Quasilinear Elliptic Problem with Singular Lower Order Term and $L^1$ Data

Abstract

In this paper, we are interested in the existence result of solutions for the nonlinear Dirichlet problem of the type:

$$\begin{aligned} \left\{ \begin{aligned}&-\mathrm{div} (M(x) \nabla u )+ \gamma u^p= B \frac{|\nabla u|^q}{u^\theta }+f\ \ \mathrm{in}\ \Omega ,\\&u> 0\ \ \mathrm{in}\ \Omega ,\\&u=0\ \ \mathrm{on}\ {\partial \Omega },\\ \end{aligned} \right. \end{aligned}$$

where $\Omega$ is a bounded open subset of $\mathbb {R}^N$, $N>2$, M(x) is a uniformly elliptic and bounded matrix, $\gamma > 0$, $B> 0$, $1\le q<2$, $0<\theta \le 1$, and the source f is a nonnegative (not identically zero) function belonging to $L^1(\Omega )$.

1 Introduction

In this paper, we deal with the following kind of problem:

$$\begin{aligned} \left\{ \begin{aligned}&-\mathrm{div} (M(x) \nabla u )+ \gamma u^p= B \frac{|\nabla u|^q}{u^\theta }+f\ \ \mathrm{in}\ \Omega ,\\&u> 0\ \ \mathrm{in}\ \Omega ,\\&u=0\ \ \mathrm{on}\ {\partial \Omega },\\ \end{aligned} \right. \end{aligned}$$

(1.1)

where $\Omega$ is any bounded open subset of $\mathbb {R}^N$, $N>2$, $\displaystyle {M: \Omega \times \mathbb {R}\rightarrow \mathbb {R}^{N^2}}$ is a bounded and measurable matrix such that there exist $\alpha , \beta > 0$ satisfying:

$$\begin{aligned} \alpha |\xi |^2 \le M(x)\xi \xi , \quad |M(x)|\le \beta \quad \text {a.e.} \ x \in \Omega , \ \forall \xi \in \mathbb {R}^N. \end{aligned}$$

(1.2)

We assume that:

$$\begin{aligned} 0<\theta \le 1, \ \gamma> 0, \ B> 0, \ 1\le q<2, \ \text {and} \ p> \frac{q}{2-q}. \end{aligned}$$

(1.3)

We suppose moreover that f is a nonnegative function, such that $f\not \equiv 0$ and that:

$$\begin{aligned} f \in L^1(\Omega ). \end{aligned}$$

(1.4)

The main features in dealing with this class of problems are the facts that the lower order term has a singularity in u and the data f belong to $L^1(\Omega )$. For $B<0$, $\gamma =0$ and the lower order term satisfies a quadratic growth assumption with respect to the gradient and depends continuously on u, the problems like (1.1) have been exhaustively studied in the literature, see for instance [4, 7, 10], while the case of singular lower order term at $u=0$ having a quadratic growth with respect to the gradient was considered in [3, 5, 14]. We refer also to [2] where the problem (1.1) is deeply studied in the case $\gamma =0$, $q=2$, $\theta =1$, and f belongs to $L^m(\Omega )$, $m>1$.

The existence and regularity results of problems (1.1) have been proved in [12] in the case where the lower order term is nonsingular and the growth of the gradient is superlinear and f belongs to $L^m(\Omega )$, $m \ge 1$ (see also [15, 16]). In addition, the problem of existence and regularity of solutions have been investigated in [11] when $\gamma =0$, $q=1$ and f belongs to $L^m(\Omega )$, $m>1$. We also quote the paper [1] in which the existence and nonexistence results related to (1.1) have been obtained using a comparison and a priori estimates if $\gamma =0$, $1<q \le 2$, and f belongs to $L^m(\Omega )$, $m \ge 1$.

Our aim is to prove the existence of solution of problem (1.1) under our assumptions (1.2)–(1.4). Note that the presence of the lower order term $\displaystyle {\gamma u^p}$ is crucial in the sense that it guarantees the existence of solution when the data f belong only in $L^1(\Omega )$. Our main result is the following

Theorem 1.1

Assume that (1.2)–(1.4) hold true. Then, there exists a solution u for (1.1), with $u>0$ in $\Omega$, in the sense that:

$u \in W^{1, r}_0(\Omega ) \cap L^p(\Omega )$ for every $1\le r<\max {\{\frac{N}{N-1},\frac{2p}{p+1} \}}$, $\frac{|\nabla u|^q}{u^\theta } \in L_\mathrm{loc}^1(\Omega )$ and that:

$$\begin{aligned} \int _{\Omega }M(x)\nabla u \nabla \varphi + \gamma \int _{\Omega } u^p \varphi = \int _{\Omega } B \frac{|\nabla u|^q}{u^\theta }\varphi +\int _{\Omega }f \varphi , \quad \forall \varphi \in C^1_c(\Omega ).\quad \end{aligned}$$

(1.5)

Moreover, if $0<\theta \le \frac{q}{q+1}$, then $\frac{|\nabla u|^q}{u^\theta }$ belongs to $L^1(\Omega )$.

