Global $C^{1,\alpha }$ regularity and existence of multiple solutions for singular p(x)-Laplacian equations

Abstract

In this paper we study the following singular p(x)-Laplacian problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} - \text{ div } \left( |\nabla u|^{p(x)-2} \nabla u\right) =\frac{ \lambda }{u^{\beta (x)}}+u^{q(x)}, &{} \text{ in }\quad \Omega , \\ u>0, &{} \text{ in }\quad \Omega , \\ u=0, &{} \text{ on }\quad \partial \Omega , \end{array}\right. \end{aligned}$$

where $\Omega$ is a bounded domain in $\mathbb {R}^N$, $N\ge 2$, with smooth boundary $\partial \Omega$, $\beta \in C^1(\bar{\Omega })$ with $0< \beta (x) <1$, $p\in C^1(\bar{\Omega })$, $q \in C(\bar{\Omega })$ with $p(x)>1$, $p(x)< q(x) +1 <p^*(x)$ for $x \in \bar{\Omega }$, where $p^*(x)= \frac{Np(x)}{N-p(x)}$ for $p(x) <N$ and $p^*(x)= \infty$ for $p(x) \ge N$. We establish $C^{1,\alpha }$ regularity of weak solutions of the problem and strong comparison principle. Based on these two results, we prove the existence of multiple (at least two) positive solutions for a certain range of $\lambda$.

Appendix

We show the Gâteaux differentiability of the energy functional $E_\lambda$ at a point $u \in W_0^{1,p(x)} (\Omega )$ satisfying $u \ge \epsilon \phi _1$ in $\Omega$ with a constant $\epsilon >0$. The result for the singular term $\int _\Omega u^{-\beta (x)}v ~dx$ is proved in a similar way as in [18].

Lemma 6.1

Let $p(x)\le q(x)+1<p^*(x)$ for all $x \in \bar{ \Omega }$. Assume that $u,v \in W_0^{1,p(x)} (\Omega )$ and u satisfies $u \ge \epsilon \phi _1$ in $\Omega$ with a constant $\epsilon >0$. Then we have

$$\begin{aligned} \lim _{t\rightarrow 0} \frac{E_\lambda (u+tv)-E_\lambda (u)}{t} = \int _\Omega |\nabla u|^{p(x)-2}\nabla u \cdot \nabla v~dx -\lambda \int _\Omega u^{-\beta (x)}v~dx -\int _\Omega u^{q(x)}v~dx. \end{aligned}$$

(6.1)

Proof

First, we easily see that

$$\begin{aligned} \frac{E_\lambda (u+tv)-E_\lambda (u)}{t}= & {} \int _\Omega \frac{1}{p(x)} \frac{ |\nabla u +t\nabla v|^{p(x) }-|\nabla u|^{p(x)} }{t}~dx \\&- \lambda \int _\Omega \frac{ 1}{1-\beta (x)} \frac{ ((u+tv)^+) ^{1-\beta (x)}-(u^+)^{1-\beta (x)}}{t}~dx\\&-\int _\Omega \frac{1}{1+q(x)} \frac{ (u+tv)^{q(x)+1}-u^{q(x)+1}}{t}~dx. \end{aligned}$$

For $0< |t| <1$, by the mean value theorem, there exists $\theta \in \mathbb {R}$ with $0<|\theta |< |t|$ such that

$$\begin{aligned} \frac{1}{p(x)} \frac{ |\nabla u +t\nabla v|^{p(x) }-|\nabla u|^{p(x)} }{t}= & {} |\nabla u + \theta \nabla v|^{p(x)-2} (\nabla u +\theta \nabla v)\cdot \nabla v \\ \nonumber\le & {} |\nabla u +\theta \nabla v|^{p(x)-1} |\nabla v|\\ \nonumber\le & {} (|\nabla u|+ |\nabla v|)^{p(x)-1} |\nabla v|. \end{aligned}$$

(6.2)

