1 Introduction

Let \(\mathbb{R}^{n}_{+} = \{ x = (x_{1},x_{2}, \ldots,x_{n}) \in \mathbb{R}^{n} : x_{i} \geq0\ \forall i=1,2, \ldots, n \}\) and \(\mathbb{R}^{n}_{++} = \{ x = (x_{1},x_{2}, \ldots,x_{n}) \in\mathbb {R}^{n} : x_{i} > 0 \ \forall i=1,2, \ldots, n \}\). The variational inequality problem is to find

$$ x\in\Omega:= \bigl\{ (u,v) : u\in\mathbb{R}_{+}^{n}, v\in \mathbb{R}_{+}^{m}, A_{1}u + A_{2}v = b \bigr\} $$

such that

$$ \bigl(x'- x\bigr)^{T} F(x) \ge0, \quad \forall x'\in\Omega, $$
(1.1)

with

$$ x=\left ( \textstyle\begin{array}{c} u \\ v \end{array}\displaystyle \right ) \quad \mbox{and} \quad F(x)=\left ( \textstyle\begin{array}{c} f_{1}(u)\\ f_{2}(v) \end{array}\displaystyle \right ), $$
(1.2)

where \(A_{1}\in\mathbb{R}^{l\times n}\), \(A_{2} \in\mathbb{R}^{l\times m}\) are given matrices, \(b\in\mathbb{R}^{l}\) is a given vector, and \(f_{1}:\mathbb{R}_{+}^{n}\to \mathbb{R}^{n}\), \(f_{2}:\mathbb{R}_{+}^{m}\to\mathbb{R}^{m}\) are given monotone operators. For further study and applications of such problems, we refer to [1–11] and the references therein. By attaching a Lagrange multiplier vector \(\lambda\in\mathbb{R}^{l}\) to the linear constraints \(A_{1}u + A_{2}v =b\), the problem (1.1)-(1.2) can be explained in terms of finding \(z \in\mathcal{Z}' := \mathbb{R}_{+}^{n} \times\mathbb{R}_{+}^{m} \times \mathbb{R}^{l}\) such that

$$ \bigl(z'-z\bigr)^{\top} Q(z) \geq0, \quad \forall z' \in\mathcal{Z}', $$
(1.3)

where

$$ z=\left ( \textstyle\begin{array}{c} u \\ v\\ \lambda \end{array}\displaystyle \right ), \qquad Q(z)= \left ( \textstyle\begin{array}{c} f_{1}(u)-A_{1}^{\top} \lambda\\ f_{2}(v)-A_{2}^{\top} \lambda\\ A_{1}u + A_{2}v-b \end{array}\displaystyle \right ), $$
(1.4)

and \(A_{1}^{\top}\) denotes the transpose of the matrix \(A_{1}\). The problem (1.3)-(1.4) is referred as a structured variational inequality problem (in short, SVI).

Yuan and Li [12] developed the following logarithmic-quadratic proximal (LQP)-based decomposition method by applying the LQP terms to regularize the ADM subproblems: For a given \(z^{k}=(u^{k},v^{k},\lambda^{k})\in\mathbb{R}^{n}_{++}\times\mathbb {R}^{m}_{++}\times\mathbb{R}^{l}\), and \(\mu\in(0,1)\), the new iterative \((u^{k+1}, v^{k+1}, \lambda^{k+1})\) is obtained via solving the following system:

$$\begin{aligned}& f_{1}(u) - A_{1}^{\top} \bigl[ \lambda^{k} - H \bigl(A_{1} u + A_{2} v^{k} - b\bigr) \bigr] + R \bigl[ \bigl(u-u^{k}\bigr) + \mu \bigl(u^{k} - U_{k}^{2} u^{-1}\bigr) \bigr] = 0, \end{aligned}$$
(1.5)
$$\begin{aligned}& f_{2}(v) - A_{2}^{\top} \bigl[ \lambda^{k} - H (A_{1} {u} + A_{2} v - b) \bigr] + S \bigl[ \bigl(v-v^{k}\bigr) + \mu\bigl(v^{k} - V_{k}^{2} v^{-1}\bigr) \bigr] = 0, \end{aligned}$$
(1.6)
$$\begin{aligned}& \lambda^{k+1} = \lambda^{k} - H \bigl(A_{1} u^{k} + A_{2} v^{k} - b\bigr), \end{aligned}$$
(1.7)

where \(H \in\mathbb{R}^{l \times l}\), \(R \in\mathbb{R}^{n \times n}\), and \(S \in\mathbb{R}^{m \times m}\) are symmetric positive definite.

Later, some LQP alternating direction methods have been proposed to make the LQP alternating direction method more practical, see, for example, [13–17] and the references therein. Each iteration of these methods contains a prediction and a correction, the predictor is obtained via solving (1.5)-(1.7) and the new iterate is obtained by a convex combination of the previous point and the one generated by a projection-type method along a descent direction. The main disadvantage of the methods proposed in [12–17] is that solving equation (1.6) requires the solution of equation (1.5). To overcome with this difficulty, Bnouhachem and Hamdi [18] proposed a parallel descent LQP alternating direction method for solving SVI.

In this paper, we propose a parallel descent LQP alternating direction method for solving the following structured variational inequality with three separable operators: Find \(y\in\Omega:= \{ (u,v,w) : u\in\mathbb {R}_{+}^{n_{1}}, v\in\mathbb{R}_{+}^{n_{2}}, w\in\mathbb{R}_{+}^{n_{2}}, A_{1}u + A_{2}v +A_{3}w= b \}\) such that

$$ \bigl(y'- y\bigr)^{\top} F(y) \ge0, \quad \forall y'\in\Omega, $$
(1.8)

with

$$ y=\left ( \textstyle\begin{array}{c} u \\ v\\ w \end{array}\displaystyle \right ),\qquad F(y)= \left ( \textstyle\begin{array}{c} f_{1}(u)\\ f_{2}(v)\\ f_{3}(w) \end{array}\displaystyle \right ), $$
(1.9)

where \(A_{1}\in\mathbb{R}^{m\times n_{1}}\), \(A_{2} \in\mathbb{R}^{m\times n_{2}}\), \(A_{3}\in\mathbb{R}^{m\times n_{3}}\) are given matrices, \(b\in\mathbb{R}^{m}\) is a given vector, and \(f_{1}:\mathbb {R}_{+}^{n_{1}}\to\mathbb{R}^{n_{1}}\), \(f_{2}:\mathbb{R}_{+}^{n_{2}}\to\mathbb{R}^{n_{2}}\), \(f_{3}:\mathbb {R}_{+}^{n_{3}}\to\mathbb{R}^{n_{3}}\) are given monotone operators. By attaching a Lagrange multiplier vector \(\lambda\in\mathbb{R}^{m}\) to the linear constraints \(A_{1}u + A_{2}v +A_{3}w= b\), the problem (1.8)-(1.9) can be explained in terms of finding \(z \in\mathcal{Z} := \mathbb{R}_{+}^{n_{1}} \times\mathbb {R}_{+}^{n_{2}} \times\mathbb{R}_{+}^{n_{3}} \times\mathbb{R}^{m}\) such that

$$ \bigl(z'-z\bigr)^{\top} Q(z) \geq0, \quad \forall z' \in \mathcal{Z}, $$
(1.10)

where

$$ z=\left ( \textstyle\begin{array}{c} u\\ v\\ w\\ \lambda \end{array}\displaystyle \right ) \quad \mbox{and} \quad Q(z)=\left ( \textstyle\begin{array}{c} f_{1}(u)-A_{1}^{\top} \lambda\\ f_{2}(v)-A_{2}^{\top} \lambda\\ f_{3}(w)-A_{3}^{\top} \lambda\\ A_{1}u + A_{2}v +A_{3}w- b \end{array}\displaystyle \right ). $$
(1.11)

The problem (1.10)-(1.11) is referred as \(\mathrm{SVI}_{3}\).

