My Math 322 Applied Mathematical Analysis Fall 2016, University of Wisconsin, Madison

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1 My Math 3 Applied Mathematical Analysis Fall 16, University of Wisconsin, Madison Nasser M. Abbasi Fall 16 compiled on Sunday December 5, 16 at 7:3 PM [public]

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3 Contents 1 Introduction syllabus HWs 7.1 HW Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem HW Summary table section.3.1 problem section.3. problem section.3.3 problem section.3.4 problem section.3.5 problem section.3.7 problem section.3.8 problem section.3.1 problem section.4.1 problem section.4. problem section.4.3 problem section.4.6 problem HW Problem.5.1e problem Problem.5. problem Problem.5.5c,d problem Problem.5.8b problem Problem.5.14 problem Problem.5. problem Problem.5.4 problem HW Problem Problem 3.. b,d Problem Problem 3.3. d Problem b Problem Problem Problem Problem HW Problem Problem Problem Problem Problem Problem Problem Problem Problem

4 Contents.6 HW Problem Problem Problem Problem b,d,g Problem Problem Problem HW Problem a Problem Problem Problem b Problem HW Problem Problem 5.1. b Problem Problem Problem Problem Problem HW Problem 8..1 a,b Problem 8.. a,d Problem Problem Problem Problem Problem b Problem Problem Problem b HW Problem Problem Problem Problem HW Problem Problem HW Problem Problem Problem Problem Problem study notes Heat PDE inside disk Mean value principle steady state, heat PDE, disk Maximum value principle steady state, heat PDE, disk Minimum value principle steady state, heat PDE, disk Heat PDE Outside disk Easy way to get the signs for Newton s cooling law

5 Chapter 1: Introduction inks: 1. piazza page requires login. Professor eslie Smith page Office hrs: MW from 1:15-:15 in Van Vleck syllabus Math 3-1 Syllabus Applied Mathematical Analysis Introduction to Partial Differential Equations MWF 1:5-1:55 in Van Hise 115 Textbook: Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, Richard Haberman, 5th Edition, Pearson. Pre-requisites: Math 319; Math 31; recommended: Math 34 or Math 3. Professor: eslie Smith, Departments of Mathematics and Engineering Physics, Office Hours in Van Vleck 85 MW 1:15-:15, lsmith. Midterm Exams: There will be two in-class exams: Monday October 1 and Monday November 14. Please plan accordingly. Each exam is 5% of the final grade. Final Exam: Saturday December 17, 5:5-7:5 PM, 35% of grade. Piazza: There will be a Piazza course page to facilitate peer-group discussions. Please consider this resource mainly as a discussion among students. The instructor will check in a few times per week. Piazza Sign-Up Page: piazza.com/wisc/fall16/math3 Piazza Course Page: piazza.com/wisc/fall16/math3/home Weekly Problem Sets: Homework is due at the beginning of class, normally on Friday. Homework problems will be selected from the book, and will be available on-line at lsmith approximately one week prior to the due date. Please write your name clearly on each homework set, stapled please! Unstapled homework will not be accepted. Grading of Homework: A grader will grade a subset of the homework problems given out each week, with some points also given for completeness. The homework scores will count for 15% of the grade. The lowest homework score will be dropped. ate Policy: Homework turned in after the beginning of class will be considered late and will be graded at 8% credit. ate homework will be accepted until 5 PM on the due date no credit thereafter, no exceptions. The policy is intended to keep everyone as current as possible. Please the instructor directly before the due time/day to make arrangements regarding late homework submission. Expectations In Class: You are required to come to class. Some classes may involve student participation such as discussion, group work, student presentation of material, etc. If you should need to miss a class for any reason, please let me know ahead of time, and make sure that you get notes and other important information from a classmate. No cell phones, ipods, computers or other electronic devices may be used in class. In particular, please refrain from texting during class. Find my mistakes in class, get brownie points! Expectations Outside of Class: In order to fully understand the material and do well in the course, it is vital that you stay on top of your reading and homework assignments. The 5

6 1 Introduction six hours minimum of work outside class includes but is not limited to reading the texts before and after the material is covered in lecture, completing/writing homework problems, and reviewing for exams. In addition, be prepared to work additional problems as needed, to formulate coherent questions for me and for your classmates, and to prepare material for discussion and or student presentation. Grading Scale for Final Grade: 9-1 A, AB, 8-88 B, BC, 7-78 C, 6-69 D, 59 and below F Course description: This is a first course in Partial Differential Equations. We will focus on the physical phenomena represented by three canonical equations the Heat Equation, aplace s Equation and the Wave Equation and learn the mathematical solution techniques. A basic starting point for these linear equations is Separation of Variables, and we will learn how to construct Eigenfunction Solutions, starting in one space dimension and then in two and three dimensions. More advanced topics include Green s function solutions, Fourier Transform solutions, and the Method of Characteristics. Course outline: The course covers most of the material in Chapters 1-5, and selected material from Chapters 7-1, 1time permitting. The topics are listed below with corresponding chapter. Chapter 1: The Heat Equation Chapter : Method of Separation of Variables Chapter 3: Fourier Series Chapter 4: Wave Equation: Vibrating Strings and Membranes Chapter 5: Sturm-iouville Eigenvalue Problems Chapter 7: Higher-Dimensional Partial Differential Equations Chapter 8: Non-homogeneous Problems Chapter 9: Green s Functions for Time Independent Problems Chapter 1: Infinite Domain Problems: Fourier Transform Solutions of Partial Differential Equations Chapter 1: The Method of Characteristics 6

7 l.~. Heat Equation in Two or Three Dimensions Chapter : HWs x 9.1 HW 1 Reference table used in HW Area magnified φ flux class uses q vector field. thermal energy per unit time per unit area. [ M φ ˆn flux Flux component that is outward normal to the surface [ M Q heat source heat energy generated per unit volume per unit time. [ M e thermal energy density. Scalar field. [ ] M T ρ density Figure mass density of Spherical materialcoordinates. which heat flows in. [ M c specific heat energy to raise temp. of unit mass by one degree Kelvin. EXERCISES ] ] T] 3 T 3 ] T 3 [ T k o ] k Thermal conductivity Used in flux equation q k u, where u is temperature. [ M κ Thermal diffusivity Used in heat equation u t κ u + Q. Where κ k ρc, u is temperature et cx, y, z, t denote tht' concentration d conservation of energy dt V e x, t dv of a pollutant A q ˆn da + the amount per unit [ ] V Qdv. Each term has units M volume. T 3 Fourier law φ a What is an expression k u. for Relates the total flux amount to temperature of pollutant gradient. in the region Divergence operator R? A vector operator. x, y, x b Suppose that the flow J of the pollutant is proportional to the gradient.1.1 Problem of the concentration. Is this reasonable? Express conservation of the pollutant. c Derive the partial differential equation governing the diffusion of the pollutant. *1.5.. For conduction of thermal energy, the heat flux vector is <p -KoVu. If in addition the molecules move at an average velocity V, a process called convection, then briefly explain why <p -KoVu + cpuv. Derive the corresponding equation for heat flow, including both conduction and convection of thermal energy assuming constant thermal properties with no sources Consider the polar coordinates x rcoso u Fourier law is used to relate the flux to the temperature u by φ k x for 1D or φ k u in general. y rsino. In addition to conduction, there is convection present. This implies there is physical material mass flowing outa of the Since control r volume x + carrying y, show thermal that ~ energy coso, with it in~ addition sino, to the process of conduction. Hencecos 8 and 88 -sin8 r the' flux is 8% adjustedr by this extra amount of thermal energy motion. The amount of mass that b Show flows out that ofr the cos surface oi + per sin unit 3 and timeii per unit - sin area oi + is cos vρ 3. [ ] [ M T M ] 1 3 T. Where ρ [ ] [ M is the mass density of the material and v ] c Using the chain rule, show that V r tr + 1i~1B and hence Vu 3 T is velocity vector of material flow at the surface. Amount of thermal~r+ energy~~ii. that vρ contains is given by vρ cu where c is the specific heat and u is the temperature. d If Therefore A Arr vρ + A81i, cu is the show additional that V A flux due ~1.:rAr to convection + ~1BA8, part. Total sinct' flux becomes 8rj8 Ii and 8lij8 -r follows from part b. φ k u + vρcu 1 Starting from first principles. Using conservation of thermal energy given by e t φ Where e is thermal energy density in the control volume. In this problem Q no energy source. The integral form of the above is d e x, t dv φ ˆn da dt V The dot product with the unit normal vector ˆn was added to indicate the normal component of φ at the surface. Since e x, t ρcu and by using divergence theorem the above is written as d ρcudv φ dt dv V V S T 3 k ] 7

8 l.~. Heat Equation in Two or Three Dimensions 9 HWs x Area Using 1 in the above and moving magnified the time derivative inside the integral which becomes partial derivative results in ρc u t dv k u vρcu dv V Moving all terms under one integral sign [ ρc u V t k u vρcu ] dv Figure Spherical coordinates. Since this is zero for all control volumes, therefore the integrand is zero EXERCISES 1.5 V ρc u t k u vρcu et cx, y, z, t denote tht' concentration of a pollutant the amount per unit Assuming κ volume. k, the above simplifies to ρc a What is an expression for the total amount of pollutant in the region R? t u κ u vu b Suppose that the flow J of the pollutant is proportional to the gradient Applying to theof property the concentration. of divergenceis ofthis the reasonable? product of scalar Express and conservation a vector givenof by the pollutant. c Derive the partial vu differential u equation v + v u governing the diffusion of the pollutant. Equation becomes *1.5.. For conduction of thermal u t energy, κ u u the v heat flux u vector is <p -KoVu. If in addition the molecules move at an average + v velocity V, a process called convection, then briefly explain why <p -KoVu + cpuv. Derive the corresponding equation for heat.1. Problem flow, including both conduction and convection of thermal energy assuming constant thermal properties with no sources Consider the polar coordinates x rcoso y rsino. a Since r x + y, show that ~ coso, ~ sino, and 88 -sin8 cos 8 r ' 8% r b Show that r cos oi + sin 3 and Ii - sin oi + cos 3. c Using the chain rule, show that V r tr + 1i~1B and hence Vu ~r+ ~~Ii. d If A Arr + A81i, show that V A ~1.:rAr + ~1BA8, sinct' 8rj8 Ii and 8lij8 -r follows from part b. 3 Chapter 1. Heat Equation part a Using Exercise 1.5.3a and the chain rule for partial derivatives, derive the special case of Exercise 1.5.3e if ur only Assume that the temperature is circularly symmetric: u ur, t, where x r cos θ 1 r x + y. We will derive the heat equation for this problem. Conside y r sin θ any circular annulus a ~ r ~ b. a Show that the total heat energy is 11" J: cpur dr. since r x + y then taking derivative w.r.t. x b Show that the How of heat energy per unit time out of the annulus at r b is -11"bKoau/ar I,b. A similar result holds at r a. r r x x c Use parts a r and b to derive the circularly symmetric heat equation without sources: au k a au at ;:ar rar. x x r r cos θ r cos θ Modify Exercise if the thermal properties depend on r Derive the heat equation in two dimensions by using Green's theorem , the two-dimensional form of the divergence theorem.

9 .1 HW 1 And taking derivative w.r.t. y Now taking derivative w.r.t. y of gives r r y y r y y r r sin θ r sin θ 4 From 4 r y Hence sin θ and sin θ y cos θ 1 r y θ y sin θ + r sin θ y. Therefore the above becomes 1 sin θ + r cos θ θ y 1 sin θ r cos θ cos θ r cos θ θ y cos θ r θ y Similarly, taking derivative w.r.t. x of 1 gives 1 r cos θ cos θ + r x x From 3, r cos θ x cos θ and x sin θ θ x, Therefore the above becomes θ 1 cos θ r sin θ x Hence θ x 1 cos θ r sin θ sin θ r sin θ θ x sin θ r Part b By definition of unit vector ˆr r r cos θ î + r sin θ ĵ r r cos θî + sin θĵ To find ˆθ, two relations are used. ˆθ 1 by definite of unit vector. Also ˆθ ˆr since these are orthogonal vectors basis vectors. Assuming that ˆθ c 1 î + c ĵ, the two equations generated are ˆθ 1 c 1 + c 1 ˆθ ˆr cos θî + sin θĵ c 1 î + c ĵ c 1 cos θ + c sin θ From, c 1 c sin θ cos θ. Substituting this into 1 gives c sin θ 1 + c cos θ c sin θ cos θ + c 9

10 HWs Solving for c gives cos θ c sin θ + cos θ c cos θ Since c is now known, c 1 is found from Hence c 1 sin θ. Therefore c 1 cos θ + c sin θ c 1 cos θ + cos θ sin θ c 1 cos θ sin θ cos θ ˆθ sin θî + cos θĵ Part c Since x x r, θ, y y r, θ, then xî + xĵ 1 x r r x + θ θ x y r r y + θ θ y Equation 1 becomes r r x + θ θ r î + x r y + θ θ ĵ y Using result found in a, the above becomes r cos θ + θ Collecting on r, θ gives sin θ r r cos θî + sin θĵ + θ r cos θî + sin θĵ + 1 r Using result from b, the above simplifies to ˆr r + ˆθ 1 r î + r sin θ + θ sin θ r î + cos θ ĵ r sin θî + cos θĵ θ θ cos θ ĵ r Hence u ˆr r u + ˆθ 1 r θ u Part d Hence A ˆr r + ˆθ 1 r θ ˆr r A rˆr + A A rˆr + A θ ˆθ ˆr r + ˆθ 1 r θ A rˆr + A θ ˆθ + ˆr r A θ ˆθ ˆθ 1 r θ A rˆr + ˆθ 1 r θ A θ ˆθ 1 1

11 .1 HW 1 But And ˆr r A rˆr ˆr r A rˆr Ar ˆr r ˆr + A ˆr r r A r r ˆr ˆr + A r A r r 1 + A r A r r ˆr r A θ ˆθ ˆr A θ ˆθ r ˆr A θ r ˆr ˆr r A θ r ˆθ + A ˆθ θ r ˆr ˆθ + A θ ˆr ˆθ r And A θ r + A θ 3 ˆθ 1 r θ A rˆr ˆθ 1 r θ A rˆr 1 r ˆθ Ar θ ˆr + A ˆr r θ 1 A r ˆθ ˆr + 1 r θ r A r ˆθ ˆr θ 1 A r r θ + 1 r A r ˆθ ˆθ Since ˆr θ ˆθ. Therefore And finally ˆθ 1 r Substituting,3,4,5 into 1 gives ˆθ 1 r θ A θ ˆθ ˆθ 1 r 1 r ˆθ θ A rˆr 1 r A r 4 A θ ˆθ θ A θ θ ˆθ + A ˆθ θ θ A θ θ ˆθ ˆθ + 1 r A θ ˆθ ˆθ θ 1 r 1 A θ r θ r A θ ˆθ ˆr 1 A θ r θ r A θ 1 A θ r θ A A r r r A r + 1 r 1 r A r + A r r + 1 A θ r θ A θ θ Add since r ra r A r + r Ar r, the above can also be written as A 1 r 1 r A r + r A r r r ra r + 1 r + 1 A θ r θ A θ θ 5 11

12 HWs Part e From part c, it was found that But ˆr r + ˆθ 1 r θ ˆr r + ˆθ 1 r ˆr θ r + ˆθ 1 r r ra r + 1 r Using result of part d, which says that A 1 r A r r, A θ 1 r θ 1 r r r r r θ r 1 r + 1r r r r θ θ A θ θ, the above becomes where now θ Hence u 1 r r u + 1 u r r r θ.1.3 Problem Chapter 1. Heat Equation Using Exercise 1.5.3a and the chain rule for partial derivatives, derive the special case of Exercise 1.5.3e if ur only Assume that the temperature is circularly symmetric: u ur, t, where r x + y. We will derive the heat equation for this problem. Consider et u u r. any From circular problem annulus part a ~ a r it ~ was b. found that a Show that the total heat x energy r cos θis 11" J: cpur dr. b Show that the How of heat y energy r sin θ per unit time out of the annulus at r b is -11"bKoau/ar ri,b. A similar result holds at r a. c Use parts a and b x to derive cos θ the circularly symmetric heat equation without sources: r y au sin θ k a au θ at ;:ar rar Modify Exercise if the y cos θ thermal r properties depend on r. θ Derive the heat equation x in sin θ two dimensions r by using Green's theorem, , the two-dimensional form of the divergence theorem. And If aplace's equation is satisfied u in u three dimensions, show that x If + u y 1 But V'uon ds u for any dosed surface. Hint: Use the divergence theorem. Give a physical x interpretation u x x of this result in the context of heat flow. u r Determine the equilibrium x r x temperature distribution inside a circular annulus rl ~ r ~ r: u *a if the outer xradius r cos is at θ temperature T and the inner at Tl b if the outer radius u is cos insulated θ + u cos θ and the inner radius is at temperature Tl x r r x u r Determine the equilibrium temperature distribution inside a circle r ~ r cos θ + u sin θ θ x r x ro if t.he boundary is fixed at temperature To. u * Consider r cos θ cos θ + u sin θ sin θ r r au ~~ rau a<r<b u at rar ar r cos θ + 1 r sin θ u r subject. to 1 au au ur.o fr, ar a, t,b, and ar b, t 1. Using physical reasoning. for what values of {3 does an equilibrium tem

13 .1 HW 1 And Substituting,3 into 1 gives Which can be written as u y u y y u r y r y u y r sin θ u sin θ + u sin θ y r r y u r r sin θ + u y r u r sin θ sin θ + u r u r sin θ + 1 r cos θ u r u u r cos θ + 1 r sin θ u r u r + 1 r u r + 1 u r r cos θ θ y cos θ u + sin θ u r + cos θ u r u 1 r r cos θ r r sin θ + 1 r cos θ u r r u r 3 Which is the special case of problem e u 1 3 r r r u r + 1 Chapter when u1. Heat u r only. Equation.1.4 Problem u r θ Using Exercise 1.5.3a and the chain rule for partial derivatives, derive the special case of Exercise 1.5.3e if ur only Assume that the temperature is circularly symmetric: u ur, t, where r x + y. We will derive the heat equation for this problem. Consider any circular annulus a ~ r ~ b. a Show that the total heat energy is 11" J: cpur dr. b Show that the How of heat energy per unit time out of the annulus at r b is -11"bKoau/ar I,b. A similar result holds at r a. c Use parts a and b to derive the circularly symmetric heat equation without sources: au k a au at ;:ar rar Modify Exercise if the thermal properties depend on r Derive the heat equation in two dimensions by using Green's theorem, Parta , the two-dimensional form of the divergence theorem. Considering the If thermal aplace's energy equation in a annulus is satisfied as shown in three dimensions, show that If V'uon ds for any dosed surface. Hint: Use the divergence theorem. Give a physical interpretation of this result in the context of heat flow Determine the equilibrium temperature distribution inside a circular annulus rl ~ r ~ r: *a if the outer radius is at temperature T and the inner at Tl b if the outer radius is insulated and the inner radius is at temperature Tl Determine the equilibrium temperature distribution inside a circle r ~ ro if t.he boundary is fixed at temperature To. * Consider 13

14 HWs Amount of thermal energy in unit thickness volume is rdrdθ cuρ dr rdθ Amount of thermal energy in annulus is π b cρu rdrdθ a r dθ b a dr Integrating gives total thermal energy Part b Using Fourier law, Since symmetric in θ, then u θ π b a cρu rdrdθ π π dθ b a b a cρu rdr φ k u k ˆr u r + ˆθ 1 u r θ and the above reduces to φ k ˆr u r Hence the heat flow per unit time through surface at r b is π π φ ˆn ds k u ˆr ˆn rdθ r But ˆn ˆr since radial unit vector. The above becomes At r b the above becomes Similarly at r a π u k r rdθ πk r u r πk b u r πk a u r rb ra cρu rdr Part c Applying that the rate of time change of total energy equal to flux through the boundaries gives d b π cρu rdr πk a u dt a r + πk b u ra r rb b πk r u dr r r a Moving d dt inside the first integral, it become partial π b a cρ u b rdr πk r u dr t a r r 14

15 .1 HW 1 Moving everything under one integral 3 Chapter 1. Heat Equation b a [ cρ u t r k r r u r ] dr Hence, since this is valid for any annulus, then the integrand is zero Using Exercise 1.5.3a and the chain rule for partial derivatives, derive the cρ u r k r u special case of Exercise 1.5.3e if ur only. t r r Assume that the temperature u is t k circularly 1 r u symmetric: u ur, t, where r x + y. We will derive the cρ heat r requation for this problem. Consider any circular annulus a ~ r ~ b. Hence u r u Where κ k cρ. a Show that the total heat energy is 11" J: cpur dr. t κ r r c Use parts a and b to derive the circularly symmetric heat equation.1.5 Problem without sources: au k a au at ;:ar rar. r b Show that the How of heat energy per unit time out of the annulus at r b is -11"bKoau/ar I,b. A similar result holds at r a Modify Exercise if the thermal properties depend on r Derive the heat equation in two dimensions by using Green's theorem, , the two-dimensional form of the divergence theorem. The earlier problem is now repeated but in this problem c c r, k k r and ρ ρ r. These are the thermal properties in the problem. Parta If aplace's equation is satisfied in three dimensions, show that π b If V'uon ds for any dosed c r surface. ρ r u rdrdθ Hint: Use dθthe c divergence r ρ r u rdr theorem. Give a physical interpretation a of this result in the context a of heat flow. π c r ρ r u rdr Determine the equilibrium temperature a distribution inside a circular annulus rl ~ r ~ r: Part b *a if the outer radius is at temperature T and the inner at Tl b if the outer radius φ is kinsulated r ˆr u r and the inner radius is at temperature The heat flow per unit Tl time through surface at r is therefore Determine π the equilibrium π temperature distribution inside a circle r ~ ro φ ˆn ds k if t.he boundary is fixed at temperature r ˆr u ˆn rdθ r To. * Consider But ˆn ˆr since radial therefore subject. to At r b the above becomes Similarly at r a Part c π π b b au ~~ rau at rar ar k r u r rdθ πk r r u r π k rb b u a<r<b au au ur.o fr, ar ra, t,b, and ar b, t 1. rb Using physical reasoning. π k for what values of {3 does an equilibrium temperature distribution exist? r ra a u ra Applying that the rate of time change of total energy equal to flux through the boundaries gives d π dt b a c r ρ r u rdr π k ra a u r + π k rb b u ra r b π k r r u dr a r r rb 15

16 HWs Using Exercise 1.5.3a and the chain rule for partial derivatives, derive the special case of Exercise 1.5.3e if ur only. Moving d dt inside Assume the first that integral, the temperature it become partial is circularly symmetric: u ur, t, where r x + y. b We will derive π c r ρ r u the heat equation b for this rdr π k r r u problem. Consider any circular annulus a ~ r ~ b. dr a t a r r a Show that the total heat energy is 11" J: cpur dr. Moving everything under one integral b Show that the How of heat energy per unit time out of the annulus at r b b is [-11"bKoau/ar c r ρ r u I,b. r A similar k r r u result ] holds at r a. dr c Use parts a a and b t to derive r the circularly r symmetric heat equation without sources: Since this is valid for any annulus then the integrand au k isa zero au c r ρ r u at r ;:ar rar k r r u Modify Exercise if t the thermal r properties r depend on r. Therefore, the heat Derive equation the heat whenequation the thermal in two properties dimensions depends by on using r becomes Green's theorem, , the two-dimensional form of the divergence theorem. ur,t t If aplace's equation is satisfied ρrcr r r in three k r rdimensions, ur,t r show that.1.6 Problem If V'uon ds for any dosed surface. Hint: Use the divergence theorem. Give a physical interpretation of this result in the context of heat flow Determine the equilibrium temperature distribution inside a circular annulus rl ~ r ~ r: *a if the outer radius is at temperature T and the inner at Tl b if the outer radius is insulated and the inner radius is at temperature Tl Determine the equilibrium temperature distribution inside a circle r ~ ro if t.he boundary is fixed at temperature To. Part a * Consider The heat equation is u au ~~ rau a<r<b t κ r r r u r. At steady state u at rar ar t. And since circular region, symmetry in θ is assumed and therefore temperature u depends only on r only. This means u r is the same at any angle θ subject. for that to specific r. This becomes a second order ODE au κ d au ur.o fr, ar a, r du t,b, and ar b, t 1. r dr dr Using physical reasoning. κ for du what values of {3 does an equilibrium temperature distribution exist? r dr + r d u dr Since κ r Integrating. Assuming du dr d u dr + 1 du r dr v r, the above becomes dv dr + 1 r v dv dr 1 r v dv v dr r ln v ln r + c 1 Where c e c 1. Since du dr v, then du v e ln r+c 1 c e ln r c 1 r dr c 1 r 1 du c r dr 16

17 .1 HW 1 Integrating u r c ln r + c 3 When r r 1, u T 1, and when r r, u T, therefore T 1 c ln r 1 + c 3 T c ln r + c 3 From first equation, c 3 T 1 c ln r 1. Substituting in second equation gives Therefore T c ln r + T 1 c ln r 1 c ln r ln r 1 + T 1 c T T 1 ln r ln r 1 Hence c 3 T 1 T T 1 ln r ln r 1 ln r 1. Therefore the steady state solution becomes Hence u r c ln r + c 3 T T 1 ln r + T 1 T T 1 ln r 1 3 ln r ln r 1 ln r ln rchapter 1 1. Heat Equation T 1 + T T 1 ln r T T 1 ln r 1 ln r ln r 1 T 1 + T T 1 ln r ln r Using Exercise 1.5.3a and lnthe r chain ln r 1 rule for partial derivatives, derive the special case of Exercise 1.5.3e if ur ln r only. r Assume that the Ttemperature 1 + T T 1 is circularly ln r symmetric: u ur, t, where r x + y. We will derive the heat r 1 equation for this problem. Consider any circular annulus a ~ r ~ b. r r 1 r1 a Show that the total heat energy is 11" u r T 1 + T T 1 J: ln cpur dr. b Show that the How of heat energy per r ln unit time out of the annulus at r b is -11"bKoau/ar I,b. A similar result holds at r a. c Use parts a and b to derive the circularly symmetric heat equation Part b without sources: Insulated condition implies du dr. So the above au is repeated, k a au but this new boundary condition is at ;:ar rar. now used at r. Starting from the general solution found in part a Modify Exercise if the thermal properties depend on r. u r c ln r + c Derive the heat equation in two dimensions When r r 1, u T 1 and when r r, du du dr. But dr c by using Green's r. Hence r r gives c theorem, , the two-dimensional form of the divergence theorem. r or c. Therefore the solution is If aplace's equation is satisfied u r in cthree 3 dimensions, show that When r r 1, u T 1, hence c 3 T 1. The solution becomes If V'uon ds u r T 1 for any dosed surface. Hint: Use the divergence theorem. Give a physical The temperature interpretation is T 1 everywhere. of this This result makes in sense the context as this is of steady heat flow. state, and no heat escapes to the outside Determine the equilibrium temperature distribution inside a circular annulus rl ~ r ~ r:.1.7 Problem *a if the outer radius is at temperature T and the inner at Tl b if the outer radius is insulated and the inner radius is at temperature Tl Determine the equilibrium temperature distribution inside a circle r ~ ro if t.he boundary is fixed at temperature To. * Consider au ~~ rau a<r<b ast problem found the solution to at the heat rar equation ar in polar coordinates with symmetry in θ to be subject. to u r c ln r + c 3 au au c must be zero since at r ur.o the temperature fr, ar must a, t be finite.,b, and The ar solution b, t 1. becomes Using physical reasoning. for u r what cvalues 3 of {3 does an equilibrium temperature distribution exist? 17

18 1.5.16, the two-dimensional form of the divergence theorem If aplace's equation is satisfied in three dimensions, show that HWs If V'uon ds Applying the boundary for any dosed conditions surface. at r Hint: r Use the divergence theorem. Give a physical interpretation of this result in the context of heat flow. T c Determine the equilibrium temperature distribution inside a circular annulus rl ~ r ~ r: Therefore, u r T The temperature *a everywhere if the outer is the radius the same is at temperature as on the edge. T and the inner at Tl b if the outer radius is insulated and the inner radius is at temperature.1.8 Problem Tl Determine the equilibrium temperature distribution inside a circle r ~ ro if t.he boundary is fixed at temperature To. * Consider subject. to au ~~ rau at rar ar a<r<b au au ur.o fr, ar a, t,b, and ar b, t 1. Using physical reasoning. for what values of {3 does an equilibrium temperature distribution exist? For equilibrium the total rate of heat flow at r a should be the same as at r b. Circumference at r a is πa and total rate of flow at r a is given by β. Hence total heat flow rate at r a is given by πa u r πaβ ra Similarly, total heat flow rate at r b is given by πb u r πb rb Therefore πaβ πa or β a b.1.9 Problem Heat Equation in Two or Three Dimensions Assume that the temperature is spherically symmetric, 1.1 ur, t, where r is the distance from a fixed point r x + y + z. Consider the heat flow without sources between any two concentric spheres of radii a and b. a Show that the total heat energy is 411" J: cpur dr. b Show that the flow of heat energy per unit time out of the spherical shell at r b is -411"b Ko 8u/ar Ir:b. A similar result holds at r a. c Use parts a and b to derive the spherically symmetric heat equation 811. ~~ r 8U at r ar 8r. * Determine the steady-state temperature distribution between two concentric spheres with radii 1 and 4, respectively, if the temperature of the outer Part a sphere is maintained at 8 and the inner sphere at see Exercise Total heat energy Isobars is, by are definition lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated. E cρudv Derive the heat equation in three V dimensions assuming constant thermal Volume v of sphere properties of radius and rno is sources. v 4 3 πr3. Hence Express the integral conservation dv law for any three-dimensional object. Assume there are no sources. dr Also 4πr assume the heat flow is specified, Vu i& g:.:, 11, z, on the entire boundary dv 4πr and dr does not depend on time. By integrating with respect to time, determine the total thermal energy. Hint: Use the initial condition Derive the integral conservation law for any three dimensional object with constant thermal properties by integrating the heat equation as

19 .1 HW 1 Equation 1 becomes, where now the r limits are from a to b Part b By definition, the flux at r b is E b 4π cρu 4πr dr a b a cρur dr u φ b k r The above is per unit area. At r b, the surface area of the sphere is 4πb. Therefore, the total energy per unit time is φ b 4πb or 4πb u k r Similarly for r a. Partc By conservation of thermal energy d dt E u 4πa k r + 4πb u k ra r d b b 4π cρur dr 4πk r u dr dt a a r r b cρ u b a t r dr k r u dr a r r Moving everything into one integral b a [ cρ u t r k r rb rb r u ] dr r Since this is valid for any limits the integrand must be zero cρ u t r k r u 1.5. Heat Equation in Two or Three r Dimensions r u t k 1 cρ r r u Assume that the temperature is spherically rsymmetric, u ur, t, where r Therefore is the distance from a fixed point r x + y + z. Consider the heat flow without sources between any two concentric spheres of radii a and b. r u u t κ r r a Show that the total heat energy is 47r fo cpur dr. Where κ k cρ b Show that the flow of heat energy per unit time out of the spherical shell at r b is -4irbKo 8u/8r Irb. A similar result holds at r a..1.1 Problemc 1 Use parts a and b to derive the spherically symmetric heat equation r 8u k 8 T8u 8t r 8r C? * Determine the steady-state temperature distribution between two concentric spheres with radii 1 and 4, respectively, if the temperature of the outer sphere is maintained at 8 and the inner sphere at see Exercise Isobars are lines of constant temperature. Show that isobars are perpendicular to any part of the boundary that is insulated. The heat equation Derive isthe u t heat κ r r equation r u r. in For three steady dimensions state u t assuming and assuming constant symmetry thermal in θ, the heat equation properties becomesand no ODE sources. in r Express the integral conservation κ d law for any three-dimensional object. Assume there are no sources. r r Also du dr assume dr the heat flow is specified, gx, y, z, on the entire boundary d and does not depend on time. By integrating with respect to time, rdetermine du dr dr the total thermal energy. Hint: Use the initial condition. r du Derive the integral conservation dr + r d u law dr for any three dimensional object with constant thermal properties by integrating the heat equation assuming no sources. Show that the result is equivalent to Orthogonal curvilinear coordinates. A coordinate system u, v, w may be introduced and defined by x xu, v, w, y yu, v, w and. rb 19

20 HWs For r et du dr Integrating v r, hence r d u dr + du dr r dv dr + v dv dr v r dv v dr r ln v ln r + c ln r+c v e ln r c 1 e 1 c 1 r Therefore, since du dr v r then du dr c 1 1 r du c 1 dr r Integrating u r c 1 r + c When r 1, u and when r 4, u 8, hence c 1 + c 8 c c From first equation, c 1 c, and from second equation 8 c c 1, hence 3 4 c 1 8 or c Therefore, the general solution becomes or u r r u r r

21 . HW. HW..1 Summary table For 1D bar eft Right λ λ > u x, t λ u u No n nπ, n 1,, 3, X n B n sin λ n x B n sin λ n x e kλnt u u x No u x u No u u x u + u x u x λ X A λ X A For periodic conditions u u and u x nπ λ n u x, t λ {}}{ a + λ n nπ, n 1, 3, 5, X n B n sin λ n x λ n nπ, n 1, 3, 5, X n A n cos λ n x tan λn λ n X λ B λ sin λ n x λ n nπ, n 1,, 3, X n A n cos λ n x u x, n 1,, 3, λ> {}}{ A n cos λn x e kλnt + B n sin λn x e kλnt,3,5, Bn sin λ n x e kλnt,3,5 An cos λ n x e kλnt A + B n sin λ n x e kλnt A + A n cos λ n x e kλnt Note on notation When using separation of variables T t is used for the time function and X x, R r, Θ θ etc. for the spatial functions. This notation is more common in other books and easier to work with as the dependent variable T, X, and the independent variable t, x, are easier to match one is upper case and is one lower case and this produces less symbols to remember and less chance of mixing wrong letters. 1

22 HWs.. section.3. Heat.3.1 Equation problem With Zero 1 Temperature Ends 55 part a et Then And EXERCISES For the following partial differential equations, what ordinary differential equations are implied by the method of separation of variables? a au ka ru * at r ar & au au * C 9x + ft o *e &U at 4U au U k c a 44 * f ate ax.3.. Consider the differential equation z +A. Determine the eigenvalues 1 u \ and corresponding eigenfunctions if satisfies the following boundary kconditions. t 1 Analyze r u r r r three cases.\ >, A, A <. You may assume that the eigenvalues are real. a and -,r u t, r T t R r *b and 51 c!o and O u If necessary, see Sec t T t R r *d and O e O and O r u u *f Oa and rob r You r may + r u assume r that A >. T R g and O r + rt R r + cb If necessary, see Sec Hence becomes Consider the heat equation 1 8U k T t R r 1 T R r + rt R r rou at - kax Note From now on T t is written as just T and similarly for R r R and R r R to simplify notations subject and make to the itboundary easier and conditions more clear to read. The above is reduced to u,t 1 and u,t. Solve the initial value problem k T R 1 if r the T R + T R temperature is initially Dividing throughout 1 by T t R r gives a ux, 6 sin s b ux, 3 sin i - sin i 1 * c ux, cos lme d ux, k T 1 r R + R R T R b d -` k 9x 1 < x < / /<x< - v ax Since each side in the above depends on a different independent variable and both are equal to each others, then each side is equal to the same constant, say λ. Therefore The following differential equations are obtained In expanded form, the above is 1 T k T 1 R r R + R R λ T + λkt rr + R + rλr dt dt + λkt t r d R dr + dr + rλr r dt 1 1 T t R r can not be zero, as this would imply that either T t or R r or both are zero, in which case there is only the trivial solution.

