6.4. Chapter 6 Exercise

You should try the following exercise questions first, then check with the answers.

• Handwritten solutions for Exercises 6.1 and 6.2

Exercise 6.1

Find the first and second characteristic polynomials and investigate the stability of the following multistep methods.

1. \(y_{j+1} ~=~ y_j + \frac{h}{2}(3f_j - f_{j-1})\)

2. \(y_{j+1} + 9y_j - 9y_{j-1} - y_{j-2} ~=~ 6h\,(f_j + f_{j-1})\)

3. \(y_{j+1} ~=~ 4y_j - 3y_{j-1} - 2hf_{j-1}\)

4. \(y_{j+1} ~=~ 2y_{j-1} - y_j + \frac{h}{2}(5f_j + f_{j-1})\)

Solution to

1. Roots are 0 and 1

2. Roots are 1, -0.1, -9.9

3. Roots are 1 and 3

4. Roots are 1 and -2

Exercise 6.2

Consider the multistep methods in Exercise 6.1, determine their order of accuracy and error constants using the techniques covered in Order and Error Constants.

Solution to

1. \(2^\text{nd}\) order, \(\frac{5}{12}\),

2. \(4^\text{th}\) order, \(\frac{1}{10}\)

3. \(2^\text{nd}\) order, \(\frac{2}{3}\),

4. \(2^\text{nd}\) order, \(\frac{1}{4}\)

Exercise 6.3

Investigate the stability of the following explicit 2-step method (i.e. mid-point rule):

\[\begin{aligned} y_{j+1} ~=~ y_{j-1}+2hf_{j} \end{aligned}\]

Use this method to approximate the solution to the following initial value problem:

\[\begin{aligned} && y'(x) \,&=\, \lambda y & y(0)\,&=\,1 & 0 &\leq \,x\, \leq 5 && \end{aligned}\]

using \(\lambda = -2\), with \(h = 0.02\) and \(h = 0.05\). Use the analytic solution \(y_{ex} = e^{-2x}\) to find the starting values, and plot the solution curves. Next, solve the same problem with the same parameters using the two-step AB method. By considering the stability of the methods and the errors in the calculations, compare the two solution methods and comment on your results.

Solution to

Re-arranging the equation gives

\[ - y_{j-1} + 0 y_j + y_{j+1} = h (0 f_{j-1} + 2 f_j + 0 f_{j+1}), \]

so the coefficients are

\[ \alpha_0 = -1, \quad \alpha_1=0, \quad \alpha_2=1, \quad \beta_0=0, \quad \beta_1 = 2, \quad \beta_2=0, \]

then the first characteristic polynomial is

\[ \rho(z) = \alpha_0 z^0 + \alpha_1 z^1 + \alpha^2 z^2 = -1 + z^2, \]

with two distinct real roots

\[ z_1 = 1, \quad z_2 = -1. \]

Thus the method satisfies the root condition, so it is zero-stable.

• Matlab solution for Exercise 6.3

Exercise 6.4

Assess the zero-stability of the multi-step method given in Example 6.1

\[ y_n = -4 y_{n-1} + 5 y_{n-2} + 4 h f_{n-1} + 2h f_{n-2}. \]

Solution to

Re-arranging the equation gives

\[ -5 y_{n-2} + 4 y_{n-1} + y_n = h \left( 2 f_{n-2} + 4 f_{n-1} + 0 f_n \right), \]

so the first characteristic polynomial is

\[ \rho(z) = -5 + 4z + z^2, \]

and it has two roots

\[ z_1 = 1, \quad z_2 = -5. \]

The method doesn’t satisfy the root condition, so it is not zero-stable.