Applying boundary conditions on an almost spherical body

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This discussion focuses on solving the Laplace equation in three dimensions, specifically under azimuthal symmetry using coordinates V=V(r,θ). The boundary conditions provided are V(R(t)+εf(t,θ),θ)=1 and V→0 as r→∞, where ε is a small perturbation. The conversation explores the applicability of separation of variables and suggests that a Green's function approach may be more suitable. The use of perturbation theory is recommended for approximating solutions, leading to a series of boundary value problems that must be solved sequentially.

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hunt_mat
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I am solving the Laplace equation in 3D:
\nabla^{2}V=0
I am considering azumuthal symmetry, so using the usual co-ordinates V=V(r,\theta). Now suppose I have two boundary conditions for [V, which are:
V(R(t)+\varepsilon f(t,\theta),\theta)=1,\quad V\rightarrow 0\quad\textrm{as}\quad r\rightarrow\infty
where \varepsilon\ll1. The boundary condition does lead itself to separation of variables does it? Would a more general Green function approach be more suitable?
 
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Since ##t## is a parameter I will ignore it. One classic way to try and find an approximate solution is to use perturbation theory. Your boundary condition can be Taylor expanded, $$
1 = V(R + \epsilon \, f(\theta),\theta) = V(R, \theta) + \epsilon \, f(\theta) V_r(R, \theta) + \frac{1}{2}\left( \epsilon \, f(\theta) \right)^2 V_{rr}(R, \theta) + \ldots $$
where ##V_r = \partial V / \partial r ##, etc, and look for a solution of the complete boundary value problem of the form,
$$ V(r,\theta) = V^{(0)}(r,\theta) + \epsilon V^{(1)}(r,\theta) + \epsilon^2 V^{(2)}(r,\theta) + \ldots $$
Here ##V^{(k)}(r,\theta)## is simply a function, and the number ##k## tells us what power of ##\epsilon## is in front of it. We plug the expansion for ##V(r,\theta)## into Laplace's equation and collect terms with the same power of ##\epsilon## to find, that for all ##k## we must have ##\nabla^2 V^{(k)}(r,\theta)=0##. Likewise, we plug the expansion for ##V(r,\theta)## into the Taylor expanded boundary condition and collect terms with the same power of ##\epsilon##. This leads to a series of boundary value problems. The first few boundary conditions are
$$
\begin{eqnarray}
V^{(0)}(R,\theta) & = & 1 \\
V^{(1)}(R,\theta) & = & -f(\theta) V^{(0)}_r(R,\theta)\\
V^{(2)}(R,\theta) & = & -f(\theta) V^{(1)}_r(R,\theta) - \frac{1}{2}f^2(\theta) V^{(0)}_r(R,\theta)\\
\vdots
\end{eqnarray}
$$

That is, you first solve ##\nabla^2 V^{(0)} = 0## with ##V^{(0)}(R,\theta) = 1##, then solve ##\nabla^2 V^{(1)} = 0## with ##V^{(1)}(R,\theta) = - f(\theta) V^{(0)}_r(R,\theta)##, etc.

You might be able to use separation of variables for each of these. If ##\epsilon## is small enough you probably just need to solve the first few problems, then add up the solutions accordingly.

Jason
 
Last edited:
Forgot to address this above.

hunt_mat said:
The boundary condition does lead itself to separation of variables does it?
Unless you are able to define a curvilinear coordinate system such that your boundary is a surface of constant coordinate (such as a sphere in spherical coordinates, ellipse in ellipsoidal coordinates, etc.), then the problem is not separable.
 

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