Remark 1.2

If $0<\theta \le \frac{q}{q+1}$, then we can use, as test functions in the formulation (1.5), not only functions in $C^1_c(\Omega )$, but also functions in $W^{1, s}_0(\Omega )$, $s>\max {\{N,\frac{2p}{p-1} \}}$.

2 Approximation of Problem (1.1)

We will prove the existence of solutions of problem (1.1) by a standard approximation procedure which avoids singularities. To this end, we consider for $n \in \mathbb {N}$ the following approximate problem:

$$\begin{aligned} \left\{ \begin{aligned}&-\mathrm{div} (M(x)\nabla u_n )+\gamma |u_n|^{p-1}u_n=B\frac{|\nabla u_n|^q}{(1+\frac{1}{n}|\nabla u_n|^q)(|u_n|+\frac{1}{n})^\theta }+f_n\ \ \mathrm{in}\ \Omega ,\\&u_n\ge 0\ \ \mathrm{in}\ \Omega ,\\&u_n=0\ \ \mathrm{on}\ {\partial \Omega },\\ \end{aligned} \right. \nonumber \\ \end{aligned}$$

(2.1)

where $f_n =\frac{f}{1+\frac{1}{n}f}$. Since the right-hand side of (2.1) is bounded, the existence of a bounded (see [18]) weak solutions $u_n \in H_0^{1}(\Omega )$ is a consequence of the classical results in [17] (see also [13]). Moreover, $u_n \ge 0$, since the right-hand side of (2.1) is positive. Now, let z be the unique weak solution of:

$$\begin{aligned} \left\{ \begin{aligned}&-\mathrm{div} (M(x)\nabla z )+\gamma z^p=\frac{f}{1+f}\ \ \mathrm{in}\ \Omega ,\\&z=0\ \ \mathrm{on}\ {\partial \Omega }.\\ \end{aligned} \right. \end{aligned}$$

(2.2)

Since $\frac{f}{1+f} \ge 0$ and $\int _{0} (s^{p+1})^{\frac{-1}{2}}= \infty$ with $p \ge 1$, the strong maximum principle (see [19]) implies that for every $\omega \subset \subset \Omega$, there exists $c_\omega$, such that the solution z of (2.2) satisfies $z \ge c_{\omega }> 0 \ \text {in} \ \omega$. Since $\frac{f}{1+\frac{1}{n}f}\ge \frac{f}{1+f}$, one has:

$$\begin{aligned} -\mathrm{div} (M(x)\nabla u_n )+\gamma u_n^p \ge -\mathrm{div} (M(x)\nabla z )+\gamma z^p; \end{aligned}$$

(2.3)

so, using $-(u_n-z)^{-}$ as test function in (2.3) with $(u_n-z)^{-}= -(u_n-z)\chi _{\{u_n<z\}}$, we obtain:

$$\begin{aligned} \int _{\{u_n<z\}} |\nabla (u_n-z)|^2- \gamma \int _{\Omega } (u_n^p-z^p) (u_n-z)^{-} \le 0. \end{aligned}$$

(2.4)

Dropping the second nonnegative term in the left-hand side of (2.4), it follows that $u_n \ge z$. Therefore:

$$\begin{aligned} u_n \ge c_{\omega }> 0 \ \text {in} \ \omega , \quad \forall \omega \subset \subset \Omega . \end{aligned}$$

(2.5)

Hereafter, we will make use of two truncation functions $T_k$ and $G_k$: for every $k \ge 0$ and $r \in \mathbb {R}$, let:

$$\begin{aligned} T_k(r)=\left\{ \begin{array}{ll} r &{} \mathrm{if} \ \ |r|\le k,\\ k\frac{r}{|r|} &{} \mathrm{if} \ \ |r|>k,\\ \end{array} \right. \quad G_k(r)=r-T_k(r). \end{aligned}$$

For the sake of simplicity, we will use when referring to the integrals the following notation:

$$\begin{aligned} \int _\Omega f = \int _\Omega f(x) \, \mathrm{d}x. \end{aligned}$$

Finally, throughout this paper, C will indicate any positive constant which depends only on data and whose value may change from line to line and sometimes in the same line.

3 Existence Result

In the next lemma, we state some a priori estimates on the solution $u_n$ and on the lower order term of the approximate problem (2.1).

