The beam under investigation consists of an elastic core sandwiched between two piezoelectric face sheets polarized through their thickness, as shown in Figure 1 . Euler-Bernoulli beam theory is applied to the face sheets, which are assumed to resist membrane and bending stresses. Timoshenko beam theory is adopted for the core, which is assumed to also resist transverse shear stress. The piezoelectric layers are fully covered on their top and bottom surfaces with electrodes. The elastic and piezoelectric materials are considered orthotropic, with their orthotropy axes aligned with those of the sandwich beam. All layers are assumed to be perfectly bonded. The transverse normal stress is considered negligible compared to the other stress components [21] - [24] . The length, width, and thickness of the beam are denoted by L, H, and h, respectively. The subscripts a, s, and c refer to the quantities associated with the bottom and top face sheets and the core, respectively.

According to the classical laminate theory based on the aforementioned assumptions, the displacement fields are described in [1] [11] [13] [24] .

$\{\begin{array}{l}u\left(x,z,t\right)=u\left(x,t\right)-z{w}_{,x}\left(x,t\right)\\ v\left(x,z,t\right)=0\\ w\left(x,z,t\right)=w\left(x,t\right)\end{array}$ (1)

In the theory of beams undergoing large deformations, the Green-Lagrange strain tensor is considered without linearization and is simplified according to the Von Kármán assumptions.

$u$: longitudinal displacement;

$w$: transverse displacement;

$z$: coordinate along the beam thickness (thickness direction).

Thus, the strain in the x-direction is given by:

$\{\begin{array}{l}\epsilon ={\epsilon }_0-z{w}_{,xx}\\ {\epsilon }_0=u_{,x}+\frac{1}{2}{w}_{,x}^2\end{array}$ (2)

It is well known that in piezoelectric materials, the electric field and strain mutually influence each other. This property enables the use of piezoelectric materials as sensors and actuators for vibration control. More specifically, this relationship can be described by constitutive equations that characterize the coupling effects between mechanical and electrical properties [1] [24] .

$\{\begin{array}{l}\sigma =c\epsilon -e^{t}E\\ D=e\epsilon +ϵE\end{array}$ (3)

$\sigma$: Stress tensor;

$\epsilon$: Strain tensor;

$D$: Electric displacement vector;

$E$: Electric field vector;

$c$: Elasticity matrix;

$e$: Piezoelectric matrix;

$ϵ$: Dielectric permittivity.

The constitutive equations of piezoelectric materials can be written in the following expanded form:

$\left\{\begin{array}{l}\left\{\begin{array}{l}{\sigma }_1\\ {\sigma }_2\\ {\sigma }_3\\ {\sigma }_4\\ {\sigma }_5\\ {\sigma }_6\end{array}\right\}\\ \left\{\begin{array}{l}{D}_1\\ D_2\\ D_3\end{array}\right\}\end{array}\right\}=\left[\begin{array}{cc}\left[\begin{array}{cccccc}{c}_{11}& c_{12}& c_{13}& 0& 0& 0\\ c_{21}& c_{22}& c_{23}& 0& 0& 0\\ c_{31}& c_{32}& c_{33}& 0& 0& 0\\ 0& 0& 0& c_{44}& 0& 0\\ 0& 0& 0& 0& c_{55}& 0\\ 0& 0& 0& 0& 0& c_{66}\end{array}\right]& \left[\begin{array}{ccc}0& 0& -e_{13}\\ 0& 0& -e_{13}\\ 0& 0& -e_{13}\\ 0& -e_{13}& 0\\ -e_{13}& 0& 0\\ 0& 0& 0\end{array}\right]\\ \left[\begin{array}{cccccc}0& 0& 0& 0& e_{15}& 0\\ 0& 0& 0& e_{24}& 0& 0\\ e_{31}& e_{32}& e_{33}& 0& 0& 0\end{array}\right]& \left[\begin{array}{ccc}{ϵ}_{11}& 0& 0\\ 0& {ϵ}_{22}& 0\\ 0& 0& {ϵ}_{33}\end{array}\right]\end{array}\right]\left\{\begin{array}{l}\left\{\begin{array}{l}{\epsilon }_1\\ {\epsilon }_2\\ {\epsilon }_3\\ {\epsilon }_4\\ {\epsilon }_5\\ {\epsilon }_6\end{array}\right\}\\ \left\{\begin{array}{l}{E}_1\\ E_2\\ E_3\end{array}\right\}\end{array}\right\}$ (4)

Displacements are considered independent of y and zero along the y-direction; the stress tensor is uniaxial, and the directions of the vectors D and E are parallel to the z-axis. Thus, the reduced constitutive relations are given by:

$\{\begin{array}{l}{\epsilon }_2=0\\ {\sigma }_3=0\end{array}\Rightarrow {\epsilon }_3=\frac{1}{c_{33}}\left(e_{33}{E}_3-c_{13}\epsilon \right)$ (5)

It follows that:

$\left(\begin{array}{l}{\sigma }_1\\ D_3\end{array}\right)=\left[\begin{array}{cc}{c}_{11}^{*}& -e_{31}^{*}\\ e_{31}^{*}& {ϵ}_{33}^{*}\end{array}\right]\left(\begin{array}{l}\epsilon \\ E_3\end{array}\right)$ (6)

With: ${ϵ}_{33}^{*}={ϵ}_{33}+\frac{e_{33}^2}{c_{33}};e_{31}^{*}=e_{31}-\frac{c_{13}}{c_{33}}{e}_{33};c_{11}^{*}=c_{11}-\frac{c_{13}^2}{c_{33}}$.

