An approximate solution of a differential inclusion with maximal monotone operator

Abstract

The theory of differential inclusions has played a central role in many areas as biological systems, physical problems and population dynamics. The principle aim of our work is to compute explicitly the discrete approximate solution of a differential inclusion including a maximal monotone operator. Also we present a numerical application of our results for showing how to compute the discrete approximate solution of its corresponding differential inclusion.

Keywords:

• Differential inclusion
• maximal monotone operator
• discrete approximate solution
• Gronwall's lemma

2010 Mathematics Subject Classifications:

• 40A25
• 45L05
• 4A60

1. Introduction

The differential inclusion has been studied by S. Adly, A. Hantoute, T. Nguyen, M. Thera and L. B. Khiet in [1–3], and by Y. Shang [4–6] in the study of fixed-time stability in the field of control engineering. We refer the interested reader to [7, 8] for more details regarding related applications (mathematical questions tied to analysis dealing with porous media or biological systems, respectively).

We consider, in this work, the following differential inclusion: (1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) where $B:{\mathbb{R}}^N\rightrightarrows {\mathbb{R}}^N$ (resp. $g:{\mathbb{R}}^N\to {\mathbb{R}}^N$) is a maximal monotone operator (resp. a k-Lipschitz function), ${\mathbb{R}}^N$ is the Euclidean N-space, and $\mathrm{dom}B$ is the domain of B which is given by $\mathrm{dom}B:=\{x\in {\mathbb{R}}^N:Bx\ne \mathrm{\varnothing }\}.$ In 2006, Mordukhovich [9] contemplated the accompanying differential inclusion ${\nu }^{\mathrm{\prime }}(s)\in G(\nu (s))$ a.e $s\in [0,b]$ where G is a Lipschitz multifunction. In 2019, he used in the work [10] the same method to study the following differential inclusion: $\nu (s)\in -N_{C\left(s\right)}(\nu (s))\mathrm{a}.\mathrm{e}0\le s\le t,\nu (0)\in C(0),$ where $C(s)$ is the convex variable set continuously evolves in time and $N_{C(s)}(\cdot )$ means the normal cone operator of $C(s).$

We present in this work a method for obtaining the best approximate solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ). The existence of the solution of this inclusion is a classical result proved by Brezis in [11].

Let n be a positive integer, and $s_i=\frac{i}{n}$ for all $i=0,\dots ,n$. Consider the following approximate problem for (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ) (2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) such that for all $0\le i\le n-1$, ${\nu }_n^{\mathrm{\prime }}(s_i)=n({\nu }_n(s_{i+1})-{\nu }_n(s_i))$ and ${\nu }_n^{\mathrm{\prime }}(s)={\nu }_n^{\mathrm{\prime }}(s_i)\mathrm{for}\mathrm{all}s$ in the interval $[s_i,s_{i+1}[.$

This problem has many applications in optimization and fixed-point theory, and there are many papers that studied the exact and approximate solution (see, e.g. [12–14]). Our idea to prove the main result of this paper is to use this problem to find an approximate solution of our principle problem (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ).

Recall that, the unique discrete trajectory ${\nu }_n(s_i),$ $i=0,\dots ,n,$ of (2(2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) ) is given by $\begin{array}{rl}& {\nu }_n(s_0)={\nu }_0,{\nu }_n(s_{i+1})=\frac{1}{n}g({\nu }_n(s_i))+J_{\frac{1}{n}}^B{\nu }_n(s_i)\\ & \mathrm{for}\mathrm{all}0\le i\le n-1.\end{array}$ We say that ${\nu }_n(\cdot )$ is a piecewise extension of the unique discrete trajectory of (2(2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) ) which is defined on the sub-interval $[s_i,s_{i+1}[$ by the following relation: ${\nu }_n(s)={\nu }_n(s_i)-(s-s_i){\nu }_n^{\mathrm{\prime }}(s_i).$ We will prove that if ${\nu }_n(\cdot )$ is a piecewise extension of the unique discrete trajectory ${\nu }_n(s_i)$ of (2(2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) ) and $\nu (\cdot )$ is a solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ), then ${\nu }_n(\cdot )$ converges uniformly to $\nu (\cdot )$.

2. Background

In this section, we give some definitions and notations that are needed throughout this article.

The orthogonal projection mapping ${\mathrm{\Pi }}_X$ onto a set X is defined as ${\mathrm{\Pi }}_X\left(s\right)=\{z\in X:\Vert z-s\Vert =\underset{y\in X}{inf}\Vert y-s\Vert \}.$ Let B be an operator on ${\mathbb{R}}^N$. The domain (resp. the graph) of B are given respectively by $\begin{array}{rl}\mathrm{dom}B:& =\{x\in {\mathbb{R}}^N:Bx\ne \mathrm{\varnothing }\},\\ \mathrm{gph}B:& =\{(x,y)\in {\mathbb{R}}^N\times {\mathbb{R}}^N:y\in Bx\}.\end{array}$ The operator B is said to be monotone if $⟨y_1-y_2,x_1-x_2⟩\ge 0\mathrm{for}\mathrm{all}(x_i,y_i)\in \mathrm{gph}B,i=1,2,$ and we said that B is maximal monotone, if B is monotone and its graph does not have an extension to a graph of another monotone operator.

For $\lambda >0$, the resolvent of B is given by $J_{\lambda }^{B}x:=(\mathrm{id}+\lambda B{)}^{-1}\left(x\right),$ and the Yoshida approximation of B is given by $B_{\lambda }x:=\frac{1}{\lambda }(\mathrm{id}-J_{\lambda }^B)\left(x\right),$ where $\mathrm{id}:$ ${\mathbb{R}}^N$ $\to$ ${\mathbb{R}}^N$ stands for the identity mapping.

