Find the eigenvalues and eigenfunctions of integral operator

$$Ku(x)= \int_0^\pi \sin(x)\sin(2y)u(y)\,\mathrm dy $$

I tried to use separable kernel to solve this, but I get a zero matrix $A$, please help, thanks!


1 Answer 1


Assuming that we are in the Hilbert space $H = L^2(0,\pi)$:

At first step we can rewrite the operator $K$ as $$ Ku(x) = \sin(x) \int_{0}^{\pi} \sin(2y)u(y)\,\mathrm{d}y . $$ From that we can conclude that $\operatorname{ran} K = \operatorname{span}\{ \sin\}$ and that $$ \ker K = \{\sin(2y)\}^{\perp} = \overline{\operatorname{span} \{\sin(nx):n\in\mathbb{N}\backslash\{2\}\} \cup \{\cos(nx):n\in\mathbb{N}\cup\{0\}\}} $$ Clearly every $u \in \ker K$ is a eigenfunction with eigenvalue $0$. For the remaining direction $u(x)=\sin(2x)$ we get $$ Ku(x) = \sin(x) \int_0^\pi \sin^2(2y) \,\mathrm{d}y = \frac{\pi}{2} \sin(x) $$ So this is not an eigenfunction.

  • $\begingroup$ Thanks a lot! So $\lambda=0$ is the only eigenvalue, and the eigenfunctions would be the functions orthogonal to the span, but why do you need to check sin(2x)? $\endgroup$
    – Areedd
    Aug 20, 2017 at 2:53
  • $\begingroup$ You are right it is clear that only functions which are in the $\operatorname{ran} K$ can be eigenfunctions to an eigenvalue $\lambda \neq 0$. So this wasn't nessacery. $\endgroup$ Aug 20, 2017 at 8:49
  • 1
    $\begingroup$ you could accept the answer if it answers your question ;) $\endgroup$ Aug 20, 2017 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.