|
Lesson 33, 34: The Interior and Exterior Dirichlet Problem for a Circle
- It is interesting to note that from this formula, it can easily be seen that
$$
u(0,0)=\frac{1}{2 \pi} \int_0^{2 \pi} g(\alpha) d \alpha .
$$ - That is, the value of the solution at the center of the circle is equal to the average value of $g$ on the boundary.
- Furthermore, by the Law of Cosines, the denominator in the integrand is the square of the length of the side opposite the angle $|\theta-\alpha|$ of the triangle with vertices $(0,0),(r, \theta)$, and $(R, \alpha)$.
- For a BVP defined on a non-circular domain, this formula can be applied by first conformally mapping the domain to a circle, using the formula on the transformed problem, and then transforming back to the original domain.
|
|