If B is maximal monotone, then $B_{\lambda }$ is Lipschitz continuous with constant $\frac{1}{\lambda }$ and maximal monotone. In this case, we have also for all $x\in {\mathbb{R}}^N,B_{\lambda }(x)\in B(J_{\lambda }^{B}x)$ and if $x\in \mathrm{dom}B$, then $\Vert B_{\lambda }(x)\Vert \le \Vert B^0(x)\Vert$ where $B^{0}x:=\{x^{\ast }\in Bx:\Vert x^{\ast }\Vert =\underset{z\in Bx}{min}\Vert z\Vert \}$ is the set of points of minimal norm in Bx.

A function w from an interval $[a,b]$ into ${\mathbb{R}}^N$ is called absolutely continuous, if for all positive real number $\epsilon >0,$ there exists a $\delta >0$ such that for every finite collection $\{[a_i,b_i]\}$ of disjoint sub-intervals of $[a,b]$, we have $\sum _i|a_i-b_i|\le \delta ⟹\sum _i|w\left(a_i\right)-w\left(b_i\right)|\le \epsilon .$ Finally, we recall the Gronwall's lemma [15, Lemma 4.1].

Lemma 2.1

Let b be a positive real number and $f(\cdot ),$ $g(\cdot )$ two functions in $L^1([0,b],\mathbb{R})$ such that the function $g(\cdot )$ is positive on $[0$,$b]$, and let w be an absolutely continuous function from $[0,b]$ into ${\mathbb{R}}^{+}$ such that $(1-\lambda )w^{\mathrm{\prime }}(t)\le f(t)w(t)+g(t)w^{\lambda }(t)\mathrm{a}.\mathrm{e}.t\in [0,b]$ with $0\le \lambda <1$. Then the following inequality is satisfied for all $0\le t\le b$ $\begin{array}{rl}& w^{1-\lambda }(t)\le w^{1-\lambda }(0)\exp \left({\int }_0^{t}f\left(\alpha \right)\mathrm{d}\alpha \right)\\ & +{\int }_0^t\exp \left({\int }_k^{t}f\left(\alpha \right)\mathrm{d}\alpha \right)g\left(s\right)\mathrm{d}s.\end{array}$

Proof.

Let ϵ be a positive real number. Put $w_{\epsilon }\left(s\right)={(w\left(s\right)+\epsilon )}^{(1-\lambda )}\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )$ The first derivative of $w_{\epsilon }$ is $\begin{array}{rl}& w_{\epsilon }^{\mathrm{\prime }}\left(s\right)\\ & =-f\left(s\right)\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha ){(w\left(s\right)+\epsilon )}^{(1-\lambda )}\\ & +\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )(1-\lambda )w^{\mathrm{\prime }}\left(s\right){(w\left(s\right)+\epsilon )}^{-\lambda }\end{array}$ After simplification, we obtain $\begin{array}{rl}{w}_{\epsilon }^{\mathrm{\prime }}\left(s\right)& =\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )\\ & \left[\frac{-f\left(s\right)w\left(s\right)+(1-\lambda )w^{\mathrm{\prime }}\left(s\right)-\epsilon f\left(s\right)}{{(w\left(s\right)+\epsilon )}^{\lambda }}\right].\end{array}$ Using the fact that $(1-\lambda )w^{\mathrm{\prime }}(t)\le f(t)w(t)+g(t)w^{\lambda }(t)\mathrm{a}.\mathrm{e}t\in [0,b]$, we get $\begin{array}{rl}{w}_{\epsilon }^{\mathrm{\prime }}\left(s\right)& \le \exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )\frac{{\left(w\left(s\right)\right)}^{\lambda }}{{(w\left(s\right)+\epsilon )}^{\lambda }}g\left(s\right)\\ & -\frac{\epsilon f\left(s\right)}{{(w\left(s\right)+\epsilon )}^{\lambda }}\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )\\ & \le \exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )\frac{{\left(w\left(s\right)\right)}^{\lambda }}{{(w\left(s\right)+\epsilon )}^{\lambda }}g\left(s\right)\\ & +\frac{\epsilon \left|f\left(s\right)\right|}{{(w\left(s\right)+\epsilon )}^{\lambda }}\exp \left({\int }_0^s\left|f\left(\alpha \right)\right|\mathrm{d}\alpha \right).\end{array}$ Hence $\begin{array}{rl}{w}_{\epsilon }^{\mathrm{\prime }}\left(s\right)& \le \exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )g\left(s\right)+{\epsilon }^{1-\lambda }\left|f\left(s\right)\right|\\ & \times \exp \left({\int }_0^s\left|f\left(\alpha \right)\right|\mathrm{d}\alpha \right)\end{array}$ Integrating this inequality from s = 0 to s = t, we get $\begin{array}{rl}& w_{\epsilon }\left(t\right)-w_{\epsilon }\left(0\right)\\ & \le {\int }_0^t\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )g\left(s\right)\mathrm{d}s\\ & +{\epsilon }^{1-\lambda }{\int }_0^t\left|f\left(s\right)\right|\exp \left({\int }_0^s\left|f\left(\alpha \right)\right|\mathrm{d}\alpha \right)\mathrm{d}s\\ & \le {\int }_0^t\exp \left({\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha \right)g\left(s\right)\mathrm{d}s\\ & +{\epsilon }^{1-\lambda }(\exp \left({\int }_0^t\left|f\left(\alpha \right)\right|\mathrm{d}\alpha \right)-1)\end{array}$ Taking $\epsilon ⟶0$ in the last inequality, we obtain $\begin{array}{rl}& w^{1-\lambda }\left(t\right)\exp (-{\int }_0^{t}f\left(\alpha \right)\mathrm{d}\alpha )-w^{1-\lambda }\left(0\right)\\ & \le {\int }_0^t\exp (-{\int }_0^{s}f\left(\alpha \right)\mathrm{d}\alpha )g\left(s\right)\mathrm{d}s.\end{array}$ Hence $\begin{array}{rl}& w^{1-\lambda }(t)\le w^{1-\lambda }(0)\exp \left({\int }_0^{t}f\left(\alpha \right)\mathrm{d}\alpha \right)\\ & +{\int }_0^t\exp \left({\int }_s^{t}f\left(\alpha \right)\mathrm{d}\alpha \right)g\left(s\right)\mathrm{d}s.\end{array}$

3. Main results

We present in this section our principle result (see Theorem 3.1) for which we give an approximate solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ).

