Nonlinear parabolic inequalities in Lebesgue-Sobolev spaces with variable exponent

Abstract

We give an existence result of the obstacle parabolic problem:

$$\begin{aligned} \left\{ \begin{array}{ll} u\ge \psi &{}\ \ \text{ a.e. } \text{ in } \ \Omega \times (0,T)\\ \displaystyle \frac{\partial b(x,u)}{\partial t} -div(a(x,t,u,\nabla u))+div(\phi (x,t,u)) =f &{}\ \ \text{ in }\ \Omega \times (0,T) .\\ \end{array} \right. \end{aligned}$$

The main contribution of our work is to prove the existence of an entropy solution without the coercivity condition on $\phi$. The second term f belongs to $L^{1}(\Omega \times (0,T))$ and $b(.,u_0)\in L^1(\Omega )$. The proof is based on the penalization methods.

Appendix

Now we state an embedding theorem, well-known Gagliardo-Nirenberg embedding theorem, in the framework of Lebesgue-Sobolev generalized, that will play a central role in our work (see [20] for the case where p is constant).

Lemma 3.6

(Gagliardo-Nirenberg generalized) Let v be a function in $W^{1,q(.)}_{0}(\Omega )\cap L^{\rho (.)}(\Omega )$ with $q,\rho \in P^{log}(\Omega ),1<q^{-}\le q(x)\le q^{+}< N, 1<\rho ^{-}\le \rho (x)\le \rho ^{+}< N$. Then there exists a positive constant C, depending on $N,\ q(x)$ and $\rho (x)$, such that

$$\begin{aligned} \parallel v\parallel _{L^{\gamma (.)}(\Omega )}\le C\parallel \nabla v\parallel _{(L^{q(.)}(\Omega ))^{N}}^{\theta }\parallel v\parallel ^{1-\theta }_{L^{\rho (.)}(\Omega )} \end{aligned}$$

for every $\theta$ and $\gamma (.)$ satisfying

$$\begin{aligned} 0\le \theta \le 1,\quad 1\le \gamma (.)\le +\infty ,\quad \frac{1}{\gamma (.)}= \theta \left( \frac{1}{q(.)}-\frac{1}{N}\right) +\frac{1-\theta }{\rho (.)} \end{aligned}$$

Proof

We prove the interpolation:

$$\begin{aligned} ||u||_{L^{p(.)}}\le C ||u||^{\theta }_{L^{p_{1}(.)}}|| u||^{1-\theta }_{L^{p_{2}(.)}} \, \text{ with } \quad \frac{1}{p(.)}=\frac{\theta }{p_{1}(.)}+\frac{1-\theta }{p_{2}(.)} \end{aligned}$$

By Hölder inequality, where $\displaystyle q(.)=\frac{p_{1}(.)}{(1-\theta )p(.)}$ and $\displaystyle q'(.)=\frac{p_{2}(.)}{\theta p(.)}$, one has

$$\begin{aligned}&\displaystyle \int _{\Omega }\Bigg |\frac{u(x)}{||u(x)||^{\theta }_{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{p(x)}\,dx\\&\quad \le C_{1} \Bigg \Vert \Bigg |\frac{u(x)}{||u(x)||^{\theta }_{L^{p_{1}(.)}}|| u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{(1-\theta )p(x)}\Bigg \Vert _{L^{q(.)}}\Bigg \Vert \Bigg |\frac{u(x)}{||u(x)||^{\theta }_{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{\theta p(x)} \Bigg \Vert _{L^{q'(.)}} \nonumber \\&\quad \le C_{1} \Bigg \Vert \Bigg |\frac{u}{||u||_{L^{p_{1}(.)}}} \Bigg |^{(1-\theta )p(x)}\Bigg |\frac{||u||_{L^{p_{1}(.)}}}{||u||_{L^{p_{2}(.)}}} \Bigg |^{(1-\theta )^{2}p(x)} \Bigg \Vert _{L^{q(.)}}\Bigg \Vert \nonumber \\&\qquad \times \Bigg |\frac{u}{||u||_{L^{p_{2}(.)}}} \Bigg |^{\theta p(x)}\Bigg |\frac{||u||_{L^{p_{2}(.)}}}{||u||_{L^{p_{1}(.)}}} \Bigg |^{\theta ^{2}p(x)} \Bigg \Vert _{L^{q'(.)}} \end{aligned}$$

