Positive solutions for a class of biharmonic problems: existence, nonexistence and multiplicity

Abstract

The main objective of this article is to consider a biharmonic problem with Navier boundary conditions. Among others, some criteria for the existence, multiplicity and nonexistence of positive solutions are established by employed fixed point theorems in a cone. In addition, we not only consider the sublinear case, but also we will study the case of appropriate combinations of superlinearity and sublinearity.

1 Introduction and main results

Biharmonic elliptic equations with various boundary conditions come from the study of traveling waves in suspension bridges [5] and static deflection of a bending beam [16], and have attracted the interest of many researchers. Some classical methods have been widely used to study biharmonic elliptic equations: the Pohozaev identities and decay estimates, see Guo-Liu [12] and Guo-Wei [13]; comparison principles, see Cosner-Schaefer [6] and Mareno [22]; degree argument, see Tarantello [28]; perturbation theory, see Wang-Shen [29]; bifurcation theory, see Lazer-McKenna [19]; the method of upper and lower solutions, see Ferrero-Warnault [10] and Pao [25]; computational methods for numerical solutions, see Pao [26] and Pao-Lu [27]; phase space analysis, see Chang-Chen [4], Díaz-Lazzo-Schmidt [9]; fixed point theorems, see Kusano-Naito-Swanso [18]; variational method, see Micheletti-Pistoia [23, 24], Xu-Zhang [31], Zhang [33], Zhou-Wu [36] and Ye-Tang [32]; Morse index, see Li-Zhang [35], Davila-Dupaigne-Wang-Wei [7], Khenissy [17] and Wei-Ye [30], and the moving-plane method, see Lin [20] and Guo-Huang-Zhou [14].

We recall some recent results of Abid-Baraket [1], Guo-Wei-Zhou [15], Arioli-Gazzola-Grunau-Mitidieri [3] and Liu-Wang [21]. In [1], Abid-Baraket applied the maximum principle to analyze the existence of singular solution to the following biharmonic elliptic problem

$$\begin{aligned} \left\{ \begin{array}{l} \Delta ^2 u=u^p \ \ \text {in} \ \Omega ,\\ u=\Delta u=0\ \ \text {on} \ \partial \Omega . \end{array} \right. \end{aligned}$$

(1.1)

Recently, Guo-Wei-Zhou [15] employed the entire radial solutions of a equation with supercritical exponent and the Kelvin’s transformation to obtain positive singular radial entire solutions of the biharmonic equation with subcritical exponent. Then, they constructed solutions with a prescribed singular set for problem (1.1) by using the expansions of such singular radial solutions at the singular point 0.

In [3], Arioli-Gazzola-Grunau-Mitidieri studied the boundary value problem

$$\begin{aligned} \left\{ \begin{array}{l} \Delta ^2 u=\lambda e^u \ \ \text {in} \ \Omega ,\\ u=\frac{\partial u}{\partial \textbf{n} }=0\ \ \text {on} \ \partial \Omega , \end{array} \right. \end{aligned}$$

(1.2)

where $\lambda \ge 0$ is a parameter, $\Omega$ is the unit ball in ${\mathbb {R}}^n\ (n\ge 5)$ and $\frac{\partial u}{\partial \textbf{n} }$ denotes the differentiation with respect to the exterior unit normal. They proved the existence of singular solutions for problem (1.2) by means of computer assistance when $5\le n\le 16$.

In [21], Liu-Wang employed a variant version of Mountain Pass Theorem to study the existence and nonexistence of positive solution to the biharmonic problem

$$\begin{aligned} \left\{ \begin{array}{l} \Delta ^2 u= f(x,u) \ \ \text {in} \ \Omega ,\\ u=\Delta u=0\ \ \text {on} \ \partial \Omega , \end{array} \right. \end{aligned}$$

where $\Omega$ is a smooth bounded domain in ${\mathbb {R}}^n\ (n> 4)$, and f satisfies

$(C_1)$ $f(x,t)\in C({{\bar{\Omega }}}\times {\mathbb {R}});\ f(x,0)=0,\ \forall x\in {{\bar{\Omega }}};\ f(x,t)\ge 0,\ \forall t\ge 0,\ x\in {{\bar{\Omega }}}\ \text {and}\ f(x,t)\equiv 0,\ \forall t\le 0,\ x\in {{\bar{\Omega }}};$

$(C_2)$ $\lim \limits _{t\rightarrow 0}\frac{f(x,t)}{t}=p(x),\ \lim \limits _{t\rightarrow +\infty }\frac{f(x,t)}{t}=l\ (0<l\le +\infty )$ uniformly in a.e. $x\in \Omega$ where $|p(x)|_\infty <\Lambda _1$, $\Lambda _1$ is the first eigenvalue of $(\Delta ^2,H^2(\Omega )\cap H_0^1(\Omega ))$;

$(C_3)$ for a.e. $x\in \Omega , \ \lim \limits _{t\rightarrow 0}\frac{f(x,t)}{t}$ is nondecreasing with respect to $t>0$.

However, to our best knowledge, in the literature, there are almost no papers using the fixed point theory in cons for completely continuous operators to study the existence, nonexistence and multiplicity of positive solutions for analogous biharmonic elliptic problems. More precisely, the study is still open for the Navier boundary value problem

$$\begin{aligned} \left\{ \begin{array}{l} \Delta ^2 u=\lambda f(x,u) \ \ \text {in} \ \Omega ,\\ u=\Delta u=0\ \ \text {on} \ \partial \Omega , \end{array} \right. \end{aligned}$$

(1.3)

where $\lambda \ne 0$ is a parameter, $\Omega$ is a smooth bounded domain in ${\mathbb {R}}^n\ (n\ge 2)$, and the nonlinearity f satisfies:

$\mathbf (f_{1})$ $f\in C({{\bar{\Omega }}}\times [0,+\infty ),[0,+\infty ))$.

If $\lambda =1$ and $f(x,u)=u^p$, then problem (1.3) reduces to the problem studied by Abid-Baraket [1] and Guo-Wei-Zhou [15].

Let

$$\begin{aligned} f^{0}:=\lim \limits _{u\rightarrow 0^+}\frac{f(x,u)}{u}\ \text {uniformly for}\ x\in {{\bar{\Omega }}}; \\ f^{\infty }:=\lim \limits _{u\rightarrow +\infty }\frac{f(x,u)}{u}\ \text {uniformly for}\ x\in {{\bar{\Omega }}}; \\ f_\infty :=\varliminf \limits _{u\rightarrow +\infty }f(x,u)\ \text {uniformly for}\ x\in {{\bar{\Omega }}}. \end{aligned}$$

The main results of this paper are the following theorems.

Theorem 1.1

Under condition $\mathbf (f_{1})$, if $f^{0}=0,\ f^{\infty }=0$ and $f_\infty \in (0,+\infty ]$, then, for any given $\tau >0$, there exists $\xi >0$ so that, for $\lambda >\xi$, problem (1.3) admits at least two positive solutions $u_{\lambda }^{(1)}(x),\ \ u_{\lambda }^{(2)}(x)$ and $\max \limits _{x\in {{\bar{\Omega }}}}u_{\lambda }^{(1)}(x)>\tau$.

Remark 1.2

One of the contributions of Theorem 1.1 is to use a simpler method, i.e. index theory of fixed points on cones to prove the multiplicity of positive solutions for biharmonic problems.

Remark 1.3

The approach used in Theorem 1.1 is completely different from those used in Abid-Baraket [1], Guo-Wei-Zhou [15], Arioli-Gazzola-Grunau-Mitidieri [3], Liu-Wang [21] and other related papers. In particular, comparing with Liu-Wang [21], the main difficulties of Theorem 1.1 lie in three main directions:

• (1) $\lambda >0$ is considered;

• (2) multiple positive solutions are obtained;

• (3) in the proof process, we do not need the monotonicity condition $(C_3)$.

