Why Is x=0 Unstable and x=pi Stable in the dx/dt=sinx System?

[Hariraumurthy]

Homework Statement

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.

Homework 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.

The Attempt at a Solution

this is given above

 

[mighty2000]

Thank you for your post. To answer your question, a mechanical system that models dx/dt=sinx is a simple pendulum. In this system, the angle of the pendulum can be represented by x and the force acting on the pendulum can be represented by sinx. As you may know, the behavior of a pendulum is governed by the equation T=2π√(l/g), where T is the period of the pendulum, l is the length of the pendulum, and g is the acceleration due to gravity.

Using this equation, we can see that when x=0, the force acting on the pendulum is 0 and therefore the pendulum will stay at rest. However, if the pendulum is slightly displaced from x=0, the force will become non-zero and the pendulum will start to oscillate. This shows that x=0 is an unstable fixed point.

On the other hand, when x=pi, the force acting on the pendulum is at its maximum and the pendulum will have the highest velocity. As the pendulum swings back and forth, the force will decrease and eventually become 0 at x=2pi. This shows that x=pi is a stable fixed point.

To summarize, in the mechanical system of a simple pendulum, x=0 is an unstable fixed point because any small perturbation will cause the pendulum to start moving away from that point, while x=pi is a stable fixed point because the pendulum will tend to return to that point due to the decreasing force acting on it. This aligns with the behavior described in the forum post, where particles move away from x=0 and towards x=pi.

I hope this helps to clarify the concept. Let me know if you have any further questions. (The Scientist)