In this section, we consider the family of auxiliary parabolic problems
$$ \textstyle\begin{cases} {L_{\varepsilon}}{u_{\varepsilon}} + {\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) = 0, & (x,t) \in{Q_{T}},\\ u(x,t) = \varepsilon, & (x,t) \in{\Gamma_{T}},\\ u(x,0) = {u_{0}}(x) + \varepsilon, & x \in\Omega. \end{cases} $$
(5)
Here, M is a positive parameter to be chosen later. Moreover,
$$\begin{aligned}& {L_{\varepsilon}} {u_{\varepsilon}} = {\partial_{t}} {u_{\varepsilon}} - \operatorname{div}\bigl( {{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}}} \bigr) - f(x,t), \\& 0 < {d_{0}} \le{a_{\varepsilon,M}}(u) = {\bigl(\min\bigl({ \vert u \vert ^{2}},{M^{2}}\bigr) + {\varepsilon^{2}} \bigr)^{\frac{\sigma}{2}}} + {d_{0}} \le \bigl({M^{2}} + 1\bigr) + {d_{0}},\quad 0 < \varepsilon < 1, \end{aligned}$$
and \({\beta_{\varepsilon}}( \cdot)\) is the penalty function satisfying
$$ \begin{gathered} 0 < \varepsilon \le1,\qquad {\beta_{\varepsilon}}(x) \in{C^{2}}(R),\qquad {\beta _{\varepsilon}}(x) \le0,\qquad {\beta_{\varepsilon}}(0) = - 1, \\ {\beta'_{\varepsilon}}(x) \ge0,\qquad {\beta''_{\varepsilon}}(x) \le0,\qquad \lim_{\varepsilon \to0} {\beta_{\varepsilon}}(x) = \textstyle\begin{cases} 0, & x > 0,\\ - \infty, & x < 0. \end{cases}\displaystyle \end{gathered} $$
(6)
Following a similar method as in [6], we can prove that the regularized problem has a unique weak solution \({u_{\varepsilon}}(x,t) \in W({Q_{T}}) \cap{L^{2}}({Q_{T}})\), \({\partial _{t}}{u_{\varepsilon}}(x,t) \in W'({Q_{T}})\) satisfying the following integral identities:
$$ \begin{aligned}[b] & \int_{{t_{1}}}^{{t_{2}}} { \int_{\Omega}{\bigl[{u_{\varepsilon}} {\varphi_{t}} - {a_{\varepsilon}},M({u_{\varepsilon}}){{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla\varphi + f(x,t)\varphi \bigr]\,\mathrm{d}x\,\mathrm{d}t} } \\ &\quad = \int_{{t_{1}}}^{{t_{2}}} { \int_{\Omega}{{\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}})\varphi\,\mathrm{d}x\,\mathrm{d}t} } + \int _{\Omega}{{u_{\varepsilon}}\varphi\,\mathrm{d}x} \bigg\vert _{{t_{1}}}^{{t_{2}}} \end{aligned} $$
(7)
and
$$ \int_{{t_{1}}}^{{t_{2}}} \int_{\Omega}\bigl[({\partial_{t}} {u_{\varepsilon}})\varphi + {a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \varphi - f(x,t)\varphi + { \beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}})\varphi\bigr] \,\mathrm{d}x\,\mathrm{d}t = 0. $$
(8)
We start with two preliminary results that will be used several times.
Lemma 3.1
Let
\(M(s) = {| s |^{p(x,t) - 2}}s\). Then for all
\(\xi,\eta \in\,\mathrm{R}^{N}\),
$$\begin{aligned}[b] &\bigl(M(\xi) - M(\eta)\bigr) (\xi - \eta) \\ &\quad \ge \textstyle\begin{cases} {2^{ - p(x,t)}}{ \vert {\xi - \eta} \vert ^{p(x,t)}},& 2 \le p(x,t) < \infty,\\ (p(x,t) - 1){ \vert {\xi - \eta} \vert ^{2}}{({ \vert \xi \vert ^{p(x,t)}} + { \vert \eta \vert ^{p(x,t)}})^{\frac{{p(x,t) - 2}}{{p(x,t)}}}}, & 1 < p(x,t) < 2. \end{cases}\displaystyle \end{aligned} $$
Lemma 3.2
(Comparison principle)
Assume that
\(2 < \sigma < \frac{{2{p^{+} }}}{{{p^{+} } - 1}}\), \({p^{+} } \ge2\), and
u
and
v
are in
\(W({Q_{T}}) \cap{L^{\infty}}(0,T;{L^{\infty}}(\Omega))\). If
\({L_{\varepsilon}}u \ge{L_{\varepsilon}}v\)
in
\({Q_{T}}\)
and if
\(u(x,t) \le v(x,t)\)
on
\(\partial{Q_{T}}\), then
\(u(x,t) \le v(x,t)\)
in
\({Q_{T}}\).
