through visosity solutions' methods
Guy Barles, Samuel Biton, Mariane Bourgoing and Olivier Ley
Laboratoire de Mathematiques et Physique Theorique
Universite de Tours
Par de Grandmont, 37200 Tours, Frane
Email: barlesuniv-tours.fr
Abstrat
In this artile, we are interested in uniqueness results for visosity solutions
of a general lass of quasilinear, possibly degenerate, paraboli equations set in
IR N
. Using lassial visosity solutions'methods, we obtain a general omparison
result for solutions with polynomial growths but with a restrition on the growth
of the initial data. The main appliation is the uniqueness of solutions for the
meanurvature equationforgraphswhihwasonlyknowninthelassofuniformly
ontinuous funtions. An appliationto themean urvature owis given.
Key-words: Quasilinear paraboli equations, solutions with polynomial growth, om-
parisonresults, mean urvature equation,visosity solutions.
AMS subjet lassiations: 35A05, 35B05,35D05,35K15, 35K55,53C44
1 Introdution
This artile is a ontinuation of the program started in [3℄ (see [2℄ for an introdutory
paper) whih aim is the study of the uniqueness properties for unbounded visosity so-
lutions of quasilinear,possibly degenerate, paraboliequationsset inIR N
. This program
wasmotivatedbythe followingsurprisingresultofEkerandHuisken[12℄: foranyinitial
data u
0 2W
1;1
lo (IR
N
), thereexists a smooth solutionof the equation
8
<
: u
t
u+ hD
2
uDu;Dui
1+jDuj 2
=0 inIR N
(0;+1);
u(x;0)=u
0
(x) in IR N
:
(1)
ThisworkwaspartiallysupportedbytheTMRprogram\Visositysolutionsandtheirappliations."
Here andbelowthe solutionuisareal-valuedfuntions, DuandD u denoterespetively
itsgradientand Hessian matrixwhile jjand h;istandfor the lassialEulideannorm
and inner produt inIR N
.
The very non-standard feature of this result is that no assumption on the behavior
of the initial data at innity is imposed and therefore the solutions may have also any
possible behavior atinnity.
A natural question is then whether suh a solution is unique or not. It is a very
intriguingandhallenging questionsine onehas totakeinaountthe lakofrestrition
on the behavior of the solutions at innity, an unusual fat. As far as we know, this
question in its full generality is still open in IR N
for N > 1 while for N = 1, the result
was proved independently and by dierent methodsby Chouand Kwong [8℄ and Barles,
Biton and Ley [4℄.
In a series of papers ([3℄, [4℄ and [5℄), we address the more general question of the
uniqueness of unbounded visosity solutions, not only for (1) but also for more general
quasilineardegenerate paraboliequationslike
8
<
: u
t Tr
b(x;t;Du)D 2
u
+H(x;t;u;Du)=0 inIR N
(0;T);
u(x;0)=u
0
(x) inIR N
;
(2)
where T > 0 is any positive onstant and b is a funtion taking values in the set of
nonnegativesymmetri matrix and H 2C(IR[0;T℄IRIR N
):
When one wants to prove suh uniqueness results, the rst key diÆulty is the one
we point out above i.e. the a priori unboundedness of solutions or more generally the
fat that solutions may have any behavior at innity. But the gradient dependene in
the diusion matrix b above is also a major diÆulty: beause of that, it does not seem
possible to obtain uniqueness results through some kind of linearization proedure and
even the uniqueness of smooth solutions isfar from being obvious, exept if one imposes
restritions on D 2
u at innity, a type of assumptions that we want to avoid. Forresults
obtained by linearization proedure, we refer the reader for example to Crandall and
Lions [10℄, Ishii[16℄ orLey [18℄where the uniqueness of solutionsof rst-order equations
were obtained using \nite speed of propagation" type properties, to Barles [1℄ where
optimal uniqueness results for solutions with exponential growth of (stationary) rst-
order equations were proved or to Barles, Bukdahn and Pardoux [6℄ for solutions with
exponentialgrowths of a system of Hamilton-Jaobi-Bellmantype Equations.
Tothebest ofourknowledge,themostgeneraluniqueness resultsforquasilinearequa-
tions{i.e. forequationsinvolvingtheabovementioneddiÆultyontheDu-dependene{
onernonlyuniformlyontinuousvisositysolutions: wereferthereaderthe\Users'guide
of visosity solutions" of Crandall, Ishii and Lions [9℄ and toGiga, Goto, Ishii and Sato
[15℄ for results inthis diretion.
The aim of this artile is to push as far as possible the lassial arguments used for
proving omparison results for visosity solutions in order to obtain suh results for the
In the general omparison theorem we are able to prove by using this approah (see
Theorem2.1),the onditionswehavetoimposeontheequationandthebehaviorofsolu-
tionatinnity,namelytohaveapolynomialgrowth,seemratherreasonable. Surprisingly
the main restrition onerns the initial data whih has to satisfy the following, rather
unnatural, ondition: there exists a modulus of ontinuity m y
and 0 < (1+ p
5)=2
suh that, forany x;y 2IR N
,
ju
0
(x) u
0
(y)jm((1+jxj+jyj)
jx yj) : (3)
Unfortunatelysuhtypeofrestritionseemstobeanunavoidableartefatofthismethod.
We do not know how optimal is this result and in partiular the limiting exponent
(1+ p
5)=2. In fat, sine this result applies not only to the mean urvature equation
(1) but alsotoa large lass of degenerate and nondegenerate equations,we do not think
thatthisexponentouldhaveageometrialinterpretation. Ontheotherhand,webelieve
that the result is true for any but we were unable toproveit.
