Existence, Uniqueness and Asymptotic Behavior
In the previous sections of the course we have developed information about particular
solutions to various specific equations. In this, the final section, we will prove qualitative
results about more general solutions to some nonlinear equations. As is typical for
nonlinear problems, none of these results is a result about nonlinear problems in general
but each applies to an equation of a specific form but with general ingredients for equations
of that form. Since most of the results make use of one or another of a large set of
inequalities, we will begin by collecting some of the more useful of these.
1. Inequalities
Lemma 1.1- For a, b 5 R and P > 0
1
2
a2 +
1
2
i)
ab ²
ii)
ab ² P a 2 + 1 b 2
4P
b2
Proof: The first result follows from the observation that for any real numbers a, b
1
2
Ýa ? bÞ 2 =
1
2
a 2 ? ab +
1
2
b 2 ³ 0.
The second result follows from
1
2
2P a ?
2
b
2P
= P a 2 ? ab +
1 b 2 ³ 0.n
4 2P
Lemma 1.2 (Young’s Inequality) For 1 < p, q < K, 1p + 1q = 1, and a, b > 0,
p
q
ab ² ap + bq .
Proof- Note that the function fÝxÞ = e x is convex, which means that
fÝūÞ = f 1p x + 1q y
Then write
² 1p fÝxÞ + 1q fÝyÞ = fÝuÞ
1
ab = e ln a+ln b = e p
ln a p +
1
q
ln b q
-x, y
= fÝūÞ
a p + b q = 1 e ln a p + 1 e ln b q = fÝuÞ.n
p
q
p
q
and
Lemma 1.3 (Holder’s inequality) For 1 < p, q < K, 1p + 1q = 1, and functions u and v,
measurable on U Ð R n such that,
|| u|| p =
XU |u| p dx
1/p
,
|| v|| q =
XU |v| q dx
1/q
,
we have
XU | u v| dx ² || u|| p || v|| q .
1
Proof- WLOG suppose || u|| p = 1, || v|| q = 1. Then, using Young’s inequality,
XU | u v| dx ² XU Ý 1p | u| p + 1q | v| q Þ dx = 1p + 1q = || u|| p || v|| q .n
These inequalities, along with others to be developed later, will be used to prove results of
existence, uniqueness and asymptotic behavior of solutions to partial differential equations.
These arguments will be carried out in the setting of a linear space of functions. These
function spaces will be equipped with a norm and often with an inner product as well.
The function space norms can be defined in various ways, leading then to different linear
spaces of function. For functions defined on a bounded open set U in R n , we may define
for u 5 CÝŪÞ
|| u|| K = sup Ū | uÝxÞ|
for u 5 MÝUÞ
|| u|| L p =
1/p
XU |u| p dx
, 1 ² p < K.
MÝUÞ = measurable functions .
In particular, when p = 2, we will write || u|| L 2 = || u|| 0 for the Hilbert norm associated with
the inner product
Ýu, vÞ 0 = XU u v dx,
|| u|| 0 = Ýu, uÞ 1/2
0 .
We define
H 1 ÝUÞ = u 5 L 2 ÝUÞ | / x j u 5 L 2 ÝUÞ, 1 ² j ² n
distributional derivative and
where / x j u denotes the
|| u|| 21 = || u|| 20 + || 4u 6 4u|| 0
is then a Hilbert norm for the inner product,
n
Ýu, vÞ 1 = Ýu, vÞ 0 + > Ý / x j u, / x j vÞ 0 .
j=1
In the case n = 1 the functions in H 1 ÝUÞ = H 1 ßa, bà are Lipschitz continuous.
Lemma 1.4 If u 5 H 1 ßa, bà then
ÝaÞ
x, y 5 ßa, bà, a ² x < y ² b,
y
uÝyÞ ? uÝxÞ = X u v ÝzÞ dz;
x
i.e., .u is absolutely continuous on ßa, bà
ÝbÞ
u 5 Cßa, bà (in fact u is Holder continuous with exponent 1/2).
Proof (a) Let ßJ, Kà denote an arbitrary subinterval of ßa, bà. Since we can define H 1 ßJ, Kà as
the completion in the ||6|| 1 norm of C 1 ßJ, Kà, it follows that there is a sequence
u n ÝxÞ 5 C 1 ßJ, Kà such that u n converges to u in the ||6|| 1 norm. This implies
||u n ? u|| 0 í 0
Then for each n
Write this as
and
||u vn ? v|| 0 í 0 as n tends to K
x
u n ÝxÞ = u n ÝJÞ + X u vn .
J
2
x
u n ÝJÞ = u n ÝxÞ ? X u vn .
J
x
x
Now u n ÝxÞ ? X u vn converges in L 2 ßJ, Kà to the limit uÝxÞ ? X v and this implies, in turn,
J
J
that u n ÝJÞ must converge (as a sequence of real numbers) to a limit we will denote by C.
Then it follows that
x
uÝxÞ = C + X v
J
for almost every x in ßJ, Kà. By modifying u on a set of measure zero, if needed, we can
assume this last equality holds at every point of ßJ, Kà. Then u is an absolutely continuous
function on the interval ßJ, Kà, and C = uÝJÞ; i.e.,
x
uÝxÞ = uÝJÞ + X v
-x 5 ßJ, Kà.
J
Since ßJ, Kà denotes an arbitrary subinterval of ßa, bà, the point x can be taken as close as
we like to either a or to b and we can conclude that uÝxÞ is absolutely continuous on the
whole interval ßa, bà.
(b) For arbitrary x, y 5 ßa, bà, a ² x < y ² b,
y
uÝyÞ ? uÝxÞ = X u v ÝzÞ dz.
x
Then
y
y
| uÝyÞ ? uÝxÞ| ² X x | u v ÝzÞ| dz = X x 1 | u v ÝzÞ| dz
²
y
X x 1 2 dz
1/2
y
X x | u v ÝzÞ| 2 dz
1/2
²
y
X x 1 2 dz
1/2
b
X a | u v ÝzÞ| 2 dz
1/2
² Ýy ? xÞ 1/2 || u v || 0 ² Ýy ? xÞ 1/2 || u|| 1 n
Lemma 1.5 (Poincare Inequality) If u 5 H 1 ßa, bà then
i) || u|| 0 ² 2Ýb ? aÞ || u v || 0
if uÝbÞ = 0
? a || u v ||
ii) || u|| 0 ² b ^
0
or uÝaÞ = 0
if uÝaÞ = uÝbÞ = 0
Proof- Suppose u 5 H 1 ßa, bà and uÝaÞ = 0. Then
b
b
b d
dx
X a uÝxÞ 2 dx = X a 1 uÝxÞ 2 dx = X a
Ýx ? aÞ uÝxÞ 2 dx
b
v
= Ýx ? aÞ uÝxÞ 2 | x=b
x=a ? X 2 Ýx ? aÞ uÝxÞ u ÝxÞ dx
a
and
b
b
X a | u| 2 dx ² 2Ýb ? aÞ X a | u | | u v | dx ² 2Ýb ? aÞ || u|| 0 || u v || 0 .
If we have uÝaÞ = 0 instead of
Then it follows that
|| u|| 0 ² 2Ýb ? aÞ || u v || 0 .
uÝbÞ = 0, then we use x ? b in place of Ýx ? aÞ in the first line of the
proof. If uÝaÞ = uÝbÞ = 0 then this estimate can be sharpened as follows. Let
d n ÝxÞ = c n sin n^Ýx?aÞ
denote the orthonormal family of eigenfunctions for the self adjoint
b?a
operator LuÝxÞ = ?u”ÝxÞ on u 5 H 1 ßa, bà | uÝaÞ = uÝbÞ = 0 . Then for any u 5 H 1 ßa, bà we
can write
3
uÝxÞ = >Ýu, d n Þ 0 d n ÝxÞ.
n
In general the convergence of this series is L 2 convergence but if u 5 H 1 ßa, bà and, in
addition, uÝaÞ = uÝbÞ = 0, then the convergence is uniform and we are entitled to write
u v ÝxÞ = >Ýu, d n Þ 0 d vn ÝxÞ.
n
where this series now converges in L 2 ßa, bà. Then
|| u v || 20 =
But
>Ýu, d n Þ 0 d vn ÝxÞ, >Ýu, d m Þ 0 d vm ÝxÞ
n
m
0
= >Ýu, d n Þ 0 >Ýu, d m Þ 0 Ýd vn , d vm Þ 0 .
n
m
Ýd vn , d vm Þ 0 = ?Ýd n ”, d m Þ 0 = V n Ýd n , d m Þ 0
hence
|| u v || 20 =
>Ýu, d n Þ 0 d vn ÝxÞ, >Ýu, d m Þ 0 d vm ÝxÞ
n
m
³ V 1 > Ýu, d n Þ 20 = V 1 || u|| 20 =
n
0
= > V n Ýu, d n Þ 20
n
^ || u|| 2
0
b?a
n
Corollary A norm is defined on X = u 5 H 1 ßa, bà | uÝbÞ = 0 by | u| 1 = || u v || 0 . and this
norm is equivalent to the usual H 1 ? norm,
|| u|| 1 = Ýu, uÞ 0 + Ýu v , u v Þ 0 . on H 1 ßa, bà.
Note that | u| 1 is not a norm on the space H 1 ßa, bà. It is only when the functions vanish at
least at one end of the interval that the Poincare inequality holds and | u| 1 defines a norm.
The reason is that when the function is ”starting from zero”, || u|| 0 can be estimated in terms
of the rate of growth, || u v || 0 .
Lemma 1.6 If u 5 H 1 ßa, bà then u 5 Cßa, bà and || u|| K ² C 0 || u|| 1 .
Proof- Assume that u 5 H 1 ßa, bà. We already showed that If u 5 H 1 ßa, bà then u 5 Cßa, bà
and hence there exist points x 0 , x 1 5 ßa, bà where u assumes its minimum and maximum
values on ßa, bà; i.e., uÝx 0 Þ ² uÝxÞ ² uÝx 1 Þ, a ² x ² b. Since we have assumed u 5 H 1 ßa, bà,
we can write
x1
uÝx 1 Þ = uÝx 0 Þ + X u v ÝxÞ dx.
x0
x1
b
Then
| uÝx 1 Þ| ² | uÝx 0 Þ| + X x | u v ÝxÞ| dx ² | uÝx 0 Þ| + X a | u v ÝxÞ| dx
and
| uÝx 1 Þ| = || u|| K ²
0
²
i.e.,
1
b?a
1
b?a
b
b
b
X a | uÝxÞ| dx + X a | u v ÝxÞ| dx
X a 1 2 dx
1/2
b
X a | uÝxÞ| 2 dx
1/2
b
+ X 1 2 dx
a
1/2
b
X a | u v ÝxÞ| 2 dx
1/2
|| u|| K ² C 0 || u|| 1 .n
4
Example 1.1 A Uniform Bound on the Solution of a Reaction Diffusion Equation
Consider the initial boundary value problem
/ t uÝx, tÞ ? / xx uÝx, tÞ = uÝx, tÞÝ1 ? uÝx, tÞÞ,
uÝx, 0Þ = fÝxÞ,
0 < x < 1, t > 0,
0 < x < 1,
/ x uÝ0, tÞ = 0 = / x uÝ1, tÞ,
Ý1.1Þ
t > 0.
We will show that if this problem has a smooth solution, (i.e., u, / t u and / xx u are all
continuous on ß0, 1à × ß0, KÞ ) and if it is true that
P ² fÝxÞ ² 1 + P, for 0 ² x ² 1,
then it follows that
P ² uÝx, tÞ ² 1 + P, for 0 ² x ² 1, t ³ 0.
To prove this result, we will suppose that it is false and show that this assumption leads to a
contradiction. Suppose uÝx, tÞ exceeds the value 1 + P. Then there must exist some time
t 0 > 0 and point x 0 5 ß0, 1à where we have
/ xx uÝx 0 , t 0 Þ ² 0,
/ t uÝx 0 , t 0 Þ ³ 0,
uÝx 0 , t 0 Þ = 1 + P,
Then the equation implies
/ t uÝx 0 , t 0 Þ = / xx uÝx 0 , t 0 Þ + Ý1 + PÞÝ1 ? Ý1 + PÞÞ ² ?PÝ1 + PÞ < 0
and this contradiction implies there can be no such point Ýx 0 , t 0 Þ. In the same way, we can
show uÝx, tÞ never drops below the value P. Thus if a smooth solution exists, it must be
bounded between P and 1 + P if this is true initially.
