Numerical Analysis - Forward Euler Method

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@FussCraft 2 жыл бұрын

really good video! whould be nice to cover stability of FE method, in order to determine appropriate time step when you dont have an analytical solution to compare to

@moritzpfurtscheller4248 2 жыл бұрын

Great explanation THANKS ALOT!

@ABHISHEKJAIN1170 4 жыл бұрын

How do we know which Euler scheme (Forward, backward, Euler-Chromer ) is to be used in solving a particular ODE ?

@pipertripp 4 жыл бұрын

It really depends on what you're after, what sort of error you can tolerate, and the nature of the equation you're trying to approximate. Forward Euler is conditionally stable. This means that if you're time step is too large you have stability issues. h = 2/a (where a is the coefficient on the dependent variable term) is the max time step you can use to ensure stability with fwd Euler. Many times this is OK, esp is a is small. If a is really large though, you have to use a very small time step to have a stable solution (note, stability and accuracy are not the same thing). Backward Euler is unconditionally stable but it's an implicit method, which is more expensive computationally. Since it's unconditionally stable, you can generally use larger time steps which reduces computations... but since it's implicit, you have to solve your function every time step. If the function is complicated, this can be time consuming. Final thought... Backward Euler will work much better for "stiff" ODEs. In general explicit methods like fwd Euler or Runge-Kutta methods don't handle stiff ODEs well. Backward differentiation methods are much better in these situations. Backward Euler is the simplest of these kinds of methods. Adams-Moulton is an example of a more sophisticated (4th order) backward differentiation method.

@KrzychVEVO 3 жыл бұрын

i dont understand what is this f(t,y), what is this? some unkown function f which takes t and y(t) as parameters? how do i know what this function will do with those parameters?

@melom806 3 жыл бұрын

f(x,y) is just a function of two variables. For example, f(x,y)=x^(2)y+xy, where as what you might be use to is f(x), which is just a function of a single variable, f(x) = x^2. I think what might make sense is v = v(0) + at. So if lets say initial velocity is 0, f(a,t) = at just like f(x,y) = x^2 + y^2. Applications of maths are important to its understanding, hence solving problems are important.