charro colorado peral

7/27/2019 Charro Colorado Peral

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MULTIPLICITY OF SOLUTIONS TO UNIFORMLY ELLIPTIC

FULLY NONLINEAR EQUATIONS WITH CONCAVE-CONVEX

RIGHT HAND SIDE

FERNANDO CHARRO, EDUARDO COLORADO, AND IRENEO PERAL

Abstract. We deal with existence, nonexistence and multiplicity of solutionsto the model problem

(P)

8 0, in ,u = 0, on ,

where Rn is a smooth bounded domain, F is a 1-homogeneous fullynonlinear uniformly elliptic operator, > 0, 0 < q < 1 < r < r and r thecritical exponent in a sense to be made precise.

We set up a general framework for Fin which there exists a positive thresh-old for existence and nonexistence. Moreover a result on multiplicity is ob-tained for 0< 0, and considerF : Rn Sn Rsatisfying the following structural hypothesis,

(F1) Uniform ellipticity: There exist constants 0 < such that for allX, Y Sn withY 0,

trace(Y) F(, X+ Y) F(, X) trace(Y)

for every Rn.

(F2) Homogeneity: F(t,tX) = t F(, X) for all t > 0. We further assumeF(0, 0) = 0.

(F3) Structure condition: There exists > 0 such that, for all X, Y Sn, and1, 2 R

n,

P,(X Y) |1 2| F(1, X) F(2, Y) P+,(X Y) + |1 2|,

2000 Mathematics Subject Classification. 35J60, 35B45, 35B33.Key words and phrases. Fully nonlinear, Uniformly elliptic, Viscosity solutions, Non-proper,

Multiplicity of solutions, Degree theory, Critical exponent, A-priori estimates.F.C. and I.P. are partially supported by projects MTM2007-65018, MEC, Spain.E.C. is partially supported by the MEC Spanish projects MTM2006-09282 and RYC-2007-

04136.1


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2 FERNANDO CHARRO, EDUARDO COLORADO, AND IRENEO PERAL

whereP, are the extremal Pucci operators defined as

P

+

,(X) =i>0

i(X) i0

i(X) i 0, in ,u = 0, on ,

with 0 < q < 1 < r has at least one nontrivial viscosity solution for < andno nontrivial solution for > . Notice that the result in [13] holds without anyrestriction on the size ofr .

Definition 2. Given F : Rn Sn R, define G(X) =F(0, X). We will say thatthe operator G : Sn R satisfies a Liouville-type result in Rn up to s whenever

v 0 is the unique non-negative viscosity solution of(2) G(D2v) = vr, in Rn,

for any 1< r < s. Finally, we define thecritical exponentfor problem (1) as,

r= sup{s | s R and G satisfies a Liouville-type result in Rn for any 1< r < s }.

Notice that (F1) and (F2) imply that G is uniformly elliptic with constants0< 0 and Rn+ = {(x1, . . . , xn) Rn :xn > 0}. Consequently, we will refer

toG as the blow-up operator hereafter.

Our goal in the present work will be to study the existence of a second nontrivialsolution for every (0, ) provided r < r. The proof involves uniform L

estimates and topological arguments.We assume the following extra hypotheses onFin order to get the L estimates.

(F4) G(QtXQ) =G(X) whereG(X) = F(0, X) as above and Q O(n) ={QSn : Q Qt =I d}.

(F5) Problems (3), (4) have no positive solution.


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MULTIPLICITY OF SOLUTIONS TO UNIFORMLY ELLIPTIC FULLY NONLINEAR. . . 3

The main result of the paper is the following abstracttheorem.

Theorem 3. ConsiderF :R

n

S

n

R

satisfying (F1) (F5), and let0< q 0, in ,u = 0, on ,

(i) has no positive solution for >,(ii) has at least one positive solution for= ,

(iii) has at least two positive solutions for every (0, ).

In some sense, these results bring to the fully nonlinear framework the well-known results on global existence and multiplicity of solutions in[2] and [3, 20](and the references therein) for the semilinear and quasilinear setting, respectively.

The proof of Theorem 3 follows some ideas of Ambrosetti et al. in [3] and involvesthe mentioned uniform L a-priori estimates.

These kind of bounds are obtained following the blow-up technique of Gidas-Spruck in [21] which leads to a contradiction with (F5) (see also [18], where differenttechniques involving topological and variational arguments are developed to get apriori uniformL estimates). As a general fact, it is difficult to verify (F5); in thisdirection we have the following results.

Theorem 4. Letv be a nontrivial non-negative bounded (viscosity) solution of

(5)

G(D2v) = f(v), in Rn+v 0, in Rn+v = 0, on Rn+,

where f is locally Lipschitz continuous function, f(0) 0 and G : Sn R isuniformly elliptic with constants 0 < < and 1-homogeneous. Furthermore,suppose that

G(QtXQ) =G(X) forQ= (qij)1i,jn, a matrix with

qij =ij if neither i nor j = n, andqij =ij otherwise.(6)

Then, v is monotonic in thexn variable:

v

xn>0 inRn+.

The non-existence of solutions to problem (4) follows from the above theorem(see Section 4 and [5], [27]).

Concerning (3), the following result for general uniformly elliptic nonlinearitiesGis proved in [16].

Theorem 5(Theorem 4.1 in [16]). LetG : Sn Rbe a uniformly elliptic operatorin the sense that there exist constants0< such that for allX, Y Sn withY 0,

trace(Y) G(X+ Y) G(X) trace(Y).

Assume G(0) = 0 and =

(n 1) + 1 > 2, and let v C(Rn) be a viscositysolution of

G(D2v) vr, in Rn,v 0, in Rn.

