1 Introduction

In this paper, we are concerned with the existence of positive solutions for the following prescribed mean curvature problem with Dirichlet boundary condition

$$\begin{aligned} \left\{ \begin{array}{l} -\text{ div } \left( \displaystyle \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) = f(u) \ \text{ in } \ \ \Omega , \ \ \\ u=0 \ \text{ on } \ \ \partial \Omega , \end{array} \right. \quad \quad \quad \quad \quad \quad \quad \quad {{(P)}} \end{aligned}$$

where \(\Omega \subset {\mathbb {R}}^{2}\) is a bounded smooth domain. The function f is given, and we seek a solution u satisfying (P).

Since the left-hand side is the mean curvature of the graph of u, problem (P) is a prescribed mean curvature equation whose prescription depends on the location of the graph. Problems of this type have been studied starting with the pioneering contributions of Gethardt [15] and Miranda [25] who constructed \(H^{1,1}\) solutions, respectively BV solutions of the prescribed mean curvature equation with Dirichlet boundary condition. We also refer to the seminal paper by [18], where there are established necessary and sufficient conditions for the existence of solutions in a particular case but without boundary conditions. Moreover, prescribed mean curvature equation has been the object of extensive studies in the past due to arises from some problems associated with differential geometry and physics such as combustible gas dynamics [4,5,6,7, 10, 14, 19, 31] and also due to the close connection with the capillary problem. For example, radial solution of (P) in \({\mathbb {R}}^{N}\) when f(u) is replaced by \(\kappa u\) has been studied in the context of the analysis of capillary surfaces, as can be seen in [9, 13, 16, 20, 22, 28] and [32].

Recently, by using variational methods, Obersnel and Omari [29] have considered the existence and multiplicity of positive solutions to problem (P) with respect to the behavior of the nonlinearity near the origin and at infinity. In the references of [29], the reader will find different contributions to the study of the prescribed mean curvature equation.

To state our main result, we need some hypotheses. The hypotheses on the continuous function f are the following:

\({(f_{1}})\):

There exists \(\alpha _0 > 0\) such that the function f(t) satisfies

$$\begin{aligned}\displaystyle \lim _{t \rightarrow \infty }\frac{f(t)}{\exp (\alpha |t|^{2})} = 0 \,\,\, \text{ for } \,\,\, \alpha > \alpha _{0} \ \ \text{ and } \ \ \displaystyle \lim _{t \rightarrow \infty }\frac{f(t)}{\exp (\alpha |t|^{2})} = \infty \,\,\, \text{ for } \,\,\, \alpha < \alpha _{0}. \end{aligned}$$
\({(f_{2}})\):

The following limit holds:

$$\begin{aligned}\displaystyle \lim _{t \rightarrow 0^+} \frac{f(t)}{ t} =0.\end{aligned}$$

Moreover, \(f(t)=0\) for all \(t\le 0\).

\({(f_{3}})\):

The function    \(t \mapsto \displaystyle \frac{f(t)}{t}\)    is increasing in   \((0, +\infty )\).

\({(f_{4}})\):

There exist \(r>{\frac{32}{7}\sqrt{2}}\) and \(\tau >\tau ^*\) such that

$$\begin{aligned} f(t) \ge \tau t^{r-1}, \end{aligned}$$

for all \(t \ge 0\), where

$$\begin{aligned} \tau >\tau ^* := \max \biggl \{\biggl [\frac{16r\sqrt{2}}{{7r-32\sqrt{2}}}\frac{c_r \alpha _0}{\pi }\biggl ]^{r-2/2}, \frac{K_2}{\delta } \biggl [\frac{32r\sqrt{2}}{{7r-32\sqrt{2}}}c_r\biggl ]^{r-2/2}, {1}\biggl \} \end{aligned}$$

and the constant \(c_r\) will appear in the Sect. 4, \(K_2>0\) will appear in Lemma 5.1, and \(\delta >0\) will appear in (3.4).

\({(f_{5}})\):

The following inequality holds:

$$\begin{aligned} 0< r F(t)\le f(t)t , \end{aligned}$$

for all \(t > 0\), where \(F(t):=\int _0^tf(s)\mathrm{d}s\).

The main result of this paper establishes the following existence and regularity property.

Theorem 1.1

Assume that conditions \((f_{1})- (f_{5})\) hold. Then, problem (P) has a positive solution \(u \in C^{1}(\overline{\Omega })\).

Hypothesis \({(f_{1}})\) is closely related to the Trudinger–Moser inequality and establishes that the function f has an exponential critical growth in \({{\mathbb {R}}}^2\).

We would like to highlight that our theorem can be applied for the model nonlinearity

$$\begin{aligned} {f(t)= \tau t^{r-1}\exp (\alpha _0 t^{2}) \ \ \text{ for } \text{ all } \ \ t\ge 0 \ \ \text{ and }\ \ f(t)=0, \ \ \text{ for } \text{ all } \ \ t\le 0}, \end{aligned}$$
(1.1)

where \(\tau\) and r are the constants in \((f_4)\) and \(\alpha _0\) is the constant in \((f_1)\).

Nonlinear problems with exponential growth have been considered recently by Alves and de Freitas [1], Alves and Santos [2], Ambrosio [3], Figueiredo and Severo [12], Li, Santos and Yang [23], Medeiros, Severo and Silva [24], etc.

There are some recent papers to prescribe mean curvature problem in two-dimensional case. In [27] the authors studied the prescribed mean curvature problem with nonhomogeneous boundary condition. More precisely, the authors investigate the boundary behavior of variational solutions of problem (P) at smooth boundary points where certain boundary curvature conditions are satisfied. In [11] the authors show a nonexistence result. To the best of our knowledge, the main result in this paper is the first work on the problem of medium curvature in dimension two and non-linearity with critical exponential growth.

The plan of the paper is as follows. We first associate to problem (P) a related nonhomogeneous auxiliary problem with Dirichlet boundary condition. In Sect. 3 we study the variational structure of this auxiliary nonlinear problem and we establish several qualitative properties of the associated energy functional. The key abstract tools in these arguments are the Trudinger–Moser inequality and the Nehari manifold method. Next, minimizing the energy function on the Nehari manifold, we prove the existence of solutions to the auxiliary problem. In the final section of this paper, we prove that the solution of the auxiliary problem is a solution of the original problem. This is essentially done by using the Moser iteration method and Stampacchia’s estimates. We refer to the recent monograph by Papageorgiou, Rădulescu and Repovš [30] for some of the abstract methods used in this paper.

2 An auxiliary problem

Consider the following auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\text{ div } \left( a(|\nabla u|^2)\nabla u\right) = f(u) &{} \text {in}\ \ \Omega ,\\ u=0 &{} \text {on}\ \ \partial \Omega , \end{array}\right. \quad \quad \quad \quad \quad \quad \quad \quad {(Aux)} \end{aligned}$$

where

$$\begin{aligned} {a(s)=\left\{ \begin{array}{ll} \displaystyle \frac{1}{\displaystyle \sqrt{1+s}} &{} s\in [0,1],\\ \displaystyle \frac{(s-2)^2+7}{8\sqrt{2}} &{} s\in [1,2),\\ \displaystyle \frac{7}{8\sqrt{2}} &{} s\in [2,\infty ). \end{array} \right. } \end{aligned}$$

Lemma 2.1

The function \(a:{\mathbb {R}}_{+}\rightarrow {\mathbb {R}}_{+}\) is decreasing and of \(C^{1}\) class. Moreover, it satisfies the following conditions:

\((a_{1})\) :

\(\displaystyle \frac{7}{8\sqrt{2}}\le a(s^{2}) \le 1, \ \ \text{ for } \text{ all } \ \ s\ge 0.\)

\((a_{2})\) :

\(\displaystyle a'(s)s\le 0< a(s), \ \ \text{ for } \text{ all } \ \ s\ge 0.\)

\((a_{3})\) :

The function

$$\begin{aligned} s\mapsto A(s^{2}) \ \ \text{ is } \text{ convex } \text{ for } \ \ s\ge 0, \end{aligned}$$

where \(A(s)=\displaystyle \int ^{s}_{0}a(t)\mathrm{d}t\).

