ON THE CURVATURE OF THE LEVEL SETS OFSOLUTIONS OF SOME NONLINEAR ELLIPTIC

EQUATIONS

FRANCESCA GLADIALI AND MASSIMO GROSSI

Abstract. In this paper we study the convexity of the level setsof solutions of the problem

(0.1)

u = f(u) in u > 0 in u = 0 on ,

where f is a superlinear function with subcritical or critical growth.Under some assumptions on the shape of we prove that the levelsets of the solution of (0.1) are strictly convex.

1. Introduction

Let us consider the problem

(1.1)

u = f(u) in u > 0 in

u = 0 on

extensive literature. In this paper we are interested in convexityresults.The convexity of the level sets of u has been investigate by many

author since 1950. It is well known that the result is true in the linearcase f(t) = 1t where 1 is the first eigenvalue of the laplacian in and u = 1 is the first eigenfunction. In [1], Brascamp and Lieb haveshown that 1 is log concave if is convex, and so it has convex levelsets. The proof of this result was rather involved and uses the fact thatthe parabolic operator

t under homogenous Dirichlet boundary

condition preserves the log concavity of the positive initial data. Lionsin [16] extended this method to other nonlinearities and showed thatif f is of the type f(t) = t tp with p > 1, , > 0, and > 1,then u is log concave. The proof needs that each solution of (1.1) canbe obtained as the limit as t of the solution of the parabolicassociated operator with some log concave initial condition.

1991 Mathematics Subject Classification. 35B05,35B50.Supported by MIUR, national project Variational Methods and Nonlinear Dif-

ferential Equations.1

2 GLADIALI AND GROSSI

To study the convexity of a solution of some elliptic and parabolicequations Korevaar in [13] introduced a concavity function

C(u, x, y, ) = u(x+ (1 )y) u(x) (1 )u(y) on

where 0 1. The function C measures how much the functionu fails to be concave. Of course u is concave if and only if C 0 on. Korevaar used a maximum principle and some boundary pointlemmas to show that if u is a solution to the generalized eigenfunctionequation then v = log u is concave. Subsequently Caffarelli and Spruckin [2] proved the log concavity of solutions of an eigenvalue problemwith weight, as well as further application, simplifying the Braschampand Lieb result, and those of Korevaar also.Kawohl in [11], [12], exploited the concavity function and showed thatif u is a solution of (1.1) with f 0, f strictly decreasing and fharmonic-concave (i.e. 1

fis convex) then the solution u is concave. In

[12] he used this result to show that if u is a solution of (1.1), with

0 < p < 1, then the function v = u1p2 is concave if is convex and

sufficiently smooth. In particular the level sets of u are convex. But ifthe exponent p is greater than 1 this method cannot be applied.Other important results are due to Gabriel who first shown the con-vexity of the level sets for the Greens function of in . In [4] and[5] he showed that the level sets are strictly convex if N = 3, and theresult extend to all dimensions. His method was generalized by Lewisin [14] where he applied it to the p-capacitary functions and by manyauthors (see for example[2], [3], [14]), to show the convexity of levelsets in ring shaped domains. For other results in this topic see [11]and references therein. But to our knowledge none of this works whenp > 1.In this work we give two example of solutions which have convex levelsets for some p > 1.

The first example deals with the following perturbed critical problem

(1.2)

u = N(N 2)up in u > 0 in

u = 0 on

where RN is a bounded smooth domain, N 3, and p =N+2N2 for > 0. In [7] Grossi and Molle have shown that the levelsets of a solution u are strictly starshaped. Here we extend their resultsto the strictly convexity. Our main result is the following

CONVEXITY 3

THEOREM 1.1. Let u be a solution of (1.2) such that

(1.3) lim0

|u|2(

u2) 22

= S

where S is the best constant in the Sobolev embedding and 2 = 2NN2 .

If has strictly positive Gauss curvature at any point p , thenthere exists such that, for every 0 the level sets of u havestrictly positive Gauss curvature at any point which is not the maximumone. In particular the level sets are strictly convex;

The second example is the following subcritical perturbed problem

(1.4)

2u+ u = up in u > 0 in

u = 0 on

where RN is a convex domain, N 2, > 0 and 1 < p < N+2N2

ifN > 3. Again we find that the level sets are convex

THEOREM 1.2. Let u a family of single peak solution of (1.4). If is convex then there exists > 0 such that for each 0 < < , uhas convex level sets.

The proof requires the understanding of problem (1.2) and (1.4)as 0. In the case of problem (1.4) only the behavior near themaximum point is needed to handle the convexity result. In case ofproblem (1.2) otherwise the exact behavior of the solution inside isinvestigate and the proof considers the deep work of [15] and [9] instudying this problem.We think that this technic can be applied to other perturbed prob-

lem when the asymptotic behavior is known.

The paper is organized as follows. In section 2 we state some knownresult and give some notation. The proof of Theorem 1.1 is given insection 3. In section 4 we discuss the subcritical case. Finally section5 is devoted to some technically estimates.

2. Some preliminaries

In this section we state some known facts about problem (1.2) thatare necessary in the proof of Theorem 1.1. We also recall the notion ofGauss curvature and normal curvature and their relationship with theconvexity of a surface.

4 GLADIALI AND GROSSI

Let u be a solution of problem (1.2) that satisfies (1.3). Let x a point such that u(x) = u. The asymptotic behavior of thesolution u was studied by Han in [9] and by Rey in [18]. They provedthe following convergence result

THEOREM 2.1 (A). Let be a smooth bounded domain in RN ,N 3. Denoting by x0 = lim0 x, we have x0 and

uu(x)[N(N 2)

S

]N4 (N 2)G(x, x0)

2|g|x \{x0}, where G(x, x0) is the Greens function for the laplacian

in with zero Dirichlet boundary condition and |g| = g(x0, x0) whereg(x, y) is the regular part of the Greens function.

Grossi and Molle in [7] have shown the following

THEOREM 2.2. If is convex then there exists such that, forevery 0 < < it occurs:

(2.1) (x x) u(x) < 0 x \ {x}.In particular, the maximum point x is the only critical point and

the superlevels are strictly starshaped.

Let us introduce some notations. Let Sf (t) the level set of a functionf , so

Sf (t) = {x : f(x) = t}.We suppose f at least C2 and |f | 6= 0.Then the level set is the surface Sf (t) = f

1(t), and the normalvector field to Sf (t) in the point x, is given by nx =

f(x)|f(x)| . We indicate

by Tx the tangent space to Sf (t) attached at the point x Sf (t). Then Tx if and only if nx = 0. If Txthe value of the Second fundamental form of Sf (t) at x on is given

by

Sx() = 1|f(x)|N

i,j=1

2f

xixj(x)ij = 1|f(x)|

tHf (x)

where Hf (x) denotes the Hessian matrix of f at the point x. If Txis a unit vector = 1, the normal curvature of Sf (t) at x in the

direction is

CONVEXITY 5

kf,x() = Sx() = 1|f(x)|tHf (x).

The Gauss-Kronecker curvature of Sf (t) at x, denoted by Kf (x)is the product of the principal curvatures of Sf (t) at x, which are sta-

tionary values of the normal curvature on the tangent space Tx. Whilethe normal curvature depends on the tangent vector the Gauss cur-vature is intrinsic.

We also mention the following well known facts

THEOREM 2.3. Let S be a compact connected oriented (N 1)-surface in RN whose Gauss-Kronecker curvature is nowhere zero. Then

i) The Gauss map N : S SN1 is one to one and onto,ii) S is strictly convex.

of the surface. Infact the convexity of a compact surface S impliesthat in each point of S the Second fundamental form is semi definite,but there may be points at which the Gauss curvature is zero.

If f is a radial function, then its level sets are spheres and kf,x() =1|x| has the same value for all tangent direction at any point of thesphere x 6= 0. Then the Gauss curvature

Kf (x) =1

|x|N1 x 6= 0for all radial function.For references about curvatures and Second fundamental form look

at [19].

