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I want to solve the equation: $$ \begin{cases} \nabla \times (A \mathbf v)=f, \quad x\in \Omega \\ \operatorname{div}(\mathbf v)=0, \end{cases} $$ where $\Omega \subset\mathbb{R}^n$, is an open set, $A$ is a $n\times n$, with $n=2,3$ and $f$ is a smooth function.

Question: Assume that A is a singular matrix. What can we say about the well-posedness of such an equation? Are there any references about such a system? I appreciate any help you can provide.

$\textbf{Edited}$: the matrix $A$ for the original is defined as $A\mathbf{v}=F\times \mathbf{v}$ where $F$ is some constant vector field. What can we say about this particular case?

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    $\begingroup$ You make no mention of x other than to state that it is a member of 𝛀. It would help me understand the question if you clearly distinguished between what is the unknown that you are trying to solve for, and what is given. $\endgroup$
    – Daniel Asimov
    Commented Dec 2 at 2:17
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    $\begingroup$ If A can be any singular matrix, it could be the zero matrix. In that case, there are no solutions if $f\neq 0$, and lots of solutions if $f=0$. In the general case, a clearly necessary condition for solutions to exist is ${\rm div}\,f=0$. $\endgroup$
    – Michael Renardy
    Commented Dec 2 at 3:19
  • $\begingroup$ @DanielAsimov $A$ can be taken constant matrix, $\boldsymbol{f}(x)$ is given, $\mathbf{v}(x)$ is the unknown. $x$ is the variable. $\endgroup$
    – Gustave
    Commented Dec 2 at 6:08
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    $\begingroup$ the singular matrix will restrict the dimensionality of the solution; as a simple case, consider $n=3$, $A=\text{diagonal}\,(1,1,0)$, $f=(0,0,F)$, with $F$ a function of $x,y$ only; then the solution is of the form $v=(v_x,v_y,0)$ with $v_x,v_y$ a function of $x,y$ only. $\endgroup$
    – Carlo Beenakker
    Commented Dec 2 at 10:52
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    $\begingroup$ that case is answered at mathoverflow.net/a/481828/11260 $\endgroup$
    – Carlo Beenakker
    Commented Dec 3 at 16:10

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