Periodic boundary value problems for nonlinear first-order impulsive dynamic equations on time scales

Abstract

By using the classical fixed point theorem for operators on cone, in this article, some results of one and two positive solutions to a class of nonlinear first-order periodic boundary value problems of impulsive dynamic equations on time scales are obtained. Two examples are given to illustrate the main results in this article.

Mathematics Subject Classification: 39A10; 34B15.

1 Introduction

Let T be a time scale, i.e., T is a nonempty closed subset of R. Let 0, T be points in T, an interval (0, T) _T denoting time scales interval, that is, (0, T) _T : = (0, T) ⋂ T. Other types of intervals are defined similarly.

The theory of impulsive differential equations is emerging as an important area of investigation, since it is a lot richer than the corresponding theory of differential equations without impulse effects. Moreover, such equations may exhibit several real world phenomena in physics, biology, engineering, etc. (see [1–3]). At the same time, the boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [4–18]. On the other hand, recently, the theory of dynamic equations on time scales has become a new important branch (see, for example, [19–21]). Naturally, some authors have focused their attention on the boundary value problems of impulsive dynamic equations on time scales [22–36]. However, to the best of our knowledge, few papers concerning PBVPs of impulsive dynamic equations on time scales with semi-position condition.

In this article, we are concerned with the existence of positive solutions for the following PBVPs of impulsive dynamic equations on time scales with semi-position condition

$$\left\{\begin{array}{c}{x}^{\Delta }\left(t\right)+f\left(t,x\left(\sigma \left(t\right)\right)\right)=0,t\in J:={\left[0,T\right]}_{\mathbf{T}},t\ne t_k,k=1,2,\dots ,m,\hfill \\ x\left(t_k^{+}\right)-x\left(t_k^{-}\right)=I_k\left(x\left(t_k^{-}\right)\right),k=1,2,\dots ,m,\hfill \\ x\left(0\right)=x\left(\sigma \left(T\right)\right),\hfill \end{array}\right.$$

(1.1)

where T is an arbitrary time scale, T > 0 is fixed, 0, T ∈ T, f ∈ C (J × [0, ∞), (-∞, ∞)), I _k ∈ C([0, ∞), [0, ∞)), t _k ∈ (0, T) _T , 0 < t₁ < ⋯ < t _m < T, and for each k = 1, 2,..., m, $x\left(t_k^{+}\right)=\underset{h\to 0^{+}}{\text{lim}}x\left(t_k+h\right)$ and $x\left(t_k^{-}\right)=\underset{h\to 0^{-}}{\text{lim}}x\left(t_k+h\right)$ represent the right and left limits of x(t) at t = t _k . We always assume the following hypothesis holds (semi-position condition):

• (H) There exists a positive number M such that

  $$Mx-f\left(t,x\right)\ge 0\mathsf{\text{for}}x\in \left[0,\infty \right),t\in {\left[0,T\right]}_{\mathbf{T}}.$$

By using a fixed point theorem for operators on cone [37], some existence criteria of positive solution to the problem (1.1) are established. We note that for the case T = R and I _k (x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem studied by [38] and for the case I _k (x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem (in the one-dimension case) studied by [39].

In the remainder of this section, we state the following fixed point theorem [37].

Theorem 1.1. Let X be a Banach space and K ⊂ X be a cone in X. Assume Ω₁, Ω₂ are bounded open subsets of X with $0\in {\Omega }_1\subset {\bar{\Omega }}_1\subset {\Omega }_2$ and Φ: $K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)\to K$ is a completely continuous operator. If

• (i) There exists u ₀ ∈ K\{0} such that u - Φu ≠ λu ₀, u ∈ K ⋂ ∂ Ω₂, λ ≥ 0; Φu ≠ τu, u ∈ K ⋂ ∂Ω₁, τ ≥ 1, or

• (ii) There exists u ₀ ∈ K\{0} such that u - Φu ≠ λu ₀, u ∈ K ⋂ ∂Ω₁, λ ≥ 0; Φu ≠ τu, u ∈ K ⋂ ∂Ω₂, τ ≥ 1.

Then Φ has at least one fixed point in $K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)$.

2 Preliminaries

Throughout the rest of this article, we always assume that the points of impulse t _k are right-dense for each k = 1, 2,...,m.

