A degenerate parabolic equation

arising in image processing

G.Citti -M.Manfredini

1 Introduction

We prove here an existence result for solutions for a parabolic equation,with non

local coecients arising in image processing.An image is a bounded function

u:D!R dened on a rectangular region D.If the function u is not regular,

the image is noisy and it is not possible to use it directly in applications,but

is necessary to smooth it by means of a nonlinear evolution problem,with

Neumann boundary data.To this end dierent model have been proposed.

Perona and Malik proposed in [PM] the following anisotropic diusion model:

@

t

u = div(f(jDuj)Du) in D [0;T];

with a suitable decreasing function f.Even though numerical experiments

provide the desired regularization eect,the problem can be ill posed from

an analytic point of view for particular choice of the function f,and really few

is known about its solutions (see [KK],[K]).Then the model was modied in

dierent ways:in [CLMC] the following equation was proposed,

@

t

u = div(f(jDG

uj)Du) in D [0;T];

where G

is a Gaussian kernel depending on a parameter .For the associated

problem with L

2

initial datum,also existence and uniqueness was proved in

[CLMC].In [ALM] and [AE] non divergence versions of the same operator was

proposed,whose simplest form is

@

t

u = f(jDG

uj)jDujdiv

Du

jDuj

+g(u) in D [0;T]:

The existence of solutions was proved with viscosity solutions methods.Equa-

tions of this type has received a lot of attention because of its geometrical

interpretation:models dened in terms of motion by mean curvature have been

proposed by [OS],[S],and model related to properties of the principal curvatures

Dipartimento di Matematica,Univ.di Bologna,P.zza di Porta S.Donato 5,40127,

Bologna,ITALY,e-mail citti@dm.unibo.it,manfredi@dm.unibo.it.Investigation supported

by University of Bologna.Founds for selected research topics.

1

are due to [CS],[ST],[SOL] We also refer to [ES],[CGG],[GGIS],[IS],[S],[GG],

for the application of viscosity methods to mean curvature equations.Similar

techniques can be applied to the study of movies,which can be considered as

family (u

)

2[0;1]

of images.

In [AGLM] the authors introduced a new model in an axiomatic way,requir-

ing that the solutions satisfy maximum and comparison primciple,and are in-

variant with respect to suitable groups of transformations.The resulting model

- which has a viscosity solution by construction - and it is the following one

@

t

u = jDuj

sign(curv(u))acc(u)

+

sign(curv(u)) u(;0) = u

0

;

where curv(u) is the mean curvature of the graph of u,and the acc(u) repre-

sents the acceleration of the movie in the direction of the spatial gradient.By

simplicity of notations we will denote

clt(u) = jDuj

sign(curv(u))acc(u)

+

:

In [G] it is proved that clt(u) has the following discretisation,which we will use

here as a denition of it.

clt(u)(x;;t) = min

1

2A

+

x

;

2

2A

x

n

ju(x +

1

; +;t) u(x;;t)j+ (1)

ju(x

2

; ;t) u(x;;t)j +j < DG

u;

1

2

> j

o

where (x;) 2 R

n

R.

A

+

x

= f 2 R

n

:x + 2

;jj 2rg

A

x

= f 2 R

n

:x 2

;jj 2rg:

A new model was introduced in [SMS]:

@

t

u = h(clt(u))div

x

(f(jD

x

Guj)D

x

u) in D[0;R][0;T] u(;;0) = u

0

;(2)

where h is of class C

1

([0;1[;R),f is of class C

2

([0;1[;R) and nonnegative.

Besides

h is nondecreasing and satisfies h(0) = 0;

f is decreasing and satisfies f(0) = 1:

In their paper the authors provide a numerical discretisation of the equation,

and some numerical experiments - see also [LS],[ZSL],[MSL].Here we provide

a rst existence result under the simplied assumptions that

infclt(u

0

) m> 0;and sup

[m;1]

h

0

(s) inf

[m;1]

h(s);(3)

where is small.A possible choice of h is the following:

h(s) =

s

2

+s

2

;

2

where is suciently small.Note that,also in this simplied assumptions,the

equation is degenerate,since its second order termonly depends on the variables

x.Besides the coecients of the equation are non local.We refer the author to

[ALM],[BN],[C],for other results concerning parabolic dierential equations

with nonlocal coecients.

