This is related to my question here Lax-Wendroff scheme stability analysis for a linear system of conservation laws , I hope it will reach more readers.
Consider the numerical scheme given by the finite difference equation:
\begin{align*}
Q_{i,j,k}^{n+1}&= \Big(I- \frac{\Delta t^2}{\Delta x^2}A^2-\frac{\Delta t^2}{\Delta y^2}B^2-\frac{\Delta t^2}{\Delta z^2}C^2\Big)Q_{i,j,k}^{n} \\
& \ +(\frac{\Delta t^2}{2\Delta x^2}A^2- \frac{\Delta t}{2\Delta x}A) Q_{i+1, j, k}^n- (\frac{\Delta t^2}{2\Delta x^2}A^2- \frac{\Delta t}{2\Delta x}A)Q_{i-1, j, k}^n \\
&\ +(\frac{\Delta t^2}{2\Delta y^2}B^2 - \frac{\Delta t}{2\Delta y}B) Q_{i, j+1, k}^n- ( \frac{\Delta t^2}{2\Delta y^2}B^2- \frac{\Delta t}{2\Delta y}B)Q_{i, j-1, k}^n \\
&\ + (\frac{\Delta t^2}{2\Delta z^2}C^2 - \frac{\Delta t}{2\Delta z}C)Q_{i, j, k+1}^n- (\frac{\Delta t^2}{2\Delta z^2}C^2- \frac{\Delta t}{2\Delta z}C)Q_{i, j, k-1}^n \\
&\quad + \frac{\Delta t^2}{8\Delta x\Delta y}(AB + BA)(Q_{i+1, j+1, k}^n- Q_{i-1, j+1, k}^n)- (Q_{i+1, j-1, k}^n- Q_{i-1, j-1, k}^n) \\
& \quad + \frac{\Delta t^2}{8\Delta x\Delta z}(AC + CA)(Q_{i+1, j, k+1}^n- Q_{i-1, j, k+1}^n)- (Q_{i+1, j, k-1}^n- Q_{i-1, j, k-1}^n) \\
&\quad + \frac{\Delta t^2}{8\Delta z\Delta y}(CB + BC)(Q_{i, j+1, k+1}^n- Q_{i, j+1, k-1}^n)- (Q_{i, j-1, k+1}^n- Q_{i, j-1, k-1}^n)
\end{align*}
Where $Q_{i,j,k}^{n+1}\in \mathbb{R}^4$ is a vector of $4$ components, at the updated one-step time $t_{n+1}= (n+1)\Delta t$
$Q_{i,j,k}^{n+1}$ is given as linear combination of values of a $19$-points stencil in the space, and $A, B, C$ are constant $4\times 4$ matrices, making this system linear.
Is this numerical scheme stable ? for what (sufficient) condition ?
Is it convergent ?
I appreciate any hints or directions I should look for.
Thanks to @editPiAf I can attempt the following, but I still need to make sense of it:
Fix a unit vector $n=(n_x, n_y, n_z)$ and note $I= \sqrt{-1}$ , we transform $$Q^{n}_{i,j,k}= \hat{Q}e^{I(wt_{n}-\kappa n.x)}= \hat{Q}e^{I(wt_{n}-\kappa(n_xi\Delta x+ n_yj\Delta y+ n_zk\Delta z))}$$
$\textbf{this is Fourier transform (one term?) but how are we allowed to do this?}$
where $t_n= n\Delta t$ and $x= (i\Delta x,j\Delta y,k\Delta z)$ is the position vector. Here $w, \kappa$ represent the angular frequency and wave number, respectively.
