Stability for a Class of Differential Set-Valued Inverse Variational Inequalities in Finite Dimensional Spaces

Abstract

In this paper, we introduce and study a new class of differential set-valued inverse variational inequalities in finite dimensional spaces. By applying a result on differential inclusions involving an upper semicontinuous set-valued mapping with closed convex values, we first prove the existence of Carathéodory weak solutions for differential set-valued inverse variational inequalities. Then, by the existence result, we establish the stability for the differential set-valued inverse variational inequality problem when the constraint set and the mapping are perturbed by two different parameters. The closedness and continuity of Carathéodory weak solutions with respect to the two different parameters are obtained.

1. Introduction

Let K be a nonempty closed convex set of $R^n$ and $F:R^n\to 2^{R^n}$ be a set-valued mapping. A set-valued inverse variational inequality, denoted by SIVI(K, F), is to find $u\in R^n$ and $u^{*}\in F\left(u\right)\cap K$ such that

$$\begin{array}{c}\hfill \langle y-u^{*},u\rangle \ge 0,\forall y\in K.\end{array}$$

(1)

The solution to this problem is denoted by SOL(K, F). We write $\dot{x}:=\frac{dx}{dt}$ for the time-derivative of function $x\left(t\right)$. In this paper, we study the following initial-value differential set-valued inverse variational inequality (denoted by DSIVI):

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}\left(t\right)=f(t,x\left(t\right))+B(t,x\left(t\right))u\left(t\right)\hfill \\ u\left(t\right)\in SOL(K,G(t,x\left(t\right))+F(·\left)\right)\hfill \\ x\left(0\right)=x_0,\hfill \end{array}\right.\end{array}$$

(2)

where $\mathrm{\Omega }:=[0,T]\times R^m$, $(f,B,G):=\mathrm{\Omega }\to R^m\times R^{m\times n}\times R^n$, $F:=R^n\to 2^{R^n}$. Time-dependent functions $x\left(t\right)$ and $u\left(t\right)$ satisfy (2) in the weak sense of Carathéodory for $t\in [0,T]$ means that x is an absolutely continuous function on $[0,T]$ , $x\left(t\right)$ satisfies the differential equation for almost all $t\in [0,T]$ and the initial-value condition. Moreover, u is an integrable function on $[0,T]$ and $u\left(t\right)\in SOL(K,G(t,x\left(t\right))+F)$ for almost all $t\in [0,T]$.

Differential variational inequalities (DVIs) arise in some applied problems such as, for example, differential Nash games, operations research, physical sciences, and structural dynamics [1,2]. DVIs were first systematically studied in finite dimensional spaces by Pang and Stewart [1] in 2008 and gained much more attention to theoretical results, numerical algorithms, and applications. Stewart [3] investigated the uniqueness for a class of index-one DVIs in finite dimensional spaces. Li et al. [4,5] researched differential mixed variational inequalities and impulsive differential variational inequalities in finite dimensional spaces and obtained some existence results and numerical methods by using some results on differential inclusions and discrete Euler time-dependent procedures. Li et al. [6] proved the existence of the Carathéodory weak solutions for differential inverse variational inequalities in finite dimensional spaces and gave an application on the time-dependent price equilibrium problem. In [7], Liu et al. first explored partial differential variational inequalities in Banach spaces and proved the nonemptiness and compactness of the solution set. For more related work about DVIs, see [8,9,10,11,12].

The inverse variational inequality, like the variational inequality, has broad applications in optimization, engineering, economics, mechanics, and transportation [13,14,15,16,17,18,19,20]. Very recently, Luo [21] studied the stability for the set-valued inverse variational inequality (1) on Banach spaces. If F is single-valued, the set-valued inverse variational inequality (1) can be reduced to the singe-valued inverse variational inequality in [13]. Furthermore, if F is single-valued and inverse, the set-valued inverse variational inequality (1) can be transformed into the classical variational inequality. However, the above transforms both failed if F is set-valued.

The stability analysis of a DVI with perturbed data is very helpful in identifying sensitive parameters, predicting the coming changes of the equilibria as a result of the changes in the governing system, and providing helpful information for designing different equilibrium systems. Gwinner [22] researched stability of the solution set for a DVI and obtained a novel upper set convergence result with respect to perturbations in the data. When the mapping and the constraint set are perturbed by different parameters, Wang et al. [23] studied the stability for a class differential mixed variational inequality in finite dimensional spaces. To the best of our knowledge, there are some results about the existence of solutions for differential variational and inverse variational inequalities in finite dimensional spaces. However, there are very few results about the existence of solutions for differential set-valued inverse variational inequalities and the stability for differential single-valued or set-valued inverse variational inequalities in finite dimensional spaces. Motivated by the aforementioned work, in this paper we are devoted to stability analysis for the DSIVI (2) in finite dimensional spaces.

The goal of this paper is to study the existence of the Carathéodory weak solutions and the stability for DSIVI (2) in finite dimensional spaces with the constraint set K and the set-valued mapping F being perturbed by two different parameters. Our results about the existence of the Carathéodory weak solutions for DSIVI (2) generalize the corresponding results in [6]. Our stability results about the differential set-valued inverse variational inequality are very new. We also give an example of a time-dependent price equilibrium control problem influenced by the seasons to show that the realistic problem can be transformed into the stability for the differential inverse variational inequality.

The paper is organized as follows. Section 2 contains some useful definitions and lemmas. In Section 3, the existence and uniqueness results of Carathéodory solutions for DSIVI (2) are considered. Furthermore, the closedness and continuity of Carathéodory solution set with respect to the perturbed data in the constraint set K and the set-valued mapping F are obtained.

2. Preliminaries

In this section, we will introduce some basic notations and preliminary results.

Definition 1

([24]). Let X and Y be two metric spaces; $Y^{*}$ is the dual space of Y. We say a set-valued mapping $F:X\to 2^Y$ is

(i) 

Upper semicontinuous at $x\in X$ if and only if for any neighborhood U of $F\left(x\right)$, there exists the neighborhood $B(x,\eta )$ of x with $\eta >0$ such that

$$\forall x^{\prime }\in B(x,\eta ),F\left(x^{\prime }\right)\subset U;$$

(ii) 

Lower semicontinuous at $x\in X$ if and only if for any $y\in F\left(x\right)$ and for any sequence of elements $x_n\in X$ converging to x, there exists a sequence of elements $y_n\in F\left(x_n\right)$ converging to y;

(iii) 

Upper hemicontinuous at $x\in X$ if and only if for any $r\in Y^{*}$, the function $x\mapsto \underset{y\in F\left(x\right)}{sup}\langle r,y\rangle$ is upper semicontinuous at x.

Definition 2

([23,25]). The set-valued mapping $F:R^n\to 2^{R^n}$ is said to be

(i) 

Strictly monotone on set $L\subset R^n$ iff for any $x,y\in L$, $x\ne y$, $x^{*}\in F\left(x\right)$, $y^{*}\in F\left(y\right)$, we have

$$\langle x^{*}-y^{*},x-y\rangle >0;$$

(ii) 

Strongly monotone with modulus $\mu >0$, if for any $x,y\in R^n$ and $x^{*}\in F\left(x\right)$, $y^{*}\in F\left(y\right)$, we have

$$\langle x^{*}-y^{*},x-y\rangle \ge \mu {\parallel x-y\parallel }^2.$$

Definition 3.

A mapping $f:\mathrm{\Omega }\to R^m$ (respectively, $B:\mathrm{\Omega }\to R^{m\times n}$) is said to be Lipschitz continuous if there exists a constant $L_f>0$ (respectively, $L_B>0$) such that, for any $(t_1,x),(t_2,y)\in \mathrm{\Omega }$, we have

$$\parallel f(t_1,x)-f(t_2,y)\parallel \le L_f\left(\right|t_1-t_2|+\parallel x-y\parallel ),$$

$$(respectively,\parallel B(t_1,x)-B(t_2,y)\parallel \le L_B\left(\right|t_1-t_2|+\parallel x-y\parallel )).$$

Lemma 1

([26], Lemma 1). Let X and Y be metric spaces. If a set-valued mapping $F:X\to P_f\left(Y\right):=\{D\subset Y:Disnonempty,closed\}$ is upper semicontinuous, then F is closed.

Lemma 2

([27], Theorem 5.1). Let $\mathbb{F}:\mathrm{\Omega }\to 2^{R^m}$ be an upper semicontinuous set-valued mapping with nonempty closed convex values. Suppose that there exists a scalar ${\rho }^{\mathbb{F}}>0$ satisfying

$$\begin{array}{c}\hfill sup\{\parallel y\parallel :y\in \mathbb{F}(t,x)\}\le {\rho }^{\mathbb{F}}(1+\parallel x\parallel ),\forall (t,x)\in \mathrm{\Omega }.\end{array}$$

(3)

Then for every $x_0\in R^m$, the DI:

$$\dot{x}\in \mathbb{F}(t,x),x\left(0\right)=x_0$$

has a weak solution in the sense of Carathéodory.

Lemma 3

([1], Lemma 6.3). Let $h:\mathrm{\Omega }\times R^n\to R^m$ be a continuous function and $U:\mathrm{\Omega }\to 2^{R^n}$ be a closed set-valued map such that for some constant ${\eta }_U>0,$

$$\underset{u\in U(t,x)}{sup}\parallel u\parallel \le {\eta }_U(1+\parallel x\parallel ),\forall (t,x)\in \mathrm{\Omega }.$$

Let $v:[0,T]\to R^m$ be a measurable function and $x:[0,T]\to R^m$ be a continuous function satisfying $v\left(t\right)\in h(t,x(t),U(t,x\left(t\right)\left)\right)$ for almost all $t\in [0,T]$. There exists a measurable function $u:[0,T]\to R^n$ such that $u\left(t\right)\in U(t,x(t\left)\right)$ and $v\left(t\right)=h(t,x(t),U(t\left)\right)$ for almost all $t\in [0,T].$

Throughout the rest of this paper, let $K\subset R^n$ be a nonempty, closed, and convex subset. The symbols $“{\rightharpoonup }^{”}$ and $“{\to }^{”}$ are used to denote the weak convergence and strong convergence. Let the barrier cone of K be denoted by

$$barr\left(K\right):=\{y\in R^n:\underset{x\in K}{sup}\langle y,x\rangle <\infty \}.$$

The recession cone of $K,$ denoted by $K_{\infty },$ is defined by

$$K_{\infty }:=\{d\in R^n:\exists t_n\to 0,\exists x_n\in K,t_n{x}_n\rightharpoonup d\}.$$

The negative polar cone of the nonempty set $D\subset R^n,$ denoted by $D^{-},$ is defined by

$$D^{-}:=\{y\in R^n:\langle y,x\rangle \le 0,\forall x\in D\}.$$

Lemma 4

([21], Theorem 4.2). Let $L:Z_1\to 2^{R^n}$ be a continuous set-valued mapping; $p_0\in Z_1$, ${\lambda }_0\in Z_2$ are given points; $F:R^n\times Z_2\to 2^{R^n}$ is a set-valued mapping and lower semicontinuous on $Z_2$. Suppose that there exists a neighborhood of $P\times \Lambda$ of $(p_0,{\lambda }_0)$, such that $L\left(p\right)$ has nonempty, closed, and convex values for any $p\in P$, and $F(x,\lambda )$ has nonempty closed values for every $x\in R^n$ and $\lambda \in \Lambda$. Moreover, for each $\lambda \in \Lambda$ and $q\in G\left(\mathrm{\Omega }\right)$, the mapping $x\mapsto q+F(x,\lambda )$ is upper hemicontinuous and monotone. If

$${\left(L\left(p_0\right)\right)}_{\infty }\cap {\{x\in R^n:q+F(x,{\lambda }_0)\cap L\left(p_0\right)\ne \varnothing \}}^{-}=\left\{0\right\},$$

then there exists a neighborhood $P^{\prime }\times {\Lambda }^{\prime }$ of $(p_0,{\lambda }_0)$ with $P^{\prime }\times {\Lambda }^{\prime }\subset P\times \Lambda$, such that for every $(p,\lambda )\in P^{\prime }\times {\Lambda }^{\prime }$ , the set $SOL\left(L\right(p),q+F(·,\lambda \left)\right)$ is nonempty and bounded.

In the rest of this paper, we assume $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$ hold.

$\left(\mathrm{A}\right)$$f,B$ and G are Lipschitz continuous functions on $\mathrm{\Omega }$ with Lipschitz constants $L_f>0$, $L_B>0$ and $L_G>0$, respectively;

$\left(\mathrm{B}\right)$B is bounded on $\mathrm{\Omega }$ with ${\delta }_{\mathrm{B}}:=\underset{(t,x)\in \mathrm{\Omega }}{sup}\Vert B(t,x)\Vert$.

Remark 1.

If $f:\mathrm{\Omega }\to R^m$ is a Lipschitz continuous function on Ω, we obtain that there exists a constant ${\rho }_f>0$, for any $(t,x)\in \mathrm{\Omega }$, such that

$$\begin{array}{ccc}\hfill \parallel f(t,x)\parallel & =& \parallel f(t,x)-f(t{}_0,0)+f(t{}_0,0)\parallel \hfill \\ & \le & \parallel f(t,x)-f(t{}_0,0)\parallel +\parallel f(t{}_0,0)\parallel \hfill \\ & \le & L_f\left(\right|t-t_0|+\parallel x\parallel )+\parallel f(t{}_0,0)\parallel \hfill \\ & \le & L_f(2T+\parallel x\parallel )+\parallel f(t{}_0,0)\parallel \hfill \\ & \le & {\rho }_f(1+\parallel x\parallel ),\hfill \end{array}$$

where $t_0\in [0,T],{\rho }_f=max\{L_f,2L_{f}T+\parallel f(t{}_0,0)\parallel \}$. Similarly, $G:\mathrm{\Omega }\to R^n$ is a Lipschitz continuous function on Ω, so there exists a constant ${\rho }_G>0$ such that $\parallel G(t,x)\parallel \le {\rho }_G(1+\parallel x\parallel )$ for any $(t,x)\in \mathrm{\Omega }$.

3. Existence and Uniqueness of Solutions for DSIVI (2)

In this section, we will show the existence and uniqueness of Carathéodory weak solutions for DSIVI (2) by applying Lemmas 2 and 3. For this purpose, we define a set-valued mapping $\mathbb{F}:\mathrm{\Omega }\to 2^{R^n}$ as follows:

$$\begin{array}{c}\hfill \mathbb{F}(t,x):=\left\{f\right(t,x)+B(t,x)u:u\in SOL(K,G(t,x)+F(·)\left)\right\}.\end{array}$$

(4)

The following lemma presents some properties of the set-valued mapping $\mathbb{F}$ defined by (4) under the hypotheses (A) and (B).

In the following, $L^2([0,T];R^n)$ is the set of all measurable functions $u:[0,T]\to R^n$, that satisfies ${\int }_0^T{\parallel u\left(t\right)\parallel }^{2}dt<+\infty$. The norm of $\parallel u\parallel$ is defined by

$$\parallel u\parallel :=({\int }_0^T{\parallel u\left(t\right)\parallel }^2{dt)}^{\frac{1}{2}}.$$

Lemma 5.

Let $(f,G,B)$ satisfy conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$. Let $K\subset R^n$ be a nonempty, bounded, closed, and convex set and $F:R^n\to 2^{R^n}$ be an upper semicontinuous set-valued mapping with nonempty closed convex values. Suppose that there exists a constant $\rho >0$ such that, for all $q\in G\left(\mathrm{\Omega }\right)$,

$$\begin{array}{c}\hfill sup\{\parallel u\parallel :u\in SOL(K,q+F(·)\left)\right\}\le \rho (1+\parallel q\parallel ).\end{array}$$

(5)

Then, there exists a constant ${\rho }^{\mathbb{F}}>0$ such that (3) holds for the mapping $\mathbb{F}$. Hence, $\mathbb{F}$ is an upper semicontinuous closed-valued mapping on Ω.

Proof. 

