# A New Approach for Studying Nonlinear Dynamic Response of a Thin Plate with Internal Resonance in a Fractional Viscoelastic Medium

Authors: Yury A. Rossikhin; Marina V. Shitikova; Jean Claude Ngenzi
Journal: Shock and Vibration
 (2015)

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## Abstract

In the previous analysis, the dynamic behaviour of a nonlinear plate embedded into a fractional derivative viscoelastic medium has been studied by the method of multiple time scales under the conditions of the internal resonances two-to-one and one-to-one, as well as the internal combinational resonances for the case when the linear parts of nonlinear equations of motion occur to be coupled. A new approach proposed in this paper allows one to uncouple the linear parts of equations of motion of the plate, while the same method, the method of multiple time scales, has been utilized for solving nonlinear equations. The influence of viscosity on the energy exchange mechanism between interacting nonlinear modes has been analyzed. It has been shown that for some internal resonances there exist such particular cases when it is possible to obtain two first integrals, namely, the energy integral and the stream function, which allows one to reduce the problem to the calculation of elliptic integrals. The new approach enables one to solve the problems of vibrations of thin bodies more efficiently.


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## 1. Introduction

It is well known that the nonlinear vibrations of plates are an important area of applied mechanics, since plates are used as structural elements in many fields of industry and technology. Different methods such as analytical [1–4], numerical [5, 6], and experimental methods [5, 7] could be employed to investigate the nonlinear vibrations of plates. Extensive review of recent research developments in the field could be found in Amabili [8, 9] and Sathyamoorthy [10].The study of free undamped [6, 9], as well as damped [2, 3, 11–13], nonlinear systems is important because the dynamics characteristics of the system—defined by the amplitude-frequency relations and modes of vibrations—are determined [6]. Moreover, nonlinear vibrations could be accompanied by such a phenomenon as the internal resonance, resulting in multimode response with a strong interaction of the modes involved [14] accompanied by the energy exchange phenomenon. The internal resonance could be found within some combination of natural frequencies of one and the same type of vibrations. As an example, a 1:3 internal resonance was discovered in [6], when the fourth natural frequency of the out-of-plane vibrations was three times larger than the fundamental frequency of the out-of-plane vibrations of the plate. A 1:1 internal resonance was investigated in [11, 15] applying a two-mode Galerkin procedure for the plate deflection in their analysis.Another type of the internal resonance was investigated in [2, 3], when one frequency of the in-plane vibrations was equal to (a 1:1 internal resonance) or twice as large as (a 1:2 internal resonance) some frequency of the out-of-plane vibrations. The combinational resonances of the additive and difference type were also investigated in [3]. These types of the internal resonance result in the energy exchange between two or three subsystems. Investigations in this field were initiated by Witt and Gorelik [16], who were among the first to show theoretically and experimentally the two-to-one internal resonance with the energy exchange from one subsystem to another using the simplest two-degree-of-freedom mechanical system as an example, and the interest of researchers in the problems of the internal resonance in mechanical systems does not relax.In the given paper, nonlinear free vibrations of a thin plate described by three coupled nonlinear differential equations are considered when the plate is being under the conditions of the internal resonance resulting in the interaction of modes corresponding to the mutually orthogonal displacements. The displacement functions are determined in terms of eigenfunctions of linear vibrations. The procedure resulting in decoupling linear parts of equations is proposed with further utilization of the method of multiple scales for solving nonlinear governing equations of motion; in so doing the amplitude functions are expanded into power series in terms of the small parameter and depend on different time scales. It is shown that the phenomenon of internal resonance can be very critical, since in a thin plate internal resonance of the types two-to-one and one-to-one as well as additive and difference combinational resonances are always present.To describe the nonlinear damped vibrations of the thin plate under consideration, the fractional derivative Kelvin-Voigt model is used, since this model has an advantage over the conventional Kelvin-Voigt model, because its prediction is in a good compliance with experimental data. Thus, for example, the experimental data obtained by Abdel-Ghaffar and Housner [17] and Abdel-Ghaffar and Scanlan [18] during ambient vibration studies of the Vincent-Thomas Suspension Bridge and the Golden Gate Bridge, respectively, show that different vibrational modes feature different amplitude damping factors, and the order of smallness of these coefficients highlights the low damping capacity of suspension combined systems, resulting in prolonged energy transfer from one partial subsystem to another. Besides, as natural frequencies of vibrations increase, the corresponding damping ratios decrease. To lead the theoretical investigations in line with the experiment, fractional derivatives were introduced by Rossikhin and Shitikova [19] for describing the processes of internal friction proceeding in suspension combined systems at nonlinear free vibrations, which allowed the authors to obtain the damping coefficients dependent on the natural frequency of vibrations. Moreover, the good agreement between the theoretical results and the experimental data has been found through the appropriate choice of the fractional parameter (the order of the fractional derivative) and the viscosity coefficient [20].


## 2. Problem Formulation and the Method of Solution

Let us consider the dynamic behavior of a free supported nonlinear thin rectangular plate (Figure1), vibrations of which in a viscoelastic medium are described in the Cartesian system of coordinates by the following three differential equations written in the dimensionless form [2, 3, 21]:(1)uxx+1-ν2β12uyy+1+ν2β1vxy+wxwxx+1-ν2β12wyy+1+ν2β12wywxy=u¨+æ1Dγu,(2)β12vyy+1-ν2vxx+1+ν2β1uxy+β1wyβ12wyy+1-ν2wxx+1+ν2β1wxwxy=v¨+æ2Dγv,(3)β2212wxxxx+2β12wxxyy+β14wyyyy-wxxux+νβ1vy-wxuxx+νβ1vxy-1-ν2β1wxyβ1uy+vx+wyβ1uxy+vxx-β12wyyνux+β1vy+wyνuxy+β1vyy-1-ν2β1wxyβ1uy+vx+wxβ1uyy+vxy=-w¨-æ3Dγw,subjected to the initial (4)u|t=0=v|t=0=w|t=0=0,(5)u˙|t=0=εU0x,y,v˙|t=0=εV0x,y,w˙|t=0=εW0x,y,as well as the boundary conditions(6)w|x=0=w|x=1=0,v|x=0=v|x=1=0,uxx=0=uxx=1=0,wxxx=0=wxxx=1=0,w|y=0=w|y=1=0,u|y=0=u|y=1=0,vyy=0=vyy=1=0,wyyy=0=wyyy=1=0,where u=u(x,y,t), v=v(x,y,t), and w=w(x,y,t) are the displacements of points located in the plate’s middle surface in the x-, y-, and z-directions, respectively; ν is Poisson’s ratio; β1=a/b and β2=h/a are the parameters defining the dimensions of the plate; a and b are the plate’s dimensions along the x- and y-axes, respectively; h is the thickness; t is the time; an overdot denotes the time-derivative; lower indices label the derivatives with respect to the corresponding coordinates; U0(x,y), V0(x,y), and W0(x,y) are functions describing the distribution of the initial velocities of the points lying in the middle surface of the plate; ε is a small value.Figure 1
Scheme of a freely supported rectangular plate.In (1)–(6), the dimensionless values are introduced similarly as it has been done in [2]:(7)u=u*a,v=v*a,w=w*a,x=x*a,y=y*b,t=t*aEρ1-ν2,where dimensional values are marked by asterisks, and E is the Young modulus of the plate material.In (1)–(3), the terms æ1Dγu, æ2Dγv, and æ3Dγw represent the viscous resistance forces initiated when the plate vibrates in the surrounding viscous medium, where æi (i=1,2,3) are damping coefficients and Dγ is the Riemann-Liouville fractional derivative of the γ-order [22]: (8)DγF=ddt∫0t‍F(t-t′)Γ(1-γ)t′γdt′.But distinct to the traditional modeling the viscous resistance forces via first-order time-derivatives [23], in the present research we adopt the fractional order time-derivatives Dγ, which, as it will be shown below, will allow us to obtain the damping coefficients dependent on the natural frequency of vibrations. It has been demonstrated in [19, 20] that such an approach for modeling the damped nonlinear vibrations of thin bodies provides the good agreement between the theoretical results and the experimental data through the appropriate choice of the fractional parameter (the order of the fractional derivative) and the viscosity coefficient.It has been noted in [2, 19] that a fractional derivative is the immediate extension of an ordinary derivative. In fact, when δ→1, Dγx tends to x˙; that is, at γ→1 the fractional derivative goes over into the ordinary derivative, and the mathematical model of the viscoelastic plate under consideration transforms into the Kelvin-Voigt model, wherein the elastic element behaves nonlinearly, but the viscous element behaves linearly. When γ→0, the fractional derivative Dγx tends to x(t). To put it otherwise, the introduction of the new fractional parameter along with the parameters æi allows one to change not only the magnitude of viscosity at the cost of an increase or decrease in the parameters æi, but also the character of viscosity at the sacrifice of variations in the fractional parameter.Reference to (5) shows that free vibrations are excited by weak disturbance from the equilibrium position.We seek the solution of (1)–(3) in the form (9)ux,y,t=∑m=1∞‍∑n=1∞‍x1mntη1mnx,y,vx,y,t=∑m=1∞‍∑n=1∞‍x2mntη2mnx,y,wx,y,t=∑m=1∞‍∑n=1∞‍x3mntη3mnx,y,where x1mn(t), x2mn(t), and x3mn(t) are the generalized displacements corresponding to the displacements in the plane of the plate and to its deflection, respectively, but the natural functions satisfying the boundary conditions (6) have the form (10)η1mnx,y=cos⁡⁡πmxsin⁡πny,η2mnx,y=sin⁡πmxcos⁡⁡πny,η3mnx,y=sin⁡πmxsin⁡πny,and m and n are integers.Linear undamped natural modes of flexural in-plane and out-of-plane vibrations of the plate are the solution of eigenvalue problem(11)-x¨1mnη1mn+x1mnη1mn,xx+1-ν2β12η1mn,yy+1+ν2β1η2mn,xyx2mn=0,(12)-x¨2mnη2mn+x2mnβ12η2mn,yy+1-ν2η2mn,xx+1+ν2β1η1mn,xyx1mn=0,(13)x¨3mnη3mn+β2212η3mn,xxxx+2β12η3mn,xxyy00000000000000+β14η3mn,yyyyx3mn=0.The set of equations (11) and (12) has the characteristic equation (14)ωmn4-(S11mn+S22mn)ωmn2+S11mnS22mn-S12mnS21mn=0,the roots of which are the dimensionless natural frequencies of the flexural in-plane vibrations of the plate(15)ω1mn2=π2(m2+β12n2),ω2mn2=1-ν2ω1mn2,where (16)S11mn=π2m2+1-ν2β12n2,S12mn=S21mn=1+ν2β1π2mn,S22mn=π21-ν2m2+β12n2.From (13), the natural frequency of the flexural out-of-plane vibrations of the plate could be obtained: (17)ω3mn2=π4β2212m2+β12n22.Substituting (9) into (1)–(3), multiplying (1), (2), and (3) by η1lk, η2lk, and η3lk, respectively, integrating over x and y, and using the orthogonality conditions for linear modes within the domains of 0≤x,y≤1, we are led to an infinite number of sets each involving a system of three coupled nonlinear ordinary differential equations of the second order in ximn(i=1,2,3): (18)x¨αmn+æαDγxαmn+Sαβmnxβmn=-Fαmnα,β=1,2,(19)x¨3mn+æ3Dγx3mn+ω3mn2x3mn=-F3mn,where the summation is carried out over two repeated indices and the elements of the matrix Sijmn are defined as (16).The linear parts of nonlinear equations (18) and (19) have been studied in time and frequency domains in detail in [24] using the Laplace integral transform method as a method of solution.The nonlinear partsFimn of (18) and (19) have the form (20)F1mn=4∑m1‍∑n1‍∑m2‍∑n2‍x3m1n1x3m2n2A1mnm1n1m2n2,F2mn=4∑m1‍∑n1‍∑m2‍∑n2‍x3m1n1x3m2n2A2mnm1n1m2n2,F3mn=-4∑m1‍∑n1‍∑m2‍∑n2‍x3m1n1x1m2n2Cmnm1n1m2n2000000000000000000+x3m1n1x2m2n2Dmnm1n1m2n2,where (21)A1mnm1n1m2n2=π3m1m22+1-ν2β12n22a1mnm1n1m2n2-1+ν2π3β12n1m2n2a2mnm1n1m2n2,A2mnm1n1m2n2=π3β1n1β12n22+1-ν2m22a3mnm1n1m2n2-1+ν2π3β1m1m2n2a4mnm1n1m2n2,Cmnm1n1m2n2=π3m2m12+νβ12n12a5mnm1n1m2n2+1-νπ3β12m1n1n2a6mnm1n1m2n2-π3m1m22+1-ν2β12n22a7mnm1n1m2n2-1+ν2β12π3n1m2n2a8mnm1n1m2n2,Dmnm1n1m2n2=π3β1n2νm12+β12n12a5mnm1n1m2n2+1-νβ1π3m1n1m2a6mnm1n1m2n2-1+ν2β1π3m1m2n2a7mnm1n1m2n2-π3β1n11-ν2m22+β12n22a8mnm1n1m2n2,and the coefficients akmnm1n1m2n2(k=1,2,…,8) depending on the combinations of sine and cosine functions entering into the eigenfunctions (10) are presented in the appendix.Since the second-rank tensorSαβmn is symmetric, then it has two real eigenvalues ωαmn defined in (15) which are in correspondence with two mutually orthogonal eigenvectors (22)L→mnIL1mnI=πmω1mn,L2mnI=πβ1nω1mn,L→mnIIL1mnII=πβ1nω1mn,L2mnII=-πmω1mn,that is,(23)LαmnILαmnI=LαmnIILαmnII=1,LαmnILαmnII=0,where here and below summation is carried out over the repeated index α, while there is no summation over the repeated indices m and n.Thus, the matrixSαβmn and the generalized displacements xαmn entering in (18) and (19) could be expanded in terms of the vectors (22) [25] as (24)Sαβmn=ω1mn2LαmnILβmnI+ω2mn2LαmnIILβmnII,(25)xαmn=X1mnLαmnI+X2mnLαmnII.Substituting (24) and (25) in (18) and (19) and then multiplying (18) successively by LimnI and LimnII with due account for (23), we obtain the following three equations: (26)X¨1mn+æ1DγX1mn+ω1mn2X1mn=-FαmnLαmnI,X¨2mn+æ2DγX2mn+ω2mn2X2mn=-FαmnLαmnII,X¨3mn+æ3DγX3mn+ω3mn2X3mn=-F3mn,where X3mn=x3mn.It should be emphasized that left-hand side parts of (26) are linear and independent of each other, while (26) are coupled only by nonlinear terms in their right-hand sides.In order to show the influence of the initial conditions (4) and (5) on the solution to be constructed, let us expand the desired functions Ximn(i=1,2,3) in a series in terms of the small parameter ε:(27)Ximn=εXimn0+ε2Ximn1+⋯i=1,2,3.Substituting (27) in the set of equations (26) and restricting ourselves by the terms of the order of ε, we are led to a linear homogeneous set of differential equations(28)X¨imn0+ωimn2Ximn0=0i=1,2,3.The solution of (28) has the form (29)Ximn0=Aimnεtexp⁡iωimnt+A-imnεtexp⁡-iωimnti=1,2,3,where Aimn(εt) and A-imn(εt) are complex conjugate functions to be found.Representing functionsAimn(εt) in the polar form (30)Aimnεt=aimnεteiψimnεti=1,2,3,the solution (27) is reduced to (31)Ximn=εXimn0=2εaimnεtcos⁡ωimnt+ψimnεti=1,2,3,where aimn(εt) and ψimn(εt) are the amplitudes and phases of nonlinear vibrations, respectively.Differentiating (31) with respect to time t and ignoring the terms of the order of ε2, we obtain (32)X˙imn=-2εaimnεtωimnsinωimnt+ψimnεti=1,2,3.Now substituting (25) in (9) with due account for (31), we have(33)u(x,y,t)=2ε∑m=1∞‍∑n=1∞‍a1mnεtcos⁡ω1mnt+ψ1mnεtL1mnI00000000000+a2mnεtcos⁡ω2mnt+ψ2mnεtL1mnII·η1mnx,y,vx,y,t=2ε∑m=1∞‍∑n=1∞‍a1mn(εt)cos⁡ω1mnt+ψ1mnεtL2mnI000000000000+a2mnεtcos⁡ω2mnt+ψ2mnεtL2mnII·η2mnx,y,wx,y,t=2ε∑m=1∞‍∑n=1∞‍a3mnεt000000000000000000·cos⁡ω3mnt+ψ3mnεtη3mnx,y.Using the initial conditions (4) and (5) with due account for relationships (32), from (33) at t=0 we obtain an infinite set of algebraic equations for determining aimn(0) and ψimn(0) (i=1,2,3):(34)a1mn(0)L1mnIcos⁡ψ1mn(0)+a2mn(0)L1mnIIcos⁡ψ2mn(0)=0,a1mn(0)L2mnIcos⁡ψ1mn(0)+a2mn(0)L2mnIIcos⁡ψ2mn(0)=0,-a1mn0ω1mnL1mnIsinψ1mn0-a2mn(0)ω2mnL1mnIIsinψ2mn(0)=æ1mn,-a1mn0ω1mnL2mnIsinψ1mn0-a2mn0ω2mnL2mnIIsinψ2mn0=æ2mn,a3mn0cos⁡ψ3mn0=0,-a3mn(0)ω3mnsinψ3mn(0)=æ3mn,where (35)æ1mn=∫01∫01U0x,yη1mnx,ydxdy∫01∫01η1mn2x,ydxdy,æ2mn=∫01∫01V0x,yη2mnx,ydxdy∫01∫01η2mn2x,ydxdy,æ3mn=∫01∫01W0x,yη3mnx,ydxdy∫01∫01η3mn2x,ydxdy.All subsequent approximations are determined from an inhomogeneous set of differential equations with known right parts. Since the general solution of such a system is the sum of two solutions, a particular solution of the inhomogeneous system and a general solution of the corresponding homogeneous system, then arbitrary constants could be chosen in such a way that the initial conditions of all subsequent approximations would be zero ones.It is known [14, 26–28] that during nonstationary excitation of thin bodies not all possible modes of vibration would be excited. Moreover, the modes which are strongly coupled by any of the so-called internal resonance conditions are initiated and dominate in the process of vibration, resulting in the energy transfer from one subsystem to another between the coupled modes; in so doing the types of modes to be excited are dependent on the character of the external excitation. As a result of damping, the modes that are not excited directly by an external excitation or through an internal resonance will die out, and only three modes will dominate the motion of the plate in three directions.Assume hereafter that the vibration process occurs in such a way that only three natural modes corresponding to the generalized displacementsX1s1s2, X2l1l2, and X3k1k2 are excited and dominate over other natural modes. In this case, the right parts of (26) are significantly simplified, and equations of free vibrations (26) take the form(36)X¨1s1s2+æ1DγX1s1s2+ω1s1s22X1s1s2=-ζ1s1s2k1k2X3k1k22,X¨2l1l2+æ2DγX2l1l2+ω2l1l22X2l1l2=-ζ2l1l2k1k2X3k1k22,X¨3k1k2+æ3DγX3k1k2+ω3k1k22X3k1k2=-ζ13k1k2s1s2X1s1s2X3k1k2-ζ23k1k2l1l2X2l1l2X3k1k2,where (37)ζ1s1s2k1k2=4A1s1s2k1k2k1k2L1s1s2I+A2s1s2k1k2k1k2L2s1s2I,(38)ζ2l1l2k1k2=4A1l1l2k1k2k1k2L1l1l2II+A2l1l2k1k2k1k2L2l1l2II,(39)ζ13k1k2s1s2=4Ck1k2k1k2s1s2L1s1s2I+Dk1k2k1k2s1s2L2s1s2I,(40)ζ23k1k2l1l2=4Ck1k2k1k2l1l2L1l1l2II+Dk1k2k1k2l1l2L2l1l2II,(41)A1s1s2k1k2k1k2=π3k1k12+1-ν2β12k22a1s1s2k1k2k1k2-1+ν2π3β12k1k22a2s1s2k1k2k1k2,(42)A2l1l2k1k2k1k2=π3β1k2β12k22+1-ν2k12a3l1l2k1k2k1k2-1+ν2π3β1k12k2a4l1l2k1k2k1k2,(43)Ck1k2k1k2s1s2=π3s1k12+νβ12k22a5k1k2k1k2s1s2+1-νπ3β12k1k2s2a6k1k2k1k2s1s2-π3k1s12+1-ν2β12s22a7k1k2k1k2s1s2-1+ν2β12π3s1s2k2a8k1k2k1k2s1s2,(44)Dk1k2k1k2l1l2=π3β1l2νk12+β12k22a5k1k2k1k2l1l2+1-νβ1π3l1k1k2a6k1k2k1k2l1l2-1+ν2β1π3k1l1l2a7k1k2k1k2l1l2-π3β1k21-ν2l12+β12l22a8k1k2k1k2l1l2,and the coefficients akmnm1n1m2n2(k=1,2,…,8) involving in (41)–(44) are presented in the appendix.From relationships (37)–(44) we could calculate all coefficients entering in (36).Omitting hereafter the subindicess1s2, k1k2, and l1l2 for ease of presentation, (36) could be rewritten as (45)X¨1+æ1DγX1+ω12X1+ζ1X32=0,X¨2+æ2DγX2+ω22X2+ζ2X32=0,X¨3+æ3DγX3+ω32X3+X3ζ13X1+ζ23X2=0.


