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Unformatted text preview: The Wave Equation
1. Acoustic Waves
We consider a general conservation statement for a region U Ð R 3 containing a fluid which
3 3x
is flowing through the domain U with velocity field V = VÝ3, t Þ . Let _ = _Ý3, t Þ denote the
x
3 = FÝ3, t Þ denote the fluid flux at Ý3, t Þ. Then
3x
(scalar) fluid density at Ý3, t Þ, and let F
x
x
3x
FÝ3, t Þ = _Ý3, t Þ VÝ3, t Þ describes the direction and speed of the fluid flow at Ý3, t Þ. Proceeding
x 3x
x
as we have in previous examples, we obtain the following equation asserting that the fluid
mass is conserved during the flow
x
x 3x
/ t _Ý3, t Þ + div _Ý3, t ÞVÝ3, t Þ =0 for all Ý3, t Þ 5 U × Ý0, T Þ
x This is another special case of the equation / t u + div F ? s = 0 we have seen before, this
3
3
time with u = _, F = _V, and s = 0. This is one equation for four unknowns,
3
_ and the 3 components of V. An additional equation is obtained from the assertion that
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,
d/dt X _Ý3, t Þ VÝ3, t Þ dx = ? X p 3 dSÝ3Þ,
x 3x
n
x
B /B where B denotes an arbitrary ball in U and p = pÝ3, t Þ denotes the scalar pressure field in the
x
fluid. Then by an integral identity that is related to the divergence theorem, X/B p 3 dSÝ3Þ = XB 4p dx,
n
x
we arrive at
d
dt _Ý3, t Þ VÝ3, t Þ
x 3x 3
= / t _Ý3, t Þ VÝ3, t Þ + V 6 4 _Ý3, t Þ VÝ3, t Þ
x 3x
x 3x = ?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 = f Ý_ Þ,
where f denotes a fluid dependent function relating pressure to density.
In one dimension, this system has the form 1 / t _Ýx, t Þ + / x Ý_Ýx, t ÞVÝx, t ÞÞ = 0, (1.1) / t Ý_VÞ + V/ x Ý_ V Þ + / x p = 0,
p = f Ý_ 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, t Þ = 0 + fÝx, t Þ where _Ýx, t Þ = _ 0 Ý1 + dÝx, t ÞÞ f  << 1 where _ 0 = const and d  << 1 pÝx, t Þ = fÝ_ Þ = fÝ_ 0 Þ + f v Ý_ 0 ÞÝ_ ? _ 0 Þ = p 0 + _ 0 f v Ý_ 0 Þ dÝx, t Þ.
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 Ý1.1 Þ and neglecting any terms that
involve products of perturbations, leads to
/ t dÝx, t Þ + / x fÝx, t Þ = 0
/ t fÝx, t Þ + f v Ý_ 0 Þ / x dÝx, t Þ = 0.
Then it is easy to show that both dÝx, t Þ and fÝx, t Þ satisfy the same second order equation,
/ tt uÝx, t Þ ? a 2 / xx uÝx, t Þ = 0, Ý1.2 Þ 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 wavelike behavior. 2. Electromagnetic Waves
In a region U Ð R 3 with no charges present and no currents, the electric force field
3 3x
3 3x
E = EÝ3, t Þ, and the magnetic force field, B = BÝ3, t Þ, satisfy Maxwell’s equations 2 3x
iÞ 0 = div EÝ3, t Þ,
3x
3x
iiÞ 0 = curl EÝ3, t Þ + 1 / t BÝ3, t Þ
C
3x
3x
iiiÞ 0 = curl BÝ3, t Þ ? 1 / t EÝ3, t Þ
C
3x
ivÞ 0 = divBÝ3, t Þ
Apply the curl operator to ii) and recall that
3
3
3x
curl curl EÝ3, t Þ = grad div E ? div 4 E ,
to obtain 3
3
grad div E ? div 4 E + 1
C 3
/ t curl B = 0. Then it follows from i) and iii) that
3x
? 4 2 EÝ3, t Þ +
and similarly, 1
C2 3x
/ tt EÝ3, t Þ = 3,
0 3x
? 4 2 BÝ3, t Þ + 1
C2 3x
/ tt BÝ3, t Þ = 3.
