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Localization and calculation for C-eigenvalues of a piezoelectric-type tensor


  • The largest C-eigenvalue of a piezoelectric tensor determines the highest piezoelectric coupling constant. In this paper, we first provide a new C-eigenvalue localization set for a piezoelectric-type tensor and prove that it is tighter than some existing sets. And then, we present a direct method to find all C-eigentriples of a piezoelectric-type tensor of dimension 3. Finally, we show the effectiveness of the direct method by numerical examples.

    Citation: Shunjie Bai, Caili Sang, Jianxing Zhao. Localization and calculation for C-eigenvalues of a piezoelectric-type tensor[J]. Electronic Research Archive, 2022, 30(4): 1419-1441. doi: 10.3934/era.2022074

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  • The largest C-eigenvalue of a piezoelectric tensor determines the highest piezoelectric coupling constant. In this paper, we first provide a new C-eigenvalue localization set for a piezoelectric-type tensor and prove that it is tighter than some existing sets. And then, we present a direct method to find all C-eigentriples of a piezoelectric-type tensor of dimension 3. Finally, we show the effectiveness of the direct method by numerical examples.



    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 A=(aijk)Rn×n×n be a third-order n-dimensional tensor. If the latter two indices of A are symmetric, i.e., aijk=aikj, where i,j,k[n]:={1,2,,n}, then A is called a piezoelectric-type tensor. When n=3, A is called a piezoelectric tensor.

    Definition 1.2. [12,Definition 2.2] Let A=(aijk)Rn×n×n be a piezoelectric-type tensor. If there is a real number λR, two vectors x:=(x1,x2,,xn)Rn{0} and y:=(y1,y2,,yn)Rn{0} such that

    {Ayy=λx,(1.1)xAy=λy,(1.2)xx=1,(1.3)yy=1,(1.4)

    where

    (Ayy)i:=j,k[n]aijkyjyk,(xAy)k:=i,j[n]aijkxiyj,

    then λ is called a C-eigenvalue of A, and x and y are its associated left and right C-eigenvectors, respectively. Then, (λ,x,y) is called a C-eigentriple of A and σ(A) is used to denote the spectrum of A, which is a set of all C-eigenvalues of 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 A=(aijk)Rn×n×n be a piezoelectric-type tensor.

    (a) The C-eigenvalues of A and their associated left and right C-eigenvectors always exist.

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

    λ=xAyy:=i,j,k[n]aijkxiyjyk.

    Furthermore, (λ,x,y), (λ,x,y) and (λ,x,y) are also C-eigentriples of A.

    (c) Let λ be the largest C-eigenvalue of A. Then

    λ=max{xAyy:xx=1,yy=1}.

    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 A=(aijk)Rn×n×n be a piezoelectric-type tensor. Then

    σ(A)Γ(A):=j[n]Γj(A),

    where

    Γj(A):={zR:|z|Rj(A)}andRj(A):=l,k[n]|alkj|.

    Definition 1.2 and Property 1.1 indicate that a C-eigenvalue λ is real and both λ and λ 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 σ(A)[ρΓ,ρΓ], where ρΓ=maxj[n]{Rj(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.

    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=(x1,x2,,xn)Rn and y=(y1,y2,,yn)Rn. Then

    (i[n]xiyi)2i[n]x2ii[n]y2i.

    Theorem 2.1. Let A=(aijk)Rn×n×n be a piezoelectric-type tensor. Then

    σ(A)Ω(A):=(i[n]ˆΩi(A))(i,j[n],ij(˜Ωi,j(A)Ki(A))),

    where

    ˆΩi(A):={zR:|z|¯rii(A)},˜Ωi,j(A):={zR:(|z|¯rii(A))(|z|˜rij(A))˜rii(A)¯rij(A)},Ki(A):={zR:|z|ri(A)},

    and

    ¯rij(A):=l[n]a2lij,˜rij(A):=l[n](k[n],ki|alkj|)2,ri(A):=l[n](k[n]|alki|)2,i,j[n].

    Proof. Let (λ,x,y) be a C-eigentriple of A. Let |yt||ys|maxk[n],kt,ks{|yk|}. Then 0|yt|1. From (1.2), we have

    λyt=l,k[n]alktxlyk=l[n]alttxlyt+l,k[n],ktalktxlyk.

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

    |λ||yt|l,k[n]|alkt||xl||yt|=|yt|l[n](|xl|k[n]|alkt|)|yt|l[n]|xl|2l[n](k[n]|alkt|)2=|yt|l[n](k[n]|alkt|)2,

    i.e.,

    |λ|l[n](k[n]|alkt|)2=rt(A),

    which implies that λKt(A), and

    |λ||yt|l[n]|altt||xl||yt|+l,k[n],kt|alkt||xl||yk|l[n]|altt||xl||yt|+l,k[n],kt|alkt||xl||ys|=|yt|l[n]|altt||xl|+|ys|l[n](|xl|k[n],kt|alkt|)|yt|l[n]a2lttl[n]x2l+|ys|l[n]x2ll[n](k[n],kt|alkt|)2=|yt|l[n]a2ltt+|ys|l[n](k[n],kt|alkt|)2=¯rtt(A)|yt|+˜rtt(A)|ys|,

    i.e.,

    (|λ|¯rtt(A))|yt|˜rtt(A)|ys|. (2.1)

    If |ys|=0 in (2.1), we have |λ|¯rtt(A), which implies that λˆΩt(A)Ω(A).

