Exact Solutions of the Cubic-Quintic Duffing Equation Using Leaf Functions
EE XACT S OLUTIONS OF THE C UBIC -Q UINTIC D UFFING E QUATION U SING L EAF F UNCTIONS
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Kazunori Shinohara ∗ Department of Mechanical Systems EngineeringDaido University10-3 Takiharu-cho, Minami-ku, Nagoya 457-8530, Japan [email protected]
February 5, 2021 A BSTRACT
The exact solutions of both the cubic Duffing equation and cubic-quintic Duffing equation arepresented by using only leaf functions. In previous studies, exact solutions of the cubic Duffingequation have been proposed using functions that integrate leaf functions in the phase of trigonometricfunctions. Because they are not simple, the procedures for transforming the exact solutions arecomplicated and not convenient. The first derivative of the leaf function sleaf ( l ) can be derivedas the root of − (sleaf ( l )) . This derivative can be factored, and the factors are ( l ) , − sleaf ( l ) , and ( l )) . These factors or multiplications of factors are exact solutionsto the Duffing equation. Some of these exact solutions are of the same type as the cubic Duffingequation reported in previously. Some of these exact solutions satisfy the exact solutions of thecubic–quintic Duffing equations with high nonlinearity. These rules can also be applied to the leaffunction cleaf ( l ) . In this study, the relationship between the parameters of these exact solutions andthe coefficients of the terms of the Duffing equation is clarified. We numerically analyze these exactsolutions, plot the waveform, and discuss the periodicity and amplitude of the waveform. K eywords Leaf functions · Hyperbolic leaf functions · Lemniscate functions · Jacobi elliptic functions · Ordinarydifferential equations · Duffing equation · Nonlinear equations.
The Duffing equation is a nonlinear second-order differential equation. The equation consists of the first derivative,the second derivative, the polynomial x , and the trigonometric function (the external force term). The behavior ofthe solution of the Duffing equation easily changes depending on the initial value and the polynomial coefficients,and it is difficult to predict its solution. To clarify the behavior of the solution, research based mainly on numericalanalysis with high-precision calculation is conducted. Different forms of the fractional Duffing equation have beenanalyzed by using the different parameters of the equation and its fractional derivative orders [1]. The efficient multistepdifferential transform method has been applied to obtain accurate approximate solutions [2]. An analytical approximatetechnique combining both the homotopy perturbation method and variational formulation has also been presented [3].Lai et al. proposed a method of linearized harmonic balance with Newton’s method for solving accurate analyticalapproximations [4]. Das et al. proposed some modifications of a domain decomposition method to obtain accurateclosed-form approximate solutions of Duffing- and Liénard-type nonlinear ordinary differential equations [5]. In [1],it was demonstrated that exact solutions of some of these nonlinear differential equations do not exist. Therefore,investigating approximate solutions of these types of equations can play a vital role. However, exact analytical solutionsor exact solutions of the Duffing equations have been discussed. To construct a transformation that converts the n thpower to an m th power Duffing-type ordinary differential equation, a systematic analytical approach has been presented ∗ a r X i v : . [ m a t h . G M ] F e b PREPRINT - F
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5, 2021[6]. Elías-Zúñiga has attempted to find only one set of exact solutions by using the Jacobi elliptic function [7]. Razzakobtained the exact solution of the cubic–quintic Duffing equation. However, Razzak’s solution procedure is complicatedand contains a set of algebraic equations with Jacobian elliptic functions that are not easily solved [3].In this study, by using leaf functions, an exact solution of the cubic–quintic Duffing equation is proposed. The cubicDuffing equation and cubic–quintic Duffing equation are represented by the following equations, respectively: d x ( t )d t + αx ( t ) + βx ( t ) = 0 , (1) d x ( t )d t + αx ( t ) + βx ( t ) + γx ( t ) = 0 . (2)By using leaf functions, exact solutions of the cubic Duffing equation were proposed for a free oscillation system, adivergence system, and a damping system [8] [9]. Because these exact solutions are expressed by using trigonometricfunctions for the phases of leaf functions, the expansions of exact solutions become complicated and the usabilityof applications becomes low. In this study, by using only leaf functions without trigonometric functions, the exactsolutions of the cubic Duffing equation are presented for a free-oscillation system. For example, the following formulais obtained by differentiating the leaf function sleaf ( x ) : (cid:112) − (sleaf ( t )) = (cid:112) { ( t ) }{ − sleaf ( t ) }{ ( t )) } . (3)As shown in the above equation, a factor obtained by factoring the equation is assumed to be the solution x ( t ) . Forexample, these solutions are as follows: x ( t ) = (cid:112) ( t ) , (cid:112) − sleaf ( t ) , (cid:112) ( t )) , (cid:112) − (sleaf ( t )) . (4)A factor obtained by factoring the derivative of the leaf function or a part of the multiplication of these factors satisfiesthe cubic Duffing equation or the cubic–quintic Duffing equation. Although the exact solution of the cubic Duffingequation has been discussed, this study is the first to describe the exact solution of the cubic–quintic Duffing equation.In this study, the exact solutions of the cubic–quintic Duffing equation are presented. The exact solutions are graphed,and the waveforms of the exact solutions are visualized. The exact solutions of the cubic Duffing equation were obtained by using the leaf functions. There are eight types ofexact solutions.