Theorem 1.4

Under condition $\mathbf (f_{1})$, (i) if $0<f^{0}<+\infty$, then there are $l_{0}>0$ and $\lambda _0>0$ such that, for every $0<r<l_{0}$, problem (1.3) admits a positive solution $u_{r}$ satisfying $\Vert u_{r}\Vert _C=r$ associated with

$$\begin{aligned} \lambda =\lambda _{r}\in (0,\lambda _{0}]. \end{aligned}$$

(1.4)

(ii) if $f^{0}=+\infty$, then there are $l^*>0$ and $\lambda ^*>0$ such that, for any $0<r^{*}<l^*$, problem (1.3) admits a positive solution $u_{ r^{*}}$ satisfying $\Vert u_{ r^{*}}\Vert _C=r^{*}$ for any

$$\begin{aligned} \lambda =\lambda _{ r^{*}}\in (0,\lambda ^{*}]. \end{aligned}$$

(iii) if $f^{0}<+\infty$ and $f^{\infty }<+\infty$, then there exists ${\underline{\lambda }}>0$ such that problem (1.3) admits no positive solutions for $\lambda \in ({\underline{\lambda }},\infty )$.

Corollary 1.5

Under condition $\mathbf (f_{1})$, if $f^{0}=0$ and $f^{\infty }=0$, then problem (1.3) admits no positive solution for sufficiently large $\lambda$.

Next, in Theorems 1.6–1.9 and Theorem 1.11, we will employ some techniques different from that used in Theorem 1.1 to prove some existence and multiplicity results. Conclusions to be demonstrated in Theorems 1.6–1.9 and Theorem 1.11 are true for any positive parameter $\lambda$. We hence may suppose that $\lambda =1$ in problem (1.3) for simplicity.

We introduce the following notations.

$$\begin{aligned} f^{\gamma }=\limsup \limits _{u \rightarrow \gamma }\max \limits _{x\in {{\bar{\Omega }}}} \frac{f(x,u)}{u^{\alpha }},\ \ \ f_{\gamma }=\liminf \limits _ {u \rightarrow \gamma }\min \limits _{x\in {{\bar{\Omega }}}} \frac{f(x,u)}{u^{\beta }}, \end{aligned}$$

where $\gamma$ denotes $0^+$ or $+\infty$, $\alpha , \beta \in (0,+\infty )$.

We consider the following three cases for $\alpha , \beta \in (0,+\infty ):$

$$\begin{aligned} \alpha =1; 0<\alpha , \beta <1\ \text {and}\ \alpha >1. \end{aligned}$$

Case $\alpha =1$ is treated in Theorems 1.6–1.7.

Theorem 1.6

Under condition $\mathbf (f_{1})$, if $f^0=0\ \text {or}\ f^\infty =0,$ and there exist $\eta >0$ and $l>0$ so that $u\ge \eta$ and $x\in {{\bar{\Omega }}}$ implies

$$\begin{aligned} f(x,u)\ge l, \end{aligned}$$

(1.5)

then problem (1.3) possesses at least one positive solution.

Theorem 1.7

Under condition $\mathbf (f_{1})$, if $f^0=0\ \text {and}\ f^\infty =0,$ and there exist $\eta >0$ and $l>0$ such that $u\ge \eta$ and $x\in {{\bar{\Omega }}}$ implies (1.5) holds, then problem (1.3) possesses at least two positive solutions $u^*$ and $u^{**}$ with

$$\begin{aligned} 0<\Vert u^*\Vert _C<\eta <\Vert u^{**}\Vert _C, \end{aligned}$$

where $\Vert \cdot \Vert _C$ denotes the norm of real Banach space $C({{\bar{\Omega }}})$.

Theorems 1.8–1.9 deal with the case $0<\alpha <1$ and $0<\beta <1$.

Theorem 1.8

Under condition $\mathbf (f_{1})$, if

$$\begin{aligned} f_0=\infty \ \text {and}\ f^\infty =0, \end{aligned}$$

then problem (1.3) possesses at least one positive solution.

Theorem 1.9

Under condition $\mathbf (f_{1})$, if $f^\infty =0$ and there exist $\eta >0$ and $l>0$ such that $u\ge \eta$ and $x\in {{\bar{\Omega }}}$ implies (1.5) holds, then problem (1.3) possesses at least one positive solution.

Remark 1.10

The method applied in the proof of Theorems 1.6–1.7 is invalid when we employ it to demonstrate Theorems 1.8–1.9 for the case $0<\alpha <1$ and $0<\beta <1$. Therefore we need to introduce a different technique to verify Theorems 1.8–1.9.

Next we study the case $\alpha >1$ in Theorem 1.11.

Theorem 1.11

Under condition $\mathbf (f_{1})$, if $f^0=0$ and there exist $\eta >0$ and $l>0$ such that $u\ge \eta$ and $x\in {{\bar{\Omega }}}$ implies (1.5) holds, then problem (1.3) possesses at least one positive solution.

Remark 1.12

The fixed point index theorem on cones is valid in the proof of Theorem 1.11, but the approach used in the proof of Theorems 1.6–1.9 is invalid.

The rest of the paper is organized as follows. In Sect. 2, we first apply an idea from Guo-Huang-Zhou [14] to transfer problem (1.3) into a second-order elliptic system. Then, we obtain the Green’s function of problem (1.3) by means of the Green’s function of the corresponding second-order elliptic boundary value problem. Consequently we get the expression of the solution for problem (1.3). In Sect. 3, we apply index theory of fixed points for completely continuous operators to study the existence, nonexistence and multiplicity of positive solutions to problem (1.3). Sections 4–6 are, respectively, devoted to the study of existence and multiplicity of positive solutions to problem (1.3) under the case $\lambda =1$.

2 Second-order elliptic system

In this section, we first apply an idea from Guo-Huang-Zhou [14] to transfer problem (1.3) into a second order elliptic system. Then, we obtain the Green’s function of problem (1.3) by means of the Green’s function of the corresponding second order elliptic boundary value problem. Consequently, we get the expression of the solution for problem (1.3).

Let $-\Delta u=\omega$. Then, we can transfer the biharmonic problem (1.3) into the following second order elliptic system

$$\begin{aligned} \left\{ \begin{array}{l} -\Delta u=\omega \ \ \text {in} \ \Omega ,\\ -\Delta \omega =\lambda f(x,u) \ \ \text {in} \ \Omega ,\\ u=0=\omega \ \ \text {on} \ \partial \Omega . \end{array} \right. \end{aligned}$$

(2.1)

It follows from (2.1) that

$$\begin{aligned} u(x)=\int _{{{\bar{\Omega }}}}G^*(x,y)\omega (y)\textrm{d}y, \end{aligned}$$

(2.2)

$$\begin{aligned} \omega (x)=\lambda \int _{{{\bar{\Omega }}}}G^*(x,y)f(y,u(y))\textrm{d}y, \end{aligned}$$

(2.3)

where $G^*(x,y)$ is the Green’s function of $-\Delta$ on $\Omega$, which verifies

$$\begin{aligned} 0\le G^*(x,y)\le C|x-y|^{2-n}, \end{aligned}$$

where $n\ge 3$, the constant C depends only on $\Omega$.

Moreover, for $x,y\in \Omega , x\ne y$, one finds that

$$\begin{aligned} 0\le G^*(x,y)\le \frac{1}{4\pi |x-y|},\ \ n=3, \\ 0\le G^*(x,y)\le \frac{1}{2\pi } \ln \frac{d}{|x-y|},\ \ n=2, \end{aligned}$$

where d denotes the diameter of $\Omega$.