We define

$$\begin{array}{l}PC=\{x\in {[0,\sigma (T)]}_{\mathbf{T}}\to R:x_k\in C(J_k,R),k=0,1,2,\dots ,m\text{ and there exist}\\ x(t_k^{+})\text{and}x(t_k^{-})\text{with}x(t_k^{-})=x(t_k),k=1,2,\dots ,m\},\end{array}$$

where x _k is the restriction of x to J_k = (t _k , t_k+1] _T ⊂ (0, σ(T)] _T , k = 1, 2,..., m and J₀ = [0, t₁] _T , t_{m +1}= σ(T).

Let

$$X=\left\{x:x\in PC,x\left(0\right)=x\left(\sigma \left(T\right)\right)\right\}$$

with the norm $∥x∥=\underset{t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}}{\text{sup}}\left|x\left(t\right)\right|$, then X is a Banach space.

Lemma 2.1. Suppose M > 0 and h: [0, T] _T → R is rd-continuous, then x is a solution of

$$x\left(t\right)=\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)I_k\left(x\left(t_k\right)\right),t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}},$$

where $G\left(t,s\right)=\left\{\begin{array}{cc}\frac{e_M\left(s,t\right)e_M\left(\sigma \left(T\right),0\right)}{e_M\left(\sigma \left(T\right),0\right)-1},\hfill & \hfill 0\le s\le t\le \sigma \left(T\right),\hfill \\ \frac{e_M\left(s,t\right)}{e_M\left(\sigma \left(T\right),0\right)-1},\hfill & \hfill 0\le t<s\le \sigma \left(T\right),\hfill \end{array}\right.$

if and only if x is a solution of the boundary value problem

$$\left\{\begin{array}{c}{x}^{\Delta }\left(t\right)+Mx\left(\sigma \left(t\right)\right)=h\left(t\right),t\in J:={\left[0,T\right]}_{\mathbf{T}},t\ne t_k,k=1,2,\dots ,m,\hfill \\ x\left(t_k^{+}\right)-x\left(t_k^{-}\right)=I_k\left(x\left(t_k^{-}\right)\right),k=1,2,\dots ,m,\hfill \\ x\left(0\right)=x\left(\sigma \left(T\right)\right).\hfill \end{array}\right.$$

Proof. Since the proof similar to that of [34, Lemma 3.1], we omit it here.

Lemma 2.2. Let G(t, s) be defined as in Lemma 2.1, then

$$\frac{1}{e_M\left(\sigma \left(T\right),0\right)-1}\le G\left(t,s\right)\le \frac{e_M\left(\sigma \left(T\right),0\right)}{e_M\left(\sigma \left(T\right),0\right)-1}\mathsf{\text{for}}\mathsf{\text{all}}t,s\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}.$$

Proof. It is obviously, so we omit it here.

Remark 2.1. Let G(t, s) be defined as in Lemma 2.1, then ${\int }_0^{\sigma \left(T\right)}G\left(t,s\right)\Delta s=\frac{1}{M}$.

For u ∈ X, we consider the following problem:

$$\{\begin{array}{l}{x}^{\Delta }(t)+Mx(\sigma (t))=Mu(\sigma (t))-f(t,u(\sigma (t)),t\in {[0,T]}_T,t\ne t_k,k=1,2,\dots ,m,\hfill \\ x(t_k^{+})-x(t_k^{-})=I_k(x(t_k^{-})),k=1,2,\dots ,m,\hfill \\ x(0)=x(\sigma (T)).\hfill \end{array}$$

(2.1)

It follows from Lemma 2.1 that the problem (2.1) has a unique solution:

$$x\left(t\right)=\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h_u\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)I_k\left(x\left(t_k\right)\right),t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}},$$

where h _u (s) = Mu(σ(s)) - f(s, u(σ(s))), s ∈ [0, T] _T .

We define an operator Φ: X → X by

$$\Phi \left(u\right)\left(t\right)=\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h_u\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)I_k\left(u\left(t_k\right)\right),t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}.$$

It is obvious that fixed points of Φ are solutions of the problem (1.1).

Lemma 2.3. Φ: X → X is completely continuous.

Proof. The proof is divided into three steps.

Step 1: To show that Φ: X → X is continuous.

