Global Higher Integrability of the Gradient of Weak Solutions of a Quasilinear Elliptic Equation

Abstract

In this paper, we establish global higher integrability of the gradient of the solution of the quasilinear elliptic equation $\Delta _A u=\text {div}\left( \frac{a(|F|)}{|F|}F\right)$ in ${\mathbb {R}}^n$, where $\Delta _A$ is the so called A-Laplace operator.

1 Introduction

We are interested in higher integrability of the gradient of the solution of the following problem:

$$\begin{aligned} (P){\left\{ \begin{array}{ll} &{}~~ u\in W^{1,A}({\mathbb {R}}^n),\\ &{}~~\displaystyle {\Delta _A u=\text {div}\left( \Theta (F)\right) }\qquad \text {in }{\mathbb {R}}^n \end{array}\right. } \end{aligned}$$

where $n\geqslant 2$, $F=(F_1,\ldots ,F_n)\in L^{A}({\mathbb {R}}^n)$, $\Delta _A u={\text {div}\left( \Theta (\nabla u)\right) }$, $\Theta (X)={{a(|X|)}\over {|X|}}X$, and ${A(t)=\int _0^t a(s)\mathrm{d}s}$, with a a function in $C^1( (0,\infty ))\cap C^0([0,\infty ))$ satisfying $a(0)=0$, and the condition

$$\begin{aligned} a_0\leqslant {{ta'(t)}\over {a(t)}}\leqslant a_1 \quad \forall t>0, \quad a_0, a_1 \hbox { positive constants}. \end{aligned}$$

(1.1)

Without loss of generality, we shall assume that $a_0<1<a_1$.

We call a solution of problem (P) any function $u\in W^{1,A}({\mathbb {R}}^n)$ that satisfies

$$\begin{aligned} \int _{{\mathbb {R}}^n} \Theta (\nabla u).\nabla \varphi \mathrm{d}x=\int _{{\mathbb {R}}^n} \Theta (F).\nabla \varphi \mathrm{d}x\quad \forall \varphi \in \mathcal {D}({\mathbb {R}}^n) \end{aligned}$$

We recall the definition of the Orlicz space $L^A({\mathbb {R}}^n)$ and its norm (see [10])

$$\begin{aligned}&L^A({\mathbb {R}}^n)=\left\{ \,u\in L^1({\mathbb {R}}^n)\,:\,\int _{{\mathbb {R}}^n} A(|u(x)|)\mathrm{d}x<\infty \,\right\} , \\&||u||_A=\inf \left\{ k>0\,:\,\int _{{\mathbb {R}}^n} A\Big ({{|u(x)|}\over k}\Big )\mathrm{d}x\le 1\,\right\} \end{aligned}$$

The dual space of $L^A({\mathbb {R}}^n)$ is the Orlicz space $L^{\widetilde{A}}({\mathbb {R}}^n)$, where $\widetilde{A}(t)=\int _{0}^{t} a^{-1}(s) \mathrm{d}s$ and $a^{-1}$ is the inverse function of a.

The Orlicz–Sobolev space $W^{1,A}({\mathbb {R}}^n)$ and its norm are given by

$$\begin{aligned} W^{1,A}({\mathbb {R}}^n)=\left\{ \,u\in L^A({\mathbb {R}}^n)\,:\,|\nabla u|\in L^A({\mathbb {R}}^n)\,\right\} ,\quad ||u||_{1,A}=||u||_A+||\nabla u||_A. \end{aligned}$$

Both $L^A({\mathbb {R}}^n)$ and $W^{1,A}({\mathbb {R}}^n)$ are reflexive Banach spaces.

The following useful inequalities can be easily deduced from (1.1) (see [9])

$$\begin{aligned}&\displaystyle {t a(t) \over 1+ a_1} \le A(t) \le t a(t)\quad \forall t \ge 0, \end{aligned}$$

(1.2)

$$\begin{aligned}&\displaystyle s a(t) \le sa(s) + t a(t)\quad \forall s,t \ge 0, \end{aligned}$$

(1.3)

$$\begin{aligned}&\displaystyle \min ( s^{a_0},s^{a_1}) a(t) \le a(st) \le \max ( s^{a_0},s^{a_1}) a(t)\quad \forall s,t \ge 0, \end{aligned}$$

(1.4)

$$\begin{aligned}&\displaystyle \min ( s^{1+a_0},s^{1+ a_1}) {A(t) \over 1+ a_1} \!\le \!A(st) \!\le \! {(1+ a_1)}\max ( s^{1+a_0},s^{1+ a_1}) A(t) \quad \!\forall s,t \!\ge \! 0.\nonumber \\ \end{aligned}$$

(1.5)

Using (1.5) and the convexity of A, we obtain

$$\begin{aligned} A(s+t)= & {} A\left( 2.\left( \frac{s+t}{2}\right) \right) \le (1+ a_1)2^{1+a_1} A\left( \frac{s+t}{2}\right) \nonumber \\\le & {} (1+ a_1)2^{a_1}(A(s)+A(t))\quad \forall s,t \ge 0 \end{aligned}$$

(1.6)

We also recall the following monotonicity inequality (see [3])

$$\begin{aligned} \Big (\Theta (X)-\Theta (Y)\Big ). (X-Y)\geqslant & {} C(A,n)|X-Y|^2{ a\big ((|X|^2+|Y|^2)^{1/2}\big )\over {(|X|^2+|Y|^2)^{1/2}}} \nonumber \\&\qquad \forall (X,Y)\in {\mathbb {R}}^{2n}{\setminus }\{ 0\} \end{aligned}$$

(1.7)

where C(A, n) is a positive constant depending only on A and n.

There is a wide range of functions a(t) satisfying (1.1). In particular, we observe that we have $a_0=a_1=p-1$ if and only if $a(t)=ct^{a_0}$ for some positive constant c. In this case, we have $A(t)={{ct^p}\over p}$ and $\Delta _A=c\Delta _p$, where $\Delta _p$ is the $p-$Laplace operator. We refer to [4] for more examples of these functions.

Definition 1.1

We say that a function $A: [0,\infty )\rightarrow [0,\infty )$ is an $N-$function if $A'\in C^1((0,\infty ))\cap C^0([0,\infty ))$, $A'(0)=0$, and $A'$ satisfies (1.1).

Here is the main result of this paper.

Theorem 1.1

Let $B: [0,\infty )\rightarrow [0,\infty )$ be a function such that $B\circ A^{-1}$ is an $N-$function and let $u\in W^{1,A}({\mathbb {R}}^n)$ be a solution of problem (P). If $F\in L^B({\mathbb {R}}^n)$, then we have $|\nabla u|\in L^B({\mathbb {R}}^n)$ with

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u|)\mathrm{d}x\le C\int _{{\mathbb {R}}^n} B(|F|) \mathrm{d}x \end{aligned}$$

where C is a positive constant depending only on n, A, and B.

Remark 1.1

We would like to mention that this regularity is well known for the Laplace equation (see [8]). For the p-Laplace equation, we refer to [8] and to [5] for systems.

Example 1.1

Assume that $\Phi$ is an $N-$function with $\Phi '=\varphi$ and let $B=\Phi \circ A$. Then, B is an $N-$function as a compose of two $N-$functions, namely $\Phi$ and A. Moreover, $B\circ A^{-1}=\Phi$ is also an $N-$function. Hence, the following functions satisfy the assumption of Theorem 1.1:

• $B(t)=(A(t))^p$ with $p>1$, ($\Phi (t)=t^p$).

• $B(t)=(A(t))^\alpha \ln (\beta A(t)+\gamma )$ with $\alpha , \beta , \gamma >0$, ($\Phi (t)=t^\alpha \ln (\beta t+\gamma )$).

• $B(t)=(A(t))^{p(A(t))}$, ($\Phi (t)=t^{p(t)}$), provided p(t) satisfies for all $t>0$: $\alpha _0\leqslant t\ln (t)p'(t)+p(t)-1\leqslant \alpha _1$, for some positive constants $\alpha _0$ and $\alpha _1$.

Corollary 1.1

Under the assumption of Theorem 1.1, if moreover, the integral ${\int _1^\infty \frac{B^{-1}(t)}{t^{\frac{n+1}{n}}}\mathrm{d}t}$ is finite, then we have $u\in C_{\mu }({\mathbb {R}}^n)$, where ${\mu (t)=\int _t^\infty \frac{B^{-1}(s)}{s^{\frac{n+1}{n}}}\mathrm{d}s}$ and $C_{\mu }({\mathbb {R}}^n)$ is the space of continuous functions with $\mu$ as a modulus of continuity.

Proof

From Theorem 1.1, we know that $u\in W^{1,B}({\mathbb {R}}^n)$. Since the integral ${\int _1^\infty \frac{B^{-1}(t)}{t^{\frac{n+1}{n}}}\mathrm{d}t}$ is finite, the result follows from the embedding $W^{1,B}({\mathbb {R}}^n)\subset C_{\mu }({\mathbb {R}}^n)$ (see [1, Theorem 8.40]).$\square$

Remark 1.2

Since $(B\circ A^{-1})'$ satisfies (1.1), using (1.5) with $t=1$ and ${s=\frac{1}{t}}$, we see that there exist two positive constants $\mu >1$ and K such that

$$\begin{aligned} 0\le B\circ A^{-1}\left( \frac{1}{t}\right) \le \frac{K}{t^\mu } \quad \text {for all }~ t \ge 1. \end{aligned}$$

Therefore, the integral ${\int _1^\infty B\circ A^{-1}\left( \frac{1}{t}\right) dt}$ is finite. This property will be used in the proof of Theorem 1.1 in Sect. 3.

In the sequel, we will denote by u a solution of problem (P). In Sect. 2, we recall some well known results about $A-$harmonic functions and establish a few Lemmas to pave the way for the proof of Theorem 1.1 which will be given in Sect. 3.

2 Some Auxiliary Lemmas

Let $x_0\in {\mathbb {R}}^n$ and $R>0$. For each open ball $B_R(x_0)$ in ${\mathbb {R}}^n$ of center $x_0$ and radius R, let v be the unique solution of the following problem:

$$\begin{aligned} (P_0){\left\{ \begin{array}{ll} &{} \quad v\in W^{1,A}(B_R(x_0)),\\ &{}\quad \Delta _A v=0\quad \text {in }B_R(x_0), \\ &{} \quad v=u\quad \text {in }\partial B_R(x_0) \end{array}\right. } \end{aligned}$$

First, we recall some properties of the solution of problem $(P_0)$.

Lemma 2.1

$$\begin{aligned} \int _{B_R(x_0)} a(|\nabla v|) |\nabla v| \le 2^{a_1+2}\int _{B_R(x_0)} a(|\nabla u|) |\nabla u|\mathrm{d}x \end{aligned}$$

Proof

See [3], Proof of Lemma 3.1. $\square$

Remark 2.1

Using (1.2) and Lemma 2.1, we obtain

$$\begin{aligned} \int _{B_R(x_0)} A(|\nabla v|)\mathrm{d}x\le & {} \int _{B_R(x_0)} a(|\nabla v|) |\nabla v|\mathrm{d}x\nonumber \\\le & {} 2^{a_1+2}\int _{B_R(x_0)} a(|\nabla v|) |\nabla v|\mathrm{d}x\nonumber \\\le & {} (1+a_1)2^{a_1+2}\int _{B_R(x_0)} A(|\nabla u|)\mathrm{d}x \end{aligned}$$

Lemma 2.2

There exists a positive constant $C_1=C_1(n,a_1)$ such that

$$\begin{aligned} \sup _{ B_{R/2}(x_0)} A(|\nabla v|)\le {C_1\over R^n} \int _{B_{R}(x_0)} A(|\nabla v|)\mathrm{d}x \end{aligned}$$

Proof

See [9], Lemma 5.1.

