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

  • 2
    $\begingroup$ 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 $\endgroup$ Oct 13, 2010 at 7:09


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