The Root Locus Method for Finding the Interval of Absolute Stability

7.3. The Root Locus Method for Finding the Interval of Absolute Stability#

This method consists of repeatedly solving the characteristic polynomial for various values of \(h\lambda\) in the neighbourhood of the origin. Note that for simple linear and quadratic cases the resulting characteristic polynomial can be solved and a table or plots of values of \(\,h\lambda\,\) against roots \(|z_1|\) and \(|z_2|\) allows us to determine the interval of absolute stability. Any numerical method such as Newton Raphson iteration, or mathematics software packages like Matlab may be used for the approximate solution of the characteristic polynomial.

Example 7.2

Find the interval of absolute stability of the two-step explicit AB method,

\[ \begin{aligned} y_{j+1} ~=~ y_j + \frac{h}{2}(3f_j - f_{j-1}) \end{aligned} \]

when applied to \(y' = \lambda y\).

Solution

The characteristic equation is

\[ \begin{aligned} \L (z,h\lambda) ~=~ z^2 - \left(1 + \frac{3}{2}h\lambda\right)z + \frac{1}{2}h\lambda ~=~ 0 \end{aligned} \]

and its roots are

\[\begin{split} \begin{aligned} z_1 ~&=~ \frac{1}{2} \left( 1 + \frac{3}{2}h\lambda + \sqrt{1 + h\lambda + \frac{9}{4}\left(h\lambda\right)^2} \right) \\[1ex] z_2 ~&=~ \frac{1}{2} \left( 1 + \frac{3}{2}h\lambda - \sqrt{1 + h\lambda + \frac{9}{4}\left(h\lambda\right)^2} \right) \end{aligned} \end{split}\]

The following table gives the calculated values of \(|z_1|\) and \(|z_2|\) for various \(h\lambda\).

\(h\lambda\)

-1.1

-1.0

-0.67

-0.50

0.0

0.50

1.0

|\(z_1\)|

0.48

0.50

0.58

0.39

1.00

0.16

0.22

|\(z_2\)|

1.13

1.00

0.58

0.64

0.00

1.59

2.28

By definition the interval of absolute stability is the set of \(h\lambda\) for which

\[ \begin{aligned} \left| z_1(h\lambda) \right| < 1, \qquad \left| z_2(h\lambda) \right| < 1\,. \end{aligned} \]

For \(\lambda\) real, the acceptable values of \(\,h\lambda\,\) are \(-1 \leq h\lambda \leq 0\,\); i.e. the method is stable inside the interval \(\,[-1,0]\,\).

This result implies that if the 2-step AB method is used to solve the differential equation \(\,y' = -100y\), with \(\lambda = -100\), and \(\,h\lambda > -1\), then \(\,h < \dfrac{1}{100}\) for the method to be absolutely stable.

Example 7.3

Find the interval of absolute stability for the implicit, two-step, Adams method

\[ \begin{aligned} y_{j+1} ~=~ y_j + \frac{h}{12}(5f_{j+1} + 8f_j - f_{j-1}) \end{aligned} \]

when applied to \(\,y' = \lambda y\).

Solution

The characteristic equation is

\[ \begin{aligned} \L (z,h\lambda) ~=~ \left(1 - \frac{5}{12}h\lambda\right)z^2 - \left(1 + \frac{8}{12}h\lambda\right)z + \frac{1}{12}h\lambda ~=~ 0 \end{aligned} \]

Let \(a=\dfrac{h\lambda}{12} \) and its roots are

\[ \begin{aligned} z_{1,2} = \frac{1 + 8a \pm\sqrt{\left(1 + 8a\right)^2 - 4a\left(1 - 5a\right)}}{2(1 - 5a)} \end{aligned} \]

The following table gives the calculated values of \(|z_1|\) and \(|z_2|\) for various \(h\lambda\). Note that the interval \(-6 \leq h\lambda \leq 0\) is an interval of absolute stability for the method which becomes partially unstable when \(h > \dfrac{6}{|\lambda|}\).

\(h\lambda\)

-7.0

-6.0

-5.0

-4.0

-3.0

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

|\(z_1\)|

0.14

0.14

0.15

0.16

0.18

0.22

0.28

0.15

0.06

1.0

0.03

0.05

|\(z_2\)|

1.07

1.00

0.91

0.78

0.62

0.41

0.28

0.39

0.61

0.0

1.65

2.81

Note

Note that the characteristic polynomials (particularly those with higher degree) have very often complex roots and therefore it is appropriate to use a mathematics package (e.g. Matlab) for the computation of the roots. To illustrate this, a Matlab program is attached that shows a simple way of finding the interval of absolute stability using a software package.

The interval of absolute stability for various order Adams methods is shown in the following Table.

Order

Admas-Bashforth

Adams-Moulton

1

\([ -2.00, 0 ]\)

\([-\infty, 0 ]\cup [2.00, \infty]\)

2

\([ -1.00, 0 ]\)

\([ -\infty, 0 ]\)

3

\([ -0.55, 0 ]\)

\([ -6.00, 0 ]\)

4

\([ -0.30, 0 ]\)

\([ -3.00, 0 ]\)

5

\([ -0.16, 0 ]\)

\([ -1.84, 0 ]\)

6

\([ -0.09, 0 ]\)

\([ -1.19, 0 ]\)

Note

For all above methods as order increases, the interval of absolute stability decreases.