Hybrid differential equations with maxima via Picard operators theory

Abstract

The aim of this paper is to discuss some basic problems (existence and uniqueness, data dependence) of the Cauchy problem for a hybrid differential equation with maxima using weakly Picard operators technique.

Authors

D. Otrocol
(Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy,

Keywords

Differential equations with maxima, Cauchy problem, data dependence, weakly Picard operators.

Cite this paper as:

D. Otrocol, Hybrid differential equations with maxima via Picard operators theory, Stud. Univ. Babes-Bolyai Math., 61 (2016) no. 4, pp. 421-428.

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About this paper

Journal

Studia Universitatis Babes-Bolyai Mathematica

Publisher Name

Univ. Babes-Bolyai, Cluj-Napoca, Romania

DOI
Print ISSN

0252-1938

Online ISSN

2065-961X

MR

MR3583207

ZBL

1397.34108

Google Scholar

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[14] Rus, I.A., Generalized contractions and applications, Cluj University Press, 2001.

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Hybrid differential equations with maxima via Picard operators theory

Diana Otrocol ‘’T. Popoviciu” Institute of Numerical Analysis,
Romanian Academy
P.O.Box. 68-1, 400110, Cluj-Napoca,
Romania
dotrocol@ictp.acad.ro
Abstract.

The aim of this paper is to discuss some basic problems (existence and uniqueness, data dependence) of the Cauchy problem for a hybrid differential equation with maxima using weakly Picard operators technique.

Key words and phrases:
differential equations with maxima, Cauchy problem, data dependence, weakly Picard operators
1991 Mathematics Subject Classification:
47H10, 34K05

1. Introduction

Recently, the interest in differential equations with “maxima” has increased exponentially. Such equations model real world problems whose present state depends significantly on its maximum value on a past time interval. For example, many problems in the control theory correspond to the maximal deviation of the regulated quantity. Some qualitative properties of the solutions of ordinary differential equations with “maxima” can be found in [1, 2, 5], [16, 17] and the references therein.

The main goal of the presented paper is to study a hybrid differential equation with maxima, using the theory of weakly Picard operators. The theory of Picard operators was introduced by I. A. Rus (see [12], [14] and their references) to study problems related to fixed point theory. This abstract approach is used by many mathematicians and it seemed to be a very useful and powerful method in the study of integral equations and inequalities, ordinary and partial differential equations (existence, uniqueness, differentiability of the solutions), etc.

In this paper we consider the following hybrid differential equation with maxima

x(t)=f(t,x(t))+g(t,maxaξtx(ξ)),x^{\prime}(t)=f(t,x(t))+g(t,\underset{a\leq\xi\leq t}{\max}x(\xi)), (1.1)

with initial condition

x(a)=x0,x(a)=x_{0}, (1.2)

where t[a,b],a,b,x0m,f,g:[a,b]×mm.t\in[a,b],\ a,b\in\mathbb{R},x_{0}\in\mathbb{R}^{m},\ f,g:[a,b]\times\mathbb{R}^{m}\mathbb{\rightarrow R}^{m}.

We use the terminologies and notations from [12] and [14]. For the convenience of the reader we recall some of them.

Let (X,d)(X,d) be a metric space and A:XXA:X\rightarrow X an operator. We denote by A0:=1X,A1:=A,An+1:=AnA,nA^{0}:=1_{X},\;A^{1}:=A,\ A^{n+1}:=A^{n}\circ A,\;\;n\in\mathbb{N}, the iterate operators of the operator AA. We also have:

P(X)\displaystyle P(X) :={YX|Yϕ},\displaystyle:=\{Y\subseteq X\ |\ Y\neq\phi\},
FA\displaystyle F_{A} :={xXA(x)=x},\displaystyle:=\{x\in X\mid A(x)=x\},
I(A)\displaystyle I(A) :={YXA(Y)Y,Y}.\displaystyle:=\{Y\subset X\mid A(Y)\subset Y,Y\neq\emptyset\}.
Definition 1.1.

Let (X,d)(X,d) be a metric space. An operator A:XXA:X\rightarrow X is a Picard operator (PO) if there exists xXx^{\ast}\in X such that FA={x}F_{A}=\{x^{\ast}\} and the sequence (An(x0))n(A^{n}(x_{0}))_{n\in\mathbb{N}} converges to xx^{\ast} for all x0Xx_{0}\in X.

