Global structure of sign-changing solutions for discrete Dirichlet problems

Definition 1.1

Suppose that a function y : T ˆ → ℝ . If y ( t 0 ) = 0 , then t 0 is a zero of y. If y ( t 0 ) = 0 or y ( t 0 ) y ( t 0 + 1 ) < 0 for some t 0 ∈ { 1 , … , T − 1 } , then y has a generalized zero at t 0 ∈ T .

Definition 1.2

A function y : T ˆ → ℝ is said to have a simple generalized zero at t 0 ∈ T provided that one of the following conditions is satisfied:

1. y ( t 0 ) = 0 , y ( t 0 − 1 ) y ( t 0 + 1 ) < 0 ;

2. y ( t 0 ) y ( t 0 + 1 ) < 0 .

It is well-known that the eigenvalue problem

has a finite sequence of simple eigenvalues

and the eigenfunction ϕ k corresponding to μ k has exactly k − 1 simple generalized zeros in T (see [1], and for other spectral results of related problems we refer to [2]). Let

be the generalized zeros of ϕ k .

Theorem 1.3

For each fixed integer k ∈ { 1 , 2 , … , T } and each fixed ν ∈ { + , − } , let (A1)–(A5) hold. Then, there exist λ ⁎ ∈ ( 0 , μ k / f 0 ) and λ ⁎ > μ k / f 0 , such that

1. (1.1) has at least one S k ν solution if λ = λ ⁎ ;

2. (1.1) has at least two S k ν solutions if λ ⁎ < λ ≤ μ k / f 0 ;

3. (1.1) has at least three S k ν solutions if μ k / f 0 < λ < λ ⁎ ;

4. (1.1) has at least two S k ν solutions if λ = λ ⁎ ;

5. (1.1) has at least one S k ν solution if λ > λ ⁎ .

Remark 1.4

Existence of c ˆ in (1.4) can be guaranteed by (A2). For example, let T = 100 , m 2 = 5 . By using Mathematica 9.0, the distance between two consecutive generalized zeros of ψ m 2 ( t ) was found to be 20, which means c ⁎ = 20 . Thus, it is easy to check that c ˆ = 10 by the definition of c ˆ .

Lemma 2.1

Suppose that ( λ , u ) is a nontrivial solution of ( 1.1 ). Then, there exists k 0 ∈ { 1 , 2 , … , T } , such that

Proof

Suppose on the contrary that for every k ∈ { 1 , 2 , … , T } ,

Then, there exists t 0 ∈ T such that

Since u ≢ 0 on T ˆ , we may assume that

On the other hand, it follows from equation (1.1) and f ( 0 ) = 0 that

which implies that

However, by (2.2) and the fact u ( t 0 ) = 0 , we get

which contradicts (2.1).□

Lemma 2.2

Assume that (A1), (A3) and (A4) hold. Let (A2) be satisfied for some k ∈ ℕ . Then for each ν ∈ { + , − } , there exists an unbounded subcontinuum C k ν which is branching from ( μ k / f 0 , 0 ) for ( 1.1 ). Moreover, if ( λ , u ) ∈ C k ν , then u is a S k ν solution of ( 1.1 ).

Lemma 2.3

Assume that (A1), (A3) and (A4) hold. Let (A2) be satisfied for some k ∈ ℕ . Let u be a S k ν solution of ( 1.1 ). Then, there exists a constant C > 0 independent of u such that

Proof

According to discrete Rolle’s theorem (see [1]), there exists x 0 ∈ T , such that Δ u ( x 0 ) = 0 or Δ u ( x 0 − 1 ) Δ u ( x 0 ) < 0 . Then by a direct computation, it is easy to see that

Noting that (A3) and (A4) imply that

for some f ⁎ > 0 , it follows from (2.5) that

1. if Δ u ( x 0 ) = 0 , then

   | Δ u ( x ) | = λ ∑ s = x + 1 x 0 | h ( s ) f ( u ( s ) ) | ≤ λ f ⁎ ∥ u ∥ ∑ s = x + 1 x 0 h ( s ) ≤ λ f ⁎ ∑ s = 0 T + 1 h ( s ) ∥ u ∥ ;