For each function $f: {\mathbb {R}}^n \rightarrow {\mathbb {R}}~({\mathbb {R}}^n)$, let and ${(f)_{r}=(f)_{0,r} }$. Then we have the following property of the function v. $\square$

Lemma 2.3

There exist two positive constants $\alpha =\alpha (n,a_1)<1$ and $C_2=C_2(n,a_1)$ such that we have for any $r\in (0,R)$

Proof

See [9], Lemma 5.1. $\square$

Remark 2.2

Using (1.6), Remark 2.1, and Jensen’s inequality, we obtain

(2.1)

Combining (2.1) and Lemma 2.3, we obtain

The next lemma is the key tool in the proof of Theorem 1.1.

Lemma 2.4

There exist two positive constants $\gamma =\gamma (n,a_0,a_1)$ and $m=m(\alpha ,n,a_0,a_1)$ such that for each $\delta \in (0,1)$, we have for any $x_0\in {\mathbb {R}}^n$, $R>0$ and $r\in (0,R)$:

The proof of Lemma 2.4 requires several lemmas. $\square$

Lemma 2.5

Let $G\,:\, {\mathbb {R}}^{2n}{\setminus }\{0\}\,\rightarrow \, {\mathbb {R}}$ defined by

$$\begin{aligned} G(\xi ,\zeta )=\int _0^1 {{a(|\theta _t|)}\over {|\theta _t|}}dt,\quad \theta _t=t\xi +(1-t)\zeta . \end{aligned}$$

Then there exists two positive constants $c_{a_1,n}$ depending only on n and $a_1$, and $C_{a_0,n}$ depending only on n and $a_0$ such that:

$$\begin{aligned}\forall (\xi ,\zeta )\in {\mathbb {R}}^{2n}{\setminus }\{0\},\quad c_{a_1,n} {{a(|\xi |+|\zeta |)}\over {|\xi |+|\zeta |}}\le G(\xi ,\zeta ) \le C_{a_0,n} {{a(|\xi |+|\zeta |)}\over {|\xi |+|\zeta |}} \end{aligned}$$

The proof of Lemma 2.5 is based on the following lemma proved in [2] for $n=2$ and whose proof extends easily to $n\ge 3$.

Lemma 2.6

Let $n\geqslant 2$, $p>1$ and let $F_p\,:\, {\mathbb {R}}^{2n}{\setminus }\{0\}\,\rightarrow \, {\mathbb {R}}$ defined by

$$\begin{aligned} F_p(\xi ,\zeta )=\int _0^1 |t\xi +(1-t)\zeta |^{p-2}dt \end{aligned}$$

Then, there exists two positive constants $c(p,n)<C(p,n)$ depending only on p and n such that:

$$\begin{aligned} \forall (\xi ,\zeta )\in {\mathbb {R}}^{2n}{\setminus }\{0\},\quad c(p,n)\big (|\xi |^2+|\zeta |^2\big )^{{p-2}\over 2}\le F_p(\xi ,\zeta ) \le C(p,n)\big (|\xi |^2+|\zeta |^2\big )^{{p-2}\over 2} \end{aligned}$$

Proof of Lemma 2.5

For $(\xi ,\zeta )\in {\mathbb {R}}^{2n}{\setminus }\{0\}$, we set

$$\begin{aligned} X_0={\xi \over {(|\xi |^2+|\zeta |^2)^{1/2}}},\quad X_1={\zeta \over {(|\xi |^2+|\zeta |^2)^{1/2}}}. \end{aligned}$$

It follows that

$$\begin{aligned} \theta _t=(|\xi |^2+|\zeta |^2)^{1/2}(tX_0+(1-t)X_1),\quad |\theta _t|=(|\xi |^2+|\zeta |^2)^{1/2}|tX_0+(1-t)X_1|. \end{aligned}$$

Then, we have by inequality (1.4)

$$\begin{aligned}&a(|\theta _t|)\le \max \big (|tX_0+(1-t)X_1|^{a_0},|tX_0+(1-t)X_1|^{a_1}\big )a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big ) \\&\min \big (|tX_0+(1-t)X_1|^{a_0},|tX_0+(1-t)X_1|^{a_1}\big )a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big ) \le a(|\theta _t|) \end{aligned}$$

Note that since $|tX_0+(1-t)X_1|\le t|X_0|+(1-t)|X_1|\le t+(1-t)=1$ and $0< a_0 \le a_1$, we have $|tX_0+(1-t)X_1|^{a_1} \le |tX_0+(1-t)X_1|^{a_0}.$ Hence we get

$$\begin{aligned}&|tX_0+(1-t)X_1|^{a_1}a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big ) \le a(|\theta _t|)\\&\quad \le |tX_0+(1-t)X_1|^{a_0}a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big ) \end{aligned}$$

which leads to

$$\begin{aligned} F_{a_1+1}(X_0,X_1){{a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big )}\over {(|\xi |^2+|\zeta |^2)^{1/2}}}\le G(\xi ,\zeta ) \le F_{a_0+1}(X_0,X_1){{a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big )}\over {(|\xi |^2+|\zeta |^2)^{1/2}}}. \end{aligned}$$