The main aim of this paper is to present the parallel descent LQP alternating direction method for solving \(\mathrm{SVI}_{3}\) and to investigate the convergence rate of this method. We show that the proposed method has the \(O(1/t)\) convergence rate. The iterative algorithm and results presented in this paper generalize, unify, and improve the previously known results in this area.

2 The proposed method

For any vector \(u\in\mathbb{R}^{n}\), \(\Vert u\Vert _{\infty}=\max\{\vert u_{1}\vert , \ldots ,\vert u_{n}\vert \}\). Let \(D \in\mathbb{R}^{n \times n}\) be a symmetry positive definite matrix, we denote the D-norm of u by \(\Vert u\Vert _{D}^{2}=u^{T}Du\).

The following lemma provides a basic property of projection operator onto a closed convex subset Ω of \(\mathbb{R}^{l}\). We denote by \(P_{\Omega,D}(\cdot)\) the projection operator under the D-norm, that is,

$$P_{\Omega,D}(v)=\operatorname{argmin} \bigl\{ \Vert v-u\Vert _{D} : u \in\Omega\bigr\} . $$

Lemma 2.1

Let D be a symmetry positive definite matrix and Ω be a nonempty closed convex subset of \(\mathbb{R}^{l}\). Then

$$ \bigl(z-P_{\Omega,D}[z]\bigr)^{\top} D \bigl(P_{\Omega,D}[z]-v\bigr) \geq0, \quad \forall z \in\mathbb{R}^{l}, v\in\Omega. $$
(2.1)

We make the following standard assumptions.

Assumption 2.1

\(f_{1}\) is monotone with respect to \(\mathbb{R}^{n_{1}}_{+}\), that is, \((f_{1}(x)-f_{1}(y))^{T}(x-y)\ge0\), \(\forall x,y\in\mathbb{R}^{n_{1}}_{+}\), \(f_{2}\) is monotone with respect to \(\mathbb{R}^{n_{2}}_{+}\), and \(f_{3}\) is monotone with respect to \(\mathbb{R}^{n_{3}}_{+}\).

Assumption 2.2

The solution set of \(\mathrm{SVI}_{3}\), denoted by \(\mathcal {Z}^{*}\), is nonempty.

We propose the following parallel LQP alternating direction method for solving \(\mathrm{SVI}_{3}\):

Algorithm 2.1

  1. Step 0.

    Given \(\varepsilon>0\), \(\mu\in(0,1)\), \(\beta\in ( \frac {\sqrt{3}}{2},1 )\), \(\gamma\in(0,2)\) and \(z^{0}=(u^{0}, v^{0},w^{0}, \lambda^{0}) \in\mathbb{R}^{n_{1}}_{++}\times \mathbb{R}^{n_{2}}_{++}\times\mathbb{R}^{n_{3}}_{++}\times\mathbb{R}^{m}\). Set \(k=0\).

  2. Step 1.

    Compute \(\tilde{z}^{k}=(\tilde{u}^{k},\tilde{v}^{k},\tilde{w}^{k},\tilde {\lambda}^{k})\in\mathbb{R}^{n_{1}}_{++}\times \mathbb{R}^{n_{2}}_{++}\times\mathbb{R}^{n_{3}}_{++}\times\mathbb {R}^{m}\) by solving the following system:

    $$\begin{aligned}& f_{1}(u) - A_{1}^{\top} \bigl[\lambda^{k} - H \bigl(A_{1} u + A_{2} v^{k} +A_{3} w^{k}- b\bigr) \bigr] \\& \quad {} + R_{1}\bigl[\bigl(u-u^{k}\bigr)+ \mu\bigl(u^{k} - U_{k}^{2} u^{-1}\bigr)\bigr]=0, \end{aligned}$$
    (2.2)
    $$\begin{aligned}& f_{2}(v) - A_{2}^{\top} \bigl[ \lambda^{k} - H \bigl(A_{1} u^{k} + A_{2} v +A_{3} w^{k}- b\bigr)\bigr] \\& \quad {}+ {R_{2}}\bigl[ \bigl(v-v^{k}\bigr) + \mu\bigl(v^{k} - V_{k}^{2} v^{-1}\bigr)\bigr]=0, \end{aligned}$$
    (2.3)
    $$\begin{aligned}& f_{3}(w) - A_{3}^{\top} \bigl[ \lambda^{k} - H \bigl(A_{1} u^{k} + A_{2} v^{k}+A_{3} w - b\bigr)\bigr] \\& \quad {}+ {R_{3}}\bigl[ \bigl(w-w^{k}\bigr) + \mu\bigl(w^{k} - W_{k}^{2} w^{-1}\bigr)\bigr]=0, \end{aligned}$$
    (2.4)
    $$\begin{aligned}& \tilde{\lambda}^{k} = \lambda^{k} - \beta H \bigl(A_{1} \tilde{u}^{k} + A_{2} \tilde{v}^{k}+A_{3} \tilde{w}^{k} - b\bigr) , \end{aligned}$$
    (2.5)

    where \(H \in\mathbb{R}^{m \times m}\), \({R_{1}} \in\mathbb{R}^{n_{1} \times n_{1}}\), \({R_{2}} \in\mathbb{R}^{n_{2} \times n_{2}}\) and \({R_{3}} \in\mathbb{R}^{n_{3} \times n_{3}}\) are symmetric positive definite matrices. \(U_{k}\), \(V_{k}\), and \(W_{k}\) are positive definite diagonal matrices defined by \(U_{k} =\operatorname{diag} ( u^{k}_{1}, \ldots,u^{k}_{n} )\), \(V_{k} = \operatorname{diag} ( v^{k}_{1}, \ldots,v^{k}_{n} )\), \(W_{k} = \operatorname{diag} ( w^{k}_{1}, \ldots,w^{k}_{n} )\).

  3. Step 2.

    If \(\max\{\Vert u^{k}-\tilde{u}^{k}\Vert _{\infty}, \Vert v^{k}-\tilde{v}^{k}\Vert _{\infty}, \Vert w^{k}-\tilde{w}^{k}\Vert _{\infty}, \Vert \lambda^{k}-\tilde{\lambda}^{k}\Vert _{\infty}\}<\epsilon\), then stop.

  4. Step 3.

    The new iterate \(z^{k+1}(\tau_{k})=(u^{k+1},v^{k+1},w^{k+1},\lambda ^{k+1})\) is given by

    $$ z^{k+1}(\tau_{k})= (1-\sigma) z^{k}+ \sigma P_{\mathcal{Z},G}\bigl[z^{k}-\gamma \tau_{k}G^{-1}g \bigl(z^{k},\tilde{z}^{k}\bigr)\bigr], \quad\sigma\in(0,1), $$
    (2.6)