The exact solution is x ( t ) = A (cid:112) ( ωt )) . (5)The ordinary differential equation is d x ( t )d t − ω x ( t ) + 2 (cid:16) ωA (cid:17) x ( t ) = 0 . (6) The exact solution is x ( t ) = A (cid:112) − (cleaf ( ωt )) . (7)The ordinary differential equation is d x ( t )d t + 3 ω x ( t ) − (cid:16) ωA (cid:17) x ( t ) = 0 . (8) The exact solution is x ( t ) = A (cid:112) ( ωt )) . (9)The ordinary differential equation is d x ( t )d t − ω x ( t ) + 2 (cid:16) ωA (cid:17) x ( t ) = 0 . (10)2 PREPRINT - F
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5, 2021
The exact solution is x ( t ) = A (cid:112) − (sleaf ( ωt )) . (11)The ordinary differential equation is d x ( t )d t − ω x ( t ) + 2 (cid:16) ωA (cid:17) x ( t ) = 0 . (12) The exact solution is x ( t ) = A (cid:112) ( ωt )) + A (cid:112) ( ωt )) . (13)The ordinary differential equation is d x ( t )d t − ω (2 √ x ( t ) + 2 (cid:16) ωA (cid:17) x ( t ) = 0 . (14) The exact solution is x ( t ) = A (cid:112) ( ωt )) − A (cid:112) ( ωt )) . (15)The ordinary differential equation is d x ( t )d t − ω (2 √ − x ( t ) + 2 (cid:16) ωA (cid:17) x ( t ) = 0 . (16) The exact solution is x ( t ) = A (cid:112) (cleafh ( ωt )) + 1 . (17)The ordinary differential equation is d x ( t )d t + 3 ω x ( t ) − (cid:16) ωA (cid:17) x ( t ) = 0 . (18) The exact solution is x ( t ) = A (cid:112) (cleafh ( ωt )) − . (19)The ordinary differential equation is d x ( t )d t − ω x ( t ) − (cid:16) ωA (cid:17) x ( t ) = 0 . (20) The exact solutions of the cubic–quintic Duffing equation are obtained by using the leaf functions. There are six typesof exact solutions.
The exact solution is x ( t ) = A (cid:112) ( ωt ) . (21)The ordinary differential equation is d x ( t )d t + 32 ω x ( t ) − ω A x ( t ) + 34 ω A x ( t ) = 0 . (22)3 PREPRINT - F
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The exact solution is x ( t ) = A (cid:112) − sleaf ( ωt ) . (23)The ordinary differential equation that satisfies the above exact solution is the same as Eq. (22). The exact solution is x ( t ) = A (cid:112) ( ωt ) . (24)The ordinary differential equation that satisfies the above exact solution is the same as Eq. (22). The exact solution is x ( t ) = A (cid:112) − cleaf ( ωt ) . (25)The ordinary differential equation that satisfies the above exact solution is the same as Eq. (22). The exact solution is x ( t ) = A (cid:112) cleafh ( ωt ) + 1 . (26)The ordinary differential equation is d x ( t )d t − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) = 0 . (27) The exact solution is x ( t ) = A (cid:112) cleafh ( ωt ) − . (28)The ordinary differential equation is d x ( t )d t − ω x ( t ) − ω A x ( t ) − ω A x ( t ) = 0 . (29) The graphs in the section are the waveforms or the divergence obtained by numerical analysis based on differentialequations and initial conditions. The waveforms or divergences can also be obtained by using the exact solution. Thewaveforms or the divergences with respect to the exact solutions are plotted for each graph. The vertical and horizontalaxes represent the variables x ( t ) and t , respectively.Fig. 1 (exact solution 1), Fig. 3 (exact solution 2), Fig. 5 (exact solution 3), Fig. 7 (exact solution 4), Fig. 9 (exactsolution 5), Fig. 11 (exact solution 6), Fig. 13 (exact solution 7), Fig. 15 (exact solution 8), Fig. 17 (exact solution 9),Fig. 19 (exact solution 10), Fig. 21 (exact solution 11), Fig. 23 (exact solution 12), Fig. 25 (exact solution 13), andFig. 27 (exact solution 14), show the graphs when the period parameter ω changes, with the amplitude parameter set to A = 1 . From the graphs, one can observe that the amplitude of the wave, x ( t ) , does not change even if the parameter ω changes. The range of x ( t ) with respect to time t satisfies the following inequalities:(i) For Figs. 1 and 5 (exact solutions 1 and 5), (cid:53) x ( t ) (cid:53) √ . (30)(ii) For Figs. 3 and 7 (exact solutions 2 and 4), (cid:53) x ( t ) (cid:53) . (31)4 PREPRINT - F
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5, 2021(iii) For Fig. 9 (exact solution 5), ( (cid:59) . · · · ) (cid:53) x ( t ) (cid:53) √ (cid:59) . · · · ) . (32)(iv) For Fig. 11 (exact solution 6), − ( √ − (cid:59) − . · · · ) (cid:53) x ( t ) (cid:53) √ − (cid:59) . · · · ) . (33)(v) For Fig. 13 (exact solution 7), . (cid:53) x ( t ) . (34)(vi) For Figs. 15 and 27 (exact solution 8 and 14), . (cid:53) x ( t ) . (35)(vii) For Figs. 17, 19, 21, and 23 (exact solutions 9, 10, 11, and 12), . (cid:53) x ( t ) (cid:53) √ . (36)(viii) For Fig. 25 (exact solution 13), √ (cid:53) x ( t ) . (37)When the absolute value of the parameter ω increases, the period of the waveform x ( t ) decreases with constantamplitude. In contrast, when the absolute value of the parameter ω decreases, the period of the waveform x ( t ) increaseswith constant amplitude. The period of wave x ( t ) is defined as T . The relationships with the parameter ω are as follows:(i) For Figs. 1, 3, 5, 7, and 11 (exact solutions 1, 2, 3, 4, and 6) T = π | ω | . (38)(ii) For Fig. 9 (exact solution 5), T = π | ω | . (39)(iii) For Figs. 13, 15, 25, and 27, (exact solutions 7, 8, 13, and 14) T = 2 η | ω | . (40)(iv) For Figs. 17, 19, 21, and 23, (exact solution 9, 10, 11, and 12) T = 2 π | ω | . (41)Figs. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20,22, 24, 26, and 28 show the variations of the amplitude parameter A with ω = 1 .Because the parameter ω is fixed at ω = 1 , the wave periods are T = π (Figs. 2, 4, 6, 8, and 12), T = π (Fig. 10), T = 2 η (Figs. 14, 16, 26, and 28), and T = 2 π (Figs. 18, 20, 22, and 24). As the absolute value | A | increases, thewave undulation increases with the period kept constant. As the absolute value | A | decreases, the wave undulationdecreases with the period kept constant. The range of variable x ( t ) satisfies the following inequalities:(i) For Figs. 2 and 6 (exact solutions 1 and 3), | A | (cid:53) x ( t ) (cid:53) √ | A | . (42)(ii) For Figs. 4 and 8 (exact solutions 2 and 4)s (cid:53) x ( t ) (cid:53) | A | . (43)(iii) For Fig. 10 (exact solution 5), | A | (cid:53) x ( t ) (cid:53) (1 + √ | A | (44)5 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 1: Waves of exact solution 1 with respect to variation of the parameter ω related to the period ( A = 1 )(iv) For Fig. 12 (exact solution 6), − ( √ − | A | (cid:53) x ( t ) (cid:53) ( √ − | A | . (45)(v) For Figs. 14 and 26 (exact solution 7 and 13), √ A (cid:53) x ( t ) ( A (cid:61) , (46) x ( t ) (cid:53) √ A ( A < . (47)(vi) For Figs. 18, 20, 22, and 24 (exact solutions 9, 10, 11, and 12), (cid:53) x ( t ) (cid:53) √ | A | . (48)(vii) For Figs. 16 and 28 (exact solutions 8 and 14), (cid:53) x ( t ) ( A (cid:61) , (49) x ( t ) (cid:53) A < . (50) In a previous study, the exact solutions of the cubic Duffing equation were presented by using the leaf function. However,the calculations of the exact solutions are complicated because they are described by using the leaf function for thephase of the trigonometric function. In this study, the exact solutions of the cubic Duffing equation are presented byusing only leaf functions without using trigonometric functions. Furthermore, the exact solutions are derived for thecubic–quintic Duffing equation, which has strong nonlinearity.The conclusions are as follows:• The exact solutions of the cubic Duffing equation and cubic–quintic Duffing equation are presented by usingleaf functions. When the first derivatives of the leaf function are derived, the first derivatives (cid:112) − (sleaf ( l )) and − (cid:112) − (cleaf ( l )) are obtained. Some of these combinations of factors based on factorization of thefirst derivatives can be applied as the exact solutions of the cubic-quintic Duffing equation.6 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 2: Waves of exact solution 1 with respect to variation of the parameter A related to the amplitude ( ω = 1 )• The exact solutions have two parameters: parameter A related to the amplitude and parameter ω related to theperiod. By using parameter A , the amplitude of the wave according to the cubic-quintic Duffing equation canbe adjusted while the period of the wave is constant. By using parameter ω , the wave period can be adjustedwhile the wave amplitude is constant. The relations between these parameters and the coefficient of the termof the Duffing equation are clarified. Appendix A (Symbols)
The parameter m represents an integer. The symbols η and π represent constants (see Appendices P and Q). The sign ± of the derivative ± (cid:112) − (sleaf ( l )) of the leaf function sleaf ( l ) depends on the domain of the phase l [10]. Theleaf function cleaf ( l ) and the hyperbolic leaf function cleafh ( l ) are the same [11] [12]. Therefore, the differentiationsare derived by dividing them into regions.The symbol lim t ↓ a f ( t ) represents the right-side limit. As the variable t decreases, it approaches point a from right toleft on the t axis of the graph. The symbol lim t ↑ a represents the left-side limit. As the variable t increases, it approachespoint a from left to right on the t axis of the graph. Appendix B (Exact solution 1)
Here we prove that exact solution 1 satisfies the Duffing equation. The first derivatives of exact solution 1 are obtainedas follows:(i) In the case where (2 m − π (cid:53) ωt (cid:53) mπ , d x ( t )d t = A ( ωt ) (cid:16) ω (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) ( ωt )) = Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (51)7 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 3: Waves of exact solution 2 with respect to variation of the parameter ω related to the period ( A = 1 )(ii) In the case where mπ (cid:53) ωt (cid:53) (2 m + 1) π , d x ( t )d t = A ( ωt ) (cid:16) − ω (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) ( ωt )) = − Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (52)The second derivatives of exact solution 1 are obtained as follows:(i) In the case where (2 m − π (cid:53) ωt (cid:53) mπ , d x ( t )d t = Aω (cid:16)(cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) − (cleaf ( ωt )) + Aω cleaf ( ωt ) (cid:16) − ω cleaf ( ωt ) (cid:112) ( ωt )) (cid:17) = Aω (cid:8) − ( ωt )) (cid:9) (cid:112) ( ωt )) . (53)(ii) In the case where mπ (cid:53) ωt (cid:53) (2 m + 1) π , d x ( t )d t = − Aω (cid:16) − (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) − (cleaf ( ωt )) − Aω cleaf ( ωt ) (cid:16) cleaf ( ωt ) (cid:112) ( ωt )) (cid:17) = Aω (cid:8) − ( ωt )) (cid:9) (cid:112) ( ωt )) . (54)By substituting Eq. (5) into Eq. (53) (or Eq. (54)), Eq. (6) of the Duffing equation is derived. The initial conditions areas follows: x (0) = A (cid:112) (0)) = A (cid:112) = √ A, (55) d x (0)d t = ± Aω cleaf (0) (cid:112) − (cleaf (0)) = ± Aω (cid:112) − (1) = 0 . (56)8 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 4: Waves of exact solution 2 with respect to variation of the parameter A related to the amplitude ( ω = 1 ) Appendix C (Exact solution 2)
