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The problem is related to this question: How to find eigenfunctions of a linear operator (follow-up question) I posted earlier.

Suppose I want to solve the following integral equation: $\int_0^1 K(x,t)y(t)dt=\sqrt{2x^2-2x+1}$ where $K(x,t)=\max((1-x)t,(1-t)x),0

Eigenfunctions of $ K(x,t)$ was found by @oenamen in the answer to the above-mentioned question. I thought one should be able to use the eigenfunctions to find a solution to the above equation but I fail to see if this is the case. I would appreciate any suggestions or comments on this.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3179/discussion-between-mikael-anderson-and-oenamen)2012-04-20

1 Answers 1

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Explicit solution

Take the second derivative of each side of the integral equation, $y = f''.$ Plugging this back into the integral equation we find that $f$ must satisfy the Robin boundary conditions obeyed by the eigenfunctions, $\begin{eqnarray*} f'(0) + f(0) &=& 0 \\ f'(1) - f(1) &=& 0. \end{eqnarray*}$ (We should expect to be able to expand a function obeying these boundary conditions in terms of the eigenfunctions.) Since $f(x) = \sqrt{2x^2-2x+1}$ satisfies these boundary conditions the solution exists. Thus, $y = \frac{1}{(2x^2-2x+1)^{3/2}}.$

Eigenfunction expansion

We write the integral equation schematically as $f = K y.$ Let $y_n$ denote the $n$th eigenfunction of $K$, $y_n = \lambda_n K y_n.$ These have been found for the operator pertinent to this question here. The eigenfunctions are orthogonal and we assume they have been normalized $y_m \cdot y_n = \delta_{mn}.$ (The inner product is $f\cdot g = \int_0^1 d t\, f(t)g(t)$.)

Picard's theorem mentioned in the comments states that $f$ can be expanded in terms of the eigenfunctions, $f = \sum f_n y_n$ where $f_n = y_n \cdot f$. Then $\begin{eqnarray*} y &=& K^{-1} \sum f_n y_n \\ &=& \sum \lambda_n f_n y_n \\ &=& \sum c_n y_n \end{eqnarray*}$ Thus, the coefficients of the expansion for $y$ are $c_n = \lambda_n f_n$.

Some details

Define $y_0 = A_0(\sqrt{\lambda_0} \cosh\sqrt{\lambda_0} x - \sinh\sqrt{\lambda_0} x)$ and $y_n = A_n(\sqrt{\mu_n} \cos\sqrt{\mu_n} x - \sin\sqrt{\mu_n} x)$ for $n\ge 1$. Note that $\lambda_n = -\mu_n$ for $n\ge 1$. Normalizing we find $A_0 \approx 0.769,\hspace{3ex} A_1 \approx 0.672, \hspace{3ex} A_2 \approx 0.241, \hspace{3ex} A_3 \approx 0.154, \hspace{3ex} \ldots$ The coefficients $c_n = \lambda_n \int_0^1 d t\, y_n(t) f(t)$ are $c_0 \approx 1.94, \hspace{3ex} c_1 \approx 0, \hspace{3ex} c_2 \approx -0.757, \hspace{3ex} c_3 \approx 0, \hspace{3ex}\ldots $ The function $y = \sum_{n=0}^3 c_n y_n$ already provides a very good approximate solution to the integral equation.

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    To @oenamen: Hi oenamen. Many thanks for your comments and the references. Very useful indeed.2012-04-26