A standard procedure for nding solutions of the Cauchy problem for a

parabolic equation on a square with Neumann boundary conditions is to extend

the initial datum u

0

on all the space by re ections and periodicity and prove

that the resulting Cauchy problem on all the space has periodic solutions.We

then prove the following:

Theorem 1.1

Let u

0

be a periodic,Lipschitz continuous function dened in

R

n

R,satisfying condition (3) and 0 u

0

(x;) 1:Then there exists a

constant T > 0,and a periodic viscosity solution u of problem (2),dened on

R

n

R[0;T],Lipschitz continuous in (x;) and Holder continuous in t.For

any xed ,and any xed 2 [0;1] the function

(x;t)!u

(x;t) = u(x;;t) (4)

is of class C

2+;1+=2

(R

n

]0;T[) in the variables x and t.Besides,there exist

constants K

1

and K

2

such that if u

0

and v

0

are bounded by 1,periodic and

Lipschitz continuous functions on R

n

R,the corresponding solutions u and v

satisfy

jju vjj

L

1

(R

n+1

[0;T])

K

1

e

K

2

T

jju

0

v

0

jj

L

1

(R

n+1

)

:(5)

The structure of the second order termof the operator in (2) can be described

as follows.We call Lip(D) the set of Lipschitz continuous functions on a set D,

Bd(D) theset of bounded functions,and

Lu =

n

X

i=1

a

i

(u)(x;;t)@

i;i

u +

n

X

i=1

b

i

(u)(x;;t)D

i

u;(6)

where

a

i

;b

i

:Lip(D)\Bd(D)!Lip(D)\Bd(D);

for every compact D in R

n

R.There exist constants C

0

;C

1

such that for every

D,for every u 2 Lip(D)\Bd(D),for every i = 1; n,for every (x;;t) 2 D

C

0

a

i

(u)(x;;t) C

1

(jjujj

1

+1);b

i

(u)(x;;t) C

1

(jjujj

1

+1):(7)

The following condition is satised:

j@

h

(a

i

(u))j +j@

h

(b

i

(u))j C

1

supj@

h

uj +C

1

;(8)

where is a suitably small constant,satisfying

2

=

2

exp(6

2

)

C

2

0

128C

2

1

and

2

=

C

0

64

1 +

4C

2

1

C

0

1

:

3

For every Lipschitz continuous functions u and v

ja

i

(u)(x;;t) a

i

(v)(x;;t)j +jb

i

(u)(x;;t) b

i

(v)(x;;t)j C

1

jjuvjj

1

:(9)

Finally,if D = R

n

R,a

i

is invariant with respect to translations:for every

xed we call

a

i

(u(; +

0

;))(x;;t) = a

i

(u)(x; +

0

;t):(10)

Theorem 1.1 is then a consequence of the following more general result:

Theorem 1.2

Let u

0

be a Lipschitz continuous function dened in R

n

R,

satisfying condition 0 u

0

(x;) 1:Then there exists a bounded,Lipschitz

continuous viscosity solution u of problem

u

t

= Lu in R

n+1

[0;T[

u(x;;0) = u

0

(x;) in R

n+1

:

(11)

For any xed ,and any xed 2 [0;1] the function u

dened in (4) is of

class C

2+;1+=2

in the variables x and t.Besides,the stability condition (5) is

veried.

The proof of this result follows essentially the main ideas of the classical exis-

tence results,as nd in the [LUS],or the user guide,but,due to the degeneracy

of the operator,we have to organize in an new way the estimate of the gradient

(D

x

u;@

u) of the solution.Indeed,using the Bernstein method,we rst prove

an a priori bound only for the spatial gradient D

x

u.Using this estimate,we

obtain a stability inequality for the solutions,from which we deduce the esti-

mate of @

u.The fact that the coecients depend globally on the unknown u

introduce some additional technical diculties.Indeed,even though we use an

elliptic regularisation,and we always work with regular functions,we are forced

to use an approach proposed by [CLM] for studying the viscosity solutions.

Let us give a more detailed sketch of the proof.In Section 2 we consider the

elliptic regularisation of the operator L:

L

"

u =

n

X

i=1

a

i

@

i;i

u +"

2

@

;

u +

n

X

i=1

b

i

(u)@

i

u;(12)

and we rst prove the existence of solutions of the Cauchy problem

@

t

u = L

"

u in Q

u(x;;t) = u

0

(x;) in @

Q

(13)

on the bounded cylinder Q = B

R

[0;T[,with parabolic boundary @

Q.In

particular,with a suitable modication of the Bernstein method we prove that

the gradient of the solution satises the following estimate

jjD

x

ujj

L

1

(Q

R

)

+"jj@

ujj

L

1

(Q

R

)

C

for a constant C independent of R and".Letting R!1we nd a solution on

all Q = R

n+1

[0;T].

4

In Section 3,using the fact that the estimate of the x- gradient is independent

of",we prove an uniqueness and stability result for solutions of (13) on R

n+1

[0;T]:

Theorem There exist constants K

1

and K

2

such that if u and v are two

solutions Lipschitz continuous and bouded of (13) with initial data u

0

and v

0

respectively,then

jju vjj

L

1

(R

n+1

[0;T])

K

1

e

K

2

T

jju

0

v

0

jj

L

1

(R

n+1

)

:

Let us note explicitly that this results holds even if the comparison principle

is not satised.Finally we deduce the boundeness of @

u from this estimate.

