Spectrum of Discrete Second-Order Difference Operator with Sign-Changing Weight and Its Applications

Theorem A. Problem (4) has an infinite sequence of simple eigenvalues

and the eigenfunction corresponding to μ_k,± has exactly k − 1 simple zeros in (0, 1).

Theorem B. Problem (9) has exactly T real and simple eigenvalues λ_k, k ∈ 𝕋, which satisfies

and the eigenfunction corresponding to λ_k has exactly k − 1 simple generalized zeros.

Theorem 1. Let n be the number of elements in 𝕋₊. Let (H0) hold and let ν ∈ {+, −}. Then

• (a)

  if 1 ≤ n ≤ T − 1, then (1) and (2) have T real and simple eigenvalues, which can be ordered as follows:

  $$ ()

• (b)

  every eigenfunction ψ_k,ν corresponding to the eigenvalue λ_k,ν has exactly k − 1 simple generalized zeros.

Remark 2. There is also much literature dealing with difference equations similar to (12) subject to various boundary value conditions. We refer to [15–22] and the references therein. However, the weight m(t) > 0 in these papers.

Definition 3 (see [22].)Suppose that a function $$. If y(t) = 0, then t is a zero of y. If y(t) = 0 and y(t − 1)y(t + 1) < 0 for some t ∈ {2, …, T − 1}, then t is a simple zero of y. If y(t)y(t + 1) < 0 for some t ∈ {1, …, T − 1}, then y has a node at the point

The nodes and simple zeros of y are called the simple generalized zeros of y.

Lemma 4 (Lagrange-type identities). For t ∈ 𝕋,

Proof. We write (15) for the two arguments in full, giving

Multiplying, respectively, by u(t, μ),   u(t, λ) and subtracting, we have and putting t = 1 and recalling that u(0, λ) = u(0, μ) = 0, we derive (17) with t = 1. Induction over t then yields (17) from (19) in the general case.

Corollary 5. For t ∈ 𝕋,

Proof. Dividing (17) by μ − λ and making μ → λ for fixed λ, then we get the desired result.

Corollary 6. For t ∈ 𝕋, and complex λ,

Proof. Set $$ in (17). Then (21) is obtained.

Lemma 7. For t ∈ 𝕋, if (λ, u) is a solution of

satisfying u(0) = 0, u(t)u(t + 1) ≤ 0, and u≢0 on 𝕋, then

Proof. It is easy to see that λ ≠ 0. Now, multiplying (22) by u(s), then we get that

which implies the desired result.

Lemma 8. For t ∈ {1,2, …, T + 1}, the polynomial u(t, λ) has precisely t − 1 real and simple zeros.

Proof. This proof is divided into two steps.

Step 1 (each zero of u(t, λ) is real). Suppose on the contrary that u(t, λ) has a complex zero λ₀; then $$. Furthermore,

On the other hand, if λ₀ is a zero of u(t, λ), then λ₀ is an eigenvalue of the linear eigenvalue problem Now, by Lemma 7 and Corollary 6, we get that the above determinant does not equal zero, which is a contradiction. Hence, the zeros of u(t, λ) are all real for t ∈ 𝕋.

Step 2 (all of the zeros of u(t, λ) are simple). Suppose on the contrary that u(t, λ) has a multiple zeros λ_*, necessarily real. Then u(t, λ_*) = 0 and u^′(t, λ_*) = 0. Moreover,

However, by Lemma 7 and Corollary 5, we get that the above determinant does not equal zero, a contradiction.

Lemma 9 (see [14].)Let n ∈ {1, …, T − 1}. Then (1) and (2) have two principal eigenvalues λ_1,+ > 0 > λ_1,− and the corresponding eigenfunctions do not change their sign.

Lemma 10. Let ν ∈ {+, −}. Then dim⁡  ker⁡(L − λ_k,νmI) = 1, where I denotes the identity operator.

