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I am trying to solve a 4th order nonlinear PDE for a real function $u(x,y)$ of two variables. It is too complicated to reproduce here but it exhibits the following two very nice properties:

1) if $u(x,y)$ is a solution, then $f(u(x,y))$ is also a solution for any function $f$.

2) if we think of the equation in the complex plane $z=x+iy$, then it is satisfied by every (anti-)holomorphic function, that is, $g(z)$ and $g(\bar{z})$ are solutions for any function $g$.

Has anyone ever encountered such a nonlinear PDE? If so, does it have a special name, and what are its known properties?

I am interested in finding as many real solutions as possible -- due to the nonlinearity, property 2) is not very useful in that regard. However, I've found a handful of solutions by inspection, and I suspect that the equation may in fact be integrable (though I am not sure how to test this), partly because of the above properties, and partly because of physical reasons.

Note: Property 1) has a nice geometric origin. Every function $u(x,y)$ defines a foliation of the plane by the level sets $u(x,y)=$ constant. We can then think of $u$ as a coordinate labeling the "leaves" in the foliation, and the change $u\to f(u)$ is just a relabeling of coordinates (as long as $f$ is monotonic).

In other words, this PDE can be thought of as an equation for a foliation. Any given foliation can be represented by many functions $u(x,y)$, all of which are functions of each other, and property 1) is just the statement of reparameterization invariance.

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It's easy to derive a third-order (nonlinear) differential equation for $u(x,y)$ that satisfies your conditions (1) and (2): Namely, set $\theta(x,y) = \arctan\bigl(u_y(x,y)/u_x(x,y)\bigr)$ and then consider $$ \theta_{xx} + \theta_{yy} = 0. $$ When one expresss this equation explicitly in terms of the partial derivatives of $u$, one obtains a third-order nonlinear polynomial PDE whose solutions have exactly the properties (1) and (2). (However, I'm actually interpreting (2) to mean that the real part of $g$ satisfies your equation, since you started out stating that the equation was for a real function $u(x,y)$, while a nonconstant holomorphic function $g(z)$ is never real-valued. Maybe you have a typo somewhere? Clarification of this point would help.)

The general solution of this third order equation depends on three arbitrary functions of one variable.

Now, your fourth-order equation could be a (combination of) derivative(s) of this third-order equation. If it is suitably non-degenerate its general solution will depend on four functions of one variable, so it will have solutions that don't satisfy this third-order equation, in which case, there will be solutions that aren't of the special form you have found so far.

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  • $\begingroup$ Thank you for your response, Robert. I did not mean that the real part of $g(z)$ satisfies my equation for any function $g$, otherwise I would be able to get infinitely many real solutions (which is what I care about). I really meant: my equation is automatically satisfied if we allow $u(x,y)$ to be complex and take it to be of the form $u(x,y)=g(x\pm iy)$ for any $g$. If the equation were linear, then the superposition $g(z)+g(\bar{z})$ would be a real solution, as desired, but this trick fails for nonlinear equations. $\endgroup$ – Alex Lupsasca Apr 11 '16 at 19:05
  • $\begingroup$ Your example of a PDE does indeed exhibit the two properties I listed and is extremely interesting. Does this equation have a name or did you just come up with it on the spot? $\endgroup$ – Alex Lupsasca Apr 11 '16 at 19:13
  • $\begingroup$ @AlexLupsasca: So you didn't really mean to assume that $u(x,y)$ is real; that confused me. Is your fourth-order equation a real polynomial in $u(x,y)$ and its derivatives? Also: Presumably, you don't really mean any $g:\mathbb{C}\to\mathbb{C}$; you require that $g$ be either holomorphic or anti-holomorphic in order for it to satisfy your equation, right?. It's entirely possible that someone has named the third-order equation, but, in fact, I just wrote it down based on your description of the properties of the solution. I don't have a reference. $\endgroup$ – Robert Bryant Apr 11 '16 at 19:30
  • $\begingroup$ The 4th order equation is indeed a real polynomial in $u(x,y)$ and its derivatives, and I am only interested in real solutions $u(x,y)$. But if I allow complex solutions, then as you wrote, any (anti-)holomorphic $g$ will do the job. Kudos on coming up with that diff. eq. -- it seems very nontrivial to me. I guess I should try variations on that theme and see if my equation can be written in terms of derivatives of a 3rd order equation of that ilk. Thank you! $\endgroup$ – Alex Lupsasca Apr 12 '16 at 2:54
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All foliations are locally diffeomorphic. So your equation depends on more than just a foliation. It also uses, most likely, the ambient Riemannian metric. You need to let us known what else it depends on (only the metric, perhaps only the affine connection of the metric, perhaps only the projective connection of the metric, perhaps only the conformal structure). Then we might recognize it.

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  • $\begingroup$ This equation is obtained by dimensional reduction from a physical equation in 4 dimensions. Working in cylindrical coordinates $(t,\rho,z,\phi)$, we take the 4-dimensional metric to be block diagonal in $(t,\phi)$ and $(\rho,z)$, with coefficients only functions of $(\rho,z)$. We can then project onto the $(x,y)\equiv(\rho,z)$ plane. The resulting equation in the plane remembers $g_{t,\phi}(x,y)$ and $g_{\rho,z}(x,y)$. I suppose this is an "ambient Riemannian metric". $\endgroup$ – Alex Lupsasca Apr 11 '16 at 19:19

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