Paper 1, Section II, C

Numerical Analysis | Part IB, 2019

(a) An ss-step method for solving the ordinary differential equation

dydt=f(t,y)\frac{d \mathbf{y}}{d t}=\mathbf{f}(t, \mathbf{y})

is given by

l=0sρlyn+l=hl=0sσlf(tn+l,yn+l),n=0,1,\sum_{l=0}^{s} \rho_{l} \mathbf{y}_{n+l}=h \sum_{l=0}^{s} \sigma_{l} \mathbf{f}\left(t_{n+l}, \mathbf{y}_{n+l}\right), \quad n=0,1, \ldots

where ρl\rho_{l} and σl(l=0,1,,s)\sigma_{l}(l=0,1, \ldots, s) are constant coefficients, with ρs=1\rho_{s}=1, and hh is the time-step. Prove that the method is of order p1p \geqslant 1 if and only if

ρ(ez)zσ(ez)=O(zp+1)\rho\left(e^{z}\right)-z \sigma\left(e^{z}\right)=O\left(z^{p+1}\right)

as z0z \rightarrow 0, where

ρ(w)=l=0sρlwl,σ(w)=l=0sσlwl\rho(w)=\sum_{l=0}^{s} \rho_{l} w^{l}, \quad \sigma(w)=\sum_{l=0}^{s} \sigma_{l} w^{l}

(b) Show that the Adams-Moulton method

yn+2=yn+1+h12(5f(tn+2,yn+2)+8f(tn+1,yn+1)f(tn,yn))\mathbf{y}_{n+2}=\mathbf{y}_{n+1}+\frac{h}{12}\left(5 \mathbf{f}\left(t_{n+2}, \mathbf{y}_{n+2}\right)+8 \mathbf{f}\left(t_{n+1}, \mathbf{y}_{n+1}\right)-\mathbf{f}\left(t_{n}, \mathbf{y}_{n}\right)\right)

is of third order and convergent.

[You may assume the Dahlquist equivalence theorem if you state it clearly.]

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