Existence and Multiplicity of Solutions for a Class of Particular Boundary Value Poisson Equations

Abstract

In this paper, a special class of boundary value problems, $-▵u=\lambda u^q+u^r,\mathrm{in}\mathsf{\Omega },$$u>0$, $\mathrm{in}\mathsf{\Omega },$$\mathbf{n}·\nabla u+g\left(u\right)u=0,\mathrm{on}\partial \mathsf{\Omega },$ where ${\displaystyle 0<q<1<r<\frac{N+2}{N-2}}$ and $g:[0,\infty )\to (0,\infty )$ is a nondecreasing $C^1$ function. Here, $\mathsf{\Omega }\subset {\mathbb{R}}^N$($N\ge 3$) is a bounded domain with smooth boundary $\partial \mathsf{\Omega }$ and $\lambda >0$ is a parameter. The existence of the solution is verified via sub- and super-solutions method. In addition, the influences of parameters on the minimum solution are also discussed. The second positive solution is obtained by using the variational method.

1. Introduction

This paper deals with the nonlinear boundary value problems:

$$\left\{\begin{array}{cc}-▵u=\lambda u^q+u^r\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ u>0\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u+g\left(u\right)u=0}\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

(1)

where $\mathsf{\Omega }\subset {\mathbb{R}}^N$($N\ge 3$) is a bounded domain with smooth boundary $\partial \mathsf{\Omega }. \mathsf{▵}$ and $\lambda$ are the Laplace operator and the real parameter, respectively. This problem arises in thermal explosion theory. In recent years, this kind of problem has no longer been limited to mathematical research. It involves many fields, such as physics, biology, environmental systems and economic systems (see [1,2,3,4] and the references therein). The nonlinear boundary condition is inspired by the following Dirichlet boundary problem. For example, Rey in [5] proved the existence of the solution of

$$\left\{\begin{array}{cc}{\displaystyle -▵v=\epsilon f(x,v)+{\left|v\right|}^{\frac{4}{N-2}}v}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v=0\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

(2)

where $\mathsf{\Omega }\subset {\mathbb{R}}^N$ is a bounded domain. In addition, $f(x,v)$ is a term of lesser order than $v^{\frac{N+2}{N-2}}$. When $\epsilon$ tends to zero, the asymptotic behavior of the solution of (2) is obtained. In [6], Tarantello showed the non-uniqueness of solutions for

$$\left\{\begin{array}{cc}-▵v=f+{\left|v\right|}^{p-2}v\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v=0\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

and ${\displaystyle p=\frac{2N}{N-2}}$ ($N\ge 3$), $f\ne 0$. Denote by ${\left(H_0^1\left(\mathsf{\Omega }\right)\right)}^{-1}$ the dual space of $H_0^1\left(\mathsf{\Omega }\right)$; then, $f\in {\left(H_0^1\left(\mathsf{\Omega }\right)\right)}^{-1}$ will be

$${\int }_{\mathsf{\Omega }}fv\le c_N{(\parallel \nabla v\parallel }_2{)}^{\frac{N+2}{2}},\mathrm{for}\forall v\in H_0^1\left(\mathsf{\Omega }\right),{\parallel v\parallel }_p=1,$$

where ${\displaystyle c_N=\frac{4}{N-2}{\left(\frac{N-2}{N+2}\right)}^{\frac{N+2}{4}}}$. When $N=3$, Huang [7] proved that the problem

$$\left\{\begin{array}{cc}{\displaystyle -{▵}_{p}v={\lambda \left|v\right|}^{s-2}v+{\left|v\right|}^{t-2}v}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v=0\hfill & \mathrm{on}\partial \mathsf{\Omega }\hfill \end{array}\right.$$

(3)

has a positive solution, where ${\lambda }^{*}<\lambda <{\lambda }_1$ and $1<s<p<t$ (${\displaystyle t\le \frac{NP}{N-p}}$), when $N>3$ and ${\displaystyle 2\le s<p<t\le \frac{Np}{N-p}}$. For the case of $0<\lambda <{\lambda }_1$, Huang also proved the existence of the solution of (3).

In addition, Ambrosetti et al. [8] discussed the existence of the below question.

$$\left\{\begin{array}{cc}-▵v=\lambda v^q+v^p\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v=0\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

with ${\displaystyle 0<q<1<p\le \frac{N+2}{N-2}}$.

Some other studies of the existence of Dirichlet boundary value problems can be found in [4,9,10,11,12,13,14] and the references therein. For the Poisson equations with nonlinear boundary conditions, we recall the following works presented in the literature (see [15,16,17,18] and the references therein). In [15], Garcia-Azorero and others discussed the concave–convex problem with the nonlinear boundary conditions.

$$\left\{\begin{array}{cc}{\displaystyle -▵v+v={\left|v\right|}^{p-2}v}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla v={\lambda \left|v\right|}^{q-2}v}\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

where ${\displaystyle q\in (1,\frac{2(N-1)}{N-2})}$, ${\displaystyle p\in (1,\frac{2N}{N-2}]}$ and $\lambda \in (0,+\infty )$. If $1<q<2<p$ and $\lambda$ is small, there exist two positive solutions, and for large $\lambda$ there is no positive solution.

In thermal explosion theory, Ko and Prashanth [17] proved that the two-dimensional elliptic equations

$$\left\{\begin{array}{cc}{\displaystyle -▵v=\lambda e^{v^{\alpha }}}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla v+g\left(v\right)v=0}\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

have a positive solution which is not unique, for $\alpha \in (0,2]$. In [18], Yu and Yan showed that there is a positive solution of the problem

$$\left\{\begin{array}{cc}{\displaystyle -▵v+\frac{K\left(x\right)}{v^{\alpha }}=\lambda v^p}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v>0\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla v+g\left(v\right)v=0}\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

where $\alpha ,p\in (0,1)$. Among them, the authors discuss three cases of $K\left(x\right)$ (positive function, negative function and sign changing function).

Gordon et al. [16] proved the uniqueness and variety of positive solutions for the problem below.

$$\left\{\begin{array}{cc}{\displaystyle -▵v=\lambda f\left(v\right)}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla v+c\left(v\right)v=0}\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

where $f:[0,\infty )\to (0,\infty )\in C^1$, $\lambda >0$ is a parameter and $c\in C^{1,\gamma }$ is a non subtractive function defined on $[0,\infty )$, which satisfies $c\left(0\right)>0$. In addition, ${\displaystyle \underset{s\to \infty }{lim}\frac{f\left(s\right)}{s}=0}$.

Differently from the above papers, consider the problem (1) in which $f(x,u)=\lambda u^q+u^r$, with ${\displaystyle 1<q<1<r<\frac{N+2}{N-2}}$ and $g\left(u\right)$ satisfies the following assumptions.

Hypothesis 1

(H1). $g:[0,\infty )\to (0,\infty )$ is a increasing $C^1$ function and satisfying $g\left(0\right)=g_0>0$.

Hypothesis 2

(H2). There exists $C_0$ such that for any $y>0$, we have

$${\int }_0^{y}g\left(s\right)sds\ge \frac{1}{r+1}g\left(y\right)y^2+C_0{y}^2.$$

Hypothesis 3

(H3). g satisfying

$$\underset{s\to +\infty }{lim}\frac{g\left(s\right)}{s^{r-1}}=0.$$

Remark 1.(H3) indicates that the highest power of g is less than $r-1$. The function g satisfying this assumption exists. For example, $g\left(s\right)=s^k+1$ with $k<r-1$, so there exists ${\displaystyle 0<C_0<\frac{r-1}{2(r+1)}}$ such that

$$\begin{array}{c}{\displaystyle {\int }_0^{y}g\left(s\right)sds=\frac{y^2}{2}+\frac{y^{k+2}}{k+2}\ge \frac{y^2}{2}+\frac{y^{k+2}}{r+1}}\hfill \\ {\displaystyle =C_0{y}^2+\frac{y^2}{2}-C_0{y}^2+\frac{y^{k+2}}{r+1}>C_0{y}^2+(\frac{1}{2}-\frac{r-1}{2(r+1)})y^2+\frac{y^{k+2}}{r+1}}\hfill \\ {\displaystyle =\frac{1}{r+1}{y}^2+\frac{y^{k+2}}{r+1}+C_0{y}^2}\hfill \\ {\displaystyle =\frac{1}{r+1}g\left(y\right)y^2+C_0{y}^2}\hfill \end{array}$$

is true.

It is well known that the sub- and super-solutions method is an important tool for solving the existence of initial and boundary value problems (see [19,20,21,22,23]). In this paper, using the sub- and super-solutions method, we present some new results on the existence of positive solutions for problem (1).

The definition of energy functional corresponding to the problem (1) is introduced.

$$I_{\lambda }\left(u\right)=\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u|}^{2}dx-{\int }_{\mathsf{\Omega }}F\left(u\right)dx+{\int }_{\partial \mathsf{\Omega }}G\left(u\right)d\sigma ,$$

(4)

with ${\displaystyle 0<q<1<r<\frac{N+2}{N-2}}$. ${\displaystyle F\left(u\right):={\int }_0^u\lambda s^q+s^{r}ds}$ and ${\displaystyle G\left(u\right):={\int }_0^{u^{+}}g\left(s\right)sds}$, where the symbol $d\sigma$ denotes the surface measure of $\partial \mathsf{\Omega }$.

Definition 1.

$u\in H^1\left(\mathsf{\Omega }\right)$is a weak solution of (1) if it satisfies

$${\int }_{\mathsf{\Omega }}\nabla u·\nabla \varphi dx={\int }_{\mathsf{\Omega }}(\lambda u^q\varphi +u^r\varphi )dx-{\int }_{\partial \mathsf{\Omega }}g\left(u\right)u\varphi d\sigma ,$$

for any $\varphi \in H^1\left(\mathsf{\Omega }\right)$.

More precisely, $u\in H^1\left(\mathsf{\Omega }\right)$ is a weak solution of (1) if and only if $u\in H^1\left(\mathsf{\Omega }\right)$ is a critical point of $I_{\lambda }$ and u is a positive solution.

Finally, the following results are obtained.

Theorem 1.

Let ${\displaystyle q\in (0,1)}$, ${\displaystyle r\in (1,\frac{N+2}{N-2})}$ and g satisfy (H1), (H2) and (H3)

(i) There exists $\mathsf{\Lambda }\in \mathbb{R}$ and $\mathsf{\Lambda }>0$, such that (1) has at least one positive solution for $\lambda =\mathsf{\Lambda }$. There is no positive solution in (1), for $\lambda \in (\mathsf{\Lambda },+\infty )$. There are at least two positive solutions in (1), for $0<\lambda <\mathsf{\Lambda }$.