In Section 4 we see that equation (2) satises these assumptions,and we

conclude the proof of Theorem 1.1,with an other regularisation procedure.

Acknoledgment

We are deeply indebted with F.Sgallari for bringing the problem to our

attention,and with A.Sarti for many useful conversations on the subject of

their model.

2 A priori bound of the spatial gradient

In this section we prove the existence of a solution of the initial value problem

(13).The proof is based on an a priori estimate of the spatial gradient.For

simplicity we will introduce the following notation:

e

@

h

= @

h

;h = 1 ;n;

e

@

n+1

="@

:(14)

Then this operator in (12) will be written as

L

"

=

n+1

X

i=1

a

i

e

@

i;i

u +

n

X

i=1

b

i

(u)

e

@

i

u;(15)

where a

n+1

= 1 and also this last coecients satises assumptions (7),(8),(9)

with the same constants C

0

and C

1

independent of".We consider the Cauchy

problem (13) on the bounded cylinder Q

R

= B

R

[0;T],with initial datum

0 u

0

(x;) 1 in B

R

:(16)

We rst note that the solutions of (13) satisfy the maximum principle,so

that (16) implies

0 u(x;;t) 1 8(x;;t) 2 B

R

[0;T]:

As we noted in the introduction the classical gradient estimate can not be

applied directly,since the coecients depend globally on u.Hence we suitably

modify the Bernstein method in order to apply it to our situation.

5

Theorem 2.1

If u 2 C

2

(Q)\Lip(

Q) is a solution of problem (13) on the

bounded cylinder Q = B

R

[0;T],then there exists a constant

e

C

1

independent

of R and"such that

jjD

x

ujj

L

1

(Q

R

)

+"jj@

ujj

L

1

(Q

R

)

e

e

C

1

T

jjD

x

u

0

jj

L

1

(B

R

)

+"jj@

ujj

L

1

(B

R

)

;

where D

x

is the gradient with respect to the xvariable.A possible choice of

e

C

1

is

e

C

1

= 4

2C

2

1

+1

C

0

;

where C

0

and C

1

are dened in (7) and (8).

Proof If is an increasing function to be chosen later,we can always represent

u in the form:u = (u).Then the function u is a solution of

@

t

u =

n+1

X

i=1

a

i

(u)

e

@

i;i

u +

n+1

X

i=1

a

i

(u)

00

(u)

0

(u)

(

e

@

i

u)

2

+

n

X

i=1

b

i

(u)

e

@

i

u:

Let be a nonnegative function in C

1

0

(Q

R

).Multiplying by

e

@

h

(

e

@

h

u) we get

Z

@

t

u

e

@

h

(

e

@

h

u)dxddt =

Z

n+1

X

i=1

a

i

(u)

e

@

i;i

u

e

@

h

(

e

@

h

u)dxddt+ (17)

+

Z

n+1

X

i=1

a

i

(u)

00

(u)

0

(u)

(

e

@

i

u)

2

e

@

h

(

e

@

h

u)dxddt +

Z

n

X

i=1

b

i

(u)

e

@

i

u

e

@

h

(

e

@

h

u)dxddt:

Let us consider one term at a time.Integrating by parts the rst one we get

Z

@

t

u

e

@

h

(

e

@

h

u)dxddt =

1

2

Z

(

e

@

h

u)

2

@

t

dxddt:(18)

The second becomes

Z

n+1

X

i=1

a

i

(u)

e

@

i;i

u

e

@

h

(

e

@

h

u)dxddt = (19)

Z

n+1

X

i=1

e

@

h

a

i

e

@

i;i

u

e

@

h

u dxddt +

Z

n+1

X

i=1

e

@

i

a

i

e

@

i;h

u

e

@

h

u dxddt+

+

Z

n+1

X

i=1

a

i

e

@

i;h

u

2

dxddt +

1

2

Z

n+1

X

i=1

a

i

e

@

i

(

e

@

h

u)

2

e

@

i

dxddt:

Hence inserting (18) and (19) in (17),summing over h,and denoting

v =

n+1

X

i=1

(

e

@

h

u)

2

;

6

where

e

@

h

is dened in (14),we obtain

1

2

Z

v@

t

+

1

2

Z

n+1

X

i=1

a

i

e

@

i

v

e

@

i

=

Z

F ;

where

F =

n+1

X

i=1

n+1

X

h=1

a

i

e

@

i;h

u

2

+

e

@

h

a

i

e

@

i;i

u

e

@

h

u

e

@

i

a

i

e

@

i;h

u

e

@

h

u+

+

e

@

h

a

i

(u)