Proof

Since

$$\begin{aligned} {a'(s)=\left\{ \begin{array}{ll} \displaystyle \frac{-1}{2\displaystyle \sqrt{(1+s)^{3}}} &{} s\in [0,1],\\ \displaystyle \frac{s-2}{4\sqrt{2}} &{} s\in [1,2),\\ 0 &{} s\in [2,\infty ), \end{array} \right. } \end{aligned}$$

the items \((a_1)\) and \((a_2)\) follow by straightforward computation. For item \((a_3)\) note that

$$\begin{aligned} (A(s^{2}))'' = 2[ a(s^{2})+2s^2 a'(s^2)]. \end{aligned}$$

If we define \(b(s):=a(s)+2s a'(s)\), we can prove that b is strictly decreasing in \([0,\frac{6}{5}]\), strictly increasing in \([\frac{6}{5}, 2]\) and constant in \([2, +\infty )\). Then,

$$\begin{aligned} (A(s^{2}))'' \ge b\left(\frac{6}{5}\right)=\frac{19}{40\sqrt{2}}>0, \end{aligned}$$

for all \(s \ge 0\) and this completes the proof. \(\square\)

In this section, we prove some auxiliary results which will be very useful throughout the paper.

Lemma 2.2

If \((a_{1})-(a_2)\) are true, then:

(i):

The function \(s\mapsto a(s^2) s\) is increasing.

(ii):

For all \(x,y \in {\mathbb {R}}^{2}\), we have

$$\begin{aligned} \langle a(|x|^{2})x- a(|y|^{2})y, x-y\rangle \ge \frac{7}{8\sqrt{2}} |x-y|^{2}. \end{aligned}$$
(2.1)

Proof

In order to prove (i), note that using \((a_{1})-(a_2)\), we get

$$\begin{aligned} (a(s^2)s)'= 2s^{2}a'(s^2)+a(s^2) = b(s^2) >0.\end{aligned}$$

Let us prove (ii). Firstly, note that for \(z\in {\mathbb {R}}^{2}\), we have

$$\begin{aligned} \displaystyle \frac{\partial }{\partial z_{i}}(a(|z|^{2})z_{j})=a(|z|^{2})\delta _{ij}+2a'(|z|^{2})z_{i}z_{j}, \end{aligned}$$

where we have denoted \(\delta _{ij}\) by the Kronecker delta. Hence, for all \(z, \xi \in {\mathbb {R}}^{2}\) we get

$$\begin{aligned} \sum ^{2}_{i,j=1}\displaystyle \frac{\partial }{\partial z_{i}}(a(|z|^{2})z_{j})\xi _{i}\xi _{j}= a(|z|^{2})|\xi |^{2}+2a'(|z|^{2})\sum ^{2}_{i,j=1}z_{i}z_{j}\xi _{i}\xi _{j}. \end{aligned}$$
(2.2)

Since

$$\begin{aligned} \sum ^{2}_{i,j=1}z_{i}z_{j}\xi _{i}\xi _{j}=\left( \sum ^{2}_{j=1}z_{j}\xi _{j}\right) ^{2}, \end{aligned}$$
(2.3)

we have

$$\begin{aligned} \sum ^{2}_{i,j=1}z_{i}z_{j}\xi _{i}\xi _{j}=|z|^{2}|\xi |^2\cos ^2(\theta ) \quad \text{ for } \text{ some } \theta \in [0,2\pi )\text{. } \end{aligned}$$

Thus, using (2.2) we deduce that

$$\begin{aligned} \sum ^{2}_{i,j=1}\displaystyle \frac{\partial }{\partial z_{i}}(a(|z|^{2})z_{j})\xi _{i}\xi _{j}&= |\xi |^{2}[a(|z|^{2})+2a'(|z|^{2})|z|^{2}\cos ^2(\theta )]\\&= |\xi |^{2}[a(|z|^{2})+2a'(|z|^{2})|z|^{2} +2a'(|z|^{2})|z|^{2}(\cos ^2(\theta )-1). \end{aligned}$$

From \((a_2)\), we have that

$$\begin{aligned} 2a'(|z|^{2})|z|^{2}(\cos ^2(\theta )-1)\ge 0. \end{aligned}$$

Hence,

$$\begin{aligned} \sum ^{2}_{i,j=1}\displaystyle \frac{\partial }{\partial z_{i}}(a(|z|^{2})z_{j})\xi _{i}\xi _{j}&\ge |\xi |^{2}[a(|z|^{2})+2a'(|z|^{2})|z|^{2}]\nonumber \\&= b(|z|^2)|\xi |^{2} \ge \frac{7}{8\sqrt{2}}|\xi |^{2} \ \ \text{ for } \text{ all } \ \ z, \xi \in {\mathbb {R}}^{2}. \end{aligned}$$
(2.4)

Now for \(z=y+t(x-y)\), \(t \in [0,1]\) and \(\xi =x-y\), we have

$$\begin{aligned} \langle a(|x|^{2})x-a(|y|^{2})y, x-y\rangle= & {} \displaystyle \sum ^{2}_{j=1}(a(|x|^{2})x_{j}-a(|y|^{2})y_{j})(x_{j}-y_{j})\\= & {} \displaystyle \int ^{1}_{0}\sum ^{2}_{i,j=1}\displaystyle \frac{\partial }{\partial z_{i}}(a(|z|^{2})z_{j})\xi _{i}\xi _{j} \mathrm{d}t. \end{aligned}$$

Finally, using (2.4) we get (2.1). \(\square\)

The next result due to Stampacchia [33] will be useful in the arguments used in this paper.

Lemma 2.3

Let \(B(\eta )\) be a given \(C^1\) vector field in \({\mathbb {R}}^2\) and f(x, s) a bounded Carathéodory function in \(\Omega \times {\mathbb {R}}\). Let \(u \in H^1_0(\Omega )\) be a solution of

$$\begin{aligned} -\mathrm{div}\,(B(\nabla u))=f(x,u)\quad \text{ in } \Omega \text{, }\qquad u=0\quad \text{ on } \partial \Omega \text{, } \end{aligned}$$

that is,

$$\begin{aligned} \int _\Omega B(\nabla u)\cdot \nabla \varphi = f(x,u)\varphi \quad \quad \forall \varphi \in H^1_0(\Omega ). \end{aligned}$$

Assume that there exists \(0< \nu < K\) such that

$$\begin{aligned} \nu |\xi |^2 \le \sum _{i,j=1}^2\frac{\partial B_i}{\partial \eta _j}(\nabla u)\xi _i\xi _j \quad \text{ and } \quad \left| \frac{\partial B_i}{\partial \eta _j}(\nabla u)\right| \le K, \end{aligned}$$
(2.5)

for all \(i,j = 1, 2\) and \(\xi \in {\mathbb {R}}^2\). Then, \(u \in W^{2,p}(\Omega ) \cap C^{1,\alpha }(\overline{\Omega })\), for all \(\alpha \in (0,1)\) and for all \(p \in (1,\infty )\). Moreover,

$$\begin{aligned} \Vert u\Vert _{1,\alpha } \le O(\nu , K, \Omega , \Vert f(\cdot ,u)\Vert _\infty ). \end{aligned}$$
(2.6)

In the following result, we show that the differential operator involved in (Aux) verifies conditions (2.5).

Lemma 2.4

Assume that hypotheses \((a_{1})-(a_2)\) are fulfilled. Then, for all \(u \in H^{1}_{0}(\Omega )\), the second-order differential operator \(B(\nabla u)= a(|\nabla u|^{2})\nabla u\) satisfies (2.5) of Lemma 2.3.

Proof

Note that

$$\begin{aligned} \frac{\partial B_i}{\partial \eta _j}(\eta )= \frac{\partial }{\partial \eta _j}(a(|\eta |^{2})\eta _i)=a(|\eta |^{2})\delta _{ij}+ 2 a'(|\eta |^{2})\eta _{i}\eta _{j}, \end{aligned}$$

and then

$$\begin{aligned} \displaystyle \sum ^{2}_{i,j=1} \frac{\partial B_i}{\partial \eta _j}(\nabla u)\xi _i\xi _j=a(|\nabla u|^{2})|\xi |^2+ 2 a'(|\nabla u|^{2}) \displaystyle \sum ^{2}_{i,j=1}\frac{ \partial u}{\partial x_i}\frac{ \partial u}{\partial x_j}\xi _{i}\xi _{j}. \end{aligned}$$

Now, observing (2.2), we can repeat the reasoning of the proof of Lemma 2.2, and using \((a_{1})-(a_2)\), we conclude that

$$\begin{aligned} {\displaystyle \sum ^{2}_{i,j=1} \frac{\partial B_i}{\partial \eta _j}(\nabla u)\xi _i\xi _j \ge \frac{7}{8\sqrt{2}} |\xi |^{2}.} \end{aligned}$$

On the other hand, using \((a_{1})-(a_2)\), we get

$$\begin{aligned} \left| \frac{\partial B_i}{\partial \eta _j}(\nabla u)\right|&= \left| a(|\nabla u|^{2})\delta _{ij}+ 2 a'(|\nabla u|^{2})\frac{ \partial u}{\partial x_i}\frac{ \partial u}{\partial x_j}\right| \\&\le a(|\nabla u|^{2})+|a'(|\nabla u|^{2})||\nabla u|^{2} \le K, \end{aligned}$$

for some positive constant \(K>0\). The proof is now complete. \(\square\)

3 The variational framework and some technical lemmas

Note that, by the hypothesis \((a_{1})\), we have that the functional \(I: H_0^{1}(\Omega )\rightarrow {\mathbb {R}}\) given by

$$\begin{aligned} I(u) = \frac{1}{2}\displaystyle \int _{\Omega }A(|\nabla u|^{2}) \mathrm{d}x - \displaystyle \int _{\Omega }F(u) \mathrm{d}x \end{aligned}$$

is well defined, where \(F(t)=\displaystyle \int ^{t}_{0}f(s) \mathrm{d}s\).

