This paper is concerned with the existence of positive solutions of the following elliptic mixed boundary value problem:

{ − Δ u = f ( x , u ) , x ∈ Ω , u = 0 , x ∈ σ , ∂ u ∂ ν = g ( x , u ) , x ∈ Γ ,

where Ω is a bounded domain in R n with Lipschitz boundary ∂Ω, σ ∪ Γ = ∂ Ω , σ ∩ Γ = Ø , Γ is a sufficiently smooth ( n − 1 ) -dimensional manifold, and ν is the outward normal vector on ∂Ω. We assume f : Ω × R → R , g : Γ × R → R are continuous and satisfy

(S1) f ( x , t ) ≥ 0 , ∀ t ≥ 0 , x ∈ Ω , f ( x , 0 ) = 0 . f ( x , t ) ≡ 0 , ∀ t < 0 , x∈Ω.

(S2) For almost every x∈Ω, f ( x , t ) t is nondecreasing with respect to t > 0 .

(S3) lim t → 0 f ( x , t ) t = p ( x ) , lim t → + ∞ f ( x , t ) t = q ( x ) ≢ 0 uniformly in a.e. x∈Ω, where ∥ p ( x ) ∥ ∞ < λ 1 , λ 1 is the first eigenvalue of (2), 0 ≤ p ( x ) , q ( x ) ∈ L ∞ ( Ω ) .

(S4) There exists c 1 , c 2 > 0 such that | f ( x , t ) | ≤ c 1 + c 2 | t | p − 1 for some p ∈ ( 2 , 2 n n − 2 ) as n ≥ 3 and p ∈ ( 2 , + ∞ ) as n = 1 , 2 .

The eigenvalue problem of (1) is studied by Liu and Su in 1

{ − Δ u = λ u in  Ω , u = 0 on  σ , ∂ u ∂ ν = λ u on  Γ .

There exists a set of eigenvalues { λ k } and corresponding eigenfunctions { u k } which solve problem (2), where 0 ≤ λ 1 ≤ λ 2 ≤ ⋯ ≤ λ k ≤ ⋯ , λ k → ∞ as k → ∞ , λ 1 = inf 0 ≠ u ∈ V ∫ Ω | ∇ u | 2 d x ∫ Ω | u | 2 d x + ∫ Γ | u | 2 d s .

There have been many papers concerned with similar problems at resonance under the boundary condition; see 2345678910. Moreover, some multiplicity theorems are obtained by the topological degree technique and variational methods; interested readers can see 11121314151617. Problem (1) is different from the classical ones, such as those with Dirichlet, Neuman, Robin, No-flux, or Steklov boundary conditions.

In this paper, we assume V : = { v ∈ H 1 ( Ω ) : v | σ = 0 } is a closed subspace of H 1 ( Ω ) . We define the norm in V as ∥ u ∥ 2 = ∫ Ω | ∇ u | 2 d x + ∫ Γ | γ u | 2 d s , ∥ ⋅ ∥ L p ( Ω ) is the L p ( Ω ) norm, ∥ ⋅ ∥ L p ( Γ ) is the L p ( Γ ) norm, γ : V → L 2 ( Γ ) is the trace operator with γ u = u Γ for all u ∈ H 1 ( Ω ) , that is continuous and compact (see 18). Furthermore, we define g = γ f , 0 ≤ g ( x , t ) ≤ | γ f ( x , t ) | for t>0 (see 1). Then, by (S3), we obtain

lim t → + ∞ g ( x , t ) t ≤ lim t → + ∞ | γ f ( x , t ) | t = q ( x ) ≢ 0 , a.e.  x ∈ Ω ¯ .

Let Ω be a bounded domain with a Lipschitz boundary; there is a continuous embedding V ↪ L y ( Ω ) for y ∈ [ 2 , 2 n n − 2 ] when n≥3, and y ∈ [ 2 , + ∞ ) when n = 1 , 2 . Then there exists γ y > 0 , such that

∥ u ∥ L y ( Ω ) ≤ γ y ∥ u ∥ , ∀ u ∈ V .

