# Integral Curves of Vector Fields with Zero Divergence or Zero Curl

Say we've got some vector field which at every point indicates the instantaneous velocity of a particle moving through that point.

I'm trying to gain some intuition for what the possible trajectories for particles would look like in the cases that this vector field has zero divergence, zero curl, or both...

There's GOT to be something special about the trajectories in such vector fields... I'm still having trouble wrapping my head around divergence and curl...although, I think intuitively understanding possible trajectories will help.

As for my background...I have not taken differential equations, and am just about done with an introductory course on multi-variable calculus.

Thanks again!

Edit:

After receiving some answers and hints, I'd like to write out some of my thoughts...

Say we have the vector-field $$\vec{F} = (2x,-2y)$$. This vector-field has both zero divergence and zero curl.

There are many possible ways to interpret such a vector-field: ie, an acceleration, a force...etc...but, lets focus on these two interpretations:

First, lets consider it the gradient of a function. In this case, the function would be $$f(x,y)=x^2-y^2$$, and thus $$\nabla(f)=\vec{F}$$.

As I understand it, the fact that $$\vec{F}$$ has no curl means that it can be the gradient of a function in the first place, because a line-integral in a closed circle is zero.

As for what it means for the divergence to be zero everywhere...well, the divergence of the gradient is the Laplacian of the original function. It means that the function is harmonic, so that at each point, the "bending" is equal in all directions. If this function represented some stretched surface, no point would have any reason to bend, because the force on any point would cancel out from the points around it.

Those are both beautiful and intuitive results...but, that wasn't what my question was driving it.

I want to interpret $$\vec{F}$$ as a velocity-field. That is, at each point, $$2x=\frac{dx}{dt}$$ and $$2y=\frac{dy}{dt}$$.

In that case, integral curves (flow-lines) starting from some point $$(x_0,y_0)$$ would look like:

$$(x_0e^{2t}, y_0e^{-2t})$$

And as for those flow-lines...I have absolutely no intuition what's special about them, coming from the fact that the divergence is zero, or that the curl is zero. These are what I wish to understand!

Thanks!

Here are a few things for you to prove to yourself:

(1) If $$\vec F$$ is conservative (i.e., a gradient field), then the flow lines (these are your trajectories) cannot be closed curves. Why? Could I deduce from this that if $$\text{curl }\vec F = \vec 0$$, then $$\vec F$$ has no closed flow lines?

(2) If you're in the plane and $$\text{div }\vec F>0$$ (or, similarly, $$<0$$) in a region, then once again $$\vec F$$ has no closed flow lines in that region.

small HINTS:

(1) Fundamental Theorem of Calculus for line integrals. (2) Divergence Theorem in two dimensions. For my follow-up question in (1), remember that if $$\vec F$$ is (continuously differentiable and) conservative, then $$\text{curl }\vec F = \vec 0$$. But there is a famous example where the converse of this fails.

• Hi Ted! Thanks again for answering another of my questions! Hmm...okay, well, for number 1...the gradient of a function points in the local uphill direction. If that gradient represents a velocity, then that means that at any point, you're always moving uphill on the function we got the gradient from. We can't go around in a closed loop and end up at the same point if we're always moving uphill. However...that only shows that if F is a gradient, it has no closed flow lines...as to why it leads to the more general idea that the curl being zero means no closed flow lines...I'm not yet sure... Jun 2 '20 at 21:40
• As for the second question...if the divergence is positive everywhere in the region, that means there must be a preferred direction in which the velocity is increasing...hmm, I guess we could once again think of the function from which $\vec{F}$ came from. It means that everywhere in a region, there's more "uphill" than "downhill". So, if the instantaneous velocity is "uphill" everywhere in the region...I'm not sure where to go from here... Jun 2 '20 at 21:44
• I think part of what's confusing for me is that when I think of vector fields, I'm thinking of rates of change of each of the coordinates with respect to time (hence, velocity fields) while when we think of gradients, its the slope of some other coordinate ($z$) with respect to $x, y$ (as an example)...and, the distinction is confusing. Jun 2 '20 at 21:46
• You're confusing matters (perhaps this happened the last time?) when you think about "uphill." That is with reference to the graph of the function. I've edited with some gentle hints. Jun 2 '20 at 22:12
• I've update my question a bit, I think I've explained my confusion, and the specifics of what I want to understand a little better now. As for the hints...okay, from the fundamental theorem, that means that the curl of any gradient is zero, right? Because we can't go around in a closed-loop, integrate the gradient, and have a non-zero integral, since that would break the vertical line test and the function wouldn't be a function to begin with. As for your second hint....😔...I'm not sure what to do with it! Jun 3 '20 at 16:03

This is more of a long comment than an answer:

For the divergence free case, there is an important consecuente, although is not a property of any integral curve in particular, but of them as a whole. Lets consider $$\Psi(t,x)$$ the flow of your vector field, that is $$\partial_t \Psi(t,x) = u(\Psi(t,x))$$ $$\Psi(0,x) = x$$ so if we fix $$x$$, $$t\rightarrow \Psi(t,x)$$ is the integral curve of $$u$$ starting at $$x$$. Then the condition of $$u$$ being divergence free is equivalent to $$\Phi(t,\cdot)$$ being measure presserving, for every $$t$$ where it is defined.

To see this we can calculate the volume of $$V_t = \Psi(t, \cdot) V$$ for an arbitrary measurable set $$V$$. Using the change of variables formula $$\mathop{vol}(V_t) = \int_{\Psi(t, \cdot) V} dx = \int_V \det(D_x\Psi(t,x))dx,$$ so $$\Psi$$ would be volume presserving if and only if (under suitable regularity assumptions) $$\det(D_x\Psi(t,x)) = 1.$$ Differentiating with respect to $$t$$ this is equivalent to $$\mathop{div} u = tr(D u(t,x)) = 0$$

Regarding to the zero curl case, i would just ad that regardless of the topology, we can always say that at least locally $$u$$ is the potential of some function $$\phi$$ (because we have simply connected neighbourhoods), alternatively we can find a global potential but it will be multi valued.