Related to the question : Eigenvalues of the circle over the Laplacian operator, how is it possible to find $c_1$ and $c_2$ related the explicit function $g(x)=c_1 \cos (\mu x)+ c_2 \sin (\mu x)$?
Someone told me that since $μ$ is a nonnegative integer, the eigenspace corresponding to $μ$ is the span of $\{\cos(μx),\sin(μx)\}$. So there is no "finding" of $c_1$,$c_2$ to be done. You can identify a basis of eigenfunctions by choosing particular sets of values of $c_1$, $c_2$.
The question is : The spectrum of the circle is it the set of k's for which there is a solution?
Ultimately, I want to find the spectrum of the circle. To do that, I have to find an explicit function for the D.E $$-g'' = \lambda g.$$
To get more information, I think the page 7 of this link could be interesting.
Any solution of $-g''=\lambda g$ is a linear combination of $\cos\sqrt\lambda x$ and $\sin\sqrt\lambda x$. Because you want your eigenfunctions to be functions on the circle, you need the period to be an integer: $$ \cos\sqrt\lambda (x+2\pi)=\cos\sqrt\lambda x, \ \ \ \ \ \ \sin\sqrt\lambda (x+2\pi)=\sin\sqrt\lambda x. $$ For this you need $\sqrt\lambda\in\mathbb N$, so you get $\lambda=k^2$, $k\in\mathbb N$.
Then, for the eigenvalue $k^2$, you have eigenfunctions $\cos kx$ and $\sin kx$. A linear combination of eigenvectors (for the same eigenvalue) is again an eigenvector, so $g(x)=c_1\,\cos kx+c_2\,\sin kx$ will be an eigenvector for $k^2$, for any choice of $c_1$ and $c_2$.
You would determine $c_1$ and $c_2$ if you had initial conditions