13.2.nb - H*Here we solve the partial differential equation...

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H * Here we solve the partial differential equation for steady state temperature distribution in the semi - infinite slab with the Laplace equation 2 T @ x,y D = 0 with the Boundary Conditions H BC L T H x,0 L = T 0 = 100, T @ x, D = 0, T @ 0,y D = 0, T @ L,y D = 0 where L = 10. Assuming a solution of the form T @ x,y D = X @ x D * Y @ y D and using the 2D Laplacian 2 = 2 x 2 + 2 y 2 which gives X" @ x D Y @ y D + X @ x D Y" @ y D = 0. Dividing by X @ x D Y @ y D 0 we get X" @ x X @ x D + Y" @ y Y @ y D = 0, which can only be satisfied if both terms are the same constant or X" @ x X @ x D =- Y" @ y Y @ y D = - k 2 , where k is the separation constant. We can solve these two diffy - Q's using DSolve. * L DSolve @ X'' @ x D + k^2 * X @ x D ã 0, X @ x D , x D 88 X @ x D Ø C @ 1 D Cos @ k x D + C @ 2 D Sin @ k x D<< DSolve @ Y'' @ y D - k^2 * Y @ y D ã 0, Y @ y D , y D 99 Y @ y D Ø ‰ k y C @ 1 D + ‰ - k y C @ 2 D== H * That is X @ x D =+ A * Sin @ k * x D + B * Cos @ k * x D and Y @ y D = C * Exp @ k * y D + D * Exp @ - k * y D , as in the notes. Demanding the BC's X @ 0 D = 0, X @ L D = 0 gives B = 0 and k = k n = n p ê L H the "quantization condition"
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13.2.nb - H*Here we solve the partial differential equation...

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