Let us now consider an arbitrary piezoelectric layer, actuator, or sensor, placed between $z_{-},z_{+}$ ( $z_{-}\text{}<z_{+}$), with center $z_c=\left(z_{-}\text{}+z_{+}\right)/2$ and thickness h. The electrostatic equilibrium equation, assuming no volume charge density, is given by:

$\frac{\partial D_3}{\partial z}=0$ (7)

Using the boundary conditions, $D\left(z_{-}\right)=0$ or $D\left(z_{+}\right)=0$, we have:

$D_3\left(z\right)=0$ (8)

Thus, the electric field in the sensor, as a function of displacement, is given by:

$E_3\left(z\right)=-\frac{e_{31}^{*}}{{ϵ}_{33}^{*}}\epsilon =-\frac{e_{31}^{*}}{{ϵ}_{33}^{*}}\left(u_{,x}+\frac{1}{2}{w}_{,x}^2-z{w}_{,xx}\right)$ (9)

Since the electric field is derived from a potential, we have:

$E_3=-\frac{\partial \varphi }{\partial z}$ (10)

Consequently, the potential difference is given by:

$\Delta \varphi =\varphi \left(z_{+}\right)-\varphi \left(z_{-}\right)=-{\displaystyle {\int }_{z_{-}}^{z_{+}}{E}_3\left(z\right)\text{d}z}=\frac{e_{31}^{*}}{{\in }_{33}^{*}}{h}_i\left(u_{,x}+\frac{1}{2}{w}_{,x}^2-z_i{w}_{,xx}\right)$ (11)

With: $z_{-}=z_i-\frac{h_i}{2}$ and ${z_{+}=z_i+\frac{h_i}{2}|}_{i=a,s}$.

From Equations (9) and (11), we have:

$E_3\left(z\right)=-\frac{\Delta \varphi }{h_i}+\frac{e_{31}^{*}}{{ϵ}_{33}^{*}}\left(z-z_i\right)w_{,xx}$ (12)

The core of the beam is assumed to be conductive with a uniform potential set to zero. The sensor potential, denoted ${\varphi }_s\left(x\right)$, is then given by:

${\varphi }_s=\Delta {\varphi }_s=\frac{e_{31}^{*}}{{ϵ}_{33}^{*}}{h}_s\left(u_{,x}+\frac{1}{2}{w}_{,x}^2-z_s{w}_{,xx}\right)$ (13)

The actuator potential ${\varphi }_a\left(x\right)$ depends on the sensor output potential ${\varphi }_s\left(x\right)$ through a proportional-derivative control law described as follows:

${\varphi }_a=G_p{\varphi }_s+G_d{\stackrel{\dot{}}{\varphi }}_s$ (14)

In this formulation, G_p represents the proportional gain, which acts directly on the sensor displacement and thus modifies the effective stiffness perceived by the structure. By adjusting G_p, the stiffening or softening behavior of the system can be influenced, and the effective natural frequency can be shifted. G_d, on the other hand, is the derivative (or velocity) gain, which acts on the sensor velocity and contributes to the effective damping of the system, reducing vibration amplitudes and the hysteresis associated with nonlinear responses.

Using Equations (12) and (13), the electric fields in the sensor and actuator are, respectively, given by:

$E_3^s\left(z\right)=-\frac{{\varphi }_s}{h_s}+\frac{e_{31}^{*}}{{ϵ}_{33}^{*}}\left(z-z_s\right)w_{,xx}$ (15)

$E_3^a\left(z\right)=\frac{{\varphi }_a}{h_a}+\frac{e_{31}^{*}}{{ϵ}_{33}^{*}}\left(z-z_a\right)w_{,xx}$ (16)

The potentials ${\varphi }_a$ and ${\varphi }_s$ are independent of z and $z_s=\left(h_c+h_s\right)/2$ et $z_a=-\left(h_c+h_s\right)/2$.

The direct and inverse piezoelectric coefficients were taken into account in these formulations and will be involved in the dynamic behavior of the beam [1] [12] [25] [26] .

To determine the dynamic equation of the beam, we use the variational formulation, Hamilton’s principle, and Equations (5), (14), and (15). The beam is subjected to axial and transverse excitations F_x and F_z [1] [24] [27] [28] .

