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Nonlinear Dynamics Strogatz(m

 
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Jul8-11, 06:10 PM   #1
 

Nonlinear Dynamics Strogatz(m


1. The problem statement, all variables and given/known data
2.1.5(A mechanical Analog)
a) Find a mechanical system approximately governed by dx/dt=sinx
b)Using your physical intuition explain why it becomes obvious that x*=0 is an unstable fixed point and x*=pi is a stable fixed point.
(note*this is exactly how it appears in strogatz)

I have given a solution using the flow on the line below.

2. Relevant equations
If you plot dx/dt on the axis normally labled y and x on the x-axis as usual, if you take x=0.1, sinx will be greater than 0 and therefore dx/dt will be greater than 0 and therefore if a particle starts from x=0.1, it will move more to the right.

Likewise if a particle starts from x=-0.1, sin(x) will be negative and therefore dx/dt will be negative and therefore the particle will move to the left.

Therefore the particle will move away from x=0 in both directions.

But if you take x=pi, a little before x=pi, sin(x) will be positive and therefore dx/dt is greater than 0 and therefore from the left of x=pi the particle will move to the right.

If you take a point a little to the right of x=pi like x=3.2, then sin(x) will be negative, dx/dt will be negative and the particle will move to the left(from the right.

The particles movement will be as follows(this follows from what I said before)
------<--0-->---->--pi--<----<---2pi--->-->- ect.

to sum up, from the picture it is clear that for x around 0, the particle will go away from x=0 and hence x=0 is unstable and is therefore an unstable fixed point(it is still a fixed point because sin(x)=0 here and therefore dx/dt=0 here. If a particle was to be exactly at x=0, it would stay there until the slightest disturbance came.

for x around pi, the particles are moving towards x=pi.

Therefore without the mechanical analog, it is quite clear that x=0 is an unstable fixed point and x=pi is a stable fixed point.

MY QUESTION IS WHAT IS A SYSTEM THAT MODELS THIS AND USING THIS HOW DOES IT BECOME OBVIOUS THAT x*=0 is an unstable fixed point and x*=pi is a stable fixed point.


A more quantitative approach to this that you are probably familiar with is if you have a system dx/dt=f(x),

for a small pertubation N, if Nf'(x*)(x* is a fixed point)>0, then x* is unstable. If Nf'(x*)<0, then x* is a fixed point.


3. The attempt at a solution

this is given above
 
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