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Multiple sign-changing solutions to semilinear elliptic resonant problems
Nonlinear Analysis 72 (2010) 3820–3827
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Multiple sign-changing solutions to semilinear elliptic resonant problemsI Fuyi Li ∗ , Yuhua Li School of Mathematical Sciences, Shanxi University, Taiyuan 030006, People’s Republic of China
article
abstract
info
Article history: Received 7 September 2009 Accepted 12 January 2010
In this paper, we use the topological degree theory and the critical groups to investigate the elliptic equation −∆u = f (x, u) in Ω and subject to u = 0 on ∂ Ω , and establish a multiple solutions theorem which guarantees that this problem has at least six nontrivial solutions under some resonant conditions. If this problem has only finitely many solutions then, of these solutions, there are two positive solutions, two negative solutions and two sign-changing solutions. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Sign-changing solution Elliptic resonant problem Topological degree Critical group
1. Introduction In this paper, we discuss the existence of positive solutions, negative solutions and sign-changing solutions to the following semilinear elliptic Dirichlet boundary value problem (BVP):
−∆u(x) = f (x, u(x)), u|∂ Ω = 0.
x ∈ Ω,
(1.1)
Here, Ω ⊂ RN is a bounded domain with smooth boundary and f ∈ C 1 (Ω × R1 , R1 ) satisfies the subcritical growth condition
|ft0 (x, t )| 6 C (1 + |t |p−2 ),
( x, t ) ∈ Ω × R 1 ,
(1.2)
where C > 0 is a constant, p ∈ (2, 2 ) and 2 = 2N /(N − 2) if N > 3; 2 = ∞ if N = 1, 2. Let λ1 < λ2 6 λ3 6 · · · 6 λn 6 · · · be all the eigenvalues of −∆u = λu in Ω and subject to u = 0 on ∂ Ω , counting with their multiplicity, the corresponding eigenfunctions are denoted by e1 , e2 , . . . , en , . . .. Since e1 has constant sign, we may let e1 > 0 in Ω . We now assume that the nonlinearity f satisfies the following conditions. ∗
∗
∗
(f1 ) f ∈ C 1 (Ω × R1 , R1 ), and f (x, t )t > 0 for all (x, t ) ∈ Ω × R1 ; (f2 ) ft0 (x, 0) = λ2n0 +1 for some n0 > 1 with λ2n0 < λ2n0 +1 and for all x ∈ Ω . Moreover, there exists δ > 0 such that f (x, t )t 6 λ2n0 +1 t 2 for all (x, t ) ∈ Ω × [−δ, δ]; (f3 ) there exist n1 > 1 with λ2n1 < λ2n1 +1 , C1 > 0, and α ∈ (0, 1) such that lim|t |→∞ f (x, t )/t = λ2n1 +1 exists uniformly for x ∈ Ω , and 1 1 2 |f (x, t ) − λ2n1 +1 t | 6 C1 (1 + |t |α ), (x, t ) ∈ Ω × R1 , lim F ( x , t ) − λ t = +∞ 2n + 1 1 |t |→∞ |t |2α 2 Rt uniformly for x ∈ Ω , where F (x, t ) = 0 f (x, s)ds for all (x, t ) ∈ Ω × R1 ; I Project supported by the National Natural Science Foundation of China (Grant No. 10771128) and the NSF of Shanxi Province (2006011002).
∗
Corresponding author. E-mail address: [email protected] (F. Li).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.01.018
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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(f4 ) there exists b > 0 such that |f (x, t )| < b for all (x, t ) ∈ Ω × [−bc , bc ], where c = maxx∈Ω e(x), and e is the solution of the BVP: −∆u = 1 in Ω and subject to u = 0 on ∂ Ω . The following theorem is the main result of this paper. Theorem 1.1. Suppose that the conditions (f1 )–(f4 ) hold. Then the BVP (1.1) has at least six nontrivial solutions. Moreover, if (1.1) has only finitely many solutions then, of these solutions, there are two positive solutions, two negative solutions and two sign-changing solutions. In recent years there have been many papers studying sign-changing solutions [1–9]. For example, by using the Morse theory and the topological degree theory, Liu and Sun in [8] obtained one positive solution and two sign-changing solutions to the elliptic problem (1.1) with jumping nonlinearities, where f satisfied the following assumptions: (f1 0 ) (f2 0 ) (f3 0 ) (f4 0 )
f ∈ C 1 (Ω × R1 , R1 ); f (x, 0) = 0 and λi < ft0 (x, 0) < λi+1 for some i > 2 and all x ∈ Ω ; lim supt →+∞ f (x, t )/t < λ1 uniformly in x ∈ Ω ; λ1 < lim inft →−∞ f (x, t )/t 6 lim supt →−∞ f (x, t )/t < +∞ uniformly in x ∈ Ω .
In [5], Bartsch, Chang and Wang assumed the following hypotheses on f with f (x, t ) = f (t ). (i) f ∈ C 1 (R1 , R1 ), f (0) = 0; (ii) lim sup|t |→∞ f (t )/t < λ1 ; (iii) lim|t |→∞ f 0 (t ) = ω. If ω = λk with eigenspace V = {u ∈ C0∞ (Ω ) : −∆u = ωu} then f (t ) − ωt is bounded, and
Z Ω
F (u(x)) −
1
F (u(x)) −
1
2
ωu2 (x) dx → ∞ for u ∈ V , kuk → ∞
or
Z Ω
2
ωu2 (x) dx → −∞ for u ∈ V , kuk → ∞.
They used the Morse theory to obtain that when (i) and (ii), or (i) and (iii) with ω < λ2 hold, and that f 0 (0) > λ2 , the BVP (1.1) has a sign-changing solution. Dancer and Du [2], used a Conley index to prove the BVP (1.1) with f (x, t ) = f (t ) has one positive solution, one negative solution and one sign-changing solution under some assumptions and the following conditions: (H0 ) limt →0+ f (t )/t = a > λ1 , limt →0− f (t )/t = b > λ1 and (H∞ ) lim sup|t |→∞ f (t )/t < λ1 ; or (H0 ) limt →0 f (t )/t < λ1 and (H∞ ) limt →+∞ f (t )/t = a > λ1 , limt →−∞ f (t )/t = b > λ1 . The authors [10,1,3,4,6] all proved that the BVP (1.1) with f (x, t ) = f (t ) has at least two sign-changing solutions provided that f is C 1 , f 0 (0) ∈ (λi , λi+1 ) for some i > 2 and lim sup|t |→∞ f (t )/t < λ1 . In [11], the authors used the topological degree theory to study the fourth-order ordinary differential equation boundary value problems, and they obtained six solutions, of which two are positive solutions, two negative solutions and two signchanging solutions. In the above papers [10,1–4,6,8], our case was not considered, f is not resonant at 0 or ∞. In the condition (iii), f may be resonant at ∞ to λk . However, they assumed that f (t ) − λk t is bounded. We, in this paper, consider the case that f is resonant at 0 and ∞, and f (t ) − λk t may not be bounded. The conditions (f2 ) and (f3 ) imply that f is resonant at 0 and ∞. We add the compressing condition (f4 ) to obtain more solutions. Moreover, the condition (f4 ) is valid (see [12]) and easily satisfied. In this paper, we combine the topological degree theory with the Morse theory to study the BVP (1.1) and obtain the same result with [11] in which they considered the case that f crosses even eigenvalue at 0 and ∞. We not only extend the result in [11] to the elliptic problems, but also extend it to the resonant case at 0 and ∞. Moreover, we prove that the BVP (1.1) has at least five solutions when the functional may not satisfy the (PS) condition. Of them, two are positive solutions, two negative solutions and one sign-changing solution. See Theorem 3.1 in Section 3. The present paper is organized as follows: In the Section 2, we give some lemmas and preliminaries which are useful to our main result. We prove the main theorems in Section 3. At last, we give an example to show that our work is significant. 2. Lemmas and preliminaries In the following, we first recall some lemmas and preliminaries about the critical group and the topological degree. Let H be a real Hilbert space, f ∈ C 1 (H , R1 ), and let K = {u ∈ H : f 0 (u) = 0} denote the critical set of f , fa = {u ∈ H : f (u) 6 a} denote the level sets of f for all a ∈ R1 .
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F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
Definition 2.1 ([12, Definition 4.1, p. 32][13]). Let p be an isolated critical point of f , and let c = f (p). We call Cq (f , p) = Hq (fc ∩ Up , (fc \ {p}) ∩ Up ; G) the qth critical group, with coefficient group G of f at p for all q ∈ N0 = {0, 1, . . .}, where Up is a neighborhood of p such that K ∩ (fc ∩ Up ) = {p}, and H∗ (X , Y ; G) stands for the singular relative homology groups with the Abelian coefficient group G. According to the excision property of the singular homology theory, the critical groups are well defined; i.e., they do not depend on a special choice of the neighborhood Up . If f satisfies the (PS) condition and the critical values of f are bounded from below by some a ∈ R1 , then the critical groups of f at infinity are defined by Bartsch and Li [14] as Cq (f , ∞) = Hq (H , fa ),
q ∈ N0 .
