On an indefinite, non-Lipschitz semilinear elliptic problem: Compactly supported solutions, multiplicity and uniqueness

J. Math. Anal. Appl. 389 (2012) 569–590

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On an indefinite, non-Lipschitz semilinear elliptic problem: Compactly supported solutions, multiplicity and uniqueness Qiuping Lu 1 Centro de Modelamiento Matemático, Av. Blanco Encalada 2120, Santiago, Chile

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 1 August 2011 Available online 7 December 2011 Submitted by M. del Pino

For a sign-changing function a(x), we consider solutions of the following semilinear elliptic problem in R N with N  3:

  −u = γ a+ − a− uq + u p ,

Keywords: Non-Lipschitz semilinear elliptic PDE Concave-convex nonlinearities Compactly supported solutions

where

γ > 0 and 0 < q < 1 < p <





u  0 and u ∈ D 1,2 R N , N +2 . N −2

We show that all solutions are compactly

supported if lim inf|x|→∞ a− (x) > 0. When Ω + = {x ∈ R N | a(x) > 0} has several connected components, we prove that there exists an interval on γ , in which two solutions exist and are positive in Ω + . We also give a uniqueness result for solution with small L ∞ -norm. In the end if a(x) = a(|x|) and a(x) is strictly decreasing in |x|, we show that all solutions are radially symmetric and are decreasing in |x|. © 2011 Elsevier Inc. All rights reserved.

1. Introduction For a locally Hölder continuous and sign-changing function a(x) in R N , we study the following elliptic problem in R N with N  3:

⎧ ⎨ −u = aγ (x)uq + u p in R N , 0 < q < 1 < p < N + 2 , N −2   ⎩ u0 in R N , u ∈ D1,2 R N

(1.1)

where γ > 0, aγ (x) = γ a+ (x) − a− (x), a+ (x) = max(0, a(x)) and a− (x) = max(0, −a(x)). By D 1,2 (R N ) we mean the com pletion of C 0∞ (R N ) under the Dirichlet semi-norm, ( RN |∇ u |2 dx)1/2 . The following assumption is always made on a− (x) throughout this paper,

lim inf a− (x) > 0. |x|→∞

Notice that this assumption includes the case lim inf|x|→∞ a− (x) = ∞. Let Ω + = {x ∈ R N | a(x) > 0}, Ω 0+ = {x ∈ R N | a(x)  0} and Ω − = {x ∈ R N | a(x) < 0}. Since a(x) is sign-changing, Ω + and Ω 0+ are bounded and not empty. Equations of this type (1.1) arise as stationary solutions to degenerate reaction–diffusion equations introduced by Gurtin and MacCamy in [12,13] to model the evolution of a biological population (also see [2]). To emphasize the dependence on λ, Eq. (1.1) is often referred as (1.1)γ (the subscript γ is omitted if no confusion arises). The important feature of this equation

1

E-mail address: [email protected]. Supported by Proyecto Fondecyt Posdoctorado No. 3100050.

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2011.12.008

©

2011 Elsevier Inc. All rights reserved.

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

is that it not only combines a non-Lipschitz nonlinearity u q with a sign-changing coefficient a(x), but also exhibits combined effects of concave and convex nonlinearities in Ω + . Such “concave plus convex” nonlinearities in a bounded domain have been studied by Ambrosetti, Brezis and Cerami [4] (see also [1,9]), so we expect some similar results. It was originally observed by Schatzman [20] that solutions could vanish on large sets and in fact that, under appropriate hypotheses on a(x), there exist solutions with compact support. We show more: Theorem 1.1. Every weak solution of (1.1) is compactly supported. So this problem (1.1) indeed is a free boundary problem. This theorem is very useful in the sense that it helps simplify the calculations for the proof of Theorem 2.5 and Lemma 5.7, which are crucial parts of this paper. The sublinear term u q , 0 < q < 1, is essential for this phenomena to happen. If instead considering the same Eq. (1.1) with q  1, then a simple application of classical strong maximum principle shows that a nonnegative solution must be strictly positive in R N , so the existence of compactly supported solutions would be impossible. In another paper [2] a similar equation −u = a(x)u q + b(x)u p with b(x)  0 was studied, and it was shown that all the nonnegative solutions in D 1,2 (R N ) have compact support and the support of these solutions stay inside a big ball with radius depending on a(x) and independent of any particular solution. Comparing to [2], the size of the support of solutions to (1.1) cannot be controlled. Actually an important result by Cortázar, Elgueta and Felmer [7] shows that the equation − v = v p − v q in R N has a unique compactly supported radial solution, which suggests that (1.1)γ could have a solution with support lying completely in Ω − . For example, Example 1.2. Let Ω +  B (0, r ) and a(x) ≡ −1 in R N − B (0, r ) for some r > 0. From [7], we may construct solution of (1.1)γ with arbitrarily large support by gluing together the compactly supported solutions of − v = v p − v q in disjoint balls in R N − B (0, r ). We want to study the structure of solution set of (1.1) in case that Ω + has several components. We make the following assumption on Ω + :

⎧ k ⎪ ⎪ ⎨ Ω + has k < ∞ connected components with Ω + = Ω +, i

⎪ ⎪ ⎩

+

(1.2)

i =1

and each connected component Ωi satisfies an interior ball condition.

Set M = {1, 2, 3, . . . , k}. Under (1.2), for any solution u (x) of (1.1) and any i ∈ M, by strong maximum principle and Hopf’s

lemma u (x) is either completely positive in Ωi+ or completely vanishes in Ωi+ . To organize the set of solutions of (1.1)γ according to the pattern of their support, we define the following classes of solutions: Definition 1.3. (1) For any nonempty I ⊂ M, denote by S I ,γ the class of solutions of (1.1)γ which are positive in Ω I+ = (2) N I ,γ denotes the set {u ∈ S I ,γ | u ≡ 0 in Ω + − Ω I+ }.



i∈ I

Ωi+ .

When γ > 0 is small, we show that there exists a “small” solution, which is a local minimizer (see Theorem 2.5), but for γ large there is no solution at all. Theorem 1.4. For any nonempty I ⊂ M, there exists 0 < Γ I < ∞ such that: 1. S I ,γ = ∅ when 0 < γ  Γ I and S I ,γ = ∅ when γ > Γ I ; 2. S I ,γ has a minimal element u I ,γ for all 0 < γ  Γ I ; 3. u I ,γ  L ∞ (RN ) → 0 as γ → 0+ . Actually the L ∞ -norm of all the solutions of (1.1)γ with nonpositive energy uniformly goes to zero as (see Proposition 2.11). The existence of a least energy solution for (1.1) is also shown in Proposition 2.12. Definition 1.5. We say that a(x) is admissible if (1.2) holds and: 1. Ω 0+ also has exactly k connected components with Ω 0+ = 2. for each i ∈ M,

Ωi0+

satisfies an interior ball condition,

3. Ωi+ ⊂ Ωi0+ for i ∈ M and dist(Ωi0+ , Ω 0j + ) > 0 for i = j.

k

0+ i =1 Ωi ,

γ tends to zero

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

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To state the following theorem we also need another condition on a(x):





there exists C > 0 such that a+ (x)  C dist x, ∂Ω 0+ for any x ∈ R N .

(1.3)

We should mention that if Ω +  Ω 0+ or a+ is locally Lipschitz in R N , then (1.3) is satisfied. We are going to give some existence and uniqueness results for N I ,γ . Let supp(u ) denote the support of u in R N . Theorem 1.6. 1. Assume a(x) is admissible, given δ > 0 small, there exists A 1 > 0 depending on δ , independent of γ , such that solution u of (1.1) satisfies supp(u ) ⊂ {x ∈ R N | dist(x, Ω 0+ )  2δ} if u  L ∞ (RN )  A 1 . In particular N I ,γ = ∅ for small γ . 2. Assume a(x) is admissible and (1.3) holds, then there exists A 2 > 0, independent of γ , such that there is at most one element u in N I ,γ with u  L ∞ (RN )  A 2 . Unlike the results in [2], solution of (1.1) is not unique in general. In fact there are at least two elements in N M ,γ = S M ,γ . To study multiplicity of solutions, we adopt a variational framework for this problem. As mentioned in [1], variational analysis of solutions in N I ,γ , I = M is difficult since these solutions have infinite dimensional negative spaces associated to them (see Remark 2.8). Therefore we will only consider the solutions u ∈ S M ,γ , that is u (x) > 0 for all x ∈ Ω + . Consider the Banach space







E = v ∈ D1,2 R N

a− | v |q+1 dx < ∞



RN

endowed with the norm

 v E =

 12   q+1 1 |∇ v |2 dx + a− | v |q+1 dx .

RN

RN

Actually since we always assume that lim inf|x|→∞ a− (x) > 0, E is a subspace of D 1,2 (R N ) ∩ L q+1 (R N ) and they are the same if lim sup|x|→∞ a− (x) < ∞. Define the energy functional I γ : E → R associated with (1.1)γ as

Iγ (v ) =

1

|∇ v |2 dx +

2 RN

1 q+1





a− | v |q+1 − γ a+ v +

q+1 

RN

dx −

1 p+1



v+

 p +1

dx.

RN

Basically following the arguments in [8] we see that I γ is C 1 from E to R. For convenience we denote by Γ = Γ M and U γ the minimal element in S M ,γ for 0 < γ  Γ . We study the following minimization problem in a convex constraint set





Inf I γ ( v ) v ∈ Y



and

Y = { v ∈ E | 0  v  U Γ a.e.}.

As in Lemma 2.7 the infimum is attained at some function v γ in Y and v γ ∈ S M ,γ . We show that v γ actually is a local minimizer of I γ in E: Theorem 1.7. For 0 < γ < Γ , v γ is a local minimizer for I γ in E; that is, there exists δ > 0 such that

I γ ( v γ )  I γ ( v ) for all v ∈ E with  v − v γ  E < δ. Recall that Brezis and Nirenberg [5] first observed that minimization in the C 1 -topology (for example, the sub- and super-solution construction above) yields local minimizers in the weaker H 1 -topology for a large class of subcritical elliptic variational problems. See also [3] for remarks on supercritical problems. Given that we have a local minimizer for γ ∈ (0, Γ ), we expect a second solution by using the mountain pass theorem of Ambrosetti and Rabinowitz [19]. As the embedding E → L p +1 (R N ) is not compact we always expect the Palais–Smale condition to be an important issue in variational problems posed on R N . To illustrate how compactness may break down for these specific problems, we may return to Example 1.2, let w be the compactly supported radial solution of − v = v p − v q and take a sequence {xn } with limn→∞ |xn | = ∞, then we see that I γ ( w (x − xn )) is bounded and I γ ( w (x − xn )) = 0, but w (x − xn ) converges to zero weakly in E. The strategy we usually use to eliminate this loss of compactness is either to consider a(x) with radial symmetry and restrict our attention to the class of radial functions, or to assume some properties for a(x) at infinity and apply the Concentration-Compactness arguments in [15]. The radial setting for this problem has been done in [16], and it basically shows that if a(x) = a(|x|), then (1.1) has at least two solutions with radial symmetry.

