A variational problem involving a negative fractional Soboblev norm. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:22:43Z http://mathoverflow.net/feeds/question/80357 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80357/a-variational-problem-involving-a-negative-fractional-soboblev-norm A variational problem involving a negative fractional Soboblev norm. Dorian 2011-11-08T04:06:27Z 2011-12-05T21:53:18Z <p>I've run into the problem of trying to evaluate the following:</p> <p>$\max_{ \xi} \iint_{\partial B \times \partial B} \xi(x) \Phi(|x-y|) \xi(y) dS(x)dS(y)$ </p> <p>subject to $\int_{\partial B} \xi(y)dS(y) = 0$ and $\int_{\partial B}\xi^2(y)dS(y)=1$ where $B \subset \mathbb{R}^3$ is a ball of radius $1$ and $\Phi(|x-y|)=\frac{1}{|x-y|}$ is the Newtonian potential.</p> <p>This seems to resemble an inverse fractional Soblev norm such as $H^{-1}$ and moreover appears to be related to the problem of finding an optimal Poincare constant.</p> <p>My guess is that the maximum is obtained for $\xi=+1$ on the upper half and $\xi=-1$ on the lower half. Given this however, I still cannot do an explicit calculation to determine this quantity. Is there a standard reference for such problems arising in Potential Theory perhaps which will allow one to evaluate (even approximately) such expressions?</p> <p>For instance I know I can rewrite the above as: $\int_{\partial B} |\nabla w|^2$ where $-\Delta w = \mu$ and $\mu(x) = \xi(x)dS(x)$ but I'm not sure how this can help me to evaluate such an expression.</p> <p>To summarize, I would like to try to evaluate the above double integral for the particular function $\xi = +1$ on the upper half of the ball and $\xi=-1$ on the lower half. Being able to solve explicitly the above maximization problem would be a bonus.</p> http://mathoverflow.net/questions/80357/a-variational-problem-involving-a-negative-fractional-soboblev-norm/82741#82741 Answer by Ray Yang for A variational problem involving a negative fractional Soboblev norm. Ray Yang 2011-12-05T21:53:18Z 2011-12-05T21:53:18Z <p>The potential of $\xi$ is a harmonic function $u$ on the unit ball in the obvious way. $u(x) = \int_{\partial B} \frac{\xi(y)}{|x-y|} dS(y).$</p> <p>Consider the spherical harmonics decomposition of $u$, given by</p> <p>$u = \sum_{l=0}^\infty f^l r^l Y^l (\theta, \phi)$</p> <p>The energy given above reduces to $C_n \int_{\partial B} u u_\nu dS = \sum l |f^l|^2$, while the normalization condition you have given reduces to $\int_{\partial B} |u_\nu|^2 dS = 1 = \sum l^2 |f^l|^2$. Ignoring the case $l=0$ (which has zero energy), you get the maximum of the two ratios at $l=1$ or $f^l = \delta_{1l}$. </p>