Theory and Modern Applications

# 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 [13]). At the same time, the boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [418]. On the other hand, recently, the theory of dynamic equations on time scales has become a new important branch (see, for example, [1921]). Naturally, some authors have focused their attention on the boundary value problems of impulsive dynamic equations on time scales [2236]. 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,\phantom{\rule{1em}{0ex}}t\in J:={\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}t\ne {t}_{k},\phantom{\rule{1em}{0ex}}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),\phantom{\rule{1em}{0ex}}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):

1. (H)

There exists a positive number M such that

$Mx-f\left(t,x\right)\ge 0\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,\infty \right),\phantom{\rule{1em}{0ex}}t\in {\left[0,T\right]}_{\mathbf{T}}.$

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 $0\in {\Omega }_{1}\subset {\stackrel{̄}{\Omega }}_{1}\subset {\Omega }_{2}$ and Φ: $K\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)\to K$ is a completely continuous operator. If

1. (i)

There exists u 0 K\{0} such that u - Φu ≠ λu 0, u K ∂ Ω2, λ ≥ 0; Φuτu, u K ∂Ω1, τ ≥ 1, or

2. (ii)

There exists u 0 K\{0} such that u - Φu ≠ λu 0, u K ∂Ω1, λ ≥ 0; Φuτu, u K ∂Ω2, τ ≥ 1.

Then Φ has at least one fixed point in $K\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)$.

## 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

where x k is the restriction of x to Jk = (t k , tk+1] T (0, σ(T)] T , k = 1, 2,..., m and J0 = [0, t1] T , tm +1= σ(T).

Let

$X=\left\{x:x\in PC,\phantom{\rule{1em}{0ex}}x\left(0\right)=x\left(\sigma \left(T\right)\right)\right\}$

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

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

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

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),\phantom{\rule{1em}{0ex}}t\in J:={\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}t\ne {t}_{k},\phantom{\rule{1em}{0ex}}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),\phantom{\rule{1em}{0ex}}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

$\frac{1}{{e}_{M}\left(\sigma \left(T\right),0\right)-1}\le G\left(t,s\right)\le \frac{{e}_{M}\left(\sigma \left(T\right),0\right)}{{e}_{M}\left(\sigma \left(T\right),0\right)-1}\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}t,s\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}.$

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:

$\left\{\begin{array}{l}{x}^{\Delta }\left(t\right)+Mx\left(\sigma \left(t\right)\right)=Mu\left(\sigma \left(t\right)\right)-f\left(t,u\left(\sigma \left(t\right)\right),\phantom{\rule{0.1em}{0ex}}t\in {\left[0,T\right]}_{T},\phantom{\rule{0.1em}{0ex}}t\ne {t}_{k},\phantom{\rule{0.1em}{0ex}}k=1,2,\dots ,m,\hfill \\ x\left({t}_{k}^{+}\right)-x\left({t}_{k}^{-}\right)={I}_{k}\left(x\left({t}_{k}^{-}\right)\right),\phantom{\rule{0.1em}{0ex}}k=1,2,\dots ,m,\hfill \\ x\left(0\right)=x\left(\sigma \left(T\right)\right).\hfill \end{array}$
(2.1)

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

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

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

We define an operator Φ: XX by

$\Phi \left(u\right)\left(t\right)=\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right){h}_{u}\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,{t}_{k}\right){I}_{k}\left(u\left({t}_{k}\right)\right),\phantom{\rule{1em}{0ex}}t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}.$

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

Lemma 2.3. Φ: XX is completely continuous.

Proof. The proof is divided into three steps.

