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Given $M$ a Riemannian manifold and $\Omega\subset M$ the capacity of $\Omega$ is defined as $$ \mathrm{cap}(\Omega)=\inf \int_{M\setminus\Omega}{|\mathrm{grad} \varphi|^2 dV} $$ where $\varphi$ ranges over all continuous, compactly supported functions on $M\setminus\Omega$ which are $C^{\infty}$ on $M\setminus\overline{\Omega}$ and which are equal to 1 on $\partial\Omega$.

Is it known what is the capacity of a ball of radius $r$ in the $n$-th dimensional hyperbolic space?

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2 Answers 2

Sounds like a home work; here are some hints:

The capacity is equal to the integral of |gradient|$^2$ of spherically symmetric harmonic function with 1 on the bry of the ball and zero at infinity. The function $f$ depends only on the radius, say $r$. You can cook an ODE for $f$, something like $$f''(r)+\frac{(n-1){\cdot}\cosh r}{\sinh r}{\cdot}f'(r)=0.$$ Then you should solve it and integrate $$\mathrm{vol}\\, S^{n-1}{\cdot}\int\limits_R^\infty(f')^2{\cdot}(\sinh r)^{n-1}\\, dr.$$

(I do not know if will get the answer in a simple form, but it will be good for all practical purposes...)

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up vote 3 down vote accepted

The capacity of a set $\Omega$ it is known to be $$ \mathrm{cap}(\Omega)=-\int_{\partial\Omega}{\frac{\partial \Phi}{\partial \nu}dA} $$ where $\Phi$ is an harmonic function with $\Phi|\partial\Omega=1$ and $A$ is the $(n-1)$ area of $\partial\Omega$ and $\frac{\partial}{\partial\nu}$ is the normal derivative along $\partial\Omega$ exterior to $\Omega$.

The Laplacian in the hyperbolic space in polar coordinates has the form: $$ \Delta_{H^{n}} f(t,\xi) = \sinh^{1-n}t \frac{\partial}{\partial t}\left(\sinh^{n-1}t\frac{\partial f}{\partial t}\right) + \sinh^{-2}t\Delta_\xi f $$ where $\Delta\xi$ is the Laplace–Beltrami operator on the ordinary unit $n-1$-sphere.

Therefore, if $f$ is a radial harmonic function then: $$ \Delta_{H^{n}}(f)=0 \iff (n-1)\cosh(t)f'(t)+\sinh(t)f''(t)=0 $$ and $$ \mathrm{cap}(B(x,r))=-\mathrm{vol}(S^{n-1})f'(r) $$

I'm in a rush I hope it makes sense!

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