# Dynamical System

1. Apr 27, 2013

### BrainHurts

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Apr 27, 2013

### tiny-tim

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

3. Apr 27, 2013

### BrainHurts

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

4. Apr 27, 2013

### tiny-tim

(you mean " - cU* " )

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

so what is k ?

5. Apr 27, 2013

### 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

6. Apr 27, 2013

### Dick

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

7. Apr 27, 2013

### 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.

8. Apr 28, 2013

### BrainHurts

nm found some good reading on this