S-shaped connected component of positive solutions for second-order discrete Neumann boundary value problems

Lemma 1.1

[15] Problem (1.5) possesses N real and simple eigenvalues μ k ( q ) , which can be ordered as

and the corresponding eigenfunction ϕ k changes its sign exactly k − 1 times on [ 1 , N ] Z . For simplicity, denote μ 1 ( q ) to be μ 1 , and the associated eigenfunction ϕ is positive.

Remark 1.1

The eigenvalue μ 1 is the minimum of the “Rayleigh quotient,” that is

and

Furthermore, we assume that

(H1) there exist α > 0 , f 0 > 0 and f 1 > 0 such that lim s → 0 + f ( s ) − f 0 s s 1 + α = − f 1 ;

(H2) f ∞ ≔ lim s → ∞ f ( s ) s = 0 .

Let η 1 be the principal eigenvalue of the eigenvalue problem

and the corresponding eigenfunction ω 1 is positive, N 2 ≤ t ˆ ≤ N + 1 2 .

Furthermore, we assume that

(H3) there exists s 0 > 0 such that

where q ˆ = max s ∈ [ 0 , N + 1 ] Z q ( s ) , 0 < σ < 1 be as in (2.2).

It is easy to find that if (H1) holds, then

Theorem 1.1

(see Figure 1) Assume that (H1)–(H3) hold. Then there exist λ ∗ ∈ ( 0 , μ 1 f 0 ) and λ ∗ > μ 1 f 0 such that

1. (1.1) has at least one positive solution if λ = λ ∗ ;

2. (1.1) has at least two positive solutions if λ ∗ < λ ≤ μ 1 f 0 ;

3. (1.1) has at least three positive solutions if μ 1 f 0 < λ < λ ∗ ;

4. (1.1) has at least two positive solutions if λ = λ ∗ ;

5. (1.1) has at least one positive solution if λ > λ ∗ .

Lemma 2.1

[7, Lemma 2.1] Let A = 1 2 ( q ( n ) + 2 + q 2 ( n ) + 4 q ( n ) ) , ρ = ( A N − A − N ) ( A 2 − 1 ) , then the unique solution of (2.1) can be expressed by

where

In addition, G ( n , s ) > 0 for all ( n , s ) ∈ [ 0 , N + 1 ] Z × [ 1 , N ] Z .

Denote

Obviously, 0 < m < M and 0 < σ < 1 (see [7]).

We extend f to a continuous function f ˜ defined by

Obviously, within the context of positive solutions, problem (1.1) is equivalent to the same problem with f replaced by f ˜ . Furthermore, f ˜ is an odd function for s ∈ ℝ . In the sequel of the proof we shall replace f with f ˜ . However, for the sake of simplicity, the modified function f ˜ will still be denoted by f.

Lemma 2.2

( μ 1 f 0 , 0 ) is a bifurcation point of problem (1.1) and the corresponding bifurcation branch C in ℝ × W whose closure contains ( μ 1 f 0 , 0 ) is either unbounded or contains a pair ( λ ¯ , 0 ) , where λ ¯ is an eigenvalue of problem (1.5) and λ ¯ ≠ μ 1 .

Proof

Let ξ , ζ ∈ C ( ℝ , ℝ ) be such that f ( u ) = f 0 u + ξ ( u ) , f ( u ) = f ∞ u + ζ ( u ) , clearly

Let us consider

as a bifurcation problem from the trivial solution u ≡ 0 .

By Lemma 2.1, problem (2.3) can be equivalently written as

where

Then it is well known that L and H : ℝ × E → E are completely continuous (see [8]).

By a simple calculation, we have

with respect to λ varying in bounded intervals.

Denote by S the closure in ℝ × W of the set of all non-trivial solutions ( λ , u ) of (2.3) with λ > 0 . Theorem 1.3 in [19] yields the existence of a maximal closed connected set C in S such that ( μ 1 f 0 , 0 ) ∈ C and at least one of the following conditions holds:

1. C is unbounded in ℝ × W ;

2. there exists a characteristic value of ℒ , with μ ˆ ≠ μ 1 , such that ( μ ˆ , 0 ) ∈ C . The proof is complete.□

Lemma 2.3

Let C ⊂ K ∪ { ( μ 1 f 0 , 0 ) } , then the last alternative of Lemma 2.2 is impossible.