2. if Δ u ( x 0 − 1 ) Δ u ( x 0 ) < 0 , then

Lemma 2.4

Assume that (A1), (A3) and (A4) hold. Let (A2) be satisfied for some k ∈ ℕ . Let { ( λ n , u n ) } be a sequence of S k ν solutions to ( 1.1 ), which satisfies ∥ u n ∥ → 0 and λ n → μ k / f 0 . Let ϕ k ( x ) be the kth eigenfunction of ( 1.2 ), which satisfies ∥ ϕ k ∥ = 1 . Then, there exists a subsequence of { u n } , again denoted by { u n } , such that u n / ∥ u n ∥ converges uniformly to ϕ k on T .

Proof

Set v n ≔ u n / ∥ u n ∥ . Then, it is easy to see that ∥ v n ∥ = 1 . It follows from Lemma 2.3 that ∥ Δ v n ∥ is bounded, so there is a subsequence of v n uniformly convergent to a limit v. Furthermore, there exists a subsequence of it such that Δ v n ( 0 ) converges to some constant c. We again denote by v n the subsequence. We note that v ∈ Y , v ( 0 ) = v ( T + 1 ) = 0 and ∥ v ∥ = 1 . Rewriting equation (1.1) with ( λ , u ) = ( λ n , u n ) , we obtain

Dividing both sides of (2.7) by ∥ u n ∥ , we get

Noting that u n ( x ) → 0 for all x ∈ T ˆ , so for each fixed t ∈ T ˆ , one has f ( u n ( t ) ) u n ( t ) → f 0 . It follows from (2.6) that w n ( x ) converges to

for each fixed x ∈ T ˆ . Therefore, by recalling (2.8), one has

The fact coupled with (2.9) yields that v n ( x ) converges to

which implies that v is a nontrivial solution of (1.2) with λ = μ k , and hence v ≡ ϕ k .□

Lemma 2.5

Assume that (A1), (A3) and (A4) hold. Let (A2) be satisfied for some k ∈ ℕ . Let C k ν be as in Lemma 2.2. Then there exists δ > 0 such that for ( λ , u ) ∈ C k ν and | λ − μ k / f 0 | + ∥ u ∥ ≤ δ , one has λ > μ k / f 0 .

Proof

Suppose on the contrary that there exists a sequence { ( λ n , u n ) } such that ( λ n , u n ) ∈ C k ν , which satisfies λ n → μ k / f 0 , ∥ u n ∥ → 0 and λ n ≤ μ k / f 0 . According to Lemma 2.4, there exists a subsequence of { u n } , for convenience denote it by { u n } , such that u n ∥ u n ∥ converges uniformly to ϕ k on T ˆ , where ϕ k ( x ) ∈ S k ν is the kth eigenfunction of (1.2) with ∥ ϕ k ∥ = 1 . Multiplying the equation of (1.1) with ( λ , u ) = ( λ n , u n ) by u n and by a direct computation, one has

and accordingly

It follows from Lemma 2.4 that after taking a subsequence and relabeling if necessary, u n ∥ u n ∥ converges to ϕ k in Y.

then coupled with (2.10), one has

with a function ζ : ℕ → ℝ satisfying

That is,

From condition (A3), we have

and

This contradicts λ n ≤ μ k / f 0 .□

Lemma 3.1

Assume that (A2) holds for some k ∈ ℕ . Let u be a S k ν solution of (1.1). Then there exist two generalized zeros α u and β u , such that

Moreover,

where σ = min { min J u β u − α u , β u − max J u β u − α u } .

Proof

It is an immediate consequence of the fact that u is concave down in T ˆ .□

Lemma 3.2

Assume that (A1) and (A5) hold. Let σ be as in Lemma 3.1 and u be a S k ν solution of (1.1) with ∥ u ∥ = 1 σ s 0 . Then λ < μ k / f 0 .

Proof

Let u be a S k ν solution of (1.1). It follows from Lemma 3.1 that

We note that u is a solution of

Lemma 4.1

Assume that (A3) and (A4) hold. Let ( λ , u ) be a S k ν solution of (1.1). Then, there exists λ ⁎ > 0 such that λ ≥ λ ⁎ .

