Positive solutions of 1-D half Laplacian equation with singular and exponential nonlinearity

Abstract

In this article, we study the existence, multiplicity, regularity and asymptotic behavior of the positive solutions to the problem of half-Laplacian with singular and exponential growth nonlinearity in one dimension (see below $(P_{λ})$ ). We prove two results regarding the existence and multiplicity of solutions to the problem $(P_{λ})$ . In the first result, existence and multiplicity have been proved for classical solutions via bifurcation theory while in the latter result multiplicity has been proved for critical exponential nonlinearity by variational methods. An independent question of symmetry and monotonicity properties of classical solution has been answered in the paper. To characterize the behavior of large solutions, we further study isolated singularities for the singular semi linear elliptic equation in $Ω \subset R^{n}$ , $n ⩾ 2 s$ involving exponential growth nonlinearities in the more general framework of ${(- Δ)}^{s}$ operator and for all $0 < s < 1$ (see below $(P_{s})$ ).

Keywords

Singular problem exponential nonlinearity half-Laplacian asymptotic behavior moving plane method

1. Introduction and main results

The purpose of the article is to study the existence, multiplicity and regularity of solutions to the following problem: $\begin{array}{l} (P_{λ}) \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = λ (\frac{1}{u^{δ}} + f (u)), u > 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) \end{matrix} \end{array}$ where $f (u) = h (u) e^{u^{α}}$ , $1 ⩽ α ⩽ 2$ , $δ > 0$ , $λ ⩾ 0$ and $h (t)$ is assumed to be a smooth perturbation of $e^{t^{α}}$ as $t \to \infty$ . Here, the fractional Laplacian ${(- Δ)}^{s}$ is defined as $\begin{matrix} {(- Δ)}^{s} u (x) = 2 P. V. C_{s} \int_{R^{n}} \frac{u (x) - u (y)}{| x - y |^{n + 2 s}} d y, \end{matrix}$ where P.V. denotes the Cauchy principal value, $s \in (0, 1)$ , $n ⩾ 2 s$ and $C_{s} = π^{- \frac{n}{2}} 2^{2 s - 1} s \frac{Γ (\frac{n + 2 s}{2})}{Γ (1 - s)}$ , Γ being the Gamma function. When $s = \frac{1}{2}$ , $C_{s} = \frac{1}{2 π}$ . The fractional Laplacian operator is the infinitesimal generator of Lévy stable diffusion process and has been a classic topic in Fourier analysis and nonlinear partial differential equations for a long time due to its vast applications in real-world problems such as finance, optimization, thin obstacle problem, quasi-geostrophic flow. The study and understanding of existence, multiplicity, and regularity of solutions to elliptic equations have been a matter of intensive research. Recently, equations involving fractional powers of Laplacian attracted a large number of researchers, especially in the study of existence and multiplicity of solutions (see [16]). This interest has been provoked and sustained by the applications of such results in mathematical physics and geometry. There is ample amount of papers on the existence and regularity of solutions involving ${(- Δ)}^{s}$ operator with singular nonlinearity. For instance, Adimurthi, Giacomoni, and Santra [2] studied the existence and regularity of weak solutions to the following purely singular problem $\begin{array}{l} {(- Δ)}^{s} u = u^{- δ}, u > 0 in Ω, u = 0 in R^{n} ∖ Ω, \end{array}$ where Ω is smooth bounded domain in $R^{n}$ , $n > 2 s$ , $s \in (0, 1)$ and $δ > 0$ such that $δ (2 s - 1) < 1 + 2 s$ . Moreover, here authors studied the existence of a global branch of solutions of the following problem: $\begin{array}{l} (1.1) & {(- Δ)}^{s} u = λ (u^{- δ} + f (u)), u > 0 in Ω, u = 0 in R^{n} ∖ Ω, \end{array}$ where $λ > 0$ and the function f is of subcritical growth. Recently, Giacomoni, Mukherjee and Sreenadh [13] proved the global multiplicity result for the nonlocal singular problem (1.1) with Sobolev critical nonlinearity $f (u) = u^{\frac{N + 2 s}{N - 2 s}}$ . For more results on nonlocal problems with singular tnonlinearity, interested readers can refer to [10,12,17] and references therein.

From the above results, a critical question originates from the case of critical dimension $n = 1$ when $s = \frac{1}{2}$ . That is what happens to the existence, multiplicity, asymptotic behavior and regularity of the solutions to problem $(P_{λ})$ in case $s = \frac{1}{2}$ . The critical profile for existence of solutions has been established in the works [10,11] with regular nonlinearities. In this paper, we answer the questions of existence, bifurcation, multiplicity and regularity of solutions with singular nonlinear terms as in $(P_{λ})$ . We remark that in contrast to higher dimensions, there is no restriction on δ is required in dimension one. Before stating the results, let us recall some definitions of function spaces from the work of [16] and define the notion of (very) weak solutions. Define $\begin{array}{l} X : = {u : R \to R | measurable, u |_{(- 1, 1)} \in L^{2} ((- 1, 1)) and \frac{(u (x) - u (y))}{| x - y |} \in L^{2} (Q)} \end{array}$ where $Q = R^{2} ∖ {(- 1, 1)}^{c} \times {(- 1, 1)}^{c}$ and ${(- 1, 1)}^{c} = R ∖ (- 1, 1)$ endowed with the norm $\begin{array}{l} ‖ u ‖_{X} = ‖ u ‖_{L^{2} ((- 1, 1))} + C_{s} {(\int_{Q} \frac{| u (x) - u (y) |^{2}}{| x - y |^{2}} d x d y)}^{\frac{1}{2}} . \end{array}$ Define the Hilbert space $X_{0}$ as $\begin{array}{l} X_{0} : = {u \in X : u = 0 in R ∖ (- 1, 1)} \end{array}$ equipped with the inner product $\begin{array}{l} ⟨ u, v ⟩ = C_{s} \int_{Q} \frac{(u (x) - u (y)) (v (x) - v (y))}{| x - y |^{2}} d x d y . \end{array}$ As in [3] we have the following definition of weak solutions to problem $(P_{λ})$ .

Definition 1.1.
A function $u \in L^{1} (R)$ with $u \equiv 0$ on $R ∖ (- 1, 1)$ is said to be a weak solution of $(P_{λ})$ if ${inf}_{K} u > 0$ for any compact set $K \subset (- 1, 1)$ and for any $ϕ \in σ$ , $\begin{array}{l} (1.2) & \int_{- 1}^{1} u {(- Δ)}^{\frac{1}{2}} ϕ = C_{s} \int_{Q} \frac{(u (x) - u (y)) (ϕ (x) - ϕ (y))}{| x - y |^{2}} d x d y = λ \int_{- 1}^{1} (\frac{1}{u^{δ}} + h (u) e^{u^{α}}) ϕ d x, \end{array}$ where $\begin{matrix} σ = & {ψ ∣ ψ : R \to R, measurable, {(- Δ)}^{\frac{1}{2}} ψ \in L^{\infty} ((- 1, 1)) and \\ ϕ has compact support in (- 1, 1)} . \end{matrix}$ Using the regularity theory of fractional Laplacian we define the set of classical solutions of $(P_{λ})$ : Definition 1.2.
The set of classical solutions to $(P_{λ})$ is defined as $\begin{array}{l} S = {(λ, u) \in R^{+} \times C_{0} ([- 1, 1]) : u is a weak solution to (P_{λ}) in X_{0}} . \end{array}$
Remark 1.3.
Regularity of a classical solution u (proved later in Theorem 1.9) for the problem $(P_{λ})$ implies $u \in C_{ϕ_{δ}}^{+} ((- 1, 1))$ (defined below). Then by using Hardy’s inequality (see [14, Corollary 1.4.4.10, p. 23]) in (1.2) together with the fact that $C_{c}^{\infty} ((- 1, 1))$ is dense in $X_{0}$ , we obtain that $\frac{1}{u^{δ}}$ belongs to dual space of $X_{0}$ for all $δ > 0$ and hence (1.2) holds for all $ϕ \in X_{0}$ .

Definition 1.4.
For $ϕ \in C_{0} ([- 1, 1])$ with $ϕ > 0$ in $(- 1, 1)$ , the set $C_{ϕ} ((- 1, 1))$ is defined as $\begin{array}{l} C_{ϕ} ((- 1, 1)) = & {u \in C_{0} ([- 1, 1]) : there exists c ⩾ 0 such that | u (x) | ⩽ c ϕ (x), \\ for all x \in (- 1, 1)}, \end{array}$ endowed with the natural norm $‖ \frac{u}{ϕ} ‖_{L^{\infty} ((- 1, 1))}$ .
Definition 1.5.
The positive cone of $C_{ϕ} ((- 1, 1))$ is the open convex subset of $C_{ϕ} ((- 1, 1))$ defined as $\begin{array}{l} C_{ϕ}^{+} ((- 1, 1)) = {u \in C_{ϕ} ((- 1, 1)) : inf_{x \in (- 1, 1)} \frac{u (x)}{ϕ (x)} > 0} . \end{array}$

To analyze the existence and regularity of solutions of $(P_{λ})$ , the key ingredient is to study the boundary behavior of the weak solution of the following problem: $\begin{matrix} (1.3) & (P) \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = \frac{1}{d {(x)}^{α} {log}^{β} (\frac{A}{d (x)})} & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{matrix}$ For the operator ${(- Δ)}^{s}$ with $n > 2 s$ , Abatangelo [1] studied the boundary behavior of the corresponding problem like (1.3) with $β = 0$ and $0 < α < 1 + s$ . The case $n = 1$ and $s = \frac{1}{2}$ has been left open as the Green function for the half-Laplacian in one dimension is different (see [7]) from that of ${(- Δ)}^{s}$ with $n > 2 s$ . In this article we explore this case and prove the following theorem. We point out that this result is new and of independent interest.
Theorem 1.6.
Let A be a positive constant such that $A ⩾ 2$ . Then the weak solution of (1.3) satisfies $\begin{matrix} (1.4) & \begin{array}{l} c_{1} d {(x)}^{\frac{1}{2}} ⩽ u (x) ⩽ c_{2} d {(x)}^{\frac{1}{2}} for 0 < α < \frac{1}{2} and β = 0, \\ c_{3} d {(x)}^{1 - α} ⩽ u (x) ⩽ c_{4} d {(x)}^{1 - α} for \frac{1}{2} < α < \frac{3}{2} and β = 0, \\ c_{5} d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{A}{d (x)}) ⩽ u (x) ⩽ c_{6} d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{A}{d (x)}) for α = \frac{1}{2} and 0 ⩽ β < 1 \end{array} \end{matrix}$ where $c_{i}$ , $i = 1, 2, \dots, 5$ are constants.

To get appropriate sub and supersolutions of the problem $(P_{λ})$ , we now turn our attention to the following pure singular problem $(P_{δ})$ . $\begin{array}{l} (P_{δ}) \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = \frac{1}{u^{δ}}, u > 0, & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{array}$ The barrier function $ϕ_{δ}$ is defined as follows: $\begin{matrix} (1.5) & \begin{matrix} ϕ_{δ} = \{\begin{matrix} ϕ_{1} & if 0 < δ < 1, \\ ϕ_{1} {(log (\frac{2}{ϕ_{1}}))}^{\frac{1}{2}} & if δ = 1, \\ ϕ_{1}^{\frac{2}{δ + 1}} & if δ > 1, \end{matrix} \end{matrix} \end{matrix}$ where $ϕ_{1}$ is the normalized ( $‖ ϕ_{1} ‖_{L^{\infty} (Ω)} = 1$ ) eigen function corresponding to the smallest eigen value of ${(- Δ)}^{\frac{1}{2}}$ on $X_{0}$ . We recall that $ϕ_{1} \in C^{\frac{1}{2}} (R)$ and $ϕ_{1} \in C_{d^{\frac{1}{2}}}^{+} ((- 1, 1))$ (see Proposition 1.1 and Theorem 1.2 of [19]). For the problem $(P_{δ})$ , we are concerned about the existence, the asymptotic behavior and the regularity of the solution. In this regard we have the following result:
Theorem 1.7.

For all $δ > 0$ , there exists a unique $u \in C_{0} ([- 1, 1])$ classical solution of $(P_{δ})$ . Moreover, $u \in X_{0} \cap C_{ϕ_{δ}}^{+} ((- 1, 1))$ where $ϕ_{δ}$ is defined in (1.5).

The classical solution u to $(P_{δ})$ belongs to $C^{γ} (R)$ with $\begin{matrix} (1.6) & γ = \{\begin{matrix} \frac{1}{2} & if δ < 1, \\ \frac{1}{2} - ϵ & if δ = 1, for all ϵ > 0 small enough, \\ \frac{1}{δ + 1} & if δ > 1 . \end{matrix} \end{matrix}$

Now we will state some assumptions on the function h:
$h : [0, \infty) \to R$ is a positive function of class $C^{2}$ in $(0, \infty)$ with $h (0) = 0$ and such that the map $t \to t^{- δ} + h (t) e^{t^{α}}$ is convex for all $t > 0$ .

For any $ϵ > 0$ , ${lim}_{t \to \infty} h (t) e^{- ϵ t^{α}} = 0$ and ${lim}_{t \to \infty} h (t) e^{ϵ t^{α}} = \infty$ .
First, we recall the definition of an asymptotic bifurcation point and then state the result regarding existence of a global branch of classical solutions to $(P_{λ})$ for $1 < α ⩽ 2$ .
Definition 1.8.
A point $Λ_{a} \in [0, \infty)$ is said to be an asymptotic bifurcation point, if there exists a sequence $(λ_{n}, u_{n}) \in S$ such that $λ_{n} \to Λ_{a}$ and $‖ u_{n} ‖_{L^{\infty} ((- 1, 1))} \to \infty$ as $n \to \infty$ .

To study the existence, multiplicity of solutions to $(P_{λ})$ , we seek assistance of global bifurcation theory due to P. H. Rabinowitz [18] and proved the following result.
Theorem 1.9.
Let h satisfy the hypothesis (H1) and (H2) and $δ > 0$ . Then the following holds:
There exists $Λ \in (0, + \infty)$ and $γ > 0$ such that $S \subset [0, Λ] \times (X_{0} \cap C_{ϕ_{δ}}^{+} ((- 1, 1)) \cap C^{γ} (R))$ , where γ is defined in Theorem 1.7 and $ϕ_{δ}$ is defined in (1.5).

There exists a connected unbounded branch $C$ of solutions to $(P_{λ})$ in $R^{+} \times C_{0} ([- 1, 1])$ , emanating from $(0, 0)$ such that for any $λ \in (0, Λ)$ , there exists $(λ, u_{λ}) \in C$ with $u_{λ}$ being minimal solution to $(P_{λ})$ . Furthermore, as $λ \to Λ^{-}$ , $u_{λ} \to u_{Λ}$ in $X_{0}$ , where $u_{Λ}$ is a classical solution to $(P_{Λ})$ .

The curve $(0, Λ) ∋ λ \to u_{λ} \in C_{0} ([- 1, 1])$ is of class $C^{2}$ .

