Suggested Solution to Assignment 7
|
|
- Harold Pope
- 5 years ago
- Views:
Transcription
1 MATH 422 (25-6) partial diferential equations Suggested Solution to Assignment 7 Exercise 7.. Suppose there exists one non-constant harmonic function u in, which attains its maximum M at x. Then by the mean value property, we have u(x ) uds, B(x, r) B(x,r) for all B(x, r). maximum. Thus u(x) M for all x B(x, r) since u is continuous and M is its Now since u is not a constant, there exists x such that u(x ) u(x ). Since is a region, we can find a continuous a curve γ(t) ( t ) such that γ() x and γ() x. Set E : { t u(γ(t)) M}, then E is closed since u and γ are both continuous. Let t E, then by the above result, u(x) M for all x B(γ(t ), r). So E is open. Hence, E [, since E and [, is connected. But this contradicts to u(x ) u(x ) and then we complete the proof. 2. Suppose u and u 2 are solutions of the problem. Then u u u 2 satisfies u in, on. Thus, by the Green s first identity for v u, we have u ds u 2 dx + Hence, Therefore, u and then u(x) is a constant.. u 2 dx. u udx. 3. Suppose u and u 2 are solutions of the problem. Then u u u 2 satisfies u in, + au on. Thus, by the Green s first identity for v u, we have u ds u 2 dx + u udx. Hence, u 2 dx au 2 ds. Since a(x) >, u and then u(x) is a constant C. So by the Robin boundary conditions, we have Ca(x). This shows that C and u.. 4. Suppose u and u 2 are solutions of the problem. Then u u u 2 satisfies u t k u in (, ), u on (, ), u(x, ) in. Thus, by the Green s first identity for v u, we have u ds u 2 dx + u udx.
2 MATH 422 (25-6) partial diferential equations Therefore, by the diffusion equation u 2 dx + uu t dx k u 2 dx + d u 2 dx. 2k dt efine E(t) : u2 dx, then the above equality implies d dte(t). Note that E() and E(t), we have E(t). So we obtain u and the uniqueness for the diffusion equation with irichlet boundary conditions is proved.. 5. Suppose u(x) minimizes the energy. Let v(x) be any function and ɛ be any constant, then E[u E[u + ɛv E[u + ɛ u vdx ɛ hvds + ɛ 2 v 2 dx. Hence, by calculus for any function v. By Green s first identity, u vdx + u vdx hvds v ds hvds. Noting v is an arbitrary function, let v vanish on firstly and then the above equality implies u in. So the above equality changes to v ds hvds. Since v is arbitrary, h on and then u(x) is a solution of the following Neumann problem u in, h on bdy. Note that the Neumann problem has a unique solution up to constant and the functional E[w does not change if a constatn is added to w, thus it is the only funtion that can minimize the energy up to a constant. Note that here we assume functions u, v and the domain are smooth enough, at least under which the Green s first identity can be hold. 6. (a) Suppose u and u 2 are solutions of the problem. Then u u u 2 tends to zero at infinity, and is constant on A and constant on B, and satisfies ds ds. A Suppose that u, then without loss of generality, we assume that there exists x such that u(x ) >. Since u tends to zero at infinity, so there exists R such that u(x) u(x ) 2 if x R. Then u is a harmonic function in B(, R) and u. Thus, the B max u u(x ) 2 u(x ) max B(,R) B(,R) maximum is attained on A or B by the Maximum Principle. WOLG, we assume that u attain its maximum on A and then by any point of A is a maximum point since u is constant on A. Hence, by the Hopf maximum principle, / > on A, but this contradicts to the condition Therefore, u and then the solution is unique. A 2 ds.
3 MATH 422 (25-6) partial diferential equations (b) If not, then u(x) have a negative minimum. As above, by the Minimum Principle, we get that u attains its minimum on A or B. WOLG, we assume that u attains its minimum on A, then < by the Hopf principle since u is not a constant. But this contradicts to A ds Q >. So u in, (c) Suppose that there exists x such that u(x ). Then by (b) and the Strong Minimum Principle, u is a constant in. But this contradicts to A ds Q >. So u > in. (Actually, we can prove that u > on A and B by the Hopf principle again as in (b)). 7. (Extra Problem ) Suppose u and u 2 are solutions of the problem. Then (u u 2 ) u 3 u 3 2 (u u 2 )(u 2 + u u 2 + u 2 2) in Thus, (u u 2 ) + a(x)(u u 2 ) on (u u 2 ) 2 (u 2 + u u 2 + u 2 2) (u u 2 ) (u u 2 ) (u u 2 ) (u u 2 ) u u 2 2 a(x)(u u 2 ) 2 u u 2 2 since a(x). (u u 2 ) 2 (u 2 + u u 2 + u 2 2 ) u u 2 in 8. (Extra Problem 2) (a) E[u : 2 ( u 2 + b(x)u 2 ) + f(x)u (b)if u is a solution, then v C 2(), v u + b(x)uv + f(x)v u v + b(x)uv + f(x)v w C 2 (), w h on, take v w u, E[w 2 v u 2 + v u + 2 b(x)v2 + 2 b(x)u2 + b(x)uv + f(x)v + f(x)u 2 v b(x)v2 + E[u since b(x). E[u Conversely, v C 2(), t R, w : u + tv C2, w u h on d dt E[u + tv t u v + b(x)uv + f(x)v v( u + b(x)u + f(x)), v C 2 () u + b(x)u + f(x) in 3
