If you consider the Potential Equation in one dimension on some Interval $(0,R)$ and look at the eigenvalue problem: $$-u'' = \lambda \ u \ \ \text{ on } (0,R)$$ with zero-"boundary conditions": $$u(0)=u(R)=0$$ and $$\lambda>0$$
It's clear that the functions $\varphi(x)= \sin \left( \frac{n \pi}{R} x\right) $ for all $n \in \mathbb{N}\setminus\{0\}$ are eigenfunctions to the corresponding eigenvalues $\left( \frac{n \pi}{R} \right)^2$.
But how do you know that these are indeed all eigenfunctions? (except the ones you get by multiplying a constant factor)
In finite dimensional spaces i would use some dimension argument since eigenfunctions of different eigenvalues are linear independent, but since here the eigenfunctions are spanning some Sobolevspace $H^{1}_0\left( \left( 0,R \right)\right)$, this argument doesn't work anymore and it's difficult to see that these functions are indeed spanning the whole space and form some kind of basis.
I would be grateful for help!
Thank's!
Assume for discussion that $\lambda$ is real and non-zero. Classical existence and uniqueness theorems tell you that the general solution to $$ u''+\lambda u = 0, $$ has the form $$ u(x)=A\sin\sqrt{\lambda}x+B\cos\sqrt{\lambda}x, $$ where $A$ and $B$ are arbitrary constants. In order for $u(0)=0$, you must have $B=0$. Then $u(x)=A\sin\sqrt{\lambda}x$ is a solution of $u(0)=u(R)$ iff either (a) $A=0$ or (b) $\sqrt{\lambda}R = \pm \pi,\pm 2\pi,\pm 3\pi,\cdots$. You want (b) because you're not interested in $0$ solutions. The case where $\lambda=0$ can be handled similarly.
NOTE: There are no solutions for non-real complex $\lambda$.