# Bifurcation Analysis of Composite Laminated Piezoelectric Rectangular Plate Structure in the Case of 1:2 Internal Resonance

Authors: Y. H. Guo; W. Zhang
Journal: Mathematical Problems in Engineering
 (2019)

---

## Abstract

In this paper, the authors study the bifurcation problems of the composite laminated piezoelectric rectangular plate structure with three bifurcation parameters by singularity theory in the case of 1:2 internal resonance. The sign function is employed to the universal unfolding of bifurcation equations in this system. The proposed approach can ensure the nondegenerate conditions of the universal unfolding of bifurcation equations in this system to be satisfied. The study presents that the proposed system with three bifurcation parameters is a high codimensional bifurcation problem with codimension 4, and 6 forms of universal unfolding are given. Numerical results show that the whole parametric plane can be divided into several persistent regions by the transition set, and the bifurcation diagrams in different persistent regions are obtained.


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## Body

## 1. Introduction

The bifurcation equations with multiple variables can be reduced to the bifurcation equation with single state variable by elimination method, but the work [1] showed that some bifurcation characteristics of system would be vanished. In addition, the bifurcation property of system with multiple bifurcation parameters is more complex than single parametric bifurcation feature [1]. Therefore, considering the system with all variables and parameters is close to the real physics problem. When an original steady-state system is subjected to small perturbations, all possible bifurcation behaviors can be revealed from the universal unfolding of bifurcation equations.The singularity theory unifies the treatment of many diverse problems in steady-state bifurcation; such unification has the obvious advantage of elegance. The singularity theory was introduced to study the bifurcation problems by Golubitsky and Schaeffer [2] in 1985; the singularity theory has been greatly developed. Golubitsky and Keyfitz [3] studied the steady-state solutions for a continuous flow stirred tank chemical reactor by the singularity theory and obtained the qualitative bifurcation diagrams. Schaeffer and Golubitsky [4] applied the singularity theory to investigate the bifurcation phenomena of a model chemical reaction. Futer [5] and Sitta [6] considered symmetry of bifurcation parameters and studied the classification of bifurcation problems of codimension not greater than 1. Lavassani et al. [7] studied equivariant multiple parameter problems by singularity theorem and gave finite definite theorem and normal form of more parameters problem. Chen and Colleagues [8, 9] firstly proposed the C–L method to study the bifurcation of the periodic solution of nonlinear dynamical equation undergoing parameter excitation by combining the L–S method and the singularity theory. Seyranian and Mailybaev [10] gave multiple parameter stability theory and discussed the influence of system parameters on stability when parameters change. Jin and Zou [11] applied singularity theory to a restrained pipe conveying fluid and obtained the dynamical behavior in different persistent regions. Chen et al. [12] gave bifurcation analysis of an arch structure of parametric and forced excitation with codimension 5. Li et al. [13] extended Lyapunov-Schmidt reduction to fractional ordinary differential systems (FODSs) with Caputo derivatives. Li et al. [14, 15] gave bifurcation analysis of the Kuramoto-Sivashinsky equation in one-spatial dimension with three kinds of boundary value conditions and studied symmetry breaking bifurcation in O
2
×
O
2-symmetric nonlinear large problems. In the applications of the two state variable singularity theory, the galloping of transmission conductor was studied. Nigol [16, 17] gave torsional galloping mechanism under galloping excitation. Yu [18] gave torsional feedback mechanism galloping excitation.Piezoelectric materials are new type functional materials in engineering applications and can be utilized as the actuators and sensors in engineering structures; for example, it was used to construct morphing structures or morphing wings which can change the aircraft shape [19–21]. Many researchers have given considerable attention to the nonlinear dynamics of composite laminated plates. Pai and Nayfeh [22] presented a general nonlinear theory for research on the nonlinear dynamics of elastic composite plates undergoing moderate-rotation oscillations by considering the geometric nonlinearities. Shukla et al. [23] gave an analytical approach to analyze the nonlinear dynamic responses of a laminated composite plate with spatially oriented short fibers in each layer of the composite. Ye et al. [24, 25] dealt with the nonlinear dynamic characteristics of a parametrically excited, simply supported, symmetric cross-ply composite laminated rectangular thin plate, and a simply supported antisymmetric cross-ply composite laminated rectangular thin plate under parametric excitation. Lee and Reddy [26] investigated the nonlinear dynamic responses of laminated composite plates under thermostatically loading. Zhang et al. [27] investigated the nonlinear oscillations and chaotic dynamics of a parametrically excited simply supported symmetric cross-ply laminated composite rectangular thin plate with the geometric nonlinearity and nonlinear damping. Hao et al. [28] analyzed the nonlinear oscillations, bifurcations, and chaos of a functionally graded materials plate. Zhang et al. [29] established the governing equations of motion for the nonlinear oscillations of a simply supported symmetric cross-ply composite laminated piezoelectric plate subjected to the transverse, in-plane excitations, and the excitation loaded by piezoelectric layers and studied the periodic and chaotic dynamics of the composite laminated piezoelectric plate. Zhang et al. [30] studied the motion equations of simply supported flexible beam with five nonlinear terms under parametric excitation and analyzed the different forms of bifurcated response curve when the parameters are located in different regions. Zhang et al. [31, 32] studied the global nonlinear dynamics of thin plates under parametric excitation. The research on mechanic-electric couple responses of piezoelectric composite laminates plates has important theoretical significance and engineering application value. In the study of one-way mechanic-electric couple responses, Crawley et al. [33] presented the development and experimental verification of the induced strain actuation of plate components of an intelligent structure. Equations related to the actuation strains, created by induced strain actuators, to the strains induced in the actuator/substrate system were derived for isotropic and anisotropic plates. Plate strain energy relations are also developed. Wang et al. [34] analyzed laminated plates with distributed piezoelectric actuators by classical laminated plate theory. In studying of two-way mechanic-electric couple responses, Allik et al. [35] derived the finite element equations of piezoelectric body vibration. Lim et al. [36] studied the dynamic characteristics of isotropic plates with piezoelectric sensors and actuators by a 20-node block element, a 13-node transition element, and a 9-node plate element. Wang et al. [37] studied the optimization of thickness and embedding position of a piezoelectric actuator in composite laminates and analyzed an isotropic plate with isotropic actuator and a composite plate with anisotropic actuator.The singularity theory of asymmetry dynamical systems with multiple state variables and single bifurcation parameter had been well developed by Golubitsky and Schaeffer. The universal unfolding of bifurcation equations was obtained [2], and it showed that the number of codimensions was equal to the number of auxiliary parameters. However, in asymmetry dynamical systems with multiple bifurcation parameters, the number of auxiliary parameters is less than codimension, which is contradiction to the previous conclusion. In this paper, the sign function is introduced to solve this problem, which leads to the fact that nondegenerate conditions of universal unfolding of the bifurcation equations are satisfied. It may determine the summation of the number of sign functions and the number of auxiliary parameters is equal to codimension. For the dynamical equations of the composite laminated piezoelectric rectangular plate structure in the case of principal parametric resonance and 1/2 subharmonic resonance for the first-order mode and primary resonance for the second-order mode [29], the four-dimensional nonlinear averaged equation is obtained by the Galerkin approach. By employed sign function, the universal unfoldings of the 1:2 internal resonance bifurcation equation are derived. The transition sets in the parameters plane are calculated and the bifurcation diagrams are depicted.


