lajos/p3s2

David mentions his calculations show that the CPC-conjecture is false in the \(p=3\), \(s=2\) case, for example. I can confirm this, namely, for the Ferracina-Spijker (2008, Section 3.3.1.) optimal example the maximal SSP coefficient of the RK-method is \(1+\sqrt{3}\) (this is their \(c(K)\)), while \(MonRad\)(numerator of the stability function corresponding to those \(\kappa_{i,j}\)’s) is only \(\frac{3+\sqrt{3}}{2}\) being \(<1+\sqrt{3}\).

Nevertheless I proved with Mma the following. Consider all \((2,2)-\)rational functions with approximation order \(p=3\). Then we have only one free real parameter \(\alpha\):

\[ \frac{1+(\alpha +1) z+\left(\frac{1}{3}+\frac{\alpha }{2}\right) z^2}{1+\alpha z+\left(-\frac{\alpha }{2}-\frac{1}{6}\right)z^2}. \]

Since this method has a non-negative \(A\) matrix, according to the corresponding trace inequality we also have \(\alpha ^2-4 \left(-\frac{\alpha }{2}-\frac{1}{6}\right)\ge 0\). It is then easy to prove with Mma that \(MonRad\)(the above one-parameter family of rational functions with trace inequality imposed)\(\le 1+\sqrt{3}\), further, equality holds if and only if \(\alpha=-1+\frac{1}{\sqrt{3}}\), that is, the optimal stability function is the one corresponding to matrix \(A\) given by Ferracina-Spijker.

Notice that we proved this result in the general \(p=3\), \(s=2\) RK-class, and not only in the SDIRK class.

We can even strengthen the above result. First notice that for each real \(\alpha\), the resultant of \(1+(\alpha +1) z+\left(\frac{1}{3}+\frac{\alpha }{2}\right) z^2\) and \(1+\alpha z+\left(-\frac{\alpha }{2}-\frac{1}{6}\right)z^2\) is \(\frac{1}{12}\), meaning that the numerator and the denominator have no common zeros in the above class of rational functions. We can also easily prove that the trace inequality this time is equivalent to the fact that a rational function of this form has at least one positive real pole. But according to Corollary 3.4 of the Griend-Kraaijevanger article (1986, p. 416), having a positive real pole is necessary to have \(MonRad\)(a rational function with numerator 1+… and denominator 1+… and without common factors\()>0\). So the above computations imply that the maximal radius of absolute monotonicity of all \((2,2)-\)rational functions with approximation order \(p=3\) is \(\le 1+\sqrt{3}\), and equality implies uniqueness. We prove this last statement below.

Proof. For an arbitrary real \(\alpha\), set \(\psi(z):= \frac{1+(\alpha +1) z+\left(\frac{1}{3}+\frac{\alpha }{2}\right) z^2}{1+\alpha z+\left(-\frac{\alpha }{2}-\frac{1}{6}\right)z^2}\). The resultant condition mentioned above (\(=\frac{1}{12}\ne 0\)) is directly verified. Secondly, as cited above, \(MonRad(\psi) > 0\) implies the denominator having a (positive) real root, so a necessary condition is that the discriminant of the denominator should be \(\alpha ^2-4 \left(-\frac{\alpha }{2}-\frac{1}{6}\right)\ge 0\).

Then one easily computes that the inequalities \(\psi(-1-\sqrt{3})\ge 0\), \(\psi\prime(-1-\sqrt{3})\ge 0\) and \(\alpha ^2-4 \left(-\frac{\alpha }{2}-\frac{1}{6}\right)\ge 0\) imply \(\alpha=-1+\frac{1}{\sqrt{3}}\). For this unique \(\alpha\), the corresponding \(\psi_*\) has \({\psi\prime}_{*}(-1-\sqrt{3})=0\), so \(MonRad\le 1+\sqrt{3}\) by Griend-Kraaijevanger’s Lemma 4.5 with \(\ell=1\). We finally verify that \(MonRad(\psi_{*})=1+\sqrt{3}\).

We have \[\psi_*(z)=\frac{\left(2+\sqrt{3}\right) \left(\left(\sqrt{3}-1\right) z^2+2 \sqrt{3} z+6\right)}{\left(\sqrt{3}+3-z\right)^2},\] from which one can prove recursively for \(n\ge 1\) that \[{{\psi_{*}^{(n)}}}(z)=\frac{6 n! \left((12 n-3)+\sqrt{3} (7 n-2)+\left(3+2 \sqrt{3}\right) z\right)}{\left(\sqrt{3}+3-z\right)^{n+2} },\] yielding for \(n=0\) \[ {{\psi_{*}}}(-1-\sqrt{3})=1-\frac{\sqrt{3}}{2}>0 \] and for \(n\ge 1\) \[ {{\psi_{*}^{(n)}}}(-1-\sqrt{3})=\frac{3 \left(12+7 \sqrt{3}\right) n!(n-1)}{2^{n+1}\left(2+\sqrt{3}\right)^{n+2}}\ge 0, \] so \(\psi_{*}\) is absolutely monotonic at \(z=-1-\sqrt{3}\). Therefore we can apply Griend-Kraaijevanger’s Lemma 4.3 with \(B(\psi_{*})=\infty\), \(\ell=0\), \(x=-1-\sqrt{3}\) and \(\alpha_0=3+\sqrt{3}\), showing the absolute monotonicity of \(\psi_{*}\) on \((-x,0]\) as well. (Of course, the above explicit formulae for \(\psi_{*}^{(n)}\) easily permit us to directly verify \(\psi_{*}^{(n)}(z)\ge 0\) for all \(n\ge 0\) and all \(-1-\sqrt{3}\le z\le0\).)

\(\bullet\) Note: similarly to the “\(r\le 2s\)”-conjecture, the denominator of the optimal \(\psi_*\) is a perfect power, i.e., its pole has (the maximal possible) order \(s=2\), however, this time the numerator does not have multiple roots.

\(\bullet\) Note: the above proof provides us with a formula for the second to last row in the table on the last page of Griend-Kraaijevanger’s article, i.e., for \(R_{2/2,3}\).

\(\bullet\) If we assumed \(A\ge 0\), then we’d have \(-\alpha=\tau:=\mathrm{tr}(A)\ge 0\) and the trace inequality \(s\tau_2\ge \tau^2\). But in the proof we do not assume \(A\ge 0\). Nevertheless, if \(s=2\), then “the denominator of the stability function has a real root” (i.e. the discriminant \(\ge 0\) condition) is equivalent to this trace inequality (being the consequence of \(A\ge 0\)).

For \(s=3\), the discriminant condition is more complicated than the corresponding trace inequalities and I have found no relations between the sign of the discriminant and the validity/violation of the trace inequalities.