Paper 1, Section II, B

Numerical Analysis | Part IB, 2021

For the ordinary differential equation

y=f(t,y),y(0)=y~0,t0\boldsymbol{y}^{\prime}=\boldsymbol{f}(t, \boldsymbol{y}), \quad \boldsymbol{y}(0)=\tilde{\boldsymbol{y}}_{0}, \quad t \geqslant 0

where y(t)RN\boldsymbol{y}(t) \in \mathbb{R}^{N} and the function f:R×RNRN\boldsymbol{f}: \mathbb{R} \times \mathbb{R}^{N} \rightarrow \mathbb{R}^{N} is analytic, consider an explicit one-step method described as the mapping

yn+1=yn+hφ(tn,yn,h)\boldsymbol{y}_{n+1}=\boldsymbol{y}_{n}+h \varphi\left(t_{n}, \boldsymbol{y}_{n}, h\right)

Here φ:R+×RN×R+RN,n=0,1,\varphi: \mathbb{R}_{+} \times \mathbb{R}^{N} \times \mathbb{R}_{+} \rightarrow \mathbb{R}^{N}, n=0,1, \ldots and tn=nht_{n}=n h with time step h>0h>0, producing numerical approximations yn\boldsymbol{y}_{n} to the exact solution y(tn)\boldsymbol{y}\left(t_{n}\right) of equation ()(*), with y0\boldsymbol{y}_{0} being the initial value of the numerical solution.

(i) Define the local error of a one-step method.

(ii) Let \|\cdot\| be a norm on RN\mathbb{R}^{N} and suppose that

φ(t,u,h)φ(t,v,h)Luv,\|\boldsymbol{\varphi}(t, \boldsymbol{u}, h)-\boldsymbol{\varphi}(t, \boldsymbol{v}, h)\| \leqslant L\|\boldsymbol{u}-\boldsymbol{v}\|,

for all h>0,tR,u,vRNh>0, t \in \mathbb{R}, \boldsymbol{u}, \boldsymbol{v} \in \mathbb{R}^{N}, where LL is some positive constant. Let t>0t^{*}>0 be given and e0=y0y(0)\boldsymbol{e}_{0}=\boldsymbol{y}_{0}-\boldsymbol{y}(0) denote the initial error (potentially non-zero). Show that if the local error of the one-step method ( \uparrow ) is O(hp+1)\mathcal{O}\left(h^{p+1}\right), then

maxn=0,,t/hyny(nh)etLe0+O(hp),h0\max _{n=0, \ldots,\left\lfloor t^{*} / h\right\rfloor}\left\|\boldsymbol{y}_{n}-\boldsymbol{y}(n h)\right\| \leqslant e^{t^{*} L}\left\|\boldsymbol{e}_{0}\right\|+\mathcal{O}\left(h^{p}\right), \quad h \rightarrow 0

(iii) Let N=1N=1 and consider equation ()(*) where ff is time-independent satisfying f(u)f(v)Kuv|f(u)-f(v)| \leqslant K|u-v| for all u,vRu, v \in \mathbb{R}, where KK is a positive constant. Consider the one-step method given by

yn+1=yn+14h(k1+3k2),k1=f(yn),k2=f(yn+23hk1).y_{n+1}=y_{n}+\frac{1}{4} h\left(k_{1}+3 k_{2}\right), \quad k_{1}=f\left(y_{n}\right), \quad k_{2}=f\left(y_{n}+\frac{2}{3} h k_{1}\right) .

Use part (ii) to show that for this method we have that equation (††) holds (with a potentially different constant LL ) for p=2p=2.

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