I am trying to integrate a distribution, but I am having some problems understanding how. I am trying to replicate the following exercise:
and the solution is the one at the bottom.
The problem is that it does not explains how to find the values for $\lambda $ and I cant't really understand how it has been found. I have been on this for long, and I have posed already a question on this, which is How do I find the integral of a distribution? , but I have deicided to pose another question, since I have not understood how to solve this kind of problems, sorry for that.
So, the original distribution is:
$\Delta =span\begin{Bmatrix} \begin{pmatrix} x_1\\ 1\\ 0\\ x_3 \end{pmatrix} & \begin{pmatrix} e^{x^2}\\ 0\\ 0\\ 0 \end{pmatrix} \end{Bmatrix}$
and so, I have to find he values of $\lambda $ for which:
$\begin{pmatrix} \frac{d\lambda }{dx_1} &\frac{d\lambda }{dx_2} &\frac{d\lambda }{dx_3} & \frac{d\lambda }{dx_4} \end{pmatrix}\cdot \begin{pmatrix} x_1 & e^{x^2}\\ 1& 0\\ 0& 0\\ x_3& 0 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix}$
Can somebody please help me understand?
You're looking, locally, for level hypersurfaces of the functions $\lambda_1,\lambda_2$ whose intersection will give the distribution. Thus, the normal vectors $\nabla\lambda_1$ and $\nabla\lambda_2$ must be orthogonal to the vectors spanning the distribution.
There's not much point having the $e^{x_2}$ (note your typo) in the second vector. The vector $(1,0,0,0)$ will do just fine. So $\lambda_1$ and $\lambda_2$ should be independent of the variable $x_1$. Orthogonality to the vector $(x_1,1,0,x_3)$ then reduces to orthogonality to the vector $(0,1,0,x_3)$. Linear algebra tells us that the gradient vectors must therefore be in the span of $(0,0,1,0)$ and $(0,-x_3,0,1)$, so we take $\lambda_1(x) = x_3$ and $\lambda_2(x)=x_4-x_2x_3$ (Hint: Change the second vector so that mixed partials will match).
REMARK: My favorite way to approach these integrability problems is to use the differential forms version of Frobenius integrability. In particular, the distribution is given by $\omega_1=0$, $\omega_2=0$, where $\omega_1 = dx_3$ and $\omega_2 = x_3\,dx_2-dx_4$. Since $d\omega_1=0$ and $d\omega_2 = dx_3\wedge dx_2 = \omega_1\wedge dx_2$, the system is integrable. Indeed, if we consider $\tilde\omega_2 = \omega_2+x_2\omega_1 = x_3\,dx_2 - dx_4 + x_2\,dx_3 = d(x_2x_3-x_4)$, we see that the distribution is given by taking level surfaces of $\lambda_1=x_3$ and $\lambda_2 = x_2x_3-x_4$.