Lemma 3.1

Assume (1.2)–(1.4). Then, the sequence $u_n$ is bounded in $W^{1, r}_0(\Omega ) \cap L^p(\Omega )$ for every $1\le r<\max {\{\frac{N}{N-1},\frac{2p}{p+1} \}}$ and $\frac{|\nabla u_n|^q}{u_n^\theta }$ is bounded in $L_\mathrm{loc}^1(\Omega )$.

Proof

Let $\lambda >1$, and choose $1-(1+G_1(u_n))^{1-\lambda }$ as test function in (2.1), using (1.2) and the fact that $f_n \le f$, we thus have:

$$\begin{aligned}&(\lambda -1) \alpha \int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda }+ \gamma \int _{\{ u_n\ge 1\}} u_n^p(1-u_n^{1-\lambda })\nonumber \\&\quad \le B\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^q}{u_n^\theta }+ \int _{\Omega }f, \end{aligned}$$

(3.1)

using Young inequality, we obtain:

$$\begin{aligned}&B\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^q}{u_n^\theta }\le B\int _{\{ u_n\ge 1\}}|\nabla u_n|^q= B\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^q}{u_n^{\frac{q\lambda }{2}}} u_n^{\frac{q\lambda }{2}}\\&\quad \le \frac{(\lambda -1) \alpha }{2}\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda }+C\int _{\{ u_n\ge 1\}} u_n^{\frac{q\lambda }{2-q}}, \end{aligned}$$

which implies from (3.1) that:

$$\begin{aligned}&\frac{(\lambda -1) \alpha }{2}\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda }+\gamma \int _{\{ u_n\ge 1\}} u_n^p(1-u_n^{1-\lambda })\nonumber \\&\quad \le C\int _{\{ u_n\ge 1\}} u_n^{\frac{q\lambda }{2-q}}+\int _{\Omega }f. \end{aligned}$$

(3.2)

Next, we choose $T> 1$, such that $1-T^{1-\lambda }= \frac{1}{2}$, we have:

$$\begin{aligned} \frac{1}{2}\int _{\{ u_n> T\}} u_n^p \le \int _{\{ u_n> T\}} u_n^p(1-u_n^{1-\lambda }) \le \int _{\{ u_n\ge 1\}} u_n^p(1-u_n^{1-\lambda }), \end{aligned}$$

that is:

$$\begin{aligned} \frac{1}{2}\int _{\{ u_n\ge 1\}} u_n^p= & {} \frac{1}{2}\int _{\{ 1 \le u_n\le T\}} u_n^p+\frac{1}{2}\int _{\{ u_n> T\}} u_n^p\\\le & {} \frac{1}{2} T^p \mathrm{meas}(\Omega )+\int _{\{ u_n\ge 1\}} u_n^p(1-u_n^{1-\lambda }). \end{aligned}$$

Hence, from (3.2), we deduce that:

$$\begin{aligned}&\frac{(\lambda -1) \alpha }{2}\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda }+\frac{\gamma }{2} \int _{\{ u_n\ge 1\}} u_n^p\\&\quad \le C\int _{\{ u_n\ge 1\}} u_n^{\frac{q\lambda }{2-q}}+C. \end{aligned}$$

Choosing $\lambda$ such that $\frac{q\lambda }{2-q}<p$, that is $1<\lambda < \frac{p(2-q)}{q}$. Note that this choice of $\lambda$ is possible, since $\frac{p(2-q)}{q}>1$. Therefore:

$$\begin{aligned} \int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda } \le C, \end{aligned}$$

(3.3)

and

$$\begin{aligned} \int _{\Omega } u_n^p \le C. \end{aligned}$$

(3.4)

Now, we choose $\varepsilon < \frac{1}{n}$ and use $(T_1(u_n)+\varepsilon )^\theta -\varepsilon ^\theta$ as test function, dropping the positive term and using (1.2), we have:

$$\begin{aligned}&\alpha \theta \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{1-\theta }} \le B\int _{\Omega }\frac{|\nabla u_n|^q}{(u_n+\frac{1}{n})^\theta }(T_1(u_n)+\varepsilon )^\theta \nonumber \\&\qquad +\int _{\Omega } f_n (T_1(u_n)+\varepsilon )^\theta \nonumber \\&\quad \le B\int _{\{ u_n\ge 1\}} |\nabla u_n|^q+ B\int _{\{ u_n<1\}}|\nabla u_n|^q+ (1+\varepsilon )^\theta \int _{\Omega } f. \end{aligned}$$

(3.5)

Using Young inequality together with (3.3) and (3.4) and the fact that $\frac{q\lambda }{2-q}<p$ yield that:

$$\begin{aligned} \int _{\{ u_n\ge 1\}} |\nabla u_n|^q= & {} \int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^q}{u_n^{\frac{\lambda q}{2}}} u_n^{\frac{\lambda q}{2}}\\\le & {} C\int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda }+ C\int _{\Omega } u_n^p \le C. \end{aligned}$$

Then, from (3.5) and the above estimate, using again young inequality, we obtain:

$$\begin{aligned}&\alpha \theta \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{1-\theta }} \le B \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{\frac{q}{2}(1-\theta )}}(T_1(u_n)+\varepsilon )^{\frac{q}{2}(1-\theta )}\\&\qquad +(1+\varepsilon )^\theta \int _{\Omega } f+C \\&\quad \le \frac{\alpha \theta }{2} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{1-\theta }}+C(1+\varepsilon )^{\frac{q}{2-q}(1-\theta )}+(1+\varepsilon )^\theta \int _{\Omega } f+C, \end{aligned}$$

and so

$$\begin{aligned} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{1-\theta }}\le C\Bigg ((1+\varepsilon )^{\frac{q}{2-q}(1-\theta )}+(1+\varepsilon )^\theta \Bigg ), \end{aligned}$$

which gives:

$$\begin{aligned} \int _{\Omega }|\nabla T_1(u_n)|^2= & {} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{1-\theta }}(T_1(u_n)+\varepsilon )^{1-\theta }\\\le & {} C (1+\varepsilon )^{1-\theta }\Bigg ((1+\varepsilon )^{\frac{q}{2-q}(1-\theta )}+(1+\varepsilon )^\theta \Bigg ). \end{aligned}$$

Letting $\varepsilon$ tends to 0 yields that:

$$\begin{aligned} \int _{\Omega }|\nabla T_1(u_n)|^2 \le C. \end{aligned}$$

(3.6)

Combining (3.3) and (3.6), we have:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla u_n|^2}{(1+u_n)^\lambda } \le \int _{\{ u_n\ge 1\}}\frac{|\nabla u_n|^2}{u_n^\lambda }+\int _{\Omega }|\nabla T_1(u_n)|^2 \le C, \end{aligned}$$

(3.7)

which holds for every $\lambda >1$. Now, we proceed as in [12] (Proposition A.1 and A.2 in the Appendix A). Let $1 \le r<2$, using the estimate (3.7) together with Hölder inequality, we obtain:

$$\begin{aligned} \int _{\Omega }|\nabla u_n|^r \le \int _{\Omega }\frac{|\nabla u_n|^r}{(1+u_n)^{\frac{r\lambda }{2}}}(1+u_n)^{\frac{r\lambda }{2}}\le C \Bigg (\int _{\Omega }(1+u_n)^{\frac{r\lambda }{2-r}}\Bigg )^{1-\frac{r}{2}}. \end{aligned}$$

(3.8)

Sobolev inequality implies that:

$$\begin{aligned} \Bigg (\int _{\Omega } u_n^{r^*}\Bigg )^{\frac{r}{r^*}}\le C\Bigg (\int _{\Omega }(1+u_n)^{\frac{r\lambda }{2-r}}\Bigg )^{1-\frac{r}{2}}. \end{aligned}$$

Choosing r such that $r^*=\frac{rN}{N-r}=\frac{r\lambda }{2-r}$ gives that $r=\frac{N(2-\lambda )}{N-\lambda }$. Notice that $\frac{r}{r^*} >1-\frac{r}{2}$ (which is equivalent to $N>2$), and since $\lambda >1$, it follows that:

$$\begin{aligned} \int _{\Omega }|\nabla u_n|^r\le C, \quad \forall \ 1 \le r< \frac{N}{N-1}. \end{aligned}$$

(3.9)

On the other hand, starting from (3.8) and thanks to (3.4), noticing that $\frac{r\lambda }{2-r}\le p$ is equivalent to $r\le \frac{2p}{p+\lambda }$, and since $\lambda >1$, we thus obtain:

$$\begin{aligned} \int _{\Omega }|\nabla u_n|^r\le C, \quad \forall \ 1 \le r< \frac{2p}{p+1}. \end{aligned}$$

(3.10)

Combining (3.9) and (3.10) to conclude that:

$$\begin{aligned} \int _{\Omega }|\nabla u_n|^r\le C, \quad \forall \ 1 \le r< \max \Bigg \{\frac{N}{N-1}, \ \frac{2p}{p+1} \Bigg \}. \end{aligned}$$

(3.11)

Recalling (2.5), estimate (3.11) and by means of Hölder inequality, it follows for every $\omega \subset \subset \Omega$ that:

$$\begin{aligned} \int _{\omega }\frac{|\nabla u_n|^q}{u_n^\theta } \le \frac{\mathrm{meas}(\Omega )^{\frac{r-q}{r}}}{c_\omega ^\theta }\Vert u_n \Vert ^q_{W_0^{1, r}(\Omega )} \le C. \end{aligned}$$

(3.12)

$\square$

Now, we prove the global boundedness of $\frac{|\nabla u_n|^q}{u_n^\theta }$ in $L^1(\Omega )$ in the case where $0<\theta \le \frac{q}{q+1}$.