## 3. Method of Solution

An approximate solution of (45) for small but finite amplitudes weakly varying with time can be represented by a third-order uniform expansion in terms of different time scales in the following form [29]:(46)Xi=εXi1T0,T1,T2,…+ε2Xi2T0,T1,T2,…+ε3Xi3T0,T1,T2,…+⋯,where i=1,2,3, ε is a small dimensionless parameter of the same order of magnitude as the amplitudes, and Tn=εnt are new independent variables; among them, T0=t is a fast scale characterizing motions with the natural frequencies and T1=εt and T2=ε2t are slow scales characterizing the modulation of the amplitudes and phases of the modes with nonlinearity.Recall that the first and the second time-derivatives are defined, respectively, as follows:(47)ddt=D0+εD1+ε2D2+⋯,(48)d2dt2=D02+2εD0D1+ε2D12+2D0D2+⋯.Further as a fractional derivative we will utilize the Gru¨nvald-Letnikov definition which is defined on a semiaxis [30](49)ddtγφt=limh→0Δhγφthγ0<γ<1,where (50)Δhγφt=∑k=0∞‍-1kγkφt-khis the γth-order finite difference of the function φ(t) with the step h and with the point t as a center and (γk) are the binomial coefficients which are determined by the formula (51)γk=-1k-1γΓk-γΓ1-γΓk+1.Substituting (47) in (49), we obtain following Rossikhin and Shitikova [19] that (52)ddtγ=D0+εD1+ε2D2+⋯γ=D0γ+εγD0γ-1D1+12ε2γγ-1D0γ-2D12+2D0γ-1D2+⋯,where Dn=∂/∂Tn and D0γ,D0γ-1,D0γ-2,… are the Gru¨nvald-Letnikov fractional derivatives, which for “reasonably good” functions, such as exponential functions e±iωt, coincide with the Riemann-Liouville fractional derivatives in time t defined on a semiaxis [30].Considering that viscosity of the medium surrounding the plate under consideration is small, that is,(53)æi=εkμiτiγ,where τi is the relaxation time of the ith generalized displacement, μi is a finite value, and the choice of k depends on the order of smallness of the viscosity coefficients æi, substituting (46)–(52) in (45), after equating the coefficients at like powers of ε to zero, we are led to a set of recurrence equations to various orders:to orderε(54)D02X11+ω12X11=0,D02X21+ω22X21=0,D02X31+ω32X31=0;to orderε2(55)D02X12+ω12X12=-2D0D1X11-ζ1X312-μ12-kτ1γD0γX11;D02X22+ω22X22=-2D0D1X21-ζ2X312-μ22-kτ2γD0γX21;D02X32+ω32X32=-2D0D1X31-ζ13X11X31-ζ23X21X31-μ32-kτ3γD0γX31;to orderε3(56)D02X13+ω12X13=-2D0D1X12-D12+2D0D2X11-2ζ1X31X32-μ12-kτ1γ·D0γX12+D0γ-1D1X11-μ1k-1τ1γD0γX11,(57)D02X23+ω22X23=-2D0D1X22-D12+2D0D2X21-2ζ2X31X32-μ22-kτ2γ·D0γX22+D0γ-1D1X21-μ2k-1τ2γD0γX21,(58)D02X33+ω32X33=-2D0D1X32-D12+2D0D2X31-ζ13X11X32+X12X31-ζ23X21X32+X22X31-μ32-kτ3γD0γX32+D0γ-1D1X31-μ3k-1τ3γD0γX31.In order to construct the uniformly valid solution, it is necessary on each step to use the solution from the preceding step and to eliminate secular terms during integration [29].We will seek the solution of (54) in the form (59)Xj1=AjT1,T2exp⁡⁡iωjT0+A-jT1,T2exp⁡⁡-iωjT0j=1,2,3,where Aj(T1,T2)(j=1,2,3) are unknown complex functions and A-j(T1,T2) are the complex conjugates of Aj(T1,T2).To solve the sets of (55) and (56)–(58), it is necessary to specify the action of the fractional derivative D0γ  (8) on the functions Xj1, that is, to calculate D0γeiωjt. It has been shown in [31] that (60)D0γeiωjt=iωjγeiωjt+sin⁡πγπ∫0∞‍uγu+iωje-utdu.The first term in (60) is equivalent to the action of the fractional derivative with the low limit tending to -∞:(61)D0γx(t)=1Γ1-γddt∫-∞t‍xsdst-sγ,the application of which for the exponential function is reduced to (62)D0γeiωjt=iωjγeiωjt,while the second term of (60), as it has been proved in [32], does not influence the solution constructed via the method of multiple time scales restricted to the zeroth- and first-order approximations.In other words, even the utilization of exact formula (60) in the problem under consideration produces completely equivalent results given by the approximate formula (62) if the solution is constructed via the method of multiple time scales within the considered orders of approximation. Thus, in further analysis we will utilize formula (62).For further analysis it is also a need to specify the order of weak damping.


### 3.1. Viscosity of the Order ofε

Let us first consider the case of viscosity of the order ofε. Then at k=1  (55) are reduced to(63)D02X12+ω12X12=-2D0D1X11-ζ1X312-μ1τ1γD0γX11,D02X22+ω22X22=-2D0D1X21-ζ2X312-μ2τ2γD0γX21,D02X32+ω32X32=-2D0D1X31-X31ζ13X11+ζ23X21-μ3τ3γD0γX31.Substituting (59) in the right-hand side of (63) with due account for (62) yields (64)D02X12+ω12X12=-2iω1D1A1T1exp⁡iω1T0-ζ1A32exp⁡⁡2iω3T0+A3A-3-μ1τ1γA1iω1γexp⁡iω1T0+cc,D02X22+ω22X22=-2iω2D1A2T1exp⁡iω2T0-ζ2A32exp⁡(2iω3T0)+A3A-3-μ2τ2γA2iω2γexp⁡(iω2T0)+cc,D02X32+ω32X32=-2iω3D1A3T1exp⁡iω3T0-μ3τ3γA3iω3γexp⁡iω3T0-ζ13A-3A1A3exp⁡iω1+ω3T000000000+A1A-3exp⁡iω1-ω3T0-ζ23A-3A2A3exp⁡iω2+ω3T00000000+A2A-3exp⁡iω2-ω3T0+cc,where cc is the complex conjugate part to the preceding terms.Reference to (64) shows that the following types of the internal resonance could occur on this step:(1)
the two-to-one internal resonance, when one natural frequency is twice the other natural frequency,(65)ω1=2ω3ω2≠ω1,ω2≠2ω3,(66)ω2=2ω3ω1≠ω2,ω1≠2ω3;(2)
the one-to-one-to-two internal resonance, that is,(67)ω1=ω2=2ω3,or1:1:2.


### 3.2. Viscosity of the Order ofε2

Now let us consider vibrations of a nonlinear plate puttingk=2 in its equations of motion (55): (68)D02X12+ω12X12=-2D0D1X11-ζ1X312,D02X22+ω22X22=-2D0D1X21-ζ2X312,D02X32+ω32X32=-2D0D1X31-X31ζ13X11+ζ23X21.Substituting (59) in the right-hand side of (68) yields (69)D02X12+ω12X12=-2iω1D1A1T1exp⁡iω1T0-ζ1A32exp⁡2iω3T0+A3A-3+cc,(70)D02X22+ω22X22=-2iω2D1A2T1exp⁡iω2T0-ζ2A32exp⁡2iω3T0+A3A-3+cc,(71)D02X32+ω32X32=-2iω3D1A3T1exp⁡iω3T0-ζ13A-3A1A3exp⁡iω1+ω3T000000000+A1A-3exp⁡iω1-ω3T0-ζ23A-3A2A3exp⁡iω2+ω3T00000000+A2A-3exp⁡iω2-ω3T0+cc.Assume that the natural frequenciesωj (j=1,2,3) possess such magnitudes that any of the combinations from (65) to (67) could not occur. Then the functions exp⁡⁡(±iωjT0) entering into the right-hand side of (69) produce secular terms.To eliminate circular terms in (69), it is necessary to vanish to zero the coefficients standing at exp⁡±iωjT0, that is,(72)D1AjT1,T2=0,whence it follows that Aj (j=1,2,3) are T1-independent.Considering (72), solution of (69)–(71) at j=1,2,3 has the form(73)X12=F1T2exp⁡⁡iω1T0+k1A32exp⁡2iω3T0+k2A3A-3+cc,X22=F2T2exp⁡⁡iω2T0+k3A32exp⁡2iω3T0+k4A3A-3+cc,X32=F3(T2)exp⁡(iω3T0)+k5A1A3exp⁡i(ω1+ω3)T0+k6A2A3exp⁡iω2+ω3T0+k7A3A-1exp⁡iω3-ω1T0+k8A3A-2exp⁡iω3-ω2T0+cc,where Fj(T2)  (j=1,2,3) are new functions to be determined and coefficients ki (i=1,2,…,8) have the form(74)k1=ζ14ω32-1,k2=-ζ1ω12,k3=ζ24ω32-1,k4=-ζ2ω22,k5=ζ13ω1ω1+2ω3,k6=ζ23ω2ω2+2ω3,k7=ζ13ω1ω1-2ω3,k8=ζ23ω2ω2-2ω3.Substituting (59) and (73) with due account for (72) in (56)–(58) at k=2, we have(75)D02X13+ω12X13=-2iω1D2A1+μ1τ1γiω1γA1+2ζ1k5A1A3A-3·exp⁡⁡iω1T0-2ζ1F3A3exp⁡⁡2iω3T0+F-3A300000000+k5A1A32exp⁡iω1+2ω3T000000000+k6A2A32exp⁡iω2+2ω3T000000000+k6A2A3A-3exp⁡⁡iω2T000000000+k7A-1A32exp⁡i2ω3-ω1T000000000+k8A32A-2exp⁡i2ω3-ω2T0+cc,D02X23+ω22X23=-2iω2D2A2+μ2τ2γiω2γA2+2ζ2k6A2A3A-3·exp⁡iω2T0-2ζ2F3A3exp⁡2iω3T0+F-3A3000000000+k5A1A32exp⁡iω1+2ω3T0000000000+k6A2A32exp⁡iω2+2ω3T0000000000+k5A1A3A-3exp⁡iω1T0000000000+k7A-1A32exp⁡i2ω3-ω1T0000000000+k8A32A-2exp⁡i2ω3-ω2T0+cc,D02X33+ω32X33=-2iω3D2A3+μ3τ3γiω3γA30000000+ζ13k2+ζ23k4A32A-30000000+ζ13k7A1A-1A3+ζ23k8A2A-2A3·exp⁡(iω3T0)-ζ13k1+ζ23k3A33exp⁡3iω3T0-ζ13A-1F3A1+F1A3exp⁡iω3+ω1T00000000000+A3F-1+A-1F3exp⁡iω3-ω1T0-ζ23A-2F3A2+F2A3exp⁡iω3+ω2T0000000000+A3F-2+A-2F3exp⁡iω3-ω2T0-ζ13k5A12A3exp⁡i(2ω1+ω3)T00000000000+k7A12A-3exp⁡i(2ω1-ω3)T0-ζ23k6A22A3exp⁡i(2ω2+ω3)T00000000000+k8A22A-3exp⁡i2ω2-ω3T0-ζ13k6+ζ23k5A1A2A3exp⁡i(ω1+ω2+ω3)T0-ζ13k6+ζ23k7A-1A2A3exp⁡i(ω2+ω3-ω1)T0-ζ13k8+ζ23k5A1A-2A3exp⁡i(ω1-ω2+ω3)T0-ζ13k8+ζ23k7A-1A-2A3·exp⁡i(ω3-ω1-ω2)T0+cc.Reference to (75) shows that the following types of the internal resonance could occur on this step:(1)
the one-to-one internal resonance, when one natural frequency is nearly equal to the other natural frequency,(76)ω1=ω2ω3≠ω1,ω3≠ω2,(77)ω1=ω3ω2≠ω1,ω2≠ω3,(78)ω2=ω3ω1≠ω2,ω1≠ω3;(2)
the one-to-one-to-one internal resonance(79)ω1=ω2=ω3;(3)
the combinational resonance of the additive-difference type, that is,(80)ω1=ω2+2ω3,orω2=ω1-2ω3,or2ω3=ω1-ω2,(81)ω1=2ω3-ω2,orω2=2ω3-ω1,or2ω3=ω1+ω2,(82)ω1=ω2-2ω3,orω2=2ω3+ω1,or2ω3=ω2-ω1.


## 3.1. Viscosity of the Order ofε

Let us first consider the case of viscosity of the order ofε. Then at k=1  (55) are reduced to(63)D02X12+ω12X12=-2D0D1X11-ζ1X312-μ1τ1γD0γX11,D02X22+ω22X22=-2D0D1X21-ζ2X312-μ2τ2γD0γX21,D02X32+ω32X32=-2D0D1X31-X31ζ13X11+ζ23X21-μ3τ3γD0γX31.Substituting (59) in the right-hand side of (63) with due account for (62) yields (64)D02X12+ω12X12=-2iω1D1A1T1exp⁡iω1T0-ζ1A32exp⁡⁡2iω3T0+A3A-3-μ1τ1γA1iω1γexp⁡iω1T0+cc,D02X22+ω22X22=-2iω2D1A2T1exp⁡iω2T0-ζ2A32exp⁡(2iω3T0)+A3A-3-μ2τ2γA2iω2γexp⁡(iω2T0)+cc,D02X32+ω32X32=-2iω3D1A3T1exp⁡iω3T0-μ3τ3γA3iω3γexp⁡iω3T0-ζ13A-3A1A3exp⁡iω1+ω3T000000000+A1A-3exp⁡iω1-ω3T0-ζ23A-3A2A3exp⁡iω2+ω3T00000000+A2A-3exp⁡iω2-ω3T0+cc,where cc is the complex conjugate part to the preceding terms.Reference to (64) shows that the following types of the internal resonance could occur on this step:(1)
the two-to-one internal resonance, when one natural frequency is twice the other natural frequency,(65)ω1=2ω3ω2≠ω1,ω2≠2ω3,(66)ω2=2ω3ω1≠ω2,ω1≠2ω3;(2)
the one-to-one-to-two internal resonance, that is,(67)ω1=ω2=2ω3,or1:1:2.


## 3.2. Viscosity of the Order ofε2

Now let us consider vibrations of a nonlinear plate puttingk=2 in its equations of motion (55): (68)D02X12+ω12X12=-2D0D1X11-ζ1X312,D02X22+ω22X22=-2D0D1X21-ζ2X312,D02X32+ω32X32=-2D0D1X31-X31ζ13X11+ζ23X21.Substituting (59) in the right-hand side of (68) yields (69)D02X12+ω12X12=-2iω1D1A1T1exp⁡iω1T0-ζ1A32exp⁡2iω3T0+A3A-3+cc,(70)D02X22+ω22X22=-2iω2D1A2T1exp⁡iω2T0-ζ2A32exp⁡2iω3T0+A3A-3+cc,(71)D02X32+ω32X32=-2iω3D1A3T1exp⁡iω3T0-ζ13A-3A1A3exp⁡iω1+ω3T000000000+A1A-3exp⁡iω1-ω3T0-ζ23A-3A2A3exp⁡iω2+ω3T00000000+A2A-3exp⁡iω2-ω3T0+cc.Assume that the natural frequenciesωj (j=1,2,3) possess such magnitudes that any of the combinations from (65) to (67) could not occur. Then the functions exp⁡⁡(±iωjT0) entering into the right-hand side of (69) produce secular terms.To eliminate circular terms in (69), it is necessary to vanish to zero the coefficients standing at exp⁡±iωjT0, that is,(72)D1AjT1,T2=0,whence it follows that Aj (j=1,2,3) are T1-independent.Considering (72), solution of (69)–(71) at j=1,2,3 has the form(73)X12=F1T2exp⁡⁡iω1T0+k1A32exp⁡2iω3T0+k2A3A-3+cc,X22=F2T2exp⁡⁡iω2T0+k3A32exp⁡2iω3T0+k4A3A-3+cc,X32=F3(T2)exp⁡(iω3T0)+k5A1A3exp⁡i(ω1+ω3)T0+k6A2A3exp⁡iω2+ω3T0+k7A3A-1exp⁡iω3-ω1T0+k8A3A-2exp⁡iω3-ω2T0+cc,where Fj(T2)  (j=1,2,3) are new functions to be determined and coefficients ki (i=1,2,…,8) have the form(74)k1=ζ14ω32-1,k2=-ζ1ω12,k3=ζ24ω32-1,k4=-ζ2ω22,k5=ζ13ω1ω1+2ω3,k6=ζ23ω2ω2+2ω3,k7=ζ13ω1ω1-2ω3,k8=ζ23ω2ω2-2ω3.Substituting (59) and (73) with due account for (72) in (56)–(58) at k=2, we have(75)D02X13+ω12X13=-2iω1D2A1+μ1τ1γiω1γA1+2ζ1k5A1A3A-3·exp⁡⁡iω1T0-2ζ1F3A3exp⁡⁡2iω3T0+F-3A300000000+k5A1A32exp⁡iω1+2ω3T000000000+k6A2A32exp⁡iω2+2ω3T000000000+k6A2A3A-3exp⁡⁡iω2T000000000+k7A-1A32exp⁡i2ω3-ω1T000000000+k8A32A-2exp⁡i2ω3-ω2T0+cc,D02X23+ω22X23=-2iω2D2A2+μ2τ2γiω2γA2+2ζ2k6A2A3A-3·exp⁡iω2T0-2ζ2F3A3exp⁡2iω3T0+F-3A3000000000+k5A1A32exp⁡iω1+2ω3T0000000000+k6A2A32exp⁡iω2+2ω3T0000000000+k5A1A3A-3exp⁡iω1T0000000000+k7A-1A32exp⁡i2ω3-ω1T0000000000+k8A32A-2exp⁡i2ω3-ω2T0+cc,D02X33+ω32X33=-2iω3D2A3+μ3τ3γiω3γA30000000+ζ13k2+ζ23k4A32A-30000000+ζ13k7A1A-1A3+ζ23k8A2A-2A3·exp⁡(iω3T0)-ζ13k1+ζ23k3A33exp⁡3iω3T0-ζ13A-1F3A1+F1A3exp⁡iω3+ω1T00000000000+A3F-1+A-1F3exp⁡iω3-ω1T0-ζ23A-2F3A2+F2A3exp⁡iω3+ω2T0000000000+A3F-2+A-2F3exp⁡iω3-ω2T0-ζ13k5A12A3exp⁡i(2ω1+ω3)T00000000000+k7A12A-3exp⁡i(2ω1-ω3)T0-ζ23k6A22A3exp⁡i(2ω2+ω3)T00000000000+k8A22A-3exp⁡i2ω2-ω3T0-ζ13k6+ζ23k5A1A2A3exp⁡i(ω1+ω2+ω3)T0-ζ13k6+ζ23k7A-1A2A3exp⁡i(ω2+ω3-ω1)T0-ζ13k8+ζ23k5A1A-2A3exp⁡i(ω1-ω2+ω3)T0-ζ13k8+ζ23k7A-1A-2A3·exp⁡i(ω3-ω1-ω2)T0+cc.Reference to (75) shows that the following types of the internal resonance could occur on this step:(1)
the one-to-one internal resonance, when one natural frequency is nearly equal to the other natural frequency,(76)ω1=ω2ω3≠ω1,ω3≠ω2,(77)ω1=ω3ω2≠ω1,ω2≠ω3,(78)ω2=ω3ω1≠ω2,ω1≠ω3;(2)
the one-to-one-to-one internal resonance(79)ω1=ω2=ω3;(3)
the combinational resonance of the additive-difference type, that is,(80)ω1=ω2+2ω3,orω2=ω1-2ω3,or2ω3=ω1-ω2,(81)ω1=2ω3-ω2,orω2=2ω3-ω1,or2ω3=ω1+ω2,(82)ω1=ω2-2ω3,orω2=2ω3+ω1,or2ω3=ω2-ω1.


## 4. Governing Nonlinear Differential Equations Describing Amplitude-Phase Modulation for Different Types of the Internal Resonance of the Order ofε

To deduce the nonlinear differential equations describing the modulation of amplitudes and phases of the nonlinear plate under consideration, we should consider separately each type of the internal resonance which could occur with due account for weak damping of the orderε.


### 4.1. Internal Resonance 2:1

Let us consider case (65), when ω1=2ω3, while ω2≠ω1 and ω2≠2ω3. Case (66) could be treated similarly.Then eliminating secular terms in (64), we obtain the following solvability equations: (83)2iω1D1A1(T1)+μ1iω1τ1γA1+ζ1A32=0,(84)2iω2D1A2(T1)+μ2iω2τ2γA2=0,(85)2iω3D1A3(T1)+μ3iω3τ3γA3+ζ13A1A-3=0.Reference to the set of equations (83)–(85) shows that its second equation is independent of other two, while the first and third ones represent a set of two nonlinear equations. The similar situation was noted in [2] for a fractionally damped nonlinear plate in the case of the two-to-one internal resonance, when equations describing plate’s in-plane motion are coupled.Let us multiply (83)–(85), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that (86)Ai=aieiφii=1,2,3,as a result we have (87)a12·+s1a12+ω1-1ζ1a1a32sin⁡δ=0,a32·+s3a32-ω3-1ζ13a1a32sin⁡δ=0,φ˙1-12σ1-12ζ1ω1-1a32a1-1cos⁡⁡δ=0,φ˙3-12σ3-12ζ13ω3-1a1cos⁡⁡δ=0,a22·+s2a22=0,φ˙2-12σ2=0,where a dot denotes differentiation with respect to T1, δ=2φ3-φ1, and(88)si=μiτiγωiγ-1sinψ,σi=μiτiγωiγ-1cos⁡ψ,ψ=12πγi=1,2,3.The set of nonlinear equations (87)–(88) was solved in [2]. It has been found that during free vibrations of the plate being under the conditions of the two-to-one internal resonance three regimes can be observed: stationary (absence of damping at γ=0 and μ=0), quasi-stationary (damping is defined by an ordinary derivative at γ=1), and transient (damping is defined by a fractional derivative at 0<γ<1).