Now, if we apply Lemma 2.6 with $p=a_0+1$ and $p=a_1+1$, we get since $|X_0|^2+|X_1|^2=1$

$$\begin{aligned}&F_{a_0+1}(X_0,X_1)\le C(a_0+1,n)\big (|X_0|^2+|X_1|^2\big )^{{a_0-1}\over 2}=C(a_0+1,n) \\&F_{a_1+1}(X_0,X_1)\ge c_{a_1+1,n}\big (|X_0|^2+|X_1|^2\big )^{{a_1-1}\over 2}=c(a_1+1,n) \end{aligned}$$

Finally, we obtain

$$\begin{aligned} c(a_1+1,n){{a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big )}\over {(|\xi |^2+|\zeta |^2)^{1/2}}}\leqslant G(\xi ,\zeta ) \leqslant C(a_0+1,n){{a\big ((|\xi |^2+|\zeta |^2)^{1/2}\big )}\over {(|\xi |^2+|\zeta |^2)^{1/2}}} \end{aligned}$$

Since the two norms $|\xi |+|\zeta |$ and $(|\xi |^2+|\zeta |^2)^{1/2}$ are equivalent, the lemma follows using (1.4) $\square$

Lemma 2.7

For any $X, Y\in {\mathbb {R}}^n$, we have:

$$\begin{aligned} A(|X|)\ge A(|Y|)+\left<\Theta (Y),X-Y\right> \end{aligned}$$

Proof

Let $X, Y\in {\mathbb {R}}^n$. First, we have

$$\begin{aligned} A(|X|)-A(|Y|)= & {} \int _0^1 \frac{d}{dt}[A(|tX+(1-t)Y|)]dt\\= & {} \int _0^1 A'(|tX+(1-t)Y|).\frac{<tX+(1-t)Y,X-Y>}{|tX+(1-t)Y|}dt \\= & {} \int _0^1 a(|tX+(1-t)Y|).\frac{<tX+(1-t)Y,X-Y>}{|tX+(1-t)Y|}dt \\= & {} \int _0^1 \left<\Theta (tX+(1-t)Y),X-Y\right>dt \end{aligned}$$

Next, using (1.7), this leads to

$$\begin{aligned} A(|X|)-A(|Y|)= & {} \int _0^1 \frac{1}{t}\left<\Theta (tX+(1-t)Y),t(X-Y)\right>dt\\\ge & {} \int _0^1 \frac{1}{t}\left<\Theta (Y),t(X-Y)\right>dt\nonumber \\= & {} \left<\Theta (Y),X-Y\right> \end{aligned}$$

which achieves the proof. $\square$

Lemma 2.8

For any $\delta \in (0,1)$, $X, Y\in {\mathbb {R}}^n$, we have:

$$\begin{aligned} |A(|X|)-A(|Y|)|\le \frac{C_3}{\delta ^{a_1(1+a_1)}}A(|X-Y|)+C_4\delta ^{a_0+1}(A(|X|)+A(|Y|)) \end{aligned}$$

where $C_3=(1+a_1)C_{a_0,n}$ and $C_4=(1+a_1)2^{a_1}C_3$

Proof

Let $\delta \in (0,1)$, $X, Y\in {\mathbb {R}}^n$. First, we have

$$\begin{aligned} |A(|X|)-A(|Y|)|= & {} \left| \int _0^1 \frac{d}{dt}[A(|tX+(1-t)Y|)]dt\right| \nonumber \\= & {} \left| \int _0^1 A'(|tX+(1-t)Y|).\frac{<tX+(1-t)Y,X-Y>}{|tX+(1-t)Y|}dt\right| \nonumber \\= & {} \left| \int _0^1 a(|tX+(1-t)Y|).\frac{<tX+(1-t)Y,X-Y>}{|tX+(1-t)Y|}dt\right| \nonumber \\\le & {} |X-Y|.(|X|+|Y|).\int _0^1 \frac{a(|tX+(1-t)Y|)}{|tX+(1-t)Y|}dt \end{aligned}$$

Next, we get by using Lemma 2.5

$$\begin{aligned} |A(|X|)-A(|Y|)|\le & {} C_{a_0,n}|X-Y|.(|X|+|Y|). {{a(|X|+|Y|)}\over {|X|+|Y|}}\nonumber \\= & {} C_{a_0,n}|X-Y|.a(|X|+|Y|) \end{aligned}$$

(2.2)

We observe that we have by (1.2)–(1.4), for $s, t\ge 0$ since $\delta \in (0,1)$

$$\begin{aligned} s a(t)= & {} \frac{s}{\delta ^{a_1}}.\delta ^{a_1}a(t)\le \frac{s}{\delta ^{a_1}}.a(\delta t) \le \frac{s}{\delta ^{a_1}}.a\left( \frac{s}{\delta ^{a_1}}.\right) +\delta t a(\delta t)\nonumber \\\le & {} \frac{s}{\delta ^{a_1}}.\frac{1}{(\delta ^{a_1})^{a_1}}a(s) +\delta t \delta ^{a_0}a(t)= \frac{sa(s)}{\delta ^{a_1(1+a_1)}}+\delta ^{a_0+1}ta(t)\nonumber \\\le & {} \frac{1+a_1}{\delta ^{a_1(1+a_1)}}A(s)+(1+a_1)\delta ^{a_0+1}A(t) \end{aligned}$$

(2.3)

Using (2.2) and (2.3), with $s=|X-Y|$ and $t=|X|+|Y|$, we get

$$\begin{aligned} |A(|X|)-A(|Y|)|\le & {} \frac{(1+a_1)C_{a_0,n}}{\delta ^{a_1(1+a_1)}}A(|X-Y|)+(1+a_1)C_{a_0,n}\delta ^{a_0+1}\nonumber \\&A(|X|+|Y|) \end{aligned}$$

