Solve Dynamical System: Find Fixed Points and k Value for Origin

[BrainHurts]

Homework Statement

$\frac{dU}{dz} = V, \frac{dV}{dz}=k+cU-6U^{2}$ c $\in$ ℝ

Find the fixed points of the system (these are solutions U=U*, V=V* where U*,V* $\in$ ℝ) and determine the value of k so that the origin is a fixed point of the system

Homework Equations

The Attempt at a Solution

I'm not too familiar with dynamical systems but I believe that we want

$\frac{dU}{dz}|_{V=V^{*}} = V^{*} = 0$

and

$\frac{dV}{dz}| _{U=U^{*}} = k+cU-6U^{2} = k+cU^{*}-6(U^{*})^{2} = 0$

So we want a value of k so that the origin is a fixed point of the system. I'm not quite sure what this means.

 

[tiny-tim]

Hi BrainHurts!

So we want a value of k so that the origin is a fixed point of the system. I'm not quite sure what this means.

It means dU/dz and dV/dz are both 0 at (U,V) = (0,0)

 

[BrainHurts]

So the origin is a fixed point of the system when k=6(U*)² + cU*?

 

[tiny-tim]

So the origin is a fixed point of the system when k=6(U*)² + cU*?

(you mean " - cU* " )

that's a strange way of putting it, but yes …

so what is k ? ​

 

[BrainHurts]

yes haha sorry, k is some arbitrary real number, but for (U,V)=(0,0) being a fixed point implies that

V*=0

and 6U*²-cU*-k=0

this gives us

U*= $\frac{c\pm\sqrt{c^{2}+24k}}{12}$, but we want U*=0 which means that means that k=0 right?

$0=\frac{c\pm\sqrt{c^{2}-24k}}{12} \Rightarrow -c=\pm\sqrt{c^{2}-24k} \Rightarrow c^{2}=c^{2} - 24k$ and we get k = 0

 

[Dick]

yes haha sorry, k is some arbitrary real number, but for (U,V)=(0,0) being a fixed point implies that

V*=0

and 6U*²-cU*-k=0

this gives us

U*= $\frac{c\pm\sqrt{c^{2}+24k}}{12}$, but we want U*=0 which means that means that k=0 right?

$0=\frac{c\pm\sqrt{c^{2}-24k}}{12} \Rightarrow -c=\pm\sqrt{c^{2}-24k} \Rightarrow c^{2}=c^{2} - 24k$ and we get k = 0

Sure, k=0. You didn't really need the quadratic equation to draw that conclusion, but ok.

 

[BrainHurts]

Ok going on to the next part of this question

c) Find the equations for the orbits of the system, these are the trajectories in the UV-plane determined by functions of the form V=V(U) and show that a first integral of the orbit equation is

V²=2kU + cU²-4U³+h

So in doing so I got

$\frac{dV}{dU}=\frac{\frac{dV}{dz}}{\frac{dU}{dz}}=\frac{k+cU-6U^{2}}{V} \Rightarrow VdV = (k+cU-6U^{2})dU$ and this gives us the desired result.

So one place where I read what orbits are are objects like V(U), V(V(U))... so I'm not really sure when to stop.

 

[BrainHurts]

nm found some good reading on this

http://math.colorado.edu/~jkeller/math4430fall10/Orbits.pdf