An Elliptic Problem with Unbounded Coefficients and Two Singularities

Abstract

In this paper, we study the existence and regularity of solutions to quasilinear elliptic equations having quadratic growth with respect to the gradient and two singularities.

1 Introduction

In this paper, we prove the existence and regularity of nonnegative solutions to the following boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+\vert u\vert ^q]\nabla u\right) =g(u)|\nabla u|^2+h(u)f(x) &{} \text {in} &{} \Omega , \\ u=0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(1.1)

where $\Omega$ is an open and bounded set of of ${\mathbb {R}}^{N}$, with $N\ge 2$, $\gamma ,q>0$, f is a nonnegative integrable function, and $a: \Omega \rightarrow {\mathbb {R}}$ is a measurable function. Moreover, g and h are nonnegative functions which possibly satisfy $g(0)=\infty$ and/or $h(0)=\infty$.

The above problem (1.1) concerns quasilinear elliptic equations having a lower-order term with quadratic growth respecting the gradient. The interest for studying these types of problems lies in the calculation of variations (see, e.g. [1, 12, 17]). For example, if we consider the functional

$$\begin{aligned} J(v)=\frac{1}{2}\int _{\Omega }[a(x)+\vert v\vert ^{1-\theta }]\vert \nabla v\vert ^2-\int _{\Omega }f(x)v, \end{aligned}$$

(1.2)

the Euler-Lagrange equation associated to the functional J is

$$\begin{aligned} -\mathrm{{div}}\left( [a(x)+\vert v\vert ^{1-\theta }]\nabla v\right) +\frac{1-\theta }{2}\frac{\vert \nabla v\vert ^2}{\vert v\vert ^{\theta }}\text {sign}(v)=f. \end{aligned}$$

Let us now consider the following boundary value problem

$$\begin{aligned} \left\{ \begin{array}{cc} -2\Delta u+\frac{|\nabla u|^{2}}{ u}=f(x) &{} x\in \Omega , \\ u(x)=0 &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$

(1.3)

if we define $v= 2\frac{u}{\sqrt{\vert u\vert }}$, then the function v is a solution of the following semilinear problem

$$\begin{aligned} \left\{ \begin{array}{cc} -\Delta v=\frac{f(x)}{\vert v\vert } &{} x\in \Omega , \\ v(x)=0 &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$

(1.4)

which is singular on the right hand side. Let us note that, in the case of nonnegative f, in [7], the authors considered the elliptic semilinear problems whose simplest model is

$$\begin{aligned} \left\{ \begin{array}{cc} -\Delta u=\frac{f}{\vert u\vert ^{\gamma }} &{} x\in \Omega , \\ u=0 &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$

(1.5)

where $\gamma >0$. More precisely, they showed that the term $\frac{f}{\vert u\vert ^{\gamma }}$ has a regularizing effect on the solutions u. In another work [11], the author studied the existence of solutions to the following elliptic problem having degenerate coercivity:

$$\begin{aligned} \left\{ \begin{array}{cc} -\mathrm{{div}}\left( \frac{\nabla u}{(1+|u|)^p}\right) =\frac{f}{\vert u\vert ^{\gamma }} &{} x\in \Omega , \\ u=0 &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$

(1.6)

where $p,\gamma >0$.

Recently, for $q, \gamma > 0$, we study existence and regularity of positive solutions for unbounded elliptic problems whose model is (see [8] and [14]):

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}[(1+|u|^q)\nabla u]=\frac{f}{|u|^{\gamma }} &{} \text {in} &{} \Omega , \\ u=0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(1.7)

where $f\in L^{m}(\Omega )$, $m\ge 1$.

Several articles deal with the existence of solutions to elliptic problems with a singular right hand side (for example [2, 4, 5] and [9]), namely with the model problem

$$\begin{aligned} \left\{ \begin{array}{cc} -\text {div}\left( M(x)|\nabla u|^{p-2}\nabla u\right) =F(u)+\frac{f(x)}{u^{\gamma }} &{} x\in \Omega , \\ u(x)=0 &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$

(1.8)

under various hypotheses on $1<p<\infty$, $\gamma >0$, $M: \Omega \rightarrow {\mathbb {R}}$ and $F: {\mathbb {R}}\rightarrow {\mathbb {R}}$. Indeed, the existence of positive solutions for (1.8) has been shown in [2] for $p=2$, $M(x)=1$, $f=\lambda >0$ and $F(\xi )=\xi ^r$ where $0<r<(N+2)/(N-2)$. Again, under the assumptions $p=2$, $f=\lambda >0$ and $F(\xi )=\xi ^r$ where $r>0$, the existence of positive solutions of (1.8) is shown in [4] provided M is a bounded uniformly elliptic matrix and $\lambda$ small. Later, in [9], it is proved the uniqueness (and existence) of positive solution for (1.8) where $p>1$, $M(x)=1$, $F(\xi )=0$ and $f \ge 0$ in $\Omega$ (not identically zero). Recently, in [5], the authors have studied the existence and regularity of positive solutions to the problem (1.8) under the assumption that $p>1$, $0<\gamma <1$, $F(\xi )=-\xi \vert \xi \vert ^{r-1}$ with $r>0$, f is a positive function in $L^{1}(\Omega )$ and M is a continuous Lipschitz function and that there exist $0< \alpha < \beta$ such that $\alpha \le M(x) \le \beta$ a.e. $\Omega$.

Without forgetting that in [16], it has been proved the existence of a nonnegative weak solution to a class of singular elliptic problems with a general measure as source term whose simplest model is $-\Delta u=\frac{f(x)}{u^{\gamma }}+\mu$, where $\mu$ is a nonnegative bounded Radon measure on $\Omega$. In addition, several authors have studied the existence and regularity of p(x)-Laplacian Dirichlet problems (see [13, 18, 19]).

This work is motivated by the results of [15], where the author studied the existence and uniqueness of nonnegative solutions to the following singular elliptic problem

$$\begin{aligned} \left\{ \begin{array}{cc} -\Delta _{p}u =g(u)|\nabla u|^p+h(u)f(x) &{} x\in \Omega , \\ u(x)=0 &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$

(1.9)

where $\Omega$ is an open bounded subset of ${\mathbb {R}}^{N}$ ($N\ge 2$), $1< p <N$, $\Delta _{p}u=\text {div}\left( |\nabla u|^{p-2}\nabla u\right)$ is the p−Laplace operator, g and h are nonnegative functions which possibly satisfy $g(0)=\infty$ and/or $h(0)=\infty$, and the datum $f \in L^1(\Omega )$ is nonnegative.

It was natural to ask this question: can we study the existence of solutions to problems (1.9) if we consider that the main operator is not of Leray–Lions type? The purpose of this article is to answer this question in the case $p = 2$, more precisely we replaced in problems (1.9) the Laplace operator by an elliptic operator with unbounded coefficients. The main difficulties posed by this problem were that the principal part of the differential operator $\text {div}((a(x)+|u|^{q})\nabla u)$ is not well defined on the whole $H_{0}^{1}(\Omega )$, the solutions not belong, in general, to $H^{1}_{0}(\Omega )$ and the two terms on the right side have a singularity at $u = 0$.

2 Hypotheses and Main Results

2.1 Notation

Before stating more precisely our problem, we introduce some notation. We recall that $\Omega$ is a bounded open set of ${\mathbb {R}}^{N}$ with $N \ge 2$. For any $p > 1$, $p' =\frac{p}{p-1}$ will be the Hölder conjugate exponent of p, and if $1 \le p < N$, we will denote by $p^* =\frac{Np}{N-p}$ its Sobolev conjugate exponent of p. We will also denote $\vert E\vert$ the Lebesgue measure of a measurable set E in ${\mathbb {R}}^{N}$. As usual, let us denote by ${\mathcal {P}}$ the Poincare constant, i.e.

$$\begin{aligned} {\mathcal {P}}=\inf _{u\in H^{1}_{0}(\Omega )-\{0\}}\frac{\Vert \nabla u\Vert _{2}^{2}}{\Vert u\Vert _{2}^{2}}. \end{aligned}$$

For a fixed $k>0$, we recall the definition of a truncated function $T_k(s)$ defined by

$$\begin{aligned} T_k(s)=\max \lbrace \min \lbrace k,s\rbrace ,-k\rbrace . \end{aligned}$$

We also consider

$$\begin{aligned} G_k(s)=s-T_k(s). \end{aligned}$$

As usual, we define the positive and negative part of a measurable function u(x) by

$$\begin{aligned} u(x)=u^{+}(x)-u^{-}(x) \ \ \text {where} \ \ u^{+}(x)=u(x)\chi _{\lbrace u\ge 0\rbrace } \ \ \text {and} \ \ u^{-}(x)=-u(x)\chi _{\lbrace u<0\rbrace }. \end{aligned}$$

Moreover, we will also use the auxiliary functions defned by

$$\begin{aligned} V_{k,\delta } (s)= \left\{ \begin{array}{rl} 1 &{} \text {if } s \le k,\\ \frac{k+\delta -s}{\delta } &{} \text {if } k<s<k+\delta ,\\ 0 &{} \text {si } s \ge k+\delta , \end{array} \right. \end{aligned}$$

(2.1)

for every $k,\delta >0$, and

$$\begin{aligned} S_{\delta } (s)=1-V_{\delta ,\delta } (s). \end{aligned}$$

(2.2)

Throughout this paper, C will always denote a positive constant which only depends on the parameters of our problem; its value can change from line to line and, sometimes, on the same line.

2.2 Assumptions and Main Results

For an open, bounded set $\Omega \subset {\mathbb {R}}^{N}$ and $q>0$, we are interested in studying the following problem:

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+\vert u\vert ^q]\nabla u\right) =g(u)|\nabla u|^2+h(u)f(x) &{} \text {in} &{} \Omega , \\ u\ge 0 &{} \text {on} &{} \Omega ,\\ u=0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(2.3)

under the following hypotheses:

($H_1$):

f is a positive function in $L^1(\Omega )$ (that is $f(x) \ge 0$ and not zero a.e. $\Omega$).