## 2. The Motion Equations and Perturbation Analysis

In this section, we summarize on modeling of composite laminated piezoelectric rectangular plate simply supported at four-edge in the literature [29]. The mechanical modal of the composite laminated piezoelectric rectangular plate is shown in Figure 1, where the edge lengths are α and β, respectively, and thickness is h. The composite laminated piezoelectric rectangular plate is considered as regular symmetric cross-ply laminates with n layers. Some of these layers are made of the PVDF piezoelectric materials as actuators, while others are made of fiber-reinforced composite materials. It is assumed that different layers of the symmetric cross-ply composite laminated piezoelectric rectangular plate are perfectly bonded with each other and with piezoelectric actuator layers embedded in the plate. A Cartesian coordinate system O
x
y
z is located in the middle surface of the composite laminated piezoelectric rectangular plate. In this study, that u
,
v
,
w and u
0
,
v
0
,
w
0 are assumed to represent the displacements of an arbitrary point and a point in the middle surface of the composite laminated piezoelectric rectangular plate in the x, y, and z directions, respectively. It is also assumed that the in-plane excitations of the composite laminated piezoelectric rectangular plate are loaded along the y direction at x
=
0 and the x direction at y
=
0 with the form of q
0
+
q
x
cos
⁡
Ω
1
t and q
1
+
q
y
cos
⁡
Ω
2
t, respectively. The transverse excitation, which loads to the composite laminated piezoelectric rectangular plate, is represented by q
=
q
3
cos
⁡
Ω
3
t. The dynamic electrical loading is expressed as E
z
=
E
z
cos
⁡
Ω
4
t.Figure 1
The simplified model of piezoelectric plate structure.According to literature [29], applying the Reddy’s third-order shear deformation theory (TSDT), the Hamilton’s principle, the Galerkin method, and the boundary condition, a two degree-of-freedom nonlinear ordinary differential equation of the composite laminated piezoelectric rectangular plate is obtained as follows:(1a)
w
¨
1
+
ω
1
2
w
1
+
μ
1
w
˙
1
+
α
2
cos
⁡
Ω
1
t
+
α
3
cos
⁡
Ω
2
t
+
α
4
cos
⁡
Ω
4
t
w
1
+
α
5
w
1
2
w
2
+
α
6
w
2
2
w
1
+
α
7
w
1
3
+
α
8
w
2
3
=
F
1
cos
⁡
Ω
3
t
,
(1b)
w
¨
2
+
ω
2
2
w
2
+
μ
2
w
˙
2
+
β
2
cos
⁡
Ω
1
t
+
β
3
cos
⁡
Ω
2
t
+
β
4
cos
⁡
Ω
4
t
w
1
+
β
5
w
2
2
w
1
+
β
6
w
1
2
w
2
+
β
7
w
2
3
+
β
8
w
1
3
=
F
2
cos
⁡
Ω
3
t
,where all coefficients in (1a) and (1b) are presented in Appendix.To obtain a system suitable for the application of the method of multiple scales, the scales transformations may be introduced as(2)
μ
1
→
ε
μ
1
,
μ
2
→
ε
μ
2
,
α
2
→
ε
α
2
,
α
3
→
ε
α
3
,
α
4
→
ε
α
4
,
α
5
→
ε
α
5
,
α
6
→
ε
α
6
,
α
7
→
ε
α
7
,
α
8
→
ε
α
8
,
β
2
→
ε
β
2
,
β
3
→
ε
β
3
,
β
4
→
ε
β
4
,
β
5
→
ε
β
5
,
β
6
→
ε
β
6
,
β
7
→
ε
β
7
,
β
8
→
ε
β
8
,
F
1
→
ε
F
1
,
F
2
→
ε
F
2
,where ε is a small parameter.Substituting (2) into (1a) and (1b), we obtain the following dimensionless two degree-of-freedom nonlinear vibratory systems:(3a)
w
¨
1
+
ω
1
2
w
1
+
ε
μ
1
w
˙
1
+
ε
α
2
cos
⁡
Ω
1
t
+
ε
α
3
cos
⁡
Ω
2
t
+
ε
α
4
cos
⁡
Ω
4
t
w
1
+
ε
α
5
w
1
2
w
2
+
ε
α
6
w
2
2
w
1
+
ε
α
7
w
1
3
+
ε
α
8
w
2
3
=
ε
F
1
cos
⁡
Ω
3
t
,
(3b)
w
¨
2
+
ω
2
2
w
2
+
ε
μ
2
w
˙
2
+
ε
β
2
cos
⁡
Ω
1
t
+
ε
β
3
cos
⁡
Ω
2
t
+
ε
β
4
cos
⁡
Ω
4
t
w
1
+
ε
β
5
w
2
2
w
1
+
ε
β
6
w
1
2
w
2
+
ε
β
7
w
2
3
+
ε
β
8
w
1
3
=
ε
F
2
cos
⁡
Ω
3
t
.The method of multiple scales is used to find the uniform solutions of (3a) and (3b) in the following form:(4a)
w
1
t
,
ε
=
x
0
T
0
,
T
1
+
ε
x
1
T
0
,
T
1
+
⋯
,
(4b)
w
2
t
,
ε
=
y
0
T
0
,
T
1
+
ε
y
1
T
0
,
T
1
+
⋯
,where T
0
=
t and T
1
=
ε
tThe time derivative used in the method of multiple scales is given as(5a)
d
d
t
=
∂
∂
T
0
+
∂
∂
T
1
∂
T
1
∂
t
+
⋯
=
D
0
+
ε
D
1
+
⋯
,
(5b)
d
2
d
t
2
=
D
0
+
ε
D
1
+
⋯
2
=
D
0
2
+
2
ε
D
0
D
1
+
⋯
,where D
0
=
∂
/
∂
T
0 and D
1
=
∂
/
∂
T
1.We consider the case of principal parametric resonance and 1:2 internal resonance. In this resonant case, there are the following resonant relations:(6)
ω
1
2
=
1
4
ω
2
+
ε
σ
1
,
ω
2
2
=
ω
2
+
ε
σ
2
,
Ω
1
=
ω
,
Ω
2
=
Ω
3
=
Ω
4
=
ω
,
ω
2
≈
2
ω
1
,where ω
1 and ω
2 are two different linear natural frequencies, σ
1 and σ
2 are two detuning parameters.Substituting (4a) and (4b)–(6) into (3a) and (3b) and balancing the coefficients of corresponding powers of ε on the left-hand and right-hand sides of equations, the differential equations are obtained as follows:Orderε
0:(7a)
D
0
2
x
0
+
1
4
ω
2
x
0
=
0
,
(7b)
D
0
2
y
0
+
ω
2
y
0
=
0
.Orderε
1:(8a)
D
0
2
x
1
+
1
4
ω
2
x
1
=
μ
1
D
0
x
0
+
α
2
+
α
3
+
α
4
cos
⁡
ω
t
x
0
-
σ
1
x
0
+
α
5
x
0
2
y
0
,
+
α
6
x
0
y
0
2
+
α
7
x
0
3
+
α
8
y
0
3
+
F
1
cos
⁡
ω
t
+
2
D
0
D
1
x
0
,
(8b)
D
0
2
y
1
+
ω
2
y
1
=
μ
2
D
0
y
0
+
β
2
+
β
3
+
β
4
cos
⁡
ω
t
y
0
-
σ
2
y
0
+
β
5
y
0
2
x
0
,
+
β
6
y
0
x
0
2
+
β
7
y
0
3
+
β
8
x
0
3
+
F
2
cos
⁡
ω
t
+
2
D
0
D
1
y
0
.The solution of (7a) and (7b) in the complex form can be written as(9a)
x
0
=
A
T
1
e
1
/
2
ω
T
0
+
A
-
T
1
e
-
1
/
2
ω
T
0
,
(9b)
y
0
=
B
T
1
e
ω
T
0
+
B
-
T
1
e
-
ω
T
0
,where A
- and B
- are the complex conjugates of A and B, respectively.Substituting (9a) and (9b) into (8a) and (8b) yields(10a)
D
0
2
x
1
+
1
4
ω
2
x
1
=
-
σ
1
A
+
1
2
α
2
+
α
3
+
α
4
A
-
-
i
ω
D
1
A
+
i
1
2
μ
1
ω
A
+
2
α
6
A
B
B
-
+
3
a
7
A
2
A
-
e
i
1
/
2
ω
T
0
+
c
c
+
N
S
T
,
(10b)
D
0
2
y
1
+
ω
2
y
1
=
-
σ
2
B
-
2
i
ω
D
1
B
+
i
μ
2
ω
B
+
2
β
6
A
A
-
B
+
1
2
F
2
+
3
β
7
B
2
B
-
e
i
ω
T
0
+
c
c
+
N
S
T
,where c
c represents the parts of the complex conjugates of the function on the right-hand side of (10a) and (10b) and N
S
T represents the terms that do not produce secular terms.Eliminating the secular terms from (10a) and (10b) yields(11a)
D
1
A
=
-
1
2
μ
1
A
+
i
1
2
σ
1
A
-
i
a
6
A
B
B
-
-
i
1
4
α
2
+
α
3
+
α
4
A
-
-
i
3
2
α
7
A
2
A
-
,
(11b)
D
1
B
=
-
1
2
μ
2
B
+
i
1
4
σ
2
B
-
i
1
2
β
6
A
A
-
B
-
i
3
4
β
7
B
2
B
-
-
i
1
8
F
2
.The functionsA and B may be expressed in the polar form(12)
A
=
1
2
a
1
e
i
φ
1
B
=
1
2
a
2
e
i
φ
2where φ
1 and φ
2 are the phase angles. Substituting (12) into (11a) and (11b) and separating the real and imaginary part, the four-dimensional averaged equation in polar form is obtained as(13a)
a
˙
1
=
-
1
2
μ
1
a
1
-
1
4
α
2
+
α
3
+
α
4
a
1
sin
⁡
2
φ
1
,
(13b)
a
1
φ
˙
1
=
1
2
σ
1
a
1
-
1
4
α
6
a
1
a
2
2
-
3
8
α
7
a
1
3
-
1
4
α
2
+
α
3
+
α
4
a
1
cos
⁡
2
φ
1
,
(13c)
a
˙
2
=
-
1
2
μ
2
a
2
-
1
4
F
2
sin
⁡
φ
2
,
(13d)
a
2
φ
˙
2
=
1
4
σ
2
a
2
-
1
8
β
6
a
1
2
a
2
-
3
16
β
7
a
2
3
-
1
4
F
2
cos
⁡
φ
2
.Making the left-hand of (13a), (13b), (13c), and (13d) equal to zero, eliminating 2
φ
1 in (13a) and (13b), and eliminating φ
2 in (13c) and (13d) by using the relations between trigonometric functions, we obtained the frequency-response functions of the system as follows:(14a)
1
2
μ
1
a
1
2
+
1
2
σ
1
a
1
-
1
4
α
6
a
1
a
2
2
-
3
8
α
7
a
1
3
2
=
1
4
α
2
+
α
3
+
α
4
a
1
2
,
(14b)
1
2
μ
2
a
2
2
+
1
4
σ
2
a
2
-
1
8
β
6
a
1
2
a
2
-
3
16
β
7
a
2
3
2
=
1
4
F
2
2
.Expanding (14a) and (14b) leads to the following bifurcation equations:(15a)
k
11
a
1
6
+
k
12
a
1
4
a
2
2
+
k
13
a
1
2
a
2
4
+
k
14
σ
1
a
1
4
+
k
15
σ
1
a
1
2
a
2
2
+
σ
1
2
a
1
2
+
k
16
a
1
2
=
0
,
(15b)
k
21
a
2
6
+
k
22
a
1
4
a
2
2
+
k
23
a
1
2
a
2
4
+
k
24
σ
2
a
2
4
+
k
25
σ
2
a
1
2
a
2
2
+
σ
2
2
a
2
2
+
4
μ
2
2
a
2
2
-
F
2
2
=
0
,where(16)
k
11
=
9
16
α
7
2
,
k
12
=
3
4
α
6
α
7
,
k
13
=
1
4
α
6
2
,
k
14
=
-
3
2
α
7
,
k
15
=
-
α
6
,
k
16
=
μ
1
2
-
1
4
α
2
+
α
3
+
α
4
2
,
k
21
=
9
16
β
7
2
,
k
22
=
1
4
β
6
2
,
k
23
=
3
4
β
6
β
7
,
k
24
=
-
3
2
β
7
,
k
25
=
-
β
6
.


## 3. Singularity Analysis

We know that the universal unfolding of the bifurcation equations can reveal all possible bifurcation behaviors when the original system is subject to small perturbations. In this section, we will use the introduced sign function in Proposition4 to establish the universal unfolding of the bifurcation equations (15a) and (15b).Let(17)
g
=
g
1
,
g
2
∈
E
→
z
,
λ
,where(18a)
g
1
=
k
11
a
1
6
+
k
12
a
1
4
a
2
2
+
k
13
a
1
2
a
2
4
+
k
14
σ
1
a
1
4
+
k
15
σ
1
a
1
2
a
2
2
+
σ
1
2
a
1
2
+
k
16
a
1
2
,
(18b)
g
2
=
k
21
a
2
6
+
k
22
a
1
4
a
2
2
+
k
23
a
1
2
a
2
4
+
k
24
σ
2
a
2
4
+
k
25
σ
2
a
1
2
a
2
2
+
σ
2
2
a
2
2
+
4
μ
2
2
a
2
2
-
F
2
2
,
(18c)
z
=
a
1
,
a
2
,
λ
=
σ
1
,
σ
2
,
F
2
.In the following discussion,Γ
=
1 is adopted in R
T
g
,
Γ.


### 3.1. The Restricted Tangent Space

Proposition 1.
The restricted tangent spaceR
T
g
,
1 of the germ g
z
,
λ is given(19)
R
T
g
,
1
=
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2Proof.
According to Proposition 1.4 (see pages 169 in Golubitsky and Schaeffer II (1985)),R
T
g
,
1 is generated (as module over E
z
,
λ) by the fourteen mappings:(20)
g
1
,
0
,
g
2
,
0
,
0
,
g
1
,
0
,
g
2
,
a
1
g
1
,
a
1
,
g
2
,
a
1
,
a
2
g
1
,
a
1
,
g
2
,
a
1
,
σ
1
g
1
,
a
1
,
g
2
,
a
1
,
σ
2
g
1
,
a
1
,
g
2
,
a
1
,
F
2
g
1
,
a
1
,
g
2
,
a
1
,
a
1
g
1
,
a
2
,
g
2
,
a
2
,
a
2
g
1
,
a
2
,
g
2
,
a
2
,
σ
1
g
1
,
a
2
,
g
2
,
a
2
,
σ
2
g
1
,
a
2
,
g
2
,
a
2
,
F
2
g
1
,
a
2
,
g
2
,
a
2
.
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2 is generated by the twenty mappings(21)
a
1
2
,
0
,
a
1
a
2
,
0
,
a
2
2
,
0
,
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
,
0
,
F
2
2
,
0
,
0
,
a
1
2
,
0
,
a
1
a
2
,
0
,
a
2
2
,
0
,
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
,
0
,
F
2
2
.
By the following proof that there exists a reversible matrixA between (20) and (21) when a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
=
0,0
,
0,0
,
0, we observe that(22)
u
1
u
2
=
A
v
1
v
2
=
A
1
A
2
A
3
A
4
v
1
v
2
,where(23a)
u
1
=
g
1
,
0
g
2
,
0
0
,
g
1
0
,
g
2
,
a
1
g
1
,
a
1
,
g
2
,
a
1
a
2
g
1
,
a
1
,
g
2
,
a
1
σ
1
g
1
,
a
1
,
g
2
,
a
1
1
×
7
T
,
(23b)
u
2
=
σ
2
g
1
,
a
1
,
g
2
,
a
1
F
2
g
1
,
a
1
,
g
2
,
a
1
a
1
g
1
,
a
2
,
g
2
,
a
2
a
2
g
1
,
a
2
,
g
2
,
a
2
,
σ
1
g
1
,
a
2
,
g
2
,
a
2
σ
2
g
1
,
a
2
,
g
2
,
a
2
F
2
g
1
,
a
2
,
g
2
,
a
2
1
×
7
T
,
(23c)
v
1
=
a
1
2
,
0
F
2
2
,
0
0
,
a
1
2
0
,
F
2
2
a
2
2
,
0
a
1
a
2
,
0
σ
1
a
1
,
0
σ
2
a
1
,
0
F
2
a
1
,
0
σ
1
a
2
,
0
1
×
10
T
,
(23d)
v
2
=
σ
2
a
2
,
0
F
2
a
2
,
0
0
,
a
1
a
2
0
,
a
2
2
0
,
σ
1
a
1
0
,
σ
2
a
1
0
,
F
2
a
1
0
,
σ
1
a
2
0
,
σ
2
a
2
0
,
F
2
a
2
1
×
10
T
,
(23e)
A
1
=
C
1
0
0
0
0
0
0
0
0
0
0
-
1
0
0
C
2
0
0
0
0
0
0
0
C
1
0
0
0
0
0
0
0
0
0
0
-
1
0
0
0
0
0
0
C
3
0
C
4
0
0
0
0
0
0
0
0
0
0
0
0
C
3
0
0
0
0
0
0
0
0
0
0
C
3
0
0
0
7
×
10
,
(23f)
A
2
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
4
0
0
0
0
0
0
0
0
0
0
0
C
4
0
0
0
0
0
7
×
10
,
(23g)
A
3
=
0
0
0
0
0
0
0
C
3
0
0
0
0
0
0
0
0
0
0
C
3
0
C
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
5
0
0
0
0
0
0
0
0
0
0
C
5
0
0
0
0
0
0
0
0
0
0
C
5
0
0
0
0
0
0
0
0
0
0
C
5
0
7
×
10
,
(23h)
A
4
=
0
0
0
0
0
C
4
0
0
0
0
0
0
0
0
0
0
C
4
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
C
6
7
×
10
,where (24a)
C
1
=
k
11
a
1
4
+
k
12
a
1
2
a
2
2
+
k
13
a
2
4
+
k
14
σ
1
a
1
2
+
k
15
σ
1
a
2
2
+
σ
1
2
+
k
16
,
(24b)
C
2
=
k
21
a
2
4
+
k
22
a
1
4
+
k
23
a
1
2
a
2
2
+
k
24
σ
2
a
2
2
+
k
25
σ
2
a
1
2
+
σ
2
2
+
4
μ
2
2
,
(24c)
C
3
=
6
k
11
a
1
4
+
4
k
12
a
1
2
a
2
2
+
2
k
13
a
2
4
+
4
k
14
σ
1
a
1
2
+
2
k
15
σ
1
a
2
2
+
2
σ
1
2
+
2
k
16
,
(24d)
C
4
=
2
k
12
a
1
2
+
4
k
13
a
2
2
+
2
k
15
σ
1
a
1
a
2
,
(24e)
C
5
=
4
k
22
a
1
2
+
2
k
23
a
2
2
+
2
k
25
σ
2
a
2
2
,
(24f)
C
6
=
6
k
21
a
2
4
+
2
k
22
a
1
4
+
4
k
23
a
1
2
a
2
2
+
4
k
24
σ
2
a
2
2
+
2
k
25
σ
2
a
1
2
+
2
σ
2
2
+
8
μ
2
2
.
In (15a) and (15b), if k
16
=
0, bifurcation analysis is meaningless, so k
16
≠
0 will be studied in this paper. When a
1
=
a
2
=
σ
1
=
σ
2
=
F
2
=
0, substituting μ
2
>
0 into (24a), (24b), (24c), (24d), (24e), and (24f), it can be obtained that(25)
C
1
=
k
16
≠
0
,
C
2
=
4
μ
2
2
>
0
,
C
3
=
2
k
16
≠
0
,
C
4
=
0
,
C
5
=
0
,
C
6
=
8
μ
2
2
>
0
.
This matrixA equals(26)
A
1
′
A
2
′
A
3
′
A
4
′
,where(27a)
A
1
′
=
k
16
0
0
0
0
0
0
0
0
0
0
-
1
0
0
4
μ
2
2
0
0
0
0
0
0
0
k
16
0
0
0
0
0
0
0
0
0
0
-
1
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
2
k
16
0
0
0
7
×
10
,
(27b)
A
2
′
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
μ
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
×
10
,
(27c)
A
3
′
=
0
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
×
10
,
(27d)
A
4
′
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
7
×
10
,
In the14
×
20 matrix A, it is easy to observe that the following rows and columns are linearly dependent: the 1th and the 5th rows; the 2th and 5th columns; the 10th, 11th, 12th, 15th, 16th, and 17th columns are identically zero. Therefore, the 5th row and the 5th, 10th, 11th, 12th, 15th, 16th, and 17th columns are eliminated. The original 14
×
20 matrix has rank 13. The proof is completed.


