STRICT CONVEXITY OF 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 suitable function with subcritical or critical growth.Under some assumptions on the Gauss curvature of we provethat the level sets of the solution of (0.1) are strictly convex.

1. Introduction

In this paper we study the shape of the level sets of the problem

(1.1)

u = f(u) in u > 0 in

u = 0 on

where is a (strictly) convex domain of RN , N 2 and f is a smoothnonlinearity.In particular we are interested to establish the convexity of the level

set.This problem has been investigate by many author since 1950. A

first important results concerns the linear case f(t) = 1t where 1 isthe first eigenvalue of the laplacian in and u = 1 is the first eigen-function. In [1], Brascamp and Lieb showed that 1 is log concaveif is convex, and so it has convex level sets. The proof of this re-sult was rather involved and uses the fact that the parabolic operatort under homogeneous Dirichlet boundary condition preserves the

log concavity of the positive initial data. Lions in [15] extended thismethod to other nonlinearities and he showed that if f is of the typef(t) = t tp with p > 1, , > 0, and > 1, then u is log con-cave. The proof needs that each solution of (1.1) can be obtained asthe limit as t of the solution of the parabolic associated operator

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

ferential Equations.1

2 GLADIALI AND GROSSI

with some log concave initial condition.

To study the convexity of a solution of some elliptic and parabolicequations Korevaar in [11] 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 Brascampand Lieb result, and those of Korevaar also.Kawohl in [9], [10], exploited the concavity function and showed

that if u is a solution of (1.1) with f 0, f strictly decreasing andf harmonic-concave (i.e. 1

fis convex) then the solution u is concave.

In [10] he used this result to show that if u is a solution of (1.1), with

f(t) = tp, 0 < p < 1, then the function v = u1p2 is concave if is

convex and sufficiently smooth. In particular the level sets of u areconvex. But if the exponent p is greater than 1 this method cannot beapplied.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 [12] where he applied it to the p-capacitary functions and by manyauthors (see for example[2], [3], [12]), to show the convexity of levelsets in ring shaped domains. For other results in this topic see [9] andreferences therein. We point out that none of this works if f(t) = tp

with p > 1.To our knowledge the only result on the strict convexity of the level

set for the nonlinearity f(t) = tp is due to C. S. Lin ([14]). In this paperthe author proves the result for strictly convex domain of R2 and healso assume that u is the least energy solution for (1.1).In this paper we give two examples of solutions which have convex

level sets for some superlinear nonlinearities where the dimension ofthe space is greater than 1. We point out that our method can be usedto handle other singularly perturbed elliptic problems.

CONVEXITY 3

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 [6] Grossi and Molle have shown that for small enoughthe level sets of a solution u are strictly starshaped. In this paper weextend this result proving the strict convexity of the level set providedthat has positive curvature More precisely we have the followingresult:

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 , then thereexists such that, for every 0 the level sets of u have strictlypositive Gauss curvature at any point which is not the maximum one.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 if

N > 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 the maxi-mum point is needed to handle the convexity result. In case of problem(1.2) the exact behavior of the solution inside is investigate and ourproof relies on the deep works [13] and [8]. We think that this techniccan be applied to other perturbed problems when the asymptotic be-havior is known.

4 GLADIALI AND GROSSI

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.

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.Let us introduce some notations. Let Sf (t) the level set of a function

f , soSf (t) = {x : f(x) = t}.

We suppose f at least C2 and |f | 6= 0. Then the level set is thesurface Sf (t) = f

1(t) and the normal vector field to Sf (t) in the pointx 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 Tx the 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 thedirection is

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

The Gauss-Kronecker curvature of Sf (t) at x, denoted by Kf (x) is theproduct of the principal curvatures of Sf (t) at x, which are stationaryvalues of the normal curvature on the tangent space Tx. While thenormal curvature depends on the tangent vector the Gauss curvatureis intrinsic.We also mention the following well known facts

THEOREM 2.1. 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.

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= 0

CONVEXITY 5

for all radial function. For references about curvatures and Secondfundamental form look at [18].Now we state some known result about the convexity of level sets in

convex ring, see [2]

THEOREM 2.2. 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 convexC1+ hypersurfaces.

Here we quote some results on the convexity of the level set for theGreens function. In [12] Lewis generalizes the work of Gabriel in [4],[5] about the convexity of the level sets of G. The result is

THEOREM 2.3. Given a convex ring \ , let u be an harmonicfunction such that u = 0 on and u = 1 on . Then

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.Theorem 2.3 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 its 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}. 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 [8] and by Rey in [17]. They provedthe following convergence result

6 GLADIALI AND GROSSI

THEOREM 2.4. 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 laplacianin with zero Dirichlet boundary condition and |g| = g(x0, x0) whereg(x, y) is the regular part of the Greens function. Moreover there exists > 0 independent of n such that

lim0

u = 1.Grossi and Molle in [6] have shown the following

THEOREM 2.5. 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 thesuperlevels are strictly starshaped.

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 is a solution of problem (1.2) corresponding to the valuepn = p n, that satisfies (1.3) then(3.1) Kun(zn) 0 nwhere Kun(zn) is the Gauss curvature of the surface Sun(un(zn)) ={x suchthat un(x) = un(zn)} at the point zn. We use un unand xn xn denotes the unique critical point of un. Notice that from(2.1) the normal nzn at the surface Sun(un(zn)) is always defined by

nzn =un(zn)un(zn) . Up to a subsequence zn z .

