Eigenvalues and eigenfunctions of the Fredholm integral equation of the second kind

I'm trying to solve the following question:

Find the eigenfunctions and eigenvalues of the following integral operator:

$$$$Ku(x) = \int_{0}^{\pi}(\sin(x)\sin(y)+\alpha\cos(x)\cos(y))u(y) dy.$$$$

My attempted solution is as follows:

Let $$(\lambda, u)$$ be such an eigenpair. Then,

\begin{align} \lambda u(x) & = \int_{0}^{\pi}(\sin(x)\sin(y)+\alpha\cos(x)\cos(y))u(y)dy \\ & = \sin(x)\int_{0}^{\pi}\sin(y)u(y)dy + \alpha \cos(x)\int_{0}^{\pi}\sin(y)u(y)dy \\ \end{align}

We convert the above equation into a differential equation. Differentiating twice, we get,

$$$$\lambda u^{''}(x) = - u(x).$$$$

As $$\lambda = 0$$ is clearly not an eigenvalue (if not, then the only possible eigenfunction is $$u = 0$$ almost everywhere), we have the general solution is,

$$$$u(x) = A\sin\bigg(\frac{x}{\sqrt{\lambda}}\bigg) + B\cos\bigg(\frac{x}{\sqrt{\lambda}}\bigg).$$$$

We need two boundary conditions. From the eigenvalue equation above, we have,

\begin{align} u(0) & = \frac{\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\ u(\pi) & = \frac{-\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\ \end{align}

However, I don't how to proceed from here. The boundary conditions above are functions of $$u$$.

Please provide relevant hints. This is a homework problem, and I would like to finish it on my own.

• You got $u(0)=-u(\pi)$. One more boundary condition and you will be able to eliminate $A$ and $B$ in the general solution, concluding the exercise. – Giuseppe Negro Mar 11 at 10:14

There is a error, you have that

\begin{align} u(0) & = \frac{\alpha}{\lambda} \int_{0}^{\pi}\cos(y)u(y)dy \\ u(\pi) & = \frac{-\alpha}{\lambda} \int_{0}^{\pi}\cos(y)u(y)dy \\ \end{align}

So $$u(0)=-u(\pi)$$. Another condition can be

\begin{align} u(\frac{\pi}{2}) & = \frac{\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\ u(\frac{3\pi}{2}) & = \frac{-\alpha}{\lambda} \int_{0}^{\pi}\sin(y)u(y)dy \\ \end{align}

So $$u(\frac{\pi}{2})=-u(\frac{3\pi}{2})$$

But

$$u(0)=Asin(\frac{0}{\sqrt{\lambda}})+Bcos(\frac{0}{\sqrt{\lambda}})=B$$

and

$$u(\pi)= Asin(\frac{\pi}{\sqrt{\lambda}})+Bcos(\frac{\pi}{\sqrt{\lambda}})$$

So you have that

$$Asin(\frac{\pi}{\sqrt{\lambda}}) =-B(1+cos(\frac{\pi}{\sqrt{\lambda}}))$$

To the other hand

$$u(\frac{\pi}{2})=Asin(\frac{\pi}{2\sqrt{\lambda}})+Bcos(\frac{\pi}{2\sqrt{\lambda}})$$

while

$$u(\frac{3\pi}{2})= Asin(\frac{3\pi}{2\sqrt{\lambda}}) +Bcos(\frac{3\pi}{2\sqrt{\lambda}})$$

so

$$A(sin(\frac{\pi}{2\sqrt{\lambda}})+sin(\frac{3\pi}{2\sqrt{\lambda}}))=-B(cos(\frac{\pi}{2\sqrt{\lambda}}) + cos(3\frac{\pi}{2\sqrt{\lambda}})$$

• How did you get rid of the factor of $\sin$ in the equation for $u(\pi)$. We haven't assumed $\lambda$ is an integer. – user82261 Mar 11 at 10:43
• Ops, you’re right. Thanks you – Federico Fallucca Mar 11 at 10:48
• I can't quite figure out how to deal with the case $A \neq 0, B \neq 0$. In that case, I have two equations in three unknowns, $A, B$ and $\lambda$. – user82261 Mar 11 at 10:51
• I think I get you mean you mean. We can plug the general solution in the eigenvalue equation to make the above argument sufficient as well, and get the answer. – user82261 Mar 11 at 10:58