Since, for $x,y\in {{\bar{\Omega }}}\subset {\mathbb {R}}^n\ (n\ge 2)$, $G^*(x,y)$ is nonnegative, continuous (when $x\ne y$) and symmetric, there must be two points $x_0$ and $y_0$ with $x_0\ne y_0$, which are interior points of $\Omega$, such that

$$\begin{aligned} G^*(x_0,y_0)=G^*(y_0,x_0)>0. \end{aligned}$$

Thus, there are $\tau _1, \tau _2, \tau _3>0$ and two disjoint small closed balls $B_1, B_2, B_3\in \Omega$ such that

$$\begin{aligned} \left\{ \begin{array}{l} G^*(x,y)\ge \tau _1,\ \ \forall (x,y)\in (B_{1}\times B_{2})\cup (B_{2}\times B_{1}),\\ G^*(y,z)\ge \tau _2,\ \ \forall (y,z)\in (B_{2}\times B_{3})\cup (B_{3}\times B_{2}),\\ G^*(x,z)\ge \tau _3,\ \ \forall (x,z)\in (B_{1}\times B_{3})\cup (B_{3}\times B_{1}), \end{array} \right. \end{aligned}$$

(2.4)

where

$$\begin{aligned} B_1=\{x\in \Omega :|x-x_0|\le \delta \}, \\ B_2=\{x\in \Omega :|x-y_0|\le \delta \}, \\ B_3=\{x\in \Omega :|x-z_0|\le \delta \}. \end{aligned}$$

It is easy to see that

$$\begin{aligned} \text {mes} B_1=\text {mes} B_2=\text {mes} B_3. \end{aligned}$$

On the other hand, from (2.2) and (2.3), we have

$$\begin{aligned} \begin{array}{ll} u(x)=-\int _{{{\bar{\Omega }}}}G^*(x,y)\omega (y)\textrm{d}y\\ \ \ \ \ \ \ \ =\lambda \int _{{{\bar{\Omega }}}}G^*(x,y)\int _{{{\bar{\Omega }}}}G^*(y,z) f(z,u(z))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ =\lambda \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,y)G^*(y,z) f(z,u(z))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ =\lambda \int _{{{\bar{\Omega }}}}G(x,z) f(z,u(z))\textrm{d}z, \end{array} \end{aligned}$$

(2.5)

where

$$\begin{aligned} G(x,z)=\int _{{{\bar{\Omega }}}}G^*(x,y)G^*(y,z)\textrm{d}y. \end{aligned}$$

(2.6)

Thus, we give the expression of Green’s function for problem (1.3). Obviously, $G(x,z)=G(z,x)$ and $G(x,z)\ge 0$ for $x,z\in {{\bar{\Omega }}}$.

3 Proof of Theorems 1.1 and 1.4

In this section, we first consider the multiplicity of positive solutions for problem (1.3) by using the fixed point index in a cone, which is used in Zhang [34].

Lemma 3.1

([8]) Let E be a real Banach space and K be a cone in E. For $r>0$, define $K_{r}=\{x\in K: \Vert x\Vert <r \}$. Assume that $T: {{\bar{K}}}_{r}\rightarrow K$ is completely continuous such that $Tx\ne x$ for $x\in \partial K_{r}=\{x\in K:\Vert x\Vert =r\}$.

(i) If $\Vert Tx\Vert \ge \Vert x\Vert$ for $x\in \partial K_{r}$, then $i(T, K_{r}, K)=0$.

(ii) If $\Vert Tx\Vert \le \Vert x\Vert$ for $x\in \partial K_{r}$, then $i(T, K_{r}, K)=1$.

Let $E=C({{\bar{\Omega }}})$ be the real Banach space with supremum norm $\Vert \cdot \Vert _C$, and define a cone K in E as

$$\begin{aligned} K=\{u:u\in E,\ u(x)\ge 0,\ x\in {{\bar{\Omega }}}\}. \end{aligned}$$

(3.1)

For $\varrho >0$, we also define

$$\begin{aligned} D_\varrho =\{u: u\in E,\ \Vert u\Vert _C<\varrho \}, \end{aligned}$$

and

$$\begin{aligned} \partial K_\varrho = K\cap \partial D_{\varrho }=\{u\in K: \Vert u\Vert _C=\varrho \}. \end{aligned}$$

For $u\in K$, we define $T_\lambda : K\rightarrow E$ to be

$$\begin{aligned} T_\lambda u(x)=\lambda \int _{{{\bar{\Omega }}}}G(x,y)f(y, u(y))\textrm{d}y, \end{aligned}$$

(3.2)

where G(x, y) is defined in (2.6).

When $\mathbf (f_1)$ hold, it is well known that $T_\lambda : K\rightarrow E$ is completely continuous.

Proof of Theorem 1.1

For any given $\tau >0$, it follows from $f_\infty \in (0,+\infty ]$ that there exist $\eta >0$ and $l>\tau$ such that

$$\begin{aligned} f(x,u)\ge \eta ,\ \forall x\in {{\bar{\Omega }}}, \ \ u\ge l. \end{aligned}$$

(3.3)

Letting $\xi =(\tau _1\tau _2\eta (\text {mes}B_2)(\text {mes}B_3))^{-1}l$, then for $0<\lambda <\xi$, one can prove that ${{\hat{T}}}_{\lambda }: K\rightarrow K$ is completely continuous.

Considering $f^0=0$, there exists $0<r<l$ such that

$$\begin{aligned} f(x,u)\le \varepsilon _{1}u,\ \forall x\in {{\bar{\Omega }}},\ 0\le u\le r, \end{aligned}$$

(3.4)

where $\varepsilon _{1}>0$ satisfies

$$\begin{aligned} \lambda \varepsilon _{1}\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C<1, \end{aligned}$$

(3.5)

and for $i\in \{1,2\}$, $\phi _i\in C^2({{\bar{\Omega }}})$ gratify

$$\begin{aligned} \left\{ \begin{array}{l} -\Delta \phi _i=1 \ \ \text {in} \ \Omega ,\\ \phi _i=0\ \ \text {on} \ \partial \Omega . \end{array} \right. \end{aligned}$$

(3.6)

So, for $u\in K\cap \partial D_r$, we have from (2.5), (2.6), (3.2), (3.4) and (3.5) that

$$\begin{aligned} \begin{array}{ll} \Vert T_\lambda u\Vert _C=\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}G(x,y)f(y, u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y, u(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)f(y, u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \lambda \varepsilon _{1}\Vert u\Vert _C\int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \lambda \varepsilon _{1}\Vert u\Vert _C\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \ < \Vert u\Vert _C. \end{array} \end{aligned}$$

It hence follows from (ii) of Lemma 3.1 that

$$\begin{aligned} i(T_{\lambda }, K_{r}, K)=1. \end{aligned}$$

(3.7)

Now turning to $f^\infty =0$, there exists $\sigma >0$ so that

$$\begin{aligned} f(x,u)\le \varepsilon _{2}u,\ \ \forall x\in {{\bar{\Omega }}},\ \ u>\sigma , \end{aligned}$$

where $\varepsilon _{2}>0$ satisfies

$$\begin{aligned} 2\lambda \varepsilon _{2}\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\le 1. \end{aligned}$$

(3.8)

We hence have

$$\begin{aligned} 0\le f(x,u)\le \varepsilon _{2}u+{\mathfrak {M}}_{\sigma },\ \ \forall x\in {{\bar{\Omega }}},\ \ u\ge 0, \end{aligned}$$

(3.9)

where

$$\begin{aligned} {\mathfrak {M}}_{\sigma }=\max \limits _{x\in {{\bar{\Omega }}},\ 0\le u\le \sigma }f(x,u)+1>0. \end{aligned}$$

Let

$$\begin{aligned} R>\max \bigg \{l,2\lambda {\mathfrak {M}}_{\sigma }\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\bigg \}. \end{aligned}$$

(3.10)

Thus, for $u\in K\cap \partial D_{R}$, we derive from (2.5), (2.6), (3.2), (3.8), (3.9) and (3.10) that

$$\begin{aligned} \begin{array}{ll} \Vert T_\lambda u\Vert _C=\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}G(x,y)f(y, u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y, u(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)f(y, u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \lambda (\varepsilon _{2}\Vert u\Vert _C+{\mathfrak {M}}_{\sigma })\int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \lambda (\varepsilon _{2}\Vert u\Vert _C+{\mathfrak {M}}_{\sigma })\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \ < \frac{\Vert u\Vert _C}{2}+\frac{R}{2}=\Vert u\Vert _C. \end{array} \end{aligned}$$

It hence follows from (ii) of Lemma 3.1 that

$$\begin{aligned} i(T_{\lambda }, K_{R}, K)=1. \end{aligned}$$

(3.11)