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

Theorem 1.8. Assume either lim inf|x|→∞ a− (x) = ∞, or there exist a∞ > 0 and R > 0 such that lim|x|→∞ a− (x) = a∞ and a− (x) < a∞ for all |x|  R. Then, S M ,γ contains at least two elements v γ and V γ for all γ ∈ (0, Γ ) with V γ  v γ . With the help of this theorem and Theorem 1.6, we can look deep inside S M ,γ . Notice that the minimal element U γ and the local minimizer v γ both stay in S M ,γ . Proposition 1.9. Assume Theorem 1.8 holds. If a(x) is admissible and (1.3) holds, then for very small with I γ ( v γ ) < 0, and I γ ( V γ ) > 0.

γ > 0, v γ coincides with U γ

Let us denote the mountain pass solution by V γ , we could not rule out the possibility that supp( v γ )∩ supp( V γ − v γ ) = ∅, which means that V γ and v γ may coincide in the region Ω + . But under special circumstance we can eliminate this possibility. Assume now a(x) = a(|x|), let r1 = sup{r  0 | a(r )  0}, then 0 < r1 < ∞ because Ω 0+ is bounded and nonempty. We show the following theorem via the moving planes method of Gidas, Ni and Nerenberg [11]. Theorem 1.10. If a(r ) is decreasing in [0, r1 ] and strictly decreasing in [r1 , ∞), any nonzero solution of (1.1) is radially symmetric and decreases as r increases. In particular all solutions of (1.1) have connected support. From time to time we use C , C 1 and C 2 to denote some generic constants. We use B (x, r ) to denote the ball in R N with center x and radius r. Theorem 1.1 and Theorem 1.4 are proved in Section 2, Theorem 1.6 is proved in Section 4, Theorem 1.7 is proved in Section 3, Theorem 1.8 and Proposition 1.9 are shown in Section 5, finally Theorem 1.10 is done in Section 6. 2. Compact support and existence In this section we first use a very simple comparison argument to prove Theorem 1.1. The regularity of solutions of (1.1) follows from standard bootstrap arguments, see Appendix B in Struwe [21], and standard elliptic theory, see [22]. Since we do not assume lim sup|x|→∞ a− (x) < ∞, the estimate offered by Cortázar, Elgueta and Felmer [7] is not enough to guarantee solution u of (1.1) goes to zero at infinity. We introduce another estimate from Theorem 1.1 of Chapter 4 in [14]. Taking ρ > 0 such that Ω 0+ ⊂ B (0, ρ ), let u be a solution of (1.1), then for any ball B (x, 1) ⊂ B (x, 2) ⊂ RN − B (0, ρ ), x ∈ RN , we have: Lemma 2.1. 1. For γ  1, there exists a continuous function h : R+ ∪ {0} → R, with h(0) = 0, such that

  u L ∞ ( B (0,ρ +4))  K h u  H 1 ( B (0,ρ +5)) , where K and h are independent of particular γ and u. 2. Let c ( y ) ∈ L t ( B (x, 2)) for some t > N2 with c ( y ) Lt ( B (x,2))  1. Suppose u (x)  0 is a sub-solution in the following sense

  ∇ u ∇φ + cu φ  0 for any φ ∈ H 01 B (x, 2) and φ  0 in B (x, 2).

B (x,2)

Then u  L ∞ ( B (x,1))  C u  L 2∗ ( B (x,2)) for some positive constant C independent of γ and u. Proof. First part is a simple application of Lemma 2.1 in [7] and second part is a simple application of Theorem 1.1 of Chapter 4 in [14]. 2 Lemma 2.2. lim|x|→∞ u (x) = 0. Moreover if γ  1 and u  E goes to 0, then u  L ∞ (RN ) goes to zero uniformly in γ . Proof. For x ∈ R N − B (0, ρ + 3), we have B (x, 1) ⊂ B (x, 2) ⊂ R N − B (0, ρ ) and −u ( y ) − (u p −1 ( y ))u ( y )  0 for any p −1 , we have that u  L 2∗ ( B (x,2)) and c  N +δ for y ∈ B (x, 2). Since ( p − 1) N2 < 2∗ = N2N −2 and u ∈ E, taking c = u L

2

( B (x,2))

some small δ > 0, both go to zero, as |x| goes to infinity. Applying above lemma we have lim|x|→∞ u (x) = 0. For the second part, repeating previous argument we have that u  L ∞ (RN − B (0,ρ +3)) goes to zero as u  E goes to zero. Next since u ∈ E, then u  H 1 ( B (0,ρ +5)) goes to zero as u  E goes to zero. Applying previous lemma we know that u  L ∞ ( B (0,ρ +4)) goes to zero as u  E goes to zero. 2 Now we begin the proof of Theorem 1.1.

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

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Proof of Theorem 1.1. Since lim inf|x|→∞ a− (x) > 0, there exist R > 0 and 1 > a0 > 0 such that a− (x)  2a0 for all x ∈ a0 2 ( 2 + N −2) 1−q 1−q

R N − B (0, R ). Let A = 12 [

1

] 1−q , then it is easy to see that A p −q < a0 . Since lim|x|→∞ u (x) = 0, we can pick R

even larger so that u (x) < A for any x ∈ R N − B (0, R ). Now for any y ∈ R N − B (0, R + 2), let V (x) = [

1

a0 2

( 2 + N −2) 1−q 1−q

2

] 1−q |x − y | 1−q , we have

  − V (x) = −a0 V (x)q in B ( y , 1) and V ∂ B ( y , 1) > A . If we have that V (x)  u (x) in B ( y , 1), then u ( y ) = 0 for all y ∈ R N − B (0, R + 2). Therefore supp(u ) ⊂ B (0, R + 2). Otherwise there exists x0 ∈ B ( y , 1) so that V (x0 ) < u (x0 ). We may assume V − u attains minimal value at x0 , hence 0  −( V − u )(x0 ) = (a− u q − a0 V q − u p )(x0 )  a0 (u q − V q )(x0 ) + (a0 u q − u p )(x0 )  a0 (u q − V q )(x0 ) > 0, which is a contradiction. 2 Remark 2.3. When 1 < p <

N +2 , N −2

any weak solution u (x) of (1.1) is classical, and from Lemma 2.2 we have that u (x)

goes to 0 uniformly as |x| tends to ∞. But when p >

N +2 , N −2

from the fact that u is a weak solution of (1.1) with u ∈ D (R ) ∩ L (R ), we cannot derive that u (x) is classical and u (x) goes to 0 uniformly as |x| → ∞. If we assume that u (x) goes to 0 uniformly as |x| → ∞, then we know that u (x) is smooth in R N − B (0, r ) for some r > 0, therefore we can follow the proof of Theorem 1.1 and conclude that u (x) has compact support. Please also see Remark 2.13 in the end of this 1, 2

N

p +1

N

section. Now we turn to the existence of minimal element in S I ,γ if it is not empty. We have the following theorem, which is the second part of Theorem 1.4. Theorem 2.4. If I = ∅ and S I ,γ = ∅, then there exists a minimal element u I ,γ in S I ,γ . Proof. This theorem is done by Monotone Iteration process and the crucial step is to find a starting sub-solution to iterate. Actually let u 1 be the smallest solution among all solutions of −u = aγ (x)u q , which are at least positive in Ω I+ . We can q q p start the iteration process with −un+1 + a− un+1 = γ a+ un + un . Since the proof is already presented in [16], we omit it. 2 Next we begin to study the existence of (1.1)γ in S I ,γ . First let us define for nonempty I ⊂ M := {1, 2, . . . , k} (recall that k denotes the number of connected components of Ω + ):

  Γ I ≡ sup γ > 0 S I ,γ = ∅ for (1.1)γ . In the following theorem we are going to show the existence of a local minimizer for small γ in E and the first part of Theorem 1.4 except the existence at Γ I is an immediate consequence of the following theorem. Theorem 2.5. For small γ > 0, there exists v¯ γ ∈ S M ,γ with  v¯ γ  E < Moreover limγ →0  v¯ γ  L ∞ (RN ) = 0.

γ such that I γ ( v¯ γ ) = inf{ I γ (u ) | u  E < γ } = αγ < 0.

1   1 Proof. Let us denote ∇ u 2 = ( RN |∇ u |2 dx) 2 and u a− = ( RN a− |u |q+1 dx) q+1 . Now for very small 1 > u  E = γ , we have

I γ (u ) 

1 2 1

1

q +1

∇ u 22 + u a2− − C 1 γ ∇ u 2 2

q +1

 u 2E − C 1 γ u  E 4

p +1

− C 2 u  E

p +1

− C 2 u  E 1

1

4

8

 (γ )2 − C 1 (γ )q+2 − C 2 (γ ) p +1  γ 2 .

Let αγ = inf{ I γ (u ) | u  E  γ }, it is easy to see that αγ → 0 as γ → 0. Now we want to show that 0  φ ∈ C 0∞ (Ωi+ ) for some i ∈ M. Hence for very small t > 0, we have t φ E < γ and

1

αγ  I γ (t φ) = t 2 2

Ωi+

|∇φ|2 dx −

1 q+1

γ > 0, letting

t q +1 γ

Ωi+

a+ φ q+1 dx −

t p +1 p+1

αγ < 0. Take φ with

φ p +1 dx < 0.

Ωi+

We expect to find a local minimizer v¯ γ ∈ S M ,γ with  v¯ γ  E < γ and I γ ( v¯ γ ) = αγ . Let B γ = {u ∈ E | u  E < γ }. By Ekeland’s variational principle, there exists a sequence {un } ⊂ B γ such that I γ (un ) → αγ and I γ (un ) → 0. Notice that we may assume un  0 and v¯ γ converges weakly to v¯ γ in E. We will show that un converges strongly to v¯ γ in E after the

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

proof of Lemma 5.7. Let us assume that this is true for the time being, that is I γ ( v¯ γ ) = αγ with  v¯ γ  E < γ . Moreover we see that limγ →0  v¯ γ  E = 0, then from Lemma 2.2 we have limγ →0  v¯ γ  L ∞ (RN ) = 0. Finally we want to show that v¯ γ ∈ S M ,γ . Indeed suppose for some i ∈ M, v¯ γ ≯ 0 in Ωi+ , then the strong maximum

principle and Hopf’s lemma imply that v¯ γ ≡ 0 over Ωi+ . Like in the previous part we take φ with 0  φ ∈ C 0∞ (Ωi+ ), then for very small t > 0, we have t φ + v¯ γ  < γ and I γ (t φ + v¯ γ ) = I γ (t φ) + I γ ( v¯ γ ) < I γ ( v¯ γ ) = αγ . This is a contradiction. 2 Recall that the minimal element in S I ,γ is denoted as u I ,γ . The following theorem is a part of Theorem 1.4 except the existence at Γ I . Proposition 2.6. Assume that (1.2) holds, then 0 < Γ I < ∞, u I ,γ exists and is increasing in γ , that is

u I ,γ1  u I ,γ2

for 0 < γ1 < γ2 < Γ I .

