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Given $X_1=\frac{\partial}{\partial x}-yz\frac{\partial}{\partial z}$ and $X_2=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}$ vector fields in $\mathbb{R}^3$, I know the integral curves through $(x_0,y_0,z_0)$ are respectively:

$X_1$: $x(t)=x_0+t$, $y(t)=y_0$, $z(t)=z_0 e^{-y_0 t}$

$X_2$: $x(t)=x_0 e^s$, $y(t)=y_0 e^{-s}$, $z(t)=z_0$.

On the other hand, since the distribution generated by both vector fields is involutive, I'm sure there is an integral manifold passing through each point of $\mathbb{R}^3$, which in particular contains the integral curves of both fields passing through that same point (right?). How may I, if not intuitively, obtain the expression for them?

In this case I obtained them in a non-self-convincing way, by solving a PDE system, and they are of the form $z=\frac{z_0}{e^{-x_0 y_0}} e^{-xy}$, condition which is indeed satisfied by both families of integral curves above.

The question is: how could I have obtained this last result in a smarter way? Any geometrical interpretations will do too, of course.

Thank you.

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up vote 3 down vote accepted

Starting from $(x_0,y_0,z_0)$, follow the flow of $X_2$ for $s$ units of time and then the flow of $X_1$ for $t$ units of time. This takes you to the point $$ (x,y,z) = (x_0 e^s + t, y_0 e^{-s}, z_0 \exp(- y_0 e^{-s} t)), $$ which lies on the integral surface through $(x_0,y_0,z_0)$, of course. Now eliminate $s$ and $t$ from this: \begin{multline*} z = z_0 \exp(- y_0 e^{-s} t) = z_0 \exp(-y t) = z_0 \exp(- y (x - x_0 e^{s})) \\ = z_0 \exp(-yx + x_0 y e^s) = z_0 \exp(-xy + x_0 y_0) . \end{multline*}

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Just what I was looking for. Thanks. – sheriff Jan 10 '13 at 16:55

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