### 3.2. The Recognition Problem

Proposition 2.
Let(28)
g
z
,
λ
=
h
z
,
λ
+
q
z
,
λwhere(29a)
h
=
k
16
a
1
2
,
4
μ
2
2
a
2
2
-
F
2
2
,
(29b)
q
=
k
11
a
1
6
+
k
12
a
1
4
a
2
2
+
k
13
a
1
2
a
2
4
+
k
14
σ
1
a
1
4
+
k
15
σ
1
a
1
2
a
2
2
+
σ
1
2
a
1
2
,
k
21
a
2
6
+
k
22
a
1
4
a
2
2
+
k
23
a
1
2
a
2
4
+
k
24
σ
2
a
2
4
+
k
25
σ
2
a
1
2
a
2
2
+
σ
2
2
a
2
2
.Ifq
∈
M
3
+
M
2
σ
1
,
σ
2
,
F
2
+
M
F
2
2
E
→
z
,
λ is a higher order term, then g is strongly equivalent to h.Proof.
By Proposition1, a higher order term of M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2 is (30)
M
3
+
M
2
σ
1
,
σ
2
,
F
2
+
M
F
2
2
.
For polynomialg, it is observed that (31a)
k
11
a
1
6
,
0
,
k
12
a
1
4
a
2
2
,
0
,
k
13
a
1
2
a
2
4
,
0
,
0
,
k
21
a
2
6
,
0
,
k
22
a
1
4
a
2
2
a
n
d
0
,
k
23
a
1
2
a
2
4
b
e
l
o
n
g
t
o
M
3
E
→
z
,
λ
.
(31b)
k
14
σ
1
a
1
4
,
0
,
k
15
σ
1
a
1
2
a
2
2
,
0
,
σ
1
2
a
1
2
,
0
,
0
,
k
24
σ
2
a
2
4
,
0
,
k
25
σ
2
a
1
2
a
2
2
a
n
d
0
,
σ
2
2
a
2
2
b
e
l
o
n
g
t
o
M
2
σ
1
,
σ
2
,
F
2
E
→
z
,
λ
.
Thus(32)
q
∈
M
3
+
M
2
σ
1
,
σ
2
,
F
2
+
M
F
2
2
E
→
z
,
λ
.
By applying Theorem 1.3 (see pages 168 in Golubitsky and Schaeffer II (1985)), we can obtain thatg is strongly equivalent to h. The proof is completed.According to Proposition2, polynomial g will be replaced by polynomial h in the following discussion.Proposition 3.
The nondegenerate conditions ofh satisfy(33)
D
>
0
;
h is equivalent to(34)
±
a
1
2
,
a
2
2
-
F
2
2
.Proof.
According to(
2.6
), (
2.7
), and (
2.8
) (see pages 402 in Golubitsky and Schaeffer I (1985)), we have(35)
Q
h
=
2
k
16
a
1
0
0
8
μ
2
2
a
2
=
16
k
16
μ
2
2
a
1
a
2
,and(36)
D
=
b
2
-
4
a
c
=
256
k
16
2
μ
2
4
.
Sincek
16
≠
0, μ
2
>
0, we obtain(37)
D
>
0
.
We rescalea
1 and a
2; the normal form of the germ h can be expressed as(38)
±
a
1
2
,
a
2
2
-
F
2
2
.
The proof is completed. It can be observed from (38) that its universal unfolding needs to complement the linear term about the state variables and constant term.We improve Theorem 3.1 (see pages 409 in Golubitsky and Schaeffer I (1985)) by introducing a sign functionε. Proposition 4 is given as follows.Proposition 4.
Let(39)
H
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
h
1
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
,
h
2
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γbe a four-parameter unfolding of bifurcation problem h in two state variables which satisfies the nondegeneracy conditions Proposition 3. Then H is a universal unfolding of h if and only if(40)
H
a
1
,
a
2
,
F
2
,
0,0
,
0
=
h
1
a
1
,
a
2
,
F
2
,
0,0
,
0
,
h
2
a
1
,
a
2
,
F
2
,
0,0
,
0
,and(41)
d
e
t
P
=
d
e
t
0
h
1
,
a
1
a
1
h
1
,
a
1
a
2
0
h
2
,
a
1
a
1
h
2
,
a
1
a
2
0
h
1
,
a
2
a
1
h
1
,
a
2
a
2
0
h
2
,
a
2
a
1
h
2
,
a
2
a
2
h
1
,
σ
1
h
1
,
σ
1
a
1
h
1
,
σ
1
a
2
h
2
,
σ
1
h
2
,
σ
1
a
1
h
2
,
σ
1
a
2
h
1
,
σ
2
h
1
,
σ
2
a
1
h
1
,
σ
2
a
2
h
2
,
σ
2
h
2
,
σ
2
a
1
h
2
,
σ
2
a
2
h
1
,
F
2
h
1
,
F
2
a
1
h
1
,
F
2
a
2
h
2
,
F
2
h
2
,
F
2
a
1
h
2
,
F
2
a
2
h
1
,
γ
h
1
,
γ
a
1
h
1
,
γ
a
2
h
2
,
γ
h
2
,
γ
a
1
h
2
,
γ
a
2
≠
0
,where σ
1, σ
2, and F
2 are bifurcation parameters and γ is auxiliary parameter when evaluated at a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0.Proof.
MatrixP can be expressed by six vectors(42)
P
=
α
1
,
α
2
,
α
3
,
α
4
,
α
5
,
α
6
T
,in which(43a)
α
1
=
0
,
h
1
,
a
1
a
1
,
h
1
,
a
1
a
2
,
0
,
h
2
,
a
1
a
1
,
h
2
,
a
1
a
2
,
(43b)
α
2
=
0
,
h
1
,
a
2
a
1
,
h
1
,
a
2
a
2
,
0
,
h
2
,
a
2
a
1
,
h
2
,
a
2
a
2
,
(43c)
α
3
=
h
1
,
σ
1
,
h
1
,
σ
1
a
1
,
h
1
,
σ
1
a
2
,
h
2
,
σ
1
,
h
2
,
σ
1
a
1
,
h
2
,
σ
1
a
2
,
(43d)
α
4
=
h
1
,
σ
2
,
h
1
,
σ
2
a
1
,
h
1
,
σ
2
a
2
,
h
2
,
σ
2
,
h
2
,
σ
2
a
1
,
h
2
,
σ
2
a
2
,
(43e)
α
5
=
h
1
,
F
2
,
h
1
,
F
2
a
1
,
h
1
,
F
2
a
2
,
h
2
,
F
2
,
h
2
,
F
2
a
1
,
h
2
,
F
2
a
2
,
(43f)
α
6
=
h
1
,
γ
,
h
1
,
γ
a
1
,
h
1
,
γ
a
2
,
h
2
,
γ
,
h
2
,
γ
a
1
,
h
2
,
γ
a
2
.
Substituting (34) into (43a), (43b), (43c), (43d), (43e), and (43f),(44)
α
1
=
0
,
±
2,0
,
0,0
,
0
,
α
2
=
0,0
,
0,0
,
0,2
,
α
3
=
0,0
,
0,0
,
0,0
,
α
4
=
0,0
,
0,0
,
0,0
,
α
5
=
0,0
,
0,0
,
0,0
,
α
6
=
0,0
,
0,0
,
0,0
.
In (44), only two vectors α
1 and α
2 are linearly independent. Since H is a four-parameter universal unfolding of h, there exist four complementary linearly independent vectors for h. It is easy to find a complementary subspace to h, spanned by (45)
λ
a
2
,
0
,
0
,
λ
′
a
1
,
λ
′
′
,
λ
′
′
,
-
γ
,
γ
,where the values of λ, λ
′, and λ
′
′ are one of ε
1
σ
1, ε
2
σ
2, and ε
3
F
2, which are the values different from each other, where ε
i
i
=
1,2
,
3 are sign function, ε
i
=
+
1
,
0
,
-
1.
The introducedε
i can satisfy (40) and (41). It is obtained from (45) (46)
H
=
±
a
1
2
+
λ
a
2
+
λ
′
′
-
γ
,
a
2
2
-
F
2
2
+
λ
′
a
1
+
λ
′
′
+
γ
.
Obviously, whenε
i
=
γ
=
0, we have(47)
H
a
1
,
a
2
,
F
2
,
0,0
,
0
=
h
1
a
1
,
a
2
,
F
2
,
0,0
,
0
,
h
2
a
1
,
a
2
,
F
2
,
0,0
,
0
.
Substituting (46) into (41), when ε
i
≠
0 and a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0, we can get(48)
d
e
t
P
≠
0
.
Conversely, only two vectorsα
1 and α
2 in (44) are linearly independent. Because d
e
t
P
≠
0, the linearly independent vectors that complement h need four. It can be adopted that(49)
λ
a
2
,
0
,
0
,
λ
′
a
1
,
λ
′
′
,
λ
′
′
,
-
γ
,
γ
,
ThusH
=
±
a
1
2
+
λ
a
2
+
λ
′
′
-
γ
,
a
2
2
-
F
2
2
+
λ
′
a
1
+
λ
′
′
+
γ will be obtained. The proof is completed.


### 3.3. The Polynomial Space

Proposition 5.
The polynomial space is given by(50)
R
d
h
z
,
λ
Y
1
,
…
,
d
h
z
,
λ
Y
m
,
h
λ
,
λ
h
λ
,
λ
2
h
λ
,
…
=
R
a
1
,
0
,
0
,
a
2
,
0
,
F
2
.Proof.
To derive the above equality, it should be noted that(51)
u
1
′
u
2
′
=
B
v
1
′
v
2
′
=
B
1
B
2
B
3
B
4
v
1
′
v
2
′
,where(52a)
u
1
′
=
h
1
,
a
1
,
0
h
1
,
a
2
,
0
h
1
,
σ
1
,
0
h
1
,
σ
2
,
0
h
1
,
F
2
,
0
1
×
5
Τ
,
(52b)
u
2
′
=
0
,
h
2
,
a
1
0
,
h
2
,
a
2
0
,
h
2
,
σ
1
0
,
h
2
,
σ
2
0
,
h
2
,
F
2
1
×
5
Τ
,
(52c)
v
1
′
=
a
1
,
0
a
2
,
0
σ
1
,
0
σ
2
,
0
F
2
,
0
1
×
5
Τ
,
(52d)
v
2
′
=
0
,
a
1
0
,
a
2
0
,
σ
1
0
,
σ
2
0
,
F
2
1
×
5
Τ
,
(52e)
B
1
=
D
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
B
2
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
(52f)
B
3
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
B
4
=
0
0
0
0
0
0
D
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
D
3
,where(53)
D
1
=
±
2
,
D
2
=
2
,
D
3
=
-
2
.
It is found thatu
1
′
T
u
2
′
T can be expressed by three vectors a
1
,
0, 0
,
a
2, and 0
,
F
2. Hence it can be obtained that(54)
R
d
h
z
,
λ
Y
1
,
…
,
d
h
z
,
λ
Y
m
,
h
λ
,
λ
h
λ
,
λ
2
h
λ
,
…
=
R
a
1
,
0
,
0
,
a
2
,
0
,
F
2
.
The proof is completed.