    If |ys|>0 in (2.1), then |λ|¯rtt(A)+˜rtt(A)rt(A), which implies that λKt(A). Now, suppose that λˆΩt(A), i.e., |λ|>¯rtt(A). The s-th equation of (1.2) is

    λys=l,k[n]alksxlyk=l[n]altsxlyt+l,k[n],ktalksxlyk,

    which implies that

    |λ||ys|l[n]|alts||xl||yt|+l,k[n],kt|alks||xl||yk|l[n]|alts||xl||yt|+l,k[n],kt|alks||xl||ys|=|yt|l[n]|alts||xl|+|ys|l[n](|xl|k[n],kt|alks|)|yt|l[n]a2ltsl[n]x2l+|ys|l[n]x2ll[n](k[n],kt|alks|)2=|yt|l[n]a2lts+|ys|l[n](k[n],kt|alks|)2=¯rts(A)|yt|+˜rts(A)|ys|,

    i.e.,

    (|λ|˜rts(A))|ys|¯rts(A)|yt|. (2.2)

    By multiplying (2.1) and (2.2) and eliminating |ys||yt|>0, we have

    (|λ|¯rtt(A))(|λ|˜rts(A))˜rtt(A)¯rts(A),

    which implies that λ(˜Ωt,s(A)Kt(A))Ω(A), and consequently, σ(A)Ω(A) by the arbitrariness of λ.

    Next, we discuss the relationship between the set Ω(A) in Theorem 2.1 and the set Γ(A) in Theorem 1.1.

    Theorem 2.2. Let A=(aijk)Rn×n×n be a piezoelectric-type tensor. Then

    Ω(A)Γ(A).

    Proof. Let zΩ(A). By Theorem 2.1, there exists i,j[n] such that zˆΩi(A) or z(˜Ωi,j(A)Ki(A)).

    Case Ⅰ. If zˆΩi(A), then

    |z|¯rii(A)Ri(A),

    which implies that zΓ(A).

    Case Ⅱ. If z(˜Ωi,j(A)Ki(A)), then z˜Ωi,j(A) and zKi(A). If z˜Ωi,j(A), then

    (|z|¯rii(A))(|z|˜rij(A))˜rii(A)¯rij(A).

    If ˜rii(A)¯rij(A)=0, then

    |z|¯rii(A)Ri(A)or|z|˜rij(A)Rj(A),

    which implies that zΓ(A).

    If ˜rii(A)¯rij(A)>0, then

    |z|¯rii(A)˜rii(A)|z|˜rij(A)¯rij(A)1.

    Therefore,

    |z|¯rii(A)˜rii(A)1or|z|˜rij(A)¯rij(A)1,

    i.e.,

    |z|¯rii(A)+˜rii(A)Ri(A)or|z|˜rij(A)+¯rij(A)Rj(A),

    and consequently zΓ(A).

    When zKi(A), we have

    |z|ri(A)Ri(A),

    then zΓ(A). Hence, Ω(A)Γ(A).

    Similarly, we can write the set Ω(A) as [ρΩ,ρΩ], where

    ρΩ:=max{maxi[n]{¯rii(A)},maxi,j[n],ijmin{12νi,j(A),ri(A)}}, (2.3)

    and

    νi,j(A):=˜rij(A)+¯rii(A)+(˜rij(A)¯rii(A))2+4˜rii(A)¯rij(A).

    Theorem 2.2 indicates that ρΩρΓ and [ρΩ,ρΩ][ρΓ,ρΓ].

    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 A=(aijk)Rn×n×n be a piezoelectric-type tensor.Then all C-eigentriples are given as follows:

    (a) If a211=a311=a112=a113=0, then (a111,(1,0,0),(±1,0,0)) and (a111,(1,0,0),(±1,0,0)) are four C-eigentriples of A.

    (b) (λ,x,y) is a C-eigentriple of A, where

    λ=a111x1+a211x2+a311x3,

    and

    x=(x1,x2,x3),y=(±1,0,0),

    x1, x2 and x3 are the real roots of the following equations:

    {a211x1a111x2=0,(3.1)a311x1a111x3=0,(3.2)a112x1+a212x2+a312x3=0,a113x1+a213x2+a313x3=0,x21+x22+x23=1.

    (c) (λ,x,y) and (λ,x,y) are two C-eigentriples of A, where

    λ=a111y21+a122y22+a133y23+2a112y1y2+2a113y1y3+2a123y2y3,

    and

    x=(1,0,0),y=(y1,y2,y3),

    y1, y2 and y3 are the real roots of the following equations:

    {a222y22+a211y21+a233y23+2a221y2y1+2a232y3y2+2a231y3y1=0,a333y23+a311y21+a322y22+2a331y3y1+2a323y2y3+2a321y2y1=0,a112y21+(a122a111)y1y2+a132y1y3a121y22a131y2y3=0,(3.3)a113y21+(a133a111)y1y3+a123y1y2a121y2y3a131y23=0,(3.4)y21+y22+y23=1.

    (d) (λ,x,y) is a C-eigentriple of A, where

    λ=a112x1t+a122x1+a222x2+a212x2t+a322x3+a312x3t,

    and

    x=(x1,x2,x3),y=±(t,1,0)t2+1,

    x1, x2 and x3 and t are the real roots of the following equations:

    {a222x1+a211x1t2+2a221x1ta111x2t2a122x22a112tx2=0,(3.5)a311t2x1+a322x1+2a321x1ta111x3t2a122x32a112x3t=0,(3.6)a112x1t2+(a122a111)x1t+(a222a211)x2t+(a322a311)x3t+a212x2t2+a312x3t2a321x3a221x2a121x1=0,(3.7)a113x1t+a123x1+a223x2+a213x2t+a313x3t+a323x3=0,x21+x22+x23=1.

