1.1. The method of characteristics 3
Example 1.1.1. If c ∈ R is constant, then u ∈
C∞([0,T]
× R) satisfies
(1.1.2) ∂tu + c ∂xu = 0
if and only if there is an f ∈ C∞(R) so that u = f(x − ct). The function f
is uniquely determined.
Proof. For constant c the characteristics along which u is constant are the
lines (t, x + ct). Therefore, u(t, x) = u(0,x − ct) proving the result with
f(x) := u(0,x).
This shows that the Cauchy problem consisting of (1.1.2) together with the
initial condition u|t=0 = f is uniquely solvable with solution f(x − ct). The
solutions are waves translating rigidly with velocity equal to c.
Exercise 1.1.1. Find an explicit solution formula for the solution of the
Cauchy problem
∂tu + c ∂xu + z(t, x)u = 0, u|t=0 = g,
where z ∈
C∞.
Example 1.1.2 (D’Alembert’s formula). If c ∈ R \ 0, then u ∈
C∞([0,T]
×
R) satisfies
(1.1.3) utt −
c2
uxx = 0
if and only if there are smooth f, g ∈
C∞(R)
so that
(1.1.4) u = f(x − ct) + g(x + ct) .
The set of all pairs
˜
f, ˜ g so that this is so is of the form
˜
f = f + b , ˜ g = g − b
with b ∈ C.
Proof. Factor
∂t
2
−
c2∂x 2
= (∂t − c∂x) (∂t + c∂x) = (∂t + c∂x) (∂t − c∂x) .
Conclude that
u+ := ∂tu − c ∂xu and u− := ∂tu + c ∂xu
satisfy
(1.1.5)
(
∂t ± c ∂x
)
u± = 0 .
Example 1.1.1 implies that there are smooth F± so that
(1.1.6) u± = F±(x ∓ ct) .
In order for (1.1.4) and (1.1.6) to hold, one must have
(1.1.7) F+ = (∂t − c∂x)u = (1 +
c2)f
, F− = (∂t + c∂x)u = (1 +
c2)g
.
