Fredholm integral equation

Question

How can I solve the following fredholm integral equation $$ψ(x)=x+λ\int_{0}^{2π}|x-t|ψ(t)dt$$ The kernel contains absolute value

Answer 1

First we express it in the form \begin{align} (I-\lambda K)\psi = x \end{align} where \begin{align} K\psi(x):= \int_{0}^{2\pi} |x-t|\psi(t)\, \mathrm{d}t \end{align} Existence and uniqueness is guaranteed if $|\lambda| < \frac{1}{\left\|K\right\|}$, so assuming $K\colon C[0,2\pi] \to C[0,2\pi]$ we can estimate \begin{align} \left\|K\psi\right\|_{\infty} &\le \max_{x\in [0,2\pi]} \int_{0}^{2\pi} |x-t| \, \mathrm{d}t \left\|\psi\right\|_{\infty} \\ &= \max_{x\in [0,2\pi]} (2\pi^2 - 2 \pi x + x^2) \left\|\psi\right\|_{\infty} \\ &= 2\pi^{2}\left\|\psi\right\|_{\infty} \end{align} So assuming $\lambda \in \left(-\frac{1}{2\pi^{2}}, \frac{1}{2\pi^{2}}\right)$, we have convergence of the Neumann series \begin{align} (I-\lambda K)^{-1} = \sum_{i=0}^{\infty} \lambda^{i}K^{i}. \end{align} If $f(x):=x$, then \begin{align} \psi &= \sum_{i=0}^{\infty} \lambda^{i}K^{i}f \\ &= f + \lambda K[f] + \lambda^{2}K^{2}[f] + \cdots \end{align} Let's work out a couple of these integrals just for fun: \begin{align} K[f](x) &= \int_{0}^{2\pi} |x-t|t \, \mathrm{d}t \\ &= \frac{1}{3} (8\pi^{3} -6\pi^2x + x^{3}) \end{align} And: \begin{align} K^{2}[f](x) &= \int_{0}^{2\pi} |x-t|\left( \frac{1}{3} (8\pi^{3} -6\pi^2t + t^{3}) \right) \, \mathrm{d}t \\ &= \frac{1}{3} \left(x^5-2 \pi x^4-4 \pi ^2 x^3+20 \pi ^3 x^2-28 \pi ^4 x+16 \pi ^5\right) \end{align} To avoid pain in the ass, I calculated these terms via the following Mathematica code:

Assuming[x > 0 && x < 2 \[Pi] , Integrate[Abs[x - t]*t, {t, 0, 2*\[Pi]}]]

For $\lambda = \frac{1}{3\pi^{2}}$, the partial sums up to $\mathcal{O}(\lambda^{6})$ are shown below: