Barycentric rational interpolation method for solving KPP equation

1.   Introduction

A piezoelectric-type tensor is a third-order tensor, which plays an important role in physics ^{[1,2,3,4,5,6]} and engineering ^{[7,8,9,10,11]}. In particular, the largest C-eigenvalue of a piezoelectric tensor and its associated left and right C-eigenvectors play an important role in the piezoelectric effect and the converse piezoelectric effect in the solid crystal ^[12,13]. Moreover, in the process of manufacturing and developing micro/nano-electromechanical devices, the development of new multifunctional intelligent structures needs consideration piezoelectric effect ^[14,15]. In order to explore more information about piezoelectric-type tensors, Chen et al. ^[12] introduced piezoelectric-type tensors and their C-eigentriples.

Definition 1.1. [12,Definition 2.1] Let $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{n\times n\times n}$ be a third-order $n$-dimensional tensor. If the latter two indices of $\mathcal{A}$ are symmetric, i.e., $a_{ijk} = a_{ikj}$, where $i, j, k\in[n]: = \{1, 2, \ldots, n\}$, then $\mathcal{A}$ is called a piezoelectric-type tensor. When $n = 3$, $\mathcal{A}$ is called a piezoelectric tensor.

Definition 1.2. [12,Definition 2.2] Let $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor. If there is a real number $\lambda\in\mathbb{R}$, two vectors $x: = (x_{1}, x_{2}, \ldots, x_{n})^{\top}\in\mathbb{R}^{n}\backslash\{0\}$ and $y: = (y_{1}, y_{2}, \ldots, y_{n})^{\top}\in\mathbb{R}^{n}\backslash\{0\}$ such that

where

then $\lambda$ is called a C-eigenvalue of $\mathcal{A}$, and $x$ and $y$ are its associated left and right C-eigenvectors, respectively. Then, $(\lambda, x, y)$ is called a C-eigentriple of $\mathcal{A}$ and $\sigma(\mathcal{A})$ is used to denote the spectrum of $\mathcal{A}$, which is a set of all C-eigenvalues of $\mathcal{A}$.

Chen et al. provided the following results for the C-eigenvalues and C-eigenvectors of a piezoelectric-type tensor.

Property 1.1. [12,Theorem 2.3] Let $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor.

(a) The C-eigenvalues of $\mathcal{A}$ and their associated left and right C-eigenvectors always exist.

(b) Let $\lambda$ be a C-eigenvalue and $(x, y)$ be its associated left and right C-eigenvectors. Then

Furthermore, $(\lambda, x, -y)$, $(-\lambda, -x, y)$ and $(-\lambda, -x, -y)$ are also C-eigentriples of $\mathcal{A}$.

(c) Let $\lambda^{*}$ be the largest C-eigenvalue of $\mathcal{A}$. Then

As we all know, the largest C-eigenvalue of a piezoelectric-type tensor and its associated C-eigenvectors constitute the best rank-one piezoelectric-type approximation. In view of this, Liang and Yang ^[16,17] designed two methods to calculate the largest C-eigenvalue of a piezoelectric-type tensor. Later, Zhao and Luo ^[18] provided a method to calculate all C-eigentriples of a piezoelectric-type tensor by converting the C-eigenvalue problem to the Z-eigenvalue problem of tensors. Moreover, many researchers considered the C-eigenvalue localization problem and provided many C-eigenvalue localization sets ^{[19,20,21,22,23,24]}. For instance, Che et al. ^[25] presented the following Geršgorin-type C-eigenvalue localization set.

Theorem 1.1. [25,Theorem 2.1] Let $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor. Then

where

Definition 1.2 and Property 1.1 indicate that a C-eigenvalue $\lambda$ is real and both $\lambda$ and $-\lambda$ are C-eigenvalues, which implies that a C-eigenvalue localization set is always symmetric with respect to the origin. Therefore, the result of Theorem 1.1 is equivalent to $\sigma(\mathcal{A})\subseteq[-\rho_\Gamma, \rho_\Gamma]$, where $\rho_\Gamma = \max\limits_{j\in[n]}\{R_j(\mathcal{A})\}$.

The remainder of the paper is organized as follows. In Section 2, we construct a new C-eigenvalue localization set and prove that it is sharper than some existing sets. In Section 3, we provide a direct method to find all C-eigentriples when $n = 3$. In Section 4, we reviewed the practical application of C-eigentriples of a piezoelectric-type tensor to the piezoelectric effect and converse piezoelectric effect. In Section 5, we verify the effectiveness of obtained results by numerical examples. In Section 6, we give a summary of this paper.

