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I would appreciate it if Someone would be kind enough to share some insights about the following question:

Suppose $(M,g)$ is a 3 dimensional Riemannian manifold. Consider the following system of pdes: $<\nabla^g \phi_1 , \nabla^g u>_g =0$ and $ <\nabla^g \phi_2 , \nabla^g u>_g=0$

Geometrically this is equivalent to trying to find a function whose level sets are perpendicular to level sets of $\phi_1$ and $\phi_2$.

Can this always be done at least locally? Is the Frobenius integrability condition satisfied?

Thanks,

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In any dimension it appears to me that what happens depends on the smallest "Lie algebra" that contains the two vector fields $V = \nabla\phi_1$ and $W = \nabla\phi_2$. If the iterated Lie brackets of $V$ and $W$ generate a sub-bundle (i.e, the rank of the subspace at each point is a constant independent of the point), then that sub-bundle is integrable. Then $u$ most be constant along the integral submanifolds.

ADDED: In dimension 3 there are only two nonsingular possibilities: Either $[V,W]$ always lies in the span of $V$ and $W$ or $[V,W]$ sis always transverse to the $2$-plane spanned by $V$ and $W$. In the first case, the span of $V$ and $W$ is integrable and $u$ can be any function that is constant along the integral surfaces. In the latter, $u$ must be constant on the whole manifold.

Other cases are more difficult to analyze.

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  • $\begingroup$ I see. That means in dimension 3, such a u always exists locally though right? $\endgroup$
    – Ali
    Commented Nov 30, 2014 at 21:02
  • $\begingroup$ Oh I guess such a u might not exist in the singular cases $\endgroup$
    – Ali
    Commented Nov 30, 2014 at 21:08
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    $\begingroup$ Yes, in the sense that the constant function always satisfies those two equations. As Thomas Richard said, a nonconstant solution exists only if the Frobenius condision holds. This is not necessarily so. $\endgroup$
    – Deane Yang
    Commented Nov 30, 2014 at 21:09
  • $\begingroup$ My bad. Of course the constant solution is not interesring. Thanks $\endgroup$
    – Ali
    Commented Nov 30, 2014 at 21:26
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    $\begingroup$ An example where Frobenius does not hold is: $$\phi_1 = x + z \text{ and }\phi_2 = y + z^2$$ $\endgroup$
    – Deane Yang
    Commented Nov 30, 2014 at 21:39
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Set $V$ to be the subbundle of $TM$ whose fiber at $p$ is the span of $\nabla\phi_1(p)$ and $\nabla\phi_2(p)$. Your equation on $u$ is equivalent to requiring that $du$ vanishes on $V$. Frobenius integrability theorem tells you it has (EDIT: non constant) local solutions if and only if $V$ is integrable (i.e. it is stable under Lie bracket). It is not too hard to find functions $\phi_1$ and $\phi_2$ such that $[\nabla\phi_1(p),\nabla\phi_2(p)]\notin V$.

EDIT: This example is actually false, the one provided by Deane Yang works.
For instance on $\mathbb{R}^3$, take $\phi_1(x,y,z)=x$ and $\phi_2(x,y,z)=xyz$.

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  • $\begingroup$ I am not sure what I am missing here...I agree with the condition you have mentioned but I am not sure the example works.. to make this clear take $ u(x_1,x_2,x_3) = e^{\frac{x_3^2-x_2^2}{2}}$ $\endgroup$
    – Ali
    Commented Nov 29, 2014 at 1:34
  • $\begingroup$ It is clear that u solves the aforementioned system of odes in the question in euclidean space $\endgroup$
    – Ali
    Commented Nov 29, 2014 at 1:35
  • $\begingroup$ I'll have a look but I'm skeptical. $\endgroup$
    – Thomas Richard
    Commented Nov 29, 2014 at 7:26
  • $\begingroup$ Ok, I did the computation: if $u$ solves the problem, so does any $f(u)$ for $f$ a smooth function from $\mathbb{R}$ to $\mathbb{R}$. So I'll take $u(x,y,z)=z^2-y^2$ instead of yours. Then $\nabla\phi_1=\partial_x$ and $\nabla\phi_2=yz\partial_x+xz\partial_y+xy\partial_z$, while $\nabla u= 2z\partial_z-2y\partial_y$. On can check that the inner products between the gradients don't vanish everywhere. $\endgroup$
    – Thomas Richard
    Commented Dec 2, 2014 at 9:14
  • $\begingroup$ :your example simply does not work as one can clearly see that the inner products DO vanish everywhere actually as I also mentioned above. Anyways one can indeed find many examples where the system has no solution. Thanks for your comment $\endgroup$
    – Ali
    Commented Dec 3, 2014 at 5:35

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