Lipschitz stability of differential equations with noninstantaneous impulses
 Snezhana Hristova^{1}Email authorView ORCID ID profile and
 Radoslava Terzieva^{1}
https://doi.org/10.1186/s1366201610456
© The Author(s) 2016
Received: 17 October 2016
Accepted: 28 November 2016
Published: 9 December 2016
Abstract
Nonlinear differential equations with noninstantaneous impulses are studied. The impulses start abruptly at some points and their actions continue on given finite intervals. We pursue the study of Lipschitz stability using Lyapunov functions. Some sufficient conditions for Lipschitz stability, uniform Lipschitz stability, and uniform global Lipschitz stability are obtained. Examples are given to illustrate the results.
Keywords
noninstantaneous impulses differential equations Lipschitz stabilityMSC
34A37 34D201 Introduction
The problems of stability of solutions of differential equations via Lyapunov functions have been successfully investigated in the past. One type of stability, very useful in real world problems, is the socalled Lipschitz stability. Dannan and Elaydi [1] introduced the notion of Lipschitz stability for ordinary differential equations. As is mentioned in [1] this type of stability is important only for nonlinear problems, since it coincides with uniform stability in linear systems.

The duration of these changes is relatively short compared to the overall duration of the whole process and the changes turn out to be irrelevant to the development of the studied process. The mathematical models in such cases can be adequately created with the help of impulsive equations (see, for example, [2–5], the monographs [6, 7] and the references therein).

The duration of these changes is not negligible short, i.e. these changes start impulsively at arbitrary fixed points and remain active on finite initially time intervals. The model of this situation is the noninstantaneous impulsive differential equation. Hernandez and O’Regan [8] introduced this new class of differential equations where the impulses are not instantaneous and they investigated the existence of mild and classical solutions. We refer the reader for some recent results such as existence to [9, 10], to stability [11–16], to periodic boundary value problems [17, 18].
Some examples of such processes can be found in physics, biology, population dynamics, ecology, pharmacokinetics, and others.
In this paper Lipschitz stability of solutions of nonlinear noninstantaneous impulsive differential equations is defined and studied. Several sufficient conditions for Lipschitz stability, uniform Lipschitz stability, and global uniform Lipschitz stability are obtained. Some examples illustrating the results are given. Note that noninstantaneous impulsive differential equations are natural generalizations of impulsive differential equations and some of the obtained sufficient conditions are a generalization of some results in [19]. Also, Lipschitz stability of impulsive functionaldifferential equations is studied in [20].
2 Preliminaries
In this paper we assume two increasing sequences of points \(\{t_{i}\} _{i=1}^{\infty}\) and \(\{s_{i}\}_{i=0}^{\infty}\) are given such that \(0 < s_{0}<t_{i}<s_{i}<t_{i+1}<s_{i+1}\), \(i=1,2, \ldots\) , and \(\lim_{k\rightarrow \infty}t_{k}=\infty\).
Let \(t_{0} \in\bigcup_{k=0}^{\infty}[s_{k},t_{k+1})\) be a given arbitrary point. Without loss of generality we will assume that \(t_{0} \in[0,s_{0})\).
Remark 1
The functions \(\Psi_{k}\) are called impulsive functions and the intervals \((s_{k},t_{k+1}]\), \(k=0, 1,2,\ldots\) are called intervals of noninstantaneous impulses.
Remark 2
In the partial case \(s_{k}=t_{k+1}\), \(k=0,1,2,\ldots\) each interval of noninstantaneous impulses is reduced to a point, and the problem (1) is reduced to an IVP for an impulsive differential equation with points of jump \(t_{k}\) and impulsive condition \(x(t_{k}+0)=I_{k}(x(t_{k}0))\equiv\Psi_{k}(t_{k},x(t_{k}0),x(t_{k}0))\).