23 . HW Part b et Then And 1 u k t u x v u k x u x, t T X u t T X u x X T u x X T Substituting these in 1 gives 1 k T X X T v k X T Dividing throughout by T X gives 1 1 T k T X X v X k X Since each side in the above depends on a different independent variable and both are equal to each others, then each side is equal to the same constant, say λ. Therefore The following differential equations are obtained 1 T k T X X v X k X λ The above in expanded form is T + λkt X v k X + λx Part d et Then And Substituting these in 1 gives dt + λkt t dt d X dx v dx + λx x k dx 1 u k t 1 r r u r r u t, r T R u t T R r u r u r r r + r u r rt R + r T R 1 k T R 1 r rt R + r T R r T R + T R 1 Dividing throughout by T R gives 1 T k T R r R + R R 3

24 HWs Since each side in the above depends on a different independent variable and both are equal to each others, then each side is equal to the same constant, say λ. Therefore The following differential equations are obtained 1 T k T R r R + R R λ.3. Heat Equation With Zero Temperature T + λkt Ends EXERCISES.3 rr + R + λrr The above.3.1. in expanded For the form following is partial differential equations, what ordinary differential equations are implied by the method of separation of variables? dt + λkt t a au ka ru dt b * at r ar -` k 9x - v r d & R ax dr + dr + λrr r au au dr * C 9x + ft o d..3 section.3. problem &U 4U au U *e k c at a 44 * f ate ax Part d.3.. Consider the differential equation z +A. Determine the eigenvalues \ and corresponding eigenfunctions if satisfies the following boundary conditions. Analyze three cases.\ >, A, A <. You may assume that the eigenvalues are real. a and -,r *b and 51 c!o and O If necessary, see Sec *d and O e O and O *f Oa and Ob You may assume that A >. g and O + cb If necessary, see Sec Consider the heat equation OU 8U at - kax subject to the boundary conditions u,t d φ and u,t. dx Solve the initial value problem + λφ if the temperature is initially φ a ux, 6 sin s dφ b ux, 3 sin i - sin i dx 1 < x < / * c ux, cos lme d ux, Substituting an assumed solution of the form φ Ae rx in the above /<x< ODE and simplifying gives the characteristic equation r + λ r λ r ± λ Assuming λ is real. The following cases are considered. case λ < In this case, λ and also λ, are positive. Hence both roots ± λ are real and positive. et λ s Where s >. Therefore the solution is φ x Ae sx + Be sx dφ dx Asesx Bse sx 55 4

25 . HW Applying the first boundary conditions B.C. gives Applying the second B.C. gives φ A + B dφ dx As Bs s A B A B The last step above was after dividing by s since s. Therefore, the following two equations are solved for A, B A + B A B The second equation implies A B and the first gives A or A. Hence B. Therefore the only solution is the trivial solution φ x. λ < is not an eigenvalue. case λ In this case the ODE becomes The solution is Applying the first B.C. gives d φ dx φ x Ax + B dφ dx A φ B Applying the second B.C. gives dφ dx A Hence A, B are both zero in this case as well and the only solution is the trivial one φ x. λ is not an eigenvalue. case λ > In this case, λ is negative, therefore the roots are both complex. Hence the solution is r ±i λ φ x Ae i λx + Be i λx Which can be writing in terms of cos, sin using Euler identity as φ x A cos λx + B sin λx Applying first B.C. gives φ A cos + B sin A λx The solution now is φ x B sin. Hence dφ dx λb cos λx 5

26 HWs Applying the second B.C. gives dφ dx λb cos λ λb cos λ λ Since λ then either B or cos. But B gives trivial solution, therefore cos λ This implies nπ λ n 1, 3, 5, In other words, for all positive odd integers. n < can not be used since λ is assumed positive. λ nπ n 1, 3, 5, The eigenfunctions associated with these eigenvalues are nπ φ n x B n sin x n 1, 3, 5, Part f d φ dx + λφ φ a φ b It is easier to solve this if one boundary condition was at x. So that one constant drops out. et τ x a and the ODE becomes where now the independent variable is τ d φ τ dτ + λφ τ 1 With the new boundary conditions φ and φ b a. Assuming the solution is φ Ae rτ, the characteristic equation is r + λ r λ r ± λ Assuming λ is real and also assuming λ > per the problem statement then λ is negative, and both roots are complex. r ±i λ This gives the solution Applying first B.C. φ τ A cos λτ + B sin λτ φ A cos + B sin A λτ Therefore the solution is φ τ B sin. Applying the second B.C. φ b a B sin λ b a λ B leads to trivial solution. Choosing sin b a gives λn b a nπ λn nπ b a n 1,, 3 6

27 . HW Or λ n nπ b a n 1,, 3, The eigenfunctions associated with these eigenvalue are φ n τ B n sin λn τ nπ B n sin b a τ Transforming back to x nπ φ n x B n sin x a b a Part g d φ dx + λφ φ dφ + φ dx Assuming solution is φ Ae rx, the characteristic equation is r + λ r λ r ± λ The following cases are considered. case λ < In this case λ and also λ are positive. Hence the roots ± λ are both real. et λ s Where s >. This gives the solution φ x A e sx + B e sx Which can be manipulated using sinh sx esx e sx, cosh sx esx +e sx to the following φ x A cosh sx + B sinh sx Where A, B above are new constants. Applying the left boundary condition gives φ A The solution becomes φ x B sinh sx and hence dφ dx conditions gives s cosh sx. Applying the right boundary φ + dφ dx B sinh s + Bs cosh s B sinh s + s cosh s But B leads to trivial solution, therefore the other option is that But the above is sinh s + s cosh s tanh s s Since it was assumed that s > then the RHS in the above is a negative quantity. However the tanh function is positive for positive argument and negative for negative argument. The above implies then that s <. Which is invalid since it was assumed s > and is the length of the bar. Hence B is the only choice, and this leads to trivial solution. λ < is not an eigenvalue. case λ 7

28 HWs In this case, the ODE becomes The solution is Applying left B.C. gives d φ dx φ x c 1 x + c φ c The solution becomes φ x c 1 x. Applying the right B.C. gives φ + dφ dx c 1 + c 1 c Since c 1 leads to trivial solution, then 1 + is the only other choice. But this invalid since > length of the bar. Hence c 1 and this leads to trivial solution. λ is not an eigenvalue. case λ > This implies that λ is negative, and therefore the roots are both complex. This gives the solution Applying first B.C. gives r ±i λ φ x Ae i λx + Be i λx A cos λx + B sin λx φ A cos + B sin A λx The solution becomes φ x B sin and Applying the second B.C. dφ dx λb cos λx dφ dx λb cos + φ λ + B sin λ λ λ λ Dividing 1 by cos, which can not be zero, because if cos, then B sin from above, and this means the trivial solution, results in B λ + tan λ But B, else the solution is trivial. Therefore tan λ λ The eigenvalue λ is given by the solution to the above nonlinear equation. The text book, in section 5.4, page 196 gives the following approximate asymptotic solution which becomes accurate only for large n and not used here λn π n 1 Therefore the eigenfunction is φ λ x B sin λx λ Where λ is the solution to tan λ. 1 8

29 c!o and O If necessary, see Sec *d and O e O and O..4 section.3.3 *f Oa problem and 3 Ob You may assume that A >. g and O + cb If necessary, see Sec HW.3.3. Consider the heat equation OU 8U at - kax subject to the boundary conditions u,t and u,t. Solve the initial value problem if the temperature is initially a ux, 6 sin s b ux, 3 sin i - sin i * c ux, cos lme d ux, 1 < x < / /<x< Part b u t k u x et u x, t T t X x, and the PDE becomes 1 k T X X T Dividing by XT 1 T k T X X Since each side depends on different independent variable and both are equal, they must be both equal to same constant, say λ where λ is assumed to be real. The two ODE s are 1 T k T X X λ T + kλt 1 X + λx Starting with the space ODE equation, with corresponding boundary conditions dx dx, dx dx. Assuming the solution is X x erx, Then the characteristic equation is r + λ r λ r ± λ The following cases are considered. case λ < In this case, λ and also λ are positive. Hence the roots ± λ are both real. et λ s Where s >. This gives the solution X x A cosh sx + B sinh sx Applying the left B.C. gives X B cosh B The solution becomes X x A cosh sx. Applying the right B.C. gives X A cosh sx 9

30 HWs But cosh sx, hence A and the trivial solution. λ < is not an eigenvalue. case λ The ODE becomes d X dx The solution is X x c 1 x + c Applying left boundary conditions gives X c Hence the solution becomes X x c 1 x. Applying the right B.C. gives X c 1 Hence c 1. Hence trivial solution. λ is not an eigenvalue. case λ > Hence λ is negative, and the roots are both complex. r ±i λ The solution is X x A cos λx + B sin λx The boundary conditions are now applied. The first B.C. gives X A λx The ODE becomes X x B sin. Applying the second B.C. gives X B sin λ B else the solution is trivial. Therefore taking sin λ λn nπ n 1,, 3, Hence eigenvalues are λ n n π n 1,, 3, The eigenfunctions associated with these eigenvalues are nπ X n x B n sin x The time domain ODE is now solved. T + kλ n T has the solution T n t e kλnt For the same set of eigenvalues. Notice that there is no need to add a new constant in the above as it will be absorbed in the B n when combined in the following step below. The solution to the PDE becomes u n x, t T n t X n x But for linear system the sum of eigenfunctions is also a solution, therefore u x, t u n x, t nπ B n sin x e k nπ t 3

31 . HW Initial conditions are now applied. Setting t, the above becomes u x, 3 sin πx sin 3πx nπ B n sin x As the series is unique, the terms coefficients must match for those shown only, and all other B n terms vanish. This means that by comparing terms πx 3πx πx 3π 3 sin sin B 1 sin + B 3 sin x Therefore And all other B n. The solution is Part d Part b found the solution to be B 1 3 B 3 1 π u x, t 3 sin x e k π t sin u x, t The new initial conditions are now applied. 3π x nπ B n sin x e k nπ t e k 3π t f x nπ B n sin x 1 Where f x { 1 < x / / < x < Multiplying both sides of 1 by sin mπ x and integrating over the domain gives mπ [ sin x mπ nπ ] f x dx B n sin x sin x dx Interchanging the order of integration and summation mπ sin x f x dx [B n mπ nπ ] sin x sin x dx But sin mπ x sin nπ x dx for n m, hence only one term survives mπ sin x f x dx B m sin mπ x dx 31

32 HWs Renaming m back to n and since sin mπ x dx nπ sin x f x dx B n Hence the solution is nπ sin x f x dx B n cos nπ x nπ nπ nπ nπ nπ sin x f x dx + nπ sin x dx + nπ cos nπ x [ nπ cos the above becomes sin + cos nπ sin x f x dx nπ x dx + cos nπ x nπ nπ x + cos ] [ + cos nπ + cos nπ nπ cos + 1 cos nπ + cos nπ cos + 1 cos nπ u x, t nπ B n sin x e k nπ t With B n nπ cos cos nπ + 1 nπ 1 1 n nπ + cos nπ 56 Chapter. Method of Separation of Variables..5 section.3.4 problem 4 [Your answer in part c may involve certain integrals that do not need to be evaluated.] Part a.3.4. Consider k, subject to u, t, u, t, and ux, f x. *a What is the total heat energy in the rod as a function of time? b What is the flow of heat energy out of the rod at x? at x? *c What relationship should exist between parts a and b?.3.5. Evaluate be careful if n m nzrx m7rx sin sin dx forn>,m>. By definition the total heat energy is Use the trigonometric identity E ρcu x, t dv sin asin b [cosa - b - cosa + b]. Assuming constant cross section area A, the above becomes assuming all thermal properties are *.3.6. Evaluate constant n7rx m7rx cog cc dx for n > O, m >. E ρcu x, t Adx Use the trigonometric identity But u x, t was found to be cos a cos b [cosa + b + cosa - b]. nπ u x, t B n sin Be careful if a - b or a + b. x e k nπ t V.3.7. Consider the following boundary value problem if necessary, see Sec..4.1: nπ ] 3 k 8U with au, to, au, t, and ux, f x. at ax ax ax a Give a one-sentence physical interpretation of this problem.

33 . HW For these boundary conditions from problem.3.3. Where B n was found from initial conditions. Substituting the solution found into the energy equation gives nπ E ρca B n sin x e k nπ t dx Part b ρca ρca ρca B n e k nπ B n e k nπ t nπ sin x dx t cos nπ x nπ B n e k nπ t cos nπ ρca B n e k nπ t 1 cos nπ nπ ρca [ Bn nπ 1 cos nπ e k π n nπ + cos By definition, the flux is the amount of heat flow per unit time per unit area. Assuming the area is A, then heat flow at x into the rod per unit time call it H x is H x A φ x Ak u x x Similarly, heat flow at x out of the rod per unit time is t ] H x A φ x Ak u x x To obtain heat flow at x leaving the rod, the sign is changed and it becomes Ak u x x. u x, t B n sin nπ x e k nπ t then u x B n nπ cos nπ x Then at x then heat flow leaving of the rod becomes Ak u x Ak x And at x, the heat flow out of the bar Ak u x Ak x Ak Ak nπ B ne k e k nπ t nπ t B n nπ cos nπ B n nπ cos nπ e κ 1 n B n nπ e κ e κ nπ t nπ t nπ t Since Part c Total E inside the bar at time t is given by initial energy E t and time integral of flow of heat energy into the bar. Since from part a Then initial energy is E ρca π E t ρca π B n nπ t n e k 1 cos nπ B n n 1 cos nπ 33

34 HWs And total heat flow into the rod per unit time is Ak u x x + Ak u x x, therefore ρca B n nπ t t π n e k 1 cos nπ Ak u x + Ak u x x dx x t u u Ak dx x x But Hence ρca π u u x x π nb n 1 n e k nπ t π nb n e k nπ 56 Chapter. Method of Separation t of Variables π [Your answer in part c may involve certain nb n 1 n e k nπ t integrals that nb n e k nπ t do not need to be evaluated.] B n.3.4. Consider nπ t n exp k 1 cos nπ Akπ t k, nb n 1 n e k nπ subject to u, t, u, t, and ux, f x. t nb n e k nπ t dx *a What is the total heat energy in the rod as a function of time?..6 section.3.5 problem 5 b What is the flow of heat energy out of the rod at x? at x? *c What relationship should exist between parts a and b?.3.5. Evaluate be careful if n m nzrx Use the trigonometric identity m7rx sin sin dx forn>,m>. sin asin b [cosa - b - cosa + b]. *.3.6. Evaluate n7rx m7rx cog nπx cc mπx I sin sin dx Considering first the Use case the m trigonometric n. The integral identity becomes dx for n > O, m >. cos a cos b I sin nπx [cosa dx + b + cosa - b]. For the case wherebe n careful m, using if a - b or a + b Consider the following boundary value problem if necessary, see Sec..4.1: sin a sin b 1 cos a b cos a + b k 8U The integral I becomes with au, to, au, t, and ux, f x. at ax ax ax I 1 nπx a cosgive a one-sentence physical interpretation of this problem. mπx nπx cos + mπx dx b Solve by the method of separation of variables. First show that there 1 are πx no n separated m solutions πx n which + m cos cos exponentially dx grow in time. [Hint: The answer is 1 sin πxn m 1 ux, sin πxn+m t Ao + > cos nix. πn m πn+m What is An? πx n m sin π n m π n + m sin But πx n m sin sin π n m sin And since n m is integer, then sin π n m, therefore sin πx n + m sin sin π n + m sin πx n + m 1 πxn m. Similarly Note that the term n m showing in the denominator is not a problem now, since this is the case where n m. 34

35 sin sin dx forn>,m>. Use the trigonometric identity Since n + *.3.6. m is integer Evaluate then sin π n + m and n7rx m7rx cog nπx cc mπx sin sin dx Use the trigonometric identity sin asin b [cosa - b - cosa + b]. sin πxn+m. Therefore { n m otherwise dx for n > O, m >.. HW..7 section.3.7 problem 6 cos a cos b [cosa + b + cosa - b]. Be careful if a - b or a + b Consider the following boundary value problem if necessary, see Sec..4.1: k 8U with au, to, au, t, and ux, f x. at ax ax ax a Give a one-sentence physical interpretation of this problem. b Solve by the method of separation of variables. First show that there are no separated solutions which exponentially grow in time. [Hint: The answer is What is An? ux, t Ao + > cos nix. part a This PDE describes how temperature u changes in a rod of length as a function of time t and location x. The left and right end are insulated, so no heat escapes from these boundaries. Initially at t, the temperature distribution in the rod is described by the function f x. Part b u t k u x et u x, t T t X x, then the PDE becomes 1 k T X X T Dividing by XT 1 T k T X X Since each side depends on different independent variable and both are equal, they must be both equal to same constant, say λ. Where λ is assumed real. The two ODE s generated are 1 T k T X X λ T + kλt 1 X + λx Starting with the space ODE equation, with corresponding boundary conditions dx dx, dx dx. Assuming the solution is X x erx, Then the characteristic equation is r + λ r λ r ± λ The following cases are considered. case λ < In this case, λ and also λ are positive. Hence the roots ± λ are both real. et λ s 35

36 HWs Where s >. This gives the solution Applying the left B.C. gives X x A cosh sx + B sinh sx dx A sinh sx + B cosh sx dx dx dx B cosh B The solution becomes X x A cosh sx and hence dx dx gives dx dx A sinh s A sinh sx. Applying the right B.C. A result in trivial solution. Therefore assuming sinh s implies s which is not valid since s > and. Hence only trivial solution results from this case. λ < is not an eigenvalue. case λ The ODE becomes d X dx The solution is Applying left boundary conditions gives X x c 1 x + c dx dx c 1 dx dx c 1 Hence the solution becomes X x c. Therefore dx dx. Applying the right B.C. provides no information. Therefore this case leads to the solution X x c. Associated with this one eigenvalue, the time equation becomes dt dt hence T is constant, say α. Hence the solution u x, t associated with this λ is u x, t X T c α A where constant c α was renamed to A to indicate it is associated with λ. λ is an eigenvalue. case λ > Hence λ is negative, and the roots are both complex. The solution is Applying the left B.C. gives X x A cos λx r ±i λ + B sin λx dx dx A λ sin λx + B λ cos λx dx dx B λ cos B λ 36

37 λx Therefore B as λ >. The solution becomes X x A cos and dx Applying the right B.C. gives dx dx A λ sin λ λ A gives a trivial solution. Selecting sin gives λ nπ n 1,, 3, dx A λ sin. HW λx. Or nπ λ n n 1,, 3, Therefore the space solution is nπ X n x A n cos x n 1,, 3, The time solution is found by solving This has the solution dt n dt + kλ nt n T n t e kλnt e k nπ t n 1,, 3, For the same set of eigenvalues. Notice that no need to add a constant here, since it will be absorbed in the A n when combined in the following step below. Since for λ the time solution was found to be constant, and for λ > the time solution is e k nπ t, then no time solution will grow with time. Time solutions always decay with time as the exponent k nπ t is negative quantity. The solution to the PDE for λ > is u n x, t T n t X n x n, 1,, 3, But for linear system sum of eigenfunctions is also a solution. Hence Part c u x, t u λ x, t + A + From the solution found above, setting t gives Therefore, f x must satisfy the above Part d u x, A + f x A + u n x, t nπ A n cos x e k nπ t nπ A n cos x nπ A n cos x Multiplying both sides with cos mπ x where in this problem m, 1,, since there was an eigenvalue associated with λ, and integrating over the domain gives mπ f x cos x dx mπ cos x A + mπ A cos x dx + mπ A cos x dx + nπ A n cos x dx mπ cos x A n cos nπ A n cos x dx mπ x cos nπ x dx 37

38 HWs Interchanging the order of summation and integration mπ f x cos x dx mπ A cos x dx + A n case m When m then cos mπ x 1 and the above simplifies to f x dx A dx + But cos nπ x dx and the above becomes f x dx A n A A dx mπ nπ cos x cos x dx 1 nπ cos x dx Therefore A 1 f x dx case m > From 1, one term survives in the integration when only n m, hence mπ f x cos x dx A But cos mπ x dx and the above becomes Therefore For n 1,, 3, Part e The solution was found to be mπ cos x dx + A m mπ f x cos x dx A m A n f x cos nπ x dx cos mπ x dx nπ u x, t A + A n cos x e k nπ t.3. Heat Equation With Zero Temperature Ends 57 In the limit as t the term e k c Show that the initial nπ t condition,. What ux, is left f is x, A is. satisfied But A if 1 f x dx from above. This quantity is the average of the initial temperature. f x Ao + E A. cos..8 section.3.8 d Using problem Exercise 7.3.6, solve for AO and Ann > 1. e What happens to the temperature distribution as t -+ oo? Show that it approaches the steady-state temperature distribution see Sec *.3.8. Consider 8u u & kax - au. This corresponds to a one-dimensional rod either with heat loss through the lateral sides with outside temperature a >, see Exercise 1..4 or with insulated lateral sides with a heat sink proportional to the temperature. Suppose that the boundary conditions are u,t and u,t. a What are the possible equilibrium temperature distributions if a >? b Solve the time-dependent problem [ux, f x] if a >. Analyze the temperature for large time t --+ oo and compare to part a. *.3.9. Redo Exercise.3.8 if a <. [Be especially careful if -a/k n7r/.].3.1. For two- and three-dimensional vectors, the fundamental property of dot products, A B IAI[BI cos9, implies that IA - BI < IAIIBI In this exercise we generalize this to n-dimensional vectors and functions, in which case.3.44 is known as Schwarz's inequality. [The names of Cauchy and Buniakovsky are also associated with.3.44.] a Show that IA - -ybi > implies.3.44, where ry A B/B B.

39 . HW part a Equilibrium is at steady state, which implies u t and the PDE becomes an ODE, since u u x at steady state. Hence d u dx α k u The characteristic equation is r α k or r ± α k. Since α > and k > then the roots are real, and the solution is α k u A e x + B e α k x This can be rewritten as Applying left B.C. gives α α u x A cosh k x + B sinh k x u A cosh A The solution becomes u x B sinh α k x. Applying the right boundary condition gives u B sinh α k B leads to trivial solution. Setting sinh α k implies α k. But this is not possible since. Hence the only solution possible is u x Part b u t k u x αu u t + αu k u x Assuming u x, t X x T t and substituting in the above gives Dividing by kxt XT + αxt kt X T kt + α k X X Since each side depends on different independent variable and both are equal, they must be both equal to same constant, say λ. Where λ is assumed real. The two ODE s are Or 1 T k T + α k X X λ 1 T k T + α k λ X X λ T + α + λk T X + λx The solution to the space ODE is the familiar where λ > is only possible case, As found in problem.3.3, part d. Since it has the same B.C. nπ X n B n sin x n 1,, 3, 39

40 HWs Where λ n nπ. The time ODE is now solved. This has the solution dt n dt + α + λ nk T n T n t e α+λnkt e αt e nπ For the same eigenvalues. Notice that no need to add a constant here, since it will be absorbed in the B n when combined in the following step below. Therefore the solution to the PDE is kt u n x, t T n t X n x But for linear system sum of eigenfunctions is also a solution. Hence u x, t u n x, t nπ B n sin x e αt e nπ kt nπ e αt B n sin x e nπ kt Where e αt was moved outside since it does not depend on n. From initial condition nπ.3. Heat Equation With u x, Zero Temperature f x Ends B n sin x 57 c Show that the initial condition, ux, f x, is satisfied if Applying orthogonality of sin as before to find B n results in B n f x Ao + E nπ A. cos sin x f x dx d Using Exercise.3.6, solve for AO and Ann > 1. Hence the solutione becomes What happens to the temperature distribution as t -+ oo? Show that it approaches the steady-state temperature distribution see Sec. u x, t [ nπ ] 1.4. nπ e αt sin x f x dx sin x e *.3.8. Consider nπ kt 8u u This corresponds to a one-dimensional rod either with heat loss through the Hence it is clear that in the limit as t becomes large u x, t since the sum is multiplied by e αt lateral sides with outside temperature a >, see Exercise 1..4 or with and α > insulated lateral sides with a heat sink proportional to the temperature. Suppose that the boundary limconditions u x, t are & kax - au. t u,t and u,t. This agrees with part a a What are the possible equilibrium temperature distributions if a >?..9 section.3.1 b Solve problem the time-dependent 8 problem [ux, f x] if a >. Analyze the temperature for large time t --+ oo and compare to part a. 4 *.3.9. Redo Exercise.3.8 if a <. [Be especially careful if -a/k n7r/.].3.1. For two- and three-dimensional vectors, the fundamental property of dot products, A B IAI[BI cos9, implies that IA - BI < IAIIBI In this exercise we generalize this to n-dimensional vectors and functions, in which case.3.44 is known as Schwarz's inequality. [The names of Cauchy and Buniakovsky are also associated with.3.44.] a Show that IA - -ybi > implies.3.44, where ry A B/B B. b Express the inequality using both b Cn *c Generalize.3.44 to functions. [Hint: et A A. B mean the integral J AxBx dx.] Solve aplace's equation inside a rectangle: V U + 8y axe u u n. subject to the boundary conditions u,y gy ux, u, y ux, H. Hint: If necessary, see Sec..5.1.

41 . HW A γb A γb A γb Since A γb then A γb A γb Expanding A A γ A B γ B A + γ B B But A B B A, hence A A γ A B + γ B B Using the definition of γ A B B B A A A B into the above gives B B A B + A B B B B B A A A B A B B B + B B A B A A B B A A B B A B A A B B A B But A B A B since A B is just a number. The above becomes A A B B A B And A A A and B B B by definition as well. Therefore the above becomes A B A B Taking square root gives A B A B Which is Schwarz s inequality. Part b From the norm definition Then A x + y + z A A A x + y + z Hence And A B A B a n b n a n b n Therefore the inequality can be written as A B A B Part c a n b n a n Using A B for functions to mean A x B x dx then inequality for functions becomes A x B x dx A x dx B x dx b n 41

42 ux,6+4cos31rx HWs Coefficients Bn 1 A I!: dx /r` O / fz.in n- dx Z nxs An J /z toy dz O an - En I f.- nrs dz I. /!ssfn nva ds 1..1 section.4.1 problem 9 EXERCISES.4 *.4.1. Solve the heat equation 8u/8t k8u/8x, < x <, t >, subject to 8xO,t t>, t t>. a ux, x < / 1 x>/ b c ux, - sin d ux, -3 cos jx The same boundary conditions was encountered in problem.3.7, therefore the solution used here starts from the same general solution already found, which is Part b λ nπ λ n n 1,, 3, nπ u x, t A + A n cos x e k nπ t u x, cos 3πx Comparing terms with the general solution at t which is results in u x, A + And all other A n. Hence the solution is Part c Hence A 6 A 3 4 u x, t cos nπ A n cos x 3π x u x, sin πx sin πx A + e k 3π t nπ A n cos x Multiplying both sides of 1 by cos mπ x and integrating gives πx mπ mπ mπ sin cos x dx A cos x + cos x mπ A cos x dx + Interchanging the order of integration and summation sin πx mπ cos x dx mπ A cos x dx + A n cos nπ x dx 1 mπ nπ A n cos x cos x dx A n mπ nπ cos x cos x dx 4

43 . HW Case m The above becomes sin But cos nπ x dx hence sin πx dx πx dx A dx + A dx A n πx A sin A cos πx π π cos dx π nπ cos x dx + cos π π π Hence Case m > A 4 π sin πx mπ cos x dx One term survives the summation resulting in πx mπ sin cos x dx 4 π But cos mπ x dx and cos mπ But Therefore sin Hence the solution becomes πx sin cos mπ A cos x dx + A n mπ cos x x dx, therefore mπ x dx A m A n 4 sin dx + A m mπ nπ cos x cos x dx πx nπ cos x dx πx nπ cos x 1 + cos nπ dx π n cos nπ A n 4 π n 1 4 1n + 1 π n 1 u x, t 4 π + 4 π..11 section.4. problem 1 n 1,, 3, 1 n + 1 n 1 cos nπ x e k nπ t cos mπ x dx 7 Chapter. Method of Separation of Variables *.4.. Solve z k8-z with 8, t u, t ux, fx For this problem you may assume that no solutions of the heat equation exponentially grow in time. You may also guess appropriate orthogonality conditions for the eigenfunctions. *.4.3. Solve the eigenvalue problem subject to d, - _AO dx 43

44 HWs u t u κ x et u x, t T t X x, then the PDE becomes 1 κ T X X T Dividing by XT 1 T κ T X X Since each side depends on different independent variable and both are equal, they must be both equal to same constant, say λ. Where λ is real. The two ODE s are 1 T κ T X X λ T + kλt 1 X + λx Per problem statement, λ, so only two cases needs to be examined. Case λ The space equation becomes X with the solution X Ax + b Hence left B.C. implies X or A. Therefore the solution becomes X b. The right B.C. implies X or b. Therefore this leads to X as the only solution. This results in trivial solution. Therefore λ is not an eigenvalue. Case λ > Starting with the space ODE, the solution is X x A cos λx + B sin λx dx dx A λ sin λx + B λ cos λx eft B.C. gives dx dx B λ Hence B since it is assumed λ and λ >. Solution becomes X x A cos λx Applying right B.C. gives X A cos λ λ A leads to trivial solution. Therefore cos or Hence Therefore λ nπ λ n And the corresponding time solution n 1, 3, 5, nπ n 1, 3, 5, nπ X n x A n cos x T n e k nπ t n 1, 3, 5, 44

45 . HW Hence From initial conditions u n x, t X n T n u x, t f x,3,5,,3,5, Multiplying both sides by cos mπ x and integrating mπ f x cos x dx nπ A n cos x e k nπ t,3,5, nπ A n cos x mπ nπ A n cos x cos x dx Interchanging order of summation and integration and applying orthogonality results in mπ f x cos x dx A m 7 Chapter A n. Method nπ of Separation of Variables f x cos x dx *.4.. Solve Therefore the solution is z k8-z with 8, t u x, t [ nπ ] nπ f x cos x dx cos x e k nπ t u, t,3,5, ux, fx..1 section.4.3 For this problem problem 11 you may assume that no solutions of the heat equation exponentially grow in time. You may also guess appropriate orthogonality conditions for the eigenfunctions. *.4.3. Solve the eigenvalue problem subject to d, - _AO dx 7r and ;jjo 1r. dx.4.4. Explicitly show that there are no negative eigenvalues for do x _ -A subject to dz and. dφ.4.5. This problem presents dx an + λφ alternative derivation of the heat equation for a thin wire. The equation φ for a circular φ π wire of finite thickness is the twodimensional heat equation dφ in polar dφ coordinates. Show that this reduces to.4.5 if the temperature dx does dx π not depend on r and if the wire is very thin. First solution using transformation et τ.4.6. x π, hence Determine the above the system equilibrium becomes temperature distribution for the thin circular ring of Section.4.: dφ a Directly from the dτequilibrium + λφ problem see Sec. 1.4 b By computing the φ limit π as t φ -π oo of the time-dependent problem dφ dφ.4.7. Solve aplace's equation π inside dτ a dτ circle π of radius a, The characteristic equation is r + λ I.9U V U or r ± λ. Assuming λ is real. r Or r 8r + rz 9, There are three cases to consider. Case λ < et s λ > subject to the boundary condition ua,9 f9. φ τ c 1 cosh sτ + c sinh sτ Hint: If necessary, φ τ see sc 1 Sec. sinh.5.. sτ + sc cosh sτ 45

46 HWs Applying first B.C. gives Applying second B.C. gives φ π φ π c 1 cosh sπ c sinh sπ c 1 cosh sπ + c sinh sπ c sinh sπ c sinh sπ 1 φ π φ π sc 1 sinh sπ + sc cosh sπ sc 1 sinh sπ + sc cosh sπ c 1 sinh sπ c 1 sinh sπ Since sinh sπ is zero only for sπ and sπ is not zero because s >. Then the only other option is that both c 1 and c in order to satisfy equations 1. Hence trivial solution. Hence λ < is not an eigenvalue. Case λ The space equation becomes dφ with the solution φ τ Aτ + B. Applying the first B.C. dτ gives φ π φ π Aπ + B Aπ + B Aπ Hence A. The solution becomes φ τ B. And φ τ. The second B.C. just gives. Therefore the solution is φ τ C Where C is any constant. Hence λ is an eigenvalue. Case λ > Applying first B.C. gives Applying second B.C. gives φ τ c 1 cos λτ + c sin λτ φ τ c 1 λ sin λτ + c λ cos λτ φ π φ π c 1 cos λπ c sin λπ c 1 cos λπ + c sin λπ c sin λπ c sin λπ 3 φ π φ π c 1 λ sin λπ + c λ cos λπ c 1 λ sin λπ + c λ cos λπ c 1 λ sin λπ c 1 sin λπ Both 3 and can be satisfied for non-zero λπ. The trivial solution is avoided. Therefore the eigenvalues are sin λπ λn π nπ n 1,, 3, Hence the corresponding eigenfunctions are { } cos λn τ, sin λn τ λ n n n 1,, 3, {cos nτ, sin nτ} 46

47 . HW Transforming back to x using τ x π {cos n x π, sin n x π} {cos nx nπ, sin nx nπ} But cos x π cos x and sin x π sin x, hence the eigenfunctions are { cos nx, sin nx} The signs of negative on an eigenfunction or eigenvector do not affect it being such as this is just a multiplication by 1. Hence the above is the same as saying the eigenfunctions are Summary {cos nx, sin nx} eigenfunctions λ arbitrary constant λ > {cos nx, sin nx} for n 1,, 3 Second solution without transformation note: Using transformation as shown above seems to be easier method than this below. The characteristic equation is r + λ or r ± λ. Assuming λ is real. There are three cases to consider. Case λ < In this case λ is positive and the roots are both real. Assuming λ s where s >, then the solution is First B.C. gives The second B.C. gives φ x Ae sx + Be sx φ x Ase sx Bse sx φ φ π A + B Ae sπ + Be sπ A 1 e sπ + B 1 e sπ 1 φ φ π As Bs Ase sπ Bse sπ A 1 e sπ + B 1 + e sπ After dividing by s since s. Now a by system is setup from 1, 1 e sπ 1 e sπ 1 e sπ A 1 + e sπ B Since this is Mx b with b then for non-trivial solution M must be zero. Checking the determinant to see if it is zero or not: 1 e sπ 1 e sπ 1 se sπ 1 + se sπ 1 e sπ 1 + e sπ 1 e sπ 1 e sπ 1 + e sπ + e sπ 1 1 e sπ e sπ e sπ + e sπ e sπ + e sπ e sπ + e sπ 4 + e sπ + e sπ cosh sπ Hence for the determinant to be zero so that non-trivial solution exist then cosh sπ or cosh sπ 1 which has the solution sπ. Which means s. But the assumption was that s >. This implies only a trivial solution exist and λ < is not an eigenvalue. case λ The space equation becomes dφ with the solution φ x Ax + B. Applying the first B.C. dx gives B Aπ + B Aπ 47

48 HWs Hence A. The solution becomes φ x B. And φ x. The second B.C. just gives. Therefore the solution is φ x C Where C is any constant. Hence λ is an eigenvalue. Case λ > In this case the solution is Applying first B.C. gives φ x A cos λx φ x A λ sin A 1 cos π λ B sin π λ Applying second B.C. gives A λ sin π λ A sin + B λ λ cos π λ + B 1 cos + B sin λx λx + B λ cos φ φ π A A cos φ φ π π λ B λ A λ sin π λ π λ λx + B sin π λ π λ + B λ cos π λ Therefore 1 cos π λ sin π λ sin π λ 1 cos π λ A B 3 Setting M to obtain the eigenvalues gives 1 cos π λ 1 cos π λ + sin π λ sin π 7 Chapter. Method of Separation of λ Variables 1 cos π *.4.. Solve λ Hence Or cos π λ 1 π λ n nπ n, 4, For this problem you may assume λn n that no solutions of the heat equation exponentially grow in time. You may also n guess, 4, appropriate orthogonality conditions for the eigenfunctions. *.4.3. Solve the eigenvalue problem λn n n 1,, 3, λ n n d, - _AO dxn 1,, 3, Therefore the eigenfunctions subject to are φ n x 7r {cos and nx,;jjo sin nx} 1r. dx Summary.4.4. Explicitly show that there are no negative eigenvalues for do eigenfunctions _ -A subject to dz and. λ x arbitrary constant λ > {cos nx, sin nx} for n 1,, This problem presents an alternative derivation of the heat equation for a thin wire. The equation for a circular wire of finite thickness is the two-.4.6 problem heat equation 1 in polar coordinates. Show that this reduces..13 sectiondimensional to.4.5 if the temperature does not depend on r and if the wire is very thin. z k8-z with 8, t u, t ux, fx.4.6. Determine the equilibrium temperature distribution for the thin circular ring of Section.4.: 48 a Directly from the equilibrium problem see Sec. 1.4 b By computing the limit as t - oo of the time-dependent problem.4.7. Solve aplace's equation inside a circle of radius a, V U subject to the boundary condition I.9U r Or r 8r + rz 9,

49 . HW The PDE for the thin circular ring is Part a At equilibrium u t and the PDE becomes u t k u x u, t u, t u, t u, t t t u x, f x u x As it now has one independent variable, it becomes the following ODE to solve Solution to d u dx is d u x dx u u du du dx dx u x c 1 x + c Where c 1, c are arbitrary constants. From the first B.C. Hence the solution becomes u u c 1 + c c 1 + c c 1 c 1 u x c The second B.C. adds nothing as it results in. Hence the solution at equilibrium is u x c This means at equilibrium the temperature in the ring reaches a constant value. Part b The time dependent solution was derived in problem.4.3 and also in section.4, page 6 in the book, given by u x, t a + a n cos nπx e k nπx t + a n sin As t the terms e k nπx t and the above reduces to u x, a Since a is constant, this is the same result found in part a. nπx e k nπx t 49

50 Using the two-dimensional divergence theorem, we conclude that see Exercise i Vu-ft ds HWs Since is proportional to the heat flow through the boundary,.5.61 implies that the net heat flow through the boundary must be zero in order for a steady state to exist. This is clear physically, because otherwise there would be a change.3 HWin time 3 of the thermal energy inside, violating the steady-state assumption. Equation.5.61 is called the solvability condition or compatibility condition for.3.1 Problem aplace's.5.1e equation. problem 1 EXERCISES Solve aplace's equation inside a rectangle < x <, < y < H, with the following boundary conditions:,y,,y, *a ax- Tx- b Tx- O, y 9y, TX-, y, *c "',y, u,y 9y, d uo,y 9y, u,y, *e u,y, u,y, ux,, ux,, ux,, x,, Fyux, - x,, ux, H f x ux, H ux, H ux, H ux, H f x f uo, y f y, u, y, x,, TV- g TX-, y, YX-, y, ux / x > / au et u x, y X x Y x. Substituting this into the PDE u yy- x, H x < /' and simplifying gives u x y "u x, H Consider ux, y satisfying aplace's X equation inside a rectangle < x < Y, < y < H subject to the boundary X Yconditions Each side depends on different independent,y Ou x, Yxvariable and they are equal, therefore they must be equal to same constant.,y X x, H f x. Y Without solving this X problem, Y briefly ±λ explain the physical condition Since the boundary conditions under which along there the is a xsolution direction to this are problem. the homogeneous ones, λ is selected in the above. Two ODE s b Solve 1, this are problem obtained by the asmethod followsof separation of variables. Show that the method works only under the condition of part a. X + λx 1 With the boundary conditions X X And With the boundary conditions Y λy Y Y Y H f x In all these cases λ will turn out to be positive. This is shown for this problem only and not be repeated again. The solution to 1 is Case λ < X Ae λx + Be λx X A cosh λx + B sinh λx λx λ At x, the above gives A. Hence X B sinh. At x this gives X B sinh. λ But sinh only at and λ, therefore B and this leads to trivial solution. Hence λ < is not an eigenvalue. Case λ X Ax + B Hence at x this gives B and the solution becomes X B. At x, B. Hence the trivial solution. λ is not an eigenvalue. Case λ > Solution is X A cos λx + B sin λx 5