### 4.2. Internal Resonance 1:1:2

Let us consider case (67), when ω1=ω2=2ω3. Then eliminating secular terms in (64), we obtain the following solvability equations: (89)2iω1D1A1T1+μ1iω1τ1γA1+ζ1A32=0,2iω2D1A2T1+μ2iω2τ2γA2+ζ2A32=0,2iω3D1A3T1+μ3iω3τ3γA3+ζ13A1A-3+ζ23A2A-3=0.Reference to the set of equations (89) shows that, in contrast to (83)–(85) corresponding to case (65) when ω1=2ω3, while ω2≠ω1 and ω2≠2ω3, now we obtain the set of three coupled nonlinear equations.Representing functionsAi entering in (89) in the polar form (86) and applying the same procedure as it has been carried out above for the 2:1 internal resonance, we have (90)a12·+s1a12+ω1-1ζ1a1a32sin⁡2φ3-φ1=0,(91)a22·+s2a22+ω2-1ζ2a2a32sin⁡2φ3-φ2=0,(92)a32·+s3a32-ω3-1ζ13a1a32sin2φ3-φ1-ω3-1ζ23a2a32sin2φ3-φ2=0,(93)φ˙1-12σ1-12ζ1ω1-1a32a1-1cos⁡2φ3-φ1=0,(94)φ˙2-12σ2-12ζ2ω2-1a32a2-1cos⁡2φ3-φ2=0,(95)φ˙3-12σ3-12ζ13ω3-1a1cos⁡2φ3-φ1-12ζ23ω3-1a2cos⁡2φ3-φ2=0.The nonlinear set of equations (90)–(95) with the corresponding initial conditions completely describes the vibrational process of the mechanical system being investigated under the condition of the internal resonance 1:1:2 and could be solved numerically. This case has been discussed in [33].Introducing new functionsξ1(T1), ξ2(T1), and ξ3(T1) such that (96)a12=ζ1ω3ζ13ω1ξ1e-s1T1,a22=ζ2ω3ζ23ω2ξ2e-s2T1,a32=ξ3e-s3T1and adding (90)–(95) with due account for (96), we obtain (97)ξ˙1e-s1T1+ξ˙2e-s2T1+ξ˙3e-s3T1=0.Equation (97) describes the law of energy variation for this case of the internal resonance.


## 4.1. Internal Resonance 2:1

Let us consider case (65), when ω1=2ω3, while ω2≠ω1 and ω2≠2ω3. Case (66) could be treated similarly.Then eliminating secular terms in (64), we obtain the following solvability equations: (83)2iω1D1A1(T1)+μ1iω1τ1γA1+ζ1A32=0,(84)2iω2D1A2(T1)+μ2iω2τ2γA2=0,(85)2iω3D1A3(T1)+μ3iω3τ3γA3+ζ13A1A-3=0.Reference to the set of equations (83)–(85) shows that its second equation is independent of other two, while the first and third ones represent a set of two nonlinear equations. The similar situation was noted in [2] for a fractionally damped nonlinear plate in the case of the two-to-one internal resonance, when equations describing plate’s in-plane motion are coupled.Let us multiply (83)–(85), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that (86)Ai=aieiφii=1,2,3,as a result we have (87)a12·+s1a12+ω1-1ζ1a1a32sin⁡δ=0,a32·+s3a32-ω3-1ζ13a1a32sin⁡δ=0,φ˙1-12σ1-12ζ1ω1-1a32a1-1cos⁡⁡δ=0,φ˙3-12σ3-12ζ13ω3-1a1cos⁡⁡δ=0,a22·+s2a22=0,φ˙2-12σ2=0,where a dot denotes differentiation with respect to T1, δ=2φ3-φ1, and(88)si=μiτiγωiγ-1sinψ,σi=μiτiγωiγ-1cos⁡ψ,ψ=12πγi=1,2,3.The set of nonlinear equations (87)–(88) was solved in [2]. It has been found that during free vibrations of the plate being under the conditions of the two-to-one internal resonance three regimes can be observed: stationary (absence of damping at γ=0 and μ=0), quasi-stationary (damping is defined by an ordinary derivative at γ=1), and transient (damping is defined by a fractional derivative at 0<γ<1).


## 4.2. Internal Resonance 1:1:2

Let us consider case (67), when ω1=ω2=2ω3. Then eliminating secular terms in (64), we obtain the following solvability equations: (89)2iω1D1A1T1+μ1iω1τ1γA1+ζ1A32=0,2iω2D1A2T1+μ2iω2τ2γA2+ζ2A32=0,2iω3D1A3T1+μ3iω3τ3γA3+ζ13A1A-3+ζ23A2A-3=0.Reference to the set of equations (89) shows that, in contrast to (83)–(85) corresponding to case (65) when ω1=2ω3, while ω2≠ω1 and ω2≠2ω3, now we obtain the set of three coupled nonlinear equations.Representing functionsAi entering in (89) in the polar form (86) and applying the same procedure as it has been carried out above for the 2:1 internal resonance, we have (90)a12·+s1a12+ω1-1ζ1a1a32sin⁡2φ3-φ1=0,(91)a22·+s2a22+ω2-1ζ2a2a32sin⁡2φ3-φ2=0,(92)a32·+s3a32-ω3-1ζ13a1a32sin2φ3-φ1-ω3-1ζ23a2a32sin2φ3-φ2=0,(93)φ˙1-12σ1-12ζ1ω1-1a32a1-1cos⁡2φ3-φ1=0,(94)φ˙2-12σ2-12ζ2ω2-1a32a2-1cos⁡2φ3-φ2=0,(95)φ˙3-12σ3-12ζ13ω3-1a1cos⁡2φ3-φ1-12ζ23ω3-1a2cos⁡2φ3-φ2=0.The nonlinear set of equations (90)–(95) with the corresponding initial conditions completely describes the vibrational process of the mechanical system being investigated under the condition of the internal resonance 1:1:2 and could be solved numerically. This case has been discussed in [33].Introducing new functionsξ1(T1), ξ2(T1), and ξ3(T1) such that (96)a12=ζ1ω3ζ13ω1ξ1e-s1T1,a22=ζ2ω3ζ23ω2ξ2e-s2T1,a32=ξ3e-s3T1and adding (90)–(95) with due account for (96), we obtain (97)ξ˙1e-s1T1+ξ˙2e-s2T1+ξ˙3e-s3T1=0.Equation (97) describes the law of energy variation for this case of the internal resonance.


## 5. Governing Nonlinear Differential Equations Describing Amplitude-Phase Modulation for Different Types of the Internal Resonance of the Order ofε2

To deduce the nonlinear differential equations describing the modulation of amplitudes and phases of the nonlinear plate under consideration, we should consider separately each type of the internal resonance which could occur with due account for weak damping of the orderε2.


### 5.1. Internal Resonance 1:1

The one-to-one internal resonance could be of two types: when the natural frequencies of two modes of in-plane vibrations are close to each other, that is, whenω1=ω2=ω, or when the natural frequency of the out-of-plane mode is close to the natural frequency of one of the in-plane modes, that is, ω3=ω1=ω or ω3=ω2=ω.


#### 5.1.1. Internal Resonanceω1=ω2=ω

Now let us first consider case (76) of the one-to-one internal resonance, when ω1=ω2=ω, but ω3≠ω.Then eliminating secular terms in (75), we obtain the following solvability equations: (98)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A3A-3=0,(99)2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A3A-3=0,(100)2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k5A1A-2A3+ζ13k6+ζ23k7A-1A2A3=0.Let us multiply (98)–(100), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that the functions Ai(T2) could be represented in the polar form (86), where ai and φi(i=1,2,3) are T2-dependent, as a result we have (101)a12·+s1a12=-2ω-1ζ1k6a1a2a32sinδ,(102)a22·+s2a22=2ω-1ζ2k5a1a2a32sinδ,(103)a32·+s3a32=-ω3-1ζ13k6-k8+ζ23k7-k5·a1a2a32sinδ,(104)φ˙1=12σ1+ω-1ζ1k5a32+ω-1ζ1k6a1-1a2a32cos⁡δ,(105)φ˙2=12σ2+ω-1ζ2k6a32+ω-1ζ2k5a1a2-1a32cos⁡δ,(106)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k6-k8+ζ23k7-k5·a1a2cos⁡δ,where the phase difference has the form δ=φ2-φ1 and an overdot denotes the differentiation with respect to T2.It could be shown that the expression in square brackets in (103) and (106) is vanished to zero when ω1=ω2=ω. Then (103) turns out to be uncoupled with respect to all other equations from (101) to (106): (107)a32·+s3a32=0,the solution of which has the form (108)a32=c3e-s3T2,where c3 is a constant of integration to be determined from the initial conditions, while (106) is reduced to (109)φ˙3=12σ3+12ω3-1·ζ13k7a12+ζ23k8a22+ζ13k2+ζ23k4a32,whence it follows that the phase φ3 is dependent only on the squared values of all amplitudes of the three interacting modes of vibrations.Introducing new functionsξ1(T2) and ξ2(T2) such that (110)a12=ζ1k6ωξ1exp⁡-s1T2,a22=ζ2k5ωξ2exp⁡-s2T2and adding (101) and (102) with due account for (110) yield (111)ξ˙2+ξ˙1e(s2-s1)T2=0,while subtracting (104) from (105) we obtain (112)δ˙=φ˙2-φ˙1=Σ+ω-1ζ2k6-ζ1k5a32+ω-1ζ2k5a12-ζ1k6a22a32a1a2cos⁡δ,where Σ=(1/2)(σ2-σ1).From (101) and (102) we could find that (113)a2a32a1=-12ωζ1k61sinδξ˙1ξ1,a1a32a2=12ωζ2k51sinδξ˙2ξ2.Considering (110) and (113), (112) could be reduced to (114)δ˙-Σ=Ke-s3T2+12ξ˙1ξ1+ξ˙2ξ2cotδ,where K=ω-1ζ2k6-ζ1k5c3.Equations (101) and (112) could be rewritten in another form if we substitute (110) and (108) in all its terms: (115)ξ˙1=-2bξ1ξ2e-(s3+1/2s2-1/2s1)T2sinδ,(116)δ˙-Σ=Ke-s3T2+bξ1e-s1T2-ξ2e-s2T2·cos⁡δξ1ξ2e-(s3+1/2s2+1/2s1)T2,where b=ω-1ζ1ζ2k5k6.The nonlinear set of equations (111), (115), and (116), or (114), with the initial conditions (117)ξ1T2=0=ξ10,ξ2T2=0=ξ20,ξ3T2=0=ξ30,δT2=0=δ0,completely describes the vibrational process of the mechanical system being investigated under the condition of the one-to-one internal resonance ω1=ω2 and could be solved numerically.It should be noted that the equation defining the stream function could be obtained from (114), if we neglect the first term in (114) which decays rather rapidly as compared with its last term.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (111) has the form (118)ξ˙2+ξ˙1=0,whence it follows that (119)ξ2+ξ1=E0,(120)ξ1=E0ξ,ξ2=E01-ξ,where E0 is the initial energy.Considering (120), (115) and (116), wherein the term rapidly decaying with time has been neglected, are reduced to the following: (121)ξ˙=-2bξ1-ξe-sT2sinδ,(122)δ˙=-b-ξ1/21-ξ-1/2+ξ-1/21-ξ1/2·cos⁡δe-sT2=-b1-2ξξ1-ξcos⁡δe-sT2,while (114), wherein its first term rapidly decaying with time has been neglected, takes the form (123)δ˙tanδ≈12ξ˙2ξ2+12ξ˙1ξ1,and it could be integrated, resulting in its first integral (124)Gξ,δ=ξ1/21-ξ1/2cos⁡δ=G0ξ0,δ0.The second first integral (124) defines the stream function G(ξ,δ) such that (125)vξ=ξ˙=2b∂G∂δe-sT2,vδ=δ˙=-2b∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (121) and (124) and integrating over T2, we have (126)∫ξ0ξ‍dξξ1-ξ-G02=2bse-sT2-1.The integral in the left-hand side of (126) can be transformed to an incomplete elliptic integral of the first kind and can be easily calculated using special tables [34].The solution of (126) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (121) as (127)δ˙=dδdT2=dδdξξ˙=-2bdδdξξ1/21-ξ1/2e-sT2sinδand retain all terms in (116) with due account for (120), then (116) could be rewritten as follows: (128)dcos⁡δdξ+121-2ξξ1-ξcos⁡δ=K2bξ1-ξ.Integrating (128) yields (129)G1ξ,δ=ξ1/21-ξ1/2cos⁡δ-12Kb-1ξ=G10ξ0,δ0.Note that relationships (125) and (126) are valid in this case as well.


#### 5.1.2. Internal Resonanceω1=ω3=ω

Now let us consider case (77), when ω1=ω3=ω, but ω2≠ω. Note that case (78), when ω2=ω3=ω, but ω1≠ω, could be treated similarly.Then eliminating secular terms in (75), we obtain the following solvability equations: (130)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k7A-1A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k7A12A-3=0.Representing functionsAi entering in (130) in the polar form and applying the same procedure as it has been carried out above for the other case of the 1:1 internal resonance, as a result we have (131)a12·+s1a12=-2ω-1ζ1k7a12a32sinδ,(132)a22·+s2a22=0,(133)a32·+s3a32=ω-1ζ13k7a12a32sinδ,(134)φ˙1=12σ1+ω-1ζ1k5a32+ω-1ζ1k7a32cos⁡δ,(135)φ˙2=12σ2+ω2-1ζ2k6a32,(136)φ˙3=12σ3+12ω-1ζ13k7a12+12ω-1ζ23k8a22+12ω-1ζ13k2+ζ23k4a32+12ω-1ζ13k7a12cos⁡δ,where the phase difference has the form δ=2(φ3-φ1).Since (132) is independent of others from (131) to (136), its solution has the form (137)a22=c2e-s2T2,where c2 is a constant of integration to be determined from the initial conditions.Introducing new functionsξ1(T2) and ξ3(T2) such that (138)a12=ζ1k7ωξ1exp⁡-s1T2,a32=ζ13k7ωξ3exp⁡-s3T2and adding (131) and (133) with due account for (138) yield (139)ξ˙1e-s1T2+2ξ˙3e-s3T2=0,while subtracting (134) from (136) we obtain (140)δ˙=2φ˙3-φ˙1=Σ+ω-1ζ13k2+ζ23k4-2ζ1k5a32hhhhhhhhhhhhh+ω-1ζ13k7a12+ω-1ζ23k8a22hhhhhhhhhhhhh+ω-1k7ζ13a12-2ζ1a32cos⁡δ,where Σ=σ3-σ1.From (131) and (133) we could find that (141)-2ζ1k7ωa32=ξ˙1ξ11sinδ,ζ13k7ωa12=ξ˙3ξ31sinδ.Considering (138) and (141), (131) and (140) could be reduced to (142)ξ˙1=-2bξ1ξ3e-s3T2sinδ,δ˙-Σ=bξ1e-s1T2+K2e-s2T2+K3ξ3e-s3T2+ξ˙1ξ1+ξ˙3ξ3cotδ=bξ1e-s1T2+K2e-s2T2+K3ξ3e-s3T2+bξ1e-s1T2-2ξ3e-s3T2cos⁡δ,where (143)b=ω-2k72ζ1ζ13,K2=ω-1ζ23k8c2,K3=ω-2ζ13k7ζ13k2+ζ23k4-2ζ1k5.The nonlinear set of equations (139) and (142) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the one-to-one internal resonance ω1=ω3 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (139) has the form (144)ξ˙1+2ξ˙3=0,whence it follows that (145)ξ1+2ξ3=E0,(146)ξ1=E0ξ,ξ3=12E01-ξ.Considering (146), (142) are reduced to (147)ξ˙=-bE0ξ1-ξe-sT2sinδ,(148)δ˙=bE0ξ+K2+12K3E0(1-ξ)e-sT2+ξ˙1ξ1+ξ˙3ξ3cotδ=bE0ξ+K2+12K3(1-ξ)e-sT2-bE01-2ξe-sT2cos⁡δ,while (148), wherein its first term rapidly decaying with time has been neglected, takes the form (149)δ˙tanδ≈ξ˙1ξ1+ξ˙3ξ3,and it could be integrated, resulting in its first integral (150)Gξ,δ=ξ1-ξcos⁡δ=G0ξ0,δ0.This first integral (150) defines the stream function G(ξ,δ) such that (151)vξ=ξ˙=bE0∂G∂δe-sT2,vδ=δ˙=-bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (147) and (150) and integrating over T2, we have (152)∫ξ0ξ‍dξξ21-ξ2-G02=bE0se-sT2-1.The solution of (152) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (147) as (153)δ˙=dδdT2=dδdξξ˙=-bE0dδdξξ1-ξe-sT2sinδ=bE0dcos⁡δdξξ1-ξe-sT2and retain all terms in (148) with due account for (146), then (148) could be rewritten as follows: (154)dcos⁡δdξ+1-2ξξ1-ξcos⁡δ=K3bξ+11-ξ+K2bE0ξ1-ξ.Integrating (154) yields(155)G1ξ,δ=ξ1-ξcos⁡δ+12K3b-11-ξ2-12ξ2-K2bE0-1ξ=G10ξ0,δ0.Note that relationships (151) and (152) are valid in this case as well.Thus, during free vibrations of the plate being under the conditions of the one-to-one internal resonance three regimes can be observed: stationary (absence of damping atγ=0), quasi-stationary (damping is defined by an ordinary derivative at γ=1), and transient (damping is defined by a fractional derivative at 0<γ<1).


#### 5.1.3. Internal Resonance 1:1:1

Now let us consider case (79), when ω1=ω2=ω3=ω.Then eliminating secular terms in (75), we obtain the following solvability equations: (156)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A3A-3+2ζ1k7A-1A32+2ζ1k8A-2A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A3A-3+2ζ2k7A-1A32+2ζ2k8A-2A32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k7A12A-3+ζ23k8A22A-3+ζ13k6+ζ23k7A-1A2A3+ζ13k8+ζ23k5A1A-2A3=0.Representing functionsAi entering in (151) in the polar form and applying the same procedure as it has been carried out above for the cases of the 1:1 internal resonance, as a result we have (157)a12·+s1a12=-2ω-1ζ1·k7a12sin2φ3-φ100000+k6a1a2sinφ2-φ100000+k8a1a2sin⁡2φ3-φ2-φ1a12a32,a22·+s2a22=-2ω-1ζ2·k8a22sin2φ3-φ200000-k5a1a2sin⁡φ2-φ100000+k7a1a2sin2φ3-φ2-φ1a22a32,a32·+s3a32=ω-1ζ13k7a12sin2φ3-φ10000000+ζ23k8a22sin2(φ3-φ2)a32,φ˙1=12σ1+ω-1ζ1·a1-1k5+k7cos⁡2φ3-φ10000+k6a1-1a2cos⁡φ2-φ10000+k8a1-1a2cos⁡(2φ3-φ2-φ1)a32,φ˙2=12σ2+ω2-1ζ2·a2-1k6+k8cos⁡2φ3-φ20000+k5a1a2-1cos⁡φ2-φ10000+k7a1a2-1cos⁡(2φ3-φ2-φ1)a32,φ˙3=12σ3+12ω-1·ζ13k7a12+ζ23k8a22+ζ13k2+ζ23k4a32+12ω-1ζ13k7a12cos⁡2φ3-φ100000000000+ζ23k8a22cos⁡2φ3-φ2-23ω-3ζ13ζ23a1a2cos⁡φ2-φ1.The nonlinear set of equations (157) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the internal resonance 1:1:1 and could be solved numerically.


### 5.2. Combinational Resonances of the Additive-Difference Type

Reference to (75) shows that three cases of the combinational internal resonance of the additive-difference type (80)–(82) could occur for the plate under consideration.


#### 5.2.1. Combinational Resonance:2ω3=ω1+ω2

Let us first consider case (81), when 2ω3=ω1+ω2.Then eliminating secular terms in (75), we obtain the following solvability equations:(158)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k8A-2A32=0,(159)2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k7A-1A32=0,(160)2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k7A1A2A-3=0.Let us multiply (158)–(160), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that the functions Ai(T2) could be represented in the polar form (86), where ai and φi(i=1,2,3) are T2-dependent, as a result we have (161)a12·+s1a12=-2ω1-1ζ1k8a1a2a32sinδ,(162)a22·+s2a22=-2ω2-1ζ2k7a1a2a32sinδ,(163)a32·+s3a32=ω3-1ζ13k8+ζ23k7a1a2a32sinδ,(164)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k8a1-1a2a32cos⁡δ,(165)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k7a1a2-1a32cos⁡δ,(166)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k8+ζ23k7a1a2cos⁡δ,where the phase difference has the form δ=2φ3-φ2-φ1.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (167)a12=ζ1k8ω1ξ1exp⁡-s1T2,a22=ζ2k7ω2ξ2exp⁡-s2T2,a32=ζ13k8+ζ23k7ω3ξ3exp⁡-s3T2and adding (161)–(163) with due account for (167) yield (168)ξ˙1e-s1T2+ξ˙2e-s2T2+4ξ˙3e-s3T2=0,while subtracting (164) and (165) from the doubled (166) we obtain (169)δ˙=2φ˙3-φ˙1-φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000-ω1-1ζ1k5-ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k8+ζ23k7a1a2-ω1-1ζ1k8a2a32a1000000-ω2-1ζ2k7a1a32a2cos⁡δ,where 2Σ=2σ3-σ1-σ2.From (161)–(163) we could find that (170)-2ζ1k8ω1a2a32a1=ξ˙1ξ11sinδ,-2ζ2k7ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k8+ζ23k7ω3a1a2=ξ˙3ξ31sinδ.Then (161), (163), and (169) could be rewritten in the following form: (171)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(172)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(173)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ10000000-ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T20000000-ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where (174)b=ζ13k8+ζ23k7ω3ζ1ζ2k7k8ω1ω2,K1=ζ1ζ13k7k8ω1ω3,K2=ζ2ζ23k7k8ω2ω3,K3=ζ13k8+ζ23k72ω3ζ13k2+ζ23k4ω3-ζ1k5ω1-ζ2k6ω2.The nonlinear set of equations (168), (171), (172), and (173) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω1+ω2 and could be solved numerically. This case has been discussed in [35].Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (168) has the form (175)ξ˙1+ξ˙2+4ξ˙3=0,whence it follows that (176)ξ1+ξ2+4ξ3=E0,(177)ξ1=2E0c1-ξ,ξ2=2E0c2-ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (178)2c1+2c2+4c3=1.Considering (177), (171) and (173) are reduced to (179)ξ˙=2bE0(c1-ξ)(c2-ξ)(c3+ξ)e-sT2sinδ,(180)δ˙=2E0K1c1-ξ+K2c2-ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1-ξ+K2c2-ξ+K3c3+ξe-sT2-bE0c1-ξc2-ξ-c3+ξc2-ξc1-ξ000000000-c3+ξc1-ξc2-ξe-sT2cos⁡δ,while (180), wherein its first term rapidly decaying with time has been neglected, takes the form (181)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (182)Gξ,δ=c3+ξc1-ξc2-ξcos⁡δ=G0ξ0,δ0.This first integral (182) defines the stream function G(ξ,δ) such that (183)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (179) and (182) and integrating over T2, we have (184)∫ξ0ξ‍dξc3+ξ2c1-ξc2-ξ-G02=2bE0s1-e-sT2.The solution of (184) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (179) as(185)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1-ξc2-ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1-ξc2-ξc3+ξe-sT2and retain all terms in (180) with due account for (177), then (180) could be rewritten as follows: (186)-dcos⁡δdξ+121c1-ξ+121c2-ξ-1c3+ξcos⁡δ=K1c1-ξbc3+ξc2-ξ+K2c2-ξbc3+ξc1-ξ+K3bc1-ξc2-ξ.Integrating (186) yields (187)G1ξ,δ=c3+ξc1-ξc2-ξcos⁡δ-12K1b-1c1-ξ2-12K2b-1c2-ξ2+12K3b-1c3+ξ2=G10ξ0,δ0.Note that relationships (183) and (184) are valid in this case as well.


