I'm trying to find the eigenvalues and eigenfunctions for the integral operator $Ku=\displaystyle \int_{-1}^1 (1-|x-y|) \,u(y) \, dy$

Since I want to find $\mu,u$ such that $Ku=\mu u$, we get the equation

$\mu u(x) = \displaystyle \int_{-1}^x \,u(y) \, dy - x \displaystyle \int_{-1}^x \,u(y) \, dy + \displaystyle \int_{-1}^x y \,u(y) \, dy - \displaystyle \int_{1}^x \,u(y) \, dy + \displaystyle \int_{1}^x y \,u(y) \, dy - x \displaystyle \int_{1}^x \,u(y) \, dy$

Taking derivatives on both sides I get:

$\mu u'(x)= - \displaystyle \int_{-1}^x \,u(y) \, dy - \displaystyle \int_{1}^x \,u(y) \, dy$.

Taking derivatives again we get:

$\mu u''(x)=-2 u(x)$ whose solutions depend on the sign of $\mu$.

If $\mu>0, u(x)=A \cos(\omega x) + B \sin(\omega x) $ and

If $\mu<0, u(x)=A e^{\omega x} + B e^{- \omega x}$, where $\omega=\sqrt{-2/\mu}$

Now I need help trying to determine the parameters $A,B$. I tried to find initial conditions that would help. For instance, $u(1)=\displaystyle \int_{-1}^1 y \,u(y) \, dy$ and $u'(1)=\displaystyle \int_{-1}^1 \,u(y) \, dy$.

How can I use these conditions to get the values of A and B?

If it helps, according to the book I am using (J.P. Kenner, Principles of Applied Mathematics) the solution says $\lambda_n=\dfrac{8}{n^2 \pi^2} $ is a double eingenvalue with corresponding eigenfunctions $\phi_n(x)=\sin(n\pi x/2)$ and $\psi_n(x)=\cos(n \pi x/2)$ for $n$ odd.

Any help will be greatly appreciate it. Thanks!


If you go back to the two equations you derived, you obtain that $$u (1)+u (-1)=u '(1)+u'(-1)=0. $$ This discards the case $\mu <0$ and, when $\mu>0$, you need (with $\omega=\sqrt {2/\mu} $) $$ 0=A\cos \omega+B\sin\omega+A\cos\omega-B\sin\omega=2A\cos\omega $$ and $$ 0=-A\sin (\omega)+B\cos (\omega)-A\sin (-\omega)+B\cos (-\omega)=2B\cos\omega. $$ One of $A,B $ has to be nonzero so we need $\cos\omega=0$. That is $$\sqrt {\frac2\mu}=\omega=\left (\frac\pi2+n\pi\right)=\frac {(2n+1)\pi}2. $$Thus $$\mu_n=\frac {8}{(2n+1)^2\pi^2}. $$

As we are free to choose either $A $ or $B $ nonzero, we get the two linearly independent eigenvectors $$\sin\frac {(2n+1)\pi x}2,\ \ \ \cos\frac {(2n+1)\pi x}2. $$

  • $\begingroup$ Thanks a lot for the help Martin! $\endgroup$ – user569959 Nov 4 '18 at 20:25
  • $\begingroup$ My pleasure. I've always found it lovely how eigenvalues pop up in these kind of problems. $\endgroup$ – Martin Argerami Nov 4 '18 at 20:27

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