The Wave Equation
1. Acoustic Waves
We consider a general conservation statement for a region U Ð R 3 containing a fluid which
3 =V
3 Ýx3, tÞ . Let _ = _Ýx3, tÞ denote the
is flowing through the domain U with velocity field V
3
3
(scalar) fluid density at Ýx3, tÞ, and let F = FÝx3, tÞ denote the fluid flux at Ýx3, tÞ. Then
3 Ýx3, tÞ = _Ýx3, tÞ V
3 Ýx3, tÞ describes the direction and speed of the fluid flow at Ýx3, tÞ. Proceeding
F
as we have in previous examples, we obtain the following equation asserting that the fluid
mass is conserved during the flow
3 Ýx3, tÞ
/ t _Ýx3, tÞ + div _Ýx3, tÞV
=0
for all Ýx3, tÞ 5 U × Ý0, TÞ
This is another special case of the equation / t u + div F ? s = 0 we have seen before, this
3 = _V
3 , and s = 0. This is one equation for four unknowns,
time with u = _, F
3 . An additional equation is obtained from the assertion that
_ and the 3 components of V
momentum is conserved during the flow. This conservation statement, that the time rate of
change of momentum equals the sum of the applied forces, can be expressed in terms of
the state variables by the vector equation,
3 Ýx3, tÞ dx = ? X p 3
d/dt X _Ýx3, tÞ V
n dSÝx3Þ,
B
/B
where B denotes an arbitrary ball in U and p = pÝx3, tÞ denotes the scalar pressure field in the
fluid. Then by an integral identity that is related to the divergence theorem,
X/B p 3n dSÝx3Þ = XB 4p dx,
we arrive at
d
dt
3 Ýx3, tÞ
_Ýx3, tÞ V
3 Ýx3, tÞ + V
3 6 4 _Ýx3, tÞ V
3 Ýx3, tÞ
= / t _Ýx3, tÞ V
= ?4p.
This adds three equations to the system but also adds a new unknown, p, so the unknowns
now consist of _, V 1 , V 2 , V 3 , and p. To complete the system we add the so called equation of
state, a constitutive equation which asserts that
p = fÝ_Þ,
where f denotes a fluid dependent function relating pressure to density.
In one dimension, this system has the form
1
/ t _Ýx, tÞ + / x Ý_Ýx, tÞVÝx, tÞÞ = 0,
(1.1)
/ t Ý_VÞ + V/ x Ý_ VÞ + / x p = 0,
p = fÝ_Þ
This is a system of nonlinear first order equations. The solution of this system is, in general,
quite difficult, even in one dimension. Therefore we consider the simpler problem of
modeling the propagation of acoustic waves in the fluid. Acoustic waves are small
amplitude perturbations in the density field in a quiescent fluid. That is,
VÝx, tÞ = 0 + fÝx, tÞ
where
_Ýx, tÞ = _ 0 Ý1 + dÝx, tÞÞ
|f| << 1
where _ 0 = const
and
|d| << 1
pÝx, tÞ = fÝ_Þ = fÝ_ 0 Þ + f v Ý_ 0 ÞÝ_ ? _ 0 Þ = p 0 + _ 0 f v Ý_ 0 Þ dÝx, tÞ.
These equations express that the unperturbed velocity and density fields are equal to zero
and _ 0 = const, respectively, while the perturbations in these fields, f and d, are much less
than 1 in magnitude. The perturbation in the pressure field is determined from the density
perturbation and the equation of state.
Substituting these expressions into the equations Ý1.1Þ and neglecting any terms that
involve products of perturbations, leads to
/ t dÝx, tÞ + / x fÝx, tÞ = 0
/ t fÝx, tÞ + f v Ý_ 0 Þ / x dÝx, tÞ = 0.
Then it is easy to show that both dÝx, tÞ and fÝx, tÞ satisfy the same second order equation,
/ tt uÝx, tÞ ? a 2 / xx uÝx, tÞ = 0,
Ý1.2Þ
where f v Ý_ 0 Þ = a 2 in this case. Equation (1.2) is referred to as the wave equation due to
the fact that its solutions exhibit wave-like behavior.
2. Electromagnetic Waves
In a region U Ð R 3 with no charges present and no currents, the electric force field
3 =E
3 Ýx3, tÞ, and the magnetic force field, B
3 =B
3 Ýx3, tÞ, satisfy Maxwell’s equations
E
2
3 Ýx3, tÞ,
iÞ 0 = div E
3 Ýx3, tÞ
3 Ýx3, tÞ + 1 / t B
iiÞ 0 = curl E
C
3 Ýx3, tÞ
3 Ýx3, tÞ ? 1 / t E
iiiÞ 0 = curl B
C
3 Ýx3, tÞ
ivÞ 0 = divB
Apply the curl operator to ii) and recall that
3 ? div 4 E
3 ,
3 Ýx3, tÞ = grad div E
curl curl E
to obtain
3 ? div 4 E
3 +
grad div E
1
C
3 = 0.