#### 5.2.2. Combinational Resonance:2ω3=ω1−ω2

Now let us consider case (80), when 2ω3=ω1-ω2.Then eliminating secular terms in (75), we obtain the following solvability equations:(188)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k7A1A-32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k6+ζ23k7A1A-2A-3=0.Applying to (188) the same procedure as it has been done above for (158)–(160), as a result we have (189)a12·+s1a12=-2ω1-1ζ1k6a1a2a32sinδ,(190)a22·+s2a22=2ω2-1ζ2k7a1a2a32sinδ,(191)a32·+s3a32=ω3-1ζ13k6+ζ23k7a1a2a32sinδ,(192)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k6a1-1a2a32cos⁡δ,(193)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k7a1a2-1a32cos⁡δ,(194)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k6+ζ23k7a1a2cos⁡δ,where the phase difference has the form δ=2φ3+φ2-φ1.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (195)a12=ζ1k6ω1ξ1exp⁡-s1T2,a22=ζ2k7ω2ξ2exp⁡-s2T2,a32=ζ13k6+ζ23k7ω3ξ3exp⁡-s3T2and adding (189)–(191) with due account for (195) yield (196)2ξ˙1e-s1T2+ξ˙2e-s2T2+2ξ˙3e-s3T2=0,while subtracting (192) from the sum of (193) and the doubled (194) we obtain (197)δ˙=2φ˙3-φ˙1+φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000-ω1-1ζ1k5+ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k6+ζ23k7a1a2-ω1-1ζ1k6a2a32a1000000+ω2-1ζ2k7a1a32a2cos⁡δ,where 2Σ=2σ3-σ1+σ2.From (189)–(191) we could find that (198)-2ζ1k6ω1a2a32a1=ξ˙1ξ11sinδ,2ζ2k7ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k6+ζ23k7ω3a1a2=ξ˙3ξ31sinδ.Then (189), (191), and (197) could be rewritten in the following form: (199)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(200)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(201)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ100000000-ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T200000000+ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where(202)b=ζ13k6+ζ23k7ω3ζ1ζ2k6k7ω1ω2,K1=ζ1ζ13k6k7ω1ω3,K2=ζ2ζ23k7k8ω2ω3,K3=ζ13k6+ζ23k72ω3ζ13k2+ζ23k4ω3-ζ1k5ω1+ζ2k6ω2.The nonlinear set of equations (196), (199), (200), and (201) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω1-ω2 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (196) has the form (203)2ξ˙1+ξ˙2+2ξ˙3=0,whence it follows that (204)2ξ1+ξ2+2ξ3=E0,(205)ξ1=2E0c1-ξ,ξ2=2E0c2+ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (206)4c1+2c2+2c3=1.Considering (205), (199) and (201) are reduced to (207)ξ˙=2bE0c1-ξc2+ξc3+ξe-sT2sinδ,(208)δ˙=2E0K1c1-ξ+K2c2+ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1-ξ+K2c2+ξ+K3c3+ξe-sT2-bE0c1-ξc2+ξ-c3+ξc2+ξc1-ξ00000000+c3+ξc1-ξc2+ξe-sT2cos⁡δ,while (208), wherein its first term rapidly decaying with time has been neglected, takes the form (209)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (210)Gξ,δ=c3+ξc1-ξc2+ξcos⁡δ=G0ξ0,δ0.This first integral (210) defines the stream function G(ξ,δ) such that (211)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (207) and (210) and integrating over T2, we have (212)∫ξ0ξ‍dξc3+ξ2(c1-ξ)(c2+ξ)-G02=2bE0s1-e-sT2.The solution of (212) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (207) as (213)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1-ξc2+ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1-ξc2+ξc3+ξe-sT2,and retain all terms in (208) with due account for (205), then (208) could be rewritten as follows: (214)-dcos⁡δdξ+121c1-ξ-121c2+ξ-1c3+ξcos⁡δ=K1c1-ξbc3+ξc2+ξ+K2c2+ξbc3+ξc1-ξ+K3bc1-ξc2+ξ.Integrating (214) yields (215)G1ξ,δ=c3+ξc1-ξc2+ξcos⁡δ-12K1b-1c1-ξ2+12K2b-1c2+ξ2+12K3b-1c3+ξ2=G10ξ0,δ0.Note that relationships (211) and (212) are valid in this case as well.


#### 5.2.3. Combinational Resonance:2ω3=ω2-ω1

Now let us consider the third possible case of the combinational resonance (82), that is, when 2ω3=ω2-ω1.Then eliminating secular terms in (75), we obtain the following solvability equations: (216)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k8A2A-32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k5A-1A2A-3=0,reducing to six differential equations in amplitudes and phases of the three interacting modes:(217)a12·+s1a12=2ω1-1ζ1k8a1a2a32sinδ,(218)a22·+s2a22=-2ω2-1ζ2k5a1a2a32sinδ,(219)a32·+s3a32=ω3-1ζ13k8+ζ23k5a1a2a32sinδ,(220)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k8a1-1a2a32cos⁡δ,(221)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k5a1a2-1a32cos⁡δ,(222)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k8+ζ23k5a1a2cos⁡δ,where the phase difference has the form δ=2φ3+φ1-φ2.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (223)a12=ζ1k8ω1ξ1exp⁡-s1T2,a22=ζ2k5ω2ξ2exp⁡-s2T2,a32=ζ13k8+ζ23k5ω3ξ3exp⁡-s3T2,and adding (217)–(219) with due account for (223) yield (224)ξ˙1e-s1T2+2ξ˙2e-s2T2+2ξ˙3e-s3T2=0,while subtracting (221) from the sum of (220) and the doubled (222) we obtain (225)δ˙=2φ˙3+φ˙1-φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000+ω1-1ζ1k5-ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k8+ζ23k5a1a2+ω1-1ζ1k8a2a32a100000-ω2-1ζ2k5a1a32a2cos⁡δ,where 2Σ=2σ3+σ1-σ2.From (217)–(219) we could find that (226)2ζ1k8ω1a2a32a1=ξ˙1ξ11sinδ,-2ζ2k5ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k8+ζ23k5ω3a1a2=ξ˙3ξ31sinδ.Then (217), (219), and (225) could be rewritten in the following form: (227)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(228)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(229)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ100000000+ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T200000000-ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where (230)b=ζ13k8+ζ23k5ω3ζ1ζ2k5k8ω1ω2,K1=ζ1ζ13k7k8ω1ω3,K2=ζ2ζ23k5k8ω2ω3,K3=ζ13k8+ζ23k52ω3ζ13k2+ζ23k4ω3+ζ1k5ω1-ζ2k6ω2.The nonlinear set of equations (224), (227), (228), and (229) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω2-ω1 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (224) has the form (231)ξ˙1+2ξ˙2+2ξ˙3=0,whence it follows that (232)ξ1+2ξ2+2ξ3=E0,(233)ξ1=2E0c1+ξ,ξ2=2E0c2-ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (234)2c1+4c2+2c3=1.Considering (233), (227) and (229) are reduced to (235)ξ˙=2bE0c1+ξc2-ξc3+ξe-sT2sinδ,(236)δ˙=2E0K1c1+ξ+K2c2-ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1+ξ+K2c2-ξ+K3c3+ξe-sT2-bE0c1+ξc2-ξ+c3+ξc2-ξc1+ξ0000000000-c3+ξc1+ξc2-ξe-sT2cos⁡δ,while (236), wherein its first term rapidly decaying with time has been neglected, takes the form (237)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (238)Gξ,δ=c3+ξc1+ξc2-ξcos⁡δ=G0ξ0,δ0.This first integral (238) defines the stream function G(ξ,δ) such that (239)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (235) and (238) and integrating over T2, we have (240)∫ξ0ξ‍dξc3+ξ2c1+ξc2-ξ-G02=2bE0s1-e-sT2.The solution of (240) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (235) as (241)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1+ξc2-ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1+ξc2-ξc3+ξe-sT2and retain all terms in (236) with due account for (233), then (236) could be rewritten as follows: (242)-dcos⁡δdξ+-121c1+ξ+121c2-ξ-1c3+ξcos⁡δ=K1c1+ξbc3+ξc2-ξ+K2c2-ξbc3+ξc1+ξ+K3bc1+ξc2-ξ.Integrating (242) yields (243)G1ξ,δ=c3+ξc1+ξc2-ξcos⁡δ+12K1b-1c1+ξ2-12K2b-1c2-ξ2+12K3b-1c3+ξ2=G10(ξ0,δ0).Note that relationships (239) and (240) are valid in this case as well.


## 5.1. Internal Resonance 1:1

The one-to-one internal resonance could be of two types: when the natural frequencies of two modes of in-plane vibrations are close to each other, that is, whenω1=ω2=ω, or when the natural frequency of the out-of-plane mode is close to the natural frequency of one of the in-plane modes, that is, ω3=ω1=ω or ω3=ω2=ω.


### 5.1.1. Internal Resonanceω1=ω2=ω

Now let us first consider case (76) of the one-to-one internal resonance, when ω1=ω2=ω, but ω3≠ω.Then eliminating secular terms in (75), we obtain the following solvability equations: (98)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A3A-3=0,(99)2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A3A-3=0,(100)2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k5A1A-2A3+ζ13k6+ζ23k7A-1A2A3=0.Let us multiply (98)–(100), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that the functions Ai(T2) could be represented in the polar form (86), where ai and φi(i=1,2,3) are T2-dependent, as a result we have (101)a12·+s1a12=-2ω-1ζ1k6a1a2a32sinδ,(102)a22·+s2a22=2ω-1ζ2k5a1a2a32sinδ,(103)a32·+s3a32=-ω3-1ζ13k6-k8+ζ23k7-k5·a1a2a32sinδ,(104)φ˙1=12σ1+ω-1ζ1k5a32+ω-1ζ1k6a1-1a2a32cos⁡δ,(105)φ˙2=12σ2+ω-1ζ2k6a32+ω-1ζ2k5a1a2-1a32cos⁡δ,(106)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k6-k8+ζ23k7-k5·a1a2cos⁡δ,where the phase difference has the form δ=φ2-φ1 and an overdot denotes the differentiation with respect to T2.It could be shown that the expression in square brackets in (103) and (106) is vanished to zero when ω1=ω2=ω. Then (103) turns out to be uncoupled with respect to all other equations from (101) to (106): (107)a32·+s3a32=0,the solution of which has the form (108)a32=c3e-s3T2,where c3 is a constant of integration to be determined from the initial conditions, while (106) is reduced to (109)φ˙3=12σ3+12ω3-1·ζ13k7a12+ζ23k8a22+ζ13k2+ζ23k4a32,whence it follows that the phase φ3 is dependent only on the squared values of all amplitudes of the three interacting modes of vibrations.Introducing new functionsξ1(T2) and ξ2(T2) such that (110)a12=ζ1k6ωξ1exp⁡-s1T2,a22=ζ2k5ωξ2exp⁡-s2T2and adding (101) and (102) with due account for (110) yield (111)ξ˙2+ξ˙1e(s2-s1)T2=0,while subtracting (104) from (105) we obtain (112)δ˙=φ˙2-φ˙1=Σ+ω-1ζ2k6-ζ1k5a32+ω-1ζ2k5a12-ζ1k6a22a32a1a2cos⁡δ,where Σ=(1/2)(σ2-σ1).From (101) and (102) we could find that (113)a2a32a1=-12ωζ1k61sinδξ˙1ξ1,a1a32a2=12ωζ2k51sinδξ˙2ξ2.Considering (110) and (113), (112) could be reduced to (114)δ˙-Σ=Ke-s3T2+12ξ˙1ξ1+ξ˙2ξ2cotδ,where K=ω-1ζ2k6-ζ1k5c3.Equations (101) and (112) could be rewritten in another form if we substitute (110) and (108) in all its terms: (115)ξ˙1=-2bξ1ξ2e-(s3+1/2s2-1/2s1)T2sinδ,(116)δ˙-Σ=Ke-s3T2+bξ1e-s1T2-ξ2e-s2T2·cos⁡δξ1ξ2e-(s3+1/2s2+1/2s1)T2,where b=ω-1ζ1ζ2k5k6.The nonlinear set of equations (111), (115), and (116), or (114), with the initial conditions (117)ξ1T2=0=ξ10,ξ2T2=0=ξ20,ξ3T2=0=ξ30,δT2=0=δ0,completely describes the vibrational process of the mechanical system being investigated under the condition of the one-to-one internal resonance ω1=ω2 and could be solved numerically.It should be noted that the equation defining the stream function could be obtained from (114), if we neglect the first term in (114) which decays rather rapidly as compared with its last term.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (111) has the form (118)ξ˙2+ξ˙1=0,whence it follows that (119)ξ2+ξ1=E0,(120)ξ1=E0ξ,ξ2=E01-ξ,where E0 is the initial energy.Considering (120), (115) and (116), wherein the term rapidly decaying with time has been neglected, are reduced to the following: (121)ξ˙=-2bξ1-ξe-sT2sinδ,(122)δ˙=-b-ξ1/21-ξ-1/2+ξ-1/21-ξ1/2·cos⁡δe-sT2=-b1-2ξξ1-ξcos⁡δe-sT2,while (114), wherein its first term rapidly decaying with time has been neglected, takes the form (123)δ˙tanδ≈12ξ˙2ξ2+12ξ˙1ξ1,and it could be integrated, resulting in its first integral (124)Gξ,δ=ξ1/21-ξ1/2cos⁡δ=G0ξ0,δ0.The second first integral (124) defines the stream function G(ξ,δ) such that (125)vξ=ξ˙=2b∂G∂δe-sT2,vδ=δ˙=-2b∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (121) and (124) and integrating over T2, we have (126)∫ξ0ξ‍dξξ1-ξ-G02=2bse-sT2-1.The integral in the left-hand side of (126) can be transformed to an incomplete elliptic integral of the first kind and can be easily calculated using special tables [34].The solution of (126) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (121) as (127)δ˙=dδdT2=dδdξξ˙=-2bdδdξξ1/21-ξ1/2e-sT2sinδand retain all terms in (116) with due account for (120), then (116) could be rewritten as follows: (128)dcos⁡δdξ+121-2ξξ1-ξcos⁡δ=K2bξ1-ξ.Integrating (128) yields (129)G1ξ,δ=ξ1/21-ξ1/2cos⁡δ-12Kb-1ξ=G10ξ0,δ0.Note that relationships (125) and (126) are valid in this case as well.


### 5.1.2. Internal Resonanceω1=ω3=ω

Now let us consider case (77), when ω1=ω3=ω, but ω2≠ω. Note that case (78), when ω2=ω3=ω, but ω1≠ω, could be treated similarly.Then eliminating secular terms in (75), we obtain the following solvability equations: (130)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k7A-1A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k7A12A-3=0.Representing functionsAi entering in (130) in the polar form and applying the same procedure as it has been carried out above for the other case of the 1:1 internal resonance, as a result we have (131)a12·+s1a12=-2ω-1ζ1k7a12a32sinδ,(132)a22·+s2a22=0,(133)a32·+s3a32=ω-1ζ13k7a12a32sinδ,(134)φ˙1=12σ1+ω-1ζ1k5a32+ω-1ζ1k7a32cos⁡δ,(135)φ˙2=12σ2+ω2-1ζ2k6a32,(136)φ˙3=12σ3+12ω-1ζ13k7a12+12ω-1ζ23k8a22+12ω-1ζ13k2+ζ23k4a32+12ω-1ζ13k7a12cos⁡δ,where the phase difference has the form δ=2(φ3-φ1).Since (132) is independent of others from (131) to (136), its solution has the form (137)a22=c2e-s2T2,where c2 is a constant of integration to be determined from the initial conditions.Introducing new functionsξ1(T2) and ξ3(T2) such that (138)a12=ζ1k7ωξ1exp⁡-s1T2,a32=ζ13k7ωξ3exp⁡-s3T2and adding (131) and (133) with due account for (138) yield (139)ξ˙1e-s1T2+2ξ˙3e-s3T2=0,while subtracting (134) from (136) we obtain (140)δ˙=2φ˙3-φ˙1=Σ+ω-1ζ13k2+ζ23k4-2ζ1k5a32hhhhhhhhhhhhh+ω-1ζ13k7a12+ω-1ζ23k8a22hhhhhhhhhhhhh+ω-1k7ζ13a12-2ζ1a32cos⁡δ,where Σ=σ3-σ1.From (131) and (133) we could find that (141)-2ζ1k7ωa32=ξ˙1ξ11sinδ,ζ13k7ωa12=ξ˙3ξ31sinδ.Considering (138) and (141), (131) and (140) could be reduced to (142)ξ˙1=-2bξ1ξ3e-s3T2sinδ,δ˙-Σ=bξ1e-s1T2+K2e-s2T2+K3ξ3e-s3T2+ξ˙1ξ1+ξ˙3ξ3cotδ=bξ1e-s1T2+K2e-s2T2+K3ξ3e-s3T2+bξ1e-s1T2-2ξ3e-s3T2cos⁡δ,where (143)b=ω-2k72ζ1ζ13,K2=ω-1ζ23k8c2,K3=ω-2ζ13k7ζ13k2+ζ23k4-2ζ1k5.The nonlinear set of equations (139) and (142) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the one-to-one internal resonance ω1=ω3 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (139) has the form (144)ξ˙1+2ξ˙3=0,whence it follows that (145)ξ1+2ξ3=E0,(146)ξ1=E0ξ,ξ3=12E01-ξ.Considering (146), (142) are reduced to (147)ξ˙=-bE0ξ1-ξe-sT2sinδ,(148)δ˙=bE0ξ+K2+12K3E0(1-ξ)e-sT2+ξ˙1ξ1+ξ˙3ξ3cotδ=bE0ξ+K2+12K3(1-ξ)e-sT2-bE01-2ξe-sT2cos⁡δ,while (148), wherein its first term rapidly decaying with time has been neglected, takes the form (149)δ˙tanδ≈ξ˙1ξ1+ξ˙3ξ3,and it could be integrated, resulting in its first integral (150)Gξ,δ=ξ1-ξcos⁡δ=G0ξ0,δ0.This first integral (150) defines the stream function G(ξ,δ) such that (151)vξ=ξ˙=bE0∂G∂δe-sT2,vδ=δ˙=-bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (147) and (150) and integrating over T2, we have (152)∫ξ0ξ‍dξξ21-ξ2-G02=bE0se-sT2-1.The solution of (152) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (147) as (153)δ˙=dδdT2=dδdξξ˙=-bE0dδdξξ1-ξe-sT2sinδ=bE0dcos⁡δdξξ1-ξe-sT2and retain all terms in (148) with due account for (146), then (148) could be rewritten as follows: (154)dcos⁡δdξ+1-2ξξ1-ξcos⁡δ=K3bξ+11-ξ+K2bE0ξ1-ξ.Integrating (154) yields(155)G1ξ,δ=ξ1-ξcos⁡δ+12K3b-11-ξ2-12ξ2-K2bE0-1ξ=G10ξ0,δ0.Note that relationships (151) and (152) are valid in this case as well.Thus, during free vibrations of the plate being under the conditions of the one-to-one internal resonance three regimes can be observed: stationary (absence of damping atγ=0), quasi-stationary (damping is defined by an ordinary derivative at γ=1), and transient (damping is defined by a fractional derivative at 0<γ<1).


### 5.1.3. Internal Resonance 1:1:1

Now let us consider case (79), when ω1=ω2=ω3=ω.Then eliminating secular terms in (75), we obtain the following solvability equations: (156)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A3A-3+2ζ1k7A-1A32+2ζ1k8A-2A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A3A-3+2ζ2k7A-1A32+2ζ2k8A-2A32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k7A12A-3+ζ23k8A22A-3+ζ13k6+ζ23k7A-1A2A3+ζ13k8+ζ23k5A1A-2A3=0.Representing functionsAi entering in (151) in the polar form and applying the same procedure as it has been carried out above for the cases of the 1:1 internal resonance, as a result we have (157)a12·+s1a12=-2ω-1ζ1·k7a12sin2φ3-φ100000+k6a1a2sinφ2-φ100000+k8a1a2sin⁡2φ3-φ2-φ1a12a32,a22·+s2a22=-2ω-1ζ2·k8a22sin2φ3-φ200000-k5a1a2sin⁡φ2-φ100000+k7a1a2sin2φ3-φ2-φ1a22a32,a32·+s3a32=ω-1ζ13k7a12sin2φ3-φ10000000+ζ23k8a22sin2(φ3-φ2)a32,φ˙1=12σ1+ω-1ζ1·a1-1k5+k7cos⁡2φ3-φ10000+k6a1-1a2cos⁡φ2-φ10000+k8a1-1a2cos⁡(2φ3-φ2-φ1)a32,φ˙2=12σ2+ω2-1ζ2·a2-1k6+k8cos⁡2φ3-φ20000+k5a1a2-1cos⁡φ2-φ10000+k7a1a2-1cos⁡(2φ3-φ2-φ1)a32,φ˙3=12σ3+12ω-1·ζ13k7a12+ζ23k8a22+ζ13k2+ζ23k4a32+12ω-1ζ13k7a12cos⁡2φ3-φ100000000000+ζ23k8a22cos⁡2φ3-φ2-23ω-3ζ13ζ23a1a2cos⁡φ2-φ1.The nonlinear set of equations (157) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the internal resonance 1:1:1 and could be solved numerically.


