0 linearizing eulers equations near this explicit

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Unformatted text preview: = X (a) = 0, Y (0) = Y (b) = 0. The only nontrivial solutions are with µ = −(mπ/a)2 , X (x) = sin(mπx/a), m = 1, 2, 3, . . . , and with ν = −(nπ/b)2 , Y (y ) = sin(nπy/b), n = 1, 2, 3, . . . . The corresponding solutions of the Helmholtz equation are fmn (x, y ) = sin(mπx/a) sin(nπy/b), with λmn = −(mπ/a)2 − (nπ/b)2 . For any given choice of m and n, the corresponding solution to (5.9) is g (t) = e−c 2 λmn t = e−c 2 ((mπ/a)2 +(nπ/b)2 )t . Hence for each choice of m and n, we obtain a product solution to the heat equation with Dirichlet boundary condition: um,n (x, y, t) = sin(mπx/a) sin(nπy/b)e−c 2 125 ((mπ/a)2 +(nπ/b)2 )t . The general solution to the heat equation with Dirichlet boundary conditions is an arbitrary superposition of these product solutions, ∞ bmn sin(mπx/a) sin(nπy/b)e−c 2 u(x, y, t) = ((mπ/a)2 +(nπ/b)2 )t . (5.11) m,n=1 To find the constants bmn appearing in (5.11), we need to apply the initial condition u(x, y, 0) = h(x, y ). The result is ∞ h(x, y ) = bmn sin(mπx/a) sin(nπy/b), (5.12) m,n=1 expressing the fact that the bmn ’s are the coefficients of what is called the d...
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This document was uploaded on 01/12/2014.

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