 # Let x′(t)=f(t,x(t)),t in (0,T) with x(0)=x_0 f satifies the Lipschitz-condition f(t,x)−f(t,y)<=L|x−y| h in (0,1/L) is the step size and the approximation x_k for x(t_k)=hk is given by x_k=x_(k−1)+hf(t_k,x_k). Now I would be very interested how to derive the error |x_k−x(t_k)|<=(1)/(1−Lh)(|x_(k−1)−x(t_(k−1))|+(h^2)/(2)max_(s in [0,T])|x′′(s)|) I tried to look up it up in some numerical analysis books but it is always different Julia Chang 2022-09-16 Answered
Let
${x}^{\prime }\left(t\right)=f\left(t,x\left(t\right)\right),t\in \left(0,T\right)$ with $x\left(0\right)={x}_{0}$
$f$ satifies the Lipschitz-condition $f\left(t,x\right)-f\left(t,y\right)\le L|x-y|$
$h\in \left(0,\frac{1}{L}\right)$ is the step size and the approximation ${x}_{k}$ for $x\left({t}_{k}\right)=hk$ is given by ${x}_{k}={x}_{k-1}+hf\left({t}_{k},{x}_{k}\right)$.
Now I would be very interested how to derive the error
$|{x}_{k}-x\left({t}_{k}\right)|\le \frac{1}{1-Lh}\left(|{x}_{k-1}-x\left({t}_{k-1}\right)|+\frac{{h}^{2}}{2}\underset{s\in \left[0,T\right]}{max}|{x}^{″}\left(s\right)|\right)$
I tried to look up it up in some numerical analysis books but it is always different
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First, we get local truncation error.
$x\left({t}_{k+1}\right)=x\left({t}_{k}\right)+hf\left({t}_{k},x\left({t}_{k}\right)\right)+{\tau }_{k}$
${\tau }_{k}=x\left({t}_{k+1}\right)-x\left({t}_{k}\right)-hf\left({t}_{k},x\left({t}_{k}\right)\right)=\frac{{h}^{2}}{2}{x}^{″}\left(\eta \right)$. Where $\eta \in \left({t}_{k},{t}_{k+1}\right)$.
Then we get the bound,
$|x\left({t}_{k+1}\right)-{x}_{k+1}|\le \left(1+hL\right)|x\left({t}_{k}\right)-{x}_{k}|+|{\tau }_{k}|$
$\le \left(1+hL\right)|x\left({t}_{k}\right)-{x}_{k}|+\frac{{h}^{2}}{2}\underset{s\in \left(0,T\right)}{max}|{x}^{″}\left(s\right)|$
$\le \frac{1}{1-hL}|x\left({t}_{k}\right)-{x}_{k}|+\frac{{h}^{2}}{2}\underset{s\in \left(0,T\right)}{max}|{x}^{″}\left(s\right)|$
Where the last step is from geometric series. Maybe it would be helpful if you listed the other results you are talking about, and then we can show that they're equivalent.