(Bending and local multiplicity near λ) $λ = Λ$ is a bifurcation point, that is, there exists a unique $C^{2}$ -curve $(λ (s), u (s)) \in C$ , where the parameter s varies in an open interval about the origin in $R$ , such that $\begin{array}{l} λ (0) = Λ, u (0) = u_{Λ}, λ^{'} (0) = 0, λ^{″} (0) < 0 . \end{array}$

(Asymptotic bifurcation point) $C$ admits an asymptotic bifurcation point $Λ_{a}$ satisfying $0 ⩽ Λ_{a} ⩽ Λ$ .

Now we study the qualitative properties of solutions for the problem $(P_{λ})$ . In light of the maximum principle (see [15]) and the moving plane method, we derive the radial symmetry and monotonicity properties of the weak solutions with respect to $| x |$ . More precisely, we prove the following result:
Theorem 1.10.
For $1 ⩽ α ⩽ 2$ , $δ > 0$ , let h satisfies (H1)–(H2), f is Lipschitz function. Then every positive solution $(λ, u) \in S$ of $(P_{λ})$ is symmetric and strictly decreasing in $| x |$ i.e. $u (x) > u (y)$ for all $| x | < | y |$ and $x, y \in (- 1, 1)$ .

To prove Theorem 1.10, we have used tools of maximum principle proved in [15]. In particular we have used Lemma 3.2, Proposition 3.5 and a combination of both gives rise to a small volume maximum principle Proposition 3.6 of [15].

Assertion $(v)$ in Theorem 1.9 conclude that the connected branch admits at least one asymptotic bifurcation point. To characterize the blow up behavior at $λ = Λ_{a}$ we study the behavior of isolated singularities as in Brezis–Lions problem (see [4]) for the fractional Laplacian operator.

We consider the following problem: $\begin{matrix} (P_{s}) \{\begin{matrix} {(- Δ)}^{s} u = g (u), u ⩾ 0 & in Ω^{'}, \\ u = 0 & in R^{n} ∖ Ω, \\ u \in L^{1} (Ω), & g (u) \in L_{loc}^{t} (Ω^{'}), \end{matrix} \end{matrix}$ where $0 < s < 1$ , $t > \frac{n}{2 s} ⩾ 1$ , $Ω \subset R^{n}$ be a bounded domain with $0 \in Ω$ and $Ω^{'} = Ω ∖ {0}$ . The notion of distributional solution for $(P_{s})$ is defined as follows:
Definition 1.11.
A function u is said to be a distributional solution of $(P_{s})$ if $u \in L^{1} (Ω)$ such that $g (u) \in L_{loc}^{1} (Ω^{'})$ and $\begin{matrix} \int_{Ω} u (x) {(- Δ)}^{s} ϕ d x = \int_{Ω} g (u) ϕ d x \end{matrix}$ for all $ϕ \in C_{c}^{\infty} (Ω)$ with $supp (ϕ) \subset Ω^{'}$ .

In [5], authors have studied the problem $(P_{s})$ by assuming the existence of classical solution u of $(P_{s})$ with polynomial type nonlinearity. In the next theorem, we extend the result of Chen and Quaas [5] for the problem $(P_{s})$ satisfying weaker assumption of distributional solution and for a larger class of nonlinearities (in particular exponential growth nonlinearity).
Theorem 1.12.
For $0 < s < 1$ , let u be non-negative distributional solution of $(P_{s})$ such that $u \in L^{1} (Ω)$ , $g (u) \in L_{loc}^{t} (Ω^{'})$ for $t > \frac{n}{2 s} ⩾ 1$ . Then there exists $k ⩾ 0$ such that u is distributional solution of $\begin{matrix} \{\begin{matrix} {(- Δ)}^{s} u = g (u) + k δ_{0}, u ⩾ 0, & in Ω, \\ u = 0 & in R^{n} ∖ Ω, \\ g (u) \in L^{1} (Ω), \end{matrix} \end{matrix}$ i.e. $\begin{matrix} \int_{Ω} u {(- Δ)}^{s} ϕ - g (u) ϕ d x = k ϕ (0) for all ϕ \in C_{c}^{\infty} (Ω) . \end{matrix}$

As an application of Theorem 1.12, we characterize the asymptotic behavior of large solutions and prove the following result:
Theorem 1.13.
For $1 < α ⩽ 2$ , $δ > 0$ , assume $Λ_{a} > 0$ be an asymptotic bifurcation point as in Definition 1.8. Then, for any sequence $(λ_{k}, u_{k}) \in S \cap ({λ_{k} ⩾ Λ_{a}} \times C_{0} ([- 1, 1]))$ such that $λ_{k} \to Λ_{a}$ and $‖ u_{k} ‖_{L^{\infty} ((- 1, 1))} \to \infty$ , the following assertions holds:
$0 \in Ω$ is the only blow up point for a sequence $u_{k}$ .

$u_{k} \to u$ in $C_{loc}^{s} ((- 1, 1) ∖ {0})$ where u is a weak (singular) solution to $(P_{λ})$ . Moreover, $u (0) = \infty$ , $u \in L^{p} ((- 1, 1))$ for any $1 ⩽ p < \infty$ , $u \notin X_{0}$ and $\frac{1}{u^{δ}} + f (u) \in L^{1} ((- 1, 1))$ .

We have the following remark about the above theorem.
Remark 1.14.

Under the conditions of Theorem 1.13 in assertion (i), we expect concentration phenomena to hold for large solutions as $λ \to 0$ .

Let G be the primitive of g defined as $g (t) = \frac{1}{t^{δ}} + f (t)$ and assume that $G (t) = O (g (t))$ as $t \to \infty$ . If the sequence of large solutions, say $u_{k}$ , have bounded energy i.e. $\begin{matrix} J (u_{k}) = \frac{1}{2} ‖ u_{k} ‖_{X_{0}}^{2} - \int_{- 1}^{1} G (u_{k}) d x < C \end{matrix}$ where C is independent of k, then assertion (ii) cannot hold (see proof in the Appendix).

From the above remark we expect the existence and multiplicity of bounded energy solutions to $(P_{λ})$ for $α = 2$ and λ small. We notice that the growth $α = 2$ is critical in the sense of Trudinger–Moser inequality (see Theorem 4.1). Precisely, the embedding $u ∋ X_{0} \mapsto e^{| u |^{α}} \in L^{1} ((- 1, 1))$ is compact for all $α \in (1, 2)$ and is continuous for $α = 2$ . Via Variational techniques, peculiarly, we proved the global multiplicity result to the problem $(P_{λ})$ for all $δ > 0$ , under the following assumptions on the function f.

$h \in C^{1} (R^{+})$ , $h (0) = 0$ , $h (t) > 0$ for $t > 0$ and $f (t) = h (t) e^{t^{2}}$ is nondecreasing in t.

For any $ϵ > 0$ , ${lim}_{t \to \infty} (h (t) + h^{'} (t)) e^{- ϵ t^{2}} = 0$ and ${lim}_{t \to \infty} h (t) t e^{ϵ t^{q}} = \infty$ for some $0 ⩽ q < 1$ .

There exists $M_{1}, M_{2}, K > 0$ such that $F (t) = \int_{0}^{t} h (s) e^{s^{2}} d s < M_{1} (f (t) + 1)$ and $f^{'} (t) ⩾ K f (t) - M_{2}$ for all $t > 0$ .
Example 1.
An example of the function h satisfying the above conditions is $h (x) = x^{k} e^{x^{γ}}$ , $k > 0$ , $0 ⩽ γ < 2$ .

We prove the following multiplicity theorem.
Theorem 1.15.

If f satisfies the assumption (K1)–(K5). There exists a $Λ > 0$ such that

For every $λ \in (0, Λ)$ the problem $(P_{λ})$ admits two solutions in $X_{0} \cap C_{ϕ_{δ}}^{+} ((- 1, 1))$ .

For $λ = Λ$ there exists a solution in $X_{0} \cap C_{ϕ_{δ}}^{+} ((- 1, 1))$ .

For $λ > Λ$ , there exists no solution.

Let $u \in X_{0}$ be any positive solution to $(P_{λ})$ where $λ \in (0, Λ]$ , $δ > 0$ . Then $u \in C^{γ} (R)$ where γ is defined (1.6).

To prove Theorem 1.15, we followed the approach of [13]. To obtain the first solution, we use the standard Perron’s method on the functional $J_{λ}$ (see (4.1)). To get a second solution, we use the assumption (K2) to guarantee that the energy level of the Palais Smale sequence is below the first critical level. For that we seek help of Moser functions (see [22]) and then by using mountain-pass Lemma we prove the existence of a second solution. Notice that Theorem 1.15 shows the existence of solution in the energy space $X_{0}$ . We remark that the Hölder regularity proved in Theorem 1.15 is the optimal due to the behavior of the solution near the points $- 1$ and 1. We also remark that Theorem 1.15 holds for the subcritical nonlinearities like $f (t) = h (t) e^{t^{α}}$ with $1 ⩽ α < 2$ as well. We left the calculations for the readers. To the best of our knowledge, there is no work which deals the singular problems for the half-Laplacian operator in one dimension. In this regard, the results proved in the present paper are completely new.

Turning to the layout of the paper: In Section 2, we give proofs of Theorem 1.6, Theorem 1.7 and Theorem 1.9. In Section 3, we provide the details to the proofs of Theorem 1.10, Theorem 1.12 and Theorem 1.13. In Section 4, we give the proof of Theorem 1.15.
2. Global bifurcation result

In this section we first study the boundary behavior of the weak solution of (1.3). We further studied the pure singular problem $(P_{δ})$ and prove Theorem 1.7 which deals with the existence and regularity of solutions of $(P_{δ})$ . In a same flow, we establish a global branch of classical solutions to $(P_{λ})$ .

Proposition 2.1 ([7]).

The Green function $G (x, y)$ associated to ${(- Δ)}^{\frac{1}{2}}$ is the following: $\begin{array}{l} G (x, y) ≍ log (1 + \frac{d {(x)}^{\frac{1}{2}} d {(y)}^{\frac{1}{2}}}{| x - y |}) for all (x, y) \in (- 1, 1) \times (- 1, 1) . \end{array}$

Proof of Theorem 1.6.
Using the fact that $\frac{1}{d {(y)}^{α} {log}^{β} (A / d (y))} \in L^{1} (d x, d^{\frac{1}{2}})$ , we have the following integral representation formula for the solution u to (1.3) $\begin{array}{l} u (x) = \int_{- 1}^{1} \frac{G_{B} (x, y)}{d {(y)}^{α} {log}^{β} (A / d (y))} d y . \end{array}$ Therefore, from Proposition 2.1, up to multiplicative constants, $\begin{array}{l} (2.1) & \frac{u (x)}{d {(x)}^{\frac{1}{2}}} ⩽ \int_{- 1}^{1} log (1 + \frac{\sqrt{d (x) d (y)}}{| x - y |}) \frac{d y}{d {(x)}^{\frac{1}{2}} d {(y)}^{α} {log}^{β} (A / d (y))} . \end{array}$ Without loss of generality, we can assume $x \in [0, 1]$ . Set $ε = d (x)$ , $r = d (y)$ and $x = (1 - ε)$ . Observe that the integral in (2.1) is symmetric around 0. Thus it is enough to consider the case $y \in [0, 1]$ , from above transformations, we have $y = 1 - r$ . To prove (1.4) we divide the proof in several steps.

Step 1: When $α < \frac{1}{2}$ and $β = 0$ . We rewrite $\begin{matrix} (2.2) & \begin{matrix} \frac{u (x)}{d {(x)}^{\frac{1}{2}}} & ⩽ \int_{0}^{1} log (1 + \frac{\sqrt{ε r}}{| r - ε |}) \frac{d r}{ε^{\frac{1}{2}} r^{α}} = \int_{0}^{\frac{1}{ε}} \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t \\ = (\int_{0}^{\frac{1}{2}} + \int_{\frac{1}{2}}^{\frac{3}{2}} + \int_{\frac{3}{2}}^{\frac{1}{ε}}) \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t . \end{matrix} \end{matrix}$ For the first integral we have $\begin{array}{l} \int_{0}^{\frac{1}{2}} \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ C \int_{0}^{\frac{1}{2}} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ C, \end{array}$ for some positive constant $c_{2}$ . For the second integral we have $\begin{array}{l} \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ C \int_{\frac{1}{2}}^{\frac{3}{2}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t < 2 C . \end{array}$ For the third integral we have $\begin{array}{l} \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ C \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{ε^{\frac{1}{2} - α}}{t^{α + \frac{1}{2}}} d t ⩽ C . \end{array}$ It implies that there exists a positive constant $c_{2}$ (large enough) such that $u (x) ⩽ c_{2} d {(x)}^{\frac{1}{2}}$ . This affirms an upper bound of the solutions. For lower bound of the solutions, notice that the integrals in (2.2) works as a lower bound of $\frac{u (x)}{d {(x)}^{\frac{1}{2}}}$ (up to constants). Now we divide the proof in two cases:
Case 1: If $ε ⩾ \frac{1}{3}$ then $\begin{array}{l} \frac{u (x)}{d {(x)}^{\frac{1}{2}}} & ⩾ \int_{0}^{\frac{1}{3}} \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩾ {(\frac{1}{3})}^{\frac{1}{2} - α} C \int_{0}^{\frac{1}{3}} t^{\frac{1}{2} - α} d t = \frac{C}{\frac{3}{2} - α} {(\frac{1}{3})}^{2 - 2 α} > \frac{2 C}{27} . \end{array}$

Case 2: If $ε < \frac{1}{3}$ then $\begin{array}{l} \frac{u (x)}{d {(x)}^{\frac{1}{2}}} & ⩾ \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{ε^{\frac{1}{2} - α}}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩾ C \int_{\frac{1}{2 ε}}^{\frac{1}{ε}} \frac{ε^{\frac{1}{2}}}{\sqrt{t}} d t = 2 (1 - 1 / \sqrt{2}) . \end{array}$
It implies that there exists a positive constants $c_{1}$ (small enough) such that $c_{1} d {(x)}^{\frac{1}{2}} ⩽ u (x)$ .