Crandall and Lions [11℄ obtained related results under similar assumptions but for
fully nonlinear pdes whose Hamiltonians depend only on the seond derivatives of the
solutionu. Theirtehniques donot apply toour ase. Comparing tothe existeneresult
for the mean urvature mentioned above, ondition (3) may appear restritive but we
point out that our proof works under a general struture assumption on b and not only
for the mean urvature equation. We refer to Setion 2 for a preise statement of the
assumptions onb and H and toSetion 3for adetailedtreatmentof the mean urvature
equation for graphs.
Unfortunatelythe proofof thisomparisonresult isvery tehnial (seeSetion4)and
reliesonthe use ofatriky test-funtionthat, aswealreadymentioneditabove,wetried
to buildin anoptimal way.
This result yieldsaomparisonpriniplebetween semiontinuous sub- and supersolu-
tions whihisa key toolforobtainingexisteneresultsthrough the Perron's method(see
[17℄, [9℄, et.). It thereforeallows us toshow suh an existeneresult of solutionswith a
suitablegrowth for (2).
The paper is divided as follows: in the next setion, we start by setting the problem
and the assumptions we will use. Then we state a omparison priniple (Theorem 2.1)
whih is the main result of the paper. As an immediate onsequene, we obtain the ex-
istene and uniqueness of a ontinuous visosity solution toour problem (Corollary2.1).
We end this setion with some examples of equations whih are inluded in our study.
Setion3dealswiththefundamentalexampleofthe meanurvature equationforgraphs.
Weprovideanappliationoftheuniqueness theoremtothemeanurvatureowofentire
graphs. The lastsetions4 and 5 are devoted to the proofs of the main theorems.
y
Afuntion m:IR +
!IR +
issaidtobeamodulusofontinuityifm(0
+
):=lim
s!0+
m(s)=0and
m(t+s)m(t)+m(s)foranys;t2IR +
:
position at the University of Padova. He would like to thank the department of mathe-
matis and espeiallyMartino Bardifor their kind hospitality and fruitfulexhanges.
2 Statement of the results and examples
Before stating the problem and our results, we introdue some notations. In the sequel,
M
N;M
is the set of N M{matries and S
N
(respetively S +
N
) denotes the set of the
symmetri (respetively symmetri nonnegative) matries. For every A 2 M
N;M , A
T
denotes the transpose of A. Finally, we introdue the spae C
poly
of loally bounded,
possibly disontinuous, funtions on IR N
[0;T℄ whih have a polynomial growth with
respet to the spae variable. More preisely, u 2 C
poly
if there exists a onstant k > 0
suh that
u(x;t)
1+jxj k
!
jxj!+1
0; uniformlywith respet to t2[0;T℄:
We willuse the followingassumptions for equation (2):
(H1) Thereexistsaontinuousfuntion :IR N
[0;T℄IR N
!M
N;M
andsomeonstant
~
C suh that
b(x;t;p)=(x;t;p)(x;t;p) T
;
k(x;t;p) (y;t;p)k
~
Cjx yj;
k(x;t;p) (x;t;q)k
~
C
jp qj
1+jpj+jqj :
(H2) The funtion H isontinuous onIR N
[0;T℄IRIR N
, u7!H(x;t;u;p)is nonde-
reasingforevery(x;t;p)andthere exists amodulusof ontinuitym~ suhthat, forevery
x;y;p;q2IR N
;u2IR and t2 [0;+1);
jH(x;t;u;p) H(y;t;u;p)jm~ ((1+jpj)jx yj);
and
jH(x;t;u;p) H(x;t;u;q)jm~((1+jxj)jp qj) :
Finally we say that a funtion ! : IR N
!IR satises the assumption (H3-!)() if there
exists amodulus of ontinuity m suh that, forevery x;y2IR N
;
j!(x) !(y)jm((1+jxj+jyj)
jx yj):
Our main result is the following
poly
C
poly
) bea upper-semiontinuous(respetivelylower-semiontinuous) visositysubsolution
(respetively visosity supersolution) of (2). If
u(x;0)u
0
(x)v(x;0) in IR N
;
where u
0
is a funtion whih satises (H3-u
0
)() with 0<(1+ p
5)=2, then
uv in IR N
[0;T℄:
The proof is postponed toSetion4. An immediateonsequene of Theorem 2.1is the
Corollary 2.1 Assume that(H1)and(H2)holdandletu
0
beafuntionwhih satises
(H3-u
0
)(),with0<(1+ p
5)=2. InC
poly
, there existsaunique ontinuousvisosity
solution u of (2). Moreover, there exists C >0 suh that
ju(x;t)jC(1+jxj +1
):
TheexisteneofasolutionisaonsequeneoftheomparisonresultbyusingPerron's
method. A priori,this existene resultstakesplae inthe lass offuntions with polyno-
mialgrowthbut we prove that the solutioninheritsthe growth ofthe initialdatum. The
proof willbe given inSetion 5.
Remark 2.1 : Conerningtheexisteneoflassialsolutionstotheseequations,werefer
toChouand Kwong [8℄who providedW 1;1
loalboundsfor thesolutionsof alargelass
of quasilinearequationswithoutgrowth restritiononthe initialdata. The existeneofa
smooth solutionto the mean urvature equation for graphs, also in the ase when there
isnorestrition onthe initialdata, wasrst shown by Eker and Huisken [12℄and Evans
and Spruk [14℄ (see also[3℄).
