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:C→R or equivalently f:R2→R. We denote by ¯∂ and ∂ the usual operators
¯∂=12(∂x+i∂y)
and
∂=12(∂x−i∂y).
The system is:
¯∂n1(z)=(1+|z|2)−1n2(z)
∂n2(z)=−(1+|z|2)−1n1(z)
and the question is as follows. Suppose that
lim|z|→∞|z|n1(z)=lim∥z|→∞|z|n2(z)=0
Can one prove that n1(z)=n2(z)=0 if one assumes a priori that n1 and n2 belong to Lp(R2) for all p>2 (including p=∞)? 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