# backward euler method error

Let

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

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## 1 Answer

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.

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