($H_2$):

$a: \Omega \rightarrow {\mathbb {R}}$ is a measurable function satisfying the following conditions:

$$\begin{aligned} \alpha \le a(x)\le \beta , \end{aligned}$$

(2.4)

for almost every $x\in \Omega$, where $\alpha$ and $\beta$ are positive constant.

($H_3$):

$g : [0, \infty ) \longmapsto [0, \infty ]$ and $h : [0, \infty ) \longmapsto [0, \infty ]$ are continuous and possibly singular functions, finite outside the origin with $g(0) \ne 0$ and $h(0) \ne 0$ and such that

$$\begin{aligned} \begin{array}{ccccc} \exists c_1, \theta , s_1>0&\text {such that}&g(s)\le \frac{c_1}{s^{\theta }}&\text {if}&s\le s_1, \end{array} \end{aligned}$$

(2.5)

and

$$\begin{aligned} \begin{array}{ccccc} \exists c_2, \gamma , s_2>0&\text {such that}&h(s)\le \frac{c_2}{s^{\gamma }}&\text {if}&s\le s_2. \end{array} \end{aligned}$$

(2.6)

($H_4$):

We suppose that $\theta <1$, then we also assume that following growth relation between g and h is satisfied:

$$\begin{aligned} \limsup _{s\longrightarrow \infty } e^{\Gamma (s)}h(s)<\infty , \end{aligned}$$

(2.7)

such that the function $\Gamma$ is defined by

$$\begin{aligned} \Gamma (s)=\frac{1}{\alpha }\int _{0}^{s}g(t)dt. \end{aligned}$$

(2.8)

In the sequel, we will use the auxiliary function that is defined by, for $s\ge 0$

$$\begin{aligned} \Psi (s)=\int _{0}^{s}e^{\Gamma (t)}dt, \end{aligned}$$

(2.9)

Our main existence results are as follows.

Theorem 2.1

Let $q>0$. Assume that ($H_1$)- ($H_4$) holds true. If $\gamma \le 1$, then there exists a solution u of (2.3), strictly positive in $\Omega$, in the sense that $T_k(u) \in H_{0}^{1}(\Omega )$ for any $k > 0$,

$$\begin{aligned}{} & {} u^{q+1}, \Psi (u)\in W_{0}^{1,\sigma } (\Omega ) \quad \text {for every} \quad \sigma <\frac{N}{N-1}, \end{aligned}$$

(2.10)

$$\begin{aligned}{} & {} g(u)\vert \nabla u\vert ^2\varphi , h(u)f\varphi \in L^{1}(\Omega ), \end{aligned}$$

(2.11)

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u^q]\nabla u\nabla \varphi = \int _{\Omega }g(u)\vert \nabla u\vert ^2\varphi + \int _{\Omega }h(u)f\varphi \end{aligned}$$

(2.12)

for every $\varphi$ in $W_{0}^{1,p}(\Omega )\cap L^{\infty }(\Omega )$, $p>N$.

Remark 2.2

Note that, the regularity given on $\Psi (u)$ is also valid for u itself. Indeed, by definition of $\Psi$, one has that both $\Psi (u)\ge u$ and $|\nabla \Psi (u)| \ge |\nabla u|$ hold.

Before giving the second result, let us start by giving the definition of the weak solution of our problem:

Definition 2.3

We say that a positive and measurable function u is a weak solution to problem (2.3) if $T_k(G_{\epsilon }(u))\in H_{0}^{1}(\Omega )$ for any $k, \epsilon >0$, if $u^q\vert \nabla u\vert ,$ $g(u)\vert \nabla u\vert ^2, h(u)f\in L^{1}_{\text {loc}}(\Omega )$ and if

$$\begin{aligned} \int _{\Omega }(a(x)+ u^q)\nabla u\nabla \varphi =\int _{\Omega } g(u)|\nabla u|^2\varphi +\int _{\Omega }h(u)f(x)\varphi \end{aligned}$$

(2.13)

for every $\varphi \in C_{c}^{1}(\Omega )$

Theorem 2.4

Let $0<q<1/(N-1)$. Under the assumptions $(H_1)-(H_4).$ If $\gamma >1$, then there exists a weak solution u of (2.3) such that $\Psi (u)\in W_{\text {loc}}^{1,\sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}.$ Moreover, $u\in W^{r,1}_{\text {loc}}(\Omega )$ for every $r<\frac{N(q+1)}{N+q-1},$ and $T_k(u)\in H_{\text {loc}}^{1}(\Omega )$.

3 The Approximated Problem

Let us consider for $n\in {\mathbb {N}}^{*},$ the following approximating problems:

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+\vert u_n\vert ^q]\nabla u_n\right) =g_n(u_n)|\nabla u_n|^2+h_n(u_n)f_n &{} \text {in} &{} \Omega , \\ u_n=0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(3.1)

where $g_n(s)=T_n(g(s)),$ $h_n(s)=T_n(h(s))$ and $f_n=T_n(f).$

We are now going to prove the existence of a weak solution of problem (3.1) for every fixed $n\in {\mathbb {N}}^{*},$ and for this, we will need to give the following truncations of $\Gamma$, $\Psi$ defned in (2.8) and (2.9)

$$\begin{aligned} \Gamma _n(s)=\frac{1}{\alpha }\int _{0}^{s}g_n(t)dt. \end{aligned}$$

(3.2)

and

$$\begin{aligned} \Psi _n(s)=\int _{0}^{s}e^{\Gamma _n(t)}dt. \end{aligned}$$

(3.3)

Lemma 3.1

Let l be positive function belonging to $L^{\infty }(\Omega )$. Assume that the assumptions ($H_2$), ($H_3$) and ($H_4$) hold true. For $q>0$, then there exists a positive solution $u_n\in H^{1}_{0}(\Omega )\cap L^{\infty }(\Omega )$ of the problem

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+\vert u_n\vert ^q]\nabla u_n\right) =g_n(u_n)|\nabla u_n|^2+h_n(u_n)l(x) &{} \text {in} &{} \Omega , \\ u_n=0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(3.4)

Proof

This proof is based on standard Schauder’s fixed point argument. Let $v\in L^2(\Omega )$, and define $w = S(v)$ to be the unique solution of

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+\vert T_n(w)\vert ^{q}]\nabla w\right) =g_n(w)|\nabla w|^2+h_n(v)l(x) &{} \text {in} &{} \Omega , \\ w =0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(3.5)

From the results of [6], the operator S is well defined, i.e. for any nonnegative v belonging to $L^2(\Omega )$, there exists a nonnegative solution w to (3.5). Furthermore, w is bounded by the results of [20]. Let $c_n$ be a positive constant such that $\Vert w\Vert _{L^{\infty }(\Omega )}< c_n$. We take $e^{\Gamma _n(w)}w$ as a test function in (3.1), and we use the Hölder’s inequality and the formula (2.4) to get that

$$\begin{aligned} \alpha \int _{\Omega }|\nabla w|^{2}\le & {} \int _{\Omega }[a(x)+\vert T_n(w)\vert ^{q}]|\nabla w|^{2}e^{\Gamma _n(w)}\\= & {} \int _{\Omega }h_n(w)e^{\Gamma _n(w)}l(x),\\\le & {} ne^{\frac{nc_n}{\alpha }}\Vert l\Vert _{L^{\infty }(\Omega )}\sqrt{\vert \Omega \vert }\Vert w\Vert _{L^2(\Omega )}. \end{aligned}$$

We apply the Poincaré’s inequality in the previous formula, we have

$$\begin{aligned} \Vert w\Vert _{L^2(\Omega )}\le \frac{ne^{\frac{nc_n}{\alpha }}\Vert l\Vert _{L^{\infty }(\Omega )}\sqrt{\vert \Omega \vert }}{\alpha {\mathcal {P}}}=r_n, \end{aligned}$$

which implies that the ball of $L^2(\Omega )$ of radius $r_n$ is invariant for S. Moreover, from the $H^{1}_{0}(\Omega )\hookrightarrow L^2(\Omega )$ embedding, it is easy to prove that S is continuous and compact. Thus, the Schauder theorem shows that S has a fixed point or equivalently there exists a non negative solution $u_n \in H^{1}_{0}(\Omega )$ to problems

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+\vert T_n(u_n)\vert ^{q}]\nabla u_n\right) =g_n(u_n)|\nabla u_n|^2+h_n(u_n)l(x) &{} \text {in} &{} \Omega , \\ u_n =0 &{} \text {on} &{} \partial \Omega , \end{array} \right. \end{aligned}$$

(3.6)

Now, for $k> 0$, we take $G_k(u_n)e^{\Gamma _n(u_n)}$ as a test function in the weak formulation of (3.6) and use (2.4) and (2.7) to obtain

$$\begin{aligned} \alpha \int _{A_k}\vert \nabla G_k(u_n)) \vert ^2\le \sup _{s\ge k}[h(s)e^{\Gamma (s)}]\int _{A_k} G_k(u_n)l(x) \end{aligned}$$

where $A_k=\{x\in \Omega : |u_n|>k\}$. We use the method of Stampacchia ( see [20]) to deduce that the sequence $u_n$ is bounded in $L^{\infty }(\Omega )$. Supposing that $u_n$ is bounded by $d_n$ in $L^{\infty }(\Omega )$, we have that $u_n:=u_{n+[d_n]+1}\in L^{\infty }(\Omega )\cap H^{1}_{0}(\Omega )$ is a solution of (3.4). $\square$

In addition, use the results of lemma 3.1 to conclude that there is a solution $u_n\in L^{\infty }(\Omega )\cap H^{1}_{0}(\Omega )$ of problems (3.1).

The following lemma will be very useful, as it gives us an a priori estimate on the summability of $\Psi _n(u_n)$ in some Sobolev spaces, where $u_n$ is a solution of problem (3.1) and $\Psi _n$ is the function defined by (3.3). Note again that, since $\Psi _n(u_n) \ge u_n$ and $|\nabla \Psi _n(u_n)| \ge |\nabla u_n|$, one gets that all the next estimates still hold for $u_n$ in place of $\Psi _n(u_n)$.

Lemma 3.2

Let $q>0$. Suppose that the assumptions ($H_1$)-($H_4$) are satisfied. If $u_n$ be a solution to (3.1), then

• (i) If $0<\gamma \le 1$, then $\lbrace \Psi _n(u_n)\rbrace$ is uniformly bounded in $W_{0}^{1, \sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$. Furthermore, for any $k > 0$, the sequences $\lbrace T_k(\Psi _n(u_n))\rbrace$ and $\lbrace T_k(u_n)\rbrace$ are uniformly bounded in $H_{0}^{1}(\Omega )$. As a result, there exists a subsequence (not relabeled) and a function u such that $u_n$ almost everywhere converges to u, and $T_k(u_n)$ weakly converges to $T_k(u)$ in $H_{0}^{1}(\Omega )$.

• (ii) If $\gamma >1$, then $\lbrace \Psi _n(u_n)\rbrace$ is uniformly bounded in $W_{\text {loc}}^{1, \sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$. Moreover, for all $k > 0$, the sequences $\lbrace T_k(\Psi _n(u_n))\rbrace$ and $\lbrace T_k(u_n)\rbrace$ are uniformly bounded in $H_{\text {loc}}^{1}(\Omega )$. Finally, there exists a subsequence (not relabeled) and a function u such that $u_n$ almost everywhere converges to u, and $T_k(u_n)$ weakly converges to $T_k(u)$ in $H_{\text {loc}}^{1}(\Omega )$.

Proof

• (i) Let $0<\gamma \le 1$, by definition of the function $\Psi$, it is easy to show that $\Psi (s)s^{-\gamma }$ is finite in zero. Take $k>0$, the use of $e^{\Gamma _n(u_n)}T_k(\Psi _n(u_n))$ as test function in (3.1), (2.6) and (2.7) implies that

  $$\begin{aligned}{} & {} \int _{\lbrace \vert \Psi _n(u_n)\vert \le k\rbrace }[a(x)+ u_{n}^{q}]\nabla \Psi _n(u_n) \nabla u_n e^{\Gamma _n(u_n)} \\{} & {} \quad \le \int _{\Omega }h_n(u_n) f_n e^{\Gamma _n(u_n)}T_k(\Psi _n(u_n)), \\{} & {} \quad \le c_2\max _{s\in [0,s_2]}[s^{-\gamma }\Psi (s)e^{\Gamma (s)}]\Vert f\Vert _{L^1(\Omega )}+ k \sup _{s>s_2}[h(s)e^{\Gamma (s)}] \Vert f\Vert _{L^1(\Omega )}. \end{aligned}$$

  By (2.4), it follows:

  $$\begin{aligned} \int _{\Omega }\vert \nabla T_k(\Psi _n(u_n))\vert ^2\le C(k+1) \end{aligned}$$

  (3.7)

  Now, with a similar reasoning, we can choose $e^{\Gamma (u_n)} T_k(u_n)$ ($k > 0$) as the test function to get

  $$\begin{aligned} \int _{\Omega }\vert \nabla T_k(u_n)\vert ^2\le C(k+1). \end{aligned}$$

  (3.8)

  We apply Lemmas 4.1 and 4.2 of [3] to deduce that $\Psi _n(u_n)$ is bounded in $W_{0}^{1, \sigma }$ with $\sigma <\frac{N}{N-1}$ respect to n. Hence, there exists a function u such that, up to subsequences, $u_n \longrightarrow u$ a.e. in $\Omega$. Moreover, using (3.8), we get that, up to subsequences, $T_k(u_n) \longrightarrow T_k(u)$ weakly in $H_{0}^{1}(\Omega )$.

• (ii) Let $k, \epsilon >0$. Choosing $e^{\Gamma _n(u_n)}T_k(G_{\epsilon }(\Psi _n(u_n)))$ as test function in (3.1), using (2.4), (2.6) and (2.7), we obtain

  $$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u_{n}^{q}]\vert \nabla T_k(G_{\epsilon }(\Psi _n(u_n)))\vert ^2 \\{} & {} \quad \le \int _{\lbrace \epsilon<\Psi _n(u_n), u_n<s_2\rbrace }h_n(u_n) f_n e^{\Gamma _n(u_n)}T_k(G_{\epsilon }(\Psi _n(u_n))) \\{} & {} \qquad + \int _{\lbrace u_n>s_2 \rbrace } h_n(u_n) f_n e^{\Gamma _n(u_n)}T_k(G_{\epsilon }(\Psi _n(u_n))),\\{} & {} \quad \le c_2e^{(1+\gamma )s_2}\epsilon ^{1-\gamma }\Vert f\Vert _{L^1(\Omega )}+ k \sup _{s>s_2}[h(s)e^{\Gamma (s)}] \Vert f\Vert _{L^1(\Omega )}. \end{aligned}$$

  So that

  $$\begin{aligned} \int _{\Omega }\vert \nabla T_k(G_{\epsilon }(\Psi _n(u_n)))\vert ^2\le C_{\epsilon }(k+1) \end{aligned}$$

  (3.9)

  we also use Lemmas 4.1 and 4.2 of [3] to deduce that for any $\epsilon >0$, $G_{\epsilon }(\Psi _n(u_n))$ is bounded in $W_{0}^{1, \sigma }$ with $\sigma <\frac{N}{N-1}$ respect to n. It remains to analyse the behaviour of $\nabla \Psi _n(u_n)$ on $\lbrace \Psi _n(u_n) \le \epsilon \rbrace$. Let $\varphi$ be a function in $C^{1}_{c}(\Omega )$, and let $\omega$ be the support of $\varphi$. We take $(T_{\epsilon }(\Psi _n(u_n))-\epsilon )\varphi ^2$ as a test function in (3.1) and use (2.4) to get that

  $$\begin{aligned}{} & {} \alpha \int _{\Omega }e^{-\Gamma _n(u_n)} \vert \nabla T_{\epsilon }(\Psi _n(u_n)) \vert ^2\varphi ^2 \\{} & {} \quad +2\int _{\Omega } [a(x)+u_{n}^{q}]e^{-\Gamma _n(u_n)}\nabla T_{\epsilon }(\Psi _n(u_n))\varphi \nabla \varphi (T_{\epsilon }(\Psi _n(u_n))-\epsilon )\le 0. \end{aligned}$$

  We can use Young’s inequality with $\epsilon ^{'}$, to obtain

  $$\begin{aligned}{} & {} \alpha e^{-\Gamma (\epsilon )}\int _{\Omega } \vert \nabla T_{\epsilon }(\Psi _n(u_n)) \vert ^2\varphi ^2 \\{} & {} \quad \le 2\epsilon ^{'}\int _{\Omega } \vert \nabla T_{\epsilon }(\Psi _n(u_n)) \vert ^2\varphi ^2+ C(\omega ,\epsilon ^{'}, \epsilon )\int _{\Omega } \vert \nabla \varphi \vert ^2. \end{aligned}$$

  Letting $\epsilon ^{'}=\alpha e^{-\Gamma (\epsilon )}/4$, we get

  $$\begin{aligned} \int _{\Omega }\vert \nabla T_{\epsilon }(\Psi _n(u_n)) \vert ^2\varphi ^2\le C. \end{aligned}$$

  (3.10)

  This last estimate implies that $\lbrace u_n\rbrace$ is uniformly bounded in $W_{\text {loc}}^{1, \sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$. Furthermore, by a similar reasoning, we use $(T_{k}(u_n)-k)\varphi ^2$ as a test function in the weak formulation of (3.1), to show that $T_{\epsilon }(u_n)$ is uniformly bounded in $H_{\text {loc}}^{1}(\Omega )$. Consequently, there exists a function u such that, up to a subsequence, $u_n \longrightarrow u$ a.e. in $\Omega$, and up to subsequences, $T_k(u_n) \rightharpoonup T_k(u)$ weakly in $H_{\text {loc}}^{1}(\Omega )$.

$\square$

4 First Result

4.1 A Priori Estimates

Lemma 4.1

Under the assumptions of Theorem 2.1, let $\lbrace u_n\rbrace$ be the sequence of solutions to problems (3.1) and u the function given by Lemma 3.2, then the following holds:

• (1) The sequence $\lbrace u_{n}^{q+1}\rbrace$ is uniformly bounded in $W_{0}^{1,\sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$.

• (2) For any $k>0$, the sequence $\lbrace T_k(u_n)\rbrace$ strongly converges to $T_k(u)$ in $H_{0}^{1}(\Omega )$.

Proof

• (1) Take $k>0$, and choose $\varphi =T_k(u_{n}^{q+1})e^{\Gamma _n(u_n)}$ as a test function in the weak formulation of (3.1) to get that

  $$\begin{aligned} \int _{\Omega }u_{n}^{q}\nabla u_n \nabla T_k(u_{n}^{q+1})e^{\Gamma _n(u_n)}\le \int _{\Omega } h_n(u_n)f_n T_k(u_{n}^{q+1})e^{\Gamma _n(u_n)}. \end{aligned}$$

  Observe that $u_{n}^{q}\nabla u_n =\frac{1}{q+1}\nabla u_{n}^{q+1}$. Therefore, by hypothesis (2.7), we have

  $$\begin{aligned}{} & {} \frac{1}{q+1}\int _{\Omega } \vert \nabla T_k(u_{n}^{q+1})\vert ^2 \nonumber \\{} & {} \quad \le \int _{\lbrace u_n\le s_2\rbrace }h_n(u_n)f_n T_k(u_{n}^{q+1})e^{\Gamma _n(u_n)}+\int _{\lbrace u_n> s_2\rbrace }h_n(u_n)f_n T_k(u_{n}^{q+1})e^{\Gamma _n(u_n)} \nonumber \\{} & {} \quad \le c_2s_{2}^{q+1-\gamma }e^{\Gamma (s_2)}\Vert f\Vert _{L^{1}(\Omega )}+k\Vert f\Vert _{L^{1}(\Omega )}\sup _{s>s_2}h(s)e^{\Gamma (s)}\le C(1+k). \end{aligned}$$

  (4.1)

  Hence, the previous estimate allows to deduce that $u_{n}^{q+1}$ is uniformly bounded in $W_{0}^{1,\sigma }(\Omega )$ with $\sigma <\frac{N}{N-1}$ (see [3]), and Fatou Lemma implies $u^{q+1}$ belongs to $W_{0}^{1,\sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$.

• (2) we consider the following function

  $$\begin{aligned} w_{n,k,h}=T_{2k}(u_n-T_h(u_n)+T_k(u_n)-T_k(u)), \ \ 0<k<h, \end{aligned}$$

  (4.2)

  and we take $w_{n,k,h}^{+}e^{\Gamma _n(u_n)}$ as a test function in the weak formulation of (3.1), we obtain

  $$\begin{aligned} \int _{\Omega }[a(x)+u_{n}^{q}]\nabla u_n \nabla w_{n,k,h}^{+}e^{\Gamma _n(u_n)}\le \int _{\Omega } h_n(u_n)f_n e^{\Gamma _n(u_n)}w_{n,k,h}^{+}. \end{aligned}$$

  (4.3)

  Now, for the right hand side, we take $\delta < k$ and we can write

  $$\begin{aligned} \int _{\Omega } h_n(u_n)f_n e^{\Gamma _n(u_n)}w_{n,k,h}^{+}= & {} \int _{\lbrace u_n\le \delta \rbrace } h_n(u_n)f_ne^{\Gamma _n(u_n)} (u_n-T_k(u))^+\nonumber \\{} & {} +\int _{\lbrace u_n> \delta \rbrace } h_n(u_n)f_ne^{\Gamma _n(u_n)} w_{n,k,h}^{+}. \end{aligned}$$

  (4.4)

  In addition, since $\gamma \le 1$, we have

  $$\begin{aligned} h_n(u_n)f_ne^{\Gamma _n(u_n)} (u_n-T_k(u))^+\chi _{\lbrace u_n\le \delta \rbrace }\le \max _{s\in [0,\delta ]}[h(s)e^{\Gamma (s)}s]f\in L^1(\Omega ) \end{aligned}$$

  Observe that $(u_n-T_k(u))^+\chi _{\lbrace u_n\le \delta \rbrace }$ converges almost everywhere to zero as $n \longrightarrow \infty$. Using the Lebesgue dominated convergence theorem, we deduce that

  $$\begin{aligned} \lim _{n\longrightarrow \infty } \int _{\lbrace u_n\le \delta \rbrace } h_n(u_n)f_ne^{\Gamma _n(u_n)} (u_n-T_k(u))^+=0. \end{aligned}$$

  (4.5)

  On the other hand, since

  $$\begin{aligned} h_n(u_n)f_ne^{\Gamma _n(u_n)} w_{n,k,h}^{+}\chi _{\lbrace u_n> \delta \rbrace }\le \sup _{s>\delta }[h(s)e^{\Gamma (s)}]fw_{n,k,h}^{+}, \end{aligned}$$

  we derive that

  $$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\lbrace u_n> \delta \rbrace } h_n(u_n)f_ne^{\Gamma _n(u_n)} w_{n,k,h}^{+}\le C\int _{\Omega }f T_{2k}(u-T_h(u))^{+}, \end{aligned}$$

  and taking $h\longrightarrow \infty$ in the previous one has

  $$\begin{aligned} \lim _{h\longrightarrow \infty }\lim _{n\longrightarrow \infty } \int _{\lbrace u_n> \delta \rbrace } h_n(u_n)f_ne^{\Gamma _n(u_n)} w_{n,k,h}^{+}=0. \end{aligned}$$

  (4.6)

  Thus, thanks to both (4.5) and (4.6), we have

  $$\begin{aligned} \lim _{h\longrightarrow \infty }\lim _{n\longrightarrow \infty } \int _{\Omega } h_n(u_n)f_ne^{\Gamma _n(u_n)} w_{n,k,h}^{+}=0. \end{aligned}$$

  (4.7)

  For the left hand side of (4.3), we remark that $\nabla w_{n,k,h}=0,$ if $u_n\ge h+2k={\mathcal {K}}$. Thus

  $$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u_{n}^{q}]\nabla w_{n,k,h}^{+} \nabla u_n e^{\Gamma _n(u_n)} \nonumber \\{} & {} \quad \ge \int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla u_n e^{\Gamma _n(u_n)}\nonumber \\{} & {} \qquad - \int _{\lbrace k<u_n<h+2k \rbrace } [a(x)+u_{n}^{q}]\nabla T_k(u) \nabla T_{{\mathcal {K}}}(u_n) e^{\Gamma _n(u_n)}, \end{aligned}$$

  (4.8)

  Using the weak convergence of $\nabla T_k(u_n)$ in $H_{0}^{1}(\Omega )$ for any $k >0$, we deduce that

  $$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\lbrace k<u_n<h+2k \rbrace } [a(x)+u_{n}^{q}]\nabla T_k(u) \nabla T_{{\mathcal {K}}}(u_n) e^{\Gamma _n(u_n)}=0. \end{aligned}$$

  (4.9)

  Passing to the limit in (4.3) first with respect to n and after as h tends to infinity, we deduce from (4.7), (4.8) and (4.9) that

  $$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla u_n e^{\Gamma _n(u_n)}=0. \end{aligned}$$

  In addition, it is easy to notice that

  $$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla T_k(u) e^{\Gamma _n(u_n)}=0, \end{aligned}$$

  which implies that

  $$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\Omega }\vert \nabla (T_k(u_n)-T_k(u))^{+}\vert ^2=0. \end{aligned}$$

  (4.10)

  Similarly, using $w_{n,k,h}^{-}e^{\Gamma _n(u_n)}$, it is possible to show that

  $$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\Omega }\vert \nabla (T_k(u_n)-T_k(u))^{-}\vert ^2=0, \end{aligned}$$

  (4.11)

  which implies the strong convergence of $\lbrace T_k(u_n)\rbrace$ in $H_{0}^{1}(\Omega )$. Moreover, we can also deduce that, up to a subsequence, $\nabla T_k(u_n) \longrightarrow \nabla T_k(u)$ a.e. in $\Omega$. This in turn implies that, up to a subsequence,

  $$\begin{aligned} \nabla u_n \longrightarrow \nabla u \ \ \ \hbox { a.e. in}\ \Omega \end{aligned}$$

$\square$

4.2 Passing to the Limit

Our aim is to show that the weak limit u given by Lemma 3.2 is a solution of the problem (2.3). We recall that by Lemma 3.2, we have $\Psi (u)\in W_{0}^{1,\sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$ and $T_k(u) \in H_{0}^{1}(\Omega )$ for any $k > 0$. Moreover, $u^{q+1}$ belongs to $W_{0}^{1,\sigma }(\Omega )$ for every $\sigma <\frac{N}{N-1}$ by Lemma 4.1. Before passing to the limit in (3.1), we will show first that $g(u)\vert \nabla u\vert ^2\varphi$ and $h(u)f\varphi$ both belong to $L^{1}(\Omega )$ for every $\varphi \in W_{0}^{1,p}(\Omega )\cap L^{\infty }(\Omega )$ with $p>N$. To prove the claim, we choose $\varphi$ as a test function in the weak formulation of (3.1) and we obtain, by (2.4), that

$$\begin{aligned} \int _{\Omega }(g_n(u_n)\vert \nabla u_n\vert ^2+h_n(u_n)f_n)\varphi \le \beta \int _{\Omega }\vert \nabla u_n\vert \vert \nabla \varphi \vert +\int _{\Omega }u_{n}^{q}\vert \nabla u_n\vert \vert \nabla \varphi \vert . \end{aligned}$$

By Young’s inequality with exponent p, we find

$$\begin{aligned}{} & {} \int _{\Omega }(g_n(u_n)\vert \nabla u_n\vert ^2+h_n(u_n)f_n)\varphi \nonumber \\{} & {} \quad \le \frac{\beta }{p^{'}}\int _{\Omega }\vert \nabla u_n\vert ^{p^{'}}+\frac{\beta }{p}\int _{\Omega }\vert \nabla \varphi \vert ^{p}+\frac{1}{p^{'}(q+1)} \int _{\Omega }\vert \nabla u_{n}^{q+1}\vert ^{p^{'}}+\frac{1}{p(q+1)} \int _{\Omega }\vert \nabla \varphi \vert ^{p}.\nonumber \\ \end{aligned}$$

(4.12)

Observe that $p^{'}<\frac{N}{N-1}$. Thanks to lemma 3.2 and 4.1, we have the boundedness $(g_n(u_n)\vert \nabla u_n\vert ^2+h_n(u_n)f_n)\varphi$ in $L^1(\Omega )$, which implies by the Fatou Lemma that $(g(u)\vert \nabla u\vert ^2+h(u)f)\varphi$ belongs to $L^1(\Omega )$. Indeed, if $\varphi$ is nonnegative, it is easy to demonstrate the previous result. In the other case, we use the decomposition $\varphi =\varphi ^+-\varphi ^-$.

Now, we pass to the limit in (3.1). For this reason, let us take $V_{\delta ,\delta }(u_n)\varphi$ as test function in (3.1), where $V_{\delta ,\delta }$ is defined in (2.1) and $\varphi$ be a nonnegative function in $W_{0}^{1,p}(\Omega )\cap L^{\infty }(\Omega )$, we obtain

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u_{n}^{q}]\nabla u_n \nabla \varphi V_{\delta ,\delta }(u_n)-\frac{1}{\delta } \int _{\lbrace \delta<u_n<2\delta \rbrace } [a(x)+u_{n}^{q}]\vert \nabla u_n\vert ^2\varphi \nonumber \\{} & {} \quad =\int _{\Omega } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi . \end{aligned}$$

(4.13)

Recalling that $V_{\delta ,\delta }$ is bounded, and using the weak convergence of $u_n$ and $u_{n}^{q+1}$ in $W_{0}^{1,p^{'}}(\Omega )$ (see Lemma 3.2 and Lemma 4.1), to deduce that

$$\begin{aligned}{} & {} \lim _{n\longrightarrow \infty }\int _{\Omega }[a(x)+u_{n}^{q}]\nabla u_n \nabla \varphi V_{\delta ,\delta }(u_n) \\{} & {} \quad = \lim _{n\longrightarrow \infty }\int _{\Omega }a(x)\nabla u_n \nabla \varphi V_{\delta ,\delta }(u_n) +\frac{1}{q+1}\lim _{n\longrightarrow \infty }\int _{\Omega }\nabla u_{n}^{q+1} \nabla \varphi V_{\delta ,\delta }(u_n)\\{} & {} \quad = \int _{\Omega }a(x)\nabla u \nabla \varphi V_{\delta ,\delta }(u) +\frac{1}{q+1}\int _{\Omega }\nabla u^{q+1}\nabla \varphi V_{\delta ,\delta }(u)\\{} & {} \quad = \int _{\Omega }[a(x)+u^{q}]\nabla u \nabla \varphi V_{\delta ,\delta }(u). \end{aligned}$$

In addition, using again that $V_{\delta ,\delta }$ is bounded, and we also have that both u and $u^{q+1}$ belong to $L^1(\Omega )$, then

$$\begin{aligned} \lim _{\delta \longrightarrow \infty }\lim _{n\longrightarrow \infty }\int _{\Omega }[a(x)+u_{n}^{q}]\nabla u_n \nabla \varphi V_{\delta ,\delta }(u_n)=\int _{\Omega }[a(x)+u^{q}]\nabla u \nabla \varphi . \end{aligned}$$

(4.14)

For the second integral of the left hand side of (4.13), we take $\varphi = e^{\Gamma _n(u_n)}S_{\delta }(u_n)$ where $\delta >0$ and $S_{\delta }$ is the function given by (2.2), as test function in the weak formulation of (3.1), we get

$$\begin{aligned} \frac{1}{\delta }\int _{\lbrace \delta< u_n<2\delta \rbrace } [a(x)+u_{n}^{q}]\vert \nabla u_n\vert ^2e^{\Gamma _n(u_n)}\le \int _{\Omega }h_n(u_n)f_ne^{\Gamma _n(u_n)}S_{\delta }(u_n) \end{aligned}$$

Passing to the limit in the previous inequality as n tends to infinity, and using Fatou Lemma on the integral of the left hand side, we have

$$\begin{aligned} \frac{1}{\delta }\int _{\lbrace \delta< u<2\delta \rbrace } [a(x)+u^{q}]\vert \nabla u\vert ^2e^{\Gamma (u)}\le \int _{\Omega }h(u)fe^{\Gamma (u)}S_{\delta }(u). \end{aligned}$$

Note that, by construction, we have $S_{\delta }(u) \longrightarrow 0$ as $\delta \longrightarrow \infty$. Hence, using Fatou lemma and the assumption (2.7), we conclude

$$\begin{aligned} \lim _{\delta \longrightarrow \infty } \frac{1}{\delta }\int _{\lbrace \delta< u<2\delta \rbrace } [a(x)+u^{q}]\vert \nabla u\vert ^2e^{\Gamma (u)}=0. \end{aligned}$$

(4.15)

Moreover, for the right hand side of (4.13), let $t > 0$ such that

$t \notin \left\{ \eta : \vert \lbrace u = \eta \rbrace \vert > 0\right\}$, and we write

$$\begin{aligned}{} & {} \int _{\Omega } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi \\{} & {} \quad = \int _{\lbrace u_n\le t \rbrace } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi \\{} & {} \qquad +\int _{\lbrace u_n> t \rbrace } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi . \end{aligned}$$

First, since $T_k(u_n)\longrightarrow T_k(u)$ strongly in $H_{0}^{1}(\Omega )$ for all $k>0$, we show by a similar argument to the one used in the proof of Theorem 2.5 of [15], that

$$\begin{aligned}{} & {} \lim _{t\longrightarrow 0}\lim _{n\longrightarrow \infty } \int _{\lbrace u_n> t \rbrace } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi \nonumber \\{} & {} \quad =\int _{\Omega } \left( g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace }+ h(u)f\right) V_{\delta ,\delta }(u)\varphi . \end{aligned}$$

(4.16)

Secondly, we choose $V_{\delta ,\delta }(u_n)V_{t,t}(u_n)\varphi$ where $0\le \varphi \in W_{0}^{1,p}(\Omega )\cap L^{\infty }(\Omega )$

$(p>N)$, as test function in (3.1), we obtain

$$\begin{aligned}{} & {} \int _{\lbrace u_n\le t \rbrace } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi \\{} & {} \quad \le \int _{\Omega } [a(x)+u_{n}^{q}]\nabla u_n\nabla \varphi V_{\delta ,\delta }(u_n) V_{t,t}(u_n). \end{aligned}$$

Observe that $0\le V_{\delta ,\delta }\le 1$ for all $\delta >0$. Using (2.4) and Young’s inequality, it follows that

$$\begin{aligned}{} & {} \int _{\lbrace u_n\le t \rbrace } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi \\{} & {} \quad \le \frac{\beta }{p^{'}}\int _{\Omega }\vert \nabla u_n\vert ^{p^{'}}V_{t,t}(u_n)+\frac{\beta }{p}\int _{\Omega }\vert \nabla \varphi \vert ^{p}V_{t,t}(u_n)+\frac{1}{p^{'}(q+1)} \int _{\Omega }\vert \nabla u_{n}^{q+1}\vert ^{p^{'}}V_{t,t}(u_n) \\{} & {} \qquad +\frac{1}{p(q+1)}\int _{\Omega }\vert \nabla \varphi \vert ^{p}V_{t,t}(u_n). \end{aligned}$$

Passing to the limit in the previous one first with respect to n and after when t tends to 0, we find

$$\begin{aligned}{} & {} \lim _{t\longrightarrow 0}\limsup _{n\longrightarrow \infty }\int _{\lbrace u_n\le t \rbrace } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) V_{\delta ,\delta }(u_n)\varphi \nonumber \\{} & {} \quad \le \frac{\beta }{p^{'}}\int _{\lbrace u=0\rbrace }\vert \nabla u\vert ^{p^{'}}+\frac{\beta }{p}\int _{\lbrace u=0\rbrace }\vert \nabla \varphi \vert ^{p}+\frac{1}{p^{'}(q+1)} \int _{\lbrace u=0\rbrace }\vert \nabla u^{q+1}\vert ^{p^{'}}\nonumber \\{} & {} \qquad +\frac{1}{p(q+1)}\int _{\lbrace u=0\rbrace }\vert \nabla \varphi \vert ^{p}=0. \end{aligned}$$

(4.17)

So, using (4.16) and (4.17), we deduce

$$\begin{aligned}{} & {} \lim _{n\longrightarrow \infty } \int _{\Omega } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) \varphi V_{\delta ,\delta }(u_n) \\{} & {} \quad =\int _{\Omega } \left( g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace }+ h(u)f\right) \varphi V_{\delta ,\delta }(u). \end{aligned}$$

Moreover, since $[g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace }+ h(u)f]\varphi$ belongs to $L^1(\Omega )$, we have

$$\begin{aligned}{} & {} \lim _{\delta \longrightarrow \infty }\lim _{n\longrightarrow \infty } \int _{\Omega } \left( g_n(u_n)\vert \nabla u_n\vert ^2+ h_n(u_n)f_n \right) \varphi V_{\delta ,\delta }(u_n) \nonumber \\{} & {} \quad =\int _{\Omega } \left( g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace }+ h(u)f\right) \varphi . \end{aligned}$$

(4.18)

Taking into account (4.14), (4.15) and (4.18), and passing to the limit in (4.13) first with respect to n and after as $\delta$ tends to $+\infty$, we get

$$\begin{aligned} \int _{\Omega }[a(x)+u^{q}]\nabla u \nabla \varphi = \int _{\Omega } g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace }\varphi +\int _{\Omega } h(u)f\varphi , \end{aligned}$$

(4.19)

for every $\varphi$ in $W_{0}^{1,p}(\Omega )\cap L^{\infty }(\Omega )$ with $p>N$.

Note that, we have to show that $u>0$ almost everywhere in $\Omega$ which means, along with the previous formulation, that (2.12) holds. To prove this claim, because of density, we now take $\varphi$ a nonnegative function belongs to $C_{0}^{1}(\Omega )$ and choose $V_{k,\delta }(u)\varphi$ ($k,\delta >0$) as test function in (4.19), we get

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u^{q}]\nabla u \nabla \varphi V_{k,\delta }(u)-\frac{1}{\delta } \int _{\lbrace k<u<k+\delta \rbrace } [a(x)+u^{q}]\vert \nabla u\vert ^2\varphi \\{} & {} \quad =\int _{\Omega } g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace } V_{k,\delta }(u)\varphi +\int _{\Omega } h(u)f V_{k,\delta }(u)\varphi . \end{aligned}$$

Dropping the second term of the previous formula because it is positive and we can pass to the limit as $\delta \longrightarrow 0$, using Fatou lemma and Lebesgue theorem, one obtains

$$\begin{aligned} \int _{\lbrace u\le k\rbrace }[a(x)+T_k(u)^{q}]\nabla u \nabla \varphi \ge \int _{\lbrace u\le k\rbrace } g(u)\vert \nabla u\vert ^2\chi _{\lbrace u>0\rbrace } \varphi +\int _{\lbrace u\le k\rbrace } h(u)f \varphi . \end{aligned}$$

(4.20)

Furthermore, since

$$\begin{aligned} \alpha \le a(x)+T_k(u)^{q}\le \beta +k^{q}, \end{aligned}$$

and $-\text {div}[a(x)+T_k(u)^{q}]\ge 0$ and not identically zero, then the strong maximum principle (see Theorem 1.2 of [21]) implies that $u>0$ almost everywhere in $\Omega$. This completes the proof.

5 Second Result

5.1 A Priori Estimates

Here, we are going to prove that the sequence $\lbrace u_n\rbrace$ is not 0 in $\Omega$. For this, we are going to show that it is uniformly away from zero in every compact set in $\Omega$.

Lemma 5.1

Under the same assumptions of Lemma 3.2. If $u_n$ is the solution of problem (3.1), then for every $\omega \subset \subset \Omega$, there exists a constant $c_{\omega } > 0$ such that

$$\begin{aligned} u_n\ge c_{\omega } > 0 \ \ \text {a.e. in} \ \ \omega . \end{aligned}$$

Proof

Note that, according to reference [10], there is a function ${\tilde{h}}$ everywhere finite, nonincreasing and such that ${\tilde{h}}(s) \le h_n(s)$ for all $s \ge 0$. Let us observe that it follows from the previous Lemma that there exists a nonnegative solution $v_n\in L^{\infty }(\Omega )\cap H^{1}_{0}(\Omega )$ to the following problems

$$\begin{aligned} \left\{ \begin{array}{ccc} -\mathrm{{div}}\left( [a(x)+ v_{n}^{q}]\nabla v_n\right) ={\tilde{h}}(v_n)f_n &{} \text {in} &{} \Omega , \\ v_n =0 &{} \text {on} &{} \partial \Omega . \end{array} \right. \end{aligned}$$

(5.1)

Choosing $(v_n-v_{n+1})^+$ as test function in (5.1), recalling that ${\tilde{h}}$ is nonincreasing, we have

$$\begin{aligned} \int _{\Omega }[a(x)+ v_{n}^{q}]\nabla v_n\nabla (v_n-v_{n+1})^+= & {} \int _{\Omega }{\tilde{h}}(v_n)f_n [v_n-v_{n+1}]^+,\\\le & {} \int _{\Omega }{\tilde{h}}(v_{n+1})f_n [v_n-v_{n+1}]^+. \end{aligned}$$

Since $0 \le f_n \le f_{n+1}$, one has

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+ v_{n}^{q}]\nabla v_n\nabla (v_n-v_{n+1})^+\le \int _{\Omega }{\tilde{h}}(v_{n+1})f_{n+1} [v_n-v_{n+1}]^+\\{} & {} \quad =\int _{\Omega }[a(x)+ v_{n+1}^{q}]\nabla v_{n+1}\nabla (v_n-v_{n+1})^+,\\{} & {} \quad \le \int _{\Omega }[a(x)+ v_{n}^{q}]\nabla v_{n+1}\nabla (v_n-v_{n+1})^+. \end{aligned}$$

So, by (2.4), we deduce that

$$\begin{aligned} \alpha \int _{\Omega }\vert \nabla (v_n-v_{n+1})^+\vert ^2\le & {} \int _{\Omega }[a(x)+ v_{n}^{q}]\vert \nabla (v_n-v_{n+1})^+\vert ^2\le 0. \end{aligned}$$

Consequently, we have $\int _{\Omega }\vert \nabla (v_n-v_{n+1})^+\vert ^2=0$, and so, using Poincare’s inequality, we get $(v_n-v_{n+1})^+=0$ a.e. $x\in \Omega$. This implies that $v_n\le v_{n+1}$ a.e. $x\in \Omega$.

We recall that $v_1$ is bounded, indeed, $v_1 \le c$, for some positive constant c. Then, it follows that

$$\begin{aligned} -\mathrm{{div}}\left( [a(x)+v_{1}^{q}]\nabla v_1\right) \ge f_1 {\tilde{h}}(c) \ \ x\in \Omega . \end{aligned}$$

Thanks to (2.4), we have $\alpha \le a(x)+v_{1}^{q}\le \beta +c^q$. Thus, we infer that $v_{1}$ is a supersolution of a linear Dirichlet problem with a strictly positive and bounded, measurable coefficient. The strong maximum principle implies that $v_1>0$ a.e. $x\in \Omega$. Moreover, Harnack’s inequality gives the stronger conclusion: for every $\omega \subset \subset \Omega$ there exists $c_{\omega }$ such that $v_1 \ge c_{\omega }$ a.e. in $\omega$. Therefore, using that the sequence $\lbrace v_n\rbrace$ is increasing, one deduces that

$$\begin{aligned} v_n \ge c_{\omega } \ \ \text {a.e. in} \ \ \omega \ \ \text {for every} \ \ n \in {\mathbb {N}}^{*}. \end{aligned}$$

(5.2)

Now, let us take $(v_n-u_{n})^+$ as a test function in the weak formulation of (5.1). Thus, recalling that the function ${\tilde{h}}$ is decreasing and verifies the following inequality ${\tilde{h}}\le h_n$, we deduce

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+ v_{n}^{q}]\nabla v_n\nabla (v_n-u_{n})^+ \\{} & {} \quad =\int _{\Omega }{\tilde{h}}(v_n)f_n [v_n-u_{n}]^+\le \int _{\Omega }{\tilde{h}}(u_{n})f_n [v_n-u_{n}]^+\le \int _{\Omega }h_n(u_{n})f_n [v_n-u_{n}]^+\\{} & {} \quad =\int _{\Omega }[a(x)+ u_{n}^{q}]\nabla u_n\nabla (v_n-u_{n})^{+}-\int _{\Omega }g_n(u_n)|\nabla u_n|^2(v_n-u_{n})^{+},\\{} & {} \quad \le \int _{\Omega }[a(x)+ v_{n}^{q}]\nabla u_n\nabla (v_n-u_{n})^{+}-\int _{\Omega }g_n(u_n)|\nabla u_n|^2(v_n-u_{n})^{+}. \end{aligned}$$

Therefore, taking into account $g_n\ge 0$, we obtain

$$\begin{aligned}{} & {} \alpha \int _{\Omega }\vert \nabla (v_n-u_n)^+\vert ^2 \le \int _{\Omega }[a(x)+ v_{n}^{q}]\vert \nabla (v_n-u_n)^+\vert ^2, \\{} & {} \quad \le -\int _{\Omega }g_n(u_n)|\nabla u_n|^2(v_n-u_{n})^{+} \le 0. \end{aligned}$$

Hence, $(v_n-u_n)^+ = 0$ a.e. in $\Omega$, which implies that $v_n\le u_n$. Finally, from (5.2), we conclude that $u_n \ge v_n>c_{\omega }$ a.e. in $\omega$ for every $n \in {\mathbb {N}}^{*}$. $\square$

Lemma 5.2

Under the hypotheses of theorem 2.4, if $u_n$ is a solution to problem (3.1), then the sequence $\lbrace u_{n}^{q}|\nabla u_n|\rbrace$ is uniformly bounded in $L^{\sigma }_{\text {loc}}(\Omega ),$ for every $\sigma <\frac{N}{N-1}.$ Moreover, the sequence $\lbrace u_n\rbrace$ is uniformly bounded in $W^{1, r}_{\text {loc}}(\Omega ),$ for every $r<\frac{N(q+1)}{N+q-1}.$

Proof

The proof is divided into three steps:

Step 1. We want to prove that for every $\lambda >1,$ $(1+u_{n})^{q-\lambda }\vert \nabla u_n\vert ^2\in L^{1}_{\text {loc}}(\Omega ).$ Indeed, let $0\le \varphi \in C^{1}_{c}(\Omega )$ and $\omega = \text {Supp}\varphi$ is the support of $\varphi$. Let us take $\left[ 1-\frac{1}{(1 +u_{n})^{\lambda -1}}\right] e^{\Gamma _n(u_n)}\varphi ^2$, with $\lambda > 1$, as a test function in the weak formulation of (3.1), using (2.4), (2.7) and Lemma 5.1, we obtain

$$\begin{aligned}{} & {} (\lambda -1)\int _{\Omega }e^{\Gamma _n(u_n)}\frac{|\nabla u_{n}|^{2}}{ (1+u_{n})^{\lambda -q }}\varphi ^2+2\int _{\Omega }[a(x)+u_{n}^{q}][1-(u_{n}+1)^{1-\lambda }]\nabla \Psi _n(u_{n})\nabla \varphi \varphi \nonumber \\{} & {} \quad \le \sup _{s>c_{\omega }}[h(s)e^{\Gamma (s)}] \Vert f\Vert _{L^1(\Omega )}\Vert \varphi ^2\Vert _{L^{\infty }(\Omega )}. \end{aligned}$$

(5.3)

Note that, since $0<q<\frac{1}{N-1}$, we have $q+1<\frac{N}{N-1}$. Let us choose $\sigma > 0$ such that

$$\begin{aligned} q+1<\sigma <\frac{N}{N-1}. \end{aligned}$$

(5.4)

We observe that the choice of $\sigma$ is equivalent to require

$$\begin{aligned} qN<q\sigma ^{'}<q+1. \end{aligned}$$

(5.5)

We can use Young’s inequality on the second term on the left side of (5.3), Lemma 3.2 and both (5.4) and (5.5) to obtain

$$\begin{aligned}{} & {} (\lambda -1)\int _{\Omega }\frac{|\nabla u_{n}|^{2}}{ (1+u_{n})^{\lambda -q }}\varphi ^2 \nonumber \\{} & {} \quad \le C+\frac{2}{\sigma }\int _{\Omega }\vert \nabla \Psi _n(u_{n})\vert ^{\sigma }\varphi ^{\sigma }+\frac{2}{\sigma ^{'}}\int _{\Omega } [a(x)+u_{n}^{q}]^{\sigma ^{'}}[1-(u_{n}+1)^{1-\lambda }]^{\sigma ^{'}}\vert \nabla \varphi \vert ^{\sigma ^{'}}, \nonumber \\{} & {} \quad \le C+\frac{2}{\sigma }\int _{\Omega }\vert \nabla \Psi _n(u_{n})\vert ^{\sigma }\varphi ^{\sigma }+C \int _{\Omega } (1+u_{n})^{q\sigma ^{'}}\vert \nabla \varphi \vert ^{\sigma ^{'}}, \nonumber \\{} & {} \quad \le C+\frac{2}{\sigma }\int _{\Omega }\vert \nabla \Psi _n(u_{n})\vert ^{\sigma }\varphi ^{\sigma }+C \int _{\Omega }u_{n}^{q\sigma ^{'}}\vert \nabla \varphi \vert ^{\sigma ^{'}}\le C. \end{aligned}$$