Then, if0< r /( 2), we havev 0.


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Notice that, even though the exponent /( 2) could not be maximal for aprecise nonlinearityG, Theorem 5 provides a lower bound for the critical exponent,

valid for the whole class of uniformly elliptic operators. Further restrictions on Fallow one to improve the range of exponents. For example, the maximal range forthe linear equation is known to be

1< r 1, hence without restriction on the growth ofr. This motivatesthe fact that the critical exponent in Definition 2 comes from the Liouville resultfor (3) alone.

Remark 7. Hypotheses (F4), (F5) are only used in the proof of the uniform L

estimates, Proposition 12, in order to carry out the blow-up argument.

The paper is organized as follows. Section 2 is devoted to some preliminaryresults that will be used in the sequel. Then, in Section 3, the proof of Theorem 3is given. The blow-up argument is presented there and uniform L a-priori boundsare stated under hypotheses (F1) (F5). Next, the existence of a second solutionto (1) is proved using a-priori bounds and theoretical degre arguments.

Section 4 is devoted to prove Theorem 4 following [5] and [27].Finally, Section 5 is devoted to applications of the abstract framework described

in Sections 3 and 4. The examples considered fulfill hypotheses (F1) (F5) andinclude the model case whereGis a Pucci extremal operator (subsection 5.1) whichinclude the laplacian as a particular case and then, concave (convex) operators(subsection 5.2) and a class of Isaacs operators, which are neither concave norconvex (subsection 5.3). We impose hypothesis (F4) in all the cases except whenGis a Pucci operator, where the condition is built-in.

2. Preliminaries

2.1. Eigenvalues. Under hypotheses (F1), (F2) and (F3), Theorem 8 in [6] holds(hereC2 regularity of is required). Hence, we know that there exists a principaleigenvalue 1 forFdefined as

1= sup{ | v >0 in s.t. F(v, D2v) v}.

in the sense that 1< and there exists a nontrivial solution (eigenfunction) to

(7)

F(v, D2v) = v, in ,

v >0, in ,v= 0, on .

Moreover, by definition of1, we know that for every > 1 problem (7) does nothave strictly positive solutions.

Other references for the existence of eigenvalues in the fully nonlinear setting are[27], [8] and the references therein.

The existence of such a principal eigenvalue and eigenfuntion is used in the proofof Theorem 3.

2.2. Hopfs Lemma for uniformly elliptic equations. We recall here the Hopfboundary lemma. An adaptation of the proof in [22, Section 3.2] can be found in[13]. For further refinements, see [25] (see also [4] and [28]).


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Proposition 8 (Hopfs Lemma). Let be a bounded domain and u a viscositysolution of

F(u, D2u) 0 in,whereF satisfies(F1), (F2), and(F3). In addition, letx0 satisfy

i) u(x0)> u(x) for allx .ii) satisfies an interior sphere condition atx0.

Then, for every nontangential directionpointing into ,

limt0+

u(x0+t) u(x0)

t 0 in. ConsiderF : Rn Sn R verifying(F1), (F2) and(F3). Finally, letu, v C()such that

F(u, D2u) f(x), and F(v, D2v) g(x), in,

in the viscosity sense. Assumeu v on, thenu v in. Furthermore, iff < gin, we haveu < v in.

Proof. 1. We can assume v > 0 in , since adding a constant to both u and vdoes not affect the problem. Now, let

v(x) = (1 +) v(x).

Indeed, by homogeneity,

(8) F(v, D2

v) (1 + ) g(x), and u v 0 on ,for small enough. Now, we argue by contradiction. Suppose that there existsx0 such that

(u v)(x0) = max

(u v)> 0.

Then, (8) implies x0 /. Now define

w(x, y) =u(x) v(y)

2|x y|2

and denote (x, y) such that w(x, y) = max w(x, y). Such pairs (x, y)satisfy

(1) lim

|x y|2 = 0.

(2) lim

w(x, y) = w(x,x) = max

(u v), whenever (x,x) is an accumulation

point of (x, y).

Properties (1) and (2) are well known and can be found, for example, in Lemma3.1 in [15].

Hence, we can assumex, yx0as hereafter without loss of generality.As a consequence, x, y for every large enough; applying the MaximumPrinciple for semicontinuous functions (see for instance, [14] and [15]), there existtwo symmetric matricesX, Y such that

(x y), X

J2+

u(x), and

(x y), Y

J2

v(y),

and X Yin the sense of matrices. By definition of viscosity sub- and superso-lutions (see [15]), we have

F

(x y), X

f(x),


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and

F(x y), Y (1 + )g(y) (1 + )f(y).Then, by degenerate ellipticity, we get

(1 +)f(y) f(x) F

(x y), Y

F

(x y), X

0.

Letting , by continuity we arrive at

0< f(x0) 0,

a contradiction. Thus, u v in , and, letting 0, we find u v in .

2. We assume in the sequel thatf < g in . Ifu =v we have to proveu < vin .Since we have already proved that u v in , suppose to the contrary that thereexists x0 such that u(x0) = v(x0), that is, x0 is a maximum point of u v.Consequently, x0 is the only maximum point ofu(x) v(x) |x x0|

4.

Considerw(x, y) =u(x) v(y) |x x0|

4 2

|x y|2

and (x, y) such thatw(x, y) = max w(x, y) as before. By properties (1) and(2) above, we have x, y x0 as . As a consequence, x, y for every large enough. Reasoning as above, we can find two symmetric matrices X, Ysuch that

(x y), X

J2+

u(x) |x x0|4

, and

(x y), Y

J2

v(y),

and XYin the sense of matrices. As a consequence,(x y) + 4|x x0|

2(x x0),

X+ 4|x x0|2Id+ 4(x x0) (x x0) J

2+u(x),

By definition of viscosity sub- and supersolution (see [15]), we have

g(x) f(y) F((x y), Y)

F

(x y) + 4|x x0|2(x x0),

X+ 4|x x0|2Id+ 4(x x0) (x x0)

P,(Y X) +O(|x x0|

2) O(|x x0|2).

as 0. Letting 0 we get 0< g(x0) f(x0) 0 by hypothesis, and we aredone.

