hw11_sol - Solutions to Homework 11 May 23, 2011 9.52 The...

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Unformatted text preview: 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, < 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, < 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 = m L , m = 1 , 2 , 3 ,... d 2 = 0 , l n = m L , n = 1 , 2 , 3 ,... 1 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 2 L 2 + m 2 2 L 2 , or c 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( mx L ) sin( ny L )( a mn cos( radicalbig m 2 + n 2 ct L + b mn sin( radicalbig m 2 + n 2 ct L )) as requested (noting that c 2 = T )....
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This note was uploaded on 08/30/2011 for the course CGN 2200 taught by Professor Glagola during the Spring '11 term at University of Florida.

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hw11_sol - Solutions to Homework 11 May 23, 2011 9.52 The...

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