Let ${\left\{u_n\right\}}_{n=1}^{\infty }$ be a sequence such that u _n → u (n → ∞) in X. Since f(t, u) and I _k (u) are continuous in x, we have

$$\begin{array}{c}\left|h_{un}\left(t\right)-h_u\left(t\right)\right|=\left|M\left(u_n-u\right)-\left(f\left(t,u_n\right)-f\left(t,u\right)\right)\right|\to 0\left(n\to \infty \right),\\ \left|I_k\left(u_n\left(t_k\right)\right)-I_k\left(u\left(t_k\right)\right)\right|\to 0\left(n\to \infty \right).\end{array}$$

So

$$\begin{array}{c}\left|\Phi \left(u_n\right)\left(t\right)-\Phi \left(u\right)\left(t\right)\right|\\ =\left|\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)\left[h_{u_n}\left(s\right)-h_u\left(s\right)\right]\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)\left[I_k\left(u_n\left(t_k\right)\right)-I_k\left(u\left(t_k\right)\right)\right]\right|\\ \le \frac{e_M\left(\sigma \left(T\right),0\right)}{e_M\left(\sigma \left(T\right),0\right)-1}\left[\underset{0}{\overset{\sigma \left(T\right)}{\int }}\left|h_{u_n}\left(s\right)-h_u\left(s\right)\right|\Delta s+\sum _{k=1}^m\left|I_k\left(u_n\left(t_k\right)\right)-I_k\left(u\left(t_k\right)\right)\right|\right]\to 0\left(n\to \infty \right),\end{array}$$

which leads to ||Φu _n - Φu|| → 0 (n → ∞). That is, Φ: X → X is continuous.

Step 2: To show that Φ maps bounded sets into bounded sets in X.

Let B ⊂ X be a bounded set, that is, ∃ r > 0 such that ∀ u ∈ B we have ||u|| ≤ r. Then, for any u ∈ B, in virtue of the continuities of f(t, u) and I _k (u), there exist c > 0, c _k > 0 such that

$$\left|f\left(t,u\right)\right|\le c,\left|I_k\left(u\right)\right|\le c_k,k=1,2,\dots ,m.$$

We get

$$\begin{array}{ll}\hfill \left|\Phi \left(u\right)\left(t\right)\right|& =\left|\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h_u\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)I_k\left(u\left(t_k\right)\right)\right|\\ \le \underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)\left|h_u\left(s\right)\right|\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)\left|I_k\left(u\left(t_k\right)\right)\right|\\ \le \frac{e_M\left(\sigma \left(T\right),0\right)}{e_M\left(\sigma \left(T\right),0\right)-1}\left[\sigma \left(T\right)\left(Mr+c\right)+\sum _{k=1}^m{c}_k\right].\end{array}$$

Then we can conclude that Φu is bounded uniformly, and so Φ(B) is a bounded set.

Step 3: To show that Φ maps bounded sets into equicontinuous sets of X.

Let t₁, t₂ ∈ (t _k , t_k+1] _T ⋂ [0, σ(T)] _T , u ∈ B, then

$$\begin{array}{l}\left|\Phi \left(u\right)\left(t_1\right)-\Phi \left(u\right)\left(t_2\right)\right|\\ \le \underset{0}{\overset{\sigma \left(T\right)}{\int }}\left|G\left(t_1,s\right)-G\left(t_2,s\right)\right|\left|h_u\left(s\right)\right|\Delta s+\sum _{k=1}^m\left|G\left(t_1,t_k\right)-G\left(t_2,t_k\right)\right|\left|I_k\left(u\left(t_k\right)\right)\right|.\end{array}$$

The right-hand side tends to uniformly zero as |t₁ - t₂| → 0.

Consequently, Steps 1-3 together with the Arzela-Ascoli Theorem shows that Φ: X → X is completely continuous.

Let

$$K=\left\{u\in X:u\left(t\right)\ge \delta ∥u∥,t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}\right\},$$

where $\delta =\frac{1}{e_M\left(\sigma \left(T\right),0\right)}\in \left(0,1\right)$. It is not difficult to verify that K is a cone in X.

From condition (H) and Lemma 2.2, it is easy to obtain following result:

Lemma 2.4. Φ maps K into K.