2.   A new C-eigenvalue localization set

In this section, we provide a new C-eigenvalue localization set and prove that it is sharper than the set in Theorem 1.1. Before that, the following lemma is needed.

Lemma 2.1. [26,pp. 10, Cauchy-Schwarz inequality] Let $x = (x_1, x_2, \ldots, x_n)^{\top}\in\mathbb{R}^{n}$ and $y = (y_1, y_2, \ldots, y_n)^{\top}\in\mathbb{R}^{n}$. Then

Theorem 2.1. Let $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor. Then

where

and

Proof. Let $(\lambda, x, y)$ be a C-eigentriple of $\mathcal{A}$. Let $|y_{t}|\geq |y_{s}| \geq \max\limits_{k\in [n], k\neq t, k\neq s} \{|y_{k}|\}$. Then $0\leq |y_{t}|\leq 1$. From (1.2), we have

Taking the modulus in above equation and using the triangle inequality and Lemma 2.1, we have

i.e.,

which implies that $\lambda\in\mathcal{K}_t(\mathcal{A})$, and

i.e.,

If $|y_{s}| = 0$ in (2.1), we have $|\lambda|\leq \overline{r}_{t}^t(\mathcal{A})$, which implies that $\lambda\in\widehat{\Omega}_{t}(\mathcal{A})\subseteq\Omega(\mathcal{A})$.

If $|y_{s}| > 0$ in (2.1), then $|\lambda|\leq \overline{r}_{t}^t(\mathcal{A})+\widetilde{r}_{t}^t(\mathcal{A})\leq r_t(\mathcal{A})$, which implies that $\lambda\in\mathcal{K}_{t}(\mathcal{A})$. Now, suppose that $\lambda\notin\widehat{\Omega}_{t}(\mathcal{A})$, i.e., $|\lambda| > \overline{r}_{t}^t(\mathcal{A})$. The $s$-th equation of (1.2) is

which implies that

i.e.,

By multiplying (2.1) and (2.2) and eliminating $|y_{s}||y_{t}| > 0$, we have

which implies that $\lambda\in\Big(\widetilde{\Omega}_{t, s}(\mathcal{A})\cap\mathcal{K}_{t}(\mathcal{A})\Big)\subseteq\Omega(\mathcal{A})$, and consequently, $\sigma(\mathcal{A})\subseteq \Omega(\mathcal{A})$ by the arbitrariness of $\lambda$.

Next, we discuss the relationship between the set $\Omega(\mathcal{A})$ in Theorem 2.1 and the set $\Gamma(\mathcal{A})$ in Theorem 1.1.

Theorem 2.2. Let $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor. Then

Proof. Let $z\in\Omega(\mathcal{A})$. By Theorem 2.1, there exists $i, j\in[n]$ such that $z\in\widehat{\Omega}_{i}(\mathcal{A})$ or $z\in\Big(\widetilde{\Omega}_{i, j}(\mathcal{A})\cap\mathcal{K}_{i}(\mathcal{A})\Big)$.

Case Ⅰ. If $z\in\widehat{\Omega}_{i}(\mathcal{A})$, then

which implies that $z\in\Gamma(\mathcal{A})$.

Case Ⅱ. If $z\in\Big(\widetilde{\Omega}_{i, j}(\mathcal{A})\cap\mathcal{K}_{i}(\mathcal{A})\Big)$, then $z\in\widetilde{\Omega}_{i, j}(\mathcal{A})$ and $z\in\mathcal{K}_{i}(\mathcal{A})$. If $z\in\widetilde{\Omega}_{i, j}(\mathcal{A})$, then

If $\widetilde{r}_{i}^i(\mathcal{A})\overline{r}_{j}^i(\mathcal{A}) = 0$, then

which implies that $z\in\Gamma(\mathcal{A})$.

If $\widetilde{r}_{i}^i(\mathcal{A})\overline{r}_{j}^i(\mathcal{A}) > 0$, then

Therefore,

i.e.,

and consequently $z\in\Gamma(\mathcal{A})$.

When $z\in\mathcal{K}_{i}(\mathcal{A})$, we have

then $z\in\Gamma(\mathcal{A})$. Hence, $\Omega(\mathcal{A})\subseteq\Gamma(\mathcal{A})$.

Similarly, we can write the set $\Omega(\mathcal{A})$ as $[-\rho_{\Omega}, \rho_{\Omega}]$, where

and

Theorem 2.2 indicates that $\rho_{\Omega}\leq\rho_\Gamma$ and $[-\rho_{\Omega}, \rho_{\Omega}]\subseteq[-\rho_\Gamma, \rho_\Gamma]$.