51 .3 HW 3 λx At x this gives A and the solution becomes X B sin. At x B sin λ λ For non-trivial solution sin or λ nπ where n 1,, 3,, therefore Eigenfunctions are nπ λ n n 1,, 3, nπ X n x B n sin x For the Y ODE, the solution is nπ nπ Y n C n cosh y + D n sinh y Y n nπ nπ C n sinh y nπ nπ + D n cosh y Applying B.C. at y gives The eigenfunctions Y n are Now the complete solution is produced u n x, y Y n X n Y Y nπ C n cosh D n cosh nπ C n D n Y n D n nπ cosh nπ y D n nπ cosh nπ y + sinh D n nπ cosh nπ y + sinh n 1,, 3, 3 nπ + D n sinh y nπ y nπ nπ y B n sin x et D n B n B n since a constant. no need to make up a new symbol. nπ nπ nπ nπ u n x, y B n cosh y + sinh y sin x Sum of eigenfunctions is the solution, hence nπ nπ nπ nπ u x, y B n cosh y + sinh y sin x The nonhomogeneous boundary condition is now resolved. Therefore f x u x, H f x At y H nπ nπ nπ nπ B n cosh H + sinh H sin x Multiplying both sides by sin mπ x and integrating gives mπ mπ f x sin x dx sin x nπ nπ nπ nπ B n cosh H + sinh H sin x dx Hence nπ nπ nπ B n cosh H + sinh H mπ mπ mπ B m cosh H + sinh H nπ mπ sin x sin x dx B n f x sin nπ x dx nπ cosh nπ H + sinh nπ 4 H This completes the solution. In summary u x, y nπ nπ nπ nπ B n cosh y + sinh y sin x With B n given by 4. The following are some plots of the solution above for different f x. 51

52 HWs Figure.1: Solution using fx x, 1, H 1 Figure.: Solution using fx sin1x, 1, H 1 Figure.3: Solution using fx cos4x, 1, H 1 Figure.4: Solution using fx sin3x cosx, 5, H 1 5

53 *c "',y, u,y 9y, d uo,y 9y, u,y, *e u,y, u,y, f uo, y f y, u, y,.3. Problem.5. problem g TX-, y, YX-, y, ux,, x,, Fyux, - x,, x,, TVux / ux, H ux, H ux, H f x.3 HW 3 "u x, H x > / au yy- x, H x < /'.5.. Consider ux, y satisfying aplace's equation inside a rectangle < x <, < y < H subject to the boundary conditions b,y Yx-,y Ou x, x, H f x. Without solving this problem, briefly explain the physical condition under which there is a solution to this problem. Solve this problem by the method of separation of variables. Show that the method works only under the condition of part a. 86 Chapter. Method of Separation of Variables c The solution [part b] has an arbitrary constant. Determine it by consideration of the time-dependent heat equation subject to the initial condition ux,y, gx,y *.5.3. Solve aplace's equation outside a circular disk r > a subject to the boundary condition part a a ua, 9 In + 4 cos 39 At steady state, b ua,9 there will f9 be no heat energy flowing across the boundaries. Which implies the flux is zero. Three of the boundaries are already insulated and hence the flux is zero at those boundaries as given. Therefore, You may the assume flux should that ur, also 9 be remains zero atfinite the top as r boundary - oo. at steady state. *.5.4. By definition, For aplace's the flux isequation φ k u inside ˆn. Direction a circular of disk fluxr vector < a, isusing from.5.45 hot to cold. and At the top boundary,.5.47, this becomes show that φ k u x, H 1 y ur,9 f6 +Eancosn9-81 db. Therefore, For the condition of a solution, a total flux on the boundary is zero, or n_ Using cos z Re [ei], sum the resulting φdx geometric series to obtain Poisson's integral formula. Using 1 in the above gives.5.5. Solve aplace's equation inside the quarter-circle of radius 1 < <- 7r/, < r < 1 subject to the boundary u conditions k x, H dx y * a r,, u r,, u1, f O u x, H dx b Ou r,, 6u r, z y, u1, f But u y Part b x, H f x and the above becomes * c ur,, u r, z, Ou 1, 9 f O d r, o o, r, o, 1, e ge f x dx Show that the solution [part d] exists only if fo g9 d9. Explain this condition physically. Using.5.6. separation Solve aplace's of variables equation results inside the following a semicircle two of ODE s radius a < r < a, < 9 < a subject to the boundary conditions X + λx *a u on the diameter and ua, X 9 g9 X b the diameter is insulated and ua, g9 And.5.7. Solve aplace's equation inside a 6 wedge of radius a subject to the boundary conditions Y λy a ur,, u r, a Y, ua, 9 f Y f x * b r,, r, 3, ua, 9 f The solution to the X x ODE has been obtained before as X n A + A n cos λn x n 1,, 3, X n A n cos λn x n, 1,, 3, 1 53

54 HWs Where λ n nπ. In this ODE λ is applicable as well as λ >. As found in last HW. Now the Y y ODE is solved for same set of eigenvalues. For λ the ODE becomes Y and solution is Y Cy + D. Hence Y C and since Y then C. Hence the solution is Y C, where C is some new constant. For λ >, the solution is Y n C n cosh λn y + D n sinh λn y n 1,, 3, Y n C n λn sinh λn y + D n λn cosh λn y At y Y n D n λn n 1,, 3, Since λ n > for n 1,, 3, then D n and the Y y solution becomes Y n C + C n cosh λn y n 1,, 3, Y n C n cosh λn y n, 1,, 3, Combining 1 and gives u n x, y X n Y n A n cos λn x C n cosh λn y A n cos λn x cosh λn y n, 1,, 3, n, 1,, 3, Where A n C n above was combined and renamed to A n No need to add new symbol. Hence by superposition the solution becomes u x, y n A n cos λn x cosh λn y Since λ and cos λ x cosh λ y 1, the above can be also be written as u x, y A + At y H, it is given that u y x, H f x. But At y H the above becomes nπ nπ A n cos x cosh y u y nπ nπ nπ A n cos x sinh y f x To verify part a by integrating both sides f x dx But cos nπ x dx, hence nπ nπ nπ A n cos x sinh H nπ nπ nπ A n cos x sinh H dx nπ nπ A n sinh H f x dx nπ cos x dx The verification is completed. Now back to 4 and multiplying by cos mπ x and integrating mπ nπ f x cos x dx A n cos x nπ λn sinh H dx nπ nπ A n sinh H cos x λn dx A m sinh mπ H

55 .3 HW 3 Hence A n f x cos nπ x dx sinh nπ n 1,, 3, H Therefore the solution now becomes from 3 u x, y A + f x cos nπ x dx sinh nπ H Only A remains to be found. This is done in next part. Part c nπ nπ cos x cosh y Since at steady state, total energy is the same as initial energy. Initial temperature is given as g x, y, therefore initial thermal energy is found by integrating over the whole domain. This is D, therefore H ρcg x, y da ρc g x, y dydx H Setting the above to ρc u x, y dydx found in last part, gives one equation with one unknown, which is A to solve for. Hence But ρc H H H H g x, y dydx ρc g x, y dydx H H g x, y dydx A H + A + A dydx + A n nπ nπ A n cos x cosh y dydx H H cos nπ nπ H cos x cosh y dydx cosh nπ nπ A n cos x cosh y dydx nπ nπ x cosh y dydx 5 nπ y nπ cos x dx dy Where cos nπ x dx. Hence the whole sum vanish. Therefore 5 reduces to H Summary The complete solution is u x, y 1 H H g x, y dydx A H g x, y dydx + A 1 H The following are some plots of the solution. H g x, y dydx f x cos nπ x dx sinh nπ H nπ nπ cos x cosh y Figure.5: Solution using gx, y xy, fx sin3x, 5, H 1 55

56 HWs Figure.6: Solution using gx, y siny cosxy, fx x, 5, H 1 86 Chapter. Method of Separation of Variables c The solution [part b] has an arbitrary constant. Determine it by consideration of the time-dependent heat equation subject to the initial condition ux,y, gx,y *.5.3. Solve aplace's equation outside a circular disk r > a subject to the boundary condition a ua, 9 In + 4 cos 39 b ua,9 f9 You may assume that ur, 9 remains finite as r - oo. *.5.4. For aplace's equation inside a circular disk r < a, using.5.45 and Figure.7:.5.47, Solution show using that gx, y y siny cosxy, fx sin1x, 1, H 1 ur,9 f6 +Eancosn9-81 db..3.3 Problem.5.5c,d problem 3 Using cos z Re [ei], sum the resulting geometric series to obtain Poisson's integral formula. Part c a n_.5.5. Solve aplace's equation inside the quarter-circle of radius 1 < <- 7r/, < r < 1 subject to the boundary conditions * a r,, u r,, u1, f O b Ou r,, 6u r, z, u1, f * c ur,, u r, z, Ou 1, 9 f O d r, o o, r, o, 1, e ge Show that the solution [part d] exists only if fo g9 d9. Explain this condition physically Solve aplace's equation inside a semicircle of radius a < r < a, < 9 < a subject to the boundary conditions *a u on the diameter and ua, 9 g9 b the diameter is insulated and ua, g9 The aplace PDE in polar coordinates is.5.7. Solve aplace's equation inside a 6 wedge of radius a subject to the boundary conditions r u r + r u r + u θ A a ur,, u r, a, ua, 9 f With boundary conditions * b r,, r, 3, ua, 9 f u r, u r, π B Assuming the solution can be written as u 1, θ f θ u r, θ R r Θ θ 56

57 .3 HW 3 And substituting this assumed solution back into the A gives Dividing the above by RΘ gives r R Θ + rr Θ + RΘ r R R + r R R + Θ Θ r R R + r R R Θ Θ Since each side depends on different independent variable and they are equal, they must be equal to same constant. say λ. r R R + r R R Θ Θ λ This results in the following two ODE s. The boundaries conditions in B are also transferred to each ODE. This gives And Θ + λθ Θ 1 π Θ r R + rr λr R < Starting with 1. Consider the Case λ <. The solution in this case will be Θ A cosh λθ + B sinh λθ λθ Applying first B.C. gives A. The solution becomes Θ B sinh. Applying second B.C. gives B sinh π λ But sinh is zero only when λ π which is not the case here. Therefore B and hence trivial solution. Hence λ < is not an eigenvalue. Case λ The ODE becomes Θ with solution Θ Aθ + B. First B.C. gives B. The solution becomes Θ Aθ. Second B.C. gives A π, hence A and trivial solution. Therefore λ is not an eigenvalue. Case λ > The ODE becomes Θ + λθ with solution Θ A cos λθ + B sin λθ The first B.C. gives A. The solution becomes Θ B sin λθ And the second B.C. gives B sin λ π For non-trivial solution sin or λ π And the eigenfunctions are λn n π λ nπ for n 1,, 3,. Hence the eigenvalues are λ n 4n n 1,, 3, Θ n θ B n sin nθ n 1,, 3, 3 Now the R ODE is solved. There is one case to consider, which is λ > based on the above. The ODE is r R + rr λ n R r R + rr 4n R n 1,, 3, 57

58 HWs This is Euler ODE. et R r r p. Then R pr p 1 and R p p 1 r p. This gives r p p 1 r p + r pr p 1 4n r p p p r p + pr p 4n r p r p p pr p + pr p 4n r p p 4n p ±n Hence the solution is R r Cr n + D 1 r n Applying the condition that R < implies D, and the solution becomes Using 3,4 the solution u n r, θ is R n r C n r n n 1,, 3, 4 u n r, θ R n Θ n C n r n B n sin nθ B n r n sin nθ Where C n B n was combined into one constant B n. No need to introduce new symbol. The final solution is u r, θ u n r, θ B n r n sin nθ Now the nonhomogeneous condition is applied to find B n. Hence r u 1, θ f θ becomes r u r, θ B n n r n 1 sin nθ f θ B n n sin nθ Multiplying by sin mθ and integrating gives When n m then π π f θ sin mθ dθ π sin mθ sin nθ dθ sin mθ nb n π π π B n n sin nθ dθ sin mθ sin nθ dθ 5 sin nθ dθ 1 1 [ sin 4nθ 1 [θ] π 1 π 4 1 8n cos 4nθ dθ ] π 4n sin 4n π sin And since n is integer, then sin 4n π sin nπ and the above becomes π 4. 58

59 .3 HW 3 Now for the case when n m using sin A sin B 1 cos A B cos A + B then π sin mθ sin nθ dθ 1 π π π 1 cos mθ nθ cos mθ + nθ dθ cos mθ nθ dθ 1 π π cos mθ + nθ dθ 1 cos m n θ dθ 1 cos m + n θ dθ 1 [ ] π sin m n θ 1 [ ] π sin m + n θ m n m + n 1 4 m n [sin m n θ] π 1 4 m + n [sin m + n θ] π 1 [ sin m n π ] 1 [ sin m + n π 4 m n 4 m + n ] Since m n π π m n which is integer multiple of π and also m + n π of π then the whole term above becomes zero. Therefore 5 becomes is integer multiple Hence π f θ sin mθ dθ mb m π 4 B n πn π f θ sin nθ dθ Summary: the final solution is u r, θ B n r n sin nθ B n πn The following are some plots of the solution π f θ sin nθ dθ Figure.8: Solution using fθ θ sin3θ Figure.9: Solution using fθ θ 59

60 HWs Part d The aplace PDE in polar coordinates is With boundary conditions r u r + r u r + u θ Assuming the solution is Substituting this back into the PDE gives Dividing by RΘ gives u r, u r, π u 1, θ f θ u r, θ R r Θ θ r R Θ + rr Θ + RΘ r R R + r R R + Θ Θ r R R + r R R Θ Θ Since each side depends on different independent variable and they are equal, they must be equal to same constant. say λ. r R R + r R R Θ Θ λ This results in two ODE s with the following boundary conditions And Θ + λθ Θ 1 Θ π r R + rr λr R < Starting with 1. Consider Case λ < The solution will be Θ A cosh λθ + B sinh λθ And Θ A λ sinh λθ + B λ cosh λθ Applying first B.C. gives B λθ λ, therefore B and the solution becomes A cosh and Θ A λθ λ sinh. Applying second B.C. gives A λ π λ π λ sinh. But sinh since λ, therefore A and the trivial solution results. Hence λ < is not an eigenvalue. Case λ The ODE becomes Θ With solution Θ Aθ + B And Θ A. First B.C. gives A. Hence Θ B. Second B.C. produces no result and the solution is constant. Hence Θ C Where C is constant. Therefore λ is an eigenvalue. Case λ > The ODE becomes Θ + λθ with solution Θ A cos λθ + B sin λθ Θ A λ sin λθ + B λ cos λθ 6

61 .3 HW 3 The first B.C. gives B λ or B. The solution becomes Θ A cos λθ And Θ A λθ λ sin. The second B.C. gives A λ sin λ π For non-trivial solution sin or λ π And the eigenfunction is λn n Now the R ODE is solved. The ODE is π λ nπ for n 1,, 3,. Hence the eigenvalues are λ n 4n n 1,, 3, Θ n θ A n cos nθ n 1,, 3, 3 r R + rr λr Case λ The ODE becomes r R + rr. et v r R r and the ODE becomes r v + rv Dividing by r v r + 1 r v r Using integrating factor e 1 r dr e ln r r. Hence Hence d dr rv rv A v r A r But since v r R r then R c 1 r. The solution to this ODE Is A R r r dr + B Therefore, for λ the solution is Since Then A and the solution is just a constant Case λ > The ODE is R r A ln r + B r lim R r < r R r B r R + rr 4n R n 1,, 3, The et R r r p. Then R pr p 1 and R p p 1 r p. This gives r p p 1 r p + r pr p 1 4n r p p p r p + pr p 4n r p r p p pr p + pr p 4n r p p 4n p ±n 61

62 HWs Hence the solution is The condition that Implies D, Hence the solution becomes R r Cr n + D 1 r n lim R r < r Now the solutions are combined. For λ the solution is R n r C n r n n 1,, 3, 4 u r, θ C B Which can be combined to one constant B. Hence And for λ > the solution is u n r, θ R n Θ n u B 5 C n r n A n cos nθ B n r n cos nθ Where C n A n are combined into one constant B n. Hence u n r, θ B n r n cos nθ 6 Equation 5 and 6 can be combined into one this now includes eigenfunctions for both λ and λ > u r, θ B + B n r n cos nθ 7 Where B represent the products of the eigenfunctions for R and Θ for λ. Now the nonhomogeneous condition is applied to find B n. Hence r u 1, θ g θ becomes r u r, θ B n n r n 1 cos nθ g θ B n n cos nθ 8 Multiplying by cos mθ and integrating gives π g θ cos mθ dθ π cos mθ nb n π B n n cos nθ dθ cos mθ cos nθ dθ 9 As in the last part, the integral on right gives π 4 when n m and zero otherwise, hence π π g θ cos nθ dθ nb n 4 B n πn π Therefore the final solution is from 7 and 9 u r, θ B + B + B n r n cos nθ πn π g θ cos nθ dθ n 1,, 3, g θ cos mθ dθ r n cos nθ 6

63 .3 HW 3 The unknown constant B can be found if given the initial temperature as was done in problem.5. part c. To answer the last part. Using 8 and integrating But π π g θ dθ π nb n cos nθ dθ nb n π cos nθ dθ [ ] π sin nθ cos nθ dθ n 1 sin n n π 1 sin nπ n Since n is an integer. This condition physically means the same as in part b problem.5.. Which is, since at steady state the flux must be zero on all boundaries, and g θ represents the flux over the surface of the quarter circle, then the integral of the flux must be zero. This means there is no thermal energy flowing across the boundary..3.4 Problem.5.8b problem 4.5. aplace's Equation Solve aplace's equation inside a circular annulus a < r < b subject to the boundary conditions * a ua, 9 f O, ub, 9 g9 b a,, ub, g9 67 c a, Wr- f, b, g 3T If there is a solvability condition, state it and explain it physically. *.5.9. Solve aplace's equation inside a 9 sector of a circular annulus a < r < b, < < ir/ subject to the boundary conditions The aplace PDE in polar coordinates is a ur,, ur, it/, ua, 9, ub, f b ur,, r ur,ir/ u fr, ua,, ub,9 r + r u r + u θ With.5.1. Using the maximum principles for aplace's equation, prove that the solution of Poisson's equation, Vu gx, subject to u f x on the boundary, is unique. u a, θ Do Exercise r u b, θ g θ.5.1. a Using the divergence theorem, determine an alternative expression for Assuming the solutionffuudxdydz. can be written as b Using part a, prove that the solution of aplace's equation Vu with u given on the u r, boundary θ R r is unique. Θ θ c Modify part b if on the boundary. And substituting this assumed solution back into the A gives d Modify part b if on the boundary. Show that Newton's law of cooling corresponds to h <. r R Θ + rr Θ + RΘ Prove that the temperature satisfying aplace's equation cannot attain its Dividing the above minimum by RΘin gives the interior Show that the "backward" heat equation r R R + r R au R + Θ Θ u -k 8x, r R at R + r R R Θ Θ subject to u, t u, t and ux, f x, is not well posed. Hint: Since each side depends Show that onif different the data are independent changed an variable arbitrarily andsmall theyamount, are equal, for example, they must be equal to same constant. say λ. 1 f x -' f x + n srn _ r R R + r R R Θ Θ λ for large n, then the solution ux, t changes by a large amount.] Solve aplace's equation inside a semi-infinite strip < x < oo, < y < H subject to the boundary conditions A B 63

64 HWs This results in the following two ODE s. The boundaries conditions in B are also transferred to each ODE. This results in And Θ + λθ 1 Θ π Θ π Θ π Θ π r R + rr λr R a Starting with 1 Case λ < The solution is Θ θ A cosh λθ + B sinh λθ First B.C. gives Θ π Θ π A cosh λπ + B sinh λπ A cosh λπ A cosh λπ B sinh λπ A cosh B sinh λπ + B sinh λπ λπ λπ + B sinh λπ But sinh only at zero and λ, hence B and the solution becomes Applying the second B.C. gives Θ θ A cosh λθ Θ θ A λ cosh λθ Θ π Θ π A λ cosh λπ A λ cosh λπ A λ cosh λπ A λ cosh λπ A λ cosh λπ λπ But cosh hence A. Therefore trivial solution and λ < is not an eigenvalue. Case λ The solution is Θ Aθ + B. Applying the first B.C. gives Θ π Θ π Aπ + B πa + B πa A And the solution becomes Θ B. A constant. Hence λ is an eigenvalue. Case λ > The solution becomes Applying first B.C. gives Θ A cos λθ Θ A λ sin + B sin λθ λθ + B λ cos λθ Θ π Θ π A cos λπ + B sin λπ A cos λπ + B sin λπ A cos λπ B sin λπ A cos λπ + B sin λπ B sin λπ 3 64

65 .3 HW 3 Applying second B.C. gives Θ π Θ π A λ sin λπ + B λ cos λπ A λ sin λπ + B λ cos λπ A λ sin λπ + B λ cos λπ A λ sin λπ + B λ cos λπ A λ sin λπ A λ sin λπ A sin λπ 4 Equations 3,4 can be both zero only if A B which gives trivial solution, or when λπ λπ sin. Therefore taking sin gives a non-trivial solution. Hence Hence the solution for Θ is Now the R equation is solved The case for λ gives Θ A + λπ nπ n 1,, 3, λ n n n 1,, 3, A n cos nθ + B n sin nθ 5 r R + rr R + 1 r R r As was done in last problem, the solution to this is R r A ln r + C Since r > no need to keep worrying about r and is removed for simplicity. Applying the B.C. gives Evaluating at r a gives Hence A, and the solution becomes Which is a constant. Case λ > The ODE in this case is et R r p, the above becomes R A 1 r A 1 a R r C r R + rr n R n 1,, 3, r p p 1 r p + rpr p 1 n r p p p 1 r p + pr p n r p p p 1 + p n p n p ±n Hence the solution is R n r Cr n + D 1 r n n 1,, 3, Applying the boundary condition R a gives R n r nc n r n 1 nd n 1 r n+1 R n a nc n a n 1 nd n 1 a n+1 nc n a n nd n C n a n D n D n C n a n 65

66 HWs The solution becomes R n r C n r n + C n a n 1 r n n 1,, 3, C n r n + an r n Hence the complete solution for R r is R r C + C n r n + an r n 6 Using 5,6 gives u n r, θ R n Θ n [ u r, θ C + D + ] [ C n r n + an r n A + A n cos nθ C n r n + an r n ] A n cos nθ + B n sin nθ + B n sin nθ C n r n + an r n Where D C A. To simplify more, A n C n is combined to A n and B n C n is combined to B n. The full solution is u r, θ D + A n r n + an r n cos nθ + B n r n + an r n sin nθ The final nonhomogeneous B.C. is applied. u b, θ g θ g θ D + For n, integrating both sides give π A n b n + an b n cos nθ + π g θ dθ π D 1 π D dθ π π π g θ dθ For n >, multiplying both sides by cos mθ and integrating gives π π g θ cos mθ dθ + π π π π π + π D cos mθ dθ B n b n + an b n sin nθ A n b n + an b n cos mθ cos nθ dθ B n b n + an b n cos mθ sin nθ dθ Hence But π π g θ cos mθ dθ π π π π + + π π D cos mθ dθ π A n b n + an b n B n b n + an b n π cos mθ cos nθ dθ π cos mθ sin nθ dθ 7 π cos mθ cos nθ dθ π n m cos mθ cos nθ dθ n m 66

67 .3 HW 3 And And Then 7 becomes π π π π cos mθ sin nθ dθ π π D cos mθ dθ g θ cos nθ dθ πa n b n + an A n 1 π π π Again, multiplying both sides by sin mθ and integrating gives Hence But And And π π π Then 9 becomes π g θ sin mθ dθ + g θ sin mθ dθ π π π π π π π π π π π + π + + π π D sin mθ dθ b n g θ cos nθ dθ 8 b n + an b n A n b n + an b n sin mθ cos nθ dθ B n b n + an b n D sin mθ dθ π A n b n + an b n B n b n + an b n sin mθ sin nθ dθ π sin mθ cos nθ dθ π sin mθ sin nθ dθ 9 π sin mθ sin nθ dθ π n m sin mθ sin nθ dθ π This complete the solution. Summary u r, θ D + D 1 π A n 1 π B n 1 π π π π π π π π π n m sin mθ cos nθ dθ π D sin mθ dθ g θ sin nθ dθ πb n b n + an B n 1 π A n r n + an r n g θ dθ g θ cos nθ dθ b n + an b n g θ sin nθ dθ b n + an b n The following are some plots of the solution. π π cos nθ + b n g θ sin nθ dθ b n + an b n B n r n + an r n sin nθ 67

68 HWs Figure.1: Solution using fθ sin3θ, a.3, b.5 Figure.11: Solution using fθ 3θ, a.3, b.6 Figure.1: Solution using fθ 1θ, a.1, b.4 68

69 ffuudxdydz. b Using part a, prove that the solution of aplace's equation Vu with u given on the boundary is unique. c Modify part b if on the boundary..3 HW 3 d Modify part b if on the boundary. Show that Newton's law of cooling corresponds to h <..3.5 Problem.5.14 problem Prove that the temperature satisfying aplace's equation cannot attain its minimum in the interior Show that the "backward" heat equation au at u -k 8x, subject to u, t u, t and ux, f x, is not well posed. Hint: Show that if the data are changed an arbitrarily small amount, for example, 1 srn _ f x -' f x + n for large n, then the solution ux, t changes by a large amount.] Solve aplace's equation inside a semi-infinite strip < x < oo, < y < H subject to the boundary conditions 1 u k t u x u, t u, t Assume u x, t XT. Hence the PDE becomes Hence, forλ real u x, f x 1 k T X X T 1 k T T X X 1 k T X X λ The space ODE was solved before. Only positive eigenvalues exist. The solution is X x B n sin λn x The time ODE becomes With solution T nπ λ n n 1,, 3, T n λ n T n T n λ n T n T n t A n e λnt T t A n e λnt For the same eigenvalues. Therefore the full solution is nπ u x, t C n sin x e Where C n A n B n. Applying initial conditions gives nπ f x C n sin x Multiplying by sin mπ x and integrating results in f x sin mπ x dx mπ sin C n sin C m x nπ t 1 nπ C n sin x dx mπ x sin nπ x dx 69

70 HWs Therefore The solution 1 becomes u x, t C n nπ f x sin x dx nπ nπ f x sin dx x sin x e Assuming initial data is changed to f x + 1 n sin nπ x then f x + 1 m sin mπ x Multiplying both sides by sin mπ x and integrating mπ f x sin x dx + 1 m sin mπ x dx nπ C n sin x mπ f x sin x dx + 1 m C m C n mπ sin nπ t x nπ C n sin x nπ f x sin x dx + 1 n Therefore, the new solution is nπ ũ x, t f x sin x dx + 1 nπ sin n x nπ t e 88 mπ f x sin x Chapter nπ. Method dx sin x nπ of t Separation e + 1 nπ n sin of x Variables nπ t e a 8' x,, "' x, mπ H, u, nπ y f y f x sin x dx sin x nπ t 1 nπ e + n sin x nπ e b ux,, ux, H, u,y fy But f c x sin ux, mπ x, dx sin ux, nπ x H e nπ, t,y fy u x, t, therefore the above can be written as d x,, Ou x, H, a:, y f y 1 nπ ũ x, t u x, t + n sin x nπ t e Show that the solution [part d] exists only if f H f y dy Consider aplace's equation inside a rectangle < x For large n, the difference between initial data f x and f x + 1 < n sin, nπ< x y < H, with is very small, since the boundary conditions 1 n. However, the effect in the solution above, due to the presence of e nπ t is that 1 n e nπ t 8u8x, 8u8yx, increases now for large n, y since, the auexponential 8x, y gy, is to the positive, power, &" x, and H itf grows x at a faster rate than 1 8y n grows small as n increases, with the net effect that the produce blow up for large n. a What is the solvability condition and its physical interpretation? This is because the power of the exponential is positive and not negative is normally would be the b Show that ux, y case. Also by looking at the series of e nπax t - y is a which is 1 + solution nπ 4 t + if f nπ x 6 and t 3 3! + gy, then are 1 n e nπ t expands to 1 n + 1 constants nπ 4 [under t n + 1 nπ 6 the conditions of part a]. t c Under the 3 nconditions 3! + of part which a, becomes solve the very general large case for[nonconstant large n. In the normal PDEf x case, and thegy]. above[hints: solutionuse would part have b and instead the been fact that the following f x f + [f x - f.,.], where f.,. f f 1x dx.] nπ ũ x, t u x, t + n sin x e nπ t Show that the mass density px, t satisfies k + V pu due to conservation of And now as n then mass. 1 n sin nπ x e nπ t as well. Notice that sin nπ If the mass density is constant, using the result of Exercise.5.17, show x term is not important for that this analysis, as its value oscillates between 1 and Show that the streamlines are parallel to the fluid velocity..3.6 Problem.5. problem Show that anytime there is a stream function, V x u From u and v-,derive u,-rue-.5.. Show the drag force is zero for a uniform flow past a cylinder including circulation Consider the velocity ug at the cylinder. Where do the maximum and minimum occur? The force exerted by the fluid on the cylinder is given by equation.5.56, page 77 of the text as.5.4. Consider.the velocity ue at the cylinder. If the circulation is negative, show that the velocity will be larger π above the cylinder than below. F p cos θ, sin θ adθ.5.5. A stagnation point is a place where u. For what values of the circulation Where a is thedoes cylinder a stagnation radius, point p is the exist fluid on the pressure. cylinder? This vector has components. The x component.5.6. is the For drag what force values andof the will y component u,. off is the the cylinder? left forcefor asthese illustrated 6, where by this for diagram. what values of r will ue also? Show that r/ a 81T B satisfies aplace's equation. Show that the streamlines are circles. Graph the streamlines. t

71 .3 HW 3 ift force F y y direction π p sinθa dθ radius a r θ F x π p cosθa dθ Drag force x direction Therefore the drag force per unit length is π F x p cos θadθ 1 Now the pressure p needs to be determined in order to compute the above. The fluid pressure p is related to fluid flow velocity by the Bernoulli condition p + 1 ρ u C Where C is some constant and ρ is fluid density and u is the flow velocity vector. Hence in order to find p, the fluid velocity is needed. But the fluid velocity is given by u u rˆr + u θ ˆθ 1 Ψ Ψ ˆr r θ r ˆθ Since the radial component of the fluid velocity is zero at the surface if the cylinder This is one of the boundary conditions used to derive the solution, then only the tangential component comes into play. Hence u Ψ r but r Ψ r, θ c 1 ln + u r a a sin θ r Therefore And hence Ψ r c 1 r + u 1 + a r sin θ u Ψ r c 1 r + u 1 + a r sin θ At the surface r a, hence u c 1 a + u sin θ Substituting this into in order to solve for pressure p gives p + 1 ρ c 1 a + u sin θ C p C 1 ρ c 1 a + u sin θ Substituting the above into 1 in order to solve for the drag gives π F x [C 1 ρ c ] 1 a + u sin θ cos θadθ The above is the quantity that needs to be shown to be zero. F x ac π cos θdθ a ρ π c 1 a + u sin θ cos θdθ 71

72 HWs 88 Chapter. Method of Separation of Variables But π cos θdθ a hence 8' x, the above, simplifies "' x, H to, u, y f y b F x a ux, π ρ c, ux, H 1, u,y fy a + u sin θ cos θdθ c ux,, ux, H,,y fy a π ρ c 1 a cos θ + 4u sin θ cos θ 4 c 1 d x,, Ou x, H, a: a u, siny θ cos f y θdθ a [ c π π ρ 1 a cos θdθ + 4u sin θ cos θdθ 4 c π ] Show that the solution [part d] exists only if f H f y 1 a u dy. sin θ cos θdθ Consider aplace's equation inside a rectangle < x <, < y < H, with But the boundary π cos θdθ and conditions π 8u sin θ cos θdθ hence the above reduces to 8x, 8u8yx, y, au 8x, y gy,, &" x, H f x π 8y a What is the solvability F x 4aρu condition sin and θits cos physical θdθ interpretation? b Show that ux, y Ax - y is a solution if f x and gy are But sin θ 1 1 cos constants θ and[under the above the conditions becomes of part a]. c Under the conditions of part a, solve the general case [nonconstant π 1 F x 4aρu 1 f x and gy]. [Hints: Use part b and the fact that f x f + cos θ cos θdθ [f x - f.,.], where f.,. f f x dx.] 1 π 4aρu cos θdθ 1 π Show that the mass density px, t satisfies k + V pu due to conservation of mass. cos θ cos θdθ But π If the mass density is constant, using the result cos θdθ and by orthogonality of cos function π of Exercise.5.17, show that cos θ cos θ dθ as well. Therefore the above reduces to Show that the streamlines are parallel to the fluid velocity. F x.5.. Show that anytime there is a stream function, V x u. The drag force x component of the force exerted by fluid on the cylinder is zero just outside the surface of.5.1. the surface From uof the cylinder. and v- Which,derive is whatu,-rue- the question asks to show..5.. Show the drag force is zero for a uniform flow past a cylinder including.3.7 Problemcirculation..5.4 problem Consider the velocity ug at the cylinder. Where do the maximum and minimum occur?.5.4. Consider.the velocity ue at the cylinder. If the circulation is negative, show that the velocity will be larger above the cylinder than below A stagnation point is a place where u. For what values of the circulation does a stagnation point exist on the cylinder? Introduction For The what stream values velocity of will u in u,. Cartesian off the coordinates cylinder? is For these 6, where for what values of r will ue also? u uî + vĵ.5.7. Show that r/ a 81T B satisfies aplace's equation. Show that the streamlines are circles. Graph the streamlines. Ψ y î Ψ x ĵ 1 Where Ψ is the stream function which satisfies aplace PDE in D Ψ. In Polar coordinates the above becomes The solution to Ψ was found under the following conditions u u rˆr + u θ ˆθ 1 Ψ Ψ ˆr r θ r ˆθ 1. When r very large, or in other words, when too far away from the cylinder or the wing, the flow lines are horizontal only. This means at r the y component of u in 1 is zero. This means Ψx,y x. Therefore Ψ x, y u y where u is some constant. In polar coordinates this implies Ψ r, θ u r sin θ, since y r sin θ.. The second condition is that radial component of u is zero. In other words, 1 Ψ r θ when r a, where a is the radius of the cylinder. 3. In addition to the above two main condition, there is a condition that Ψ at r Using the above three conditions, the solution to Ψ was derived in lecture Sept. 3, 16, to be r Ψ r, θ c 1 ln + u r a a sin θ r 7

73 .3 HW 3 Using the above solution, the velocity u can now be found using the definition in as follows 1 Ψ r θ 1 r u r a cos θ r Hence, in polar coordinates u 1 r u Ψ r c 1 r + u r a r 1 + a r sin θ cos θ ˆr c1r + u 1 + a sin θ r ˆθ 3 Now the question posed can be answered. The circulation is given by Γ π u θ rdθ But from 3 u θ c1r + u 1 + a sin θ, therefore the above becomes r Γ At r a the above simplifies to But π sin θdθ, hence π Γ π π c1 r + u π 1 + a r c1 a + u sin θ adθ c 1 au sin θdθ sin θ rdθ π c 1 dθ au sin θdθ π Γ c 1 dθ c 1 π Since Γ <, then c 1 >. Now that c 1 is known to be positive, then the velocity is calculated at θ π and then at θ +π to see which is larger. Since this is calculated at r a, then the radial velocity is zero and only u θ needs to be evaluated in 3. At θ π 1 + a At r a At θ +π At r a c1 u π c1 sin r + u r r u 1 + a r 1 + a c1 r u r sin π π c1 u π a u c 1 a + u 4 c1 u +π r + u c1 r + u 1 + a r 1 + a r π sin c1 u π a + u c 1 a u 5 Comparing 4,5, and since c 1 >, then the magnitude of u θ at π is larger than the magnitude of u θ at π. Which implies the stream flows faster above the cylinder than below it. 73

