N1 3 v by the initial conditions u x 0 f x and ut

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Unformatted text preview: (t) + λc2 T (t) = 0 II. From the SLP analysis, Xn (x) = βn sin (nx) and λn = n2 , where n = 1, 2, . . . III. Solving for the T (T ), T (t) = an cos (nct) + bn sin (nct) . IV. The solution becomes, ∞ u (x, t) = sin (nx) [An cos (nct) + Bn sin (nct)] . n=1 3 V. By the initial conditions u (x, 0) = f (x) and ut (x, 0) = g (x), ∞ An sin (nx) = πx − x2 ; u (x, 0) = n=1 ∞ ut (x, 0) = ncBn sin (nx) = 0. n=1 VI. Find An and Bn , 2π π x − x2 sin (nx) dx π0 2 −1 −1 = cos (nx) − (π − 2x) π x − x2 π n n2 2 −2 4 2 = cos (nπ ) + 3 = 3 [1 − (−1)n ] 3 πn n nπ 8 , if n is odd, = n3 π ; 0, if n is even. π 2 Bn = (0) sin (nx) dx = 0. nπc 0 An = sin (nx) + (−2) 1 n3 π cos (nx) Therefore, the solution is ∞ u (x, t) = n=1 n is odd 8 sin (nx) cos (nct) . n3 π Laplace’s Equation 1. I. Let the solution be u (x, y ) = X (x) Y (y ), then uxx = ∂2 ∂2 [X (x) Y (y )] = X ′′ (x) Y (y ) and uyy = 2 [X (x) Y (y )] = X (x) Y ′′ (y ) . ∂x2 ∂y The PDE becomes uxx + uyy = 0 =⇒ X ′′ (x) Y (y ) + X (x) Y ′′ (y ) = 0 =⇒ Y ′′ (y ) X ′′ (x) =− = −λ, X (x) Y (y ) where λ is a constant. Form the boundary conditions, u (0, y ) =X (0) Y (y ) = 0 =⇒ X (0) = 0 u (a, y ) =X (a) Y (y ) = 0 =⇒ X (a) = 0 since we do not want Y (y ) = 0. Then we have two differential equations: X ′′ (x) + λX (x) = 0, X (0) = X (a) = 0 · · · SLP Y ′′ (y ) − λY (y ) = 0 4 0 II. From the SLP analysis that we have done before, nπ x a Xn (x) = β...
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This document was uploaded on 01/23/2014.

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