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If a function h(x) is defined for positive x, then sign(x)h(|x|) is the odd extension of
the function. If h(x) is defined for x ∈ (−L . . . L) then its periodic extension is

x + L
h x − 2L .
2L
We combine these two formulas to form odd periodic extensions.
x + L x + L

˜
f(x) = sign x − 2L f x − 2L
2L 2L

x + L x + L
˜ g(x) = sign x − 2L g x − 2L
2L 2L
Now we can write the solution for the vibrations of a string with fixed ends.
Z x+ct
1 1
˜
˜
u(x, t) = f(x − ct) + f(x + ct) + ˜ g(ξ) dξ
2 2c x−ct
The case of a finite string is more complex. There we encounter the problem that
even though f and g are only known for 0 < x < a, x ± ct can take any value from −∞
to ∞. So we have to figure out a way to continue the function beyond the length of the
string. The way to do that depends on the kind of boundary conditions: Here we shall
only consider a string fixed at its ends.

u(0, t) = u(a, t) = 0,

∂u
u(x, 0) = f(x), (x, 0) = g(x).
∂t
Initially we can follow the approach for the infinite string as sketched above, and we find
that
1
F(x) = (f(x) + Γ(x) + C),
2
1
G(x) = (f(x) − Γ(x) − C).
2
Look at the boundary condition u(0, t) = 0. It shows that

1 1
(f(ct) + f(−ct)) + (Γ(ct) − Γ(−ct)) = 0.
2 2
Now we understand that f and Γ are completely arbitrary functions — we can pick any
form for the initial conditions we want. Thus the relation found above can only hold when
both terms are zero
f(x) = −f(−x), Γ(x) = Γ(x). (6.8)
Now apply the other boundary condition, and find

f(a + x) = −f(a − x), Γ(a + x) = Γ(a − x). (6.9)

The reflection conditions for f and Γ are similar to those for sines and cosines, and both
f and Γ have period 2a.

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