(5.6)

Step 2. Here, we show that $\lbrace u_{n}^{q}|\nabla u_n|\rbrace$ is uniformly bounded in $L^{r}_{\text {loc}}(\Omega )$ for every $r<\frac{N}{N-1}$. For this, let $\sigma <2$ and $0\le \varphi \in C^{1}_{c}(\Omega )$, then we use Hölder inequality with exponent $\frac{2}{\sigma }$ and step 1 to obtain

$$\begin{aligned}{} & {} \int _{\Omega }u_{n}^{q\sigma }|\nabla u_{n}|^{\sigma }\varphi ^{\sigma } \\{} & {} \quad \le \int _{\Omega }\frac{|\nabla u_{n}\vert ^{\sigma }}{ (1+u_{n})^{\frac{\sigma (\lambda -q)}{2}}}\varphi ^{\frac{\sigma ^2(N-2)}{2(N-\sigma )}}(1+u_{n})^{\frac{ \sigma (\lambda +q)}{2}}\varphi ^{\frac{N\sigma (2-\sigma )}{2(N-\sigma )}}, \\{} & {} \quad \le \left( \int _{\Omega }\frac{|\nabla u_{n}\vert ^{2}}{ (1+u_{n})^{\lambda -q}}\varphi ^{\frac{\sigma (N-2)}{N-\sigma }}\right) ^{\frac{\sigma }{2}}\left( \int _{\Omega }(1 +u_{n})^{\frac{\sigma (\lambda +q)}{2-\sigma }}\varphi ^{\sigma ^*}\right) ^{\frac{2-\sigma }{2}},\\{} & {} \quad \le C\left( \int _{\Omega }(1 +u_{n})^{\frac{\sigma (\lambda +q)}{2-\sigma }}\varphi ^{\sigma ^*}\right) ^{\frac{2-\sigma }{2}}. \end{aligned}$$

Using the Sobolev inequality, we get

$$\begin{aligned} \left( \int _{\Omega }u_{n}^{(q+1)\sigma ^{*}}\varphi ^{\sigma ^*}\right) ^{\frac{ \sigma }{\sigma ^{*}}}\le C\left( \int _{\Omega }(1 +u_{n})^{\frac{ \sigma (\lambda +q)}{2-\sigma }}\varphi ^{\sigma ^*}\right) ^{\frac{2-\sigma }{2}}+C. \end{aligned}$$

Noticing that $\frac{ \sigma }{\sigma ^{*}}>\frac{2-\sigma }{2},$ and choosing $\sigma$ such that $(q+1)\sigma ^{*}=\frac{\sigma (\lambda +q)}{2-\sigma },$ yields $\sigma =\frac{N(2+q-\lambda )}{N(q+1)-(\lambda +q)}$. Hence, using Young’s inequality with $\epsilon$, we have

$$\begin{aligned} \left( \int _{\Omega }u_{n}^{(q+1)\sigma ^{*}}\varphi ^{\sigma ^*}\right) ^{\frac{ \sigma }{\sigma ^{*}}}\le \epsilon \left( \int _{\Omega }(1 +u_{n})^{(q+1)\sigma ^{*}}\varphi ^{\sigma ^*}\right) ^{\frac{ \sigma }{\sigma ^{*}}}+C_{\epsilon }. \end{aligned}$$

(5.7)

It is easy to check that the assumption $\lambda >1$ implies that $\sigma< \frac{N}{N-1} < 2$.

Step 3. Finally, we show that $\lbrace u_{n}\rbrace$ is uniformly bounded in $W^{1, r}_{\text {loc}}(\Omega )$ for every $r<\frac{N(q+1)}{N+q-1}$. For the proof of this step, let us take $\sigma <2$ and $0\le \varphi \in C^{1}_{c}(\Omega )$, then as before, we have

$$\begin{aligned}{} & {} \int _{\Omega }|\nabla u_{n}|^{r}\varphi ^{r}\le \int _{\Omega }\frac{|\nabla u_{n}\vert ^{r}}{ (1+u_{n})^{\frac{r(\lambda -q)}{2}}}\varphi ^{\frac{r^2(N-2)}{2(N-r)}}(1+u_{n})^{\frac{ r(\lambda -q)}{2}}\varphi ^{\frac{Nr(2-r)}{2(N-r)}}, \\{} & {} \quad \le \left( \int _{\Omega }\frac{|\nabla u_{n}\vert ^{2}}{ (1+u_{n})^{\lambda -q}}\varphi ^{\frac{r(N-2)}{N-r}}\right) ^{\frac{r}{2}}\left( \int _{\Omega }(1 +u_{n})^{\frac{r(\lambda -q)}{2-r}}\varphi ^{r^*}\right) ^{\frac{2-r}{2}},\\{} & {} \quad \le C\left( \int _{\Omega }(1 +u_{n})^{\frac{r(\lambda -q)}{2-r}}\varphi ^{r^*}\right) ^{\frac{2-r}{2}}. \end{aligned}$$

We now choose r such that $r^{*}=\frac{r(\lambda -q)}{2-r}$, that is, $r=\frac{N(2+q-\lambda )}{N+q-\lambda }$. Remark that since $\lambda>$, we have $r<\frac{N(q+1)}{N+q-1}<2$. Using the Sobolev inequality, one has

$$\begin{aligned} \left( \int _{\Omega }u_{n}^{r^*}\varphi ^{r^*}\right) ^{\frac{r}{r^{*}}}\le C\left( \int _{\Omega }(1+u_{n})^{r^*}\varphi ^{r^*}\right) ^{\frac{\lambda -q}{r^*}}+C. \end{aligned}$$

(5.8)

We observe that $\lambda -q<r$, and so, the previous formula implies that $\lbrace u_n\rbrace$ is uniformly bounded in $L_{\text {loc}}^{r^*}(\Omega )$. This boundedness then implies the boundedness of $\lbrace u_n\rbrace$ in $W^{1, r}_{\text {loc}}(\Omega )$. $\square$

Lemma 5.3

Again under the assumptions of Theorem 2.1, if $u_n$ is the solution of problem (3.1), then $\lbrace T_k(u_n)\rbrace$ strongly converges to $T_k(u)$ in $H^{1}_{\text {loc}}(\Omega )$ as $n \longrightarrow \infty$ for any $k > 0$.

Proof

Reasoning as in the lemma 4.1, we choose $e^{\Gamma _n(u_n)}w_{n,k,h}^{+}\varphi ^2$ as the test function in the weak formulation of (3.1), where $\varphi$ is a positive function in $C_{c}^{1}(\Omega )$, where $\omega$ is the support of $\varphi$, and $w_{ n ,k,h}$ is the function defined in (4.2). We have

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u_{n}^{q}]\nabla w_{n,k,h}^{+} \nabla u_n e^{\Gamma _n(u_n)}\varphi ^2 +2\int _{\Omega }[a(x)+u_{n}^{q}] \nabla \Psi _n(u_n) \nabla \varphi \varphi w_{n,k,h}^{+} \nonumber \\{} & {} \quad \le \int _{\Omega }h_n f_ne^{\Gamma _n(u_n)}w_{n,k,h}^{+}\varphi ^2. \end{aligned}$$

(5.9)

Hence, using (2.7) and Lemma 5.1, we get

$$\begin{aligned} h_n f_ne^{\Gamma _n(u_n)}w_{n,k,h}^{+}\varphi ^2\le \Vert \varphi \Vert _{L^{\infty }(\Omega )}^{2} \sup _{s>c_{\omega }}[h(s)e^{\Gamma (s)}] fw_{n,k,h}^{+}. \end{aligned}$$

Recalling that $w_{n,k,h}^{+}$ converges to $T_{2k}(G_h(u))$ as $n \longrightarrow \infty$, and then, it converges to zero once $h \longrightarrow \infty$; both convergences are *-weakly in $L^{\infty }(\Omega )$, then we deduce that

$$\begin{aligned} \lim _{h\longrightarrow \infty }\lim _{n\longrightarrow \infty }\int _{\Omega }h_n f_ne^{\Gamma _n(u_n)}w_{n,k,h}^{+}\varphi ^2=0. \end{aligned}$$

(5.10)

In addition, we also choose $\sigma$ such that $q+1<\sigma <\frac{N}{N-1}$. Using Young’s inequality with exponents $\sigma$, we have

$$\begin{aligned}{} & {} \vert \int _{\Omega }[a(x)+u_{n}^{q}] \nabla \Psi _n(u_n) \nabla \varphi \varphi w_{n,k,h}^{+}\vert \\{} & {} \quad \le C \int _{\Omega } (1+u_{n})^{q\sigma ^{'}}\vert \nabla \varphi \vert ^{\sigma ^{'}}w_{n,k,h}^{+} +C \int _{\Omega }\vert \nabla \Psi _n(u_n)\vert ^{\sigma }\varphi ^{\sigma }w_{n,k,h}^{+}, \end{aligned}$$

which implies that

$$\begin{aligned} \lim _{h\longrightarrow \infty }\lim _{n\longrightarrow \infty } \int _{\Omega }[a(x)+u_{n}^{q}] \nabla \Psi _n(u_n) \nabla \varphi \varphi w_{n,k,h}^{+}=0, \end{aligned}$$

(5.11)

because $(1+u_{n})^{q\sigma ^{'}}\vert \nabla \varphi \vert ^{\sigma ^{'}}\longrightarrow (1+u)^{q\sigma ^{'}}\vert \nabla \varphi \vert ^{\sigma ^{'}}$ strongly in $L^1(\Omega )$ as $n \longrightarrow \infty$, $\vert \nabla \Psi _n(u_n)\vert ^{\sigma }\varphi ^{\sigma }\longrightarrow \vert \nabla \Psi (u)\vert ^{\sigma }\varphi ^{\sigma }$ strongly in $L^1(\Omega )$ as $n \longrightarrow \infty$, and $w_{n,k,h}^{+}$ converges to $T_{2k}(G_h(u))$ as $n \longrightarrow \infty$, and then, it converges to zero once $h \longrightarrow \infty$; both convergences are *-weakly in $L^{\infty }(\Omega )$.

