# Positive solutions to PBVPs for nonlinear first-order impulsive dynamic equations on time scales

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## Abstract

By using the classical fixed point theorem for operators on a cone, in this paper, 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.

## Introduction

The theory of dynamic equations on time scales has been a new important mathematical branch  since it was initiated by Hilger . At the same time, the boundary value problems of impulsive dynamic equations on time scales have received considerable attention  since the theory of impulsive differential equations is a lot richer than the corresponding theory of differential equations without impulse effects .

In this paper, we concerned with the existence of positive solutions for the following PBVPs of impulsive dynamic equations on time scales

$$\left \{ \begin{array}{@{}l} x^{\bigtriangleup}(t)+p(t)x(\sigma(t))=f(t,x(\sigma(t))),\quad t\in J:= [ 0,T ] _{\mathbb{T}}, t\neq t_{k}, \\ x(t_{k}^{+})-x(t_{k}^{-})=I_{k}(x(t_{k}^{-})),\quad k=1,2,\ldots,m, \\ x(0)=x(\sigma(T)), \end{array} \right .$$
(1.1)

where $$\mathbb{T}$$ is an arbitrary time scale, $$T>0$$ is fixed, 0, $$T\in\mathbb {T}$$, $$f\in C ( J\times[ 0,\infty ) , [ 0,\infty ) )$$, $$I_{k}\in C ( [ 0,\infty ) , [ 0,\infty ) )$$, $$p: [ 0,T ] _{\mathbb{T}}\rightarrow(0,\infty)$$ is right-dense continuous, $$t_{k}\in ( 0,T ) _{\mathbb{T}}$$, $$0< t_{1}<\cdots <t_{m}<T$$, and, for each $$k=1,2,\ldots,m$$, $$x(t_{k}^{+})=\lim_{h\rightarrow 0^{+}}x(t_{k}+h)$$ and $$x(t_{k}^{-})=\lim_{h\rightarrow0^{-}}x(t_{k}+h)$$ represent the right and left limits of $$x(t)$$ at $$t=t_{k}$$.

By using the Guo-Krasnoselskii fixed point theorem, Wang  considered the existence of one or two positive solutions to the problem (1.1).

In , by using the Schaefer fixed point theorem, Wang and Weng obtained the existence of at least one solution to the problem (1.1).

When $$I_{k}(x)\equiv0$$, $$k=1,2,\ldots,m$$, [25, 26] considered the existence of solutions to the problem (1.1) by means of the Schaefer fixed point theorem; when $$p(t)=0$$, the problem (1.1) reduces to the problem studied by [12, 19].

Motivated by the results mentioned above, in this paper, we shall obtain the existence of one and two solutions to the problem (1.1) by means of a fixed point theorem in cones. The results obtained in this paper improve the results in  intrinsically.

Throughout this work, we assume knowledge of time scales and the time-scale notation, first introduced by Hilger . For more on time scales, please see the texts by Bohner and Peterson [2, 3].

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

### Theorem 1.1

()

Let X be a Banach space and $$K\subset X$$ be a cone in X. Assume $$\Omega_{1}$$, $$\Omega_{2}$$ are bounded open subsets of X with $$0\in\Omega_{1}\subset\overline{\Omega}_{1}\subset \Omega_{2}$$ and $$\Phi:K\cap(\overline{\Omega}_{2}\backslash\Omega _{1})\to K$$ is a completely continuous operator. If

1. (i)

there exists $$u_{0}\in K\backslash\{0\}$$ such that $$u-\Phi u\neq \lambda u_{0}$$, $$u\in K\cap\partial\Omega_{2}$$, $$\lambda\geq0$$; $$\Phi u\neq\tau u$$, $$u\in K\cap\partial\Omega_{1}$$, $$\tau\geq1$$, or

2. (ii)

there exists $$u_{0}\in K\backslash\{0\}$$ such that $$u-\Phi u\neq \lambda u_{0}$$, $$u\in K\cap\partial\Omega_{1}$$, $$\lambda\geq0$$; $$\Phi u\neq\tau u$$, $$u\in K\cap\partial\Omega_{2}$$, $$\tau\geq1$$,

then Φ has at least one fixed point in $$K\cap(\overline{\Omega}_{2}\backslash\Omega_{1})$$.

## Preliminaries

Throughout the rest of this paper, we always assume that the points of impulse $$t_{k}$$ are right-dense for each $$k=1,2,\ldots,m$$.

We define

\begin{aligned} PC={}&\bigl\{ x\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}\rightarrow R:x_{k}\in C(J_{k},R), k=0,1,2,\ldots,m \mbox{ and there exist}\\ &{} x\bigl(t_{k}^{+}\bigr)\mbox{ and }x \bigl(t_{k}^{-}\bigr)\mbox{ with }x \bigl(t_{k}^{-}\bigr)=x(t_{k}), k=1,2,\ldots,m \bigr\} , \end{aligned}

where $$x_{k}$$ is the restriction of x to $$J_{k}= \left.(t_{k},t_{k+1}]\right._{\mathbb{T}}\subset \left.(0,\sigma(T)]\right._{\mathbb{T}}$$, $$k=1,2,\ldots,m$$, and $$J_{0}=[0,t_{1}]_{\mathbb{T}}$$, $$t_{m+1}=\sigma(T)$$.

Let

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

with the norm $$\Vert x\Vert =\sup_{ t\in[0,\sigma (T)]_{\mathbb{T}}}\vert x(t)\vert$$, then X is a Banach space.

### Lemma 2.1

Suppose $$M>0$$ and $$h:[0,T]_{\mathbb{T}}\rightarrow R$$ is rd-continuous, then x is a solution of

$$x(t)=\int_{0}^{\sigma(T)}G(t,s)h(s)\triangle s+\sum _{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k})\bigr),\quad t\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}},$$

where

$$G(t,s)=\left \{ \begin{array}{@{}l@{\quad}l} \frac{e_{M}(s,t)e_{M}(\sigma(T),0)}{e_{M}(\sigma(T),0)-1}, & 0\leq s\leq t\leq \sigma(T), \\ \frac{e_{M}(s,t)}{e_{M}(\sigma(T),0)-1}, & 0\leq t< s\leq\sigma(T), \end{array} \right .$$

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

$$\left \{ \begin{array}{@{}l} x^{\bigtriangleup}(t)+Mx(\sigma(t))=h(t), \quad t\in J, t\neq t_{k},\\ x(t_{k}^{+})-x(t_{k}^{-})=I_{k}(x(t_{k}^{-})),\quad k=1,2,\ldots,m,\\ x(0)=x(\sigma(T)). \end{array} \right .$$

### Proof

Since the proof is similar to that of , Lemma 3.1, we omit it here. □

### Lemma 2.2

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

$$\frac{1}{e_{M}(\sigma(T),0)-1}\leq G(t,s)\leq\frac{e_{M}(\sigma(T),0)}{e_{M}(\sigma(T),0)-1}\quad\textit{for all } t,s\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}.$$

### Proof

It is obvious, so we omit it here. □

### Remark 2.1

Let $$G(t,s)$$ be defined as in Lemma 2.1, then $$\int_{0}^{\sigma(T)}G(t,s)\triangle s=\frac{1}{M}$$.

Let $$m=\min_{t\in [ 0,T ] _{\mathbb{T}}}p(t)$$, $$M=\max_{t\in [ 0,T ] _{\mathbb{T}}}p(t)$$, then $$0< m\leq M<\infty$$. For $$u\in X$$, we consider the following problem:

$$\left \{ \begin{array}{l} x^{\bigtriangleup}(t)+Mx(\sigma(t))=Mu(\sigma(t))-p(t)u(\sigma (t))+f(t,u(\sigma(t))),\quad t\in J, t\neq t_{k}, \\ x(t_{k}^{+})-x(t_{k}^{-})=I_{k}(x(t_{k}^{-})),\quad k=1,2,\ldots,m, \\ x(0)=x(\sigma(T)). \end{array} \right .$$
(2.1)

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

$$x(t)=\int_{0}^{\sigma(T)}G(t,s)h_{u}(s) \triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k})\bigr),\quad t\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}},$$

where $$h_{u}(s)=Mu(\sigma(s))-p(t)u(\sigma(t))+f(s,u(\sigma(s)))$$, $$s\in [0,T]_{\mathbb{T}}$$.

We define the operator $$\Phi:X\rightarrow X$$ by

$$\Phi(u) (t)=\int_{0}^{\sigma(T)}G(t,s)h_{u}(s) \triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(u(t_{k})\bigr),\quad t\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}.$$

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

### Lemma 2.3

$$\Phi:X\rightarrow X$$ is completely continuous.

### Proof

Since the proof is similar to that of , Lemma 3.3, we omit it here. □

Let

$$K= \bigl\{ u\in X:u(t)\geq\delta \Vert u\Vert , t\in \bigl[0,\sigma (T) \bigr]_{\mathbb{T}} \bigr\} ,$$

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

From Lemma 2.2, it is easy to obtain the following result.

Φ maps K into K.

## Main results

For convenience, we denote

\begin{aligned}& f^{0}=\lim_{ u\rightarrow0^{+}}\sup\max_{t\in [ 0,T ] _{\mathbb{T}}} \frac{f(t,u)}{u},\qquad f^{\infty}=\lim_{ u\rightarrow \infty }\sup\max _{t\in [ 0,T ] _{\mathbb{T}}}\frac{f(t,u)}{u}, \\& f_{0}=\lim_{ u\rightarrow0^{+}}\inf\min_{t\in [ 0,T ] _{\mathbb{T}}} \frac{f(t,u)}{u},\qquad f_{\infty}=\lim_{ u\rightarrow \infty }\inf\min _{t\in [ 0,T ] _{\mathbb{T}}}\frac{f(t,u)}{u}, \end{aligned}

and

$$I_{0}=\lim_{ u\rightarrow0^{+}}\frac{I_{k}(u)}{u},\qquad I_{\infty}=\lim_{ u\rightarrow\infty}\frac{I_{k}(u)}{u}.$$

Now we state our main results.

### Theorem 3.1

Suppose that

(H1):

$$f_{0}>M$$, $$f^{\infty}< m$$, $$I_{\infty}=0$$ for any k; or

(H2):

$$f_{\infty}>M$$, $$f^{0}< m$$, $$I_{0}=0$$ for any k.

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

### Proof

Firstly, we assume (H1) holds. Then there exist $$\varepsilon>0$$ and $$\beta>\alpha>0$$ such that

\begin{aligned}& f(t,u)\geq(M+\varepsilon)u,\quad t\in [ 0,T ] _{\mathbb {T}}, u\in ( 0,\alpha ], \end{aligned}
(3.1)
\begin{aligned}& I_{k}(u)\leq\frac{[e_{M}(\sigma(T),0)-1]\varepsilon}{2Mme_{M}(\sigma (T),0)}u,\quad u\in [ \beta, \infty )\mbox{ for any }k, \end{aligned}
(3.2)

and

$$f(t,u)\leq(m-\varepsilon)u,\quad t\in [ 0,T ] _{\mathbb {T}}, u\in [ \beta,\infty ) .$$
(3.3)

Let $$\Omega_{1}= \{ u\in X:\Vert u\Vert < r_{1} \}$$, where $$r_{1}=\alpha$$. Choose $$u_{0}=1$$, then $$u_{0}\in K\backslash\{0\}$$. We assert that

$$u-\Phi u\neq\lambda u_{0},\quad u\in K\cap\partial \Omega_{1},\lambda\geq0.$$
(3.4)

Suppose on the contrary that there exist $$\overline{u}\in K\cap\partial \Omega_{1}$$ and $$\overline{\lambda}\geq0$$ such that

$$\overline{u}-\Phi\overline{u}=\overline{\lambda}u_{0}.$$

Let $$\zeta=\min_{t\in[0,\sigma(T)]_{\mathbb{T}}}\overline{u}(t)$$, then $$\zeta\geq\delta \Vert \overline{u}\Vert =\delta r_{2}=\beta$$, and we have from (3.1)

\begin{aligned} \overline{u}(t) =&\Phi(\overline{u}) (t)+\overline{\lambda} \\ =&\int_{0}^{\sigma(T)}G(t,s)h_{\overline{u}}(s)\triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(\overline{u}(t_{k})\bigr)+\overline{\lambda} \\ \geq&\int_{0}^{\sigma(T)}G(t,s)\bigl[M-p(t)+M+\varepsilon \bigr]u\bigl(\sigma (t)\bigr)\triangle s+\overline{\lambda} \\ \geq&\frac{(M+\varepsilon)}{M}\zeta+\overline{\lambda},\quad t\in \bigl[0,\sigma(T) \bigr]_{\mathbb{T}}. \end{aligned}

Therefore,

$$\zeta=\min_{t\in[0,\sigma(T)]_{\mathbb{T}}}\overline{u}(t)\geq\frac{(M+\varepsilon)}{M}\zeta+ \overline{\lambda}>\zeta,$$

On the other hand, let $$\Omega_{2}= \{ u\in X:\Vert u\Vert < r_{2} \}$$, where $$r_{2}=\frac{\beta}{\delta}$$.