Lemma 3.2

Assume (1.2)–(1.4). If $0<\theta \le \frac{q}{q+1}$, then $\frac{|\nabla u_n|^q}{u_n^\theta }$ is bounded in $L^1(\Omega )$.

Proof

To prove Lemma 3.2 we have to distinguish between three cases.

First case. Assume that $0<\theta <\frac{1}{2}$. We choose $\varepsilon < \frac{1}{n}$ and use $(T_1(u_n)+\varepsilon )^{1-2\theta }-\varepsilon ^{1-2\theta }$ as test function in (2.1), using (1.2), dropping the positive term, we have:

$$\begin{aligned}&\alpha (1-2\theta ) \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{2\theta }} \le B(1+\varepsilon )^{1-2\theta }\\&\quad \times \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{q\theta }}(T_1(u_n)+\varepsilon )^{(q-1)\theta }\\&\quad +B(1+\varepsilon )^{1-2\theta }\int _{\{ u_n\ge 1\}} |\nabla u_n|^q+(1+\varepsilon )^{1-2\theta }\int _{\Omega } f, \end{aligned}$$

using (3.11) and Young inequality, we obtain:

$$\begin{aligned} \frac{\alpha (1-2\theta )}{2} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{2\theta }} \le C(\varepsilon ), \end{aligned}$$

where $C(\varepsilon )$ is such that ${\lim _{\varepsilon \rightarrow 0}} \ C(\varepsilon ) <+\infty .$ That is:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{2\theta }} \le C. \end{aligned}$$

(3.13)

Using (3.11), (3.13), and Hölder inequality, we obtain:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla u_n|^q}{(u_n+\varepsilon )^{\theta }}\le & {} \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{\theta }}+\int _{\{ u_n\ge 1\}} |\nabla u_n|^q \\\le & {} C+\int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{q\theta }}(T_1(u_n)+\varepsilon )^{(q-1)\theta } \\\le & {} C+(1+\varepsilon )^{(q-1)\theta }\int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{q\theta }} \\\le & {} C+\Bigg ( \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(u_n+\varepsilon )^{2\theta }} \Bigg )^{\frac{q}{2}} (1+\varepsilon )^{(q-1)\theta } meas(\Omega )^{\frac{2-q}{q}} \le C. \end{aligned}$$

Thanks to Fatou’s Lemma, letting $\varepsilon$ tends to zero, we deduce that:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla u_n|^q}{u_n^{\theta }} \le C. \end{aligned}$$

Second case. Assume that $\theta =\frac{1}{2}$, we choose $\varepsilon < \frac{1}{n}$ and use $\log (T_1(u_n)+\varepsilon )-\log (\varepsilon )$ as test function, using (1.2), dropping the positive term, we obtain:

$$\begin{aligned} \alpha \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{T_1(u_n)+\varepsilon }\le & {} B (1+\varepsilon )^{\frac{1}{2}}\int _{\Omega }|\nabla T_1(u_n)|^q \\&+ B \log (1+\varepsilon )\int _{\Omega } |\nabla G_1(u_n)|^q+\log (1+\varepsilon )\int _{\Omega } f, \end{aligned}$$

Using (3.3) and (3.6), we thus have:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{T_1(u_n)+\varepsilon } \le C ((1+\varepsilon )^{\frac{1}{2}}+\log (1+\varepsilon )), \end{aligned}$$

and by Hölder inequality, it follows that:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{\frac{1}{2}}}= & {} \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{\frac{q}{2}}} (T_1(u_n)+\varepsilon )^{\frac{q-1}{2}}\\\le & {} (1+\varepsilon )^{\frac{q-1}{2}} \mathrm{meas}(\Omega )^{\frac{2-q}{q}} \Bigg (\int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{T_1(u_n)+\varepsilon } \Bigg )^{\frac{q}{2}} \\\le & {} C(1+\varepsilon )^{\frac{q-1}{2}} \mathrm{meas}(\Omega )^{\frac{2-q}{q}} \Bigg ( (1+\varepsilon )^{\frac{1}{2}}+\log (1+\varepsilon )\Bigg )^{\frac{q}{2}}. \end{aligned}$$

Thanks to Fatou’s Lemma, letting $\varepsilon$ tends to zero, we deduce that:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla u_n|^q}{u_n^{\theta }} \le C. \end{aligned}$$

Third case. Assume that $\frac{1}{2}<\theta <\frac{q}{q+1}$, we choose $\varepsilon < \frac{1}{n}$, using $(T_1(u_n)+\varepsilon )^{2\theta -1}-\varepsilon ^{2\theta -1}$ and reasoning as above, we obtain:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{2(1-\theta )}} \le C; \end{aligned}$$

(3.14)

using Hölder inequality together with (3.14) yields:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{\theta }}= & {} \int _{\Omega }\frac{|\nabla T_1(u_n)|^q}{(T_1(u_n)+\varepsilon )^{q(1-\theta )}} (T_1(u_n)+\varepsilon )^{q(1-\theta )-\theta } \\\le & {} (1+\varepsilon )^{q-(q+1)\theta } \mathrm{meas}(\Omega )^{\frac{2-q}{q}} \Bigg (\int _{\Omega }\frac{|\nabla T_1(u_n)|^2}{(T_1(u_n)+\varepsilon )^{2(1-\theta )}}\Bigg )^{\frac{q}{2}}\\\le & {} C (1+\varepsilon )^{q-(q+1)\theta }. \end{aligned}$$

By Fatou’s Lemma, letting $\varepsilon$ tends to zero, we deduce that:

$$\begin{aligned} \int _{\Omega }\frac{|\nabla u_n|^q}{u_n^{\theta }} \le C. \end{aligned}$$

$\square$

Remark 3.3

Thanks to estimates (3.4) and (3.11), it follows that there exists a function $u \in W^{1, r}_0(\Omega ) \cap L^p(\Omega )$, such that up to a subsequence, $u_n$ converges to u weakly in $W^{1, r}_0(\Omega )$ and $L^p(\Omega )$ and a.e. in $\Omega$.

We prove now the following convergence result.

Proposition 3.4

Under assumptions (1.2)–(1.4), we have:

$$\begin{aligned} u_n^p \rightarrow u^p \ \ \text {strongly in} \ \ L^1(\Omega ). \end{aligned}$$

Proof

We take $T_1(u_n-T_h(u_n))$ as test function in (2.1), dropping the positive term, using (1.2), and we then have:

$$\begin{aligned}&\alpha \int _{\{ h \le u_n\le h+1\}}|\nabla u_n|^2+\gamma \int _{\{ u_n \ge h+1\}} u_n^p\\&\quad \le B\int _{\{ h \le u_n\le h+1\}}|\nabla u_n|^q+B\int _{\{ u_n > h+1\}}|\nabla u_n|^q+\int _{\{ u_n \ge h\}} f, \end{aligned}$$

which implies using (3.11), Young together with Hölder inequalities that:

$$\begin{aligned}&\frac{\alpha }{2} \int _{\{ h \le u_n\le h+1\}}|\nabla u_n|^2+\gamma \int _{\{ u_n \ge h+1\}} u_n^p \\&\quad \le C \mathrm{meas}{\{ u_n> h\}}^{1-\frac{q}{2}}+B\Vert u_n \Vert ^q_{W_0^{1, r}(\Omega )} \mathrm{meas}{\{ u_n> h\}}^{\frac{r-q}{r}}+\int _{\{ u_n \ge h\}} f, \\&\quad \le C \mathrm{meas}{\{ u_n> h\}}^{1-\frac{q}{2}}+C \mathrm{meas}{\{ u_n > h\}}^{\frac{r-q}{r}}+\int _{\{ u_n \ge h\}} f. \end{aligned}$$

Letting $n \rightarrow +\infty$ and then $h \rightarrow +\infty$, we obtain:

$$\begin{aligned} \int _{\{ u_n \ge h+1\}} u_n^p \le w(n, h), \end{aligned}$$

(3.15)

where w(n, h) tends to zero when $n \rightarrow +\infty$ and $h \rightarrow +\infty$.

Let E be a measurable subset of $\Omega$, and we have:

$$\begin{aligned} \int _{E}u_n^p \le \int _{\{ u_n > h\}}u_n^p+h^p \mathrm{meas}(E). \end{aligned}$$

Then, thanks to (3.15), we take the limit as $\mathrm{meas}(E)$ tends to zero, h tends to infinity, and since $u_n^{p}$ converges to $u^p$ almost everywhere, we easily conclude by Vitali’s theorem the proof of Proposition 3.4.

To conclude the proof of Theorem 1.1, we need to prove the strong convergence of $T_k(u_n)$ to $T_k(u)$ in $W_\mathrm{loc}^{1, r}(\Omega )$. First, we choose $(T_k(u_n)+\varepsilon )^\theta -\varepsilon ^\theta$ as test function in (2.1) with $\varepsilon < \frac{1}{n}$, using (1.2), dropping the positive term and by means of Hölder inequality and (3.11), we get:

$$\begin{aligned}&\alpha \int _{\Omega } \frac{|\nabla T_k(u_n)|^2}{(T_k(u_n)+\varepsilon )^{1-\theta }} \le B\int _{\Omega } |\nabla u_n|^q+ (k+\varepsilon )^\theta \int _{\Omega } f_n\\&\quad \le C\Vert u_n \Vert ^q_{W_0^{1, r}(\Omega )}+(k+\varepsilon )^\theta \Vert f \Vert _{L^1(\Omega )}. \end{aligned}$$

Then, from the previous inequality, we deduce that:

$$\begin{aligned}&\int _{\Omega } |\nabla T_k(u_n)|^2=\int _{\Omega } \frac{|\nabla T_k(u_n)|^2}{(T_k(u_n)+\varepsilon )^{1-\theta }} (T_k(u_n)+\varepsilon )^{1-\theta }\\&\quad \le (k+\varepsilon )^{1-\theta } (C\Vert u_n \Vert ^q_{W_0^{1, r}(\Omega )}+(k+\varepsilon )^\theta \Vert f \Vert ^q_{L^1(\Omega )}). \end{aligned}$$

Letting $\varepsilon$ goes to zero:

$$\begin{aligned} \int _{\Omega } |\nabla T_k(u_n)|^2 \le C_ k, \end{aligned}$$

(3.16)

which implies that:

$$\begin{aligned} T_k(u_n) \ \text {converges to} \ T_k(u) \ \text {weakly} \ \text {in} \ H_0^{1}(\Omega ) \ \text {and a.e. in} \ \Omega . \end{aligned}$$

(3.17)

Then, thanks to (3.12), the right-hand side of (2.1) is bounded in $L_\mathrm{loc}^1(\Omega )$, and by the result of [8] (see also [6, 9]), we deduce that, up to subsequences (not relabeled), $\nabla u_n$ converges to $\nabla u$ a.e. in $\Omega$ which, in turn, implies that $\frac{|\nabla u|^q}{u^\theta }$ belongs to $L_\mathrm{loc}^1(\Omega )$ and that $\frac{|\nabla u|^q}{u^\theta }$ belongs to $L^1(\Omega )$ if $0<\theta \le \frac{q}{q+1}$.

Next, let $\varphi \in C^1_c(\Omega )$, $\varphi \ge 0$, $\varphi \equiv 1$ on $\omega \subset \subset \Omega$ and use $T_h(u_n-T_k(u))\varphi$ as test function in (2.1), we thus have thanks to (1.2), (2.5), and (3.11):

$$\begin{aligned}&\alpha \int _{\Omega } |\nabla T_h(u_n-T_k(u))|^2 \ \varphi \le C h \Vert \nabla \varphi \Vert _{L^{\infty }(\Omega )}\Vert u_n \Vert _{W_0^{1, r}(\Omega )}\\&\quad + h \Vert \varphi \Vert _{L^{\infty }(\Omega )} \ \frac{C}{c_\omega ^\gamma }\Vert u_n \Vert ^q_{W_0^{1, r}(\Omega )} \\&\quad +h \Vert \varphi \Vert _{L^{\infty }(\Omega )} \int _{\Omega } f-\int _{\Omega } M(x) \nabla T_k(u) \nabla T_h(u_n-T_k(u)) \ \varphi . \end{aligned}$$

Since $T_h(u_n-T_k(u))$ converges to $T_h(u-T_k(u))$ weakly in $H^1_0(\Omega )$, by (1.2) and using the fact that $\nabla T_k(u) \nabla T_h(u-T_k(u))\equiv 0$, we obtain:

$$\begin{aligned} \int _{\Omega } M(x) \nabla T_k(u) \nabla T_h(u_n-T_k(u)) \ \varphi =\omega (n). \end{aligned}$$

Hence:

$$\begin{aligned} \alpha \limsup _{n \rightarrow \infty } \int _{\Omega }|\nabla T_h(u_n-T_k(u))|^2 \ \varphi \le Ch. \end{aligned}$$

(3.18)

Let now r be such that $r<2$, where r is as in the statement of Lemma 3.1, we can write:

$$\begin{aligned}&\int _{\omega }|\nabla T_k(u_n)-\nabla T_k(u)|^r\le \int _{\Omega }|\nabla T_k(u_n)-\nabla T_k(u)|^r \varphi \nonumber \\&\quad =\int _{\{ |u_n-u| \le h, u_n\le k\}}|\nabla u_n-\nabla T_k(u)|^r\varphi \nonumber \\&\qquad +\int _{\{ |u_n-u| \le h, u _n>k\}}|\nabla T_k(u)|^r\varphi \nonumber \\&\qquad +\int _{\{ |u_n-u| > h\}}|\nabla T_k(u_n)-\nabla T_k(u)|^r\varphi . \end{aligned}$$

(3.19)

