# Iterative trapezoidal method for differential equations

I am studying numerical methods for differential equations. I came accros the trapezoidal method in two forms, an explicit and an iterative one. I would like to know the advantages and disadvantages of each of those methods. Furthermore, how can I study the stability for the iterative method? Which stability definition is the better one and why?. I explain both methods below.

Consider an initial value problem given by $y' = f(t, y)$ and $y(a) = t_0$, where $f$ is defined in $[a, b]\times[\alpha, \beta]$ and satisfies the Lipschitz property with Lipschitz's constant $L$.

Given a natural number $n$ and $h = \frac{b-a}{n}$, we are trying to approximate the unique solution of the problem at $t_i = a + ih \ \forall i = 0, 1 \ldots n$. If $y$ is the solution, we call $y_i = y(t_i)$ and we denote $w_i$ to the approximations obtained by the applied method.

One of the methods studied is the explicit trapezoidal method. It follows the following rule:

$w_{i+1} = w_i + \frac{h}{2} \left[f(t_i,w_i) + f(t_{i}+h, w_i + h f(t_i,w_i))\right]$

We have proved that it has a local error of order 3 and, hence, a global error of order 2.

Then, reading some books I came accros the iterative trapezoidal method, which solves the following implicit equation:

$w_{i}= w_{i-1} + \frac{h}{2} \left[f(t_{i-1}, w_{i-1}) + f(t_i, w_{i})\right]$

The idea is taking an initial approximation $w_i^{(0)}$ and defining the following sequence:

$w_{i} ^{(j+1)} = w_{i-1} + \frac{h}{2} \left[f(t_{i-1}, w_{i-1}) + f(t_i, w_{i}^{(j)})\right]$

The limit of that sequence is taken as $w_i$. I have proved that the sequence converges if $hL/2 < 1$ and that if we use $w_i$, then the local error is $O(h^3)$. However, why is this method useful and how can I study its stability?

3. The stability of these method are studied by examining their behavior when applied to the simple test equation $y' = \lambda y$. When the real part of $\lambda$ is strictly less than zero, then the exact solution decays to zero as $t$ tends to infinity. The implicit method will reproduce this behavior regardless of the time step $h$. The explicit method will only produce this behavior for sufficient small values of $h$.