0 Evidently, each component of the electric and magnetic fields satisfy the 3dimensional
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)
x
x
x
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = FÝ3, t Þ 3 5 Rn, 0 < t < T
x uÝ3, 0 Þ = fÝ3Þ,
x
x 35R
x / t uÝ3, 0 Þ = gÝ3Þ,
x
x (3.1) 3 5 Rn.
x (b) InitialBoundary Value Problem
x
x
x
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = FÝ3, t Þ n 3 5 Rn, 0 < t < T
x uÝ3, 0 Þ = fÝ3Þ,
x
x 3 5 Rn
x / t uÝ3, 0 Þ = gÝ3Þ,
x
x (3.2) 3 5 Rn.
x BCßuÝ3, t Þà = hÝ3, t Þ,
x
x 3 5 /U, 0 < t < T
x # 3 where BCßuÝ3, t Þà denotes one of the types of boundary conditions we have discussed.
x
(c) Dirichlet Problem for the Wave Equation
x
x
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 3 5 U Ð Rn, 0 < t < T
x uÝ3, 0 Þ = fÝ3Þ,
x
x 35UÐR
x uÝ3, T Þ = gÝ3Þ,
x
x 3 5 U Ð Rn
x uÝ3, t Þ = 0,
x (3.3) 3 5 /U, 0 < t < T
x 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 1dimensional wave equation,
/ tt uÝx, t Þ ? / xx uÝx, t Þ = 0
uÝ x , 0 Þ = 0
uÝx, T Þ = 0,
uÝ0, t Þ = uÝL, t Þ = 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, t Þ = C sinÝn^x/L Þ sinÝm^t/T Þ ,
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 1dimensional wave equation. Recall
that for arbitrary differentiable functions of one variable, F and G,
Ý/ t ? a / x ÞFÝx + at Þ = 0, and Ý/ t + a / x ÞGÝx ? at Þ = 0. This implies
Ý/ tt ? a 2 / xx ÞßFÝx + at Þ + GÝx ? at Þà = Ý/ t + a / x ÞÝ/ t ? a / x ÞßFÝx + at Þ + GÝx ? at Þà = 0
i.e., uÝx, t Þ = FÝx + at Þ + GÝx ? at Þ solves Ý/ tt ? a 2 / xx Þ uÝx, t Þ = 0 Now consider the initial value problem Ý3.1 Þ for n = 1,
/ tt uÝx, t Þ ? a 2 / xx uÝx, t Þ = 0
uÝx, 0 Þ = fÝxÞ, x 5 R, 0 < t < T,
x 5 R, Ý4.1 Þ 4 / t uÝx, 0 Þ = gÝx Þ,
Then x 5 R. uÝx, t Þ = FÝx + at Þ + GÝx ? at Þ
uÝx, 0 Þ = FÝx Þ + GÝx Þ = fÝx Þ
/ t uÝx, 0 Þ = F v Ýx Þ Ýa Þ + G v Ýx Þ Ý?a Þ = gÝx Þ.
FÝx Þ + GÝx Þ = fÝx Þ,
a ßF v Ýx Þ ? G v Ýx Þ à = gÝx Þ or This leads to
and
1
2 F Ýx Þ = x
fÝx Þ + 1 X gÝs Þ ds,
0
2a Then
uÝx, t Þ =
= 1
2 x
FÝx Þ ? GÝx Þ = 1 X gÝs Þ ds
a0 GÝx Þ = 1
2 x
fÝx Þ ? 1 X gÝs Þ ds..
