Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 341 / 1

Along an exact autonomous-ODE solution, $f(y)=y^{\prime }$ andMove every term to the left and substitute the Taylor expansions about $t_n$. The coefficients of $h^j{y}^{(j)}(t_n)$ vanish for $0\le j\le 4$, while the first nonzero coefficient isThus the local defect is $O(h^5)$. At $h=0$ the first characteristic polynomial iswhich satisfies the root condition for a multistep method. The method is therefore zero-stable and has

Apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and set $z=h\lambda$. Its amplification roots satisfyAt $z=0$ the roots are $1$ and $5/11$. A root can leave the unit disk only through $\xi =e^{i\phi }$. Substitution gives the boundary equationDirect separation into real and imaginary parts shows that both branches satisfy $\mathrm{Re}z\ge 0$; equality occurs on the branch through $z=0$. Hence no root crosses the unit circle in $\mathrm{Re}z<0$. Moreover both roots tend to zero as $z\to \infty$ in the left half-plane. Consequently the method is A-stable, and it also strongly damps the infinitely stiff limit.