Example 1.2 Asymptotic Behavior of the Solution of a Reaction Diffusion Equation
We continue to consider the initial boundary value problem (1.1) and we will prove some
results on the solution behavior as t tends to infinity. First we define
1
FÝtÞ = X / x uÝx, tÞ 2 dx,
0
t ³ 0.
Note that if we differentiate the equation (1.1) with respect to x, we see that
vÝx, tÞ = / x uÝx, tÞ satisfies
/ t vÝx, tÞ ? / xx vÝx, tÞ = vÝx, tÞÝ1 ? 2uÝx, tÞÞ,
vÝ0, tÞ = 0 = vÝ1, tÞ,
0 < x < 1, t > 0,
t > 0.
5
Then
1
1
0
0
F v ÝtÞ = X 2 vÝx, tÞ / t vÝx, tÞ dx = X 2 vÝx, tÞ ß / xx v + vÝ1 ? 2uÞà dx
1
1
0
0
2
2
= 2 v / x v | x=1
x=0 ? 2 X / x vÝx, tÞ dx + 2 X v Ý1 ? 2uÞ dx
Since u is positive on the interval ß0, 1à, this reduces to
1
F v ÝtÞ ² 2 X Ýv 2 ? / x v 2 Þdx = 2|| v|| 20 ? 2|| / x v|| 20 .
0
Now using the sharp version of the Poincare inequality, we obtain
F v ÝtÞ ² 2|| v|| 20 ? 2^ 2 || v|| 20 . = ?2Ý^ 2 ? 1Þ FÝtÞ
This implies
d e2
dt
which leads to
or
^ 2 ?1 t
FÝtÞ ² e ?2
FÝtÞ
^ 2 ?1 t
||/ x uÝ6, tÞ|| 20 ² e ?2
²0
FÝ0Þ
^ 2 ?1 t
|| f v || 20 .
This estimate implies that / x uÝ6, tÞ converges to zero in L 2 ß0, 1à as t tends to infinity. This
does not mean u tends to zero and it does not necessarily mean u tends to a constant but it
does mean that as t tends to infinity, uÝx, tÞ ¸ uÝtÞ; i.e., u becomes independent of x.
In order to analyze the asymptotic behavior of the solution, define
1
EÝtÞ = X ÝuÝx, tÞ ? 1Þ 2 dx
0
Then
for t ³ 0.
1
1
E v ÝtÞ = 2 X ÝuÝx, tÞ ? 1Þ / t uÝx, tÞ dx = 2 X Ýu ? 1Þß/ xx u + uÝ1 ? uÞà dx
0
0
1
2
2
= 2Ýu ? 1Þ / x u | x=1
x=0 ? 2 X ß/ x u + uÝ1 ? uÞ à dx
0
1
1
² ?2 X ßuÝ1 ? uÞ 2 à dx ² ?2P X Ý1 ? uÞ 2 dx = ?2P EÝtÞ.
0
0
This implies
or
EÝtÞ ² e ?2Pt EÝ0Þ,
|| uÝ6, tÞ ? 1|| 20 ² e ?2Pt || fÝ6Þ ? 1|| 20 ¸ 0 as t ¸ K.
The solution tends to the constant 1 as t tends to infinity.
Example 1.3 A More General Uniform Bound on the Solution of a Reaction Diffusion
Equation
Consider the following slightly more general reaction diffusion equation,
/ t uÝx, tÞ ? / xx uÝx, tÞ = FÝuÝx, tÞÞ,
uÝx, 0Þ = fÝxÞ,
uÝ0, tÞ = 0 = uÝ1, tÞ,
0 < x < 1, t > 0,
(1.2)
0 < x < 1,
t > 0.
where we suppose F 5 CÝR 1 Þ.