(ii) For $0<\lambda <\mathsf{\Lambda }$, (1) has a minimal positive solution $u_{\lambda }$, and the map $\lambda \mapsto u_{\lambda }(·)$ is increasing. Moreover, for $\lambda \in (0,\mathsf{\Lambda })$, $I_{\lambda }$ has a local minimum near zero.

This paper is divided into the following sections. In the second section, we list and show several lemmas that can be widely applied. The Lemmas proposed in the third and fourth part are proved under the condition of Theorem 1 and prepare us for the proof of Theorem 1. The fifth part focuses on proving our results.

2. Preliminaries

In this section, we rephrase problem (1) in the general form

$$\left\{\begin{array}{cc}{\displaystyle -▵u=f(x,u)}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ u>0\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u+g\left(u\right)u=0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

(5)

The corresponding definitions of sub-solution and super-solution are given as follows:

Definition 2.

A function ${\chi }_2:\overline{\mathsf{\Omega }}\to \mathbb{R}$ is called a super-solution of (5) if ${\chi }_2\in C^2\left(\mathsf{\Omega }\right)\bigcap C\left(\overline{\mathsf{\Omega }}\right)$ and

$$\left\{\begin{array}{cc}{\displaystyle ▵{\chi }_2+f(x,{\chi }_2)\le 0}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla {\chi }_2+g\left({\chi }_2\right){\chi }_2\ge 0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

Definition 3.

A function ${\chi }_1:\overline{\mathsf{\Omega }}\to \mathbb{R}$ is called a sub-solution of (5) if ${\chi }_1\in C^2\left(\mathsf{\Omega }\right)\bigcap C\left(\overline{\mathsf{\Omega }}\right)$ and

$$\left\{\begin{array}{cc}{\displaystyle ▵{\chi }_1+f(x,{\chi }_1)\ge 0}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla {\chi }_1+g\left({\chi }_1\right){\chi }_1\le 0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

Lemma 1

(see [24]). Let $f\in \overline{\mathsf{\Omega }}\times \mathbb{R}$ and ${\displaystyle \frac{\partial f}{\partial u}}$ be continuous and g satisfy $\left(H1\right)$. If μ and χ are the sub- and super-solutions of problem (5), respectively, such that $\mu \le \chi$, then problem (5) has at least one solution u satisfying

$$\mu \le u\le \chi ,\mathrm{on}\overline{\mathsf{\Omega }}.$$

From the above lemma, it can be seen that if you want to obtain the solution by the sub- and super-solution method, you must prove that the sub-solution is less than or equal to the super-solution.

In order to compare the sub- and the super-solution more conveniently, the following comparison lemma is proposed.

Lemma 2

(see [17]). Let $w_1,w_2\in C^{2,\beta }\left(\mathsf{\Omega }\right)\bigcap C^{1,\beta }\left(\overline{\mathsf{\Omega }}\right)$ satisfy $-▵w_1\le -▵w_2$, in $\mathsf{\Omega }$, ${\displaystyle \mathbf{n}·\nabla w_1+g\left(w_1\right)w_1\le \mathbf{n}·\nabla w_2+g\left(w_2\right)w_2}$, on $\partial \mathsf{\Omega }$. Then $w_1<w_2$ in $\overline{\mathsf{\Omega }}$.

Lemma 3

(see [18]). Let $f:\overline{\mathsf{\Omega }}\times (0,\infty )\to \mathbb{R}$ be a continuous function such that ${\displaystyle \frac{f(x,s)}{s}}$ is strictly decreasing in $(0,\infty )$. Let ${\varpi }_1,{\varpi }_2\in C^2\left(\mathsf{\Omega }\right)\bigcap C\left(\overline{\mathsf{\Omega }}\right)$ satisfy:

(a) 

${\displaystyle ▵{\varpi }_2+f(x,{\varpi }_2)\ge 0\ge ▵{\varpi }_1+f(x,{\varpi }_1),\mathrm{in}\mathsf{\Omega }}$;

(b) 

${\displaystyle \mathbf{n}·\nabla {\varpi }_1+g\left({\varpi }_1\right){\varpi }_1\ge c\ge \mathbf{n}·\nabla {\varpi }_2+g\left({\varpi }_2\right){\varpi }_2,on\partial \mathsf{\Omega }}$ with c a nonnegative constant and ${\varpi }_1,{\varpi }_2>0\mathrm{in}\mathsf{\Omega }$;

(c) 

$▵{\varpi }_2\in L^1\left(\mathsf{\Omega }\right)$. Then ${\varpi }_1\ge {\varpi }_2$ in $\overline{\mathsf{\Omega }}$.

Let $H^1\left(\mathsf{\Omega }\right)=\{u:u\in L^2(\partial \mathsf{\Omega }),\nabla u\in L^2\left(\mathsf{\Omega }\right)\}$. We have the following norm:

$${\parallel u\parallel }_{H^1\left(\mathsf{\Omega }\right)}^2:={\int }_{\mathsf{\Omega }}{|\nabla u|}^{2}dx+b{\int }_{\partial \mathsf{\Omega }}{\left|u\right|}^{2}d\sigma ,$$

where b is a positive constant. For convenience, take $b=1$ in the following proof process.

Remark 2.

Thanks to the trace imbedding and the imbedding of Cherrier (see [25,26,27]), it follows that ${\parallel ·\parallel }_{H^1}$ is indeed an equivalent norm in $H^1\left(\mathsf{\Omega }\right)$. In other words, there are $M_1$ and $M_2>0$ such that

$$M_1{\parallel \nu \parallel }_{H_0^1\left(\mathsf{\Omega }\right)}\le {\parallel \nu \parallel }_{H^1\left(\mathsf{\Omega }\right)}\le M_2{\parallel \nu \parallel }_{H_0^1\left(\mathsf{\Omega }\right)},\forall \nu \in H^1\left(\mathsf{\Omega }\right)$$

where ${\displaystyle {\parallel \nu \parallel }_{H_0^1\left(\mathsf{\Omega }\right)}^2={\int }_{\mathsf{\Omega }}{\left(\right|\nabla \nu |}^2+{\left|\nu \right|}^2)dx}$ and $H_0^1\left(\mathsf{\Omega }\right)=\{v:v\in L^2\left(\mathsf{\Omega }\right),\nabla v\in L^2\left(\mathsf{\Omega }\right)\}$.

In the proof, we will apply the next result.

Lemma 4

(see [28]). (Rellich–Kondrachov Compactness Theorem) Assume $\mathsf{\Omega }$ is a bounded open subset of ${\mathbb{R}}^N$ and $\partial \mathsf{\Omega }$ is $C^1$. Suppose $1\le p<N$. Then,

$$W^{1,p}\left(\mathsf{\Omega }\right)\subset \subset L^q\left(\mathsf{\Omega }\right),$$

for each ${\displaystyle 1\le q<p\ast =\frac{Np}{N-p}}$.

When studying the nonlinear problems on the boundary, we should also pay attention to the following embedding conditions on the boundary.

Lemma 5

(see [29]). Let $\mathsf{\Omega }$ be a smooth bounded domain in ${\mathbb{R}}^N$, $N\ge 2$. For any $p>1$, with ${\displaystyle p\le \frac{N}{N-2}}$ if $N\ge 3$, we have the validity of the Sobolev trace embedding of $H^1\left(\mathsf{\Omega }\right)$ into $L^p(\partial \mathsf{\Omega })$; namely, there exists a positive constant S such that

$${S\parallel u\parallel }_{L^{p+1}(\partial \mathsf{\Omega })}\le {\parallel u\parallel }_{H^1\left(\mathsf{\Omega }\right)},$$

for all $u\in H^1\left(\mathsf{\Omega }\right)$.

In order to construct a sub-solution, the following boundary value problem will be used.

$$\left\{\begin{array}{cc}-▵v=\lambda v^q\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v>0\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla v+g\left(v\right)v=\rho }\hfill & \mathrm{on}\partial \mathsf{\Omega }\hfill \end{array}\right.$$

(6)

where $q\in (0,1)$ and $\rho \ge 0$.

Lemma 6

(see [18]). Let $q\in (0,1)$, $\lambda >0$. The problem (6) has only solution $\vartheta \in C\left(\overline{\mathsf{\Omega }}\right)\cap C^{2,\beta }\left(\mathsf{\Omega }\right)$ and $\vartheta >0$ on $\partial \mathsf{\Omega }$.

The second solution of (1) is proved by variational method. The following lemma will be used.

Lemma 7

(see [30,31]). Let F be a functional on a Banach space X, $F\in C^1(X,\mathbb{R})$. Let us assume that there exists $r,R>0$ such that

(i)

$F\left(u\right)>r$and$\forall u\in X$with$\parallel u\parallel =R$;

(ii)

$F\left(0\right)=0$and$F\left(w_0\right)<r$for some$w_0\in X$with$\parallel w_0\parallel >R$.

Let us define $\mathsf{\Gamma }=\gamma \in C([0,1],X):\gamma \left(0\right)=0,\gamma \left(l\right)=w_0$ and

$${\displaystyle c=\underset{\gamma \in \mathsf{\Gamma }}{inf}\underset{t\in [0,1]}{max}F\left(\gamma \left(t\right)\right).}$$

Then, there exists a sequence $\left\{u_j\right\}\in X$ such that $F\left(u_j\right)\to c$ and $F^{\prime }\left(u_j\right)\to 0$ in $X^{*}$ (dual of X).

3. Constraints of $\mathbf{\lambda }$ When Solutions Exist

We define

$$\mathsf{\Lambda }=sup\{\lambda >0:(1\left)\mathrm{has}\mathrm{a}\mathrm{positive}\mathrm{solution}\right\}.$$

(7)

Lemma 8.

$0<\mathsf{\Lambda }<\infty$.

Proof. 

Let e be a solution of

$$\left\{\begin{array}{cc}-▵u=1\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u+g_{0}u=0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

Since $0<q<1<r$, we can seek out ${\lambda }_0$ such that for all $0<\lambda \le {\lambda }_0$ there exists $M=M\left(\lambda \right)>0$ satisfying

$${\displaystyle M\ge \lambda M^q{e}^q+M^r{e}^r.}$$

Then, the function $Me>0$ verifies

$${\displaystyle -▵\left(Me\right)=M\ge \lambda {\left(Me\right)}^q+{\left(Me\right)}^r,\mathrm{in}\mathsf{\Omega }}$$

and

$${\displaystyle \mathbf{n}·\nabla Me+g\left(Me\right)Me\ge M(\mathbf{n}·\nabla e+g_{0}e)=0,\mathrm{on}\partial \mathsf{\Omega }.}$$

It guarantees that $Me$ is a super-solution of (1).

