#  On an Anisotropic $p$-Laplace equation with variable singular exponent

Kaushik Bal, Prashanta Garain and Tuhina Mukherjee

###### Abstract.

In this article, we study the following anisotropic p-Laplacian equation with
variable exponent given by

$(P)\left\\{\begin{split}-\Delta_{H,p}u&=\frac{\lambda
f(x)}{u^{q(x)}}+g(u)\text{ in }\Omega,\\\ u&>0\text{ in }\Omega,\,u=0\text{ on
}\partial\Omega,\end{split}\right.$

under the assumption $\Omega$ is a bounded smooth domain in $\mathbb{R}^{N}$
with $p,N\geq 2$, $\lambda>0$ and $0<q\in C(\bar{\Omega})$. For the purely
singular case that is $g\equiv 0$, we proved existence and uniqueness of
solution. We also demonstrate the existence of multiple solution to $(P)$
provided $f\equiv 1$ and $g(u)=u^{r}$ for $r\in(p-1,p^{*}-1)$.

Key words: Anisotropic $p$-Laplace equation, Singular nonlinearity, Variable
exponent, Approximation method, Variational approach.

2010 Mathematics Subject Classification: 35D30, 35J62, 35J75.

##  1\. Introduction and Main results

In this article, we establish the existence of unique weak solution for the
problem

(1.1) $\left\\{\begin{split}-\Delta_{H,p}u&=\frac{f(x)}{u^{q(x)}}\text{ in
}\Omega,\\\ u&>0\text{ in }\Omega,\,u=0\text{ on
}\partial\Omega,\end{split}\right.$

and existence of at least two weak solutions for

(1.2)
$\left\\{\begin{split}-\Delta_{H,p}u&=\frac{\lambda}{u^{q(x)}}+u^{r}\text{ in
}\Omega,\\\ u&>0\text{ in }\Omega,\,u=0\text{ on
}\partial\Omega,\end{split}\right.$

where $\Omega$ is a bounded smooth domain in $\mathbb{R}^{N}$ with $N\geq 2$
and $0<q\in C(\bar{\Omega})$. Our other important assumptions for (1.1) are
$2\leq p<\infty$ and $f\in L^{1}(\Omega)\setminus\\{0\\}$ is a nonnegative
function (see Theorem 1.4-1.5 for precise assumptions). For the problem (1.2),
we assume the parameter $\lambda$ to be positive and $r\in(p-1,p^{*}-1)$,
where $p^{*}=\frac{Np}{N-p}$ is the critical Sobolev exponent for $1<p<N$
(refer Theorem 1.6-1.7). The key operator in our problems, that is the Finsler
p-Laplacian is defined as

$\Delta_{H,p}u:=\text{div}\big{(}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\big{)},$

where $\nabla_{\eta}$ denotes the gradient operator with respect to the $\eta$
variable and $H:\mathbb{R}^{N}\to[0,\infty)$ is known as the Finsler-Minkowski
norm satisfying the following hypothesis:

  1. (H0)

$H(x)\geq 0$, for every $x\in\mathbb{R}^{N}$.

  2. (H1)

$H(x)=0$, if and only if $x=0$.

  3. (H2)

$H(tx)=|t|H(x)$, for every $x\in\mathbb{R}^{N}$ and $t\in\mathbb{R}$.

  4. (H3)

$H\in C^{\infty}\left(\mathbb{R}^{N}\setminus\\{0\\}\right)$.

  5. (H4)

the Hessian matrix $\nabla_{\eta}^{2}\big{(}\frac{H^{2}}{2}\big{)}(x)$ is
positive definite for all $x\in\mathbb{R}^{N}\setminus\\{0\\}$.

The dual $H_{0}:\mathbb{R}^{N}\to[0,\infty)$ of $H$ is defined by

(1.3) $H_{0}(\xi):=\sup_{x\in\mathbb{R}^{N}\setminus\\{0\\}}\frac{\langle
x,\xi\rangle}{H(x)}.$

More details on this can be found in Bao-Chern-Shen [7], Xia [47] and
Rockafellar [45].

Anisotropic $p$-Laplace equations has been a topic of considerable attention
in the recent years. We refer to Alvino-Ferone-Trombetti-Lions [3], Belloni-
Ferone-Kawohl [8], Belloni-Kawohl-Juutinen [9], Bianchini-Giulio [10], Xia
[47], Cianchi-Salani[18], Ferone-Kawohl [25], Kawohl-Novaga [33] and the
references therein, for detailed analysis of the operator and its important
features. We present some remarks and examples now to give a little more
insight on the anisotropic p-Laplace operator and the adjoined norm.

######  Remark 1.1.

Since all norms in $\mathbb{R}^{N}$ are equivalent, there exist positive
constants $C_{1},C_{2}$ such that

$C_{1}|x|\leq H(x)\leq C_{2}|x|,\quad\forall\,x\in\mathbb{R}^{N}.$

We are able to define an equivalent norm on the Sobolev space
$W_{0}^{1,p}(\Omega)$ as

(1.4) $\|u\|_{X}:=\left(\int_{\Omega}H(\nabla
u)^{p}\,dx\right)^{\frac{1}{p}}.$

Examples: Let $x=(x_{1},x_{2},\ldots,x_{N})\in\mathbb{R}^{N}$,

  1. (i)

then, for $t>1$, we define

(1.5) $H_{t}(x):=\Big{(}\sum_{i=1}^{N}|x_{i}|^{t}\Big{)}^{\frac{1}{t}};$

  2. (ii)

for $\lambda,\mu>0$, we define

(1.6)
$H_{\lambda,\mu}(x):=\sqrt{\lambda\sqrt{\sum_{i=1}^{N}x_{i}^{4}}+\mu\sum_{i=1}^{N}x_{i}^{2}}.$

Then, one may check that the functions
$H_{t},H_{\lambda,\mu}:\mathbb{R}^{N}\to[0,\infty)$ given by (1.5) and (1.6)
satisfies all the hypothesis from $(H0)-(H4)$ or refer Mezei-Vas [36].

######  Remark 1.2.

For $i=1,2$ if $\lambda_{i},\mu_{i}$ are positive real numbers such that
$\frac{\lambda_{1}}{\mu_{1}}\neq\frac{\lambda_{2}}{\mu_{2}}$, then
$H_{\lambda_{1},\mu_{1}}$ and $H_{\lambda_{2},\mu_{2}}$ given by (1.6) defines
two non-isometric norms in $\mathbb{R}^{N}$.

######  Remark 1.3.

Moreover, for $H=H_{t}$ given by (1.5) we have

(1.7) $\Delta_{H_{t},p}u=\begin{cases}\Delta_{p}u:=\text{div}(|\nabla
u|^{p-2}\nabla u),\,(p\text{-Laplacian})\,\text{if }t=2,\,1<p<\infty,\\\
\mathcal{S}_{p}=\sum_{i=1}^{N}\frac{\partial}{\partial
x_{i}}\Big{(}|u_{i}|^{p-2}u_{i}\Big{)},\,\text{if
}t=p\in(1,\infty),\end{cases}$

where $u_{i}:=\frac{\partial u}{\partial x_{i}}$, for $i=1,2,\ldots,N$.

This last remark puts an important touch to our problem stating that Finsler
p-Laplacian generalizes the $p$-Laplacian. So our result are true for a
general set of quasilinear operators and we strongly point out here that (1.1)
and (1.2) are new even for Laplacian.

The equations (1.1) and (1.2) are singular in the sense that the
nonlinearities in our consideration blow up near the origin, due to the
presence of the term $u^{-q(x)}$ where $q$ is a positive function. The
fundamental feature of our article is the singular variable exponent
$q:\overline{\Omega}\to(0,\infty)$ which is a continuous function up to the
boundary and it is positive. Due to the blow up phenomenon, critical point
theory fails here and the standard technique of approximating the problem with
a non singular set up comes to our rescue. Another striking feature is the
singular exponent being a function here which covers both $0<q(x)<1$ and
$q(x)\geq 1$ cases together. In the $0<q(x)<1$ for all $x\in\Omega$, we may
employ some well known techniques to derive the weak solution. But when $q$ is
not bounded by $1$, under an additional assumption of restricting $q$ in
$(0,1]$ near the boundary of $\Omega$, we were able to show existence of
unique weak solution to (1.1) i.e. purely singular case. For (1.2), fixing
$0<q(x)<1$ we show that the variational techniques using the Gateaux
differentiability of corresponding energy functional can be advantageous to
show existence of at least two weak solutions.

When $q(x)=q$ is a fixed positive constant, singular problems has been
investigated in many different contexts in the recent years. In particular,
the $p$-Laplace equation

(1.8) $-\Delta_{p}u=h(x,u)\text{ in }\Omega,\,u>0\text{ in }\Omega,\,u=0\text{
on }\partial\Omega,$

has been studied concerning the existence, uniqueness and multiplicity results
widely for any $1<p<\infty$, including both purely singular and perturbed
singular nonlinearity $h$.

In the purely singular case, we refer the reader to Crandall-Rabinowitz-Tartar
[19], Boccardo-Orsina [13] for the semilinear case and De Cave [20], Canino-
Sciunzi-Trombetta [15] for the quasilinear case respectively. Whereas for the
perturbed singular case, one can look into Arcoya-Boccardo [4] for the
semilinear case and Giacomoni-Schindler-Takáč [31], Bal-Garain [5] for the
quasilinear case respectively. One may also refer to Garain-Mukherjee [26, 28]
and Leggat-Miri [35], dos Santos-Figueiredo-Tavares [21], Bal-Garain [6, 27]
for the study of singular problems in the context of weighted and anisotropic
$p$-Laplace operator respectively. Furthermore, we motivate readers to read
Ghergu-Rădulescu [30], Oliva-Orsina-Petitta [39, 40, 41, 42] for an extensive
study of singular elliptic problems.

As stated earlier, our main concern in this article is to investigate the
existence results in the presence of a variable singular exponent $q$. To the
best of our knowledge, the first article to deal with such phenomenon was by
Carmona-Aparicio [16], where the authors studied the following singular
Laplace equation

(1.9) $\begin{split}-\Delta u=u^{-q(x)}\text{ in }\Omega,\,u>0\text{ in
}\Omega,\,u=0\text{ on }\partial\Omega.\end{split}$

In addition to (1.9), for purely singular nonlinearity, Chu-Gao-Gao [17]
studied a more general class of second order elliptic equation; Miri [37]
studied anisotropic $p$-Laplace equations; Zhang [49] studied $p(x)$-Laplace
equation accomodating variable singular exponent. Alves-Santos-Siqueira [2]
obtained existence and uniqueness for a more general $p(x)$-Laplace equation.
For the existence results of a system of $p(x)$-Laplace equation, we refer
Alves-Moussaoui [1]. In the purturbed case, Byun-Ko [14] discussed
multiplicity for a class of $p(x)$-Laplace equation. Papageorgiou-Scapellato
[43] obtained existence results for a purely singular $(p(x),q(x))$-Laplace
equation. In the nonlocal setting, problem (1.9) has been investigated
recently by Garain-Mukherjee in [29]. For an extensive literature of variable
exponent problems, interested readers can go through Rădulescu-Repovš [46].