Define
6
1
EÝtÞ = X ß 12 / x uÝx, tÞ 2 ? GÝuÝx, tÞÞà dx
for t ³ 0,
0
where
u
GÝuÞ = X FÝsÞ ds; i.e. G v ÝuÞ = FÝuÞ -u.
0
Then
1
E v ÝtÞ = X ß/ x uÝx, tÞ / xt uÝx, tÞ ? G v ÝuÝx, tÞÞ / t uÝx, tÞà dx
0
1
= X ß? / xx uÝx, tÞ / t uÝx, tÞ ? FÝuÝx, tÞÞ / t uÝx, tÞà dx
0
where we integrated by parts and used the BC’s to alter the first term in the integral. Then if
u = uÝx, tÞ solves (1.2), this reduces to
1
1
0
0
E v ÝtÞ = ? X ß / xx uÝx, tÞ + FÝuÝx, tÞÞ à / t uÝx, tÞ dx = ? X / t uÝx, tÞ 2 dx ² 0
for t ³ 0.
Then EÝtÞ ² EÝ0Þ for t ³ 0. Now suppose F satisfies
lim sup
|u|¸K
FÝuÞ
u ²0
Ý1.3Þ
Note that this condition is satisfied by FÝuÞ = ? u 2k+1 but it is not satisfied for
FÝuÞ = ± u 2k . We will discuss this condition further when we present the existence proof for
this problem. The condition implies that for P > 0, there exists a constant C P such that
u
u
0
0
GÝuÞ ? P u 2 = X FÝsÞ ds ? P u 2 = X
Then
1
EÝtÞ = X ß 12 / x uÝx, tÞ 2 ? GÝuÝx, tÞÞà dx ³
0
FÝsÞ
s ? 2P s ds ² C P .
1
2
1
1
X 0 / x uÝx, tÞ 2 dx ? P X 0 u 2 dx ? C P .
In view of the boundary conditions the sharp Poincare inequality implies that for any
R, 0 < R < 12 ,
1
2
1
1
1
X 0 / x uÝx, tÞ 2 dx = Ý 12 ? RÞ X 0 / x uÝx, tÞ 2 dx + R X 0 / x uÝx, tÞ 2 dx
1
1
0
0
³ Ý 12 ? RÞ X / x uÝx, tÞ 2 dx + R^ 2 X uÝx, tÞ 2 dx
and then
1
1
0
0
EÝtÞ ³ Ý 12 ? RÞ X / x uÝx, tÞ 2 dx + ÝR^ 2 ? PÞ X uÝx, tÞ 2 dx ? C P
1
1
³ C R X / x uÝx, tÞ 2 dx + X uÝx, tÞ 2 dx ? C P = C R || u|| 21 ? C P
0
where R is chosen so that Ý
0
1
2
? RÞ = ÝR^ 2 ? PÞ = C R . Then we have
|| u|| 21 ² 1 ÝEÝtÞ + C P Þ ² 1 ÝEÝ0Þ + C P Þ.
CR
CR
Ý1.4aÞ
1
But
EÝ0Þ = X ß 12 f v ÝxÞ 2 ? GÝfÝxÞÞà dx depends only on the initial data so (1.4) may be
0
termed an a-priori estimate for the solution; i.e., it is an estimate on the H 1 ? norm of the
solution that is based only on the initial data and the function F from the equation. It follows
from lemma 1.6 that
7
C0
ÝEÝ0Þ + C P Þ .
CR
|| u|| K ²
Ý1.4bÞ
We can now use this estimate to prove the existence of a solution to (1.2).
Example 1.4 Existence of a Global Solution for a Reaction Diffusion Equation
We will prove the existence of a global solution for (1.2) under the assumptions
i)
f 5 Cß0, 1à
ii) FÝuÞ is Lipschitz on R; i.e., for each K > 0, 0M K > 0, such that
Ý1.5Þ
|FÝuÞ ? FÝvÞ| ² M K | u ? v| for all u, v 5 R, |u|, |v| < K..
iii)
lim sup
|u|¸K
FÝuÞ
u ²0
We have to recall that the unique solution for the linear problem
/ t vÝx, tÞ ? / xx vÝx, tÞ = pÝx, tÞ,
vÝx, 0Þ = fÝxÞ,
0 < x < 1, t > 0,
0 < x < 1,
vÝ0, tÞ = 0 = vÝ1, tÞ,
t > 0.
can be written as
1
t
1
0
0 0
vÝx, tÞ = X GÝx, y, tÞ fÝyÞ dy + X X GÝx, y, t ? sÞ pÝy, sÞ dyds.