Step 2: When $α > \frac{1}{2}$ and $β = 0$ . We rewrite $\begin{matrix} (2.3) & \frac{u (x)}{d {(x)}^{1 - α}} ⩽ \int_{0}^{\frac{1}{ε}} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t = (\int_{0}^{\frac{1}{2}} + \int_{\frac{1}{2}}^{\frac{3}{2}} + \int_{\frac{3}{2}}^{\frac{1}{ε}}) \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t . \end{matrix}$ For the first integral, $\begin{array}{l} \int_{0}^{\frac{1}{2}} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ \int_{0}^{\frac{1}{2}} t^{\frac{1}{2} - α} d t < C . \end{array}$ By using the same estimation as in Step 1, $\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ C$ . For the third integral $\begin{array}{l} \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ \int_{\frac{3}{2}}^{\infty} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩽ C \int_{\frac{3}{2}}^{\infty} \frac{1}{t^{\frac{1}{2} + α}} d t < C . \end{array}$ Observe that from the estimation of first and third integral is valid only when $\frac{1}{2} < α < \frac{3}{2}$ . For the lower bound, notice that integrals in (2.3) serve as lower bound as well. Hence $\begin{array}{l} \frac{u (x)}{d {(x)}^{1 - α}} ⩾ \int_{0}^{\frac{1}{2}} \frac{1}{t^{α}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t ⩾ \int_{0}^{\frac{1}{2}} t^{\frac{1}{2} - α} d t > C . \end{array}$ Thus we can choose appropriate positive constant $c_{3}$ and $c_{4}$ such that $\begin{array}{l} c_{3} d {(x)}^{1 - α} ⩽ u (x) ⩽ c_{4} d {(x)}^{1 - α} . \end{array}$ Step 3: When $α = \frac{1}{2}$ and $0 ⩽ β < 1$ . Clearly we can take $A = 2$ . Then $\begin{array}{l} \frac{u (x)}{d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{d (x)})} ⩽ (\int_{0}^{\frac{1}{2}} + \int_{\frac{1}{2}}^{\frac{3}{2}} + \int_{\frac{3}{2}}^{\frac{1}{ε}}) \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{t^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{ε}) {log}^{β} (\frac{2}{ε t})} d t . \end{array}$ The first integral $\begin{array}{l} \int_{0}^{\frac{1}{2}} \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{t^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{ε}) {log}^{β} (\frac{2}{ε t})} d t & ⩽ \frac{C}{log 2} \int_{0}^{\frac{1}{2}} \frac{d t}{{(log (\frac{1}{ε}) + log (\frac{2}{t}))}^{β}} \\ ⩽ C \int_{0}^{\frac{1}{2}} \frac{d t}{log {(\frac{2}{t})}^{β}} ⩽ C \int_{0}^{\frac{1}{2}} \frac{d t}{log 4^{β}} ⩽ \frac{C}{log 4} . \end{array}$ For the second integral $\begin{array}{l} \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{t^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{ε}) {log}^{β} (\frac{2}{ε t})} d t & C \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{{log}^{β} (\frac{2}{t})} d t ⩽ C \int_{\frac{1}{2}}^{\frac{3}{2}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t < 2 C . \end{array}$ For the third integral, $\begin{array}{l} \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{t^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{ε}) {log}^{β} (\frac{2}{ε t})} d t & ⩽ \frac{C}{{log}^{1 - β} (\frac{2}{ε})} \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{d t}{t {(- log (\frac{ε t}{2}))}^{β}} \\ ⩽ \frac{C}{{log}^{1 - β} (\frac{2}{ε})} {(log \frac{4}{3 ε})}^{1 - β} < C . \end{array}$ For the lower bound, we again divide it in two cases:
Case 1: If $ε > \frac{1}{3}$ then $\begin{array}{l} \frac{u (x)}{d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{d (x)})} ⩾ \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{t^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{ε}) {log}^{β} (\frac{2}{ε t})} d t ⩾ C \int_{\frac{1}{2}}^{\frac{3}{2}} log (1 + \frac{\sqrt{t}}{| t - 1 |}) d t > C . \end{array}$

Case 2: If $ε < \frac{1}{3}$ then $\begin{array}{l} \frac{u (x)}{d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{d (x)})} & ⩾ \int_{\frac{3}{2}}^{\frac{1}{ε}} \frac{log (1 + \frac{\sqrt{t}}{| t - 1 |})}{t^{\frac{1}{2}} {log}^{1 - β} (\frac{2}{ε}) {log}^{β} (\frac{2}{ε t})} d t ⩾ \frac{C}{{log}^{1 - β} (\frac{2}{ε})} \int_{log \frac{3 ε}{4}}^{log \frac{1}{2}} \frac{d z}{{(- z)}^{β}} > C > 0 . \end{array}$
It implies there exists suitable positive constant $c_{5}$ and $c_{6}$ such that $\begin{array}{l} c_{5} d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{A}{d (x)}) ⩽ u (x) ⩽ c_{6} d {(x)}^{\frac{1}{2}} {log}^{1 - β} (\frac{A}{d (x)}) . \end{array}$ □

We now study the pure singular problem.
Proof of Theorem 1.7.
(i) The proof goes along the lines of [2, Theorem 1.2] for $s = \frac{1}{2}$ . For the sake of completeness, we will give a short brief of the proof. Let us first consider the case $δ < 1$ . We introduce the following approximated problem: $\begin{array}{l} (P_{δ}^{ϵ}) \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = \frac{1}{{(u + ϵ)}^{δ}}, u > 0, & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{array}$ Following the same arguments and assertions as in proof of [2, Theorem 1.2], there exists a unique weak solution $u_{ϵ} \in X_{0} \cap C^{\frac{1}{2}} (R)$ to $(P_{δ}^{ϵ})$ , $u_{ϵ}$ is a monotone increasing sequence as $ϵ \to 0^{+}$ and there exists a constant $c > 0$ such that $u_{ϵ} ⩾ c ϕ_{1}$ . Moreover, $\begin{array}{l} sup_{ϵ > 0} \int_{R} {({(- Δ)}^{1 / 4} u_{ϵ})}^{2} d x < \infty . \end{array}$ To get an upper bound on $u_{ϵ}$ , we will use the integral representation and the Green’s function $G (x, y)$ . Clearly, $\begin{array}{l} u_{ϵ} (x) = \int_{- 1}^{1} \frac{G (x, y)}{{(u_{ϵ} (y) + ϵ)}^{δ}} d y . \end{array}$ Then for a suitable positive constant C independent of ϵ, we have $\begin{array}{l} \frac{u_{ϵ} (x)}{d {(x)}^{\frac{1}{2}}} ⩽ C \int_{- 1}^{1} \frac{log (1 + \frac{d {(x)}^{\frac{1}{2}} d {(y)}^{\frac{1}{2}}}{| x - y |})}{d {(x)}^{\frac{1}{2}} {(u_{ϵ} (y) + ϵ)}^{δ}} d y ⩽ C \int_{- 1}^{1} \frac{log (1 + \frac{d {(x)}^{\frac{1}{2}} d {(y)}^{\frac{1}{2}}}{| x - y |})}{d {(x)}^{\frac{1}{2}} d {(y)}^{\frac{δ}{2}}} d y . \end{array}$ Utilizing the fact that $δ < 1$ and the proof of Theorem 1.6 Step 1, we obtain that $\begin{array}{l} u_{ϵ} (x) / d {(x)}^{\frac{1}{2}} < \infty, for all x \in (- 1, 1) . \end{array}$ Thus, we infer that $u = {lim}_{ϵ \to 0^{+}} u_{ϵ} ⩽ c ϕ_{1}$ and u is the unique weak solution to $(P_{δ})$ . Also, $\begin{array}{l} c ϕ_{1} ⩽ u ⩽ C ϕ_{1} \end{array}$ for some suitable constants c, C. This completes the proof of the theorem in case of $δ < 1$ . In a similar manner, for the case $δ ⩾ 1$ , we will follow the proof of Theorem 1.2 of [2] coupled with Theorem 1.6. Precisely, we will get unique solution of $(P_{δ})$ such that $\begin{array}{l} \frac{1}{C_{1}} ϕ_{1} {log}^{\frac{1}{2}} (\frac{2}{ϕ_{1}}) ⩽ u ⩽ C_{1} ϕ_{1} {log}^{\frac{1}{2}} (\frac{2}{ϕ_{1}}), if δ = 1, \\ \frac{1}{C_{2}} ϕ_{1}^{\frac{2}{δ + 1}} ⩽ u ⩽ C_{2} ϕ_{1}^{\frac{2}{δ + 1}}, if δ > 1, \end{array}$ for some appropriate positive constant $C_{1}$ and $C_{2}$ . For the Part (ii), the proof follows from Theorem 1.4 of [2]. We remark that all classical solutions belong to space $X_{0}$ as well. □

Define $Λ : = sup {λ > 0 : (P_{λ}) has a weak solution}$ . Lemma 2.2.
It holds $0 < Λ < \infty$ .
Proof.
Let u be the solution of $(P_{δ})$ given by Theorem 1.7 then ${\underline{u}}_{λ} = λ^{\frac{1}{δ + 1}} u$ is a solution of $\begin{array}{l} \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = \frac{λ}{u^{δ}}, u > 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{array}$ Moreover, ${\underline{u}}_{λ}$ is a strict subsolution of $(P_{λ})$ . Also, let ${\overline{u}}_{λ} = {\underline{u}}_{λ} + M U$ for some $M > 1$ , where U is a solution of $\begin{array}{l} \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = 1, u > 0, & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{array}$ There exists $λ_{0} > 0$ such that ${\overline{u}}_{λ}$ is a supersolution of $(P_{λ})$ for all $λ ⩽ λ_{0}$ . Now we define the following iterative scheme for all $λ ⩽ λ_{0}$ , starting with $u_{0} = {\underline{u}}_{λ}$ and ( $n ⩾ 1$ ) $\begin{array}{l} \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u_{n} + λ C u_{n} - \frac{λ}{u_{n}^{δ}} = λ C u_{n - 1} + λ f (u_{n - 1}), u > 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1), \end{matrix} \end{array}$ where $C = C (λ_{0}) > 0$ is large enough such that $t \to C t + f (t)$ is non decreasing in $(0, ‖ {\overline{u}}_{λ_{0}} ‖_{L^{\infty}})$ . Taking into account monotonicity of the operator ${(- Δ)}^{\frac{1}{2}} u - λ u^{- δ}$ , using the Comparison Principle [13, Lemma 2.2] and the proof of Theorem 1.7, we have that ${u_{n}}$ is increasing and ${u_{n}} \subset C^{\frac{1}{2}} (R) \cap C_{ϕ_{δ}}^{+} ((- 1, 1))$ . Furthermore, for all $λ ⩽ λ_{0}$ , ${\underline{u}}_{λ} ⩽ u_{n} ⩽ {\overline{u}}_{λ}$ . Using Theorem 1.7, we have ${sup}_{n \in N} ‖ u_{n} ‖_{C^{γ} (R)} ⩽ C_{0}$ for some $C_{0} = C_{0} (δ, λ_{0})$ large enough and γ is defined in Theorem 1.7. Hence $u_{n} \to u$ in $C (R)$ and u satisfies $\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = \frac{λ}{u^{δ}} + f (u) \end{matrix}$ in the sense of distributions. Hence from the above arguments we get $Λ > 0$ . From the superlinear behavior of $f (t)$ at infinity, we obtain that $Λ < \infty$ . □
Proof of Theorem 1.9.
The proof follows from Theorem 1.6 of [2] (see also [8]). □
Remark 2.3.
Consider the problem $\begin{array}{l} (P_{λ}^{K}) \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = λ (\frac{K (x)}{u^{δ}} + f (u)), u > 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) \end{matrix} \end{array}$ where $K \in C_{loc}^{ν} ((- 1, 1))$ , $ν \in (0, 1)$ such that ${inf}_{x \in (- 1, 1)} K (x) > 0$ and satisfies for some $0 ⩽ β < 1$ and $c_{1}, c_{2} > 0$ such that $c_{1} d {(x)}^{- β} ⩽ K (x) ⩽ c_{2} d {(x)}^{- β}$ , for all $x \in (- 1, 1)$ . By modifying our barrier function $ϕ_{δ}$ (see (1.5)) with respect to the growth of $K (x)$ , we can prove Theorem 1.7. Subsequently, we can also prove Theorem 1.9 for the problem $(P_{λ}^{K})$ (same as [2, Theorem 1.6]).

3. Study of isolated singularities and qualitative properties

In this section, we study the qualitative properties as symmetry, monotonicity of solutions to the problem $(P_{λ})$ and asymptotic behavior of the connected branch $C$ . In order to describe the asymptotic behavior of large solutions, we first study Brezis–Lions problem in the setting of fractional Laplacian operator. In the spirit of Brezis and Lions work (see [4]), we classify the singularities of non-negative distributional solutions of fractional semilinear elliptic equation $(P_{s})$ for $n ⩾ 2 s$ . We assume that $g : R^{+} \to R^{+}$ is continuous function with $g (0) = 0$ and Ω be a smooth bounded domain in $R^{n}$ .

Theorem 3.1.
Let u be nonnegative distributional solution of $(P_{s})$ in the sense of Definition 1.11 and $g (u) \in L_{loc}^{t} (Ω^{'})$ for $t > \frac{n}{2 s} ⩾ 1$ . Then $g (u) \in L^{1} (Ω)$ .
Proof.
We follow the approach as in [5]. Suppose by contradiction that $g (u) \notin L^{1} (Ω)$ . Since $g (u) \in L_{loc}^{t} (Ω^{'})$ then for any small $r > 0$ there exists a sequence ${R_{m}}_{m \in N} \in (0, r)$ such that $R_{m} \to 0$ and $\begin{matrix} (3.1) & \int_{B_{r} (0) ∖ B_{R_{m}} (0)} g (u) d x = m . \end{matrix}$ Consider the problem $\begin{matrix} (3.2) & \{\begin{matrix} {(- Δ)}^{s} u_{m} = χ_{Ω ∖ B_{R_{m}} (0)} g (u), u ⩾ 0 & in Ω, \\ u_{m} = 0 & in R^{n} ∖ Ω . \end{matrix} \end{matrix}$ Since $χ_{Ω ∖ B_{R_{m}} (0)} g (u) \in L^{t} (Ω)$ for $t > \frac{n}{2 s}$ , then there exists a sequence of classical solutions ${u_{m}}$ solving (3.2) such that $u_{m} \in L^{\infty} (Ω) \cap C^{β} (\overline{Ω})$ for some $β \in (0, 1)$ (see Proposition 1.4 in [20]).

Let Φ be the fundamental solution of ${(- Δ)}^{s}$ . i.e. ${(- Δ)}^{s} Φ = δ_{0}$ in $D^{'} (R^{n})$ , where $\begin{matrix} Φ (x) = \{\begin{matrix} \frac{Γ (\frac{n}{2} - s)}{2^{2 s} π^{n / 2} Γ (s)} \frac{1}{| x |^{n - 2 s}} & if n \neq 2 s \\ \frac{- 1}{π} log (| x |) & if n = 2 s . \end{matrix} \end{matrix}$ Since $u ⩾ 0$ and $u_{m}$ is bounded in $L^{\infty} (Ω)$ therefore ${lim}_{x \to 0} (u + Φ) (x) = + \infty$ and for each $m \in N$ , there exists $r_{m} > 0$ such that $u + Φ ⩾ u_{m}$ in $B_{r_{m}} (0) ∖ {0}$ . Then by the weak comparison principle we obtain $u + Φ ⩾ u_{m}$ in $R^{n} ∖ {0}$ .

