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This means that if g is close to f then the integral of .
u
v
/
2
, at any time t>0,
must be small. This, in turn, implies that the region where
v
differs significantly from
u
must be small (see Exercise
8.13
). Hence, we conclude that minor changes in the
initial condition of (
8.1
)-(
8.3
) will not alter its solution significantly. The problem
is stable!
8.1.4
Uniqueness
We will now prove that (
8.1
)-(
8.3
) can have at most one smooth solution. Assume
that both
u
and
v
are smooth solutions of this problem; that is,
u
solves (
8.1
)-(
8.3
)
and
v
solves (
8.12
)-(
8.14
), with g.x/
D
f.x/for all x in .0; 1/.Then(8.15) implies
that
Z
1
.
u
.x; t /
v
.x; t //
2
dx
D
0
for t>0:
0
Since
u
and
v
are both smooth, it follows that the function .
u
v
/
2
is continuous.
Furthermore,
.
u
.x; t /
v
.x; t //
2
0
for all x
2
Œ0; 1 and t
0;
and we can therefore conclude that
u
.x; t /
D
v
.x; t /
for all x
2
Œ0; 1 and t
0;
see Exercise
8.12
.
The problem (
8.1
)-(
8.3
) can have at most one smooth solution.
8.1.5
Maximum Principles
Consider a steel rod positioned along the unit interval Œ0; 1. Suppose that the tem-
perature of this rod at any position x
2
.0; 1/ is given by the function f.x/ at time
t
D
0.Thatis,f represents the initial temperature distribution in the rod. Further-
more, assume that the temperature at the endpoints x
D
0 and x
D
1 of the rod at
time t>0is determined by the functions g
1
D
g
1
.t / and g
2
D
g
2
.t /, respectively,
and that the rod is isolated elsewhere.
9
Under these circumstances the temperature
u
.x; t / is, as we have seen in Chap. 7, governed by the following set of equations:
9
Heat can only leave and enter the rod through its endpoints.