Moreover, there is a continuous boundary trace embedding V ↪ L z ( Γ ) for z ∈ [ 2 , 2 ( n − 1 ) n − 2 ] when n≥3, and z ∈ [ 2 , + ∞ ) when n=1,2. Then there exists k z > 0 , such that

∥ u ∥ L z ( Γ ) ≤ k z ∥ u ∥ , ∀ u ∈ V .

It is well known that to seek a nontrivial weak solution of problem (1) is equivalent to finding a nonzero critical value of the C 1 functional

J ( u ) = 1 2 ∫ Ω | ∇ u | 2 d x − ∫ Ω F ( x , u ) d x − ∫ Γ G ( s , u ) d s ,

where u ∈ V , F ( x , u ) = ∫ 0 u f ( x , t ) d t , G ( x , u ) = ∫ 0 u g ( x , t ) d t . Moreover, by (S1) and the Strong maximum principle, a nonzero critical point of J is in fact a positive solution of (1). In order to find critical points of the functional (6), one often requires the technique condition, that is, for some μ > 2 , ∀ | u | ≥ M > 0 , x∈Ω,

0 < μ F ( x , u ) ≤ u f ( x , u ) , F ( x , u ) = ∫ 0 u f ( x , t ) d t .

It is easy to see that the condition (AR) implies that lim u → + ∞ F ( x , u ) u 2 = + ∞ , that is, f ( x , u ) must be superlinear with respect to u at infinity. In the present paper, motivated by 19 and 20, we study the existence and nonexistence of positive solutions for problem (1) with the asymptotic behavior assumptions (S3) of f at zero and infinity. Moreover, we also study superlinear of f at infinity with q ( x ) ≡ + ∞ in (S3), which is weaker than the (AR) condition, that is the (AR) condition does not hold.

In order to get our conclusion, we define the minimization problem

Λ = inf { ∫ Ω | ∇ u | 2 d x : u ∈ V , ∫ Ω q ( x ) u 2 d x + ∫ Γ q ( s ) u 2 d s = 1 } ,

then Λ > 0 , which is achieved by some φ Λ ∈ V with φ Λ ( x ) > 0 a.e. in Ω; see Lemma 1.

We denote by c, c 1 , c 2 universal constants unless specified otherwise. Our main results are as follows.

Theorem 1 Let conditions (S1) to (S3) hold, then:

(i) If Λ > 1 , then the problem (1) has no any positive solution in V.

(ii) If Λ < 1 , then the problem (1) has at least one positive solution in V.

(iii) If Λ = 1 , then the problem (1) has one positive solution u ( x ) ∈ V if and only if there exists a constant c > 0 such that u ( x ) = c φ Λ ( x ) and f ( x , u ) = q ( x ) u ( x ) , g ( x , u ) = q ( x ) u ( x ) a.e. x∈Ω, where φΛ(x)>0 is the function which achieves Λ.

Corollary 2 Let conditions (S1) to (S3) with q ( x ) ≡ l > 0 hold, then:

(i) If l < λ 1 , then the problem (1) has no any positive solution in V.

(ii) If λ 1 < l < + ∞ , then the problem (1) has at least one positive solution in V.

(iii) If l = λ 1 , then the problem (1) has one positive solution u(x)∈V if and only if there exists a constant c>0 such that u ( x ) = c φ 1 ( x ) and f ( x , u ) = λ 1 u ( x ) , g ( x , u ) = λ 1 u ( x ) a.e. x∈Ω, where φ 1 ( x ) > 0 is the eigenfunction of the λ1.

Theorem 3 Let conditions (S1) to (S4) with q(x)≡+∞ hold, then the problem (1) has at least one positive solution in V.

We need the following lemmas.

Lemma 1 If q(x)∈L∞(Ω), q ( x ) ≥ 0 , q ( x ) ≢ 0 , then Λ>0 and there exists φ Λ ( x ) ∈ V such that Λ = ∫ Ω | ∇ φ Λ | 2 d x and ∫ Ω q ( x ) φ Λ 2 d x + ∫ Γ q ( s ) φ Λ 2 d s = 1 . Moreover, φΛ(x)>0 a.e. in V.