Since ${lim}_{y \to x} G (x, y) = + \infty$ , there exists $r_{1} > 0$ such that $G (x, y) ⩾ 1$ in $x, y \in B_{r_{1}} (0)$ . Now by using (3.1) we obtain that, $\begin{matrix} u_{m} (x) & = \int_{Ω} G (x, y) χ_{Ω ∖ B_{R_{m}} (0)} g (u) d y = \int_{Ω ∖ B_{R_{m}} (0)} G (x, y) g (u) d y \\ ⩾ \int_{B_{r_{1}} (0) ∖ B_{R_{m}} (0)} g (u) = m \to \infty \end{matrix}$ which implies $u + Φ = + \infty$ in $K ⋐ B_{r_{1}} (0) ∖ {0}$ , which is not possible. Therefore $g (u) \in L^{1} (Ω)$ . □

Let $ξ : R^{n} \to [0, 1]$ be radially symmetric increasing function such that $ξ \in C^{\infty} (R^{n})$ and $\begin{matrix} ξ (x) = \{\begin{matrix} 1 & if x \in R^{n} ∖ B_{2} (0), \\ 0 & if x \in B_{1} (0) . \end{matrix} \end{matrix}$ Define $u_{ϵ} = u ξ_{ϵ}$ where $ξ_{ϵ} (x) = ξ (\frac{x}{ϵ})$ for all $x \in R^{n}$ . Then for any $x \in Ω ∖ {0}$ , $\begin{matrix} (3.3) & \begin{matrix} {(- Δ)}^{s} u_{ϵ} (x) & = C_{n, s} \int_{R^{n}} \frac{u_{ϵ} (x) - u_{ϵ} (y)}{| x - y |^{n + 2 s}} d y \\ = C_{n, s} P. V. \int_{R^{n}} \frac{u (x) ξ_{ϵ} (x) - u (y) ξ_{ϵ} (y)}{| x - y |^{n + 2 s}} d y \\ = ξ_{ϵ} (x) {(- Δ)}^{s} u (x) + u (x) {(- Δ)}^{s} ξ_{ϵ} (x) \\ - C_{n, s} P. V. \int_{R^{n}} \frac{(u (x) - u (y)) (ξ_{ϵ} (x) - ξ_{ϵ} (y))}{| x - y |^{n + 2 s}} d y . \end{matrix} \end{matrix}$ Now we prove the following result:
Theorem 3.2.
Let $P : C_{c}^{\infty} (Ω) \to R$ be the operator such that $\begin{matrix} P (ϕ) = \int_{Ω} u {(- Δ)}^{s} ϕ - g (u) ϕ d x for all ϕ \in C_{c}^{\infty} (Ω) \end{matrix}$ where $u \in L^{1} (Ω)$ is a non-negative distributional solution of $(P_{s})$ and $g (u) \in L^{1} (Ω)$ . Then
$P (ϕ) = 0$ for any $ϕ \in C_{c}^{\infty} (Ω)$ with $supp (ϕ) \subset Ω ∖ {0}$ .

There exists constants $c_{a}$ such that $\begin{matrix} P (ϕ) = \sum_{| a | = 0}^{\infty} c_{a} D^{a} ϕ (0) \end{matrix}$ where $a = (a_{1}, a_{2}, \dots, a_{n})$ with $a_{i} \in N$ , $| a | = \sum_{i = 1}^{n} a_{i}$ , $D^{a} = (\partial^{a_{1}} ϕ, \partial^{a_{2}} ϕ, \dots, \partial^{a_{n}} ϕ)$ .

Proof.
Let us prove assertion 1 and consider $ϵ > 0$ small enough then we have $\begin{matrix} | \int_{Ω} (u {(- Δ)}^{s} ϕ - g (u) ϕ) | d x & ⩽ \int_{Ω} | (u_{ϵ} {(- Δ)}^{s} ϕ - g (u) ϕ) d x | \\ + | \int_{Ω} u (1 - ξ_{ϵ} (x)) {(- Δ)}^{s} ϕ d x | . \end{matrix}$ Since $ϕ \in C_{c}^{\infty} (Ω)$ with $supp (ϕ) \subset Ω ∖ {0}$ , then there exists $r > 0$ such that $ϕ = 0$ in $B_{r} (0)$ . Then by using integration by parts formula with ${(- Δ)}^{s} u_{ϵ} \in L^{\infty} (Ω)$ (Lemma 2.2 in [6]) and (3.3) we obtain, $\begin{matrix} | \int_{Ω} (u {(- Δ)}^{s} ϕ - g (u) ϕ) d x | \\ ⩽ | \int_{Ω} ϕ {(- Δ)}^{s} u_{ϵ} - g (u) ϕ) d x | + | \int_{B_{2 ϵ} (0)} u {(- Δ)}^{s} ϕ d x | \\ ⩽ | \int_{B_{2 ϵ} (0)} u {(- Δ)}^{s} ϕ d x | + | \int_{Ω ∖ B_{r} (0)} (ξ_{ϵ} (x) {(- Δ)}^{s} u (x) + u (x) {(- Δ)}^{s} ξ_{ϵ} (x) - g (u)) ϕ d x | \\ + C_{n, s} | \int_{Ω ∖ B_{r} (0)} ϕ (x) \int_{R^{n}} \frac{(u (x) - u (y)) (ξ_{ϵ} (x) - ξ_{ϵ} (y))}{| x - y |^{n + 2 s}} d y d x | . \end{matrix}$ For $x \in Ω ∖ B_{r} (0)$ , $y \in B_{ϵ} (0)$ and $ϵ < \frac{r}{4}$ , we have $\begin{matrix} | {(- Δ)}^{s} ξ_{ϵ} (x) | = C_{n, s} \int_{B_{2 ϵ} (0)} \frac{1 - ξ_{ϵ} (y)}{| x - y |^{n + 2 s}} d y ⩽ C \frac{| B_{2 ϵ} (0) |}{{(r - ϵ)}^{n + 2 s}} . \end{matrix}$ Therefore, $\begin{matrix} (3.4) & lim_{ϵ \to 0} \int_{Ω ∖ B_{r} (0)} u (x) | ϕ | | {(- Δ)}^{s} ξ_{ϵ} (x) | = 0 . \end{matrix}$ Also, $\begin{matrix} (3.5) & \begin{matrix} | \int_{Ω ∖ B_{r} (0)} ϕ (x) \int_{R^{n}} \frac{(u (x) - u (y)) (ξ_{ϵ} (x) - ξ_{ϵ} (y)}{| x - y |^{n + 2 s}} d y d x | \\ ⩽ ‖ ϕ ‖_{L^{\infty} (Ω)} | \int_{Ω ∖ B_{r} (0)} u (x) \int_{B_{2 ϵ} (0)} \frac{| 1 - ξ_{ϵ} (y) |}{| x - y |^{n + 2 s}} d y d x \\ - \int_{Ω ∖ B_{r} (0)} (\int_{B_{2 ϵ} (0)} \frac{u (y) | 1 - ξ_{ϵ} (y) |}{| x - y |^{n + 2 s}} d y) d x | \\ ⩽ \frac{‖ ϕ ‖_{L^{\infty} (Ω)}}{{(r - ϵ)}^{n + 2 s}} (C ϵ^{n} \int_{Ω ∖ B_{r} (0)} u (x) d x + | Ω ∖ B_{r} (0) | \int_{B_{2 ϵ} (0)} u (y) d y) \to 0 as ϵ \to 0 . \end{matrix} \end{matrix}$ Since $ξ_{ϵ} (x) = 1$ in $Ω ∖ B_{r} (0)$ and u is distributional solution of $(P_{s})$ we obtain, $\begin{matrix} (3.6) & \int_{Ω ∖ B_{r} (0)} (ξ_{ϵ} (x) {(- Δ)}^{s} u (x) - g (u)) ϕ d x = 0 . \end{matrix}$ Therefore, by combining (3.4), (3.5) and (3.6) we obtain $P (ϕ) = 0$ for all $ϕ \in C_{c}^{\infty} (Ω)$ with $supp (ϕ) \subset Ω ∖ {0}$ . Since $u \in L^{1} (Ω)$ and $g (u) \in L^{1} (Ω)$ , then P is a bounded linear functional on $C_{c}^{\infty} (Ω)$ . Therefore by using Theorem XXXV in [21], we obtain $\begin{matrix} P = \sum_{| a | ⩽ m} c_{a} D^{a} δ_{0} \end{matrix}$ where $c_{a} \in R$ and $δ_{0}$ denotes the Dirac mass at origin. i.e. for all $ϕ \in C_{c}^{\infty} (Ω)$ $\begin{matrix} (3.7) & P (ϕ) = \sum_{| a | ⩽ m} c_{a} D^{a} ϕ (0) . \end{matrix}$ □
Theorem 3.3.
Let P be a bounded linear functional satisfying (3.7). Then $\begin{matrix} c_{a} = 0 for any | a | ⩾ 1 . \end{matrix}$
Proof.
Let $η \in C^{\infty} (R^{n})$ with $supp (η) \subset \overline{B_{1} (0)}$ and $| a | ⩾ 1$ such that $D^{a} η (0) = c_{a}$ for every $| a | ⩽ m$ (see [4]). Define $η_{ϵ} (x) = η (\frac{x}{ϵ})$ for $x \in R^{n}$ , then from (3.7) we obtain, $\begin{matrix} (3.8) & P (η_{ϵ}) = \sum_{| a | ⩽ m} c_{a} D^{a} η_{ϵ} (0) = C \frac{c_{a}^{2}}{ϵ^{| a |}} = \int_{Ω} (u {(- Δ)}^{s} η_{ϵ} - g (u) η_{ϵ}) d x . \end{matrix}$ Let $r > 0$ and divide the integral in (3.8) into two parts: $\begin{matrix} | \int_{Ω} u (x) {(- Δ)}^{s} η_{ϵ} (x) | & ⩽ \int_{Ω ∖ B_{r} (0)} u (x) | {(- Δ)}^{s} η_{ϵ} (x) | d x + \int_{B_{r} (0)} u (x) | {(- Δ)}^{s} η_{ϵ} (x) | d x . \end{matrix}$ For $x \in Ω ∖ B_{r} (0)$ with ϵ small enough, we obtain $\begin{matrix} (3.9) & \begin{matrix} | {(- Δ)}^{s} η_{ϵ} (x) | & = | P. V. \int_{B_{ϵ} (0)} \frac{η (\frac{x}{ϵ}) - η (\frac{y}{ϵ})}{| x - y |^{n + 2 s}} d y | ⩽ \frac{2 ‖ η ‖_{L^{\infty} (B_{1} (0))} | B_{ϵ} (0) |}{{(r - ϵ)}^{n + 2 s}} \to 0 as ϵ \to 0 \end{matrix} \end{matrix}$ and for $x \in B_{r} (0)$ we obtain, $\begin{matrix} (3.10) & \begin{matrix} | \int_{B_{r} (0)} u (x) {(- Δ)}^{s} η (\frac{x}{ϵ}) d x | & ⩽ {‖ {(- Δ)}^{s} η ‖}_{L^{\infty} (B_{r} (0))} \int_{B_{r} (0)} u (x) d x \to 0 as r \to 0 . \end{matrix} \end{matrix}$ independently of ϵ. Therefore by combining (3.9) and (3.10) with $ϵ < r$ , we obtain $\begin{matrix} (3.11) & \begin{matrix} | \int_{Ω} u (x) {(- Δ)}^{s} η_{ϵ} (x) d x | & ⩽ o (1) as ϵ \to 0 . \end{matrix} \end{matrix}$ Also, the second integral in (3.8) satisfies $\begin{matrix} (3.12) & \begin{matrix} \int_{Ω} g (u) η_{ϵ} d x & ⩽ ‖ η ‖_{L^{\infty} (Ω)} \int_{B_{ϵ} (0)} g (u) d x \to 0 as ϵ \to 0 . \end{matrix} \end{matrix}$ From (3.8), (3.11) and (3.12), $c_{a}^{2} ⩽ C_{1} ϵ^{| a |} o (1)$ as $ϵ \to 0$ . Therefore we have $c_{a} = 0$ , for all $| a | ⩾ 1$ since ϵ is arbitrary. □
Proof of Theorem 1.12.
Follows from combining Theorem 3.2 and Theorem 3.3. □

Now we prove Theorems 1.10 and 1.13 concerning the qualitative properties of classical solution and asymptotic behavior of large solution for half Laplacian operator and $n = 1$ :
Proof of Theorem 1.10.
With the assistance of maximum principle in narrow domains (see [15]) and moving plane method, we prove the monotonicity and radial symmetry of classical solutions in Ω. Without loss of generality, we assume $Ω = (- 1, 1)$ and $(λ, u)$ be classical solution of $(P_{λ})$ for $λ ⩾ λ_{0}$ (obtained from Theorem 1.9).

Define $R_{h} (x) : = (2 h - x)$ be the reflection of the point x about h and $\begin{matrix} v_{h} (x) : = u_{h} (x) - u (x) where u_{h} (x) = u (R_{h} (x)) . \end{matrix}$

Step 1: Positivity of $v_{h}$ near $- 1$ and 1:

Clearly for $| h |$ sufficiently large, $v_{h} (x) ⩾ 0$ . Now we prove that $v_{h} (x) ⩾ 0$ in $(- 1, h) \cap H_{h}^{-}$ if $h ⩽ 0$ and in $(h, 1) \cap H_{h}^{+}$ if $h > 0$ where $H_{h}^{\pm} = {x \in R : x ≷ h}$ and h lies in the neighborhood of $x_{0} \in \partial Ω = {- 1, 1}$ . Suppose that $v_{h} < 0$ in $K \subset (- 1, h) \cap H_{h}^{-}$ for some $h ⩽ 0$ . Since f is Lipschitz and noting that $supp ({(- v_{h})}^{+}) \subset (- 1, 2 h + 1)$ , we have $\begin{matrix} ⟨ {(- Δ)}^{1 / 2} (- v_{h}), {(- v_{h})}^{+} ⟩ & = λ \int_{- 1}^{2 h + 1} (\frac{1}{u^{δ}} - \frac{1}{{(u_{h})}^{δ}} + f (u) - f (u_{h})) {(- v_{h})}^{+} d x \\ ⩽ C \int_{K} {({(u - u_{h})}^{+})}^{2} d x . \end{matrix}$ Then by Poincaré inequality, we obtain $\begin{matrix} \int_{R} {({(- Δ)}^{\frac{1}{4}} {(u - u_{h})}^{+})}^{2} ⩽ C \int_{- 1}^{2 h + 1} {({(- v_{h})}^{+})}^{2} d x ⩽ C (diam (K)) \int_{R} {({(- Δ)}^{\frac{1}{4}} {(u - u_{h})}^{+})}^{2} . \end{matrix}$ Then by choosing h close enough to $- 1$ we get, $C (diam (K)) < 1$ and then ${(- v_{h})}^{+} = {(u - u_{h})}^{+} = 0$ . Similarly in the case of $(h, 1) \cap H_{h}^{+}$ for $h > 0$ . Now by moving the point in the neighborhood of $- 1$ and 1 we obtain there exists $T > 0$ independent of u such that $\begin{matrix} (3.13) & \{\begin{matrix} u (x - t) is non-increasing & \forall (t, x) \in [0, T] \times (- 1, h) if h ⩽ 0, \\ u (x - t) is non-decreasing & \forall (t, x) \in [0, T] \times (h, 1) if h ⩾ 0 . \end{matrix} \end{matrix}$

Step 2: Positivity of $v_{h}$ in interior of $(- 1, 1)$ :

In Step 1, we have proved that $v_{h} ⩾ 0$ in the neighborhood of $- 1$ and 1. So, without loss of generality we can assume that $h ⩾ 0$ be the smallest value such that $v_{h} ⩾ 0$ in $(h, 1)$ . Then the mean value Theorem implies $v_{h}$ satisfies the following for some $θ \in (0, 1)$ $\begin{matrix} (3.14) & {(- Δ)}^{\frac{1}{2}} v_{h} + \frac{δ v_{h}}{{(θ u + (1 - θ) u_{h})}^{δ + 1}} = f (u_{h}) - f (u) in (h, 1) . \end{matrix}$
Claim 1: For every compact subset $K \subset (h, 1)$ , $ess {inf}_{K} v_{h} > 0$ .

