Slight Perturbations — An Introduction to Perturbation Methods (2)

轻微的扰动——摄动法简介(2) 苏剑林 · 2013-02-06 · Mathematics approximation perturbation perturbation

To help everyone get more familiar with the basic steps of perturbation methods, this post walks through another example: solving an algebraic equation using perturbation methods. This one actually came out of real research work:

$$\begin{eqnarray*} x=\frac{k(1+k^2+k^4+l^2)}{2(1+k^2)^2} \\ k=\frac{dy}{dx}\end{eqnarray*} $$

This is a differential equation. To solve it, the best approach is of course to first solve the first equation for the form of $k=k(x)$, and then integrate. However, since quintic equations have no general explicit solution, we're forced to look for approximate solutions instead. Of course, anyone familiar with Mathematica would probably just go straight to numerical computation. Here I'll only consider the perturbation method.

Let's rewrite the original equation into the following form:

$$x=\frac{k}{2}[1+\frac{l^2}{(1+k^2)^2}]$$more

It's not hard to see that when $l=0$, the original equation has a simple solution $k=2x$. Starting from this solution, we look for an approximation valid when l is small. Of course, given the complexity of the original equation, I'll only consider the first-order approximation. Let the approximate solution be $k=2x+l^2 p$. Substituting into the original equation, we get

$$x=\frac{2x+l^2 p}{2}[1+\frac{l^2}{(1+(2x+l^2 p)^2)^2}]$$

Keeping only the first-order approximation gives:

$$x=\frac{2x+l^2 p}{2}[1+\frac{l^2}{(1+(2x)^2)^2}]$$

Expanding this out, we obtain:

$$\frac{x l^2}{(1+4x^2)^2}+\frac{l^2 p}{2}=0$$

That is,

$$p=-\frac{2x}{(1+4x^2)^2}$$

That is,

$$k=2x-l^2\frac{2x}{(1+4x^2)^2}=\frac{dy}{dx}$$

Integrating gives

$$y=x^2+\frac{l^2}{4(1+4x^2)}$$

This is the approximate solution to the original differential equation. It doesn't diverge at x=0, and the approximation gets better the larger |x| is, which suggests this solution has quite a nice shape overall. Of course, l needs to be fairly small.