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

regatamin2 2022-01-22 Answered
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.
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Robert Pina
Answered 2022-01-22 Author has 42 answers
For the method that you wrote down, you indeed have
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}
So the stability region does not depend on the step size h.
In the particular case you are asking about, you can use that
0.25w2+w+1=(0.5w+1)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 0.25w2+w+1 is also real, so the condition |0.25w2+w+1|<1 simplifies to 1<0.25w2+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
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.
Not exactly what you’re looking for?
Ask My Question
Carl Swisher
Answered 2022-01-23 Author has 28 answers
I think you mean to type(?)
{zC|0.25h2z2+hz+1<1}
Why to impose this condition? Consider the following at time step n:
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.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-09-07
Find the laplace transform of f(t)=(t3)u2(t)(t2)u3(t)
asked 2022-05-22
I'm currently stuck with a problem where I'm supposed to find all solutions that are asymptotic to the line y = 3 t when t . This is the demand, from here I'm supposed to create a first order linear differential equation. Can someone help me get started with this problem? Unsure of how to start....
Asymptotic would mean that for example x ( t ) = y ( t ) 1 / t would satisfy the given demand, not sure how to go further with this although.
A first order linear differential equation means I should have some sort of connection between my function and the derivative of the function, I cant make that connection....
It's my first time using this forum so I've probably made every mistake you can make, hopefully my question is still relevant...
asked 2021-11-15
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the slotuion.
(x2)y+y+(x2)(tanx)y=0
y(3)=1
y(3)=6
asked 2020-11-17
Solve the initial value problem below using the method of Laplace transforms
y"35y=144t366t
y(0)=0
y(0)=47
asked 2021-12-29
Solve for the differential equations and get the general solution. Simplify your answer free from In. dy=tanxtanydx
asked 2021-11-20
Show that y=x2sin(x) and y=0 are both solutions of
x2y4xy+(x2+6)y=0
and that both satisfy the conditions y(0)=0 and y(0)=0. Does this theorem contradict Theorem A? If not, why not?
asked 2022-06-21
I need to find a series solution to the following simple differential equation
x 2 y = y
Assuming the solution to be of the form y = a n x n and equating the coefficients on both the sides, all the coefficients turn out to be zero which is definitely wrong.
Any help is appreciated.