74 HWs.4 HW If the mass density is constant, using the result of Exercise.5.17, show that Show that the streamlines are parallel to the fluid velocity..5.. Show that anytime there is a stream function, V x u From u and v-,derive u,-rue-.5.. Show the drag force is zero for a uniform flow past a cylinder including.4.1 Problemcirculation Consider the velocity ug at the cylinder. Where do the maximum and minimum occur?.5.4. Consider.the velocity ue at the cylinder. If the circulation is negative, show that the velocity will be larger above the cylinder than below A stagnation point is a place where u. For what values of the circulation does a stagnation point exist on the cylinder? Introduction. The stream velocity u in Cartesian coordinates is.5.6. For what values of will u,. off the cylinder? For these 6, where for what values of r will ue u also? uî + vĵ.5.7. Show that r/ a 81T B satisfies aplace's equation. Show that the streamlines are circles. Graph the streamlines. Ψ y î Ψ x ĵ 1 Where Ψ is the stream function which satisfies aplace PDE in D Ψ. In Polar coordinates the above becomes The solution to Ψ was found under the following conditions u u rˆr + u θ ˆθ 1 Ψ Ψ ˆr r θ r ˆθ 1. When r very large, or in other words, when too far away from the cylinder or the wing, the flow lines are horizontal only. This means at r the y component of u in 1 is zero. This means Ψx,y x. Therefore Ψ x, y u y where u is some constant. In polar coordinates this implies Ψ r, θ u r sin θ, since y r sin θ.. The second condition is that radial component of u is zero. In other words, 1 Ψ r θ when r a, where a is the radius of the cylinder. 3. In addition to the above two main condition, there is a condition that Ψ at r Using the above three conditions, the solution to Ψ was derived in lecture Sept. 3, 16, to be r Ψ r, θ c 1 ln + u r a a sin θ r Using the above solution, the velocity u can now be found using the definition in as follows 1 Ψ r θ 1 r u r a cos θ r Hence, in polar coordinates u 1 r u Ψ r c 1 r + u r a r 1 + a r sin θ cos θ ˆr c1r + u 1 + a sin θ r ˆθ 3 Now the question posed can be answered. The circulation is given by Γ π u θ rdθ But from 3 u θ c1r + u 1 + a sin θ, therefore the above becomes r Γ At r a the above simplifies to π Γ π π c1 r + u π 1 + a r c1 a + u sin θ adθ c 1 au sin θdθ sin θ rdθ π c 1 dθ au sin θdθ 74

75 .4 HW 4 But π sin θdθ, hence π Γ c 1 dθ c 1 π Since Γ <, then c 1 >. Now that c 1 is known to be positive, then the velocity is calculated at θ π and then at θ +π to see which is larger. Since this is calculated at r a, then the radial velocity is zero and only u θ needs to be evaluated in 3. At θ π At r a c1 u π r + u 1 + a r π sin π sin 3.. Convergence Theorem c1 r u 1 + a r n7rx c1 r u 1 + a n7rx n7rx an f x cos dx cos dx J r nir sin / u π c1 a u c 1 a + u 4 bn 1 At θ +π f x sin nirx 1 dx J -1 nirx sin n dx - cos / x nir / c1 u +π r + u 1 + a π nn r sin Cos - - cos n7r 3..8 n7r c1 We omit simplifications that arise r by + noting u 1 + that a r sinn7r, cosnir -1", and so on. At r a 1 sinn7r - sin nir nir EXERCISES 3. c1 u π a + u For the following functions, sketch c the Fourier series off x on the interval 1 - < x <. Compare f x to its Fourier a u series: 5 Comparing 4,5, a andfx since 1 c 1 >, then the magnitude of *b u θ at fx π is larger x than the magnitude of u θ at π. Which c implies fx1+x the stream flows faster above the *d cylinder fx than ex below it..4. Problem 3.. b,d g fx x e f x { x x > * f f x 1+x x < / x > / x > O 3... For the following functions, sketch the Fourier series of f x on the interval - < x < and determine the Fourier coefficients: I/ *a fxx *c fx sin e fx I jxj < / jxi > / b fx e-x d fx * f f x l x < l x x> x< x> g fx I 1 x < l x> Part b The following is sketch of periodic extension of e x from x for 1 for illustration. The function will converge to e x between x and between x 3 and between x 3 and so on. But at the jump discontinuities which occurs at x, 3,,, 3, it will converge to the average shown as small circles in the sketch. 75

76 HWs o o 1.5 o o Periodic extension of exp-x from By definitions, a 1 T a n 1 T/ b n 1 T/ T/ T/ T/ T/ T/ T/ f x dx f x cos n f x sin n The period here is T, therefore the above becomes a 1 a n 1 b n 1 These are now evaluated for f x e x a 1 e x dx 1 Now a n is found e x 1 a n 1 f x dx f x cos π T π T n π x dx f x sin n π x dx x dx x dx 1 e x 1 e e e e e x cos n π x dx This can be done using integration by parts. udv uv vdu. et I e x cos n π x dx and u cos n π x, dv e x, du nπ sin n π x, v e x, therefore I [uv] vdu [ e x cos n π ] x nπ [ e cos n π + e cos [ e cos nπ + e cos nπ ] nπ e x sin n π x dx n π ] nπ e x sin e x sin n π x dx n π x dx Applying integration by parts again to e x sin n π x dx where now u sin n π x, dv e x 76

77 .4 HW 4 du nπ cos n π x, v e x, hence the above becomes I [ e cos nπ + e cos nπ ] nπ uv vdu [ e cos nπ + e cos nπ ] nπ {}}{ [ e x sin n π ] x + nπ [ e cos nπ + e cos nπ ] nπ nπ [ e cos nπ + e cos nπ ] nπ e x cos e x cos n π x dx n π x dx But e x cos n π x dx I and the above becomes nπ I e cos nπ + e cos nπ I Simplifying and solving for I nπ I + I cos nπ e e nπ I 1 + cos nπ e e + n π I cos nπ e e I + n π cos nπ e e Hence a n becomes But cos nπ 1 n hence Similarly for b n a n 1 + n π cos nπ e e a n 1 n e n π + e b n 1 e x sin n π x dx This can be done using integration by parts. udv uv vdu. et I e x sin n π x dx and u sin n π x, dv e x, du nπ cos n π x, v e x, therefore I [uv] vdu { [ }} { e x sin nπ n π ] x + nπ e x cos n π x dx e x cos n π x dx e x cos n π x dx Applying integration by parts again to e x cos n π x dx where now u cos n π x, dv e x du nπ sin n π x, v e x, hence the above becomes I nπ uv vdu nπ [ e x cos n π ] x nπ e x sin n π x dx nπ nπ e cos n π + e cos cos nπ e e nπ n π nπ e x sin n π x dx e x sin n π x dx 77

78 HWs But e x cos n π x dx I and the above becomes Simplifying and solving for I I I I nπ nπ nπ I + I nπ n π cos nπ e e nπ I I nπ cos nπ e e nπ I cos nπ e e nπ cos nπ e e nπ cos nπ e e I + n π nπ cos nπ e e Hence b n becomes But cos nπ 1 n hence b n 1 + n π nπ + n π nπ cos nπ e e cos nπ e e b n 1 n nπ e + n π e Summary a e e a n 1 n e n π + e b n 1 n nπ e + n π e f x a + a n cos n a + π x + b n sin n T a n cos n π x + b n sin n π x π x T The following shows the approximation f x for increasing number of terms. Notice the Gibbs phenomena at the jump discontinuity Fourier series approximation, number of terms 3 Showing 3 periods extenstion of -.., with 1 fx x 78

79 .4 HW 4 Fourier series approximation, number of terms 1 Showing 3 periods extenstion of -.., with 1 fx x Fourier series approximation, number of terms 5 Showing 3 periods extenstion of -.., with fx x Part d The following is sketch of periodic extension of f x from x for 1 for illustration. The function will converge to f x between x and between x 3 and between x 3 and so on. But at the jump discontinuities which occurs at x, 3,,, 3, it will converge to the average 1 shown as small circles in the sketch o o o o Showing 3 periods extenstion of fx between -.., with 1 By definitions, a 1 T a n 1 T/ b n 1 T/ T/ T/ T/ T/ T/ T/ f x dx f x cos n f x sin n π T π T x dx x dx 79

80 HWs The period here is T, therefore the above becomes a 1 a n 1 b n 1 These are now evaluated for given f x Now a n is found a n a x f x dx f x cos n π x dx f x sin n π x dx f x dx f x dx + f x cos n π x dx f x cos x cos n π x dx xdx n π x dx + f x dx f x cos n π x dx Integration by parts. et u x, du 1, dv cos n π x, v sin n π x n π, hence the above becomes Now b n is found a n 1 {}}{ nπ x sin n π x 1 nπ 1 nπ 1 b n sin n π x dx cos n π x n π cos n π nπ x n n π cos π x [ n π cos n π ] 1 n π [ 1n 1] f x sin n π x dx f x sin x sin n π x dx n π x dx + sin n π x n π dx f x sin n π x dx 8

81 .4 HW 4 Integration by parts. et u x, du 1, dv sin n π x, v cos n π x n π, hence the above becomes b n 1 nπ x cos n π x cos n π + x n π dx nπ cos n π 1 + nπ {}}{ 1 [ nπ 1n + sin n π x ] nπ n π cos n π x dx Summary nπ 1n 1 n+1 nπ a 4 a n n π [ 1n 1] b n 1 n+1 nπ f x a + a n cos a + n π x + b n sin n T a n cos n π x + b n sin n π x π x T The following shows the approximation f x for increasing number of terms. Notice the Gibbs phenomena at the jump discontinuity..8.6 Fourier series approximation, number of terms 3 Showing 3 periods extenstion of -.., with 1 fx x Fourier series approximation, number of terms 1 Showing 3 periods extenstion of -.., with fx x 81

82 HWs Fourier series approximation, number of terms 5 Showing 3 periods extenstion of -.., with fx x 96 Chapter 3. Fourier Series.4.3 Problem Show 3..4that the Fourier series operation is linear: that is, show that the Fourier series of c1 f x + cgx is the sum of cl times the Fourier series of f x and c times the Fourier series of gx Suppose that f x is piecewise smooth. What value does the Fourier series of f x converge to at the endpoint x -? at x? 3.3 Fourier Cosine and Sine Series It will converge to the average value of the function at the end points after making periodic extensions In of this the section function. we show Specifically, that the series at x of sines theonly Fourier and series the series will of converge cosines to only are special cases of a Fourier series. 1 f + f Fourier Sine Series Odd functions. An odd function is a function with the property f -x And at x it will converge to - f x. The sketch of an odd function 1 for x < will be minus the mirror image of f + f f x for x >, as illustrated in Fig Examples of odd functions are f x x3 Notice that in 114if fact, f any has odd same power value and as f f x, sin then 4x. The there integral will not Chapter of be an a odd 3. jump Fourier function discontinuity Series over when a symmetric interval is zero any contribution from x > will be canceled by a periodic extension are made, and the above formula simply gives the value of the function at either contribution from x <. end, since it is thea same fx value. 1 b fx1+x c.4.4 Problem 3.3. d fx _ { 1 + x x > *d fx ex e fx e-x x > For the following functions, sketch the Fourier sine series of f x and determine its Fourier coefficients. Figure An odd function. Fourier a b series of odd functions. et us calculate f x 3 [Verify formula ] the Fourier /6 < x coeffi- < / x > / cients of an odd function: c fx ao x x < / 1 f 1 J_f x dx * x > / d f x For the following functions, sketch the Fourier sine series of f x. Also, roughly sketch the sum of a finite number of nonzero terms at least the first two of the Fourier sine series: { Both are zero because the integrand, f x 1 cos x nirx/, < f x is odd being the product of an even function a f x cos cos n7rx/ irx/ [Use and an formula odd function ] x > f x. Since an, all the cosine functions which are even will not appear in the Fourier series of an odd function. The Fourier b series fx _ of { an 1 x < / The first step is to sketch f x over. This is the result for 1 as an example. odd x > function / is an infinite series of odd functions sines: an ffxcosdx. c f x x [Use formula ] f x - n1x, bn sin Sketch the Fourier cosine series of f x sin irx/. Briefly discuss. 1 x < /6 1 x < / x > / For the following functions, sketch the Fourier cosine series of f x and determine its Fourier coefficients: 1 x < /6 a f x x b f x 3 /6 < x < / c f x x > / fx x > / For the following functions, sketch the Fourier cosine series of f x. Also, roughly sketch the sum of a finite number of nonzero terms at least the first two of the Fourier cosine series: a f x x [Use formulas 3.3. and ]

83 .4 HW 4 Original fx function defined for The second step is to make an odd extension of f x over. This is the result. odd extension of fx defined for The third step is to extend the above as periodic function with period as normally would be done and mark the average value at the jump discontinuities. This is the result 1. o.5 o o o o o o Showing 3 periods extenstion of fx between -.., with 1 o o Now the Fourier sin series is found for the above function. Since the function f x is odd, then only b n will exist f x π b n sin n x b n sin n π x Where b n 1 π f x sin n x dx 1 f x sin n π x dx 83

84 HWs Since f x sin n π x is even, then the above becomes Therefore b n / / f x sin n π x dx [ cos n π x [ nπ nπ nπ nπ 1 sin n π x dx + sin n π x dx n π ] / cos n π ] / x [ cos n π ] 1 [ nπ ] cos 1 [ nπ ] 1 cos f x nπ 1 cos nπ / sin n π x dx sin n π x The following shows the approximation f x for increasing number of terms. Notice the Gibbs phenomena at the jump discontinuity. Fourier series approximation, number of terms 3 Showing 3 periods extenstion of -.., with fx x Fourier series approximation, number of terms 1 Showing 3 periods extenstion of -.., with fx x 84

85 .4 HW 4 Fourier series approximation, number of terms 5 Showing 3 periods extenstion of -.., with Chapter 3. Fourier Series a fx.5 1 b fx1+x c fx fx _ {. 1 + x x > *d fx ex e fx -.5 e-x x > For the following functions, sketch the Fourier sine series of f x and determine its Fourier coefficients Problem b a [Verify formula ] c fx x x < / x > / x 1 x < /6 b f x 3 /6 < x < / x > / * d f x 1 x < / x > / For the following functions, sketch the Fourier sine series of f x. Also, roughly sketch the sum of a finite number of nonzero terms at least the first two of the Fourier sine series: a f x cos irx/ [Use formula ] b fx _ { 1 x < / x > / c f x x [Use formula ] Sketch the Fourier cosine series of f x sin irx/. Briefly discuss For the following functions, sketch the Fourier cosine series of f x and This is the same problem as 3.3. part d. But it asks to plot for n 1 and n in the sum. determine its Fourier coefficients: The sketch of the Fourier sin series was done above in solving 3.3. partd and will not be repeated again. From above, it was found that 1 x < /6 a f x x b f x 3 /6 < x < / c f x x > / fx x > / f x B n sin n π x For the following functions, sketch the Fourier cosine series of f x. Also, roughly sketch the sum of a finite number of nonzero terms at least the Where B n [ nπfirst 1 two cos nπ ] of the. Fourier The following cosine series: is the plot for n 1 1. a f x x [Use formulas 3.3. and ] b f x c f x x</ 1 x > 1 _ x < / 1 x > / [Use carefully formulas 3..6 and 3..7.] [Hint: Add the functions in parts b and c.] Show that ex is the sum of an even and an odd function. 85

86 HWs Fourier series approximation, number of terms 1 Showing 3 periods extenstion of -.., with 1 Fourier series approximation, number of terms Showing 3 periods extenstion of -.., with fx. fx x x Fourier series approximation, number of terms 3 Showing 3 periods extenstion of -.., with 1 Fourier series approximation, number of terms 4 Showing 3 periods extenstion of -.., with fx. fx x x Fourier series approximation, number of terms 5 Showing 3 periods extenstion of -.., with 1 Fourier series approximation, number of terms 6 Showing 3 periods extenstion of -.., with fx. fx x x Fourier series approximation, number of terms 7 Showing 3 periods extenstion of -.., with 1 Fourier series approximation, number of terms 8 Showing 3 periods extenstion of -.., with fx. fx x x Fourier series approximation, number of terms 9 Showing 3 periods extenstion of -.., with 1 Fourier series approximation, number of terms 1 Showing 3 periods extenstion of -.., with fx. fx x x.4.6 Problem Cosine and Sine Series a Determine formulas for the even extension of any f x. Compare to the formula for the even part of f x. b Do the same for the odd extension of f x and the odd part of f x. c Calculate and sketch the four functions of parts a and b if J x x> fx x x <. Graphically add the even and odd parts of f x. What occurs? Similarly, add the even and odd extensions. What occurs then? What is the sum of the Fourier sine series of f x and the Fourier cosine series of f x? [What is the sum of the even and odd extensions of f x?] * If f x e_z x >, what are the even and odd parts of f x? Given a sketch of fx, describe a procedure to sketch the even and odd parts of f x a Graphically show that the even terms n even of the Fourier sine series

87 .4 HW 4 Part a The even extension of f x is But the even part of f x is { f x x > f x x < 1 f x + f x Part b The odd extension of f x is While the odd part of f x is { f x x > f x x < 1 f x f x Part c First a plot of f x is given f x { x x > x x < 1. Plot of fx fx x A plot of even extension and the even part for f x Is given below Plot of even extension Plot of even part fx.4 fx x x A plot of odd extension and the odd part is given below 1. Plot of odd extension.1 Plot of odd part of fx.5.5 function x function x Adding the even part and the odd part gives back the original function 87

88 HWs 1. Plot of odd +even parts of fx.8 function x Plot of adding the even extension and the odd extension is below. Plot of even extension+odd extension of fx 1.5 function x.4.7 Problem Part a 3.4. Term-by-Term Differentiation Suppose that f x is continuous [except for a jump discontinuity at x x i f xa a and f xo 1 and df /dx is piecewise smooth. *a Determine the Fourier sine series of df /dx in terms of the Fourier cosine series coefficients of f x. b Determine the Fourier cosine series of df /dx in terms of the Fourier sine series coefficients of fx Suppose that f x and df /dx are piecewise smooth. a Prove that the Fourier sine series of a continuous function f x can only be differentiated term by term if f and f. Fourier sin series b of f Prove x is that given the Fourier by, assuming cosine period series of isa continuous function f x can be differentiated term by term Using determine f x the Fourier b n cosine n π series of sin 7rx/. x There are some things wrong in the following demonstration. Find the Where mistakes and correct them. In this exercise we b n attempt to f obtain x sinthe n Fourier π cosine coefficients of ex: x dx Applying integration by parts. et f x nirx exa +E dv, u A.cos sin n π r. x, then v f x, du 3.4. nπ cos nπ x. Since v f x has has jump discontinuity at x as described, and assuming x >, then, and using sin n π x Differentiating at x yields b nir n udv nirx e-, [ ] the [uv] Fourier x + [uv] sine x + series of ex. vdu Differentiating again yields O n7r [ sin n π ] x [ ex - An cos nx x f x + sin n π ] x f x nπ nπ f x cos x + x dx sin Since n π equations 3.4. and give the Fourier cosine series of ex, they must x f x sin n π be identical. Thus, x+ f x + nπ nπ f x cos x dx 1 88 `4 obviously wrong!. An By correcting the mistakes, you should be able to obtain A and An without

89 .4 HW 4 In the above, sin n π x and at x was used. But f x α f x + β sin n π x+ And since sin is continuous, then sin n π x to b n α β sin n π x nπ On the other hand, the Fourier cosine series for f x is given by Where f x a + a 1 a n f x dx a n cos n π x f x cos n π x dx Therefore f x cos n π x dx a n. Substituting this into gives b n α β sin n π x nπ a n n α β sin π x nπ a n Hence Summary the Fourier sin series of f x is sin n π x. Equation 1 simplifies nπ f x cos x dx b n sin n π x α β nπ a n 3 f x b n sin n π x With b n given by 3. The above is in terms of a n, which is the Fourier cosine series of f x, which is what required to show. In addition, the cos series of f x can also be written in terms of sin series of f x. From 3, solving for a n a n nπ b n nπ sin f x a + 1 n α β n π x π b n π sin n π x nπ α β cos x This shows more clearly that the Fourier series of f x has order of convergence in a n as 1 n as expected. Part b Fourier cos series of f x is given by, assuming period is Where a f x x n f x dx a n cos n π x f x dx + x + [f x] x + [f x] x + f x dx 1 [α f ] + [f β] α β f + f + 89

90 HWs And for n > a n f x cos n π x dx Applying integration by parts. et f x dv, u cos n π x, then v f x, du nπ Since v f x has has jump discontinuity at x as described, then a n udv [ [uv] x + [uv] x + [ cos n π ] x x f x ] vdu [ cos n π ] x f x x nπ cos n π x f x f + cos nπ f cos n π x+ nπ f x sin x dx f x + nπ + sin nπ x. nπ f x sin x dx But f x α f x + β And since cos is continuous, then cos n π x cos n π x+ cos n π x, therefore 1 becomes a n cos nπ f f + cos n π x α β + nπ nπ f x sin x dx On the other hand, the Fourier sin series for f x is given by Where f x b n n b n sin n π x f x sin n π x dx Therefore f x sin n π x dx b n. Substituting this into gives a n cos nπ f f + cos n π x α β + nπ cos nπ f f + cos n π x 1n f f + cos n π x b n α β + α β + nπ b n nπ b n 1 Hence a n 1n f f + cos n π x α β + nπ b n 3 Summary the Fourier cos series of f x is f x n a n cos n π x α β f + f a + a n 1n f f + n cos π x α β + nπ b n The above is in terms of b n, which is the Fourier sin series of f x, which is what required to show. 9

91 Consider subject to au _ au at kaxe au/ax,t, au/ax,t, and ux, fx..4.8 Problem Solve 3.4.9in the following way. ook for the solution as a Fourier cosine series. Assume that u and au/ax are continuous and au/axe and au/at are piecewise smooth. Justify all differentiations of infinite series. *3.4.9 Consider the heat equation with a known source qx, t: kjx + qx, t with u, t and u, t. Assume that qx, t for each t > is a piecewise smooth function of x. Also assume that u and au/ax are continuous functions of x for t > and au/axe and au/at are piecewise smooth. Thus, ux, t _ What ordinary differential equation does differential equation. N,t sin.x. satisfy? Do not solve this Modify Exercise if instead au/ax, t and au/ax, t..4 HW 4 The PDE is Consider the nonhomogeneous heat equation with a steady heat source: u t k u + q x, t 1 x kax +gx. Since the boundary conditions are homogenous at Dirichlet conditions, then the solution can be written down as Solve this equation with the initial condition u x, t ux, b n t sinfx n π x and the boundary conditions Since the solution is assumed to be continuous with continuous derivative, then term by term differentiation is allowed w.r.t. x uo,t and u, t. Assume that a continuous u solution exists with continuous derivatives. [Hints: Expand x the n π solution as b n t cos n π a Fourier sine x series i.e., use the method of eigenfunction expansion. Expand gx as a Fourier sine series. Solve for the Fourier sine u series of the solution. Justify all differentiations with respect to x.] x nπ bn t sin n π x Also using * assumption Solve the that following u is smooth, nonhomogeneous then problem: t k5- + u e-t + e- t cos 3x [assume that # k37r/] t db n t sin n π dt x Substituting,3 into 1 gives db n t sin n π dt x k nπ Expanding q x, t as Fourier sin series in x. Hence nπ q x, t q n t sin x Where now q n t are time dependent given by by orthogonality q n t nπ q x, t sin x Hence 4 becomes db n t dt sin n π x nπ k bn t sin n π x 3 bn t sin n π x + q x, t 4 + nπ q t n sin x Applying orthogonality the above reduces to one term only db n t sin n π nπ dt x k bn t sin n π nπ x + q t n sin x Dividing by sin n π x db n t dt db n t nπ k bn t + q n t dt nπ + k Bn t q n t 5 The above is the ODE that needs to be solved for b n t. It is first order inhomogeneous ODE. The question asks to stop here. 91

92 Also assume that u and au/ax are continuous functions of x for t > and au/axe and au/at are piecewise smooth. Thus, HWs ux, t _.4.9 Problem What ordinary differential equation does differential equation. N,t sin.x. satisfy? Do not solve this Modify Exercise if instead au/ax, t and au/ax, t Consider the nonhomogeneous heat equation with a steady heat source: kax +gx. at Solve this equation with the initial condition and the boundary conditions ux, fx uo,t and u, t. Assume that a continuous solution exists with continuous derivatives. [Hints: Expand the solution as a Fourier sine series i.e., use the method of eigenfunction expansion. Expand gx as a Fourier sine series. Solve for the Fourier sine series of the solution. Justify all differentiations with respect to x.] * Solve the following nonhomogeneous problem: The PDE is k5- + e-t + e- t cos 3x [assume that # k37r/] u t k u + g x 1 x Since the boundary conditions are homogenous Dirichlet conditions, then the solution can be written down as u x, t b n t sin n π x Since the solution is assumed to be continuous with continuous derivative, then term by term differentiation is allowed w.r.t. x u x n π b n t cos n π x u x nπ bn t sin n π x Also using assumption that u t is smooth, then u t db n t dt sin n π x 3 Substituting,3 into 1 gives db n t dt sin n π x k nπ bn t sin n π x + g x 4 Using hint given in the problem, which is to expand g x as Fourier sin series. Hence Where Hence 4 becomes db n t dt sin n π x g x g n nπ g n sin x nπ g x sin x nπ k bn t sin n π x Applying orthogonality the above reduces to one term only db n t dt + nπ g n sin x sin n π nπ x k bn t sin n π nπ x + g n sin x 9

93 .4 HW 4 Dividing by sin n π x db n t dt db n t nπ k bn t + g n dt nπ + k bn t g n 5 This is of the form y + ay g n, where a k nπ. This is solved using an integration factor µ e at, where d dt e at y e at g n, giving the solution y t 1 µ µg n dt + c µ Hence the solution to 5 is b n t e k nπ t b n t e k nπ e k nπ t g n dt + c t e k nπ t b n t kn π g n + c kn π g n + ce k nπ t Where c above is constant of integration. Hence the solution becomes u x, t At t, u x, f x, therefore Therefore Solving for c gives b n t sin n π x kn π g n + ce k f x kn π g n + c c nπ t sin n π x kn π g n + c sin n π x f x sin n π x dx f x sin n π x dx kn π g n This completes the solution. Everything is now known. Summary u x, t b n t sin n π x b n t kn π g n + ce k g n c nπ t nπ g x sin x f x sin n π x dx kn π g n 93

94 EXERCISES 3.5 HWs.5 HW Consider xbnsinnrx a Determine bn from , 3.3.1, and Problem 3.5. b Fdr what values of x is an equality? *c Derive the Fourier cosine series for x3 from For what values of x will this be an equality? a Using and 3.3.1, obtain the Fourier cosine series of x. b From part a, determine the Fourier sine series of x Generalize Exercise 3.5., in order to derive the Fourier sine series of xm, m odd. Part a Suppose that cosh x - F_', b,, sin nirx/. Equation , page 1 is the Fourier sin series of x a Determine bn by correctly differentiating this series twice. b Determine x bn by Bintegrating n sin n π x this series twice. < x < Where Show that Bn in satisfies Bn an/n7r/, where a is defined by B n nπ 1n Evaluate The goal is to find the Fourier cos series of 1 x Since x x tdt, then x x tdt. Hence from [ x by evaluating x at x. B n sin n π ] t dt Interchanging * the Evaluate order of summation and integration the above becomes T3- x + 73 x B n sin n π t dt using Complex Form of Fourier n π Series cos n π B t n [ With periodic boundary conditions, we have found the theory of Fourier series to be quite useful: nπ B n cos n π ] x t [ nirx nlrx f x. ac nπ + F, B n cos n π ] an cos f- x 1 + bn sin nπ B n cos n π x + nπ B n But a Fourier cos series has the form Comparing 1 and gives Using for B n the above becomes And x nπ B n cos n π x + x A + A A n nπ B n A n nπ A n cos n π x nπ 1n+1 1 n nπ B n nπ 4 π nπ 1n+1 1 n+1 1 n nπ B n nπ 1 94

95 .5 HW 5 But 1n+1 1 π, hence the above becomes n Summary The Fourier cos series of x is Part b Since 1 x A A 4 π π 1 3 A n cos n π x 1 n cos n π nπ x x 3 3 x t dt Then, using result from part a for Fourier cos series of t results in [ x x 3 3 A + A n cos n π ] t dt x 3 3 dt + 3 x 1 n nπ t x n x nπ [ x n sin n π t nπ n π x + 3 nπ 1n nπ x cos n π t dt cos n π t dt ] x [ sin n π t ] x 1 n 3 sin n π nπ x Using which is x B n sin n π x, with B n nπ 1n+1 the above becomes x 3 Combining all above terms [ x 3 nπ 1n+1 sin n π x Will try to simplify more to obtain B n x 3 nπ 1n n 1 n 3 nπ 1 n 3 nπ [ ] nπ [ ] nπ 1 n 3 sin n π nπ x ] 3 sin n π nπ x sin n π x sin n π x Comparing the above to the standard Fourier sin series x 3 B n sin n π x then the above is the required sin series for x 3 with [ B n 1 n 3 ] sin n π nπ nπ x Expressing the above using B n from x 1 to help find recursive relation for next problem. 95

96 HWs Will now use the notation i B n to mean the B n for x i. Then since 1 B n nπ 1n+1 1 n nπ for x, then, using 3 B n as the B n for x 3, the series for x 3 can be written 3.6. Complex Form of Fourier Series 131 [ x 3 1 n nπ + 6 ] n π sin n π EXERCISES 3.5 x [1 1 n B n + 6 ] n π sin n π Consider x xbnsinnrx Where now [ 3 B n 1 n Bn ] a Determine bn from , 3.3.1, and n π b Fdr what values of x is an equality? The above will help in the next problem in order to find recursive relation. *c Derive the Fourier cosine series for x3 from For what values of x will this be an equality?.5. Problem a Using and 3.3.1, obtain the Fourier cosine series of x. b From part a, determine the Fourier sine series of x Generalize Exercise 3.5., in order to derive the Fourier sine series of xm, m odd Consider Suppose that cosh x - F_', b,, sin nirx/. Result from ast a problem Determine showed bn by that correctly differentiating this series twice. b Determine bn by integrating this series twice. x Bn 1 sin n π a Determine bn from , Show that Bn in satisfies Bn an/n7r/, x 3.3.1, and where a is defined by b Fdr what values of x is B n 1 n an equality? Evaluate nπ And 3.6. Complex Form of Fourier Series EXERCISES 3.5 This suggests * that Evaluate [1 by evaluating x n at x B n. + 3 ] n π sin n π x using T3- + [3 1 n B n n π 3.6 Complex Form 1 B n + 6of nfourier π Series 3 B n 1 n [ ] sin n π x And in general With periodic boundary conditions, we have found the theory of Fourier series to be quite useful: [m x m 1 n B n + m! ] n π sin n π b Determine bn by integrating this series twice. x nirx nlrx f x. ac + F, an cos f- + bn sin Where [ m B n 1 n m 4 B n + m! ] n π The above is a recursive definition to find x m Fourier series for m odd..5.3 Problem ] 131 xbnsinnrx *c Derive the Fourier cosine series for x3 from For what values of x will this be an equality? a Using and 3.3.1, obtain the Fourier cosine series of x. b From part a, determine the Fourier sine series of x Generalize Exercise 3.5., in order to derive the Fourier sine series of xm, x 5 m odd. Suppose that cosh x - F_', b,, sin nirx/. a Determine bn by correctly differentiating this series twice Show that Bn in satisfies Bn an/n7r/, where a is defined by Evaluate by evaluating at x * Evaluate T3- + using Complex Form of Fourier Series Equation is x 4 π 3 With periodic x boundary conditions, we have found the theory of Fourier series to sin be quite useful: πx 3πx 5πx 7πx sin sin sin nirx nlrx f x. ac + F, an cos f- + bn sin 3.6.1

97 .5 HW 5 etting x Hence Or in gives sin π π 3 + sin 3π sin 5π sin 7π π 3 sin π + sin 3 π sin 5 π sin 7 π π π π 3 π π Problem Chapter 3. Fourier Series EXERCISES 3.6 * Consider x < xo f x 1/ xo <X < xo + x>xo+. Assume that xo > - and xo + A <. Determine the complex Fourier coefficients c,, If f x is real, show that c_,a c,,. The function defined above is the Dirac delta function. in the limit, as. Now c n 1 1 f x e in π x dx x + 1 x ] e in π x x + in π x ein π x dx [ [e in π x] x + inπ x 1 in π e in π x + e in π x Since eiz e iz i sin z. The denominator above has i in it. Factoring out e in π x + from the above gives 1 c n in π ein 1 n π ein π x + π x + e in π e in π e in π e in π Now the form is sin z is obtained, hence it can be written as π x + c n ein n π i sin n π Or c n cos n π x + + i sin n π x + sin n π nπ 97

98 5j Ox, 4..9 where c Tolpox. Equation 4..9 is called the one-dimensional wave equation. The notation c is introduced because To/pox has the dimensions of velocity HWs squared. We will show that c is a very important velocity. For a uniform string, c is constant..5.5 Problem 4..1 EXERCISES a Using Equation 4..7, compute the sagged equilibrium position uex if Qx, t -g. The boundary conditions are uo and u. b Show that vx, t ux, t - uex satisfies Show that c has the dimensions of velocity squared. Part a Consider a particle whose x-coordinate in horizontal equilibrium is designated by a. If its vertical and horizontal displacements are u and v, Equation 4..7 isrespectively, determine its position x and y. Then show that ρ x u t dy T u + Q x, t ρ x 4..7 x 8u/8a Replacing Q x, t by g dx v/8a' ρ x u Derive equations for horizontal t T u gρ x and xvertical displacements without ignoring string v. Assume is sagged that the butstring is notis moving. perfectly flexible and that the tension is At equilibrium, the determined by an experimental law Derive the partial y differential equation for a vibrating string in the simplest possible manner. You may assume the string has constant mass density po, you may assume the tension To is constant, and you may assume small displacements with small slopes. g u E x This is equilibrium position. Sagged due to only weight of string Therefore u E t. The above becomes T u E x gρ x This is now partial differential equation in only x. It becomes an ODE d u E dx gρ x T With boundary conditions u E, u E. By double integration the solution is found. Integrating once gives du x E dx gρ s ds + c 1 T Integrating again x s gρ z u E dz + c 1 ds + c T x s gρ z x dz ds + c 1 ds + c g T x T s ρ z dzds + c 1 x + c 1 Equation 1 is the solution. Applying B.C. to find c 1, c. At x the above gives The solution 1 becomes And at x the above becomes u E g T x g T c 1 g T c s s s ρ z dzds + c 1 x ρ z dzds + c 1 ρ z dzds 98

99 .5 HW 5 Substituting this into gives the final solution u E g x s T If the density was constant, 3 reduces to u E gρ T g ρ z dz ds + T x gρ T x gρ T gρ sds + T T x gρ x x s sds x ρ z dz ds x 3 Here is a plot of the above function for g 9.8, 1, T 1, ρ.1 for verification. ue sagged equilibrium position ue x Part b Equation 4..9 is Since And ρ x u E t Then by subtracting from 1 u t T ρ x u x 4..9 ρ x u t T u + Q x, t ρ x 1 x u E T + Q x, t ρ x x ρ x u t ρ x u E u t T x + Q x, t ρ x T u E x u ρ x t u E u t T x u E x Q x, t ρ x Since v x, t u x, t u E x, t then v t equation becomes u u E t t and v x u u E x x, therefore the above Which is QED. ρ x v t T v x v t T v ρ x x c v x 99

100 HWs ignated by a. If its vertical and horizontal displacements are u and v, respectively, determine its position x and y. Then show that dy 8u/8a dx v/8a'.5.6 Problem Derive 4..5 equations for horizontal and vertical displacements without ignoring v. Assume that the string is perfectly flexible and that the tension is determined by an experimental law Derive the partial differential equation for a vibrating string in the simplest possible manner. You may assume the string has constant mass density po, you may assume the tension To is constant, and you may assume small displacements with small slopes. et us consider a small segment of the string of length x from x to x + x. The mass of this segment is ρ x, where ρ is density of the string per unit length, assumed here to be constant. et the angle that the string makes with the horizontal at x and at x + x be θ x, t and θ x + x, t respectively. Since we are only interested in the vertical displacement u x, t of the string, the vertical force on this segment consists of two parts: Its weight acting downwards and the net tension resolved in the vertical direction. et the total vertical force be F y. Therefore net tension on segment in vertical direction weight {}}{{}}{ F y ρ xg + T x + x, t sin θ x + x, t T x, t sin θ x, t Applying Newton s second law in the vertical direction F y ma y where a y ux,t t gives the equation of motion of the string segment in the vertical direction and m ρ x, ρ x u x, t t Dividing both sides by x ρ xg + T x + x, t sin θ x + x, t T x, t sin θ x, t ρ u x, t t Taking the limit x ρg + T x + x sin θ x + x, t T x sin θ x, t x ρ u x, t t ρg + T x, t sin θ x, t x Assuming small angles then u u x x giving ρ u x, t sin θ tan θ cos θ sin θ, then we can replace sin θ in the above with t ρg + T x, t x u x, t x Assuming tension T x, t is constant, say T then the above becomes ρ u x, t t ρg + T x u x, t t T u x, t ρ x ρg u x, t x Setting T ρ c then the above becomes u x, t t c u x, t x ρg Note: In the above g gravity acceleration was used instead of Q x, t as in the book to represent the body forces. In other words, the above can also be written as u x, t t c u x, t x + ρq x, t This is the required PDE, assuming constant density, constant tension, small angles and small vertical displacement. 1

