**4. The solution of the forced undamped Duffing-Helmholtz equation**

Suppose that the physical system to be studied is under the influence of some constant external/excitation force, so the standard Duffing-Helmholtz equation can be reformulated to the following constant forced Duffing-Helmholtz i.v.p.

$$\begin{cases} \ddot{\mathbf{x}} + p\mathbf{x} + q\mathbf{x}^2 + r\mathbf{x}^3 = F, \\ \varkappa(\mathbf{0}) = \varkappa\_0 \mathbf{and} \varkappa'(\mathbf{0}) = \dot{\varkappa}\_0. \end{cases} \tag{62}$$

For solving the i.v.p. (62), the following assumption is introduced

$$\mathbf{x}(t) = \mathbf{y}(t) + \boldsymbol{\zeta},\tag{63}$$

where *ζ* is a solution to the cubic algebraic equation

$$r\zeta^3 + q\zeta^2 + p\zeta - F = \mathbf{0}.\tag{64}$$

Substituting Eq. (63) into the i.v.p. (62), we have

$$(p''(t) + (p + 2q\zeta + 3r\zeta^2)y(t) + (q + 3r\zeta)y(t)^2 + r\gamma(t)^3 = 0. \tag{65}$$

Note that the constant forced Duffing-Helmholtz Eq. (62) has been reduced to the standard Duffing-Helmholtz Eq. (65) with the following new initial conditions

$$
\mathbf{y}(\mathbf{0}) = \mathbf{x}\_0 - \boldsymbol{\zeta} \,\, \mathbf{\&} \mathbf{y}'(\mathbf{0}) = \dot{\mathbf{x}}\_0. \tag{66}
$$

### **Example 6.**

Suppose that we have the following i.v.p. and we want to solve it

$$\begin{cases} \ddot{v} + 2v - 12v^2 + v^3 = 4, \\ v(0) = 1 \text{ \& } v'(0) = 1. \end{cases} \tag{67}$$

It is clear that the i.v.p. (67) is a constant forced Duffing-Helmholtz equation. The solution of this problem is given by

$$v(t) = 15.8046 - \frac{15.7714}{1 + 0.0322539 \wp(0.761045 - t; -9.41667, 47.287)}.\tag{68}$$

The comparison between the solution (68) and the RK4 solution is introduced in **Figure 6**.

**Figure 6.** *A comparison between the solution (68) and the RK4 solution.*

The periodicity of solution (68) is given by

$$T = 2\int\_{1.93657}^{\infty} \frac{1}{\sqrt{4\pi^3 + 9.41667\varkappa - 47.287}}d\varkappa = 1.68202... $$