3. The abstract result. Proof of Theorem 3

For the readers convenience we recall the main result already stated in theintroduction.

Theorem 3. ConsiderF : Rn Sn R satisfying (F1) (F5), and let0< q 0, in ,u = 0, on ,

(i) has no positive solution for >,(ii) has at least one positive solution for= ,

(iii) has at least two positive solutions for every (0, ).

The proof is divided into several steps, organized as subsections.


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3.1. Non-existence for large .

Theorem 10. For large enough, problem (9) has no solution in the viscositysense.

Proof. Fix > 1 and consider

0= rqr1 (r 1)

(1 q)1q

(r q)rq

1r1

.

In order to reach a contradiction, suppose that there exists > 0 such that theproblem (9) has a solution u. Then, we have

(10) F(u, D2u) = u

q+ u

r> u in ,

in the viscosity sense. In fact, it is enough to demonstrate that

mintR+

(t)> where (t) = tq1 + tr1.

It is easy to check that

d

dt(t) = 0 t =

(1 q)

(r 1)

1rq

,

which, indeed, is a minimum. Since (t) both as t 0 and t , it is aglobal minimum. Then,

(t) =r1rq

(r q) (1 q)q1rq

(r 1)r1rq

>

by our selection of. On the other hand, define = 1, where 1 is a solutionof (7). Hopfs Lemma (Proposition 8) implies that there exists > 0 such that u. Then,

(11) F(, D2) = 1 < u in ,

by definition of.By construction, we have 0 < u, where and u satisfy (10) and (11).

Hence, we can apply the iteration method to getv, satisfying v u, a viscositysolution of

F(v, D2v) = v.

Hence, v is a positive solution of (7), which is a contradiction with the definitionof1.

3.2. Existence of one solution for (9) with 0 < . In [13] it is proved

that there exists a threshold > 0 such that problem (9) has at least one positivesolution for 0 < < and no positive solution for > . Here we extend theresult to the critical value = .

Proposition 11. Let Rn be a smooth bounded domain, and suppose thatF : Rn Sn R satisfies(F1), (F2) and(F3). Let0< q 0.Then there exists a number > 0 such that problem (9) has at least one solutionfor every (0, ].

Proof. Given 0 be the solution to (9) found in [13]. As is pointedout in Remark 1, it is possible to follow the arguments in Proposition 4.10 in [10] inorder to get Krylov-Safonov C estimates. Namely, there exists a positive constantC such that

uC() C, uniformly in ,


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since the right hand side of the equation in (9) is uniformly bounded by Proposition12,

uq+u

r u

q+ u

r C.

Consequently, applying the Arzela-Ascoli Theorem, we can find a sequence {uj}with j as j which converges uniformly to some u. Such uj areviscosity solutions to problem (9) with = j ; hence, we can pass to the limit inthe viscosity sense to find that u is a solution to problem (9) with = .

In order to prove that u > 0, notice that by construction uj > c >0 uniformly in j, so u > 0. Then we get u > 0 from the weak Harnackinequality (see [10])

Thus, statements (i) and (ii) in Theorem 3 are proved. The rest of this sectionis devoted to prove the existence of a second solution in (0, ).

3.3. Existence of a second solution in (0, ).

3.3.1. L estimates: Blow-up argument. We present first the blow-up method in[21] adapted to the viscosity setting. We summarize the arguments in the followingresult.

Proposition 12. LetF : RnSn Rsatisfy(F1)(F5)and letu be a nontrivialviscosity solution of problem (9) with 0 < q < 1 < r < r. Then there exists aconstantC >0 independent of andu such thatuL C.

For the proof, we proceed by contradiction. Since we aim to prove that u(x) Cwith C = C(r, ) independent of u, suppose to the contrary that there exists asequence{uk}k of positive solutions to

(12)

F(uk, D2uk) = u

qk+ u

rk, in ,

uk(x) = 0, on ,

and a sequence of points {zk}k such that

Mk = sup

uk = uk(zk) as k .

Without loss of generality we can assume zk z as k . There are twocases to be considered, either z or z .

CASE 1: z . Let 2d= dist(z, ) in the sequel. Now, consider

y=x zk

k, x= zk+ ky,

where

2

r1

k Mk = 1.

We can then define

(13) vk(y) = 2

r1

k uk(x).

Lemma 13. Fork large enough, the functionvk(y) defined in (13) is a viscositysolution of

(14) F

kyvk(y), D2yvk(y)

=

2(rq)r1

k vqk(y) +v

rk(y) inBd/k(0).

Proof. In order to prove that vk is a viscosity solution of (14), we treat first thesubsolution case.

Consider C2, y0 Bd/k(0) such that vk has a local maximum at y0.Indeed, we can suppose without loss of generality that touches vk from above aty0, that is,

(vk )(y) (vk )(y0) = 0,


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for all y in a neighborhood ofy0. Define

(x) =

2r1

k x zk

k

.

Then, touchesukfrom above atx0= zk+ky0 (it is here wherey0 Bd/k(0)plays a role), namely

uk(x0) =uk(zk+ky0) = 2r1

k vk(y0) =2r1

k (y0) = (x0),

and

uk(x) = uk(zk+ky) = 2r1

k vk(y) 2r1

k (y) = (x),

for all x in a neighborhood ofx0.We can compute the derivatives of (x) in terms of those of(y)

x(x0) = r1r1

k y(y0),

D2x(x0) =

2rr1

k D2y(y0).

