On a generalization of the Poincaré lemma to equations of the type \(dw+a\wedge w=f\). (Q1746001)

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: \textbf{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, \begin{itemize}\item[(i)] any \(w\in C^{1}\left(\Omega,\Lambda^{k}\right)\) solution of \[dw+a\wedge w=f,\tag{P}\] is a solution of \[da\wedge w=df+a\wedge f;\tag{0.1}\]\item[(ii)] if rank\([da]\) \(\geq 2(k+1)\), (P) and (0.1) have at most one solution;\item[(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 \(w\in C^{r};\)\item[(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);\item[(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 \(w\in C^{r}\).\end{itemize} 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.