Step 1: To show that Φ: XX 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

$\begin{array}{c}\left|{h}_{un}\left(t\right)-{h}_{u}\left(t\right)\right|=\left|M\left({u}_{n}-u\right)-\left(f\left(t,{u}_{n}\right)-f\left(t,u\right)\right)\right|\to 0\left(n\to \infty \right),\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|{I}_{k}\left({u}_{n}\left({t}_{k}\right)\right)-{I}_{k}\left(u\left({t}_{k}\right)\right)\right|\to 0\left(n\to \infty \right).\end{array}$

So

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\Phi \left({u}_{n}\right)\left(t\right)-\Phi \left(u\right)\left(t\right)\right|\\ =\left|\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)\left[{h}_{{u}_{n}}\left(s\right)-{h}_{u}\left(s\right)\right]\Delta s+\sum _{k=1}^{m}G\left(t,{t}_{k}\right)\left[{I}_{k}\left({u}_{n}\left({t}_{k}\right)\right)-{I}_{k}\left(u\left({t}_{k}\right)\right)\right]\right|\\ \le \frac{{e}_{M}\left(\sigma \left(T\right),0\right)}{{e}_{M}\left(\sigma \left(T\right),0\right)-1}\left[\underset{0}{\overset{\sigma \left(T\right)}{\int }}\left|{h}_{{u}_{n}}\left(s\right)-{h}_{u}\left(s\right)\right|\Delta s+\sum _{k=1}^{m}\left|{I}_{k}\left({u}_{n}\left({t}_{k}\right)\right)-{I}_{k}\left(u\left({t}_{k}\right)\right)\right|\right]\to 0\left(n\to \infty \right),\end{array}$

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

$\left|f\left(t,u\right)\right|\le c,\phantom{\rule{1em}{0ex}}\left|{I}_{k}\left(u\right)\right|\le {c}_{k},\phantom{\rule{1em}{0ex}}k=1,2,\dots ,m.$

We get

$\begin{array}{ll}\hfill \left|\Phi \left(u\right)\left(t\right)\right|& =\left|\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right){h}_{u}\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,{t}_{k}\right){I}_{k}\left(u\left({t}_{k}\right)\right)\right|\phantom{\rule{2em}{0ex}}\\ \le \underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right)\left|{h}_{u}\left(s\right)\right|\Delta s+\sum _{k=1}^{m}G\left(t,{t}_{k}\right)\left|{I}_{k}\left(u\left({t}_{k}\right)\right)\right|\phantom{\rule{2em}{0ex}}\\ \le \frac{{e}_{M}\left(\sigma \left(T\right),0\right)}{{e}_{M}\left(\sigma \left(T\right),0\right)-1}\left[\sigma \left(T\right)\left(Mr+c\right)+\sum _{k=1}^{m}{c}_{k}\right].\phantom{\rule{2em}{0ex}}\end{array}$

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 , tk+1] T [0, σ(T)] T , u B, then

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\left|\Phi \left(u\right)\left({t}_{1}\right)-\Phi \left(u\right)\left({t}_{2}\right)\right|\phantom{\rule{2em}{0ex}}\\ \le \underset{0}{\overset{\sigma \left(T\right)}{\int }}\left|G\left({t}_{1},s\right)-G\left({t}_{2},s\right)\right|\left|{h}_{u}\left(s\right)\right|\Delta s+\sum _{k=1}^{m}\left|G\left({t}_{1},{t}_{k}\right)-G\left({t}_{2},{t}_{k}\right)\right|\left|{I}_{k}\left(u\left({t}_{k}\right)\right)\right|.\phantom{\rule{2em}{0ex}}\end{array}$

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: XX is completely continuous.

Let

$K=\left\{u\in X:u\left(t\right)\ge \delta ∥u∥,\phantom{\rule{1em}{0ex}}t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}\right\},$

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

$\begin{array}{ll}\hfill {f}^{0}& =\underset{u\to {0}^{+}}{\text{lim}}\text{sup}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{max}}\frac{f\left(t,u\right)}{u},\phantom{\rule{1em}{0ex}}{f}^{\infty }=\underset{u\to \infty }{\text{lim}}\text{sup}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{max}}\frac{f\left(t,u\right)}{u},\phantom{\rule{2em}{0ex}}\\ \hfill {f}_{0}& =\underset{u\to {0}^{+}}{\text{lim}}\text{inf}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{min}}\frac{f\left(t,u\right)}{u},\phantom{\rule{1em}{0ex}}{f}_{\infty }=\underset{u\to \infty }{\text{lim}}\text{inf}\underset{t\in {\left[0,T\right]}_{\mathbf{T}}}{\text{min}}\frac{f\left(t,u\right)}{u}.\phantom{\rule{2em}{0ex}}\end{array}$

and

${I}_{0}=\underset{u\to {0}^{+}}{\text{lim}}\frac{{I}_{k}\left(u\right)}{u},\phantom{\rule{1em}{0ex}}{I}_{\infty }=\underset{u\to \infty }{\text{lim}}\frac{{I}_{k}\left(u\right)}{u}.$

