1. Dynamical Systems

Hi, I am studying dynam systems and have missed some classes and am not sure if i'm on the right track or not,

Dynamical system;

dx/dt = rx - ( x/(1 + (x^2)))

when asked to calc the fixed points for this system i figured to set the RHS to equal 0 and solve for x which gave me;

x = (+/-) sqrt((1/r) - 1 )

Have i done this correctly?

It then asks me to determine the stability of the fixed points and i'm unsure how to proceed.

Any help will be greatly appreciated am really struggling with this subject.

Many thanks.

2. Originally Posted by monster
Hi, I am studying dynam systems and have missed some classes and am not sure if i'm on the right track or not,

Dynamical system;

dx/dt = rx - ( x/(1 + (x^2)))

when asked to calc the fixed points for this system i figured to set the RHS to equal 0 and solve for x which gave me;

x = (+/-) sqrt((1/r) - 1 )

Have i done this correctly?
yes

It then asks me to determine the stability of the fixed points and i'm unsure how to proceed.

Any help will be greatly appreciated am really struggling with this subject.

Many thanks.
Let $x_0$ denote a fixed point, and put $x=x_0+\varepsilon$, now use $\varepsilon$ as the new variable, and if when the DE is linearised:

$\frac{d\varepsilon}{dt}<0$ when $\varepsilon>0$ and $\frac{d\varepsilon}{dt}>0$ when $\varepsilon<0$ the fixed point is stable.

That is a positive departure from the fixed point gives a negative rate of change of departure, and a negative departude gives a positive rate of change of departure.

and so on ..

CB

3. dont forget the trivial x=0 fixed point

4. Dynamical systems cont.

Many thanks for help above!

Have come across another problem of similar nature where i am again to calculate the fixed points and thier stability but the equation has me stumped.

dx/dt = (x^3) - ((r+1)x^2 ) + 2rx - r^2

obviously a cubic but am unsure how to calculate the fixed points of this system.

5. Originally Posted by monster
Many thanks for help above!

Have come across another problem of similar nature where i am again to calculate the fixed points and thier stability but the equation has me stumped.

dx/dt = (x^3) - ((r+1)x^2 ) + 2rx - r^2

obviously a cubic but am unsure how to calculate the fixed points of this system.
By inspection $x=r$ is a root of the cubic:

$x^3-(r+1)x^2+2rx-r^2=0$

and as:

$x^3-(r+1)x^2+2rx-r^2=(x-r)(x^2-x+r)$

this is the only real root and so the only real fixed point.

CB

6. Cheers! helps heaps,

Have gone ahead and completed my original questing and calculated the stability of the fixed points,

I said that;
$
f(x) = rx - \frac{x}{1+x^2}
$

and then that the derivative;
$
f ' (x) = r - \frac{1-x^2}{(1+x^2)^2}
$

and i then substituted equation of fixed points in for x,
which gave a messy function of r which i simplified down to;
$
f ' (x_0) = 2r - 2r^2
$

Where $x_0$denotes fixed points.

And considering that the question stated r > 0
i figure that;

fixed points where 0 < r < 1 are unstable
fixed points where r > 1 are stable
and when r=1 fixed points inconclusive?

I hope i did all the algebra in between steps correctly, but have i gone about this the right way am trying to follow text book but a little hard to understand without examples.

Many thanks for help, is VERY helpful.

7. Originally Posted by monster
Cheers! helps heaps,

Have gone ahead and completed my original questing and calculated the stability of the fixed points,

I said that;
$
f(x) = rx - \frac{x}{1+x^2}
$

and then that the derivative;
$
f ' (x) = r - \frac{1-x^2}{(1+x^2)^2}
$

and i then substituted equation of fixed points in for x,
which gave a messy function of r which i simplified down to;
$
f ' (x_0) = 2r - 2r^2
$

Where $x_0$denotes fixed points.

And considering that the question stated r > 0
i figure that;

fixed points where 0 < r < 1 are unstable
fixed points where r > 1 are stable
and when r=1 fixed points inconclusive?

I hope i did all the algebra in between steps correctly, but have i gone about this the right way am trying to follow text book but a little hard to understand without examples.

Many thanks for help, is VERY helpful.
Putting $x=r+\varepsilon$ we get:

$
\frac{d \varepsilon}{dt}=r^2 \varepsilon + O(\varepsilon^2)
$

So if $r \ne 0$ a small enough positive deviation from the fixed point gives a positive error rate and a small enough negative error gives a negative error rate. Thus the fixed point is unstable for non-zero $r$, (when $r=0$ we would have to look at higher order terms in $\varepsilon$ to investigate stability)

Alternativly solve the linearised equation for $\varepsilon$ :

$
\frac{d \varepsilon}{dt}=r^2 \varepsilon
$

to get:

$\varepsilon = \varepsilon_0 e^{r^2 t}$

which for $r \ne 0$ shows the error grows.

CB