# Is there a solution to this integral equation?

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 $$K(x,t)=\max((1-x)t,(1-t)x),0<x<1,0<t<1.$$

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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Why not take the second derivative of both sides and see what happens ... – user26872 Apr 20 '12 at 4:36
To @oenamen: There is a theorem by Picard which states that equation of the above type has a solution if and only if the right hand side can be expanded into a mean-square convergent series with respect to eigenfunctions of the kernel $K(x, t)$. I was trying to use that instead of trying to solve the equation by differentiation. – Mikael Anderson Apr 20 '12 at 12:12
Would not your argument imply that the equation in my question does not have a solution since Picard's theorem is formulated as if and only if''? – Mikael Anderson Apr 20 '12 at 20:13
To @oenamen: I do not know how I can start a chat but if you have a minute to chat please let me know. I just finished my calculations in Mathematica and I am puzzled by your previous comment that $y=-f''$ because I think $y=f''$ was correct. – Mikael Anderson Apr 20 '12 at 22:25
– Mikael Anderson Apr 20 '12 at 22:27

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: Many thanks for taking the time to provide the detailed answer. I appreciate all your help. – Mikael Anderson Apr 21 '12 at 11:38
@MikaelAnderson: Glad to help, Mikael. – user26872 Apr 21 '12 at 15:58
To @oenamen: I have followed all the details of the calculations and I get the same results as in your answer above. For $n=3$, I get the following approximation of the solution: $$y(t)= 1.49516 (1.5434 \cosh (1.5434 t)-\sinh (1.5434 t))-0.182402 (5.95017 \cos (5.95017 t)-\sin (5.95017 t))$$ However, when I plot this together with the exact answer $\frac{1}{\left(2 t^2-2 t+1\right)^{3/2}}$ I see great difference between the exact and approximation close to the boundaries at 0 and 1. I wonder if you get the same answer as above for $n=3$. – Mikael Anderson Apr 22 '12 at 18:12
Further, I thought one could improve the approximation by increasing $n$ to say 50 or 100 but something which puzzles me is that the approximation gets worse by increasing $n$. This does not sound right so I would appreciate to hear your comments on this. – Mikael Anderson Apr 22 '12 at 18:14
@MikaelAnderson: Hi Mikael. Your $y$ looks fine. At $x=0$ and $1$ I find the fit and exact solution disagree by about $20\%$. However, you should base the goodness of fit on how well $y$ solves the integral equation. On that basis the fit is very good indeed. I recommend doing the integral to see this. – user26872 Apr 22 '12 at 19:11