4 MATH 422 (25-6) partial diferential equations Exercise Let r x and ɛ : {x ɛ < r < R}, where R is large enough such that φ outside r < R/2. Using the Green s second identity, [ ɛ x φ(x)dx x φ φ ds x x R x ɛ ɛ [ x φ φ x [ ɛ φ + φ c 2 ds + x ɛ ds 4φ 4ɛ φ, where φ denotes the average value of φ on the sphere {r ɛ}, and φ this sphere. Since φ is continuous and φ is bounded, Hence, 4φ 4ɛ φ B(x,R) φ() x <R [ x φ φ 4φ() as ɛ. x φ(x)dx 4. x ds denotes the average value of φ on 3. Choosing B(x, R) in the representation formula () and using the divergence theorem, [ u(x ) u(x) ( x x ) + x x ds 4 x x R 4R 2 4R 2 [ R 2 u(x) + ds R 4 x x R x x R uds + 4R uds. x x <R 4. (Extra Problem) φ(x) Cc 2 (R 2 ), we need to show that φ() log x φ(x) dx R 2 2 udx proof: Let r x and ε : x : ε < r < R, where R is large enough such that φ if r R/2. Using the Green s identity, log x φ(x)dx [log x φ ε ε φ log x ds [log x φ x ε φ log x ds [log ε φ φ ε ds x ε 2 φ 2ε log ε φ 4
5 MATH 422 (25-6) partial diferential equations where φ, bounded, Exercise 7.3 φ φ denote the average value of φ, φ on the sphere x ε. Since φ is continuous and 2 φ φ 2ε log ε 2φ(), as ε. Suppose G and G 2 both are the Green s functions for the operator and the domain at the point x d. By (i) and (iii), we obtain that G (x) + 4 x x and G 2(x) + both are harmonic 4 x x functions in. Thus, by (ii) and the uniqueness theorem of harmonic function, we have G (x) + that is, the Green s function is unique. 3. In the textbook, 4 x x G 2(x) + 4 x x, A ɛ x a ɛ where u(x) G(x, a) and v(x) G(x, b). Note that (u v v )ds, u(x) G(x, a) + H(x, a) 4 x a, where H(x, a) is a harmonic function throughout the domain. What s more, v(x) G(x, b) is a harmonic function in { x a < ɛ}, then we have [ v A ɛ ( + H(x, a)) x a ɛ 4 x a v ( + H(x, a)) ds 4 x a [ v x a ɛ 4 x a + v ( 4 x a ) ds [ + H(x, a) v x a ɛ v H(x, a) ds v(a) [H(x, a) v v H(x, a) dx v(a). x a <ɛ is Here we use the representation formula (7.2.) and. So lim A ɛ v(a) G(a, b). ɛ Exercise 7.4. By (i), we know that G(x) is a linear function in [, x and [x, l. Since (ii) and G(x) is continuous at x, we have { kx, < x x ; G(x) kx x l (x l), x < x < l. Hence, G(x) + 2 x x is harmonic at x if and only if if and only if (G(x) + 2 x x ) x (G(x) + 2 x x ) x, + k l x. l 5
6 MATH 422 (25-6) partial diferential equations So the one-dimensional Green s function for the interval (, l) at the point x (, l) is G(x) { l x l x, < x x ; x l (x l), x < x < l. 2. Assume that h(x, y) is a continuous function that vanished outside {(x, y) x 2 +y 2 R 2 } and h(x, y) M. Then by the formula (3), we have u(x, y, z ) M [(x x ) 2 + (y y ) 2 + z dxdy {(x,y) x 2 +y 2 R 2 } MR 2 2( x 2 + y2 +, when z2 R)3 x 2 + y2 + z2 > R. Therefore, u satisfies the condition at infinity: u(x), as x. 3. From (3), we have u(x, y, z ) z 2 z 2 [(x x ) 2 + (y y ) 2 + z h(x, y)dxdy [x 2 + y 2 + z h(x + x, y + y )dxdy. Now we change variables such that x z s cos θ, y z s sin θ. Then Since u(x, y, z ) z (z 2 s 2 + z 2 ) 3 2 h(z s cos θ + x, z s sin θ + y )z 2 s dsdθ s(s 2 + ) 3 2 h(z s cos θ + x, z s sin θ + y ) dsdθ. s(s 2 + ) 3 2 ds (s 2 + ) 3 2, lim u(x, y, z ) h(x, y ), z if the limit can be taken in the integration, for example, when h(x, y) is bounded. 5. Here is one of suggested explains. Since the half-plane {y > } is not bounded, this only means that the solution is not unique under the (uncompleted) boundary condition or it is not a well-posed problem. But generally it will be a well-posed problem if we add another boundary condition, such as u(x), as x. 6. (a) Using the method of reflection, as in the dimension three, we have where x (x, y), x (x, y ), x (x, y ). G(x, x ) 2 log x x 2 log x x, 6
7 MATH 422 (25-6) partial diferential equations (b) Since G y y + y 2[(x x ) 2 + (y + y ) 2 y y 2[(x x ) 2 + (y y ) 2 y [(x x ) 2 + y 2 on y, the solution is given by u(x, y ) y h(x) (x x ) 2 + y 2 dx. (c) By (b), we have u(x, y ) y h(x) (x x ) 2 + y 2 dx x 2 dx (a) If u(x, y) f( x ), then y (b) So f has to satisfy the following OE where a, b are contants. (c) Using the polar coordinate, u x y f ( x y ), u y x y 2 f ( x y ). u xx + u yy y 2 f ( x y ) + x2 y 4 f ( x y ) + 2x y 3 f ( x y ). f(t) ( + t 2 )f (t) + 2tf (t). t a ds + b a arctan t + b, + s2 x x + y y y f ( x y )x r x y 2 f ( x y )y r. v(r, θ) cθ + d, where c, d are constants. So in the (x, y)-coordinate, we have v(x, y) c arctan y x + d. (d) By (a), So u(x, y) f( x y ) a arctan x y + b. h(x) lim y u(x, y) 2 a + b, x > ; b, x ; a + b, x <. 2 (e) By (d), we see that h(x) is not continuous unless u is a constant. This agrees with the condition that h(x) is continuous in Exercise 6. 7
