# Power series solution to integral equation

Hi guys i'm reading a paper in which the authors have two coupled integral equation for the function $f(x)$ and $g(x)$, in order to solve this problem they employ a power series expansion of these function in powers of a small parameter $\epsilon$ $f(x)=\epsilon f^{(1)}(x)+\epsilon^{2} f^{(2)}(x)+O(\epsilon^{3})$ and the same for $g(x)$. I don't understand how to proceed to the solution once you take this approximation, so if you could help me with a simplified problem to let me understand how to proceed it would be great...

Let's consider the following equation (which represent a simplified version of my problem)

$K = \frac{4}{\epsilon}f(x)+f(x)\ln(1-\frac{x^{2}}{l^{2}})+\int_{-l}^{l}(f(x)-f(s))(\frac{2}{(x-s)^{2}})ds$

Now if i take the approximation of the function as a power series and i truncate to second order for example, how do i proceed? (K and are real positive numbers)

By direct substitution we have $$\begin{array}{l}K = \frac{4}{\varepsilon }\left( {\varepsilon {f^{(1)}}(x) + {\varepsilon ^2}{f^{(2)}}(x) + O({\varepsilon ^3})} \right) + \\\left( {\varepsilon {f^{(1)}}(x) + {\varepsilon ^2}{f^{(2)}}(x) + O({\varepsilon ^3})} \right)\ln (1 - \frac{{{x^2}}}{{{l^2}}}) + \int\limits_{ - l}^{ + l} {\left( \begin{array}{l}\varepsilon {f^{(1)}}(x) + {\varepsilon ^2}{f^{(2)}}(x) + O({\varepsilon ^3}) - \\\varepsilon {f^{(1)}}(s) - {\varepsilon ^2}{f^{(2)}}(s) - O({\varepsilon ^3})\end{array} \right)\frac{2}{{{{(x - s)}^2}}}ds} \end{array}$$Now, separating the zero and first order terms in $\varepsilon$ yields $$\left\{ \begin{array}{l}K = 4{f^{(1)}}(x)\\0 = 4{f^{(2)}}(x) + {f^{(1)}}(x)\ln (1 - \frac{{{x^2}}}{{{l^2}}}) + \int\limits_{ - l}^{ + l} {\left( {{f^{(1)}}(x) - {f^{(1)}}(s)} \right)\frac{2}{{{{(x - s)}^2}}}ds} \end{array} \right.$$ which is easily solved to yield $$\left\{ \begin{array}{l}{f^{(1)}}(x) = \frac{K}{4}\\{f^{(2)}}(x) = - \frac{K}{{16}}\ln (1 - \frac{{{x^2}}}{{{l^2}}})\end{array} \right.$$Hope it helps ;)
What is being performed here is the so called ¨naive expansion¨. What you have to do is to substitute for $f(x)$ and collect all terms of equal order in $\epsilon$ and solve them separately.
For example, upon substitution, the only $\mathcal{O}(1)$-term on the right hand side is $4f^{(1)}(x)$. You have to set it equal to the $\mathcal{O}(1)$-component in $K$ and solve the equation. When you collect all the $\mathcal{O}(\epsilon)$-terms on both sides, you shall see that there is a $f^{(1)}(x)$-dependence in the equation that you acquire. Therefore, you have to have $f^{(1)}(x)$ at your disposal to solve for $f^{(2)}(x)$. In general you may need all the $f^{(n)}(x)$ with $n \leq m-1$ in order to solve for $f^{(m)}(x)$.
So, if you are only interested in the solution up to $\mathcal{O}(\epsilon^2)$, you can simply stop after $f^{(2)}(x)$.