To establish our claim, we follow the proof of Proposition 3.6 in [15]. Since $v_{h} ≢ 0$ in $(h, 1)$ then for $x^{} \in (h, 1)$ , it is enough to prove that $ess {inf}_{B_{r} (x^{})} v_{h} > 0$ for r sufficiently small. From Step 1, $v_{h} ⩾ 0$ and $v_{h} (x) = - v_{h} (R_{h} (x))$ in $H_{h}^{+}$ , there exists a bounded set $B \subset H_{h}^{+}$ with $x^{} \notin \overline{B}$ and $\begin{matrix} (3.15) & \tilde{μ} : = inf_{B} v_{h} > 0 . \end{matrix}$ Using Lemma 2.1 in [15], we fix r such that $U = B_{2 r} (x^{})$ and $\begin{matrix} (3.16) & 0 < r < \frac{1}{4} dist (x^{}, B \cup (R ∖ H_{h}^{+})) and λ_{1} (U) ⩾ C_{L} (f) \end{matrix}$ where $C_{L} (f)$ is the Lipschitz constant of f and $λ_{1} (U)$ is the first eigenvalue of ${(- Δ)}^{s}$ in U. Now, in order to apply Proposition 3.5 in [15], we construct a subsolution of ${(- Δ)}^{\frac{1}{2}} \tilde{v} = c (x) \tilde{v}$ in U where $\begin{array}{l} c (x) = \{\begin{matrix} \frac{f (u_{h}) - f (u)}{v_{h}} - \frac{δ}{{(θ u + (1 - θ) u_{h})}^{δ + 1}} & if v_{h} \neq 0, \\ 0 & if v_{h} = 0 . \end{matrix} \end{array}$ Define $\begin{matrix} k : R \to R, k (x) = m (x) - m (R_{h} (x)) + a [1_{B} (x) - 1_{B} (R_{h} (x))] \end{matrix}$ where a will be determined later and $m \in C_{c}^{2} (R)$ such that $0 ⩽ m ⩽ 1$ on $R$ and $\begin{array}{l} m (x) = \{\begin{matrix} 1 & if | x - x^{} | ⩽ r, \\ 0 & if | x - x^{} | ⩾ 2 r, \end{matrix} \end{array}$ and satisfies $k (R_{h} (x)) = - k (x)$ on $H_{h}^{+}$ , $k = 0$ in $H_{h}^{+} ∖ (U \cup B)$ and $k = a$ on B. Then by Proposition 2.3 in [15] we obtain, $\begin{matrix} \int_{R} \int_{R} \frac{(m (x) - m (y)) (ϕ (x) - ϕ (y))}{| x - y |^{2}} d x d y ⩽ C \int_{U} ϕ (x) d x \end{matrix}$ for $ϕ \in τ$ , $ϕ ⩾ 0$ and $C = C (m)$ independent of ϕ. Since $ϕ = 0$ in $R ∖ U$ , $(U \cap B) \cup (U \cap R_{h} (B)) = \emptyset$ and $\begin{matrix} m (R_{h} (x)) ϕ (x) = 1_{B} (x) ϕ (x) = 1_{R_{h} (B)} (x) ϕ (x) = 0 in R . \end{matrix}$ Then we have $\begin{array}{l} \int_{R} \int_{R} \frac{(k (x) - k (y)) (ϕ (x) - ϕ (y))}{| x - y |^{2}} d x d y ⩽ C_{a} \int_{U} ϕ (x) d x \end{array}$ where $\begin{matrix} C_{a} : = C + sup_{x \in U} \int_{R_{h} (U)} \frac{1}{| x - y |^{2}} d y - a inf_{x \in U} \int_{B} (\frac{1}{| x - y |^{2}} - \frac{1}{| x - R_{h} (y) |^{2}}) d y . \end{matrix}$ Since $| x - y | ⩽ | x - R_{h} (y) |$ for all $x, y \in H_{h}^{+}$ , $\overline{U} \subset H_{h}^{+}$ and continuity of the function $x \mapsto \int_{B} (\frac{1}{| x - y |^{2}} - \frac{1}{| x - R_{h} (y) |^{2}}) d y$ implies $\begin{matrix} inf_{U} \int_{B} (\frac{1}{| x - y |^{2}} - \frac{1}{| x - R_{h} (y) |^{2}}) d y > 0 . \end{matrix}$ Now by taking a sufficiently large enough such that $C_{a} ⩽ - C_{L} (f)$ and using $v_{h} ⩾ 0$ in U, we obtain $\begin{matrix} \int_{R} \int_{R} \frac{(k (x) - k (y)) (ϕ (x) - ϕ (y))}{| x - y |^{2}} d x d y \\ ⩽ - C_{L} (f) \int_{U} ϕ (x) d x \\ ⩽ \int_{U} λ (\frac{f (u_{h}) - f (u)}{v_{h}} - \frac{δ}{{(θ u + (1 - θ) u_{h})}^{δ + 1}}) k (x) ϕ (x) d x . \end{matrix}$ Then by using (3.15), (3.16) and Proposition 3.5 in [15], we obtain ${\tilde{v}}_{h} (x) : = v_{h} (x) - \frac{\tilde{μ}}{a} k (x) ⩾ 0$ a.e. in U so that $v_{h} (x) ⩾ \frac{\tilde{μ}}{a} k (x) = \frac{\tilde{μ}}{a} > 0$ a.e. in $B_{r} (x^{})$ and which completes the Claim 1.
Claim 2: $h = 0$ .

We argue by contradiction and suppose $h > 0$ . Since h is the smallest value such that $v_{h} ⩾ 0$ in $(h, 1)$ , so we claim that for a small $ϵ > 0$ we have $v_{h - ϵ} ⩾ 0$ in $(h - ϵ, 1)$ and thus get a contradiction that h is the smallest value. For this claim we follow the proof of Proposition 3.5 in [15]. Fix γ (to be determined later) and let $K ⋐ (h, 1)$ such that $| (h, 1) ∖ K | ⩽ \frac{γ}{2}$ . Then by using Claim 1, $v_{h} ⩾ r > 0$ in K and then by continuity $v_{h - ϵ} > 0$ in K for ϵ small enough. Since $v_{h - ϵ}$ satisfies (3.14) in $(h - ϵ, 1) ∖ K$ and by taking $w : = 1_{H_{h - ϵ}^{+}} v_{h - ϵ}^{-}$ such that $supp (w) \subset (h - ϵ, 1) ∖ K$ as a test function, we obtain $\begin{matrix} (3.17) & \begin{matrix} ⟨ {(- Δ)}^{\frac{1}{2}} v_{h - ϵ}, w ⟩ = \int_{(h - ϵ, 1) ∖ K} (\frac{- δ v_{h - ϵ}}{{(θ u + (1 - θ) u_{h - ϵ})}^{δ + 1}} + f (u_{h - ϵ}) - f (u)) w d x . \end{matrix} \end{matrix}$ We observe that $\begin{matrix} [w + v_{h - ϵ}] w = [1_{H_{h - ϵ}^{+}} v_{h - ϵ}^{+} + 1_{R ∖ H_{h - ϵ}^{+}} v_{h - ϵ}] 1_{H_{h - ϵ}^{+}} v_{h - ϵ}^{-} = 0 in R \end{matrix}$ and therefore $\begin{matrix} (3.18) & \begin{matrix} {[w (x) - w (y)]}^{2} + [v_{h - ϵ} (x) - v_{h - ϵ} (y)] [w (x) - w (y)] \\ = - (w (x) [w (x) + v_{h - ϵ} (y)] + w (y) [w (x) + v_{h - ϵ} (x)]) . \end{matrix} \end{matrix}$ Now using $| x - y | ⩽ | x - R_{h - ϵ} (y) |$ for all $x, y \in H_{h - ϵ}^{+}$ , $R_{h - ϵ} (R ∖ H_{h - ϵ}^{+}) = H_{h - ϵ}^{+}$ and from (3.18), we obtain $\begin{matrix} (3.19) & \begin{matrix} ⟨ {(- Δ)}^{\frac{1}{2}} w, w ⟩ + ⟨ {(- Δ)}^{\frac{1}{2}} v_{h - ϵ}, w ⟩ \\ = - 2 \int_{H_{h - ϵ}^{+}} \int_{R} \frac{w (x) [w (y) + v_{h - ϵ} (y)]}{| x - y |^{2}} d y d x \\ = - 2 \int_{H_{h - ϵ}^{+}} \int_{R} \frac{w (x) [1_{H_{h - ϵ}^{+}} v_{h - ϵ}^{+} (y) + 1_{R ∖ H_{h - ϵ}^{+}} v_{h - ϵ} (y)]}{| x - y |^{2}} d y d x \\ = - 2 \int_{H_{h - ϵ}^{+}} \int_{H_{h - ϵ}^{+}} w (x) (\frac{v_{h - ϵ}^{+} (y)}{| x - y |^{2}} - \frac{v_{h - ϵ} (y)}{| x - R_{h - ϵ} |^{2}}) d y d x ⩽ 0 . \end{matrix} \end{matrix}$ Let $λ_{1}$ be the first eigenvalue of ${(- Δ)}^{s}$ in $(h - ϵ, 1) ∖ K$ and then by combining (3.17) and (3.19) we obtain, $\begin{matrix} λ_{1} ((h - ϵ, 1) ∖ K) \int_{(h - ϵ, 1) ∖ K} {| v_{h - ϵ}^{-} |}^{2} d x \\ ⩽ ⟨ {(- Δ)}^{\frac{1}{2}} w, w ⟩ ⩽ - ⟨ {(- Δ)}^{\frac{1}{2}} v_{h - ϵ}, w ⟩ \\ = \int_{(h - ϵ, 1) ∖ K} \frac{δ v_{h - ϵ} 1_{(h - ϵ, 1) ∖ K} v_{h - ϵ}^{-}}{{(θ u + (1 - θ) u_{h})}^{δ + 1}} d x + \int_{(h - ϵ, 1) ∖ K} (- f (u_{h - ϵ}) + f (u)) 1_{(h - ϵ, 1) ∖ K} v_{h - ϵ}^{-} d x \\ ⩽ C_{L} \int_{(h - ϵ, 1) ∖ K} {| v_{h - ϵ}^{-} |}^{2} d x . \end{matrix}$ Since $λ_{1} (Ω) \to \infty$ when $| Ω | \to 0$ (see Lemma 2.1 in [15]) then by choosing γ small enough we get $v_{h - ϵ} ⩾ 0$ in $(h - ϵ, 1)$ , which is a contradiction. Therefore $h = 0$ i.e. $u (- x) ⩾ u (x)$ and then by repeating the same proof for largest value of h over $(- 1, h)$ we obtain $u (x) = u (- x)$ for all $x \in (- 1, 1)$ . Since $h = 0$ , therefore (3.13) and Claim 1 imply u is strictly decreasing in $| x |$ . □

Now we prove result describing the asymptotic behavior of connected branch $C$ : Proof of Theorem 1.13.
Let $(λ, u) \in S \cap ({λ ⩾ Λ_{a}} \times C_{0} ([- 1, 1]))$ be the solution of the problem $(P_{λ})$ . Then from Theorem 1.10 we obtain, u is decreasing with respect to $| x |$ then for every $ϵ > 0$ there exists $β_{1} > 0$ such that for any $x \in (- 1, - ϵ) \cup (ϵ, 1)$ , we have a measurable set $M_{ϵ}$ satisfying
$| M_{ϵ} | ⩾ β_{1}$ and $M_{ϵ} \subset (- 1 + ϵ, 1 - ϵ)$ .

$u (y) ⩾ u (x)$ , $\forall y \in M_{ϵ}$ .
Then by multipying $ϕ_{1}$ to the equation satisfied by u, we obtain $\begin{matrix} λ_{1} \int_{- 1}^{1} u ϕ_{1} = λ (\int_{- 1}^{1} \frac{ϕ_{1}}{u^{δ}} d x + \int_{- 1}^{1} f (u) ϕ_{1} d x) \end{matrix}$ and for any $m ⩾ \frac{λ_{1}}{Λ_{a}}$ , there exists a $C > 0$ , $\begin{matrix} m t - C ⩽ \frac{1}{t^{δ}} + f (t), t \in R^{+} . \end{matrix}$ Then by using $u (y) ⩾ u (x)$ , $\forall y \in M_{ϵ}$ we obtain for $C_{1} > 0$ large enough, $\begin{matrix} u (x) \int_{M_{ϵ}} ϕ_{1} d x ⩽ (m - \frac{λ_{1}}{Λ_{a}}) \int_{- 1}^{1} u ϕ_{1} d x ⩽ C_{1} . \end{matrix}$ Together with $| M_{ϵ} | ⩾ β_{1}$ it implies that $u (x) ⩽ C_{2}$ for all $x \in (- 1, - ϵ) \cup (ϵ, 1)$ . So, $\begin{matrix} (3.20) & sup_{u \in S \cap ({λ ⩾ λ_{0}} \times C_{0} ([- 1, 1]))} ‖ u ‖_{L^{\infty} ((- 1, 1) ∖ [ϵ, ϵ])} ⩽ c_{ϵ} < \infty . \end{matrix}$ Suppose there exists a sequence $(λ_{k}, u_{k})$ of solutions in $S \cap ({Γ ⩾ λ ⩾ Λ_{a}} \times C_{0} ([- 1, 1]))$ such that $λ_{k} \to Λ_{a}$ and $‖ u_{k} ‖_{L^{\infty}} \to \infty$ as $k \to \infty$ , then (3.20) implies “0” is the blow up point. Hence by regularity of $u_{k}$ and compact embedding we obtain $u_{k} \to u$ uniformly on compact subsets of $(- 1, 1) ∖ {0}$ . Since $u_{k}$ satisfies (1.2), then from the proof of Remark 1.14, we obtain $‖ g (u_{k}) ‖_{L^{1}} ⩽ C_{2}$ , where $C_{2}$ is independent of k. Then Fatou’s lemma and Vitali’s convergence theorem give $u \in L^{p} ((- 1, 1))$ and $‖ u_{k} - u ‖_{L^{p} ((- 1, 1))} \to 0$ for any $1 ⩽ p < \infty$ . Now, by passing to the limit as $k \to \infty$ we obtain u satisfies (in the sense of Definition 1.11): $\begin{matrix} \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = Λ_{a} g (u) & in (- 1, 1) ∖ {0}, \\ u ⩾ 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1), \end{matrix} \end{matrix}$ with $g (u) \in L^{1} (Ω)$ . Then by Theorems 3.1 and 1.12 there exists $μ ⩾ 0$ such that u satisfies (in the sense of Definition 1.1) $\begin{matrix} (3.21) & \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = Λ_{a} g (u) + μ δ_{0} & in (- 1, 1), \\ u ⩾ 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{matrix}$ Suppose $μ \neq 0$ . Hence we have $u (x) = Λ_{a} g (u) * Φ (x) + μ Φ (x) + l (x)$ where l is a s-harmonic function in $(- 1, 1)$ and $Φ (x) = \frac{- 1}{π} log (| x |)$ . Therefore $u (x) ⩾ log (| x |^{- μ / π}) - C$ and since $α > 1$ , $\begin{matrix} f (u) ⩾ h (log (| x |^{- μ / π}) - C) {exp}^{{(log (| x |^{- μ / π}) - C)}^{α}} ⩾ h (log (| x |^{- μ / π}) - C) | x |^{- μ p / π} \end{matrix}$ for all $p > 1$ , $0 < | x | ⩽ | x_{ρ} |$ and $| x_{ρ} |$ small. Then by integrating $f (u)$ over a small ball B around 0, we obtain $\int_{B} f (u) = \infty$ which contradicts $f (u) \in L^{1} ((- 1, 1))$ . Therefore $μ = 0$ . This completes the proof of Theorem 1.13. □

4. Global multiplicity result via variational method

In this section, we will show the existence and multiplicity of solutions of $(P_{λ})$ by using variational methods. The energy functional corresponds to problem $(P_{λ})$ is defined as $\begin{array}{l} (4.1) & J_{λ} (u) = \frac{1}{2} ‖ u ‖^{2} - λ \int_{- 1}^{1} (G (u) + F (u)) d x, \end{array}$ where $\begin{array}{l} G (u) = \{\begin{matrix} 0 & if u ⩽ 0 and δ > 0, \\ \frac{u^{1 - δ}}{1 - δ} & if u > 0 and δ \neq 1, \\ ln u & if u > 0 and δ = 1 . \end{matrix} \end{array}$ We will state the Trudinger–Moser inequality in case of half-Laplacian which is recently proved by Takahashi [22].

Theorem 4.1.
Let Ω be an open bounded interval in $R$ . Then it holds $\begin{matrix} π = max {c : sup_{‖ u ‖_{X_{0}} ⩽ 1} \int_{Ω} e^{c u^{2}} d x < \infty} . \end{matrix}$

Using the above theorem one can see that the functional $J_{λ}$ is well defined. Lemma 4.2.
For each $λ \in (0, Λ]$ , $(P_{λ})$ admits a weak solution provided (K1) and (K2) holds.
Proof.
We use the classical Perron’s method to proof the existence of a solution. Let $\underline{u} = {\underline{u}}_{λ}$ where ${\underline{u}}_{λ}$ is defined in Lemma 2.2. Then $\underline{u}$ is a strict subsolution of $(P_{λ})$ . Let $λ^{'} \in (0, Λ)$ then it is easy to see that $u_{λ^{'}}$ is a solution of $(P_{λ^{'}})$ and forms a supersolution of $(P_{λ})$ . Note that such a $λ^{'}$ exists because of definition of Λ. Let $\overline{u} = u_{λ^{'}}$ and $M : = {u \in X_{0} ∣ \underline{u} ⩽ u ⩽ \overline{u}}$ . Then M is closed, convex and $J_{λ}$ is coercive and weakly semi lower continuous on M. It implies that $u_{n}$ is a sequence in M such that $J_{λ} (u_{n}) \to {inf}_{u \in M} J_{λ} (u) > - \infty$ when $n \to \infty$ and $u_{n} ⩽ \overline{u}$ . It implies ${u_{n}}$ is bounded in $X_{0}$ . Then there exists $u_{λ} \in M$ such that (up to a subsequence) $u_{n} ⇀ u_{λ}$ weakly in $X_{0}$ .
Claim: $u_{λ}$ is weak solution of $(P_{λ})$ .

For $ϕ \in X_{0}$ and $ε > 0$ small enough, define $v_{ε} = u_{λ} + ε ϕ - ϕ^{ε} + ϕ_{ε} \in M$ , where $\begin{array}{l} ϕ^{ε} = {(u_{λ} + ε ϕ - \overline{u})}^{+} and ϕ_{ε} = {(u_{λ} + ε ϕ - \underline{u})}^{-} \end{array}$ By construction $ϕ^{ε}, ϕ_{ε} \in X_{0} \cap L^{\infty} ((- 1, 1))$ and $u_{λ} + t (v_{ε} - u_{λ}) \in M$ for each $t \in (0, 1)$ , we have $\begin{array}{l} 0 & ⩽ lim_{t \to 0^{+}} \frac{J_{λ} (u_{λ} + t (v_{ε} - u_{λ})) - J_{λ} (u_{λ})}{t} \\ = \int_{Q} (v_{ε} - u_{λ}) {(- Δ)}^{\frac{1}{2}} u_{λ} d x - \int_{- 1}^{1} u_{λ}^{- δ} (v_{ε} - u_{λ}) d x - \int_{- 1}^{1} f (u_{λ}) (v_{ε} - u_{λ}) d x . \end{array}$ Now using the same arguments as in proof of [13, Proposition 3.2] coupled with Lemma 4.3, we have desired result. For the case $λ = Λ$ using the same assertions as in [13, Theorem 3.4], we have desired result. □
Lemma 4.3.
Let $λ \in (0, Λ)$ and $u_{λ}$ denotes the weak solution of $(P_{λ})$ obtained in Lemma 4.2. Then $u_{λ}$ is a local minimum of the functional $J_{λ}$ .
Proof.
The proof follows by using the same arguments as in [13, Lemma 3.3](see [9]), one can proof that $u_{λ}$ is local minimum of the functional $J_{λ}$ in $X_{0}$ topology. We left the calculations for the readers. □
Lemma 4.4.
There exists a positive weak solution of $(P_{Λ})$ and any weak solution of $(P_{λ})$ for $λ \in (0, Λ]$ , belongs to $L^{\infty} ((- 1, 1)) \cap C_{ϕ_{δ}}^{+} ((- 1, 1))$ where $ϕ_{δ}$ is defined in (1.5). Moreover,
Proof.
See [13, Theorem 3.4, Proposition 4.1]. □

4.1. Existence of second solution for $(P_{λ})$ , $0 < λ < Λ$

The concern of this section is to prove the existence of a second solution for $(P_{λ})$ . Let $u_{λ}$ is the first weak solution of $(P_{λ})$ in $X_{0}$ topology obtained in Lemma 4.2. Now, consider the following problem, which is $(P_{λ})$ translated by $u_{λ}$ : $\begin{array}{l} (\tilde{P_{λ}}) \{\begin{matrix} {(- Δ)}^{\frac{1}{2}} u = λ ({(u + u_{λ})}^{- δ} - {(u_{λ})}^{- δ} + f (u + u_{λ}) - f (u_{λ})), u > 0 & in (- 1, 1), \\ u = 0 & in R ∖ (- 1, 1) . \end{matrix} \end{array}$ For $x \in (- 1, 1)$ , define $\begin{array}{l} \tilde{g} (x, s) = ({(s + u_{λ} (x))}^{- δ} - {(u_{λ} (x))}^{- δ}) χ_{R^{+}} (s), \tilde{G} (x, t) = \int_{0}^{t} \tilde{g} (x, s) d s, \\ \tilde{f} (x, s) = (f (s + u_{λ} (x)) - f (u_{λ} (x))) χ_{R^{+}} (s), \tilde{F} (x, t) = \int_{0}^{t} \tilde{f} (x, s) d s . \end{array}$ Let ${\tilde{J}}_{λ} : X_{0} \to R$ be the energy functional associated with $(\tilde{P_{λ}})$ defined as $\begin{array}{l} {\tilde{J}}_{λ} (u) = \frac{‖ u ‖^{2}}{2} - λ \int_{- 1}^{1} \tilde{G} (x, u (x)) d x - λ \int_{- 1}^{1} \tilde{F} (x, u (x)) d x . \end{array}$

Remark 4.5.

By Theorem 4.1, it can be easily shown that the map $X_{0} ∋ u \to \frac{1}{2} ‖ u ‖^{2} - λ \int_{- 1}^{1} \tilde{F} (x, u (x)) d x \in R$ is a $C^{1}$ map. The map $X_{0} ∋ u \to λ \int_{- 1}^{1} \tilde{G} (x, u (x)) d x \in R$ is locally Lipschitz. Therefore, ${\tilde{J}}_{λ}$ is a sum of a $C^{1}$ and a Lipschitz functional. Hence, the generalized derivative of ${\tilde{J}}_{λ}$ exist for all $u \in X_{0}$ and given by $\begin{array}{l} {\tilde{J}}_{λ}^{0} (u, ϕ) = lim_{h \to 0} sup_{t \to 0} \frac{{\tilde{J}}_{λ} (u + h + t ϕ) - {\tilde{J}}_{λ} (u + h)}{t}, ϕ \in X_{0} . \end{array}$ We say u is a generalized critical point if ${\tilde{J}}_{λ}^{0} (u, ϕ) ⩾ 0$ for all $ϕ \in X_{0}$ .

For any $u \in X_{0}$ , $\begin{array}{l} J_{λ} (u^{+} + u_{λ}) = J_{λ} (u_{λ}) + {\tilde{J}}_{λ} (u) - \frac{‖ u^{-} ‖^{2}}{2} - 4 \int_{R} \int_{R} \frac{u^{+} (x) u^{-} (y)}{| x - y |^{2}} d x d y . \end{array}$ Since $u_{λ}$ is a local minimum of $J_{λ}$ , it follows that 0 is a local minimum of ${\tilde{J}}_{λ}$ in $X_{0}$ -topology.

One can easily prove that if $u ⩾ 0$ then $\begin{array}{l} {\tilde{J}}_{λ}^{0} (u, ϕ) ⩽ ⟨ u_{λ} + u, ϕ ⟩ - λ \int_{- 1}^{1} {(u_{λ} + u)}^{- δ} ϕ d x - λ \int_{- 1}^{1} f (u_{λ} + u) ϕ d x . \end{array}$

Now we will use the machinery of mountain pass Lemma and Ekeland variational principle to prove the existence of second solution. We will show the existence of solution in the following cone: $\begin{array}{l} T = {u \in X_{0} : u ⩾ 0 a.e. in (- 1, 1)} . \end{array}$ Since 0 is local minimum of ${\tilde{J}}_{λ}$ in $X_{0}$ topology, there exists a $ρ_{0} > 0$ such that ${\tilde{J}}_{λ} (0) ⩽ {\tilde{J}}_{λ} (u)$ provided $‖ u ‖ < ρ_{0}$ . We distinguish two cases:

(Zero Altitude): $inf {{\tilde{J}}_{λ} (u) : u \in T, ‖ u ‖ = ρ} = {\tilde{J}}_{λ} (0) = 0$ for all $ρ \in (0, ρ_{0})$ .

(Mountain Pass): There exists $ρ_{1} \in (0, ρ_{0})$ such that $inf {{\tilde{J}}_{λ} (u) : u \in T ‖ u ‖ = ρ_{1}} > {\tilde{J}}_{λ} (0)$ .

Lemma 4.6.

Let (ZA) holds for some $λ \in (0, Λ)$ . Then there exists a non-trivial generalized critical point $v_{λ} \in T$ for ${\tilde{J}}_{λ}$ .

Proof.

