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.


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

$$f(t_{n},y_{n}) = \lambda y_{n}$$


$$f(t_{n+1},y_{n+1}) = \lambda y_{n+1}$$

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 $\mathbb{R_{-}}$.

  • $\begingroup$ Oh phenomenal, I kind of feel stupid for not having attempted this. In a first semester course, we called this "finding the symbol" for a method, and it was exactly what we had done for single-step methods. It hadn't occurred to me to attempt the same thing here. Sometimes wading through a new text and notation, you forget what you already know. Thank you for taking the time to write out the details! $\endgroup$ – cnolte Aug 1 '16 at 3:39
  • $\begingroup$ I know what you mean, I personally found numerical methods quite confusing when I started learning about them. If you need anymore help, let me know. $\endgroup$ – mattos Aug 1 '16 at 4:07
  • $\begingroup$ Agreed, especially considering how similar so much of the information is. I had a quick question regarding the Adams-Bashforth method given above. Since I will have more than just $y_{n+1}$ and $y_n$, I will end up with two coefficients, one for $y_n$, the other for $y_{n-1}$. In this case, do I need to set each to be less than one, and let the intersection of those two regions be my area of stability, or should I write $\phi(z)$ as the combination of the two, and still set $\phi(z) < 1$?. Both appear valid to me at first, as the former would (I think) ensure that each term stays stable, while $\endgroup$ – cnolte Aug 1 '16 at 15:40
  • $\begingroup$ while the later ensures that, as a whole, $y_{n+1}$ remains stable. My gut, though, is the later. Also, it's hardly an issue, but for completeness, in your definition of $S$, you might require that Re$(z) < 0$ since it is allowed to be complex. Is that right? $\endgroup$ – cnolte Aug 1 '16 at 15:42
  • $\begingroup$ See here, section $4.4$, for a slightly different approach to getting the stability region for the Adams-Bashforth. It may help your overall understanding. An alternative approach is to use induction on the method with the standard test problem and note that $y_{0}$ is a given constant. $\endgroup$ – mattos Aug 2 '16 at 1:34

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