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In compact surfaces, there are no solution to the following equation:

$$ \nabla^2 \phi(\sigma)=\delta^2(\sigma) $$ since there is no place to go for the electric field lines $E_i=-\partial_i\phi$.

This seems to forbid some charge configurations to happen in electrostatic on a compact surface. Is this theory ill defined? What kind of mechanism protect the theory of dangerous charges configuration?

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  • $\begingroup$ What do you mean by compact surfaces in electrostatics? In electrostatics, we are either interested in the surfaces of conductors or insulators. In the case of conductors, the electric field inside the conductor is zero and just outside it, $E^{\perp} = \sigma/\varepsilon_0$ and $E^{\parallel} = 0$. In the case of insulators, there are boundary conditions to be satisfied also. Otherwise, the electric field will `begin' on the charges producing the field and decrease to zero far away from the field source or end on any conductors or insulators or -ve charges. $\endgroup$ – Physics_Et_Al Jan 13 '18 at 19:36
  • $\begingroup$ Are you referring to the notion of compact support? $\endgroup$ – Physics_Et_Al Jan 13 '18 at 19:38
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    $\begingroup$ I am considering Electrostatic in a two dimensional compact space. A sphere $S_2$ for example. Is not the usal 3 dimensional eletrostatic. $\endgroup$ – Nogueira Jan 13 '18 at 19:47
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The problem is Poincare' s lemma joined with compactness (without boundary) of the 2-manifold $M$ we are considering. Due to them, the flux of the electrical field should be simultaneously zero and $1$ for a $1$-surface (a closed curve) sourronding the support of the delta function, since this curve can viewed as the boundary of a region including the charge but also the boundary of its complement. This property must be valid for every density of charge over the manifold: the total charge must be zero. As a consequence, there is no solution of the eqution you wrote since you are dealing with a density of charge whose integral is different from zero.

There are however some ways out. The simplest one is obtained by adding a negative constant to the right hand side of your equation whose integral on the whole compact surface cancel the integral of the delta. This is a continuous density of charge that compensates the localized charge arising from the delta function.
As a matter of fact the fundamental solution must satisfy $$\Delta_x G(x,y) = \delta(x,y) - S^{-1}\tag{1},$$ where $S$ is the $2$-volume of the whole manifold $M$.

This procedure works for a flat 2- torus in particular, where this sort of foundamental solution can be explicitly computed by means of a double Fourier series (try and pay attention to the zero mode).

The obtaioned fundamental solution produces, using the standard convolution procedure, the potential field (satisfying Poisson equation) generated by a smooth density of charge $\rho$ provided the total (integrated) charge vanishes as it is required. $$\varphi(x) = \int_M G(x,y) \rho(y) dy\tag{2}.$$ This is evident from (1), passing the Laplacian under the sign of integration.

All that works also in compact $n$ dimensional manifold (without boundary) referring to Poisson equation constructed out of a smooth Riemannian metric defined on it.

It is interesting to notice that outside the support of the delta function, the said fundamental solutions which are distributions, are however smooth functions as a consequence of elliptic regularity theorems. For the 2-torus the Fourier series you find has to be considered a series of distributions. However it weakly converges to a smooth function outside the support of the delta.

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