6
Wave Equation
Pinchover and Rubinstein, Chapter 4.
We consider the homogeneous wave equation in one-dimension,
utt − c2 uxx = 0,
−∞ ≤ a < x < b ≤ ∞, t > 0
(6.1)
To find the general solution of (6.1), we can proceed as follows.
We introduce the new coordinates which transform (6.1) into its canonical
form. Let
ξ = x + ct, η = x − ct
an set w(ξ, η) = u(x(ξ, η), t(ξ, η)). Applying the chain rule to u(x, y) =
w(ξ(x, t), η(x, t)), we find that
ξx = 1 = ηx
and
ξt = c = −ηt
and
ut = wξ ξt + wη ηt = c(wξ − ηξ )
ux = wξ ξx + wη ηx = wξ + wη
utt = c2 (wξξ − 2wξη + wηη )
uxx = wξξ + 2wξη + wηη .
Substituting into (6.1), we get
0 = utt − c2 uxx = −4c2 wξη = 0
so that wRξη = 0. Since (wξ )η = 0, it follows that wξ = f (ξ) from which we
get w = f (ξ) dξ + G(η). Hence, the general solution of wξη = 0 is given
by
w(ξ, η) = F (ξ) + G(η),
where F and G are two arbitrary C 2 -function.
w(ξ(x, t), η(x, t)), we find that
Then, since u(x, y) =
u(x, y) = F (x + ct) + G(x − ct).
(6.2)
Conversely, if F and G are of class C 2 , then u defined by u(x, t) = F (x +
ct) + G(x − ct) is a classical solution of (6.1). T The families of lines
x − ct = constant
and
42
x + ct = constant,
on (x, t) plane are called the characteristics of the wave equation. These
are straight lines with slopes ±1/c. The function G(x − ct) represents a
wave moving to the right with the spped c and it is called the forward wave
while F (x + ct) is a wave moving to the left with the speed c, it is called the
backward wave. The constant c is called the wave speed.
Next we extend the notion of a solution. Consider two piecewise continuous functions F and G. Then formula (6.2) for u defines a piecewise
continuous function which is a superposition of the forward wave F (x − ct)
and the the backward wave G(x+ct). It is possible to find two sequences (Fn )
and (Gn ) of smooth functions such that Fn (x)) → F (x) and Gn (x) → G(x)
ate each point x, and, in addition, (Fn ) and (Gn ) converge to F and G,
respectively, on any closed and bounded interval not containing points of
discontinuity. The functions un (x, t) = Fn (x − ct) + Gn (x + ct) are classical
solutions of (6.1), however, the limit function u(x, t) = F (x−ct)+g(x+ct) is
not a classical solution. We call u defined by (6.2) in which F and/or G are
piecewise continuous functions a generalized solution of the wave equation.
Next assume that for a given time t0 > 0 the solution u of (6.1) is smooth
except at the point (x0 , t0 ). By the formula (6.2), either F is not smooth at
x0 − ct0 and/or G is not smooth at x0 + ct0 . If F is not smooth at x0 − ct0 ,
then F is not smooth along the characteristic x − ct = x0 − ct0 and if G
is not smooth at x0 + ct0 , then it is not smooth along the characteristic
x + ct = x0 + ct0 . This means that the singularities (points were u is not
smooth) are traveling along characteristics. This is a typical phenomena for
hyperbolic equations in contrast to parabolic equations.
6.1
The Cauchy problem and the d’Alambert’s formula
The initial value problem (Cauchy problem) for the 1-dimenisonal wave
equation is given by
utt − c2 uxx = 0,
(x, t) ∈ R × (0, ∞)
u(x, 0) = φ(x)
(6.3)
ut (x, 0) = ψ(x).
The general solution of the wave equation is given by
u(x, t) = F (x + ct) + G(x − ct)
(6.4)
where F and G are two arbitrary functions of class C 2 . Hence at t = 0,
F (x) + G(x) = φ(x).
