Figure 4: Error in $L_\infty$-norm compared to the analytical solution for advection $N_\theta = 2000$, $q = \sqrt{2}$, $n = 24$ and $m = -34$ (left), $m = -33$ (right)
on $[0, 2\pi]^2$ with $A_1 = 1$ and $A_2 = q$. So $r, R, v$ can be chosen so that $\tilde{v} = q$, that is
In this way of presentation, we remark that we have three zones: a first zone, when the error does not decrease, a second zone where the error decreases according to the order of interpolation and then a third zone when the error does no more decrease which depends on $N_\theta/|m|$. Note that for a fixed degree and a given $N_\theta/|m|$, the error corresponding to the standard or aligned method lies on the same curve. But for a given $N_\phi$, there is shift in abscissa which corresponds to the distance between $N_\varphi/|n|$ and $N_\varphi/|\hat{k}_\parallel|$. So when the error of the standard method is in the left of the curve, the error of the new method lie on the right of it. So, for example, in order to have an error around the error of discretization in $\theta$ (when the error begins to saturate: beginning of the third zone), as we are on the same curve, we typically need $\frac{N_\varphi}{|\hat{k}_\parallel|}$ points for the aligned method instead of $\frac{N_\varphi}{|n|}$ points. The factor of gain is thus
On Figure 7, we fix $m = -34$ and change the values of $n$. Now we consider as abscissa only $N_\varphi/|m|$. We then see the same effect. We can remark that when $n$ is smaller than $|\hat{k}_\parallel|$ (here $n=5$ and $\hat{k}_\parallel \simeq -19$), the classical method is more accurate. For $n$ and $\hat{k}_\parallel$ similar, both methods have the same accuracy ($n=12$ and $\hat{k}_\parallel \simeq -12$). Then increasing $n$, $|\hat{k}_\parallel|$ is diminished ($n=23$ and $\hat{k}_\parallel \simeq -1$) and the aligned method becomes more and more efficient. For $\hat{k}_\parallel \simeq 0$, we have the best result (this corresponds to the previous curve: with $n=24$). Increasing again $n$, the results are still better but less and less ($n=30$ and $\hat{k}_\parallel \simeq 6$). For $n=100$ (not shown), we approach again the curve corresponding to $n=12$. We have considered here LAG9, $N_\theta = 200$, except for $n=23$, where $N_\theta = 400$. This to see, that the saturation error diminishes with $N_\theta$ (on the previous plots, we had seen that the error increases with $m$).
On Figures 5,6,7, we have considered the following values for $N_\varphi$: