Taylor M.E. Partial Differential Equations III: Nonlinear Equations
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368 15. Nonlinear Parabolic Equations
for r<2;2 .0; 1/. It follows that
(6.45) fv
j
W j 2 Z
C
g is compact in C
1=2ı
Œ0; T ; C
1
.I /
; for all ı>0:
Consequently, (6.41) is sharpened to
(6.46) f
j
W j 2 Z
C
g is compact in C
3=2ı
Œ0; T
;
for all ı>0. It follows that fv
j
g has a limit point
(6.47) v 2
\
0<<1;ı>0
C
Œ0; T ; C
2ı2
.I /
;
and f
j
g has a limit point
(6.48) 2
\
ı>0
C
3=2ı
Œ0; T
:
It remains to show that such v; are unique and give a solution to the Stefan
problem.
To investigate this, choose
0
.t/ satisfying
(6.49)
0
.0/ D 1;
P
0
.0/ D2af
0
.1/;
P
0
0I
define w
1
.t; x/ to solve (6.5) with D
0
;thenset
1
.t/ D 1 2a
Z
t
0
@
x
w
1
.; 1/ d ;
and continue, obtaining a sequence w
j
;
j
;j2 Z
C
, in a fashion similar to that
used to get the sequence v
j
;
j
. As in Lemma 6.1, we see that each
j
satisfies
(6.12). Now we want to compare the differences v
j
w
j
and
j
j
with v
j C1
w
j C1
and
j C1
j C1
.SetV D v
j C1
w
j C1
. Thus V satisfies the PDE
(6.50)
@V
@t
D
1
j
V
xx
C
1
2
P
j
j
xV
x
C
1
j
1
j
@
xx
w
j C1
C
1
2
P
j
j
P
j
j
!
x@
x
w
j C1
;
together with
(6.51) V.0;x/ D 0; V .t; 0/ D V.t;1/ D 0:
6. The Stefan problem 369
Note that the analysis above also gives uniform estimates of the form (6.40)–
(6.46)onw
j
and
j
.Now,forV we have the integral equation
(6.52)
V.t/ D
Z
t
0
ˇ
j
./e
˛
j
.t;/
x@
x
V./
d
C
Z
t
0
1
j
./
1
j
./
e
˛
j
.t;/
@
xx
w
j C1
./ d
C
1
2
Z
t
0
P
j
./
j
./
P
j
./
j
./
!
e
˛
j
.t;/
x@
x
w
j C1
./
d
D S
1
C S
2
C S
3
;
where ˇ
j
./ and ˛
j
.t; / are as in (6.16). As in (6.21), we replace the integrand
in S
1
by
(6.53) ˇ
j
./
@
x
e
˛
j
.t;/
N
M
x
e
˛
j
.t;/
V./:
Thus
(6.54)
kS
1
k
C
1
A
Z
t
0
ˇ
j
./˛
j
.t; /
1=2
kV./k
C
1
d
B
j
.t/
1=2
"
sup
0t
P
j
./
#
Z
t
0
.t /
1=2
kV./k
C
1
d
Ct
1=2
sup
0t
kV./k
C
1
;
provided 0 t T , with T small enough that the uniform estimates on
j
and
P
j
apply.
