Existence and regularity of solutions to a quasilinear elliptic problem involving variable sources

Ying Chu; Ruimei Gao; Yan Sun

doi:10.1186/s13661-017-0888-4

Research
Open Access

Existence and regularity of solutions to a quasilinear elliptic problem involving variable sources

Ying Chu¹,
Ruimei Gao¹Email author and
Yan Sun¹

Boundary Value Problems20172017:155

https://doi.org/10.1186/s13661-017-0888-4

Received: 13 September 2017
Accepted: 17 October 2017
Published: 24 October 2017

Abstract

The authors of this paper prove the existence and regularity results for the homogeneous Dirichlet boundary value problem to the equation $-\operatorname{div}(|\nabla u|^{p-2}\nabla u)=\frac{f(x)}{u ^{\alpha(x)}}$ with $f\in L^{m}(\Omega)$ ($m\geqslant1$) and $\alpha(x)>0$. Due to the nonlinearity of a p-Laplace operator and the anisotropic variable exponent $\alpha(x)$, some classical methods may not directly be applied to our problem. In this paper, we construct a suitable test function and apply the Leray-Schauder fixed point theorem to prove the existence of positive solutions with necessary a priori estimate and compact argument. Furthermore, we also discuss the relationship among the regularity of solutions, the summability of f and the value of $\alpha(x)$.

Keywords

quasilinear elliptic problem
nonlinear singular term
existence
regularity
variable exponent

1 Introduction

In this paper, we study the existence of solutions for the following quasilinear elliptic problem with nonlinear singular terms and variable exponent:

$$ \textstyle\begin{cases} -\operatorname {div}(\vert \nabla u\vert ^{p-2}\nabla u)= \frac{f(x)}{u^{\alpha(x)}}, &x\in\Omega, \\ u=0, &x\in \partial\Omega, \end{cases} $$

(1.1)

where Ω is a bounded domain in $R^{N}$ ($N\geqslant p$) with smooth boundary ∂Ω, $p>2$, $\alpha(x)$ is a continuous function on Ω̅, $\alpha(x)>0$, $\alpha^{+}= \sup_{x\in\overline{\Omega}}\alpha(x)$, $\alpha^{-}=\inf_{x\in\overline {\Omega}}\alpha(x) $, f is a nonnegative function belonging to the Lebesgue space $L^{m}(\Omega)$ for some suitable $m\geqslant1$.

Problem (1.1) has been widely applied in many areas such as the contexts of chemical heterogeneous catalysts, non-Newtonian fluids and also the theory of heat conduction in electrically conducting materials, see [1–4] for detailed discussion.

For constant-exponent cases, Lazer and Mckenna in [5] discussed the case when $p=2$ and f is a positive regular function in Ω̅. They proved that the solution was in $H_{0}^{1}( \Omega)$ if and only if $\alpha<3$, while it was not in $C^{1}(\overline{ \Omega})$ if $\alpha>1$.

Lair and Shaker in [6] improved the results of [5]. More specially speaking, they proved that this problem with $0<\alpha<1$ has a unique weak positive solution in $H_{0}^{1}( \Omega)$ if $f(x)$ is a nonnegative nontrivial function in $L^{2}(\Omega)$.

In 2004, the results of Lair and Shaker were generalized by Zhang and Cheng (see [7]) to the following problem:

$$ \textstyle\begin{cases} -\Delta u=f(x)g(u), &x\in\Omega, \\ u=0, &x\in\partial\Omega, \end{cases} $$

(1.2)

where $g(s)$ is singular near $s=0$. They proved the existence and uniqueness of classical solutions under the assumption that $f(x)\in C^{\alpha}(\Omega)$.

Recently, Boccardo and Orsina in [8] studied the existence, regularity and nonexistence of solutions for the following problem:

$$ \textstyle\begin{cases} -\operatorname {div}(M(x)\nabla u)= \frac{f(x)}{u^{\alpha}}, &x\in\Omega, \\ u=0, &x\in \partial\Omega. \end{cases} $$

(1.3)

They discussed how the summability of f and the value of α affected the existence and regularities of solutions to the above problems. For other results of the related problems, the interested readers may refer to [9, 10] and the references therein. Problem (1.3) was discussed and extended to the more general problem of which the right-hand side is $f(x)/u^{\alpha(x)}$ in [11]. For more related questions, refer to [12–14].

In this paper, we generalize the results in [11] to the case when the left-hand side is a p-Laplace operator. Due to the nonlinearity of a p-Laplace operator and the anisotropic variable exponent $\alpha(x)$, some classical methods may not directly be applied to our problem. We apply the method of regularization and the Schauder fixed point theorem, construct a suitable test function as well as a necessary compactness argument to overcome the difficulties arising from a variable exponent and a nonlinear differential operator and give an almost complete classification of coefficient m and variable exponent $\alpha(x)$, then we prove the existence and regularity of solutions.

2 Preliminaries

Firstly, we give the definition of weak solutions to problem (1.1).

Definition 2.1

A function $u\in W^{1,p}_{0}(\Omega)$ is called a weak solution of problem (1.1) if the following identity holds:

$$ \int_{\Omega} \vert \nabla u\vert ^{p-2}\nabla u \nabla \varphi \,\mathrm{dx}= \int_{\Omega}\frac{f}{u^{\alpha(x)}}\varphi \,\mathrm{dx}, \quad\forall\varphi \in C_{0}^{\infty}(\Omega). $$

(2.1)

In order to prove our results, we will consider the following approximation problem:

$$ \textstyle\begin{cases} -\operatorname {div}(\vert \nabla u_{n}\vert ^{p-2}\nabla u_{n})= \frac{f_{n}}{(u_{n}+\frac{1}{n})^{\alpha(x)}}, &x\in\Omega, \\ u_{n}=0, &x\in \partial\Omega, \end{cases} $$

(2.2)

where $f_{n}(x)=\min\{f(x),n\}$, $n\in N$.

Lemma 2.1

Problem (2.2) has a nonnegative solution $u_{n}$ in $W_{0}^{1,p}( \Omega)\bigcap L^{\infty}(\Omega)$.

Proof

Let $n\in N$ be fixed, and ω be a function in $L^{p}(\Omega)$. It is not difficult to prove that the following problem has a unique solution $v\in W_{0}^{1,p}(\Omega)\bigcap L^{\infty}(\Omega)$ (see [15, 16]):

$$ \textstyle\begin{cases} -\operatorname {div}(\vert \nabla v\vert ^{p-2}\nabla v)= \frac{f_{n}}{(\vert \omega \vert +\frac{1}{n})^{\alpha(x)}}, &x\in\Omega, \\ v=0, &x\in \partial\Omega. \end{cases} $$

(2.3)

So, for any $\omega\in L^{p}(\Omega)$, we define the mapping $\Gamma: L^{p}(\Omega)\rightarrow L^{p}(\Omega)$ as $\Gamma( \omega)=v$. Taking v as a test function for (2.3), we have

$$ \int_{\Omega} \vert \nabla v\vert ^{p}\,\mathrm{dx}= \int_{\Omega}\frac{f_{n}}{(\vert \omega \vert +\frac{1}{n})^{\alpha(x)}}v\,\mathrm{dx}\leqslant \int_{ \Omega}\frac{n}{(\frac{1}{n})^{\alpha^{+}}}v\,\mathrm{dx}\leqslant n ^{\alpha^{+}+1} \int_{\Omega} \vert v\vert \,\mathrm{dx}. $$

