# What is known about the equation $u_{xy}+u_{yx}=0$?

I was thinking a bit about PDEs and realized that I haven't seen any PDEs whose solutions possess non-equal mixed partial derivatives or where this possibility is at least taken seriously. So, I was wondering:

What is known about such equations? Is there a general theory of them?

A reference would be very welcome. (If such equations make sense, of course.)

In particular, I was wondering what can be said about the equation

$$u_{xy}+u_{yx}=0$$

Are there any interesting solutions to this equation, in any sense?

I am mostly interested in functions $u:\Omega\to\Bbb R$, where $\Omega\subseteq\Bbb R^2$ is a domain and preferably $u_{xy}\neq 0$, but weak solutions of any kind are also welcome.

Thanks in advance.

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shouldn't $u_{xy} = u_{yx}$ ? – Santosh Linkha Sep 9 '12 at 20:21
looks like out of my scope :( – Santosh Linkha Sep 9 '12 at 20:30
The only example I know of, where you have a function with $u_{xy}\neq u_{yx}$, the inequality is valid in a given point. I suppose it would be very difficult to build an example valid in an open set. – enzotib Sep 9 '12 at 20:36
In the sense of distributions, $u_{xy} = u_{yx}$ always. So you won't find any interesting weak solutions. – Robert Israel Sep 9 '12 at 20:54
I'm sure there is no general theory for this sort of thing. Linear PDE are best understood in terms of distributions (which sometimes happen to be nice functions), and as @RobertIsrael pointed out, the distributional derivatives always commute. It does make an entertaining real analysis question: if $u_{xy}+u_{yx}=0$ holds in a domain, does it follow that $u_{xy}\equiv u_{yx}\equiv 0$? I would guess that the answer is yes... – user31373 Sep 10 '12 at 0:31

## 1 Answer

I think I may have some rough idea about this type of problem, probably I am wrong though, should be a comment really.

If in $\mathbb{R}^2$, $u_{xy}\neq u_{yx}$, while the first derivatives exists, in weak sense, $u\in H^1$ but $u\notin H^2$, and the problem can be rewritten as: $$\tag{1}\mathrm{div}(R \nabla u) = 0$$ where $R = \begin{pmatrix}0&1 \\1&0\end{pmatrix}$ is the reflection transformation with respect to the line $y=x$, multiply both sides of (1) by a smooth test function $v\in C^{\infty}_0$, suppose $u$ is a piecewisely defined function:

$$\int_{\mathbb{R}^2} \mathrm{div}(R \nabla u) v = -\int_{\mathbb{R}^2} R \nabla u\cdot \nabla v + \sum_{\Omega_i}\int_{\partial \Omega_i} vR\nabla u\cdot \nu$$ The first term is gone if $u$ is a weak solution, for second term, if the reflection transformation would make the normal component across the boundary still continuous, ie $(vR\nabla u\cdot \nu|_{\partial \Omega^+} + vR\nabla u\cdot \nu|_{\partial \Omega^-})$ be zero.

A possible candidate $u$ would a piecewise linear function that is $1$ on $y=x$ and $0$ on $y = x\pm 1$, has linear decay within the stripe and $0$ elsewhere, $\mathbb{R}^2$ is being divided into 2 parts, where a.e. $u_{xy} = 0 = u_{yx}$, while the normal component of $R\nabla u$ is continuous across $y=x$, hence we have $\mathrm{div}(R\nabla u)=0$.

I tried for an hour and can't think of an explicit example of $u_{xy} = -u_{yx}$ on a positive Lebesgue measure set, maybe it is as LVK suggested.

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