I'm new to the idea of solving ODEs using the backward Euler. I have a system which I solve using the Backward Euler (actually backward Euler + Newton's method since I can't find a closed form solution).
I understand the math involved in solving it, i.e. how to rearrange the equations to get $\vec{x}_{n+1}$, the position at time $n+1$. However, I'm not sure WHY it works.
Given the equation:
$\vec{x}_{n+1} = \vec{x}_n + \Delta t f(t, \vec{x}_{n+1})$ (backward Euler)
compared to
$\vec{x}_{n+1} = \vec{x}_n + \Delta t f(t, \vec{x}_{n})$ (forward Euler)
Why is backward Euler more stable? why does solving for $\vec{x}_{n+1}$ implicitly like that, searching for an equilibrium, produce unlimited stability?
EDIT:
The ODE looks like this:
$\vec{P}_{n+1}^i = \vec{P}_n^i + \Delta t [ \sum_{j \not = i} K (\vec{P}_{n}^j – \vec{P}_n^i) + \sum_{j \not = i} \frac{G(\vec{P}_{n}^i – \vec{P}_n^j)}{|\vec{P}_{n}^j – \vec{P}_n^i|^2}]$
For a bunch of particles (each being $i$) in two dimensions (so each position $\vec{P} = [x, y]^T$).
So basically for each particle I have an attraction force from all other particles and a repulsion force from all other particles. Closed form solution doesn't seem possible for this when I substitute all the $\vec{P}_n^i$ with $\vec{P}_{n+1}^i$ (except the first one, of course).
Also, I'm aware to make it a true Backward Euler I'd have to solve the state of all the system, otherwise my stability isn't really unconditional.
However, if I were to solve this as a system, why does using the backward Euler give increased stability?
Best Answer
There is plenty of literature out there discussing the stability of ODE integrators. A good intro to the topic might be these lecture notes, which consider linear ODEs first, introducing the concept of A and L stability.
For a nonlinear ODE like yours, you could start by looking into this site. There, you find the concept of B and algebraic stability. This can lead you to this article, which presents conditions under which different integrators work or not.