101 shape and the other to the right at velocity c with a different fixed shape. We are claiming that the solution to the one-dimensional wave equation can be written as ux, t Rx - ct + Sx + ct,.5 HW 5 even if the boundary conditions are not fixed at x and x. We will show.5.7 Problem and discuss this further in the Exercises and in Chapter 1. EXERCISES Consider vibrating strings of uniform density po and tension To. *a What are the natural frequencies of a vibrating string of length fixed at both ends? *b What are the natural frequencies of a vibrating string of length H, which is fixed at x and "free" at the other end [i.e., Ou/8xH, t 1? Sketch a few modes of vibration as in Fig c Show that the modes of vibration for the odd harmonics i.e., n 1, 3, 5,... of part a are identical to modes of part b if H /. Verify that their natural frequencies are the same. Briefly explain using symmetry arguments In Sec. 4. it was shown that the displacement u of a nonuniform string satisfies u 9u Part a Po To 8x + Q, The natural frequencies where Q of represents vibrating the string vertical of component length with of the fixed body ends, force isper given unit by length. equation in the book, which If Q is the, the solution partial to differential the stringequation wave equation is homogeneous. A slightly different homogeneous equation occurs if Q au. a u Show x, t that if sin a < n, the π body force is restoring toward u. Show that if a >, the body x A n cos n πc force tends to t + B n sin n πc push the string further t away from its unperturbed position u. The frequency of the time solution part of the PDE is given by the arguments of eigenfucntions b Separate variables if pox and ax but To is constant for physical A n cos n πc t + B n sin reasons. n πc t Analyze. Therefore the time-dependent n πc represents ordinary the circular differential frequency equation. ω Comparing general form of cos ωt with cos n *c Specialize part πc b t we see that each mode n has circular frequency given by to the constant coefficient case. Solve the initial value problem if a < : ω n n πc u, t ux, For n 1,, 3,. In cycles per seconds Hertz, and since ω πf, then πf n πc. Solving for f gives u, t 5 x, f W. Where c T ρ Part b What are the frequencies f n of vibration? n πc π n c in all of the above. Equation above was for a string with fixed ends. Now the B.C. are different, so we need to solve the spatial equation again to find the new eigenvalues. Starting with u X x T t and substituting this in the PDE ux,t t c ux,t x T X c T X 1 T c T with < x < H gives X X λ Where both sides are set equal to some constant λ. We now obtain the two ODE s to solve. The spatial ODE is And the time ODE is X + λx X X H T + λc T The eigenvalues will always be positive for the wave equation. Taking λ > the solution to the space ODE is X x A cos λx + B sin λx Applying first B.C. gives A 11

102 HWs λx Hence X x B sin and X x B λx λ cos. Applying second B.C. gives B λ cos λh Therefore for non-trivial solution, we want λh n π for n 1, 3, 5, or written another way λh n 1 π n 1,, 3, Therefore λ n n 1 π n 1,, 3, H These are the eigenvalues. Now that we know what λ n is, we go back to the solution found before, which is u x, t sin λn x A n cos λn ct + B n sin λn ct And see now that the circular frequency ω n is given by ω n λ n c n 1 π c n 1,, 3, H In cycles per second, since ω πf then n 1 πf n π c H n 1 f n c n 1,, 3, H The following are plots for n 1,, 3, 4, 5 for t 3 seconds by small time increments. *solution for HW 5, problem 4.4.1* f[x_, n_, t_] : Module[{H 1, c 1, lam}, lam n - 1/ Pi/H; Sin[lam x] Sin[lam c t] ] ; Table[Plot[f[x, 1, t], {x,, 1}, AxesOrigin -> {, }], {t,,3,.5}]; p abeled[show[ Mode vibration, from t to t3 seconds Mode n vibration, from t to t3 seconds 1

103 .5 HW Mode n3 vibration, from t to t3 seconds Mode n4 vibration, from t to t3 seconds Mode n5 vibration, from t to t3 seconds Part c For part a, the harmonics had circular frequency ω n nπ c. Hence for odd n, these will generate π c, 3 π c, 5 π c, 7 π c, 1 For part b, ω n modes gives n 1 π H c. When H, this becomes ω n n 1 π c. ooking at the first few c, 1 π c, 3 1 π c, 4 1 π c, 1 1 π π c, 3π c, 5π c, 7π c, Comparing 1 and we see they are the same. Which is what we asked to show. 13

104 HWs.5.8 Problem Chapter 4. Wave Equation Consider a slightly damped vibrating string that satisfies 11 Po & T o a Briefly explain why /3 >. *b Determine the solution by separation of variables that satisfies the boundary conditions and the initial conditions u, t and u, t ux, fx and 8tx, gx- You can assume that this frictional coefficient Q is relatively small 3 < 4rrpoTo/ Redo Exercise 4.4.3b by the eigenfunction expansion method Redo Exercise 4.4.3b if 4rrpoTo/ < p < 16rrpoTo/. Part a For , from show that u ux, ρ t Rx - ct + Sx + ct, t T u x β u t The term β u where R and S are some functions. t is the force that acts on the spring segment due to damping. This is the Viscous damping force which If a vibrating is proportional string satisfying to speed, where β represents is initially viscous at rest, damping gx coefficient., This damping force always show that opposes the direction of the motion. Hence if u u t > then β t should come out to be negative. This occurs if β ux, >. t OnI [Fx the other - ct + hand, Fx + ifct], u u t < then β t should now be positive. Which means again that β must be positive quantity. Hence only case were the damping force always opposes where the Fx motion is the odd of the periodic string extension is when β of > f. x. Hints. 1. For all x, Fx _ An sin!. Part b. sin a cos b [sina + b + sina - b]. Starting with u Comment: X x T t This and result substituting shows that this the inpractical the above difficulty PDE with of summing < x < an gives infinite number of terms of a Fourier series may be avoided for the onedimensional wave equation. ρ T X T T X βt X If a vibrating string ρ satisfying T is initially unperturbed, f x _, with the initial velocity T T + β T given, T T show X X that λ Ect Hence we obtain two ODE s. The space ODE is ux, t 1 Gx dam, c X + λx t where Gx is the odd periodic extension X of gx. Hints: And the time ODE is 1. For all x, Gx _ O_1 X nir-c sin nt T + c βt + c λt T f x T g x The eigenvalues will always be positive for the wave equation. Hence taking λ > the solution to the space ODE is X x A cos λx + B sin λx Applying first B.C. gives A λx Hence X B sin. Applying the second B.C. gives Therefore B sin λ λ nπ n 1,, 3, nπ λ n 1,, 3, 14

105 .5 HW 5 Hence the space solution is X nπ b n sin x Now we solve the time ODE. This is second order ODE, linear, with constant coefficients. 1 ρ T T T + β T T T λ ρ T + β T + λt T T T + β ρ T + T ρ λt Where in the above λ λ n for n 1,, 3,. The characteristic equation is r + c βr + c λ. The roots are found from the quadratic formula Replacing λ nπ, gives r 1, B ± B 4AC A β ρ ± β ρ 4 T ρ λ β ± 1 β 4 T λ ρ ρ r 1, β ± 1 β ρ ρ β β ρ ± 1 β ρ ± 1 ρ ρ 4 T nπ ρ ρ 4 T n π ρ β n π 4ρ T We are told that β π < 4ρ T, what this means is that β n π 4ρ T <, since n >. This means we will get complex roots. et n π 4ρ T β Hence the roots can now be written as r 1, β ± i ρ ρ Therefore the time solution is T n t e β ρ t A n cos t + B n sin t ρ ρ This is sinusoidal damped oscillation. Therefore T t e β t ρ A n cos t + B n sin t ρ ρ Combining 1 and, gives the total solution nπ u x, t sin x e β t ρ A n cos t + B n sin t ρ ρ 3 Where b n constants for space ODE merged with the constants A n, B n for the time solution. Now we are ready to find A n, B n from initial conditions. At t f x nπ sin x A n 15

106 HWs Multiplying both sides by sin mπ x and integrating gives mπ f x sin x dx Changing the order of integration and summation Hence mπ f x sin x dx A n A n A m mπ nπ sin x sin x A n dx mπ nπ sin x sin x dx nπ f x sin x dx To find B n, we first take time derivative of the solution above in 3 which gives t u x, t nπ sin x e β t ρ A n sin t + B n cos t ρ ρ ρ ρ β nπ sin ρ x e β t ρ A n cos t + B n sin t ρ ρ At t, using the second initial condition gives g x nπ sin x B n β nπ A n sin ρ ρ x Multiplying both sides by sin mπ x and integrating gives mπ g x sin x dx Changing the order of integration and summation mπ g x sin x dx B n ρ mπ nπ sin x sin x β mπ nπ B n dx A n sin ρ ρ x sin x B m ρ β A n ρ B m β A n ρ ρ mπ nπ sin x sin x dx β ρ A n mπ nπ sin x sin x Hence B m ρ β ρ A n B m mπ g x sin g x sin x dx mπ x dx + β A n ρ ρ This completes the solution. Summary of solution u x, t nπ sin x e β t ρ A n cos t ρ A n B n n π 4ρ T nπ f x sin x dx mπ g x sin β x + B n sin dx + β ρ A n ρ t ρ 16

107 4.5. Vibrating Membrane Problem sin a sin b 1 [cosa - b cosa + b]. See the comment after Exercise From 4.4.1, derive conservation of energy for a vibrating string,.5 HW 5 Hence de '9U &U wt -c8x8to, where the total energy E is the sum of the kinetic energy, defined by f 8u dx, and the potential energy, defined by f z & dx What happens to the total energy E of a vibrating string see Exercise a If u, T and u, t b If Ou,t and u,t E 1 u dx + c u dx c If u, t and Ou t, t -ryu, t with xy > d If y < in part c Show that de the potential and kinetic energies defined in Exercise are equal for dt a traveling 1 d u dx + c d u dx dt wave, t u Rx - ct. dt x Moving d dt inside Using the integral, , itprove becomes that the partial solution derivative of is unique a Using , calculate the energy of one normal mode. de b Show that the total energy, when ux, t satisfies , is the sum of the dt 1 u dx + c u dx 1 energies contained t t in each mode. t x But 4.5 Vibrating u Membrane u u u u t t t t t t t + u u u t t u t t The heat equation in one spatial dimension is 8u/8t k8u/8x. In two or three And dimensions, the temperature satisfies 8u/8t kvu. In a similar way, the vibration of a string one u dimension can u u be extended u to uthe vibration of a membrane two dimensions. t x t x x x t x + u u x x t u u x x t The vertical displacement of a vibrating string satisfies the one-dimensional wave Substituting,3 into 1 gives equation 8u 8u de c 8x There are important dt 1 u u physical problems t t dx + c u u that solve x x t dx u u,9 cvu, t t dx + c u u x x t dx But u t known c u as then the the two- above or three-dimensional becomes wave equation. An example of a physical problem x that satisfies a two-dimensional wave equation is the vibration of a highly stretched membrane. This can be thought of as a two-dimensional vibrating string. We will give a brief de [ ] derivation u in the manner described by Kaplan [1981], omitting dt c u t x dx + c u u x x t dx u c u t x dx + c u u x x t dx u c u u u t x + x x t But since the integrand in 4 can also be written as Then 4 becomes x Which is what we are asked to show. u t u x u u x t x + u u t x de dt c x c u t QED. u x u t u x 3 dx 4 dx 17

108 HWs 168 Chapter 5. Sturm-iouville Eigenvalue Problems.6 HW 6 EXERCISES Problem 5.3. Part a * Do Exercise 4.4.b. Show that the partial differential equation may be put into Sturm-iouville form Consider U 11u 8u To 8x +au+/3. a Give a brief physical interpretation. What signs must a and 'o have to be physical? b Allow p, a, /3 to be functions of x. Show that separation of variables works only if Q cp, where c is a constant. c If cp, show that the spatial equation is a Sturm-iouville differential equation. Solve the time equation. * Consider the non-sturm-iouville differential equation dx + ax dx + [AQx + -Yxlo. Multiply this equation by Hx. Determine Hx such that the equation ρ u may be reduced to the tstandard T u u + αu + β t Sturm-iouville t form: The PDE equation represents the vertical d displacement u x, t of the string as a function of time and horizontal position. This is 1Ddx wave [px equation. d-j + [Ao,x The + term qx1 q5 β u. t represents the damping force can be due to motion of the string in air or fluid. The damping coefficient β must be negative to make β u Given ax, 3x, and -yx, what are px, ax, and qx? t opposite to direction of motion. Damping force is proportional to velocity and acts opposite to direction Consider of motion. heat flow with convection see Exercise 1.5.: The term αu represents the stiffness in the system. This is a restoring force, and acts also U opposite to direction of motion and is proportional 19U to current 49U displacement from equilibrium position. Hence α < also. cat ka - VoOx a Show that the spatial ordinary differential equation obtained by separation of variables is not in Sturm-iouville form. Part b *b Sore the initial boundary value problem et u X x T t. Substituting this into the above PDE gives u,t u, t ux, f x. ρt X T X T + αxt + βt X Dividing by XT c Solve the initial boundary value problem ρ T T T X O,t X + α + β T T ρ T, t T β T TX_ X Tzux, T T X + α fx. To make each side depends on one variable only, we move ρ x, β x to the right side since these depends on x. Then dividing by ρ x gives T T β T ρ T T X ρx + α ρ If βx βx ρx c is constant, then we see the equations have now been separated, since ρx on x any more and the above becomes do not depend T T ct T T X ρx + α x ρ x Now we can say that both side is equal to some constant λ giving the two ODE s Or T T ct T λ T X ρx + α ρ λ T ct + λt α X + X + λ ρ T T 18

109 .6 HW 6 Part c From above, the spatial ODE is α X + X + λ ρ 1 T T Comparing to regular Sturm iouville RS form, which is d px + qx + λσx dx px + p X + q + λσ X Comparing 1 and we see that p 1 q α T σ ρ T To solve the time ODE T ct +λt, since this is second order linear with constant coefficients, then the characteristic equation is r cr + λ r B B A ± 4AC A c c ± 4λ Hence the 168 two solutions are Chapter 5. Sturm-iouville Eigenvalue Problems EXERCISES 5.3 T 1 t e c + c 4λ c c 4λ * Do Exercise 4.4.b. Show that the partial t differential equation may be put into Sturm-iouville T form. t e The general solution Consider is linear combination of the above two solution, therefore final solution is 11u U c T t c 1 e + c 8u To 4λ c t + c e c 4λ 8x +au+/3. t a Give a brief physical interpretation. What signs must a and 'o have to Where c 1, c are arbitrary be physical? constants of integration. b Allow p, a, /3 to be functions of x. Show that separation of variables.6. Problem 5.3.3works only if Q cp, where c is a constant. c If cp, show that the spatial equation is a Sturm-iouville differential equation. Solve the time equation. * Consider the non-sturm-iouville differential equation t dx + ax dx + [AQx + -Yxlo. Multiply this equation by Hx. Determine Hx such that the equation may be reduced to the standard Sturm-iouville form: d dx [px d-j + [Ao,x + qx1 q5. Given ax, 3x, and -yx, what are px, ax, and qx? Consider heat flow with convection see Exercise 1.5.: 19U d φ dφcat + α x + λβ ka x - VoOx + γ x φ a Show that dx the spatial dx ordinary differential equation obtained by separation gives of variables is not in Sturm-iouville form. Multiplying by H x *b H x Sore φ the x initial + H x boundary α x φ value x + problem H x λβ x + γ x φ 1 Comparing 1 to Sturm iouville form, which is U 49U u,t u, t ux, d f x. pφ + qφ + λσφ dx c Solve p the x initial φ x boundary + p x φ value x + problem q + λσ φ x O,t TX_, t Tzux, fx. 19

110 HWs 5.3. Sturm-iouville Eigenvalue Problems 169 Then we need to satisfy For the Sturm-iouville eigenvalue H x P problem, x H x α x P x x + AO with and, Therefore, by combining the above, we obtain one ODE equation to solve for H x verify the following general properties: dx dx a There is an infinite H number x H of x eigenvalues α x with a smallest but no largest. This is first order separable ODE. H H α or ln H αdx + c or b The nth eigenfunction has n - 1 zeros. c The eigenfunctions H are complete Ae αxdx and orthogonal. d What does the Rayleigh quotient say concerning negative and zero Where A is some constant. eigenvalues? By comparing 1, again, we see that Redo Exercise q λσ for λβ the x Sturm-iouville H x + γ x H eigenvalue x problem Summary of solution QED Which of statements 1-5 σ x of the β x theorems H x of this section are valid for the following eigenvalue q x problem? γ x H x.6.3 Problem Part a dx + A with and. I P x H x H x Ae + AO with αxdx O d- d Show that A > for the eigenvalue problem d + a - x with, 1. Is A an eigenvalue? Consider the eigenvalue problem xdx + x + AO with 1O, and bo a Show that multiplying by 1/x puts this in the Sturm-iouville form. This multiplicative factor is derived in Exercise b Show that A >. *c Since is an equidimensional equation, determine all positive eigenvalues. Is A an eigenvalue? Show that there is an infinite number of eigenvalues with a smallest, but no largest. d The eigenfunctions are orthogonal with what weight according to Sturm-iouville theory? Verify the orthogonality using properties of integrals. e Show that the nth eigenfunction has n - 1 zeros Reconsider Exercise with the boundary conditions dx 1 and b. x φ + xφ + λφ 1 φ 1 φ b Multiplying 1 by 1 x where x gives xφ + φ + λ x φ Comparing to Sturm-iouville form Then pφ + p φ + q + λσ φ 3 p x q σ 1 x And since the given boundary conditions also satisfy the Sturm-iouville boundary conditions, then is a regular Sturm-iouville ODE. 11

111 .6 HW 6 Partb Using equation in page 16 of text called Raleigh quotient, which applies to regular Sturm- iouville ODE, which relates the eigenvalues to the eigenfunctions λ [pφφ ] xb x1 + b 1 p φ qφ dx b 1 φ σdx [p b φ b φ b p 1 φ 1 φ b] + b 1 p φ qφ dx b 1 φ σdx Using p x, q, σ 1 x and using φ 1, φ b, then the above simplifies to λ b 1 p φ dx φ x dx b 1 The integrands in the numerator and denominator can not be negative, since they are squared quantities, and also since x > as the domain starts from x 1, then RHS above can not be negative. This means the eigenvalue λ can not be negative. It can only be λ. QED. Partc The possible values of λ > are determined by trying to solve the ODE and seeing which λ produces non-trivial solutions given the boundary conditions. The ODE to solve is 1 above. Here it is again We know λ, so we do not need to check for negative λ. Case λ. Equation 1 becomes x φ + xφ + λφ 1 x φ + xφ xφ + φ d xφ dx Hence xφ c 1 where c 1 is constant. Therefore d dx φ c 1 x or At x 1, φ 1, hence φ c 1 1 x dx + c c 1 ln x + c c 1 ln 1 + c But ln 1, therefore c. The solution now becomes At the right end, x b, φ b, therefore φ c 1 ln x c 1 ln b But since b > 1 the above implies that c 1. This gives trivial solution. Therefore λ is not an eigenvalue. Case λ > x φ + xφ + λφ This is non-constant coefficients, linear, second order ODE. et φ x x p. Equation 1 becomes x p p 1 x p + xpx p 1 + λx p Dividing by x p gives the characteristic equation p p 1 x p + px p + λx p p p 1 + p + λ p p + p + λ p λ 111

112 HWs Since λ then p is complex. Therefore the roots are Therefore the two solutions eigenfunctions are p ±i λ φ 1 x x i λ φ x x i λ To more easily use standard form of solution, the standard trick is to rewrite these solution in exponential form φ 1 x e i λ ln x φ x e i λ ln x The general solution to 1 is linear combination of these two solutions, therefore φ x c 1 e i λ ln x + c e i λ ln x Since λ > then the above can be written using trig functions as φ x c 1 cos λ ln x + c sin λ ln x We are now ready to check for allowed values of λ by applying B.C s. The first B.C. gives c 1 cos λ ln 1 + c sin λ ln 1 c 1 cos + c sin c 1 Hence the solution now simplifies to Applying the second B.C. gives φ x c sin λ ln x c sin λ ln b For non-trivial solution we want λ ln b nπ n 1,, 3, nπ λ ln b nπ λ n n 1,, 3, ln b Therefore, there are infinite numbers of eigenvalues. The smallest is when n 1 given by π λ 1 ln b Part d From Equation 5.3.6, page 159 in textbook, the eigenfunction are orthogonal with weight function σ x b In this problem, the weight σ 1 x Now we can verify the orthogonality b 1 a φ n x φ m x σ x dx Using the substitution z ln x, then dz then the above integral becomes I φ n x φ m x σ x dx n m and the solution eigenfuctions were found above to be φ n x sin λn ln x zln b z ln b dx 1 x sin xb x1 nπ sin nπ mπ 1 sin ln b ln x sin ln b ln x x dx. When x 1, z ln 1 and when x b, z ln b, ln b z nπ ln b z sin mπ sin dz ln b z dx dx mπ ln b z dz But sin nπ ln b z and sin mπ ln b z are orthogonal functions now with weight 1. Hence the above gives when n m using standard orthogonality of the sin functions we used before many times. QED. 11

113 .6 HW 6 Parte The n th eigenfunction is nπ φ n x sin ln b ln x Here, the zeros are inside the interval, not counting the end points x 1 and x b. And nπ ln b ln x x1 nπ ln b nπ ln b ln x xb nπ ln b ln b nπ Hence for n 1, The domain of φ 1 x is π. And there are no zeros inside this for sin function not counting the end points. For n, the domain is π and sin has one zero inside this at π, not counting end points. And for n 3, the domain is 3π and sin has two zeros inside this at π, π, not counting end points. And so on. Hence φ n x has n 1 zeros not counting the end points..6.4 Problem 5.5. Self-Adjoint b,d,g Operators 181 EXERCISES A Sturm-iouville eigenvalue problem is called self-adjoint if p udv du dx - vdx b since then fq [uv - vu] dx for any two functions u and v satisfying the boundary conditions. Show that the following yield self-adjoint problems. a O and O b V. and O c a - hq5 and d d ta b and pa 1 a pb - b e a b and lka _ b [self-adjoint only if pa pb] f q and [in the situation in which p ] bounded and lim;r.o px- *g Under what conditions is the following self-adjoint if p is constant? + a + Q d!k+-r,+&- d Prove that the eigenfunctions corresponding to different eigenvalues of the following eigenvalue problem are orthogonal: The Sturm-iouville ODE is d dx [px pφ d_] + qφ 4x λσφ + Aox dx Or in operator form, with the defining boundary conditions d dx p d dx + q, becomes The operator is self adjoined when What is the weighting function? b [φ] λσφ Consider the eigenvalue u problem [v] dx v -avx46, [u] dx subject to a given set of homogeneous boundary conditions. Suppose that a For the above to work out, we need tojb show that [uv - vu] dx p uv vu b a for all functions u and v satisfying the same set of boundary conditions. And this is what Prove we will that do eigenfunctions now. corresponding to different eigenvalues are orthogonal with what weight?. b a a 113

114 HWs Partb Here a and b. p uv vu b a p u dv dx v du dx [ p Substituting u v and dv dx given gives Part d p uv vu b a [p u dv du v dx dx [ ] p u dv ] du v dx dx du dx into the above since there are the B.C. dv ] du dx dx p u v p uv vu b a p u dv dx v du a dx b [ p a u a v a v a u a p b u b v b v b u b ] p a u a v a p a v a u a p b u b v b + p b v b u b 1 We are given that u a u b and v a v b and p a u a p b u b and p a v a p b v b. We start by replacing u a by u a and replacing v a by v b in 1, this gives p uv vu b a p a u b v a p a v b u a p b u b v b + p b v b u b u b p a v a p b v b + v b p b u b p a u a Now using p a u a p b u b and p a v a p b v b in the above gives Part g p uv vu b a u b p b v b p b v b + v b p b u b p b u b p is constant. Hence We are given that And u b + v b p uv vu p u dv dx v du dx p [ u v v u u v v u ] 1 u + αu + βu u + γu + δu 3 v + αv + βv 4 v + γv + δv 5 From, From 3 From 4 From 5 u αu βu u γu δu v αv βv v γv δv 114

115 .6 HW 6 Using these 4 relations in equation 1 gives where p is removed out, since it is constant, to simplify the equations uv vu u v v u u v + v u 5.5. Self-Adjoint Operators αu βu γv δv 181 αv βv γu δu EXERCISES 5.5 u v + v u A Sturm-iouville eigenvalue problem is called self-adjoint if Simplifying uv vu du αu γv + αu δv p udv b + βu γv + βu δv dx - vdx a αv γu + αv δu + βv γu + βv δu since u v then fq [uv + v u - vu] dx for any two functions u and v satisfying the boundary conditions. Show that the following yield self-adjoint αu problems. γv + αu δv + βu γv + βu δv αv a γu O and αv O δu βv γu βv δu u v + v u Collecting b V. and O uv vu αδ c a u v - hq5 and v u d + βδ ta u v b and v pa 1 u a d pb - b e a b and lka _ b [self-adjoint only if pa pb] + αγ u v v u f + βγ q u and v v [in the situation u in which p ] bounded and lim;r.o px- u *g v + v u αδ Under what conditions is u v v u the following + βγ self-adjoint if p u v v u is constant? u v v u αδ u v v u βγ v u u v u v v u et u v v u then we!k+-r,+&- see that the above is just d Prove that the uv eigenfunctions vu corresponding αδ βγ to different eigenvalues of the following eigenvalue problem are orthogonal: αδ βγ 1 Hence, for uv vu, we need dx [px d_] + 4x + Aox with the boundary conditions αδ βγ Problem What is the weighting function? + a + Q d Consider the eigenvalue problem -avx46, subject to a given set of homogeneous boundary conditions. Suppose that jb [uv - vu] dx for all functions u and v satisfying the same set of boundary conditions. Prove that eigenfunctions corresponding to different eigenvalues are orthogonal with what weight?. But We are given that b a u [v] v [u] dx 1 [v] λ v σ x v [u] λ u σ x u 3 Where σ x is the weight function of the corresponding Sturm-iouville ODE that u, v are its solution eigenfunctions. Substituting,3 into 1 gives b a u λ v σ x v v λ u σ x u dx b a λ v σ x uv + λ u σ x uvdx λ u λ v b a σ x uvdx 115

116 c If z depends on x and a parameter r, show that HWs d Using part c, evaluate 8z/8r if z e''x. Since u, v are different e Obtain eigenfunctions, a second solution then the of y λ u λ, using v as part these d. are different eigenvalues. There is one eigenfunction Prove that if corresponding x is a root of to a sixth-order each eigenvalue. polynomial Therefore with real thecoefficients, above says that then a is also a root. b σ x u x v x dx For a Hence different eigenfunctions u x, v x are orthogonal to each others. The weight is σ x..6.6 Problem with p and q real, carefully show that arz5 d pd Consider a fourth-order linear differential operator, d a Show that vv - vu is an exact differential. b Evaluate fo [uv - vu] dx in terms of the boundary data for any functions u and v. c Show that fo [uv - vu] dx if u and v are any two functions satisfying the boundary conditions 46 1 d d 1. d Give another example of boundary conditions such that f 1 [uv - vu] dx Self-Adjoint Operators 183 e For the eigenvalue problem [using the boundary conditions in part c] d44 + aez, dx4 show that the eigenfunctions corresponding to different eigenvalues are orthogonal. What is the weighting function? * For the eigenvalue problem Part a dx + Aexb d4 subject to the boundary conditions dx 4 O1 ad 1, -, show that the eigenvalues u [v] are v less [u] than u d4 v or equal to zero A <. Don't worry; in a physical context that is exactly dx 4 what v d4 u dx is 4 expected. Is A an eigenvalue? uv 4 vu 4 We want to obtaina expression Show that of 5.5. form yields d if at least one of the boundary conditions is of the regular Sturm-iouville such that it comes type. out to be uv4 vu 4. If we can do dx this, then it is exact differential. Now, since b Do part a if one boundary condition is of the singular type. d *a Suppose that uv u v u v + uv 4 u v u v 1 dx And px + rx + qx. d Consider vu v u v u + vu 4 v u v u dx b Then 1- gives 1 vu dx. a d uv u v d By vu repeated v u integration u v by + parts, uv 4 determine u v the u v adjoint v operator u + vu ' 4 v u v u dx dx such that b b u 1 [u'v v + uv - vuj 4 u dx v u Hx v v u vu 4 + v u + v u auv 4 vu 4 What is Hx? Under what conditions does ', the self-adjoint Hence we found that case? [Hint: Show that d uv u v vu d p + v \ u uv 4 vu 4 \ 1 dx ' pdx + TX 4 u dx[v] dx v [u] Therefore u [v] b v If[u] is exact differential. u and + u, 116 d Pdr JJ what boundary conditions should vx satisfy for HxI a, called the adjoint boundary conditions?

117 .6 HW 6 Part b I b a b a u [v] v [u] dx d uv u v vu + v u dx dx uv u v vu + v u b a u b v b u b v b v b u b + v b u b u a v a u a v a v a u a + v a u a Or I u b v b u b v b v b u b+v b u b u a v a+u a v a+v a u a v a u a Part c From partb, I Since we are given that 1 u [v] v [u] dx uv u v vu + v u 1 1 φ φ φ 1 φ 1 The above will give u v u v u 1 v 1 u 1 v 1 Substituting these into 1 gives 1 u [v] v [u] dx u 1 v 1 u 1 v 1 v 1 u 1 + v 1 u 1 u v + u v + v u v u Therefore 1 u [v] v [u] dx v 1 u 1 + v + + u Part d Any boundary conditions which makes uv u v vu + v u 1 will do. For example, φ φ φ 1 φ 1 The above will give u v u v u 1 v 1 u 1 v 1 117

118 HWs Substituting these into 1 gives 1 u [v] v [u] dx u 1 v 1 u 1 v 1 v 1 u 1 + v 1 u 1 u v + u v + v u v u v 1 v 1 u 1 + u 1 v + v + u u Part e Given Therefore Therefore, for eigenfunctions u, v we have d 4 dx 4 φ + λex φ [φ] λe x φ [u] λ u e x u [v] λ v e x v Where λ u, λ v are the eigenvalues associated with eigenfunctions u, v and they are not the same. Hence now we can write Self-Adjoint Operators u [v] v [u] dx e For the eigenvalue u problem λ v e x [using v v the λboundary u e x u dx conditions in part c] 1 1 λ v e x d44 uv + λ + u e aez x uvdx dx4 show that the eigenfunctions corresponding λ u λ v e x to different eigenvalues are orthogonal. What is the weighting uv function? dx * For the eigenvalue problem λ u λ v Since λ u λ v then 1 1 e x uv dx dx + Aexb subject to the boundary conditions e x uv dx O1 Hence u, v are orthogonal to each others with weight function e x. 1, ad -,.6.7 Problemshow that the eigenvalues are less than or equal to zero A <. Don't worry; in a physical context that is exactly what is expected. Is A an eigenvalue? Parta a Show that 5.5. yields if at least one of the boundary conditions is of the regular Sturm-iouville type. b Do part a if one boundary condition is of the singular type *a Suppose that Equation 5.5. is Consider px b + rx + qx. 1 vu dx., p φ 1 φ φ φ 1 a constant 5.5. ooking at boundaryby conditions repeated atintegration one end, sayby atparts, x a determine left end, the andadjoint let theoperator boundary' conditions there be such that b b β 1 [u'v φ a + β- φ vuj a dx Hx a Therefore for eigenfunctions φ 1, φ we obtain 118 b If JJ d Pdr 4 What is Hx? Under what conditions does ', the self-adjoint case? [Hint: Show β 1 that φ 1 a + β φ 1 a 1 βd 1 φ a + pβ φ \a \ 1 ' pdx + TX dx dx u and + u, what boundary conditions should vx satisfy for HxI a, called

119 .6 HW 6 From 1, From φ 1 a β 1 β φ 1 a 3 φ a β 1 β φ a 4 Substituting 3,4 into φ 1 φ φ φ 1 gives, at end point a, the following φ 1 a φ a φ a φ 1 a φ 1 a β 1 φ a φ a β 1 φ 1 a β β β 1 β φ a φ 1 a + β 1 β φ a φ 1 a In the above, we evaluated φ 1 φ φ φ 1 at one end point, and found it to be zero. But φ 1φ φ φ 1 is the Wronskian W x. It is known that if W x at just one point, then it is zero at all points in the range. Hence we conclude that φ 1 φ φ φ 1 For all x. This also means the eigenfunctions φ 1, φ are linearly dependent. This gives equation QED. Partb Equation 5.5. is p φ 1 φ φ φ 1 constant 5.5. At one end, say end x a, is where the singularity exist. This means p a. Now to show that p φ 1 φ φ φ 1 at x a, we just need to show that φ 1φ φ φ 1 is bounded. Since in that case, we will have A, where A is some value which is φ 1 φ φ φ 1. But boundary conditions at x 1 must be φ a < and also φ a <. This is always the case at the end where p. Then let φ a c 1 and φ a c, where c 1, c are some constants. Then we write Hence it follows immediately that φ 1 a c 1 φ 1 a c φ a c 1 φ a c φ 1 φ φ φ 1 c 1 c c c 1 Hence we showed that φ 1 φ φ φ 1 is bounded. Then p φ 1φ φ φ 1. QED. 119

120 461, since a1 f, uola dx/ fa la dx. We now proceed in a similar way. Since 1\ < An for n >, it follows that RQ[u] > A, HWs and furthermore the equality holds only if an for n > [i.e., u axj since al already. We have just proved the following theorem: The minimum.7 HW value 7for all continuous functions ux that are orthogonal to the lowest eigenfunction and satisfy the boundary conditions is the next-to-lowest eigenvalue. Further.7.1 Problem generalizations also a follow directly from EXERCISES Use the Rayleigh quotient to obtain a reasonably accurate upper bound for the lowest eigenvalue of a + - x with and 1 b d +ax4with d and 1+-1 *c St +A. with - and!ttl See Exercise a Consider the eigenvalue problem part a subject to and 1. Show that A > be sure to show that d φ A. dx + λ x φ Prove that is valid in the following φ way. Assume u/a is piecewise smooth so that φ 1 u _ E bnnx Putting the equation in the form d Determine bn. [Hint: Using φ Green's formula 5.5.5, show that bn -anan if u and du/dx are continuous dx and x φ λφ if u satisfies the same homogeneous And comparingboundary it to the conditions standard Sturm-iouville as the eigenfunctions formonx.] Shows that p d φ dφ + p + qφ λσφ dx dx p 1 q x σ 1 Now the Rayleigh quotient is λ pφφ p φ qφ dx 1 σφ dx Substituting known values, and since φ, φ 1 the above simplifies to λ 1 φ + x φ dx 1 φ dx Now we can say that 1 λ min λ 1 φ + x φ dx 1 1 φ dx We now need a trial solution φ trial to use in the above, which needs only to satisfy boundary conditions to use to estimate lowest λ min. The simplest such function will do. The boundary conditions are φ,φ 1. We see for example that φ trial x x 1 works, since 1

121 194 Chapter 5. Sturm-iouville Eigenvalue Problems.7 HW 7 φ trial x x, We and thus have φ trial a minimization and φ theorem for the lowest eigenvalue A1. We can ask if there are corresponding trial theorems 1 1for the 1 higher. So will eigenvalues. use this in Interesting 1 generalizations immediately follow from λ min λ 1 x + x If we x 1 insist that al, then dx F O O_ anan 1 x f 6 1 n a dx RQ[u] dx 1 x + x b x 4 x En an fn 'Vna dx dx This means that in addition we are 1 restricting x4 x our + 1 function dx u to be orthogonal to 461, since a1 f, uola dx/ fa 1 la dx. 4x + x 6 We now x 4 proceed + x in a similar way. Since dx 1\ < An for n >, it follows that 1 x4 x + 1 dx 1 3x RQ[u] + x 6 > dx A, and furthermore the equality 1 holds x4 x only + if 1dx an for n > [i.e., u axj since al already. We have just proved x x7 1the following theorem: The minimum value for all continuous functions ux that are 1 5 x5 3 x3 + x orthogonal 7 1 to the lowest eigenfunction and satisfy the boundary conditions is the next-to-lowest eigenvalue. Further generalizations also follow directly 15 from EXERCISES Use the Rayleigh quotient to obtain a reasonably accurate upper bound Hence for the lowest eigenvalue of λ a + - x with and 1.7. Problem 5.6.d b +ax4with d and 1+-1 *c St +A. with - and!ttl See Exercise a Consider the eigenvalue problem subject to A. and 1. Show that A > be sure to show that Prove that is valid in the following way. Assume u/a is piecewise smooth so that u _ E bnnx d φ Determine bn. [Hint: Using dx + λ x φ Green's formula 5.5.5, show that bn -anan if u and du/dx are continuous and φ if u satisfies the same homogeneous boundary conditions as the eigenfunctions φ 1 Onx.] Putting the equation in the form d φ dx x φ λφ And comparing it to the standard Sturm-iouville form Shows that p d φ dφ + p + qφ λσφ dx dx p 1 q x σ 1 Now the Rayleigh quotient is λ pφφ p φ qφ dx 1 σφ dx Substituting known values, and since φ, φ 1 the above simplifies to λ 1 φ + x φ dx 1 φ dx 11

