# How to Choose Basis Solutions for Eigenvalue Problem

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.

• There is no need to take real and imaginary parts. It only makes things more complicated. – Robert Israel Mar 22 at 18:40
• 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? – swygerts Mar 22 at 19:34