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Suppose $\mathcal{L}$ is a bounded linear operator and I have the solution to Eigenvalue problem

$\mathcal{L} \phi + \lambda \phi = 0$

wish to solve the following PDE

$\left(-\partial_t + \mathcal{L}\right)u = 0$.

If the spectrum of $\mathcal{L}$ is continuous or discrete, then a general solution to the PDE is

$\int C_q e^{- \lambda_q t} \phi_q dq$

or

$\sum_q C_q e^{- \lambda_q t} \phi_q$,

where the $C$'s are constants.

But, what if the spectrum of $\mathcal{L}$ is mixed and has a continuous part, a discrete part, and a singular part? Is there a general way to write the solution to the above PDE if I do not know the spectrum of $\mathcal{L}$?

This has come up in my research because I need to work with the $e^{- \lambda_q t}$ but I do not know what the spectrum of $\mathcal{L}$?

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Well, I have to confess that I am not sure about your goals. In which space are you working? But there is always a way to represent the solution as a generazied exponential function, called operator semigroup. If your operator $\mathcal L$ is indeed bounded, then you can represent the solution via Dunford-Riesz calculus. A good reference for this and much more general cases is Engel-Nagel: A short course on operator semigroups, Springer.

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  • $\begingroup$ I'll check out the "short course on operator semigroups" as you suggested. In the mean-time, I think I have found an answer to my quandary (on wikipedia embarrassingly). I can just write my solution as $\int e^{-\lambda t} \phi_\lambda \mu( d\lambda )$ where $\mu$ is the spectral measure. At least, I think this works (I'm a physicist ... we aren't normally taught things like this in our coursework--at least I never was--so we need to learn this as we go in research). In any case, thank you very much for your reply to my question. $\endgroup$
    – psyduck
    Commented Feb 15, 2011 at 1:42
  • $\begingroup$ Another good sourse to your problem might be the book Weidmann: Linear operators in Hilbert spaces, Springer. I belive, in research we all have to learn all the time. $\endgroup$
    – András Bátkai
    Commented Feb 15, 2011 at 8:31

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