Consider the following ODE-system \begin{align} \dot{x}(t)&=f(x(t),z(t))\\ \dot{y}(t)&=g(x(t),y(t),z(t)) \\ \dot{z}(t)&=h((x(t),y(t),z(t))) \end{align} with the following initial/terminal conditions: $\lim\limits_{t\rightarrow \infty} x(t) = x_\infty, y(0)=y_0, z(0)=0$. Moreover, it is known that $\lim_{t\rightarrow \infty}y(t)=0$ and $\lim_{t\rightarrow \infty} z(t)=z_\infty>0$. The solution of the system must therefore converge to a point $(x_\infty,0,z_\infty)$, where $x(\infty)$ is known but $z(\infty)$ is not. Assume that $f,g$, and $h$ are Lipschitz continuous.

Under which conditions does this system have a unique solution? I understand that if I had an initial condition for $x(0)$, standard theorems like Picard-Lindeloef would apply. However, without this initial condition but a boundary condition at infinity I do not know whether there exist theorems to apply.


1 Answer 1


I'll assume that $f,g,h$ are Lipschitz continuous so you have local existence and uniqueness.

In general it is unlikely that there are any solutions to your problem, but there are some cases where this will happen. Since you don't say what $y$ and $z$ are supposed to be doing as $t \to \infty$ while $x \to x_\infty$, there are many scenarios, but the simplest is that there is an isolated fixed point $P = (x_\infty, y_\infty, z_\infty)$ that is the limit of your solution as $t \to \infty$. The next question is: what is the dimension of the stable manifold for that fixed point?

If that dimension is $0$, the only way to approach it is to start there, i.e. you must have $y_0 = y_\infty$ and $0 = z_\infty$, and the unique solution is the constant solution $x = x_\infty, y=y_\infty, z = z_\infty$.

If the dimension is $1$, there are two trajectories approaching $P$ from opposite directions, and your problem has a unique solution if one of these happens to intersect the line $y=y_\infty, z= 0$ in exactly one point and the other does not intersect it.

If the dimension is $2$, i.e. the stable manifold is a surface, you want that surface to intersect the line $y=y_\infty, z=0$ in exactly one point. This may be the most likely scenario.

If the dimension is $3$, you won't have a unique solution: the set of initial conditions that end up approaching the fixed point is an open set, and if it intersects your line $y=y_\infty, z=0$ it does so in an open interval.

  • $\begingroup$ Thank you for your reply - I added the Lipschitz continuity to the question. Regarding your next point: I know that $\lim_{t\rightarrow \infty} y(t) = 0$, $z(t)$ also converges to some positive number $\lim_{t\rightarrow \infty} z(t)=z(\infty)>0$. Finally, I am not aware of the concept of stable manifolds - could you please point my to a reference or briefly let me know what that means in this context? $\endgroup$
    – ctmntlds
    Nov 13, 2020 at 15:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .