PHYSICS 116C Homework 4 Solutions
|
|
- Brice Thomas
- 5 years ago
- Views:
Transcription
1 PHYSICS 116C Homework 4 Solutions 1. ( Simple hrmonic oscilltor. Clerly the eqution is of the Sturm-Liouville (SL form with λ = n 2, A(x = 1, B(x =, w(x = 1. Legendre s eqution. Clerly the eqution is of the SL form with λ = n(n+1, A(x = 1 x 2, B(x =, w(x = 1. Bessel s eqution. This is not obviously in the SL form. To mke it so, multiply by f(x nd choose f(x such tht the coefficient of y is the derivtive of the coefficient of y, i.e. or which hs solution f(x = 1/x nd so xy +y + [c which is of the SL for with d ( x 2 f(x = xf(x dx f (x = f(x x 2nx ν2 x ] y =, λ = c 2 n, A(x = x, B(x = ν 2 /x, w(x = x. Hermite s eqution. Proceeding s for Bessel s eqution we multiply by f(x nd require which hs solution nd so which is of the SL form with df dx = 2xf(x f(x = exp( x 2, e x2 y 2xe x2 y +2ne x2 y =, λ = 2n, A(x = e x2, B(x =, w(x = e x2. Lguerre s eqution. Proceeding s for Bessel s eqution we multiply by f(x nd require or which hs solution f = e x nd so which is of the SL form with d (xf = (1 xf(x dx f = f, xe x y +(1 xe x y +ne x y =, λ = n, A(x = xe x, B(x =, w(x = e x. 1
2 (b The equtions for y 1 nd y 2 re d [ ] A(xy dx 1 +[λ1 w(x+b(x]y 1 = (1 d [ ] A(xy dx 2 +[λ2 w(x+b(x]y 2 =. (2 Multiply Eq. (1 by y 2 nd Eq. (2 by y 1, subtrct, nd integrte from to b. This gives ( d [ ] y 2 A(xy d [ ] b dx 1 y1 A(xy dx 2 dx+(λ 1 λ 2 w(xy 1 (xy 2 (xdx =. (3 The integrnd in the first term cn be written s d [ ] y 2 A(xy d [ ] dx 1 y1 A(xy dx 2 Eq. (3 therefore becomes [ A(x(y 1 y 2 y 2y 1 ] b +(λ 1 λ 2 = d [ A(x(y dx 1 y 2 y 2y 1 ] ( A(xy 1y 2 A(xy 2y 1 = d [ A(x(y dx 1 y 2 y 2y 1 ]. Hence, for λ 1 λ 2 we obtin the orthogonlity condition provided the boundry term vnishes, i.e. w(xy 1 (xy 2 (xdx =. w(xy 1 (xy 2 (xdx =, (4 [ A(x(y 1 y 2 y 2y 1 ] b =. (5 (c Simple hrmonic oscilltor. =,b = 2π. The y n re sinnx nd cosnx which re periodicwithperiod2π. A(x = 1ndso the contributions from the two limits in Eq. (5 cncel. The orthogonlity condition, Eq.(4, is just the fmilir orthogonlity of sines nd cosines over the intervl from to 2π. Legendre s eqution. = 1,b = 1. The y n re Legendre polynomils, P n (x. Eq. (5 is stisfied becuse A(x = 1 x 2 vnishes t x = ±1. The orthogonlity condition, Eq. (4, is 1 1 P n (xp m (xdx = (n m, which ws shown in clss. Bessel s eqution. =,b = 1. The y n re Bessel functions J ν (c n x. In Eq. (5, the contribution t x = vnishes becuse A(x = x nd the contribution t x = 1 vnishes becuse c n is zero of the Bessel function nd hence y(1 =. The orthogonlity condition, Eq. (4, is 1 which ws shown in clss. xj ν (c n xj ν (c m xdx = (n m, 2
3 Hermite s eqution. =,b =. The y n re Hermite polynomils H n (x. Eq. (5 is stisfied becuse A(x = exp( x 2 vnishes for x ±. The orthogonlity condition, Eq. (4, is e x2 H n (xh m (xdx = (n m, which ws stted in clss. Lguerre s eqution. =,b =. The y n re Lguerre polynomils. L n (x. Eq. (5 is stisfied becuse A(x = xe x vnishes t x = nd. The orthogonlity condition, Eq. (4, is which ws stted in clss. e x L n (xl m (xdx = (n m, (d Assuming completeness of the y n we cn expnd function f(x in terms of them s f(x = m m y m (x. (6 Multiplying by y n (x, integrting from to b, nd using the orthogonlity reltion, Eq. (4, gives w(xy n (xf(xdx = n w(xyn(xdx, 2 which gives where n = 1 I n I n = w(xy n (xf(xdx. (7 w(xy 2 n(xdx. (8 (e For the simple hrmonic oscilltor eqution, =,b = 2π, nd the solutions re cosnx nd sinnx. Hence Eq. (7 is just the fmilir Fourier series f(x = m ( m cosmx+b m sinmx. I n in Eq. (8 is or nd hence Eq. (7 corresponds to 2π 2π cos 2 nxdx = π sin 2 nxdx = π, n = 1 π b n = 1 π 2π 2π f(xcosnxdx f(xsinnxdx, 3
4 which re indeed the coefficients in Fourier series. 2. ( Using the divergence theorem FdV = R B F ds with F = u u gives ( u 2 u+ u 2 dv = R B u u ds = B u ds, (9 n where we used (ua = u A+A u, nd / n is the derivtive of u in the direction of the (outwrd norml to the surfce. (b Consider two solutions of Lplce s eqution u 1 nd u 2 which hve the sme vlues on the boundry B. Denote the difference u 1 u 2 by u. This stisfies Lplce s eqution, 2 u =, nd hs the vlue u = on the boundry. Hence inserting these properties into Eq. (9 gives u 2 dv =. (1 R Hence u =, so u is constnt, in the volume V, but since u = on the boundry S we must hve u = everywhere in V. Hence u 1 = u 2 nd so the solution to Lplce s eqution whose vlue is specified on the boundry is unique. (c Repet the rgument of the previous section but with / n = on the surfce. This gin yields u 2 dv =, (11 R showing tht u = so u is constnt. Since we do not know the vlue of u u 1 u 2 t ny point, ll we cn sy is tht the solution is determined up to n rbitrry constnt. 3. ( In one dimension, the diffusion eqution is t = u D 2 x 2. (12 Using we hve u(x,t = 1 exp ( x2 t1/2 4Dt t = 1 2t (13 x2 u+ u, (14 4Dt2 where the first term comes from differentiting 1/t 1/2 nd the second from differentiting the 1/t in the exponentil. We lso hve x = x 2Dt u, 4
5 nd differentiting gin gives x 2 = 1 ( x 2 2Dt u+ u. 2Dt We see tht this is just D 1 times Eq. (14, nd hence the diffusion eqution, Eq. (12, hs solution given by by u(x,t in Eq. (13. (b In sphericl polr coordintes in three dimensions, the diffusion eqution is t = D 1 ( r 2 r 2, (15 r r where we ssumed sphericl symmetry. Tking u(r,t = 1 exp ( r2 t3/2 4Dt (16 we hve We lso hve r = r 2Dt u, nd tking the pproprite second derivtive gives t = 3 r2 u+ u, (17 2t 4Dt2 ( r 2 ( = 3r2 r 2 r r 2Dt u+r2 u. 2Dt Thisisjustr 2 /D timeseq.(17, ndhencethediffusionequtioninsphericlpolrs, Eq.(15, hs solution given by u(r,t in Eq. ( The wve eqution in two dimensions is x u y 2 = 1 v 2 t 2. (18 Tking where k 2 = k 2 x +k 2 y, we hve nd similrly u(x,y,t = sin(k x xsin(k y ysin(kvt, (19 y 2 = k2 yu, x 2 = k2 xu, t 2 = k2 v 2. Hence the wve eqution, Eq. (18 hs solution given by Eq. (19. 5
6 5. We hve to find the solution of Lplce s eqution in the rectngulr region x, y b which stisfies the boundry conditions u(,y = u(,y = u(x, =, nd u(x,b = u sin(πx/. We look for seprble solution of Lplce s eqution, i.e. u(x, y = X(xY(y. Substituting nd dividing by u gives X X + Y Y =. Since the first term is only function of x nd the second term only function of y ech term must be constnt in order for the eqution to be true for ll x nd y. Setting the first term to equl k 2 we hve X +k 2 X =, Y k 2 Y =. A solution which stisfies the boundry conditions u(,y = u(x, = is X(x = sin(kx,y(y = sinh(ky. To stisfy X( = we need k = 2πn/ (n = 1,2,3,, nd so we form the liner combintion. u(x,y = C n sin sinh ( nπy. WedeterminethecoefficientsC n byrequiringthtthelstboundrycondition,u(x,b = u sin(πx/, is stisfied. By inspection we hve C n = (n 1, nd C 1 = u /sinh(πb/, so u(x,y = u sinh( πb sin ( πx sinh ( πy. Note: One could work out ll the Fourier coefficients explicitly nd show tht they re ll zero except for C 1, but this is unnecessry work. If you cn see the solution by inspection, like here, then you cn use this informtion (unless you re expressly told to go through the derivtion. 