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where the arbitrary constant b n is the product of the two arbitrary constants A n and a n
in equations (9.5) and (9.6). Note that each u n satisfies the two boundary conditions

u n (0, t) = u n (l, t) = 0 for all t > 0.
So far we have not used the initial condition.
Because of linearity of a heat equation, linear combinations of solutions are also solu-
tions. If we let
∞
2 2 2
X πnx π n c
u n (x, t) = b n sin e − l 2 t ,
l
n=1
then setting t = 0 and using the initial condition u(x, 0) = f(x) results in the requirement
that
∞
X πnx
u(x, 0) = b n sin ≡ f(x), 0 ≤ x ≤ l.
l
n=1
Recognizing that this is just a Fourier Sine Series for the function f(x) on 0 ≤ x ≤ l,
the constants b n must be the coefficients of that Sine Series. Therefore, the complete
solution of our heat equation is given by
l
∞ Z
2 2 2
X πnx π n c 2 πnx
u n (x, t) = b n sin e − l 2 t , b n = f(x) sin dx. (9.7)
l l l
n=1
0
0
Example 6.1 A rod of length 10 meters is initially heated to a temperature of 100 C
0
on the left half 0 ≤ x ≤ 5 and to a temperature of 40 C on the right half 5 ≤ x ≤ 10.
If the rod is completely insulated except at the two ends, and the temperature at both
0
ends is held constant at 0 C for all t > 0, find the temperature u(x, t) in the rod for
K
2
0 ≤ x ≤ 10, t > 0. Assume the constant c = = 4.
sρ
We know that the general solution is given by
∞
2 2
X πnx 4π n
t
u(x, t) = b n sin e − 100 .
10
n=1
Writing the piecewise continuous function f(x) in the form

100 for 0 ≤ x ≤ 5
f(x) =
40 for 5 < x ≤ 10
The coefficients b n are given by
10 5 10
Z Z Z
2 πnx 1 πnx 1 πnx
b n = f(x) sin dx = 100 sin dx + 40 sin dx =
10 10 5 10 5 10
0 0 5
200 nπ 80 nπ
= 1 − cos + cos − cos (nπ)
nπ 2 nπ 2
Even though forty terms were used in the Fourier Series it can be seen that the
approximation of the initial temperature function at time t = 0 is very rough. For t > 0
the exponential functions tend to smooth the solution. Note that, as expected, the
0
temperature along the entire rod tends to 0 C as t → ∞.

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