Definition 1.2.

Let (X,d)(X,d) be a metric space. An operator A:XXA:X\rightarrow X is a weakly Picard operator (WPO) if the sequence (An(x))n(A^{n}(x))_{n\in\mathbb{N}} converges for all xXx\in X, and its limit (which may depend on xx) is a fixed point of AA.

Definition 1.3.

If AA is weakly Picard operator then we consider the operator AA^{\infty} defined by A:XX,A(x):=limnAn(x)A^{\infty}:X\rightarrow X,\ A^{\infty}(x):=\underset{n\rightarrow\infty}{\lim}A^{n}(x).

Obviously, A(X)=FAA^{\infty}(X)=F_{A}. Moreover, if AA is a PO and we denote by xx^{\ast} its unique fixed point, then A(x)=xA^{\infty}(x)=x^{\ast}, for each xXx\in X.

2. Existence and uniqueness

We prove the existence and uniqueness for the solution of the problem (1.1)-(1.2) using the Perov’s Theorem as in [7]. For standard techniques, when it is used the Banach contraction principle, see [13], [9] and [10].

Theorem 2.1.

(Perov’s fixed point theorem) Let (X,d)(X,d) with d(x,y)md(x,y)\in\mathbb{R}^{m}, be a complete generalized metric space and A:XXA:X\rightarrow X an operator. We suppose that there exists a matrix QMm×m(+)Q\in M_{m\times m}(\mathbb{R}_{+}), such that

  • (i)

    d(A(x),A(y))Qd(x,y)d(A(x),A(y))\leq Qd(x,y), for all x,yX;x,y\in X;

  • (ii)

    Qn0Q^{n}\rightarrow 0, as nn\rightarrow\infty.

Then

  • (a)

    FA={x},F_{A}=\{x^{\ast}\},

  • (b)

    An(x)xA^{n}(x)\rightarrow x^{\ast}, as nn\rightarrow\infty and

    d(An(x),x)(IQ)1Qnd(x0,A(x0)),xX,n;d(A^{n}(x),x^{\ast})\leq(I-Q)^{-1}Q^{n}d(x_{0},A(x_{0})),\ \forall x\in X,\forall n\in\mathbb{N}^{\ast};
  • (c)

    d(x,x)(IQ)1d(x,A(x)),xX.d(x,x^{\ast})\leq(I-Q)^{-1}d(x,A(x)),\ \forall x\in X.

We consider on m\mathbb{R}^{m} the following vectorial norm

|x|:=(|x1||xm|).\left|x\right|:=\left(\begin{array}[c]{c}\left|x_{1}\right|\\ \vdots\\ \left|x_{m}\right|\end{array}\right).

We have the following result:

Theorem 2.2.

We assume that:

  • (i)

    f,gC([a,b]×m,m);f,g\in C([a,b]\times\mathbb{R}^{m},\mathbb{R}^{m});

  • (ii)

    there exist LfL_{f} and LgL_{g} nonnegative matrices such that

    |f(t,u1)f(t,u2)|\displaystyle\left|f(t,u^{1})-f(t,u^{2})\right| Lf|u1u2|,\displaystyle\leq L_{f}\left|u^{1}-u^{2}\right|,
    |g(t,v1)g(t,v2)|\displaystyle\left|g(t,v^{1})-g(t,v^{2})\right| Lg|v1v2|,\displaystyle\leq L_{g}\left|v^{1}-v^{2}\right|,

    t[a,b]\forall t\in[a,b] and u1=(u11,,um1),u2=(u12,,um2),v1=(v11,,vm1),v2=(v12,,vm2)m;u^{1}=(u_{1}^{1},\ldots,u_{m}^{1}),u^{2}=(u_{1}^{2},\ldots,u_{m}^{2}),v^{1}=(v_{1}^{1},\ldots,v_{m}^{1}),v^{2}=(v_{1}^{2},\ldots,v_{m}^{2})\in\mathbb{R}^{m};

  • (iii)

    the matrix

    Q:=(ba)(Lf+Lg)Q:=(b-a)(L_{f}+L_{g}) (2.1)

    is convergent to 0, i.e. Qn0Q^{n}\rightarrow 0, as n.n\rightarrow\infty.