/ t curl B
Then it follows from i) and iii) that
and similarly,
3 Ýx3, tÞ +
? 42E
1
C2
3 Ýx3, tÞ = 3
/ tt E
0,
3 Ýx3, tÞ +
? 42B
1
C2
3 Ýx3, tÞ = 3
/ tt B
0.
Evidently, each component of the electric and magnetic fields satisfy the 3-dimensional
wave equation. As we have seen in the past, very different physical phenomena can be
modelled by the same mathematical description.
3. Some Problems for the Wave Equation
We can add various auxiliary conditions to the wave equation to try to get a well posed
problem.
(a) Pure Initial Value Problem (Cauchy Problem)
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = FÝx3, tÞ
3
x 5 Rn, 0 < t < T
uÝx3, 0Þ = fÝx3Þ,
3
x5R
/ t uÝx3, 0Þ = gÝx3Þ,
3
x 5 Rn.
(b) Initial-Boundary Value Problem
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = FÝx3, tÞ
3
x 5 Rn, 0 < t < T
uÝx3, 0Þ = fÝx3Þ,
3
x 5 Rn
/ t uÝx3, 0Þ = gÝx3Þ,
3
x 5 Rn.
BCßuÝx3, tÞà = hÝx3, tÞ,
(3.1)
n
3
x 5 /U, 0 < t < T
(3.2)
#
3
where BCßuÝx3, tÞà denotes one of the types of boundary conditions we have discussed.
(c) Dirichlet Problem for the Wave Equation
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = 0
3
x 5 U Ð Rn, 0 < t < T
uÝx3, 0Þ = fÝx3Þ,
3
x5UÐR
uÝx3, TÞ = gÝx3Þ,
3
x 5 U Ð Rn
uÝx3, tÞ = 0,
3
x 5 /U, 0 < t < T
(3.3)
n
Problems (a) and (b) are examples of well posed problems for the wave equation, while (c)
is not well posed.
Problem 23 Consider the Dirichlet problem for the 1-dimensional wave equation,
/ tt uÝx, tÞ ? / xx uÝx, tÞ = 0
uÝx, 0Þ = 0
uÝx, TÞ = 0,
uÝ0, tÞ = uÝL, tÞ = 0,
0 < x < L, 0 < t < T,
0 < x < L,
0 < x < L,
0 < t < T.
Show that if T/L = m/n, for integers m and n, then the problem has infinitely many solutions
u mn Ýx, tÞ = C sinÝn^x/LÞ sinÝm^t/TÞ ,
but if T/L is irrational, then the trivial solution is the only solution. This shows that the
solution does not depend continuously on the data, which in this case is the shape
(dimensions) of the domain, 0 < x < L, 0 < t < T .
4. The One Dimensional Wave Equation
We will begin by considering the simplest case, the 1-dimensional wave equation. Recall
that for arbitrary differentiable functions of one variable, F and G,
Ý/ t ? a / x ÞFÝx + atÞ = 0,
and
Ý/ t + a / x ÞGÝx ? atÞ = 0.
This implies
Ý/ tt ? a 2 / xx ÞßFÝx + atÞ + GÝx ? atÞà = Ý/ t + a / x ÞÝ/ t ? a / x ÞßFÝx + atÞ + GÝx ? atÞà = 0
i.e.,
uÝx, tÞ = FÝx + atÞ + GÝx ? atÞ solves Ý/ tt ? a 2 / xx Þ uÝx, tÞ = 0
Now consider the initial value problem Ý3.1Þ for n = 1,
/ tt uÝx, tÞ ? a 2 / xx uÝx, tÞ = 0
uÝx, 0Þ = fÝxÞ,
x 5 R, 0 < t < T,
x 5 R,
Ý4.1Þ
4
/ t uÝx, 0Þ = gÝxÞ,
Then
x 5 R.
uÝx, tÞ = FÝx + atÞ + GÝx ? atÞ
uÝx, 0Þ = FÝxÞ + GÝxÞ = fÝxÞ
/ t uÝx, 0Þ = F v ÝxÞ ÝaÞ + G v ÝxÞ Ý?aÞ = gÝxÞ.
FÝxÞ + GÝxÞ = fÝxÞ,
a ßF v ÝxÞ ? G v ÝxÞ à = gÝxÞ or
This leads to
and
1
2
FÝxÞ =
x
fÝxÞ + 1 X gÝsÞ ds,
0
2a
Then
uÝx, tÞ =
=
1
2
x
FÝxÞ ? GÝxÞ = 1a X gÝsÞ ds
0
GÝxÞ =
1
2
x
fÝxÞ ? 1 X gÝsÞ ds..