Moreover limγ →0 u I ,γ  L ∞ (RN ) = 0. Proof. From above theorem we see that for small γ > 0, we have v¯ γ ∈ S M ,γ ⊂ S I ,γ . So we conclude that Γ I > 0. Now we want to show that Γ I < ∞. Suppose otherwise Γ I = ∞. For each Ωi+ , i ∈ I , taking a small ball B i such that B i  Ωi+ , let a = infx∈ i∈ I B i a(x), then we have a > 0. Let ϕi , λi to be the first positive eigenfunction and eigenvalue of the following problem:

−ϕi = λi ϕi in B i and ϕi = 0 on ∂ B i . Notice that



(ϕi u − u ϕi ) dx = Bi



∂ ϕi ∂u ∂ ϕi ϕi dS = u− u dS  0, ∂n ∂n ∂n

∂ Bi

∂ Bi

where n is the outer unit normal vector of ∂ B i . Therefore we have



u ϕi dx =

λi Bi

that is



−ϕi u dx  Bi

−u ϕi dx = Bi

a+ u q ϕi dx +

Bi

u p ϕi dx, Bi

  λi u − γ auq − u p ϕi dx  0 for i ∈ I and γ > 0.

Bi



By assumption u is positive in Ω I+ = i ∈ I Ωi+ , but λi t − γ at q − t p = t q (λi t 1−q − γ a − t p −q ) < 0 for all t > 0 if γ is sufficiently large, this is a contradiction. We must have Γ I < ∞. Since Γ I > 0, we know that S I ,γ is not empty, from Theorem 2.4, the minimal element u I ,γ exists in S I ,γ . It is easy to see that u I ,γ2 acts naturally as a super-solution for (1.1)γ1 . Noticing that u I ,γ2 has compact support, with proper small subsolution which is supported at each Ωi+ for i ∈ I , (1.1)γ1 has a compactly supported solution u in S I ,γ1 such that u  u I ,γ2 by the sub- and super-solution method. Since u I ,γ1 is the minimal element in S I ,γ1 , we have u I ,γ1  u I ,γ2 . Finally from above theorem we see that for small γ > 0, we have v¯ γ ∈ S M ,γ ⊂ S I ,γ . So v¯ γ  u I ,γ1 , from above theorem again we have limγ →0 u I ,γ  L ∞ (RN ) = 0. 2 So far we have established an interval of existence for (1.1)γ , γ ∈ (0, Γ I ), in the class S I ,γ , where I ⊂ M indicates the components of Ω + in which these solutions must be positive. Now we assert that a solution of class S I ,γ must exist at the endpoint of the maximal interval of existence, γ = Γ I . This is the “extremal solution” for this family (see [6]). Lemma 2.7. Suppose u ∈ N I ,γ for some I γ (u γ ) < 0.

γ > 0, then N I ,γ admits an element u γ for every 0 < γ  γ . Moreover u γ  u and

Proof. For 0 < γ  γ , u is a super-solution for Eq. (1.1)γ and 0 is a sub-solution. We consider the following minimization problem in a convex constraint set





Inf I γ ( v ) v ∈ X



and

X = { v ∈ E | 0  v  u a.e.}.

Notice that u has compact support, so following the arguments as in Struwe [21], the infimum is achieved at some u γ ∈ X and I γ (u γ )φ = 0 for all ϕ ∈ C 0∞ (R N ), and by routine regularity arguments u γ is a solution to (1.1)γ . Since u γ ∈ X , it



vanishes on the components Ω + − i ∈ I Ωi+ . Actually as in the proof of Theorem 2.5, we can show that I γ (u γ ) < 0 and u γ does not vanish in Ωi+ for each i ∈ I . Therefore we have u γ ∈ N I ,γ and I γ (u γ ) < 0. 2

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

575

Remark 2.8. Given the variational formulation of the problem as an infimum, it is natural to ask whether the solutions obtained by above lemma are local minimizers of I γ in any sense. Notice this cannot be the case when I = M. Indeed following the arguments of the last part of the proof of Theorem 2.5, we can decrease the value of I γ near such solution , where j ∈ / I . So the existence of a second solution in the classes N I ,γ remains an open by small perturbations in each Ω + j question. Corollary 2.9. For 0 < γ < Γ I , I γ (u I ,γ ) < 0, where u I ,γ is the minimum element in S I ,γ . Proof. We just simply apply above lemma with u = u I ,γ , γ = γ and some J ⊂ M such that I ⊂ J and u I ,γ ∈ N J ,γ . Hence by above lemma we get a solution u γ ∈ N J ,γ ⊂ S I ,γ such that

0  u γ  u I ,γ .

I γ (u γ ) < 0 and

Since u I ,γ is the minimal element in S I ,γ , we must have u γ = u I ,γ .

2

In order to show the existence at Γ I , we need to prove some estimates.

  q+1 Lemma 2.10. RN |∇ u I ,γ |2 dx + RN a− u I ,γ dx  C (a+ )γ

1 q+1

1

Since q+1 >



1 2

−

1 2

−

1



q +1

a− u I ,γ dx +

2



1

−

2

RN

>

1

p +1

1 p+1



1 p+1

, where C (a+ ) is some constant depending on a+ and Ω + .

1 I ( u I ,γ ) u γ 2 γ

Proof. From above corollary we know that I γ (u I ,γ ) −



p +1 p −q

p +1

< 0, that is



u I ,γ dx < γ RN

1 q+1

−

1



2

q +1

a+ u I ,γ dx.

RN

, from the above inequality we have





p +1

u I ,γ dx < γ

1 q+1

RN

−

1



2

q +1

a+ u I ,γ dx.

(2.1)

RN

Since a+ is compactly supported, we get

q +1





a+ u I ,γ dx  a+  L ∞ (RN )



supp(a+ )

RN

where C (a+ ) is some constant depending on a+ and

  p −q u I ,γ L p+1 (RN )  C a+ γ



1 q+1

−

Ω + . Putting this back to (2.1), we find

   −1  1  a+  ∞ N 1 − 1 . L (R ) 2 2 p+1

With this and (2.1), noticing that I γ (u I ,γ ) < 0, i.e.

1





q +1 q +1 u I ,γ dx  C a+ a+  L ∞ (RN ) u I ,γ  L p+1 (RN ) ,

|∇ u I ,γ |2 dx +

2 RN

we conclude this lemma.

1 q+1

q +1

a− u I ,γ dx <

RN

γ q+1

q +1

a+ u I ,γ dx +

RN

1 p+1

p +1

u I ,γ dx, RN

2

Now we are ready to complete the proof of Theorem 1.4. Proof of Theorem 1.4. Taking an increasing sequence {γn } with limit Γ I , from the above lemma, we know that u I ,γn  E is uniformly bounded. Hence there exists u Γ I ∈ E such that

u I ,γn u Γ I













weakly in D1,2 R N , L p +1 R N and L q+1 R N .

Moreover we may assume u I ,γn → u Γ I a.e. in R N . From Proposition 2.6 we know that u I ,γn is increasing in n, so by monotone convergence theorem

u I ,γn → u Γ I









strongly in L p +1 R N and L q+1 R N ,

which implies that u Γ I = 0. Next since I γ n (u I ,γn ) = 0, taking any

(2.2)

ϕ ∈ C 0∞ (RN ), we have I γ (u I ,γn )ϕ = 0. By (2.2), passing the

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

limit on n, we have



∇ u Γ I ∇ ϕ dx = RN

q

aΓ I u Γ I ϕ +

RN

p

uΓI ϕ .

RN

Therefore u Γ I is a weak solution of (1.1)Γ I . By routine regularity arguments, u Γ I is a classical solution. Next we want to show that u Γ I is the minimal element in S I ,Γ I , i.e. u Γ I = u I ,Γ I . Actually from above we see that S I ,Γ I is not empty. Picking any U ∈ S I ,Γ I , we just need to apply Lemma 2.7 to Eq. (1.1)γ with u = U , γ = Γ I and some J ⊂ M such that I ⊂ J and U ∈ N J ,Γ J . We get a solution u γ to (1.1)γ such that u γ ∈ S I ,γ and we also have U  u γ  u I ,γ . Since limγn →Γ − u I ,γn = u Γ I , we have U  u Γ I . 2 I

Now let us take a look at the nonpositive energy solution u γ of (1.1)γ . Proposition 2.11. Suppose for all small γ > 0, I γ (u γ )  0, then u γ  L ∞ (RN ) → 0 as γ → 0.

  q+1 Proof. Since I γ (u γ )  0 and I γ (u γ ) = 0, using the same proof of Lemma 2.10, we get RN |∇ u γ |2 dx + RN a− u γ dx  C (a+ )γ

p +1 p −q

, which implies that u γ  E → 0 as

γ → 0. From Lemma 2.2, we have u γ L ∞ (RN ) → 0 as γ → 0. 2

In the end of this section we want to point out the existence of a least energy solution for (1.1). Proposition 2.12. Suppose S I ,γ = ∅ for some inf{ I γ (u ) | I γ (u ) = 0 and u  0}.

γ > 0 and I ⊂ M, then there exists a solution vˆ γ of (1.1) such that I γ ( vˆ γ ) =

ˆ γ = inf{ I γ (u ) | I γ (u ) = 0 and u  0}. Since S I ,γ = ∅ for some Proof. Let α αˆ γ < 0. Now we have a sequence {un } with the following properties:

ˆ γ + o(1) 0 > I γ (u n ) = α

and

γ > 0 and I ⊂ M, from Corollary 2.9 we have

I γ (un ) = 0.





q+1

p +1

Using the same proof of Lemma 2.10, we get RN |∇ un |2 dx + RN a− un dx  C (a+ )γ p−q , which implies that un  E is ˆ γ and vˆ γ is compactly bounded. Therefore we may assume that un → vˆ γ weakly in E, then we have I γ ( vˆ γ ) = 0, I γ ( vˆ γ )  α supported. We will show that un converges strongly to vˆ γ in E after the proof of Lemma 5.7. 2 Remark 2.13. We want to point out that Lemma 2.2 is still true when p = N +2 , N −2

N +2 , N −2

therefore with minor changes all the results

+2 in this section hold for p = which implies that Theorem 1.4 holds for p = N , for details please see [17]. N −2 We use variational arguments to prove the existence Theorem 2.5, so it is not clear that we can extend the existence re+2 sult for p > N . Actually for all p > 1, we can modify the proof of Lemma 3.2 in [16] and construct a smooth super-solution N −2

K γ > 0 in R N for (1.1) such that K γ (x) goes to 0 as |x| → ∞ and limγ →0  K γ  L ∞ (RN ) = 0 when γ is small. Therefore for all p > 1 (1.1) has a smooth compactly supported solution u γ ∈ S M ,γ when γ is small, moreover limγ →0 u γ  L ∞ (RN ) = 0. For later on denote Γ M by Γ , u M ,γ by U γ and u M ,ΓM by U Γ . 3. Existence of local minimizers Brezis and Nirenberg [5] first observed that minimization in the C 1 -topology yields local minimizers in the weaker H -topology for a large class of subcritical elliptic variational problems. Now we employ a similar idea from [1] to prove Theorem 1.7. Recall that U γ represents the minimal element in S M ,γ for 0 < γ  Γ , here Γ = Γ M . Now consider the following minimization problem in a convex constraint set 1





Inf I γ ( v ) v ∈ Y



and

Y = { v ∈ E | 0  v  U Γ a.e.}.