3.   A direct method to calculate all C-eigentriples when ${\textit n}$ = 3

By the idea to find all M-eigenpairs of a fourth-order tensor in Theorem 7 of ^[27], we in this section present a direct method to find all C-eigenpair when $n = 3$.

Theorem 3.1. Let $\mathcal{A} = (a_{ijk})\in \mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor.Then all C-eigentriples are given as follows:

(a) If $a_{211} = a_{311} = a_{112} = a_{113} = 0$, then $(a_{111}, (1, 0, 0)^\top, (\pm 1, 0, 0)^{\top})$ and $(-a_{111}, (-1, 0, 0)^\top, (\pm 1, 0, 0)^\top)$ are four C-eigentriples of $\mathcal{A}$.

(b) $(\lambda, x, y)$ is a C-eigentriple of $\mathcal{A}$, where

and

$x_{1}$, $x_{2}$ and $x_{3}$ are the real roots of the following equations:

(c) $(\lambda, x, y)$ and $(-\lambda, -x, y)$ are two C-eigentriples of $\mathcal{A}$, where

and

$y_1$, $y_2$ and $y_3$ are the real roots of the following equations:

(d) $(\lambda, x, y)$ is a C-eigentriple of $\mathcal{A}$, where

and

$x_1$, $x_2$ and $x_3$ and $t$ are the real roots of the following equations:

(e) $(\lambda, x, y)$ and $(-\lambda, -x, y)$ are two C-eigentriples of $\mathcal{A}$, where

and

$y_1$, $y_2$, $y_3$ and $t$ are the real roots of the following equations:

(f) $(\lambda, x, y)$ is a C-eigentriple of $\mathcal{A}$, where

and

$u_{1}$, $u_{2}$, $v_{1}$ and $v_{2}$ are the real roots of the following equations:

Proof. When $n = 3$, the specific form of (1.1)–(1.4) is

To proceed, we break the arguments into six Cases.

(a) Assume that $x = (1, 0, 0)^{\top}$ and $y = (\pm1, 0, 0)^{\top}$. Then (3.18)–(3.25) becomes $\lambda = a_{111}$ and $a_{211} = a_{311} = a_{112} = a_{113} = 0$.

Assume that $x = (-1, 0, 0)^{\top}$ and $y = (\pm1, 0, 0)^{\top}$. Then (3.18)–(3.25) becomes $\lambda = -a_{111}$ and $a_{211} = a_{311} = a_{112} = a_{113} = 0$.

Hence, if $a_{211} = a_{311} = a_{112} = a_{113} = 0$, then $(a_{111}, (1, 0, 0)^\top, (\pm 1, 0, 0)^{\top})$ and $(-a_{111}, (-1, 0, 0)^\top, (\pm 1, 0, 0)^\top)$ are four C-eigentriples of $\mathcal{A}$.

(b) Assume that $y = (\pm1, 0, 0)^{\top}$. Then (3.18)–(3.25) becomes

By (3.26) and (3.27), we have (3.1). By (3.26) and (3.28), we have (3.2). Next, solving (3.1), (3.2), (3.30), (3.31) and (3.24), we can obtain $x_1$, $x_2$ and $x_3$, which implies that $x = (x_{1}, x_{2}, x_{3})^{\top}$ and $y = (\pm1, 0, 0)^{\top}$ are a pair of C-eigenvectors. Furthermore, by (3.29), we can get a C-eigenvalue $\lambda$ of $\mathcal{A}$.

(c) Assume that $x = (1, 0, 0)^{\top}$. Then (3.18)–(3.25) becomes

By (3.35) and (3.36), we have (3.3). By (3.35) and (3.37), we have (3.4). Next, solving (3.3), (3.4), (3.25), (3.33) and (3.34), we can obtain $y_1$, $y_2$ and $y_3$, which implies that $x = (1, 0, 0)^\top$ and $y = (y_1, y_2, y_3)^{\top}$ are a pair of C-eigenvectors. Furthermore, by (3.32), we can get a C-eigenvalue $\lambda$ of $\mathcal{A}$.

Assume that $x = (-1, 0, 0)^{\top}$. Similarly, we have

with its a pair of C-eigenvectors are $x = (-1, 0, 0)^{\top}$ and $y = (y_1, y_2, y_3)^{\top}$, which also satisfies (3.3), (3.4), (3.25), (3.33) and (3.34).