It does not seem feasible to get a good estimate on S
2
in terms of the C
1
-norm
of w
j C1
, but we do have the following:
(6.55)
kS
2
k
C
1
A
Z
t
0
j
j
./
j
./j ˛
j
.t; /
.2/=2
kw
j C1
./k
C
1C
d
C sup
0t
j
j
./
j
./j sup
0t
kw
j C1
./k
C
1C
t
=2
;
for any 2 .0; 1/. Finally,
(6.56) kS
3
k
C
1
C sup
0t
ˇ
ˇ
ˇ
ˇ
ˇ
P
j
./
j
./
P
j
./
j
./
ˇ
ˇ
ˇ
ˇ
ˇ
sup
0t
kw
j C1
./kt
1=2
:
370 15. Nonlinear Parabolic Equations
Consequently, for t T sufficiently small, 0<<1,wehave
(6.57)
1
2
sup
0t
kV./k
C
1
C sup
0t
j
j
./
j
./j sup
0t
kw
j C1
./k
C
1C
t
=2
C C sup
0t
ˇ
ˇ
ˇ
ˇ
ˇ
P
j
./
j
./
P
j
./
j
./
ˇ
ˇ
ˇ
ˇ
ˇ
sup
0t
kw
j C1
./k
C
1
t
1=2
:
Therefore
(6.58)
sup
0t
j
P
j C1
./
P
j C1
./j
C sup
0t
j
j
./
j
./j sup
0t
kw
j C1
./k
C
1C
t
=2
C C sup
0t
j
P
j
./
P
j
./j sup
0t
kw
j C1
./k
C
1
t
1=2
:
It follows easily that, for T small enough,
(6.59) k
j
j
k
C
1
.Œ0;T /
! 0 and kv
j
w
j
k
C.Œ0;T ;C
1
.I //
! 0;
as j !1. Thus we have the following short-time existence result:
Proposition 6.2. Given f 2 C
1
.I /; f 0; f .0/ D f.1/ D 0, there are a
T>0and a unique solution v; to (6.5)–(6.6), satisfying (6.47)–(6.48). Hence
there is a unique solution u;s to (6.1)–(6.2)on0 t T , satisfying
(6.60) u 2 C
Œ0; T ; C
r2
.I /
;s2 C
3=2ı
Œ0; T
; Ps 0;
for all 2 Œ0; 1/; r < 2; ı > 0.
We want to improve this to a global existence theorem. To do this, we need
further estimates on the local solution v; s. First, it will be useful to have some
regularity results on v not given by (6.60).
Lemma 6.3. The solution v of Proposition 6.2 satisfies
(6.61) v 2 C
Œ0; T ; H
r
.I /
\ C
1
Œ0; T ; H
r2
.I /
; for all r<
5
2
:
Proof. This follows by arguments similar to those used above, plus the following
variant of (6.18):
(6.62) O W H
s
.I / ! H
s
.S
1
/; for 0 s<
1
2
:
6. The Stefan problem 371
We know from (6.60)thatxv
x
./ is continuous in 2 Œ0; T with values in
C
1"
.I / H
12"
.I / H
1
2
"
.I /, so use of the integral equation (6.7)gives
(6.61).
We now take a look at
(6.63) h.t/ D
Z
1
0
v.t; x/ dx;
which, by (6.14), is kv.t/k
L
1
.Wehave
(6.64)
dh
dt
D
1
Z
1
0
@
2
x
v.t; x/ dx C
1
2
P
Z
1
0
x@
x
v.t; x/ dx;
and integrations by parts give for the right side:
(6.65)
1
j
P
2a
@
x
v.t; 0/
!
1
2
P
Z
1
0
v.t; x/ dx
D
1
a
Ps
s
1
s
2
@
x
v.t; 0/
Ps
s
h.t/:
Consequently,
(6.66)
d
dt
sh
D
Ps
a
1
s
@
x
v.t; 0/;
and integration of this gives
(6.67) s.t/h.t/ C
1
a
s.t/ C
Z
t
0
1
s./
@
x
v.; 0/ d D
1
a
C
Z
1
0
f.x/dx:
From (6.14), @
x
v.t; 0/ 0, so each term on the left side of (6.67) is positive. This
gives the upper bound in the two-sided bound,
(6.68) 1 s.t/ 1 C a
Z
1
0
f.x/dx D A;
the lower bound following from the monotonicity Ps 0. Thus 1 .t/ A
2
.
Hence, in (6.7)–(6.9), we have
(6.69) A
2
.t / ˛.t; / t :
We can likewise examine .d=dt/kv.t/k
2
, but since that analysis won’t be cru-
cial for our existence theorem, we leave it to the exercises.