Moreover, we have

$$\begin{aligned} I'(u)\phi =\displaystyle \int _{\Omega }a(|\nabla u|^{2})| \nabla u\nabla \phi \ \mathrm{d}x - \displaystyle \int _{\Omega }f(u)\phi \ \mathrm{d}x, \end{aligned}$$

for all \(\phi \in H_0^{1}(\Omega )\). Thus, I is a \({{\mathcal {C}}}^1\) functional on \(H_0^{1}(\Omega )\) and its critical points are weak solution of problem (Aux).

The Nehari manifold associated to the functional I is given by

$$\begin{aligned} \mathcal {N}= \{u \in H^{1}_{0}(\Omega )\backslash \{0\}:\ J(u)=0\}, \end{aligned}$$

where \(J(u)=I'(u)u\) for \(u \in H^{1}_{0}(\Omega )\). Let us start with the following important result due to Trudinger [34] and Moser [26].

Theorem 3.1

For every \(u \in H^{1}_{0}(\Omega )\) and \(\alpha >0\)

$$\begin{aligned} \exp (\alpha u^{2}) \in L^{1}(\Omega ) \end{aligned}$$
(3.1)

and there is a constant \(M>0\) such that

$$\begin{aligned} \sup _{\Vert u\Vert _{H^{1}_{0}(\Omega )}\le 1} \int _{\Omega }\exp (\alpha u^{2})\mathrm{d}x \le M, \end{aligned}$$
(3.2)

for every \(\alpha \le 4\pi .\)

Moreover, if \(\alpha > 4\pi\), then

$$\begin{aligned} \sup _{\Vert u\Vert _{H^{1}_{0}(\Omega )}\le 1} \int _{\Omega }\exp (\alpha u^{2})\mathrm{d}x =\infty . \end{aligned}$$
(3.3)

Note that, from \((f_{2})\), for any \(\varepsilon >0\), there exists \(\delta >0\) such that

$$\begin{aligned} |f(t)|\le \varepsilon |t| \end{aligned}$$
(3.4)

and

$$\begin{aligned} |F(t)|\le \frac{1}{2}\varepsilon |t|^{2}, \end{aligned}$$
(3.5)

for all \(0< t\le \delta\).

Furthermore, from \((f_{1})\), given \(\alpha >\alpha _0\), there exists \(K>0\) such that

$$\begin{aligned} |f(t)|\le \varepsilon \exp (\alpha t^{2} ), \end{aligned}$$

for all \(t\ge K\). In particular, we get

$$\begin{aligned} {|f(t)|\le \frac{ \varepsilon }{K} t\exp (\alpha t^{2} ),} \end{aligned}$$
(3.6)

with implies

$$\begin{aligned} |f(t)t|\le \frac{\varepsilon }{K} t^{2}\exp (\alpha t^{2} )\le \frac{\varepsilon }{K^{q-2}} |t|^{q}\exp (\alpha t^{2} )= C_{\varepsilon }|t|^{q}\exp (\alpha t^{2} ), \end{aligned}$$
(3.7)

where \(C_{\varepsilon }=\frac{\varepsilon }{K^{q-2}}\). Moreover, from (3.6), we get

$$\begin{aligned} |F(t)|\le \frac{\varepsilon }{\alpha K}\exp (\alpha t^{2} )\le \frac{\varepsilon }{\alpha _0 K^{q+1}}|t|^{q}\exp (\alpha t^{2} )=\tilde{C_{\varepsilon }} |t|^{q}\exp (\alpha t^{2} ), \end{aligned}$$
(3.8)

for all \(t\ge K\), for all \(\alpha >\alpha _0\), for all \(q\ge 0\) with \(\tilde{C_{\varepsilon }}= \frac{\varepsilon }{\alpha _0 K^{q+1}}\).

Consequently, using (3.4), (3.5), (3.7) and (3.8), for all \(\varepsilon >0\) and for all \(\alpha >\alpha _0\), there exists \(C_{\varepsilon }>0\) such that

$$\begin{aligned} \int _\Omega f(u)u \mathrm{d}x\le \varepsilon \int _\Omega |u|^2 \mathrm{d}x +C_{\varepsilon }\int _\Omega |u|^{q}\exp (\alpha |u|^{2} )\mathrm{d}x \end{aligned}$$
(3.9)

and

$$\begin{aligned} \int _\Omega F(u)\mathrm{d}x\le \frac{\varepsilon }{2} \int _\Omega |u|^2 \mathrm{d}x+\tilde{C_\varepsilon }\int _\Omega |u|^{q}\exp (\alpha |u|^{2} )\mathrm{d}x, \end{aligned}$$
(3.10)

for all \(u \in H^{1}_{0}(\Omega )\). In particular, in this paper, we will use \(q>2\).

In the next result, we prove that \(\mathcal {N}\) is not empty and that I restricted to \(\mathcal {N}\) is bounded from below.

Lemma 3.1

For each \(u \in H^{1}_{0}(\Omega ) \setminus \{0\}\), there exists a unique \(t>0\) such that \(tu \in \mathcal {N}\). Moreover, \(I(u)> 0\) for every \(u \in \mathcal {N}\).

Proof

Given \(u \in H^{1}_{0}(\Omega ) \setminus \{0\}\), let \(\mathcal {T}_u(t)=I(tu)\) for \(t>0\). Then, \(tu \in \mathcal {N}\) if and only if \(\mathcal {T}_{u}'(t)=0\). Note that, taking \(\varepsilon >0\) sufficiently small in (3.9) and using \((a_{1})\) and Sobolev embedding, there exists \(C>0\) such that

$$\begin{aligned} \mathcal {T}_{u}(t)&= \frac{1}{2}\int _{\Omega }A(|t \nabla u|^{2})\mathrm{d}x - \int _{\Omega }F(tu)\mathrm{d}x\\&\ge \frac{7/8\sqrt{2}-C\varepsilon }{2} t^2\Vert u\Vert ^2 - t^{q}\tilde{C_\varepsilon }\int _\Omega |u|^{q}\exp ( \alpha | t u|^{2} )\mathrm{d}x. \end{aligned}$$

Using Hölder’s inequality with \(s',s>1\), we get

$$\begin{aligned} \mathcal {T}_{u}(t)\ge \frac{7/8\sqrt{2}-C\varepsilon }{2} t^2\Vert u\Vert ^2 - t^{q}\tilde{C_\varepsilon }\biggl (\int _\Omega |u|^{q s'}\mathrm{d}x\biggl )^{1/s'} \biggl (\int _\Omega \exp \left( \alpha s \Vert t u\Vert ^{2}\left(\frac{u}{\Vert u\Vert }\right)^{2} \right) \mathrm{d}x\biggl )^{1/s}. \end{aligned}$$

Choosing \(\alpha > \alpha _0\) and \(t_1>0\) such that \(\alpha s\Vert t_1 u\Vert ^{2}< 4\pi\), using (3.2) we obtain

$$\begin{aligned} \mathcal {T}_u(t)\ge D_1 t^{2}_{1} - D_2 t^{q}_{1} \end{aligned}$$

for some \(D_1,D_2>0\) and for all \(0\le t\le t_1\). Thus, since \(2<q\), there exists \(0<t^{*}\le t_1\) such that \(\mathcal {T}_u(t)>0\) for all \(0<t< t^{*}\le t_1\).

Now, from \((a_1)\) and \((f_{4})\), we have

$$\begin{aligned} \frac{\mathcal {T}_u(t)}{t^2}\le \frac{1}{2}\Vert u\Vert ^2 -\frac{\tau }{r} t^{r-2}\int _{\Omega ^{+}_{u}} u^{r}\mathrm{d}x, \end{aligned}$$

where \(\Omega ^{+}_{u}=\{x\in \Omega : u(x)>0\}\). Therefore, since \(r>2\), we conclude \(\displaystyle \lim _{t \rightarrow +\infty } \mathcal {T}_u(t)=- \infty\). Consequently, there exists at least one \(t(u)>0\) such that \(\mathcal {T}_u'(t(u))=0\), that is, \(t(u)u \in \mathcal {N}\).