0
2a x+at
x?at
ßfÝx + at Þ + fÝx ? at Þà+ 1 X gÝs Þ ds ? 1 X gÝs Þ ds
0
0
2a
2a
1
2 x+at
ßfÝx + at Þ + fÝx ? at Þà + 1 X gÝs Þ ds
x?at
2a Ý4.2 Þ 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, t Þ, t > 0, only when the point Ýx, t Þ 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 ÝR Þ, and g 5 C 1 ÝR Þ, then the D’Alembert solution, Ý4.2 Þ, solves the Cauchy
problem Ý4.1 Þ. In addition, it is evident from Ý4.2 Þ 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,t Þ Ý4.3 Þ where for each t 0 > 0, the solution max is taken over the interval
áÝx, t Þ : 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, t Þ is a solution for Ý4.1 Þ for smooth data fÝx Þ, gÝx Þ 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, t Þ vanishes for
x  > C + at. Now, for t ³ 0, let EÝt Þ be given by
E Ýt Þ = 1
2 XR ß / t uÝx, t Þ 2 + a 2 / x uÝx, t Þ 2 à dx Ý5.1 Þ Then
0 ² E Ý0 Þ = 1
2 XR ß gÝx Þ 2 + a 2 f v Ýx Þ 2 à dx = 1
2 C X ?C ß gÝx Þ 2 + a 2 f v Ýx Þ 2 à dx < K. Moreover,
E v Ýt Þ = X ß / t uÝx, t Þ / tt uÝx, t Þ + a 2 / x uÝx, t Þ / tx uÝx, t Þà dx
R But XR / x uÝx, t Þ / tx uÝx, t Þ dx = / x uÝx, t Þ / t uÝx, t Þ  K ? XR / xx uÝx, t Þ / t uÝx, t Þ dx
?K and since u vanishes for x  > C + at, 6 E v Ýt Þ = X / t uÝx, t Þ ß / tt uÝx, t Þ ? a 2 / xx uÝx, t Þà dx = 0.
R
Then EÝt Þ is a nonnegative constant. Physically, EÝt Þ is related to the total energy in the
system. In particular, / t uÝx, t Þ 2 is proportional to the kinetic energy in the system, while
/ x uÝx, t Þ 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, t Þ ? a 2 / xx uÝx, t Þ + b 2 / t uÝx, t Þ = 0, for x 5 R, t > 0, then, in the same way, we find
E v Ýt Þ = X / t uÝx, t Þ ß / tt uÝx, t Þ ? a 2 / xx uÝx, t Þà dx =
R = X / t uÝx, t Þ ß ? b 2 / t uÝx, t Þà dx ² 0.
R .
Evidently, the additional term dissipates energy, causing E(t) to decrease as long as
/ t uÝx, t Þ is different from zero. Since E(t) is nonnegative, 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, t Þ solves the equation
/ tt uÝx, t Þ ? a 2 / xx uÝx, t Þ + b 2 uÝx, t Þ = 0,
then for x 5 R, t > 0, E v Ýt Þ = X / t uÝx, t Þ ß / tt uÝx, t Þ ? a 2 / xx uÝx, t Þà dx =
R = X / t uÝx, t Þ ß ? b 2 uÝx, t Þà dx = ? 1 b 2
2
R This implies E D Ýt Þ = E Ýt Þ + 1
2 d
dt XR uÝx, t Þ 2 dx. b 2 X uÝx, t Þ 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, t Þ ? a 2 / xx vÝx, t Þ = FÝx, t Þ, x 5 R, t > 0, (6.1) vÝx, 0 Þ = / t vÝx, 0 Þ = 0, x 5 R, 7 where FÝx, t Þ is given. Then, for a fixed s 5 R, let wÝx, t; s Þ be given by
wÝx, t; s Þ = 1
2a x+at X x?at FÝz, s Þ dz (6.2) It is easy to check that wÝx, t; s Þ solves (just note that wÝx, t; s Þ = PÝx + atÞ ? PÝx ? atÞ),
/ tt wÝx, t Þ ? a 2 / xx wÝx, t Þ = 0,
wÝx, 0 Þ = 0, / t wÝx, 0 Þ = FÝx, s Þ, x 5 R, t > 0,
x 5 R, and
v Ýx , t Þ = t X 0 wÝx, t ? s; s Þ ds = 1
2a t x+aÝt?sÞ X 0 X x?aÝt?sÞ FÝz, s Þ 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, t Þ = wÝx, 0; t Þ + X / t wÝx, t ? s; s Þ ds
0
t t / tt vÝx, t Þ = / t wÝx, 0; t Þ + X / tt wÝx, t ? s; s Þ ds = FÝx, t Þ + X a 2 / xx wÝx, t ? s; s Þ ds
0
0
t = FÝx, t Þ + a 2 / xx X wÝx, t ? s; s Þ ds = FÝx, t Þ + a 2 / xx vÝx, t Þ.