Before giving examples of equations to whih these results apply, let us make some
ommentsabouttheassumption(H1)and(H3-!)(),(H2)beinganaturalandlassial
assumption. In(H1),thetworstonditionsare lassial;notethattherstoneimplies
thatthe equationisdegenerateparaboli. Ofourse,thethird oneisthe mostinteresting
sine it onerns the behavior of b in the gradient variable and we reall that it is a key
diÆultyhere. Wehave hosenthis typeofassumptionfor inpbeausethis isthe type
of dependene we have for (1). We have deided not to onsider in this paper dierent
hoiesof p{dependene inordertokeep thelengthof thispaperreasonable. They would
lead to resultswith, in partiular, dierent limitationonthe growth of the initialdata.
It is anyway worth pointingout that, with suh typeof assumptions on; thereis no
hope to prove a better result than a omparison priniple in C
poly
or for solutions with
exponentialgrowth: indeed the heat equationsatises (H1). In order togo further, one
hastotakeinaountsomedegenerayoftheequation(typially(p)p!0asjpj!+1)
but our proof does not see atallthis kindof property.
that, for every (x;t;p)2IR N
[0;T℄IR N
;
k(x;t;p)kC(1+jxj); (4)
a key fat inthe proof.
A more readable versionof assumption (H3-!)()onthe initialdata ofthe equation
is when! is loallyLipshitz ontinuous, then (H3-!) holds if
jD!(x)jC(1+jxj
) for almost every x2IR N
:
We end this setion by a listof equations forwhih the previous results apply.
1. Equations of the form
u
t
a(x;t)
u
(1+jDuj 2
)
+H(x;t;u;Du)=0:
The above results apply when > 0, H satises (H2) and a 2 C(IR N
[0;+1)) is a
bounded nonnegative funtion suh that p
a is Lipshitz ontinuous in x,uniformlywith
respet tot: Sine it isnot obvious, wehek the thirdstatement of (H1). Without loss
of generality, we an suppose a 1; the dimension of the spae N = 1 and q p 0:
We have
(p) (q)=
1
(1+p 2
)
=2 1
1+ p
2
q 2
1+q 2
=2
!
:
From the inequality 1 (1+x)
=2
x for 1x0; we get
(p) (q)
1
(1+p 2
)
=2
(q p)(p+q)
1+q 2
2q
1+q 2
(p q):
Sine we assumed 0 p q; we an nd a onstant C independent of p;q suh that
2q=(1+q 2
)C=(1+p+q):It givesthe onlusion.
2. Nongeometri urvature ow:
u
t div
Du
p
1+jDuj 2
=0:
3. Consider
u
t
Tr[A(x;t)
I
DuDu
1+jDuj 2
A(x;t)D 2
u℄+H(x;t;u;Du)=0;
If A is a bounded ontinuous funtion from IR [0;+1) into M
N
, Lipshitz in x
(uniformly int) and H satises (H2),then our results apply. In partiular,
u
t
a(x;t) p
1+jDuj 2
div
Du
p
1+jDuj 2
+(x;t)
!
=0
fulllstheassumptionsassoonasthe ontinuous funtionsa;arebounded, aisnonneg-
ative and p
a; are Lipshitz ontinuous in x uniformlyin t: If a 1 and is onstant,
thenwereognizethegeometrieikonalequation. Themeanurvatureequationforgraphs
(a1 and 0) isstudied ingreat details inthe following setion.
4.
u
t
sup
2A Tr[b
(x;t;Du)D 2
u℄+sup
2B H
(x;t;u;Du)=0;
when (H1)holds for b
and (H2) holds for H
with onstants independent of ;:
5. Withminor adaptations inthe proof of the omparison result we an deal with
u
t
f Tr[b(x;t;Du)D 2
u℄
+H(x;t;u;Du)=0;
under (H1) and (H2) for a nondereasing nonnegative funtion f whih satises, for
every z;z 0
2IR ;
f(z) f(z 0
)Cjz 0
zj
;
with 2[0;1℄:
3 Appliation to the mean urvature equation
Ourmainmotivationistoprovesomeuniquenessresultsforthe meanurvatureequation
for graphs. In this setion, we reall some fats about this equation and show that it
enters theframeworkof Theorem2.1. Wethenapply these resultstothemean urvature
ow forentire graphsin IR N+1
:
Wean write the meanurvatureequation indierentforms. To followthe notations
of the previous setion, wewill write itas
u
t Tr
b
(Du)D 2
u
=0; (5)
where, forevery p2 IR N
,
b
(p)=I
pp
1+jpj 2
: (6)
The same equation isoften written
u
t
u+ hD
2
uDu;Dui
1+jDuj 2
=0 inIR N
(0;+1); (7)
u
t p
1+jruj 2
div
ru
p
1+jruj 2
!
=0: (8)
The operator b
dened by (6)maps IR N
into S +
N
. For every p2 IR N
; we an dene
the positive symmetri square root b 1=2
(p) of b
(p) 2 S +
N
: The following lemma provides
some propertiesof b 1=2
.
Lemma 3.1 For every p2IR N
;
b 1=2
(p)=I
1
p
1+jpj 2
(1+ p
1+jpj 2
) pp:
Moreover, b 1=2
isboundedandLipshitzontinuousinIR N
; thereexistsa positiveonstant
C suh that, for every p;q2IR N
;
kb 1=2
(p) b 1=2
(q)k
Cjp qj
1+jpj+jqj
: (9)
Proof of Lemma 3.1. We rst ompute b 1=2
. For every q 2 (Spanp)
?
; b
(p)q = q and
b
(p)p=p=(1+jpj 2
). Itfollowsthatweanlookforb 1=2
intheformb 1=2
(p)=I f(p)pp.
Then, aneasy alulation gives the result. Lookingat the formula, it islear that b 1=2
is
ontinuous and bounded.