It follows from the deformation lemma that the critical groups at infinity do not depend on the choice of a. Using the two concepts, we have the following famous Morse inequality [12,13]: ∞ ∞ X X (−1)q Mq = (−1)q βq , q =0
where Mq =
q =0
P
rank Cq (f , p), and βq = rank Cq (f , ∞) for all q ∈ N0 .
p∈K
Lemma 2.1 ([12, Theorem 3.2, p. 100]). Let H be a real Hilbert space, and let f ∈ C 2 (H , R1 ) satisfy the (PS) condition. Assume that f 0 (x) = x − Ax,
x ∈ H,
where A is a compact mapping, and that p0 is an isolated critical point of f . Then we have ind(f 0 , p0 ) =
∞ X
(−1)q rank Cq (f , p0 ).
q =0
The following lemma establishes the relationship of fixed point index and topological degree. As X = P is a cone, the idea of this result was given by Liu [15]. See also [11,16]. Let X be a retract of a real Banach space E, U be a relatively open subset of X , A : closX U → X be a completely continuous operator. Suppose that A has not fixed points on ∂X U and that the fixed point set of A is bounded. Then we have the next lemma. Lemma 2.2. If any fixed point of A in U is the interior point of X , then there exists a bounded open subset O of E with O ⊂ U such that O contains all fixed points of A in U and deg(I − A, O, 0) = i(A, U , X ). Proof. Let S = {u ∈ U : u = Au}. Then S ⊂ X ◦ , where X ◦ denotes the interior of X . Thus,Sfor any u ∈ S, there exists δu ∈ (0, 1) such that B(u, δu ) ⊂ X and B(u, δu ) ∩ X ⊂ U, and then B(u, δu ) ⊂ U. Now let O = u∈S B(u, δu ). Then O ⊂ U is a bounded open subset of E. By the excision property of the fixed point index, we have i(A, U , X ) = i(A, O, X ). By the definition of the fixed point index, we have i(A, O, X ) = deg(I − A ◦ r , B(0, R) ∩ r −1 (O), 0), where r : E → X is an arbitrary retraction and R is a large enough positive number such that S ⊂ B(0, R). Now we assume that u∗ ∈ B(0, R) ∩ r −1 (O) such that u∗ = A ◦ r (u∗ ). Since r : E → X and A : closX U → X , it follows that u∗ ∈ X and u∗ = r (u∗ ) ∈ O. Thus, u∗ ∈ O whenever u∗ ∈ B(0, R) ∩ r −1 (O) is a fixed point of A ◦ r. By the excision property of the topological degree, we have deg(I − A ◦ r , B(0, R) ∩ r −1 (O), 0) = deg(I − A ◦ r , O, 0) = deg(I − A, O, 0). We complete the proof.
Here, the fixed point index can be defined on an unbounded set. See [17, p. 238]. Now we turn to consider the BVP (1.1). Let H01 (Ω ) be the Sobolev space with the inner product
(u, v) =
Z Ω
∇ u(x)∇v(x)dx
and the corresponding norm kuk = (u, u)1/2 for all u, v ∈ H01 (Ω ).
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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Define the functional by J (u) =
1
Z
2
Ω
|∇ u(x)|2 dx −
Z Ω
F (x, u(x))dx,
u ∈ H01 (Ω ).
By the growth condition (1.2), J is well defined and of C 2 , and J 0 = I − (−∆)−1 f. It is well known that the solution of the BVP (1.1) is equivalent to the critical point of the functional J. Moreover, the critical point of the functional J is exactly the solution of the operator equation u = (−∆)−1 fu := Au. By the Sobolev theorem, A : H01 (Ω ) → H01 (Ω ) is completely continuous. And it follows from [18, Theorem 4.2] that A : C01 (Ω ) → C01 (Ω ) is also completely continuous. Remark 2.1. Let O be a bounded open subset of H01 (Ω ), there be no zero points of I − A on ∂ O. Since C01 (Ω ) can be compactly embedded into H01 (Ω ), it follows from the bootstrap argument and the definition of Leray–Schauder degree that degH 1 (Ω ) (I − A, O, 0) = degC 1 (Ω ) (I − A, C01 (Ω ) ∩ O, 0). 0 0 In what follows, degC 1 (Ω ) is denoted simply by deg. 0
Remark 2.2. Remark 2.1 implies that two topological degrees in both H01 (Ω ) and C01 (Ω ) are same. Combining with Lemma 2.1, we can obtain the connection between the topological degree in C01 (Ω ) and the critical group, deg(I − A, O, 0) =
∞ X X (−1)q rank Cq (J , p0 ). p0 ∈K ∩O q=0
Here and in what follows, we always assume that J has only finitely many critical points. The next lemma is mainly used to obtain the (PS) condition and the critical groups of J at ∞. Lemma 2.3. Suppose that (f3 ) holds. Then J satisfies the (PS) condition and Cq (J , ∞) = δq,2n1 G for all q ∈ N0 . Proof. Let B = I − λ2n1 +1 (−∆)−1 , and G(u) = Ω [F (x, u(x)) − 21 λ2n1 +1 u2 (x)]dx for all u ∈ H01 (Ω ). Then B is a bounded selfadjoint operator defined on H01 (Ω ), and G ∈ C 1 (H01 (Ω ), R1 ) has a compact differential G0 = (−∆)−1 f − λ2n1 +1 (−∆)−1 . According to the spectral decomposition of the operator B, we have H01 (Ω ) = H− ⊕ H0 ⊕ H+ , where H± , H0 are invariant subspaces corresponding to the positive/negative, and zero spectrum of B respectively. In fact, H0 = ker B, H− = span{e1 , e2 , . . . , e2n1 }, and H+ = (H− ⊕ H0 )⊥ . Consequently, B± := B|H± has a bounded inverse on H± , and dim(H− ⊕ H0 ) < ∞. By the condition (f3 ), we have
R
Z |(G (u), v)| = [f (x, u(x)) − λ2n1 +1 u(x)]v(x)dx ZΩ 6 C1 (1 + |u(x)|α )|v(x)|dx 0
Ω
6 C1 [kvk1 + kukαα 2∗ kvk2N /(N +2) ] 6 C2 (1 + kukα )kvk,
u, v ∈ H01 (Ω ),
where C2 > 0 is a constant. Thus, kG0 (u)k 6 C2 (1 + kukα ) for all u ∈ H01 (Ω ). On the other hand, it follows from the condition (f3 ) that for any R > 0 there exists C3 > 0 such that F (x, t ) −
1 2
λ2n1 +1 t 2 > R|t |2α − C3 ,
( x, t ) ∈ Ω × R 1 .
Since all norms are equivalent on H0 , there exists C4 > 0 such that C4 kuk2α 6 G(u) > R
Z Ω
|u(x)|2α dx − C3 m(Ω ) > RC4 kuk2α − C3 m(Ω ),
R
Ω
|u(x)|2α dx for all u ∈ H0 . Thus,
u ∈ H0 ,
and G(u) → +∞, kuk2α
u ∈ H0 , kuk → ∞.
According to the proposition in [19], also see [20, Proposition 2.1], the conclusions hold.
Lemma 2.4. If the conditions (f2 ) and (f3 ) hold, then Cq (J , 0) = δq,2n0 G for all q ∈ N0 . Proof. By the condition (f3 ) and Lemma 2.3, we get that J satisfies the (PS) condition. It is easy to see that 0 is a degenerate critical point of J with the Morse index 2n0 . By the shifting theorem [12,13], Cq (J , 0) = Cq−2n0 (e J , 0) for all q ∈ N0 , wheree J is
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F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
the restriction of J to N := ker J 00 (0). By the condition (f2 ), we have F (x, t ) 6
λ2n0 +1 t 2 for all (x, t ) ∈ Ω × [−δ, δ]. For this δ > 0, there exists δ1 > 0 such that |u(x)| 6 δ for all x ∈ Ω and u ∈ Bδ1 ∩ N, where Bδ1 = {u ∈ H01 (Ω ) : kuk < δ1 }. Thus, Z Z 1 1 1 (2.1) F (x, u(x))dx > kuk2 − λ2n0 +1 |u(x)|2 dx = 0, u ∈ Bδ1 ∩ N . J (u) = kuk2 − 2
2
Ω
2
1 2
Ω
If there exists r ∈ (0, δ1 ] such that for any u ∈ (Br \{0})∩ N, there exists x0 ∈ Ω satisfying f (x0 , u(x0 ))u(x0 ) < λ2n0 +1 |u(x0 )|2 , then F (x0 , u(x0 )) =
u(x0 )
Z
f (x0 , s)ds <
0
1 2
λ2n0 +1 |u(x0 )|2 .