### 3.4. The Universal Unfolding

Proposition 6.
The codimension ofh in E
→
z
,
λ is 4, and the universal unfolding of the normal form (34) is given as(55a)
G
1
=
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
,
(55b)
G
2
=
±
a
1
2
+
2
ε
22
σ
2
a
2
+
2
ε
21
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
23
σ
1
a
1
+
γ
,
(55c)
G
3
=
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
,
(55d)
G
4
=
±
a
1
2
+
2
ε
42
F
2
a
2
+
ε
41
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
43
σ
2
a
1
+
ε
41
σ
1
+
γ
,
(55e)
G
5
=
±
a
1
2
+
2
ε
52
σ
1
a
2
+
ε
51
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
53
F
2
a
1
+
ε
51
σ
2
+
γ
,
(55f)
G
6
=
±
a
1
2
+
2
ε
62
F
2
a
2
+
ε
61
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
63
σ
1
a
1
+
ε
61
σ
2
+
γ
.where γ is auxiliary parameter and ε
i
j are sign functions, ε
i
j
=
+
1
,
0
,
-
1; i
=
1
,
…
,
6, j
=
1,2
,
3.Proof.
According to Theorem 2.1 and equation 2.7 (see pages 211 in Golubitsky and Schaeffer II (1985)), it can be obtained that(56)
T
h
,
1
=
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2
E
→
z
,
λ
+
R
a
1
,
0
,
0
,
a
2
,
0
,
F
2
.
It is not difficult to see that the largest intrinsic idealI
t
r
T
h
,
1 contained in T
h
,
1 is just(57)
I
t
r
T
h
,
1
=
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2
E
→
z
,
λ
,and(58)
I
t
r
T
h
,
1
⊥
=
R
a
1
,
0
,
a
2
,
0
,
σ
1
,
0
,
σ
2
,
0
,
F
2
,
0
,
-
1,0
,
0
,
a
1
,
0
,
a
2
,
0
,
σ
1
,
0
,
σ
2
,
0
,
F
2
,
0,1
,whose dimension is 12.
Clearly, the dimension ofR
a
1
,
0
,
0
,
a
2
,
0
,
F
2 is 3; hence there exists a basis for a subspace of I
t
r
T
h
,
1
⊥ which is complementary to R
a
1
,
0
,
0
,
a
2
,
0
,
F
2.
Consequently, it is easy to find a complementary subspace toT
h
,
1, spanned by (59)
a
2
,
0
,
0
,
a
1
,
F
2
,
0
,
σ
1
,
0
,
0
,
σ
1
,
σ
2
,
0
,
0
,
σ
2
,
-
1,0
a
n
d
0,1
.
According to Proposition4, simplified (59),(60)
a
2
,
0
,
0
,
a
1
,
F
2
,
0
,
σ
1
,
σ
1
,
σ
2
,
σ
2
,
-
1,1
.
It can be observed thatM
σ
1
,
σ
2
,
F
2
E
→
z
,
λ is generated by the twelve mappings(61)
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
,
0
,
0
,
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
.
Equation (60) makes two auxiliary parameters β
1 and β
2 for a
2
,
0 and 0
,
a
1, then β
1
a
2
,
0 and 0
,
β
2
a
1 will occur. According to Proposition 4 and (61), β
1
a
2
,
0 and one of three vectors σ
1
a
2
,
0, σ
2
a
2
,
0, and F
2
a
2
,
0 are linearly independent. 0
,
β
2
a
1 and one of three vectors 0
,
σ
1
a
1, 0
,
σ
2
a
1, and 0
,
F
2
a
1 are linearly independent. Therefore, a
2
,
0 and 0
,
a
1 will be eliminated.
Three sign functionsε
1, ε
2, and ε
3 and an auxiliary parameter γ in universal unfolding of h in (60), such as ε
1
F
2
,
0, ε
2
σ
1
,
ε
2
σ
1, ε
3
σ
2
,
ε
3
σ
2, and -
γ
,
γ, will occur in universal unfolding of h. Since F
2, σ
1, σ
2, and γ are all parameters, according to Proposition 4, substituting four vectors ε
1
F
2
,
0, ε
2
σ
1
,
ε
2
σ
1, ε
3
σ
2
,
ε
3
σ
2, and -
γ
,
γ into (41), only two vectors are linearly independent. By (41), it is easy to find -
γ
,
γ will necessarily occur in matrix P.
Thus,ε
1
F
2
,
0
&
-
γ
,
γ or ε
2
σ
1
,
ε
2
σ
1
&
-
γ
,
γ or ε
3
σ
2
,
ε
3
σ
2
&
-
γ
,
γ will be chosen as one of complementary bases of T
h
,
1 in E
→
z
,
λ.
In order to simplify the notation,ε
2
σ
1
,
ε
2
σ
1 is denoted as ε
1
σ
1
,
ε
1
σ
1 and ε
3
σ
2
,
ε
3
σ
2 is denoted as ε
1
σ
2
,
ε
1
σ
2.
According to Proposition4 and (61), there exist two linearly independent vectors as complementary basis of R
T
h
,
1 in T
h
,
1. They are σ
1
a
2
,
0
&
0
,
σ
2
a
1 or σ
2
a
2
,
0
&
0
,
σ
1
a
1 when ε
1
F
2
,
0 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ. They are F
2
a
2
,
0
&
0
,
σ
2
a
1 or σ
2
a
2
,
0
&
0
,
F
2
a
1 when ε
1
σ
1
,
ε
1
σ
1 and -
γ
,
γ are chosen as one of complementary bases of T
h
,
1 in E
→
z
,
λ. They are F
2
a
2
,
0
&
0
,
σ
1
a
1 or σ
1
a
2
,
0
&
0
,
F
2
a
1 when ε
1
σ
2
,
ε
1
σ
2 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ.
There exist two sign functionsε
2 and ε
3, such that ε
2
σ
1
a
2
,
0
&
0
,
ε
3
σ
2
a
1 or ε
2
σ
2
a
2
,
0
&
0
,
ε
3
σ
1
a
1 will occur when ε
1
F
2
,
0 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ. ε
2
F
2
a
2
,
0
&
0
,
ε
3
σ
2
a
1 or ε
2
σ
2
a
2
,
0
&
0
,
ε
3
F
2
a
1 will occur when ε
1
σ
1
,
ε
1
σ
1 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ. ε
2
F
2
a
2
,
0
&
0
,
ε
3
σ
1
a
1 or ε
2
σ
1
a
2
,
0
&
0
,
ε
3
F
2
a
1 will occur when ε
1
σ
2
,
ε
1
σ
2 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ.
According to the analysis above, the universal unfolding ofh is given by(62a)
G
1
=
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
,
(62b)
G
2
=
±
a
1
2
+
2
ε
22
σ
2
a
2
+
2
ε
21
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
23
σ
1
a
1
+
γ
,
(62c)
G
3
=
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
,
(62d)
G
4
=
±
a
1
2
+
2
ε
42
F
2
a
2
+
ε
41
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
43
σ
2
a
1
+
ε
41
σ
1
+
γ
,
(62e)
G
5
=
±
a
1
2
+
2
ε
52
σ
1
a
2
+
ε
51
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
53
F
2
a
1
+
ε
51
σ
2
+
γ
,
(62f)
G
6
=
±
a
1
2
+
2
ε
62
F
2
a
2
+
ε
61
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
63
σ
1
a
1
+
ε
61
σ
2
+
γ
.
When sign functionsε
i
j and auxiliary parameter γ equal zero, namely, ε
i
j
=
0 (i
=
1
,
…
,
6; j
=
1,2
,
3) and γ
=
0, (62a), (62b), (62c), (62d), (62e), and (62f) can be obtained such that(63)
G
i
a
1
,
a
2
,
F
2
,
0,0
,
0,0
=
h
a
1
,
a
2
,
F
2
.
Using Proposition4, when ε
i
j
≠
0, it is obtained from (62a), (62b), (62c), (62d), (62e), and (62f) that(64)
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
2
ε
12
0
0
0
0
0
0
0
2
ε
13
0
2
ε
11
0
0
0
0
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
0
0
2
ε
23
0
0
0
2
ε
22
0
0
0
2
ε
21
0
0
0
0
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
ε
31
0
0
ε
31
0
0
0
0
2
ε
32
0
0
0
0
0
0
0
2
ε
33
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
ε
41
0
0
ε
41
0
0
0
0
0
0
2
ε
43
0
0
0
2
ε
42
0
0
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
2
ε
52
0
0
0
ε
51
0
0
ε
51
0
0
0
0
0
0
2
ε
53
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
0
0
2
ε
63
0
ε
61
0
0
ε
61
0
0
0
0
2
ε
62
0
0
0
-
1
0
0
1
0
0
≠
0
.
The proof is completed. It can be found from (55a), (55b), (55c), (55d), (55e), and (55f) that codimension of h in E
→
z
,
λ is equal to the summation of the number sign functions and the number of auxiliary parameter.


