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Fundamental Solution

Now, let's consider the ``fundamental solution'' u*i, which is the solution to

 \begin{displaymath}
\nabla^2u^*_i+\Delta^i=0,
\end{displaymath} (6.4)

where $\Delta^i$ is the Dirac delta function at point i, i.e.

\begin{displaymath}\int_\Omega\Delta^idV=
\left\{\begin{array}{l}1,\ i\in\Omega\\ 0\ {\rm otherwise}\end{array}\right..
\end{displaymath} (6.5)

We know that

\begin{displaymath}\int_\Omega u(\nabla^2u^*_i+\Delta^i)dV=\int_\Omega u\nabla^2u^*_i dV+u^i,
\end{displaymath} (6.6)

so since u*i satisfies equation 6.4, equation 6.3 becomes

 \begin{displaymath}
u^i +\int_{\partial\Omega_2}u\nabla u^*_i\cdot\vec{dA}
+\int...
...i\bar{q}dA
+\int_{\partial\Omega_1}u^*_i\nabla u\cdot\vec{dA}.
\end{displaymath} (6.7)

Incidentally, this allows the calculation of u at any internal point igiven the u and $\nabla u\cdot\hat{n}$ values on the boundary, according to

\begin{displaymath}u^i = \int_{\partial\Omega}u^*_i\nabla u\cdot\vec{dA}
-\int_{\partial\Omega}u\nabla u^*_i\cdot\vec{dA}.
\end{displaymath} (6.8)

The ``fundamental solutions'' u*i that fit in here are

\begin{displaymath}u^*_i=\frac{1}{2\pi}\ln\left(\frac{1}{r}\right)
\end{displaymath} (6.9)

in 2-D (r is the distance from point i), and

\begin{displaymath}u^*_i=\frac{1}{4\pi r}
\end{displaymath} (6.10)

in 3-D. For a point i on the surface, because the singularity intersects the integration path, we modify equation 6.7 to read:

 \begin{displaymath}
\frac{u^i}{2}+\int_{\partial\Omega_2}u\nabla u^*_i\cdot\vec{...
...i\bar{q}dA
+\int_{\partial\Omega_1}u^*_i\nabla u\cdot\vec{dA}.
\end{displaymath} (6.11)

(Note: if the surface isn't smooth at point i, ui shouldn't be multiplied by $\frac{1}{2}$ but by something else.) This is valid for both 2-D and 3-D cases.


next up previous contents
Next: Boundary elements Up: Theory Previous: Basics
Adam Clayton Powell IV
1999-07-23