# Analytic continuation of an NLS soliton

The attractive nonlinear Schroedinger equation $$i \partial_t \psi = - \frac {1} {2} \Delta \psi - |\psi|^{p-1} \psi$$ in the $$H^1$$-subcritical case $$1 < p$$, $$\frac {d} {2} + \frac {2} {p-1} < 1$$ has a soliton solution

$$\psi (t, x) = e^{i t/2} Q (|x|)$$

where $$Q : [0, \infty) \to \mathbb {R}$$ is a solution to the ordinary differential equation

$$(1) \qquad Q'' (r) + \frac {d-1} {r} Q' (r) + 2 Q (r)^p - Q (r) = 0$$

satisfying $$Q' (0) = 0$$, where $$Q (0) > 0$$ is the unique initial value such that the solution satisfies $$Q > 0$$ and $$\lim _{x \to \infty} Q (x) = 0$$.

I want to evaluate oscillatory integrals involving $$Q$$, in particular in order to understand the asymptotics of the Fourier transform of $$\psi (t, \cdot)$$. Using the saddle-point method, this requires understanding analytic continuations of $$Q$$. What is known about this analytic continuation? In particular, what is the point where it can't be analytically extended which has the smallest imaginary part, and what is the nature of the singularity there?

Here's what I figured out:

Obvious $$Q$$ can be extended to an even function on all of $$\mathbb {R}$$.

An analytic continuation of $$Q$$ is still a solution to the equation (1). By the existence theory of complex ODEs, $$Q$$ can be analytically extended except at points where $$Q \to 0$$, $$Q \to \infty$$, and possibly at $$t = 0$$. $$Q$$ is smooth at $$t = 0$$, and although I haven't checked this, I'd be amazed if the power series doesn't converge.

Moreover, rewriting $$Q$$ as $$Q (r) = u (r) r^{-(d-1)/2} e^{-r}$$, we get the equation

$$u'' (r) - 2 u' (r) + \frac {(d-1) (d-3)} {4 r^2} u (r) + 2 r^{- (p-1) (d-1)/2} e^{- (p-1)r} u (r)^p = 0$$

Since $$0 < \lim_{r \to +\infty} u (r) < \infty$$, the nonlinear term becomes negligible as $$r \to \infty$$, which leads me to expect that the radius of converge of $$u$$ (and hence $$Q$$) around $$r$$ goes to infinity as $$r$$ goes to infinity. Therefore I expect that $$Q$$ can be analytically extended to some strip $$|\mathrm {Im} (z)| < c$$, which implies that the Fourier transform of $$\psi$$ decays exponentially.

The function $$v (x) = Q (ix)$$ is real-valued and satifies

$$v'' (x) = \frac {d-1} {x} v' (x) + 2 v (x)^p - v (x)$$

It can be shown that $$v (0) = Q (0) > (1/2)^{1/p}$$, and so $$v'' (0) > 0$$. By continuity $$v (x), v' (x) > 0$$ for all $$x \geq 0$$ which in turn implies $$v (x) \to \infty$$ as $$x \to T^{-}$$ for some finite $$T$$. In particular $$Q$$ cannot be extended to an entire function.

• In the case $d=1$, $p=3$, $Q$ is basically the hyperbolic secant function (up to some normalising constants). For $d=1$ and other values of p one gets some fractional power of sech (again up to some normalising factors I am too lazy to work out here). I would expect the analyticity properties of the $d>1$ case to be similar to that of the $d=1$ case. – Terry Tao Oct 2 '18 at 18:02
• see e.g. annals.math.princeton.edu/wp-content/uploads/… for the general explicit form of the soliton in the d=1 case. – Terry Tao Oct 2 '18 at 18:04
• The $d=1$ case has two interesting behaviors: All the singularities are of the form $Q \to \infty$, none are $Q \to 0$, and these singularities are entirely on the imaginary axis. However, I don't see a reason to expect these two facts to generalize to higher $d$. – Itai Bar-Natan Oct 3 '18 at 21:51