    (e) (λ,x,y) and (λ,x,y) are two C-eigentriples of A, where

    λ=t2+1(a222y22+a211y22+a233y23+2a221y1y2+2a232y2y3+2a231y1y3), (3.8)

    and

    x=(t,1,0)t2+1,y=(y1,y2,y3),

    y1, y2, y3 and t are the real roots of the following equations:

    {a222y22t+a211y21t+a233y23t+2a221y2y1t+2a232y3y2t+2a231y3y1ta111y21a122y22a133y232a112y1y22a113y1y32a123y2y3=0,(3.9)a333y23+a311y21+a322y22+2a331y3y1+2a323y2y3+2a321y2y1=0,a112ty21+(a122a111)ty1y2+a132ty1y3+(a222a211)y2y1+a212y21+a232y1y3a121ty22a131y2ty3a221y22a231y2y3=0,(3.10)a113ty21+(a133a111)ty1y3+a123ty1y2+a223y1y2+a213y21+(a233a211)y1y3a121ty2y3a131ty23a221y2y3a231y23=0,(3.11)y21+y22+y23=1.

    (f) (λ,x,y) is a C-eigentriple of A, where

    λ=±(a113u1v1+a123u1v2+a133u1+a213u2v1+a223u2v2+a233u2+a313v1+a323v2+a333)/u21+u22+1, (3.12)

    and

    x=±(u1,u2,1)u21+u22+1,y=±(v1,v2,1)v21+v22+1, (3.13)

    u1, u2, v1 and v2 are the real roots of the following equations:

    {a222u1v22+a211v21u1+a233u1+2a221v1u1v2+2a232u1v2+2a231v1u1a111u2v21a122u2v22a133u22a112u2v1v22a113u2v12a123u2v2=0,(3.14)a333u1+a311u1v21+a322u1v22+2a331u1v1+2a323u1v2+2a321u1v1v2a111v21a122v22a1332a112v1v22a113v12a123v2=0,(3.15)a112u1v21+(a122a111)u1v1v2+a132v1u1+(a222a211)u2v1v2+a212u2v21+a232v1u2+a332v1+(a322a311)v1v2+a312v21a121u1v22a131u1v2a221u2v22a231v2u2a331v2a321v22=0,(3.16)a113u1v21+(a133a111)u1v1+a123v1u1v2+a223v1u2v2+a323v1v2+(a233a211)u2v1+a213u2v21+(a333a311)v1+a313v21a121u1v2a131u1a221u2v2a231u2a331a321v2=0.(3.17)

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

    {a111y21+a122y22+a133y23+2a112y1y2+2a113y1y3+2a123y2y3=λx1,(3.18)a222y22+a211y21+a233y23+2a221y2y1+2a232y3y2+2a231y3y1=λx2,(3.19)a333y23+a311y21+a322y22+2a331y3y1+2a323y2y3+2a321y2y1=λx3,(3.20)a111x1y1+a121x1y2+a131x1y3+a221x2y2+a211x2y1+a231x2y3+a331x3y3+a311x3y1+a321x3y2=λy1,(3.21)a112x1y1+a122x1y2+a132x1y3+a222x2y2+a212x2y1+a232x2y3+a332x3y3+a322x3y2+a312x3y1=λy2,(3.22)a113x1y1+a133x1y3+a123x1y2+a223x2y2+a233x2y3+a213x2y1+a333x3y3+a313x3y1+a323x3y2=λy3,(3.23)x21+x22+x23=1,(3.24)y21+y22+y23=1.(3.25)

    To proceed, we break the arguments into six Cases.

    (a) Assume that x=(1,0,0) and y=(±1,0,0). Then (3.18)–(3.25) becomes λ=a111 and a211=a311=a112=a113=0.

    Assume that x=(1,0,0) and y=(±1,0,0). Then (3.18)–(3.25) becomes λ=a111 and a211=a311=a112=a113=0.

    Hence, if a211=a311=a112=a113=0, then (a111,(1,0,0),(±1,0,0)) and (a111,(1,0,0),(±1,0,0)) are four C-eigentriples of A.

    (b) Assume that y=(±1,0,0). Then (3.18)–(3.25) becomes

    {a111=λx1,(3.26)a211=λx2,(3.27)a311=λx3,(3.28)a111x1+a211x2+a311x3=λ,(3.29)a112x1+a212x2+a312x3=0,(3.30)a113x1+a213x2+a313x3=0,(3.31)x21+x22+x23=1.

    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 x1, x2 and x3, which implies that x=(x1,x2,x3) and y=(±1,0,0) are a pair of C-eigenvectors. Furthermore, by (3.29), we can get a C-eigenvalue λ of A.

    (c) Assume that x=(1,0,0). Then (3.18)–(3.25) becomes

    {a111y21+a122y22+a133y23+2a112y1y2+2a113y1y3+2a123y2y3=λ,(3.32)a222y22+a211y21+a233y23+2a221y2y1+2a232y3y2+2a231y3y1=0,(3.33)a333y23+a311y21+a322y22+2a331y3y1+2a323y2y3+2a321y2y1=0,(3.34)a111y1+a121y2+a131y3=λy1,(3.35)a112y1+a122y2+a132y3=λy2,(3.36)a113y1+a133y3+a123y2=λy3,(3.37)y21+y22+y23=1.

    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 y1, y2 and y3, which implies that x=(1,0,0) and y=(y1,y2,y3) are a pair of C-eigenvectors. Furthermore, by (3.32), we can get a C-eigenvalue λ of A.

    Assume that x=(1,0,0). Similarly, we have

    λ=(a111y21+a122y22+a133y23+2a112y1y2+2a113y1y3+2a123y2y3),

    with its a pair of C-eigenvectors are x=(1,0,0) and y=(y1,y2,y3), which also satisfies (3.3), (3.4), (3.25), (3.33) and (3.34).

