# Uniqueness of Positive Solutions for Fractional Schrödinger Equations with
General Nonlinearities

Xinyu Lia and Linjie Songa

aDepartment of Mathematics, Tsinghua University, Beijing, 100084, PR China
Email<EMAIL_ADDRESS>is supported by "Shuimu Tsinghua Scholar
Program". Email<EMAIL_ADDRESS>
###### Abstract

In 2013, R.L. Frank and E. Lenzmann [R.L. Frank, E. Lenzmann, Uniqueness of
non-linear ground states for fractional Laplacians in $\mathbb{R}$, Acta Math.
210 (2) (2013) 261-318] study the following problem:

$\displaystyle(-\Delta)^{s}u+u=u^{p-1}\quad\text{in }\mathbb{R}^{N},$

where $s\in(0,1)$, $N=1$, $p\in(2,2_{s}^{*})$, and $2_{s}^{*}$ is the critical
fractional Sobolev exponent. They proved that the ground state is unique (up
to translations). Then in 2016, they, together with L. Silvestre [R.L. Frank,
E. Lenzmann, L. Silvestre, Uniqueness of radial solutions for the fractional
Laplacian, Comm. Pure Appl. Math. 69 (9) (2016) 1671-1726] showed similar
uniqueness results for high dimensions ($N\geqslant 2$), in which they
proposed a challenging open problem to extend their results about non-
degeneracy and uniqueness of ground states to nonlinearities $f(u)$ beyond the
pure-power case. To the best of our knowledge, this question is still
unresolved so far. In this paper, we aim to give a full affirmative answer to
this open issue for a large class of convex nonlinearities. In fact, we prove
the uniqueness of the positive solution.

Keywords: Fractional Schrödinger equation; uniqueness; non-degeneracy; general
nonlinearities.

2020 MSC: 35A02, 35B09, 35R11.

Data availability statement: Our manuscript has no associate data.

###### Contents

  1. 1 Introduction and Overview on Main Results
  2. 2 Preliminaries
    1. 2.1 Existence of the Positive Solution
    2. 2.2 Definitions of the Non-degeneracy and Morse Index
    3. 2.3 An Oscillation Estimate of the Second Radial Eigenfunction
    4. 2.4 Variational Characterization of the Second Eigenvalue
  3. 3 Non-degeneracy of the Positive Solution
    1. 3.1 Non-degeneracy in the Non-Radial Sector
    2. 3.2 Polarization and Non-degeneracy in the Radial Sector
    3. 3.3 Completing the Proof of Theorem 2.5
  4. 4 Uniqueness of the Positive Solution
    1. 4.1 Construction of the Global Branch
    2. 4.2 Uniqueness of the Global Branch
    3. 4.3 Proof of Theorem 1.1

##  1 Introduction and Overview on Main Results

Fractional differential operators and the fractional Sobolev spaces attract
the attention of many researchers, both from a purely mathematical standpoint
and from a standpoint of various applications, since these operators naturally
arise in many different contexts, such as anomalous diffusion and quasi-
geostrophic flows, turbulence and water waves, molecular dynamics, and
relativistic quantum mechanics of stars (see [5, 8, 12, 29] and the references
therein). It also has various applications in probability and finance (see [2,
4, 7]). The fractional Schrödinger equation discovered by Laskin [23, 24, 25]
is a fundamental equation of fractional quantum mechanics. Here, what played a
central role within these models is often so-called ground state solutions, or
simply ground states. By this, we mean non-trivial, non-negative, and radial
functions $u=u(|x|)\geqslant 0$ that vanish at infinity and satisfy (in the
distributional sense) an equation of the form

$\displaystyle(-\Delta)^{s}u+F(u)=0\quad\text{in }\mathbb{R}^{N}.$ (1.1)

In most examples of interest, the existence of ground states follows from
variational arguments. Based on this variational approach, it is natural in
these cases to require that a ground state is also a minimizer for some
related variational problem. After the existence of ground states is
established, some properties, such as the uniqueness and the non-degeneracy
(see Theorem 2.5 below for the definition) play an important role in many
applications, for example, they are of fundamental importance in the stability
and blow-up analysis for the corresponding solitary waves.

However, in striking contrast to the issue of existence, it seems fair to say
that extremely little is known about the uniqueness of ground states for
problems like (1.1), until the outstanding papers [18, 19]. For the
"classical" limiting case with $s=1$, uniqueness can be addressed by standard
ordinary differential equation (ODE) methods. When $s\in(0,1)$, the equation
is nonlocal and ODE techniques fail to work. Before [18, 19], the known
nonperturbative uniqueness results are for equations with special forms and
summarized as follows. (See [16, 21] for some perturbative uniqueness results
when $s$ is close to 1.)

  * •

Ground state solitary waves for the Benjamin-Ono equation in $N=1$ dimension;
see [1].

  * •

Optimizers for fractional Sobolev inequalities in $N\geqslant 1$ dimensions;
see [26] and [11].

In [18] and [19] the authors studied

$(-\Delta)^{s}u+u-u^{p-1}=0\quad\text{in }\mathbb{R}^{N},$ (1.2)

and proved uniqueness and non-degeneracy of ground states for all $s\in(0,1)$,
in arbitrary dimension $N\geqslant 1$, where $2<p<2^{*}_{s}$ and the critical
fractional Sobolev exponent $2^{*}_{s}$ will given below. In [19], the authors
proposed a challenging open problem:

  * •

to extend the results about non-degeneracy and uniqueness of ground states
obtained in [18, 19] to nonlinearities $f(u)$ beyond the pure-power case.

In this paper, we aim to give a full affirmative answer to the above open
question for a large class of convex nonlinearities. More precisely, we
consider the following nonlocal problem:

$\left\\{\begin{aligned} &(-\Delta)^{s}u+\lambda u=f(u)\quad\text{in
}\mathbb{R}^{N},\\\ &u>0\quad\text{in }\mathbb{R}^{N},\quad u(x)\to
0\quad\text{as }|x|\to\infty,\end{aligned}\right.$ (1.3)

where $s\in(0,1)$, $\lambda>0$ and $(-\Delta)^{s}$ denotes the fractional
Laplacian. Throughout this paper the following hypotheses on
$f:[0,\infty)\to\mathbb{R}$ will be assumed.

  * $(f_{1})$

$f\in C^{1,\gamma}(\mathbb{R}^{+},\mathbb{R})$ for some
$\gamma>\max\\{0,1-2s\\}$, and $f(0)=0$.

  * $(f_{2})$

$f^{\prime}(t)$ is non-decreasing and $f^{\prime}(t)t>f(t)$, $\forall t>0$.

  * $(f_{3})$

There exist $q,r\in(2,2^{*}_{s})$ such that

$\displaystyle 0<qF(t)\leqslant f(t)t\leqslant rF(t),\quad\forall t\geqslant
0.$ (1.4)

where $F(t)=\int_{0}^{t}f(s)ds$, $2^{*}_{s}=\frac{2N}{N-2s}\text{ if
}2s<N,\text{ and }2^{*}_{s}=\infty\text{ if }2s\geqslant N.$

  * $(f_{4})$

For some $a_{0}>0$ and $r\in(2,2_{s}^{*})$, we have $f(t)=a_{0}t^{r-1}+g(t)$
and there exist $q\in(2,r)$ and $q_{0}\in(q,r)$ such that

$\displaystyle 0\leqslant g(t)\leqslant C(t^{q-1}+t^{q_{0}-1}),\quad\forall
t\geqslant 0,$ (1.5)

where $C$ is a positive constant. Without loss of generality, in the
following, we always assume $a_{0}=1$.

We remark that $(f_{2})$ is a convexity condition. A classical example
satisfying $(f_{1})-(f_{4})$ is the sum of pure-power nonlinearities
$f(t)=\sum^{k}_{i=1}t^{p_{i}-1}$ with $2<p_{i}<2_{s}^{*}$, $1\leqslant
i\leqslant k$. Moreover, our conditions cover more cases such as
$F(t)=\frac{1}{r}t^{r}+\frac{t^{q}}{1+t}$ with $r\in(2,2_{s}^{*})$ and
$q\in(3,r)$.

In the recent work [17], the authors proved that positive solutions to the
equation (1.2) with the pure-power nonlinearity are unique up to translations
and non-degenerate, for all $s\in(0,1),N\geqslant 1$ and $2<p<2^{*}_{s}$. This
generalizes the result of Frank, Lenzmann, and Silvestre [19], where the same
uniqueness and non-degeneracy result was proven for positive solutions whose
Morse index is equal to $1$.

Inspired by [17], we study the uniqueness problem for positive solutions of
the equation (1.3) with general nonlinearities. Our main result is the
following.

######  Theorem 1.1.

Assume that $(f_{1})-(f_{4})$ hold. Let $N\geqslant 1,s\in(0,1)$, and
$\lambda>0$. Then the positive solution $u=u(|x|)>0$ of (1.3) is unique (up to
translations) and it is non-degenerate (see Theorem 2.5 below for the
definition).

We remark that our result in the above Theorem 1.1 also holds in the case
$s=1$, and covers some new results which were not obtained by ODE techniques.
In the local case when $s=1$, the uniqueness of the positive solution for
problem (1.2) with all $p\in(2,2^{*})$ was established in a celebrated paper
by Kwong [22] (see also [27]). Our method provides a new proof for this pure-
power case by taking $p$ as a parameter and extending it to $p\searrow 2$.
Things become difficult for a more complicated case with double-power
nonlinearities, i.e., $f(u)=u^{p-1}+u^{q-1}$ with $2<p<q<2^{*}$. Indeed, in
$N\geqslant 3$ dimensions, [9] established the uniqueness of the positive
solution covering examples $f(u)=u^{p-1}+u^{q-1}$ with
$2<p<q\leqslant\frac{2N-2}{N-2}$. Note that there is a gap between the
exponent $\frac{2N-2}{N-2}$ in [9] and the critical exponent
$2^{*}=\frac{2N}{N-2}$. There seems to be no progress to fill this gap since
1991, [9]. We remark that our result solves completely this problem. In fact,
$(f_{1})-(f_{4})$ in the case $s=1$ cover the example
$f(u)=\sum^{k}_{i=1}u^{p_{i}-1}$ for the full range $2<p_{i}<2^{*}$,
$1\leqslant i\leqslant k$, and in arbitrary dimension $N\geqslant 1$.

Sketch of Main Ideas.  To prove Theorem 1.1, we need to overcome two main
challenges caused by the general nonlinearities. The first one is to obtain
the non-degeneracy of the positive solution, i.e., to prove that the kernel of
the self-adjoint operator $L_{+}:=(-\Delta)^{s}+\lambda-f^{\prime}(u)$ acting
on $L^{2}(\mathbb{R}^{N})$ is exactly
$\operatorname{span}\\{\partial_{x_{i}}u:i=1,\cdots,N\\}$. In [18, 19], the
non-degeneracy of the positive solution with Morse index 1 was established by
using a concrete construction depending on the pure-power nonlinearities. It
seems to be hard to find a similar construction for nonlinearities beyond the
pure-power case. Next, we state our key steps to establish the non-degeneracy.
Firstly, we show that the kernel of the linearized operator $L_{+}$ in the
non-radial sector is exactly
$\operatorname{span}\\{\partial_{x_{i}}u:i=1,\cdots,N\\}$. It is easy to see
that $\operatorname{span}\\{\partial_{x_{i}}u:i=1,\cdots,N\\}$ belongs to the
kernel of $L_{+}$. To rule out further elements in the kernel in the non-
radial sector, we decompose the action of $L_{+}$ using spherical harmonics.
In fact, this latter argument is based in spirit on an argument in [19]. In
the second step, we assume that the Morse index $u$ is equal to 1, and suppose
by contradiction that there exists a radial eigenfunction $w$ with respect to
the second eigenvalue $\mu_{2}=0$ of $L_{+}$. By the oscillation result in
[19] (see Proposition 2.7), $w$ changes its sign exactly once. Then we exhibit
a non-radial second eigenfunction $w_{a}$, by using the polarization
techniques. We conclude that $w_{a}$ must be a linear combination of
$w,\partial_{x_{1}}u,\cdots,\partial_{x_{N}}u$. Then we show that $w_{a}$ must
be $k_{0}w$ for some $k_{0}\in\mathbb{R}$, which establishes the triviality of
the kernel of the linearized operator $L_{+}$ in the radial sector. This shows
that the positive solution whose Morse index is equal to 1 is non-degenerate.
Finally, inspired by [17], we prove that any positive solution of (1.3) has
Morse index 1 and complete the proof of non-degeneracy. In fact, the topic of
using polarization techniques to obtain non-degeneracy of positive solutions
with Morse index 1 was recently touched by [13] and [14], for equation (1.2)
on the balls in $\mathbb{R}^{N}$ with pure-power nonlinearities. For the
problem on the balls, the kernel of $L_{+}$ is empty while it is not empty on
the entire space. The non-empty kernel makes new difficulties in our
arguments.