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

$f\left(t,u\right)\ge \epsilon u,\phantom{\rule{1em}{0ex}}t\in {\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}u\in \left(0,\alpha \right],$
(3.1)
${I}_{k}\left(u\right)\le \frac{\left[{e}_{m}\left(\sigma \left(T\right),0\right)-1\right]\epsilon }{2Mm{e}_{M}\left(\sigma \left(T\right),0\right)}u,u\in \left(0,\alpha \right],\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{any}}\phantom{\rule{2.77695pt}{0ex}}k,$
(3.2)

and

$f\left(t,u\right)\le -\epsilon u,\phantom{\rule{1em}{0ex}}t\in {\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}u\in \left[\beta ,\infty \right).$
(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

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

which yields ||Φ(u)|| < ||u||.

Therefore

$\Phi u\ne \tau u,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{1},\phantom{\rule{1em}{0ex}}\tau \ge 1.$
(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

$u-\Phi u\ne \lambda {u}_{0},\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{2},\phantom{\rule{1em}{0ex}}\lambda \ge 0.$
(3.5)

Suppose on the contrary that there exist $ū\in K\cap \partial {\Omega }_{2}$ and $\stackrel{̄}{\lambda }\ge 0$ such that

$ū-\Phi ū=\stackrel{̄}{\lambda }{u}_{0}.$

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

$\begin{array}{ll}\hfill ū\left(t\right)& =\Phi \left(ū\right)\left(t\right)+\stackrel{̄}{\lambda }\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right){h}_{ū}\left(s\right)\Delta s+\sum _{k=1}^{m}G\left(t,{t}_{k}\right){I}_{k}\left(ū\left({t}_{k}\right)\right)+\stackrel{̄}{\lambda }\phantom{\rule{2em}{0ex}}\\ \ge \underset{0}{\overset{\sigma \left(T\right)}{\int }}G\left(t,s\right){h}_{ū}\left(s\right)\Delta s+\stackrel{̄}{\lambda }\phantom{\rule{2em}{0ex}}\\ \ge \frac{\left(M+\epsilon \right)}{M}\varsigma +\stackrel{̄}{\lambda },\phantom{\rule{1em}{0ex}}t\in {\left[0,\sigma \left(T\right)\right]}_{\mathbf{T}}.\phantom{\rule{2em}{0ex}}\end{array}$

Therefore,

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

It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point ${u}^{*}\in K\cap \left({\stackrel{̄}{\Omega }}_{2}\{\Omega }_{1}\right)$, 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

$f\left(t,u\right)\ge {\epsilon }^{\prime }u,\phantom{\rule{1em}{0ex}}t\in {\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}u\in \left[{\beta }^{\prime },\infty \right),$
(3.6)
${I}_{k}\left(u\right)\le \frac{\left[{e}_{M}\left(\sigma \left(T\right),0\right)-1\right]{\epsilon }^{\prime }}{2Mm{e}_{M}\left(\sigma \left(T\right),0\right)}u,\phantom{\rule{1em}{0ex}}u\in \left[{\beta }^{\prime },\infty \right)\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{any}}\phantom{\rule{2.77695pt}{0ex}}k,$
(3.7)

and

$f\left(t,u\right)\le -{\epsilon }^{\prime }u,\phantom{\rule{1em}{0ex}}t\in {\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}u\in \left(0,{\alpha }^{\prime }\right].$
(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

$u-\Phi u\ne \lambda {u}_{0},\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{3},\phantom{\rule{1em}{0ex}}\lambda \ge 0.$
(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

