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1. Setting the Scene
16
solves (1.37) for any smooth functions f and g. In fact, all solutions of
(1.37) can be written in the form (1.38); see Exercise 1.12. Now it follows
from (1.36) that
u(x, t)= f(x + t)+ g(x − t)
(1.39)
solves (1.33) for any smooth f and g. This can be verified by direct deriva-
tion:

u tt = f + g

=⇒ u tt = u xx .
u xx = f + g

Next we turn our attention to the initial data (1.33) and (1.34). We want
to determine the functions f and g in (1.39) such that (1.33) and (1.34)
are satisfied. Of course, φ and ψ are supposed to be given functions.
By (1.39) we have
u(x, t)= f(x + t)+ g(x − t)
and

u t (x, t)= f (x + t) − g (x − t).
Inserting t = 0, (1.34) and (1.35) imply that
φ(x)= f(x)+ g(x) (1.40)
and
ψ(x)= f (x) − g (x). (1.41)

By differentiating (1.40) with respect to x,weget

φ(x)= f (x)+ g (x). (1.42)
Combining (1.41) and (1.42) yields
1

f = φ + ψ

2
and
1

g = φ − ψ ,
2
and thus, by integration, we have
1 1 s
f(s)= c 1 + φ(s)+ ψ(θ)dθ (1.43)
2 2
0