Sinceuk is a viscosity subsolution of (12), by homogeneity, we get

F

ky(y0), D2y(y0)

2(rq)r1

k vqk(y0) +v

rk(y0),

which is what we aimed for. The supersolution case is analogous.

We can fixR >0 and suppose without loss of generality (taking k large enough)thatBR(0) Bd/k(0).

Our hypotheses on the uniform ellipticity and structure ofF imply that

P,(D2vk) k|vk|

2(rq)r1

k vqk(y) +v

rk(y), in Bd/k(0),

and

P+,(D2vk) +k|vk|

2(rq)r1

k vqk(y) +v

rk(y), in Bd/k(0).

In particular, sincevkL(BR) = 1 by construction, and k 0 as k , wecan fix >0 and find for k large enough that

P,(D2vk) |vk| 1 + , in Bd/k(0),

and

P+,(D2vk) +|vk| (1 +), in Bd/k(0).

Hence, from the Harnack inequality (which follows from the ABP estimate, availableby (F3), see [10] and [12]), we get uniform C estimates (see [10]),

(15) vkC(BR/2) C(n,R,,, 1 +),

for some 0< R and apply the arguments above to the subse-quencevkj in BR1(0). Then, we get a new subsequence vkj1 such that

limkj1

vkj1 =v, uniformly in BR1(0), andv(0) = 1.

Notice that, since {vkj1 }kj1 {vkj}kj the limits of both subsequences coincide in

BR(0).


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We can consider an increasing sequence of radii {Rj}j and iterate this procedureto get a diagonal subsequence vk such that

(16) limk

vk = v, uniformly in BR(0)R >0, andv(0) = 1.

Finally, we take limits in the viscosity sense in (14), which is the content of thefollowing lemma.

Lemma 14. The limitv(y) in (16) is a viscosity solution of G

D2yv(y)

=vr(y), in Rn,

0 v(y) v(0) = 1, in Rn,

whereG(X) =F(0, X).

Proof. Consider C2 and y0 such that v has a strict local maximum at y0,that is,

(v )(y)< (v )(y0),for all y =y0 in a neighborhood ofy0.

Fix R > 0 such that y0 BR(0). By uniform convergence in compact sets, wededuce that there exist a sequence of points yk y0 as k such that vk has a local maximum at yk, that is,

(vk )(y) (vk )(yk),

for all y = yk near yk. Without loss of generality, we can further suppose yk BR(0) Bd/k(0) for every k > k0.

Then, sincevk is a viscosity solution of (14), we have

(17) F

k(yk), D2(yk)

2(rq)r1

k vqk(yk) +v

rk(yk).

Lettingk in (17) we arrive atF

0, D2(y0)

vr(y0),

and we have proved that v is a viscosity subsolution. The supersolution case issymmetric.

We conclude the argument in this case pointing out that the statement in Lemma14 contradicts our assumption (F5).

CASE 2: z . In this case the reduction argument leads to a problem in eitherRn or a half space Rn+. Without loss of generality, we can assume that z = 0. Inthis way, the tangent space to at the origin is given byx, = 0, for some fixed Rn. Moreover, after a rotation, we can suppose that = (0, ..., 0, 1). Considerk defined as in Case 1, see (13), and define the scaled function

vk(y) = 2

r1

k u(zk+ky

, z(n)k +ky

(n))

wherezk = (zk, z

(n)k ),y = (y

, y(n)), with z k, y Rn1 andz

(n)k , y

(n) R.

Considerdk =z(n)k /k+o(1) ask , which correspond to the distance from

the maximum ofvk to the boundary of k = 1k

( (zk, z(n)k )).

Then, we have to consider the following alternatives depending on the behaviorofdk:

(1) {dk} is unbounded. In this case, passing to the limit in a similar way tothe Case 1, we arrive to the equation G

D2yv(y)

= vr(y) in Rn, with

0 v(y) v(0) = 1, and we reach a contradiction.(2) {dk} is bounded. We can take a subsequence, if necessary, such that dk

s 0.


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MULTIPLICITY OF SOLUTIONS TO UNIFORMLY ELLIPTIC FULLY NONLINEAR.. . 11

In the second case there are two alternatives to be considered. Ifs = 0 we geta contradiction with the continuity of the limit function v, since on the one hand

v(0) = 1, and on the other hand v(y) = 0 for any y Rn verifyingy, = 0, inparticular for y = 0.

Ifs >0, we reach the problem

(18)

G(D2yv(y)) = vr(y), v 0, in Rn+,

0 v(y) v(0,..., 0, s) = 1, in Rn+,v(x, 0) = 0, x Rn1.

Then, by construction, v in (18) has a maximum at (0, . . . , 0, s). This impliesv(0, . . . , 0, s) = 0 and in particular

v

xn(0, . . . , 0, s) = 0,

which contradicts our assumption (F5) and concludes the proof of Proposition 12.

3.3.2. Theoretical degree arguments. OnceL estimates have been proved, we pro-ceed to the proof of the existence of a second solution using degree theoretic argu-ments.

Fix (0, ) and consider 0 < m < < M < . Define vM, wm to beviscosity solutions to(19)

F(vM, D2vM) = M in ,

vM > 0 in ,vM = 0 on .

F(wm , D2wm) = m d(x), in ,

wm > 0, in ,wm = 0, on ,

respectively, whered(x) is the normalized distance to the boundary, that is,

d(x) = dist(x, )

dist(, ) .

