Proof of Uniqueness + Problems on p. 38

In the spirit of understanding complex and abstract concepts as simply as possible, I am going to attempt recreating/re-explaining Arnold’s proof of the uniqueness of the solution for $\dot x = v(x)$ with a smooth RHS.

We suppose $\phi_0(t) = x_0$ an equilibrium position and $\phi(t)$ is a solution that satisfies $\phi(t_0) = x_0$ and $\phi(t_1) = x_1$ is not an equilibrium position. What is going to go wrong?

To start, we need to consider the equilibrium position that is closest to $\phi(t_1) = x_1$ and this might not be $x_0$ . Let the closest one be $\phi(t_2) = x_2$ . We’re going to show that it is going to infinitely long for a particle to travel from $x_1$ to $x_2$ (Note: recall that the particle would be moving along a 1-dim. line and $x_1$ and $x_2$ are two points on that line).

Why?

The key point is that the slope of each tangent line at each point of the integral represents the velocity of the particle. So, as the particle gets closer to $x_2$ , its velocity is also slowing down to 0 (and this slowing down is going to be faster than the rate at which it is approaching $x_2$ so that it’s never going to reach it in finite time).

The difference in time comes from Barrow’s formula:

Now, since $\latex v(x)$ is smooth, within a neighbourhood of $\latex x_2$, its slope is bounded:

i.e. there exists k such that $|v(\xi)| < k|\xi - x_2|$ . Hence,

We can always compare this motion with a faster linear motion. But since the linear motion is going to take infinite time, this motion will also take infinite time.

First, we need to ask what does it mean for something to converge faster than exponentially?

Given two integral curves $f_1(t)$ and $f_2(t)$ . My understanding is that eventually, for some large M, for all t > M, $|f_1(t) - f_2(t)| < k*e^{-mt}$ (where k, m are positive integers).

When one of the curves is an equilibrium position, like Arnold had said previously, the motion of the a particle described by the second integral curve can be compared to a suitable and more rapid motion in a linear field, and we know that the motion in the linear field converges exponentially.

But now, if both integral curves are not equilibrium positions, in what situation will they converge to each other faster than exponentially? Note that the RHS of $\dot x = v(x)$ is smooth.

I had originally came up with this field of integral curves:

Here, the curves get mapped to each other under translation along the t-axis and they converge to each other faster than exponentially. However, if we are to express this field using an autonomous equation, the RHS will not be smooth due to the degree 3. And this will be true as long as we try to decide to have a polynomial in the exponent.

But, what if we are to do something like the one below?

If we are to restrict our equation to x > 0, then we would have $\dot x = -ln(x)x$ .

Key point: as long as we can compare the motion to a suitable motion in a more rapid linear field, then we’re good.

Since the derivative of v exists on the interval of definition, then in the neighborhood of x₂, there exists k>0 such that $|v'(\xi)| < k$ for all $\xi$ in the neighborhood. Hence, it follows that $|v(\xi)| < k|\xi - x_2|$ . We can compare the motion of the particle approaching x₂ to a linear motion.

By Barrow’s formula,

$t - t_1 = \int_{x_1}^{x} \frac{1}{v(\xi)} \mathrm{d}\xi$ , where $\phi(t_1) = x_1$ .

Suppose $\phi(t_0) = x_0$ , then as x₁ approaches x₀, $t - t_1 = \int_{x_1}^{x} \frac{1}{v(\xi)} \mathrm{d}\xi$ goes to infinity. It takes infinite time for the particle to reach x₀.

Suppose v(x₀) = 0, then for any $\xi \in I$ (where I is the interval on which v(x) is defined),

$|v(\xi)| \leq k|\xi - x_0|$ , which is what we want.

The underlying concept is that because of smoothness, we can locally see $\bar{\phi_1}(t)$ as a solution of an autonomous equation (that is, we can find a local inverse for $\bar{\phi_1}(t)$ ).

Proof of Uniqueness + Problems on p. 38

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