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Give a physical explanation for why the Neumann Problem $$ U_{xx}+U_{yy}=q(x,y) $$ $$ \nabla U(p)\cdot n(p)=g(p) \quad \forall p\in C $$ on $D$ for Poissons equation, has no solution, unless we assume the compatibility condition $$ \iint q(x,y)dxdy=\int g(p(s))ds, $$ Where $s$ denotes the arc length parameter along the boundary $C$ of the region $D$.

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Should that be $U_{xx}+U_{yy}=q(x,y)$? –  Matt Apr 10 '13 at 22:12

2 Answers 2

If you think of $U$ is the temperature, then you have a stationary heat equation. The right-hand side $-q$ is an energy-density (heat load). The integral $$-\int\int q(x,y)\,dx\,dy$$ describes the amount of energy, which is feeded into the system over the area $D$. An analogue holds for $g$ which is a energy-density on the boundary $C$. Hence, the integral $$ \int g \, ds$$ equals the amount of energy, which is feeded into the system over the boundary $C$. Hence, if $$-\int\int q(x,y)\,dx\,dy + \int g \, ds \ne 0$$ you have a positive (or negative) energy flow. Hence, there exist no stationary temperature (the temperature would tend to $\pm \infty$) which could solve the heat equation.

Note that the equation is typically written as $$- ( U_{xx} + U_{yy} ) = -q.$$ Therefore, my explanation contains $-q$ instead of $q$.

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Here I will give an electrostatic potential version of answer:

For any conservative electric field $\mathbf{E}$ (not caused by changing magnetic field), $\mathbf{E} = -\nabla U$, where $U$ is an electrostatic potential. By Gauss Law: $$ \nabla \cdot \mathbf{E} = q $$ where $q$ is the average charge density in the region $D$ at a certain point.

Now consider a thin wire with zero width sitting on the boundary curve $C$ of $D$, with charge density $g$ at point $p\in C$.

Then the electrostatic potential of this wire is $U$, and the electric field of this wire at any point $(x,y)\in D$ is $\mathbf{E}$, then $$ -\Delta U = \nabla \cdot \mathbf{E} = q $$ Hence the total charge in this region is: $$ \iint_D q(x,y) \,dxdy \tag{1} $$

On the other hand, the thin wire sits only on the boundary curve, with charge density $g$, this translates to electric field $\mathbf{E}$'s boundary condition, is: $$ \mathbf{E}\cdot n = -\nabla U \cdot n = -g $$ The total charge of the wire is: $$ -\int_C g \,ds \tag{2} $$

(1) and (2) must be the same for we have no other charged objects in this model. Notice the sign difference, for the positive charge is assume to have the electric field lines pointing away from the charge.


Remark 1: if you like to think in Coulumb's law then a thought experiment can be you put a test wire with charge density 1 along the wire so instantaneously that the test wire's electric field does not interfere with the original field, then the Coulumb force's by two ways of thinking should be the same.

Remark 2: Mathematically it is just divergence theorem, in that the right hand side of Poisson equation must be lying in the range of the operator $\Delta$.

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