Step 1Firstly we suppose z 6= x0, so z \B(x0) for some > 0. In this

domain we have from Theorem 2.4 that gn(x)[N(N2)

S

]N4 (N2)G(x,x0)

2|g|

in C2( \B(x0)

), where gn = unun. From (3.1) we haveKgn(zn) =

Kun(zn) 0 n and passing to the limitlimn

Kgn(zn) = KG(z) 0.If z is an interior point this is impossible from Theorem 2.3 andCorollary 2.1. If otherwise z , this is impossible since we supposethe Gauss curvature of is strictly positive.

CONVEXITY 7

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

unpn12

),

for someR > 0. Consider the function un(x) =1

unun

(xn +

x

unpn12

).

By standard blow-up results, see [8] for example, un converges to

U(x) =(

11+|x|2

)N22

in C2(B(0, R)), for all R R. If zn Bnthen zn = un

pn12 (zn xn) B(0, R) and up to a subsequence

zn z B(0, R). Suppose z 6= 0. From (3.1) we haveKun(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 curvature of U in 0 since SU(1) = {0}. But zn 6= 0 n andKun(zn) 0 implies there exists at least a tangent direction n Tznsuch that kun,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 degen-erate maximum point for the function U(x) and then tHU(0) > 0and so a contradiction follows.

Step 3Finally we consider the case where zn x0 but zn / Bn. From

equation (1.2) we have

(3.2) un(x) = N(N 2)

G(x, y)upnn (y)dy

where G(x, y) = 1(N2)n

1|xy|N2g(x, y) and g(x, y) is the regular part

of the Greens function of the laplacian in . Here we use some ideasby [13]. Let rn = |zn xn|; we have 1 rn R

unpn12

, and rn 0 as

n . Let vn(x) = r2

pn1n un(xn + rnx) for x n = xnrn . Then by

(3.2)

vn(x) = N(N 2)r2

pn1n

G(xn + rnx, y)upnn (y)dy.

Letting y = xn +z

unpn12

we get

vn(x) = N(N 2) r2

pn1n

unpn12

N

n

G

(xn + rnx, xn +

z

unpn12

)

upnn

(xn +

z

unpn12

)dz

8 GLADIALI AND GROSSI

where n = unpn12 ( xn). Multiplying by vn = r

2pn1n un

and recalling the definition of un we get

(3.3)

vnvn(x) = Nn

r4(N2)

4(N2)nn

unN22

n

n

1

rN2nx z

rnunpn12

N2 upnn (z)dz

N(N 2) r4(N2)

4(N2)nn

unN22

n

n

g

(xn + rnx, xn +

z

unpn12

)upnn (z)dz.

In [8] it is proved that

0 un(z) CU(z) = C(

1

1 + |z|2)N2

2

in RN .

From this estimate we deriven

upnn (z)dz RN

(1

1+|z|2)N+2

2dz = n

N.

Since g(x, ) is a bounded function we getn

g

(xn + rnx, xn +

z

unpn12

)upnn (z)dz g(x0, x0)

RN

(1

1 + |z|2)N+2

2

.

Now we consider the first integral in (3.3) i.e.

N

n

r (N2)2n

4(N2)nn

unN22

n

n

1x zrnun

pn12

N2 upnn (z)dz.By the definition of rn we have rnun

pn12 > R

unpn12un

pn12 = R for

each R > 0, and so rnunpn12 as n . If n is large enough

and x 6= 0 then 1x zrnun

pn12

< C|x| and we can pass to the limit asn. Moreover we know from Theorem 2.4 that limn unn =1. Then also limn rnn = 1. We can pass to the limit in (3.3) getting

(3.4) vnvn(x) 1|x|N2 in C( \ {0})

It is easily seen that the convergence in (3.4) is C2(K), for any compactset K \ {0}. Now let zn = znxnrn . Then zn z, where |z| = 1.From (3.1) and the C2 convergence of vnvn to V (x) = 1|x|N2 we get

K||vn||vn(zn) = rN1n Kun(zn) 0 n

CONVEXITY 9

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

1|z|N1 > 0. This finishes the proof of Theorem 1.1.

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:

(4.1)

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 decreasing

DEFINITION 4.1. Let u C2() C1() a family of solutions of(1.4), for small , and let x be such that u = u(x). Thenu is said a family of single-peak solutions near a point x0 if

i) x x0 as 0,ii) N

[12

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

up+1

] E as 0.

The following lemma is well known (see [16] for example)

LEMMA 4.1. If u is a family of single-peak solutions of (1.4), then,for sufficiently small, x is the only local maximum point and

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

the problem (1.4) (see for example [16], [7] and the references therein).We are in position to give the proof of Theorem 1.2.

Proof of Theorem 1.2. Let v(y) = u(x + y), where x are the max-imum points 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 seethat v W in C2loc(RN), and W satisfies (4.1). Furthermore 0 is anondegenerate maximum point for W . We want to show that in eachball B = B(x, R) for R R the function u has strictly convex levelsets. This follows from the convergence of v W as in Theorem 1.1.Let us suppose by contradiction that there exists a sequence n 0,and points zn Bn Bn , zn 6= xn, with Gauss curvature less orequal than zero. Then the same is true for the function vn, so that

10 GLADIALI AND GROSSI

Kvn(zn) = N1n Kun(zn) 0 where zn = znxnn and zn B(0, R). Up to

a subsequence zn z B(0, R), and if z 6= 0 then limnKvn(zn) =KW (z) 0. But this is not possible since W is radial and its Gausscurvature is strictly positive if z 6= 0.Now consider the case z = 0; the condition Kvn(zn) 0 implies

that there exists at least a unit vector n Tzn and kvn,zn(n) = tnHvn(zn)n 0. Up to a subsequence n , = 1 and tnHvn(zn)n tHW (0) 0 and this is not possible since 0 isa non degenerate maximum point for W , and its Hessian matrix isnegative defined in 0.Reasoning as in [16], > 0, R we can find a ball B such

that u(x) in \B. 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