Proof

Suppose on the contrary, if there exists ( λ n , u n ) → ( λ ¯ , 0 ) , when n → + ∞ with ( λ n , u n ) ∈ C , u n ≢ 0 , λ ¯ ≠ μ 1 / f 0 . Let v n ≔ u n / ∥ u n ∥ , then ∥ v n ∥ = 1 and v n is the solution of the problem:

Because v n is bounded, so there is a subsequence of v n , again denote by { v n } , such that v n → v 0 as n → + ∞ . It is easy to see that ( λ ¯ , v 0 ) verifies problem (1.5). On the other hand, we can easily show that the bifurcation points must be eigenvalues. Thus, C does not join to (0, 0) because 0 is not the eigenvalue of problem (1.5). Clearly, (0, 0) is the only solution of problem (1.1) for λ = 0 . Hence, we have C ∩ ( { 0 } × W ) = ∅ . It follows that λ ¯ ≠ μ 1 / f 0 . Lemma 1.1 follows v 0 must change its sign, and as a consequence for some n large enough, u n must change sign. This is a contradiction.□

Lemma 2.4

Assume (H1). Let { ( λ n , u n ) } be a sequence of positive solutions to (1.1) which satisfies λ n → μ 1 f 0 and ∥ u n ∥ → 0 . Let ϕ be the positive eigenfunction corresponding to μ 1 , which satisfies ∥ ϕ ∥ = 1 . Then there exists a subsequence of { u n } , again denoted by { u n } , such that u n ∥ u n ∥ converges uniformly to ϕ on [ 0 , N + 1 ] Z .

Proof

Set v n ≔ u n ∥ u n ∥ , since v n is bounded, a subsequence of { v n } uniformly converges to a limit v ∈ W with ∥ v ∥ = 1 , and we again denote by { v n } the subsequence.

For every ( λ n , u n ) , we have

Dividing the both sides of (2.4) by ∥ u n ∥ , we get

Since u n ( t ) → 0 for all t ∈ [ 0 , N + 1 ] Z , we conclude that ξ ( u n ( t ) ) ∥ u n ∥ → 0 for each fixed t ∈ [ 0 , N + 1 ] Z .

Lebesgue’s dominated convergence theorem shows that

for each fixed t ∈ [ 0 , N + 1 ] Z , which means that v is a nontrivial solution of (1.5) with λ = μ 1 , and hence v ≡ ϕ .□

Lemma 2.5

Let u , φ ∈ W . Then

Proof

Since Δ u ( 0 ) = Δ u ( N ) = 0 , we have

Lemma 2.6

Assume (H1). Then there exists δ > 0 such that for all ( λ , u ) ∈ C and | λ − μ 1 f 0 | + ∥ u ∥ ≤ δ imply λ > μ 1 f 0 .

Proof

Assume to the contrary that there exists a sequence { ( λ n , u n ) } ⊂ C such that λ n → μ 1 f 0 , ∥ u n ∥ → 0 and λ n ≤ μ 1 f 0 . By Lemma 2.4, there exists a subsequence of { u n } , again denoted by { u n } , such that u n ∥ u n ∥ converges uniformly to ϕ on [ 0 , N + 1 ] Z , where ϕ is the positive eigenfunction corresponding to μ 1 , which satisfies ∥ ϕ ∥ = 1 . Multiplying equation (1.1) with ( λ , u ) = ( λ n , u n ) by u n and summation it over [ 1 , N ] Z , we obtain

By simple computation and using the definition of μ 1 in Remark 1.1, we get

that is,

Lebesgue’s dominated convergence theorem, condition (H1) and Lemma 2.4 imply that

and

This contradicts λ n < μ 1 f 0 .□

Definition 2.1

[20] We say that a solution y ( t ) of problem (1.1) has a generalized zero at t 0 provided that y ( t 0 ) = 0 if t 0 = 0 and if t 0 > 0 either y ( t 0 ) = 0 or y ( t 0 − 1 ) y ( t 0 ) < 0 .