8 MATH 422 (25-6) partial diferential equations 9. Here the methof of reflection is used. Let x (x, y, z ) such that ax + by + cz >, then its reflection point x (x, y, z ) about the hyperplane {ax + by + cz } satisfies x 2a x a 2 + b 2 + c 2 (ax + by + cz ), y 2b y a 2 + b 2 + c 2 (ax + by + cz ), z 2c z a 2 + b 2 + c 2 (ax + by + cz ). So the Green s function for the tilted half-space {(x, y, z) : ax + by + cz > } is given by G(x, x ) 4 x x + 4 x x.. In case x, the formula for the Green s function is G(x, ) 4 x + 4a, since 4 x + 4a C2 and is harmonic in B a (), except at the point x ; [ 4 x + 4a x a ; and [ 4 x + 4a + 4 x is harmonic at. 3. Let x, then we have G(x, x ) 4 x x + 4 r x/a ax /r is the Green s function at x for the whole ball, where a is the radius and r x. Let x (x, y, z ), then G(x, x ) 4 x x + 4 r x/a ax /r is the Green s function at x for the whole ball. Note that G(x, x ) G(x, x ) on and G(x, x ) is a harmonic function on. Hence, G(x, x ) G(x, x ) is the Green s function for by the uniqueness of the Green s function. 5. (a) If v(x, y) is harmonic and u(x, y) v(x 2 y 2, 2xy), then Thus u x 2xv x + 2yv y, u y 2yv x + 2xv y. u xx + u yy 2v x + 4x 2 v xx + 4xyv xy + 4xyv yx + 4y 2 v yy 2v x + 4y 2 v xx 4xyv xy 4xyv yx + 4x 2 v yy 4(x 2 + y 2 )(v xx + v yy ). (b) Using the polar coordinates, the transformation si (r cos θ, r sin θ) (r 2 cos 2θ, r 2 sin 2θ). Therefore, it maps the first quadrant onto the half-plane {y > }. Note that the transformation is one-one which also maps the x-positive-axis to x-positive-axis and y-positive-axis to x-negative-axis. 8
9 MATH 422 (25-6) partial diferential equations 7. (a) Here the method of reflection is used and you can also use the result of Exercise 55 to find the Green s function for the quadrant Q. Let x (x, y ) Q, x y ( x, y ), x x (x, y ) and x ( x, y ). Then it is easy to verify that G(x, x ) 2 log x x 2 log x xy 2 log x xx + 2 log x x is the Green s function at x for the quadrant Q. (b) (Extra Problem 3) By formula () in Section 7.3 in the textbook, u(x ) u(x) G(x, x ) ds bdy x G G x,y> x x ( x x 2 x x ) 2 G G x>,y y y ( x x 2 x x ) 2 g(y)( x x 2 x x )dy + y 2 h(x)( x x 2 x x )dx 2 9. Consider the four-dimensional laplacian u u xx + u yy + u zz + u ww. Show that its fundamental solution is r 2, where r 2 x 2 + y 2 + z 2 + w 2. proof: irect computation according to its definition. 2. Using the conclusion in 7 and 2, we have the singular part of the Green function is 8ω 4 x x 2, where ω 4 is the volume of S 3, x (x, y, z, w) and x (x, y, z, w ). So the Green function is 8ω 4 x x 2 + 8ω 4 x x 2, where x (x, y, z, w ). 2. By the formula (2)(3) in Section7.3, we get u(x ) bdy (u N N) bdy N since N on the boundary. Thus we proved the following theorem: If N(x, x ) is the Neumann function, then the solution of the Neumann problem is given by the formula u(x ) 22. (Extra Problem 4)By the formula (4) in Section7.4, u(x ) z 2 z bdy N x 2 +y 2 + x x 3 ds. If the bounded condition is dropped, then we have v u + az where a is arbitrary real number is also a solution. 9
MATH 425, FINAL EXAM SOLUTIONS
MATH 425, FINAL EXAM SOLUTIONS Each exercise is worth 50 points. Exercise. a The operator L is defined on smooth functions of (x, y by: Is the operator L linear? Prove your answer. L (u := arctan(xy u
More informationMath 342 Partial Differential Equations «Viktor Grigoryan
Math 342 Partial ifferential Equations «Viktor Grigoryan 3 Green s first identity Having studied Laplace s equation in regions with simple geometry, we now start developing some tools, which will lead
More informationMATH 124B Solution Key HW 05
7.1 GREEN S FIRST IENTITY MATH 14B Solution Key HW 05 7.1 GREEN S FIRST IENTITY 1. erive the 3-dimensional maximum principle from the mean value property. SOLUTION. We aim to prove that if u is harmonic
More informationGREEN S IDENTITIES AND GREEN S FUNCTIONS
GREEN S IENTITIES AN GREEN S FUNCTIONS Green s first identity First, recall the following theorem. Theorem: (ivergence Theorem) Let be a bounded solid region with a piecewise C 1 boundary surface. Let
More informationu xx + u yy = 0. (5.1)
Chapter 5 Laplace Equation The following equation is called Laplace equation in two independent variables x, y: The non-homogeneous problem u xx + u yy =. (5.1) u xx + u yy = F, (5.) where F is a function
More informationApplications of the Maximum Principle
Jim Lambers MAT 606 Spring Semester 2015-16 Lecture 26 Notes These notes correspond to Sections 7.4-7.6 in the text. Applications of the Maximum Principle The maximum principle for Laplace s equation is
More informationMultivariable Calculus
2 Multivariable Calculus 2.1 Limits and Continuity Problem 2.1.1 (Fa94) Let the function f : R n R n satisfy the following two conditions: (i) f (K ) is compact whenever K is a compact subset of R n. (ii)
More informationMATH COURSE NOTES - CLASS MEETING # Introduction to PDEs, Fall 2011 Professor: Jared Speck
MATH 8.52 COURSE NOTES - CLASS MEETING # 6 8.52 Introduction to PDEs, Fall 20 Professor: Jared Speck Class Meeting # 6: Laplace s and Poisson s Equations We will now study the Laplace and Poisson equations
More informationMATH COURSE NOTES - CLASS MEETING # Introduction to PDEs, Spring 2018 Professor: Jared Speck
MATH 8.52 COURSE NOTES - CLASS MEETING # 6 8.52 Introduction to PDEs, Spring 208 Professor: Jared Speck Class Meeting # 6: Laplace s and Poisson s Equations We will now study the Laplace and Poisson equations
More informationFinal Exam May 4, 2016
1 Math 425 / AMCS 525 Dr. DeTurck Final Exam May 4, 2016 You may use your book and notes on this exam. Show your work in the exam book. Work only the problems that correspond to the section that you prepared.