Tohek (9), we write
kb 1=2
(p) b 1=2
(q)k = kf(q)qq f(p)ppk
= kq(q p)f(q)+(q p)pf(q)+pp(f(q) f(p))k
(jqjf(q)+jpjf(q))jp qj+jpj 2
jf(p) f(q)j:
Without lossof generality, wean takejqjjpj. On the one hand,
jqjf(q)+jpjf(q) 2jqjf(q)
2jqj
p
1+jqj 2
(1+ p
1+jqj 2
)
2
1+ p
1+jqj 2
2
1+jqj
2
1+ 1
2
(jpj+jqj)
4
1+jpj+jqj
: (10)
On the otherhand, by setting g(p)= p
1+jpj 2
;
jpj 2
jf(q) f(p)j jpj 2
f(p)f(q)jg(p)(1+g(p)) g(q)(1+g(q))j
jpj 2
f(p)f(q)j1+g(p)+g(q)jjg(p) g(q)j:
(p;q)7!jpj 2
f(p)f(q)j1+g(p)+g(q)j(1+jpj+jqj)
is ontinuous and bounded on the set f(p;q) 2 IR 2N
: jpj jqjg thus there exists a
onstant C
1
>0 suhthat for every p;q 2IR N
; jpjjqj; we have
jpj 2
f(p)f(q)j1+g(p)+g(q)jjg(p) g(q)j C
1 jp qj
1+jpj+jqj
: (11)
Combining(10) and (11), we obtain the result with C =4+C
1
: 2
Thanks tothe previous lemmaand the existeneof asmooth solutionto (7)(f. [12℄,
[14℄ and [3℄) weget
Theorem 3.1 Assumethatu
0
2C(IR N
)satises(H3-u
0
)(),with0 <(1+ p
5)=2.
Then (7) (or equivalently (5) or (8)) has a unique solution u 2 C(IR N
[0;+1))\
C 1
(IR N
(0;+1)) in C
poly :
Weturntoageometrialappliationtotheso-alledlevel-setapproahtothe generalized
evolution of hypersurfaes by their mean urvature. This method, introduedfor numer-
ial omputations by Osher and Sethian [19℄, was developed theoretially by Evans and
Spruk [13℄ and Chen,Giga and Goto [7℄.
Let us reall briey the level-set approah in the ase of the mean urvature motion
of entire graphs. We onsider the graph of the initial datum u
0
2 C(IR ) of (7) as an
hypersurfae
0
=f(x;y)2IR N
IR:u
0
(x)=ygofIR N+1
. Dene
0
=f(x;y)2IR N+1
:
u
0
(x)<yg. Wetake auniformlyontinuous funtion v
0 :IR
N+1
!IR suh that
0
=f(x;y)2IR N+1
:v
0
(x;y)=0g and
0
=f(x;y)2IR N+1
:v
0
(x;y)>0g (12)
(hoose the signed-distane to
0
for instane). Next,we onsider a funtion v :IR N+1
(0;+1) ! IR suh that v(x;u(x;t);t) =0 where u is a solution of (7). Formally, v has
to satisfythe well-known geometrialmean urvature equation
8
<
: v
t
v+ hD
2
vDv;Dvi
jDvj 2
in IR N+1
(0;T);
v(;;0)=v
0
in IR N+1
:
This equation admits a unique visosity solution v 2 UC(IR N+1
[0;+1)) for every
initial datum v
0
2 UC(IR N+1
), where UC denotes the uniformly ontinuous funtions.
Moreover, the level-set approahworks: it meansthat we an dene, for every t2[0;T℄;
t
=f(x;y)2IR N+1
:v(x;y;t)=0g and
t
=f(x;y)2IR N+1
:v(x;y;t)>0g;
t t t t 0 0
of v
0
: The family (
t )
t
is alled the generalized evolution by mean urvature of the graph
0 and
t
is alled the front. A natural issue is the onnetion between this generalized
evolution and the lassial motionby meanurvature of the graphof u
0
. Note that
t is
dened as the 0-level-set of a uniformly ontinuous funtion; it may be very irregular in
generaland an even develop an interior inIR N+1
. In our ontext, we have
Theorem 3.2 If u
0
2C(IR N
) satises (H3-u
0
)(), with 0 <(1+ p
5)=2, then, for
every t 2[0;T℄, the set
t
isa entire smooth graph, namely
t
=f(x;y)2IR N+1
:y=u(x;t)g;
whereuistheuniquesmoothsolutionof(7)withinitialdatumu
0
. Moreover,theevolution
of
t
agrees with the lassial motion by mean urvature in the sense of the dierential
geometry.
Proof of Theorem3.2. From[3℄, weknowthat, if westart withanhypersurfae
0
whih
is anentire ontinuous graph inIR N
IR , then, forevery t2[0;T℄,
t
=f(x;y)2IR N+1
:u (x;t)y u +
(x;t)g;
whereu and u +
are respetivelythe minimaland themaximal (disontinuous)visosity
solution of (7). In the speial ase of the mean urvature equation, we proved that the
boundary of the front
t
is smooth. It follows that u and u +
are smooth. At this step,
we aimat applying Theorem 3.1to showthat u =u +
. To doit, we need to know that
u ;u +
2C
poly
: NotethatCorollary2.1isnot suÆientbeausewesuppose, apriori,that
the solutionhas a polynomialgrowth. To overome this diÆulty, we invoke a L 1
loal
bound for(7)(see [3℄): thereexists aonstantC suhthat, forevery(x;t)2IR N
[0;T℄;
ju(x;t)jmaxfju
0 (y)j+
p
2Ct :y2
B(x;
p
2Ct)g:
From (H3-u
0
)(),thereexists apositive onstant
~
C suh thatju
0 (y)j
~
C(1+jyj 1+
):It
follows
ju(x;t)j
~
C(1+(jxj+ p
2Ct) 1+
)+ p
2CT;
whih proves the laim.
