Existence of solution for a general class of elliptic equations with exponential growth

Abstract

We use Galerkin approximations to show the existence of solution for a class of elliptic equations on bounded domains in $\mathbb {R}^2$ with subcritical or critical exponential nonlinearities. We are able to solve the problem under more general assumptions usually assumed in the variational the approach, but not in our paper.

1 Introduction

We prove the existence of solution of the problem

$$\begin{aligned} \left\{ \begin{array}{lll} -\Delta v = \lambda v^{q} + f(v) &{}\hbox {in}&{} \Omega \\ v>0 &{}\hbox {in}&{} \Omega \\ v=0 &{}\hbox {on}&{} \partial \Omega , \end{array} \right. \end{aligned}$$

(1)

where $\Omega \subset \mathbb {R}^2$ is a bounded domain with smooth boundary, $\lambda >0$ is a parameter, $0<q<1$, $f: [0, \infty ) \rightarrow \mathbb {R}$ is a continuous function, and

$$\begin{aligned} 0\le f(s)s\le C|s|^p\exp (\alpha \,s^2) \end{aligned}$$

(2)

where $2< p<+\infty$ and $\alpha >0$.

We state our main result.

Theorem 1.1

Suppose that $f:[0,\infty ) \rightarrow \mathbb {R}$ is a continuous function satisfying (2). Then there exists $\lambda ^*>0$ such that for every $\lambda \in (0,\lambda ^*)$, the problem (1) has a positive weak solution $u \in H^1_0(\Omega ) \cap H^2(\Omega )$.

Elliptic problems of the type

$$\begin{aligned} \left\{ \begin{array}{lll} -\Delta v = g(x,v) &{}\hbox {in}&{} \Omega \\ v=0 &{}\hbox {on}&{} \partial \Omega , \end{array} \right. \end{aligned}$$

(3)

in $\Omega \subset \mathbb {R}^2$ where g(x, v) is continuous and behaves like $\exp (\alpha |v|^{2})$ as $|v| \rightarrow +\infty$ have been studied by many authors, see [6, 10–12, 16, 19]. One of the main ingredients is the Trudinger–Moser inequality introduced in [18, 21], namely. Given $u \in H^1_0(\Omega )$, then

$$\begin{aligned} e^{\sigma |u|^{2}} \in L^1(\Omega )\,\,\,\text{ for } \text{ every }\,\,\,\sigma >0, \end{aligned}$$

(4)

and there exists a positive constant L such that

$$\begin{aligned} \sup _{\Vert u\Vert _{H^1_0(\Omega )}\le 1}\int _{\Omega }e^{\sigma |u|^{2}}{\hbox {d}}x \le L\,\,\,\text{ for } \text{ every }\,\,\,\sigma \le 4\pi . \end{aligned}$$

(5)

We say that g has subcritical growth at $+\infty$ if for every $\sigma >0$

$$\begin{aligned} \lim _{s \rightarrow +\infty }\frac{|g(x,s)|}{e^{\sigma s^2}}=0 \end{aligned}$$

and g has critical growth at $+\infty$ if there exists $\sigma _0>0$ such that

$$\begin{aligned} \lim _{s \rightarrow +\infty }\frac{|g(x,s)|}{e^{\sigma s^2}}=0 \, \forall \, \sigma >\sigma _0 \quad \text{ and } \,\, \lim _{s \rightarrow +\infty }\frac{|g(x,s)|}{e^{\sigma s^2}}=+\infty \, \forall \, \sigma <\sigma _0. \end{aligned}$$

The only assumptions we assume are that $0<q<1$, f is continuous and satisfies the growth assumption (2), and thus the nonlinearity $g(s)=\lambda s^q + f(s)$ of problem (1) can have subcritical or critical behavior at $+\infty$.

Most papers treat problem (3) by means of variational methods, and then usually it is assumed that g has subcritical or critical growth and sometimes $g(s) \ge c |s|^p$, where $c>0$ is a constant, see [12]. Another common assumption on g is the so-called Ambrosetti–Rabinowitz condition

$$\begin{aligned} \exists R>0 \text{ and } \theta > 2 \text{ such } \text{ that } 0< \theta G(x,s) \le s g(x,s) \,\, \forall |s| \ge R \text{ and } x \in \Omega , \end{aligned}$$

where $G(s)=\int _0^s g(t){\hbox {d}}t$, see [10–12].

Even when the Ambrosetti–Rabinowitz can be dropped, some conditions have to be assumed to give compactness of the Palais-Samle sequences or Cerami sequences, see for instance [16] where they assume

$$\begin{aligned}&g: \overline{\Omega }\times \mathbb {R} \text{ is } \text{ continuous } \text{ and } g(x,0) = 0;\\&\quad \exists t_0 > 0 \text{ and } M > 0 \text{ such } \text{ that } 0 < G(x,s) \le M g(x,s) \,\, \forall |s| \ge t_0 \text{ and } x \in \Omega ;\\&0 < 2 G(x,s) \le s g(x,s) \,\, \forall |s| \ge 0 \text{ and } x \in \Omega . \end{aligned}$$

A problem in $\Omega = \mathbb {R}^2$ without Ambrosetti–Rabinowitz condition and exponential growth on g different from (2) has been addressed in [15].