    (d) Assume that y=(y1,y2,0), where y20. Then (3.18)–(3.25) becomes

    {a111y21+a122y22+2a112y1y2=λx1,a222y22+a211y21+2a221y2y1=λx2,a311y21+a322y22+2a321y2y1=λx3,a111x1y1+a121x1y2+a221x2y2+a211x2y1+a311x3y1+a321x3y2=λy1,a112x1y1+a122x1y2+a222x2y2+a212x2y1+a322x3y2+a312x3y1=λy2,a113x1y1+a123x1y2+a223x2y2+a213x2y1+a313x3y1+a323x3y2=0,x21+x22+x23=1,y21+y22=1.

    Let t=y1y2. Then the above equations become

    {(a111t2+a122+2a112t)y22=λx1,(3.38)(a222+a211t2+2a221t)y22=λx2,(3.39)(a311t2+a322+2a321t)y22=λx3,(3.40)a111x1t+a121x1+a221x2+a211x2t+a311x3t+a321x3=λt,(3.41)a112x1t+a122x1+a222x2+a212x2t+a322x3+a312x3t=λ,(3.42)a113x1t+a123x1+a223x2+a213x2t+a313x3t+a323x3=0,(3.43)x21+x22+x23=1.

    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 x1, x2, x3 and t, which leads to its a pair of C-eigenvectors x=(x1,x2,x3) and y=±(t,1,0)t2+1. Furthermore, by (3.42), we can get a C-eigenvalue λ of A.

    (e) Assume that x=(x1,x2,0), where x20. Then (3.18)–(3.25) becomes

    {a111y21+a122y22+a133y23+2a112y1y2+2a113y1y3+2a123y2y3=λx1,a222y22+a211y21+a233y23+2a221y2y1+2a232y3y2+2a231y3y1=λx2,a333y23+a311y21+a322y22+2a331y3y1+2a323y2y3+2a321y2y1=0,a111x1y1+a121x1y2+a131x1y3+a221x2y2+a211x2y1+a231x2y3=λy1,a112x1y1+a122x1y2+a132x1y3+a222x2y2+a212x2y1+a232x2y3=λy2,a113x1y1+a133x1y3+a123x1y2+a223x2y2+a233x2y3+a213x2y1=λy3,x21+x22=1,y21+y22+y23=1.

    Let t=x1x2. Then the above equations become

    {a111y21+a122y22+a133y23+2a112y1y2+2a113y1y3+2a123y2y3=λtx2,(3.44)a222y22+a211y21+a233y23+2a221y2y1+2a232y3y2+2a231y3y1=λx2,(3.45)a333y23+a311y21+a322y22+2a331y3y1+2a323y2y3+2a321y2y1=0,(3.46)a111tx2y1+a121tx2y2+a131tx2y3+a221x2y2+a211x2y1+a231x2y3=λy1,(3.47)a112tx2y1+a122tx2y2+a132tx2y3+a222x2y2+a212x2y1+a232x2y3=λy2,(3.48)a113tx2y1+a133tx2y3+a123tx2y2+a223x2y2+a233x2y3+a213x2y1=λy3,(3.49)y21+y22+y23=1.

    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 y1, y2, y3 and t. And then, by x21+x22=1, we have x2=±1t2+1. Furthermore, by (3.45), we can get a C-eigenvalue λ of A in (3.8) and its a pair of C-eigenvectors x=±(t,1,0)t2+1 and y=(y1,y2,y3).

    (f) Assume that x=(x1,x2,x3) and y=(y1,y2,y3), where x30 and y30. Let

    u1=x1x3,u2=x2x3,v1=y1y3,v2=y2y3.

    Then (3.18)–(3.25) becomes

    {a111v21y23+a122v22y23+a133y23+2a112v1v2y23+2a113v1y23+2a123v2y23=λu1x3,(3.50)a222v22y23+a211v21y23+a233y23+2a221v1v2y23+2a232v2y23+2a231v1y23=λu2x3,(3.51)a333y23+a311v21y23+a322v22y23+2a331v1y23+2a323v2y23+2a321v1v2y23=λx3,(3.52)a111u1v1x3+a121u1v2x3+a131u1x3+a221u2v2x3+a211u2v1x3+a231u2x3+a331x3+a311v1x3+a321v2x3=λv1,(3.53)a112u1v1x3+a122u1v2x3+a132u1x3+a222u2v2x3+a212u2v1x3+a232u2x3+a332x3+a322v2x3+a312v1x3=λv2,(3.54)a113u1v1x3+a133u1x3+a123u1v2x3+a223u2v2x3+a233u2x3+a213u2v1x3+a333x3+a313v1x3+a323v2x3=λ.(3.55)

    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 u1, u2, v1 and v2. Furthermore, by x21+x22+x23=1 and y21+y22+y23=1, we can get x and y in (3.13); by (3.55), we can get a C-eigenvalue λ of 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

    f(x)=a0xn+a1xn1++an,a00,n>0.g(x)=b0xm+b1xm1++bm,b00,m>0.

    Then the resultant of f(x) and g(x) is represented by Resx(f,g), that is, the determinant of (n+m)×(n+m) is as follows:

    Resx(f,g)=det(a0a1a0a2a1a2a0ama1ama2amncolumnsd0d1d0d2d1d2b0dnb1dnb2bn)mcolumns,

    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 u1, u2, v1 and v2, 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 f1(u1,u2,v1,v2), f2(u1,u2,v1,v2), f3(u1,u2,v1,v2) and f4(u1,u2,v1,v2) in turn, and regard f1, f2, f3 and f4 as functions with u1 as a variable and u2, v1 and v2 as their coefficients. If Resu1(f1,f2)=0, which is a function with u2, v1 and v2 as three variables, then f1 and f2 have a common root. Similarly, Resu1(f1,f3)=0 and Resu1(f1,f4)=0 can be obtained and they are functions with u2, v1 and v2 as three variables.