3 Main results

For convenience, we denote

$$\begin{array}{ll}\hfill f^0& =\underset{u\to 0^{+}}{\text{lim}}\text{sup}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{max}}\frac{f\left(t,u\right)}{u},f^{\infty }=\underset{u\to \infty }{\text{lim}}\text{sup}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{max}}\frac{f\left(t,u\right)}{u},\\ \hfill f_0& =\underset{u\to 0^{+}}{\text{lim}}\text{inf}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{min}}\frac{f\left(t,u\right)}{u},f_{\infty }=\underset{u\to \infty }{\text{lim}}\text{inf}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{min}}\frac{f\left(t,u\right)}{u}.\end{array}$$

and

$$I_0=\underset{u\to 0^{+}}{\text{lim}}\frac{I_k\left(u\right)}{u},I_{\infty }=\underset{u\to \infty }{\text{lim}}\frac{I_k\left(u\right)}{u}.$$

Now we state our main results.

Theorem 3.1. Suppose that

(H₁) f₀ > 0, f^∞ < 0, I₀ = 0 for any k; or

(H₂) f_∞ > 0, f⁰ < 0, I_∞ = 0 for any k.

Then the problem (1.1) has at least one positive solutions.

Proof. Firstly, we assume (H₁) holds. Then there exist ε > 0 and β > α > 0 such that

$$f\left(t,u\right)\ge \epsilon u,t\in {\left[0,T\right]}_{\mathbf{T}},u\in \left(0,\alpha \right],$$

(3.1)

$$I_k\left(u\right)\le \frac{\left[e_m\left(\sigma \left(T\right),0\right)-1\right]\epsilon }{2Mm{e}_M\left(\sigma \left(T\right),0\right)}u,u\in \left(0,\alpha \right],\mathsf{\text{for}}\mathsf{\text{any}}k,$$

(3.2)

and

$$f\left(t,u\right)\le -\epsilon u,t\in {\left[0,T\right]}_{\mathbf{T}},u\in \left[\beta ,\infty \right).$$

(3.3)

Let Ω₁ = {u ∈ X: ||u|| < r₁}, where r₁ = α. Then u ∈ K ⋂ ∂Ω₁, 0 < δα = δ ||u|| ≤ u(t) ≤ α, in view of (3.1) and (3.2) we have

$$\begin{array}{ll}\hfill \Phi \left(u\right)\left(t\right)& =\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h_u\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)I_k\left(u\left(t_k\right)\right)\\ \le \underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)\left(M-\epsilon \right)u\left(\sigma \left(s\right)\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)\frac{\left[e_M\left(\sigma \left(T\right),0\right)-1\right]\epsilon }{2Mm{e}_M\left(\sigma \left(T\right),0\right)}u\left(t_k\right)\\ \le \frac{\left(M-\epsilon \right)}{M}∥u∥+\frac{e_M\left(\sigma \left(T\right),0\right)}{e_M\left(\sigma \left(T\right),0\right)-1}\sum _{k=1}^m\frac{\left[e_M\left(\sigma \left(T\right),0\right)-1\right]\epsilon }{2Mm{e}_M\left(\sigma \left(T\right),0\right)}∥u∥\\ =\frac{\left(M-\frac{\epsilon }{2}\right)}{M}∥u∥\\ <∥u∥,t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}},\end{array}$$

which yields ||Φ(u)|| < ||u||.

Therefore

$$\Phi u\ne \tau u,u\in K\cap \partial {\Omega }_1,\tau \ge 1.$$

(3.4)

On the other hand, let Ω₂ = {u ∈ X: ||u|| < r₂}, where $r_2=\frac{\beta }{\delta }$.

Choose u₀ = 1, then u₀ ∈ K\{0}. We assert that

$$u-\Phi u\ne \lambda u_0,u\in K\cap \partial {\Omega }_2,\lambda \ge 0.$$

(3.5)

Suppose on the contrary that there exist $ū\in K\cap \partial {\Omega }_2$ and $\bar{\lambda }\ge 0$ such that

$$ū-\Phi ū=\bar{\lambda }{u}_0.$$

Let $\varsigma =\underset{t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}}{\text{min}}ū\left(t\right)$, then $\varsigma \ge \delta ∥ū∥=\delta r_2=\beta$, we have from (3.3) that

$$\begin{array}{ll}\hfill ū\left(t\right)& =\Phi \left(ū\right)\left(t\right)+\bar{\lambda }\\ =\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h_{ū}\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,t_k\right)I_k\left(ū\left(t_k\right)\right)+\bar{\lambda }\\ \ge \underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)h_{ū}\left(s\right)\Delta s+\bar{\lambda }\\ \ge \frac{\left(M+\epsilon \right)}{M}\varsigma +\bar{\lambda },t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}.\end{array}$$

Therefore,

$$\varsigma =\underset{t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}}{\text{min}}ū\left(t\right)\ge \frac{\left(M+\epsilon \right)}{M}\varsigma +\bar{\lambda }>\varsigma ,$$

which is a contradiction.