We first prove that there is a constant ${\rho }^{\mathbb{F}}>0$ such that (3) holds for the mapping $\mathbb{F}$. For any $y\in \mathbb{F}(t,x)$, from the definition of $\mathbb{F},$ we know there exists $u\in SOL(K,G(t,x)+F(·\left)\right)$ such that $y=f(t,x)+B(t,x)u.$ From conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$, it is easy to see that there exists positive constants ${\rho }_f$ and ${\rho }_G$ such that

$$\begin{array}{c}\hfill \parallel f(t,x)\parallel \le {\rho }_f(1+\parallel x\parallel ),\end{array}$$

(6)

and

$$\begin{array}{c}\hfill \parallel G(t,x)\parallel \le {\rho }_G(1+\parallel x\parallel ).\end{array}$$

(7)

Applying (5), we obtain

$$\begin{array}{c}\hfill \parallel y\parallel \le \parallel f(t,x)\parallel +\parallel B(t,x)u\parallel \le ({\rho }_f+{\delta }_B\rho +{\delta }_B\rho {\rho }_G)(1+\parallel x\parallel ).\end{array}$$

If we let ${\rho }^{\mathbb{F}}={\rho }_f+{\delta }_B\rho +{\delta }_B\rho {\rho }_G,$ then (3) holds.

Next, we prove that $\mathbb{F}$ is upper semicontinuous. We note that under the linear growth condition (3), the upper semicontinuity of F holds if $\mathbb{F}$ is closed. Therefore, we need to prove $\mathbb{F}$ is closed on $\mathrm{\Omega }$. Let $\left\{(t_n,x_n)\right\}\subset \mathrm{\Omega }$ be a sequence converging to some vector $(t_0,x_0)\in \mathrm{\Omega }$ and $\{f(t_n,x_n)+B(t_n,x_n)u_n\}\subset \mathbb{F}(t_n,x_n)$ converging to $z_0$, where $u_n\in SOL(K,G(t_n,x_n)+F(·))$. It follows that sequence $\left\{u_n\right\}$ is bounded by (5). Therefore, $\left\{u_n\right\}$ has a convergent subsequence, denoted again by $\left\{u_n\right\}$, with a limit point $u_0\in R^n$. According to $u_n\in SOL(K,G(t_n,x_n)+F(·))$, it is easy to see that there exists $u_n^{*}\in F\left(u_n\right)$ and $G(t_n,x_n)+u_n^{*}\in$K such that

$$\langle y-G(t_n,x_n)-u_n^{*},u_n\rangle \ge 0,\forall y\in K.$$

By the boundedness of $K,$ we get $\left\{u_n^{*}\right\}$ is bounded and has a convergent subsequence with a limit $u_0^{*}$, as the set-valued mapping F is upper semicontinuous with nonempty closed convex values. By Lemma 1, we obtain that F is closed, which means $Graph\left(F\right):=\{(x,y)\in R^n\times R^n:y\in F\left(x\right)\}$ is closed. We know $(u_n,u_n^{*})\in Graph\left(F\right)$ since $u_n^{*}\in F\left(u_n\right)$. Therefore, $u_0^{*}\in F\left(u_0\right)$. Since G is Lipschitz continuous, it follows $\left\{G(t_n,x_n)\right\}$ converges to $G(t_0,x_0)$. Since K is closed, it follows that $G(t_0,x_0)+u_0^{*}\in (G(t_0,x_0)+F\left(u_0\right))\cap K$ and

$$\langle y-G(t_0,x_0)-u_0^{*},u_0\rangle \ge 0,\forall y\in K.$$

That means $u_0\in SOL(K,G(t_0,x_0)+F(·))$ and so

$$f(t_n,x_n)+B(t_n,x_n)u_n\to z_0=f(t_0,x_0)+B(t_0,x_0)u_0\in \mathbb{F}(t_0,x_0).$$

Therefore, $\mathbb{F}$ is closed. This completes the proof. □

Remark 2.

We would like to point out that Lemma 5 extends Lemma 2.5 in [6].

Theorem 1.

Let $(f,G,B)$ satisfy conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$. Let $K\subset R^n$ be a nonempty, bounded, closed, and convex set and $F:R^n\to 2^{R^n}$ be an upper semicontinuous set-valued mapping with nonempty closed convex values. Suppose for any $q\in G\left(\mathrm{\Omega }\right)$, there exists a constant $\rho >0$ such that (5) holds and the set $SOL(K,q+F(·\left)\right)$ is nonempty, closed, and convex. Then, DSIVI (2) has a Carathéodory weak solution.

Proof. By Lemma 5, we obtain that there exists a constant ${\rho }^{\mathbb{F}}>0$ such that (3) holds for the mapping $\mathbb{F}$ defined by (4) and $\mathbb{F}$ is an upper semicontinuous closed-valued mapping on $\mathrm{\Omega }$. Next, for any $(t,x)\in \mathrm{\Omega },$ we prove $\mathbb{F}(t,x)$ is convex. Since for any $q\in G\left(\mathrm{\Omega }\right)$, $SOL(K,q+F(·\left)\right)$ is nonempty, it is easy to see that $\mathbb{F}(t,x)$ is nonempty. However, for any $f(t,x)+B(t,x)u_1,f(t,x)+B(t,x)u_2\in \mathbb{F}(t,x)$, where $u_1,u_2\in SOL(K,G(t,x)+F(·))$, by the convex of $SOL(K,G(t,x)+F(·\left)\right)$, we know that there exists a constant $\eta \in (0,1)$ such that

$$\begin{array}{ccc}& & \eta (f(t,x)+B(t,x)u_1)+(1-\eta )(f(t,x)+B(t,x)u_2)\hfill \\ & =& f(t,x)+B(t,x)(\eta u_1+(1-\eta )u_2)\in \mathbb{F}(t,x),\hfill \end{array}$$

where $\eta u_1+(1-\eta )u_2\in SOL(K,G(t,x)+F(·)).$ This means $\mathbb{F}(t,x)$ is convex.

Because $\mathbb{F}$ is an upper semicontinuous set-valued mapping with nonempty closed convex values and there exists a constant ${\rho }^{\mathbb{F}}>0$ such that (3) holds for the mapping $\mathbb{F}$, by Lemma 2, we obtain that the following differential inclusion $\dot{x}\in \mathbb{F}(t,x),x\left(0\right)=x_0$ has a Carathéodory weak solution $x\left(t\right)$. Thus, we have for any $t\in [0,T],$

$$\begin{array}{ccc}\hfill {\int }_0^t\dot{x}\left(s\right)ds& =& {\int }_0^t[f(s,x\left(s\right))+B(s,x\left(s\right))u\left(s\right)]ds,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel x\left(t\right)\parallel & \le & \parallel x_0\parallel +{\int }_0^t{\rho }^{\mathbb{F}}(1+\parallel x\left(s\right)\parallel )ds.\hfill \end{array}$$

Then, by the Gronwall inequality, we obtain

$$\begin{array}{c}\hfill \parallel x\left(t\right)\parallel \le (\parallel x_0\parallel +{\rho }^{\mathbb{F}}T)e^{{\rho }^{\mathbb{F}}T}.\end{array}$$

(8)

Therefore, from the above two inequalities we can obtain that $x\left(t\right)$ is absolutely continuous on $[0,T].$ Let $U(t,x):=SOL(K,G(t,x)+F(·\left)\right)$ and $h(t,x,u):=f(t,x)+B(t,x)u$. We conclude by Lemma 3 that there exists a measurable function $u\left(t\right)$ such that $u\left(t\right)\in SOL(K,G(t,x\left(t\right))+F(·\left)\right)$ and $\dot{x}\left(t\right)=f(t,x)+B(t,x)u\left(t\right)$ for almost all t. By Lemma 6, it follows that for almost all $t\in [0,T]$, there exists $\rho >0$ such that

$$\parallel u\left(t\right)\parallel \le \rho (1+\parallel G(t,x\left(t\right))\parallel ),$$

where $u\left(t\right)\in SOL(K,G(t,x\left(t\right))+F(·\left)\right).$ From (6) and (8), it follows from the above inequality that for almost all $t\in [0,T],$

$$\parallel u\left(t\right)\parallel \le \rho (1+{\rho }_G(1+(\parallel x_0\parallel +{\rho }^{\mathbb{F}}T)e^{{\rho }^{\mathbb{F}}T}\left)\right).$$

Therefore, $u\left(t\right)$ is integrable on $[0,T]$. This completes the proof. □

Lemma 6.

