A proximal alternating minimization algorithm for the largest C-eigenvalue of piezoelectric-type tensors

C-eigenvalues of piezoelectric-type tensors play an important role in piezoelectric effect and converse piezoelectric effect. While the largest C-eigenvalue of a given piezoelectric-type tensor has concrete physical meaning which determines the highest piezoelectric coupling constant. In this paper, we focus on computing the maximum C-eigenvalue of piezoelectric-type tensors which is a third degree polynomial problem. To do that, we first establish the equivalence between the proposed polynomial optimization problem (POP) and a multi-linear optimization problem (MOP) under conditions that the original objective function is concave. Then, an augmented POP (which can also be regarded as a regularized POP) is introduced for the purpose to guarantee the concavity of the underlying objective function. Theoretically, both the augmented POP and the original problem share the same optimal solutions when the compact sets are specified as unit spheres. By exploiting the multi-block structure of the resulting MOP, we accordingly propose a proximal alternating minimization algorithm to get an approximate optimal value of the maximum C-eigenvalue. Furthermore, convergence of the proposed algorithm is established under mild conditions. Finally, some preliminary computational results on synthetic data sets are reported to show the efficiency of the proposed algorithm.

The authors would like to thank the editor and the two anonymous referees for their valuable comments, which greatly helped us to improve the quality of this paper. H. Chen was supported in part by National Natural Science Foundation of China (NSFC) (No. 12071249), and Shandong Province Natural Science Foundation of Distinguished Young Scholars (No. ZR2021JQ01). Y. Wang was supported in part by NSFC (No. 12071250).

Authors and Affiliations

1. School of Management Science, Qufu Normal University, Rizhao, Shandong, People’s Republic of China

   Wenjie Wang, Haibin Chen & Yiju Wang

2. Department of Mathematics and Statistics, Curtin University, Perth, WA, Australia

   Guanglu Zhou

Corresponding author

Correspondence to Haibin Chen.

Publisher's Note

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Appendix

In this appendix, we present detailed proofs for Theorem 5.

Proof of Theorem 5

1. (1)

   Since \(f_{\alpha }(\mathbf{x}, \mathbf{y}, \mathbf{y})\) is a linear function with respect to \(\mathbf{x}\), we know that, for \(k\in N\), it follows that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})-f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})=\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), \mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\rangle . \end{aligned}$$

   (20)

   If \(\nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})=\mathbf{0}\), then \(\mathbf{x}^{(k+1)}=\mathbf{x}^{(k)}\) from Algorithm 2 and

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})=f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}). \end{aligned}$$

   If \(\nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\ne \mathbf{0}\), by (20), it follows that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})-f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})&=\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), ~\mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\rangle \\&=\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}),\\&\quad \,-\frac{\nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})}{\Vert \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\Vert }-\mathbf{x}^{(k)}\rangle \\&=-\Vert \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\Vert \\&\quad \,-\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), \mathbf{x}^{(k)}\rangle . \end{aligned}$$

   Combining this with the Cauchy-schwarz inequality and \(\Vert \mathbf{x}^{(k)}\Vert =1\), we know that

   $$\begin{aligned} \langle \nabla _\mathbf{x}f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), \mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\rangle \le 0, \end{aligned}$$

   which implies that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\le f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}). \end{aligned}$$

   (21)

   On the other hand, we can prove similarly that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})\le f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}). \end{aligned}$$

   (22)

   Combining (21) with (22), that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})\le f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}). \end{aligned}$$

   On the other hand, if there exists a k such that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})= f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), \end{aligned}$$

   then, by (21)–(22), we have

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})= f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})= f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}). \end{aligned}$$

   From (20) and the fact that \(\Vert \mathbf{x}^{(k+1)}\Vert =\Vert \mathbf{x}^{(k)}\Vert =1\), we know that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})< f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}) \Leftrightarrow \mathbf{x}^{(k+1)}\ne \mathbf{x}^{(k)}. \end{aligned}$$

   Similarly, we can obtain that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})< f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}) \Leftrightarrow \mathbf{y}^{(k+1)}\ne \mathbf{y}^{(k)}, \end{aligned}$$

   which means that \(\{f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})\}\) is strictly decreasing.