By the Poincaré inequality (on the left-hand side) and the Sobolev embedding theorem on the right-hand side ($W^{1,p}(\Omega )\hookrightarrow L^{1}(\Omega)$), we get that

$$\begin{aligned} \Vert v\Vert ^{p}_{W^{1,p}}\leqslant Cn^{\alpha^{+}+1} \Vert v\Vert _{W^{1,p}}, \end{aligned}$$

this implies that

$$ \Vert v\Vert _{W^{1,p}}\leqslant Cn^{\frac{\alpha^{+}+1}{p-1}}. $$

Since the embedding $W^{1,p}(\Omega)\hookrightarrow L^{p}(\Omega)$ is compact, we obtain that Γ is a compact operator. Moreover, if $u=\lambda\Gamma u$ for some $0<\lambda\leqslant1$, then $\Gamma u=\frac{u}{\lambda}$ and hence $\Vert u\Vert _{L^{p}(\Omega)}\leqslant \Vert u\Vert _{W^{1,p}(\Omega)}\leqslant C$ for a constant C independent of λ. Then, by Schauder’s fixed point theorem, we know that there exists $u_{n}\in W_{0}^{1,p}(\Omega)$ such that $u_{n}=\Gamma(u_{n})$, i.e., problem (2.2) has a solution. Since $\frac{f_{n}}{(u_{n}+ \frac{1}{n})^{\alpha(x)}}\geqslant0$, the maximum principle in [17, 18] shows that $u_{n}\geqslant0$, $u _{n}\in L^{\infty}(\Omega)$. □

Lemma 2.2

The sequence $\{u_{n}\}$ is increasing with respect to n, $u_{n}>0$ in Ω, and for every $\Omega'\subset\subset\Omega $, there exists $C_{\Omega'}>0$ (independent of n) such that

$$ u_{n}(x)\geqslant C_{\Omega'}>0 \quad \textit{for every }x\in \Omega', \textit{for every }n\in N. $$

(2.4)

Proof

Due to $0\leqslant f_{n}\leqslant f_{n+1}$ and $\alpha(x)>0$, we have that

$$\begin{aligned}& -\operatorname {div}\bigl(\vert \nabla u_{n}\vert ^{p-2}\nabla u_{n}\bigr)=\frac {f_{n}}{(u_{n}+ \frac{1}{n})^{\alpha(x)}}\leqslant\frac{f_{n+1}}{(u_{n}+ \frac{1}{n+1})^{\alpha(x)}}, \\& -\operatorname {div}\bigl(\vert \nabla u_{n+1}\vert ^{p-2}\nabla u_{n+1}\bigr)=\frac{f_{n+1}}{(u _{n+1}+\frac{1}{n+1})^{\alpha(x)}}, \end{aligned}$$

so that

$$ -\operatorname {div}\bigl(\vert \nabla u_{n} \vert ^{p-2}\nabla u_{n}-\vert \nabla u_{n+1} \vert ^{p-2}u _{n+1}\bigr) =f_{n+1} \frac{(u_{n+1}+\frac{1}{n+1})^{\alpha(x)}-(u_{n}+ \frac{1}{n+1})^{\alpha(x)}}{(u_{n}+\frac{1}{n+1})^{\alpha(x)}(u_{n+1}+ \frac{1}{n+1})^{\alpha(x)}}. $$

Choosing $(u_{n}-u_{n+1})_{+}=\max\{u_{n}-u_{n+1},0\}$ as a test function and observing that

$$\begin{aligned} &\bigl(\vert \nabla u_{n}\vert ^{p-2}\nabla u_{n}-\vert \nabla u_{n+1}\vert ^{p-2}\nabla u_{n+1}\bigr) \nabla(u_{n}-u_{n+1})_{+} \geqslant0, \\ & \biggl(\biggl(u_{n+1}+\frac{1}{n+1}\biggr)^{\alpha(x)}- \biggl(u_{n}+\frac{1}{n+1}\biggr)^{ \alpha(x)} \biggr) (u_{n}-u_{n+1})_{+}\leqslant0, \end{aligned}$$

we get that

$$ 0\leqslant \int_{\Omega}\bigl(\vert \nabla u_{n}\vert ^{p-2}\nabla u_{n}-\vert \nabla u _{n+1}\vert ^{p-2}\nabla u_{n+1}\bigr)\nabla (u_{n}-u_{n+1})_{+} \,\mathrm{dx}\leqslant0, $$

which implies that $(u_{n}-u_{n+1})_{+}=0$ a.e. in Ω, that is, $u_{n}\leqslant u_{n+1}$ for every $n\in N$. Since the sequence $\{u_{n}\}$ is increasing with respect to n, we only need to prove that (2.4) holds for $u_{1}$. Using Lemma 2.1, we know that $u_{1}\in L^{\infty}(\Omega)$, i.e., there exists a constant C (depending only on Ω and N) such that

$$ \Vert u_{1}\Vert _{L^{\infty}(\Omega)}\leqslant C\Vert f_{1}\Vert _{L^{\infty}( \Omega)}\leqslant C, $$

then

$$ -\operatorname {div}\bigl(\vert \nabla u_{1}\vert ^{p-2}\nabla u_{1}\bigr)=\frac {f_{1}}{(u_{1}+1)^{ \alpha(x)}}\geqslant\frac{f_{1}}{(C+1)^{\alpha(x)}}. $$

Since $\frac{f_{1}}{(C+1)^{\alpha(x)}}\geqslant0$, $\frac{f_{1}}{(C+1)^{ \alpha(x)}}\not\equiv0$, the strong maximum principle implies that $u_{1}>0$ in Ω and (2.4) holds for $u_{1}$. Because of the monotonicity of $u_{n}$, (2.4) holds for $u_{n}$. □

Remark 2.1

If $u_{n}$ and $v_{n}$ are two solutions of (2.2), following the lines of the proof of the first part in Lemma 2.2, we may show that $u_{n}\leqslant v_{n}$. By symmetry, this implies that the solution of (2.2) is unique.

Lemma 2.3

The solution $u_{1}$ to problem (2.2) with $n=1$ satisfies

$$ \int_{\Omega}u_{1}^{-r}\,\mathrm{dx}< \infty, \quad \forall r< 1. $$

(2.5)

Proof

By $\frac{\min\{f(x), 1\}}{(u_{1}+1)^{\alpha(x)}}\leqslant1$ and Lemma 2.2 in [17], we know that there exists $0<\beta<1$ such that $u_{1}\in C^{1,\beta}(\overline{\Omega})$ and $\Vert u_{1}\Vert _{C ^{1,\beta}}\leqslant C$, which implies that the gradient of $u_{1}$ exists everywhere, then the Hopf lemma in [19] shows that $\frac{\partial u_{1}(x)}{\partial\nu}>0 $ in Ω̅, where ν is the outward unit normal vector of ∂Ω at x. Moreover, following the lines of the proof of lemma in [5], we get that

$$ \int_{\Omega}u_{1}^{r}\,\mathrm{dx}< \infty \quad \mbox{if and only if} \quad r>-1. $$

□

We know clearly that the estimates on $u_{n}$ depend on f and $\alpha(x)$, we will discuss this in different cases.

3 The case $0<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha^{+}<1$

In this case, we obtain a priori estimates on $u_{n}$ in $H_{0}^{1}( \Omega)$ only if f is more regular than $L^{1}(\Omega)$. We have the following results.

Lemma 3.1

([20])

Let $u_{n}$ be the solution of (2.2) with $0<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha^{+}<1$, and suppose that $f\in L^{m}(\Omega)$ with $m=\frac{Np}{Np-N+p+(N-p)\alpha^{-}}= (\frac{p^{*}}{1-\alpha^{-}} )^{\prime}$. Then the sequence $\{u_{n}\}$ is bounded in $W_{0}^{1,p}(\Omega)$.

Once we have the boundedness of $u_{n}$, we can prove an existence result for (1.1).

Theorem 3.1

([20])

Suppose that f is a nonnegative function in $L^{m}(\Omega)$ ($f\not\equiv0$), with $m=\frac{Np}{Np-N+p+(N-p)\alpha ^{-}}= (\frac{p^{*}}{1-\alpha^{-}} )'$, $f\not\equiv0$, and let $0<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha^{+}<1$. Then problem (1.1) has a solution $u\in W_{0}^{1,p}(\Omega)$ satisfying (2.1).

The summability of u depends on the summability of f, which is proved in the next lemma.

