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17
1.4 Cauchy Problems
and
1
1
g(s)= c 2 + φ(s) −
2
2
where c 1 and c 2 are constants of integration. From (1.40) we note that
φ(x)= f(x)+ g(x),
and thus by adding (1.43) and (1.44), we observe that
c 1 + c 2 =0. 0 s ψ(θ)dθ, (1.44)
Putting s = x + t in (1.43) and s = x − t in (1.44), it follows from (1.39)
that
1 x+t 1 x−t
1
u(x, t)= φ(x + t)+ φ(x − t) + ψ(θ)dθ − ψ(θ)dθ,
2 2 2
0 0
or
x+t

1
1
u(x, t)= φ(x + t)+ φ(x − t) + ψ(θ)dθ. (1.45)
2 2 x−t
This formula is referred to as the d’Alembert solution. Let us use it to
compute the solution of one Cauchy problem.
Example 1.4 Consider the Cauchy problem
u tt = u xx , x ∈ R ,t > 0 ,
u(x, 0)=0 , x ∈ R , (1.46)
u t (x, 0) = cos(x) , x ∈ R .
Since φ(x)=0 and ψ(x) = cos(x), it follows by (1.45) that
1 x+t
u(x, t)= cos(θ)dθ
2 x−t
x+t
1
= sin(θ)
2 x−t
1
= sin(x + t) − sin(x − t) ,
2
so
u(x, t) = cos(x) sin(t). (1.47)

It is straightforward to check by direct computation that (1.47) in fact
solves (1.46).