most recent 30 from http://mathoverflow.net 2013-05-19T15:56:35Z http://mathoverflow.net/feeds/question/41933 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41933/uniqueness-for-solution-of-a-d-dbar-system-related-to-davey-stewartson-solitons Peter Perry 2010-10-12T19:05:31Z 2010-10-13T07:57:05Z

<p>This question concerns a system of equations that arise in the study of one-soliton solutions to the Davey-Stewartson equation. </p> <p>In what follows, $f(z)$ denotes a function which depends smoothly (but not necessarily analytically!) on $z=x+iy$. Thus $f:\mathbb{C} \rightarrow \mathbb{R}$ or equivalently $f:\mathbb{R}^2 \rightarrow \mathbb{R}$. We denote by $\overline{\partial}$ and $\partial$ the usual operators $$ \overline{\partial} = \frac{1}{2} \left( \partial_x + i \partial_y \right) $$ and $$\partial = \frac{1}{2} \left( \partial_x - i \partial_y \right). $$</p> <p>The system is:</p> <p>$$\overline{\partial} n_1(z) = (1+|z|^2)^{-1} n_2(z)$$ $$\partial n_2(z) = -(1+|z|^2)^{-1} n_1(z)$$</p> <p>and the question is as follows. Suppose that </p> <p>$$\lim_{|z|\rightarrow \infty} |z| n_1(z) = \lim_{\|z| \rightarrow \infty} |z| n_2(z) = 0$$</p> <p>Can one prove that $n_1(z)=n_2(z)=0$ if one assumes <em>a priori</em> that $n_1$ and $n_2$ belong to $L^p(R^2)$ for all $p>2$ (including $p=\infty$)? For this purpose one can assume that the limits above exist.</p> <p>Thanks in advance for any help.</p> <p>Peter Perry, University of Kentucky</p>