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13
consequently
u(x, t)= φ(x − at).
We conclude that the problem (1.24) is solved by the formula (1.26) for
any smooth φ and constant a. It is straightforward to check that (1.26)
actually solves (1.24);
u(x, 0) = φ(x),
and

u t = −aφ (x − at) 1.4 Cauchy Problems (1.26)
=⇒ u t + au x =0.

u x = φ (x − at)
Hence both the initial condition and the differential equation are fulfilled.

Example 1.2 Consider the Cauchy problem
u t + xu x =0, x ∈ R,t > 0,
(1.27)
u(x, 0) = φ(x),x ∈ R.
Now the characteristics are defined by

x (t)= x(t), x(0) = x 0
so
−t
x(t)= x 0 e t and x 0 = xe .
Since

u x(t),t = φ(x 0 )
(see (1.23)), we get

−t
u(x, t)= φ xe . (1.28)
As above, it is a straightforward task to check that (1.28) solves (1.27).

1.4.2 First-Order Nonhomogeneous Equations

The method of characteristics can also be utilized for nonhomogeneous
problems. Consider the Cauchy problem

u t + a(x, t)u x = b(x, t), x ∈ R,t > 0,
(1.29)
u(x, 0) = φ(x), x ∈ R.