Question 5.57: Use the Numerov method with h = 0.2, to determine y(0.6), wh......
Use the Numerov method with h = 0.2, to determine y(0.6), where y(x) denotes the solution of the initial value problem
y″ + xy = 0, y(0) = 1, y′(0) = 0.
The Numerov method is given by
\begin{aligned} y_{n+1}-2 y_{n}+y_{n-1} & =\frac{h^{2}}{12}\left(y_{n+1}^{\prime \prime}+10 y_{n}^{\prime \prime}+y_{n-1}^{\prime \prime}\right), n \geq 1 .\\ & =-\frac{h^{2}}{12}\left[x_{n+1} y_{n+1}+10 x_{n} y_{n}+x_{n-1} y_{n-1}\right] \end{aligned}
Solving for y_{n+1}, we get
\left[1+\frac{h^{2}}{12} x_{n+1}\right] y_{n+1}=2 y_{n}-y_{n-1}-\frac{h^{2}}{12}\left[10 x_{n} y_{n}+x_{n-1} y_{n-1}\right].
Here, we require the values y_{0} and y_{1} to start the computation. The Numerov method has order four and we use a fourth order single step method to determine the value y_{1}. The Taylor series method gives
y(h)=y(0)+h y^{\prime}(0)+\frac{h^{2}}{2} y^{\prime \prime}(0)+\frac{h^{3}}{6} y^{\prime \prime \prime}(0)+\frac{h^{4}}{24} y^{(4)}(0) .
We have y(0)=1, y^{\prime}(0)=0, y^{\prime \prime}(0)=0, y^{\prime \prime \prime}(0)=-1, y^{(4)}(0)=0,
y^{(5)}(0)=0, y^{(6)}(0)=4.
Hence, we obtain
y(h)=1-\frac{h^{3}}{6}+\frac{h^{6}}{180}+\ldots
For h=0.2, we get
y(0.2) \approx y_{1}=1-\frac{(0.2)^{3}}{6}+\ldots \approx 0.9986667.
We have the following results, using the Numerov method.
n=1: \quad\left[1+\frac{h^{2}}{12}(0.4)\right] y_{2}=2 y_{1}-y_{0}-\frac{h^{2}}{12}\left[10(0.2) y_{1}+0\right]
or \quad y_{2}=\frac{1}{1.0013333}\left[2(0.9986667)-1-\frac{0.04}{12}\{2(0.9986667)\}\right]=0.9893565.
n=2:\left[1+\frac{h^{2}}{12}(0.6)\right] y_{3}=2 y_{2}-y_{1}-\frac{h^{2}}{12}\left[10(0.4) y_{2}+0.2 y_{1}\right]
=0.9642606.