Then if u solves (1.2) it follows that
1
t
0
0 0
1
uÝx, tÞ = X GÝx, y, tÞ fÝyÞ dy + X X GÝx, y, t ? sÞ FÝuÝy, sÞÞ dyds
Ý1.6Þ
Then (1.6) is a nonlinear integral equation with the propert that a solution of (1.2) is a
solution of (1.6). If u solves (1.6) then u solves (1.2), at least in some sense, but it may not
be a classical solution.
We define a nonlinear mapping
1
t
0
0 0
1
®ßuà = X GÝx, y, tÞ fÝyÞ dy + X X GÝx, y, t ? sÞ FÝuÝy, sÞÞ dyds
on the linear space X = Cß0, T 0 : Cß0, 1àà. Here X is the space of continuous functions of t,
0 ² t ² T 0 , which take their values in the space of continuous functions of x, 0 ² x ² 1. This
space is a complete normed linear space when equipped with the norm
|| u|| K = max |uÝx, tÞ| : 0 ² x ² 1, 0 ² t ² T 0 .
Here T 0 denotes a positive time to be specified later. Then for u,v in X, we have
t
1
| ®ßuà ? ®ßvà| = | X X GÝx, y, t ? sÞÝ FÝuÝy, sÞÞ ? FÝvÝy, sÞÞ Þ dyds |
0 0
t
1
² X X | GÝx, y, t ? sÞ| | FÝuÝy, sÞÞ ? FÝvÝy, sÞÞ | dyds
0 0
8
hence
t
1
|| ®ßuà ? ®ßvà|| K ² ||FÝuÝy, sÞÞ ? FÝvÝy, sÞÞ|| K X 0 X 0 | GÝx, y, t ? sÞ| dyds
t
1
² M K ||u ? v|| K X X | GÝx, y, t ? sÞ| dyds ² C T 0 M K ||u ? v|| K .
0 0
Here we used
t
1
X 0 X 0 | GÝx, y, t ? sÞ| dyds ² C T 0
0 ² t ² T0.
Now recall that for F satisfying (1.5iii), the solutions of (1.2) satisfy || u|| K ² K for some
constant K depending only on the initial data. Then the constant M K in (1.5ii) depends on
the initial data as well but the estimate
||FÝuÝy, sÞÞ ? FÝvÝy, sÞÞ|| K ² M K || u ? v|| K
holds uniformly for all solutions u, v of equation (1.2). Then the time T 0 is chosen such that
CM K T 0 = J < 1. In this case ® is a contraction mapping of X into X and there exists then a
unique fixed point which must be a solution of (1.6). This fixed point is a local (and possibly
weak) solution to the IBVP in the sense that it is valid only on the strip
0 ² x ² 1, 0 ² t ² T 0 . However, since condition (1.5iii) implies that condition (1.5ii) is
uniform in u,v, we can make the local solution into a global one. We use the local solution
as the initial condition to solve, by the same technique, a second IBVP, one that begins at
t = T 0 , with u = Ýx, t 0 Þ. Since the constants K, M K do not depend on T 0 the solution to the
new problem is valid on an inteval of the same width, namely, ß0, 1à × ßT 0 , 2T 0 à. This process
can be repeated over and over, to obtain a solution valid on ß0, 1à × ßT 0 , nT 0 à for any integer,
n. This extending of a local solution into a global one can be carried out as long as F is
uniformly Lipschitz. The combination of (1.5ii) and (1.5iii) result in the needed uniformity. It
could also result from an F that is uniformly Lipschitz on R without reference to condition
2
(1.5iii). For example FÝuÞ = e ?u is is uniformly Lipschitz on R but FÝuÞ = u m for m > 0 is
not.
Note that FÝuÞ = ?u 3 satisfies (1.5iii) but FÝuÞ = ± u 2 does not. Then (1.2) has a global
solution for the first choice of F but only a local solution for the second choice. The next
example will show why this is the case.