00

(u)

0

(u)

(

e

@

i

u)

2

e

@

h

u

!

n

X

i=1

n+1

X

h=1

e

@

h

b

i

(u)

e

@

i

u

e

@

h

u:

Let us estimate F

F

n+1

X

i=1

n+1

X

h=1

a

i

e

@

i;h

u

2

+j

e

@

i;i

uj

2

+

1

j

e

@

h

a

i

j

2

j

e

@

h

uj

2

+j

e

@

i;h

uj

2

+

1

j

e

@

i

a

i

j

2

j

e

@

h

uj

2

+

+j

e

@

h

a

i

j

2

j

e

@

i

uj

2

+

00

(u)

0

(u)

2

(

e

@

i

u)

2

(

e

@

h

u)

2

+

+a

i

(u)

00

(u)

0

(u)

0

(

e

@

i

u)

2

(

e

@

h

u)

2

+(

e

@

i;h

u)

2

+

a

2

i

(u)

00

(u)

0

(u)

2

(

e

@

i

u)

2

(

e

@

h

u)

2

+

+

n

X

i=1

n+1

X

h=1

j

e

@

h

(b

i

(u))

e

@

i

u

e

@

h

uj +j

e

@

i;h

uj

2

+

1

jb

i

(u)j

2

j

e

@

h

uj

2

(if = C

0

=4,where C

0

is dened in (7) and we set b

n+1

= 0 for simplicity of

notations)

n+1

X

i=1

n+1

X

h=1

1

(

e

@

h

a

i

)

2

+(

e

@

i

a

i

)

2

+(

e

@

i

a

i

)

2

+j

e

@

h

b

i

(u)j

2

j@

h

uj

2

+

+

n+1

X

i=1

1

a

2

i

+jb

i

(u)j

2

+1

v +

n+1

X

i=1

00

(u)

0

(u)

2

+a

i

00

(u)

0

(u)

0

+

a

2

i

00

(u)

0

(u)

2

!

v

2

:

If we choose

00

(u)

0

(u)

0

0;

and use the assumptions (7),(8),(9),then the estimate for F becomes:

F

8(n +1)

C

2

1

2

supv(

0

)

2

+C

2

1

v +

(2C

2

1

+1)v

+

+

n+1

X

i=1

00

(u)

0

(u)

2

+C

0

00

(u)

0

(u)

0

+

C

2

1

00

(u)

0

(u)

2

!

v

2

7

e

C

1

v+(n+1)

32

C

0

C

2

1

2

(

0

)

2

+

1+

4C

2

1

C

0

00

(u)

0

(u)

2

v sup v+C

0

(n+1)

00

(u)

0

(u)

0

v

2

;

for a suitable constant

e

C

1

= 4

2C

2

1

+1

C

0

only dependent on the assumptions.We can now make the same choice of as

in [LUS].We set

:[;2]!R (x) =

Z

2

exp(

q

)d

1

Z

2

exp(

q

)d;(20)

where is dened in (8).The assumption made on assure the existence of

constants

e

C

2

and

e

C

3

such that

(n +1)

32

C

0

C

2

1

2

(

0

)

2

+

1 +

4C

2

1

C

0

00

(u)

0

(u)

2

e

C

2

;(21)

2

e

C

2

e

C

3

and

e

C

3

C

0

(n +1)

00

(u)

0

(u)

0

(for reader convenience the computations are collected in Remark 2.1 below).

Then

F

e

C

1

v +

e

C

2

v supv

e

C

3

v

2

:

The estimate for the gradient is a consequence of the following lemma.

Lemma 2.1

Let v be a nonnegative solution of class C

0

(

Q

R

)\C

1

(Q

R

) of the

following nonlinear equation:

1

2

Z

v@

t

dxddt +

1

2

Z

n+1

X

i=1

a

i

@

i

v@

i

dxddt =

Z

F(v) dxddt;

with

F

e

C

1

v +

e

C

2

v supv

e

C

3

v

2

;

and

2

e

C

1

<

e

C

2

.Then

sup

Q

R

v e

e

C

1

T

sup

@

(Q

R

)

v:

Proof If we set!(x;;t) = v(x;;t)e

e

C

1

t

;the function!is a solution of

1

2

Z

!@

t

+

1

2

Z

n+1

X

i=1

a

i

@

i

!@

i

=

Z

e

F ;

with

e

F 2 L

1

and

e

F

e

C

2

!sup!

e

C

3

!

