# Is there a vector field that is equal to its own curl?

I was wondering if there is a vector field that satisfies the following condition:

I’d like to offer a high-brow answer to this question. While I realize that this solution is likely to be above the OP’s level, I’m hoping it’ll be of some interest to others.

Claim: Let $u(x,y)$, $v(x,y)$, $w(x)$ be any real-analytic functions (defined on a neighborhood of $0$). Then there exists a unique real-analytic vector field $F = (F^1, F^2, F^3)$ that satisfies $\text{curl}\,F = F$ and

We’ll show this by applying the Cauchy-Kovalevskaya Theorem twice. Here I should acknowledge that I first saw this techinque applied to $\text{curl}\,F = F$ in an exercise in Robert Bryant’s “Nine Lectures on Exterior Differential Systems.”

Cauchy-Kovalevskaya Theorem: If $\mathbf{H}$ and $\phi$ are real-analytic functions near the origin, then there is a neighborhood of the origin on which there exists a unique real-analytic solution $\mathbf{g}(\mathbf{x}, t)$ to

A PDE system of this form will be called a “Cauchy problem.”

## Setup

The equation $\text{curl}\,F = F$ can be viewed as an overdetermined system of PDEs — in fact, a system of 4 first-order quasilinear equations for 3 unknown functions. That is, if we write $F = (F^1, F^2, F^3)$, then the condition $\text{curl}\,F = F$ becomes:

The fourth equation is a hidden “integrability condition” of sorts: namely, any solution $F$ to $\text{curl}\,F = F$ must satisfy $\text{div}\,F = 0$. This gives us a fourth (first-order quasilinear) equation

## The Cauchy Problems

We can write this system as a sequence of two Cauchy problems as follows. Let $u(x,y)$, $v(x,y)$, and $w(x)$ be arbitrary real-analytic functions.

We first consider the problem of finding $g(x,y)$ such that

By Cauchy-Kovalevskaya$^\dagger$, there exists a unique real-analytic solution $g(x,y)$.

Second, letting $g(x,y)$ be a solution as above, we consider the problem of finding $F^1, F^2, F^3$ such that

By Cauchy-Kovalevskaya again, there exists a unique real-analytic solution $F = (F^1, F^2, F^3)$.

By construction, this solution $(F^1, F^2, F^3)$ satisfies equations (1), (2) and (4). One can check (exercise!$^*$) that this $F$ necessarily satisfies equation (3) as well. This completes the proof. $\lozenge$

## End notes

$^\dagger$ Admittedly overkill here

$^*$ Hint for Exercise: Consider $E(x,y,z) := \frac{\partial F^1}{\partial y} - \frac{\partial F^2}{\partial x} - F^3$. Check that $E(x,y,0) = 0$ and $\frac{\partial E}{\partial z} = 0$, and then apply uniqueness in Cauchy-Kovalevskaya.

Note that this exercise shows why we needed the first Cauchy problem at all: that is, we couldn’t just choose $F^1(x,y,0) = g(x,y)$ completely arbitrarily!