It implies u ;u +
2 C
poly
and from Theorem 3.1, we obtain u = u +
:= u. Finally,
t
=Graph(u(;t))isasmooth submanifoldof IR N+1
(inpartiular,
t
never fattens). In
this ase, the generalized evolution ts inwith the lassial evolution by mean urvature
(see Evans and Spruk [13℄ and [14℄ for the agreement with an alternative generalized
motion). 2
Wereferto[3℄ and[5℄for some moregeneralgeometrialmotions. Inthese works,we
assoiate a geometrial motion to some quasilinear equations of the type (2) for whih
the above tehniques apply.
1. We argueby ontradition assumingthat there exists (~x;
~
t)2IR N
[0;T℄ suh that
u(~x;
~
t)>v(~x;
~
t): (13)
For ";>0;pk >maxf2;+1g tobe hosen later on, we introduethe funtions
dened by : for every x 0
;y 0
2IR N
,
K(x 0
):=1+jx 0
j k
; '(y 0
):=
jy 0
j p
"
p
and (x 0
;y 0
):=K(x 0
)('(y 0
)+ ):
Then we onsider the test-funtion given by : for everyx;y2IR N
;
(x;y;t):=e Lt
(x+y;x y)+t;
where L; >0.
2. The rst step of the proof is the
Lemma 4.1 Under the assumptions of Theorem 2.1, there exists a onstant C >0 suh
that, for any x2IR N
and t2[0;T℄;
u(x;t)C(1+jxj +1
) and v(x;t) C(1+jxj +1
): (14)
Moreover, if pk >+1, the
sup
(x;y;t)2(IR N
) 2
[0;T℄
u(x;t) v(y;t) (x;y;t)
is nite and is ahieved at a point (x ;y;
t). Finally, we an hoose the parameters ,
and " smallenough in order to have a positive supremum and jx yj 1.
We postponethe proofofthis lemmatotheend ofthe setion. Fromnowon, wesuppose
that the supremum ispositiveand that jx yj 1.
3. The idea of the proof isthe following : the seven next steps are devoted to show that
we an x the parameters in in order to fore the maximum to be ahieved at time
t =0. The last step deals with the ase t =0. We prove that the partiular form of the
test-funtion andthe assumptionabout the modulusof ontinuity of theinitialdatalead
to aontradition with (13) whihwillbethe end of the proof.
4. We rst onsider the ase when the supremum is ahieved ata point suh that
t>0.
By applying the fundamental result of the Users' guide to visosity solutions ([9,
Theorem 8.3℄),we knowthat, forevery >0,thereexista
1
;a
2
2IR andX;Y 2S
N suh
that
(a
1
;D
x
(x ;y;
t);X)2
P 2;+
(u)(x;
t);
(a
2
; D
y
(x ; y;
t);Y)2
P (v)(y;
t);
a
1 a
2
=
t (x;y;
t);
and
X 0
0 Y
A+A 2
where A=D 2
(x ; y;
t): (15)
Therefore, sine u and v are respetively sub- and supersolution of (2), we have
a
1
Tr[ b(x;
t;D
x
(x ; y;
t))X℄+H(x;
t;u(x;
t);D
x
(x ; y;
t)) 0 (16)
and
a
2
Tr[b(y;
t; D
y
(x ;y;
t))Y℄+H(y;
t;v(y;
t); D
y
(x;y;
t))0: (17)
By subtrating(17) from (16), we obtain
+L(x ;y;
t) Tr[b(x ;
t;D
x
(x ;y;
t))X℄ Tr[b(y;
t; D
y
(x ; y;
t))Y℄
+H(y;
t;v(y;
t); D
y
(x ; y;
t)) H(x;
t;u(x;
t);D
x
(x ; y;
t)): (18)
In order to show that the maximum annotbe ahieved for
t>0, we have to prove that
this inequalityannot holdfor asuitablehoieof the parameters and, todoso,we have
to estimatethe right-handside of this inequality.
5. Forthe sake of notational simpliity,we write
x
=(x;
t;D
x
(x; y;
t)) and
y
=(y;
t; D
y
(x ;y;
t))
and omit in the following omputations the dependene on (x ;y;
t) when there is no
ambiguity. Forany orthonormal basis (e
i )
1iN=1 of IR
N
, we have
Tr[b(x;
t;D
x
(x ; y;
t))X b(y;
t; D
y
(x ; y;
t))Y℄=Tr
T
x X
x
T
y Y
y
= N
X
i=1 hX
x e
i
;
x e
i
i hY
y e
i
;
y e
i
i: (19)
On another hand, wean write(15) as,for all; 2IR N
;
hX;i hY;ie L
t
hD 2
xx
( +);+i+2e L
t
hD 2
xy
(+); i
+e L
t
hD 2
yy
( ); i+hA 2
(;);(;)i:
x i y i
beomes
+L A
1 +A
2 +A
3 +A
4
; (20)
where
8
>
>
>
>
>
>
<
>
>
>
>
>
>
: A
1
=e L
t
kD 2
xx k k
x +
y k
2
;
A
2
=2e L
t
kD 2
xy k k
x +
y kk
x
y k;
A
3
=e L
t
kD 2
yy k k
x
y k
2
;
A
4
=H(y;
t;v(y;
t); D
y
) H(x;
t;u(x;
t);D
x ):
In the fourth next steps, we estimate the A
i
;1 i 4. These estimates are heavy to
obtainbutthebasiideaiseasy: wewanttoprovethat,forasuitablehoieofparameters,
eah jA
i
j is bounded by
C+, where
C is a xed onstant and is suÆiently small.
