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

→F=∇×→F

**Answer**

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=(F1,F2,F3) that satisfies curlF=F and

F1(x,0,0)=w(x)F2(x,y,0)=u(x,y)F3(x,y,0)=v(x,y).

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

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

∂g∂t(x,t)=H(x,t,g(x,t),∂g∂xi)g(x,0)=ϕ(x)

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

## Setup

The equation curlF=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=(F1,F2,F3), then the condition curlF=F becomes:

∂F2∂z−∂F3∂y=F1∂F3∂x−∂F1∂z=F2∂F1∂y−∂F2∂x=F3.

The fourth equation is a hidden “integrability condition” of sorts: namely, any solution F to curlF=F must satisfy divF=0. This gives us a fourth (first-order quasilinear) equation

∂F1∂x+∂F2∂y+∂F3∂z=0.

## 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

∂g∂y=∂u∂x+v(x,y)g(x,0)=w(x).

By Cauchy-Kovalevskaya†, 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 F1,F2,F3 such that

∂F1∂z=−F2+∂F3∂x∂F2∂z=F1+∂F3∂y∂F3∂z=−∂F1∂x−∂F2∂yF1(x,y,0)=g(x,y)F2(x,y,0)=u(x,y)F3(x,y,0)=v(x,y)

By Cauchy-Kovalevskaya again, there exists a unique real-analytic solution F=(F1,F2,F3).

By construction, this solution (F1,F2,F3) satisfies equations (1), (2) and (4). One can check (exercise!∗) that this F necessarily satisfies equation (3) as well. This completes the proof. ◊

## End notes

† Admittedly overkill here

∗ Hint for Exercise: Consider E(x,y,z):=∂F1∂y−∂F2∂x−F3. Check that E(x,y,0)=0 and ∂E∂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 F1(x,y,0)=g(x,y) completely arbitrarily!

**Attribution***Source : Link , Question Author : Community , Answer Author : Jesse Madnick*