## 5.1.1. Internal Resonanceω1=ω2=ω

Now let us first consider case (76) of the one-to-one internal resonance, when ω1=ω2=ω, but ω3≠ω.Then eliminating secular terms in (75), we obtain the following solvability equations: (98)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A3A-3=0,(99)2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A3A-3=0,(100)2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k5A1A-2A3+ζ13k6+ζ23k7A-1A2A3=0.Let us multiply (98)–(100), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that the functions Ai(T2) could be represented in the polar form (86), where ai and φi(i=1,2,3) are T2-dependent, as a result we have (101)a12·+s1a12=-2ω-1ζ1k6a1a2a32sinδ,(102)a22·+s2a22=2ω-1ζ2k5a1a2a32sinδ,(103)a32·+s3a32=-ω3-1ζ13k6-k8+ζ23k7-k5·a1a2a32sinδ,(104)φ˙1=12σ1+ω-1ζ1k5a32+ω-1ζ1k6a1-1a2a32cos⁡δ,(105)φ˙2=12σ2+ω-1ζ2k6a32+ω-1ζ2k5a1a2-1a32cos⁡δ,(106)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k6-k8+ζ23k7-k5·a1a2cos⁡δ,where the phase difference has the form δ=φ2-φ1 and an overdot denotes the differentiation with respect to T2.It could be shown that the expression in square brackets in (103) and (106) is vanished to zero when ω1=ω2=ω. Then (103) turns out to be uncoupled with respect to all other equations from (101) to (106): (107)a32·+s3a32=0,the solution of which has the form (108)a32=c3e-s3T2,where c3 is a constant of integration to be determined from the initial conditions, while (106) is reduced to (109)φ˙3=12σ3+12ω3-1·ζ13k7a12+ζ23k8a22+ζ13k2+ζ23k4a32,whence it follows that the phase φ3 is dependent only on the squared values of all amplitudes of the three interacting modes of vibrations.Introducing new functionsξ1(T2) and ξ2(T2) such that (110)a12=ζ1k6ωξ1exp⁡-s1T2,a22=ζ2k5ωξ2exp⁡-s2T2and adding (101) and (102) with due account for (110) yield (111)ξ˙2+ξ˙1e(s2-s1)T2=0,while subtracting (104) from (105) we obtain (112)δ˙=φ˙2-φ˙1=Σ+ω-1ζ2k6-ζ1k5a32+ω-1ζ2k5a12-ζ1k6a22a32a1a2cos⁡δ,where Σ=(1/2)(σ2-σ1).From (101) and (102) we could find that (113)a2a32a1=-12ωζ1k61sinδξ˙1ξ1,a1a32a2=12ωζ2k51sinδξ˙2ξ2.Considering (110) and (113), (112) could be reduced to (114)δ˙-Σ=Ke-s3T2+12ξ˙1ξ1+ξ˙2ξ2cotδ,where K=ω-1ζ2k6-ζ1k5c3.Equations (101) and (112) could be rewritten in another form if we substitute (110) and (108) in all its terms: (115)ξ˙1=-2bξ1ξ2e-(s3+1/2s2-1/2s1)T2sinδ,(116)δ˙-Σ=Ke-s3T2+bξ1e-s1T2-ξ2e-s2T2·cos⁡δξ1ξ2e-(s3+1/2s2+1/2s1)T2,where b=ω-1ζ1ζ2k5k6.The nonlinear set of equations (111), (115), and (116), or (114), with the initial conditions (117)ξ1T2=0=ξ10,ξ2T2=0=ξ20,ξ3T2=0=ξ30,δT2=0=δ0,completely describes the vibrational process of the mechanical system being investigated under the condition of the one-to-one internal resonance ω1=ω2 and could be solved numerically.It should be noted that the equation defining the stream function could be obtained from (114), if we neglect the first term in (114) which decays rather rapidly as compared with its last term.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (111) has the form (118)ξ˙2+ξ˙1=0,whence it follows that (119)ξ2+ξ1=E0,(120)ξ1=E0ξ,ξ2=E01-ξ,where E0 is the initial energy.Considering (120), (115) and (116), wherein the term rapidly decaying with time has been neglected, are reduced to the following: (121)ξ˙=-2bξ1-ξe-sT2sinδ,(122)δ˙=-b-ξ1/21-ξ-1/2+ξ-1/21-ξ1/2·cos⁡δe-sT2=-b1-2ξξ1-ξcos⁡δe-sT2,while (114), wherein its first term rapidly decaying with time has been neglected, takes the form (123)δ˙tanδ≈12ξ˙2ξ2+12ξ˙1ξ1,and it could be integrated, resulting in its first integral (124)Gξ,δ=ξ1/21-ξ1/2cos⁡δ=G0ξ0,δ0.The second first integral (124) defines the stream function G(ξ,δ) such that (125)vξ=ξ˙=2b∂G∂δe-sT2,vδ=δ˙=-2b∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (121) and (124) and integrating over T2, we have (126)∫ξ0ξ‍dξξ1-ξ-G02=2bse-sT2-1.The integral in the left-hand side of (126) can be transformed to an incomplete elliptic integral of the first kind and can be easily calculated using special tables [34].The solution of (126) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (121) as (127)δ˙=dδdT2=dδdξξ˙=-2bdδdξξ1/21-ξ1/2e-sT2sinδand retain all terms in (116) with due account for (120), then (116) could be rewritten as follows: (128)dcos⁡δdξ+121-2ξξ1-ξcos⁡δ=K2bξ1-ξ.Integrating (128) yields (129)G1ξ,δ=ξ1/21-ξ1/2cos⁡δ-12Kb-1ξ=G10ξ0,δ0.Note that relationships (125) and (126) are valid in this case as well.


## 5.1.2. Internal Resonanceω1=ω3=ω

Now let us consider case (77), when ω1=ω3=ω, but ω2≠ω. Note that case (78), when ω2=ω3=ω, but ω1≠ω, could be treated similarly.Then eliminating secular terms in (75), we obtain the following solvability equations: (130)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k7A-1A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k7A12A-3=0.Representing functionsAi entering in (130) in the polar form and applying the same procedure as it has been carried out above for the other case of the 1:1 internal resonance, as a result we have (131)a12·+s1a12=-2ω-1ζ1k7a12a32sinδ,(132)a22·+s2a22=0,(133)a32·+s3a32=ω-1ζ13k7a12a32sinδ,(134)φ˙1=12σ1+ω-1ζ1k5a32+ω-1ζ1k7a32cos⁡δ,(135)φ˙2=12σ2+ω2-1ζ2k6a32,(136)φ˙3=12σ3+12ω-1ζ13k7a12+12ω-1ζ23k8a22+12ω-1ζ13k2+ζ23k4a32+12ω-1ζ13k7a12cos⁡δ,where the phase difference has the form δ=2(φ3-φ1).Since (132) is independent of others from (131) to (136), its solution has the form (137)a22=c2e-s2T2,where c2 is a constant of integration to be determined from the initial conditions.Introducing new functionsξ1(T2) and ξ3(T2) such that (138)a12=ζ1k7ωξ1exp⁡-s1T2,a32=ζ13k7ωξ3exp⁡-s3T2and adding (131) and (133) with due account for (138) yield (139)ξ˙1e-s1T2+2ξ˙3e-s3T2=0,while subtracting (134) from (136) we obtain (140)δ˙=2φ˙3-φ˙1=Σ+ω-1ζ13k2+ζ23k4-2ζ1k5a32hhhhhhhhhhhhh+ω-1ζ13k7a12+ω-1ζ23k8a22hhhhhhhhhhhhh+ω-1k7ζ13a12-2ζ1a32cos⁡δ,where Σ=σ3-σ1.From (131) and (133) we could find that (141)-2ζ1k7ωa32=ξ˙1ξ11sinδ,ζ13k7ωa12=ξ˙3ξ31sinδ.Considering (138) and (141), (131) and (140) could be reduced to (142)ξ˙1=-2bξ1ξ3e-s3T2sinδ,δ˙-Σ=bξ1e-s1T2+K2e-s2T2+K3ξ3e-s3T2+ξ˙1ξ1+ξ˙3ξ3cotδ=bξ1e-s1T2+K2e-s2T2+K3ξ3e-s3T2+bξ1e-s1T2-2ξ3e-s3T2cos⁡δ,where (143)b=ω-2k72ζ1ζ13,K2=ω-1ζ23k8c2,K3=ω-2ζ13k7ζ13k2+ζ23k4-2ζ1k5.The nonlinear set of equations (139) and (142) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the one-to-one internal resonance ω1=ω3 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (139) has the form (144)ξ˙1+2ξ˙3=0,whence it follows that (145)ξ1+2ξ3=E0,(146)ξ1=E0ξ,ξ3=12E01-ξ.Considering (146), (142) are reduced to (147)ξ˙=-bE0ξ1-ξe-sT2sinδ,(148)δ˙=bE0ξ+K2+12K3E0(1-ξ)e-sT2+ξ˙1ξ1+ξ˙3ξ3cotδ=bE0ξ+K2+12K3(1-ξ)e-sT2-bE01-2ξe-sT2cos⁡δ,while (148), wherein its first term rapidly decaying with time has been neglected, takes the form (149)δ˙tanδ≈ξ˙1ξ1+ξ˙3ξ3,and it could be integrated, resulting in its first integral (150)Gξ,δ=ξ1-ξcos⁡δ=G0ξ0,δ0.This first integral (150) defines the stream function G(ξ,δ) such that (151)vξ=ξ˙=bE0∂G∂δe-sT2,vδ=δ˙=-bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (147) and (150) and integrating over T2, we have (152)∫ξ0ξ‍dξξ21-ξ2-G02=bE0se-sT2-1.The solution of (152) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (147) as (153)δ˙=dδdT2=dδdξξ˙=-bE0dδdξξ1-ξe-sT2sinδ=bE0dcos⁡δdξξ1-ξe-sT2and retain all terms in (148) with due account for (146), then (148) could be rewritten as follows: (154)dcos⁡δdξ+1-2ξξ1-ξcos⁡δ=K3bξ+11-ξ+K2bE0ξ1-ξ.Integrating (154) yields(155)G1ξ,δ=ξ1-ξcos⁡δ+12K3b-11-ξ2-12ξ2-K2bE0-1ξ=G10ξ0,δ0.Note that relationships (151) and (152) are valid in this case as well.Thus, during free vibrations of the plate being under the conditions of the one-to-one internal resonance three regimes can be observed: stationary (absence of damping atγ=0), quasi-stationary (damping is defined by an ordinary derivative at γ=1), and transient (damping is defined by a fractional derivative at 0<γ<1).


## 5.1.3. Internal Resonance 1:1:1

Now let us consider case (79), when ω1=ω2=ω3=ω.Then eliminating secular terms in (75), we obtain the following solvability equations: (156)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A3A-3+2ζ1k7A-1A32+2ζ1k8A-2A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A3A-3+2ζ2k7A-1A32+2ζ2k8A-2A32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k7A12A-3+ζ23k8A22A-3+ζ13k6+ζ23k7A-1A2A3+ζ13k8+ζ23k5A1A-2A3=0.Representing functionsAi entering in (151) in the polar form and applying the same procedure as it has been carried out above for the cases of the 1:1 internal resonance, as a result we have (157)a12·+s1a12=-2ω-1ζ1·k7a12sin2φ3-φ100000+k6a1a2sinφ2-φ100000+k8a1a2sin⁡2φ3-φ2-φ1a12a32,a22·+s2a22=-2ω-1ζ2·k8a22sin2φ3-φ200000-k5a1a2sin⁡φ2-φ100000+k7a1a2sin2φ3-φ2-φ1a22a32,a32·+s3a32=ω-1ζ13k7a12sin2φ3-φ10000000+ζ23k8a22sin2(φ3-φ2)a32,φ˙1=12σ1+ω-1ζ1·a1-1k5+k7cos⁡2φ3-φ10000+k6a1-1a2cos⁡φ2-φ10000+k8a1-1a2cos⁡(2φ3-φ2-φ1)a32,φ˙2=12σ2+ω2-1ζ2·a2-1k6+k8cos⁡2φ3-φ20000+k5a1a2-1cos⁡φ2-φ10000+k7a1a2-1cos⁡(2φ3-φ2-φ1)a32,φ˙3=12σ3+12ω-1·ζ13k7a12+ζ23k8a22+ζ13k2+ζ23k4a32+12ω-1ζ13k7a12cos⁡2φ3-φ100000000000+ζ23k8a22cos⁡2φ3-φ2-23ω-3ζ13ζ23a1a2cos⁡φ2-φ1.The nonlinear set of equations (157) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the internal resonance 1:1:1 and could be solved numerically.


## 5.2. Combinational Resonances of the Additive-Difference Type

Reference to (75) shows that three cases of the combinational internal resonance of the additive-difference type (80)–(82) could occur for the plate under consideration.


### 5.2.1. Combinational Resonance:2ω3=ω1+ω2

Let us first consider case (81), when 2ω3=ω1+ω2.Then eliminating secular terms in (75), we obtain the following solvability equations:(158)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k8A-2A32=0,(159)2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k7A-1A32=0,(160)2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k7A1A2A-3=0.Let us multiply (158)–(160), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that the functions Ai(T2) could be represented in the polar form (86), where ai and φi(i=1,2,3) are T2-dependent, as a result we have (161)a12·+s1a12=-2ω1-1ζ1k8a1a2a32sinδ,(162)a22·+s2a22=-2ω2-1ζ2k7a1a2a32sinδ,(163)a32·+s3a32=ω3-1ζ13k8+ζ23k7a1a2a32sinδ,(164)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k8a1-1a2a32cos⁡δ,(165)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k7a1a2-1a32cos⁡δ,(166)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k8+ζ23k7a1a2cos⁡δ,where the phase difference has the form δ=2φ3-φ2-φ1.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (167)a12=ζ1k8ω1ξ1exp⁡-s1T2,a22=ζ2k7ω2ξ2exp⁡-s2T2,a32=ζ13k8+ζ23k7ω3ξ3exp⁡-s3T2and adding (161)–(163) with due account for (167) yield (168)ξ˙1e-s1T2+ξ˙2e-s2T2+4ξ˙3e-s3T2=0,while subtracting (164) and (165) from the doubled (166) we obtain (169)δ˙=2φ˙3-φ˙1-φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000-ω1-1ζ1k5-ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k8+ζ23k7a1a2-ω1-1ζ1k8a2a32a1000000-ω2-1ζ2k7a1a32a2cos⁡δ,where 2Σ=2σ3-σ1-σ2.From (161)–(163) we could find that (170)-2ζ1k8ω1a2a32a1=ξ˙1ξ11sinδ,-2ζ2k7ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k8+ζ23k7ω3a1a2=ξ˙3ξ31sinδ.Then (161), (163), and (169) could be rewritten in the following form: (171)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(172)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(173)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ10000000-ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T20000000-ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where (174)b=ζ13k8+ζ23k7ω3ζ1ζ2k7k8ω1ω2,K1=ζ1ζ13k7k8ω1ω3,K2=ζ2ζ23k7k8ω2ω3,K3=ζ13k8+ζ23k72ω3ζ13k2+ζ23k4ω3-ζ1k5ω1-ζ2k6ω2.The nonlinear set of equations (168), (171), (172), and (173) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω1+ω2 and could be solved numerically. This case has been discussed in [35].Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (168) has the form (175)ξ˙1+ξ˙2+4ξ˙3=0,whence it follows that (176)ξ1+ξ2+4ξ3=E0,(177)ξ1=2E0c1-ξ,ξ2=2E0c2-ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (178)2c1+2c2+4c3=1.Considering (177), (171) and (173) are reduced to (179)ξ˙=2bE0(c1-ξ)(c2-ξ)(c3+ξ)e-sT2sinδ,(180)δ˙=2E0K1c1-ξ+K2c2-ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1-ξ+K2c2-ξ+K3c3+ξe-sT2-bE0c1-ξc2-ξ-c3+ξc2-ξc1-ξ000000000-c3+ξc1-ξc2-ξe-sT2cos⁡δ,while (180), wherein its first term rapidly decaying with time has been neglected, takes the form (181)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (182)Gξ,δ=c3+ξc1-ξc2-ξcos⁡δ=G0ξ0,δ0.This first integral (182) defines the stream function G(ξ,δ) such that (183)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (179) and (182) and integrating over T2, we have (184)∫ξ0ξ‍dξc3+ξ2c1-ξc2-ξ-G02=2bE0s1-e-sT2.The solution of (184) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (179) as(185)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1-ξc2-ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1-ξc2-ξc3+ξe-sT2and retain all terms in (180) with due account for (177), then (180) could be rewritten as follows: (186)-dcos⁡δdξ+121c1-ξ+121c2-ξ-1c3+ξcos⁡δ=K1c1-ξbc3+ξc2-ξ+K2c2-ξbc3+ξc1-ξ+K3bc1-ξc2-ξ.Integrating (186) yields (187)G1ξ,δ=c3+ξc1-ξc2-ξcos⁡δ-12K1b-1c1-ξ2-12K2b-1c2-ξ2+12K3b-1c3+ξ2=G10ξ0,δ0.Note that relationships (183) and (184) are valid in this case as well.


### 5.2.2. Combinational Resonance:2ω3=ω1−ω2

Now let us consider case (80), when 2ω3=ω1-ω2.Then eliminating secular terms in (75), we obtain the following solvability equations:(188)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k7A1A-32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k6+ζ23k7A1A-2A-3=0.Applying to (188) the same procedure as it has been done above for (158)–(160), as a result we have (189)a12·+s1a12=-2ω1-1ζ1k6a1a2a32sinδ,(190)a22·+s2a22=2ω2-1ζ2k7a1a2a32sinδ,(191)a32·+s3a32=ω3-1ζ13k6+ζ23k7a1a2a32sinδ,(192)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k6a1-1a2a32cos⁡δ,(193)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k7a1a2-1a32cos⁡δ,(194)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k6+ζ23k7a1a2cos⁡δ,where the phase difference has the form δ=2φ3+φ2-φ1.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (195)a12=ζ1k6ω1ξ1exp⁡-s1T2,a22=ζ2k7ω2ξ2exp⁡-s2T2,a32=ζ13k6+ζ23k7ω3ξ3exp⁡-s3T2and adding (189)–(191) with due account for (195) yield (196)2ξ˙1e-s1T2+ξ˙2e-s2T2+2ξ˙3e-s3T2=0,while subtracting (192) from the sum of (193) and the doubled (194) we obtain (197)δ˙=2φ˙3-φ˙1+φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000-ω1-1ζ1k5+ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k6+ζ23k7a1a2-ω1-1ζ1k6a2a32a1000000+ω2-1ζ2k7a1a32a2cos⁡δ,where 2Σ=2σ3-σ1+σ2.From (189)–(191) we could find that (198)-2ζ1k6ω1a2a32a1=ξ˙1ξ11sinδ,2ζ2k7ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k6+ζ23k7ω3a1a2=ξ˙3ξ31sinδ.Then (189), (191), and (197) could be rewritten in the following form: (199)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(200)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(201)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ100000000-ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T200000000+ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where(202)b=ζ13k6+ζ23k7ω3ζ1ζ2k6k7ω1ω2,K1=ζ1ζ13k6k7ω1ω3,K2=ζ2ζ23k7k8ω2ω3,K3=ζ13k6+ζ23k72ω3ζ13k2+ζ23k4ω3-ζ1k5ω1+ζ2k6ω2.The nonlinear set of equations (196), (199), (200), and (201) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω1-ω2 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (196) has the form (203)2ξ˙1+ξ˙2+2ξ˙3=0,whence it follows that (204)2ξ1+ξ2+2ξ3=E0,(205)ξ1=2E0c1-ξ,ξ2=2E0c2+ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (206)4c1+2c2+2c3=1.Considering (205), (199) and (201) are reduced to (207)ξ˙=2bE0c1-ξc2+ξc3+ξe-sT2sinδ,(208)δ˙=2E0K1c1-ξ+K2c2+ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1-ξ+K2c2+ξ+K3c3+ξe-sT2-bE0c1-ξc2+ξ-c3+ξc2+ξc1-ξ00000000+c3+ξc1-ξc2+ξe-sT2cos⁡δ,while (208), wherein its first term rapidly decaying with time has been neglected, takes the form (209)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (210)Gξ,δ=c3+ξc1-ξc2+ξcos⁡δ=G0ξ0,δ0.This first integral (210) defines the stream function G(ξ,δ) such that (211)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (207) and (210) and integrating over T2, we have (212)∫ξ0ξ‍dξc3+ξ2(c1-ξ)(c2+ξ)-G02=2bE0s1-e-sT2.The solution of (212) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (207) as (213)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1-ξc2+ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1-ξc2+ξc3+ξe-sT2,and retain all terms in (208) with due account for (205), then (208) could be rewritten as follows: (214)-dcos⁡δdξ+121c1-ξ-121c2+ξ-1c3+ξcos⁡δ=K1c1-ξbc3+ξc2+ξ+K2c2+ξbc3+ξc1-ξ+K3bc1-ξc2+ξ.Integrating (214) yields (215)G1ξ,δ=c3+ξc1-ξc2+ξcos⁡δ-12K1b-1c1-ξ2+12K2b-1c2+ξ2+12K3b-1c3+ξ2=G10ξ0,δ0.Note that relationships (211) and (212) are valid in this case as well.