2. Monotonicity Methods
Suppose u = uÝx, tÞ, v = vÝx, tÞ both satisfy
/ t uÝx, tÞ ? / xx uÝx, tÞ = FÝuÝx, tÞÞ,
0 < x < 1, t > 0,
/ x uÝ0, tÞ = 0 = / x uÝ1, tÞ,
Ý2.1Þ
t > 0,
and
uÝx, 0Þ ² vÝx, 0Þ
Then wÝx, tÞ = uÝx, tÞ ? vÝx, tÞ,
0 < x < 1.
Ý2.2Þ
satisfies
/ t wÝx, tÞ ? / xx wÝx, tÞ = FÝuÞ ? FÝvÞ,
wÝx, 0Þ ² 0,
/ x wÝ0, tÞ = 0 = / x wÝ1, tÞ,
0 < x < 1, t > 0,
0 < x < 1,
t > 0,
9
But, for F 5 C 1 ÝRÞ,
u
1
FÝuÞ ? FÝvÞ = X F v ÝsÞ ds = Ýu ? vÞ X F v Ýv + aÝu ? vÞÞ da = pÝx, tÞ wÝx, tÞ
v
0
where
1
pÝx, tÞ = X F v ÝvÝx, tÞ + aÝuÝx, tÞ ? vÝx, tÞÞÞ da, 0 < x < 1, t > 0.
0
Then
/ t wÝx, tÞ ? / xx wÝx, tÞ = pÝx, tÞ wÝx, tÞ
and it follows from the initial and boundary conditions and a maximum principle arguments,
that
wÝx, tÞ ² 0 for 0 ² x ² 1, t ³ 0.
That is, Ý2.1Þ and Ý2.2Þ imply that
uÝx, tÞ ² vÝx, tÞ for 0 ² x ² 1, t ³ 0. This result is
called a ”comparison principle” and it asserts that for solutions of the equation and
conditions in (2.1), the initial ordering (2.2) is preserved as time progresses.
Now suppose u = uÝx, tÞ satisfies
/ t uÝx, tÞ ? / xx uÝx, tÞ = FÝuÝx, tÞÞ,
uÝx, 0Þ = fÝxÞ,
0 < x < 1, t > 0,
0 < x < 1,
/ x uÝ0, tÞ = 0 = / x uÝ1, tÞ,
Ý2.3Þ
t > 0,
and v = vÝtÞ solves the ode problem,
v v ÝtÞ = FÝvÝtÞÞ,
t > 0,
vÝ0Þ = C.
In particular, let v m ÝtÞ, v M ÝtÞ denote the solutions in the cases C = min x5ß0,1à fÝxÞ = f m , and
C = max x5ß0,1à fÝxÞ = f M , respectively. Then it follows from the comparison principle that
v m ÝtÞ ² uÝx, tÞ ² v M ÝtÞ
for 0 ² x ² 1, t ³ 0.
Ý2.4Þ
Note that vÝtÞ satisfies the derivative boundary conditions in Ý2.3Þ since vÝtÞ it is independent
of x .
If FÝuÞ = u 2 then vÝtÞ =
C
and
1 ? Ct
fM
fm
² uÝx, tÞ ²
1 ? fmt
1 ? fMt
If f m > 0, this result implies that uÝx, tÞ ¸ K as t ¸ 1/f m > 0; i.e., the solution ”blows up” at a
finite time. On the other hand, if f m ² f M < 0, then uÝx, tÞ ¸ 0 as t ¸ K. Note that for either of
the functions, FÝuÞ = ± u 2 , the solution of the initial boundary value problem could
experience blow up at a finite positive time depending on the sign of the initial data.
C
If FÝuÞ = ?u 3 then vÝtÞ =
and
1 + 2C 2 t
fm
² uÝx, tÞ ²
1 + 2f 2m t
fM
1 + f 2M t
10
In this case uÝx, tÞ ¸ 0 as t ¸ K, whatever the sign of the initial state.
If FÝuÞ = uÝ1 ? uÞ then vÝtÞ =
C
and
C + Ý1 ? CÞe ?t
fM
fm
² uÝx, tÞ ²
f m + Ý1 ? f m Þe ?t
f M + Ý1 ? f M Þe ?t
.
