I would like to diagonalize the differential operator $D=-\partial^2_t+a^2$ with Dirichlet boundary conditions $x(0)=x(T)=0.$
So far I have tried to find the eigenfunctions of $D$, $$Df = \lambda f$$ by considering cases when $\lambda>0, \lambda=0,$ and $\lambda<0.$ However, it seems like for each case, the only function that satisfies this is the zero vector, which isn't an eigenfunction.
Is this differential operator diagonalizable?
Assuming we are working with only ODEs, then we are essentially solving \begin{align} f''(t)-(a^2-\lambda)f(t) = 0 \end{align} with boundary conditions $f(0) = f(T) = 0$.
We see that this only has solution provided $a^2-\lambda<0$, which means the solution to the ODE is given by \begin{align} f(t) = C_1 \sin \sqrt{\lambda-a^2} t + C_2 \cos\sqrt{\lambda-a^2}t. \end{align} Using the fact that $f(0) = 0$, then it follows \begin{align} f(t) = C_1\sin\sqrt{\lambda-a^2} t. \end{align} Using the second condition yields that \begin{align} \sqrt{\lambda-a^2}T = n\pi \ \ \Rightarrow \ \ \lambda= \frac{n^2\pi^2}{T^2}+a^2. \end{align}