Using a curve as an attracting manifold

28 Apr, 2026

This is a math post.

I was wondering how easy it is to build a dynamical system with a user-defined attracting curve, e.g., a sine wave. You pick a point and if it starts on the curve, then it remains there, otherwise it smoothly gets attracted to the curve as $t$ increases.

I don't recall where I first saw this, but I think it's a standard construction.

Let's say $f:ℝ\to ℝ$ is our function and the curve of interest is $(x,y)$, where $y=f(x)$. Since we want to build a dynamical system, we will assume that $x$ and $y$ depend on $t\ge 0$. To simplify matters, we assume $x^{\prime }=1$, i.e., constant speed on the x axis and movement from left to right. Now we just need to find the right behavior for $y$.

How to get $y$ to behave the way we want?

We can first define $u:=y(t)-f(x(t))$. We would want $u(t)$ to be decreasing when $u(t)>0$ and increasing when $u(t)<0$. An easy way to do this: fix a $k>0$ and require that

which is an ODE with solution $u(t)=\exp (-kt)u(0)$. This immediately gives

As time passes, $y(t)$ "forgets" the initial deviation and behaves like $f(x)$. How quickly that happens depends on the magnitude of $k$.

And so we indeed have a deviation that keeps

with target manifold $y=f(x)$ (here $f(x)=\sin x+\frac{1}{2}\sin (2x)$).

In the plot above, the white curve is $f$. The rest of the colorful curves are the trajectories of random points -- over time they converge to $f(x)$ (as they should, by construction). You can also see the vector field as little arrows -- this shows how a point should move over time.

Extending to variable gain $k(x)$

We can also let $k$ be a function of $x$:

The same deviation variable satisfies

so

which means:

In this case the memory of the initial deviation does not have to go away, i.e., we could pick $k$ so that we never quite stabilize.

Extending to generic curves in ${ℝ}^n$

As long as the curve is reasonable (smooth and without sharp corners), we can define a vector field that has the curve as a stable manifold. That more or less follows the previous ideas.

Find the manifold given the vector field

What if we have a vector field $F:{ℝ}^n\to {ℝ}^n$ and we want to find the attracting manifold? We will assume $F$ is a smooth vector field, e.g., $C^1$.

First of all, given a point $z_0\in {ℝ}^n$, $z$ will evolve according to the ODE

This uniquely identifies the trajectory $z(t)$.

We now want to find $u:{ℝ}^n\to ℝ$ such that $u(z)=0$ can identify the attracting manifold. To do this, we first set $u$ to behave like before on the trajectories, i.e., pick $k>0$ and

We would like to incorporate $F$ in this (cause it controls how the trajectories move). If we use chain rule on $du/dt$ (and since $z^{\prime }=F$) and re-arrange:

This is a transport PDE for $u$! We can solve it with the method of characteristics. The solution is

where $\xi$ identifies the characteristic we are on and $\Phi$ is some arbitrary function, e.g., $\Phi (\xi )=\xi$.

At this point, I asked ChatGPT to write the TeX for the previous example, see below:

Example from before

For our vector field (from before), the characteristics satisfy

Since $\dot{x}=1$, we can use $x$ as the time variable. Then

Define the residual along the characteristic,

Then

So

Equivalently,

This quantity labels the characteristic. Therefore the general solution of the transport PDE is

where $\Phi$ is arbitrary.

Choosing $\Phi (\xi )=\xi$ gives

and therefore the attracting manifold is