From physics, just to use a well known example, we know that the relationship between the magnetic induction $\mathbf{B}$ and the potential vector $\mathbf{A}$ is given by:
$$\mathbf{B} = \nabla\times\mathbf{A}$$
My question is: could/does exist an operator $\mathrm{\hat{O}}$ (or with a bad notation: "$(\nabla\times)^{-1}"$ such that
$$(\nabla\times)^{-1}\mathbf{B} = \mathbf{A}$$
I mean: knowing the magnetic field $\mathbf{B}$, is there some operator $\mathrm{O}$ such that $\mathrm{\hat{O}} \mathbf{B} = \mathbf{A}$?
Best Answer
First, note that we expect such an inverse will be nonlocal, since the inverse of the standard differential operator $d/dx$ is the integral $\int_a^x \, dx'$.
Suppose in the simplest case we have the simply connected domain $\mathbb{R}^3$ (with enough decay at infinity that whatever integrals we write down will converge), and we are trying to solve $$ \nabla \times A = B $$ for $A$. Taking another curl gives $$ \nabla \times (\nabla \times A) = \nabla \times B, $$ and it looks like I've made things worse. But we have $$ \nabla \times (\nabla \times A) = \nabla(\nabla \cdot A)-\nabla^2 A, $$ where the last term is the vector Laplacian. Now, if we can say that $\nabla \cdot A=0$ (which it's admittedly not clear is possible; let's come back to that), then we need to solve the equation $$ -\nabla^2 A = \nabla \times B. $$ But if we use Cartesian coordinates, the vector Laplacian acts like the ordinary Laplacian on each component of $A$; therefore, we can invert it using the Green's function for the Laplacian (if you haven't met this, it's given by the solution to $-\nabla_x^2 G(x-y) = \delta(x-y)$ satisfying the right boundary conditions), which in this case is $-1/(4\pi |x-y|)$. Then we define $$ A_B(x) = \int_{\mathbb{R}^3} G(x-y) (\nabla \times B)(y) \, dy=\int B(y) \times [\nabla G(x-y)] \, dy, $$ integrating by parts. Does this work? Well, $$ \nabla \times (X \times a) = (a\cdot \nabla)X-(\nabla \cdot X)a, $$ so we have $$ \nabla \times A_B = \int (B(y) \cdot \nabla)\nabla G(x-y) \, dy + \int B(y) (-\nabla^2 G(x-y)) \, dy; $$ the former term is zero because if we integrate it by parts, we get a $\nabla \cdot B$, which is zero since $B$ is supposed to be the curl of something. The second term is just $B(x)$ by the definition of the Green's function!
(Some more care is needed in the above: deciding how to actually turn the $\nabla_x$ into a $\nabla_y$ and so on, but that's the right idea.)
Okay, that works. Now let's tidy up. We firstly want to show that we can take $\nabla \cdot A=0$. Suppose we define $\Lambda$ so that $-\nabla^2\Lambda=\nabla \cdot A$ (easy enough, using the Green's function). But then $A_{\Lambda}=A+\nabla \Lambda$ also solves $\nabla \times A_{\Lambda} = B$, and has zero divergence. This also tells us how to get from our $A_B$ to a more general $A$ that does not have $\nabla \cdot A=0$: add on a gradient of something.