We are able to solve (1) under weaker assumptions by using the Galerkin method. For that matter, we approximate f by Lipschitz functions in Sect. 2. We solve the approximating problems (11) in Sect. 3. Section 4 is devoted to prove Theorem 1.1, and in doing so, we show that the solutions $v_n$ of problem (11) are bounded away from zero and converge to a positive solution of (1).

Problem (1) is also studied in $\mathbb {R}^2$, see for instance [1, 2, 4, 8, 22]. Problems with nonlinearities with exponential growth are also important in conformal geometry [9, 17].

2 Approximating functions

To prove Theorem 1.1, we approximate f by Lipschitz functions $f_k:\mathbb {R} \rightarrow \mathbb {R}$ defined by

$$\begin{aligned} f_k(s)=\displaystyle \left\{ \begin{array}{lll} -k\left[ G\left( -k-\frac{1}{k}\right) - G(-k)\right] , &{}\hbox {if}&{} s\le -k\\ -k\left[ G\left( s-\frac{1}{k}\right) - G(s)\right] , &{}\hbox {if}&{} -k\le s \le -\frac{1}{k}\\ k^2s\left[ G\left( -\frac{2}{k}\right) - G\left( -\frac{1}{k}\right) \right] , &{}\hbox {if}&{} -\frac{1}{k}\le s\le 0\\ k^2s\left[ G\left( \frac{2}{k}\right) - G\left( \frac{1}{k}\right) \right] , &{}\hbox {if}&{} 0\le s\le \frac{1}{k}.\\ k\left[ G\left( s+\frac{1}{k}\right) - G(s)\right] , &{}\hbox {if}&{} \frac{1}{k}\le s \le k\\ k\left[ G\left( k+\frac{1}{k}\right) - G(k)\right] , &{}\hbox {if}&{} s\ge k.\\ \end{array} \right. \end{aligned}$$

(6)

where $G(s)=\int _0^sf(\xi ){\hbox {d}}\xi$.

The following approximation result was proved in [20].

Lemma 2.1

Let $f:\mathbb {R} \rightarrow \mathbb {R}$ be a continuous function such that $sf(s)\ge 0$ for every $s \in \mathbb {R}$. Then there exists a sequence $f_k:\mathbb {R} \rightarrow \mathbb {R}$ of continuous functions satisfying

• (i) $sf_k(s)\ge 0$ for every $s \in \mathbb {R}$;

• (ii) $\forall \, k \in \mathbb {N}$ $\exists c_k>0$ such that $|f_k(\xi ) - f_k(\eta )|\le c_k|\xi - \eta |$ for every $\xi , \eta \in \mathbb {R}$;

• (iii) $f_k$ converges uniformly to f in bounded subsets of $\mathbb {R}$.

The sequence $f_k$ of the previous lemma has some additional properties.

Lemma 2.2

Let $f: \mathbb {R} \rightarrow \mathbb {R}$ be a continuous function satisfying (2) for every $s \in \mathbb {R}$. Then the sequence $f_k$ of Lemma 2.1 satisfies

• (i) $\forall \, k \in \mathbb {N}$, $0\le sf_k(s) \le C_1|s|^p\exp (4\alpha \,s^2)$ for every $|s|\ge \frac{1}{k}$;

• (ii) $\forall \, k \in \mathbb {N}$, $0\le sf_k(s) \le C_2|s|^2\exp (4\alpha \,s^2)$ for every $|s|\le \frac{1}{k}$,

where $C_1$ and $C_2$ are positive constants independent of k.

Proof of Lemma 2.2

Everywhere in this proof the constant C is the one of (2).

First step. Suppose that $-k\le s \le -\frac{1}{k}$.

By the mean value theorem, there exists $\eta \in \left( s-\frac{1}{k},s \right)$ such that

$$\begin{aligned} f_k(s)=-k\left[ G\left( s-\frac{1}{k}\right) - G(s)\right] =-kG'(\eta )\left( s-\frac{1}{k}-s\right) =f(\eta ) \end{aligned}$$

and

$$\begin{aligned} sf_k(s)=sf(\eta ). \end{aligned}$$

Since $s-\frac{1}{k}<\eta <s<0$ and $f(\eta )<0$, we have $sf(\eta ) \le \eta f(\eta )$. Therefore,

$$\begin{aligned} sf_k(s) \le \eta f(\eta )\le & {} C|\eta |^p\exp (\alpha \,|\eta |^2)\\\le & {} C|s-\frac{1}{k}|^p\exp \left( \alpha \,|s-\frac{1}{k}|^2\right) \\\le & {} C\left( |s| +\frac{1}{k}\right) ^p\exp \left( \alpha \,\left( |s| +\frac{1}{k}\right) ^2\right) \\\le & {} C(2|s|)^p\exp (\alpha \,(2|s|)^2)\\= & {} C2^p|s|^p\exp (4\alpha \,|s|^2). \end{aligned}$$

Second step. Assume $\frac{1}{k}\le s \le k$.