Proof By the Sobolev embedding function V ↪ L 2 ( Ω ) and Fatou’s lemma, it is easy to know that Λ>0 and there exists φΛ(x)∈V, which satisfies Λ, that is, ∫ Ω q ( x ) φ Λ 2 d x + ∫ Γ q ( s ) φ Λ 2 d s = 1 . Furthermore, we assume φ Λ ( x ) ≥ 0 , then φ Λ ( x ) could replace by | φ Λ ( x ) | . By the Strong maximum principle, we know φΛ(x)>0 a.e. in V. □

Lemma 2 If conditions (S1) to (S3) hold, then there exists β , ρ > 0 such that J | ∂ B ρ ( 0 ) ≥ β , ∀ u ∈ V , ∥ u ∥ = ρ .

Proof By condition (S3), there exists δ > 0 , ε > 0 such that f ( x , u ) u ≤ λ 1 − ε , g ( x , u ) u ≤ γ f ( x , u ) u ≤ λ 1 − ε as 0 < | u | ≤ δ . Which implies that F ( x , u ) ≤ 1 2 ( λ 1 − ε ) u 2 + c | u | y , G ( x , u ) ≤ 1 2 ( λ 1 − ε ) u 2 + c | u | z .

By (4) and (5), we obtain

J ( u ) = 1 2 ∥ ∇ u ∥ L 2 ( Ω ) 2 − ∫ Ω F ( x , u ) d x − ∫ Γ G ( s , u ) d s ≥ 1 2 ∥ ∇ u ∥ L 2 ( Ω ) 2 + 1 2 ∥ γ u ∥ L 2 ( Γ ) 2 − 1 2 ∥ γ u ∥ L 2 ( Γ ) 2 − 1 2 ( λ 1 − ε ) ∥ u ∥ L 2 ( Ω ) 2 − c ∥ u ∥ L y ( Ω ) y − 1 2 ( λ 1 − ε ) ∥ u ∥ L 2 ( Γ ) 2 − c ∥ u ∥ L z ( Γ ) z ≥ 1 2 ∥ u ∥ 2 − 1 2 ( λ 1 − ε ) 1 λ 1 ∥ u ∥ 2 − c γ y y ∥ u ∥ y − 1 2 ( λ 1 − ε + 1 ) 1 λ 1 + 1 ∥ u ∥ 2 − c k z z ∥ u ∥ z = [ ε ( 2 λ 1 + 1 ) 2 λ 1 ( λ 1 + 1 ) − 1 2 ] ∥ u ∥ 2 − c γ y y ∥ u ∥ y − c k z z ∥ u ∥ z .

Hence, y , z > 2 ; we take ε which satisfies ε ( 2 λ 1 + 1 ) 2 λ 1 ( λ 1 + 1 ) − 1 2 > 0 , that is, ε > λ 1 ( λ 1 + 1 ) 2 λ 1 + 1 . Then we take a positive constant β such that J|∂Bρ(0)≥β as ∥u∥=ρ, and is small enough. □

Lemma 3 If conditions (S1) to (S3) hold, Λ<1, φΛ(x)>0 is defined by Lemma 1, then J ( t φ Λ ( x ) ) → − ∞ as t → + ∞ .

Proof If Λ<1, φΛ(x)>0 is defined by Lemma 1, by Fatou’s lemma, and (S3), we have

So, J(tφΛ(x))→−∞ as t→+∞. □

Lemma 4 Let conditions (S1) and (S2) hold. If a sequence { u n } ⊂ V satisfies 〈 J ′ ( u n ) , u n 〉 → 0 as n → + ∞ , then there exists a subsequence of { u n } , still denoted by {un} such that J ( t u n ) ≤ 1 + t 2 2 n + J ( u n ) for all t>0, n ≥ 1 .

Proof Since 〈 J ′ ( u n ) , u n 〉 → 0 as n→+∞, for a subsequence, we may assume that

− 1 n < 〈 J ′ ( u n ) , u n 〉 = ∥ ∇ u n ∥ L 2 ( Ω ) 2 − ∫ Ω f ( x , u n ) u n d x − ∫ Γ g ( s , u n ) u n d s < 1 n , ∀ n ≥ 1 .

For any fixed x∈Ω and n≥1, set

ψ 1 ( t ) = t 2 2 f ( x , u n ) u n − F ( x , t u n ) , ψ 2 ( t ) = t 2 2 g ( s , u n ) u n − G ( s , t u n ) .

