A Nonlinear Parabolic Initial Boundary Value Problem
For U Ð R n , open, bounded with smooth boundary @, consider the IBVP
/ t uÝx, tÞ ? divÝaÝuÝx, tÞÞ4uÝx, tÞÞ = fÝx, tÞ
in U × Ý0, TÞ
uÝx, tÞ = 0
on @ × Ý0, TÞ
uÝx, 0Þ = u 0 ÝxÞ
in U
where
a:R¸R
belongs to L K ÝRÞ
and
0 < C 0 ² aÝsÞ ² C 1 -s 5 R.
V = H 10 ÝUÞ Ð H = L 2 ÝUÞ Ð V v = H ?1 ÝUÞ
Let
and define
Evidently,
aÝu, vÞ = X aÝuÞ 4u 6 4v dx
U
for u, v 5 V
| aÝu, vÞ| ² C 1 | XU 4u 6 4v dx| ² C 1 || u|| V || v|| V
and it follows that AÝuÞ defined by
ÖAÝuÞ, v× V v ×V = aÝu, vÞ
for u, v 5 V
defines a nonlinear map from V into V’ such that for all u 5 V,
||AÝuÞ|| V v ² C 1 || u|| V
Then we define uÝtÞ 5 L 2 ß0, T : Và to be a weak solution of the IBVP if
(1)
Ýu v ÝtÞ, vÞ H + aÝuÝtÞ, vÞ = ÝfÝtÞ, vÞ H
forall v 5 V,
uÝ0Þ = u 0
Alternatively, uÝtÞ must satisfy
(2)
u v ÝtÞ + AÝuÝtÞÞ = fÝtÞ,
uÝ0Þ = u 0 .
1
Existence and Uniqueness Theorem
Suppose f 5 L 2 ß0, T : Hà and u 0 5 H. Then there exists a unique weak solution
for the IBVP. This weak solution has the following additional smoothness,
u v ÝtÞ 5 L 2 ß0, T : V v à
and
uÝtÞ 5 Cß0, T : Hà
Proof- Since the embedding of V in H is compact, it follows that there is an orthonormal basis for H, áw k â, which is simultaneously an orthogonal basis for V; i.e.,
Ýw j , w k Þ H =
0
if
j®k
V j Ýw j , w j Þ V
if
j=k
Then, for N = 1, 2, ... let
N
u N ÝtÞ = > k=1 c k,N ÝtÞ w k
satisfy,
(3)
Ýu vN ÝtÞ, w j Þ H + aÝu N ÝtÞ, w j Þ = ÝfÝtÞ, w j Þ H
Ýu N Ý0Þ ? u 0 , w j Þ H = 0
for j = 1, ..., N
for j = 1, ..., N
This is a system of nonlinear ODE’s for the coefficients ác j,N ÝtÞâ,
N
c vj,N ÝtÞ + > k=1 B j,k Ýc 1 , ..., c N Þ c k,N ÝtÞ = f j ÝtÞ
c j,N Ý0Þ = Ýu 0 , , w j Þ H
where
B j,k Ýc 1 , ..., c N Þ = X aÝu N Þ 4w j 6 4w k dx
U
For each fixed N, it is well known that this nonlinear IVP has a unique solution on an interval
ß0, T N à with T N ² T. In order to have T N < T the solution would have to become infinite as t
tends to T N . However, the a-priori estimates we are about to prove will show that such
unbounded behavior for u N ÝtÞ is not possible. Therefore, for each N, an approximate
solution u N ÝtÞ exists on ß0, Tà.
a-priori estimates
It follows from (3) that
(4)
d
dt
|| u N ÝtÞ|| 2H + 2aÝ u N ÝtÞ, u N ÝtÞÞ = 2ÝfÝtÞ, u N ÝtÞÞ H .
