# Forward Euler Method: how to derive global error

I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.

I'm given a system such that: $$\frac{du}{dt} = v ~ ~ \& ~ ~ \frac{dv}{dt} = -f(u)$$

with Hamiltonian $$H = \frac{1}{2} \left(\frac{du}{dt}\right)^2 + \int f du.$$

I have to show that using the forward Euler method leads to a global error for $$H$$ that grows like $$nh^2$$ for step size $$h$$ and number of steps $$n$$.

I know that the global error can be calculated via $$\epsilon = |U^n - U(T)|$$ but I'm not sure how to apply it in this case.

Thanks for any help!

EDIT: So if I understand correct, I have to calculate $$|H_{n+1}-H_n|$$

## 1 Answer

If you take the example $$f(u)=u$$, then you can interpret the Hamilton system in the complex plane as $$\dot z=\dot u+i\dot v = -i(u+iv)=-iz.$$

The Euler forward iteration thus produces elements $$z_{k+1}=z_k+h(-iz_k)=(1-ih)z_k\implies z_n=(1-ih)^nz_0.$$

The Hamilton function is $$H=\frac12(v^2+u^2)=\frac12|z|^2$$, which gives on the Euler solution $$H_n=\frac12|z_n|^2=\frac12(1+h^2)^n|z_0|^2=e^{nh^2+O(nh^4)}H_0$$ so that indeed $$H_n-H_0=nh^2e^{\frac12nh^2+O(nh^4)}H_0$$.

Under more general conditions one gets \begin{align} H_{k+1}&=\frac12(v_k-hf(u_k))^2+F(u_k+hv_k) \\ &=H_k-hv_kf(u_k)+\frac12h^2f(u_k)^2 ~ + ~ f(u_k)(hv_k)+\frac12f'(u_k)(hv_k)^2+O(h^3) \\ &=H_k+\frac12h^2\bigl[f(u_k)^2+f'(u_k)v_k^2\bigr]+O(h^3) \end{align} Now you have to argue that the coefficient of the $$h^2$$ term remains bounded and the claim of the task follows.