Multiple solutions for inhomogeneous critical semilinear elliptic problems

Nonlinear Analysis 68 (2008) 2569–2593 www.elsevier.com/locate/na Multiple solutions for inhomogeneous critical semilinear elliptic problems Youyan W...

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Nonlinear Analysis 68 (2008) 2569–2593 www.elsevier.com/locate/na

Multiple solutions for inhomogeneous critical semilinear elliptic problems Youyan Wan a,b , Jianfu Yang c,∗ a Wuhan Institute of Physics and Mathematics, The Chinese Academy of Sciences, P.O. Box 71010, Wuhan, Hubei, 430071, PR China b The College of Mathematics and Computer, Jianghan University, Wuhan, Hubei, 430056, PR China c Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, PR China

Received 20 May 2005; accepted 6 February 2007

Abstract In this paper, we consider the existence of multiple positive solutions for an inhomogeneous critical semilinear elliptic problem. We show that the problem possesses at least four positive solutions. c 2007 Elsevier Ltd. All rights reserved.

Keywords: Multiple positive solution; Critical exponent; Existence

1. Introduction This paper is concerned with the existence of multiple positive solutions of the following nonhomogeneous critical semilinear elliptic equations:  ∗ −∆u + a(x)u = u 2 −1 + f (x) x ∈ Ω , (1.1) u>0 x ∈ Ω,  u ∈ H01 (Ω ), where Ω ⊂ R N is a bounded domain, N ≥ 3, 2∗ = It is well known that the critical problem  ∗ −∆u = u 2 −1 x ∈ Ω , u>0 x ∈ Ω,  u ∈ H01 (Ω )

2N N −2 ,

and f (x) ∈ H −1 (Ω ), f (x) ≥ 0.

(1.2)

has no solution via Pohozaev identity if Ω ⊂ R N is a bounded star-shaped domain. However, the situation is different if there is a lower perturbation term f (x, u): ∗ Corresponding author.

E-mail address: [email protected] (J. Yang). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.02.005

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 ∗ −∆u = u 2 −1 + f (x, u) u>0  u ∈ H01 (Ω ).

x ∈ Ω, x ∈ Ω,

(1.3)

In [5], Brezis and Nirenberg have made great contributions on this aspect; they obtained one solution of (1.3) using the mountain pass theorem. For the nonhomogeneous critical semilinear elliptic equation, ∗ −∆u = |u|2 −2 u + f (x) x ∈ Ω , (1.4) u ∈ H01 (Ω ). Tarantello [9] showed that problem (1.4) has at least two solutions. The main idea is to divide the Nehari manifold Λ = {u ∈ H01 (Ω ); hI 0 (u), ui = 0} into three parts, Λ+ , Λ− and Λ0 , and to use the Ekeland variational principle to get one solution for Λ+ and another solution for Λ− . The effect of a(x) for the existence of solutions was studied by Benci, Cerami [4] and Passaseo [6]. In [4], Benci and Cerami considered the existence of solutions for the problem  ∗ −∆u + a(x)u = |u|2 −2 u x ∈ R N , (1.5) u>0 x ∈ RN ,  1,2 N u ∈ D (R ). They proved that problem (1.5) has at least one positive solution under some assumptions on a(x). In [6], Passaseo considered the problem  ∗ −∆u + a(x)u = u 2 −1 x ∈ Ω , u>0 x ∈ Ω,  u ∈ H01 (Ω ), N

where a(x) = α(x) ¯ + λ2 α(λ(x − x0 )), α, ¯ α ∈ L 2 (R N ), α, ¯ α ≥ 0, kαk kαk L

N 2

(R N )

< S(2

2 N

L

N 2

(R N )

(1.6)

6= 0, x0 is a fixed point of Ω ,

− 1), and λ is a positive parameter. Passaseo showed that problem (1.6) has at least two positive

solutions if λ > λ¯ for some λ¯ > 0. The arguments used are the quantitative deformation lemma and the comparison N N of levels of the energy functional. Furthermore, the energy of these two solutions belongs to ( N1 S 2 , N2 S 2 ), where S is the best Sobolev constant. On the other hand, for the subcritical problem in the whole space, Bahri and Li showed in [3] that the problem  p−1 u = 0 x ∈ RN , −∆u + u − q(x)|u| u>0 x ∈ RN ,  1 N u ∈ H (R ), +2 ∞ N where 1 < p < N N −2 if N ≥ 3; 1 < p < +∞ if N = 1, 2, and q ∈ L (R ), has a positive solution provided that q(x) satisfies further conditions. Meanwhile, in [11] Zhu obtained two positive solutions for the problem  p N −∆u + u = u + f (x) x ∈ R , N u>0 x ∈R ,  u ∈ H 1 (R N ), N +2 where f (x) ∈ L 2 (R N ), f (x) ≥ 0, f (x) 6≡ 0, 1 < p < N −2 if N ≥ 3; 1 < p < +∞ if N = 2. One of the solutions in [11] is a local minimum of the energy functional; the other one was obtained using the mountain pass theorem. Recently, Adachi and Tanaka [2] considered the following problem:  p N −∆u + u = a(x)u + f (x) x ∈ R , (1.7) u>0 x ∈ RN ,  1 N u ∈ H (R ),

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+2 −1 N where 1 < p < N N −2 if N ≥ 3; 1 < p < +∞ if N = 1, 2, f (x) ∈ H (R ), f (x) ≥ 0, f (x) 6≡ 0, and N a(x) ∈ C(R ). They obtained four positive solutions for problem (1.7). The first solution is a local minimum of the corresponding energy functional near zero; the second and the third solutions are obtained using the Ljusternik–Schnirelman category, and the last one is found using the method of Bahri and Li in [3]. In this paper we consider the existence of multiple positive solutions of problem (1.1); it involves the critical exponent, the effect of the coefficient a(x), and the perturbation term f (x). Assume that a(x) is a non-negative function satisfying the following conditions:

(A1 ) a(x) ∈ L ∞ (Ω ); (A2 ) there exists Ω1 ⊂⊂ Ω such that a(x) = 0 for x ∈ Ω1 ; N (A3 ) a(x) = α(x) ¯ + λ2 α(λ(x − x0 )), where α, ¯ α ∈ L 2 (R N ), α, ¯ α ≥ 0, kαk and λ is a positive parameter.

L

N 2

(R N )

6= 0, x0 is a fixed point in Ω ,

Our main result is as follows. Theorem 1.1. Suppose k f k H −1 (Ω ) is sufficiently small. (i) If (A1 ) holds, then (1.1) has at least a solution u loc . (ii) Furthermore, if f ∈ L p (Ω ), p > N2 and (A2 ) holds, then (1.1) has another two solutions u 2 , u 3 . (iii) Moreover, if we assume further (A3 ) and 2

kαk L

N 2

(R N )

< S(2 N − 1),

(1.8)

then there exists λ¯ > 0 such that, for λ > λ¯ , (1.1) has one more solution u 4 . Consequently, problem (1.1) has at least four positive solutions under the hypotheses (A1 )–(A3 ). Solutions of (1.1) will be found as critical points of the functional Z Z Z 1 1 1 ∗ a(x)u 2 dx − ∗ u 2+ dx − I f (u) = kuk2H 1 (Ω ) + f udx, 2 2 Ω 2 Ω 0 Ω where Z u + = max{u(x), 0},

kuk H 1 (Ω ) = 0

2

1 2

|∇u| dx Ω

.

To obtain critical points of I f by critical point theory, one needs to verify the (PS) condition for I f . A sequence {u n } ⊂ H01 (Ω ) is said to satisfy the (PS)c if I f (u n ) → c and I 0f (u n ) → 0 imply that {u n } contains a convergent ∗ subsequence. In the critical case, the (PS) condition is not valid since the imbedding of H01 (Ω ) into L 2 (Ω ) is not compact. However, the (PS)c condition holds for c belonging to certain intervals. We shall find solutions in the first and second intervals, as stated below. The first solution u loc (x) will be found as a local minimum of I f (u) near zero. If f (x) 6≡ 0, then u loc (x) is a positive solution of (1.1) and I f (u loc ) < 0. Let J f (v) = max I f (tv) t>0

be a functional defined on Σ+ , where Σ+ = {v ∈ Σ ; v+ 6≡ 0}, Σ = {v ∈ H01 (Ω ); kvk H 1 (Ω ) = 1}. We shall prove 0

that the critical points of I f : H01 (Ω ) → R and J f : Σ+ → R are related to solutions of (1.1). To get the second and third solutions u 2 and u 3 of (1.1), we need a result of global compactness, namely that there exists a K ∈ N such that the (PS)c condition for I f (u) and J f (v) breaks down if and only if c = N I f (u 0 (x)) + Nk S 2 , k = 1, 2, . . . , K , where u 0 (x) is a critical point of I f (u). Furthermore, we see that I f (u) and N

N

N

J f (v) satisfy the (PS)c condition for c 0 and k f k H −1 (Ω ) are small enough. Next, denote [J f ≤ c] = {u ∈ Σ+ ; J f (u) ≤ c}. We then show that [J f ≤ I f (u loc (x)) +

