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Given vector fields $Y$, $Z$ on a (possibly compact) manifold $M$, I would like to know about the existence of solutions $X$ to the differential equation $$ \nabla_Y X + a \cdot \mathrm{div}(Y)\cdot X + b X = Z$$ where $\nabla$ denotes the Levi-Civita connection.

What about local and/or global solutions? What if $Y = \mathrm{grad} \phi$ for some function $\phi$?

In the case that Y is the radial vector field around some special point, this is easily transferred into a normal ODE. But I don't really see what to do in a more general situation.

\Edit: So, as I should have seen myself, existence of a local solution is quite clear because this is just a system of ODEs along the integral curves of V. Furthermore, you can say basically nothing about global solutions. So, let us concentrate on local solutions.

Another question though:

It turns out, I have to solve the equation locally, but around the points where $Y$ vanishes. And now it doesn't seem clear at all, what happens.

Because, if $Y$ behaves like the radial vector field, then you basically have to solve singular ODEs, but if $Y$ rotates around $p$ (like $- \mathrm{sin}(x^1) \partial_1 + \mathrm{cos}(x^2) \partial_2$ around $0$), then it is not clear at all to me how I can ensure that the solutions fit together smoothly at $p$.

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This is just a family of linear ODE's along the vector field $Y$. So you can start with any value of $X$ along any hypersurface that is transversal to $Y$ and integrate it on integral curves of $Y$ until two distinct integral curves of $Y$ cross. –  Deane Yang Nov 26 '11 at 0:24
If $Y=\mathop{\rm grad} \phi$ and $\phi$ has more then one critical point then for generic $Z$ there is no solution. That follows from Deane's comment. –  Anton Petrunin Nov 26 '11 at 3:53
Thank you very much. But does that give me any information about global solutions? –  Kofi Nov 26 '11 at 12:17
I don't see how to say anything about global solutions without more assumptions. –  Deane Yang Nov 26 '11 at 15:31
As Anton mentioned, global solutions often do not exist. And, if they do, it's not easy to figure out when. –  Deane Yang Nov 26 '11 at 20:30
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