(3.1)

Like in Lemma 2.7, the infimum is attained at some function in Y , say v γ , and v γ ∈ S M ,γ . Moreover I γ ( v γ ) < 0. Lemma 3.1. Each connected component of the set {x ∈ R N | U γ > 0} contains at least one connected component of Ω + . Proof. We know that U γ ∈ S M ,γ , by definition of S M ,γ , U γ (x) > 0 for all x ∈ Ω + . If there is one connected component S 0 of {x ∈ R N | U γ > 0} such that S 0 ∩ Ω + ≡ ∅, then we have another solution U¯ γ ∈ S M ,γ with U¯ γ (x) = U γ (x) for all x ∈ R N − S 0 and U¯ γ (x) ≡ 0 for all x ∈ S 0 . Therefore U γ is not the minimal element in S M ,γ , it is a contradiction. 2

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

Lemma 3.2. For 0 < {x ∈ R N | U Γ (x) > 0}.

577

γ < Γ , let u be a solution to (1.1)γ such that 0  u  U Γ in RN , then u (x) < U Γ (x) for all x ∈ A =

Proof. Let v = U Γ − u  0 in R N . We have

−U Γ + a− U Γ = Γ a+ U Γ + U Γ  γ a+ uq + u p = −u + a− uq . q

q

p

So we obtain that in R N , −(U Γ − u ) + a− (U Γ − u q )  0. We may rewrite it as q

− v + a−



q

U Γ − uq UΓ − u



v  0.

Let A i be one of the connected components of A. Suppose that for some x0 ∈ A i , v (x0 ) = 0. Let S i = {x ∈ A i | v (x) = 0}, then S i is not empty since x0 ∈ S i . Since U Γ (x0 ) > 0, we may take a small ball B = B (x0 , r ) such that B  A i . Over B we have q

0

q

U Γ − uq



UΓ − u

UΓ UΓ

,

q

U −u q

which implies that UΓ −u uniformly bounded in B. Since v  0 in R N , then by strong maximum principle we have v ≡ 0 Γ in B, which means S i is open in A i . By continuity S i is also close in A i . Since S i is not empty, then S i = A i . From previous lemma we see that A i contains some connected component of Ω + . This leads to a contradiction when comparing the equations satisfied by u and U Γ . 2 Now we restate Theorem 1.7 and give a detailed proof. Theorem 3.3. For 0 < γ < Γ , v γ is a local minimizer for I γ in E, that is, there exists δ > 0 such that

I γ ( v γ )  I γ ( v ) for all v ∈ E with  v − v γ  E  δ. Proof. We prove this theorem by contradiction. Suppose there exists a sequence {un } ⊂ E such that un → v γ strongly in E and I γ (un ) < I γ ( v γ ). It is easy to see that un → v γ in D 1,2 (R N ) ∩ L q+1 (R N ). Let





un− = max{−un , 0} and

v n = max 0, min{un , U Γ } ,

w n = (un − U Γ )+ .

So we have un = v n − un− + w n , v n ∈ Y and w n ∈ E. Moreover w n is uniformly bounded in E, un− and w n have disjoint supports. We notice that for some x ∈ R N , if U Γ (x) = 0, then v n (x) = 0. Define the measurable sets



T n = supp un−

S n = supp( w n ),









R n = x ∈ R N 0  un  U Γ .

and

For any x ∈ S n , v n (x) = U Γ (x) and un (x)  U Γ (x)  v γ (x). Therefore for any x ∈ R N





0  w n (x) = (un − U Γ )+ (x)  (un − v γ )+ (x)  un (x) − v γ (x) , ∗

which implies that w n → 0 in L q+1 (R N ) ∩ L 2 (R N ). Recall that A = {x ∈ R N | U Γ (x) > 0}, let B = R N − A − Ω 0+ and B 0 = (R N − A ) ∩ Ω 0+ , thus R N = A ∪ B ∪ B 0 . We also have that A and B 0 are bounded in R N , and B ⊂ Ω − by the strong maximum principle and Hopf’s lemma. 1 1 a ( v )q+1 + p + ( v ) p +1 and K 1 (x) = H (x, U Γ + w n ) − H (x, U Γ ) − For convenience for v  0, set H (x, v ) = q+ 1 γ 1



H v (x, U Γ ) w n , where H v (x, v ) = aγ v q + v p ; set K 2 = RN in S n , noticing that supp( v n ) ⊂ R n ∪ S n , we have

I γ (u n ) =

1



|∇ un |2 dx −

2 RN

 =





H x, un+ dx +

RN

1 2

Rn

= Rn ∪ Sn

RN

1 |∇ un− |2 2



1 q+1

+

1 a− |un− |q+1 dx. q+1

q+1

a− un−

Since un = U Γ + w n and v n = U Γ

dx

 

 1 |∇ v n |2 − H (x, v n ) dx + |∇ un |2 − H (x, un ) dx + K 2



2

Sn



  1 1 2 2 |∇ v n | − H (x, v n ) dx + K 2 + |∇ w n | + ∇ U Γ ∇ w n dx − H (x, U Γ + w n ) − H (x, U Γ ) dx 2

2

= Iγ (vn) + K 2 +

Sn

1 2

Sn

|∇ w n |2 + ∇ U Γ ∇ w n − H v (x, U Γ ) w n − K 1 (x) dx.

Sn

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

Since U Γ is a super-solution with respect to (1.1)γ , we have



∇ U Γ ∇ wn  RN

q

p

aγ U Γ w n + U Γ w n dx =

RN

H v (x, U Γ ) w n dx.

RN

Therefore we have

I γ (u n )  I γ ( v n ) + K 2 +



1

|∇ w n |2 dx −

2 RN

K 1 (x) dx. Sn

Since I γ ( v n )  I γ ( v γ ) and I γ ( v γ ) > I γ (un ), we get I γ ( v n ) > I γ (un ). Noticing that aγ = 0 and U Γ = 0 in B 0 , we find

0 > K2 +



1

|∇ w n |2 dx −

2

= K2 +



1

K 1 (x) dx S n ∩( A ∪ B ∪ B 0 )

RN

2

|∇ w n | dx −

2

K 1 (x) dx − Sn ∩B

Sn ∩ A

RN

K 1 (x) dx − Sn ∩B 0

1 p+1

p +1

wn

dx.

(3.2)

2∗

Since w n → 0 in L q+1 (R N ) ∩ L (R N ) and B 0 is bounded, we have

p +1

wn

 dx =

B0

 pp−+11 

p +1

wn

dx

p +1

wn

B0

B0

Now we are going to estimate each term in



  2 K 1 (x) dx  o(1) |∇ w n | dx ,

Sn ∩ A



K 1 (x)  −C

Sn ∩B

RN q +1

a− w n

2

RN



Sn ∩ B

K 1 (x) dx and show that

 |∇ w n |2 dx ,

(3.5)

RN

1

0 > I γ (u n ) − I γ ( v γ )  K 2 + 1

(3.3)

(3.4)



where C = q+1 − p +1 . Therefore we have

   o(1) |∇ w n |2 dx .

K 1 (x) dx and

dx + o(1)

Sn ∩B 1



Sn ∩ A

 p+2 1 dx

1

2

 

q +1 |∇ w n |2 dx − o(1) |∇ w n |2 dx + C a− w n dx

RN

RN



  − 2 2 ∇ u + a− u − q+1 dx + 1 − o(1) |∇ w n | dx  0, n n

Sn ∩B

2

RN

RN

which is a contradiction  for n large enough. This theorem is done if (3.4) and (3.5) are true. Now we estimate S ∩ A K 1 (x) dx. First we want to show that | A ∩ S n | → 0 as n → ∞. Indeed, let n δ > 0, set

E n = {x ∈ A | un > U Γ > v γ + δ} and

F n = {x ∈ A | un > U Γ and U Γ  v γ + δ}.

It is clear that A ∩ S n ⊂ E n ∪ F n . From Lemma 3.2, we see that 0 = |{x ∈ A | U Γ  v γ }| = |

∞

j =1 {x ∈ + δ1 }| < 12

1 }|. j

 > 0 be given. For

A | U Γ  v γ + 1j }| =

lim j →∞ |{x ∈ A | U Γ  v γ + Hence there exists δ1 > 0 so that | F n |  |{x ∈ A | U Γ  v γ  for all n. But on the other hand, since un → v γ strongly in L q+1 (R N ), there exists n1 > 0 such that for all n  n1

1 2

q +1

δ1



(un − v γ )q+1 dx 

 RN

q +1

δ1

q +1

dx = δ1

| E n |,

En

1 2

so we have | E n | <  , which implies that | A ∩ S n |  | E n | + | F n | <  . Since | A ∩ S n | → 0 as n → ∞, we have

w n2 dx A∩ Sn

 | A ∩ Sn|

2 n

 RN

∗ w n2

 n−n 2 dx

   o(1) |∇ w n |2 dx . RN

(3.6)

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

579

Similarly we also have

p +1 w n dx

 pp−+11 



p +1 w n dx

=

A∩ Sn

p +1

wn

A∩ Sn

 p+2 1

   o(1) |∇ w n |2 dx .

dx

A∩ Sn

(3.7)

RN

0+

Since U Γ > 0 in Ω − B 0 , by Hopf’s lemma there exists l > 0 such that U Γ  l for all x ∈ Ω 0+ − B 0 . Noticing that A ∩ B 0 = ∅, we have from (3.6)

  1 1 q +1 q q +1 aγ (x) (U Γ + w n ) − − U Γ w n dx U q+1 q+1 Γ

0 A ∩ S n ∩Ω 0+

   Γ a+ 

L ∞ (R N ) A∩ Sn

On A ∩

1 2

∩Ω 0+

ql

q −1

w n2 dx

  2  o(1) |∇ w n | dx .

S n ∩ Ω − we have



aγ A ∩ S n ∩Ω −

1 q+1

(3.8)

RN

1

(U Γ + w n )q+1 −

q +1

q+1

UΓ

 q − U Γ w n dx  0.

(3.9)

To estimate the other term, we notice that there exists θ = θ(x) ∈ (0, 1)

1

0

p+1

1

( U Γ + w n ) p +1 −

p +1

UΓ

p+1

 1 p p −1  2 − U Γ w n = p (U Γ + θ w n ) p −1 w n2  C 1 + w n wn , 2

as a consequence, from (3.6) and (3.7)

1

A∩ Sn

p+1

1

( U Γ + w n ) p +1 −

p +1

p+1

UΓ

  p − U Γ w n dx  o(1) |∇ w n |2 dx .

(3.10)

RN

Combining (3.8),(3.9) and (3.10), we get (3.4). We estimate S ∩ B K 1 (x) dx. To estimate the terms on the set B, we should notice that B ⊂ Ω − . Since U Γ = 0 on B, then n we have that

0 Sn ∩B

 

1 1 q +1 q q +1 aγ (x) (U Γ + w n ) − − U Γ w n dx = − U q+1 q+1 Γ

Sn ∩B

a−

q +1

q+1

wn

dx.

