0
$\begingroup$

The electrostatic Laplace problem for the exterior of a disk can be solved in a straightforward manner using separation of variables. Suppose we have a unit disk $\Omega$ with a charge density of $f$ on the boundary. Then the solution to the exterior problem, \begin{align} & \Delta v(x) = 0 \quad \quad x\in \mathbb{R}^2\setminus \overline{\Omega}, \\ & v |_{\partial D}(x) = f, \end{align} is given through separation of variables as $$ v(x) = v(r,\theta) = \sum_{n=0}^\infty r^{-n}(a_n \cos(n\theta)+b_n \sin(n\theta)), $$ where $$ a_n = \frac{1}{2\pi}\int_0^{2\pi} f(\theta)\cos(n\theta) d\theta, \\ b_n = \frac{1}{2\pi}\int_0^{2\pi} f(\theta)\sin(n\theta) d\theta. $$ But what about the case where we have two disks $\Omega_1$ and $\Omega_2$, and we want to determine the solution to the exterior Laplace problem for a charge density $f$, with $f$ defined on the union of the disks $D=\Omega_1\cup \Omega_2$? Say for example the disks both have radius $1$ and are symmetric on the $x_1$ axis with a distance of $6$ between them.

Can we just take define two new origins $O_1$ and $O_2$ at the center of the disks, with associated polar coordinates $(r_1,\theta_1)$ and $(r_2,\theta_2)$, and then say the following: $$ v(x) = \sum_{n=0}^\infty r_1^{-n}(a_n \cos(n\theta_1)+b_n \sin(n\theta_1)) + \sum_{n=0}^\infty r_2^{-n}(a_n \cos(n\theta_2)+b_n \sin(n\theta_2)). $$ Is this a valid solution? If not, why not?

$\endgroup$
0
$\begingroup$

This is indeed a solution due to the linearity of the Lapacian. If I have 2 charge distributions $\rho_1$ and $\rho_2$ and

\begin{align} \nabla^2 v_1 &= \rho_1\\ \nabla^2 v_2 &= \rho_2 \end{align} then we have that \begin{align} \nabla^2 v_1 + \nabla^2 v_2 &= \rho_1 + \rho_2\\ &= \nabla^2( v_1 + v_2)\;. \end{align}

This is known as the principle of superposition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.