On the other hand, for $u \in {{\bar{K}}}_{l}^{R}=\{u\in K: \Vert u\Vert _C\le R, \min \limits _{x\in B_3}u(x)\ge l\}$, (2.5), (2.6), (3.2), (3.8), (3.9) and (3.10) yield that

$$\begin{aligned} \Vert T_\lambda u\Vert _C<R. \end{aligned}$$

Furthermore, for $u \in {{\bar{K}}}_{l}^{R}$, from (2.4) (2.5), (2.6), (3.2), and (3.3), we obtain that

$$\begin{aligned} \begin{array}{l} \min \limits _{x\in B_1}( T_\lambda u)(x)=\lambda \min \limits _{x\in B_1}\int _{{{\bar{\Omega }}}}G(x,y)f(y,u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge \lambda \min \limits _{x\in B_1}\int _{B_2}\int _{B_3}G^*(x,z)G^*(z,y)f(y, u(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge \lambda \tau _1\tau _2\eta (\text {mes}B_2)(\text {mes}B_3)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ >\xi \tau _1\tau _2\eta (\text {mes}B_2)(\text {mes}B_3)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =l. \end{array} \end{aligned}$$

Letting $u_{0}\equiv \frac{l+R}{2}$ and $H(t,u)=(1-t)T_{\lambda }u+tu_{0}$, then $H:[0,1]\times {{\bar{K}}}_{l}^{R}\rightarrow K$ is completely continuous, and from the analysis above, we obtain for $(t,u)\in [0,1]\times {{\bar{K}}}_{l}^{R}$

$$\begin{aligned} H(t,u)\in K_{l}^{R}. \end{aligned}$$

(3.12)

Therefore, for $t\in [0,1], u\in \partial K_{l}^{R}$, we have $H(t,u)\ne u.$ Hence, by the normality property and the homotopy invariance property of the fixed point index, we obtain

$$\begin{aligned} i( T_{\lambda }, K_{l}^{R}, K)=i(u_{0}, K_{l}^{R}, K)=1. \end{aligned}$$

(3.13)

Consequently, by the solution property of the fixed point index, $T_{\lambda }$ admits a fixed point $u_{\lambda }^{(1)}$ with $u_{\lambda }^{(1)} \in K_{l}^{R}$, and

$$\begin{aligned} \max \limits _{x\in {{\bar{\Omega }}}}u_{\lambda }^{(1)}(x)\ge \min \limits _{x\in B_3}u_{\lambda }^{(1)}(x)>l>\tau . \end{aligned}$$

On the other hand, it follows from (3.7), (3.11) and (3.13) together with the additivity of the fixed point index that

$$\begin{aligned} \begin{array}{ll} i(T_{\lambda }, K_{R}\backslash ({{\bar{K}}}_{r}\cup {{\bar{K}}}_{l}^{R}), K) \\ = i(T_{\lambda }, K_{R}, K)-i(T_{\lambda }, K_{l}^{R}, K)-i(T_{\lambda }, K_{r}, K) \\ =1-1-1=-1. \end{array} \end{aligned}$$

(3.14)

According to the solution property of the fixed point index, $T_{\lambda }$ so possesses a fixed point $u_{\lambda }^{(2)}$ with $u_{\lambda }^{(2)} \in K_{R}\backslash ({{\bar{K}}}_{r}\cup \bar{K}_{l}^{R})$. It is easy to see that $u_{\lambda }^{(1)}\ne u_{\lambda }^{(2)}$. This finishes the proof of Theorem 1.1. $\square$

Next, we will prove the existence and nonexistence of positive solution to problem (1.3). To this goal, we need to state one well-known result of the fixed point index on cones for completely continuous operators, which is the base of our approaches.

Lemma 3.2

(Corollary 2.3.1, Guo and Lakshmikantham [11]) Let K be a cone in a real Banach space E and let $\Omega$ be a bounded open set of E. Assume that the operator $A:K\cap {{\bar{\Omega }}} \rightarrow K$ is completely continuous. If there exists a $u_{0}>0$ such that

$$\begin{aligned} u-Au\ne tu_{0},\ \ \forall u\in K\cap \partial \Omega ,\ \ t\ge 0, \end{aligned}$$

then

$$\begin{aligned} i(A, K\cap \Omega , K)=0. \end{aligned}$$

Proof of Theorem 1.4

It is well known that problem (1.3) is equivalent to the following nonlinear integral equation

$$\begin{aligned} u(x)=\lambda \int _{{{\bar{\Omega }}}}G(x,y)f(y,u(y))\textrm{d}y, \end{aligned}$$

(3.15)

where G(x, y) is defined in (2.6).

Consider the operator

$$\begin{aligned} {{\hat{T}}} u(x)=\int _{{{\bar{\Omega }}}}G(x,y)f(y, u(y))\textrm{d}y. \end{aligned}$$

(3.16)

Since G(x, y) and f(x, u) are nonnegative, it is easy to see that ${{\hat{T}}}: K\rightarrow K$ is completely continuous.

Part (i). It follows from $0<f^{0}<+\infty$ that there exist $0<l_{1}<l_{2}$ and $\mu >0$ such that

$$\begin{aligned} l_{1} u< f(x,u)< l_{2} u\ \ (\forall x\in {{\bar{\Omega }}},\ \ 0\le u\le \mu ). \end{aligned}$$

(3.17)

Let $\lambda _{0}=(l_{1}\tau _1\tau _2\text {mes} B_2\text {mes} B_3)^{-1}$ and $l_{0}=\mu$. We now demonstrate that $\lambda _0$ and $l_0$ are the numbers to be required.

On one hand, for $u\in K\cap \partial D_{r}$, we have

$$\begin{aligned} 0\le u(x)\le r<l_0=\mu ,\ \ x\in {{\bar{\Omega }}}. \end{aligned}$$

On the other hand, we may suppose that

$$\begin{aligned} u-\lambda _0 {{\hat{T}}}u\ne 0\ \ (\forall u\in K\cap \partial D_{r}). \end{aligned}$$

(3.18)

If not, then there is $u_{r}\in K\cap \partial D_{r}$ such that $\lambda _0 {{\hat{T}}} u_{r}=u_{r}$ and so (1.4) already holds for $\lambda _{r}=\lambda _{0}$.

Define $\psi (x)\equiv 1$ for $x\in {{\bar{\Omega }}}$. Then, $\psi \in K$ gratifying $\Vert \psi \Vert _C\equiv 1$. We now demonstrate that

$$\begin{aligned} u-\lambda _0 {{\hat{T}}}u\ne \zeta \psi \ \ (\forall u\in K\cap \partial D_{r}, \ \zeta \ge 0). \end{aligned}$$

(3.19)

Assume that there are $u_{1}\in K\cap \partial D_{r}$ and $\zeta _{1}\ge 0$ such that $u_{1}-\lambda _0 \hat{T}u_{1}=\zeta _{1}\psi$, then (3.18) indicates that $\zeta _{1}>0$, and $u_{1}=\zeta _{1}\psi +\lambda _0 {{\hat{T}}}u_{1}\ge \zeta _{1}\psi$. Let $\zeta ^{*}=\sup \{\zeta |u_{1}\ge \zeta \psi \}$. Then $\zeta _{1}\le \zeta ^{*}<+\infty$ and $u_{1}\ge \zeta ^{*}\psi$. Therefore,

$$\begin{aligned} \zeta ^{*}=\zeta ^{*}\Vert \psi \Vert _C\le \Vert u_{1}\Vert _C=r. \end{aligned}$$

(3.20)

Consequently, for any $x\in B_{1}$, we derive from (2.4), (2.5), (2.6), (3.16), and (3.17), that

$$\begin{aligned} \begin{array}{ll} u_{1}(x)=\lambda _0\int _{{{\bar{\Omega }}}}G(x,y)f(y,u_1(y))\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ =\lambda _0\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*{z,y}f(y,u_1(y))\textrm{d}z\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \lambda _0\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*{z,y}l_1u_1(y)\textrm{d}z\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \lambda _0\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*{z,y}l_1\zeta ^*\psi (z)\textrm{d}z\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \lambda _0l_1\zeta ^*\int _{B_2}G^*(x,z)\textrm{d}z\int _{B_3}G^*{z,y}\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \lambda _0l_1\zeta ^*\tau _1\tau _2\text {mes} B_2\text {mes} B_3+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ = \zeta ^{*}+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ = (\zeta ^{*}+\zeta _{1})\psi (x), \end{array} \end{aligned}$$

which indicates that $u_{1}(x)\ge (\zeta ^{*}+ \zeta _{1}) \psi (x)$ for $x\in B_1$. This contradicts the definition of $\zeta ^{*}$. Thus, (3.19) holds and hence it yields from Lemma 3.2 that

$$\begin{aligned} i(\lambda _0 {{\hat{T}}},K\cap D_{r},K)=0. \end{aligned}$$

(3.21)

It is widely known that

$$\begin{aligned} i(\theta ,K\cap D_{r},K)=1, \end{aligned}$$

(3.22)

where $\theta$ denotes the zero operator.