For the variational principle, we have:

$\begin{array}{c}{\displaystyle \underset{V}{\int }{\sigma }_1\delta \epsilon \text{d}v}={\displaystyle \underset{V_s}{\int }{\sigma }_1\delta \epsilon \text{d}{v}_s}+{\displaystyle \underset{V_c}{\int }{\sigma }_1\delta \epsilon \text{d}{v}_c}+{\displaystyle \underset{V_a}{\int }{\sigma }_1\delta \epsilon \text{d}{v}_a}\\ ={\displaystyle {\int }_0^L\left(N\delta {\epsilon }_0+M\delta w_{,xx}\right)\text{d}x}\\ ={\displaystyle {\int }_0^L\left(F_X\delta u+F_Z\delta w\right)\text{d}x}-{\left(\rho S\right)}_{*}{\displaystyle {\int }_0^L\left(\ddot{u}\delta u+\ddot{w}\delta w\right)\text{d}x}\end{array}$ (17)

With: $\{\begin{array}{l}{\left(\rho S\right)}_{*}={\rho }_s{S}_s+{\rho }_c{S}_c+{\rho }_a{S}_a\\ \delta {\epsilon }_0=\delta u_{,x}+w_{,x}\delta w_{,x}\end{array}$

According to Hamilton’s principle, we have:

$\{\begin{array}{l}{N}_{xx}={\displaystyle \underset{s}{\int }{\sigma }_{xx}\text{d}s}=ES\left[\frac{\partial u}{\partial x}+\frac{1}{2}{\left(\frac{\partial w}{\partial x}\right)}^2\right]\\ M_{xx}={\displaystyle \underset{s}{\int }z{\sigma }_{xx}\text{d}s}=-EI\frac{{\partial }^{2}w}{\partial x^2}\end{array}$ (18)

If we integrate over the entire thickness and width, and assume that the piezoelectric layers are symmetrical ( $h_a=h_s$), the axial force N and the bending moment M are determined from the previous equation as follows:

$N={\sigma }_1^a{S}_a+{\sigma }_1^c{S}_c+{\sigma }_1^s{S}_s=\left({\sigma }_1^a+{\sigma }_1^s\right)S_s+E_c{S}_c\epsilon$ (19)

Using Equation (5), we obtain:

${\sigma }_1^a+{\sigma }_1^s=2c_{11}^{*}\epsilon -e_{31}^{*}\left(E_3^a+E_3^s\right)$ (20)

$E_3^a+E_3^s=\frac{{\varphi }_a-{\varphi }_s}{h_s}$ (21)

Using Equations (2), (13)-(16), we obtain:

$\begin{array}{l}N=\left(2c_{11}^{*}{S}_s+E_c{S}_c+\left(1-G_p\right)S_s\frac{{\left(e_{31}^{*}\right)}^2}{{ϵ}_{33}^{*}}\right){\epsilon }_0-\left(1-G_p\right)S_s\frac{{\left(e_{31}^{*}\right)}^2}{{ϵ}_{33}^{*}}{z}_s{w}_{,xx}\\ -G_d{S}_s\frac{{\left(e_{31}^{*}\right)}^2}{{ϵ}_{33}^{*}}\left({\stackrel{\dot{}}{u}}_{,x}+w_{,x}{\stackrel{\dot{}}{w}}_x-z_s{w}_{,xx}\right)\end{array}$ (22)

We also obtain the moment in the same manner; thus, we have:

$\{\begin{array}{l}N={\left(ES\right)}_{*}{\epsilon }_0-B_N{w}_{,xx}-{\left(ES\right)}_{pe}{G}_d\left({\stackrel{\dot{}}{u}}_{,x}+w_{,x}{\stackrel{\dot{}}{w}}_{,x}-{\stackrel{\dot{}}{w}}_{,xx}{z}_s\right)\\ M=-B_M{\epsilon }_0+{\left(EI\right)}_{*}{w}_{,xx}-{\left(ES\right)}_{pe}{z}_s{G}_d\left({\stackrel{\dot{}}{u}}_{,x}+w_{,x}{\stackrel{\dot{}}{w}}_{,x}-{\stackrel{\dot{}}{w}}_{,xx}{z}_s\right)\end{array}$ (23)

With: $\{\begin{array}{l}{\left(ES\right)}_{*}=E_c{S}_c+2c_{11}^{*}{S}_s+{\left(ES\right)}_{pe}\left(1-G_p\right);\\ {\left(ES\right)}_{pe}=S_s\frac{{\left(e_{31}^{*}\right)}^2}{{ϵ}_{33}^{*}};\\ B_N={\left(ES\right)}_{pe}\left(1-G_p\right)z_s;\\ B_M={\left(ES\right)}_{pe}\left(1+G_p\right)z_s;\\ {\left(EI\right)}_{*}=E_c{I}_c+2c_{11}^{*}\left(I_s+S_s{z}_s^2\right)+\frac{{\left(ES\right)}_{pe}}{S_s}\left(2I_s+\left(1+G_p\right)z_s^2{S}_s\right).\end{array}$

By applying the variational principle to the displacements u and w, and integrating by parts once for the terms in $\delta u_{,x}$ and $\delta w_{,x}$, and twice for the terms in $\delta w_{,xx}$, we obtain the following partial differential equations [1] [12] - [15] .