Fix $ρ \in (0, ρ_{0})$ . By using the definition of infimum of there exist ${u_{n}} \subseteq T$ with $‖ u_{n} ‖ = ρ$ and ${\tilde{J}}_{λ} (u_{n}) ⩽ 1 / n$ . Let $0 < σ < \frac{1}{2} min {ρ_{0} - ρ, ρ}$ and define the set $\begin{array}{l} A = {u \in T : ρ - σ ⩽ ‖ u ‖ ⩽ ρ + σ} \end{array}$ which is closed in $X_{0}$ and ${\tilde{J}}_{λ}$ is continuous on A. Now using the Ekeland Variational principle, we obtain the existence of a sequence ${v_{n}} \in A$ such that $\begin{matrix} (4.2) & \begin{matrix} \{\begin{matrix} {\tilde{J}}_{λ} (v_{n}) ⩽ {\tilde{J}}_{λ} (u_{n}) ⩽ \frac{1}{n}, ‖ u_{n} - v_{n} ‖ ⩽ \frac{1}{n}, \\ {\tilde{J}}_{λ} (v_{n}) ⩽ {\tilde{J}}_{λ} (z) + \frac{1}{n} ‖ z - v_{n} ‖ for all z \in A . \end{matrix} \end{matrix} \end{matrix}$ For a given $z \in T$ , we can choose $ε > 0$ such that $v_{n} + ε (v - v_{n}) \in A$ . From (4.2), we obtain that $\begin{array}{l} \frac{{\tilde{J}}_{λ} (v_{n} + ε (z - v_{n})) - {\tilde{J}}_{λ} (v_{n})}{ε} ⩾ - \frac{1}{n} ‖ z - v_{n} ‖ . \end{array}$ Taking $ε \to 0^{+}$ we get $\begin{array}{l} {\tilde{J}}_{λ}^{0} (v_{n}, z - v_{n}) ⩾ - \frac{1}{n} ‖ z - v_{n} ‖ for all z \in A . \end{array}$ From Remark 4.5, we deduce that for any $z \in A$ , $\begin{array}{l} (4.3) & \begin{matrix} ⟨ u_{λ} + v_{n}, z - v_{n} ⟩ - λ \int_{- 1}^{1} {(u_{λ} + v_{n})}^{- δ} (z - v_{n}) d x - λ \int_{- 1}^{1} f (u_{λ} + v_{n}) (z - v_{n}) d x \\ ⩾ - \frac{1}{n} ‖ z - v_{n} ‖ . \end{matrix} \end{array}$ Since $v_{n}$ is a bounded sequence in $X_{0}$ therefore, there exists $v_{λ} \in X_{0}$ such that $v_{n} ⇀ v_{λ}$ weakly in $X_{0}$ as well as almost everywhere in $(- 1, 1)$ . We claim that $v_{λ}$ is a weak solution of $(\tilde{P_{λ}})$ . For any $ψ \in C_{c}^{\infty} ((- 1, 1))$ , set $\begin{array}{l} ψ_{n, ε} = {(v_{n} + ε ψ)}^{-} and z = v_{n} + ε ψ + ψ_{n, ε} = {(v_{n} + ε ψ)}^{+} \in T . \end{array}$ Hence as a result of (4.3) and the choice of z, we have $\begin{matrix} ⟨ u_{λ} + v_{n}, z - v_{n} ⟩ - λ \int_{- 1}^{1} {(u_{λ} + v_{n})}^{- δ} (ε ψ + ψ_{n, ε}) d x - λ \int_{- 1}^{1} f (u_{λ} + v_{n}) (ε ψ + ψ_{n, ε}) d x \\ ⩾ - \frac{1}{n} ‖ ε ψ + ψ_{n, ε} ‖ . \end{matrix}$ Observe that $ψ_{n, ε} \to ψ_{ε} = {(v_{λ} + ε ψ)}^{-}$ a.e. in $(- 1, 1)$ , $| ψ_{n, ε} | ⩽ ε | ψ |$ in $(- 1, 1)$ and by using dominated convergence theorem one can easily show that $ψ_{n, ε} \to ψ_{ε}$ in $L^{m} ((- 1, 1))$ for all $m > 1$ . Moreover, $ψ_{n, ε} ⇀ ψ_{ε}$ weakly in $X_{0}$ . Using the same arguments as in [13, Lemma 4.2], we have $\begin{array}{l} (4.4) & ⟨ u_{λ} + v_{n}, ε ψ + ψ_{n, ε} ⟩ ⩽ ⟨ u_{λ} + v_{λ}, ε ψ + ψ_{ε} ⟩ + o_{n} (1) . \end{array}$ By Hardy’s Inequality (see [14, Corollary 1.4.4.10, p. 23]) and dominated convergence theorem, $\begin{array}{l} (4.5) & \int_{- 1}^{1} {(u_{λ} + v_{n})}^{- δ} (ε ψ + ψ_{n, ε}) d x \to \int_{- 1}^{1} {(u_{λ} + v_{λ})}^{- δ} (ε ψ + ψ_{ε}) d x . \end{array}$ Taking into account the hypothesis (K2), Theorem 4.1 and Vitali’s convergence theorem, we get $\begin{array}{l} \int_{- 1}^{1} f (u_{λ} + v_{n}) ψ d x \to \int_{- 1}^{1} f (u_{λ} + v_{λ}) ψ d x . \end{array}$ Using the mean value Theorem, definition of $ψ_{n, ε}$ and the fact that $f^{'} > 0$ , we deduce that $\begin{array}{l} f (u_{λ} + v_{n}) ψ_{n, ε} ⩽ (f (u_{λ} - ε ψ) + f^{'} (ξ_{n}) (v_{n} + ε ψ)) ψ_{n, ε} \\ f (u_{λ} - ε ψ) ψ_{n, ε} ⩽ f (u_{λ} - ε ψ) ε | ψ | \in L^{1} ((- 1, 1)) . \end{array}$ This on using dominated convergence theorem gives $\begin{array}{l} (4.6) & \int_{- 1}^{1} f (u_{λ} + v_{n}) (ε ψ + ψ_{n, ε}) d x \to \int_{- 1}^{1} f (u_{λ} + v_{λ}) (ε ψ + ψ_{ε}) d x . \end{array}$ Using (4.4), (4.5) and (4.6), we obtain that $\begin{array}{l} ⟨ u_{λ} + v_{λ}, ε ψ + ψ_{ε} ⟩ - \int_{- 1}^{1} {(u_{λ} + v_{λ})}^{- δ} (ε ψ + ψ_{ε}) d x - \int_{- 1}^{1} f (u_{λ} + v_{λ}) (ε ψ + ψ_{ε}) d x ⩾ 0 . \end{array}$ Employing the fact that $u_{λ}$ is a weak solution of $(P_{λ})$ , we get $\begin{array}{rcl} (4.7) & \begin{matrix} ⟨ u_{λ} + v_{λ}, ψ ⟩ - \int_{- 1}^{1} {(u_{λ} + v_{λ})}^{- δ} ψ d x - \int_{- 1}^{1} f (u_{λ} + v_{λ}) ψ d x \\ ⩾ - \frac{1}{ε} ⟨ v_{λ}, ψ_{ε} ⟩ + \frac{1}{ε} \int_{- 1}^{1} ({(u_{λ} + v_{λ})}^{- δ} - {(u_{λ})}^{- δ}) ψ_{ε} d x + \frac{1}{ε} \int_{- 1}^{1} (f (u_{λ} + v_{λ}) - f (u_{λ})) ψ_{ε} d x \\ ⩾ - \frac{1}{ε} ⟨ v_{λ}, ψ_{ε} ⟩ + \frac{1}{ε} \int_{Ω_{ε}} ({(u_{λ} + v_{λ})}^{- δ} - {(u_{λ})}^{- δ}) ψ_{ε} d x \end{matrix} \end{array}$ where the last inequality follows using the fact that f is an increasing function and $v_{λ} ⩾ 0$ and $supp (ψ_{ε}) = : Ω_{ε} \subset (- 1, 1)$ . Keeping in mind that $u_{λ}^{- δ} ϕ \in L^{1} ((- 1, 1))$ , $\begin{array}{l} \frac{1}{ε} \int_{Ω_{ε}} ({(u_{λ} + v_{λ})}^{- δ} - {(u_{λ})}^{- δ}) ψ_{ε} d x ⩽ \frac{2}{ε} \int_{- 1}^{1} u_{λ}^{- δ} ψ = o (1) . \end{array}$ Furthermore, trivial calculations gives $\begin{array}{l} - \frac{1}{ε} ⟨ v_{λ}, ψ_{ε} ⟩ & ⩾ \int_{Ω_{ε}} \int_{Ω_{ε}} \frac{(v_{λ} (x) - v_{λ} (y)) (ψ (x) - ψ (y))}{| x - y |^{2}} d x d y \\ + 2 \int_{Ω_{ε}^{c}} \int_{Ω_{ε}} \frac{(ψ (x) - ψ (y)) (v_{λ} + ψ) (y)}{| x - y |^{2}} d x d y . \\ + 2 \int_{Ω_{ε}} \int_{Ω_{ε}^{c}} \frac{v_{λ} (x) ψ (x)}{| x - y |^{2}} d x d y . \end{array}$ Letting $ε \to 0$ in (4.7), we deduce that, for all $ψ \in C_{c}^{\infty} ((- 1, 1))$ , $\begin{array}{l} ⟨ u_{λ} + v_{λ}, ψ ⟩ - \int_{- 1}^{1} {(u_{λ} + v_{λ})}^{- δ} ψ d x - \int_{- 1}^{1} f (u_{λ} + v_{λ}) ψ d x ⩾ 0 . \end{array}$ It implies that $v_{λ}$ is a generalized critical point. Now we will show that $v_{λ} ≢ 0$ . Note that $‖ v_{n} ‖ ⩾ max {2 ρ - ρ_{0}, 0} ⩾ 0$ , so it enough to show that $v_{n} \to v_{λ}$ strongly in $X_{0}$ . Let $z = v_{λ}$ in (4.3), $\begin{array}{l} ‖ v_{λ} - v_{n} ‖^{2} & ⩽ ⟨ u_{λ} + v_{λ}, v_{λ} - v_{n} ⟩ + \frac{1}{n} ‖ v_{λ} - v_{n} ‖ - λ \int_{- 1}^{1} {(u_{λ} + v_{n})}^{- δ} (v_{λ} - v_{n}) d x \\ - λ \int_{- 1}^{1} f (u_{λ} + v_{n}) (v_{λ} - v_{n}) d x . \end{array}$ Observe that ${(u_{λ} + v_{n})}^{- δ} (v_{λ} - v_{n}) \to 0$ as $n \to \infty$ a.e. on $(- 1, 1)$ , $u_{λ} \sim ϕ_{δ}$ in the neighborhood of $- 1$ and 1. In consequence of Hardy’s inequality and Hölder inequality, for any measurable set $E \subset (- 1, 1)$ and $δ > 1$ , we have $\begin{array}{l} \int_{E} {(u_{λ} + v_{n})}^{- δ} (v_{λ} - v_{n}) d x & ⩽ \int_{E} u_{λ}^{- δ} | v_{λ} - v_{n} | d x ⩽ C \int_{E} ϕ_{δ}^{\frac{- 2 δ}{δ + 1}} | v_{λ} - v_{n} | d x \\ ⩽ C \int_{E} d^{\frac{1 - δ}{2 (δ + 1)}} \frac{| v_{λ} - v_{n} |}{d^{\frac{1}{2}}} d x ⩽ C ‖ v_{λ} - v_{n} ‖_{X_{0}} {‖ d^{\frac{1 - δ}{2 (δ + 1)}} ‖}_{L^{2} (E)} . \end{array}$ Thus in a consequence of Vitali’s convergence theorem $\int_{- 1}^{1} {(u_{λ} + v_{n})}^{- δ} (v_{λ} - v_{n}) d x \to 0$ . Rewrite $\begin{array}{l} f (u_{λ} + v_{n}) (v_{λ} - v_{n}) = f (u_{λ} + v_{n}) (u_{λ} + v_{λ}) - f (u_{λ} + v_{n}) (u_{λ} + v_{n}) . \end{array}$ Using the same arguments used for (4.6), one can easily show that $\begin{array}{l} \int_{- 1}^{1} f (u_{λ} + v_{n}) (u_{λ} + v_{λ}) d x \to \int_{- 1}^{1} f (u_{λ} + v_{λ}) (u_{λ} + v_{λ}) d x . \end{array}$ Let $z_{n} = u_{λ} + v_{n}$ and $z_{λ} = u_{λ} + v_{λ}$ then $f (z_{n}) z_{n} \to f (z_{λ}) z_{λ}$ a.e. in $(- 1, 1)$ . Let k be any integer such that $k > ‖ u_{λ} ‖_{\infty}$ . Using (K2), $\begin{array}{l} \int_{{z_{n} ⩾ k}} f (z_{n}) z_{n} d x ⩽ C \int_{{z_{n} ⩾ k}} e^{\frac{3 z_{n}^{2}}{2}} z_{n} d x ⩽ C \int_{{z_{n} ⩾ k}} e^{2 z_{n}^{2}} z_{n} d x ⩽ C e^{- k^{2}} \int_{- 1}^{1} e^{3 z_{n}^{2}} d x . \end{array}$ By means of the Hölder inequality and the relation $z_{n}^{2} ⩽ 2 (u_{λ}^{2} + {(z_{n} - u_{λ})}^{2})$ , we deduce that $\begin{array}{l} (4.8) & \int_{{z_{n} ⩾ k}} f (z_{n}) z_{n} d x & ⩽ C e^{- k^{2}} \int_{- 1}^{1} e^{6 u_{λ}^{2}} e^{6 v_{n}^{2}} d x ⩽ C e^{- k^{2}} {‖ e^{6 u_{λ}^{2}} ‖}_{L^{p} ((- 1, 1))} {‖ e^{6 v_{n}^{2}} ‖}_{L^{p^{'}} ((- 1, 1))} . \end{array}$ Now for $ρ_{0}$ small enough, we can choose $p^{'} > 1$ such that $6 p^{'} ‖ v_{n} ‖ ⩽ 12 p^{'} ρ_{0} < π$ . With the help of Trudinger–Moser inequality and (4.8), we have $\int_{{z_{n} ⩾ k}} f (z_{n}) z_{n} d x ⩽ C e^{- k^{2}}$ where C is independent of n. Hence for k large enough, $\begin{array}{l} \int_{- 1}^{1} f (z_{n}) z_{n} d x ⩽ \int_{{z_{n} ⩽ k}} f (z_{n}) z_{n} d x + C e^{- k^{2}} . \end{array}$ Letting $n \to \infty$ and $k \to \infty$ , ${lim sup}_{n \to \infty} \int_{- 1}^{1} f (z_{n}) z_{n} d x ⩽ \int_{- 1}^{1} f (z_{λ}) z_{λ} d x$ . Using the fact that $v_{n} ⇀ v_{λ}$ weakly in $X_{0}$ , we get $⟨ u_{λ} + v_{λ}, v_{λ} - v_{n} ⟩ \to 0$ as $n \to \infty$ . Therefore, from all the calculations, we obtain that $‖ v_{λ} - v_{n} ‖ \to 0$ as $n \to \infty$ . □

Now we will prove the existence of second solution if (MP) holds. Before this we will prove some preliminary results. We recall the definition of Moser function $ω_{n}$ for half-Laplacian, which is recently given by Takahashi [22]. $\begin{array}{l} ω_{n} (x) = \frac{1}{\sqrt{π}} \{\begin{matrix} {(log n)}^{\frac{1}{2}} & if | x | ⩽ \frac{1}{n}, \\ - log {(n)}^{- \frac{1}{2}} log | x | & if \frac{1}{n} ⩽ | x | ⩽ 1, \\ 0 & if | x | ⩾ 1 . \end{matrix} \end{array}$ Fix $x_{0} \in R$ and $r > 0$ such that $ω_{n}^{r} (x) = ω_{n} (\frac{x - x_{0}}{r})$ has support in $(- 1, 1)$ . Note that $‖ ω_{n}^{r} ‖ = 1$ .

Lemma 4.7.

Assume (K1)–(K3). Then the following holds.

${\tilde{J}}_{λ} (t ω_{n}^{r}) \to - \infty$ as $t \to \infty$ .

For a suitable $x_{0} \in (- 1, 1)$ and $r > 0$ small enough, ${sup}_{t > 0} {\tilde{J}}_{λ} (t ω_{n}^{r}) < \frac{π}{2}$ .

Proof.

(i) Using (K2), there exist positive constants $C_{1}$ and $C_{2}$ such that $\begin{array}{l} \tilde{F} (x, t) ⩾ C_{1} e^{\frac{1}{2} {(u_{λ} + t)}^{2}} - C_{2} - f (u_{λ}) t for t ⩾ 0 . \end{array}$ Hence for some $C > 0$ , $\begin{array}{l} {\tilde{J}}_{λ} (t ω_{n}^{r}) & ⩽ \frac{t^{2}}{2} - C λ \int_{B_{\frac{r}{n}} (x_{0})} e^{\frac{t^{2}}{2} | ω_{n}^{r} |^{2}} d x = \frac{t^{2}}{2} - \frac{C λ r}{n} e^{\frac{t^{2} log n}{π}} \to - \infty as t \to \infty . \end{array}$