Theorem 3.1

Let ${\nu }_n(\cdot )$ be a piecewise extension of the unique discrete trajectory ${\nu }_n(s_i),$ $i=0,\dots ,n,$ of (2(2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) ) such that ${\nu }_n(0)={\nu }_0$. Then ${\nu }_n(\cdot )$ converges uniformly to the solution $\nu (\cdot )$ of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ).

We need the following propositions to prove Theorem 3.1.

Proposition 3.2

If $\nu (\cdot )$ is the solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ), then the following inequalities are satisfied for every real number s in the interval $[0,1].$

1. $\Vert {\nu }^{\mathrm{\prime }}(s)\Vert \le e^k\Vert {\nu }^{\mathrm{\prime }}(0)\Vert$ and $\Vert \nu (s)\Vert \le e^k\Vert {\nu }^{\mathrm{\prime }}(0)\Vert +\Vert {\nu }_0\Vert ,$

2. $\Vert g(\nu (s))\Vert \le k{e}^k\Vert {\nu }^{\mathrm{\prime }}(0)\Vert +2k\Vert {\nu }_0\Vert +\Vert g({\nu }_0)\Vert ,$

3. $\Vert b(\nu (s))\Vert \le (k+1)e^k\Vert {\nu }^{\mathrm{\prime }}(0)\Vert +2k\Vert {\nu }_0\Vert +\Vert g({\nu }_0)\Vert ,$

where ${\nu }^{\mathrm{\prime }}(s)=g(\nu (s))-b(\nu (s))$ a.e. $s\in [0,1],$ $b(\nu (s))\in B(\nu (s)).$

Proof.

Since $\nu (\cdot )$ is a solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ), we have ${\nu }^{\mathrm{\prime }}(s)=g(\nu (s))-b(\nu (s))$ a.e. $s\in [0,1]$. So $\mathrm{\forall }t\in ]0,1[$ such that $s+t\le 1$, we have ${\nu }^{\mathrm{\prime }}(s+t)=g(\nu (s+t))-b(\nu (s+t))$ a.e. $s\in [0,1]$. This equation with the initial condition $\nu (0)={\nu }_0\in \mathrm{dom}B$ has a unique solution $\nu (\cdot )$ which is absolutely continuous on $[0,1]$ (see, e.g. [11]).

Fix $s\in ]0,1[.$ As g is k-Lipschitz function and B is a maximal monotone operator, we have $\begin{array}{rl}& \frac{\mathrm{d}}{\mathrm{d}t}{\Vert \nu (s+t)-\nu \left(s\right)\Vert }^2\\ & =2⟨{\nu }^{\mathrm{\prime }}(s+t)-{\nu }^{\mathrm{\prime }}\left(s\right),\nu (s+t)-\nu \left(s\right)⟩\\ & =2⟨g\left(\nu (s+t)\right)-b\left(\nu (s+t)\right)-g\left(\nu \left(s\right)\right)-b\left(\nu \left(s\right)\right),\\ & \nu (s+t)-\nu \left(s\right)⟩\\ & =2⟨g\left(\nu (s+t)\right)-g\left(\nu \left(s\right)\right),\nu (s+t)-\nu \left(s\right)⟩\\ & -2⟨b\left(\nu (s+t)\right)-b\left(\nu \left(s\right)\right),\nu (s+t)-\nu \left(s\right)⟩\\ & \le 2k{\Vert \nu (s+t)-\nu \left(s\right)\Vert }^2.\end{array}$ Using Gronwall Lemma (see Lemma 2.1), we get $\begin{array}{rl}& {\Vert \nu (t+s)-\nu \left(s\right)\Vert }^2\le {\Vert \nu \left(t\right)-\nu \left(0\right)\Vert }^2{e}^{2ks}\\ & \mathrm{for}\mathrm{all}0\le t\le 1.\end{array}$ Therefore $\Vert \nu (t+s)-\nu \left(s\right)\Vert \le \Vert \nu \left(t\right)-\nu \left(0\right)\Vert e^{ks}.$ Dividing the last inequality by $\Vert t\Vert$ and taking $t\to 0$, we obtain that $\Vert {\nu }^{\mathrm{\prime }}\left(s\right)\Vert \le \Vert {\nu }^{\mathrm{\prime }}\left(0\right)\Vert {\mathrm{e}}^{ks}.$ As ${\nu }^{\mathrm{\prime }}(0)=g({\nu }_0)-b({\nu }_0)$, we get $\Vert {\nu }^{\mathrm{\prime }}\left(s\right)\Vert \le \Vert g\left({\nu }_0\right)-b\left({\nu }_0\right)\Vert e^k.$ Using the fact that $\nu (s)={\int }_0^s{\nu }^{\mathrm{\prime }}(t)\mathrm{d}t+{\nu }_0,$ we get $\begin{array}{rl}\Vert \nu \left(s\right)\Vert & \le \Vert {\int }_0^s{\nu }^{\mathrm{\prime }}\left(t\right)\mathrm{d}t\Vert +\Vert {\nu }_0\Vert \\ & \le {\int }_0^s\Vert {\nu }^{\mathrm{\prime }}\left(t\right)\Vert \mathrm{d}t+\Vert {\nu }_0\Vert \\ & \le e^k\Vert {\nu }^{\mathrm{\prime }}\left(0\right)\Vert +\Vert {\nu }_0\Vert .\end{array}$ By the Lipschitzianity of g, we get for all $s\in [0,1]$ : $\begin{array}{rl}\Vert g\left(\nu \left(s\right)\right)\Vert & =\Vert g\left(\nu \left(s\right)\right)-g\left({\nu }_0\right)+g\left({\nu }_0\right)\Vert \\ & \le \Vert g\left(\nu \left(s\right)\right)-g\left({\nu }_0\right)\Vert +\Vert g\left({\nu }_0\right)\Vert \\ & \le k\Vert \nu \left(s\right)-{\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert \\ & \le k\Vert \nu \left(s\right)\Vert +k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert \\ & \le k{e}^k\Vert {\nu }^{\mathrm{\prime }}\left(0\right)\Vert +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert .\end{array}$ Since ${\nu }^{\mathrm{\prime }}(s)=g(\nu (s))-b(\nu (s))$ a.e. $s\in [0,1]$, we obtain $b(\nu (s))=-{\nu }^{\mathrm{\prime }}(s)+g(\nu (s))$ a.e. $s\in [0,1]$. Therefore $\begin{array}{rl}& \Vert b\left(\nu \left(s\right)\right)\Vert \\ & =\Vert g\left(\nu \left(s\right)\right)-{\nu }^{\mathrm{\prime }}\left(s\right)\Vert \\ & \le \Vert g\left(\nu \left(s\right)\right)\Vert +\Vert {\nu }^{\mathrm{\prime }}\left(s\right)\Vert \\ & \le k{e}^k\Vert {\nu }^{\mathrm{\prime }}\left(0\right)\Vert +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert +e^k\Vert {\nu }^{\mathrm{\prime }}\left(0\right)\Vert \\ & \le (k+1)e^k\Vert {\nu }^{\mathrm{\prime }}\left(0\right)\Vert +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert .\end{array}$

