Particle Motion Analysis

The motion of the particle in one dimension is described by the velocity function:

\( \displaystyle v(x) = \frac{dx}{dt} = v_0 \sin \left( \frac{\pi x}{x_0} \right) \)

where \(v_0\) and \(x_0\) are positive constants.

Finding Fixed Points

Fixed points are positions where the velocity is zero, meaning the particle would stay there if placed at that point. We find these by setting \(v(x) = 0\):

\( v_0 \sin \left( \frac{\pi x}{x_0} \right) = 0 \)

Since \(v_0\) is a positive constant, we must have:

\( \sin \left( \frac{\pi x}{x_0} \right) = 0 \)

This occurs when the argument of the sine function is an integer multiple of \(\pi\):

\( \frac{\pi x}{x_0} = n\pi \)

where \(n\) is an integer (\(n = 0, \pm 1, \pm 2, \ldots\)). Solving for \(x\), we get the fixed points:

\( x_n = nx_0 \)

So the fixed points are at \(x = 0, x_0, -x_0, 2x_0, -2x_0, \ldots\).

Analyzing Stability of Fixed Points

To understand the long-time behavior, we need to determine whether these fixed points are stable or unstable. A stable fixed point attracts nearby particles, while an unstable fixed point repels them.

We can analyze stability by looking at the sign of the velocity \(v(x)\) in the regions around each fixed point or by calculating the derivative of \(v(x)\) with respect to \(x\), \(dv/dx\), at each fixed point.

The derivative is:

\( \displaystyle \frac{dv}{dx} = \frac{d}{dx} \left( v_0 \sin \left( \frac{\pi x}{x_0} \right) \right) = v_0 \left( \frac{\pi}{x_0} \right) \cos \left( \frac{\pi x}{x_0} \right) \)

Evaluating this at the fixed points \(x_n = nx_0\):

\( \left. \frac{dv}{dx} \right|_{x=nx_0} = v_0 \left( \frac{\pi}{x_0} \right) \cos (n\pi) = v_0 \left( \frac{\pi}{x_0} \right) (-1)^n \)

For a fixed point to be stable, \(dv/dx\) at that point must be negative. For it to be unstable, \(dv/dx\) must be positive.

• At \(x=0\) (\(n=0\)): \( \left. \frac{dv}{dx} \right|_{x=0} = v_0 \left( \frac{\pi}{x_0} \right) (-1)^0 = v_0 \frac{\pi}{x_0} \). Since \(v_0, x_0 > 0\), this is positive. Thus, \(x=0\) is an unstable fixed point.
• At \(x=x_0\) (\(n=1\)): \( \left. \frac{dv}{dx} \right|_{x=x_0} = v_0 \left( \frac{\pi}{x_0} \right) (-1)^1 = -v_0 \frac{\pi}{x_0} \). This is negative. Thus, \(x=x_0\) is a stable fixed point.
• At \(x=2x_0\) (\(n=2\)): \( \left. \frac{dv}{dx} \right|_{x=2x_0} = v_0 \left( \frac{\pi}{x_0} \right) (-1)^2 = v_0 \frac{\pi}{x_0} \). This is positive. Thus, \(x=2x_0\) is an unstable fixed point.
• At \(x=-x_0\) (\(n=-1\)): \( \left. \frac{dv}{dx} \right|_{x=-x_0} = v_0 \left( \frac{\pi}{x_0} \right) (-1)^{-1} = -v_0 \frac{\pi}{x_0} \). This is negative. Thus, \(x=-x_0\) is a stable fixed point.

The stable fixed points are at odd multiples of \(x_0\) (\(x = x_0, -x_0, 3x_0, -3x_0, \ldots\)), and the unstable fixed points are at even multiples of \(x_0\) (\(x = 0, 2x_0, -2x_0, \ldots\)).

Particle's Long-Time Behavior

The particle starts at \(x/x_0 = \varepsilon\), where \(|\varepsilon| \ll 1\). This means the initial position \(x = \varepsilon x_0\) is very close to the unstable fixed point at \(x=0\).

An unstable fixed point repels nearby trajectories. If the particle starts exactly at an unstable fixed point, it stays there. However, any tiny deviation will cause it to move away.

• If the particle starts slightly to the right of \(x=0\) (i.e., \(\varepsilon > 0\), so \(x = \varepsilon x_0\) is small and positive), it is in the region \(0 < x < x_0\). In this region, \(0 < \pi x/x_0 < \pi\), so \(\sin(\pi x/x_0) > 0\). This means \(v(x) > 0\), and the particle moves towards increasing \(x\). The nearest stable fixed point in this direction is \(x_0\).
• If the particle starts slightly to the left of \(x=0\) (i.e., \(\varepsilon < 0\), so \(x = \varepsilon x_0\) is small and negative), it is in the region \(-x_0 < x < 0\). In this region, \(-\pi < \pi x/x_0 < 0\), so \(\sin(\pi x/x_0) < 0\). This means \(v(x) < 0\), and the particle moves towards decreasing \(x\). The nearest stable fixed point in this direction is \(-x_0\).

The question asks about the long-time behavior starting near \(x=0\) (\(|\varepsilon| \ll 1\)). Since \(x=0\) is unstable, the particle will move away from it towards a stable fixed point. Given the options and the nature of stable points at \(x_0, -x_0\), etc., the particle will tend towards the nearest stable point depending on which side of the unstable point \(x=0\) it starts.

The correct option states the particle tends towards \(x = x_0\). This implies the particle must start in the basin of attraction for \(x_0\), which, for a starting point very close to \(x=0\), means it starts slightly to the right of \(x=0\) (i.e., \(\varepsilon > 0\)). In this case, the velocity is positive, and the particle moves towards \(x_0\), which is a stable fixed point. As it approaches \(x_0\), the velocity decreases, and in the long time, the particle tends towards \(x_0\).

Thus, starting near the unstable point \(x=0\), if the initial perturbation is such that the particle moves into the region \(0 < x < x_0\), it will be attracted to the stable fixed point \(x=x_0\).

The final answer is tends towards x = x₀.