### 3.5. The Transition Set

The bifurcation behaviors of (55a), (55b), (55c), (55d), (55e), and (55f) are discussed in the following analysis. The transition set Σ of the universal unfolding G
:
R
n
×
R
×
R
k
→
R
n is formulated as follows (see pages 409 in Golubitsky and Schaeffer I (1985)).The bifurcation variety set is(65a)
B
=
α
∈
R
k
:
∃
z
,
λ
s
u
c
h
t
h
a
t
G
z
,
λ
,
α
=
0
,
a
n
d
r
a
n
k
D
G
z
,
λ
,
α
≤
1
.The hysteresis variety set is(65b)
H
=
α
∈
R
k
:
∃
z
,
λ
,
v
s
u
c
h
t
h
a
t
G
z
,
λ
,
α
=
0
,
d
G
z
,
λ
,
α
·
v
=
0
a
n
d
d
2
G
z
,
λ
,
α
v
,
v
∈
r
a
n
g
e
d
G
z
,
λ
,
α
.The double limit point variety set is(65c)
D
=
α
∈
R
k
:
∃
z
1
,
z
2
,
λ
s
u
c
h
t
h
a
t
z
1
≠
z
2
,
G
z
i
,
λ
,
α
=
0
,
a
n
d
det
⁡
d
G
z
,
λ
,
α
=
0
,
i
=
1,2
.Then the transition set isΣ
=
B
∪
H
∪
D.The equations of bifurcation points for (55a) are obtained as(66a)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
=
0
,
(66b)
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
=
0
,
(66c)
±
a
1
ε
13
σ
2
=
ε
12
σ
1
a
2
=
ε
12
a
2
0
,
o
r
±
a
1
ε
13
σ
2
=
ε
12
σ
1
a
2
=
0
ε
13
a
1
,
o
r
±
a
1
ε
13
σ
2
=
ε
12
σ
1
a
2
=
ε
11
-
F
2
.Adding (66a) to (66b) obtains (67)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
+
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
=
0
.By the third equation of (66c), it can be obtained that(68)
±
a
1
=
ε
11
ε
13
σ
2
-
F
2
,and(69)
a
2
=
-
ε
12
σ
1
F
2
ε
11
.Substituting (68) and (69) into (67), bifurcation occurs when (70)
ε
11
σ
1
2
-
ε
11
F
2
4
+
2
-
2
σ
1
2
F
2
3
-
2
σ
2
2
F
2
±
ε
11
σ
2
2
=
0
.By the first and second equation of (66c), bifurcation occurs when(71)
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0
.The equations of hysteresis points for (55a) are obtained as follows:(72a)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
=
0
,
(72b)
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
=
0
,
(72c)
±
a
1
a
2
-
ε
12
ε
13
σ
1
σ
2
=
0
,
(72d)
v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G
,where v
=
v
1
,
v
2.Now we assumeσ
1
≠
0 and σ
2
≠
0. Then, it can be taken that(73)
v
=
±
a
1
,
ε
12
σ
1
,and it is noted that v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G if and only if (74)
v
1
2
,
v
2
2
·
ε
12
σ
1
,
a
2
=
0
.Equation (74) implies that (75)
ε
12
a
1
2
+
σ
1
a
2
=
0
.Multiplying (75) by a
1 and substituting it into (72c) we obtain(76)
a
1
=
∓
ε
13
σ
1
2
/
3
σ
2
1
/
3
.Using (72c) and (76), we obtain(77)
a
2
=
-
ε
12
σ
1
1
/
3
σ
2
2
/
3
.Substituting (76) and (77) into (67), hysteresis occurs when(78)
F
2
2
-
2
ε
11
F
2
+
σ
1
4
/
3
σ
2
2
/
3
+
σ
1
2
/
3
σ
2
4
/
3
=
0
,and(79)
F
2
2
-
2
ε
11
F
2
+
3
σ
1
4
/
3
σ
2
2
/
3
-
3
σ
1
2
/
3
σ
2
4
/
3
=
0
.The equations of the double limit points for (55a) are obtained as(80a)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
=
0
,
(80b)
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
=
0
,
(80c)
a
1
a
2
-
ε
12
ε
13
σ
1
σ
2
=
0
,
(80d)
a
11
,
a
21
≠
a
12
,
a
22
,
w
h
e
n
F
2
=
c
o
n
s
t
.Using (70), double limit points occur when(81)
σ
1
=
±
1
,
F
2
=
±
ε
11
2
a
n
d
σ
2
=
0
,
F
2
=
2
ε
11
.B, H, and D have equations as follows:(82a)
B
:
ε
11
σ
1
2
-
ε
11
F
2
4
+
2
-
2
σ
1
2
F
2
3
-
2
σ
2
2
F
2
±
ε
11
σ
2
2
=
0
,
(82b)
H
:
F
2
2
-
2
ε
11
F
2
+
σ
1
4
/
3
σ
2
2
/
3
+
σ
1
2
/
3
σ
2
4
/
3
=
0
;
F
2
2
-
2
ε
11
F
2
+
3
σ
1
4
/
3
σ
2
2
/
3
-
3
σ
1
2
/
3
σ
2
4
/
3
=
0
,
(82c)
D
:
σ
1
=
±
1
,
F
2
=
±
ε
11
2
;
σ
2
=
0
,
F
2
=
2
ε
11
.In the same way, transition setΣ of the universal unfolding (55b), (55c), (55d), (55e), and (55f) may be obtained.For (55c), a short calculation shows that bifurcation occurs when(83a)
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
=
0
,
(83b)
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
=
0
,
(83c)
±
a
1
ε
33
F
2
=
ε
32
σ
2
a
2
=
ε
31
ε
31
o
r
±
a
1
ε
33
F
2
=
ε
32
σ
2
a
2
=
ε
32
a
2
0
o
r
±
a
1
ε
33
F
2
=
ε
32
σ
2
a
2
=
0
-
F
2
+
ε
33
a
1
.Adding (83a) to (83b) obtains(84)
a
2
2
±
a
1
2
+
2
ε
32
σ
2
a
2
+
2
ε
33
F
2
a
1
-
F
2
2
+
2
ε
31
σ
1
=
0
.By the first equation of (83c) and (84), a short calculation shows that bifurcation occurs when(85)
σ
1
=
-
3
2
ε
31
σ
2
2
-
±
3
-
1
2
ε
31
F
2
2
.By the second and third equation of (83c), bifurcation occurs when(86)
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0
.The equations of hysteresis points for (55c) are(87a)
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
=
0
,
(87b)
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
=
0
,
(87c)
±
a
1
a
2
-
ε
32
ε
33
F
2
σ
2
=
0
,
(87d)
v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G
,where v
=
v
1
,
v
2.Now assumeσ
1
≠
0 and σ
2
≠
0, then it may be taken that(88)
v
=
±
a
1
,
ε
32
σ
2and note that v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G if and only if(89)
v
1
2
,
v
2
2
·
ε
32
σ
2
,
a
2
=
0
.Equation (89) implies that (90)
ε
32
a
1
2
+
σ
2
a
2
=
0
,Next multiply (90) by a
1 and substitute (87c) to obtain(91)
a
1
=
∓
ε
33
F
2
1
/
3
σ
2
2
/
3
.Using (91) and (87c) obtains(92)
a
2
=
-
ε
32
F
2
2
/
3
σ
2
1
/
3
.Finally, substituting (91) and (92) into (84), hysteresis occurs when(93)
σ
1
=
1
2
ε
31
F
2
2
/
3
σ
2
4
/
3
+
F
2
2
/
3
σ
2
2
/
3
+
F
2
4
/
3
,and(94)
σ
1
=
1
2
ε
31
F
2
2
/
3
3
σ
2
4
/
3
-
3
F
2
2
/
3
σ
2
2
/
3
+
F
2
4
/
3
.The equations of double limit points for (55c) are(95a)
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
=
0
,
(95b)
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
=
0
,
(95c)
±
a
1
a
2
-
ε
32
ε
33
F
2
σ
2
=
0
,
(95d)
a
11
,
a
21
≠
a
12
,
a
22
,
w
h
e
n
σ
1
=
c
o
n
s
t
.By (85), a more involved calculation implies that double limit points occur for (96a)
σ
2
=
0
,
σ
1
≤
0
a
n
d
F
2
=
0
,
σ
1
≤
0
w
h
e
n
ε
31
=
1
a
n
d
t
a
k
e
n
t
h
e
p
l
u
s
s
i
g
n
i
n
85
,
(96b)
σ
2
=
0
,
σ
1
≥
0
a
n
d
F
2
=
0
,
σ
1
≥
0
w
h
e
n
ε
31
=
-
1
a
n
d
t
a
k
e
n
t
h
e
p
l
u
s
s
i
g
n
i
n
85
,
(96c)
σ
2
=
0
,
σ
1
≥
0
a
n
d
F
2
=
0
,
σ
1
≤
0
w
h
e
n
ε
31
=
1
a
n
d
t
a
k
e
n
t
h
e
m
i
n
u
s
s
i
g
n
i
n
85
,
(96d)
σ
2
=
0
,
σ
1
≤
0
a
n
d
F
2
=
0
,
σ
1
≥
0
w
h
e
n
ε
31
=
-
1
a
n
d
t
a
k
e
n
t
h
e
m
i
n
u
s
s
i
g
n
i
n
85
.We defineρ, θ, and δ as follows:(97)
F
2
=
ρ
3
cos
3
⁡
θ
,
σ
2
=
ρ
3
sin
3
⁡
θ
,
σ
1
=
ρ
6
δ
.Thusρ
,
θ
,
δ are a singular modification of cylindrical coordinates in f
T
σ
2
σ
1
-space.In these coordinates, substituting (97) into (85), (93), and (94), B, H, and D have equations as follows: (98a)
B
:
δ
=
-
3
2
ε
31
sin
6
⁡
θ
-
±
3
-
1
2
ε
31
cos
6
⁡
θ
,
(98b)
H
:
δ
=
1
2
ε
31
cos
2
⁡
θ
sin
4
⁡
θ
+
cos
2
⁡
θ
sin
2
⁡
θ
+
cos
4
⁡
θ
;
δ
=
1
2
ε
31
cos
2
⁡
θ
3
sin
4
⁡
θ
-
3
cos
2
⁡
θ
sin
2
⁡
θ
+
cos
4
⁡
θ
,
(98c)
D
:
θ
=
0
,
π
2
,
π
,
3
π
2
;
δ
≤
0
;
ε
31
=
1
.
o
r
θ
=
0
,
π
2
,
π
,
3
π
2
;
δ
≥
0
;
ε
31
=
-
1From the above analysis, we deduce that (82a), (82b), (82c) and (98a), (98b), (98c) represent the relationship between two detuning parameters and a transverse excitation when the composite laminated piezoelectric rectangular plate structure subjected to small perturbations shows the bifurcation, hysteresis, and double limit points. If the disturbed system is equivalent to the original system, the space made up by these perturbations is the restricted tangent space. We know that the items σ
1
a
2 and σ
2
a
1 of complemented to h in (55a) belong to the restricted tangent space and the variation of parameters σ
1 and σ
2 make no difference in the bifurcation property of the original system, while the variation of parameter F
2 of complemented to h in (55a) has a qualitative influence on the bifurcation property of the original system. The similar analysis can be made in (55b), (55c), (55d), (55e), and (55f).


## 3.1. The Restricted Tangent Space

Proposition 1.
The restricted tangent spaceR
T
g
,
1 of the germ g
z
,
λ is given(19)
R
T
g
,
1
=
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2Proof.
According to Proposition 1.4 (see pages 169 in Golubitsky and Schaeffer II (1985)),R
T
g
,
1 is generated (as module over E
z
,
λ) by the fourteen mappings:(20)
g
1
,
0
,
g
2
,
0
,
0
,
g
1
,
0
,
g
2
,
a
1
g
1
,
a
1
,
g
2
,
a
1
,
a
2
g
1
,
a
1
,
g
2
,
a
1
,
σ
1
g
1
,
a
1
,
g
2
,
a
1
,
σ
2
g
1
,
a
1
,
g
2
,
a
1
,
F
2
g
1
,
a
1
,
g
2
,
a
1
,
a
1
g
1
,
a
2
,
g
2
,
a
2
,
a
2
g
1
,
a
2
,
g
2
,
a
2
,
σ
1
g
1
,
a
2
,
g
2
,
a
2
,
σ
2
g
1
,
a
2
,
g
2
,
a
2
,
F
2
g
1
,
a
2
,
g
2
,
a
2
.
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2 is generated by the twenty mappings(21)
a
1
2
,
0
,
a
1
a
2
,
0
,
a
2
2
,
0
,
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
,
0
,
F
2
2
,
0
,
0
,
a
1
2
,
0
,
a
1
a
2
,
0
,
a
2
2
,
0
,
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
,
0
,
F
2
2
.
By the following proof that there exists a reversible matrixA between (20) and (21) when a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
=
0,0
,
0,0
,
0, we observe that(22)
u
1
u
2
=
A
v
1
v
2
=
A
1
A
2
A
3
A
4
v
1
v
2
,where(23a)
u
1
=
g
1
,
0
g
2
,
0
0
,
g
1
0
,
g
2
,
a
1
g
1
,
a
1
,
g
2
,
a
1
a
2
g
1
,
a
1
,
g
2
,
a
1
σ
1
g
1
,
a
1
,
g
2
,
a
1
1
×
7
T
,
(23b)
u
2
=
σ
2
g
1
,
a
1
,
g
2
,
a
1
F
2
g
1
,
a
1
,
g
2
,
a
1
a
1
g
1
,
a
2
,
g
2
,
a
2
a
2
g
1
,
a
2
,
g
2
,
a
2
,
σ
1
g
1
,
a
2
,
g
2
,
a
2
σ
2
g
1
,
a
2
,
g
2
,
a
2
F
2
g
1
,
a
2
,
g
2
,
a
2
1
×
7
T
,
(23c)
v
1
=
a
1
2
,
0
F
2
2
,
0
0
,
a
1
2
0
,
F
2
2
a
2
2
,
0
a
1
a
2
,
0
σ
1
a
1
,
0
σ
2
a
1
,
0
F
2
a
1
,
0
σ
1
a
2
,
0
1
×
10
T
,
(23d)
v
2
=
σ
2
a
2
,
0
F
2
a
2
,
0
0
,
a
1
a
2
0
,
a
2
2
0
,
σ
1
a
1
0
,
σ
2
a
1
0
,
F
2
a
1
0
,
σ
1
a
2
0
,
σ
2
a
2
0
,
F
2
a
2
1
×
10
T
,
(23e)
A
1
=
C
1
0
0
0
0
0
0
0
0
0
0
-
1
0
0
C
2
0
0
0
0
0
0
0
C
1
0
0
0
0
0
0
0
0
0
0
-
1
0
0
0
0
0
0
C
3
0
C
4
0
0
0
0
0
0
0
0
0
0
0
0
C
3
0
0
0
0
0
0
0
0
0
0
C
3
0
0
0
7
×
10
,
(23f)
A
2
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
4
0
0
0
0
0
0
0
0
0
0
0
C
4
0
0
0
0
0
7
×
10
,
(23g)
A
3
=
0
0
0
0
0
0
0
C
3
0
0
0
0
0
0
0
0
0
0
C
3
0
C
5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C
5
0
0
0
0
0
0
0
0
0
0
C
5
0
0
0
0
0
0
0
0
0
0
C
5
0
0
0
0
0
0
0
0
0
0
C
5
0
7
×
10
,
(23h)
A
4
=
0
0
0
0
0
C
4
0
0
0
0
0
0
0
0
0
0
C
4
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
C
6
0
0
0
0
0
0
0
0
0
0
C
6
7
×
10
,where (24a)
C
1
=
k
11
a
1
4
+
k
12
a
1
2
a
2
2
+
k
13
a
2
4
+
k
14
σ
1
a
1
2
+
k
15
σ
1
a
2
2
+
σ
1
2
+
k
16
,
(24b)
C
2
=
k
21
a
2
4
+
k
22
a
1
4
+
k
23
a
1
2
a
2
2
+
k
24
σ
2
a
2
2
+
k
25
σ
2
a
1
2
+
σ
2
2
+
4
μ
2
2
,
(24c)
C
3
=
6
k
11
a
1
4
+
4
k
12
a
1
2
a
2
2
+
2
k
13
a
2
4
+
4
k
14
σ
1
a
1
2
+
2
k
15
σ
1
a
2
2
+
2
σ
1
2
+
2
k
16
,
(24d)
C
4
=
2
k
12
a
1
2
+
4
k
13
a
2
2
+
2
k
15
σ
1
a
1
a
2
,
(24e)
C
5
=
4
k
22
a
1
2
+
2
k
23
a
2
2
+
2
k
25
σ
2
a
2
2
,
(24f)
C
6
=
6
k
21
a
2
4
+
2
k
22
a
1
4
+
4
k
23
a
1
2
a
2
2
+
4
k
24
σ
2
a
2
2
+
2
k
25
σ
2
a
1
2
+
2
σ
2
2
+
8
μ
2
2
.
In (15a) and (15b), if k
16
=
0, bifurcation analysis is meaningless, so k
16
≠
0 will be studied in this paper. When a
1
=
a
2
=
σ
1
=
σ
2
=
F
2
=
0, substituting μ
2
>
0 into (24a), (24b), (24c), (24d), (24e), and (24f), it can be obtained that(25)
C
1
=
k
16
≠
0
,
C
2
=
4
μ
2
2
>
0
,
C
3
=
2
k
16
≠
0
,
C
4
=
0
,
C
5
=
0
,
C
6
=
8
μ
2
2
>
0
.
This matrixA equals(26)
A
1
′
A
2
′
A
3
′
A
4
′
,where(27a)
A
1
′
=
k
16
0
0
0
0
0
0
0
0
0
0
-
1
0
0
4
μ
2
2
0
0
0
0
0
0
0
k
16
0
0
0
0
0
0
0
0
0
0
-
1
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
2
k
16
0
0
0
7
×
10
,
(27b)
A
2
′
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
μ
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
×
10
,
(27c)
A
3
′
=
0
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
2
k
16
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7
×
10
,
(27d)
A
4
′
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
0
0
0
0
0
0
0
0
0
0
8
μ
2
2
7
×
10
,
In the14
×
20 matrix A, it is easy to observe that the following rows and columns are linearly dependent: the 1th and the 5th rows; the 2th and 5th columns; the 10th, 11th, 12th, 15th, 16th, and 17th columns are identically zero. Therefore, the 5th row and the 5th, 10th, 11th, 12th, 15th, 16th, and 17th columns are eliminated. The original 14
×
20 matrix has rank 13. The proof is completed.


