Exact BDF stability angles with maple

Abstract

BDF formulas are among the most efficient methods for numerical integration, in particular of stiff equations (see e.g. Gear in Numerical initial value problems in ordinary differential equations, Prentice Hall, Upper Saddle River, 1971). Their excellent stability properties are known for precisely half a century, from the first calculation of their angles of $A(\alpha )$-stability by Nørsett (BIT, 9:259–263, 1969). Later, more insight was gained and more precise values were calculated numerically (see for example Hairer and Wanner in Solving ordinary differential equations, Springer, New York, 1996, Sect. V.2). This was the state-of-the-art, when Akrivis and Katsoprinakis (BIT, 2019. https://doi.org/10.1007/s10543-019-00768-1) discovered exact values for these angles. In this note we simplify the derivation and results by using Maple.

Let $z=x+iy=e^{i\varphi }$ lie on the unit circle, then the root locus curve for k-step BDF is

$$\begin{aligned} w=u+iv=\sum _{j=1}^k \frac{1}{j} (1-z^{-1})^j\,, \qquad (0\le \varphi \le 2\pi ) \end{aligned}$$

(see e.g., [3, Sect. V.1, Eq. (1.17)]). It describes the boundary of the stability domain (see Fig. 1).

Fig. 1

Root locus curve and stability domain for BDF4

We use the classical parametrization of the unit circle

$$\begin{aligned} x=\frac{1-t^2}{1+t^2}\,,\quad y=\frac{2t}{1+t^2}\,,\quad t=\tan \frac{\varphi }{2}\,, \qquad ( -\infty<t<+\infty ) \end{aligned}$$

(see e.g., Euler’s Calculus Integralis, Caput V, § 261, 1768). The angle $\alpha$ of $A(\alpha )$-stability is then computed by the following Maple commands:

Here, $p=\tan \alpha =-\frac{v}{u}$, and pdn is the numerator of $\frac{dp}{dt}$, a polynomial in $t^2$ due to symmetry, whose largest root $t_0^2$ we have to compute (corresponding to the extremal value $w_0$ furthest away from the origin). For $k=3,4$ and 6, this $t_0^2$ is rational, which leads to expressions for $p_0=\tan \alpha$ containing only one square root (with Digits:=30):

k	$t_0^2$	$p_0=\tan \alpha$	$\alpha$
3	$\frac{9}{35}$	$\frac{329}{135}\sqrt{35}$	$86.032366860211647332387423479^{\circ }$
4	$\frac{2}{3}$	$\frac{699}{512}\sqrt{6}$	$73.351670474578482110409536864^{\circ }$
5	$\times$	$\times$	$51.839755836049910391602721533^{\circ }$
6	$\frac{15}{13}$	$\frac{45503}{1974375}\sqrt{195}$	$17.839777792245700101632480553^{\circ }$

The case $k=5$ is more complicated, since here¹

$$\begin{aligned} pdn=15(248t^4-275t^2+25)(t^2+1)^4,~~\hbox {largest root:}~~ \textstyle t_0^2=\frac{275}{496}+\frac{5}{496}\sqrt{2033}\,. \end{aligned}$$

Therefore we have to simplify the expression for $p_0$ by using the field structure of the set $\{ a+b\sqrt{2033}~;~ a,b \hbox { rationals}\}$. The command evala(simplify(p0)) in Maple leads to

$$\begin{aligned} p_0= \left( -\frac{51844971}{14086400}+\frac{5765167}{70432000}\sqrt{2033}\right) \sqrt{8525+155\sqrt{2033}}. \end{aligned}$$

FormalPara Remark

By differentiating x (instead of p), we obtain exact values for the values of D in Gear’s definition of stiff stability (see [2] or [3, p. 250]) as

k	3	4	5	6
D	$\displaystyle \frac{1}{12}$	$\displaystyle \frac{2}{3}$	$\displaystyle \frac{93}{80}+\frac{25}{48}\sqrt{5}$	$\displaystyle \frac{243}{40}$