, 17 min read

Stability Mountain for Cryer's Formula

Colin Cryer in his article on $A_0$-stable methods mentioned in passing that there is a method, which is $A_0$-stable but not $A[\alpha]$-stable for any $\alpha$. For the terminology see Tendler-like Formulas for Stiff ODEs. This post shows the stability mountain of this method:

$$ y_{n+2} - y_{n+1} = \frac{h}{4}\left( f_{n+2} + 2 f_{n+1} + f_n \right). $$

It has generating polynomials

$$ \rho(\zeta) = \zeta^2 - \zeta, \qquad \sigma(\zeta) = (\zeta+1)^2/4. $$

It is a two-step method of consistency order only of one. So, it has no practical relevance, for example, in comparison to the implicit Euler-method.

References.

  1. Cryer, Colin W.: “A new class of highly-stable methods: $A_0$-stable methods”, BIT 13, 153-159(1973)
  2. Jeltsch, Rolf: “Stiff Stability and its Relation to $A_0$- and $A(0)$-Stability”, SIAM J. Num. Anal., Vol 13, No. 1, March 1976

A method is called $A_0$-stable if the stability region contains the entire set

$$ \{z\in\mathbb{C}: \mathop{\rm Re}(z)\le0 \land \mathop{\rm Im}(z)=0 \}, $$

i.e., the entire left real axis. Obviously, every $A[\alpha]$-stable method is $A_0$-stable. Above example shows, that the opposite statement is not true.

This means that the $A_0$-stable methods are a larger class than the $A[\alpha]$-stable methods.

The output of stabregion4 -df Cryer2:

Cryer2, p=1, k=2, l=1
                 0.0000
                -4.0000
                 4.0000
                 1.0000
                 2.0000
                 1.0000
rho_0              0.000000000
rho_1              0.000000000
rho_2              0.500000000 <-----

The matrices in question are given below.

k=2, l=1, rest=1, n=2, nrest=1, nsq=4, colLen=6
Cryer2
A1
              1.00000      0.00000
              0.00000      4.00000
A0
              0.00000     -1.00000
              0.00000     -4.00000
B1
              0.00000      0.00000
              0.00000      1.00000
B0
              0.00000      0.00000
              1.00000      2.00000
parasitic roots of Cryer2
        nr      real                    imag                    abs                     1-th root
            0      1.00000000         0.00000000              1.00000000                1.00000000
            1      0.00000000        -0.00000000              0.00000000                0.00000000
radius at infinity of Cryer2
        nr      real                    imag                    abs                     1-th root
            0      0.00000000         0.00000000              0.00000000                0.00000000
            1      0.00000000         0.00000000              0.00000000                0.00000000
            3      0.00000000        -0.00000000              0.00000000                0.00000000

1. Stability region

Below is the output of:

stabregion4 -aguptri -f Cryer2 -oj -L100:-10:1:4 -r150

One clearly sees the parabolic nature of the border of the stability region. Every ray from the origin starting above the parabola will eventually cross the parabola. So, you can decrease the step-size under all circumstances without losing stability. But increasing the step-size can lead into an unstable region.

Therefore the above values for delta and alpha are wrong. The method is not $A[\alpha]$-stable nor $S[\delta]$-stable.

2. Stability mountain

Below is the output of:

stabregion4 -f Cryer2 -o3 -L100:-10:1:4 -r110

You can rotate the graphic around any axis.

3. Cryer's k-step methods

Mapping the unit circle to the left half-plane:

$$ z = {\zeta+1\over\zeta-1}, \qquad \zeta={z+1\over z-1}. $$

The generating polynomials then become:

$$ r(z) = \left({z-1\over 2}\right)^k \rho\left({z+1\over z-1}\right) = \sum_{j=0}^k a_j z^k, $$

and

$$ s(z) = \left({z-1\over2}\right)^2 \sigma\left({z+1\over z-1}\right) = \sum_{j=0}^k b_j z^k. $$

Cryer (1973) further proves the following.

1. An $A_0$-stable method must be implicit.

2. The $k$-step Adams-Moulton method with $k\ge2$ is not $A_0$-stable.

4. The $k$-step method of order $k$ corresponding to $s(z)=(z+d)^k$ is $A_0$-stable if $d\ge2^{k+1}$.

4. The $k$-step method of order $k$ corresponding to $s(z)=(z+d)^k$ is not $A_0$-stable if $k\ge8$ and $d\le\frac{1}{2}k^{1/2}-1$.

A method which is $A[\alpha]$-stable for some $\alpha\gt 0$ is called $A(0)$-stable.

Jeltsch (1976) proves the following.

5. Cryer's $k$-step method with $\sigma(\zeta)=(\zeta+d)^k$ is $A(0)$-stable and stiffly-stable if

$$ -1 \lt d\le -1 + {2\over 1+2^{k+1}}. $$

6. To any positive constant $D$ any positive integer $k$ there exists a stiffly-stable $k$-step method of order $k$ which contains the entire left half-plane left from $-D$.