1 N

N

S 2 − ε] is not empty and

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cat

J f ≤ I f (u loc (x)) +

1 N S2 −ε N

≥2

(1.9)

for ε > 0 small, where cat stands for Ljusternik–Schnirelman category. So we obtain another two positive solutions N u 2 and u 3 with 0 < I f (u i ) < I f (u loc (x)) + N1 S 2 , i = 2, 3. In order to prove (1.9), inspired by [2], we need to construct two mappings, 1 N N −1 N F:S = {y ∈ R ; |y| = 1} → J f ≤ I f (u loc (x)) + S 2 − ε , N 1 N G : J f ≤ I f (u loc (x)) + S 2 − ε → S N −1 N such that F ◦ G is homotopic to the identity. Finally, the last solution u 4 will be obtained by using the quantitative deformation lemma and comparing the levels of the energy functional on the manifold M = {u ∈ H01 (Ω ); kuk2∗ = 1}, which is the argument used by Passaseo [6]. The paper is organized as follows. In Section 1, we show the existence of a local minimum of I f and prove some preliminary results; the second and third solutions will be found in Section 2 by Ljusternik–Schnirelman theory; Finally, we show that there is a high-energy solution in Section 4. 2. Preliminaries and local minimum In this section, we collect preliminary facts for future reference, and we show the existence of a local minimum for the functional I f related to (1.1). Denote by E = H01 (Ω ) the usual Sobolev space and by E ∗ the dual space of E. Lemma 2.1. Assume f ∈ E ∗ and (A1 ). Then: (i) I f ∈ C 2 (E, R) and Z

Z Z 2∗ −1 a(x)uϕdx − u+ ϕdx − f ϕdx, Ω Ω Ω Z Z 2∗ −2 2 00 2 2 ∗ I f (u)(ϕ, ϕ) = kϕk E + a(x)ϕ dx − (2 − 1) u+ ϕ dx

I 0f (u)ϕ = hu, ϕi E +

Ω

(2.1) (2.2)

Ω

for all ϕ ∈ E. (ii) Assume that u ∈ E is a critical point of I f (u). Then u is a non-negative solution of (1.1). Moreover, u is a positive solution of (1.1) if u 6≡ 0 or f 6≡ 0. Proof. The proof of (i) is standard. Now we prove (ii). Let u ∈ E be a critical point of I f , then Z Z 2 2 (|∇u − | + a(x)u − )dx = f u − dx ≤ 0 Ω

Ω

since f (x) ≥ 0. This implies that u− = 0

a.e in Ω ,

thus u ≥ 0. The positivity of u in the cases u(x) 6≡ 0 or f (x) 6≡ 0 follows from the maximum principle.

We show in the following lemma the existence of a local minimum of I f . Lemma 2.2. Assume (A1 ) and let f ∈ E ∗ . Then there exist r > 0 and d > 0 such that: (i) I f (u) is strictly convex in Br = {u ∈ E; kuk E < r }; (ii) If k f k E ∗ ≤ d, then we have infkuk E =r I f (u) > 0. Moreover, I f (u) has a unique critical point u loc ( f ; x) in Br , that is I f (u loc ( f ; x)) = inf I f (u). u∈Br

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

Proof. Since I 00f (u)(ϕ, ϕ)

≥

− (2 − 1)

kϕk2E

∗

= [1 − (2 − 1)S ∗

Z

2∗

2∗ −2 Z ∗ 2

|u| dx Ω ∗

− 22

∗

|ϕ|2 dx

2 2∗

Ω

∗ kuk2E −2 ]kϕk2E , 2∗

1

for all ϕ ∈ E, I 00f (u) is positive definite in u ∈ Br provided that r < [(2∗ − 1) · S − 2 ]− 2∗ −2 , and then I f is strictly convex in Br , (i) follows. Next, we prove (ii). For kuk E = r , there holds ∗ 1 2 S −2 /2 2∗ 1 1 ∗ r 2 − r k f k E∗ . I f (u) ≥ r − r − r k f kE = − 2 2∗ 2 2∗ (2∗ − 1) Hence, there exists d > 0 such that infkuk E =r I f (u) > 0 if k f k E ∗ ≤ d. Since I f (u) is strictly convex in Br and infkuk E =r I f (u) > I f (0) = 0, we see that I f (u) has a unique critical point u loc ( f ; x) in Br satisfying I f (u loc ( f ; x)) =

inf

kuk E
I f (u).

We remark that u loc (0; x) = 0. The following lemma shows that u loc ( f ; x) actually bifurcates from the trivial solution if k f k E ∗ is the parameter. Lemma 2.3. ku loc ( f ; x)k E → 0 as k f k E ∗ → 0. Proof. We derive from I 0f (u loc ) = 0 that Z Z N +2 1 1 a(x)u 2loc dx − f u loc dx. I f (u loc ) = ku loc k2E + N N Ω 2N Ω Since I f (u loc ) ≤ 0, we have (N + 2) k f k E ∗ ku loc ( f ; x)k E 2

ku loc ( f ; x)k2E ≤ which yields

(N + 2) k f kE ∗ . 2 The assertion follows. ku loc ( f ; x)k E ≤

To find the second and third solutions of (1.1), we shall use the theory of Ljusternik–Schnirelman category on manifolds. We then need to exhibit properties of the functional J f . Recall that the best Sobolev constant S can be characterized as kuk2E

S := inf

06=u∈E

kuk22∗

,

1 R where kuk p = ( Ω |u| p dx) p . It is well known that

U (x) =

[N (N − 2)] [1 + |x|2 ]

N −2 4

N −2 2

solves ∗ −1

−∆u = u 2

in R N

Z

Z

(2.3)

and |∇U |2 dx = RN

∗

RN

N

U 2 dx = S 2 .

(2.4)

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

See, for instance, Theorem 1.42 and Lemma 1.46 of [10]. Let ξ(x) ∈ C0∞ (R N ) be a cut-off function such that 0 ≤ ξ ≤ 1, ξ(x) ≡ 1 ∀x ∈ B(0, η), ξ(x) ≡ 0 ∀x ∈ B(0, 2η)c . For > 0, x ∈ R N , we define 2−N x−z , if z ∈ R N ; u ε,z (x) = ξz (x)Uε,z (x) if z ∈ Ωη , Uε,z (x) = ε 2 U ε where Ωη := {x ∈ Ω , dist(x, ∂Ω ) > 2η, η > 0}, ξa (x) = ξ(x − a). If z = 0, we denote u ε = u ε,0 , Uε = Uε,0 . It is known that (see, for example, Lemma 1.46 of [10]) Z Z N 2 |∇Uε |2 dx + O(ε N −2 ) = S 2 + O(ε N −2 ), (2.5) |∇u ε | dx = RN Ω Z Z N ∗ ∗ |Uε |2 dx + O(ε N ) = S 2 + O(ε N ). |u ε |2 dx = (2.6) RN

Ω

Now we define the functional I0 : E → R by Z Z 1 1 1 ∗ a(x)u 2 dx − ∗ u 2 dx. I0 (u) = kuk2E + 2 2 Ω 2 Ω + Let J0 (v) = maxt>0 I0 (tv). Lemma 2.4. Assume (A1 ). Then infv∈Σ+ J0 (v) is not achieved and infv∈Σ+ J0 (v) = Proof. First, we show that infv∈Σ+ J0 (v) = ! ∗1 R 1 + Ω a(x)v 2 dx 2 −2 R 2∗ Ω v+ dx

t0 =

or

1 N

N

S 2 . Indeed, for any v ∈ Σ+ , solving

1 N

N

S2.

dI0 (tv) dt

= 0, we get

t = 0.

Since I0 (t0 v) > I0 (0) = 0 and I0 (tv) → −∞, as t → +∞, we have R N 1 1 + Ω a(x)v 2 dx 2 J0 (v) = I0 (t0 v) = . R 2∗ N 2−2 N v dx Ω +

(2.7)

By the Sobolev inequalities, J0 (v) ≥

1 N

Z

∗

Ω

2 v+ dx

− N −2 2

≥

1 N

Z

∗

|v|2 dx Ω

− N −2 2

≥

1 N S2, N

(2.8)

namely, inf J0 (v) ≥

v∈Σ+

Let v =

1 N S2. N

(2.9)

u ε,z ku ε,z k E ,

J0

where z ∈ Ω is fixed and B(z, 2η) ⊂ Ω . By (2.5)–(2.7), u ε,z u ε,z = max I0 t t>0 ku ε,z k E ku ε,z k E !N R 2 dx 2 a(x)u 1 N ε,z S 2 + O(ε N −2 ) 1+ Ω . = N ku ε,z k2E

Since Z

p

RN

Uε,z dx = O(ε N −

(N −2) p 2

)

(2.10)

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593 p for all p ∈ ( NN−2 , N2N −2 ) and a ∈ L (Ω ) for all 1 ≤ p < ∞, choosing p > deduce Z N −N +2 a(x)u 2ε,z dx ≤ kak p kUε,z k L 2 p0 (R N ) = O(ε p0 ),

N 2

if N > 3 and

3 2

< p < 3 if N = 3, we (2.11)

Ω

p , 0 < Np0 − N + 2 < N − 2. Whence, by (2.10), we have where p0 = p−1 N u ε,z 1 N −N +2 J0 ). ≤ S 2 + O(ε p0 ku ε,z k E N

(2.12)

Hence, inf J0 (v) =

v∈Σ+

1 N S2. N

Secondly, suppose, by contradiction, that there were v ∈ Σ+ satisfying 1 N S2. N We see from (2.8) that Z − N −2 2 1 N 1 1 N ∗ S2 ≤ ≤ S2. |v|2 dx N N N Ω J0 (v) =

This implies that Z − N −2 · 2 N 2 kvk2E 2∗ S= = |v| dx . kvk22∗ Ω This contradicts the fact that S is not achieved on a bounded domain in R N . The conclusion follows.