Then $$u\in K\cap\partial\Omega_{2}$$, $$0<\delta\beta=\delta \Vert u\Vert \leq u(t)\leq\beta$$, and in view of (3.2) and (3.3) we have

\begin{aligned} \Phi(u) (t) =&\int_{0}^{\sigma(T)}G(t,s)h_{u}(s) \triangle s+\sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(u(t_{k})\bigr) \\ \leq&\int_{0}^{\sigma(T)}G(t,s)\bigl[M-p(t)+m-\varepsilon \bigr]u\bigl(\sigma (s)\bigr)\triangle s\\ &{}+\sum_{k=1}^{m}G(t,t_{k}) \frac{[e_{M}(\sigma (T),0)-1]\varepsilon}{2Mme_{M}(\sigma(T),0)}u(t_{k}) \\ \leq&\frac{(M-\varepsilon)}{M}\Vert u\Vert +\frac{e_{M}(\sigma (T),0)}{e_{M}(\sigma(T),0)-1}\sum _{k=1}^{m}\frac{[e_{M}(\sigma(T),0)-1]\varepsilon }{2Mme_{M}(\sigma(T),0)}\Vert u\Vert \\ =&\frac{(M-\frac{\varepsilon}{2})}{M}\Vert u\Vert \\ < &\Vert u\Vert , \quad t\in\bigl[0,\sigma(T)\bigr]_{\mathbb{T}}, \end{aligned}

which yields $$\Vert \Phi(u)\Vert <\Vert u\Vert$$.

Therefore

$$\Phi u\neq\tau u,\quad u\in K\cap\partial\Omega_{1},\tau \geq1.$$
(3.5)

It follows from (3.4), (3.5), and Theorem 1.1 that Φ has a fixed point $$u^{*}\in K\cap(\overline{\Omega}_{2}\backslash\Omega_{1})$$, and $$u^{*}$$ is the desired positive solution of the problem (1.1).

Next, suppose that (H2) holds. Then we can choose $$\varepsilon ^{\prime }>0$$ and $$\beta^{\prime}>\alpha^{\prime}>0$$ such that

\begin{aligned}& f(t,u)\geq\bigl(M+\varepsilon^{\prime}\bigr)u,\quad t\in [ 0,T ] _{\mathbb{T}}, u\in \bigl[ \beta^{\prime},\infty \bigr) , \end{aligned}
(3.6)
\begin{aligned}& I_{k}(u)\leq\frac{[e_{M}(\sigma(T),0)-1]\varepsilon^{\prime}}{2Mme_{M}(\sigma(T),0)}u,\quad u\in \bigl( 0, \alpha^{\prime} \bigr] \mbox{ for any }k, \end{aligned}
(3.7)

and

$$f(t,u)\leq\bigl(m-\varepsilon^{\prime}\bigr)u,\quad t\in [ 0,T ] _{\mathbb{T}}, u\in \bigl( 0,\alpha^{\prime} \bigr] .$$
(3.8)

Let $$\Omega_{3}= \{ u\in X:\Vert u\Vert < r_{3} \}$$, where $$r_{3}=\alpha^{\prime}$$. Then for any $$u\in K\cap\partial\Omega_{3}$$, $$0<\delta \Vert u\Vert \leq u(t)\leq \Vert u\Vert =\alpha ^{\prime }$$.

It is similar to the proof of (3.5), by (3.7) and (3.8) we have

$$\Phi u\neq\tau u,\quad u\in K\cap\partial\Omega_{4},\tau \geq1.$$
(3.9)

Let $$\Omega_{4}= \{ u\in X:\Vert u\Vert < r_{4} \}$$, where $$r_{4}=\frac{\beta^{\prime}}{\delta}$$. Then for any $$u\in K\cap\partial \Omega _{4}$$, $$u(t)\geq\delta \Vert u\Vert =\delta r_{4}=\beta^{\prime}$$, by (3.6) , it is easy to obtain

$$u-\Phi u\neq\lambda u_{0},\quad u\in K\cap\partial \Omega_{3},\lambda\geq0.$$
(3.10)

It follows from (3.9), (3.10), and Theorem 1.1 that Φ has a fixed point $$u^{*}\in K\cap(\overline{\Omega}_{4}\backslash\Omega_{3})$$, and $$u^{*}$$ is the desired positive solution of the problem (1.1). □

In particular, we have the following results, which are main results of .