    where

    $$\begin{aligned}& \tau_{k}=\frac{\varphi(z^{k},\tilde{z}^{k})}{\Vert z^{k}- \tilde{z}^{k}\Vert ^{2}_{G}}, \end{aligned}$$
    (2.7)
    $$\begin{aligned}& \varphi\bigl(z^{k},\tilde{z}^{k}\bigr)=\bigl\Vert z^{k}- \tilde{z}^{k}\bigr\Vert _{M}^{2} \\& \hphantom{ \varphi\bigl(z^{k},\tilde{z}^{k}\bigr)=}{}+ \frac {1}{\beta}\bigl(\lambda^{k}-\tilde{\lambda}^{k} \bigr)^{T} \bigl(A_{1}\bigl(u^{k}- \tilde{u}^{k}\bigr) +A_{2}\bigl(v^{k}- \tilde{v}^{k}\bigr)+A_{3}\bigl(w^{k}- \tilde{w}^{k}\bigr) \bigr), \end{aligned}$$
    (2.8)
    $$\begin{aligned}& g(z^{k},\tilde{z}^{k}) \\& \quad = \left ( \textstyle\begin{array}{c} f_{1}(\tilde{u}^{k}) - A_{1}^{\top} \tilde{\lambda}^{k}+A_{1}^{\top} H[A_{1}(u^{k}-\tilde{u}^{k}) +A_{2}(v^{k}-\tilde{v}^{k})+A_{3}(w^{k}-\tilde{w}^{k}) +\frac{1-\beta}{\beta }H^{-1}(\lambda^{k}-\tilde{\lambda}^{k})]\\ f_{2}(\tilde{v}^{k}) - A_{2}^{\top} \tilde{\lambda}^{k}+ A_{2}^{\top} H[A_{1}(u^{k}-\tilde{u}^{k})+A_{2}(v^{k}-\tilde{v}^{k})+A_{3}(w^{k}-\tilde{w}^{k}) +\frac{1-\beta}{\beta}H^{-1}(\lambda^{k}-\tilde{\lambda}^{k})]\\ f_{3}(\tilde{w}^{k}) - A_{3}^{\top} \tilde{\lambda}^{k}+ A_{3}^{\top} H[A_{1}(u^{k}-\tilde{u}^{k})+A_{2}(v^{k}-\tilde{v}^{k})+A_{3}(w^{k}-\tilde{w}^{k}) +\frac{1-\beta}{\beta}H^{-1}(\lambda^{k}-\tilde{\lambda}^{k})]\\ A_{1}\tilde{u}^{k}+A_{2}\tilde{v}^{k}+A_{3}\tilde{w}^{k}-b \end{array}\displaystyle \right ), \\& \\& G=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} (1+\mu){R_{1}}+A_{1}^{\top} HA_{1} & 0& 0 &0 \\ 0 & (1+\mu){R_{2}} +A_{2}^{\top} HA_{2} & 0&0 \\ 0 &0 &(1+\mu){R_{3}} +A_{3}^{\top} HA_{3} & 0 \\ 0 & 0 &0 & \frac{1}{\beta}H^{-1} \end{array}\displaystyle \right ), \end{aligned}$$
    (2.9)

    and

    $$M=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} {R_{1}} +A_{1}^{\top}HA_{1} & 0& 0 &0 \\ 0 & {R_{2}} +A_{2}^{\top} HA_{2} & 0&0 \\ 0 &0 & {R_{3}} +A_{3}^{\top} HA_{3} & 0 \\ 0 & 0 &0 & \frac{1}{\beta}H^{-1} \end{array}\displaystyle \right ). $$

Set \(k:= k + 1\) and go to Step 1.

Remark 2.1

As special cases, we can obtain some new LQP alternating methods as follows:

  1. (a)

    If \(u^{k+1}=\tilde{u}^{k}\), \(v^{k+1}=\tilde{v}^{k}\), \(w^{k+1}=\tilde{w}^{k}\) and \(\lambda^{k+1}=\tilde{\lambda}^{k}\) in (2.2), (2.3), (2.4) and (2.5), respectively, we obtain a new method which can be viewed as an extension of that proposed in [10] for solving structured variational inequality with three separable operators in a parallel way.

  2. (b)

    If \(u^{k+1}=\tilde{u}^{k}\), \(v^{k+1}=\tilde{v}^{k}\), \(w^{k+1}=\tilde{w}^{k}\), \(\lambda^{k+1}=\tilde{\lambda}^{k}\), and \(\beta=1\) in (2.2), (2.3), (2.4) and (2.5), respectively, we obtain a new method which can be viewed as an extension of that proposed in [12] for solving structured variational inequality with three separable operators in a parallel wise.

  3. (c)

    If \(\beta=1\), the proposed method can be viewed as an extension of that proposed in [18] for solving structured variational inequality with three separable operators.

We need the following result in the convergence analysis of the proposed method.

Lemma 2.2

[12]

Let \(q(u)\in\mathbb{R}^{n}\) be a monotone mapping of u with respect to \(\mathbb{R}^{n}_{+}\) and \(R\in\mathbb{R}^{n\times n}\) be a positive definite diagonal matrix. For a given \(u^{k} >0\), if \(U_{k}:= \operatorname {diag}(u^{k}_{1},u^{k}_{2},\ldots, u^{k}_{n})\) (the diagonal matrix with elements \(u^{k}_{1}\), \(u^{k}_{2}\), … , \(u^{k}_{n}\)) and \(u^{-1}\) be an n-vector whose jth element is \(1/u_{j}\), then the equation

$$ q(u) + R\bigl[\bigl(u -u^{k}\bigr) + \mu\bigl( u^{k} - U_{k}^{2} u^{-1}\bigr)\bigr]=0 $$
(2.10)

has a unique positive solution u. Moreover, for any \(v\geq0\), we have

$$ (v-u)^{\top} q(u) \geq{ \frac{1+\mu}{2} } \bigl(\Vert u - v \Vert _{R}^{2} - \bigl\Vert u^{k} - v \bigr\Vert _{R}^{2} \bigr) + { \frac{1-\mu}{2} }\bigl\Vert u^{k} - u \bigr\Vert _{R}^{2}. $$
(2.11)

The next theorem is useful for the convergence analysis.

Theorem 2.1

For given \(z^{k}\in \mathbb{R}^{n_{1}}_{++}\times\mathbb{R}^{n_{2}}_{++}\times \mathbb{R}^{n_{3}}_{++}\times\mathbb{R}^{m}\), let \(\tilde{z}^{k}\) be generated by (2.2)-(2.5). Then

$$\begin{aligned} \varphi\bigl(z^{k},\tilde{z}^{k}\bigr)\geq \frac{2\beta-\sqrt{3}}{2\beta} \bigl\Vert z^{k}- \tilde{z}^{k}\bigr\Vert ^{2}_{G} \end{aligned}$$
(2.12)

and

$$\begin{aligned} \tau_{k}\geq\frac{2\beta-\sqrt{3}}{2\beta}. \end{aligned}$$
(2.13)

Proof

It follows from (2.8) that

$$\begin{aligned} \varphi\bigl(z^{k},\tilde{z}^{k}\bigr) = & \bigl\Vert z^{k}- \tilde{z}^{k}\bigr\Vert _{M}^{2}+\frac{1}{\beta}\bigl(\lambda^{k}-\tilde {\lambda}^{k}\bigr)^{\top}\bigl(A_{1} \bigl(u^{k}-\tilde{u}^{k}\bigr) +A_{2} \bigl(v^{k}-\tilde{v}^{k}\bigr)+A_{3} \bigl(w^{k}-\tilde{w}^{k}\bigr)\bigr) \\ = & \bigl\Vert u^{k}-\tilde{u}^{k}\bigr\Vert _{R_{1}}^{2}+\bigl\Vert A_{1}u^{k}-A_{1} \tilde{u}^{k}\bigr\Vert _{H}^{2}+\bigl\Vert v^{k}-\tilde{v}^{k}\bigr\Vert _{R_{2}}^{2} +\bigl\Vert A_{2}v^{k} -A_{2} \tilde{v}^{k} \bigr\Vert _{H}^{2} \\ &{}+\bigl\Vert w^{k}-\tilde{w}^{k}\bigr\Vert _{R_{3}}^{2}+\bigl\Vert A_{3}w^{k} -A_{3}\tilde{w}^{k}\bigr\Vert _{H}^{2} +\frac{1}{\beta}\bigl\Vert \lambda^{k}-\tilde{\lambda}^{k} \bigr\Vert _{H^{-1}}^{2} \\ &{}+\frac{1}{\beta}\bigl(\lambda^{k}-\tilde{\lambda}^{k} \bigr)^{\top }\bigl(A_{1}\bigl(u^{k}- \tilde{u}^{k}\bigr)+A_{2}\bigl(v^{k}- \tilde{v}^{k}\bigr)+A_{3}\bigl(w^{k}- \tilde{w}^{k}\bigr)\bigr). \end{aligned}$$
(2.14)