Proof. Suppose that φ_k,ν ∈ X and ϕ_k,ν ∈ X are two eigenfunctions corresponding to λ_k,ν. Then φ_k,ν(0) = ϕ_k,ν(0) = 0 and there exists a constant c ∈ ℝ such that φ_k,ν(1) = cϕ_k,ν(1) ≠ 0. Now, by the recurrence relation (15), we get that φ_k,ν(t) = cϕ_k,ν(t) for $$.

Lemma 11. For fixed real λ, the zeros of u(x, λ), 0 ≤ x ≤ T + 1 are simple.

Proof. Suppose that x^′ is a zero of u(x, λ). Now, the proof can be divided into two cases.

Case 1. If x₀ ∈ 𝕋, then by virtue of (15), we get that u(x₀ + 1, λ)u(x₀ − 1, λ) < 0.

Case 2. If x₀ ∈ (t, t + 1) for some t ∈ 𝕋. Then (∂/∂x)u(x, λ) exists at x = x₀ and is not zero by the definition of u(x, λ).

Lemma 12. Let x(λ) be the zeros of u(t + 1, λ). Then x(λ) is a continuous function of λ.

Lemma 13. For λ > 0, the zeros, x(λ), of u(x, λ), x ∈ (0, T + 1] is a decreasing function of λ. For λ < 0, x(λ) is an increasing function of λ for x ∈ (0, T + 1].

Proof. Let the zero x(λ) occur in (t, t + 1]. Since u(x, λ) is linear in (t, t + 1], the location of this zero is given by

conversely, x(λ) as given by this equation will actually be a zero of u(x, λ) if u(t, λ) ≠ u(t + 1, λ) and if x(λ) so the given falls in the interval (t, t + 1]. If we differentiate the right of (31), with respect to λ, the result is found to be and this does not equal zero by Lemma 7 and Corollary 5. Furthermore, for λ > 0, x^′(λ) < 0 and for λ < 0, x^′(λ) > 0. This combines with the continuity of x(λ), and we get the desired result.

Lemma 14. Let ν ∈ {+, −}. The sequence

exhibits for λ_1,− ≤ λ ≤ λ_1,+ no changes of sign; for |λ_k,ν| < λ ≤ |λ_k+1,ν|, exactly k changes sign; for λ > λ_p,+, exactly p changes sign; for λ < λ_T−p,−, exactly T − p changes sign.

Proof. To prove the times of changes of sign of (33), it is equivalent to find the number of zeros of u(x, λ), x ∈ (1, T + 1). We only deal with the case that ν = +; the case ν = − is similar.

By Lemma 12, for k = 1,2, 3, …, p − 1, there exist p − 1 functions x_k,+(λ), which satisfy x_k,+(λ_k,+) = T + 1 and are all decreasing function of λ; moreover, for fixed λ, there are the zeros of u(x, λ). Since u(0, λ) = 0 and u(1, λ) = 1, it follows that x_k,+(λ)∈(1, T + 1) for λ > λ_k,+.

For 0 ≤ λ ≤ λ_1,+, by Lemma 8, we get that u(x, λ) does not have a zero in (1, T + 1).

Let λ_⋄ ∈ (λ_1,+, λ_2,+] be arbitrary. Since x_1,+(λ) is continuous and decreasing, x_1,+(λ) will intersect with λ = λ_⋄ at (λ_⋄, x_1,+(λ_⋄)). Moreover, x_1,+(λ_1,+) = T + 1 and λ_⋄ > λ_1,+, which implies that x_1,+(λ_⋄) < T + 1. Thus, for λ = λ_2,+, (33) changes its sign exactly one time.

Now, we claim that for the same λ, x_k,+(λ) and x_k+1,+(λ) have no common zero for each k ∈ {1,2, …, p − 1}.

Suppose the contrary, then there exists λ^* ∈ (λ_2,+, λ_p,+) such that

Let Then, for λ ≤ λ^*, $$ is the k^*th zero of u(x, λ).

Case 1. If there exists t₀ ∈ 𝕋 such that $$, then by the definition of u(x, λ), we obtain that the signs of u(t₀, λ^*) and u(t₀ + 1, λ^*) are opposite. Without loss of generality, suppose that u(t₀, λ^*) > 0 and u(t₀ + 1, λ^*) < 0.

Now, consider the variation of $$ when λ varies. Take ε > 0 sufficiently small, by the continuity of $$ with respect to λ, for λ ∈ (λ^* − ε, λ^*), $$, and also u(t₀, λ) > 0, u(t₀ + 1, λ) < 0 since u(t, λ) is a continuous function of λ. However, for λ ∈ (λ^* − ε, λ^*), $$ is the (k^* + 1)th zero of u(x, λ), which implies that u(t₀, λ) < 0, u(t₀ + 1, λ) > 0, a contradiction.

Case 2. If there exists t₀ ∈ 𝕋 such that $$, then we consider the sign of u(t₀ − 1, λ^*) and u(t₀ + 1, λ^*), and we can get the similar contradiction as in Case 1.

This claim implies that, for the same λ,   x_k,+(λ),   k = 1,2, …, p − 1 do not intersect with each other. Thus, for λ_k−1,+ < λ_⋄ ≤ λ_k,+, there are k − 1 functions: x_j,+, j = 1,2, …, k − 1 which intersect with λ = λ_⋄ in (1, T + 1) at k − 1 different points; that is, for λ_k−1,+ < λ_⋄ ≤ λ_k,+, u(x, λ_⋄) has exactly k − 1 zeros.

This completes the proof.

Lemma 15. Problems (1) and (2) have exactly n positive eigenvalues and exactly T − n negative eigenvalues.

Proof. First we show that ψ_p,+ changes its sign exactly n − 1 times.

Let us consider the following T + 2 ordered polynomials:

where Q_j(λ) is a polynomial of degree precisely j − 1 of λ.

Observation 1. Consider

where NSC(λ) is the number of sign changes of

Observation 2. For j ∈ {2, …, T}, denoted by Γ(j), the number of the elements in the set

Then Γ(T) = n. Now if λ > 0 is large enough. Since changes sign exactly Γ(T) times, it follows that changes sign exactly Γ(T) times as λ > 0 is large enough; that is, This together with (37) implies that NSC(λ_p,+) = n. So, ψ_p,+ changes its sign exactly n times.

Next, using the result of first step and Lemma 14, it follows that p = n.

Proof of Theorem 1. From Lemma 7–Lemma 15, the results of Theorem 1 hold.

Theorem 16. Suppose that (H0), (H1), and (H2) hold. Assume that f₀ < f_∞. Then

• (i)

  if

  $$ ()

•  

  problems (12) and (13) have at least four sign-changing solutions $$, $$, $$, and $$;

• (ii)

  if

  $$ ()

•  

  problems (12) and (13) have at least four sign-changing solutions $$, $$, $$, and $$.

Moreover, for r ∈ (λ_k,+/f_∞, λ_k,+/f₀], there also exist at least two sign-changing solutions $$ and $$; meanwhile, for r ∈ [λ_l,−/f₀, λ_l,−/f_∞), there also exist at least two sign-changing solutions $$ and $$.

Theorem 17. Suppose that (H0), (H1), and (H2) hold. Assume that f_∞ < f₀. Then

• (i)

  if

  $$ ()

•   

  problems (12) and (13) have at least four sign-changing solutions $$, $$, $$, and $$;

• (ii)

  if

  $$ ()

•  

  problems (12) and (13) have at least four sign-changing solutions $$, $$, $$, and $$.

Moreover, for r ∈ (λ_k,+/f₀, λ_k,+/f_∞], there exist at least two sign-changing solutions $$ and $$; meanwhile, for r ∈ [λ_l,−/f_∞, λ_l,−/f₀), there also exist at least two sign-changing solutions $$ and $$.

