# Physical interpretation of the integral formula for the solution of Laplace equation with Dirichlet/Neumann boundary condition

Suppose we have a bounded domain $D$ with smooth boundary, with $G(x,y)$ being the Green's function for the Poisson equation on $D$, i.e. $G(x,\cdot)=0$ on $\partial D$ and $\Delta_y G(x,y)=\delta_x(y)$. We know that the solution of $\Delta u=f, u|_{\partial D}=0$ is given by $u(x)=\int_D{G(x,y)\,f(y)\mathrm dy}$. Its physical interpretation is exactly the Coulomb's law and the superposition principle.

Also we have a formula for the Dirichlet problem: $$u(x)=\int_{\partial D}{K(x,y)u(y)\mathrm dy}, \text{where }K(x,y)=\frac{\partial}{\partial n_y}G(x,y)$$

Realizing that the normal derivative of electric potential is the surface charge density on a conductor, I think there might be some physical interpretation of this using the Coulomb's law involving the surface charge. However I failed to find one because $\partial D$ cannot be a conductor if $u|_{\partial D}$ is nonconstant.

The Neumann condition is trickier. Consider a different Green's function $G_N(x,y)$ such that $\dfrac{\partial}{\partial n}G(x,\cdot)=c$ on $\partial D$ and $\Delta_y G(x,y)=\delta_x(y)$. The constant $c$ must be $1$ over the area of $\partial D$ by compatibility. (I read it from here.) By Green's representation formula, one can get $$u(x)=C-\int_{\partial D}{G_N(x,y)\frac{\partial u}{\partial n_y}\mathrm dy},$$where $C$ is a constant.

Again, is there any interpretation here? In this case a basic and not necessarily physical interpretation also helps because I know too few about Neumann condition. (I don't know the physical background of Neumann condition, nor can I find $G_N$ on simple domains like balls and half planes.)

## 1 Answer

There is an interpretation for the Dirichlet condition, i.e. $u(x)=\int_{\partial D}{K(x,y)u(y)\mathrm dy}, \text{where }K(x,y)=\frac{\partial}{\partial n_y}G(x,y)$, due to a physics friend of mine, H. Lam.

Given a charge distribution $Q$, we write the potential of these charges given by Coulomb's Law as $U(Q)$, a function on $\mathbb R^n$. Given a potential function $u$, we write the potential energy of the charge distribution $Q$ in $u$ as $uQ$. We have $U(Q_1)Q_2=U(Q_2)Q_1$. In other words $U(\cdot)\cdot$ is a symmetric bilinear functional on the space of charge distributions.

Now given $x\in D$. Consider the first scenario where $\mathbb R^n-D$ is a conductor and there is a unit point charge $Q$ at $x$. The induced charge $Q_i$ lies in $\partial D$ and we have $G(x,\cdot)=U(Q+Q_i)$ (here $G(x,\cdot)$ are defined to be $0$ outside $D$). By a property of conductors, $K(x,y)=\dfrac{\partial}{\partial n_y}G(x,y)$ is minus the surface charge density of $Q_i$.

Consider another scenario where the potential inside $D$ is given by $u$. Let $Q_u$ be a possible charge distribution that realizes this scenario. As $u$ is harmonic, $Q_u$ lies in $\mathbb R^n-D$.

Back to the formula. We have $u(x)=u(Q)=U(Q_u)Q$ and $RHS=\int_{\partial D}{K(x,y)u(y)\mathrm dy}=-U(Q_u)Q_i$. Then $LHS-RHS=U(Q_u)(Q+Q_i)=U(Q+Q_i)Q_u=G(x,\cdot)Q_u=0$ as $Q_u$ is complete outside $D$.

In spirit this is the same as the mathematical proof, as $U(Q_1)Q_2=\langle U(Q_1),\Delta U(Q_2) \rangle_{L^2(\mathbb R^n)}$, which can be proved to be symmetric by integration by parts.

As a side observation, this method can also give an interpretation of the symmetry of Green's function, or Principle of Reciprocity as is referred to in electrostatics.

Let $x,y\in D$, we want to prove $G(x,y)=G(y,x)$. Suppose the outside of $D$ is a conductor. In one scenario, there is a unit point charge $Q_x$ at $x$. The induced charge on $\partial D$ is denoted by $Q_x^i$. We have $U(Q_x+Q^i_x)=G(x,\cdot)$. In other scenario, we similarly define $Q_y$ and $Q^i_y$. Note that $U(Q_x+Q^i_x)Q^i_y=G(x,\cdot)Q^i_y=0$ as $Q^i_y$ lies on $\partial D$. Therefore

$$G(x,y)=U(Q_x+Q^i_x)Q_y=U(Q_x+Q^i_x)(Q_y+Q^i_y)=U(Q_y+Q^i_y)(Q_x+Q^i_x)=G(y,x)$$