After establishing the non-degeneracy, we use a continuation argument to prove
the uniqueness. In contrast to the discussions in [18, 19], the general
nonlinearities bring new difficulties. Here, we use a homotopic argument by
continuation in $\lambda$. More precisely, starting from some solution $u_{0}$
of

$(-\Delta)^{s}u_{0}+\lambda_{0}u_{0}=f(u_{0})\quad\text{in }\mathbb{R}^{N},$

we seek to construct a branch $u_{\lambda}$ with $\lambda$ stemming from
$\lambda_{0}$ and solving the problem

$(-\Delta)^{s}u+\lambda u=f(u)\quad\text{in }\mathbb{R}^{N}.$

The local existence and uniqueness (in suitable function space) for
$u_{\lambda}$ with $u_{\lambda_{0}}=u_{0}$ follows from an implicit function
argument, provided the linearization around $u_{0}$ is invertible, which is
the first key step in the argument. The second key point is now to derive
suitable apriori bounds that guarantee that the branch $u_{\lambda}$ can be
extended to $\lambda\nearrow+\infty$. Finally, we aim to prove that the
positive solution is unique when $\lambda$ is large enough. To do it, we use
scaling arguments and define a one-to-one transform $T_{\lambda}$ from
$H^{s}(\mathbb{R}^{N})$ to $H^{s}(\mathbb{R}^{N})$ by

$T_{\lambda}u=\lambda^{-\frac{1}{r-2}}u(\frac{x}{\lambda^{\frac{1}{2s}}}),\quad\forall
u\in H^{s}(\mathbb{R}^{N}).$

Then $u$ is a weak solution of (1.3), if and only if $v$ is a solution of the
following equation:

$(-\Delta)^{s}v+v=\mu^{1-r}f(\mu v)\quad\text{in }\mathbb{R}^{N},$ (1.6)

where $v=T_{\lambda}u,$ and $\mu=\lambda^{\frac{1}{r-2}}$. Hence, in order to
prove the uniqueness of the positive solution of (1.3) for large $\lambda$, it
suffices to show the uniqueness of the positive solution of (1.6) when $\mu$
is large enough. By deriving some priori bounds and decay estimates, we get
that $v_{\mu}$, the positive solution of (1.6), converges as $\mu\to\infty$ to
a positive solution of a limit equation. Then using the uniqueness and non-
degeneracy of the positive solution of this limit equation, we complete the
proof.

Plan of the Paper. We organize this paper as follows. In Section 2 we give
some preliminary results. The Section 3 is devoted to the proof of Theorem 2.5
(non-degeneracy result). The proof of Theorem 1.1 (uniqueness result) will be
given in Section 4.

Notation and Conventions. Throughout this paper, we employ the common abuse of
notation by writing both $f=f(|x|)$ and $f=f(r)$ for any radial functions $f$
on $\mathbb{R}^{N}$. We use standard notation for $L^{p}$ and Sobolev spaces
and $L_{\mathrm{rad}}^{p}(\mathbb{R}^{N})$ denotes the space of radial and
square-integrable functions on $\mathbb{R}^{N}$. For $k\in\mathbb{N}$ and
$0<\gamma\leqslant 1$ the Hölder space $C^{k,\gamma}(\mathbb{R}^{N})$ is
equipped with the norm $\|u\|_{C^{k,\gamma}}=\sum_{|\alpha|\leqslant
k}\left\|\partial_{x}^{\alpha}u\right\|_{L^{\infty}}+\|u\|_{C^{\gamma}}$,
where $\|u\|_{C^{\gamma}}=\sup_{x\neq y}\frac{|u(x)-u(y)|}{|x-y|^{\gamma}}$.
We often write $L^{p}$ instead of $L^{p}(\mathbb{R}^{N})$ etc. We denote
$\langle\cdot,\cdot\rangle_{L^{2}}$ and $|\cdot|_{2}$ the inner product and
norm on $L^{2}(\mathbb{R}^{N})$, respectively. For a function $v$ we denote

$v^{+}:=\max\\{v,0\\}\geqslant 0\qquad\mbox{ and }\qquad
v^{-}:=\max\\{-v,0\\}\geqslant 0,$ (1.7)

hence $v=v^{+}-v^{-}$.

We employ the following convention for constants in this paper: Unless
otherwise explicitly stated, we write

$X\lesssim_{a,b,c,\ldots}Y$

to denote that $X\leqslant CY$ with some constant $C>0$ that only depends on
the quantities $a,b,c,\ldots$ and the space dimension $N\geqslant 1$.
Moreover, we require that $C>0$ can be chosen uniform if $a,b,c,\ldots$ range
in some fixed compact set. We write $X\lesssim Y$ for simplicity without
causing ambiguity. We denote by $X\sim Y$ if $X\lesssim Y$ and $Y\lesssim X$.

##  2 Preliminaries

Let $s\in(0,1)$ and $(-\Delta)^{s}$ denotes the fractional Laplacian given by

$(-\Delta)^{s}u(x)=c_{N,s}\
P.V.\int_{\mathbb{R}^{N}}\frac{u(x)-u(y)}{|x-y|^{N+2s}}dy,\quad
c_{N,s}:=4^{s}\pi^{-\frac{N}{2}}\frac{\Gamma(\frac{N}{2}+s)}{\Gamma(2-s)}s(1-s).$

We define $H^{s}(\mathbb{R}^{N})$ the fractional Sobolev space,

$\displaystyle H^{s}(\mathbb{R}^{N})=\left\\{u\in
L^{2}(\mathbb{R}^{N}):\frac{|u(x)-u(y)|}{|x-y|^{\frac{N}{2}+s}}\in
L^{2}(\mathbb{R}^{N}\times\mathbb{R}^{N})\right\\},$ (2.1)

endowed with the (equivalent) norm

$\displaystyle\|u\|^{2}:=\lambda|u|_{2}^{2}+\mathcal{E}(u,u),$ (2.2)

where, for $u,v\in H^{s}(\mathbb{R}^{N})$,

$\displaystyle\mathcal{E}(u,v):=\dfrac{c_{N,s}}{2}\iint_{\mathbb{R}^{N}\times\mathbb{R}^{N}}\dfrac{(u(x)-u(y))(v(x)-v(y))}{|x-y|^{N+2s}}dx\,dy.$

###  2.1 Existence of the Positive Solution

In this paper, we focus on real-valued solutions. A natural approach to
construct nontrivial solutions $u$ for equation (1.3) is given by minimizing
$J$ on the Nehari manifold $\mathcal{N}$, where
$J:H^{s}(\mathbb{R}^{N})\rightarrow\mathbb{R}$ is the energy functional given
by

$J(v):=\frac{1}{2}\mathcal{E}(v,v)+\frac{\lambda}{2}|v|_{2}^{2}-\int_{\mathbb{R}^{N}}F(v)dx,\quad
F(t):=\int_{0}^{t}f(s)ds,$

and the set $\mathcal{N}$ is the Nehari manifold

$\mathcal{N}:=\left\\{v\in
H^{s}(\mathbb{R}^{N})\backslash\\{0\\}:\mathcal{E}(v,v)+\lambda|v|_{2}^{2}-\int_{\mathbb{R}^{N}}f(v)vdx=0\right\\}.$

Assumptions $(f_{1})-(f_{3})$ guarantee, from standard variational arguments,
existence of a ground state of (1.3), which is positive and radially
symmetric, for any $\lambda>0$.

We summarize the following existence results along with fundamental properties
of positive solutions for equation (1.3).

######  Proposition 2.1.

Let $N\geqslant 1,s\in(0,1)$, $\lambda>0$ and assumptions $(f_{1})-(f_{3})$
hold true. Then the following holds.
$(i)$ Existence : Equation (1.3) admits a solution $0<u\in
H^{s}(\mathbb{R}^{N})$.
$(ii)$ Symmetry, regularity, and decay : If $u\in H^{s}(\mathbb{R}^{N})$ with
$u>0$ solves (1.3), then there exists some $x_{0}\in\mathbb{R}^{N}$ such that
$u\left(\cdot-x_{0}\right)$ is radial, positive, and strictly decreasing in
$\left|x-x_{0}\right|$. Moreover, the function $u$ belongs to
$H^{2s+1}(\mathbb{R}^{N})\cap C^{\max\\{1,2s+\alpha\\}}(\mathbb{R}^{N})$ for
some $\alpha>0$ and it satisfies

$\frac{C_{1}}{1+|x|^{N+2s}}\leqslant
u(x)\leqslant\frac{C_{2}}{1+|x|^{N+2s}}\quad\text{ for }x\in\mathbb{R}^{N},$

with some constants $C_{2}\geqslant C_{1}>0$ depending on $s,N,\lambda$, and
$u$.

######  Remark 2.2.

See [30, Chapter 4] for the existence result in Proposition 2.1 (i). As for
the results in (ii), we can apply the moving plane method for the fractional
Laplacian in [10] to obtain the radial symmetry for the positive solutions and
use arguments modified by [19, Proposition 3.1 and Appendix D] to get the
regularity and decay. See also in [6] for some results about the existence,
regularity, and decay.

###  2.2 Definitions of the Non-degeneracy and Morse Index

The solution obtained by Proposition 2.1 (i) has more properties, for example,
its Morse index is equal to 1 and it is non-degenerate. For further
explanations, we introduce the following notions.

######  Definition 2.3 (Morse index).

Let $u\in H^{s}(\mathbb{R}^{N})$ be a solution of (1.3), and we define the
linearized operator

$L_{+}=(-\Delta)^{s}+\lambda-f^{\prime}(u).$

Then, $L_{+}$ is a self-adjoint operator on $L^{2}(\mathbb{R}^{N})$ with
quadratic-form domain $H^{s}(\mathbb{R}^{N})$ and operator domain
$H^{2s}(\mathbb{R}^{N})$. The Morse index of $u$ is defined as

$\mu\left(u\right):=\sharp\left\\{e<0:e\text{ is an eigenvalue of
}L_{+}\right\\},$

where multiplicities of eigenvalues are taken into account. Noticing that
$L_{\text{rad }}^{2}(\mathbb{R}^{N})$ is an invariant subspace of $L_{+}$, we
also introduce the Morse index of $u$ in the sector of radial functions by
defining

$\mu_{\text{rad }}\left(u\right):=\sharp\left\\{e<0:e\text{ is an eigenvalue
of }L_{+}\text{ restricted to }L_{\text{rad
}}^{2}(\mathbb{R}^{N})\right\\}\text{. }$

######  Remark 2.4.

If $u$ is the solution of (1.3) obtained by Proposition 2.1 (i) such that
$J(u)=\inf_{\mathcal{N}}J(v)$, then $\mu\left(u\right)=\mu_{\text{rad
}}\left(u\right)=1$. In fact, on the one hand, by $(f_{2})$,

$\left\langle
L_{+}u,u\right\rangle_{L^{2}}=\int_{\mathbb{R}^{N}}f(u)u-f^{\prime}(u)u^{2}dx<0$

Thus, by the min-max principle, the operator $L_{+}$ has at least one negative
eigenvalue, i.e. $\mu\left(u\right)\geqslant 1$. Since $u$ is radial, then
$\mu_{\text{rad }}\left(u\right)\geqslant 1$ by applying min-max principle in
$L_{rad}^{2}(\mathbb{R}^{N})$. On the other hand, since $u$ is a minimum point
in the Nehari manifold whose codimension is $1$, it is easy to verify that
$\mu_{\text{rad }}\left(u\right)\leqslant$ $\mu\left(u\right)\leqslant 1$.

Now we state our result on the non-degeneracy, whose definition will also be
given.

######  Theorem 2.5 (Non-degeneracy).

Let $N\geqslant 1,s\in(0,1)$, $\lambda>0$, and assumptions $(f_{1})-(f_{3})$
hold true. Suppose that $u\in H^{s}(\mathbb{R}^{N})$ is a positive solution of
(1.3). Then $\mu(u)=\mu_{rad}(u)=1$ and $u$ is non-degenerate, in the sense
that

$\ker
L_{+}=\operatorname{span}\left\\{\partial_{x_{1}}u,\ldots,\partial_{x_{N}}u\right\\},$

where $L_{+}$ is defined by Definition 2.3.

###  2.3 An Oscillation Estimate of the Second Radial Eigenfunction

In this subsection, we collect some preliminary information, which is useful
to prove Theorem 2.5. We start with the following result from [19].

Let $N\geqslant 1$ and $s\in(0,1)$ be given. We consider the linear equation

$(-\Delta)^{s}u+Vu=0\quad\text{in }\mathbb{R}^{N}.$ (2.3)

We require that the potential $V:\mathbb{R}^{N}\rightarrow\mathbb{R}$
satisfies the following conditions.

  * (V1)

$V=V(|x|)$ is radial and non-decreasing in $|x|$.

  * (V2)

$V\in C^{0,\gamma}(\mathbb{R}^{N})$ for some $\gamma>\max\\{0,1-2s\\}$.

######  Proposition 2.6 (Frank-Lenzmann-Silvestre, [19], Corollary 1).

Suppose $N\geqslant 1,s\in(0,1)$, and let $V$ satisfy (V1) and (V2). Consider
$H=(-\Delta)^{s}+V$ acting on $L_{\mathrm{rad}}^{2}(\mathbb{R}^{N})$. Then all
eigenvalues of $H$ are simple.

