does this PDE have a name? My students found an interesting PDE today,
$$ u_{xxx}+u_{yyy}+u_{zzz}-3u_{xyz} = 0. $$
Does anyone happen to know if this PDE has a name?
Incidentally, one curious solution I just found is $u(x,y,z) = e^{x+y-2z}$. 
Connection with other question: in my other question "Is this a wave equation"
I asked about solving a system of first order ODEs in dependent variables $u,v,w$ and dependent variables $x,y,z$. I believe that solutions of that system are likewise solutions of this system. However, the curious solution (and many others Will has elucidated) are not solutions to the first order system. The third order PDE considered in this question has additional, and from my viewpoint, extraneous solutions.
Moreover, the connection with circulant matrices is not at accidental! This PDE was constructed from the left-regular representation of the group algebra of the cyclic group of order $3$. In view of this origin, I find the comments here delightful.
 A: I don't know if it has a name but here are a bunch of solutions. Write $D_x$ for the operator which partially differentiates by $x$, etc. Then the equation can be written 
$$(D_x^3 + D_y^3 + D_z^3 - 3D_x D_y D_z) u = 0.$$
This operator happens to be a product of linear factors as was observed in the comments. Letting $\omega$ be a primitive third root of unity we can write it as
$$(D_x + D_y + D_z)(D_x + \omega D_y + \omega^2 D_z)(D_x + \omega^2 D_y + \omega D_z) u = 0.$$
Consequently a solution to any of the equations
$$u_x + u_y + u_z = 0$$
$$u_x + \omega u_y + \omega^2 u_z = 0$$
$$u_x + \omega^2 u_y + \omega u_z = 0$$
gives a solution to the original equation, and one might hope that linear combinations of such solutions give all solutions to the original equation (I don't know if this is true). I believe the solutions to the first equation are all linear combinations of solutions of the form
$$u(x, y, z) = f(ax + by + cz)$$
whenever $a + b + c = 0$. If we want $u$ to be real-valued, then the second two equations are equivalent but impose two conditions on $u$ by taking real and imaginary parts, namely
$$u_x - \frac{u_y + u_z}{2} = 0$$
$$u_y - u_z = 0.$$
This is equivalent to requiring $u_x = u_y = u_z$, which I believe is equivalent to 
$$u(x, y, z) = g(x + y + z).$$
A: Here is a general solution,
$$ u \left( x,y,z \right) ={\it F_1} \left( y-x,z-x \right) +{\it F_2}
 \left( y- \left( -\frac{1}{2}+\frac{i\sqrt {3}}{2} \right) x,z- \left( -\frac{1}{2}-\frac{i
\sqrt {3}}{2} \right) x \right) + {\it F_3} \left( y- \left( -\frac{1}{2}-\frac{i
\sqrt {3}}{2} \right) x, z- \left( -\frac{1}{2}+\frac{i\sqrt {3}}{2} \right) x \right) \,.$$
Note that your solution is a special case of this solution.