0
2a
x+at
x?at
ßfÝx + atÞ + fÝx ? atÞà+ 1 X gÝsÞ ds ? 1 X gÝsÞ ds
0
0
2a
2a
1
2
x+at
ßfÝx + atÞ + fÝx ? atÞà + 1 X gÝsÞ ds
x?at
2a
Ý4.2Þ
This is referred to as the D’Alembert solution of the wave equation. Note that for a fixed
Ýx 0 , t 0 Þ, the solution value uÝx 0 , t 0 Þ depends only on the data values in the interval
DÝx 0 , t 0 Þ = ßx 0 ? at 0 , x 0 + at 0 à . This interval is referred to as the domain of dependence for
the point Ýx 0 , t 0 Þ. Data values outside this interval have no influence on the value of u at
the point Ýx 0 , t 0 Þ.
2
time axis
x + t = 2, x ? t = ?2
-2
0
x axis
2
Domain of Dependence
The values of the data at a point x 0 have an effect on the value of the solution at a point
Ýx, tÞ, t > 0, only when the point Ýx, tÞ lies inside the wedge shaped region between the lines
x + at = x 0 and x ? at = x 0 . This region is referred to as the domain of influence for the
point x 0 . For a fixed time t 1 > 0, the data values at x 0 influence the solution values uÝx, t 1 Þ
for all x in the interval ßx 0 ? at 1 , x 0 + at 1 à. At the later time t 2 > t 1 , the data values at x 0
influence the solution values uÝx, t 2 Þ for all x in the larger interval ßx 0 ? at 2 , x 0 + at 2 à. In the
amount of time from t 1 to t 2 , the interval of influence expands by the amount aÝt 2 ? t 1 Þ,
which is to say the domain of influence is expanding at the rate a.
5
1
time axis
-2
x + t = ?1, x ? t = ?1
0
-1
x axis
Domain of Influence
If f 5 C 2 ÝRÞ, and g 5 C 1 ÝRÞ, then the D’Alembert solution, Ý4.2Þ, solves the Cauchy
problem Ý4.1Þ. In addition, it is evident from Ý4.2Þ that if u 1 , u 2 are solutions corresponding
to data pairs áf 1 , g 1 â and áf 2 , g 2 â, then
max | u 1 ? u 2 | ² max
| f 1 ? f 2 | + t 0 max
|g 1 ? g 2 |
x
x
Ýx,tÞ
Ý4.3Þ
where for each t 0 > 0, the solution max is taken over the interval
áÝx, tÞ : x 1 ? at ² x ² x 2 + at, 0 ² t ² t 0 â the data max is taken over the interval ßx 1 , x 2 à. Note
that the estimate (4.3) implies both the uniqueness of the solution and the continuous
dependence on the data.
5. Energy Integrals
Suppose uÝx, tÞ is a solution for Ý4.1Þ for smooth data fÝxÞ, gÝxÞ and assume that the data
vanishes for |x| > C where C denotes some positive but finite constant. Then for each finite
positive t it follows from the discussion of the domain of influence, that uÝx, tÞ vanishes for
|x| > C + at. Now, for t ³ 0, let EÝtÞ be given by
EÝtÞ =
1
2
XR ß / t uÝx, tÞ 2 + a 2 / x uÝx, tÞ 2 à dx
Ý5.1Þ
Then
0 ² EÝ0Þ =
1
2
XR ß gÝxÞ 2 + a 2 f v ÝxÞ 2 à dx =
1
2
C
X ?C ß gÝxÞ 2 + a 2 f v ÝxÞ 2 à dx < K.
Moreover,
E v ÝtÞ = X ß / t uÝx, tÞ / tt uÝx, tÞ + a 2 / x uÝx, tÞ / tx uÝx, tÞà dx
R
But
XR / x uÝx, tÞ / tx uÝx, tÞ dx = / x uÝx, tÞ / t uÝx, tÞ | K?K ? XR / xx uÝx, tÞ / t uÝx, tÞ dx
and since u vanishes for |x| > C + at,
6
E v ÝtÞ = X / t uÝx, tÞ ß / tt uÝx, tÞ ? a 2 / xx uÝx, tÞà dx = 0.
R
Then EÝtÞ is a non-negative constant. Physically, EÝtÞ is related to the total energy in the
system. In particular, / t uÝx, tÞ 2 is proportional to the kinetic energy in the system, while
/ x uÝx, tÞ 2 is proportional to the potential energy stored in the system. As the system evolves
in time,the energy simply changes from kinetic to potential and back again. This is because
the equation in (4.1) contains no term which dissipates energy. Note that if (4.1) is modified
as follows,
/ tt uÝx, tÞ ? a 2 / xx uÝx, tÞ + b 2 / t uÝx, tÞ = 0,
for x 5 R, t > 0,
then, in the same way, we find
E v ÝtÞ = X / t uÝx, tÞ ß / tt uÝx, tÞ ? a 2 / xx uÝx, tÞà dx =
R
= X / t uÝx, tÞ ß ? b 2 / t uÝx, tÞà dx ² 0.
