lajos/p2s3 Miscellaneous

In this note I’ll collect some observations concerning the \(p=2\), \(s=3\) case.

By applying the formula for the derivative of the determinant once more, we see that the stability function takes the following form:

\[ \psi(z)=\frac{1+(1-\tau)z+\frac{1}{2}\left((1-\tau)^2-\tau_2\right)z^2+\frac{1}{6}((1-\tau)^3-1+3(\tau-1)\tau_2-2\tau_3+6\gamma)z^3}{1-\tau z+\frac{1}{2}(\tau^2-\tau_2)z^2-\frac{1}{6}(\tau^3-3\tau\cdot\tau_2+2\tau_3)z^3} \]

with \(\gamma:=\textrm{tr}(A^2 eb^T)\). Alternatively, by also introducing \(\sigma\) as the second invariant of the matrix \(A\) and \(\delta:=\det(A)\) (in other words, if \(\lambda_k\) denote the three eigenvalues of \(A\), then \(\tau=\sum_k \lambda_k\), \(\sigma=\lambda_1 \lambda_2+\lambda_1 \lambda_3+\lambda_2 \lambda_3\), and \(\delta=\prod_k \lambda_k\)), we can have a slightly less non-linearity in the parameters as

\[ \psi(z)=\frac{1+(1-\tau)z+\left(\frac{1}{2}-\tau+\sigma\right)z^2+\left(-\frac{\tau}{2}+\sigma-\delta+\gamma\right)z^3}{1-\tau z+\sigma z^2-\delta z^3}. \]

For the conjectured maximal SSP coefficient 6, we can assume that the numerator and denominator have no common roots, since otherwise the stability function would be either a constant or by the Griend-Kraaijevanger article (1986) the maximal SSP coefficient \(\le R_{m/n,2}\) (with \(m\le 3\) and \(n\le 2\)) \(\le 3+\sqrt{2\cdot 3}<6\). But how can we generalize this “no common root condition” for \(s>3\)?

The trace inequalities in the \(s=3\) case \(3\tau_2\ge \tau^2\) and \(9\tau_3\ge \tau^3\) imply

\[ \tau^2\ge 3\sigma \quad \textrm{and} \quad 27\delta+8\tau^3\ge 27\sigma\tau. \]

As for \(\gamma\), we have the trivial lower estimate \(\gamma\ge 0\).

There are some other constraints however that are implied by conditions (6.35a\(-\)b.) For example, it is easy to see that \(MinRowSum(A)\le \frac{1}{2}\le MaxRowSum(A)\).

I could prove that \(\gamma=\)tr\((A^2 e b^T)\in\left[\frac{1}{2}\cdot MinRowSum(A),\frac{1}{2}\cdot MaxRowSum(A)\right]\), but I could not make much use of this fact so far.

A simple observation: if we assume \(MinRowSum(A)=0\), then \(A\) has a zero row, so \(\delta=0\), so by the Griend-Kraaijevanger article (1986) the maximal SSP coefficient would be \(<6\). Therefore, we can assume \(MinRowSum(A)>0\), implying \(\gamma>0\).

There are other inequalities (in terms of the row sums of \(A\)) that are consequences of the non-negativity of the components of vector \(b\). WRITE THESE UP, if necessary.

Since we can assume that the numerator and denominator have no common roots, the numerator and the denominator both should be positive on \(-6 < z\le 0\) by continuity. (This observation generalizes my earlier conjecture in the \(s=2\) case that \(\det(I+rA)=\det(I+rK) > 0\), with the notation of (6.35a\(-\)d.))

The numerator being non-negative at \(z=-6\) implies

\[ \gamma\le \delta -\frac{5}{6}\sigma +\frac{13}{36}\tau+\frac{13}{216}. \]

I’ve found the following statement in the book of Hairer-Wanner (Solving Ordinary Differential Equations II., p. 179.): “Let \(r\) denote the threshold factor of the rational function \(R\). Then its stability domain includes the disc {$ z : |z+r| <= r $}.” Can this statement have any significance for us?

It would be nice to figure out whether it is possible to generalize https://mathwiki.kaust.edu.sa/lajos/Bounds%20on%20the%20coefficients%20of%20a%20polynomial%20having%20all%20derivatives%20non-negative, if, instead of a finite sum, we have an infinite series, being the Taylor series of the stability function (as a rational function) around \(z=-2s\):

\[ \psi(z)=\sum_{k=0}^{\infty}\gamma_k (1+\frac{z}{2s})^k. \]

The Perron-Frobenius theorem or Griend-Kraaijevanger (1986) implies that there is a positive real pole \(\alpha_0 > 0\) of \(\psi\), and there are no other poles of \(\psi\) within the open disc centered at the origin with radius \(\alpha_0\). Moreover, Griend-Kraaijevanger say that \(MonRad(\psi)\le B(\psi)\). So in order to have \(MonRad\ge 2s\), we need \(B\ge 2s\). But this implies by the definition of \(B(\psi)\) that the open disc with center \(-2s\) and radius \(2s+\alpha_0\) does not contain any other poles.

This justifies the existence of the above Taylor series of \(\psi\). Moreover, we know that the radius of convergence of a Taylor series is equal to the distance from the center to the nearest singularity, that is, \(2s+\alpha_0\).

(Does this explain the “no common root condition” in general?)

The \(p=2\) order conditions \(1=\psi(0)=\psi\prime(0)=\psi\prime\prime(0)\) imply that

\[ \sum_{k=0}^{\infty}\gamma_k=1, \]

\[ \sum_{k=0}^{\infty}k \gamma_k=2s, \] and \[ \sum_{k=0}^{\infty}k(k-1)\gamma_k=(2s)^2. \] From \(\psi^{(k)}(-2s)\ge 0\) (for integer \(k\ge 0\)) we get \(\gamma_k\ge 0\).

How can we show that \(\gamma_0=\gamma_1=\ldots=\gamma_{s-1}=0\), implying that \(\psi\) has a root at \(z=-2s\) with multiplicity \(s\)? How should we take into account the effect of \(A\ge 0\) on the \(\gamma_k\) coefficients?