Here we prove that exact solution 2 satisfies the Duffing equation The first derivatives of exact solution 1 are obtainedas follows:(i) In the case where mπ (cid:53) ωt (cid:53) (2 m + 1) π , d x ( t )d t = A − ( ωt ) (cid:16) ω (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) − (cleaf ( ωt )) = − Aω cleaf ( ωt ) (cid:112) ( ωt )) . (57)(ii) In the case where (2 m − π (cid:53) ωt (cid:53) mπ , d x ( t )d t = A − ( ωt ) (cid:16) − ω (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) − (cleaf ( ωt )) = Aω cleaf ( ωt ) (cid:112) ( ωt )) . (58)The second derivatives of exact solution 2 are obtained as follows:(i) In the case where mπ (cid:53) ωt (cid:53) (2 m + 1) π , d x ( t )d t = − Aω (cid:16) ω (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) ( ωt )) − Aω cleaf ( ωt ) (cid:16) ω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) (cid:17) = − Aω (cid:8) ( ωt )) (cid:9) (cid:112) − (cleaf ( ωt )) . (59)9 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 5: Waves of exact solution 3 with respect to variation of the parameter ω related to the period ( A = 1 )(ii) In the case where (2 m − π (cid:53) ωt (cid:53) mπ , d x ( t )d t = Aω (cid:16) − (cid:112) − (cleaf ( ωt )) (cid:17) (cid:112) ( ωt )) + Aω cleaf ( ωt ) (cid:16) − cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) (cid:17) = − Aω (cid:8) ( ωt )) (cid:9) (cid:112) − (cleaf ( ωt )) . (60)By substituting Eq. (7) into Eq. (59) (or Eq. (60)), Eq. (8) of the Duffing equation is obtained. The initial condition ofthe differential equation, Eq. (8), is as follows: x (0) = A (cid:112) − (cleaf (0)) = A (cid:112) − (1) = 0 . (61)Exact solution 2 is a continuous function. However, it cannot be partially differentiated. Exact solution 2 is notdifferentiable at t = m π ω : lim t ↑ m π ω d x ( t )d t = −√ Aω, (62) lim t ↓ m π ω d x ( t )d t = √ Aω. (63)
Appendix D (Exact solution 3)
Here we prove that exact solution 3 satisfies the Duffing equation. The first derivatives of exact solution 3 are obtainedas follows:(i) In the case where m − π (cid:53) ωt (cid:53) m +12 π , d x ( t )d t = A ( ωt ) (cid:16) ω (cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) ( ωt )) = Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) . (64)10 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 6: Waves of exact solution 3 with respect to variation of the parameter A related to the amplitude ( ω =1)(ii) In the case where m +12 π (cid:53) ωt (cid:53) m +32 π , d x ( t )d t = A ( ωt ) (cid:16) − ω (cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) ( ωt )) = − Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) . (65)The second derivatives of exact solution 3 are obtained as follows:(i) In the case where m − π (cid:53) ωt (cid:53) m +12 π , d x ( t )d t = Aω (cid:16)(cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) − (sleaf ( ωt )) + Aω sleaf ( ωt ) (cid:16) − sleaf ( ωt ) (cid:112) ( ωt )) (cid:17) = Aω (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) . (66)(ii) In the case where m +12 π (cid:53) ωt (cid:53) m +32 π , d x ( t )d t = − Aω (cid:16) − ω (cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) − (sleaf ( ωt )) − Aω sleaf ( ωt ) (cid:16) ω sleaf ( ωt ) (cid:112) ( ωt )) (cid:17) = Aω (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) . (67)By substituting Eq. (9) into Eq. (66) (or Eq. (67)), Eq. (10) of the Duffing equation is obtained. The initial conditionsof the differential equation, Eq. (10), are as follows: x (0) = A (cid:112) (0)) = A (cid:112) = A, (68) d x (0)d t = ± Aω sleaf (0) (cid:112) − (sleaf (0)) = ± Aω · · (cid:112) − (0) = 0 . (69)11 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 7: Waves of exact solution 4 with respect to variation of the parameter ω related to the period ( A = 1 ) Appendix E (Exact solution 4)
Here we prove that exact solution 4 satisfies the Duffing equation The first derivatives of exact solution 4 are obtainedas follows:(i) In the case where m − π (cid:53) ωt (cid:53) m +12 π , d x ( t )d t = A ( ωt ) (cid:16) ω (cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) − (sleaf ( ωt )) = Aω sleaf ( ωt ) (cid:112) ( ωt )) . (70)(ii) In the case where m +12 π (cid:53) ωt (cid:53) m +32 π , d x ( t )d t = A ( ωt ) (cid:16) − ω (cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) − (sleaf ( ωt )) = − Aω sleaf ( ωt ) (cid:112) ( ωt )) . (71)The second derivatives of exact solution 4 are obtained as follows:(i) In the case where m − π (cid:53) ωt (cid:53) m +12 π , d x ( t )d t = Aω (cid:16)(cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) ( ωt )) + Aω sleaf ( ωt ) (cid:16) sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) (cid:17) = Aω (cid:8) ( ωt )) (cid:9) (cid:112) − (sleaf ( ωt )) . (72)12 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 8: Waves of exact solution 4 with respect to variation of the parameter A related to the amplitude ( ω = 1 )(ii) In the case where m +12 π (cid:53) ωt (cid:53) m +32 π , d x ( t )d t = − Aω (cid:16) − ω (cid:112) − (sleaf ( ωt )) (cid:17) (cid:112) ( ωt )) − Aω sleaf ( ωt ) (cid:16) − ω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) (cid:17) = Aω (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) . (73)By substituting Eq. (11) into Eq. (72) (or Eq. (73)), Eq. (12) of the Duffing equation is obtained. The initial conditionof the differential equation, Eq. (12), is as follows: x (0) = A (cid:112) − (sleaf (0)) = A (cid:112) − (0) = A. (74)Exact solution 4 is a continuous function. However, it cannot be partially differentiated. Exact solution 4 is notdifferentiable at t = m − π ω : lim t ↑ m − π ω d x ( t )d t = −√ Aω, (75) lim t ↓ m − π ω d x ( t )d t = √ Aω. (76)
Appendix F (Exact solution 5)
