## 2 Problem Formulation and Lipschitz Constant Estimates
: _Assume that \(F:[0,T]\times\mathbb{R}^{n}\to\mathbb{R}^{n}\) is uniformly \(\alpha\)-strongly monotone on \(\mathbb{R}^{n}\) and \(F\) is Lipschitz continuous on \([0,T]\times\mathbb{R}^{n}\) with constant \(L\), i.e.,_  
\[\|F(t_{1},x_{1})-F(t_{2},x_{2})\|\leq L(|t_{1}-t_{2}|+\|x_{1}-x_{2}\|)\,,\qquad \forall x_{1},x_{2}\in\mathbb{R}^{n},\ \forall\ t_{1},t_{2}\in[0,T].\]  
_Moreover, we assume that \(K(t)\) is a closed and convex set for any \(t\in[0,T]\), and there exists \(M\geq 0\) such that_  
\[\|p_{K(t_{1})}(x)-p_{K(t_{2})}(x)\|\leq M|t_{1}-t_{2}|,\qquad\forall\ x\in \mathbb{R}^{n},\ \forall\ t_{1},t_{2}\in[0,T], \tag{3}\]  
_where \(p_{K(t)}(x)\) denotes the projection of \(x\) on the closed convex set \(K(t)\)._  
_Then, for any \(t\in[0,T]\), there exists a unique solution \(x(t)\) of the \(VI(F,K(t))\) and \(x(t)\) is Lipschitz continuous on \([0,T]\), with an estimated constant equal to_  
\[\Lambda_{1}= \begin{cases}\inf_{z\in(0,2\delta)}\biggl{[}\frac{M}{1-s}+\frac{z (1+z)}{s(1-s)}\biggr{]},&\text{if }\widetilde{\alpha}<1,\\ M+2\sqrt{2M}+1,&\text{if }\widetilde{\alpha}=1,\end{cases} \]  
_where \(\widetilde{\alpha}=\alpha/L\) and \(s=\sqrt{z^{2}-2\widetilde{\alpha}z+1}\)._  
Proof. : The existence and uniqueness of the solution \(x(t)\) follows from Theorem 1. Given \(t_{1},t_{2}\in[0,T]\), we denote \(x_{1}:=x(t_{1})\) and \(x_{2}:=x(t_{2})\). We know that  
\[x_{1}=p_{K(t_{1})}(x_{1}-\lambda F(t_{1},x_{1})),\qquad x_{2}=p_{K(t_{2})}(x_ {2}-\lambda F(t_{2},x_{2})),\]  
hold for any \(\lambda>0\), due to the characterization of the projection onto a closed convex set and the Brower fixed point theorem (see, e.g., [8]). Hence, we obtain the following estimate:  
\[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\] \[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_ {2}-\lambda F(t_{2},x_{2}))\|\] \[\qquad+\|p_{K(t_{1})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})} (x_{1}-\lambda F(t_{1},x_{1}))\|\] \[\leq M|t_{1}-t_{2}|+\|(x_{2}-\lambda F(t_{2},x_{2}))-(x_{1}- \lambda F(t_{1},x_{1}))\|, \tag{4}\]  
where the second inequality follows from assumption (3) and the non-expansiveness of the projection map. Moreover, we have  
\[\|(x_{2}-\lambda F(t_{2},x_{2}))-(x_{1}-\lambda F(t_{1},x_{1}))\| ^{2}\] \[\qquad=\|x_{2}-x_{1}\|^{2}+\lambda^{2}\|F(t_{2},x_{2})-F(t_{1},x _{1})\|^{2}-2\lambda\langle x_{2}-x_{1},F(t_{2},x_{2})-F(t_{1},x_{1})\rangle\] \[\qquad\leq\|x_{2}-x_{1}\hold for any \(x,y\in\mathbb{R}^{n}\) and \(t\in[0,T]\). Thus, we obtain  
\[1+\lambda^{2}L^{2}-2\lambda\alpha\geq 1+\lambda^{2}L^{2}-2\lambda L=(1-\lambda L)^{ 2}\geq 0,\]  
hence \(s:=\sqrt{1+\lambda^{2}L^{2}-2\lambda\alpha}\) is well defined. Notice that if \(\alpha<L\), then \(s>0\) for any \(\lambda>0\), while if \(\alpha=L\) then \(s>0\) for any \(\lambda\in(0,1/L)\cup(1/L,+\infty)\). Moreover, we have  
\[1+\lambda L=\sqrt{1+\lambda^{2}L^{2}+2\lambda L}>\sqrt{1+\lambda^{2}L^{2}-2 \lambda\alpha}=s,\]  
thus, if \(s>0\), we have  
\[\lambda^{2}L^{2}|t_{2}-t_{1}|^{2}\leq\bigg{[}\lambda L\bigg{(}\frac{1+\lambda L }{s}\bigg{)}|t_{2}-t_{1}|\bigg{]}^{2}. \tag{6}\]  
It follows from (5) and (6) that  
\[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\]0 \[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\]1 \[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\]2  
hence  
\[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\]3  
It follows from (4) and (7) that  
\[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\]4  
Notice that \(1-s>0\) if and only if \(\lambda\in(0,2\alpha/L^{2})\).