It is easy to check (see [13]) that, for t >0 sufficiently small, the function

u= t wm

is a viscosity solution to

(20)

F(u, D2u) m uq + ur, in ,u > 0, in ,u = 0, on .

In addition, there existsT(M)> 0 such that

u= T(M) vM,

is a viscosity solution to

F(u, D2u) Muq + ur, in ,

u > 0, in ,u = 0, on .

We can assume without loss of generality that t < T(M). Indeed, since in theviscosity sense,

F(u, D2u) t m< T(M) M =F(u, D2u),

we can apply Proposition 9 in order to get

u < u in .

We define

X :={v C1() : v= 0 on , v >0 in }


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endowed with the C1 topology, and

K(v) := F1(f(v)),

wheref(v) =vq + vr for simplicity.

The main properties ofK are the following:

(1) K : XX(see [10] and Remark 1).(2) u < K(u) and u > K(u) in .

Moreover ifv K(u), we have

F(u, D2u) m uq +ur < uq + ur =F(v, D2v)

in the viscosity sense. By proposition 9, u < v =K(u) in . The secondinequality, u > K(u), follows in a similar way.

(3) K is compact in C1 (see [10] and Remark 1).

Now, define X as

= {v X :u v u}.Indeed,K: . To see this, letv , that is, v X such thatu v u. Weare going to show that u < K(v)< u. Let w = K(v); then,

F(u, D2u) m uq +ur < vq +vr =F(w, D2w).

Again, Proposition 9 impliesu < w in and henceu < K(v). The other inequalityfollows analogously.

By theC1, estimates in [10] (see Remark 1) and the above computations, we see

thatKis compact. MoreoverK() is a compact set inX. Thus, the Schauderfixed point theorem implies that there existsu such that K(u) = u, i.e., asolution to problem (9) with = .

Ifu is not the unique fixed point in we are done. Otherwise, by Proposition

9, it is easy to show that u < u< u in . Moreover, we have the following result.

Lemma 15. There exists > 0 such that u+ B1(0) , whereB1(0) denotesthe unit ball centered at 0 inX.

Proof. For >0 sufficiently small, we define

:={x | dist(x, )< }.

Previously we have proved that u < u< u; then using theC1, estimates given in

[10] (see also remark 1), there exists a constantC >0 such thatu(x)< Cdist(x, )for any x . By Hopfs Boundary Lemma 8 it is easy to conclude the existenceof a constantc >0 such thatc dist(x, )< u(x) for anyx (for >0 possiblysmaller than before, depends on the geometric properties of the domain).

Let x0 and x0

= (x0) the outward unitary normal to at x0. By theaforementionedC1, estimates, we obtain(21)

t0= tx0 >0 such that u(x0 tx0)< Cdist(x0 tx0 , ), 0< t < t0.

Observe thatx0 is arbitrary, so we can cover by

x

Btx(x), fortxdefined

as in (21). By the compactness of, there exists m N, xj and tj = txj ,forj = 1,...,m, verifying (21) and such that

mj=1 Btj (xj). As a consequence,

there existst >0 such that

(22) u(x tx)< Cdist(x tx, ), 0< t


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necessary. These arguments prove that there exists a positive constant Cverifying

(23) u(x)< Cdist(x, ), x .

Arguing in a similar way, using the Hopf Boundary Lemma 8, instead of the gradientestimates, we obtain the existence of a constant c >0 such that

(24) c dist(x, )< u(x) x .

Using (23), (24) and C1, estimates, one can interpolate the distance times a smallpositive constant in the inequalitiesu < u < u, in the following sense: there exists0< 0 such that w(x0 tx0) 0 suchthatw(x)< 0dist(x, ) for allx B0(x0) . Using thatx0 is arbitrarytogether with compactness arguments as before, we conclude w(x)< dist(x, )for all x for some >0 small enough. By similar arguments we can prove thesecond inequality in (25).

It only remains to prove the claim. To this end, we adapt the proof method ofTheorem 5.3 in [10].

Fix H andH1 such thatH1 HH. Let us denote

u(x) =u(x) and v(x) = u(x)

for simplicity. Consider their sup- and inf-convolutions (see for instance chapter 5in [10]), respectively,

u(x) = supyH

u(y) +

1

|y x|2

, for x H,

and

v(x) = infyH

v(y) +

1

|y x|2

, forx H.

Indeed,u, v are, respectively, viscosity solutions to

(27) F(u, D2u) fmu+ c2 +o() in H1

and

(28) F(v, D2v) f

v c2 + o()

in H1

wherec = max{u, v} (recall that u, v C1,).We show the details in the case ofu. Let C2, and x H such that (u

)(x) (u )(x) = 0 for all x B = {y H : |y x| < dist(x, H)}, withx H such that

u(x) = supyH

u(y) +

1

|y x|2

=u(x) +

1

|x x|2.

Define (y) =(y+ x x) + 1|x x|2 . Then, u has a local maximum

at x, that is, (u )(y) (u )(x) = 0. Notice that

(29) |x x| u


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and consequently x H for small enough. Since x B, we then have y =x x + x H.

Hence, since u satisfiesF(u, D2u) fm(u) in the viscosity sense, we have bythe definition of viscosity solution

F((x), D2(x)) fm

u(x)

and, by the definition of ,

F((x), D2(x)) fm

u(x)

.

Finally, sinceu C1,, its Taylors expansion and (29) yields

u(x) u(x) + u |x x| +o(|x x|) u(x) +c2+o().

