Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 8D (Numerical Analysis)

Apply the numerical method to the scalar test equation $y^{\prime }=\lambda y$, with $z=h\lambda$. Its linear stability domain consists of those $z\in \mathbb{C}$ for which the discrete solution stays bounded as the step count tends to infinity. For a one-step method with stability function $R$, this isIf asymptotic decay is required, use $|R(z)|<1$ instead; this only changes the boundary conventions in part (b). A method is A-stable if its stability domain contains the left half-plane, so every decaying linear test equation is stable for every positive step size. For a multistep method, the corresponding definition uses the root condition for a multistep method for every characteristic root, including simplicity of unit-modulus roots.

Expand the second stage by the Taylor theorem, writing $f=f(t_n,y_n)$ and $D_{y}f$ for the derivative in the state variable:The update therefore agrees with the exact solution's Taylor series:Thus the one-step local truncation error is $O(h^3)$, and the method has order at least two under the usual smoothness and Lipschitz continuity assumptions.

For $f=\lambda y$, the two-stage Runge-Kutta method givesOn the real axis, $R(z)=\frac{1}{2}(z+1{)}^2+\frac{1}{2}\ge \frac{1}{2}$, so $|R(z)|\le 1$ is equivalent to $(z+1{)}^2\le 1$. HenceThe decay convention gives $(-2,0)$. Since negative real values such as $z=-3$ lie outside the linear stability domain, the method is not A-stable. Its order is exactly two, since $R(z)$ lacks the cubic term of $e^z$.