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Comparison of formulations for the double-layer potential

In this section we compare our FMM formulation and Fujiwara's one for the double-layer potential in elastodynamics to clarify the difference between these formulations. In this section S denotes a surface in a domain and $\mbox{\boldmath$\space n $ }$ the unit vector normal to S.

First, we describe our formulation for the double-layer potential. The double-layer potential is written as

 
$\displaystyle W_i(\mbox{\boldmath$ x $ })=\int_{S_y} C_{jkpl}\frac{\partial}{\p...
...\boldmath$ y $ }) u_j(\mbox{\boldmath$ y $ })
n_k(\mbox{\boldmath$ y $ }) dS_y.$     (3.160)

where ui is the displacement vector.

Substituting (3.142) into (3.162), we obtain

\begin{eqnarray*}W_i(\mbox{\boldmath$ x $ })=
\sum_{n=0}^{\infty}\sum_{m=-n}^{n}...
...overrightarrow{O\mbox{\boldmath$ x $ }})
\widehat{M}^L_{n,m}(O)
\end{eqnarray*}


where $\widehat{M}^T_{r,n,m}(O)$ and $\widehat{M}^L_{n,m}(O)$ are the same as (3.147) and (3.148). In our formulation the number of the multipole moments is 4. Similarly, one can easily find that the number of the multipole moments for the single-layer potential is also 4.

Second, we describe Fujiwara's formulation for the double-layer potential.

Fujiwara's starting point is to write the fundamental solution (3.139) as

 
$\displaystyle \Gamma_{ip}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ }) =
\fra...
..._T^2} \frac{\partial^2}{\partial y_i \partial y_p}
\frac{e^{i k_L R}}{R}\right)$     (3.161)

Substituting (3.120) into (3.163), one obtains
 
$\displaystyle \Gamma_{ip}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ }) =
\sum...
...tialy_p} \overline{{\cal I}}_n^m(k_L,\overrightarrow{O\mbox{\boldmath$ y $ }}),$     (3.162)

where functions $F^T_{ip,n,m}(\overrightarrow{O\mbox{\boldmath$\space x $ }})$ and $F^L_{ip,n,m}(\overrightarrow{O\mbox{\boldmath$\space x $ }})$ are defined as

\begin{eqnarray*}F^T_{n,m}(\overrightarrow{O\mbox{\boldmath$ x $ }}) &=& \frac{i...
...u }
{\cal O}_n^m(k_L,\overrightarrow{O\mbox{\boldmath$ x $ }}).
\end{eqnarray*}


Substituting (3.164) into (3.162), one obtains

\begin{eqnarray*}W_i(\mbox{\boldmath$ x $ })=
\sum_{n=0}^{\infty}\sum_{m=-n}^{n}...
...rightarrow{O\mbox{\boldmath$ x $ }})
\widetilde{M}^L_{i,n,m}(O)
\end{eqnarray*}


where $\widetilde{M}^T_{i,n,m}(O)$ and $\widetilde{M}^L_{i,n,m}(O)$ are defined as

\begin{eqnarray*}\widetilde{M}^T_{i,n,m}(O) &=&
\int_{S_y} C_{jkpl}\left(\delta_...
...})
u_j(\mbox{\boldmath$ y $ }) n_k(\mbox{\boldmath$ y $ }) dS_y
\end{eqnarray*}


Both $\widetilde{M}^T_{p,n,m}(O)$ and $\widetilde{M}^L_{p,n,m}(O)$ have three components. Therefore, the total number of the multipole moment for the double-layer potential is 6(=3$\times$2) for a given pair of n and m.
next up previous contents
Next: Concluding remarks Up: Three-dimensional scattering of elastic Previous: Numerical examples
Ken-ichi Yoshida
2001-07-28