In addition, in order to apply Lemma 1, the existence of sub-solutions needs to be confirmed. For $\epsilon >0$ small enough, the above discussion can deduce

$${\displaystyle \lambda {\left(\epsilon \vartheta \right)}^q={\left(\epsilon \vartheta \right)}^q=-▵\left(\epsilon \vartheta \right)\le \lambda {\epsilon }^q{\vartheta }^q+{\epsilon }^r{\vartheta }^r.}$$

On $\partial \mathsf{\Omega }$, since g is a nondecreasing $C^1$ function,

$${\displaystyle \mathbf{n}·\nabla \epsilon \vartheta +g\left(\epsilon \vartheta \right)\epsilon \vartheta \le \epsilon (\mathbf{n}·\nabla \vartheta )+g\left(\vartheta \right)\epsilon \vartheta =\epsilon (\mathbf{n}·\nabla \vartheta +g(\vartheta \left)\vartheta \right)=0.}$$

Therefore, $\epsilon \vartheta$ is a sub-solution of problem (1).

Let $\epsilon$ be sufficiently small to satisfy $\epsilon \vartheta <Me$. Therefore, by Lemma 1, problem (1) admits a positive solution u such that

$$\epsilon \vartheta \le u\le Me,$$

whenever $\lambda \le {\lambda }_0$ and thus $\mathsf{\Lambda }\ge {\lambda }_0$.

Next, prove that $\mathsf{\Lambda }$ is finite; namely, there is a positive constant $\overline{\lambda }$ such that $\mathsf{\Lambda }<\overline{\lambda }$.

The following eigenvalue problem,

$$\left\{\begin{array}{cc}-▵v=\lambda v\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ v=0\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

(8)

and ${\lambda }_1$ and ${\phi }_1$ are the corresponding minimum eigenvalue and eigenfunction respectively. If u is a positive solution of (1) corresponding to parameter $\lambda$, then

$${\displaystyle {\int }_{\mathsf{\Omega }}-▵u·{\phi }_{1}dx={\int }_{\mathsf{\Omega }}\lambda u^q{\phi }_{1}dx+{\int }_{\mathsf{\Omega }}{u}^r{\phi }_{1}dx,}$$

where ${\phi }_1$ is solution of (8).

Through the computations and ${\phi }_1=0$ on $\partial \mathsf{\Omega }$, obtain

$$\begin{array}{c}{\displaystyle {\int }_{\mathsf{\Omega }}\lambda u^q{\phi }_{1}dx+{\int }_{\mathsf{\Omega }}{u}^r{\phi }_{1}dx}\hfill \\ {\displaystyle ={\int }_{\mathsf{\Omega }}-▵u·{\phi }_{1}dx}\hfill \\ {\displaystyle ={\int }_{\partial \mathsf{\Omega }}-\frac{\partial u}{\partial \mathbf{n}}·{\phi }_{1}d\sigma +{\int }_{\mathsf{\Omega }}\nabla u\nabla {\phi }_{1}dx}\hfill \\ {\displaystyle =0+{\int }_{\partial \mathsf{\Omega }}u\frac{\partial {\phi }_1}{\partial \mathbf{n}}d\sigma -{\int }_{\mathsf{\Omega }}u▵{\phi }_{1}dx}\hfill \\ {\displaystyle ={\int }_{\partial \mathsf{\Omega }}u\frac{\partial {\phi }_1}{\partial \mathbf{n}}d\sigma +{\int }_{\mathsf{\Omega }}u{\lambda }_1{\phi }_{1}dx.}\hfill \end{array}$$

By taking into account that ${\displaystyle \frac{\partial {\phi }_1}{\partial \mathbf{n}}<0}$, we have

$$\begin{array}{cc}{\displaystyle {\lambda }_1{\int }_{\mathsf{\Omega }}u{\phi }_{1}dx}\hfill & {\displaystyle =\lambda {\int }_{\mathsf{\Omega }}{u}^q{\phi }_{1}dx+{\int }_{\mathsf{\Omega }}{u}^r{\phi }_{1}dx-{\int }_{\partial \mathsf{\Omega }}u\frac{\partial {\phi }_1}{\partial \mathbf{n}}d\sigma ,}\hfill \\ \hfill & {\displaystyle \ge \lambda {\int }_{\mathsf{\Omega }}{u}^q{\phi }_{1}dx+{\int }_{\mathsf{\Omega }}{u}^r{\phi }_{1}dx.}\hfill \end{array}$$

(9)

Let $\overline{\lambda }$ satisfy

$${\displaystyle \overline{\lambda }{t}^q+t^r>{\lambda }_{1}t,\forall t\in \mathbb{R},t>0.}$$

Since ${\phi }_1>0$ in $\mathsf{\Omega }$,

$${\displaystyle {\int }_{\mathsf{\Omega }}(\overline{\lambda }{u}^q+u^r){\phi }_{1}dx>{\int }_{\mathsf{\Omega }}{\lambda }_{1}u{\phi }_{1}dx.}$$

(10)

The previous relations (9) and (10) imply that

$${\displaystyle {\int }_{\mathsf{\Omega }}(\overline{\lambda }{u}^q+u^r){\phi }_{1}dx>{\int }_{\mathsf{\Omega }}{\lambda }_{1}u{\phi }_{1}dx\ge \lambda {\int }_{\mathsf{\Omega }}{u}^q{\phi }_{1}dx+{\int }_{\mathsf{\Omega }}{u}^r{\phi }_{1}dx}$$

and $\lambda <\overline{\lambda }$. Hence, $\mathsf{\Lambda }\le \overline{\lambda }$ and $0<\mathsf{\Lambda }<\infty$. □

Lemma 9.

Under the assumptions of Theorem 1, the solution of problem (1) exists for all $0<\lambda <\mathsf{\Lambda }$.

Proof. 

Given $\lambda <\mathsf{\Lambda }$, from the definition of upper bound, there exists ${\lambda }_0>0$ such that $\lambda <{\lambda }_0<\mathsf{\Lambda }$ when $\lambda ={\lambda }_0$.

Since

$${\displaystyle -▵u_{{\lambda }_0}={\lambda }_0{u}_{{\lambda }_0}^q+u_{{\lambda }_0}^r>\lambda u_{{\lambda }_0}^q+u_{{\lambda }_0}^r,\mathrm{in}\mathsf{\Omega }}$$

and

$${\displaystyle \mathbf{n}·\nabla u_{{\lambda }_0}+g\left(u_{{\lambda }_0}\right)u_{{\lambda }_0}=0,\mathrm{on}\partial \mathsf{\Omega };}$$

therefore, $u_{{\lambda }_0}$ is a super-solution for (1) when the parameter is $\lambda$.

As $\epsilon >0$ is small enough to ensure $\epsilon \vartheta <u_{{\lambda }_0}$, by exploiting Lemma 1, there is a positive solution u of (1) satisfying $\epsilon \vartheta \le u\le u_{{\lambda }_0}$ for all $0<\lambda <\mathsf{\Lambda }$. □

When using variational method to solve such problems, we can usually refer to weak solutions and to the energy functional (4) associated with problem (1).

It is easy to verify $I_{\lambda }\in C^1(H^1\left(\mathsf{\Omega }\right),\mathbb{R})$ and

$$\begin{array}{cc}(I_{\lambda }^{\prime }\left(u\right),v)\hfill & {\displaystyle ={\int }_{\mathsf{\Omega }}\nabla u·\nabla vdx-{\int }_{\mathsf{\Omega }}(\lambda u^q+u^r)vdx+{\int }_{\partial \mathsf{\Omega }}g\left(u\right)uvd\sigma ,}\hfill \end{array}$$

for any $u,v\in H^1\left(\mathsf{\Omega }\right)$. Here $\lambda \in (0,\mathsf{\Lambda })$, $\mathsf{\Lambda }$ is given by the definition (7).

Lemma 10.

Suppose that the minimal positive solution of problem (1) exists. Then, $u_1<u_2$ for $l_1<l_2$ and $l_1,l_2\in (0,\mathsf{\Lambda })$. Here, $u_1$ is the minimal positive solution of problem (1) for $\lambda =l_1$.

Proof. 

Indeed, if $l_1<l_2$ then

$$-▵u_2=l_2{u}_2^q+u_2^r\ge l_1{u}_2^q+u_2^r,\mathrm{in}\mathsf{\Omega }$$

and $u_2$ is a sub-solution of (1) satisfying $\epsilon \vartheta <u_2$ for a small enough $\epsilon >0$. By Lemma 1, problem (1) has a positive solution. We obtain $\epsilon \vartheta \le u_1\le u_2$ by Lemma 1.

Hence, we get $u_1<u_2$ by $u_1\not\equiv u_2$ and the strong maximum principle.

From

$$-▵u_2=l_2{u}_2^q+u_2^r\ge l_1{u}_1^q+u_1^r=-▵u_1,\mathrm{in}\mathsf{\Omega }$$

and

$$\mathbf{n}·\nabla u_2+g\left(u_2\right)u_2=0=\mathbf{n}·\nabla u_1+g\left(u_1\right)u_1,\mathrm{on}\partial \mathsf{\Omega },$$

it can be deduced that $u_1<u_2$ in $\overline{\mathsf{\Omega }}$ by Lemma 2. □

Lemma 11.

For all $\lambda \in (0,\mathsf{\Lambda })$, problem (1) has a positive solution u. Thus, $I_{\lambda }$ obtains a local minimum in the $C^1$ topology.

Proof. 

There exists ${\lambda }_1$ such that ${\displaystyle \underline{\lambda }<{\lambda }_1<\mathsf{\Lambda }}$ and the minimal positive solution $u_1=u_{{\lambda }_1}$ defined in Lemma 10 and $u_1>0$ in $\overline{\mathsf{\Omega }}$ by (7). Let ${\displaystyle u_{\underline{\lambda }}}$ be the unique positive solution of (6) with $\rho =0$

$$\left\{\begin{array}{cc}{\displaystyle -▵u=\underline{\lambda }{u}^q}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u+g\left(u\right)u=0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

Since ${\displaystyle \frac{s^q}{s}=s^{q-1}}$ is strictly decreasing for each $0<s<\infty$, we have ${\displaystyle ▵u_{\underline{\lambda }}+\underline{\lambda }{u}_{{\underline{\lambda }}^q}=0\ge -u_1^r={\lambda }_1{u}_1^q+▵u_1}$ in $\mathsf{\Omega }$ and ${\displaystyle \mathbf{n}·\nabla u_{\underline{\lambda }}+g\left(u_{\underline{\lambda }}\right)u_{\underline{\lambda }}=0=\mathbf{n}·\nabla u_1+g\left(u_1\right)u_1}$ on $\partial \mathsf{\Omega }$. Therefore, ${\displaystyle u_{\underline{\lambda }}\le u_1}$ in $\overline{\mathsf{\Omega }}$ by the Hopf maximum principle and Lemma 3.

