Generalized Boundary Integral Equation Method for Boundary Value Problems of Two-D Isotropic Lattice Laplacian

Abstract

A generalized boundary integral equation method for boundary value problems of two-dimensional isotropic lattice Laplacian is proposed in this paper. The proposed method is an extension of the classical boundary integral equation method with notable advantage. By utilizing the asymptotic expression of the fundamental solution at infinity, this method effectively addresses the challenge of numerical integration involving singular integral kernels. The introduction of Green’s formulas, Dirichlet and Neumann traces, and other tools which are parallel to the traditional integral equation method, form a solid foundation for the development of the generalized boundary integral equation method. The solvability of boundary integral equations and the solvability of lattice interface problem are important guarantees for the feasibility of this method, and these are emphasized in this paper. Subsequently, the generalized boundary integral equation method is applied to boundary value problems equipped with either Dirichlet or Neumann boundary conditions. Simple numerical examples demonstrate the accuracy and effectiveness of the generalized boundary integral equation method.

Appendix A. Derivation of Asymptotic Expression of the Fundamental Solution

In the appendix, we give the derivation of the asymptotic expression for the fundamental solution at infinity.

Given generating vectors $\xi =(\xi _1, \xi _2)$ and $\eta =(\eta _1, \eta _2)$, we know that the fundamental solution is defined as

$$\begin{aligned} G(t)=\frac{1}{|2\pi T|}\int _{2\pi T}\frac{[\exp (it\cdot k)-1]dk}{2\sum _{s\in \varPi }\sin ^2\left( \frac{s\cdot k}{2}\right) \gamma (s)},\ \forall \, t \in \varPi . \end{aligned}$$

Writing t and k in component form $t = (a,b)^{T}$, $k = (x,y)^{T}$, and using the variable substitutions

$$\begin{aligned} X=\xi _1 x+\xi _2 y, \quad Y=\eta _1 x+\eta _2 y, \end{aligned}$$

i.e.,

$$\begin{aligned} x=\alpha _1 X+\beta _1 Y, \quad y=\alpha _2 X+\beta _2 Y, \end{aligned}$$

where

$$\begin{aligned} \alpha _1=\frac{\eta _2}{\eta _2 \xi _1-\eta _1 \xi _2}, \alpha _2=\frac{-\eta _1}{\eta _2 \xi _1-\eta _1 \xi _2}, \beta _1=\frac{-\xi _2}{\eta _2 \xi _1-\eta _1 \xi _2}, \beta _2=\frac{\xi _1}{\eta _2 \xi _1-\eta _1 \xi _2}, \end{aligned}$$

we know

$$\begin{aligned} 2\pi G(a,b)= T_1(F), \end{aligned}$$

where $T_1$ is defined in article [8], and

$$\begin{aligned} F(X,Y)=\frac{1}{2\sum _{s\in \varPi }\sin ^2\left( \frac{s\cdot k}{2}\right) \gamma (s)}\sim \frac{1}{\lambda _\gamma (x^2+y^2)} \quad \textrm{as} \quad (x,y) \rightarrow (0,0). \end{aligned}$$

In the above equation,

$$\begin{aligned} \lambda _\gamma (x^2+y^2)&=\lambda _\gamma ((\alpha _1 X+\beta _1 Y)^2+(\alpha _2 X+\beta _2 Y)^2) \\&=\lambda _\gamma ((\alpha _1^2+\beta _1^2)X^2+(\alpha _2^2+\beta _2^2)Y^2+2(\alpha _1 \beta _1+\alpha _2 \beta _2)XY). \end{aligned}$$

Let

$$\begin{aligned}{} & {} g=\left( \begin{array}{cc} \lambda _\gamma (\alpha _1^2+\beta _1^2) &{} \lambda _\gamma (\alpha _1 \beta _1+\alpha _2 \beta _2)\\ \lambda _\gamma (\alpha _1 \beta _1+\alpha _2 \beta _2) &{} \lambda _\gamma (\alpha _2^2+\beta _2^2) \end{array}\right) ,\\{} & {} ({\tilde{X}},{\tilde{Y}})^{T}=g^{\frac{1}{2}}(X,Y)^{T}, \end{aligned}$$

then we know

$$\begin{aligned} 2\pi G(a,b)=\textrm{det}g^{-\frac{1}{2}}T_1(F\circ g^{-\frac{1}{2}})({\tilde{a}},{\tilde{b}}), \end{aligned}$$

where

$$\begin{aligned} ({\tilde{a}},{\tilde{b}})^{T}= g^{-\frac{1}{2}} (a,b)^{T}. \end{aligned}$$

Following Corollary 2 in [8], we have

$$\begin{aligned} T_1(F\circ g^{-\frac{1}{2}})=-\textrm{log}{\tilde{t}}+\textrm{log}2-\varUpsilon +T(F_1)\\ -\underset{\epsilon \rightarrow 0}{\textrm{lim}}\left( \textrm{log}\epsilon +(2\pi )^{-1} \int _{\epsilon }^{\infty }rdr\int _{0}^{2\pi }F \circ g^{-\frac{1}{2}} d\theta \right) \end{aligned}$$

where ${\tilde{t}}=({\tilde{a}}^2+{\tilde{b}}^2)^{\frac{1}{2}}$, $F_1=F-{\tilde{R}}^{-2}S({\tilde{R}})$, ${\tilde{R}}=({\tilde{X}}^2+{\tilde{Y}}^2)^{\frac{1}{2}}$. S is a mollifier mentioned in [8].

The limit in the above equation is a definite value that can be calculated, so it remains to determine $T(F_1)$. Using Taylor expansion and following Corollary 1, Theorem 2 in [8], we obtain that

$$\begin{aligned} T(F_1)&\sim \frac{1}{{\tilde{t}}^2}\left( { C_1 \left( 6-\frac{24{\tilde{a}}^2}{{\tilde{t}}^2}+\frac{16{\tilde{a}}^4}{{\tilde{t}}^4} \right) + C_2 \left( 6-\frac{24{\tilde{b}}^2}{{\tilde{t}}^2}+\frac{16{\tilde{b}}^4}{{\tilde{t}}^4} \right) } \right. \nonumber \\&\quad \left. +C_3 \left( -2+\frac{16{\tilde{a}}^2{\tilde{b}}^2}{{\tilde{t}}^4}\right) + C_4\left( -\frac{12{\tilde{a}}{\tilde{b}}}{{\tilde{t}}^2}+\frac{16{\tilde{a}}^3{\tilde{b}}}{{\tilde{t}}^4} \right) + C_5 \left( -\frac{12{\tilde{a}}{\tilde{b}}}{{\tilde{t}}^2}+\frac{16{\tilde{b}}^3{\tilde{a}}}{{\tilde{t}}^4} \right) \right) , \end{aligned}$$

where $C_i, i=1,2,3,4,5$ are constants depend on g. Then we have that

$$\begin{aligned} T(F_1) = O\left( \frac{1}{{\tilde{t}}^2}\right) . \end{aligned}$$

So from all of this, we get (6).