Then we get

F((x), D2(x)) fm

u(x) +c2+ o()

.

and (27) is proved.Continuing with the proof of the claim, we aim to prove that for small enough,

(30) P,

D2(u v)

|(u v)| 0 in H1,

in the viscosity sense. Then, since H1 is arbitrary, and u v converges

uniformly to u v (see [10]), we can pass to the limit in the viscosity sense in (30)to get (26).

For > 0 small enough, we can fix 0 > 0 sufficiently small to ensure that forany 0< < 0,

(31) fm

u(x) +c2 +o()

f

v(x) c2 +o()

, x H

wheref(t) =tq + tr. Clearly, 0 depends on u,v, u, v and H.

Then, for < 0 we consider a paraboloid P touching u v from above atx0 H1. More precisely, we consider Pverifying

(u v) P

(x)

(u v) P

(x0) = 0, x Br(x0),

for r >0 to be fixed. We want to prove P,

D2P(x0)

|P(x0)| 0. To thisend, take >0 and define

w(x) =v(x) u(x) + P(x) + |x x0|

2 r2.

The regularity ofu, v, imply

(32) |w(x)|= |w(x) w(x0)| C |x x0| < C r x Br(x0)

for some positive constants and Cdepending on u, v,1 andP. Hence, sinceP

is fixed, we may assume that r is small enough to fulfill(33) D2P r 0 independent of such that, every u > 0 solution to problem (9)satisfies

uC1 C.

Take > C. Clearly, there are no solution u to problem (9) with uC1 =. Bythe homotopy invariance of the Leray-Schauder degree, we get

deg(I K, B1(0), 0) = deg(I K+, B1(0), 0) = 0.


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Now, by the excision property and (35) we deduce

degI K, B1(0) \ {u+B1(0)}, 0= deg

I K, B1(0), 0

deg

I K, u+B1(0), 0

=1.

Hence,K has another fixed point u B1(0) \ {u+B1(0)}.It remains to show that the trivial solution u = 0 has degree 0, i.e., deg

I

K, B1(0), 0

= 0 for any >0 sufficiently small.

To this aim, notice that there exists such that for any >0 the problem

(36)

F(u, D2u) = uq + ur +, u >0, in ,

u = 0, on

does not have any positive solution. Indeed, arguing as in the proof of Theorem3.12 in [13], for a given >0 it is easy to find >0 such that

uq + ur >(1+ )u,

and hence, obviously,

uq +ur + >(1+)u,

for any > 0. Following the arguments in the proof of Theorem 3.12 in [13], weget a positive eigenfunctionassociated to 1+ , a contradiction with the resultson existence of eigenvalues in [6], already discussed in Subsection 2.1.

It follows that the homotopy

H(, u) = u F1

uq +ur +

is admissible and hence

deg

I K, B1(0), 0

= deg

H(0, ), B1(0), 0

= deg

H(1, ), B1(0), 0

= 0

for all >0. Then, again by homotopy, one finds

deg

I K, B1(0), 0

= deg

I K, B1(0), 0

= 0.

As a consequence, there exists > 0 such that u = 0 is the unique nonnegativesolution of (9) in B1(0), which yields (iii) and finish the proof of Theorem 3.

4. A monotonicity property. Proof of Theorem 4

Here we provide the proof of Theorem 4, a monotonicity result in the spirit ofTheorem 3.1 in [27] and Corollary 1.3 in [5]. We recall the statement of the Theoremfor the readers convenience.

Theorem 4. Letv be a nontrivial non-negative bounded (viscosity) solution of

(37)

G(D2v) = f(v), in Rn+v 0, on Rn

+v = 0, in Rn+,

where f is a locally Lipschitz continuous function, f(0) 0 and G : Sn Ris uniformly elliptic with constants 0 < < and 1-homogeneous. Furthermore,suppose that

G(QtXQ) = G(X) for Q= (qij)1i,jn a matrix with

qij =ij if neither i nor j = n and qij =ij otherwise.(38)

Then, v is monotonic in thexn variable:

v

xn>0 inRn+.

Before going into the proof, let us recall some well-known general results in the

form needed below (see for instance [9]).


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Proposition 16 (Strong Maximum Principle). Let be a regular domain and letv be a non-negative viscosity solution to P+,(D

2v) c(x)v in withc(x) L.

Then, eitherv vanishes identically in orv(x)> 0 for allx . Moreover, in thelatter case for anyx0 such thatv(x0) = 0,

limsupt0

v(x0 t) v(x0)

t 0.

We are going to use the moving plane method as in [27]. For each , we define,

as usual,T ={x R

n+: xn = }, ={x R

n+: 0< xn < },

and the functions

v(x) = v(y, 2 xn), w(x) =v(x) v(x), x= (y, xn) Rn1 R+,

defined in .First, we point out that

G

D2v(x)

=f

v(x)

in ,

in the viscosity sense. Let us consider for example the subsolution case. Take C2

and x0 = (y0, x0n) such that v has a local maximum at x0. Define

(x) = (y, 2xn). It is easy to see that v has a local maximum at

(y0, 2 x0n). Then, D 2(y, xn) =QD2(y, 2 xn)Q where Q is a matrix withelements qij = ij if neither i nor j = n and qij = ij otherwise. Finally, bydefinition ofv , we get

f

v(y0, x0)

G

D2(y0, 2x0n)

=G

QD2(y0, x0n)Q

=G

D2(y0, x

0n)

in ,

which is what we aimed for.

Next, we have to show that w =v v satisfies

(39) P+,

D2w(x)

c(x)w(x),

in the viscosity sense, where

c(x) =

f

v(x)

f

v(x)

v(x) v(x) , ifv(x)=v(x)

0, otherwise.


18/25


Notice that, again sincef is Lipschitz, we havec(x) L. The proof follows the

ideas in [17].

