Page 86 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

76 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
We now proceed by separating variables. Let

U(τ, ξ, η) = T(τ)X(ξ)Y (η) (5.4)

Differentiating and inserting into (5.3) and dividing by (5.4) we find

2
T X Y + r Y X

= (5.5)
T XY
where the primes indicate differentiation with respect to the variable in question and r = a/b.
Since the left-hand side of (5.5) is a function only of τ and the right-hand side is only a function
of ξ and η both sides must be constant. If the solution is to be finite in time we must choose
2
2
the constant to be negative, –λ . Replacing T /T by –λ and rearranging,

X Y
2
−λ − = r (5.6)
X Y
Once again we see that both sides must be constants. How do we choose the signs? It should be
clear by now that if either of the constants is positive solutions for X or Y will take the form of
hyperbolic functions or exponentials and the boundary conditions on ξ or η cannot be satisfied.
Thus,

T
=−λ 2 (5.7)
T
X 2
=−β (5.8)
X
Y
r 2 =−γ 2 (5.9)
Y
Note that X and Y are eigenfunctions of (5.8) and (5.9), which are Sturm–Liouville equations
and β and γ are the corresponding eigenvalues.
Solutions of (5.7), (5.8), and (5.9) are

2
T = A exp(−λ τ) (5.10)
X = B 1 cos(βξ) + B 2 sin(βξ) (5.11)
Y = C 1 cos(γη/r) + C 2 sin(γη/r) (5.12)

Applying the first homogeneous boundary condition, we see that X(0) = 0, so that B 1 = 0.
Applying the third homogeneous boundary condition we see that Y (0) = 0, so that C 1 = 0.
The second homogeneous boundary condition requires that sin(β) = 0, or β = nπ.The last
homogeneous boundary condition requires sin(γ/r) = 0, or γ = mπr. According to (5.6),
2
2
2
λ = β + γ . Combining these solutions, inserting into (5.4) we have one solution in the