Numerical analysis - Error analysis for the second order Taylor method

Question

I'm trying to prove the following result for the second order Taylor method:

1. $f$ is continuous and satisfies a Lipschitz condition with constant L on $D=\{(t,y)\mid a\leq t \leq b, -\infty<y<\infty\}.$
2. There exists a constant $M$ with $|y'''(t)|\leq M$ for $t∈[a,b],$ where $y(t)$ denotes the unique solution to the IVP $y'=f(t,y)$ , $a\leq t \leq b$ , $y(a)=\alpha.$
3. Let $w_0,w_1,...,w_N$ be the approximation generated by the second order Taylor method for some integer $N.$ Then, $|y(t_i)-w_i|\leq \frac{h^2M}{6L}[e^{L(t_i-a)}-1]$ for each $i=0,1,...,N$.

First, let $T_i=\frac{y_{i+1}-y_i-hy'_i-\frac{h^2}{2}y''_i}{h}$ and $T=max(|T_1|,...,|T_N|).$ Note that $T_i$ is the $i$th Local truncation error. Then, by the triangle inequality, \begin{align} |e_{i+1}|&=|y_{i+1}-w_{i+1}| \\ &=|\color{red}{T_ih+y_i+hy'_i}\color{blue}{+\frac{h^2}{2}y''_i}\color{red}{-w_i-hf(t_i,w_i)}\color{blue}{-\frac{h^2}{2}(f_t(t_i,w_i)+f(t_i,w_i)f_y(t_i,w_i))}| \\ &\leq \color{red}{hT+|e_i|+hL|e_i|}\color{blue}{+|\frac{h^2}{2}y''_i-\frac{h^2}{2}(f_t(t_i,w_i)+f(t_i,w_i)f_y(t_i,w_i))|}. \end{align} Here's where I'm stuck. For the first order method, the $\color{blue}{blue}$ part doesn't appear and only the $\color{red}{red}$ part appears. So I could obtain the error bound for the first order case using the simple geometric series argument and the fact that $1+x\leq e^x$ for all real $x$. But, for the second order case, I couldn't figure out how to control the $\color{blue}{blue}$ part. Any help would be appreciated!!

Answer 1

You need, as a first step, to use that also for the exact solution $$ y''(t)=f_t(t,y(t))+f_y(t,y(t))\,f(t,y(t)) \\ \implies y_i''=f_t(t_i,y_i)+f_y(t_i,y_i)\,f(t_i,y_i). $$ From this the formula for the Taylor method was derived, after all.

---

(2/28/2021) But one does not need to go into these details after all. Let $y_k(t)$, $k=0,...,n$ (different from the $y_i=y(t_i)$ in the question) be the exact ODE solution to the initial condition $y_k(t_k)=w_i$. Then by construction $$w_{k+1}=y_k(t_k)+hy_k'(t_k)+\frac{h^2}2y''_k(t_k).$$ Thus using the usual Taylor remainder term $$ w_{k+1}-y_k(t_k+h)=-\frac{h^3}6y_k'''(t_k+\theta_kh), ~~ \theta_k\in(0,1) $$ Using the Grönwall lemma one gets for $t\ge t_{k+1}$ $$\begin{align} |y_{k+1}(t)-y_k(t)|&\le e^{L(t-t_{k+1})}|y_{k+1}(t_{k+1})-y_k(t_{k+1})| \\ &=e^{L(t-t_{k+1})}|w_{k+1}-y_k(t_k+h)|\le e^{L(t-t_{k+1})}\frac{h^3}{6}M_3 \end{align}$$ Summing up at $t=t_n$ up to $k=n-1$ gives via triangle inequality and geometric sums \begin{align} |w_n-y_0(t_n)|&\le\sum_{k=0}^{n-1}|y_{k+1}(t)-y_k(t)| \\ &\le \sum_{k=0}^{n-1} e^{Lh(n-1-k)}\frac{h^3}{6}M_3 \\ &=\frac{e^{L(t_n-t_0)}-1}{e^{Lh}-1}\frac{h^3}{6}M_3 <\frac{e^{L(t_n-t_0)}-1}{L}\frac{h^2}{6}M_3 \end{align}