Then,
d
dt
|| u N ÝtÞ|| 2H + 2C 0 || u N ÝtÞ|| 2V ² 2 || fÝtÞ || H || u N ÝtÞ || H
2
² Ý1/C 0 Þ || fÝtÞ || 2H + C 0 || u N ÝtÞ || 2V
or,
(5)
d
dt
|| u N ÝtÞ|| 2H + C 0 || u N ÝtÞ|| 2V ² Ý1/C 0 Þ || fÝtÞ || 2H
d
dt
|| u N ÝtÞ|| 2H ² Ý1/C 0 Þ || fÝtÞ || 2H
Then, clearly,
Integrate this last expression from t = 0 to t = b, to get
T
|| u N ÝbÞ|| 2H ² || u N Ý0Þ|| 2H + Ý1/C 0 Þ X 0 || fÝtÞ || 2H dt ² ||u 0 || 2H + || f || 2L 2 ß0,T,Hà
Since this holds for all b 5 ß0, Tà it follows that
(6)
|| u N || 2L K ß0,T:Hà ² ||u 0 || 2H + || f || 2L 2 ß0,T,Hà = M 1
This shows that the sequence of approximate solutions is a bounded infinite set
in L K ß0, T : Hà
Integrating (5) from 0 to T, leads to
T
T
|| u N ÝTÞ|| 2H + C 0 X 0 || u N ÝtÞ|| 2V dt ² || u N Ý0Þ|| 2H + Ý1/C 0 Þ X 0 || fÝtÞ || 2H dt
i.e.,
(7)
|| u N || 2L 2 ß0,T:Và ² Ý1/C 0 Þ || u 0 || 2H + Ý1/C 0 Þ 2 || f || 2L 2 ß0,T:Hà = M 2
This shows that the sequence of approximate solutions is a bounded infinite set
in L 2 ß0, T : Và.
Next, for each N, let P N : V ¸ V N = spanáw 1 , ..., w N â denote the projection from
V into V N . Then it follows from (3) that for each N,
Öu vN ÝtÞ + AÝ u N ÝtÞÞ ? fÝtÞ, P N v× V v ×V = 0
-v 5 V
This is equivalent to
ÖP fN áu vN ÝtÞ + AÝ u N ÝtÞÞ ? fÝtÞâ, v× V v ×V = 0
-v 5 V
where P fN : V vN ¸ V v But P fN áu vN ÝtÞ = u vN ÝtÞ, so
|| u vN ÝtÞ|| L 2 ß0,T:V v à =
T
X 0 Ýsup ||v|| V=1 | Ö u vN ÝtÞ, v× V v ×V |Þ 2 dt
1/2
3
|| u vN ÝtÞ|| L 2 ß0,T:V v à = ||P fN áAÝ u N ÝtÞÞ ? fÝtÞâ|| L 2 ß0,T:V v à
² || AÝ u N ÝtÞÞ ? fÝtÞ|| L 2 ß0,T:V v à
² || AÝ u N Ý6ÞÞ|| L 2 ß0,T:V v à + || f || L 2 ß0,T:V v à
² C 1 || u N || L 2 ß0,T:Và + || f || L 2 ß0,T:V v à ;
i.e.,
|| u vN ÝtÞ|| L 2 ß0,T:V v à ² M 1 C 1 + || f || L 2 ß0,T:V v à = M 3
(8)
This shows that the sequence of derivatives approximate solutions is
a bounded infinite set in L 2 ß0, T : V v à
passing to the limit
The a-priori estimates (7) and (8) imply the existence of a subsequence of
á u N ÝtÞâ such that
(a) u N ÝtÞ ¸ uÝtÞ weakly in L 2 ß0, T : Và
(b) u vN ÝtÞ ¸ vÝtÞ weakly in L 2 ß0, T : V v à
The usual distributional argument is used to show that vÝtÞ = u v ÝtÞ.
It follows from ÝaÞ and the fact that V is compactly embedded in H that
there is a further subsequence (still denoted by á u N ÝtÞâ) such that
(c) u N ÝtÞ ¸ uÝtÞ strongly in L 2 ß0, T : Hà
It also follows from (7) and the assumptions on aÝ6Þ, that
|| AÝ u N Ý6ÞÞ|| L 2 ß0,T:V v à ² C 1 M 2
which leads to
(d) AÝ u N Ý6ÞÞ ¸ B 1 weakly in L 2 ß0, T : V v à
Now let
u
bÝuÞ = X aÝsÞ ds;
0
i.e., b v ÝuÞ = aÝuÞ, bÝ0Þ = 0, and bÝuÞ is continuous on R.
Also,
C 0 |u| ² | bÝuÞ| ² C 1 |u|
-u 5 R
4
Since 4bÝuÞ = aÝuÞ4u, it follows from (6) and (7) that
||bÝ u N Ý6ÞÞ|| L 2 ß0,T:Và ² C 1 M 1
-N.
