Question 11.2.1: Apply the Shooting method with Newton’s Method to the bounda...
Apply the Shooting method with Newton’s Method to the boundary-value problem
y^{\prime \prime}=\frac{1}{8}\left(32+2 x^{3}-y y^{\prime}\right), \quad \text { for } 1 \leq x \leq 3, \text { with } y(1)=17 \text { and } y(3)=\frac{43}{3}
Use N = 20, M = 10, and T O L=10^{-5} , and compare the results with the exact solution y(x)=x^{2}+16 / x .
We need approximate solutions to the initial-value problems
y^{\prime \prime}=\frac{1}{8}\left(32+2 x^{3}-y y^{\prime}\right), \quad \text { for } 1 \leq x \leq 3, \text { with } y(1)=17 \text { and } y^{\prime}(1)=t_{k}
and
z^{\prime \prime}=\frac{\partial f}{\partial y} z+\frac{\partial f}{\partial y^{\prime}} z^{\prime}=-\frac{1}{8}\left(y^{\prime} z+y z^{\prime}\right), \quad \text { for } 1 \leq x \leq 3 \text {, with } z(1)=0 \text { and } z^{\prime}(1)=1,
at each step in the iteration. If the stopping technique in Algorithm 11.2 requires
\left|w_{1, N}\left(t_{k}\right)-y(3)\right| \leq 10^{-5} ,
then we need four iterations and t_{4}=-14.000203 . The results obtained for this value of t are shown in Table 11.2.
Table 11.2
\begin{array}{lccl}\hline x_{i} & w_{1, i} & y\left(x_{i}\right) & \left|w_{1, i}-y\left(x_{i}\right)\right| \\\hline 1.0 & 17.000000 & 17.000000 & \\1.1 & 15.755495 & 15.755455 & 4.06 \times 10^{-5} \\1.2 & 14.773389 & 14.773333 & 5.60 \times 10^{-5} \\1.3 & 13.997752 & 13.997692 & 5.94 \times 10^{-5} \\1.4 & 13.388629 & 13.388571 & 5.71 \times 10^{-5} \\1.5 & 12.916719 & 12.916667 & 5.23 \times 10^{-5} \\1.6 & 12.560046 & 12.560000 & 4.64 \times 10^{-5} \\1.7 & 12.301805 & 12.301765 & 4.02 \times 10^{-5} \\1.8 & 12.128923 & 12.128889 & 3.14 \times 10^{-5} \\1.9 & 12.031081 & 12.031053 & 2.84 \times 10^{-5} \\2.0 & 12.000023 & 12.000000 & 2.32 \times 10^{-5} \\2.1 & 12.029066 & 12.029048 & 1.84 \times 10^{-5} \\2.2 & 12.112741 & 12.112727 & 1.40 \times 10^{-5} \\2.3 & 12.246532 & 12.246522 & 1.01 \times 10^{-5} \\2.4 & 12.426673 & 12.426667 & 6.68 \times 10^{-6} \\2.5 & 12.650004 & 12.650000 & 3.61 \times 10^{-6} \\2.6 & 12.913847 & 12.913845 & 9.17 \times 10^{-7} \\2.7 & 13.215924 & 13.215926 & 1.43 \times 10^{-6} \\2.8 & 13.554282 & 13.554286 & 3.46 \times 10^{-6} \\2.9 & 13.927236 & 13.927241 & 5.21 \times 10^{-6} \\3.0 & 14.333327 & 14.333333 & 6.69 \times 10^{-6} \\\hline\end{array}