# Find eigenvalues and eigenfunctions for integral operator

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!

## 1 Answer

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 $$\sin\frac {(2n+1)\pi x}2,\ \ \ \cos\frac {(2n+1)\pi x}2.$$

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