It follows from (2.1) that J (u) < 0. Thus, 0 is a local minimum ofe J, and Cq (e J , 0) = δq0 G. Therefore, J , 0) = δq,2n0 G, Cq (J , 0) = Cq−2n0 (e
q ∈ N0 .
If not, there exists {un } ⊂ (Bδ1 \ {0}) ∩ N such that f (x, un (x)) = λ2n0 +1 un (x) for all x ∈ Ω and un → 0 as n → ∞, then every un is a solution to the BVP (1.1), and the BVP (1.1) has infinitely many solutions. This contradicts the assumption. The proof is completed. By Lemma 2.4 and Remark 2.2, it is easy to see that Lemma 2.5. Suppose that (f2 ) and (f3 ) hold. Then there exists r0 > 0 such that for all r ∈ (0, r0 ], deg(I − A, B(0, r ), 0) = 1, where B(0, r ) = {u ∈ C01 (Ω ) : kuk < r } denotes an open ball in C01 (Ω ). The following two lemmas deal with the fixed point index. Let X = {u ∈ C01 (Ω ) : −be(x) 6 u(x) 6 be(x), x ∈ Ω }, P = {u ∈ C01 (Ω ) : u(x) > 0, x ∈ Ω }. Then X and P are nonempty closed convex subsets of C01 (Ω ), and P is also a cone in C01 (Ω ). Lemma 2.6. Suppose that (f1 ) and (f2 ) hold. There exists r1 ∈ (0, r0 ] such that for all r ∈ (0, r1 ], i(A, P ∩ B(0, r ), P ) = 0,
i(A, (−P ) ∩ B(0, r ), −P ) = 0.
Proof. By the condition (f1 ), A : P → P is completely continuous. It follows from the condition (f2 ) that there exist δ1 > 0 and ρ1 ∈ (0, r0 ] such that f (x, t ) > λ1 (1 + δ1 )t ,
(x, t ) ∈ Ω × [0, ρ1 ].
(2.2)
If u = Au + µe1 for some µ > 0 and u ∈ P ∩ ∂ B(0, r ), where r ∈ (0, ρ1 ] is a positive number; that is
−∆u(x) = f (x, u(x)) + µλ1 e1 (x), u|∂ Ω = 0,
x ∈ Ω,
(2.3)
then we have from (2.2) that
−∆u(x) > λ1 (1 + δ1 )u(x),
x ∈ Ω.
Thus,
λ1
Z Ω
u(x)e1 (x)dx > λ1 (1 + δ1 )
Z Ω
u(x)e1 (x)dx,
and this is a contradiction. Therefore, according to the property of fixed point index [21, Corollary 2.3.1, p. 90], we get i(A, P ∩ B(0, r ), P ) = 0. Similarly, we can also show that there exists ρ2 ∈ (0, r0 ] such that i(A, (−P ) ∩ B(0, r ), −P ) = 0 for all r ∈ (0, ρ2 ]. Let r1 = min{ρ1 , ρ2 }. Then the conclusion holds. Lemma 2.7. Suppose that (f1 ) and (f4 ) hold. Then there exists a bounded open subset O of C01 (Ω ) with 0 ∈ O ⊂ X such that O contains all fixed points of A in X , and i(A, P ∩ O, P ) = 1, deg(I − A, O, 0) = 1.
i(A, (−P ) ∩ O, −P ) = 1, (2.4)
Proof. It follows from the condition (f4 ) that A(X ) ⊂ X . Since the fixed point set of A in X is bounded, i(A, X , X ) = 1. If u = Au for some u ∈ X , then u satisfies (1.1). By the condition (f4 ) and the strong maximum principle, we have u ∈ X ◦ .
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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Thus, Lemma 2.2 implies that there exists a bounded open subset O of C01 (Ω ) with 0 ∈ O ⊂ X such that O contains all the fixed points of A in X , and deg(I − A, O, 0) = i(A, X , X ) = 1. In the same way, if there are µ ∈ [0, 1] and u ∈ P ∩ X such that u = µAu, then u ∈ X ◦ . According to the homotopy invariance of the fixed point index, we have i(A, P ∩ X ◦ , P ) = 1. Since any fixed point of A in P ∩ X ◦ belongs to P ∩ O, it follows from the excision property of the fixed point index that i(A, P ∩ X ◦ , P ) = i(A, P ∩ O, P ) = 1. Similarly, we can also obtain that i(A, (−P ) ∩ O, −P ) = 1. Thus, we complete the proof. Since X is not bounded in C01 (Ω ), it follows X 6⊂ B(0, R) for any R > 0. We, however, convert the unbounded set X into a bounded set O by Lemma 2.7. Therefore, the proof of our main theorem becomes more easy. Lemma 2.8. Let (f1 ) hold. Suppose that limt →+∞ f (x, t )/t > λ1 and limt →−∞ f (x, t )/t > λ1 uniformly for x ∈ Ω , and that there are C1 > 0 and C2 > 0 such that |f (x, t )| 6 C1 |t | + C2 for all (x, t ) ∈ Ω × R1 . Then there exists R0 > r0 such that for all R > R0 , i(A, P ∩ B(0, R), P ) = 0,
i(A, (−P ) ∩ B(0, R), −P ) = 0.
Proof. We only prove that i(A, P ∩ B(0, R), P ) = 0. Since limt →+∞ f (x, t )/t > λ1 uniformly for x ∈ Ω , there exist δ > 0 and C3 > 0 such that f (x, t ) > λ1 (1 + δ)t − C3 ,
(x, t ) ∈ Ω × [0, ∞).
(2.5)
Define H (µ, u) = Au + µe1 for all (µ, u) ∈ [0, ∞) × P. If u ∈ P and µ > 0 satisfy H (µ, u) = u, then u and µ satisfy (2.3). Consequently, it follows from (2.5) that
−∆u(x) > λ1 (1 + δ)u(x) − C3 + µλ1 e1 (x),
x ∈ Ω,
and
λ1
Z Ω
Z
u(x)e1 (x)dx > λ1 (1 + δ)
Ω
u(x)e1 (x)dx − C3
Z Ω
e1 (x)dx + µλ1
Z Ω
e21 (x)dx.
This implies µλ1 Ω e21 (x)dx 6 C3 Ω e1 (x)dx. Thus, there is a µ0 > 0 such that for µ > µ0 , H (µ, ·) has no fixed points in P. To show the existence of such a constant R1 > 0 which is able to insure that kuk < R1 if u ∈ P is a solution of u = H (µ, u) for some µ > 0, we adopt an indirect argument. Suppose that there exists a sequence {(µn , un )} ⊂ [0, ∞) × P with kun k → ∞ satisfying un = H (µn , un ). Then (µn , un ) satisfies the following BVP
R
R
−∆un (x) = f (x, un (x)) + µn λ1 e1 (x), un |∂ Ω = 0.
x ∈ Ω,
(2.6)
Since f (x, t ) 6 C1 t + C2 for all (x, t ) ∈ Ω × [0, ∞), we have from (2.6) that f (x, un )un
Z 1= Ω
k un k
2
dx + µn λ1
Z
Z
e 1 un Ω
k un k
dx 6 C1 2
Ω
vn2 dx + C2
vn
Z Ω
k un k
dx + µn λ1
e1 vn
Z Ω
k un k
dx,
(2.7)
where vn = un /kun k for all n = 1, 2, . . .. Since H01 (Ω ) can be compactly embedded into L2 (Ω ), there exists a converging subsequence {vnk } of {vn } in L2 (Ω ), and we let vnk → v in RL2 (Ω ). It follows from Riesz’s theorem that v(x) > 0, a.e. x ∈ Ω . By µn 6 µ0 and kun k → ∞, we get from (2.7) that 1 6 C1 Ω v 2 dx. This implies that v > 0, a.e. x ∈ Ω , and v 6= 0. On the other hand, we have from (2.5) and (2.6) that
λ1
Z Ω
vnk e1 dx > λ1 (1 + δ)
Z Ω
vnk e1 dx − C3
Z
e1 Ω
kunk k
dx.