It willlead to aontradition with (20) if we take L large enough.
In allthe estimates below, C willdenoteaonstant whih mayvaryfromlinetoline,
may depend onk, p but not on", and L.
6. Estimateof A
1 .
We have
kD
xx
kk(k 1)jx+yj k 2
('+ ):
From (4), we get
A
1
2e
L
t
C 2
k(k 1)jx+yj k 2
((1+jxj) 2
+(1+jyj) 2
)('+ ):
But, sinejx yj 1, wean ompare both jxjand jyj withjx+yj. Itfollows thatthere
exists aonstant C =C(k)suh that
A
1 Ce
L
t
K('+ ): (21)
7. Estimateof A
2 .
An expliit omputation gives D 2
xy
= DK D'. We may assume without loss of
generality that x 6=yor equivalently D' 6=0 sine otherwise A
2
would be 0 and auses
noproblem.
From inequalities (4)and jx yj 1, we thenobtain
A
2
2Ce L
t
jDKjjD'j(3+jx+yj)k
x
y
k: (22)
Using (H1),we have
k
x
y
k k(x;
t;D
x
) (y;
t;D
x
)k+k(y;
t;D
x
) (y;
t; D
y )k
Cjx yj +C jD
x +D
y j
1+jD
x
j+jD
y j
:
But D
x =e
Lt
(DK('+ )+KD') and D
y =e
Lt
(DK('+ ) KD'). Weobtain
k
x
y
k Cjx yj +C
2e L
t
jDKj('+ )
1+ 1
2 (jD
x +D
y
j+jD
x
D
y j)
Cjx yj +2C
jDKj('+ )
KjD'j
: (23)
Combiningthis lastinequality with (22) and using that
jDKj=kjx+yj k 1
and jD'j=p
jx yj p 1
"
p
;
showthat thereexists apositiveonstant C =C(k;p) suh that
A
2 Ce
L
t
K('+ ): (24)
8. Estimateof A
3 .
This step is the most tehnial one. First,the omputationof D 2
yy
gives
A
3 e
L
t
KkD 2
'k k
x
y k
2
and we have
k
x
y k
2
= k(x ;
t;D
x
) (y;
t;D
x )k
2
+2k(x;
t;D
x
) (y;
t;D
x
)kk(y;
t;D
x
) (y;
t; D
y )k
+k(y;
t;D
x
) (y;
t; D
y )k
2
:
It follows
jA
3
j 2Ke L
t
k(x;
t;D
x
) (y;
t;D
x )k
2
kD 2
'k
+2Ke L
t
k(y;
t;D
x
) (y;
t; D
y )k
2
kD 2
'k: (25)
We estimateseparately the two terms whih appear inthe right-handside (25). For the
rst one, using (H1), we obtain
2Ke L
t
k(x ;
t;D
x
) (y;
t;D
x )k
2
kD 2
'k2C 2
p(p 1)e L
t
K' (26)
beause jx yj 2
kD 2
'kp(p 1)'.
The estimate of the seond term is the hardest one. Again we may assume without
seond term =2Ke L
t
k(y;
t;D
x
) (y;
t; D
y )k
2
kD 2
'k
2CKe L
t
jx yj p 2
"
p
CjD
x +D
y j
1+jD
x
j+jD
y j
2
2CKe L
t
jx yj p 2
"
p
jD
x +D
y j
1+ 1
2 (jD
x +D
y
j+jD
x
D
y j)
2
2CKe L
t
jx yj p 2
"
p
2e L
t
('+ )jDKj
1+e L
t
('+ )jDKj+e L
t
KjD'j
2
2CKe L
t
jx yj p 2
"
p
e L
t
('+ )jDKj
2
(e L
t
('+ )jDKj) 2
(e L
t
KjD'j) 2
for every positive; suh that +1.
We nowdistinguish two ases.
|1 st
ase. If ' , then taking =0 and =1, we get
seondterm 2CKe L
t
jx yj p 2
"
p e
2L
t
' 2
jDKj 2
e 2L
t
K 2
jD'j 2
2CKe L
t
'
jDKj
K
2
2CKe L
t
': (27)
|2 nd
ase. Otherwise, '. Then
seond term 2CKe L
t
jx yj p 2
"
p
e L
t
jDKj
2
(e L
t
jDKj) 2
e L
t
K pjx yj
p 1
"
p
2
2CK 1 2
e L
t (3 2 2)
jx yj
p 2 2(p 1)
"
p+2p
( jDKj) 2 2
:
We hoose =
p 2
2(p 1)
and this yields
seondterm2Ce 3L
t
"
p
p 1
2 2
K 1
p 1
jDKj 2 2
:
From now on, we take
p>k+1 (28)
and we hoose = 1
2
2 with
=
p k 1
(p 1)(k 1)
: (29)
Usingthat jDKjkK k
; we obtain
seondterm 2Ce 3L
t
"
p
p 1
2 2
K 1
p 1 +
(2 2)(k 1)
k
2Ce 3L
t
"
p
p 1
K : (30)
Combining(27) and (30), we obtain
seond termmax n
2Ce L
t
K'; 2Ce 3L
t
"
p=(p 1)
K o
; (31)
and from (26) and (31), we get
jA
3
j Ce
L
t
K'+CKe L
t
('+ )
+max n
2Ce L
t
K'; 2Ce 3L
t
"
p=(p 1)
K o
Cmax n
e L
t
K('+ ); e 3L
t
"
p=(p 1)
K o
; (32)
where C is a positive onstant independent of ", , L and .
