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Nonlinear Dynamics Strogatz(m |
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| Jul8-11, 06:10 PM | #1 |
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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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