As, $\displaystyle p^{-}\le p(.)\le p^{+}$ that is $\displaystyle \theta ^{2}p^{-}\le \theta ^{2}p(.)\le \theta ^{2}p^{+}$ and $\displaystyle (1-\theta )^{2}p^{-}\le (1-\theta )^{2}p(.)\le (1-\theta )^{2}p^{+}$, then we have:

$$\begin{aligned} \Bigg |\frac{||u||_{L^{p_{1}(.)}}}{||u||_{L^{p_{2}(.)}}} \Bigg |^{(1-\theta )^{2}p(x)}\le max \left( \Bigg |\frac{||u||_{L^{p_{1}(.)}}}{||u||_{L^{p_{2}(.)}}} \Bigg |^{(1-\theta )^{2}p^{+}};\Bigg |\frac{||u||_{L^{p_{1}(.)}}}{||u||_{L^{p_{2}(.)}}} \Bigg |^{(1-\theta )^{2}p^{-}} \right) \end{aligned}$$

which implies

$$\begin{aligned}&\displaystyle \int _{\Omega }\Bigg |\frac{u(x)}{||u(x)||^{\theta }_{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{p(x)}\,dx\\&\quad \le C_{2} \Bigg \Vert \Bigg |\frac{u}{||u||_{L^{p_{1}(.)}}} \Bigg |^{(1-\theta )p(x)}\Bigg \Vert _{L^{q(.)}}\Bigg \Vert \Bigg |\frac{u}{||u||_{L^{p_{2}(.)}}} \Bigg |^{\theta p(x)}\Bigg \Vert _{L^{q'(.)}}\\&\quad \le C_{2} max\left( \Bigg (\int _{\Omega } \Bigg |\frac{u}{||u||_{L^{p_{1}(.)}}} \Bigg |^{p_{1}(x)}\,dx \Bigg )^{\frac{1}{q^{-}}};\, \Bigg (\int _{\Omega } \Bigg |\frac{u}{||u||_{L^{p_{1}(.)}}} \Bigg |^{p_{1}(x)}\,dx \Bigg )^{\frac{1}{q^{+}}}\right) \\&\qquad \times \, max\left( \Bigg (\int _{\Omega } \Bigg |\frac{u}{||u||_{L^{p_{2}(.)}}} \Bigg |^{p_{2}(x)}\,dx \Bigg )^{\frac{1}{q'^{-}}};\Bigg (\int _{\Omega } \Bigg |\frac{u}{||u||_{L^{p_{2}(.)}}} \Bigg |^{p_{2}(x)}\,dx \Bigg )^{\frac{1}{q'^{+}}}\right) \end{aligned}$$

Due to definition of the Luxambourg’s norm, and the fact that $\displaystyle \frac{1}{q^{+}}\le \frac{1}{q^{-}},\quad \displaystyle \frac{1}{q'^{+}}\le \frac{1}{q'^{-}}$

We obtain

$$\begin{aligned} \displaystyle \int _{\Omega }\Bigg |\frac{u(x)}{||u(x)||^{\theta } _{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{p(x)}\,dx\le & {} C_{2} \Bigg (\int _{\Omega }\Bigg |\frac{u}{||u||_{L^{p_{1}(.)}}} \Bigg |^{p_{1}(x)}\,dx \Bigg )^{\frac{1}{q^{+}}}\nonumber \\&\,\times \Bigg (\int _{\Omega }\Bigg |\frac{u}{||u||_{L^{p_{2}(.)}}} \Bigg |^{p_{2}(x)}\,dx \Bigg )^{\frac{1}{q'^{+}}}\le C_{3}. \end{aligned}$$

Firstly: If $C_{3} \le 1$, Then $\displaystyle \int _{\Omega }\Bigg |\frac{u(x)}{||u(x)||^{\theta }_{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{p(x)}\,dx\le 1,$ which give

$$\begin{aligned} \displaystyle \Vert u\Vert _{p(.)}\le \Vert u\Vert ^{\theta }_{p_{1}(.)}\Vert u\Vert ^{1-\theta }_{p_{2}(.)} \end{aligned}$$