2

exp(

e

C

1

t):

8

Let (x

0

;

0

;t

0

) a maximum point for!in B

R

[0;T],and assume by contradic-

tion that

M

0

=!(x

0

;

0

;t

0

) > max

@

Q

R

!= M:

Then we can choose such that

!(x

0

;

0

;t

0

) > M;2 M

0

:

Let us denote (! M

0

+ )

+

its positive part.Let F

j

be a sequence in C

1

converging to

e

F as j!+1,and let!

j

the corresponding solution.Then!

j

uniformly converges to!,and a simple integration by parts ensures that for

every j

Z

!

j

M

0

n+1

X

i=1

a

i

(u)(@

i

!

j

)

2

dxddt =

Z

F

j

(!

j

M

0

+)

+

dxddt:

Letting j go to 1 we obtain

Z

!M

0

n+1

X

i=1

a

i

(u)(@

i

!)

2

dxddt

e

C

2

Z

M

0

!(!M

0

+)

+

e

e

C

1

t

dxddt

Z

e

C

3

!

2

(!M

0

)

+

e

e

C

1

t

dxddt

(since!(x;;t) > M

0

> M

0

=2)

(2

e

C

2

e

C

3

)

Z

!

2

(!M

0

)

+

e

e

C

1

t

dxddt < 0

This contradiction proves the assertion.

For reader convenience we compute explicitly the derivative of the function

introduced in (20),showing that the relation (21) is satised:

Remark 2.1

Let

:[;2]!R

be the function dened in (20).Then

0

=

Z

2

exp(s

2

)ds

1

exp(x

2

);

00

(u)

0

(u)

= 2x:

We can choice

e

C

3

= (n +1)C

0

00

(u)

0

(u)

0

= 2C

0

(n +1):

Since is dened in (8) as

2

=

C

0

64

1 +

4C

2

1

C

0

1

;

9

then

(n +1)

1 +

4C

2

1

C

0

00

(u)

0

(u)

2

(n +1)

1 +

4C

2

1

C

0

16

2

(n +1)C

0

4

=

e

C

2

2

By assumption (8)

2

=

2

exp(6

2

)

C

2

0

128C

2

1

and by a direct computation

(

0

)

2

exp(6

2

)

2

then

(n +1)

32

C

0

C

2

1

2

(

0

)

2

(n +1)

32

C

0

C

2

1

2

exp(6

2

)

2

(n +1)C

0

4

=

e

C

2

2

:

Relation (21) is proved.

It is standard to prove the existence of a solution of problem (13) on the

cylinder Q = B

R

[0;T],using the estimate of the gradient just established.

We refer for example to [LSU] Theorem 1.1 cap VI x1 and cap V x6.Letting

R!1,we immediately deduce

Theorem 2.2

Let u

0

be a Lipschitz continuous function dened in R

n

R,

satisfying condition 0 u

0

(x;) 1:Then for every T > 0 there exists a

solution u of class C

2+;1+=2

in the variables (x;) and t of the problem

u

t

= L

"

u in R

n+1

[0;T[

u(x;;0) = u

0

(x;) in R

n+1

;

(22)

which satises

jjD

x

ujj

1

+"jj@

ujj

1

e

e

C

1

T

jjD

x

u

0

jj

1

+"jj@

ujj

1

;

and

ju(x;;t) u(x;;t

0

)j

e

C

1

jt t

0

j

1=2

;

for every (x;;t),(x;;t

0

),with a constant

e

C

1

independent of".

3 Stability inequality

In this section we prove that the solution found in Theorem 2.2 is unique,and

conclude the proof of Theorem 1.2.Even though the solutions are regular,we

are forced to use a technique introduced for studying the viscosity solutions in

[ALM].However the choice of the main parameters is dierent here,because we

do not have and estimate of the complete gradient,and we do not yet assume

that the solution is periodic.

10

Theorem 3.1

Let u

0

and v

0

be bounded and Lipschitz continuous on R

n

R.

Let u and v be the correspondent viscosity solutions of problem (22).There

exists a constant K such that

jju vjj

L

1

(R

n+1

[0;T])

Kjju

0

v

0

jj

L

1

(R

n+1

)

:

Proof Let and constants to be xed later,and dependent only on jjujj

1

,

jjvjj

1

,jjD

x

ujj

1

and let

(x;y;;t) = u(x;;t) v(y;;t)

jx yj

4

4

t

jxj

2

+jyj

2

+jj

2

R

;(23)

with R > 0.Since u and v are bounded,then has a maximum,at a point

say (x

0

;y

0

;

0

;t

0

).We can always assume that u(0;0;0) v(0;0;0),so that the

maximum of is nonnegative:

(x

0

;y

0

;

0

;t

0

) (0;0;0;0) = u(0;0;0) v(0;0;0) 0:

Let us rst assume that t

0

> 0.Since all the considered functions are of class

C

2

,at the point (x

0

;y

0

;