More informationMath The Laplacian. 1 Green s Identities, Fundamental Solution
Math. 209 The Laplacian Green s Identities, Fundamental Solution Let be a bounded open set in R n, n 2, with smooth boundary. The fact that the boundary is smooth means that at each point x the external
More informationMath 4263 Homework Set 1
Homework Set 1 1. Solve the following PDE/BVP 2. Solve the following PDE/BVP 2u t + 3u x = 0 u (x, 0) = sin (x) u x + e x u y = 0 u (0, y) = y 2 3. (a) Find the curves γ : t (x (t), y (t)) such that that
More informationLecture No 1 Introduction to Diffusion equations The heat equat
Lecture No 1 Introduction to Diffusion equations The heat equation Columbia University IAS summer program June, 2009 Outline of the lectures We will discuss some basic models of diffusion equations and
More informationMATH 124B Solution Key HW 03
6.1 LAPLACE S EQUATION MATH 124B Solution Key HW 03 6.1 LAPLACE S EQUATION 4. Solve u x x + u y y + u zz = 0 in the spherical shell 0 < a < r < b with the boundary conditions u = A on r = a and u = B on
More information[2] (a) Develop and describe the piecewise linear Galerkin finite element approximation of,
269 C, Vese Practice problems [1] Write the differential equation u + u = f(x, y), (x, y) Ω u = 1 (x, y) Ω 1 n + u = x (x, y) Ω 2, Ω = {(x, y) x 2 + y 2 < 1}, Ω 1 = {(x, y) x 2 + y 2 = 1, x 0}, Ω 2 = {(x,
More informationPartial Differential Equations
Part II Partial Differential Equations Year 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2015 Paper 4, Section II 29E Partial Differential Equations 72 (a) Show that the Cauchy problem for u(x,
More informationContents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9
MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Contents Multiple Integrals 3 2 Vector Fields 9 3 Line and Surface Integrals 5 4 The Classical Integral Theorems 9 MATH 32B-2 (8W)
More informationSummer 2017 MATH Solution to Exercise 5
Summer 07 MATH00 Solution to Exercise 5. Find the partial derivatives of the following functions: (a (xy 5z/( + x, (b x/ x + y, (c arctan y/x, (d log((t + 3 + ts, (e sin(xy z 3, (f x α, x = (x,, x n. (a
More informationFinal: Solutions Math 118A, Fall 2013
Final: Solutions Math 118A, Fall 2013 1. [20 pts] For each of the following PDEs for u(x, y), give their order and say if they are nonlinear or linear. If they are linear, say if they are homogeneous or
More informationUNIVERSITY OF MANITOBA
Question Points Score INSTRUCTIONS TO STUDENTS: This is a 6 hour examination. No extra time will be given. No texts, notes, or other aids are permitted. There are no calculators, cellphones or electronic
More informationSolutions: Problem Set 4 Math 201B, Winter 2007
Solutions: Problem Set 4 Math 2B, Winter 27 Problem. (a Define f : by { x /2 if < x
More informationM545 Homework 8. Mark Blumstein. December 9, [f (x)] 2 dx = 1. Find such a function if you can. If it cannot be found, explain why not.
M545 Homework 8 Mark Blumstein ecember 9, 5..) Let fx) be a function such that f) f3), 3 [fx)] dx, and 3 [f x)] dx. Find such a function if you can. If it cannot be found, explain why not. Proof. We will
More informationMath 127C, Spring 2006 Final Exam Solutions. x 2 ), g(y 1, y 2 ) = ( y 1 y 2, y1 2 + y2) 2. (g f) (0) = g (f(0))f (0).