Let us define the following cut-off nonlinear function:

$${\widehat{f}}_{\lambda }(x,t)=\left\{\begin{array}{cc}{\displaystyle \lambda {\left(u_{\underline{\lambda }}\left(x\right)\right)}^q+{\left(u_{\underline{\lambda }}\left(x\right)\right)}^r}\hfill & \mathrm{if}t\le u_{\underline{\lambda }}\left(x\right),\hfill \\ {\displaystyle \lambda t^q+t^r}\hfill & \mathrm{if}{u}_{\underline{\lambda }}\left(x\right)<t<u_1\left(x\right).\hfill \\ {\displaystyle \lambda {\left(u_1\left(x\right)\right)}^q+{\left(u_1\left(x\right)\right)}^r}\hfill & \mathrm{if}t\ge u_1\left(x\right)\hfill \end{array}\right.$$

and ${\displaystyle {\widehat{F}}_{\lambda }(x,u)={\int }_0^u{\widehat{f}}_{\lambda }(x,t)dt}$, $x\in \mathsf{\Omega }$. Then, ${\widehat{I}}_{\lambda }:H^1\left(\mathsf{\Omega }\right)\to \mathbb{R}$ is given by

$${\widehat{I}}_{\lambda }\left(u\right)=\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u|}^{2}dx-{\int }_{\mathsf{\Omega }}{\widehat{F}}_{\lambda }(x,u)dx+{\int }_{\partial \mathsf{\Omega }}G\left(u\right)d\sigma .$$

This functional is coercive and bounded from below. Obviously, ${\widehat{I}}_{\lambda }\left(u\right)=I_{\lambda }\left(u\right)$ when $u_{\underline{\lambda }}\left(x\right)<u\left(x\right)<u_1\left(x\right)$. Let $u_{\lambda }$ be a global minimizer of ${\widehat{I}}_{\lambda }$ on $H^1\left(\mathsf{\Omega }\right)$. Then, $u_{\lambda }$ is the solution of

$$\left\{\begin{array}{cc}{\displaystyle -▵u={\widehat{f}}_{\lambda }(x,u)}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u+g\left(u\right)u=0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

Through the define of ${\widehat{f}}_{\lambda }(x,t)$, we obtain ${\displaystyle {\widehat{f}}_{\lambda }(x,u_{\underline{\lambda }})\le {\widehat{f}}_{\lambda }(x,u_{\lambda })\le {\widehat{f}}_{\lambda }(x,u_1)}$ and ${\displaystyle u_{\underline{\lambda }}<u_{\lambda }<u_1}$ in $\overline{\mathsf{\Omega }}$ by Lemma (2). In addition, $u_{\lambda }$ is a solution of (3).

Let ${\displaystyle \Xi =min\{\underset{x\in \overline{\mathsf{\Omega }}}{min}|u_{\underline{\lambda }}\left(x\right)-u_{\lambda }\left(x\right)|,\underset{x\in \overline{\mathsf{\Omega }}}{min}|u_1\left(x\right)-u_{\lambda }\left(x\right)|\}}$. On the set ${\displaystyle \{u:\parallel u-u_{\lambda }{\parallel }_{C^1}<\frac{\Xi }{2}\}}$, ${\widehat{I}}_{\lambda }=I_{\lambda }$. Hence, $u_{\lambda }$ is a local minimal for $I_{\lambda }$. □

Remark 3.

We observe that $u_{\lambda }$ has negative energy $I_{\lambda }\left(u_{\lambda }\right)<0$.

In fact, $I_{\lambda }\left(u_{\lambda }\right)\le I\left(u_1\right)$, ${\widehat{I}}_{\lambda }\left(u_1\right)=I\left(u_1\right)$ and

$$\begin{array}{c}{\displaystyle ({\widehat{I}}_{\lambda }^{\prime }\left(u_1\right),u_1)={\int }_{\mathsf{\Omega }}{|\nabla u_1|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}g\left(u_1\right)u_1^{2}d\sigma -{\int }_{\mathsf{\Omega }}\lambda u_1^{q+1}+u_1^{r+1}dx=0.}\hfill \end{array}$$

Hence we have

$$\begin{array}{cc}{\displaystyle I_{\lambda }\left(u_{\lambda }\right)}\hfill & \le I_{\lambda }\left(u_1\right)={\widehat{I}}_{\lambda }\left(u_1\right)\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u_1|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}{\int }_0^{u_1}g\left(s\right)sdsd\sigma -{\int }_{\mathsf{\Omega }}{\int }_0^{u_1}\lambda u_1^q+u_1^{r}dsdx}\hfill \\ \hfill & {\displaystyle \le \frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla u_1{|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}g\left(\right|u_1\left|\right)\frac{u_1^2}{2}d\sigma -{\int }_{\mathsf{\Omega }}\lambda u_1^{q+1}+u_1^{r+1}dx}\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}\lambda u_1^{q+1}+u_1^{r+1}dx-{\int }_{\mathsf{\Omega }}\lambda u_1^{q+1}+u_1^{r+1}dx}\hfill \\ \hfill & {\displaystyle =-\frac{1}{2}{\int }_{\mathsf{\Omega }}\lambda u_1^{q+1}+u_1^{r+1}dx<0.}\hfill \end{array}$$

4. The Second Solution

The proof of the existence of the second solution is very long. For the convenience of readers, it will be proved separately. Next, let us prove an important result about bounded $PS$ sequences.

Lemma 12.

Let $\left\{u_n\right\}\subset H^1\left(\mathsf{\Omega }\right)$ be a bounded (PS) sequence for ${\overline{I}}_{\lambda }$ which is defined by Equation ((12)). Then, $u_n\to u$ in $H^1\left(\mathsf{\Omega }\right)$.

Proof. 

Indeed, the (PS) sequence $\left\{u_n\right\}$ of ${\overline{I}}_{\lambda }$ satisfies

$$|{\overline{I}}_{\lambda }\left(u_n\right)|\le C\mathrm{and}{\overline{I}}_{\lambda }^{\prime }\left(u_n\right)\to 0,\mathrm{as}n\to \infty .$$

Since $\left\{u_n\right\}\subset H^1\left(\mathsf{\Omega }\right)$ is bounded, we get that $\left\{u_n\right\}$ weakly converges to u; i.e., $u_n\rightharpoonup u$, and

$$({\overline{I}}_{\lambda }^{\prime }\left(u\right),u_n-u)\to 0,\mathrm{as}n\to \infty .$$

Obviously, it is

$$({\overline{I}}_{\lambda }^{\prime }\left(u_n\right),u_n-u)\to 0,\mathrm{as}n\to \infty .$$

Combined with the previous two equations, there are

$$({\overline{I}}_{\lambda }^{\prime }\left(u_n\right)-{\overline{I}}_{\lambda }^{\prime }\left(u\right),u_n-u)\to 0,\mathrm{as}n\to \infty .$$

$\parallel u_n{-u\parallel }_{L^2\left(\mathsf{\Omega }\right)}\to 0$ and $\parallel u_n{-u\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}\to 0$ since ${\displaystyle 2<r+1<\frac{2N}{N-2}}$.

For $u_0$ as the fixed parameter defined by Lemma 11, a fixed parameter $\lambda$ and $|\lambda u^q|\le \lambda (\left|u\right|+1)$,

$$\begin{array}{c}{\displaystyle |{\int }_{\mathsf{\Omega }}\left(\lambda \right|u_n+u_0{|}^q-\lambda |u+u_0{|}^q\left)(u_n-u)dx\right|}\hfill \\ {\displaystyle \le \lambda {\int }_{\mathsf{\Omega }}|u_n-u\left|\right(|u_n+u_0{|}^q+|u+u_0{|}^q)dx}\hfill \\ {\displaystyle \le \lambda {\int }_{\mathsf{\Omega }}|u_n-u\left|\right(2+|u_n+u_0|+|u+u_0\left|\right)dx}\hfill \\ {\displaystyle \le 2\lambda ·meas\left(\mathsf{\Omega }\right)\parallel u_n{-u\parallel }_{L^2\left(\mathsf{\Omega }\right)}+\lambda ({\int }_{\mathsf{\Omega }}|u_n+u_0{|}^2{dx)}^{\frac{1}{2}}({\int }_{\mathsf{\Omega }}|u_n-u{{|}^{2}dx)}^{\frac{1}{2}}}\hfill \\ {\displaystyle +\lambda ({\int }_{\mathsf{\Omega }}|u+u_0{|}^2{dx)}^{\frac{1}{2}}({\int }_{\mathsf{\Omega }}|u_n-u{{|}^{2}dx)}^{\frac{1}{2}}}\hfill \\ {\displaystyle =2\lambda ·meas\left(\mathsf{\Omega }\right)\parallel u_n{-u\parallel }_{L^2\left(\mathsf{\Omega }\right)}+\lambda (\parallel u_n+u_0{\parallel }_{L^2\left(\mathsf{\Omega }\right)}+\parallel u+u_0{\parallel }_{L^2\left(\mathsf{\Omega }\right)})\parallel u_n{-u\parallel }_{L^2\left(\mathsf{\Omega }\right)}}\hfill \\ \to 0,\mathrm{as}n\to \infty \hfill \end{array}$$

by $H\ddot{o}lder$ inequality. Thus, we get

$${\int }_{\mathsf{\Omega }}\left(\lambda \right|u_n+u_0{|}^q-\lambda |u+u_0{|}^q)(u_n-u)dx=o\left(1\right),\mathrm{as}n\to \infty .$$

Similarly, it can be inferred that

$$\begin{array}{c}{\displaystyle |{\int }_{\mathsf{\Omega }}|u_n+u_0{|}^r-|u+u_0{|}^r\left)(u_n-u)dx\right|}\hfill \\ {\displaystyle \le (\parallel u_n+u_0{\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}^r+\parallel u+u_0{\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}^r)\parallel u_n{-u\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}}\hfill \\ \to 0,\mathrm{as}n\to \infty .\hfill \end{array}$$