Letting k → ∞, we have that Ω v e1 dx 6 0, which is impossible. Thus, there exists R1 > 0 such that kuk 6 R1 for any (µ, u) ∈ [0, ∞) × P with u = H (µ, u), and a standard bootstrap argument implies that there exists R01 > 0 such that kukC 1 (Ω ) 6 R01 for all such u. Therefore, if R > R01 , then u 6= H (µ, u) for all u ∈ P ∩ ∂ B(0, R) and µ > 0, and the property of
R
0
fixed point index [21] implies that i(A, P ∩ B(0, R), P ) = 0. Similarly, there exists R02 > 0 such that i(A, (−P ) ∩ B(0, R), −P ) = 0 for all R > R02 . Let R0 = max{R01 , R02 }. Then the conclusion is true. Remark 2.3. In the proof of Lemma 2.8, the following property does not hold
(−∆)−1 u > e0 (x)k(−∆)−1 ukC 1 (Ω ) , 0
u ∈ P,
where e0 (x) > 0 for x ∈ Ω is some function. However, we still prove that the conclusion holds.
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F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
3. Proof of main theorems The Proof of Theorem 1.1. The proof of Lemma 2.8 implies that there exists R > 0 large enough such that O ⊂ B(0, R) and B(0, R) contains all positive and negative fixed points of A. It follows from Lemmas 2.7 and 2.8 that i(A, P ∩ (B(0, R) \ O), P ) = 0 − 1 = −1,
(3.1)
i(A, (−P ) ∩ (B(0, R) \ O), −P ) = 0 − 1 = −1.
(3.2)
Consequently, the BVP (1.1) has at least a positive solution u1 ∈ P ∩ (B(0, R) \ O) and a negative solution u2 ∈ (−P ) ∩ (B(0, R) \ O). If u ∈ P \ {0} is a solution of the BVP (1.1), then the condition (f1 ) and the strong maximum principle imply that u ∈ P ◦ . Similarly, if u ∈ −P \ {0} is a solution of the BVP (1.1), then u ∈ (−P )◦ . It follows from Lemma 2.2, (3.1) and (3.2) that there exist two bounded open subsets U1 ⊂ P ∩ (B(0, R) \ O) containing all positive fixed points of A in P ∩ (B(0, R) \ O) and U2 ⊂ (−P ) ∩ (B(0, R) \ O) containing all negative fixed points of A in (−P ) ∩ (B(0, R) \ O) such that deg(I − A, U1 , 0) = −1,
deg(I − A, U2 , 0) = −1.
If there are no more critical points of J in Remark 2.2, (2.4) and (3.3) that 1 = (−1)2n1 =
∞ X
C01
(3.3)
(Ω ) \ (O ∪ U 1 ∪ U 2 ), then it follows from the Morse inequality, Lemma 2.3,
(−1)q rank Cq (J , ∞) = 1 + (−1) + (−1) = −1,
q =0
which is impossible. Thus, there exists a solution u3 ∈ C01 (Ω ) \ (O ∪ U 1 ∪ U 2 ), and this is a sign-changing solution. On the other hand, by Lemmas 2.6 and 2.7, we have that for all r ∈ (0, r0 ] satisfying B(0, r ) ⊂ O, i(A, P ∩ (O \ B(0, r )), P ) = 1,
i(A, (−P ) ∩ (O \ B(0, r )), −P ) = 1,
where B(0, r ) denotes B(0, r ). Thus, the BVP (1.1) has at least a positive solution u4 ∈ P ∩(O \ B(0, r )) and a negative solution u5 ∈ (−P ) ∩ (O \ B(0, r )). From again Lemma 2.2, there exist U3 ⊂ P ∩ (O \ B(0, r )) containing all positive solutions in O \ B(0, r ) and U4 ⊂ (−P ) ∩ (O \ B(0, r )) containing all negative solutions in O \ B(0, r ), and deg(I − A, U3 , 0) = 1,
deg(I − A, U4 , 0) = 1.
(3.4)
It follows from Lemma 2.5 that for all r ∈ (0, r0 ] with B(0, r ) ⊂ O, deg(I − A, B(0, r ), 0) = 1.
(3.5)
If A has no fixed points in O \ (B(0, r ) ∪ U 3 ∪ U 4 ), then the additivity property of the topological degree implies that deg(I − A, O, 0) = deg(I − A, B(0, r ), 0) + deg(I − A, U3 , 0) + deg(I − A, U4 , 0). By (2.4), (3.4) and (3.5), we have 1 = 1 + 1 + 1. This is a contradiction. Thus, the BVP (1.1) has at least a solution u6 ∈ O \ (B(0, r ) ∪ U 3 ∪ U 4 ). This is a sign-changing solution. The proof is completed. Theorem 3.1. Let (f1 ), (f2 ) and (f4 ) hold. Suppose that limt →+∞ f (x, t )/t > λ1 and limt →−∞ f (x, t )/t > λ1 uniformly for x ∈ Ω , and that there are two constants C1 > 0 and C2 > 0 such that |f (x, t )| 6 C1 |t | + C2 for all (x, t ) ∈ Ω × R1 . Then the BVP (1.1) has at least five solutions. Of them, two positive solutions, two are negative solutions and one sign-changing solution. Proof. When the condition (f3 ) does not hold, J may not satisfy the (PS) condition. Therefore, we define g ∈ C 1 (Ω × R1 , R1 ) as g (x, t ) = ω(t )f (x, t ) for all (x, t ) ∈ Ω × R1 , where ω ∈ C0∞ (R1 ) satisfies ω(t ) = 1 for all t ∈ [−bc , bc ], ω(t ) = 0 for all |t | > 2bc, and |ω(t )| 6 1 for all t ∈ R1 . Let J 1 ( u) =
1 2
where G(x, t ) =
kuk2 −
Z Ω
Rt 0
G(x, u(x))dx,
u ∈ H01 (Ω ),
g (x, s)ds for all (x, t ) ∈ Ω × R1 . It follows from the definition of g that there is a constant C > 0 such
that |g (x, t )| 6 C for all (x, t ) ∈ Ω × R1 . Thus, J1 satisfies the (PS) condition and the critical points of J1 in X are all the critical points of J in X . The definition of g implies that g satisfies the conditions (f1 ), (f2 ) and (f4 ). Therefore, we have from the proof of Theorem 1.1 that J1 has at least three critical points in O ⊂ X , of them, there are one positive critical point, one negative critical point and one sign-changing critical point. Therefore, the BVP (1.1) has at least three solutions in O; one positive solution, one negative solution and one sign-changing solution. By Lemmas 2.7 and 2.8, there exists R > 0 with O ⊂ B(0, R) such that i(A, P ∩ (B(0, R) \ O), P ) = 0 − 1 = −1, i(A, (−P ) ∩ (B(0, R) \ O), −P ) = 0 − 1 = −1. Thus, the BVP (1.1) has at least one positive solution in P ∩(B(0, R)\ O) and one negative solution in (−P )∩(B(0, R)\ O).
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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Example 3.1. Let n be a positive integer such that λ2n0 +1 c /n < 1, f (x, t ) = f (t ) be an odd function defined as follows
2 − 2n t 2 λ2n0 +1 te c 2 , c 2nt − c − 21 f (t ) = λ2n0 +1 e 1 + cos π , 4n 2nc − c p 3√ 2 λ2n1 +1 (t − c ) 1 − e−(t −c ) + 3 t − c ln 1 + 3 (t − c )4 , 4
h
c
, h c 2n t ∈ ,c , t ∈ 0,
2n
t ∈ [c , ∞).