9. Estimateof A
4 :
Sine we have
0<(x ;y;
t)u(x;
t) v(y;
t)
and sine u7!H(;;u;)is nondereasing, we rst get
A
4
= H(y;
t;v(y;
t); D
y
) H(x;
t;u(x;
t);D
x )
H(y;
t;u(x;
t); D
y
) H(x;
t;u(x;
t);D
x ):
Then, using (H2),it follows that
A
4
jH(y;
t;u(x;
t); D
y
) H(y;
t;u(x;
t);D
x )j
+jH(y;
t;u(x;
t);D
x
) H(x;
t;u(x;
t);D
x )j
m((1+jyj)jD
x +D
y
j)+m((1+jD
x
j)jx yj) :
Beause of the properties of a modulus of ontinuity, for every r > 0 and every > 0,
there exists a positive onstant C() suh that m(r) +C()r: Taking = =2 and
up to replaeat eah step C() by alarger one whihdepends only on, we obtain
A
4
2
+C()
(1+jyj)jD
x +D
y
j+(1+jD
x
j)jx yj
2
+C()
e L
t
(1+jyj)jDKj(' + )+(1+e L
t
(jDKj('+ )+KjD'j))jx yj
2
+C()
e L
t
K('+ )+jx yj
:
jx yj
jx yj
"
p
+"
q
K('+ )+"
q
and nally, we obtain
A
4
+C()
e L
t
K('+ )+"
q
; (33)
where C()is a positive onstantindependent of ;" and L.
10. End of the ase
t>0.
Pluggingestimates (21), (24), (32) and (33) in(20) yields
+L
2
+(C+C())+Ce 3L
t
"
p=(p 1)
K+C()"
q
;
where C isa positiveonstant whihis independent of ", , and L. Thus, we an rst
hoose a onstant L large enough, namely L() C+C()+1; suh that the previous
inequalitybeomes
2
+Ce 3L
t
"
p=(p 1)
K+C()"
q
; (34)
for every positive"; and :Then we take="
`
, hoosing ` largeenough suh that
` >
p
p 1
: (35)
Thus (34) reads
2 +e
L
t
K('+ )Ce 3LT
"
` p=(p 1)
K+C()"
q
:
Taking " small enough suh that C()"
q
< =2 and Ce 2LT
"
` p=(p 1)
1; we obtain a
ontradition. Finally, the onlusion of the preeding steps is: if we hoose L;p and k
as above and take = "
`
(where ` satises (35)), then neessarily
t = 0 for suÆiently
small".
11. From the previous step, we know that the maximum of the funtion u v is
ahieved at a point (x;y; 0) whih implies, from (13) and Lemma 4.1, that there exists
Æ>0 suh that
0<Æu(~x;
~
t) v(~x;
~
t) "
`
e L
~
t
K(2~x)
~
tu(x;0) v(y;0) K(x+y)(' +"
`
);
for " and smallenough. Sineu(x;0) u
0
v(y;0),this inequality leadsto
Æ u
0
(x) u
0
(y) K(x+y)(' +"
`
):
0
Æ m((1+jxj+jyj)
jx yj) K(x+y)(' +"
`
)
Æ
2
+C(Æ)(1+jxj+jyj)
jx yj K(x+y)('( x y) +"
`
): (36)
Sinejx yj 1,thereexists
~
C suhthat (1+jxj +jyj)
~
C(1+jx+yj
)andtherefore
(36) yields
Æ
2
C(Æ)
~
C(1+jx+yj
)jx yj (1+jx+yj k
)
jx yj p
"
p
+"
`
: (37)
Inordertostudytheright-handsideofthisinequality,weintroduethefuntiong dened
by
g(r;s)=C(Æ)
~
C(1+r
)s (1+r k
)
s p
"
p +"
`
;
for every r;s 0: One easily shows that there exists a onstant C depending only on
C(Æ);
~
C and psuh that
g(r;s)C (1+r
) p
p 1
(1+r k
) 1
p 1
"
p
p 1
"
`
(1+r k
): (38)
Then,using that,forevery r>0;
(1+r
) p
p 1
(1+r k
) 1
p 1
~
C(1+r p k
p 1
),with
~
C =
~
C(N; ;p;k),we
obtain from(38)
g(r;s) f(r):=C
~
C
1+r p k
p 1
"
p
p 1
(1+r k
)"
`
; (39)
for every r;s 0:Wetakep large enoughsuh that
p k >0: (40)
Sine, in addition, we have already hosen k > +1 at the beginning, the funtion f
ahieves a maximumat
r=C 0
"
`(p 1) p
p( k )
;
where the C 0
depends only on C;
~
C;p; ;k: Replaing r by this value, we obtain that for
every r 0;
f(r)C
~
C"
p
p 1
+
C"
; (41)
where
C is a onstant depending onC;
~
C;p; ;k and
=
kp `(p k)
p(k ) :
Æ
2 C
"
p
p 1
+"
with C isa onstant dependingonly onÆ; ;p;k and N.
In order to onlude through the above inequality by letting " tend to 0, we need to
have > 0 i.e. to be able to hoose the parameters in suh a way to have `<
kp
p k
sine p k >0from(40). Todoso,we reallthat wehavehosenk >maxf2;+1g at
the beginning, p>k+1 (see (28)) and `>
p
(p 1)
(see (35)) where =
p k 1
(p 1)(k 1)
(see (29)). Thus, we need to nd `;k and psuh that
k>maxf2;+1g; p>k+1; p>k and
p(k 1)
p k 1
<`<
kp
p k
: (42)
Tofulll the lastondition in(42), itis suÆienttohave p( k( 1))>2k: Itfollows
that, if wean nd a suitablek; up totakeplarge enough, then allthe onditions would
be satised. Wedistinguish two ases:
|1 st
ase. If 1; then we take k >2 and we are done.
|2 nd
ase. If >1,then wehave tond k suh that
+1<k <
1
:
It leads to aondition on ;namely 2
<+1 whih is automatiallysatised provided
<
1+ p
5
2
as wesupposed it.