(d) Assume that $y = (y_1, y_2, 0)^\top$, where $y_2\neq 0$. Then (3.18)–(3.25) becomes

Let $t = \frac{y_{1}}{y_{2}}$. Then the above equations become

By (3.38) and (3.39), we have (3.5). By (3.38) and (3.40), we have (3.6). By (3.4) and (3.42), we have (3.7). Next, solving (3.5)–(3.7), (3.24) and (3.43), we can obtain $x_1$, $x_2$, $x_3$ and $t$, which leads to its a pair of C-eigenvectors $x = (x_1, x_2, x_3)^\top$ and $y = \pm\frac{(t, 1, 0)^{\top}}{\sqrt{t^2+1}}$. Furthermore, by (3.42), we can get a C-eigenvalue $\lambda$ of $\mathcal{A}$.

(e) Assume that $x = (x_1, x_2, 0)^\top$, where $x_2\neq 0$. Then (3.18)–(3.25) becomes

Let $t = \frac{x_{1}}{x_{2}}$. Then the above equations become

By (3.44) and (3.45), we have (3.9). By (3.47) and (3.48), we have (3.10). By (3.47) and (3.49), we have (3.11). Next, solving (3.9)–(3.11), (3.25) and (3.46), we can obtain $y_1$, $y_2$, $y_3$ and $t$. And then, by $x_1^2+x_2^2 = 1$, we have $x_2 = \pm \frac{1}{\sqrt{t^{2}+1}}$. Furthermore, by (3.45), we can get a C-eigenvalue $\lambda$ of $\mathcal{A}$ in (3.8) and its a pair of C-eigenvectors $x = \pm\frac{(t, 1, 0)^\top}{\sqrt{t^{2}+1}}$ and $y = (y_1, y_2, y_3)^{\top}$.

(f) Assume that $x = (x_1, x_2, x_3)^{\top}$ and $y = (y_1, y_2, y_3)^{\top}$, where $x_3\neq 0$ and $y_3\neq 0$. Let

Then (3.18)–(3.25) becomes

By (3.50) and (3.51), we have (3.14). By (3.50) and (3.52), we have (3.15). By (3.53) and (3.54), we have (3.16). By (3.53) and (3.55), we have (3.17). Next, solving (3.14)–(3.17), we can obtain $u_1$, $u_2$, $v_1$ and $v_2$. Furthermore, by $x_1^2+x_2^2+x_3^2 = 1$ and $y_1^2+y_2^2+y_3^2 = 1$, we can get $x$ and $y$ in (3.13); by (3.55), we can get a C-eigenvalue $\lambda$ of $\mathcal{A}$ in (3.12).

Remark 3.1. ⅰ) Solving the system of nonlinear equations in (b) of Theorem 3.1, we can choose any three of the first four equations to form a system of linear equations. If the determinant of the coefficient matrix of the system of linear equations is not equal to 0, then the system of linear equations has only zero solution and obviously it is not the solution of the last equation in the system of nonlinear equations, which implies that the system of nonlinear equations in (b) has no solution. If the determinant is equal to 0, then we need to verify that whether this solution satisfies the other two equations.

ⅱ) Solving these system of nonlinear equations in (c)–(f) of Theorem 3.1, we can use the resultant method in algebraic geometry ^[28] to find their solutions, which has been verified and is feasible in finding all M-eigenpairs of an order 4 dimension 2 tensor in Theorem 7 of ^[27].

Given two polynomials $f(x)$ and $g(x)$, where

Then the resultant of $f(x)$ and $g(x)$ is represented by ${\text{Res}}_x(f, g)$, that is, the determinant of $(n + m)\times(n + m)$ is as follows:

where the blank spaces are filled with zeros. It is showed in [28,pp. 78] that over the complex field $f(x)$ and $g(x)$ have a common root if and only if their resultant is zero.

Solving the systems (3.14)–(3.17) with four variables $u_1$, $u_2$, $v_1$ and $v_2$, we can obtain its solutions by successive elimination of variables as follows:

Step 1: Mark the four equations in the systems (3.14)–(3.17) as $f_{1}(u_{1}, u_{2}, v_{1}, v_{2})$, $f_{2}(u_{1}, u_{2}, v_{1}, v_{2})$, $f_{3}(u_{1}, u_{2}, v_{1}, v_{2})$ and $f_{4}(u_{1}, u_{2}, v_{1}, v_{2})$ in turn, and regard $f_{1}$, $f_{2}$, $f_{3}$ and $f_{4}$ as functions with $u_1$ as a variable and $u_2$, $v_1$ and $v_2$ as their coefficients. If ${\text{Res}}_{u_1}(f_1, f_2) = 0$, which is a function with $u_2$, $v_1$ and $v_2$ as three variables, then $f_1$ and $f_2$ have a common root. Similarly, ${\text{Res}}_{u_1}(f_1, f_3) = 0$ and ${\text{Res}}_{u_1}(f_1, f_4) = 0$ can be obtained and they are functions with $u_2$, $v_1$ and $v_2$ as three variables.