Then by Lemma 5.4,

$$\begin{aligned} \int _\Omega \left( |\nabla u|+ |\nabla v|\right) ^{p\left( x\right) -1} |\nabla v| ~dx\le & {} 2 ~\Vert \left( |\nabla u|+|\nabla v|\right) ^{p\left( x\right) -1} \Vert _{p'\left( x\right) } \Vert \nabla v\Vert _{p\left( x\right) }\\= & {} 2 ~\Vert |\nabla u|+|\nabla v| \Vert _{p\left( x\right) } \Vert \nabla v\Vert _{p\left( x\right) }\\\le & {} 2^{p_+} \left( \Vert \nabla u\Vert _{p\left( x\right) }+\Vert \nabla v\Vert _{p\left( x\right) }\right) \Vert \nabla v\Vert _{p\left( x\right) }, \end{aligned}$$

where $p'(x)= \frac{p(x)}{p(x)-1}$. Thus, $(|\nabla u|+ |\nabla v|)^{p(x)-1} |\nabla v| \in L^1(\Omega )$. Now it follows from the dominated convergence theorem that

$$\begin{aligned}&\lim _{t\rightarrow 0} \int _\Omega \frac{1}{p(x)} \frac{ |\nabla u +t\nabla v|^{p(x) }-|\nabla u|^{p(x)} }{t}~dx\\&\quad = \int _\Omega \lim _{\theta \rightarrow 0} \left( |\nabla u + \theta \nabla v|^{p(x)-2} \left( \nabla u +\theta \nabla v\right) \cdot \nabla v \right) dx\\&\quad =\int _\Omega |\nabla u|^{p(x)-2}\nabla u \cdot \nabla v~dx. \end{aligned}$$

Likewise, we have that

$$\begin{aligned} \lim _{t\rightarrow 0 }\int _\Omega \frac{1}{1+q(x)} \frac{ (u+tv)^{q(x)+1}-u^{q(x)+1}}{t}~dx = \int _\Omega u^{q(x)}v~dx. \end{aligned}$$

To prove the result for the singular term $\int _\Omega u^{-\beta (x) }v ~dx$, let us define

$$\begin{aligned} F(u)= \int _\Omega \frac{1}{1-\beta (x)} (u^+)^{1-\beta (x)} ~dx,~~ \text{ for }~u \in W^{1,p(x)}_0 (\Omega ). \end{aligned}$$

For $\xi \in \mathbb {R}{\setminus } \{0\}$, we define

$$\begin{aligned} z(\xi ) = \frac{1}{1-\beta (x)} \frac{d}{d\xi }(\xi ^+)^{1-\beta (x)}= \left\{ \begin{array}{l@{\quad }l} \xi ^{-\beta (x)} &{} \text{ if }\quad \xi >0,\\ 0&{} \text{ if }\quad \xi <0. \end{array}\right. \end{aligned}$$

Consequently, we find

$$\begin{aligned} \frac{F(u+tv)-F(u)}{t}= \int _\Omega \left( \int _0^1 z(u+stv)~ds \right) v~ dx. \end{aligned}$$

(6.3)

Notice that for almost everywhere $x \in \Omega$, we have $u(x)>0$ and

$$\begin{aligned} \int _0^1 z(u(x)+stv(x))ds \rightarrow z(u(x))= u(x)^{-\beta (x)}~\text{ as }~t\rightarrow 0. \end{aligned}$$

Moreover,

$$\begin{aligned} \int _0^1 z(u(x)+stv(x))~ds\le & {} \int _0^1 |u(x)+stv(x) |^{-\beta (x)}ds\nonumber \\\le & {} c (\max _{0\le s \le 1 } |u(x)+stv(x)|)^{-\beta (x)}\nonumber \\\le & {} c u^{-\beta (x)} \le C_\beta (\epsilon \phi _1)^{-\beta (x)} \le c \phi _1^{-\beta (x)}, \end{aligned}$$

(6.4)

with a constant $c=c(\beta ,\epsilon )$ independent of $x \in \Omega$. Indeed, for the second inequality in (6.4), we refer to the following inequality (Lemma A.1 in [24]): if $0<\delta <1$, then there exists $c=c(\delta )>0$ such that

$$\begin{aligned} \int _0^1 |a+sb|^{-\delta } ds \le c \left( \max _{0\le s\le 1} |a+sb|\right) ^{-\delta }, \end{aligned}$$

for all $a,b \in \mathbb {R}$ with $|a|+|b|>0$. Finally, we observe from Hölder’s inequality and Lemma 2.7 that

$$\begin{aligned} \int _\Omega v\phi _1^{-\beta (x)} dx \le C_\epsilon \int _\Omega \frac{v}{d(x)} d(x)^{1-\beta } dx \le 2C_\epsilon \left\| \frac{v}{d(x)} \right\| _{p(x)} \Vert d(x)^{1-\beta (x)}\Vert _{p'(x)} \le \tilde{C} \Vert \nabla v \Vert _{p(x)}, \end{aligned}$$

to discover that $v\phi _1 \in L^1(\Omega )$. Hence, from the dominated convergence theorem, we get

$$\begin{aligned} \lim _{t\rightarrow 0} \int _\Omega \frac{ 1}{1-\beta (x)} \frac{ ((u+tv)^+)^{1-\beta (x)}-(u^+)^{1-\beta (x)}}{t}~dx= \int _\Omega u^{-\beta (x)}v~dx. \end{aligned}$$

