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

Question

I'm trying to solve the following question:

Find the eigenfunctions and eigenvalues of the following integral operator:

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

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,

\begin{equation} \lambda u^{''}(x) = - u(x). \end{equation}

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,

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

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.

Answer 1

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}}) $

Answer 2

Since $$ Ku(x)=\sin(x)\int_{0}^{\pi}\sin(y)u(y)\,dy + \alpha \cos(x)\int_{0}^{\pi}\cos(y)u(y)\,dy, \tag{1} $$ the eigenfunctions of $K$ associated with nonzero eigenvalues must be linear combinations of $\sin(x)$ and $\cos(x)$. Thus,

\begin{align} Ku(x)&=K(A\sin(x)+B\cos(x)) \\ &=\int_0^{\pi}(\sin(x)\sin(y)+\alpha\cos(x)\cos(y))(A\sin(y)+B\cos(y))dy \\ &=\frac{\pi}{2}A\sin(x)+\frac{\alpha\pi}{2}B\cos(x), \tag{2} \end{align} which implies that $\sin(x)$ and $\cos(x)$ are eigenfuntions of $K$ with eigenvalues $\frac{\pi}{2}$ and $\frac{\alpha\pi}{2}$, respectively.

In addition, any function that is orthogonal to both $\sin(x)$ and $\cos(x)$ in the interval $[0,\pi]$ is an eigenfunction of $K$ with eigenvalue $0$. Examples: $\sin(2n+1)x$ and $\cos(2n+1)x$ for $n\in\mathbb{N}_{>0}$.