R
.
Evidently, the additional term dissipates energy, causing E(t) to decrease as long as
/ t uÝx, tÞ is different from zero. Since E(t) is non-negative, E(t) will decrease steadily toward
zero. The added term here has the physical interpretation of being related to friction. Adding
other lower order terms to the wave equation does not produce the same dissipative effect.
For example, if uÝx, tÞ solves the equation
/ tt uÝx, tÞ ? a 2 / xx uÝx, tÞ + b 2 uÝx, tÞ = 0,
then
for x 5 R, t > 0,
E v ÝtÞ = X / t uÝx, tÞ ß / tt uÝx, tÞ ? a 2 / xx uÝx, tÞà dx =
R
= X / t uÝx, tÞ ß ? b 2 uÝx, tÞà dx = ? 12 b 2
R
This implies
E D ÝtÞ = EÝtÞ +
1
2
d
dt
XR uÝx, tÞ 2 dx.
b 2 X uÝx, tÞ 2 dx = constant.
R
Note that the energy integral provides an additional method for proving uniqueness of the
solution to the Cauchy problem for the wave equation. To use the arguments presented
here, however, we have to suppose that the data f and g have compact support.
6. The Inhomogeneous Wave Equation
Consider the problem with homogeneous initial conditions but inhomogeneous equation,
/ tt vÝx, tÞ ? a 2 / xx vÝx, tÞ = FÝx, tÞ,
x 5 R, t > 0,
(6.1)
vÝx, 0Þ = / t vÝx, 0Þ = 0, x 5 R,
7
where FÝx, tÞ is given. Then, for a fixed s 5 R, let wÝx, t; sÞ be given by
wÝx, t; sÞ = 1
2a
x+at
X x?at FÝz, sÞ dz
(6.2)
It is easy to check that wÝx, t; sÞ solves (just note that wÝx, t; sÞ = PÝx + atÞ ? PÝx ? atÞ),
/ tt wÝx, tÞ ? a 2 / xx wÝx, tÞ = 0,
wÝx, 0Þ = 0, / t wÝx, 0Þ = FÝx, sÞ,
x 5 R, t > 0,
x 5 R,
and
vÝx, tÞ =
t
X 0 wÝx, t ? s; sÞ ds =
1
2a
t
x+aÝt?sÞ
X 0 X x?aÝt?sÞ FÝz, sÞ dz ds.
(6.3)
To see that v(x,t) must be given by (6.3), differentiate (6.3) with respect to t,
t
/ t vÝx, tÞ = wÝx, 0; tÞ + X / t wÝx, t ? s; sÞ ds
0
t
t
0
0
/ tt vÝx, tÞ = / t wÝx, 0; tÞ + X / tt wÝx, t ? s; sÞ ds = FÝx, tÞ + X a 2 / xx wÝx, t ? s; sÞ ds
t
= FÝx, tÞ + a 2 / xx X wÝx, t ? s; sÞ ds = FÝx, tÞ + a 2 / xx vÝx, tÞ.
0
It follows now that the solution of
/ tt uÝx, tÞ ? a 2 / xx uÝx, tÞ = FÝx, tÞ,
x 5 R, t > 0
uÝx, 0Þ = fÝxÞ,
x 5 R,
/ t uÝx, 0Þ = gÝxÞ,
x 5 R,
(6.4)
is given by
uÝx, tÞ = 1 ßfÝx + atÞ + fÝx ? atÞà + 1
2a
2
x+at
X x?at gÝsÞ ds +
1
2a
t
x+aÝt?sÞ
X 0 X x?aÝt?sÞ FÝz, sÞ dz ds
(6.5)
7. A Fundamental Solution for the Wave Equation
We can show that a fundamental solution for the 1-dimensional wave equation is given by
uÝx, tÞ = 1 HÝat ? |x|Þ.
2a
This means, that uÝx, tÞ solves
L W uÝx, tÞ = Ý/ tt ? a 2 / xx ÞuÝx, tÞ = NÝx, tÞ,
? K < x < K, t > 0.
8
Here a > 0 and HÝ6Þ denotes the Heaviside step function that is zero when its argument is
negative and is one when the argument is positive. Since the delta notation is only formal at
this point, we restate the condition defining the fundamental solution as follows,
K
K
X ?K X 0 dÝx, tÞL W uÝx, tÞdtdx = dÝ0, 0Þ
Ý7.1Þ
for all smooth functions d which are zero for |x|, t sufficiently large. To show that u does, in
fact, satisfy (7.1), we first use integration by parts to write
K
K
K
K
X ?K X 0 dÝx, tÞL W uÝx, tÞdtdx = X ?K X 0 uÝx, tÞ L W dÝx, tÞdtdx.