Lemma 3.2

Suppose that $f\in L^{m}(\Omega)$, $m\geqslant\frac{Np}{Np-N+p+(N-p) \alpha^{-}}$, and let $0<\alpha^{-}\leqslant\alpha(x)\leqslant \alpha^{+}<1$. Then the solution u of (1.1) given by Theorem 3.1 is such that:

(i)

if $m>\frac{N}{p}$, then $u\in L^{\infty}(\Omega)$;
(ii)

if $\frac{Np}{Np-N+p+(N-p)\alpha^{-}}\leqslant m<\frac {N}{p}$, then $u\in L^{s}(\Omega)$, $s=\frac{Nm(\alpha^{-}+p-1)}{N-pm}$.

Proof

To prove (i), let $k>1$ and define $G_{k}(s)=(s-k)_{+}$. Taking $G_{k}(u_{n})$ as a test function in (2.2), we obtain

$$ \int_{\Omega}\bigl\vert \nabla G_{k}(u_{n}) \bigr\vert ^{p}\,\mathrm{dx}\leqslant \int_{ \Omega}\bigl(\bigl\vert \nabla G_{k}(u_{n}) \bigr\vert ^{p-2}\nabla G_{k}(u_{n})\bigr)\cdot \nabla G_{k}(u_{n})\,\mathrm{dx}= \int_{\Omega}\frac{f_{n}G_{k}(u_{n})}{(u _{n}+\frac{1}{n})^{\alpha(x)}}\,\mathrm{dx}. $$

Since $G_{k}(u_{n})\neq0$, it implies that

$$ \int_{\Omega}\bigl\vert \nabla G_{k}(u_{n}) \bigr\vert ^{p}\,\mathrm{dx}\leqslant \int_{ \Omega}fG_{k}(u_{n})\,\mathrm{dx}. $$

(3.1)

Starting from inequality (3.1), Theorem 4.2 in [21] shows that there exists a constant C (independent of n) such that

$$ \Vert u_{n}\Vert _{L^{\infty}(\Omega)}\leqslant C\Vert f\Vert _{L^{m}(\Omega)}, $$

which implies that u belongs to $L^{\infty}(\Omega)$.

To prove (ii), noting that if $m=\frac{Np}{Np-N+p+(N-p)\alpha^{-}}$, $s=\frac{Np}{N-p}=p^{*}$, since $u\in W_{0}^{1,p}(\Omega)$, the result when $m=\frac{Np}{Np-N+p+(N-p)\alpha^{-}}$ is true by the Sobolev embedding theorem. If $\frac{Np}{Np-N+p+(N-p)\alpha^{-}}< m< \frac{N}{p}$, letting $\delta>1$ and choosing $u_{n}^{p\delta-p+1}$ as a test function in (2.2), using Hölder’s inequality, we get that

$$ \begin{aligned}[b] &(p\delta-p+1) \int_{\Omega} \vert \nabla u_{n}\vert ^{p}u_{n}^{p\delta-p} \,\mathrm{dx}\\ &\quad \leqslant \int_{\{x\in\Omega, u_{n}\geqslant1\}}\frac{fu_{n}^{p \delta-p+1}}{u_{n}^{\alpha^{-}}}\,\mathrm{dx}+ \int_{\{x\in\Omega, u_{n}< 1\}}\frac{fu_{n}^{p\delta-p+1}}{u_{n} ^{\alpha^{+}}}\,\mathrm{dx}\\ &\quad = \Vert f\Vert _{L^{m}(\Omega)} \biggl( \biggl( \int_{\Omega }u_{n}^{(p\delta-p+1- \alpha^{-})m'}\,\mathrm{dx}\biggr)^{\frac{1}{m'}} \\ &\quad\quad{} + \vert \Omega \vert ^{ \frac{\alpha^{+}-\alpha^{-}}{(p\delta-p+1-\alpha^{-})m'}} \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'}\,\mathrm{dx}\biggr)^{\frac{p \delta-p+1-\alpha^{+}}{(p\delta-p+1-\alpha^{-})m'}} \biggr). \end{aligned} $$

(3.2)

By the Sobolev inequality (on the left-hand side), we have that

$$ \int_{\Omega} \vert \nabla u_{n}\vert ^{p}u_{n}^{p\delta -p}\,\mathrm{dx}=\frac{1}{ \delta^{p}} \int_{\Omega}\bigl\vert \nabla u_{n}^{\delta} \bigr\vert ^{p}\,\mathrm{dx}\geqslant\frac{S}{\delta^{p}} \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}}, $$

(3.3)

where S is the constant of the Sobolev embedding theorem. Combining with (3.2) and (3.3), we have that

$$ \begin{aligned}[b] & \frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega}u_{n}^{p^{*} \delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \\ & \quad \leqslant \Vert f\Vert _{L^{m}( \Omega)} \biggl( \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}} \\ &\quad\quad {} + \vert \Omega \vert ^{ \frac{\alpha^{+}-\alpha^{-}}{(p\delta-p+1-\alpha^{-})m'}} \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'}\,\mathrm{dx}\biggr)^{\frac{p \delta-p+1-\alpha^{+}}{(p\delta-p+1-\alpha^{-})m'}} \biggr). \end{aligned} $$

(3.4)

We choose δ in such a way that $p^{*}\delta=(p\delta-p+1- \alpha^{-})m'$, i.e.,

$$ \delta=\frac{(\alpha^{-}+p-1)m(N-p)}{p(N-mp)}, $$

which yields that $\delta>1$ if and only if $\frac{Np}{Np-N+p+(N-p) \alpha^{-}}< m<\frac{N}{p}$, and that $p^{*}\delta=\frac{Nm(\alpha ^{-}+p-1)}{N-pm}=s$. Therefore, (3.4) becomes

$$ \begin{aligned} \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} &\leqslant \frac{\delta^{p}}{S(p\delta-p+1)}\Vert f\Vert _{L^{m}(\Omega)} \biggl( \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{1}{m'}} \\ &\quad{} +\vert \Omega \vert ^{ \frac{\alpha^{+}-\alpha^{-}}{(p\delta-p+1-\alpha^{-})m'}} \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{p\delta-p+1-\alpha ^{+}}{(p\delta-p+1-\alpha^{-})m'}} \biggr), \end{aligned} $$

which implies that

$$ \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{p-1+\alpha^{+}}{p ^{*}\delta}} \leqslant\frac{\delta^{p}}{S(p\delta-p+1)}\Vert f\Vert _{L ^{m}(\Omega)} \biggl( \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{ \alpha^{+}-\alpha^{-}}{p^{*}\delta}} +\vert \Omega \vert ^{\frac{\alpha^{+}- \alpha^{-}}{p^{*}\delta}} \biggr). $$

(3.5)

Using Young’s inequality on the right-hand side in (3.5), we have that

$$ \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{p-1+\alpha^{+}}{p ^{*}\delta}} \leqslant\frac{\delta^{p}}{S(p\delta-p+1)}\Vert f\Vert _{L ^{m}(\Omega)} \biggl(\varepsilon \biggl( \int_{\Omega}u_{n}^{s} \,\mathrm{dx}\biggr)^{\frac{p-1+\alpha^{+}}{p^{*}\delta}} + \varepsilon^{-\frac{\alpha^{+}-\alpha^{-}}{p-1+\alpha^{-}}}+\vert \Omega \vert ^{\frac{ \alpha^{+}-\alpha^{-}}{p^{*}\delta}} \biggr), $$

where $\varepsilon=\frac{S(p\delta-p+1)}{2\delta^{p}\Vert f\Vert _{L^{m}( \Omega)}}$. Thus, we get that

$$ \biggl( \int_{\Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{p-1+\alpha^{+}}{p ^{*}\delta}} \leqslant\frac{2\delta^{p}\Vert f\Vert _{L^{m}(\Omega)}}{S(p \delta-p+1)} \biggl( \biggl( \frac{2\delta^{p}\Vert f\Vert _{L^{m}(\Omega)}}{S(p \delta-p+1)} \biggr)^{\frac{\alpha^{+}-\alpha^{-}}{p-1+\alpha^{-}}} +\vert \Omega \vert ^{\frac{\alpha^{+}-\alpha^{-}}{p^{*}\delta}} \biggr). $$

(3.6)

Therefore, we know that $u_{n}$ is bounded in $L^{s}(\Omega)$, so is $u\in L^{s}(\Omega)$. □

Theorem 3.2

Suppose that $f\in L^{m}(\Omega)$, $\frac{(p-1+\alpha^{+})N}{(\alpha ^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}\leqslant m<\frac{Np}{Np-N+p+(N-p) \alpha^{-}}$, and $0<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha ^{+}<1$. Then problem (1.1) has a solution u in $W_{0}^{1,q}( \Omega)$, $q=\frac{Nm(\alpha^{-}+p-1)}{N-m(1-\alpha^{-})}$.