Suppose, by contradiction, that there are \(t>0\) and \(\widetilde{t}>0\) such that

$$\begin{aligned} \int _\Omega a(|t \nabla u|^{2})| \nabla u|^{2}\mathrm{d}x = \int _\Omega \frac{f(tu)}{t}u\mathrm{d}x \end{aligned}$$

and

$$\begin{aligned} \int _\Omega a(|\widetilde{t} \nabla u|^{2})| \nabla u|^{2}\mathrm{d}x = \int _\Omega \frac{f(\widetilde{t}u)}{\widetilde{t}}u \mathrm{d}x. \end{aligned}$$

Then,

$$\begin{aligned} \int _\Omega \left[ a(|t \nabla u|^{2})| \nabla u|^{2} - a(|\widetilde{t} \nabla u|^{2})| \nabla u|^{2}\right] \mathrm{d}x= \int _\Omega \left[ \frac{f(tu)}{t}- \frac{f(\widetilde{t}u)}{\widetilde{t}} \right] u \mathrm{d}x. \end{aligned}$$

If \(t>\widetilde{t}\), from Lemma 2.1 and \((f_{3})\), we have

$$\begin{aligned} 0> \int _\Omega \left[ a(|t \nabla u|^{2})| \nabla u|^{2} - a(|\widetilde{t} \nabla u|^{2})| \nabla u|^{2}\right] \mathrm{d}x= \int _\Omega \left[ \frac{f(tu)}{t}- \frac{f(\widetilde{t}u)}{\widetilde{t}} \right] u \mathrm{d}x>0, \end{aligned}$$

which is a contradiction. In the same way, we obtain that we cannot have the case \(t<\widetilde{t}\). We conclude that there is a unique parameter \(t>0\) such that \(t(u)u \in \mathcal {N}\). Note, in particular, that t(u) is a global maximum point of \(\mathcal {T}_u\) and \(\mathcal {T}_u(t(u))>0\), i.e. \(I(t(u)u)>0\). Since \(t(u)=1\) if \(u \in \mathcal {N}\), we deduce that \(I(u)>0\) for every \(u \in \mathcal {N}\). \(\square\)

In the next result we prove that sequences in \(\mathcal {N}\) cannot converge to 0.

Lemma 3.2

There exists a constant \(C>0\) such that \(0<C\le \Vert u\Vert\), for every \(u \in \mathcal {N}\).

Proof

Suppose, by contradiction, that there is \((u_n) \subset \mathcal {N}\) such that

$$\begin{aligned} u_n \rightarrow 0 \ \ \text{ in } \ \ H^{1}_{0}(\Omega ). \end{aligned}$$
(3.11)

Then, using (3.10), we have

$$\begin{aligned} \displaystyle \int _{\Omega }a(|\nabla u_n|^{2})|\nabla u_n|^{2} \mathrm{d}x= \displaystyle \int _{\Omega }f(u_n)u_n \mathrm{d}x\le \varepsilon \int _\Omega |u_n|^2 \mathrm{d}x +C_{\varepsilon }\int _\Omega |u_n|^{q}\exp (\alpha |u_n|^{2} )\mathrm{d}x. \end{aligned}$$

Now, from \((a_1)\) we get

$$\begin{aligned} {\displaystyle \frac{7}{8\sqrt{2}}}\displaystyle \int _{\Omega }|\nabla u_n|^{2} \mathrm{d}x \le \varepsilon \int _\Omega |u_n|^2 \mathrm{d}x+C_{\varepsilon }\int _\Omega |u_n|^{q}\exp (\alpha |u_n|^{2} )\mathrm{d}x. \end{aligned}$$

Using Sobolev embedding, there exists \(C>0\) such that

$$\begin{aligned} {\left(\displaystyle \frac{7}{8\sqrt{2}} - C \varepsilon \right)}\Vert u_n\Vert ^{2} \le C_{\varepsilon }\int _\Omega |u_n|^{q}\exp \left(\alpha \Vert u_n\Vert ^{2}\left(\frac{|u_n|}{\Vert u_n\Vert }\right)^{2} \right)\mathrm{d}x. \end{aligned}$$

Using Hölder’s inequality with \(s',s>1\), we have

$$\begin{aligned} {\left(\displaystyle \frac{7}{8\sqrt{2}} - C \varepsilon \right)}\Vert u_n\Vert ^{2} \le C_\varepsilon \biggl (\int _\Omega |u_n|^{q s'}\mathrm{d}x\biggl )^{1/s'} \biggl (\int _\Omega \exp \left( \alpha s \Vert u_n\Vert ^{2}\left(\frac{u_n}{\Vert u_n\Vert }\right)^{2} \right)\mathrm{d}x \biggl )^{1/s}. \end{aligned}$$
(3.12)

Note that by (3.11), there is \(n_0 \in \mathbb {N}\) such that

$$\begin{aligned} \Vert u_n\Vert ^{2}< \frac{4\pi }{ \alpha s} \end{aligned}$$

for all \(n \ge n_0\) and for some \(\alpha >\alpha _0\). Then, from (3.2) and Sobolev embedding again, we have

$$\begin{aligned} {\left(\displaystyle \frac{7}{8\sqrt{2}} - C \varepsilon \right)}\Vert u_n\Vert ^{2} \le MC_\varepsilon \biggl (\int _\Omega |u_n|^{q s\prime}\mathrm{d}x\biggl )^{1/s\prime}\le MC_\varepsilon C \Vert u_n\Vert ^{q}. \end{aligned}$$

This inequality implies

$$\begin{aligned} \frac{{\left(\displaystyle \frac{7}{8\sqrt{2}} - C \varepsilon \right)} }{MC_\varepsilon C } \le \Vert u_n\Vert ^{q-2}. \end{aligned}$$

Since \(q>2\), the above inequality contradicts (3.11) and the lemma is proved. \(\square\)

We set \(c:=\inf _{\mathcal {N}}I\), and in the next result we will prove that minimizing sequence for c are bounded.

Lemma 3.3

If \((u_n) \subset \mathcal {N}\) is a minimizing sequence for c, then \((u_n)\) is bounded.

Proof

Suppose, by contradiction, that up to a subsequence, \(\Vert u_{n}\Vert \rightarrow \infty\) and consider \(v_{n}=\frac{u_{n}}{\Vert u_{n}\Vert }\rightharpoonup v_{0}\). If \(v_{0}=0\), then for all \(t>0\), from \((a_1)\) we obtain

$$\begin{aligned} c+o_{n}(1)= I(u_{n})&= I(\Vert u_{n}\Vert v_{n})\ge I(tv_{n})\\&\ge {\displaystyle \frac{7}{16\sqrt{2}}}t^{2} - \displaystyle \int _{\Omega }F(tv_{n}) \ \mathrm{d}x, \end{aligned}$$

where \(o_{n}(1)\rightarrow 0\) as \(n\rightarrow +\infty\). Since \(v_n \rightarrow 0\) in \(L^{q}(\Omega )\) and \(\Vert v_n\Vert < 4\pi\), using (3.10), Hölder inequality as in (3.12) and Theorem 3.1, we get

$$\begin{aligned} \displaystyle \int _{\Omega }F(tv_{n}) \ \mathrm{d}x\rightarrow 0. \end{aligned}$$

But this last convergence implies

$$\begin{aligned} c\ge {\displaystyle \frac{7}{16\sqrt{2}}}t^{2}, \ \ \text{ for } \text{ all } \ \ t>0, \end{aligned}$$

which is a contradiction.

Suppose now that \(v_{0}\ne 0\). Then,

$$\begin{aligned} \frac{1}{\Vert u_{n}\Vert ^{2}}I(u_{n})=\frac{c}{\Vert u_{n}\Vert ^{2}}= o_{n}(1), \end{aligned}$$

where \(o_{n}(1)\rightarrow 0\) as \(n\rightarrow +\infty\). Hence, using \((a_1)\) and \((f_4)\), we get

$$\begin{aligned} \frac{\tau }{r}\Vert u_n\Vert ^{r-2}\displaystyle \int _{\Omega } |v_n|^{r} \mathrm{d}x \le \displaystyle \int _{\Omega }\frac{F(\Vert u_{n}\Vert v_{n})}{\Vert u_{n}\Vert ^{2}}= \frac{1}{2\Vert u_{n}\Vert ^{2}}\displaystyle \int _{\Omega } A(|\nabla u_n|^{2} \mathrm{d}x +o_{n}(1)\le \frac{{1}}{2}+o_{n}(1). \end{aligned}$$

Since \(v_n \rightarrow v\) in \(L^{r}(\Omega )\) and \(\Vert u_n\Vert \rightarrow +\infty\), we have a contradiction. \(\square\)

To end up this section, let us prove that if the minimum of I over \(\mathcal {N}\) is achieved in some \(u_0 \in \mathcal {N}\), then in fact \(u_0\) is a critical point of I. This follows from some arguments used in [8, Proof of Theorem 1.1 (completed)].