Here we used the fact that dÝx, tÞ is zero for |x|, t sufficiently large to eliminate all the
boundary terms in the integration by parts. Now we note that
HÝat ? |x|Þ =
1 if ?K < x < 0, t > ? ax
1 if 0 < x < K, t > ax
0 if otherwise
in order to write
K
K
K K
X ?K X 0 uÝx, tÞ L W dÝx, tÞdtdx = 1 X 0 X x Ý/ t ? a / x ÞÝ/ t + a / x ÞdÝx, tÞdtdx
2a
a
0
K
+ 1 X X x Ý/ t + a / x ÞÝ/ t ? a / x ÞdÝx, tÞdtdx.
2a ?K ? a
In the first of these integrals, let t = b + ax so as t varies over ax , K , b varies over ß0, Kà
and / t = / b .In the second integral, let t = b ? ax so as t varies over ? ax , K , b varies over
ß0, Kà and / t = / b . Then
K
K
K K
X ?K X 0 uÝx, tÞ L W dÝx, tÞdtdx = 1 X 0 X 0 Ý/ b ? a / x ÞÝ/ b + a / x Þd x, b + ax dbdx
2a
0
K
+ 1 X X Ý/ b + a / x ÞÝ/ b ? a / x Þd x, b ? ax dbdx.
?K
0
2a
But
so
d d x, b +
dx
d d x, b ?
dx
x
a
x
a
= / x d + 1a / b d,
= /xd ? 1
a / b d,
K
K
K K
X ?K X 0 uÝx, tÞ L W dÝx, tÞdtdx = 1 X 0 X 0 d Ý/ b ? a / x Þd x, b + ax dxdb
2
dx
9
K 0
d Ý/ b + a / x Þd x, b ? x dxdb,
?1 X X
a
0
?K dx
2
K
K
= ? 1 X Ý/ b ? a/ x ÞdÝ0, bÞ db ? 1 X Ý/ b + a / x ÞdÝ0, bÞdb,
2 0
2 0
K
= ? X / b dÝ0, bÞ db = dÝ0, 0Þ.
0
Using this fundamental solution, we can proceed as in section 9 of the previous chapter and
construct Green’s functions for the wave operator. In particular, we can show that (6.5) can
be interpreted in terms of the Green’s function for the 1-d wave equation.
8. The Wave Equation in R n
In R n , the wave equation has the form
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = 0
for x 5 R n , t > 0.
We will consider several simple solutions for this equation.
Plane Wave Solutions
c63
x ? at = C, is the
Let 3
c denote a fixed unit vector in R n . Then for each constant, C, 3
n
equation of a plane in R having 3
c as its normal vector. As t varies, the plane moves in the
direction of the normal vector with speed equal to a. Now for F a smooth function of one
variable, let uÝx3, tÞ = FÝc3 6 3
x ? atÞ and note that
¾
x ? atÞ a 2 ? a 2 F ”Ýc3 6 3
x ? atÞ 3
c63
c = 0.
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = F ”Ýc3 6 3
Then uÝx3, tÞ = FÝc3 6 3
x ? atÞ is called a plane wave solution to the wave equation. If the wave
operator is changed to the related operator, / tt uÝx3, tÞ ? / f A /uÝx3, tÞ = 0, where A denotes a
symmetric n by n matrix, then uÝx3, tÞ = FÝc3 6 3
x ? atÞ solves the new equation if
x ? atÞ a 2 ? F ”Ýc3 6 3
x ? atÞ 3
cfA 3
c = 0;
/ tt uÝx3, tÞ ? / f A /uÝx3, tÞ = F ”Ýc3 6 3
cfA 3
c = 0. It is evident that plane wave solutions for this equation propagate
i.e., if a 2 ? 3
with a different speed in every direction 3
c.
Note that uÝx3, tÞ = FÝc3 6 3
x ? atÞ + GÝc3 6 3
x + atÞ is not the general solution to the wave
n
equation in R since it is clear that any unit vector 3
c 5 R n produces a solution to the PDE,
1
whereas in R the only unit vector is c = 1.
Spherical Wave Solutions
In spherical coordinates,
¾
10
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = / tt uÝx3, tÞ ? r 1?n / r Ýr n?1 / r uÞ ? r ?2 C g u = 0
and for spherically symmetric waves, u = uÝr, tÞ this reduces to
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = / tt uÝx3, tÞ ? r 1?n / r Ýr n?1 / r uÞ = 0.
For n = 3, this is,
/ tt uÝx3, tÞ ? r ?2 / r Ýr 2 / r uÞ = / tt uÝr, tÞ ? / rr u ? 2r / r u = 0.
Let wÝr, tÞ = r uÝr, tÞ, so that / r w = r / r u + u, and / rr w = r / rr u + 2 / r u. Then if uÝr, tÞ is any
spherically symmetric solution of the wave equation in R 3 ,
r / tt uÝr, tÞ ? r / rr u ? 2 / r u = / tt w ? / rr w = 0.