### 5.2.3. Combinational Resonance:2ω3=ω2-ω1

Now let us consider the third possible case of the combinational resonance (82), that is, when 2ω3=ω2-ω1.Then eliminating secular terms in (75), we obtain the following solvability equations: (216)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k8A2A-32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k5A-1A2A-3=0,reducing to six differential equations in amplitudes and phases of the three interacting modes:(217)a12·+s1a12=2ω1-1ζ1k8a1a2a32sinδ,(218)a22·+s2a22=-2ω2-1ζ2k5a1a2a32sinδ,(219)a32·+s3a32=ω3-1ζ13k8+ζ23k5a1a2a32sinδ,(220)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k8a1-1a2a32cos⁡δ,(221)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k5a1a2-1a32cos⁡δ,(222)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k8+ζ23k5a1a2cos⁡δ,where the phase difference has the form δ=2φ3+φ1-φ2.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (223)a12=ζ1k8ω1ξ1exp⁡-s1T2,a22=ζ2k5ω2ξ2exp⁡-s2T2,a32=ζ13k8+ζ23k5ω3ξ3exp⁡-s3T2,and adding (217)–(219) with due account for (223) yield (224)ξ˙1e-s1T2+2ξ˙2e-s2T2+2ξ˙3e-s3T2=0,while subtracting (221) from the sum of (220) and the doubled (222) we obtain (225)δ˙=2φ˙3+φ˙1-φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000+ω1-1ζ1k5-ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k8+ζ23k5a1a2+ω1-1ζ1k8a2a32a100000-ω2-1ζ2k5a1a32a2cos⁡δ,where 2Σ=2σ3+σ1-σ2.From (217)–(219) we could find that (226)2ζ1k8ω1a2a32a1=ξ˙1ξ11sinδ,-2ζ2k5ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k8+ζ23k5ω3a1a2=ξ˙3ξ31sinδ.Then (217), (219), and (225) could be rewritten in the following form: (227)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(228)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(229)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ100000000+ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T200000000-ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where (230)b=ζ13k8+ζ23k5ω3ζ1ζ2k5k8ω1ω2,K1=ζ1ζ13k7k8ω1ω3,K2=ζ2ζ23k5k8ω2ω3,K3=ζ13k8+ζ23k52ω3ζ13k2+ζ23k4ω3+ζ1k5ω1-ζ2k6ω2.The nonlinear set of equations (224), (227), (228), and (229) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω2-ω1 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (224) has the form (231)ξ˙1+2ξ˙2+2ξ˙3=0,whence it follows that (232)ξ1+2ξ2+2ξ3=E0,(233)ξ1=2E0c1+ξ,ξ2=2E0c2-ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (234)2c1+4c2+2c3=1.Considering (233), (227) and (229) are reduced to (235)ξ˙=2bE0c1+ξc2-ξc3+ξe-sT2sinδ,(236)δ˙=2E0K1c1+ξ+K2c2-ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1+ξ+K2c2-ξ+K3c3+ξe-sT2-bE0c1+ξc2-ξ+c3+ξc2-ξc1+ξ0000000000-c3+ξc1+ξc2-ξe-sT2cos⁡δ,while (236), wherein its first term rapidly decaying with time has been neglected, takes the form (237)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (238)Gξ,δ=c3+ξc1+ξc2-ξcos⁡δ=G0ξ0,δ0.This first integral (238) defines the stream function G(ξ,δ) such that (239)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (235) and (238) and integrating over T2, we have (240)∫ξ0ξ‍dξc3+ξ2c1+ξc2-ξ-G02=2bE0s1-e-sT2.The solution of (240) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (235) as (241)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1+ξc2-ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1+ξc2-ξc3+ξe-sT2and retain all terms in (236) with due account for (233), then (236) could be rewritten as follows: (242)-dcos⁡δdξ+-121c1+ξ+121c2-ξ-1c3+ξcos⁡δ=K1c1+ξbc3+ξc2-ξ+K2c2-ξbc3+ξc1+ξ+K3bc1+ξc2-ξ.Integrating (242) yields (243)G1ξ,δ=c3+ξc1+ξc2-ξcos⁡δ+12K1b-1c1+ξ2-12K2b-1c2-ξ2+12K3b-1c3+ξ2=G10(ξ0,δ0).Note that relationships (239) and (240) are valid in this case as well.


## 5.2.1. Combinational Resonance:2ω3=ω1+ω2

Let us first consider case (81), when 2ω3=ω1+ω2.Then eliminating secular terms in (75), we obtain the following solvability equations:(158)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k8A-2A32=0,(159)2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k7A-1A32=0,(160)2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k7A1A2A-3=0.Let us multiply (158)–(160), respectively, by A-1, A-2, and A-3 and find their complex conjugates. Adding every pair of the mutually adjoint equations with each other and subtracting one from another and considering that the functions Ai(T2) could be represented in the polar form (86), where ai and φi(i=1,2,3) are T2-dependent, as a result we have (161)a12·+s1a12=-2ω1-1ζ1k8a1a2a32sinδ,(162)a22·+s2a22=-2ω2-1ζ2k7a1a2a32sinδ,(163)a32·+s3a32=ω3-1ζ13k8+ζ23k7a1a2a32sinδ,(164)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k8a1-1a2a32cos⁡δ,(165)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k7a1a2-1a32cos⁡δ,(166)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k8+ζ23k7a1a2cos⁡δ,where the phase difference has the form δ=2φ3-φ2-φ1.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (167)a12=ζ1k8ω1ξ1exp⁡-s1T2,a22=ζ2k7ω2ξ2exp⁡-s2T2,a32=ζ13k8+ζ23k7ω3ξ3exp⁡-s3T2and adding (161)–(163) with due account for (167) yield (168)ξ˙1e-s1T2+ξ˙2e-s2T2+4ξ˙3e-s3T2=0,while subtracting (164) and (165) from the doubled (166) we obtain (169)δ˙=2φ˙3-φ˙1-φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000-ω1-1ζ1k5-ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k8+ζ23k7a1a2-ω1-1ζ1k8a2a32a1000000-ω2-1ζ2k7a1a32a2cos⁡δ,where 2Σ=2σ3-σ1-σ2.From (161)–(163) we could find that (170)-2ζ1k8ω1a2a32a1=ξ˙1ξ11sinδ,-2ζ2k7ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k8+ζ23k7ω3a1a2=ξ˙3ξ31sinδ.Then (161), (163), and (169) could be rewritten in the following form: (171)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(172)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(173)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ10000000-ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T20000000-ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where (174)b=ζ13k8+ζ23k7ω3ζ1ζ2k7k8ω1ω2,K1=ζ1ζ13k7k8ω1ω3,K2=ζ2ζ23k7k8ω2ω3,K3=ζ13k8+ζ23k72ω3ζ13k2+ζ23k4ω3-ζ1k5ω1-ζ2k6ω2.The nonlinear set of equations (168), (171), (172), and (173) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω1+ω2 and could be solved numerically. This case has been discussed in [35].Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (168) has the form (175)ξ˙1+ξ˙2+4ξ˙3=0,whence it follows that (176)ξ1+ξ2+4ξ3=E0,(177)ξ1=2E0c1-ξ,ξ2=2E0c2-ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (178)2c1+2c2+4c3=1.Considering (177), (171) and (173) are reduced to (179)ξ˙=2bE0(c1-ξ)(c2-ξ)(c3+ξ)e-sT2sinδ,(180)δ˙=2E0K1c1-ξ+K2c2-ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1-ξ+K2c2-ξ+K3c3+ξe-sT2-bE0c1-ξc2-ξ-c3+ξc2-ξc1-ξ000000000-c3+ξc1-ξc2-ξe-sT2cos⁡δ,while (180), wherein its first term rapidly decaying with time has been neglected, takes the form (181)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (182)Gξ,δ=c3+ξc1-ξc2-ξcos⁡δ=G0ξ0,δ0.This first integral (182) defines the stream function G(ξ,δ) such that (183)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (179) and (182) and integrating over T2, we have (184)∫ξ0ξ‍dξc3+ξ2c1-ξc2-ξ-G02=2bE0s1-e-sT2.The solution of (184) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (179) as(185)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1-ξc2-ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1-ξc2-ξc3+ξe-sT2and retain all terms in (180) with due account for (177), then (180) could be rewritten as follows: (186)-dcos⁡δdξ+121c1-ξ+121c2-ξ-1c3+ξcos⁡δ=K1c1-ξbc3+ξc2-ξ+K2c2-ξbc3+ξc1-ξ+K3bc1-ξc2-ξ.Integrating (186) yields (187)G1ξ,δ=c3+ξc1-ξc2-ξcos⁡δ-12K1b-1c1-ξ2-12K2b-1c2-ξ2+12K3b-1c3+ξ2=G10ξ0,δ0.Note that relationships (183) and (184) are valid in this case as well.


## 5.2.2. Combinational Resonance:2ω3=ω1−ω2

Now let us consider case (80), when 2ω3=ω1-ω2.Then eliminating secular terms in (75), we obtain the following solvability equations:(188)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k6A2A32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k7A1A-32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k6+ζ23k7A1A-2A-3=0.Applying to (188) the same procedure as it has been done above for (158)–(160), as a result we have (189)a12·+s1a12=-2ω1-1ζ1k6a1a2a32sinδ,(190)a22·+s2a22=2ω2-1ζ2k7a1a2a32sinδ,(191)a32·+s3a32=ω3-1ζ13k6+ζ23k7a1a2a32sinδ,(192)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k6a1-1a2a32cos⁡δ,(193)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k7a1a2-1a32cos⁡δ,(194)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k6+ζ23k7a1a2cos⁡δ,where the phase difference has the form δ=2φ3+φ2-φ1.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (195)a12=ζ1k6ω1ξ1exp⁡-s1T2,a22=ζ2k7ω2ξ2exp⁡-s2T2,a32=ζ13k6+ζ23k7ω3ξ3exp⁡-s3T2and adding (189)–(191) with due account for (195) yield (196)2ξ˙1e-s1T2+ξ˙2e-s2T2+2ξ˙3e-s3T2=0,while subtracting (192) from the sum of (193) and the doubled (194) we obtain (197)δ˙=2φ˙3-φ˙1+φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000-ω1-1ζ1k5+ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k6+ζ23k7a1a2-ω1-1ζ1k6a2a32a1000000+ω2-1ζ2k7a1a32a2cos⁡δ,where 2Σ=2σ3-σ1+σ2.From (189)–(191) we could find that (198)-2ζ1k6ω1a2a32a1=ξ˙1ξ11sinδ,2ζ2k7ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k6+ζ23k7ω3a1a2=ξ˙3ξ31sinδ.Then (189), (191), and (197) could be rewritten in the following form: (199)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(200)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(201)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ100000000-ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T200000000+ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where(202)b=ζ13k6+ζ23k7ω3ζ1ζ2k6k7ω1ω2,K1=ζ1ζ13k6k7ω1ω3,K2=ζ2ζ23k7k8ω2ω3,K3=ζ13k6+ζ23k72ω3ζ13k2+ζ23k4ω3-ζ1k5ω1+ζ2k6ω2.The nonlinear set of equations (196), (199), (200), and (201) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω1-ω2 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (196) has the form (203)2ξ˙1+ξ˙2+2ξ˙3=0,whence it follows that (204)2ξ1+ξ2+2ξ3=E0,(205)ξ1=2E0c1-ξ,ξ2=2E0c2+ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (206)4c1+2c2+2c3=1.Considering (205), (199) and (201) are reduced to (207)ξ˙=2bE0c1-ξc2+ξc3+ξe-sT2sinδ,(208)δ˙=2E0K1c1-ξ+K2c2+ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1-ξ+K2c2+ξ+K3c3+ξe-sT2-bE0c1-ξc2+ξ-c3+ξc2+ξc1-ξ00000000+c3+ξc1-ξc2+ξe-sT2cos⁡δ,while (208), wherein its first term rapidly decaying with time has been neglected, takes the form (209)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (210)Gξ,δ=c3+ξc1-ξc2+ξcos⁡δ=G0ξ0,δ0.This first integral (210) defines the stream function G(ξ,δ) such that (211)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (207) and (210) and integrating over T2, we have (212)∫ξ0ξ‍dξc3+ξ2(c1-ξ)(c2+ξ)-G02=2bE0s1-e-sT2.The solution of (212) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (207) as (213)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1-ξc2+ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1-ξc2+ξc3+ξe-sT2,and retain all terms in (208) with due account for (205), then (208) could be rewritten as follows: (214)-dcos⁡δdξ+121c1-ξ-121c2+ξ-1c3+ξcos⁡δ=K1c1-ξbc3+ξc2+ξ+K2c2+ξbc3+ξc1-ξ+K3bc1-ξc2+ξ.Integrating (214) yields (215)G1ξ,δ=c3+ξc1-ξc2+ξcos⁡δ-12K1b-1c1-ξ2+12K2b-1c2+ξ2+12K3b-1c3+ξ2=G10ξ0,δ0.Note that relationships (211) and (212) are valid in this case as well.


## 5.2.3. Combinational Resonance:2ω3=ω2-ω1

Now let us consider the third possible case of the combinational resonance (82), that is, when 2ω3=ω2-ω1.Then eliminating secular terms in (75), we obtain the following solvability equations: (216)2iω1D2A1+μ1iω1τ1γA1+2ζ1k5A1A3A-3+2ζ1k8A2A-32=0,2iω2D2A2+μ2iω2τ2γA2+2ζ2k6A2A3A-3+2ζ2k5A1A32=0,2iω3D2A3+μ3iω3τ3γA3+ζ13k2+ζ23k4A32A-3+ζ13k7A1A-1A3+ζ23k8A2A-2A3+ζ13k8+ζ23k5A-1A2A-3=0,reducing to six differential equations in amplitudes and phases of the three interacting modes:(217)a12·+s1a12=2ω1-1ζ1k8a1a2a32sinδ,(218)a22·+s2a22=-2ω2-1ζ2k5a1a2a32sinδ,(219)a32·+s3a32=ω3-1ζ13k8+ζ23k5a1a2a32sinδ,(220)φ˙1=12σ1+ω1-1ζ1k5a32+ω1-1ζ1k8a1-1a2a32cos⁡δ,(221)φ˙2=12σ2+ω2-1ζ2k6a32+ω2-1ζ2k5a1a2-1a32cos⁡δ,(222)φ˙3=12σ3+12ω3-1ζ13k7a12+12ω3-1ζ23k8a22+12ω3-1ζ13k2+ζ23k4a32+12ω3-1ζ13k8+ζ23k5a1a2cos⁡δ,where the phase difference has the form δ=2φ3+φ1-φ2.Introducing new functionsξ1(T2), ξ2(T2), and ξ3(T2) such that (223)a12=ζ1k8ω1ξ1exp⁡-s1T2,a22=ζ2k5ω2ξ2exp⁡-s2T2,a32=ζ13k8+ζ23k5ω3ξ3exp⁡-s3T2,and adding (217)–(219) with due account for (223) yield (224)ξ˙1e-s1T2+2ξ˙2e-s2T2+2ξ˙3e-s3T2=0,while subtracting (221) from the sum of (220) and the doubled (222) we obtain (225)δ˙=2φ˙3+φ˙1-φ˙2=Σ+ω3-1ζ13k2+ζ23k40000000+ω1-1ζ1k5-ω2-1ζ2k6a32+ω3-1ζ13k7a12+ω3-1ζ23k8a22+ω3-1ζ13k8+ζ23k5a1a2+ω1-1ζ1k8a2a32a100000-ω2-1ζ2k5a1a32a2cos⁡δ,where 2Σ=2σ3+σ1-σ2.From (217)–(219) we could find that (226)2ζ1k8ω1a2a32a1=ξ˙1ξ11sinδ,-2ζ2k5ω2a1a32a2=ξ˙2ξ21sinδ,ζ13k8+ζ23k5ω3a1a2=ξ˙3ξ31sinδ.Then (217), (219), and (225) could be rewritten in the following form: (227)ξ˙1=-2bξ1ξ2ξ3e-(s3+1/2s2-1/2s1)T2sinδ,(228)ξ˙3=bξ1ξ2ξ3e-1/2(s2+s1)T2sinδ,(229)δ˙-Σ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=K1ξ1e-s1T2+K2ξ2e-s2T2+K3ξ3e-s3T2+bξ1ξ2e-1/2(s1+s2)T2ξ3ξ2ξ100000000+ξ3ξ2ξ1e-(s3+1/2s2-1/2s1)T200000000-ξ3ξ1ξ2e-(s3+1/2s1-1/2s2)T2cos⁡δ,where (230)b=ζ13k8+ζ23k5ω3ζ1ζ2k5k8ω1ω2,K1=ζ1ζ13k7k8ω1ω3,K2=ζ2ζ23k5k8ω2ω3,K3=ζ13k8+ζ23k52ω3ζ13k2+ζ23k4ω3+ζ1k5ω1-ζ2k6ω2.The nonlinear set of equations (224), (227), (228), and (229) with the initial conditions (117) completely describes the vibrational process of the mechanical system being investigated under the condition of the combinational internal resonance 2ω3=ω2-ω1 and could be solved numerically.Particular Case. In the particular case at Σ=0 and s1=s2=s3=s, (224) has the form (231)ξ˙1+2ξ˙2+2ξ˙3=0,whence it follows that (232)ξ1+2ξ2+2ξ3=E0,(233)ξ1=2E0c1+ξ,ξ2=2E0c2-ξ,ξ3=E0c3+ξ,where c1, c2, and c3 are constants of integration such that (234)2c1+4c2+2c3=1.Considering (233), (227) and (229) are reduced to (235)ξ˙=2bE0c1+ξc2-ξc3+ξe-sT2sinδ,(236)δ˙=2E0K1c1+ξ+K2c2-ξ+K3c3+ξe-sT2+12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3cotδ=2E0K1c1+ξ+K2c2-ξ+K3c3+ξe-sT2-bE0c1+ξc2-ξ+c3+ξc2-ξc1+ξ0000000000-c3+ξc1+ξc2-ξe-sT2cos⁡δ,while (236), wherein its first term rapidly decaying with time has been neglected, takes the form (237)δ˙tanδ≈12ξ˙1ξ1+12ξ˙2ξ2+ξ˙3ξ3,and it could be integrated, resulting in its first integral (238)Gξ,δ=c3+ξc1+ξc2-ξcos⁡δ=G0ξ0,δ0.This first integral (238) defines the stream function G(ξ,δ) such that (239)vξ=ξ˙=-2bE0∂G∂δe-sT2,vδ=δ˙=2bE0∂G∂ξe-sT2,which describes steady-state vibrations of an elastic plate decaying with time.Eliminating the variableδ from (235) and (238) and integrating over T2, we have (240)∫ξ0ξ‍dξc3+ξ2c1+ξc2-ξ-G02=2bE0s1-e-sT2.The solution of (240) allows one to find the value ξ(T2) and, thus, to solve the problem under consideration.If we representδ˙ considering (235) as (241)δ˙=dδdT2=dδdξξ˙=2bE0dδdξc1+ξc2-ξc3+ξe-sT2sinδ=-2bE0dcos⁡δdξc1+ξc2-ξc3+ξe-sT2and retain all terms in (236) with due account for (233), then (236) could be rewritten as follows: (242)-dcos⁡δdξ+-121c1+ξ+121c2-ξ-1c3+ξcos⁡δ=K1c1+ξbc3+ξc2-ξ+K2c2-ξbc3+ξc1+ξ+K3bc1+ξc2-ξ.Integrating (242) yields (243)G1ξ,δ=c3+ξc1+ξc2-ξcos⁡δ+12K1b-1c1+ξ2-12K2b-1c2-ξ2+12K3b-1c3+ξ2=G10(ξ0,δ0).Note that relationships (239) and (240) are valid in this case as well.


## 6. Numerical Investigations

As examples, let us carry out the qualitative analysis of the cases of one-to-one internal resonance (76), when two different modes of in-plane vibrations are coupled, and (77), when one mode of in-plane vibrations is coupled with a certain mode of out-of-plane vibrations.


### 6.1. One-to-One Internal Resonanceω1=ω2 (76)

For this case, the stream functionG(ξ,δ) is defined by relationship (129), and the phase portrait to be constructed according to (129) depends essentially on the magnitudes of the coefficient K.


#### 6.1.1. Case (76) at K=0

For this case, the stream functionG(ξ,δ) is constructed according to (124) and the streamlines of the phase fluid in the phase plane ξ-δ are presented in Figure 2. Magnitudes of G are indicated by digits near the curves which correspond to the streamlines; the flow directions of the phase fluid elements are shown by arrows on the streamlines.Figure 2
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at K=0.Reference to Figure2 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π/2)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=0 and inside it the value G preserves its sign. The function G attains its extreme magnitudes ±0.5 at the points with the coordinates ξ=1/2, δ=±πn(n=0,1,2,…).The transition of fluid elements from the pointsξ=0, δ=π/2±2πn to the points ξ=0, δ=-π/2±2πn(n=0,1,2,…) and from the points ξ=1, δ=-π/2±2πn to the points ξ=1,δ=π/2±2πn(n=0,1,2,…) proceeds instantly, because according to the distribution of the phase velocity along the section δ=0 the magnitude of v tends to infinity as ξ→0 and ξ→1, respectively. The distribution of the velocity along the vertical lines δ=±πn has the aperiodic character, while in the vicinity of the line ξ=1/2 it possesses the periodic character.The transition of fluid elements from the pointsξ=1, δ=-π/2±2πn to the points ξ=1,δ=π/2±2πn(n=0,1,2,…) proceeds instantly, because according to the distribution of the phase velocity along the section δ=0 the magnitude of v tends to infinity as ξ→1.Along the linesδ=±(π/2)±2πn(n=0,1,2,…) pure amplitude modulated aperiodic motions are realized, since with an increase in time t the value ξ increases from ξ0 to 1 (along the line δ=-π/2) or decreases from ξ0 to 0 (along the line δ=π/2), and solution (126) could be written in the form of a soliton-like solution: (244)-arcsin⁡1-2ξξ0ξ=2bse-sT2-1,(245)δ(T2)=δ0=π2±πn,n=0,1,2,….Streamlines give a pictorial estimate of the connection ofG with all types of the energy-exchange mechanism. Thus, in the case of undamped vibrations, that is, when the damping coefficient is equal to zero and s=0, the points with the coordinates ξ0=1/2, δ0=±πn(n=0,1,2,…) correspond to the stationary regime, since ξ˙=0 and δ˙=0 according to (125). The stationary points ξ0=1/2, δ0=±πn are centers; as with a small deviation from a center, a phase element begins to move around the stationary point along a closed trajectory. Closed streamlines correspond to the periodic change of both amplitudes and phases.TheT2-dependence of the values ξ and 1-ξ which are proportional to the square of the amplitudes a1 and a2, respectively, at δ0=π/4 and ξ0=0.5 is presented in Figure 2, wherein the solid and dashed lines correspond, respectively, to the cases of damped and undamped vibrations. Reference to Figure 3 shows the damping of the energy exchange between two subsystems. Moreover, the points of touching of solid curves, that is, the points of tangency of the envelopes of vibrations (ξi=1-ξi,T2i), allow one to determine the value characterizing the damping of energy during vibrations via the formula(246)ln⁡ai+1/aiT2(i-1)-T2i,where ai are defined by (110) with account for (120).Figure 3
TheT2*-dependence of the dimensionless amplitudes of vibrations for the case of 1:1 internal resonance.


