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CHAPTER 11
InitialBoundary Value Problems
T
he infnite spatial domains considered in the previous chapter give insight
regarding the behavior oF waves and diFFusions. However, since such
domains are not physically realistic, we need to develop new techniques For
solving
pde
s on bounded domains. As a frst step towards solving the heat and
wave equations over fnite spatial domains (such as the interval 0
≤
x
≤
L
in
one space dimension), we will solve these equations on “semiinfnite” domains
whose boundaries consist oF one point.
11.1. Heat and Wave Equations on a HalfLine
We begin by solving the homogeneous Dirichlet problem For the heat equation
on the interval 0
≤
x
<
∞
; that is,
u
t
=
κ
u
xx
(
0
<
x
<
∞
)
(
11
.
1
)
u
(
x
,0
)=
φ
(
x
)(
0
<
x
<
∞
)
(
11
.
2
)
u
(
0,
t
0
(
t
≥
0
)
.(
11
.
3
)
The homogeneous boundary condition is quite important For the solution tech
nique that Follows. In the context oF heat transFer within a “onedimensional”
wire, this Dirichlet boundary condition is analogous to immersing the
x
=
0 end
oF the wire in a bath oF ice water with temperature zero degrees Celsius.
We will solve the homogeneous Dirichlet problem (
11
.
1
)–(
11
.
3
) using a
refection
method
, temporarily extending our spatial domain to the entire real line and
solving a Cauchy problem instead. By quoting the Formula For the solution
297
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heat and wave equations on a half

line
φ
odd
(x)
φ
(x)
xx
Figure
11
.
1
.
Illustration of the odd extension of a function
φ
(
x
)
.
of the Cauchy problem in the preceding chapter, we will obtain the solution
of (
11
.
1
)–(
11
.
3
) by restricting ourselves to the original spatial domain.
First, recall that a function
f
(
x
)
of a single variable is called
odd
if it has the
property that
f
(

x
)=

f
(
x
)
for all real
x
. Examples of odd functions include
sin
(
x
)
and
x
3
. If
f
is an odd function, notice that
f
(
0
f
(

0

f
(
0
)
, which
implies that
f
(
0
0. Now, referring to the initial condition (
11
.
2
) above, we
de±ne the
odd extension
of
φ
(
x
)
as
φ
odd
(
x
φ
(
x
)
if
x
>
0

φ
(

x
)
if
x
<
0
0
if
x
=
0.
By construction,
φ
odd
is an odd function and is de±ned for all real
x
(see Fig
ure
11
.
1
). Now consider the
Cauchy
problem
v
t
=
κ
v
(

∞
<
x
<
∞
)
v
(
x
,0
φ
odd
(
x
)(

∞
<
x
<
∞
)
.
From the previous chapter, we know that the solution is given by the convolution
of the heat kernel
S
(
x
,
t
)
with the initial condition:
v
(
x
,
t
) = (
S
±
φ
odd
x
,
t
#
∞

∞
S
(
x

y
,
t
)
φ
odd
(
y
)
d
y
.
We claim that the restriction of
v
(
x
,
t
)
to the domain
x
≥
0 is the solution of
the Dirichlet problem (
11
.
1
)–(
11
.
3
). To see why, we need to verify that all three
conditions of our Dirichlet problem are satis±ed. Certainly
v
(
x
,
t
)
satis±es the
initial

boundary value problems
299
same
pde
as
u
(
x
,
t
)
on the domain
x
>
0. The initial conditions also match on
that domain, because
v
(
x
,0
)=
φ
(
x
u
(
x
)
whenever
x
>
0. Checking the
boundary condition requires a bit more care. As an exercise, you should verify
that since the initial condition for
v
(
x
,
t
)
is odd, then the solution
v
(
x
,
t
)
will
remain odd for all
t
>
0. That is,
v
(

x
,
t

v
(
x
,
t
)
for all
t
≥
0. By our earlier
remarks on odd functions, this implies that
v
(
0,
t
0 for all
t
≥
0. It follows
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