# Doubt regarding question on Bogomolny equations

### PROBLEM:

Let $$\bf E$$ and $$\bf B$$ be static, and source free electric and magnetic fields in $$\mathbb R^3,$$ respectively which are infinitely many times differentiable. Also assume that they satisfy either of the following : $$\mathbf E = \mathbf B ~\textrm{or}~ \mathbf E = -\mathbf B.$$ (Bogomolny equations). Prove that as long as the equations describe a finite energy configuration, $$\mathbf E = \mathbf B = 0.$$

### MY ATTEMPT:

Consider a very large boundary $$S,$$ compared to the region under consideration. Since $$\int \mathbf E^2~\mathrm dv$$ all over space is a finite value(since the energy configuration described is finite), the integrand must go to zero at infinity. Or, at my chosen surface $$S,~~ \mathbf E = -\textrm{grad}(V) = 0.$$ Thus we see that the potential $$V$$ is the same everywhere on the surface $$S$$ (since $$\bf E$$ is zero everywhere on the surface). Let $$V(S) = a.$$ So the problem reduces to the boundary problem, $$V(S) = a$$ at all points on $$S$$ and also the laplacian of $$V$$ is zero everywhere inside $$S$$ (since it is source free). One evident solution to this is $$\bf E = 0,$$ (as a result of which $$\bf B = 0,$$ from the Bogomolny equations). Hence, this must be the only solution by uniqueness.

### MY DOUBT:

Is it okay to start working in the boundary conditions using this surface $$S$$ as I defined it? If not, can the definition of $$S$$ be changed to satisfy the rest of the solution? If that cannot be done, please suggest other solutions.

PS: This is a problem from RUDOLF ORTVAY COMPETITION IN PHYSICS (1997) problem 28. I do not consider this as a homework question and was asked by no one to get this solved. This is out of my interest only. If you feel like suggesting the homework tag please do so.

• Lelouch, you know very well that the tag is "homework-and-exercises"! This is certainly an exercise. You have definitely shown an attempt to solve the problem. The issue is whether you are asking a conceptual question, or merely asking "Is my solution correct?" I think it is the latter. If you disagree, please can you make clear what your "conceptual doubt" is? Commented Aug 20, 2016 at 15:31
• What i don't understand is whether it is meaningful, mathematically / physically to work in the limiting case mentioned here. If it is, i am almost sure the rest is correct. Commented Aug 20, 2016 at 15:46
• I think this is a perfectly fine question. Not every question involving proving an equation is an 'exercise'. There are plenty of pure conceptual issues here to deal with. Commented Aug 20, 2016 at 22:24

It is false that if $\int_{\mathbb R^n} |f|^2~\mathrm dx^n <+ \infty$ then necessarily $f(x) \to 0$ as $|x| \to +\infty$, so your approach cannot work.

The only proof I see is however based on a non-trivial property of harmonic functions whose proof is a bit technical as it relies upon some properties of subharmonic functions and Hoelder inequality:

Proposition. If $g : \mathbb R^n \to \mathbb R$ is harmonic and $$\int_{\mathbb R^n} |g|^p~\mathrm dx^n <+\infty$$ for some $p$ with $1\leq p < +\infty$, then $g=0$ everywhere.

Next the way is easy. As ${\bf E}$ (supposed to be $C^1$) is static, $\nabla \wedge {\bf E}=0$ everywhere in $\mathbb R^3$ which is simply connected and thus ${\bf E} = -\nabla f$ for some $C^2$ scalar function $f$. Since ${\bf E} = \pm {\bf B}$, also $\nabla \cdot {\bf E}=0$, so that $\Delta f=0$. In other words $f$ is harmonic on the whole $\mathbb R^3$. In particular $f$ is also $C^\infty$.

On the other hand we know that $$\int_{\mathbb R^n} {\bf E}^2~\mathrm dx^3 <+ \infty$$ which means $$\int_{\mathbb R^n} g_k^2~\mathrm dx^3 <+ \infty$$ for $g_k := \frac{\partial f}{\partial x_k}$ where $k=1,2,3$ and every $g_k$ is harmonic as well obviously. Applying Proposition, we immediately have that $g_k=0$ everywhere in $\mathbb R^3$ for $k=1,2,3$. In other words is $f$ is constant (since $\mathbb R^3$ is connected). We have eventually obtained that ${\bf E} = -\nabla f=0$ everywhere as wanted.

ADDENDUM. I constructed a short and quite elementary proof of Proposition for the case $p=2$, the only relevant here.

If $g$ is harmonic and $x$ is a point in its domain, the theorem of average value of harmonic functions states that $$g(x) = \frac{\displaystyle\int_{B_R} g ~\mathrm dx^n}{\textrm{Vol}(B_R)}\tag{1}$$ where $B_R$ is a closed ball of finite radius $R$ centered on $x$ completely included in the domain of $g$. The Cauchy-Schwartz inequality says that $$\left|\int_{B_R} g~\mathrm dx^n\right|= \left|\int_{B_R} 1 \cdot g~\mathrm dx^n\right| \leq \sqrt{\int_{B_R} 1^2 ~\mathrm dx^n}\sqrt{\int_{B_R} g^2~\mathrm dx^n}= \sqrt{\textrm{Vol}(B_R)}\sqrt{\int_{B_R} g^2~\mathrm dx^n}\:.$$ Inserting this result in (1), for $g$ everywhere harmonic in $\mathbb R^n$ we have $$0\leq |g(x)| \leq \sqrt{\frac{\displaystyle\int_{B_R} g^2~\mathrm dx^n}{\textrm{Vol}(B_R)}} \to 0 \quad \mbox{for R \to +\infty}\:.$$ The limit can be computed because (a) $R$ can be taken arbitrarily large since $g$ is everywhere defined in $\mathbb R^n$, (b) $\textrm{Vol}(B_R) = C_n R^n \to +\infty$ and (c) $\int_{B_R} g^2~\mathrm dx^n \to \int_{\mathbb R^n} g^2 ~\mathrm dx^n <+\infty$ (e.g., using the dominate convergence theorem). We conclude that $g(x)=0$ for every $x \in \mathbb R^n$.

• Can you please provide a counterexample to disprove the necessity of E going to zero as x goes to infinity ? Commented Aug 20, 2016 at 15:58
• for continuos , differentiable E. Commented Aug 20, 2016 at 16:09
• f(x) = 1 in [0,1], 2 in [2, 2+1/2^3], 3 in [3, 3+ 1/3^4], 4 in [4, + 1/4^5],... 0 in all other points Commented Aug 20, 2016 at 16:10
• You can make as smooth this example as you want obtaining a $C^\infty$ function... Commented Aug 20, 2016 at 16:11
• Indeed the point is not the smoothness of $f$, but the fact that it is harmonic which prevents these pathologic structures. In $\mathbb R$ the only harmonic functions are of the form $ax+b$ and thus my counterexample does not apply, in more dimension the properties of harmonic functions are much more complicated but they are enough to get rid of crazy oscillating functions in view (essentially by means of the so called maximum principle and related results)... Commented Aug 20, 2016 at 16:20