$\{\begin{array}{l}-N_{,x}+{\left(\rho S\right)}_{*}\ddot{u}=F_X\\ M_{,xx}-{\left(N{w}_{,x}\right)}_{,x}+{\left(\rho S\right)}_{*}\ddot{w}=F_Z\end{array}$ (24)

Assuming that the axial force and the axial displacement inertia are negligible, system (24) becomes:

$\{\begin{array}{l}-N_{,x}=0\\ M_{,xx}-{\left(N{w}_{,x}\right)}_{,x}+{\left(\rho S\right)}_{*}\ddot{w}=F_Z\end{array}$ (25)

The axial force depends only on time $\left(N\left(x,t\right)=N\left(t\right)\right)$, so system (25) becomes:

$\{\begin{array}{l}N=\frac{1}{2}{\left(ES\right)}_{*}{w}_{,x}^2-B_N{w}_{,xx}-{\left(ES\right)}_{pe}{G}_d\left(w_{,x}{\stackrel{\dot{}}{w}}_{,x}-{\stackrel{\dot{}}{w}}_{,xx}{z}_s\right)\\ M=\frac{1}{2}{B}_M{w}_{,x}^2+{\left(EI\right)}_{*}{w}_{,xx}-{\left(ES\right)}_{pe}{G}_d{z}_s\left(w_{,x}{\stackrel{\dot{}}{w}}_{,x}-{\stackrel{\dot{}}{w}}_{,xx}{z}_s\right)\end{array}$ (26)

En intégrant la première équation du système (26) entre (O et L), on obtient:

$N\left(t\right)=\frac{1}{2L}{\left(ES\right)}_{*}{\displaystyle {\int }_0^L{w}_{,x}^2\text{d}x}-\frac{B_N}{L}{\displaystyle {\int }_0^L{w}_{,xx}\text{d}x}-\frac{{\left(ES\right)}_{pe}}{L}{G}_d{\displaystyle {\int }_0^L\left(w_{,x}{\stackrel{\dot{}}{w}}_{,x}-{\stackrel{\dot{}}{w}}_{,xx}{z}_s\right)\text{d}x}$ (27)

By substituting (27) into the second equation of system (26), the following dynamic equation is obtained:

$\begin{array}{l}{\left(\rho S\right)}_{*}\ddot{w}+{\left(EI\right)}_{*}{w}_{,xxxx}-N\left(t\right)w_{,xx}-B_M\left(w_{,xx}^2+w_{,x}{w}_{,xxx}\right)\\ -{\left(\rho S\right)}_{pe}{G}_d{z}_s\left({\stackrel{\dot{}}{w}}_{,x}{w}_{,xxx}+2{\stackrel{\dot{}}{w}}_{,xx}{w}_{,xx}+w_{,x}{\stackrel{\dot{}}{w}}_{,xxx}-z_s{\stackrel{\dot{}}{w}}_{,xxxx}\right)=F_Z\end{array}$ (28)

This differential equation describes the transverse dynamic behavior of the piezoelectric-elastic-piezoelectric beam subjected to a transverse excitation and active control based on the feedback control law when the axial force and axial displacement effects are neglected. The free and forced nonlinear vibrations, as well as the active control of the beam, can be analyzed by solving Equation (25) or (28) [12] .

In this work, the axial effects are neglected; therefore, only Equation (28), which describes the dynamic behavior of the beam without axial effects, is solved.

According to the principle of the method of averaging, Equation (30) can be rewritten by introducing the perturbation parameter; thus, we have:

$\ddot{q}+{\omega }_L^{2}q=-\epsilon \left[2\mu \stackrel{\dot{}}{q}+{\alpha }_2{q}^2+{\alpha }_3{q}^3+{\alpha }_{4}q\stackrel{\dot{}}{q}+{\alpha }_5{q}^2\stackrel{\dot{}}{q}-F_1\cos \left(\omega t\right)\right]$ (32)

If $\epsilon =0$, the general solution of Equation (31) is given by:

$q=a\cos \left({\omega }_{L}t+\beta \right)$ (33)

Since $a$ and $\beta$ are constants, the derivative of Equation (32) is:

$\stackrel{\dot{}}{q}=-a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)$ (34)

If $\epsilon \ne 0$, the solution of Equation (32) takes the form of Equation (34), but with $a$ and $\beta$ now varying with time. Differentiating Equation (33) then yields:

$\stackrel{\dot{}}{q}=-a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)+\stackrel{\dot{}}{a}\cos \left({\omega }_{L}t+\beta \right)-a\stackrel{\dot{}}{\beta }\sin \left({\omega }_{L}t+\beta \right)$ (35)

By comparing Equations (34) and (35), we obtain:

$\stackrel{\dot{}}{a}\cos \left({\omega }_{L}t+\beta \right)-a\stackrel{\dot{}}{\beta }\sin \left({\omega }_{L}t+\beta \right)=0$ (36)

Let us differentiate Equation (33) with respect to time:

$\ddot{q}=-\stackrel{\dot{}}{a}{\omega }_L\sin \left({\omega }_{L}t+\beta \right)-a{\omega }_L^2\cos \left({\omega }_{L}t+\beta \right)-a\stackrel{\dot{}}{\beta }{\omega }_L\cos \left({\omega }_{L}t+\beta \right)$ (37)

By substituting (34), (35), and (37) into (32), we obtain:

