Let $D$ be the distribution in $\mathbb{R}^3$ generated by the vector fields $X=ye^x \ \partial_y-\partial_z$, $Y=\partial_x$.
- Find an integral submanifold for $D$ passing for $(1,0,0)$
- Say whether $D$ is integrable or not.
- Say whether an integral submanifold for $D$ passing for $(1,1,1)$ exists. If so, write it explicitly.
Since $[X,Y]=-ye^x\ \partial_y$ which is not, in general, a linear composition of $X$ and $Y$, $D$ is not involutive, hence not integrable. Some points of $\mathbb{R}^3$ are not contained in an integral submanifold.
General PDE's system
Now, for a fixed $q \in \mathbb{R}^3$ I'm looking for a 2-dimensional immersed submanifold of $\mathbb{R}^3$ passing for $q$, i.e. the image of an injective immersion $\sigma$, whose tangent space is described everywhere by $X$ and $Y$. The problem can be expressed in terms of a PDE's system: $$\sigma: A \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3, \quad \sigma(t,s)=(\sigma^1,\sigma^2,\sigma^3)(t,s), \quad S=\sigma(A) \subset \mathbb{R}^3$$ $$T_{\sigma(t,s)}S\equiv <\frac{\partial \sigma}{\partial t}(t,s),\frac{\partial \sigma}{\partial s}(t,s)>=D_{\sigma(t,s)}\equiv <X \left( \sigma(t,s) \right),Y \left( \sigma(t,s) \right)>$$ $$\sigma(t_0,s_0)=q $$
which I don't have any experience with.
Simple solution for $(1,0,0)$
Without using the system, the solution for the point $q=(1,0,0)$ is simple. In general $$D_{(a,b,c)}=<(0,be^a,-1),(1,0,0) > = \{(x,y,z):y+be^az=0 \}$$ I'm asking
- $q \in S$
- $T_pS=D_p \quad \forall p \in S$
Then in particular $T_qS=D_q$, where $D_{(1,0,0)}$ is just the plane $\{y=0\}$.
Exceptionally $q$ belongs to this plane, which then is a good candidate to be the integral submanifold, fulfilling condition 1. We have to check condition 2 for $S=\{y=0\}$: since $S$ is a plane passing for the origin, $T_pS=S$. What is more, for $p \in S$ , $p=(a,0,c)$, and then $D_p=\{y=0\}=S$. So both the conditions are fulfilled and $\{y=0\}$ is in integral submanifold of $D$ passing for $(1,0,0)$.
General solution
The same trick won't work for $q=(1,1,1)$ since $D_{(1,1,1)}=\{y+ez=0\}$ which does not contain $q$.
Is there a general way to continue from this point, or a test so say whether this integral submanifold exists or not?
What I would try is follow the fluxes of $X$ and $Y$ from $(1,1,1)$ and see what happens. If the submanifold for $(1,1,1)$ exists, I should be able to parametrize it this way. This works in the involutive case (Frobenius theorem), but in this non-involutive case the fluxes do not commute, so which should be followed first?
The flux of $X$ is $\theta_t(x,y,z)= \left( x,ye^{te^x},-t+z \right)$, the flux of $Y$ is $\alpha_s(x,y,z)=\left(s+x,y,z \right)$. $$\theta_t \cdot \alpha_s (1,1,1)=\left(s+1,e^{te^{s+1}},-t+1 \right)$$ $$\alpha_s \cdot \theta_t (1,1,1)=\left(s+1,e^{te},-t+1 \right)$$
Case 1
Using $\theta_t \cdot \alpha_s$ we obtain the submanifold parametrized by $(t,s) \mapsto (x,y,z)=\left(s+1,e^{te^{s+1}},-t+1 \right)$, i.e. $s=x-1,t=1-z$ and $$S= \{y=e^{(1-z)e^x}\}$$ $$T_{(a,b,c)}S=<\left(0,be^a,-1 \right),\left(1,b(1-c)e^a,0 \right)>$$ where of course $(a,b,c) \in S$ and $b=e^{(1-c)e^a}$. I would like this space to be $D_{(a,b,c)}$, so these two vectors should fulfill $y+be^az=0$, which does not happen (the first one is ok, but the second is not).
Case 2
Using $\alpha_s \cdot \theta_t$ we obtain the submanifold parametrized by $(t,s) \mapsto (x,y,z)=\left(s+1,e^{te},-t+1 \right)$, i.e. $s=x-1,t=1-z$ and $$S= \{y=e^{(1-z)e}\}$$ $$T_{(a,b,c)}S=<\left(0,be,-1 \right),\left(1,0,0 \right)>$$ $b=e^{(1-c)e^a}$. Again these two vectors should fulfill $y+be^az=0$, which does not happen (the second one is ok, but the first is not).
Does this mean the integral submanifold for $(1,1,1)$ does not exist?