Then (f1 )– (f4 ) hold for α = 5/8 and b = 1. According to Theorem 1.1, the BVP (1.1) has at least six solutions. References [1] K.-C. Chang, S. Li, J. Liu, Remarks on multiple solutions for asymptotically linear elliptic boundary value problems, Topol. Methods Nonlinear Anal. 3 (1994) 179–187. [2] E. Dancer, Y. Du, Multiple solutions of some semilinear elliptic equations via the generalized Conley index, J. Math. Anal. Appl. 189 (1995) 848–871. [3] T. Bartsch, Z.-Q. Wang, On the existence of sign-changing solutions for semilinear Dirichlet problems, Topol. Methods Nonlinear Anal. 7 (1996) 115–131. [4] E. Dancer, Y. Du, The generalized Conley index and multiple solutions of some semilinear elliptic problems, Abstract Appl. Anal. 1 (1996) 103–135. [5] T. Bartsch, K.-C. Chang, Z.-Q. Wang, On the Morse indices of sign changing solutions of nonlinear elliptic problems, Math. Z. 233 (2000) 655–677. [6] S. Li, Z.-Q. Wang, Mountain pass theorem in order intervals and multiple solutions for semilinear elliptic Dirichlet problems, J. Anal. Math. 81 (2000) 373–396. [7] S. Li, Z.-Q. Wang, Ljusternik–Schnirelman theory in partially ordered Hilbert spaces, Trans. Amer. Math. Soc. 354 (2002) 3207–3227. [8] Z. Liu, J. Sun, An elliptic problem with jumping nonlinearities, Nonlinear Anal. 63 (2005) 1070–1082. [9] W. Zou, On finding sign-changing solutions, J. Funct. Anal. 234 (2006) 364–419. [10] H. Hofer, Variational and topological methods in partially order Hilbert spaces, Math. Ann. 261 (1982) 493–514. [11] C. Pang, W. Dong, Z. Wei, Multiple solutions for fourth-order boundary value problems, J. Math. Anal. Appl. 314 (2006) 464–476. [12] K.-C. Chang, Infinite Dimensional Morse Theory and Multiple Solution Problems, Birkhäuser, Boston, 1993. [13] J. Mawhin, M. Willem, Critical Point Theory and Hamiltonian Systems, Springer, New York, 1989. [14] T. Bartsch, S. Li, Critical point theory for asymptotically quadratic functionals and applications to problems with resonance, Nonlinear Anal. 28 (1997) 419–441. [15] Z. Liu, On Dancer’s conjecture and multiple solutions of elliptic differential equations, Northeast. Math. J. 9 (1993) 388–394. [16] X. Xu, Multiple sign-changing solutions for some m-point boundary value problems, Electron. J. Differential Equations 89 (2004) 1–14. [17] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, Berlin Heidelberg, 1985. [18] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Rev. 18 (1976) 620–709. [19] S. Li, J. Liu, Computations of critical groups at degenerate critical point and applications to nonlinear differential equations with resonance, Houston J. Math. 25 (1999) 563–582. [20] W. Zou, S. Li, J. Liu, Nontrivial solutions for resonant cooperative elliptic systems via computations of critical groups, Nonlinear Anal. 38 (1999) 229–247. [21] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988.
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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Multiple sign-changing solutions to semilinear elliptic resonant problemsI Fuyi Li ∗ , Yuhua Li School of Mathematical Sciences, Shanxi University, Taiyuan 030006, People’s Republic of China
article
abstract
info
Article history: Received 7 September 2009 Accepted 12 January 2010
In this paper, we use the topological degree theory and the critical groups to investigate the elliptic equation −∆u = f (x, u) in Ω and subject to u = 0 on ∂ Ω , and establish a multiple solutions theorem which guarantees that this problem has at least six nontrivial solutions under some resonant conditions. If this problem has only finitely many solutions then, of these solutions, there are two positive solutions, two negative solutions and two sign-changing solutions. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Sign-changing solution Elliptic resonant problem Topological degree Critical group
1. Introduction In this paper, we discuss the existence of positive solutions, negative solutions and sign-changing solutions to the following semilinear elliptic Dirichlet boundary value problem (BVP):
−∆u(x) = f (x, u(x)), u|∂ Ω = 0.
x ∈ Ω,
(1.1)
Here, Ω ⊂ RN is a bounded domain with smooth boundary and f ∈ C 1 (Ω × R1 , R1 ) satisfies the subcritical growth condition
|ft0 (x, t )| 6 C (1 + |t |p−2 ),
( x, t ) ∈ Ω × R 1 ,
(1.2)
where C > 0 is a constant, p ∈ (2, 2 ) and 2 = 2N /(N − 2) if N > 3; 2 = ∞ if N = 1, 2. Let λ1 < λ2 6 λ3 6 · · · 6 λn 6 · · · be all the eigenvalues of −∆u = λu in Ω and subject to u = 0 on ∂ Ω , counting with their multiplicity, the corresponding eigenfunctions are denoted by e1 , e2 , . . . , en , . . .. Since e1 has constant sign, we may let e1 > 0 in Ω . We now assume that the nonlinearity f satisfies the following conditions. ∗
∗
∗
(f1 ) f ∈ C 1 (Ω × R1 , R1 ), and f (x, t )t > 0 for all (x, t ) ∈ Ω × R1 ; (f2 ) ft0 (x, 0) = λ2n0 +1 for some n0 > 1 with λ2n0 < λ2n0 +1 and for all x ∈ Ω . Moreover, there exists δ > 0 such that f (x, t )t 6 λ2n0 +1 t 2 for all (x, t ) ∈ Ω × [−δ, δ]; (f3 ) there exist n1 > 1 with λ2n1 < λ2n1 +1 , C1 > 0, and α ∈ (0, 1) such that lim|t |→∞ f (x, t )/t = λ2n1 +1 exists uniformly for x ∈ Ω , and 1 1 2 |f (x, t ) − λ2n1 +1 t | 6 C1 (1 + |t |α ), (x, t ) ∈ Ω × R1 , lim F ( x , t ) − λ t = +∞ 2n + 1 1 |t |→∞ |t |2α 2 Rt uniformly for x ∈ Ω , where F (x, t ) = 0 f (x, s)ds for all (x, t ) ∈ Ω × R1 ; I Project supported by the National Natural Science Foundation of China (Grant No. 10771128) and the NSF of Shanxi Province (2006011002).
∗
Corresponding author. E-mail address: [email protected] (F. Li).
0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.01.018
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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(f4 ) there exists b > 0 such that |f (x, t )| < b for all (x, t ) ∈ Ω × [−bc , bc ], where c = maxx∈Ω e(x), and e is the solution of the BVP: −∆u = 1 in Ω and subject to u = 0 on ∂ Ω . The following theorem is the main result of this paper. Theorem 1.1. Suppose that the conditions (f1 )–(f4 ) hold. Then the BVP (1.1) has at least six nontrivial solutions. Moreover, if (1.1) has only finitely many solutions then, of these solutions, there are two positive solutions, two negative solutions and two sign-changing solutions. In recent years there have been many papers studying sign-changing solutions [1–9]. For example, by using the Morse theory and the topological degree theory, Liu and Sun in [8] obtained one positive solution and two sign-changing solutions to the elliptic problem (1.1) with jumping nonlinearities, where f satisfied the following assumptions: (f1 0 ) (f2 0 ) (f3 0 ) (f4 0 )
f ∈ C 1 (Ω × R1 , R1 ); f (x, 0) = 0 and λi < ft0 (x, 0) < λi+1 for some i > 2 and all x ∈ Ω ; lim supt →+∞ f (x, t )/t < λ1 uniformly in x ∈ Ω ; λ1 < lim inft →−∞ f (x, t )/t 6 lim supt →−∞ f (x, t )/t < +∞ uniformly in x ∈ Ω .
In [5], Bartsch, Chang and Wang assumed the following hypotheses on f with f (x, t ) = f (t ). (i) f ∈ C 1 (R1 , R1 ), f (0) = 0; (ii) lim sup|t |→∞ f (t )/t < λ1 ; (iii) lim|t |→∞ f 0 (t ) = ω. If ω = λk with eigenspace V = {u ∈ C0∞ (Ω ) : −∆u = ωu} then f (t ) − ωt is bounded, and
Z Ω
F (u(x)) −
1
F (u(x)) −
1
2
ωu2 (x) dx → ∞ for u ∈ V , kuk → ∞
or
Z Ω
2
ωu2 (x) dx → −∞ for u ∈ V , kuk → ∞.
They used the Morse theory to obtain that when (i) and (ii), or (i) and (iii) with ω < λ2 hold, and that f 0 (0) > λ2 , the BVP (1.1) has a sign-changing solution. Dancer and Du [2], used a Conley index to prove the BVP (1.1) with f (x, t ) = f (t ) has one positive solution, one negative solution and one sign-changing solution under some assumptions and the following conditions: (H0 ) limt →0+ f (t )/t = a > λ1 , limt →0− f (t )/t = b > λ1 and (H∞ ) lim sup|t |→∞ f (t )/t < λ1 ; or (H0 ) limt →0 f (t )/t < λ1 and (H∞ ) limt →+∞ f (t )/t = a > λ1 , limt →−∞ f (t )/t = b > λ1 . The authors [10,1,3,4,6] all proved that the BVP (1.1) with f (x, t ) = f (t ) has at least two sign-changing solutions provided that f is C 1 , f 0 (0) ∈ (λi , λi+1 ) for some i > 2 and lim sup|t |→∞ f (t )/t < λ1 . In [11], the authors used the topological degree theory to study the fourth-order ordinary differential equation boundary value problems, and they obtained six solutions, of which two are positive solutions, two negative solutions and two signchanging solutions. In the above papers [10,1–4,6,8], our case was not considered, f is not resonant at 0 or ∞. In the condition (iii), f may be resonant at ∞ to λk . However, they assumed that f (t ) − λk t is bounded. We, in this paper, consider the case that f is resonant at 0 and ∞, and f (t ) − λk t may not be bounded. The conditions (f2 ) and (f3 ) imply that f is resonant at 0 and ∞. We add the compressing condition (f4 ) to obtain more solutions. Moreover, the condition (f4 ) is valid (see [12]) and easily satisfied. In this paper, we combine the topological degree theory with the Morse theory to study the BVP (1.1) and obtain the same result with [11] in which they considered the case that f crosses even eigenvalue at 0 and ∞. We not only extend the result in [11] to the elliptic problems, but also extend it to the resonant case at 0 and ∞. Moreover, we prove that the BVP (1.1) has at least five solutions when the functional may not satisfy the (PS) condition. Of them, two are positive solutions, two negative solutions and one sign-changing solution. See Theorem 3.1 in Section 3. The present paper is organized as follows: In the Section 2, we give some lemmas and preliminaries which are useful to our main result. We prove the main theorems in Section 3. At last, we give an example to show that our work is significant. 2. Lemmas and preliminaries In the following, we first recall some lemmas and preliminaries about the critical group and the topological degree. Let H be a real Hilbert space, f ∈ C 1 (H , R1 ), and let K = {u ∈ H : f 0 (u) = 0} denote the critical set of f , fa = {u ∈ H : f (u) 6 a} denote the level sets of f for all a ∈ R1 .
