1
$\begingroup$

The spectrum$(\mathbb{S}^1)=\{\lambda_k=k^2\ : k \in \mathbb{N}\}$, and the eigenfunctions $\mu_k(t)$ associated to the eigenvalues $\lambda_k$ are $a_k \cos kt + b_k \sin kt$ under the Laplacian operator. A things I know is $\mu_k$ is a periodic function on the interval $[0, 2 \pi)$, and I would like to find the set of normalized eigenfunction. How do I find out if I don't know the explicit values of $a_k$ and $b_k$

$\endgroup$

closed as unclear what you're asking by user7530, Claude Leibovici, user91500, John B, Davide Giraudo May 24 '16 at 9:53

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ I am a bit confused. You talk about the spectrum of a set but then talk about eigenvalues (presumably of some differential operator). Can you please be more clear? $\endgroup$ – Cameron Williams May 24 '16 at 0:41
  • $\begingroup$ @CameronWilliams Is it more clearer? $\endgroup$ – user332990 May 24 '16 at 0:44
  • $\begingroup$ So basically you just need to find values for $a_k$ and $b_k$? Try computing the integrals $\int_0^{2\pi} cos^2(kt)\,dt$ and likewise for $\sin$. (If I'm understanding you correctly.) You should have that $a_k^2\int_0^{2\pi}\cos^2(kt)\,dt = 1$. $\endgroup$ – Cameron Williams May 24 '16 at 0:46
  • $\begingroup$ @CameronWilliams Is it the property of normalization? Could you explain a bit more? $\endgroup$ – user332990 May 24 '16 at 0:51
  • 1
    $\begingroup$ Yes it is. You want to have that $\int_0^{2\pi} (a_k\cos^2(kt))^2\,dt = 1$ and likewise for $\sin$. This is what it means for the functions to be $L^2$-normalized. $\endgroup$ – Cameron Williams May 24 '16 at 0:52