(3.11)

1

Similar to estimate | A ∩ S n |, we can show that | P n | = |{x ∈ Ω − | w n  (a− (x)) p−q }| → 0 as n → ∞. Therefore

B ∩ Sn

1 p+1

p +1

wn

1

dx = B ∩ Sn − P n



1 p+1

p +1

wn

p+1

a

−

1

dx + B ∩ Sn ∩ P n

p +1

p+1

wn



q +1 w n dx + o(1)

B ∩ Sn

dx

 2

|∇ w n | dx . RN

Hence from above and (3.11) we have (3.5):





K 1 (x) dx = Sn ∩B

aγ Sn ∩B





1 p+1

1

q +1

q+1

−



wn

1 q+1



dx + Sn ∩B q +1

a− w n

Sn ∩B

1 p+1

p +1

wn

dx

 dx + o(1)

 |∇ w n |2 dx .

2

RN

Remark 3.4. With almost identical proof we can show that Theorem 3.3 holds also for p = N +2 . N −2

(3.12)

N +2 , N −2

for details see [17].

Theorem 3.3 is not true in general for p > We are going to give an example for this situation. Recall from Remark 2.13, we know that (1.1) has a compactly supported smooth solution when γ is small, so we may assume (1.1)γ1 has a compactly supported smooth solution at γ1 , say u γ1 . Now we may replace U Γ in (3.1) by u γ1 , by sub- and sup-solution method we have a compactly supported smooth solution for (1.1)γ for γ < γ1 , let us still denote it by v γ .

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

+2 We will show that when p > N , for any solution v γ of (1.1) with v γ ∈ E ∩ L p +1 (R N ) and supp( v γ ) ⊂ B (0, r ) for some N −2 r > 0, v γ is not a local minimizer in E. In fact for  > 0, we provide two types of test function φ ∈ E such that

φ  E → 0 as  → 0 and I γ ( v γ + φ ) < I γ ( v γ ) for all small  > 0. Take x0 ∈ R N such that B (x0 , 1) ∩ B (0, r ) = ∅ and infx∈ B (x0 ,r ) a− (x) > 0. Let η be the cut-off function with the properties that supp(η) ⊂ B (x0 , 1) and η ≡ 1 around x0 . Now set φ (x) = η|x − x0 |−α , where α > 0 and 2(α + 1) < N. Immediately



p +1

= ∞, hence we know that φ ∈ E and φ  E → 0 as  → 0. Now for large p with ( p + 1)α > n, we know that B (x ,1) φ 0 I γ ( v γ + φ ) = −∞ < I γ ( v γ ). This is some extreme type of test function. Next we give another type of test function, which is originated from [3]. For above x0 and B (x0 , 1), let φ =  −σ φ1 (x/ ),  p +1 2 2 N < σ < min( N − , p+ ). It is easy to see that B (x ,1) φ dx = O ( N −σ ( p +1) ) and where 0  φ1 ∈ C 0∞ ( B (x0 , 1)) and p − 1 2 1



  |∇φ |2 dx = O  N −2−2σ ,

B (x0 ,1)

0

q +1

a − φ

dx = O





 N −σ (q+1) .

B (x0 ,1)

2 N < σ < min( N − , p+ ), we have N − σ (q + 1) > N − 2 − 2σ > N − σ ( p + 1) > 0. Hence φ  E → 0 2 1 as  → 0 and I γ ( v γ + φ ) − I γ ( v γ ) = I γ (φ )  −c  N −σ ( p +1) (−1 + o(1)) < 0 for some c > 0 and all small  > 0.

Since p >

N +2 N −2

and

2 p −1

To this end for small γ > 0, we have the minimal element U γ ∈ S M ,γ , local minimizer v γ ∈ S M ,γ , another local minimizer v¯ γ ∈ S M ,γ from Theorem 2.5 and least energy solution vˆ γ from Proposition 2.12, all of them have nonpositive energy. From Proposition 2.11, we see that their L ∞ -norm tends to zero as γ → 0. In the next section we are going to show that for small γ they are in fact the same if a(x) is admissible and (1.3) holds. 4. Uniqueness of the small solution In this section we give the proof of Theorem 1.6. Here is the proof for the first part, which is very similar to the proof of Theorem 1.1. Proof of Theorem 1.1. Since a(x) is admissible, we can take δ > 0 small so that

δ

1 10









min dist Ωi0+ , Ω 0j + 1  i = j  k > 0.

+ 0+ 0+ N Denote Ωi0,m δ = {x ∈ R | dist(x, Ωi )  mδ} with 1  i  k and m  1, and Ωmδ = it is clear that 0 < aδ → 0 as δ → 0. So we may assume that aδ < 1 and δ < 1.

Let A 1 = 12 [

aδ

1

2 ( 2 +n−2) 1−q 1−q

2

p −q

] 1−q δ 1−q , then we have A 1



i∈M

+ Ωi0,m δ , let aδ =

1 2

minx∈RN −Ω 0+ a− (x), δ

< aδ . Now for any x ∈ R N − Ωδ0+ and u  L ∞ (RN )  A 1 , we have

  −u = aγ uq + u p = −a− uq + u p  −aδ uq + u p − aδ uq  −aδ uq . Next for any y ∈ R N − Ω20δ+ , let V (x) = [

aδ

2 ( 2 +n−2) 1−q 1−q

1

2

] 1−q |x − y | 1−q , then we have

  − V = −aδ V q in B ( y , δ) and V ∂ B ( y , δ) > A 1 . So if we have that V (x)  u (x) in B ( y , δ), then u ( y ) = 0 for all y ∈ R N − Ω20δ+ . Therefore supp(u ) ⊂ Ω20δ+ . Otherwise there exists x0 ∈ B ( y , δ) so that V (x0 ) < u (x0 ). We may assume V − u attain minimal value at x0 , hence 0  −( V − u )(x0 ) = aδ (u q − V q )(x0 ) > 0, which is a contradiction. Finally from Theorem 2.5, we know that for small γ > 0 there exists v¯ γ ∈ S M ,γ with limγ →0  v¯ γ  L ∞ (RN ) = 0. Therefore we conclude that N I ,γ is not empty for small γ > 0. 2 We are going to start the proof for the second part of Theorem 1.6. When a(x) is admissible and L ∞ -norm of solution u is small, the first part of Theorem 1.6 says that solution u is somehow “localized”, which means that we only need to prove the second part when Ω 0+ is connected. We follow the arguments in Ambrosetti, Brezis and Cerami [4]. We begin with the following lemma. Lemma 4.1. Let z denote the unique solution satisfying

−z = a+ (x) zq , z > 0 in Ω 0+ ,

x ∈ Ω 0+ , z = 0 on ∂Ω 0+ .

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

581

Then there exists β > 0 such that for any φ ∈ H 01 (Ω 0+ )



|∇φ|2 − qzq−1 a+ φ 2 dx  β

Ω 0+

φ 2 dx.

Ω 0+

Proof. Let us recall that z can be obtained by the following variational problem:

 min

1 2

|∇ u |2 − Ω 0+

1 q+1







a+ |u |q+1 u ∈ H 01 Ω 0+



 .

Ω 0+

As a consequence we have Ω 0+ |∇φ|2 − qzq−1 a+ φ 2 dx  0 for φ ∈ H 01 (Ω 0+ ). We want to point out that this expression  q−1 + 2 a φ dx makes sense by Hardy inequality (see [4]). Namely we have the first eigenvalue λ1 {− − qa+ zq−1 }  0. Ω 0+ qz Suppose that λ1 {− − qa+ zq−1 } = 0, then there exists φ ∈ H 01 (Ω 0+ ) and φ > 0 in Ω 0+ such that

−φ − qa+ zq−1 φ = 0. 







Hence we have Ω 0+ ∇φ∇ z dx = q Ω 0+ a+ zq φ . On the other hand from the equation we have Ω 0+ ∇φ∇ z dx = Ω 0+ a+ zq φ , this is a contradiction because q < 1. So we have λ1 {− − qa+ zq+1 } > 0. 2 Next we enlarge the domain Ω 0+ a little. Let δ > 0 small, denote the set {x ∈ R N | dist(x, Ω 0+ ) < 2δ} by Ω20δ+ . We have the following theorem. Lemma 4.2. Assume Ω 0+ is connected and (1.3) holds. Then there exist δ0 > 0 small and β1 > 0 such that for any φ ∈ H 01 (Ω20δ+ )

|∇φ|2 − qzq−1 a+ φ 2 dx  β1

Ω20δ+

0

φ 2 dx.

Ω20δ+

0

0

Proof. By Hopf’s lemma there exist C 1 > 0 and C 2 > 0 such that







C 1 dist x, ∂Ω 0+  z(x)  C 2 dist x, ∂Ω 0+



for any x ∈ Ω 0+ ,

which means that there exists C > 0, ( zq−1 a+ )(x)  C for any x ∈ R N . Therefore λ1 {− − qa+ zq−1 } is well defined in H 01 (Ω20δ+ ). This lemma is concluded by the continuous dependence of the first eigenvalue on the domain. 2 0

The condition (1.3) says that: there exists C > 0 such that a+ (x)  C dist(x, ∂Ω 0+ ) for any x ∈ R N . This condition is not hard to be satisfied by a, for example Ω +  Ω 0+ or a+ is a Lipschitz function in R N . Indeed if a+ is Lipschits in R N , then for any x ∈ Ω 0+ , there exists y ∈ ∂Ω 0+ such that |x − y | = dist(x, ∂Ω 0+ ), hence a+ (x) = |a+ (x) − a+ ( y )|  C |x − y | = C dist(x, ∂Ω 0+ ). Theorem 4.3. If Ω 0+ is connected and (1.3) holds, then there exists a constant A > 0 such that (1.1) has at most one solution u with u  L ∞ (RN )  A. Proof. For δ0 > 0 small, from the first part of Theorem 1.6 we see that there exists A 1 > 0 such that any solution u of (1.1) with u  L ∞ (RN )  A 1 satisfies supp(u ) ⊂ Ω0+ with  = 2δ0 . For this δ0 , making it smaller if necessary, Lemma 4.2 says that there exists β1 > 0 such that for any φ ∈ H 01 (Ω0+ ), we have

|∇φ|2 − qzq−1 a+ φ 2 dx  β1

Ω0+

φ 2 dx.

(4.1)

Ω0+ β

1

p −1

Now we take A 2 = min{ A 1 , ( 2p1 ) p−1 }, then p A 2

< β1 .

0+

Notice that we assume Ω is connected, so there is no confusion that we denote u γ to be the minimal solution of (1.1)γ with u γ  L ∞ (RN )  A 2 . Suppose, by contradiction, there is another solution w = u γ + v and  w  L ∞ (RN )  A 2 , then we have supp( v ) ⊂ Ω0+ and v ∈ H 01 (Ω0+ ). Moreover v  0 in Ω0+ and  v  H 1 (Ω 0+ ) > 0. Let ξ = γ

1 1−q

0

z, then we have

−ξ = γ a+ ξ q in Ω 0+ and ξ > 0 in Ω 0+ .



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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

We also have −u γ = aγ u γ + u γ  aγ u γ = γ a+ u γ in Ω 0+ . By uniqueness and sub- and super-solution method we have q

1 1−q

uγ  ξ = γ

p

q

q

in Ω 0+ .

z

We know that v satisfies

   q q p − v  − v + a− (u γ + v )q − u γ = γ a+ (u γ + v )q − u γ + (u γ + v ) p − u γ . Hence we have



Ω0+











Ω0+

γ a+ v (u γ + v )q − uqγ dx +

= Ω 0+

q−1

q



γ a+ vquqγ−1 v dx +

|∇ v |2 dx  Ω0+

Ω 0+ q−1

We also have u γ



p (u γ + v ) p − u γ v dx.