(ii) On the contrary, suppose that there exists a subsequence of $N$ such that ${sup}_{t > 0} {\tilde{J}}_{λ} (t ω_{n}^{r}) ⩾ \frac{π}{2}$ . That is, $\frac{t_{n}^{2}}{2} - λ \int_{- 1}^{1} \tilde{G} (x, t_{n} ω_{n}) d x - λ \int_{- 1}^{1} \tilde{F} (x, t_{n} ω_{n}) d x ⩾ \frac{π}{2}$ . Using (K1), we have $\begin{array}{l} \frac{t_{n}^{2}}{2} - λ \int_{- 1}^{1} \tilde{G} (x, t_{n} ω_{n}) d x ⩾ \frac{π}{2} . \end{array}$ Clearly, $\tilde{g} (x, s) ⩽ 0$ for $x \in (- 1, 1)$ and $s > 0$ . For and $x \in B_{r} (x_{0})$ , applying Taylor’s expansion, $\tilde{g} (x, s) = \frac{δ s}{u_{λ}^{δ + 1}} + o (s^{2})$ . It implies that $\begin{array}{l} \int_{- 1}^{1} \tilde{G} (x, t_{n} ω_{n}) d x = \int_{B_{r} (x_{0})} \tilde{G} (x, t_{n} ω_{n}) d x ⩽ \int_{B_{r} (x_{0})} \int_{0}^{t_{n} ω_{n}} C s d s d x ⩽ C t_{n}^{2} O ({(log n)}^{- 1}) . \end{array}$ As a result, we get $\begin{array}{l} (4.9) & t_{n}^{2} ⩾ π - O ({(log n)}^{- 1}) . \end{array}$ Since $\frac{d}{d t} \tilde{J} (t ω_{n}^{r}) |_{t = t_{n}} = 0$ , we get $t_{n}^{2} - λ \int_{- 1}^{1} \tilde{g} (x, t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x = λ \int_{- 1}^{1} \tilde{f} (x, t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x$ . Again using the fact that $| \tilde{g} (x, s) | ⩽ C s$ , we have $\begin{array}{l} (4.10) & \int_{- 1}^{1} \tilde{f} (x, t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x ⩽ t_{n}^{2} [1 + O ({(log n)}^{- 1})] . \end{array}$ By definition of $\tilde{f}$ and the fact that $‖ ω_{n}^{r} ‖_{L^{1} ((- 1, 1))} = O ({(log n)}^{- \frac{1}{2}})$ , we have $\begin{array}{l} \int_{- 1}^{1} \tilde{f} (x, t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x = \int_{- 1}^{1} f (u_{λ} + t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x - t_{n} O ({(log n)}^{- \frac{1}{2}}) . \end{array}$ Now we will estimate $\int_{- 1}^{1} f (u_{λ} + t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x$ from below. Let $μ = {min}_{B_{r} (x_{0})} u_{λ}$ . Taking into account (K1), (K2), definition of $ω_{n}^{r}$ and (4.9), we have $\begin{matrix} \int_{- 1}^{1} f (u_{λ} + t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x \\ ⩾ \int_{B_{r} (x_{0})} h (u_{λ} + t_{n} ω_{n}^{r}) e^{{(u_{λ} + t_{n} ω_{n}^{r})}^{2}} t_{n} ω_{n}^{r} d x \\ ⩾ C h (μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}}) e^{{(μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}})}^{2}} t_{n} {(log n)}^{\frac{1}{2}} \int_{B_{\frac{r}{n}} (x_{0})} d x \\ ⩾ \frac{C r}{n} e^{- {(μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}})}^{q}} e^{2 (μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}})} e^{t_{n}^{2} \frac{log n}{π}} t_{n} {(log n)}^{\frac{1}{2}} {(μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}})}^{- 1} \\ ⩾ C e^{2 (μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}}) - {(μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}})}^{q}} . \end{matrix}$ Thus we obtain $\begin{array}{l} (4.11) & \int_{- 1}^{1} \tilde{f} (x, t_{n} ω_{n}^{r}) t_{n} ω_{n}^{r} d x = C e^{2 (μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}}) - {(μ + \frac{t_{n}}{\sqrt{π}} {(log n)}^{\frac{1}{2}})}^{q}} - t_{n} O ({(log n)}^{- \frac{1}{2}}) . \end{array}$ Since $(log n) \to \infty$ and $t_{n}$ is bounded away from 0 as $n \to \infty$ , we obtain a contradiction from (4.10) and (4.11). □

Lemma 4.8.

Let ${u_{k} : ‖ u_{k} ‖ = 1}$ be a sequence of $X_{0}$ functions converging weakly to a nonzero function u. Then for all $p < {(1 - ‖ u ‖)}^{- 1}$ , $\begin{array}{l} sup_{k \in N} \int_{- 1}^{1} e^{π p | u_{k} |^{2}} d x < \infty . \end{array}$

Proof.

See [11, Lemma 4.4]. □

Lemma 4.9.

Assume (K1)–(K3) and fix $λ \in (0, Λ)$ . Let (MP) holds then there exists a nontrivial generalized critical point $v_{λ}$ of ${\tilde{J}}_{λ}$ .

Proof.

Define the complete metric space $\begin{array}{l} Y = {η \in (C [0, 1], T) : η (0) = 0, ‖ η (1) ‖ > ρ_{1}, {\tilde{J}}_{λ} (η (1)) < {\tilde{J}}_{λ} (0) = 0}, \end{array}$ with metric space defined as $d (η, η_{'}) = {max}_{t \in [0, 1]} {‖ η (t) - η^{'} (t) ‖}$ for all $η, η^{'} \in Y$ . Fix $x_{0} \in (- 1, 1)$ and $r > 0$ such that Lemma 4.7 (ii) holds. Now choose $t_{0} > 1$ such that ${\tilde{J}}_{λ} (t_{0} ω_{n}^{r}) < 0$ . Note that existence of $t_{0}$ holds by Lemma 4.7 (i). Let $η (t) = t t_{0} ω_{n}^{r}$ , $t > 0$ . Then $η \in Y$ . Define the mountain-pass critical level $\begin{array}{l} γ_{0} = inf_{η \in Y} max_{t \in [0, 1]} {\tilde{J}}_{λ} (η (t)) . \end{array}$ From Lemma 4.7, we have $0 < γ_{0} < \frac{π}{2}$ . Define $Ψ : X \to R$ as $Ψ (η) = {max}_{t \in [0, 1]} {\tilde{J}}_{λ} (η (t))$ , $η \in Y$ . Applying the Ekeland’s variational principle, we get a sequence ${η_{k}} \subset Y$ such that $\begin{array}{l} Ψ (η_{k}) < γ_{0} + \frac{1}{k} and Ψ (η_{k}) < Ψ (η) + \frac{1}{k} ‖ η - η_{k} ‖, for all η \in Y . \end{array}$ Denote $Z_{k} = {t \in (0, 1) : {\tilde{J}}_{λ} (η_{k} (t)) = {max}_{s \in [0, 1]} {\tilde{J}}_{λ} (η_{k} (s))}$ . Now using the arguments and assertions as in [13, Lemma 4.4], there exist $t_{k} \in Z_{k}$ such that if $v_{k} = η_{k} (t_{k})$ , then

${\tilde{J}}_{λ}^{0} (v_{k}; \frac{w - v_{k}}{max {1, ‖ w - v_{k} ‖}}) ⩾ - \frac{1}{k}$ for all $w \in T$ ,

${\tilde{J}}_{λ} (v_{k}) \to γ_{0}$ as $k \to \infty$ .

Taking

w = u_{λ} + 2 v_{k}

in (i), we obtain,

\begin{array}{l} (4.12) & - \frac{1}{k} max {1, ‖ u_{λ} + v_{k} ‖} ⩽ ‖ u_{λ} + v_{k} ‖^{2} - λ \int_{- 1}^{1} ({(u_{λ} + v_{k})}^{1 - δ} + f (u_{λ} + v_{k}) (u_{λ} + v_{k})) d x . \end{array}

Fix

ε > 0

, using (K4), then there exists

C_{ε} > 0

such that

\begin{array}{l} \tilde{F} (x, v_{k}) & ⩽ \int_{0}^{u_{λ} + v_{k}} f (s) d s - f (u_{λ}) v_{k} ⩽ M (f (u_{λ} + v_{k}) + 1) - f (u_{λ}) v_{k} \\ ⩽ ε f (u_{λ} + v_{k}) (u_{λ} + v_{k}) + C_{ε} - f (u_{λ}) v_{k} . \end{array}

Now using the fact that

\tilde{G} (x, v_{k}) ⩽ 0

, we have

\begin{array}{l} (4.13) & \frac{1}{2} ‖ v_{k} ‖^{2} - λ ε f (u_{λ} + v_{k}) (u_{λ} + v_{k}) - C_{ε} + f (u_{λ}) v_{k} ⩽ γ_{0} + o_{k} (1) . \end{array}

From (4.12) and (4.13), we have

\begin{array}{l} (\frac{1}{2} - ε) ‖ v_{k} ‖^{2} < C + λ \int_{- 1}^{1} u_{λ}^{- δ} v_{k} d x + ‖ u_{λ} ‖^{2} + \frac{ε}{k} max {1, ‖ u + λ + v_{k} ‖} . \end{array}

With the help of Hardy’s inequality, we obtain that

{v_{k}}

is a bounded sequence in T. Therefore, there exists a

v_{λ} \in T

such that

v_{k} ⇀ v_{λ}

X_{0}

. From (i), we have

\begin{array}{l} (4.14) & {\tilde{J}}_{λ}^{0} (v_{k}; w - v_{k}) ⩾ - \frac{1}{k} (1 + ‖ w ‖), \end{array}

for all

w \in T

. Using the same assertions and arguments as in proof of Lemma 4.6, one can easily prove that

v_{λ}

is a generalized critical point of

(\tilde{P_{λ}})

. From (4.12),

{lim sup}_{k \to \infty} \int_{- 1}^{1} f (u_{λ} + v_{k}) (u_{λ} + v_{k}) < \infty

. Hence by Vitali’s convergence theorem,

\int_{- 1}^{1} f (u_{λ} + v_{k}) d x \to \int_{- 1}^{1} f (u_{λ} + v_{λ}) d x

. Now using (K4) and genralized dominated convergence theorem

\int_{- 1}^{1} \tilde{F} (x, v_{k}) d x \to \int_{- 1}^{1} \tilde{F} (x, v_{λ}) d x

. Using the fact that

u_{λ} \sim ϕ_{δ}

, Hardy’s inequality and similar arguments used above we can easily prove that

\int_{- 1}^{1} \tilde{G} (x, v_{k}) d x \to \int_{- 1}^{1} \tilde{G} (x, v_{λ}) d x

. Since

v_{k} ⇀ v_{λ}

weakly in

X_{0}

\begin{array}{l} \tilde{J} (v_{λ}) ⩽ \underset{k \to \infty}{lim inf} {\tilde{J}}_{λ} (v_{k}) = γ_{0} . \end{array}

Since

{lim}_{k \to \infty} {\tilde{J}}_{λ} (v_{k}) = γ_{0}

and if

v_{k} \to v_{λ}

strongly in

X_{0}

then

0 < γ_{0} = {\tilde{J}}_{λ} (v_{λ})

implies

v_{λ} \neq 0

. Therefore, to show

v_{λ} \neq 0

, it is enough to show that

v_{k} \to v_{λ}

strongly in

X_{0}

. Let if possible then

v_{k} ↛ v_{λ}

X_{0}

then

{\tilde{J}}_{λ} (v_{λ}) < γ_{0}

, we can assume

{\tilde{J}}_{λ} (v_{λ}) = 0

otherwise

v_{λ} \neq 0

. From Remark 4.5, we have

J_{λ} (u_{λ} + v_{λ}) = J_{λ} (u_{λ})

. We can choose

ε > 0

small enough so that

\begin{array}{l} (4.15) & (γ_{0} + J_{λ} (u_{λ} + v_{λ}) - J_{λ} (u_{λ})) = γ_{0} (1 + ε) < \frac{π}{2} . \end{array}

Define

Θ = λ \int_{- 1}^{1} F (u_{λ} + v_{λ}) + G (u_{λ} + v_{λ}) d x

. Using Remark 4.5, we have

{\tilde{J}}_{λ} (v_{k}) = J_{λ} (u_{l} a + v_{k}) - J - λ (u_{λ})

. Therefore,

2 (γ_{0} + Θ + J_{λ} (u_{λ})) = {lim}_{k \to \infty} ‖ u_{λ} + v_{k} ‖^{2}

. Since

{\tilde{J}}_{λ} (v_{λ}) < γ_{0}

then

‖ u_{λ} + v_{λ} ‖^{2} < {lim}_{k \to \infty} ‖ u_{λ} + v_{k} ‖^{2}

. It gives that

\begin{array}{l} (4.16) & 0 < ‖ u_{λ} + v_{λ} ‖^{2} < 2 (γ_{0} + Θ + J_{λ} (u_{λ})) . \end{array}

Taking into account (4.15), (4.16) and the fact that

J_{λ} (u_{λ} + v_{λ}) = \frac{1}{2} ‖ u_{λ} + v_{λ} ‖^{2} - Θ

, we deduce that

\begin{array}{l} (1 + ε) ‖ u_{λ} + v_{k} ‖^{2} & < \frac{π (γ_{0} + Θ + J_{λ} (u_{λ}))}{γ_{0} + Θ + J_{λ} (u_{λ}) + \frac{1}{2} ‖ u_{λ} + v_{λ} ‖^{2}} = π {(1 - \frac{‖ u_{λ} + v_{λ} ‖^{2}}{2 (γ_{0} + Θ + J_{λ} (u_{λ}))})}^{- 1} . \end{array}

Now taking into mind (4.16), we can choose

p > 1

such that

\begin{array}{l} \frac{(1 + ε) ‖ u_{λ} + v_{k} ‖^{2}}{p} ⩽ p < {(1 - \frac{‖ u_{λ} + v_{λ} ‖^{2}}{2 (γ_{0} + Θ + J_{λ} (u_{λ}))})}^{- 1} . \end{array}

Therefore, from Lemma 4.8,

{lim sup}_{k \to \infty} \int_{- 1}^{1} e^{(1 + ε) {(u_{λ} + v_{k})}^{2}} d x ⩽ {lim sup}_{k \to \infty} \int_{- 1}^{1} e^{\frac{π p {(u_{λ} + v_{k})}^{2}}{‖ u_{λ} + v_{k} ‖^{2}}} d x < \infty

. We write

\begin{array}{l} \int_{- 1}^{1} f (u_{λ} + v_{k}) v_{k} d x = \int_{- 1}^{1} f (u_{λ} + v_{k}) (u_{λ} + v_{k}) d x - \int_{- 1}^{1} f (u_{λ} + v_{k}) u_{λ} d x . \end{array}

From (K2), given

ε_{1} < ε

and

N \in N

, for some

C > 0

, we have

\begin{array}{l} \int_{- 1}^{1} f (u_{λ} + v_{k}) (u_{λ} + v_{k}) d x & = (\int_{v_{k} ⩽ N} + \int_{v_{k} > N}) f (u_{λ} + v_{k}) (u_{λ} + v_{k}) d x \\ ⩽ \int_{v_{k} ⩽ N} f (u_{λ} + v_{k}) (u_{λ} + v_{k}) d x + C \int_{v_{k} > N} e^{(1 + ε_{1}) {(u_{λ} + v_{k})}^{2}} d x \\ ⩽ \int_{v_{k} ⩽ N} f (u_{λ} + v_{k}) (u_{λ} + v_{k}) d x + C e^{(ε_{1} - ε) N} . \end{array}

Now letting

k \to \infty

and then

N \to \infty

, we obtain,

\begin{array}{l} \underset{k \to \infty}{lim sup} \int_{- 1}^{1} f (u_{λ} + v_{k}) (u_{λ} + v_{k}) d x ⩽ \int_{- 1}^{1} f (u_{λ} + v_{λ}) (u_{λ} + v_{λ}) d x . \end{array}

Hence

\begin{array}{l} (4.17) & \underset{k \to \infty}{lim sup} \int_{- 1}^{1} f (u_{λ} + v_{k}) (v_{k} - v_{λ}) d x ⩽ 0 . \end{array}

On the other hand, since we assume

v_{k} ↛ v_{λ}

then by using Remark 4.5, (4.14) with

w = v_{λ}

and the fact that

v_{k} ⇀ v_{λ}

, we have

\begin{array}{l} (4.18) & 0 < ν ⩽ ‖ v_{λ} - v_{k} ‖^{2} ⩽ o (1) - λ \int_{- 1}^{1} f (u_{λ} + v_{k}) (v_{λ} - v_{k}) d x . \end{array}

From (4.17) and (4.18), we obtain contradiction. Therefore,

v_{λ} \neq 0

. □

Proof of Theorem 1.15.

The proof follows from Lemma 4.2, Lemma 4.4, Lemma 4.6 along with Lemma 4.9. The proof of Hölder regularity follows straightaway from Lemma 4.4 and [2, Theorem 1.2] with $β = 0$ . □

Footnotes

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