Here we prove that exact solution 5 satisfies the Duffing equation. The first derivative of exact solution 5 is obtained byEq. (51) (or Eq. (52)) and Eq. (64) (or Eq. (65)).(i) In the case where m − π (cid:53) ωt (cid:53) m π , d x ( t )d t = Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) + Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (77)13 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 9: Waves of exact solution 5 with respect to variation of the parameter ω related to the period ( A = 1 )(ii) In the case where m π (cid:53) ωt (cid:53) m +12 π , d x ( t )d t = Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) − Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (78)(iii) In the case where m +12 π (cid:53) ωt (cid:53) m +22 π , d x ( t )d t = − Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) − Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (79)(iv) In the case where m +22 π (cid:53) ωt (cid:53) m +32 π , d x ( t )d t = − Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) + Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (80)With respect to all domains in (i)–(iv), the second derivative of exact solution 5 is obtained as follows: d x ( t )d t = Aω (cid:8) − ( ωt )) (cid:9) (cid:112) ( ωt )) + Aω (cid:8) − ( ωt )) (cid:9) (cid:112) ( ωt )) . (81)By using Eq. (148), the relational expression is obtained by the following transformation of the equation: (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) = 1 + (cleaf ( ωt )) + 1 − (cleaf ( ωt )) ( ωt )) (cid:112) ( ωt )) = 2 (cid:112) ( ωt )) = √ (cid:112) ( ωt )) . (82)14 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 10: Waves of exact solution 5 with respect to variation of the parameter A related to the amplitude ( ω = 1 )Similarly, the following relational expression is obtained: (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) = 1 + (sleaf ( ωt )) + 1 − (sleaf ( ωt )) ( ωt )) (cid:112) ( ωt )) = 2 (cid:112) ( ωt )) = √ (cid:112) ( ωt )) . (83)Next, the following equation is derived: ( x ( t )) = A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) + 3 A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) + 3 A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) + A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) = A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) + 3 A √ (cid:112) ( ωt )) + 3 A √ (cid:112) ( ωt )) + A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) = A (sleaf ( ωt )) (cid:112) ( ωt )) + A (cleaf ( ωt )) (cid:112) ( ωt )) + A (3 √ (cid:112) ( ωt )) + A (3 √ (cid:112) ( ωt )) . (84)By using Eq. (81), the above equation is transformed as follows: (sleaf ( ωt )) (cid:112) ( ωt )) + (cleaf ( ωt )) (cid:112) ( ωt )) = ( x ( t )) A − (3 √ x ( t ) A . (85)By substituting Eq. (82) into Eq. (81), Eq. (14) of the Duffing equation is obtained. The initial conditions of Eq. (14)are as follows: x (0) = A (cid:112) (0)) + A (cid:112) (0)) = (1 + √ A, (86)15 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 11: Waves of exact solution 6 with respect to variation of the parameter ω related to the period ( A = 1 ) d x (0)d t = 0 . (87) Appendix G (Exact solution 6)
Here we prove that exact solution 6 satisfies the Duffing equation.The first derivative of exact solution 6 is obtained by Eq. (51) (or Eq. (52)) and Eq. (64) (or Eq. (65)).(i) In the case where m − π (cid:53) ωt (cid:53) m π , d x ( t )d t = Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) − Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (88)(ii) In the case where m π (cid:53) ωt (cid:53) m +12 π , d x ( t )d t = Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) + Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (89)(iii) In the case where m +12 π (cid:53) ωt (cid:53) m +22 π , d x ( t )d t = − Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) + Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (90)(iv) In the case where m +22 π (cid:53) ωt (cid:53) m +32 π , d x ( t )d t = − Aω sleaf ( ωt ) (cid:112) − (sleaf ( ωt )) − Aω cleaf ( ωt ) (cid:112) − (cleaf ( ωt )) . (91)With respect to all domains in (i)–(iv), the second derivative of exact solution 6 is obtained as follows: d x ( t )d t = Aω (cid:8) − ( ωt )) (cid:9) (cid:112) ( ωt )) − Aω (cid:8) − ( ωt )) (cid:9) (cid:112) ( ωt )) . (92)16 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 12: Waves of exact solution 6 with respect to variation of the parameter A related to the amplitude ( ω = 1 )Next, the following equation is derived: ( x ( t )) = A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) + 3 A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) − A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) − A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) = A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) + 3 A √ (cid:112) ( ωt )) − A √ (cid:112) ( ωt )) − A (cid:8) ( ωt )) (cid:9) (cid:112) ( ωt )) = A (sleaf ( ωt )) (cid:112) ( ωt )) − A (cleaf ( ωt )) (cid:112) ( ωt )) + A (3 √ − (cid:112) ( ωt )) − A (3 √ − (cid:112) ( ωt )) . (93)By using Eq. (93), the above equation is transformed as follows: (sleaf ( ωt )) (cid:112) ( ωt )) − (cleaf ( ωt )) (cid:112) ( ωt )) = ( x ( t )) A − (3 √ − x ( t ) A . (94)By substituting Eq. (91) into Eq. (92), Eq. (16) of the Duffing equation is obtained. The initial conditions of Eq. (16)are as follows: x (0) = A (cid:112) (0)) − A (cid:112) (0)) = (1 − √ A, (95) d x (0)d t = 0 . (96)17 PREPRINT - F
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5, 2021 ‐ ‐ ‐ (cid:3404) (cid:3399) A (cid:3404) (cid:3399) A (cid:3404) (cid:3399) Figure 13: Divergences of exact solution 7 with respect to variation of the parameter ω related to the period ( A = 1 ) Appendix H (Exact solution 7)
