# Uniqueness for solution of a d-dbar system related to Davey-Stewartson Solitons

+ 4 like - 0 dislike
3799 views

This question concerns a system of equations that arise in the study of one-soliton solutions to the Davey-Stewartson equation.

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 $$\partial = \frac{1}{2} \left( \partial_x - i \partial_y \right).$$

The system is:

$$\overline{\partial} n_1(z) = (1+|z|^2)^{-1} n_2(z)$$ $$\partial n_2(z) = -(1+|z|^2)^{-1} n_1(z)$$

and the question is as follows. Suppose that

$$\lim_{|z|\rightarrow \infty} |z| n_1(z) = \lim_{\|z| \rightarrow \infty} |z| n_2(z) = 0$$

Can one prove that $n_1(z)=n_2(z)=0$ if one assumes a priori 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.

Thanks in advance for any help.

Peter Perry, University of Kentucky

This post imported from StackExchange MathOverflow at 2014-09-20 22:30 (UCT), posted by SE-user Peter Perry
Did you try with some Pohozaev-type inequality? e.g. like this: 1) apply $\partial$ to the first equation so it becomes $\Delta n_1+(1+|z|^2)^{-1}n_1=g$ where $g$ decays at infinity faster than $z^{-3}$ 2) multiply by $\overline{z}n_1$ and integrate over an annulus. Typically you obtain quite good information on the behavior of the gradient with this method
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.