Singular boundary method using time-dependent fundamental solution for scalar wave equations

Abstract

This study makes the first attempt to extend the meshless boundary-discretization singular boundary method (SBM) with time-dependent fundamental solution to two-dimensional and three-dimensional scalar wave equation upon Dirichlet boundary condition. The two empirical formulas are also proposed to determine the source intensity factors. In 2D problems, the fundamental solution integrating along with time is applied. In 3D problems, a time-successive evaluation approach without complicated mathematical transform is proposed. Numerical investigations show that the present SBM methodology produces the accurate results for 2D and 3D time-dependent wave problems with varied velocities c and wave numbers k.

Appendix: The detailed derivation of Eq. (13)

The derivation of Eq. (13) will be provided via a circle domain with uniformly distributed points. It is beyond our capability to derive this formula in more general situations. However, our numerical experiments show that Eq. (13) can be extended to irregular domains with non-equally distributed points.

Consider a 2D Laplace equation on Dirichlet boundary condition

$$\begin{aligned}&\nabla ^{2}u\left( x \right) =0,\quad \quad x\in \Omega , \end{aligned}$$

(31)

$$\begin{aligned}&u\left( x \right) =\bar{{u}}\left( x \right) ,\quad \quad x\in \Gamma _D , \end{aligned}$$

(32)

where $\nabla ^{2}$ denotes the Laplacian operator, u(x) represents the potentials in domain $\Omega$, $\bar{{u}}(x)$ is the known function, $\Gamma _D$ represents the Dirichlet boundary conditions.

The formulation of the SBM can be stated as

(33)

where $G_0 (x_m ,s_n )=\ln (r_{mn} )$ is the fundamental solution of 2D Laplace equations, $U_0^m$ the source intensity factors, $\alpha _n$ the unknown coefficient.

Consider $\Omega$ a circle with radius R. $\left\{ {x_m } \right\} _{m=1}^N$ denote the collocation and source points uniformly placed on the boundary $\partial \Omega$. $L_n$ represents the length of the auxiliary line in 2D as shown in Fig. 5.

For $x_m \in \partial \Omega$, it is easy to verify that the following contour integral is zero.

$$\begin{aligned} \oint {\ln (r(x_m ,s)/R)} d\Gamma _s =0, \end{aligned}$$

(34)

where r is the distance between $x_m$ and source points s on the circle. Eq. (34) can be discretized as

$$\begin{aligned} \sum _{n=1}^N {\ln (r(x_m ,x_n )/R)} =0. \end{aligned}$$

(35)

Due to the symmetry property of the circle, the derivation is independent of the indices of the collocation points. $m=1$ is used in the following derivation. Hence, we have

$$\begin{aligned} \sum _{n=1}^N {\ln (r(x_1 ,x_n )/R)} =0, \end{aligned}$$

(36)

where $x_1 =(1,0)$. Notice that

$$\begin{aligned} r(x_1 ,x_n )=2R\sin \left( \frac{n-1}{N}\pi \right) . \end{aligned}$$

(37)

Substituting Eq. (37) into Eq. (36), and define the desingularized value of the natural logarithm function as a,

$$\begin{aligned} a=\ln R-\ln \left( 2^{N-1}\prod _{n=2}^N {\sin \left( \frac{n-1}{N}\pi \right) } \right) . \end{aligned}$$

(38)

The key issue is to determine the value of $\prod \nolimits _{n=2}^N {\sin \left( \frac{n-1}{N}\pi \right) }$. Let $\omega =\cos (2\pi /N)+i\sin (2\pi /N)=\exp (2\pi i/N)$. Note that $\omega$ is a solution of the equation

$$\begin{aligned} z^{N}-1=0. \end{aligned}$$

(39)

Similarly, $\omega ^{2},\omega ^{3},...,\omega ^{N-1}$ are also the solution of Eq. (39). After some algebraic manipulations, Eq. (39) is equivalent to

$$\begin{aligned}&z^{N-1}+z^{N-2}+\cdots +z+1\nonumber \\&\quad =(z-\omega )(z-\omega ^{2})\cdots (z-\omega ^{N-1}), \end{aligned}$$

(40)

For $z=1$, we obtain

$$\begin{aligned} N=(1-\omega )(1-\omega ^{2})\cdots (1-\omega ^{N-1}). \end{aligned}$$

(41)

It follows that

$$\begin{aligned} N=\left| {1-\omega } \right| \left| {1-\omega ^{2}} \right| \cdots \left| {1-\omega ^{N-1}} \right| . \end{aligned}$$

(42)

Since

$$\begin{aligned} 1-\omega ^{k}=2\sin \frac{k\pi }{N}\left( \sin \frac{k\pi }{N}-i\cos \frac{k\pi }{N}\right) . \end{aligned}$$

(43)

We have

$$\begin{aligned} \left| {1-\omega ^{k}} \right| =2\sin \frac{k\pi }{N}. \end{aligned}$$

(44)

Therefore,

$$\begin{aligned} \prod _{n=2}^N {\sin \left( \frac{n-1}{N}\pi \right) =\frac{N}{2^{N-1}}} . \end{aligned}$$

(45)

From Eqs. (38) and (45), we have

$$\begin{aligned} a=\ln (R/N). \end{aligned}$$

(46)

It follows that $U_0^m$ for the natural logarithm function is

$$\begin{aligned} U_0^m =\ln \left( \frac{2\pi R}{2\pi N}\right) =\ln \left( \frac{L_m }{2\pi }\right) , \end{aligned}$$

(47)

where $L_m =2\pi R/N$.