Question 5.31: Consider an implicit two-step method yn + 1 – (1 + a) yn + a......
Consider an implicit two-step method
y_{n+1}-(1+a) y_{n}+a y_{n-1}=\frac{h}{12}\left[(5+a) y_{n+1}^{\prime}+8(1-a) y_{n}^{\prime}-(1+5 a) y_{n-1}^{\prime}\right]
where -1 \leq a<1, for the solution of the initial value problem y^{\prime}=f(x, y), y\left(x_{0}\right)=y_{0}.
(i) Show that the order of the two-step method is 3 if a \neq-1 and is 4 if a=-1.
(ii) Prove that the interval of absolute stability is (-6(a+1) /(a-1), 0) and that the interval of relative stability is (3(a+1) /(2(a-1)), \infty).
(i) The truncation error of the two-step method is given by
\begin{aligned} T_{n+1}= & y\left(x_{n+1}\right)-(1+a) y\left(x_{n}\right)+a y\left(x_{n-1}\right) \\ & -\frac{h}{12}\left[(5+a) y^{\prime}\left(x_{n+1}\right)+8(1-a) y^{\prime}\left(x_{n}\right)-(1+5 a) y^{\prime}\left(x_{n-1}\right)\right] \\ = & C_{0} y\left(x_{n}\right)+C_{1} h y^{\prime}\left(x_{n}\right)+C_{2} h^{2} y^{\prime \prime}\left(x_{n}\right)+C_{3} h^{3} y^{\prime \prime \prime}\left(x_{n}\right)+C_{4} h^{4} y^{(4)}\left(x_{n}\right)+\ldots \ldots . \end{aligned}
where C_{0}=0, C_{1}=0, C_{2}=0, C_{3}=0, \quad C_{4}=-(1+a) / 24.
Hence, the truncation error is
T_{n+1}=-\frac{1}{24}(1+a) h^{4} y^{(4)}\left(x_{n}\right)+\left(\frac{a-1}{90}\right) h^{5} y^{(5)}\left(x_{n}\right)+O\left(h^{6}\right) .
Therefore, the two-step method has order 3 if a \neq-1 and order 4 if a=-1.
(ii) The characteristic equation of the method is given by
\left(1-\frac{\bar{h}}{12}(5+a)\right) \xi^{2}-\left((1+a)+\frac{2}{3} \bar{h}(1-a)\right) \xi+\left(a+\frac{\bar{h}}{12}(1+5 a)\right)=0 .
Absolute Stability : Setting \xi=(1+z) /(1-z), the transformed characteristic equation is obtained as
\left(2(1+a)+\frac{\bar{h}}{3}(1-a)\right) z^{2}+2\left((1-a)-\frac{\bar{h}}{2}(1+a)\right) z-\bar{h}(1-a)=0.
The Routh-Hurwitz criterion is satisfied if
\begin{aligned} 2(1+a) & +\frac{\bar{h}}{3}(1-a)>0, \\ (1-a)- & \frac{\bar{h}}{2}(1+a)>0, \\ -\bar{h}(1-a) & >0. \end{aligned}
For \bar{h}<0 and -1 \leq a<1, the conditions will be satisfied if \bar{h} \in(-6(1+a) /(1-a), 0).
Hence, the interval of absolute stability is \bar{h} \in(-6(1+a) /(1-a), 0).
Relative Stability : It is easily verified that the roots of the characteristic equation are real and distinct for all \bar{h} and for all a. The end points of the interval of relative stability are given by \xi_{1 h}=\xi_{2 h} and \xi_{1 h}=-\xi_{2 h}. The first condition is never satisfied that is, the interval extends to +\infty, ; the second condition gives
(1+a)+\frac{2}{3} \bar{h}(1-a)=0 \quad \text { or } \quad \bar{h}=\frac{3(a+1)}{2(a-1)} .
Hence, the interval of relative stability is
\bar{h} \in\left(\frac{3(a+1)}{2(a-1)}, \infty\right).