Math Fall 2006 Sample problems for the final exam: Solutions

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1 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 on, : fx = c n e inx = fx 2 dx = c n 2. Fourier sine nd cosine trnsforms of the second derivtive: S[f ]ω = 2 f ω ω2 S[f]ω, C[f ]ω = 2 f ω 2 C[f]ω. Lplce s opertor in polr coordintes r, θ: 2 u = 2 u r 2 + r r + 2 u r 2 θ. 2 Any nonzero solution of regulr Sturm-Liouville eqution stisfies the Ryleigh quotient reltion pφ + qφ + λσφ = < x < b λ = pφφ b + b b pφ 2 qφ 2 dx. φ 2 σ dx Some tble integrls: x x 2 e ix 2 dx = i + 2x 2 e ix + C, ; 2 i 3 e αx2 e iβx dx = e α x e iβx dx = α e β2 /4α, α >, β R; 2α, α >, β R. α 2 + β2

2 Problem Let fx = x 2. i Find the Fourier series complex form of fx on the intervl,. The required series is c n e inx, c n fxe inx dx. In prticulr, c fx dx x 2 dx x 3 3 x= 3 3 = 2 3. If n then we hve to integrte by prts twice: c n x 2 e inx dx x 2 e inx dx in 2xe inx dx in = e in + e in + 2xe inx in 2 dx 2e inx in 2 x 2 e inx in 2xe inx in 2 2 n dx = + + 2xe inx dx in + 2e inx in 3 To sve time, we could insted use the tble integrl x x 2 e ix 2 dx = i + 2x 2 2 i 3 e ix + C,. 2e inx in 2 = 2 n. dx According to this integrl, c n x 2 e inx dx x2 in + 2x + 2 in 3 e inx e in + e in = 2 n. Thus x <n< n 2 n e inx. ii Rewrite the Fourier series of fx in the rel form <n< n 2 n e inx = n e inx + e inx = n cos nx. Thus x n cos nx. iii Sketch the function to which the Fourier series converges. 2

3 The series converges to the -periodic function tht coincides with fx for x. The sum is continuous nd piecewise smooth hence the convergence is uniform. The derivtive of the sum hs jump discontinuities t points + 2k, k Z. The grph is sclloped curve. iv Use Prsevl s equlity to evlute n 4. In our cse, Prsevl s equlity cn be written s f, f = g, h = f, φ n 2 φ n, φ n, gxhx dx nd φ n x = e inx. Since c n = f,φn φ nd φ n,φ n n, φ n = for ll n Z, it cn be reduced to n equivlent form fx 2 dx = c n 2. Now Therefore It follows tht fx 2 dx = x 4 dx = x5 5 c n 2 = n = n 4. n = = 5 5, Problem 2 Solve Lplce s eqution in disk, 2 u = r <, u, θ = fθ. Lplce s opertor in polr coordintes r, θ: 2 u = 2 u r 2 + r r + 2 u r 2 θ 2. We serch for the solution of the boundry vlue problem s superposition of solutions ur, θ = hrφθ < r <, < θ < with seprted vribles of Lplce s eqution in the disk. Solutions with seprted vribles stisfy periodic boundry conditions nd the singulr boundry condition ur, = ur,, u, θ <. r, = r, θ θ 3