Now we state some known result about the convexity of level sets inconvex ring, see [2]

THEOREM 2.4. Let u be the unique solution ofu = f(u) in \ u = 0 on

u = 1 on

with f(u) continuous and nonincreasing in u, f(0) = 0, and

bounded convex sets in RN . Then the level surface of u are convex C1+hypersurfaces.

Here we quote some results on the convexity of the level set for theGreens function. In [14] Lewis generalizes the work of Gabriel in [4],[5] about the convexity of the level sets of G, the result is

THEOREM 2.5. Given a convex ring \ , let u be an harmonicfunction such that u = 0 on and u = 1 on . Then

6 GLADIALI AND GROSSI

i) the set {x : u(x) > t} is convex for 0 t < 1,ii) if u 6= 0 and x \ , then all the normal curvature at x of

the level surface {y : u(y) = u(x)} are positive.The Theorem 2.5 proves the convexity of the level sets SG(t) of the

Greens function in an annular convex domain, and shows that all thenormal curvature of SG(t) are positive. This imply also that the Gausscurvature of SG(t) is positive.

COROLLARY 2.1. Let be a convex bounded domain in RN , N 3and let x0 . Then the level sets of Greens function G(x, x0) for theLaplacian in with Dirichlet boundary condition have strictly positiveGauss curvature at any point.

Proof. The result is proved by Gabriel in [4] for N = 3, and generalizeto all dimension. The idea is that in a neighborhood of x0 the Greensfunction behave essentially as 1

n(N2)|xx0|N2 , where N is the area ofthe unit sphere in RN , and so the normal curvature of his level sets isessentially the normal curvature of a sphere which is strictly positive,and increasing as x x0. Hence in a small neighborhood of x0 theGauss curvature of SG(t) is strictly positive and the level sets of G arestrictly convex. Now we can apply the Theorem of Lewis at G in \where is a strictly convex level set near x0 to have the positivenessof the Gauss curvature of SG(t) in all \ {x0}.

3. The main result

Proof of Theorem 1.1. We argue by contradiction. Let us suppose thatthere exists a sequence n > 0, n 0 and points zn \ {xn} suchthat, if un un is a solution of problem (1.2) corresponding to thevalue n, that satisfies (1.3) then

(3.1) Kun(zn) 0 nwhere Kun(zn) is the Gauss curvature of the surface Sun(zn) = {x

s.t. un(x) = un(zn)}. We use un un and xn xn denotes theunique critical point of un. Notice that from (2.1) the normal nzn at

the surface Sun(zn) is always defined by nzn =un(zn)un(zn) . Up to a sub-

sequence zn z .

Step 1Firstly we suppose z 6= x0, so z \B(x0) for some > 0.In this domain we have from Theorem 3.4 that gn(x)

[N(N2)

S

]N4 (N2)G(x,x0)

2|g|

in C2( \B(x0)

), where gn = unun.

CONVEXITY 7

From (3.1) we have Kgn(zn) = Kun(zn) 0 n and passing to thelimit

limn

Kgn(zn) = KG(z) 0.If z is an interior point this is impossible from Theorem 2.5. If

otherwise z , this is impossible since we suppose the Gauss cur-vature of is strictly positive.

Step 2Now zn x0. We first consider the case where zn Bn = B(xn, R

unpn12

),

for some R > 0.

Consider the function un(x) =1

unun

(xn +

x

unpn12

). By stan-

dard blow-up results, see [10] for example, un converges to U(x) =(1

1+|x|2)N2

2in C2(B(0, R)), for all R R. If zn Bn then zn =

unpn12 (zn xn) B(0, R) and up to a subsequence zn z

B(0, R). Suppose z 6= 0.From (3.1) we have

Kun(zn) =1

unpn12

(N1)

Kun(zn) 0 n

and passing to the limit KU(z) 0. But this is not possible sinceKU(z) =

1|z|N1 > 0.

