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sol12

# sol12 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 12...

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CAAM 336 · DIFFERENTIAL EQUATIONS Problem Set 12 · Solutions Posted Friday 26 November 2010. Due Friday 3 December 2010, 5pm. This problem set counts for 100 points, plus a 20 point bonus. 1. [50 points: 15 points for (a); 10 points for (b); 5 points for (c); 20 points for (d)] On Problem Set 10, you solved the heat equation on a two-dimensional square domain. Now we will investigate the wave equation on the same domain, a model of a vibrating membrane stretched over a square frame—that is, a square drum: u tt ( x, y, t ) = u xx ( x, y, t ) + u yy ( x, y, t ) , with 0 x 1, and 0 y 1, and t 0. Take homogeneous Dirichlet boundary conditions u ( x, 0 , t ) = u ( x, 1 , t ) = u (0 , y, t ) = u (1 , y, t ) = 0 for all x and y such that 0 x 1 and 0 y 1 and all t 0, and consider the initial conditions u ( x, y, 0) = u 0 ( x, y ) = X j =1 X k =1 a j,k (0) ψ j,k ( x, y ) , u t ( x, y, 0) = v 0 ( x, y ) = X j =1 X k =1 b j,k (0) ψ j,k ( x, y ) . Here ψ j,k ( x, y ) = 2 sin( jπx ) sin( kπy ), for j, k 1, are the eigenfunctions of the operator Lu = - ( u xx + u yy ) , with homogeneous Dirichlet boundary conditions given in Problem Set 10. You may use without proof that these eigenfunctions are orthogonal, and use the eigenvalues λ j,k = ( j 2 + k 2 ) π 2 computed for Problem Set 10. (a) We wish to write the solution to the wave equation in the form u ( x, y, t ) = X j =1 X k =1 a j,k ( t ) ψ j,k ( x, y ) . Show that the coefficients a j,k ( t ) obey the ordinary differential equation a 00 j,k ( t ) = - λ j,k a j,k ( t ) with initial conditions a j,k (0) , a 0 j,k (0) = b j,k (0) derived from the initial conditions u 0 and v 0 . (b) Write down the solution to the differential equation in part (a). (c) Use your solution to part (b) to write out a formula for the solution u ( x, y, t ). (d) Suppose the drum begins with zero velocity, v 0 ( x, y ) = 0, and displacement u 0 ( x, y ) = 200 xy (1 - x )(1 - y )( x - 1 / 4)( y - 1 / 4) = X j =1 X k =1 100(5 + 7( - 1) j )(5 + 7( - 1) k ) j 3 k 3 π 6 ψ j,k ( x, y ) . Submit surface (or contour) plots of the solution at times t = 0 , 0 . 5 , 1 . 0 , 1 . 5 , 2 . 5, using j = 1 , . . . , 10 and k = 1 , . . . , 10 in the series.

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Solution. This question follows the same pattern as the first problem on this problem set. (a) Substitute the formula u ( x, y, t ) = X j =1 X k =1 a j,k ( t ) ψ j,k ( x, y ) . into the two dimensional wave equation to obtain X j =1 X k =1 d 2 a j,k dt 2 ( t ) ψ j,k ( x, y ) = X j =1 X k =1 a j,k ( t ) 2 ∂x 2 + 2 ∂y 2 ψ j,k ( x, y ) , which implies X j =1 X k =1 d 2 a j,k dt 2 ( t ) ψ j,k ( x, y ) = X j =1 X k =1 - a j,k ( t ) λ j,k ψ j,k ( x, y ) . Take the inner product of both sides with ψ m,n and use the orthogonality of the eigenfunctions to obtain a 00 j,k ( t ) = - λ j,k a j,k ( t ) . The initial value a j,k (0) is simply the ( j, k ) coefficient in the expansion of the initial condition, u ( x, y, 0) = u 0 ( x, y ) = X j =1 X k =1 a j,k (0) ψ j,k ( x, y ) . As the differential equation describing a j,k is of second order, we require a second initial condition to determine a unique solution. Since u t ( x, y, 0) = X j =1 X k =1 a 0 j,k (0) ψ j,k ( x, y ) , and v 0 ( x, y ) = X j =1 X k =1 b j,k ψ j,k ( x, y ) , the second initial condition we require is simply the ( j, k ) coefficient of v 0 : a 0 j,k (0) = b j,k .
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