## 3.2. The Recognition Problem

Proposition 2.
Let(28)
g
z
,
λ
=
h
z
,
λ
+
q
z
,
λwhere(29a)
h
=
k
16
a
1
2
,
4
μ
2
2
a
2
2
-
F
2
2
,
(29b)
q
=
k
11
a
1
6
+
k
12
a
1
4
a
2
2
+
k
13
a
1
2
a
2
4
+
k
14
σ
1
a
1
4
+
k
15
σ
1
a
1
2
a
2
2
+
σ
1
2
a
1
2
,
k
21
a
2
6
+
k
22
a
1
4
a
2
2
+
k
23
a
1
2
a
2
4
+
k
24
σ
2
a
2
4
+
k
25
σ
2
a
1
2
a
2
2
+
σ
2
2
a
2
2
.Ifq
∈
M
3
+
M
2
σ
1
,
σ
2
,
F
2
+
M
F
2
2
E
→
z
,
λ is a higher order term, then g is strongly equivalent to h.Proof.
By Proposition1, a higher order term of M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2 is (30)
M
3
+
M
2
σ
1
,
σ
2
,
F
2
+
M
F
2
2
.
For polynomialg, it is observed that (31a)
k
11
a
1
6
,
0
,
k
12
a
1
4
a
2
2
,
0
,
k
13
a
1
2
a
2
4
,
0
,
0
,
k
21
a
2
6
,
0
,
k
22
a
1
4
a
2
2
a
n
d
0
,
k
23
a
1
2
a
2
4
b
e
l
o
n
g
t
o
M
3
E
→
z
,
λ
.
(31b)
k
14
σ
1
a
1
4
,
0
,
k
15
σ
1
a
1
2
a
2
2
,
0
,
σ
1
2
a
1
2
,
0
,
0
,
k
24
σ
2
a
2
4
,
0
,
k
25
σ
2
a
1
2
a
2
2
a
n
d
0
,
σ
2
2
a
2
2
b
e
l
o
n
g
t
o
M
2
σ
1
,
σ
2
,
F
2
E
→
z
,
λ
.
Thus(32)
q
∈
M
3
+
M
2
σ
1
,
σ
2
,
F
2
+
M
F
2
2
E
→
z
,
λ
.
By applying Theorem 1.3 (see pages 168 in Golubitsky and Schaeffer II (1985)), we can obtain thatg is strongly equivalent to h. The proof is completed.According to Proposition2, polynomial g will be replaced by polynomial h in the following discussion.Proposition 3.
The nondegenerate conditions ofh satisfy(33)
D
>
0
;
h is equivalent to(34)
±
a
1
2
,
a
2
2
-
F
2
2
.Proof.
According to(
2.6
), (
2.7
), and (
2.8
) (see pages 402 in Golubitsky and Schaeffer I (1985)), we have(35)
Q
h
=
2
k
16
a
1
0
0
8
μ
2
2
a
2
=
16
k
16
μ
2
2
a
1
a
2
,and(36)
D
=
b
2
-
4
a
c
=
256
k
16
2
μ
2
4
.
Sincek
16
≠
0, μ
2
>
0, we obtain(37)
D
>
0
.
We rescalea
1 and a
2; the normal form of the germ h can be expressed as(38)
±
a
1
2
,
a
2
2
-
F
2
2
.
The proof is completed. It can be observed from (38) that its universal unfolding needs to complement the linear term about the state variables and constant term.We improve Theorem 3.1 (see pages 409 in Golubitsky and Schaeffer I (1985)) by introducing a sign functionε. Proposition 4 is given as follows.Proposition 4.
Let(39)
H
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
h
1
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
,
h
2
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γbe a four-parameter unfolding of bifurcation problem h in two state variables which satisfies the nondegeneracy conditions Proposition 3. Then H is a universal unfolding of h if and only if(40)
H
a
1
,
a
2
,
F
2
,
0,0
,
0
=
h
1
a
1
,
a
2
,
F
2
,
0,0
,
0
,
h
2
a
1
,
a
2
,
F
2
,
0,0
,
0
,and(41)
d
e
t
P
=
d
e
t
0
h
1
,
a
1
a
1
h
1
,
a
1
a
2
0
h
2
,
a
1
a
1
h
2
,
a
1
a
2
0
h
1
,
a
2
a
1
h
1
,
a
2
a
2
0
h
2
,
a
2
a
1
h
2
,
a
2
a
2
h
1
,
σ
1
h
1
,
σ
1
a
1
h
1
,
σ
1
a
2
h
2
,
σ
1
h
2
,
σ
1
a
1
h
2
,
σ
1
a
2
h
1
,
σ
2
h
1
,
σ
2
a
1
h
1
,
σ
2
a
2
h
2
,
σ
2
h
2
,
σ
2
a
1
h
2
,
σ
2
a
2
h
1
,
F
2
h
1
,
F
2
a
1
h
1
,
F
2
a
2
h
2
,
F
2
h
2
,
F
2
a
1
h
2
,
F
2
a
2
h
1
,
γ
h
1
,
γ
a
1
h
1
,
γ
a
2
h
2
,
γ
h
2
,
γ
a
1
h
2
,
γ
a
2
≠
0
,where σ
1, σ
2, and F
2 are bifurcation parameters and γ is auxiliary parameter when evaluated at a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0.Proof.
MatrixP can be expressed by six vectors(42)
P
=
α
1
,
α
2
,
α
3
,
α
4
,
α
5
,
α
6
T
,in which(43a)
α
1
=
0
,
h
1
,
a
1
a
1
,
h
1
,
a
1
a
2
,
0
,
h
2
,
a
1
a
1
,
h
2
,
a
1
a
2
,
(43b)
α
2
=
0
,
h
1
,
a
2
a
1
,
h
1
,
a
2
a
2
,
0
,
h
2
,
a
2
a
1
,
h
2
,
a
2
a
2
,
(43c)
α
3
=
h
1
,
σ
1
,
h
1
,
σ
1
a
1
,
h
1
,
σ
1
a
2
,
h
2
,
σ
1
,
h
2
,
σ
1
a
1
,
h
2
,
σ
1
a
2
,
(43d)
α
4
=
h
1
,
σ
2
,
h
1
,
σ
2
a
1
,
h
1
,
σ
2
a
2
,
h
2
,
σ
2
,
h
2
,
σ
2
a
1
,
h
2
,
σ
2
a
2
,
(43e)
α
5
=
h
1
,
F
2
,
h
1
,
F
2
a
1
,
h
1
,
F
2
a
2
,
h
2
,
F
2
,
h
2
,
F
2
a
1
,
h
2
,
F
2
a
2
,
(43f)
α
6
=
h
1
,
γ
,
h
1
,
γ
a
1
,
h
1
,
γ
a
2
,
h
2
,
γ
,
h
2
,
γ
a
1
,
h
2
,
γ
a
2
.
Substituting (34) into (43a), (43b), (43c), (43d), (43e), and (43f),(44)
α
1
=
0
,
±
2,0
,
0,0
,
0
,
α
2
=
0,0
,
0,0
,
0,2
,
α
3
=
0,0
,
0,0
,
0,0
,
α
4
=
0,0
,
0,0
,
0,0
,
α
5
=
0,0
,
0,0
,
0,0
,
α
6
=
0,0
,
0,0
,
0,0
.
In (44), only two vectors α
1 and α
2 are linearly independent. Since H is a four-parameter universal unfolding of h, there exist four complementary linearly independent vectors for h. It is easy to find a complementary subspace to h, spanned by (45)
λ
a
2
,
0
,
0
,
λ
′
a
1
,
λ
′
′
,
λ
′
′
,
-
γ
,
γ
,where the values of λ, λ
′, and λ
′
′ are one of ε
1
σ
1, ε
2
σ
2, and ε
3
F
2, which are the values different from each other, where ε
i
i
=
1,2
,
3 are sign function, ε
i
=
+
1
,
0
,
-
1.
The introducedε
i can satisfy (40) and (41). It is obtained from (45) (46)
H
=
±
a
1
2
+
λ
a
2
+
λ
′
′
-
γ
,
a
2
2
-
F
2
2
+
λ
′
a
1
+
λ
′
′
+
γ
.
Obviously, whenε
i
=
γ
=
0, we have(47)
H
a
1
,
a
2
,
F
2
,
0,0
,
0
=
h
1
a
1
,
a
2
,
F
2
,
0,0
,
0
,
h
2
a
1
,
a
2
,
F
2
,
0,0
,
0
.
Substituting (46) into (41), when ε
i
≠
0 and a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0, we can get(48)
d
e
t
P
≠
0
.
Conversely, only two vectorsα
1 and α
2 in (44) are linearly independent. Because d
e
t
P
≠
0, the linearly independent vectors that complement h need four. It can be adopted that(49)
λ
a
2
,
0
,
0
,
λ
′
a
1
,
λ
′
′
,
λ
′
′
,
-
γ
,
γ
,
ThusH
=
±
a
1
2
+
λ
a
2
+
λ
′
′
-
γ
,
a
2
2
-
F
2
2
+
λ
′
a
1
+
λ
′
′
+
γ will be obtained. The proof is completed.


## 3.3. The Polynomial Space

Proposition 5.
The polynomial space is given by(50)
R
d
h
z
,
λ
Y
1
,
…
,
d
h
z
,
λ
Y
m
,
h
λ
,
λ
h
λ
,
λ
2
h
λ
,
…
=
R
a
1
,
0
,
0
,
a
2
,
0
,
F
2
.Proof.
To derive the above equality, it should be noted that(51)
u
1
′
u
2
′
=
B
v
1
′
v
2
′
=
B
1
B
2
B
3
B
4
v
1
′
v
2
′
,where(52a)
u
1
′
=
h
1
,
a
1
,
0
h
1
,
a
2
,
0
h
1
,
σ
1
,
0
h
1
,
σ
2
,
0
h
1
,
F
2
,
0
1
×
5
Τ
,
(52b)
u
2
′
=
0
,
h
2
,
a
1
0
,
h
2
,
a
2
0
,
h
2
,
σ
1
0
,
h
2
,
σ
2
0
,
h
2
,
F
2
1
×
5
Τ
,
(52c)
v
1
′
=
a
1
,
0
a
2
,
0
σ
1
,
0
σ
2
,
0
F
2
,
0
1
×
5
Τ
,
(52d)
v
2
′
=
0
,
a
1
0
,
a
2
0
,
σ
1
0
,
σ
2
0
,
F
2
1
×
5
Τ
,
(52e)
B
1
=
D
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
B
2
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
(52f)
B
3
=
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
B
4
=
0
0
0
0
0
0
D
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
D
3
,where(53)
D
1
=
±
2
,
D
2
=
2
,
D
3
=
-
2
.
It is found thatu
1
′
T
u
2
′
T can be expressed by three vectors a
1
,
0, 0
,
a
2, and 0
,
F
2. Hence it can be obtained that(54)
R
d
h
z
,
λ
Y
1
,
…
,
d
h
z
,
λ
Y
m
,
h
λ
,
λ
h
λ
,
λ
2
h
λ
,
…
=
R
a
1
,
0
,
0
,
a
2
,
0
,
F
2
.
The proof is completed.


