The double-power nonlinear Schrödinger equation and its generalizations: uniqueness, non-degeneracy and applications

Abstract

In this paper we first prove a general result about the uniqueness and non-degeneracy of positive radial solutions to equations of the form $\Delta u+g(u)=0$. Our result applies in particular to the double power non-linearity where $g(u)=u^q-u^p-\mu u$ for $p>q>1$ and $\mu >0$, which we discuss with more details. In this case, the non-degeneracy of the unique solution $u_\mu$ allows us to derive its behavior in the two limits $\mu \rightarrow 0$ and $\mu \rightarrow \mu _*$ where $\mu _*$ is the threshold of existence. This gives the uniqueness of energy minimizers at fixed mass in certain regimes. We also make a conjecture about the variations of the $L^2$ mass of $u_\mu$ in terms of $\mu$, which we illustrate with numerical simulations. If valid, this conjecture would imply the uniqueness of energy minimizers in all cases and also give some important information about the orbital stability of $u_\mu$.

Appendix A: Non-degeneracy of \(u_0\)

Let $d\geqslant 3$ and $p>q>1$. Let $u_0$ be the unique positive radial solution to the equation

$$\begin{aligned} -\Delta u_0+u_0^p-u_0^q=0 \end{aligned}$$

which decays like $u_0(r)\sim C_0r^{2-d}$ at infinity [7, 31, 46, 47]. Define

$$\begin{aligned} \mathcal {L}_0=-\Delta +p(u_0)^{p-1}-q(u_0)^{q-1} \end{aligned}$$

the corresponding linearized operator.

Lemma A.1

(Non-degeneracy of $u_0$) Let v be the unique solution to

$$\begin{aligned}{\left\{ \begin{array}{ll} v''+(d-1)\frac{v'}{r}+qu_0^{q-1}v-pu_0^{p-1}v=0,\\ v(0)=1,\\ v'(0)=0. \end{array}\right. } \end{aligned}$$

Then we have $v\notin L^2({\mathbb R }^d)$, so that

$$\begin{aligned} \ker (\mathcal {L}_0)_\mathrm{rad}=\{0\},\qquad \ker (\mathcal {L}_0)=\left\{ \partial _{x_1}u_0,\ldots ,\partial _{x_d}u_0\right\} . \end{aligned}$$

Proof

We assume by contradiction that $v\in L^2({\mathbb R }^d)$. Then we have the bounds

$$\begin{aligned} |v(r)|\leqslant \frac{C}{r^{d-2}},\qquad |v'(r)|\leqslant \frac{C}{r^{d-1}},\qquad |v''(r)|\leqslant \frac{C}{r^{d}}\qquad \forall r\geqslant 1. \end{aligned}$$

We have proved in Lemma 7.3 that the similar function $v_\mu$ at $\mu >0$ vanishes only once over $(0,\infty )$ and diverges to $-\infty$. From the convergence $g'_\mu (u_\mu )\rightarrow g_0'(u_0)$, it follows that $v_\mu \rightarrow v$ locally. In particular, v vanishes at most once over $(0,\infty )$. On the other hand, since we have assumed that $v\in L^2({\mathbb R }^d)$, it has to vanish at least once. This is because we know that $\mathcal {L}_0$ admits a negative eigenvalue with a positive eigenfunction and that v has to be orthogonal to this eigenfunction. Hence v vanishes exactly once, at some $r_*\in (0,\infty )$. Next we follow step by step the argument of Lemma 7.3 and use the Wronskian identity with $f=u_0+cru_0'$ with $c=-\frac{u_0(r_*)}{r_*u_0'(r_*)}$. This gives

$$\begin{aligned} (r^{d-1}(v f' - fv'))'=r^{d-1}v I_\lambda (u_0) \end{aligned}$$

with $\lambda =1+2c$. Here

$$\begin{aligned} I_\lambda (u)=(q-\lambda )u^q-(p-\lambda )u^p \end{aligned}$$

vanishes at most once over $(0,\infty )$. Since $r^{d-1}(v f' - fv'_0)$ vanishes both at 0 and $r_*$, this proves that $I_\lambda$ must vanish on the left of $r_*$, hence has a constant sign on $(r_*,\infty )$. One difference with the case $\mu >0$ is that this sign is unknown, it depends whether $\lambda$ is on the left or right of q. In any case, we obtain that $r^{d-1}(v f' - fv')$ is either increasing or decreasing over $(r_*,\infty )$, and vanishes at $r_*$. This cannot happen because v and f decay like $r^{2-d}$ at infinity, whereas their derivatives $v'$ and $f'$ decay like $r^{1-d}$, so that $r^{d-1}(v f' - fv')$ always tends to 0 at infinity.

The rest of the argument for the sectors of positive angular momentum is identical to that of Lemma 7.5, of which we use the notation. We know that $\ker (A^{(1)})=\mathrm{span}\{u'_0\}$ and this corresponds to $\partial _{x_1}u_0,\ldots ,\partial _{x_d}u_0$ being in the kernel of $\mathcal {L}_0$. On the other hand, we have $A^{(\ell )}=A^{(1)}+\frac{\ell (\ell +d-2)-d+1}{r^2}> A^{(1)}$ for $\ell \geqslant 2$, which proves that $\ker (A^{(\ell )})=\{0\}$ for $\ell \geqslant 2$. $\square$