By the mean value theorem, there exists $\eta \in \left( s,s+\frac{1}{k}\right)$ such that

$$\begin{aligned} f_k(s)=k\left[ G\left( s+\frac{1}{k}\right) - G(s)\right] =kG'(\eta )\left( s+\frac{1}{k}-s\right) =f(\eta ) \end{aligned}$$

and

$$\begin{aligned} sf_k(s)=sf(\eta ). \end{aligned}$$

Since $0<s <\eta <s+\frac{1}{k}$ and $f(\eta )>0$, we have $sf(\eta ) \le \eta f(\eta )$. Therefore,

$$\begin{aligned} sf_k(s) \le \eta f(\eta )\le & {} C|\eta |^p\exp (\alpha \,|\eta |^2)\\\le & {} C|s+\frac{1}{k}|^p\exp \left( \alpha \,|s+\frac{1}{k}|^2\right) \\\le & {} C(2|s|)^p\exp (\alpha \,(2|s|)^2)\\= & {} C2^p|s|^p\exp (4\alpha \,|s|^2). \end{aligned}$$

Third step. Suppose that $|s|\ge k$, then

$$\begin{aligned} f_k(s)=\displaystyle \left\{ \begin{array}{lcc} -k\left[ G\left( -k-\frac{1}{k}\right) - G(-k)\right] , &{}\hbox {if}&{} s\le -k\\ k[G\left( k+\frac{1}{k}\right) - G(k)], &{}\hbox {if}&{} s\ge k.\\ \end{array} \right. \end{aligned}$$

(7)

If $s\le -k$, by the mean value theorem, there exists $\eta \in \left( -k-\frac{1}{k},-k\right)$ such that

$$\begin{aligned} f_k(s)=k\left[ G\left( -k-\frac{1}{k}\right) - G(-k)\right] =-kG'(\eta )\left( -k-\frac{1}{k}-(-k)\right) =f(\eta ) \end{aligned}$$

and

$$\begin{aligned} sf_k(s)=sf(\eta ). \end{aligned}$$

Since $-k-\frac{1}{k} <\eta <-k<0$ and $k<|\eta | < k + \frac{1}{k}$, we conclude that

$$\begin{aligned} sf_k(s)=\frac{s}{\eta }\eta f(\eta )\le & {} \frac{|s|}{|\eta |}C|\eta |^p\exp (\alpha \,|\eta |^2) =C|s||\eta |^{p-1}\exp (\alpha \,|\eta |^2)\nonumber \\\le & {} C|s|\left( k + \frac{1}{k}\right) ^{p-1}\exp \left( \alpha \,\left( k + \frac{1}{k}\right) ^2\right) \nonumber \\\le & {} C|s|\left( |s| + \frac{1}{k}\right) ^{p-1}\exp \left( \alpha \,\left( |s| + \frac{1}{k}\right) ^2\right) \nonumber \\\le & {} C|s|(2|s|)^{p-1}\exp (\alpha \,(2|s|)^2)\nonumber \\\le & {} C2^{p-1}|s|^{p}\exp (4\alpha \,|s|^2). \end{aligned}$$

(8)

If $s\ge k$, by the mean value theorem, there exists $\eta \in \left( k,k + \frac{1}{k}\right)$ such that

$$\begin{aligned} f_k(s)=k\left[ G\left( k+\frac{1}{k}\right) - G(k)\right] =kG'(\eta )\left( k+\frac{1}{k}-k\right) =f(\eta ). \end{aligned}$$

By computations similar to conclude (8) one has

$$\begin{aligned} sf_k(s)=sf(\eta )=\frac{s}{\eta }\eta f(\eta ) \le \frac{|s|}{|\eta |}C|\eta |^p \exp (\alpha \,|\eta |^2) \le C2^{p-1}|s|^{p}\exp (4\alpha \,|s|^2). \end{aligned}$$

Fourth step. Assume $-\frac{1}{k}\le s\le \frac{1}{k}$, then

$$\begin{aligned} f_k(s)=\displaystyle \left\{ \begin{array}{lcc} k^2s\left[ G\left( -\frac{2}{k}\right) - G\left( -\frac{1}{k}\right) \right] , &{}\hbox {if}&{} -\frac{1}{k}\le s\le 0\\ k^2s\left[ G\left( \frac{2}{k}\right) - G\left( \frac{1}{k}\right) \right] , &{}\hbox {if}&{} 0\le s\ge \frac{1}{k}.\\ \end{array} \right. \end{aligned}$$

(9)

If $-\frac{1}{k}\le s\le 0$, by the mean value theorem, there exists $\eta \in (-\frac{2}{k},-\frac{1}{k})$ such that

$$\begin{aligned} f_k(s)=k^2s\left[ G\left( -\frac{2}{k}\right) - G\left( -\frac{1}{k}\right) \right] =k^2sG'(\eta )\left( -\frac{2}{k}-\left( -\frac{1}{k}\right) \right) =-ksf(\eta ). \end{aligned}$$