43
(6.5)
Differentiating (6.4) with respect to t, we get
ut (x, t) = cF ′ (x + ct) − cG′ (x − ct)
so that at t = 0, we have
cF ′ (x) − cG′ (x) = ψ
which after integrating gives
F (x) − G(x) =
1
c
Z
x
ψ(τ ) dτ + C
(6.6)
0
where C = F (0) − G(0). Solving the system of two linear equation (6.5) and
(6.6). we find that
Z
1
1 x
C
F (x) = φ(x) +
(6.7)
ψ(τ ) dτ +
2
c 0
2
Z
1 x
C
1
(6.8)
ψ(τ ) dτ − .
G(x) = φ(x) −
2
c 0
2
Consequently, in view of (6.4),
u(x, t) =
1
φ(x + ct) + φ(x − ct)
+
2
2c
Z
x+ct
ψ(τ ) dτ.
(6.9)
x−ct
This is so called the d’Alambert’s formula for the solution of the Cauchy
problem (6.1).
Theorem 6.1. Let T > 0 be fixed. The Cauchy problem (6.1) in the domain
R × [0, T ] is well-posed for φ ∈ C 2 and φ ∈ C 1 .
Proof. The existence and the uniqueness of the solution of (6.1) follows from
the d’Alambert’s formula. Moreover, if f ∈ C 2 and g ∈ C 1 , then u is of
class C 2 on R × (0, ∞) and of class C 1 on R × [0, ∞). So, u given by he
d’Alambert’s formula is the classical solution of (6.1). Next we have to show
the stability of the Cauchy problem. That is, we have to show that small
changes in the initial conditions lead to small change in the solution. So, let
u1 and u2 be two solutions of (6.1) with initial conditions given by (φ1 , ψ1 )
and (φ2 , ψ2 ). If
|φ1 (x) − φ2 (x)| < δ
and
44
|ψ1 (x) − ψ2 (x)| < δ
for all x ∈ R, the for (x, t) ∈ R × [0, T ],
φ1 (x + ct) − φ2 (x + ct) φ1 (x − ct) − φ2 (x − ct) +
|u1 (x, t) − u2 (x, t)| ≤ 2
2
Z x+ct
1
+
|ψ1 (τ ) − ψ2 (τ )| dτ.
2c x−ct
1
1
≤ (δ + δ) + (2ct)δ = (1 + t)δ ≤ (1 + T )δ.
2
2c
Now, given ε > 0, take δ > 0 so that 0 < δ < ε/(1 + T ). Then
|u1 (x, t) − u2 (x, t)| ≤ (1 + T )δ < ε
for all (x, t) ∈ R × [0, T ].
Example 6.2. Solve the following Cauchy problem
utt − uxx = 0,
(x, t) ∈ R × (0, ∞)
u(x, 0) = cos x
ut (x, 0) = x.
By (6.11),
cos(x + t) + cos(x − t) 1
+
u(x, t) =
2
2
= cos x cos t + xt
Z
x+t
y dy
x−t
We have used the formula cos(a + b) = cos a cos b − sin a sin b.
6.2
Domain of dependence and region of influence
Fix a point (x0 , t0 ) and consider two points characteristics
x − ct = x0 − ct0
and
x + ct = x0 + ct0
passing through the point (x0 , t0 ). The triangle formed by those lines and
the interval [x0 − ct0 , x0 + ct0 ] is called the characteristic triangle and will
be denoted in the following by △ = △x0 t0 .
According to the d’Alambert’s formula
Z
ϕ(x0 + ct0 ) + ϕ(x0 − ct0 )
1 x0 +ct0
u(x0 , t0 ) =
+
ψ(y) dy.
