Higher-Order Dynamic Delay Differential Equations on Time Scales

Theorem 1.1 (see [18].)Suppose E is a real Banach space, K ⊂ E is a cone, let Ω_r = {u ∈ K : ∥u∥ ≤ r}. Let operator T:Ω_r → K be completely continuous and satisfy Tx ≠ x, for all x ∈ ∂Ω_r. Then

• (i)

  if ∥Tx∥ ≤ ∥x∥, for all x ∈ ∂Ω_r, then i(T, Ω_r, K) = 1;

• (ii)

  if ∥Tx∥ ≥ ∥x∥, for all x ∈ ∂Ω_r, then i(T, Ω_r, K) = 0.

Definition 2.1. u(t) is called a solution of SBVP (1.7) and (1.8) if it satisfies the following:

• (1)

  $$;

• (2)

  u(t) > 0 for all t ∈ (0, T) and satisfy conditions (1.8);

• (3)

  $$ hold for t ∈ (0, T).

In the rest of the paper, we also make the following assumptions:

•  

  H₁ f ∈ C_ld([0,+∞)ⁿ, [0, +∞));

•  

  H₂ g(t) ∈ C_ld((0, T), [0, +∞)) and there exists t₀ ∈ (0, T), such that

•  

  H₃ ζ(t) ∈ C([−τ, 0], ζ(t) > 0 on [−τ, 0) and ζ(0) = 0.

It is easy to check that condition (H₂) implies that

We can easily get the following lemmas.

Lemma 2.2. Suppose condition (H₂) holds. Then there exists a constant θ ∈ (0, 1/2) satisfing

Furthermore, the function is positive continuous functions on [θ, T − θ]; therefore, A(t) has minimum on [θ, T − θ]. Hence we suppose, that there exists L > 0 such that A(t) ≥ L, t ∈ [θ, T − θ].

Proof. At first, it is easily seen that A(t) is continuous on [θ, T − θ]. Next, let

Then, from condition (H₂), we have that the function A₁(t) is strictly monotone nondecreasing on [θ, T − θ] and A₁(θ) = 0, the function A₂(t) is strictly monotone nonincreasing on [θ, T − θ] and A₂(T − θ) = 0, which implies L = min _{t∈[θ,T−θ]} A(t) > 0. The proof is complete.

Lemma 2.3. Let u ∈ K and θ of Lemma 2.2, then

The proof of the above lemma is similar to the proof in [17, Lemma 2.2], so we omit it.

Lemma 2.4. Suppose that conditions (H₁),   (H₂),   (H₃) hold, $$ is a solution of the following boundary value problems:

where Then, $$, −τ ≤ t ≤ T is a positive solution to the SBVP (1.7) and (1.8).

Proof. It is easy to check that $$ satisfies (1.7) and (1.8).

Lemma 2.5. Suppose that conditions (H₁),   (H₂),   (H₃) hold, $$ is a solution of boundary value problems (2.13), (2.14) if and only if u(t) ∈ B is a solution of the following integral equation:

where

Proof. Necessity. Obviously, for t ∈ (−τ, 0), we have u(t) = ζ(t). If t ∈ (0,1), by the equation of the boundary condition, we have $$, $$, then there exists a constant σ ∈ [ξ, η]⊂(0, T) such that $$.

Firstly, by integrating the equation of the problems (2.13) on (σ, T), we have

then thus

By $$ and condition (2.18), let t = η on (2.18), we have

By the equation of the boundary condition (2.14), we have then Then, by (2.20) and leting t = T on (2.20), we have Then Then, by integrating (2.25) for n − 2 times on (0, T), we have Similarly, for t ∈ (0, σ), by integrating the equation of problems (2.13) on (0, σ), we have Therefore, for any t ∈ [0, T], u(t) can be expressed as equation where w(t) is expressed as (2.17). Then the results of Lemma 2.3 hold.

Sufficiency. Suppose that $$, 0 ≤ t ≤ T. Then by (2.17), we have

So, $$, 0 < t < T. These imply that (2.13) holds. Furthermore, by letting t = 0 and t = T on (2.17) and (2.29), we can obtain the boundary value equations of (2.14). The proof is complete.

Lemma 2.6. T : K → K is completely continuous.

Proof. Because

is continuous, decreasing on [0, T] and satisfies $$, then, Tu ∈ K for each u ∈ K and $$. This shows that TK ⊂ K. Furthermore, it is easy to check by Arzela-ascoli Theorem that T : K → K is completely continuous.

Lemma 2.7. Suppose that conditions (H₁),   (H₂),   (H₃) hold, the solution u(t) of problem (2.13), (2.14) satisfies

and for θ ∈ (0, T/2) in Lemma 2.2, one has

Proof. Firstly, we can have

Next, if u(t) is the solution of problem (2.13), (2.14), then $$ is concave function, and $$, t ∈ [0, T]. Thus, we have

that is, $$, t ∈ [0, T].