It now follows that
(e) bÝ u N Ý6ÞÞ ¸ B 2
weakly in L 2 ß0, T : Và
and the compactness of the embedding V Ð H then implies that
bÝ u N Ý6ÞÞ ¸ B 2 strongly in L 2 ß0, T : Hà . But bÝ6Þ is continuous on R
and u N ÝtÞ ¸ uÝtÞ strongly in L 2 ß0, T : Hà . Then it follows that
bÝ u N Ý6ÞÞ ¸ bÝuÞ strongly in L 2 ß0, T : Hà , hence bÝuÞ = B 2 .
aÝu, vÞ = X aÝuÞ 4u 6 4v dx = X 4bÝuÞ 6 4v dx
Now
U
U
= ÝbÝuÞ, vÞ V ? ÝbÝuÞ, vÞ H
hence for arbitrary v 5 V
T
T
X 0 Ö AÝ u N ÝtÞÞ, v× V v ×V dt = X 0 áÝbÝu N Þ, vÞ V ? ÝbÝu N Þ, vÞ H â dt.
Then we have
T
T
X 0 áÝbÝu N Þ, vÞ V ? ÝbÝu N Þ, vÞ H â dt ¸ X 0 áÝbÝuÞ, vÞ V ? ÝbÝuÞ, vÞ H â dt
T
T
and
X 0 Ö AÝ u N ÝtÞÞ, v× V v ×V dt ¸ X 0 Ö B 1 , v× V v ×V dt.
Also
X 0 áÝbÝuÞ, vÞ V ? ÝbÝuÞ, vÞ H â dt = X 0 aÝu, vÞ dt = X 0 Ö AÝ uÝtÞÞ, v× V v ×V dt
T
T
T
and together these results imply that AÝ uÝtÞÞ = B 1 ; i.e., AÝ u N Ý6ÞÞ ¸ AÝ uÝtÞÞ,
weakly in L 2 ß0, T : V v à. Then we can pass to the limit in (3) to see that uÝtÞ
is a weak solution of the partial differential equation. Then u v ÝtÞ = fÝtÞ ? AÝuÝtÞÞ
belongs to L 2 ß0, T : V v à and it follows that uÝtÞ 5 L 2 ß0, T : VàVCß0, T : Hà. In addition,
we can show that uÝ0Þ = u 0 .
To prove uniqueness, suppose that u 1 ÝtÞ, u 2 ÝtÞ 5 L 2 ß0, T : Và are two weak solutions
of (1). Then wÝtÞ = u 1 ÝtÞ ? u 2 ÝtÞ solves
Ö/ t wÝtÞ, v× V v ×V + ÖAÝu 1 Þ ? AÝu 2 Þ, v× V v ×V = 0
-v 5 V,
wÝ0Þ = 0.
5
But
ÖAÝu 1 Þ ? AÝu 2 Þ, v× V v ×V = XU 4ÝbÝu 1 Þ ? bÝu 2 ÞÞ 6 4v dx
= ÝbÝu 1 Þ ? bÝu 2 Þ, vÞ V ? ÝbÝu 1 Þ ? bÝu 2 Þ, vÞ H
and
ÝbÝu 1 Þ ? bÝu 2 Þ, vÞ V = ÖbÝu 1 Þ ? bÝu 2 Þ, Jv× V×V v = ÝbÝu 1 Þ ? bÝu 2 Þ, JvÞ H
where J denotes the isomorphism that associates v 5 V with a unique element
Jv 5 V v as prescribed by the fact that H is the pivot space between V and V v If we
choose v 5 V such that Jv = w, then
ÝbÝu 1 Þ ? bÝu 2 Þ, vÞ V = ÝbÝu 1 Þ ? bÝu 2 Þ, wÞ H
ÖAÝu 1 Þ ? AÝu 2 Þ, v× V v ×V = ÝbÝu 1 Þ ? bÝu 2 Þ, wÞ H ? ÝbÝu 1 Þ ? bÝu 2 Þ, vÞ H .
and
It follows that
Ö/ t wÝtÞ, v× V v ×V + ÝbÝu 1 Þ ? bÝu 2 Þ, wÞ H = ÝbÝu 1 Þ ? bÝu 2 Þ, vÞ H .
and
Ö/ t wÝtÞ, J ?1 w× V v ×V + C 0 || w|| 2H ² C 1 XU |wÝxÞ| |J ?1 wÝxÞ| dx.
²
Note also, that
and
Then from
it follows that
1
2
2
C 0 ||w|| 2H + C 2 ||J ?1 wÝxÞ|| H .
Ö/ t wÝtÞ, J ?1 w× V v ×V = Ý/ t wÝtÞ, wÞ V v =
1 d
2 dt
||wÝtÞ|| 2V v
2
||J ?1 wÝxÞ|| H = ÝJ ?1 wÝxÞ, J ?1 wÝxÞÞ H = Ýw, wÞ V v .
d
dt
||wÝtÞ|| 2V v ² 2C 2 ||wÝtÞ|| 2V v
||wÝtÞ|| V v = 0
wÝ0Þ = 0,
so the solution is unique.
6