Ω0+

By concavity we have (u γ + v )q  u γ + qu γ



γ a+ v (u γ + v )q − uqγ + (u γ + v ) p − u γp v dx.

|∇ v |2 dx 

v in Ω 0+ , so we have



p (u γ + v ) p − u γ v dx.

Ω0+ p −1

 γ −1 zq−1 in Ω 0+ and (u γ + v ) p − u γ  p A 2 p

|∇ v |2 − qa+ zq−1 v 2 dx 

Ω0+

v in Ω0+ , therefore we get

p −1 2

p A2

v dx,

Ω0+

which is a contradiction with (4.1) because  v  H 1 (Ω 0+ ) > 0. 0



2

One immediate corollary from Theorem 1.6 and Theorem 2.11 is the following: Corollary 4.4. If a(x) is admissible and (1.3) holds, then for small γ > 0, there is only one element in S M ,γ with nonpositive energy. If a(x) is admissible and (1.3) holds, for small γ > 0, from the first part of Theorem 1.6 and Lemma 2.7 we see that the least energy solution vˆ γ from Proposition 2.12 has to stay inside S M ,γ , therefore this corollary says that the minimal element U γ ∈ S M ,γ , local minimizer v γ ∈ S M ,γ , another local minimizer v¯ γ ∈ S M ,γ and least energy solution vˆ γ are the same. 5. The second solution From the previous sections we know that there exists a family of local minimizers v γ , γ ∈ (0, Γ ), for the energy functional I γ . Now we want to seek a second solution in the form u = v γ + v with v  0 by means of mountain-pass theorem. We define for any v ∈ E,

J γ (v ) =

1

|∇ v |2 dx +

2 RN

where H (x, v ) =

J γ (v ) =

v 0

a− | v |q+1 −

1 q+1



a− v +

q+1

− H (x, v ) dx,

h(x, s) ds with h(x, v ) = aγ [( v γ + v + )q − v γ ] + [( v γ + v + ) p − v γ ]. Therefore we see that

1 q+1



1 q+1

q

a

−

RN

= Iγ vγ + v



v

 +

 − q +1

dx +

p

1

|∇ v |2 − H (x, v ) dx

2 RN

1

2 − I γ ( v γ ) + ∇ v −  L 2 (RN ) + 2

1 q+1



a− v −

q+1

dx.

RN

Since v γ is a local minimizer due to Theorem 3.3, from the above formula we immediately conclude that v = 0 is a local minimizer of J γ . Lemma 5.1. For any fixed 0 < γ < Γ , there exists δ1 > 0 such that J γ ( v )  J γ (0) = 0 with  v  E < δ1 .

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

583

The following lemma deals with the boundedness of Palais–Smale sequence. For simplification let us denote



−

q +1

a |v |

 q+1 1 dx

RN

by  v  L q+1 (RN ) . a

Lemma 5.2. For γ > 0, suppose { v n } is a sequence in E with J γ ( v n ) → c γ and J γ ( v n ) → 0, then { v γ + v n+ } is uniformly bounded in E. q+1

Proof. First notice that J γ ( v n ) v n− = −(∇ v n− 2L 2 (RN ) +  v n−  q+1 N ), then we have L (R ) a

 − 2 ∇ v  2 n

L (R N )

 q+1 +  v −  q +1 n

La

        J γ ( v n ) ∇ v n−  L 2 (RN ) +  v n− Lq+1 (RN ) (R N ) a  − q+1  − 2       o(1) ∇ v n L 2 (RN ) + v n q+1 N + O (1) . La

So we have

q+1 (1 − o(1))(∇ v n− 2L 2 (RN ) +  v n−  q+1 N ) L a (R )

 o(1), that is v n− → 0 in E.

v n+ , then we reach that

Therefore we may take un = v γ +

I γ (u n ) → I γ ( v γ ) + c γ

(R )

I γ (un ) → 0.

and

It is quite standard to derive the boundedness for this Palais–Smale sequence, so we will be brief. Pick θ such that 2 < θ < 1 p + 1, then p + < θ1 < 12 < q+1 1 . Since I γ ( v γ ) < 0, we have I γ (un ) − θ1 I γ (un )un  c γ + o(1)un  E , that is 1



1 2

−

1



2

|∇ un | dx +

θ RN

 γ

1 q+1

−

1



θ

− q +1

a un





dx +

1

−

θ

RN q +1

a+ u n



1 p+1

p +1

un

dx

RN

dx + c γ + o(1)un  E .

RN

From Young’s inequality we have + q +1

a un Pick

 =



1 +  q +1  q + 1  q+1  qp++11 p −q a  un   un +  p+1 p+1



1 + a

( pp+−1q ) .



 small and C large so that Γ

q+1 p+1



p +1 q +1



1 q+1

−

1

 

θ

1 2



1

θ

−

Overall we reach



1 2

−

1



|∇ un |2 dx +

θ RN

q +1

a− u n

and

p+1

 dx +

RN

which implies that un  E is bounded.



1

1 2



1

θ

−

Γ

p −q



p+1

1 p+1

1 q+1



p +1

un

−

1

 

θ RN

1 + a



( pp+−1q )

 C.

dx  c γ + C + o(1)un  E ,

(5.1)

RN

2

Sobolev embedding from E to L p +1 (R N ) is not compact, so we want to use Concentration-Compactness arguments in [15] to derive an alternative for the classical Palais–Smale condition. We recall the hypothesis on a(x) from the statement of Theorem 1.8:

lim a− (x) = a∞ > 0 and a− (x) < a∞

|x|→∞

for x ∈ R N − B (0, R ),

(5.2)

or

lim inf a− (x) = ∞. |x|→∞

(5.3)

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

When (5.2) holds, we define the following energy functional I ∞ ( v ) for v ∈ E,

I∞(v ) =

1

|∇ v |2 dx +

2 RN

a∞ q+1

| v |q+1 dx −

RN

1 p+1



v+

 p +1

dx.

RN

Lemma 5.3. The following problem

− w = w p − a∞ w q in R N

w  0 in R N , w ∈ E ,

and

(5.4)

has a unique compactly supported solution radial W (x) = W (|x|) and all the other nonzero solutions are combinations of the translation of W . Moreover I ∞ ( w )  I ∞ ( W ) > 0 for any nonzero solution w of (5.4). Proof. See Cortázar, Elgueta and Felmer [7].

2

Now pick some x0 in R N so that supp( W (x − x0 )) ∩ supp( v γ (x)) = ∅ and supp( W (x − x0 ))  Ω − . If (5.2) holds, we also make sure supp( W (x − x0 )) ⊂ R N − B (0, R ). Denote W (x − x0 ) by W 0 . Since p > 1, there exists T > 0, J γ (t W 0 ) < 0 for t  T . Now let

c γ = inf max J γ σ ∈ S γ s∈[0,1]





σ (s) ,

where S γ = {σ ∈ C ([0 1], E ) | σ (0) = 0 and

σ (1) = T W 0 }. Lemma 5.1 asserts cγ  0.

Lemma 5.4. If (5.2) holds, c γ < I ∞ ( W 0 ). Proof. Consider

Jγ





σ (s) = sW 0 , s ∈ [0, T ], noticing that a− (x) < a∞ in RN − B (0, R ), we have

σ (s) = I γ (sW 0 + v γ ) − I γ ( v γ ) = I γ (sW 0 ) + I γ ( v γ ) − I γ ( v γ ) = I γ (sW 0 ) < I ∞ (sW 0 ). p

q

Since − W 0 = W 0 − a∞ W 0 in R N , it is not difficult to see that I ∞ (sW 0 ) achieves its maximum value at s = 1.

2

Proposition 5.5. 1. Assume (5.2) holds, if there exists u γ ∈ S M ,γ so that c γ  I γ (u γ ) − I γ ( v γ ) + I ∞ ( W 0 ), then u γ ∈ / Y . In particular u γ = v γ . 2. Assume (5.2) holds, for γ > 0 small, c γ > 0, and any u γ ∈ S M ,γ , c γ < I γ (u γ ) − I γ ( v γ ) + I ∞ ( W 0 ) if a(x) is admissible and (1.3) holds. Proof. Since c γ  I γ (u γ ) − I γ ( v γ ) + I ∞ ( W 0 ) and c γ < I ∞ ( W 0 ), we have I γ (u γ ) < I γ ( v γ ). If u γ ∈ Y , then I γ (u γ )  I γ ( v γ ) = inf v ∈Y I γ ( v ), it is a contradiction. So u γ ∈ / Y and u γ = v γ since v γ ∈ Y . If a(x) is admissible and (1.3) holds, by Corollary 4.4 we have v γ = v¯ γ , where v¯ γ is from the proof of Theorem 2.5. Recall from Theorem 2.5 there is a small sphere {u ∈ E | u  = γ } around v¯ γ , on which the energy I γ (u ) is positive. So by the definition of c γ we see that c γ > 0 for small γ . Suppose for γ > 0 small, there exists u γ ∈ S M ,γ so that c γ  I γ (u γ ) − I γ ( v γ ) + I ∞ ( W 0 ), then like previous we have I γ (u γ ) < I γ ( v γ ). Noticing that I γ ( v γ ) < 0, Corollary 4.4 asserts that u γ = v γ , which means that c γ  I ∞ ( W 0 ), it is a contradiction. 2 Since our goal is find a second solution to (1.1)γ , by above proposition we may assume that for any u ∈ S M ,γ we have

c γ < I γ (u ) − I γ ( v γ ) + I ∞ ( W 0 ). Under the assumption (5.5), we will show the compactness of Palais–Smale sequence. Lemma 5.6. Assume that un is uniformly bounded in L q+1 (R N ), ∇ un is also uniformly bounded in L 2 (R N ) and

sup y∈ R N

|un |q+1 dx → 0 as n → ∞,

B ( y ,1)

then un → 0 in L α (R N ) for α ∈ (q + 12∗ ) as n → ∞. Proof. See Lions [15].

2

For simplicity we ignore “dx” during integration and omit the sentence “as n → ∞”.

(5.5)

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

585

Lemma 5.7. Suppose γ > 0, { v n } is a sequence in E such that J γ ( v n ) → c γ and J γ ( v n ) → 0, then { v n } contains a strongly convergent subsequence in E. Moreover if v n → v 0  0, then u 0 = v γ + v 0 is a solution to (1.1)γ . Proof. In view of Lemma 5.2, taking un = v n+ + v γ , we have

I γ (u n ) → I γ ( v γ ) + c γ

I γ (un ) → 0.

and

q+1

For simplicity, we denote I γ as I and γ a+ as a+ , still keep I ∞ and c γ . We also use L a q+1 L a (R N ), L q+1 (R N ), L 2 (R N ) and L p +1 (R N ). Therefore we have

I (u n ) → I ( v γ ) + c γ

, L q+1 , L 2 and L p +1 instead of

I ( u n ) → 0.

and

Again from Lemma 5.2, we have that un  E is uniformly bounded, restricting to a subsequence if necessary, there exists u 0 ∈ E so that

un → u 0

weakly in E .