Then uÝx, tÞ ¸ 1 as t ¸ K, , uniformly in x.
Recall the condition (1.3) from the previous section. Functions FÝuÞ which satisfy this
condition behave like ?u 2k+1 for large values of |u| . As is illustrated by the example
FÝuÞ = ?u 3 , and as we proved in the previous section, solutions to the initial boundary
value problem for FÝuÞ satisfying the condition (1.3) remain bounded for all
t > 0, independent of the sign of the initial data. That this condition is sufficient but not
necessary for global boundedness is illustrated by the example, FÝuÞ = uÝ1 ? uÞ, whose
solution is also global in t.
Arguments of this type can be extended to obtain existence results. For convenience,
we will write the following IBVP
/ t uÝx, tÞ ? 4 2 uÝx, tÞ = FÝuÝx, tÞÞ
in U T = á5 U, 0 < t < Tâ
uÝx, tÞ = gÝx, tÞ
x 5 /U, 0 < t < T,
uÝx, 0Þ = u 0 ÝxÞ,
x 5 U,
as
/ t uÝx, tÞ ? 4 2 uÝx, tÞ = FÝuÝx, tÞÞ
uÝx, tÞ = gÝx, tÞ
in U T = á5 U, 0 < t < Tâ
Ýx, tÞ 5 /U T
by letting gÝx, 0Þ = u 0 ÝxÞ, x 5 U and letting /U T denote the parabolic boundary of U T . We
will call the function U 0 Ýx, tÞ an upper solution of the IBVP if
/ t U 0 Ýx, tÞ ? 4 2 U 0 Ýx, tÞ ? FÝU 0 Ýx, tÞÞ > 0
in U T = á5 U, 0 < t < Tâ
U 0 Ýx, tÞ ³ gÝx, tÞ
Ýx, tÞ 5 /U T
Similarly, V 0 Ýx, tÞ, will be called a lower solution to the IBVP if
/ t V 0 Ýx, tÞ ? 4 2 V 0 Ýx, tÞ ? FÝV 0 Ýx, tÞÞ < 0
in U T = á5 U, 0 < t < Tâ
V 0 Ýx, tÞ ² gÝx, tÞ
Ýx, tÞ 5 /U T
Assume that upper and lower solutions, U 0 and V 0 are known and suppose further that
FÝuÞ is smooth on the interval J = ámin V 0 ² u ² max U 0 â. Then choose C > 0 such that
F v ÝuÞ + C > 0
for all u in J.
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Now let U 1 = U 1 Ýx, tÞ denote the solution of the linear problem
/ t U 1 Ýx, tÞ ? 4 2 U 1 Ýx, tÞ + CU 1 Ýx, tÞ = FÝU 0 Ýx, tÞÞ + CU 0 Ýx, tÞ
U 1 Ýx, tÞ = gÝx, tÞ
in U T = á5 U, 0 < t < Tâ
Ýx, tÞ 5 /U T
Since U 1 is completely determined by U 0 , we will write, U 1 = GßU 0 à. Note that
/ t ÝU 0 ? U 1 Þ ? 4 2 ÝU 0 ? U 1 Þ + CÝU 0 ? U 1 Þ ³ 0
U0 ? U1 ³ 0
in U T = á5 U, 0 < t < Tâ
Ýx, tÞ 5 /U T
This implies that U 0 ? U 1 > 0 inside U T ; i.e., U 0 > GßU 0 à. Similarly, if V 1 = GßV 0 à, we can
show V 1 > V 0 . More generally, we can show
Lemma 2.1 If u ² v then Gßuà ² Gßvà
This allows us to define sequences, U n+1 = GßU n à and V n+1 = GßV n à, n = 0, 1, ... The
lemma implies that these sequences are monotone decreasing and monotone increasing,
respectively. If V 0 ² U 0 , these sequences converge to a common limit, which is then the
solution to the original IBVP. Of course there are numerous technical details to be checked,
including whether the sense of the convergence is sufficiently strong to imply that the limit
function satisfies the pde in the classical sense.
12