Page 35 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

THE FOURIER METHOD: SEPARATION OF VARIABLES 25
2.1.21 Separating Variables
We now solve for V using separation of variables.

V = P(τ)Q(ξ) (2.87)
P τ Q ξξ 2
= =±λ (2.88)
P Q
We must choose the minus sign once again (see Problem 1 above) to have a negative exponential
2
for P(τ). (We will see later that it’s not always so obvious.) P = exp(−λ τ).
The solution for Q is once again sines and cosines.

Q = A cos(λξ) + B sin(λξ) (2.89)
The boundary condition V (τ, 0) = 0 requires that Q(0) = 0. Hence, A = 0. The boundary
condition V (τ, 1) = 0 requires that Q(1) = 0. Since B cannot be zero, sin(λ) = 0sothatour
eigenvalues are λ = nπ and our eigenfunctions are sin(nπξ).

2.1.22 Superposition
Once again using linear superposition,
∞
2 2
V = B n exp(−n π τ)sin(nπξ) (2.90)
n=0
Applying the initial condition
∞
1
ξ(ξ − 1) = B n sin(nπξ) (2.91)
2
n=1
This is a Fourier sine series representation of 1 ξ(ξ − 1). We now use the orthogonality of
2
the sine function to obtain the coefficients B n .

2.1.23 Orthogonality
Using the concept of orthogonality again, we multiply both sides by sin(mπξ)dξ and integrate
over the space noting that the integral is zero if m is not equal to n. Thus, since

1
1
2
sin (nπξ)dξ = (2.92)
2
0
1

B n = ξ(ξ − 1) sin(nπξ)dξ (2.93)
0