In particular, if $E_{1}^{\mathrm{rad}}\leqslant E_{2}^{\mathrm{rad}}\leqslant
E_{3}^{\mathrm{rad}}\leqslant\cdots<\inf\sigma_{\mathrm{ess}}(H)$ denote (with
counting multiplicity) the discrete eigenvalues of $H$ acting on
$L_{\mathrm{rad}}^{2}(\mathbb{R}^{N})$, then we have strict inequalities

$E_{1}^{\mathrm{rad}}<E_{2}^{\mathrm{rad}}<E_{3}^{\mathrm{rad}}<\cdots<\inf\sigma_{\mathrm{ess}}(H).$

As the last main result for the linear equation (2.3), we need the following
sharp oscillation estimate for the second eigenfunction of $H$, which provides
us with a bound in agreement with Sturm-Liouville theory for ODE.

######  Proposition 2.7 (Frank-Lenzmann-Silvestre, [19], Theorem 2).

Let $H=(-\Delta)^{s}+V$ be as the one given by Proposition 2.6 above. Suppose
that $H$ has at least two radial eigenvalues
$E_{1}^{\mathrm{rad}}<E_{2}^{\mathrm{rad}}<\inf\sigma_{\mathrm{ess}}(H)$. Let
$\psi\in L^{2}(\mathbb{R}^{N})$ denote the radial eigenfunction of $H$ for the
second radial eigenvalue $E_{2}^{\mathrm{rad}}$. Then $\psi=\psi(|x|)$ changes
its sign exactly once for $|x|=r\in(0,+\infty)$. By this, we mean that there
is some $\rho>0$ such that (after multiplying $\psi$ with $-1$ if necessary)
we have

$\psi(r)\leqslant 0\text{ for }r\in\left[0,\rho\right)\text{ and
}\psi(r)>0\text{ for }r\in\left(\rho,+\infty\right),$

and $\psi\not\equiv 0$ on both intervals $\left[0,\rho\right)$ and
$\left[\rho,+\infty\right)$. Note also that $\psi(0)<0$.

######  Remark 2.8.

For the applications in the next sections, the potential $V$ is
$\lambda-f^{\prime}(u)$, where $u$ is a positive solution of (1.3). By
Proposition 2.1, we know that (V1) and (V2) hold.

###  2.4 Variational Characterization of the Second Eigenvalue

We introduce the eigenvalue problem associated to the linearized equation:

$(-\Delta)^{s}\phi+\lambda\phi-f^{\prime}(u)\phi=\mu_{i}\phi\quad\text{in
}\mathbb{R}^{N}.$ (2.4)

We denote by $\mu_{i}$ the $i$-eigenvalue and by $\phi_{i}$ the associated
eigenfunction.

Defining the bilinear form $\mathcal{B}:H^{s}(\mathbb{R}^{N})\times
H^{s}(\mathbb{R}^{N})\to\mathbb{R}$ as

$\mathcal{B}(\phi,\psi):=\mathcal{E}(\phi,\psi)+\lambda\int_{\mathbb{R}^{N}}\phi\psi
dx-\int_{\mathbb{R}^{N}}f^{\prime}(u)\phi\psi dx,$ (2.5)

one has that

$\mu_{1}=\min\\{\frac{\mathcal{B}(\varphi,\varphi)}{\int_{\mathbb{R}^{N}}\varphi^{2}dx}:\varphi\in
H^{s}(\mathbb{R}^{N})\setminus\\{0\\}\\},$ (2.6)

and

$\mu_{2}=\min\\{\frac{\mathcal{B}(\varphi,\varphi)}{\int_{\mathbb{R}^{N}}\varphi^{2}dx}:\varphi\in
H^{s}(\mathbb{R}^{N})\setminus\\{0\\},\int_{\mathbb{R}^{N}}\phi_{1}\varphi
dx\\}.$ (2.7)

Let $\phi_{2}$ be a second eigenfunction of the eigenvalue problem (2.4), then
$\phi_{2}^{\pm}\in H^{s}(\mathbb{R}^{N})\setminus\\{0\\}$ and so it can be
used as test functions in the weak formulation of the eigenvalue problem with
$\mu=\mu_{2}$, obtaining

$\mu_{2}\int_{\mathbb{R}^{N}}(\phi^{+}_{2})^{2}dx=\mathcal{B}(\phi_{2},\phi^{+}_{2})\quad\text{
and
}\quad\mu_{2}\int_{\mathbb{R}^{N}}(\phi^{-}_{2})^{2}dx=-\mathcal{B}(\phi_{2},\phi^{-}_{2}).$
(2.8)

Inspired by the ideas in [3], we introduce the following characterization for
the second eigenvalue $\mu_{2}$ of the eigenvalue problem (2.4).

######  Lemma 2.9.

Let $v\in H^{s}(\mathbb{R}^{N})$ be such that $v^{\pm}\neq 0$ in
$\mathbb{R}^{N}$.

$\mu_{2}\int_{\mathbb{R}^{N}}(v^{+})^{2}dx\geqslant\mathcal{B}(v,v^{+})\quad\text{and}\quad\mu_{2}\int_{\mathbb{R}^{N}}(v^{-})^{2}dx\geqslant-\mathcal{B}(v,v^{-}),$
(2.9)

where $\mathcal{B}$ is the bilinear form in (2.5) and $v^{\pm}$ are positive
and negative part of $v$. Then $v$ is an eigenfunction associated with the
eigenvalue $\mu_{2}$ for the linearized problem (2.4) at $u$ and equalities
hold in (2.9).

###### Proof.

The proof is similar to the one of [3, Lemma 2.1]. We omit the details. ∎

##  3 Non-degeneracy of the Positive Solution

This section is devoted to the proof of Theorem 2.5 that establishes the non-
degeneracy of solutions $u>0$ for equation (1.3). By Proposition 2.1, we can
assume that $u=u(|x|)>0$ is radial without loss of generality. The proof of
Theorem 2.5 will be divided into three main steps as follows.

###  3.1 Non-degeneracy in the Non-Radial Sector

In the first step, we show that the kernel of the linearized operator $L_{+}$
in the non-radial sector is exactly
$\operatorname{span}\\{\partial_{x_{i}}u:i=1,\cdots,N\\}$.

Since $u=u(|x|)$ is a radial function, the operator
$L_{+}=(-\Delta)^{s}+\lambda-f^{\prime}(u)$ commutes with rotations on
$\mathbb{R}^{N}$. Using the decomposition in terms of spherical harmonics

$L^{2}(\mathbb{R}^{N})=\bigoplus_{\ell\geqslant 0}\mathcal{H}_{\ell},$ (3.1)

we find that $L_{+}$ acts invariantly on each subspace

$\mathcal{H}_{\ell}=L^{2}\left(\mathbb{R}^{+},r^{N-1}dr\right)\otimes\mathcal{Y}_{\ell}.$
(3.2)

Here $\mathcal{Y}_{\ell}=\operatorname{span}\left\\{Y_{\ell,m}\right\\}_{m\in
M_{\ell}}$ denotes space of the spherical harmonics of degree $\ell$ in space
dimension $N$. Note that the index set $M_{\ell}$ depends on $\ell$ and $N$.
Recall also that
$-\Delta_{\mathbb{S}^{N-1}}Y_{\ell,m}=\ell(\ell+N-2)Y_{\ell,M}$. For each
$\ell\geqslant 0$, the action of $L_{+}$on the radial factor in
$\mathcal{H}_{\ell}$ is given by

$\displaystyle\left(L_{+,\ell}g\right)(r)=\left(\left(-\Delta_{\ell}\right)^{s}g\right)(r)+\lambda
g(r)-f^{\prime}(u)g(r),$ (3.3)

for $g\in C_{0}^{\infty}(\mathbb{R}^{+})\subset
L^{2}(\mathbb{R}^{+},r^{N-1}dr)$. Here $\left(-\Delta_{\ell}\right)^{s}$ is
given by spectral calculus and the known formula

$\displaystyle-\Delta_{\ell}=-\frac{\partial^{2}}{\partial
r^{2}}-\frac{N-1}{r}\frac{\partial}{\partial r}+\frac{\ell(\ell+N-2)}{r^{2}}.$
(3.4)

Technically, we consider (3.4) as a self-adjoint operator in
$L^{2}(\mathbb{R}^{+})$ defined as the Friedrichs extension of the
corresponding differential expression acting on
$C_{0}^{\infty}(\mathbb{R}^{+})$.

We now derive the following result inspired by [19].

######  Lemma 3.1.

Suppose $(f_{1})-(f_{3})$. Then we have
$\operatorname{ker}L_{+,1}=\operatorname{span}\left\\{\partial_{r}u\right\\}\text{
and }\operatorname{ker}L_{+,\ell}=\\{0\\}\text{ for }\ell\geqslant 2.$

###### Proof.

By differentiating equation (1.3), we see that $L_{+}\partial_{x_{i}}u=0$ for
$i=1,\cdots,N$, here $\partial_{x_{i}}u=\frac{\partial u}{\partial
x_{i}}=u^{\prime}(r)\frac{x_{i}}{r}\in\mathcal{H}_{1}$. Hence,
$L_{+,1}u^{\prime}(r)=0$. By standard arguments (because of $u^{\prime}<0$ and
the Perron-Frobenius property of $L_{+,1}$), we deduce that $0$ is the lowest
and hence non-degenerate eigenvalue of $L_{+,1}$. Thus we conclude that

$\operatorname{ker}L_{+,1}=\operatorname{span}\\{\partial_{r}u\\}.$

Finally, by [19, Lemma C.5], we have the strictly inequality
$L_{+,\ell}>L_{+,1}$ in the sense of quadratic forms for any $\ell\geqslant
2$. Since $0$ is the lowest eigenvalue of $L_{+,1},$ this implies that $0$
cannot be an eigenvalue of $L_{+,\ell}$ for any $\ell\geqslant 2$, which
completes the proof.

∎

###  3.2 Polarization and Non-degeneracy in the Radial Sector

In the second step, we assume that the Morse index of $u$ is equal to 1. Then
using polarization techniques to construct a non-radial element in $\ker
L_{+}$ and some novel ideas to address the nontrivial kernel of $L_{+}$, we
prove that any positive solution with Morse index 1 is non-degenerate.

Taking $a\in\mathbb{R}$, we consider the hyperplane

$H_{a}:=\\{x\in\mathbb{R}^{N}:x_{1}=a\\},$

and let $\overline{x}=\sigma_{a}(x):=(2a-x_{1},x_{2},\cdots,x_{N})$ be the
reflection of $x=(x_{1},x_{2},\cdots,x_{N})$ with respect to $H_{a}$. Let us
define the halfspaces

$\Sigma_{a}^{-}:=\\{x\in\mathbb{R}^{N}:x_{1}\leqslant
a\\}\quad\text{and}\quad\Sigma_{a}^{+}:=\\{x\in\mathbb{R}^{N}:x_{1}\geqslant
a\\}.$

For $w\in H^{s}(\mathbb{R}^{N})$, we define the polarization of $w$ with
respect to $H_{a}$ as the function

$\displaystyle w_{a}(x):=\begin{cases}\min\\{w(x),w(\overline{x})\\},&\quad
x\in\Sigma_{a}^{+},\\\ \max\\{w(x),w(\overline{x})\\},&\quad
x\in\Sigma_{a}^{-}.\end{cases}$ (3.5)

Let us recall that

$[w_{a}^{\pm}]_{H^{s}(\mathbb{R}^{N})}\leqslant[w^{\pm}]_{H^{s}(\mathbb{R}^{N})}\quad\text{and}\quad|w_{a}^{\pm}|_{2}=|w^{\pm}|_{2},$
(3.6)

where $[v]^{2}_{H^{s}(\mathbb{R}^{N})}=\varepsilon(v,v)$ (see, for instance,
[3, Lemma 2.3] ).

######  Lemma 3.2.

Suppose $(f_{1})-(f_{3})$. Let $w\in H^{s}(\mathbb{R}^{N})$, radially
symmetric and such that $w(x)>0$ in $\\{x\in\mathbb{R}^{N}:\rho>|x|\\},$ for
certain $\rho\in(0,+\infty)$. Then, for $a\in(0,\rho)$,

$\mathcal{B}(w,w^{+})\geqslant\mathcal{B}(w_{a},w_{a}^{+})$ (3.7)

and

$-\mathcal{B}(w,w^{-})\geqslant\mathcal{B}(w_{a},w_{a}^{-}).$ (3.8)

###### Proof.

Note that $w_{a}\in H^{s}(\mathbb{R}^{N})$ by (3.6) and from [3, Lemma 2.3],
we know that

$\displaystyle\mathcal{E}(w,w^{+})\geqslant\mathcal{E}(w_{a},w_{a}^{+})\quad\text{
and }\quad-\mathcal{E}(w,w^{-})\geqslant-\mathcal{E}(w_{a},w_{a}^{-}).$

Hence, using (3.6),

$\displaystyle\lambda\int_{\mathbb{R}^{N}}ww^{+}=\lambda|w^{+}|^{2}_{2}=\lambda|w_{a}^{+}|^{2}_{2}=\lambda\int_{\mathbb{R}^{N}}w_{a}w_{a}^{+},$

and

$\displaystyle-\lambda\int_{\mathbb{R}^{N}}ww^{-}=\lambda|w^{-}|^{2}_{2}=\lambda|w_{a}^{-}|^{2}_{2}=-\lambda\int_{\mathbb{R}^{N}}w_{a}w_{a}^{-}.$

Then, by the definition of the bilinear form $\mathcal{B}$ in (2.5), in order
to conclude the proof it is enough to show that

$\int_{\mathbb{R}^{N}}f^{\prime}(u)(w^{+})^{2}dx\leqslant\int_{\mathbb{R}^{N}}f^{\prime}(u)(w_{a}^{+})^{2}dx,$
(3.9)

and

$\int_{\mathbb{R}^{N}}f^{\prime}(u)(w^{-})^{2}dx\geqslant\int_{\mathbb{R}^{N}}f^{\prime}(u)(w_{a}^{-})^{2}dx,$
(3.10)

hold.