4 Substituting ur, θ = hrφθ into Lplce s eqution, we obtin h rφθ + r h rφθ + r 2 hrφ θ =, r 2 h r + rh r hr = φ θ φθ. Since the left-hnd side does not depend on θ while the right-hnd side does not depend on r, it follows tht r 2 h r + rh r = φ θ hr φθ = λ, λ is constnt. Then r 2 h r + rh r = λhr, φ = λφ. Conversely, if functions h nd φ re solutions of the bove ODEs for the sme vlue of λ, then ur, θ = hrφθ is solution of Lplce s eqution in polr coordintes. Substituting ur, θ = hrφθ into the periodic nd singulr boundry conditions, we get hrφ = hrφ, hrφ = hrφ, hφθ <. It is no loss to ssume tht neither h nor φ is identiclly zero. Then the boundry conditions re stisfied if nd only if φ = φ, φ = φ, h <. To determine φ, we hve n eigenvlue problem φ = λφ, φ = φ, φ = φ. This problem hs eigenvlues λ n =, n =,, 2,.... The eigenvlue λ = is simple, the others re of multiplicity 2. The corresponding eigenfunctions re φ, φ n θ = cos nθ nd ψ n θ = sin nθ for n. The function h is to be determined from the eqution r 2 h + rh = λh nd the boundry condition h <. We my ssume tht λ is one of the bove eigenvlues so tht λ. If λ > then the generl solution of the eqution is hr = c r µ + c 2 r µ, µ = λ nd c, c 2 re constnts. In the cse λ =, the generl solution is hr = c + c 2 log r, c, c 2 re constnts. In either cse, the boundry condition h < holds if c 2 =. Thus we obtin the following solutions of Lplce s eqution in the disk: u, u n r, θ = r n cos nθ, ũ n r, θ = r n sin nθ, n, 2,... A superposition of these solutions is series ur, θ = α + rn α n cos nθ + β n sin nθ, α, α,... nd β, β 2,... re constnts. Substituting the series into the boundry condition u, θ = fθ, we get fθ = α + n α n cos nθ + β n sin nθ. The right-hnd side is Fourier series on the intervl,. Therefore the boundry condition is stisfied if the right-hnd side coincides with the Fourier series A + of the function fθ on,. Hence A n cos nθ + B n sin nθ α = A, α n = n A n, β n = n B n, n, 2,... 4

5 nd A fθ dθ, A n ur, θ = A + r nan cos nθ + B n sin nθ, fθ cos nθ dθ, B n fθ sin nθ dθ, n, 2,... Problem 3 Find Green s function for the boundry vlue problem d 2 u dx 2 u = fx < x <, u = u =. The Green function Gx, x should stisfy 2 G x 2 G = δx x, Since 2 G x 2 G = for x < x nd x > x, it follows tht Gx, x = G x, x = G x, x =. { e x + be x for x < x, ce x + de x for x > x, constnts, b, c, d my depend on x. Then { G e x x x, x be x for x < x, = ce x de x for x > x. The boundry conditions imply tht = b nd ce = de. Now impose the gluing conditions t x = x, tht is, continuity of the function nd jump discontinuity of the first derivtive: The two conditions imply tht Since b = nd d = ce 2, we get Gx, x x=x = Gx, x x=x +, G x x=x + G x x=x. e x + be x = ce x + de x, ce x de x e x be x. e x = ce x + e 2 x, ce x e 2 x e x e x. Then Therefore e x = ce x e 2 x e x e x e x e x = ce x e 2 x e x ce x + e 2 x e x e x = 2c e 2. c = ex 2 e 2, = c ex + e 2 x e x = ex + e 2 x 2 e 2, d = ce 2 = ex 2e 2, b = = ex + e 2 x 2 e 2. 5

6 Finlly, Gx, x = Observe tht Gx, x = Gx, x. e x + e 2 x e x + e x 2 e 2 e x e x + e 2 x 2 e 2 for x < x, for x > x. Problem 4 Solve the initil-boundry vlue problem for the het eqution, t = 2 u < x <, t >, x 2 ux, = fx < x <, u, t =,, t + 2u, t =. x In the process you will discover sequence of eigenfunctions nd eigenvlues, which you should nme φ n x nd λ n. Describe the λ n qulittively e.g., find n eqution for them but do not expect to find their exct numericl vlues. Also, do not bother to evlute normliztion integrls for φ n. We serch for the solution of the initil-boundry vlue problem s superposition of solutions ux, t = φxgt with seprted vribles of the het eqution tht stisfy the boundry conditions. Substituting ux, t = φxgt into the het eqution, we obtin φxg t = φ xgt, g t gt = φ x φx. Since the left-hnd side does not depend on x while the right-hnd side does not depend on t, it follows tht g t gt = φ x φx = λ, λ is constnt. Then g = λg, φ = λφ. Conversely, if functions g nd φ re solutions of the bove ODEs for the sme vlue of λ, then ux, t = φxgt is solution of the het eqution. Substituting ux, t = φxgt into the boundry conditions, we get φgt =, φ gt + 2φgt =. It is no loss to ssume tht g is not identiclly zero. Then the boundry conditions re stisfied if nd only if φ =, φ + 2φ =. To determine φ, we hve n eigenvlue problem φ = λφ, φ =, φ + 2φ =. This is regulr Sturm-Liouville eigenvlue problem. If φ is n eigenfunction corresponding to n eigenvlue λ, then the Ryleigh quotient reltion holds: φφ λ = + φ x 2 dx. φx 2 dx 6