Secondly: If $C_{3} \ge 1$, we have $\displaystyle 1\le p(.) \Rightarrow \frac{1}{ C^{p(.)}_{3}}\le \frac{1}{ C_{3}}$ which implies

$$\begin{aligned} \displaystyle \int _{\Omega }\Bigg |\frac{u(x)}{C_{3}||u(x)||^{\theta }_{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{p(x)}\,dx\le & {} \int _{\Omega }\frac{1}{ C_{3}}\Bigg |\frac{u(x)}{||u(x)||^{\theta }_{L^{p_{1}(.)}}||u(x)||^{1-\theta }_{L^{p_{2}(.)}}}\Bigg |^{p(x)}\,dx\nonumber \\\le & {} 1. \end{aligned}$$

which yields

$$\begin{aligned} \Vert u\Vert _{p(.)}\le C_{3} \Vert u\Vert ^{\theta }_{p_{1}(.)}\Vert u\Vert ^{1-\theta }_{p_{2}(.)} \end{aligned}$$

(3.74)

Finally

$$\begin{aligned} \Vert u\Vert _{p(.)}\le C \Vert u\Vert ^{\theta }_{p_{1}(.)}\Vert u\Vert ^{1-\theta }_{p_{2}(.)} ,\quad \text{ With }\quad C=\max (C_{3},1) \end{aligned}$$

(3.75)

Let $p\in P^{log}(\Omega )$ and $\gamma (.)\in C({\overline{\Omega }})$, with $1\le p_{-}\le p^{+}< N$ and $p(.)\le \gamma (.)\le q^{*}(.)$

According to Lemma 2.1 and interpolation (3.75), we get

$$\begin{aligned} \displaystyle \frac{1}{\gamma (.)}= & {} \frac{\theta }{q^{*}(.)}+\frac{(1-\theta )}{p(.)}= \theta \left( \frac{1}{q(.)}-\frac{1}{N}\right) +\frac{(1-\theta )}{p(.)}\\ \displaystyle ||u||_{\gamma (.)}\le & {} C ||u||^{1-\theta }_{p(.)}||u||^{\theta }_{q^{*}(.)}\\\le & {} C||u||^{1-\theta }_{p(.)}||\nabla u||^{\theta }_{q(.)} \end{aligned}$$

$\square$

We give a corollary which is the parabolic version of Lemma 3.6.

Corollary 3.1

Let $v\in L^{q^{-}}((0,T),W_{0}^{1,q(.)}(\Omega ))\cap L^{\infty }((0,T),L^{2}(\Omega ))$, with $q\in P^{log}(\Omega )$ and  $1<q^{-}\le q(x)\le q^{+}< N$, then $v\in L^{\sigma (.)}(\Omega )$ with $\sigma (.)=q(.)(\frac{N+2}{N})$ and

$$\begin{aligned} \int _{Q_{T}}|v|^{\sigma (x)}\,dx\,dt&\le C\max \Bigg (\parallel v\parallel ^{\frac{2 q^{+}}{N}}_{L^{\infty }(0,T,L^{2}(\Omega ))};\parallel v\parallel ^{\frac{2 q^{-}}{N}}_{L^{\infty }(0,T,L^{2}(\Omega ))}\Bigg )\\&\quad \times \max \left( \Bigg (\int _{Q_{T}}\Big |\nabla v|^{q(x)}dxdt\Bigg )^{\frac{q^{+}}{q^{-}}};\Bigg (\int _{Q_{T}}|\nabla v|^{q(x)}dxdt\Bigg )^{\frac{q^{-}}{q^{+}}}\right) \end{aligned}$$

Proof

We choose $\theta =\frac{N}{N+2}$ in Lemma 3.6, then $\sigma (.)=q(.)(\frac{N+2}{N})$ and

$$\begin{aligned} \parallel v\parallel _{L^{\sigma (.)}(\Omega )}\le C\parallel v\parallel ^{(1-\theta )}_{L^{2}(\Omega )} \parallel v\parallel ^{\theta }_{L^{q(.)}(\Omega )} \end{aligned}$$

By Lemma 2.1 we have

$$\begin{aligned} \int _{\Omega }|v|^{\sigma (x)}dx\le \max \Big (\parallel v\parallel ^{\sigma ^{+}}_{L^{\sigma (.)}(\Omega )} ; \parallel v\parallel ^{\sigma ^{-}}_{L^{\sigma (.)}(\Omega )}\Big ) \end{aligned}$$