2
2c x0 −ct0
45
t
(x0 , t0 )
△x0 t0
x
(x0 − ct0 , 0)
(x0 , 0)
(x0 + ct0 , 0)
Figure 1: The characteristic triangle △x0 t0
We see that the value u(x0 , t0 ) is determined by the values of ϕ at the base of
the triangle and by the values of ψ along the base. In other words, u(x0 , t0 )
depends only on the initial data given on the interval [x0 −ct0 , x0 +ct0 ]. This
interval is called the domain of dependence of u at (x0 , t0 ). If we change the
initial conditions outside of this interval, the value of u at (x0 , t0 ) will not
change. In particular, if the initial condition on this interval are smooth,
then the solution u stays smooth in the characteristics triangle.
The dual question is which are the points (x, t) which are influenced by
the initial conditions on a fixed interval [a, b]? The set of such points is called
the region of influence of the interval [a, b]. From the previous discussion it
follows that the points of [a, b] influence value of u at the point (x0 , t0 ) if
and only if [x0 − ct0 , x0 + ct0 ] ∩ [a, b] 6= ∅. Thus, the initial conditions along
[a, b] influence those points (x, t) in the (x, t)-plane which satisfy
x − ct ≤ b
and
x + ct ≥ a.
These points form a truncated cone with base [a, b] and the edges given by
x + ct = a and x − ct = b. In particular, if the initial data ϕ ≡ 0 and ψ ≡ 0
outside of [a, b], then the solution is identically equal to 0 to the left of the
line x + ct = a and to the right of the line x − ct = b.
46
t
x + ct = a
x − ct = b
a
b
x
Figure 2: Region of inflence of the interval [a, b]
6.3
The Cauchy problem for the nonhomogeneous wave equation
We now consider the following Cauchy Problem
utt − c2 uxx = f (x, t),
(x, t) ∈ R × (0, ∞)
u(x, 0) = ϕ(x)
(6.10)
ut (x, 0) = ψ(x).
Theorem 6.3. Let ϕ ∈ C 2 , ψ ∈ C 1 , and f ∈ C 1 . Then the Cauchy problem
(6.10) has a unique classical solution u which is given by
Z
Z
1 x0 +ct
1
ϕ(x + ct) + ϕ(x − ct)
+
f (y, s) dyds,
ψ(y) dy +
u(x, t) =
2
2c x−ct
2c △xt
(6.11)
where △xt at the point (x, t).
The last integral is equal to
Z tZ
Z
f (y, s) dyds =
0
△xt
x+c(t−s)
f (y, s) dyds.
x−c(t−s)
Proof. Using the Green’s Theorem Assume that u is a solution of (6.10)
and let (x0 , t0 ) be a fix point in the (x, t)-plane and let △ be the characteristic
triangle with the vertex at (x0 , t0 ). Integrating the equation over △, we get
Z
Z
(utt − c2 uxx dxdt.
f (y, s) dyds =
△
△
47
Now, the Green, s theorem tells that given two C 1 functions P and Q,
Z
Z
P dt + Qdx.
(Px − Qt ) dxdt =
∂△
△
t
(x0 , t0 )
L3
L2
△x0 t0
x
(x0 − ct0 , 0)
(x0 + ct0 , 0)
L1
Figure 3:
We use the Green’s theorem with P = −c2 ux and Q = −ut . Then
denoting by L1 , L2 , and L3 the sides of △ we get that
Z
Z
(−c2 ux dt − ut ) dx.
(Px − Qt ) dxdt =
L1 ∪L2 ∪L3
△
• On L1 , dt = 0 so that
Z
Z
2
−c ux dt − ut dx = −
x0 +ct0
ut (y, 0) dy = −
x0 −ct0
L1
Z
x0 +ct0
ψ(y) dy
x0 −ct0
• On L2 , we have dx = −cdt and thus
Z
Z
(cux dx) + cut dt)
−c2 ux dt − ut dx =
L2
L2
= c[u(x0 , t0 ) − ϕ(x0 + ct0 )]
• On L3 , dx = cdt and so,
Z
Z
Z
2
(−cux dx − cut dt) = −c
−c ux dt − ut dx =
L3
L2
= c[u(x0 , t0 ) − ϕ(x0 − ct0 )].