$\square$

We now know that the energy functional $E_\lambda :W_0^{1,p(x)}(\Omega ) \rightarrow \mathbb {R}$ is Gâteaux-differentiable at every point $u \in W_0^{1,p(x)}(\Omega )$ satisfying $u \ge \epsilon \phi _1$ in $\Omega$ with a constant $\epsilon >0$. Its Gâteaux derivative $E'_\lambda (u)$ at u is given by

$$\begin{aligned} <E'_\lambda (u),v>= & {} \int _\Omega |\nabla u|^{p(x)-2}\nabla u \cdot \nabla v~dx -\lambda \int _\Omega u^{-\beta (x)}v~dx\nonumber \\&-\int _\Omega u^{q(x)}v~dx~~~\text{ for } ~v\in W^{1,p(x)}_0(\Omega ). \end{aligned}$$

(6.5)

Now we prove the $C^1$ differentiability of the cut off energy functional. The proof is similar to the one in [18]. For the reader’s convenience, we here provide details of its proof.

Lemma 6.2

Let $p(x) \le q(x)+1<p^*(x)$ for all $x \in \bar{ \Omega }$ and $w \in W_0^{1,p(x)} (\Omega )$ such that $w \ge \epsilon \phi _1$ in $\Omega$ for some $\epsilon >0$. Define $f_\lambda : \Omega \times \mathbb {R} \rightarrow \mathbb {R}$ such that

$$\begin{aligned} f_\lambda (x,s)= \left\{ \begin{array}{l@{\quad }l} \lambda s^{-\beta (x)}+ s^{q(x)}&{}\text{ if }\quad s\ge w(x),\\ \lambda w(x)^{-\beta (x)} +w(x)^{q(x)}&{}\text{ if }\quad s< w(x), \end{array} \right. \end{aligned}$$

(6.6)

$F_\lambda (x,s) =\int _0^s f_\lambda (x,t)~dt$, and for $u \in W_0^{1,p(x)}(\Omega )$,

$$\begin{aligned} \overline{E}_\lambda (u) =\int _\Omega \frac{1}{p(x)} |\nabla u|^{p(x)} dx - \int _\Omega F_\lambda (x,u) ~dx. \end{aligned}$$

Then $\overline{E}_\lambda$ belongs to $C^1(W_0^{1,p(x)}(\Omega ),\mathbb {R})$.

Proof

We prove the result only for the singular term, as the two other terms are treated in a standard way. Set

$$\begin{aligned} h(x,t)= {\left\{ \begin{array}{ll} t^{-\beta (x)},\quad \text{ if }\quad t \ge w(x),\\ w(x)^{-\beta (x)},\quad \text{ if }\quad t< w(x), \end{array}\right. } \end{aligned}$$

$H(x,t)= \int _0^t h(x,s)ds$, and $G(u)= \int _\Omega H(x,u)dx$. As in Lemma 6.1, we have that for all $u \in W^{1,p(x)}_0(\Omega )$, Gâteaux derivative $G'(u)$ exists and is given by

$$\begin{aligned} <G'(u),v>= \int _\Omega (\max \{u(x),w(x) \})^{-\beta (x)}v(x) dx. \end{aligned}$$

Let $u_k,u_0 \in W^{1,p(x)}_0(\Omega )$ such that $u_k \rightarrow u_0$ in $W^{1,p(x)}_0(\Omega ).$ Then

$$\begin{aligned} |<G'(u_k)-G'(u_0),v>|= & {} \left| \int _\Omega [ (\max \{u_k(x),w(x) \}) ^{-\beta (x)}v(x) \right. \\&\left. - (\max \{u_0(x),w(x) \})^{-\beta (x)}v(x) ] ~dx \right| \\\le & {} 2 \int _\Omega w^{-\beta (x)} |v|dx\\\le & {} 2 C_\epsilon \int _\Omega \phi _1^{-\beta (x)} |v|dx, \end{aligned}$$

for all $v\in W^{1,p(x)}_0(\Omega )$. Again as in Lemma 6.1, we get $\phi _1^{-\beta (x)} v \in L^1(\Omega )$, and so by Lebesgue’s dominated convergence theorem, we conclude that the Gâteaux derivative of G is continuous, which implies that $G \in C^1(W_0^{1,p(x)}(\Omega ),\mathbb {R})$. $\square$