A degenerate parabolic equation arising in image processing

paradepetΤεχνίτη Νοημοσύνη και Ρομποτική

5 Νοε 2013 (πριν από 3 χρόνια και 11 μήνες)

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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 coecients arising in image processing.An image is a bounded function
u:D!R dened 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 dierent model have been proposed.
Perona and Malik proposed in [PM] the following anisotropic diusion 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 eect,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 modied in
dierent 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 dened 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 denition 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


;jj  2rg
A

x
= f 2 R
n
:x  2


;jj  2rg:
A new model was introduced in [SMS]:
@
t
u = h(clt(u))div
x
(f(jD
x
Guj)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 simplied 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 suciently small.Note that,also in this simplied assumptions,the
equation is degenerate,since its second order termonly depends on the variables
x.Besides the coecients of the equation are non local.We refer the author to
[ALM],[BN],[C],for other results concerning parabolic dierential equations
with nonlocal coecients.
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 dened 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),dened on
R
n
R[0;T],Lipschitz continuous in (x;) and Holder 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 satised:
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
jjuvjj
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 dened 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

dened in (4) is of
class C
2+;1+=2
in the variables x and t.Besides,the stability condition (5) is
veried.
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 coecients depend globally on the unknown u
introduce some additional technical diculties.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 modication of the Bernstein method we prove that
the gradient of the solution satises 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 satised.Finally we deduce the boundeness of @

u from this estimate.
In Section 4 we see that equation (2) satises 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 coecients satises 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 coecients 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 xvariable.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 dened 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)dxddt =
Z
n+1
X
i=1
a
i
(u)
e
@
i;i
u
e
@
h
(
e
@
h
u)dxddt+ (17)
+
Z
n+1
X
i=1
a
i
(u)

00
(u)

0
(u)
(
e
@
i
u)
2
e
@
h
(
e
@
h
u)dxddt +
Z
n
X
i=1
b
i
(u)
e
@
i
u
e
@
h
(
e
@
h
u)dxddt:
Let us consider one term at a time.Integrating by parts the rst one we get
Z
@
t
u
e
@
h
(
e
@
h
u)dxddt =
1
2
Z
(
e
@
h
u)
2
@
t
dxddt:(18)
The second becomes
Z
n+1
X
i=1
a
i
(u)
e
@
i;i
u
e
@
h
(
e
@
h
u)dxddt = (19)

Z
n+1
X
i=1
e
@
h
a
i
e
@
i;i
u
e
@
h
u dxddt +
Z
n+1
X
i=1
e
@
i
a
i
e
@
i;h
u
e
@
h
u dxddt+
+
Z
n+1
X
i=1
a
i

e
@
i;h
u

2
dxddt +
1
2
Z
n+1
X
i=1
a
i
e
@
i
(
e
@
h
u)
2
e
@
i
dxddt:
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 dened 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 dened 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 dened 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
dxddt +
1
2
Z
n+1
X
i=1
a
i
@
i
v@
i
dxddt =
Z
F(v) dxddt;
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
dxddt =
Z
F
j
(!
j
M
0
+)
+
dxddt:
Letting j go to 1 we obtain
Z
!M
0

n+1
X
i=1
a
i
(u)(@
i
!)
2
dxddt 
e
C
2
Z
M
0
!(!M
0
+)
+
e
e
C
1
t
dxddt

Z
e
C
3
!
2
(!M
0
)
+
e
e
C
1
t
dxddt 
(since!(x;;t) > M
0
 > M
0
=2)
 (2
e
C
2

e
C
3
)
Z
!
2
(!M
0
)
+
e
e
C
1
t
dxddt < 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 satised:
Remark 2.1
Let
:[;2]!R
be the function dened 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 dened 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 dened 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 satises
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 dierent 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
+jj
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
+jj
2
)
R

 sup
n
u
0
(x;) v
0
(y;) 
jx yj
4


(jxj
2
+jyj
2
+jj
2
)
R
o
:
If x = y we get
u
(
x;;t
)

v
(
x;;t
)

T
+
2jxj
2
+jj
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
+jj
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
+jj
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 suciently 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 coecients.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 satises 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) dened 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 denition,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 denition 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 satised 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
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