735
Internat. J. Math. & Math. Sci.
VOL. 12 NO. 4 (1989) 735-740
UNIQUENESS AND STABILITY OF SOLUTIONS FOR
A TYPE OF PARABOLIC BOUNDARY VALUE PROBLEM
GONZALEZ-VELASCO
Department of Mathematics
ENRIQUE A.
University of Lowell
Lowell, Massachusetts, U.S.A.
(Received June 25, 1986 and in revised form October i0, 1986)
We consider a boundary value problem consisting of the one-dimensional
ABSTRACT.
+ q, where g, h and q are functions of x, subject to
By developing a maximum principle for the boundary
(hUx)
parabolic equation gu
t
x
some general boundary conditions.
rather than the equation,
value problem,
we prove the uniqueness of a nonnegative
solution that depends continuously on boundary values.
KEY WORDS AND PHRASES.
Parabolic equations, maximum principles.
1980 AMS SUBJECT CLASSIFICATION CODE. 35K20.
I.
INTRODUCTION.
{(x,t) ]R2:0 < x < a, t > 0} and
with
shall
be
concerned
0}, we
If a is a positive constant, D
D
11t2
{(x,t)
0
a,
IR of the parabolic boundary value problem
x
t
nonnegative solutions u:D
c
c
f(x)
u(x,O)
(1.1)
in D
(hUx) x +q
alU(0,t blUx(O,t
a2u(a,t) + b2ux(a,t)
gut
t
2
continuous
0
(1.2)
(1.3)
t
)
0
0
(
x
(
a
(1.4)
where we assume that
(i) g, h and q are nonnegatlve continuous functions on [O,a], g and h never vanish
and h has a continuous derivative,
(ii)
(iii)
al, a2, bl, b2,
a
+ b
>
c
0 and a
and c
2
+ b
2
2
are nonnegative constants,
>
0, so that none of the boundary conditions is
missing,
(iv)
ala 2 >
O, and
(v) f is a nonnegative continuous function with a piecewise continuous derivative
and satisfies (1.2) and (1.3).
E. A. GONZALEZ-VELASCO
736
A continuous solution on D must be bounded below, and we are assuming without
loss of generality that this bound is zero.
and c
on the constants c
This is in agreement with the constraints
2.
The question of existence of solutions is easily answered by finding first the
steady state solution and then using separation of variables and the Sturm-Liouville
theory
to
the existence of
prove
a
transient
solution.
It can be shown that a
nonnegative steady state solution always exists under the stated conditions while, if
[I].
When
ala 2 >
0 it
may
For proof see
it may exist but with probability zero.
condition (iv) fails to hold,
still
not
exist
if
either b
<
0 or b
2
<
0, as in the
problem
U
t
U
xx
u(0,t) + u X (0,t)
2u(l ,t)
which admits
no solution independent of
time.
In fact, under these conditions it
cannot even be guaranteed that the solution will be nonnegative, as in the problem
U
t
U
xx
u(0,t)
u (0,t)
u(l,t)
Ux(l,t)
3
X
whose only solution independent of time is U(x)
2x.
Uniqueness of solutions is sometimes obtained from a strong maximum principle of
A. Friedman [2], for a general type of parabolic equation.
In this paper we shall be
able to give a much shorter proof by developing, instead, a type of maximum principle
for the boundary value problem, rather than for the equation alone.
This will provide
the added benefits of enabling us to prove that the unique solution is nonnegative and
In the simpler case in which
(I.I) becomes the heat equation, all of these facts are easily concluded from the
0.
usual maximum principle of E.E. Levi, but such a theorem is not valid when q
depends continuously on the boundary and initial values.
However, the following starting statement about the minimum of a solution will be
sufficient for our purposes.
2. MAI N RE SULTS.
THEOREM I.
Let T be an arbitrary positive number and define the sets
D
D
T
{(x,t) e D: t
T
{(x,t) e D:
4
T}
t 4 T}
and
"(T
{(x,t) e
DT:
(x,t)
DT}.
737
UNIQUENESS AND STABILITY OF SOLUTIONS FOR BOUNDARY VALUE PROBLEM
T
If u:
IR
(l.l) in DT, then u attains its
is a continuous function that satisfies
minimum value on
YT"
The proof is a trivial modif[cation of that of Levi’s principle.
THEOREM 2.
Let
T’ YT
DT,
and define the sets
and u be as in Theorem
Y0
{(x,t) e
YT:
x
0, t
>
0}
Ya
{(x,t) e
T:
x
a,
>
0}.
and
If u has a negative minimum on D
T
and if
I.
u satisfies
2.
u satisfies (1.3), then it attains mlmimum on
(1.2), then it attains its minimum on
By Theorem I, u attains
Assume
contrary
that,
PROOF.
I.
some t
o >
>
>
blux(0,t 0)
else alu(0,t 0)
alu(0,t 0)
b
M
0 while u
0, or
YT
its minimum value m
the
to
YT
0)
Y0"
alto <
conclusion, u(0,t
desired
Then u (0,t
x
0, which is impossible if c
m on
YT Y0"
Ya"
< 0 on T"
0)
m for
0 and
>
0.
Thus, c
0 and then
0, contradicting the value of c I.