In the same context, we believe that Anisotropic $p$-Laplace equations with
singular nonlinearities have been very less understood and considered. We
refer to the recent articles by Biset-Mebrate-Mohammed [11], Farkas-Winkert
[24] and Farkas-Fiscella-Winkert [23] in this regard. We seek some ideas of
approximation from Boccardo-Orsina [13] and Carmona-Aparicio [16] for the
problem (1.1). On the other side, we obtain multiplicity results for the
problem (1.2) using the variational approach from Arcoya-Boccardo [4].

Notation: We gather below all the notations that has been frequently used in
our article-

  * •

$X:=W_{0}^{1,p}(\Omega)$.

  * •

For $u\in X$, we denote by $\|u\|$ to mean $\|u\|_{X}$ which is defined by
(1.4).

  * •

For a given constant $c$ and a set $S$, by $u\geq c$ in $S$, we mean $u\geq c$
almost everywhere in $S$. Moreover, we write $|S|$ to denote the Lebesgue
measure of $S$.

  * •

$\langle,\rangle$ denotes the standard inner product in $\mathbb{R}^{N}$.

  * •

The conjugate exponent of $\theta>1$ is given by
$\theta^{\prime}:=\frac{\theta}{\theta-1}$.

  * •

For $1<p<N$, we denote by $p^{*}:=\frac{Np}{N-p}$ to mean the critical Sobolev
exponent.

  * •

For $a\in\mathbb{R}$, we denote by $a^{+}:=\max\\{a,0\\}$ and
$a_{-}:=\min\\{a,0\\}$.

  * •

We write by $c,C$ or $C_{i}$ for $i\in\mathbb{N}$ to mean a positive constant
which may vary from line to line or even in the same line. If a constant $C$
depends on $r_{1},r_{2},\ldots$, we denote it by $C(r_{1},r_{2},\ldots)$.

Now, we define the notion of weak solutions as follows.
Weak solution: We say that $u\in X$ is a weak solution of the problem (1.1),
if $u>0$ in $\Omega$ and for every $\omega\Subset\Omega$, there exists a
positive constant $c_{\omega}$ such that $u\geq c_{\omega}>0$ in $\omega$,
satisfying

(1.10) $\int_{\Omega}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\nabla\phi\,dx=\int_{\Omega}\frac{f(x)}{u^{q(x)}}\phi\,dx,\quad\forall\,\phi\in
C_{c}^{1}(\Omega).$

Analogously, we say that $u\in X$ is a weak solution of the problem (1.2), if
$u>0$ in $\Omega$ and for every $\omega\Subset\Omega$, there exists a positive
constant $c_{\omega}$ such that $u\geq c_{\omega}>0$ in $\omega$, satisfying

(1.11) $\int_{\Omega}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\nabla\phi\,dx=\int_{\Omega}\Big{(}\frac{\lambda}{u^{q(x)}}+u^{r}\Big{)}\phi\,dx,\quad\forall\,\phi\in
C_{c}^{1}(\Omega).$

Our main results in this article reads includes the following.

######  Theorem 1.4.

Let $2\leq p<\infty$ and $q\in C(\overline{\Omega})$ be positive.

  1. (a)

Uniqueness: Then for any nonnegative $f\in L^{1}(\Omega)\setminus\\{0\\}$, the
problem (1.1) admits at most one weak solution in $X$.

  2. (b)

Existence: Assume that there exists a constant $\delta>0$ such that
$0<q(x)\leq 1$ in $\Omega_{\delta}$, where

$\Omega_{\delta}:=\\{x\in\Omega:\text{dist}\,(x,\partial\Omega)<\delta\\}.$

Then the problem $(\ref{maineqn})$ admits a unique weak solution $u\in X$, for
any nonnegative $f\in L^{m}(\Omega)\setminus\\{0\\}$, where

$m=\begin{cases}(p^{*})^{\prime},\text{ if }2\leq p<N,\\\ >1,\text{ if
}p=N,\\\ 1,\text{ if }p>N.\end{cases}$

In particular, we have the following existence and uniqueness result for any
$1<p<\infty$.

######  Theorem 1.5.

If

$\Delta_{H,p}\equiv\Delta_{p}\text{ or }\mathcal{S}_{p},$

as given by (1.7), then Theorem 1.4 holds for any $1<p<\infty$.

######  Theorem 1.6.

Let $2\leq p<N$ and $q\in C(\overline{\Omega})$ be such that $0<q(x)<1$ for
all $x\in\overline{\Omega}$. Then there exists $\Lambda>0$ such that for all
$\lambda\in(0,\Lambda)$, the problem $\eqref{maineqn1}$ admits at least two
distinct weak solutions in $X$, for any $r\in(p-1,p^{*}-1)$.

In particular, we have the following multiplicity result for any $1<p<N$.

######  Theorem 1.7.

If

$\Delta_{H,p}\equiv\Delta_{p}\text{ or }\mathcal{S}_{p},$

as given by (1.7), then Theorem 1.6 holds for any $1<p<N$.

We have segmented our paper as follows: In Section $2$, we discuss some
auxiliary results in our functional setting. In Section $3$ and Section $4$,
we obtain some preliminary results for the proof of Theorem 1.4, Theorem 1.5
and Theorem 1.6, Theorem 1.7 respectively. Finally, the proof of all main
results has been collected in Section $5$.

##  2\. Auxiliary results

This section is devoted to preliminary ideas which will be helpful to advance
towards establishing our principle results. We fix our assumption to
$1<p<\infty$, unless otherwise mentioned. We state some results below which
are taken from Farkas-Winkert [24, Proposition 2.1] and Xia [47, Proposition
1.2].

######  Lemma 2.1.

For every $x\in\mathbb{R}^{N}\setminus\\{0\\}$ and
$t\in\mathbb{R}\setminus\\{0\\}$ we have

  1. (A)

$x\cdot\nabla_{\eta}H(x)=H(x)$.

  2. (B)

$\nabla_{\eta}H(tx)=\text{sign}(t)\nabla_{\eta}H(x)$.

  3. (C)

$|\nabla_{\eta}H(x)|\leq C$, for some positive constant $C$.

  4. (D)

$H$ is stricly convex.

######  Corollary 2.2.

On account of Lemma 2.1 and Remark 1.1, we have the following relations-

(2.1) $H(x)^{p-1}\nabla_{\eta}H(x)\cdot x=H(x)^{p}\geq
C_{1}|x|^{p},\quad\forall\,x\in\mathbb{R}^{N},$ (2.2)
$\big{|}H(x)^{p-1}\nabla_{\eta}H(x)\big{|}\leq
C_{2}|x|^{p-1},\quad\forall\,x\in\mathbb{R}^{N}\text{ and}$ (2.3)
$H(tx)^{p-1}\nabla_{\eta}H(tx)=|t|^{p-2}tH(x)^{p-1}\nabla_{\eta}H(x),\quad\forall\,x\in\mathbb{R}^{N}\text{
and }t\in\mathbb{R}\setminus\\{0\\}.$

The following result has been adapted from Ohta [38, Proposition 4.6].

######  Lemma 2.3.

For every $x,y\in\mathbb{R}^{N}$, there exists a constant $C\geq 1$ such that

(2.4)
$H\left(\frac{x+y}{2}\right)^{2}+\frac{1}{4C^{2}}H(x-y)^{2}\leq\frac{H(x)^{2}+H(y)^{2}}{2}.$

Lemma 2.4-2.5 given below follows from arguments given in the proof of Xia
[48, Lemma $3.1-3.2$].

######  Lemma 2.4.

Let $2\leq p<\infty$. Then, for every $x,y\in\mathbb{R}^{N}$,

(2.5) $H(x)^{p}\geq H(y)^{p}+pH(y)^{p-1}\nabla_{\eta}H(y)(x-y).$

Moreover, there exists a positive constant $c=c(C,p)$, where $C$ is given by
Lemma 2.3 such that for every $x,y\in\mathbb{R}^{N}$,

(2.6) $\begin{split}H(x)^{p}&\geq
H(y)^{p}+pH(y)^{p-1}\nabla_{\eta}H(y)(x-y)+cH(x-y)^{p}.\end{split}$

###### Proof.

Firstly, we observe that $H^{p}$ is convex. Indeed, for any
$x,y\in\mathbb{R}^{N}$ and $s\in[0,1]$, by the convexity of $H$ (follows from
Lemma 2.1), we have

$\begin{split}H(sx+(1-s)y)^{p}&\leq\left(sH(x)+(1-s)H(y)\right)^{p}\\\ &\leq
sH(x)^{p}+(1-s)H(y)^{p},\end{split}$

where in the final step above, we have used the convexity of $|\cdot|^{p}$.
Hence the estimate (2.5) follows. Next we prove (2.6). For any $a,b\geq 0$ and
$p\geq 2$, we have the following elementary inequality (see Xia [48])

(2.7) $(a^{p}+b^{p})\leq(a^{2}+b^{2})^{\frac{p}{2}}.$

Let $C\geq 1$ be given by Lemma 2.3. Choosing $a=H\big{(}\frac{x+y}{2}\big{)}$
and $b=H\big{(}\frac{x-y}{2C}\big{)}$ in (2.7) and using Lemma 2.3 we get

(2.8)
$\begin{split}H\left(\frac{x+y}{2}\right)^{p}+H\left(\frac{x-y}{2C}\right)^{p}&\leq\left(\frac{H(x+y)^{2}}{4}+\frac{1}{4C^{2}}H(x-y)^{2})\right)^{\frac{p}{2}}\\\
&\leq\left(\frac{H(x)^{2}+H(y)^{2}}{2}\right)^{\frac{p}{2}}\leq\frac{H(x)^{p}}{2}+\frac{H(y)^{p}}{2},\end{split}$

where in the last line above we have used the convexity of
$|\cdot|^{\frac{p}{2}}$ for $p\geq 2$. Moreover, by (2.5), we have

(2.9) $H\left(\frac{x+y}{2}\right)^{p}\geq
H(y)^{p}+\frac{p}{2}H(y)^{p-1}\nabla_{\eta}H(y)(x-y).$

Hence using (2.9) in (2.8) the estimate (2.6) follows. ∎

Our next Lemma is a well know identity for the $p$-Laplacian operators, we
prove it for Finsler $p$-Laplacian (a general case).

######  Lemma 2.5.

Let $2\leq p<\infty$. Then, for every $x,y\in\mathbb{R}^{N}$, there exists a
positive constant $c=c(C,p)$, where $C$ is given by Lemma 2.3 such that

(2.10) $\begin{split}\langle
H(x)^{p-1}\nabla_{\eta}H(x)-H(y)^{p-1}\nabla_{\eta}H(y),x-y\rangle&\geq
cH(x-y)^{p}.\end{split}$

###### Proof.

From (2.6) in Lemma 2.4, for every $x,y\in\mathbb{R}^{N}$ we have

(2.11) $\begin{split}H(x)^{p}&\geq
H(y)^{p}+pH(y)^{p-1}\nabla_{\eta}H(y)(x-y)+cH(x-y)^{p},\end{split}$

and

(2.12) $\begin{split}H(y)^{p}&\geq
H(x)^{p}+pH(x)^{p-1}\nabla_{\eta}H(x)(y-x)+cH(y-x)^{p}.\end{split}$

Adding (2.11), (2.12) and using $(H2)$ the estimate (2.10) follows. ∎

######  Remark 2.6.

When $2\leq p<\infty$ and $H(x)=H_{2}(x)=|x|$ as given by (1.5), Lemma 2.5
coincides with the well-known algebraic inequality from Lemma 2.12.

######  Corollary 2.7.

As a consequence of Lemma 2.5, we have

(2.13) $\langle
H(x)^{p-1}\nabla_{\eta}H(x)-H(y)^{p-1}\nabla_{\eta}H(y),x-y\rangle>0,\quad\forall\,x\neq
y\in\mathbb{R}^{N}.$

The following inequalities has been borrowed from[47, Proposition $1.3$ and
$(1.2)$].

######  Lemma 2.8.

  1. (D1)

Cauchy-Schwartz inequality:

(2.14) $\langle x,\xi\rangle\leq
H(x)H_{0}(\xi),\quad\forall\,x,\xi\in\mathbb{R}^{N}.$

  2. (D2)

For all $x\in\mathbb{R}^{N}\setminus\\{0\\}$, it holds

(2.15) $H_{0}\big{(}\nabla H(x)\big{)}=1.$

Below is an important output of above inequalities.

######  Corollary 2.9.

For any $x,y\in\mathbb{R}^{N}$, it holds that

(2.16) $\langle
H(x)^{p-1}\nabla_{\eta}H(x)-H(y)^{p-1}\nabla_{\eta}H(y),x-y\rangle\geq\left(H(x)^{p-1}-H(y)^{p-1}\big{)}\big{(}H(x)-H(y)\right).$

###### Proof.

Using Lemma 2.8 we observe that

(2.17) $H(x)^{p-1}\langle\nabla_{\eta}H(x),y\rangle\leq
H(x)^{p-1}H_{0}(\nabla_{\eta}H(x))H(y)=H(x)^{p-1}H(y)\text{ and}$ (2.18)
$H(y)^{p-1}\langle\nabla_{\eta}H(y),x\rangle\leq
H(y)^{p-1}H_{0}(\nabla_{\eta}H(y))H(x)=H(y)^{p-1}H(x),$

holds for every $x,y\in\mathbb{R}^{N}$. Hence using (2.17) and (2.18), for all
$x,y\in\mathbb{R}^{N}$ we obtain

$\begin{split}&\langle
H(x)^{p-1}\nabla_{\eta}H(x)-H(y)^{p-1}\nabla_{\eta}H(y),x-y\rangle\\\
&=H(x)^{p}+H(y)^{p}-H(x)^{p-1}\langle\nabla_{\eta}H(x),y\rangle-H(y)^{p-1}\langle\nabla_{\eta}H(y),x\rangle\\\
&\geq H(x)^{p-1}+H(y)^{p-1}-H(x)^{p-1}H(y)-H(y)^{p-1}H(x)\\\
&=\big{(}H(x)^{p-1}-H(y)^{p-1}\big{)}\big{(}H(x)-H(y)\big{)}.\end{split}$

∎

The following result follows from Belloni-Ferone-Kawohl [8, Theorem 3.1].

######  Lemma 2.10.

There exists a positive eigenfunction $e_{1}\in X\cap L^{\infty}(\Omega)$ such
that $\|e_{1}\|_{L^{\infty}(\Omega)}=1$ corresponding to the first eigenvalue
$\lambda_{1}>0$ satisfying the equation

(2.19) $-\Delta_{H,p}v=\lambda_{1}|v|^{p-2}v\text{ in }\Omega,\;\;v=0\text{ in
}\partial\Omega.$

For the following Sobolev embedding, see Evans [22].

######  Lemma 2.11.

The inclusion map

$W_{0}^{1,p}(\Omega)\hookrightarrow\begin{cases}L^{t}(\Omega),\text{ for
}t\in[1,p^{*}],\text{ if }1<p<N,\\\ L^{t}(\Omega),\text{ for
}t\in[1,\infty),\text{ if }p=N,\\\ C(\overline{\Omega}),\text{ if
}p>N,\end{cases}$

is continuous. Moreover, the above mapping is compact except for $r=p^{*}$,
when $1<p<N$.

Next, we state the algebraic inequality from Peral [44, Lemma A.0.5].

######  Lemma 2.12.

For any $a,b\in\mathbb{R}^{N}$, there exists a constant $C=C(p)>0$, such that

(2.20)
$\langle|a|^{p-2}a-|b|^{p-2}b,a-b\rangle\geq\begin{cases}C|a-b|^{p},\text{ if
}2\leq p<\infty,\\\ C\frac{|a-b|^{2}}{(|a|+|b|)^{2-p}},\text{ if
}1<p<2.\end{cases}$

Our next result ensures that test functions in (1.10) and (1.11) can be chosen
from the space $X$ itself.

######  Lemma 2.13.

Let $u\in X$ be a weak solution of the problem (1.1) or (1.2), then (1.10) or
(1.11) holds, for every $\phi\in X$ respectively.

###### Proof.

Let $u\in X$ solves the problem $\eqref{maineqn}$ or (1.2). Suppose

$g(x,u)=f(x)u^{-q(x)}\text{ or }\lambda u^{-q(x)}+u^{r}.$

Therefore, for every $\phi\in C_{c}^{1}(\Omega)$, from (1.10) and (1.11) we
have

(2.21) $\int_{\Omega}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\nabla\phi\,dx=\int_{\Omega}g(x,u)\phi\,dx.$

Now for every $\psi\in X$, there exists a sequence of function
$0\leq\psi_{n}\in C_{c}^{1}(\Omega)\to|\psi|$ strongly in $X$ as $n\to\infty$
and pointwise almost everywhere in $\Omega$. We observe that

(2.22)
$\begin{split}\Big{|}\int_{\Omega}g(x,u)\psi\,dx\Big{|}&\leq\int_{\Omega}g(x,u)|\psi|\,dx\leq\liminf_{n\to\infty}\int_{\Omega}g(x,u)\psi_{n}\,dx\\\
&=\liminf_{n\to\infty}<-\Delta_{H,p}u,\psi_{n}>\leq
C\|u\|^{p-1}\lim_{n\to\infty}\|\psi_{n}\|\\\ &\leq C\|u\|^{p-1}\||\psi|\|\leq
C\|u\|^{p-1}\|\psi\|,\end{split}$

for some positive constant $C$. Let $\phi\in X$, then there exists a sequence
$\\{\phi_{n}\\}\subset C_{c}^{1}(\Omega)$ which converges to $\phi$ strongly
in $X$. We claim that

$\lim_{n\to\infty}\int_{\Omega}g(x,u)\phi_{n}\,dx=\int_{\Omega}g(x,u)\phi\,dx.$

Indeed, using $\psi=\phi_{n}-\phi$ in (2.22), we obtain

(2.23)
$\lim_{n\to\infty}\Big{|}\int_{\Omega}g(x,u)(\phi_{n}-\phi)\,dx\Big{|}\leq
C||u||^{p-1}\lim_{n\to\infty}||\phi_{n}-\phi||=0.$

Again, since $\phi_{n}\to\phi$ strongly in $X$ as $n\to\infty$, we have

(2.24) $\lim_{n\to\infty}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\nabla(\phi_{n}-\phi)\,dx=0.$

Hence, using (2.23) and (2.24) in (2.21) the result follows. ∎

##  3\. Preliminaries for the proof of Theorem 1.4 and Theorem 1.5

We present proof of our first main result in this section. Here again, we
assume $1<p<\infty$, unless otherwise mentioned. For $n\in\mathbb{N}$ and
nonnegative $f\in L^{1}(\Omega)\setminus\\{0\\}$, let us define
$f_{n}(x):=\min\\{f(x),n\\}$ and consider the following approximated problem

(3.1) $\left.\begin{aligned}
-\Delta_{H,p}u&=\frac{f_{n}(x)}{(u^{+}+\frac{1}{n})^{q(x)}}\text{ in
}\Omega,\\\ u&=0\text{ on }\partial\Omega.\end{aligned}\right\\}$

First, we prove the following useful result.

######  Lemma 3.1.

Let $g\in L^{\infty}(\Omega)\setminus\\{0\\}$ be a nonnegative function in
$\Omega$. Then there exists a unique solution $u\in X\cap L^{\infty}(\Omega)$
of the problem

(3.2) $\begin{split}-\Delta_{H,p}u=g\text{ in }\Omega,\,u>0\text{ in
}\Omega,\,u=0\text{ on }\partial\Omega,\end{split}$

such that for every $\omega\Subset\Omega$, there exists a constant
$c_{\omega}$ satisfying $u\geq c_{\omega}>0$ in $\omega$.