But this implies wÝr, tÞ = FÝr ? tÞ + GÝr + tÞ for smooth functions F and G and then
uÝr, tÞ = 1r FÝr ? tÞ + 1r GÝr + tÞ;
i.e., uÝr, tÞ is the sum of a contracting spherical wave and an expanding spherical wave. For
n = 2, the spherically symmetric wave equation has the form
/ tt uÝr, tÞ ? 4 2 uÝr, tÞ = / tt uÝr, tÞ ? r ?1 / r Ýr / r uÞ = / tt uÝr, tÞ ? / rr u ? 1r / r u = 0
and there is no simple analogue of the trick that worked for n = 3. Solutions to the wave
equation are very sensitive to changes in dimension. As another illustration of this fact, note
that in 1-dimension, the solution of the Cauchy problem
/ tt uÝx, tÞ ? / xx uÝx, tÞ = 0,
is given by
uÝx, tÞ = X
x+t 1
x?t 2
Then
/ t uÝx, tÞ =
1
2
and
/ tt uÝx, tÞ =
1
2
uÝx, 0Þ = 0,
/ t uÝx, 0Þ = gÝxÞ,
gÝsÞ ds.
ßgÝx + tÞ ? gÝx ? tÞà
ßg v Ýx + tÞ + g v Ýx ? tÞà
so that if g is continuous with a continuous derivative, then / tt uÝx, tÞ is continuous. On the
other hand, in 3-dimensions the solution of
/ tt uÝr, tÞ ? 4 2 uÝr, tÞ = 0,
11
uÝr, 0Þ = 0,
is given by
Ý1 ? r 2 Þ
/ t uÝr, 0Þ =
3/2
if r 2 < 1
= gÝrÞ,
if r 2 > 1
0
uÝr, tÞ = 1r FÝt + rÞ ? 1r FÝt ? rÞ, for F an even function.
Note that if F is even, then
uÝr, 0Þ = 1r FÝrÞ ? 1r FÝ?rÞ = 0,
and
/ t uÝr, 0Þ = 1r F v ÝrÞ ? 1r F v Ý?rÞ = 2r F v ÝrÞ.
Note that since F is even, it follows that the derivative is odd, so F v ÝrÞ ? F v Ý?rÞ = 2F v ÝrÞ.
Then the initial condition implies F v ÝrÞ = Ýr/2Þ gÝrÞ. In addition,
lim r¸0 1r FÝt + rÞ ? 1r FÝt ? rÞ
= 2F v ÝtÞ,
which leads finally to the conclusion
v
lim r¸0 uÝr, tÞ = uÝ0, tÞ = 2F ÝtÞ = t gÝtÞ =
tÝ1 ? t 2 Þ 3/2
if t 2 < 1
0
if t 2 > 1
.
Evidently, even though u, / t u, and / tt u are all continuous at t = 0 for all r ³ 0, it is still the
case that when t = 1 and r = 0 then u, and / t u, are continuous but / tt u is discontinuous.
This singularity in the second derivative is a dimension dependent phenomenon referred to
as ”focusing”.
9. Energy Integrals in R n
Let U denote a bounded open set in R n and consider the following initial boundary value
problem
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = 0
uÝx, 0Þ = / t uÝx, 0Þ = 0
uÝx, tÞ = 0
Let
EÝtÞ = X
2
U
x 5 U, t > 0,
x 5 U,
x 5 /U, t > 0.
/ t uÝx3, tÞ + | 4uÝx3, tÞ| 2 dx
for t ³ 0.
Then EÝ0Þ = 0 and
E v ÝtÞ = X 2ß/ t uÝx3, tÞ / tt uÝx3, tÞ + 4uÝx3, tÞ 6 4/ t uÝx3, tÞà dx.
U
But
XU 4uÝx3, tÞ 6 4/ t uÝx3, tÞ dx = X/U / t uÝx3, tÞ 4uÝx3, tÞ 6 3n dS ? XU 4 2 uÝx3, tÞ / t uÝx3, tÞdx
12
hence
E v ÝtÞ = X 2/ t uÝx3, tÞ ß / tt uÝx3, tÞ ? 4 2 uÝx3, tÞ à dx. + X
U
/U
/ t uÝx3, tÞ / N uÝx3, tÞ dS.
Then it follows from the wave equation and the boundary condition that E v ÝtÞ = 0. This, in
turn, implies that / t u and |4u| are both zero in Ū × ß0, KÞ, which is to say uÝx, tÞ is constant
on the same set. But uÝx, 0Þ = 0 for x 5 U, hence we conclude that the solution vanishes
identically in Ū × ß0, KÞ. This is the essential part of the proof that the solution to the IBVP for
the wave equation is unique.
The energy integral EÝtÞ can also be used to show that solutions to the n-dimensional wave
equation obey the principle of causality.