Lemma 3.4

If \(u_0 \in \mathcal {N}\) is such that

$$\begin{aligned} I(u_0) = \min _{\mathcal {N}}I=c, \end{aligned}$$

then \(I'(u_0) = 0\).

Proof

Suppose, by contradiction, that c is achieved by \(u_0\) and this one is not a critical point of I. Then, there exists \(\phi \in H^{1}_{0}(\Omega )\) such that

$$\begin{aligned} I'(u_0)\phi <0. \end{aligned}$$

By the continuity of \(I'\), we can find \(\widetilde{\varepsilon }, \widetilde{\delta } >0\) small such that

$$\begin{aligned} I'(t(u_0+ \sigma \phi ))\phi <0,\;\;\; \text{ for } \ \ t \in [1-\widetilde{\varepsilon }, 1+ \widetilde{\varepsilon }] \ \ \text{ and } \ \ \sigma \in [-\widetilde{\delta }, \widetilde{\delta }]. \end{aligned}$$
(3.13)

Moreover, since the map \(t\mapsto I(tu_0)\) attains its maximum at \(t=1\) as shown in the proof of Lemma 3.1, we have

$$\begin{aligned} I'((1-\widetilde{\varepsilon })u_0)u_0=\mathcal {T}_{u}^{\prime}(1-\widetilde{\varepsilon })>0>\mathcal {T}_{u}^{\prime}(1+\widetilde{\varepsilon })=I'((1+\widetilde{\varepsilon })u_0)u_0. \end{aligned}$$

Then, again by the continuity of \(I'\), there exists \(\overline{\sigma } \in (0,\widetilde{\delta })\) such that

$$\begin{aligned}I'((1-\widetilde{\varepsilon })(u_0+\overline{\sigma }\phi )(u_0+\overline{\sigma }\phi )>0>I'((1+\widetilde{\varepsilon })(u_0+\overline{\sigma }\phi )(u_0+\overline{\sigma }\phi ),\end{aligned}$$

i.e. \(\mathcal {T}_{u+\overline{\sigma }\phi }^{\prime}(1-\widetilde{\varepsilon })>0>\mathcal {T}_{u+\overline{\sigma }\phi }'(1+\widetilde{\varepsilon })\). It follows that

$$\begin{aligned} \overline{t}(u_0+\overline{\sigma }\phi ) \in \mathcal {N}, \ \ \text{ for } \text{ some } \ \ \overline{t} \in (1-\widetilde{\varepsilon }, 1+\widetilde{\varepsilon }). \end{aligned}$$
(3.14)

From (3.13), we have

$$\begin{aligned} I(\overline{t}(u_0+ \overline{\sigma } \phi )) - I(u_0)\le I(\overline{t}(u_0+ \overline{\sigma } \phi )) - I(\overline{t}u_0)= \overline{t}\int ^{\overline{\sigma }}_{0} I'(\overline{t}(u_0+ \sigma \phi ))\phi d\sigma <0, \end{aligned}$$

so that

$$\begin{aligned} I(\overline{t}(u_0+ \overline{\sigma } \phi )) < I(u_0) = c, \end{aligned}$$

which contradicts (3.14) [also, because \(I(u_0)=\min _{\mathcal {N}}I\)]. Therefore, \(I'(u_0)=0\) and the proof is complete. \(\square\)

4 Existence of solution to the auxiliary problem

In this section, in order to prove the existence of result in the exponential critical case, we consider the auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{l} -\Delta u = |u|^{r-2}u \ \text{ in } \ \ \Omega , \ \ \\ u \in H^{1}_{0}(\Omega ), \end{array} \right. \quad \quad \quad \quad \quad \quad \quad \quad {(A)} \end{aligned}$$

where r is the constant that appear in the hypothesis \((f_4)\).

The energy functional associated to problem (A) is defined by

$$\begin{aligned} I_r(u)=\frac{1}{2}\displaystyle \int _{\Omega }|\nabla u|^{2}\mathrm{d}x- \frac{1}{r}\displaystyle \int _{\Omega }|u|^{r}\mathrm{d}x. \end{aligned}$$

We also define the Nehari manifold

$$\begin{aligned} \mathcal {N}_r=\{u \in H^{1}_{0}(\Omega ); u\ne 0: I'_r(u)u=0\}. \end{aligned}$$

Since the embedding \(H^{1}_{0}(\Omega )\hookrightarrow L^{r}(\Omega )\) is compact, using the mountain pass theorem and the classical maximum principle, we can prove that there exists a positive solution to problems (A) given by \(w_r \in H^{1}_{0}(\Omega )\) such that

$$\begin{aligned} I_r(w_r)=c_r, \ \ I'_r(w_r)=0 \end{aligned}$$

and

$$\begin{aligned} c_r=\frac{r-2}{2r}\displaystyle \int _{\Omega }|w_r|^{r} \mathrm{d}x, \end{aligned}$$
(4.1)

where \(c_r=\displaystyle \inf _{\mathcal {N}_r}I_r\). The next result is an estimate to \(c=\displaystyle \inf _{\mathcal {N}}I\).

Lemma 4.1

The value \(c=\displaystyle \inf _{\mathcal {N}}I\) satisfies

$$\begin{aligned} c \le \frac{c_r}{\tau ^{2/(r-2)}}. \end{aligned}$$

Proof

Note that, by the hypotheses \((a_1)\) and \((f_4)\), we have

$$\begin{aligned} \displaystyle \int _{\Omega }a(|\nabla w_r|^{2})|\nabla w_r|^{2} \mathrm{d}x \le \displaystyle \int _{\Omega }|\nabla w_r|^{2} \mathrm{d}x = \displaystyle \int _{\Omega }|w_r|^{r}\le \displaystyle \int _{\Omega }f(w_r)w_r \mathrm{d}x. \end{aligned}$$

This inequality implies that \(I'(w_r)w_r \le 0\). Then, from \((f_3)\), there exists \(\beta \in (0,1{]}\) such that \(\beta w_r \in \mathcal {N}\). Using \((a_1)\) and \((f_4)\) again, we obtain

$$\begin{aligned} c\le I(\beta w_r) \le \frac{\beta ^{2}}{2}\displaystyle \int _{\Omega }|\nabla w_r|^{2} \mathrm{d}x - \frac{\tau }{r}\beta ^{r} \displaystyle \int _{\Omega }| w_r|^{r} \mathrm{d}x. \end{aligned}$$

Since \(I'_r(w_r)=0\), we conclude that

$$\begin{aligned} c\le \biggl [\frac{\beta ^{2}}{2}- \tau \frac{\beta ^{r}}{r}\biggl ]\displaystyle \int _{\Omega }| w_r|^{r} \mathrm{d}x. \end{aligned}$$

Using (4.1), we have

$$\begin{aligned} c\le \biggl [\frac{\beta ^{2}}{2}- \tau \frac{\beta ^{r}}{r}\biggl ]\frac{2rc_r }{(r-2)} \le \displaystyle \max _{s\ge 0}\biggl [\frac{s^{2}}{2}- \tau \frac{s^{r}}{r}\biggl ]\frac{2rc_r }{(r-2)}=\frac{(r-2)}{2r}\frac{1}{\tau ^{2/(r-2)}}\frac{2rc_r }{(r-2)}=\frac{c_r}{\tau ^{2/(r-2)}}. \end{aligned}$$

The proof is now complete. \(\square\)

Lemma 4.2

If \((u_n)\subset \mathcal {N}\) is a minimizing sequence for c, then

$$\begin{aligned} \displaystyle \limsup _{n \rightarrow \infty }\Vert u_n\Vert ^{2}\le \frac{2\pi }{\alpha _0}. \end{aligned}$$

Proof

Note that

$$\begin{aligned} c+ o_{n}(1)&= I(u_n) - {\frac{1}{r}} I'(u_n)u_n = \frac{1}{2}\displaystyle \int _{\Omega }A(|\nabla u_n|^{2})\mathrm{d}x - {\frac{1}{r}}\displaystyle \int _{\Omega }a(|\nabla u_n|^{2})|\nabla u_n|^{2}\mathrm{d}x\\&\quad+ {\frac{1}{r}}\displaystyle \int _{\Omega }f(u_n)u_n \mathrm{d}x - \displaystyle \int _{\Omega }F(u_n)\mathrm{d}x. \end{aligned}$$

