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

regatamin2

regatamin2

Answered question

2022-01-22

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=ky, where k is a complex number.

Answer & Explanation

Robert Pina

Robert Pina

Beginner2022-01-22Added 42 answers

You have, in fact, used the technique that you described
yn+1=(0.25h2z2+hz+1)yn
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 ={wC|0.25w2+w+1<1} 
Therefore, the stability region is independent of step size h.
You can utilize that in the specific situation you are requesting
0.25w2+w+1=(0.5w+1)2
You should be able to find the region from this. 
Regarding how to tell if a method is A-stable, you are correct. Given that explicit methods are never A-stable, you should discover that the technique is not 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 0.25w2+w+1 is also real, so the condition |0.25w2+w+1|<1 simplifies to 1<0.25w2+w+1<1. However, in actual practice, plotting the area on a computer is what most people do.
A final note: Are you sure you copied the method correctly? The method 
yn+1=yn+0.5h(f(tn,yn)+f(tn+1,yn+hf(tn,yn))  
with no factor 0.5 in front of the last h, is more popular.

Carl Swisher

Carl Swisher

Beginner2022-01-23Added 28 answers

I think you mean to type(?)
{zC|0.25h2z2+hz+1<1}
Why enforce this requirement? At time step n, think about the following:
yn=(h2z24+hz+1)ny0
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 y0ϵ=y0+ϵ, then
|ynϵyn|=|h2z24+hz+1|nϵ<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.

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