May 26, 2020 APM346 – Week 4 Justin Ko
1 Properties of the Wave Equation on R
Recall that the solution to
2
utt − c uxx = f (x, t)
x ∈ R, t > 0,
u|t=0 = g(x) x ∈ R, (1)
ut |t=0 = h(x) x ∈ R.
is given by d’Alembert’s formula
x+ct Z tZ x+c(t−s)
g(x + ct) + g(x − ct)
Z
1 1
u(x, t) = + h(s) ds + f (y, s) dyds. (2)
2 2c x−ct 2c 0 x−c(t−s)
We can use this formula to derive several nice properties satisfied by the solutions to (1).
1.1 Causality
When f ≡ 0, d’Alembert’s formula implies that waves propagate at speed c.
1. Domain of Influence: If f ≡ 0, g(x0 ) and h(x0 ) only affect the solution to (1) on the sector
4+ (x0 , 0) = {(x, t) : x0 − ct ≤ x ≤ x0 + ct}.
2. Domain of Dependence: The solution to (1) at (x0 , t0 ) only depends on f, g, h on the triangle
4(x0 , t0 ) = {(x, t) ∈ R × R+ : x0 − c(t0 − t) ≤ x ≤ x0 + c(t0 − t)}.
t t
x = x0 + c(t0 − t) x = x0 − c(t0 − t)
∆+ (x0 , t0 )
x = x0 − ct x = x0 + ct ∆
x x
x0
Proposition 1 (Causality )
Suppose f (x, t) ≡ 0.
1. If g(x) and h(x) vanish for |x| > R, then u(x, t) = 0 for x, t such that |x| > R + ct.
2. If g(x) and h(x) vanish for |x − x0 | < R then u(x, t) = 0 for all x, t ∈ 4(x0 , Rc ).
Proof. When f ≡ 0, the wave equation is determined by the characteristic lines x − ct = C1 and
x + ct = C2 . The domain of influence and domain of dependence are simply the regions separated by
the characteristic lines through the points x0 and t0 . Part (2) follows by revisting the derivation of
the first two terms in d’Alembert’s formula and part (1) is the “inverse” restatement of part (2).
Remark 1. In higher dimensions, the domains of influence and dependence are cones.
Remark 2. The propagation speed of the wave equation can also be proved independently of D’Alembert’s
formula using a local energy argument.
Page 1 of 11
, May 26, 2020 APM346 – Week 4 Justin Ko
1.2 Well-Posed
If we assume some regularity properties on the initial conditions, then d’Alembert’s formula implies
there exists a C 2 (R) solution to (1). In fact, with a bit of work, we can show that the solutions are
also unique and stable.
Proposition 2
If g ∈ C 2 (R), h ∈ C(R), and f is integrable then the IVP (1) is well-posed.
Proof. We check the three conditions that define a well-posed solution.
Existence: The existence of a solution with continuous second derivative follows immediately from
d’Alembert’s formula (2).
Uniqueness: We use an energy argument to prove uniqueness.
Difference of Solutions: Suppose u1 and u2 are C 2 (R) solutions to (1). By linearity, v = u1 − u2
solves
2
vtt − c vxx = 0,
x ∈ R, t > 0
v|t=0 = 0 x∈R (3)
vt |t=0 = 0 x ∈ R.
To prove uniqueness, it suffices to show that v ≡ 0 on the domain of the solution.
Show the Energy is Zero: We consider the energy of the solution v to (3),
1 ∞ 2
Z
v + c2 vx2 dx.
E(t) =
2 −∞ t
Since the initial conditions are 0, the energy is finite so we differentiate under the integral sign with
respect to t to see
Z ∞
E 0 (t) = vt vtt + c2 vx vxt dx
−∞
Z ∞
2
vtt − c2 vxx = 0
=c vt vxx + vx vxt dx
−∞
Z ∞
2
=c (vx vt )x dx (vx vt )x = vxx vt + vx vxt
−∞
= lim vx (y, t)vt (y, t) − vx (−y, t)vt (−y, t)
y→∞
= 0,
since the wave equation has finite propagation speed Proposition 1 (the initial conditions vanish outside
for |x| > 1), so limy→±∞ vx (y, t) = 0 for every fixed t. Since E 0 (t) = 0, we can conclude that E(t) is
constant by the mean value theorem. Furthermore, the initial conditions imply
1 ∞ 2
Z
vt (x, 0) + c2 vx2 (x, 0) dx = 0
E(0) =
2 −∞
because vt (x, 0) = 0 and v(x, 0) = 0 =⇒ vx (x, 0) = 0. Combined with the fact E(t) is constant, we
have
E(t) = 0 for all t.
