Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 17B (Numerical Analysis)

Apply a one-step numerical method to the scalar test ordinary differential equationIf one step has the formthen $R$ is the stability function and the linear stability domain isThe method is A-stable whenso it does not amplify any mode that the exact differential equation does not amplify.

For the test equation $f(y)=\lambda y$, put $z=h\lambda$. The stage equations of this implicit Runge-Kutta method areSubstitution into the update gives the stability functionFor $z=x+iy$, direct expansion givesThis is nonnegative whenever $x\le 0$, and hence $|R(z)|\le 1$ throughout the closed left half-plane, provided the denominator has no zero there.

If $a=0$, the denominator vanishes only at $z=2$. If $a\ne 0$, its zeros solveReal roots are positive, while a complex-conjugate pair has positive real part because the sum of the roots is $1/(2a^2)>0$. Thus there are no poles in the closed left half-plane for any real $a$. Thereforeis the complete set of A-stable parameters, as summarized by the A-stability of a symmetric two-stage implicit Runge-Kutta family.