Math 27C, Spring 26 Final Exam Solutions. Define f : R 2 R 2 and g : R 2 R 2 by f(x, x 2 (sin x 2 x, e x x 2, g(y, y 2 ( y y 2, y 2 + y2 2. Use the chain rule to compute the matrix of (g f (,. By the chain
More informationMath 350 Solutions for Final Exam Page 1. Problem 1. (10 points) (a) Compute the line integral. F ds C. z dx + y dy + x dz C
Math 35 Solutions for Final Exam Page Problem. ( points) (a) ompute the line integral F ds for the path c(t) = (t 2, t 3, t) with t and the vector field F (x, y, z) = xi + zj + xk. (b) ompute the line
More informationMathematical Methods - Lecture 9
Mathematical Methods - Lecture 9 Yuliya Tarabalka Inria Sophia-Antipolis Méditerranée, Titane team, http://www-sop.inria.fr/members/yuliya.tarabalka/ Tel.: +33 (0)4 92 38 77 09 email: yuliya.tarabalka@inria.fr
More informationINTRODUCTION TO PDEs
INTRODUCTION TO PDEs In this course we are interested in the numerical approximation of PDEs using finite difference methods (FDM). We will use some simple prototype boundary value problems (BVP) and initial
More informationMath 185 Homework Exercises II
Math 185 Homework Exercises II Instructor: Andrés E. Caicedo Due: July 10, 2002 1. Verify that if f H(Ω) C 2 (Ω) is never zero, then ln f is harmonic in Ω. 2. Let f = u+iv H(Ω) C 2 (Ω). Let p 2 be an integer.
More informationLeplace s Equations. Analyzing the Analyticity of Analytic Analysis DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING. Engineering Math 16.
Leplace s Analyzing the Analyticity of Analytic Analysis Engineering Math 16.364 1 The Laplace equations are built on the Cauchy- Riemann equations. They are used in many branches of physics such as heat
More informationLeplace s Equations. Analyzing the Analyticity of Analytic Analysis DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING. Engineering Math EECE
Leplace s Analyzing the Analyticity of Analytic Analysis Engineering Math EECE 3640 1 The Laplace equations are built on the Cauchy- Riemann equations. They are used in many branches of physics such as
More informationAero III/IV Conformal Mapping
Aero III/IV Conformal Mapping View complex function as a mapping Unlike a real function, a complex function w = f(z) cannot be represented by a curve. Instead it is useful to view it as a mapping. Write
More informationCALCULUS JIA-MING (FRANK) LIOU
CALCULUS JIA-MING (FRANK) LIOU Abstract. Contents. Power Series.. Polynomials and Formal Power Series.2. Radius of Convergence 2.3. Derivative and Antiderivative of Power Series 4.4. Power Series Expansion
More informationALGEBRAIC GEOMETRY HOMEWORK 3
ALGEBRAIC GEOMETRY HOMEWORK 3 (1) Consider the curve Y 2 = X 2 (X + 1). (a) Sketch the curve. (b) Determine the singular point P on C. (c) For all lines through P, determine the intersection multiplicity
More informationPartial Differential Equations
M3M3 Partial Differential Equations Solutions to problem sheet 3/4 1* (i) Show that the second order linear differential operators L and M, defined in some domain Ω R n, and given by Mφ = Lφ = j=1 j=1
More informationTEST CODE: MIII (Objective type) 2010 SYLLABUS
TEST CODE: MIII (Objective type) 200 SYLLABUS Algebra Permutations and combinations. Binomial theorem. Theory of equations. Inequalities. Complex numbers and De Moivre s theorem. Elementary set theory.
More informationAP Calculus Testbank (Chapter 9) (Mr. Surowski)
AP Calculus Testbank (Chapter 9) (Mr. Surowski) Part I. Multiple-Choice Questions n 1 1. The series will converge, provided that n 1+p + n + 1 (A) p > 1 (B) p > 2 (C) p >.5 (D) p 0 2. The series
More informationIn this chapter we study elliptical PDEs. That is, PDEs of the form. 2 u = lots,
Chapter 8 Elliptic PDEs In this chapter we study elliptical PDEs. That is, PDEs of the form 2 u = lots, where lots means lower-order terms (u x, u y,..., u, f). Here are some ways to think about the physical
More informationSeries Solution of Linear Ordinary Differential Equations
Series Solution of Linear Ordinary Differential Equations Department of Mathematics IIT Guwahati Aim: To study methods for determining series expansions for solutions to linear ODE with variable coefficients.
More informationMath 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3
Math 201 Solutions to Assignment 1 1. Solve the initial value problem: x 2 dx + 2y = 0, y(0) = 2. x 2 dx + 2y = 0, y(0) = 2 2y = x 2 dx y 2 = 1 3 x3 + C y = C 1 3 x3 Notice that y is not defined for some
More informationUNIVERSITY OF HOUSTON HIGH SCHOOL MATHEMATICS CONTEST Spring 2018 Calculus Test
UNIVERSITY OF HOUSTON HIGH SCHOOL MATHEMATICS CONTEST Spring 2018 Calculus Test NAME: SCHOOL: 1. Let f be some function for which you know only that if 0 < x < 1, then f(x) 5 < 0.1. Which of the following
More informationComplex Variables. Chapter 2. Analytic Functions Section Harmonic Functions Proofs of Theorems. March 19, 2017
Complex Variables Chapter 2. Analytic Functions Section 2.26. Harmonic Functions Proofs of Theorems March 19, 2017 () Complex Variables March 19, 2017 1 / 5 Table of contents 1 Theorem 2.26.1. 2 Theorem
More informationMATH 126 FINAL EXAM. Name:
MATH 126 FINAL EXAM Name: Exam policies: Closed book, closed notes, no external resources, individual work. Please write your name on the exam and on each page you detach. Unless stated otherwise, you
More informationFirst order wave equations. Transport equation is conservation law with J = cu, u t + cu x = 0, < x <.