Proposition 9.1 Suppose uÝx3, tÞ satisfies
/ tt uÝx3, tÞ ? 4 2 uÝx3, tÞ = 0
x 5 R n , t > 0.
Suppose, in addition, for some fixed, but arbitrary 3
x 0 5 R n , t 0 > 0, we have
uÝx, 0Þ = / t uÝx, 0Þ = 0 for
x?3
x0 | ² t0.
|3
Then uÝx3.tÞ = 0 for all Ýx3, tÞ in the light cone
x?3
x0 | ² t0 ? t .
Ýx3, tÞ : 0 ² t ² t 0 , | 3
C=
Remark: Let UÝtÞ denote the n-ball of radius t 0 ? t with center at x 0 ; i.e.,
UÝtÞ =
3
x : |3
x?3
x0 | ² t0 ? t
Ð Rn.
Then the proposition asserts that the value of uÝx30 , t 0 Þ depends only on the values of the
data inside UÝ0Þ, and if the data is zero in this ball, then the solution is zero everywhere in
the cone that is obtained by drawing all the backward characteristic lines through Ýx30 , t 0 Þ.
This cone formed in this way is called the light cone at Ýx30 , t 0 Þ.
Proof- Let
EÝtÞ = X
UÝtÞ
/ t uÝx3, tÞ 2 + | 4uÝx3, tÞ| 2 dx
for 0 ² t ² t 0
Then
.E v ÝtÞ = X
UÝtÞ
2ß/ t uÝx3, tÞ / tt uÝx3, tÞ + 4uÝx3, tÞ 6 4/ t uÝx3, tÞà dx
?X
2
/UÝtÞ
/ t uÝx3, tÞ + | 4uÝx3, tÞ| 2 dSÝxÞ.
Here we made use of the fact that
d X
fÝxÞ dx
dr B r Ýx 0 Þ
r
= d X X
fÝr, gÞ dg dr
dr 0 /B r Ýx 0 Þ
=X
/B r Ýx 0 Þ
fÝr, gÞ dg
13
and B r Ýx30 Þ = UÝtÞ = B t 0 ?t Ýx 0 Þ. Now
XUÝtÞ 4uÝx3, tÞ 6 4/ t uÝx3, tÞ dx = X/UÝtÞ / t uÝx3, tÞ 4uÝx3, tÞ 6 3n dS ? XUÝtÞ 4 2 uÝx3, tÞ / t uÝx3, tÞdx
hence
E v ÝtÞ = X
UÝtÞ
2/ t uÝx3, tÞß / tt uÝx3, tÞ ? 4 2 uÝx3, tÞà dx
+X
2
/UÝtÞ
2/ t u / N u ? Ý/ t uÝx3, tÞ + | 4uÝx3, tÞ| 2 Þ dSÝxÞ.
Notice that
n | ² 2 |/ t u| |4u| ² / t uÝx3, tÞ 2 + | 4uÝx3, tÞ| 2
| 2/ t u / N u| = 2 | / t u 4u 6 3
from which it follows that E v ÝtÞ ² 0. Since EÝ0Þ = 0, we conclude EÝtÞ is identically zero
which implies that the solution is constant inside the light cone. Since the solution is zero on
the ”floor” of the light cone, the solution must be zero throughout the light cone.n
This is referred to as the ”causality” principle because it asserts that the effects from a
cause cannot be felt at a point before the effects have had time to reach the point, travelling
from the source to the point with the finite speed dictated by the governing partial
differential equation. Another way to say this is that the value of the solution to the wave
equation at a specified point Ýx30 , t 0 Þ is not influenced by data values that lie outside the
domain of dependence for the point Ýx30 , t 0 Þ. This domain of dependence is just the floor of
the light cone obtained by drawing all backward characteristics through Ýx30 , t 0 Þ.
10. Wave-Like Evolution
We use the term diffusion-like evolution to describe the properties of the solutions to the
heat equation. These properties include
¾
1.
2.
3.
4.
5.
instantaneous smoothing of the data
infinite speed of propagation
time irreversible
existence of a Max-min principle
monotone (non-oscillatory) evolution
The wave equation also described the time evolution of a system but the properties of
solutions to the wave equation are rather different from those of solutions to the heat
equation. Solutions of the wave equation exhibit the following properties
¾
1.
2.
3.
no smoothing of the data
finite speed of propagation
time reversible
14
no Max-min principle
oscillatory time behavior
4.
5.
Points 1 and 2 are clearly illustrated by the D’Alembert solution (4.2) for the Cauchy
problem (4.1). Consider the special case where g = 0 and f is given by
fÝxÞ =
1 ? | x|
0
if | x| ² 1
if
| x| > 1
.
Then f is a continuous but not differentiable triangular pulse centered at x = 0. In this case
the solution uÝx, tÞ has exactly the same smoothness as the data since,
uÝx, tÞ = ßfÝx + atÞ + fÝx ? atÞà/2.