Here we prove that exact solution 7 satisfies the Duffing equation The first derivatives of exact solution 7 are obtainedas follows:(i) In the case where mη (cid:53) t (cid:53) (2 m + 1) η , d x ( t )d t = A ( ωt ) (cid:16) ω (cid:112) (cleafh ( ωt )) − (cid:17) (cid:112) (cleafh ( ωt )) + 1= Aω cleafh ( ωt ) (cid:112) (cleafh ( ωt )) − . (97)(ii) In the case where (2 m + 1) η (cid:53) t (cid:53) (2 m + 2) η , d x ( t )d t = A ( ωt ) (cid:16) − ω (cid:112) (cleafh ( ωt )) − (cid:17) (cid:112) (cleafh ( ωt )) + 1= − Aω cleafh ( ωt ) (cid:112) (cleafh ( ωt )) − . (98)With respect to all domains in (i) and (ii), the second derivative of exact solution 7 is obtained as follows: d x ( t )d t = Aω (cid:8) ( ωt )) − (cid:9) (cid:112) (cleafh ( ωt )) + 1 . (99)By substituting Eq. (17) into Eq. (99), Eq. (18) of the Duffing equation is obtained. The initial conditions of Eq. (18)are as follows: x (0) = A (cid:112) (cleafh (0)) + 1 = √ A, (100) d x (0)d t = 0 . (101)18 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 14: Divergences of exact solution 7 with respect to variation of the parameter A related to the amplitude ( ω = 1 ) Appendix I (Exact solution 8)
Here we prove that exact solution 8 satisfies the Duffing equation The first derivatives of exact solution 8 are obtainedas follows:(i) In the case where mη (cid:53) t (cid:53) (2 m + 1) η , d x ( t )d t = A ( ωt ) (cid:16) ω (cid:112) (cleafh ( ωt )) − (cid:17) (cid:112) (cleafh ( ωt )) − Aω cleafh ( ωt ) (cid:112) (cleafh ( ωt )) + 1 . (102)(ii) In the case where (2 m + 1) η (cid:53) t (cid:53) (2 m + 2) η , d x ( t )d t = A ( ωt ) (cid:16) − ω (cid:112) (cleafh ( ωt )) − (cid:17) (cid:112) (cleafh ( ωt )) − − Aω cleafh ( ωt ) (cid:112) (cleafh ( ωt )) + 1 . (103)With respect to all domains in (i) and (ii), the second derivative of the exact solution 8 is obtained as follows: d x ( t )d t = Aω (cid:8) ( ωt )) + 1 (cid:9) (cid:112) (cleafh ( ωt )) − . (104)By substituting Eq. (19) into Eq. (104) (or Eq. (105)), Eq. (20) of the Duffing equation is obtained. The initial conditionof Eq. (20) is as follows: x (0) = A (cid:112) (cleafh (0)) − . (105)Exact solution 8 is a continuous function. However, it cannot be partially differentiated. Exact solution 8 is notdifferentiable at t = mη ω : lim t ↑ mη ω d x ( t )d t = −√ Aω, (106)19
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5, 2021 ‐ ‐ ‐ (cid:3404) (cid:3399) A (cid:3404) (cid:3399) A (cid:3404) (cid:3399) Figure 15: Divergences of exact solution 8 with respect to variation of the parameter ω related to the period ( A = 1 ) lim t ↓ mη ω d x ( t )d t = √ Aω. (107)
Appendix J (Exact solution 9)
Here we prove that exact solution 9 satisfies the Duffing equation. The first derivatives of exact solution 9 are obtainedas follows:(i) In the case where m − π (cid:53) t (cid:53) m +12 π , d x ( t )d t = Aω (cid:112) − (sleaf ( ωt )) (cid:112) ( ωt )= Aω (cid:112) − sleaf ( ωt ) + (sleaf ( ωt )) − (sleaf ( ωt )) . (108)(ii) In the case where m +12 π (cid:53) t (cid:53) m +32 π , d x ( t )d t = Aω − (cid:112) − (sleaf ( ωt )) (cid:112) ( ωt )= − Aω (cid:112) − sleaf ( ωt ) + (sleaf ( ωt )) − (sleaf ( ωt )) . (109)The second derivatives of exact solution 9 are obtained as follows:20 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 16: Divergences of exact solution 8 with respect to variation of the parameter A related to the amplitude ( ω = 1 )(i) In the case where m − π (cid:53) t (cid:53) m +12 π , d x ( t )d t = Aω {− ( ωt ) − ( ωt )) } (cid:112) − (sleaf ( ωt )) (cid:112) − sleaf ( ωt ) + (sleaf ( ωt )) − (sleaf ( ωt )) = Aω {− ( ωt ) − ( ωt )) } (cid:112) ( ωt )= Aω (cid:40) − (cid:18) x ( t ) A − (cid:19) − (cid:18) x ( t ) A − (cid:19) (cid:41) x ( t ) A = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (110)(ii) In the case where m +12 π (cid:53) t (cid:53) m +32 π , d x ( t )d t = − Aω {− ( ωt ) − ( ωt )) } ( − (cid:112) − (sleaf ( ωt )) ) (cid:112) − sleaf ( ωt ) + (sleaf ( ωt )) − (sleaf ( ωt )) = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (111)Eq. (22) of the Duffing equation is obtained. The initial condition of the differential equation, Eq. (22), is as follows: x (0) = A (cid:112) (0) = A √ A. (112)Exact solution 9 is a continuous function. However, it cannot be partially differentiated. Exact solution 9 is notdifferentiable at t = m − π ω : lim t ↑ m − π ω d x ( t )d t = − Aω, (113) lim t ↓ m − π ω d x ( t )d t = Aω. (114)21
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5, 2021 ‐ ‐ ‐ ‐ 𝜔 (cid:3404) 𝜔 (cid:3404) (cid:3398) 𝜔 (cid:3404) 𝜔 (cid:3404) -3 𝜔 (cid:3404) -1 𝜔 (cid:3404) Figure 17: Waves of exact solution 9 with respect to variation of the parameter ω related to the period ( A = 1 ) Appendix K (Exact solution 10)
Here we prove that exact solution 10 satisfies the Duffing equation. The first derivatives of exact solution 10 are obtainedas follows:(i) In the case where m − π (cid:53) t (cid:53) m +12 π , d x ( t )d t = − Aω (cid:112) − (sleaf ( ωt )) (cid:112) − sleaf ( ωt )= − Aω (cid:112) ( ωt ) + (sleaf ( ωt )) + (sleaf ( ωt )) . (115)(ii) In the case where m +12 π (cid:53) t (cid:53) m +32 π , d x ( t )d t = − Aω − (cid:112) − (sleaf ( ωt )) (cid:112) − sleaf ( ωt )= Aω (cid:112) ( ωt ) + (sleaf ( ωt )) + (sleaf ( ωt )) . (116)The second derivatives of exact solution 10 are obtained as follows:(i) In the case where m − π (cid:53) t (cid:53) m +12 π , d x ( t )d t = − Aω { ( ωt ) + 3(sleaf ( ωt )) } (cid:112) − (sleaf ( ωt )) (cid:112) ( ωt ) + (sleaf ( ωt )) + (sleaf ( ωt )) = − Aω { ( ωt ) + 3(sleaf ( ωt )) } (cid:112) − sleaf ( ωt )= − Aω (cid:40) (cid:18) − x ( t ) A (cid:19) + 3 (cid:18) − x ( t ) A (cid:19) (cid:41) x ( t ) A = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (117)22 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 18: Waves of exact solution 9 with respect to variation of the parameter A related to the amplitude ( ω = 1 )(ii) In the case where m +12 π (cid:53) t (cid:53) m +32 π , d x ( t )d t = Aω { ( ωt ) + 3(sleaf ( ωt )) } ( − (cid:112) − (sleaf ( ωt )) ) (cid:112) ( ωt ) + (sleaf ( ωt )) + (sleaf ( ωt )) = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (118)Eq. (22) of the Duffing equation is obtained. The initial condition of the differential equation, Eq. (22), is as follows: x (0) = A (cid:112) − sleaf (0) = A √ − A. (119)Exact solution 10 is a continuous function. However, it cannot be partially differentiated. Exact solution 10 is notdifferentiable at t = m +12 π ω : lim t ↑ m +12 π ω d x ( t )d t = − Aω, (120) lim t ↓ m +12 π ω d x ( t )d t = Aω. (121)