Then, the problem (1.1)-(1.2) has a unique solution xC([a,b],m).x^{\ast}\in C([a,b],\mathbb{R}^{m}).

Proof.

We consider the generalized Banach space X=(C([a,b],m),)X=(C([a,b],\mathbb{R}^{m}),\left\|\cdot\right\|) where \left\|\cdot\right\| is the norm,

x:=(maxatb|x1(t)|maxatb|xm(t)|).\left\|x\right\|:=\left(\begin{array}[c]{c}\underset{a\leq t\leq b}{\max}\left|x_{1}(t)\right|\\ \vdots\\ \underset{a\leq t\leq b}{\max}\left|x_{m}(t)\right|\end{array}\right). (2.2)

The problem (1.1)-(1.2), xC1([a,b],m)x\in C^{1}([a,b],\mathbb{R}^{m}) is equivalent with the following fixed point equation

x(t)=x0+atf(s,x(s))𝑑s+atg(s,maxaξsx(ξ))𝑑s,t[a,b].x(t)=x_{0}+\int\nolimits_{a}^{t}f(s,x(s))ds+\int\nolimits_{a}^{t}g(s,\underset{a\leq\xi\leq s}{\max}x(\xi))ds,\ t\in[a,b]. (2.3)

We consider the operator A:XXA:X\rightarrow X, where

A(x)(t)=x0+atf(s,x(s))𝑑s+atg(s,maxaξsx(ξ))𝑑s.A(x)(t)=x_{0}+\int\nolimits_{a}^{t}f(s,x(s))ds+\int\nolimits_{a}^{t}g(s,\underset{a\leq\xi\leq s}{\max}x(\xi))ds. (2.4)

It is easy to see that if xFAx^{\ast}\in F_{A} then xx^{\ast} is a solution of (1.1)-(1.2).

Conditions (ii) implies that

|A(x)(t)A(y)(t)|\displaystyle\left|A(x)(t)-A(y)(t)\right|\leq
at|f(s,x(s))f(s,y(s))|𝑑s+at|g(s,maxaξsx(ξ))g(s,maxaξsy(ξ))|𝑑s\displaystyle\leq\int\nolimits_{a}^{t}\left|f(s,x(s))-f(s,y(s))\right|ds+\int\nolimits_{a}^{t}\left|g(s,\underset{a\leq\xi\leq s}{\max}x(\xi))-g(s,\underset{a\leq\xi\leq s}{\max}y(\xi))\right|ds
(ba)Lf(maxasb|x1(s)y1(s)|maxasb|xm(s)ym(s)|)\displaystyle\leq(b-a)L_{f}\left(\begin{array}[c]{c}\underset{a\leq s\leq b}{\max}\left|x_{1}(s)-y_{1}(s)\right|\\ \vdots\\ \underset{a\leq s\leq b}{\max}\left|x_{m}(s)-y_{m}(s)\right|\end{array}\right)
+(ba)Lg(maxasb|maxaξsx1(s)maxaξsy1(s)|maxasb|maxaξsxm(s)maxaξsym(s)|).\displaystyle\quad+(b-a)L_{g}\left(\begin{array}[c]{c}\underset{a\leq s\leq b}{\max}\left|\underset{a\leq\xi\leq s}{\max}x_{1}(s)-\underset{a\leq\xi\leq s}{\max}y_{1}(s)\right|\\ \vdots\\ \underset{a\leq s\leq b}{\max}\left|\underset{a\leq\xi\leq s}{\max}x_{m}(s)-\underset{a\leq\xi\leq s}{\max}y_{m}(s)\right|\end{array}\right).

But

maxasb|maxaξsxi(s)maxaξsyi(s)|maxasb|xi(s)yi(s)|.\underset{a\leq s\leq b}{\max}\left|\underset{a\leq\xi\leq s}{\max}x_{i}(s)-\underset{a\leq\xi\leq s}{\max}y_{i}(s)\right|\leq\underset{a\leq s\leq b}{\max}\left|x_{i}(s)-y_{i}(s)\right|.

So,

A(x)A(y)Qxy.\left\|A(x)-A(y)\right\|\leq Q\left\|x-y\right\|.