0

;t

0

) we have

@

t

u(x

0

;

0

;t

0

) @

t

v(y

0

;

0

;t

0

);(24)

@

i

u(x

0

;

0

;t

0

) =

jx

0

y

0

j

2

(x

0

y

0

)

i

+

2(x

0

)

i

R

;

@

i

v(y

0

;

0

;t

0

) =

jx

0

y

0

j

2

(x

0

y

0

)

i

2(y

0

)

i

R

;

and

0

@

D

2

x

u 0 0

0 D

2

y

v 0

0 0 @

;

(u v)

1

A

D

2

x;y;

jx

0

y

0

j

4

4

+t

0

+

jx

0

j

2

+jy

0

j

2

+j

0

j

2

R

:

(25)

If we denote A the right hand side of (25) we have

A =

jx

0

y

0

j

2

4

0

@

I I 0

I I 0

0 0 0

1

A

+

2

R

0

@

I 0 0

0 I 0

0 0 1

1

A

+

+

2

0

@

(x

0

y

0

)

N

(x

0

y

0

) (x

0

y

0

)

N

(x

0

y

0

) 0

(x

0

y

0

)

N

(x

0

y

0

) (x

0

y

0

)

N

(x

0

y

0

) 0

0 0 0

1

A

and

tr(A) C

jx

0

y

0

j

2

+

1

R

:(26)

Multiplying (25) on the right by the matrix

(u) =

0

@

diag

a

1

(u);:::;a

n

(u)

diag

p

a

1

(u)a

1

(v);:::;

p

a

n

(u)a

n

(v)

0

diag

p

a

1

(u)a

1

(v);:::;

p

a

n

(u)a

n

(v)

diag

a

1

(v);:::;a

n

(v)

0

0 0"

2

1

A

11

and considering the trace we get

n+1

X

i=1

a

i

(u)

e

@

i;i

u

n+1

X

i=1

a

i

(v)

e

@

i;i

v

n

X

i=1

a

i

(u)

1=2

a

i

(v)

1=2

2

tr(A) +

"

2

R

(27)

C

jju vjj

1

+jx

0

y

0

j

2

jx

0

y

0

j

2

+

1

R

+

"

2

R

;

where we have used (26) to estimate the the trace of A and the fact that

ja

1=2

i

(u)(x

0

;

0

;t

0

) a

1=2

i

(v)(y

0

;

0

;t

0

)j

(by (7)

C

1

0

ja

i

(u)(x

0

;

0

;t

0

) a

i

(v)(y

0

;

0

;t

0

)j

jju vjj

1

+jjD

x

ujjjx

0

y

0

j

2

;

where jjD

x

ujj

1

is uniformly bounded.Analogously

b

i

(u)@

i

u(x

0

;

0

;t

0

) b

i

(v)@

i

v(y

0

;

0

;t

0

) = (28)

=

b

i

(u)(x

0

;

0

;t

0

)b

i

(v)(y

0

;

0

;t

0

)

4jx

0

y

0

j

2

(x

0

y

0

)

i

+b

i

(u)

2(x

0

)

i

R

+b

i

(v)

2(y

0

)

i

R

jju vjj

1

+jx

0

y

0

j

jx

0

y

0

j

3

+

jx

0

j +jy

0

j

R

:

By (24) we have:

@

t

u(x

0

;

0

;t

0

) @

t

v(y

0

;

0

;t

0

) =

n+1

X

i=1

a

i

(u)

e

@

i;i

u +b

i

(u)

e

@

i

u

n+1

X

i=1

a

i

(v)

e

@

i;i

v b

i

(v)@

i

v

by (27) and (28)

C

jju vjj

2

1

+jx

0

y

0

j

2

jx

0

y

0

j

2

+

1

R

+

"

2

R

+

jju vjj

1

+jx

0

y

0

j

jx

0

y

0

j

3

+

jx

0

j +jy

0

j

R

:

Since (x

0

;

0

;t

0

) 0;for every R > 0 we have

jx

0

y

0

j

4

4

+

jx

0

j

2

+jy

0

j

2

+j

0

j

2

R

u(x

0

;

0

;t

0

) v(y

0

;

0

;t

0

)

jjujj

L

1

(R

n+1

[0;T[)

+jjvjj

L

1

(R

n+1

[0;T[)

e

C:

Hence

jx

0

j

2

+jy

0

j

2

+j

0

j

2

CR;

jx

0

y

0

j

4

4

e

C:(29)

12

Since (x

0

;y

0

;

0

;t

0

) is a maximum point for ,

u(x

0

;

0

;t

0

) v(y

0

;

