Euler Method

Consider an approximate solution of (1.8) over a small interval $\delta t = t_{n+1} - t_n$ by writing the integral as

$$\int_{t_n}^{t_{n+1}}\bi{f}(\bi{y}(t'), t')\,\d t' \approx \delta t\,\bi{f}(\bi{y}(t_n), t_n).$$ (1.10)

to obtain

$$\bi{y}(t_{n+1}) = \bi{y}(t_n) - \delta t\, \bi{f}(\bi{y}(t_n), t_n).$$

or, in a more concise notation,

$$\bi{y}_{n+1} = \bi{y}_n - \delta t\, \bi{f}(\bi{y}_n, t_n) = \bi{y}_n - \delta t\, \bi{f}_n.$$ (1.11)

We can integrate over any larger interval by subdividing the range into sections of width $\delta t$ and repeating (1.11) for each part. Equivalently we can consider that we have approximated the derivative with a forward difference

$$\left.\d\bi{y}\over\d t\right\vert _n \approx {\bi{y}_{n+1} - \bi{y}_n\over\delta t}.$$ (1.12)

We will also come across centred and backward differences,

$$\left.\d\bi{y}\over\d t\right\vert _n \approx \begin{case... ...\bi{y}_n - \bi{y}_{n-1}\over\delta t}& backward \end{cases}$$ (1.13)

respectively. Here we have used a notation which is very common in computational physics, in which we calculate $\bi{y}_n = \bi{y}(t_n)$ at discrete values of $t$ given by $t_n = n \delta t$, and $\bi{f}_n = \bi{f}(\bi{y}_n, t_n)$. In what follows we will drop the vector notation $\bi{y}$ except when it is important for the discussion.