We rewrite $\mathbb{R}^{N}=A\cup(\mathbb{R}^{N}\setminus A)$, where

$A:=\\{x\in\mathbb{R}^{N}:w_{a}(x)=w(x)\\}.$

Then

$\int_{A}f^{\prime}(u)(w^{+})^{2}dx=\int_{A}f^{\prime}(u)(w_{a}^{+})^{2}dx.$
(3.11)

Let

$\mathbb{R}^{N}\setminus
A=\\{x\in\mathbb{R}^{N}:w_{a}(x)=w(\overline{x})\\}=K^{+}\cup K^{-},$

where $\overline{x}=\sigma_{a}(x)=(2a-x_{1},x_{2},\cdots,x_{N}),$

$\displaystyle K^{+}:=(\mathbb{R}^{N}\setminus
A)\cap\Sigma^{+}_{a}=\\{x\in\mathbb{R}^{N}:w_{a}(x)=w(\overline{x})<w(x)\\}\subseteq\Sigma^{+}_{a},$
$\displaystyle K^{-}:=(\mathbb{R}^{N}\setminus
A)\cap\Sigma^{-}_{a}=\\{x\in\mathbb{R}^{N}:w_{a}(x)=w(\overline{x})>w(x)\\}\subseteq\Sigma^{-}_{a}.$

Note that

$\sigma_{a}(K^{\pm})=K^{\mp}.$

In order to prove (3.9) we split the integral and make a change of variable,
that is,

$\displaystyle\int_{\mathbb{R}^{N}\setminus
A}[f^{\prime}(u)(w_{a}^{+})^{2}-f^{\prime}(u)(w^{+})^{2}]dx$ (3.12)
$\displaystyle=$
$\displaystyle\int_{K^{-}}f^{\prime}(u(x))[(w_{a}^{+})^{2}(x)-(w^{+})^{2}(x)]dx+\int_{K^{+}}f^{\prime}(u(x))[(w_{a}^{+})^{2}(x)-(w^{+})^{2}(x)]dx$
$\displaystyle=$
$\displaystyle\int_{K^{-}}f^{\prime}(u(x))[(w^{+})^{2}(\bar{x})-(w^{+})^{2}(x)]dx+\int_{K^{+}}f^{\prime}(u(x))[(w^{+})^{2}(\bar{x})-(w^{+})^{2}(x)]dx$
$\displaystyle=$
$\displaystyle\int_{K^{-}}f^{\prime}(u(x))[(w^{+})^{2}(\bar{x})-(w^{+})^{2}(x)]dx+\int_{K^{-}}f^{\prime}(u(\bar{x}))[(w^{+})^{2}(x)-(w^{+})^{2}(\bar{x})]dx$
$\displaystyle=$
$\displaystyle\int_{K^{-}}[f^{\prime}(u(x))-f^{\prime}(u(\bar{x}))][(w^{+})^{2}(\bar{x})-(w^{+})^{2}(x)]dx$
$\displaystyle\geqslant$ $\displaystyle 0,$

where the last inequality follows since $f^{\prime}(t)$ is non-decreasing for
any $t>0$ and $u(x)>u(\bar{x})>0$ for $x\in K^{-}$, since $|x|<|\bar{x}|$ and
$u$ is radial and strictly decreasing in $r\in(0,+\infty)$. Then (3.9) follows
from (3.11) and (3.12).

The proof of (3.10) follows by similar arguments and we complete the proof.

∎

######  Lemma 3.3.

Suppose $(f_{1})-(f_{3})$. If the Morse index of $u$ is 1, then we have
$(\operatorname{ker}L_{+})\cap L^{2}_{rad}(\mathbb{R}^{N})=\\{0\\}.$

###### Proof.

Suppose by contradiction that there exists $w\in(\operatorname{ker}L_{+})\cap
L^{2}_{rad}$ and $w\neq 0$. Since the Morse index of $u$ is 1, then $0$ is the
second eigenvalue of $L_{+}$ with eigenfunction $w$. From Proposition 2.7, we
conclude that there is some $\rho>0$ such that (after multiplying $w$ with
$-1$ if necessary)

$w(r)\leqslant 0\text{ for }r\in\left[0,\rho\right)\text{ and }w(r)>0\text{
for }r\in\left(\rho,+\infty\right),$

and $w\not\equiv 0$ on both intervals $\left[0,\rho\right)$ and
$\left[\rho,+\infty\right)$. Note also that $w(0)<0$ and $w(x)=0$ if
$|x|=\rho$. Let us fix $a\in(0,\rho)$.

Next, by combining Lemma 2.9 and Lemma 3.2, one gets that $w_{a}$, the
polarization of $w$ with respect to $H_{a}$, is a second eigenfunction.
Further, $w_{a}$ is non-radial. Indeed, following [3] we take

$x^{*}:=(\rho+2a,0,\cdots,0)\in\Sigma^{+}_{a},$

so $|x^{*}|=\rho+2a\in(\rho,+\infty).$ Then, $w(x^{*})>0$. Moreover,
$\overline{x^{*}}=(-\rho,0,\cdots,0)\in\Sigma^{-}_{a}$ and
$|\overline{x^{*}}|=\rho$. Hence, $w(\overline{x}^{*})=0$, and so

$w_{a}(x^{*})=\min\\{w(x^{*}),w(\overline{x^{*}})\\}=0.$

Furthermore, $-x^{*}=(-\rho-2a,0,\cdots,0)\in\Sigma^{-}_{a}$, so
$|-x^{*}|=\rho+2a\in(\rho,+\infty)$, hence $w(-x^{*})>0$. And
$\overline{-x^{*}}=(\rho+3a,0,\cdots,0)\in\Sigma^{+}_{a}$, then
$w(\overline{-x^{*}})\geqslant 0$. As a consequence,

$w_{a}(-x^{*})=\max\\{w(-x^{*}),w(\overline{-x^{*}})\\}>0,$

and so

$w_{a}(x^{*})\neq w_{a}(-x^{*}).$

Finally, we will find a contradiction, and, it follows that
$(\operatorname{ker}L_{+})\cap L^{2}_{rad}(\mathbb{R}^{N})=\\{0\\}.$

We first consider the case when $N\geqslant 2$, and show that $w_{a}=k_{0}w$,
which will yield a contradiction since $w_{a}$ is non-radial. Proposition 2.6
gives us that $\operatorname{dim}(\operatorname{ker}L_{+}\cap
L^{2}_{rad}(\mathbb{R}^{N}))=1$. So
$\operatorname{ker}L_{+}=\operatorname{span}\\{\partial_{x_{1}}u,\cdots,\partial_{x_{N}}u,w\\},$
where $w$ is radial and sign-changing. Hence, there exist
$k_{0},k_{1},\ldots,k_{N}$ such that

$\displaystyle w_{a}=k_{0}w+\sum_{i=1}^{N}k_{i}\partial_{x_{i}}u.$ (3.13)

Consider

$y_{j,\pm}=(\underbrace{2a,\cdots,0}_{j-1},\pm\rho,0,\cdots,0),\quad\overline{y_{j,\pm}}=(\underbrace{0,\cdots,0}_{j-1},\pm\rho,0,\cdots,0),\quad
2\leqslant j\leqslant N.$

Note $y_{j,\pm}\in\Sigma_{a}^{+}$. We recall that $a\in(0,\rho)$, $w(\rho)=0$
and $w(r)>0$ for $r\in\left(\rho,+\infty\right)$. Then direct computations
show us that $w_{a}(y_{j,\pm})=0$ for $2\leqslant j\leqslant N$. Set
$c=|y_{j,\pm}|=\sqrt{4a^{2}+\rho^{2}}$, $2\leqslant j\leqslant N$, then

$0=k_{0}w(c)+k_{1}u^{\prime}(c)\frac{2a}{c}+k_{j}u^{\prime}(c)\frac{\pm\rho}{c},\quad
2\leqslant j\leqslant N.$ (3.14)

Since $u^{\prime}(c)<0$, we conclude that $k_{j}=0$ for all $2\leqslant
j\leqslant N$, and so

$w_{a}(x)=k_{0}w(x)+k_{1}u^{\prime}(|x|)\frac{x_{1}}{|x|}.$ (3.15)

Take $x_{1}$ such that $a<x_{1}<\rho$ and so $0<2a-x_{1}<x_{1}$. Set
$x^{*}=(x_{1},z)$, $\overline{x^{*}}=(2a-x_{1},z)$, where
$z\in\mathbb{R}^{N-1}$ such that $|x^{*}|=\rho$. Since
$|\overline{x^{*}}|<|x^{*}|=\rho$, there hold $w(x^{*})=w(\rho)=0$ and
$w(\overline{x^{*}})\leqslant 0$. Then
$w_{a}(x^{*})=\min\\{w(x^{*}),w(\overline{x^{*}})\\}\leqslant 0$. By (3.15),

$w_{a}(x^{*})=k_{1}u^{\prime}(\rho)\frac{x_{1}}{\rho}\leqslant 0,$ (3.16)

implying that $k_{1}\geqslant 0$.

Similarly, take $t_{1}$ such that $0<t_{1}<a$ and so $t_{1}<2a-t_{1}$. Set
$x^{**}=(t_{1},h)$, $\overline{x^{**}}=(2a-t_{1},h)$, where
$h\in\mathbb{R}^{N-1}$ such that $|x^{**}|=\rho$. Since
$|\overline{x^{**}}|>|x^{**}|=\rho$, there hold $w(x^{**})=w(\rho)=0$ and
$w(\overline{x^{**}})\geqslant 0$. Then
$w_{a}(x^{**})=\max\\{w(x^{**}),w(\overline{x^{**}})\\}\geqslant 0$. By
(3.15),

$w_{a}(x^{**})=k_{1}u^{\prime}(\rho)\frac{t_{1}}{\rho}\geqslant 0.$ (3.17)

implying that $k_{1}\leqslant 0$.

Thus $w_{a}=k_{0}w$, a contradiction!

Next, we consider the case when $N=1$. Here

$w_{a}(x)=k_{0}w(x)+k_{1}u^{\prime}(|x|)\frac{x}{|x|}.$ (3.18)

Choose $a>0$ such that $a<\rho<2a$. There hold $w(0)<0,w(2a)>0$. Take $x=0$
then $x<a$ and $\rho<2a=\overline{x}$. We have
$w_{a}(0)=\max\\{w(0),w(2a)\\}>0$. Then

$\displaystyle w_{a}(0)$ $\displaystyle=$ $\displaystyle
k_{0}w(0)+k_{1}u^{\prime}(0)$ (3.19) $\displaystyle=$ $\displaystyle
k_{0}w(0)$

since $u^{\prime}(0)=0$, implying that $k_{0}<0$. Moreover, by taking
$x=\rho$, then $x>a$ and $\overline{x}=2a-\rho$ satisfies
$|\overline{x}|<\rho$. Note that $w(2a-\rho)\leqslant 0$. We have
$w_{a}(\rho)=\min\\{w(0),w(2a-\rho)\\}<0$. Then, by

$\displaystyle
w_{a}(\rho)=k_{0}w(\rho)+k_{1}u^{\prime}(\rho)=k_{1}u^{\prime}(\rho)<0,$
(3.20)

one get $k_{1}>0$ since $u^{\prime}(\rho)<0$. On the other hand, we take
$x=2a+\rho$, then $x>a$ and $\overline{x}=-\rho$. Since $w(-\rho)=0$ and
$w(2a+\rho)\geqslant 0$, we have
$w_{a}(2a+\rho)=\min\\{w(-\rho),w(2a+\rho)\\}=0$. Then we deduce a self-
contradictory inequality

$\displaystyle 0=w_{a}(2a+\rho)$ $\displaystyle=$ $\displaystyle
k_{0}w(2a+\rho)+k_{1}u^{\prime}(2a+\rho)$ (3.21) $\displaystyle=$
$\displaystyle k_{0}w(2a+\rho)+k_{1}u^{\prime}(2a+\rho)$ $\displaystyle<$
$\displaystyle 0.$

This is a contradiction and we complete the proof. ∎

###  3.3 Completing the Proof of Theorem 2.5

Finally, inspired by [17], we prove that any positive solution of (1.3) has
Morse index 1 and complete the proof of non-degeneracy.

The following Piccone-type result will be crucial in the following proof.