Step 2: Let

and

and $g_1$, $g_2$ and $g_3$ be regarded as functions with $u_2$ as a variable and $v_1$ and $v_2$ as their coefficients. If ${\text{Res}}_{u_2}(g_1, g_2) = 0$, which is a function with $v_1$ and $v_2$ as two variables, then $g_1$ and $g_2$ have a common root. Similarly, ${\text{Res}}_{u_2}(g_1, g_3) = 0$ can be obtained and it is a function with $v_1$ and $v_2$ as two variables.

Step 3: Let

and $h_1$ and $h_2$ be regarded as functions with $v_1$ as a variable and $v_2$ as a coefficient. If ${\text{Res}}_{v_1}(h_1, h_2) = 0$, which is a function with $v_2$ as a variable, then $h_1$ and $h_2$ have a common root.

Step 4: Solving the function ${\text{Res}}_{v_1}(h_1, h_2) = 0$ with $v_2$ as a variable by Matlab command '$\mathtt {solve}$', its all real solutions $v_2$ can be obtained.

Step 5: Substituting $v_2$ to $h_1(v_1, v_2) = 0$ and $h_2(v_1, v_2) = 0$ to find all their real solutions $v_1$. And then, substituting $v_2$ and $v_1$ to $g_1(u_2, v_1, v_2) = 0$, $g_2(u_2, v_1, v_2) = 0$ and $g_3(u_2, v_1, v_2) = 0$ to find all their real solutions $u_2$. Furthermore, substituting $v_2$, $v_1$ and $u_2$ to $f_i(u_1, u_2, v_1, v_2) = 0$ for $i\in [4]$ to find all their real solutions $u_1$. Then, all real roots $u_1$, $u_2$, $v_1$ and $v_2$ of (3.14)–(3.17) are obtained. Finally, by (3.12) and (3.13), we can find all C-eigentriples of $\mathcal{A}$ in the Case (f) in Theorem 3.1.

4.   Applications

It is shown in ^[12,29] that the largest C-eigenvalue $\lambda^*$ of a piezoelectric tensor determines the highest piezoelectric coupling constant, and its corresponding C-eigenvector $y^*$ is the corresponding direction of the stress where this appears. In this section, let's review its physical background, which is shown in ^[29].

In physics, for non-centrosymmetric materials, we can write the linear piezoelectric equation as

where $\mathcal{A} = (a_{ijk})\in\mathbb{R}^{3\times3\times3}$ is a piezoelectric tensor, $T = (T_{jk})\in\mathbb{R}^{3\times3}$ is the stress tensor, and $p = (p_i) \in \mathbb{R}^3$ is the electric change density displacement.

Now, it is worth considering, under what conditions can the maximal piezoelectricity be triggered under a unit uniaxial stress? In this case, the stress tensor $T$ can be rewritten as $T = yy^{\top}$ with $y^{\top}y = 1$. Then, this maximal piezoelectricity problem can be formulated into an optimization model

By a dual norm, $\|p\|_2 = \max\limits_{x^{\top}x = 1}x^{\top}p = \max\limits_{x^{\top}x = 1}x\mathcal{A}yy$ is derived and hence the above optimization model is converted to the following optimization problem

If $(x^*, y^*)$ is an optimal solution of the above optimization problem, then $\lambda^* = x^*\mathcal{A}y^*y^*$ is the largest C-eigenvalue of $\mathcal{A}$ and $y^*$ is the unit uniaxial direction that the maximal piezoelectric effect take place along.

Theorem 4.1. [29,Theorem 7.12] Let $\lambda^*$ be the largest C-eigenvalue, $x^*$ and $y^*$ be the associated C-eigenvectors of a piezoelectric tensor $\mathcal{A}$.Then, $\lambda^*$ is the maximum value of the 2-norm of the electric polarization under a unit uniaxial stress along the optimal axial direction $y^*$.

Moreover, the linear equation of the inverse piezoelectric effect is

where $S = (S_{jk})\in\mathbb{R}^{3\times3}$ is the strain tensor and $e = (e_i) \in \mathbb{R}^3$ is the electric field strength. Next, the following maximization problem is considered:

By $\|S\|_2 = \max\limits_{y^{\top}y = 1}y^{\top}Sy = \max\limits_{y^{\top}y = 1}e\mathcal{A}yy$, the above maximization problem is rewritten as

If $(e^*, y^*)$ is an optimal solution of the above optimization problem, then $\lambda^* = e^*\mathcal{A}y^*y^*$ is the largest C-eigenvalue of $\mathcal{A}$, $e^*$ and $y^*$ are its associated C-eigenvectors.