###### Proof.

Existence: We define the energy functional $J:X\to\mathbb{R}$ by

$J(u):=\frac{1}{p}\int_{\Omega}H(\nabla u)^{p}\,dx-\int_{\Omega}gu\,dx.$

Noting Lemma 2.1 and Lemma 2.11, it can be easily seen that $J$ is coercive,
weakly lower semicontinuous and a strictly convex functional. Therefore, $J$
has a unique minimizer $u$ in $X$. Since $J\in C^{1}(X)$, so $u\in X$ is the
unique solution of the equation

(3.3) $-\Delta_{H,p}u=g\text{ in }\Omega,\,u=0\text{ on }\partial\Omega.$

Noting $g\geq 0$ and choosing $u_{-}:=\min\\{u,0\\}$ as a test function in
(3.3), by (2.1) we obtain

$\begin{split}C_{1}\int_{\Omega}|\nabla
u_{-}|^{p}\,dx&\leq\int_{\Omega}H\big{(}\nabla
u_{-}\big{)}^{p}\,dx=\int_{\Omega}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\nabla u_{-}\,dx=\int_{\Omega}gu_{-}\,dx\leq 0,\end{split}$

which gives, $u\geq 0$ in $\Omega$.
Boundedness: For any $k\geq 1$, we define a subset of $\Omega$ given by

$A(k):=\\{x\in\Omega:u(x)\geq k\text{ in }\Omega\\}.$

Choosing $\phi_{k}:=(u-k)^{+}=\max\\{u-k,0\\}$ as a test function in (3.3) and
using the continuity of the mapping $X\hookrightarrow L^{l}(\Omega)$ for some
$l>p$ from Lemma 2.11 along with (2.1) we obtain

$\begin{split}\int_{\Omega}H(\nabla\phi_{k})^{p}\,dx=\int_{\Omega}g\phi_{k}\leavevmode\nobreak\
dx\leq\|g\|_{L^{\infty}(\Omega)}\int_{A(k)}(u-k)\,dx\leq
C_{0}\|g\|_{L^{\infty}(\Omega)}|A(k)|^{\frac{l-1}{l}}\|\phi_{k}\|,\end{split}$

where $C_{0}$ is the Sobolev constant. Hence, we have

(3.4) $\|\phi_{k}\|^{p-1}\leq C|A(k)|^{\frac{l-1}{l}},$

for some positive constant $C$ depending on $\|g\|_{L^{\infty}(\Omega)}$. We
choose $h$ such that $1<k<h$ then using (3.4), $(u(x)-k)^{p}\geq(h-k)^{p}$ on
$A(h)$ and $A(h)\subset A(k)$ we get

$\displaystyle(h-k)^{p}|A(h)|^{\frac{p}{l}}$
$\displaystyle\leq\left(\int_{A(h)}(u(x)-k)^{l}\,dx\right)^{\frac{p}{l}}\leq\left(\int_{A(k)}(u(x)-k)^{l}\leavevmode\nobreak\
dx\right)^{\frac{p}{l}}$ $\displaystyle\leq\|\phi_{k}\|^{p}\leq
C|A(k)|^{\frac{p(l-1)}{l(p-1)}}.$

Hence, we obtain

$|A(h)|\leq\frac{C}{(h-k)^{l}}|A(k)|^{\frac{l-1}{p-1}}.$

Therefore, observing that $\frac{l-1}{p-1}>1$, using Kinderlehrer-Stampacchia
[34, Lemma B.1] we have

$\|u\|_{L^{\infty}(\Omega)}\leq C,$

for some positive constant $C$ depending on $\|g\|_{L^{\infty}(\Omega)}$.
Positivity: Moreover since $g\neq 0$, we have $u\neq 0$ in $\Omega$. Noting
Corollary 2.2 and Corollary 2.7, we can apply Heinonen-Kilpeläine-Martio [32,
Theorem 3.59] so that for every $\omega\Subset\Omega$, there exists a constant
$c_{\omega}>0$ such that $u\geq c_{\omega}>0$ in $\Omega$. Thus $u>0$ in
$\Omega$. ∎

Then, we have the following result concerning the problem (3.1).

######  Lemma 3.2.

Let $2\leq p<\infty$. Then for every $n\in\mathbb{N}$, there exists a unique
positive solution $u_{n}\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$ to the
problem (3.1) such that $u_{n+1}\geq u_{n}$ in $\Omega$, for every
$n\in\mathbb{N}$. Moreover for every $\omega\Subset\Omega$, there exists a
constant $c_{\omega}>0$ (independent of $n$) such that $u_{n}\geq
c_{\omega}>0$ in $\omega$.

###### Proof.

Step $1$. (Existence) We observe that for every $n\in\mathbb{N}$ and $v\in
L^{p}(\Omega)$, the function

$g_{n}(x):=\frac{f_{n}(x)}{\big{(}v^{+}+\frac{1}{n}\big{)}^{q(x)}}\in
L^{\infty}(\Omega)\setminus\\{0\\}$

is nonnegative. Therefore, by Lemma 3.1 we can define the map
$T:L^{p}(\Omega)\to L^{p}(\Omega)$ by

$T(v)=w_{n},$

where $w_{n}\in X\cap L^{\infty}(\Omega)$ is the unique solution to the
problem

(3.5) $\left.\begin{aligned}
-\Delta_{H,p}w_{n}&=\frac{f_{n}(x)}{(v^{+}+\frac{1}{n})^{q(x)}}\text{ in
}\Omega,\\\ w_{n}&>0\text{ in }\Omega,\,w_{n}=\text{ on
}\partial\Omega,\end{aligned}\right\\}$

such that for every $\omega\Subset\Omega$ there exists a positive constant
$c_{\omega}$ satisfying $w_{n}\geq c_{\omega}>0$ in $\omega$. Now, using
$w_{n}$ as a test function in the problem (3.5) and the fact that $q\in
C(\overline{\Omega})$, we obtain

(3.6) $\|w_{n}\|\leq Cn^{\frac{\|q\|_{L^{\infty}(\Omega)}+1}{p-1}},$

for some constant $C=C(p,N,\Omega)$. We define

$S:=\\{v\in L^{p}(\Omega):\lambda T(v)=v\text{ for }0\leq\lambda\leq 1\\}.$

Indeed, let $v_{1},v_{2}\in S$. Then using (3.6) and Lemma 2.11

$\begin{split}\|v_{1}-v_{2}\|_{L^{p}(\Omega)}&=\lambda\|T(v_{1})-T(v_{2})\|_{L^{p}(\Omega)}\\\
&\leq 2\lambda CC_{1}n^{\frac{\|q\|_{L^{\infty}(\Omega)}+1}{p-1}},\end{split}$

where $C=C(p,N,\Omega)$ is given by (3.6) and $C_{1}$ is the Sobolev constant.
Therefore, the set $S$ is bounded in $L^{p}(\Omega)$. To prove the continuity
of $T$, let $v_{k}\to v$ strongly in $L^{p}(\Omega)$ and $T(v_{k})=w_{n,k}$.
Then, we have

(3.7) $\int_{\Omega}H(\nabla w_{n,k})^{p-1}\nabla_{\eta}H(\nabla
w_{n,k})\nabla(w_{n,k}-u_{n})\,dx=\int_{\Omega}\frac{f_{n}(x)}{(v_{k}^{+}+\frac{1}{n})^{q(x)}}(w_{n,k}-w_{n})\,dx,$

and

(3.8) $\int_{\Omega}H(\nabla w_{n,k})^{p-1}\nabla_{\eta}H(\nabla
w_{n,k})\nabla(w_{n,k}-w_{n})\,dx=\int_{\Omega}\frac{f_{n}(x)}{(v^{+}+\frac{1}{n})^{q(x)}}(w_{n,k}-w_{n})\,dx.$

Subtracting (3.8) from (3.7), since $2\leq p<\infty$, we can use Lemma 2.5 to
obtain

(3.9)
$\begin{split}&\int_{\Omega}f_{n}(x)\left\\{\left(v_{k}^{+}+\frac{1}{n}\right)^{-q(x)}-\left(v^{+}+\frac{1}{n}\right)^{-q(x)}\right\\}(w_{n,k}-w_{n})\,dx\\\
&=\int_{\Omega}\left\\{H(\nabla w_{n,k})^{p-1}\nabla_{\eta}H(\nabla
w_{n,k})-H(\nabla w_{n})^{p-1}\nabla_{\eta}H(\nabla
w_{n})\right\\}\nabla(w_{n,k}-w_{n})\,dx\\\ &\geq
c\int_{\Omega}H\left(\nabla(w_{n,k}-w_{n})\right)^{p}\,dx,\end{split}$

for some positive constant $c$. From Hölder’s inequality, we obtain

(3.10)
$\begin{split}&\int_{\Omega}f_{n}(x)\left\\{\big{(}v_{k}^{+}+\frac{1}{n}\big{)}^{-q(x)}-\left(v^{+}+\frac{1}{n}\right)^{-q(x)}\right\\}(w_{n,k}-w_{n})\,dx\\\
&\leq
n\left(\int_{\Omega}\left|{\left(v_{k}^{+}+\frac{1}{n}\right)^{-q(x)}}-\left(v^{+}+\frac{1}{n}\right)^{-q(x)}\right|^{\frac{p}{p-1}}\,dx\right)^{\frac{p-1}{p}}\|w_{n,k}-w_{n}\|_{L^{p}(\Omega)}.\end{split}$

Since $v_{k}\to v$ strongly in $L^{p}(\Omega)$, it follows that upto a
subsequence $v_{k}\to v$ almost everywhere in $\Omega$. Also, it is easy to
observe that

$\Big{|}\big{(}v_{k}^{+}+\frac{1}{n}\big{)}^{-q(x)}-\big{(}v^{+}+\frac{1}{n}\big{)}^{-q(x)}\Big{|}\leq
2n^{||q||_{L^{\infty}(\Omega)}}.$

Since the limit is independent of the choice of the subsequence, applying the
Lebsegue dominated convergence theorem, we obtain

(3.11)
$\lim_{k\to\infty}\int_{\Omega}\left|\left(v_{k}^{+}+\frac{1}{n}\right)^{-q(x)}-\left(v^{+}+\frac{1}{n}\right)^{-q(x)}\right|^{\frac{p}{p-1}}\,dx=0.$

Thus using (3.9) along with (3.10) and (3.11) it follows that $w_{n,k}\to
w_{n}$ strongly in $X$. As a consequence, the mapping $T$ is continuous. The
compactness of $T$ follows from the fact (3.6) and the compactness of the
mapping $X\hookrightarrow L^{p}(\Omega)$ which follows from Lemma 2.11. Thus
by the Schauder fixed point theorem, $T$ has a fixed point, say $u_{n}\in X$
which solves the problem (3.1). Note that $u_{n}\geq 0$ in $\Omega$.
Step $2$. (Monotonicity) Choosing
$(u_{n}-u_{n+1})^{+}:=\max\\{u_{n}-u_{n+1},0\\}$ as a test function in (3.1)
we have