We may assume un → u 0 strongly in L q+1 ( B (0, r )) for any r > 0 and un → u 0 a.e. in R N . By interpolation we have un → u 0 strongly in L t ( B (0, r )) for any r > 0 and q + 1  t < 2∗ . It is easy to see that u 0 is a solution to (1.1)γ with u 0 ∈ S M ,γ , so u 0 has compact support. Moreover u 0  v γ . Now let us consider un1 = un − u 0 , noticing that u 0 is compactly supported, so (un − u 0 )− has bounded support. By q+1

compactness (un − u 0 )− → 0 in L p +1 ∩ L a



2





. Since I (u 0 ) = 0, then ( I (un ) − I (u 0 ))(un − u 0 )− → 0, that is

I (un ) − I (u 0 ) (un − u 0 )− = −∇(un − u 0 )−  L 2 −



q

a un − u 0 (un − u 0 )− − q

RN



p

un − u 0 (un − u 0 )− p

RN

2  = −∇(un − u 0 )− L 2 + o(1),

which implies ∇(un − u 0 )−  L 2 → 0, that is (un − u 0 )− → 0 in E. We now may assume un1 = un − u 0  0 in R N . Since u 0 has compact support, it is easy to see that

a RN



−



q un

q − u 0 (u n

p

p

u n − u 0 (u n − u 0 ) =

RN

RN

(un − u 0 ) p +1 + o(1),

RN

which implies that



a− (un − u 0 )q+1 + o(1),

− u0 ) =



I (un ) − I (u 0 ) (un − u 0 ) = I (un − u 0 )(un − u 0 ) + o(1).

Since ( I (un ) − I (u 0 ))(un

− u 0 ) = o(1), we get

∇(un − u 0 ) 2 +

RN

−

q +1

a (u n − u 0 )

RN

−

(5.6)

(un − u 0 ) p +1 = o(1).

(5.7)

RN

Again since u 0 has compact support, we have I (un ) = I (u 0 ) + I (un − u 0 ) + o(1). With the help of (5.7) we get



I (u n ) = I (u 0 ) + Let us define

1 2

δ = lim sup n→∞

y ∈R N

−

1 p+1



RN

∇(un − u 0 ) 2 +



1 q+1

−

1 p+1



a− (un − u 0 )q+1 + o(1).

(5.8)

RN

1 q+1 u dx. n

B ( y ,1)

  Now if δ = 0, from Lemma 5.6 we have un − u 0  L p+1 → 0, in turn with (5.7) we get RN |∇(un − u 0 )|2 + RN a− (un − u 0 )q+1 → 0, which means that un → u 0 strongly in E. Otherwise we have δ > 0. There exists a sequence { yn1 } ⊂ R N such that

1

|un − u 0 |q+1 dx > δ. 2

B ( yn1 ,1)

Since un → u 0 strongly in L q+1 ( B (0, r )) for any r > 0, we may assume | yn1 | → ∞.

586

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

Now if (5.3) holds, i.e. lim inf|x|→∞ a− (x) = ∞, we have un1  L q+1 → ∞, which is impossible because un  E is uniformly a

bounded. So if (5.3) holds, we must have δ = 0, which means un → u 0 strongly in E. From now on we always assume that (5.2) holds. Since a− is bounded, then E is equivalent to D 1,2 (R N ) ∩ L q+1 . We will use the equivalent norm ∇ u  L 2 + u  L q+1 instead of ∇ u  L 2 + u  L q+1 . a

(u 1 ) → 0. This is quite standard and we will be brief. First we want to show that I (un ) = I (u 0 ) + I ∞ (un1 ) + o(1) and I ∞ n In fact since lim|x|→∞ a− = a∞ , u 0 has compact support and un → u 0 strongly in L q+1 ( B (0, r )) ∩ L p +1 ( B (0, r )) for any r > 0, we can easily get I (un ) = I (u 0 ) + I ∞ (un1 ) + o(1). As for the second part, since u 0 has compact support, we can pick r > 0 such that supp(u 0 )  B (0, r ). For any ϕ ∈ C 0∞ (R N ), we have



I∞ un1







ϕ − I (un )ϕ − I (u 0 )ϕ =





q



q 

a∞ (un − u 0 )q − a− un − u 0 + a+ un − u 0

B (0,r )





q

a∞ − a− un ϕ dx −

+

q

R N − B (0,r )



q

ϕ dx

p p (un − u 0 ) p + u 0 − un ϕ dx.

B (0,r )

Since lim|x|→∞ a− = a∞ , we have



I∞ un1







ϕ − I (un )ϕ − I (u 0 )(ϕ ) = o(1)ϕ  E .

(u 1 ) → 0. Noticing that I (un )ϕ = o(1)ϕ  E and I (u 0 ) = 0, we have I ∞ n Let us define v n1 (x) = un1 (x + yn1 ) = un1 ( . + yn1 ). It is easy to see that









I ∞ v n1 = I ∞ un1 = I (un ) − I (u 0 ) + o(1)

Assume v n1 → v 1 weakly in E for some v 1 ∈ E, since





I∞ v n1 → 0.

and 

B (0,1)

(5.9)

| v n1 |q+1 dx > 12 δ , it follows that

1 q+1 1 v dx  δ, 2

B (0,1)

which implies that v 1 = 0. It is easy to see that v 1 is a solution to (5.4), so v 1 has compact support. Moreover I ∞ ( v 1 )  I ∞ ( W 0 ). As we show (un − u 0 )− → 0 in E, in the same way we can that ( v n1 − v 1 )− → 0 in E . So we may assume v n1  v 1 . Since v 1 has compact support, similar to (5.6) we have









I∞ v n1 − I ∞ v1







v n1 − v 1 = I ∞ v n1 − v 1

( v 1 ) − I ( v 1 ))( v 1 − v 1 ) = o (1), we get Since ( I ∞ n n ∞

 1  ∇ v − v 1 2 +

n

RN



a∞ v n1 − v 1

q+1

−

RN







v n1 − v 1 + o(1).

v n1 − v 1

 p +1

= o(1).

RN

Again since v 1 has compact support, we have I ∞ ( v n1 ) = I ∞ ( v 1 ) + I ∞ ( v n1 − v 1 ) + o(1). Therefore we get









I ∞ v n1 = I ∞ v 1 +

which implies that

I ∞ ( v n1 )



1 2

−

1



p+1

 1  ∇ v − v 1 2 + n

RN



1 q+1

−

1 p+1





a∞ v n1 − v 1

q+1

+ o(1),

RN

 I ∞ ( v ) + o(1). Overall from (5.9) we have 1













I (un ) − I (u 0 ) + o(1) = I ∞ un1 = I ∞ v n1  I ∞ v 1 + o(1)  I ∞ ( W 0 ) + o(1). Since I (un ) = I ( v γ ) + c γ + o(1), we have

c γ + I ( v γ )  I (u 0 ) + I ∞ ( W 0 ), which contradicts the assumption (5.5). So we must have δ = 0, which means that un → u 0 strongly in E. Since c γ < I ∞ ( W 0 ), we see that u 0 is different from v γ . 2 Now we want to complete the proof of Theorem 2.5 and Proposition 2.12, that is to show the compactness of the Palais–Smale sequence.

Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

587

Proof of Theorem 2.5 and Proposition 2.12. Recall that from the proof of Theorem 2.5, for

 αγ = Inf I γ (u ) u  E  γ 

We also know that I γ (u )  such that

I γ (un ) → αγ

and

1 8

γ > 0 small,

 B γ = u ∈ E u  E < γ . 

and

γ 2 on ∂ B γ . Therefore by Ekeland’s variational principle, there exists a sequence {un } ⊂ B γ

I γ (un ) → 0.

Notice that we may assume un  0. Since un  E < γ , we may assume un converges weakly to v¯ γ in E with  v¯ γ  E  γ . Moreover we have I γ ( v¯ γ )  αγ . v¯ γ is a solution to (1.1) and v¯ γ has compact support. Now we repeat the first part of the proof of Lemma 5.7, such as (5.6), (5.7) and (5.8), we may assume un  v¯ γ and obtain



αγ + o(1) = I γ (un ) = I γ ( v¯ γ ) +

1 2

1

−

p+1





∇(un − v¯ γ ) 2 +



RN

1 q+1

−



1 p+1



a− (un − v¯ γ )q+1 + o(1).

RN

1 1 Since I γ ( v¯ γ )  αγ , we have o(1)  ( 12 − p + ) RN |∇(un − v¯ γ )|2 + ( q+1 1 − p + ) RN a− (un − v¯ γ )q+1 , which implies un → v¯ γ 1 1 strongly in E. As the proof of Proposition 2.12 we can do the same thing and we derive that



αˆ γ + o(1) = I γ ( vˆ γ ) +

1 2

−

1 p+1



∇(un − vˆ γ ) 2 +

RN



1 q+1

−



1 p+1





a− (un − vˆ γ )q+1 .

RN

1 1 ˆ γ , we have o(1)  ( 12 − p + Since I γ ( vˆ γ )  α ) RN |∇(un − vˆ γ )|2 + ( q+1 1 − p + ) RN a− (un − vˆ γ )q+1 , which implies un → vˆ γ 1 1 strongly in E. 2

From the proof above we see that the compactness of the support of u 0 and v 1 dramatically reduce the complexity of the proof. In order to prove Theorem 1.8, we need the following theorem. Theorem 5.8. Assume (5.5) holds, suppose c γ = 0 and there exists ηγ > 0 such that for any ρ ∈ [0, ηγ ]







inf J γ ( v )  v  E = ρ = 0. Then for each ρ ∈ (0, ηγ ), the problem (1.1)γ has a solution with u − v γ  E = ρ . Proof. Since (5.5) holds, from Lemma 5.7 we know that Palais–Smale sequence is compact. Now for any fixed ρ ∈ (0, ηγ ), the set F = ∂ B (0, ρ ) in E satisfies the hypothesis of theorem (1) in Ghoussoub and Preiss [10], their theorem (1.bis) asserts the existence of a solution for each ρ ∈ (0, ηγ ) with the compactness of the Palais–Smale sequence. 2 We want to point out that, by Proposition 5.5, c γ > 0 for proof of Theorem 1.8.

γ > 0 small if a(x) is admissible and (1.3) holds. Here is the

Proof of Theorem 1.8. If (5.5) does not hold, by Proposition 5.5 we already have a second solution. Therefore for the following proof we assume (5.5) holds. If there exists some ρ < δ1 such that inf{ J γ ( v ) |  v  E = ρ } > 0, we have c γ > 0. By the mountain-pass theorem of Ambrosetti and Rabinowitz, there exists a solution V γ of (1.1)γ with J γ ( V γ ) > 0, i.e. I γ ( V γ ) > I γ ( v γ ), which implies that V γ is different from v γ . If not, but c γ > 0, we still can extract a Palais–Smale sequence by applying Theorem 4.3 from Mawhin and Willem [18] and have the same result like above. If not and c γ = 0, then for all ρ ∈ [0, δ1 ), we have inf{ J γ ( v ) |  v  E = ρ } = 0, then from Theorem 5.8 we see that there are infinite many solutions of (1.1)γ . 2 In the end we point out that Proposition 1.9 is the result of Corollary 4.4, so we omit the proof. 6. Radial symmetry In this section we are going to use the method of moving plane by Gidas, Ni and Nerenberg [11] to prove the radial symmetry of the solutions of (1.1). Since γ is irrelevant in this section, we omit it. First assume a(x) = a(|x|), then let us denote r1 = sup{r  0 | a(r )  0}. We should point out that we do not assume here that Ω 0+ is not empty, so r1 could be

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zero. We make the following assumption on a:

a(r ) is decreasing in [0, r1 ] and is strictly decreasing in [r1 , ∞).