Using (iii), we get that the operator A:XXA:X\rightarrow X is a QQ-contraction, so FA=(x1,,xm)=xF_{A}=(x_{1}^{\ast},\ldots,x_{m}^{\ast})=x^{\ast} is the unique solution of (1.1)-(1.2). ∎

The equation (1.1) is equivalent with

x(t)=x(a)+atf(s,x(s))𝑑s+atg(s,maxaξsx(ξ))𝑑s,t[a,b],x(t)=x(a)+\int\nolimits_{a}^{t}f(s,x(s))ds+\int\nolimits_{a}^{t}g(s,\underset{a\leq\xi\leq s}{\max}x(\xi))ds,\ t\in[a,b], (2.5)

xC([a,b],m).x\in C([a,b],\mathbb{R}^{m}).

In what follows we consider the operator B:XXB:X\rightarrow X defined by B(x)(t):=B(x)(t):=the right hand side of (2.5). For x0mx_{0}\in\mathbb{R}^{m}, we consider Xx0:={xC([a,b],m)|x(a)=x0}X_{x_{0}}:=\{x\in C([a,b],\mathbb{R}^{m})|\ x(a)=x_{0}\}. It is clear that

X=x0mXx0X=\underset{x_{0}\in\mathbb{R}^{m}}{\cup}X_{x_{0}}

is a partition of X.X.\ We have

Lemma 2.3.

We suppose that the condition (C1)(C_{1}) is satisfied. Then

  • (a)

    A(X)Xx0A(X)\subset X_{x_{0}} and A(Xx0)Xx0;A(X_{x_{0}})\subset X_{x_{0}};

  • (b)

    A|Xx0=B|Xx0.A|_{X_{x_{0}}}=B|_{X_{x_{0}}}.

Remark 2.4.

From Theorem 2.2 we have that the operator AA is PO. Because A|Xx0=B|Xx0,X:=C([a,b],m)=x0mXx0,Xx0I(B)A|_{X_{x_{0}}}=B|_{X_{x_{0}}},\ X:=C([a,b],\mathbb{R}^{m})=\underset{x_{0}\in\mathbb{R}^{m}}{\cup}X_{x_{0}},X_{x_{0}}\in I(B) it follows that the operator BB is WPO and

FBXx0={x},x0m,F_{B}\cap X_{x_{0}}=\{x^{\ast}\},\ \forall x_{0}\in\mathbb{R}^{m},

where xx^{\ast} is the unique solution of the problem (1.1)-(1.2).

3. Data dependence: comparison results

Now we consider the operators AA and BB on the ordered Banach space (C([a,b],m),,)(C([a,b],\mathbb{R}^{m}),\left\|\cdot\right\|,\leq) where the order relation on m\mathbb{R}^{m} is given by: xyxiyi,i=1,m¯x\leq y\Leftrightarrow x_{i}\leq y_{i},\ i=\overline{1,m}.

In order to establish the Čaplygin type inequalities we need the following abstract result.

Lemma 3.1.

(see [14]) Let (X,d,)(X,d,\leq) be an ordered metric space and A:XXA:X\rightarrow X an operator. Suppose that AA is increasing and WPO. Then the operator AA^{\infty} is increasing.

We have the following result

Theorem 3.2.

Suppose that:

  • (a)

    the conditions of Theorem 2.2 are satisfied;

  • (b)

    f(t,):mm,g(t,):mmf(t,\cdot):\mathbb{R}^{m}\rightarrow\mathbb{R}^{m},g(t,\cdot):\mathbb{R}^{m}\rightarrow\mathbb{R}^{m} are increasing, t[a,b].\forall t\in[a,b].

Let xx^{\ast} be a solution of equation (1.1) and yy^{\ast} a solution of the inequality

y(t)f(t,y(t))+g(t,maxaξty(ξ)),t[a,b].y^{\prime}(t)\leq f(t,y(t))+g(t,\underset{a\leq\xi\leq t}{\max}y(\xi)),\ t\in[a,b].

Then y(a)x(a)y^{\ast}(a)\leq x^{\ast}(a) implies that yxy\leq x.

Proof.