122 HWs Since eigenfunction φ can not be identically zero, the denominator in the above expression can only be positive, since the integrand is positive. So we need now to consider the numerator term only: 1 φ 1 dx + x φ dx For the second term, again, this can only be positive since φ can not be zero. For the first term, there are two cases. If φ zero or not. If it is not zero, then the term is positive and we are done. This means λ >. if φ then 1 φ dx and also conclude λ > thanks to the second term 1 x φ being positive. So we conclude that λ can only be positive..7.3 Problem Problem Consider eigenvalue problem d dr also assume dφ dφ bounded. And r dφ dr λrφ, < r < 1 subject to B.C. φ < you may dr dr 1. a prove that λ. b Solve the boundary value problem. You may assume eigenfunctions are known. Derive coefficients using orthogonality. Notice: Correction was made to problem per class . Book said to show that λ > which is error changed to λ. Part a From the problem we see that p r, q, σ r. The Rayleigh quotient is The term rφφ 1 expands to λ pφφ p φ qφ dr 1 σφ dr rφφ r φ dr 1 rφ dr 1 φ 1 φ 1 φ φ 1 Since φ 1, the above is zero and Equation1 reduces to 1 λ r φ dr 1 rφ dr The denominator above can only be positive, as an eigenfunction φ can not be identically zero. For the numerator, we have to consider two cases. case 1 If φ then we are done. The numerator is positive and we conclude that λ >. case If φ then φ is constant and this means λ is possible hence λ. Now we need to show φ being constant is also possible. Since φ 1, then for φ to be true everywhere, it should also be φ which means φ is some constant. We are told that φ <. Hence means φ is constant is possible value since bounded. Hence φ is possible. Therefore λ. QED. Part b The ODE is rφ + φ + λrφ < r < 1 1 φ < φ 1 In standard form the ODE is φ + 1 r φ + λφ. This shows that r is a regular singular point. Therefore we try φ r a n r n+α Hence φ r φ r n n + α a n r n+α 1 n n + α n + α 1 a n r n+α n 1

123 .7 HW 7 Substituting back into the ODE gives r n + α n + α 1 a n r n+α + n + α a n r n+α 1 + λr a n r n+α n n n n + α n + α 1 a n r n+α 1 + n + α a n r n+α 1 + λ a n r n+α+1 n To make all powers of r the same, we subtract from the power of last term, and add to the index, resulting in n + α n + α 1 a n r n+α 1 + n + α a n r n+α 1 + λ a n r n+α 1 n For n we obtain n α α 1 a r α 1 + α a r α 1 α α 1 a + α a a α α + α a α But a we always enforce this condition in power series solution, which implies Now we look at n 1, which gives α a 1 r n+α a 1 r n+α 1 For n, now all terms join in, and we get a recursive relation a 1 n n 1 a n r n 1 + n a n r n 1 + λa n r n 1 n n 1 a n + n a n + λa n λa n a n n n 1 + n λ n a n For example, for n, we get a λ a All odd powers of n result in a n. For n 4 a 4 λ 4 a λ λ 4 a λ 4 a And for n 6 a 6 λ 6 a 4 λ λ 6 4 a And so on. The series is φ r a n r n n n n λ a a + + a r + + a 4 r a 6 r 6 + a λr a + λ r 4 4 a λ 3 r a + λr λr 4 λr 6 a From tables, Bessel function of first kind of order zero, has series expansion given by 1 n z n J o z n! n z 1 z 4 1 z z z z6 + 1 z + z4 4 z n 13

124 HWs By comparing,3 we see a match between J o z and φ r, if we let z λr we conclude that φ 1 r a J λr We can now normalized the above eigenfunction so that a 1 as mentioned in class. But it is not needed. The above is the first solution. We now need second solution. For repeated roots, the second solution will be But α, hence Hence the solution is φ r φ 1 r ln r + r α φ r φ 1 r ln r + n b n r n b n r n n φ r c 1 φ 1 r + c φ r Since φ is bounded, then c since ln not bounded at zero, and the solution becomes where a is now absorbed with the constant c 1 φ r c 1 φ 1 r cj λr The boundedness condition has eliminated the second solution altogether. Now we apply the second boundary conditions φ 1 to find allowed eigenvalues. Since φ r cj 1 λr Then φ 1 implies cj 1 λ The zeros of this are the values of λ. Using the computer, these are the first few such values. λ λ λ or λ λ λ Hence φ n r c n J λn r φ r φ n r c n J λn r 4 To find c n, we use orthogonality. Per class discussion, we can now assume this problem was part of initial value problem, and that at t we had initial condition of f r, therefore, we now write f r c n J λn r 14

125 5.9. arge Eigenvalues Asymptotic Behavior 15 Here px 1, ax 1 + x, qx, 1. Our asymptotic formula for the eigenvalues is.7 HW 7 n7r n7r nir Multiplying both sides by J λm r A... σ_ and integrating lgives - where σ r / 1 JO 1 + xo11 dx 1 + xo3/i11 3/ _ f r J λm r rdr c n J λm r J λn r rdr In Table we compare numerical results using an accurate numerical scheme on 1 the computer with the asymptotic formula. Equation is even a reasonable approximation if n 1. The percent or relative c m J error λm of r the rdr asymptotic formula improves as n increases. However, the error stays about the same though small. There are improvements to that account 1 c m J for the approximately constant λm r rdr error. c m Ω Table 5.9.: Eigenvalues \n Where Ω is some constant. Numerical Therefore answer* Asymptotic formula n assumed accurate 1 c n f r J λm r rdr Error Ω And φ r c n J λn r With the eigenvalues 7 given as above, which have to be computed for each.1993 n using the computer. *Courtesy of E. C. Gartland, Jr..7.4 Problem b EXERCISES Estimate to leading order the large eigenvalues and corresponding eigenfunctions for pcx + [,\vx + qxq if the boundary conditions are a 4 and *b Consider and d c and + hcb dxj + A1 + xo From textbook, equation 5.9.8, we are given that for large λ large. Take into account amplitude and period variations. φ x σp 1 4 exp ±i σ t λ p t dt λ x σp 1 σ t λ x 4 c 1 cos p t dt + c sin subject to and 461. Roughly sketch the eigenfunctions for A x σ t p t dt 1 Where c 1, c are the two constants of integration since this is second order ODE. For φ, the integral x σt pt dt σt pt dt and the above becomes Hence c 1 and 1 reduces to φ σp 1 4 c1 cos + c sin c 1 σp 1 4 φ x c σp 1 4 sin λ x σ t p t dt Hence φ x c σp 1 4 cos λ x c σp 1 4 cos λ x c σp 1 4 λσ p cos σ t p t dt d x λ dx λ p t dt σ x p x σ t λ x σ t p t dt σ t p t dt 15

126 error. Table 5.9.: Eigenvalues \n HWs Numerical answer* Asymptotic formula n assumed accurate Error Since φ 1then c σp 1 λσ λ p cos σ t p t dt Which means, for non-trivial solution, that σ t λn p t dt n *Courtesy of E. C. Gartland, π Jr. Therefore, for large λ, i.e. large n the estimate is EXERCISES 5.9 n 1 λn π Estimate to leading order the large σt eigenvalues pt dt and corresponding eigenfunctions for pcx n + [,\vx 1 λ n π + qxq if the boundary conditions are σt pt dt a 4 and.7.5 Problem 5.9. *b Consider and d c and + hcb dxj + A1 + xo subject to and 461. Roughly sketch the eigenfunctions for A large. Take into account amplitude and period variations. φ + λ 1 + x φ Comparing the above to Sturm-iouville form pφ + qφ λσφ Shows that p 1 q σ 1 + x Now, from textbook, equation 5.9.8, we are given that for large λ φ x σp 1 4 exp ±i x σ t λ p t dt σp 1 4 c 1 cos λ x σ t p t dt + c sin λ x σ t p t dt 1 Where c 1, c are the two constants of integration since this is second order ODE. For φ, the integral x σt pt dt σt pt dt and the above becomes Hence c 1 and 1 reduces to φ σp 1 4 c1 cos + c sin c 1 σp 1 4 φ x c σp 1 4 sin λ x σ t p t dt Applying the second boundary condition φ 1 on the above gives φ 1 c σp 1 4 sin λ 1 σ t p t dt 16

127 .7 HW 7 Hence for non-trivial solution we want, for large positive integer n 1 σ t λ dt nπ p t nπ λ 1 σt pt dt nπ tdt But tdt , hence nπ λ n λ 6.644n Therefore, solution for large λ is φ x c σp 1 4 sin λ x c σp 1 4 sin.5773n σ t x c 1 + x 1 4 sin.5773n p t dt x σ t p t dt 1 + tdt c 1 + x 1 4 sin.5773n x 3 To plot this, let us assume c 1 we have no information given to find c. What value of n to use? Will use different values of n in increasing order. So the following is plot of φ x 1 + x 1 4 sin.5773n x 3 For x 1 and for n 1,, 3,, 8. 17

128 HWs Plot of ϕx for n 1 Plot of ϕx for n ϕx. ϕx x Plot of ϕx for n 3 x Plot of ϕx for n ϕx. ϕx x Plot of ϕx for n 5 x Plot of ϕx for n ϕx. ϕx x Plot of ϕx for n 7 x Plot of ϕx for n ϕx. ϕx x x 18

129 where I is the length of nx: fa 1 dx..8 HW 8.8 HW 8 Parseval's equality simply states that the length of f squared, fa f a dx, equals the sum of squares of the components of f using an orthogonal basis of functions.8.1 Problem of unit length, 5.1.1an1 an fa b na dx. EXERCISES Consider the Fourier sine series for f x 1 on the interval < x <. How many terms in the series should be kept so that the mean-square error is 1% of f f a dx? Obtain a formula for an infinite series using Parseval's equality applied to the The Fourier sin series of f x 1 on x is given by a Fourier sine series of f x 1 on the interval < x < *b Fourier cosine series of f x x on the nπ interval < x < f x b n sin c Fourier sine series of f x x on the interval x < x < Where Consider any function f x defined for a < x < b. Approximate this function by a constant. Show that the best such constant in the mean-square sense, b n i.e., 1 minimizing the nπ f x sin mean-square deviation is the constant equal to the average of f x over the interval x dx a < x < b a Using 1 Parseval's equality, nπ express the error terms nπ 1 sin of the tail of a series. x dx + +1 sin x dx b Redo part a for a Fourier sine series on the interval < x <. 1 [ nπ nπ ] c If f x is piecewise cos smooth, x + [ nπ nπ ] estimate the tail cos in part x b. Hint: Use integration by parts. 1 [ nπ ] cos Show that if nπ x [ nπ ] cos nπ x 1 f_-p f+qf, nπ [cos cos nπ] [cos nπ cos ] nπ then b + \ -f'f 1 nπ [1 cos nπ] [cos nπ 1] f dx -pf dx nπ l dx _ gfj dx J. a [p \ a We see that b n for n, 4, 6,, and b n odd for n 1, 3, 5, so we can simplify the above to be if f and df /dx are continuous Assuming that the operations of summation and integration can be interchanged, show bthat n if 1 nπ [1 1] [ 1 1] nπ f 1 and 9 Nn n, nπ [] nπ [ ] 1 4 nπ 4 nπ Equation 1 becomes f x,3,5, 4 nπ nπ sin x mean-square error is, from textbook, page 13, is given by equation E f x σ x dx,3,5, α n 1 φ nσ x dx In this problem, φ n sin nπ x and α n a n 4 nπ. The above equation becomes E For σ 1 we know that Hence E becomes f x σ x dx f x σdx E,3,5,,3,5, 16 n π 4 nπ sin nπ x σdx f x σdx sin nπ x σ x dx sin nπ x σdx,3,5, 16 n π 19

130 HWs But f x σdx for σ 1 is just 1 dx, and the above becomes E 8 π 16 π,3,5,,3,5, We need to find N so that E.1. The above becomes We need now to solve for N in the above.1 8 π.1 8 π π N 1 n,3,5, N,3,5, N,3,5, N 1 n 1 n,3,5, A small Mathematica program written which prints the RHS sum for each n, and was visually checked when it reached 1.14, here is the result 1 n 1 n 1 n In[53]: data Table[{i, Sum[1 / n^, {n, 1, i, }]}, {i, 1, 5, }] // N; Grid[Join[{{"n", "sum"}}, data], Frame All] Out[54] n sum Counting the number of terms needed to reach 1.14, we see there are 1 terms 1 rows in the table, since only odd entries are counted, the table above skips the even n values in the sum since these are all zero. 13

131 Parseval's equality simply states that the length of f squared, fa f a dx, equals the sum of squares of the components of f using an orthogonal basis of functions of unit length, an1 an fa b na dx. EXERCISES Problem Consider 5.1. b the Fourier sine series for f x 1 on the interval < x <. How many terms in the series should be kept so that the mean-square error is 1% of f f a dx? Obtain a formula for an infinite series using Parseval's equality applied to the a Fourier sine series of f x 1 on the interval < x < *b Fourier cosine series of f x x on the interval < x < c Fourier sine series of f x x on the interval < x <.8 HW Consider any function f x defined for a < x < b. Approximate this function by a constant. Show that the best such constant in the mean-square Parseval s equality sense, is i.e., given minimizing by equation the mean-square , page deviation 14 in textbook is the constant equal to the average of f x over b the interval a < x < b. b a Using Parseval's equality, f σdx express a n the φerror nσdx in terms of the tail of a a series. a The books uses αb n instead Redo part of aa n, for buta it Fourier is the sine same, series these on are the the interval coefficients < x < in. the Fourier series for f x. We nowc need If f x to find is piecewise the cosine smooth, Fourier estimate series for the f tail x in part x. This b. Hint: is givenuse by integration by parts. nπ Show that if f x a + a n cos x Where But Hence then -f'f nπ b f f x dx cos -pf x dx dx f_-p f+qf, + a \ n 1 l dx _ gfj dx J. a [p \ a if f and df 1 [ nπ nπ ] /dx are x continuous. cos x dx + +x cos x dx Assuming that the operations of summation and integration can be interchanged, show 1 [ nπ nπ ] that if x cos x dx + x cos x dx f and 9 Nn n, nπ x cos x 1 + 1n dx n π nπ x cos x 1 + 1n dx n π a n 1 ooking at few terms to see the pattern Therefore, we can write a n as And a n π ] 1 + 1n [ n π n π n { 1 },,,, 3 5, a n 4 π n n 1, 3, 5, a 1 [ f x dx x dx + [ ] x ] ] +x dx ] [ [ x + [ ] [ ]] [ + [ ] + 131

132 HWs Hence the Fourier series is f x +,3,5, 4 nπ π n cos x We now go back to equation butchapter need to5. addsturm-iouville a to it, sinceeigenvalue there is this Problems extra term with cosine Fourier series where I is b the length of nx: b b f σdx a 1 dx + a n φ nσdx a x dx dx + [ x 3 of unit length, an1 an fa b + na dx. 3 4 a fa a 1 dx. 4 π n cos nπ x dx Parseval's equality simply states that,3,5, the length of f squared, fa f a dx, equals the sum of squares ] of the components of f using an orthogonal basis of functions,3,5, 16 π 4 n 4 cos nπ x Since cos nπ x dx the above becomes Consider the Fourier 3 sine series for f x 1 on the interval < x <. How many terms in 3 the series should be kept 16 so that the mean-square error + is 1% of f f a dx? 4 π 4 n 4,3,5, Obtain a formula for 3 an infinite series using 3 8 Parseval's 3 equality applied to the + 4 π 4 n 4,3,5, a Fourier sine series of f x 1 on the interval < x < *b Fourier cosine series of f x x on the interval < x < π 4 n c Fourier sine series of f x 4,3,5, x on the interval < x < Simplifying Consider any function f x defined for a < x < b. Approximate this function by a constant. Show that the best such constant in the mean-square 1 sense, i.e., minimizing the mean-square deviation is the constant equal to the average of f x over the interval πa 4 < n 4,3,5, x < b. 1 1 n π 4 a Using Parseval's equality, express the error in terms of the tail of a series. 4 8,3,5, b Redo part a for a Fourier sine series on the interval < x <. 1 c If f x is piecewise smooth, n 4 π4 estimate the tail in part b. Hint: Use integration by parts. 96 Hence EXERCISES Show that if,3,5, π 4 f_-p f+qf, Which agrees withen the book solution given in back of book. b + \ -f'f.8.3 Problem f dx -pf dx l dx _ gfj dx J. a [p \ a if f and df /dx are continuous Assuming that the operations of summation and integration can be interchanged, show that if f and 9 Nn n, dx 5.1. Approximation Properties 1 then for normalized eigenfunctions fb fgv dx E " an/jne a generalization of Parseval's equality Using Exercises and 5.1.6, prove that b Az df -unan b -pfdx a fbp r ll z-9fi dx b fgσdx α n φ n β n φ n σdx a a [Hint et g f, assuming that term-by-term differentiation is justified.] b According to Schwarz's α 1 φinequality 1 + α φ proved + βin 1 φexercise 1 + β φ.3.1, + σdx the absolute 1 value of the pointwise a error satisfies 13 M _ z flx - E anon anon -_ > IAnlan 1/ ao InM+1 nm+1 " nm+1 A'll z 11,

133 .8 HW 8 But α 1 φ 1 + α φ + β 1 φ 1 + β φ + α 1 β 1 φ 1 + α 1 β φ 1 φ + α 1 β 3 φ 1 φ α β 1 φ φ 1 + α β φ + α β 3 φ 1 φ α 3 β 1 φ 3 φ 1 + α 3 β φ 3 φ + α 3 β 3 φ 3 + Which means when expanding the product of the two series, only the terms on the diagonal the terms with α i β j φ i φ j with i j will survive. This due to orthogonality. To show this more clearly, we put the above expansion back into the integral 1 and break up the integral into sum of integrals b a fgσdx The above simplifies to b a fgσdx b a. + + b a b a b a α 1 β 1 φ 1σdx + α 1 β 1 φ 1σdx + b α β 1 φ φ 1 σdx + α 3 β 1 φ 3 φ 1 σdx + b a a b α 1 β φ 1 φ σdx + a b a α β φ σdx + α β φ σdx + b a b α 3 β φ 3 φ σdx + b a a b α 1 β 3 φ 1 φ 3 σdx + α β 3 φ 1 φ 3 σdx + a α 3 β 3 φ 3σdx + α 3 β 3 φ 3σdx + b a α n β n φ nσdx + Since all other terms vanish due to orthogonality of eigenfunctions. The above simplifies to b a fgσdx b a α n β n φ nσdx b α n β n a φ nσdx Because the 7.3. eigenfunctions Vibrating Rectangular are normalized, Membrane then b a φ nσdx 1 and the above reduces 87 to the result needed Solve b fgσ dx z α n β z 9U. n a k, + k at axe on a rectangle < x <, < y < H subject to.8.4 Problem O, y, t x,, t ux,y, fx,y, y, t x, H, t. ZTY u Consider the wave equation for a vibrating rectangular membrane < x <, <y<h at - C subject to the initial conditions C7u 'U aul 8x + ay J ux,y, and 5 x,y, fx,y. Solve the initial value problem if a u, y, t, u, y, t, ay x,, t, au x, H, t * b, y, t,, y, t, x,, t, x, H, t Consider z z atz c C k On with k >. parta a Give a brief physical interpretation of this equation. et u X x Y y T t. Substituting this back into the PDE gives b Suppose that ux, y, t fxgyht. What ordinary differential equations are satisfied by f, g, and h? T XY c X Y T + Y XT Consider aplace's equation Dividing by XY T gives 1 T u axe + ay? + az c T X X + Y Y in a right cylinder whose base is arbitrarily shaped see Fig The top is z H and the bottom is z. Assume that a ux, y, 133

134 HWs Since left side depends on t only and right side depends on x, y only, then both must be equal to some constant, say λ We obtain the following 1 T c T X X + Y Y T + c λt X X λ λ Y Again, looking at the second ODE above, we see that the left side depends on x only, and the right side on y only. Then they must be equal to some constant, say µ and we obtain X λ X Y µ Y Which results in two ODE s. The first is X + µx X X Y And the second is λ + Y Y µ Y + Y λ µ Y Y µ λy With B.C. Y Y H Starting with the X ODE since it is simpler, the solution is X c 1 cos µx + c sin µx Applying X gives Hence solution is Applying X gives For non-trivial solution And the eigenfunctions are We now solve the Y ODE. c 1 X c sin µx c sin µ nπ µ n n 1,, 3, nπ X n x sin x Y + λ µ n Y Assuming that λ µ > for all λ, µ, we know this is the only case, since only positive λ µ n will be possible when B.C. are homogeneous Dirichlet. Then, for λ µ >, the solution is Y y c 1 cos λ µn y + c sin λ µn y Y c 1 λ µn sin λ µn y + c λ µn cos λ µn y Applying B.C. Y the above becomes c λ µn Hence c and the solution becomes Y c 1 cos λ µn y Y c 1 λ µn sin λ µn y 134

135 .8 HW 8 Applying second B.C. Y H gives c 1 λ µn sin λ µn H For non-trivial solution we want sin λ µn H λnm µ n m π H λ nm µ n m π H λ nm m π H + µn m, 1,, Hence the eigenfunctions are Y nm cos m π H y m, 1,,, n 1,, 3, For each n, m, we find solution of T + c λ nm T.The solution is T nm t A nm cos c λ nm t + B nm sin c λ nm t Putting all these results together gives u x, y, t + m1 m1 T nm t X nm x Y nm y m1 m1 [ A nm cos c λ n,m t + B nm sin c ] nπ λ n,m t sin x cos m π H y A nm cos c nπ λ nm t sin x cos m π H y B nm sin c nπ λ nm t sin x cos m π H y We now apply initial conditions to find A nm, B nm. At t Hence And the solution becomes u x, y, t u x, y, m m nπ A nm sin x cos m π H y A nm Taking derivative of the solution w.r.t. time t gives t u x, y, t At t the above becomes m α x, y B nm sin c nπ λ nm t sin x cos m π H y c λ nm B nm cos c nπ λ nm t sin x cos m π H y m c nπ λ nm B nm sin x cos m π H y Multiplying both sides by sin nπ x cos m π H y and integrating gives H nπ α x, y sin x cos m π H y dxdy c λ nm B nm c λ nm B nm c λ nm B nm m H H H sin nπ x cos m π H y sin nπ x cos m π H y dxdy dxdy 135

136 HWs Hence 4 H nπ B nm Hc α x, y sin λ nm x cos m π H y dxdy Summary of solution nπ X n x sin x n 1,, 3, nπ µ n n 1,, 3, Y nm y cos m π H y m, 1,, λ nm µ n m π m, 1,,, n 1,, 3, H T nm t B nm sin c λ nm t u x, y, t B nm sin c nπ λ nm t sin x cos m π H y Part b In this case we have m 4 B nm Hc λ nm H nπ α x, y sin x cos m π H y dxdy X + µx X X And the second spatial ODE is λ + Y Y µ Y + Y λ µ Y Y µ λy With B.C. Y Y H Starting with the X ODE. The solution is X c 1 cos µx + c sin µx X c 1 µ sin µx + c µ cos µx First B.C. gives Hence c and the solution becomes c µ X c 1 cos µx X c 1 µ sin µx Second B.C. gives Hence c 1 µ sin µ µ nπ nπ µ n, 1,, Now for the Y solution. This is the same as part a. Y nm y cos m π H y λ nm µ n m π H λ nm m π + µn H m π nπ + m, 1,,, n, 1,, H 136

137 .8 HW 8 For each n, m, we find solution of T + c λ nm T.When n, m, λ nm and the ODE becomes T With solution And total solution is T At + B u x, y, t T nm t X nm x Y nm y T t X x Y y At + B Since X x 1 and Y y 1. Applying initial conditions gives u x, y, B Therefore the solution is u x, y, t At. Applying second initial conditions gives Hence the time solution for n m is A α x, y T tα x, y For each n, m, other than n m, the time solution of T + c λ nm T is T nm t A nm cos c λ nm t + B nm sin c λ nm t Putting all these results together, we obtain u x, y, t + n m n m T nm t X nm x Y nm y n m n m [ A nm cos c λ nm t + B nm sin c ] nπ λ nm t cos x cos m π H y A nm cos c nπ λ nm t cos x cos m π H y B nm sin c nπ λ nm t cos x cos m π H y The difference in partb from parta, is that the space solutions eigenfunctions are now all cosine instead of cosine and sine. When the eigenfunction is cos the sum starts from zero. When eigenfunction is sin the sum starts from 1. Now initial conditions are applied as in part a. Hence A nm. u x, y, n m And the solution becomes u x, y, t n m Taking derivative of the solution w.r.t. time t u x, y, t At t the above becomes n m α x, y nπ A nm cos x cos m π H y B nm sin c nπ λ nm t cos x cos m π H y c λ nm B nm cos c nπ λ nm t cos x cos m π H y n m c nπ λ nm B nm cos x cos m π H y Multiplying both sides by cos nπ x cos m π H y and integrating gives H nπ α x, y cos x cos m π H y dxdy c λ nm B nm c λ nm B nm n m H H cos nπ x cos m π H y dxdy 137

138 HWs Hence Summary of solution Part c 4 B nm Hc λ nm H nπ α x, y cos x cos m π H y dxdy nπ X n x cos x n, 1,, nπ µ n n, 1,, Y nm y cos m π H y m, 1,, λ nm µ n m π m, 1,,, n, 1,, { H tα x, y n m T nm t B nm sin c λ nm t otherwise { tα x, y n m u x, y, t m1 B nm sin c λ nm t cos nπ x cos m π H y otherwise 4 H nπ B nm Hc α x, y cos λ nm x cos m π H y dxdy Same problem, but using the following boundary conditions u, y, t u, y, t u x,, t u x, H, t Since the boundary conditions are homogeneous Dirichlet then the X x ODE solution is nπ X n sin x nπ µ n 1,, 3, And Y y ODE solution is mπ Y nm y sin H y λ nm m π + µn H m π nπ + H m 1,, 3,, n 1,, 3, And the time solution is T nm t A nm cos c λ nm t + B nm sin c λ nm t m 1,, 3,, n 1,, 3, Hence the total solution is At t u x, y, t + m1 m1 T nm t X nm x Y nm y m1 m1 A nm cos c nπ mπ λ nm t sin x sin H y B nm sin c nπ mπ λ nm t sin x sin H y Hence A nm and the solution becomes u x, y, t m1 nπ mπ A nm sin x sin H y B nm sin c nπ mπ λ nm t sin x sin H y 138

139 .8 HW 8 Taking derivative t u x, y, t B nm c λ nm cos c nπ mπ λ nm t sin x sin H y 7.3. Vibrating Rectangular m1 Membrane At t α x, y B nm c nπ mπ Solve λ z nm sin z x sin H y 9U. k, + k m1 at axe Therefore, using on orthogonality a rectangle in D, < x < we, find < y < H subject to H 4 B nm Hc α x, y sin nπ mπ O, y, t λ nm x x,, t sin H y ux,y, fx,y dxdy, y, t x, H, t. ZTY u Summary of solution Consider the wave equation for a vibrating rectangular membrane < x <, <y<h nπ X n x sin x C7u n 1,, 'U 3, aul nπ at - C 8x + ay J subject µ n to the initial conditions n 1,, 3, Y nm y cos m π ux,y, H y m and 1, 5, x,y, 3, fx,y. λ nm µ n m π Solve the initial value problem m 1, if, 3,, n 1,, 3, H T nm t B nm sin c a u, y, t, u, λ nm y, t t, ay x,, t, au x, H, t u x, y, t B nm sin c nπ λ nm t sin x sin m π * b, y, t,, y, t, x,, t, x, H y H, t m Consider 4 B nm Hc λ nm z H atz c C k On z nπ mπ α x, y sin with x sin k >. H y dxdy a Give a brief physical interpretation of this equation..8.5 Problem b Suppose that ux, y, t fxgyht. What ordinary differential equations are satisfied by f, g, and h? Consider aplace's equation 87 u axe + ay? + az in a right cylinder whose base is arbitrarily shaped see Fig The top is z H and the bottom is z. Assume that a ux, y, ux, y, H f x, y and u on the "lateral" sides. a Separate the z-variable in general. *b Solve for ux, y, z if the region is a rectangular box, < x <, < y<w,<z<h. 88 Chapter 7. Higher Dimensional PDEs V Figure Part a If possible, solve aplace's equation u _ 8x + 8y + az, u u in a rectangular-shaped region, x + u y< x + u < z, < y < W, < z < H, subject et u XY Z where to the X boundary X x, conditions Y Y y, Z Z z. Substituting this back in the above gives a Tx-, y, z X, Y Z + Y XZ ux,,z + Z XY,, y, z, ux, W, z, ux,y, fx,y ux, y, H b u,y,z, u, y, z, ux,, z, ux, W, z fx,z, ux,y,, ux, y, H 139

140 HWs Dividing by XY Z gives X X + Y Y + Z Z X X + Y Y Z Z Since the left side depends on x, y only and the right side depends on z only and they are equal, they must both be the same constant. Say λ, and we write X X + Y λ 1 Y Z Z λ The problem asks to separate the z variable, then the ODE for this variable is With boundary conditions Partb Z λz Z Z H f x, y We will continue separation from parta. From 1 in part a We now need to separate X, Y. Therefore X X + Y Y X X λ λ Y As the left side depends on x only and right side depends on y only and both are equal, then they are equal to some constant, say µ The x ODE becomes And the y ODE becomes With B.C. Y X X µ λ Y Y µ X + µx 1 X X Y Y µ + λ Y + λ µ Y Y Y W Now that we have the three ODE s we start solving them. Starting with the x ODE 1. The solution is nπ X n sin x nπ µ n 1,, 3, For each n there is solution for the y ODE mπ Y nm sin W y mπ λ nm µ n m 1,, 3, W 14

141 .8 HW 8 Or mπ nπ λ nm + n 1,, 3,, m 1,, 3, W And for each n and for each m there is a solution for the z ODE we found in part a, which is The solution is, since λ nm > is Z c 1 cosh λnm z Applying B.C. Z gives Z c 1 λnm sinh λnm z Z λ nm Z Z + c sinh λnm z c λnm Hence c and the solution becomes Z c nm cosh λnm z Putting all these solutions together, we obtain u x, y, z m1 + c λnm cosh λnm z nπ mπ c nm sin x sin W y cosh λnm z Only now we apply the last boundary condition u x, y, H f x, y to find c nm. nπ mπ f x, y c nm sin x sin W y cosh λnm H 94 Chapter 7. Higher Dimensional PDEs m1 Applying D orthogonality this case, , givesthe generalized Fourier coefficient a,,,,, can be evaluated in W two equivalent nπ ways: mπ W f x, y sin x sin W y dxdy c nm cosh λnm H sin nπ x sin mπ a Using one two-dimensional orthogonality formula for the eigenfunctions W y dxdy of V +4 W b Using two one-dimensional orthogonality c nm cosh formulas λnm H 4b. Convergence. As with any Sturm-iouville eigenvalue problem see Sec. Hence 5.1, a finite series of the eigenfunctions of V + ai may be used to approximate W a function f x, y sin f x, nπ c nm x y. sin In particular, we could show that if we mπ measure error in the W y dxdy cosh mean-square λ nm H sense, W rr 4 W nπ mπ W cosh E λ nm H //.i' - dx dy, f x, y sin x sin W y / dxdy with weight function 1, then this mean-square error is minimized by the coefficients a., being chosen by , the generalized Fourier coefficients. It Summary of solution is known that the approximation improves as the number of terms increases. Furthermore, nπ E -' as all the eigenfunctions are included. We say that the series X n Eaaa,, sin converges x in the mean to f. mπ EXERCISES Y nm sin 7.4 W y mπ nπ Consider the eigenvalue problem λ nm + n 1,, 3,, m 1,, 3, W +AO nπ mπ u x, y, z c nm sin, y x sin W y cosh λnm z Ox, m1 4 W nπ mπ c nm W cosh, y λ nm H Ox, H. f x, y sin x sin W y dxdy *a Show that there is a doubly infinite set of eigenvalues. b If H, show that most eigenvalues have more than one eigenfunction..8.6 Problem 7.4. c Derive that the eigenfunctions are orthogonal in a two-dimensional sense using two one-dimensional orthogonality relations Without using the explicit solution of 7.4.7, show that A > from the Rayleigh quotient, If necessary, see Sec. 7.5: a Derive that ff uvv - vvu dx dy fuvv - vvu. fn ds. b From part a, derive Derive If necessary, see Sec [Hint: Multiply by and integrate.] 141

142 HWs Equation is And is two equivalent ways: to approximate a function f φ x, + y. λφ In particular, we could show that if we measure error in the mean-square φ, y sense, φ, y rr E // φ x,.i' - dx dy, φ x, H with weight function 1, then this mean-square error is minimized by the coefficients a., being chosen by , the generalized Fourier coefficients. It is known that the approximation φ φ ˆnds + improves φ as dxdy the number of terms increases. Furthermore, E -' as all the eigenfunctions R are included. We say that the λ R φ dxdy φ φ ˆnds Consider as we arethe toldeigenvalue φ on problem the boundary and this integration is for the boundary only. Hence λ simplifies to a Using one two-dimensional orthogonality formula for the eigenfunctions of V +4 b Using two one-dimensional orthogonality formulas 4b. Convergence. As with any Sturm-iouville eigenvalue problem see Sec. 5.1, a finite series of the eigenfunctions of V + ai may be used series Eaaa,, converges in the mean to f. EXERCISES 7.4 λ R φ dxdy φ dxdy R *a Show that there is a doubly infinite set of eigenvalues. / +AO, y Ox,, y Ox, H. The numerator can b not If be negative, H, show that sincemost the eigenvalues integrand φ have ismore not negative. than one eigenfunction. integrand, because φ can not be identically zero, as it is an eigenfunction. Similarly, the denominator has positive Hence we conclude that λ. c Derive that the eigenfunctions are orthogonal in a two-dimensional sense using two one-dimensional orthogonality relations..8.7 Problem Without using the explicit solution of 7.4.7, show that A > from the Rayleigh quotient, part a If necessary, see Sec. 7.5: a Derive that ff uvv - vvu dx dy fuvv - vvu. fn ds. b From part a, derive Derive If necessary, see Sec [Hint: Multiply by and integrate.] Equation 1- leads to u v u v + u v 1 v u v u + v u u v v u u v + u v v u + v u u v v u u v v u + u v v u But u v v u so the above reduces to u v v u u v v u Therefore u v v u dxdy u v v u dxdy 3 But the RHS of the above is of the form A dxdy where A u v v u here. Which we can apply divergence theorem on it and obtain A ˆn ds. Therefore, using divergence theorem on the RHS of 3, then 3 can be written as u v v u dxdy Which is what is required to show. u v v u ˆnds 14

143 .8 HW 8 Partb Equation is From part a, we found u v v u dxdy R φ λ1 φ λ dxdy if λ 1 λ u v v u ˆnds 1 But we know that, since both u, v satisfy the multidimensional eigenvalue problem on same domain, then And similarly v + λ v v β 1 v + β v ˆn 3 u + λ u u 4 β 1 u + β u ˆn 5 Now we will use,3,4,5 into 1 to obtain From, we see that v λ v v and from 4 u λ u u and from 3 v ˆn β 1 β v and from 5 u ˆn β 1 β u. Substituting all of these back into 1 gives u λ v v v λ u u dxdy u v ˆn v u ˆn ds λ v uv + λ u vu dxdy u β 1 v v β 1 u ds β β β1 λ u λ v uv dxdy [ uv + uv] ds β λ u λ v uv dxdy 6 We now use 6 the above to show that is correct. In 6, if we replace u φ λ1, v φ λ1 and λ u λ 1, λ v λ then 6 becomes λ 1 λ φ λ1 φ λ1 dxdy We see now that for λ 1 λ, then φ λ1 φ λ1 dxdy. Which is what we asked to show. 143

144 In general, the partial differential equation for vx, t is of the same type as for ux, t, but with a different nonhomogeneous term, since rx, t usually does not satisfy the homogeneous heat equation. The initial condition is also usually altered: HWs vx, f x - rx, f x - A - B - A] 9x HW 9 It can be seen that in general only the boundary conditions have been made homogeneous. In Sec. 8.3 we will develop a method to analyze nonhomogeneous problems.9.1 Problem with homogeneous 8..1 a,b boundary conditions. EXERCISES Solve the heat equation with time-independent sources and boundary conditions k 8U + Qx ux, fx if an equilibrium solution exists. Analyze the limits as t - oo. If no equilibrium exists, explain why and reduce the problem to one with homogeneous boundary conditions but do not solve. Assume * a Qx, u,t A, Tax-,t B b Qx,, t, ai, t B 96 c Qx,,t A 96, r,t A * d Qx k, u, t A, u, t B e Qx k, U, t, au, t f Qx sin -i-, a!i, t, Ou, t 8... Consider the heat equation with time-dependent sources and boundary conditions: Part a 9U 8t k8x + Qx, t et ux, fx. u x, t v x, t + u E x 1 Reduce the problem to one with homogeneous boundary conditions if Since Q x in this problem is zero, we can look for u E x which is the steady state solution that satisfies the non-homogenous * a 8u,t boundary At conditions. and &u,t If Q was Bt present, and if it also was time dependent, then we replace u b u, t E x by r x, t which becomes a reference function that only needs At and, t Bt D_X to satisfy the non-homogenous boundary conditions and not the PDE itself at steady state. In 1 * c, t At and u, t Bt TX_ v x, t satisfies the PDE itself but with homogenous boundary conditions. The first step is to find d uo, t and Ou, t + hu, t - Bt u E x. We use the equilibrium solution in this case. At equilibrium u Ex,t e,t and t and hence the, t + hu, t - Bt '67X '67X solution is given d u E or x u E x c 1 x + c At x, u E x A, Hence c A And solution becomes u E x c 1 x + A. at x, u Ex x c 1 B, Therefore u E x Bx + A Now we plug-in 1 into the original PDE, this gives v x, t v x, t k t x + u E x x But u E x x, hence we need to solve v x, t t k v x, t x for v x, t u x, t u E x with homogenous boundary conditions v, t, v,t t initial conditions and v x, u x, u E x f x Bx + A This PDE we already solved before in earlier HW s and we know that it has the following solution v x, t,3,5, b n sin λn x e kλnt nπ λ n n 1, 3, 5, 144

