## 3 Main Results  
In this section, we construct two methods for approximating the solution of the variational inequality problem over the solution set of the monotone inclusion problem. We establish the strong convergence results of the methods in the settings of real Hilbert spaces. The following assumptions will be useful in achieving our main results:  
**Assumption 1**. : _(A\({}_{1}\)) Operator \(D:\mathcal{H}\to\mathcal{H}\) is monotone and \(L_{2}\)-Lipschitz continuous, and \(E:\mathcal{H}\to 2^{\mathcal{H}}\) is a maximal monotone operator._  
_The solution set denoted by \(\Omega=\{p^{\star}\in(D+E)^{-1}(0):\langle\mathcal{F}p^{\star},q-p^{\star} \rangle\geq 0,\forall q\in(D+E)^{-1}(0)\}\
eq\emptyset\)._  
_\(\mathcal{F}:\mathcal{H}\to\mathcal{H}\) is \(\eta\)-strongly monotone and \(L_{1}\)-Lipschitz continuous._  
_\(\{\alpha_{k}\}\subset(0,1)\), such that \(\lim_{k\to\infty}\alpha_{k}=0\) and \(\sum_{k=1}^{\infty}\alpha_{k}=\infty\). The positive sequence \(\{\epsilon_{k}\}\) satisfies \(\lim_{k\to\infty}\frac{\epsilon_{k}}{\alpha_{k}}=0\)._  
_Let \(0<s<s^{\prime}<1\), \(\{t_{k}\}\subset[0,\infty)\) with \(\lim_{k\to\infty}t_{k}=0\), \(\{s_{k}\}\subset[0,\infty)\) with \(\lim_{k\to\infty}s_{k}=0\), and \(q_{k}\subset[0,\infty)\) with \(\sum_{k=0}^{\infty}q_{k}<\infty\)._  
**Remark 1**. : _From (9) and assumption \((A_{4})\), it is not hard to see that_  
\[\lim_{k\to\infty}\phi_{k}\|p_{k}-p_{k-1}\|=0\text{ and }\lim_{k\to\infty} \frac{\phi_{k}}{\alpha_{k}}\|p_{k}-u_{k-1}\|=0.\]  
**Remark 2**. : _Obviously, the step size (16) properly contains some well-known step sizes considered in [12; 18; 32] and many others._  
**Lemma 6**. : _Assume that Assumption 1 holds and \(\{p_{k}\}\) is the sequence generated by Algorithm 1, then \(\{\lambda_{k}\}\) defined by (16) is well-defined, and \(\lim_{k\to\infty}\lambda_{k}=\lambda>0\)._  
Proof. : Since \(D\) is \(L_{2}\)-Lipschitz continuous, such that \(L_{1}>0\), \(s_{k}\geq 0\), then by (16), if \(Dw_{k}\
eq Dv_{k}\) for all \(k\geq 1\), we obtain  
\[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]  
The remaining part of the proof of this lemma is similar to that in [33], so we omit it here. **Lemma 7**. : _If assumption \((A_{5})\) is performed, then a positive integer \(K\) exists, such that_  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]  
Proof. : Since \(0<c_{1}<c_{1}^{\prime}<2\) and \(\lim_{k\to\infty}t_{k}=0\), then a positive integer \(K_{1}\) exists, such that  
\[0<c_{1}+t_{k}\leq c_{1}^{\prime}<2,\;\forall k\geq K_{1}.\]  
For \(0<s<s^{\prime}\leq 1\), \(\lim_{k\to\infty}s_{k}=0\) and \(\lim_{k\to\infty}\lambda_{k}=\lambda\), we have  
\[\lim_{k\to\infty}\biggl{(}1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}\biggr{)} =1-s>1-s^{\prime}>0,\]  
and this means that a positive integer \(K_{2}\) exists, such that  
\[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]Setting \(K=\max\{K_{1},K_{2}\}\), which means that  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}\in( 0,1),\;\forall k\geq K.\]  
**Lemma 8**. : _Suppose that Assumption 1 holds and \(\{z_{k}\}\) is the sequence generated from Algorithm 1. Then, for \(p^{\star}\in\Omega\), the following inequality holds:_  
\[\|z_{k}-p^{\star}\|^{2}\leq\|w_{k}-p^{\star}\|^{2}+\left(1-\frac{2}{c_{1}+t_{k }}\right)\|z_{k}-w_{k}\|^{2},\;\forall k\geq 1. \tag{17}\]  
Proof. : Since \(v_{k}=(I+\lambda_{k}E)^{-1}(I-\lambda_{k}D)w_{k}\), we have that \(w_{k}-\lambda_{k}Dw_{k}-v_{k}\in\lambda_{k}Ev_{k}\).