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F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
Definition 2.1 ([12, Definition 4.1, p. 32][13]). Let p be an isolated critical point of f , and let c = f (p). We call Cq (f , p) = Hq (fc ∩ Up , (fc \ {p}) ∩ Up ; G) the qth critical group, with coefficient group G of f at p for all q ∈ N0 = {0, 1, . . .}, where Up is a neighborhood of p such that K ∩ (fc ∩ Up ) = {p}, and H∗ (X , Y ; G) stands for the singular relative homology groups with the Abelian coefficient group G. According to the excision property of the singular homology theory, the critical groups are well defined; i.e., they do not depend on a special choice of the neighborhood Up . If f satisfies the (PS) condition and the critical values of f are bounded from below by some a ∈ R1 , then the critical groups of f at infinity are defined by Bartsch and Li [14] as Cq (f , ∞) = Hq (H , fa ),
q ∈ N0 .
It follows from the deformation lemma that the critical groups at infinity do not depend on the choice of a. Using the two concepts, we have the following famous Morse inequality [12,13]: ∞ ∞ X X (−1)q Mq = (−1)q βq , q =0
where Mq =
q =0
P
rank Cq (f , p), and βq = rank Cq (f , ∞) for all q ∈ N0 .
p∈K
Lemma 2.1 ([12, Theorem 3.2, p. 100]). Let H be a real Hilbert space, and let f ∈ C 2 (H , R1 ) satisfy the (PS) condition. Assume that f 0 (x) = x − Ax,
x ∈ H,
where A is a compact mapping, and that p0 is an isolated critical point of f . Then we have ind(f 0 , p0 ) =
∞ X
(−1)q rank Cq (f , p0 ).
q =0
The following lemma establishes the relationship of fixed point index and topological degree. As X = P is a cone, the idea of this result was given by Liu [15]. See also [11,16]. Let X be a retract of a real Banach space E, U be a relatively open subset of X , A : closX U → X be a completely continuous operator. Suppose that A has not fixed points on ∂X U and that the fixed point set of A is bounded. Then we have the next lemma. Lemma 2.2. If any fixed point of A in U is the interior point of X , then there exists a bounded open subset O of E with O ⊂ U such that O contains all fixed points of A in U and deg(I − A, O, 0) = i(A, U , X ). Proof. Let S = {u ∈ U : u = Au}. Then S ⊂ X ◦ , where X ◦ denotes the interior of X . Thus,Sfor any u ∈ S, there exists δu ∈ (0, 1) such that B(u, δu ) ⊂ X and B(u, δu ) ∩ X ⊂ U, and then B(u, δu ) ⊂ U. Now let O = u∈S B(u, δu ). Then O ⊂ U is a bounded open subset of E. By the excision property of the fixed point index, we have i(A, U , X ) = i(A, O, X ). By the definition of the fixed point index, we have i(A, O, X ) = deg(I − A ◦ r , B(0, R) ∩ r −1 (O), 0), where r : E → X is an arbitrary retraction and R is a large enough positive number such that S ⊂ B(0, R). Now we assume that u∗ ∈ B(0, R) ∩ r −1 (O) such that u∗ = A ◦ r (u∗ ). Since r : E → X and A : closX U → X , it follows that u∗ ∈ X and u∗ = r (u∗ ) ∈ O. Thus, u∗ ∈ O whenever u∗ ∈ B(0, R) ∩ r −1 (O) is a fixed point of A ◦ r. By the excision property of the topological degree, we have deg(I − A ◦ r , B(0, R) ∩ r −1 (O), 0) = deg(I − A ◦ r , O, 0) = deg(I − A, O, 0). We complete the proof.
Here, the fixed point index can be defined on an unbounded set. See [17, p. 238]. Now we turn to consider the BVP (1.1). Let H01 (Ω ) be the Sobolev space with the inner product
(u, v) =
Z Ω
∇ u(x)∇v(x)dx
and the corresponding norm kuk = (u, u)1/2 for all u, v ∈ H01 (Ω ).
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
3823
Define the functional by J (u) =
1
Z
2
Ω
|∇ u(x)|2 dx −
Z Ω
F (x, u(x))dx,
u ∈ H01 (Ω ).
By the growth condition (1.2), J is well defined and of C 2 , and J 0 = I − (−∆)−1 f. It is well known that the solution of the BVP (1.1) is equivalent to the critical point of the functional J. Moreover, the critical point of the functional J is exactly the solution of the operator equation u = (−∆)−1 fu := Au. By the Sobolev theorem, A : H01 (Ω ) → H01 (Ω ) is completely continuous. And it follows from [18, Theorem 4.2] that A : C01 (Ω ) → C01 (Ω ) is also completely continuous. Remark 2.1. Let O be a bounded open subset of H01 (Ω ), there be no zero points of I − A on ∂ O. Since C01 (Ω ) can be compactly embedded into H01 (Ω ), it follows from the bootstrap argument and the definition of Leray–Schauder degree that degH 1 (Ω ) (I − A, O, 0) = degC 1 (Ω ) (I − A, C01 (Ω ) ∩ O, 0). 0 0 In what follows, degC 1 (Ω ) is denoted simply by deg. 0
Remark 2.2. Remark 2.1 implies that two topological degrees in both H01 (Ω ) and C01 (Ω ) are same. Combining with Lemma 2.1, we can obtain the connection between the topological degree in C01 (Ω ) and the critical group, deg(I − A, O, 0) =
∞ X X (−1)q rank Cq (J , p0 ). p0 ∈K ∩O q=0
Here and in what follows, we always assume that J has only finitely many critical points. The next lemma is mainly used to obtain the (PS) condition and the critical groups of J at ∞. Lemma 2.3. Suppose that (f3 ) holds. Then J satisfies the (PS) condition and Cq (J , ∞) = δq,2n1 G for all q ∈ N0 . Proof. Let B = I − λ2n1 +1 (−∆)−1 , and G(u) = Ω [F (x, u(x)) − 21 λ2n1 +1 u2 (x)]dx for all u ∈ H01 (Ω ). Then B is a bounded selfadjoint operator defined on H01 (Ω ), and G ∈ C 1 (H01 (Ω ), R1 ) has a compact differential G0 = (−∆)−1 f − λ2n1 +1 (−∆)−1 . According to the spectral decomposition of the operator B, we have H01 (Ω ) = H− ⊕ H0 ⊕ H+ , where H± , H0 are invariant subspaces corresponding to the positive/negative, and zero spectrum of B respectively. In fact, H0 = ker B, H− = span{e1 , e2 , . . . , e2n1 }, and H+ = (H− ⊕ H0 )⊥ . Consequently, B± := B|H± has a bounded inverse on H± , and dim(H− ⊕ H0 ) < ∞. By the condition (f3 ), we have
R
Z |(G (u), v)| = [f (x, u(x)) − λ2n1 +1 u(x)]v(x)dx ZΩ 6 C1 (1 + |u(x)|α )|v(x)|dx 0
Ω
6 C1 [kvk1 + kukαα 2∗ kvk2N /(N +2) ] 6 C2 (1 + kukα )kvk,
u, v ∈ H01 (Ω ),
where C2 > 0 is a constant. Thus, kG0 (u)k 6 C2 (1 + kukα ) for all u ∈ H01 (Ω ). On the other hand, it follows from the condition (f3 ) that for any R > 0 there exists C3 > 0 such that F (x, t ) −
1 2
λ2n1 +1 t 2 > R|t |2α − C3 ,
( x, t ) ∈ Ω × R 1 .
Since all norms are equivalent on H0 , there exists C4 > 0 such that C4 kuk2α 6 G(u) > R
Z Ω
|u(x)|2α dx − C3 m(Ω ) > RC4 kuk2α − C3 m(Ω ),
R
Ω
|u(x)|2α dx for all u ∈ H0 . Thus,
u ∈ H0 ,
and G(u) → +∞, kuk2α
u ∈ H0 , kuk → ∞.