If we now define
x
f
(x)
0
and a new function v: D
T
/R
by
M
m
H(x)
2H(a)
then v is continuous and satlsifies (I.I) because
v(x,t)
gvt
u(x,t)
(hvx)x
gut
(hUx)x
lh
gu
(hu)
q
t
M- )xl
m
According to Theorem I, v also attains its minimum on
at
(O,t0)
because if (x,t)
v(x,t)
while v
-=
u on
)
YT
Y0
then
M-m
u(x,t)
2
)
M
0 and b
v (0 t o
x
>
u (0,t
x
a
b- u(O’t0)
its minimum at this point.
0"
M+m
2
2
>
m
0, differentiation of v with respect to x gives
0)
M
m
2H(a)h(0)
M-m
2H(a)h(0)
<
0
Then v is decreasing in the x direction at
T
M-m
Y0"
Now, since c
on
In fact, this minimum is
YT"
(0,t0),
contradicting the fact that it has
This contradiction shows that u nst attain its minimum
738
E. A. GONZALEZ-VELASCO
2.
The proof is analogous to the one above, but based on the choice
v(x t)
M- m
2H(a) iN(a)
u(x t)
H(x)]
Q.E.D.
These theorems have two corollaries that are applicable to solutions of
gu
(hu)
t
in D
X X
alu(0,t)
a2u(a,t)
+
b2uX (0 t)
b2ux(a,t)
(2.1)
0
t
)
0
(2.2)
0
t
)
0
(2.3)
and we shall use them to complete the discussion of our boundary value problem.
COROLLARY I.
A continuous function u:
its maximum value on
YT"
T
]R that satisfies
(2.2), then it attains its maximum on
u satisfies (2.3), then it attains its maximum on
u satisfies
|.
2.
PROOF.
(2.1) in D
attains
T
If this maximum value is positive and if
YT
YT
0"
Ya"
If u satisfies any of the last three equations, so does -u, and thus -u
hypotheses
the
satisfies
of
Theorems
2
and
with q =_ 0, and c
O.
c
2
Accordingly, the statements made by these theorems about the minimum of -u translate
into the statements made here about the maximum of u, Q.E.D.
COROLLARY
(2.3).
If M
0
Let u: D /IR be
2.
and m
0}
function
4
u(x,t)
>
0 be arbitrary.
minimum value m on
m
left.
M on
0
and u(x,t)
Either M
By Theorem I, the restriction of u to
0, and then u(x,t)
O, or m < 0.
T In
max {0, M
YT
O}
YT"
Either m
m
0.
Since
T is arbltrary,
< 0, and then u(x,t)
by Corollary I, u attains its maximum on
u(x,t) < MO.
(2.1)
then
this
>
0, or M
YT
the
attains
this
and thus
inequality on the
attains its maximum value
In this second case, and again
}, and thus M M and
0.
{Y0 Ya
U
T
{YoU Ya
YT
proves
Similarly, by Corollary I, the restriction of u to D
YT.
satlsifes
{YOU Ya }’
second case, Theorem 2 implies that u attains its minimum on
m
that
5.
Let T
PROOF.
its
continuous
denote its maximum and minimum values on
0
mln {0, m
for all (x,t) e
a
0
Since T is arbitrary, this proves the inequality on the right, Q.E.D.
We now return to the original boundary value problem.
THEOREM 3.
The boundary value problem (I.I)
(1.4) has a unique continuous
]R that depends continuously on initial values, and this solution is
solution u: D
nonnegt ive.
PROOF.
If u and u are solutions of (I.I)
(1.4) corresponding to initial
2
functions f and g, then u -u satisfies (2.1)
(2.2)
and the new initial condition
2
(u
-u2)(x,O)
f(x)
g(x).
UNIQUENESS AND STABILITY OF SOLUTIONS FOR BOUNDARY VALUE PROBLEM
If M
0
and m
0
.
(u
0}
mln {0, m
for all (x,t)
zero
and
g, then Corollary 2 implies
denote the maximum and minimum values of f
that
so
Therefore,
is u
u
2.
max {0,
-u2)(x,t)
if
[O,a], M0 and m 0 are close
the continuity of
proves
This
initial conditions and, taking f
M0}
close to g on
is
739
solutions
with
respect
to
to
g, proves uniqueness.
Finally, if u has negative values then, according to Theorem 2, it attains its
negative minimum on
f
{YOU ya },
YT
where it equals f.
But this is impossible because
0. Q.E.D.
To prove the continuity of the solution with respect to both boundary and initial
conditions, notice that the steady state solutlon U as found in [I],
U(x)
+
+
(blC2h(a)
where
[(alc 2 a2c
A
0
+
b2Clh(0)
+
+ a a2
blb2Q(a)
h(a) +
+
alb2Q(a)]
a2Clh(0)h(a)H(a)
+
a2blh(a) Of
x
Q(x)
f
q
0
and
A
[a h(0)H(a) + b
]a2h(a)
+ a
b2h(0)
depend continuously on the values of the constants.
of (I.I)
(1.4) and we define v
Then, if u is the unique solution
u- U, ths function satisfies (2.1)
(2.3).
As
above, v depends continuously on f- U, which, in turn, depends continuously on f and
the constants.
That is u is stable with respect to initial and boundary values.
We
can s up our results as follows
THEOREM 4.
The boundary value problem (I.I)
(1.4) is well posed in the sense of
Had area r d.
REFERENCES
I.
2.
GONZALEZ-VELASCO, E.A., The Existence of a Steady State Solution for a Type of
Parabolic Boundary Value Problem, Int. J. Math. Educ. in Scl. & Technol.
19(3), (1988), 413-419.
PROTTER, M.H. and WEINBERGER, H.F., Maximum Principles in Differential
Equatlons, Prentice Hall, 1967.