#### 6.1.2. Case (76) at (1/2)Kb-1=0.1

In this case, the stream functionG(ξ,δ) (129) is defined as (247)Gξ,δ=ξ1-ξcos⁡δ-0.1ξ=Gξ0,δ0,and Figure 4 shows the streamlines of the phase fluid in the phase plane.Figure 4
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=0.1.As in the previous case, the phase fluid flows in an infinitely long channel, the boundaries of which are the straight linesξ=0 and ξ=1, corresponding to the phase modulated motions. In one part the streamlines are nonclosed, which corresponds to the periodic change of amplitudes and the aperiodic change of phases; in another part they are closed, which corresponds to the periodic change of both amplitudes and phases. The aperiodic regime lines are the boundaries of the closed and unclosed streamline areas. From the phase portrait in Figure 4 it is seen that the circulation zones are located in a staggered arrangement by the right and left channel sides (this configuration resembles that of von Kármán staggered vortex tracks).Each zone by the sideξ=1 is surrounded by a line with the value G=-0.1. However, only those parts of the line G=-0.1 which bound these zones from below and come closer to the side ξ=1 at the points ξ=1, δ=π/2±πn belong to the domain of the fluid flow. The transition of fluid elements from the points ξ=1, δ=(3π/2)±πn to the points ξ=0, δ=(π/2)±πn proceeds instantly. The soliton-like solution along the separatrix is written as(248)11.01arcsin1-2.021-ξ1-ξ01-ξ=2bse-sT2-1,cos⁡δ=-0.11-ξξ,wherein ξ varies from ξmin⁡=0.0099 to ξmax⁡=1.The circulation zones by the sideξ=0 are surrounded by the line with the value G=0. However, only those parts of the line G=0 which bound these zones from above and come closer to the side ξ=0 at the points ξ=0, δ=π/2±πn belong to the domain of the fluid flow. The transition of fluid elements from the points ξ=0, δ=(π/2)±πn to the points ξ=0, δ=(-π/2)±πn proceeds instantly. The soliton-like solution along the separatrix line G=0 has the form (249)-11.01arcsin1-2.02ξξ0ξ=2bse-sT2-1,cos⁡δ=0.1ξ1-ξ,wherein ξ varies from ξmin⁡=0 to ξmax⁡=0.99.Inside the two circulation zones there are points with the extreme values of the stream function: maximalGmax⁡=0.4525 and minimal Gmin⁡=-0.5525, respectively. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.45, δ=δ0=±2πn and ξ=ξ0=0.55, δ=δ0=π±2πn, respectively.Between the lines corresponding toG=0 and G=-0.1, unclosed streamlines are located which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


#### 6.1.3. Case (76) at (1/2)Kb-1=±1

In this case, the stream functionG(ξ,δ) (129) is defined as (250)Gξ,δ=ξ1-ξcos⁡δ∓ξ=Gξ0,δ0,and Figures 5 and 6 show the streamlines of the phase fluid in the phase plane, respectively.Figure 5
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=1.Figure 6
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=-1.From the comparison of Figures5 and 6 it is evident that with the increase of the parameter K the zone of the unclosed streamlines becomes wider while the circulation zones locating in a staggered arrangement by the right and left channel sides ξ=0 and ξ=1 turn out to be more narrow.The soliton-like solution along the separatrix lineG=0 near the bank ξ=0 is found to be as follows: (251)-12arcsin1-4ξξ0ξ=2bse-sT2-1,cos⁡δ=ξ1-ξ,wherein ξ varies from ξmin⁡=0 to ξmax⁡=0.5, while along the separatrix G=-1 attached to the side ξ=1 it has the form (252)12arcsin1-41-ξ1-ξ01-ξ=2bse-sT2-1,cos⁡δ=-1-ξξ,wherein ξ varies from ξmin⁡=0.5 to ξmax⁡=1.Inside the circulation zones there are the stationary points, that is, the centers corresponding to the stable stationary regimes, where the stream function takes the extreme values:Gmax⁡=0.207 at the points with the coordinates ξ=ξ0=0.1465, δ=δ0=±2πn and Gmin⁡=-1.207 at the points with the coordinates ξ=ξ0=0.8535, δ=δ0=π±2πn.At the negative value of parameterKb-1, the phase fluid flow portrait shown in Figure 5 is shifted along the δ-axis on the value π, the flow is opposite in direction (see Figure 6), and the maxima and minima change places. In so doing the solutions along the curve separatrices with G=0 and G=1 are written according to (251) and (252), respectively.


### 6.2. One-to-One Internal Resonanceω1=ω3 (77)

For this case, the stream functionG(ξ,δ) is defined by relationship (155), and the phase portrait to be constructed according to (155) depends essentially on the magnitudes of the coefficients K2 and K3.


#### 6.2.1. Case (77) at K2=K3=0

For this case, the stream functionG(ξ,δ) is constructed according to (150) and the streamlines of the phase fluid in the phase plane ξ-δ are presented in Figure 7. Magnitudes of G are indicated by digits near the curves which correspond to the streamlines; the flow directions of the phase fluid elements are shown by arrows on the streamlines.Figure 7
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2=K3=0.Reference to Figure7 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π/2)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=0 and inside it the value G preserves its sign. The function G attains its extreme magnitudes ±0.25 at the points with the coordinates ξ=1/2, δ=±πn(n=0,1,2,…).Along the boundariesξ=0 and ξ=1 the phase modulated regimes are realized because the particles of the phase fluid flow with finite velocities (as distinct to the similar case shown in Figure 2 for ω1=ω2, where velocities of fluid particles tend to infinity as ξ→0 and ξ→1), while along the lines δ=±(π/2)±2πn(n=0,1,2,…) there exist pure amplitude modulated aperiodic motions, since with an increase in time t the value ξ increases from ξ0 to 1 (along the line δ=-π/2) or decreases from ξ0 to 0 (along the line δ=π/2), and the solution (152) could be written in the form of a soliton-like solution: (253)ln⁡1-ξξξ0ξ=bE0s1-e-sT2,δT2=δ0=π2±πn,n=0,1,2,….


#### 6.2.2. Case (77) at K2(bE0)-1=-1 and K3b-1=1/2

In this particular case, the stream functionG(ξ,δ) (150) is defined as (254)Gξ,δ=ξ1-ξcos⁡⁡δ+14+12ξ-14ξ2=G0ξ0,δ0,and Figure 8 shows the streamlines of the phase fluid in the phase plane.Figure 8
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=-1 and K3b-1=1/2.As in the previous case, the phase fluid flows in an infinitely long channel, the boundaries of which are the straight linesξ=0 and ξ=1, corresponding to the phase modulated motions. In one part the streamlines are nonclosed, which corresponds to the periodic change of amplitudes and the aperiodic change of phases; in another part they are closed, which corresponds to the periodic change of both amplitudes and phases. The aperiodic regime lines are the boundaries of the closed and unclosed streamline areas. From the phase portrait in Figure 8 it is seen that the circulation zones are located in a staggered arrangement by the right and left channel sides (this configuration resembles that of von Kármán staggered vortex tracks).Each zone by the sideξ=1 is surrounded by a line with the value G=0.5. This line consists of two parts connected with each other at the branch points with the coordinates ξ=1, δ=π/2±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=1, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0.2 to ξmax⁡=1. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix with G=0.5 is written as (255)ln⁡81-ξξ-15ξ+13-41-ξ+41-ξ01-ξ=bE0se-sT2-1,cos⁡δ=1-ξ4ξ,15≤ξ≤1.The circulation zones by the sideξ=0 are surrounded by the line with the value G=0.25. This line consists of two parts connected with each other at the branch points with the coordinates ξ=0, δ=2π/3±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=0, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0 to ξmax⁡=2/3. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix line G=0.25 has the form (256)23ln⁡4ξ23-ξ65-ξ-7ξ+6ξ0ξ=bE0s1-e-sT2,cos⁡δ=-2-ξ41-ξ,0≤ξ≤23.Inside the two circulation zones there are points with the extreme values of the stream function: maximalGmax⁡=0.7 and minimal Gmin⁡=1/6, respectively. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.6, δ=δ0=±2πn and ξ=ξ0=1/3, δ=δ0=π±2πn, respectively.Between the lines corresponding toG=0.25 and G=0.5, unclosed streamlines are located which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


#### 6.2.3. Case (77) at K2(bE0)-1=-1 and K3b-1=-1

In this case, the stream functionG(ξ,δ) (155) is defined as (257)Gξ,δ=ξ1-ξcos⁡δ-12+2ξ-ξ2=G0ξ0,δ0,and Figure 9 shows the streamlines of the phase fluid in the phase plane.Figure 9
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=-1 and K3b-1=-1.From Figure9 it is seen that, unlike the previous case presented in Figure 8, the circulation zones by the side ξ=0 and the aperiodic regime disappear. If ξ→0, then the streamlines level off and tend to the line ξ=0 with G=Gmin⁡=-0.5, where the phase modulated regime is realized.Each circulation zone by the sideξ=1 is surrounded by a line with the value G=0.5. This line consists of two parts connected with each other at the branch points with the coordinates ξ=1, δ=π/2±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=1, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0.5 to ξmax⁡=1. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix with G=0.5 is written as (258)ln⁡2ξ-1-12ξ-1+11-ξ01-ξ=bE0s1-e-sT2,cos⁡δ=1-ξξ,12≤ξ≤1.Inside the circulation zones there are points with the maximal valueGmax⁡=0.625. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.75, δ=δ0=±2πn. The lines corresponding to -0.5<G<0.5 are unclosed streamlines which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


#### 6.2.4. Case (77) at K2(bE0)-1=1 and K3b-1=1

In this case, the stream functionG(ξ,δ) (155) is defined as (259)Gξ,δ=ξ1-ξcos⁡δ+12-2ξ=G0ξ0,δ0,and Figure 10 shows the streamlines of the phase fluid in the phase plane.Figure 10
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=1 and K3b-1=1.Figure10 illustrates the phase portrait with only unclosed phase fluid streamlines along which the fluid flows in the direction of a decrease in δ. With ξ→0 and ξ→1, the streamlines level off and tend, respectively, to the lines ξ=0 with G=Gmax⁡=0.5 and ξ=1 with G=Gmin⁡=-0.5, where the boundary phase modulated regimes are realized.


#### 6.2.5. Case (77) at K2=0 and K3b-1=1

In this particular case, the stream functionG(ξ,δ) (155) is defined as (260)Gξ,δ=ξ1-ξcos⁡δ-12+ξ-ξ2=G0ξ0,δ0,and Figure 11 shows the streamlines of the phase fluid in the phase plane.Figure 11
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2=0 and K3b-1=1.Reference to Figure11 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=-0.5, and inside it the value -0.5<G<0 attaining its maximal magnitude Gmax⁡=0 at the points with the coordinates ξ=1/2, δ=±2πn(n=0,1,2,…).Along the boundariesξ=0 and ξ=1 the phase modulated regimes are realized because the particles of the phase fluid flow with finite velocities: (261)vδ=∓bE01+cos⁡δe-sT2,in so doing (262)tanδ2δ0δ=∓bE0s1-e-sT2,where the signs “−” and “+” refer to the boundaries ξ=0 and ξ=1, respectively.Along the vertical linesδ=δ0=const=π±2πn(n=1,2,…) there exist stationary regimes for all 0≤ξ0≤1 attenuating with time.


## 6.1. One-to-One Internal Resonanceω1=ω2 (76)

For this case, the stream functionG(ξ,δ) is defined by relationship (129), and the phase portrait to be constructed according to (129) depends essentially on the magnitudes of the coefficient K.


### 6.1.1. Case (76) at K=0

For this case, the stream functionG(ξ,δ) is constructed according to (124) and the streamlines of the phase fluid in the phase plane ξ-δ are presented in Figure 2. Magnitudes of G are indicated by digits near the curves which correspond to the streamlines; the flow directions of the phase fluid elements are shown by arrows on the streamlines.Figure 2
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at K=0.Reference to Figure2 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π/2)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=0 and inside it the value G preserves its sign. The function G attains its extreme magnitudes ±0.5 at the points with the coordinates ξ=1/2, δ=±πn(n=0,1,2,…).The transition of fluid elements from the pointsξ=0, δ=π/2±2πn to the points ξ=0, δ=-π/2±2πn(n=0,1,2,…) and from the points ξ=1, δ=-π/2±2πn to the points ξ=1,δ=π/2±2πn(n=0,1,2,…) proceeds instantly, because according to the distribution of the phase velocity along the section δ=0 the magnitude of v tends to infinity as ξ→0 and ξ→1, respectively. The distribution of the velocity along the vertical lines δ=±πn has the aperiodic character, while in the vicinity of the line ξ=1/2 it possesses the periodic character.The transition of fluid elements from the pointsξ=1, δ=-π/2±2πn to the points ξ=1,δ=π/2±2πn(n=0,1,2,…) proceeds instantly, because according to the distribution of the phase velocity along the section δ=0 the magnitude of v tends to infinity as ξ→1.Along the linesδ=±(π/2)±2πn(n=0,1,2,…) pure amplitude modulated aperiodic motions are realized, since with an increase in time t the value ξ increases from ξ0 to 1 (along the line δ=-π/2) or decreases from ξ0 to 0 (along the line δ=π/2), and solution (126) could be written in the form of a soliton-like solution: (244)-arcsin⁡1-2ξξ0ξ=2bse-sT2-1,(245)δ(T2)=δ0=π2±πn,n=0,1,2,….Streamlines give a pictorial estimate of the connection ofG with all types of the energy-exchange mechanism. Thus, in the case of undamped vibrations, that is, when the damping coefficient is equal to zero and s=0, the points with the coordinates ξ0=1/2, δ0=±πn(n=0,1,2,…) correspond to the stationary regime, since ξ˙=0 and δ˙=0 according to (125). The stationary points ξ0=1/2, δ0=±πn are centers; as with a small deviation from a center, a phase element begins to move around the stationary point along a closed trajectory. Closed streamlines correspond to the periodic change of both amplitudes and phases.TheT2-dependence of the values ξ and 1-ξ which are proportional to the square of the amplitudes a1 and a2, respectively, at δ0=π/4 and ξ0=0.5 is presented in Figure 2, wherein the solid and dashed lines correspond, respectively, to the cases of damped and undamped vibrations. Reference to Figure 3 shows the damping of the energy exchange between two subsystems. Moreover, the points of touching of solid curves, that is, the points of tangency of the envelopes of vibrations (ξi=1-ξi,T2i), allow one to determine the value characterizing the damping of energy during vibrations via the formula(246)ln⁡ai+1/aiT2(i-1)-T2i,where ai are defined by (110) with account for (120).Figure 3
TheT2*-dependence of the dimensionless amplitudes of vibrations for the case of 1:1 internal resonance.


### 6.1.2. Case (76) at (1/2)Kb-1=0.1

In this case, the stream functionG(ξ,δ) (129) is defined as (247)Gξ,δ=ξ1-ξcos⁡δ-0.1ξ=Gξ0,δ0,and Figure 4 shows the streamlines of the phase fluid in the phase plane.Figure 4
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=0.1.As in the previous case, the phase fluid flows in an infinitely long channel, the boundaries of which are the straight linesξ=0 and ξ=1, corresponding to the phase modulated motions. In one part the streamlines are nonclosed, which corresponds to the periodic change of amplitudes and the aperiodic change of phases; in another part they are closed, which corresponds to the periodic change of both amplitudes and phases. The aperiodic regime lines are the boundaries of the closed and unclosed streamline areas. From the phase portrait in Figure 4 it is seen that the circulation zones are located in a staggered arrangement by the right and left channel sides (this configuration resembles that of von Kármán staggered vortex tracks).Each zone by the sideξ=1 is surrounded by a line with the value G=-0.1. However, only those parts of the line G=-0.1 which bound these zones from below and come closer to the side ξ=1 at the points ξ=1, δ=π/2±πn belong to the domain of the fluid flow. The transition of fluid elements from the points ξ=1, δ=(3π/2)±πn to the points ξ=0, δ=(π/2)±πn proceeds instantly. The soliton-like solution along the separatrix is written as(248)11.01arcsin1-2.021-ξ1-ξ01-ξ=2bse-sT2-1,cos⁡δ=-0.11-ξξ,wherein ξ varies from ξmin⁡=0.0099 to ξmax⁡=1.The circulation zones by the sideξ=0 are surrounded by the line with the value G=0. However, only those parts of the line G=0 which bound these zones from above and come closer to the side ξ=0 at the points ξ=0, δ=π/2±πn belong to the domain of the fluid flow. The transition of fluid elements from the points ξ=0, δ=(π/2)±πn to the points ξ=0, δ=(-π/2)±πn proceeds instantly. The soliton-like solution along the separatrix line G=0 has the form (249)-11.01arcsin1-2.02ξξ0ξ=2bse-sT2-1,cos⁡δ=0.1ξ1-ξ,wherein ξ varies from ξmin⁡=0 to ξmax⁡=0.99.Inside the two circulation zones there are points with the extreme values of the stream function: maximalGmax⁡=0.4525 and minimal Gmin⁡=-0.5525, respectively. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.45, δ=δ0=±2πn and ξ=ξ0=0.55, δ=δ0=π±2πn, respectively.Between the lines corresponding toG=0 and G=-0.1, unclosed streamlines are located which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


### 6.1.3. Case (76) at (1/2)Kb-1=±1

In this case, the stream functionG(ξ,δ) (129) is defined as (250)Gξ,δ=ξ1-ξcos⁡δ∓ξ=Gξ0,δ0,and Figures 5 and 6 show the streamlines of the phase fluid in the phase plane, respectively.Figure 5
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=1.Figure 6
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=-1.From the comparison of Figures5 and 6 it is evident that with the increase of the parameter K the zone of the unclosed streamlines becomes wider while the circulation zones locating in a staggered arrangement by the right and left channel sides ξ=0 and ξ=1 turn out to be more narrow.The soliton-like solution along the separatrix lineG=0 near the bank ξ=0 is found to be as follows: (251)-12arcsin1-4ξξ0ξ=2bse-sT2-1,cos⁡δ=ξ1-ξ,wherein ξ varies from ξmin⁡=0 to ξmax⁡=0.5, while along the separatrix G=-1 attached to the side ξ=1 it has the form (252)12arcsin1-41-ξ1-ξ01-ξ=2bse-sT2-1,cos⁡δ=-1-ξξ,wherein ξ varies from ξmin⁡=0.5 to ξmax⁡=1.Inside the circulation zones there are the stationary points, that is, the centers corresponding to the stable stationary regimes, where the stream function takes the extreme values:Gmax⁡=0.207 at the points with the coordinates ξ=ξ0=0.1465, δ=δ0=±2πn and Gmin⁡=-1.207 at the points with the coordinates ξ=ξ0=0.8535, δ=δ0=π±2πn.At the negative value of parameterKb-1, the phase fluid flow portrait shown in Figure 5 is shifted along the δ-axis on the value π, the flow is opposite in direction (see Figure 6), and the maxima and minima change places. In so doing the solutions along the curve separatrices with G=0 and G=1 are written according to (251) and (252), respectively.


## 6.1.1. Case (76) at K=0

For this case, the stream functionG(ξ,δ) is constructed according to (124) and the streamlines of the phase fluid in the phase plane ξ-δ are presented in Figure 2. Magnitudes of G are indicated by digits near the curves which correspond to the streamlines; the flow directions of the phase fluid elements are shown by arrows on the streamlines.Figure 2
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at K=0.Reference to Figure2 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π/2)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=0 and inside it the value G preserves its sign. The function G attains its extreme magnitudes ±0.5 at the points with the coordinates ξ=1/2, δ=±πn(n=0,1,2,…).The transition of fluid elements from the pointsξ=0, δ=π/2±2πn to the points ξ=0, δ=-π/2±2πn(n=0,1,2,…) and from the points ξ=1, δ=-π/2±2πn to the points ξ=1,δ=π/2±2πn(n=0,1,2,…) proceeds instantly, because according to the distribution of the phase velocity along the section δ=0 the magnitude of v tends to infinity as ξ→0 and ξ→1, respectively. The distribution of the velocity along the vertical lines δ=±πn has the aperiodic character, while in the vicinity of the line ξ=1/2 it possesses the periodic character.The transition of fluid elements from the pointsξ=1, δ=-π/2±2πn to the points ξ=1,δ=π/2±2πn(n=0,1,2,…) proceeds instantly, because according to the distribution of the phase velocity along the section δ=0 the magnitude of v tends to infinity as ξ→1.Along the linesδ=±(π/2)±2πn(n=0,1,2,…) pure amplitude modulated aperiodic motions are realized, since with an increase in time t the value ξ increases from ξ0 to 1 (along the line δ=-π/2) or decreases from ξ0 to 0 (along the line δ=π/2), and solution (126) could be written in the form of a soliton-like solution: (244)-arcsin⁡1-2ξξ0ξ=2bse-sT2-1,(245)δ(T2)=δ0=π2±πn,n=0,1,2,….Streamlines give a pictorial estimate of the connection ofG with all types of the energy-exchange mechanism. Thus, in the case of undamped vibrations, that is, when the damping coefficient is equal to zero and s=0, the points with the coordinates ξ0=1/2, δ0=±πn(n=0,1,2,…) correspond to the stationary regime, since ξ˙=0 and δ˙=0 according to (125). The stationary points ξ0=1/2, δ0=±πn are centers; as with a small deviation from a center, a phase element begins to move around the stationary point along a closed trajectory. Closed streamlines correspond to the periodic change of both amplitudes and phases.TheT2-dependence of the values ξ and 1-ξ which are proportional to the square of the amplitudes a1 and a2, respectively, at δ0=π/4 and ξ0=0.5 is presented in Figure 2, wherein the solid and dashed lines correspond, respectively, to the cases of damped and undamped vibrations. Reference to Figure 3 shows the damping of the energy exchange between two subsystems. Moreover, the points of touching of solid curves, that is, the points of tangency of the envelopes of vibrations (ξi=1-ξi,T2i), allow one to determine the value characterizing the damping of energy during vibrations via the formula(246)ln⁡ai+1/aiT2(i-1)-T2i,where ai are defined by (110) with account for (120).Figure 3
TheT2*-dependence of the dimensionless amplitudes of vibrations for the case of 1:1 internal resonance.