Therefore,

$$\begin{aligned} sf_k(s)= & {} -ks^2f(\eta )=-k\frac{s^2}{\eta }\eta f(\eta ) \le k\frac{s^2}{|\eta |}\eta f(\eta ) \nonumber \\&\le Ck|s|^2|\eta |^{p-1}\exp (\alpha \,|\eta |^2) \le Ck|s|^2(\frac{2}{k})^{p-1}\exp (\alpha \,|\eta |^2)\nonumber \\&\le C2^{p-1}|s|^2\exp \left( \alpha \,\left( \frac{2}{k}\right) ^2\right) \nonumber \\&\le C2^{p-1}|s|^2\exp (4\alpha ) \le C2^{p-1}\exp (4\alpha )|s|^2\exp (4\alpha \,|s|^2). \end{aligned}$$

(10)

If $0\le s \le \frac{1}{k}$, by the mean value theorem, there exists $\eta \in (\frac{1}{k},\frac{2}{k})$ such that

$$\begin{aligned} f_k(s)=k^2s[G(\frac{2}{k}) - G(\frac{1}{k})]=k^2sG'(\eta )(\frac{2}{k}-\frac{1}{k})=ksf(\eta ). \end{aligned}$$

By similar computations to conclude (10) one obtains

$$\begin{aligned} sf_k(s)=ks^2f(\eta )=k\frac{s^2}{|\eta |}\eta f(\eta ) \le C2^{p-1}\exp (4\alpha )|s|^2\exp (4\alpha \,|s|^2). \end{aligned}$$

The proof of the lemma follows by taking $C_1=C2^p$ ad $C_2=C2^{p-1}$, where C is given in (2). $\square$

3 Approximate equation

To prove Theorem 1.1, we first show the existence of a solution of the following auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{lcc} -\Delta v = \lambda v^{q} + f_n(v) + \frac{1}{n} &{}\hbox {in}&{} \Omega \\ v>0 &{}\hbox {in}&{} \Omega \\ v=0 &{}\hbox {on}&{} \partial \Omega , \end{array} \right. \end{aligned}$$

(11)

where $f_n$ are given by Lemma 2.1 and Lemma 2.2.

We will use the Galerkin method together with the following fixed point theorem, see [20] and [14, Theorem5.2.5]. A similar approach was already used in [3].

Lemma 3.1

Let $F: \mathbb {R}^d \rightarrow \mathbb {R}^d$ be a continuous function such that $\left\langle F(\xi ),\xi \right\rangle \ge 0$ for every $\xi \in \mathbb {R}^d$ with $|\xi |=r$ for some $r>0$. Then, there exists $z_0$ in the closed ball $\overline{B}_r(0)$ such that $F(z_0)=0$.

The main result in this section is the following.

Lemma 3.2

There exist $\lambda ^*>0$ and $n^* \in \mathbb {N}$ such that (11) has a weak nonnegative and nontrivial solution for every $\lambda \in (0,\lambda ^*)$ and $n\ge n^*$.

Proof of Lemma 3.2

Let $\mathcal {B}=\{w_1,w_2,\dots ,w_m,\dots \}$ be an orthonormal basis of $H_0^1(\Omega )$ and define

$$\begin{aligned} W_m=[w_1,w_2,\dots ,w_m], \end{aligned}$$

to be the space generated by $\{w_1,w_2,\dots ,w_m\}$. Define the function $F:\mathbb {R}^m \rightarrow \mathbb {R}^m$ such that $F(\xi )=(F_1(\xi ),F_2(\xi ),\dots , F_m(\xi ))$, where $\xi =(\xi _1, \xi _2, ..., \xi _m) \in \mathbb {R}^m$,

$$\begin{aligned} F_j(\xi )=\int _{\Omega }\nabla v\nabla w_j - \lambda \int _{\Omega }(v_+)^{q}w_j - \int _{\Omega }f_n(v_+)w_j - \frac{1}{n}\int _{\Omega }w_j, \,\,\,j=1,2,\dots ,m \end{aligned}$$

and $v=\sum _{i=1}^m\xi _i w_i$ belongs to $W_m$. Therefore,

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle =\int _{\Omega }|\nabla v|^2 - \lambda \int _{\Omega }(v_+)^{q+1} - \int _{\Omega }f_n(v_+)v_+ - \frac{1}{n}\int _{\Omega }v, \end{aligned}$$

(12)

where $v_+ = \max \{v,0\}$ and $v_- = v_+ - v$.

Given $v \in W_m$, we define

$$\begin{aligned} \Omega ^+_n=\left\{ x \in \Omega : |v(x)|\ge \frac{1}{n}\right\} \end{aligned}$$

and

$$\begin{aligned} \Omega ^-_n=\left\{ x \in \Omega : |v(x)|< \frac{1}{n}\right\} . \end{aligned}$$

Thus, we rewrite (12) as

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle = \left\langle F(\xi ),\xi \right\rangle _P + \left\langle F(\xi ),\xi \right\rangle _{N}, \end{aligned}$$

where

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle _P=\int _{\Omega ^+_n}|\nabla v|^2 - \lambda \int _{\Omega ^+_n}(v_+)^{q+1} - \int _{\Omega ^+_n}f_n(v_+)v_+ - \frac{1}{n}\int _{\Omega ^+_n}v \end{aligned}$$

and

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle _{N}=\int _{\Omega ^-_n}|\nabla v|^2 - \lambda \int _{\Omega ^-_n}(v_+)^{q+1} - \int _{\Omega ^-_n}f_n(v_+)v_+ - \frac{1}{n}\int _{\Omega ^-_n}v. \end{aligned}$$