48
du
L2
Putting all of those expressions together we find out that
Z
Z
f (y, s) dyds = 2cu(x0 , t0 ) − c[ϕ(x0 + ct0 ) + ϕ(x0 − ct0 )] −
△
x0 +ct0
ψ(y) dy
x0 −ct0
which gives the required form
ϕ(x0 + ct0 ) + ϕ(x0 − ct0 ) 1
u(x0 , t0 ) =
+
2
2c
Z
x0 +ct0
x0 −ct0
1
ψ(y) dy+
2c
Z
f (y, s) dyds.
△
It remains to show that the above formula is indeed a solution of (6.10). In
view of the superposition principle, we have only check that
Z
1
v(x, t) =
f (y, s) dyds
2c △
is a solution of the following Cauchy problem,
utt − c2 uxx = f (x, t)
(x, t) ∈ R × (0, ∞)
(6.12)
u(x, 0) = 0
ut (x, t) = 0
First observe that
1
v(x, t) =
2c
Z
1
f (y, s) dyds =
2c
△
Z tZ
0
x+c(t−s)
f (y, s)dy ds
x−c(t−s)
from which we obtain v(x, 0) = 0. Next we take derivatives of v. To do this
we use the formula
Z b(t)
Z
d b(t)
Gt (τ, t) dτ.
G(τ, t)dτ = G(b(t), t)b′ (t) − G(a(t), t)a′ (t) +
dt a(t)
a(t)
Thus,
1
vt (x, t) =
2
Z
t
[f (x + c(t − s), s) + f (x − c(t − s), s)]ds
0
showing that vt (x, 0) = 0. Differentiating again with respect to t,
c
vtt (x, t) =
2
Z
t
[fx (x + c(t − s), s) − fx (x + c(t − s), s)] ds.
0
49
The derivatives with respect to x are given by
Z
1 t
vx (x, t) =
[f (x + c(t − s), s) − f (x − c(t − s), s)] ds
2c 0
Z
1 t
[fx (x + c(t − s), s) − fx (x − c(t − s), s)] ds.
vxx (x, t) =
2c 0
from which we get that vtt − c2 vxx = f (x, t) as claimed.
Theorem 6.4. Let T > 0 be fixed. The Cauchy problem (6.10) on R × [0, T ]
for f ∈ C 1 , ϕ ∈ C 2 , and ψ ∈ C 1 .
Corollary 6.5. Assume that ϕ and ψ are even function f (·, t) is even for
every t ≥ 0.If u is the solution of the Cauchy problem (6.10), then u(·, t) is
even function for every t ≥ 0. If ϕ and ψ are odd (periodic with period L)
and f (·, t) is odd (periodic with period L) for every t ≥ 0, then u(·, t) is odd
(period with period L) for all t ≥ 0.
Example 6.6. Solve the following Cauchy problem
utt − 9uxx = ex − e−x ,
(x, t) ∈ R × (0, ∞)
u(x, 0) = x
ut (x, 0) = sin x.
We use ex − e−x = 2 sinh x and the formula cosh(a + b) = cosh a cosh b +
sinh a sinh b. Then by (6.11),
Z
(x + 3t) + (x − 3t) 1 x+3t
+
sin y dy
u(x, t) =
2
6 x−3t
Z Z
1 t x+3(t−s) y
+
(e − e−y )dyds
6 0 x−3(t−s)
Z
1
1 t y
x+3(t−s)
= x − cos y|x+3t
+
(e + e−y )|x−3(t−s) ds
x−3t
6
6 0
1
= x + [cos(x − 3t) + cos(x + 3t)]
6
Z
1 t
x+3(t−s)
cosh s|x−3(t−s) ds
+
3 0
1
2
2
= x + sin x sin 3t − sinh x + sinh x cosh 3t.