######  Lemma 3.4 ([17], Lemma 2.1).

Let $\Omega$ be an open set of $\mathbb{R}^{N}$ and $\alpha>0$. Let $V\in
L^{1}_{loc}(\Omega)$ and $v\in C^{2s+\alpha}(\Omega)\cap
L^{1}(\mathbb{R}^{N};(1+|x|)^{-N-2s})$ satisfy

$(-\Delta)^{s}v=Vv\quad\textrm{in $\Omega$}.$ (3.22)

Let $e$ be a unit vector of $\mathbb{R}^{N}$ and let $\sigma_{e}$ denote the
reflection with respect to $\\{x\in\mathbb{R}^{N}\,:\,x\cdot e=0\\}$. Suppose
that $v>0$ on $\\{x\in\mathbb{R}^{N}\,:\,x\cdot e>0\\}$ and
$v=-v\circ\sigma_{e}$ on $\mathbb{R}^{N}$. Then for any $w\in
H^{s}(\mathbb{R}^{N})\cap C_{c}(\Omega)$ satisfying $w=-w\circ\sigma_{e}$ on
$\mathbb{R}^{N}$ and $w/v\in C_{c}(\Omega)$, we have

$[w]_{s}^{2}-\int_{\mathbb{R}^{N}}Vw^{2}dx=\int_{\\{x\cdot
e>0\\}}\int_{\\{y\cdot e>0\\}}H_{w,v}^{e}(x,y)dxdy,$ (3.23)

where $H_{w,v}^{e}\geqslant 0$ on $\\{x\cdot e>0\\}\cap\\{y\cdot e>0\\}$ is
given by

$H_{w,v}^{e}(x,y):=c_{N,s}{v(x)v(y)}\left[w(x)/v(x)-w(y)/v(y)\right]^{2}\times\left(\frac{1}{|x-y|^{N+2s}}-\frac{1}{|\sigma_{e}(x)-y|^{N+2s}}\right).$
(3.24)

######  Lemma 3.5.

Suppose $(f_{1})-(f_{3})$. Then any positive solution $u$ of (1.3) has the
Morse index equal to $1$.

###### Proof.

Note that $\mu_{1}<0$ follows by $(f_{2})$ where $\mu_{1}$ is the first
eigenvalue of $L_{+}$. Let $\mu_{2}$ be the second eigenvalue of $L_{+}$. It
suffices to prove that $\mu_{2}=0$. Note first that $\mu_{2}\leqslant 0$,
which follows that $\partial_{x_{i}}u\in\ker L_{+}$ for any
$i\in\\{1,\cdots,N\\}$. We then consider $w$ an eigenfunction corresponding to
$\mu_{2}$ with $|w|_{2}=1$. By elliptic regularity we have $w\in
C^{2}(\mathbb{R}^{N})$. Similar to the proof of Lemma 3.3, we may assume that
$w$ is non-radial. Hence, there exists a unit vector $\nu\in\mathbb{R}^{N}$
such that $w^{\nu}:=\frac{w-w\circ\sigma_{\nu}}{2}\in
H^{s}(\mathbb{R}^{N})\cap C^{2}(\mathbb{R}^{N})$ is not zero, here we define
the hyperplane $T_{\nu}:=\\{x\cdot\nu=0\\}$ and denote by $\sigma_{\nu}$ the
reflection with respect to $T_{\nu}$. Clearly since $u$ is radial, then
$w^{\nu}$ solves $(-\Delta)^{s}w^{\nu}+\lambda
w^{\nu}-f^{\prime}(u)w^{\nu}=\mu_{2}w^{\nu}$ in $\mathbb{R}^{N}$. Moreover, we
have $v:=-\nabla u\cdot\nu\in H^{s}(\mathbb{R}^{N})$ solves the equation
$(-\Delta)^{s}v+\lambda v=f^{\prime}(u)v$ in $\mathbb{R}^{N}$, and $v$ is odd
with respect to the reflection at $\nu$ and $v>0$ on $\\{x\cdot\nu>0\\}$. By
elliptic regularity, we have $v\in C^{2}(\mathbb{R}^{N})$. Applying Hopf’s
Lemma for the fractional Laplacian (see [15, Corollary 3.3] and [17, Lemma
5.3]), we have $\nabla v\cdot\nu>0$ on $\\{x\cdot\nu=0\\}$. Therefore,

$x\mapsto\frac{v(x)}{x\cdot\nu}\qquad\text{extends to a positive continuous
function on $\mathbb{R}^{N}$.}$ (3.25)

We let

$\chi\in C^{\infty}_{c}(-2,2),\quad\textrm{ with $0\leqslant\chi\leqslant 1$
on $\mathbb{R}$ and $\chi\equiv 1$ on $(-1,1)$}.$ (3.26)

For $R\in\mathbb{N}$, we define $w_{R}^{\nu}(x)=w^{\nu}(x)\chi(|x|/R)$ for all
$x\in\mathbb{R}^{N}$. By (3.25), we have $w_{R}/v\in C_{c}(\mathbb{R}^{N})$.
Applying Lemma 3.4, we get

$\displaystyle\int_{\mathbb{R}^{N}}(|(-\Delta)^{s/2}w_{R}^{\nu}|^{2}+\lambda|w_{R}^{\nu}|^{2}-f^{\prime}(u)|w_{R}^{\nu}|^{2})dx=\int_{\\{x\cdot\nu>0\\}\cap\\{y\cdot\nu>0\\}}H_{w_{R}^{\nu},v}^{\nu}(x,y)dxdy.$
(3.27)

It is not difficult to see that
$\|w_{R}^{\nu}\|_{H^{s}(\mathbb{R}^{N})}\to\|w^{\nu}\|_{H^{s}(\mathbb{R}^{N})}$
as $R\to\infty$. On the other hand by the dominated convergence theorem, as
$R\to\infty$,

$\int_{\mathbb{R}^{N}}(w_{R}^{\nu})^{2}dx\to\int_{\mathbb{R}^{N}}(w^{\nu})^{2}dx\qquad\textrm{and}\qquad\int_{\mathbb{R}^{N}}f^{\prime}(u)(w_{R}^{\nu})^{2}dx\to\int_{\mathbb{R}^{N}}f^{\prime}(u)(w^{\nu})^{2}dx.$

By Fatou’s lemma, we obtain that

$\mu_{2}\int_{\mathbb{R}^{N}}|w^{\nu}|^{2}dx\geqslant\int_{\\{x\cdot\nu>0\\}\cap\\{y\cdot\nu>0\\}}H_{w^{\nu},v}^{\nu}(x,y)dxdy\geqslant
0.$ (3.28)

Note that $w^{\nu}\not\equiv 0$. Hence, we have $\mu_{2}\geqslant 0$.
Combining with $\mu_{2}\leqslant 0$, we conclude that $\mu_{2}=0$. ∎

###### Proof of Theorem 2.5.

By Lemma 3.5, any positive solution $u$ of (1.3) has the Morse index equal to
$1$. Then by Lemma 3.3, we have $(\operatorname{ker}L_{+})\cap
L^{2}_{rad}(\mathbb{R}^{N})=\\{0\\}$. Together with Lemma 3.1, we conclude
that

$\ker
L_{+}=\operatorname{span}\left\\{\partial_{x_{1}}u,\ldots,\partial_{x_{N}}u\right\\}.$

∎

##  4 Uniqueness of the Positive Solution

In this section, we give the proof of Theorem 1.1 by using a continuation
argument. In contrast to the discussions in [18, 19], the general
nonlinearities bring new difficulties. Here, inspired by [20], we use a
homotopic argument by continuation in $\lambda$. The first step is to
construct a global branch. More precisely, we need to prove the following
result.

######  Proposition 4.1.

Assume that $(f_{1})-(f_{3})$ hold. Let $N\geqslant 1,s\in(0,1)$, and $u_{0}$
be a positive solution of (1.3) with $\lambda=\lambda_{0}>0$. Then there is a
map $u\in C^{1}([\lambda_{0},\infty),H_{\text{rad}}^{s})$ such that
$u(\lambda)$ is positive and solves (1.3) with $u(\lambda_{0})=u_{0}$.

We will prove Proposition 4.1 in Subsection 4.1. Thanks to the non-degeneracy
result of Theorem 2.5, we can construct a local branch stemming from
$\lambda=\lambda_{0}$ by using implicit function arguments. Then, in order to
extend this branch to $\infty$, we give some results about uniform bounds and
decay w.r.t. $\lambda$, see details in Subsection 4.1.

Next, we aim to show that the branch constructed by Proposition 4.1 is unique.
In fact, we define a one-to-one transform $T_{\lambda}$ from
$H^{s}(\mathbb{R}^{N})$ to $H^{s}(\mathbb{R}^{N})$:

$T_{\lambda}u=\lambda^{-\frac{1}{r-2}}u(\frac{x}{\lambda^{1/2s}}).$ (4.1)

Then we get a branch $v\in C^{1}([\mu_{0},\infty),H_{\text{rad}}^{s})$ such
that $v(\lambda):=T_{\lambda}u(\lambda)$ is positive and solves

$(-\Delta)^{s}v+v=\mu^{1-r}f(\mu v)\quad\text{in }\mathbb{R}^{N},$ (4.2)

with $v(\mu_{0})=v_{0}:=T_{\lambda_{0}}u_{\lambda_{0}}$, where
$\mu=\lambda^{\frac{1}{r-2}}$. Then, it suffices to prove the uniqueness of
the branch $v(\mu)$. As $\mu\to\infty$, we will provide the uniform bounds and
decay of $v_{\mu}$, and then to show that $v_{\mu}$ converges to the positive,
unique (up to translations) and non-degenerate solution of the limit equation
$(-\Delta)^{s}v_{*}+v_{*}-\left|v_{*}\right|^{r-2}v_{*}=0$ in
$\mathbb{R}^{N}$. Then by using implicit function arguments again, we can
prove the following result.

######  Proposition 4.2.

Assume that $(f_{1})-(f_{4})$ hold. Let $N\geqslant 1,s\in(0,1)$, and for
$i=1,2$, let $v_{i}\in C^{1}([\mu_{0},\infty),H_{\text{rad}}^{s})$ be the
global branch such that $v_{i}(\mu)>0$ solves (4.2), then $v_{1}\equiv v_{2}$.

See Subsection 4.2 for the proof of Proposition 4.2. Finally in Subsection
4.3, we will complete the proof of Theorem 1.1.

###  4.1 Construction of the Global Branch

Consider:

$\displaystyle G:\ $ $\displaystyle
H^{s}_{rad}(\mathbb{R}^{N})\times[\lambda_{0},\infty)\to
H^{-s}(\mathbb{R}^{N}),$ $\displaystyle G(u,\lambda)=(-\Delta)^{s}u+\lambda
u-f(u).$

Note that $u$ solves (1.3) is equivalent to $G(\lambda,u)=0$. In this
subsection, we set

$L_{+,\lambda}:=(-\Delta)^{s}+\lambda-f^{\prime}(u).$

######  Lemma 4.3.

Suppose $(f_{1})-(f_{3})$. Let $u_{0}\in H_{\text{rad }}^{s}(\mathbb{R}^{N})$
be a positive solution of $G(u_{0},\lambda_{0})=0$. Then for some $\delta>0$,
there exists a map $u(\lambda)\in
C^{1}(\left[\lambda_{0},\lambda_{0}+\delta),H_{\text{rad}}^{s}\right)$ such
that the following holds true, where we denote $u_{\lambda}=u(\lambda)$ in the
sequel.

(i) $u_{\lambda}$ solves $G(u,\lambda)=0$ for all
$\lambda\in[\lambda_{0},\lambda_{0}+\delta)$.

(ii) There exists $\epsilon>0$ such that $u_{\lambda}$ is the unique solution
of $G(u,\lambda)=0$ for $\lambda\in[\lambda_{0},\lambda_{0}+\delta)$ in the
neighborhood $\left\\{u\in H_{\text{rad
}}^{s}:\left\|u-u_{\lambda}\right\|_{H_{\text{rad }}^{s}}<\epsilon\right\\}$.
In particular, we have that $u_{\lambda_{0}}=u_{0}$ holds.

###### Proof.

We use an implicit function argument for the map

$\Psi:H_{\text{rad }}^{s}\times[\lambda_{0},\lambda_{0}+\delta)\rightarrow
H_{\text{rad
}}^{s},\quad\Psi(u,\lambda)=u-\frac{1}{(-\Delta)^{s}+\lambda}f(u).$

It is not difficult to see that $\Psi$ is a well-defined map of class $C^{1}$.
Furthermore, $\Psi\left(u_{0},\lambda_{0}\right)=0$ and $\Psi(u,\lambda)=0$ if
and only if $u\in H_{\text{rad }}^{s}$ solves $G(u,\lambda)=0$. Next, we
consider the Fréchet derivative

$\partial_{u}\Psi\left(u_{0},\lambda_{0}\right)=1-K,K=\frac{1}{(-\Delta)^{s}+\lambda_{0}}f^{\prime}(u_{0})$

By Theorem 2.5, $0$ is not an eigenvalue of
$L_{+,\lambda_{0}}=(-\Delta)^{s}+\lambda_{0}-f^{\prime}(u_{0})$ constrained on
$L_{rad}^{2}$. Note that the operator $K$ is compact on $L_{\text{rad }}^{2}$
and we have that $1\notin\sigma(K)$. Since $f^{\prime}(u_{0})\varphi\in
H^{-s}$ if $\varphi\in H_{\text{rad }}^{s}$, we have that $K$ maps
$H_{\text{rad }}^{s}$ to $H_{\text{rad }}^{s}$. Thus the bounded inverse
$(1-K)^{-1}$ exists on $H_{\text{rad }}^{s}$, i.e., the Fréchet derivative
$\partial_{u}\Psi$ has a bounded inverse on $H_{\text{rad }}^{s}$ at
$\left(u_{0},\lambda_{0}\right)$. By the implicit function theorem, we deduce
that the assertions $(i)$ and $(ii)$ in Lemma 4.3 hold for some $\delta>0$ and
$\epsilon>0$ sufficiently small. ∎

We now turn to the global extension of the locally unique branch $u_{\lambda}$
constructed in the above Lemma 4.3. First, we will give some uniform priori
estimates. We write $a\lesssim b$ to denote that $a\leqslant Cb$ with some
constant $C>0$ that may change from line to line and only depends on $N$,
$s,\lambda$ and $u_{0}$. Furthermore, we use $a\sim b$ to denote that both
$a\lesssim b$ and $b\lesssim a$ hold. In what follows, let us assume that
$N>2s$ holds. The arguments can be adapted with some modifications to the case
$N\leqslant 2s$.