Then (S2) implies that

ψ 1 ′ ( t ) = t f ( x , u n ) u n − f ( x , t u n ) u n = t u n [ f ( x , u n ) − f ( x , t u n ) t ] = { ≥ 0 , 0 < t ≤ 1 ; ≤ 0 , t > 1 .

It implies that ψ 1 ( t ) ≤ ψ 1 ( 1 ) , ∀ t > 0 . Following the same procedures, we obtain ψ 2 ( t ) ≤ ψ 2 ( 1 ) , ∀t>0.

For all t>0 and positive integer n, by (8), we have

J ( t u n ) = t 2 2 ∥ ∇ u n ∥ L 2 ( Ω ) 2 − ∫ Ω F ( x , t u n ) d x − ∫ Γ G ( s , t u n ) d s ≤ t 2 2 [ 1 n + ∫ Ω f ( x , u n ) u n d x + ∫ Γ g ( s , u n ) u n d s ] − ∫ Ω F ( x , t u n ) d x − ∫ Γ G ( s , t u n ) d s ≤ t 2 2 n + ∫ Ω [ 1 2 f ( x , u n ) u n − F ( x , u n ) ] d x + ∫ Γ [ 1 2 g ( s , u n ) u n − G ( s , u n ) ] d s .

On the other hand, by (8), one has

J ( u n ) = 1 2 ∥ ∇ u n ∥ L 2 ( Ω ) 2 − ∫ Ω F ( x , u n ) d x − ∫ Γ G ( s , u n ) d s ≥ 1 2 [ − 1 n + ∫ Ω f ( x , u n ) u n d x + ∫ Γ g ( s , u n ) u n d s ] − ∫ Ω F ( x , u n ) d x − ∫ Γ G ( s , u n ) d s = − 1 2 n + ∫ Ω [ 1 2 f ( x , u n ) u n − F ( x , u n ) ] d x + ∫ Γ [ 1 2 g ( s , u n ) u n − G ( s , u n ) ] d s .

One has

∫ Ω [ 1 2 f ( x , u n ) u n − F ( x , u n ) ] d x + ∫ Γ [ 1 2 g ( s , u n ) u n − G ( s , u n ) ] d s ≤ J ( u n ) + 1 2 n .

Combining (9) and (10), we have J ( t u n ) ≤ 1 + t 2 2 n + J ( u n ) . □

Lemma 5 (see 21)

Suppose E is a real Banach space, J ∈ C 1 ( E , R ) satisfies the following geometrical conditions:

(i) J ( 0 ) = 0 ; there exists ρ > 0 such that J | ∂ B ρ ( 0 ) ≥ r > 0 ;

(ii) There exists e ∈ E ∖ B ρ ( 0 ) ¯ such that J ( e ) ≤ 0 . Let Γ 1 be the set of all continuous paths joining 0 and e:

Γ 1 = { h ∈ C ( [ 0 , 1 ] , E ) | h ( 0 ) = 0 , h ( 1 ) = e } ,

and

c = inf h ∈ Γ 1 max t ∈ [ 0 , 1 ] J ( h ( t ) ) .

Then there exists a sequence { u n } ⊂ E such that J ( u n ) → c ≥ β and ( 1 + ∥ u n ∥ ) × ∥ J ′ ( u n ) ∥ E ∗ → 0 .

Proof of Theorem 1 (i) If u∈V is one positive solution of problem (1), by (3), one has

0 = 〈 J ′ ( u ) , u 〉 = ∫ Ω | ∇ u | 2 d x − ∫ Ω f ( x , u ) u d x − ∫ Γ g ( s , u ) u d s .

That is,

∫ Ω | ∇ u | 2 d x = ∫ Ω f ( x , u ) u d x + ∫ Γ g ( s , u ) u d s ≤ ∫ Ω q ( x ) u 2 d x + ∫ Γ q ( s ) u 2 d s = 1 .

It implies that Λ ≤ 1 . This completes the proof of Theorem 1(i).