By using the Cauchy-Schwarz inequality, we have

$$\begin{aligned}& \bigl(\lambda^{k}-\tilde{\lambda}^{k} \bigr)^{\top}\bigl(A_{1}\bigl(u^{k}- \tilde{u}^{k}\bigr)\bigr) \geq-\frac{1}{2} \biggl(\sqrt{3}\bigl\Vert A_{1}\bigl(u^{k}-\tilde{u}^{k}\bigr)\bigr\Vert _{H}^{2} +\frac{1}{\sqrt{3}}\bigl\Vert \lambda^{k}-\tilde{\lambda}^{k}\bigr\Vert _{H^{-1}}^{2} \biggr), \end{aligned}$$
(2.15)
$$\begin{aligned}& \bigl(\lambda^{k}-\tilde{\lambda}^{k} \bigr)^{\top}\bigl(A_{2}\bigl(v^{k}- \tilde{v}^{k}\bigr)\bigr) \geq-\frac{1}{2} \biggl(\sqrt{3}\bigl\Vert A_{2}\bigl(v^{k}-\tilde{v}^{k}\bigr)\bigr\Vert _{H}^{2} +\frac{1}{\sqrt{3}}\bigl\Vert \lambda^{k}-\tilde{\lambda}^{k}\bigr\Vert _{H^{-1}}^{2} \biggr) , \end{aligned}$$
(2.16)

and

$$\begin{aligned} \bigl(\lambda^{k}-\tilde{\lambda}^{k} \bigr)^{\top}\bigl(A_{3}\bigl(w^{k}-\tilde{w}^{k} \bigr)\bigr) \geq-\frac{1}{2} \biggl(\sqrt{3}\bigl\Vert A_{3} \bigl(w^{k}-\tilde{w}^{k}\bigr)\bigr\Vert _{H}^{2} +\frac{1}{\sqrt{3}}\bigl\Vert \lambda^{k}-\tilde{\lambda}^{k}\bigr\Vert _{H^{-1}}^{2} \biggr). \end{aligned}$$
(2.17)

Substituting (2.15), (2.16), and (2.17) into (2.14), we get

$$\begin{aligned} \varphi\bigl(z^{k},\tilde{z}^{k}\bigr) \geq & \frac{2\beta-\sqrt{3}}{2\beta} \bigl(\bigl\Vert A_{1}u^{k}-A_{1} \tilde{u}^{k}\bigr\Vert _{H}^{2} +\bigl\Vert A_{2}v^{k} -A_{2}\tilde{v}^{k} \bigr\Vert _{H}^{2}+\bigl\Vert A_{3}w^{k} -A_{3}\tilde{w}^{k} \bigr\Vert _{H}^{2} \bigr)\\ &{} +\frac{2-\sqrt{3}}{2\beta}\bigl\Vert \lambda^{k}-\tilde{ \lambda}^{k}\bigr\Vert _{H^{-1}}^{2} +\bigl\Vert u^{k}-\tilde{u}^{k}\bigr\Vert _{R_{1}}^{2}+\bigl\Vert v^{k}- \tilde{v}^{k}\bigr\Vert _{R_{2}}^{2} +\bigl\Vert w^{k}-\tilde{w}^{k}\bigr\Vert _{R_{3}}^{2} \\ {\geq}& \frac{2\beta-\sqrt{3}}{2\beta} \biggl(\bigl\Vert A_{1}u^{k}-A_{1} \tilde {u}^{k}\bigr\Vert _{H}^{2}+\bigl\Vert A_{2}v^{k} -A_{2}\tilde{v}^{k} \bigr\Vert _{H}^{2}\\ &{}+\bigl\Vert A_{3}w^{k}-A_{3}\tilde{w}^{k}\bigr\Vert _{H}^{2} + \frac{1}{\beta}\bigl\Vert \lambda^{k}-\tilde{\lambda}^{k} \bigr\Vert _{H^{-1}}^{2} \biggr) \\ &{}+\frac{2\beta-\sqrt{3}}{2\beta} \bigl(\bigl\Vert u^{k}-\tilde{u}^{k} \bigr\Vert _{R_{1}}^{2} +\bigl\Vert v^{k}- \tilde{v}^{k}\bigr\Vert _{R_{2}}^{2}+\bigl\Vert w^{k}-\tilde{w}^{k}\bigr\Vert _{R_{3}}^{2} \bigr) \\ =&\frac{2\beta-\sqrt{3}}{2\beta} \bigl(\bigl\Vert z^{k}-\tilde{z}^{k} \bigr\Vert ^{2}_{G}+(1-\mu)\bigl\Vert u^{k}- \tilde{u}^{k}\bigr\Vert _{R_{1}}^{2}\\ &{} +(1-\mu)\bigl\Vert v^{k}-\tilde{v}^{k}\bigr\Vert _{R_{2}}^{2}+(1- \mu)\bigl\Vert w^{k}-\tilde {w}^{k}\bigr\Vert _{R_{3}}^{2} \bigr) \\ \geq &\frac{2\beta-\sqrt{3}}{2\beta}\bigl\Vert z^{k}-\tilde {z}^{k}\bigr\Vert ^{2}_{G}. \end{aligned}$$

Therefore, it follows from (2.7) and (2.12) that

$$ \tau_{k}\geq\frac{2\beta-\sqrt{3}}{2\beta}, $$
(2.18)

and this completes the proof. □

3 Convergence rate

Recall that \(\mathcal{Z}^{*}\) can be characterized as (see (2.3.2) in p.159 of [19])

$$\mathcal{Z}^{*}=\bigcap_{z\in\mathcal{Z}} \bigl\{ \hat{z}\in \mathcal{Z} : (z-\hat{z})^{\top} Q(z)\geq0 \bigr\} . $$

This implies that ẑ is an approximate solution of \(\mathrm{SVI}_{3}\) with the accuracy \(\epsilon>0\) if it satisfies

$$ \hat{z}\in\mathcal{Z}\quad\hbox{and}\quad\sup_{z\in\mathcal {Z}} \bigl\{ (z-\hat{z})^{\top} Q(z)\bigr\} \leq\epsilon. $$
(3.1)

Now we show that after t iterations of the proposed method, we can find a \(\hat{z}\in\mathcal{Z}\) such that (3.1) is satisfied with \(\epsilon= O(1/t)\).

We introduce the following matrices,

$$ N=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} I & 0& 0 &0 \\ 0 & I & 0& 0 \\ 0 & 0 & I& 0 \\ -\beta HA_{1} & -\beta HA_{2} &-\beta HA_{3} & \beta I \end{array}\displaystyle \right ) $$
(3.2)

and

$$ J=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} (1+\mu){R_{1}}+A_{1}^{\top} HA_{1} & 0& 0& 0 \\ 0 &(1+\mu){R_{2}}+A_{2}^{\top} H A_{2} & 0& 0 \\ 0 &0 & (1+\mu){R_{3}}+A_{3}^{\top} H A_{3} & 0 \\ -A_{1} & -A_{2} &-A_{3} & H^{-1} \end{array}\displaystyle \right ). $$
(3.3)

By simple manipulations, we can find that \(J= GN\).

Our analysis needs a new sequence defined by

$$ \hat{z}^{k}= \left ( \textstyle\begin{array}{c} \hat{u}^{k} \\ \hat{v}^{k}\\ \hat{w}^{k}\\ \hat{\lambda}^{k} \end{array}\displaystyle \right )=\left ( \textstyle\begin{array}{c} \tilde{u}^{k}\\ \tilde{v}^{k}\\ \tilde{w}^{k}\\ {\lambda}^{k}-H(A_{1}u^{k}+A_{2}v^{k}+A_{3}w^{k}-b) \end{array}\displaystyle \right ). $$
(3.4)

Based on (3.2) and (3.4), we can easily have

$$ z^{k}-\tilde{z}^{k}=N\bigl(z^{k}- \hat{z}^{k}\bigr). $$
(3.5)

Using (1.11), (2.9), and (3.4), we obtain

$$ g\bigl(z^{k},\tilde{z}^{k}\bigr)=Q\bigl( \hat{z}^{k}\bigr). $$
(3.6)

Lemma 3.1

Let \(\hat{z}^{k}\) be defined by (3.4), \(z\in\mathcal{Z}\), and the matrix J be given by (3.3). Then

$$ \bigl(z-\hat{z}^{k}\bigr)^{\top} \bigl(Q\bigl( \hat{z}^{k}\bigr)-J\bigl(z^{k}-\hat{z}^{k}\bigr) \bigr) \geq-\mu\bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2}-\mu\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2}-\mu\bigl\Vert w^{k} - \hat{w}^{k} \bigr\Vert _{R_{3}}^{2}. $$
(3.7)

