david/step-size-strat

Let \[\Sigma = \sum_{j=0}^{k-1} h_{n-k+j}\] and \[\mu = \min_{0\le j \le k-1} h_{FE}(u_{n-k+j}).\]

Then \[\Omega_{k-1} = \Sigma/h_n\]

For the third-order methods, the greedy algorithm in the first case gives the step size

\[h_n = \frac{\Sigma \mu}{\Sigma + 2\mu}.\] Then the condition \[ \Omega_{k-1}\ge 2\] is automatically satisfied, while the condition \[\Omega_{k-1} \le 2 + 2\sqrt{2}\] leads to \[\Sigma \le 2 \sqrt{2} \mu.\]

In the second case, the greedy algorithm gives the step size (corrected by Lajos) \[h_n = \Sigma \frac{3\mu - \Sigma}{\Sigma - 2\mu}\] as long as \[ 2\mu \le \Sigma \le 3\mu.\] Then the condition \[\Omega_{k-1} > 2 + 2\sqrt{2}\] leads to \[\Sigma > 2\sqrt{2} \mu.\] Together, this means the second greedy choice works if \[ 2\sqrt{2}\mu \le \Sigma \le 3\mu.\] If \(\Sigma > 3\mu\), it seems there is no SSP step size.

Note that typically (when \(h_{FE}\) varies slowly) we have \[\Sigma \approx (k-3)\mu,\] which gives \(\mu\) for \(k=4\) and \(2\mu\) for \(k=5\). Thus we will use the first choice except when \(h_{FE}\) changes quite rapidly.