Step 1. Since $0<q<1$, then

$$\begin{aligned} \int _{\Omega ^+_n}(v_+)^{q+1} \le \int _{\Omega }|v|^{q+1} = \Vert v\Vert ^{q+1}_{L^{q+1}(\Omega )}\le C_1\Vert v\Vert ^{q+1}_{H^1_0(\Omega )}. \end{aligned}$$

(13)

By virtue of Lemma 2.2 (i), we get

$$\begin{aligned} \displaystyle \int _{\Omega ^+_n}f_n(v_+)v_+\le & {} \displaystyle C_1\int _{\Omega ^+_n}|v_+|^p\exp (4\alpha |v_+|^2){\hbox {d}}x\nonumber \\\le & {} \displaystyle C_1 \left( \int _{\Omega }|v_+|^{p+1}\right) ^{\frac{p}{p+1}}\left( \int _{\Omega }\exp (4\alpha (p+1)|v_+|^2){\hbox {d}}x\right) ^{\frac{1}{p+1}}\nonumber \\\le & {} \displaystyle C_1\Vert v\Vert _{L^{p+1}(\Omega )}^{p}\left( \int _{\Omega }\exp (4\alpha (p+1)|v|^2){\hbox {d}}x\right) ^{\frac{1}{p+1}}. \end{aligned}$$

(14)

It follows from (13) and (14) that

$$\begin{aligned} \begin{array}{rcl} \left\langle F(\xi ),\xi \right\rangle _P &{}\ge &{}\displaystyle \int _{\Omega ^+_n}|\nabla v|^2 - \lambda C_0\Vert v\Vert ^{q+1}_{H^1_0(\Omega )}\\ &{}-&{} \displaystyle C_1\Vert v\Vert ^{p}_{H^1_0(\Omega )}\left( \int _{\Omega }\exp (4\alpha (p+1)|v|^2){\hbox {d}}x\right) ^{\frac{1}{p+1}} - \frac{C_3}{n}\Vert v\Vert _{H^1_0(\Omega )}, \end{array} \end{aligned}$$

(15)

where $C_0$, $C_1$, and $C_3$ are constants depending only on C, p, and $|\Omega |$.

Step 2. Since $0<q<1$, then

$$\begin{aligned} \int _{\Omega ^-_n}(v_+)^{q+1} \le \int _{\Omega ^-_n}|v|^{q+1} \le |\Omega |\frac{1}{n^{q+1}}. \end{aligned}$$

(16)

By virtue of Lemma 2.2 (ii), we get

$$\begin{aligned} \int _{\Omega ^-_n}f_n(v_+)v_+ \le C_2 \int _{\Omega ^-_n}|v_+|^2\exp (4\alpha |v_+|^2){\hbox {d}}x\le C_2 \exp (4\alpha )|\Omega |\frac{1}{n^2}. \end{aligned}$$

(17)

It follows from (16) and (17) that

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle _N \ge \int _{\Omega ^-_n}|\nabla v|^2 - \lambda |\Omega |\frac{1}{n^{q+1}} - C_2 \exp (4\alpha )|\Omega |\frac{1}{n^2} - |\Omega |\frac{1}{n^2}. \end{aligned}$$

(18)

Thus, (15) and (18) imply

$$\begin{aligned} \begin{array}{rcl} \displaystyle \left\langle F(\xi ),\xi \right\rangle &{}\ge &{} \displaystyle \Vert v\Vert ^2_{H^1_0(\Omega )} - \lambda C_0\Vert v\Vert ^{q+1}_{H^1_0(\Omega )} - C_1\Vert v\Vert ^{p}_{H^1_0(\Omega )}\left( \int _{\Omega }\exp (4\alpha (p+1)|v|^2){\hbox {d}}x\right) ^{\frac{1}{p+1}} \\ &{} - &{}\displaystyle \frac{C_3}{n}\Vert v\Vert _{H^1_0(\Omega )} - \lambda |\Omega |\frac{1}{n^{q+1}} - C_2 \exp (4\alpha )|\Omega |\frac{1}{n^2} - |\Omega |\frac{1}{n^2}. \end{array} \end{aligned}$$

(19)

Assume now that $\Vert v\Vert _{H^1_0(\Omega )}=r$ for some $r>0$ to be chosen later. We have

$$\begin{aligned} \int _{\Omega }\exp (4\alpha (p+1)|v|^2){\hbox {d}}x = \int _{\Omega }\exp \left( 4\alpha (p+1)r^2\left( \frac{v}{\Vert v\Vert _{H^1_0(\Omega )}}\right) ^2\right) {\hbox {d}}x \end{aligned}$$

(20)

and in order to apply the Trudinger–Moser inequality (5), we must have $4\alpha (p+1)r^2 \le 4\pi$. Consequently,

$$\begin{aligned} r\le \left( \frac{\pi }{\alpha (p+1)}\right) ^{\frac{1}{2}}. \end{aligned}$$