It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point $u^{*}\in K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)$, and u* is a desired positive solution of the problem (1.1).

Next, suppose that (H₂) holds. Then we can choose ε' > 0 and β' > α' > 0 such that

$$f\left(t,u\right)\ge {\epsilon }^{\prime }u,t\in {\left[0,T\right]}_{\mathbf{T}},u\in \left[{\beta }^{\prime },\infty \right),$$

(3.6)

$$I_k\left(u\right)\le \frac{\left[e_M\left(\sigma \left(T\right),0\right)-1\right]{\epsilon }^{\prime }}{2Mm{e}_M\left(\sigma \left(T\right),0\right)}u,u\in \left[{\beta }^{\prime },\infty \right)\mathsf{\text{for}}\mathsf{\text{any}}k,$$

(3.7)

and

$$f\left(t,u\right)\le -{\epsilon }^{\prime }u,t\in {\left[0,T\right]}_{\mathbf{T}},u\in \left(0,{\alpha }^{\prime }\right].$$

(3.8)

Let Ω₃ = {u ∈ X: ||u|| < r₃}, where r₃ = α'. Then for any u ∈ K ⋂ ∂Ω₃, 0 < δ ||u|| ≤ u(t) ≤ ||u|| = α'.

It is similar to the proof of (3.5), we have

$$u-\Phi u\ne \lambda u_0,u\in K\cap \partial {\Omega }_3,\lambda \ge 0.$$

(3.9)

Let Ω₄ = {u ∈ X: ||u|| < r₄}, where $r_4=\frac{{\beta }^{\prime }}{\delta }$. Then for any u ∈ K ⋂ ∂Ω₄, u(t) ≥ δ ||u|| = δr₄ = β', by (3.6) and (3.7), it is easy to obtain

$$\Phi u\ne \tau u,u\in K\cap \partial {\Omega }_4,\tau \ge 1.$$

(3.10)

It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point $u^{*}\in K\cap \left({\bar{\Omega }}_4\backslash {\Omega }_3\right)$, and u* is a desired positive solution of the problem (1.1).

Theorem 3.2. Suppose that

(H₃) f⁰ < 0, f^∞ < 0;

(H₄) there exists ρ > 0 such that

$$\text{min}\left\{f\left(t,u\right)-u|t\in {\left[0,T\right]}_{\mathbf{T}},\delta \rho \le u\le \rho \right\}>0;$$

(3.11)

$$I_k\left(u\right)\le \frac{\left[e_M\left(\sigma \left(T\right),0\right)-1\right]}{Mm{e}_M\left(\sigma \left(T\right),0\right)}u,\delta \rho \le u\le \rho ,\mathsf{\text{for}}\mathsf{\text{any}}k.$$

(3.12)

Then the problem (1.1) has at least two positive solutions.

Proof. By (H₃), from the proof of Theorem 3.1, we should know that there exist β" > ρ > α" > 0 such that

$$u-\Phi u\ne \lambda u_0,u\in K\cap \partial {\Omega }_5,\lambda \ge 0,$$

(3.13)

$$u-\Phi u\ne \lambda u_0,u\in K\cap \partial {\Omega }_6,\lambda \ge 0,$$

(3.14)

where Ω₅ = {u ∈ X: ||u|| < r₅}, Ω₆ = {u ∈ X: ||u|| < r₆}, $r_5={\alpha }^{″},r_6=\frac{{\beta }^{″}}{\delta }.$

By (3.11) of (H₄), we can choose ε > 0 such that

$$f\left(t,u\right)\ge \left(1+\epsilon \right)u,t\in {\left[0,T\right]}_{\mathbf{T}},\delta \rho \le u\le \rho .$$

