Problems from Sections 2.2 - 2.5 in Logan

one family of such curves that carry signals and provide a distinguished
these roles. Even first order PDEs, which are actually wave-like have
perbolic problems there is always a set of characteristic curves that play
influence and
domain of
dependence.
—c2U, = 0, x > 0, —
< t < oo; u(0, t) = s(t), —
< t <
3. Solve the outgoing signal problem
tions F and Gin the general solution (2.11) using the initial conditions (2.13).
impulse. Contrast the solution with the case when f 0 and g = 0.
Region of
the solution surface and discuss the effect of giving a string at rest an initial
Figure 2.3.
displacement isf(x) = 0, and the initial velocity isg(x) = 1/(1 + 0.25x
2). Plot
2. Calculate the exact solution to the Cauchy problem when c = 2, the initial
1. Derive d’Alembert’s formula (2.14) by determining the two arbitrary func
So the signal breaks into two pieces, and they propagate in opposite direc
So the signal stays at the same place, but it spreads out and decreases in
ill + 4kt
u(x, t) = e2’(1+41t)
1
via
u(x, t) =
u(x, t) = O5(e_(_Ct)
2 + ex4ct)
2). Think of this signal as being a bit of information.
case. For example, suppose the initial signal is a Gaussian function or bellshaped
curve exp(—x
gate information is to determine how a signal is propagated in a special
is retained. Parabolic, or diffusion, equations propagate signals at infinite
herency in the wave form as it propagates, and therefore information
parabolic and hyperbolic problems. Hyperbolic, or wave-like, equations
The convection equation Ut + CU = 0, which is a wave-like equation,
amplitude. Any information in the signal is eventually lost.
:ion at (xo, to). This
rm u 0. In hy
Finally, we point out again the important differences between
speed; because the signals diffuse or smear out, there is a gradual loss of
tions at speed c. The diffusion equation Ut = ku propagates the signal
Section 1.2).
coordinate sytem where the problem simplifies (recall the examples in
propagate signals at a finite speed along characteristics; there is co
information. A good way to understand how different equations propa
propagates this signal via
That is, it moves it at speed c without distortion. The wave equation
=c2u moves it via
2).
)lution outside this
the wave equation
Exercises
fine a special coor
ward in time to the
regarding the charsignals
forward in
)und by tracing the
y the initial values
one. Looking at the
itive characteriSticS,
then the region of
constant are paths
affect the solution
—time along which
ence of the interval
a and x —
ct
always zero outside
:hat interval, so the
n surface.
:h velocity c. These
z.z. The
Jnbounded Domains Exercises 57

ct)/p
68
2. Partial Differcntial Equations on Unbounded Domains
4.
where s(t) is a known signal. Hint: Look for a right-traveling wave solution.
The three-dimensional wave equation is
5. Solve the Cauchy problem
—c2Au = 0,
where is = u(x, y, z, t) and A is the Laplacian operator. For waves with spher
ical symmetry, is = u(p, ,
where p = .Jx2 + y2 + z2. In this special case
the Laplacian is given by (Section 1.8) Au = u + u. By introducing a
change of dependent variable U = pu, show that the general solution for the
spherically symmetric wave equation
is
1
u= —(F(p—ct)+G(p+cfl).
p
Why do you think an outward-moving wave is = F(p —
amplitude?
u1—c
2u=0, xcR, t>0,
2
=c2(u + u)
decays in
-i
u(x, 0) =e1, Ut(x, 0) = cosx, x € R.
Use a computer algebra program to graph the wave profile at t = 1, 2, 3. Take
c= 1
6. In Section 1.7 we showed that any solution to Laplace’s equation has the
property that its value at a point is approximately the average of four nearby
values surrounding the point. Can we make a statement about solutions to
the wave equation? Consider any characteristic parallelogram (see Figure
2.4) whose sides are positive and negative characteristics, and let A, B, C, D
be the vertices as shown. Show that any solution to the wave equation satisfies
the relation
u(A) + u(C) = u(B) + u(D).
Figure 2.4. Characteristic
parallelogram.