Finally,in any ase, itispossibletofulll onditions (42);thus the proofof the theo-
rem is omplete. 2
Nowwe turn to the proof of the Lemma.
Proof of Lemma 4.1. We rst prove (14). We are going to doit only for the subsolution
u, the proof for the supersolution v being analogous.
To do so, we introdue for C, L, " > 0; k 2 and the sequene of smooth funtions
(
"
)
"
dened by
"
(x;t)=e Lt
C(1+jxj 2
) +1
2
+"(1+jxj k
)
:
Tedious but straightforward omputations show that, for any C, and k, if L is hosen
large enough, then
"
isa strit supersolution of (2)for any " smallenough.
On an other hand, sine u(x;0) u
0
(x) in IR N
and sine u
0
satises (H3-u
0
)(), it
is lear that if C is hosen large enough, then u(x;0)
"
(x;0)in IR N
.
Finally sine u is in C
poly
, we have, for k large enough u(x;t)=(1+jxj) ! 0 when
jxj!+1uniformlyfor t 2[0;T℄and therefore u(x;t)
"
(x;t) for jxj large enough.
Using these three properties, one shows easily that u(x;t)
"
(x;t) in IR N
[0;T℄:
indeed,beauseof the lastone, u
"
ahieves itsmaximum on IR N
[0;T℄but sine u
is a visosity subsolution of (2) and
"
is a strit smooth supersolution of this equation,
this maximum annot be ahieved for t > 0. Therefore it is ahieved for t = 0 and the
maximum is thereforenegative,proving the laim. Letting" tend to 0 yields(14).
Nowweprovetheseondpartofthelemma. Sineu(x;t) v(y;t) (x;y;t)isupper-
semiontinuous, inordertoprove thatthe supremum isahieved, itsuÆes toprovethat
thisfuntiontendsto 1whenjxj;jyj!+1(uniformlywithrespettot2[0;T℄). But
e Lt
K(x+y)('(x y)+ )+t
jx yj p
"
p
+ (1+jx+yj k
)
min
; 1
"
p
jx yj p
+1+jx+yj k
: (43)
Sine pk>maxf2;+1g;using the onvexity of r 7!r k
,we have
jx yj p
+1+jx+yj k
jx yj k
+jx+yj k
jxj k
+jyj k
:
From (43) and (14), it follows
u(x;t) v(y;t) (x;y;t) u(x;t) v(y;t) min
; 1
"
p
jxj k
+jyj k
C 1+jxj +1
+jyj +1
min
; 1
"
p
jxj k
+jyj k
whih proves the laim sine +1<k.
On another hand, wehave
u(x;
t) v(y;
t) (x ; y;
t)u(~x;
~
t) v(~x;
~
t) e L
~
t
K(2~x)
~
t:
From (13), the right-hand side is positiveif and are suÆiently small.
For and suÆiently small, we have
0<u(x;
t) v(y;
t) (x ;y;
t)C 1+jxj +1
+jyj +1
e L
t
K(x+y)(' + )
t:
By using the onvexity of r7!r +1
, it follows
1+jx+yj k
jx yj p
"
p
+
C 1+jx+yj +1
+jx yj +1
:
Sine +1<k, we obtain
jx yj p
"
p
C 1+jx yj +1
whih impliesthat jx yj 1 for " smallenough sine p >+1. And the proof of the
lemma isomplete. 2
The uniqueness part in the statement of Corollary 2.1 is an immediate onsequene of
Theorem 2.1. In this setion, we prove the existene result using Perron's method. For
the desription of this method in the ontext of visosity solutions, we refer to Ishii [17℄
and Crandall, Ishii and Lions [9℄.
A straightforward omputation shows that, if C > 0 and L are large enough, then
u(x;t) = Ce Lt
(1+jxj 2
) +1
2
and u(x;t) = Ce Lt
(1+jxj 2
) +1
2
are respetively visosity
sub- and supersolution of (2).
Wedenethe setS inthefollowingway : anloallybounded, possibly disontinuous,
funtion w dened on IR N
[0;T℄ is in S if u w u on IR N
[0;T℄ and if w
is a
visosity subsolution of (2) satisfyingthe initialondition in the visosity sense, i.e.
min
w
t Tr
b(x;0;Dw)D 2
w
+H(x;0;w;Dw); w
(x;0) u
0 (x)
0 (44)
inIR N
f0g. Then weset
u(x;t)=sup
w2S
w(x;t):
Classial arguments show that u is stilla subsolution. Moreover, when a subsolution
wsatisesthe initialondition(44), itiswell-known(see [1℄forinstane) thatw
(x;0)
u
0
(x)for allx2IR N
:From the denition ofu; itfollowsu
(x;0)u
0
(x)for allx2IR N
:
From Perron's method, u is a supersolution whih satises the initialondition (44)
where we replae \min" by \max", \w
" by \w
" and \" by \". As above, itfollows
u
(x;0)u
0
(x) for allx2IR N
:
Finally, we have u
(;0) u
0 u
(;0) in IR N
: Applying our omparison result
(Theorem 2.1) to u
and u
; we obtain u
u
whih means that u is a ontinuous
visosity solutionof (2) with initialdatum u
0
: 2
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