######  Lemma 4.4 (Pohozaev identity, [20], Lemma 4.5).

Let $u$ be a weak solution of (1.3), then the following integral identity
holds true:

$(N-2s)\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}u|^{2}dx+N\lambda\int_{\mathbb{R}^{N}}u^{2}dx=2N\int_{\mathbb{R}^{N}}F(u)dx.$
(4.3)

######  Lemma 4.5.

Suppose $(f_{1})-(f_{3})$. Then we have the following power estimates

$\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\sim\lambda\int_{\mathbb{R}^{N}}u_{\lambda}^{2}dx\sim\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}u_{\lambda}|^{2}dx\sim
1,$

for all $\lambda\in[\lambda_{0},\lambda_{*})$, where $u_{\lambda}$ is given in
Lemma 4.3 and

$\displaystyle\lambda_{*}=\sup\\{\hat{\lambda}>0:\ $ $\displaystyle
u_{\lambda}\in C^{1}([\lambda_{0},\lambda_{*}),H^{s}_{rad}(\mathbb{R}^{N})),$
$\displaystyle u_{\lambda}\text{ satisfies the assumptions of Lemma
\ref{prop4.15}}\text{ for}\lambda\in[\lambda_{0},\hat{\lambda})\\}.$

###### Proof.

We now proceed in two steps.

Step 1. We claim $\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\gtrsim 1$
for $\lambda\in(\lambda_{0},\lambda_{*})$.

By integrating $G(u_{\lambda},\lambda)=0$ after a multiplication by
$u_{\lambda}$, we have

$\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}u_{\lambda}|^{2}dx+\lambda\int_{\mathbb{R}^{N}}u_{\lambda}^{2}dx=\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx.$
(4.4)

Consequently, (4.3) and (4.4) together, give

$2s\lambda\int_{\mathbb{R}^{N}}u_{\lambda}^{2}dx=2\int_{\mathbb{R}^{N}}NF(u_{\lambda})dx-(N-2s)\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx,$
(4.5)
$2s\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}u_{\lambda}|^{2}dx=N\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx-2N\int_{\mathbb{R}^{N}}F(u_{\lambda})dx.$
(4.6)

By assumption $(f_{3})$, we have $F(t)\sim f(t)t$, therefore we derive

$\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\sim\lambda\int_{\mathbb{R}^{N}}u_{\lambda}^{2}dx\sim\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}u_{\lambda}|^{2}dx.$
(4.7)

Since $2<q<r<\frac{2N}{N-2s}$, from the following fractional Gagliardo-
Nirenberg-Sobolev inequality

$\int_{\mathbb{R}^{n}}|u|^{q}dx\leqslant
C_{opt}\left(\int_{\mathbb{R}^{n}}|(-\Delta)^{s/2}u|^{2}dx\right)^{\frac{N(q-2)}{4s}}\left(\int_{\mathbb{R}^{n}}|u|^{2}dx\right)^{\frac{q}{2}-\frac{N(q-2)}{4s}},\quad\forall
u\in H^{s}(\mathbb{R}^{N}),$ (4.8)

where $C_{opt}$ is the best constant and depends on $s,N,q$, we derive

$\displaystyle\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{R}^{N}}u_{\lambda}^{q}+u_{\lambda}^{r}dx$ (4.9)
$\displaystyle\lesssim$
$\displaystyle\left(\int_{\mathbb{R}^{n}}|(-\Delta)^{s/2}u|^{2}dx\right)^{\frac{N(q-2)}{4s}}\left(\int_{\mathbb{R}^{n}}|u|^{2}dx\right)^{\frac{q}{2}-\frac{N(q-2)}{4s}}$
$\displaystyle+\left(\int_{\mathbb{R}^{n}}|(-\Delta)^{s/2}u|^{2}dx\right)^{\frac{N(r-2)}{4s}}\left(\int_{\mathbb{R}^{n}}|u|^{2}dx\right)^{\frac{r}{2}-\frac{N(r-2)}{4s}}$
$\displaystyle\lesssim$
$\displaystyle\left(\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\right)^{q/2}\lambda^{\frac{N(q-2)}{4s}-\frac{q}{2}}$
$\displaystyle+\left(\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\right)^{r/2}\lambda^{\frac{N(r-2)}{4s}-\frac{r}{2}}.$

Thus $\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\gtrsim 1$ for
$\lambda\in(\lambda_{0},\lambda_{*})$.

Step 2. Next we show that
$\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\lesssim 1$ for
$\lambda\in(\lambda_{0},\lambda_{*})$.
Set

$g(\lambda)=\int_{\mathbb{R}^{N}}[f(u_{\lambda})u_{\lambda}-2F(u_{\lambda})]dx.$

Assumption $(f_{3})$ and (4.7), yield that

$\lambda\int_{\mathbb{R}^{N}}u^{2}_{\lambda}dx\sim\int_{\mathbb{R}^{N}}f(u_{\lambda})u_{\lambda}dx\sim
g(\lambda).$ (4.10)

Therefore, it suffices to show that $g(\lambda)$ has uniform upper bound on
$(\lambda_{0},\lambda_{*})$.
Differentiating $G(u_{\lambda},\lambda)=0$ with resect to $\lambda$, we get

$(-\Delta)^{s}\frac{d}{d\lambda}u_{\lambda}=-u_{\lambda}-\lambda\frac{d}{d\lambda}u_{\lambda}+f^{\prime}(u_{\lambda})\frac{d}{d\lambda}u_{\lambda},$
(4.11)

hence

$L_{+,\lambda}(\frac{d}{d\lambda}u_{\lambda})=-u_{\lambda}.$ (4.12)

Note that

$L_{+,\lambda}u_{\lambda}=f(u_{\lambda})-f^{\prime}(u_{\lambda})u_{\lambda},$

and this implies, via (4.12),

$\displaystyle g^{\prime}(\lambda)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}f^{\prime}(u_{\lambda})\frac{d}{d\lambda}u_{\lambda}\cdot
u_{\lambda}+f(u_{\lambda})\frac{d}{d\lambda}u_{\lambda}-2f(u_{\lambda})\frac{d}{d\lambda}u_{\lambda}dx$
(4.13) $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}f^{\prime}(u_{\lambda})u_{\lambda}\frac{d}{d\lambda}u_{\lambda}-f(u_{\lambda})\frac{d}{d\lambda}u_{\lambda}dx$
$\displaystyle=$ $\displaystyle-\langle
L_{+,\lambda}u_{\lambda},\frac{d}{d\lambda}u_{\lambda}\rangle_{L^{2}}$
$\displaystyle=$ $\displaystyle-\langle
u_{\lambda},L_{+,\lambda}\frac{d}{d\lambda}u_{\lambda}\rangle_{L^{2}}$
$\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{N}}u_{\lambda}^{2}dx.$

Via (4.10) and (4.13), we obtain the differential inequality

$\displaystyle g^{\prime}(\lambda)$ $\displaystyle\lesssim$
$\displaystyle\frac{1}{\lambda}g(\lambda),$ (4.14)

which by integration yields the uniform upper bound
$g(\lambda)\lesssim\lambda^{C}\lesssim 1$ for
$\lambda\in[\lambda_{0},\lambda_{*})$ where $C$ is a positive constant. The
proof of Lemma 4.5 is now complete. ∎

######  Lemma 4.6.

Suppose $(f_{1})-(f_{3})$ and that $u_{\lambda_{0}}(x)>0$ is positive. Let
$\lambda_{*}$ is given by Lemma 4.5. Then, for all
$\lambda\in\left[\lambda_{0},\lambda_{*}\right)$, we have

  * (i)

$u_{\lambda}(x)>0$ for all $x\in\mathbb{R}^{N}$;

  * (ii)

(uniform decay estimate) $u_{\lambda}(x)\leqslant C|x|^{-(N+2s)}$ for $|x|>R$,
where $R$ is some constant independent of $\lambda$.

###### Proof.

Define the linearized operator

$L_{-,\lambda}=(-\Delta)^{s}+\lambda-\frac{f(u)}{u}$ (4.15)

acting on $L^{2}(\mathbb{R}^{N})$. Obviously, if
$\lambda\to\tilde{\lambda}\in[\lambda_{0},\lambda_{*})$, then
$L_{\pm,\lambda}\to L_{\pm,\tilde{\lambda}}$ in norm-resolvent sense.

The proof of $(i)$ is divided into two steps. First, we show that if
$u_{\tilde{\lambda}}>0$ for some
$\tilde{\lambda}\in[\lambda_{0},\lambda_{*})$, then $u_{\lambda}>0$ for
$\lambda$ close to $\tilde{\lambda}$. Note that $L_{-,\lambda}u_{\lambda}=0$,
i.e., $0$ is an eigenvalue of $L_{-,\lambda}$ and $u_{\lambda}$ is an
eigenfunction with respect to $0$. Since
$\sigma_{ess}(L_{-,\lambda})=[\lambda,+\infty)$, we know that
$\inf\sigma(L_{-,\lambda})$ is an eigenvalue. [19, Lemma C.4] yields that the
first eigenvalue is simple and the corresponding eigenfunction can be chosen
strictly positive. Thus $\lambda_{1}(L_{-,\tilde{\lambda}})=0$ is simple and
we choose $\psi_{1,\tilde{\lambda}}=u_{\tilde{\lambda}}>0$ as the
corresponding eigenfunction, where $\lambda_{1}(L_{-,\tilde{\lambda}})$ means
the first eigenvalue of $L_{-,\tilde{\lambda}}$. Since $L_{-,\lambda}\to
L_{-,\tilde{\lambda}}$ in norm-resolvent sense when
$\lambda\to\tilde{\lambda}\in[\lambda_{0},\lambda_{*})$, we derive that
$\lambda_{1}(L_{-,\tilde{\lambda}})\to 0$. Since
$\lambda_{1}(L_{-,\tilde{\lambda}})$ is simple and isolate, we assume that
$\sigma(L_{-,\lambda})\cap(-\varepsilon,\varepsilon)=\\{\lambda_{1}(L_{-,\tilde{\lambda}})\\}$
for some small $\varepsilon$ and $\lambda$ close enough to $\tilde{\lambda}$.
Furthermore, $0$ is an eigenvalue of $L_{-,\lambda}$, showing
$\lambda_{1}(L_{-,\lambda})=0$. Since $u_{\lambda}\to u_{\tilde{\lambda}}>0$
in $H^{s}$, we obtain $u_{\lambda}>0$ for $\lambda$ close enough to
$\tilde{\lambda}$.

The second step is to show that if
$\\{\lambda_{n}\\}_{n=1}^{+\infty}\subset[\lambda_{0},\lambda_{*})$ is a
sequence with $\lambda_{n}\to\hat{\lambda}$ such that $u_{\lambda_{n}}>0$ for
some $\hat{\lambda}\in[\lambda_{0},\lambda_{*})$, then $u_{\hat{\lambda}}>0$.
Obviously, $u_{\hat{\lambda}}\geqslant 0$ since $u_{\lambda_{n}}\to
u_{\hat{\lambda}}$ in $H^{s}$. Thus $0$ is the first eigenvalue of
$L_{-,\hat{\lambda}}$ and so $u_{\hat{\lambda}}>0$. The proof of $(i)$ is
complete.

Next, we prove $(ii)$. Since $u_{\lambda}$ is monotonically decreasing with
respect to $r=|x|$, we have $|x|^{N}u^{2}_{\lambda}(x)\lesssim
u^{2}_{\lambda}(x)\int_{B_{|x|}}dy\lesssim\int_{B_{|x|}}u^{2}_{\lambda}(y)dy\lesssim
1.$ We use the uniform priori estimate shown in Lemma 4.5 in the final
inequality. Thus $u_{\lambda}(x)\lesssim|x|^{-\frac{N}{2}}$.

