Is it possible to recover a vector field from its divergence equation?

Question

I have the following vector field $p(x,y):\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ $$\nabla\cdot p=1$$ Is it somehow possible to deduce the vector field function $p$ just from the divergence equation?

Answer 1

You can't determine the constants, because taking derivatives destroys information; that's why we need initial conditions when working backwards.

One example of a solution (in three dimensions) to what you wrote would be $<x/3, y/3, z/3>$. But you could just as well say $<x/3 + f(y,z), y/3+g(x,z), z/3 + h(x,y)>$ and your arbitrary constants can actually be functions of the other two variables! So there are a lot of ways to get answers that solve your equation. It's definitely not unique.

In two dimensions, any vector function of the form $<x/2 + f(y), y/2 + g(x)>$ will satisfy your equation. Adding a constraint that $|p| = 1$ still leaves a lot of freedom of choice.

Answer 2

Far from it: Given any solution ${\bf p}$ of $\nabla \cdot {\bf p} = 1$ and any divergence-free vector field $\bf v$ (so that $\nabla \cdot {\bf v} = 0$), $$\nabla \cdot ({\bf p} + {\bf v}) = \nabla \cdot {\bf p} + \nabla \cdot {\bf v} = 1 + 0 = 1,$$ that is, ${\bf p} + {\bf v}$ also has divergence $1$.

There are many divergence-free vector fields: For any $C^2$ function $f$, we have $$\nabla \cdot (-f_y \partial_x + f_x \partial_y) = -f_{yx} + f_{xy} = 0 .$$