Proof

Applying Lemma 2.2 to (2.2), we get

$$\begin{aligned}& \bigl(u - \tilde{u}^{k}\bigr)^{\top} \bigl\{ f_{1}\bigl(\tilde{u}^{k}\bigr) -A_{1}^{\top} \bigl[\lambda^{k} - H\bigl(A_{1} \tilde{u}^{k} + A_{2}{v}^{k}+ A_{3}{w}^{k} -b\bigr) \bigr] \bigr\} \\& \quad \geq\frac{1+\mu}{2} \bigl(\bigl\Vert \tilde{u}^{k}-u \bigr\Vert _{R_{1}}^{2} - \bigl\Vert u^{k} -u\bigr\Vert _{R_{1}}^{2} \bigr) + \frac{1-\mu}{2} \bigl\Vert u^{k} - \tilde{u}^{k} \bigr\Vert _{R_{1}}^{2}. \end{aligned}$$
(3.8)

Since

$$\begin{aligned} \bigl\Vert u^{k}-u\bigr\Vert _{R_{1}}^{2}=\bigl\Vert u^{k}-\tilde{u}^{k}\bigr\Vert _{R_{1}}^{2}+ \bigl\Vert \tilde{u}^{k}-u\bigr\Vert _{R_{1}}^{2}+2 \bigl(\tilde{u}^{k}-u\bigr)^{T}R_{1} \bigl(u^{k}-\tilde{u}^{k}\bigr), \end{aligned}$$

we have

$$ \bigl(u-\tilde{u}^{k}\bigr)^{\top} R_{1}\bigl(u^{k}-\tilde{u}^{k}\bigr) = \frac{1}{2} \bigl(\bigl\Vert \tilde{u}^{k}-u\bigr\Vert _{R_{1}}^{2}-\bigl\Vert u^{k}-u\bigr\Vert _{R_{1}}^{2} \bigr) +{\frac{1}{2}}\bigl\Vert u^{k}-\tilde{u}^{k}\bigr\Vert _{R_{1}}^{2}. $$
(3.9)

Adding (3.8) and (3.9), we obtain

$$\begin{aligned}& \bigl(u - \tilde{u}^{k}\bigr)^{\top} \bigl\{ (1+ \mu)R_{1}\bigl(u^{k}-\tilde{u}^{k}\bigr)- f_{1}\bigl(\tilde{u}^{k}\bigr) + A_{1}^{\top} \bigl[\lambda^{k} - H\bigl(A_{1}u^{k} + A_{2}{v}^{k}+ A_{3}{w}^{k} -b\bigr) \bigr] \\& \quad {}+A_{1}^{\top} HA_{1}\bigl(u^{k}- \tilde{u}^{k}\bigr) \bigr\} \leq\mu\bigl\Vert u^{k} - \tilde {u}^{k} \bigr\Vert _{R_{1}}^{2}. \end{aligned}$$
(3.10)

Similarly, applying Lemma 2.2 to (2.3), we get

$$\begin{aligned}& \bigl(v - \tilde{v}^{k}\bigr)^{\top} \bigl\{ f_{2}\bigl(\tilde{v}^{k}\bigr) - A_{2}^{\top} \bigl[\lambda^{k} - H \bigl(A_{1}u^{k} + A_{2}\tilde{v}^{k}+ A_{3}{w}^{k} -b \bigr)\bigr] \bigr\} \\& \quad \geq\frac{1+\mu}{2} \bigl(\bigl\Vert \tilde{v}^{k}-v \bigr\Vert _{R_{2}}^{2} - \bigl\Vert v^{k} -v\bigr\Vert _{R_{2}}^{2} \bigr) + \frac{1-\mu}{2} \bigl\Vert v^{k} - \tilde{v}^{k} \bigr\Vert _{R_{2}}^{2}. \end{aligned}$$
(3.11)

Similar to (3.9), we have

$$ \bigl(v-\tilde{v}^{k}\bigr)^{\top} {R_{2}}\bigl(v^{k}-\tilde{v}^{k}\bigr)={ \frac{1}{2}} \bigl(\bigl\Vert \tilde{v}^{k}-v\bigr\Vert _{R_{2}}^{2}-\bigl\Vert v^{k}-v\bigr\Vert _{R_{2}}^{2} \bigr) +{\frac{1}{2}}\bigl\Vert v^{k}-\tilde{v}^{k}\bigr\Vert _{R_{2}}^{2}. $$
(3.12)

Adding (3.11) and (3.12), we have

$$\begin{aligned}& \bigl(v - \tilde{v}^{k}\bigr)^{\top} \bigl\{ (1+ \mu){R_{2}}\bigl(v^{k}-\tilde{v}^{k}\bigr)- f_{2}\bigl(\tilde{v}^{k}\bigr) + A_{2}^{\top} \bigl[\lambda^{k} -H\bigl(A_{1}u^{k} + A_{2}{v}^{k}+ A_{3}{w}^{k} -b\bigr) \bigr] \\& \quad {}+A_{2}^{\top} HA_{2}\bigl(v^{k}- \tilde{v}^{k}\bigr) \bigr\} \leq\mu\bigl\Vert v^{k} - \tilde {v}^{k} \bigr\Vert _{R_{2}}^{2}. \end{aligned}$$
(3.13)

Similarly, we have

$$\begin{aligned}& \bigl(w - \tilde{w}^{k}\bigr)^{\top} \bigl\{ (1+ \mu){R_{3}}\bigl(w^{k}-\tilde{w}^{k}\bigr)- f_{3}\bigl(\tilde{w}^{k}\bigr) + A_{3}^{\top} \bigl[\lambda^{k} -H\bigl(A_{1}u^{k} + A_{2}{v}^{k}+ A_{3}{w}^{k} -b\bigr) \bigr] \\& \quad {}+A_{3}^{\top} HA_{3}\bigl(w^{k}- \tilde{w}^{k}\bigr) \bigr\} \le\mu\bigl\Vert w^{k} - \tilde {w}^{k} \bigr\Vert _{R_{3}}^{2}. \end{aligned}$$
(3.14)

Then, by using the notation of \(\hat{z}^{k}\) in (3.4), (3.10), (3.13), and (3.14) can be written as

$$\begin{aligned}& \bigl(u - \hat{u}^{k}\bigr)^{\top} \bigl\{ (1+\mu)R_{1}\bigl(u^{k}-\hat{u}^{k}\bigr)- f_{1}(\hat {u}^{k})+A_{1}^{\top} \hat{\lambda}^{k} +A_{1}^{\top} HA_{1}\bigl(u^{k}-\hat{u}^{k}\bigr) \bigr\} \\& \quad \leq\mu\bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2}, \end{aligned}$$
(3.15)
$$\begin{aligned}& \bigl(v - \hat{v}^{k}\bigr)^{\top} \bigl\{ (1+ \mu){R_{2}}\bigl(v^{k}-\hat{v}^{k}\bigr)- f_{2}\bigl(\hat{v}^{k}\bigr)+A_{2}^{\top} \hat{\lambda}^{k} +A_{2}^{\top} HA_{2} \bigl(v^{k}-\hat{v}^{k}\bigr) \bigr\} \\& \quad \le\mu\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2}, \end{aligned}$$
(3.16)

and

$$\begin{aligned}& \bigl(w - \hat{w}^{k}\bigr)^{\top} \bigl\{ (1+ \mu){R_{3}}\bigl(w^{k}-\hat{w}^{k}\bigr)- f_{3}\bigl(\hat{w}^{k}\bigr)+A_{3}^{\top} \hat{\lambda}^{k} +A_{3}^{\top} HA_{3} \bigl(w^{k}-\hat{w}^{k}\bigr) \bigr\} \\& \quad \le\mu\bigl\Vert w^{k} - \hat{w}^{k} \bigr\Vert _{R_{3}}^{2}. \end{aligned}$$
(3.17)