From Remark 2.4 we have that BB is WPO. On the other hand, from the condition (b) and Lemma 3.1 we get that the operator BB^{\infty} is increasing. If x0mx_{0}\in\mathbb{R}^{m}, then we denote by x~0\widetilde{x}_{0} the following function

x~0:[a,b]m,x~0(t)=x0,t[a,b].\widetilde{x}_{0}:[a,b]\rightarrow\mathbb{R}^{m},\widetilde{x}_{0}(t)=x_{0},\ \forall t\in[a,b].

Hence yB(y)B2(y)B(y)=B(y~(a))B(x~(a))=xy^{\ast}\leq B(y^{\ast})\leq B^{2}(y^{\ast})\leq\ldots\leq B^{\infty}(y^{\ast})=B^{\infty}(\widetilde{y^{\ast}}(a))\leq B^{\infty}(\widetilde{x^{\ast}}(a))=x^{\ast}. ∎

In order to study the monotony of the solution of the problem (1.1)–(1.2) with respect to x0,f,gx_{0},f,g we need the following result from WPOs theory.

Lemma 3.3.

(Abstract comparison lemma, [15]) Let (X,d,)(X,d,\leq) be an ordered metric space and A,B,C:XXA,B,C:X\rightarrow X be such that:

  • (i)

    the operator A,B,CA,B,C are WPOs;

  • (ii)

    ABC;A\leq B\leq C;

  • (iii)

    the operator BB is increasing.

Then xyzx\leq y\leq z imply that A(x)B(y)C(z).A^{\infty}(x)\leq B^{\infty}(y)\leq C^{\infty}(z).

From this abstract result we obtain the following result:

Theorem 3.4.

Let fj,gjC([a,b]×m,m),j=1,3¯,f^{j},g^{j}\in C([a,b]\times\mathbb{R}^{m},\mathbb{R}^{m}),j=\overline{1,3}, and suppose that conditions the conditions from Theorem 2.2 hold. Furthermore suppose that:

  • (i)

    f1f2f3,g1g2g3;f^{1}\leq f^{2}\leq f^{3},g^{1}\leq g^{2}\leq g^{3};

  • (ii)

    f2(t,):mm,g2(t,):mmf^{2}(t,\cdot):\mathbb{R}^{m}\rightarrow\mathbb{R}^{m},g^{2}(t,\cdot):\mathbb{R}^{m}\rightarrow\mathbb{R}^{m} are increasing.

Let xjx^{\ast j} be a solution of the equation

xj(t)=fj(t,x(t))+gj(t,maxaξtx(ξ)),t[a,b] and j=1,3¯.x^{j\prime}(t)=f^{j}(t,x(t))+g^{j}(t,\underset{a\leq\xi\leq t}{\max}x(\xi)),\ t\in[a,b]\text{ and }j=\overline{1,3}.

Then x1(a)x2(a)x3(a)x^{\ast 1}(a)\leq x^{\ast 2}(a)\leq x^{\ast 3}(a), implies x1x2x3x^{\ast 1}\leq x^{\ast 2}\leq x^{\ast 3}, i.e. the unique solution of the problem (1.1)–(1.2) is increasing with respect to x0,fx_{0},f and g.g.

Proof.

From Remark 2.4, the operators Bj,j=1,3¯,B_{j},j=\overline{1,3},\ are WPOs. From the condition (ii) the operator B2B_{2} is monotone increasing. From the condition (i) it follows that B1B2B3B_{1}\leq B_{2}\leq B_{3}. Let x~j(a)(C[a,b],m)\widetilde{x}^{j}(a)\in(C[a,b],\mathbb{R}^{m}) be defined by x~j(a)=xj(a),t[a,b].\widetilde{x}^{j}(a)=x^{j}(a),\ \forall t\in[a,b].\ We notice that

x~1(a)(t)x~2(a)(t)x~3(a)(t),t[a,b].\widetilde{x}^{1}(a)(t)\leq\widetilde{x}^{2}(a)(t)\leq\widetilde{x}^{3}(a)(t),\ \forall t\in[a,b].

From Lemma 3.3 we have that B1(x~1(a))B2(x~2(a))B3(x~3(a))B_{1}^{\infty}(\widetilde{x}^{\ast 1}(a))\leq B_{2}^{\infty}(\widetilde{x}^{\ast 2}(a))\leq B_{3}^{\infty}(\widetilde{x}^{\ast 3}(a)). But xj=Bj(x~j(a))x^{\ast j}=B_{j}^{\infty}(\widetilde{x}^{\ast j}(a)), so x1x2x3x^{\ast 1}\leq x^{\ast 2}\leq x^{\ast 3}. ∎

4. Data dependence: continuity

In this section we prove the continuous dependence of the solution for equation (1.1) and suppose the conditions of the Theorem 2.2 are satisfied.

