## 2 Problem Formulation and Lipschitz Constant Estimates
It is sufficient to prove that if \(\widetilde{\alpha}<1\), then  
\[\inf_{z\in(0,2\widetilde{\alpha})}\frac{z(1+z)}{s(1-s)}=\frac{1}{\widetilde {\alpha}}.\]  
Since \(\widetilde{\alpha}>0\), we have  
\[(1+\widetilde{\alpha})z^{2}-2\widetilde{\alpha}z+1+\widetilde{\alpha}>0, \qquad\forall\,z\in\mathbb{R},\]  
thus  
\[(1+\widetilde{\alpha})z^{2}\Big{[}(1+\widetilde{\alpha})z^{2}-2\widetilde{ \alpha}z+1+\widetilde{\alpha}\Big{]}>0,\qquad\forall\,z>0,\]  
On the other hand, we have  
\[(1+\widetilde{\alpha})z^{2}\Big{[}(1+\widetilde{\alpha})z^{2}-2 \widetilde{\alpha}z+1+\widetilde{\alpha}\Big{]}\] \[\qquad=(1+\widetilde{\alpha})z^{2}4-2\widetilde{\alpha}(1+ \widetilde{\alpha})z^{3}+(1+\widetilde{\alpha})z^{2}z^{2}\] \[\qquad=(1+\widetilde{\alpha})z^{2}4+\widetilde{\alpha}^{2}z^{2}+ 1-2\widetilde{\alpha}(1+\widetilde{\alpha})z^{3}+2(1+\widetilde{\alpha})z^{2} -2\widetilde{\alpha}z-z^{2}+2\widetilde{\alpha}z-1\] \[\qquad=[(1+\widetilde{\alpha})z^{2}-\widetilde{\alpha}z+1]^{2}-z ^{2}+2\widetilde{\alpha}z-1\] \[\qquad=[(1+\widetilde{\alpha})z^{2}-\widetilde{\alpha}z+1]^{2}-z ^{2},\]hence  
\[[(1+\widetilde{\alpha})z^{2}-\widetilde{\alpha}z+1]^{2}>s^{2},\qquad\forall\;z>0.\]  
Since \(s>0\) and \((1+\widetilde{\alpha})z^{2}-\widetilde{\alpha}z+1>0\) hold for any \(z>0\), we obtain  
\[(1+\widetilde{\alpha})z^{2}-\widetilde{\alpha}z+1>s,\qquad\forall\;z>0.\]  
The latter inequality is equivalent to  
\[(1+\widetilde{\alpha})z^{2}-2\widetilde{\alpha}z+1+\widetilde{\alpha}>0, \qquad\forall\,z\in\mathbb{R},\]0  
Since \(s>0\) and \(1-s>0\) hold for any \(z\in(0,2\widetilde{\alpha})\), we obtain  
\[(1+\widetilde{\alpha})z^{2}-2\widetilde{\alpha}z+1+\widetilde{\alpha}>0, \qquad\forall\,z\in\mathbb{R},\]1  
On the other hand, we have  
\[(1+\widetilde{\alpha})z^{2}-2\widetilde{\alpha}z+1+\widetilde{\alpha}>0, \qquad\forall\,z\in\mathbb{R},\]2  
Therefore, the proof is complete. **Theorem 3** (Separable operator). : _Assume that \(F:[0,T]\times\mathbb{R}^{n}\to\mathbb{R}^{n}\) is separable, i.e,_  
\[(1+\widetilde{\alpha})z^{2}-2\widetilde{\alpha}z+1+\widetilde{\alpha}>0, \qquad\forall\,z\in\mathbb{R},\]3  
_where \(G:\mathbb{R}^{n}\to\mathbb{R}^{n}\) is \(\alpha\)-strongly monotone on \(\mathbb{R}^{n}\) and Lipschitz continuous on \(\mathbb{R}^{n}\) with constant \(L_{x}\), and \(H:[0,T]\to\mathbb{R}^{n}\) is Lipschitz continuous on \(\mathbb{R}^{n}\) with constant \(L_{t}\). 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 (3) holds._  
_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_  
\[\Lambda_{2}= \begin{cases}\inf_{z\in(0,2\hat{\alpha})}\biggl{[}\frac{M}{1-\hat {s}}+\frac{\hat{L}z(1+z)}{s(1-\hat{s})}\biggr{]},&\text{if }\hat{\alpha}<1,\\ M+2\sqrt{2M\hat{L}}+\hat{L},&\text{if }\hat{\alpha}=1,\end{cases} \]  
_where \(\hat{\alpha}=\alpha/L_{x}\), \(\hat{L}=L_{t}/L_{x}\) and \(\hat{s}=\sqrt{z^{2}-2\hat{\alpha}z+1}\)._  
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 \(x_{2}:=x(t_{2})\).