$\Phi u\ne \tau u,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{4},\phantom{\rule{1em}{0ex}}\tau \ge 1.$
(3.10)

It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point ${u}^{*}\in K\cap \left({\stackrel{̄}{\Omega }}_{4}\{\Omega }_{3}\right)$, 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

$\text{min}\left\{f\left(t,u\right)-u|t\in {\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}\delta \rho \le u\le \rho \right\}>0;$
(3.11)
${I}_{k}\left(u\right)\le \frac{\left[{e}_{M}\left(\sigma \left(T\right),0\right)-1\right]}{Mm{e}_{M}\left(\sigma \left(T\right),0\right)}u,\phantom{\rule{1em}{0ex}}\delta \rho \le u\le \rho ,\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{any}}\phantom{\rule{2.77695pt}{0ex}}k.$
(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

$u-\Phi u\ne \lambda {u}_{0},\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{5},\phantom{\rule{1em}{0ex}}\lambda \ge 0,$
(3.13)
$u-\Phi u\ne \lambda {u}_{0},\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{6},\phantom{\rule{1em}{0ex}}\lambda \ge 0,$
(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

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

$\Phi u\ne \tau u,\phantom{\rule{1em}{0ex}}u\in K\cap \partial {\Omega }_{7},\phantom{\rule{1em}{0ex}}\tau \ge 1.$
(3.16)

By Theorem 1.1, we conclude that Φ has two fixed points ${u}^{**}\in K\cap \left({\stackrel{̄}{\Omega }}_{6}\{\Omega }_{7}\right)$ and ${u}^{***}\in K\cap \left({\stackrel{̄}{\Omega }}_{7}\{\Omega }_{5}\right)$, 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

$\text{max}\left\{f\left(t,u\right)|t\in {\left[0,T\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}\delta \rho \le u\le \rho \right\}<0.$

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,\phantom{\rule{1em}{0ex}}t\in {\left[0,3\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}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

$Mx-f\left(t,x\right)=\left(t+1\right){x}^{2}\ge 0\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,\infty \right),\phantom{\rule{1em}{0ex}}t\in {\left[0,3\right]}_{\mathbf{T}},$

and

${f}_{0}\ge 1,\phantom{\rule{1em}{0ex}}{f}^{\infty }=-\infty ,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{and}}\phantom{\rule{2.77695pt}{0ex}}{I}_{0}=0.$

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,\phantom{\rule{1em}{0ex}}t\in {\left[0,3\right]}_{\mathbf{T}},\phantom{\rule{1em}{0ex}}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 $T=3,f\left(t,x\right)=4{e}^{1-4{e}^{2}}x-\left(t+1\right){x}^{2}{e}^{-x}$, and I(x) = x2e-x.

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

$\begin{array}{ll}\hfill Mx-f\left(t,x\right)& =x\left(1-4{e}^{1-4{e}^{2}}\right)+\left(t+1\right){x}^{2}{e}^{-x}\ge 0\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,\infty \right),\phantom{\rule{1em}{0ex}}t\in {\left[0,3\right]}_{\mathbf{T}},\phantom{\rule{2em}{0ex}}\\ \hfill {f}_{0}& \ge 4{e}^{1-4{e}^{2}}>0,\phantom{\rule{1em}{0ex}}{f}_{\infty }\ge 4{e}^{1-4{e}^{2}}>0,\phantom{\rule{1em}{0ex}}{I}_{0}=0\phantom{\rule{1em}{0ex}},{I}_{\infty }=0,\phantom{\rule{2em}{0ex}}\end{array}$

and

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

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

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## Acknowledgements

The author thankful to the anonymous referee for his/her helpful suggestions for the improvement of this article. This work is supported by the Excellent Young Teacher Training Program of Lanzhou University of Technology (Q200907)

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Wang, DB. Periodic boundary value problems for nonlinear first-order impulsive dynamic equations on time scales. Adv Differ Equ 2012, 12 (2012). https://doi.org/10.1186/1687-1847-2012-12