0
It follows now that the solution of
/ tt uÝx, t Þ ? a 2 / xx uÝx, t Þ = FÝx, t Þ, x 5 R, t > 0 uÝx, 0 Þ = fÝx Þ, x 5 R, / t uÝx, 0 Þ = gÝx Þ, (6.4) x 5 R, is given by
uÝx, t Þ = 1 ßfÝx + at Þ + fÝx ? at Þà + 1
2a
2 x+at X x?at gÝs Þ ds + 1
2a t x+aÝt?sÞ X 0 X x?aÝt?sÞ FÝz, s Þ dz ds (6.5) 7. A Fundamental Solution for the Wave Equation
We can show that a fundamental solution for the 1dimensional wave equation is given by
uÝx, t Þ = 1 HÝat ? x  Þ.
2a
This means, that uÝx, t Þ solves
L W uÝx, t Þ = Ý/ tt ? a 2 / xx ÞuÝx, t Þ = NÝx, t Þ, ? 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, t ÞL W uÝx, t Þdtdx = dÝ0, 0 Þ Ý7.1 Þ 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, t ÞL W uÝx, t Þdtdx = X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx.
Here we used the fact that dÝx, t Þ 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 = x
1 if ?K < x < 0, t > ? a
x
1 if 0 < x < K, t > a
0 if otherwise in order to write
K
K
KK
X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx = 1 X 0 X x Ý/ t ? a / x ÞÝ/ t + a / x ÞdÝx, t Þdtdx
2a a 0
K
+ 1 X X x Ý/ t + a / x ÞÝ/ t ? a / x ÞdÝx, t Þdtdx.
2a ?K ? a x
x
In the first of these integrals, let t = b + a so as t varies over a , K , b varies over ß0, K à
x
x
and / t = / b .In the second integral, let t = b ? a so as t varies over ? a , K , b varies over
ß0, K à and / t = / b . Then
K
K
KK
x
X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx = 1 X 0 X 0 Ý/ b ? a / x ÞÝ/ b + a / x Þd x, b + a dbdx 2a 0
K
x
+ 1 X X Ý/ b + a / x ÞÝ/ b ? a / x Þd x, b ? a dbdx.
?K 0
2a But
so d d x, b +
dx
d d x, b ?
dx x
a
x
a = / x d + 1 / b d,
a
= / x d ? 1 / b d,
a K
K
KK
x
X ?K X 0 uÝx, t Þ L W dÝx, t Þdtdx = 1 X 0 X 0 d Ý/ b ? a / x Þd x, b + a dxdb 2 dx 9 K0
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, b Þ db ? 1 X Ý/ b + a / x ÞdÝ0, b Þdb,
20
20
K = ? X / b dÝ0, b Þ db = dÝ0, 0 Þ.
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 1d wave equation. 8. The Wave Equation in R n
In R n , the wave equation has the form
x
x
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 for x 5 R n , t > 0. We will consider several simple solutions for this equation.
Plane Wave Solutions
cx
Let 3 denote a fixed unit vector in R n . Then for each constant, C, 3 6 3 ? at = C, is the
c
equation of a plane in R n having 3 as its normal vector. As t varies, the plane moves in the
c
direction of the normal vector with speed equal to a. Now for F a smooth function of one
variable, let uÝ3, t Þ = FÝ3 6 3 ? at Þ and note that
x
cx
¾ x
x
cx
cx
cc
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = F ”Ý3 6 3 ? at Þ a 2 ? a 2 F ”Ý3 6 3 ? at Þ 3 6 3 = 0.
