For $s=1$, imposing order two gives
$$
\rho_0=-1,
\qquad \sigma_0=\sigma_1=\frac12.
$$
The method is therefore the <trapezoidal rule>
$$
y_{n+1}-y_n=\frac h2(f_{n+1}+f_n).
$$
Its first <characteristic polynomials of a linear multistep method>[characteristic polynomial] is $\rho(\zeta)=\zeta-1$, so it satisfies the <root condition for a multistep method>.

For $s=2$, the four conditions through order three give
$$
\rho_0=-\frac15,
\qquad \rho_1=-\frac45,
\qquad \sigma_1=\frac45,
\qquad \sigma_2=\frac25,
$$
and hence
$$
y_{n+2}-\frac45y_{n+1}-\frac15y_n
=h\left(\frac25f_{n+2}+\frac45f_{n+1}\right).
$$
Here
$$
\rho(\zeta)=\zeta^2-\frac45\zeta-\frac15
=(\zeta-1)\left(\zeta+\frac15\right),
$$
so the <root condition for a multistep method> again holds. Both methods are consistent and zero-stable, and the <Dahlquist equivalence theorem> therefore proves that both are convergent.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-341/section-a/1/b/solution