I been working on Kuramoto-Sivashinsky Equation. In the process of analysis, I need to solve the following eigenvalues problem
$$ -u_{xxxx}-\lambda u_{xx}=\beta(\lambda)u $$
where $\lambda$ is a positive parameter and subject to periodic boundary condition $u(x+2k\pi)=u(x), k\in \mathbb{Z}$. Also, we required $\int_0^{2\pi}u(x,t)dx=0$.
This eigenvalues problem can be easily solved. We have
$$\beta_1(\lambda)=\beta_2(\lambda), \cdots, \beta_{2n-1}(\lambda)=\beta_{2n}(\lambda)=n^2(\lambda-n^2)$$
and
$$e_1=\sin x, e_2=\cos x \dots ,e_{2n-1}=\sin nx, e_{2n}=\cos nx$$
However, when working with generalized Kuramoto-Sivashinsky Equation, the eigenvalues problem become
$$ -u_{xxxx}-\lambda \nu u_{xxx}-\lambda u_{xx}=\beta(\lambda)u \ \ \ \ \ \ \ (\star) $$
Where $\lambda$ and $\nu$ are positive parameters with $\nu$ fixed. The periodic boundary condition above still applied.
I am stuck with the extra term $\lambda \nu u_{xxx}$ in equation $\star$. I need help to solve this eigenvalues problem. Also I am not sure we still have eigenvector which consists of trigonometric functions.
Thanks