$\begin{array}{l}\stackrel{\dot{}}{a}{\omega }_L\sin \left({\omega }_{L}t+\beta \right)+a\stackrel{\dot{}}{\beta }{\omega }_L\cos \left({\omega }_{L}t+\beta \right)\\ =-2\mu \epsilon \left(a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)\right)+{\alpha }_2\epsilon \left(a^2{\cos }^2\left({\omega }_{L}t+\beta \right)\right)\\ +{\alpha }_3\epsilon \left(a^3{\cos }^3\left({\omega }_{L}t+\beta \right)\right)-{\alpha }_4\epsilon \left(a^2{\omega }_L\cos \left({\omega }_{L}t+\beta \right)\sin \left({\omega }_{L}t+\beta \right)\right)\\ -{\alpha }_5\epsilon \left(a^3{\omega }_L{\cos }^2\left({\omega }_{L}t+\beta \right)\sin \left({\omega }_{L}t+\beta \right)\right)-\epsilon F_1\cos \left(\omega t\right)\end{array}$ (38)

By using (36) and (38), we have:

$\begin{array}{c}\stackrel{\dot{}}{a}=\frac{\epsilon }{{\omega }_L}\{-\mu a{\omega }_L+\mu a{\omega }_L\cos \left(2{\omega }_{L}t+2\beta \right)+{\alpha }_2\left(\frac{1}{4}{a}^2\sin \left({\omega }_{L}t+\beta \right)+\frac{1}{4}{a}^2\sin \left(3{\omega }_{L}t+3\beta \right)\right)\\ +{\alpha }_3\left(\frac{1}{4}{a}^3\sin \left(2{\omega }_{L}t+2\beta \right)+\frac{1}{8}{a}^3\sin \left(4{\omega }_{L}t+4\beta \right)\right)-\frac{1}{2}{F}_1\sin \left(\left({\omega }_L-\omega \right)t+\beta \right)\\ -{\alpha }_4\left(\frac{1}{4}{a}^2{\omega }_L\cos \left({\omega }_{L}t+\beta \right)-\frac{1}{4}{a}^2{\omega }_L\cos \left(3{\omega }_{L}t+3\beta \right)\right)-\frac{1}{2}{F}_1\sin \left(\left({\omega }_L+\omega \right)t+\beta \right)\\ -{\alpha }_5\left(\frac{1}{8}{a}^3{\omega }_L-\frac{1}{8}{a}^3{\omega }_L\cos \left(4{\omega }_{L}t+4\beta \right)\right)\}\end{array}$ (39)

By substituting (39) into (36), we obtain:

$\begin{array}{c}a\stackrel{\dot{}}{\beta }=\frac{\epsilon }{{\omega }_L}\{-\mu a{\omega }_L\sin \left(2{\omega }_{L}t+2\beta \right)+{\alpha }_2\left(\frac{3a^2}{4}\cos \left({\omega }_{L}t+\beta \right)+\frac{a^2}{4}\cos \left(3{\omega }_{L}t+3\beta \right)\right)\\ +{\alpha }_3\left(\frac{3a^3}{8}+\frac{a^3}{2}\cos \left(2{\omega }_{L}t+2\beta \right)+\frac{a^3}{8}\cos \left(4{\omega }_{L}t+4\beta \right)\right)\\ -{\alpha }_4\left(\frac{a^2}{4}{\omega }_L\sin \left({\omega }_{L}t+\beta \right)+\frac{a^2}{4}{\omega }_L\sin \left(3{\omega }_{L}t+3\beta \right)\right)-{\frac{F_1}{2}}_1\cos \left(\left({\omega }_L-\omega \right)t+\beta \right)\\ -{\alpha }_5\left(\frac{a^3}{4}{\omega }_L\sin \left(2{\omega }_{L}t+2\beta \right)+\frac{a^3}{8}{\omega }_L\sin \left(4{\omega }_{L}t+4\beta \right)\right)-\frac{F_1}{2}\cos \left(\left({\omega }_L+\omega \right)t+\beta \right)\}\end{array}$ (40)

In primary resonance, $\omega \approx {\omega }_L$ and the expressions in $\sin \left(\left({\omega }_L-\omega \right)t+\beta \right)$ and $\cos \left(\left({\omega }_L-\omega \right)t+\beta \right)$ vary slowly with respect to time in Equations (39) and (40), respectively. We then have:

$\{\begin{array}{l}\stackrel{\dot{}}{a}=-\mu a\epsilon -{\alpha }_5\epsilon \frac{a^3}{8}-\frac{\epsilon F_1}{2{\omega }_L}\sin \left(\left({\omega }_L-\omega \right)t+\beta \right)\\ a\stackrel{\dot{}}{\beta }={\alpha }_3\epsilon \frac{3a^3}{8{\omega }_L}-\frac{\epsilon F_1}{2{\omega }_L}\cos \left(\left({\omega }_L-\omega \right)t+\beta \right)\end{array}$ (41)

By setting $\gamma =\left({\omega }_L-\omega \right)t+\beta$ and $\omega ={\omega }_L+\epsilon \sigma$, the system (41) becomes:

