Page 146 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

136 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
We first solve the problem

P t = P xx + xe −λ

P(0, t) = P(1, t) = 0
P(x, 0) = 0

while holding λ constant.
Recall from Chapter 2 that one technique in this case is to assume a solution of the form

P(x,λ, t) = X(x) + W(x,λ, t)

so that

W t = W xx
W(0,λ, t) = W(1,λ, t) = 0
W(x,λ, 0) =−X(x,λ)

and

X xx + xe −λ = 0

X(0) = X(1) = 0

Separating variables in the equation for W(x, t), we find that for W(x,λ, t) = S(x)Q(t)
Q t S xx 2
= =−β
Q S
The minus sign has been chosen so that Q remains bounded. The boundary conditions
on S(x) are as follows:

S(0) = S(1) = 0

The solution gives

S = A sin(β x) + B cos(βx)
Q = Ce −β t

Applying the boundary condition at x = 0 requires that B = 0 and applying the boundary
condition at x = 1 requires that sin(β) = 0or β = nπ.
Solving for X(x) and applying the boundary conditions gives
x 2 −λ
X = (1 − x )e =−W(x,λ, 0)
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