Download Solutions to Various Partial Differential Equations - Prof. L. Norris and more Exams Differential Equations in PDF only on Docsity! MA 401.001 Fall 2005 Test #3 LK Norris 1. (10 points) Solve Poisson’s equation ∂ 2u ∂x2 + ∂ 2u ∂y2 = f(x, y) on the rectangle 0 < x < a and 0 < y < b with zero boundary data, i.e. u(0, y) = 0 = u(a, y) , 0 < y < b and u(x, 0) = 0 = u(x, b) , 0 < x < a Solution: As discussed in class and in the textbook we assume a solution of the form φ(x, y) = ∑∞ n=1 ∑∞ m=1 Enm sin( nπx z ) sin( mπy b ) so that the zero boundary conditions are satis- fied automatically. To find the coefficients Enm we insert this into the ODE and find f(x, y) = ∞∑ n=1 ∞∑ m=1 (−n 2 a2 − m 2 b2 )π2Enm sin( nπx a ) sin( mπy b ) We recognize this as a double Fourier series for the function f(x, y), so the coefficients are determined by Enm = 4 ((−n2 a2 − m2 b2 )abπ2) ∫ a 0 ∫ b 0 f(x, y) sin( nπx a ) sin( mπy b )dy dx 2. (25 points) The general solution of the ODE y′′ + ω2y = 0 is y = A sin (ωx) + B cos(ωx) Use the power series method to derive this general solution. [HINT: The Taylor series for the basic solutions are sin (ωx) = ∑∞ n=0 (−1)n(ωx)2n+1 (2n+1)! and cos(ωx) = ∑∞ n=0 (−1)n(ωx)2n (2n)! Solution: We assume a solution of the form y = ∑∞ n=0 anx n so that y′ = ∑∞ n=1 nanx n−1 and y′′ = ∑∞ n=2 n(n− 1)anxn−2. Substituting into the ODE yields ∞∑ n=2 n(n− 1)anxn−2 + ∞∑ n=0 ω2anx n = 0 Shifting indices in the first summation (m=n-2 =⇒ n=m+2) we can rewrite this as ∞∑ n=0 (n + 2)(n + 1)an+2xn + ∞∑ n=0 ω2anx n = 0 Combining the two summations (no terms need to be extracted) and setting the coefficients of the various powers of x equal to zero we find (n + 2)(n + 1)an+2xn + ω2an = 0 =⇒ an+2 = − ω2an (n + 2)(n + 1) , n >= 0. We compute a few terms to see the pattern: a2 = − ω2 2 a0 , a4 = − ω2 4 · 3 a2 = (−1)2 ω4 4! a0 , a6 = − ω2 6 · 5 a4 = (−1)3 ω6 6! a0 Hence the even indexed terms have the general form a2n = (−1)n ω 2n (2n)!a0 Similarly for the odd indexed terms we find: a3 = − ω2 3 · 2 a1 , a5 = − ω2 5 · 4 a3 = (−1)2 ω4 5! a1 , a7 = − ω2 7 · 6 a5 = (−1)3 ω6 7! a1 Since a1 is arbitrary I will replace it with ωā1 so that the powers of ω and the factorial in the denominator are the same. Thus for the odd indexed coefficients we find the general term Hence the even indexed terms have the general form a2n+1 = (−1)n ω 2n+1 (2n+1)! ā1 Hence for the solution we find y = ∞∑ n=0 anx n = ∞∑ n=0 even anx n + ∞∑ n=0 n odd anx n = ∞∑ n=0 a2nx 2n + ∞∑ n=0 a2n+1x 2n+1 = a0 ∞∑ n=0 (−1)n ω 2n (2n)! x2n + ā1 ∞∑ n=0 (−1)n ω 2n+1 (2n + 1)! x2n+1 = a0 cos(ωx) + ā1 sin(ωx) 3. (25 points) The Legendre polynomials Pn(x) are the bounded solutions on the interval [−1, 1] of the equation (1 − x2)y′′ − 2xy′ + n(n + 1)y = 0. The first few are P0(x) = 1, P1(x) = x, P2(x) = 12(3x 2 − 1), P3(x) = 12(5x 3 − 3x) and P4(x) = 18(35x 4 − 30x2 + 3). (a) The Legendre polynomials are known to be orthogonal with weight function 1 on the interval [−1, 1]. Show by direct integration that P2(x) and P4(x) are orthogonal with weight function 1 on the interval [−1, 1]. Solution:∫ 1 −1 P2(x)P4(x)dx = ∫ 1 −1 1 2 (3x2 − 1)18(35x4 − 30x2 + 3)dx = 19 ∫ 1 −1 (105x6 − 90x4 + 9x2 − 35x4 + 30x2 − 3)dx = 38 ∫ 1 0 (105x6 − 90x4 − 35x4 + 39x2 − 3)dx = 38 ( 105 7 − 90 5 − 35 5 + 39 3 − 3 ) = 38 (15− 18− 7 + 13− 3) = 0 (b) Use the formula ∫ 1 −1(Pn(x)) 2dx = 22n+1 to derive the formula for the coefficients cn in the Legendre expansion f(x) = ∞∑ n=0 cnPn(x)