Let $(f,G,B)$ satisfy conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$. Let $K\subset R^n$ be a nonempty, bounded, closed, and convex set. Suppose the following statements hold:

(i) 

$F:R^n\to 2^{R^n}$ is strictly monotone and upper hemicontinuous on $R^n$;

(ii) 

For any $q\in G\left(\mathrm{\Omega }\right)$, $K_{\infty }\cap {\{x\in R^n:q+F\left(x\right)\cap K\ne \varnothing \}}^{-}=\left\{0\right\};$

(iii) 

The interior of $barr\left(K\right)$ is nonempty.

Then, $SOL(K,q+F(·\left)\right)$ is a singleton for any $q\in G\left(\mathrm{\Omega }\right)$. Moreover, there exists a constant $\rho >0$ such that (5) holds for any $q\in G\left(\mathrm{\Omega }\right)$.

Proof. Using conditions (i)–(iii) and according to Theorem 3.2 in [21], we can obtain that $SOL(K,q+F(·\left)\right)\ne \varnothing$ for any $q\in G\left(\mathrm{\Omega }\right)$. Next, we show $SOL(K,q+F(·\left)\right)$ is a singleton for any $q\in G\left(\mathrm{\Omega }\right).$ We assume $u_1,u_2\in SOL(K,q+F(·))$ and $u_1\ne u_2$, and we have

$$\begin{array}{c}\hfill q+u_1^{*}\in (q+F\left(u_1\right))\cap K,\langle y-q-u_1^{*},u_1\rangle \ge 0,\forall y\in K,\end{array}$$

(9)

and

$$\begin{array}{c}\hfill q+u_2^{*}\in (q+F\left(u_2\right))\cap K,\langle y-q-u_2^{*},u_2\rangle \ge 0,\forall y\in K.\end{array}$$

(10)

Letting $y=q+u_2^{*}$ in (9), we have

$$\langle u_2^{*}-u_1^{*},u_1\rangle \ge 0.$$

Letting $y=q+u_1^{*}$ in (10), we have

$$\langle u_1^{*}-u_2^{*},u_2\rangle \ge 0.$$

It follows from the above two inequalities that

$$\begin{array}{c}\hfill \langle u_2^{*}-u_1^{*},u_1-u_2\rangle \ge 0.\end{array}$$

(11)

Since F is strictly monotone, $u_1\ne u_2$, $u_1^{*}\in F\left(u_1\right)$, $u_2^{*}\in F\left(u_2\right)$, we obtain

$$\langle u_2^{*}-u_1^{*},u_2-u_1\rangle >0,$$

which contradicts (11). That means $SOL(K,q+F(·\left)\right)$ is a singleton for any $q\in G\left(\mathrm{\Omega }\right)$ and so there exists a constant $\rho >0$ such that (5) holds for any $q\in G\left(\mathrm{\Omega }\right)$. This completes the proof. □

Theorem 2.

Let $K\subset R^n$ be a nonempty, bounded, closed, and convex set. Let $(f,G,B)$ satisfy conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$. Suppose the following statements hold:

(i) 

$F:R^n\to 2^{R^n}$ is strictly monotone and upper hemicontinuous on $R^n$;

(ii) 

$F:R^n\to 2^{R^n}$ is an upper semicontinuous set-valued map with nonempty closed convex values;

(iii) 

For any $q\in G\left(\mathrm{\Omega }\right)$, $K_{\infty }\cap {\{x\in R^n:q+F\left(x\right)\cap K\ne \varnothing \}}^{-}=\left\{0\right\};$

(iv) 

The interior of $barr\left(K\right)$ is nonempty.

Then, DSIVI (2) has a Carathéodory weak solution.

Proof. It follows from conditions (i), (iii), (iv), and Lemma 6 that (5) holds. By condition (ii) and Lemma 5, we know there exists a constant ${\rho }^{\mathbb{F}}>0$ such that (3) holds, where $\mathbb{F}$ is defined by (4). Applying Theorem 1, we get DSIVI (2) has a Carathéodory weak solution. This completes the proof. □

Remark 3.

From the above proof, it is easy to see that $u\in L^2([0,T];R^n)$.

Theorem 3.

Assume conditions $\left(ii\right)$–$\left(iv\right)$ in Theorem 2 hold and $F:R^n\to 2^{R^n}$ is strongly monotone and upper hemicontinuous on $R^n$. Then, DSIVI (2) has a unique Carathéodory weak solution $(x,u)\in C([0,T];R^m)\times L^2([0,T];R^n)$.

Proof. By Theorem 2, we know DSIVI (2) has Carathéodory weak solutions. Now, we only need to prove the uniqueness of the Carathéodory weak solution for DSIVI (2). For this purpose, we let $(x_1,u_1)$ and $(x_2,u_2)$ be the Carathéodory weak solutions for DSIVI (2). Therefore,

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_1\left(t\right)=x_0+{\int }_0^{t}f(\tau ,x_1\left(\tau \right))+B(\tau ,x_1\left(\tau \right))u_1\left(\tau \right)d\tau ,\mathrm{for} \mathrm{any}t\in [0,T]\hfill \\ u_1\left(t\right)\in SOL(K,G(t,x_1\left(t\right))+F(·)),\mathrm{for} \mathrm{almost} \mathrm{all}t\in [0,T]\hfill \\ x_1\left(0\right)=x_0,\hfill \end{array}\right.\end{array}$$

(12)

and

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_2\left(t\right)=x_0+{\int }_0^{t}f(\tau ,x_2\left(\tau \right))+B(\tau ,x_2\left(\tau \right))u_2\left(\tau \right)d\tau ,\mathrm{for} \mathrm{any}t\in [0,T]\hfill \\ u_2\left(t\right)\in SOL(K,G(t,x_2\left(t\right))+F(·)),\mathrm{for} \mathrm{almost} \mathrm{all}t\in [0,T]\hfill \\ x_2\left(0\right)=x_0.\hfill \end{array}\right.\end{array}$$

(13)

For almost all $t\in [0,T]$, it is easy to see $u_1\left(t\right)\in SOL(K,G(t,x_1\left(t\right))+F(·))$ and $u_2\left(t\right)\in SOL(K,G(t,x_2\left(t\right))+F(·))$. Therefore, there exists a measurable E on $[0,T]$ with $mE=0$ ($mE$ denotes the Lebesgue measure of the set E on $[0,T]$) such that for any $t\in [0,T]\setminus E,$ there exists $u_1^{*}\left(t\right)\in F\left(u_1\left(t\right)\right)$ and $G(t,x_1\left(t\right))+u_1^{*}\left(t\right)\in$K such that

$$\begin{array}{c}\hfill \langle y-G(t,x_1\left(t\right))-u_1^{*}\left(t\right),u_1\left(t\right)\rangle \ge 0,\forall y\in K\end{array}$$

(14)

and there exists $u_2^{*}\left(t\right)\in F\left(u_2\left(t\right)\right)$ and $G(t,x_2\left(t\right))+u_2^{*}\left(t\right)\in$K such that

$$\begin{array}{c}\hfill \langle y-G(t,x_2\left(t\right))-u_2^{*}\left(t\right),u_2\left(t\right)\rangle \ge 0,\forall y\in K.\end{array}$$

(15)

For $t\in [0,T]\setminus E,$ letting $y=G(t,x_2\left(t\right))+u_2^{*}\left(t\right)$ in (14), we get

$$\langle G(t,x_2\left(t\right))+u_2^{*}\left(t\right)-G(t,x_1\left(t\right))-u_1^{*}\left(t\right),u_1\left(t\right)\rangle \ge 0.$$

For $t\in [0,T]\setminus E,$ letting $y=G(t,x_1\left(t\right))+u_1^{*}\left(t\right)$ in (15), we get

$$\langle G(t,x_1\left(t\right))+u_1^{*}\left(t\right)-G(t,x_2\left(t\right))-u_2^{*}\left(t\right),u_2\left(t\right)\rangle \ge 0.$$

Therefore, for $t\in [0,T]\setminus E,$ one has

$$\langle G(t,x_1\left(t\right))+u_1^{*}\left(t\right)-G(t,x_2\left(t\right))-u_2^{*}\left(t\right),u_2\left(t\right)-u_1\left(t\right)\rangle \ge 0,$$

and

$$\langle G(t,x_1\left(t\right))-G(t,x_2\left(t\right)),u_2\left(t\right)-u_1\left(t\right)\rangle \ge \langle u_1^{*}\left(t\right)-u_2^{*}\left(t\right),u_1\left(t\right)-u_2\left(t\right)\rangle .$$

