**6. Approximate analytic solution of the damped and trigonometric forced Duffing-Helmholtz equation**

Let us define the following new i.v.p.

$$\begin{cases} \ddot{\mathbf{x}} + 2\boldsymbol{\varepsilon}\,\dot{\mathbf{x}} + p\boldsymbol{\varkappa} + q\mathbf{x}^2 + r\mathbf{x}^3 = F\cos\left(\boldsymbol{\varkappa}t\right), \\\\ \boldsymbol{\varkappa}(\mathbf{0}) = \boldsymbol{\varkappa}\_0 \,\boldsymbol{\&}\,\mathbf{x}'(\mathbf{0}) = \dot{\boldsymbol{\varkappa}}\_0. \end{cases} \tag{82}$$

We suppose that *<sup>q</sup>*<sup>2</sup> � <sup>4</sup>*pr*<0, and the following residual is defined

$$R(t) \equiv \ddot{\varkappa}(t) + 2\varepsilon \dot{\varkappa}(t) + p\varkappa(t) + q\varkappa^2(t) + r\varkappa^3(t) - F\cos(\alpha t). \tag{83}$$

Let us define the solution of i.v.p. (82) as follows

$$\mathbf{x}(t) = \exp\left(-\rho t\right)\mathbf{y}(t) + c\_1 \cos\left(a\mathbf{t}\right) + c\_2 \sin\left(a\mathbf{t}\right),\tag{84}$$

where

$$\begin{aligned} &9F^2c\_1^3 + 96e^2Fro^2c\_1^2 + \\ &4\left(64e^4o^4 + 16e^2o^6 + 3F^2pr - 3F^2ro^2 + 16e^2p^2o^2 - 32e^2po^4\right)c\_1 \\ &- 4F\left(-16e^2o^4 + 3F^2r + 16e^2po^2\right) = 0. \\ &- 6144e^3Fro^3 + 432F^2r^3c\_2^3 + 2304eFr^2o(p - o^2)c\_2^2 \\ &+ 3072e^2ro^2\left(4e^2o^2 + p^2 - 2po^2 + o^4\right)c\_2 = 0. \end{aligned} \tag{86}$$

The function *y* � *y t*ð Þ is a solution to the i.v.p.

$$\begin{cases} \mathcal{y}''(t) + 2\epsilon \mathbf{y}'(t) + \bar{p}\mathbf{y}(t) + q\mathbf{y}(t)^2 + r\mathbf{y}(t)^3 = \mathbf{0}, \\\ \mathcal{y}(\mathbf{0}) = (\mathbf{x}\_0 - \mathbf{c}\_1) \otimes \mathbf{y}'(\mathbf{0}) = (\dot{\mathbf{x}}\_0 - a\mathbf{c}\_2). \end{cases} \tag{87}$$

where *<sup>p</sup>*<sup>~</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> <sup>2</sup>*<sup>p</sup>* <sup>þ</sup> <sup>3</sup>*rc*<sup>2</sup> <sup>1</sup> <sup>þ</sup> <sup>3</sup>*rc*<sup>2</sup> <sup>2</sup> � <sup>4</sup>*ερ* <sup>þ</sup> <sup>2</sup>*ρ*<sup>2</sup> � �. The value of *ρ* can be determined from the following equation

$$\begin{aligned} \left(4\rho^3 - 12\varepsilon\rho^2 + \left(4p + 8\varepsilon^2 - 5qc\_1 + 12rc\_1^2 + 6rc\_2^2 + 5q\mathbf{x}\_0 - 12rc\_1\mathbf{x}\_0 + 6r\mathbf{x}\_0^2\right)\rho \\ -2\varepsilon\left(2p - 2qc\_1 + 5rc\_1^2 + 3rc\_2^2 + 2q\mathbf{x}\_0 - 4rc\_1\mathbf{x}\_0 + 2r\mathbf{x}\_0^2\right) = 0. \end{aligned} \tag{88}$$

**Example 8.**

Let

$$\begin{cases} \ddot{\mathbf{x}} + \mathbf{0}.2 \ \dot{\mathbf{x}} + \mathbf{1} \mathbf{3} \mathbf{x} + \mathbf{x}^2 + \mathbf{x}^3 = \mathbf{0}.25 \cos \left( \mathbf{0}.5t \right), \\\\ \mathbf{x}(\mathbf{0}) = \mathbf{0} \ \mathbf{8} \mathbf{x}'(\mathbf{0}) = -\mathbf{0}.2. \end{cases} \tag{89}$$

The approximate analytic solution of the i.v.p. (89) is given by

$$\begin{aligned} \text{x}\_{\text{app}}(t) &= e^{-0.100096t} \begin{pmatrix} 0.0587764- \\ 0 \\ \hline 1+0.914739 \wp(0.529144 - 13766.9(1 - e^{0.000072639}); 14.0404, 10.1967) \end{pmatrix} \\ &+ 0.000153771 \sin(0.5t) + 0.0196062 \cos(0.5t) \end{aligned}$$

(90)

The distance error according to the RK4 numerical solution is calculated as

$$\max\_{0 \le t \le 60} \left| \boldsymbol{\kappa}\_{\text{app}}(t) - \boldsymbol{\kappa}\_{\text{RK4}}(t) \right| = \mathbf{0}.000671928. \tag{91}$$

Moreover, solution (90) is compared with RK4 solution as shown in **Figure 8**.

*Analytical Solutions of Some Strong Nonlinear Oscillators DOI: http://dx.doi.org/10.5772/intechopen.97677*

**Figure 8.** *A comparison between solution (90) and RK4 solution.*
