I have a quadratic eigenvalue problem of the form:
$$(A_2 s^2 + A_1 s + A_0)\hat{v} = 0$$
where $s$ is the eigenvalue. The matrices $A_i$ contain derivatives up to order six, and I have six boundary conditions:
$$ \hat{v}(0) = \hat{v}(1) = 0 \\ \hat{v}^{ii}(0) = \hat{v}^{ii}(1) = 0 \\ \hat{v}^{iv}(0) = \hat{v}^{iv}(1) = 0 $$
If the eigenvalue problem was of "standard" form $A\hat{v}=\lambda \hat{v}$ then I would simply modify the first and last rows of the matrix $A$ to impose the boundary conditions. But for my problem I don't know if I should impose the boundary conditions for all three matrices, or just for $A_0$ (which is not multiplied by $s$), in which case I would set the corresponding terms of $A_2$ and $A_1$ to zero. Actually I tried to use this last approach but it didn't work (I am calculating the eigenvalues using MATLAB's polyeig
).
More specifically, the differential equation is:
$$ \left [\left(D^2 - K^2 \right)s^2 - 2 \left(D^2 - K^2\right)^2s + \left(D^2 - K^2 \right)^3 + Ra K^2 \right] \hat{v} = 0 $$
where $D = \frac{\partial^2}{\partial y^2}$, $K$ is a constant (so numerically it is a diagonal matrix), and $Ra$ is a constant scalar. The differentiation operator $D$ is implemented using Chebyshev collocation, and the higher-order derivatives can be evaluated as powers of $D$.