Combining Gagliardo-Nirenberg inequalities generalized (see Lemmas  3.6) and (2.1), one has

$$\begin{aligned} \parallel v\parallel ^{\sigma ^{+}}_{L^{\sigma (.)}(\Omega )}\le & {} C\parallel v\parallel ^{(1-\theta )\sigma ^{+}}_{L^{2}(\Omega )}\nonumber \\&\times \max \left( \Bigg (\int _{\Omega }|\nabla v|^{q(x)}\,dx\Bigg )^ {\frac{\theta \sigma ^{+}}{q^{+}}}, \Bigg (\int _{\Omega }|\nabla v|^{q(x)}\,dx\Bigg )^ {\frac{\theta \sigma ^{+}}{q^{-}}}\right) \end{aligned}$$

and

$$\begin{aligned} \parallel v\parallel ^{\sigma ^{-}}_{L^{\sigma (.)}(\Omega )}\le & {} C\parallel v\parallel ^{(1-\theta )\sigma ^{-}}_{L^{2}(\Omega )(\Omega )}\nonumber \\&\times \max \left( \Bigg (\int _{\Omega }|\nabla v|^{q(x)}\,dx\Bigg )^ {\frac{\theta \sigma ^{-}}{q^{+}}} , \Bigg (\int _{\Omega }|\nabla v|^{q(x)}\,dx\Bigg ) ^{\frac{\theta \sigma ^{-}}{q^{-}}}\right) \end{aligned}$$

Moreover, $\sigma (.)=q(.)(\frac{N+2}{N})$ implies $\frac{\theta \sigma ^{+}}{q^{+}}= \frac{\theta \sigma ^{-}}{q^{-}}=1.$ Then

$$\begin{aligned} \parallel v\parallel ^{\sigma ^{+}}_{L^{\sigma (.)}(\Omega )}\le & {} C\parallel v\parallel ^{2\frac{q^{+}}{N}}_{L^{2}(\Omega )}\max \left( \int _{\Omega }|\nabla v|^{q(x)}\,dx , \Bigg (\int _{\Omega }|\nabla v|^{q(x)}\,dx\Bigg )^ {\frac{q^{+}}{q^{-}}}\right) \\ \parallel v\parallel ^{\sigma ^{-}}_{L^{\sigma (.)}(\Omega )}\le & {} C\parallel v\parallel ^{2\frac{q^{-}}{N}}_{L^{2}(\Omega )}\max \left( \int _{\Omega }|\nabla v|^{q(x)}\,dx , \Bigg (\int _{\Omega }|\nabla v|^{q(x)}\,dx\Bigg )^{\frac{q^{-}}{q^{+}}}\right) \end{aligned}$$

Finally

$$\begin{aligned}&\max \Big (\parallel v\parallel ^{\sigma ^{+}}_{L^{\sigma (.)}(\Omega )}, \parallel v\parallel ^{\sigma ^{-}}_{L^{\sigma (.)}(\Omega )}\Big ) \\&\quad \le C\max \Big (\parallel v\parallel ^{\frac{2 q^{+}}{N}}_{L^{2}(\Omega )},\parallel v\parallel ^{\frac{2 q^{-}}{N}} _{L^{2}(\Omega )}\Big )\nonumber \\&\qquad \times \max \left( \Bigg (\int _{\Omega }\Big |\nabla v|^{q(x)}dx\Bigg )^{\frac{q^{+}}{q^{-}}},\Bigg (\int _{\Omega }|\nabla v|^{q(x)}dx\Bigg )^{\frac{q^{-}}{q^{+}}}\right) \end{aligned}$$

and so we can conclude the Proof of corollary 3.1 . $\square$

In order to prove the existence results, we prove a technical lemma (we follow the same method used in [8], and in [13] in the case where the function p(.) is constant), that yields two estimates for $|u|^{p(x)-1}$ and $|\nabla u|^{p(x)-1}$ in the Lorentz spaces with variable exponents $L^{\frac{p(.)(N+1)-N}{N(p(.)-1)},\infty }(Q_{T})$ and $L^{\frac{p(.)(N+1)-N}{(N+1)(p(.)-1)},\infty }(Q_{T})$ respectively.

