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Direct computation

Discretising the regularised integral equation (3.141) with piecewise constant shape functions, we obtain
 
$\displaystyle { t_{a}^{I}(\mbox{\boldmath$ x $ })=
n_{b}(\mbox{\boldmath$ x $ }...
..._n}\Gamma_{mp}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ })
dy_r \phi_{j}^L }$
    $\displaystyle - n_{b}(\mbox{\boldmath$ x $ }) C_{ablm} \rho \omega^2 \sum_{L}\i...
...ath$ x $ }-\mbox{\boldmath$ y $ }) n_l(\mbox{\boldmath$ y $ }) dS_{y} \phi_j^L,$ (3.155)

where SJ is a plane element in Sy and $\mbox{\boldmath$\space \phi $ }^J$ represents $\mbox{\boldmath$\space \phi $ }$ on SJ. In the evaluation of the integral in (3.157) we divide (3.139) into the static part (the fundamental solution of the equation of elastostatics) and the residual part as follows:

\begin{eqnarray*}\Gamma_{ij}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ }) = \G...
...amma^{R}_{ij}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ }),\\
\end{eqnarray*}


where $\Gamma^S_{ij}(\mbox{\boldmath$\space x $ }-\mbox{\boldmath$\space y $ })$ and $\Gamma^R_{ij}(\mbox{\boldmath$\space x $ }-\mbox{\boldmath$\space y $ })$ are given by

\begin{eqnarray*}\Gamma_{ij}^S(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ })=\f...
...rac{\partial}{\partialy_i}
\frac{\partial}{\partialy_j}\right)R,
\end{eqnarray*}



 
$\displaystyle \Gamma_{ij}^R(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ }) =
\f...
..._T^2}\right)
\frac{\partial}{\partialy_i}\frac{\partial}{\partialy_j} R\right),$     (3.156)

where $R=\vert\mbox{\boldmath$\space x $ }-\mbox{\boldmath$\space y $ }\vert$. Namely, we treat four integrals to compute (3.157) as follows:
 
$\displaystyle { t_{a}^{I}(\mbox{\boldmath$ x $ })=
n_{b}(\mbox{\boldmath$ x $ }...
..._n}\Gamma_{mp}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ })
dy_r \phi_{j}^L }$
    $\displaystyle - n_{b}(\mbox{\boldmath$ x $ }) C_{ablm} \rho \omega^2 \sum_{L}\i...
...math$ x $ }-\mbox{\boldmath$ y $ }) n_l(\mbox{\boldmath$ y $ }) dS_{y} \phi_j^L$  
    $\displaystyle + n_{b}(\mbox{\boldmath$ x $ }) C_{ablm}
\sum_{L} \oint_{\partial...
..._n}\Gamma^R_{mp}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ })
dy_r \phi_{j}^L$  
    $\displaystyle - n_{b}(\mbox{\boldmath$ x $ }) C_{ablm} \rho \omega^2 \sum_{L}\i...
...math$ x $ }-\mbox{\boldmath$ y $ }) n_l(\mbox{\boldmath$ y $ }) dS_{y} \phi_j^L$ (3.157)

In the computation of the integrals related to $\Gamma^R_{ij}(\mbox{\boldmath$\space x $ }-\mbox{\boldmath$\space y $ })$ we expand (3.158) into the following form:

 
    $\displaystyle \Gamma^R_{ij}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ })$  
    $\displaystyle = \frac{1}{4\pi\mu}
\biggl\{\delta_{ij}\left[\frac{e^{i k_T R}}{R...
...
+\left(\frac{i}{k_T R} - \frac{1}{(k_T R)^2}\right)\frac{e^{ik_TR}}{R}
\right.$  
    $\displaystyle \left. -\left(\frac{k_L}{k_T}\right)^2
\left(\frac{i}{k_T R} - \f...
...left(1+\frac{3i}{k_T R} - \frac{3}{(k_T R)^2}\right)\frac{e^{ik_TR}}{R}
\right.$  
    $\displaystyle \left. -\left(\frac{k_L}{k_T}\right)^2
\left(1+\frac{3i}{k_T R} -...
...}}{R}
+\frac{1}{2}\left(1-\left(\frac{k_L}{k_T}\right)^2\right)\right]
\biggr\}$ (3.158)

When kTR is small we use the following series instead of (3.160)
$\displaystyle \Gamma^R_{ij}(\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ })$ = $\displaystyle \frac{1}{4\pi\mu}\biggl\{\delta_{ij}
\sum_{n=0}^{\infty}
\frac{(ik_T)^{n+1}R^n}{(n+1)!(n+3)}
\left(n+2+\left(\frac{k_L}{k_T}\right)^{n+3}\right)$  
    $\displaystyle -R_{,i}R_{,j}\sum_{n=0}^{\infty}
\frac{(ik_T)^{n+1}n R^n}{(n+1)!(n+3)}
\left(1-\left(\frac{k_L}{k_T}\right)^{n+3}\right)
\biggr\}$ (3.159)

We compute the integrals related to the static part in (3.159) analytically and the rest numerically with Gaussian quadrature.
next up previous contents
Next: Algorithm Up: Numerical procedure Previous: Truncation of the infinite
Ken-ichi Yoshida
2001-07-28