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5) The eigen-solutions are the product of the solutions of the ordinary differential
equations. φ n = X n Y n Z n · · · . The solution of the partial differential equation is a
linear combination of the eigen-solutions.

X
u(x, y, z, . . .) = a n φ n

6) Solve for the coefficients, a n using the inhomogeneous boundary conditions.

8.6 Separation of variables for nonhomogeneous

equations

It is possible to upgrade the method of separation of variables to a method for solving non-
homogeneous PDEs. This technique is called also the method of eigenfunction expansion.
For example, consider the problem

u tt − u xx = cos 2πx cos 2πt, 0 < x < 1, t > 0, (8.43)
u x (0, t) = u x (1, t) = 0, t ≥ 0,
2
u(x, 0) = f(x) = cos πx, 0 ≤ x ≤ 1,
2
u t (x, 0) = g(x) = 2 cos πx, 0 ≤ x ≤ 1.
In the previous section we found the system of all eigenfunctions and the corresponding
eigenvalues of the homogeneous problem. They are

2
X n (x) = cos nπx, λ n = (nπ) , n = 0, 1, 2, . . .

Recall Fourier’s claim that any reasonable function satisfying the boundary conditions can
be uniquely expanded into (generalized) Fourier series with respect to the system of the
eigenfunctions of the problem. Since the solution u(x, t) of the problem (8.43) is a twice
differentiable function satisfying the boundary conditions, it follows that for a fixed t the
solution u can be represented as

∞
1 X
u(x, t) = T 0 (t) + T n (t) cos πnx, (8.44)
2
n=1
where T n (t) are the (time dependent) Fourier coefficients of the function u(·, t). Hence,
we need to find these coefficients.
Substituting (8.44) into the wave equation (8.43) and differentiating the series term-
by-term implies that

∞
1 X
00
2 2
00
T + (T + n π T n ) cos πnx = cos 2πt cos 2πx. (8.45)
2 0 n
n=1
Note that in the current example, the right hand side of the equation is already given in
the form of a Fourier series. The uniqueness of the Fourier expansion implies that the

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