Lemma 3.7

Assume that $\Omega$ is an open set of $\mathbb {R}^{N}$ of finite measure and $1<p^{-}\le p(x)\le p^{+}<N$. Let u be a measurable function satisfying $T_{k}(u)\in V\cap L^{\infty }((0,T);L^{2}(\Omega ))$ for every k and such that:

$$\begin{aligned} \sup _{t\in (0,T)}\int _{\Omega }|T_{k}(u)|^{2}\,dx+\int _{Q_{T}}|\nabla T_{k}(u)|^{p(x)}\,dx\,dt\le Mk\quad \forall \ k>0, \end{aligned}$$

(3.76)

where M is a positive constant. Then

$$\begin{aligned} \Vert |u|^{p(x)-1}\Vert _{L^{\frac{p(.)(N+1)-N}{N(p(.)-1)},\infty }(Q_{T})}\le & {} C \left[ M^{ \left( \frac{p^{-}}{N}+1\right) \frac{N}{N+(p^{-})^{'}}}+(\text{ meas }(Q_{T}))^{\frac{N}{N+(p^{-})^{'}}}\right] \nonumber \\\end{aligned}$$

(3.77)

$$\begin{aligned} \Vert |\nabla u|^{p(x)-1}\Vert _{L^{\frac{p(.)(N+1)-N}{(N+1)(p(.)-1)},\infty }(Q_{T})}\le & {} C\left[ M^{\frac{(N+2)(p^{-}-1)}{p^{-}(N+1)-N}}+(\text{ meas }(Q_{T}))^{\frac{N+1}{N+(p^{-})^{'}}}\right] \end{aligned}$$

(3.78)

where C is a constant depend only on N, $p^{-}$ and $p^{+}$.

Proof

In the case of $k>1$, we have

$$\begin{aligned} \displaystyle \int _{\{|u|>k\}} k^{\frac{p^{-}(N+2)}{N}} \,dx\,dt\le \displaystyle \int _{\{|u|>k\}} k^{\frac{p(x)(N+2)}{N}} \,dx\,dt \le \displaystyle \int _{Q_{T}}|T_{k}u|^{\frac{p(x)(N+2)}{N}}\,dx\,dt \end{aligned}$$

That is

$$\begin{aligned} \displaystyle k^{\frac{p^{-}(N+2)}{N}}meas \{ (x,t)\in Q_{T} \,:\, |u|> k\} < \int _{Q_{T}}|T_{k}u|^{\frac{p(x)(N+2)}{N}}\,dx\,dt \end{aligned}$$

By Gagliardo-Niremberg (see Corollary 3.1), we have

$$\begin{aligned}&\displaystyle k^{\frac{p^{-}(N+2)}{N}} meas \{ (x,t)\in Q_{T} \,:\, |u|> k\}\\&\quad \le \int _{Q_{T}}|T_{k}(u)|^{\frac{p(x)(N+2)}{N}}\,dx\,dt\\&\quad \le C\max \left( \sup _{t\in (0,T)}\Bigg (\int _{\Omega }| T_{k}(u)|^{2}dx\Bigg ) ^{\frac{p^{+}}{N}} ;\sup _{t\in (0,T)}\Bigg (\int _{\Omega }| T_{k}(u)|^{2}dx\Bigg )^{\frac{p^{-}}{N}}\right) \\&\qquad \times \max \left( \Bigg (\int _{Q_{T}}|\nabla T_{k}u|^{p(x)}dxdt \Bigg )^{\frac{p^{+}}{p^{-}}};\Bigg (\int _{Q_{T}}|\nabla T_{k}u|^{p(x)} dxdt\Bigg )^{\frac{p^{-}}{p^{+}}}\right) \end{aligned}$$

Moreover, $|T_{k}(u)|\le k$ implies $\displaystyle \int _{\Omega }|T_{k}(u)|^{2}dx\le k^{2}\text{ meas }(\Omega ),$ then $\frac{1}{k^{2}\text{ meas }(\Omega )}\displaystyle \int _{\Omega }|T_{k}(u)|^{2}dx\le 1$.