In addition, it follows from (3.4) that

$$\begin{aligned}& A_{1}\hat{u}^{k}+ A_{2} \hat{v}^{k}+A_{3}\hat{w}^{k}-b +H^{-1} \bigl(\hat{\lambda }^{k}-{\lambda}^{k}\bigr) \\& \quad {}-A_{1} \bigl(\hat{u}^{k}-u^{k}\bigr)-A_{2}\bigl( \hat{v}^{k}-v^{k}\bigr)-A_{3}\bigl( \hat{w}^{k}-w^{k}\bigr)=0. \end{aligned}$$
(3.18)

Combining (3.15)-(3.18), we get

$$\begin{aligned}& \left ( \textstyle\begin{array}{c} u - \hat{u}^{k} \\ v - \hat{v}^{k}\\ w - \hat{w}^{k}\\ \lambda- \hat{\lambda}^{k} \end{array}\displaystyle \right )^{\top} \left ( \textstyle\begin{array}{c} f_{1}(\hat{u}^{k})- A_{1}^{\top} \hat{\lambda}^{k}- ((1+\mu ){R_{1}}+A^{T}HA_{1} )(u^{k}-\hat{u}^{k})\\ f_{2}(\hat{v}^{k}) - A_{2}^{\top} \hat{\lambda}^{k}- ((1+\mu ){R_{2}}+A_{2}^{\top}HA_{2} )(v^{k}-\hat{v}^{k})\\ f_{3}(\hat{w}^{k}) - A_{3}^{\top} \hat{\lambda}^{k}- ((1+\mu ){R_{3}}+A_{3}^{\top}HA_{3} )(w^{k}-\hat{w}^{k})\\ A_{1}\hat{u}^{k}+A_{2}\hat{v}^{k}+A_{3}\hat{w}^{k} -b +A_{1}(u^{k}-\hat{u}^{k})+A_{2}(v^{k}-\hat{v}^{k})+A_{3}(w^{k}-\hat {w}^{k})-H^{-1}(\lambda^{k}-\hat{\lambda}^{k}) \end{array}\displaystyle \right ) \\& \quad \geq-\mu\bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2}-\mu\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2}-\mu\bigl\Vert w^{k} - \hat{w}^{k} \bigr\Vert _{R_{3}}^{2}. \end{aligned}$$
(3.19)

Recall the definition of J in (3.3), we obtain the assertion (3.7). The proof is completed. □

Lemma 3.2

For given \(z^{k}\in \mathbb{R}^{n_{1}}_{++}\times\mathbb{R}^{n_{2}}_{++}\times \mathbb{R}^{n_{3}}_{++}\times\mathbb{R}^{m}\) and \(z_{*}^{k} := P_{\mathcal{Z},G}[z^{k}-\tau_{k} G^{-1}g(z^{k},\tilde{z}^{k})]\), we have

$$ \gamma\tau_{k}\bigl(z-\hat{z}^{k} \bigr)^{\top} Q(z)+\frac{1}{2}\bigl(\bigl\Vert z-z^{k} \bigr\Vert _{G}^{2}-\bigl\Vert z-z_{*}^{k} \bigr\Vert _{G}^{2}\bigr)\geq\frac{1}{2} \gamma(2-\gamma) \tau_{k}^{2}\bigl\Vert z^{k}-\tilde{z}^{k} \bigr\Vert _{G}^{2}. $$
(3.20)

Proof

Since \(z_{*}^{k}\in\mathcal{Z}\), substituting \(z=z_{*}^{k}\) in (3.7), we get

$$\begin{aligned}& \gamma\tau_{k}\bigl(z_{*}^{k}- \hat{z}^{k}\bigr)^{\top}Q\bigl(\hat{z}^{k}\bigr) \end{aligned}$$
(3.21)
$$\begin{aligned}& \quad \geq \gamma\tau_{k}\bigl(z_{*}^{k}- \hat{z}^{k}\bigr)^{\top} J\bigl(z^{k}- \hat{z}^{k}\bigr) - \mu\gamma\tau_{k}\bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2}- \mu\gamma\tau_{k}\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2} \\& \qquad {} -\mu\gamma\tau_{k}\bigl\Vert w^{k} - \hat{w}^{k}\bigr\Vert _{R_{3}}^{2} \\& \quad = \gamma\tau_{k}\bigl(z^{k}-\hat{z}^{k} \bigr)^{\top} J\bigl(z^{k}-\hat{z}^{k}\bigr)+\gamma \tau _{k}\bigl(z_{*}^{k}-z^{k} \bigr)^{\top} J\bigl(z^{k}-\hat{z}^{k}\bigr) \\& \qquad {} -\mu\gamma\tau_{k}\bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2}-\mu\gamma\tau _{k}\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2} -\mu\gamma\tau_{k}\bigl\Vert w^{k} - \hat{w}^{k}\bigr\Vert _{R_{3}}^{2} \\& \quad = \gamma\tau_{k}\bigl(z^{k}-\tilde{z}^{k} \bigr)^{\top} \bigl(N^{-1}\bigr)^{\top} JN^{-1}\bigl(z^{k}-\tilde{z}^{k}\bigr) +\gamma \tau_{k}\bigl(z_{*}^{k}-z^{k} \bigr)^{\top} JN^{-1}\bigl(z^{k}-\tilde{z}^{k} \bigr) \\& \qquad {} -\gamma\tau_{k}\mu\bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2} -\gamma \tau_{k}\mu\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2}-\mu\gamma\tau_{k}\bigl\Vert w^{k} - \hat{w}^{k}\bigr\Vert _{R_{3}}^{2} \\& \quad = \gamma\tau_{k}\bigl(z^{k}-\tilde{z}^{k} \bigr)^{\top} \bigl(N^{-1}\bigr)^{\top} G \bigl(z^{k}-\tilde{z}^{k}\bigr) -\gamma\tau_{k}\mu \bigl\Vert u^{k} - \hat{u}^{k} \bigr\Vert _{R_{1}}^{2} -\gamma\tau_{k}\mu\bigl\Vert v^{k} - \hat{v}^{k} \bigr\Vert _{R_{2}}^{2} \\& \qquad {} -\mu\gamma\tau_{k}\bigl\Vert w^{k} - \hat{w}^{k}\bigr\Vert _{R_{3}}^{2} +\gamma \tau_{k}\bigl(z_{*}^{k}-z^{k} \bigr)^{\top} G\bigl(z^{k}-\tilde{z}^{k}\bigr) \\& \quad = \gamma\tau_{k}\bigl(z^{k}-\tilde{z}^{k} \bigr)^{\top} \bigl(N^{-1}\bigr)^{\top} M \bigl(z^{k}-\tilde{z}^{k}\bigr) +\gamma\tau_{k} \bigl(z_{*}^{k}-z^{k}\bigr)^{\top} G \bigl(z^{k}-\tilde{z}^{k}\bigr) \\& \quad \geq \gamma\tau_{k}\varphi\bigl(z^{k},\tilde{z}^{k} \bigr)+\gamma\tau _{k}\bigl(z_{*}^{k}-z^{k} \bigr)^{\top} G\bigl(z^{k}-\tilde{z}^{k}\bigr) \\& \quad \geq \gamma\tau_{k}\varphi\bigl(z^{k},\tilde{z}^{k} \bigr)-\frac{1}{2}\bigl\Vert z^{k}-z_{*}^{k}\bigr\Vert _{G}^{2} -\frac{1}{2}\gamma^{2} \tau_{k}^{2}\bigl\Vert z^{k}-\tilde{z}^{k} \bigr\Vert _{G}^{2} \\& \quad = \frac{1}{2} \gamma(2-\gamma)\tau_{k}^{2}\bigl\Vert z^{k}-\tilde{z}^{k}\bigr\Vert _{G}^{2}- \frac{1}{2}\bigl\Vert z^{k}-z_{*}^{k}\bigr\Vert _{G}^{2}. \end{aligned}$$
(3.22)

