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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.
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 ) =
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
x Am cosh[
(b − y )] + Bm sinh[
a Using the boundary conditions on the horizontal sides, starting with uy (x, b) =
0, we ﬁnd that
∞ 0 = B0 +
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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.
- Fall '11