Since F is strongly monotone on $R^n$, it yields for almost all $t\in [0,T],$

$$\begin{array}{c}\hfill \langle u_1^{*}\left(t\right)-u_2^{*}\left(t\right),u_1\left(t\right)-u_2\left(t\right)\rangle \ge \mu {\parallel u_1\left(t\right)-u_2\left(t\right)\parallel }^2.\end{array}$$

(16)

From the Cauchy–Schwarz inequality, we know that

$$\begin{array}{ccc}\hfill & & \langle G(t,x_1\left(t\right))-G(t,x_2\left(t\right)),u_2\left(t\right)-u_1\left(t\right)\rangle \hfill \\ & \le & \parallel G(t,x_1\left(t\right))-G(t,x_2\left(t\right))\parallel \parallel u_1\left(t\right)-u_2\left(t\right)\parallel .\hfill \end{array}$$

(17)

Therefore, combining (16) and (17), we get for almost all $t\in [0,T],$

$$\begin{array}{ccc}\hfill \mu \parallel u_1\left(t\right)-u_2\left(t\right)\parallel & \le & \parallel G(t,x_1\left(t\right))-G(t,x_2\left(t\right))\parallel \hfill \\ & \le & L_G\parallel x_1\left(t\right)-x_2\left(t\right)\parallel \hfill \end{array}$$

(18)

and

$$\begin{array}{ccc}\hfill \parallel u_1\left(t\right)-u_2\left(t\right)\parallel & \le & \frac{L_G}{\mu }\parallel x_1\left(t\right)-x_2\left(t\right)\parallel .\hfill \end{array}$$

(19)

Furthermore, from (12), (13), and (18), we infer that for any $t\in [0,T],$

$$\begin{array}{ccc}& & \parallel x_1\left(t\right)-x_2\left(t\right)\parallel \hfill \\ & =& {\int }_0^t\parallel f(\tau ,x_1\left(\tau \right))+B(\tau ,x_1\left(\tau \right))u_1\left(\tau \right)-f(\tau ,x_2\left(\tau \right))+B(\tau ,x_2\left(\tau \right))u_2\left(\tau \right)\parallel d\tau \hfill \\ & \le & {\int }_0^t\parallel f(\tau ,x_1\left(\tau \right))-f(\tau ,x_2\left(\tau \right))\parallel d\tau +{\delta }_B{\int }_0^t\parallel u_1\left(\tau \right)-u_2\left(\tau \right)\parallel d\tau \hfill \\ & \le & L_f{\int }_0^t\parallel x_1\left(\tau \right)-x_2\left(\tau \right)\parallel d\tau +{\delta }_B\frac{L_G}{\mu }{\int }_0^t\parallel x_1\left(\tau \right)-x_2\left(\tau \right)\parallel d\tau \hfill \\ & \le & (L_f+{\delta }_B\frac{L_G}{\mu }){\int }_0^t\parallel x_1\left(\tau \right)-x_2\left(\tau \right)\parallel d\tau .\hfill \end{array}$$

Apparently, there exists a constant $C=L_f+{\delta }_B\frac{L_G}{\mu }>0$ such that

$$\parallel x_1\left(t\right)-x_2\left(t\right)\parallel \le C{\int }_0^t\parallel x_1\left(\tau \right)-x_2\left(\tau \right)\parallel d\tau .$$

According to the Gronwall inequality, we get $x_1\left(t\right)=x_2\left(t\right)$ for all $t\in [0,T]$, so $x_1=x_2$ in $C([0,T],R^m)$. From (18), we have $u_1\left(t\right)=u_2\left(t\right)$ in $R^n$ for almost all $t\in [0,T]$. This means $u_1=u_2$ in $L^2([0,T],R^n)$. This completes the proof. □

4. Stability for DSIVI (2)

In this section, we aim to study the stability for DSIVI (2) in finite dimensional spaces when both the mapping and the constraint set are perturbed by two different parameters. For this purpose, we consider the parametric DSIVI, denoted by DSIVI ($L\left(p\right),$$G(t,x(t\left)\right)+F(·,\lambda )$), as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}\left(t\right)=f(t,x\left(t\right))+B(t,x\left(t\right))u\left(t\right)\hfill \\ u\left(t\right)\in SOL\left(L\right(p),G(t,x\left(t\right))+F(·,\lambda \left)\right)\hfill \\ x\left(0\right)=x_0,\hfill \end{array}\right.\end{array}$$

(20)

where $(Z_1,d_1)$ and $(Z_2,d_2)$ are two metric spaces. The nonempty closed convex subset K of $R^n$ in DSIVI (2) is perturbed by a parameter p, which varies over $(Z_1,d_1)$. Therefore, K is a perturbed set. That means $L:Z_1\to 2^{R^n}$ is a set-valued mapping with nonempty closed convex values. The mapping $F:R^n\to 2^{R^n}$ is a set-valued mapping that is perturbed by a parameter $\lambda$, and $\lambda$ varies over $(Z_2,d_2)$. That is to say, $F:R^n\times Z_2\to 2^{R^n}$. In what follows, to simplify notation, we let $S(p,\lambda )$ denote the Carathéodory weak solution for DSIVI (20). Next, we will establish the closedness and continuity of the mapping $(p,\lambda )\to S(p,\lambda )$.

Theorem 4.

Let $(f,G,B)$ satisfy conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right),$ $p_0\in Z_1$, ${\lambda }_0\in Z_2$ be two given points. Assume the following conditions hold.

(i) 

$L:Z_1\to 2^{R^n}$ is a continuous set-valued mapping with nonempty bounded closed convex values and $\bigcup _{p\in U\left(p_0\right)}L\left(p\right)$ is compact, where $U\left(p_0\right)$ is a neighborhood of $p_0$;

(ii) 

$F:R^n\times Z_2\to 2^{R^n}$ is an upper semicontinuous set-valued mapping with nonempty closed convex values on $R^n\times Z_2$ and lower semicontinuous on $Z_2$;

(iii) 

There exists a neighborhood Λ of ${\lambda }_0,$ for each $\lambda \in \Lambda$, the mapping $x\mapsto q+F(x,\lambda )$ is upper hemicontinuous and monotone for any $q\in G\left(\mathrm{\Omega }\right)$;

(iv) 

The set $SOL(L\left(p_0\right),q+F(·,{\lambda }_0))$ is nonempty and bounded for any $q\in G\left(\mathrm{\Omega }\right)$;

(v) 

$F:R^n\times Z_2\to 2^{R^n}$ is strictly monotone and upper hemicontinuous on $R^n$ .

Then, $S(p,\lambda )$ is closed at $(p_0,{\lambda }_0)\in Z_1\times Z_2$.

Proof. From Theorem 3.2 in [21], we know that condition $\left(iv\right)$ is equivalent to conditions $\left(iii\right)$ and $\left(iv\right)$ in Theorem 2. By conditions $\left(i\right)$–$\left(iv\right)$, it follows from Lemma 4 that there exists a neighborhood $P^{\prime }\times {\Lambda }^{\prime }$ of $(p_0,{\lambda }_0)$, $P^{\prime }\times {\Lambda }^{\prime }\subset P\times \Lambda$, such that for each $(p,\lambda )\in P^{\prime }\times {\Lambda }^{\prime }$, the set $SOL\left(L\right(p),q+F(·,\lambda \left)\right)$ is nonempty and bounded. It follows from Lemma 6 that there exists a constant $\rho >0$ such that (5) holds for any $q\in G\left(\mathrm{\Omega }\right)$. It is obvious that DSIVI (20) has solutions by Theorem 2.