(3.15)

Let Ω₇ = {u ∈ X: ||u|| < ρ}, for any u ∈ K ⋂ ∂Ω₇, δρ = δ ||u|| ≤ u(t) ≤ ||u|| = ρ, from (3.12) and (3.15), it is similar to the proof of (3.4), we have

$$\Phi u\ne \tau u,u\in K\cap \partial {\Omega }_7,\tau \ge 1.$$

(3.16)

By Theorem 1.1, we conclude that Φ has two fixed points $u^{**}\in K\cap \left({\bar{\Omega }}_6\backslash {\Omega }_7\right)$ and $u^{***}\in K\cap \left({\bar{\Omega }}_7\backslash {\Omega }_5\right)$, and u** and u*** are two positive solution of the problem (1.1).

Similar to Theorem 3.2, we have:

Theorem 3.3. Suppose that

(H₄) f₀ > 0, f_∞ > 0, I₀ = 0, I_∞ = 0;

(H₅) there exists ρ > 0 such that

$$\text{max}\left\{f\left(t,u\right)|t\in {\left[0,T\right]}_{\mathbf{T}},\delta \rho \le u\le \rho \right\}<0.$$

Then the problem (1.1) has at least two positive solutions.

4 Examples

Example 4.1. Let T = [0, 1] ∪ [2, 3]. We consider the following problem on T

$$\left\{\begin{array}{c}{x}^{\Delta }\left(t\right)+f\left(t,x\left(\sigma \left(t\right)\right)\right)=0,t\in {\left[0,3\right]}_{\mathbf{T}},t\ne \frac{1}{2},\hfill \\ x\left({\frac{1}{2}}^{+}\right)-x\left({\frac{1}{2}}^{-}\right)=I\left(x\left(\frac{1}{2}\right)\right),\hfill \\ x\left(0\right)=x\left(3\right),\hfill \end{array}\right.$$

(4.1)

where T = 3, f(t, x) = x - (t + 1)x², and I(x) = x²

Let M = 1, then, it is easy to see that

$$Mx-f\left(t,x\right)=\left(t+1\right)x^2\ge 0\mathsf{\text{for}}x\in \left[0,\infty \right),t\in {\left[0,3\right]}_{\mathbf{T}},$$

and

$$f_0\ge 1,f^{\infty }=-\infty ,\mathsf{\text{and}}{I}_0=0.$$

Therefore, by Theorem 3.1, it follows that the problem (4.1) has at least one positive solution.

Example 4.2. Let T = [0, 1] ∪ [2, 3]. We consider the following problem on T

$$\left\{\begin{array}{c}{x}^{\Delta }\left(t\right)+f\left(t,x\left(\sigma \left(t\right)\right)\right)=0,t\in {\left[0,3\right]}_{\mathbf{T}},t\ne \frac{1}{2},\hfill \\ x\left({\frac{1}{2}}^{+}\right)-x\left({\frac{1}{2}}^{-}\right)=I\left(x\left(\frac{1}{2}\right)\right),\hfill \\ x\left(0\right)=x\left(3\right),\hfill \end{array}\right.$$

(4.2)

where $T=3,f\left(t,x\right)=4e^{1-4e^2}x-\left(t+1\right)x^2{e}^{-x}$, and I(x) = x²e^-x.

Choose M = 1, ρ = 4e², then $\delta =\frac{1}{2e^2}$, it is easy to see that

$$\begin{array}{ll}\hfill Mx-f\left(t,x\right)& =x\left(1-4e^{1-4e^2}\right)+\left(t+1\right)x^2{e}^{-x}\ge 0\mathsf{\text{for}}x\in \left[0,\infty \right),t\in {\left[0,3\right]}_{\mathbf{T}},\\ \hfill f_0& \ge 4e^{1-4e^2}>0,f_{\infty }\ge 4e^{1-4e^2}>0,I_0=0,I_{\infty }=0,\end{array}$$

and

$$\max (f(t,u)\left|t\in {[0,T]}_T,\delta \rho \le u\le \rho \}=\max \{f(t,u)\right|t\in {[0,3]}_T,2\le u\le 4e^2\}=16e^{3-4e^2}(1-e)<0.$$

Therefore, together with Theorem 3.3, it follows that the problem (4.2) has at least two positive solutions.