Therefore, it derives from (3.21) and (3.22) and the homotopy invariance that there are $u_{r}\in K\cap \partial D_{r}$ and $0<\nu _{r}<1$ so that $\nu _{r} \lambda _0{{\hat{T}}}u_{r}=u_{r}$, which indicates that

$$\begin{aligned} 0<\lambda _{r}=\lambda _{0}\nu _{r}<\lambda _{0}. \end{aligned}$$

This gives the proof of (1.4).

Part (ii). If $f^{0}=+\infty$, then there are $l_3>0$ and $\mu ^{*}>0$ so that

$$\begin{aligned} f(x,u)> l_3 u,\ \ \forall x\in {{\bar{\Omega }}},\ \ 0\le u\le \mu ^{*}. \end{aligned}$$

Next, we verify that $l^*=\mu ^{*}$ and $\lambda _{*}=(l_{3}\tau _1\tau _2\text {mes} B_2\text {mes} B_3)^{-1}$ are required. Thus, for $u\in K\cap \partial D_{r^*}$, we derive that

$$\begin{aligned} 0\le u(x)\le r^*<l^*=\mu ^*,\ \ x\in {{\bar{\Omega }}}. \end{aligned}$$

Similar to the proof of (i), replacing (3.18), one can suppose that

$$\begin{aligned} u-\lambda _*{{\hat{T}}}u\ne 0\ \ (\forall u\in K\cap \partial D_{r^*}), \end{aligned}$$

and replacing (3.19) we can demonstrate that

$$\begin{aligned} u-\lambda _*{{\hat{T}}}u\ne \zeta \psi \ \ (\forall u\in K\cap \partial D_{r^*}, \ \zeta \ge 0). \end{aligned}$$

It follows from Lemma 3.2 that $i(\lambda _*{{\hat{T}}},K\cap D_{r^*},K)=0.$ Seeing that $i(\theta ,K\cap D_{r},K)=1$, one can easily demonstrate that there are $u_{r^*}\in K\cap \partial D_{r^*}$ and $0<\nu _{r^*}<1$ so that $\nu _{r^*} \lambda _*{{\hat{T}}}u_{r^*}=u_{r^*}$. So, Theorem 1.4 (ii) holds for $\lambda _{r^*}= \lambda _{*}\nu _{r^*}<\lambda _{*}$.

Part (iii). If $f^{0}<\infty$ and $f^{\infty }<\infty$, then there are positive numbers $\eta _{1}>0,\ \ \eta _{2}>0,\ \ h_{1}>0$ and $h_{2}>0$ so that $h_{1}<h_{2}$ and for $x \in {{\bar{\Omega }}},\ \ 0<u\le h_{1}$, we derive that

$$\begin{aligned} f(x,u)\le \eta _{1}u, \end{aligned}$$

(3.23)

and for $x \in {{\bar{\Omega }}},\ \ u\ge h_{2}$, we derive that

$$\begin{aligned} f(x,u)\le \eta _{2}u. \end{aligned}$$

(3.24)

Set

$$\begin{aligned} \eta ^{*}=\max \bigg \{\eta _{1},\eta _{2}, \ \ \max \bigg \{\frac{f(x,u)}{u}: x \in {{\bar{\Omega }}},\ h_{1}\le u \le h_{2}\bigg \}\bigg \}>0. \end{aligned}$$

Then, we derive that

$$\begin{aligned} f(x,u)\le \eta ^{*} u,\ x \in {{\bar{\Omega }}},\ u \in [0,\infty ). \end{aligned}$$

(3.25)

Assume that $v\in K$ is a positive solution to problem (1.3). We will demonstrate that this leads to a contradiction for $\lambda < {\underline{\lambda }}=(\eta ^{*}\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C)^{-1}$.

In fact, for $\lambda <{\underline{\lambda }}$, we derive from (2.5), (2.6), (3.2), and (3.25) that

$$\begin{aligned} \begin{array}{ll} \Vert T_\lambda v\Vert _C=\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}G(x,y)f(y, v(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y, v(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\lambda \int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)f(y, v(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \lambda \eta ^{*}\Vert v\Vert _C\int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \lambda \eta ^{*}\Vert v\Vert _C\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \ < {\underline{\lambda }}\eta ^{*}\Vert v\Vert _C\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \ =\Vert v\Vert _C, \end{array} \end{aligned}$$

which indicates that

$$\begin{aligned} \Vert v\Vert _C=\Vert T_\lambda v\Vert _C< \Vert v\Vert _C. \end{aligned}$$

This is a contradiction. $\square$

Proof of Corollary 1.5

The proof of Corollary 1.1 is a direct consequence of the proof for Theorem 1.4 (iii). Under the conditions of Corollary 1.5, we can obtain the intervals of $\lambda$ so that problem (1.3) admits no positive solutions. $\square$

Remark 3.3

If we consider the following Navier boundary value problem

$$\begin{aligned} \left\{ \begin{array}{l} \lambda \Delta ^2 u= f(x,u) \ \ \text {in} \ \Omega ,\\ u=\Delta u=0\ \ \text {on} \ \partial \Omega , \end{array} \right. \end{aligned}$$

(3.26)

where $\lambda \ne 0$ is a parameter, $\Omega$ is a bounded domain in ${\mathbb {R}}^n\ (n\ge 2)$, and the nonlinearity f satisfies $\mathbf (f_{1})$, then we have the following conclusions.

Theorem 3.4

Under condition $\mathbf (f_{1})$ holds, if $0<f^{0}<+\infty$, then there exists $l_{0}>0$ such that, for every $0<r<l_{0}$, problem (3.26) admits a positive solution $u_{r}$ satisfying $\Vert u_{r}\Vert _C=r$ associated with

$$\begin{aligned} \lambda =\lambda _{r}\in [\lambda _{0},{{\bar{\lambda }}}_{0}], \end{aligned}$$

where $\lambda _{0}$ and ${{\bar{\lambda }}}_{0}$ are two positive finite numbers.

Proof

The proof is similar to that of Theorem 1.4 (i). We hence omit it here. $\square$

4 Proof of Theorem 1.6 and Theorem 1.7

In this section, we will prove Theorem 1.6 and Theorem 1.7. To achieve this goal, we first state a well-known result of the fixed point, which is the base of our approaches.

Lemma 4.1

(Theorem 2.3.3, Guo and Lakshmikantham [11]) Let $\Omega _{1}$ and $\Omega _{2}$ be two bounded open sets in a real Banach space E such that $\theta \in \Omega _{1}$ and ${{\bar{\Omega }}}_{1}\subset \Omega _{2}$. Let P be a cone in E and let operator $A:P\cap ({{\bar{\Omega }}}_{2}\backslash \Omega _{1})\rightarrow P$ be completely continuous. Suppose that one of the following two conditions

(a) $Ax \not \ge x,\forall \ \ x\in P\cap \partial \Omega _{1}$ and $Ax \not \le x,\forall \ x\in P\cap \partial \Omega _{2}$

and

(b) $Ax \not \le x,\forall \ x\in P\cap \partial \Omega _{1}$ and $Ax \not \ge x,\forall \ \ x\in P\cap \partial \Omega _{2}$

is satisfied. Then, A has at least one fixed point in $P\cap (\Omega _{2}\backslash {{\bar{\Omega }}}_{1})$.

Remark 4.2

It is clear to see that the fixed point of A in Lemmas 4.1 can not reach the boundary of $\Omega _1$ and $\Omega _2$.

Proof of Theorem 1.6

It is well known that problem (1.3) is equivalent to the following nonlinear integral equation

$$\begin{aligned} u(x)=\int _{{{\bar{\Omega }}}}G(x,y)f(y,u(y))\textrm{d}y \end{aligned}$$

(4.1)

when $\lambda =1$, where G(x, y) is defined in (2.6).

Consider the operator

$$\begin{aligned} T_1u(x)=\int _{{{\bar{\Omega }}}}G(x,y)f(y, (u(y))\textrm{d}y, \end{aligned}$$

(4.2)

It is generally known that $T_1$ maps K into K is a completely continuous operator.

Case (1), $f^0=0$.

Considering $f^0=0$, there exists $r>0$ such that

$$\begin{aligned} f(x,u)\le \varepsilon r,\ \forall x\in {{\bar{\Omega }}},\ 0\le u\le r, \end{aligned}$$

(4.3)

where $\varepsilon >0$ satisfy $\varepsilon \Vert \phi _1\Vert _C\Vert \phi _2\Vert _C<1$, and $\phi _i$ are defined in (3.6) for $i\in \{1,2\}$.

We can prove that

$$\begin{aligned} Au\not \ge u,\ \ u\in K,\ \ \Vert u\Vert _{C}=r. \end{aligned}$$

(4.4)

In fact, if there exists $u_{1}\in K\cap \partial D_{r}$ such that $T_1u_{1}\ge u_{1}$, then from (2.5), (2.6), (3.6), (4.2) and (4.3) we have

$$\begin{aligned} \begin{array}{ll} 0\le u_{1}(x)\le T_1u_1(x)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}G(x,y)f(y, u_1(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y, u_1(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)f(y, u_1(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \varepsilon r\int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \varepsilon r\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \ \ \ <r=\Vert u_{1}\Vert _{C}. \end{array} \end{aligned}$$

This leads to $\Vert u_{1}\Vert _{C}<\Vert u_{1}\Vert _{C}$, which is a contraction. It so follows that (4.4) holds.

Case (2), $f^\infty =0$.

Considering $f^{\infty }=0$, there exists ${{\bar{R}}}>r>0$ such that

$$\begin{aligned} f(x,u)\le {{\bar{\varepsilon }}} u,\ \forall x\in {{\bar{\Omega }}},\ u\ge {{\bar{R}}}, \end{aligned}$$

where ${{\bar{\varepsilon }}}>0$ satisfies $\frac{1}{2}{{\bar{\varepsilon }}}\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C<1$.

We hence have

$$\begin{aligned} 0\le f(x,u)\le {{\bar{\varepsilon }}} u+{\mathfrak {M}},\ \ \forall x\in {{\bar{\Omega }}},\ \ u\ge 0, \end{aligned}$$

(4.5)

where

$$\begin{aligned} {\mathfrak {M}}=\max \limits _{x\in {{\bar{\Omega }}},\ 0\le u\le {{\bar{R}}}}f(x,u)+1>0. \end{aligned}$$

Let

$$\begin{aligned} R>\max \bigg \{{{\bar{R}}},2{\mathfrak {M}}\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\bigg \}. \end{aligned}$$

(4.6)

Then, one can prove that

$$\begin{aligned} T_1u\not \ge u,\ \ u\in K,\ \ \Vert u\Vert _{C}=R. \end{aligned}$$

(4.7)

In fact, if there exists $u_{2}\in K$ with $\Vert u_{2}\Vert _{C}=R$ so that $T_1u_{2}\ge u_{2}$, then it follows from (2.5), (2.6), (3.6), (4.5) and (4.6) that

$$\begin{aligned} \begin{array}{ll} 0\le u_{2}(x)\le T_1u_2(x)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}G(x,y)f(y, u_2(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y, u_2(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)f(y, u_2(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \varepsilon ({{\bar{\varepsilon }}} \Vert u_2\Vert _C+{\mathfrak {M}})\int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \le ({{\bar{\varepsilon }}} \Vert u_2\Vert _C+{\mathfrak {M}})\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \ \ \ <\frac{\Vert u_2\Vert _C}{2}+\frac{R}{2}=\Vert u_{2}\Vert _{C}, \end{array} \end{aligned}$$

which leads to $\Vert u_{2}\Vert _{C}<\Vert u_{2}\Vert _{C}$. This is a contraction. It so follows that (4.7) holds.

Next, we demonstrate that

$$\begin{aligned} T_1u\not \le u,\ \ u\in K,\ \ \Vert u\Vert _{C}=\eta , \end{aligned}$$

(4.8)

where ${{\bar{R}}}<\eta <R$.

In reality, if there is $u_{0}\in K$ with $\Vert u_{0}\Vert _{C}=\eta$ so that $T_1u_{0}\le u_{0}$.

It so follows from (1.5), when a $\eta$ is fixed, there is a $l> 0$ so that

$$\begin{aligned} f(x,u_1)\ge l>\frac{\eta }{\tau _1\tau _2\text {mes} B_2\text {mes} B_3},\ \forall x\in {{\bar{\Omega }}},\ u\ge \eta . \end{aligned}$$

(4.9)

We notice that $u\ge \eta$ implies that $\Vert u\Vert _C\ge \eta$. Therefore, for $u\in \partial K_\eta$, we derive from (2.4), (2.5), (2.6), (3.2) and (4.9) that

$$\begin{aligned} \begin{array}{ll} x\in B_{1}\Longrightarrow u_0(x)\ge T_1u_0(x)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}G(x,y)f(y,u_0(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge \int _{B_2}\int _{B_3}G^*(x,z)G^*(z,y)f(y, u_0(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge \tau _1\tau _2l(\text {mes}B_2)(\text {mes}B_3)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ >\eta . \end{array} \end{aligned}$$

This is a contraction. It so follows that (4.8) holds.

Applying Lemma 4.1 to (4.4) and (4.8) or (4.7) and (4.8) yields that operator $T_1$ possesses one fixed point u with $u\in K_{r,\eta }$ or $u\in K_{\eta ,R}$. It hence follows that problem (1.3) admits at least one positive solutions u with $r<\Vert u\Vert _{C}<\eta$ or $\eta<\Vert u\Vert _{C}<R$. This gives the proof of Theorem 1.6. $\square$

Proof of Theorem 1.7

Take $0<\eta _1<\eta <\eta _2$. When $f^0=0$, similar to the proof of (4.4), one can demonstrate that

$$\begin{aligned} T_1u\not \ge u,\ \ u\in K,\ \ \Vert u\Vert _{C}=\eta _1. \end{aligned}$$

(4.10)

When $f^\infty =0$, similar to the proof of (4.7), we derive that

$$\begin{aligned} T_1u\not \ge u,\ \ u\in K,\ \ \Vert u\Vert _{C}=\eta _2. \end{aligned}$$

(4.11)

Under condition (1.5), similar to the proof of (4.8), one can prove that

$$\begin{aligned} T_1u\not \le u,\ \ u\in K,\ \ \Vert u\Vert _{C}=\eta . \end{aligned}$$

(4.12)

Therefore, from (4.10), (4.11) and (4.12), Lemma 4.1 yields that $T_1$ possesses two fixed point $u^{*}, u^{**}$ gratifying that $u^{*}\in K_{\eta _1,\eta }, u^{**}\in K_{\eta ,\eta _2}$. It so follows that problem (1.3) possesses at least two positive solutions $u^{*}, u^{**}$ gratifying that

$$\begin{aligned} 0<\Vert u^{*}\Vert _{C}<\eta <\Vert u^{**}\Vert _{C}. \end{aligned}$$

This completes the proof of Theorem 1.7. $\square$

5 Proof of Theorems 1.8 and 1.9

In this section, we will employ the following fixed point theorems on cones to prove Theorem 1.8 and Theorem 1.9 for the case $0<\alpha <1$ and $0<\beta <1$.

