Question 5.71: Consider the homogeneous boundary value problem y″ + Λ y = 0......
Consider the homogeneous boundary value problem
\begin{gathered} y^{\prime \prime}+ \Lambda y=0, \\ y(0)=y(1)=0. \end{gathered}
(a) Show that the application of the fourth order Numerov method leads to the system
\left[\mathrm{J}-\frac{\lambda}{1+\lambda / 12} \mathrm{I}\right] \mathrm{y}=\mathrm{0}
where \lambda=h^{2} \Lambda.
(b) Show that the approximation to the eigenvalues by the second and fourth order methods are given by 2(1-\cos n \pi h) / h^{2} and 12(1-\cos n \pi h) /\left[h^{2}(5+\cos n \pi h)\right], 1 \leq n \leq N -1 , respectively, where h=1 / N.
(c) Noticing that \Lambda_{n}=n^{2} \Pi^{2}, show that the relative errors
\frac{\Lambda_{n}-h^{-2} \lambda_{n}}{ \Lambda_{n}}
for the second and the fourth order methods are given by \Lambda_{n} h^{2} / 12 and \Lambda_{n}^{2} h^{4} / 240, respectively, when terms of higher order in h are neglected.
(a) The application of the Numeröv method leads to the system
\begin{aligned} & y_{n+1}-2 y_{n}+y_{n-1}+\frac{\lambda}{12}\left(y_{n+1}+10 y_{n}+y_{n-1}\right)=0, \\ & n=1,2, \ldots N-1, \end{aligned}
or -y_{n+1}+2 y_{n}-y_{n-1}-\frac{\lambda}{1+\lambda / 12} y_{n}=0 ,
n=1,2, \ldots, N-1,
where \lambda=h^2 \Lambda .
Incorporating the boundary conditions we obtain
\left(\mathrm{J}-\frac{\lambda}{1+\lambda / 12} \mathrm{I}\right) \mathrm{y}=\mathrm{0}.
(b) For the second order method, we have
\begin{array}{r} y_{n+1}-2\left[1-(\lambda / 2] y_{n}+y_{n-1}=0\right., \\ y_{0}=y_{N}=0. \end{array}
The characteristic equation of the difference equation is given by
\xi^{2}-2[1-(\lambda / 2)] \xi+1=0.
Substitute \cos \theta=1-(\lambda / 2). Then the roots of the characteristic equation are given by \xi=\cos \theta \pm i \sin \theta=e^{ \pm i \theta}. The solution of the difference scheme becomes
y_{n}=C_{1} \cos n \theta+C_{2} \sin n \theta .
Boundary conditions y(0)=0, y(1)=0 lead to C_{1}=0, C_{2} \sin (N \theta)=0, or \theta=n \pi / N. Since h=1 / N, we have
1-\frac{1}{2} \lambda_n=\cos \theta=\cos (n \pi h) \text {, or } \quad \lambda_n=2[1-\cos (n \pi h)] \text {, }
or \Lambda_n=\frac{2}{h^2}[1-\cos (n \pi h)] .
Similarly, for the Numeröv method, we substitute
\frac{1-5 \lambda / 12}{1+\lambda / 12}=\cos \theta
and find that \theta=n \pi / N=n \pi h.
The eigenvalue is given by
\left\{\left[1-\frac{5}{12} \lambda_{n}\right] /\left[1+\frac{1}{12} \lambda_{n}\right]\right\}=\cos (n \pi h), \text { or } \quad \lambda_{n}=\frac{12[1-\cos (n \pi h)]}{5+\cos (n \pi h)}
or \Lambda_{n}=\frac{12}{h^{2}}\left[\frac{1-\cos (n \pi h)}{5+\cos (n \pi h)}\right] .
(c) The analytical solution of the eigenvalue problem gives \Lambda_{n}=n^{2} \pi^{2}.
For the second order method, we obtain
\begin{aligned} \frac{1}{h^{2}} \lambda_{n} & =\frac{2}{h^{2}}[1-\cos (n \pi h)] \\ & =n^{2} \pi^{2}\left[1-\frac{1}{12} n^{2} \pi^{2} h^{2}+O\left(h^{4}\right)\right], \end{aligned}
Thus, the relative error in the eigenvalue is given by
\frac{\Lambda_{n}-h^{-2} \lambda_{n}}{\Lambda_{n}}=\frac{\Lambda_{n}}{12} h^{2}+O\left(h^{4}\right).
We have
\begin{aligned} & {[5+\cos (n \pi h)]^{-1} }=\left[6-\frac{1}{2} n^{2} \pi^{2} h^{2}+\frac{1}{24} n^{4} \pi^{4} h^{4}+O\left(h^{6}\right)\right]^{-1} \\ &=\frac{1}{6}\left[1-\left\{\frac{1}{12} n^{2} \pi^{2} h^{2}-\frac{1}{144} n^{4} \pi^{4} h^{4}+O\left(h^{6}\right)\right\}\right]^{-1} \\ &=\frac{1}{6}\left[1+\frac{1}{12} n^{2} \pi^{2} h^{2}+O\left(h^{6}\right)\right] \\ & \frac{[1-\cos (n \pi h)]}{[5+\cos (n \pi h)]}=\left(\frac{1}{6}\right)\left(\frac{1}{2} n^{2} \pi^{2} h^{2}\right)\left[1-\frac{1}{12} n^{2} \pi^{2} h^{2}+\frac{1}{360} n^{4} \pi^{4} h^{4}+O\left(h^{6}\right)\right] \\ & \times[1+\left.\frac{1}{12} n^{2} \pi^{2} h^{2}+O\left(h^{6}\right)\right] \\ &=\frac{1}{12} n^{2} \pi^{2} h^{2}\left[1-\frac{1}{240} n^{4} \pi^{4} h^{4}+O\left(h^{6}\right)\right] \\ & \frac{1}{h^{2}} \lambda_{n}=\frac{12}{h^{2}} \frac{[1-\cos (n \pi h)]}{[5+\cos (n \pi h)]}=n^{2} \pi^{2}\left[1-\frac{1}{240} n^{4} \pi^{4} h^{4}+O\left(h^{6}\right)\right] \end{aligned}
Therefore, for the Numeröv method, the relative error is given by
\frac{\Lambda_{n}-h^{-2} \lambda_{n}}{\Lambda_{n}}=\frac{1}{240} n^{2} \pi^{2} h^{4}+O\left(h^{6}\right)=\frac{1}{240} \Lambda_{n}^{2} h^{4}+O\left(h^{6}\right).