Dear Chang,

the theorem actually concerns the collection of the vector fields that form the distribution and not the objects to which vector-fields are applied.

So, if you have vector-valued functions and each component of the vector admits its own distribution, the integrability can be computed by component-wise application of the theorem.

Concerning the rank of the Jacobian of the vector - valued function, it is given by applying the Frobenius theorem to the merge of the distributions of all of the function's components. Actually the rank of the Jacobian is the dimension of the space minus the rank of the solutions of the resulting system.

Concerning the PDE that you refer to, it does not define the distribution but the Jacobian of a vector-valued function, actually it asks for J(u)=transpose(J(u)) which I think implies that J(u)=constant.

In general, as it is referred in the wikipedia page, in case that you have the 1-forms that define a base for the Jacobian, the corresponding distribution is integrable iff the interior products of the 1-forms and of their exterior derivative with the vector fields of the distribution vanish.

I think this interpretes as: if A=[a1,...,ak] is the base of the Jacobian, the distributions Δ1,...,Δk consist of vector fields that correspond to vectors Vi : (J(ai) - transpose(J(ai))) Vi = 0, i=1...k.

Best,

Dimitris

Dear Arabadjis,

Thank you very much for your detailed answer. I get something, still, not quite understood and hope your further explanation.

1. I think distribution is precisely defined, they're {d/dx_i}, irrelevant to pde system.

2. It's right that Frobenius theorem could be applied component-wise if NO two vector fields mix together. Obviously, this pde system doesn't belong to this class and some transformation that changes for a proper coordinate is required if it exists.

3. Where does this come from "in case that you have the 1-forms that define a base for the Jacobian, the corresponding distribution is integrable iff the interior products of the 1-forms and of their exterior derivative with the vector fields of the distribution vanish."

Dear Chang,

Concerning the distributions, {d/dxi} is the linear base for the vector fields, but each distribution is itself a collection of vector fields that, also, belong to the tangent space of the considered manifold.

In the case of Frobenius theorem, this manifold is defined implicitly via the level sets of the scalar functions, over which the considered distribution vanishes. Specifically, if a set of vector fields {Vi} belongs to the tangent space of a manifold (thus forming a distribution), then one wants to know the rank of the gradient fields of the scalar functions that remain constant on that manifold (i.e. u : {Vi(u)=0}) and that is what Frobenius theorem offers. Thus, Frobenius theorem concerns the distribution (the considered subspace of the tangent space) and not the actual form of the object to which the distribution applies.

For 2. I think you mean that no two components of the vector valued function should be coupled in the PDEs. Yes, this does not allow for component-wise application of the theorem but the main reason is that the resulting system of equations for each component is inhomogenous and, thus, it cannot define a distribution (the vector fields don't belong to the tangent space of the implicit manifold).

For the case of 1-forms you can consider the subsection "Frobenius' theorem in modern language" > "Differential forms formulation" and mainly its 3rd paragraph in the wikipedia page you mentioned. The phrase in question just evaluates the differential ideal and its closedness condition under exterior derivative.

Thank you very much and now I need more reading on that