372 15. Nonlinear Parabolic Equations
We next look at the rate of change of k@
x
v.t/k
2
L
2
.Since@
t
v D 0 for x D 0; 1,
we have
(6.70)
1
2
d
dt
@
x
v.t/
2
L
2
D .@
x
@
t
v; @
x
v/ D.@
t
v; @
2
x
v/
D
1
@
2
x
v
2
L
2
1
2
P
x@
x
v; @
2
x
v
;
and an integration by parts, plus use of @
x
v.t; 1/ D
P
.t/=2a,gives
(6.71)
d
dt
@
x
v.t/
2
L
2
D
2
@
2
x
v
2
L
2
P
P
2
8a
2
C
1
2
P
@
x
v
2
L
2
or, equivalently,
(6.72) s
d
dt
1
s
@
x
v
2
L
2
D
2
@
2
x
v
2
L
2
P
P
2
8a
2
:
Since the right side of (6.72)is 0, this gives, upon integration,
(6.73) k@
x
v.t/k
2
L
2
s.t/k@
x
f k
2
L
2
:
Note that, by (6.68), the right side is Ak@
x
f k
2
L
2
, which is independent of t.
Using (6.73)and(6.7), we have, for r 2 Œ1; 2/,
(6.74) kv.t/k
H
r
ke
˛.t;0/
f k
H
r
C C k@
x
f k
L
2
Z
t
0
ˇ./˛.t;/
r=2
d;
and the first term on the right is kf k
H
r
,since
f 2 H
r
.I / \ H
1
0
.I / H) Of 2 H
r
.S
1
/;
for r<5=2.By(6.69), we deduce that, for r 2 Œ1; 2/,
(6.75) kv.t/k
H
r
C
1
C C
2
Z
t
0
Ps./ .t /
r=2
d;
with C
j
D C
j
.r/kf k
H
r
independent of t.Takingr 2 .3=2; 2/, and using
Ps.t/ Dv
x
.t; 1/=2a,whichis C kvk
H
r
, we deduce that
(6.76) Ps.t/ K
1
C K
2
Z
t
0
Ps. /.t /
r=2
d:
6. The Stefan problem 373
From this we can establish the following important estimate:
Lemma 6.4. If v solves (6.3)for0 t T and satisfies (6.61), then
(6.77) sup
0tT
Ps.t/ K
0
;
where K
0
D K
0
.kf k
H
r
/.rD 3=2 C "/ is independent of T .
Proof. Pick >0small enough that
R
0
t
r=2
dt 1=2K
2
. Thus, writing the
interval Œ0; t as Œ0; t [ Œt ; t,wehave
(6.78) K
2
Z
t
0
Ps./ .t /
r=2
d
1
2
sup
t
Ps./ C C
2
r=2
s.t / 1
:
We conclude from (6.76)and(6.68)that
(6.79) sup
0tT
Ps.t/
1
2
sup
0tT
Ps.t/ C K
1
C K
2
r=2
akf k
L
1
;
which gives (6.77).