    Step 2: Let

    g1(u2,v1,v2):=Resu1(f1,f2)=0,g2(u2,v1,v2):=Resu1(f1,f3)=0,

    and

    g3(u2,v1,v2):=Resu1(f1,f4)=0,

    and g1, g2 and g3 be regarded as functions with u2 as a variable and v1 and v2 as their coefficients. If Resu2(g1,g2)=0, which is a function with v1 and v2 as two variables, then g1 and g2 have a common root. Similarly, Resu2(g1,g3)=0 can be obtained and it is a function with v1 and v2 as two variables.

    Step 3: Let

    h1(v1,v2):=Resu2(g1,g2)=0andh2(v1,v2):=Resu2(g1,g3)=0.

    and h1 and h2 be regarded as functions with v1 as a variable and v2 as a coefficient. If Resv1(h1,h2)=0, which is a function with v2 as a variable, then h1 and h2 have a common root.

    Step 4: Solving the function Resv1(h1,h2)=0 with v2 as a variable by Matlab command 'solve', its all real solutions v2 can be obtained.

    Step 5: Substituting v2 to h1(v1,v2)=0 and h2(v1,v2)=0 to find all their real solutions v1. And then, substituting v2 and v1 to g1(u2,v1,v2)=0, g2(u2,v1,v2)=0 and g3(u2,v1,v2)=0 to find all their real solutions u2. Furthermore, substituting v2, v1 and u2 to fi(u1,u2,v1,v2)=0 for i[4] to find all their real solutions u1. Then, all real roots u1, u2, v1 and v2 of (3.14)–(3.17) are obtained. Finally, by (3.12) and (3.13), we can find all C-eigentriples of A in the Case (f) in Theorem 3.1.

    It is shown in [12,29] that the largest C-eigenvalue λ 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

    pi=j,k[3]aijkTjk,

    where A=(aijk)R3×3×3 is a piezoelectric tensor, T=(Tjk)R3×3 is the stress tensor, and p=(pi)R3 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 with yy=1. Then, this maximal piezoelectricity problem can be formulated into an optimization model

    {maxp2s.t.p=Ayy,yy=1.

    By a dual norm, p2=maxxx=1xp=maxxx=1xAyy is derived and hence the above optimization model is converted to the following optimization problem

    maxxAyys.t.xx=1,yy=1.

    If (x,y) is an optimal solution of the above optimization problem, then λ=xAyy is the largest C-eigenvalue of A and y is the unit uniaxial direction that the maximal piezoelectric effect take place along.

    Theorem 4.1. [29,Theorem 7.12] Let λ be the largest C-eigenvalue, x and y be the associated C-eigenvectors of a piezoelectric tensor A.Then, λ 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

    Sjk=iaijkei,

    where S=(Sjk)R3×3 is the strain tensor and e=(ei)R3 is the electric field strength. Next, the following maximization problem is considered:

    {maxS2:=maxyy=1ySys.t.Sjk=i[3]eiaijk,j,k[3],ee=1.

    By S2=maxyy=1ySy=maxyy=1eAyy, the above maximization problem is rewritten as

    max{eAyy:ee=1,yy=1}.

    If (e,y) is an optimal solution of the above optimization problem, then λ=eAyy is the largest C-eigenvalue of A, e and y are its associated C-eigenvectors.

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

    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 AVFeSb with its nonzero entries

    a123=a213=a312=3.68180667.

    (b) The second piezoelectric tensor ASiO2 with its nonzero entries

    a111=a122=a212=0.13685,a123=a213=0.009715.

    (c) The third piezoelectric tensor ACr2AgBiO8 with its nonzero entries

    a123=a213=0.22163,a113=a223=2.608665,a311=a322=0.152485,a312=0.37153.

    (d) The fourth piezoelectric tensor ARbTaO3 with its nonzero entries

    a113=a223=8.40955,a311=a322=4.3031,a222=a212=a211=5.412525,a333=5.14766.

    (e) The fifth piezoelectric tensor ANaBiS2 with its nonzero entries

    a113=8.90808,a223=0.00842,a311=7.11526,a322=0.6222,a333=7.93831.

    (f) The sixth piezoelectric tensor ALiBiB2O5 with its nonzero entries

    a112=0.34929,a211=0.16101,a222=0.12562,a312=2.57812,a123=2.35682,a213=0.05587,a233=0.1361,a323=6.91074.

    (g) The seventh piezoelectric tensor AKBi2F7 with its nonzero entries

    a111=12.64393,a211=2.59187,a311=1.51254,a123=1.59052,a122=1.08802,a212=0.10570,a312=0.08381,a233=0.81041,a113=1.96801,a213=0.71432,a313=0.39030,a333=0.23019,a112=0.22465,a222=0.08263,a322=0.68235,a323=0.19013,a133=4.14350,a223=0.51165.

    (h) The eigth piezoelectric tensor ABaNiO3 with its nonzero entries

    a113=0.038385,a223=0.038385,a311=a322=6.89822,a333=27.4628.

    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.