for any \(k=0,1,2,\ldots\) the function \(X_{k}(t)\), \(t\in[t_{k},s_{k}]\) is a solution of the initial value problem for ODE \(x'=f(t,x)\), \(x(t_{k})=x(t_{k};t_{0},x_{0})\), respectively;

on any interval \((s_{k},t_{k+1}]\), \(k=0,1,2,\ldots\) the solution \(x(t;t_{0},x_{0})\) satisfies the algebraic equation \(x(t;t_{0},x_{0})=\Psi _{k}(t,x(t;t_{0},x_{0}), X_{k}(t_{1}0))\).
Remark 3
According to the above description any solution of (1) might have a discontinuity at any point \(s_{k}\), \(k =0, 1,2, \ldots\) .
Now we will illustrate the influence of the impulsive condition on the behavior of the solution.
Example 1
Therefore, if any impulsive function \(\Psi_{k}(t,x,y)\) do not depend on y the solution does not depend on the initial value \(x_{0}\) for \(t>s_{k}\) (see Cases 2 and 3).
Remark 4
Note in some papers (see, for example [14]) the functions of noninstantaneous impulses are given in the form \(g_{k}(t,x(t))\), i.e. they do not depend on the value of the solution before the jump \(x(s_{k}0)\). Then the solution will depend on the initial value only on the interval \([t_{0},s_{0}]\). Then the meaning of the stability as well dependence of the solution on the initial value is lost.
 (H1)
The function \(f \in C(\bigcup_{k=0}^{\infty}[t_{k}, s_{k}]\times\mathbb{R}^{n}, \mathbb{R}^{n})\) and \(f(t,0)\equiv0\).
 (H2)
For any \(k=0,1,2,\ldots\) and any fixed \(t\in [s_{k},t_{k+1}]\) and \(y\in\mathbb{R}^{n}\) the algebraic equation \(x=\Psi _{k}(t,x,y)\) has unique solution \(x=\phi_{k}(t,y)\) with \(\phi_{k} \in C([s_{k},t_{k+1}] \times\mathbb{R}^{n}, \mathbb{R}^{n})\) and \(\phi_{k}(t,0)\equiv0\).
Remark 5
The function \(a(u)=K_{1}u\), \(K_{1}\in(0,1]\) is from the class \(K(\mathbb{R}_{+})\) with \(q(u)\equiv u\). The function \(a(u)=K_{1} u^{2}\), \(K_{1}>0\) is from the class \(M([0,1])\).
We will use the class Λ of Lyapunovlike functions, defined and used for impulsive differential equations in [7].
Definition 1

the function \(V(t,x)\) is a continuous on \(J/\{t_{k} \in J\}\times\Delta\) and it is locally Lipschitz with respect to its second argument;

for each \(s_{k} \in J\) and \(x \in\Delta\) there exist finite limits$$ V(s_{k},x)=V(s_{k}0,x)= \lim_{t \uparrow s_{k}}V(t,x) \quad \text{and}\quad V(s_{k}+0,x)= \lim_{t \downarrow s_{k}}V(t,x). $$
3 Main results
We define Lipschitz stability [1] of systems of differential equations with noninstantaneous impulses.
Definition 2
Lipschitz stability

Lipschitz stable if there exists \(M \geq1\) and for every \(t_{0} \geq0\) there exists \(\delta=\delta(t_{0})>0\) such that, for any \(x_{0} \in\mathbb{R}^{n}\), the inequality \(x_{0}<\delta\) implies \(x(t;t_{0},x_{0}) \leq Mx_{0}\) for \(t \geq t_{0}\);

uniformly Lipschitz stable if there exist \(M \geq1\) and \(\delta>0\) such that for any \(t_{0}\geq0\) and \(x_{0} \in\mathbb{R}^{n}\) the inequality \(x_{0}<\delta\) implies \(x(t;t_{0},x_{0}) \leq Mx_{0}\) for \(t \geq t_{0}\);