(3.12) $\int_{\Omega}H(\nabla u_{n})^{p-1}\nabla_{\eta}H(\nabla
u_{n})\nabla(u_{n}-u_{n+1})^{+}\,dx=\int_{\Omega}\frac{f_{n}(x)}{\left(u_{n}+\frac{1}{n}\right)^{q(x)}}(u_{n}-u_{n+1})^{+}\,dx,$

and

(3.13) $\int_{\Omega}H(\nabla u_{n+1})^{p-1}\nabla_{\eta}H(\nabla
u_{n+1})\nabla(u_{n}-u_{n+1})^{+}\,dx=\int_{\Omega}\frac{f_{n+1}(x)}{\left(u_{n+1}+\frac{1}{n+1}\right)^{q(x)}}(u_{n}-u_{n+1})^{+}\,dx.$

Subtracting (3.13) from (3.12), since $2\leq p<\infty$, using Lemma 2.5 we
obtain

(3.14)
$\begin{split}&\int_{\Omega}\left\\{\frac{f_{n}(x)}{\left(u_{n}+\frac{1}{n}\right)^{q(x)}}-\frac{f_{n+1}(x)}{\left(u_{n+1}+\frac{1}{n+1}\right)^{q(x)}}\right\\}(u_{n}-u_{n+1})^{+}\,dx\\\
&=\int_{\Omega}\left\\{H(\nabla u_{n})^{p-1}\nabla_{\eta}H(\nabla
u_{n})-H(\nabla u_{n+1})^{p-1}\nabla_{\eta}H(\nabla
u_{n+1})\right\\}\nabla(u_{n}-u_{n+1})^{+}\,dx\\\ &\geq
c\int_{\Omega}H\left(\nabla(u_{n}-u_{n+1})^{+}\right)^{p}\,dx,\end{split}$

for some positive constant $c$. Since $f_{n}(x)\leq f_{n+1}(x)$ for almost
every $x\in\Omega$, we get

(3.15)
$\begin{split}&\int_{\Omega}\left\\{\frac{f_{n}(x)}{\big{(}u_{n}+\frac{1}{n}\big{)}^{q(x)}}-\frac{f_{n+1}(x)}{\big{(}u_{n+1}+\frac{1}{n+1}\big{)}^{q(x)}}\right\\}(u_{n}-u_{n+1})^{+}\,dx\\\
&\leq\int_{\Omega}f_{n+1}(x)\frac{\big{(}u_{n+1}+\frac{1}{n+1}\big{)}^{q(x)}-\big{(}u_{n}+\frac{1}{n}\big{)}^{q(x)}}{\big{(}u_{n}+\frac{1}{n}\big{)}^{q(x)}\big{(}u_{n+1}+\frac{1}{n+1}\big{)}^{q(x)}}(u_{n}-u_{n+1})^{+}\,dx\leq
0.\end{split}$

Hence from (3.14) and (3.15), we have $u_{n+1}\geq u_{n}$ in $\Omega$.
Step $3$. (Uniqueness) Let $n\in\mathbb{N}$ and $u_{n},v_{n}\in X\cap
L^{\infty}(\Omega)$ solves the problem (3.1). Then choosing
$(u_{n}-v_{n})^{+}:=\max\\{u_{n}-v_{n},0\\}$ as a test function in (3.1) and
proceeding similarly as the proof of Step $2$ above, we get $u_{n}\geq v_{n}$
in $\Omega$. Choosing $(v_{n}-u_{n})^{+}:=\max\\{v_{n}-u_{n},0\\}$ as a test
function in (3.1), we get $v_{n}\geq u_{n}$ in $\Omega$. This gives that
$u_{n}=v_{n}$ in $\Omega$.
Step $4$. (Positivity) From Step $1$, we infer that there exists a constant
$c_{\omega}>0$ such that $u_{1}\geq c_{\omega}>0$, for every
$\omega\Subset\Omega$. Hence taking into account Step $2$, we have $u_{n}\geq
c_{\omega}>0$ in $\omega$ for every $\omega\Subset\Omega$, where
$c_{\omega}>0$ is independent of $n$. Thus $u_{n}>0$ in $\Omega$ for all
$n\in\mathbb{N}$, from which the result follows. ∎

######  Remark 3.3.

We emphasize that if $\Delta_{H,p}=\Delta_{p}$ or $\mathcal{S}_{p}$ as given
by (1.7), then noting Lemma 2.12, the results in Lemma 3.2 will hold for any
$1<p<\infty$.

##  4\. Preliminaries for the proof of Theorem 1.6 and Theorem 1.7

Throughout this section, we assume $1<p<N$, unless otherwise mentioned. Here
we prove some preliminary results required to prove Theorem 1.6-1.7.

Let us denote the energy functional
$I_{\lambda}:X\to\mathbb{R}\cup\\{\pm\infty\\}$ corresponding to the problem
(1.2) by

$I_{\lambda}(u):=\frac{1}{p}\int_{\Omega}H(\nabla
u)^{p}\,dx-\lambda\int_{\Omega}\frac{(u^{+})^{1-q(x)}(x)}{1-q(x)}\leavevmode\nobreak\
dx-\frac{1}{r+1}\int_{\Omega}(u^{+})^{r+1}\leavevmode\nobreak\ dx.$

Now for $\epsilon>0$, we consider the following approximated problem

$(P_{\lambda,\epsilon})\left\\{\begin{aligned}
-\Delta_{H,p}u&=\frac{\lambda}{(u^{+}+\epsilon)^{q(x)}}+(u^{+})^{r}\;\text{in}\;\Omega,\\\
u&=0\;\text{ on }\;\partial\Omega,\end{aligned}\right.$

for which the corresponding energy functional is given by

$I_{\lambda,\epsilon}(u)=\frac{1}{p}\int_{\Omega}H(\nabla
u)^{p}\,dx-\lambda\int_{\Omega}\frac{[(u^{+}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}]}{1-q(x)}\leavevmode\nobreak\
dx-\frac{1}{r+1}\int_{\Omega}(u^{+})^{r+1}\leavevmode\nobreak\ dx.$

It is easy to verify that $I_{\lambda,\epsilon}\in C^{1}(X,\mathbb{R})$,
$I_{\lambda,\epsilon}(0)=0$ and $I_{\lambda,\epsilon}(v)\leq
I_{0,\epsilon}(v)$, for all $v\in X$.

Our next Lemma states that $I_{\lambda,\epsilon}$ satisfies the Mountain Pass
Geometry.

######  Lemma 4.1.

There exists $R>0,\,\rho>0$ and $\Lambda>0$ depending on $R$ such that

$\inf\limits_{\|v\|\leq
R}I_{\lambda,\epsilon}(v)<0\;\text{and}\;\inf\limits_{\|v\|=R}I_{\lambda,\epsilon}(v)\geq\rho,\text{
for }\lambda\in(0,\Lambda).$

Moreover, there exists $T>R$ such that $I_{\lambda,\epsilon}(Te_{1})<-1$ for
$\lambda\in(0,\Lambda)$, where $e_{1}$ is given by Lemma 2.10.

###### Proof.

We fix $l=|\Omega|^{\frac{1}{\left(\frac{p^{*}}{r+1}\right)^{\prime}}}$. Then
using Hölder’s inequality and Lemma 2.11, for any $v\in X$ we get

$\int_{\Omega}(v^{+})^{r+1}\leavevmode\nobreak\
dx\leq\left(\int_{\Omega}|v|^{p^{*}}\right)^{\frac{r+1}{p*}}|\Omega|^{\frac{1}{(\frac{p^{*}}{r+1})^{\prime}}}\leq
Cl\|v\|^{r+1},$

for some positive constant $C$ independent of $v$. Observing

$\lim_{t\to
0}\frac{I_{\lambda,\epsilon}(te_{1})}{t}=-\lambda\int_{\Omega}\epsilon^{-q(x)}e_{1}\,dx<0,$

we can choose $k\in(0,1)$ sufficiently small and set
$\|v\|=R:=k(\frac{r+1}{pCl})^{\frac{1}{r+1-p}}$ such that

$\inf\limits_{\|v\|\leq R}I_{\lambda,\epsilon}(v)<0.$

Moreover, since $R<(\frac{r+1}{pCl})^{\frac{1}{r+1-p}}$, we obtain

$\displaystyle
I_{0,\epsilon}(v)\geq\frac{R^{p}}{p}-\frac{ClR^{r+1}}{r+1}:=2\rho\,(\text{say})>0.$

We define

$\Lambda:=\frac{\rho}{\sup\limits_{\|v\|=R}\left(\displaystyle\frac{1}{1-q(x)}\int_{\Omega}|v|^{1-q(x)}\leavevmode\nobreak\
dx\right)},$

which is a positive constant and since $\rho,R$ depends on $k,r,p,|\Omega|,C$
so does $\Lambda$. We know that

(4.1) $(v^{+}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}\leq(v^{+})^{1-q(x)},$

which gives

$\displaystyle I_{\lambda,\epsilon}(v)$
$\displaystyle\geq\frac{1}{p}\int_{\Omega}H(\nabla
v)^{p}\,dx-\frac{1}{r+1}\int_{\Omega}(v^{+})^{r+1}\,dx-\frac{\lambda}{1-q(x)}\int_{\Omega}(v^{+})^{1-q(x)}\,dx$
$\displaystyle=I_{0,\epsilon}(v)-\frac{\lambda}{1-q(x)}\int_{\Omega}(v^{+})^{1-q(x)}\,dx.$

Therefore

$\displaystyle\inf\limits_{\|v\|=R}I_{\lambda,\epsilon}(v)$
$\displaystyle\geq\inf\limits_{\|v\|=R}I_{0,\epsilon}(v)-\lambda\sup\limits_{\|v\|=R}\left(\frac{1}{1-q(x)}\int_{\Omega}|v|^{1-q(x)}\leavevmode\nobreak\
dx\right)$ $\displaystyle\geq
2\rho-\lambda\sup\limits_{\|v\|=R}\left(\frac{1}{1-q(x)}\int_{\Omega}|v|^{1-q(x)}\leavevmode\nobreak\
dx\right)\geq\rho,$

if $\lambda\in(0,\Lambda).$ Lastly, it is easy to see that
$I_{0,\epsilon}(te_{1})\to-\infty$, as $t\to+\infty$ which implies that we can
choose $T>R$ such that $I_{0,\epsilon}(Te_{1})<-1$. Hence

$I_{\lambda,\epsilon}(Te_{1})\leq I_{0,\epsilon}(Te_{1})<-1$

which completes the proof. ∎

As a consequence of Lemma 4.1, we have

$\inf\limits_{\|v\|=R}I_{\lambda,\epsilon}(v)\geq\rho\,\max\\{I_{\lambda,\epsilon}(Te_{1}),I_{\lambda,\epsilon}(0)\\}=0.$

Our next Lemma ensures that $I_{\lambda,\epsilon}$ satisfies the Palais Smale
$(PS)_{c}$ condition.

