<s>[INST] The order of accuracy of the implicit Euler method is equal to $1$ I want to show that the order of accuracy of the implicit Euler method is equal to $1$.
That's what I have tried:
We have the initial value problem
$\left\{\begin{matrix}
y'(t)=f(t,y(t)) &, a \leq t \leq b \\ 
y(a)=y_0 & 
\end{matrix}\right.$
Using Taylor's expasion, we have $y(t^{n+1})=y(t^n)+hy'(t^n)+\frac{h^2}{2}y''(\xi_n)$.
From the implicit Euler method we have $y^{n+1}=y^n+hf(t^{n+1},y^{n+1})$.
Thus:
$$|y(t^{n+1})-y^{n+1}|=|y(t^n)-y^n+h(f(t^n,y(t^n))-f(t^{n+1},y^{n+1}))+ \frac{h^2}{2}y''(\xi_n)| \leq |y(t^n)-y^n|+h|y(t^n)-y^{n+1}|+ \frac{h^2}{2}|y''(\xi_n)|=|y(t^n)-y^n|+h|y^{n+1}-y(t^n)|+ \frac{h^2}{2}|y''(\xi_n)| \leq |y(t^n)-y^n|+h|y^{n}-y(t^n)|+ \frac{h^2}{2}|y''(\xi_n)|=(1+h)|y(t^n)-y^n|+ \frac{h^2}{2}|y''(\xi_n)| \\\leq (1+h)^{n+1} |y(t^0)-y(0)|+ \sum_{i=0}^n (1+h)^i \frac{h^2}{2}|y''(\xi_{n-i})|=\frac{h^2}{2} \sum_{i=0}^n (1+h)^i |y''(\xi_{n-i})| \overset{M:=\max_{i=0, \dots, n} |y''(\xi_i)|}{\leq } \frac{Mh^2}{2} \sum_{i=0}^n (1+h)^i= \frac{Mh^2}{2} \left( \frac{(1+h)^{n+1}-1}{h}\right)=\frac{Mh}{2}((1+h)^{n+1}-1) \leq \frac{Mh}{2}(e^{(n+1)h}-1) \overset{(\star)}{\leq } \frac{Mh}{2} (e^{b-a}-1)=CM \frac{h}{2}$$
$(\star): t^{n+1}=a+(n+1)h \\ t^{n+1} \leq b \\ \Rightarrow a+(n+1)h \leq b \Rightarrow (n+1)h \leq b-a$
Therefore, the order of accuracy is at least $1$.
We consider the initial value problem:
$\left\{\begin{matrix}
y'(t)=2t &, 0 \leq t \leq 1 \\ 
y(0)=0 & 
\end{matrix}\right.$
Its real solution is $y(t)=t^2, 0 \leq t \leq 1$.
$y^{n+1}=y^n+2ht^{n+1}=y^n+2h (n+1)h=y^n+ 2h^2(n+1) $
$y^1=2h^2=1 \cdot 2 \cdot h^2$
$y^2=2h^2+4h^2=6h^2= 3 \cdot 2 \cdot h^2$
$y^3=6h^2+6h^2=12h^2=3 \cdot 4 \cdot h^2$
$y^4= 12h^2+8h^2=20h^2=4 \cdot 5 \cdot h^2$
We see that $y^n=n(n+1)h^2$
For $n=N$ we have $y^N=N(N+1)h^2=(N+1)h=1+h$.
$\epsilon^N=|y(t^N)-y^N|=|y(1)-(1+h)|=|1-1-h|=h$.
So the order of accuracy is exactly $1$.
Could you tell me if it is right?
 [/INST] Your first line should be, using $y(t^n)=y(t^{n+1})-h·f(t^{n+1}, y(t^{n+1}))+\frac{h^2}{2}·y''(\xi_n)$ as the appropriate Taylor expansion,
$$
|y(t^{n+1})-y^{n+1}|
=\left|y(t^n)-y^n+h·\bigl(f(t^{n+1},y(t^{n+1}))-f(t^{n+1},y^{n+1})\bigr)+ \frac{h^2}{2}·y''(\xi_n)\right| 
\\
\leq |y(t^n)-y^n|+h·L·|y(t^{n+1})-y^{n+1}|+ \frac{h^2}{2}·|y''(\xi_n)|
$$
where $L$ is the Lipschitz constant in $y$-direction of $f$. So that
$$
|y(t^{n+1})-y^{n+1}|\leq \frac{1}{1-L·h}·\left(|y(t^n)-y^n|+ \frac{h^2}{2}·|y''(\xi_n)|\right)
$$
Everything else looks correct at least in principle.
</s>