Let $\Omega \in \mathbb{R}^d$ be an open bounded set with smooth boundary. Consider $$-\bigtriangleup q(x) = \lambda q(x)$$ with either Dirichlet or Neumann boundary conditions $$q(x)=0, x \in \partial \Omega$$or $$\bigtriangledown q(x) \centerdot n(x) = 0, x \in \partial \Omega$$ I need to show that for any eigenvalue, we can find a real-valued eigenfunction. I have shown that the eigenvalues are real by taking the conjugate of the pde, multiplying the conjugate one by $q$ and the non-conjugate one by the conjugate of $q$ and integrating. I've tried subtracting the conjugate equation from the original and integrating but that didn't get me anywhere. The best I can get from doing any integration is that $\frac{q(x)}{\bigtriangleup q(x)}$ is real but that follows immediately from the pde if you know $\lambda$ is real. I've been stuck for a while and don't know what to do. Thanks.

  • $\begingroup$ Multiply by $\bar{q}$, integrate over the domain. Then integrate by parts the LHS. That should do the trick. $\endgroup$ – maxmilgram Apr 13 at 19:15
  • $\begingroup$ @maxmilgram what am I supposed to get from doing that? The boundary term vanishes and the other term is a dot product of two gradients. If i integrate that term by parts I get another vanishing boundary term and the conjugate of what I started with which is the RHS if you substitute the pde back in. $\endgroup$ – N Dizzle Apr 13 at 20:20
  • $\begingroup$ You get $||\nabla q||_{L^2}^2=\lambda||q||_{L^2}^2$, don't you? $\endgroup$ – maxmilgram Apr 14 at 6:52
  • $\begingroup$ Yeah that's what I got, I guess I just don't see how that means $q$ is real. $\endgroup$ – N Dizzle Apr 14 at 18:04
  • $\begingroup$ Oh! Really sorry my bad, i missread that as "eigenvalues". So I gave a proof for eigenvalues being real. I dont think your statement about eigenfunctions is actually true. Take a real eigenfunction $q(x)$. Then $i\cdot q(x)$ is an eigenfunction as well. $\endgroup$ – maxmilgram Apr 14 at 18:12

Your Answer

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

Browse other questions tagged or ask your own question.