Proof

The lines of our proof are that if we can prove that $u_{n}$ is bounded in $W_{0}^{1,q}(\Omega)$ (with q as in the statement), the existence of a solution u in $W_{0}^{1,q}(\Omega)$ of (1.1) will be proved by passing to the limit in (2.2) as in the proof of Theorem 3.1. To prove that $u_{n}$ is bounded in $W_{0}^{1,q}( \Omega)$, we begin by proving that it is bounded in $L^{s}(\Omega)$, with $s=\frac{Nm(\alpha^{-}+p-1)}{N-pm}$. To attain this goal, we choose $u_{n}^{p\delta-p+1}$ as a test function in (2.2) as in the statement of Lemma 3.2, where $\frac{p-1+\alpha^{+}}{p} \leqslant\delta<1$; however, $\nabla u_{n}^{p\delta-p+1}$ will be singular at $u_{n}=0$. Therefore, we choose $(u_{n}+\varepsilon)^{p \delta-p+1}-\varepsilon^{p\delta-p+1}$ as a test function in (2.2), where $\varepsilon<\frac{1}{n}$ for n fixed. We have that

$$ (p\delta-p+1) \int_{\Omega} \vert \nabla u_{n}\vert ^{p}(u_{n}+\varepsilon)^{p \delta-p}\,\mathrm{dx}\leqslant \int_{\Omega}\frac{f_{n}(u_{n}+ \varepsilon)^{p\delta-p+1}}{(u_{n}+\varepsilon)^{\alpha(x)}} \,\mathrm{dx}. $$

Since $f_{n}\leqslant f$, we have that

$$ \begin{aligned}[b] &(p\delta-p+1) \int_{\Omega} \vert \nabla u_{n}\vert ^{p}(u_{n}+\varepsilon)^{p \delta-p}\,\mathrm{dx}\\ &\quad \leqslant \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1-\alpha ^{-}} \,\mathrm{dx}+ \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1- \alpha^{+}} \,\mathrm{dx}. \end{aligned} $$

(3.7)

By the Sobolev embedding theorem ($W_{0}^{1,p}(\Omega)\hookrightarrow L^{p^{*}}(\Omega)$) on the left-hand side, it follows that

$$ \begin{aligned}[b] \int_{\Omega} \vert \nabla u_{n}\vert ^{p}(u_{n}+\varepsilon)^{p\delta-p} \,\mathrm{dx}&= \int_{\Omega}\frac{\vert \nabla ((u_{n}+\varepsilon)^{ \delta}-\varepsilon^{\delta} )\vert ^{p}}{\delta^{p}}\,\mathrm{dx}\\ &\geqslant \frac{S}{\delta^{p}} \biggl( \int_{\Omega} \bigl((u_{n}+\varepsilon )^{\delta}-\varepsilon^{\delta} \bigr)^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p ^{*}}}, \end{aligned} $$

(3.8)

where S is the best constant of the Sobolev embedding theorem. Combining (3.7) with (3.8), we have that

$$ \begin{aligned}[b] &\frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega} \bigl((u_{n}+ \varepsilon)^{\delta}- \varepsilon^{\delta} \bigr)^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \\ &\quad \leqslant \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1-\alpha ^{-}} \,\mathrm{dx}+ \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1- \alpha^{+}} \,\mathrm{dx}. \end{aligned} $$

(3.9)

Using Hölder’s inequality on the right-hand side, we get

$$ \begin{aligned} &\frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega} \bigl((u_{n}+ \varepsilon)^{\delta}- \varepsilon^{\delta} \bigr)^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \\ &\quad \leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \int_{\Omega}(u_{n}+\varepsilon)^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}} \\ &\quad\quad {} + \vert \Omega \vert ^{ \frac{\alpha^{+}-\alpha^{-}}{(p\delta-p+1-\alpha^{-})m'}}\Vert f\Vert _{L ^{m}(\Omega)} \biggl( \int_{\Omega}(u_{n}+\varepsilon)^{(p\delta-p+1- \alpha^{-})m'}\,\mathrm{dx}\biggr) ^{\frac{p\delta-p+1-\alpha^{+}}{(p \delta-p+1-\alpha^{-})m'}}. \end{aligned} $$

Letting $\varepsilon\rightarrow0$, we get (3.4), i.e.,

$$ \begin{aligned} \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p ^{*}}} &\leqslant \frac{\delta^{p}}{S(p\delta-p+1)}\Vert f\Vert _{L^{m}( \Omega)} \biggl( \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}} \\ &\quad{} + \vert \Omega \vert ^{ \frac{\alpha^{+}-\alpha^{-}}{(p\delta-p+1-\alpha^{-})m'}} \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'}\,\mathrm{dx}\biggr)^{\frac{p \delta-p+1-\alpha^{+}}{(p\delta-p+1-\alpha^{-})m'}} \biggr), \end{aligned} $$

where δ is chosen in such a way that $p^{*}\delta=(p\delta-p+1- \alpha^{-})m'$, i.e.,

$$ \delta=\frac{(\alpha^{-}+p-1)(N-p)m}{p(N-mp)}. $$

If $m= \frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}$, we choose $\delta=\frac{p-1+\alpha^{+}}{p}$ in (3.9), and letting $\varepsilon\rightarrow0$, we have that

$$ \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p ^{*}}}\leqslant\frac{\delta^{p}}{S(p\delta-p+1)} \biggl( \int_{\Omega }fu_{n}^{p\delta-p+1-\alpha^{-}}\,\mathrm{dx}+ \int_{\Omega}f \,\mathrm{dx}\biggr). $$

Using Hölder’s inequality and Young’s inequality, we get that

$$ \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p ^{*}}} \leqslant \frac{\delta^{p}\Vert f\Vert _{L^{m}(\Omega)}}{S(p\delta -p+1)} \biggl(\varepsilon \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}}+ \varepsilon^{- \frac{p^{*}}{pm'-p^{*}}}+\vert \Omega \vert ^{\frac{1}{m'}} \biggr), $$

where $\varepsilon=\frac{S(p\delta-p+1)}{2\delta^{p}\Vert f\Vert _{L^{m}( \Omega)}}$. Thus we have that

$$ \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p ^{*}}} \leqslant \frac{2\delta^{p}\Vert f\Vert _{L^{m}(\Omega)}}{S(p\delta -p+1)} \biggl( \biggl( \frac{2 \delta^{p}\Vert f\Vert _{L^{m}(\Omega)}}{S(p\delta -p+1)} \biggr)^{\frac{p^{*}}{pm'-p ^{*}}}+\vert \Omega \vert ^{\frac{1}{m'}} \biggr). $$

Therefore we obtain that $u_{n}$ is bounded in $L^{\frac{N(p-1+\alpha ^{+})}{N-p}}(\Omega)$, where $\frac{N(p-1+\alpha^{+})}{N-p}$ is the value of s for $m=\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+ \alpha^{+})}$.

If $\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha ^{+})}< m<\frac{Np}{Np-N+p+(N-p) \alpha^{-}}$, it is clear that the inequality on m holds true if and only if $\frac{p-1+\alpha^{+}}{p}<\delta<1$, starting from (3.4) and arguing as in the proof of Lemma 3.2, we also get that $u_{n}$ is bounded in $L^{s}(\Omega)$ with $s=\frac{Nm( \alpha^{-}+p-1)}{N-pm}$.

