Differential equation, limit cycles, stability, phase plane. By 11/12

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Homework Help Overview

The discussion revolves around a system of differential equations describing the trajectory of an arrow in a phase plane. Participants are tasked with deriving an ordinary differential equation (ODE) for the radial coordinate and analyzing limit cycles and their stability.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive an ODE for the radial coordinate and analyze stability points. Some participants question the correctness of the equations and the independence of certain terms. Others discuss the implications of the derived ODE and the behavior of the system near critical points.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into stability analysis and the behavior of the system. There is a recognition of potential typographical errors in the equations, which has led to further clarification and confirmation of results.

Contextual Notes

There are noted typographical errors in the original equations that may affect the analysis. Participants are also considering the implications of these errors on the stability and behavior of the system.

debian
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Homework Statement



The trajectory of an arrow in space obeys the following system of equations:

\dot{x} = y+(x^2+y^2-3)^2 (x^3-x+xy^2)

\dot{y} = y+(x^2+y^2-3)^2 (y^3-y+x^2y)

1. Questions

a) Derive an ODE for the radial coordiante r(t) = \sqrt[]{x^2(t)+y^2(t)}

b) Show that the system admits two limit cycles and classify their stability

c) The target is standing at the origin (0,0) of the phase plane. Let Po=(x(0),y(0))\neq (0,0) denote the initial position of the arrow at the time t = 0. Determine and plot the region in the (x,y) plane containing all the points Po such that the arrow will hit the target at some instant of time t > 0


The Attempt at a Solution



I derived ODE in radial coordiante and got:

\dot{r}=(r^2-3)^2(r^2-1)r

So we have 5 points to check the stability: 0, 1,-1,\sqrt{3},-\sqrt{3}

Than I calculate the derivative of \dot{r} and evaluate in those 5 points: I get five zeros so they are undecided.

Then I daw the plot of \dot{r} and if the \dot{r}>0 it does mean that arrow on the phase line is to the right and if \dot{r}<0 the arrow is to the left.

So The phase line looks like this: <- (-\sqrt{3}) <- (-1) -> (0) <- (1) -> (\sqrt{3}) ->

Is that correct? What should I do next?
 
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debian said:

Homework Statement



The trajectory of an arrow in space obeys the following system of equations:

\dot{x} = y+(x^2+y^2-3)^2 (x^3-x+xy^2)

\dot{y} = y+(x^2+y^2-3)^2 (y^3-y+x^2y)

Are you sure these are correct? Because they tell me that
r\dot r = x\dot x + y\dot y
will not be independent of \theta, because the linear terms give me
xy + y^2 = r^2(\cos\theta\sin\theta + \sin^2\theta).

Assuming that you are correct that
\dot r = r(r^2 - 3)^2(r^2 - 1) = f(r)
then since by convention r \geq 0, there is a fixed point at the origin and two orbits, at r = \sqrt 3 and r = 1.

To determine the stability, you look at the sign of f&#039;(r). If it's negative then the point is stable, and if it's positive then the point is unstable. If it's zero then it's inconclusive and you have to look at higher derivatives.

Alternatively, in one dimension r is increasing if \dot r &gt; 0 and decreasing if \dot r &lt; 0. By inspection, \dot r &lt; 0 if 0 &lt; r &lt; 1 and \dot r &gt; 0 when 1 &lt; r &lt; \sqrt 3 and r &gt; \sqrt 3. So the origin is stable, the orbit at 1 is unstable, and the orbit at \sqrt 3 is stable from below and unstable from above.

For part (c) you need to think about what trajectories eventually end up at the origin.
 
There is a typo: shoul be -x instead of one of y

\dot{x} = y+(x^2+y^2-3)^2 (x^3-x+xy^2)

\dot{y} = -x+(x^2+y^2-3)^2 (y^3-y+x^2y)
 
debian said:
There is a typo: shoul be -x instead of one of y

\dot{x} = y+(x^2+y^2-3)^2 (x^3-x+xy^2)

\dot{y} = -x+(x^2+y^2-3)^2 (y^3-y+x^2y)

Then your result for \dot r is correct.
 

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