According to the proposition in [19], also see [20, Proposition 2.1], the conclusions hold.
Lemma 2.4. If the conditions (f2 ) and (f3 ) hold, then Cq (J , 0) = δq,2n0 G for all q ∈ N0 . Proof. By the condition (f3 ) and Lemma 2.3, we get that J satisfies the (PS) condition. It is easy to see that 0 is a degenerate critical point of J with the Morse index 2n0 . By the shifting theorem [12,13], Cq (J , 0) = Cq−2n0 (e J , 0) for all q ∈ N0 , wheree J is
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F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
the restriction of J to N := ker J 00 (0). By the condition (f2 ), we have F (x, t ) 6
λ2n0 +1 t 2 for all (x, t ) ∈ Ω × [−δ, δ]. For this δ > 0, there exists δ1 > 0 such that |u(x)| 6 δ for all x ∈ Ω and u ∈ Bδ1 ∩ N, where Bδ1 = {u ∈ H01 (Ω ) : kuk < δ1 }. Thus, Z Z 1 1 1 (2.1) F (x, u(x))dx > kuk2 − λ2n0 +1 |u(x)|2 dx = 0, u ∈ Bδ1 ∩ N . J (u) = kuk2 − 2
2
Ω
2
1 2
Ω
If there exists r ∈ (0, δ1 ] such that for any u ∈ (Br \{0})∩ N, there exists x0 ∈ Ω satisfying f (x0 , u(x0 ))u(x0 ) < λ2n0 +1 |u(x0 )|2 , then F (x0 , u(x0 )) =
u(x0 )
Z
f (x0 , s)ds <
0
1 2
λ2n0 +1 |u(x0 )|2 .
It follows from (2.1) that J (u) < 0. Thus, 0 is a local minimum ofe J, and Cq (e J , 0) = δq0 G. Therefore, J , 0) = δq,2n0 G, Cq (J , 0) = Cq−2n0 (e
q ∈ N0 .
If not, there exists {un } ⊂ (Bδ1 \ {0}) ∩ N such that f (x, un (x)) = λ2n0 +1 un (x) for all x ∈ Ω and un → 0 as n → ∞, then every un is a solution to the BVP (1.1), and the BVP (1.1) has infinitely many solutions. This contradicts the assumption. The proof is completed. By Lemma 2.4 and Remark 2.2, it is easy to see that Lemma 2.5. Suppose that (f2 ) and (f3 ) hold. Then there exists r0 > 0 such that for all r ∈ (0, r0 ], deg(I − A, B(0, r ), 0) = 1, where B(0, r ) = {u ∈ C01 (Ω ) : kuk < r } denotes an open ball in C01 (Ω ). The following two lemmas deal with the fixed point index. Let X = {u ∈ C01 (Ω ) : −be(x) 6 u(x) 6 be(x), x ∈ Ω }, P = {u ∈ C01 (Ω ) : u(x) > 0, x ∈ Ω }. Then X and P are nonempty closed convex subsets of C01 (Ω ), and P is also a cone in C01 (Ω ). Lemma 2.6. Suppose that (f1 ) and (f2 ) hold. There exists r1 ∈ (0, r0 ] such that for all r ∈ (0, r1 ], i(A, P ∩ B(0, r ), P ) = 0,
i(A, (−P ) ∩ B(0, r ), −P ) = 0.
Proof. By the condition (f1 ), A : P → P is completely continuous. It follows from the condition (f2 ) that there exist δ1 > 0 and ρ1 ∈ (0, r0 ] such that f (x, t ) > λ1 (1 + δ1 )t ,
(x, t ) ∈ Ω × [0, ρ1 ].
(2.2)
If u = Au + µe1 for some µ > 0 and u ∈ P ∩ ∂ B(0, r ), where r ∈ (0, ρ1 ] is a positive number; that is
−∆u(x) = f (x, u(x)) + µλ1 e1 (x), u|∂ Ω = 0,
x ∈ Ω,
(2.3)
then we have from (2.2) that
−∆u(x) > λ1 (1 + δ1 )u(x),
x ∈ Ω.
Thus,
λ1
Z Ω
u(x)e1 (x)dx > λ1 (1 + δ1 )
Z Ω
u(x)e1 (x)dx,
and this is a contradiction. Therefore, according to the property of fixed point index [21, Corollary 2.3.1, p. 90], we get i(A, P ∩ B(0, r ), P ) = 0. Similarly, we can also show that there exists ρ2 ∈ (0, r0 ] such that i(A, (−P ) ∩ B(0, r ), −P ) = 0 for all r ∈ (0, ρ2 ]. Let r1 = min{ρ1 , ρ2 }. Then the conclusion holds. Lemma 2.7. Suppose that (f1 ) and (f4 ) hold. Then there exists a bounded open subset O of C01 (Ω ) with 0 ∈ O ⊂ X such that O contains all fixed points of A in X , and i(A, P ∩ O, P ) = 1, deg(I − A, O, 0) = 1.
i(A, (−P ) ∩ O, −P ) = 1, (2.4)
Proof. It follows from the condition (f4 ) that A(X ) ⊂ X . Since the fixed point set of A in X is bounded, i(A, X , X ) = 1. If u = Au for some u ∈ X , then u satisfies (1.1). By the condition (f4 ) and the strong maximum principle, we have u ∈ X ◦ .
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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Thus, Lemma 2.2 implies that there exists a bounded open subset O of C01 (Ω ) with 0 ∈ O ⊂ X such that O contains all the fixed points of A in X , and deg(I − A, O, 0) = i(A, X , X ) = 1. In the same way, if there are µ ∈ [0, 1] and u ∈ P ∩ X such that u = µAu, then u ∈ X ◦ . According to the homotopy invariance of the fixed point index, we have i(A, P ∩ X ◦ , P ) = 1. Since any fixed point of A in P ∩ X ◦ belongs to P ∩ O, it follows from the excision property of the fixed point index that i(A, P ∩ X ◦ , P ) = i(A, P ∩ O, P ) = 1. Similarly, we can also obtain that i(A, (−P ) ∩ O, −P ) = 1. Thus, we complete the proof. Since X is not bounded in C01 (Ω ), it follows X 6⊂ B(0, R) for any R > 0. We, however, convert the unbounded set X into a bounded set O by Lemma 2.7. Therefore, the proof of our main theorem becomes more easy. Lemma 2.8. Let (f1 ) hold. Suppose that limt →+∞ f (x, t )/t > λ1 and limt →−∞ f (x, t )/t > λ1 uniformly for x ∈ Ω , and that there are C1 > 0 and C2 > 0 such that |f (x, t )| 6 C1 |t | + C2 for all (x, t ) ∈ Ω × R1 . Then there exists R0 > r0 such that for all R > R0 , i(A, P ∩ B(0, R), P ) = 0,
i(A, (−P ) ∩ B(0, R), −P ) = 0.
Proof. We only prove that i(A, P ∩ B(0, R), P ) = 0. Since limt →+∞ f (x, t )/t > λ1 uniformly for x ∈ Ω , there exist δ > 0 and C3 > 0 such that f (x, t ) > λ1 (1 + δ)t − C3 ,
(x, t ) ∈ Ω × [0, ∞).
(2.5)
Define H (µ, u) = Au + µe1 for all (µ, u) ∈ [0, ∞) × P. If u ∈ P and µ > 0 satisfy H (µ, u) = u, then u and µ satisfy (2.3). Consequently, it follows from (2.5) that
−∆u(x) > λ1 (1 + δ)u(x) − C3 + µλ1 e1 (x),
x ∈ Ω,
and
λ1
Z Ω
Z
u(x)e1 (x)dx > λ1 (1 + δ)
Ω
u(x)e1 (x)dx − C3
Z Ω
e1 (x)dx + µλ1
Z Ω
e21 (x)dx.
