Motivation: if we are given

$$ u_x=v_y \qquad v_x=u_y $$

then it follows $u_{xx}=u_{yy}$ and $v_{xx}=v_{yy}$. If we think of $x$ as time then these are one-dimensional wave equations for $u$ and $v$.

Question: suppose $u,v,w$ are functions dependent on $x,y,z$ such that $$ u_x=v_y=w_z, \qquad \& \qquad v_x=w_y=u_z, \qquad \& \qquad w_x=u_y=v_z. $$ Do the conditions above imply wave equations for $u,v,w$?

Something I read seems to imply these equations produce a three-dimensional wave equation. But this seems wrong since one of the variables counting as time only leaves two independent spatial variables. For example, $u_{xx}=u_{yy}+u_{zz}$ is a two-dimensional wave equation. But, perhaps the three-dimensionality refers to the dependent variables as a triple $(u,v,w)$. Maybe there is the same wave equation for each component in $(u,v,w)$? I do not insist that $x$ be the "time" in the equation, I cannot judge from the motivating example if $x$ (or $y$) plays a special role.

Thanks in advance for your insights.


These equations imply

$$ u_{xx} = v_{zz} = w_{yy} (= u_{yz} = v_{xy} = w_{xz}) $$ $$ u_{yy} = v_{xx} = w_{zz} (= u_{xz} = v_{yz} = w_{xy}) $$ $$ u_{zz} = v_{yy} = w_{xx} (= u_{xy} = v_{xz} = w_{yz}) $$

let's go ahead and change $x$ to $t$ to be suggestive. We can write (for instance) $$ u_{tt} + v_{tt} + w_{tt} = (w_{yy} + u_{yy} + v_{yy}) $$

$$ (u + v + w)_{tt} = (u + v + w)_{yy} $$ or, alternately, $$ (u + v + w)_{tt} = (u + v + w)_{zz} $$ or even $$ (u + v + w)_{tt} = \frac{1}{2}((u + v + w)_{yy} + (u + v + w)_{zz}) $$ So you can get wave-equations for $\widetilde{u} = u + v + w$ in a few different ways, though it isn't really like the standard 2D wave equation, it seems, more like a 1D equation extended to the plane via a symmetry condition.

  • $\begingroup$ this looks promising. I'll look back at this tomorrow. If nobody gets a better result then you certainly have answered the question. $\endgroup$ – James S. Cook Sep 6 '12 at 5:20
  • $\begingroup$ Alright. I'll think about it some more in the meantime also. My guess is that you cannot tease out obviously wave-like equations for $u,v,w$ alone, but I could be very wrong. $\endgroup$ – BaronVT Sep 6 '12 at 5:33
  • $\begingroup$ Thanks BaronVT, if you find anything else feel free to email me... or post here again. I really want PDEs for $u,v,w$ alone... but the nature of the problem where these arise tends to make for these sort of ambiguous patterns. $\endgroup$ – James S. Cook Sep 6 '12 at 18:20

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