145 .9 HW 9 With b n found from orthogonality using initial conditions v x, f x Bx + A Hence v x, f x Bx + A sin λm x dx,3,5, f x Bx + A sin λm x dx b m b n Therefore, from 1 the solution is u x, t,3,5, b n sin λn x b n sin λn x sin λm x dx f x Bx + A sin λn x dx n 1, 3, 5, 3,3,5, b n sin λn x e kλnt + With b n given by 3 and eigenvalues λ n given by. Part b et u E x {}}{ Bx + A u x, t v x, t + r x 1 Since Q x in this problem is zero, we can look for r x, since unique equilibrium solution is not possible due to both boundary conditions being insulated. The idea is that, if we can find u E then we use that, else we switch to reference function r x which only needs to satisfy the non-homogenous boundary condition u E x but does not have to satisfy equilibrium solution. et At x, second equation above reduces to r x c 1 x + c x r x c 1 + c x c 1 Hence r x c x. Now r x c x. At x, this gives c B or c B, therefore r x B x The above satisfies the non-homogenous B.C. at the right, and also satisfies the homogenous B.C. at the left. Now we plug-in 1 into the original PDE, this gives Hence We now treat k B v x, t t v x, t t v x, t k x v x, t k x v x, t t k v x, t x + u E x x + B + k B k v x, t x + kb as forcing function. So the above can be written as v x, t t k v x, t x + Q The above is now solved using eigenfunction expansion, since no steady state equilibrium solution exist. et v x, t a n t φ n x 3 n 145

146 HWs Where the index starts from zero, since there is a zero eigenvalue, due to B.C. being Neumann. φ n x are the eigenfunctions of the corresponding homogenous PDE vx,t t k vx,t x with homogenous BC v,t t, v,t t. This we solved before. The eigenfunctions are nπ φ n x cos x With eigenvalues λ n n π n, 1,, Notice that λ. Substituting 3 into gives a n t φ n x k a n t d φ n x dx + Q n n Term by term differentiation is justified, since v x, t and φ n x both solve the same homogenous B.C. problem. Since d φ nx λ dx n φ n x the above equation reduces to a n t φ n x k a n t λ n φ n x + Q n Now we expand Q, which gives n By orthogonality case n a n t φ n x k a n t λ n φ n x + q n φ n x n n n a n t + ka n t λ n q n a t + ka t λ q But λ a t q But since Q kb is constant, then kb n q nφ n x implies that kb q φ x. But φ x 1 for this problem. Hence q kb and the ODE becomes a t kb Hence a t kb t + c 1 case n > a n t + ka n t λ n q n Since all q n for n > the above becomes a n t + ka n t λ n Integrating factor is µ e kλnt. Hence d dt an t e kλnt or a n t c e kλnt Therefore the solution from 3 becomes v x, t kb t + c 1 + c e kλnt cos λn x 4 Now we find the initial conditions on v x, t. Since u x, v x, + r x then Hence equation 4 at t becomes v x, f x B x f x B x c 1 + c cos λn x 146

147 .9 HW 9 We now find c 1, c by orthogonality. case n f x B x cos λ x dx c 1 cos λ x dx But λ f x B 35 Chapter 8. Nonhomogeneous Problems x dx c 1 dx In general, the partial differential f x B equation for vx, t is of the same type as for ux, t, but with a different nonhomogeneous x dx c 1 term, since rx, t usually does not satisfy the homogeneous heat equation. The initial condition is also usually altered: c 1 1 f x B vx, f x - rx, f x - A - B - A] x 9x. dx 8..9 case n > It can be seen that in general only the boundary conditions have been made homogeneous. In Sec. 8.3 we will develop a method to analyze nonhomogeneous problems with homogeneous boundary conditions. f x B x cos λm x dx cos λn x cos λm x dx EXERCISES 8. c Solve the heat equation with time-independent c sources and boundary conditions c f B k 8U + Qx x cos λn x dx Therefore the solution for v x, t is now complete from 4. Hence if an equilibrium solution exists. Analyze the limits as t - oo. If no equilibrium u x, exists, t explain v x, t why + r x and reduce the problem to one with homogeneous boundary conditions but do not solve. Assume kb t + c 1 + e kλnt cos λn x + B x c * a Qx, u,t A, ux, fx Tax-,t B b Qx,, t, ai, t B 96 Where c 1, c are given c by Qx above, result. This completes,t A 96 the, solution. r,t A * d Qx k, u, t A, u, t B.9. Problem 8.. e Qx a,d k, U, t, au, t f Qx sin -i-, a!i, t, Ou, t 8... Consider the heat equation with time-dependent sources and boundary conditions: 9U 8t k8x + Qx, t ux, fx. Reduce the problem to one with homogeneous boundary conditions if * a 8u,t At and &u,t Bt b u, t At and, t Bt D_X * c, t At and u, t Bt TX_ d uo, t and Ou, t + hu, t - Bt e,t and, t + hu, t - Bt '67X '67X Part a et u x, t v x, t + r x, t 1 Since the problem has time dependent source function Q x, t then r x, t is now a reference function that only needs to satisfy the non-homogenous boundary conditions which in this problem are at both ends and v x, t has homogenous boundary conditions. The first step is to find r x, t. et r x, t c 1 t x + c t x Then r x, t x c 1 t + c t x 147

148 HWs At x And at x Solving for c 1, c gives Replacing 1 into ux,t t k ux,t x A t c 1 t B t c 1 t + c t c t B t c 1 t r x, t A t x + + Q x, t gives Bt At x But r x v x, t r x, t k v x, t r x, t + Q x, t t x v t r t k v x k r + Q x, t x Bt At, hence the above reduces to et v t k v B t A t + Q x, t k x Q x, t Q x, t + r t k Bt At + r t then becomes v t k v x + Q x, t The above PDE now has homogenous boundary conditions And initial condition is v, t t v, t t v x, u x, r x, B A f x A x + x The problem does not ask us to solve it. So will stop here. Part d et u x, t v x, t + r x, t 1 Since the problem has time dependent source function Q x, t then r x, t is now a reference function that only needs to satisfy the non-homogenous boundary conditions which in this problem are at both ends and v x, t has homogenous boundary conditions. The boundary condition r x, t need to satisfy is et r, t + hr, t hb t x r, t r x, t c 1 t x + c t Since r, t then c. Now we use the right side non-homogenous B.C. to solve for c 1. Plugging the above into the right side B.C. gives c 1 + hc 1 hb t c 1 hb t 1 + h 148

149 .9 HW 9 Hence r x, t hbt 1+h x 3 The rest is very similar to what we did in part a. Replacing 1 into the original PDE ux,t k ux,t x But r x + Q x, t gives v x, t r x, t k v x, t r x, t + Q x, t t x v t r t k v x k r + Q x, t x hence the above reduces to v t k v r + Q x, t + x t et 8.3. Eigenfunction Expansion with Q x, Homogeneous t Q x, t + BCs r 353 t Then 4 becomes Solve the two-dimensional heat equation with circularly symmetric timeindependent sources, boundary v t k conditions, v + Q x, t and initial conditions inside a x circle: The above PDE now has homogeneous boundary conditions rc 8r r 8 + Qr with v, t ur, v, f r t and ua, t T. t Solve the two-dimensional heat equation with time-independent boundary And initial condition conditions: is au _ k 8U + au v x, at u x, ax r x, W subject to the boundary conditions hb f x 1 + h x u, y, t ux,, t The problem does not ask us to solveu, it. y, Sot will stop ux, here. H, t gx and the initial condition.9.3 Problem 8..5 Analyze the limit as t -' oo. ux,y, fx,y Solve the initial value problem for a two-dimensional heat equation inside a circle of radius a with time-independent boundary conditions: au kvu at ua,,t ur, 9, g9 fr, Solve the wave equation with time-independent sources, t au 8u at u r, θ, t uc ax + Qx ux, k fx t r + 1 u r r + 1 u r θ a ux, gx, u, θ, t < θ if an "equilibrium" solution exists. Analyze the behavior for large t. If no equilibrium exists, u a, θ, explain t g θ why and reduce the problem to one with homogeneous boundary u r, π, t conditions. u r, π, t Assume that u u * a Qx, r, π, u, t t A, r, π, t u, t B θ θ b Qx 1, U, t, u,t With initial conditions c Qx u r, θ, 1, f r, u, θ. Since t A, the boundary u, t conditions B are not homogenous, and since there are no time[hint: dependent Add problems sources, then a and in this b.] case we look for u E r, θ which is solution at steady state which : d needs Qx to satisfy sin the nonhomogeneous, u, t, u, B.C., t where u r, θ, t v r, θ, t + u E r, θ and v r, θ, t solves the PDE but with homogenous B.C. Therefore, we need to find equilibrium solution for aplace PDE on disk, that only needs to satisfy the nonhomogeneous B.C. u E 4 u E r + 1 r u E r + 1 r u E θ 149

150 HWs With boundary condition u E, θ < θ u E a, θ g θ u E r, π u E r, π u E θ r, π u E r, π θ But this PDE we have already solved before. But to practice, will solve it again. et u E r, θ R r Θ θ Where R r is the solution in radial dimension and Θ θ is solution in angular dimension. Substituting u E r, θ in the PDE gives Dividing by R r Φ θ R Θ + 1 r R Θ + 1 r Θ R R R + 1 R r R + 1 Θ r Θ r R R + r R R Θ Θ Hence each side is equal to constant, say λ and we obtain Or r R R + r R R λ Θ Θ λ We start with Φ ODE. The boundary conditions on 3 are r R + rr λr 1 Θ θ Θ + λθ Θ π Θ π π Θ θ π case λ The solution is Φ c 1 θ + c. Hence we obtain, from first initial conditions πc 1 + c πc 1 + c c 1 Second boundary conditions just says that c c, so any constant will do. Hence λ is an eigenvalue with constant being eigenfunction. case λ > The solution is Θ θ c 1 cos λθ + c sin λθ The first boundary conditions gives c 1 cos λπ + c sin λπ c 1 cos λπ + c sin λπ c 1 cos λπ c sin λπ c 1 cos λπ + c sin λπ c sin λπ 3 From second boundary conditions we obtain Θ θ λc 1 sin λθ + c λ cos λθ Therefore λc 1 sin λπ + c λ cos λπ λc 1 sin λπ + c λ cos λπ λc1 sin λπ + c λ cos λπ λc 1 sin λπ + c λ cos λπ λc1 sin λπ λc 1 sin λπ c 1 sin λπ 4 15

151 .9 HW 9 Both 3 and 4 are satisfied if Therefore Θ n θ λπ nπ n 1,, 3, λ n n 1,, 3, λ {}}{ à + à n cos nθ + B n sin nθ 5 I put tilde on top of these constants, so not confuse them with constants used for v r, θ, t found later below. Now we go back to the R ODE given by r R + rr λ n R and solve it. This is Euler PDE whose solution is found by substituting R r r α. The solution comes out to be ecture 9 R n r c + c n r n 6 Combining 5,6 we now find u E as u En r, θ R n r Θ n θ u E r, θ à + à n cos nθ r n + B n sin nθ r n n à n cos nθ r n + B n sin nθ r n 7 Where c was combined with A. Now the above equilibrium solution needs to satisfy the nonhomogenous B.C. u E a, θ g θ. Using orthogonality on 7 to find A n, B n gives g θ à n cos nθ a n + B n sin nθ a n π For n For n > g θ cos n θ dθ n π n à n cos nθ cos n θ a n dθ + n π à n cos nθ cos n θ a n dθ + π Ãn cos n θ a n dθ π π g θ dθ à à 1 π g θ cos nθ dθ Ãn à n 1 π π π π π dθ g θ dθ π B n sin nθ cos n θ a n dθ {}}{ π B n sin nθ cos n θ a n dθ n cos nθ a n dθ g θ cos nθ dθ Similarly, we apply orthogonality to find B n which gives for n > only B n 1 π π g θ sin nθ dθ Therefore, we have found u E r, θ completely now. It is given by u E r, θ à + à n cos nθ r n + B n sin nθ r n à 1 π à n 1 π B n 1 π π π π g θ dθ g θ cos nθ dθ g θ sin nθ dθ 151

152 HWs The above satisfies the non-homogenous B.C. u E a, θ g θ. Now, since u r, θ, t v r, θ, t + u E r, θ, then we need to solve now for v r, θ, t specified by v r, θ, t t v, θ, t < θ v a, θ, t v k r + 1 v r r + 1 v r θ v r, π, t v r, π, t v v r, π, t r, π, t θ θ et v r, θ, t R r Θ θ T t. Substituting into 8 gives T RΘ k R T Θ + 1 r R T Θ + 1 r Θ RT Dividing by R r Θ θ T t gives 1 T k T R R + 1 R r R + 1 Θ r Θ et first separation constant be λ, hence the above becomes 8 Or 1 T k T λ R R + 1 R r R + 1 Θ r Θ λ T + λkt r R R + r R R + r λ Θ Θ We now separate the second equation above using µ giving Or r R R + r R R + r λ µ Θ Θ µ R + 1 r R + R λ µ r 9 Θ + µθ 1 Equation 9 is Sturm-iouville ODE with boundary conditions R a and bounded at r and 1 has periodic boundary conditions as was solved above. The solution to 1 is given in 5 above, no change for this part. Θ n θ λ {}}{ A + n A n cos nθ + B n sin nθ A n cos nθ + B n sin nθ 11 Therefore 9 becomes R + 1 r R + R λ n with n, 1,,. We found the solution to r this Sturm-iouville before, it is given by R nm r J n λnm r n, 1,,, m 1,, 3, 1 a Where λ nm z nm where a is the radius of the disk and z nm is the m th zero of the Bessel function of order n. This is found numerically. We now just need to find the time solution from T + λ nm kt. This has solution T nm t e kλ nmt 13 15

153 .9 HW 9 Now we combine 11,1,13 to find solution for v r, θ, t v nm r, θ, t Θ n θ R nm r T nm t v r, θ, t A n cos nθ J n λnm r e kλ nmt + n m1 m1 B n sin nθ J n λnm r e kλ nmt We now need to find A n, B n, which is found from initial conditions on v r, θ, which is given by Hence from 14, at t f r, θ u E r, θ n m1 v r, θ, u r, θ, u E r, θ f r, θ u E r, θ A n cos nθ J n λnm r + m1 B n sin nθ J n λnm r For each n, inside the m sum, cos nθ and sin nθ will be constant. So we need to apply orthogonality twice in order to remove both sums. Multiplying 15 by cos n θ and integrating gives π f r, θ u E r, θ cos n θ π dθ A n J n λnm r cos nθ cos n θ dθ π π + π π n m1 m1 B n J n λnm r sin nθ cos n θ The second sum in the RHS above goes to zero due to π π sin nθ cos n θ dθ and we end up with π π π f r, θ u E r, θ cos nθ dθ A n cos nθ π m1 J n λnm r dθ We now apply orthogonality again, but on Bessel functions and remember to add the weight r. The above becomes a π a π f r, θ u E r, θ cos nθ J n λnm r rdθdr A n cos nθ J n λnm r J n λnm r rd Hence π A n A n a a π π f r, θ u E r, θ cos nθ J n λnm r rdθdr a π π cos nθ Jn λnm r rdθdr π π π m cos nθ Jn λnm r rdθdr n, 1,,, m 1,, 3, We will repeat the same thing to find B n. The only difference now is to use sin nθ. repeating these steps gives a π π B n f r, θ u E r, θ sin nθ J n λnm r rdθdr a π π sin nθ Jn λnm r n, 1,,, m 1,, 3, rdθdr This complete the solution. Summary of solution u r, θ, t v r, θ, t + u E r, θ A n cos nθ J n λnm r e kλ nmt + Where n m1 u E r, θ Ã + Ã 1 π Ã n 1 π B n 1 π m1 Ã n cos nθ r n + B n sin nθ r n π π π g θ dθ g θ cos nθ dθ g θ sin nθ dθ B n sin nθ J n λnm r e kλ nmt + u E r, θ 153

154 HWs And And A n B n Where λ nm of order n. a π π f r, θ u E r, θ cos nθ J n λnm r rdθdr a π π cos nθ Jn λnm r rdθdr a π π f r, θ u E r, θ sin nθ J n λnm r rdθdr a π π sin nθ Jn λnm r rdθdr a z nm.9.4 Problem Problem Solve the initial value problem n, 1,,, m 1,, 3, n, 1,,, m 1,, 3, where a is the radius of the disk and z nm is the m th zero of the Bessel function cρ u t x u K + qu + f x, t 1 x Where c, ρ, K, q are functions of x only, subject to conditions u, t, u, t, u x, g x. Assume that eigenfunctions are know. Hint: let d d dx K dx + q solution Because this problem has homogeneous B.C. but has time dependent source i.e. non-homogenous in the PDE itself, then we will use the method of eigenfunction expansion. In this method, we first need to find the eigenfunctions φ n x of the associated PDE without the source being present. Then use these φ n x to expand the source f x, t as generalized Fourier series. We now switch to the associated homogenous PDE in order to find the eigenfunctions. This the same as above, but without the source term. u t 1 cρ u, t u, t x u x, g x u K + q x cρ u We are told to assume the eigenfunctions φ n x are known. But it is better to do this explicitly, also needed to find the weight. et u X x T t. Then becomes Dividing by XT gives T X 1 cρ K X T + 1 cρ K X T + q cρ XT T T 1 X cρ K X + 1 cρ K X X + q cρ As the right side depends on x only, and the left side depends on t only, we can now separate them. Using λ as separation constant gives T + λt And for the x part 1 X cρ K X + 1 cρ K X X + q cρ λ K X + K X + qx λcρx K X + qx λcρx A We now see this is Sturm-iouville ODE, with p K q q σ cρ And [X] d dx d dx dx K + qx dx dx K + q dx 154

155 .9 HW 9 Where [X] λcρx The solution to S-, with homogeneous B.C. is given as X x a n φ n x When we plug-in this back into, and incorporate the time solution from T + λ n T, we end up with solution for as u x, t a n t φ n x 3 Where now the Fourier coefficients became time dependent. We now substitute this back into the original PDE 1 with the source present the nonhomogeneous PDE and obtain cρ a n t φ n x a n t [φ n x] + f x, t 4 We now expand f x, t using same eigenfunctions found from the homogeneous PDE solution we can do this, since eigenfunctions found from Sturm-iouville can be used to expand any piecewise continuous function. et f x, t f n t φ n x 5 Hence 4 becomes cρ a n t φ n x a n t [φ n x] + f n t φ n x 6 But from above, we know that [φ n x] λ n cρφ n x, hence 6 becomes cρ a n t φ n x cρ λ n a n t φ n x + f n t φ n x cρa n t φ n x + cρλ n a n t φ n x a n t + λ n a n t cρφ n x f n t φ n x f n t φ n x By orthogonality, weight is cρ then from the above we obtain The solution to the above is a n t + λ n a n t f n t t a n t e λnt f n s e λns ds + ce λnt To find constant of integration c in the above, we use initial conditions. At t Hence the solution becomes c a n t a n t e λnt f n s e λns ds + a n e λnt e λnt a n + To find a n, from 3, putting t gives Applying orthogonality g x t a n φ n x g x φ n x dx a n a n f n s e λns ds φ n x cρdx g x φ n x dx φ n x cρdx 155

156 HWs And finally, to find f n t, which is the generalized Fourier coefficient of the expansion of the source in 5 above, we also use orthogonality f x, t φ n x dx f n t f n t φ n x cρdx f x, t φ n x dx φ n x cρdx Summary of solution The solution to cρ u t u x K x + qu + f x, t is given by u x, t a n t φ n x Where a n t is the solution to 8.4. Eigenfunction Expansion a Using n t + λgreen's n a n t Formula f n t 359 Given by Where And where c, p, KO, and q are functions of x t only, subject to the conditions a n t e a λncρt n + f n s e λncρs ds u, t, u, t, and ux, gx. Assume that the eigenfunctions f n t f x, are t known. φ [Hint: let n x dx Consider φ n x cρdx 11 f KoxYX-J Ko > O,o > 8t ax a and with the boundary conditions a n g x initial φ n x conditions: dx φ n x cρdx ux, gx, u, t A, and u, t B..9.5 Problem *a Find a time-independent solution, uox. b Show that limi_.,.. ux, t f x independent of the initial conditions. [Show that f x uox.] * Solve Solve 8u kvu+fr,t 8t inside the circle r < a with u at r a and initially u. d: KoA +q] 8u _ 8zu + sin 5x a-ze Since this problem has homogeneous B.C. c?tbut8x has time dependent source i.e. non-homogenous in the PDE itself, subject thento weu, willt use 1, the u7r, method t, of and eigenfunction ux,. expansion. In this method, we first find* the eigenfunctions Solve φ n x of the associated homogenous PDE without the source being present. Then use these φ n x to expand the source 8uu _ f x, t as generalized Fourier series. We now switch to the associated homogenous PDE in order 8tto find 8xthe eigenfunctions. u u r, t. There is no θ. Hence subject to u, t, u, t t, and ux,. u r, t u k t r + 1 u 1 r r 8.4 Method of Eigenfunction Expansion u a, t Using Green's Formula u, t < With or Without u r, Homogeneous Boundary Conditions We need to solve the above in order to find the eigenfunctions φ n r. et u R r T t. Substituting In this thissection back into we reinvestigate 1 gives problems that may have nonhomogeneous boundary conditions. We still use the method of eigenfunction expansion. For example, consider T R k R T + 1 r R T Dividing by RT PDE: kaxz + Qx, t T at k T R R + 1 R r R et separation constant be λ. We obtain z T + kλt 156

157 .9 HW 9 And R R + 1 R r R λ R + 1 r R λr rr + R + λrr This is a singular Sturm-iouville ODE. Standard form is rr λrr Hence p r q σ r We solved R + 1 r R + λr before. The solution is R n r J λn r Where λ n is found by solving J λn a. Now that we know what the eigenfunctions are, then we write u r, t a n t J λn r Where a n t is function of time since it includes the time solution in it. Now we use the above in the original PDE with the source in it u r, t t k u + f r, t 3 Where u λu. Substituting into 3, and using f r, t f n t J λn r gives a n t J λn r k a n t + kλ n a n t J λn r λ n a n t J λn r + f n t J λn r Applying orthogonality, the above simplifies to The solution is a n t + kλ n a n t f n t t a n t e kλnt f n s e kλns ds + ce kλnt To find constant of integration c in the above, we use initial conditions. At t Hence the solution becomes c a n t a n t e kλnt f n s e kλns ds + a n e kλnt e kλnt a n + To find a n, from, putting t gives Hence a n. Therefore a n t becomes. t f n s e kλns ds a n J λn r t a n t e kλnt f n s e kλns ds f n t J λn r 157

158 HWs And finally, to find f n t, which is the generalized Fourier coefficient of the expansion of the source in 3 above, we also use orthogonality a f r, t J λn r rdr f n t a J λn r rdr f n t f r, t J λn r 8.4. Eigenfunction Expansion Using Green's rdr a Formula J λn r 359 rdr Summary of solution where c, p, KO, and q are functions of x only, subject to the conditions The solution to ur,t t k + 1 u u, r r t +, f u, r, t t is given, and by ux, gx. u r Assume that the eigenfunctions are known. [Hint: let u r, t a n t J λn r Consider 11 f KoxYX-J Ko > O,o > Where a n t is the solution to 8t ax a a and n t + kλ n a n t f n t with the boundary conditions initial conditions: Given by ux, gx, u, t t A, and u, t B. a n t e kλnt f n s e kλns ds *a Find a time-independent solution, uox. Where a f n t f r, t J b Show that limi_.,.. ux, t f x λn independent r rdr of the initial conditions. [Show that f x uox.] a λn r rdr * Solve.9.6 Problem Solve * Solve J a 8u kvu+fr,t 8t inside the circle r < a with u at r a and initially u. 8u _ 8zu c?t 8x + sin 5x a-ze subject to u, t 1, u7r, t, and ux,. 8uu _ d: KoA +q] This problem has nonhomogeneous B.C. and non-homogenous 8t 8x in the PDE itself source present. First step is to usesubject reference to function u, t to, remove u, t the t, nonhomogeneous and ux,. B.C. then use the method of eigenfunction expansion on the resulting problem. et r x c 8.4 Method of Eigenfunction 1 x + c Expansion Using Green's Formula With or Without r x Homogeneous 1 Boundary Conditions x π At x, r x 1, hence 1 c and at x π, r x, hence c 1 π + 1 or c 1 1 π, hence Therefore In this section we reinvestigate u x, t problems v x, t that + r may x have nonhomogeneous boundary conditions. We still use the method of eigenfunction expansion. For example, Where v x, consider t solution for the given PDE but with homogeneous B.C., therefore v x, t v x, t t PDE: v, t v π, t x + e t sin 5x z 1 at v x, u x, r x kaxz + Qx, t x x π π 1 We now solve 1. This is homogeneous in the PDE itself. To solve, we first solve the nonhomogeneous PDE in order to find the eigenfunctions. Hence we need to solve This has solution v x, t t v x, t v x, t x a n t φ n x 158

159 .9 HW 9 With φ n x sin λn x λ n n n 1,, 3 n 1,, 3 Plug-in back into 1 gives a n t φ n x a n t φ n x + e t sin 5x But x φ n x λ n φ n nφ n, hence the above becomes a n t x φ n x + e t sin 5x a n t φ n x + n a n t φ n x e t sin 5x a n t + n a n t sin nx e t sin 5x Therefore, since Fourier series expansion is unique, we can compare coefficients and obtain { a n t + n e t n 5 a n t n 5 For the case n 5 Hence a 5 t + 5a 5 t e t d a5 t e 5t e 3t dt a 5 t e 5t e 3t dt + c e3t 3 + c a 5 t e t 3 + ce 5t At t, a c, hence c a And the solution becomes a 5 t 1 3 e t + a 5 1 e 5t 3 For the case n 5 At t, a n c, hence Therefore a n t a n t + n a n t d a n t e n t dt a n t e nt c a n t ce n t a n t a n e nt { 1 3 e t + a e 5t n 5 a n e n t n 5 To find a n we use orthogonality. Since u x, t v x, t + r x, then u x, t a n t sin nx + 1 x π 159

160 HWs And at t the above becomes a n sin nx Or Applying orthogonality π Therefore a n π x π 1 sin nx dx π π + 1 x π x π 1 a n sin nx x π 1 sin n x π dx a n sin n x dx x π π 1 sin nx dx [ π sin nx dx + 1 π ] x sin nx dx π π [ cos nx π + 1 sin nx x cos nx π ] π n π n n [ cos nπ sin nπ π cos nπ sin π n n π n n n [ 1 n π n π ] 8.4. Eigenfunction Expansion Using 1n Green's Formula 363 n π n [ 1 n 1 ] even if the boundary π n n conditions 1n are homogeneous. In fact, it is this derivation that justifies the differentiation nof infinite series of eigenfunctions used in Sec We now have two procedures to solve nonhomogeneous partial differential equations nπ with nonhomogeneous boundary conditions. By subtracting any function that just solves the nonhomogeneous boundary conditions, we can solve a related problem Therefore with a 5 homogeneous 5π. Hence boundary conditions by the eigenfunction expansion method. Alternatively, we can solve directly the original { 1 a n t 3 e t + problem with nonhomogeneous boundary conditions by the method of eigenfunction 5π 3 1 expansions. e 5t n In 5both cases we need the eigenfunction expansion of some function nπ e n t wx, t: n 5 Where wx, t E a,, tnx u x, t v x, t + r x If wx, t satisfies the same homogeneous boundary conditions as nx, then we claim that this series will converge reasonably fast. However, a n t sin nx + 1 x if wx, t satisfies nonhomogeneous boundary conditions, then not only will the π series not satisfy the boundary conditions at x and x, but the series will converge more slowly everywhere. Thus, the advantage of reducing a problem to homogeneous boundary.9.7 Problem conditions is that b the corresponding series converges fasts r. EXERCISES 8.4 ] cos n In these exercises, do not make a reduction to homogeneous boundary conditions. Solve the initial value problem for the heat equation with timedependent sources au ka +Qx,t ux, fx subject to the following boundary conditions: a u,t At, * b, t At, TX_, t Bt YX_ 16 et Use the method of cigenfunction expansions to solve, without reducing to homogeneous boundary conditions: Consider 8u _ 8u u x, t atb n t k 8x φ n x 1 n u O, t ux, fx cxpx at ax [Kox _ l 1 A }constants. B + 9xu + f x, t

161 .9 HW 9 Where in this problem φ n x are the eigenfunctions of the corresponding homogenous PDE, which due to having both sides insulated, we know they are given by φ n x cos nπ x where now n, 1,, and λ n nπ. That is why the sum above starts from zero and not one. We now substitute 1 back into the given PDE, but remember not to do term-by-term differentiation on the spatial terms. b n t φ n x k x + Q x, t u n But Q x, t i q n t φ n x so the above becomes b n t φ n x k u x + q n t φ n x n Multiplying both sides by φ m x and integrating b n t φ n x φ m x dx n Applying orthogonality b n t φ n x dx Dividing both sides by φ n x dx gives We now use Green s formula to simplify then n k u x φ m x dx + k u x φ n x dx + q n t q n t φ n x φ m x dx n φ n x dx b n t k u φ x n x dx + q n t 1A φ n x dx But we know from Green s formula that uφ x n x dx. We rewrite u u x φ n x dx v [u] u [v] dx p v [u] dx v du dx u dv dx x [u] and let φ n x v, In this problem p 1, so we solve for v [u] dx which is really all what we want from the above and obtain v [u] dx u [v] dx v du dx u dv dx v [u] dx v du dx u dv + u [v] dx dx Since we said φ n x v, then we replace these back into the above to make it more explicit u x φ n x dx φ n x du dx udφ n x + u [φ n x] dx dx But [φ n x] λ n φ n x and above becomes u x φ n x dx φ n x du dx udφ n x λ n uφ n x dx dx We are now ready to substitute boundary conditions. In this problem we know that du, t B t dx dφ n, t d nπ dx dx cos x x nπ nπ sin x nπ φ n, t cos x cos nπ 1 n x dφ n, t d nπ dx dx cos x x nπ φ n, t cos x 1 du, t A t dx x x 161

162 HWs Now we have all the information to evaluate u x φ n x dx φ n du dx u dφ n dx λ n uφ n x dx Which becomes u x φ n x dx 1 n B t A t λ n φ n du dx u dφ n dx uφ n x dx 1 n B t A t λ n uφ n x dx 3 Now we need to sort out the uφ n x dx term above, since u x, t is unknown, so we can t leave the above as is. But we know from u x, t n b n t φ n x that b n t uφnxdx by φ nxdx orthogonality. Hence uφ n x dx b n t φ n x dx. Using this in 3, we finally found the result for u φ x n x dx u x φ n x dx 1 n B t A t λ n b n t But φ n x dx cos nπ x dx hence Substituting the above in 1A gives u x φ n x dx 1 n B t A t λ n b n t b n t k 1 n B t A t λ n b n t b n t k 1 n B t A t λ n b n t + q n t φ n x dx + q n t k 1n B t A t kλ n b n t + q n t 4 Or b n t + kλ n b n t q n t + k 1n B t A t Now that we found the differential equation for b n t we solve it. The integrating factor is µ e kλnt, hence the solution is Integrating d dt µb n t µq n t + µ k 1n B t A t µb n t µq n t dt + µ k 1n B t A t dt + c Or b n t e kλnt e kλnt q n t dt + e kλnt k 1n B t A t dt + ce kλnt The constant of integration c is b n, therefore b n t e kλnt e kλnt q n t dt + e kλnt k 1n B t A t dt + b n e kλnt The above could also be written as t b n t e kλnt e kλns q n s ds + t e kλns k 1n B s A s ds + b n e kλnt Now that we found b n t, the last step is to determine b n. This is done from initial conditions By orthogonality b n u x, b n φ n x n f x φ n x dx φ n x dx nπ f x cos x dx 16

163 claim that this series will converge reasonably fast. However, if wx, t satisfies nonhomogeneous boundary conditions, then not only will the series not satisfy the boundary conditions at x and x, but the series will converge more slowly everywhere. Thus, the advantage of reducing a problem to homogeneous boundary.9 HW 9 conditions is that the corresponding series converges fasts r. This complete the solution. Summary of result The solution is In these exercises, do not make a reduction to homogeneous boundary conditions. Solve the u initial x, t value bproblem n t φ n x for the heat equation with timedependent sources n Where t au t b n t e kλnt e kλns q n s ds + e kλns ka +Qx,t k 1n B s A s ds + b n e kλnt Where EXERCISES 8.4 ux, fx subject to the following b n boundary conditions: nπ f x cos x dx a u,t At, And * b, t At, q n t, t Bt TX_ YX_ nπ Q x, t cos x dx Use the method of cigenfunction expansions to solve, without reducing to And homogeneous boundary conditions: nπ λ n n, 1,, 3, 8u _ 8u at k 8x.9.8 Problem u O, t Consider ux, fx cxpx at ax [Kox _ l 1 A }constants. B + 9xu + f x, t 364 Chapter 8. Nonhomogeneous Problems ux, gx u, t at u, t,qt. Assume that the eigenfunctions,ax of the related homogeneous problem are known. a Solve without reducing to a problem with homogeneous boundary conditions. b Solve by first reducing to a problem with homogeneous boundary conditions Reconsider z kz + Qx, t Part a ux, fx uo,t From problem 8.3.3, we found the eigenfunctions φ n x u, from t the Sturm-iouville. to have weight Assume that the solution ux, t has the appropriate smoothness, so that it may be represented by a Fourier σ cosine cρ series et ux, t _ cnt cos nx. u x, t n-b n t φ n x Solve for dcn/dt. Show that c satisfies a first-order nonhomogeneous ordinary abovedifferential in the PDEequation, gives but part of the nonhomogeneous term is not Substituting the known. Make a brief philosophical conclusion. σ b n t φ n x [u] + f x, t 8.5 Forced Vibrating Membranes and Resonance K i1 Where The method of eigenfunction expansion may also be applied to nonhomogeneous partial x differential x + q. Following same procedure using Green s formula on page 35, we obtain equations with more than two independent variables. An interesting example is a vibrating membrane of arbitrary shape. In our previous analysis of membranes, σ db n t + kλ vibrations n b n t were f n t caused + k λ by n α t 1 the initial conditions. n β t Another mechanism that will dt 1 put a membrane into motion is an φ external n x σdxforce. The linear nonhomogeneous partial differential equation that describes a vibrating membrane Where is u f x, t cvzu f n + t Qx, φ n y, x t, c3tz where Qx, y, t represents a time- and f spatially dependent external force. To be completely general, there should n t f x, t φ n x σdx be some boundary condition along the boundary of the membrane. However, it is more usual φ n x σdx for a vibrating membrane to be fixed with zero vertical displacement. Thus, we will specify this homogeneous boundary condition, u,

164 HWs The solution to 1 is found using integrating factor. db n t dt + λn b n t 1 σ σ f n t + Hence µ e λn cρ t and the solution becomes b n t e λn σ t 1 σ Where c is found from e λn σ t f n t dt + And b n is found from initial conditions This complete the solution. Summary Solution is Where b n t e λn σ t 1 σ b n Part b g x φ n x σdx φ n x σdx σ cρ e λn cρ t f n t dt + g x b n k σ λn φ n x σdx b n c k σ λn α t 1 n β t φ n x σdx b n φ n x u x, t g x φ n x σdx φ n x σdx k σ λn b n t φ n x φ n x σdx e λn σ t α t 1 n β t dt e λn σ t α t 1 n β t dt + ce λn σ t + b n e λn σ t The first step is to obtain a reference function r x, t where u x, t v x, t+r x, t. The reference function only needs to satisfy the nonhomogeneous B.C. We see that β t α t r x, t α t + x does the job. Now we solve the following PDE cρ v t x v, t v π, t v x, g x Using Green s formula, starting with K v x v x, t + q x v + f x, t α + b n t φ n x i1 β α x Where we used instead of above now, since both v x, t and φ n x satisfy the homogenous B.C., and where b n t satisfies the ODE Where σ cρ and σ db n t dt f x, t f n t + λ n b n t f n t 1 f n t φ n x f x, t φ n x σdx φ n x σdx 164

