I have a real, fourth order linear operator $L$ and want to solve the eigenvalue problem \begin{equation*} Lv = \lambda v, \end{equation*} where $\lambda \in \mathbb{C}$. I further want to impose periodic boundary conditions, but I will mention that later. I know that I will find roots $r_j$ of the characteristic polynomial of $L - \lambda$, these roots $r_j$ will give me exponential functions as solutions so that the general solution is \begin{equation*} v(x) = \sum_{j=1}^4 c_j e^{r_j x}. \end{equation*} Now, I want to impose periodic boundary conditions \begin{equation*} v^{(k)}(0) = v^{(k)}(L), \ 0 \leq k \leq 3. \end{equation*} for some $L > 0$, in order to determine which values of $\lambda \in \mathbb{C}$ give nontrivial $v$.

The whole point of my question is this: is $\{ e^{r_jx} \}_{j=1}^4$ a suitable basis for finding the $\lambda$ that work, or should I convert to real and complex parts (as one does when $\lambda$ is real meaning the the roots $r_j$ come in complex pairs) in order to find a basis? This is also confusing to me because in the latter case, I will have between four and eight basis functions depending on roots being real or complex.

  • $\begingroup$ There is no need to take real and imaginary parts. It only makes things more complicated. $\endgroup$ – Robert Israel Mar 22 at 18:40
  • $\begingroup$ But I should at least write the complex exponentials out as $e^{a_jx}( \cos (b_jx) + i \sin (b_jx))$ in order to find suitable $\lambda$, yes? $\endgroup$ – swygerts Mar 22 at 19:34

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