3
9
9
50
Here is another way of deriving the formula for the solution of (6.10)
Operator Method. We start with the motivation.Consider the first-order
inhomogeneous ODE
u′ (t) + au(t) = f (t)
(6.13)
u(0) = ϕ
Multiplying the equations by eat we get, using the fundamental theorem of
calculus,
eat u′ (t) + aeat u(t) = (eat u(t))′ = eat f (t)
so that after integrating over [0, t] we get
at
a0
e u(t) − e u(0) =
Z
t
eas f (s) ds.
0
So,
−at
u(t) = e
−at
ϕ+e
Z
t
as
−at
e f (s) ds = e
ϕ+
Z
t
e−a(t−s) f (s) ds.
0
0
Introducing the solution operator
S(t)ϕ = e−at ϕ,
we note that the solution of the homogoeneous Cauchy problem (f=0) is
given by u(t) = S(t)ϕ while the solution of the nonhomogoeneous Cauchy
problem (6.13) is given by
u(t) = S(t)ϕ +
Z
t
S(t − s)ϕ ds.
0
Now consider the Cauchy problem for the second order O.D.E.,
u′′ (t) + a2 u(t) = f (t)
(6.14)
u(0) = ϕ
′
u (0) = ψ
Introducing v = a1 u′ we can write (6.15) as a system of two equations,
u′ = av
1
v ′ = −au + f
a
51
or in the matrix form as
′ 0 a
u
u
0
+
·
= 1
−a 0
v
v
af
Setting
u
U=
,
v
0 a
u
A=
·
,
−a 0
v
0
F = 1 ,
af
ϕ
Φ=
,
ψ
the (6.15) can be written as
U ′ + AU = F
U (0) = Φ.
P
n n
Denoting by eAt = ∞
n=1 A t /n! and multiplying both sides of the equation
AT
by e
the equation becomes
(eAt U )′ = eAt F (t)
which after integration over [0, t] gives
−At
U (t) = e
Φ+
Z
t
e−A(t−s) F (s) ds.
0
Introducing the solution operator S(t)Ψ = e−At Ψ we note as before that the
solution of the homogeneous Cuachy problem (F = 0) is given by U (t) =
S(t)Φ while the solution of the inhomogeneous Cuachy problem is given by
U (t) = S(t)Φ +
Z
t
S(t − s)F (s) ds.
0
Next we consider the inhomogeneous Cauchy problem (6.10). Introducing v = ut , it can be written as the system
ut − v = 0
vt − c2 uxx = f
or in the matrix form as
u
0
−1
u
0
+
·
=
.
v t
−c2 ∂x2
0
v
f
52
(6.15)
Now setting
u
U=
,
v
0 −1
A=
,
−c2
0
0
F =
,
f
ϕ
Φ=
,
ψ
(6.10) can be written as
U ′ + AU = F
U (0) = Φ.
Using the previous cases a hint, introduce the solution operator S(t) so that
U (x, t) = S(t)Φ is a solution of the homogeneous Cauchy Problem, (6.10)
can be written as
U ′ + AU = 0
U (0) = Φ.
In view of the d’Alambert’s formula
ϕ(x + ct) + ϕ(x − ct)
1
u(x, t) =
+
2
2c
Z
x+tc
ψ(y) dy,
x−ct
U has to be of the form

ϕ(x+ct)+ϕ(x−ct)
2
+

U (x, t) = 
′
′ (x−ct)
c ϕ (x+ct)−ϕ
+
2
1
2c
R x+tc
x−ct
ψ(y)
ψ(x+ct)+ψ(x−ct)
2



Thus, we may define the solution operator S(t) by


R x+tc
ϕ(x+ct)+ϕ(x−ct)
1
ψ(y)
+
2
2c x−ct
ϕ


S(t)Φ = S(t)
=

ψ
′
′
(x−ct)
ψ(x+ct)+ψ(x−ct)
c ϕ (x+ct)−ϕ
+
2
2
So, one could expect that
Z
U (x, t) = S(t)Φ +
t
S(t − s)F (s) ds
0
is the solution of the inhomogeneous Cauchy problem. The first component
of U (x, t) is given by
ϕ(x + ct) + ϕ(x − ct) 1
u(x, t) =
+
2
2c
Z
x+tc
x−ct
1
ψ(y)+
2c
Z tZ
0
which is exactly the same formula as obtained before.