(2.4)

Using (1.6), we get from (2.4)

$$\begin{aligned} |A(|X|)-A(|Y|)|\le & {} \frac{C_3}{\delta ^{a_1(1+a_1)}}A(|X-Y|)+(1+a_1)^22^{a_1}C_{a_0,n}\delta ^{a_0+1}\nonumber \\&(A(|X|)+A(|Y|)) \end{aligned}$$

Hence, the lemma follows. $\square$

Lemma 2.9

There exist two positive constants ${C_5=\frac{(1+a_1)2^{\frac{a_1}{2}}}{C(A,n)} }$ and

${C_6=(1+a_1)2^{a_1}\left( 1+(1+a_1)2^{a_1+2}\right) \left( C_5+\frac{1+a_1}{4}\right) }$ such that we have for each $\eta \in (0,1)$ , $x_0\in {\mathbb {R}}^n$, $R>0$ and $r\in (0,R)$:

Proof

We observe that by translation, it is enough to prove the lemma for $x_0=0$. Using $w=(u-v)\chi _{B_R}$ as a test function for problems (P) and $(P_0)$ and subtracting the two equations and using (2.3), we get for $\eta \in (0,1)$:

$$\begin{aligned}&\int _{B_R}\big (\Theta (\nabla u)-\Theta (\nabla v)\big )(\nabla u-\nabla v)\mathrm{d}x\nonumber \\&\qquad =\int _{B_R}\frac{a(|F|)}{|F|}F.\nabla w \mathrm{d}x \le \int _{B_R} a(|F|).|\nabla w| \mathrm{d}x\nonumber \\&\qquad \le ~ \frac{1+a_1}{\eta ^{a_1(1+a_1)}}\int _{B_R}A(|F|)\mathrm{d}x+(1+a_1)\eta ^{a_0+1}\int _{B_R}A(|\nabla w|)\mathrm{d}x \end{aligned}$$

(2.5)

Using (2.5) and (1.7), we get

$$\begin{aligned} \frac{C(A,n)}{2^{\frac{a_1}{2}}}\int _{B_R}{{a(|\nabla u|+|\nabla v|)}\over {|\nabla u|+|\nabla v|}}.|\nabla w|^2\mathrm{d}x\le & {} \frac{1+a_1}{\eta ^{a_1(1+a_1)}}\int _{B_R}A(|F|)\mathrm{d}x\\&+(1+a_1)\eta ^{a_0+1}\int _{B_R}A(|\nabla w|)\mathrm{d}x \end{aligned}$$

or

$$\begin{aligned}&\int _{B_R}{{a(|\nabla u|+|\nabla v|)}\over {|\nabla u|+|\nabla v|}}.|\nabla w|^2\mathrm{d}x\nonumber \\&\qquad \le ~ \frac{C_5}{\eta ^{a_1(1+a_1)}}\int _{B_R}A(|F|)\mathrm{d}x+C_5\eta ^{a_0+1}\int _{B_R}A(|\nabla w|)\mathrm{d}x \end{aligned}$$

(2.6)

By (1.2), we can write

$$\begin{aligned} \int _{B_r}A(|\nabla w|)\mathrm{d}x\le & {} \int _{B_r}|\nabla w|a(|\nabla w|)\mathrm{d}x\nonumber \\= & {} \int _{E_{1,r}}|\nabla w|a(|\nabla w|)\mathrm{d}x+\int _{E_{2,r}}|\nabla w|a(|\nabla w|)\mathrm{d}x \end{aligned}$$

(2.7)

where $E_{1,r}=B_r\cap \{~(|\nabla u|+|\nabla v|)a(|\nabla w|)\le |\nabla w|a(|\nabla u|+|\nabla v|)~\}$ and $E_{2,r}=B_r\cap \{~(|\nabla u|+|\nabla v|)a(|\nabla w|)>|\nabla w|a(|\nabla u|+|\nabla v|)~\}$.

From the definition of $E_{1,r}$, we see that

$$\begin{aligned} \int _{E_{1,r}}|\nabla w|a(|\nabla w|)\mathrm{d}x\le \int _{E_{1,r}}{{a(|\nabla u|+|\nabla v|)}\over {|\nabla u|+|\nabla v|}}.|\nabla w|^2\mathrm{d}x \end{aligned}$$

(2.8)

Using the monotonicity of a, the definition of $E_{2,r}$, and Young’s inequality, we get

$$\begin{aligned}&\int _{E_{2,r}}|\nabla w|a(|\nabla w|)\mathrm{d}x~\le ~\int _{E_{2,r}}a(|\nabla u|+|\nabla v|).|\nabla w|\mathrm{d}x\nonumber \\&\quad \le \int _{E_{2,r}}\left( \frac{a(|\nabla u|+|\nabla v|)}{|\nabla u|+|\nabla v|}.|\nabla w|^2\right) ^{\frac{1}{2}} .\left( (|\nabla u|+|\nabla v|)a(|\nabla u|+|\nabla v|)\right) ^{\frac{1}{2}}\mathrm{d}x\nonumber \\&\quad \le \frac{1}{\eta }\int _{E_{2,r}}{{a(|\nabla u|+|\nabla v|)}\over {|\nabla u|+|\nabla v|}}.|\nabla w|^2\mathrm{d}x +\frac{\eta }{4}\int _{E_{2,r}}(|\nabla u|+|\nabla v|)a(|\nabla u|+|\nabla v|)\mathrm{d}x\nonumber \\&\quad \le \frac{1}{\eta }\int _{E_{2,r}}{{a(|\nabla u|+|\nabla v|)}\over {|\nabla u|+|\nabla v|}}.|\nabla w|^2\mathrm{d}x +\frac{(1+a_1)\eta }{4}\int _{B_r}A(|\nabla u|+|\nabla v|)\mathrm{d}x\nonumber \\ \end{aligned}$$

(2.9)

Combing (2.6), (2.7), (2.8), and (2.9), and using (1.6) and Remark 2.1, we arrive at

$$\begin{aligned}&\int _{B_r}A(|\nabla w|)\mathrm{d}x\\&\quad \le \frac{1}{\eta }\int _{B_r}{{a(|\nabla u|+|\nabla v|)}\over {|\nabla u|+|\nabla v|}}.|\nabla w|^2\mathrm{d}x +\frac{(1+a_1)\eta }{4}\int _{B_R}A(|\nabla u|+|\nabla v|)\mathrm{d}x\nonumber \\&\quad \le \frac{C_5}{\eta ^{a_1(1+a_1)+1}}\int _{B_R}A(|F|)\mathrm{d}x+C_5\eta ^{a_0}\int _{B_R}A(|\nabla w|)\mathrm{d}x\nonumber \\&\qquad +\frac{(1+a_1)\eta }{4}\int _{B_R}A(|\nabla u|+|\nabla v|)\mathrm{d}x\nonumber \\&\quad \le \frac{C_5}{\eta ^{a_1(1+a_1)+1}}\int _{B_R}A(|F|)\mathrm{d}x+\left( C_5\eta ^{a_0} +\frac{(1+a_1)\eta }{4}\right) \int _{B_R}A(|\nabla u|+|\nabla v|)\mathrm{d}x\nonumber \\&\quad \le (1+a_1)2^{a_1}\left( C_5\eta ^{a_0} +\frac{(1+a_1)\eta }{4}\right) \left( \int _{B_R}A(|\nabla u|)\mathrm{d}x+\int _{B_R}A(|\nabla v|)\mathrm{d}x\right) \nonumber \\&\quad \qquad +\frac{C_5}{\eta ^{a_1(1+a_1)+1}}\int _{B_R}A(|F|)\mathrm{d}x\nonumber \\&\quad \le (1+a_1)2^{a_1}\left( C_5 +\frac{1+a_1}{4}\right) (1+(1+a_1)2^{a_1+2})\eta ^{a_0}\int _{B_R}A(|\nabla u|)\mathrm{d}x\nonumber \\&\qquad \qquad +\frac{C_5}{\eta ^{a_1(1+a_1)+1}}\int _{B_R}A(|F|)\mathrm{d}x \end{aligned}$$

Hence, the lemma follows. $\square$

Proof of Lemma 2.4

Let $\delta \in (0,1)$, $R>0$ and $r\in (0,R)$. First, we have by Lemma 2.8 used for $X=\nabla u(x)$ and $Y=\nabla u(y)$

This leads by (1.6) to

(2.10)

Using (1.6) again, we get

(2.11)

Using Jensen’s inequality, we derive

(2.12)

From (2.11) and (2.12), we obtain

(2.13)

Combining (2.10) and (2.13), we obtain

(2.14)

where ${C_7=2C_3(1+ a_1)^22^{2a_1}(1+(1+ a_1)2^{a_1}) }$ and ${C_8=2C_3(1+ a_1)^32^{3a_1} }$.

Finally, by taking into account Remark 2.2 and Lemma 2.8, we obtain from (2.14) for any $\eta \in (0,1)$

To conclude the proof, we let $\gamma =\max (C_5C_7,C_6C_7+C_2C_8(1+a_1)^22^{2a_1+3},2C_4)$, ${m=n+\frac{(\alpha +n)(1+a_1(1+a_1))}{a_0} }$, and choose $\eta$ such that ${\eta ^{a_0}=\left( {r\over R}\right) ^{\alpha +n} }$ or $\eta =\left( {r\over R}\right) ^{\frac{\alpha +n}{a_0}}$. Then we obtain

$\square$

3 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1. First, we recall that for each function $f\in L^1({\mathbb {R}}^n)$, the Hardy–Littlewood maximal function associated with f is given by

and the sharp maximal function associated with f is defined by

Proof of Theorem 1.1

We deduce from Lemma 2.4 that we have for every $\delta \in (0,1)$, $r>0$, ${R=\frac{r}{\delta ^{\frac{a_1(1+a_1)+a_0+1}{\alpha }} } }$, ${\kappa =a_1(1+a_1)+\frac{m(a_1(1+a_1)+a_0+1)}{\alpha } }$, and $x_0\in {\mathbb {R}}^n$

which leads to

$$\begin{aligned} (A(|\nabla u|))^{\sharp }(x_0)~\le ~\frac{\gamma }{\delta ^{\kappa }}M[A(|F|)](x_0) +2\gamma \delta ^{a_0+1} M[A(|\nabla u|)](x_0) \end{aligned}$$

(3.1)

Now, let B be a function such that $B\circ A^{-1}$ is an $N-$function and assume that $F\in C_0^\infty ({\mathbb {R}}^n)$ and $|\nabla u|\in L^B({\mathbb {R}}^n)$. Then by using (1.6), we obtain from (3.1) (see [6]) for $\gamma '=\gamma (1+c_1)2^{c_1}$, where $c_1$ is the equivalent to the constant $a_1$ for the function $B\circ A^{-1}$

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u|)\mathrm{d}x\le & {} \int _{{\mathbb {R}}^n}B\left( A^{-1}(A(|\nabla u|))^{\sharp }(x)\right) \mathrm{d}x \nonumber \\\le & {} 2\gamma '\delta ^{1+a_0} \int _{{\mathbb {R}}^n}B\left( A^{-1}M[A(|\nabla u|)](x)\right) \mathrm{d}x\nonumber \\&+\frac{\gamma '}{\delta ^{\kappa }}\int _{{\mathbb {R}}^n}B\left( A^{-1}M[A(|F|)](x)\right) \mathrm{d}x \end{aligned}$$

(3.2)

By the Hardy–Littlewood maximal theorem (see [7]), we have for some positive constant $C_9$ depending only upon n, A and B

$$\begin{aligned} \int _{{\mathbb {R}}^n}B\left( A^{-1}M[A(|F|)](x)\right) \mathrm{d}x\le & {} C_9\int _{{\mathbb {R}}^n}B(|F|)\mathrm{d}x \end{aligned}$$

(3.3)

$$\begin{aligned} \int _{{\mathbb {R}}^n}B\left( A^{-1}M[A(|\nabla u|)](x)\right) \mathrm{d}x\le & {} C_9\int _{{\mathbb {R}}^n}B(|\nabla u|)\mathrm{d}x \end{aligned}$$

(3.4)

We deduce then from (3.2), (3.3) and (3.4) that

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u|)\mathrm{d}x \le 2\gamma ' C_9\delta ^{1+a_0} \int _{{\mathbb {R}}^n}B(|\nabla u|)\mathrm{d}x +\frac{\gamma ' C_9}{\delta ^{\kappa }}\int _{{\mathbb {R}}^n}B(|F|)\mathrm{d}x \end{aligned}$$

(3.5)

Now, if we choose $\delta$ such that $\delta \in \left( 0, \min \left( 1,\left( 2\gamma ' C_9\right) ^{-\frac{1}{1+a_0}}\right) \right)$, we get

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u|)\mathrm{d}x \le \frac{\gamma ' C_9}{\delta ^{\kappa }(1-2\gamma ' C_9\delta ^{1+a_0})}\int _{{\mathbb {R}}^n}B(|F|)\mathrm{d}x \end{aligned}$$

(3.6)

This completes the proof when $F\in C_0^\infty ({\mathbb {R}}^n)$ and $|\nabla u|\in L^B({\mathbb {R}}^n)$. Next, we will establish it when $F\in L^{B}({\mathbb {R}}^n)$. To do that, we consider a sequence $(F_k)_k$ of vector functions in $C_0^\infty ({\mathbb {R}}^n)$ that converges to F in $L^{B}({\mathbb {R}}^n)$. We denote by $(u_k)_k$ the sequence of unique solutions of the following problem

$$\begin{aligned} (P_k){\left\{ \begin{array}{ll} &{}u_k\in W^{1,A}({\mathbb {R}}^n),\\ &{}{\text {div}\big (\Theta (\nabla u_k)\big )=\text {div}\left( \Theta (F_k)\right) }\qquad \text {in }{\mathbb {R}}^n \end{array}\right. }\end{aligned}$$

Since $F_k$ has compact support in ${\mathbb {R}}^n$, there exists $l_k>0$ such that $\text {suppt}(F_k)\subset B_{l_k}$, and therefore we have for all k

$$\begin{aligned} \text {div}\big (\Theta (\nabla u_k)\big )=0\quad \text {in }{\mathbb {R}}^n{\setminus } B_{l_k} \end{aligned}$$

(3.7)

We will prove that $|\nabla u_k|\in L^B({\mathbb {R}}^n)$. Since $u_k\in C^1({\mathbb {R}}^n)$, it is enough to show that ${\int _{\{|x|>2l_k\}} B(|\nabla u_k(x)|)\mathrm{d}x<\infty }$. We observe that because of (3.7), we can apply Lemma 2.3 in $B_{|x|-l_k}(x)$ for each $x\in {\mathbb {R}}^n{\setminus } B_{2l_k}$

$$\begin{aligned} A(|\nabla u_k(x)|)\le {C_1\over (|x|-l_k)^n} \int _{B_{|x|-l_k}(x)} A(|\nabla u_k|)\mathrm{d}y \end{aligned}$$

(3.8)

As observed in Remark 1.2, there exists two positive constants $\mu$ and K such that

$$\begin{aligned} 0\le B\circ A^{-1}(st) \le K s^\mu B\circ A^{-1}(t) \qquad \forall s, t \ge 0. \end{aligned}$$

(3.9)

Using (3.8) and (3.9), and taking into account Remark 1.2, we get

$$\begin{aligned}&\int _{\{|x|>2l_k\}} B(|\nabla u_k(x)|)\mathrm{d}x\nonumber \\&\quad \le K\left( C_1\int _{B_{|x|-l_k}(x)}A(|\nabla u_k|)\mathrm{d}y\right) ^\mu . \int _{\{|x|>2l_k\}}B\circ A^{-1}\left( {{1}\over (|x|-l_k)^n}\right) \mathrm{d}x \nonumber \\&\quad \le K\omega _n\left( C_1\int _{{\mathbb {R}}^n}A(|\nabla u_k|)\mathrm{d}y\right) ^\mu . \int _{2l_k}^\infty r^{n-1}B\circ A^{-1}\left( {{1}\over (r-l_k)^n}\right) dr \nonumber \\&\quad \le K\omega _n2^{n-1}\left( C_1\int _{{\mathbb {R}}^n}A(|\nabla u_k|)\mathrm{d}y\right) ^\mu . \int _{2l_k}^\infty (r-l_k)^{n-1}B\circ A^{-1}\left( {{1}\over (r-l_k)^n}\right) dr \nonumber \\&\quad =\frac{K\omega _n 2^{n-1}}{n}\left( C_1\int _{{\mathbb {R}}^n}A(|\nabla u_k|)\mathrm{d}y\right) ^\mu . \int _{l_k^n}^\infty B\circ A^{-1}\left( {{1}\over t}\right) dt <\infty \end{aligned}$$