Theorem 4.1.

Let x0j,fj,gj,j=1,2x_{0}^{j},f^{j},g^{j},j=1,2 satisfy the conditions from Theorem 2.2. Furthermore we suppose there exist η1,η2,η3+m,\eta^{1},\eta^{2},\eta^{3}\in\mathbb{R}_{+}^{m}, such that

  1. (i)

    |x0jx0j|η1\left|x_{0}^{j}-x_{0}^{j}\right|\leq\eta^{1};

  2. (ii)

    |f1(t,u)f2(t,u)|η2,|g1(t,v)g2(t,v)|η3,tC[a,b],u,vm.\left|f^{1}(t,u)-f^{2}(t,u)\right|\leq\eta^{2},\left|g^{1}(t,v)-g^{2}(t,v)\right|\leq\eta^{3},\ \forall t\in C[a,b],u,v\in\mathbb{R}^{m}.

Then

x(t;x01,f1,g1)x(t;x02,f2,g2)(IQ)1(η1+(ba)(η2+η3)),\left\|x^{\ast}(t;x_{0}^{1},f^{1},g^{1})-x^{\ast}(t;x_{0}^{2},f^{2},g^{2})\right\|\leq(I-Q)^{-1}(\eta^{1}+(b-a)(\eta^{2}+\eta^{3})),

where x(t;x0j,fj,gj)x^{\ast}(t;x_{0}^{j},f^{j},g^{j}) are the solutions of the problem (1.1)–(1.2) with respect to x0j,fj,gj,j=1,2.x_{0}^{j},f^{j},g^{j},j=1,2.

Proof.

Consider the operator Ax0j,fj,gj,j=1,2.A_{x_{0}^{j},f^{j},g^{j}},j=1,2. From Theorem 2.2 it follows that

Ax01,f1,g1(x)Ax01,f1,g1(y)Qxy,x,yX.\left\|A_{x_{0}^{1},f^{1},g^{1}}(x)-A_{x_{0}^{1},f^{1},g^{1}}(y)\!\right\|\leq Q\left\|x-y\right\|,\forall x,y\in X.

Additionally

Ax01,f1,g1(x)Ax02,f2,g2(x)(η1+(ba)(η2+η3)).\left\|A_{x_{0}^{1},f^{1},g^{1}}(x)-A_{x_{0}^{2},f^{2},g^{2}}(x)\!\right\|\leq(\eta^{1}+(b-a)(\eta^{2}+\eta^{3})).

Then

x(t;x01,f1,g1)x(t;x02,f2,g2)=\displaystyle\left\|x^{\ast}(t;x_{0}^{1},f^{1},g^{1})-x^{\ast}(t;x_{0}^{2},f^{2},g^{2})\right\|=
=Ax01,f1,g1(x(t;x01,f1,g1))Ax02,f2,g2(x(t;x02,f2,g2))\displaystyle=\left\|A_{x_{0}^{1},f^{1},g^{1}}(x^{\ast}(t;x_{0}^{1},f^{1},g^{1}))-A_{x_{0}^{2},f^{2},g^{2}}(x^{\ast}(t;x_{0}^{2},f^{2},g^{2}))\!\right\|
Ax01,f1,g1(x(t;x01,f1,g1))Ax01,f1,g1(x(t;x02,f2,g2))\displaystyle\leq\left\|A_{x_{0}^{1},f^{1},g^{1}}(x^{\ast}(t;x_{0}^{1},f^{1},g^{1}))-A_{x_{0}^{1},f^{1},g^{1}}(x^{\ast}(t;x_{0}^{2},f^{2},g^{2}))\!\right\|
+Ax01,f1,g1(x(t;x02,f2,g2))Ax02,f2,g2(x(t;x02,f2,g2))\displaystyle\quad+\left\|A_{x_{0}^{1},f^{1},g^{1}}(x^{\ast}(t;x_{0}^{2},f^{2},g^{2}))-A_{x_{0}^{2},f^{2},g^{2}}(x^{\ast}(t;x_{0}^{2},f^{2},g^{2}))\!\right\|
Qx(t;x01,f1,g1)x(t;x02,f2,g2)+(η1+(ba)(η2+η3)).\displaystyle\leq Q\left\|x^{\ast}(t;x_{0}^{1},f^{1},g^{1})-x^{\ast}(t;x_{0}^{2},f^{2},g^{2})\right\|+(\eta^{1}+(b-a)(\eta^{2}+\eta^{3})).