53
x+c(t−s)
f (y, s) dyds
x−c(t−s)
6.4
The wave equation on a half-line
Consider the Dirichlet problem on the half-line,
utt − c2 uxx = 0,
x > 0, t > 0
x≥0
u(x, 0) = φ(x),
x≥0
ut (x, 0) = ψ(x),
u(0, t) = 0,
t > 0.
with φ of class C 2 on [0, ∞) and ψ of class C 1 on [0, ∞) satisfying the
compatibility condition φ(0) = φ′′ (0) = ψ(0) = 0.
In order to find the solution we extend φ and ψ to all of R by odd
reflection. Namely, we set
(
(
ψ(x)
x≥0
φ(x)
x
≥
0
e
e
and
ψ(x)
=
φ(x)
=
−ψ(−x) x ≥ 0.
−φ(−x) x ≥ 0
Then φe and ψe are odd functions on R. The d’Alambert’s formula defines
the solution u
e on R × (0, ∞) of the initial value problem,
u
ett − c2 u
exx = 0,
e
u
e(x, 0) = φ(x),
e
u
et (x, 0) = ψ(x),
x > 0, t > 0
x≥0
x≥0
Then we set u(x, t) = u
e(x, t). Clearly, u solves the equation on (0, ∞) ×
e
(0, ∞). Moreover, for x > 0, u(x, 0) = u
e(x, 0) = φ(x)
= φ(x) and similarly
e
ut (x, 0) = u
et = ψ(x) = ψ(x). It remains to show that u(0, t) = 0. For this
it suffices to show that u
e(·, t) is an odd function for all t > 0. Indeed, set
v(x, t) = −e
u(−x, t). Then
vt (x, t) = −e
ut (−x, t)
vx (x, t) = u
ex (−x, t)
vxx (x, t) = −e
uxx (−x, t)
so that
vtt (x, t) = −e
utt (−x, t),
vtt − c2 vxx = − u
ett − c2 u
exx = 0
e
e
and v(x, 0) = −e
u(−x, t) = −φ(−x)
= φ(x)
and vt (x, 0) = −e
ut (−x, 0) =
e
e
−ψ(−x) = ψ(x) . By Proposition 6.3, v(x, t) = u(x, t), i.e., u(x, t) =
−u(−x, t) as claimed.
54
So what is the solution u in terms of φ and ψ. Recall that in view of the
d’Alambert’s formula
Z
e + ct) + φ(x
e − ct)
φ(x
1 x+ct e
u
e(x, t) =
+
ψ(y) dy.
2
2c x−ct
e ± ct) = φ(x + ±ct) and
Then, if t > 0 and x − ct >, we have φ(x
Z
φ(x + ct) + φ(x − ct)
1
u(x, t) =
+
2
2c
x+ct
ψ(y) dy.
x−ct
e − ct) = −φ(−(x − ct)) = −φ(ct − x) and
If t > 0 and x − ct < 0, then φ(x
e
ψ(y)
= −ψ(−y) for y < 0. So,
Z
x+ct
x−ct
e
ψ(y)
dy =
Z
0
e
ψ(y)
dy +
x−ct
Z ct−x
Z
x+ct
0
ψ(y) dy +
=−
0
Z ct+x
ψ(y) dy
=
Z
e
ψ(y)
dy
x+ct
ψ(y) dy
0
ct−x
and
1
φ(x + ct) − φ(ct − x)
+
u(x, t) =
2
2c
55
Z
ct+x
ψ(y) dy.
ct−x