Lemma 2.7

Let P k ≥ p k for k ∈ [ m , n + 1 ] Z . Also let y ( k ) , z ( k ) be solutions of the following difference equations:

respectively. If y ( m ) = y ( n + 1 ) = 0 but without any generalized zeros in [ m + 1 , n ] Z , then either there exists τ ∈ [ m + 1 , n ] Z such that τ is a generalized zero of z or P k = p k and Δ y ( k ) y ( k ) = Δ z ( k ) z ( k ) .

Lemma 3.1

Assume (H3). Let u be a positive solution of (1.1) with ∥ u ∥ = s 0 , then λ < μ 1 f 0 .

Proof

Let u be a positive solution of (1.1) with ∥ u ∥ = s 0 . Without loss of generality, assume that there exists x 0 ∈ [ 0 , N + 1 ] Z such that ∥ u ∥ = u ( x 0 ) . This together with the fact

implies that

We note that u is a solution of

Now we assume λ ≥ μ 1 f 0 . Then for t ∈ [ 0 , t ˆ ] Z , by (H3), we get

Since η 1 is the principal eigenvalue of the eigenvalue problem (1.6). By Lemma 2.7, we know that u has at least one generalized zero on [ 0 , t ˆ ] Z . This contradicts the fact that u ( t ) > 0 on [ 0 , t ˆ ] Z .□

Lemma 3.2

Assume that (H1)–(H3) hold. Then C joins ( μ 1 f 0 , 0 ) to ( − ∞ , + ∞ ) .

Proof

Assume on the contrary that there exists λ M a blow up point and λ M < + ∞ . Then there exists a sequence { ( λ n , u n ) } satisfying λ n + ∥ u n ∥ → + ∞ , such that lim n → ∞ λ n = λ M and lim n → ∞ ∥ u n ∥ = + ∞ . We note that λ n > 0 for all n ∈ ℕ , since ( 0 , 0 ) is the only solution of problem (2.3) for λ = 0 and C ∩ ( { 0 } × W ) = ∅ .

Since ( λ n , u n ) ∈ C , we divide the equation

by ∥ u n ∥ and set ω n ≔ u n ∥ u n ∥ . Since ω n is bounded, after taking a subsequence if necessary, we have that ω n → ω for some ω ∈ W with ∥ ω ∥ = 1 . Let ζ ˜ ( u ) = max u ≤ s ≤ 2 u | ζ ( s ) | , then ζ ˜ is nondecreasing and

Since

Then

Combining the fact that f ∞ = 0 , we have

again choosing a subsequence and relabeling if necessary. This contradicts with ∥ ω ∥ = 1 .□

Proof of Theorem 1.1

By Lemmas 2.3 and 2.4, C is bifurcating from ( μ 1 f 0 , 0 ) and goes rightward. By Lemma 3.2, there exists ( λ 0 , u 0 ) ∈ C such that ∥ u 0 ∥ = s 0 and λ 0 < μ 1 f 0 . By Lemmas 2.4, 3.1 and 3.2, C passes through some points ( μ 1 f 0 , v 1 ) and ( μ 1 f 0 , v 2 ) with ∥ v 1 ∥ < s 0 < ∥ v 2 ∥ , and there exist λ ̲ and λ ̲ which satisfy 0 < λ ̲ < μ 1 f 0 < λ ̲ and both (i) and (ii):

1. if λ ∈ ( μ 1 f 0 , λ ̲ ] , then there exist u and v such that ( λ , u ) , ( λ , v ) ∈ C and ∥ u ∥ < ∥ v ∥ < s 0 ;

2. if λ ∈ ( λ ̲ , μ 1 f 0 ] , then there exist u and v such that ( λ , u ) , ( λ , v ) ∈ C and ∥ u ∥ < s 0 < ∥ v ∥ .