Then uÝ3, t Þ = FÝ3 6 3 ? at Þ is called a plane wave solution to the wave equation. If the wave
x
cx
x
x
operator is changed to the related operator, / tt uÝ3, t Þ ? / f A /uÝ3, t Þ = 0, where A denotes a
symmetric n by n matrix, then uÝ3, t Þ = FÝ3 6 3 ? at Þ solves the new equation if
x
cx
x
x
cx
cx
cc
/ tt uÝ3, t Þ ? / f A /uÝ3, t Þ = F ”Ý3 6 3 ? at Þ a 2 ? F ”Ý3 6 3 ? at Þ 3f A 3 = 0;
cc
i.e., if a 2 ? 3f A 3 = 0. It is evident that plane wave solutions for this equation propagate
with a different speed in every direction 3.
c
Note that uÝ3, t Þ = FÝ3 6 3 ? at Þ + GÝ3 6 3 + at Þ is not the general solution to the wave
x
cx
cx
equation in R n since it is clear that any unit vector 3 5 R n produces a solution to the PDE,
c
whereas in R 1 the only unit vector is c = 1.
Spherical Wave Solutions
In spherical coordinates,
¾ 10 / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = / tt uÝ3, t Þ ? r 1?n / r Ýr n?1 / r u Þ ? r ?2 C g u = 0
x
x
x
and for spherically symmetric waves, u = uÝr, t Þ this reduces to
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = / tt uÝ3, t Þ ? r 1?n / r Ýr n?1 / r u Þ = 0.
x
x
x
For n = 3, this is,
/ tt uÝ3, t Þ ? r ?2 / r Ýr 2 / r u Þ = / tt uÝr, t Þ ? / rr u ? 2 / r u = 0.
x
r
Let wÝr, t Þ = r uÝr, t Þ, so that / r w = r / r u + u, and / rr w = r / rr u + 2 / r u. Then if uÝr, t Þ is any
spherically symmetric solution of the wave equation in R 3 ,
r / tt uÝr, t Þ ? r / rr u ? 2 / r u = / tt w ? / rr w = 0.
But this implies wÝr, t Þ = FÝr ? t Þ + GÝr + t Þ for smooth functions F and G and then
uÝr, t Þ = 1 FÝr ? t Þ + 1 GÝr + t Þ;
r
r
i.e., uÝr, t Þ 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, t Þ ? 4 2 uÝr, t Þ = / tt uÝr, t Þ ? r ?1 / r Ýr / r u Þ = / tt uÝr, t Þ ? / rr u ? 1 / r u = 0
r
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 1dimension, the solution of the Cauchy problem
/ tt uÝx, t Þ ? / xx uÝx, t Þ = 0,
is given by uÝx, t Þ = X x+t 1
x?t 2 Then / t uÝx, t Þ = 1
2 and / tt uÝx, t Þ = 1
2 uÝx, 0 Þ = 0, / t uÝx, 0 Þ = gÝx Þ, gÝs Þ ds. ßgÝx + t Þ ? gÝx ? t Þà ßg v Ýx + t Þ + g v Ýx ? t Þà so that if g is continuous with a continuous derivative, then / tt uÝx, t Þ is continuous. On the
other hand, in 3dimensions the solution of
/ tt uÝr, t Þ ? 4 2 uÝr, t Þ = 0, 11 uÝr, 0 Þ = 0, is given by Ý1 ? r 2 Þ / t uÝr, 0 Þ = 3/2 if r 2 < 1 = gÝr Þ, if r 2 > 1 0 uÝr, t Þ = 1 FÝt + r Þ ? 1 FÝt ? r Þ, for F an even function.
r
r Note that if F is even, then
uÝr, 0 Þ = 1 FÝr Þ ? 1 FÝ?r Þ = 0,
r
r and / t uÝr, 0 Þ = 1 F v Ýr Þ ? 1 F v Ý?r Þ = 2 F v Ýr Þ.
r
r
r Note that since F is even, it follows that the derivative is odd, so F v Ýr Þ ? F v Ý?r Þ = 2F v Ýr Þ.