######  Proposition 4.2.

Let $2\leq p<N$, then $I_{\lambda,\epsilon}$ satisfies the $(PS)_{c}$
condition, for any $c\in\mathbb{R}$ that is if $\\{u_{k}\\}\subset X$ is a
sequence satisfying

(4.2) $I_{\lambda,\epsilon}(u_{k})\to
c\;\text{and}\;I_{\lambda,\epsilon}^{\prime}(u_{k})\to 0$

as $k\to\infty$, then $\\{u_{k}\\}$ contains a strongly convergent subsequence
in $X$.

###### Proof.

Let $\\{u_{k}\\}\subset X$ satisfies (4.2) then we claim that $\\{u_{k}\\}$
must be bounded in $X$. To see this using (4.1), we obtain

(4.3)
$\begin{split}I_{\lambda,\epsilon}(u_{k})-\frac{1}{r+1}I_{\lambda,\epsilon}^{\prime}(u_{k})u_{k}&=\left(\frac{1}{p}-\frac{1}{r+1}\right)\int_{\Omega}H(\nabla
u_{k})^{p}\,dx-{\lambda}\int_{\Omega}\frac{(u_{k}^{+}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx\\\
&\quad+\frac{\lambda}{r+1}\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q(x)}u_{k}\leavevmode\nobreak\
dx\\\ &\geq\left(\frac{1}{p}-\frac{1}{r+1}\right)\int_{\Omega}H(\nabla
u_{k})^{p}\,dx-\lambda\int_{\Omega}\frac{(u_{k}^{+})^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx\\\
&\quad\quad+\frac{\lambda}{r+1}\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q(x)}u_{k}\leavevmode\nobreak\
dx\\\ &\geq\left(\frac{1}{p}-\frac{1}{r+1}\right)\int_{\Omega}H(\nabla
u_{k})^{p}\,dx-\lambda\int_{\Omega}\frac{(u_{k}^{+})^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx-\frac{\lambda C}{\epsilon(r+1)}\|u_{k}\|,\end{split}$

for some positive constant $C$ (independent of $k$) where we have used Lemma
2.11 and the fact $0<q(x)<1$ in $\overline{\Omega}$. Due to the same
reasoning, we obtain

(4.4)
$\begin{split}-\int_{\Omega}\frac{(u_{k}^{+})^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx&\geq-\int_{\Omega}\frac{|u_{k}|^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\ dx\\\
&\geq\frac{-1}{1-\|q\|_{L^{\infty}(\Omega)}}\left(\int_{\Omega\cap\\{|u_{k}|\geq
1\\}}|u_{k}|^{1-q(x)}dx+\int_{\Omega\cap\\{|u_{k}|<1\\}}|u_{k}|^{1-q(x)}dx\right)\\\
&\geq\frac{-1}{1-\|q\|_{L^{\infty}(\Omega)}}\left(\int_{\Omega\cap\\{|u_{k}|\geq
1\\}}|u_{k}|dx+\int_{\Omega\cap\\{|u_{k}|<1\\}}|u_{k}|^{1-\|q\|_{L^{\infty}(\Omega)}}dx\right)\\\
&\geq-C\left(\|u_{k}\|+\|u_{k}\|^{1-\|q\|_{L^{\infty}(\Omega)}}\right),\end{split}$

for some positive constant $C$ independent of $k$. Thus inserting (4.4) into
(4.3) and using the fact $r+1>p$ along with Remark 1.1, we get

(4.5)
$I_{\lambda,\epsilon}(u_{k})-\frac{1}{r+1}I_{\lambda,\epsilon}^{\prime}(u_{k})u_{k}\geq
C_{1}\|u_{k}\|^{p}-C\left(\|u_{k}\|+\|u_{k}\|^{1-\|q\|_{L^{\infty}(\Omega)}}\right),$

for some positive constant $C_{1}$ (independent of $k$). Also from (4.2) it
follows that for $k$ large enough

(4.6)
$\left|I_{\lambda,\epsilon}(u_{k})-\frac{1}{r+1}I_{\lambda,\epsilon}^{\prime}(u_{k})u_{k}\right|\leq
C+o(\|u_{k}\|),$

for some positive constant $C$ (independent of $k$). Combining (4.5) and
(4.6), our claim follows since $p>1$. By reflexivity of $X$, there exists
$u_{0}\in X$ such that up to a subsequence, $u_{k}\rightharpoonup u_{0}$
weakly in $X$ as $k\to\infty$.
Claim: $u_{k}\to u_{0}$ strongly in $X$ as $k\to\infty$.
By (4.2), we have that

$\lim_{k\to\infty}\left(\mathcal{\int}_{\Omega}H(\nabla
u_{k})^{p-1}\nabla_{\eta}H(\nabla u_{k})\nabla
u_{0}\,dx-\lambda\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q(x)}u_{0}\leavevmode\nobreak\
dx-\int_{\Omega}(u_{k}^{+})^{r}u_{0}\leavevmode\nobreak\ dx\right)=0$

and

$\lim_{k\to\infty}\left(\mathcal{\int}_{\Omega}H(\nabla
u_{k})^{p-1}\nabla_{\eta}H(\nabla u_{k})\nabla
u_{k}\,dx-\lambda\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q(x)}u_{k}\leavevmode\nobreak\
dx-\int_{\Omega}(u_{k}^{+})^{r}u_{k}\leavevmode\nobreak\ dx\right)=0.$

Now

(4.7) $\begin{split}&\lim\limits_{k\to\infty}\int_{\Omega}\big{(}H(\nabla
u_{k})^{p-1}\nabla_{\eta}H(\nabla u_{k})-H(\nabla
u_{0})^{p-1}\nabla_{\eta}H(\nabla u_{0})\big{)}.\nabla(u_{k}-u_{0})\,dx\\\
&=\lim\limits_{k\to\infty}\left(\lambda\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q}u_{k}\leavevmode\nobreak\
dx+\int_{\Omega}(u_{k}^{+})^{r}u_{k}\leavevmode\nobreak\
dx-\lambda\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q}u_{0}\leavevmode\nobreak\
dx-\int_{\Omega}(u_{k}^{+})^{r}u_{0}\leavevmode\nobreak\ dx\right)\\\
&\quad-\lim_{k\to\infty}\left(\int_{\Omega}H(\nabla
u_{0})^{p-1}\nabla_{\eta}H(\nabla u_{0}).\nabla u_{k}\leavevmode\nobreak\
dx-\int_{\Omega}H(\nabla u_{0})^{p}\leavevmode\nobreak\ dx\right).\end{split}$

From weak convergence of $\\{u_{k}\\}$ we get

(4.8) $\lim_{k\to\infty}\left(\int_{\Omega}H(\nabla
u_{0})^{p-1}\nabla_{\eta}H(\nabla u_{0})\nabla u_{k}\leavevmode\nobreak\
dx-\int_{\Omega}H(\nabla u_{0})^{p}\leavevmode\nobreak\ dx\right)=0.$

Also, we have

$\displaystyle\left|(u_{k}^{+}+\epsilon)^{-q(x)}u_{0}\right|\leq\epsilon^{-q(x)}u_{0}\text{
and
}\int_{\Omega}\left|\epsilon^{-q(x)}u_{0}\right|dx\leq\|\epsilon^{-q(x)}\|_{L^{\infty}(\Omega)}\int_{\Omega}|u_{0}|\leavevmode\nobreak\
dx<+\infty.$

Thus Lebesgue Dominated convergence theorem gives that

(4.9)
$\lim_{k\to\infty}\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q(x)}u_{0}\leavevmode\nobreak\
dx=\int_{\Omega}(u_{0}^{+}+\epsilon)^{-q(x)}u_{0}\leavevmode\nobreak\ dx.$

Since $u_{k}\to u_{0}$ pointwise in $\Omega$, for any measurable subset $E$ of
$\Omega$ we have

$\begin{split}\int_{E}|(u_{k}^{+}+\epsilon)^{-q(x)}u_{k}|\leavevmode\nobreak\
dx&\leq\int_{E}\epsilon^{-q(x)}u_{k}\leavevmode\nobreak\
dx\leq\|\epsilon^{-q(x)}\|_{L^{\infty}(\Omega)}\|u_{k}\|_{L^{p^{*}}(\Omega)}|E|^{\frac{p^{*}-1}{p^{*}}}\leq
C(\epsilon)|E|^{\frac{p^{*}-1}{p^{*}}},\end{split}$

so from Vitali convergence theorem, it follows that

(4.10)
$\lim\limits_{k\to\infty}\lambda\int_{\Omega}(u_{k}^{+}+\epsilon)^{-q(x)}u_{k}\leavevmode\nobreak\
dx=\lambda\int_{\Omega}(u_{0}^{+}+\epsilon)^{-q(x)}u_{0}\leavevmode\nobreak\
dx.$

Similarly, due to $r+1<p^{*}$, we have

$\int_{E}|(u_{k}^{+})^{r}u_{0}|\leavevmode\nobreak\
dx\leq\|u_{0}\|_{L^{p^{*}}(\Omega)}\left(\int_{E}(u_{k}^{+})^{rp^{*}{{}^{\prime}}}\leavevmode\nobreak\
dx\right)^{\frac{1}{p^{*}{{}^{\prime}}}}\leq C_{3}|E|^{\alpha}$

and

$\int_{E}|(u_{k}^{+})^{r}u_{k}|\leavevmode\nobreak\
dx\leq\|u_{k}\|_{L^{p^{*}}(\Omega)}\left(\int_{E}(u_{k}^{+})^{rp^{*}{{}^{\prime}}}\leavevmode\nobreak\
dx\right)^{\frac{1}{p*{{}^{\prime}}}}\leq C_{4}|E|^{\beta}$

for some positive constants $C_{3},C_{4},\alpha$ and $\beta$. Therefore Vitali
convergence theorem gives

(4.11) $\lim_{k\to\infty}\int_{\Omega}(u_{k}^{+})^{r}u_{0}\leavevmode\nobreak\
dx=\int_{\Omega}(u_{0}^{+})^{r}u_{0}\leavevmode\nobreak\ dx,$

and

(4.12) $\lim_{k\to\infty}\int_{\Omega}(u_{k}^{+})^{r}u_{k}\leavevmode\nobreak\
dx=\int_{\Omega}(u_{0}^{+})^{r}u_{0}\leavevmode\nobreak\ dx.$