2. (2)

   If Algorithm 2 terminates in finite steps, without loss of generality, suppose it stops at the t-th iterative step. From the proof of (1), it holds that

   $$\begin{aligned} f_{\alpha }(\mathbf{x}^{(t+1)}, \mathbf{y}^{(t+1)}, \mathbf{y}^{(t+1)})=f_{\alpha }(\mathbf{x}^{(t)}, \mathbf{y}^{(t)}, \mathbf{y}^{(t)}) \quad and \quad \mathbf{x}^{(t+1)}=\mathbf{x}^{(t)},~ \mathbf{y}^{(t+1)}=\mathbf{y}^{(t)}. \end{aligned}$$

   Then, by Algorithm 2, it follows that

   $$\begin{aligned} \nabla _\mathbf{x}f_{\alpha }(\mathbf{x}^{(t)}, \mathbf{y}^{(t)}, \mathbf{y}^{(t)})=0,~~~\nabla _\mathbf{y}f_{\alpha }(\mathbf{x}^{(t)}, \mathbf{y}^{(t)}, \mathbf{y}^{(t)})=0. \end{aligned}$$

   Thus, for any \(\mathbf{x}, \mathbf{y}\in D\), it holds that

   $$\begin{aligned} \langle \mathbf{x}-\mathbf{x}^{(t)}, \nabla _\mathbf{x}f_{\alpha }(\mathbf{x}^{(t)}, \mathbf{y}^{(t)}, \mathbf{y}^{(t)})\rangle +\langle \mathbf{y}-\mathbf{y}^{(t)}, \nabla _\mathbf{y}f_{\alpha }(\mathbf{x}^{(t)}, \mathbf{y}^{(t)}, \mathbf{y}^{(t)})\rangle =0, \end{aligned}$$

   which implies that \((\mathbf{x}^{(t)}, \mathbf{y}^{(t)})\) is a stationary point of (6).

3. (3)

   In this case, from the proof of (1), we know that

   $$\begin{aligned} \begin{aligned}&f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})-f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\\&\quad \le f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})-f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\\&\quad =\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), \mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\rangle \le 0. \end{aligned} \end{aligned}$$

   Since the sequence \(\{f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\}\) is bounded and monotonic, it follows that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\langle \nabla _\mathbf{x}f_{\alpha }(\mathbf{x}^{(k)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)}), \mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\rangle =0. \end{aligned}$$

   (23)

   On the other hand, since the generated sequence \(\{(\mathbf{x}^{(k)}, \mathbf{y}^{(k)})\}\) is bounded, it has at least one accumulation point. Without loss of generality, assume \((\hat{\mathbf{x}}, \hat{\mathbf{y}})\) is an accumulation point of the subsequence \(\{(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)})\}\) such that \(\Vert \hat{\mathbf{x}}\Vert =\Vert \hat{\mathbf{y}}\Vert =1.\) By the continuity of \(\nabla _{\mathbf{x}}f_{\alpha }(\mathbf{x}, \mathbf{y}, \mathbf{y})\), if it holds that

   $$\begin{aligned} \lim \limits _{j\rightarrow \infty }\nabla _\mathbf{x}f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)})=\nabla _\mathbf{x}f_{\alpha }(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{y}})=\mathcal {A}\hat{\mathbf{y}}\hat{\mathbf{y}}-\alpha \hat{\mathbf{x}}=0, \end{aligned}$$

   we obtain that

   $$\begin{aligned} {\mathcal {A}}\hat{\mathbf{y}}\hat{\mathbf{y}}=\alpha \hat{\mathbf{x}} \end{aligned}$$

   (24)

   If \(\lim \limits _{j\rightarrow \infty }\nabla _\mathbf{x}f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)})\ne 0\), by Algorithm 2, (20) and (23) it follows that

   $$\begin{aligned} \begin{aligned} 0&=\lim _{j\rightarrow \infty }\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)}), ~\mathbf{x}^{(k_j+1)}-\mathbf{x}^{(k_j)}\rangle \\&=\lim _{j\rightarrow \infty }\left\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)}), -\frac{\nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)})}{\Vert \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)})\Vert }-\mathbf{x}^{(k_j)}\right\rangle \\&=\lim _{j\rightarrow \infty }\left( -\Vert \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)})\Vert -\langle \nabla _\mathbf{x} f_{\alpha }(\mathbf{x}^{(k_j)}, \mathbf{y}^{(k_j)}, \mathbf{y}^{(k_j)}), \hat{\mathbf{x}}\rangle \right) \\&=-\Vert \nabla _\mathbf{x} f_{\alpha }(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{y}})\Vert -\langle \nabla _\mathbf{x} f_{\alpha }(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{y}}), \hat{\mathbf{x}}\rangle . \end{aligned} \end{aligned}$$

