Can you find a vector field that equals its own curl?

[math man]

I was looking through a calculus book doing some of the practice problems where I was asked to calculate the curl of a few functions. One of them got me thinking, is there a function whose curl is itself? Much like how e^{x} is it's own derivative, is there a vector field that is it's own curl?
∇×F = F
Are there any non-zero general solutions to this problem? This will expand to a system of 3 linear PDE's with 3 unknowns. The problem is, I'm new to PDE's and don't know how to solve them. I am familiar with ODE's to some extent, and I understand that the goal is to simplify a PDE into an ODE. So my question... how would I go about solving this problem? Do you know any good online references that would teach me? Thank you for your time.

 

[Marioeden]

There are certainly several.

One example that I can think of off the top of my head is:
F = (0, cos(x), -sin(x))

 

[math man]

Good, so it is solvable. Following your example...
<0, sin(x), cos(x)>
<cos(y), 0, sin(y)>
<-sin(y), 0, cos(y)>
<sin(z), cos(z), 0>
<cos(z), -sin(z), 0>

Are all solutions to my problem. Is there a general method to solve for these, a way to combine them into a single general form from which they can all be derived?

 

[hunt_mat]

This will give you three equations for three unknowns, cross differentiate to obtain three separate equations.

 

[math man]

How do you cross differentiate something?

 

[chiro]

Use the normal formula for cross product (like say in determinant form) and then set-up three equations corresponding to each component (as pointed out by hunt_mat above).

If you know how to calculate the curl component-wise for an unspecified F, then the rest is equating each element of F to the respective element of curl(F).

 

[HallsofIvy]

\nabla f\vec{i}+ g\vec{j}+ h\vec{k}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}- \left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+\left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}

So \nabla \vec{F}= \vec{F} gives the three partial differential equations
\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}= f
\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}= g
\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}= h

You can reduce those to higher derivative equations in just one unknown function similarly to how you can reduce systems of numerical equations. To make it easier to see what you are doing, rewrite those in "operator form", writing "D_x" for the partial derivative with respect to x, etc.
D_yh- D_zg= f
D_zf- D_xh= g
D_xg- D_yf= h

You can treat those derivatives like numerical coefficients and multiply to get like coefficients, etc. then add or subtract to reduce. (I said like numerical coefficients. "Multiplying" on the right will be applying the derivative to that function.)

 

[math man]

Thank you HallsofIvy. I had a hunch that you might solve for the operators and not for the components of F, but I didn't think to treat them like numbers. After a few hours of trying random things, I finally was able to get it down to 1 PDE by using matrix row operations. Thank you everyone for your posts.