    Table 1.  Comparison among ϱ, ϱmin, ˜ϱmin, ρΓ, ρL, ρM, ρΥ, ργ, ρΩS, ρC, ρG, ρB, ρmin, ρΨ, ρΩ and λ.
    AVFeSb ASiO2 ACr2AgBiO8 ARbTaO3 ANaBiS2 ALiBiB2O5 AKBi2F7 ABaNiO3
    ϱ 7.3636 0.2882 5.6606 30.0911 17.3288 15.2911 22.6896 33.7085
    ϱmin 7.3636 0.2834 5.6606 23.5377 16.8548 12.3206 20.2351 27.5396
    ˜ϱmin 7.3636 0.2393 4.6717 22.7163 14.5723 12.1694 18.7025 27.5396
    ρΓ 7.3636 0.2834 5.6606 23.5377 16.8548 12.3206 20.2351 27.5396
    ρL 7.3636 0.2744 4.8058 23.5377 16.5640 11.0127 18.8793 27.5109
    ρM 7.3636 0.2834 4.7861 23.5377 16.8464 11.0038 19.8830 27.5013
    ρΥ 7.3636 0.2834 4.7335 23.5377 16.8464 10.9998 19.8319 27.5013
    ργ 7.3636 0.2744 4.7860 23.0353 16.4488 10.2581 18.4090 27.5013
    ρΩS 7.3636 0.2744 4.2732 23.0353 16.4486 10.2581 17.7874 27.4629
    ρC 6.3771 0.1943 3.7242 16.0259 11.9319 7.7540 13.5113 27.4629
    ρG 6.3771 0.2506 4.0455 21.5313 13.9063 9.8718 14.2574 29.1441
    ρB 5.2069 0.2345 4.0026 19.4558 13.4158 10.0289 15.3869 27.5396
    ρmin 6.5906 0.1942 3.5097 18.0991 11.9324 8.1373 14.3299 27.4725
    ρΨ 6.5906 0.1942 4.2909 18.9140 11.9319 8.1501 14.0690 27.4629
    ρΩ 5.2069 0.2005 3.5097 19.2688 11.9319 8.6469 13.6514 27.4629
    λ 4.2514 0.1375 2.6258 12.4234 11.6674 7.7376 13.5021 27.4628

     | Show Table
    DownLoad: CSV

    In Table 1, λ is the largest C-eigenvalue of a piezoelectric tensor; ϱ and ϱmin are respectively the right boundaries of the interval [ϱ,ϱ] and [ϱmin,ϱmin] obtained by Theorems 1 and 2 in [20]; ˜ϱmin are respectively the right boundary of the interval [˜ϱmin,˜ϱmin] obtained by Theorem 2.1 of [13]; ρL and ρM are respectively the right boundaries of the intervals [ρL,ρL] and [ρM,ρM] obtained by 2.2 and 2.4 in [25]; ρΥ is the right boundary of the interval [ρΥ,ρΥ] obtained by Theorem 2.1 of [24]; ργ is the right boundary of the interval [ργ,ργ] obtained by Theorem 2.1 of [23]; ρΩS is the right boundary of the interval [ρΩS,ρΩS] obtained by Theorem 5 of [19]; ρC is the right boundary of the interval [ρC,ρC] obtained by Theorem 7 of [21]; ρG, ρB and ρmin are respectively the right boundaries of the intervals [ρG,ρG], [ρB,ρB] and [ρmin,ρmin]obtained by Theorems 2.3–2.5 in [18]; ρΨ is the right boundary of the interval [ρΨ,ρΨ] obtained by Theorem 2.1 of [22]; ρΓ is the right boundary of the interval [ρΓ,ρΓ] obtained by Theorem 1.1;ρΩ is the right boundary of the interval [ρΩ,ρΩ] obtained by Theorem 2.1.

    From Table 1, it can be seen that:

    ⅰ) ρΩ is smaller thanϱ, ϱmin, ˜ϱmin, ρΓ, ρL, ρM, ρΥ, ργfor the eight piezoelectric tensors.

    ⅱ) ρΩρΩS, ρΩρB for the eight piezoelectric tensors.

    ⅲ) For some tensors, ρΩ is smaller than ρC, ρG, ρmin and ρΨ.For the other tensors, ρΩ is bigger than or equal to ρC, ρG, ρmin and ρΨ.For examples, for AVFeSb, ρΩ<ρC, ρΩ<ρG, ρΩ<ρmin and ρΩ<ρΨ; For ASiO2, ρΩ>ρC, ρΩ>ρmin and ρΩ>ρΨ; For ABaNiO3, ρΩ>ρG.

    Ⅱ. Calculation for all C-eigentriples of the seventh piezoelectric tensor AKBi2F7 by Theorem 3.1.

    All C-eigentriples of AKBi2F7 are obtained by Theorem 3.1 and are showen in Table 2. And the calculation process is shown in Appendix.

    Table 2.  All C-eigentriples of AKBi2F7.
    λ x1 x2 x3 y1 y2 y3
    13.50214 0.97050 0.20974 0.11890 0.97226 0.05065 0.22836
    13.50214 0.97050 0.20974 0.11890 0.97226 0.05065 0.22836
    4.46957 0.98196 0.18905 0.00362 0.22771 0.41491 0.88091
    4.46957 0.98196 0.18905 0.00362 0.22771 0.41491 0.88091
    0.54486 0.75981 0.36879 0.53544 0.06168 0.87047 0.48833
    0.54486 0.75981 0.36879 0.53544 0.06168 0.87047 0.48833
    0.54486 0.75981 0.36879 0.53544 0.06168 0.87047 0.48833
    0.54486 0.75981 0.36879 0.53544 0.06168 0.87047 0.48833
    4.46957 0.98196 0.18905 0.00362 0.22771 0.41491 0.88091
    4.46957 0.98196 0.18905 0.00362 0.22771 0.41491 0.88091
    13.50214 0.97050 0.20974 0.11890 0.97226 0.05065 0.22836
    13.50214 0.97050 0.20974 0.11890 0.97226 0.05065 0.22836

     | Show Table
    DownLoad: CSV

    Let ARn×n×n be a piezoelectric-type tensor. In this paper, we in Theorem 2.1 constructed a C-eigenvalue interval Ω(A) to locate all C-eigenvalues of 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 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.

