# Solving the nonlinear differential equation $m \ddot x +\alpha x + \beta x^3 = 0$

As the header says: I want to solve the differential equation $$m \ddot x +\alpha x + \beta x^3 = 0$$, with initial conditions $$x(0) = -x_0$$, $$\dot x(0)=0$$. It comes up in the solution to the equations of motion of the so-called undamped duffing oscillator, an undamped oscillator driven by nonlinear force $$-\alpha x - \beta x^3$$. My research has shown that the solution involves elliptic integrals, which I have never seen before.

Here is my approach so far: I use conservation of energy. The total Energy should be $$E_{tot} = T + V = \frac{1}{2}m \dot x^2 +\alpha \frac{x^2}{2}+ \beta \frac{x^4}{4} = E_{t= 0} = \alpha \frac{x_0^2}{2}+ \beta \frac{x_0^4}{4}$$. Now I would transfrom this equation into $$\dot x = \sqrt{\frac{2}{m}(\alpha \frac{x^2-x_0^2}{2}+ \beta \frac{x_0^4-x^4}{4})}$$ and then obtain by seperation of variables $$dt = \frac{dx}{\sqrt{\frac{2}{m}(\alpha \frac{x^2-x_0^2}{2}+ \beta \frac{x_0^4-x^4}{4})}}$$. By integrating both sides, I'm guessing the integral on the right $$\int \frac{dx}{\sqrt{\frac{2}{m}(\alpha \frac{x^2-x_0^2}{2}+ \beta \frac{x_0^4-x^4}{4})}}$$ is the elliptic integral.

Is my reasoning correct so far? And if yes, how do you solve (analytically) the integral?

Your reasoning is correct. Analytical solving confirmes the result : $$mx''=-\alpha x-\beta x^3$$ $$2mx''x'=-2\alpha xx'-2\beta x^3x'$$ $$m(x')^2=-\alpha x^2-\frac12\beta x^4+C$$ With conditions $$x(0)=-x_0$$ and $$x'(0)=0$$ we get $$C=\alpha x_0^2+\frac12\beta x_0^4$$ $$m(x')^2=-\alpha (x^2-x_0^2)-\frac12\beta (x^4-x_0^4)$$ $$x'=\pm\sqrt{\frac{1}{m}\left(\alpha (x_0^2-x^2)+\frac12\beta (x_0^4-x^4)\right)}$$ $$t=\int \frac{dx}{\sqrt{\frac{\alpha}{m} (x_0^2-x^2)+\frac{\beta}{2m} (x_0^4-x^4)}}+c$$ With conditions $$x(0)=-x_0$$ : $$t=\int_{-x_0}^x \frac{d\xi}{\sqrt{\frac{\alpha}{m} (x_0^2-\xi^2)+\frac{\beta}{2m} (x_0^4-\xi^4)}}$$

This is an elliptic integral of the first kind.

The inverse function $$x(t)$$ cannot be expressed with a finite number of elementary functions. A closed form involves the $$sn$$ Jacobi elliptic function.

So, the analytic solving is possible in terms of Jacobi elliptic function, but rather arduous. In practice solving thanks to numerical calculus would be easier.

Answer to the further question raised in comments : "What if the task is to solve the approximate answer for $$x(t)\simeq −x_0$$ ? "
A Taylor approximate is : $$x(t)\simeq x(0)+x'(0)t+\frac12 x''(0)t^2$$ $$x(0)=-x_0\quad;\quad x'(0)=0$$
$$x''(0)=\frac{1}{m}\left(-\alpha x(0)-\beta x(0)^3 \right)=\frac{1}{m}\left(\alpha x_0+\beta x_0^3 \right)$$
$$x(t)\simeq -x_0+\frac{1}{2m}\left(\alpha x_0+\beta x_0^3 \right)t^2$$
• what if the task is to solve the approximate answer for $x(t) \approx -x_0$? – ghthorpe Dec 13 '18 at 6:49