## 3.4. The Universal Unfolding

Proposition 6.
The codimension ofh in E
→
z
,
λ is 4, and the universal unfolding of the normal form (34) is given as(55a)
G
1
=
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
,
(55b)
G
2
=
±
a
1
2
+
2
ε
22
σ
2
a
2
+
2
ε
21
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
23
σ
1
a
1
+
γ
,
(55c)
G
3
=
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
,
(55d)
G
4
=
±
a
1
2
+
2
ε
42
F
2
a
2
+
ε
41
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
43
σ
2
a
1
+
ε
41
σ
1
+
γ
,
(55e)
G
5
=
±
a
1
2
+
2
ε
52
σ
1
a
2
+
ε
51
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
53
F
2
a
1
+
ε
51
σ
2
+
γ
,
(55f)
G
6
=
±
a
1
2
+
2
ε
62
F
2
a
2
+
ε
61
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
63
σ
1
a
1
+
ε
61
σ
2
+
γ
.where γ is auxiliary parameter and ε
i
j are sign functions, ε
i
j
=
+
1
,
0
,
-
1; i
=
1
,
…
,
6, j
=
1,2
,
3.Proof.
According to Theorem 2.1 and equation 2.7 (see pages 211 in Golubitsky and Schaeffer II (1985)), it can be obtained that(56)
T
h
,
1
=
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2
E
→
z
,
λ
+
R
a
1
,
0
,
0
,
a
2
,
0
,
F
2
.
It is not difficult to see that the largest intrinsic idealI
t
r
T
h
,
1 contained in T
h
,
1 is just(57)
I
t
r
T
h
,
1
=
M
2
+
M
σ
1
,
σ
2
,
F
2
+
F
2
2
E
→
z
,
λ
,and(58)
I
t
r
T
h
,
1
⊥
=
R
a
1
,
0
,
a
2
,
0
,
σ
1
,
0
,
σ
2
,
0
,
F
2
,
0
,
-
1,0
,
0
,
a
1
,
0
,
a
2
,
0
,
σ
1
,
0
,
σ
2
,
0
,
F
2
,
0,1
,whose dimension is 12.
Clearly, the dimension ofR
a
1
,
0
,
0
,
a
2
,
0
,
F
2 is 3; hence there exists a basis for a subspace of I
t
r
T
h
,
1
⊥ which is complementary to R
a
1
,
0
,
0
,
a
2
,
0
,
F
2.
Consequently, it is easy to find a complementary subspace toT
h
,
1, spanned by (59)
a
2
,
0
,
0
,
a
1
,
F
2
,
0
,
σ
1
,
0
,
0
,
σ
1
,
σ
2
,
0
,
0
,
σ
2
,
-
1,0
a
n
d
0,1
.
According to Proposition4, simplified (59),(60)
a
2
,
0
,
0
,
a
1
,
F
2
,
0
,
σ
1
,
σ
1
,
σ
2
,
σ
2
,
-
1,1
.
It can be observed thatM
σ
1
,
σ
2
,
F
2
E
→
z
,
λ is generated by the twelve mappings(61)
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
,
0
,
0
,
σ
1
a
1
,
0
,
σ
2
a
1
,
0
,
F
2
a
1
,
0
,
σ
1
a
2
,
0
,
σ
2
a
2
,
0
,
F
2
a
2
.
Equation (60) makes two auxiliary parameters β
1 and β
2 for a
2
,
0 and 0
,
a
1, then β
1
a
2
,
0 and 0
,
β
2
a
1 will occur. According to Proposition 4 and (61), β
1
a
2
,
0 and one of three vectors σ
1
a
2
,
0, σ
2
a
2
,
0, and F
2
a
2
,
0 are linearly independent. 0
,
β
2
a
1 and one of three vectors 0
,
σ
1
a
1, 0
,
σ
2
a
1, and 0
,
F
2
a
1 are linearly independent. Therefore, a
2
,
0 and 0
,
a
1 will be eliminated.
Three sign functionsε
1, ε
2, and ε
3 and an auxiliary parameter γ in universal unfolding of h in (60), such as ε
1
F
2
,
0, ε
2
σ
1
,
ε
2
σ
1, ε
3
σ
2
,
ε
3
σ
2, and -
γ
,
γ, will occur in universal unfolding of h. Since F
2, σ
1, σ
2, and γ are all parameters, according to Proposition 4, substituting four vectors ε
1
F
2
,
0, ε
2
σ
1
,
ε
2
σ
1, ε
3
σ
2
,
ε
3
σ
2, and -
γ
,
γ into (41), only two vectors are linearly independent. By (41), it is easy to find -
γ
,
γ will necessarily occur in matrix P.
Thus,ε
1
F
2
,
0
&
-
γ
,
γ or ε
2
σ
1
,
ε
2
σ
1
&
-
γ
,
γ or ε
3
σ
2
,
ε
3
σ
2
&
-
γ
,
γ will be chosen as one of complementary bases of T
h
,
1 in E
→
z
,
λ.
In order to simplify the notation,ε
2
σ
1
,
ε
2
σ
1 is denoted as ε
1
σ
1
,
ε
1
σ
1 and ε
3
σ
2
,
ε
3
σ
2 is denoted as ε
1
σ
2
,
ε
1
σ
2.
According to Proposition4 and (61), there exist two linearly independent vectors as complementary basis of R
T
h
,
1 in T
h
,
1. They are σ
1
a
2
,
0
&
0
,
σ
2
a
1 or σ
2
a
2
,
0
&
0
,
σ
1
a
1 when ε
1
F
2
,
0 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ. They are F
2
a
2
,
0
&
0
,
σ
2
a
1 or σ
2
a
2
,
0
&
0
,
F
2
a
1 when ε
1
σ
1
,
ε
1
σ
1 and -
γ
,
γ are chosen as one of complementary bases of T
h
,
1 in E
→
z
,
λ. They are F
2
a
2
,
0
&
0
,
σ
1
a
1 or σ
1
a
2
,
0
&
0
,
F
2
a
1 when ε
1
σ
2
,
ε
1
σ
2 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ.
There exist two sign functionsε
2 and ε
3, such that ε
2
σ
1
a
2
,
0
&
0
,
ε
3
σ
2
a
1 or ε
2
σ
2
a
2
,
0
&
0
,
ε
3
σ
1
a
1 will occur when ε
1
F
2
,
0 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ. ε
2
F
2
a
2
,
0
&
0
,
ε
3
σ
2
a
1 or ε
2
σ
2
a
2
,
0
&
0
,
ε
3
F
2
a
1 will occur when ε
1
σ
1
,
ε
1
σ
1 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ. ε
2
F
2
a
2
,
0
&
0
,
ε
3
σ
1
a
1 or ε
2
σ
1
a
2
,
0
&
0
,
ε
3
F
2
a
1 will occur when ε
1
σ
2
,
ε
1
σ
2 and -
γ
,
γ are chosen as one of complementary basis of T
h
,
1 in E
→
z
,
λ.
According to the analysis above, the universal unfolding ofh is given by(62a)
G
1
=
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
,
(62b)
G
2
=
±
a
1
2
+
2
ε
22
σ
2
a
2
+
2
ε
21
F
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
23
σ
1
a
1
+
γ
,
(62c)
G
3
=
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
,
(62d)
G
4
=
±
a
1
2
+
2
ε
42
F
2
a
2
+
ε
41
σ
1
-
γ
,
a
2
2
-
F
2
2
+
2
ε
43
σ
2
a
1
+
ε
41
σ
1
+
γ
,
(62e)
G
5
=
±
a
1
2
+
2
ε
52
σ
1
a
2
+
ε
51
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
53
F
2
a
1
+
ε
51
σ
2
+
γ
,
(62f)
G
6
=
±
a
1
2
+
2
ε
62
F
2
a
2
+
ε
61
σ
2
-
γ
,
a
2
2
-
F
2
2
+
2
ε
63
σ
1
a
1
+
ε
61
σ
2
+
γ
.
When sign functionsε
i
j and auxiliary parameter γ equal zero, namely, ε
i
j
=
0 (i
=
1
,
…
,
6; j
=
1,2
,
3) and γ
=
0, (62a), (62b), (62c), (62d), (62e), and (62f) can be obtained such that(63)
G
i
a
1
,
a
2
,
F
2
,
0,0
,
0,0
=
h
a
1
,
a
2
,
F
2
.
Using Proposition4, when ε
i
j
≠
0, it is obtained from (62a), (62b), (62c), (62d), (62e), and (62f) that(64)
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
2
ε
12
0
0
0
0
0
0
0
2
ε
13
0
2
ε
11
0
0
0
0
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
0
0
2
ε
23
0
0
0
2
ε
22
0
0
0
2
ε
21
0
0
0
0
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
ε
31
0
0
ε
31
0
0
0
0
2
ε
32
0
0
0
0
0
0
0
2
ε
33
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
ε
41
0
0
ε
41
0
0
0
0
0
0
2
ε
43
0
0
0
2
ε
42
0
0
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
2
ε
52
0
0
0
ε
51
0
0
ε
51
0
0
0
0
0
0
2
ε
53
0
-
1
0
0
1
0
0
≠
0
,
d
e
t
0
±
2
0
0
0
0
0
0
0
0
0
2
0
0
0
0
2
ε
63
0
ε
61
0
0
ε
61
0
0
0
0
2
ε
62
0
0
0
-
1
0
0
1
0
0
≠
0
.
The proof is completed. It can be found from (55a), (55b), (55c), (55d), (55e), and (55f) that codimension of h in E
→
z
,
λ is equal to the summation of the number sign functions and the number of auxiliary parameter.


## 3.5. The Transition Set

The bifurcation behaviors of (55a), (55b), (55c), (55d), (55e), and (55f) are discussed in the following analysis. The transition set Σ of the universal unfolding G
:
R
n
×
R
×
R
k
→
R
n is formulated as follows (see pages 409 in Golubitsky and Schaeffer I (1985)).The bifurcation variety set is(65a)
B
=
α
∈
R
k
:
∃
z
,
λ
s
u
c
h
t
h
a
t
G
z
,
λ
,
α
=
0
,
a
n
d
r
a
n
k
D
G
z
,
λ
,
α
≤
1
.The hysteresis variety set is(65b)
H
=
α
∈
R
k
:
∃
z
,
λ
,
v
s
u
c
h
t
h
a
t
G
z
,
λ
,
α
=
0
,
d
G
z
,
λ
,
α
·
v
=
0
a
n
d
d
2
G
z
,
λ
,
α
v
,
v
∈
r
a
n
g
e
d
G
z
,
λ
,
α
.The double limit point variety set is(65c)
D
=
α
∈
R
k
:
∃
z
1
,
z
2
,
λ
s
u
c
h
t
h
a
t
z
1
≠
z
2
,
G
z
i
,
λ
,
α
=
0
,
a
n
d
det
⁡
d
G
z
,
λ
,
α
=
0
,
i
=
1,2
.Then the transition set isΣ
=
B
∪
H
∪
D.The equations of bifurcation points for (55a) are obtained as(66a)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
=
0
,
(66b)
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
=
0
,
(66c)
±
a
1
ε
13
σ
2
=
ε
12
σ
1
a
2
=
ε
12
a
2
0
,
o
r
±
a
1
ε
13
σ
2
=
ε
12
σ
1
a
2
=
0
ε
13
a
1
,
o
r
±
a
1
ε
13
σ
2
=
ε
12
σ
1
a
2
=
ε
11
-
F
2
.Adding (66a) to (66b) obtains (67)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
+
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
=
0
.By the third equation of (66c), it can be obtained that(68)
±
a
1
=
ε
11
ε
13
σ
2
-
F
2
,and(69)
a
2
=
-
ε
12
σ
1
F
2
ε
11
.Substituting (68) and (69) into (67), bifurcation occurs when (70)
ε
11
σ
1
2
-
ε
11
F
2
4
+
2
-
2
σ
1
2
F
2
3
-
2
σ
2
2
F
2
±
ε
11
σ
2
2
=
0
.By the first and second equation of (66c), bifurcation occurs when(71)
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0
.The equations of hysteresis points for (55a) are obtained as follows:(72a)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
=
0
,
(72b)
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
=
0
,
(72c)
±
a
1
a
2
-
ε
12
ε
13
σ
1
σ
2
=
0
,
(72d)
v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G
,where v
=
v
1
,
v
2.Now we assumeσ
1
≠
0 and σ
2
≠
0. Then, it can be taken that(73)
v
=
±
a
1
,
ε
12
σ
1
,and it is noted that v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G if and only if (74)
v
1
2
,
v
2
2
·
ε
12
σ
1
,
a
2
=
0
.Equation (74) implies that (75)
ε
12
a
1
2
+
σ
1
a
2
=
0
.Multiplying (75) by a
1 and substituting it into (72c) we obtain(76)
a
1
=
∓
ε
13
σ
1
2
/
3
σ
2
1
/
3
.Using (72c) and (76), we obtain(77)
a
2
=
-
ε
12
σ
1
1
/
3
σ
2
2
/
3
.Substituting (76) and (77) into (67), hysteresis occurs when(78)
F
2
2
-
2
ε
11
F
2
+
σ
1
4
/
3
σ
2
2
/
3
+
σ
1
2
/
3
σ
2
4
/
3
=
0
,and(79)
F
2
2
-
2
ε
11
F
2
+
3
σ
1
4
/
3
σ
2
2
/
3
-
3
σ
1
2
/
3
σ
2
4
/
3
=
0
.The equations of the double limit points for (55a) are obtained as(80a)
±
a
1
2
+
2
ε
12
σ
1
a
2
+
2
ε
11
F
2
-
γ
=
0
,
(80b)
a
2
2
-
F
2
2
+
2
ε
13
σ
2
a
1
+
γ
=
0
,
(80c)
a
1
a
2
-
ε
12
ε
13
σ
1
σ
2
=
0
,
(80d)
a
11
,
a
21
≠
a
12
,
a
22
,
w
h
e
n
F
2
=
c
o
n
s
t
.Using (70), double limit points occur when(81)
σ
1
=
±
1
,
F
2
=
±
ε
11
2
a
n
d
σ
2
=
0
,
F
2
=
2
ε
11
.B, H, and D have equations as follows:(82a)
B
:
ε
11
σ
1
2
-
ε
11
F
2
4
+
2
-
2
σ
1
2
F
2
3
-
2
σ
2
2
F
2
±
ε
11
σ
2
2
=
0
,
(82b)
H
:
F
2
2
-
2
ε
11
F
2
+
σ
1
4
/
3
σ
2
2
/
3
+
σ
1
2
/
3
σ
2
4
/
3
=
0
;
F
2
2
-
2
ε
11
F
2
+
3
σ
1
4
/
3
σ
2
2
/
3
-
3
σ
1
2
/
3
σ
2
4
/
3
=
0
,
(82c)
D
:
σ
1
=
±
1
,
F
2
=
±
ε
11
2
;
σ
2
=
0
,
F
2
=
2
ε
11
.In the same way, transition setΣ of the universal unfolding (55b), (55c), (55d), (55e), and (55f) may be obtained.For (55c), a short calculation shows that bifurcation occurs when(83a)
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
=
0
,
(83b)
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
=
0
,
(83c)
±
a
1
ε
33
F
2
=
ε
32
σ
2
a
2
=
ε
31
ε
31
o
r
±
a
1
ε
33
F
2
=
ε
32
σ
2
a
2
=
ε
32
a
2
0
o
r
±
a
1
ε
33
F
2
=
ε
32
σ
2
a
2
=
0
-
F
2
+
ε
33
a
1
.Adding (83a) to (83b) obtains(84)
a
2
2
±
a
1
2
+
2
ε
32
σ
2
a
2
+
2
ε
33
F
2
a
1
-
F
2
2
+
2
ε
31
σ
1
=
0
.By the first equation of (83c) and (84), a short calculation shows that bifurcation occurs when(85)
σ
1
=
-
3
2
ε
31
σ
2
2
-
±
3
-
1
2
ε
31
F
2
2
.By the second and third equation of (83c), bifurcation occurs when(86)
a
1
,
a
2
,
σ
1
,
σ
2
,
F
2
,
γ
=
0,0
,
0,0
,
0,0
.The equations of hysteresis points for (55c) are(87a)
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
=
0
,
(87b)
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
=
0
,
(87c)
±
a
1
a
2
-
ε
32
ε
33
F
2
σ
2
=
0
,
(87d)
v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G
,where v
=
v
1
,
v
2.Now assumeσ
1
≠
0 and σ
2
≠
0, then it may be taken that(88)
v
=
±
a
1
,
ε
32
σ
2and note that v
1
2
,
v
2
2
∈
r
a
n
g
e
d
G if and only if(89)
v
1
2
,
v
2
2
·
ε
32
σ
2
,
a
2
=
0
.Equation (89) implies that (90)
ε
32
a
1
2
+
σ
2
a
2
=
0
,Next multiply (90) by a
1 and substitute (87c) to obtain(91)
a
1
=
∓
ε
33
F
2
1
/
3
σ
2
2
/
3
.Using (91) and (87c) obtains(92)
a
2
=
-
ε
32
F
2
2
/
3
σ
2
1
/
3
.Finally, substituting (91) and (92) into (84), hysteresis occurs when(93)
σ
1
=
1
2
ε
31
F
2
2
/
3
σ
2
4
/
3
+
F
2
2
/
3
σ
2
2
/
3
+
F
2
4
/
3
,and(94)
σ
1
=
1
2
ε
31
F
2
2
/
3
3
σ
2
4
/
3
-
3
F
2
2
/
3
σ
2
2
/
3
+
F
2
4
/
3
.The equations of double limit points for (55c) are(95a)
±
a
1
2
+
2
ε
32
σ
2
a
2
+
ε
31
σ
1
-
γ
=
0
,
(95b)
a
2
2
-
F
2
2
+
2
ε
33
F
2
a
1
+
ε
31
σ
1
+
γ
=
0
,
(95c)
±
a
1
a
2
-
ε
32
ε
33
F
2
σ
2
=
0
,
(95d)
a
11
,
a
21
≠
a
12
,
a
22
,
w
h
e
n
σ
1
=
c
o
n
s
t
.By (85), a more involved calculation implies that double limit points occur for (96a)
σ
2
=
0
,
σ
1
≤
0
a
n
d
F
2
=
0
,
σ
1
≤
0
w
h
e
n
ε
31
=
1
a
n
d
t
a
k
e
n
t
h
e
p
l
u
s
s
i
g
n
i
n
85
,
(96b)
σ
2
=
0
,
σ
1
≥
0
a
n
d
F
2
=
0
,
σ
1
≥
0
w
h
e
n
ε
31
=
-
1
a
n
d
t
a
k
e
n
t
h
e
p
l
u
s
s
i
g
n
i
n
85
,
(96c)
σ
2
=
0
,
σ
1
≥
0
a
n
d
F
2
=
0
,
σ
1
≤
0
w
h
e
n
ε
31
=
1
a
n
d
t
a
k
e
n
t
h
e
m
i
n
u
s
s
i
g
n
i
n
85
,
(96d)
σ
2
=
0
,
σ
1
≤
0
a
n
d
F
2
=
0
,
σ
1
≥
0
w
h
e
n
ε
31
=
-
1
a
n
d
t
a
k
e
n
t
h
e
m
i
n
u
s
s
i
g
n
i
n
85
.We defineρ, θ, and δ as follows:(97)
F
2
=
ρ
3
cos
3
⁡
θ
,
σ
2
=
ρ
3
sin
3
⁡
θ
,
σ
1
=
ρ
6
δ
.Thusρ
,
θ
,
δ are a singular modification of cylindrical coordinates in f
T
σ
2
σ
1
-space.In these coordinates, substituting (97) into (85), (93), and (94), B, H, and D have equations as follows: (98a)
B
:
δ
=
-
3
2
ε
31
sin
6
⁡
θ
-
±
3
-
1
2
ε
31
cos
6
⁡
θ
,
(98b)
H
:
δ
=
1
2
ε
31
cos
2
⁡
θ
sin
4
⁡
θ
+
cos
2
⁡
θ
sin
2
⁡
θ
+
cos
4
⁡
θ
;
δ
=
1
2
ε
31
cos
2
⁡
θ
3
sin
4
⁡
θ
-
3
cos
2
⁡
θ
sin
2
⁡
θ
+
cos
4
⁡
θ
,
(98c)
D
:
θ
=
0
,
π
2
,
π
,
3
π
2
;
δ
≤
0
;
ε
31
=
1
.
o
r
θ
=
0
,
π
2
,
π
,
3
π
2
;
δ
≥
0
;
ε
31
=
-
1From the above analysis, we deduce that (82a), (82b), (82c) and (98a), (98b), (98c) represent the relationship between two detuning parameters and a transverse excitation when the composite laminated piezoelectric rectangular plate structure subjected to small perturbations shows the bifurcation, hysteresis, and double limit points. If the disturbed system is equivalent to the original system, the space made up by these perturbations is the restricted tangent space. We know that the items σ
1
a
2 and σ
2
a
1 of complemented to h in (55a) belong to the restricted tangent space and the variation of parameters σ
1 and σ
2 make no difference in the bifurcation property of the original system, while the variation of parameter F
2 of complemented to h in (55a) has a qualitative influence on the bifurcation property of the original system. The similar analysis can be made in (55b), (55c), (55d), (55e), and (55f).


