Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 38A (Numerical Analysis)

Use the stage solution continuous from $k_2=f(y)$ as $h\to 0$. Its displacement is $\delta =h[(1-a)f+a{k}_2]=hf+a{h}^2{f}^{\prime }f+O(h^3)$. Taylor expansion of $f(y+\delta )$ givesConsequently the Runge-Kutta method advances byThe exact solution has third-order term $h^3(f^{\prime 2}f+f^{\prime \prime }{f}^2)/6$. The first two orders agree for every real $a$, but the $f^{\prime \prime }{f}^2$ coefficient is always $1/4$ instead of $1/6$. Therefore the general nonlinear method has $\boxed{\text{order }2}$ for every fixed real $a$. The special value $a=1/3$ matches the linear test equation at third order but does not raise the nonlinear order. This is the two-stage Runge-Kutta family with an implicit second stage.

For the linear test equation $f(y)=\lambda y$ put $z=h\lambda$. The second stage solves $(1-az)k_2=\lambda y[1+(1-a)z]$, and substitution into the update gives the stability functionA-stability requires no poles and $|R(z)|\le 1$ throughout the closed left half-plane. If $a\ne 1/2$, the quadratic numerator gives unbounded magnitude as $z\to -\infty$; when $a=0$ the denominator is constant and the same conclusion holds. Thus $a=1/2$ is necessary. At this value $R(z)=(1+z/2)/(1-z/2)$, whose only pole is at positive $z=2$. Moreoveron the left half-plane. Hence the method is A-stable exactly for $\boxed{a=1/2}$. Its limit at negative infinity is $-1$, so it is not L-stable.