First order wave equations Transport equation is conservation law with J = cu, u t + cu x = 0, < x
More informationMATH20411 PDEs and Vector Calculus B
MATH2411 PDEs and Vector Calculus B Dr Stefan Güttel Acknowledgement The lecture notes and other course materials are based on notes provided by Dr Catherine Powell. SECTION 1: Introctory Material MATH2411
More informationHOMEWORK 8 SOLUTIONS
HOMEWOK 8 OLUTION. Let and φ = xdy dz + ydz dx + zdx dy. let be the disk at height given by: : x + y, z =, let X be the region in 3 bounded by the cone and the disk. We orient X via dx dy dz, then by definition
More informationLecture notes: Introduction to Partial Differential Equations
Lecture notes: Introduction to Partial Differential Equations Sergei V. Shabanov Department of Mathematics, University of Florida, Gainesville, FL 32611 USA CHAPTER 1 Classification of Partial Differential
More information1MA6 Partial Differentiation and Multiple Integrals: I
1MA6/1 1MA6 Partial Differentiation and Multiple Integrals: I Dr D W Murray Michaelmas Term 1994 1. Total differential. (a) State the conditions for the expression P (x, y)dx+q(x, y)dy to be the perfect
More information0, otherwise. Find each of the following limits, or explain that the limit does not exist.
Midterm Solutions 1, y x 4 1. Let f(x, y) = 1, y 0 0, otherwise. Find each of the following limits, or explain that the limit does not exist. (a) (b) (c) lim f(x, y) (x,y) (0,1) lim f(x, y) (x,y) (2,3)
More informationSolutions of Math 53 Midterm Exam I
Solutions of Math 53 Midterm Exam I Problem 1: (1) [8 points] Draw a direction field for the given differential equation y 0 = t + y. (2) [8 points] Based on the direction field, determine the behavior
More informationMATH 31BH Homework 5 Solutions
MATH 3BH Homework 5 Solutions February 4, 204 Problem.8.2 (a) Let x t f y = x 2 + y 2 + 2z 2 and g(t) = t 2. z t 3 Then by the chain rule a a a D(g f) b = Dg f b Df b c c c = [Dg(a 2 + b 2 + 2c 2 )] [
More informationELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS EXERCISES I (HARMONIC FUNCTIONS)
ELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS EXERCISES I (HARMONIC FUNCTIONS) MATANIA BEN-ARTZI. BOOKS [CH] R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. Interscience Publ. 962. II, [E] L.
More informationPDEs, part 1: Introduction and elliptic PDEs
PDEs, part 1: Introduction and elliptic PDEs Anna-Karin Tornberg Mathematical Models, Analysis and Simulation Fall semester, 2013 Partial di erential equations The solution depends on several variables,
More informationPRACTICE PROBLEMS. Please let me know if you find any mistakes in the text so that i can fix them. 1. Mixed partial derivatives.
PRACTICE PROBLEMS Please let me know if you find any mistakes in the text so that i can fix them. 1.1. Let Show that f is C 1 and yet How is that possible? 1. Mixed partial derivatives f(x, y) = {xy x
More informationElectromagnetism HW 1 math review
Electromagnetism HW math review Problems -5 due Mon 7th Sep, 6- due Mon 4th Sep Exercise. The Levi-Civita symbol, ɛ ijk, also known as the completely antisymmetric rank-3 tensor, has the following properties:
More informationSection Taylor and Maclaurin Series
Section.0 Taylor and Maclaurin Series Ruipeng Shen Feb 5 Taylor and Maclaurin Series Main Goal: How to find a power series representation for a smooth function us assume that a smooth function has a power
More informationJUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 13 (Second moments of a volume (A)) A.J.Hobson
JUST THE MATHS UNIT NUMBER 13.13 INTEGRATION APPLICATIONS 13 (Second moments of a volume (A)) by A.J.Hobson 13.13.1 Introduction 13.13. The second moment of a volume of revolution about the y-axis 13.13.3
More informationMATH 220: Problem Set 3 Solutions
MATH 220: Problem Set 3 Solutions Problem 1. Let ψ C() be given by: 0, x < 1, 1 + x, 1 < x < 0, ψ(x) = 1 x, 0 < x < 1, 0, x > 1, so that it verifies ψ 0, ψ(x) = 0 if x 1 and ψ(x)dx = 1. Consider (ψ j )
More informationModule 7: The Laplace Equation
Module 7: The Laplace Equation In this module, we shall study one of the most important partial differential equations in physics known as the Laplace equation 2 u = 0 in Ω R n, (1) where 2 u := n i=1
More information1 Review of di erential calculus
Review of di erential calculus This chapter presents the main elements of di erential calculus needed in probability theory. Often, students taking a course on probability theory have problems with concepts
More informationPreliminary Exam 2018 Solutions to Morning Exam
Preliminary Exam 28 Solutions to Morning Exam Part I. Solve four of the following five problems. Problem. Consider the series n 2 (n log n) and n 2 (n(log n)2 ). Show that one converges and one diverges
More informationBoundary conditions. Diffusion 2: Boundary conditions, long time behavior
Boundary conditions In a domain Ω one has to add boundary conditions to the heat (or diffusion) equation: 1. u(x, t) = φ for x Ω. Temperature given at the boundary. Also density given at the boundary.
More informationApplied Math Qualifying Exam 11 October Instructions: Work 2 out of 3 problems in each of the 3 parts for a total of 6 problems.
Printed Name: Signature: Applied Math Qualifying Exam 11 October 2014 Instructions: Work 2 out of 3 problems in each of the 3 parts for a total of 6 problems. 2 Part 1 (1) Let Ω be an open subset of R
More informationMATH 131P: PRACTICE FINAL SOLUTIONS DECEMBER 12, 2012
MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, This is a closed ook, closed notes, no calculators/computers exam. There are 6 prolems. Write your solutions to Prolems -3 in lue ook #, and your solutions to
More informationFinal Exam 2011 Winter Term 2 Solutions
. (a Find the radius of convergence of the series: ( k k+ x k. Solution: Using the Ratio Test, we get: L = lim a k+ a k = lim ( k+ k+ x k+ ( k k+ x k = lim x = x. Note that the series converges for L
More informationLOWELL JOURNAL. MUST APOLOGIZE. such communication with the shore as Is m i Boimhle, noewwary and proper for the comfort
- 7 7 Z 8 q ) V x - X > q - < Y Y X V - z - - - - V - V - q \ - q q < -- V - - - x - - V q > x - x q - x q - x - - - 7 -» - - - - 6 q x - > - - x - - - x- - - q q - V - x - - ( Y q Y7 - >»> - x Y - ] [
More informationMathematics of Physics and Engineering II: Homework problems
Mathematics of Physics and Engineering II: Homework problems Homework. Problem. Consider four points in R 3 : P (,, ), Q(,, 2), R(,, ), S( + a,, 2a), where a is a real number. () Compute the coordinates
More informationReview For the Final: Problem 1 Find the general solutions of the following DEs. a) x 2 y xy y 2 = 0 solution: = 0 : homogeneous equation.
Review For the Final: Problem 1 Find the general solutions of the following DEs. a) x 2 y xy y 2 = 0 solution: y y x y2 = 0 : homogeneous equation. x2 v = y dy, y = vx, and x v + x dv dx = v + v2. dx =
More informationis a weak solution with the a ij,b i,c2 C 1 ( )
Thus @u @x i PDE 69 is a weak solution with the RHS @f @x i L. Thus u W 3, loc (). Iterating further, and using a generalized Sobolev imbedding gives that u is smooth. Theorem 3.33 (Local smoothness).
More informationImplicit Functions, Curves and Surfaces
Chapter 11 Implicit Functions, Curves and Surfaces 11.1 Implicit Function Theorem Motivation. In many problems, objects or quantities of interest can only be described indirectly or implicitly. It is then
More informationChapter 4. Inverse Function Theorem. 4.1 The Inverse Function Theorem
Chapter 4 Inverse Function Theorem d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d dd d d d d This chapter
More informationSuggested Solution to Assignment 6
MATH 4 (6-7) partial diferential equations Suggested Solution to Assignment 6 Exercise 6.. Note that in the spherical coordinates (r, θ, φ), Thus, Let u = v/r, we get 3 = r + r r + r sin θ θ sin θ θ +
More informationGeorgia Tech PHYS 6124 Mathematical Methods of Physics I
Georgia Tech PHYS 612 Mathematical Methods of Physics I Instructor: Predrag Cvitanović Fall semester 2012 Homework Set #5 due October 2, 2012 == show all your work for maximum credit, == put labels, title,
More information1. The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions.
Complex Analysis Qualifying Examination 1 The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions 2 ANALYTIC FUNCTIONS:
More informationAP Calculus 2004 AB FRQ Solutions
AP Calculus 4 AB FRQ Solutions Louis A. Talman, Ph. D. Emeritus Professor of Mathematics Metropolitan State University of Denver July, 7 Problem. Part a The function F (t) = 8 + 4 sin(t/) gives the rate,
More informationMath 31CH - Spring Final Exam
Math 3H - Spring 24 - Final Exam Problem. The parabolic cylinder y = x 2 (aligned along the z-axis) is cut by the planes y =, z = and z = y. Find the volume of the solid thus obtained. Solution:We calculate
More informationChain Rule. MATH 311, Calculus III. J. Robert Buchanan. Spring Department of Mathematics
3.33pt Chain Rule MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Spring 2019 Single Variable Chain Rule Suppose y = g(x) and z = f (y) then dz dx = d (f (g(x))) dx = f (g(x))g (x)
More informationMATH 124B: INTRODUCTION TO PDES AND FOURIER SERIES
MATH 24B: INTROUCTION TO PES AN FOURIER SERIES SHO SETO Contents. Fourier Series.. Even, Odd, Period, and Complex functions 5.2. Orthogonality and General Fourier Series 6.3. Convergence and completeness
More informationJim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt
Jim Lambers MAT 28 ummer emester 212-1 Practice Final Exam olution 1. Evaluate the line integral xy dx + e y dy + xz dz, where is given by r(t) t 4, t 2, t, t 1. olution From r (t) 4t, 2t, t 2, we obtain
More informationJUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 10 (Second moments of an arc) A.J.Hobson
JUST THE MATHS UNIT NUMBER 13.1 INTEGRATION APPLICATIONS 1 (Second moments of an arc) by A.J.Hobson 13.1.1 Introduction 13.1. The second moment of an arc about the y-axis 13.1.3 The second moment of an
More informationThe first order quasi-linear PDEs
Chapter 2 The first order quasi-linear PDEs The first order quasi-linear PDEs have the following general form: F (x, u, Du) = 0, (2.1) where x = (x 1, x 2,, x 3 ) R n, u = u(x), Du is the gradient of u.
