Grid Refinement

$ \def \half {\frac{1}{2}} \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} $ So far so good for grid coarsening. Let's consider instead the inverse of grid coarsening, which is grid refinement. Now it is quite clear that - with DoubleGrid as well as with TripleGrid - all iterations are exactly the reverse of the above.
Thus, for DoubleGrid refinement: $$ y_k = (y_{k+1} - 2)^2 $$ This actually means that we have to solve a quadratic equation for $y_{k+1}$ : $$ y_{k+1}^2 - 4 y_{k+1} + 4 - y_k = 0 $$ The discriminant $D$ of a quadratic equation $a.x^2 + b.x + c = 0$ is very well known: $$ D = b^2 - 4.a.c $$ Hence, for the problem at hand, the discriminant is: $$ D = 16 - 4(4-y_k) = 4 y_k $$ And it is positive for $y_k \ge 0$. The solutions of the quadratic equation are then: $$ y = 2 \pm \sqrt{y_k} $$ But these solutions may both serve again as input for a next iteration, associated with DoubleGrid mesh refinement. This means that $\sqrt{y_k} \le 2$ and therefore $y_k \le 4$. Thus we have restricted the possible values of $y_k$ to the interval $\left[0,4\right]$. (The inverse of) this interval is known in MultiGrid (read: DoubleGrid) Calculus as the Dangerous Interval. The whole danger being a division by zero in the accompanying Gaussian elimination process. We can see that the restriction $0 \le y_k \le 4$ is both necessary and sufficient. Because it is impossible for the iterands $y_k$ to "escape" from that interval.
If, on the other hand $y_k > 4$, then one of the roots becomes negative, while the other remains positive and becomes greater than $4$ again. With the next iteration, only the positive root can be used, resulting again in a negative and a positive root. The latter becomes larger and larger. This form of DoubleGrid refinement has been advertized as the Safe Interval in Multigrid Calculus.
But, as before, it should have been easier to employ the angles instead: $$ \cos(\phi_k) = \cos(2\phi_{k+1}) \slechts \phi_{k+1} = \left\{\: \frac{\phi_k}{2} \:,\: \pi - \frac{\phi_k}{2} \:\right\} $$ Now we will start with the value that is certainly dangerous, namely $y_0 = 0$. Because that value will give rise to a division by zero during the Gaussian elimination process. Then we have: $$ \cos(\phi_0) = \half y_0 - 1 = - 1 \hieruit \phi_0 = \pi $$ The next step is: $$ \phi_1 = \left\{\: \frac{\pi}{2} \:,\: \pi - \frac{\pi}{2} \:\right\} = \pi \left\{\: \frac{1}{2} \:\right\} $$ Add to this the initial value $\pi$ , then we have in total for the dangerous values so far: $$ \phi_1 \cup \phi_0 = \pi\,\left\{\: \half \:,\: 1 \: \right\} $$ The next step is: $$ \phi_2 = \left\{\: \frac{\pi}{4} \:,\: \pi - \frac{\pi}{4} \:\right\} = \pi\,\left\{\: \frac{1}{4} \:,\: \frac{3}{4} \:\right\} $$ Add to this the other iterands, then we have in total for the dangerous values so far: $$ \phi_2 \cup \phi_1 \cup \phi_0 = \pi \,\left\{\:\frac{1}{4}\:,\:\half\:,\:\frac{3}{4}\:,\:1\:\right\} = \pi\, \left\{\:\frac{1}{4}\:,\:\frac{2}{4}\:,\:\frac{3}{4}\:,\:\frac{4}{4}\:\right\} $$ Now it's not difficult anymore to make an educated guess for the next steps. And it is conjectured that: $$ \phi_n \cup \phi_{n-1} \cup ... \cup \phi_2 \cup \phi_1 \cup \phi_0 = \frac{\pi}{2^n}\,\left\{\:1\:,\:2\:,\:3\:,\:.\,.\,.\:,2^n\:\right\} $$ The physical meaning of this is that (the angles of the) dangerous points are arbitrarily dense in the domain $0 < y_n \le 4$ . Sooner or later, any number in the dangerous domain will become a candidate for division by zero in the (iterative version of the) Gaussian elimination process. But the above is only a replay of results that have been derived, more thouroughly, in the Multigrid Calculus document. Let's proceed now with TripleGrid and make a fresh start.