The right-hand side of (3.7) is bounded with respect to n (and ε, which we take smaller than 1) by using the estimate on $u_{n}$ in $L^{s}(\Omega)$ and the choice of δ.

Since $\delta<1$,

$$ \int_{\Omega}\frac{\vert \nabla u_{n}\vert ^{p}}{(u_{n}+\varepsilon)^{p-p \delta}} \,\mathrm{dx}= \int_{\Omega} \vert \nabla u_{n}\vert ^{p}(u_{n}+\varepsilon )^{p\delta-p}\,\mathrm{dx}\leqslant C. $$

If $q=\frac{Nm(\alpha^{-}+p-1)}{N-m(1-\alpha^{-})}< p$, by Hölder’s inequality, we have that

$$ \begin{aligned}[b] \int_{\Omega} \vert \nabla u_{n}\vert ^{q}\,\mathrm{dx}&= \int _{\Omega}\frac{\vert \nabla u_{n}\vert ^{q}}{(u_{n}+\varepsilon)^{(1-\delta )q}}(u_{n}+\varepsilon )^{(1-\delta)q}\,\mathrm{dx}\\ &\leqslant \biggl( \int_{\Omega}\frac{\vert \nabla u_{n}\vert ^{p}}{(u_{n}+ \varepsilon)^{p(1-\delta)}}\,\mathrm{dx}\biggr)^{\frac{q}{p}} \biggl( \int_{\Omega}(u_{n}+\varepsilon)^{\frac{pq(1-\delta)}{p-q}} \,\mathrm{dx}\biggr)^{1-\frac{q}{p}} \\ &\leqslant C \biggl( \int_{\Omega}(u_{n}+\varepsilon)^{\frac{pq(1- \delta)}{p-q}}\,\mathrm{dx}\biggr)^{1-\frac{q}{p}}. \end{aligned} $$

(3.10)

The choice of δ and the value of q are such that $\frac{pq(1- \delta)}{p-q}=s$, so that the right-hand side of (3.10) is bounded with respect to n and ε. Hence, $u_{n}$ is bounded in $W_{0}^{1,q}(\Omega)$. □

Theorem 3.3

Suppose that $f\in L^{m}(\Omega)$, $\frac{1}{2-p-\alpha^{+}+p\delta }< m<\frac{(p-1+ \alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}$ ($\frac{p-1+ \alpha^{-}}{p}<\delta<\frac{p-1+\alpha^{+}}{p}$), and $0<\alpha ^{-}\leqslant\alpha(x)\leqslant\alpha^{+}<1$. Then problem (1.1) has a solution u in $W_{0}^{1,q}(\Omega)$, $q=\frac{Nm( \alpha^{-}+p-1)}{N-m(1-\alpha^{-})}$.

Proof

The lines of our proof are similar to those in the proof of Theorem 3.2. We also begin by proving that $u_{n}$ is bounded in $L^{s}(\Omega)$, with $s=\frac{Nm(\alpha^{-}+p-1)}{N-pm}$. To this aim, we also choose $(u_{n}+\varepsilon)^{p\delta-p+1}- \varepsilon^{p\delta-p+1}$ as a test function in (2.2), where $\frac{p-1+\alpha^{-}}{p}<\delta<\frac{p-1+\alpha^{+}}{p}$, $\varepsilon<\frac{1}{n}$ for n fixed. Since $f_{n}\leqslant f$, using the Sobolev embedding theorem ($W_{0}^{1,p}(\Omega )\hookrightarrow L^{p^{*}}(\Omega)$) on the left-hand side again, we have that

$$ \begin{aligned} &\frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega} \bigl((u_{n}+ \varepsilon)^{\delta}- \varepsilon^{\delta} \bigr)^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \\ &\quad \leqslant \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1-\alpha ^{-}} \,\mathrm{dx}+ \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1- \alpha^{+}} \,\mathrm{dx}, \end{aligned} $$

where S is the best constant of the Sobolev embedding theorem.

Using Hölder’s inequality and Lemma 2.3 on the right-hand side, we get that

$$ \begin{aligned} &\frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega} \bigl((u_{n}+ \varepsilon)^{\delta}- \varepsilon^{\delta} \bigr)^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \\ &\quad \leqslant \int_{\Omega}f(u_{n}+\varepsilon)^{p\delta-p+1-\alpha ^{-}} \,\mathrm{dx}+ \int_{\Omega}\frac{f}{u_{1}^{p-1+\alpha^{+}-p\delta }}\,\mathrm{dx}\\ &\quad \leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \int_{\Omega }(u_{n}+\varepsilon )^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}}+C\Vert f\Vert _{L^{m}(\Omega)}. \end{aligned} $$

Letting $\varepsilon\rightarrow0$, we have that

$$ \biggl( \int_{\Omega}u_{n}^{p^{*}\delta} \,\mathrm{dx}\biggr)^{\frac{p}{p ^{*}}} \leqslant\frac{\delta^{p}}{S(p\delta-p+1)}\Vert f\Vert _{L^{m}( \Omega)} \biggl( \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}}+C \biggr), $$

(3.11)

where δ is chosen in such a way that $p^{*}\delta=(p\delta-p+1- \alpha^{-})m'$, i.e.,

$$ \delta=\frac{(\alpha^{-}+p-1)(N-p)m}{p(N-mp)}. $$

If $1< m<\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha ^{+})}$, it is clear that the inequality on m holds true if and only if $\frac{p-1+\alpha^{-}}{p}<\delta<\frac{p-1+\alpha^{+}}{p}$, and arguing as to the case $m=\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+ \alpha^{+})}$ in the proof of Theorem 3.2, we also obtain that $u_{n}$ is bounded in $L^{s}(\Omega)$, with $s=\frac{Nm(\alpha ^{-}+p-1)}{N-pm}$.

Since $\delta<1$,

$$ \int_{\Omega}\frac{\vert \nabla u_{n}\vert ^{p}}{(u_{n}+\varepsilon)^{p-p \delta}} \,\mathrm{dx}= \int_{\Omega} \vert \nabla u_{n}\vert ^{p}(u_{n}+\varepsilon )^{p\delta-p}\,\mathrm{dx}\leqslant C. $$

If $q=\frac{Nm(\alpha^{-}+p-1)}{N-m(1-\alpha^{-})}< p$, similarly to the proof of Theorem 3.2, we have by Hölder’s inequality that

$$ \int_{\Omega} \vert \nabla u_{n}\vert ^{q}\,\mathrm{dx}\leqslant C \biggl( \int_{ \Omega}(u_{n}+\varepsilon)^{\frac{pq(1-\delta)}{p-q}}\,\mathrm{dx}\biggr)^{1-\frac{q}{p}}. $$

Due to the choice of δ and the value of q, the right-hand side of the above inequality is bounded with respect to n and ε. Hence, $u_{n}$ is bounded in $W_{0}^{1,q}(\Omega)$. □

4 The case $1<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha^{+}$

The case $1<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha^{+}$ has many analogies with the case $0<\alpha^{-}<\alpha^{+}<1$. In this case, we can also prove that $u_{n}$ is bounded in $W_{0}^{1,p}(\Omega)$ only if f is more regular than $L^{1}(\Omega)$ and $\alpha^{+}$ and $\alpha^{-}$ is close to 1. Hence we obtain the existence of problem (1.1).

Lemma 4.1

Suppose that $f\in L^{m} (\Omega)$ ($m>1$), let $u_{n}$ be the solution of (2.2) with $1<\alpha^{-}<\alpha^{+}<2-\frac{1}{m}$. Then $u_{n}$ is bounded in $W_{0}^{1,p}(\Omega)$.

Proof

Taking $u_{n}$ as a test function in (2.2), we obtain that

$$ \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\,\mathrm{dx}\leqslant \int_{\Omega}\frac{f}{u _{n}^{\alpha(x)-1}}\,\mathrm{dx}. $$

Using Lemma 2.1 and Lemma 2.2, we know that $u_{n}\geqslant u_{1}$ and there exists a constant $M>0$ s.t. $u_{1}\leqslant M$. Hence $(\frac{M}{u_{1}})^{\alpha(x)-1}\leqslant (\frac{M}{u_{1}})^{\alpha^{+}-1}$, and we have that

$$ \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\,\mathrm{dx}\leqslant \int_{\Omega}\frac{f}{u _{1}^{\alpha(x)-1}}\,\mathrm{dx}\leqslant \bigl(1+M^{\alpha^{+}-\alpha^{-}}\bigr) \int_{\Omega}\frac{f}{u_{1}^{\alpha^{+}-1}}\,\mathrm{dx}. $$

Using Hölder’s inequality on the right-hand side and Lemma 2.3, we obtain

$$ \begin{aligned} \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\,\mathrm{dx}\leqslant C\bigl(1+M^{\alpha ^{+}-\alpha^{-}}\bigr)\Vert f\Vert _{L^{m}(\Omega)}. \end{aligned} $$

Therefore, $u_{n}$ is bounded in $W_{0}^{1,p}(\Omega)$. □

Once we have the boundedness of $u_{n}$, we can prove the following existence theorem along the lines of Theorem 3.1.

Theorem 4.1

Suppose that $f\in L^{m}(\Omega)$ ($m>1$), $f\not\equiv0$ and $1<\alpha^{-}<\alpha^{+}<2-\frac{1}{m}$. Then problem (1.1) has a solution u in $W_{0}^{1,p}(\Omega)$.

The summability of u can be proved along the lines of Lemma 3.2 with small changes.

Lemma 4.2

Suppose that $f\in L^{m}(\Omega)$ ($m>1$) and $1<\alpha^{-}<\alpha^{+}<2- \frac{1}{m}$. Then the solution u of (1.1) given by Theorem 4.1 is such that:

(i)

if $m>\frac{N}{p}$, then $u\in L^{\infty}(\Omega)$;
(ii)

if $\frac{(p-1+\alpha^{+})N}{(\alpha ^{-}+p-1)(N-p)+p(p-1+\alpha ^{+})}\leqslant m<\frac{N}{p}$, then $u\in L^{s}(\Omega)$, $s=\frac{Nm( \alpha^{-}+p-1)}{N-pm}$.

Proof

The proof of (i) is similar to the proof of Lemma 3.2(i), we omit the details here.

To prove (ii), we choose $u_{n}^{p\delta-p+1}$ as a test function with $\delta\geqslant\frac{p-1+\alpha^{+}}{p}$ in (2.2). Similarly to the proof of Lemma 3.2, we obtain that

$$ \frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega}u_{n}^{p^{*} \delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \int_{\Omega}fu _{n}^{p\delta-p+1-\alpha^{-}}\,\mathrm{dx}+ \int_{\Omega}fu_{n}^{p \delta-p+1-\alpha^{+}}\,\mathrm{dx}. $$

(4.1)

If $m= \frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}$, choosing $\delta=\frac{p-1+\alpha^{+}}{p}$ in (4.1), by Hölder’s inequality, we get that

$$ \frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega}u_{n}^{p^{*} \delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \Vert f\Vert _{L^{m}( \Omega)} \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}}+\vert \Omega \vert ^{1-\frac {1}{m}}\Vert f\Vert _{L^{m}( \Omega)}. $$

We choose δ in such a way that $p^{*}\delta=(p\delta-p+1- \alpha^{-})m'$, i.e., $\delta=\frac{(\alpha^{-}+p-1)m(N-p)}{p(N-mp)}$. Since $m=\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha ^{+})}$, we get that $\frac{p}{p^{*}}>\frac{1}{m'}$. Because $s=p^{*}\delta$, we have the boundedness of $u_{n}$ in $L^{\frac{N(p-1+ \alpha^{+})}{N-p}}(\Omega)$, which is the value of s for $m=\frac{N(p-1+\alpha^{+})}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}$.

If $\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}< m< \frac{N}{p}$, starting from inequality (4.1), using Hölder’s inequality, we get that

$$ \begin{aligned} \frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega}u_{n}^{p^{*} \delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} &\leqslant \Vert f\Vert _{L^{m}( \Omega)} \biggl( \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'} \,\mathrm{dx}\biggr)^{\frac{1}{m'}} \\ &\quad{} + \vert \Omega \vert ^{ \frac{\alpha^{+}-\alpha^{-}}{(p\delta-p+1-\alpha^{-})m'}} \biggl( \int_{\Omega}u_{n}^{(p\delta-p+1-\alpha^{-})m'}\,\mathrm{dx}\biggr)^{\frac{p \delta-p+1-\alpha^{+}}{(p\delta-p+1-\alpha^{-})m'}} \biggr). \end{aligned} $$

We also choose δ in such a way that $p^{*}\delta=(p\delta-p+1- \alpha^{-})m'$, which yields that $\delta>\frac{p-1+\alpha^{+}}{p}$ if and only if $m>\frac{(p-1+\alpha^{+})N}{(\alpha^{-}+p-1)(N-p)+p(p-1+ \alpha^{+})}$, and that $p^{*}\delta=s$. So, since $\frac{p}{p^{*}}> \frac{1}{m'}$ being $m<\frac{N}{p}$, we have the boundedness of $u_{n}$ in $L^{s}(\Omega)$, so does $u\in L^{s}(\Omega)$. □

Moreover, we can prove that a positive power of $u_{n}$ is bounded in $W_{0}^{1,p}(\Omega)$ only if f is more regular than $L^{1}(\Omega )$ and $\alpha^{+}$ is close to $\alpha^{-}$, and we only have the boundedness of $u_{n}$ in $W_{\mathrm{loc}}^{1,p}(\Omega)$.

Lemma 4.3

Suppose that $f\in L^{m}(\Omega)$ ($m>1$), let $u_{n}$ be the solution of (2.2) with $1<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha ^{+}$ and $\alpha^{+}-\alpha^{-}<1-\frac{1}{m}$. Then $u_{n}^{\frac{p-1+ \alpha^{-}}{p}}$ is bounded in $W_{0}^{1,p}(\Omega)$, and $u_{n}$ is bounded in $W_{\mathrm{loc}}^{1,p}(\Omega)$ and in $L^{s}(\Omega)$, with $s=\frac{N(\alpha^{-}+p-1)}{N-p}$.

Proof

Taking $u_{n}^{\alpha^{-}}$ as a test function in (2.2), since $\frac{u_{n}^{\alpha^{-}}}{(u_{n}+\frac{1}{n})^{\alpha^{-}}}\leqslant 1$ and $f_{n}\leqslant f$, by Hölder’s inequality and Lemma 2.3, we get that

$$ \begin{aligned} \alpha^{-} \int_{\Omega} \vert \nabla u_{n}\vert ^{p}u_{n}^{\alpha^{-}-1} \,\mathrm{dx}&\leqslant \int_{\Omega}\frac{fu_{n}^{\alpha^{-}}}{(u_{n}+ \frac{1}{n})^{\alpha^{-}}}\,\mathrm{dx}+ \int_{\Omega}\frac{fu_{n}^{ \alpha^{-}}}{u_{n}^{\alpha^{+}}}\,\mathrm{dx}\\ &\leqslant \int_{\Omega}f \,\mathrm{dx}+ \int_{\Omega}\frac{f}{u_{n} ^{\alpha^{+}-\alpha^{-}}}\,\mathrm{dx}\leqslant \vert \Omega \vert ^{1- \frac{1}{m}}\Vert f\Vert _{L^{m}(\Omega)}+C\Vert f\Vert _{L^{m}(\Omega)}. \end{aligned} $$

Since

$$ \int_{\Omega} \vert \nabla u_{n}\vert ^{p}u_{n}^{\alpha ^{-}-1}\,\mathrm{dx}=\frac{p ^{p}}{(\alpha^{-}+p-1)^{p}} \int_{\Omega}\bigl\vert \nabla u_{n}^{\frac {\alpha ^{-}+p-1}{p}} \bigr\vert ^{p}\,\mathrm{dx}, $$

we have that

$$ \frac{p^{p}\alpha^{-}}{(\alpha^{-}+p-1)^{p}} \int_{\Omega}\bigl\vert \nabla u _{n}^{\frac{\alpha^{-}+p-1}{p}} \bigr\vert ^{p}\,\mathrm{dx}\leqslant\bigl(C+\vert \Omega \vert ^{1-\frac{1}{m}}\bigr)\Vert f\Vert _{L^{m}(\Omega)}. $$

Thus, we have that $u_{n}^{\frac{\alpha^{-}+p-1}{p}}$ is bounded in $W_{0}^{1,p}(\Omega)$.

Applying the Sobolev embedding theorem to $u_{n}^{ \frac{\alpha^{-}+p-1}{p}}(W_{0}^{1,p}(\Omega)\hookrightarrow L^{p ^{*}}(\Omega))$, we get that

$$ S \biggl( \int_{\Omega}\bigl\vert u_{n}^{\frac{\alpha^{-}+p-1}{p}}\bigr\vert ^{p^{*}} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \int_{\Omega}\bigl\vert \nabla u _{n}^{\frac{\alpha^{-}+p-1}{p}} \bigr\vert ^{p}\,\mathrm{dx}, $$

where S is the best constant of the Sobolev embedding theorem. Since the boundedness of $u_{n}^{\frac{\alpha^{-}+p-1}{p}}$ in $W_{0}^{1,p}( \Omega)$, we thus have the boundedness of $u_{n}$ in $L^{s}(\Omega)$.

To prove the boundedness of $u_{n}$ in $W_{\mathrm{loc}}^{1,p}(\Omega)$, we choose $u_{n}\varphi^{p}$ as a test function in (2.2), where $\varphi\in C_{0}^{\infty}(\Omega)$, $\Omega'=\{x\in\Omega, \varphi\neq0\}$. By (2.4), we have that

$$ \begin{aligned} & \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\varphi ^{p}\,\mathrm{dx}+p \int_{\Omega }\bigl(\vert \nabla u_{n}\vert ^{p-2}\nabla u_{n}\bigr)\cdot\nabla \varphi u_{n} \varphi ^{p-1}\,\mathrm{dx}\\ &\quad \leqslant \int_{\Omega} \frac{f_{n}\varphi^{p}}{u_{n}^{\alpha(x)-1}}\,\mathrm{dx}\leqslant \int_{\Omega}\frac{f_{n}\varphi^{p}}{C_{\Omega'}^{\alpha(x)-1}} \,\mathrm{dx}\leqslant\frac{1}{\min\{C_{\Omega'}^{\alpha^{+}-1}, C _{\Omega'}^{\alpha^{-}-1}\}} \int_{\Omega}f_{n}\varphi^{p} \,\mathrm{dx}. \end{aligned} $$

By Young’s inequality, we have that

$$ \begin{aligned} &p \int_{\Omega}\bigl(\vert \nabla u_{n}\vert ^{p-2}\nabla u_{n}\bigr)\cdot\nabla\varphi u_{n} \varphi^{p-1}\,\mathrm{dx}\\ &\quad \leqslant \frac{p\frac{1}{2(p-1)}}{\frac{p}{p-1}} \int_{\Omega} \bigl\vert \vert \nabla u_{n}\vert ^{p-2}\nabla u_{n} \bigr\vert ^{\frac{p}{p-1}} \varphi^{p} \,\mathrm{dx}+\frac{p(2(p-1))^{p-1}}{p} \int_{\Omega} \vert \nabla\varphi \vert ^{p}u _{n}^{p}\,\mathrm{dx}\\ &\quad \leqslant \frac{1}{2} \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\varphi^{p} \,\mathrm{dx}+ \bigl(2(p-1) \bigr)^{p-1} \int_{\Omega} \vert \nabla\varphi \vert ^{p}u _{n}^{p}\,\mathrm{dx}. \end{aligned} $$

Since $u_{n}$ is bounded in $L^{s}(\Omega)$ (where $s\geqslant p$), by Hölder’s inequality, we obtain that

$$ \begin{aligned} &\frac{1}{2} \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\varphi ^{p}\,\mathrm{dx}\\ & \quad \leqslant\frac{1}{\min\{C_{\Omega'}^{\alpha^{+}-1}, C_{\Omega'} ^{\alpha^{-}-1}\}} \int_{\Omega}f_{n}\varphi^{p}\,\mathrm{dx}+ \bigl(2(p-1)\bigr)^{p-1} \int_{\Omega} \vert \nabla\varphi \vert ^{p}u_{n}^{p} \,\mathrm{dx}\\ & \quad \leqslant \frac{\Vert \varphi \Vert _{L^{\infty}(\Omega)}^{p}}{\min \{C _{\Omega'}^{\alpha^{+}-1}, C_{\Omega'}^{\alpha^{-}-1}\}} \int_{ \Omega}f \,\mathrm{dx}+\bigl(2(p-1)\bigr)^{p-1}\Vert \nabla\varphi \Vert _{L^{\infty}( \Omega)}^{p} \int_{\Omega}u_{n}^{p}\,\mathrm{dx}\\ & \quad \leqslant \frac{\vert \Omega \vert ^{1-\frac{1}{m}}\Vert \varphi \Vert _{L^{\infty}( \Omega)}^{p}\Vert f\Vert _{L^{m}(\Omega)}}{\min\{ C_{\Omega'}^{\alpha^{+}-1}, C_{\Omega'}^{\alpha^{-}-1}\}} +\bigl(2(p-1)\bigr)^{p-1}\vert \Omega \vert ^{1- \frac{p}{s}}\Vert \nabla\varphi \Vert _{L^{\infty}(\Omega )}^{p} \biggl( \int_{ \Omega}u_{n}^{s}\,\mathrm{dx}\biggr)^{\frac{p}{s}}, \end{aligned} $$

and hence $u_{n}$ is bounded in $W_{\mathrm{loc}}^{1,p}(\Omega)$. □

Once we have the boundedness of $u_{n}$, we can prove the following existence theorem along the lines of Theorem 3.1.

Theorem 4.2

Suppose that f is a nonnegative function in $L^{m}(\Omega)$ ($m>1$), ($f \not\equiv0$), $1<\alpha^{-}\leqslant\alpha(x)\leqslant\alpha^{+}$ and $\alpha^{+}-\alpha^{-}<1-\frac{1}{m}$. Then problem (1.1) has a solution u in $W_{\mathrm{loc}}^{1,p}(\Omega)$. Furthermore, $u^{\frac{ \alpha^{-}+p-1}{p}}$ belongs to $W_{0}^{1,p}(\Omega)$.

The summability of u can be proved as the following lemma, the proof is similar to the proof of Lemma 4.2

Lemma 4.4

Suppose that $f\in L^{m}(\Omega)$, $1<\alpha^{-}\leqslant\alpha(x) \leqslant\alpha^{+}$ and $\alpha^{+}-\alpha^{-}<1-\frac{1}{m}$. Then the solution u of (1.1) given by Theorem 4.2 is such that:

(i)

if $m>\frac{N}{p}$, then $u\in L^{\infty}(\Omega)$;
(ii)

if $\frac{N(p-1+\alpha^{+})}{(\alpha ^{-}+p-1)(N-p)+p(p-1+\alpha ^{+})}\leqslant m<\frac{N}{p}$, then $u\in L^{s}(\Omega)$, $s=\frac{Nm( \alpha^{-}+p-1)}{N-mp}$.

5 The case $0<\alpha^{-}<1<\alpha^{+}$

If $0<\alpha^{-}<1<\alpha^{+}$, the boundedness of $u_{n}$ in $W_{0}^{1,p}(\Omega)$ can also be obtained only if f is more regular than $L^{1}(\Omega)$. Furthermore, the existence of problem (1.1) is obtained, the proof has many analogies with the case $0<\alpha^{-}<\alpha^{+}<1$. We have the following results.

