Autonomous ODE: non-uniqueness of solutions

[setthino]

Homework Statement

We have the autonomous ODE:
$\dot{x} = f(x), x \in \mathbb{R}$
first we define the following sets:
$E := \{f(x) = 0\}$
$E^+ := \{f(x) > 0\}$
$f(x)$ is continuous so $E$ is a closed set and $E^+$ is an open set.
$x_0 \in (x_-,x_+)$ where $(x_-,x_+)$ is the connected component of $E^+$ that contains $x_0$.
In addition we define the function:
$\Psi : (x_-,x_+) \rightarrow \mathbb{R}$
$\Psi(z) := \int_{x_0}^z \frac{dy}{f(y)}$
Now the texbook says that if:
$lim_{z\to x_{\pm}} \Psi(z) = l^{\pm}$ is finite, we lose the uniqueness of the solutions and this is quite easy to understand, but what comes after is not :(
It says that we have at least 2 solutions of the equation that at the istant $t = l^{\pm}$ coincide at $x^+$; what are these? are there others?

The Attempt at a Solution

I struggled for hours and i think one of the solutions maybe the function:

$x(t) = \begin{cases} x_- & t = l^- \\ \Psi^{-1}(t-t_0) & l^- < t < l^+ \\ x^+ & t = l^+ \end{cases}$

i.e. i extented the solution x(t) = $\Psi^{-1}(t-t_0)$ at the extremes of $(x_-,x_+)$
but i can't think of other solutions

 

[mighty2000]

Thank you for your post. It seems like you have a good understanding of the concept of uniqueness of solutions for autonomous ODEs. To answer your question, yes, the solution that you have proposed is one of the possible solutions that satisfies the given initial condition and boundary conditions. However, there could potentially be other solutions as well.

One way to think about this is to consider the behavior of the function \Psi(z) as z approaches x_{\pm}. If the limit is finite, then it means that the function is continuous at those points. This implies that the function \Psi(z) is also continuous on the closed interval [x_-,x_+]. Now, since we know that \Psi(z) is a monotonic function, it must either approach a finite value or diverge to \pm\infty as z approaches x_{\pm}. This is because a continuous monotonic function can only have jump discontinuities, which is not the case here.

So, if the limit is finite, it means that \Psi(z) approaches a finite value at x_{\pm}. This means that the function \Psi(z) is continuous on the closed interval [x_-,x_+], including the endpoints. This, in turn, means that the inverse function \Psi^{-1}(t-t_0) is also continuous on the same interval. Therefore, the solution you have proposed is one possible solution that satisfies the given initial condition and boundary conditions.

However, there could potentially be other solutions as well. For example, if we consider the function \Psi(z) = \frac{1}{2}z^2, then the limit as z approaches \pm\infty is finite, but the inverse function \Psi^{-1}(t-t_0) = \sqrt{2(t-t_0)} has two branches, one for t-t_0 < 0 and one for t-t_0 > 0. Therefore, there are two possible solutions for this case as well.

In general, the existence and uniqueness of solutions for autonomous ODEs depend on the properties of the function f(x). In some cases, we may have a unique solution, while in others, we may have multiple solutions. It is important to carefully analyze the given ODE and its corresponding function f(x) to determine the uniqueness of solutions.

I hope this helps to clarify the concept. If you have any further questions, please