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Chem Differential Eq HW Solutions Fall 2011 52

Chem Differential Eq HW Solutions Fall 2011 52 - 52 Chapter...

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52 Chapter 3 Partial Differential Equations in Rectangular Coordinates Solutions to Exercises 3.10 1. We use a combination of solutions from (2) and (3) and try a solution of the form u ( x, y ) = n =1 sin mx A m cosh m (1 - y ) + B m sinh my . (If you have tried a different form of the solution, you can still do the prob- lem, but your answer may look different from the one derived here. The reason for our choice is to simplify the computations that follow.) The boundary conditions on the vertical sides are clearly satisfied. We now determine A m and B m so as to satisfy the conditions on the other sides. Starting with u (1 , 0) = 100, we find that 100 = m =1 A m cosh m sin mx. Thus A m cosh m is the sine Fourier coefficient of the function f ( x ) = 100. Hence A m cosh m = 2 π π 0 100 sin mx dx A m = 200 πm cosh m [1 - ( - 1) m ] . Using the boundary condition u y ( x, 1) = 0, we find 0 = m =1 sin
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