Returning to (6.75), we deduce that the solution to (6.3) given by Proposition
6.2 satisfies, for any r 2 Œ1; 2/,
(6.80) kv.t/k
H
r
.I /
K
r
;0 t T;
with K
r
independent of T . We know that xv
x
./ has an H
s
-bound for any s<1,
and, via (6.62), we can use such a bound on xv
x
./ for s<
1
2
, to conclude, via
(6.7), that (6.80) holds for any r 2 Œ1; 5=2/. Now familiar methods establish the
following:
Theorem 6.5. Given f 2 C
1
.I /; f 0; f .0/ D f.1/ D 0, there is a unique
solution v; to (6.5)–(6.6), defined for all t 2 Œ0; 1/, satisfying
(6.81) v 2 C
Œ0; 1/; H
r
.I /
\ C
1
Œ0; 1/; H
r2
.I /
;r<
5
2
;
and
(6.82) 2 C
Œ0; 1/
;<
3
2
:
We now tackle the task of showing that u and s, or equivalently v and ,are
smooth for t 2 .0; 1/ D J . It is convenient to set
(6.83) V.T;x/ D v.t; x/; T D ˛.t; 0/ D
Z
t
0
./
1
d;
374 15. Nonlinear Parabolic Equations
so @
T
V.T;x/ D .t/@
t
v.t; x/, and we have
(6.84)
@V
@T
D V
xx
C .T /xV
x
;.T/D
1
2
P
.t/;
with
(6.85) V.0;x/ D f.x/; V.t;x/ D 0; for x 2 @I:
Note that (6.6) is equivalent to
(6.86) .T / DaV
x
.T; 1/:
In place of (6.7), we use
(6.87) V.T/ D e
T
f C
Z
T
0
./e
.T /
xV
x
./
d:
The results (6.81)and(6.82)imply
(6.88)
V 2 C
Œ0; 1/; H
r
.I /
\ C
1
Œ0; 1/; H
r2
.I /
;2 C
1=2ı
Œ0; 1/
;
for any r<5=2;ı>0. Consequently, V 2 C
1=2ı
Œ0; 1/; H
3=2ı
.I /
for all
ı>0,so./xV
x
./ 2 C
1=2ı
Œ0; 1/; H
1=2ı
.I /
. The following lemma is
useful.
Lemma 6.6. Suppose
G.T / D
Z
T
0
e
.T /
F./ d:
Then, for any r>0;s2 .0; 1=2/,
F 2 C
Œ0; 1/; L
2
.I /
\C
r
.0; 1/; H
s
.I /
H) G 2 C
r
.0; 1/; H
sC2ı
.I /
;
for all ı>0.
The proof is straightforward. Applying this to F./ D ./xV
x
./, we deduce
that the right side of (6.87) belongs to C
1=2ı
.0; 1/; H
5=2ı
.I /
,forallı>0.
Thus, with J D .0; 1/,
(6.89) V 2 C
1=2ı
J;H
5=2ı
.I /
:
Note that this is stronger than the first inclusion in (6.88). Making use of the PDE
(6.84), we deduce that V
T
2 C
1=2ı
.J; H
1=2ı
.I //,so
6. The Stefan problem 375
(6.90) V 2 C
3=2ı
J;H
1=2ı
.I /
:
Interpolation of (6.89)and(6.90)gives
(6.91) V 2 C
1ı
J;H
3=2ı
.I /
I
hence, by (6.86),
(6.92) 2 C
1ı
.J /:
Now we have improved all three parts of (6.88), essentially increasing the degree
of regularity in T by one half, at least for T 2 J D .0; 1/. Iterating this argu-
ment, we obtain
(6.93)
V 2 C
j=2ı
.J; H
5=2ı
.I // \ C
1Cj=2ı
.J; H
1=2ı
.I //;
2 C
1=2Cj=2ı
.J /;
for each j 2 Z
C
. We are well on the way to establishing the following:
Proposition 6.7. The solutions v; of Theorem 6.5 have the property
(6.94) v 2 C
1
.0; 1/ I
;2 C
1
.0; 1/
:
Proof. That 2 C
1
.J / follows from 2 C
1
.J /. For the rest, it suffices to
show that V 2 C
1
.J I/. We get this from (6.93) together with the PDE (6.84).
In fact, this yields
(6.95) V
xx
2 C
1
.J; H
3=2ı
.I //;
hence
(6.96) V 2 C
1
.J; H
7=2ı
.I //:
Iterating this argument finishes the proof.
A number of variants of (6.1)–(6.2) are studied. For example, one often sees a
nonhomogeneous boundary condition at the left boundary:
(6.97) u.t; 0/ D g.t/:
Or the boundary condition at x D 0 may be of Neumann type:
(6.98) u
x
.0; t/ D h.t/:
376 15. Nonlinear Parabolic Equations
Both the PDE and the boundary conditions may have t-dependent coefficients.
For example, the PDE might be
(6.99) u
t
D A.t/u
xx
:
Some studies of these problems can be found in Chap. 8 of [Fr1]andin[KMP],
where particular attention is paid to the nature of the dependence of the solution
on the coefficient A.t/,assumedtobe>0.
