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The boundary and initial conditions become
v(0, t) = 0, v x (h, t) = 0, v(x, 0) = f(x) − µ(x, 0).

Thus we have a heat equation with the source s(x, t)−µ t (x, t). We could apply separation
of variables to find a solution of the form
∞ r
X 2 (2n − 1)πx
u(x, t) = µ(x, t) + u n (t) sin .
h 2h
n=1

9.5.2 Direct eigenfunction expansion.

Alternatively we could seek a direct eigenfunction expansion of u(x, t).

∞ r
X 2 (2n − 1)πx
u(x, t) = u n (t) sin .
h 2h
n=1
Note that the eigenfunctions satisfy the homogeneous boundary conditions while u(x, t)
does not. If we choose any fixed time t = t 0 and form the periodic extension of the
function u(x, t 0 ) to define it for x outside the range (0, h), then this function will have
jump discontinuities. This means that our eigenfunction expansion will not converge
uniformly. We are not allowed to differentiate the series with respect to x. We can’t just
plug the series into the partial differential equation to determine the coefficients. Instead,
we will multiply Equation 9.11, by an eigenfunction and integrate from x = 0 to x = h.
To avoid differentiating the series with respect to x, we will use integration by parts to
2
move derivatives from u(x, t) to the eigenfunction. (We will denote λ n = (2n−1)π .)
2h
r Z h r Z h
2 p 2 p
sin( λ n x)(u t − κu xx ) dx = sin( λ n x)s(x, t) dx
h 0 h 0
r r Z
2 h p i h 2 p h p
0
u (t) − κ u x sin( λ n x) + κ λ n u x cos( λ n x) dx = s n (t)
n
h 0 h 0
r r
2 2 p h p i h
0 n
u (t) − κ(−1) u x (h, t) + κ λ n u cos( λ n x) +
n
h h 0
r Z h
2 p
+ κλ n u sin( λ n x) dx = s n (t)
h 0
r r
2 2 p
0 n
u (t) − κ(−1) β(t) − κ λ n u(0, t) + κλ n u n (t) = s n (t)
n
h h
r
2 p
0
n
u (t) + κλ n u n (t) = κ λ n α(t) + (−1) β(t) + s n (t)
n
h
Now we have an ordinary differential equation for each of the u n (t). We obtain initial
conditions for them using the initial condition for u(x, t).
∞ r
X 2 p
u(x, 0) = u n (0) sin( λ n x) = f(x)
h
n=1
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