It follows that $|\nabla u_k|\in L^B({\mathbb {R}}^n)$. Therefore, (3.6) is valid for $u_k$ and we have

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u_k|)\mathrm{d}x \le \frac{\gamma ' C_9}{\delta ^{\kappa }(1-2\gamma ' C_9\delta ^{1+a_0})}\int _{{\mathbb {R}}^n}B(|F_k|)\mathrm{d}x \quad \forall k \end{aligned}$$

(3.10)

Since $F_k\rightarrow F$ strongly in $L^B({\mathbb {R}}^n)$, we deduce from (3.10) that we have for some positive integer $k_0$

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u_k|)\mathrm{d}x \le \frac{2\gamma ' C_9}{\delta ^{\kappa }(1-2\gamma ' C_9\delta ^{1+a_0})}\int _{{\mathbb {R}}^n}B(|F|)\mathrm{d}x \quad \forall k\ge k_0 \end{aligned}$$

(3.11)

Therefore, $u_k$ is uniformly bounded in $W^{1,B}({\mathbb {R}}^n)$ and consequently has a weakly convergent subsequence to some function v in $W^{1,B}({\mathbb {R}}^n)$. Using Lemma 2.7 with function B, we get

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla u_k|)\mathrm{d}x\ge \int _{{\mathbb {R}}^n} B(|\nabla v|)\mathrm{d}x+\int _{{\mathbb {R}}^n}\left<\frac{b(|\nabla v|)}{|\nabla v|}\nabla v,\nabla u_k-\nabla v\right>\mathrm{d}x \end{aligned}$$

(3.12)

Passing to the limit in (3.11) and taking into account (3.12) and the convergence for the weak topology, we infer that

$$\begin{aligned} \int _{{\mathbb {R}}^n} B(|\nabla v|)\mathrm{d}x \le \frac{2\gamma ' C_9}{\delta ^{\kappa }(1-2\gamma ' C_9\delta ^{1+a_0})}\int _{{\mathbb {R}}^n}B(|F|)\mathrm{d}x \end{aligned}$$

(3.13)

The proof of the theorem will be complete if we prove that v is a solution of problem (P). To this end, we observe that it is enough to show that $\Theta (\nabla u_k)$ has a weakly convergent subsequence in $L^{\widetilde{A}}({\mathbb {R}}^n)$ to $\Theta (\nabla v)$. The proof is well known for monotone and continuous operators. We give it here for the sake of completeness.

Using $w=u_k-v$ as a test function for problem $(P_k)$, we get

$$\begin{aligned} \int _{{\mathbb {R}}^n} \Theta (\nabla u_k).(\nabla u_k-\nabla v)\mathrm{d}x =\int _{{\mathbb {R}}^n}\Theta (F_k).(\nabla u_k-\nabla v) \mathrm{d}x \end{aligned}$$

(3.14)

Now, given that $\nabla u_k$ is uniformly bounded in $L^{A}({\mathbb {R}}^n)$, $\Theta (\nabla u_k)$ is uniformly bounded in $L^{\widetilde{A}}({\mathbb {R}}^n)$, and, therefore, has a weakly convergent subsequence to some vector function V in $L^{\widetilde{A}}({\mathbb {R}}^n)$. Moreover, $u_k\rightharpoonup v$ weakly in $W^{1,B}({\mathbb {R}}^n)$ and $F_k\rightarrow F$ strongly in $L^{B}({\mathbb {R}}^n)$. Therefore, in particular, $u_k\rightharpoonup v$ weakly in $W^{1,A}({\mathbb {R}}^n)$ and $F_k\rightarrow F$ strongly in $L^{A}({\mathbb {R}}^n)$. Hence, we obtain from (3.14)

$$\begin{aligned}&\limsup _{k\rightarrow \infty }\int _{{\mathbb {R}}^n}|\nabla u_k|a(|\nabla u_k|)\mathrm{d}x=\int _{{\mathbb {R}}^n} V.\nabla v\mathrm{d}x \end{aligned}$$

(3.15)

At this point, we use (1.7) for $X=\nabla u_k$ and a vector function $Y\in L^{A}({\mathbb {R}}^n)$

$$\begin{aligned} \int _{{\mathbb {R}}^n} \big (\Theta (\nabla u_k)-\Theta (Y)\big ). (\nabla u_k-Y) \mathrm{d}x\ge 0. \end{aligned}$$

Letting $k\rightarrow \infty$ and taking into account (3.15), one can check that

$$\begin{aligned} \int _{{\mathbb {R}}^n}\big (V-\Theta (Y)\big ). (\nabla v-Y) \mathrm{d}x\ge 0. \end{aligned}$$

Now choosing $Y=\nabla v-\lambda \vartheta$, where $\lambda$ is a positive number and $\vartheta$ is an arbitrary vector function in $\mathcal {D}({\mathbb {R}}^n)$, we obtain

$$\begin{aligned} \int _{{\mathbb {R}}^n} \big (V-\Theta (\nabla v-\lambda \vartheta )\big ). \vartheta \mathrm{d}x\ge 0. \end{aligned}$$

Letting $\lambda \rightarrow 0$, we get ${\int _{{\mathbb {R}}^n} \big (V-\Theta (\nabla v)\big ). \vartheta \mathrm{d}x\ge 0.}$ Since $\vartheta$ is arbitrary in $\mathcal {D}({\mathbb {R}}^n)$, it follows that $V=\Theta (\nabla v)$ in ${\mathbb {R}}^n$. This achieves the proof. $\square$