Appendix L (Exact solution 11)
Here we prove that exact solution 11 satisfies the Duffing equation. The first derivatives of exact solution 11 are obtainedas follows:(i) In the case where mπ (cid:53) t (cid:53) (2 m + 1) π , d x ( t )d t = Aω − (cid:112) − (cleaf ( ωt )) (cid:112) ( ωt )= − Aω (cid:112) − cleaf ( ωt ) + (cleaf ( ωt )) − (cleaf ( ωt )) . (122)23 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) -1 𝜔 (cid:3404) 𝜔 (cid:3404) -3 𝜔 (cid:3404) 𝜔 (cid:3404) 𝜔 (cid:3404) (cid:3398) Figure 19: Waves of exact solution 10 with respect to variation of the parameter ω related to the period ( A = 1 )(ii) In the case where (2 m + 1) π (cid:53) t (cid:53) (2 m + 2) π , d x ( t )d t = Aω (cid:112) − (cleaf ( ωt )) (cid:112) ( ωt )= Aω (cid:112) − cleaf ( ωt ) + (cleaf ( ωt )) − (cleaf ( ωt )) . (123)The second derivatives of exact solution 11 are obtained as follows:(i) In the case where mπ (cid:53) t (cid:53) (2 m + 1) π , d x ( t )d t = − Aω {− ( ωt ) − ( ωt )) } ( − ω (cid:112) − (cleaf ( ωt )) )2 (cid:112) − cleaf ( ωt ) + (cleaf ( ωt )) − (cleaf ( ωt )) = Aω {− ( ωt ) − ( ωt )) } (cid:112) ( ωt )= Aω (cid:40) − (cid:18) x ( t ) A − (cid:19) − (cid:18) x ( t ) A − (cid:19) (cid:41) x ( t ) A = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (124)(ii) In the case where (2 m + 1) π (cid:53) t (cid:53) (2 m + 2) π , d x ( t )d t = Aω {− ( ωt ) − ( ωt )) } ( ω (cid:112) − (cleaf ( ωt )) )2 (cid:112) − cleaf ( ωt ) + (cleaf ( ωt )) − (cleaf ( ωt )) = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (125)Eq. (22) of the Duffing equation is obtained. The initial condition of the differential equation, Eq. (22), is as follows: x (0) = A (cid:112) (0) = A √ √ A. (126)24 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 20: Waves of exact solution 10 with respect to variation of the parameter A related to the amplitude ( ω = 1 )Exact solution 11 is a continuous function. However, it cannot be partially differentiated. Exact solution 11 is notdifferentiable at t = (2 m − π ω : lim t ↑ (2 m − π ω d x ( t )d t = − Aω, (127) lim t ↓ (2 m − π ω d x ( t )d t = Aω. (128)
Appendix M (Exact solution 12)
Here we prove that exact solution 12 satisfies the Duffing equation. The first derivatives of exact solution 12 are obtainedas follows:(i) In the case where mπ (cid:53) t (cid:53) (2 m + 1) π , d x ( t )d t = Aω − ( − (cid:112) − (cleaf ( ωt )) )2 (cid:112) − cleaf ( ωt )= Aω (cid:112) ( ωt ) + (cleaf ( ωt )) + (cleaf ( ωt )) . (129)(ii) In the case where (2 m + 1) π (cid:53) t (cid:53) (2 m + 2) π , d x ( t )d t = Aω − (cid:112) − (cleaf ( ωt )) (cid:112) − cleaf ( ωt )= − Aω (cid:112) ( ωt ) + (cleaf ( ωt )) + (cleaf ( ωt )) . (130)The second derivatives of exact solution 12 are obtained as follows:25 PREPRINT - F
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5, 2021 ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 21: Waves of exact solution 11 with respect to variation of the parameter ω related to the period ( A = 1 )(i) In the case where mπ (cid:53) t (cid:53) (2 m + 1) π , d x ( t )d t = Aω { ( ωt ) + 3(cleaf ( ωt )) } ( − ω (cid:112) − (cleaf ( ωt )) )2 (cid:112) ( ωt ) + (cleaf ( ωt )) + (cleaf ( ωt )) = − Aω { ( ωt ) + 3(cleaf ( ωt )) } (cid:112) − cleaf ( ωt )= − Aω (cid:40) (cid:18) − x ( t ) A (cid:19) + 3 (cid:18) − x ( t ) A (cid:19) (cid:41) x ( t ) A = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (131)(ii) In the case where (2 m + 1) π (cid:53) t (cid:53) (2 m + 2) π , d x ( t )d t = − Aω { ( ωt ) + 3(cleaf ( ωt )) } ( ω (cid:112) − (cleaf ( ωt )) )2 (cid:112) ( ωt ) + (cleaf ( ωt )) + (cleaf ( ωt )) = − ω x ( t ) + 2 ω A x ( t ) − ω A x ( t ) . (132)Eq. (22) of the Duffing equation is obtained. The initial condition of the differential equation, Eq. (22), is as follows: x (0) = A (cid:112) − cleaf (0) = A √ − . (133)Exact solution 12 is a continuous function. However, it cannot be partially differentiated. Exact solution 12 is notdifferentiable at t = 2 m π ω : lim t ↑ m π ω d x ( t )d t = − Aω, (134) lim t ↓ m π ω d x ( t )d t = Aω. (135)26
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 22: Waves of exact solution 11 with respect to variation of the parameter A related to the amplitude ( ω = 1 ) Appendix N (Exact solution 13)