Using (4.8), (4.9), (4.10), (4.11) and (4.12) in (4.7), we obtain

$\lim\limits_{k\to\infty}\int_{\Omega}\big{(}H(\nabla
u_{k})^{p-1}\nabla_{\eta}H(\nabla u_{k})-H(\nabla
u_{0})^{p-1}\nabla_{\eta}H(\nabla u_{0})\big{)}.\nabla(u_{k}-u_{0})\,dx=0.$

Then using (2.16) from Corollary 2.9, we obtain $u_{k}\to u_{0}$ strongly in
$X$ as $k\to\infty$ which proves our claim. ∎

######  Remark 4.3.

From Lemma 4.1, Proposition 4.2 and Mountain Pass Lemma, for every
$\lambda\in(0,\Lambda)$, there exists a $\zeta_{\epsilon}\in X$ such that
$I_{\lambda,\epsilon}^{\prime}(\zeta_{\epsilon})=0$ and

$I_{\lambda,\epsilon}(\zeta_{\epsilon})=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I_{\lambda,\epsilon}(\gamma(t))\geq\rho>0,$

where

$\Gamma=\big{\\{}\gamma\in C([0,1],X):\gamma(0)=0,\gamma(1)=Te_{1}\big{\\}}.$

Using (4.4) together with Vitali convergence theorem, if $u_{k}\rightharpoonup
u_{0}$ weakly in $X$, then we have

$\lim_{k\to\infty}\int_{\Omega}\frac{(u_{k}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}}{1-q(x)}\,dx=\int_{\Omega}\frac{(u_{0}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}}{1-q(x)}\,dx.$

Therefore, $I_{\lambda,\epsilon}$ is weakly lower semicontinuous. Furthermore,
as a consequence of Lemma 4.1, since for every $\lambda\in(0,\Lambda)$ we have
$\inf\limits_{\|v\|\leq R}I_{\lambda,\epsilon}(v)<0$, there exists a nonzero
$\nu_{\epsilon}\in X$ such that $\|\nu_{\epsilon}\|\leq R$ and

(4.13) $\inf\limits_{\|v\|\leq
R}I_{\lambda,\epsilon}(v)=I_{\lambda,\epsilon}(\nu_{\epsilon})<0<\rho\leq
I_{\lambda,\epsilon}(\zeta_{\epsilon}).$

Thus, $\zeta_{\epsilon}$ and $\nu_{\epsilon}$ are two different non trivial
critical points of $I_{\lambda,\epsilon}$, provided $\lambda\in(0,\Lambda)$.

######  Lemma 4.4.

Let $2\leq p<N$, then the critical points $\zeta_{\epsilon}$ and
$\nu_{\epsilon}$ of $I_{\lambda,\epsilon}$ are nonnegative in $\Omega.$

###### Proof.

Testing $(P_{\lambda,\epsilon})$ with $\min\\{\zeta_{\epsilon},0\\}$ and
$\min\\{\nu_{\epsilon},0\\}$, proceeding similarly as in the proof of Lemma
3.1, we have $\zeta_{\epsilon},\nu_{\epsilon}\geq 0$ in $\Omega$. ∎

######  Remark 4.5.

As in Remark 3.3, if $\Delta_{H,p}=\Delta_{p}$ or $\mathcal{S}_{p}$ as given
by (1.7), then noting Lemma 2.12, we have Proposition 4.2 and Lemma 4.4 valid
for any $1<p<N$.

######  Lemma 4.6.

There exists a constant $\Theta>0$ (independent of $\epsilon$) such that
$\|v_{\epsilon}\|\leq\Theta$, where $v_{\epsilon}=\zeta_{\epsilon}$ or
$\nu_{\epsilon}$.

###### Proof.

The result trivially holds if $v_{\epsilon}=\nu_{\epsilon}$ so we deal with
the case $v_{\epsilon}=\zeta_{\epsilon}$. Recalling the terms from Lemma 4.1,
we define $A=\max\limits_{t\in[0,1]}I_{0,\epsilon}(tTe_{1})$ then

$A\geq\max_{t\in[0,1]}I_{\lambda,\epsilon}(tTe_{1})\geq\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I_{\lambda,\epsilon}(\gamma(t))=I_{\lambda,\epsilon}(\zeta_{\epsilon})\geq\rho>0>I_{\lambda,\epsilon}(\nu_{\epsilon}).$

Therefore

(4.14)
$\frac{1}{p}\int_{\Omega}H(\nabla\zeta_{\epsilon})^{p}\,dx-{\lambda}\int_{\Omega}\frac{(\zeta_{\epsilon}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx-\frac{1}{r+1}\int_{\Omega}\zeta_{\epsilon}^{r+1}\leavevmode\nobreak\ dx\leq
A.$

Choosing $\phi=-\frac{\zeta_{\epsilon}}{p+1}$ as a test function in
$(P_{\lambda,\epsilon})$ we obtain

(4.15)
$-\frac{1}{r+1}\int_{\Omega}H(\nabla\zeta_{\epsilon})^{p}\,dx+\frac{\lambda}{r+1}\int_{\Omega}\frac{\zeta_{\epsilon}}{(\zeta_{\epsilon}+\epsilon)^{q(x)}}\,dx+\frac{1}{r+1}\int_{\Omega}\zeta_{\epsilon}^{r+1}\,dx=0.$

Adding (4.14) and (4.15) we get

$\displaystyle\left(\frac{1}{p}-\frac{1}{r+1}\right)\int_{\Omega}H(\nabla\zeta_{\epsilon})^{p}\,dx$
$\displaystyle\leq{\lambda}\int_{\Omega}\frac{(\zeta_{\epsilon}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx-\frac{\lambda}{r+1}\int_{\Omega}\frac{\zeta_{\epsilon}}{(\zeta_{\epsilon}+\epsilon)^{q(x)}}\,dx+A$
$\displaystyle\leq{\lambda}\int_{\Omega}\frac{(\zeta_{\epsilon}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}}{1-q(x)}\leavevmode\nobreak\
dx+A$ $\displaystyle\leq
C(\|\zeta_{\epsilon}\|+\|\zeta_{\epsilon}\|^{1-\|q\|_{L^{\infty}(\Omega)}})+A,$

for some positive constant $C$ being independent of $\epsilon$, where the last
inequality is deduced using the estimate (4.4), Hölder inequality along Lemma
2.11. Therefore, using $r+1>p$, the sequence $\\{\zeta_{\epsilon}\\}$ is
uniformly bounded in $X$ with respect to $\epsilon$. This completes the proof.
∎

##  5\. Proof of the main results

Proof of Theorem 1.4:

  1. (a)

Uniqueness: Let $u,v\in X$ be two distinct weak solutions of
$\eqref{maineqn}$. Then by Lemma 2.13 we can choose $\phi=(u-v)^{+}$ as a test
function in (1.10) to get

$\langle-\Delta_{H,p}u+\Delta_{H,p}v,(u-v)^{+}\rangle=\int_{\Omega}f\left(\frac{1}{u^{q(x)}}-\frac{1}{v^{q(x)}}\right)(u-v)^{+}\,dx\leq
0.$

Using Lemma 2.5 we obtain $u\leq v$ in $\Omega$. Similarly, choosing
$\phi=(v-u)^{+}$ as a test function in (1.10) we get $v\leq u$ in $\Omega$.
Hence the uniqueness follows.

  2. (b)

Existence: Let $u_{n}\in X$ denotes the weak solution of (3.1) then choosing
$u_{n}$ as a test function in (3.1) and using Remark 1.1, we get

$\int_{\Omega}H(\nabla
u_{n})^{p}\,dx\leq\int_{\Omega}\frac{f_{n}(x)u_{n}}{\big{(}u_{n}+\frac{1}{n}\big{)}^{q(x)}}\,dx.$

Now denote by $\omega_{\delta}=\Omega\setminus\overline{\Omega_{\delta}}$ then
by Lemma 3.2, we get $u_{n}\geq c_{\omega_{\delta}}>0$ in $\omega_{\delta}$.
We observe that

$\displaystyle\int_{\Omega}\frac{f_{n}(x)u_{n}}{(u_{n}+\frac{1}{n})^{q(x)}}\,dx$
$\displaystyle=\int_{\overline{{\Omega}_{\delta}}}\frac{f_{n}(x)u_{n}}{(u_{n}+\frac{1}{n})^{q(x)}}\,dx+\int_{\omega_{\delta}}\frac{f_{n}(x)u_{n}}{(u_{n}+\frac{1}{n})^{q(x)}}\,dx$
$\displaystyle\leq\int_{\overline{{\Omega}_{\delta}}}f(x)u_{n}^{1-q(x)}\,dx+\int_{\omega_{\delta}}\frac{f(x)}{u_{n}^{q(x)}}u_{n}\,dx$
$\displaystyle\leq\int_{\overline{{\Omega}_{\delta}}\cap\\{u_{n}\leq
1\\}}f(x)\,dx+\int_{{\overline{{\Omega}_{\delta}}}\cap\\{{u_{n}\geq
1\\}}}f(x)u_{n}\,dx+\int_{\omega_{\delta}}\frac{f(x)}{c_{\omega_{\delta}}^{q(x)}}u_{n}\,dx$
$\displaystyle\leq\|f\|_{L^{1}(\Omega)}+\big{(}1+\|c_{\omega_{\delta}}^{-q(x)}\|_{L^{\infty}(\Omega)}\big{)}\int_{\Omega}f(x)u_{n}\,dx.$

Let $2\leq p<N$. Then, using H$\ddot{\text{o}}$lder’s inequality and Lemma
2.11 we obtain for $f\in L^{m}(\Omega)$ with $m=(p^{*})^{\prime}$

$\|u_{n}\|^{p}\leq\|f\|_{L^{1}(\Omega)}+C\|f\|_{L^{m}(\Omega)}\|u_{n}\|,$

which implies that the sequence $\\{u_{n}\\}$ is uniformly bounded in $X$.
Thus up to a subsequence, by Lemma 2.11 for $1\leq t<p^{*}$, we get

$\displaystyle u_{n}$ $\displaystyle\rightharpoonup u\text{ weakly in }X,$
$\displaystyle\,u_{n}$ $\displaystyle\to u\text{ strongly in
}L^{t}(\Omega),\text{ and }$ $\displaystyle u_{n}$ $\displaystyle\to u\text{
pointwise in }\Omega\text{ as }n\to\infty.$

