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hw11_sol - Solutions to Homework 11 9.52 The problem is 1...

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Solutions to Homework 11 May 23, 2011 9.52 The problem is w xx + w yy = 1 c 2 w tt , for 0 < x < L, 0 < y < L w ( x = 0) = w ( x = L ) = w ( y = 0) = w ( y = L ) = 0; w bounded . Looking for separable solutions by substituting w ( x, y, t ) = X ( x ) Y ( y ) T ( t ) , we get X ′′ Y T + XY ′′ T = 1 c 2 XY T ′′ , for 0 < x < l, 0 < y < l X (0) = X ( L ) = Y (0) = Y ( L ) = 0; X, Y, T bounded . or X ′′ X = T ′′ c 2 T - Y ′′ Y = - k 2 T ′′ c 2 T + k 2 = Y ′′ Y = - l 2 giving us Storm-Liouville problems for X and Y , with general solutions X ′′ + k 2 X = 0 X = c 1 sin( kx ) + c 2 cos( kx ) Y ′′ + l 2 Y = 0 Y = d 1 sin( ly ) + d 2 cos( ly ) Using the boundary conditions, we need c 2 = 0 , k m = L , m = 1 , 2 , 3 , . . . d 2 = 0 , l n = L , n = 1 , 2 , 3 , . . . 1
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For T , the general solution is T ′′ + c 2 ( k 2 + l 2 ) T = 0 T = e 1 sin( c radicalbig k 2 + l 2 t ) + e 2 cos( c radicalbig k 2 + l 2 t ) with no boundary conditions applicable. However, the requirements on k and l mean that c k 2 + l 2 will be in the form c radicalBig m 2 π 2 L 2 + m 2 π 2 L 2 , or L m 2 + n 2 . Considering a superposition of these separable solutions across positive integer m and n and consolidating the coefficients into a mn c 1 m d 1 n e 2 mn , b mn c 1 m d 1 n e 1 mn , we get w ( x, y, t ) = summationdisplay m =1 summationdisplay n =1 sin( mπx L ) sin( nπy L )( a mn cos( radicalbig m 2 + n 2 cπt L + b mn sin( radicalbig m 2 + n 2 cπt L )) as requested (noting that c 2 = T ρ ). 9.58 (a) The problem for the steady-state case is T xx = 0 T (0) = T 1 , hlT x ( L ) + T ( L ) - T 0 = 0 The general solution is T ( x ) = a + bT , and putting in the boundary condi-
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