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# Problem2.5.1e - Problem 2.5.1(e The problem statement is 2...

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Problem 2.5.1(e) The problem statement is : (1) 2 u ÅÅÅÅÅÅÅÅÅ x 2 + 2 u ÅÅÅÅÅÅÅÅÅ y 2 = 0 BC1:u H 0, y L = 0 BC2:u H L, y L = 0 BC3:u H x, 0 L - u ÅÅÅÅÅÅÅ y H x, 0 L = 0 BC4:u H x, H L = f H x L We will use separation of variables and look for a solution of the form u H x, y L = h H x L f H y L . Substituting this form for the solution into the PDE gives: (2) 1 ÅÅÅÅ h 2 h ÅÅÅÅÅÅÅÅÅ x 2 = - 1 ÅÅÅÅ f 2 f ÅÅÅÅÅÅÅÅÅ y 2 = -l The appropriate eigenvalue problem is (3) 2 h ÅÅÅÅÅÅÅÅÅ x 2 + l h = 0 BC1:h H 0 L = 0 BC2:h H L L = 0 The solution is (4) h H x L = c 1 Sin H n p x ê L L where (5) l = I n p ÅÅÅÅÅÅÅ L M 2 , n = 1, 2, 3 Next we solve for the function f H y L . The governing equation is: (6) 2 f ÅÅÅÅÅÅÅÅÅ y 2 - lf = 0 BC1: f H 0 L - „f ÅÅÅÅÅÅÅ y H 0 L = 0 The solution is (7) f H y L = c 2 è!!! l Cosh I è!!! l y M + c 2 Sinh I è!!! l y M Hence the general solution is

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(8) u H x, y L = n = 1 A n 9 è!!! l Cosh I è!!! l y M + Sinh I è!!! l y M= Sin H n p x ê L L To find the coefficients, we use the orthogonality properties of the eigenfunctions, and BC4:
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