Proposition 3.3

Let ${\nu }_n(\cdot )$ be a piecewise extension of the unique discrete trajectory ${\nu }_n(s_i)$ of (2(2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) ). Then the following inequalities are satisfied for every real number s in the interval $[0,1]$ and for all $0\le i\le n.$

1. $\Vert {\nu }_n^{\mathrm{\prime }}(s)\Vert \le e^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert$ and $\Vert {\nu }_n(s)\Vert \le e^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +\Vert {\nu }_0\Vert ,$

2. $\Vert g({\nu }_n(s))\Vert \le k{e}^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +2k\Vert {\nu }_0\Vert +\Vert g({\nu }_0)\Vert ,$

3. $\Vert B_{\frac{1}{n}}({\nu }_n(s_i))\Vert \le (k+1)e^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +2k\Vert \nu (0)\Vert +\Vert g({\nu }_0)\Vert .$

Proof.

Let ${\nu }_n(\cdot )$ be a piecewise extension of the unique discrete trajectory ${\nu }_n(s_i)$ of the following problem: $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i))-B_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0.\end{array}$ Then for all $0\le i\le n-1$, we have $\begin{array}{rl}& {\nu }_n(s_{i+1})\\ & ={\nu }_n(s_i)+{\displaystyle \frac{1}{n}}{\nu }_n^{\mathrm{\prime }}(s_i)\\ & ={\nu }_n(s_i)+{\displaystyle \frac{1}{n}}[g({\nu }_n(s_i))-B_{{\displaystyle \frac{1}{n}}}({\nu }_n(s_i))]\\ & ={\nu }_n(s_i)+{\displaystyle \frac{1}{n}}g({\nu }_n(s_i))-({\nu }_n(s_i)-J_{{\displaystyle \frac{1}{n}}}^B{\nu }_n(s_i)).\end{array}$ Therefore $\begin{array}{rl}& {\nu }_n(s_{i+1})=J_{\frac{1}{n}}^B{\nu }_n(s_i)+\frac{1}{n}g({\nu }_n(s_i))\\ & \mathrm{for}\mathrm{all}0\le i\le n-1.\end{array}$ Hence $\begin{array}{rl}& {\nu }_n(s_{i+2})-{\nu }_n(s_{i+1})\\ & =J_{\frac{1}{n}}^B{\nu }_n(s_{i+1})-J_{\frac{1}{n}}^B{\nu }_n(s_i)\\ & +\frac{1}{n}[g({\nu }_n(s_{i+1}))-g({\nu }_n(s_i))].\end{array}$ Since $J_{\lambda }^B$ (resp. g) is 1-Lipschitz (resp. k-Lipschitz) continuous function, $\begin{array}{rl}\Vert {\nu }_n(s_{i+2})-{\nu }_n(s_{i+1})\Vert & \le \Vert J_{\frac{1}{n}}^B{\nu }_n(s_{i+1})-J_{\frac{1}{n}}^B{\nu }_n(s_i)\Vert \\ & +\frac{1}{n}\Vert g({\nu }_n(s_{i+1}))-g({\nu }_n(s_i))\Vert \end{array}$ Hence $\begin{array}{rl}& \Vert {\nu }_n(s_{i+2})-{\nu }_n(s_{i+1})\Vert \\ & \le \Vert {\nu }_n(s_{i+1})-{\nu }_n(s_i)\Vert +\frac{k}{n}\Vert {\nu }_n(s_{i+1})-{\nu }_n(s_i)\Vert \end{array}$ That means $\Vert {\nu }_n(s_{i+2})-{\nu }_n(s_{i+1})\Vert \le (\frac{k}{n}+1)\Vert {\nu }_n(s_{i+1})-{\nu }_n(s_i)\Vert$ Consequently $\begin{array}{rl}\Vert {\nu }_n(s_{i+2})-{\nu }_n(s_{i+1})\Vert & \le (\frac{k}{n}+1)\Vert {\nu }_n(s_{i+1})-{\nu }_n(s_i)\Vert \\ & \le {(\frac{k}{n}+1)}^2\Vert {\nu }_n(s_i)-{\nu }_n(s_{i-1})\Vert \\ & ⋮\\ & \le {(\frac{k}{n}+1)}^{i+1}\Vert {\nu }_n(s_1)-{\nu }_n(s_0)\Vert \\ & \le {(\frac{k}{n}+1)}^n\Vert {\nu }_n(s_1)-{\nu }_n(s_0)\Vert .\end{array}$ Since $(\frac{k}{n}+1{)}^n\le e^k$, $\begin{array}{rl}& \Vert {\nu }_n(s_{i+2})-{\nu }_n(s_{i+1})\Vert \\ & \le e^k\Vert {\nu }_n(s_1)-{\nu }_n(s_0)\Vert \mathrm{for}\mathrm{all}0\le i\le n-2.\end{array}$ We deduce that $\begin{array}{rl}& \Vert n({\nu }_n(s_{i+2})-{\nu }_n(s_{i+1}))\Vert \\ & \le e^k\Vert n({\nu }_n(s_1)-{\nu }_n(s_0))\Vert \mathrm{for}\mathrm{all}0\le i\le n-2.\end{array}$ Therefore $\Vert {\nu }_n^{\mathrm{\prime }}(s_i)\Vert \le e^k\Vert {\nu }_n^{\prime }(0)\Vert \mathrm{for}\mathrm{all}1\le i\le n-1.$ Clearly the previous inequality is also true when i = 0, and since ${\nu }_n^{\prime }(s)={\nu }_n^{\prime }(s_i)$ for all $s\in [s_i,s_{i+1}[$ and $i=0,\dots ,n-1$, we get $\Vert {\nu }_n^{\mathrm{\prime }}(s)\Vert \le e^k\Vert {\nu }_n^{\prime }(0)\Vert \mathrm{for}\mathrm{all}s\in [0,1].