From \((a_1)\) and \((f_5)\), we get

$$\begin{aligned} c+ o_{n}(1)\ge & {} {\biggl (\frac{7}{32\sqrt{2}}-\frac{1}{r}\biggl )}\displaystyle \int _{\Omega }|\nabla u_n|^{2}\mathrm{d}x= {\biggl (\frac{7}{32\sqrt{2}}-\frac{1}{r}\biggl )}\Vert u_n\Vert ^{2}. \end{aligned}$$

Since \(r> \frac{32\sqrt{2}}{7}\), we obtain

$$\begin{aligned} \Vert u_n\Vert ^{2}\le \frac{32\sqrt{2}r}{(7r-32\sqrt{2})}c+ o_{n}(1) \end{aligned}$$

By the estimate on c in Lemma 4.1, we find

$$\begin{aligned} \Vert u_{n}\Vert ^{2} \le \frac{32\sqrt{2}r}{(7r-32\sqrt{2})} \frac{c_r}{\tau ^{2/(r-2)}}+o_{n}(1). \end{aligned}$$

Since \(\tau > \tau ^{*}\) in \((f_4)\), then

$$\begin{aligned} \Vert u_{n}\Vert ^{2} < \frac{2\pi }{\alpha _0}+o_n(1) \end{aligned}$$

and the result follows. \(\square\)

The next result establishes some compactness properties of minimizing sequences.

Lemma 4.3

If \((u_{n})\subset \mathcal {N}\) is a minimizing sequence for c, then there exists \(u\in H^{1}_{0}(\Omega )\) such that

$$\begin{aligned} \displaystyle \int _{\Omega }f(u_{n})u_{n}\mathrm{d}x\rightarrow \displaystyle \int _{\Omega }f(u)u \mathrm{d}x \end{aligned}$$

and

$$\begin{aligned} \displaystyle \int _{\Omega }F(u_{n})\mathrm{d}x\rightarrow \displaystyle \int _{\Omega }F(u)\mathrm{d}x. \end{aligned}$$

Proof

It is enough to prove the first limit, since the second one is analogous. By Lemma 4.2, we have

$$\begin{aligned} \displaystyle \limsup _{n \rightarrow \infty }\Vert u_n\Vert ^{2}\le \frac{2\pi }{\alpha _0} \end{aligned}$$
(4.2)

and, up to a subsequence, then there exists \(u\in H^{1}_{0}(\Omega )\) such that

$$\begin{aligned} u_{n}(x)\rightarrow u(x) \ \ \text{ a.e. } \text{ in } \ \ \Omega \end{aligned}$$

and

$$\begin{aligned} f(u_{n}(x))u_{n}(x)\rightarrow f(u(x))u(x) \ \ \text{ a.e. } \text{ in } \ \ \Omega . \end{aligned}$$

Now it is sufficient to prove that there is \(g:{\mathbb {R}} \rightarrow {\mathbb {R}}\) such that \(|f(u_n)u_n|\le g(u_n)\) with \((g(u_n))\) convergent in \(L^{1}(\Omega )\).

Note that by the inequality (3.9) we have

$$\begin{aligned} |f(u_{n}(x))u_{n}(x)|\le & {} \varepsilon |u_{n}(x)|^2 +C_{\varepsilon }|u_{n}(x)|^{q}\exp \biggl (\alpha |u_{n}(x)|^{2}\biggl ):=g(u_n(x)). \end{aligned}$$

We will prove that \((g(u_n))\) is convergent in \(L^{1}(\Omega )\). First, note that

$$\begin{aligned} \displaystyle \int _{\Omega }|u_{n}|^2 \mathrm{d}x \rightarrow \displaystyle \int _{\Omega }|u|^2 \mathrm{d}x. \end{aligned}$$
(4.3)

Considering \(s,s'>1\) such that \(\frac{1}{s} + {\frac{1}{s'}} =1\) and s close to 1,  we get

$$\begin{aligned} |u_{n}|^{q} \rightarrow |u|^{q} \ \ \text{ in } \ \ L^{s'}(\Omega ). \end{aligned}$$
(4.4)

Now choosing \(\alpha >\alpha _0\) but close to 1, we have that

$$\begin{aligned} \displaystyle \int _{\Omega }\exp \biggl (\alpha s |u_{n}(x)|^{2}\biggl )\mathrm{d}x=\displaystyle \int _{\Omega }\exp \biggl (\alpha s\Vert u_n\Vert ^{2} \biggl (\frac{|u_{n}(x)|}{\Vert u_n\Vert }\biggl )^{2}\biggl )\mathrm{d}x. \end{aligned}$$

Using Lemma 4.2, we can conclude that

$$\begin{aligned} \displaystyle \int _{\Omega }\exp \biggl (\alpha s |u_{n}(x)|^{2}\biggl )\mathrm{d}x \le \displaystyle \int _{\Omega }\exp \left(4\pi \biggl (\frac{|u_{n}(x)|}{\Vert u_n\Vert }\biggl )^{2}\right)\mathrm{d}x \end{aligned}$$

Then, it follows by Theorem 3.1 that there is \(M>0\) such that

$$\begin{aligned} \displaystyle \int _{\Omega }\exp \biggl (\alpha s |u_{n}(x)|^{2}\biggl )\mathrm{d}x \le M. \end{aligned}$$

Since

$$\begin{aligned} \exp \biggl (\alpha |u_{n}(x)|^{2} \biggl )\rightarrow \exp \biggl (\alpha |u(x)|^{2}\biggl ) \ \ \text{ a.e. } \text{ in } \ \ \Omega , \end{aligned}$$

we use [21, Lemma 4.8] and conclude that

$$\begin{aligned} \exp \biggl (\alpha |u_{n}|^{2} \biggl )\rightharpoonup \exp \biggl (\alpha |u|^{2}\biggl ) \ \ \text{ in } \ \ L^{s}(\Omega ). \end{aligned}$$
(4.5)

Now using (4.4), (4.5) and [21, Lemma 4.8] again, we conclude

$$\begin{aligned} \displaystyle \int _{\Omega }f(u_{n})u_{n}\mathrm{d}x\rightarrow \displaystyle \int _{\Omega }f(u)u \mathrm{d}x. \end{aligned}$$

The proof is complete. \(\square\)

Theorem 4.1

The auxiliary problem has a nonnegative solution \(v_0 \in H^{1}_{0}(\Omega )\).

Proof

Consider \((u_{n}) \subset \mathcal {N}\) a minimizing sequence for c. Then, by Lemma 3.3, \((u_{n})\) is bounded in \(H^{1}_{0}(\Omega )\) and, up to a subsequence,

$$\begin{aligned} u_{n}\rightharpoonup u_0 \ \ \text{ in } \ \ H^{1}_{0}(\Omega ). \end{aligned}$$

We claim that \(u_0 \not \equiv 0\). Indeed, if \(u_0 \equiv 0\), then, from \((a_{1})\) and Lemma 4.3, we get

$$\begin{aligned} {\frac{7}{8\sqrt{2}}}\Vert u_n\Vert ^2\le \displaystyle \int _{\Omega }a(|\nabla u_{n}|^{2})|\nabla u_{n}|^{2}\mathrm{d}x =\displaystyle \int _{\Omega }f(u_{n})u_{n}\mathrm{d}x \rightarrow 0, \end{aligned}$$

which implies

$$\begin{aligned} \Vert u_{n}\Vert \rightarrow 0, \end{aligned}$$

contradicting Lemma 3.2. Let \(t_{0}>0\) such that \(v_{0}:=t_{0}u_{0} \in \mathcal {N}\). Since by \((a_3)\) the function \(s\mapsto A(s^{2})\) is convex, we get \(\displaystyle \int _{\Omega }A(|\nabla tu_0|^{2}) \mathrm{d}x\le \displaystyle \liminf _{n\rightarrow \infty }\displaystyle \int _{\Omega }A(|\nabla tu_n|^{2}) \mathrm{d}x\). From Lemma 4.3, we infer that \(\displaystyle \int _{\Omega }F(tu_0) \mathrm{d}x= \displaystyle \lim _{n\rightarrow \infty }\displaystyle \int _{\Omega }F(tu_n) \mathrm{d}x\). Then,

$$\begin{aligned} c \le I(v_{0})\le \displaystyle \liminf _{n\rightarrow \infty }I(t_{0}u_{n}). \end{aligned}$$

From Lemma 3.1, we conclude that

$$\begin{aligned} \displaystyle \liminf _{n\rightarrow \infty }I(t_{0}u_{n})\displaystyle \le \liminf _{n\rightarrow \infty }\max _{t\ge 0}I(tu_{n})=\liminf _{n\rightarrow \infty }I(u_{n})=c. \end{aligned}$$

The equality \(I'(v_0)=0\) is a consequence of Lemma 3.4. Since \(f(t)=0\), for all \(t\le 0\), we get \(v_0\ge 0\) in \(\Omega\). \(\square\)

5 Proof of Theorem 1.1

We first establish some estimates on solutions of the auxiliary problem from which the existence of positive solution of problem (P) will be deduced. Let us point out that the classical elliptic regularity theory [17] cannot be applied immediately because the coefficients in the differential operator are not necessarily continuous. Throughout this section, we assume that \(\gamma =7/(8\sqrt{2})\) and \(\Gamma =1\).

