Chem Differential Eq HW Solutions Fall 2011 52

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

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Unformatted text preview: 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 ) = sin mx Am cosh m(1 − y ) + Bm sinh my . n=1 (If you have tried a different form of the solution, you can still do the problem, 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 Am and Bm so as to satisfy the conditions on the other sides. Starting with u(1, 0) = 100, we find that ∞ 100 = Am cosh m sin mx. m=1 Thus Am cosh m is the sine Fourier coefficient of the function f (x) = 100. Hence Am cosh m = 2 π π 100 sin mx dx 0 ⇒ Am = 200 [1 − (−1)m ] . πm cosh m Using the boundary condition uy (x, 1) = 0, we find ∞ 0= sin mx Am (−m) sinh[m(1 − y )] + mBm cosh my m=1 Thus y =1 . ∞ 0= mBm sin mx cosh m. m=1 By the uniqueness of Fourier series, we conclude that mBm cosh m = 0 for all m. Since m cosh m = 0, we conclude that Bm = 0 and hence u(x, y ) = = 200 π 400 π ∞ m=1 ∞ k =0 [1 − (−1)m ] sin mx cosh[m(1 − y )] m cosh m sin[(2k + 1)x] cosh[(2k + 1)(1 − y )]. (2k + 1) cosh(2k + 1) 5. We combine solutions of different types from Exercise 4 and try a solution of the form ∞ u(x, y ) = A0 + B0 y + cos m=1 mπ mπ mπ x Am cosh[ (b − y )] + Bm sinh[ y] . a a a Using the boundary conditions on the horizontal sides, starting with uy (x, b) = 0, we find that ∞ 0 = B0 + m=1 mπ mπ mπ Bm cos x cosh b. a a a ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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