# Thread: A problem about semistable equilibrium solution

1. ## A problem about semistable equilibrium solution

I was solving a ODEs problem about semistable equilibrium solution.

The equation is dy/dt=k(1-y)^2. y(0) = M (M is an arbitrary constant)

It says the y=1 is a semistable solution for this equation, which other solutions below it approach it and those above it grow father away.

However when i get the general solution which is
y=[t+M/(k-kM)]/[t+1/(k-kM)]
i found that when t approaches infinity, the y correspondingly approaches 1 no matter the initial value M is either greater than 1 or smaller than 1.

If this is the case it seems that the qualitative analysis describing the characteristic of semistable solution is contradictory with what i got, i was

thank you very much for you valuable help,
thanks alot in advance...

2. Originally Posted by rexhegemony
I was solving a ODEs problem about semistable equilibrium solution.

The equation is dy/dt=k(1-y)^2. y(0) = M (M is an arbitrary constant)

It says the y=1 is a semistable solution for this equation, which other solutions below it approach it and those above it grow father away.

However when i get the general solution which is
y=[t+M/(k-kM)]/[t+1/(k-kM)]
i found that when t approaches infinity, the y correspondingly approaches 1 no matter the initial value M is either greater than 1 or smaller than 1.

If this is the case it seems that the qualitative analysis describing the characteristic of semistable solution is contradictory with what i got, i was

thank you very much for you valuable help,
thanks alot in advance...
Look at a particular example say when k = 1 and y(0) = 2. The solution of this is

$y = \frac{t-2}{t-1}$ which if you plot this solution shows that you'll go away from the equilibrium for t > 0 and has a singularity at t = 1. The same will happen for the general solution for M > 1 (k > 0).

i know that if you plug some particular values of k and M into the general solution and get certain particular solution, that qualitative description about semistable equilibrium solution is indeed right...

But look at that general solution y=[t+M/(k-kM)]/[t+1/(k-kM)],
whats the limitation when t approaches infinity? its plausible to be 1 instead of any other figures...

this is wat i confused with...

thank you very much for your help!

4. This problem has been discussed more than two years ago in...

http://www.mathhelpforum.com/math-help/calculus/13907-semistable-equilibrium-solutions.html

... and here we found the following definition of 'semistable solution'...

… a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable…

In such case for $k>0$ forall $y$ is $y^{'}(t) \ge 0$ and that means that if $M \le 1$ the solution will converge to 1 and if $M>1$ the solution will depart from 1. The opposite is of course if $k<0$...

Kind regards

$\chi$ $\sigma$

5. ## An odd conclusion

Originally Posted by Danny
Look at a particular example say when k = 1 and y(0) = 2. The solution of this is

$y = \frac{t-2}{t-1}$ which if you plot this solution shows that you'll go away from the equilibrium for t > 0 and has a singularity at t = 1. The same will happen for the general solution for M > 1 (k > 0).
Hi thank you so much for you example.

Firstly i forgot to specify that k is indeed a positive constant in this case.

Then lets follow you example :

1. k=1, initial condition is y(0)=2, then we end up with [IMG]file:///D:/DOCUME%7E1/ZLibAuth/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]...
I use excel plot the curve it's such that when t increases
y approaches 1. so far so good..

2. k=1, initial condition is y(0)=0.5 (which in this situation initial value is smaller than 1) , it yields to be y=(t+1)/(t+2); still i plot the curve wat happens is when t increases, y approaches 1 again.... but but but according to the definition of semistable solution this is totally not supposed to be so.
(that fancy definition "… a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable…")(thanks to Chisigma)

Thank you all so much!

6. Originally Posted by chisigma
This problem has been discussed more than two years ago in...

http://www.mathhelpforum.com/math-help/calculus/13907-semistable-equilibrium-solutions.html

... and here we found the following definition of 'semistable solution'...

… a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable…

In such case for $k>0$ forall $y$ is $y^{'}(t) \ge 0$ and that means that if $M \le 1$ the solution will converge to 1 and if $M>1$ the solution will depart from 1. The opposite is of course if $k<0$...

Kind regards

$\chi$ $\sigma$
Thank you so much for your help.

but now it seems that i have a problem if we simply plug some figures into this question....