# pitchfork bifurcation and sketching full bifurcation diagram

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• October 12th 2008, 08:01 PM
tk1234
pitchfork bifurcation and sketching full bifurcation diagram and more
given system of ODE

dx/dt = ax+xy-x^3
dy/dt = b-x^2-y^2

first, show that there is saddle-node bifurcation for b=0, without using the theorem

second, determine the curve in the a-b plane for which there exists a pitchfork bifurcation without using the theorem

third, sketch the full bifurcation set in the a-b plane (including homoclinic bifurcation)

Help plz
• October 14th 2008, 06:25 AM
shawsend
I'm no expert ok, but I find these fascinating so I'd like to try: I'd first solve for the equilibrium points in terms of a and b:

$ax+xy-x^3=0$
$b-x^2-y^2=0$

These are:

$e_1=\left\{0,-\sqrt{b}\right\}$
$e_2=\left\{0,\sqrt{b}\right\}$
$e_3=\left\{-\frac{\sqrt{-1+2 a-\sqrt{1-4 a+4 b}}}{\sqrt{2}},\frac{1}{2} \left(-1-\sqrt{1-4 a+4 b}\right)\right\}$
$e_4 \left\{\frac{\sqrt{-1+2 a-\sqrt{1-4 a+4 b}}}{\sqrt{2}},\frac{1}{2} \left(-1-\sqrt{1-4 a+4 b}\right)\right\}$
$e_5=\left\{-\sqrt{-\frac{1}{2}+a+\frac{1}{2} \sqrt{1-4 a+4 b}},-\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a+4 b}\right\}$
$e_6=\left\{\sqrt{-\frac{1}{2}+a+\frac{1}{2} \sqrt{1-4 a+4 b}},-\frac{1}{2}+\frac{1}{2} \sqrt{1-4 a+4 b}\right\}
$

Fixed points are created whenever the equilibrium points are real so the root objects completely control the creation and destruction of fixed points. Immediately we see the creation of $e_1$ and $e_2$ whenever $b$ becomes non-negative. I would suspect through further analysis, i.e., linearizing, and calculating eigenvalues, we could determine that these two fixed points are indeed a saddle and node hence the saddle-node bifurcation. Probably other ways too though. In the first plot, I depicted the flow in phase space about these two points (red) when a=4 and b=1 (green section in second plot). The top fixed point looks to be a saddle (points move to then away from it). The bottom point looks a source (node). Points move away from it. That would need to be verified algebraically though.

The remaining four have real y-component whenever the quantity $b>a-1/4$ or the region above the line $b=a-1/4$. These have real x component whenever $-1/2+a+1/2\sqrt{1-4a+4b}$ or $-1+2a-\sqrt{1-4a+4b}$ are non-negative. These regions are depicted in the second plot below with the number of fixed points indicated in each.

Again, further analysis could determine the types of remaining fixed points as well as further types of bifurations such as Hopf bifurcations. This is just a start.
• October 14th 2008, 10:30 AM
shawsend
I'd like to add a little rigor to my analysis above regarding the fixed points when a=4 and b=1. In this case we have:

$x'=4x+xy-x^3$
$y'=1-x^2-y^2$

The fixed points are $e_1=(0,1)$ and $e_2=(0,-1)$. In order to linearize these, I move the points to the origin via change of variables.

For $e_1$ it's $u=x$ and $v=y-1$.

For $e_2$ it's $u=x$ and $v=y+1$.

Making these changes results in the new differential equations with fixed points at the origin. In the case of $e_1$ we have:

$u'=5u+uv-u^3$
$v'=-2v-u^2-v^2$

Since the fixed points are now at the origin, the non-linear terms have little impact for u and v values near the origin and the resulting linear system is an approximation to the non-linear system:

$u'=5v+0v$
$v'=0u-2v$

The eigen values for this one is -2 and 5 which is a saddle. Likewise, doing the same thing with the other one, I get for the linear system:

$u'=3u+0v$
$v'=0u+2v$

The eigenvalues of which are 2 and 3 which means this is a source, i.e,. a node.
• October 15th 2008, 10:51 PM
tk1234
Thanks !!!