By exploiting the above relations, the $H\ddot{o}lder$ inequality and assumption ($H_1$), we obtain

$$\left[g\left(s_1\right)-g\left(s_2\right)\right](s_1-s_2)\ge 0$$

and $\overline{g}(x,s)=0$ for $s<0$.

$$\begin{array}{c}o\left(1\right)=(I_{\lambda }^{\prime }\left(u_n\right)-I_{\lambda }^{\prime }\left(u\right),u_n-u)\hfill \\ {\displaystyle ={\int }_{\mathsf{\Omega }}{|\nabla (u_n-u)|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}(\overline{g}(x,u_n)-\overline{g}(x,u))(u_n-u)d\sigma }\hfill \\ {\displaystyle -{\int }_{\mathsf{\Omega }}\left[\lambda |u_n+u_0{|}^q-\lambda u_0^q-\left(\right|u+u_0{|}^q-\lambda u_0^q)\right](u_n-u)dx}\hfill \\ {\displaystyle -{\int }_{\mathsf{\Omega }}\left[|u_n+u_0{|}^r-u_0^r-\left(\right|u+u_0{|}^r-u_0^r)\right](u_n-u)dx}\hfill \\ {\displaystyle ={\int }_{\mathsf{\Omega }}{|\nabla (u_n-u)|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\left(g(u_n+u_0)(u_n+u_0)-g(u+u_0)(u+u_0)\right)(u_n-u)d\sigma }\hfill \\ {\displaystyle -{\int }_{\mathsf{\Omega }}\left(\lambda |u_n+u_0{|}^q-{|u+u_0|}^q\right)(u_n-u)dx-{\int }_{\mathsf{\Omega }}\left(|u_n+u_0{|}^r-{|u+u_0|}^r\right)(u_n-u)dx}\hfill \\ {\displaystyle \ge {\int }_{\mathsf{\Omega }}{|\nabla (u_n-u)|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}(g(u_n+u_0)(u_n+u_0)-g(u_n+u_0)(u+u_0))(u_n-u)d\sigma }\hfill \\ {\displaystyle +{\int }_{\partial \mathsf{\Omega }}(g(u_n+u_0)(u+u_0)-g(u+u_0)(u+u_0))(u_n-u)d\sigma }\hfill \\ {\displaystyle \ge {\int }_{\mathsf{\Omega }}{|\nabla (u_n-u)|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}{g}_0{(u_n-u)}^{2}d\sigma }\hfill \\ {\displaystyle +{\int }_{\partial \mathsf{\Omega }}(g(u_n+u_0)-g(u+u_0))\left[(u_n+u_0)-(u+u_0)\right](u+u_0)d\sigma }\hfill \\ {\displaystyle \ge min\{1,g_0\}{\parallel u_n-u\parallel }_{H^1\left(\mathsf{\Omega }\right)}^2,\mathrm{as}n\to \infty .}\hfill \end{array}$$

Thus, $u_n$ strongly converges to u in $H^1\left(\mathsf{\Omega }\right)$. The proof is concluded. □

5. Proof of Theorem

Proof of Theorem 1(i).

The first part (i) is divided into two steps: we first prove the existence of the solution, and then prove whether the solution is unique.

Through Lemmas 8–10, the solution of problem (1) exists, for any $\lambda \in (0,\mathsf{\Lambda })$.

Firstly, the following argument shows the second solution of (1) exists. Let us look for a second positive solution of the form $u=u_0+v$, where $u_0=u_{\lambda }$ is the positive solution found in Lemma 11. The function v satisfies

$$\left\{\begin{array}{cc}-▵v=\overline{f}(x,v),\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ \mathbf{n}·\nabla v+\overline{g}(x,v)=0,\hfill & \mathrm{on}\partial \mathsf{\Omega }\hfill \end{array}\right.$$

(11)

with

$${\displaystyle \overline{f}(x,s\left(x\right))=\left\{\begin{array}{cc}\lambda {(u_0+s\left(x\right))}^q+{(u_0+s\left(x\right))}^r-\lambda u_0^q-u_0^r,\hfill & s\left(x\right)\ge 0,\hfill \\ 0,\hfill & s\left(x\right)<0,\hfill \end{array}\right.}$$

$${\displaystyle \overline{g}(x,s\left(x\right))=\left\{\begin{array}{cc}g(s\left(x\right)+u_0)(s\left(x\right)+u_0)-g\left(u_0\right)u_0,\hfill & s\left(x\right)\ge 0,\hfill \\ 0,\hfill & s\left(x\right)<0\hfill \end{array}\right.}$$

and

$${\displaystyle {\overline{I}}_{\lambda }\left(u\right)=\frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla u|dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}(x,u)d\sigma -{\int }_{\mathsf{\Omega }}\overline{F}(x,u)dx,}$$

(12)

where ${\displaystyle \overline{F}(x,u)={\int }_0^u\overline{f}(x,s)ds}$ and ${\displaystyle \overline{G}(x,u)={\int }_0^u\overline{g}(x,s)ds}$. For convenience, we write $s\left(x\right)$ as s. The second component on $\overline{f}(x,u)$ and $\overline{g}(x,u)$ represents a function of $u=u\left(x\right)$.

Note that ${\overline{I}}_{\lambda }\left(0\right)=0$ and $v=0$ is a local minimum of ${\overline{I}}_{\lambda }$ in $H^1\left(\mathsf{\Omega }\right)$. Let $v^{+}$ be the positive part of v. As

$$\begin{array}{c}{\displaystyle I_{\lambda }\left(u_0\right)=\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u_0|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}G\left(u_0\right)d\sigma -{\int }_{\mathsf{\Omega }}F\left(u_0\right)dx}\hfill \end{array}$$

and

$$\begin{array}{cc}{\overline{I}}_{\lambda }\left(v\right)\hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla v^{+}{|}^{2}dx+\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{-}|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}\left(v^{+}\right)d\sigma -{\int }_{\mathsf{\Omega }}\overline{F}\left(v^{+}\right)dx}\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{+}|}^{2}dx-{\int }_{\mathsf{\Omega }}F(u_0+v^{+})dx+{\int }_{\mathsf{\Omega }}\lambda u_0^q{v}^{+}+\lambda u_0^r{v}^{+}dx}\hfill \\ \hfill & {\displaystyle +\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{-}|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}\left(v^{+}\right)d\sigma +{\int }_{\mathsf{\Omega }}F\left(u_0\right)dx.}\hfill \end{array}$$

By computation, we have

$$\begin{array}{cc}{I}_{\lambda }(u_0+v^{+})\hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla v^{+}{|}^{2}dx+\frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla u_0{|}^{2}dx+{\int }_{\mathsf{\Omega }}|\nabla u_0\left|\right|\nabla v^{+}|dx}\hfill \\ \hfill & {\displaystyle +{\int }_{\partial \mathsf{\Omega }}G(u_0+v^{+})d\sigma -{\int }_{\mathsf{\Omega }}F(u_0+v^{+})dx}\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla v^{+}{|}^{2}dx+\frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla u_0{|}^{2}dx+{\int }_{\mathsf{\Omega }}|\nabla u_0\left|\right|\nabla v^{+}|dx}\hfill \\ \hfill & {\displaystyle +{\int }_{\partial \mathsf{\Omega }}G(u_0+v^{+})d\sigma -{\int }_{\mathsf{\Omega }}F(u_0+v^{+})dx}\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{+}|}^{2}dx-{\int }_{\mathsf{\Omega }}F(u_0+v^{+})dx+{\int }_{\mathsf{\Omega }}(\lambda u_0^q+\lambda u_0^r)v^{+}dx}\hfill \\ \hfill & {\displaystyle +\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u_0|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}G(u_0+v^{+})d\sigma -{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}^{+}d\sigma .}\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}{\overline{I}}_{\lambda }\left(v\right)\hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{-}|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}\left(v^{+}\right)d\sigma +{\int }_{\mathsf{\Omega }}F\left(u_0\right)dx+I(u_0+v^{+})}\hfill \\ \hfill & {\displaystyle +{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}^{+}d\sigma -\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u_0|}^{2}dx-{\int }_{\partial \mathsf{\Omega }}{\int }_0^{u_0+v^{+}}g\left(s\right)sdsd\sigma }\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{-}|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}\left(v^{+}\right)d\sigma +{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}^{+}d\sigma +I(u_0+v^{+})}\hfill \\ \hfill & {\displaystyle +{\int }_{\mathsf{\Omega }}F\left(u_0\right)dx-\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla u_0|}^{2}dx-{\int }_{\partial \mathsf{\Omega }}{\int }_0^{u_0}g\left(s\right)sdsd\sigma -{\int }_{\partial \mathsf{\Omega }}{\int }_{u_0}^{u_0+v^{+}}g\left(s\right)sdsd\sigma }\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{-}|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}\left(v^{+}\right)d\sigma +I(u_0+v^{+})-I\left(u_0\right)}\hfill \\ \hfill & {\displaystyle +{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}^{+}d\sigma -{\int }_{\partial \mathsf{\Omega }}{\int }_{u_0}^{u_0+v^{+}}g\left(s\right)sdsd\sigma }\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v^{-}|}^{2}dx+I(u_0+v^{+})-I\left(u_0\right)+{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}^{+}d\sigma }\hfill \\ \hfill & {\displaystyle +{\int }_{\partial \mathsf{\Omega }}{\int }_0^{v^{+}}g(s+u_0)(s+u_0)-g\left(u_0\right)u_{0}dsd\sigma -{\int }_{\partial \mathsf{\Omega }}{\int }_0^{v^{+}}g(t+u_0)(t+u_0)dtd\sigma }\hfill \\ \hfill & {\displaystyle =\frac{1}{2}{\parallel \nabla v^{-}\parallel }_{L^2\left(\mathsf{\Omega }\right)}^2+I(u_0+v^{+})-I\left(u_0\right)}\hfill \end{array}$$

with $t=s-u_0$. From the Lemma 11, ${\displaystyle {\overline{I}}_{\lambda }\left(v\right)\ge \frac{1}{2}{\parallel \nabla v^{-}\parallel }_{L^1\left(\mathsf{\Omega }\right)}^2\ge 0={\overline{I}}_{\lambda }\left(0\right)}$ is obtained and $v=0$ is the local minimum of ${\overline{I}}_{\lambda }$ in $H^1\left(\mathsf{\Omega }\right)$.