globally uniformly Lipschitz stable if there exists \(M \geq1\) such that for any \(t_{0} \geq0\) and \(x_{0} \in \mathbb{R}^{n}\) the inequality \(x_{0}<\infty\) implies \(x(t;t_{0},x_{0}) \leq Mx_{0}\) for \(t \geq t_{0}\).
Example 2
The above example shows the presence of noninstantaneous impulses and the type of impulsive functions that have influence on the behavior of the solution.
 (H3)
The function \(g(t,u) \in C(\bigcup_{k=0}^{\infty }[t_{k},s_{k}] \times\mathbb{R}_{+}, \mathbb{R})\), \(g(t,0)=0\), and for any \(k=0,1,2,\ldots\) the functions \(\psi_{k}:[s_{k},t_{k+1}] \times\mathbb{R}_{+} \rightarrow\mathbb{R}_{+}\) are nondecreasing with respect to their second argument and \(\psi_{k}(t,0)=0\).
In the main study we will use the following result.
Proposition 1
Theorem 3.1.1 [21]
Let the function \(V \in C([t_{0},T]\times\mathbb{R}^{n},\mathbb{R}_{+})\) and \(V(t,x)\) be locally Lipschitz in x and \(D_{+}V(t,x)\leq g(t,V(t,x))\) for \((t,x) \in[t_{0},T]\times\mathbb{R}^{n}\), where \(g\in C([t_{0},T] \times\mathbb{R}_{+},\mathbb{R})\). Let \(\tilde {r}(t)=r(t;t_{0},u_{0})\) be the maximal solution of the scalar differential equation \(u'=g(t,u)\) with initial condition \(u(t_{0})=u_{0}\geq0\), existing on \([t_{0},T]\). If \(x(t)=x(t;t_{0},x_{0})\) is any solution of the IVP for ODE \(x'=f(t,x)\), \(x(t_{0})=x_{0}\) existing on \([t_{0},T]\) such that \(V(t_{0},x_{0})\leq u_{0}\), then the inequality \(V(t,x(t)) \leq \tilde{r}(t)\) for \(t \in[t_{0},T]\) holds.
Lemma 1
 1.
Conditions (H1), (H2), and (H3) are satisfied.
 2.
The function \(x^{*}(t)=x(t;t_{0},x_{0}) \in NPC^{1}([t_{0},T], \Delta)\) is a solution of (7), where \(T \geq t_{0}\) is a given constants, \(\Delta\subset\mathbb{R}^{n}\).
 3.The function \(V \in\Lambda([t_{0},T], \Delta)\) is such that:
 (i)
the inequality \(D_{+}V(t,x^{*}(t))\leq g(t,V(t,x^{*}(t)))\) for \(t \in[t_{0},T] \cap(\bigcup_{k=0}^{\infty}(t_{k},s_{k})) \) holds;
 (ii)for all \(k=0, 1, 2,3, \ldots\) the inequalityholds.$$V\bigl(t, \phi_{k}\bigl(t,x^{*}(s_{k}0)\bigr)\bigr)\leq \psi_{k}\bigl(t,V\bigl(s_{k}0,x^{*}(s_{k}0)\bigr) \bigr)\quad \textit{for } t \in[t_{0},T]\cap(s_{k},t_{k+1}] $$
 (i)
If \(V(t_{0},x_{0})\leq u_{0}\), then the inequality \(V(t,x^{*}(t))\leq r(t)\) for \(t \in[t_{0},T]\) holds, where \(r(t)=r(t;t_{0},u_{0})\) is the maximal solution of (12) with \(u_{0}\geq0\).
Proof
We use induction to prove Lemma 1.
Apply Proposition 1 to the interval \([t_{1},s_{1}] \cap[t_{0},T]\) with the initial value \(u_{0}=r(t_{1};t_{0},u_{0})\) and \(V(t_{1},\overline{x}(t_{1}))\leq r(t_{1};t_{0},u_{0})\) and we obtain \(V(t,\overline{x}(t))\leq r(t)\), \(t\in[t_{1},s_{1}] \cap[t_{0},T]\). Therefore \(V(t,x^{*}(t))\leq r(t;t_{0},u_{0})\), \(t\in(t_{1},s_{1}] \cap[t_{0},T]\).
Continue this process and an induction argument proves the claim of Lemma 1 is true for \(t \in[t_{0},T]\). □
Theorem 1
 1.
Conditions (H1)(H3) are fulfilled.
 2.There exists a function \(V(t,x) \in\Lambda(\mathbb{R}^{+}, \mathbb{R}^{n})\) with Lipschitz constant L in \(S_{\rho}\), \(V(t,0)=0\), and:
 (i)the inequalityholds, where \(b\in K (\mathbb{R}_{+})\);$$b\bigl(\Vert x\Vert \bigr)\leq V(t,x), \quad x \in\mathbb{R}^{n}, t \in\mathbb{R}_{+}, $$
 (ii)
the inequality \(D^{+}V(t,x) \leq g(t,V(t,x))\), \(t \in\bigcup_{k=0}^{\infty}(t_{k},s_{k}) \), \(x \in\mathbb{R}^{n}\), holds;
 (iii)for any \(k=1,2, \ldots\) the inequalityholds.$$V\bigl(t,\phi_{k}(t,y)\bigr)\leq\psi_{k} \bigl(t,V(s_{k}0,y)\bigr),\quad t \in(s_{k},t_{k+1}], y \in \mathbb{R}^{n}, $$
 (i)
 3.
The zero solution of (12) is Lipschitz stable.
Then the zero solution of (7) is Lipschitz stable.
Proof
Since \(V(t_{0},0)=0\) there exists a \(\delta_{2}=\delta_{2}(t_{0},\delta_{1})>0\) such that \(V(t_{0},x)<\delta_{1}\) for \(\Vert x\Vert <\delta_{2}\). The function \(V(t,x)\) is Lipschitz on \(S_{\rho}\) then \(\x\<\rho\) implies \(V(t,x)=V(t,x)V(t,0)\leq L\x\\).
Let \(\delta=\min \{ \delta_{1},\delta_{2},\rho \} \) and choose \(M_{1}\geq1\) such that \(M_{1}> M L\) and let \(M_{2}=q(M_{1})\). Note since \(M_{1}\geq1\) we have \(M_{2}\geq1\) and δ depends on \(t_{0}\) and M, therefore on \(M_{2}\).
Now let the initial value be such that \(\Vert x_{0}\Vert <\delta\). Consider a solution \(x(t)=x(t;t_{0},x_{0})\) of system (7). Let \(u_{0}^{\ast}=V(t_{0},x_{0})\). Then from the choice of \(x_{0}\) it follows that \(u_{0}^{\ast }=V(t_{0},x_{0})<\delta_{1}\) for \(\Vert x_{0}\Vert <\delta\). Therefore, the function \(u^{\ast}(t)\) satisfies (14) for \(t\geq t_{0}\) with \(u_{0}=u^{*}_{0}\), where \(u^{\ast }(t)=u(t;t_{0},u_{0}^{\ast})\) is a solution of (12).
Corollary 1
Let the conditions of Theorem 1 be satisfied with \(b(u)=K_{1} u\), \(K_{1}>0\).
Then the zero solution of (7) is Lipschitz stable.
Proof
The proof is similar to the one of Theorem 1 with \(M_{1}\geq1: M_{1}> M \frac{L}{K_{1}}\) and \(M_{2}=M_{1}\). □
Theorem 2
 1.
Conditions (H1)(H3) are fulfilled.
 2.There exists a function \(V(t,x) \in\Lambda(\mathbb{R}_{+}, \mathbb{R}^{n})\) and:
 (i)the inequalitiesholds, where \(b\in K ([0,\rho])\), \(a\in M ([0,\rho])\), \(\rho>0\);$$b\bigl(\Vert x\Vert \bigr)\leq V(t,x)\leq a\bigl(\Vert x\Vert \bigr), \quad x \in S_{\rho}, t \in\mathbb{R}_{+} $$
 (ii)
the inequality \(D^{+}V(t,x) \leq g(t,V(t,x))\), \(t \in\bigcup_{k=0}^{\infty}(t_{k},s_{k}) \), \(x \in S_{\rho}\) holds;
 (iii)for any \(k=0,1,2, \ldots\) the inequalityholds.$$V\bigl(t,\phi_{k}(t,y)\bigr)\leq\psi_{k} \bigl(t,V(s_{k}0,y)\bigr),\quad t \in(s_{k},t_{k+1}], y \in S_{\rho}, $$
 (i)
 3.
The zero solution of (12) is uniformly Lipschitz stable (uniformly globally Lipschitz stable).
Then the zero solution of (7) is uniformly Lipschitz stable (uniformly globally Lipschitz stable).
Proof
From the inclusions \(b\in K ([0,\rho])\) and \(a\in {M}([0,\rho])\) there exist a function \(q_{b}(u)\) and a positive constant \(K_{a}\). Choose \(M_{1}\geq1\) such that \(M_{1}>q_{b}(M) K_{a}\) and \(\delta_{2}\leq \frac{\rho}{M_{1}}\). Therefore, \(\delta_{2}\leq\rho\).