If z = 0, |U(0)| = 0 and we cannot speak of Gauss curvatureof U in 0 since SU(1) = {0}. But zn 6= 0 n and Kun(zn) 0implies there exists at least a tangent direction n Tzn such thatkun,zn(n) = 1|un(zn)| tnHun(zn)n 0. Now n with = 1,and tnHun(zn)n tHU(0) 0.But 0 is a non degenerate maximum point for the function U(x) and

then tHU(0) > 0 and so a contradiction follows.

Step 3 Finally we consider the case where zn x0 but zn / Bn. Herewe use some ideas by [15]. Let rn = |zn xn|; we have 1 rn

R

unpn12

, and rn 0 as n. Let vn(x) = r2

pn1n un(xn+ rnx). Then

vn solves

(3.2)

vn = N(N 2)vpnn in nvn 0 in nvn = 0 on n

8 GLADIALI AND GROSSI

where n =xnrn

RN , and vn = vn(0) = r2

pn1n un(xn)

R2

pn1un un(xn) > R

2pn1 . Since R can be chosen large as you want then

vn as n.We extend the function vn to zero outside of n. Integrating equation

(3.2) we get, for any C0 (RN)

(3.3)RN

(vnvn) =RNvnvn() = N(N2)

RNvnvpnn .

In the same manner as in [9] (see the appendix for details) we havethe following estimate

(3.4) 0 vn(x) vn 2N2vn

2N2 + |x|2

N22 in nusing (3.4) we get that

vnvn vn vn 2N2vn

2N2 + |x|2

N22 = 1vn

2N2 + |x|2

N22

1|x|N2

Hence vnvn is uniformly bounded in L1+loc (RN) for 1 < NN2 .Then, up to a subsequence vnvn weakly tends to V L1+loc (RN).Letting n =

pn12

and using (3.4) we get

RNvnvpnn (x)(x)dx = vn

RN

vpnn

(y

vnn

)(

yvnn

)vnNn

dy

= vn1Nn+pnRN

1

vnpn vpnn

(y

vn

)

(y

vnn

)dy =

RN

1

vnpn vpnn

(y

vn

)

(y

vnn

)dy

(0)RN

[(1

1 + |y|2)N2

2

]N+2N2

= (0)nN.

Using (3.4) we can pass to the limit in (3.3) for each C0 (RN),and we have that vnvn V that solves

CONVEXITY 9

V = N(N 2)(0) in RNin the sense of distribution. The solution is

V (x) =1

|x|N2 + wwhere w solves w = 0 in distributional sense in RN . By regularity

result of solutions for harmonic function, see [10] Corollary 4.1.2, w isa regular harmonic function in RN . From (3.4) and corollary 5.1 weget w(x) |x|N2 and so w(x) 0 as |x| . Then by LiouvilleTheorem w = 0 and V = 1|x|N2 . From (3.4) we have also that vnvnis uniformly bounded in B1(0) \B(0) and so the convergence is inC2(B1(0) \B(0)) for each > 0.Let zn =

znxnrn

. Then zn z, where |z| = 1. From (3.1) we have

Kvn(zn) = rN1n Kun(zn) 0 n

and passing to the limit KV (z) 0and this is not possible sinceKV (z) =

1|z|N1 > 0.

4. The subcritical case

We start this section by recalling the definition of single-peak solutionto the problem (1.4).Denote by W the unique solution of:

W +W = W p in RNW > 0 in RN

lim|y|W (y) = 0, W (0) = W.and define the energy of the solution W as

E =1

2

RN|W |2 + 1

2

RN

W 2 1p+ 1

RN

W p+1.

Let us recall thatW is a radial function and strictly radially decreas-ing

DEFINITION 4.1. Let u C2()C1(RN) a family of solutions of(P), for small , and let x RN be such that u = u(x). Thenu is said a family of single-peak solutions near a point x0 RN if

i) x x0 as 0,ii) N

[12

RN (

2|u|2 + u2) 1p+1RN u

p+1

] E as 0.