In addition, at positive integer values of t, t = N, the solution is seen to be composed of two
triangular pulses, one centered at x = Na, and one centered at x = ?Na, and each of the
pulses has half the amplitude of the original pulse. Evidently, the pulse at x = 0 splits into
two halves and each half pulse has travelled a distance equal to Na in time t = N, which is
to say the pulses travel with finite speed equal to a.
Points 3,4, and 5 are illustrated by the solution for the following IBVP on a bounded interval.
/ tt uÝx, tÞ = a 2 / xx uÝx, tÞ,
uÝx, 0Þ = fÝxÞ, / t uÝx, 0Þ = 0,
uÝ0, tÞ = uÝL, tÞ = 0,
0 < x < L, 0 < t,
0 < x < L,
0 < t.
It will be shown later that the solution for this problem can be written
K
uÝx, tÞ = > n=1 f n cosÝn^at/LÞ sinÝn^x/LÞ
where f n denote the Fourier coefficients for fÝxÞ; i.e.,
L
f n = 2 X fÝxÞ sinÝn^x/LÞ dx, n = 1, 2, ...
L 0
It is known from the theory of Fourier series that the series
K
f n sinÝn^x/LÞ
> n=1
converges (in some sense) to *fÝxÞ, the odd, L-periodic extension of fÝxÞ. Then, since
cosÝn^at/LÞ sinÝn^x/LÞ =
1
2
ßsinÝn^Ýx + atÞ/LÞ + sinÝn^Ýx ? atÞ/LÞà,
we see that
15
K
uÝx, tÞ = 1 > n=1 f n ßsinÝn^Ýx + atÞ/LÞ + sinÝn^Ýx ? atÞ/LÞà
2
= 1 *fÝx + atÞ + *fÝx ? atÞ .
2
Clearly this solution is oscillatory in time and, since positive time is interchangeable with
negative time, the solution is time reversible. The energy, EÝtÞ, has been shown to be
constant in time so the solution does not die out with increasing time like the solution to the
heat equation. Finally, if fÝxÞ is a triangular pulse of height one, centered at x = L/2, then
plotting uÝx, tÞ versus x for various times shows that for t = L/a, we see uÝx, L/aÞ is a
triangular pulse centered at x = L/2, but with height equal to minus one. Clearly there can
be no max-min principle for this equation.
The behavior exhibited by this solution of the IBVP problem in 1-dimension is typical of
solutions to the wave equation in other settings. Collectively we refer to this behavior as
wave-like evolution, as opposed to diffusion-like evolution, which is the behavior exhibited
by the solutions to the heat equation. The heat equation and wave equation are prototypes
for the classes of parabolic and hyperbolic partial differential equations, respectively. The
Laplace equation is the prototype for the class of elliptic partial differential equations. We
will now explain the meaning of this terminology.
11. Classification
The Laplace, heat and wave operators in n-dimensions can be written in the following way
/fIn 3
/uÝx3Þ
4 2 uÝx3Þ = 3
where 3
/ f = ß/ 1 , ..., / n à
/fIn 3
/uÝx3, tÞ
/ t uÝx3, tÞ ? 4 2 uÝx3, tÞ = / t uÝx3, tÞ ? 3
2
f
/ tt uÝx3, tÞ ? 4 uÝx3, tÞ = / tt uÝx3, tÞ ? / I n /uÝx3, tÞ
where I n denotes the n by n identity matrix. If we do not distinguish between the space and
time variables but think of these as operators in n variables (thus the time variable becomes
x n in the heat and wave operators) the second order terms in each of these operators are all
/uÝx3Þ where the n by n matrix J n is as follows,
of the form 3
/fJn 3
Laplace J n = I n ,
heat J n =
I n?1 0
0
0
,
wave J n =
I n?1
0
0
?1
Now consider the operator defined by an arbitrary symmetric n by n matrix, A.
LßuÝx3Þà= / f A n /uÝx3Þ
Recall that since A is symmetric, it has n real (but not necessarily distinct) eigenvalues.
Then we say that the operator L is:
16
elliptic
if the eigenvalues of A are all of one sign
parabolic
if zero is an eigenvalue for A of multiplicity one,
and the remaining n ? 1 eigenvalues have one sign
hyperbolic
if A has n ? 1 eigenvalues of one sign and one eigenvalue
with the opposite sign
Note that when n = 2 this classification is exhaustive in the sense that every matrix A falls
into one of these classes. If n is greater than 2 then there are matrices which are in none of
the 3 classes and, consequently, there are partial differential operators that are neither
elliptic, parabolic nor hyperbolic. On the other hand, many of the partial differential
equations that occur in applications are one of these three types.
The reason for these names is that when n = 2, the locus of points satisfying 3
x f Ax3 = C, is
an ellipse, parabola or hyperbola, respectively, when the matrix A has the properties
assigned to each of the names.
17