6. We look for seprble solution of Lplce s eqution, i.e. u(x, y = X(xY(y. Substituting nd dividing by u gives X X + Y Y =. Since the first term is only function of x nd the second term only function of y ech term must be constnt in order for the eqution to be true for ll x nd y. Setting the first term to equl k 2 we hve X +k 2 X =, Y k 2 Y =. A solution which stisfies the boundry conditions u(,y = u(x, = is X(x = sin(kx,y(y = sinh(ky. To stisfy X( = we need k = 2πn/ (n = 1,2,3,, nd so we form the liner combintion. u(x,y = C n sin sinh ( nπy. To stisfy the boundry condition t y = b we need [ ( ] nπb u x( x = C n sinh sin. 6
7 In other words we hve to determine the sine Fourier series for u x( x in the intervl x nd the n-th Fourier coefficient is C n sinh(nπb/, i.e. ( nπb 2 C n sinh = u x( x sin dx. Integrting by prts twice gives ( { nπb 2 [ ] C n sinh = u x( xcos + ( 2xcos nπ = 2u ( nπx ( 2xcos dx nπ = 2u { [ ] ( 2xsin ( 2 sin nπ nπ = 4u ( nπx (nπ 2 sin dx = 4u [ ] (nπ 2 cos = 4u nπ (nπ 2 nπ [1 ( 1n ] (n even, = 8 2 (nπ 3 u (n odd. Hence the solution is (writing n = 2m 1 with m = 1,2,3, u(x,y = u 2 ( 8 π 3 m=1 1 (2m 1 3 } dx } dx ( (2m 1πy sinh ( (2m 1πx ( sin. (2m 1πb sinh 7. We strt with [ ] sin (2n 1πx L T(x,y = 4T e (2n 1πy/L. (2n 1π Since sin[(2n 1πx/L] = Im[exp(i(2n 1πx/L] we cn write where T(x,y = 4T π Im ( z 2n 1, (2 2n 1 [ ] π(ix y z = exp = exp( yπ/l(cos(πx/l+isin(πx/l. L Usingtheseriesforln(1+z = z z 2 /2+z 3 /3, whichlsoimpliesln(1 z = z z 2 /2 z 3 /3, we see tht the series in Eq. (2 is just tht for ln(1+z ln(1 z = ln[(1+z/(1 z]. Hence T(x,y = 2T [ ( ] 1+z π Im ln. (21 1 z 7
8 For complex number w = r iθ we hve Imlnw = Imln(re iθ = Im(lnr + iθ = θ = rg(w = tn 1 (Imw/Rew. Letting z = s+it we hve 1+z 1 z = 1+s+it 1 s it = (1+s+it(1 s+it (1 s it(1 s+it = 1 s2 t 2 +2it (1 s 2 +t 2, nd so Hence Im[(1+z/(1 z] Re[(1+z/(1 z] = 2t 1 s 2 t 2 = 2sin(πx/Le πy/l 1 e 2πy/L = sin(πx/l sinh(πy/l. T(x,y = 2T ( sin(πx/l π tn 1. (22 sinh(πy/l Note: For x, sin(πx/l nd so T(,y =. For x L, sin(πx/l nd so T(L,y =. For y, sinh(πy/l nd so lim T(x,y =. y For y, sinh(πy/l, so tn 1 [sin(πx/l/sinh(πy/l] tn 1 = π/2 nd so T(x, = T. Hence the closed form expression in Eq. (22 stisfies the boundry conditions s required. 8
Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationLecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature
Lecture notes on Vritionl nd Approximte Methods in Applied Mthemtics - A Peirce UBC Lecture 6: Singulr Integrls, Open Qudrture rules, nd Guss Qudrture (Compiled 6 August 7) In this lecture we discuss the
More informationPartial Differential Equations
Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for One-Dimensionl Eqution The reen s function provides complete solution to boundry
More informationChapter 28. Fourier Series An Eigenvalue Problem.
Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationMath 5440 Problem Set 3 Solutions
Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 25 1: Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping
More informationMath Fall 2006 Sample problems for the final exam: Solutions
Mth 42-5 Fll 26 Smple problems for the finl exm: Solutions Any problem my be ltered or replced by different one! Some possibly useful informtion Prsevl s equlity for the complex form of the Fourier series
More informationMath 5440 Problem Set 3 Solutions
Mth 544 Mth 544 Problem Set 3 Solutions Aron Fogelson Fll, 213 1: (Logn, 1.5 # 2) Repet the derivtion for the eqution of motion of vibrting string when, in ddition, the verticl motion is retrded by dmping