The following lemma is well known (see [17] for example)

10 GLADIALI AND GROSSI

LEMMA 4.1. If u is a family of single-peak solutions of (P), then,for sufficiently small, x is the only local maximum point and

(4.1) u(x) 0, as 0, for any x RN \ x0;There is a rich literature on the existence of single peak solutions to

the problem (1.4) (see for example [17], [8] and the references therein).We are in position to give the proof if Theorem 1.2.

Proof of Theorem 1.2. Let v(y) = u(x+y), where x are maximumpoints of u. Then v solves

v + v = vp in v > 0 in v = 0 on

where = (x)/ and RN as 0. It is easy to see thatv W in C2loc(RN), and W satisfiesFurthermore 0 is a nondegenerate maximum point for W .We want to show that in each ball B = B(x, R) for R R the

function u has strictly convex level sets. This follows from the con-vergence of v W as in Theorem 1.1. Let suppose by contradictionthere exists a sequence n 0, and points zn Bn Bn , zn 6= xn,with Gauss curvature less or equal than zero. Then the same is true forthe function vn, so that Kvn(zn) =

N1n Kun(zn) 0 where zn = znxnn

and zn B(0, R). Up to a subsequence zn z B(0, R), and ifz 6= 0 then limnKvn(zn) = KW (z) 0. But this is not possiblesince W is radial and its Gauss curvature is strictly positive if z 6= 0.Now consider the case z = 0; Kvn(zn) 0 implies there exists at lesta unit vector n Tzn and kvn,zn(n) = tnHvn(zn)n 0. Up to asubsequence n , = 1 and tnHvn(zn)n tHW (0) 0and this is not possible since 0 is a non degenerate maximum point forW , and his Hessian matrix is negative defined in 0.

Reasoning as in [17], > 0, R we can find a ball B suchthat u(x) in \ B. This is possible since u(x) 0 in C1loc(),and v W in C2loc(RN). Now consider the level set Su(); it is convexsince Su() B. The function u satisfies

u = upu2

in \ Su()u = 0 on

u = on Su()

and \Su() is a convex ring. Taking 0 in nu = 0 on n

There exists > 0 independent of n such that

un 1.Proof. It is the same of Corollary 1 of [9]. LEMMA 5.1. Let be a domain in RN and u H10 () be a positivesmooth solution of

u = a(x)uq1where a(x) L(), 2 < q0 q 2NN2 . Then there exist 0 > 0

and r0 > 0 depending on N, a, q0 and 1 N+2N2 such that forany Q RN with

B(Q,2r) uq 0, r r0, we have

uL(+1)(

p+12 )(B(Q,r))

C

r2

+1

uL+1(B(Q,2r))with C depending on N only.

LEMMA 5.2. Let u H10 () be a positive smooth solution of

u = a(x)uwith a(x) L() for some > N

2. Then for any Q RN

supB(Q,r)

u C(

1

rN

B(Q,2r)

u2NN2

)N2N+2

where C depends on aL(), ,N.

Proof of the estimate (3.4). With the notation of Step 3 of Theorem

1.1 we have vn = r2

pn1n un(xn+ rnx), and rn >

R

upn12

. From Corollary

5.1 follows rnn Rn

unpn12

= Rnunpn12 , for some > 0, so

that limn rnn = for some 0 < 1.

We make some calculation

12 GLADIALI AND GROSSI

n

vpn+1n =

n

r2(pn+1)pn1

n (un(xn + rnx))pn+1 = r

2(pn+1)pn1 N

n

(un(y))pn+1 =

rn(N22 )

2

n

(un(y))pn+1 = (

N22 )

2

(1 + o(1))

(un(y))pn+1 ;

n

|vn|2 =n

r2( 2pn1+1)n |un(xn+rnx)|2 = r

2(pn+1)pn1 N

n

|un(y)|2 =

rn(N22 )

2

n

|un(y)|2 = (N22 )

2

(1 + o(1))

|un(y)|2.

These estimates lead to

n|vn|2(

nvpn+1n

) 2pn+1

=(

N22 )

2

(1 + o(1))|un|2

(N22 )

2 2pn+1 (1 + o(1))

(upn+1n

) 2pn+1

= (N22 )

2

(1+o(1))S.