7 Note tht φφ = φφ φφ = 2 φ 2. It follows tht λ. Moreover, λ > since constnts re not eigenfunctions. Hence ll eigenvlues re positive. For ny λ > the generl solution of the eqution φ = λφ is φx = c cos λ x + c 2 sin λ x, c, c 2 re constnts. The boundry condition φ = holds if c =. Then the condition φ + 2φ = holds if A nonzero solution exists if c 2 λ cos λ + 2 sin λ =. λ cos λ + 2 sin λ = = λ = tn λ. 2 It follows tht the eigenvlues < λ < λ 2 <... re solutions of the eqution 2 λ = tn λ, nd the corresponding eigenfunctions re φ n x = sin λ n x. The function g is to be determined from the eqution g = λg. The generl solution is gt = c e λt, c is constnt. Thus we obtin the following solutions of the het eqution tht stisfy the boundry conditions: A superposition of these solutions is series u n x, t = e λnt φ n x = e λnt sin λ n x, n, 2,... ux, t = c ne λnt φ n x = c ne λnt sin λ n x, c, c 2,... re constnts. Substituting the series into the initil condition ux, = fx, we get fx = c nφ n x. The right-hnd side is generlized Fourier series. Therefore the initil condition is stisfied if the right-hnd side coincides with the generlized Fourier series of the function f, tht is, if c n = fx φ n x dx φ n x 2 dx, n, 2,... Problem 5 By the method of your choice, solve the wve eqution on the hlf-line subject to 2 u t 2 u, t =, = 2 u x 2 < x <, < t < ux, = fx, x, = gx. t Fourier s method: In view of the boundry condition, let us pply the Fourier sine trnsform with respect to x to both sides of the eqution: [ 2 ] [ u 2 ] u S t 2 = S x 2. 7

8 Let Uω, t denote the Fourier sine trnsform of the solution ux, t: Then S [ 2 ] u t 2 = 2 U t 2, Uω, t = S[u, t]ω = 2 S ux, t sin ωx dx. [ 2 ] u x 2 = 2 u, tω ω2 Uω, t = ω 2 Uω, t. Hence 2 U t 2 = ω2 Uω, t. If ω then the generl solution of the ltter eqution is Uω, t = cos ωt+b sin ωt, = ω, b = bω. Applying the Fourier sine trnsform to the initil conditions, we obtin Uω, = F ω, U ω, = Gω, t F = S[f], G = S[g]. It follows tht ω = F ω, bω = Gω/ω. Now it remins to pply the inverse Fourier sine trnsform: ux, t = S [U, t]x = F ω cos ωt + Gω sin ωt sin ωx dω, ω F ω = 2 fx sin ωx dx, Gω = 2 gx sin ωx dx. D Alembert s method: Define fx nd gx for negtive x to be the odd extensions of the functions given for positive x, i.e., f x = fx nd g x = gx for ll x >. By d Alembert s formul, the function ux, t 2 fx + t + fx t + 2 x+t is the solution of the wve eqution tht stisfies the initil conditions ux, = fx, x t x, = gx t gx dx on the entire line. Since f nd g re odd functions, it follows tht ux, t is lso odd s function of x. As consequence, u, t = for ll t. Thus the boundry condition holds s well. Bonus Problem 6 Solve Problem 5 by distinctly different method. See bove. Bonus Problem 7 Find Green function implementing the solution of Problem 2. The solution of Problem 2: ur, θ = A + r nan cos nθ + B n sin nθ, 8

9 A fθ dθ, A n It cn be rewritten s fθ cos nθ dθ, B n ur, θ = Gr, θ; θ + Gr, θ; θ fθ dθ, r ncos nθ cos nθ + sin nθ sin nθ is the desired Green function. The expression cn be simplified: Gr, θ; θ + + n= r n cos nθ θ r n e inθ θ + e inθ θ r e iθ θ n + 2 fθ sin nθ dθ, n, 2,... r e iθ θ n r e iθ θ + r e iθ θ r e iθ θ re iθ θ + re iθ θ re iθ θ 2 r 2 re iθ θ re iθ θ 2 r 2 2 2r cosθ θ + r 2. 9