We compare J f with J0 as follows. Lemma 2.5. Assume (A1 ) and f ∈ E ∗ . Then: (i) for any v ∈ Σ+ and ε ∈ (0, 1), we have N 1 1 k f k2E∗ ≤ J f (v) ≤ (1 + ε) 2 J0 (v) + k f k2E∗ ; 2ε 2ε (ii) there exists d0 > 0 such that infv∈Σ+ J f (v) > 0 if k f k E∗ ≤ d0 . N

(1 − ε) 2 J0 (v) −

Proof. (i) For v ∈ E and ε ∈ (0, 1), Z ε 1 f vdx ≤ k f k E ∗ kvk E ≤ kvk2E + k f k2E ∗ , 2 2ε Ω therefore, ) Z Z ∗ 1 1 (tv+ )2 1 1 2 2 ktvk E + a(x)(tv) dx − ∗ dx − k f k2E ∗ ≤ max I f (tv) (1 − ε) max t>0 t>0 2 2 Ω 2 Ω 1−ε 2ε ) ( Z Z ∗ 1 1 (tv+ )2 1 1 ktvk2E + a(x)(tv)2 dx − ∗ dx + k f k2E ∗ . ≤ (1 + ε) max t>0 2 2 Ω 2 Ω 1+ε 2ε (

Let s1 =

t 1 (1−ε) 2∗ −2

and s2 =

t 1

. We see that (2.13) is true.

(1+ε) 2∗ −2

(ii) By (2.13) and Lemma 2.4, we have N

N

inf J f (v) ≥ (1 − ε) 2 J0 (v) −

v∈Σ+

N S 2 1 1 k f k2E ∗ ≥ (1 − ε) 2 − k f k2E ∗ . 2ε N 2ε

So there exists d0 > 0 such that infv∈Σ+ J f (v) > 0 if k f k E ∗ ≤ d0 .

(2.13)

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

Lemma 2.6. Assume (A1 ) and f ∈ E ∗ . Then: (i) For any v ∈ Σ+ the function t 7→ I f (tv) has at most two critical points in [0, ∞); (ii) If k f k E ∗ ≤ d0 , for any v ∈ Σ+ there exists a unique t f (v) > 0 such that I f (t f (v)v) = J f (v). Moreover, t f (v) >

! ∗1 R 1 1 + Ω a(x)v 2 dx 2 −2 2∗ R 2∗ ≥ [(2∗ − 1)S − 2 ]− 2∗ −2 > 0, ∗ (2 − 1) Ω v+ dx

(2.14)

I 00f (t f (v)v)(v, v) < 0.

(2.15)

(iii) If t 7→ I f (tv) has another critical point t¯ different from t f (v), then t¯ ∈ [0, (1 −

1 −1 ∗ 2∗ −1 ) k f k E ].

Proof. (i) Let g(t) = I f (tv) for v ∈ Σ+ . It is easy to see that g (t) > 0, 00

t < t0 :=

! ∗1 R 1 + Ω a(x)v 2 dx 2 −2 R 2∗ ; (2∗ − 1) Ω v+ dx

g 00 (t) < 0,

t > t0 .

(2.16)

Hence, g 0 (t) has at most two critical points t1 and t2 satisfying 0 ≤ t1 ≤ t0 ≤ t2 .

(2.17)

(ii) Since g(0) = 0 and g(t) → −∞ as t → +∞, then supt>0 g(t) is finite. By (ii) of Lemma 2.5, infv∈Σ+ supt>0 I f (tv) > 0 and supt>0 g(t) > 0. We see that there exists t f (v) ≥ t0 such that I f (t f (v)v) = J f (v) and t f (v) 6= t0 . In fact, otherwise we would have g 0 (t0 ) = 0. By (2.15), t0 is the maximum point of g 0 (t). Then g 0 (t) ≤ 0 for all t > 0. Since g(0) = 0, we would have g(t) ≤ 0 for t > 0. This contradicts the fact that supt>0 g(t) > 0. So (2.14) holds. Since Z Z ∗ 2∗ I 00f (tv)(v, v) = 1 + a(x)v 2 dx − (2∗ − 1)t 2 −2 v+ dx, (2.18) Ω

Ω

it follows that g 00 (t) = I 00f (tv)(v, v). Equations (2.14), (2.16) and (2.18) then give (2.15). The uniqueness of t f (v) is a conclusion of (2.14) and (2.17). (iii) Suppose that g(t) has another critical point t¯ which is different from t f (v). In view of (2.14) and (2.17), t¯ satisfies t f (v) > t0 ≥ t¯. Hence Z Z ∗ 2∗ 1+ a(x)v 2 dx − t¯2 −2 v+ dx > 0. Ω

But

Ω

g 0 (t¯)

= 0, namely, Z Z Z ∗ −1 ∗ 2 2 2 t¯ + t¯ a(x)v dx − t¯ v+ dx − Ω

Ω

Ω

f vdx = 0.

Since t0 ≥ t¯, we have R t¯ =

R

1+ Ω

Ω

a(x)v 2 dx

f vdx R 2∗ ∗ − t¯2 −2 Ω v+ dx

k f k E ∗ kvk E R 2∗ −1 R 2∗ 1 + Ω a(x)v 2 dx − 1 + Ω a(x)v 2 dx (2∗ − 1) Ω v+ dx Ω v+ dx −1 1 ≤ 1− ∗ k f kE ∗ . 2 −1 ≤

R

R

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

Lemma 2.7. Assume (A1 ). There exists d1 > 0 such that if k f k E ∗ ≤ d1 , then: (i) J f ∈ C 1 (Σ+ , R) and, for h ∈ Tv Σ+ = {h ∈ E; hh, vi E = 0}, J 0f (v)h = t f (v)I 0f (t f (v)v)h. (ii) v ∈ Σ+ is a critical point of J f if and only if t f (v)v ∈ E is a critical point of I f . (iii) Moreover, the set of critical points of I f (u) can be written as [ {t f (v)v; v ∈ Σ+ , J 0f (v) = 0} {u loc ( f ; x)}. Proof. We prove (i) first. For v ∈ Σ+ Z Z Z ∗ 2∗ v+ dx − a(x)v 2 dx − t 2 −1 I 0f (tv)v = t + t Ω

Ω

Ω

f vdx.

By Lemma 2.6, there exists a unique t f (v) > 0 such that J f (v) = I f (t f (v)v)v and I 0f (t f (v)v)v = 0. Suppose that Q ∈ Σ+ × (0, +∞) is an open subset such that (v, t f (v)) ∈ Q. By Lemma 2.1, I 0f (tu)u : Q → R and g 00 (t) = I 00f (tv)(v, v) are continuous. Moreover, we know from Lemma 2.6 that g 00 (t) = I 00f (t f (v)v)(v, v) < 0. Applying the implicit function theorem, we see that there exist a neighborhood U of v, a neighborhood W of t f (v), U × W ⊂ Q and a continuous mapping ϕ : U → W such that ϕ(v) = t f (v) ∈ C 1 (Σ+ , (0, +∞)) and I 0f (ϕ(v)v)v = 0,

(2.19)

∀v ∈ U.

Therefore, J f (v) = I f (t f (v)v) ∈ C 1 (Σ+ , R) and J 0f (v)h = I 0f (t f (v)v)(t f (v)h + (t 0f (v), h)v) = t f (v)I 0f (t f (v)v)h for h ∈ Tv Σ+ . By Lemma 2.6, t f (v) > 0. So (ii) readily follows from (i). Now we prove (iii). Suppose that u ∈ E is a critical point of I f . Let u = tv, where v ∈ Σ+ and t > 0. By (iii) of Lemma 2.6, we have either t = t f (v) or −1 1 t ≤ 1− ∗ k f kE ∗ . 2 −1 So either u ∈ E is a critical point of J f or kuk E ≤ (1 − 2∗1−1 )−1 d1 ≤ r1 , provided that d1 > 0 small enough. In the second case, by Lemma 2.2, I f has a unique critical point u loc ( f ; x) in Br1 . It is standard to prove the following result. See, for instance, the proof of Theorem 3.1 in [8]; we omit the proof. Lemma 2.8. Assume (A1 ) and f ∈ E ∗ , k f k E ∗ ≤ d1 . Then: (i) J f (v) → ∞ as dist E (v, ∂Σ+ ) ≡ inf{kv − uk E ; u ∈ ∂Σ+ } → 0. (ii) Assume that (vm )∞ m=1 ⊂ Σ+ satisfies J f (vm ) → c > 0 and kJ 0f (vm )kTv∗m Σ+ ≡ sup{J 0f (vm )h; h ∈ Tvm Σ+ , khk E = 1} → 0 j

j

j

as m → ∞. Then, there exist k ∈ N , sequences of positive numbers (εm ) and (xm ) ⊂ Ω verifying εm → 0 and j j dist(xm , ∂Ω )/εm → ∞ for 1 ≤ j ≤ k, as m → ∞; also, there exist a non-negative solution u 0 ∈ E of (1.1) and non-trivial solutions u j ∈ D 1,2 (R N ), 1 ≤ j ≤ k, of problem ∗

2 −1 , −∆u = u +

in R N

(2.20)

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

such that a subsequence of (vm ), still denoted by (vm ), satisfies

k P j

0 u (x) + u m (x)

j=1

vm −

→ 0,

k P

j 0

u m (x)

u (x) +

1,2 N j=1 (R ) D 1,2 (R N )

D

as m → ∞, where j

j

u m (x) = (εm )−

N −2 2

j

j

u j ((x − xm )/εm ),

1 ≤ j ≤ k, m ∈ N.

