# a question about solutions of a system with four equations

I am looking for solutions of this system. Is such system solvable? here $a,b,c,...$ are smooth functions and $h(x)\in C^1$ and we take $h_i=\frac{\partial \varphi}{\partial x_i}$, and also $h(x)=(h_1(x_1,x_2,x_3),h_2(x_1,x_2,x_3),h_3(x_1,x_2,x_3))$

$a(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_3^2}+\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_1}+\frac{\partial^2\varphi}{\partial x_2 \partial x_1}\frac{\partial^2\varphi}{\partial x_3 \partial x_2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}]-$

$(\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_2}+\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_1}+\frac{\partial^2\varphi}{\partial x_3^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_1})+$

$b(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2^2}-\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_1}]+c(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_3^2}-\frac{\partial^2\varphi}{\partial x_1 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_1}]+$

$d(x,h(x))[\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_3^2}-\frac{\partial^2\varphi}{\partial x_2 \partial x_3}\frac{\partial^2\varphi}{\partial x_3 \partial x_2}]+e(x,h(x))[\frac{\partial^2\varphi}{\partial x_1^2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}-\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}]-$

$f(x,h(x))[\frac{\partial^2\varphi}{\partial x_2^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_3}-\frac{\partial^2\varphi}{\partial x_1 \partial x_2}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}]+g(x,h(x))[\frac{\partial^2\varphi}{\partial x_3^2}\frac{\partial^2\varphi}{\partial x_1 \partial x_2}-\frac{\partial^2\varphi}{\partial x_1 \partial x_3}\frac{\partial^2\varphi}{\partial x_2 \partial x_3}]+$

$h'(x,h(x))\frac{\partial^2\varphi}{\partial x_1^2}-i(x,h(x))\frac{\partial^2\varphi}{\partial x_2^2}+j(x,h(x))\frac{\partial^2\varphi}{\partial x_3^2}+k(x,h(x))\frac{\partial^2\varphi}{\partial x_1 \partial x_2}-$

$l(x,h(x))\frac{\partial^2\varphi}{\partial x_2 \partial x_3}+m(x,h(x))\frac{\partial^2\varphi}{\partial x_1 \partial x_3}$

$n(x,h(x))\frac{\partial \varphi}{\partial x_1}+o(x,h(x))\frac{\partial \varphi}{\partial x_2}+p(x,h(x))\frac{\partial \varphi}{\partial x_3}+q(x,h(x))=0$

&

$\frac{\partial^2\varphi}{\partial x_1\partial x_2}=\frac{\partial^2\varphi}{\partial x_2\partial x_1}$

$\frac{\partial^2\varphi}{\partial x_1\partial x_3}=\frac{\partial^2\varphi}{\partial x_3\partial x_1}$

$\frac{\partial^2\varphi}{\partial x_2\partial x_3}=\frac{\partial^2\varphi}{\partial x_3\partial x_2}$

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Is $h(x)$ one of the unknowns? If not, the notation seems unnecessarily complicated. Are the coefficient functions real analytic? In the real analytic category, there are always local solutions for an unknown $\phi$, given all of the other functions, by the Cauchy-Kovalevskaya theorem. The last three equations follows immediately if $\phi$ is $C^2$, so in the real analytic category. If we know nothing about the coefficient functions (perhaps not even measureable functions) then I don't think anyone could tell you anything about solvability. –  Ben McKay Apr 7 at 12:17
@Hassan Jolany: By the way 'root-systems' has nothing to do with this question, so you should remove this tag. I don't think that 'ca.analysis-and-odes' has anything to do with it either. Finally, I imagine that you are asking this question because of my answer to your other question at mathoverflow.net/questions/126203. This is not the kind of four equations with no solutions that I had in mind when I wrote that comment. That problem is not one of analysis; it would persist even in the analytic category. If I have time, I'll add to my comment there to give you an illustrative example. –  Robert Bryant Apr 7 at 13:11
@Hassan Jolany: Unfortunately, your definition of 'quadratic form' is faulty because, obviously, we will have $q(\alpha,\beta)=-q(\beta,\alpha)$ for all $\alpha,\beta\in \Pi$, so $q$ is skewsymmetric rather than symmetric. As such, while it is possible for $q$ to be nondegenerate, there is no sense of 'determined' (i.e., what most people would call 'definite'). You cannot define 'elliptic' this way. It is more subtle than that. –  Robert Bryant Apr 7 at 16:22
It seems to me that if you want to assume only $C^1$ and want only a $C^1$ solution, then the usual PDE theory does not apply, especially since the equation is nonlinear. There is some chance that the $h$-principle invented by Nash and Kuiper and refined by Gromov would apply. Unfortunately, this is not an easy theory to learn. There is Gromov's book, Partial Differential Relations, but I've been told that the book by Eliashberg and Mišačev is easier to understand. I do not know enough about this to say whether it's relevant or not. –  Deane Yang Apr 7 at 16:32
Since you say now that $h$ is $C^1$ and that $h=(h_1,h_2,h_3)$ and $h_i=\partial \phi/\partial x_i$, it follows that $\phi$ is $C^2$, so the last three equations are automatically satisfied for any solution $\phi$ to the previous equations. You have a single quasilinear determined PDE in one unknown function $\phi$, and $h$ is really just a notation for $d\phi$. If the PDE is hyperbolic or elliptic, then it has local solutions, so that gets us somewhere. If the type is allowed to change, that could be very tricky. I think the mathematical community is losing its enthusiasm for this question. –  Ben McKay Apr 8 at 8:05