# Does Mixed boundary conditions change Von Neumann Stability Analysis

I'm looking at the 1D Advection-Diffusion Equation $$\frac{\partial\phi}{\partial t}+\frac{\partial}{\partial x}\left(u\phi - D\frac{\partial \phi}{\partial x}\right)=0$$ with $u$ and $D$ constant and the following Dirichlet-Neumann Boundary Conditions: $$\phi(t,0)=\alpha,\quad\text{ and }\quad\phi_{x}(t,1)=\beta$$ I have a very similar case, but for simplicity we can consider with the 2nd Order CDS Scheme $$u\frac{\phi_{i+1}-\phi_{i-1}}{2 \Delta x}-D\frac{\phi_{i+1}-2\phi_{i}+\phi_{i-1}}{\Delta x^{2}}=0$$ Does Neumann stability analysis still hold? I have not been able to find the explanation out right if why. Also does the inhomogeneous case differ?

Take $P=\frac{u\Delta x}{D}$ s.t. the 2nd Order CDS scheme becomes $$-\frac{D}{\Delta x^{2}}\left[\left(1-\frac{P}{2}\right)\phi_{i+1}-2\phi_{i}+\left(1+\frac{P}{2}\right)\phi_{i-1}\right]=0$$ where the discretization at the boundaries is done by:

Left Dirichlet BC: $\alpha = \frac{\phi_{1}+\phi_{0}}{2}\implies \phi_{0}=-\phi_{1}+2\alpha$

such that $$-\frac{D}{\Delta x^{2}}\left[\left(1-\frac{P}{2}\right)\phi_{2}-\left(3+\frac{P}{2}\right)\phi_{1}\right]=2\frac{D}{\Delta x^{2}}\left(1+\frac{P}{2}\right)\alpha$$

Right Neumann BC: $\beta = \frac{\phi_{N+1}+\phi_{N}}{\Delta x}\implies \phi_{N+1}=-\phi_{N}+\Delta x\beta$

such that

$$-\frac{D}{\Delta x^{2}}\left[\left(\frac{P}{2}-3\right)\phi_{N}+\left(1+\frac{P}{2}\right)\phi_{N-1}\right]=\frac{D}{\Delta x}\left(1-\frac{P}{2}\right)\beta$$

How would you show that Fourier Modes don't apply to the boundaries?

• Remarks: (i) Should it be $-D$ in your PDE? Otherwise the discretisation is inconsistent. (ii) Are $u$ and $D$ positive numbers? Otherwise there is no chance that the PDE is well-posed (assuming $-D$ in the PDE). (iii) Stability of course depends on the discretisation near the boundaries, which you have not included here. (iv) Von Neumann analysis assumes that the solution takes the form of a Fourier series. If you can show that the Fourier modes satisfy the boundary conditions, you can still use Von Neumann. However, generally they wont and you have to resort to other techniques. – ekkilop Aug 2 '17 at 10:25
• Apologies for the mistake, it's been corrected. So for this example how would you show that Fourier modes don't satisfy the boundary conditions? Can you cite a paper or text where this or something similar is demonstrated? – DanBernou Aug 3 '17 at 9:18
• Fourier modes are periodic by definition so they cannot satisfy the boundary conditions ($\alpha$ and $\beta$ are constants, right?) unless the entire solution is constant. This would of course mean that $\beta = 0$ and that we have a constant initial condition. For less trivial situations, stability can be analysed using e.g. the Energy method or by Normal mode analysis (Laplace transform). Both these techniques are described in many texts e.g. springer.com/us/book/9783540749929 – ekkilop Aug 3 '17 at 10:50

von Neumann analysis assumes a plane wave like solution $u = \alpha e^{i(\kappa x - \omega t )}$. So it only applies to infinite space. (I do not even think it applies to periodic space).
Sengupta, Tapan K., Anurag Dipankar, and Pierre Sagaut. "Error dynamics: beyond von Neumann analysis." Journal of Computational Physics 226.2 (2007): 1211-1218.