Using (3.6), \(z_{*}^{k}\) is the projection of \(z^{k}-\gamma\tau_{k} G^{-1}Q(\hat{z}^{k})\) on \(\mathcal{Z}\), it follows from (2.1) that

$$\bigl(z^{k}-\gamma\tau_{k} G^{-1}Q\bigl( \hat{z}^{k}\bigr)-z_{*}^{k}\bigr)^{\top} G \bigl(z-z_{*}^{k}\bigr)\leq0, \quad\forall z\in\mathcal{Z}, $$

and consequently, we have

$$\gamma\tau_{k} \bigl(z-z_{*}^{k} \bigr)^{\top} Q\bigl(\hat{z}^{k}\bigr)\geq \bigl(z^{k}-z_{*}^{k} \bigr)^{\top} G\bigl(z-z_{*}^{k}\bigr). $$

Using the identity \(x^{\top} Gy =\frac{1}{2} (\Vert x\Vert _{G}^{2}-\Vert x-y\Vert _{G}^{2}+\Vert y\Vert _{G}^{2} )\) to the right hand side of the last inequality, we obtain

$$ \gamma\tau_{k} \bigl(z-z_{*}^{k} \bigr)^{\top} Q\bigl(\hat{z}^{k}\bigr)\geq\frac{1}{2} \bigl(\bigl\Vert z-z_{*}^{k}\bigr\Vert _{G}^{2}-\bigl\Vert z-z^{k}\bigr\Vert _{G}^{2} \bigr) +\frac{1}{2}\bigl\Vert z^{k}-z_{*}^{k}\bigr\Vert _{G}^{2}. $$
(3.23)

Adding (3.21) and (3.23), we get

$$\gamma\tau_{k}\bigl(z-\hat{z}^{k}\bigr)^{\top} Q \bigl(\hat{z}^{k}\bigr)+\frac{1}{2}\bigl(\bigl\Vert z-z^{k} \bigr\Vert _{G}^{2}-\bigl\Vert z-z_{*}^{k} \bigr\Vert _{G}^{2}\bigr)\geq\frac{1}{2} \gamma(2-\gamma) \tau_{k}^{2}\bigl\Vert z^{k}-\tilde{z}^{k} \bigr\Vert _{G}^{2}, $$

and by using the monotonicity of Q, we obtain (3.20) and the proof is completed. □

Lemma 3.3

Let \(z^{k}\in \mathbb{R}^{n_{1}}_{++}\times \mathbb{R}^{n_{2}}_{++}\times\mathbb{R}^{n_{3}}_{++}\times\mathbb {R}^{m}\) and \(z^{k+1}(\tau_{k})\) be generated by (2.6). Then

$$ \gamma\sigma\tau_{k}\bigl(z-\hat{z}^{k} \bigr)^{\top} Q(z)+\frac{1}{2}\bigl(\bigl\Vert z-z^{k}\bigr\Vert _{G}^{2}-\bigl\Vert z-z^{k+1}( \tau_{k})\bigr\Vert _{G}^{2}\bigr) \geq \frac{1}{2} \sigma\gamma(2-\gamma)\tau_{k}^{2}\bigl\Vert z-\tilde{z}^{k}\bigr\Vert _{G}^{2}. $$
(3.24)

Proof

We have

$$\begin{aligned}& \bigl\Vert z-z^{k}\bigr\Vert _{G}^{2}- \bigl\Vert z-z^{k+1}(\tau_{k})\bigr\Vert _{G}^{2} \end{aligned}$$
(3.25)
$$\begin{aligned}& \quad = \bigl\Vert z^{k}-z\bigr\Vert _{G}^{2}- \bigl\Vert z^{k}-\sigma\bigl(z^{k}-z_{*}^{k} \bigr)-z\bigr\Vert _{G}^{2} \\& \quad = 2\sigma\bigl(z^{k}-z\bigr)^{\top} G\bigl(z^{k}-z_{*}^{k} \bigr)-\sigma^{2}\bigl\Vert z^{k}-z_{*}^{k} \bigr\Vert _{G}^{2} \\& \quad = 2\sigma \bigl(\bigl\Vert z^{k}-z_{*}^{k} \bigr\Vert _{G}^{2}-\bigl( z-z_{*}^{k} \bigr)^{\top} G\bigl(z^{k}-z_{*}^{k}\bigr) \bigr)-\sigma^{2}\bigl\Vert z^{k}-z_{*}^{k} \bigr\Vert _{G}^{2}. \end{aligned}$$
(3.26)

Using the identity

$$\begin{aligned} \bigl(z-z_{*}^{k}\bigr)^{\top} G \bigl(z^{k}-z_{*}^{k}\bigr) = &{ \frac{1}{2}} \bigl(\bigl\Vert z_{*}^{k}-z\bigr\Vert _{G}^{2}- \bigl\Vert z^{k}-z\bigr\Vert _{G}^{2} \bigr) +{\frac{1}{2}}\bigl\Vert z^{k}-z_{*}^{k}\bigr\Vert _{G}^{2}, \end{aligned}$$

we get

$$ \bigl\Vert z^{k}-z_{*}^{k}\bigr\Vert _{G}^{2}-2\bigl(z-z_{*}^{k} \bigr)^{\top} G\bigl(z^{k}-z_{*}^{k}\bigr)= \bigl\Vert z^{k}-z\bigr\Vert _{G}^{2}-\bigl\Vert z_{*}^{k}-z\bigr\Vert _{G}^{2}. $$
(3.27)

Substituting (3.27) into (3.25), we obtain

$$\begin{aligned} \bigl\Vert z-z^{k}\bigr\Vert _{G}^{2}- \bigl\Vert z-z^{k+1}(\tau_{k})\bigr\Vert _{G}^{2} =&\sigma\bigl(\bigl\Vert z-z^{k}\bigr\Vert _{G}^{2}-\bigl\Vert z-z_{*}^{k}\bigr\Vert _{G}^{2}\bigr) +\sigma(1-\sigma)\bigl\Vert z^{k}-z_{*}^{k}\bigr\Vert _{G}^{2} \\ \geq&\sigma\bigl(\bigl\Vert z-z^{k}\bigr\Vert _{G}^{2}-\bigl\Vert z-z_{*}^{k}\bigr\Vert _{G}^{2}\bigr). \end{aligned}$$
(3.28)

Substituting (3.28) into (3.20), we obtain (3.24), the required result. □

Theorem 3.1

Let \(z^{*}\) be a solution of \(\mathrm{SVI}_{3}\) and \(z^{k+1}(\tau_{k})\) be generated by (2.6). Then \(z^{k}\) and \(\tilde{z}^{k}\) are bounded, and

$$ \bigl\Vert z^{k+1}(\tau_{k}) - z^{*}\bigr\Vert _{G}^{2}\leq\bigl\Vert z^{k} -z^{*}\bigr\Vert _{G}^{2}-c\bigl\Vert z^{k}-\tilde {z}^{k}\bigr\Vert ^{2}_{G}, $$
(3.29)

where

$$c:={ \frac{\sigma\gamma(2-\gamma)(2\beta-\sqrt {3})^{2}}{4\beta^{2}}}>0. $$

Proof

Setting \(z=z^{*}\) in (3.24), we obtain

$$\begin{aligned} \bigl\Vert z^{k+1}(\tau_{k}) - z^{*}\bigr\Vert _{G}^{2} \leq&\bigl\Vert z^{k} -z^{*}\bigr\Vert _{G}^{2}-\sigma\gamma(2-\gamma)\tau_{k}^{2} \bigl\Vert z^{k}-\tilde {z}^{k}\bigr\Vert ^{2}_{G}+ 2\gamma\sigma\tau_{k}\bigl(z^{*}-\hat{z}^{k} \bigr)^{\top} Q\bigl(z^{*}\bigr) \\ \leq&\bigl\Vert z^{k} -z^{*}\bigr\Vert _{G}^{2}- \sigma\gamma(2-\gamma)\tau_{k}^{2}\bigl\Vert z^{k}-\tilde {z}^{k}\bigr\Vert ^{2}_{G} \\ \leq&\bigl\Vert z^{k} -z^{*}\bigr\Vert _{G}^{2}- \frac{\sigma\gamma(2-\gamma)(2\beta-\sqrt {3})^{2}}{4\beta^{2}}\bigl\Vert z^{k}-\tilde{z}^{k}\bigr\Vert ^{2}_{G}. \end{aligned}$$

Then we have

$$\bigl\Vert z^{k+1}(\tau_{k})-z^{*}\bigr\Vert _{G}\leq\bigl\Vert z^{k}-z^{*}\bigr\Vert _{G} \leq\cdots\leq\bigl\Vert z^{0}-z^{*}\bigr\Vert _{G}, $$

and thus, \(\{ z^{k}\}\) is a bounded sequence.