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More information10 Elliptic equations
1 Elliptic equtions Sections 7.1, 7.2, 7.3, 7.7.1, An Introduction to Prtil Differentil Equtions, Pinchover nd Ruinstein We consider the two-dimensionl Lplce eqution on the domin D, More generl eqution
More informationMath 473: Practice Problems for the Material after Test 2, Fall 2011
Mth 473: Prctice Problems for the Mteril fter Test, Fll SOLUTION. Consider the following modified het eqution u t = u xx + u x. () Use the relevnt 3 point pproximtion formuls for u xx nd u x to derive
More informationOrthogonal functions
Orthogonl functions Given rel vrible over the intervl (, b nd set of rel or complex functions U n (ξ, n =, 2,..., which re squre integrble nd orthonorml b U n(ξu m (ξdξ = δ n,m ( if the set of of functions
More information1 1D heat and wave equations on a finite interval
1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion
More informationc n φ n (x), 0 < x < L, (1) n=1
SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationPDE Notes. Paul Carnig. January ODE s vs PDE s 1
PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationProblem Set 3 Solutions
Msschusetts Institute of Technology Deprtment of Physics Physics 8.07 Fll 2005 Problem Set 3 Solutions Problem 1: Cylindricl Cpcitor Griffiths Problems 2.39: Let the totl chrge per unit length on the inner
More information1 E3102: a study guide and review, Version 1.0
1 E3102: study guide nd review, Version 1.0 Here is list of subjects tht I think we ve covered in clss (your milege my vry). If you understnd nd cn do the bsic problems in this guide you should be in very
More informationSection 7.1 Integration by Substitution
Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationFunctions of Several Variables
Functions of Severl Vribles Sketching Level Curves Sections Prtil Derivtives on every open set on which f nd the prtils, 2 f y = 2 f y re continuous. Norml Vector x, y, 2 f y, 2 f y n = ± (x 0,y 0) (x
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationElliptic Equations. Laplace equation on bounded domains Circular Domains
Elliptic Equtions Lplce eqution on bounded domins Sections 7.7.2, 7.7.3, An Introduction to Prtil Differentil Equtions, Pinchover nd Rubinstein 1.2 Circulr Domins We study the two-dimensionl Lplce eqution
More informationDifferential Equations 2 Homework 5 Solutions to the Assigned Exercises
Differentil Equtions Homework Solutions to the Assigned Exercises, # 3 Consider the dmped string prolem u tt + 3u t = u xx, < x , u, t = u, t =, t >, ux, = fx, u t x, = gx. In the exm you were supposed
More informationAnti-derivatives/Indefinite Integrals of Basic Functions
Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationReview SOLUTIONS: Exam 2
Review SOUTIONS: Exm. True or Flse? (And give short nswer ( If f(x is piecewise smooth on [, ], we cn find series representtion using either sine or cosine series. SOUTION: TRUE. If we use sine series,
More information3 Mathematics of the Poisson Equation
3 Mthemtics of the Poisson Eqution 3. Green functions nd the Poisson eqution () The Dirichlet Green function stisfies the Poisson eqution with delt-function chrge 2 G D (r, r o ) = δ 3 (r r o ) (3.) nd
More informationLecture 24: Laplace s Equation
Introductory lecture notes on Prtil Differentil Equtions - c Anthony Peirce. Not to e copied, used, or revised without explicit written permission from the copyright owner. 1 Lecture 24: Lplce s Eqution
