Use the non-self-starting Heun method to perform the same computations as were performed previously in Example 22.2 using Heun’s method. That is, integrate y′ = 4e^0.8t − 0.5y from t = 0 to 4 with a step size of 1. As with Example 22.2, the initial condition at t = 0 is y = 2. However, because we

Question

Use the non-self-starting Heun method to perform the same computations as were performed previously in Example 22.2 using Heun’s method. That is, integrate [latex]y′ = 4e^{0.8t} − 0.5y [/latex] from t = 0 to 4 with a step size of 1. As with Example 22.2, the initial condition at t = 0 is y = 2. However, because we are now dealing with a multistep method, we require the additional information that y is equal to –0.3929953 at t = –1.

Accepted Answer

The predictor [Eq. (23.7)] is used to extrapolate linearly from t = –1 to 1:
Predictor (Fig. 23.7a): [latex]y_{i+1}^{0}=y_{i-1}^m+f\left(t_i, y_i^m ight) 2 h[/latex] (23.7)

[latex]y_1^0=-0.3929953+\left[4e^{0.8(0)}-0.5(2) ight] 2=5.607005[/latex]
The corrector [Eq. (23.8)] is then used to compute the value:

Corrector (Fig. 23.7b): [latex] \begin{array}{r} y_{i+1}^j=y_i^m+\frac{f\left(t_i, y_i^m ight)+f\left(t_{i+1}, y_{i+1}^{j-1} ight)}{2} h \quad \quad (23.8) \ \quad \quad \quad \quad ( ext { for } j=1,2, \ldots, m) \end{array}[/latex]

[latex]y_1^1=2+\frac{4 e^{0.8(0)}-0.5(2)+4 e^{0.8(1)}-0.5(5.607005)}{2} 1=6.549331[/latex]

which represents a true percent relative error of –5.73% (true value = 6.194631). This error is somewhat smaller than the value of –8.18% incurred in the self-starting Heun.
Now, Eq. (23.8) can be applied iteratively to improve the solution:
[latex]y_1^2=2+\frac{3+4 e^{0.8(1)}-0.5(6.549331)}{2}1=6.313749[/latex]
which represents an error of –1.92%. An approximate estimate of the error can be determined using Eq. (23.9):

[latex]\left|\varepsilon_a ight|=\left|\frac{y_{i+1}^j-y_{i+1}^{j-1}} {y_{i+1}^j} ight| imes 100 \% [/latex] (23.9)

[latex]\left|\varepsilon_a ight|=\left|\frac{6.313749-6.549331}{6.313749} ight| imes 100 \%=3.7 \%[/latex]

Equation (23.8) can be applied iteratively until εa falls below a prespecified value of [latex]ε_s[/latex]. As was the case with the Heun method (recall Example 22.2), the iterations converge on a value of 6.36087 ([latex]ε_t[/latex] = –2.68%). However, because the initial predictor value is more accurate, the multistep method converges at a somewhat faster rate.
For the second step, the predictor is

[latex]y_2^{0}=2+\left[4 e^{0.8(1)}-0.5(6.36087) ight] 2=13.44346 \quad \varepsilon_t=9.43 \%[/latex]

which is superior to the prediction of 12.0826 ([latex]ε_t[/latex] = 18%) that was computed with the original Heun method. The first corrector yields 15.76693 ([latex]ε_t[/latex] = 6.8%), and subsequent iterations converge on the same result as was obtained with the self-starting Heun method: 15.30224 ([latex]ε_t[/latex] = –3.09%). As with the previous step, the rate of convergence of the corrector is somewhat improved because of the better initial prediction.

Question 23.3: Use the non-self-starting Heun method to perform the same co...

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