Then

$$\begin{aligned} \sup _{\Vert v\Vert _{H^1_0(\Omega )}\le 1}\int _{\Omega }\exp \left( 4\alpha (p+1)r^2\left( \frac{v}{\Vert v\Vert _{H^1_0(\Omega )}}\right) ^2\right) {\hbox {d}}x \le L. \end{aligned}$$

Hence,

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle \ge r^2 - \lambda C_0 r^{q+1} - C_1 L^{\frac{1}{p+1}}r^p {-} \frac{C_3}{n}r - \lambda |\Omega |\frac{1}{n^{q+1}} {-} C_2 \exp (4\alpha )|\Omega |\frac{1}{n^2} - |\Omega |\frac{1}{n^2}. \end{aligned}$$

We need to choose r such that

$$\begin{aligned} r^2 - C_1 L^{\frac{1}{p+1}}r^p\ge \frac{r^2}{2}; \end{aligned}$$

in other words,

$$\begin{aligned} r\le \frac{1}{(2 C_1 L^{\frac{1}{p+1}})^{\frac{1}{p-2}}}; \end{aligned}$$

thus, let $r=\min \left\{ \frac{1}{2(2C_1 L^{\frac{1}{p+1}})^{\frac{1}{p-2}}}, \left( \frac{\pi }{\alpha (p+1)}\right) ^{\frac{1}{2}}\right\}$, and hence

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle \ge \frac{r^2}{2} - \lambda C_0 r^{q+1} - \frac{C_3}{n}r - \lambda |\Omega |\frac{1}{n^{q+1}} - C_2 \exp (4\alpha )|\Omega |\frac{1}{n^2} - |\Omega |\frac{1}{n^2}. \end{aligned}$$

Now, defining $\rho =\frac{r^2}{2} - \lambda C_0 r^{q+1}$, we choose $\lambda ^*>0$ such that $\rho >0$ for $\lambda < \lambda ^*$. Therefore, we choose

$$\begin{aligned} \lambda ^*=\frac{r^{1-q}}{4C_0}. \end{aligned}$$

Now we choose $n^* \in \mathbb {N}$ such that

$$\begin{aligned} \frac{C_3}{n}r + \lambda |\Omega |\frac{1}{n^{q+1}} + C_2 \exp (4\alpha )|\Omega |\frac{1}{n^2} + |\Omega |\frac{1}{n^2} <\frac{\rho }{2}, \end{aligned}$$

for every $n\ge n^*$. Let $\xi \in \mathbb {R}^m$, such that $|\xi |=r$, then for $\lambda < \lambda ^*$ and $n\ge n^*$ we obtain

$$\begin{aligned} \left\langle F(\xi ),\xi \right\rangle \ge \frac{\rho }{2}>0. \end{aligned}$$

(21)

For every $n \in \mathbb {N}$, $f_n$ is a Lipschitz function, and then by Lemma 3.1 for every $m \in \mathbb {N}$ there exists $y \in \mathbb {R}^m$ with $|y|\le r$ such that $F(y)=0$, that is, there exists $v_m \in W_m$ verifying

$$\begin{aligned} \Vert v_m\Vert _{H^1_0(\Omega )}\le r \,\, \text{ for } \text{ every } \,\, m \in \mathbb {N} \end{aligned}$$

and such that

$$\begin{aligned} \int _{\Omega }\nabla v_m\nabla w = \lambda \int _{\Omega }(v_{m+})^{q}w + \int _{\Omega }f_n(v_{m+})w + \frac{1}{n}\int _{\Omega }w, \,\,\, \forall \, w \in W_m. \end{aligned}$$

(22)

Since $W_m \subset H^1_0(\Omega )$ $\forall \, m \in \mathbb {N}$ and r does not depend on m, then $(v_m)$ is a bounded sequence in $H^1_0(\Omega )$. Then, for some subsequence, there exists $v \in H^1_0(\Omega )$ such that

$$\begin{aligned} v_m \rightharpoonup v \,\,\, \text{ weakly } \text{ in } \,\,\, H^1_0(\Omega ) \end{aligned}$$

(23)

and

$$\begin{aligned} v_m \rightarrow v \,\,\, \text{ in } \,\,\, L^2(\Omega )\,\,\, \text{ and } \text{ a.e. } \text{ in }\,\,\,\Omega . \end{aligned}$$

(24)

Let $k \in \mathbb {N}$, then for every $m\ge k$ we obtain

$$\begin{aligned} \int _{\Omega }\nabla v_m\nabla w_k = \lambda \int _{\Omega }(v_{m+})^{q}w_k + \int _{\Omega }f_n(v_{m+})w_k + \frac{1}{n}\int _{\Omega }w_k, \,\,\, \forall \, w_k \in W_k. \end{aligned}$$

(25)

It follows from (23) that

$$\begin{aligned} \int _{\Omega }\nabla v_m\nabla w_k \rightarrow \int _{\Omega }\nabla v\nabla w_k \,\,\,\text{ as }\,\,\, m \rightarrow \infty \end{aligned}$$

