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:

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
gt(x,t)=H(x,t,g(x,t),gxi)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:
F2zF3y=F1F3xF1z=F2F1yF2x=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
F1x+F2y+F3z=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
gy=ux+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
F1z=F2+F3xF2z=F1+F3yF3z=F1xF2yF1(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):=F1yF2xF3. Check that E(x,y,0)=0 and Ez=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

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