Theorem 18. Let (H0), (H1), and ($$2) hold. Then (12) and (13) have a sign-changing solution in $$, (k = 1,2, …, n) if and only if r ≠ 0. Moreover, for r ∈ (−∞, λ_l,−/f₀), there exist at least two solutions $$ and $$, and there also exist at least two solutions $$ and $$ for r ∈ (λ_k,+/f₀, +∞).

Lemma 19. Suppose that (r, u) is a nontrivial solution of (12) and (13); then there exists k₀ ∈ {1,2, …, max⁡{n, T − n}} such that

Proof. Suppose on the contrary that for every k ∈ {1,2, …, max⁡{n, T − n}}

Then there exists t₀ ∈ 𝕋 such that Since u≢0 on $$, we may assume that On the other hand, it follows from (52) and f(0) = 0 that which implies that However, by (59) and the fact u(t₀) = 0, we get which contradicts (58).

Lemma 20. Suppose that (H0), (H1), and (H2) hold. Then for $$,

Proof. Suppose on the contrary that there exists $$ such that

By (H0), (H1), and (H2), there exists b ≥ 0 such that m(t)f(s) + bs is strictly increasing in s for s ∈ [0, s₁]. Then and since Δ²s₁ = 0 = f(s₁), Subtracting, we get Let w = s₁ − u, and applying boundary value problem (12) and (13), we have

Let e : 𝕋 → (0, ∞), such that

Let G₁(t, s) be the Green function of the boundary value problem From r > 0, b > 0, applying Theorem 6.8 and Corollary 6.7 of [24], we have G₁(t, s) > 0, ∀  t, s ∈ 𝕋 and G(0, s) = G(T + 1, s) = 0 for s ∈ 𝕋.

Problem (70) is equivalent to

By using the positivity of G₁(t, s) and e(s), we have that is, s₁ > u(t),   t ∈ 𝕋. This contradicts (64).

There exists $$ such that

Note that in this case r < 0, so we can choose $$ such that $$ is strictly decreasing in s for s ∈ [0, s₁]. Then and since Δ²s₁(t − 1) = 0 = f(s₁), Subtracting, we get Let w = s₁ − u, and applying boundary value condition (13), we have Similar to the above proof, we have s₁ > u(t),   t ∈ 𝕋. This contradicts (73).

Lemma 21. Suppose that (H0), (H1), and (H2) hold. Then for $$, we have

Proof. It is similar to the proof of Lemma 20, so we omit it.

Lemma 22. Suppose (H0), (H1), and (H2) hold. Then

Proof. Firstly, we will prove $$. Take Λ ⊂ ℝ as an interval such that Λ∩{(λ_j,+/f_∞) | j ∈ {1,2, …, n}} = {λ_k,+/f_∞} and ℳ is a neighborhood of (λ_k,+/f_∞, ∞) whose projection on ℝ lies in Λ and whose projection on X is bounded away from 0. Then by [25, Theorem 1.6 and Corollary 1.8], we have that either

• (1)

  $$ is bounded in ℝ × E in which case $$ meets{(λ, 0) | λ ∈ ℝ}; or

• (2)

  $$ is unbounded.

Moreover if (2) occurs and $$ has a bounded projection on ℝ, then $$ meets $$ where $$.

Obviously Lemma 21 implies that (1) does not occur. So $$ is unbounded.

Lemma 19 guarantees that $$ is a component of solutions of (80) in $$ which meets (λ_k,+/f_∞, ∞). Therefore there is no j ∈ {1,2, …, n}∖{k} such that $$ also meets (λ_j,+/f_∞, ∞). Otherwise, there will exist $$ such that y has a multiple zero point t₀ ∈ (0, T + 1); that is, y(t₀) = 0 and y(t₀ − 1)y(t₀ + 1) ≥ 0. However this contradicts with Lemma 19, and consequently $$ is unbounded. Thus

and similarly we have

Proof of Theorems 16, 17, and 18. From Lemmas 19–22 we have already completed the proof of Theorems 16 and 17. We note that if f_∞ = ∞, then λ_k,+/f_∞ = 0 and λ_l,−/f_∞ = 0 which imply that the results of Theorem 18 hold.