Moreover, J f (vm ) → I f (u 0 ) +

k X

I ∞ (u j )

j=1

as m → ∞. Lemma 2.9. Assume (A1 ). For any ε > 0 there exists a δ(ε) ∈ (0, d1 ] such that, if k f k E ∗ ≤ δ(ε): N

N

(i) N1 S 2 − ε ≤ infv∈Σ+ J f (v) ≤ N1 S 2 + ε; (ii) J f satisfies (PS)c condition for c ∈ (−∞, I f (u loc ) +

1 N

N

S2)

S (I f (u loc ) +

1 N

N

S2,

2 N

N

S 2 − ε).

Proof. By (i) of Lemma 2.5, for σ ∈ (0, 1), N

N

inf J0 (v) ≤ inf J f (v) + [(1 − σ )− 2 − 1] inf J f (v) + (1 − σ )− 2

v∈Σ+

v∈Σ+

v∈Σ+

Since infv∈Σ+ J0 (v) =

1 k f k2E ∗ . 2σ

N

1 N

S 2 , for k f k2E ∗ ≤ d0 , by Lemma 2.5 again, N 3 2 1 N 0 < inf J f (v) ≤ S 2 + d02 . 2 N v∈Σ+

Therefore, infv∈Σ+ J f (v) is bounded, provided that k f k E ∗ ≤ d0 . Choosing σ1 > 0 small enough so that N ¯ [(1 − σ )− 2 − 1] infv∈Σ J f (v) < ε for 0 < σ < σ1 and k f k E ∗ ≤ d0 , then there exists d(ε) ∈ (0, d1 ] such 2

+

N

that (1 − σ )− 2

1 2 2σ k f k E ∗

<

ε 2

¯ ¯ if k f k E ∗ ≤ d(ε). As a result, for k f k E ∗ ≤ d(ε)

1 N S 2 ≤ inf J f (v) + ε. N v∈Σ+ On the other hand, by Lemma 2.5, for σ ∈ (0, 1), N

N

inf J f (v) + [(1 + σ )− 2 − 1] inf J f (v) − (1 + σ )− 2

v∈Σ+

v∈Σ+

1 k f k2E ∗ ≤ inf J0 (v). 2σ v∈Σ+

ˆ ˆ In the same way, we may choose σ > 0 small enough and d(ε) ∈ (0, d1 ] so that, for k f k E ∗ ≤ d(ε), inf J f (v) ≤

v∈Σ+

1 N S 2 + ε. N

So (i) follows. N Next, we prove (ii). We first consider the case c ∈ (−∞, I f (u loc ( f ; x))+ N1 S 2 ). By (ii) of Lemma 2.5, I f (u 0 ) > 0. While I f (u loc ) ≤ 0, we see from (iii) of Lemma 2.7 that I f (u loc ) = inf{I f (u 0 ); u 0 is a critical point of I f (u) ∈ E}. Furthermore, by Lemma 2.8 c + o(1) = J f (vm ) → I f (u 0 ) +

k X j=1

I ∞ (u j )

(2.21)

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593 N

as m → ∞. Since I ∞ (u) ≥ N1 S 2 for the solutions u ∈ D 1,2 (R N ) of problem (2.20), we deduce that k = 0. Hence, vm → u 0 in E as m → ∞. N N If c ∈ (I f (u loc ) + N1 S 2 , N2 S 2 − ε), we see from Lemma 2.7 that, for ε > 0, there exists d(ε) ∈ (0, d1 ] such that N

u 0 = u loc (x) or I f (u 0 ) ≥ infv∈Σ+ J f (v) ≥ N1 S 2 − ε for all k f k E ∗ ≤ d(ε). If u 0 = u loc (x), since u loc is a critical point of I f (u), we have Z Z 1 N +2 1 I f (u 0 ) = I f (u loc ) = ku loc k2E + a(x)u 2loc dx − f u loc dx. N N Ω 2N Ω Again, by Lemma 2.8, (2.21) holds true. If k = 1, c = I f (u loc ) + If k = 2, then, Z f u loc dx ≤ k f k E ∗ ku loc k E ≤ r1 k f k E ∗ . −I f (u loc ) ≤

1 N

N

S 2 , which contradicts the assumption.

Ω

ε r1 , we obtain N N ∈ (I f (u loc ) + N1 S 2 , N2 S 2 − ε). In the case that u 0 6= u loc (x), we

Choosing k f k E ∗ ≤ c

c = I f (u loc ) +

2 N

N

S2 ≥

2 N

N

S 2 − ε, which is a contradiction of the fact that

deduce from (i) that, for all 1 ≤ j ≤ k, k X 1 N 2 N 1 N c = I f (u 0 ) + I ∞ (u j ) ≥ S 2 − ε + S 2 = S 2 − ε. N N N j=1

Again, we obtain a contradiction. Consequently, k = 0, provided that k f k E ∗ is small enough, and then vm → u 0 in E as m → ∞.

3. Existence of second and third positive solutions The second and third solutions are found by using the following results related to the Ljusternik–Schnirelman category. Lemma 3.1. Suppose that M is a Hilbert manifold and Ψ ∈ C 1 (M, R). Assume that, for c0 ∈ R and k ∈ N , Ψ satisfies (PS)c condition for c ≤ c0 and cat({x ∈ M; Ψ (x) ≤ c0 }) ≥ k. Then Ψ has at least k critical points in {x ∈ M; Ψ (x) ≤ c0 }. The proof of this lemma can be found, for instance, in [1] or in [7]. Lemma 3.2. Let X be a topological space. Suppose that there exist two continuous mappings, F : S N −1 = {y ∈ R N ; |y| = 1} → X,

G : X → S N −1

such that G ◦ F is homotopic to identity. Then cat(X ) ≥ 2. See [2] for the proof. Denote [J f ≤ c] = {u ∈ Σ+ ; J f (u) ≤ c}. From now on, we shall construct two mappings, 1 N 1 N F : S N −1 → J f ≤ I f (u loc ) + S 2 − ε ; G : J f ≤ I f (u loc ) + S 2 − ε → S N −1 N N such that F ◦ G is homotopic to the identity. This, in turn, implies by Lemmas 3.1 and 3.2 the existence of the second and third solutions. We first construct the mapping F. η Set Ω1 := {x ∈ Ω1 ; dist(x, ∂Ω1 ) > 2η}.

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Lemma 3.3. Assume (A1 ), (A2 ) and f (x) 6≡ 0, k f k E ∗ ≤ d1 , f ∈ L p (Ω ), p > that I f (u loc + tu ε,z ) 0 such

1 N S2, N

(3.1)

η

for 0 < ε < ε0 , t ≥ 0 and a.e z ∈ Ω1 . Proof. We may verify that Z 1 ∗ ∗ ∗ ∗ 2∗ −1 I f (u loc ( f ; x) + tu ε,z ) = I f (u loc ) + I0 (tu ε,z ) − ∗ ((u loc + tu ε,z )2 − u 2loc − t 2 u 2ε,z − 2∗ u loc tu ε,z )dx. 2 Ω By (2.5) and (2.6), R

1 max I0 (tu ε,z ) = t>0 N

1 N S2 N

= For z ∈ Z

R N |∇u ε,z |2 dx + Ω a(x)u 2ε,z dx 2 R N −2 2∗ 2 Ω u ε,z dx N Z 2 − N2 2 1+S a(x)u ε,z dx + O(ε N −2 ).

Ω

Ω

η Ω1 ,

Ω

by (A2 ) we have Z Z 2 a(x)u 2ε,z dx = a(x)ξz2 (x)Uε,z dx ≤ Ω

B(z,2η)

2 a(x)Uε,z dx = 0.

η

N

S 2 + O(ε N −2 ) for z ∈ Ω1 , and then Z 1 N 1 ∗ 2 [(u loc + tu ε,z )2 I f (u loc + tu ε,z ) ≤ I f (u loc ) + S − ∗ N 2 Ω

Thus, I0 (tu ε,z ) ≤

1 N

∗

∗

∗

∗

− u 2loc − t 2 u 2ε,z − 2∗ u 2loc−1 tu ε,z ]dx + O(ε N −2 ).