### Corollary 3.1

Suppose that

(H1):

$$f_{0}=\infty$$, $$f^{\infty}=0$$, $$I_{\infty}=0$$ for any k; or

(H2):

$$f_{\infty}=\infty$$, $$f^{0}=0$$, $$I_{0}=0$$ for any k.

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

### Theorem 3.2

Suppose that

(H3):

$$f^{0}< m$$, $$f^{\infty}< m$$, $$I_{0}=0$$, $$I_{\infty}=0$$;

(H4):

there exists $$\rho >0$$ such that

$$\min\bigl\{ f(t,u)-p(t)u\mid t\in [ 0,T ] _{\mathbb{T}}, \delta \rho \leq u\leq\rho\bigr\} >0.$$
(3.11)

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

### Proof

By (H3), from the proof of Theorem 3.1, we see that there exist $$\beta^{\prime\prime}>\rho>\alpha^{\prime\prime}>0$$ such that

\begin{aligned}& \Phi u\neq\tau u,\quad u\in K\cap\partial\Omega_{5},\tau \geq1, \end{aligned}
(3.12)
\begin{aligned}& \Phi u\neq\tau u,\quad u\in K\cap\partial\Omega_{6},\tau \geq1, \end{aligned}
(3.13)

where $$\Omega_{5}= \{ u\in X:\Vert u\Vert < r_{5} \}$$, $$\Omega _{6}= \{ u\in X:\Vert u\Vert < r_{6} \}$$, $$r_{5}=\alpha^{\prime \prime}$$, $$r_{6}=\frac{\beta^{\prime\prime}}{\delta}$$.

By (3.11) of (H4), we can choose $$\varepsilon>0$$ such that

$$f(t,u)-p(t)u\geq\varepsilon u, \quad t\in [ 0,T ] _{\mathbb {T}}, \delta\rho\leq u\leq\rho.$$
(3.14)

Let $$\Omega_{7}= \{ u\in X:\Vert u\Vert <\rho \}$$, for any $$u\in K\cap\partial\Omega_{7}$$, $$\delta\rho=\delta \Vert u\Vert \leq u(t)\leq \Vert u\Vert =\rho$$, from (3.14), it is similar to the proof of (3.4), and we have

$$u-\Phi u\neq\lambda u_{0},\quad u\in K\cap\partial \Omega_{7},\lambda\geq0.$$
(3.15)

By Theorem 1.1, from (3.12), (3.13), and (3.15) we conclude that Φ has two fixed points $$u^{**}\in K\cap(\overline{\Omega}_{6}\backslash \Omega _{7})$$ and $$u^{***}\in K\cap(\overline{\Omega}_{7}\backslash\Omega_{5})$$, and $$u^{**}$$ and $$u^{***}$$ are two positive solutions of the problem (1.1). □

Similar to Theorem 3.2, we have the following.

### Theorem 3.3

Suppose that

(H5):

$$f_{0}>M$$, $$f_{\infty}>M$$;

(H6):

there exists $$\rho>0$$ such that

\begin{aligned}& \max\bigl\{ f(t,u)-p(t)u\mid t\in [ 0,T ] _{\mathbb{T}}, \delta \rho \leq u\leq\rho\bigr\} < 0; \\& I_{k}(u)\leq\frac{[e_{M}(\sigma(T),0)-1]}{Mme_{M}(\sigma (T),0)}u, \quad\delta\rho\leq u\leq\rho\textit{ for any }k. \end{aligned}

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

### Remark 3.1

If (H3) in Theorem 3.2 is replaced by $$f^{0}=0$$, $$f^{\infty}=0$$, or if (H5) in Theorem 3.3 is replaced by $$f_{0}=\infty$$, $$f_{\infty}=\infty$$, then the results of Theorem 3.2 and Theorem 3.3 are also hold.

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## Acknowledgements

The author express her gratitude to the anonymous referee for his/her valuable suggestions. The research was supported by Natural Science Foundation of Gansu Province of China (Grant No. 1310RJYA080).

## Author information

Correspondence to Wen Guan. 