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

.