There are also two-phase Stefan problems, where the ice is not assumed to be
at temperature 0, but rather at a temperature u
i
.t; x/ 0, to be determined as
part of the problem. Furthermore, these problems are most interesting in higher-
dimensional space. More material on this can be found in [Fr2].
Exercises
1. If v solves (6.3)–(6.4), show that
d
dt
s
v.t/
2
L
2
D
2
s
v
x
.t/
2
L
2
I
hence
s.t/kv.t/k
2
L
2
C 2
Z
t
0
s./
1
v
x
./
2
L
2
d Dkf k
2
L
2
:
2. If u satisfies (6.1)–(6.2), show that
(6.100) s.t/
2
D 1 C 2a
Z
1
0
xf .x / d x 2a
Z
s.t/
0
xu.t; x/ dx:
Compare the upper bound on s.t/ this gives with (6.68).
Show that, conversely, (6.1)and(6.100)imply(6.1)–(6.2). This result, or rather its ana-
logue in the more general context of the nonhomogeneous boundary condition (6.97),
played a role in the analysis in [CH].
7. Quasi-linear parabolic equations I
In this section we begin to study the initial-value problem
(7.1)
@u
@t
D
X
j;k
A
jk
.t;x;D
1
x
u/@
j
@
k
u C B.t; x; D
1
x
u/; u.0/ D f:
Here, u takes values in R
K
, and each A
jk
can be a symmetric K K matrix;
we assume A
jk
and B are smooth in their arguments. We assume for simplicity
that x 2 M D T
n
,then-dimensional torus. Modifications for a more general,
7. Quasi-linear parabolic equations I 377
compact M will be contained in the stronger analysis made in 8. We impose the
following “strong parabolicity condition”:
(7.2)
X
j;k
A
jk
.t;x;D
1
x
u/
j
k
C
0
jj
2
I;
where to say that a pair of symmetric K K matrices S
j
satisfies S
1
S
2
is to
say that S
1
S
2
is a positive-semidefinite matrix.
We will use a “modified Galerkin method” to produce a short-time solution.
We consider the approximating equation
(7.3)
@u
"
@t
D J
"
X
A
jk
.t;x;D
1
x
J
"
u
"
/@
j
@
k
J
"
u
"
C J
"
B.t; x; D
1
x
J
"
u
"
/
D J
"
L
"
J
"
u
"
C J
"
B
"
;
u
"
.0/ D J
"
f:
Here J
"
is a Friedrichs mollifier, which we can take in the form
J
"
D '."
p
/;
with an even function ' 2 S.R/; '.0/ D 1. Equivalently, the Fourier coefficients
O
f.k/of f 2 D
0
.T
n
/ are related to those of J
"
f by
.J
"
f/O.k/ D '."jkj/
O
f.k/; k 2 Z
n
:
For any fixed ">0, the right side of (7.3) is Lipschitz in u
"
with values in
practically any Banach space of functions, so the existence of short-time solutions
to (7.3) follows by the material of Chap. 1. Our task will be to show that the
solution u
"
exists for t in an interval independent of " 2 .0; 1 and has a limit as
" & 0, solving (7.1).
To do this, we estimate the H
`
-norm of solutions to (7.3). We begin with
(7.4)
d
dt
kD
˛
u
"
.t/k
2
L
2
D 2.D
˛
J
"
L
"
J
"
u
"
;D
˛
u
"
/ C 2.D
˛
B
"
;D
˛
J
"
u
"
/:
Since J
"
commutes with D
˛
and is self-adjoint, we can write the first term on the
right as
(7.5) 2.L
"
D
˛
J
"
u
"
;D
˛
J
"
u
"
/ C 2.ŒD
˛
;L
"
J
"
u
"
;D
˛
J
"
u
"
/:
To analyze the first term in (7.5), write it as
(7.6) 2.L
"
v; v/ D2
X
j;k
.A
jk
"
@
j
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