More informationConsequently, the temperature must be the same at each point in the cross section at x. Let:
HW 2 Comments: L1-3. Derive the het eqution for n inhomogeneous rod where the therml coefficients used in the derivtion of the het eqution for homogeneous rod now become functions of position x in the
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationUnit 5. Integration techniques
18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationDiscrete Least-squares Approximations
Discrete Lest-squres Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve
More information1 The Lagrange interpolation formula
Notes on Qudrture 1 The Lgrnge interpoltion formul We briefly recll the Lgrnge interpoltion formul. The strting point is collection of N + 1 rel points (x 0, y 0 ), (x 1, y 1 ),..., (x N, y N ), with x
More informationWave Equation on a Two Dimensional Rectangle
Wve Eqution on Two Dimensionl Rectngle In these notes we re concerned with ppliction of the method of seprtion of vriles pplied to the wve eqution in two dimensionl rectngle. Thus we consider u tt = c
More informationThe Dirac distribution
A DIRAC DISTRIBUTION A The Dirc distribution A Definition of the Dirc distribution The Dirc distribution δx cn be introduced by three equivlent wys Dirc [] defined it by reltions δx dx, δx if x The distribution
More information) 4n+2 sin[(4n + 2)φ] n=0. a n ρ n sin(nφ + α n ) + b n ρ n sin(nφ + β n ) n=1. n=1. [A k ρ k cos(kφ) + B k ρ k sin(kφ)] (1) 2 + k=1
Physics 505 Fll 2007 Homework Assignment #3 Solutions Textbook problems: Ch. 2: 2.4, 2.5, 2.22, 2.23 2.4 A vrint of the preceeding two-dimensionl problem is long hollow conducting cylinder of rdius b tht
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationu t = k 2 u x 2 (1) a n sin nπx sin 2 L e k(nπ/l) t f(x) = sin nπx f(x) sin nπx dx (6) 2 L f(x 0 ) sin nπx 0 2 L sin nπx 0 nπx
Chpter 9: Green s functions for time-independent problems Introductory emples One-dimensionl het eqution Consider the one-dimensionl het eqution with boundry conditions nd initil condition We lredy know
More informationSuggested Solution to Assignment 5
MATH 4 (5-6) prti diferenti equtions Suggested Soution to Assignment 5 Exercise 5.. () (b) A m = A m = = ( )m+ mπ x sin mπx dx = x mπ cos mπx + + 4( )m 4 m π. 4x cos mπx dx mπ x cos mπxdx = x mπ sin mπx
More informationGreen function and Eigenfunctions
Green function nd Eigenfunctions Let L e regulr Sturm-Liouville opertor on n intervl (, ) together with regulr oundry conditions. We denote y, φ ( n, x ) the eigenvlues nd corresponding normlized eigenfunctions
More informationENGI 9420 Lecture Notes 7 - Fourier Series Page 7.01
ENGI 940 ecture Notes 7 - Fourier Series Pge 7.0 7. Fourier Series nd Fourier Trnsforms Fourier series hve multiple purposes, including the provision of series solutions to some liner prtil differentil
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationName Solutions to Test 3 November 8, 2017
Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier
More informationMath 3B Final Review
Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationJackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero
More informationBest Approximation in the 2-norm
Jim Lmbers MAT 77 Fll Semester 1-11 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the -norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationProblem Set 3 Solutions