Then the initial condition implies F v Ýr Þ = Ýr/2Þ gÝr Þ. In addition,
lim r¸0 1 FÝt + r Þ ? 1 FÝt ? r Þ
r
r = 2F v Ýt Þ, which leads finally to the conclusion v lim r¸0 uÝr, t Þ = uÝ0, t Þ = 2F Ýt Þ = t gÝt Þ = 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 uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0
x
x
uÝx, 0 Þ = / t uÝx, 0 Þ = 0
uÝ x , t Þ = 0
Let E Ýt Þ = X 2 U x 5 U, t > 0,
x 5 U,
x 5 /U, t > 0. / t uÝ3, t Þ +  4uÝ3, t Þ 2 dx
x
x for t ³ 0. Then EÝ0 Þ = 0 and
x
x
x
x
E v ÝtÞ = X 2ß/ t uÝ3, t Þ / tt uÝ3, t Þ + 4uÝ3, t Þ 6 4/ t uÝ3, t Þà dx.
U But XU 4uÝ3, t Þ 6 4/ t uÝ3, t Þ dx = X/U / t uÝ3, t Þ 4uÝ3, t Þ 6 3 dS ? XU 4 2 uÝ3, t Þ / t uÝ3, t Þdx
x
x
x
x
n
x
x
12 hence E v ÝtÞ = X 2/ t uÝ3, t Þ ß / tt uÝ3, t Þ ? 4 2 uÝ3, t Þ à dx. + X
x
x
x
U /U / t uÝ3, t Þ / N uÝ3, t Þ dS.
x
x 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, t Þ is constant
on the same set. But uÝx, 0 Þ = 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 EÝt Þ can also be used to show that solutions to the ndimensional wave
equation obey the principle of causality.
Proposition 9.1 Suppose uÝ3, t Þ satisfies
x
x
x
/ tt uÝ3, t Þ ? 4 2 uÝ3, t Þ = 0 x 5 R n , t > 0. Suppose, in addition, for some fixed, but arbitrary 30 5 R n , t 0 > 0, we have
x
uÝx, 0 Þ = / t uÝx, 0 Þ = 0 for xx
 3 ? 30  ² t 0 . Then uÝ3.t Þ = 0 for all Ý3, t Þ in the light cone
x
x
x
xx
Ý3, t Þ : 0 ² t ² t 0 ,  3 ? 30  ² t 0 ? t . C= Remark: Let UÝt Þ denote the nball of radius t 0 ? t with center at x 0 ; i.e.,
UÝt Þ = 3 :  3 ? 30  ² t 0 ? t
x
xx Ð Rn. Then the proposition asserts that the value of uÝ30 , t 0 Þ depends only on the values of the
x
data inside UÝ0 Þ, 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 Ý30 , t 0 Þ.
x
30 , t 0 Þ.
This cone formed in this way is called the light cone at Ýx
Proof Let E Ýt Þ = X UÝt Þ / t uÝ3, t Þ 2 +  4uÝ3, t Þ 2 dx
x
x for 0 ² t ² t 0 Then
.E v ÝtÞ = X UÝt Þ 2ß/ t uÝ3, t Þ / tt uÝ3, t Þ + 4uÝ3, t Þ 6 4/ t uÝ3, t Þà dx
x
x
x
x
?X 2 /UÝt Þ / t uÝ3, t Þ +  4uÝ3, t Þ 2 dSÝx Þ.
x
x Here we made use of the fact that
dX
fÝx Þ dx
dr B r Ýx 0 Þ r
= d XX
fÝr, g Þ dg dr
dr 0 /B r Ýx 0 Þ =X /B r Ýx 0 Þ fÝr, g Þ dg 13 x
and B r Ý30 Þ = UÝt Þ = B t 0 ?t Ýx 0 Þ. Now XUÝt Þ 4uÝ3, t Þ 6 4/ t uÝ3, t Þ dx = X/UÝt Þ / t uÝ3, t Þ 4uÝ3, t Þ 6 3 dS ? XUÝt Þ 4 2 uÝ3, t Þ / t uÝ3, t Þdx
x
x
x
x
n
x
x
hence
E v Ýt Þ = X UÝt Þ 2/ t uÝ3, t Þß / tt uÝ3, t Þ ? 4 2 uÝ3, t Þà dx
x
x
x
+X 2 /UÝt Þ 2/ t u / N u ? Ý/ t uÝ3, t Þ +  4uÝ3, t Þ 2 Þ dSÝx Þ.