Let $\phi\in C_{c}^{\infty}(\Omega)$ be such that $\mathrm{supp}\,\phi=\omega$
where $\omega\Subset\Omega$. Then by Lemma 3.2, there exists a constant
$c_{\omega}>0$ independent of $n$ such that $u_{n}\geq c_{\omega}>0$ in
$\omega$. Hence, we get

$\frac{f_{n}}{\left(u_{n}+\frac{1}{n}\right)^{q(x)}}\phi\leq
c_{\omega}^{-q(x)}f(x)\phi(x)\in L^{1}(\Omega).$

Thus we can apply Boccardo-Murat [12, Theorem 2.1] to obtain that upto a
subsequence still denoted by $\\{u_{n}\\}$, we have

$\lim_{n\to\infty}\nabla u_{n}(x)=\nabla u(x)\text{ for }x\text{ in }\Omega.$

Hence, we have

$\lim_{n\to\infty}H(\nabla u_{n}(x))^{p-1}\nabla_{\eta}H\big{(}\nabla
u_{n}(x)\big{)}=H(\nabla u(x))^{p-1}\nabla_{\eta}H\big{(}\nabla
u(x)\big{)}\text{ for }x\text{ in }\Omega.$

Moreover, by Corollary 2.2 we observe that

$\|H(\nabla u_{n})^{p-1}\nabla_{\eta}H(\nabla
u_{n})\|_{L^{\frac{p}{p-1}}(\Omega)}^{\frac{p}{p-1}}\leq\|u_{n}\|^{p}\leq C,$

for some positive constant $C$ which is independent of $n$. As a consequence,
the sequence

$H(\nabla u_{n})^{p-1}\nabla_{\eta}H(\nabla u_{n})\rightharpoonup H(\nabla
u)^{p-1}\nabla_{\eta}H(\nabla u)\text{ weakly in }X\text{ as }n\to\infty.$

Indeed, since the weak limit is independent of the subsequence, the above fact
holds for every $n$. Therefore, we have

(5.1) $\lim_{n\to\infty}\int_{\Omega}H(\nabla
u_{n}(x))^{p-1}\nabla_{\eta}H(\nabla
u_{n}(x))\nabla\phi\,dx=\int_{\Omega}H(\nabla u)^{p-1}\nabla_{\eta}H(\nabla
u)\nabla\phi\,dx,$

for every $\phi\in C_{c}^{\infty}(\Omega)$. Moreover by the Lebsegue dominated
convergence theorem, we have

(5.2)
$\lim_{n\to\infty}\int_{\Omega}\frac{f_{n}}{\big{(}u_{n}+\frac{1}{n}\big{)}^{q(x)}}\phi\,dx=\int_{\Omega}\frac{f}{u^{q(x)}}\,\phi\,dx\text{
for all }\phi\in C_{c}^{\infty}(\Omega).$

Now, we define by

$u:=\lim_{n\to\infty}u_{n}.$

Then, due to the monotonicity of $\\{u_{n}\\}$ we get $u\geq c_{\omega}>0$ for
every $\omega\Subset\Omega$. Hence from (5.1) and (5.2), we get $u$ is our
required solution and the result follows.

Noting Lemma 2.11, for $p\geq N$ the result follows similarly.

Proof of Theorem 1.5: Noting Lemma 2.12 and Remark 3.3, proceeding similarly
as in the proof of Theorem 1.4, the result follows.

Proof of Theorem 1.6: As a resultant of Lemma 4.4 and Lemma 4.6, upto a
subsequence we get that $\zeta_{\epsilon}\rightharpoonup\zeta_{0}$ and
$\nu_{\epsilon}\rightharpoonup\nu_{0}$ weakly in $X$ as $\epsilon\to 0^{+}$,
for some non negative $\zeta_{0},\nu_{0}\in X$. In the sequel, we establish
that $\zeta_{0}\neq\nu_{0}$ and forms a weak solution to our problem (1.2).
For convenience, we denote by $v_{0}$ either $\zeta_{0}$ or $\nu_{0}$. We
prove the result in the following steps.
Step $1$. Here, we prove that $v_{0}\in X$ is a weak solution to the problem
(1.2).
We observe that for any $\epsilon\in(0,1)$ and $t\geq 0$,

$\frac{\lambda}{(t+\epsilon)^{q(x)}}+t^{r}\geq\frac{\lambda}{(t+1)^{q(x)}}+t^{r}\geq\text{min}\left\\{1,\frac{\lambda}{2}\right\\}:=C>0,\text{
say}.$

As a consequence we get

$-\Delta_{H,p}\,v_{\epsilon}=\frac{\lambda}{(v_{\epsilon}+\epsilon)^{q(x)}}+v_{\epsilon}^{r}\geq
C>0.$

By Lemma 3.1, let $\xi\in X$ satisfies

$-\Delta_{H,p}\xi=C\text{ in }\Omega,\,\xi>0\text{ in }\Omega,$

such that for every $\omega\Subset\Omega$, there exists a constant
$c_{\omega}>0$ satisfying $\xi\geq c_{\omega}>0$ in $\Omega$. Then, for every
nonnegative $\phi\in X$, we have

$\displaystyle\int_{\Omega}H(v_{\epsilon})^{p-1}\nabla_{\eta}H(v_{\epsilon})\nabla\phi\,dx$
$\displaystyle=\int_{\Omega}\Big{(}\frac{\lambda}{(v_{\epsilon}+\epsilon)^{q(x)}}+v_{\epsilon}^{r}\Big{)}\phi\,dx$
$\displaystyle\geq\int_{\Omega}C\phi\,dx=\int_{\Omega}H(\nabla\xi)^{p-1}\nabla_{\eta}H(\nabla\xi)\nabla\phi\,dx.$

Now choosing $\phi=(\xi-v_{\epsilon})^{+}$, we obtain

$\int_{\Omega}\Big{\\{}H(\nabla\xi)^{p-1}\nabla_{\eta}H(\nabla\xi)-H(\nabla
v_{\epsilon})^{p-1}\nabla_{\eta}H(\nabla v_{\epsilon})\Big{\\}}\nabla(\xi-
v_{\epsilon})^{+}\,dx\leq 0.$

Now applying Lemma 2.5, we obtain $v_{\epsilon}\geq\xi$ in $\Omega$. Hence, we
get the existence of a constant $c_{\omega}>0$ (independent of $\epsilon$)
such that

(5.3) $v_{\epsilon}\geq c_{\omega}>0,\text{ for every }\omega\Subset\Omega.$

Therefore using $r+1<p^{*}$ along with Lemma 4.6 and the fact (5.3) as in the
proof of Theorem 1.4 we can apply Boccardo-Murat [12, Theorem 2.1] to obtain

$\int_{\Omega}H(\nabla v_{0})^{p-1}\nabla_{\eta}H(\nabla
v_{0})\nabla\phi\,dx=\lambda\int_{\Omega}\frac{\phi}{v_{0}^{q}(x)}\,dx+\int_{\Omega}v_{0}^{r}\phi\,dx.$

Hence the claim follows.
Step $2$. Now we are going to prove that $\zeta_{0}\neq\nu_{0}$.
Choosing $\phi=v_{\epsilon}\in X$ as a test function in
$(P_{\lambda,\epsilon})$ we get

$\int_{\Omega}H(\nabla v_{\epsilon})^{p-1}\nabla_{\eta}H(\nabla
v_{\epsilon})\nabla\phi\,dx=\lambda\int_{\Omega}\frac{v_{\epsilon}}{(v_{\epsilon}+\epsilon)^{q(x)}}\,dx+\int_{\Omega}v_{\epsilon}^{r+1}\,dx.$

Since $r+1<p^{*}$, using Lemma 2.11 we obtain

$\lim\limits_{\epsilon\to
0^{+}}\int_{\Omega}(v_{\epsilon})^{r+1}\,dx=\int_{\Omega}v_{0}^{r+1}\,dx.$

Moreover, since

$0\leq\frac{v_{\epsilon}}{(v_{\epsilon}+\epsilon)^{q(x)}}\leq
v_{\epsilon}^{1-q(x)},$

using (4.4) together with Vitali convergence theorem, we get

$\lambda\lim\limits_{\epsilon\to
0^{+}}\int_{\Omega}\frac{v_{\epsilon}}{(v_{\epsilon}+\epsilon)^{q(x)}}\,dx=\lambda\int_{\Omega}v_{0}^{1-q(x)}\,dx.$

Therefore for every $\phi\in X$, we have

$\lim_{\epsilon\to 0^{+}}\int_{\Omega}H(\nabla
v_{\epsilon})^{p-1}\nabla_{\eta}H(\nabla
v_{\epsilon})\nabla\phi\,dx=\lambda\int_{\Omega}v_{0}^{1-q(x)}\,dx+\int_{\Omega}v_{0}^{r+1}\,dx.$

Using Lemma 2.13 we can choose $\phi=v_{0}$ as a test function in (1.11) to
deduce that

$\int_{\Omega}H(\nabla
v_{0})^{p}\,dx=\lambda\int_{\Omega}v_{0}^{1-q(x)}\,dx+\int_{\Omega}v_{0}^{r+1}\,dx.$

Hence we obtain

$\lim\limits_{\epsilon\to 0}\int_{\Omega}H(\nabla
v_{\epsilon})^{p}\,dx=\int_{\Omega}H(\nabla v_{0})^{p}\,dx.$

By the Vitali convergence theorem, we get

$\lim\limits_{\epsilon\to
0}\int_{\Omega}[(v_{\epsilon}+\epsilon)^{1-q(x)}-\epsilon^{1-q(x)}]\,dx=\int_{\Omega}v_{0}^{1-q(x)}\,dx,$

which together with the strong convergence of $v_{\epsilon}$ in $X$ implies
$\lim\limits_{\epsilon\to
0}I_{\lambda,\epsilon}(v_{\epsilon})=I_{\lambda}(v_{0}).$ Hence from (4.13) we
get $\zeta_{0}\neq\nu_{0}.$

Proof of Theorem 1.7: Noting Lemma 2.12 and Remark 4.5, proceeding similarly
as in the proof of Theorem 1.6, the result follows.

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Kaushik Bal
Department of Mathematics and Statistics,
Indian Institute of Technology Kanpur,
Kanpur-208016,
Uttar Pradesh, India
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Prashanta Garain
Department of Mathematics,
Ben-Gurion University of the Negev,
P.O.B.-653,
Beer Sheva-8410501, Israel
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Tuhina Mukherjee
Department of Mathematics,
Indian Institute of Technology Jodhpur,
Jodhpur-342037,
Rajasthan, India
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