To this aim, let C2 such that w has a local minimum at some pointx0 . In other words, x0 is a local maximum of v v +. As usual in thetheory of viscosity solutions, introduce for every >0

(x, y) =v(x) v(y) +(x) |x y|2

2 |x x0|

4.

For small enough, attains a maximum in at some point (x, y) Br(x0) Br(x0) for some r >0. Since x0 is a local strict maximum of

xv(x) v(x) +(x) |x x0|4

standard results of the theory of viscosity solutions (see [15]) yieldsx, y x0and|xy|

2

2 0 as 0.

In addition, defining (x, y) = (x) +

|xy|2

2 +|x x0|

4

, the results in [15]imply that for any given >0, there exist matrices X, Y Sn such that

(x(x, y), X) J2,+

v(x)

(y(x, y), Y) J2,

v(y),(40)

and

1

+ A

I

X 0

0 Y

A+A2,

whereA = D2(x, y). From there, setting = 2, it is standard to see that

X Y D2(x) +O(2 + |x x0|

2).

By definition (see [15]) of viscosity solutions and (40), we get

G(X) f

v(x)

and G(Y) f

v(y)

,

and subtracting in the previous inequalities we obtain

f

v(y)

f

v(x)

G(Y) G(X)

G

X+D2(x) + O(2 + |x x0|

2)

G(X)

P+,(D2(x)) + O(

2 + |x x0|2).

Letting 0, we get (39).

Then,w 0 in if is small enough, sincew 0 on and hence we canapply the maximum principle in narrow domains (Proposition 17).

We define,

= sup{: w

0 in

for all < }> 0.

Using Hopfs Lemma, we conclude that w >0 in and

v

xn=

1

2

wxn

>0 on T

for every 0< . If we prove that =, we have finished.Suppose to the contrary that < . We can fix 0 small such the maximum

principle holds forG() c(x) in +0\ 0 .

Claim:There exists0 (0, 0] such that for each (0, 0) we havew+ 0in 0\ 0 .

Once the claim is proven, we can apply the maximum principle in narrow domainsto

P+,

D2w(x)

c(x)w(x) in +\ 0 0 ,


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withc as before, and conclude thatw+ 0 in + , contradicting the maxi-mality of.

Hence, it remains to prove the claim. It follows in a similar way to Lemma 3.1in [27]. We include the details for the readers convenience.

Suppose that the claim were false, that is, that there exist sequences m 0

and x(m) = (y(m), x(m)n ) 0\ 0 such that

(41) w+m(x(m))< 0.

We can suppose that x(m)n x0n [0,

0] asm .We define the functions

v(m)(y, xn) = v(y+ y(m), xn)

and respectively

w(m) (y, xn) =v

(m)(y, 2 xn) v(m)(y, xn).

Notice thatG

D2v(m)

=f

v(m)(x)

.

in the viscosity sense. Then, it is standard to show (see for instance Proposition4.11 in [10]) that there exists a subsequence and a limit v C such that v(m) vuniformly in compact sets as m and

G

D2v

=f

v(x)

in the viscosity sense.By the strong maximum principle (Proposition 16), we have that either v is

strictly positive in Rn+ or v 0 in Rn+.

Suppose first that v >0 in Rn+. By what we have already shown, we know that

w

(m)

(y, xn) =w(y+y

(m)

, xn)> 0 in for all

. Hence the limit functionw = limm w

(m) is non-negative in for all

.

So we can repeat the moving plane argument for v and get , where

is to v what is to v . Since w satisfies

P+,

D2w(x)

c(x)w(x)

we can apply the strong maximum principle and get, as before, that w >0 in for all . On the other hand, by continuity and (41), we have w

(0, x0n) = 0and x0n (0,

0], a contradiction.

Suppose next that v 0 in Rn+. We fix the rectangular domains

Q1=

x Rn+ : 1< x1< 1, . . . , 1< xn1< 1, 0< xn < 2 + 1

,

Q2=

x Rn+ : 2< x1< 2, . . . , 2< xn1< 2,02 < xn < 2 + 2

.

Sincev (m) converges uniformly to zero in Q2, we can suppose that v(m) 1 in Q2

form sufficiently small. We set

m= v(m)(0, x(m)n ) and v

(m) =v(m)

m.

Now, the function v (m) satisfies

(42) G

D2v(m)(x)

= f

v(m)(x)

v(m)(x) v(m)(x), x Q2.

The Harnack inequality (see Chapter 4 in [10]) implies

supQ1

v(m) C1

infQ1

v(m) C1

.


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Next, we recall that w

0 in , which implies

v(m)

(y, xn) v(m)

(y, 2

xn) C1, for (y, xn)

.

Thus, v(m)L(Q) C1, where

Q=

x Rn+: 1< x1< 1, . . . , 1< xn1< 1, 0< xn < 2 + 1

,

hence, our C estimates yields (up to a subsequence) that v (m) v Cuniformlyin compact sets, and v is a viscosity solution of

G

D2v

lv,

wherel = limt0

f(t)/t. By the strong maximum principle, eitherv 0 in Q or v >0

in Q. The first possibility is excluded since v (0, x0n) = 1.

We introduce the functions

z(y, xn) = v(y, 2 xn) v(y, xn)

defined in Q for all + 1/2. We have, by continuity,

z

0 and z

(0, x0n) = 0.

Since

P+,

D2z(x)

lz(x),

the strong maximum principle, implies z

0 in Q. This contradicts thefact thatv = 0 on {xn = 0} and v >0 on{xn = 2

}.

5. Applications of Theorem 3

In Section 3 we have developed an abstract framework for the global multiplicityresult, Theorem 3. To make it more explicit, we have proved Theorem 3 underhypotheses (F1) (F5).

