Is there any result (Schauder-like estimates, $L^2$ estimates or similar) to equations of the form $$ {\rm div}(Av)=f $$ where $A$ is the "unknown" (i.e. I would like estimates on $A$ depending on $f$,$v$)? Thanks!
1 Answer
Of course not, because this equation is far from being elliptic. Actually, it is even under-determined, in the sense that you have only one equation, for $n^2$ unknowns (where the matrix is $n\times n$).
Let me however give you a result in this direction, that I discovered two years ago, which has important consequences in various domains.
Let $A$ be symmetric, with entries in the space $\cal M$ of bounded measures. Suppose that $A$ is positive semi-definite. Suppose at last that ${\rm div}(A\vec e_i)\in\cal M$ for every $i\in[1,n]$. Then $(\det A)^\frac1n$, which is a priori a bounded measure, is actually an element of $L^{\frac n{n-1}}$. This qualitative result is associated with a functional inequality.
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$\begingroup$ Thanks! It is obvious indeed that one needs less generality, and I was looking for results on these lines $\endgroup$ Commented Mar 12, 2020 at 20:34