Q:

Global Error Bound for Implicit Euler

I'm trying to derive an error bound for the approximate solution obtained with Backward Euler to the IVP 
$$\frac{dy}{dt} = f(y), \hspace{.75cm} y(0) = y_0,$$
where $f \in C^2(\mathbb{R})$ with global Lipschitz constant $K$, and $t \in [0,T]$. I know that the local truncation error at each time step $\tau_n$ is bounded by $\frac{M h^2 }{2} $, where $$M = \sup_{t \in [0,T]} \left\{ f(y(t))f'(y(t)) \right\}.$$ Letting $e_n = y(t_n) - y_n$, we have that \begin{align*}
\frac{Mh^2}{2} &\geq \left|{\tau_n} \right| \\
&= \left| y(t_n) - y(t_{n-1}) - hf(y(t_n)) \right| \\
&= \left| y(t_n) - y(t_{n-1}) - hf(y(t_n)) - (y_n - y_{n-1} - hf_n) \right| \\
&= \left| e_{n}-e_{n-1} - h(f(y(t_n))-f_n) \right| \\
& \geq \left| e_n \right| - \left| e_{n-1} \right| - hK \left| e_n \right|   ,
\end{align*}
which implies that $$\left| e_n \right| \leq \frac{\left| e_{n-1} \right| + \frac{Mh^2}{2}}{1- hK}   $$
Continuing recursively, we find that \begin{align*} \left| e_n \right| &\leq \frac{Mh^2}{2(1 - hK)^n} \sum_{j = 0}^{n-1} (1- hK)^j \\
&\leq \frac{Mh^2}{2(1 - hK)^n} \frac{1}{1-(1-hK)} \\
&= \frac{Mh}{2K(1-hK)^n} \\
&= \frac{Mh}{2K(1-hK)^{t_n/h}} \\
&\leq \frac{Mh}{2K(1-hK)^{T/h}}
\end{align*}
Using L'hopital's rule, we find that $$\lim_{h \to 0} (1-hk)^{-T/h} =e^{TK},$$
which implies that $$ (1-hk)^{-T/h} = e^{TK} + O(h). $$
Therefore, $$\left| e_n \right| \leq \frac{Mhe^{TK}}{2K} + O(h^2). $$
Is this bound reasonable? If so, is there a way to make this bound tighter? Thanks in advance!

A:

This looks as tight as possible in this generality. You could use the estimate
$$
(1-x)^{-1}=\exp(-\ln(1-x))=\exp(x+\tfrac12x^2+\tfrac13x^3+...)\le\exp(x+x^2)
$$
for $0\le x\le\frac12$ to get an explicit bound in the last formula instead of the general $O(h^2)$ class.