Choose $L>0$ so that

$$0={\overline{I}}_{\lambda }\left(0\right)\le {\overline{I}}_{\lambda }\left(u\right),\mathrm{for}\mathrm{all}{\parallel u\parallel }_{H^1\left(\mathsf{\Omega }\right)}\le L.$$

Since $q<1<r$ and the highest power of g is less than the $r-1$ of (H3),

$$\begin{array}{c}{\displaystyle {\overline{I}}_{\lambda }\left(tu\right)=\frac{t^2}{2}{\int }_{\mathsf{\Omega }}{|\nabla u|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}\overline{G}\left(u\right)d\sigma -{\int }_{\mathsf{\Omega }}\overline{F}\left(u\right)dx}\\ {\displaystyle \le \frac{t^2}{2}{\int }_{\mathsf{\Omega }}{|\nabla u|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}{\int }_0^{tu}g(s+u_0)(s+u_0)dsd\sigma }\\ {\displaystyle -{\int }_{\mathsf{\Omega }}{\int }_0^{tu}\lambda {(s+u_0)}^q+{(s+u_0)}^r-\lambda u_0^q-u_0^{r}dsdx\to -\infty }\end{array}$$

(13)

as $t\to +\infty$. Fix $\mu \in H^1\left(\mathsf{\Omega }\right)-\left\{0\right\}$ such that ${\overline{I}}_{\lambda }\left(\mu \right)<0$. Necessarily, ${\parallel \mu \parallel }_{H^1\left(\mathsf{\Omega }\right)}>L$. Set

$$\mathsf{\Gamma }=\{\gamma :[0,1]\to H^1\left(\mathsf{\Omega }\right):\gamma \mathrm{is}\mathrm{continuous},\gamma \left(0\right)=0,\gamma \left(1\right)=\mu \}$$

and define the mountain-pass level

$$\beta =\underset{\gamma \in \mathsf{\Gamma }}{inf}\underset{t\in [0,1]}{sup}{\overline{I}}_{\lambda }\left(\gamma \left(t\right)\right).$$

Clearly, $\beta \ge 0$ since ${\overline{I}}_{\lambda }\left(0\right)=0$. We recall the definition of the $PS$ sequence around the closed set F.

Definition 4.

We define the closed set${\displaystyle F=\{u_n\in H^1\left(\mathsf{\Omega }\right):\parallel u{\parallel }_{H^1\left(\mathsf{\Omega }\right)}=\frac{L}{2}\}}$if$\beta =0$and$F=H^1\left(\mathsf{\Omega }\right)$if$\beta >0$.

Definition 5.

${\left(PS\right)}_{F,\beta }$means a sequence$\left\{u_n\right\}\in H^1\left(\mathsf{\Omega }\right)$such that

$$\underset{n\to \infty }{lim}dist(u_n,F)=0,\underset{n\to \infty }{lim}{\overline{I}}_{\lambda }\left(u_n\right)=\beta \mathrm{and}\underset{n\to \infty }{lim}{\parallel {\overline{I}}_{\lambda }^{\prime }\left(u_n\right)\parallel }_{{\left(H^1\left(\mathsf{\Omega }\right)\right)}^{*}}=0,$$

where${\left(H^1\left(\mathsf{\Omega }\right)\right)}^{*}$is the dual space of$H^1\left(\mathsf{\Omega }\right)$.

We have the following two cases:

(i)

${\displaystyle inf\{{\overline{I}}_{\lambda }\left(u\right):u\in H^1\left(\mathsf{\Omega }\right)\mathrm{and}\parallel u{\parallel }_{H^1\left(\mathsf{\Omega }\right)}=l\}}$ for all $l<L$.

In this case, Ghoussoub and Preiss proved the existence of such a ${\left(PS\right)}_{F,\beta }$ sequence (see [32]). Next, we just need to prove that there is also a ${\left(PS\right)}_{F,\beta }$ sequence in the following case.

(ii)

There exists $0<l_1<L$ such that

$$inf\{{\overline{I}}_{\lambda }\left(u\right):u\in H^1\left(\mathsf{\Omega }\right)\mathrm{and}\parallel u{\parallel }_{H^1\left(\mathsf{\Omega }\right)}=l_1\}>0.$$

Note that $\beta =0$ implies that (i) holds and (ii) implies $\beta >0$. In the case (ii), we can find $r>0$ such that ${\overline{I}}_{\lambda }\left(u\right)>r$, $\forall u\in H^1\left(\mathsf{\Omega }\right)$ with $\parallel u\parallel =l_1$, ${\overline{I}}_{\lambda }\left(0\right)=0$ and ${\overline{I}}_{\lambda }\left(\mu \right)<0<r$ for some $\mu \in H^1\left(\mathsf{\Omega }\right)$ with $\parallel \mu \parallel >L>l_1$. By Lemma 7, ${\left(PS\right)}_{F,\beta }$ hold and there exists a sequence $\left\{v_n\right\}\in X$ such that ${\overline{I}}_{\lambda }\left(v_n\right)\to \beta$ and ${\overline{I}}_{\lambda }^{\prime }\left(v_n\right)={\epsilon }_n\to 0$.

Hence, we have $|({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),v_n)|\le {\epsilon }_n{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}$ and $({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),v_n)\ge -{\epsilon }_n{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}$. In addition, $({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),u_0)\to 0$ for fixed $u_0\in H^1\left(\mathsf{\Omega }\right)$.

Note that $u_0\in H^1\left(\mathsf{\Omega }\right)$ is the local minimum positive solution of $I_{\lambda }$. Thus,

$$<I_{\lambda }^{\prime }\left(u_0\right),\phi >=0,\mathrm{for}\mathrm{any}\phi \in H^1\left(\mathsf{\Omega }\right)$$

and

$$\begin{array}{c}{\displaystyle {\int }_{\mathsf{\Omega }}\nabla u_0·\nabla \phi dx+{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0\phi d\sigma ={\int }_{\mathsf{\Omega }}(\lambda u_0^q+u_0^r)\phi dx\le T_1{\parallel \phi \parallel }_{L^2},}\hfill \end{array}$$

(14)

where ${\displaystyle T_1=({\int }_{\mathsf{\Omega }}\left(\lambda \right|u_0{|}^q+|u_0{|}^r{{)}^{2}dx)}^{\frac{1}{2}}>0}$.

Since $1<q+1<2$ and (14)

$$\begin{array}{c}{\displaystyle \beta +o_n\left(1\right)={\overline{I}}_{\lambda }\left(v_n\right)-\frac{1}{r+1}({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),u_0)+\frac{1}{r+1}({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),v_n+u_0)-\frac{1}{r+1}({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),v_n+u_0)}\hfill \\ {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla v_n|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}{\int }_0^{v_n}g(s+u_0)(s+u_0)-g\left(u_0\right)u_{0}dsd\sigma +\frac{1}{r+1}({\overline{I}}_{\lambda }^{\prime }\left(v_n\right),v_n)}\hfill \\ {\displaystyle -{\int }_{\mathsf{\Omega }}\frac{\lambda |v_n+u_0{|}^{q+1}-\lambda u_0^{q+1}}{q+1}-\lambda u_0^q|v_n|dx-{\int }_{\mathsf{\Omega }}\frac{|v_n+u_0{|}^{r+1}-u_0^{r+1}}{r+1}-u_0^r\left|v_n\right|dx}\hfill \\ {\displaystyle -\frac{1}{r+1}{\int }_{\mathsf{\Omega }}|\nabla v_n\left|\right|\nabla (v_n+u_0)|dx-\frac{1}{r+1}{\int }_{\partial \mathsf{\Omega }}(g(v_n+u_0)(v_n+u_0)-g\left(u_0\right)u_0)(v_n+u_0)d\sigma }\hfill \\ {\displaystyle +\frac{\lambda }{r+1}{\int }_{\mathsf{\Omega }}{(v_n+u_0)}^{q+1}-u_0^q(v_n+u_0)dx+\frac{1}{r+1}{\int }_{\mathsf{\Omega }}{(v_n+u_0)}^{r+1}-u_0^r(v_n+u_0)dx}\hfill \\ {\displaystyle \ge \frac{1}{2}{\int }_{\mathsf{\Omega }}|\nabla v_n{|}^{2}dx-\frac{1}{r+1}{\int }_{\mathsf{\Omega }}|\nabla v_n{|}^{2}dx-\frac{1}{r+1}{\int }_{\mathsf{\Omega }}|\nabla v_n\left|\right|\nabla u_0|dx}\hfill \\ {\displaystyle +{\int }_{\partial \mathsf{\Omega }}{\int }_{u_0}^{v_n+u_0}g\left(t\right)tdtd\sigma -{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}_{n}d\sigma -\frac{1}{r+1}{\epsilon }_n{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}}\hfill \\ {\displaystyle -\frac{1}{r+1}{\int }_{\partial \mathsf{\Omega }}g(v_n+u_0){(v_n+u_0)}^{2}d\sigma +\frac{1}{r+1}{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0(v_n+u_0)d\sigma }\hfill \\ {\displaystyle -\frac{\lambda }{q+1}{\int }_{\mathsf{\Omega }}|v_n+u_0{|}^{q+1}dx+\frac{\lambda }{q+1}{\int }_{\mathsf{\Omega }}{u}_0^{q+1}dx+\lambda {\int }_{\mathsf{\Omega }}{u}_0^q\left|v_n\right|dx}\hfill \\ {\displaystyle -\frac{1}{r+1}{\int }_{\mathsf{\Omega }}\lambda |v_n+u_0{|}^{r+1}dx+\frac{1}{r+1}{\int }_{\mathsf{\Omega }}{u}_0^{r+1}dx+{\int }_{\mathsf{\Omega }}{u}_0^r\left|u_n\right|dx}\hfill \\ {\displaystyle +\frac{\lambda }{r+1}{\int }_{\mathsf{\Omega }}|v_n+u_0{|}^{q+1}dx-\frac{\lambda }{r+1}{\int }_{\mathsf{\Omega }}{u}_0^{q+1}dx-\frac{\lambda }{r+1}{\int }_{\mathsf{\Omega }}{u}_0^q\left|u_n\right|dx}\hfill \\ {\displaystyle +\frac{1}{r+1}{\int }_{\mathsf{\Omega }}|v_n+u_0{|}^{r+1}dx-\frac{1}{r+1}{\int }_{\mathsf{\Omega }}{u}_0^{r+1}dx-\frac{1}{r+1}{\int }_{\mathsf{\Omega }}{u}_0^r\left|u_n\right|dx}\hfill \\ {\displaystyle \ge \frac{r-1}{2(r+1)}{\int }_{\mathsf{\Omega }}|\nabla v_n{|}^{2}dx+C_0{\int }_{\partial \mathsf{\Omega }}|v_n+u_0{|}^{2}d\sigma -{\int }_{\partial \mathsf{\Omega }}{\int }_0^{u_0}g\left(t\right)tdtd\sigma -\frac{1}{r+1}{\int }_{\mathsf{\Omega }}|\nabla v_n\left|\right|\nabla u_0|dx}\hfill \\ {\displaystyle -{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)u_0{v}_{n}d\sigma +\frac{r}{r+1}{\int }_{\mathsf{\Omega }}(\lambda u_0^q+u_0^r)|u_n|dx-\frac{{\epsilon }_n}{r+1}{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}}\hfill \\ {\displaystyle -\lambda \frac{r-q}{(q+1)(r+1)}{\int }_{\mathsf{\Omega }}|v_n+u_0{|}^{q+1}dx+\lambda \frac{r-q}{(q+1)(r+1)}{\int }_{\mathsf{\Omega }}{\left|u_0\right|}^{q+1}dx}\hfill \\ {\displaystyle \ge \frac{r-1}{2(r+1)}{\int }_{\mathsf{\Omega }}|\nabla v_n{|}^{2}dx+C_0{\int }_{\partial \mathsf{\Omega }}|v_n{|}^{2}d\sigma -{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)\frac{u_0^2}{2}d\sigma -\frac{1}{r+1}{\int }_{\mathsf{\Omega }}|\nabla v_n\left|\right|\nabla u_0|dx}\hfill \\ {\displaystyle +{\int }_{\mathsf{\Omega }}|\nabla v_n\left|\right|\nabla u_0|dx-{\int }_{\mathsf{\Omega }}(\lambda u_0^q+u_0^r)|u_n|dx+\frac{r}{r+1}{\int }_{\mathsf{\Omega }}(\lambda u_0^q+u_0^r)\left|u_n\right|dx}\hfill \\ {\displaystyle -\lambda \frac{r-q}{(q+1)(r+1)}{\int }_{\mathsf{\Omega }}max\left\{2\right|v_n|,2|u_0{\left|\right\}}^{q+1}dx-\frac{{\epsilon }_n}{r+1}{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}}\hfill \\ {\displaystyle \ge min\{\frac{r-1}{2(r+1)},C_0\}\parallel v_n{\parallel }_{H^1\left(\mathsf{\Omega }\right)}^2-{\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)\frac{u_0^2}{2}d\sigma -\frac{1}{r+1}{\int }_{\mathsf{\Omega }}(\lambda u_0^q+u_0^r)\left|u_n\right|dx}\hfill \\ {\displaystyle -\lambda \frac{2^{q+1}(r-q)}{(q+1)(r+1)}max\{\parallel v_n{\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1},\parallel u_0{\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1}\}-\frac{{\epsilon }_n}{r+1}{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}}\hfill \\ {\displaystyle \ge min\{\frac{r-1}{2(r+1)},C_0\}\parallel v_n{\parallel }_{H^1\left(\mathsf{\Omega }\right)}^2-C_2-\frac{1}{r+1}{T}_1\parallel v_n{\parallel }_{L^2\left(\mathsf{\Omega }\right)}-\frac{{\epsilon }_n}{r+1}{\parallel v_n\parallel }_{H^1\left(\mathsf{\Omega }\right)}}\hfill \\ {\displaystyle -\lambda \frac{2^{q+1}(r-q)}{(q+1)(r+1)}max\{\parallel v_n{\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1},\parallel u_0{\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1}\}}\hfill \\ {\displaystyle \ge C_1\parallel v_n{\parallel }_{H^1\left(\mathsf{\Omega }\right)}^2-C_2-C_3\parallel v_n{\parallel }_{H^1\left(\mathsf{\Omega }\right)}-C_{4}max\{\parallel v_n{\parallel }_{H^1\left(\mathsf{\Omega }\right)}^{q+1},\parallel u_0{\parallel }_{H^1\left(\mathsf{\Omega }\right)}^{q+1}\},}\hfill \end{array}$$

