Thread: Chaos theory: bifurcation diagram

Hi, I have this problem.

I need to sketch a bifurcation diagram for the fixed points of the family $\displaystyle \mu \rightarrow f_\mu$ where $\displaystyle f_\mu= \mu x^2(1-x)$.

I have that the fixed points are $\displaystyle x=0$ and $\displaystyle x=\frac{1\pm\sqrt{1-\frac{4}{\mu}}}{2} $

I know that:
When $\displaystyle \mu < 4$ $\displaystyle f_\mu$ drops below the identity mapping and so there is only one fixed point at $\displaystyle x=0$

When $\displaystyle \mu =4$ the identity mapping is tangent to $\displaystyle f_\mu$ and so there are 2 fixed points at $\displaystyle x=0 $and $\displaystyle x= 0.5$

When $\displaystyle \mu > 4$ $\displaystyle f_\mu$ rises above the identity mapping and so there are 3 one fixed points at x=0 and $\displaystyle x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$

So i can sort of picture how the diagram will look, I just can't figure out for what values of $\displaystyle \mu$ the fixed points $\displaystyle x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$ will be repelling/attracting.

I know that $\displaystyle f'_\mu(x) = 3-\frac{\mu}{2}\pm\frac{\sqrt{\mu^2 - 4\mu} }{2}$ at the last two fixed points. And that if $\displaystyle |f'_\mu(x)|<1$ then fixed point is an attractor and if $\displaystyle |f'_\mu(x)|>1$ then fixed point is a repellor. But i'm stuck here.

Please help.
Katy

hi !
Using the fixed point x=0 and $\displaystyle
f^{\prime}=2\mu x - 3\mu x^2]$

x=0 is a atracting fixed point , cause $\displaystyle
|f^{\prime}| <1$.

for the other fixed point $\displaystyle x_{2}= \frac{1+\sqrt{1-4\mu}}{2}$

evaluating , you get $\displaystyle
f^{\prime}(x_{2})=\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}$

in case you want to know the values for $\displaystyle \mu$ so that $\displaystyle x_{2}$ is an atracting fixed point you have to solve the inequality

$\displaystyle |\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|<1$
and for a repelling fixed point solve:
$\displaystyle |\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|>1$
to sketch the bifurcation, in the x -axis you plot $\displaystyle \mu$ and in the y-axis put the fixed points.

locate the coordinates $\displaystyle (\mu,1)=(\frac{1}{4},1)$, and the use the results of the inequalities to draw the behaviour your are looking for.!

Originally Posted by harkapobi

Hi, I have this problem.

I need to sketch a bifurcation diagram for the fixed points of the family $\displaystyle \mu \rightarrow f_\mu$ where $\displaystyle f_\mu= \mu x^2(1-x)$.

I have that the fixed points are $\displaystyle x=0$ and $\displaystyle x=\frac{1\pm\sqrt{1-\frac{4}{\mu}}}{2} $

I know that:
When $\displaystyle \mu < 4$ $\displaystyle f_\mu$ drops below the identity mapping and so there is only one fixed point at $\displaystyle x=0$

When $\displaystyle \mu =4$ the identity mapping is tangent to $\displaystyle f_\mu$ and so there are 2 fixed points at $\displaystyle x=0 $and $\displaystyle x= 0.5$

When $\displaystyle \mu > 4$ $\displaystyle f_\mu$ rises above the identity mapping and so there are 3 one fixed points at x=0 and $\displaystyle x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$

So i can sort of picture how the diagram will look, I just can't figure out for what values of $\displaystyle \mu$ the fixed points $\displaystyle x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$ will be repelling/attracting.

I know that $\displaystyle f'_\mu(x) = 3-\frac{\mu}{2}\pm\frac{\sqrt{\mu^2 - 4\mu} }{2}$ at the last two fixed points. And that if $\displaystyle |f'_\mu(x)|<1$ then fixed point is an attractor and if $\displaystyle |f'_\mu(x)|>1$ then fixed point is a repellor. But i'm stuck here.

Please help.
Katy

Originally Posted by Sofia

hi !
Using the fixed point x=0 and $f^{\prime}=2\mu x - 3\mu x^2$

x=0 is a atracting fixed point , cause $|f^{\prime}| <1$.

for the other fixed point $x_{2}= \frac{1+\sqrt{1-4\mu}}{2}$

evaluating , you get $f^{\prime}(x_{2})=\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}$

in case you want to know the values for $\mu$ so that $x_{2}$ is an atracting fixed point you have to solve the inequality

$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|<1$

and for a repelling fixed point solve:
$|\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|>1$

to sketch the bifurcation, in the x -axis you plot $\mu$ and in the y-axis put the fixed points.

locate the coordinates $(\mu,1)=(\frac{1}{4},1)$, and the use the results of the inequalities to draw the behaviour your are looking for.!

Use [tex][/tex]

Originally Posted by Sofia

hi !
Using the fixed point x=0 and $\displaystyle
f^{\prime}=2\mu x - 3\mu x^2]$

x=0 is a atracting fixed point , cause $\displaystyle
|f^{\prime}| <1$.

for the other fixed point $\displaystyle x_{2}= \frac{1+\sqrt{1-4\mu}}{2}$

evaluating , you get $\displaystyle
f^{\prime}(x_{2})=\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}$

in case you want to know the values for $\displaystyle \mu$ so that $\displaystyle x_{2}$ is an atracting fixed point you have to solve the inequality

$\displaystyle |\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|<1$
and for a repelling fixed point solve:
$\displaystyle |\frac{6\mu^2-\mu\sqrt{1-4\mu}+\mu}{2}|>1$
to sketch the bifurcation, in the x -axis you plot $\displaystyle \mu$ and in the y-axis put the fixed points.

locate the coordinates $\displaystyle (\mu,1)=(\frac{1}{4},1)$, and the use the results of the inequalities to draw the behaviour your are looking for.!

Hi, thanks for your help. Those inequalities are the parts i'm struggling with. I just can't solve them. Any clues on how I can do this?

Thanks

Well, you could just plot it, see what's up, then run a Solve in Mathematica on it:

Code:

Plot[(6*m^2 - m*Sqrt[1 - 4*m] + m)/2, {m, -1, 1}]
Solve[(6*m^2 - m*Sqrt[1 - 4*m] + m)/2 == 1, m]