$ For every $s\in [0,1],$ we have $y_n(s)={\int }_0^s{\nu }_n^{\mathrm{\prime }}(r)\mathrm{d}r+{\nu }_0.$ Then $\begin{array}{rl}\Vert {\nu }_n\left(s\right)\Vert & \le \Vert {\int }_0^s{\nu }_n^{\mathrm{\prime }}\left(r\right)dr\Vert +\Vert {\nu }_0\Vert \\ & \le {\int }_0^s\Vert {\nu }_n^{\mathrm{\prime }}\left(r\right)\Vert dr+\Vert {\nu }_0\Vert \\ & \le s{e}^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +\Vert {\nu }_0\Vert \\ & \le e^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +\Vert {\nu }_0\Vert .\end{array}$ By the Lipschitzianity of g, we get for every $s\in [0,1]$: $\begin{array}{rl}\Vert g\left({\nu }_n\left(s\right)\right)\Vert & =\Vert g\left({\nu }_n\left(s\right)\right)-g\left({\nu }_0\right)+g\left({\nu }_0\right)\Vert \\ & \le \Vert g\left({\nu }_n\left(s\right)\right)-g\left({\nu }_0\right)\Vert +\Vert g\left({\nu }_0\right)\Vert \\ & \le k\Vert {\nu }_n\left(s\right)-{\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert \\ & \le k\Vert {\nu }_n\left(s\right)\Vert +k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert \\ & \le k{e}^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert .\end{array}$ Since ${\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i))-B_{\frac{1}{n}}({\nu }_n(s_i))$ for all $0\le i\le n-1$, we obtain that $\begin{array}{rl}& \Vert B_{\frac{1}{n}}({\nu }_n(s_i))\Vert \\ & =\Vert g({\nu }_n(s_i))-{\nu }_n^{\mathrm{\prime }}\left(s_i\right)\Vert \\ & \le \Vert g({\nu }_n(s_i))\Vert +\Vert {\nu }_n^{\mathrm{\prime }}\left(s_i\right)\Vert \\ & \le (k+1)e^k\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert .\end{array}$

Proof

Proof of theorem 3.1

For all $s\in [s_i,s_{i+1}[$, we have $\begin{array}{rl}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}s}{\Vert {\nu }_n(s)-\nu (s)\Vert }^2& =⟨{\nu }_{{}_n}^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n(s)-\nu (s)⟩\\ & =⟨{\nu }_{{}_n}^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n(s_i)-\nu (s)⟩\\ & +⟨{\nu }_{{}_n}^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n(s)-{\nu }_n(s_i)⟩.\end{array}$ Since ${\nu }_n^{\mathrm{\prime }}(s)={\nu }_n^{\mathrm{\prime }}(s_i)$ for all $s\in [s_i,s_{i+1}[$ and $0\le i\le n-1$, we have clearly that: ${\nu }_n(s)={\nu }_n(s_i)-(s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)$ for all $s\in [s_i,s_{i+1}[$ and $0\le i\le n-1$. Therefore $\begin{array}{rl}\Vert {\nu }_n(s)-{\nu }_n(s_i)\Vert & \le (s-s_i)\Vert {\nu }_n^{\prime }(s_i)\Vert \\ & \le \frac{1}{n}\Vert {\nu }_n^{\prime }(s_i)\Vert .\end{array}$ Using the Proposition 3.3 and the fact that ${\nu }_n^{\mathrm{\prime }}(0)=g({\nu }_0)-B_{\frac{1}{n}}({\nu }_0)$ and $\Vert B_{\frac{1}{n}}({\nu }_0)\Vert$ is less than or equal $\Vert b({\nu }_0)\Vert$, we get $\begin{array}{rl}\Vert {\nu }_n(s)-{\nu }_n(s_i)\Vert & \le \frac{1}{n}{e}^k\Vert {\nu }_n^{\prime }(0)\Vert \\ & \le \frac{1}{n}{e}^k(\Vert g\left({\nu }_0\right)\Vert +\Vert b\left({\nu }_0\right)\Vert ).\end{array}$ Applying Proposition 3.2 and the last inequality, we obtain that $\begin{array}{rl}& ⟨{\nu }_{{}_n}^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n(s)-{\nu }_n(s_i)⟩\\ & \le \Vert {\nu }_{{}_n}^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s)\Vert \Vert {\nu }_n(s)-{\nu }_n(s_i)\Vert \\ & \le {\displaystyle \frac{2(m_1{)}^2}{n}},\end{array}$ where $m_1=e^k(\Vert g({\nu }_0)\Vert +\Vert b({\nu }_0)\Vert )$.