(26)

and by (24) one obtains

$$\begin{aligned} \int _{\Omega }f_n(v_{m+})w_k \rightarrow \int _{\Omega }f_n(v_{+})w_k \,\,\,\text{ as }\,\,\, m \rightarrow \infty . \end{aligned}$$

(27)

Indeed, by Lemma 2.1 (ii) it follows that $|f_n(v_{m+}) - f_n(v_{+})|\le c_n|v_{m+}-v_{+}|$; hence,

$$\begin{aligned} \left| \int _{\Omega }f_n(v_{m+})w_k - \int _{\Omega }f_n(v_{+})w_k\right| \le c_n\Vert w_k\Vert _{L^2(\Omega )}\Vert v_{m}-v\Vert _{L^2(\Omega )} \,\,\,\text{ as }\,\,\, m \rightarrow \infty \end{aligned}$$

and then (24) implies (27). By (23), (27), and Sobolev compact imbedding, letting $m \rightarrow \infty$ one has

$$\begin{aligned} \lambda \int _{\Omega }(v_{m+})^{q}w_k + \int _{\Omega }f_n(v_{m+})w_k + \frac{1}{n}\int _{\Omega }w_k \rightarrow \lambda \int _{\Omega }(v_{+})^{q}w_k + \int _{\Omega }f_n(v_{+})w_k + \frac{1}{n}\int _{\Omega }w_k. \end{aligned}$$

(28)

By (25), (26), and (28)

$$\begin{aligned} \int _{\Omega }\nabla v\nabla w_k = \lambda \int _{\Omega }(v_{+})^{q}w_k + \int _{\Omega }f_n(v_{+})w_k + \frac{1}{n}\int _{\Omega }w_k, \,\,\, \forall \, w_k \in W_k. \end{aligned}$$

(29)

Since $[W_k]_{k \in \mathbb {N}}$ is dense in $H^1_0(\Omega )$, we conclude that

$$\begin{aligned} \int _{\Omega }\nabla v\nabla w = \lambda \int _{\Omega }(v_{+})^{q}w + \int _{\Omega }f_n(v_{+})w + \frac{1}{n}\int _{\Omega }w, \,\,\, \forall \, w \in H^1_0(\Omega ). \end{aligned}$$

(30)

Furthermore, $v\ge 0$ in $\Omega$. In fact, since $v_- \in H^1_0(\Omega )$, then from (30) we obtain

$$\begin{aligned} \int _{\Omega }\nabla v\nabla v_- = \lambda \int _{\Omega }(v_{+})^{q}v_- + \int _{\Omega }f_n(v_{+})v_- + \frac{1}{n}\int _{\Omega }v_-. \end{aligned}$$

Hence,

$$\begin{aligned} - \Vert v_-\Vert ^2_{H^1_0(\Omega )} = \int _{\Omega }\nabla v\nabla v_- = \int _{\Omega }f_n(v_{+})v_- + \frac{1}{n}\int _{\Omega }v_- \ge 0, \end{aligned}$$

then $v_- \equiv 0$ a.e. in $\Omega$. $\square$

4 Proof of the main result

In this section, we prove Theorem 1.1. We will use the unique solution $\widetilde{w}$ of the problem

$$\begin{aligned} \left\{ \begin{array}{lcc} -\Delta \widetilde{w} = \widetilde{w}^{q} &{}\hbox {in}&{} \Omega \\ \widetilde{w}>0 &{}\hbox {in}&{} \Omega \\ \widetilde{w}=0 &{}\hbox {on}&{} \partial \Omega , \end{array} \right. \end{aligned}$$

(31)

for $0<q<1$, see for instance [7]. The solution $\widetilde{w}$ allows us to bound from below the solutions $v_n$ of (11).

The following lemma of [20, Theorem1.1] is used to show that $v_n$ converges to a solution v of (1).

Lemma 4.1

Let $\Omega$ be a bounded open set in $\mathbb {R}^N$, $u_k : \Omega \rightarrow \mathbb {R}$ be a sequence function, and $g_k : \mathbb {R} \rightarrow \mathbb {R}$ be a sequence of functions such that $g_k(u_k)$ are measurable in $\Omega$ for every $k \in \mathbb {N}$. Assume that $g_k(u_k) \rightarrow v$ a.e. in $\Omega$ and $\int _{\Omega }|g_k(u_k)u_k|{\hbox {d}}x<C$ for a constant C independent of k. And suppose that for every $B \subset \mathbb {R}$, B bounded, there is a constant $C_B$ depending only on B such that $|g_k(x)|\le C_B$, for all $x \in B$ and $k \in \mathbb {N}$. Then $v \in L^1(\Omega )$ and $g_k(u_k) \rightarrow v$ in $L^1(\Omega )$.

Proof of Theorem 1.1

By Lemma 3.2, equation (11) has a weak solution $v_n \in H^1_0(\Omega )$ for each $n \in \mathbb {N}$. Since $0<q<1$ and $f_n$ is Lipschitz, then $\lambda v_n^{q} + f_n(v_n) + \frac{1}{n} \in L^p(\Omega )$ with $p > 2$. Hence, $v_n \in C^{1,\alpha }(\overline{\Omega })$ with $0<\alpha <1$, see [13]. Therefore, $v_n \in H^1_0(\Omega )\cap C^{1,\alpha }(\overline{\Omega })$.