Lemma 5.1

If \(v_0 \in H^{1}_{0}(\Omega )\) is a solution of the auxiliary problem, then \(v_0 \in L^{\infty }(\Omega )\) and there exists \(K_2>0\) not depending on \(v_0\) such that

$$\begin{aligned} \Vert v_0\Vert _{\infty }\le K_{2}\biggl [ \frac{4r}{(r\gamma -4\Gamma )} \frac{c_r}{\tau ^{2/(r-2)}} \biggl ]^{2}. \end{aligned}$$

Proof

Since \(I(v_0)=c\) and \(I'(v_0)=0\), arguing as Lemma 4.2, we have

$$\begin{aligned} \Vert v_{0}\Vert ^{2} \le \frac{4r}{(r\gamma -4\Gamma )} \frac{c_r}{\tau ^{2/(r-2)}}. \end{aligned}$$
(5.1)

Considering \(\tau > \tau ^{*}\) in \((f_4)\), then

$$\begin{aligned} \Vert v_{0}\Vert ^{2} < \frac{2\pi }{\alpha _0}. \end{aligned}$$
(5.2)

In what follows, let \(R>R_{1}>0\) with \(R>1\) and take a cutoff function \(\eta _{R} \in C^{\infty }_{0}(\Omega )\) such that \(0 \le \eta _R \le 1\), \(\eta _{R} \equiv 0\) in \(B_{R}^{c}\), \(\eta _R \equiv 1\) in \(B_{R_{1}}\) and \(|\nabla \eta _{R}|\le C/R\), where \(B_{R}(0)\subset \Omega\) is a ball in \({\mathbb {R}}^2\) and \(C>0\) is a constant.

Define for \(L > 0\),

$$\begin{aligned} v_{L,0}(x) =\ \ \left\{ \begin{array}{l} v_{0}(x), \quad \text{ if } \quad v_{0}(x) \le L \\ \\ L, \quad \text{ if } \quad v_{0}(x) \ge L, \\ \end{array} \right. \\z_{L,0} = \eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}v_{0} \quad \text{ and } \quad w_{L,0} = \eta _{R}v_{0} v_{L,0}^{\sigma - 1} \end{aligned}$$

with \(\sigma > 1\) to be determined later. Taking \(z_{L,0}\) as a test function we obtain

$$\begin{aligned} I'(v_{0})z_{L,0}=0. \end{aligned}$$

In other words,

$$\begin{aligned} \int _{\Omega }a(|\nabla v_0|^{2})\nabla v_{0} \nabla z_{L,0} \mathrm{d}x=\int _{\Omega }f(v_{0})z_{L,0}\mathrm{d}x. \end{aligned}$$

By (3.9), (5.2) and Theorem 3.1, we obtain

$$\begin{aligned} \int _{\Omega }a(|\nabla v_0|^{2})\nabla v_{0} \nabla z_{L,0} \mathrm{d}x\le \epsilon \displaystyle \int _{\Omega }v_0z_{L,0}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }|v_{0}^{q-1}z_{L,0}|^{s'}\mathrm{d}x\biggl )^{1/s'}. \end{aligned}$$

Using \(z_{L,0}\) and \((a_1)\), we obtain

$$\begin{aligned} \gamma \int _{\Omega }\eta _{R}^{2}v_{L,0}^{2(\sigma -1)}|\nabla v_{0}|^{2} \ \mathrm{d}x&\le - \int _{\Omega }a(|\nabla v_0|^{2})\eta _{R}v_{0}v_{L,0}^{2(\sigma -1)}\nabla \eta _{R}\nabla v_{0} \ \mathrm{d}x \\&\quad- 2(\sigma -1)\int _{\Omega }a(|\nabla v_0|^{2})v_{L,0}^{(2\sigma -3)}v_{0}\nabla v_{0} \nabla v_{L,0} \mathrm{d}x\\&\quad + \epsilon \displaystyle \int _{\Omega }|v_0|^{2}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }|v_{0}^{q}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}|^{s'}\mathrm{d}x\biggl )^{1/s'} . \end{aligned}$$

The definition of \(v_{L,0}\) implies

$$\begin{aligned} -2(\sigma -1)\int _{\Omega }a(|\nabla v_0|^{2})v_{L,0}^{(2\sigma -3)}v_{0}\nabla v_{0} \nabla v_{L,0}\mathrm{d}x\le 0. \end{aligned}$$

Thus, by \((a_1)\) again

$$\begin{aligned} \gamma \int _{\Omega }\eta _{R}^{2}v_{L,0}^{2(\sigma -1)}|\nabla v_{0}|^{2} \ \mathrm{d}x&\le \Gamma C_{1} \int _{\Omega }\eta _{R}v_{0}v_{L,0}^{2(\sigma -1)}|\nabla \eta _{R}||\nabla v_{0}| \ \mathrm{d}x \\&\quad + \epsilon \displaystyle \int _{\Omega }|v_0|^{2}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }|v_{0}^{q}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}|^{s'}\mathrm{d}x\biggl )^{1/s'}. \end{aligned}$$

Taking \(\widetilde{\tau }>0\) and using Young’s inequality, we obtain

$$\begin{aligned} \gamma \int _{\Omega }\eta _{R}^{2}v_{L,0}^{2(\sigma -1)}|\nabla v_{0}|^{2} \ \mathrm{d}x \le&\Gamma C_{1} \int _{\Omega }\bigg (\widetilde{\tau }\eta _{R}^{2}|\nabla v_{0}|^{2}+ C_{\widetilde{\tau }}v_{0}^{2}|\nabla \eta _{R}|^{2} \biggl )v_{L,0}^{2(\sigma -1)} \ \mathrm{d}x \\&\quad +\epsilon \displaystyle \int _{\Omega }|v_0|^{2}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }|v_{0}^{q}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}|^{s'}\mathrm{d}x\biggl )^{1/s'}. \end{aligned}$$

Choosing \(\widetilde{\tau }\) sufficient small, it follows that

$$\begin{aligned} \int _{\Omega }\eta _{R}^{2}v_{L,0}^{2(\sigma -1)}|\nabla v_{0}|^{2} \ \mathrm{d}x&\le C_{2}\bigg (\int _{\Omega }v_{0}^{2}v_{L,0}^{2(\sigma -1)}|\nabla \eta _{R}|^{2} \ \mathrm{d}x \nonumber \\&\quad + \epsilon \displaystyle \int _{\Omega }|v_0|^{2}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }|v_{0}^{q}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}|^{s'}\mathrm{d}x\biggl )^{1/s'} \bigg ). \end{aligned}$$
(5.3)

On the other hand, we get

$$\begin{aligned} S_{\Upsilon } \Vert w_{L,0}\Vert ^{2}_{L^{\Upsilon }(\Omega )}\le & {} \int _{\Omega } \left| \nabla \left( \eta _{R}v_{0} v_{L,0}^{\sigma -1}\right) \right| ^2 \\\le & {} \int _{\Omega } |v_{0}|^2 v_{L,0}^{2(\sigma -1)}|\nabla \eta _{R}|^2 + \int _{\Omega } \eta _{R}^2\left| \nabla \left( v_{0} v_{L,0}^{\sigma -1}\right) \right| ^2, \end{aligned}$$

where \(S_\Upsilon\) is the best Sobolev constant of \(H^{1}_{0}(\Omega )\) in \(L^{\Upsilon }(\Omega )\) and \(\Upsilon .1\) that will fix after. But

$$\begin{aligned} \int _{\Omega } \eta _{R}^2\left| \nabla \left( v_{0} v_{L,_{0}}^{\sigma -1}\right) \right| ^2= & {} \int _{\{|v_{0}| \le L\}}\eta _{R}^2\left| \nabla \left( v_{0} v_{L,0}^{\sigma -1}\right) \right| ^2 + \int _{\{|v_{0}|> L\}} \eta _{R}^2\left| \nabla \left( v_{0} v_{L,0}^{\sigma -1}\right) \right| ^2 \\= & {} \int _{\{|v_{0}| \le L\}} \eta _{R}^2\left| \nabla v_{0}^{\sigma }\right| ^2 + \int _{\{|v_{0}| > L\}} \eta _{R}^2 L^{2(\sigma -1)}\left| \nabla v_{0}\right| ^2 \\&\le \sigma ^2 \int _{\Omega }\eta _{R}^2 v_{L,0}^{2(\sigma -1)}|\nabla v_{0}|^2, \end{aligned}$$

and therefore,

$$\begin{aligned} \Vert w_{L,0}\Vert ^{2}_{L^{\Upsilon }(\Omega )} \le C_3\sigma ^{2} \left( \int _{\Omega } |v_{0}|^{2}v_{L,0}^{2(\sigma - 1)}|\nabla \eta _{R}|^{2} + \int _{\Omega } \eta _{R}^{2} v_{L,0}^{2(\sigma - 1)}|\nabla v_{0}|^{2}\right) . \end{aligned}$$