0

;t

0

)

jx

0

y

0

j

4

4

t

0

jx

0

j

2

+jy

0

j

2

+j

0

j

2

R

=

(x

0

;y

0

;

0

;t

0

) (y

0

;y

0

;

0

;t

0

) u(y

0

;

0

;t

0

)v(y

0

;

0

;t

0

)t

0

2jy

0

j

2

+j

0

j

2

R

:

Thus

jx

0

y

0

j

4

4

u(x

0

;

0

;t

0

) v(y

0

;

0

;t

0

) +

jy

0

j

2

jx

0

j

2

R

e

Ljx

0

y

0

j +

jx

0

y

0

j(jx

0

j +jy

0

j)

R

;

where

e

L is the Lipschitz constant in x for u.In particular we deduce

jx

0

y

0

j

3

4

e

L+

jx

0

j +jy

0

j

R

(by (29))

e

L+

C

p

R

e

L+1:

If we choose

=

3

jju vjj

3

;

inserting in the estimate of we deduce

Cjju vjj

1

+ +1

+C

jju vjj

2

R

1 +

2

+

C

R

1=2

;

and this is a contradiction,if

= 2Cjju vjj

1

+ +1

+2C

jju vjj

2

R

1 +

2

+

2C

R

1=2

:(30)

Hence t

0

= 0,and for every t,for every x,y

u(x;;t) v(y;;t)

jx yj

4

t

(jxj

2

+jyj

2

+jj

2

)

R

sup

n

u

0

(x;) v

0

(y;)

jx yj

4

(jxj

2

+jyj

2

+jj

2

)

R

o

:

If x = y we get

u

(

x;;t

)

v

(

x;;t

)

T

+

2jxj

2

+jj

2

R

+

jj

u

0

v

0

jj

+

sup

r>0

n

L

0

r

r

4

4

o

;

where L

0

is the Lipschitz norm of v

0

= T +

2jxj

2

+jj

2

R

+jju

0

v

0

jj +

3

4

L

4=3

0

1=3

=

for the choice of and ,

= 2CTjju vjj

1

+ +1

+2CT

jju vjj

2

R

1 +

2

+

2CT

R

1=2

+

13

+

2jxj

2

+jj

2

R

+jju

0

v

0

jj +

3

4

L

4=3

0

jju vjj:

Since x and are xed and the constants C;T;R;L

0

; do not depend on R,

letting R go to +1 we get:

u(x;;t) v(x;;t)

2CTjju vjj

1

+ +1

+jju

0

v

0

jj +

3

4

L

4=3

0

jju vjj:

We now conclude,choosing = L

4=3

0

,and T suciently small.

Therefore,if T

1

is an arbitrary interval of time in [0;+1[,and NT T

1

we

deduce,iterating this argument that

jju vjj

1

C

T

1

jju

0

v

0

jj

1

:

for a constant C depending on jjujj

1

,jjvjj

1

,jjD

x

ujj

1

.

Proof of Theorem 1.2 By assumption u

0

is a bounded and Lipschitz

continuous function on R

n

R.For every"> 0 Theorem2.2 provides a solution

(u

"

) of the regularized problem (22),with initial condition u

0

,satisfying

jjD

x

u

"

jj

1

C;

for a constant C only dependent on u

0

and independent of".On the other side,

by (10),if we x

0

2 R,the function

v

"

(x;;t) = u

"

(x; +

0

;t)

is a solution of the same problem,with initial datum

v

0

(x;) = u

0

(x; +

0

):

Then

ju

"

(x;;t) u

"

(x; +

0

;t)j = ju

"

(x;;t) v

"

(x;;t)j

(by Theorem 3.1)

ju

0

(x;) v

0

(x;)j = ju

0

(x;) u

0

(x; +

0

)j

0

:

The Lipschitz continuity is then proved.Letting"!0 we found a viscosity

lipschitz continuous solution of (11).Keeping xed,the function u

can be

considered a solution of an uniformly parabolic equation,with Lipschitz contin-

uous coecients.Hence it belongs to C

2+;1+=2

;for every 2]0;1[,uniformly

with respect to .

Remark 3.1

If the initial datum is periodic,the solution of (12) is periodic.

Indeed if u is a solution,also u

h

= u( + h) is a solution of the same Cauchy

problem,so that it coincides with u,by the asserted uniqueness.

14

4 Application to the model

In this section we show how to apply Theorem 1.2 to equation (2) and we

conclude the proof of Theorem 1.1.

In order to write equation (2) in the nondivergence form (6) we set

a

"

i

(u) = (h(clt(u)) +")f(jDG uj) (31)

b

i

(u) = clt

2

(u)f

0

(jDG uj)

n

X

i;j=1

D

2

i;j

G u

D

j

G u

jDG uj

:(32)

Clearly (2) is obtained for (6) for"= 0.