Here we prove that exact solution 13 satisfies the Duffing equation. The first derivatives of exact solution 13 are obtainedas follows:(i) In the case where mη < t < (4 m + 1) η , d x ( t )d t = Aω (cid:112) (cleafh ( ωt )) − (cid:112) cleafh ( ωt ) + 1= Aω (cid:112) (cleafh ( ωt )) − (cleafh ( ωt )) + cleafh ( ωt ) − . (136)(ii) In the case where (4 m − η < t < mη , d x ( t )d t = A − ω (cid:112) (cleafh ( ωt )) − (cid:112) cleafh ( ωt ) + 1= − Aω (cid:112) (cleafh ( ωt )) − (cleafh ( ωt )) + cleafh ( ωt ) − . (137)With respect to all domains in (i) and (ii), the second derivative of exact solution 13 is obtained as follows: d x ( t )d t = 32 ω x ( t ) − ω A x ( t ) + 34 ω A x ( t ) . (138)Eq. (27) of the Duffing equation is obtained. Because the function cleafh ( t ) becomes negative in the domain (4 m + 1) η < t < (4 m + 3) η , exact solution 13 does not have a real number.The initial conditions of the differential equation, Eq. (27), are as follows: x (0) = A (cid:112) cleafh (0) + 1 = A √ √ A, (139) d x (0)d t . = 0 (140)27 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ 𝜔 (cid:3404) 𝜔 (cid:3404) -2 𝜔 (cid:3404) 𝜔 (cid:3404) (cid:3398) 𝜔 (cid:3404) (cid:3398) 𝜔 (cid:3404) Figure 23: Waves of exact solution 12 with respect to variation of the parameter ω related to the period ( A = 1 ) Appendix O (Exact solution 14)
Here we prove that exact solution 14 satisfies the Duffing equation. The first derivatives of exact solution 14 are obtainedas follows:(i) In the case where the domain mη < t < (4 m + 1) η , d x ( t )d t = Aω (cid:112) (cleafh ( ωt )) − (cid:112) cleafh ( ωt ) − Aω (cid:112) (cleafh ( ωt )) + (cleafh ( ωt )) + cleafh ( ωt ) + 1 . (141)(ii) In the case where the domain (4 m − η < t < mη , d x ( t )d t = Aω − (cid:112) (cleafh ( ωt )) − (cid:112) cleafh ( ωt ) − − Aω (cid:112) (cleafh ( ωt )) + (cleafh ( ωt )) + cleafh ( ωt ) + 1 . (142)With respect to all domains in (i) and (ii), the second derivative of exact solution 14 is obtained as follows: d x ( t )d t = 32 ω x ( t ) + 2 ω A x ( t ) + 34 ω A x ( t ) . (143)Eq. (29) of the Duffing equation is obtained. The initial condition of the differential equation, Eq. (29), is as follows: x (0) = A (cid:112) cleafh (0) − A √ − . (144)Exact solution 14 is a continuous function. However, it cannot be partially differentiated. Exact solution 14 is notdifferentiable at t = 4 m η ω : lim t ↑ m η ω d x ( t )d t = − Aω, (145)28
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 24: Waves of exact solution 12 with respect to variation of the parameter A related to the amplitude ( ω = 1 ) lim t ↓ m η ω d x ( t )d t = Aω. (146)
Appendix P
The limit of the hyperbolic leaf function cleafh ( l ) is calculated using the following equation [12]: η = (cid:90) ∞ √ t − dt = 1 . · · · . (147) Appendix Q
The constant π is calculated using the following equation [10] [11]: π = 2 (cid:90) √ − t dt = 2 . · · · . (148) Appendix R
The relation equations with the basis n = 2 are described. The relation equation between the leaf function sleaf ( l ) andthe leaf function cleaf ( l ) is as follows [11]: (sleaf ( l )) + (cleaf ( l )) + (sleaf ( l )) · (cleaf ( l )) = 1 . (149) References [1] P. Pirmohabbati, A. Sheikhani, H. Najafi, and Ali Abdolahzadeh Ziabari. Numerical solution of full fractionalduffing equations with cubic-quintic-heptic nonlinearities.
AIMS Mathematics , 5:1621–1641, 02 2020.29
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5, 2021 ‐ ‐ ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 25: Divergences of exact solution 13 with respect to variation of the parameter ω related to the period ( A = 1 )[2] Iman Khatami, Ehsan Zahedi, and Mohsen Zahedi. Efficient solution of nonlinear duffing oscillator. Journal ofApplied and Computational Mechanics , 6(2):219–234, 2020.[3] Md Abdur Razzak. An analytical approximate technique for solving cubic quintic duffing oscillator.
AlexandriaEngineering Journal , 55(3):2959 – 2965, 2016.[4] S.K. Lai, C.W. Lim, B.S. Wu, C. Wang, Q.C. Zeng, and X.F. He. Newton harmonic balancing approach foraccurate solutions to nonlinear cubic quintic duffing oscillators.
Applied Mathematical Modelling , 33(2):852 –866, 2009.[5] Prakash Das, Debabrata Singh, and M.M. Panja. An efficient scheme for accurate closed-form approximatesolution of some duffing-and lienard-type equations. 16:8213–8225, 02 2019.[6] Gholam-Ali Zakeri and Emmanuel Yomba. Exact solutions of a generalized autonomous duffing-type equation.
Applied Mathematical Modelling , 39(16):4607 – 4616, 2015.[7] Alex Elias-Zuniga. Exact solution of the cubic-quintic duffing oscillator.
Applied Mathematical Modelling ,37(4):2574 – 2579, 2013.[8] Kazunori Shinohara. Exact solutions of the cubic duffing equation by leaf functions under free vibration.
ComputerModeling in Engineering & Sciences , 115(2):149–215, 2018.[9] Kazunori Shinohara. Damped and divergence exact solutions for the duffing equation using leaf functions andhyperbolic leaf functions.
Computer Modeling in Engineering & Sciences , 118(3):599–647, 2019.[10] Kazunori Shinohara. Special function: Leaf function r = sleaf n ( l ) (first report). Bulletin of Daido University ,51:23–38, 2016.[11] Kazunori Shinohara. Special function: Leaf function r = cleaf n ( l ) (second report). Bulletin of Daido University ,51:39–68, 2016.[12] Kazunori Shinohara. Special function: Hyperbolic leaf function r = cleafh n ( l ) (second report). Bulletin of DaidoUniversity , 52:83–105, 2017. 30
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 26: Divergences of exact solution 13 with respect to variation of the parameter A related to the amplitude( ω = 1 ) ‐ ‐ ‐ ‐ ‐ 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) 𝜔 (cid:3404) (cid:3399) Figure 27: Divergences of exact solution 14 with respect to variation of the parameter ω related to the period ( A = 1 )31 PREPRINT - F
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5, 2021 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ (cid:3404) A (cid:3404) A (cid:3404) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) A (cid:3404) (cid:3398) Figure 28: Divergences of exact solution 14 with respect to variation of the parameter A related to the amplitude( ω = 1= 1