165 .9 HW 9 The solution to 1 is found using integrating factor µ e λn σ t, hence b n t e λn σ t 1 e λn σ t f n t dt + b n e λn σ t σ And b n is found from initial conditions v x, g x α + β α x b n b n φ n x i1 g x This complete the solution. Summary Solution is given by u x, t b n t φ n x + r x, t i1 b n t φ n x + α t + i1 α + β α φ n x σdx β t α t x x φ n x σdx Where b n t e λn σ t 1 σ e λn σ t f n t dt + b n e λn σ t And b n g x α + β α φ n x σdx x φ n x σdx And 8.5. Forced Vibrating Membranes and Resonance f n t f x, t φ n x σdx 371 φ n x σdx is a particular solution of Where σ cρ ti dty + wov f t.9.9 Problem 8.5. What is the general solution? What solution satisfies the initial conditions yo yo and g vo? Part a Consider a vibrating string with time-dependent forcing: z atz cz 8x + Qx, t u, t ux, fx u,t 5x,. a Solve the initial value problem. *b Solve the initial value problem if Qx, t gx cos wt. For what values of w does resonance occur? Consider a vibrating string with friction with time-periodic forcing -t c axz -pet + gx coswt et u, t ux, fx u,t gfx,. u x, t A n t φ n x a Solve this initial value problem if 3 is moderately small < Q < Where we used instead cir/. of above, since the PDE given has homogeneous B.C. We know that φ n x sin λ n b x Compare for n 1, this, 3, solution where to Exercise λ n 8.5.b. nπ. Substituting the above in the given PDE gives Solve the initial value problem for a vibrating string with time-dependent forcing. A n t φ n x c A n t d φ n x z dx + Q x, t at cz 8xz + Qx, t, ux, f x, x,, But Q x, t St q n t φ n x, hence the above becomes subject to the following boundary conditions. Do not reduce to homogeneous boundary conditions: A n t φ n x c A n t d φ n x dx + g n t φ n x a u, t At, u, t Bt b u, t,,t c P;, t At, u, t z Solve the initial value problem for a membrane with time-dependent forcing and fixed boundaries u, 165

166 HWs But d φ nx dx λ n φ n x, hence A n t φ n x c λ n A n t φ n x + Multiplying both sides by φ m x and integrating gives Hence g n t φ n x A n t φ m x φ n x dx c λ n A n t φ m x φ n x dx + A n t φ n x dx c λ n A n t Now we solve the above ODE. et solution be φ n x dx + g n t A n t + c λ n A n t g n t A n t A h n t + A p n t g n t φ m x φ n x dx φ n x dx Which is the sum of the homogenous and particular solutions. The homogenous solution is A h n t c 1n cos c λ n t + c n sin c λ n t And the particular solution depends on q n t. Once we find q n t, we plug-in everything back into u x, t A n t φ n x and then use initial conditions to find c 1n, c n, the two constant of integrations. We will do this in the second part. Part b Now we are given that Q x, t g x cos ωt. Hence Where g n t Q x, t φ n x dx φ n x dx γ n cos ωt g x φ n x dx cos ωt γ n φ n x dx g x φ n x dx φ n x dx is constant that depends on n. Now we use the above in result found in part a We know the homogenous solution from part a. A h n t c 1n cos c λ n t A n t + c λ n A n t γ n cos ωt 1 + c n sin c λ n t We now need to find the particular solution. Will solve using method of undetermined coefficients. Case 1 ω c λ n no resonance We can now guess A p n t z 1 cos ωt + z sin ωt Plugging this back into 1 gives Collecting terms z 1 cos ωt + z sin ωt + c λ n z 1 cos ωt + z sin ωt γ n cos ωt ωz 1 sin ωt + ωz cos ωt + c λ n z 1 cos ωt + z sin ωt γ n cos ωt ω z 1 cos ωt ω z sin ωt + c λ n z 1 cos ωt + z sin ωt γ n cos ωt cos ωt ω z 1 + c λ n z 1 + sin ωt ω z + c λ n z γn cos ωt Therefore we obtain two equations in two unknowns ω z 1 + c λ n z 1 γ n ω z + c λ n z 166

167 .9 HW 9 From the second equation, z and from the first equation Hence Therefore z 1 c λ n ω γ n γ n z 1 c λ n ω A p n t z 1 cos ωt + z sin ωt γ n c cos ωt λ n ω A n t A h n t + A p n t c 1n cos c λ n t Now we need to find c 1n, c n. Since u x, t A n t φ n x At t the above becomes + c n sin c λ n t + γ n c cos ωt λ n ω c 1n cos c λ n t + c n sin c γ n nπ λ n t + c cos ωt sin λ n ω x f x Applying orthogonality mπ f x sin x dx mπ f x sin x Rearranging dx c 1n mπ f x sin x γ n dx c λ n ω γ n nπ c 1n + c λ n ω sin x nπ c 1n sin x + γ n nπ c λ n ω sin x nπ mπ c 1n sin x sin x dx + sin nπ x γ n dx + c λ n ω sin nπ x dx c 1n c 1n γ n nπ mπ c λ n ω sin x sin x dx sin nπ x dx sin nπ x dx f x sin mπ x dx sin nπ x dx f x sin We now need to find c n. For this we need to differentiate the solution once. u x, t t mπ x dx γ n c λ n ω γ n c λ n ω c λ n c 1n sin c λ n t + c λ n c n cos c γ n nπ λ n t c ω sin ωt sin λ n ω x Applying initial conditions ux, t Hence Therefore the final solution is And gives A n t c 1n cos u x, t c nπ λ n c n sin x c n c γ n λ n t + c cos ωt λ n ω nπ A n t sin x 167

168 HWs Where c 1n mπ f x sin x γ n dx c λ n ω Case ω c λ n Resonance case. Now we can t guess A p n t z 1 cos ωt + z sin ωt so we have to use A p n t z 1 t cos ωt + z t sin ωt Substituting this in A n t + c λ n A n t γ n cos ωt gives But z 1 t cos ωt + z t sin ωt + c λ n z 1 t cos ωt + z t sin ωt γ n cos ωt z 1 t cos ωt + z t sin ωt z 1 cos ωt z 1 ωt sin ωt + z sin ωt + z ωt cos ωt z 1 ω sin ωt z 1 ω sin ωt + z 1 ω t cos ωt Hence becomes + z ω cos ωt + z ω cos ωt z ω t sin ωt z 1 ω sin ωt z 1 ω t cos ωt + z ω cos ωt z ω t sin ωt z 1 ω sin ωt z 1 ω t cos ωt+z ω cos ωt z ω t sin ωt+c λ n z 1 t cos ωt + z t sin ωt γ n cos ωt Comparing coefficients we see that z ω γ n or And z 1. Therefore Therefore z γ n ω A p n t γ n t sin ωt ω A n t A h n t + A p n t c 1n cos c λ n t + c n sin c λ n t + γ n c t sin ωt λ n We now can find c 1n, c n from initial conditions. u x, t A n t φ n x c 1n cos c λ n t + c n sin c λ n t + γ n nπ c t sin ωt sin λ n x 4 At t f x c 1n nπ c 1n sin x nπ f x sin x dx Taking time derivative of 4 and setting it to zero will give c n. Since initial speed is zero then c n. Hence A n t c 1n cos c λ n t + γ n c t sin ωt λ n This completes the solution. Summary of solution The solution is given by u x, t A n t φ n x Case ω c λ n And A n t c 1n cos c γ n λ n t + c cos ωt λ n ω c 1n mπ f x sin x γ n dx c λ n ω 168

169 Consider a vibrating string with friction with time-periodic forcing -t c axz -pet + gx coswt u, t ux, fx And u,t gfx,. γ n g x φ n x dx a Solve this initial value problem if 3 is moderately small < Q < cir/. φ n x dx And λ n nπ b Compare this solution to Exercise 8.5.b., n 1,, 3, Case ω c λ n Solve resonance the initial value problem for a vibrating string with time-dependent forcing. A n t c 1n cos c λ n t + γ n z c t sin ωt λ n at cz 8xz + Qx, t, ux, f x, x,, St And subject to the following c 1n boundary conditions. nπ Do not reduce to homogeneous boundary conditions: f x sin x dx a u, t At, u, t Bt.9.1 Problemb u, b t,,t c P;, t At, u, t Solve the initial value problem for a membrane with time-dependent forcing and fixed boundaries u, c Vu + Qx, y, t,.9 HW 9 37 Chapter 8. Nonhomogeneous Problems if the membrane is ux,y, fx,y, x,y,, a a rectangle < x <, < y < H b a circle r < a *c a semicircle < < ir,r < a d a circular annulus a < r < b Consider the displacement ur,, t of a forced semicircular membrane of radius a Fig that satisfies the partial differential equation The solution to the corresponding homogeneous PDE 1 8u 1 8 8u\ 1 8u r + 9r', t' c 8t r 8r uc 8r J + r 88 t c Is u r, θ, t a n t J n λnm r cos nθ + a n t J n λnm r sin nθ n m1 u Where λ nm are found by solving roots of J n λnm a. To make things simpler, we will write Zero displacement Figure u r, θ, t a i t Φ i r, θ with the homogeneous boundary iconditions: Where the above means the double sum of all eigenvalues λ ur,, t, ur, ir, t, i. So Φ and i r, θ represents J a,, t n λnm r {cos nθ, sin θ} combined. So double sum is implied everywhere. Given this, we now expand the source term and the initial conditions Q r, θ, t q i t Φ i r, θ ur,, Hr, i and 5 r, 9,. And the original *a PDE Assume becomes that ur,, t E F, at4r,, where r, are the eigen- of the related homogeneous problem. What initial conditions functions does a i t at Φ λ satisfy? i c What a i differential t Φ i r, equation θ + does q i t at Φ i satisfy? r, θ 1 i i i b What are the eigenfunctions? But c Solve for ur,, t. Hint: See Exercise Φ i r, θ λ i Φ i r, θ 8.6 Poisson's Equation Hence 1 becomes m1 We have applied a the method of eigenfunction expansion to nonhomogeneous timedependent boundary value problems for PDEs with or without homogeneous bound- i t Φ i r, θ + c λ i a i t Φ i r, θ q i t Φ i r, θ i i ary conditions. In each case, the method of eigenfunction expansion, a i t + c λ i a i t Φ i r, θ q i t Φ i r, θ i u E aitoi, i 169

170 HWs Applying orthogonality gives Where q i t The solution to the homogenous ODE is a h i t A i cos a i t + c λ i a i t q i t a π π Q r, θ, t Φ i r, θ rdrdθ a π π Φ i r, θ rdrdθ c λ i t + B i sin c λ i t And the particular solution is found if we know what Q r, θ, t and hence q i t. For now, lets call the particular solution as a p i t. Hence the solution for a i t is a i t A i cos c λ i t + B i sin c λ i t + a p i t Plugging the above into the u r, θ, t i a i t Φ i r, θ, gives u r, θ, t A i cos c λ i t + B i sin c λ i t + a p i t Φ i r, θ i We now find A i, B i from initial conditions. At t f r, θ i A i + a p i Φ i r, θ Applying orthogonality a π π a π π f r, θ Φ j r, θ rdrdθ Taking time derivative of u r, θ, t t a π π A i + a p i Φ i r, θ Φ j r, θ rdrdθ i π f r, θ Φ j r, θ rdrdθ A j + a p a j Φ j r, θ rdrdθ π a π A i + a p i π f r, θ Φ i r, θ rdrdθ i a π π Φ i r, θ rdrdθ A i c λ i sin c λ i t + c λ i B i cos c λ i t + dap i t Φ i r, θ dt At t i c λ i B i + dap i Φ i r, θ dt Hence B i. Therefore the final solution is u r, θ, t A i cos c λ i t + a p i t Φ i r, θ i Where This complete the solution. a π A i + a p i π f r, θ Φ i r, θ rdrdθ a π π Φ i r, θ rdrdθ 17

171 .1 HW 1.1 HW 1 Math 3 Smith: Problem Set 1 Due Wednesday Dec. 7, For the following problems, determine a representation of the solution in terms of a symmetric Green s function. Use appropriate homogeneous boundary conditions for the Green s function. Show that the boundary terms can also be understood using homogeneous solutions of the differential equation. d u dx fx, < x < 1, u A, du 1 B 1 dx d u +u fx, < x <, u A, u B, nπ dx d u dx fx, < x <, u A, du +hu 3 dx d u dx +du +u fx, < x < 1, u, u1 1 4 dx.1.1 Problem 1 d u du f x ; < x < ; u A; dx dx 1 B Note: I used for the length instead of one. Will replace by one at the very end. This makes it more clear. Compare the above to the standard form d p du + qu f x dx dx Therefore pu f x p x 1 Green function is G x, x will use x which is what the book uses, instead of a, as x is more clear. Green function is the solution to d G x, x dx δ x x G, x 1 dg, x dx Where x is the location of the impulse. Since d Gx,x d Gx,x dx dx for x x, then the solution to, which is a linear function in this case, is broken into two regions { A1 x + A G x, x < x < x B 1 x + B x < x < The first solution, using G, x gives A and the second solution using dg1,x dx B 1, hence the above reduces to { A1 x x < x G x, x x < x B gives 1 We are left with constants to A 1, B to find. The continuity condition at x x gives A 1 x B The jump discontinuity of the derivative of G x, x at x x, gives the final equation d d dx G x, x dx G x, x 1 p x 1 region region 1 171

172 HWs Since p x 1 in this problem. But Hence Therefore dg x, x dx { A1 x < x x < x A 1 1 Solving,4 gives B x. Hence the Green function is, from 1 { x x < x G x, x x x < x 3 A Gx, x x x x And dgx,x dx where now derivative is w.r.t. x, is { dg x, x x < x dx 1 x < x dgx,x dx slope at upper region x x 1 slope 1 at lower region We now have all the information needed to evaluate the solution to the original ODE. boundary terms particular solution {}}{{}}{ x [ u x G x, x f x dx + p x G x, x du x p x u x dg x, x ] x dx dx Since p x 1 then x [ y x G x, x f x dx G x, x du x u x dg x, x ] x dx dx x [ ] et u h G x, x dux dx u x dgx,x x dx, hence x But u h G x, du x dx u dg x, x dx G x, du x dx + u dg x, x dx G x, x du x B dx dg x, x dx dg x, x dx G x, 1 x 17

173 .1 HW 1 Then u h xb A We see that the boundary terms are linear in x, which is expected as the fundamental solutions for the homogenous solution as linear. The complete solution is y x x G x, x f x dx xb A G x, x f x dx + xb + A x f x dx + For example, if f x x, or f x x then y x x x 3 3 x 3 3 x x x dx x x x x x xf x dx + xb + A 1 x 1 x 1 x A 1 x + Bx x3 x dx + xb + A + xb + A + xb + A To verify the result, this was solved directly, with f x x, giving same answer as above. In[39]: DSolve[{u''[x] x, u[] A, u'[1] B}, u[x], x] Out[39] u[x] A - 3 x + 6 B x + x3 In[4]: Out[4] Expand[%] u[x] A - x + B x + x3 6 And if f x x, or f x x, then y x x x x f x dx + x 4 4 x 4 4 x x dx x x x 3 x 3 x x 1 x 1 x 1 3 x3 3 A 1 3 x + Bx x4 xf x dx + xb + A x dx + xb + A + xb + A + xb + A To verify the result, this was solved directly, with f x x, giving same answer as above. In[41]: DSolve[{u''[x] x^, u[] A, u'[1] B}, u[x], x] Out[41] u[x] A - 4 x + 1 B x + x4 In[4]: Out[4] Expand[%] u[x] A - x 3 + B x + x4 1 This shows the benefit of Green function. Once we know G x, x, then changing the source term, requires only convolution to find the new solution, instead of solving the ODE again as normally done. 173

174 HWs.1. Problem d u + u f x ; < x < ; u A; u B; nπ dx Solution Compare the above to the standard form Therefore Green function is the solution to d dx p du dx + qu f x pu + qu f x p x 1 d G x, x dx + G x, x δ x x G, x G, x Where x is the location of the impulse. Since d Gx,x d Gx,x dx + G x, x, is broken into two regions dx for x x, then the solution to G x, x { A1 cos x + A sin x x < x B 1 cos x + B sin x x < x The first boundary condition on the left gives A 1. Second boundary conditions on the right gives Hence the solution now looks like { G x, x B 1 cos + B sin B 1 B sin cos A sin x x < x sin B cos cos x + B sin x x < x But sin B cos cos x + B sin x B sin x cos cos x sin cos Using trig identity sin a b sin a cos b cos a sin b, the above can be written as hence the solution becomes { Continuity at x gives G x, x B cos A sin x x < x sin x x < x B cos sin x, A sin x B cos sin x And jump discontinuity on derivative of G gives B cos cos x A cos x 1 p x 1 B cos cos x A cos x 1 3 Now we need to solve,3 for A, B to obtain the final solution for G x, x. From, 1 Plug into 3 A B cos sin x sin x 4 B cos cos x B sin x cos x 1 cos sin x B cos cos x B cos sin x cos x 1 sin x B cos x B sin x cos x cos sin x B cos x sin x cos x cos sin x B sin x cos x cos x sin x cos sin x 174

175 .1 HW 1 But using trig identity sin a b sin a cos b cos a sin b we can write above as B sin x x cos sin x B sin cos sin x Now that we found B, we go back and find A from 4 A cos sin x sin Therefore Green function is, from 1 { Or G x, x sin x sin G x, x cos sin x sin B cos sin x sin 1 cos sin x sin x sinx sin sin x x < x sin x x < x 1 cos { sinx sin x sin sin sin x x < x sin x x < x It is symmetrical. Here is a plot of G x, x for some arbitrary x located at x.75 for Problem Green function. x Gx,x x Now comes the hard part. We need to find the solution using boundary terms particular solution {}}{{}}{ x [ y x G x, x f x dx + p x G x, x du x p x u x dg x, x ] x dx dx x 6 The first step is to find dgx,x dx. From 5, we find dg x, x dx { cosx cos x sin sin sin x x < x sin x x < x 7 Now we plug everything in 6. But remember that G, x, G, x, u A, u B. Hence [ p x G x, x du x p x u x dg x, x ] x dx dx x du dg x, du dg x, G x, u G x, + u dx dx dx dx cos x cos x B sin x + A sin x sin x sin x 1 sin x B sin sin x + A sin B sin x x + Asin sin sin 175

176 HWs But p 1, hence the above becomes du dg x, du dg x, G x, + u + G x, u dx dx dx dx cos x cos x + B sin x + A sin x sin x sin x 1 sin x + B sin sin x A sin B sin x x Asin sin sin We see that the boundary terms are linear combination of sin and cosine in x, which is expected as the fundamental solutions for the homogenous solution as linear combination of sin and cosine in x as was found initially above. Equation 6 becomes y x x G x, x f x dx + B sin x x Asin sin sin Now we can do the integration part. Therefore x x sin x G x, x f x dx sin x f x dx + sin We can test the solution to see if it correct. et f x x or f x x, hence x G x, x f x dx Hence the solution is x sin x x sin x dx + sin x x x 8 sin x sin sin x f x dx sin sin x dx x sin x sin x x sin x dx + sin x x sin x dx sin sin x sin x x cos x + sin x + sin x + x cos x sin x sin sin x cos x sin x sin x sin x + sin sin x cos x sin x sin x x cos x sin x + sin sin sin 1 sin x + x cos x sin x x cos x sin x sin u x 1 sin x + x cos x sin x x cos x sin x + sin sin x x cos x sin x sin x sin x + sin sin sin B sin x x Asin sin sin 1 sin x + x cos x sin x x cos x sin x + B sin x A sin x sin To verify, the problem is solved directly using CAS, and solution above using Green function was compared, same answer confirmed. In[81]: f x; 1; A 1; B ; computersolution u[x] /. First@DSolve[{u''[x] + u[x] f, u[] A, u[] B}, u[x], x]; mysolusinggreenfunction 1 / Sin[] - Sin[x] + x Cos[x - ] Sin[x] - x Cos[x] Sin[x - ] + B Sin[x] - A Sin[x - ]; Simplify[computerSolution - mysolusinggreenfunction] Out[87].1.3 Problem 3 d u du f x ; < x < ; u A; + hu dx dx Solution Compare the above to the standard form d p du f x dx dx pu f x 176

177 .1 HW 1 Therefore p x 1 Green function is the solution to d G x, x dx δ x x G, x d dx G, x + hg, x Where x is the location of the impulse. Since d Gx,x d Gx,x dx, is broken into two regions { A1 x + A G x, x x < x B 1 x + B x < x dx for x x, then the solution to The first boundary condition on the left gives A. Second boundary conditions on the right gives Hence the solution now looks like But Hence Continuity at x gives G x, x B 1 + h B 1 + B { B h hb hb 1+h B 1 hb 1 + h A 1 x x < x x + B x < x hb hb x + B x + B 1 + h 1 + h 1 + h 1 + h B 1 + h hx 1 + h G x, x { A 1 x x < x x < x B 1+h hx 1+h A 1 x B 1 + h hx 1 + h And jump discontinuity on derivative of G gives We solve,3 for A 1, B. From 3 B cos cos x A cos x 1 p x 1 1 hb 1 + h A 1 1 p x 1 3 Substituting in hb 1 + h A 1 1 A 1 hb 1 + h 1 hb 1 + h 1 + h 1 + h hb 1 h 1 + h hb 1 h x B 1 + h hx 1 + h 1 + h hb 1 h x B 1 + h hx hb x x hx B + hb hx B B hx 1 h + hx x + hx B 1 + h x 1 h 1 + h 1 + h x x 177

178 HWs Hence Therefore 1 becomes But G x, x x 1 + h hx 1 + h A 1 hb 1 h 1 + h h x 1 h 1 + h hx 1 h 1 + h hx 1 + h 1 + h 1 + h hx 1 + h 1 { hx 1+h 1 x x < x x 1+h hx 1+h x < x 1 h + hx x 1 + h hx x 1 + h 1 + h 1 + h hx x 1 + h 1 1 Hence 1 becomes hx 1+h G x, x 1 x x < x hx 1+h 1 x x < x We see they are symmetrical in x, x. Here is a plot of G x, x for some arbitrary x located at x.75, h 1, for 1.. Problem 3 Green function. x Gx,x x We need to find the solution using y x x [ G x, x f x dx + p x G x, x du x u x dg x, x ] x dx dx x 3 The first step is to find dgx,x dx. From, we find dg x, x dx { hx 1+h x < x hx 1+h 1 x < x 4 Now we plug everything in 3. But remember that G, x, dg,x dx hg, x, u 178

179 .1 HW 1 A, du dx hu. Hence [ G x, x du x u x dg x, x ] x dx dx x du dg x, du dg x, G x, u G x, + u dx dx dx dx dg x, G x, hu u hg x, + A dx {}}{ hx G x, hu + u hg x, + A 1 + h 1 x hx A 1 + h 1 Now we do the integration, From 3, and since p 1 then we obtain x hx y x G x, x f x dx A 1 + h 1 particular solution x boundary terms {}}{ { x }} { hx G x, x f x dx + G x, x f x dx A 1 + h 1 We see that the boundary terms are linear combination x, which is expected as the fundamental solutions for the homogenous solution as linear in x as was found initially above. Plugging values From 3 for G x, x for each region into the above gives y x x hx 1 + h 1 x f x dx + x hx hx 1 + h 1 xf x dx A 1 + h 1 This completes the solution. Now we should test it. et f x x or f x x and compare to direction solution. The above becomes hx x y x 1 + h 1 x hx hx dx + x x 1 + h 1 x dx A 1 + h 1 hx x 3 x 1 + h 1 hx + x hx 3 x 1 + h x dx A 1 + h 1 hx x 3 h 1 + h 1 x 3 + x h 3 x hx A x 1 + h 1 hx x 3 h 1 + h x h 3 h x h 3 + x hx A 1 + h 1 1 h 3 x 3 x + hx 3 + x 3 hx A h 1 + h 1 1 x h x h A 1 hx h 1 + h 1 x h x h h x + A h 1 + h To verify, the problem is solved directly, and solution above using Green function was compared, same answer confirmed. In[15]: ClearAll[x,, y, A, h] Out[156] f x; computersolution u[x] /. First@DSolve[{u''[x] f, u[] A, u'[] + h u[] }, u[x], x]; h - x mysolusinggreenfunction h x3 1 + h - x ^ h A ; 1 + h Simplify[computerSolution - mysolusinggreenfunction].1.4 Problem 4 u + u + u f x ; < x < ; u ; u 1 179

180 HWs Solution Since the coefficient on u is, then the Integrating factor is µ x e dx e x. Multiplying the ODE by µ x gives e x u + e x u + e x u e x f x d e x u + e x u e x f x dx To keep the solution consistent with the class notes, we now multiply both sides by 1 in order to obtain the same form as used in class notes. Hence our ODE is We now see from above that d e x u e x u e x f x dx p x e x Once we found p x, we now find the Green function. The Green function is the solution to d G x, x dx + dg x, x + G x, x δ x x dx G, x G, x Where x is the location of the impulse. We first need to find fundamental solutions to the homogeneous ODE. The solution to u +u +u is found by characteristic method. r +r+1, hence r + 1. Therefore the roots are r 1, double root. Hence the fundamental solutions are Therefore u 1 e x u xe x { A1 u G x, x 1 + A u x < x B 1 u 1 + B u x > x { A1 e x + A xe x x < x B 1 e x + B xe x x > x The first boundary condition on the left end gives A 1 from the first region. The second B.C. on the right end, gives B 1 e + B e B 1 B e e B Hence the above solution now reduces to { A G x, x xe x x < x B e x + B xe x x > x Simplifying B e x + B xe x B x e x, the above can be written as { A G x, x xe x x < x B x e x 1 x > x Continuity at x gives A x e x B x e x A x B x And jump discontinuity on derivative of G gives { d dx G x, x A e x xe x x < x B 1 x + e x x > x Hence important note: we use instead of + d dx p dy dx + 1 px below and not 1 px B 1 x + e x A e x x e x because we started with d dx 1 p x 1 e x e x p dy dx + 18

181 .1 HW 1 Dividing by e x to simplify gives We solve,3 for A 1, B. From 3 B 1 x + A 1 x e x 3 B e x + A 1 x 1 x + 4 Substituting in Hence, from 4 A x e x + A 1 x x 1 x + A x 1 x + e x x + A 1 x x A x 1 x + 1 x x e x x A e x x x 1 x + 1 x x 1 e x x B e x + 1 e x x 1 x 1 x + 1 x e x Therefore the solution 1 becomes { 1 G x, x e x x xe x x < x 1 x e x x e x 5 x > x Or G x, x We see they are symmetrical in x, x. x.75 for 1..4 { x xe x x x < x x x e x x x > x 5 Here is a plot of G x, x for some arbitrary x located at Problem 4 Green function. x.75.3 Gx,x x We now need to find the solution using homogeneous solution/boundary terms particular solution {}}{{}}{ x [ y x G x, x f x dx + p x G x, x du x p x u x dg x, x ] x dx dx x 6 The first step is to find dgx,x dx. From 5, we find { dg x, x dx xe x x xe x x x x < x e x x x xx e x x 7 x > x 181

182 HWs Now we plug everything in 3. But remember that G x,, G x,, u, u 1, p x e x. The following is the result of the homogeneous part [ p x G x, x du x p x u x dg x, x dx dx e G x, du e dg x, u dx dx x<x branch from 7 { }}{ e 1 xe x x xe x x x e xe x xe x e xe x xe x Now we complete the integration, From 3 u x particular {}}{ x x ] x x e G x, x + homogeneous { }} { xe x G x, x f x dx + G x, x f x dx + Plug-in in values From 5 G x, x for each region, u x x from x>x branch in 5 { }} { x x e x x g x dx + x x G x, x f x dx + xe x from x<x branch in 5 du + e dg x, u dx dx { }} { x xe x x g x dx + 1 xe x+ This completes the solution. Now we should test it. et f x x or f x x. But since we multiplied by e x integrating factor at start, we should now use g x e x x as f x below.the above becomes x x u x x e x e x x x x dx + xe x x e x x dx + xe x But x e x x Hence 8 becomes x x e x dx xe x x x x x x e x dx + xe x x e x dx + e x + x x e x x x dx e + e x + x x + x x e x u x + e x + x x xe x e + e x + x x + x + xe x Which can be simplified to For 1, the above becomes u x 1 e x 3e x + + e x x xe u x x x e x + e x + x e e x Verification To verify the above, a plot of the solution was compare to Mathematica result. Here is plot of the result. My solution gives exact plot as Mathematica. 8 18

183 .1 HW 1 In[773]: Out[775] ClearAll[, x, f] f x; computersolution u[x] /. First@DSolve[{u''[x] + u'[x] + u[x] f, u[], u[] 1}, u[x], x] e -x - e x - x + 3 e x - e x + e x x In[778]: mysolution - - x Exp[-x] / - + Exp[x] + x^ - x - x Exp[-x] / Exp[] - + Exp[x] + - x - x + x^ + x / Exp[ - x]; Simplify[mysolution] Out[779] In[78]: Out[78] e -x e x + + e x - + x - e x mysolution - computersolution // Simplify In[781]: 1; p1 Plot[computerSolution, {x,, 1}, PlotRange All, ImageSize 3, Plotabel "Mathematica answer", PlotStyle Blue, Gridines Automatic, GridinesStyle ightgray]; In[783]: p Plot[mysolution, {x,, 1}, PlotRange All, ImageSize 3, Plotabel "Manual solution, Green function method", PlotStyle {Red}, Gridines Automatic, GridinesStyle ightgray]; 1. Mathematica answer 1. Manual solution.8.8 Out[575]

184 HWs.11 HW a Use the method of images to solve Math 3 Homework 11 Due Wednesday Dec. 14, 16 ux fx in a semi-infinite D domain with boundary condition ux, hx. b Use the method of images to solve ux fx in a semi-infinite D domain with boundary condition ux, / y hx. c a Use the method of images to solve ux fx in a semi-infinite 3D domain with boundary condition ux,,z/ y hx,z.. Using the method of images, solve 1a 1b 1c ux fx in the D domain x, y withboundary conditions u,y gyand ux, hx. The following are the general steps used in all the problems below : 1. Image points were placed to satisfy homogenous boundary conditions for Green function using the solution for infinite domain.. The Green formula was applied to determine the particular solution and the boundary terms. 3. Derivative of Green function was found and used in the result found above. 4. The role of x, x was reversed in the final expression to express the final result as u x instead of u x Problem 1 Part a Green function on infinite domain, which is the solution to Is given by G x, x 1 δ x x G x, x 1 ln r π 1 π ln x x 1 x π ln x + y y 1 4π ln x x + y y By placing a negative impulse at location x x, y, the Green function for semi-infinite domain is obtained y Region where solution exist + x, y x x, y 184

185 .11 HW 11 G x, x 1 π ln r 1 1 π ln r 1 π ln x x 1 1 4π ln π ln x x x x + y y ln x x + y + y 1 4π ln x x + y y x x + y + y 1 The following is 3D plot of the above Green function, showing the image impulse and showing that G at the line y marked as red Out[4] The Green function in 1 is now used to solve u x f x, with u x, h x. Starting with Green formula for D u x G x, x G x, x u x da u x G x, x G x, x u x ˆn ds u x G x, x G x, x u x ĵ ds G x, x u x u x G x, x ĵ ds u x G x, x u x G x, x y y y But G x, x δ x, x and u x f x, therefore the above becomes dx u x δ x, x da G x, x f x da u x G x, x u x G x, x y y y dx Since u x δ x, x da u x, the above reduces to u x G x, x f x da u x G x, x u x G x, x y dy y u x G x, x f x da + And since G x, x at y, therefore dx u x G x, x u x G x, x y y y dx u x G x, x f x da + u x G x, x y y dx 185

186 HWs And since u x h x at y, then u x G x, x f x da G x, x h x y y dx 3 dg x, x dy is now evaluated to complete the solution. Using G x, x in equation 1, therefore y dg x, x dy 1 4π 1 4π d dy ln Evaluating the above at y gives dg x, x dy x x + y y ln y y x x + y y y + y x x + y + y y 1 4π 1 π x x + y + y y y x x + y x x + y y x x + y Replacing the above into 3 gives u x G x, x f x da + y π h x x x + y Using the expression for G x, x from 1, the above result becomes u x, y 1 4π x y And finally, order of x, x is reversed giving u x, y 1 4π Part b x y ln x x + y y x x + y + y f x, y dydx + y π ln x x + y y x x + y + y f x, y dy dx + y π dx h x x x + y dx h x x x + y dx This is similar to part a, and the image is placed on the same location as shown above, but now the boundary conditions are different. Starting from equation in part a u x G x, x f x da + u x G x, x u x G x, x y y y dx 1 But now G x, x y at y and not G x, x as in part a. This means the image is a positive impulse and not negative as in part a. Therefore G x, x becomes the following G x, x 1 π ln r π ln r 1 π ln x x π ln π ln x x x x + y y + ln x x + y + y The following is 3D plot of the above Green function, showing that showing that G x, x y at y marked as red Out[3] Green function, semi-infinite D, x at 1,1 and image at 1,-1 partb 186

187 .11 HW 11 Since G x, x y at y then 1 becomes u x G x, x f x da + G x, x u x y y dx But u x y h x at y, hence the above reduces to u x G x, x f x da + G x, x y h x dx 3 Evaluating G x, x at y gives ln G x, x y 1 4π 1 4π 1 4π ln x x + y y + ln ln x x + y x x + y 1 π ln x x + y Substituting the above in RHS of 3 gives u x x y Reversing the role of x, x gives Part c u x x y + ln G x, x f x, y dydx + 1 π x x + y + y y x x + y G x, x f x, y dy dx + 1 π In infinite 3D domain, the Green function for Poisson PDE is given by Where r is given by r G x, x 1 4πr x x + y y + z z ln x x + y h x dx ln x x + y h x dx x x, y, z is the location of the impulse. Since G y, then the same sign impulse is located at x x, y, z, and the Green function becomes G x, x 1 1 4πr 4πr 1 1 4π x x + y y + z z 1 x x + y + y + z z 1 Using Green formula in 3D gives u x G x, x G x, x u x dv u x G x, x G x, x u x ˆn dxdz u x G x, x G x, x u x ĵ dxdz G x, x u x u x G x, x ĵ dxdz u x, y, z G x, x u x G x, y, z, x y y y dxdz But G x, x δ x, x and u x f x, and the above becomes u x δ x, x dv G x, x f x dv u x G x, x u x G x, x y y y dxdz 187

188 HWs But u x δ x, x dv u x, hence u x Rearranging But G x, x f x dv u x G x, x f x dv + u x y h x, z and we impose G x, x y y Evaluating G x, x y gives u x G x, x f x dv + u x G x, x u x G x, x y y y u x G x, x u x G x, x y y y y, therefore the above becomes dxdz dxdz G x, x y h x, z dxdz 3 G x, x y π x x + y y + z z x x + y + y + z z π x x + y + z z x x + y + z z 1 π x x + y + z z Using the above in 3 results in y u x G x, x f x dv 1 π 1 h x, z dxdz x x + y + z z And finally reversing the role of x, x gives the final answer u x G x, x f x dv 1 π 1 h x, z dx dz x x + y + z z.11. Problem Green function in D on infinite domain, which is the solution to Is given by G x, x δ x x G x, x 1 ln r π A negative impulse is placed x 1 x, y and another negative impulse at x x, y and positive one at x 3 x, y. The following is a diagram showing the placement of images. y x, y Region where solution exist + x, y x + x, y x, y 188

189 .11 HW 11 Or The resulting Green function becomes G x, x 1 1 ln r π π ln r 1 1 π ln r + 1 π ln r 3 G x, x 1 4π ln x x + y y 1 4π ln x x + y + y 1 4π ln x + x + y y + 1 4π ln x + x + y + y 1 The following is 3D plot of the above Green function, showing the image impulse and showing that G at the line y and also at line x. ines marked as red and blue Out[5] Now that the Green function is found, it is used to solve u x f x, with u x, h x, u, y g y.starting with Green formula for D u x G x, x G x, x u x da u x G x, x G x, x u x ˆn ds s 1 + u x G x, x G x, x u x ˆn ds s To simplify the notation, from now on, G is used of G x, x, and also u instead of u x. The line s 1 in the above is the line x >, y and s is the line x, y >. Therefore the above becomes u G G u da u G G u ĵ ds + u G G u î ds s 1 s Or u G da G u da G u y u G y y dx + But G x, x δ x, x and u x f x, hence the above reduces to u x δ x, x da Gf x da But u x δ x, x da u x therefore Or u x Gf x da u x Gf x da + G u y u G y y G u y u G y y G u y u G y y dx + dx + dx + Since G x, x at y, and G x, x at x, the above becomes u x Gf x da u G y y dx G u x u G x x dy G u x u G x x G u x u G x x G u x u G x x u G x x dy dy dy dy 189

190 HWs Since u x h x at y and u x g y at x then u x, y Gf x da h x G y y dx g y G x x dy dg dy and G y x x are now evaluated to complete the solution. Using G x, x in equation 1 gives G y 1 y y 4π x x + y y 1 y + y 4π x x + y + y 1 y y 4π x + x + y y + 1 y + y 4π x + x + y + y Evaluating the above at y results in G 1 y y y 4π x x 1 y + y 4π x x + y 1 y 4π x + x + 1 y + y 4π x + x + y Or Finding G x gives G x 1 4π G y 1 y y π x + x + y 1 4π x x x x + y y x + x x + x + y y 1 4π + 1 4π 1 x x + y x x x x + y + y x + x x + x + y + y Evaluating the above at x results in G 1 x x x 4π x + y y 1 x 4π x + y + y 1 x 4π x + y y + 1 x 4π x + y + y Or G x x 1 π x π Substituting 3,4 into gives the final answer u x, y y π x π Reversing the role of x, x gives u x, y y π x π x x + y + y 1 π x + y y 1 x + y + y 1 x + y y G x, x f x dxdy 1 1 h x x + x + y x x dx + y 1 g y x + y + y 1 x + y y dy G x, x f x, y dx dy h x g y x 1 x + x + y 1 x x + y 1 x + y + y 1 x + y y dx dy

191 .11 HW 11 Where G x, x is given by equation 1. This complete the solution. The following is 3D plot of the solution for small area is first quadrant generated using Mathematica using f x e x 4 y 5 g y 1 sin 5y h x 5 cos x Out[194] This is a contour plot of the above solution Out[186]