For $f\in C([0,T]×\mathbb{R})$, let $|f(t,x)-f(t,y)|\leq L|x-y|$ for a $L\geq0$. Also suppose we have a differential equation
\begin{align}
y'(t) &= f(t,y(t))\\
y(0)&=y_0
\end{align}
with $y_1,\dots, y_N$ the solutions given by the Forward Euler method.
So far I found that a bound for the maximum error is given by
$$\max_{0\leq n \leq N} e_n \leq \frac{M}{2L}(e^{LT}-1)h$$
where $M=\max_{0\leq t \leq T}|y''(\tau)|$ for a $\tau\in[0,T]$, and $h=T/N$ the step size.
My question is: how do I find this bound for the following initial value problem:
\begin{align}
y'(t)&=\arctan(y(t))\\
y(0)&=y_0?
\end{align}
I am stuck at finding $M$ and $L$. How could I find the maximum of the double derivative of $y(t)$ if I do not know anything about the function?
2026-03-29 20:37:29.1774816649
Maximum error bound for ODE
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1
In fact, the correct expression is $\displaystyle M = \max_{t \in [0,T]}|y''(t)|$. In order to get an error estimate, you don't need to exactly compute $M$, you just need to have an upper bound: $$ |y''(t)| = |(\arctan(y(t)))'| = \left|\dfrac{y'(t)}{1+y(t)^2}\right|=\left| \dfrac{\arctan(y(t))}{1+y(t)^2}\right|\leq \frac{\pi}{2}. $$