EULER's METHOD

Euler's Method:

given $y'=f(t,y)$, with $y(t_0) = y_0$, use

$$y_{n+1} = y_n + hf(t_n,y_n), \mbox{ with } t_n = t_0+nh,$$

where $h$ is some fixed stepsize, and we want $y_n \simeq y(t_n)$. Note: we do not get $y(t)$; only $y_1, y_2, \ldots \simeq y(t_1), y(t_2), \ldots$.
Issues: accuracy? efficiency?

Motivations:

1. If $y_n \simeq y(t_n)$, then $f(t_n,y_n)$ is the slope of the tangent line to the solution at $t_n$, so if $h$ is small enough, we can approximate $y(t)$ near $t_n$ using the tangent line, and then $y(t_n) \simeq y_n + hf(t_n,y_n)$.
2. The integrated form of the DE is
   
   $$\rule{0pt}{5pt}\hspace{-4cm}y(t_{n+1}) = y(t_n) + \int_{t_n}^{t_{n+1}}f(s,y(s))ds$$
   
   
   so, using rectangle approximation for integral,
   
   $$y(t_{n+1})\simeq y(t_n)+hf(t_n,y(t_n)) \simeq y_n+hf(t_n,y_n).$$
   
   

EXISTENCE and UNIQUENESS THEORY

General DE:

$y'=f(t,y)$, given $y(0) = 0$.

Solution Theorem:

if $f$ and $f_y$ are continuous for
$R = \{ (y,t) :\vert t\vert < a, \vert y\vert < b\}$, then there exists
an $h>0$, such that, for all $\vert t\vert \leq h \leq a$, there exists
a unique solution $\phi(t)$ for $y'=f(t,y)$.

Picard's Proof

• Any solution $\phi(t)$ must satisfy
  
  $$\phi(t) = \int_0^tf(s,\phi(s))ds.$$
  
  

• Iteration: make an initial guess $\phi_0$, for $\phi$, and use
  
  $$\phi_{i+1}(t) = \int_0^tf(s,\phi_i(s))ds,$$
  
  
  for $i = 0, 1, ...,$ until approximations $\phi_i(t)$ converge.
• Need to show convergence to a unique limit $\phi(t)$.
• Key argument shows convergence of the series
  
  $$\phi_1(t) + \sum_{k=1}^{\infty}(\phi_{k+1}(t)-\phi_k(t)),$$
  
  
  using
  
  $$\vert\phi_{k+1}-\phi_k\vert=\vert f(t,\phi_{k+1})-f(t,\phi_k)\vert\leq K\vert\phi_{k+1}-\phi_k\vert,$$
  
  
  where $K = \max_D \vert f_y\vert$.