0
$\begingroup$

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,

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I see. That means in dimension 3, such a u always exists locally though right? $\endgroup$
    – Ali
    Nov 30, 2014 at 21:02
  • $\begingroup$ Oh I guess such a u might not exist in the singular cases $\endgroup$
    – Ali
    Nov 30, 2014 at 21:08
  • 1
    $\begingroup$ 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
    Nov 30, 2014 at 21:09
  • $\begingroup$ My bad. Of course the constant solution is not interesring. Thanks $\endgroup$
    – Ali
    Nov 30, 2014 at 21:26
  • 1
    $\begingroup$ An example where Frobenius does not hold is: $$\phi_1 = x + z \text{ and }\phi_2 = y + z^2$$ $\endgroup$
    – Deane Yang
    Nov 30, 2014 at 21:39
2
$\begingroup$

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 $\phi_2(x,y,z)=xyz$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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
    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
    Nov 29, 2014 at 1:35
  • $\begingroup$ I'll have a look but I'm skeptical. $\endgroup$
    – Thomas Richard
    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
    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
    Dec 3, 2014 at 5:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.