Inverting an anisotropic distribution I have come across a research problem where I need to solve an integral equation of the form
$\int A^{-1}(x,y) \nabla_z\cdot\left[\nabla_y\cdot\left(v(y) G(y-z) \right) v(z)\right]dy = \delta(x-z)$,
where $v:\mathbb{R}^d\to\mathbb{R}^d$.
What I need is an understanding of how the distribution $A^{-1}(x,y)$ changes with respect to $v$. Specifically, I need to evaluate integrals like
$$\int \frac{\partial A^{-1}(x,y)}{\partial v(x)}f(y)dy\quad\text{   and  }\iint f(x) \frac{\partial A^{-1}(x,y)}{\partial v(x)\partial v(y)} f(y)dxdy.$$
Has anybody come across anything similar?
 A: OK, so I think I have a solution. First let's start with an easier problem. Suppose we know a pseudo-differential operator $P(-\Delta)$ such that $P(-\Delta)G(x,y)=\delta(x-y)$, and we want the inverse of the operator $f(x)G(x,y)f(y)$, where $f:\mathbb{R}\to\mathbb{R}$.
We wish to find a differential operator $L(x,y)$, such that
$$ \int L(x,y)f(y)G(y,z)f(z) dy = \delta(x-z). $$
The solution 
$$
 L(x,y) g(y)=\delta(x-y)\frac{1}{f(y)} P(-\Delta_y) \frac{g(y)}{f(y)}
$$
satisfies our desired relation. To see why, let us do the calculation
\begin{align*}
\int L(x,y)f(y)G(y,z)f(z)dy &= \int \delta(x-y)\frac{1}{f(y)} P(-\Delta_y)\left[ \frac{f(y)}{f(y)} G(y,z)f(z) \right]dy \\
&=\int\delta(x-y) \frac{1}{f(y)}\delta(y-z) f(z) dy \\
&=\delta(x-z).
\end{align*}
Now returning to the original problem, I can rewrite the operator to invert as follows
$$ M(x,D_x)M(y,D_y) G(x,y), $$
where $M(x,D)=v(x)\cdot\nabla_x+ \nabla_x\cdot v(x)$. Then, similar to above, the operator
$$ L(x,y)g(y) = \delta(x-y){M(y,D_y)}^{-1}\left[P(-\Delta_y)\left({M(y,D_y)}^{-1}g(y) \right) \right] $$
is the inverse. 