## 4. Numerical Results

The transition sets of the composite laminated piezoelectric rectangular plate structure with respect to small perturbations of the bifurcation parameter are depicted by (82a)–(82c) and (98a)–(98c) in Figures 2, 3, and 6, where B implies bifurcation set, H implies hysteresis set, and D implies double limit points. It is observed from Figures 2, 3, and 6 that the amplitudes a
1 has a solution in region (1), which means that frequency-response equation of the original system has only one solution and the composite laminated piezoelectric rectangular plate structure subjected to small perturbations is in stable state. In region (2), there are multiple solutions, corresponding to the multiple solutions band of frequency-response equations of the original system. The composite laminated piezoelectric rectangular plate structure is in an unstable state under small disturbance, and there is a jumping phenomenon. There is no solution to region (3). It means that there is no solution to the frequency-response equations of the original system. The composite laminated piezoelectric rectangular plate structure with small disturbance is in a stable state.Figure 2
Three-dimensional sketch of the transitions variety for (55a) when ε
11
=
1, in which taken the plus sign in (55a), B represents the bifurcation set; H represents the hysteresis set; D represents double limit points.Figure 3
Three-dimensional sketch of the transitions variety for (55a) when ε
11
=
1, in which taken the minus sign in (55a), B represents the bifurcation set; H represents the hysteresis set; D represents double limit points.By the changing of three bifurcation parameters, to understand bifurcation properties of the composite laminated piezoelectric rectangular plate structure subjected to small perturbations, Figure4 is the bifurcation diagrams of taken the plus sign in (55a) when ε
11
=
1. In Figure 4(a), the parameters are chosen as σ
1
=
0.1, σ
2
=
2.1, γ
=
10, in which there are three limit points. In Figure 4(b), the parameters are chosen as F
2
=
1, σ
2
=
0.2, γ
=
0, in which there is one limit point. In Figure 4(c), the parameters are chosen as F
2
=
0.1, σ
1
=
0.3, γ
=
1, in which there is one limit point. Figure 5 is the bifurcation diagrams of taken the minus sign in (55a) when ε
11
=
1. In Figure 5(a), the parameters are chosen as σ
1
=
0.1, σ
2
=
2.1, γ
=
10, in which there are two limit points. In Figure 5(b), the parameters are chosen as F
2
=
1, σ
2
=
0.2, γ
=
0, in which there is one limit point. In Figure 5(c), the parameters are chosen as F
2
=
0.1, σ
1
=
0.3, γ
=
1, in which there is one limit point. By comparing the numerical results of the disturbed system (55a) above, it is easy to find that the bifurcation properties of the composite laminated piezoelectric rectangular plate structure under small disturbances in region (2) is significantly affected by the positive and negative signs in (55a).Figure 4
The bifurcation diagrams of taken the plus sign in (55a) when ε
11
=
1. Solid line represents a
1; dash line represents a
2.
(a)
(b)
(c)Figure 5
The bifurcation diagrams of taken the minus sign in (55a) when ε
11
=
1. Solid line represents a
1; dash line represents a
2.
(a)
(b)
(c)Figure 6
The plan of the transition sets for taken the plus sign in (55c) when ε
31
=
-
1, in which taken the plus sign in (55c), B represents the bifurcation set; H represents the hysteresis set; D represents double limit points.Figure7 is the bifurcation diagrams of taken the plus sign in (55c) when ε
31
=
-
1. In Figure 7(a), the parameters are chosen as σ
1
=
1.1, σ
2
=
1, γ
=
2, in which there is one limit point. In Figure 7(b), the parameters are chosen as F
2
=
2, σ
2
=
1.1, γ
=
1, in which there is one limit point. In Figure 7(c), the parameters are chosen as F
2
=
1, σ
1
=
1, γ
=
1, in which there is no limit point.Figure 7
The bifurcation diagrams of taken the plus sign in (55c) when ε
31
=
-
1. Solid line represents a
1; dash line represents a
2.
(a)
(b)
(c)It is observed that jumping phenomenon and hilltop bifurcation occur at the critical point from region (1) to region (2), and only hilltop bifurcation occurs at the critical point from region (3) to region (2).Meanwhile, by choosing different bifurcation parameters, the changing of the width of the multiple solutions band is studied. For taken the plus sign in (55a) when ε
11
=
1, detuning parameter is chosen as the controlling parameter to study the changing of the width of the multiple solutions band. In Figure 8(a), the parameters are chosen as F
2
=
0.1, γ
=
0, σ
2
=
-
1. In Figure 8(b), the parameters are chosen as F
2
=
0.5, γ
=
0, σ
2
=
-
1. In Figure 8(c), the parameters are chosen as F
2
=
1, γ
=
0, σ
2
=
-
3. In Figure 8(d), the parameters are chosen as F
2
=
1, γ
=
1, σ
2
=
-
1. It is observed from Figure 8 that parameter γ has a great influence on the width of the multiple solutions band. The parameters F
2 and σ
2 are almost negligible for the width of the multiple solutions band when γ is fixed. In Figure 9(a), the parameters are chosen as F
2
=
0.5, γ
=
0, σ
1
=
-
3. In Figure 9(b), the parameters are chosen as F
2
=
0.5, γ
=
0, σ
1
=
-
1. In Figure 9(c), the parameters are chosen as F
2
=
0.3, γ
=
1, σ
1
=
-
1. In Figure 9(d), the parameters are chosen as F
2
=
0.5, γ
=
1, σ
1
=
-
1. It is observed from Figures 9(a) and 9(b) that parameter σ
1 has a great influence on the number of solutions when F
2 and γ are fixed. The parameter F
2 has a great influence on the width of the multiple solutions band when γ and σ
1 are fixed from Figures 9(c) and 9(d).Figure 8
Frequency-response curves of the first-order mode with the different parameters for (55a).
(a)
(b)
(c)
(d)Figure 9
Frequency-response curves of the second-order mode with the different parameters for (55a).
(a)
(b)
(c)
(d)Taking the plus sign in (55c) when ε
31
=
-
1, the detuning parameter σ
1 as the controlling parameter to study the changing of the width of the multiple solutions band. In Figure 10(a), the parameters are chosen as F
2
=
1, γ
=
0, σ
2
=
1. In Figure 10(b), the parameters are chosen as F
2
=
3, γ
=
1, σ
2
=
1. In Figure 10(c), the parameters are chosen as F
2
=
3, γ
=
1, σ
2
=
2. In Figure 10(d), the parameters are chosen as F
2
=
3, γ
=
2, σ
2
=
2. It is observed from Figure 10 that parameters F
2 and γ have a great influence on the number of solutions when σ
2 is fixed.Figure 10
Frequency-response curves of the first-order mode with the different parameters for (55c).
(a)
(b)
(c)
(d)


## 5. Conclusions

In this paper, the sign function was introduced, which leads to that the nondegenerate conditions of universal unfolding of the bifurcation equations of the composite laminated piezoelectric rectangular plate structure are satisfied. Then, we proved that bifurcation of the composite laminated piezoelectric rectangular plate structure in the case of 1:2 internal resonance is a high codimensional problem with codimension 4, and the universal unfolding of the bifurcation equations of the composite laminated piezoelectric rectangular plate structure was obtained.The transition set divided the parameter spaces into three regions. The results indicated that the composite laminated piezoelectric rectangular plate structure with respect to small perturbations of the bifurcation parameter was in a stable state in regions (1) and (3) of Figures2, 3, and 6. The composite laminated piezoelectric rectangular plate structure with respect to small perturbations of the bifurcation parameter was in an unstable state in regions (2) of Figures 2, 3, and 6, with jumping phenomenon or hilltop bifurcation. The results also showed that jumping phenomenon and hilltop bifurcation occurred at the critical point from region (1) to region (2), and only hilltop bifurcation occurred at the critical point from region (3) to region (2).For this composite laminated piezoelectric rectangular plate structure, it was demonstrated that there exist abundant dynamic bifurcation patterns in a small perturbation sense. Some new dynamic bifurcation phenomena were also presented. Clearly, these results provided some inspiration and guidance for the analysis and dynamic designs of this structure.



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*Source: 6830274-2019-02-20.xml*