More informationPartial Differential Equations (TATA27)
Partial Differential Equations (TATA7) David Rule Spring 9 Contents Preliminaries. Notation.............................................. Differential equations...................................... 3
More informationNumerical Solutions to Partial Differential Equations
Numerical Solutions to Partial Differential Equations Zhiping Li LMAM and School of Mathematical Sciences Peking University Numerical Methods for Partial Differential Equations Finite Difference Methods
More informationCalculus of Variations and Computer Vision
Calculus of Variations and Computer Vision Sharat Chandran Page 1 of 23 Computer Science & Engineering Department, Indian Institute of Technology, Bombay. http://www.cse.iitb.ernet.in/ sharat January 8,
More informationMATH 411 NOTES (UNDER CONSTRUCTION)
MATH 411 NOTES (NDE CONSTCTION 1. Notes on compact sets. This is similar to ideas you learned in Math 410, except open sets had not yet been defined. Definition 1.1. K n is compact if for every covering
More informationMath Homework 2
Math 73 Homework Due: September 8, 6 Suppose that f is holomorphic in a region Ω, ie an open connected set Prove that in any of the following cases (a) R(f) is constant; (b) I(f) is constant; (c) f is
More informationHouston Journal of Mathematics. c 2015 University of Houston Volume 41, No. 4, 2015
Houston Journal of Mathematics c 25 University of Houston Volume 4, No. 4, 25 AN INCREASING FUNCTION WITH INFINITELY CHANGING CONVEXITY TONG TANG, YIFEI PAN, AND MEI WANG Communicated by Min Ru Abstract.
More informationLECTURE # 0 BASIC NOTATIONS AND CONCEPTS IN THE THEORY OF PARTIAL DIFFERENTIAL EQUATIONS (PDES)
LECTURE # 0 BASIC NOTATIONS AND CONCEPTS IN THE THEORY OF PARTIAL DIFFERENTIAL EQUATIONS (PDES) RAYTCHO LAZAROV 1 Notations and Basic Functional Spaces Scalar function in R d, d 1 will be denoted by u,
More informationNORTHEASTERN UNIVERSITY Department of Mathematics
NORTHEASTERN UNIVERSITY Department of Mathematics MATH 1342 (Calculus 2 for Engineering and Science) Final Exam Spring 2010 Do not write in these boxes: pg1 pg2 pg3 pg4 pg5 pg6 pg7 pg8 Total (100 points)
More information2.2 Separable Equations
2.2 Separable Equations Definition A first-order differential equation that can be written in the form Is said to be separable. Note: the variables of a separable equation can be written as Examples Solve
More informationConformal maps. Lent 2019 COMPLEX METHODS G. Taylor. A star means optional and not necessarily harder.
Lent 29 COMPLEX METHODS G. Taylor A star means optional and not necessarily harder. Conformal maps. (i) Let f(z) = az + b, with ad bc. Where in C is f conformal? cz + d (ii) Let f(z) = z +. What are the
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More information1. The accumulated net change function or area-so-far function
Name: Section: Names of collaborators: Main Points: 1. The accumulated net change function ( area-so-far function) 2. Connection to antiderivative functions: the Fundamental Theorem of Calculus 3. Evaluating
More informationMinimal Surfaces: Nonparametric Theory. Andrejs Treibergs. January, 2016
USAC Colloquium Minimal Surfaces: Nonparametric Theory Andrejs Treibergs University of Utah January, 2016 2. USAC Lecture: Minimal Surfaces The URL for these Beamer Slides: Minimal Surfaces: Nonparametric
More informationAdvanced Calculus Math 127B, Winter 2005 Solutions: Final. nx2 1 + n 2 x, g n(x) = n2 x
. Define f n, g n : [, ] R by f n (x) = Advanced Calculus Math 27B, Winter 25 Solutions: Final nx2 + n 2 x, g n(x) = n2 x 2 + n 2 x. 2 Show that the sequences (f n ), (g n ) converge pointwise on [, ],
More informationMathematical Analysis II, 2018/19 First semester
Mathematical Analysis II, 208/9 First semester Yoh Tanimoto Dipartimento di Matematica, Università di Roma Tor Vergata Via della Ricerca Scientifica, I-0033 Roma, Italy email: hoyt@mat.uniroma2.it We basically
More informationTEST CODE: MMA (Objective type) 2015 SYLLABUS
TEST CODE: MMA (Objective type) 2015 SYLLABUS Analytical Reasoning Algebra Arithmetic, geometric and harmonic progression. Continued fractions. Elementary combinatorics: Permutations and combinations,
More informationLaplace s Equation. Chapter Mean Value Formulas
Chapter 1 Laplace s Equation Let be an open set in R n. A function u C 2 () is called harmonic in if it satisfies Laplace s equation n (1.1) u := D ii u = 0 in. i=1 A function u C 2 () is called subharmonic
More informationMcGill University April 20, Advanced Calculus for Engineers
McGill University April 0, 016 Faculty of Science Final examination Advanced Calculus for Engineers Math 64 April 0, 016 Time: PM-5PM Examiner: Prof. R. Choksi Associate Examiner: Prof. A. Hundemer Student
More informationu(0) = u 0, u(1) = u 1. To prove what we want we introduce a new function, where c = sup x [0,1] a(x) and ɛ 0:
6. Maximum Principles Goal: gives properties of a solution of a PDE without solving it. For the two-point boundary problem we shall show that the extreme values of the solution are attained on the boundary.
More informationPROBLEM SET 6. E [w] = 1 2 D. is the smallest for w = u, where u is the solution of the Neumann problem. u = 0 in D u = h (x) on D,
PROBLEM SET 6 UE ATE: - APR 25 Chap 7, 9. Questions are either directl from the text or a small variation of a problem in the text. Collaboration is oka, but final submission must be written individuall.
More informationMath 46, Applied Math (Spring 2009): Final
Math 46, Applied Math (Spring 2009): Final 3 hours, 80 points total, 9 questions worth varying numbers of points 1. [8 points] Find an approximate solution to the following initial-value problem which
More information