Proof

From Lemma 2.3, there exists a constant C > 0 , which is independent of u such that (2.4) holds. Let ∥ u ∥ = u ( x 0 ) , then it follows from (2.4) that

that is, λ ≥ ( C T ) − 1 .□

Lemma 4.2

Assume that (A1) and (A4) hold. Let (A2) be satisfied for some k ∈ ℕ . Let u be an S k ν solution of ( 1.1 ). Then, there exists a constant C > 0 independent of u such that for f ̲ ( s ) = min σ s ≤ t ≤ s f ( t ) / t , one has λ f ̲ ( ∥ u ∥ ) ≤ C .

Proof

Let

be the generalized zeros of u. Then, by (A2) we may get

We may assume that

or

We only deal with case (4.2) since case (4.3) can be treated by the similar way.

Let

Let t ˆ ∈ T satisfy t l + 1 − t l 2 ≤ t ˆ ≤ t l + 1 − t l + 1 2 and

If u ( t l ) = u ( t l + 1 ) = 0 , then it follows from (4.2) that

where

If u ( t l ) = A ≠ 0 , u ( t l + 1 ) = B ≠ 0 , then it follows from (4.2) again that

where u ˜ ( t ˆ ) is the solution of Δ 2 u ˜ ( t ˆ − 1 ) = 0 with boundary value condition u ˜ ( t l ) = A , u ˜ ( t l + 1 ) = B . Then by the similar argument as above, we complete the proof.□

Lemma 4.3

Assume that (A1) and (A3) hold, (A2) be satisfied for some k ∈ ℕ . Let J be a compact interval in ( 0 , ∞ ) . Then, there exists a constant M J > 0 such that for all λ ∈ J and given ν ∈ { + , − } , one has all possible S k ν solutions u of (1.1) satisfy ∥ u ∥ ≤ M J .

Proof

Suppose on the contrary that there exists a sequence { u n } of S k ν solutions of (1.1) with { λ n } ⊂ J ≜ [ a , b ] and ∥ u n ∥ → ∞ as n → ∞ . Then by the same argument used in the proof of Lemma 4.2, with obvious changes, we may get

Let

where Q = ∑ s = 1 T G ( t , s ) h ( s ) . Then by (A3), there exists u ρ > 0 such that u > u ρ implies f ( u ) < ρ u .

Let m ρ ≜ max u ∈ [ 0 , u ρ ] f ( u ) and let A n ≜ { t ∈ [ t m , t m + 1 ] Z : u n ( t ) ≤ u ρ } and B n ≜ { t ∈ [ t m , t m + 1 ] Z : u n ( t ) > u ρ } . Then we have,

for 0 ≤ s ≤ δ n . Thus,

On B n , u n ( s ) > u ρ implies f ( u n ( s ) ) ∥ u n ∥ ≤ f ( u n ( s ) ) u n ( s ) ≤ ρ . Accordingly,

Since 0 < a < λ n ≤ b for all n, we have 1 b ≤ 1 λ n ≤ 1 a for all n and

According to the fact ∥ u n ∥ → ∞ as n → ∞ , we get

This contradiction completes the proof.□

Lemma 4.4

Assume that (A1), (A3) and (A4) hold. Let (A2) hold for some k ∈ ℕ . Then there exists { ( λ n , u n ) } such that ( λ n , u n ) ∈ C k ν , λ n → ∞ as n → ∞ and ∥ u n ∥ → ∞ .

Proof

We only deal with the case ν = + .

It follows from Lemma 2.2 that C k ν is unbounded, so there exists a sequence { ( λ n , u n ) } of solutions of (1.1) such that { ( λ n , u n ) } ⊂ ℝ + × S k + and | λ n | + ∥ u n ∥ → ∞ . Moreover, Lemma 4.1 implies that λ n > 0 . Suppose on the contrary that there exists a bounded subsequence { ( λ n k , u n k ) } . Then from Lemma 4.3, ∥ u n k ∥ is bounded, which contradicts the fact that | λ n | + ∥ u n ∥ → ∞ . So, λ n → ∞ . Then, Lemma 4.2 implies that f ̲ ( ∥ u n ∥ ) → 0 . It deduces from (A4) that ∥ u n ∥ → ∞ .□