Theorem 4.2. [29,Theorem 7.13] Let $\lambda^*$ be the largest C-eigenvalue and $x^*$ and $y^*$ be its associated C-eigenvectors of a piezoelectric tensor $\mathcal{A}$.Then, $\lambda^*$ is the largest spectral norm of a strain tensor generated by the converse piezoelectric effect under unit electric field strength $\|x^*\| = 1$.

5.   Numerical examples

In this section, numerical examples are given to verify the obtained theoretical results.

Example 1. Consider the eight piezoelectric tensors in [12,Examples 1-8].

(a) The first piezoelectric tensor is $\mathcal{A}_{\rm VFeSb}$ with its nonzero entries

(b) The second piezoelectric tensor $\mathcal{A}_{\rm SiO_2}$ with its nonzero entries

(c) The third piezoelectric tensor $\mathcal{A}_{\rm Cr_2AgBiO_8}$ with its nonzero entries

(d) The fourth piezoelectric tensor $\mathcal{A}_{\rm RbTaO_3}$ with its nonzero entries

(e) The fifth piezoelectric tensor $\mathcal{A}_{\rm NaBiS_2}$ with its nonzero entries

(f) The sixth piezoelectric tensor $\mathcal{A}_{\rm LiBiB_2O_5}$ with its nonzero entries

(g) The seventh piezoelectric tensor $\mathcal{A}_{\rm KBi_2F_7}$ with its nonzero entries

(h) The eigth piezoelectric tensor $\mathcal{A}_{\rm BaNiO_3}$ with its nonzero entries

Ⅰ. Localization for all C-eigenvalues of the above eight piezoelectric tensors.

Now, we use these C-eigenvalues intervals in Theorems 2.1 and 1.1, Theorems 1 and 2 of ^[20], Theorem 2.1 of ^[13], Theorems 2.2 and 2.4 of ^[25], Theorem 2.1 of ^[24], Theorem 2.1 of ^[23], Theorem 5 of ^[19], Theorem 7 of ^[21], Theorems 2.3–2.5 of ^[18] and Theorem 2.1 of ^[22] to locate all Ceigenvalues of the above eight piezoelectric tensors. Numerical results are shown in Table 1. Since these intervals are symmetric about the origin, only their right boundaries are listed in Table 1.

In Table 1, $\lambda^{*}$ is the largest C-eigenvalue of a piezoelectric tensor; $\varrho$ and $\varrho_{\min}$ are respectively the right boundaries of the interval $[-\varrho, \varrho]$ and $[-\varrho_{\min}, \varrho_{\min}]$ obtained by Theorems 1 and 2 in ^[20]; $\widetilde{\varrho}_{\min}$ are respectively the right boundary of the interval $[-\widetilde{\varrho}_{\min}, \widetilde{\varrho}_{\min}]$ obtained by Theorem 2.1 of ^[13]; $\rho_{\mathcal{L}}$ and $\rho_{\mathcal{M}}$ are respectively the right boundaries of the intervals $[-\rho_{\mathcal{L}}, \rho_{\mathcal{L}}]$ and $[-\rho_{\mathcal{M}}, \rho_{\mathcal{M}}]$ obtained by 2.2 and 2.4 in ^[25]; $\rho_{\Upsilon}$ is the right boundary of the interval $[-\rho_{\Upsilon}, \rho_{\Upsilon}]$ obtained by Theorem 2.1 of ^[24]; $\rho_{\gamma}$ is the right boundary of the interval $[-\rho_{\gamma}, \rho_{\gamma}]$ obtained by Theorem 2.1 of ^[23]; $\rho_{\Omega^{S}}$ is the right boundary of the interval $[-\rho_{\Omega^{S}}, \rho_{\Omega^{S}}]$ obtained by Theorem 5 of ^[19]; $\rho_{C}$ is the right boundary of the interval $[-\rho_{C}, \rho_{C}]$ obtained by Theorem 7 of ^[21]; $\rho_{G}$, $\rho_{B}$ and $\rho_{\min}$ are respectively the right boundaries of the intervals $[-\rho_{G}, \rho_{G}]$, $[-\rho_{B}, \rho_{B}]$ and $[-\rho_{\min}, \rho_{\min}]$obtained by Theorems 2.3–2.5 in ^[18]; $\rho_{\Psi}$ is the right boundary of the interval $[-\rho_{\Psi}, \rho_{\Psi}]$ obtained by Theorem 2.1 of ^[22]; $\rho_{\Gamma}$ is the right boundary of the interval $[-\rho_{\Gamma}, \rho_{\Gamma}]$ obtained by Theorem 1.1;$\rho_{\Omega}$ is the right boundary of the interval $[-\rho_{\Omega}, \rho_{\Omega}]$ obtained by Theorem 2.1.

