Stability Region for 2nd order Multi-Step Methods The question I am trying to solve is as follows:
What is the intersection of stability region for the 2nd order Adams-Bashforth method (given below) with the real axis
$$y_{n+1} = y_n + h \left[ \frac{3}{2} f(t_{n},y_{n}) - \frac{1}{2} f(t_{n-1},y_{n-1}) \right]$$
Similarly, the 2nd order Adams-Moulton method is the trapezoidal method
$$y_{n+1} = y_n + h \left[ \frac{1}{2} f(t_{n+1},y_{n+1}) + \frac{1}{2} f(t_{n},y_{n}) \right]$$
What is the intersection of stability region with the real axis for this method?
I really have no idea where to begin. When doing similar problems for RK methods, we usually begin with the model problem, $Y'(t) = \lambda Y(t)$ for $t \geq 0$ and $Y(0) = 1$. The solution to this is $Y(t) = e^{\lambda t}$. The text that I am using doesn't give a very detailed method on how to approach multi-step stability, but it does reference using the characteristic polynomial.
I appreciate any help or insight that anybody may have in this. Ideally, if I could see a general strategy for approaching one of the given methods, I think I should be able to figure out where to go for the second.
 A: I'm going to do a fairly detailed answer because there seems to be some confusion about applying the test problem to this particular method. For the test problem $y' = f(t,y) = \lambda y$, you have 
$$f(t_{n},y_{n}) = \lambda y_{n}$$
and 
$$f(t_{n+1},y_{n+1}) = \lambda y_{n+1}$$
If we substitute these into the trapezoidal method, we find
\begin{align}
y_{n+1} &= y_{n} + \frac{h}{2} \bigg(f(t_{n+1},y_{n+1}) + f(t_{n},y_{n}) \bigg) \\
&= y_{n} + \frac{h}{2} \bigg(\lambda y_{n+1} + \lambda y_{n} \bigg) \\
\implies y_{n+1} - \frac{h \lambda}{2} y_{n+1} &= y_{n} + \frac{h \lambda}{2} y_{n} \\
\implies \bigg(1 - \frac{h \lambda}{2} \bigg) y_{n+1} &= \bigg(1 + \frac{h \lambda}{2} \bigg) y_{n} \\
\implies y_{n+1} &= \frac{1 + z/2}{1 - z/2} y_{n}
\end{align}
where $z = h \lambda \in \mathbb{C}$. We know that the coefficient of $y_{n}$ must be less than $1$ for stability, hence we require
$$\phi(z) = \bigg \lvert \frac{1 + z/2}{1 - z/2} \bigg \rvert < 1$$
Notice that for $z > 0$, $\phi(z) > 1$ with a pole at $z = 2$. When $z < 0$, $\phi(z) < 1$. Finally, when $z = 0$, $\phi(z) = 1$. Hence, the stability region is given by
$$S = \{z \in \mathbb{C} : z < 0 \}$$
which is the left half of the complex plane. So the intersection of the stability region and the real axis, $S \cap \mathbb{R}$, is given by $\mathbb{R_{-}}$.