We have by (23) that

$$\begin{aligned} v_m \rightharpoonup v_n \,\,\, \text{ weakly } \text{ in } \,\,\, H^1_0(\Omega ) \,\, \text{ as } \,\, m \rightarrow \infty . \end{aligned}$$

(32)

Therefore,

$$\begin{aligned} \Vert v_n\Vert _{H^1_0(\Omega )} \le \liminf _{m \rightarrow \infty }\Vert v_m\Vert _{H^1_0(\Omega )}\le r, \,\, \forall \, n \in \mathbb {N}, \end{aligned}$$

and r does not depend on n. Thus, there exists $v \in H^1_0(\Omega )$ such that

$$\begin{aligned} v_n \rightharpoonup v \,\,\, \text{ weakly } \text{ in } \,\,\, H^1_0(\Omega ) \,\, \text{ as } \,\, n \rightarrow \infty . \end{aligned}$$

(33)

By Sobolev compact imbedding for $1 \le s < +\infty$,

$$\begin{aligned} v_n \rightarrow v \,\,\text{ in }\,\,L^s(\Omega ) \,\, \text{ and } \,\, \text{ a.e. } \text{ in } \,\, \Omega . \end{aligned}$$

Note that

$$\begin{aligned} \left\{ \begin{array}{lcc} -\Delta v_n \ge \lambda v_n^{q} , &{}\hbox {in}&{} \Omega \\ v_n>0 &{}\hbox {in}&{} \Omega \\ v_n=0 &{}\hbox {on}&{} \partial \Omega . \end{array} \right. \end{aligned}$$

(34)

By rescaling, thus $w_n=\lambda ^{\frac{1}{q-1}}v_n$ and we obtain

$$\begin{aligned} -\Delta \left( \frac{w_n}{\lambda ^{\frac{1}{q-1}}}\right) \ge \lambda \left( \frac{w_n}{\lambda ^{\frac{1}{q-1}}}\right) ^q \end{aligned}$$

implying

$$\begin{aligned} -\Delta \,w_n\ge w_n^q. \end{aligned}$$

(35)

By Lemma 3.3 of [5], it follows that $w_n \ge \widetilde{w}$ $\forall \, n \in \mathbb {N}$, that is,

$$\begin{aligned} v_n \ge \lambda ^{\frac{1}{1-q}}\widetilde{w} \,\,\text{ a.e. } \text{ in }\,\,\Omega , \,\, \forall \, n \in \mathbb {N}. \end{aligned}$$

(36)

Letting $n \rightarrow +\infty$ in (36), we obtain

$$\begin{aligned} v \ge \lambda ^{\frac{1}{1-q}}\widetilde{w} \,\,\text{ a.e. } \text{ in }\,\,\Omega \end{aligned}$$

showing that $v>0$ in $\Omega$.

We prove now that v is a solution of (1). Since

$$\begin{aligned} v_n \rightarrow v\,\, \text{ a.e. } \text{ in }\,\, \Omega , \end{aligned}$$

we have

$$\begin{aligned} f_n(v_n(x)) \rightarrow f(v(x))\,\, \text{ a.e. } \text{ in }\,\, \Omega , \end{aligned}$$

(37)

by the uniform convergence of Lemma 2.1 (iii).

Recall from (30) that

$$\begin{aligned} \int _{\Omega }\nabla v_n\nabla w = \lambda \int _{\Omega }(v_n)^{q}w + \int _{\Omega }f_n(v_n)w + \frac{1}{n}\int _{\Omega }w, \,\,\, \forall \, w \in H^1_0(\Omega ). \end{aligned}$$

(38)

Taking $w=v_n$ in (38) and since $v_n$ is bounded in $H_0^1(\Omega )$, we obtain

$$\begin{aligned} \int _{\Omega }f_n(v_n)v_n{\hbox {d}}x \le C, \end{aligned}$$

(39)

for every $n \in \mathbb {N}$, where $C>0$ is a constant independent of n. By (37), (39), and by the expression of $f_n$ defined in (6), the assumptions of Lemma 4.1 are satisfied, implying

$$\begin{aligned} f_n(v_n) \rightarrow f(v)\,\, \text{ in }\,\, L^1(\Omega ). \end{aligned}$$

(40)

It follows from (4) that $e^{v^2} \in L^1(\Omega )$, and in view of (2) and Hölder inequality, we conclude that $f(v) \in L^2(\Omega )$.

By (38), we have

$$\begin{aligned} \int _{\Omega }\nabla v\nabla w = \lambda \int _{\Omega }v^{q}w + \int _{\Omega }f(v)w, \,\,\, \forall \, w \in H^1_0(\Omega ). \end{aligned}$$

(41)

Since $f(v) \in L^2(\Omega )$ and $\lambda \,v^{q} \in L^2(\Omega )$, we conclude from (41) that $v \in H^2(\Omega )$ and

$$\begin{aligned} -\Delta v= \lambda \,v^{q} + f(v). \end{aligned}$$

The proof of the theorem is complete. $\square$