    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).

    The author declares there is no confict of interest.

    The following is the calculation process for all C-eigentriples of AKBi2F7 by Theorem 3.1.

    (a) Because a2110, Case (a) in Theorem 3.1 does not holds.

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

    {2.59187x112.64393x2=0,(A.1)1.51254x112.64393x3=0,(A.2)1.08802x1+0.10570x2+0.08381x3=0,(A.3)1.96801x1+0.71432x2+0.39030x3=0,x21+x22+x23=1.

    The three Eqs (A.1)–(A.3) yield a linear system of equation Ax=0, where x=(x1,x2,x3) and

    A=(2.5918712.6439301.51254012.643931.088020.105700.08381).

    From det(A)=179.00740, the solution of Ax=0 is x=(x1,x2,x3)=(0,0,0), which contradicts with x21+x22+x23=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

    {f1(y1,y2,y3)=0.08263y22+2.59187y21+0.81041y23+0.2114y1y2+1.0233y2y3+1.42864y1y3=0,f2(y1,y2,y3)=0.23019y23+1.51254y21+0.68235y22+0.7806y1y3+0.38026y2y3+0.16762y1y2=0,f3(y1,y2,y3)=0.22465y210.22465y2211.55591y1y2+1.59052y1y31.96801y2y3=0,f4(y1,y2,y3)=1.96801y211.96801y238.50043y1y3+a123y1y20.22465y2y3=0,f5(y1,y2,y3)=y21+y22+y23=1.

    We now regard fi(y1,y2,y3), i[5] as a function with y1 as a variable and y2 and y3 as two coefficients and obtain their resultants as follows:

    Resy1(f1,f2)=2.68640y421.92404y3y325.51122y22y23+2.09880y33y2+3.18879y43,Resy1(f1,f3)=27.58122y42+331.64703y3y32+134.75405y23y2255.45236y33y2+4.93315y43.

    Let g1(y2,y3):=Resy1(f1,f2), g2(y2,y3):=Resy1(f1,f3), and its resultant

    Resy2(g1,g2)=95376348203.97653y163=0.

    Then y3=0. Substituting y3=0 into g1 and g2, we have

    g1(y2,y3)=2.68640y42,g2(y2,y3)=27.58122y42.

    Let g1(y2,y3)=0 and g2(y2,y3)=0. We have y2=0. Substituting y2=0 and y3=0 into f1, we have f1(y1,y2,y3)=2.59187y21. Solving f1(y1,y2,y3)=0, we have y1=0. However, y1=y2=y3=0 is not solution of y21+y22+y23=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

    {f1(u1,u2,v1,v2):=0.08263u1v22+2.59187v21u1+0.81041u1+0.21140v1u1v2+1.02330u1v2+1.42864v1u112.64393u2v211.08802u2v224.14350u20.44930u2v1v23.93602u2v13.18104u2v2=0,f2(u1,u2,v1,v2):=0.23019u1+1.51254u1v21+0.68235u1v22+0.78060u1v1+0.38026u1v2+0.16762u1v1v212.64393v211.08802v220.44930v1v23.93602v13.18104v24.14350=0,f3(u1,u2,v1,v2):=0.22465u1v2111.55591u1v1v2+1.59052v1u12.50924u2v1v2+0.10570u2v21+0.51165v1u2+0.19013v10.83019v1v2+0.08381v210.22465u1v221.96801u1v20.10570u2v220.71432v2u20.39030v20.08381v22=0,f4(u1,u2,v1,v2):=1.96801u1v218.50043u1v1+1.59052v1u1v2+0.51165v1u2v2+0.19013v1v21.78146u2v1+0.71432u2v211.742730v1+0.39030v210.22465u1v21.96801u10.10570u2v20.71432u20.08381v20.39030=0.

    We now regard fi(u1,u2,v1,v2), i[4] as a function with u1 as a variable and u2, v1 and v2 as three coefficients and obtain their resultants as follows:

    Resu1(f1,f2)=(0.84336v20.95379+2.32838v1+3.78649v22+2.58431v32+0.74241v42+6.42916v21+15.82324v31+19.12445v41+10.62991v2v21+4.23911v22v1+2.79896v2v31+0.48895v32v1+10.34857v22v21+4.57094v2v1)u29.10936v13.95976v22v216.81799v29.81234v2v122.65734v2v213.01186v22v13.83745v2v310.26713v32v14.47928v221.37622v320.08990v4226.60934v2128.26528v3132.771423v413.35793=0,
    Resu1(f1,f3)=(8.73334v2+7.00497v18.00778v223.02304v320.25316v42+8.00778v21+22.47179v31+3.11442v4174.15833v2v2139.62519v22v1152.49246v2v3112.90368v32v18.58379v22v2153.09898v2v1)u2+0.15408v10.31630v20.38580v22v211.03583v2v12.07169v2v211.03607v22v12.13403v2v310.08632v32v10.46731v220.11801v320.00693v42+0.33955v21+0.61253v31+0.21722v41=0,

    and

    Resu1(f1,f4)=(8.733348.00778v245.43188v13.02304v220.25316v3254.00439v21103.32952v31+26.73481v41+6.67230v2v213.93604v22v1+22.47179v2v31+1.77280v32v1+3.02304v22v2123.92868v2v1)u2+0.07244v22v211.96992v10.46731v21.83150v2v1+0.08538v2v21+0.03284v22v1+0.57530v2v31+0.01571v32v10.11801v220.00693v323.18504v213.95933v31+1.01161v410.31630=0.

    Let

    g1(u2,v1,v2):=Resu1(f1,f2),g2(u2,v1,v2):=Resu1(f1,f3),g3(u2,v1,v2):=Resu1(f1,f4).