Chemistry 36 Dr Jen M Stndrd Problem Set 3 Solutions 1 Verify for the prticle in one-dimensionl box by explicit integrtion tht the wvefunction ψ ( x) π x is normlized To verify tht ψ ( x) is normlized,
More informationMathematics Higher Block 3 Practice Assessment A
Mthemtics Higher Block 3 Prctice Assessment A Red crefully 1. Clcultors my be used. 2. Full credit will be given only where the solution contins pproprite working. 3. Answers obtined from reding from scle
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationAike ikx Bike ikx. = 2k. solving for. A = k iκ
LULEÅ UNIVERSITY OF TECHNOLOGY Division of Physics Solution to written exm in Quntum Physics F0047T Exmintion dte: 06-03-5 The solutions re just suggestions. They my contin severl lterntive routes.. Sme/similr
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationAM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h
AM Mthemticl Anlysis Oct. Feb. Dte: October Exercises Lecture Exercise.. If h, prove the following identities hold for ll x: sin(x + h) sin x h cos(x + h) cos x h = sin γ γ = sin γ γ cos(x + γ) (.) sin(x
More informationMA 201: Partial Differential Equations Lecture - 12
Two dimensionl Lplce Eqution MA 201: Prtil Differentil Equtions Lecture - 12 The Lplce Eqution (the cnonicl elliptic eqution) Two dimensionl Lplce Eqution Two dimensionl Lplce Eqution 2 u = u xx + u yy
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More information1 Techniques of Integration
November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.
More informationPhys 6321 Final Exam - Solutions May 3, 2013
Phys 6321 Finl Exm - Solutions My 3, 2013 You my NOT use ny book or notes other thn tht supplied with this test. You will hve 3 hours to finish. DO YOUR OWN WORK. Express your nswers clerly nd concisely
More informationChapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1
Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4
More informationAdvanced Computational Fluid Dynamics AA215A Lecture 3 Polynomial Interpolation: Numerical Differentiation and Integration.
Advnced Computtionl Fluid Dynmics AA215A Lecture 3 Polynomil Interpoltion: Numericl Differentition nd Integrtion Antony Jmeson Winter Qurter, 2016, Stnford, CA Lst revised on Jnury 7, 2016 Contents 3 Polynomil
More informationFourier Series. with the period 2π, given that sin nx and cos nx are period functions with the period 2π. Then we.
. Definition We c the trigonometric series the series of the form + cos x+ b sin x+ cos x+ b sin x+ or more briefy + ( ncos nx+ bnsin nx) () n The constnts, n nd b, n ( n,, ) re coefficients of the series
More informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More informationThe Product Rule state that if f and g are differentiable functions, then
Chpter 6 Techniques of Integrtion 6. Integrtion by Prts Every differentition rule hs corresponding integrtion rule. For instnce, the Substitution Rule for integrtion corresponds to the Chin Rule for differentition.
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
More informationOrthogonal Polynomials and Least-Squares Approximations to Functions
Chpter Orthogonl Polynomils nd Lest-Squres Approximtions to Functions **4/5/3 ET. Discrete Lest-Squres Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny
More informationn=0 ( 1)n /(n + 1) converges, but not n=100 1/n2, is at most 1/100.
Mth 07H Topics since the second exm Note: The finl exm will cover everything from the first two topics sheets, s well. Absolute convergence nd lternting series A series n converges bsolutely if n converges.
