Lecture Six

January 27, 2021

1 The nite interval

The wave equation with xed ends:

2
vtt = c vxx
 0 < x < l,
v(x, 0) = φ(x), vt (x, 0) = ψ(x),

v(0, t) = v(l, t) = 0.


We extend them to the whole line


φ(x)
 0<x<l
φext = −φ(−x) −l < x < 0

extended to be of period 2l.


and 
ψ(x)
 0<x<l
ψext = −ψ(−x) −l < x < 0

extended to be of period 2l.


So the formula of v is
Z x+ct
1 1
v(x, t) = [φext (x − ct) + φext (x + ct)] + ψext (s)ds.
2 2c x−ct

Case (0,0), x − ct ≥ 0, x + ct ≤ l and t > 0:

Z x+ct
1 1
v(x, t) = [φ(x − ct) + φ(x + ct)] + ψ(s)ds.
2 2c x−ct

Case (0,1), x − ct ≥ 0 , x + ct ≥ l and x ≤ l:

Z l Z 2l
1 1 1
v(x, t) = [φ(x − ct) − φ(2l − x − ct)] + ψ(s)ds − ψ(2l − s)ds
2 2c x−ct 2c l
Z l Z 2l−x−ct
1 1 1
= [φ(x − ct) − φ(2l − x − ct)] + ψ(s)ds + ψ(s)ds
2 2c x−ct 2c l
Z 2l−x−ct
1 1
= [φ(x − ct) − φ(2l − x − ct)] + ψ(s)ds.
2 2c x−ct

1
 Case (1,1), −l ≤ x − ct ≤ 0 and 2l ≥ x + ct ≥ l:
Z 0 Z l Z x+ct
1 1
v(x, t) = [−φ(ct − x) − φ(2l − x − ct)] + [ −ψ(−s)ds + ψ(s)ds − ψ(2l − s)ds]
2 2c x−ct 0 l
Z 2l−x−ct
1 1
= [−φ(ct − x) − φ(2l − x − ct)] + ψ(s)ds.
2 2c ct−x

Case (1,2), −l ≤ x − ct ≤ 0 , x + ct ≥ 2l and x ≤ l:

1
v(x, t) = [−φ(ct − x) + φ(x + ct − 2l)]
2
Z 0 Z l Z 2l Z x+ct
1
= + [ −ψ(−s)ds + ψ(s)ds − ψ(2l − s)ds + ψ(s − 2l)ds]
2c x−ct 0 l 2l
Z ct−x
1 1
= [−φ(ct − x) + φ(x + ct − 2l)] − ψ(s)ds,
2 2c x+ct−2l

which depends only on the initial values on the interval [x + ct − 2l, ct − x].
Similarly, you can derive the formula of v on each domain of Chapter 3
Figure 6 in the textbook.

2 Waves with a source

We are going to solve wave equations with a source term f on the whole line

2
utt − c uxx = f (x, t), −∞ < x < ∞

u(x, 0) = φ(x), (1)

ut (x, 0) = ψ(x),


where f (x, t) is a given function. For instance, f (x, t) could be interpreted

as an external force acting on an innitely long vibrating string. This is an
inhomogeneous linear equation. If you can nd a solution uf to the equation

2
utt − c uxx = f (x, t) −∞ < x < ∞

u(x, 0) = 0

ut (x, 0) = 0,


then with the solution uhom = 12 [φ(x + ct) + φ(x − ct)] + 2c1 x−ct
x+ct
ψ for the
R

homogeneous linear wave equation with the initial dates φ and ψ , you will get
the solution u = uf + uhom for the problem (1).
We will show that
Z Z Z tZ x+c(t−s)
1 1
uf (x, t) = f= f (y, s)dyds, (2)
2c 4 2c 0 x−c(t−s)

where 4 is the domain of dependence of (x, t) (characteristic triangle).

2
 The formula illustrates the eect of a force f on u(x, t) is obtained by simply
integrating f over the past history of the point (x, t) back to the initial time
t = 0. This is another example of the causality principle.
This is a well-posed problem.
• Existence is from the explicit formula.
• The uniqueness can be deduced from the Energy identity see Lec 5, section
2.
• The stability is from: suppose u1 is the solution with data (φ1 , ψ1 , f1 ) and
u2 is the solution with data (φ2 , ψ2 , f2 ). Then the dierence u = u1 − u2
is given by the following formula due to linearity:
1
u(x, t) = [φ1 (x + ct) − φ2 (x + ct) + φ1 (x − ct) − φ2 (x − ct)]
2
Z x+ct
1
+ (ψ1 − ψ2 )
2c x−ct
Z Z
1
+ (f1 − f2 ). (3)
2c 4

Dene the uniform norm

kwk = max |w(x)|
−∞<x<∞

and
kwkT = max |w(x, t)|
−∞<x<∞,0≤t≤T

where T is xed.
So from the formula (1), we have the estimate
T2
ku1 − u2 kT ≤ kφ1 − φ2 k + T kψ1 − ψ2 k + kf1 − f2 kT .
2
If kφ1 − φ2 k, kψ1 − ψ2 k and kf1 − f2 kT are small, then ku1 − u2 kT is also
small. This proved the stability.
Now we start to prove the formula (2).
Proof. Method of Characteristic Coordinates. We intoduce coordinates ξ =
x+ct
2c , η = ct−x
2c such that

utt − c2 uxx = uξη = f (cξ − cη, ξ + η).

Integrate along η then along ξ

Z ξ Z η
u = f (cξ 0 − cη 0 , ξ 0 + η 0 )dη 0 dξ 0 ,

3
 where the lower limits of integration are arbitrary. We may make a particular
choice of the lower limits to nd a particular solution in the domain 40 =
{(ξ 0 , η 0 )|ξ 0 + η 0 ≥ 0, ξ 0 ≤ ξ, η 0 ≤ η}
Z ξ Z η
u(x, t) = f (cξ 0 − cη 0 , ξ 0 + η 0 )dη 0 dξ 0
η −ξ 0
Z Z
1
? = f (x, t)dxdt.
2c 4

Try to gure out the last equality by yourself.