**2.2 Second case: Δ < 0**

For <sup>Δ</sup> <sup>&</sup>lt;0, in this case *<sup>q</sup>*<sup>&</sup>lt; 0 and then, *<sup>δ</sup>* <sup>¼</sup> *<sup>p</sup>*2�<sup>Δ</sup> �*<sup>q</sup>* <sup>&</sup>gt; 0, *<sup>δ</sup>* <sup>¼</sup> def <sup>2</sup>*<sup>p</sup>* <sup>þ</sup> *qx*<sup>2</sup> 0 � �*x*<sup>2</sup> <sup>0</sup> <sup>þ</sup> <sup>2</sup>*x*\_ <sup>2</sup> <sup>0</sup> >0. Let us introduce the solution in the following form

$$\mathbf{x}(t) = A - \frac{2A}{\mathbf{1} + \mathbf{y}(t)},\tag{18}$$

where *y* ¼ *y t*ð Þ is a solution of some Duffing equation

$$y''(t) + my(t) + ny^3(t),\tag{19}$$

with initial conditions

$$\begin{split} \boldsymbol{\jmath}(\mathbf{0}) &= \boldsymbol{\jmath}\_{0} = \frac{2\boldsymbol{A}\dot{\boldsymbol{x}}\_{0}}{\left(\boldsymbol{A} - \boldsymbol{\varkappa}\_{0}\right)^{2}} \\ \boldsymbol{\jmath}^{\prime}(\mathbf{0}) &= \dot{\boldsymbol{\jmath}}\_{0} = \frac{\boldsymbol{A} + \boldsymbol{\varkappa}\_{0}}{\boldsymbol{A} - \boldsymbol{\varkappa}\_{0}}. \end{split} \tag{20}$$

**Figure 1.** *A comparison between the analytical solution (17) and the approximate numerical RK4.*

Inserting ansatz (18) into Eq. (4) and taking the below relation into account

$$\left(\mathbf{y}'(t)\right)^2 = \dot{\mathbf{y}}\_0 + m\mathbf{y}\_0^2 + \frac{n}{2}\mathbf{y}\_0^4 - m\mathbf{y}^2(t) - \frac{n}{2}\mathbf{y}^4(t),\tag{21}$$

we get

$$\begin{aligned} &-A\left(A^2q + 4m\dot{y}\_0^2 + 2n\dot{y}\_0^4 + p + 4\dot{\varphi}\_0^2\right) \\ &+ A\left(3A^2q - 2m - p\right)y(t) \\ &-A\left(3A^2q - 2m - p\right)y(t)^2 \\ &+ A\left(A^2q - 2n + p\right)y(t)^3 = 0. \end{aligned} \tag{22}$$

Equating the coefficients of *y <sup>j</sup>* ð Þ*t* to zero, gives an algebraic system. A solution to this system gives

$$\begin{aligned} m &= \frac{1}{2} \left( -p + 3A^2 q \right), n = \frac{1}{2} \left( p + A^2 q \right), \\ A &= \sqrt[4]{\frac{(2p + qx\_0^2)x\_0^2 + 2\dot{x}\_0^2}{-q}} = \sqrt[4]{\frac{\delta}{-q}}. \end{aligned}$$

Note that the i.v.p. (19)-(20) has a positive discriminant and it is given by

$$\left(\left(m+n\mathbf{y}\_0^2\right)^2+2n\,\dot{\mathbf{y}}\_0^2\right)=\frac{\delta(A-\varkappa\_0)^4\left(2A^4\dot{\mathbf{x}}\_0^2+\delta\left(A^2+\varkappa\_0^2\right)^2\right)}{4A^8\varkappa\_0^2}.$$

Then the problem reduces to the first case. Accordingly, the solution of the i.v.p. (4)-(5) maybe written in the form,

$$\mathbf{x}(t) = A - \frac{\mathbf{2}A}{\mathbf{1} + B \frac{b\_0 \text{cn}(\sqrt{\alpha t}|m) + b\_1 \text{sn}(\sqrt{\alpha t}|m) \text{dn}(\sqrt{\alpha t}|m)}{\mathbf{1} + b\_2 \text{sn}^2(\sqrt{\alpha t}|m)}},\tag{23}$$

where

$$\begin{aligned} m &= \frac{B^2(A^2q+p)}{2A^2(B^2+3)q + 2(B^2-1)p}, \\ m &= \frac{1}{2}(A^2(B^2+3)q + (B^2-1)p), \\ b\_0 &= \frac{A+\mathbf{x}\_0}{AB-B\mathbf{x}\_0}, b\_1 = \frac{2A\dot{\mathbf{x}}\_0}{B\sqrt{\dot{\alpha}}(A-\mathbf{x}\_0)^2}, \\ b\_2 &= -\frac{2A\mathbf{x}\_0(\mathbf{x}\_0-A)(p+q\mathbf{x}\_0^2) + \alpha(A+\mathbf{x}\_0)(A-\mathbf{x}\_0)^2 + 4A\dot{\mathbf{x}}\_0^{-2}}{2\alpha(A-\mathbf{x}\_0)^2(A+\mathbf{x}\_0)}, \\ A &= \sqrt[4]{\frac{(2p+q\mathbf{x}\_0^2)\dot{\alpha}\_0^2 + 2\dot{\mathbf{x}}\_0^2}{-q}} = \sqrt[4]{\frac{\delta}{-q}}, \\ B &= \sqrt{\frac{A\left(2\sqrt{2}\sqrt{q(A^2q-p)} - 3Aq\right) + p}{A^2q+p}}. \end{aligned} \tag{24}$$

The solution (23) is unbounded and its periodicity is given by

$$T = \left| \frac{4K(m)}{\sqrt{\alpha}} \right| = \left| \frac{4K(1-m)}{m\sqrt{\alpha}} \right|. \tag{25}$$

#### **Example 2.**

Let us assume the following i.v.p.

$$\begin{cases} \varkappa''(t) + \varkappa(t) - \varkappa^3(t) = \mathbf{0}, \\ \varkappa(\mathbf{0}) = -\mathbf{1} \otimes \varkappa'(\mathbf{0}) = -\mathbf{1}. \end{cases} \tag{26}$$

The solution of the i.v.p. (26) according to the relation (23) reads

$$x(t) = \textbf{1.31607} - \frac{\textbf{2.63215}}{\textbf{1} + \frac{0.13647 \textbf{cn} (1.7539 \textbf{e} \| 1.0035 \textbf{S}) - 0.2797 \textbf{d} \textbf{n} (1.7539 \textbf{e} \| 1.0035 \textbf{S}) \textbf{m} (1.0035 \textbf{S})}}{1 - 1.00463 \textbf{n} (1.7539 \textbf{d} \| 1.0035 \textbf{S})} \tag{27}$$

and the periodicity of this solution is given by

$$T = 9.57783. \tag{28}$$

Solution (27) is displayed in **Figure 2**.