From Table 1, it can be seen that:

ⅰ) $\rho_{\Omega}$ is smaller than$\varrho$, $\varrho_{\min}$, $\widetilde{\varrho}_{\min}$, $\rho_{\Gamma}$, $\rho_{\mathcal{L}}$, $\rho_{\mathcal{M}}$, $\rho_{\Upsilon}$, $\rho_{\gamma}$for the eight piezoelectric tensors.

ⅱ) $\rho_{\Omega}\leq \rho_{\Omega^S}$, $\rho_{\Omega}\leq \rho_B$ for the eight piezoelectric tensors.

ⅲ) For some tensors, $\rho_{\Omega}$ is smaller than $\rho_C$, $\rho_G$, $\rho_{\min}$ and $\rho_{\Psi}$.For the other tensors, $\rho_{\Omega}$ is bigger than or equal to $\rho_{C}$, $\rho_{G}$, $\rho_{\min}$ and $\rho_{\Psi}$.For examples, for $\mathcal{A}_{\rm VFeSb}$, $\rho_{\Omega} < \rho_C$, $\rho_\Omega < \rho_G$, $\rho_\Omega < \rho_{\min}$ and $\rho_\Omega < \rho_{\Psi}$; For $\mathcal{A}_{\rm SiO_2}$, $\rho_{\Omega} > \rho_C$, $\rho_\Omega > \rho_{\min}$ and $\rho_\Omega > \rho_{\Psi}$; For $\mathcal{A}_{\rm BaNiO_3}$, $\rho_{\Omega} > \rho_G$.

Ⅱ. Calculation for all C-eigentriples of the seventh piezoelectric tensor $\mathcal{A}_{\rm KBi_2F_7}$ by Theorem 3.1.

All C-eigentriples of $\mathcal{A}_{\rm KBi_2F_7}$ are obtained by Theorem 3.1 and are showen in Table 2. And the calculation process is shown in Appendix.

6.   Conclusions

Let $\mathcal{A}\in \mathbb{R}^{n\times n\times n}$ be a piezoelectric-type tensor. In this paper, we in Theorem 2.1 constructed a C-eigenvalue interval $\Omega(\mathcal{A})$ to locate all C-eigenvalues of $\mathcal{A}$ and proved that it is tighter than that in [25,Theorem 2.1]. Subsequently, we in Theorem 3.1 provided a direct method to find all C-eigentriples of $\mathcal{A}$ when $n = 3$. Although the method in Theorem 3.1 is divided into six Cases, it is indeed a little complicated, but it can be seen from Example 1 that this method is feasible.

Acknowledgments

The authors are very grateful to the five anonymous referees for their insightful comments and constructive suggestions, which considerably improve our manuscript. Caili Sang's work is supported by Science and Technology Plan Project of Guizhou Province (Grant No. QKHJC-ZK[2021]YB013). Jianxing Zhao's work is supported by Science and Technology Plan Project of Guizhou Province (Grant No. QKHJC-ZK[2022]YB215).

Conflict of interest

The author declares there is no confict of interest.

Appendix

The following is the calculation process for all C-eigentriples of $\mathcal{A}_{\rm KBi_2F_7}$ by Theorem 3.1.

(a) Because $a_{211}\neq 0$, Case (a) in Theorem 3.1 does not holds.

(b) The system in Case (b) of Theorem 3.1 is

The three Eqs (A.1)–(A.3) yield a linear system of equation $Ax = 0$, where $x = (x_{1}, x_{2}, x_{3})^{\top}$ and

From $\det(A) = 179.0074\neq 0$, the solution of $Ax = 0$ is $x = (x_1, x_2, x_3)^\top = (0, 0, 0)^\top$, which contradicts with $x_1^2+x_2^2+x_3^2 = 1$. Hence, the system in Case (b) of Theorem 3.1 has no solution.

