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I have a third order linear differential equation, with a free parameter, and boundary conditions that depend on that parameter. I don't think it is possible to obtain an analytic solution, but I would like to know if the eigenfunction, i.e. the family of solutions that correspond to different values of the free parameter, are complete. As a side note, is there any example for an equation whose eigenfunction are not complete?

user2535797
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  • Welcome to MSE! This question would be much easier to answer helpfully if you explicitly typed out your differential equation. – Ben Grossmann Aug 01 '13 at 16:16
  • Thanks for the quick response. The equation has the form

    $$ \sum_n a_n \left( x \right) \cdot y^{(n)}(x) = 0 $$

    where $a_n \left( x \right)$ are rational functions whose explicit forms are just too big to fit in this page.

     – user2535797 Aug 01 '13 at 18:00
  • Wait, when you say that you would like to know if the eigenfunction is complete, do you mean complete in the sense that there are no additional eigenfunctions (which is what I originally thought you meant) or complete in the sense of metric spaces, i.e. under the $|\cdot |_\infty$ norm? – Ben Grossmann Aug 01 '13 at 20:33
  • Complete in the sense that any continuous function can be represented as a linear superposition of the eigenfunctions. – user2535797 Aug 01 '13 at 20:54

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For your side note: if I understood what you mean correctly, here is a differential equation whose corresponding eigenfunction is not "complete":

$$ y' = y; \\ \text{Boundary condition: } y(0) = \lambda $$

Surely, we cannot express a general continuous function as superpositions of $\lambda e^x$.

If you mean an eigenfunction in the sense that $L(f) = \lambda f$, then things become different. It is necessary, at least, to consider complex eigenvalues.

Ben Grossmann
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  • The case $L \left( f \right) = \lambda f$ is more relevant to my case, and the free parameter $\lambda$ is allowed to be complex. – user2535797 Aug 02 '13 at 04:32