Then we find $$Q^{n+1}_{i,j,k}= \hat{Q}e^{Iw\Delta t+ I(wt_{n}-\kappa(n_xi\Delta x+ n_yj\Delta y+ n_zk\Delta z))}= e^{Iw\Delta t}Q^{n}_{i,j,k}$$ and $$Q^{n}_{i+\delta i ,j+\delta j,k+\delta k}= \hat{Q}e^{I(wt_{n}-\kappa n.x- \kappa n.\delta x)}= e^{-I \kappa n.\delta x}\hat{Q}e^{I(wt_{n}-\kappa n.x)}= e^{-I \kappa n.\delta x}Q^{n}_{i,j,k}$$ For example, \begin{align*} Q^{n}_{i+1 ,j-1,k}&= e^{- I\kappa (n_x\Delta x-n_y\Delta y)}Q^{n}_{i,j,k} \\ Q^{n}_{i ,j+1,k}&= e^{- I\kappa (n_y\Delta y)}Q^{n}_{i,j,k} \end{align*}
Then we obtain, by using the scheme equation, where we denote $G= e^{Iw\Delta t}$ and $I$ the identity matrix \begin{align*} 0= Q^{n}_{i,j,k}\Bigg[ -GI + \Big(I- \frac{\Delta t^2}{\Delta x^2}A^2-\frac{\Delta t^2}{\Delta y^2}& B^2-\frac{\Delta t^2}{\Delta z^2}C^2\Big) \\ \ +(\frac{\Delta t^2}{2\Delta x^2}A^2 &- \frac{\Delta t}{2\Delta x}A) e^{- I\kappa (n_x\Delta x)}- (\frac{\Delta t^2}{2\Delta x^2}A^2- \frac{\Delta t}{2\Delta x}A)e^{ I\kappa (n_x\Delta x)} \\ \ +(\frac{\Delta t^2}{2\Delta y^2}B^2 &- \frac{\Delta t}{2\Delta y}B) e^{- I\kappa (n_y\Delta y)}- (\frac{\Delta t^2}{2\Delta y^2}B^2- \frac{\Delta t}{2\Delta y}B)e^{ I\kappa (n_y\Delta y)} \\ \ + (\frac{\Delta t^2}{2\Delta z^2}C^2 &- \frac{\Delta t}{2\Delta z}C)e^{- I\kappa (n_z\Delta z)}- (\frac{\Delta t^2}{2\Delta z^2}C^2- \frac{\Delta t}{2\Delta z}C)e^{ I\kappa (n_z\Delta z)} \\ \quad + \frac{\Delta t^2}{8\Delta x\Delta y}(AB + BA)(e^{- I\kappa (n_x\Delta x+n_y\Delta y)}&- e^{- I\kappa (-n_x\Delta x+n_y\Delta y)}- e^{- I\kappa (n_x\Delta x-n_y\Delta y)}+ e^{- I\kappa (-n_x\Delta x-n_y\Delta y)}) \\ \quad + \frac{\Delta t^2}{8\Delta x\Delta z}(AC + CA)(e^{- I\kappa (n_x\Delta x+n_z\Delta z)}&- e^{- I\kappa (-n_x\Delta x+n_z\Delta z)}- e^{- I\kappa (n_x\Delta x-n_z\Delta z)}+ e^{ I\kappa (n_x\Delta x+n_z\Delta z)}) \\ \quad + \frac{\Delta t^2}{8\Delta z\Delta y}(CB + BC)(e^{- I\kappa (n_y\Delta y+n_z\Delta z)}&- e^{- I\kappa (n_y\Delta y-n_z\Delta z)}- e^{- I\kappa (-n_y\Delta y+n_z\Delta z)}+ e^{ I\kappa (n_y\Delta y+n_z\Delta z)})\Bigg] \end{align*} $\textbf{then how to go from here?}$ should we deduce that $G$ (hence $w\Delta t$) must satisfy \begin{align*} GI= \Big(I- \frac{\Delta t^2}{\Delta x^2}A^2-\frac{\Delta t^2}{\Delta y^2}& B^2-\frac{\Delta t^2}{\Delta z^2}C^2\Big) \\ \ +(\frac{\Delta t^2}{2\Delta x^2}A^2 &- \frac{\Delta t}{2\Delta x}A) e^{- I\kappa (n_x\Delta x)}- (\frac{\Delta t^2}{2\Delta x^2}A^2- \frac{\Delta t}{2\Delta x}A)e^{ I\kappa (n_x\Delta x)} \\ \ + \ ... \\ \quad + \frac{\Delta t^2}{8\Delta x\Delta y}(AB + BA)(e^{- I\kappa (n_x\Delta x+n_y\Delta y)}&- e^{- I\kappa (-n_x\Delta x+n_y\Delta y)}- e^{- I\kappa (n_x\Delta x-n_y\Delta y)}+ e^{- I\kappa (-n_x\Delta x-n_y\Delta y)}) \\ \quad + \ ... \end{align*} for all values of the unit vector $n$ ?