## On a generalization of the Poincaré lemma to equations of the type $$dw+a\wedge w=f$$.(English)Zbl 1463.35181

The paper is deal with solving the following problem. Consider $$\Omega\subset\mathbb{R}^{n}$$ an open bounded set and $$0\leq k\leq n-1$$. Find $$w:\Omega\to\Lambda^k$$ such that $L_{a}^{k}(w):=dw+a\wedge w=f,\tag{P}$ where $$a:\Omega\to\Lambda^{1}$$ and $$f:\Omega\to\Lambda^{k+1}$$ are given differential forms. When $$k=0$$, defining $$w$$, the $$0$$-form with a scalar field and $$a$$, $$f$$ the $$1$$-forms with vector fields, the equation for the gradient is taken as follow $\nabla w+w\cdot a=f.$ When $$k=1$$ and $$n=3$$, equation (P) is equivalent to the curl equation $\operatorname{curl}w+a\times w=f,$ where $$w,a$$ and $$f$$ are vector fields, and $$\times$$ represents the vector product in $$\mathbb{R}^{3}$$. If $$k=n-1$$ equation (P) is equivalent to the divergence equation $\operatorname{div}w+\langle a,w\rangle=f.$ The equation (P) is a generalization of the Poincaré lemma. If $$a$$ is exact, that is there exists $$A$$ a $$0$$-form such that $$dA=a$$, then (P) is equivalent to $d\left(e^{A}w\right)=e^{A}f.$ So, there exists a solution to (P) if and only if $$e^{A}f$$ is exact.
For the above reason, the author focuses the attention to the cases where $$a$$ is nonexact. Given an exterior $$2$$-form $$\alpha$$, naturally associate a skew symmetric square matrix $$\bar{\alpha}$$ by defining $$\bar{\alpha}_{ij}=\alpha^{ij}$$ if $$i<j$$ and $$\bar{\alpha}_{ij}=-\alpha^{ij}$$ if $$i>j$$. The rank of $$\alpha$$ noted rank$$[\alpha]$$ is then the rank of the matrix $$\bar{\alpha}$$. Here, the rank of $$da$$ is assumed to be constant on the whole of $$\Omega$$. In this work, the main achievement of this article is the following theorem:
Theorem 0.1 Let $$\Omega\subset\mathbb{R}^{n}$$ be an open bounded set, $$0\leq k\leq n-1,$$ $$r\geq 1$$ be integers, $$a\in C^{1}\left(\Omega,\Lambda^{1}\right)$$ and $$f\in C^{1}\left(\Omega,\Lambda^{k+1}\right)$$. Then,
(i)
any $$w\in C^{1}\left(\Omega,\Lambda^{k}\right)$$ solution of $dw+a\wedge w=f,\tag{P}$ is a solution of $da\wedge w=df+a\wedge f;\tag{0.1}$
(ii)
if rank$$[da]$$ $$\geq 2(k+1)$$, (P) and (0.1) have at most one solution;
(iii)
if rank$$[da]\equiv 2m\geq 2(k+2)$$, $$a\in C^{r+2}$$ and $$f\in C^{r+1}$$, then (P) and (0.1) are equivalent and admit at most one solution $$w\in C^{r};$$
(iv)
if $$k\geq1$$ and $$u\in C^{1}\left(\Omega,\Lambda^{k-1}\right)$$ is such that $$du\in C^{1}$$ and $da\wedge u=f,$ then $$w=du+a\wedge u\in C^{1}$$ is a solution of (P);
(v)
if rank$$[da]\equiv 2m\geq 2(n-k)$$, $$a\in C^{r+3}$$ and $$f\in C^{r+1}$$, (P) always admits a solution $$w\in C^{r}$$.

In additional, the notations, propositions and remarks are given for this problem. The other main theorems are proved and cited for proof of many theorems. As a conclusion of this paper, according to the states of rank$$[da]$$, the solutions of the equation are reached under some conditions.

### MSC:

 35F35 Systems of linear first-order PDEs 58A10 Differential forms in global analysis