# How to find eigenfunctions of a linear operator (follow-up question)

This is a follow up to this question. My original question was answered by @oeanamen but for convenience here I state the follow up question completely.

I am interested in calculating characteristic values and eigenfunctions of $$K(x,t)=\max((1−x)t,(1−t)x),0<x<1,0<t<1$$ and find $λ_i$ and $y_i(x)$ such that

$$y_i(x)−λ_i\int_0^1K(x,t)y_i(t)dt=0.$$

After taking the second derivative of the above equation we find $$y''=λy.$$ As oeanamen suggested in the comments to the original question a solution is $$A \left(\sqrt{\lambda } \cosh \left(\sqrt{\lambda } x\right)-\sinh \left(\sqrt{\lambda } x\right)\right).$$

I realize that this is indeed a solution but I wonder if this is the only solution. I would also like to understand the details of calculation leading to the value of $\lambda$.

My last question is what the corresponding eigenfunction is for $K(x,t)$ defined above.

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As with the previous problem we can convert the integral equation into a differential equation by taking the second derivative of the integral equation with respect to $x$. We find $$y'' - \lambda y = 0.$$ We immediately throw away the solution for $\lambda =0$ ($y = A x + B$) since it implies $y = 0$ in the integral equation. Thus, the solutions will be of the form $$y = A \cosh\sqrt\lambda x + B \sinh\sqrt\lambda x.$$ In fact, by examining the integral equation and its first derivative evaluated at $x=0$ and $x=1$ we can convince ourselves that the solutions must satisfy Robin boundary conditions $$\begin{eqnarray*} y'(0) + y(0) &=& 0 \\ y'(1) - y(1) &=& 0. \end{eqnarray*}$$ These boundary conditions make finding a closed form for the eigenvalues impossible. The solutions are peculiar. For $\lambda>0$ there is one eigenfunction. For $\lambda<0$ there is a tower of eigenfunctions. For large and negative $\lambda$ we will find approximate eigenvalues of the form $\lambda_n \approx -n^2\pi^2$.

The boundary conditions imply that $B = -A/\sqrt\lambda$ and that the eigenvalues satisfy the condition $$$$\tanh\sqrt\lambda = \frac{2\sqrt\lambda}{1+\lambda}. \tag{1}$$$$

Case I: $\lambda > 0$

There is one solution to equation (1) for $\lambda>0$. It must be found numerically. It is $\lambda_0 \approx 2.38.$ The eigenfunction is $$y_0 = A(\sqrt\lambda_0 \cosh \sqrt\lambda_0 x - \sinh\sqrt\lambda_0 x).$$

Case II: $\lambda < 0$

Define $\mu = -\lambda$. The condition on the eigenvalues becomes $$$$\tan\sqrt\mu = \frac{2\sqrt\mu}{1-\mu}. \tag{2}$$$$ There is an infinite tower of countable solutions to equation (2). We find, for example, $$\mu_1 \approx 5.43 \approx \pi^2, \hspace{5ex} \mu_2 \approx 35.4 \approx (2\pi)^2, \hspace{5ex} \mu_3 \approx 84.8 \approx (3\pi)^2.$$ In the limit of large $\mu$, the right-hand side of (2) vanishes. Thus, for large $\mu$, $\sqrt\mu = n \pi$ will be an approximate solution, where $n\in\mathbb{N}$. (These are the positive zeros of the tangent function.) That is, $\mu_n \approx n^2\pi^2$ for $n$ large. The eigenfunctions are
$$y_n = A(\sqrt\mu_n \cos \sqrt\mu_n x - \sin\sqrt\mu_n x).$$

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@ oenamen: Many thanks indeed for your detailed answer. I really appreciate that. –  Mikael Anderson Apr 17 '12 at 10:16
@MikaelAnderson: Glad to help. A related integral equation has been handled here. The solution appears to be correct. Some arguments are made initially regarding the existence and properties of the solutions using Fredholm theory. –  user26872 Apr 17 '12 at 16:35
@ oenamen: Thanks for the link. I have learned a lot from your contribution and I am grateful for that. –  Mikael Anderson Apr 17 '12 at 18:44
@ oenamen: I just wanted to test the eigenfunctions for a few values of $n$ to make sure that I understand the solution but I don't get the right answer. For instance I get $$-\pi^2\int_0^1 K(x,t)(\pi \cos (\pi x)-\sin (\pi x))dt=-2 \pi (x-1)-\sin (\pi x)+\pi \cos (\pi x).$$ Should not the result be $\pi \cos (\pi x)-\sin (\pi x)$? What am I missing here? –  Mikael Anderson Apr 18 '12 at 19:34
To @oenamen: I am probably missing something trivial here but when I calculate equation (2) for different values of $n$ it does not seem like $\mu_i$ given above solve that equation. For instance these are the values left and right hand side of equation (2) for $n=1,\ldots,10$:$$\{0.47669,-4.72624,1.1659,1.1023 4,-7.95271,0.31569,-0.197149, 0.197051,10.2372,0.546753\}$$ and $$\{-0.708395,-0.326582,-0.214623, -0.160169,-0.127842,-0.106403 ,-0.0911341,-0.0797037,-0.070 8241,-0.0637265\}.$$ –  Mikael Anderson Apr 18 '12 at 21:04