Now, we prove $S(p,\lambda )$ is closed at $(p_0,{\lambda }_0)$. Let $\left\{(p_n,{\lambda }_n)\right\}\subset P\times \Lambda$ be a given sequence with $(p_n,{\lambda }_n)\to (p_0,{\lambda }_0)$ and $(x_n,u_n)\in S(p_n,{\lambda }_n)$ with $(x_n,u_n)\to (x_0,u_0)$ in $C([0,T];R^m)\times L^2([0,T],R^n).$ Therefore,

(a)

For any $0\le s\le t\le T,$

$$x_n\left(t\right)-x_n\left(s\right)={\int }_s^{t}f(\tau ,x_n\left(\tau \right))+B(\tau ,x_n\left(\tau \right))u_n\left(\tau \right)d\tau ;$$

(b)

For almost all $t\in [0,T]$, there exists $u_n^{*}\left(t\right)\in F(u_n\left(t\right),{\lambda }_n)$ and $G(t,x_n\left(t\right))+u_n^{*}\left(t\right)\in$$L\left(p_n\right)$, for any $y_n\in L\left(p_n\right)$, such that

$$\langle y_n-G(t,x_n\left(t\right))-u_n^{*}\left(t\right),u_n\left(t\right)\rangle \ge 0;$$

(c)

The initial condition

$$x_n\left(0\right)=x_0.$$

From the convergence $u_n$ converges to $u_0$ in $L^2([0,T],R^n)$, we obtain

$${\int }_{[0,T]}{\parallel u_n\left(t\right)-u_0\left(t\right)\parallel }^{2}dt<\infty$$

and

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{\parallel u_n-u_0\parallel }_{L^2}& =& \underset{n\to \infty }{lim}({\int }_{[0,T]}\parallel u_n\left(t\right)-u_0\left(t\right){{\parallel }^{2}dt)}^{\frac{1}{2}}\hfill \\ & =& 0.\hfill \end{array}$$

Moreover, applying the Holder inequality, we know

$$\begin{array}{ccc}\hfill {\int }_{[0,T]}\parallel u_n\left(t\right)-u_0\left(t\right)\parallel dt& \le & ({\int }_{[0,T]}\parallel u_n\left(t\right)-u_0\left(t\right){{\parallel }^{2}dt)}^{\frac{1}{2}}{\left({\int }_{[0,T]}{1}^{2}dt\right)}^{\frac{1}{2}}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{\int }_{[0,T]}\parallel u_n\left(t\right)-u_0\left(t\right)\parallel dt& \le & \underset{n\to \infty }{lim}({\int }_{[0,T]}\parallel u_n\left(t\right)-u_0\left(t\right){{\parallel }^{2}dt)}^{\frac{1}{2}}=0.\hfill \end{array}$$

This means $u_n$ converges to $u_0$ in $L^1([0,T],R^n)$, which is equivalent to $\parallel u_n-u_0{\parallel }_{L^1}\to 0$. By Theorem 4.9 in [28], there exists a sequence $u_n\left(t\right)$ and a function $h\in L^1$ such that

$$\begin{array}{c}\hfill u_n\left(t\right)\to u_0\left(t\right),\mathrm{for} \mathrm{almost} \mathrm{all}t\in [0,T]\end{array}$$

(21)

and

$$\begin{array}{c}\hfill \parallel u_n\left(t\right)\parallel \le h\left(t\right),\mathrm{for} \mathrm{almost} \mathrm{all}t\in [0,T].\end{array}$$

(22)

Combining (21) and (22), by the Lebesgue control convergence theorem, we know

$$\underset{n\to \infty }{lim}{\int }_{[0,T]}{u}_n\left(t\right)dt={\int }_{[0,T]}{u}_0\left(t\right)dt.$$

However, from $\left(b\right)$, it is easy to see that $G(t,x_n\left(t\right))+u_n^{*}\left(t\right)\in L\left(p_n\right)$ for almost all $t\in [0,T]$. By condition $\left(i\right)$, there exists a neighborhood $U\left(p_0\right)$ of $p_0$ such that $\bigcup _{p\in U\left(p_0\right)}L\left(p\right)$ is compact. Therefore, $\left\{u_n^{*}\left(t\right)\right\}$ has a subsequence, denoted again by $\left\{u_n^{*}\left(t\right)\right\}$, such that $u_n^{*}\left(t\right)\to u_0^{*}\left(t\right)$. Since $(u_n,{\lambda }_n)\to (u_0,{\lambda }_0),$ it follows from Lemma 1 and condition $\left(ii\right)$ that $u_0^{*}\left(t\right)\in F(u_0\left(t\right),{\lambda }_0)$. Moreover, the lower semicontinuity of L implies that, for any $y\in L\left(p_0\right)$, there exists a sequence $\left\{y_n\right\}$ with $y_n\in L\left(p_n\right)$ such that $y_n\to y$.

Now, by $\left(a\right)$, $\left(b\right)$, and $\left(c\right)$, we have

(a′)

For any $0\le s\le t\le T,$

$$x_0\left(t\right)-x_0\left(s\right)={\int }_s^{t}f(\tau ,x_0\left(\tau \right))+B(\tau ,x_0\left(\tau \right))u_0\left(\tau \right)d\tau ;$$

(b′)

For almost all $t\in [0,T]$, there exists $u_0^{*}\left(t\right)\in F(u_0\left(t\right),{\lambda }_0)$ and $G(t,x_0\left(t\right))+u_0^{*}\left(t\right)\in$$L\left(p_0\right)$, for any $y\in L\left(p_0\right)$, such that

$$\langle y-G(t,x_0\left(t\right))-u_0^{*}\left(t\right),u_0\left(t\right)\rangle \ge 0;$$

(c′)

The initial condition

$$x_0\left(0\right)=x_0.$$

Therefore, it deduces that $(x_0,u_0)\in S(p_0,{\lambda }_0)$. This completes the proof. □

Theorem 5.

Let $(f,G,B)$ satisfy conditions $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$; $p_0\in Z_1$, ${\lambda }_0\in Z_2$ are given points. Assume the following conditions hold.

(i) 

$L:Z_1\to 2^{R^n}$ is a continuous set-valued mapping with nonempty bounded closed convex values, and there exists a neighborhood $U\left(p_0\right)$ of $p_0$ such that $\bigcup _{p\in U\left(p_0\right)}L\left(p\right)$ is compact;

(ii) 

$F:R^n\times Z_2\to 2^{R^n}$ is a upper semicontinuous set-valued mapping with nonempty closed convex values on $R^n\times Z_2$ and lower semicontinuous on $Z_2$;

(iii) 

For each $\lambda \in \Lambda$ and $q\in G\left(\mathrm{\Omega }\right)$, the mapping $x\mapsto q+F(x,\lambda )$ is upper hemicontinuous and monotone, where Λ is a neighborhood of ${\lambda }_0$;

(iv) 

There exists a neighborhood $U(p_0,{\lambda }_0)$ of $(p_0,{\lambda }_0)$ such that

$$\bigcup _{(p,\lambda )\in U(p_0,{\lambda }_0)}SOL(L\left(p\right),q+F(·,\lambda ))$$

is bounded for any $q\in G\left(\mathrm{\Omega }\right)$;

(v) 

$F:R^n\times Z_2\to 2^{R^n}$ is strongly monotone and upper hemicontinuous on $R^n$.

Then, $S(p,\lambda )$ is continuous at $(p_0,{\lambda }_0)\in Z_1\times Z_2$.

Proof. From Theorem 3.2 in [21], we know that condition $\left(iv\right)$ is equivalent to conditions $\left(iii\right)$ and $\left(iv\right)$ in Theorem 2. It follows from Theorem 3 that $S(p,\lambda )$ is a singleton by conditions $\left(i\right),\left(ii\right),\left(iv\right),$ and $\left(v\right)$. Let $S(p_n,{\lambda }_n)=(x_n,u_n)$ with $(p_n,{\lambda }_n)\to (p_0,{\lambda }_0)$. Next, we need to prove sequence $\left\{x_n\right\}$ and $\left\{u_n\right\}$ have convergent subsequences, respectively.

Step 1.$\left\{x_n\right\}$ is uniformly bounded.

It is known that $(x_n,u_n)\in S(p_n,{\lambda }_n)$. Therefore, for almost all $t\in [0,T]$,

$$\begin{array}{c}\hfill {\dot{x}}_n\left(t\right)=f(t,x_n\left(t\right))+B(t,x_n\left(t\right))u_n\left(t\right),n=1,2,\cdots .\end{array}$$

(23)

Since $\mathbb{F}(t,x_n)$ defined by (4) satisfies (3), for any $t\in [0,T]$, we have

$$\parallel x_n\left(t\right)\parallel \le \parallel x_0\parallel +{\int }_0^t{\rho }^{\mathbb{F}}(1+\parallel x_n\left(s\right)\parallel )ds.$$

Applying the Gronwall inequality, we know

$$\parallel x_n\left(t\right)\parallel \le (\parallel x_0\parallel +{\rho }^{\mathbb{F}}T)e^{{\rho }^{\mathbb{F}}T}.$$

Clearly, $\left\{x_n\right\}$ is uniformly bounded with $\parallel x\parallel =\underset{t\in [0,T]}{sup}\parallel x\left(t\right)\parallel$.

Step 2.$\left\{x_n\right\}$ is an equicontinuous family of functions. Since $(x_n,u_n)\in S(p_n,{\lambda }_n),$ for almost all $t\in [0,T],$$u_n\left(t\right)\in SOL(L\left(p_n\right),G(t,x_n\left(t\right))+F(·,{\lambda }_n)).$ By condition $\left(iv\right)$, for almost all $t\in [0,T]$ and $n=1,2,\cdots$, there exists a constant C such that $\parallel u_n\left(t\right)\parallel \le C$.

In reality, (23) means for all $0\le s\le t\le T$,

$$x_n\left(t\right)-x_n\left(s\right)={\int }_s^{t}f(\tau ,x_n\left(\tau \right))+B(\tau ,x_n\left(\tau \right))u_n\left(\tau \right)d\tau ,n=1,2,\cdots .$$

We note that f is a Lipschitz continuous function on $\mathrm{\Omega },$ so for all $(t,x)\in \mathrm{\Omega }$, there exists a constant ${\rho }_f>0$ such that

$$\begin{array}{c}\hfill \parallel f(t,x)\parallel \le {\rho }_f(1+\parallel x\parallel ).\end{array}$$

(24)

Since B is bounded on $\mathrm{\Omega }$ and $\left\{u_n\left(t\right)\right\}$ is bounded for almost all $t\in [0,T]$, by (24), we have

$$\begin{array}{ccc}& & \parallel x_n\left(t\right)-x_n\left(s\right)\parallel \hfill \\ & =& \parallel {\int }_s^{t}f(\tau ,x_n\left(\tau \right))+B(\tau ,x_n\left(\tau \right))u_n\left(\tau \right)d\tau \parallel \hfill \\ & \le & {\int }_s^t\parallel f(\tau ,x_n\left(\tau \right))\parallel d\tau +{\int }_s^t\parallel B(\tau ,x_n\left(\tau \right))u_n\left(\tau \right)\parallel d\tau \hfill \\ & \le & {\int }_s^t\parallel f(\tau ,x_n\left(\tau \right))\parallel d\tau +{\int }_s^t\parallel B(\tau ,x_n\left(\tau \right))\parallel \parallel u_n\left(\tau \right)\parallel d\tau \hfill \\ & \le & {\int }_s^t{\rho }_f(1+\parallel x_n\left(\tau \right)\parallel )d\tau +{\delta }_{B}C|t-s|\hfill \\ & \le & {\rho }_f|t-s|+{\rho }_f(\parallel x_0\parallel +{\rho }^{\mathbb{F}}T)e^{{\rho }^{\mathbb{F}}T}|t-s|+{\delta }_{B}C|t-s|\hfill \\ & \le & ({\rho }_f(1+(\parallel x_0\parallel +{\rho }^{\mathbb{F}}T)e^{{\rho }^{\mathbb{F}}T})+{\delta }_{B}C\left)\right|t-s|.\hfill \end{array}$$

Let $M={\rho }_f(1+(\parallel x_0\parallel +{\rho }^{\mathbb{F}}T)e^{{\rho }^{\mathbb{F}}T})+{\delta }_{B}C$. Therefore, there exists a constant M such that, for any $n=1,2,\cdots$,

$$\parallel x_n\left(t\right)-x_n\left(s\right)\parallel \le M|t-s|.$$

Then, sequence $\left\{x_n\right\}$ is equicontinuous. We can apply the Arzelà–Ascoli theorem to deduce that $\left\{x_n\right\}$ has a subsequence, denoted again by $\left\{x_n\right\}$, which converges to $x_0$.

Step 3.$S(p_n,{\lambda }_n)\to S(p_0,{\lambda }_0)$ in $C([0,T];R^m)\times L^2([0,T];R^n).$ We know that $u_n\left(t\right)\in SOL(L\left(p_n\right),G(t,x_n\left(t\right))+F(·,{\lambda }_n))$ for almost all $t\in [0,T]$. Then, there exists a measure E with $mE=0$ such that $u_n\left(t\right)\in SOL(L\left(p_n\right),G(t,x_n\left(t\right))+F(·,{\lambda }_n))$ for any $t\in [0,T]\setminus E.$ That is, for any $t\in [0,T]\setminus E$, there exists $u_n^{*}\left(t\right)\in F(u_n\left(t\right),{\lambda }_n)$ and $G(t,x_n\left(t\right))+u_n^{*}\left(t\right)\in$$L\left(p_n\right)$ such that

$$\begin{array}{c}\hfill \langle y-G(t,x_n\left(t\right))-u_n^{*}\left(t\right),u_n\left(t\right)\rangle \ge 0,\forall y\in L\left(p_n\right).\end{array}$$

(25)

Take any small h such that $t+h\in [0,T]\setminus E$ and $u_n(t+h)\in SOL(L\left(p_n\right),G(t+h,x_n(t+h))+F(·,{\lambda }_n))$. Then, there exists $u_n^{*}(t+h)\in F(u_n(t+h),{\lambda }_n)$ and $G(t+h,x_n(t+h))+u_n^{*}(t+h)\in$$L\left(p_n\right)$ such that, for any $t+h\in [0,T]\setminus E$,

$$\begin{array}{c}\hfill \langle y-G(t+h,x_n(t+h))-u_n^{*}(t+h),u_n(t+h)\rangle \ge 0,\forall y\in L\left(p_n\right).\end{array}$$

(26)

For any $t+h\in [0,T]\setminus E$, letting $y=G(t+h,x_n(t+h))+u_n^{*}(t+h)$ in (25), we have

$$\langle G(t+h,x_n(t+h))+u_n^{*}(t+h)-G(t,x_n\left(t\right))-u_n^{*}\left(t\right),u_n\left(t\right)\rangle \ge 0.$$

For any $t\in [0,T]\setminus E$, letting $y=G(t,x_n\left(t\right))+u_n^{*}\left(t\right)$ in (26), we have

$$\langle G(t,x_n\left(t\right))+u_n^{*}\left(t\right)-G(t+h,x_n(t+h))-u_n^{*}(t+h),u_n(t+h)\rangle \ge 0.$$

Therefore,

$$\begin{array}{ccc}& & \langle G(t+h,x_n(t+h))-G(t,x_n\left(t\right)),u_n\left(t\right)-u_n(t+h)\rangle \hfill \end{array}$$

(27)

$$\begin{array}{ccc}& \ge & \langle u_n^{*}\left(t\right)-u_n^{*}(t+h),u_n\left(t\right)-u_n(t+h)\rangle .\hfill \end{array}$$

(28)

By the monotonicity of F,

$$\langle u_n^{*}\left(t\right)-u_n^{*}(t+h),u_n\left(t\right)-u_n(t+h)\rangle \ge \mu {\parallel u_n\left(t\right)-u_n(t+h)\parallel }^2.$$

Applying the Cauchy–Schwarz inequality, we get

$$\begin{array}{ccc}& & \langle G(t+h,x_n(t+h))-G(t,x_n\left(t\right)),u_n\left(t\right)-u_n(t+h)\rangle \hfill \\ & \le & \parallel G(t+h,x_n(t+h))-G(t,x_n\left(t\right))\parallel \parallel u_n\left(t\right)-u_n(t+h)\parallel .\hfill \end{array}$$

Thus, combining the above two inequalities and applying the Lipschitz continuity of G, we obtain

$$\begin{array}{ccc}\hfill \mu \parallel u_n\left(t\right)-u_n(t+h)\parallel & \le & \parallel G(t+h,x_n(t+h))-G(t,x_n\left(t\right))\parallel \hfill \\ & \le & L_G\left(\right|h|+\parallel x_n(t+h)-x_n\left(t\right)\parallel ),\hfill \end{array}$$

which means

$$\begin{array}{ccc}\hfill \parallel u_n\left(t\right)-u_n(t+h)\parallel & \le & \frac{L_G}{\mu }\left(\right|h|+\parallel x_n(t+h)-x_n\left(t\right)\parallel )\hfill \\ & \le & \frac{L_G}{\mu }\left(\right|h|+M\left|h\right|)\hfill \\ & \le & \frac{L_G}{\mu }(M+1)\left|h\right|.\hfill \end{array}$$

