# Semi-discrete FVM space discretisation of Advection-Diffusion, stability and convergence.

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I'm looking into different schemes of the 1D constant coefficient Advection-Diffusion Equation with a cell-centered Finite volume approach. \begin{align} \frac{\partial c}{\partial t} +v\frac{\partial c}{\partial x} - D\frac{\partial^{2} c}{{\partial x}^{2}} = S\qquad x\in[0,1]\\ \end{align} Specifically, with Dirichlet-Neumann boundary conditions \begin{align} \text{BC:}\quad & \begin{cases} c(t,x=0)=g_{1}(t)\\ c_{x}(t,1)=g_{2}(t) \end{cases}\\ \text{IC:}\quad & c(0,x)=f(x) \end{align} Where I get really confused is how to compute the eigenvalues of the resulting ODE system given the boundary conditions are applied on their outer cell boundary. $$\text{Resulting ODE:}\quad Mc' = -Ac + b$$ (to simplify the problem take source term $S=0$.),[midpoint rule is used for $\frac{\partial c}{\partial t}$

Typically here I would have a D-D homogeneous BC case and substitute a solution vector $c=e^{\lambda t}v$ and get the generalised eigenvalue problem $$(A+\lambda M)v=0$$ Does the same hold for mixed BC?

Now since the schemes I'm looking at are "gnarly" let us take an example scheme similar to 2nd order Central Difference Scheme times $\alpha$, with $M=I$. So $A= -\frac{D}{\Delta x^{2}}[1,-2,1]$ with the first and last rows adjusted for the BCs. For completeness: $b$ is the boundary data where $b_{1} = \frac{2\alpha D g_{1}(t)}{\Delta x^{2}}$(by ghost point), $b_{N} = \frac{\alpha D g_{2}(t)}{\Delta x}$(averaged to boundary point) and the rest being $0$.

So now we take the Dirichlet-Neumann BC to be homogenous, reducing our ODE system to: \begin{align} &Mc'=-Ac,\qquad\qquad (M=I)\\ &\implies (A+\lambda I)v=0? \end{align} but how do we factor in these mixed boundary conditions? $c_{1}=0=\frac{\partial c_{N}}{\partial x}$ but $c_{1}\neq c_{N}\neq 0$(or at least it should right?)

Also, does Von Neumann Stability Analysis still hold for such mixed boundary conditions?

Finally more specifically, one of the schemes I'm looking at $A$ is still tridiagonal, and M is lower bidiagonal (main ($0$th) diagonal, and ($-1st$) lower diagonal), but neither is symmetric or skew (both with constant coefficients on the diagonals [except BC changes on A]). Nor are they (Semi-)Positive/Negative Definite. I'm out of ideas....