Since $k>1$ and by Lemma 2.1, we get

$$\begin{aligned} \text{ meas }(\Omega )\le \displaystyle \int _{\{|u|>k\}}|T_{k}(u)|^{p^{*}(x)}dx\le \displaystyle \int _{\Omega }|T_{k}(u)|^{p^{*}(x)}dx\le c_{1} \displaystyle \int _{\Omega }|\nabla T_{k}(u)|^{p(x)}, \end{aligned}$$

which implies

$$\begin{aligned} 1\le \frac{c_{1}}{meas(\Omega )} \displaystyle \int _{\Omega }|\nabla T_{k}(u)|^{p(x)}dx. \end{aligned}$$

Then for some constant $C>0$, the result is

$$\begin{aligned}&\displaystyle k^{\frac{p^{-}(N+2)}{N}} meas \{ (x,t)\in Q_{T} \,:\, |u|> k\} \nonumber \\&\quad \le C\sup _{t\in (0,T)} \Bigg (\int _{\Omega }| T_{k}(u)|^{2}\,dx\Bigg )^{\frac{p^{-}}{N}} \Bigg (\int _{Q_{T}}\Big |\nabla T_{k}u|^{p(x)}dxdt\Bigg )\nonumber \\&\quad \le C \big (MK\big )^{\frac{p^{-}}{N}+1} \end{aligned}$$

(3.79)

or equivalently (taking $k=h^{\frac{1}{p^{-}-1}}$), for every $h>0$

$$\begin{aligned} h^{\frac{1}{p^{-}-1}\frac{p^{-}(N+2)}{N}}\text{ meas }\{(x,t)\in Q_{T} \,:\, |u|^{p(x)-1}> h\}\le CM^{\frac{p^{-}}{N}+1}\times h^{\frac{1}{p^{-}-1} \left( \frac{p^{-}}{N}+1\right) }, \end{aligned}$$

we deduce that

$$\begin{aligned} \text{ meas }\{(x,t)\in Q_{T} \,:\, |u|^{p(x)-1}> h\}\le CM^{\frac{p^{-}}{N}+1}\times h^{-\frac{N+(p^{-})^{'}}{N}}. \end{aligned}$$

(3.80)

Using the embedding in Lorentz space with variable exponent

$$\begin{aligned} L^{\eta ^{+},\infty }(Q_T)\hookrightarrow L^{\eta (.),\infty }(Q_T) \end{aligned}$$

(3.81)

with $\eta (.)=\frac{p(.)(N+1)-N}{N(p(.)-1)}$, and the fact that $\eta ^{+}\le \frac{N+(p^{-})^{'}}{N}$, we have:

$$\begin{aligned}&\displaystyle |||u|^{p(x)-1}||_{L^{\eta (.),\infty }(Q_{T})}\le C|||u|^{p(x)-1}||_{L^{\frac{N+(p^{-})^{'}}{N},\infty }(Q_{T})}\\&\quad \le C sup_{h>0} h\,\left[ meas\{(x,t)\in Q_{T}:\,\,|u|^{p(x)-1}>h \}\right] ^{\frac{N}{N+(p^{-})^{'}}} \\&\quad \le C sup_{0<h<1} h \, \left[ meas\{(x,t)\in Q_{T}:\,\,|u|^{p(x)-1}>h \}\right] ^{\frac{N}{N+(p^{-})^{'}}}\\&\qquad +\,Csup_{1\le h <h_{0}} h \, \left[ meas\{(x,t)\in Q_{T}:\,\,|u|^{p(x)-1}>h \}\right] ^{\frac{N}{N+(p^{-})^{'}}}\\&\qquad +\,Csup_{ h>h_{0}} h \, \left[ meas\{(x,t)\in Q_{T}:\,\,|u|^{p(x)-1}>h \}\right] ^{\frac{N}{N+(p')^{+}}}\\&\quad \le 2C h_{0}(\text{ meas }(Q_{T}))^{\frac{N}{N+(p^{-})^{'}}}+ C M^{ \left( \frac{p^{-}}{N}+1\right) \frac{N}{N+(p^{-})^{'}}}. \end{aligned}$$

Taking $h_0=M^{(\frac{p^{-}}{N}+1)\frac{N}{N+(p^{-})^{'}}}+1$ we have (3.77).

Now we prove the estimate involving the gradient of u, for every $k>1$ and for every $\lambda >0$ we have:

$$\begin{aligned} meas\{(x,t)\in Q_T \,:\,|\nabla u|>\lambda \}\le & {} meas\{(x,t)\in Q_T\,:\, |\nabla u|>\lambda \,\text{ and }\,|u|\le k\}\\&+ meas\{(x,t)\in Q_T\,:\, |\nabla u|>\lambda \,\text{ and }\, |u|>k\}. \end{aligned}$$

In the case of $\lambda >1$ and by (3.76), we know that

$$\begin{aligned} \displaystyle \lambda ^{p^{-}}meas \{(x,t)\in Q_T \,:\,|\nabla u|>\lambda \, \text{ and }\, |u|\le k \}\le & {} \int \int _{|u|\le k}\lambda ^{p^{-}}\,dx\,dt\\\le & {} \int \int _{Q_T}|\nabla T_k|^{p(x)} \,dx\,dt.\\\le & {} Mk \end{aligned}$$

That is $meas \{(x,t)\in Q_T\,:\, |\nabla u|^{p(x)-1}>\lambda \, \text{ and }\, |u|\le k \}\le \frac{Mk}{\lambda ^{(p^{-})'}}$.

On the other hand by (3.79), we have:

$$\begin{aligned} meas \{(x,t)\in Q_T \,:\,|\nabla u|^{p(x)-1}>\lambda \, \text{ and }\, |u|> k \}\le C M^{\frac{p^{-}}{N}+1} k^{\frac{-p^{-}(N+2)}{N}}. \end{aligned}$$

Then

$$\begin{aligned} meas \{(x,t)\in Q_T\,:\, |\nabla u|^{p(x)-1}>\lambda \}\le & {} \frac{Mk}{\lambda ^{(p^{-})'}}+C M^{\frac{p^{-}}{N}+1} k^{\frac{-p^{-}(N+2)}{N}}\\\le & {} \frac{Mk}{\lambda ^{(p^{-})'}}+C M^{\frac{p^{-}}{N}+1} k^{-\frac{p^{-}(N+1)-N}{N}}. \end{aligned}$$

If we Take $k=M^{\frac{1}{N+1}}\lambda ^{\frac{N}{(N+1)(p^{-}-1)}},$ we have

$$\begin{aligned} meas \{(x,t)\in Q_T \,:\,|\nabla u|^{p(x)-1}>\lambda \}\le C\frac{M^{\frac{N+2}{N+1}}}{\lambda ^{\frac{P^{-}(N+1)-N}{(N+1)(p^{-}-1)}}}. \end{aligned}$$

Thanks to (3.81), we have

$$\begin{aligned}&\displaystyle |||\nabla u|^{p(x)-1}||_{L^{\frac{p(.)(N+1)-N}{(N+1)(p(.)-1)},\infty }(Q_{T})} \nonumber \\&\quad \le C sup_{\lambda >0} \lambda \, \left[ meas\{(x,t)\in Q_{T}:\,\,|\nabla u|^{p(x)-1}>\lambda \}\right] ^{\frac{N+1}{N+(p^{-})^{'}}}\nonumber \\&\quad \le C sup_{0< \lambda <1} \lambda \, \left[ meas\{(x,t)\in Q_{T}:\,\,|\nabla u|^{p(x)-1}> \lambda \}\right] ^{\frac{N+1}{N+(p^{-})^{'}}}\\&\qquad +\,sup_{1\le \lambda <\lambda _{0}} \lambda \, \left[ meas\{(x,t)\in Q_{T}:\,\,|\nabla u|^{p(x)-1}>\lambda \}\right] ^{\frac{N+1}{N+(p^{-})^{'}}}\\&\qquad +\,sup_{ \lambda > \lambda _{0}} \lambda \, \left[ meas\{(x,t)\in Q_{T}:\,\,|\nabla u|^{p(x)-1}>\lambda \}\right] ^{\frac{N+1}{N+(p^{-})^{'}}}\\&\quad \le 2C\lambda _{0}(\text{ meas }(Q_{T}))^{\frac{N+1}{N+(p^{-})^{'}}}+C M^{\frac{(N+2)(p^{-}-1)}{p^{-}(N+1)-N}}. \end{aligned}$$

Then, if we choose $\lambda _{0}= M^{\frac{(N+2)(p^{-}-1)}{p^{-}(N+1)-N}}+1$, we obtain (3.78) and the Proof of lemma 3.7 is complete. $\square$