Are there square integrable vector field solutions of curl(F) = F?

329 Views Asked by At

Are There Square Integrable Vector Field Solutions of Curl(F) = F ?

More exactly, consider three functions which satisfy the set of three simultaneous linear first order PDE's implied by the expression curl(F) = F with no finite boundary conditions. These functions for the components of F should be expressed in terms of known functions to provide some intuitive understanding of the space of solutions and provide a basis for proving some theorems about that space. Square integrable (SQI) means the integral of the norm of the solution over all space gives a finite value.

When supplied with appropriate time factors, the vector field satisfying these conditions can serve as both the electric field and the magnetic field of a solution of Maxwell's equations. This type of solution is called self dual. If the norm of the solution vector F is SQI, the solution could be considered as a clump of energy, if reasonably stable. The stability would depend on the details of the interaction of the time factors with the components of F. This (Physics Stack} question is concerned only with the components of F. They may be thought of as a kind of geometric framework of the Maxwell solution. Their calculation is already quite challenging.. The time factors and stability can be considered later.

Solutions without finite boundary conditions exist in at least three coordinate systems: rectangular, cylindrical and spherical. In rectangular coordinates (x,y,z) the solutions are of the form (0, sin(x), cos(x)). In cylindrical coordinates a solution can be obtained with Bessel functions of the radius: the r component is zero, the theta component is a first order Bessel function of r, and the r component is a zero order Bessel function of r; in symbolic form (0, J1(r), J0(r)). In spherical coordinates, solutions have been expressed in terms of spherical Bessel functions by Chubykalo and Espinoza in "Journal of Physics A Math 35" in 2002. These spherical solutions go to zero in all directions but not fast enough to localize the function enough to get an SQI solution. No solution that I know of is SQI. Are there SQI solutions (with no boundary conditions except at infinity and which can be expressed in terms of known functions) in general orthogonal curvilinear coordinate systems?

Special emphasis is on toroidal and bispherical coordinates. Definitions of these coordinate systems can be found online by using those names as search terms or in "Methods of Theoretical Physics" by Morse and Feshbach and also in "Field Theory Handbook" by Moon and Spencer. As explained by Morse and Feshbach, use of a special coordinate system is a guess that a solution can be found with integral lines that coincide with the coordinate lines thus permitting a simple expression for that solution. I have not been able to find any way to separate the variables in these coordinates for the expression curl(F) = F.

Here is something that may help in finding SQI solutions. Since F is a curl, it must be that div(F) = 0 and its integral lines are closed curves or can terminate only at a singularity or infinity. Thus the location of the singular points of the coordinate system play a strong, possibly dominant, roll in forming the shape of the solution. In the coordinate systems mentioned above there is at most one singular point or at most one singular line.

In other orthogonal curvilinear systems there are more than one singular point and/or more than one singular line which might help to localize the solution. For example, in bispherical coordinates the two foci are singular points and in an infinitesimal neighborhood around them, the metric is the same as spherical coordinates so there are solutions like those of spherical coordinates in the infinitesimal focal neighborhoods. These solutions can be continued (but numeric solutions are of little value in carrying out the objectives mentioned above.) The sphere like surfaces of fixed value of a larger size than those which surround a focal point in bispherical coordinates may form a region containing most of the value when they crunch into the corresponding spheres around the other focal point. Thus a more localized solution may be created which might be SQI.

Questions: Is this reasoning correct or are there flaws?
Are there solutions F (in terms of known functions) of the expression curl(F) = F for toroidal or bispherical coordinates? Are there SQI solutions F (in terms of known functions and filling all space) of the expression curl(F) = F for toroidal or bispherical coordinates? If so, what are they?
Are there reasons for thinking there are no such solutions?

3

There are 3 best solutions below

4
On

If $\newcommand{\bF}{\mathbf{F}} \bF$ is square-integrable its Fourier transform exists. In Fourier space, $(\nabla \times \bF)(x) = \bF(x)$ is $i\mathbf k \times \bF(\mathbf k) = \bF(\mathbf k) $. This equation has no non-trivial solutions.

1
On

For the more mathematically minded:

By taking the curl of your equation we see that any solution must satisfy $-\nabla^2 \mathbf F + \mathbf F = 0.$ Let us consider the more general equation $-\nabla^2 \mathbf F + \mathbf F = \mathbf u$ where $\mathbf u$ is a known function that vanishes at infinity. For any test function $\mathbf G$, it must then be the case that, by integration by parts, this expression is $$\int \mathbf G \cdot ((-\nabla^2 + 1) \mathbf F) = \int (\nabla_i G_j)(\nabla_i F_j) + G_i F_i= \int \mathbf G\cdot \mathbf u. $$ This idea of looking for solutions to PDE:s by moving derivatives to test functions is generally called a weak formulation. It's very important for PDE:s because for example it is the starting point for finite element methods.

Define a bilinear form $B$ on a suitable Sobolev space by $$B(\mathbf F, \mathbf G) := \int (\nabla_i G_j)(\nabla_i F_j) + G_i F_i$$ The bilinear form $B$ is bounded, $|B(\mathbf G, \mathbf F)| \le C \|\mathbf F \| \| \mathbf G\|$ by the Poincaré inequality and Cauchy-Schwarz, and clearly coercive $B(\mathbf F, \mathbf F) \ge c\| \mathbf F \|$. By the Lax-Milgram theorem, there exists a unique $\mathbf F$ such that $$B(\mathbf G, \mathbf F) = \int \mathbf G \cdot \mathbf u$$ holds for any $\mathbf G$.

This of course means that with $\mathbf u \equiv 0$, no non-trivial solution exists.

See also Example 2 on the Wikipedia page about elliptic operators. The operator $-\nabla^2 + 1 $ is clearly of that form. Thus the elliptic regularity theorem detailed in the section below applies and the weak formulation isn't too weak.

0
On

The solutions of the equation which expresses the equality of the vector product of the vector k of the Fourier transform variables and the vector F(k) of the transforms of the components of F, with that vector F(k), can only have solutions which are divisors of zero, ie., must have infinite values, and thus the corresponding components cannot be SQI. The simplest answer to the stated question is therefore NO. However some well-behaved and useful functions have transforms having infinite values, eg., Bessel functions. The question should be rephrased so as not to emphasize SQI solutions.