(c) The system in Case (c) of Theorem 3.1 is

We now regard $f_i(y_1, y_2, y_3)$, $i\in [5]$ as a function with $y_1$ as a variable and $y_2$ and $y_3$ as two coefficients and obtain their resultants as follows:

Let $g_1(y_2, y_3): = {\text{Res}}_{y_1}(f_1, f_2)$, $g_2(y_2, y_3): = {\text{Res}}_{y_1}(f_1, f_3)$, and its resultant

Then $y_3 = 0$. Substituting $y_3 = 0$ into $g_1$ and $g_2$, we have

Let $g_1(y_2, y_3) = 0$ and $g_2(y_2, y_3) = 0$. We have $y_2 = 0$. Substituting $y_2 = 0$ and $y_3 = 0$ into $f_1$, we have $f_1(y_1, y_2, y_3) = 2.59187y_1^2$. Solving $f_1(y_1, y_2, y_3) = 0$, we have $y_1 = 0$. However, $y_1 = y_2 = y_3 = 0$ is not solution of $y_1^2+y_2^2+y_3^2 = 1$. Hence, the system in Case (c) of Theorem 3.1 has no solution.

(d) Similar to solution for Case (c), the system in Case (d) of Theorem 3.1 has no solution.

(e) Similar to solution for Case (c), the system in Case (e) of Theorem 3.1 has no solution.

(f) The system in Case (f) of Theorem 3.1 is

We now regard $f_i(u_1, u_2, v_1, v_2)$, $i\in [4]$ as a function with $u_1$ as a variable and $u_2$, $v_1$ and $v_2$ as three coefficients and obtain their resultants as follows:

and

Let

Then their resultants are

and

Let

Their results are

Next, we obtain the solution of the system (f) by the following steps:

Step 1. Solving ${\text{Res}}_{v_1}(h_1, h_2) = 0$, we have

Step 2. Substituting $v_2 = -0.85035$ into $h_1(v_1, v_2)$, and letting $h_1(v_1, v_2) = 0$, it all real roots are $v_1 = -0.43450$ or $v_1 = 4.25751$. Substituting $v_2 = -0.85035$ into $h_2(v_1, v_2)$, and letting $h_2(v_1, v_2) = 0$, it all real roots are $v_1 = -0.48184$, $-0.17296$, 0.0000000000038876, or 5.04918. It is easy to see that $h_1(v_1, v_2) = 0$ and $h_2(v_1, v_2) = 0$ have no common solution, which implies that $v_2 = -0.85035$ is not a solution of the system (f).

Step 3. Substituting $v_2 = 0.22179$ into $h_1(v_1, v_2)$, and letting $h_1(v_1, v_2) = 0$, it all real roots are $v_1 = -0.43450$ or 4.25751. substituting $v_2 = 0.22179$ into $h_2(v_1, v_2)$, and letting $h_2(v_1, v_2) = 0$, it all real roots are $v_1 = -0.24034$ or 4.25751. It is easy to see that $v_1 = 4.25751$ is a common solution of $h_1(v_1, v_2) = 0$ and $h_2(v_1, v_2) = 0$.

Step 4. Substituting $v_2 = 0.22179$ and $v_1 = 4.25751$ into $g_1(u_2, v_1, v_2)$, $g_2(u_2, v_1, v_2)$ and $g_3(u_2, v_1, v_2)$, and letting $g_1(u_2, v_1, v_2) = 0$, its all real roots are $u_2 = 1.76393$; letting $g_2(u_2, v_1, v_2) = 0$, its all real roots are $u_2 = 1.76393$; letting $g_3(u_2, v_1, v_2) = 0$, its all real roots are $u_2 = 1.76393$. Hence, the common solution of $g_1(u_2, v_1, v_2) = 0$, $g_2(u_2, v_1, v_2) = 0$ and $g_3(u_2, v_1, v_2) = 0$ is $u_2 = 1.76393$.

Step 5. Substituting $v_2 = 0.22179$, $v_1 = 4.25751$ and $u_2 = 1.76393$ into $f_1$, and letting $f_1(u_1, u_2, v_1, v_2) = 0$, we can get its all real roots $u_1 = 8.16186$.

Step 6. By $v_2 = 0.22179$, $v_1 = 4.25751$, $u_2 = 1.76393$, $u_1 = 8.16186$, (3.2) and (3.3), we can get the corresponding C-eigentriples as follows:

$\bullet$ $\lambda = 13.50214$ and its C-eigenvectors are

$\bullet$ $\lambda = -13.50214$ and its C-eigenvectors are

Step 7. For other values of $v_2$, such as, $-7.79725$, $-1.78254$, $-11.53378$, $-0.75907$, $-0.70532$, 0.47100, we can also obtain their corresponding C-eigentriples by using the method similar to Steps 3–6.

Finally, we find all C-eigentriples, which is listed in Table 2.