Using again Proposition 3.3, and the fact that $\Vert {\nu }_n^{\mathrm{\prime }}(0)\Vert \le \Vert g({\nu }_0)\Vert +\Vert b({\nu }_0)\Vert$, we get for all $i=1,\dots ,n-1$ (3) $\begin{array}{rl}& \Vert B_{\frac{1}{n}}({\nu }_n(s_i))\Vert \\ & \le (k+1)e^k(\Vert g\left({\nu }_0\right)\Vert +\Vert b\left({\nu }_0\right)\Vert )\\ & +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert .\end{array}$(3) On the other hand, we have for all $s\in [s_i,s_{i+1}[$: $\begin{array}{rl}& ⟨{\nu }_n^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n\left(s_i\right)-\nu (s)⟩\\ & =⟨g\left({\nu }_n\left(s_i\right)\right)-g(\nu (s)),{\nu }_n\left(s_i\right)-\nu (s)⟩\\ & -⟨B_{\frac{1}{n}}\left({\nu }_n\left(s_i\right)\right)-b(\nu (s)),{\nu }_n\left(s_i\right)-\nu (s)⟩\\ & =⟨g\left({\nu }_n\left(s_i\right)\right)-g(\nu (s)),{\nu }_n\left(s_i\right)-\nu (s)⟩\\ & -⟨B_{\frac{1}{n}}\left({\nu }_n\left(s_i\right)\right)-b(\nu (s)),J_{\frac{1}{n}}^B{\nu }_n\left(s_i\right)-\nu (s)⟩\\ & -⟨B_{\frac{1}{n}}\left({\nu }_n\left(s_i\right)\right)-b(\nu (s)),{\nu }_n\left(s_i\right)-J_{\frac{1}{n}}^B{\nu }_n\left(s_i\right)⟩.\end{array}$ Since $B_{\frac{1}{n}}({\nu }_n(s_i))\in B(J_{\frac{1}{n}}^B{\nu }_n(s_i))$ and B  is monotone, we can show that $⟨B_{\frac{1}{n}}({\nu }_n(s_i))-b(\nu (s)),J_{\frac{1}{n}}^B{\nu }_n(s_i)-\nu (s)⟩\ge 0.$ On the other hand, we have $\begin{array}{rl}& ⟨B_{\frac{1}{n}}({\nu }_n(s_i))-b(\nu (s)),{\nu }_n(s_i)-J_{\frac{1}{n}}^B{\nu }_n(s_i)⟩\\ & \le \Vert B_{\frac{1}{n}}({\nu }_n(s_i))-b(\nu (s))\Vert \Vert {\nu }_n(s_i)-J_{\frac{1}{n}}^B{\nu }_n(s_i)\Vert \\ & \le \frac{\Vert B_{\frac{1}{n}}({\nu }_n(s_i))\Vert +\Vert b(\nu (s))\Vert }{n}\Vert B_{\frac{1}{n}}({\nu }_n(s_i))\Vert .\end{array}$

From Proposition 3.2 and the inequality (3(3) $\begin{array}{rl}& \Vert B_{\frac{1}{n}}({\nu }_n(s_i))\Vert \\ & \le (k+1)e^k(\Vert g\left({\nu }_0\right)\Vert +\Vert b\left({\nu }_0\right)\Vert )\\ & +2k\Vert {\nu }_0\Vert +\Vert g\left({\nu }_0\right)\Vert .\end{array}$(3) ), we get $⟨B_{\frac{1}{n}}({\nu }_n(s_i))-b(\nu (s)),{\nu }_n(s_i)-J_{\frac{1}{n}}^B{\nu }_n(s_i)⟩\le \frac{2{\left(m_2\right)}^2}{n}.$ where $m_2=(k+1)e^k(\Vert g({\nu }_0)\Vert +\Vert b({\nu }_0)\Vert )+2k\Vert {\nu }_0\Vert +\Vert g({\nu }_0)\Vert .$

