University:
Whitman College
Course:
MATH-367 | Introduction to Partial Differential Equations
Academic year:

2022
Views:

240

Pages:

2
Author:

sizwakalev84

Extra Practice: Solve the Wave Equation, one free, one fixed NOTE: It would be better to use d’Alembert’s solution (which is covered later). This is only for practice manipulating these series when separating variables. The PDE we consider is: utt = c2 uxx 0 0 X(x) = C1 cos( λ x) + C2 sin( λ x) Finishing up this last solution in X, T , we see that for X, X(0) = 0 ⇒ C1 = 0 For the other end, for n = 1, 2, 3, · · ·, X 0 (L) = 0 ⇒ √ √ λC2 cos( λ L) = 0 ⇒ √ λL = (2n − 1)π 2 Therefore, the eigenvalues and eigenfunctions are: λn = (2n − 1)π 2L !2 , (2n − 1)π Xn (x) = sin x 2L For time, we have: (2n − 1)πc T + 2L 00 1 !2 T =0 ! so that ! ! (2n − 1)πc (2n − 1)πc T (t) = An cos t + Bn sin t 2L 2L The general solution is: ∞ X (2n − 1)π x u(x, t) = sin 2L n=1 ! ! (2n − 1)πc (2n − 1)πc An cos t + Bn sin t 2L 2L To solve for the coefficients, we go back to the initial conditions: ∞ X ! (2n − 1)π u(x, 0) = An sin x = f (x) 2L n=1 Therefore, these coefficients are the Fourier cosine coefficients for f (x): ! 2ZL (2n − 1)π An = f (x) sin x dx, for n = 1, 2, 3, · · · L 0 2L Similarly, ∞ X ! (2n − 1)πc (2n − 1)π Bn ut (x, 0) = sin x = g(x) L 2L n=1 Therefore, ! 2ZL (2n − 1)π (2n − 1)πc = g(x) sin x dx, for n = 1, 2, 3, · · · Bn 2L L 0 2L 2 !!

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