    Then their resultants are

    Resu2(g1,g2)=106.21826v81+(5025.64756v2+839.61383)v71+(865.25666v226653.08038v2+998.05266)v61+(1083.52137v325229.62871v227381.37411v2+1094.99786)v51+(137.50780v41593.23550v324933.36887v224733.12663v2+632.08073)v41+(70.75506v52741.31965v422567.50800v324243.14169v222467.61827v2+335.99902)v31+(5.62145v6290.33840v52524.86290v421366.09759v321783.36316v22843.25156v2+90.73526)v21+(1.29518v7223.97008v62135.36791v52405.58564v42682.18237v32624.18604v22209.71678v2+23.37525)v1+(0.02790v820.72569v726.69240v6228.96790v5272.02906v42105.34644v3286.25447v2229.02430v2)=0

    and

    Resu2(g1,g3)=895.48403v81+(852.85983v22690.30193)v71+(304.63154v22+1073.45667v24096.10056)v61+(172.77376v32+134.01848v222360.05799v25632.48044)v51+(29.00602v42+203.47367v32304.10992v222878.73753v23926.74222)v41+(10.47279v52+55.51707v42189.59056v321392.51379v223083.36023v22315.99105)v31+(0.80682v62+8.30736v524.53864v42259.96426v32927.90427v221442.92349v2831.16935)v21+(0.17104v72+2.07986v622.49973v5263.10600v42263.71529v32525.06788v22551.19369v2230.96997)v1+(0.02790v720.72569v626.69240v5228.96790v4272.02906v32105.34644v2286.25447v229.02430)=0.

    Let

    h1(v1,v2):=Resu2(g1,g2)andh2(v1,v2):=Resu2(g1,g3).

    Their results are

    Resv1(h1,h2)=1.62711×1046v6421.18415×1044v6324.48673×1043v6221.18681×1041v6122.43052×1040v6024.06317×1039v5925.7302×1038v5826.93518×1037v5727.24748×1036v5626.49253×1035v5524.87257×1034v5422.98951×1033v5321.49477×1032v5226.3372×1032v5122.51516×1031v5021.06401×1030v4925.02246×1030v4822.43069×1029v4721.09655×1028v4624.44371×1028v4521.60392×1027v4425.08938×1027v4321.39566×1026v4223.2426×1026v4126.05752×1026v4027.48945×1026v392+1.29902×1026v382+4.01056×1025v372+1.382×1024v362+3.13265×1024v352+5.06222×1024v342+4.80914×1024v3321.80398×1024v3221.82876×1023v3124.38298×1023v3027.14302×1023v2928.82265×1023v2827.81232×1023v2723.29361×1023v262+1.21659×1023v2525.53171×1023v2422.91829×1022v2323.73689×1022v222+1.93914×1022v212+1.01905×1021v202+8.28494×1022v1924.6185×1022v1821.27125×1021v1729.35242×1022v1623.19849×1022v152+7.28108×1024v142+6.11682×1023v132+6.78336×1024v122+1.47291×1024v112+8.0039×1024v1022.71136×1024v921.27876×1024v82+1.35623×1024v72+3.48519×1025v62+2.48651×1026v522.61144×1026v421.87784×1027v326.6019×1029v22+2.43536×1029v26.98911×1031.

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

    Step 1. Solving Resv1(h1,h2)=0, we have

    v2=0.85035,0.22179,7.79725,1.78254,11.53378,0.75907,0.70532,0.47100.

    Step 2. Substituting v2=0.85035 into h1(v1,v2), and letting h1(v1,v2)=0, it all real roots are v1=0.43450 or v1=4.25751. Substituting v2=0.85035 into h2(v1,v2), and letting h2(v1,v2)=0, it all real roots are v1=0.48184, 0.17296, 0.0000000000038876, or 5.04918. It is easy to see that h1(v1,v2)=0 and h2(v1,v2)=0 have no common solution, which implies that v2=0.85035 is not a solution of the system (f).

    Step 3. Substituting v2=0.22179 into h1(v1,v2), and letting h1(v1,v2)=0, it all real roots are v1=0.43450 or 4.25751. substituting v2=0.22179 into h2(v1,v2), and letting h2(v1,v2)=0, it all real roots are v1=0.24034 or 4.25751. It is easy to see that v1=4.25751 is a common solution of h1(v1,v2)=0 and h2(v1,v2)=0.

    Step 4. Substituting v2=0.22179 and v1=4.25751 into g1(u2,v1,v2), g2(u2,v1,v2) and g3(u2,v1,v2), and letting g1(u2,v1,v2)=0, its all real roots are u2=1.76393; letting g2(u2,v1,v2)=0, its all real roots are u2=1.76393; letting g3(u2,v1,v2)=0, its all real roots are u2=1.76393. Hence, the common solution of g1(u2,v1,v2)=0, g2(u2,v1,v2)=0 and g3(u2,v1,v2)=0 is u2=1.76393.

    Step 5. Substituting v2=0.22179, v1=4.25751 and u2=1.76393 into f1, and letting f1(u1,u2,v1,v2)=0, we can get its all real roots u1=8.16186.

    Step 6. By v2=0.22179, v1=4.25751, u2=1.76393, u1=8.16186, (3.2) and (3.3), we can get the corresponding C-eigentriples as follows:

    λ=13.50214 and its C-eigenvectors are

    x=(0.97050,0.20974,0.11890),y=±(0.97226,0.05065,0.22836).

    λ=13.50214 and its C-eigenvectors are

    x=(0.97050,0.20974,0.11890),y=±(0.97226,0.05065,0.22836).

    Step 7. For other values of v2, 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.



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