We write $u_{\lambda}=\frac{1}{(-\Delta)^{s}+\theta}z_{\lambda}$ where
$0<\theta<\lambda$ and
$z_{\lambda}=(\theta-\lambda)u_{\lambda}+f(u_{\lambda})$. Since
$u_{\lambda}(x)\lesssim|x|^{-\frac{N}{2}}$, we can take $R$ large enough such
that $suppz_{\lambda}^{+}\subset B_{\frac{R}{2}}$ and $R$ is independent of
$\lambda$. Suppose $G_{s,\theta}(x)$ is the kernel of the resolvent
$((-\Delta)^{s}+\theta)^{-1}$ on $\mathbb{R}^{N}$. Notice that
$f(u)\lesssim(u^{q-1}+u^{r-1})$, then we obtain

$\displaystyle\int_{B_{\frac{R}{2}}}z_{\lambda}^{+}(y)dy$
$\displaystyle\lesssim$ $\displaystyle\int_{B_{\frac{R}{2}}\cap
suppz_{\lambda}^{+}}(u_{\lambda}^{q-1}+u^{r-1}_{\lambda})dy$ (4.16)
$\displaystyle\lesssim$
$\displaystyle(\int_{B_{\frac{R}{2}}}1dy)^{\frac{1}{q}}(\int_{B_{\frac{R}{2}}\cap
suppz_{\lambda}^{+}}u_{\lambda}^{q}dy)^{\frac{q-1}{q}}+(\int_{B_{\frac{R}{2}}}1dy)^{\frac{1}{r}}(\int_{B_{\frac{R}{2}}\cap
suppz_{\lambda}^{+}}u^{r}_{\lambda}dy)^{\frac{r-1}{r}}$
$\displaystyle\lesssim$ $\displaystyle R^{\frac{N}{q}}+R^{\frac{N}{r}}$

We use the uniform priori estimates shown in Lemma 4.5 in the final
inequality. Then, we can conclude that: for all $|x|>R$,

$\displaystyle u_{\lambda}(x)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}G_{s,\theta}(x-y)u_{\lambda}(y)dy\leqslant\int_{B_{\frac{R}{2}}}G_{s,\theta}(x-y)z^{+}_{\lambda}(y)dy$
(4.17) $\displaystyle\lesssim$
$\displaystyle\int_{B_{\frac{R}{2}}}|x-y|^{-(N+2s)}z^{+}_{\lambda}(y)dy\lesssim\int_{B_{\frac{R}{2}}}|x|^{-(N+2s)}z^{+}_{\lambda}(y)dy$
$\displaystyle\lesssim$ $\displaystyle|x|^{-(N+2s)}.$

∎

######  Lemma 4.7.

Suppose $(f_{1})-(f_{3})$. Let
$\\{\lambda_{n}\\}_{n=1}^{+\infty}\subset[\lambda_{0},\lambda_{*})$ be a
sequence such that
$\lambda_{n}\to\lambda_{*},u_{\lambda_{n}}=u_{\lambda_{n}}(|x|)>0$. Then,
after possibly passing to a subsequence, we have $u_{\lambda_{n}}\to
u_{\lambda_{*}}$ in $H^{s}_{rad}(\mathbb{R}^{N})$, where
$u_{\lambda_{*}}(|x|)>0$ satisfies

$(-\Delta)^{s}u_{\lambda_{*}}+\lambda_{*}u_{\lambda_{*}}=f(u_{\lambda_{*}})\quad\text{in
}\mathbb{R}^{N}.$ (4.18)

###### Proof.

By Lemma 4.5, $u_{\lambda_{n}}$ is bounded in $H^{s}$. Passing to a
subsequence if necessary, we assume that $u_{\lambda_{n}}\rightharpoonup
u_{\lambda_{*}}$ in $H^{s},u_{\lambda_{n}}\rightarrow u_{\lambda_{*}}$ in
$L_{loc}^{p}$ for $1\leqslant p<2_{s}^{*},u_{\lambda_{n}}\rightarrow
u_{\lambda_{*}}$ a.e. on $\mathbb{R}^{N}$. With the help of the uniform decay
estimate shown in Lemma 4.6 $(ii)$, we obtain that $u_{\lambda_{n}}\rightarrow
u_{\lambda_{*}}$ in $L^{p}$ for $1\leqslant p<2_{s}^{*}$. Bootstrap arguments
implies that $u_{\lambda_{n}}\rightarrow u_{\lambda_{*}}$ in $H^{s}$.
Obviously, $u_{\lambda_{*}}\geqslant 0$ satisfies equation (4.18). Thus 0 is
the first eigenvalue of $L_{-,\lambda_{*}}$, showing $u_{\lambda_{*}}>0$. ∎

###### Proof of Proposition 4.1.

It suffices to prove that $\lambda_{*}=\infty$, which is given by Lemma 4.5.
Suppose on the contrary that $\lambda_{*}<\infty$. By Lemma 4.6, we can assume
that $u(\lambda)$ is positive for all $\lambda\in[\lambda_{0},\lambda_{*})$.
Then, by Lemma 4.7, we get $u_{\lambda_{*}}$ which is a positive solution of
(4.18). Applying Theorem 2.5, one obtain that $u_{\lambda_{*}}$ is non-
degenerate. Then using implicit function arguments again (like the proof of
Lemma 4.3), we can extend the branch $u(\lambda)$ to $\lambda_{*}+\delta$,
contradicting to the definition of $\lambda_{*}$. Hence, we complete the
proof. ∎

###  4.2 Uniqueness of the Global Branch

In this subsection, we aim to prove Proposition 4.2, and, we set

$L_{+,\mu}=(-\Delta)^{s}+1-\mu^{2-r}f^{\prime}(\mu v).$

######  Lemma 4.8.

Suppose $(f_{1})-(f_{4})$. Let $v(\mu)\in
C^{1}([\mu_{0},\infty),H_{\text{rad}}^{s})$ be the global branch such that
$v_{\mu}:=v(\mu)$ is positive and solves (4.2). Then we have the priori bounds

$\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx\sim\int_{\mathbb{R}^{N}}v_{\mu}^{2}dx\sim\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx\sim
1,$

uniformly w.r.t. $\mu\in[\mu_{0},\infty)$.

###### Proof.

We divide the proof into the following two steps.

Step 1. Lower bounds.

First, we claim

$\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx\sim\int_{\mathbb{R}^{N}}v_{\mu}^{2}dx\sim\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx.$

By integrating (4.2) after a multiplication by $v_{\mu}$, we obtain

$\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx+\int_{\mathbb{R}^{N}}v_{\mu}^{2}dx=\int_{\mathbb{R}^{N}}\mu^{1-r}f(\mu
v_{\mu})v_{\mu}dx.$ (4.19)

For (4.2), we have the Pohozaev type identity

$(N-2s)\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx+N\int_{\mathbb{R}^{N}}v_{\mu}^{2}dx=2N\int_{\mathbb{R}^{N}}\mu^{-r}F(\mu
v_{\mu})dx.$ (4.20)

Via (4.19) and (4.20), we have

$2s\int_{\mathbb{R}^{N}}v_{\mu}^{2}dx=2N\int_{\mathbb{R}^{N}}\mu^{-r}F(u_{\lambda})dx-(N-2s)\int_{\mathbb{R}^{N}}\mu^{1-r}f(\mu
v_{\mu})v_{\mu}dx,$ (4.21)

and

$2s\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx=N\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx-2N\mu^{-r}\int_{\mathbb{R}^{N}}F(\mu v_{\mu})dx.$ (4.22)

Assumption $(f_{3})$ implies $F(t)\sim f(t)t$, hence

$\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx\sim\int_{\mathbb{R}^{N}}v_{\mu}^{2}dx\sim\int_{\mathbb{R}^{N}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx.$

Since $2<q<r<2^{*}_{s}=\frac{2N}{N-2s}$, from the fractional Gagliardo-
Nirenberg-Sobolev inequality (4.8), we can derive

$\displaystyle\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})v_{\mu}dx$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{R}^{N}}\mu^{q-r}v_{\mu}^{q}+\mu^{r-r}v_{\mu}^{r}dx$
(4.23) $\displaystyle\lesssim$
$\displaystyle\mu^{q-r}(\int_{\mathbb{R}^{n}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx)^{\frac{N(q-2)}{4s}}\left(\int_{\mathbb{R}^{n}}|v_{\mu}|^{2}dx\right)^{\frac{q}{2}-\frac{N(q-2)}{4s}}$
$\displaystyle+(\int_{\mathbb{R}^{n}}|(-\Delta)^{s/2}v_{\mu}|^{2}dx)^{\frac{N(r-2)}{4s}}\left(\int_{\mathbb{R}^{n}}|v_{\mu}|^{2}dx\right)^{\frac{r}{2}-\frac{N(r-2)}{4s}}$
$\displaystyle\lesssim$
$\displaystyle\mu^{q-r}(\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx)^{q/2}$ $\displaystyle+(\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx)^{r/2}.$

Thus $\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})v_{\mu}dx\gtrsim 1$ for
$\mu\in[\mu_{0},\infty)$.

Step 2. Upper bounds. Next we show that $\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx\lesssim 1$ for $\mu\in[\mu_{0},\infty)$.

Set

$k(\mu)=\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})v_{\mu}dx,$

$h(\mu)=2\mu^{-r}\int_{\mathbb{R}^{N}}F(\mu v_{\mu})dx,$

and

$B(\mu)=\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})\mu v_{\mu}-2F(\mu
v_{\mu})dx.$

By $(f_{3})$, we have $B(\mu)\sim k(\mu)$. Now it suffices to prove that
$B(\mu)\lesssim 1$.

Clearly, a brief computation gives

$\displaystyle k^{\prime}(\mu)$ $\displaystyle=$
$\displaystyle(1-r)\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx+\mu^{1-r}\int_{\mathbb{R}^{N}}f^{\prime}(\mu
v_{\mu})[v_{\mu}+\mu\frac{d}{d\mu}v_{\mu}]v_{\mu}dx$ (4.24) $\displaystyle+$
$\displaystyle\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})\frac{d}{d\mu}v_{\mu}dx$ $\displaystyle=$
$\displaystyle(1-r)\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx+\mu^{1-r}\int_{\mathbb{R}^{N}}f^{\prime}(\mu
v_{\mu})v_{\mu}^{2}+f^{\prime}(\mu v_{\mu})\mu v_{\mu}\frac{d}{d\mu}v_{\mu}dx$
$\displaystyle+$ $\displaystyle\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})\frac{d}{d\mu}v_{\mu}dx$

and

$\displaystyle h^{\prime}(\mu)$ $\displaystyle=$ $\displaystyle
2(-r)\mu^{-(r+1)}\int_{\mathbb{R}^{N}}F(\mu
v_{\mu})dx+2\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})[v_{\mu}+\mu\frac{d}{d\mu}v_{\mu}]dx.$ (4.25)

Differentiating (4.2) with respect to $\mu$, we have

$(-\Delta)^{s}\frac{d}{d\mu}v_{\mu}+\frac{d}{d\mu}v_{\mu}=(1-r)\mu^{-r}f(\mu
v_{\mu})+\mu^{1-r}f^{\prime}(\mu v_{\mu})[v_{\mu}+\mu\frac{d}{d\mu}v_{\mu}],$
(4.26)

consequently,

$L_{+,\mu}(\frac{d}{d\mu}v_{\mu})=\mu^{-r}\left[f^{\prime}(\mu v_{\mu})\mu
v_{\mu}+(1-r)f(\mu v_{\mu})\right].$ (4.27)

Note that $L_{+,\mu}v_{\mu}=\mu^{1-r}[f(\mu v_{\mu})-f^{\prime}(\mu
v_{\mu})\mu v_{\mu}],$ then we go further

$\displaystyle k^{\prime}(\mu)-h^{\prime}(\mu)$ $\displaystyle=$
$\displaystyle(1-r)\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx+\mu^{1-r}\int_{\mathbb{R}^{N}}f^{\prime}(\mu
v_{\mu})v_{\mu}^{2}+f^{\prime}(\mu v_{\mu})\mu v_{\mu}\frac{d}{d\mu}v_{\mu}dx$
(4.28) $\displaystyle+\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})\frac{d}{d\mu}v_{\mu}dx$
$\displaystyle-2(-r)\mu^{-(r+1)}\int_{\mathbb{R}^{N}}F(\mu
v_{\mu})dx-2\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})[v_{\mu}+\mu\frac{d}{d\mu}v_{\mu}]dx$ $\displaystyle=$
$\displaystyle(1-r)\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})v_{\mu}dx+\mu^{1-r}\int_{\mathbb{R}^{N}}f^{\prime}(\mu
v_{\mu})v_{\mu}^{2}dx$
$\displaystyle+\mu^{1-r}\int_{\mathbb{R}^{N}}f^{\prime}(\mu v_{\mu})\mu
v_{\mu}\frac{d}{d\mu}v_{\mu}dx-\mu^{1-r}\int_{\mathbb{R}^{N}}f(\mu
v_{\mu})\frac{d}{d\mu}v_{\mu}dx$
$\displaystyle-2(-r)\mu^{-(r+1)}\int_{\mathbb{R}^{N}}F(\mu
v_{\mu})dx-2\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})v_{\mu}dx$
$\displaystyle=$ $\displaystyle\mu^{-(r+1)}\int_{\mathbb{R}^{N}}(1-r)f(\mu
v_{\mu})\mu v_{\mu}+f^{\prime}(\mu v_{\mu})(\mu v_{\mu})^{2}dx$
$\displaystyle-\langle
L_{+,\mu}v_{\mu},\frac{d}{d\mu}v_{\mu}\rangle_{L^{2}}+2\mu^{-(r+1)}\int_{\mathbb{R}^{N}}rF(\mu
v_{\mu})-f(\mu v_{\mu})\mu v_{\mu}dx.$

Clearly,

$\displaystyle\langle L_{+,\mu}v_{\mu},\frac{d}{d\mu}v_{\mu}\rangle_{L^{2}}$
$\displaystyle=$ $\displaystyle\langle
v_{\mu},L_{+,\mu}\frac{d}{d\mu}v_{\mu}\rangle_{L^{2}}$ (4.29) $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{N}}\mu^{-r}v_{\mu}\left[f^{\prime}(\mu
v_{\mu})\mu v_{\mu}+(1-r)f(\mu v_{\mu})\right]dx,$

recalling that (4.27) holds. Going back to (4.28), we have

$\displaystyle B^{\prime}(\mu)$ $\displaystyle=$
$\displaystyle\mu^{-(r+1)}\int_{\mathbb{R}^{N}}(1-r)f(\mu v_{\mu})\mu
v_{\mu}+f^{\prime}(\mu v_{\mu})(\mu v_{\mu})^{2}dx$ (4.30) $\displaystyle-$
$\displaystyle\langle
L_{+,\mu}v_{\mu},\frac{d}{d\mu}v_{\mu}\rangle_{L^{2}}+2r\mu^{-(r+1)}\int_{\mathbb{R}^{N}}F(\mu
v_{\mu})-f(\mu v_{\mu})\mu v_{\mu}dx$ $\displaystyle=$ $\displaystyle
2\mu^{-(r+1)}\int_{\mathbb{R}^{N}}rF(\mu v_{\mu})-f(\mu v_{\mu})\mu
v_{\mu}dx.$

It is easy to see $B(\mu)>0,B^{\prime}(\mu)>0$.

