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

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
$$y_n = A(\sqrt\mu_n \cos \sqrt\mu_n x - \sin\sqrt\mu_n x).$$

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    @ oenamen: Many thanks indeed for your detailed answer. I really appreciate that.2012-04-17
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    @MikaelAnderson: Glad to help. A related integral equation has been handled [here](http://math.stackexchange.com/questions/87124/solve-fx-lambda-int-limits-01-maxx-txtftdt). The solution appears to be correct. Some arguments are made initially regarding the existence and properties of the solutions using Fredholm theory.2012-04-17
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    @ oenamen: Thanks for the link. I have learned a lot from your contribution and I am grateful for that.2012-04-17
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    @ 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?2012-04-18
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    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 $$\{-0.708395,-0.326582,-0.214623, -0.160169,-0.127842,-0.106403 ,-0.0911341,-0.0797037,-0.070 8241,-0.0637265\}.$$2012-04-18
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    @MikaelAnderson: If you read my solution carefully you will find the eigenvalue is only approximately $n^2\pi^2$ and only for $n$ very large. Certainly not for $n=1$. *The eigenvalues must be found numerically.* I give the first three eigenvalues to three significant figures in the body of the answer. The tenth eigenvalue is roughly 983 and $10^2\pi^2$ is about 987. The hundredth eigenvalue is roughly 98692 and $100^2\pi^2$ is about 98696.2012-04-18
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    @MikaelAnderson: If you put the function $y = A(\sqrt\mu \cos \sqrt\mu x - \sin\sqrt\mu x)$ into the integral equation you will find the condition that it be a solution is equation (2). Thus, any $\mu$ solving (2) will be an eigenvalue.2012-04-18
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    To @oenamen: Thanks, I appreciate your reply. Please let me know if I understand this correctly. It seems that $\mu_0=5.4341315058465565$ solves equation (2). Should this integral $$-\mu_0\int_0^1 K(x,t)(\sqrt{\mu_0} \cos (\sqrt{\mu_0 t})-\sin (\sqrt{\mu_0 t}))dt$$ be (at least approximately) equal to $$\sqrt{\mu_0}\cos (\sqrt{\mu_0 x})-\sin (\sqrt{\mu_0 x})? $$2012-04-19
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    @MikaelAnderson: Yes. And you should find that it is.2012-04-19
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    To @oenamen: Well, I calculate that integral in Mathematica and, if I have done it correctly, I get: $$2.89845 x-12. \sqrt{x} \sin \left(2.33112 \sqrt{x}\right)+(9.32449 x-5.14773) \cos \left(2.33112 \sqrt{x}\right)+6.05358.$$2012-04-19
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3152/discussion-between-oenamen-and-mikael-anderson)2012-04-19
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    To @oenamen: My sincere thanks to oenamen for finding a typo in my Mathematica program in the chat.2012-04-19
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    To @oenamen: I was reviewing your solution above and I have got stuck in one detail regarding case II: $\lambda<0$. When I simplify the eigenfunction in this case the imaginary part $i$ remains in the formula (in converting $\sinh$ to $\sin$). I wonder if you could let me know if you factorize it into the constant $A$ or am I missing something trivial here. Many thanks.2013-02-12
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    @MikaelAnderson: Welcome back. Good question. We start with $y = A \cosh\sqrt\lambda x + B \sinh\sqrt\lambda x$. Applying the first boundary condition we find $y = \frac{A}{\sqrt{\lambda}}(\sqrt\lambda \cosh\sqrt\lambda x - \sinh\sqrt\lambda x) \to A(\sqrt\lambda \cosh\sqrt\lambda x - \sinh\sqrt\lambda x)$. Then we set $\lambda = -\mu$ so $y = i A(\sqrt\mu \cos\sqrt\mu x - \sin\sqrt\mu x) \to A(\sqrt\mu \cos\sqrt\mu x - \sin\sqrt\mu x)$. We keep redefining $A$ in a convenient way so as not to introduce unnecessary symbols.2013-02-12
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    @oenamen: Thanks again for your detailed answer. So given that $A$ in case II is complex, how does it affect calculation of the normalizing constants for $y_n$? I am thinking about your calculations for $A_i, i\geq 1$ in [this question](http://math.stackexchange.com/questions/134161/is-there-a-solution-to-this-integral-equation). I mean don't we need to have $i^2=-1$ somewhere?2013-02-12
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    @MikaelAnderson: Glad to help. If you want real solutions you are free to assume $A$ is real. Without some other restrictions $A$ is complex. In this case the inner product would be modified slightly, $f\cdot g = \int_0^1 d t\, \overline{f(t)}g(t)$, where $\overline f$ is the complex conjugate of $f$.2013-02-12
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    @oenamen: So just to make sure that I understand your answer: does it mean that the calculations for $A_i, i\geq 1$ in [this question](http://math.stackexchange.com/questions/134161/is-there-a-solution-to-this-integral-equation) are still valid? Or if I want real solutions I can not use those values of $A_i, i\geq 1$? Thanks.2013-02-12
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    @MikaelAnderson: I suggest you play around with this yourself. Is the solution written in the other answer real? What about the $A$s? The $c$s? The eigenfunctions? What happens if we let $A$ be complex there? Do we get a different solution?2013-02-12