# 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?

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Perhaps this would be better on Mathematics? If you think so, don't repost it there, I can migrate it for you. – David Z Oct 22 '12 at 1:05
It's appropriate here, too. Let me ask -- what is the large |y| behavior of A^{-1}? Obviously, I have integration by parts in mind. – MarkWayne Oct 22 '12 at 1:17
@MarkWayne I'm not exactly sure what the behavior is for large |y|, G(y-z) is assumed to be any generic Green's function for a PDO, say for simplicity the negative Laplacian. – Josh Chang Oct 23 '12 at 21:42
@Qmechanic Yes, the operator results actually from two integration by parts, from the functional integral wrt $\phi$ with an action consisting of of a quadratic term like $\int\int \nabla \phi(x) v(x) G(x-y) v(y) \nabla \phi(y)dxdy$. There is also a determinant term that I need a handle on as well. – Josh Chang Oct 23 '12 at 21:45

## 1 Answer

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.

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