A vector field satisfies both $\nabla\cdot v=0$ and $\Vert v \Vert = 1$

Question

It is trivial to satisfy just one of them. In 2D for example ($r = \sqrt{x^2 + y^2}$ below), $\nabla\cdot(x,-y) = 0$. However, $\Vert v \Vert = r$, so only the divergence condition is satisfied. On the other hand for $(x/r, y/r)$, we have $\Vert v \Vert = 1$ but $\nabla\cdot v=1/r$ instead of 0.

Note that the first one is a differential constraint, whereas the second one is only an algebraic constraint. Is there a standard name for this kind of problem that involves a mixture of both algebraic and differential operations?

Edit: A uniform unit vector field is excluded because it's trivial.

Edit: Solution in 2D

Thanks to the paper @LuisFerreira pointed to me, I can construct a 2D field that satisfies the requirement, except at some singular points. The proof in the paper is beyond me so unfortunately I can not explain why such singularity exists.

First: recognize that $v = \nabla^\perp \psi$, where $\psi$ is the stream function and the operation gives a 2D vector field $(\psi_y, -\psi_x)$. Then the divergence condition is satisfied automatically

Second: the norm constraint is $\Vert v \Vert = \Vert \nabla^\perp\psi \Vert = \Vert \nabla \psi \Vert = 1$. The 2nd equality is clear if you express them in components. $\Vert \nabla \psi \Vert = 1$ is the Eikonal equation, a classical solution of which is the distance field.

Example, the distance field to the origin is $\psi(x,y)=r=\sqrt{x^2+y^2}$, so $$ v = \nabla^\perp \psi = \left( \frac{y}{\sqrt{x^2+y^2}}, -\frac{x}{\sqrt{x^2+y^2}} \right) $$

A plot

$v$ is divergence free and has unit length, but it's singular at the origin.

Comment In 3D we don't have scalar stream function so the problem can be quite different/difficult (we do have vector potential, which itself is a vector instead of scalar field, so I don't think it will help).

Edit Another 2D example based on the distance field of an ellipse. This one won't permit any analytical description.

Answer 1

These are usually called Differential algebraic equations, which involve a mixture of differential equations and algebraic constraints. Note that these are different from algebraic differential equations, which are a whole other beast.

EDIT: It seems like you were also looking for an actual solution to this equation and I missed that, so I'll try to find some direction at least: Since $\|v\|=1$, we can write $v(x,y)=(v_1(x,y),v_2(x,y))$ as $(cos(\theta(x,y)), sin(\theta(x,y)))$. Applying divergence to this expression, we arrive at $cos(\theta)\frac{\partial \theta}{\partial y}-sin(\theta)\frac{\partial \theta}{\partial x}=0$, which isn't a very nice PDE to deal with.

EDIT: This paper appears to have some interesting results, but might be too technical for what you're looking for.

Answer 2

Just take any unit vector $\mathbf{e}$, and set $v(x,y)=\mathbf{e}$ for all $x,y$.