(3.2)

Since I f (u loc + tu ε,z ) → I f (u loc ) < 0 as t → 0, and I f (u loc + tu ε,z ) → −∞ as t → +∞, there exist 0 < m < M such that I f (u loc + tu ε,z ) ≤ 0 S for t ∈ [0, m] [M, ∞). So we only need to show that (3.1) is true for t ∈ [m, M]. Since Z Z N −2 ∗ ∗ ∗ ∗ ∗ ∗ 2∗ −1 ku loc + tu ε,z k22∗ = ku loc k22∗ + t 2 ku ε,z k22∗ + 2∗ t |u loc |2 −2 u loc u ε,z dx + 2∗ t 2 −1 u ε,z u loc dx + o(ε 2 ) Ω

Ω

η

for a.e z ∈ Ω1 , we have I f (u loc + tu ε,z ) ≤ I f (u loc ) + η

Z N −2 1 N ∗ ∗ S 2 − t 2 −1 u loc u 2ε,z−1 dx + O(ε N −2 ) + o(ε 2 ) N Ω

for a.e z ∈ Ω1 . As f ∈ L p (Ω ), p > minx∈B(z,2η) u loc > 0 and

N 2,

(3.3)

we get u loc ∈ C 0 (Ω¯ ). From (A1 ) and (ii) of Lemma 2.1, we obtain that

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

t

2∗ −1

Z Ω

2∗ −1 u loc u ε,z dx

Z

u loc ξ

≥c

2∗ −1

(x − z)

B(z,2η)

≥ cε

Z

N −2 2

= O(ε

min

x∈B(z,2η) N −2 2

u loc B(0,η)

ε

N −2 2

2 (1 + | x−z ε | ) 1

(1 + |z|)

N +2 2

N +2 2

· εN

dx

dz

).

(3.4)

By (3.3) and (3.4), N −2 N −2 1 N S 2 + O(ε N −2 ) + o(ε 2 ) − O(ε 2 ). N The conclusion then follows for t ∈ [m, M], provided that > 0 sufficiently small.

I f (u loc + tu ε,z ) ≤ I f (u loc ) +

Remark 3.4. If f (x) ≡ 0, then (3.1) is not true. In fact, if f (x) ≡ 0, then u loc ≡ 0 and I f (u loc + t f (u ε,z )u ε,z ) = N I0 (t f (u ε,z )u ε,z ) > N1 S 2 . Lemma 3.5. Assume (A1 ), (A2 ) and k f k E ∗ ≤ d1 . Then there exist d2 ∈ (0, d1 ] and ε1 < ε0 such that, for η k f k E ∗ ≤ d2 , 0 < ε ≤ ε1 and a.e z ∈ Ω1 , there exists a unique s = s( f, ε, z) > 0 in a neighborhood of 1 satisfying u loc + su ε,z ku loc + su ε,z k E = t f . (3.5) ku loc + su ε,z k E Moreover, the mapping ε ∈ {ε ∈ R; 0 < ε < ε1 } → s( f, ε, z) ∈ (0, ∞) is continuous. Proof. The lemma is proved by the implicit function theorem. Set Φ(s, f, ε, z) = hI 0f (u loc + su ε,z ), u loc + su ε,z i E ∗ ,E . We claim that (3.5) holds if and only if Φ(s, f, ε, z) = 0. In fact, if (3.5) holds, then u loc + su ε,z d u loc + su ε,z =0 If tf dt ku loc + su ε,z k E ku loc + su ε,z k E namely, d (I f (u loc + su ε,z )) = 0. dt This implies that Φ(s, f, ε, z) = 0. On the other hand, supposing that Φ(s, f, ε, a) = 0, then by (ii) of Lemma 2.6, for v ∈ Σ+ there exists a unique t f (v) > 0 such that I f (t f (v)v) = J f (v) if k f k E ∗ ≤ d0 . Therefore, (3.5) holds. It may be verified that Z Z Z ∗ 2 2 Φ(1, 0, ε, z) = |∇u ε,z | dx + a(x)u ε,z dx − u 2ε,z dx → 0; Ω Ω Ω Z Z Z ∂ ∗ 2 2 ∗ Φ(s, 0, ε, z) = 2 |∇u ε,z | dx + 2 a(x)u ε,z dx − 2 u 2ε,z dx ∂s Ω Ω Ω s=1 N

→ −(2∗ − 2)S 2 < 0, as ε → 0+. The implicit function theorem then allows us to find a unique mapping s = s( f, ε, z) in a neighborhood of 1 such that Φ(s, f, ε, a) = 0, and ε 7→ s( f, ε, a) is continuous. Define a mapping Fρ (y) : S N −1 = {y ∈ R N ; |y| = 1} → Σ+ by Fρ (y) =

u loc + s( f, ε, ρy)u ε,ρy (x) , ku loc + s( f, ε, ρy)u ε,ρy (x)k E

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593 η

where z = ρy ∈ Ω1 \ {0} is determined in Lemma 3.3. Let d2 and ε1 be as in Lemma 3.5. Lemma 3.6. Assume (A1 ), (A2 ) and f ∈ L p (Ω ), p > J f (Fρ (y)) < I f (u loc ) +

N 2,

0 < k f k E ∗ ≤ d2 . Then, for 0 < ε < ε1 , we have

1 N S2. N

Proof. Since J f (v) = I f (t f (v)v), we have J f (Fρ (y)) = I f (u loc + s( f, ε, ρy)u ε,ρy (x)). The conclusion follows by Lemma 3.3.

Lemma 3.7. Under the hypothesis of Lemma 3.6, there exists δ1 (ε) > 0 such that 1 N N −1 2 Fρ (S ) ⊂ J f ≤ I f (u loc ) + S − δ1 (ε) . N Proof. By Lemma 3.6, J f (v) 0 such that J f (v) ≤ I f (u loc ) +

1 N S 2 − δ(v). N

Choose 1 δ(v) = 2

1 N I f (u loc ) + S 2 − J f (v) . N

By Lemma 2.7, δ(v) ∈ C 1 (Σ+ , R). Since Fρ (S N −1 ) is compact, there exists v0 ∈ Fρ (S N −1 ) such that 1 1 N 2 min δ(v) = I f (u loc ) + S − J f (v0 ) > 0. 2 N v∈Fρ (S N −1 ) Therefore, 1 N S 2 − min δ(v) v∈Fρ (y) N 1 N 1 1 N = I f (u loc ) + S 2 − I f (u loc ) + S 2 − J f (v0 ) . N 2 N

J f (v) ≤ I f (u loc ) +

The proof is completed by choosing 1 1 N δ1 (ε) = I f (u loc ) + S 2 − J f (v0 ) . 2 N Next, we construct the mapping G. Lemma 3.8. Assume (A1 ). Then, there exists δ0 > 0 such that if J0 (v) ≤ one has Z

1 N S 2 + δ0 , N

x |∇v|2 dx 6= 0. Ω |x|

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593 N

Proof. Suppose that v ∈ Σ+ satisfies J0 (v) ≤ N1 S 2 + δ0 , where δ0 > 0 is to be determined. By the Ekeland variational principle, there exists v˜ ∈ Σ+ such that kv˜ − vk E ≤

p

δ0 ,

kJ00 (v)k ˜ Tv˜∗ Σ+ ≤

p

δ0 ,

J0 (v) ˜ ≤

1 N S 2 + δ0 . N

Invoking (ii) of Lemma 2.8, we see that for R > 0 there exists σ = σ (R) > 0 such that if v˜ ∈ Σ+ satisfying J0 (v) ˜ ≤

1 N S 2 + σ, N

then there exist 0 < ε <

1 R

ε 1 < , dist(y, ∂Ω ) R

kJ00 (v)k ˜ Tv˜∗ Σ+ ≤ σ, and y ∈ Ω , we may assume y 6= 0, so that

2−N x−y

ε 2 U ε 1

≤ .

v˜ −

2−N

x−y

R

ε 2 U ε 1,2 N

1,2 N D (R ) D

(3.6)

(R )

Choosing δ0 ≤ {σ (2R)2 , σ (2R), 4R1 2 }, we have

Uε,y Uε,y

≤ kv − vk ˜ D 1,2 (R N ) + v˜ −

v −

kUε,y k D 1,2 (R N ) 1,2 N kUε,y k D 1,2 (R N ) D

(R )

≤

p

D 1,2 (R N )

1 1 ≤ δ0 + 2R R

as well as Z " # Z |∇Uε,y |2 |∇Uε,y |2 x 2 2 |∇v| − dx ≤ |∇v| − dx Ω |x| kUε,y k2D 1,2 (R N ) kUε,y k2D 1,2 (R N ) RN  ! 2  12 Z !2  21 Z Uε,y |∇Uε,y | 4 ≤ |∇v| + dx  ≤ . ∇ v − dx   N N kU k kU k R ε,y D 1,2 (R N ) ε,y D 1,2 (R N ) R R Thus, Z Z 4 |∇Uε,y |2 x x 2 dx − ≤ |∇v| dx . Ω |x| kUε,y k2 1,2 N R Ω |x| D (R ) We claim that Z |∇Uε,y |2 x Ω |x| kUε,y k2 1,2 D

for

N +2 N +3

(R N )

(3.7)

N −2 1 (1−α)(N −2) (α−1)N +3α−2 dx ≥ 1 − C1 ε − C2 ε − C3 R

(3.8)

< α < 1. In fact,

Z |∇Uε,y |2 y x − dx Ω |x| |y| kUε,y k2 1,2 N D (R ) Z Z |∇Uε,y |2 |∇Uε,y |2 x y y x = − dx − − dx 2 R N |x| |y| kUε,y k2 1,2 N N |x| |y| kUε,y k D 1,2 (R N ) R \Ω D (R ) Z Z |∇Uε,y |2 |∇Uε,y |2 x x y y ≤ − dx + − dx R N |x| |y| kUε,y k2 1,2 N R N \Ω |x| |y| kUε,y k2 1,2 N D

= H1 + H2 .

(R )

D

(R )

(3.9)

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

We estimate H1 and H2 separately. Z Z y y x x 1 2 2 |∇Uε,y | dx + |∇Uε,y | dx − − H1 = 2 α N α |x| |y| |x| |y| kUε,y k D 1,2 (R N ) B(y,ε ) R \B(y,ε ) =

1 (I1 + I2 ). kUε,y k2D 1,2 (R N )

(3.10)

Since y 6= 0, we have x y |x|y| − x|x| + x|x| − |x|y| ||y| − |x|| + |x − y| 2|x − y| ≤ ≤ . |x| − |y| = |x y| |y| |y|

(3.11)

Thus Z I1 ≤ for α >

B(y,εα )

2|x − y| 2ε α |∇Uε,y |2 dx ≤ |y| |y|

Z B(y,εα )

|∇Uε,y |2 dx ≤ C1 ε(α−1)N +3α−2

(3.12)

N +2 N +3 .

There also holds Z I2 ≤ 2 |∇Uε,y |2 dx ≤ C2 ε(1−α)(N −2) R N \B(y,εα )

(3.13)

for α < 1. Therefore, H1 ≤ C1 ε (α−1)N +3α−2 + C2 ε (1−α)(N −2) for

< α < 1. Choosing τ = so that B(y, 2τ ) ⊂ Ω , by (3.6) we get Z x 1 y H2 ≤ − |∇Uε,y |2 dx 2 N kUε,y k D 1,2 (R N ) R \Ω |x| |y| Z 2 ≤ |∇Uε,y |2 dx kUε,y k2D 1,2 (R N ) R N \Ω ε N −2 ≤C 2τ N −2 1 . ≤C R

Hence Z |∇Uε,y |2 x y − Ω |x| |y| kUε,y k2 1,2 D

for

(3.14)

1 2 dist(y, ∂Ω )

N +2 N +3

(R N )

(3.15)

N −2 1 dx ≤ C1 ε (α−1)N +3α−2 + C2 ε (1−α)(N −2) + C R

< α < 1. It results that Z N −2 |∇Uε,y |2 y 1 (1−α)(N −2) (α−1)N +3α−2 dx − C ε − C ε − C 1 2 2 Ω |y| kUε,y k 1,2 N R D (R ) Z |∇Uε,y |2 x dx ≤ . Ω |x| kUε,y k2 1,2 N

N +2 N +3

D

(3.16)

(R )

In the same way, we have Z R N −2 2 |∇Uε,y |2 y 1 R N \Ω |∇Uε,y | dx dx = 1 − R ≥1−C . 2 2 Ω |y| kUε,y k 1,2 N R R N |∇Uε,y | dx D (R )

(3.17)

Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

2585

In conclusion, (3.8) holds, and (3.7) yields 1 − C1 ε(α−1)N +3α−2 − C2 ε(1−α)(N −2) − C3 for

N +2 N +3

< α < 1.

Z N −2 1 4 x − ≤ |∇v|2 dx R R Ω |x|

Lemma 3.9. Assume f ∈ E ∗ and (A1 ). There exists d3 ∈ (0, d2 ] such that, if k f k E ∗ ≤ d3 , then 1 N 1 N J f < I f (u loc ) + S 2 ⊂ J0 < S 2 + δ0 . N N Proof. By (i) of Lemma 2.5, for ε ∈ (0, 1) and v ∈ Σ+ , N 1 J0 (v) ≤ (1 − ε)− 2 J f (v) + k f k2E ∗ . 2ε Since I f (u loc ) ≤ 0, we have N 2 1 1 1 N J0 (v) 0 such that, if 0 < ε < ε3 , " # N 2 δ0 1 1 N −1 S2 < . 1−ε N 2

(3.18)

(3.19)

For 0 < ε < ε3 , we choose d3 ∈ (0, d2 ] such that, if k f k E ∗ ≤ d4 ,

1 1−ε

N 2

·

1 δ0 k f k2E ∗ < . 2ε 2

(3.20)

The conclusion is completed by (3.19) and (3.20).

Define

1 N G : J f < I f (u loc ) + S 2 N

→ S N −1 ,

by Z

x G(v) = |∇v|2 dx Ω |x|

Z

x 2 |∇v| dx . Ω |x|

From Lemmas 3.8 and 3.9, we know that, if k f (x)k E ∗ ≤ d4 , Z x |∇v|2 dx 6= 0 Ω |x| for v ∈ [J f η ρy ∈ Ω1 \ {0}, the mapping G ◦ Fρ : S N −1 → S N −1 ; is homotopic to the identity.

y 7→ G(Fρ (y))

N 2.

For ε > 0, k f k E ∗ sufficiently small, and a.e

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593 u

ε,ρy N −1 , where γ (θ ) = Proof. Let γ (θ ), θ ∈ [θ1 , θ2 ] be a regular geodesic between G( ku ε,ρy 1 k E ) and G(Fρ (y)) on S

u

ε,ρy G( ku ε,ρy k E ), γ (θ2 ) = G(Fρ (y)). Set

γ (θ ) = γ (2(θ1 − θ2 )θ + θ2 ),

1 θ ∈ 0, . 2

Define the mapping ζ (θ, y) : [0, 1] × S N −1 → S N −1 by  1   γ (θ ), θ ∈ 0,   2  ζ (θ, y) = u 2(1−θ)ε,ρy (x) 1  , θ ∈ G , 1   ku 2(1−θ)ε,ρy (x)k E 2   y, θ = 1. η

First, we show that ζ (θ, y) is well defined. In fact, since a(x) ≥ 0 satisfies (A1 ), by (2.12) we have for z ∈ Ω1 that u ε,z 1 N J0 ≤ S 2 + o(1) ku ε,z k E N η

as ε → 0+. Then, for ε small enough and z ∈ Ω1 , u ε,z 1 N ≤ S 2 + δ0 , J0 ku ε,z k E N where δ0 is as in Lemma 3.8. Therefore, ζ (θ, y) is well defined. Second, we prove that ζ (θ, y) ∈ C([0, 1] × S N −1 , S N −1 ). There holds u ε,ρy 1 lim γ (θ ) = γ (θ1 ) = G =ζ ,y . ku ε,ρy k E 2 θ→ 21 We claim that u 2ε(1−θ),ρy (x) lim G = y. θ →1 ku 2ε(1−θ),ρy (x)k E

(3.21) η

In fact, it is sufficient to prove that, for z ∈ Ω1 \ {0}, Z x z |∇u ε,z |2 − dx = o(1) Ω |x| |z| ku ε,z k2E

(3.22)

η

as ε → 0+, since it implies for z ∈ Ω1 \ {0} that Z x |∇u ε,z |2 z dx = + o(1) 2 |z| Ω |x| ku ε,z k E η

as ε → 0+. So, for ρy ∈ Ω1 \ {0} and |y| = 1, lim G

θ →1

u 2ε(1−θ),ρy (x) ku 2ε(1−θ),ρy (x)k E

The claim follows.

R

2 x |∇u 2(1−θ)ε,ρy | Ω |x| ku 2(1−θ)ε,ρy k2 dx

= lim θ →1 R Ω

ρy

= |ρy| = y. ρy x |∇u 2(1−θ)ε,ρy | |ρy| |x| ku 2(1−θ)ε,ρy k2 dx E E 2

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593 η

Now we prove (3.22). For z ∈ Ω1 \ {0} there holds Z z |∇u ε,z |2 x − dx Ω |x| |z| ku ε,z k2E Z Z x z 1 2 − ≤ |∇u | dx + ε,z |x| |z| α ku k2 B(z,ε )

ε,z E

=

1 ku ε,z k2E

x z 2 − |∇u | dx ε,z |z| Ω \B(z,εα ) |x| (3.23)

[L 1 + L 2 ],

where α < 1. We may find ε2 > 0 small such that ε α < η for 0 < ε < ε2 . By (3.11), we have Z Z 2|x − z| 2ε α |∇u ε,z |2 dx = o(1) L1 ≤ |∇u ε,z |2 dx ≤ |z| |z| B(z,εα ) B(z,εα ) as ε → 0+. We may verify directly by computation that Z L2 ≤ 2 |∇u ε,z |2 dx = o(1)

(3.24)

(3.25)

Ω \B(z,εα )

as ε → 0+. Equation (3.22) then follows from (3.23)–(3.25). Lastly, by the definition of ζ (θ, y), we see that, for y ∈ S N −1 , if 0 < ε < ε2 , then ζ (0, y) = G(Fρ (y)),

ζ (1, y) = y.

Therefore, G ◦ Fρ is homotopic to the identity. The proof is complete.

The proof of the existence of the second and third positive solutions. We shall show (ii) of Theorem 1.1. By Lemmas 2.9, 3.2 and 3.9, J f (v) has at least two different critical points N u¯ 2 and u¯ 3 in [J f < I f (u loc ) + N1 S 2 ]. Let u 2 = t f (u¯ 2 )u¯ 2 and u 3 = t f (u¯ 3 )u¯ 3 . By (ii) of Lemma 2.5, we have N

0 < I f (u i ) < I f (u loc ) + N1 S 2 , i = 2, 3. Using (ii) of Lemma 2.1 and (ii) Lemma 2.7, we know that u 2 and u 3 are two positive solutions of (1.1). 4. The existence of a fourth positive solution The fourth positive solution of the problem will be found in this section on the manifold M := {u ∈ E; kuk2∗ = 1}. N N To this end, we shall find two proper positive constants A and B such that N1 S 2 + I f < A ≤ B < N1 S 2 − µ for some small µ > 0, so that there exists a critical point of I f with the energy in [A, B]. Similar to Lemmas 2.8 and 2.9, we have the following result. Lemma 4.1. Assume (A1 ). Then, for ε > 0 there exists d(ε) ∈ (0, d2 ] such that I f (u) satisfies the (PS)c condition N S N N for c ∈ (−∞, I f (u loc ) + N1 S 2 ) (I f (u loc ) + N1 S 2 , N2 S 2 − ε), provided that k f k E ∗ ≤ d(ε). Denote M(R N ) := {u ∈ H 1 (R N ); kuk L 2∗ (R N ) = 1}. We define two continuous mappings β, : M(R N ) → R N and γ : M(R N ) → R+ by Z Z x x ∗ |u(x)|2∗ dx. |u(x)|2 dx, γ (u) = − β(u) β(u) = R N 1 + |x| R N 1 + |x| Let Z

h

C(α) := inf RN

|∇u| + α(x)u 2

2

i

1 dx; u ∈ M(R ), β(u) = 0, γ (u) = . 3 N

2

We may show, as in [6], that C(α) > S. Choose σ > 0 so that S + σ < min{C(α), 2 N S}. Let ϕ ∈ H01 (B(0, 1)) be a function satisfying

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

 ϕ ∈ C ∞ (B(0, 1)), ϕ(x) > 0 ∀x ∈ B(0, 1)    ϕ Z is radial symmetric and Z|x1 | < |x2 | H⇒ ϕ(x1 ) > ϕ(x2 ), ∗   |∇ϕ|2 dx < S + σ. ϕ 2 dx = 1, S< 

(4.1)

B(0,1)

B(0,1)

Define the mapping Tε,y : M(R N ) → M(R N ) by u x−y ε . Tε,y (u) = x−y

u

2∗ N ε

L

(R )

Let Fλ (u) =

Z Ω

{|∇u|2 + [α(x) ¯ + λ2 α(λ(x − x0 ))]u 2 }dx

and βλ = β ◦ Tλ,−λx0 ,

γλ = γ ◦ Tλ,−λx0 .

Denote K = K (ε1 , ε2 , r ) = {(y, ε) ∈ R N × R; |y| ≤ r, ε1 ≤ ε ≤ ε2 } for r > 0, ε1 > 0 and ε2 > 0. It is proved in [6] that there exist r, 1 and 2 such that the following result holds. Lemma 4.2. Assume (A3 ) and kαk

2

L

N 2

(R N )

¯ there holds < S(2 N − 1). There is a λ¯ > 0 such that, if λ > λ,

S < sup{Fλ ◦ T 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ ∂ K } < S+ε

λ

< C(α) 1 ≤ inf Fλ (u); u ∈ M, βλ (u) = 0, γλ (u) = 3 ≤ sup{Fλ ◦ T 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ K } λ

2 N

< 2 S.

(4.2)

We set N −2 4

, M+ := {u ∈ M, u + 6≡ 0}, Z Z 1 1 1 ∗ u 2+ dx, a(x)u 2 dx − ∗ I0 (u) := kuk2E + 2 2 Ω 2 Ω Z Z 1 1 1 ∗ 2 2 J (u) := kuk E + a(x)u dx − ∗ |u|2 dx. 2 2 Ω 2 Ω

b := S

Lemma 4.3. Assume (A3 ) and kαk

2

J

N 2

(R N )

< S(2 N − 1). Then, there exists λ¯ > 0 such that, if λ > λ¯ ,

1 N S 2 < sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } λ N u u 1 u < inf J (u); ∈ M, βλ = 0, γλ = b b b 3 u u 1 u ≤ inf I0 (u); ∈ M+ , βλ = 0, γλ = b b b 3 ≤ sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } λ

2 N < S2. N

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

Proof. For (y, ε) ∈ K we have J (bT 1 ,x0 λ

Z 2 1 N −2 1 N 2 2 ◦ Tε,y (ϕ)) = S ∇T 1 ,x0 ◦ Tε,y (ϕ) + a(x)(T 1 ,x0 ◦ Tε,y (ϕ)) dx − ∗ S 2 . λ λ 2 2 Ω

Letting u ∈ E, satisfying

u b

(4.3)

∈ M, βλ ( ub ) = 0, γλ ( ub ) = 13 , then there holds

Z u 2 1 N 1 N −2 u 2 dx − ∗ S 2 . S 2 ∇ + a(x) 2 b b 2 Ω First, by Lemma 4.2 and (4.3), we have J (u) =

(4.4)

1 N −2 1 N S 2 sup{Fλ ◦ T 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ ∂ K } − ∗ S 2 λ 2 2 1 N −2 1 N > S 2 S− ∗S2 2 2 1 N = S2. N

sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } = λ

Next, by Lemma 4.2, (4.3) and (4.4), 1 1 N −2 S 2 inf Fλ (u); u ∈ M, βλ (u) = 0, γλ (u) = − 2 3 u u u 1 N −2 u = S 2 inf Fλ ; ∈ M, βλ = 0, γλ = 2 b b b b u u 1 u = 0, γλ = . = inf J (u); ∈ M, βλ b b b 3

sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } < λ

1 N S2 2∗ 1 1 N − ∗S2 3 2

Last, again by Lemma 4.2 and (4.3), we have 1 N −2 1 N S 2 sup{Fλ ◦ T 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ K } − ∗ S 2 λ 2 2 1 N −2 2 1 N < S 2 2N S − ∗ S 2 2 2 2 N < S2. N

sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } = λ

2

The last strict inequality is true because, if N ≥ 3, we have 2 N − 1 −

2 N

< 0.

Denote Z Z Z 1 1 1 2 2 2∗ ˆ J (u) = kuk E + a(x)u dx − ∗ |u| dx − f udx. 2 2 Ω 2 Ω Ω The constants A and B are defined in the following lemma. Lemma 4.4. Assume (A3 ) and kαk 2 N

N

2

J

N 2

(R N )

< S(2 N − 1). There exists λ¯ > 0 such that, if λ > λ¯ , for 0 < µ <

S 2 − sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K }, we have λ

1 N S 2 + I f (u loc ) < sup{ Jˆ(bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } λ N u u 1 u ˆ < inf J (u); ∈ M, βλ = 0, γλ = b b b 3 u u 1 u = 0, γλ = ≤ A := inf I f (u); ∈ M+ , βλ b b b 3

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

≤ B := sup{I f (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } λ

if k f k E ∗

2 N < S2 −µ N is small enough.

(4.5)

Proof. First, by the choice of ϕ, we have sup{ Jˆ(bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } λ

≥ sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } + inf{−bk f k E ∗ kT 1 ,x0 ◦ Tε,y (ϕ)k E ; (y, ε) ∈ ∂ K } λ

λ

> sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } − S

N −2 4

λ

1 2

(S + σ ) k f k E ∗ .

By Lemmas 2.3 and 4.3, we can choose C1 > 0 such that, if k f k E ∗ ≤ C1 , then sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } − λ

N −2 1 1 N S 2 > S 4 (S + σ ) 2 k f k E ∗ . N

Therefore, if k f k E ∗ ≤ C1 , we have sup{ Jˆ(bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } > λ

1 N S 2 + I f (u loc ). N

Second, by the choice of ϕ and Lemma 4.3, we have sup{I f (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } λ

≤ sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } + sup{bk f k E ∗ kT 1 ,x0 ◦ Tε,y (ϕ)k E ; (y, ε) ∈ K } λ

λ

≤ sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } + S

N −2 4

λ

For 0 < µ < µ+S

2 N

N −2 4

S

N 2

1

(S + σ ) 2 k f k E ∗ .

− sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K }, we choose C2 > 0 so that, if k f k E ∗ ≤ C2 , then λ

1

(S + σ ) 2 k f k E ∗ <

2 N S 2 − sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K }. λ N

Therefore sup{I f (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K } < λ

2 N S 2 − µ. N

Last, by Lemma 4.3, we can choose u u 1 1 u inf J (u); ∈ M, βλ = 0, γλ = − sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } . ν< λ 2 b b b 3 Then, there exists {u m } such that ubm ∈ M, βλ ( ubm ) = 0, γλ ( ubm ) = 31 and u u 1 u = 0, γλ = < Jˆ(u m ) inf Jˆ(u); ∈ M, βλ b b b 3 u u 1 u < inf Jˆ(u); ∈ M, βλ = 0, γλ = +ν b b b 3 2 N S 2 − µ + ν. (4.6) N Actually, (u m )m is bounded. In fact, if this is not the case, then we would have ku m k E → ∞ as m → ∞. Then, if k f k E ∗ ≤ C2 , <

∗

1 b2 Jˆ(u m ) ≥ ku m k2E − ∗ − C2 ku m k E → +∞ 2 2

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

as m → ∞. This contradicts (4.6). So, there exists C > 0 such that ku m k E ≤ C. Hence, u u 1 u inf Jˆ(u); ∈ M, βλ +ν = 0, γλ = b b b 3 > Jˆ(u m ) u u 1 u − Ck f k E ∗ . ≥ inf J (u); ∈ M, βλ = 0, γλ = b b b 3 This yields u u 1 u ˆ = 0, γλ = inf J (u); ∈ M, βλ b b b 3 u u u 1 ≥ inf J (u); ∈ M, βλ − Ck f k E ∗ − ν. = 0, γλ = b b b 3 Choose 0 < C3 < C2 such that, if k f k E ∗ ≤ C3 , then u u 1 1 u Ck f k E ∗ < inf J (u); ∈ M, βλ = 0, γλ = 2 b b b 3 − sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } . λ

Hence, if k f k E ∗ ≤ C3 , then u u 1 u ˆ inf J (u); ∈ M, βλ = 0, γλ = b b b 3 u u 1 u u 1 u u = 0, γλ = − inf J (u); ∈ M, βλ = 0, γλ = > inf J (u); ∈ M, βλ b b b 3 b b b 3 − sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } λ

= sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } λ

Z f · bT 1 ,x0 ◦ Tε,y (ϕ)dx; (y, ε) ∈ ∂ K ≥ sup{J (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K } + sup − λ λ Ω Z f · bT 1 ,x0 ◦ Tε,y (ϕ)dx; (y, ε) ∈ ∂ K ≥ sup J (bT 1 ,x0 ◦ Tε,y (ϕ)) − λ

λ

Ω

= sup{ Jˆ(bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ ∂ K }. λ

N

Therefore, for 0 < µ < N2 S 2 − sup{I0 (bT 1 ,x0 ◦ Tε,y (ϕ)); (y, ε) ∈ K }, there exists C = min{C1 , C2 , C3 } such that, λ if k f k E ∗ ≤ C, then the conclusion is true. Lemma 4.5. Assume (A3 ) and kαk

2

J

N 2

(R N )

< S(2 N − 1). Then, there exists λ¯ > 0 such that, if λ > λ¯ , k f k E ∗ ≤ C

and µ > 0 small enough, then I f has at least a critical point u 4 satisfying 2 N 1 N S 2 + I f (u loc ) < I f (u 4 ) < S 2 − µ. N N Proof. We argue by contradiction. Suppose that I f has no critical point in [A, B]. By Lemma 4.3, 1 N 2 N S 2 + I f (u loc ) < A ≤ B < S 2 − µ. N N N

We see from Lemma 4.2 that I f satisfies the (PS)c condition if c ∈ ( N1 S 2 + I f (u loc ), C ∈ (A, B) and a continuous mapping Γ : I fB × [0, 1] → I fB such that Γ (u, 0) = u,

2 N

N

S 2 − µ). Then, there exist

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Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

Γ (u, 1) ∈ I Cf Γ (u, t) = u

∀u ∈ I fB , ∀t ∈ [0, 1] ∀u ∈ I Cf ,

where M+ = {u ∈ M, u + 6≡ 0}. By Lemma 4.4, we obtain {bT 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ ∂ K } ⊆ I Cf , λ

{bT 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ K } ⊆ I fB . λ

Hence, Γ [bT 1 ,x0 ◦ Tε,y (ϕ), t] = bT 1 ,x0 ◦ Tε,y (ϕ) λ

∀(y, ε) ∈ ∂ K ∀t ∈ [0, 1];

λ

(4.7)

sup{I f ◦ Γ [bT 1 ,x0 ◦ Tε,y (ϕ), 1]; (y, ε) ∈ K } ≤ C < A.

(4.8)

λ

We define a continuous function χ : K × [0, 1] → R N × R by 1 χ (y, ε, t) = ((1 − 2t)y + 2tβ ◦ Tε,y (ϕ), (1 − 2t)ε + 2tγ ◦ Tε,y (ϕ)) ∀(y, ε) ∈ K ∀t ∈ 0, , 2 1 1 χ (y, ε, t) = βλ ◦ Γ [bT 1 ,x0 ◦ Tε,y (ϕ), 2t − 1], γλ ◦ Γ [bT 1 ,x0 ◦ Tε,y (ϕ), 2t − 1] λ λ b b 1 ,1 . ∀(y, ε) ∈ K ∀t ∈ 2 Since {bT 1 ,x0 ◦ Tε,y (ϕ); (y, ε) ∈ K } ⊆ I fB , λ

then χ is well defined. We may verify that 1 1 ∀(y, ε) ∈ ∂ K ∀t ∈ 0, . χ (y, ε, t) = ϑ(y, ε, 2t) 6= 0, 3 2 By (4.7), χ (y, ε, t) = (βλ ◦ T 1 ,x0 ◦ Tε,y (ϕ), γλ ◦ T 1 ,x0 ◦ Tε,y (ϕ)) = $ λ λ 1 ∀(y, ε) ∈ ∂ K ∀t ∈ ,1 . 2

1 y, ε, 2

1 6 = 0, 3

Thus, there exists ( y¯ , ε¯ ) ∈ K such that 1 1 1 γλ ◦ Γ [bT 1 ,x0 ◦ Tε¯ , y¯ (ϕ), 1] = . βλ ◦ Γ [bT 1 ,x0 ◦ Tε¯ , y¯ (ϕ), 1] = 0, λ λ b b 3 It results that u 1 u u I f ◦ Γ [bT 1 ,x0 ◦ Tε¯ , y¯ (ϕ), 1] ≥ inf I f (u); ∈ M+ , βλ = 0, γλ = = A > C. λ b b b 3 This contradicts (4.8). The proof is complete.

Proof of Theorem. By Lemma 2.2, if k f k E ∗ is small, then (1.1) has a solution u loc and I f (u loc ) < 0. By (ii) of Lemma 2.1, we see that u loc is a positive solution of (1.1). (i) then follows. (ii) has been proved in Section 3. Lemmas 2.1 and 4.5 give (iii). Acknowledgement This work is supported by the National Natural Sciences Foundation of China, grant no. 10571175 and 10631030. References [1] A. Ambrosetti, Critical points and nonlinear variational problems, Bull. Soc. Math. France 120 (49) (1992) Memoire. [2] S. Adachi, K. Tanaka, Four positive solutions for the semilinear elliptic equation: −∆u + u = a(x)u p + f (x) in R N , Calc. Var. 11 (2000) 63–95.

Y. Wan, J. Yang / Nonlinear Analysis 68 (2008) 2569–2593

2593

[3] A. Bahri, Y.Y. Li, On the min-max procedure for the existence of a positive solution for certain scalar field equations in R N , Rev. Mat. Iberoamericana 6 (1990) 1–15. [4] V. Benci, G. Cerami, Existence of positive solutions of the equation −∆u + a(x)u = u N +2/N −2 in R N , J. Funct. Anal. 88 (1990) 90–117. [5] H. Brezis, L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 (1983) 437–477. ∗ [6] D. Passaseo, Some sufficient conditions for the existence of positive solutions to the equation −∆u + a(x)u = u 2 −1 in bounded domains, Ann. Inst. H. Poincar´e Anal. Nonli. 13 (1996) 185–227. [7] J.T. Schwartz, Nonlinear Functional Analysis, Gordon Breach, 1969. [8] M. Struwe, Variational Methods, Springer-Verlag, 1996. [9] G. Tarantello, On nonhomogeneous elliptic equations involving critical Sobolev exponent, Ann. Inst. H. Poincar´e Anal. Nonli. 9 (1992) 281–304. [10] M. Willem, Minimax Theorem, Birkh¨auser, Berlin, 1996. [11] X.P. Zhu, A perturbation result on positive entire solutions of a semilinear elliptic equation, J. Differential Equations 92 (1991) 163–178.

Multiple solutions for inhomogeneous critical semilinear elliptic problems

Multiple solutions for inhomogeneous critical semilinear elliptic problems

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