Lemma 5.1

(See [2], Theorem 12.3) Let P be a cone in a real Banach space E. Assume $\Omega _{1}, \Omega _{2}$ are bounded open sets in E with $\theta \in \Omega _{1},\ \ {{\bar{\Omega }}}_{1}\subset \Omega _{2}$. If

$$\begin{aligned} A:P\cap ({{\bar{\Omega }}}_{2}\backslash \Omega _{1})\rightarrow P \end{aligned}$$

is completely continuous such that either

• (a) there exists a $u_{0}>0$ such that $u-Au\ne tu_{0}, \forall u\in P\cap \partial \Omega _2, t\ge 0$; $Au\ne \mu u, \forall u\in P\cap \partial \Omega _1, \mu \ge 1$, or

• (b) there exists a $u_{0}>0$ such that $u-Au\ne tu_{0}, \forall u\in P\cap \partial \Omega _1, t\ge 0$; $Au\ne \mu u, \forall u\in P\cap \partial \Omega _2, \mu \ge 1$. Then A has at least one fixed point in $P\cap (\Omega _{2}\backslash {{\bar{\Omega }}}_{1})$.

Remark 5.2

Obviously, the fixed point of A in Lemmas 5.1 can not reach the boundary of $\Omega _1$ and $\Omega _2$.

Proof of Theorem 1.8

We assume that there is $r_1>0$ so that

$$\begin{aligned} u-T_1 u\ne \theta ,\ \forall u\in K,\ 0<\Vert u\Vert _C\le r_1. \end{aligned}$$

(5.1)

If not, then there is $u_{r_1}\in K\cap \partial D_{r_1}$ so that

$$\begin{aligned} T_1u_{r_1}=u_{r_1}. \end{aligned}$$

Considering $f_0=\infty$, there is $\sigma >0$ and $r_2>0$ so that

$$\begin{aligned} f(x,u)\ge \sigma u^{\beta }\ \ (\forall x\in {{\bar{\Omega }}},\ \ 0\le u\le r_2). \end{aligned}$$

(5.2)

Let $\psi (x)\equiv 1$ for $x\in {{\bar{\Omega }}}$. Then $\psi \in K$ with $\Vert \psi \Vert _C\equiv 1$. Next, we demonstrate that

$$\begin{aligned} u-T_1u\ne \zeta \psi \ \ (\forall u\in K\cap \partial D_{r}, \ \zeta \ge 0), \end{aligned}$$

(5.3)

where

$$\begin{aligned} 0<r<\min \{r_1, r_2,(\tau \sigma \text {mes}B_{\delta })^{\frac{1}{1-\beta }}\}. \end{aligned}$$

In reality, if there are $u_{1}\in K\cap \partial D_{r}$ and $\zeta _{1}\ge 0$ such that $u_{1}-T_1u_{1}=\zeta _{1}\psi$, then (5.1) indicates that $\zeta _{1}>0$. But, $u_{1}=\zeta _{1}\psi +T_1u_{1}\ge \zeta _{1}\psi$. Set

$$\begin{aligned} \zeta ^{*}=\sup \{\zeta |u_{1}\ge \zeta \psi \}. \end{aligned}$$

Thus, we have $\zeta _{1}< \zeta ^{*}<+\infty$ and $u_{1}\ge \zeta ^{*}\psi$. So,

$$\begin{aligned} \zeta ^{*}=\zeta ^{*}\Vert \psi \Vert _C\le \Vert u_{1}\Vert _C=r\le (\tau _1\tau _2\sigma \text {mes}B_{2}\text {mes}B_{3})^{\frac{1}{1-\beta }}. \end{aligned}$$

(5.4)

Therefore, for any $x\in B_{1}$, we follow from (2.4), (2.5), (2.6), (3.2), (5.2) and (5.4) that

$$\begin{aligned} \begin{array}{ll} u_{1}(x)=\int _{{{\bar{\Omega }}}}G(x,y)f(y,u_1(y))\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \int _{{{\bar{\Omega }}}}G(x,y)\sigma u_1^{\beta }(y)\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \int _{{{\bar{\Omega }}}}G(x,y)\sigma (\zeta ^{*}\psi (y))^\beta \textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \sigma (\zeta ^{*})^\beta \int _{B_{2}}G^*(x,z)\textrm{d}z\int _{B_{3}}G^*(z,y)\textrm{d}y+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \sigma (\zeta ^{*})^\beta \tau _1\tau _2\text {mes} B_2\text {mes} B_3+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \zeta ^{*}+\zeta _{1}\psi (x)\\ \ \ \ \ \ \ \ \ = (\zeta ^{*}+\zeta _{1})\psi (x). \end{array} \end{aligned}$$

This contradicts the definition of $\zeta ^{*}$. So (5.3) holds.

Next, turning to $f^\infty =0$, then there are $l>0$ and $r_2>0$ so that

$$\begin{aligned} f(x,u)\le lu^{\alpha }(x),\ \forall u\ge r_2. \end{aligned}$$

Let

$$\begin{aligned} L=\max \limits _{x\in {{\bar{\Omega }}}, 0\le u\le r_2}f(x,u). \end{aligned}$$

Then we derive that

$$\begin{aligned} f(x,u)\le l\Vert u\Vert ^{\alpha }_C+L,\ \forall x\in {{\bar{\Omega }}}, u\in [0,+\infty ). \end{aligned}$$

(5.5)

Take R be large enough ($R>r$) such that

$$\begin{aligned} \frac{L\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C}{R}+\frac{l\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C}{R^{1-\alpha }}<1. \end{aligned}$$

(5.6)

Now, we are going to prove that

$$\begin{aligned} \forall u\in K\cap \partial D_{R},\ \mu \ge 1\Rightarrow T_1 u\ne \mu u. \end{aligned}$$

(5.7)

In fact, if there are $u_{2}\in K\cap \partial D_{R}$ and $\mu _0\ge 1$ such that $T_1u_{2}=\mu _0 u_2$, then it follows from (2.5), (2.6), (3.2), (3.6) and (5.5) that

$$\begin{aligned} \begin{array}{ll} \mu _0u_{2}(x)=\int _{{{\bar{\Omega }}}}G(x,y)f(y,u_2(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*{z,y}f(y,u_2(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le \int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)(L+lu^{\alpha }(z))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le (L+l\Vert u\Vert ^{\alpha })\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \le (L+l\Vert u\Vert ^{\alpha })\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C. \end{array} \end{aligned}$$

(5.8)

Thus it follows from (5.8) that

$$\begin{aligned} \begin{array}{ll} \mu _0 R=\mu _0 \Vert u_2\Vert _C\\ \ \ \ \ \ \ \le (L+l\Vert u\Vert ^{\alpha })\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \le (L+lR^{\alpha })\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C. \end{array} \end{aligned}$$

It hence derives from (5.6) that

$$\begin{aligned} \mu _0\le \frac{L\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C}{R}+\frac{l\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C}{R^{1-\alpha }}<1. \end{aligned}$$

This contradicts $\mu _0\ge 1$, which indicates that (5.7) holds.

Therefore, according to (b) of Lemma 5.1, it yields from (5.3) and (5.7) that operator $T_1$ possesses a fixed point u in $K\cap (D_{R}\backslash {{\bar{D}}}_{r})$ with $r< \Vert u\Vert _C< R$. This follows that problem (1.3) has at least one positive solution u with $r<\Vert u\Vert _C< R$, and so the proof of Theorem 1.8 is completed. $\square$

Proof of Theorem 1.9

Take $0<\eta <\eta _1$. When $f^\infty =0$, similar to the proof of (5.7), we can prove that

$$\begin{aligned} \forall u\in K\cap \partial D_{\eta _1},\ \mu \ge 1\Rightarrow T_1 u\ne \mu u. \end{aligned}$$

(5.9)

Let $\psi (x)\equiv 1$ for $x\in {{\bar{\Omega }}}$. Then, $\psi \in K$ with $\Vert \psi \Vert _C\equiv 1$. Next, we can prove that

$$\begin{aligned} u-T_1u\ne \zeta \psi \ \ (\forall u\in K\cap \partial D_{\eta }, \ \zeta \ge 0). \end{aligned}$$

(5.10)

In reality, if there are $u_{2}\in K\cap \partial D_{\eta }$ and $\zeta _{2}\ge 0$ such that $u_{2}-T_1u_{2}=\zeta _{2}\psi$, then (5.10) indicates that $\zeta _{2}>0$. But, $u_{2}=\zeta _{2}\psi +T_1u_{2}\ge \zeta _{2}\psi$. Set

$$\begin{aligned} \zeta ^{**}=\sup \{\zeta ^*|u_{2}\ge \zeta ^*\psi \}. \end{aligned}$$

