To help everyone become more familiar with the basic steps of the perturbation method, this article discusses another example of solving an algebraic equation using the perturbation method. This example originates from actual research:
\begin{aligned} x &= \frac{k(1+k^2+k^4+l^2)}{2(1+k^2)^2} \\ k &= \frac{dy}{dx} \end{aligned}
This is a differential equation. To solve this equation, the best method is, of course, to first solve for k=k(x) from the first equation and then integrate. However, since quintic equations do not have a general explicit solution, we are forced to consider approximate solutions. Generally speaking, those familiar with Mathematica would directly perform numerical computations. Here, I only consider the perturbation method.
We transform the original equation into the following form: x = \frac{k}{2} \left[ 1 + \frac{l^2}{(1+k^2)^2} \right]
It is not difficult to find that when l=0, the original equation has a simple solution k=2x. Starting from this solution, we look for an approximation when l is very small. Due to the complexity of the original equation, I will only consider the first-order approximation. Let the approximate solution be k=2x+l^2 p. Substituting this into the original equation, we get: x = \frac{2x+l^2 p}{2} \left[ 1 + \frac{l^2}{(1+(2x+l^2 p)^2)^2} \right]
Considering only the first-order approximation (terms involving l^2), we have: x = \frac{2x+l^2 p}{2} \left[ 1 + \frac{l^2}{(1+(2x)^2)^2} \right]
Expanding this yields: \frac{x l^2}{(1+4x^2)^2} + \frac{l^2 p}{2} = 0
Which gives: p = -\frac{2x}{(1+4x^2)^2}
Thus: k = 2x - l^2 \frac{2x}{(1+4x^2)^2} = \frac{dy}{dx}
Integrating yields: y = x^2 + \frac{l^2}{4(1+4x^2)}
This is the approximate solution to the original differential equation. It does not diverge at x=0, and the larger |x| is, the better the approximation, indicating that the behavior of this solution is quite good. Of course, l needs to be sufficiently small.
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