Now, we are ready to come to the main point of the proof of Lemma 4.8. It is
convenient to use the following notation

$M_{\mu}=\int_{\mathbb{R}^{N}}\left|v_{\mu}\right|^{2},\quad
T_{\mu}=\int_{\mathbb{R}^{N}}\left|(-\Delta)^{s/2}v_{\mu}\right|^{2},\quad
V_{\mu}=\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})\mu v_{\mu},$

for $\mu\in[\mu_{0},\infty)$. We first recall that

$M_{\mu}\sim T_{\mu}\sim V_{\mu}.$

Note that

$\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})\mu
v_{\mu}=\int_{\mathbb{R}^{N}}v_{\mu}^{r}dx+\mu^{-r}\int_{\mathbb{R}^{N}}g(\mu
v_{\mu})\mu v_{\mu}dx,$ (4.31)

by $(f_{4})$. For any $p\in(2,r)$, in light of (4.31), we have

$\displaystyle\int_{\mathbb{R}^{N}}|v_{\mu}|^{p}dx$ $\displaystyle\lesssim$
$\displaystyle(\int_{\mathbb{R}^{N}}|v_{\mu}|^{2}dx)^{\frac{\alpha
p}{2}}(\int_{\mathbb{R}^{N}}|v_{\mu}|^{r}dx)^{\frac{(1-\alpha)p}{r}}$ (4.32)
$\displaystyle\lesssim$ $\displaystyle(M_{\mu})^{\frac{\alpha
p}{2}}(V_{\mu})^{\frac{(1-\alpha)p}{r}}$ $\displaystyle\sim$ $\displaystyle
M_{\mu},$

where the last inequality holding by
$\frac{1}{p}=\frac{\alpha}{2}+\frac{1-\alpha}{r}$. Then by $(f_{4})$ again, we
derive that

$\displaystyle B^{\prime}(\mu)$ $\displaystyle=$ $\displaystyle
2\mu^{-(r+1)}\int_{\mathbb{R}^{N}}rF(\mu v_{\mu})-f(\mu v_{\mu})\mu v_{\mu}dx$
(4.33) $\displaystyle=$ $\displaystyle
2\mu^{-(r+1)}\int_{\mathbb{R}^{N}}rG(\mu v_{\mu})-g(\mu v_{\mu})\mu v_{\mu}dx$
$\displaystyle\lesssim$ $\displaystyle\mu^{-(r+1)}\int_{\mathbb{R}^{N}}(\mu
v_{\mu})^{q}+(\mu v_{\mu})^{q_{0}}dx$ $\displaystyle\lesssim$
$\displaystyle\mu^{-(r-q+1)}M_{\mu}+\mu^{-(r-q_{0}+1)}M_{\mu},$

where the last inequality holding by (4.32). Hence

$\displaystyle\frac{B^{\prime}(\mu)}{B(\mu)}$ $\displaystyle\lesssim$
$\displaystyle\frac{\mu^{-(r-q+1)}M_{\mu}+\mu^{-(r-q_{0}+1)}M_{\mu}}{V_{\mu}}$
(4.34) $\displaystyle\lesssim$ $\displaystyle\mu^{-(\theta+1)},$

where $\theta=r-q_{0}>0$. In light of (4.34), we obtain the differential
inequality

$\displaystyle(\ln B(\mu))^{\prime}$ $\displaystyle\lesssim$
$\displaystyle\mu^{-\theta-1},$ (4.35)

which by integration yields

$\displaystyle\ln B(\mu)$ $\displaystyle\lesssim$
$\displaystyle-\frac{1}{\theta}\mu^{-\theta}+C_{0},$ (4.36)

where $C_{0}>0$ is a constant independent of $\mu$. Hence

$\displaystyle B(\mu)\lesssim C_{1}e^{-\frac{1}{\theta}\mu^{-\theta}}\sim 1.$
(4.37)

which yields the uniform upper bound for $\mu\in[\mu_{0},\infty)$, The proof
of Lemma 4.8 is now complete. ∎

######  Lemma 4.9.

Suppose $(f_{1})-(f_{4})$. Then, for all $\mu\in\left[\mu_{0},\infty\right)$,
we have $v_{\mu}(x)\lesssim|x|^{-N}$ for $|x|\geqslant R$ for some $R$
independent of $\mu$.

###### Proof.

First, we claim that $v_{\mu}\in L^{1}$ holds for any fixed $\mu$. Indeed,
recalling that $u_{\lambda}\in L^{1}$ holds by the uniform decay estimate in
Lemma 4.6, we conclude $v_{\mu}\in L^{1}$ by the definition of $v_{\mu}$
directly.

Integrating equation

$(-\Delta)^{s}v_{\mu}+v_{\mu}=\mu^{1-r}f(\mu v_{\mu})$ (4.38)

over $\mathbb{R}^{N}$ and using the fact that
$\int_{\mathbb{R}^{N}}(-\Delta)^{s}v_{\mu}=0$ holds, we obtain the identity

$\int_{\mathbb{R}^{N}}v_{\mu}=\int_{\mathbb{R}^{N}}\mu^{1-r}f(\mu v_{\mu}).$
(4.39)

Notice that

$V_{\mu}=\mu^{-r}\int_{\mathbb{R}^{N}}f(\mu v_{\mu})\mu
v_{\mu}=\int_{\mathbb{R}^{N}}v_{\mu}^{r}dx+\mu^{-r}\int_{\mathbb{R}^{N}}g(\mu
v_{\mu})\mu v_{\mu}dx,$ (4.40)

by $(f_{4})$. Together with Hölder’s inequality, this gives

$\displaystyle\int_{\mathbb{R}^{N}}\mu^{1-r}f(\mu v_{\mu})dx$
$\displaystyle\lesssim$ $\displaystyle\int_{\mathbb{R}^{N}}\mu^{1-r}(\mu
v_{\mu})^{q-1}+\mu^{1-r}(\mu v_{\mu})^{r-1}dx$ (4.41) $\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{R}^{N}}\mu^{q-r}(v_{\mu})^{q-1}+\int_{\mathbb{R}^{N}}(v_{\mu})^{r-1}$
$\displaystyle\lesssim$
$\displaystyle\mu^{q-r}\left(\int_{\mathbb{R}^{N}}v_{\mu}\right)^{\frac{1}{q}}\left(\int_{\mathbb{R}^{N}}v_{\mu}^{q}\right)^{\frac{q-2}{(q-1)^{2}}}+\left(\int_{\mathbb{R}^{N}}v_{\mu}\right)^{\frac{1}{r}}\left(\int_{\mathbb{R}^{N}}v_{\mu}^{r}\right)^{\frac{r-2}{(r-1)^{2}}}$
$\displaystyle\lesssim$
$\displaystyle\mu^{q-r}\left(\int_{\mathbb{R}^{N}}v_{\mu}\right)^{\frac{1}{q}}+\left(\int_{\mathbb{R}^{N}}v_{\mu}\right)^{\frac{1}{r}},$

where the last inequality holds by using the uniform estimate proved in Lemma
4.8. Going back to (4.39), we obtain

$\displaystyle\int_{\mathbb{R}^{N}}v_{\mu}$ $\displaystyle\lesssim$
$\displaystyle\mu^{q-r}\left(\int_{\mathbb{R}^{N}}v_{\mu}\right)^{\frac{1}{q}}+\left(\int_{\mathbb{R}^{N}}v_{\mu}\right)^{\frac{1}{r}}.$
(4.42)

Since $\mu\in[\mu_{0},+\infty)$ and $q<r$, we deduce the uniform bound

$\displaystyle\int_{\mathbb{R}^{N}}v_{\mu}\lesssim 1.$ (4.43)

Since, the function $v_{\mu}(r)>0$ is decreasing in $r$. Hence, for any
$|x|>0$,

$\int_{\mathbb{R}^{N}}v_{\mu}\geqslant\int_{|y|\leqslant|x|}v_{\mu}(|y|)dy\gtrsim|x|^{N}v_{\mu}(|x|).$
(4.44)

In view of (4.43) we conclude that $v_{\mu}(x)\lesssim|x|^{-N}$ holds for
$|x|\geqslant R$ for some $R$ independent of $\mu$, completing the proof of
Lemma 4.9. ∎

######  Lemma 4.10.

Let $\\{\mu_{n}\\}_{n=1}^{\infty}\subset\left[\mu_{0},\infty\right)$ be a
sequence such that $\mu_{n}\rightarrow\infty$ and suppose that
$v_{\mu_{n}}(|x|)>0$ for $n\in\mathbb{N}$. Then, after possibly passing to a
subsequence, we have $v_{\mu_{n}}\rightarrow v_{*}$ in $H^{s}(\mathbb{R}^{N})$
as $n\rightarrow+\infty$. Moreover, the function $v_{*}(|x|)>0$ is positive
and satisfies

$(-\Delta)^{s}v_{*}+v_{*}-|v_{*}|^{r-2}v_{*}=0\quad\text{in }\mathbb{R}^{N}.$
(4.45)

###### Proof.

Let $v_{n}=v_{\mu_{n}}$ in the following. Recall the uniform bound
$\left\|v_{n}\right\|_{H^{s}}\lesssim 1$ by the priori bounds in Lemma 4.8.
Passing to a subsequence if necessary, we have $v_{n}\rightharpoonup v_{*}$
weakly in $H^{s}$. Furthermore, by local Rellich compactness, we can assume
that $v_{n}\rightarrow v_{*}$ in $L_{\text{loc }}^{p}$, $2\leqslant
p<2_{s}^{*}$ and pointwise a.e. in $\mathbb{R}^{N}$. Thanks to the uniform
decay bound in Lemma 4.9, we can upgrade this to $v_{n}\rightarrow v_{*}$
strongly in $L^{p}$, $2\leqslant p<2_{s}^{*}$. Bootstrap arguments imply that
$v_{n}\rightarrow v_{*}$ in $H^{s}$. Obviously, $v_{*}\geqslant 0$ satisfies
equation (4.45). Thus 0 is the first eigenvalue of
$L_{-,\infty}:=(-\Delta)^{s}+1-|v_{*}|^{r-2}$, showing $v_{*}>0$. ∎

###### Proof of Proposition 4.2.

Let $v_{1},v_{2}\in C^{1}([\mu_{0},\infty),H_{\text{rad}}^{s})$ be two maps
such that for $i=1,2$, $v_{i}(\mu)>0$ solves (4.2). For convenience, we let
$\kappa=1/\mu$ and $\kappa_{0}=1/\mu_{0}$. Then for $i=1,2$, we consider maps
$v_{i}(\kappa)\in C^{1}((0,\kappa_{0}],H_{\text{rad}}^{s})$, and aim to show
that $v_{1}(\kappa)=v_{2}(\kappa)$ for all $\kappa\in(0,\kappa_{0}]$. By Lemma
4.10, as $\kappa\to 0$, both $v_{1}(\kappa)$ and $v_{2}(\kappa)$ converge to
$v_{*}$ which is positive and solves (4.45). By [17], $v_{*}$ is unique (up to
translations) and non-degenerate. Then, by implicit function arguments like
the proof of Lemma 4.3, we obtain that $v_{1}(\kappa)=v_{2}(\kappa)$ for small
positive $\kappa$, and thus for all $\kappa\in(0,\kappa_{0}]$. The proof is
complete. ∎

###  4.3 Proof of Theorem 1.1

###### Proof.

Suppose on the contrary that (1.3) has two solutions $u_{1,\lambda_{0}}$ and
$u_{2,\lambda_{0}}$ for some $\lambda=\lambda_{0}>0$. Then by Proposition 4.1,
there exist two maps $u_{1},u_{2}\in
C^{1}([\lambda_{0},\infty),H_{\text{rad}}^{s})$ such that for $i=1,2$,
$u_{i}(\lambda)$ is positive and solves (1.3) with
$u_{i}(\lambda_{0})=u_{i,\lambda_{0}}$. Then, since $T(\lambda)$ is one-to-
one, Proposition 4.2 shows us that $u_{1}(\lambda)\equiv u_{2}(\lambda)$.
Hence, $u_{1}\equiv u_{2}$, and we complete the proof. ∎

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