x
x Notice that
n
x2
x
 2/ t u / N u  = 2  / t u 4u 6 3  ² 2 / t u  4u  ² / t uÝ3, t Þ +  4uÝ3, t Þ 2
from which it follows that E v ÝtÞ ² 0. Since EÝ0Þ = 0, we conclude EÝt Þ 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 Ý30 , t 0 Þ is not influenced by data values that lie outside the
x
domain of dependence for the point Ý30 , t 0 Þ. This domain of dependence is just the floor of
x
the light cone obtained by drawing all backward characteristics through Ý30 , t 0 Þ.
x 10. WaveLike Evolution
We use the term diffusionlike 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 Maxmin principle
monotone (nonoscillatory) 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 Maxmin 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
f Ýx Þ = 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 + at Þ + fÝx ? at Þà/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, t Þ = a 2 / xx uÝx, t Þ,
uÝx, 0 Þ = fÝx Þ, / t uÝx, 0 Þ = 0,
uÝ0, t Þ = uÝL, t Þ = 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, t Þ = > n=1 f n cosÝn^at/L Þ sinÝn^x/L Þ
where f n denote the Fourier coefficients for fÝx Þ; i.e.,
L
f n = 2 X fÝx Þ sinÝn^x/L Þ dx, n = 1, 2, ...
L0 It is known from the theory of Fourier series that the series
K
> n=1 f n sinÝn^x/L Þ converges (in some sense) to *Ýx Þ, the odd, Lperiodic extension of fÝx Þ. Then, since
f
cosÝn^at/L Þ sinÝn^x/L Þ = 1
2 ßsinÝn^Ýx + at Þ/L Þ + sinÝn^Ýx ? at Þ/L Þà, we see that 15 K
uÝx, t Þ = 1 > n=1 f n ßsinÝn^Ýx + at Þ/L Þ + sinÝn^Ýx ? at Þ/L Þà
2 f
= 1 *Ýx + at Þ + *Ýx ? at Þ .
f
2
Clearly this solution is oscillatory in time and, since positive time is interchangeable with
negative time, the solution is time reversible. The energy, EÝt Þ, 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 fÝx Þ is a triangular pulse of height one, centered at x = L/2, then
plotting uÝx, t Þ versus x for various times shows that for t = L/a, we see uÝx, L/a Þ is a
triangular pulse centered at x = L/2, but with height equal to minus one. Clearly there can
be no maxmin principle for this equation.
The behavior exhibited by this solution of the IBVP problem in 1dimension is typical of
solutions to the wave equation in other settings. Collectively we refer to this behavior as
wavelike evolution, as opposed to diffusionlike 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 ndimensions can be written in the following way
/ /x
x
4 2 uÝ3Þ = 3f I n 3uÝ3Þ where 3f = ß/ 1 , ..., / n à
/ x
x
x
/ /x
/ t uÝ3, t Þ ? 4 2 uÝ3, t Þ = / t uÝ3, t Þ ? 3f I n 3uÝ3, t Þ
23
x
x
x
/ tt uÝ3, t Þ ? 4 uÝx, t Þ = / tt uÝ3, t Þ ? / f I n /uÝ3, t Þ
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
/x
of the form 3f J n 3uÝ3Þ where the n by n matrix J n is as follows,
/
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.
x
LßuÝ3Þà= / f A n /uÝ3Þ
x
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 3f Ax = C, is
x3
an ellipse, parabola or hyperbola, respectively, when the matrix A has the properties
assigned to each of the names. 17 ...
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This note was uploaded on 08/30/2011 for the course ECE 3041 taught by Professor Brewer during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Brewer
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