General solution to an ultrahyperbolic PDE

Question

$\DeclareMathOperator\SO{SO}$The following PDE defined on $\mathbb{R}^2$ $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}f(x,y) = 0,$$ has solution $$f(x,y) = g(x) + h(y),$$ where $g,h : \mathbb{R} \to \mathbb{R}$ are arbitrary (nice enough) functions. I believe this to be a general solution to the equation, for any choice of boundary conditions. I don't have a proof of this and I'm happy to be shown otherwise if that's the case.

I want to solve a similar looking PDE for $(x,y) \in \mathbb{R}^{n}\times\mathbb{R}^{n},$ which is given by $$\frac{\partial}{\partial x}\cdot\frac{\partial}{\partial y}f(x,y) = 0,$$ where the dot product is with the Euclidean metric on $\mathbb{R}^{n}$. There is an $\SO(n)$ symmetry under $x \mapsto M x, y \mapsto M^T y$. I can see that there are still solutions of the form $f_1(x,y) = g(x) + h(y)$ where $g,h : \mathbb{R}^n \to \mathbb{R}$ , but that these are no longer the only solutions.

I've come up with something quite general that solves the equation;$$f_2(x,y) = \int d^na\,\left( g^x(x \wedge a) \,h^y(y\cdot a) + g^y(y \wedge a) \,h^x(x\cdot a)\right) + f_1(x,y),$$ where $g^x,g^y:\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ and $h^x,h^y:\mathbb{R} \to \mathbb{R}$ are arbitrary (sufficiently nice) functions, which can depend on $a$. I added on $f_1(x,y)$ explicility, although I think it may be possible to fiddle the form of the integral and the choice of $g^x,g^y,h^x,h^y$ to include solutions of the form $f_1(x,y)$ without adding them on separately. One feature that I like about this solution is that it respects the $\SO(n)$ symmetry of the original PDE. The integral over the vector $a$ is also natural, in the context of the physical problem where the PDE comes from.

Is anyone able to find a general solution to this equation, in a form which respects the $\SO(n)$ symmetry, possibly with some (hopefully quite general) restriction on the boundary conditions?

Is my solution general for some choice of boundary conditions?

Answer 1

The standard method of constructing solutions is the following:

First, observe that, if $(a,b)\in\mathbb{R}^n\times\mathbb{R}^n$ is any pair of vectors that satisfies $a\cdot b = 0$, and $h:\mathbb{R}\to\mathbb{R}$ is any smooth function, then $$ f(x,y) = h(\,a{\cdot}x + b{\cdot}y\,) $$ is a solution of the ultrahyperbolic equation.

Now, you want to use linearity to superimpose these solutions as $(a,b)$ vary by integrating over the space of such pairs. The way to do this is to let $$M = \{(a,b)\in\mathbb{R}^n\times\mathbb{R}^n\ |\ a{\cdot}b = 0\ \text{and}\ a{\cdot}a + b{\cdot}b = 1\,\},$$ which is a manifold of dimension $2n{-}2$. (Note that it is $4$ points when $n=1$.) Fix a positive smooth measure $d\xi$ on $M$ that is invariant under $(a,b)\mapsto(-a,-b)$ and under the $\mathrm{O}(n)$-action $A\cdot(a,b) = (Aa,Ab)$ (which is easily seen to exist). Now let $h:M\times\mathbb{R}\to\mathbb{R}$ be any smooth map that satisfies $h(-a,-b,-t)=h(a,b,t)$ and let $$ f(x,y) = \int_{(a,b)\in M} h(a,b,\,\,a{\cdot}x + b{\cdot}y\,)\,d\xi $$ Then $f$ is a solution of the ultrahyperbolic equation. (Note that, when $n=1$, the above 'integration' is just the sum of two terms, yielding the familiar solution that you have mentioned in the problem.)

More invariant descriptions are possible, based on the fact that $M$ is (up to multiples) the set of nonzero null vectors of a 'split' quadratic form on $\mathbb{R}^{2n}$, and hence the ultrahyperbolic equation is actually invariant under a much larger group than $\mathrm{O}(n)$, namely $\mathrm{CO}(n,n)\subset\mathrm{GL}(2n,\mathbb{R})$ (plus translations). When you look at the literature on the ultrahyperbolic equation, you'll see how this works, particularly when you have a look at Johns' "Plane Waves and Spherical Means".

Answer 2

This is not an answer to the main question as the one offered by Robert Bryant, but it is a comment to the statement pertaining the two-variable boundary value problem for the ultrahyperbolic equation:

I believe this to be a general solution to the equation, for any choice of boundary conditions. I don't have a proof of this and I'm happy to be shown otherwise if that's the case.

The Dirichlet problem $$ \begin{cases} \dfrac{\partial}{\partial x}\dfrac{\partial}{\partial y} f(x,y)=f_{xy}(x,y)=0 & x,y\in[0,1]\\ f(x,\alpha(x))=\varphi_1(x) & x\in [0,1]\\ f(x,\beta(x))=\varphi_2(x) & x\in [0,1] \end{cases}\label{1}\tag{1} $$ where $\alpha, \beta:[0,1] \to [0,1]$ are appropriate strictly monotonic functions, is overdetermined (this, according to [1] pp. 6, was proved by Mauro Picone in [3]). Problem \eqref{1} was thoroughly studied by Gaetano Fichera: in [2] he found necessary and sufficient conditions (compatibility conditions) on the Dirichlet datum $(\varphi_1,\varphi_2)$ for it to be solvable with $f\in C^1([0,1]^2)$. In [1] he drops these compatibility conditions in order to accurately describe the singularities developed by the general solution $$ f(x,y) =g(x)+ h(y),\label{2}\tag{2} $$ by describing the kind of singularities developed by $g(x)$ and $h(y)$ as $x,y\to 0^+$. Therefore, it is probably right that \eqref{2} is the general solution to \eqref{1}, even when the boundary data are such that the solution behaves singularly at some point of the domain where the problem is posed.

References

[1] Gaetano Fichera, "Studio delle singolarità della soluzione di un problema di Dirichlet per l'equazione $u_{xy} = 0$" (Italian), Atti della Accademia Nazionale dei Lincei, Classe di Scienze Fisiche, Matematiche e Naturali. Rendiconti Serie 8 50 (1971), fasc. n.1, pp. 6-17, MR0310434, Zbl 0231.35046.

[2] Gaetano Fichera, "Su un problema di Dirichlet per l'equazione $u_{xy}=0$". (Italian) Atti dell'Accademia delle Scienze di Torino, Classe di Scienze Fisiche Matematiche e Naturali 105 (1971), pp. 355–366, MR298228.

[3] Mauro Picone, "Sulle equazioni alle derivate parziali del second’ordine del tipo iperbolico in due variabili indipendenti" (Italian), Rendiconti del Circolo Matematico di Palermo 30, pp. 349-376 (1910), DOI: 10.1007/BF03014882, JFM 41.0424.04.