Therefore, for all real number s in the interval $[s_i,s_{i+1}]$ and $0\le i\le n-1$, $\begin{array}{rl}& ⟨{\nu }_n(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n(s_i)-\nu (s)⟩\\ & \le k\Vert {\nu }_n(s_i)-\nu (s)\Vert +{\displaystyle \frac{2(m_2{)}^2}{n}}.\end{array}$ As ${\nu }_n(s)={\nu }_n(s_i)-(s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)$ for all $s\in [s_i,s_{i+1}[$ and $i=0,\dots ,n-1$, we can obtain directly that $\begin{array}{rl}{\Vert {\nu }_n(s_i)-\nu (s)\Vert }^2& \le {\Vert {\nu }_n(s)-\nu (s)\Vert }^2+{\Vert (s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)\Vert }^2\\ & +2\Vert {\nu }_n(s)-\nu (s)\Vert \Vert (s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)\Vert .\end{array}$ Hence $\begin{array}{rl}& 2\Vert {\nu }_n(s)-\nu (s)\Vert \Vert (s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)\Vert \\ & \le {\Vert {\nu }_n(s)-\nu (s)\Vert }^2+{\Vert (s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)\Vert }^2.\end{array}$ So $\begin{array}{rl}& {\Vert {\nu }_n(s_i)-\nu (s)\Vert }^2\\ & \le 2{\Vert {\nu }_n(s)-\nu (s)\Vert }^2+2{\Vert (s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)\Vert }^2.\end{array}$ Hence for all $s\in [0,1]$, we have $\begin{array}{rl}& ⟨{\nu }_{{}_n}^{\mathrm{\prime }}(s)-{\nu }^{\mathrm{\prime }}(s),{\nu }_n(s)-\nu (s)⟩\\ & \le 2k{\Vert {\nu }_n(s)-\nu (s)\Vert }^2+2k{\Vert (s-s_i){\nu }_n^{\mathrm{\prime }}(s_i)\Vert }^2\\ & +{\displaystyle \frac{2{\left(m_2\right)}^2}{n}}+{\displaystyle \frac{2{\left(m_1\right)}^2}{n}}\\ & \le 2k{\Vert {\nu }_n(s)-\nu (s)\Vert }^2+{\displaystyle \frac{(2k+2){\left(m_1\right)}^2}{n}}+{\displaystyle \frac{2{\left(m_2\right)}^2}{n}}\\ & \le 2k{\Vert {\nu }_n(s)-\nu (s)\Vert }^2+{\displaystyle \frac{c}{n}},\end{array}$ where $c=(2k+2)(m_1{)}^2+2(m_2{)}^2.$

Consequently $\begin{array}{rl}& \frac{\mathrm{d}}{\mathrm{d}s}\frac{1}{2}\Vert {\nu }_n(s)-\nu (s){\Vert }^2\\ & \le 2k\Vert {\nu }_n(s)-\nu (s){\Vert }^2+{\displaystyle \frac{c}{n}},\mathrm{\forall }s\in [0,1].\end{array}$ From Gronwall's Lemma 2.1, we get for all $0\le s\le 1$ ${\Vert {\nu }_n(s)-\nu (s)\Vert }^2\le \left(\frac{c}{4kn}\right)e^{2kt}-\frac{c}{4kn}.$ Finally $\underset{s\in [0,1]}{sup}\Vert {\nu }_n(s)-\nu (s){\Vert }^2\le \left(\frac{c}{4kn}\right)e^{2k}-\frac{c}{4kn}$. Hence ${\nu }_n(s)\to \nu (s)$ as $n\to +\mathrm{\infty }.$

3.1. Numerical application

We can find an approximate solution of the differential inclusion (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ) by taking n large enough. Let ${\nu }_n(\cdot )$ be a piecewise extension of (2(2) $\begin{array}{l}{\nu }_n^{\mathrm{\prime }}(s_i)=g({\nu }_n(s_i)){-B}_{\frac{1}{n}}({\nu }_n(s_i))\mathrm{for}\mathrm{all}0\le i\le n-1\\ {\nu }_n(0)={\nu }_0,\end{array}$(2) ): $\begin{array}{rl}{\nu }_n(s_{i+1})& ={\displaystyle \frac{1}{n}}g({\nu }_n(s_i))+J_{{\displaystyle \frac{1}{n}}}^B{\nu }_n(s_i)\\ & \mathrm{for}\mathrm{all}0\le i\le n-1,\\ {\nu }_n(0)& ={\nu }_0.\end{array}$ We consider here $s_i:=\frac{i}{n}\in [0,1]\mathrm{for}\mathrm{all}0\le i\le n-1$. Then $\begin{array}{l}{\nu }_n(0)={\nu }_0(\mathrm{given})\\ {\nu }_1={\nu }_n\left({\displaystyle \frac{1}{n}}\right)={\displaystyle \frac{1}{n}}g({\nu }_0)+J_{{\displaystyle \frac{1}{n}}}^B({\nu }_0)\\ {\nu }_2={\nu }_n\left({\displaystyle \frac{2}{n}}\right)={\displaystyle \frac{1}{n}}g(x_1)+J_{{\displaystyle \frac{1}{n}}}^B({\nu }_1),\\ ⋮\\ {\nu }_j={\nu }_n\left({\displaystyle \frac{j}{n}}\right)={\displaystyle \frac{1}{n}}g({\nu }_j)+J_{{\displaystyle \frac{1}{n}}}^B({\nu }_j),\\ ⋮\\ {\nu }_n(1)={\displaystyle \frac{1}{n}}g({\nu }_{n-1})+J_{{\displaystyle \frac{1}{n}}}^B({\nu }_{n-1}).\end{array}$ The sequence $({\nu }_n(\cdot ){)}_n$ converges uniformally to $\nu (\cdot )$ the unique solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ). Then for n large enough, ${\nu }_n(s_i),$ $i=0,\dots ,n$ are approximate solution of (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ) in $s_i$.

We have ${\nu }_n(s)={\nu }_n(s_i)-(s-s_i){\nu }_n(s_i)$ for all $s\in [s_i,s_{i+1}[$ ${\Vert {\nu }_n(s)-\nu (s)\Vert }^2\le \left({\displaystyle \frac{c}{2kn}}\right)e^{2ks}-{\displaystyle \frac{c}{2kn}},\mathrm{\forall }s\in [0,1].$ Hence ${\nu }_n(\cdot )$ is an approximate solution to (1(1) $\begin{array}{l}{\nu }^{\mathrm{\prime }}(s)\in g(\nu (s))-B(\nu (s))\mathrm{a}.\mathrm{e}.s\in [0,1]\\ \nu (0)={\nu }_0\in \mathrm{dom}B,\end{array}$(1) ) for n large enough.

