1.2. Examples of propagation of singularities using progressing waves 13
The rays are the integral curves of
(1.2.3) ∂t ± ∂x .
Structures are rigidly transported at speeds ±1.
There is an energy law. If u is a smooth solution whose support intersects
each time slab a t b in a compact set, one has
d
dt
R
ut
2
+ ux
2
dx = ∂t(ut
2
+ ux)
2
dx
= 2ut(utt uxx) + ∂x(2ut ux) dx = 0 ,
since the first summand vanishes and the second is the x derivative of a
function vanishing outside a compact set.
The fundamental solution that solves (1.2.3) together with the initial
values
(1.2.4) u(0,x) = 0 , ut(0,x) = δ(x) ,
is given by the explicit formula
(1.2.5) u(t, x) =
sgn t
2
χ[−t,t] =
1
2
h(x t) h(x + t) ,
where h denotes Heaviside’s function, the characteristic function of ]0, ∞[.
Exercise 1.2.1. i. Derive (1.2.5) by solving the initial value problem using
the Fourier transform in x. Hint. You will likely decompose an expression
regular at ξ = 0 into two that are not. Use a principal value to justify this
step.
ii. The proof of D’Alembert’s formula (1.2.2) shows that every distribu-
tion solution of (1.2.1) is given by (1.2.2) for f, g distributions on R. Derive
(1.2.5) by finding the f, g that yield the solution of (1.2.4). Hint. You will
need to find the solutions of du/dx = δ(x).
The singularities of the solution (1.2.5) lie on the characteristic curves
through (0, 0). This is a consequence of Theorem 1.1.4. In fact, define v as
the solution of
vtt vxx = 0, v(0,x) = 0, vt(0,x) = x+/2,
2
x+ := max{x, 0} .
Introduce
V := (v1,v2,v), v1 := ∂tv ∂xv, v2 := ∂tv + ∂xv
to find
∂tV +

⎝0
1 0 0
1 0
0 0 −1


∂xV +

⎝0
0 0 1
0
0⎠
0 0 0

V = 0 .
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