$\{\begin{array}{l}\stackrel{\dot{}}{a}=-\mu a\epsilon -{\alpha }_5\epsilon \frac{a^3}{8}-\frac{\epsilon F_1}{2{\omega }_L}\sin \left(\gamma \right)\\ a\stackrel{\dot{}}{\gamma }+\epsilon a\sigma ={\alpha }_3\epsilon \frac{3a^3}{8{\omega }_L}-\frac{\epsilon F_1}{2{\omega }_L}\cos \left(\gamma \right)\end{array}$ (42)

Initially, $a$ and $\gamma$ oscillate, and as time increases, they become constants. Thus, for $\stackrel{\dot{}}{a}=0$ and $\stackrel{\dot{}}{\gamma }=0$, we have:

$\{\begin{array}{l}\frac{\mu }{{\omega }_L}+\frac{{\alpha }_5{a}^2}{8{\omega }_L}=\frac{-F_1}{2a{\omega }_L^2}\sin \left(\gamma \right)\\ \sigma -3\frac{{\alpha }_3{a}^2}{8{\omega }_L}=\frac{-F_1}{2a{\omega }_L^2}\cos \left(\gamma \right)\end{array}$ (43)

From system (43), the following equation is obtained:

${\left[\frac{\omega }{{\omega }_L}-\left(1+3\frac{{\alpha }_3{a}^2}{8{\omega }_L^2}\right)\right]}^2+{\left[\frac{\mu }{{\omega }_L}+\frac{{\alpha }_5{a}^2}{8{\omega }_L}\right]}^2={\left(\frac{F_1}{2a{\omega }_L^2}\right)}^2$ (44)

In the case of secondary resonance, Equation (31) can be rewritten by introducing the perturbation parameter in the following form:

$\ddot{q}+{\omega }_L^{2}q=-\epsilon \left[2\mu \stackrel{\dot{}}{q}+{\alpha }_2{q}^2+{\alpha }_3{q}^3+{\alpha }_{4}q\stackrel{\dot{}}{q}+{\alpha }_5{q}^2\stackrel{\dot{}}{q}\right]+F_1\cos \left(\omega t\right)$ (45)

If $\epsilon =0$ and using the principle of superposition, the general solution of Equation (45) is given by:

$q=a\cos \left({\omega }_{L}t+\beta \right)+2\Lambda \cos \left(\omega t\right)$ (46)

$2\Lambda =\frac{F_1}{{\omega }_L^2-{\omega }^2}$ (47)

The derivative of (46) is given as follows:

$\stackrel{\dot{}}{q}=-a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)-2\Lambda \omega \sin \left(\omega t\right)$ (48)

If $\epsilon \ne 0$, as in primary resonance, the derivative of (47), using the variation of constants, is given by:

$\stackrel{\dot{}}{q}=-a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)+\stackrel{\dot{}}{a}\cos \left({\omega }_{L}t+\beta \right)-a\stackrel{\dot{}}{\beta }\sin \left({\omega }_{L}t+\beta \right)-2\Lambda \omega \sin \left(\omega t\right)$ (49)

By comparing (48) and (49), we obtain:

$\stackrel{\dot{}}{a}\cos \left({\omega }_{L}t+\beta \right)-a\stackrel{\dot{}}{\beta }\sin \left({\omega }_{L}t+\beta \right)=0$ (50)

The derivative of (47) is given by:

$\ddot{q}=-a{\omega }_L^2\cos \left({\omega }_{L}t+\beta \right)-\stackrel{\dot{}}{a}{\omega }_L\sin \left({\omega }_{L}t+\beta \right)-a\stackrel{\dot{}}{\beta }{\omega }_L\cos \left({\omega }_{L}t+\beta \right)-2\Lambda {\omega }^2\cos \left(\omega t\right)$(51)

By substituting (46), (48), and (30) into (45), we obtain:

$\begin{array}{l}\stackrel{\dot{}}{a}{\omega }_L\sin \left({\omega }_{L}t+\beta \right)+a\stackrel{\dot{}}{\beta }\cos \left({\omega }_{L}t+\beta \right)=-2\mu \epsilon \left[a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)+2\Lambda \omega \sin \left(\omega t\right)\right]\\ +{\alpha }_2\epsilon {\left[a\cos \left({\omega }_{L}t+\beta \right)+2\Lambda \cos \left(\omega t\right)\right]}^2+{\alpha }_3\epsilon {\left[a\cos \left({\omega }_{L}t+\beta \right)+2\Lambda \cos \left(\omega t\right)\right]}^3\\ +{\alpha }_4\epsilon \left[\left(a\cos \left({\omega }_{L}t+\beta \right)+2\Lambda \cos \left(\omega t\right)\right)\left(-a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)-2\Lambda \omega \sin \left(\omega t\right)\right)\right]\\ +{\alpha }_5\epsilon \left[{\left(a\cos \left({\omega }_{L}t+\beta \right)+2\Lambda \cos \left(\omega t\right)\right)}^2\left(-a{\omega }_L\sin \left({\omega }_{L}t+\beta \right)-2\Lambda \omega \sin \left(\omega t\right)\right)\right]\end{array}$(52)

Using (50) and (52), we obtain:

$\begin{array}{c}\stackrel{\dot{}}{a}=\frac{\epsilon }{{\omega }_L}\left\{-\mu a{\omega }_L\left(1-\cos \left(2{\omega }_{L}t+2\beta \right)\right)+2\mu \Lambda \omega \cos \left(\left({\omega }_L+\omega \right)t+\beta \right)-2\mu \Lambda \omega \cos \left(\left({\omega }_L-\omega \right)t+\beta \right)\right\}\\ +\frac{{\alpha }_2\epsilon }{{\omega }_L}\{\left(2{\Lambda }^2+\frac{a^2}{4}\right)\sin \left({\omega }_{L}t+\beta \right)+\frac{a^2}{4}\sin \left(3{\omega }_{L}t+3\beta \right)+{\Lambda }^2\sin \left(\left({\omega }_L+2\omega \right)t+\beta \right)\\ \stackrel{}{}+{\Lambda }^2\sin \left(\left({\omega }_L-2\omega \right)t+\beta \right)+a\Lambda \sin \left(\left(2{\omega }_L+\omega \right)t+2\beta \right)+a\Lambda \sin \left(\left(2{\omega }_L-\omega \right)t+2\beta \right)\}\\ +\frac{{\alpha }_3\epsilon }{{\omega }_L}\{\left(3a{\Lambda }^2+\frac{a^3}{4}\right)\sin \left(2{\omega }_{L}t+2\beta \right)+\frac{a^3}{8}\sin \left(4{\omega }_{L}t+4\beta \right)\\ +\left(3{\Lambda }^3+\frac{3a^2\Lambda }{4}\right)\sin \left(\left({\omega }_L+\omega \right)t+\beta \right)+\left(3{\Lambda }^3+\frac{3a^2\Lambda }{4}\right)\sin \left(\left({\omega }_L-\omega \right)t+\beta \right)\\ +\frac{3a^2\Lambda }{2}\sin \left(\left(2{\omega }_L+2\omega \right)t+2\beta \right)+\frac{3a^2\Lambda }{2}\sin \left(\left(2{\omega }_L-2\omega \right)t+2\beta \right)+{\Lambda }^3\sin \left(\left({\omega }_L+3\omega \right)t+\beta \right)\\ +\frac{3a^2\Lambda }{4}\sin \left(\left(3{\omega }_L+\omega \right)t+3\beta \right)+\frac{3a^2\Lambda }{4}\sin \left(\left(3{\omega }_L-\omega \right)t+3\beta \right)+{\Lambda }^3\sin \left(\left({\omega }_L-3\omega \right)t+\beta \right)\}\\ +\frac{{\alpha }_4\epsilon }{{\omega }_L}\{-\frac{a^2{\omega }_L}{4}\cos \left({\omega }_{L}t+\beta \right)+\frac{a^2{\omega }_L}{4}\cos \left(3{\omega }_{L}t+3\beta \right)-\Lambda a{\omega }_L\cos \left(\omega t\right)\\ +\frac{a\Lambda }{2}\left({\omega }_L+\omega \right)\cos \left(\left(2{\omega }_L+\omega \right)t+2\beta \right)+\frac{a\Lambda }{2}\left({\omega }_L-\omega \right)\cos \left(\left(2{\omega }_L-\omega \right)t+2\beta \right)\end{array}$

$\begin{array}{c}\stackrel{}{}+\omega {\Lambda }^2\cos \left(\left(2{\omega }_L+\omega \right)t+2\beta \right)-\omega {\Lambda }^2\cos \left(\left(2{\omega }_L-\omega \right)t+2\beta \right)\}\\ +\frac{{\alpha }_5\epsilon }{{\omega }_L}\{-\left(a{\omega }_L{\Lambda }^2+\frac{a^3{\omega }_L}{8}\right)-a{\omega }_L{\Lambda }^2\left(\cos \left(2\omega t\right)+\cos \left(2{\omega }_{L}t+2\beta \right)\right)+\frac{a^3{\omega }_L}{8}\cos \left(4{\omega }_{L}t+4\beta \right)\\ +\left(\omega {\Lambda }^3-\frac{a^2{\omega }_L\Lambda }{2}+\frac{a^2\omega \Lambda }{4}\right)\cos \left(\left({\omega }_L+\omega \right)t+\beta \right)-\left(\omega {\Lambda }^3+\frac{a^2{\omega }_L\Lambda }{2}+\frac{a^2\omega \Lambda }{4}\right)\cos \left(\left({\omega }_L-\omega \right)t+\beta \right)\\ +\left(\frac{a^2{\omega }_L\Lambda }{2}+\frac{a^2\omega \Lambda }{4}\right)\cos \left(\left(3{\omega }_L+\omega \right)t+3\beta \right)+\left(\frac{a^2{\omega }_L\Lambda }{2}-\frac{a^2\omega \Lambda }{4}\right)\cos \left(\left(3{\omega }_L-\omega \right)t+3\beta \right)\\ +\left(a\omega {\Lambda }^2+\frac{a{\omega }_L{\Lambda }^2}{2}\right)\cos \left(\left(2{\omega }_L+2\omega \right)t+2\beta \right)+\left(-a\omega {\Lambda }^2+\frac{a{\omega }_L{\Lambda }^2}{2}\right)\cos \left(\left(2{\omega }_L-2\omega \right)t+2\beta \right)\\ \stackrel{}{}+\omega {\Lambda }^3\cos \left(\left({\omega }_L+3\omega \right)t+\beta \right)-\omega {\Lambda }^3\cos \left(\left({\omega }_L-3\omega \right)t+\beta \right)\}\end{array}$ (53)

By substituting (53) into (50), we obtain:

(54)

In secondary resonance, there are two cases: superharmonic resonance and subharmonic resonance.