(6.1)

We will present a list of lemmas to prove this theorem. Taking any nonzero solution of (1.1), say u, by Theorem 1.1 u is compact supported. Let r2 = sup{|x|  0 | u (x) > 0}, it is clear r1 < r2 < ∞ and u (∂ B (0, r2 )) = 0. Let us consider now an arbitrary direction τ , which for simplicity we can assume to be τ = e 1 . For λ  0 we define

  Σλ = (x1 , x2 , . . . , xn ) x1 > λ and T λ = ∂Σλ .

For x ∈ R N , let xλ = 2(λ − x1 )e 1 + x be the reflection of x with respect to T λ . We define the functions u λ , aλ , w λ : R N → R as

  u λ (x) = u xλ ,

  aλ (x) = a xλ ,

w λ = u λ (x) − u (x).

Any x ∈ Σλ , it is easy to see that aλ (x)  a(x). We set

  λ0 = inf λ  0 w λ (x)  0, ∀x ∈ Σλ .

Clearly since u has compact support, λ0 < ∞. Since u is nonzero, rotating the axis if necessary, with the help of strong maximum principle and Hopf’s lemma we may assume that u (r3 e 1 ) > 0 for some r3 ∈ (r1 , r2 ). Lemma 6.1. Σλ0 ∩ B (0, r2 ) = ∅, i.e. λ0 < r2 . Proof. From strong maximum principle and Hopf’s lemma, noticing that u (r3 e 1 ) > 0, by definition of r2 we see that a(r2 ) < 0. Since we also have u (∂ B (0, r2 )) = 0, then there exist δ1 > 0 and σ1 > 0 such that

a− (r )  σ1 , p −q A1

r ∈ [r2 − δ1 , r2 ] and

u (x)  A 1 ,

x ∈ Σr2 −δ1 ∩ B (0, r2 ),

σ1 q

where = p and the second part is just by continuity of u at r2 e 1 . Now if there exists x1 ∈ Σr2 −δ1 ∩ B (0, r2 ) such that w r2 −δ1 (x1 ) < 0. Since w r2 −δ1 (x)  0 on ∂[Σr2 −δ1 ∩ B (0, r2 )], we may assume w r2 −δ1 attains minimum value at x1 . Therefore we have



q

p







0  − w r2 −δ1 (x1 ) = ar2 −δ1 (x1 )u r2 −δ1 (x1 ) + u r2 −δ1 (x1 ) − a(x1 )u q (x1 ) + u p (x1 )

    q p  a(x1 )ur2 −δ1 (x1 ) + ur2 −δ1 (x1 ) − a(x1 )uq (x1 ) + u p (x1 ) > 0,

where the last two inequalities are true because ar2 −δ1 (x1 ) > a(x1 ), A 1 > u (x1 ) > u r2 −δ1 (x1 )  0 and a− (x1 )  σ1 . This is a contradiction, this lemma is proven. 2 Now if λ0 = 0, Theorem 1.10 is proved. But if λ0 > 0, by definition of λ0 , there exists a sequence {λk } such that

λk < λ0 and λk ↑ λ0 and the function w λk possesses a negative minimum value at xk ∈ Σλk ∩ B (0, r2 ). It follows that

∇ w λk (xk ) = 0 and xk ∈ Σλk ∩ B (0, r2 ).

w λk (xk ) < 0,

Consequently the sequence {xk } is bounded, restricting to a subsequence if necessary, we can assume that it converges to x¯ ∈ Σλ0 ∩ B (0, r2 ), which means that w λ0 (¯x)  0. Moreover we have

w λ0 (¯x) = 0 and

∇ w λ0 (¯x) = 0.

Lemma 6.2. If a− (¯x) > 0, then u (¯x) > 0. Proof. Let us assume otherwise u (¯x) = 0. Noticing that a− (¯x) > 0, by continuity we may assume for large k, that there exist σ2 > 0 and A 2 > 0 with A 2p−q = σp2 q such that

a− (xk )  σ2

u (xk )  A 2 .

and

Now for large k, w λk (x)  0 on ∂[Σλk ∩ B (0, r2 )]. But



q

p







0  − w λk (xk ) = aλk (xk )u λ (xk ) + u λ (xk ) − a(xk )u q (xk ) + u p (xk ) k

k

      q p since aλk (xk )  a(xk )  a(xk )u λk (xk ) + u λk (xk ) − a(xk )uq (xk ) + u p (xk )   > 0 since A 2  u (xk ) > u λk (xk )  0 and a− (xk )  σ2 .

This is a contradiction.

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

589

Lemma 6.3. If a− (¯x) > 0 and x¯ ∈ Σλ0 ∩ B (0, r2 ), then w λ0 (x) ≡ 0 for all x ∈ B (¯x,  ), where  is a small positive number. Proof. From above lemma we see that u (¯x) > 0, so for small

B (¯x,  )  Σλ0 ∩ B (0, r2 ) and

inf

x∈ B (¯x, )

 > 0 we have

u (x) > 0.

Since w λ0 (x)  0 in Σλ0 ∩ B (0, r2 ), we see that infx∈ B (¯x, ) u λ0 (x) > 0. Now in B (¯x,  ), noticing that aλ0 (x)  a(x), we get

      q p  q p  − w λ0 + aλ0 (x)uq + u p − aλ0 (x)u λ0 + u λ0  − w λ0 + a(x)uq + u p − aλ0 (x)u λ0 + u λ0 = 0. So we obtain

 q p  u q − u λ0 u p − u λ0 − w λ0 + aλ0 (x) + w λ0  0 in B (¯x,  ). u λ0 − u u λ0 − u q

Also noticing that c (x) = aλ0 (x)

u q −u λ

0

u λ0 −u

p

+

u p −u λ

0

u λ0 −u

is uniformly bounded in B (¯x,  ) and w λ0  0 in B (¯x,  ), by the strong

maximum principle w λ0 (x) ≡ 0 for all x ∈ B (¯x,  ) since w λ0 (¯x) = 0.

2

Similarly we also have the following result. Lemma 6.4. If a− (¯x) > 0 and x¯ ∈ T λ0 ∩ B (0, r2 ), then w λ0 (x) ≡ 0 for all x ∈ B (¯x,  ) ∩ Σλ0 , where  is a small positive number. Proof. Following the same steps as above, if w λ0 (x) is not identically zero inside B (¯x,  ) ∩ Σλ0 , by strong maximum principle w λ0 (x) > 0 for all x ∈ B (¯x,  ) ∩ Σλ0 . Since u (¯x) > 0, then x¯ ∈ B (0, r2 ) ∩ T λ0 . Applying Hopf’s lemma, we have which contradicts the fact that ∇ w λ0 (¯x) = 0. 2

∂ w λ0 (¯x) ∂n

< 0,

Lemma 6.5. If r1 > 0, then λ0  r1 . Proof. Suppose otherwise λ0 > r1 , then by definition of r1 we find out that a− (¯x) > 0. From Lemma 6.2 we see that u (¯x) > 0, which means that x¯ ∈ / ∂ B (0, r2 ). From Lemma 6.3 and Lemma 6.4 we see that, x¯ ∈ Σλ0 ∩ B (0, r2 ) or x¯ ∈ B (0, r2 ) ∩ T λ0 , both cases imply that w λ0 (x) ≡ 0 in Σλ0 ∩ B (¯x,  ) for some small positive  . Taking x0 ∈ B (¯x,  ) ∩ Σλ0 , making x0 close to x¯ so that u (x0 ) > 0 and a(x0 ) < 0, then we have q

p

a(x0 )u q (x0 ) + u p (x0 ) = −u (x0 ) = −u λ0 (x0 ) = aλ0 (x0 )u λ0 (x0 ) + u λ0 (x0 ). Since w λ0 (x0 ) = 0, u (x0 ) > 0 and a(x0 ) < aλ0 (x0 ), it is a contradiction. So we have λ0  r1 .

2

From above lemma, we immediately have the following interesting result: Proposition 6.6. Under the assumption (6.1), if r1 > 0, then any nonzero solution of (1.1) is positive in B (0, r1 ). Proof. From the above lemma, we see that λ0  r1 . Since u (r3 e 1 ) > 0 and r3 > r1 , we have u (r1 e 1 ) > 0. By continuity, strong maximum principle and Hopf’s lemma we have u > 0 in B (0, r1 ). 2 Lemma 6.7. If r1 = 0, then λ0 = 0. Proof. In this case a(r ) is strictly decreasing, we can repeat the proof of Lemma 6.5 to conclude that λ0 = 0.

2

Now we give the proof of Theorem 1.10. Proof of Theorem 1.10. If r1 = 0, then from previous lemma we have λ0 = 0. So we assume r1 > 0, then from Proposition 6.6 there exists δ3 > 0 such that u (x) > 0 for all x ∈ B (0, r1 + δ3 ). If a(¯x) < 0, that is a− (¯x) > 0, from Lemma 6.2 we see that u (¯x) > 0, which means that x¯ ∈ / ∂ B (0, r2 ). Now we can repeat the same arguments of Lemma 6.5 to get a contradiction. If a(¯x)  0, then u (¯x) > 0 and x¯ ∈ B (0, r1 + δ3 ) ∩ Σλ0 or x¯ ∈ B (0, r1 ) ∩ T λ0 . We can apply the same arguments of Lemma 6.3 and Lemma 6.4 over the domain B (0, r1 + δ3 ) ∩ Σλ0 , and conclude that

u (x) ≡ 0

for all x ∈ B (0, r1 + δ3 ) ∩ Σλ0 .

Hence we can pick a point x∗ ∈ B (0, r1 + δ3 ) ∩ Σλ0 in the e 1 axis such that

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Q. Lu / J. Math. Anal. Appl. 389 (2012) 569–590

  a x∗ < 0

and

    u x∗ = u λ0 x∗ > 0,

then we have aλ0 (x∗ ) > a(x∗ ), but we also have

            q   p   a x∗ u q x∗ + u p x∗ = −u x∗ = −u λ0 x∗ = aλ0 x∗ u λ0 x∗ + u λ0 x∗ , it is a contraction. Overall we have λ0 = 0.

2

In the end we want to point out that if f (s) ∈ C 1 (R+ , R+ ) with f (0) = 0, then Theorem 1.10 holds with u p replaced by f (u ). The same proof works because f (s) − sq is strictly decreasing inside a small neighborhood around 0. Acknowledgments This paper is supported by Proyecto Fondecyt Posdoctorado No. 3100050. I would like to thank Professor Juan Dávila for useful discussions during the preparation of this paper, especially for the proof of Theorem 1.6; and Professor Stanley Alama for reading the preprint of this paper and giving useful suggestions during my stay in McMaster University. I also want to give my thanks to the reviewer for the useful comments and suggestions.

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