Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

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 $$\begin{equation} \tanh\sqrt\lambda = \frac{2\sqrt\lambda}{1+\lambda}. \tag{1} \end{equation}$$

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 $$\begin{equation} \tan\sqrt\mu = \frac{2\sqrt\mu}{1-\mu}. \tag{2} \end{equation}$$ 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
$$y_n = A(\sqrt\mu_n \cos \sqrt\mu_n x - \sin\sqrt\mu_n x).$$

share|improve this answer
    
@ 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
show 13 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.