   Combining this with the fact that \(\Vert \hat{\mathbf{x}}\Vert =1\), we know that

   $$\begin{aligned} \nabla _\mathbf{x} f_{\alpha }(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{y}})=-\Vert \nabla _\mathbf{x} f_{\alpha }(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{y}})\Vert \hat{\mathbf{x}}, \end{aligned}$$

   which implies that

   $$\begin{aligned} {\mathcal {A}}\hat{\mathbf{y}}\hat{\mathbf{y}}=(\alpha -\Vert \nabla _\mathbf{x}f_{\alpha }(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{y}})\Vert )\hat{\mathbf{x}}. \end{aligned}$$

   (25)

   Combining (24) and (25), we obtain the \(\hat{\mathbf{x}}\)-part KKT system for the problem of (6). Similarly, we can obtain the \(\hat{\mathbf{y}}\)-part for the KKT system, which implies that \((\hat{\mathbf{x}}, \hat{\mathbf{y}})\) is a KKT point of problem (6).

4. (4)

   To prove

   $$\begin{aligned} \lim _{k\rightarrow \infty }\Vert (\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)})-(\mathbf{x}^{(k)}, \mathbf{y}^{(k)})\Vert =0, \end{aligned}$$

   it suffices to prove that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\Vert =0,~~\text{ and }~~\lim _{k\rightarrow \infty }\Vert \mathbf{y}^{(k+1)}-\mathbf{y}^{(k)}\Vert =0. \end{aligned}$$

   From (23), it is not difficult to show that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\left( \Vert \mathbf{x}^{(k+1)}\Vert ^{2}-\langle \mathbf{x}^{(k+1)}, \mathbf{x}^{(k)}\rangle \right) =0, \end{aligned}$$

   which implies that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\langle \mathbf{x}^{(k+1)}, \mathbf{x}^{(k)}\rangle =\lim _{k\rightarrow \infty }\Vert \mathbf{x}^{(k+1)}\Vert ^{2}=1. \end{aligned}$$

   So, it follows that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \mathbf{x}^{(k+1)}-\mathbf{x}^{(k)}\Vert =0. \end{aligned}$$

   On the other hand, if \(f_{\alpha }(\mathbf{x}, \mathbf{y}, \mathbf{y})\) is strictly concave with respect to \(\mathbf{y}\in D\), then \(\nabla ^2_{\mathbf{y}}f_{\alpha }(\mathbf{x},\mathbf{y},\mathbf{y})\) is negative definite and \(\nabla _{\mathbf{y}}f_{\alpha }(\mathbf{x},\mathbf{y},\mathbf{y})\ne \mathbf{0}\) for any \(\mathbf{y}\in D\). By the property of concave functions, it holds that

   $$\begin{aligned}&f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k+1)}, \mathbf{y}^{(k+1)})-f_{\alpha }(\mathbf{x}^{(k+1)}, \mathbf{y}^{(k)}, \mathbf{y}^{(k)})\\&\quad <\langle \nabla _{\mathbf{y}}f_{\alpha }(\mathbf{x}^{(k+1)},\mathbf{y}^{(k)},\mathbf{y}^{(k)}), \mathbf{y}^{(k+1)}-\mathbf{y}^{(k)}\rangle . \end{aligned}$$

   Due to a similar proof of (23), we obtain that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\langle \nabla _{\mathbf{y}}f_{\alpha }(\mathbf{x}^{(k+1)},\mathbf{y}^{(k)},\mathbf{y}^{(k)}), \mathbf{y}^{(k+1)}-\mathbf{y}^{(k)}\rangle =0, \end{aligned}$$

   which implies that

   $$\begin{aligned} \lim _{k\rightarrow \infty }\left( \Vert \mathbf{y}^{(k+1)}\Vert ^{2}-\langle \mathbf{y}^{(k+1)}, \mathbf{y}^{(k)}\rangle \right) =0 \end{aligned}$$

   and

   $$\begin{aligned} \lim _{k\rightarrow \infty }\langle \mathbf{y}^{(k+1)}, \mathbf{y}^{(k)}\rangle =\lim _{k\rightarrow \infty }\Vert \mathbf{y}^{(k+1)}\Vert ^{2}=1. \end{aligned}$$

   Thus, we have

   $$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \mathbf{y}^{(k+1)}-\mathbf{y}^{(k)}\Vert =0 \end{aligned}$$

   and the desired results hold. \(\square \)

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