The present section is devoted to examples in which it is possible to prove Propo-sition 12, or in other words, examples for which hypotheses (F4), (F5) hold. Themonotonicity property in Section 4 holds in all the subsequent examples.

We treat first (subsection 5.1) the important example when Gis a Pucci extremaloperator, which include the Laplacian as a particular case.

In subsections 5.2, and 5.3 we treat concave (convex) operators and, respectively,a class of Isaacs operators, which are neither concave nor convex. We will alwaysassume that the operators considered satisfy hypothesis (F4), which, in fact, isbuilt-in in the case of Pucci extremal operators.

It is worth comparing here these examples with the results in [2] where problem(1) for the operator is studied using variational methods.

The results in [2] are optimal, since the maximal range of exponents,1 < r < 2 1, is known thanks to the results in [21]. However, the arguments(in particular those leading to the L estimates) are strongly dependent on thestructure of the Laplacian.

The examples below lack important features of the Laplacian, such as the varia-tional structure or classical regularity of the solutions (which is not known in somecases), which make necessary the use of the viscosity framework in Sections 3 and

4.


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5.1. Extremal Pucci operators. In the precise context ofF involving a Pucciextremal operator, that is

F(, X) =P,(X) +H(),

withH : Rn R homogeneous and Lipschitz continuous and such that H(0) = 0,it is clear thatF satisfies (F1)-(F4).

Concerning (F5), the Liouville-type result in Rn by Cutri-Leoni (Theorem 5)applies. On the other hand, Quaas and Sirakov, following the ideas in [5], provedTheorem 4 in this case, with G() =P,().

Observe thatP, admitsC2, estimates, in the sense that if the function u is a

viscosity solution to the equation P,(D2u) =g (x) in a ball B2R and g C

for

some (0, 1), thenu C2, and moreover,

uC2,(BR) CuL(B2R)+ gC(B2R)

for some constantC >0. With the aim of these C2,-estimates the proof of Theorem4 could be simplified.

The abstract existence result (Theorem 3) holds and reads as follows.

Theorem 18. Consider Rn, n 3, a smooth bounded domain, and set

r=(n 1) +

(n 1) .

LetH : Rn R be homogeneous and Lipschitz continuous. Then, for0< q 0, in ,u = 0, on

(i) has no positive solution for >,(ii) has at least one viscosity positive solution for= ,

(iii) has at least two viscosity positive solutions for every (0, ).

As an important consequence of Theorem 4, Quaas and Sirakov ([27]) deduce aLiouville-type result in Rn+ for Pucci operators, adapting arguments in [5] to thenonlinear setting.

Theorem 19 (Theorem 1.5 in [27]). Supposen 3 and set

r=(n 2) +

(n 2) .

Then the problem

(44)

P,

D2v

= vr, in Rn+

v = 0, on Rn+

does not have a nontrivial non-negative bounded solution, provided 1 < r r (or1< r


22/25


maximal range for the Laplacian (the Pucci operator with = = 1) is known tobe

1< p max{r, 2

1}. As far as we know, to establish the Liouville result in therange r < r < r+ is an open problem.

5.2. Concave and convex operators. Let F : Rn Sn R satisfy (F1)-(F4),and suppose that the blow-up operator G(X) = F(0, X) is concave (or convex).

The hypothesis on G being concave (or convex), yields the following regularityresult by Evans and Krylov. For the proof we refer to [10], Section 8.1.

Theorem 20. LetG: Sn R be a uniformly elliptic (with constants0<


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5.3. A class of Isaacs operators. Finally we intend to apply the above analysisto a class of operators which are neither convex nor concave and for which classical

regularity is not known in general.Consider the class of Isaacs operators

(46) F(, X) = suplL

infkK

Lk,l(, X)

,

whereK andLare arbitrary sets of indexes and Lk,l are of the form

Lk,l(, X) = trace(Ak,lX) +Hk,l()

for Hk,l : Rn R 1-homogeneous and Lipschitz continuous (all with the sameconstant), such that Hk,l(0) = 0 and Ak,l is a family of matrices with the sameellipticity constants.

In addition, we assume that the symmetry hypothesis (F4) holds for the operatorG(X) = F(0, X). More precisely, we require that for every Q On, whenever

Ak,l Sn gives rise to an operator Lk,l, with (k, l) K L, the matrix QAk.lQtgives rise to Lk,l for some other pair (k, l) K L.

Classical regularity is not known in general for problems of the form (46). How-ever in [11] classical regularity is proved for Isaacs operators of the particular form

(47) F(, X) = min

F(, X), F(, X)

, Rn and XSn

with F : Rn Sn R and F : Rn Sn R uniformly elliptic operatorsrespectively convex and concave in the matrix argument. ClearlyFin (47) is neitherconcave nor convex. As before, a model operator satisfying all the hypotheses abovecould be

F(u, D2u) = min

infkK

Lku, suplL

Llu

whereLku= trace(AkD

2u) +Hk(u).

Notice that, since classical regularity is not known in general, the viscosity settingbecomes crucial in this case. Arguing as in Sections 3 and 4 we have the followingresult.

Theorem 22. Consider Rn,n 3, a smooth domain. LetF : RnSn Rbean operator of the form (46) satisfying (F1), with ellipticity constants0< ,(F2), (F3) and(F4). Finally, set

r=(n 1) +

(n 1) .

Then, for0< q


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Departamento de Matematicas, Univ. Autonoma de Madrid, 28049 Madrid, Spain.

E-mail address: [email protected], [email protected]

Departamento de Matematicas, Univ. Carlos III de Madrid, 28911 Madrid, Spain.

E-mail address: [email protected]

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