Since 1 we necessarily have = 1. Using this and equation (3.2)we have

n

|vn|2 = (1 + o(1))S(

n

vpn+1n

) 2pn+1

,

n

|vn|2 = N(N 2)n

vpn+1n

and so

limn

n

vpn+1n [

S

N(N 2)]N

2

.

Now we apply the usual blowing-up technique to the vn. Let vn(x) =1

vnvn(x

vnpn12

) then vn = N(N 2)vpnn in n = vnpn12 n.

Notice that vn(0) = 1, 0 vn(x) 1 for x n and n RN . Thusby standard elliptic theory vn U uniformly on every compact set, andU solves U = N(N 2)U N+2N2 in RN , U(0) = 1 and 0 U(x) 1in RN , so that U(x) =

(1

1+|x|2)N2

2.

CONVEXITY 13

uniformly over B1 and soB1vpn+1n

B1

(1

1+|x|2)N

= C > 0, and so

C B1

vpn+1n =

|x|1

1

vnpn+1

(vn

(x

vnpn12

))pn+1= vnpn1+

pn12

N

|y|vn

pn12

vpn+1n (y) vnnN22

n

vpn+1n

and then vnn (

CRn

vpn+1n

) 2N2

(

C

( CSN(N2))N2

) 2N2

. So it

exists > 0 such that vnn 1.To establish the estimate (3.4) is equivalent to have

(5.1) vn(x) CU(x) n.Let wn be the Kelvin transform of vn. Then wn(x) = |x|N+2vn

(x|x|2)

and wn solvewn = N(N 2)|x|(N2)nwpnn in n, wn 0, where n is the

image of n under the Kelvin transform. Notice that n is RN except

for a small region near the origin. To have (5.1) we show that

wn C in n.Since 0 vn 1 then wn |x|2N in n and so we only need to

limit wn near the origin.

Let a(x) = N(N 2)|x|(N2)n . To use lemma 5.1 we need thata(x) L(n). In fact a(x)L(n) is bounded independent of n:

a(x)L(n) = N(N 2) supxn

|x|(N2)n = N(N 2) supyn

|y|(N2)n

= N(N 2) supzn

|z|(N2)nvnpn12

(N2)n

= N(N 2)vnpn12

(N2)n supz

( |z xn|rn

)(N2)n 1

pn12

(N2)N(N 2)diam()(N2)n

(1 + o(1)) C

and C can be chosen independent of n. For 0 and r0 as in Lemma5.1, (consider wn to be zero outside

n), we have

B(0,2r)

wpn+1n 0,since vn Lpn+1(n) the same is true for wn in n, and so applyingLemma 5.1

B(0,r)

w(pn+1)

2

2n C

14 GLADIALI AND GROSSI

where C is again independent of n.

Now let b(x) = N(N 2)|x|(n2)nwpn1n ; b(x) L(n) with =(pn+1)2

21

pn1 >N2, in fact

nb(x) = (N(N 2))

(n\B(0,r)

(|x|(n2)nwpn1n )+

nB(0,r)

(|x|(n2)nwpn1n )) .We have already shown that |x|(n2)n C in n, and so

n\B(0,r)

(|x|(n2)nwpn1n ) C n\B(0,r)

w(pn+1)

2

2n

n\B(0,r)

|x| 2N2

N2 C.

Also

nB(0,r)

(|x|(n2)nwpn1n ) C nB(0,r)

w(pn+1)

2

2n C.

Hence we apply Lemma 5.2 at the equation wn = b(x)wn in n wefind wnL(B(0, r

2)) C.

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CONVEXITY 15

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Dipartimento di Matematica, Universita` di Roma La Sapienza -P.le A. Moro 2 - 00185 Roma - Italy.

E-mail address: gladiali@mat.uniroma1.it

Dipartimento di Matematica, Universita` di Roma La Sapienza -P.le A. Moro 2 - 00185 Roma - Italy.

E-mail address: grossi@mat.uniroma1.it

1. Introduction2. Some preliminaries 3. The main result 4. The subcritical case 5. Appendix References