TMA4120 Matematikk 4K Eksamen sommer 2016

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1 Norges teknisk naturvitenskapelige universitet Institutt for matematiske fag TMA40 Matematikk 4K Eksamen sommer 06 Løsningsforslag We apply the Laplace transform to the differential equation y + y = sint we get the equation s Y sf0 f 0 + Y = s +, where Y = Ly. Since f 0 = f0 = 0 we have that isolating the Y we get s Y + Y = s +, Y = s + s +. Now using fraction reduction we have that Y = s + s +, hence applying the inverse of the Laplace transform we get L Y = yt = L s + L s + = sint sin t. The Fourier series of fx = cosht on the interval [ π, π] is given by gx = π sinhπ n + + n cosnx. Since lim gx = lim gx = coshπ we have that x π + x π coshπ = π sinhπ n + + n cosnπ = π sinhπ n + + n n π sinhπ n + + n = π sinhπ + + n. Therefore, π coshπ sinhπ = + + n, + n = π cothπ. 9. august 06 Side av 6

2 3 a Stard separation of variables gives F + kf = 0, G kg = 0 for some constant k R. Three cases to be considered: i k = λ < 0. Using the boundary condition 0 = F 0 = F π we see that F is identically ero. ii k = 0. Using the boundary condition 0 = F 0 = F π, we see that F constant is the only solution. iii k = λ > 0. Using the boundary condition 0 = F 0 = F π, we see in the stard manner that λ = n Z F x = β n cosnx, n Z, for any constant β n R. Since cosinus is an even function it suffices to consider nonnegative integers. Using this result for the equation for G, we infer that Gy = A n e ny + B n e ny, n N, for any constants A n, B n R. For n = 0 we find Gy = A 0 y + B 0 for constants A 0, B 0 R. Thus the general solution of the form u = F G reads { A 0 y + B 0, for n = 0, ux, y = F xgy = u n x, y = cosnx A n e ny + B n e ny, for n N. b The general solution reads ux, y = u n x, y = A 0 y + B 0 + cosnx A n e ny + B n e ny. n=0 The boundary condition at y = 0 yields ux, 0 = B 0 + cosnx A n + B n = 0. By the uniqueness of Fourier series we conclude that B 0 = 0 A n = B n for n N. The boundary condition at y = π yields u y x, π = A 0 + cosnxna n e nπ/ + e nπ/ = + cosx = + cosx + cos x = 3 + cosx + cosx. Again by the uniqueness of Fourier series we conclude that A 0 = 3/, A e π/ + e π/ =, A e π + e π = /, while all the other constants vanish. We may write A = coshπ/, A = 8 coshπ. 9. august 06 Side av 6

3 Thus ux, y = 3 y + coshπ/ cosx e y e y + 8 coshπ cosx e y e y. 4 We take the Fourier transform Kreysig, p. 57 F f g = πff Fg = π e ω /4 d F dx e x π = /4 e ω iωf e x πi = ω e ω /4 e ω /4 4 πi = ω e ω /. 4 Here we also used the formula for the Fourier transform of the derivative Kreysig, p. 56. Using that hx = π Fhe iωx dω Kreysig, p. 54 we see that f g = π = π = πi π 4 = i 4 Ff ge iωx dω πi ω e ω / e iωx dω 4 ω e ω / e iωx dω ω e ω / e iωx dω 5 For = x + iy C \ {0} we have that f = = x + iy = x + iy = So we have that f = ux, y + ivx, y where ux, y = The Cauchy Riemann equations say that x x + y i y x + y. x x + y vx, y = y x + y. u x = v y u y = v x. 9. august 06 Side 3 av 6

4 But u x = x x x + y = x + y xx x + y = y x x + y, u y = x y x + y = 0 xy x + y = xy x + y, v x = y x x + y = 0 xy x + y = xy x + y, v y = y y x + y = x + y yy x + y = x y x + y = y x x + y. So the Cauchy-Riemann equations are trivially satisfied. Finally using Theorem in page 67 Kreysig, it follows that fx + iy = ux, y + ivx, y is analytic if u v satisfy the Cauchy-Riemann equations have continuous partial derivatives. But since our domain D is C\{0} this is true, therefore f is analytic in C \ {0}. 6 a The function f = + have singularities at = i, i. So we can write f = + = i + i = Whence = i = i are singularities of order. i + i. Observe that both singularities are inside the circle =, hence by the Residue Integration Method it follows that fd = πires =i f + Res = i f. = So we have to compute the residues of the singularities. Since both singularities have order we can use the formula Therefore Res =0 = lim 0 [ 0 f]. Res =i = lim i [ i f] = lim i + i = lim i + i + i 3 = 0. Thus, Res = i = lim [ + i i f] = lim i = i = lim i + i i 3 = 0. fd = πires =i f + Res = i f = πi0 + 0 = august 06 Side 4 av 6

5 b The function f = has only two singularities = i = i. Observe + that the distance between i i is, i.e., i i = i =. Then there are two Laurent series of f with center = i: one with convergence region 0 < i <, the other with convergence region i >. Then since i i = i = < we have that for = i the first Laurent series converges, i.e., the one with convergence region i <. 7 Let fx = x +x 4, we want to compute 0 x + x 4 dx. Let be the closed path that is the upper-half circle with center the origo radius R. Observe that we can decompose the path in two parts: the horiontal line [ R, R] the arc S R. Then R + 4 d = x + x 4 dx + R S R + 4 d. Observe that f = has singularities the solutions of 4 + = 0, that are + 4 w = e i π 4 = + i, w = e i π 4 + π = + i, w 3 = e i π 4 +π = i w 4 = e i π 4 + 3π = i. All this singularities have order. Using the Residue integration method when R > we have that + 4 d = πires =w f + Res =w f. Now we compute the residues. Since all the singularities have order we can apply the formula in Page 7 in Kreysig to get Res =w 4 + = ] ] 4 + ]=w = 4 3 Res =w 4 + = 4 + ]=w = 4 3 Then ] =w = 4 =w = 4 ] =w = 4e i π 4 =w = 4e i 3π 4 d = πi i 4 + i = πi = 4 e i π 4 = 4 i = 4 e i 3π 4 = 4 i i = π. Now using the ML-inequality we have that + 4 d Max SR 4 + Length S R R R 4 πr = π R. S R 9. august 06 Side 5 av 6

6 Then when have that hence lim R lim R S R d = lim + 4 R + 4 d = 0, R R x + x 4 dx + S R + 4 d. Thus, π = x + x 4 dx + 0. Finally, since fx = so x x 4 + is an even function we have that x π = + x 4 dx, 0 0 x + x 4 dx =. 9. august 06 Side 6 av 6

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