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There are also two derivatives along the t− direction and hence we need two further
conditions here. We need to know the initial position and the initial velocity of the string.
That is,
u(x, 0) = ϕ(x) and u t (x, 0) = ψ(x)

for some known functions ϕ(x) and ψ(x).
Thus, the following task arises:

2
u tt = c u xx , x ∈ (0; l), t > 0 (8.1)

u(x, 0) = ϕ(x), u t (x, 0) = ψ(x), x ∈ [0; l] (8.2)
u(0, t) = u(l, t) = 0, t ≥ 0 (8.3)

Solution by separation of variables is a powerful technique that is applicable to many
areas of mathematics. The idea is to look first for solutions of a particularly simple form,
then combine to obtain the most general solution.
We assume that the solution of the task (8.1) - (8.3) has the form

u(x, t) = X(x)T(t),

where X(x) and T(t) are nonzero functions. Substituting into (8.1) we get

00
00
2
X(x)T (t) = c X (x)T(t).
2
Dividing both sides of the last equality by c X(x)T(t) we obtain
00
00
T (t) X (x)
= (8.4)
2
c T(t) X(x)
Since the left side of (8.4) depends only on t and the right side depends only on x each
side of this equality can only be equal to some constant. Thus
00
00
T (t) X (x)
= = −λ, λ = constant,
2
c T(t) X(x)
or
00
2
T (t) + λc T(t) = 0 (8.5)
00
X (x) + λX(x) = 0 (8.6)
It follows from (8.3) that
X(0) = X(l) = 0 (8.7)

So we have to solve the eigenvalue problem (8.6),(8.7).
√
2
k + λ = 0, k = ±i λ.

The general solution for (8.6) is
√ √
X(x) = C 1 cos λx + C 2 sin λx.

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