(15)

with ${\displaystyle C_1=min\{\frac{r-1}{2(r+1)},C_0\}}$, ${\displaystyle C_2={\int }_{\partial \mathsf{\Omega }}g\left(u_0\right)\frac{u_0^2}{2}d\sigma }$, ${\displaystyle C_3=\frac{1}{r+1}{T}_1{S}_2+\frac{{\epsilon }_n}{r+1}}$ and ${\displaystyle C_4=\lambda \frac{2^{q+1}(r-q)}{(q+1)(r+1)}{S}_{q+1}^{q+1}}$, where $S_i$ satisfies ${\parallel u\parallel }_{L^i\left(\mathsf{\Omega }\right)}\le S_i{\parallel u\parallel }_{H^1\left(\mathsf{\Omega }\right)}$ and $C_2$ is a constant value.

Therefore, we get the bounded (PS) sequence $\left\{v_n\right\}$ of $I_{\lambda }$. Accordingly with the properties of bounded sequences and Lemma 12, we have

$$u_n\to u,\mathrm{in}{H}^1\left(\mathsf{\Omega }\right).$$

Next, let us verify the conditions of the Mountain Pass Theorem (see [33]). Obviously, ${\overline{I}}_{\lambda }\left(0\right)=0$. In the previous proof, we found $r>0$ such that ${\overline{I}}_{\lambda }\left(u\right)>r$, $\forall u\in H^1\left(\mathsf{\Omega }\right)$ with $\parallel u\parallel =l_1$.

In Equation (13),

$${\overline{I}}_{\lambda }\left(tu\right)\to -\infty ,\mathrm{as}t\to +\infty ,\mathrm{for}\mathrm{any}u\in H^1\left(\mathsf{\Omega }\right)-\left\{0\right\}.$$

Choose a sufficiently large $t=t^{\infty }$ such that ${\overline{I}}_{\lambda }\left(u^{\infty }\right){|}_{\parallel u^{\infty }{\parallel }_{H^1\left(\mathsf{\Omega }\right)}>L>l_1}<0$. For ${\displaystyle \beta =\underset{\gamma \in \mathsf{\Gamma }}{inf}\underset{t\in [0,1]}{sup}}$${\overline{I}}_{\lambda }\left(\gamma \left(t\right)\right)$ and

$$\mathsf{\Gamma }=\{\gamma :[0,1]\to H^1\left(\mathsf{\Omega }\right):\gamma \mathrm{is}\mathrm{continuous},\gamma \left(0\right)=0,\gamma \left(1\right)=\mu \}$$

and the critical value $\beta >r$ by the mountain pass theorem (see [33]).

Through the above argument, there is a solution $u_1\in H^1\left(\mathsf{\Omega }\right)$ such that

$${\overline{I}}_{\lambda }^{\prime }\left(u_1\right)=0\mathrm{and}{\overline{I}}_{\lambda }\left(u_1\right)=\beta \ge r>0$$

and $u_1$ is a critical point of functional ${\overline{I}}_{\lambda }$.

Assuming $u_1=0$, the contradiction is obtained from ${\overline{I}}_{\lambda }\left(u_1\right)=0\ne \beta ={\overline{I}}_{\lambda }\left(u_1\right)$. Since $-▵u_1=\overline{f}(x,u_1)\ge 0$ in $\mathsf{\Omega }$ and $\mathbf{n}·\nabla u_1+g\left(u_1\right)u_1\ge 0$ on $\partial \mathsf{\Omega }$, we get $u_1>0$ in $\overline{\mathsf{\Omega }}$ by Lemma 2. Therefore, $u_0+u_1$ is the second positive solution of the problem (1).

When $\lambda =\mathsf{\Lambda }$, the problem has a positive solution.

Let $\left\{{\lambda }_n\right\}$ be a sequence satisfying ${\lambda }_n\to \mathsf{\Lambda }$, where ${\lambda }_n\le \mathsf{\Lambda }$. Then there exists a sequence of solutions $\left\{u_{{\lambda }_n}\right\}\subset H^1\left(\mathsf{\Omega }\right)$ to problem (1) for ${\lambda }_n$ fulfilling

$${\displaystyle sup{I}_{{\lambda }_n}\left(u_{{\lambda }_n}\right)<+\infty ,\mathrm{and}{I}_{{\lambda }_n}^{\prime }\left(u_{{\lambda }_n}\right)=0.}$$

From Lemma 11, ${\displaystyle {\left\{u_{\underline{\lambda }}\right\}}_{0\le \underline{\lambda }}\le \mathsf{\Lambda }}$ is uniformly bounded in $C^1\left(\overline{\mathsf{\Omega }}\right)$ and $I_{\lambda }\left(u_{\lambda }\right)\le 0$. In fact, $-▵u_{\underline{\lambda }}=\underline{\lambda }{u}_{\underline{\lambda }}^q<\mathsf{\Lambda }{u}_{\mathsf{\Lambda }}^q=-▵u_{\mathsf{\Lambda }}$ in $\mathsf{\Omega }$ and $\mathbf{n}·\nabla u_{\underline{\lambda }}+g\left(u_{\underline{\lambda }}\right)u_{\underline{\lambda }}=0=\mathbf{n}·\nabla u_{\mathsf{\Lambda }}+g\left(u_{\mathsf{\Lambda }}\right)u_{\mathsf{\Lambda }}$ on $\partial \mathsf{\Omega }$, where ${\displaystyle u_{\mathsf{\Lambda }}}$ is the unique positive solution of

$$\left\{\begin{array}{cc}{\displaystyle -▵u=\mathsf{\Lambda }{u}^q}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u+g\left(u\right)u=0}\hfill & \mathrm{on}\partial \mathsf{\Omega }.\hfill \end{array}\right.$$

Hence, for all $0\le \underline{\lambda }\le \mathsf{\Lambda }$, $\left\{u_{\underline{\lambda }}\right\}$ is less than $u_{\mathsf{\Lambda }}$. $u_{{\lambda }_n}\le u_{\mathsf{\Lambda }}$, for any ${\lambda }_n\in [0,\mathsf{\Lambda }]$. Therefore, $\left\{u_{{\lambda }_n}\right\}$ is a bounded sequence in $H^1\left(\mathsf{\Omega }\right)$, so $u_{{\lambda }_n}\rightharpoonup u_{\mathsf{\Lambda }}$. For parameter $\lambda =\mathsf{\Lambda }$, it is proved that the problem has at least one positive solution.

Thirdly, accordingly to (7), there are no positive solutions for $\lambda >\mathsf{\Lambda }$.

This concludes the proof of the first part (i) of Theorem 1. □

Proof of Theorem 1(ii).

