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3. The Heat Equation
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3.3 The Principle of Superposition
In step 1 we found that the functions {u k (x, t)} given by (3.18) solve the
following problems:
(u k ) t =(u k ) xx
u k (0,t)= u k (1,t)=0,
u k (x, 0) = sin (kπx),
for k =1, 2,... . Now, we want to use these solutions to solve more general
problems of the form for x ∈ (0, 1), t > 0, (3.19)
for x ∈ (0, 1), t > 0,
u t = u xx
u(0,t)= u(1,t)=0, (3.20)
u(x, 0) = f(x).
Suppose first that the initial function f can be written as a finite linear
combination of the eigenfunctions {sin (kπx)}. Thus, there exist constants
N
{c k } such that
k=1
N

f(x)= c k sin (kπx). (3.21)
k=1
Then, by linearity, it follows that the solution of (3.20) is given by
N
2
−(kπ) t
u(x, t)= c k e sin (kπx). (3.22)
k=1
You can easily check that this is a solution by explicit differentiation.
Example 3.1 Let us look at one simple example showing some typical
features of a solution of the heat equation. Suppose

f(x) = 3 sin (πx) + 5 sin (4πx);

then the solution of (3.20) is given by
2
2 −16π t
−π t
u(x, t)=3e sin (πx)+5e sin (4πx).
This solution is graphed, as a function of x, in Fig. 3.2 for t =0, 0.01, 0.1.
Notice here that the maximum value of the solution is attained at t =0,
and that the entire solution becomes smaller as t increases. We easily see,
both from the figure and from the formulas, that this solution approaches
zero as t tends to infinity.