Example 3.4

Application to Electrical

In this example, we apply our result to the case of electrical circuits where current source is fixed. We discuss the circuit considered in [16]. $L{\nu }^{\mathrm{\prime }}(s)+R\nu (s)\in -{\mathrm{N}}_I(\nu (s))\mathrm{with}I=[c,+\mathrm{\infty }[.$ If $\nu (0)=0$ and L = R, then ${\nu }^{\mathrm{\prime }}(s)\in -\nu (s)-{\mathrm{N}}_I(\nu (s)).$

In this example, $g(\nu (s))=-\nu (s),$ $B(\nu (s))={\mathrm{N}}_I(\nu (s))$ and $J_{\frac{1}{n}}^B(\nu )={\mathrm{\Pi }}_I(\nu ).$

We have ${\nu }_n(s_i),$ $i=0,\dots ,n$ is a discrete trajectory of $\begin{array}{rl}{\nu }_n(s_{i+1})& ={\displaystyle \frac{1}{n}}g({\nu }_n(s_i))+{\mathrm{\Pi }}_I({\nu }_n(s_i)),\\ & i=0,\dots ,n-1,\\ {\nu }_n(0)& ={\nu }_0,\end{array}$ where $s_i=\frac{i}{n}\in [0,1]$ for all $i=0,\dots ,n$. Then

${\nu }_n(0)=0$ $(\mathrm{given})$

${\nu }_1={\nu }_n\left(\frac{1}{n}\right)=c$

${\nu }_2={\nu }_n\left(\frac{2}{n}\right)=-\frac{1}{n}c+{\mathrm{\Pi }}_I(c)=(1-\frac{1}{n})c,$

${\nu }_3={\nu }_n\left(\frac{3}{n}\right)=-\frac{1}{n}(1-\frac{1}{n})c+c=(1-\frac{1}{n}+\frac{1}{n^2})c,$

${\nu }_4={\nu }_n\left(\frac{4}{n}\right)=-\frac{1}{n}(1-\frac{1}{n}+\frac{1}{n^2})c+c=(1-\frac{1}{n}+\frac{1}{n^2}-\frac{1}{n^3})c,$

⋮

${\nu }_i={\nu }_n\left(\frac{i}{n}\right)=(1-\frac{1}{n}+\frac{1}{n^2}-\frac{1}{n^3}+\cdots +(-1{)}^{i-1}\frac{1}{n^{i-1}})c.$

For n large enough: $\begin{array}{rl}& {\nu }_n(s)\simeq \nu (s)\simeq \frac{n+{\left(\frac{-1}{n}\right)}^{i-1}}{n+1}c\\ & \mathrm{for}\mathrm{all}s\in [\frac{i}{n},\frac{i+1}{n}[\mathrm{and}0\le i\le n.\end{array}$

Example 3.5

We consider the differential inclusion (4) $\begin{array}{l}\dot{x}(t)\in x(t)-{\mathrm{N}}_{[0,+\mathrm{\infty })}(x(t))\mathrm{a}.\mathrm{e}t\in [0,1]\\ x(0)=1\in [0,+\mathrm{\infty }),\end{array}$(4) over the space of all absolutely continuous arcs $x:[0,1]\to \mathbb{R}.$ where ${\mathrm{N}}_{[0,+\mathrm{\infty })}(x)=\{y\in \mathbb{R}/y(x-z)\ge 0,\mathrm{\forall }z\in [0,+\mathrm{\infty })\}$

The differential inclusion (4(4) $\begin{array}{l}\dot{x}(t)\in x(t)-{\mathrm{N}}_{[0,+\mathrm{\infty })}(x(t))\mathrm{a}.\mathrm{e}t\in [0,1]\\ x(0)=1\in [0,+\mathrm{\infty }),\end{array}$(4) ) has a unique solution ${x(t)=e}^t.$

We take the following discrete approximations: (5) $\begin{array}{l}{\dot{x}}_n(t_j)=x_n(t_j),j=0,\dots ,n\\ x_n(0)=1\in [0,+\mathrm{\infty }),\end{array}$(5) where $t_j{:=jh}_n\in [0,1]$ as $j=0,\dots ,n$, $h_n=\frac{1}{n}$ such that

1. ${\dot{x}}_n(t_j)=\frac{x_n(t_{j+1})-x_n(t_j)}{h_n}$ for all $j=0,\dots ,n$.

2. ${\dot{x}}_n(t)={\dot{x}}_n(t_j),\mathrm{\forall }t\in [t_j,t_{j+1}[,$ for all $j=0,\dots ,n$.

3. $x_n(t)=x_n(t_j)-(t-t_j){\dot{x}}_n(t_j),\mathrm{\forall }t\in [t_j,t_{j+1}[,$ and $j=0,\dots ,n.$

We can write the differential inclusion (5(5) $\begin{array}{l}{\dot{x}}_n(t_j)=x_n(t_j),j=0,\dots ,n\\ x_n(0)=1\in [0,+\mathrm{\infty }),\end{array}$(5) ) as follows: $\begin{array}{l}{x}_n(t_{j+1})={\displaystyle \frac{1}{n}}{x}_n(t_j)+x_n(t_j),j=0,\dots ,n-1,\\ x_n(0)=1.\end{array}$ Clearly $x_n\left(\frac{1}{n}\right)=\frac{1}{n}+1$, $x_n(\frac{2}{n})={(\frac{1}{n}+1)}^2,\dots ,x_n\left(\frac{j}{n}\right)={(\frac{1}{n}+1)}^j$ and $x_n(1)={(\frac{1}{n}+1)}^n$, and if n = 1000, j = 500, we get $x_{1000}(0.5)={(\frac{1}{1000}+1)}^{500}\simeq 1,6483$ and $e^{0.5}\simeq 1,6486.$

Conclusion

We have studied, in this work, a differential inclusion including a maximal monotone operator, where a discrete approximate solution of an interesting problem has been given explicitly. The method used in this work can be applied in the future to a single-solution inclusions by selecting some discrete modulators.

Acknowledgments

The author express his deep gratitude to the referee for his/her meticulous reading and valuable suggestions which have definitely improved the paper.

Disclosure statement

No potential conflict of interest was reported by the author(s).