In this case, we have $\omega \approx \frac{1}{3}{\omega }_L$ et $\omega \approx \frac{1}{2}{\omega }_L$.

First case: $\omega \approx \frac{1}{3}{\omega }_L,\Lambda =\frac{9F_1}{{\omega }_L^2}$

The expressions for $\sin \left(\left({\omega }_L-3\omega \right)t+\beta \right)$ and $\cos \left(\left({\omega }_L-3\omega \right)t+\beta \right)$ vary slowly with time. Equations (53) and (54) then become:

$\{\begin{array}{l}\stackrel{\dot{}}{a}+\mu a\epsilon +\frac{{\alpha }_5\epsilon a^3}{8}+{\alpha }_5\epsilon a{\Lambda }^2=\frac{{\alpha }_3\epsilon {\Lambda }^3}{{\omega }_L}\sin \left(\left({\omega }_L-3\omega \right)t+\beta \right)-\frac{{\alpha }_5\epsilon {\Lambda }^3}{3{\omega }_L}\cos \left(\left({\omega }_L-3\omega \right)t+\beta \right)\\ a\stackrel{\dot{}}{\beta }-\frac{3{\alpha }_3\epsilon a^3}{8{\omega }_L}-\frac{3{\alpha }_3\epsilon a{\Lambda }^2}{{\omega }_L}=\frac{{\alpha }_3\epsilon {\Lambda }^3}{{\omega }_L}\cos \left(\left({\omega }_L-3\omega \right)t+\beta \right)+\frac{{\alpha }_5\epsilon {\Lambda }^3}{3{\omega }_L}\sin \left(\left({\omega }_L-3\omega \right)t+\beta \right)\end{array}$ (55)

By setting $\gamma =\left({\omega }_L-3\omega \right)t+\beta$ and $3\omega ={\omega }_L+\epsilon \sigma$, the system (55) becomes:

$\{\begin{array}{l}\stackrel{\dot{}}{a}+\mu a\epsilon +\frac{{\alpha }_5\epsilon a^3}{8}+{\alpha }_5\epsilon a{\Lambda }^2=\frac{{\alpha }_3\epsilon {\Lambda }^3}{{\omega }_L}\sin \left(\gamma \right)-\frac{{\alpha }_5\epsilon {\Lambda }^3}{3{\omega }_L}\cos \left(\gamma \right)\\ a\stackrel{\dot{}}{\gamma }+\epsilon a\sigma -\frac{3{\alpha }_3\epsilon a^3}{8{\omega }_L}-\frac{3{\alpha }_3\epsilon a{\Lambda }^2}{{\omega }_L}=\frac{{\alpha }_3\epsilon {\Lambda }^3}{{\omega }_L}\cos \left(\gamma \right)+\frac{{\alpha }_5\epsilon {\Lambda }^3}{3{\omega }_L}\sin \left(\gamma \right)\end{array}$ (56)

If the system tends toward a steady state, $\stackrel{\dot{}}{a}=0$ and $\stackrel{\dot{}}{\gamma }=0$, we have:

$\{\begin{array}{l}\mu +\frac{{\alpha }_5{a}^2}{8}+{\alpha }_5{\Lambda }^2=\frac{{\alpha }_3{\Lambda }^3}{a{\omega }_L}\sin \left(\gamma \right)-\frac{{\alpha }_5{\Lambda }^3}{3a{\omega }_L}\cos \left(\gamma \right)\\ \sigma -\frac{3{\alpha }_3{a}^2}{8{\omega }_L}-\frac{3{\alpha }_3{\Lambda }^2}{{\omega }_L}=\frac{{\alpha }_3{\Lambda }^3}{a{\omega }_L}\cos \left(\gamma \right)+\frac{{\alpha }_5{\Lambda }^3}{3a{\omega }_L}\sin \left(\gamma \right)\end{array}$ (57)

From system (57), we obtain:

${\left[\frac{\omega }{{\omega }_L}-\left(\frac{1}{3}+\frac{{\alpha }_3{a}^2}{8{\omega }_L^2}+\frac{{\alpha }_3{\Lambda }^2}{{\omega }_L^2}\right)\right]}^2+{\left[\frac{\mu }{3{\omega }_L}+\frac{{\alpha }_5{a}^2}{24{\omega }_L}+\frac{{\alpha }_5{\Lambda }^2}{3{\omega }_L}\right]}^2={\left(\frac{{\alpha }_3{\Lambda }^3}{3a{\omega }_L^2}\right)}^2+{\left(\frac{{\alpha }_5{\Lambda }^3}{9a{\omega }_L}\right)}^2$(58)