On existence and uniqueness of a

solution to the Cauchy problem for the parabolic p-Laplace

equation in the whole space.

Mikhail Surnachev Keldysh Institute of Applied Mathematics

(RAS) Joint work with V.V. Zhikov (Vladimir)

2

Setting of the problem.

2

2

div | | , , 0,

( ,0) ( ) ( ), 1.

p n

n

loc

uu u x t

t

u x f x L p

The problem was previously studied by E. DiBenedetto in the L1 setting.

Our aim was to give an independent proof in the L2 setting. These results will be used later to study stabilization properties of solutions.

(1)

3

Definition of a solution

By a solution to the problem (1) in the cylinder

2 1,(0, ; ( )) (0, ; ( ))n p p n

loc locu C T L L T W

such that for any compactly (in space) supported test-function

2 2

( , ) :

( (0, )), ( ,0) ( ), ( (0, ))n n p n

t

x t

L T x L L T

and any t1 from the interval (0,T) there holds the following integral identity:

Q (0, )n

T T

we mean a function

4

1 1( , ) ( , ) ( )

( ) ( ,0)

nT T

n

t

Q Q

u x t x t dx u dxdt A u dxdt

f x x dx

2Here ( ) for .

p nA

The solution u=u(x,t)is called a global solution to (1) if it is a solution in the cylinder QT for any T>0.

5

Main results

Two different cases: degenerate (p>2) and singular (1<p<2).

The singular case proved to be easier to handle compared with the degenerate case. (Surprisingly!)

In the singular case there are no additional conditions on the initial data to get a global existence theorem.

To get the uniqueness theorem in the singular case it is sufficient to restrict the range of the exponent p:

22

2

np

n

6

Theorem 1. Let 1<p<2. Then there exists a global solution to (1) with the following energy estimate:

2

2 2

2 2 2 2

0(0, )

sup ( , ) ( )

where ( , ) and B : .

R R R

pn

p p p

t TB B T B

n

R

u x t dx u dxdt f x dx CT R

C C n p x x R

Moreover, for the difference of two solutions with the same initial data u1(x,0)=u2(x,0)=f(x) we have

Hence, if 2n/(n+2) <p<2 a solution to (1) is unique.

(2)

2 2

2 2 2

1 20

sup ( ) ( , ) .

R

pn

p p

t TB

u u x t dx CT R

(3)

7

Indeed, note that the constant C in (3) is independent of both R,T. Note also that for 2n/(n+2)<p<2 we

have 𝑛 +2𝑝

𝑝−2< 0. It remains to send R to infinity in (3)

to get the left-hand side of this expression equal to zero.

The proof of the energy estimate in the singular case is elementary. It will be given the end of the talk.

Quite unlike what happens in the field of the regularity theory, the degenerate case (p>2) is much harder and requires a lot of technical work. Moreover, we managed to prove the uniqueness of solutions in this case in narrower classes than the existence.

8

Degenerate case (p>2).

To state our theorem we need to introduce the following (weighted energy) classes.

First, to describe the initial data we need the classes

2 2( ) : ( ) 1 | | .n

psn

s locf L f x x dx

F

To describe the behaviour at infinity of solutions we introduce the function classes

9

2

0

( )

( , ) :

.sup ( , ) (1 | |) | | (1 | |)n

T

s T

ps p ps

t TQ

B Q

u u x t

u x t x dx u x dxdt

Now we introduce two critical values of the parameter s.

22 ,

221 ,

22 2.

e u

e

if p npn n

s sp p

s if p np

The parameter se is responsible for the existence of solutions and the parameter su is responsible for the uniqueness of solutions.

10

Theorem 2. Let p>2. Assume that the initial data f

belongs to the class Fs and s<se. Then there exists a

global solution to the Cauchy problem (1) which belongs

to the class Bs(QT) for all T>0. If s<su a

solution in QT is unique.

To make the statement more precise we need to introduce weight functions which additionally depend on the parameter R.

,

1, | | ,

| || | .

pss R

if x R

xif x R

R

11

Then Theorem 2 gives the following quantitative estimate of the norm of the solution. It asserts that for any T>0 there exists R=R(f,T) such that

2

, ,0

2

,

sup ( , ) ( ) ( )

( ) ( )

nT

n

p

s R s Rt T

Q

s R

u x t x dx u x dxdt

C f x x dx

where C=C(n,p). The parameter R is chosen such that

2 2

2 2 2

, 0( ) ( )n

pn

p p

s Rf x x dx c T R

where c0= c0(n,p,s) is a positive constant. It can be easily proved that (5) holds for any R sufficiently large.

(4)

(5)

12

When 2<p<n we require in fact an additional (natural) integrability assumption on the solution in the uniqueness result. In can be easily seen that the existence part of the theorem produces solutions which satisfy this additional regularity condition.

1

0

(1 ) .n

p

T np ns n pu x dx dt

13

Proof of the energy estimate for p<2.

Let be a nonnegative “cut-off” function. In the integral identity which defines a solution choose :

0( ) ( )nx C

u

2

2 2

2 1

1( , ) ( ) | | ( )

2

1( ) ( ) | |

2

1 1( ) ( ) | | ( ) | | .

2 2

n

n

n

p

Q

p

Q

p p p

Q Q

u x x dx u x dxdt

f x x dx u u u dxdt

f x x dx u x dxdt C u dxdt

Doing the obvious cancellation and taking the supremum of the left-hand side with respect to we obtain

14

2

0

2 1

sup ( , ) ( ) | |

( ) ( ) | | .

nT

nT

p

t TQ

p p

Q

u x t x dx u dxdt

f x x dx C u dxdt

Next, use the Young inequality again to estimate

1

21 3 2

2 2 2

21 3 2

2 2 2

0

| |

1

2

1sup ( , ) ( ) .