Assuming $\rho =0$, (6) has an only positive solution $\vartheta$ by Lemma 6 with

$$-▵\vartheta =\lambda {\vartheta }^q\le \lambda {\vartheta }^q+{\vartheta }^r,\mathrm{in}\mathsf{\Omega }$$

and

$$\mathbf{n}·\nabla \vartheta +g\left(\vartheta \right)\vartheta =0,\mathrm{on}\partial \mathsf{\Omega }.$$

Since problem (1) has a positive solution for any $\lambda \in (0,\mathsf{\Lambda })$,

$${\displaystyle ▵u+\lambda u^q\le 0=▵\vartheta +\lambda {\vartheta }^r,\mathrm{in}\mathsf{\Omega },}$$

$${\displaystyle \mathbf{n}·\nabla \vartheta +g\left(\vartheta \right)\vartheta =0=\mathbf{n}·\nabla u+g\left(u\right)u,\mathrm{on}\partial \mathsf{\Omega }}$$

and $\vartheta \le u$ by Lemma 3.

Construct the following monotone iteration:

$$\left\{\begin{array}{cc}{\displaystyle -▵u_{k+1}=\lambda u_k^q+u_k^r,}\hfill & \mathrm{in}\mathsf{\Omega },\hfill \\ {\displaystyle \mathbf{n}·\nabla u_{k+1}+g\left(u_{k+1}\right)u_{k+1}=0,}\hfill & \mathrm{on}\partial \mathsf{\Omega },\hfill \end{array}\right.$$

with $u_0=\vartheta$. Let u be an arbitrary solution; then, $u\ge \vartheta$ is a super-solution of (1).

For $k=0$,

$${\displaystyle -▵u_1=\lambda u_0^q+u_0^r\ge \lambda u_0^q=-▵u_0=-▵\vartheta ,\mathrm{in}\mathsf{\Omega }}$$

and

$${\displaystyle \mathbf{n}·\nabla u_1+g\left(u_1\right)u_1=0=\mathbf{n}·\nabla u_0+g\left(u_0\right)u_0,\mathrm{on}\partial \mathsf{\Omega }.}$$

Hence, $\vartheta =u_0\le u_1$, by Lemma 2.

Since ${(\lambda s^q+s^r)}^{\prime }=q\lambda s^{q-1}+r{s}^{r-1}>0$, the function $\lambda s^q+s^r$ is strictly increasing. For $k=1$,

$${\displaystyle -▵u_2=\lambda u_1^q+u_1^r>\lambda u_0^q+u_0^r=-▵u_1,\mathrm{in}\mathsf{\Omega }}$$

and

$${\displaystyle \mathbf{n}·\nabla u_1+g\left(u_1\right)u_1=0=\mathbf{n}·\nabla u_2+g\left(u_2\right)u_2,\mathrm{on}\partial \mathsf{\Omega }.}$$

This implies by Lemma 2 that $u_1\le u_2$.

For any $k\in \mathbb{N}$,

$${\displaystyle \vartheta =u_0\le u_1\le u_2\le \cdots \le u_k\le \cdots \le u.}$$

Thus, $u_k\le u$ by iterating this process. In addition,

$${\displaystyle u_{{\lambda }_0}=\underset{k\to +\infty }{lim}{u}_k\le \underset{k\to +\infty }{lim}u=u,}$$

so $u_{{\lambda }_0}$ is a minimal positive solution of (1).

Denote by $u_{{\lambda }_0}$ the minimal positive solution of (1) for ${\lambda }_0\in (0,\mathsf{\Lambda })$. Moreover, by Lemma 10 we get the strict inequality $u_{{\lambda }_0}<u_{{\lambda }_1}$ for ${\lambda }_0<{\lambda }_1$.

Let $\lambda$ as defined above,

$$\begin{array}{cc}{I}_{\lambda }\left(tv\right)\hfill & {\displaystyle =\frac{1}{2}{\int }_{\mathsf{\Omega }}{|\nabla tv|}^{2}dx-\lambda {\int }_{\mathsf{\Omega }}dx{\int }_0^{tv}{\left|s\right|}^{q}ds-{\int }_{\mathsf{\Omega }}{\int }_0^{tv}{\left|s\right|}^{r}ds+{\int }_{\partial \mathsf{\Omega }}d\sigma {\int }_0^{tv}g\left(s\right)sds}\hfill \\ \hfill & {\displaystyle =\frac{t^2}{2}{\int }_{\mathsf{\Omega }}{|\nabla v|}^{2}dx-\frac{\lambda t^{q+1}}{q+1}{\int }_{\mathsf{\Omega }}{\left|v\right|}^{q+1}dx-\frac{t^{r+1}}{r+1}{\int }_{\mathsf{\Omega }}{\left|v\right|}^{r+1}dx+{\int }_{\partial \mathsf{\Omega }}d\sigma {\int }_0^{tv}g\left(s\right)sds,}\hfill \end{array}$$

for any $v\in H^1$. In a sufficiently small neighborhood near the zero point,

$$\begin{array}{c}{I}_{\lambda }\left(tv\right)\hfill \\ {\displaystyle =\frac{t^2}{2}{\int }_{\mathsf{\Omega }}{|\nabla v|}^{2}dx-\frac{\lambda t^{q+1}}{q+1}{\parallel v\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1}-\frac{t^{r+1}}{r+1}{\parallel v\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}^{r+1}+{\int }_{\partial \mathsf{\Omega }}d\sigma {\int }_0^{tv}(g_0+o\left(1\right))sds}\hfill \\ {\displaystyle =\frac{t^2}{2}{\int }_{\mathsf{\Omega }}{|\nabla v|}^{2}dx+\frac{g_0{t}^2}{2}{\int }_{\partial \mathsf{\Omega }}{v}^{2}d\sigma +\frac{o\left(t^2\right)}{2}{\int }_{\partial \mathsf{\Omega }}{v}^{2}d\sigma -\frac{\lambda t^{q+1}}{q+1}{\parallel v\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1}}\hfill \\ {\displaystyle -\frac{t^{r+1}}{r+1}{\parallel v\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}^{r+1}}\hfill \\ {\displaystyle \le max\{\frac{1}{2},\frac{g_0}{2}\}{t}^2({\int }_{\mathsf{\Omega }}{|\nabla v|}^{2}dx+{\int }_{\partial \mathsf{\Omega }}{v}^{2}d\sigma )+\frac{o\left(t^2\right)}{2}{\int }_{\partial \mathsf{\Omega }}{v}^{2}d\sigma -\frac{\lambda t^{q+1}}{q+1}{\parallel v\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1}}\hfill \\ {\displaystyle -\frac{t^{r+1}}{r+1}{\parallel v\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}^{r+1}}\hfill \\ {\displaystyle =max\{\frac{1}{2},\frac{g_0}{2}\}{t}^2{\parallel v\parallel }_{H^1\left(\mathsf{\Omega }\right)}^2+\frac{o\left(t^2\right)}{2}{\int }_{\partial \mathsf{\Omega }}{v}^{2}d\sigma -\frac{\lambda t^{q+1}}{q+1}{\parallel v\parallel }_{L^{q+1}\left(\mathsf{\Omega }\right)}^{q+1}-\frac{t^{r+1}}{r+1}{\parallel v\parallel }_{L^{r+1}\left(\mathsf{\Omega }\right)}^{r+1}}\hfill \\ <0,\mathrm{as}t\to 0.\hfill \end{array}$$

Accordingly to the calculation, there exists ${\displaystyle \parallel u_m{\parallel }_{H^1\left(\mathsf{\Omega }\right)}<\frac{1}{n}}$. Let ${\displaystyle \left\{u_n\right\}\subset B_{\frac{1}{n}}\left(0\right)}$ be a minimizing sequence for $I_{\lambda }$, where ${\displaystyle B_{\frac{1}{n}}\left(0\right)=:\{u\in H^1{\left(\mathsf{\Omega }\right):\parallel u\parallel }_{H^1\left(\mathsf{\Omega }\right)}\le \frac{1}{n}\}}$. Since the bounded sequence $\left\{u_n\right\}\subset H^1\left(\mathsf{\Omega }\right)$ has convergent subsequence $\{u_{n_k\}}$, there exists $u_m\in H^1\left(\mathsf{\Omega }\right)$ such that $u_{n_k}\rightharpoonup u_m$. Clearly,

$${\displaystyle {\int }_{\mathsf{\Omega }}{|\nabla u|}^{2}dx\le \underset{k\to \infty }{lim}inf{\int }_{\mathsf{\Omega }}{|\nabla u_{n_k}|}^{2}dx}$$

is established. In addition, since ${\displaystyle q,r<\frac{2N}{N-2}}$,

$${\displaystyle \lambda {\int }_{\mathsf{\Omega }}{u}_{n_k}^{q}dx\to \lambda {\int }_{\mathsf{\Omega }}{u}_m^{q}dx}$$

and

$${\displaystyle {\int }_{\mathsf{\Omega }}{u}_{n_k}^{r}dx\to {\int }_{\mathsf{\Omega }}{u}_m^{r}dx,}$$

by Lemma 5 and the convergence theorem. It also follows that ${\displaystyle {\int }_{\partial \mathsf{\Omega }}G\left(u_{n_k}\right)d\sigma \to {\int }_{\partial \mathsf{\Omega }}G\left(u_m\right)d\sigma }$, according to the compactness of the tracked embedding. Hence, we have ${\displaystyle I_{\lambda }\left(u_{\lambda }\right)\le \underset{k\to \infty }{lim}inf{I}_{\lambda }\left(u_{n_k}\right)=\underset{B_{\frac{1}{n}}\left(0\right)}{inf}{I}_{\lambda }}$. Since ${\displaystyle u_m\in B_{\frac{1}{n}}\left(0\right)}$, there must be ${\displaystyle I_{\lambda }\left(u_m\right)=\underset{B_{\frac{1}{n}}\left(0\right)}{inf}{I}_{\lambda }}$. Based on the above discussion, $u_m$ is the local minimum of $I_{\lambda }$ in the set ${\displaystyle \{u\in H^1\left(\mathsf{\Omega }\right):\parallel u{\parallel }_{H^1\left(\mathsf{\Omega }\right)}\le B_{\frac{1}{n}}\left(0\right)\}}$. Notice that because $I_{\lambda }\left(0\right)=0>I_{\lambda }\left(u_m\right)$, there is $u_m\not\equiv 0$. The local minimum value can be obtained near zero of $I_{\lambda }$. □

Therefore, Theorem 1 has been fully proved.

6. Concluding Remarks

In this paper, we did not only prove the existence of an application of the sub- and super-solutions method, but also proved the existence of the second solution via variational method. The results show that the uniqueness of the positive solution of the elliptic equation with a special boundary is related to the parameters of the internal nonlinear equation.