Let \(\delta=\min \{ \delta_{1},\delta_{2},\frac{\delta _{1}}{K_{a}} \} \). Choose the initial value \(x_{0}\): \(\Vert x_{0}\Vert <\delta\). Therefore, \(\x_{0}\<\delta\leq\delta _{2}\leq\rho\), i.e. \(x_{0}\in S_{\rho}\). Consider the solution \(x(t)=x(t;t_{0},x_{0})\) of system (7) for the chosen initial data. Let \(u_{0}^{\ast}=V(t_{0},x_{0})\). From the choice of \(x_{0}\) and the properties of the function \(a(u)\) applying condition 2(i) we get \(u_{0}^{\ast }=V(t_{0},x_{0})\leq a(\x_{0}\)\leq K_{a}\x_{0}\< K_{a} \delta\leq\delta _{1}\). Therefore, the function \(u^{\ast}(t)\) satisfies (17) for \(t\geq t_{0}\) with \(u_{0}=u^{*}_{0}\), where \(u^{\ast }(t)=u(t;t_{0},u_{0}^{\ast})\) is a solution of (12).
Assume (18) is not true. Therefore, there exists a point \(T>t_{0}\) such that \(\x(t)\\leq M_{1}\x_{0}\\) for \(t\in[t_{0},T]\), \(\x(T)\= M_{1}\x_{0}\\) and \(\x(t)\> M_{1}\x_{0}\\) for \(t\in(T,T+\epsilon]\), where \(\epsilon>0\) is a small enough number. Then for \(t\in[t_{0},T]\) the inequalities \(\x(t)\\leq M_{1}\x_{0}\< M_{1}\delta\leq M_{1}\delta_{2} \leq\rho\) hold, i.e. \(x(t)\in S_{\rho}\) for \(t\in[t_{0},T]\).
The proof of globally uniformly Lipschitz stability is analogous and we omit it. □
Corollary 2
Let (H1)(H3) and condition 2 of Theorem 2 be satisfied with \(g(t,x)\equiv0\) and \(\psi_{k}(t,x)\equiv x\).
Then the zero solution of (7) is uniformly Lipschitz stable.
Corollary 3
Then the zero solution of (7) is uniformly Lipschitz stable.
Proof
Consider the quadratic Lyapunov function \(V(t,x)=x^{2}\) for which \(D^{+}V(t,x) =2xf(t,x)\) and condition 2 of Theorem 2 is satisfied with \(K_{1}\leq1\), \(K_{2}\geq1\), \(g(t,x)=0\), and \(\psi_{k}(t,x)\equiv x\). □
Theorem 3
 2.
 (i)
the inequalities \(\lambda_{1}(t)\x\^{2}\leq V(t,x)\leq\lambda _{2}(t)\x\^{2}\), \(x \in S_{\rho}\), \(t \in\mathbb{R}^{+} \) holds, where \(\lambda _{1},\lambda_{2}\in C(\mathbb{R}_{+},(0,\infty))\) and there exist positive constants \(A_{1}\), \(A_{2}\): \(A_{1}< A_{2}\) such that \(\lambda_{1}(t)\geq A_{1}\), \(\lambda _{2}(t)\leq A_{2}\) for \(t\geq0\), and \(\rho>0\).
 (i)
If the zero solution of (12) is uniformly Lipschitz stable (uniformly globally Lipschitz stable) then the zero solution of (7) is uniformly Lipschitz stable(uniformly globally Lipschitz stable).
Proof
The proof is similar to the one of Theorem 2 where \(M_{1}=\sqrt {M\frac{A_{2}}{A_{1}}}\). □
4 Applications
Define the function \(V(t,x)=x^{2}\).
Then condition 2(i) of Theorem 3 is satisfied for \(\lambda_{1}(t)=0.5\), \(\lambda_{2}(t)=1.5\).
For any \(x: x\leq\rho\), \(\rho=\frac{b\sqrt{b^{2}+4 ac}}{2c}>0\) we have \(D^{+}V(t,x)=2x^{2} (cx^{2}bxa ) \leq0\), \(t \in\bigcup_{k=0}^{\infty}(t_{k},s_{k}) \), \(x \in S_{\rho}\). Therefore, condition 2(ii) is satisfied with \(g(t,x)\equiv0\).
The condition 2(ii) is satisfied for \(\psi_{k}(t,x)\equiv C_{k}x\).
Declarations
Acknowledgements
The research is partially supported by the Fund NPD, Plovdiv University.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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