2

T

T T

nT

p p

Q

pp p

p p

Q Q

pp p

p p

t TQ

u dxdt

u dxdt CT dxdtT

u x t x dx CT dxdt

15

2

0

21 3 2

2 2 2

sup ( , ) ( ) | | ( )

2 ( ) ( ) .

nT

nT

p

t TQ

pp p

p p

Q

u x t x dx u x dxdt

f x x dx CT dxdt

Combining the last two inequalities we easily obtain

Notice that 1/p-3/2>-1 for 1<p<2. Let ccR be a nonnegative “cut-off” function such that c

is 1 on BR , 0 outside B2R , bounded from above by 1, and with the modulus of its gradient bounded by 4/R. Taking in (6)

0 ( )nC

2 /(2 ).p p

R R c

(6)

we obtain

16

2

0

2 2

2 2 2

sup ( , ) ( ) | | ( )

2 ( ) ( ) .

nT

n

p

t TQ

pn

p p

u x t x dx u x dxdt

f x x dx CT R

Since R1 on the ball BR and R0 outside B2R we obtain (2):

2

2

0(0, )

2 2

2 2 2

sup ( , )

( ) .

R R

R

p

t TB B T

pn

p p

B

u x t dx u dxdt

f x dx CT R

17

Obtaining the energy estimate for the difference of two solutions.

Let u1 and u2 be two solutions to the problem (1). Denote w=u1-u2 . Substracting the integral identities for u1 and u2 with the test function w, we obtain

18

2

2 1 2 1

2 1

2 1 2 1

1

1( , ) ( )

2

( ) ( ) ( )

( ) ( )

1( ) ( ) ( )

2

| | .

n

Q

Q

Q

p p

Q

w x x dx

A u A u u u x dxdt

w A u A u dxdt

A u A u u u x dxdt

C w dxdt

(7)

19

Here we used the following algebraic inequality:

1( ) ( ) ( ) ( )

for any ,

p

p

n

A b A a C A b A a b a

a b

with the constant C=C(n,p). Doing the obvious cancellation in (7) yields

2

0

2 1 2 1

2 1

sup ( , ) ( )

( ) ( )

( ) ( ) | | .

n

T

nT

t T

Q

p p

Q

w x t x dx

A u A u u u dxdt

f x x dx C w dxdt

(8)

20

We proceed further exactly as before to obtain (3).

Remark: Inequality (8) was the subject of a separate lemma. Although its proof is elementary in nature we did not manage to find it in standard reference sources like the book of E. DiBenedetto or the lecture notes of P. Lindqvist. This inequality is the cornerstone of proving the uniqueness in the singular case.

21

Some notes on the degenerate case.

Here we use a special transform. Let

with ( ).u ve x

For the function v we get the modified equation

(2 ) div( ( )) ( -1) ( )

where .

p

te v A b p A b

b v v

Further on we will denote (2 ) .pe

(9)

22

Multiplying equation (9) by v and integrating by parts (these operations are formal, since in reality we work with integral identities) we obtain the following energy equality:

2

2

1( , ) ( ) | |

2

1( ,0) ( ) ( ) .

2

n

n

p

Q

Q

v x x dx b dxdt

v x x dx p v A b dxdt

(10)

Applying the Young inequality to the second term in the right-hand side and taking the supremum with respect to from 0 to T yields

23

2

0

2

sup ( , ) ( ) | |

( ,0) ( ) .

nT

nT

p

t TQ

p

Q

v x x dx b dxdt

v x x dx C v dxdt

Denote in the last inequality

2

0

2

0

sup ( , ) ( ) , | | ,

( ,0) ( ) , .

nT

nT

p

t TQ

p

Q

I v x x dx K b dxdt

I v x x dx J v dxdt

(11)

Using this notation we rewrite (11) as

24

0 .I K I CJ (12)

Now it is a good time to specify the function :

0, if | | ,

| |ln if | | .

x R

xs x R

R

Thus, the function depends on the two parameters, s and R . Using such choice of we can estimate the term J in the right-hand side of (12) for s≤se .

25

(1 /2) /2( ) .n n p pJ K C R I T (13)

Using this estimate we obtain from (12)

(1 /2) /2

0 .n n p pI K I R TC I

While it almost looks like an energy estimate it still is not one. We need to use yet one trick, which we present here in the form of a lemma.

Lemma 1. Let a positive continuous function x(t) satisfy the inequality

1

0 , 0.x x Mx

Let x(0)=x0. There exists a positive constant c0=c0 such that the inequality x0M

1/≤c0 implies

(14)

26

the estimate x(t)≤c1x0 for all t>0 with a constant c1=c1. We can take

1/

0 1

1 1( ) , ( ) .

1 1c c

Applying Lemma 1 to inequality (14) we obtain the following conditional result:

2 2

2 2 2

0 2

0

If

then

)

(

.

n

pn

p p pI f x e d c

I

x

C

R

K

T

I

(15)

27

The second lemma effectively asserts that condition (15) holds for all sufficiently large R:

Lemma 2. The following estimate is valid:

2 ( ), 0,( )

(1), 0.n

ps

p o R sf x e dx

O s

It remains to observe that

2.

2e

ps s ps n

p

to complete the proof.

28

Remark: The main (technical) problem with the uniqueness in the degenerate case: estimate of the difference of flows

2 2 2

( ) ( ) ( ) ( ), .p p

A b A a C a b A b A a b a

(Compare with (8)).