Lemma 5.1

Suppose that $f\in L^{m}(\Omega)$, with $m=\frac{Np}{Np-N+p+(N-p) \alpha^{-}}$, and let $u_{n}$ be the solution of (2.2) with $0<\alpha^{-}<1<\alpha^{+}<2-\frac{1}{m}$. Then the sequence $\{u_{n}\}$ is bounded in $W_{0}^{1,p}(\Omega)$.

Proof

We choose $u_{n}$ as a test function in (2.2), by Hölder’s inequality and Lemma 2.3, since $f_{n}\leqslant f$, we have that

$$ \begin{aligned}[b] \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\,\mathrm{dx}&\leqslant \int_{\Omega}fu _{n}^{1-\alpha^{-}}\,\mathrm{dx}+ \int_{\Omega}\frac{f}{u_{1}^{\alpha ^{+}-1}}\,\mathrm{dx}\\ & \leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \int_{\Omega }u_{n}^{(1-\alpha ^{-})m'}\,\mathrm{dx}\biggr)^{\frac{1}{m'}} +C\Vert f\Vert _{L^{m}(\Omega)}. \end{aligned} $$

(5.1)

Applying the Sobolev embedding theorem on the left-hand side, we get

$$ S \biggl( \int_{\Omega}u_{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \int_{\Omega} \vert \nabla u_{n}\vert ^{p}\,\mathrm{dx}. $$

(5.2)

Combining (5.1) with (5.2) implies that

$$ S \biggl( \int_{\Omega}u_{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \int_{\Omega }u_{n}^{(1-\alpha ^{-})m'}\,\mathrm{dx}\biggr)^{\frac{1}{m'}}+C\Vert f\Vert _{L^{m}(\Omega)}. $$

Let $p^{*}=(1-\alpha^{-})m'$, it follows that

$$ S \biggl( \int_{\Omega}u_{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \int_{\Omega }u_{n}^{p^{*}} \,\mathrm{dx}\biggr)^{\frac{1}{m'}}+C\Vert f\Vert _{L^{m}(\Omega)}. $$

By Young’s inequality, we get that

$$ S \biggl( \int_{\Omega}u_{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \varepsilon \biggl( \int_{\Omega}u _{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} + \varepsilon^{-\frac{p^{*}}{pm'-p^{*}}} \biggr)+C\Vert f\Vert _{L^{m}(\Omega)}. $$

Thus, we have that

$$ \biggl( \int_{\Omega}u_{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} \leqslant\frac{\Vert f\Vert _{L^{m}(\Omega )}}{S-\varepsilon \Vert f\Vert _{L^{m}( \Omega)}} \bigl(\varepsilon^{-\frac{p^{*}}{pm'-p^{*}}}+C \bigr). $$

We choose $\varepsilon=\frac{S}{2\Vert f\Vert _{L^{m}(\Omega)}}$ to get

$$ \biggl( \int_{\Omega}u_{n}^{p^{*}}\,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} =\frac{2 \Vert f\Vert _{L^{m}(\Omega)}}{S} \biggl( \biggl(\frac{2\Vert f\Vert _{L^{m}(\Omega)}}{S} \biggr)^{\frac{p^{*}}{pm'-p^{*}}}+C \biggr). $$

So the boundedness of $u_{n}$ in $L^{p^{*}}(\Omega)$ is obtained. Using the estimate and (5.1) again, we have the estimate of $u_{n}$ in $W_{0}^{1,p}(\Omega)$. □

Once the boundedness of $u_{n}$ in $W_{0}^{1,p}(\Omega)$ is obtained, we can prove the following existence theorem.

Theorem 5.1

Suppose that $f\in L^{m}(\Omega)$ with $m=\frac{Np}{Np-N+p+(N-p) \alpha^{-}}$, $f\not\equiv0$, and $0<\alpha^{-}<1<\alpha^{+}<2- \frac{1}{m}$. Then problem (1.1) has a solution u in $W_{0}^{1,p}(\Omega)$.

Lemma 5.2

Suppose that $f\in L^{m}(\Omega)$ with $m\geqslant\frac{Np}{Np-N+p+(N-p) \alpha^{-}}$, and $0<\alpha^{-}<1<\alpha^{+}<2-\frac{1}{m}$. Then the solution u of (1.1) given by Theorem 5.1 is such that:

(i)

if $m>\frac{N}{p}$, then $u\in L^{\infty}(\Omega)$;
(ii)

if $\frac{Np}{Np-N+p+(N-p)\alpha^{-}}\leqslant m<\frac {N}{p}$, then $u\in L^{s}(\Omega)$, $s=\frac{Nm(\alpha^{-}+p-1)}{N-mp}$.

Proof

The proof of (i) is similar to that for Lemma 3.2(i), we omit the details here.

To prove (ii), if $\frac{N(p-1+\alpha^{+})}{(\alpha^{-}+p-1)(N-p)+p(p-1+ \alpha^{+})}\leqslant m<\frac{N}{p}$, the proof is identical to that for Lemma 4.2, we also omit it here.

If $m=\frac{Np}{Np-N+p+(N-p)\alpha^{-}}$, we can prove the results by the Sobolev embedding theorem.

If $\frac{Np}{Np-N+p+(N-p)\alpha^{-}}< m<\frac{N(p-1+\alpha^{+})}{( \alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}$, we choose $1<\delta<\frac{p-1+ \alpha^{+}}{p}$, and use once again $u_{n}^{p\delta-p+1}$ as a test function in (2.2). Using $\delta>1>\frac{p-1+\alpha^{-}}{p}$, as well as Hölder’s inequality, the Sobolev embedding theorem, Lemma 2.3, we get that

$$ \begin{aligned} \frac{S(p\delta-p+1)}{\delta^{p}} \biggl( \int_{\Omega}u_{n}^{p^{*} \delta} \,\mathrm{dx}\biggr)^{\frac{p}{p^{*}}} &\leqslant \int_{\Omega}fu _{n}^{p\delta-p+1-\alpha^{-}}\,\mathrm{dx}+ \int_{\Omega}\frac{f}{u _{1}^{p-1+\alpha^{+}-p\delta}} \,\mathrm{dx}\\ &\leqslant \Vert f\Vert _{L^{m}(\Omega)} \biggl( \int_{\Omega }u_{n}^{(p\delta -p+1-\alpha^{-})m'}\,\mathrm{dx}\biggr)^{\frac{1}{m'}}+C\Vert f\Vert _{L^{m}( \Omega)}. \end{aligned} $$

The choice of δ in such a way that $p^{*}\delta=(p\delta-p+1- \alpha^{-})m'$ yields that $1<\delta<\frac{p-1+\alpha^{+}}{p}$ if and only if $\frac{Np}{Np-N+p+(N-p)\alpha^{-}}< m<\frac{N(p-1+\alpha^{+})}{( \alpha^{-}+p-1)(N-p)+p(p-1+\alpha^{+})}$, and that $p^{*}\delta=s$. The choice of $m<\frac{N}{p}$ implies that $\frac{p}{p^{*}}>\frac{1}{m'}$. Thus we have the boundedness of $u_{n}$ in $L^{s}(\Omega)$, and so does the limit u in $L^{s}(\Omega)$. □

6 Conclusions

In this paper, we study the existence and regularity of solutions to the quasilinear elliptic problem with nonlinear singular terms and variable exponent. Due to the nonlinearity of a p-Laplace operator and the anisotropic variable exponent $\alpha(x)$, some classical methods may not directly be applied to our problem. We construct a suitable test function and apply the Leray-Schauder fixed point theorem to prove the existence of positive solutions with necessary a priori estimate and compact argument. Furthermore, we prove that the existence and regularity of solutions depend on the summability of f and the value of $\alpha(x)$.

Declarations

Acknowledgements

We are very grateful to the anonymous referees for their valuable suggestions that improved the article. The work is supported by the National Natural Science Foundation of China (11501051, 11626044) and the fund of the “Thirteen Five” Scientific and Technological Research Planning Project of the Department of Education of Jilin Province in China(JJKH20170301KJ).

Authors’ contributions

All authors collaborated in all the steps concerning the research and achievements presented in the final manuscript. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)

School of Science, Changchun University of Science and Technology, Changchun, 130022, P.R. China

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