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

Question

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?

Answer 1

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.

IN ADDITION :

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 : $$x(t)\simeq x(0)+x'(0)t+\frac12 x''(0)t^2$$ $x(0)=-x_0\quad;\quad x'(0)=0$

$x''(0)=\frac{1}{m}\left(-\alpha x(0)-\beta x(0)^3 \right)=\frac{1}{m}\left(\alpha x_0+\beta x_0^3 \right)$

$$x(t)\simeq -x_0+\frac{1}{2m}\left(\alpha x_0+\beta x_0^3 \right)t^2$$

So, to answer, there was no need to explicitly solve the ODE, thus no need for elliptic integral.