(29)

Let $l=\frac{L_G}{\mu }(M+1)$. Since $\left\{x_n\right\}$ is equicontinuous, it follows from (29) that for any $ϵ>0$, there exists $\delta =min\{T,\frac{ϵ}{l\sqrt{2T}}\}$ such that, for all $n=1,2,\dots$ and all $\left|h\right|\le \delta$,

$$\begin{array}{ccc}\hfill {\int }_0^{T-h}{\parallel u_n(t+h)-u_n\left(h\right)\parallel }^{2}dt& \le & {\int }_0^{T-h}{l}^2{h}^{2}dt\hfill \\ & \le & l^2{h}^2(T-h)\hfill \\ & \le & l^2{\delta }^2(T+\delta )\hfill \\ & <& {ϵ}^2.\hfill \end{array}$$

(30)

It is known that $\parallel u_n{\parallel }_{L^2}=({\int }_{[0,T]}\parallel u_n\left(t\right){{\parallel }^{2}dt)}^{\frac{1}{2}}<\infty$, which means $\left\{u_n\right\}$ is bounded in $L^2[0,T]$. Applying inequality (30) and the boundedness of $\left\{u_n\right\}$, by Corollary 1.34 in [29], we get that the sequence $\left\{u_n\right\}$ is relatively compact in $L^2[0,T]$. We can obtain the closure of $\left\{u_n\right\}$ is compact. Therefore, $\left\{u_n\right\}$ exists a convergent subsequence, denoted again by $\left\{u_n\right\}$, which converges to $u_0$. Up to now, we get subsequence $(x_n,u_n)=S(p_n,{\lambda }_n)$ with $x_n\to x_0$ and $u_n\to u_0$. From Theorem 4, $S(p,\lambda )$ is closed at $(p_0,{\lambda }_0)$. This means $(x_n,u_n)\to (x_0,u_0)=S(p_0,{\lambda }_0)$ and so $S(p,\lambda )$ is continuous at $(p_0,{\lambda }_0).$ This completes the proof. □

5. An Example of a Time-Dependent Spatial Price Equilibrium Control Problem

In this section, we will give an example of the differential inverse variational inequality to the time-dependent spatial price equilibrium control problem. As discussed by Scrimali [15], assume that a single commodity is produced at m supply market, with typical supply market denoted by i, and is consumed at n demand markets, with typical demand market denoted by j, during the time interval $[0,T]$ with $T>0$. Let $(i,j)$ denote the typical pair of producers and consumers for $i=1,2,\cdots ,m$ and $j=1,2,\cdots ,n$. Let $S_i\left(t\right)$ be the supply of the commodity produced at supply market i at time $t\in [0,T]$ and group the supplies into a column vector

$$S\left(t\right)=(S_1\left(t\right),\cdots S_m\left(t\right))\in R^m.$$

Let $D_j\left(t\right)$ be the demand of the commodity associated with demand market j at time $t\in [0,T]$ and group the demands into a column vector

$$D\left(t\right)=(D_1\left(t\right),\cdots D_n\left(t\right))\in R^n.$$

Let $x_{ij}\left(t\right)$ be the commodity shipment from supply market i to demand market j at time $t\in [0,T]$ and group the commodity shipments into a column vector $x\left(t\right)\in R^{mn}$.

Li et al. [6] studied the time-dependent spatial price equilibrium control problem by establishing the relation between the problem and a differential inverse variational inequality. We restate it here with a concise version.

Assume that, for any $t\in [0,T]$,

$$S_i\left(t\right)=\sum _{j=1}^n{x}_{ij}\left(t\right),D_j\left(t\right)=\sum _{i=1}^m{x}_{ij}\left(t\right)$$

and resource exploitations $S\left(x\right(t),u(t\left)\right)$ at supply market and consumption $D\left(x\right(t),u(t\left)\right)$ at demands market can be controlled by adjusting the tax $u\left(t\right).$ Let

$$W(t,x(t),u(t\left)\right)=\left(S\right(x\left(t\right),u\left(t\right)),D(x\left(t\right),u\left(t\right)\left)\right),\forall t\in [0,T],$$

which can be written as

$$W(t,x(t),u(t\left)\right)=G(t,x(t\left)\right)+F\left(u\right(t\left)\right),$$

where $G(t,x)$ is a Carathéodory function with $\gamma \left(t\right)\in L^2[0,T]$ such that

$$\parallel G(t,x)\parallel \le \gamma \left(t\right)+\parallel x\parallel .$$

and F is a continuous mapping. Let

$$L=\{w\in L^2([0,T],R^{m+n}):\underline{w}\left(t\right)\le w\left(t\right)\le \overline{w}\left(t\right)\mathrm{for} \mathrm{almost} \mathrm{all}t\in [0,T]\}$$

be the set of a feasible state influenced by the adjusted taxes $u\left(t\right),$ where $\underline{w}\left(t\right)=(\underline{S}\left(t\right),\underline{D}\left(t\right))$ and $\overline{w}\left(t\right)=(\overline{S}\left(t\right),\overline{D}\left(t\right))$ denote the lower and upper capacity constraints, respectively. Under some appropriate assumptions, finding the solution of a time-dependent optimal control equilibrium problem is equivalent to finding the Carathéodory solution $\left(x\right(t),u(t\left)\right)$ for the following differential inverse variational inequality:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}\left(t\right)=f(t,x\left(t\right))+B(t,x\left(t\right))u\left(t\right)\hfill \\ u\left(t\right)\in SOL(-L,-G(t,x\left(t\right))-F(·\left)\right)\hfill \\ x\left(0\right)=x_0,\hfill \end{array}\right.\end{array}$$

For more details, we refer the reader to [6].

However, the total amount of supply for a commodity and the relevant tax adjustments policy on the markets always vary with the sales season and the off-season [21]. In real life, any minute change in the proportion of each strategy seen will lead to a change in strategy. Let 0 denote the off-season and 1 denote the sales season. During the off-season, policy-makers will motivate manufacturers to develop resources by lowering the taxes they need to bear. During the sales season, policy-makers resist more development resources by increasing taxes on manufacturers. That means the set L of a feasible state is influenced by a parameter $p,$ where $p\in \{0,1\}$. Because the supply and demand of the commodity are also influenced by the seasons, we assume the mapping F is influenced by a parameter $\lambda ,$ where $\lambda \in \{0,1\}.$ Now, the time-dependent spatial price equilibrium control problem can be transformed into the following differential inverse variational inequality including parameters:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\dot{x}\left(t\right)=f(t,x\left(t\right))+B(t,x\left(t\right))u\left(t\right)\hfill \\ u\left(t\right)\in SOL(-L(p),-G(t,x\left(t\right))-F(·,\lambda \left)\right)\hfill \\ x\left(0\right)=x_0,\hfill \end{array}\right.\end{array}$$

Therefore, the time-dependent spatial price equilibrium control problem influenced by seasons will lead to a stability problem for a class of differential inverse variational inequalities.

6. Conclusions

The paper is concerned with the stability analysis of differential set-valued inverse variational inequalities in finite dimensional spaces. First, we proved an important result about a set-valued mapping, Lemma 5, which extends Lemma 2.5 in [6] and plays an important role in proving the existence of Carathéodory weak solutions for DSIVI (2). Then we obtained the existence of Carathéodory weak solutions for DSIVI (2). Second, we established closedness and continuity for the differential set-valued inverse variational inequality problem when the constraint set and the mapping are perturbed by two different parameters. Finally, we gave an example of a time-dependent spatial price equilibrium control problem, which can be transformed into a differential inverse variational inequality in finite dimensional spaces.

For further research, we can note the following directions: First, to adapt the main methods to study the existence of Carathéodory weak solutions and stability for differential set-valued inverse mixed variational inequalities in finite dimensional spaces; second, to use the theory of semigroups, set-valued mappings, and variational inequality to study the partial differential set-valued inverse variational inequalities in Banach spaces.