It follows from (3.29) that

$$\sum_{k=0}^{\infty}c\bigl\Vert z^{k}-\tilde{z}^{k}\bigr\Vert ^{2}_{G}< + \infty, $$

which means that

$$ \lim_{k\to\infty}\bigl\Vert z^{k}- \tilde{z}^{k}\bigr\Vert _{G}=0. $$
(3.30)

Since \(\{ z^{k}\}\) is a bounded sequence, we conclude that \(\{\tilde {z}^{k}\}\) is also bounded. □

The global convergence of the proposed method can be proved by similar arguments as in [18]. Hence the proof is omitted.

Theorem 3.2

The sequence \(\{z^{k}\}\) generated by the proposed method converges to some \(z^{\infty}\) which is a solution of \(\mathrm{SVI}_{3}\).

Now, we are ready to present the \(O(1/t)\) convergence rate of the proposed method.

Theorem 3.3

For any integer \(t > 0\), we have a \(\hat{z}_{t}\in\mathcal{Z}\) which satisfies

$$(\hat{z}_{t}-z)^{\top} Q(z)\leq\frac{1}{2\gamma\sigma\Upsilon_{t}}\bigl\Vert z-z^{0}\bigr\Vert _{G}^{2},\quad\forall z\in \mathcal{Z}, $$

where

$$\hat{z}_{t}=\frac{1}{\Upsilon_{t}}\sum_{k=0}^{t} \tau_{k}\hat{z}^{k}\quad \textit{and} \quad\Upsilon_{t}= \sum_{k=0}^{t}\tau_{k}. $$

Proof

Summing the inequality (3.24) over \(k=0,\ldots,t\), we obtain

$$\Biggl( \Biggl(\sum_{k=0}^{t}\gamma\sigma \tau_{k} \Biggr) z-\sum_{k=0}^{t} \gamma\sigma\tau_{k}\hat{z}^{k} \Biggr)^{\top} Q(z)+ \frac{1}{2}\bigl\Vert z-z^{0}\bigr\Vert _{G}^{2} \geq0. $$

Using the notations of \(\Upsilon_{t}\) and \(\hat{z}_{t}\) in the above inequality, we derive

$$(\hat{z}_{t}-z)^{\top} Q(z)\leq\frac{1}{2\gamma\sigma\Upsilon_{t}}\bigl\Vert z-z^{0}\bigr\Vert _{G}^{2},\quad\forall z\in \mathcal{Z}. $$

Indeed, \(\hat{z}_{t}\in\mathcal{Z}\) because it is a convex combination of \(\hat{z}^{0}\), \(\hat{z}^{1}\), … , \(\hat{z}^{t}\). The proof is completed. □

Remark 3.1

It follows from (2.13) that

$$\Upsilon_{t}\geq\frac{2\beta-\sqrt{3}}{2\beta} (t+1 ). $$

Suppose that, for any compact set \(\mathcal {D}\subset\mathcal{Z}\), let \(d =\sup\{\Vert z-z^{0}\Vert _{G}\vert z\in\mathcal {D}\}\). For any given \(\epsilon>0\), after at most

$$t= \biggl[\frac{\beta d^{2}}{(2\beta-\sqrt{3})\gamma\sigma\epsilon } \biggr] $$

iterations, we have

$$(\hat{z}_{t}-z)^{T}Q(z)\leq\epsilon,\quad \forall z\in\mathcal {D}. $$

That is, the \(O(1/t)\) convergence rate is established in an ergodic sense.

4 Preliminary computational results

In this section, we present some numerical examples to illustrate the proposed method. We consider the following optimization problem with matrix variables, which is studied in [18]:

$$ \min \biggl\{ \frac{1}{2} \Vert U-C\Vert _{F}^{2}\Big\vert U\in S_{+}^{n} \biggr\} , $$
(4.1)

where \(\Vert \cdot \Vert _{F}\) is the matrix Fröbenius norm, i.e., \(\Vert C\Vert _{F} = ( \sum_{i=1}^{n}\sum_{j=1}^{n} \vert C_{ij}\vert ^{2} )^{1/2}\) and

$$S_{+}^{n}=\bigl\{ M\in\mathbb{R}^{n\times n} : M^{\top} =M, M \succeq 0 \bigr\} . $$

It has been shown in [18] that solving problem (4.1) is equivalent to the following variational inequality problem: Find \(X^{*}=(U^{*},V^{*},Z^{*})\in\Omega= S_{+}^{n} \times S_{+}^{n}\times \mathbb{R}^{n\times n}\) such that

$$\begin{aligned} \textstyle\begin{cases} \langle U-U^{*},(U^{*}-C)-Z^{*}\rangle\geq0, \\ \langle V-V^{*},(V^{*}-C)+Z^{*}\rangle\geq0,&\forall X=(U,V,Z)\in \Omega,\\ U^{*} - V^{*} =0. \end{cases}\displaystyle \end{aligned}$$
(4.2)

The problem (4.2) is a special case of (1.3)-(1.4) with matrix variables where \(A_{1}=I_{n\times n}\), \(A_{2}=-I_{n\times n}\), \(b=0\), \(f_{1}(U)=U-C\), \(f_{2}(V)=V-C\), and \(\mathcal{W} =S_{+}^{n} \times S_{+}^{n}\times\mathbb{R}^{n\times n}\). For simplification, we take \(R_{1}=r_{1}I_{n\times n}\), \(R_{2}=r_{2}I_{n\times n}\), and \(H=I_{n\times n}\) where \(r_{1}>0\) and \(r_{2}>0\) are scalars. In all tests we take \(\mu=0.5\), \(\beta=0.88\), \(C=\operatorname{rand}(n)\), and \((U^{0}, V^{0}, Z^{0})=(I_{n\times n},I_{n\times n},0_{n\times n})\) as the initial point in the test. The iteration is stopped as soon as

$$\max\bigl\{ \bigl\Vert U^{k}-\tilde{U}^{k}\bigr\Vert , \bigl\Vert V^{k}-\tilde{V}^{k}\bigr\Vert ,\bigl\Vert Z^{k}-\tilde{Z}^{k}\bigr\Vert \bigr\} \leq10^{-6}. $$

All codes were written in Matlab, we compare the proposed method with those in [18] and [17]. The iteration numbers, denoted by k, and the computational time for the problem (4.1) with different dimensions are given in Tables 1-2.

Table 1 Numerical results for problem ( 4.1 ) with \(\pmb{r_{1}=0.5}\) , \(\pmb{r_{2}=5}\)
Full size table
Table 2 Numerical results for problem ( 4.1 ) with \(\pmb{r_{1}=1}\) , \(\pmb{r_{2}=10}\)
Full size table

Tables 1-2 show that the proposed method is more flexible and efficient for the problem tested.

5 Conclusions

In this paper, we proposed a new modified logarithmic-quadratic proximal alternating direction method for solving structured variational inequalities with three separable operators. The prediction point is obtained by solving series of related systems of nonlinear equations in a parallel way. Global convergence of the proposed method is proved under mild assumptions. We have proved the \(O(1/t)\) convergence rate of the parallel LQP alternating direction method. Some preliminary numerical examples are reported to verify the effectiveness of the proposed method in practice.