Therefore, using Hölder inequality, we obtain:

$$\begin{aligned}&\int _{\Omega }|\nabla T_k(u_n)-\nabla T_k(u)|^r \varphi \le \int _{\Omega }|\nabla T_h(u_n-T_k(u))|^r\varphi \\&\quad +\Vert \varphi \Vert _{L^{\infty }(\Omega )} \Bigg (\int _{\{u _n>k\}}|\nabla T_k(u)|^r\\&\quad +(\mathrm{meas}{\{ |u_n-u| > h\}})^{1-\frac{r}{2}} \Bigg (\int _{\Omega }|\nabla T_k(u_n)-\nabla T_k(u)|^2\Bigg )^{\frac{r}{2}}\Bigg ) . \end{aligned}$$

Thus, combining (3.18) and (3.19) and thanks to (3.16) and Lebesgue’s convergence theorem, we obtain for every $h>0$ and every $k>0$:

$$\begin{aligned} \limsup _{n \rightarrow \infty } \int _{\Omega }|\nabla T_k(u_n)-\nabla T_k(u)|^r\ \varphi \le \Bigg (\frac{Ch}{\alpha }\Bigg )^{\frac{r}{2}} \Vert \varphi \Vert _{L^{\infty }(\Omega )} \mathrm{meas}(\Omega )^{1-\frac{r}{2}}, \end{aligned}$$

Letting h tends to zero, we finally have:

$$\begin{aligned} \limsup _{n \rightarrow \infty } \int _{\Omega }|\nabla T_k(u_n)-\nabla T_k(u)|^r\ \varphi =0. \end{aligned}$$

On the other hand, using $1-(1+G_k(u_n))^{1-\lambda }$ as test function in (2.1) with $\lambda >1$ and following the same proof of that (3.3) and using Hölder inequality, we obtain:

$$\begin{aligned} \int _{\{ u_n\ge k\}}|\nabla u_n|^r \le w(k). \end{aligned}$$

(3.20)

Hence:

$$\begin{aligned} u_n \ \text {converges to} \ u \ \text {strongly} \ \text {in} \ W_\mathrm{loc}^{1, r}(\Omega ). \end{aligned}$$

(3.21)

Now, let $E\subset \subset \omega$ be a measurable set, and using (2.5) and Hölder inequality, we obtain:

$$\begin{aligned} \int _{E}\frac{|\nabla u_n|^q}{u_n^{\theta }}\le & {} \int _{E}\frac{|\nabla u_n|^q}{u_n^{\theta }} \varphi \le \frac{\Vert \varphi \Vert _{L^{\infty }(\Omega )} }{k^\theta } \int _{\{ u_n> k\}}|\nabla u_n|^q+\frac{\Vert \varphi \Vert _{L^{\infty }(\Omega )} }{c_\omega ^\theta } \int _{E}|\nabla u_n|^q\\\le & {} \frac{\Vert \varphi \Vert _{L^{\infty }(\Omega )} }{k^\theta } (\mathrm{meas}{\{ u_n> k\}})^{1-\frac{q}{r}} \Bigg (\int _{\{ u_n > k\}}|\nabla u_n|^r\Bigg )^{\frac{q}{r}}\\&+\frac{\Vert \varphi \Vert _{L^{\infty }(\Omega )} }{c_\omega ^\theta } \mathrm{meas}(E)^{1-\frac{q}{r}} \Bigg (\int _{E}|\nabla u_n|^r\Bigg )^{\frac{q}{r}}. \end{aligned}$$

Taking the limit as $\mathrm{meas}(E)$ tends to zero, k tends to infinity, using (3.20) and (3.21), and since $\frac{|\nabla u_n|^q}{u_n^{\theta }}$ converges to $\frac{|\nabla u|^q}{u^{\theta }}$ almost everywhere, we easily verify thanks to Vitali’s theorem that:

$$\begin{aligned} \frac{|\nabla u_n|^q}{u_n^{\theta }} \text { converge strongly to} \ \ \frac{|\nabla u|^q}{u^{\theta }}\ \text {in} \ L_\mathrm{loc}^1(\Omega ). \end{aligned}$$

(3.22)

Now, we take $\varphi \in C^1_{c}(\Omega )$ as test function in (2.1), and we have:

$$\begin{aligned} \int _{\Omega }M(x)\nabla u_n \nabla \varphi + \gamma \int _{\Omega } u_n^p \varphi = B\int _{\Omega } \frac{|\nabla u_n|^q}{(\frac{1}{n}|\nabla u_n|^q+1)(u_n+\frac{1}{n}))^\theta }\varphi +\int _{\Omega }f _n\varphi ; \end{aligned}$$

using (3.22), (3.21), and Proposition 3.4, we can pass to the limit with respect to n to conclude the proof of Theorem 1.1. $\square$