# A-stability of Heun method for ODEs I'm trying to determine the

A-stability of Heun method for ODEs
I'm trying to determine the stability region of the Heun method for ODEs by using the equation $y\prime =ky$, where k is a complex number.
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Robert Pina
For the method that you wrote down, you indeed have
${y}_{n+1}=\left(0.25\cdot {h}^{2}\cdot {z}^{2}+hz+1\right){y}_{n}$.
The expression between parentheses is a function of $w=hz$, and the stability region consists of the numbers w for which the modulus of the expression between parentheses is at most one:
Stability region $=\left\{w\in C|0.25{w}^{2}+w+1\mid <1\right\}$
So the stability region does not depend on the step size h.
In the particular case you are asking about, you can use that
$0.25{w}^{2}+w+1={\left(0.5w+1\right)}^{2}$.
You should be able to find the region from this.
You are right about how to determine whether the method is A-stable. You should find that the method is not A-stable, because explicit methods are never A-stable (see also the Wikipedia page that you link to).
In general, to get a feeling for what the stability region looks like, one may start by restricting to the real axis. If w is real, then $0.25{w}^{2}+w+1$ is also real, so the condition $|0.25{w}^{2}+w+1|<1$ simplifies to $-1<0.25{w}^{2}+w+1<1$. But what people often do in practice is to plot the region using the computer.
A final note: Are you sure you copied the method correctly? The method
${y}_{n+1}={y}_{n}+0.5\cdot h\left(f\left({t}_{n},{y}_{n}\right)+f\left({t}_{n+1},{y}_{n}+h\cdot f\left({t}_{n},{y}_{n}\right)\right)$
with no factor 0.5 in front of the last h, is more popular.
###### Not exactly what you’re looking for?
Carl Swisher
I think you mean to type(?)
$\left\{z\in C|0.25{h}^{2}{z}^{2}+hz+1\mid <1\right\}$
Why to impose this condition? Consider the following at time step n:
${y}_{n}={\left(\frac{{h}^{2}{z}^{2}}{4}+hz+1\right)}^{n}{y}_{0}$
therefore having the first stability condition is the sufficient condition for a small disturbance in the initial value wouldn't get magnified marching in the time step: Consider ${y}_{0}^{ϵ}={y}_{0}+ϵ$, then
$|{y}_{n}^{ϵ}-{y}_{n}|={|\frac{{h}^{2}{z}^{2}}{4}+hz+1|}^{n}ϵ<ϵ$
and $ϵ$ can be happen at any time-step, or even could be the numerical error itself, this is why sometimes for a relatively not-so-small step-size h, after a few iterations, the numerical error gets magnified and the numerical solution blows up.