Let's consider a 2D-square with 4 equal subsquares containing different dielectrics.

Inside the square domain, the unknown electric potential function $\Phi$ satisfies the Laplace equation:


and satisfies the continuity conditions on the interfaces between different dielectrics:


$$\frac{\partial\Phi^+}{\partial n}\epsilon^+=\frac{\partial\Phi^-}{\partial n}\epsilon^-$$

If we apply Dirichlet Boundary conditions to the 4 edges of the square domain, then we can define a surface green function[1] G(x, y | $\eta$) to help solve the unknown function $\Phi$ inside the square domain:


Coordinate (x, y) denotes a point inside the square domain while $\eta$ denotes the point on the domain boundary. The $\Phi(\eta)$ is the Dirichlet Boundary Condition and is known. L is the closed boundary of the square domain.

Assuming the length of the square domain's each edge is a, by numerical method, I found a property of G(x, y | $\eta$):


where the 2 points corresponding to $\eta_i$ and $\eta_j$ are centrosymmetric in reference to the central point of the square domain. $\epsilon_i$ and $\epsilon_j$ are the dielectric value of the subdomains containing them.

If the domain is circle instead of square, the property holds applying Gauss' law, as done in [2] with integral from the center of circle to the boundary. But I dont't think this method can be used for square domain, because the normal vector of the boundary is not parallel with the radius EFI(electric intensity field) vector starting from the center $(\frac{a}{2},\frac{a}{2})$. Yet, I haven't found any other efficient method to this problem.

I'd like to invite you to help me solve the problem.


[1] Y. Le Coz and R. B. Iverson, “A stochastic algorithm for high speed capacitance extraction in integrated circuits,” Solid State Electron., vol. 35, no. 7,pp.1005–1012, Jul. 1992.

[2] R. Schlott, “A Monte Carlo method for the Dirichlet problem of dielectric wedges,” IEEE Trans. Microwave Theory Tech., vol. 36, p. 724, 1988.

  • $\begingroup$ I've worked out the proof even for 3-D version. Later this year, I'll share the paper link here. The basic idea is on random walk. $\endgroup$
    – Ming Yang
    Commented Feb 15, 2019 at 15:20
  • $\begingroup$ Please see this proof from ieeexplore.ieee.org/abstract/document/8966497 if you are interested $\endgroup$
    – Ming Yang
    Commented Sep 29, 2020 at 1:13

1 Answer 1


Please see this proof from ieeexplore.ieee.org/abstract/document/8966497 if you are interested


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