(ii) By Lemma 2, there exists β , ρ > 0 such that J|∂Bρ(0)≥β with ∥u∥=ρ. By Lemma 3, we obtain J ( t 0 φ Λ ( x ) ) < 0 as t 0 → + ∞ . Define

where φΛ(x)>0 is given by Lemma 1. Then c ≥ β > 0 and by Lemma 3, there exists {un}⊂V such that

(14) implies that

〈 J ′ ( u n ) , u n 〉 = ∥ ∇ u n ∥ L 2 ( Ω ) 2 − ∫ Ω f ( x , u n ) u n d x − ∫ Γ g ( s , u n ) u n d s = o ( 1 ) .

Here, in what follows, we use o ( 1 ) to denote any quantity which tends to zero as n→+∞.

If {un} is bounded in V, when Ω is bounded and f(x,u), g ( x , u ) are subcritical, we can get {un} has a subsequence strong convergence to a critical value of J, and our proof is complete. So, to prove the theorem, we only need show that {un} is bounded in V. Supposing that { u n } is unbounded, that is, ∥ u n ∥ → + ∞ as n→+∞. We order

t n = 2 c ∥ u n ∥ , w n = t n u n = 2 c u n ∥ u n ∥ .

Then { w n } is bounded in V. By extracting a subsequence, we suppose w n → w is a strong convergence in L 2 ( Ω ) , wn→w is a convergence a.e. x∈Ω, w n ⇀ w is a weak convergence in V.

We claim that w ≠ 0 . In fact, by (S1) and (S3), we know ∀ x ∈ Ω , u n ≥ 0 , and there exists M 1 , M 2 > 0 such that | f ( x , u n ) u n | ≤ M 1 , | g ( x , u n ) u n | ≤ M 2 . If w = 0 , w n → 0 is a strong convergence in L2(Ω), and by (15) and (16) we know

4 c = t n 2 ∥ u n ∥ 2 = t n 2 ( ∥ ∇ u n ∥ L 2 ( Ω ) 2 + ∥ γ u n ∥ L 2 ( Γ ) 2 ) = t n 2 ∫ Ω f ( x , u n ) u n d x + t n 2 ∫ Γ g ( s , u n ) u n d s + t n 2 ∥ γ u n ∥ L 2 ( Γ ) 2 + o ( 1 ) = ∫ Ω f ( x , u n ) u n w n 2 d x + ∫ Γ g ( s , u n ) u n w n 2 d s + t n 2 ∥ u n ∥ L 2 ( Γ ) 2 + o ( 1 ) ≤ M 1 ∫ Ω w n 2 d x + M 2 ∫ Γ w n 2 d s + ∥ w n ∥ L 2 ( Γ ) 2 + o ( 1 ) → 0 .

It is contradiction with c>0, so w≠0.

As follows, we prove w≠0 satisfies

∫ Ω ∇ φ ( x ) ∇ w ( x ) d x − ∫ Ω q 1 ( x ) φ ( x ) w ( x ) d x − ∫ Γ q 2 ( s ) φ ( s ) w ( s ) d s = 0 .

We order

By (S1) and (S3), there exists M 3 > 0 such that 0 ≤ p n ( x ) ≤ M 3 , 0 ≤ q n ( x ) ≤ M 3 , ∀ x ∈ Ω ¯ . We select a suitable subsequence and there exists h 1 ( x ) ∈ L 2 ( Ω ) , h 2 ( x ) ∈ L 2 ( Γ ) such that p n ( x ) → h 1 ( x ) is a strong convergence in L2(Ω), q n ( x ) → h 2 ( x ) is a strong convergence in L 2 ( Γ ) , and 0 ≤ h 1 ( x ) ≤ M 3 , 0 ≤ h 2 ( x ) ≤ M 3 , ∀x∈Ω¯.

It follows from wn→w is a strong convergence in L2(Ω) that

Hence, { p n ( x ) w n ( x ) } is bounded in L2(Ω), p n ( x ) w n ( x ) ⇀ h 1 ( x ) w + ( x ) in L2(Ω); { q n ( x ) w n ( x ) } is bounded in L2(Γ), q n ( x ) w n ( x ) ⇀ h 2 ( x ) w + ( x ) in L2(Γ).

