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Say we want to calculate the gravitational potential everywhere around(outside) a solid, circular, right cylinder. We want to use Poisson's equation for gravity for that (Laplace(U) = -4*pi*density or something like that). Since we want to know the potential outside the cyllinder this reduces to laplaces equation laplace(U) = 0

My question is: What would the boundary conditions for this equation be? Obviously one is that it decays to zero at infinity, but what about the boundary conditions on surface of the cyllinder?

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The boundary condition for this equation can be whatever you want, Dirichlet (value of $U({x})$) or Neumann (value of $\nabla U({x}).n(x)$), but you need first to compute it via the integral $$U({y}) = -\int_{cylinder} \frac{G\,\rho(x)}{|{y}-{x}|}\ dx$$ where $\rho({x})$ is the density inside your cylinder.

Note: once you are that far, this integral gives you also the value of the potential everywhere, so there is no need to solve the Laplace equation anymore.

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  • $\begingroup$ What do you mean, whatever I want? I can't just invent U(x) = tag(sin(cos(e^x))) can I? $\endgroup$
    – spyro386
    Dec 30 '16 at 23:18
  • $\begingroup$ I mean. The conditions must make sense. For example for electrostatics with conductive cyllinder it would be that potential is constant on the surface of it. $\endgroup$
    – spyro386
    Dec 30 '16 at 23:20
  • $\begingroup$ @spyro386 : by "whatever you want", I meant you could choose between Dirichlet or Neumann boundary condition. But of course you cannot put an arbitrary function, you need to compute it via the integral I mentioned, which for example, if using Dirichlet B.C., will give the value of the potential on the surface of the cylinder. Note that it gives you the potential everywhere, so solving Laplace after is not necessary. $\endgroup$
    – user130529
    Dec 31 '16 at 8:43
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Assuming there is no infinitely thin surface density, the boundary conditions are continuity of the potential and continuity of the derivative of the potential: $$ U_{in} = U_{out}, \qquad \frac{\partial U_{in}}{\partial n} = \frac{\partial U_{out}}{\partial n}, $$ at the surface, where $n$ is the normal coordinate to the surface. These are due to the gravitational force being finite and continuous.

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