Let us prove that these function satises the assumptions (7),(8),(9).

Lemma 4.1

Let Q be compact in R

n+1

[0;T],and let u be a bounded and

Lipschitz continuous function on Q.Then the function clt(u) dened in (1) is

bounded and Lipschitz continuous in

Q.Precisely

jjclt(u)jj

1

4jjujj

1

:(33)

For every (x;;t) there exists

1

,

2

2 R

n

such that j

1

j;j

2

j 1 and

j@

h

clt(u)(x;;t)j j@

h

u(x +

1

; +%;t)j +j@

h

u(x

2

; %;t)j+ (34)

+2j@

h

u(x;;t)j +jDG @

h

u(x;;t)j

for every t and a.e.(x;) 2 B

R

.Finally,if u and v are bounded and Lipschiz,

jclt(u)(x

0

;

0

;t

0

) clt(v)(y

0

;

0

;t

0

)j Cjju vjj:(35)

Proof

The estimate (33) follows directly by the denition,simply choosing

1

=

2

.

Let now v;u 2 Bd(Q)\Lip(Q),and assume that clt(u) clt(v) 0.Let

1

,

2

be such that

clt(v)(x;;t) = jv(x +

1

; +;t) v(x;;t)j+

+jv(x

2

; +;t) v(x;;t)j +j < DG v;

1

2

> j:

Then by denition of clt(u),

clt(u) clt(v) ju(x +

1

; +;t) u(x;;t)j+

+ju(x

2

; +;t) u(x;;t)j +j < DG u;

1

2

> j

jv(x +

1

; +;t) v(x;;t)j

jv(x

2

; +;t) v(x;;t)j j < DG v;

1

2

> j

ju(x +

1

; +;t) v(x +

1

; +;t)j+

+ju(x

2

; %;t) v(x

2

; %;t)j+

15

+2ju(x;;t) v(x;;t)j +jDG u DG vj:

And this implies (34).

Now we call e

h

a vector of the canonical basis,

u

;h

= u(x +e

h

;;t);for h = 1; ;n

and

u

;n+1

= u(x; +;t):

It then follows that for every 2 C

1

, 0

Z

@

h

clt(u) dxd = lim

!0

Z

clt(u) clt(u

;h

)

dxd

lim

!0

Z

ju(x +

1

; +;t) u

;h

(x +

1

; +;t)j

dxd+

+

Z

ju(x

2

; %;t) u

;h

(x

2

; %;t)j

dxd+

+2

Z

ju(x;;t) u

;h

(x;;t)j

dxd +

Z

jDG u DG u

;h

j

dxd =

=

Z

j@

h

u(x +

1

; +%;t)j +j@

h

u(x

2

; %;t)j+

+2j@

h

u(x;;t)j +jDG @

h

uj

dxd:

An analogous relation,holds for @

h

clt(u) and the thesis is proved.

Fromthis lemma,and the properties of the convolution,it is easy to recognize

that a

"

i

and b

i

satisfy assumptions (7),(8),(9).Let us now conclude the

Proof of Theorem1.1 By Theorem1.2 for every"there exists (u

"

) solution

of

u

"

t

=

n

X

i=1

a

i

(u

"

)(x;;t)@

i;i

u

"

+

u

"

X

i=1

b

i

(u

"

)(x;;t)D

i

u

"

;

satisfying condition (5),and

ju(x;;t) u(x;;0)j Ct

1=2

for a constant C independent of".If infclt(u

0

) > m> 0,

clt(u)(x;;t) clt(u)(x;;0) Ct

1=2

m

2

;

if ct

1=2

m=2.Then condition (7) is satised on [0;

m

2

4C

2

],with a constant

C

0

independent of".Letting"go to 0 we nd a solution u satisfying all the

conditions listed in the thesis.

16

5 References

[AGLM] L.Alvarez,F.Guichard,P.L.Lions,J.M.Morel,Axioms and

Fundamental Equations of Image Processing,Arch.Rat.Mech.Anal,

123,(1993),200-257.

[ALM] L.Alvarez,P.L.Lions,J.M.Morel,Image selective smoothing and edge

detection by nonlinear diusion in R

n

.II,SIAM J.of nonlinear analisys

29,3,(1992),845-866.

[AM] L.Alvarez,J.M.Morel,Formalization and computational aspect of

image analysis,Acta Num.,??,(1994),1-59.

[AE] L.Alvarez,J.Escalatin,Image equalization using reaction diusion

equations,SIAM J.of Appl.Math.,57,1,(1997),153-75.

[AMa] L.Alvarez,L.Mazorra,Signal and image restoration by using shock

lters and anisotropic diusion,SIAM J.of Appl.Math.,31,2,(1994),

590-605.

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