By (16), we have

Since wn⇀w is a weak convergence in V, we obtain

∫ Ω ∇ φ ( x ) ∇ w ( x ) d x − ∫ Ω h 1 ( x ) φ ( x ) w + ( x ) d x − ∫ Γ h 2 ( s ) φ ( s ) w + ( s ) d s = 0 , φ ∈ V .

We order φ = w − ; this yields ∥ w − ∥ 2 = 0 , so w = w + ≥ 0 . By the Strong maximum principle, we know w > 0 a.e. in Ω, so u n → ∞ a.e. in Ω. Combining (S3) and (3), we obtain

∫ Ω ∇ φ ( x ) ∇ w ( x ) d x − ∫ Ω q ( x ) φ ( x ) w ( x ) d x − ∫ Γ q ( s ) φ ( s ) w ( s ) d s = 0 , ∀ φ ∈ V .

This is a contradiction with Λ<1. This completes the proof of Theorem 1(ii).

(iii) If Λ=1, by Lemma 1, there exists some φΛ(x)>0, such that

∫ Ω ∇ v ( x ) ∇ φ Λ ( x ) d x = ∫ Ω q ( x ) v ( x ) φ Λ ( x ) d x + ∫ Γ q ( s ) v ( s ) φ Λ ( s ) d s .

If u is a positive solution of (1), for the above φΛ(x), we have

∫ Ω ∇ u ( x ) ∇ φ Λ ( x ) d x = ∫ Ω f ( x , u ( x ) ) φ Λ ( x ) d x + ∫ Γ g ( s , u ( s ) ) φ Λ ( s ) d s .

We order v = u in (17), and it follows from (18) that

∫ Ω ∇ u ( x ) ∇ φ Λ ( x ) d x = ∫ Ω q ( x ) u ( x ) φ Λ ( x ) d x + ∫ Γ q ( s ) u ( s ) φ Λ ( s ) d s = ∫ Ω f ( x , u ( x ) ) φ Λ ( x ) d x + ∫ Γ g ( s , u ( s ) ) φ Λ ( s ) d s ≤ ∫ Ω q ( x ) u ( x ) φ Λ ( x ) d x + ∫ Γ q ( s ) u ( s ) φ Λ ( s ) d s ,

which implies that ∫ Ω ( f ( x , u ) − q ( x ) u ( x ) ) φ Λ ( x ) d x + ∫ Γ ( g ( s , u ) − q ( s ) u ( s ) ) φ Λ ( s ) d s = 0 .

When φΛ(x)>0 a.e. in Ω, combining (S2), (S3), and (3), we obtain

f ( x , u ) ≤ q ( x ) u ( x ) , g ( x , u ) ≤ q ( x ) u ( x ) .

Then we must have f(x,u)=q(x)u(x), g ( x , u ) = q ( x ) u ( x ) a.e. in Ω, u ( x ) > 0 also achieves Λ (=1). When u = c φ Λ , c>0, we have ∫ Ω | ∇ φ Λ | 2 d x = ∫ Ω q ( x ) φ Λ 2 d x + ∫ Γ q ( s ) φ Λ 2 d s , which achieves Λ.

On the other hand, if for some c>0, u(x)=cφΛ(x) and f ( x , c φ Λ ( x ) ) = c q ( x ) φ Λ ( x ) , g ( x , u ) = c q ( x ) φ Λ ( x ) a.e. x∈Ω, since c φ Λ ( x ) also achieves Λ. This means u(x)=cφΛ(x) is a solution of problem (1) as Λ=1. This completes the proof of Theorem 1(iii). □

Proof of Corollary 2 Note that when q ( x ) ≡ l , then Λ = λ 1 l . The conclusion follows from Theorem 1. □

Proof of Theorem 3 When q(x)≡+∞, we can replace φ Λ by φ 1 in (11) and define c as in (12), then following the same procedures as in the proof of Theorem 1(ii), we need to show only that {un} is bounded in V. For this purpose, let {wn} be defined as in (16). If {wn} is bounded in V, we know wn→w is a strong convergence in L2(Ω), wn→w is convergence a.e. x∈Ω, wn⇀w is a weak convergence in V, and w ∈ V .