Since Qn0Q^{n}\rightarrow 0 as nn\rightarrow\infty, implies that (IQ)1Mmm(+)(I-Q)^{-1}\in M_{mm}(\mathbb{R}_{+}) and we finally obtain

x(t;x01,f1,g1)x(t;x02,f2,g2)(IQ)1(η1+(ba)(η2+η3)).\left\|x^{\ast}(t;x_{0}^{1},f^{1},g^{1})-x^{\ast}(t;x_{0}^{2},f^{2},g^{2})\right\|\leq(I-Q)^{-1}(\eta^{1}+(b-a)(\eta^{2}+\eta^{3})).

5. Remarks

In this section we emphasize some special cases of (1.1).

Let τ>0\tau>0 be a given number and we define the operator G:C([τ,),m)mG:C([-\tau,\infty),\mathbb{R}^{m})\rightarrow\mathbb{R}^{m} such that for any function xC([τ,),m)x\in C([-\tau,\infty),\mathbb{R}^{m}) and any point t+t\in\mathbb{R}_{+} there exists a point ξ[tτ,t]\xi\in[t-\tau,t] such that G(x)(t)=a(t)x(ξ)G(x)(t)=a(t)x(\xi) where aC(+,).a\in C(\mathbb{R}_{+},\mathbb{R}).

Consider the nonlinear delay functional differential equation

x(t)=f(t,x(t))+g(t,G(x)(t))x^{\prime}(t)=f(t,x(t))+g(t,G(x)(t)) (5.1)

for tt0t\geq t_{0} with initial condition

x(t+t0)=φ(t),t[τ,0],x(t+t_{0})=\varphi(t),\ t\in[-\tau,0],

where xm,f:+×mm,t0+,φ:[τ,0]m.x\in\mathbb{R}^{m},\ f:\mathbb{R}_{+}\times\mathbb{R}^{m}\rightarrow\mathbb{R}^{m},\ t_{0}\in\mathbb{R}_{+},\ \varphi:[-\tau,0]\rightarrow\mathbb{R}^{m}.

Particular cases of (1.1):

  • (i)

    For G(x)(t)=x(tτ),t+G(x)(t)=x(t-\tau),\ t\in\mathbb{R}_{+}, then (5.1) reduces to a delay differential equation (see [6], [12], [14], [15]);

  • (ii)

    For G(x)(t)=maxs[tτ,t]x(s),t+G(x)(t)=\underset{s\in[t-\tau,t]}{\max}x(s),\ t\in\mathbb{R}_{+}, then (5.1) reduces to a differential equation with maxima (see [16], [17], [9], [10], [1]);

  • (iii)

    For G(x)(t)=tτtx(s)𝑑s,t+,τ>0G(x)(t)=\int\nolimits_{t-\tau}^{t}x(s)ds,\ t\in\mathbb{R}_{+},\tau>0, then (5.1) reduces to a differential equation with distributed delay (see [11], [4]);

  • (iv)

    For g(t,G(x)(t))=h(x)(t)g(t,G(x)(t))=h(x)(t), where h:C([a,b],)C([a,b],)h:C([a,b],\mathbb{R})\rightarrow C([a,b],\mathbb{R}) is an abstract Volterra operator, then (5.1) reduces to a differential equation with abstract Volterra operator (see [8]);

  • (v)

    If x(t)f(t,x(t)):=ddt[x(t)f(t,x(t))],G(x)(t)=x(t),tt0x^{\prime}(t)-f(t,x(t)):=\frac{d}{dt}\left[\frac{x(t)}{f(t,x(t))}\right],\ G(x)(t)=x(t),\ t\geq t_{0}, then (5.1) reduces to a quadratic differential equation (see [3]).

Acknowledgements.

The author is grateful to professor I. A. Rus for his helpful comments and suggestions.

References

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