Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$x'(t)=f(t,x(t)), t\in(0,T)$ with $x(0)=x_0$

$f$ satifies the Lipschitz-condition $f(t,x)-f(t,y)\le L|x-y|$

$h\in (0,\frac{1}{L})$ 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

$$|x_k-x(t_k)|\le\frac{1}{1-Lh}\left(|x_{k-1}-x(t_{k-1})|+\frac{h^2}{2} \max_{s\in [0,T]}|x''(s)|\right)$$

I tried to look up it up in some numerical analysis books but it is always different

share|cite|improve this question

First, we get local truncation error.

$x(t_{k+1}) = x(t_k) + hf(t_k,x(t_k)) + \tau_k$

$\tau_k = x(t_{k+1}) - x(t_k) - hf(t_k,x(t_k)) = \frac{h^2}{2}x''(\eta)$. Where $\eta \in (t_k,t_{k+1})$.

Then we get the bound,

$$|x(t_{k+1}) - x_{k+1}| \le (1+hL)|x(t_{k}) - x_{k}| + |\tau_k|$$ $$\le (1+hL)|x(t_{k}) - x_{k}| + \frac{h^2}{2}\max_{s \in (0,T)}|x''(s)|$$ $$\le \frac{1}{1-hL}|x(t_{k}) - x_{k}| + \frac{h^2}{2}\max_{s \in (0,T)}|x''(s)|$$

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.