Then we derive $\zeta _{2}< \zeta ^{**}<+\infty$ and $u_{2}\ge \zeta ^{**}\psi$. We so have

$$\begin{aligned} \zeta ^{**}=\zeta ^{**}\Vert \psi \Vert _C\le \Vert u_{2}\Vert _C=\eta \le (\tau _1\tau _2\sigma \text {mes}B_{2}\text {mes}B_{3})^{\frac{1}{1-\beta }}. \end{aligned}$$

(5.11)

On the other hand, it follows from (1.5), when a $\eta$ is fixed, there exists a $l> 0$ such that

$$\begin{aligned} f(x,u_2)\ge l>\frac{\eta }{\tau _1\tau _2\text {mes}B_{2}\text {mes}B_{3}},\ \forall x\in {{\bar{\Omega }}},\ u\ge \eta . \end{aligned}$$

(5.12)

Thus, for any $x\in B_{1}$ and $u\in \partial K_\eta$, we follow from (2.4), (2.5), (2.6), (3.2), (5.11) and (5.12) that

$$\begin{aligned} \begin{array}{ll} u_{2}(x)=\int _{{{\bar{\Omega }}}}G(x,y)f(y,u_2(y))\textrm{d}y+\zeta _{2}\psi (x)\\ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y,u_2(y))\textrm{d}z\textrm{d}y+\zeta _{2}\psi (x)\\ \ \ \ \ \ \ \ \ \ge l\int _{B_2}\int _{B_3}G^*(x,zy)G^*(z,y)\textrm{d}z\textrm{d}y+\zeta _{2}\psi (x)\\ \ \ \ \ \ \ \ \ \ge l\tau _1\tau _2\text {mes}B_{2}\text {mes}B_{3}+\zeta _{2}\psi (x)\\ \ \ \ \ \ \ \ \ > \eta +\zeta _{2}\psi (x)\\ \ \ \ \ \ \ \ \ \ge \zeta ^{**}+\zeta _{2}\psi (x)\\ \ \ \ \ \ \ \ \ =(\zeta ^{**}+\zeta _{2})\psi (x). \end{array} \end{aligned}$$

This contradicts the definition of $\zeta ^{**}$. Therefore (5.10) holds. This completes the proof of Theorem 1.9. $\square$

6 Proof of Theorem 1.11

In this section, we intend to apply the following fixed point theorem on cones to demonstrate Theorem 1.11 for the case $\alpha >1$.

Lemma 6.1

(Theorem 2.3.4, Guo-Lakshmikantham [11]) Let P be a cone in a real Banach space E. Assume $\Omega _{1},\ \ \Omega _{2}$ are bounded open sets in E with $\theta \in \Omega _{1},\ \ {{\bar{\Omega }}}_{1}\subset \Omega _{2}$. If

$$\begin{aligned} A:P\cap ({{\bar{\Omega }}}_{2}\backslash \Omega _{1})\rightarrow P \end{aligned}$$

is completely continuous such that either

• (a) $\Vert Ax\Vert \le \Vert x\Vert ,\ \ \forall x\in P\cap \partial \Omega _{1}$ and $\Vert Ax\Vert \ge \Vert x\Vert ,\ \forall x\in P\cap \partial \Omega _{2},$ or

• (b) $\Vert Ax\Vert \ge \Vert x\Vert ,\ \ \forall x\in P\cap \partial \Omega _{1}$ and $\Vert Ax\Vert \le \Vert x\Vert ,\ \forall x\in P\cap \partial \Omega _{2},$ then A has at least one fixed point in $P\cap ({{\bar{\Omega }}}_{2}\backslash \Omega _{1})$.

Remark 6.2

Comparing with Lemma 4.1 and Lemma 5.1, the fixed point of A in Lemmas 6.1 can reach the boundary of $\Omega _1$ and $\Omega _2$.

Proof of Theorem 1.11

Since $f^{0}=0$, then there is $0<r<1$ such that

$$\begin{aligned} f(x,u)\le \varepsilon _{1}\Vert u\Vert ^\alpha _C,\ \forall x\in {{\bar{\Omega }}},\ 0\le u\le r, \end{aligned}$$

(6.1)

where $\varepsilon _{1}>0$ gratifies

$$\begin{aligned} \varepsilon _{1}\Vert \phi _1\Vert _C\Vert \phi _1\Vert _C\le 1, \end{aligned}$$

and $\phi _i$ are defined in (3.6) for $i\in \{1,2\}$.

Thus, for $x\in {{\bar{\Omega }}},\ u\in K\cap \partial D_{r_{1}},$ it hence follows from (3.2), (3.6), (6.1), $\alpha >1$ and $0<r=\Vert u\Vert _C<1$ that

$$\begin{aligned} \begin{array}{ll} \Vert T_1 u\Vert _C=\max \limits _{x\in {{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G(x,y)f(y,u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ =\max \limits _{x\in {{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*(z,y)f(y,u(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \le \varepsilon _{1} \Vert u\Vert ^{\alpha }_C\max \limits _{x\in {{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)\textrm{d}z\int _{{{\bar{\Omega }}}}G^*(z,y)\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \le \varepsilon _{1} \Vert u\Vert ^{\alpha }_C\Vert \phi _1\Vert _C\Vert \phi _2\Vert _C\\ \ \ \ \ \ \ \ \ \ \ \le \Vert u\Vert ^{\alpha }_C\\ \ \ \ \ \ \ \ \ \ \ < \Vert u\Vert _C. \end{array} \end{aligned}$$

This indicates that

$$\begin{aligned} \Vert T_1 u\Vert _C< \Vert u\Vert _C,\ \ \forall u\in K\cap \partial D_{r}. \end{aligned}$$

(6.2)

Take $\eta >1$. Next, we demonstrate that

$$\begin{aligned} \Vert T_1 u\Vert _C> \Vert u\Vert _C,\ \ \forall u\in K\cap \partial D_{\eta }. \end{aligned}$$

(6.3)

If (1.5) holds, when a $\eta$ is fixed, then there is a $l> 0$ so that

$$\begin{aligned} f(x,u)\ge l>\frac{\eta }{\tau _1\tau _2\text {mes}\ B_1\text {mes}\ B_2\text {mes}\ B_3(\text {mes}\ {{\bar{\Omega }}})^{-1}},\ \forall x\in {{\bar{\Omega }}},\ u\ge \eta . \end{aligned}$$

(6.4)

Therefore, for $u\in K\cap \partial D_{\eta }$, we get from (2.4), (2.5), (2.6), (3.2) and (6.4) that

$$\begin{aligned} \begin{array}{ll} (\text {mes}\ {{\bar{\Omega }}})\Vert T_1 u\Vert _{C}\ge \int _{{{\bar{\Omega }}}}(T_1 u)(x)dx\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}dx\int _{{{\bar{\Omega }}}}G(x,y)f(y,u(y))\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int _{{{\bar{\Omega }}}}dx\int _{{{\bar{\Omega }}}}\int _{{{\bar{\Omega }}}}G^*(x,z)G^*{z,y}f(y,u(y))\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge l\int _{B_1}dx\int _{B_2}\int _{B_3}G^*(x,z)G^*{z,y}\textrm{d}z\textrm{d}y\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge l\tau _1\tau _2\text {mes}\ B_1\text {mes}\ B_2\text {mes}\ B_3, \end{array} \end{aligned}$$

which indicates that

$$\begin{aligned} \begin{array}{ll} \Vert T_1 u\Vert _C \ge l\tau _1\tau _2\text {mes}\ B_1\text {mes}\ B_2\text {mes}\ B_3\ (\text {mes}{{\bar{\Omega }}})^{-1}>\eta = \Vert u\Vert _C. \end{array} \end{aligned}$$

This indicates that (6.3) is true.

Therefore, according to (a) of Lemma 6.1, (6.2) and (6.3) yields that operator $T_1$ possesses a fixed point u in $K\cap ({{\bar{D}}}_{\eta }\backslash D_{r})$ with $r\le \Vert u\Vert _C\le \eta$. This derives that problem (1.3) admits at least one positive solution u with $r\le \Vert x\Vert _C\le \eta$. $\square$