Page 2 of 11
1 Properties of the Wave Equation on R
Recall that the solution to
2
utt − c uxx = f (x, t)
x ∈ R, t > 0,
u|t=0 = g(x) x ∈ R, (1)
ut |t=0 = h(x) x ∈ R.
is given by d’Alembert’s formula
x+ct Z tZ x+c(t−s)
g(x + ct) + g(x − ct)
Z
1 1
u(x, t) = + h(s) ds + f (y, s) dyds. (2)
2 2c x−ct 2c 0 x−c(t−s)
We can use this formula to derive several nice properties satisfied by the solutions to (1).
1.1 Causality
When f ≡ 0, d’Alembert’s formula implies that waves propagate at speed c.
1. Domain of Influence: If f ≡ 0, g(x0 ) and h(x0 ) only affect the solution to (1) on the sector
4+ (x0 , 0) = {(x, t) : x0 − ct ≤ x ≤ x0 + ct}.
2. Domain of Dependence: The solution to (1) at (x0 , t0 ) only depends on f, g, h on the triangle
4(x0 , t0 ) = {(x, t) ∈ R × R+ : x0 − c(t0 − t) ≤ x ≤ x0 + c(t0 − t)}.
t t
x = x0 + c(t0 − t) x = x0 − c(t0 − t)
∆+ (x0 , t0 )
x = x0 − ct x = x0 + ct ∆
x x
x0
Proposition 1 (Causality )
Suppose f (x, t) ≡ 0.
1. If g(x) and h(x) vanish for |x| > R, then u(x, t) = 0 for x, t such that |x| > R + ct.
2. If g(x) and h(x) vanish for |x − x0 | < R then u(x, t) = 0 for all x, t ∈ 4(x0 , Rc ).
Proof. When f ≡ 0, the wave equation is determined by the characteristic lines x − ct = C1 and
x + ct = C2 . The domain of influence and domain of dependence are simply the regions separated by
the characteristic lines through the points x0 and t0 . Part (2) follows by revisting the derivation of
the first two terms in d’Alembert’s formula and part (1) is the “inverse” restatement of part (2).
Remark 1. In higher dimensions, the domains of influence and dependence are cones.
Remark 2. The propagation speed of the wave equation can also be proved independently of D’Alembert’s
formula using a local energy argument.
Page 1 of 11
, May 26, 2020 APM346 – Week 4 Justin Ko
1.2 Well-Posed
If we assume some regularity properties on the initial conditions, then d’Alembert’s formula implies
there exists a C 2 (R) solution to (1). In fact, with a bit of work, we can show that the solutions are
also unique and stable.
Proposition 2
If g ∈ C 2 (R), h ∈ C(R), and f is integrable then the IVP (1) is well-posed.
Proof. We check the three conditions that define a well-posed solution.
Existence: The existence of a solution with continuous second derivative follows immediately from
d’Alembert’s formula (2).
Uniqueness: We use an energy argument to prove uniqueness.
Difference of Solutions: Suppose u1 and u2 are C 2 (R) solutions to (1). By linearity, v = u1 − u2
solves
2
vtt − c vxx = 0,
x ∈ R, t > 0
v|t=0 = 0 x∈R (3)
vt |t=0 = 0 x ∈ R.
To prove uniqueness, it suffices to show that v ≡ 0 on the domain of the solution.
Show the Energy is Zero: We consider the energy of the solution v to (3),
1 ∞ 2
Z
v + c2 vx2 dx.
E(t) =
2 −∞ t
Since the initial conditions are 0, the energy is finite so we differentiate under the integral sign with
respect to t to see
Z ∞
E 0 (t) = vt vtt + c2 vx vxt dx
−∞
Z ∞
2
vtt − c2 vxx = 0
=c vt vxx + vx vxt dx
−∞
Z ∞
2
=c (vx vt )x dx (vx vt )x = vxx vt + vx vxt
−∞
= lim vx (y, t)vt (y, t) − vx (−y, t)vt (−y, t)
y→∞
= 0,
since the wave equation has finite propagation speed Proposition 1 (the initial conditions vanish outside
for |x| > 1), so limy→±∞ vx (y, t) = 0 for every fixed t. Since E 0 (t) = 0, we can conclude that E(t) is
constant by the mean value theorem. Furthermore, the initial conditions imply
1 ∞ 2
Z
vt (x, 0) + c2 vx2 (x, 0) dx = 0
E(0) =
2 −∞
because vt (x, 0) = 0 and v(x, 0) = 0 =⇒ vx (x, 0) = 0. Combined with the fact E(t) is constant, we
have
E(t) = 0 for all t.
Page 2 of 11
