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

Advances in Difference Equations20152015:83

https://doi.org/10.1186/s13662-015-0428-4

  • Received: 16 October 2014
  • Accepted: 26 February 2015
  • Published:

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.

Keywords

  • time scale
  • periodic boundary value problem
  • fixed point
  • impulsive dynamic equation

MSC

  • 39A10
  • 34B15

1 Introduction

The theory of dynamic equations on time scales has been a new important mathematical branch [13] since it was initiated by Hilger [4]. At the same time, the boundary value problems of impulsive dynamic equations on time scales have received considerable attention [521] since the theory of impulsive differential equations is a lot richer than the corresponding theory of differential equations without impulse effects [2224].

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 [18] considered the existence of one or two positive solutions to the problem (1.1).

In [20], 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 [18] intrinsically.

Throughout this work, we assume knowledge of time scales and the time-scale notation, first introduced by Hilger [4]. 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 [27].

Theorem 1.1

([27])

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 \(K\cap(\overline{\Omega}_{2}\backslash\Omega_{1})\).

2 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 [18], 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 [18], 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.

Lemma 2.4

Φ maps K into K.

3 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,
$$\zeta=\min_{t\in[0,\sigma(T)]_{\mathbb{T}}}\overline{u}(t)\geq\frac{(M+\varepsilon)}{M}\zeta+ \overline{\lambda}>\zeta, $$
which is a contradiction.

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 [18].

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.

Declarations

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).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu, 730050, People’s Republic of China

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© Guan; licensee Springer. 2015

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