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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?

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  • $\begingroup$ Is $f$ a two variable function? $\endgroup$ – Jacky Chong Feb 2 '17 at 5:21
  • $\begingroup$ No, it only depends on $t.$ $\endgroup$ – user352541 Feb 2 '17 at 5:26
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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}

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