oo.
Therefore, for each t > 0,
w(x, t) = f(()
lfr(y))G(x
-
y,
t)dy.
x = 0 the solution is
Exercises 71
tisfies the Cauchy
(x, t) are close. Let
her. We would like
i for Laplace’s equa
I close in the sense
b this end, consider
easily observe that
ther problems such
)n the initial and/or
(ii) the solution is
ions for various PDE
ming. We say that a
an initial boundary
) solve the problem,
eed only speci1y the
m (2J5)—(2.16) does
ty of continuous de
e get the boundary
in the boundary data
lution should depend
ehavior is disturbing,
er all, we want to be
ation (2.15)—(2.16) is
lation models steady
:he data on the bound
on Unbounded Domains
this mean with regard to stability?
showthatl u’(x,t)—u
2(x,t) I 8 +8Tforallx c R, 0 < t < T.Whatdoes
since f G(x —
f6G(x —y,t)dy = 6,
Iu(x, t) —
v(x,
t)I f
(y) — fr(y) II G(x
—
y,
y,
1. Show that the Cauchy problem for the backward diffusion equation,
2. Let u = u(x, y). Is the problem
uxy = 0, 0 < x,y < 1,
3. Consider two Cauchy problems for the wave equation with different initial
u’(x, 0) = f’(x), u(x, 0) = g’(x), x
I f’(x) —f
2(x) &, I g’(x) —g
2(x) IS 2,
indices and not exponents). If for all x E R we have
for i = 1, 2, where f’ ,f2, g’, and g2 are given functions (the superscripts are
= x E R, 0 < t < T,
data:
square, a well-posed problem? Discuss.
on the unit square, where the value of u is prescribed on the boundary of the
for large n.
u(x, t) = 1 + _en2tsinnx
is unstable by considering the solutions
u(x, 0) = ft), x € R,
u+u=0, xeRt>O,
Exercises cJ
Lc
closeness of the initial data implies closeness of the solution.
t)dy = 1. Therefore, in the sense interpreted above,
t) I
The solution formula (2.8) gives
—

led Domains Exercises 75
of an odd solution to the positive real axis is the solution to the given ini
tial boundary value problem. If this intuitive reasoning leaves the reader
perplexed, then one can always verify analytically that the solutions we
ace—time
have obtained by this reflection method are, in fact, solutions to the given
where the
problems.
2.24) is If the boundary condition (2.18) along x = 0 in the heat flow problem
ion x > Ct
(2.17)—(219) is replaced by a Neumann condition
y the
g and can U(O, t) = 0, t > 0,
embert’s
then the problem can be solved by extending the initial data to an
even function. The same is true for the wave equation. We leave these
calculations as exercises.
(2.27)
0,
(2.28)
Exercises
1. Solve the problem
< 0. = ku, x > 0, t > 0,
u(0,t)=0, t >0,
u(x, 0) = (x), x > 0,
with an insulated boundary condition by extending
an even function. The solution is
to all of the real axis as
u(x, y) = f[G(x y, t) + G(x + y, ](y)dy.
—
s)ds
0
2. Find a formula for the solution to the problem
u=ku, x>0,t>0,
u(0,t)=0, t >0,
eplaced by
may write u(x, 0) = 1, x > 0.
(s)ds,
Sketch a graph of several solution profiles with k = 0.5.
3. Find the solution to the problem
(2.29) 2
Utt = C U, x > 0, t > 0,
.24) is given u(0, t) = 0, t > 0,
)rx 0.
rn and wave
hr these two Pick c = 0.5 and sketch several time snapshots of the solution surface to
e restriction
observe the reflection of the wave from the boundary.

—
ku
— =
r)
G(X
r))f(y,
80 2. Partial Differential Equations on Unbounded Domains 2.6. i
Therefore, transformation of the dependent variable has changed the 2.6
problem into one with a homogeneous boundary condition, but a price
was paid—an inhomogeneity, or source term —g’(t), was introduced into
Laplac
the PDE. In general, we can always homogenize the boundary condi-
equati
tions in a linear problem, but the result is an inhomogeneous PDE; SO
ordina
inhomogeneous boundary conditions can be traded for inhomogeneous
an alg
PDEs.
operat
We can solve (2.49)—(2.51) for v(x, t) by formulating a Duhamel’s prin-
Let
ciple. However, in Section 2.5 we will observe that Laplace transform
grow t
methods can also be applied to find the solution.
means
the La;
Exercises
The La
1. Write a formula
given f
for the solution to the problem
U(s), ir
sin x, x E R, t > o, variabi
u(X, 0)
where
Ut(X, 0) = 0, X E R.
is calle
Graph the solution surface when c = 1.
Laplac
2. Write a formula for
many
the solution
s
to the problem
shortt
SrnX, XE R, t > 0, system
u(x,0)=0,
XER.
3. Using Duhamel’s principle, find a formula for the solution to the initial value returns
problem for the convection equation
Ut + cu = f(x, t), X E R, t > 0; u(X, 0) = 0, X E R.
Hint: Look at the problem
w(x,t;r)+cw(X,t;r)=O, XER,t >0; w(x,0;r)=flX,j, xeR.
Solve the problem
Ut + 2u = xet, X E R, t > 0; U(x, 0) = 0, x E R.
which i
use
which r
4. Formulate Duhamels principle and solve the initial boundary value problem The
that it c
Ut = kU +f(x, t), X > 0, t > 0,
transfoi
U(X,0)=0, X >0,
u(0,t)0, t >0.
The solution is
u(x, t) =
f f (G(X y,
—
t
—
—
+ y t
—
r)dydr.
Formuh
parts, ai
equatlo;