Finally, by Lemma 2.3, for t ∈ [θ, T − θ], we have $$. By $$, we have

The proof is complete.

Theorem 3.1. Suppose that conditions (H₁), (H₂), (H₃), and (A₂) hold. Assume that f also satisfies

•  

  A₃ f₀ = +∞;

•  

  A₄ f_∞ = +∞.

Then, the SBVP (2.13), (2.14) hase at last two solutions u₁, u₂ such that

Proof. For any u ∈ K, by Lemma 2.3, we have

First, by condition (A₃), for any N > 2/θL, there exists a constant ρ_* ∈ (0, R) such that Set $$. For any $$, by (3.2) we have For any $$, by (3.3) and Lemmas 2.3–2.6, we will discuss it from three perspectives.

• (i)

  If σ ∈ [θ, T − θ], we have

  $$ ()

• (ii)

  If σ ∈ (T − θ, T], we have

• (iii)

  If σ ∈ (0, θ), we have

  $$ ()
  Therefore, no matter under which condition, we all have
  $$ ()
  Then, by Theorem 1.1, we have
  $$ ()

Next, by condition (A₄), for any $$, there exists a constant ρ₀ > 0 such that We choose a constant ρ^* > max  {R, ρ₀/θ}, obviously ρ_* < R < ρ^*. Set $$. For any $$, by Lemma 2.3, we have Then, by (3.10), Lemmas 2.3–2.6 and also similar to the previous proof, we can also have from three perspectives that Then, by Theorem 1.1, we have Finally, set Ω_R = {u ∈ K : ∥u∥<R}. For any u ∈ ∂Ω_R, we have u(t) ≤ ∥u∥ = R, by (A₂) we know Thus, Then, by Theorem 1.1, we have Therefore, by (3.9), (3.13), (3.16), ρ_* < R < ρ^* we have Then T has fixed point $$ and fixed point $$. Obviously, u₁, u₂ are all positive solutions of problem (2.13), (2.14) and ρ_* < ∥u₁∥ < R < ∥u₂∥ < ρ^*. Proof of Theorem 3.1 is complete.

Theorem 3.2. Suppose that conditions (H₁), (H₂), (H₃), (A₁) hold. Assume that f also satisfies

•  

  A₅ f₀ = 0;

•  

  A₆ f_∞ = 0.

Then, the SBVP (2.13), (2.14) has at last two solutions u₁, u₂ such that 0 < ∥u₁∥ < r < ∥u₂∥.

Proof. First, by f₀ = 0, for ϵ₁ ∈ (0, θ_*), there exists a constant ρ_* ∈ (0, r) such that

Set $$, for any $$, by (3.18), we have that is, Then, by Theorem 1.1, we have

Next, let $$; note that f^*(x) is monotone increasing with respect to x ≥ 0. Then, from f_∞ = 0, it is easy to see that

Therefore, for any ϵ₂ ∈ (0, θ_*), there exists a constant ρ^* > r such that Set $$, for any $$, by (3.23), we have that is, Then, by Theorem 1.1, we have

Finally, set Ω_r = {u ∈ K : ∥u∥ < r}. For any u ∈ ∂Ω_r, by (A₁), Lemma 2.3 and also similar to the previous proof of Theorem 3.1, we can also have

Then, by Theorem 1.1, we have Therefore, by (3.21), (3.28), (3.26), ρ_* < r < ρ^*, we have Then T has fixed point $$ and fixed point $$. Obviously, u₁, u₂ are all positive solutions of problem (2.13), (2.14) and ρ_* < ∥u₁∥ < r < ∥u₂∥ < ρ^*. The proof of Theorem 3.2 is complete.

Theorem 3.3. Suppose that conditions (H₁), (H₂), (H₃), and (A₂) hold and

•  

  A₇ $$,

•  

  A₈ $$.

Then, the SBVP (2.13), (2.14) has at last two solutions u₁, u₂ such that 0 < ∥u₁∥ < R < ∥u₂∥.

Theorem 3.4. Suppose that conditions (H₁), (H₂), (H₃), and (A₁) hold and

•  

  A₉ $$;

•  

  A₁₀ $$.

Then, the SBVP (2.13), (2.14) has at last two solutions u₁, u₂ such that 0 < ∥u₁∥ < r < ∥u₂∥.

Example 4.1. Consider the following 3-order singular boundary value problem (SBVP) with p-Laplacian:

where So, by Lemma 2.4, we discuss the following SBVP: where Then, obviously, so conditions (H₁), (H₂), (H₃), (A₂), and (A₃) hold.

Next,

we choose R = 3, M = 2 and for θ = 1/4, because of the monotone increasing of f(u₁, u₂, u₃) on [0,∞)³, then Therefore, by we know so condition (A₂) holds. Then, by Theorem 3.1, SBVP (4.3) has at least two positive solutions v₁, v₂ and 0 < ∥v₁∥<3 < ∥v₂∥. Then, by Lemma 2.4, $$, $$, t ∈ (−1,1) are the positive solutions of the SBVP (4.1).