From this estimate and (5.3),

$$\begin{aligned} \Vert w_{L,0}\Vert ^{2}_{L^{\Upsilon }(\Omega )}&\le C_4\sigma ^{2} \int _{\Omega } |v_{0}|^{2}v_{L,0}^{2(\sigma - 1)}|\nabla \eta _{R}|^{2} \nonumber \\&+\sigma ^{2}C_{4}\biggl ( \epsilon \displaystyle \int _{\Omega }|v_0|^{2}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }|v_{0}^{q}\eta _{R}^{2}v_{L,0}^{2(\sigma - 1)}|^{s'}\mathrm{d}x\biggl )^{1/s'} \biggl ). \end{aligned}$$
(5.4)

for every \(\sigma >1\).

The above expression, the properties of \(\eta _{R}\) and \(v_{L,0} \le v_{0}\) imply that

$$\begin{aligned} \begin{array}{ll} \displaystyle \Vert w_{L,0}\Vert ^{2}_{L^{r}(\Omega )} &{}\displaystyle \le C_4\sigma ^{2} \int _{\Omega } |v_{0}|^{2\sigma }|\nabla \eta _{R}|^{2}\\ &\quad {}\displaystyle +\sigma ^{2}C_{4}\biggl ( \epsilon \displaystyle \int _{\Omega }|v_0|^{2\sigma }\eta ^{2}_{R}\mathrm{d}x+ MC_{\epsilon }\biggl (\int _{\Omega }\eta ^{2}_{R}||v_{0}|^{q-2}|v_{0}|^{2\sigma }|^{s'}\mathrm{d}x\biggl )^{1/s'}\biggl ). \end{array} \end{aligned}$$
(5.5)

Taking

$$\begin{aligned} t:= \frac{qq}{2(q-2)}>1,~~~~\Upsilon := \frac{2t}{t-1}, \end{aligned}$$
(5.6)

then we can apply Hölder’s inequality with exponents \(t/(t-1)\) and t in (5.5) to get

$$\begin{aligned} \begin{array}{lcl} \Vert w_{L,0}\Vert ^{2}_{L^{\Upsilon }(\Omega )} &{}\le &{} C_4\sigma ^{2} \Vert v_{0}\Vert _{L^{\sigma \Upsilon }(B_{R})}^{2\sigma } \left( \displaystyle \int _{B_{R}}|\nabla \eta _{R}|^{2t}\right) ^{1/t} \\ &{}&{}+ \displaystyle C_4\sigma ^{2} \Vert v_{0}\Vert _{L^{\sigma \Upsilon }(B_{R})}^{2\sigma } \left( \displaystyle \int _{B_{R}}| \eta _{R}|^{2t}\right) ^{1/t} \\ &{} +&{} M_{1}C_{\epsilon }C_4\sigma ^{2} \Vert v_{0}\Vert _{L^{\sigma \Upsilon }(B_{R})}^{2\sigma } \left( \displaystyle \int _{B_{R}}|v_{0}|^{qq/2}\right) ^{1/t}. \end{array} \end{aligned}$$
(5.7)

Since \(\eta _{R}\) is constant on \(B_{R_1} \cup B_{R}^c\) and \(|\nabla \eta _{R}| \le C/R\), we conclude that

$$\begin{aligned} \int _{B_{R}}|\nabla \eta _{R}|^{2t} = \int _{B_{R}\backslash B_{R_{1}}}|\nabla \eta _{R}|^{2t} \le \frac{C_5}{R^{2t-2}} \le C_5. \end{aligned}$$
(5.8)

We have used \(R>1\) and \(2t>q > 2\) in the last inequality.

Considering

$$\begin{aligned} \int _{\Omega }|v_{0}|^{qq/2}\le K, \end{aligned}$$

we can use (5.7) and (5.8) to conclude that

$$\begin{aligned} \Vert w_{L,0}\Vert ^{2}_{L^{\Upsilon }(\Omega )} \le C_6\sigma ^{2}\Vert v_{0}\Vert ^{2\sigma }_{L^{\sigma \Upsilon }(B_{R})}. \end{aligned}$$

Since

$$\begin{aligned} \Vert v_{L,0}\Vert ^{2\sigma }_{L^{\sigma \Upsilon }(B_{R_{1}})}&= \left( \int _{B_{R_{1}}}v_{L,0}^{\sigma \Upsilon }\right) ^{2/\Upsilon } \\&\le \left( \int _{\Omega } \eta _{R}^{\Upsilon } |v_{0}|^{\Upsilon }v_{L,0}^{\Upsilon (\sigma -1) }\right) ^{2/\Upsilon } \\&= \Vert w_{L,0}\Vert ^{2}_{L^{\Upsilon }(\Omega )} \le C_6\sigma ^{2}\Vert v_{0}\Vert ^{2\sigma }_{L^{\sigma \Upsilon }(\Omega )}, \end{aligned}$$

we can apply Fatou’s lemma in the variable L and Sobolev embedding to obtain

$$\begin{aligned} \Vert v_{0}\Vert _{L^{\sigma \Upsilon }(B_{R_{1}})}\le C_7^{1/\sigma }\sigma ^{1/\sigma }\Vert v_{0}\Vert . \end{aligned}$$

Here, \(C_7\) is a positive constant independent on R. Iterating this process, for each \(k\in \mathbb {N}\), it follows that

$$\begin{aligned} \Vert v_{0}\Vert _{L^{\sigma ^k \Upsilon }(B_{R{1}})} \le C_7^{\sum _{i=1}^k \sigma ^{-i}} \sigma ^{\sum _{i=1}^m i \sigma ^{-i}} \Vert v_{0}\Vert . \end{aligned}$$

Since \(\Omega\) can be covered by a finite number of balls \(B_{R_{1}}^j\), we have that

$$\begin{aligned} \Vert v_{0}\Vert _{L^{\sigma ^k \Upsilon }(\Omega )}\le \displaystyle \sum _{j}^{finite} \Vert v_{0}\Vert _{L^{\sigma ^k \Upsilon }(B_{R_{1}}^j)} \le \displaystyle \sum _j^{finite} C_7^{\sum _{i=1}^k \sigma ^{-i}} \sigma ^{\sum _{i=1}^m i \sigma ^{-i}} \Vert v_{0}\Vert . \end{aligned}$$

Using (5.1) and since \(\sigma >1\), we let \(k \rightarrow \infty\) to get \(K_{2}>0\) such that

$$\begin{aligned} \Vert v_{0}\Vert _{L^{\infty }(\Omega )} \le K_{2}\Vert v_{0}\Vert \le K_{2}\biggl [ \frac{4r}{(r\gamma -4\Gamma )} \frac{c_r}{\tau ^{2/(r-2)}} \biggl ]^{2}. \end{aligned}$$

The proof is now complete. \(\square\)

5.1 Proof of Theorem 1.1 completed

By Lemma (5.1) and \((f_4)\), we obtain

$$\begin{aligned} \Vert v_{0}\Vert _{L^{\infty }(\Omega )} \le \delta . \end{aligned}$$

Now from (3.4), given \(\epsilon >0\), we get

$$\begin{aligned} \Vert f(v_0)\Vert _{L^{\infty }(\Omega )} \le \epsilon \Vert v_{0}\Vert _{L^{\infty }(\Omega )}\le \epsilon \delta . \end{aligned}$$

Using Lemma 2.3, then \(v_{0} \in C^{1,\alpha }(\Omega )\), and for \(\epsilon >0\) sufficient small, we have

$$\begin{aligned} \Vert \nabla v_0\Vert _{L^{\infty }(\Omega )} <1. \end{aligned}$$

The proof of Theorem  1.1 is now complete. \(\square\)