On the other hand, we recall that $\nabla w_{n,k,h}=0,$ if $u_n\ge h+2k={\mathcal {K}}$. Thus,

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u_{n}^{q}]\nabla w_{n,k,h}^{+} \nabla u_n e^{\Gamma _n(u_n)}\varphi ^2 \\{} & {} \quad \ge \int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla u_n e^{\Gamma _n(u_n)}\varphi ^2\\{} & {} \qquad - \int _{\lbrace k<u_n<h+2k\rbrace } [a(x)+u_{n}^{q}]\nabla T_k(u) \nabla T_{{\mathcal {K}}}(u_n) e^{\Gamma _n(u_n)}\varphi ^2. \end{aligned}$$

Note that, the second term on the right hand side of the previous formula goes to zero as $n\longrightarrow \infty$. Therefore, we have that

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u_{n}^{q}]\nabla w_{n,k,h}^{+} \nabla u_n e^{\Gamma _n(u_n)}\varphi ^2 \nonumber \\{} & {} \quad \ge \int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla u_n e^{\Gamma _n(u_n)}\varphi ^2+\epsilon _n, \end{aligned}$$

(5.12)

where $\epsilon _n$ converges to zero as $n \longrightarrow \infty$.

Using (5.10)–(5.12) and passing to the limit in (5.9) first with respect to n and after as h tends to infinity, we obtain that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla u_n e^{\Gamma _n(u_n)}\varphi ^2=0. \end{aligned}$$

(5.13)

Moreover, it is easy to show that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\lbrace u_n \le k\rbrace }[a(x)+u_{n}^{q}]\nabla (T_k(u_n)-T_k(u))^{+} \nabla T_k(u) e^{\Gamma _n(u_n)}\varphi ^2=0, \end{aligned}$$

(5.14)

which implies from (5.13) and (5.14) that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\Omega }[a(x)+u_{n}^{q}]\vert \nabla (T_k(u_n)-T_k(u))^{+}\vert ^2 e^{\Gamma _n(u_n)}\varphi ^2=0. \end{aligned}$$

(5.15)

Finally, recall that $u_n>0$, then use (2.4) to conclude that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\Omega }\vert \nabla (T_k(u_n)-T_k(u))^{+}\vert ^2 \varphi ^2=0. \end{aligned}$$

(5.16)

A similar reasoning concerning $w_{n,k,h}^{-}$ allows us to obtain that

$$\begin{aligned} \lim _{n\longrightarrow \infty }\int _{\Omega }\vert \nabla (T_k(u_n)-T_k(u))^{-}\vert ^2 \varphi ^2=0, \end{aligned}$$

(5.17)

and thus, Lemma 5.3 is proved. $\square$

5.2 Passing to the Limit

In this section, we will give a proof of Theorem 2.4, i.e. we are going to show that the function u given by Lemma 3.2 is a weak solution to problem (2.3). First, we will show that $T_k(G_{\epsilon }(u)\in H_{0}^{1}(\Omega ),$ for every $k, \epsilon >0$. For this, we take $e^{\Gamma _n(u_n)} T_k(G_{\epsilon }(u_n))$ as a test function in the weak formulation of (3.1) yielding to

$$\begin{aligned} \int _{\Omega }[a(x)+u_{n}^{q}]\vert \nabla T_k(G_{\epsilon }(u_n))\vert ^2\le \int _{\Omega } h_n(u_n) f_n e^{\Gamma _n(u_n)} T_k(G_{\epsilon }(u_n)). \end{aligned}$$

Recall that $f\in L^1(\Omega )$, using both (2.4) and (2.7), we have

$$\begin{aligned} \int _{\Omega }\vert \nabla T_k(G_{\epsilon }(u_n))\vert ^2\le k\Vert f\Vert _{L^1(\Omega )}\sup _{s>\epsilon }h(s)e^{\Gamma (s)}. \end{aligned}$$

Thus, $\lbrace T_k(G_{\epsilon }(u_n)\rbrace$ is uniformly bounded in $H_{0}^{1}(\Omega )$. It follows, up to subsequences, that $T_k(G_{\epsilon }(u_n)) \rightharpoonup T_k(G_{\epsilon }(u))$ weakly in $H_{0}^{1}(\Omega )$.

Second, we will now show that $g(u)\vert \nabla u\vert ^2, h(u)f \in L^{1}_{\text {loc}}(\Omega )$. To prove this, we take a nonnegative $\varphi \in C_{0}^{1}(\Omega )$ and the assumption (2.4) to obtain that

$$\begin{aligned} \int _{\Omega }(g(u_n)\vert \nabla u_n\vert ^2+h(u_n)f_n)\varphi \le \beta \int _{\Omega }\nabla u_n\nabla \varphi +\int _{\Omega }u_{n}^{q}\nabla u_n\nabla \varphi . \end{aligned}$$

Thanks to lemma 5.2, we have

$$\begin{aligned} \int _{\Omega }(g_n(u_n)\vert \nabla u_n\vert ^2+h_n(u_n)f_n)\varphi \le C. \end{aligned}$$

Thus, we can use Fatou lemma with respect to n to conclude that $g(u)\vert \nabla u\vert ^2,$ $h(u)f \in L^{1}_{\text {loc}}(\Omega )$.

Finally, we pass to the limit in (3.1). To this aim, let $\delta >0$ and $\varphi$ belong to $\varphi \in C_{c}^{1}(\Omega )$, with $\varphi \ge 0$, and use $V_{\delta ,\delta }(u_n)\varphi$ as test function in (3.1), where $V_{\delta ,\delta }$ is given by (2.1), to get that

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+T_{2\delta }(u_{n})^{q}]\nabla T_{2\delta }(u_n) \nabla \varphi V_{\delta ,\delta }(u_n)-\frac{1}{\delta } \int _{\lbrace \delta<u_n<2\delta \rbrace } [a(x)+u_{n}^{q}]\vert \nabla T_{2\delta }(u_n)\vert ^2\varphi \nonumber \\{} & {} \quad =\int _{\Omega } g_n(u_n)\vert \nabla T_{2\delta }(u_n)\vert ^2V_{\delta ,\delta }(u_n)\varphi + \int _{\Omega } h_n(u_n)f_n V_{\delta ,\delta }(u_n)\varphi . \end{aligned}$$

(5.18)

Now, we let n tend to infinity; using the fact that $V_{\delta ,\delta }$ is bounded and by the strong convergence of $T_{2\delta }(u_n)$ in $H^{1}_{\text {loc}}(\Omega )$, we have

$$\begin{aligned}{} & {} \int _{\Omega }[a(x)+u^{q}]\nabla u \nabla \varphi V_{\delta ,\delta }(u)-\frac{1}{\delta } \int _{\lbrace \delta<u<2\delta \rbrace } [a(x)+u^{q}]\vert \nabla u\vert ^2\varphi \nonumber \\{} & {} \quad =\int _{\Omega } g(u)\vert \nabla u\vert ^2V_{\delta ,\delta }(u)\varphi + \int _{\Omega } h(u)f V_{\delta ,\delta }(u)\varphi . \end{aligned}$$

(5.19)

On the other hand, choosing $\varphi = e^{\Gamma _n(u_n)}S_{\delta }(u_n)$ where $\delta >0$ and $S_{\delta }$ is the function defined in (2.2), as test function in (3.1), we deduce

$$\begin{aligned} \frac{1}{\delta }\int _{\lbrace \delta< u_n<2\delta \rbrace } [a(x)+u_{n}^{q}]\vert \nabla u_n\vert ^2e^{\Gamma _n(u_n)}\le \int _{\Omega }h_n(u_n)f_ne^{\Gamma _n(u_n)}S_{\delta }(u_n) \end{aligned}$$

Passing to the inferior limit in the previous inequality as n goes to infinity, we get

$$\begin{aligned} \frac{1}{\delta }\int _{\lbrace \delta< u<2\delta \rbrace } [a(x)+u^{q}]\vert \nabla u\vert ^2e^{\Gamma (u)}\le \int _{\Omega }h(u)fe^{\Gamma (u)}S_{\delta }(u). \end{aligned}$$

Observe that by construction, we have $S_{\delta }(u) \longrightarrow 0$ when $\delta \longrightarrow \infty$. Thus, using Fatou lemma and hypothesis (2.7), we deduce

$$\begin{aligned} \lim _{\delta \longrightarrow \infty } \frac{1}{\delta }\int _{\lbrace \delta< u<2\delta \rbrace } [a(x)+u^{q}]\vert \nabla u\vert ^2e^{\Gamma (u)}=0. \end{aligned}$$

(5.20)

Note that, $V_{\delta ,\delta }(u)\longrightarrow 1$ as $\delta$ tends to $\infty$. Consequently, using that $V_{\delta ,\delta }$ is bounded, that both $g(u)\vert \nabla u\vert ^2$ and h(u)f belong to $L^{1}_{\text {loc}}(\Omega )$, the limit (5.20) and Lebesgue theorem, we can pass to the limit $(\delta \longrightarrow \infty )$ in (5.19) to conclude

$$\begin{aligned} \int _{\Omega }[a(x)+u^{q}]\nabla u \nabla \varphi =\int _{\Omega } g(u)\vert \nabla u\vert ^2\varphi + \int _{\Omega } h(u)f \varphi , \end{aligned}$$

(5.21)

for all $\varphi$ belongs to $C^{1}_{c}(\Omega )$.