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

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.

1 Introduction

The theory of dynamic equations on time scales has been a new important mathematical branch [1–3] 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 [5–21] since the theory of impulsive differential equations is a lot richer than the corresponding theory of differential equations without impulse effects [22–24].

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

• (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

• (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.