More informationEuler-Maclaurin Summation Formula 1
Jnury 9, Euler-Mclurin Summtion Formul Suppose tht f nd its derivtive re continuous functions on the closed intervl [, b]. Let ψ(x) {x}, where {x} x [x] is the frctionl prt of x. Lemm : If < b nd, b Z,
More informationPhysics 215 Quantum Mechanics 1 Assignment 2
Physics 15 Quntum Mechnics 1 Assignment Logn A. Morrison Jnury, 16 Problem 1 Clculte p nd p on the Gussin wve pcket α whose wve function is x α = 1 ikx x 1/4 d 1 Solution Recll tht where ψx = x ψ. Additionlly,
More informationOrthogonal Polynomials
Mth 4401 Gussin Qudrture Pge 1 Orthogonl Polynomils Orthogonl polynomils rise from series solutions to differentil equtions, lthough they cn be rrived t in vriety of different mnners. Orthogonl polynomils
More informationES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus
ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems
More informationSpace Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More information221B Lecture Notes WKB Method
Clssicl Limit B Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using
More informationIntroduction to Complex Variables Class Notes Instructor: Louis Block
Introuction to omplex Vribles lss Notes Instructor: Louis Block Definition 1. (n remrk) We consier the complex plne consisting of ll z = (x, y) = x + iy, where x n y re rel. We write x = Rez (the rel prt
More informationThe usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1
Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be non-negtive. By introducing
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationSturm-Liouville Eigenvalue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (a, b). Here we assume b > a. Let X C 2 1
Ch.4. INTEGRAL EQUATIONS AND GREEN S FUNCTIONS Ronld B Guenther nd John W Lee, Prtil Differentil Equtions of Mthemticl Physics nd Integrl Equtions. Hildebrnd, Methods of Applied Mthemtics, second edition
More information2. THE HEAT EQUATION (Joseph FOURIER ( ) in 1807; Théorie analytique de la chaleur, 1822).
mpc2w4.tex Week 4. 2.11.2011 2. THE HEAT EQUATION (Joseph FOURIER (1768-1830) in 1807; Théorie nlytique de l chleur, 1822). One dimension. Consider uniform br (of some mteril, sy metl, tht conducts het),
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationk and v = v 1 j + u 3 i + v 2
ORTHOGONAL FUNCTIONS AND FOURIER SERIES Orthogonl functions A function cn e considered to e generliztion of vector. Thus the vector concets like the inner roduct nd orthogonlity of vectors cn e extended
More informationSTURM-LIOUVILLE THEORY, VARIATIONAL APPROACH
STURM-LIOUVILLE THEORY, VARIATIONAL APPROACH XIAO-BIAO LIN. Qudrtic functionl nd the Euler-Jcobi Eqution The purpose of this note is to study the Sturm-Liouville problem. We use the vritionl problem s
More information8 Laplace s Method and Local Limit Theorems
8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved
More informationAPPM 4360/5360 Homework Assignment #7 Solutions Spring 2016
APPM 436/536 Homework Assignment #7 Solutions Spring 6 Problem # ( points: Evlute the following rel integrl by residue integrtion: x 3 sinkx x 4 4, k rel, 4 > Solution: Since the integrnd is even function,
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationHigher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors
Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector
More informationMath 113 Exam 1-Review
Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More informationIf u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du
Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible
More information4181H Problem Set 11 Selected Solutions. Chapter 19. n(log x) n 1 1 x x dx,
48H Problem Set Selected Solutions Chpter 9 # () Tke f(x) = x n, g (x) = e x, nd use integrtion by prts; this gives reduction formul: x n e x dx = x n e x n x n e x dx. (b) Tke f(x) = (log x) n, g (x)
More information(9) P (x)u + Q(x)u + R(x)u =0
STURM-LIOUVILLE THEORY 7 2. Second order liner ordinry differentil equtions 2.1. Recll some sic results. A second order liner ordinry differentil eqution (ODE) hs the form (9) P (x)u + Q(x)u + R(x)u =0
More informationIndefinite Integral. Chapter Integration - reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More information