This implies µλ1 Ω e21 (x)dx 6 C3 Ω e1 (x)dx. Thus, there is a µ0 > 0 such that for µ > µ0 , H (µ, ·) has no fixed points in P. To show the existence of such a constant R1 > 0 which is able to insure that kuk < R1 if u ∈ P is a solution of u = H (µ, u) for some µ > 0, we adopt an indirect argument. Suppose that there exists a sequence {(µn , un )} ⊂ [0, ∞) × P with kun k → ∞ satisfying un = H (µn , un ). Then (µn , un ) satisfies the following BVP
R
R
−∆un (x) = f (x, un (x)) + µn λ1 e1 (x), un |∂ Ω = 0.
x ∈ Ω,
(2.6)
Since f (x, t ) 6 C1 t + C2 for all (x, t ) ∈ Ω × [0, ∞), we have from (2.6) that f (x, un )un
Z 1= Ω
k un k
2
dx + µn λ1
Z
Z
e 1 un Ω
k un k
dx 6 C1 2
Ω
vn2 dx + C2
vn
Z Ω
k un k
dx + µn λ1
e1 vn
Z Ω
k un k
dx,
(2.7)
where vn = un /kun k for all n = 1, 2, . . .. Since H01 (Ω ) can be compactly embedded into L2 (Ω ), there exists a converging subsequence {vnk } of {vn } in L2 (Ω ), and we let vnk → v in RL2 (Ω ). It follows from Riesz’s theorem that v(x) > 0, a.e. x ∈ Ω . By µn 6 µ0 and kun k → ∞, we get from (2.7) that 1 6 C1 Ω v 2 dx. This implies that v > 0, a.e. x ∈ Ω , and v 6= 0. On the other hand, we have from (2.5) and (2.6) that
λ1
Z Ω
vnk e1 dx > λ1 (1 + δ)
Z Ω
vnk e1 dx − C3
Z
e1 Ω
kunk k
dx.
Letting k → ∞, we have that Ω v e1 dx 6 0, which is impossible. Thus, there exists R1 > 0 such that kuk 6 R1 for any (µ, u) ∈ [0, ∞) × P with u = H (µ, u), and a standard bootstrap argument implies that there exists R01 > 0 such that kukC 1 (Ω ) 6 R01 for all such u. Therefore, if R > R01 , then u 6= H (µ, u) for all u ∈ P ∩ ∂ B(0, R) and µ > 0, and the property of
R
0
fixed point index [21] implies that i(A, P ∩ B(0, R), P ) = 0. Similarly, there exists R02 > 0 such that i(A, (−P ) ∩ B(0, R), −P ) = 0 for all R > R02 . Let R0 = max{R01 , R02 }. Then the conclusion is true. Remark 2.3. In the proof of Lemma 2.8, the following property does not hold
(−∆)−1 u > e0 (x)k(−∆)−1 ukC 1 (Ω ) , 0
u ∈ P,
where e0 (x) > 0 for x ∈ Ω is some function. However, we still prove that the conclusion holds.
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F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
3. Proof of main theorems The Proof of Theorem 1.1. The proof of Lemma 2.8 implies that there exists R > 0 large enough such that O ⊂ B(0, R) and B(0, R) contains all positive and negative fixed points of A. It follows from Lemmas 2.7 and 2.8 that i(A, P ∩ (B(0, R) \ O), P ) = 0 − 1 = −1,
(3.1)
i(A, (−P ) ∩ (B(0, R) \ O), −P ) = 0 − 1 = −1.
(3.2)
Consequently, the BVP (1.1) has at least a positive solution u1 ∈ P ∩ (B(0, R) \ O) and a negative solution u2 ∈ (−P ) ∩ (B(0, R) \ O). If u ∈ P \ {0} is a solution of the BVP (1.1), then the condition (f1 ) and the strong maximum principle imply that u ∈ P ◦ . Similarly, if u ∈ −P \ {0} is a solution of the BVP (1.1), then u ∈ (−P )◦ . It follows from Lemma 2.2, (3.1) and (3.2) that there exist two bounded open subsets U1 ⊂ P ∩ (B(0, R) \ O) containing all positive fixed points of A in P ∩ (B(0, R) \ O) and U2 ⊂ (−P ) ∩ (B(0, R) \ O) containing all negative fixed points of A in (−P ) ∩ (B(0, R) \ O) such that deg(I − A, U1 , 0) = −1,
deg(I − A, U2 , 0) = −1.
If there are no more critical points of J in Remark 2.2, (2.4) and (3.3) that 1 = (−1)2n1 =
∞ X
C01
(3.3)
(Ω ) \ (O ∪ U 1 ∪ U 2 ), then it follows from the Morse inequality, Lemma 2.3,
(−1)q rank Cq (J , ∞) = 1 + (−1) + (−1) = −1,
q =0
which is impossible. Thus, there exists a solution u3 ∈ C01 (Ω ) \ (O ∪ U 1 ∪ U 2 ), and this is a sign-changing solution. On the other hand, by Lemmas 2.6 and 2.7, we have that for all r ∈ (0, r0 ] satisfying B(0, r ) ⊂ O, i(A, P ∩ (O \ B(0, r )), P ) = 1,
i(A, (−P ) ∩ (O \ B(0, r )), −P ) = 1,
where B(0, r ) denotes B(0, r ). Thus, the BVP (1.1) has at least a positive solution u4 ∈ P ∩(O \ B(0, r )) and a negative solution u5 ∈ (−P ) ∩ (O \ B(0, r )). From again Lemma 2.2, there exist U3 ⊂ P ∩ (O \ B(0, r )) containing all positive solutions in O \ B(0, r ) and U4 ⊂ (−P ) ∩ (O \ B(0, r )) containing all negative solutions in O \ B(0, r ), and deg(I − A, U3 , 0) = 1,
deg(I − A, U4 , 0) = 1.
(3.4)
It follows from Lemma 2.5 that for all r ∈ (0, r0 ] with B(0, r ) ⊂ O, deg(I − A, B(0, r ), 0) = 1.
(3.5)
If A has no fixed points in O \ (B(0, r ) ∪ U 3 ∪ U 4 ), then the additivity property of the topological degree implies that deg(I − A, O, 0) = deg(I − A, B(0, r ), 0) + deg(I − A, U3 , 0) + deg(I − A, U4 , 0). By (2.4), (3.4) and (3.5), we have 1 = 1 + 1 + 1. This is a contradiction. Thus, the BVP (1.1) has at least a solution u6 ∈ O \ (B(0, r ) ∪ U 3 ∪ U 4 ). This is a sign-changing solution. The proof is completed. Theorem 3.1. Let (f1 ), (f2 ) and (f4 ) hold. Suppose that limt →+∞ f (x, t )/t > λ1 and limt →−∞ f (x, t )/t > λ1 uniformly for x ∈ Ω , and that there are two constants C1 > 0 and C2 > 0 such that |f (x, t )| 6 C1 |t | + C2 for all (x, t ) ∈ Ω × R1 . Then the BVP (1.1) has at least five solutions. Of them, two positive solutions, two are negative solutions and one sign-changing solution. Proof. When the condition (f3 ) does not hold, J may not satisfy the (PS) condition. Therefore, we define g ∈ C 1 (Ω × R1 , R1 ) as g (x, t ) = ω(t )f (x, t ) for all (x, t ) ∈ Ω × R1 , where ω ∈ C0∞ (R1 ) satisfies ω(t ) = 1 for all t ∈ [−bc , bc ], ω(t ) = 0 for all |t | > 2bc, and |ω(t )| 6 1 for all t ∈ R1 . Let J 1 ( u) =
1 2
where G(x, t ) =
kuk2 −
Z Ω
Rt 0
G(x, u(x))dx,
u ∈ H01 (Ω ),
g (x, s)ds for all (x, t ) ∈ Ω × R1 . It follows from the definition of g that there is a constant C > 0 such
that |g (x, t )| 6 C for all (x, t ) ∈ Ω × R1 . Thus, J1 satisfies the (PS) condition and the critical points of J1 in X are all the critical points of J in X . The definition of g implies that g satisfies the conditions (f1 ), (f2 ) and (f4 ). Therefore, we have from the proof of Theorem 1.1 that J1 has at least three critical points in O ⊂ X , of them, there are one positive critical point, one negative critical point and one sign-changing critical point. Therefore, the BVP (1.1) has at least three solutions in O; one positive solution, one negative solution and one sign-changing solution. By Lemmas 2.7 and 2.8, there exists R > 0 with O ⊂ B(0, R) such that i(A, P ∩ (B(0, R) \ O), P ) = 0 − 1 = −1, i(A, (−P ) ∩ (B(0, R) \ O), −P ) = 0 − 1 = −1. Thus, the BVP (1.1) has at least one positive solution in P ∩(B(0, R)\ O) and one negative solution in (−P )∩(B(0, R)\ O).
F. Li, Y. Li / Nonlinear Analysis 72 (2010) 3820–3827
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Example 3.1. Let n be a positive integer such that λ2n0 +1 c /n < 1, f (x, t ) = f (t ) be an odd function defined as follows
2 − 2n t 2 λ2n0 +1 te c 2 , c 2nt − c − 21 f (t ) = λ2n0 +1 e 1 + cos π , 4n 2nc − c p 3√ 2 λ2n1 +1 (t − c ) 1 − e−(t −c ) + 3 t − c ln 1 + 3 (t − c )4 , 4
h
c
, h c 2n t ∈ ,c , t ∈ 0,
2n
t ∈ [c , ∞).
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