## 6.1.2. Case (76) at (1/2)Kb-1=0.1

In this case, the stream functionG(ξ,δ) (129) is defined as (247)Gξ,δ=ξ1-ξcos⁡δ-0.1ξ=Gξ0,δ0,and Figure 4 shows the streamlines of the phase fluid in the phase plane.Figure 4
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=0.1.As in the previous case, the phase fluid flows in an infinitely long channel, the boundaries of which are the straight linesξ=0 and ξ=1, corresponding to the phase modulated motions. In one part the streamlines are nonclosed, which corresponds to the periodic change of amplitudes and the aperiodic change of phases; in another part they are closed, which corresponds to the periodic change of both amplitudes and phases. The aperiodic regime lines are the boundaries of the closed and unclosed streamline areas. From the phase portrait in Figure 4 it is seen that the circulation zones are located in a staggered arrangement by the right and left channel sides (this configuration resembles that of von Kármán staggered vortex tracks).Each zone by the sideξ=1 is surrounded by a line with the value G=-0.1. However, only those parts of the line G=-0.1 which bound these zones from below and come closer to the side ξ=1 at the points ξ=1, δ=π/2±πn belong to the domain of the fluid flow. The transition of fluid elements from the points ξ=1, δ=(3π/2)±πn to the points ξ=0, δ=(π/2)±πn proceeds instantly. The soliton-like solution along the separatrix is written as(248)11.01arcsin1-2.021-ξ1-ξ01-ξ=2bse-sT2-1,cos⁡δ=-0.11-ξξ,wherein ξ varies from ξmin⁡=0.0099 to ξmax⁡=1.The circulation zones by the sideξ=0 are surrounded by the line with the value G=0. However, only those parts of the line G=0 which bound these zones from above and come closer to the side ξ=0 at the points ξ=0, δ=π/2±πn belong to the domain of the fluid flow. The transition of fluid elements from the points ξ=0, δ=(π/2)±πn to the points ξ=0, δ=(-π/2)±πn proceeds instantly. The soliton-like solution along the separatrix line G=0 has the form (249)-11.01arcsin1-2.02ξξ0ξ=2bse-sT2-1,cos⁡δ=0.1ξ1-ξ,wherein ξ varies from ξmin⁡=0 to ξmax⁡=0.99.Inside the two circulation zones there are points with the extreme values of the stream function: maximalGmax⁡=0.4525 and minimal Gmin⁡=-0.5525, respectively. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.45, δ=δ0=±2πn and ξ=ξ0=0.55, δ=δ0=π±2πn, respectively.Between the lines corresponding toG=0 and G=-0.1, unclosed streamlines are located which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


## 6.1.3. Case (76) at (1/2)Kb-1=±1

In this case, the stream functionG(ξ,δ) (129) is defined as (250)Gξ,δ=ξ1-ξcos⁡δ∓ξ=Gξ0,δ0,and Figures 5 and 6 show the streamlines of the phase fluid in the phase plane, respectively.Figure 5
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=1.Figure 6
Phase portrait for the case of 1:1 internal resonanceω1=ω2 at (1/2)Kb-1=-1.From the comparison of Figures5 and 6 it is evident that with the increase of the parameter K the zone of the unclosed streamlines becomes wider while the circulation zones locating in a staggered arrangement by the right and left channel sides ξ=0 and ξ=1 turn out to be more narrow.The soliton-like solution along the separatrix lineG=0 near the bank ξ=0 is found to be as follows: (251)-12arcsin1-4ξξ0ξ=2bse-sT2-1,cos⁡δ=ξ1-ξ,wherein ξ varies from ξmin⁡=0 to ξmax⁡=0.5, while along the separatrix G=-1 attached to the side ξ=1 it has the form (252)12arcsin1-41-ξ1-ξ01-ξ=2bse-sT2-1,cos⁡δ=-1-ξξ,wherein ξ varies from ξmin⁡=0.5 to ξmax⁡=1.Inside the circulation zones there are the stationary points, that is, the centers corresponding to the stable stationary regimes, where the stream function takes the extreme values:Gmax⁡=0.207 at the points with the coordinates ξ=ξ0=0.1465, δ=δ0=±2πn and Gmin⁡=-1.207 at the points with the coordinates ξ=ξ0=0.8535, δ=δ0=π±2πn.At the negative value of parameterKb-1, the phase fluid flow portrait shown in Figure 5 is shifted along the δ-axis on the value π, the flow is opposite in direction (see Figure 6), and the maxima and minima change places. In so doing the solutions along the curve separatrices with G=0 and G=1 are written according to (251) and (252), respectively.


## 6.2. One-to-One Internal Resonanceω1=ω3 (77)

For this case, the stream functionG(ξ,δ) is defined by relationship (155), and the phase portrait to be constructed according to (155) depends essentially on the magnitudes of the coefficients K2 and K3.


### 6.2.1. Case (77) at K2=K3=0

For this case, the stream functionG(ξ,δ) is constructed according to (150) and the streamlines of the phase fluid in the phase plane ξ-δ are presented in Figure 7. Magnitudes of G are indicated by digits near the curves which correspond to the streamlines; the flow directions of the phase fluid elements are shown by arrows on the streamlines.Figure 7
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2=K3=0.Reference to Figure7 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π/2)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=0 and inside it the value G preserves its sign. The function G attains its extreme magnitudes ±0.25 at the points with the coordinates ξ=1/2, δ=±πn(n=0,1,2,…).Along the boundariesξ=0 and ξ=1 the phase modulated regimes are realized because the particles of the phase fluid flow with finite velocities (as distinct to the similar case shown in Figure 2 for ω1=ω2, where velocities of fluid particles tend to infinity as ξ→0 and ξ→1), while along the lines δ=±(π/2)±2πn(n=0,1,2,…) there exist pure amplitude modulated aperiodic motions, since with an increase in time t the value ξ increases from ξ0 to 1 (along the line δ=-π/2) or decreases from ξ0 to 0 (along the line δ=π/2), and the solution (152) could be written in the form of a soliton-like solution: (253)ln⁡1-ξξξ0ξ=bE0s1-e-sT2,δT2=δ0=π2±πn,n=0,1,2,….


### 6.2.2. Case (77) at K2(bE0)-1=-1 and K3b-1=1/2

In this particular case, the stream functionG(ξ,δ) (150) is defined as (254)Gξ,δ=ξ1-ξcos⁡⁡δ+14+12ξ-14ξ2=G0ξ0,δ0,and Figure 8 shows the streamlines of the phase fluid in the phase plane.Figure 8
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=-1 and K3b-1=1/2.As in the previous case, the phase fluid flows in an infinitely long channel, the boundaries of which are the straight linesξ=0 and ξ=1, corresponding to the phase modulated motions. In one part the streamlines are nonclosed, which corresponds to the periodic change of amplitudes and the aperiodic change of phases; in another part they are closed, which corresponds to the periodic change of both amplitudes and phases. The aperiodic regime lines are the boundaries of the closed and unclosed streamline areas. From the phase portrait in Figure 8 it is seen that the circulation zones are located in a staggered arrangement by the right and left channel sides (this configuration resembles that of von Kármán staggered vortex tracks).Each zone by the sideξ=1 is surrounded by a line with the value G=0.5. This line consists of two parts connected with each other at the branch points with the coordinates ξ=1, δ=π/2±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=1, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0.2 to ξmax⁡=1. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix with G=0.5 is written as (255)ln⁡81-ξξ-15ξ+13-41-ξ+41-ξ01-ξ=bE0se-sT2-1,cos⁡δ=1-ξ4ξ,15≤ξ≤1.The circulation zones by the sideξ=0 are surrounded by the line with the value G=0.25. This line consists of two parts connected with each other at the branch points with the coordinates ξ=0, δ=2π/3±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=0, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0 to ξmax⁡=2/3. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix line G=0.25 has the form (256)23ln⁡4ξ23-ξ65-ξ-7ξ+6ξ0ξ=bE0s1-e-sT2,cos⁡δ=-2-ξ41-ξ,0≤ξ≤23.Inside the two circulation zones there are points with the extreme values of the stream function: maximalGmax⁡=0.7 and minimal Gmin⁡=1/6, respectively. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.6, δ=δ0=±2πn and ξ=ξ0=1/3, δ=δ0=π±2πn, respectively.Between the lines corresponding toG=0.25 and G=0.5, unclosed streamlines are located which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


### 6.2.3. Case (77) at K2(bE0)-1=-1 and K3b-1=-1

In this case, the stream functionG(ξ,δ) (155) is defined as (257)Gξ,δ=ξ1-ξcos⁡δ-12+2ξ-ξ2=G0ξ0,δ0,and Figure 9 shows the streamlines of the phase fluid in the phase plane.Figure 9
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=-1 and K3b-1=-1.From Figure9 it is seen that, unlike the previous case presented in Figure 8, the circulation zones by the side ξ=0 and the aperiodic regime disappear. If ξ→0, then the streamlines level off and tend to the line ξ=0 with G=Gmin⁡=-0.5, where the phase modulated regime is realized.Each circulation zone by the sideξ=1 is surrounded by a line with the value G=0.5. This line consists of two parts connected with each other at the branch points with the coordinates ξ=1, δ=π/2±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=1, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0.5 to ξmax⁡=1. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix with G=0.5 is written as (258)ln⁡2ξ-1-12ξ-1+11-ξ01-ξ=bE0s1-e-sT2,cos⁡δ=1-ξξ,12≤ξ≤1.Inside the circulation zones there are points with the maximal valueGmax⁡=0.625. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.75, δ=δ0=±2πn. The lines corresponding to -0.5<G<0.5 are unclosed streamlines which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


### 6.2.4. Case (77) at K2(bE0)-1=1 and K3b-1=1

In this case, the stream functionG(ξ,δ) (155) is defined as (259)Gξ,δ=ξ1-ξcos⁡δ+12-2ξ=G0ξ0,δ0,and Figure 10 shows the streamlines of the phase fluid in the phase plane.Figure 10
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=1 and K3b-1=1.Figure10 illustrates the phase portrait with only unclosed phase fluid streamlines along which the fluid flows in the direction of a decrease in δ. With ξ→0 and ξ→1, the streamlines level off and tend, respectively, to the lines ξ=0 with G=Gmax⁡=0.5 and ξ=1 with G=Gmin⁡=-0.5, where the boundary phase modulated regimes are realized.


### 6.2.5. Case (77) at K2=0 and K3b-1=1

In this particular case, the stream functionG(ξ,δ) (155) is defined as (260)Gξ,δ=ξ1-ξcos⁡δ-12+ξ-ξ2=G0ξ0,δ0,and Figure 11 shows the streamlines of the phase fluid in the phase plane.Figure 11
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2=0 and K3b-1=1.Reference to Figure11 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=-0.5, and inside it the value -0.5<G<0 attaining its maximal magnitude Gmax⁡=0 at the points with the coordinates ξ=1/2, δ=±2πn(n=0,1,2,…).Along the boundariesξ=0 and ξ=1 the phase modulated regimes are realized because the particles of the phase fluid flow with finite velocities: (261)vδ=∓bE01+cos⁡δe-sT2,in so doing (262)tanδ2δ0δ=∓bE0s1-e-sT2,where the signs “−” and “+” refer to the boundaries ξ=0 and ξ=1, respectively.Along the vertical linesδ=δ0=const=π±2πn(n=1,2,…) there exist stationary regimes for all 0≤ξ0≤1 attenuating with time.


## 6.2.1. Case (77) at K2=K3=0

For this case, the stream functionG(ξ,δ) is constructed according to (150) and the streamlines of the phase fluid in the phase plane ξ-δ are presented in Figure 7. Magnitudes of G are indicated by digits near the curves which correspond to the streamlines; the flow directions of the phase fluid elements are shown by arrows on the streamlines.Figure 7
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2=K3=0.Reference to Figure7 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π/2)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=0 and inside it the value G preserves its sign. The function G attains its extreme magnitudes ±0.25 at the points with the coordinates ξ=1/2, δ=±πn(n=0,1,2,…).Along the boundariesξ=0 and ξ=1 the phase modulated regimes are realized because the particles of the phase fluid flow with finite velocities (as distinct to the similar case shown in Figure 2 for ω1=ω2, where velocities of fluid particles tend to infinity as ξ→0 and ξ→1), while along the lines δ=±(π/2)±2πn(n=0,1,2,…) there exist pure amplitude modulated aperiodic motions, since with an increase in time t the value ξ increases from ξ0 to 1 (along the line δ=-π/2) or decreases from ξ0 to 0 (along the line δ=π/2), and the solution (152) could be written in the form of a soliton-like solution: (253)ln⁡1-ξξξ0ξ=bE0s1-e-sT2,δT2=δ0=π2±πn,n=0,1,2,….


## 6.2.2. Case (77) at K2(bE0)-1=-1 and K3b-1=1/2

In this particular case, the stream functionG(ξ,δ) (150) is defined as (254)Gξ,δ=ξ1-ξcos⁡⁡δ+14+12ξ-14ξ2=G0ξ0,δ0,and Figure 8 shows the streamlines of the phase fluid in the phase plane.Figure 8
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=-1 and K3b-1=1/2.As in the previous case, the phase fluid flows in an infinitely long channel, the boundaries of which are the straight linesξ=0 and ξ=1, corresponding to the phase modulated motions. In one part the streamlines are nonclosed, which corresponds to the periodic change of amplitudes and the aperiodic change of phases; in another part they are closed, which corresponds to the periodic change of both amplitudes and phases. The aperiodic regime lines are the boundaries of the closed and unclosed streamline areas. From the phase portrait in Figure 8 it is seen that the circulation zones are located in a staggered arrangement by the right and left channel sides (this configuration resembles that of von Kármán staggered vortex tracks).Each zone by the sideξ=1 is surrounded by a line with the value G=0.5. This line consists of two parts connected with each other at the branch points with the coordinates ξ=1, δ=π/2±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=1, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0.2 to ξmax⁡=1. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix with G=0.5 is written as (255)ln⁡81-ξξ-15ξ+13-41-ξ+41-ξ01-ξ=bE0se-sT2-1,cos⁡δ=1-ξ4ξ,15≤ξ≤1.The circulation zones by the sideξ=0 are surrounded by the line with the value G=0.25. This line consists of two parts connected with each other at the branch points with the coordinates ξ=0, δ=2π/3±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=0, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0 to ξmax⁡=2/3. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix line G=0.25 has the form (256)23ln⁡4ξ23-ξ65-ξ-7ξ+6ξ0ξ=bE0s1-e-sT2,cos⁡δ=-2-ξ41-ξ,0≤ξ≤23.Inside the two circulation zones there are points with the extreme values of the stream function: maximalGmax⁡=0.7 and minimal Gmin⁡=1/6, respectively. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.6, δ=δ0=±2πn and ξ=ξ0=1/3, δ=δ0=π±2πn, respectively.Between the lines corresponding toG=0.25 and G=0.5, unclosed streamlines are located which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


## 6.2.3. Case (77) at K2(bE0)-1=-1 and K3b-1=-1

In this case, the stream functionG(ξ,δ) (155) is defined as (257)Gξ,δ=ξ1-ξcos⁡δ-12+2ξ-ξ2=G0ξ0,δ0,and Figure 9 shows the streamlines of the phase fluid in the phase plane.Figure 9
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=-1 and K3b-1=-1.From Figure9 it is seen that, unlike the previous case presented in Figure 8, the circulation zones by the side ξ=0 and the aperiodic regime disappear. If ξ→0, then the streamlines level off and tend to the line ξ=0 with G=Gmin⁡=-0.5, where the phase modulated regime is realized.Each circulation zone by the sideξ=1 is surrounded by a line with the value G=0.5. This line consists of two parts connected with each other at the branch points with the coordinates ξ=1, δ=π/2±πn (n=0,1,2,…). One branch of this line corresponds to the phase modulated regime ξ=1, and the other corresponds to the aperiodic regime, wherein ξ varies from ξmin⁡=0.5 to ξmax⁡=1. At the branch point itself, the phase fluid flow velocity is equal to zero. The soliton-like solution along the separatrix with G=0.5 is written as (258)ln⁡2ξ-1-12ξ-1+11-ξ01-ξ=bE0s1-e-sT2,cos⁡δ=1-ξξ,12≤ξ≤1.Inside the circulation zones there are points with the maximal valueGmax⁡=0.625. These points are the centers corresponding to the stable stationary regimes ξ=ξ0=0.75, δ=δ0=±2πn. The lines corresponding to -0.5<G<0.5 are unclosed streamlines which are in accordance with the periodic change of the amplitudes and the aperiodic change of the phase difference.


## 6.2.4. Case (77) at K2(bE0)-1=1 and K3b-1=1

In this case, the stream functionG(ξ,δ) (155) is defined as (259)Gξ,δ=ξ1-ξcos⁡δ+12-2ξ=G0ξ0,δ0,and Figure 10 shows the streamlines of the phase fluid in the phase plane.Figure 10
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2(bE0)-1=1 and K3b-1=1.Figure10 illustrates the phase portrait with only unclosed phase fluid streamlines along which the fluid flows in the direction of a decrease in δ. With ξ→0 and ξ→1, the streamlines level off and tend, respectively, to the lines ξ=0 with G=Gmax⁡=0.5 and ξ=1 with G=Gmin⁡=-0.5, where the boundary phase modulated regimes are realized.


## 6.2.5. Case (77) at K2=0 and K3b-1=1

In this particular case, the stream functionG(ξ,δ) (155) is defined as (260)Gξ,δ=ξ1-ξcos⁡δ-12+ξ-ξ2=G0ξ0,δ0,and Figure 11 shows the streamlines of the phase fluid in the phase plane.Figure 11
Phase portrait for the case of 1:1 internal resonanceω1=ω3 at K2=0 and K3b-1=1.Reference to Figure11 shows that the phase fluid flows within the circulation zones, which tend to be located around the perimeter of the rectangles bounded by the lines ξ=0, ξ=1, and δ=±(π)±2πn(n=0,1,2,…). As this takes place, the flow in each such rectangle becomes isolated. On all four rectangle sides G=-0.5, and inside it the value -0.5<G<0 attaining its maximal magnitude Gmax⁡=0 at the points with the coordinates ξ=1/2, δ=±2πn(n=0,1,2,…).Along the boundariesξ=0 and ξ=1 the phase modulated regimes are realized because the particles of the phase fluid flow with finite velocities: (261)vδ=∓bE01+cos⁡δe-sT2,in so doing (262)tanδ2δ0δ=∓bE0s1-e-sT2,where the signs “−” and “+” refer to the boundaries ξ=0 and ξ=1, respectively.Along the vertical linesδ=δ0=const=π±2πn(n=1,2,…) there exist stationary regimes for all 0≤ξ0≤1 attenuating with time.


## 7. Conclusion

Free damped vibrations of a nonlinear plate in a fractional derivative viscoelastic medium have been investigated. Nonlinear vibrations are described by coupled three equations with respect to the three orthogonal displacements.The proposed analytical approach for investigating the damped vibrations of the nonlinear plate subjected to the conditions of the internal resonance has been possible owing to the new procedure suggested in this paper, resulting in decoupling linear parts of equations with further utilization of the method of multiple scales for solving nonlinear governing equations of motion.It has been shown that the phenomenon of internal resonance could be very critical, since in the thin plate under consideration the internal resonance is always present. Moreover, its type depends on the order of smallness of the viscosity involved into consideration. Thus, at theε-order, damped vibrations occur within the two-to-one and one-to-one-to-two internal resonance. Other types of the internal resonance, such as one-to-one, one-to-one-to-one, and combinational resonances of the additive and difference types could be found at ε2-order; that is, the type of the resonance depends on the order of smallness of the fractional derivative entering in the equations of motion of the plate.For each type of the resonance, the nonlinear sets of resolving equations in terms of amplitudes and phase differences have been obtained. It has been shown that for the one-to-one and combinational internal resonances there exist such particular cases when it is possible to obtain two first integrals, namely, the energy integral and the stream function, which allows one to reduce the problem to the calculation of elliptic integrals.The influence of viscosity on the energy exchange mechanism has been analyzed. It has been shown that each mode is characterized by its damping coefficient connected with the natural frequency by the exponential relationship with a negative fractional exponent. Thus, during free vibrations of the plate with internal resonances three regimes can be observed: stationary (absence of damping atγ=0), quasi-stationary (damping is defined by an ordinary derivative at γ=1), and transient (damping is defined by a fractional derivative at 0<γ<1).The phenomenological analysis carried out for two types of the one-to-one internal resonance with the help of the phase portraits constructed for different magnitudes of the plate parameters reveals the great variety of vibrational motions: stationary vibrations, two-sided energy exchange between two subsystems under consideration, and one-sided energy interchange resulting in the complete one-sided energy transfer. Under the presence of small viscosity all regimes attenuate with time.The analysis of the phase portraits for various oscillatory regimes cited above demonstrates that they contain closed as well as nonclosed streamlines, which are separated by the curves separatrixes. Along the separatrixes one succeeds in finding analytic solutions which determine the irreversible energy exchange from one subsystem into another and which are inherently soliton-like solutions in the theory of vibrations.Comparison of the results obtained in this paper via the new suggested procedure with those for a nonlinear plate, the motion of which is described either by one equation in terms of plate’s deflection [6] or also by three coupled nonlinear equations in terms of three displacements but without decoupling the linear parts of governing equations [2, 3], reveals the fact that the new approach allows one to discover and analyze much more diversified variety of internal resonances, including the one-to-one-to-one and one-to-one-to-two internal resonances and additional one-to-one, two-to-one, and combinational resonances (ten possible cases in all), than all other existing previously in literature approaches.It should be noted that the proposed analytical approach could be used for other cases if the matrix of the natural modes coupled by the internal resonance is Hermitian matrix; that is, it should be symmetric and all its elements should be real.



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*Source: 795606-2015-03-18.xml*