If ∥un∥→+∞, then t n → 0 and w ( x ) ≡ 0 . We set Ω 1 = { x ∈ Ω : w ( x ) = 0 } , Ω 2 = { x ∈ Ω : w ( x ) ≠ 0 } . Obviously, by (16), | u n | → + ∞ a.e. in Ω 2 . When q(x)≡+∞ in (S3), there exists K 1 , K 2 > 0 and n large enough we have | f ( x , u n ) u n | ≥ K 1 , | g ( x , u n ) u n | ≥ K 2 uniformly in x ∈ Ω 2 . Hence, by (15) and (16), we obtain

4 c = lim n → + ∞ t n 2 ∥ u n ∥ 2 = lim n → + ∞ t n 2 ( ∥ ∇ u n ∥ L 2 ( Ω ) 2 + ∥ γ u n ∥ L 2 ( Γ ) 2 ) = lim n → + ∞ t n 2 ( ∫ Ω f ( x , u n ) u n d x + ∫ Γ g ( s , u n ) u n d s + ∥ γ u n ∥ L 2 ( Γ ) 2 ) = lim n → + ∞ ( ∫ Ω f ( x , u n ) u n w n 2 d x + ∫ Γ g ( s , u n ) u n w n 2 d s + t n 2 ∥ γ u n ∥ L 2 ( Γ ) 2 ) ≥ K 1 ∫ Ω w 2 d x + K 2 ∫ Γ w 2 d s + ∥ w ∥ L 2 ( Γ ) 2 .

Noticing that w ( x ) ≠ 0 in Ω2 and K 1 , K 2 can be chosen large enough, so m Ω 2 ≡ 0 and then w(x)≡0 in Ω.

Then we know lim n → + ∞ ∫ Ω F ( x , w n ) d x + lim n → + ∞ ∫ Γ G ( s , w n ) d s = 0 , and consequently,

J ( w n ) = 1 2 ∥ ∇ w n ∥ L 2 ( Ω ) 2 + o ( 1 ) = 1 2 ∥ w n ∥ 2 − 1 2 ∥ w n ∥ L 2 ( Γ ) 2 + o ( 1 ) ≥ 1 2 ( 1 − 1 λ 1 + 1 ) ∥ w n ∥ 2 + o ( 1 ) = 2 c ( 1 − 1 λ 1 + 1 ) + o ( 1 ) .

By ∥un∥→+∞, tn→0 as n→+∞, then it follows Lemma 4 and (13), we obtain

J ( w n ) = J ( t n u n ) ≤ 1 + t n 2 2 n ≤ c .

Obviously, (19) and (20) are contradictory. So {un} is bounded in V. This completes the proof of Theorem 3. □

In this section, we give two examples on f(x,u): One satisfies (S1) to (S3) with q(x)≡+∞, but does not satisfy the (AR) condition; the other illustrates how the assumptions on the boundary are not trivial and compatible with the inner assumptions in Ω.

Example 1 Set:

f ( x , t ) = { 0 , t ≤ 0 ; t l n ( 1 + t ) , t > 0 .

Then it is easy to verify that f ( x , t ) satisfies (S1) to (S3) with p ( x ) = 0 as t → 0 and q ( x ) = + ∞ as t→+∞. In addition,

F ( x , t ) = 1 2 t 2 ln ( 1 + t ) − 1 4 t 2 + 1 2 t − 1 2 ln ( 1 + t ) .

So, for some μ>2, μ F ( x , t ) = t 2 ln ( 1 + t ) ( μ 2 − μ 4 ln ( 1 + t ) + μ 2 t l n ( 1 + t ) − μ 2 t 2 ) > t 2 ln ( 1 + t ) , for all t large.

This means f(x,t) does not satisfy the (AR) condition.

Example 2 Consider the following problem:

{ − u ″ ( x ) = α u ( x ) , 0 < x < l , u ( 0 ) = 0 , u ′ ( l ) = α u ( l ) ,

where α > 0 is a constant. It is obvious that g=γf as f ( x , u ) = α u ( x ) . Problem (21) is a case of (1); we can obtain the nontrivial solution: u ( x ) = C ˜ sin α x , C ˜ ≠ 0 .

The author declares that he has no competing interests.

Li G carried out all studies in this article.