For reference, I'm working from Ordinary Differential Equations, Third Edition by Arnol'd (page 37, Chapter 2, Section 2, "A Counterexample"). Arnol'd proposes the following theorem:
Theorem. Let $v$ be a smooth (continuously differentiable) function defined on an interval $U$ of the real axis. The solution
$\varphi$ of the equation $\dot{x} = v(x)$ with initial condition
$(t_0,x_0)$
- exists for all $t_0 \in \mathbb{R}$ and $x_0 \in U$;
- is unique in the sense that any two solutions with the same initial condition coincide in some neighborhood of the point $t_0$;
- is given by the formula:
$$t – t_0 = \int_{x_0}^{\varphi(t)} \frac{d\xi}{v(\xi)}\text{, if } v(x_0) \neq 0\text{; or }\varphi(t) \equiv x_0\text{, if }v(x_0) = 0.$$
He then goes on to give $v = x^{2/3}$ as a counterexample, stating that,
The two solutions $\varphi_1 = 0$ and $\varphi_2 = (t/3)^3$ both
satisfy the initial conditions $(0, 0)$, contrary to the assertion of
uniqueness . . . This example can be described as follows: under motion with velocity
$v(x) = x^{2/3}$ the equilibrium point $x = 0$ can be reached from
another point in finite time.
My question is: When he says, "The equilibrium point $x = 0$ can be reached from another point in finite time," what does he mean?
I have figured out that because $v = x^{2/3}$ is not differentiable at $x = 0$, it does not satisfy the smoothness hypothesis of the theorem, and that this means solutions to $v$ may not be unique. I have verified that solving the integral from the theorem using $v$, one can obtain either $\varphi_1$ or $\varphi_2$ as a valid solution. I think what is confusing me is what it physically means for two solutions to fail to be unique.
What I think Arnol'd means is that if a particle starts, for example, at $x = 0$ when $t = -1$ then, per solution $\varphi_1 = 0$, the particle will arrive at the point (0,0) one second later OR ALSO if the particle starts at $x = -\frac{1}{9}$ at $t = -1$, then per solution $\varphi_2 = (t/3)^3$, that particle will likewise arrive at (0,0) a second later. Therefore the solutions are not unique in that they "cross". Is that correct? If it is correct, does it mean that the particle following $\varphi_2$ will remain at $x = 0$ from then on? Like, will it slow to a gentle halt at $x = 0$ and then rest there for all time? Or will that particle continue past the equilibrium position per the equation $\varphi_2 = (t/3)^3$?
(Incidentally, is it OK to think of an "initial condition" as "any space/time pair that the particle will occupy at some point" since, in my example, the "initial condition" (0,0) is actually where the particles end and not where the particles "initially" start?)
I apologize for the long and detailed question! I am loving this book and this topic, but I have been stuck on this example for days at this point. I will appreciate any guidance!
Best Answer
Yes, your idea is right. If the particle moves on the curve $\varphi_1$ or $\varphi_2$, it will manage to reach the equilibrium point $x = 0$ in finite time. Note that this does not mean that the particle which moves on $\varphi_2$ remains at $x = 0$. For $t> 0$, it follows the path given by $\varphi_2$ (of course you can find more solutions for the ODE which can describe a modified movement).
Maybe it is also useful to look at the example $v(x) = -x$ which he also describes. Here $v$ tends to $0$ fast enough when $x$ tends to the equilibrium point $0$. Since the velocity is very small for $x$ sufficiently near to $0$, the solution curve $x(t) = \mathrm{e}^{-t}$ (I assume $(t_0,x_0)= (0,1)$) is unique.
I think that term "initial condition" comes from the point that you often know the start position of your object at time $t_0$, say $x(t_0) = x_0$. In many cases as in this one, it is correct that you consider it as one point which the particle goes through on its way.