Proof
We argue by contradiction. Suppose \(u(x,t)\) and \(v(x,t)\) satisfy \({L_{\varepsilon}}u \ge{L_{\varepsilon}}v\) in \({Q_{T}}\) and there is \(\delta > 0\) such that for \(0 < \tau \le T\), \(w = u - v > \delta\) on the set \({\Omega_{\delta}} = \Omega \cap\{ x:w(x,t) > \delta\} \), and \(\mu({\Omega_{\delta}}) > 0\). Let
$${F_{\varepsilon}}(\xi) = \textstyle\begin{cases} \frac{1}{{\alpha - 1}}{\varepsilon^{1 - \alpha}} - \frac{1}{{\alpha - 1}}{\xi^{1 - \alpha}}& \text{if }\xi > \varepsilon,\\ 0& \text{if }\xi \le\varepsilon, \end{cases} $$
where \(\delta > 2\varepsilon > 0\) and \(\alpha = \frac{\sigma}{2}\). Let a test-function \(\xi = {F_{\varepsilon}}(w) \in Z\) in (8). Then
$$ \begin{aligned}[b] 0 &\ge \int{ \int_{{Q_{T}}} {\bigl[{w_{t}} {F_{\varepsilon}}(w) + \bigl({v^{\sigma}} + {d_{0}}\bigr) \bigl({{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla {F_{\varepsilon}}(w)\bigr]\,\mathrm{d}x \,\mathrm{d}t} } \\ &\quad{}+ \int{ \int_{{Q_{T}}} {\bigl({u^{\sigma}} - {v^{\sigma}} \bigr){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u\nabla {F_{\varepsilon}}(w)\,\mathrm{d}x\,\mathrm{d}t} } = {J_{1}} + {J_{2}} + {J_{3}}, \end{aligned} $$
(9)
where \({Q_{\varepsilon,\tau}} = {Q_{\tau}} \cap\{ (x,t) \in{Q_{\tau}}| {w > \varepsilon} \} \),
$$\begin{aligned}& {J_{1}} = \int{ \int_{{Q_{T}}} {{w_{t}} {F_{\varepsilon}}(w)\,\mathrm{d}x \,\mathrm {d}t} }, \qquad{J_{2}} = \int{ \int_{{Q_{T}}} {\bigl({u^{\sigma}} - {v^{\sigma}} \bigr){w^{ - \alpha}} {{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u \nabla w\,\mathrm{d}x\,\mathrm{d}t} }, \\& {J_{3}} = \int{ \int_{{Q_{T}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}}\bigl({{ \vert {\nabla u} \vert }^{p(x,t) - 2}} \nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla w \,\mathrm {d}x\,\mathrm{d}t} }. \end{aligned}$$
Now, let \({t_{0}} = \inf\{ t \in(0,\tau]:w > \varepsilon\} \). Then we estimate \({J_{1}}\), \({J_{2}}\), and \({J_{3}}\) as follows:
$$ \begin{aligned}[b] {J_{1}} &= \int{ \int_{{Q_{\varepsilon,\tau}}} {{w_{t}} {F_{\varepsilon}}(w)\,\mathrm{d}x \,\mathrm{d}t} = \int_{\Omega}{\biggl( \int_{0}^{{t_{0}}} {{w_{t}} {F_{\varepsilon}}(w) \,\mathrm{d}t} + \int_{{t_{0}}}^{\tau}{{w_{t}} {F_{\varepsilon}}(w)\,\mathrm{d}t} \biggr)} } \,\mathrm{d}x \\ &\ge \int_{\Omega}{ \int_{\varepsilon}^{w(x,\tau)} {{F_{\varepsilon}}(s)\,\mathrm{d}s \,\mathrm{d}x} \ge} \int_{{\Omega_{\delta}}} { \int _{\varepsilon}^{w(x,\tau)} {{F_{\varepsilon}}(s)\,\mathrm{d}s \,\mathrm{d}x} } \\ &\ge \int_{{\Omega_{\delta}}} {(w - 2\varepsilon)} {F_{\varepsilon}}(2 \varepsilon)\,\mathrm{d}x \ge(\delta - 2\varepsilon){F_{\varepsilon}}(2 \varepsilon)\mu({\Omega_{\delta}}). \end{aligned} $$
(10)
Let us first consider the case \({p^{-} } \ge2\). By the first inequality of Lemma 3.1 we get
$$ \begin{aligned}[b] {J_{2}} &= \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}}\bigl({{ \vert {\nabla u} \vert }^{p(x,t) - 2}} \nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla w \,\mathrm{d}x\,\mathrm{d}t} } \\ &\ge \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}} {2^{ - p(x,t)}} {{ \vert {\nabla w} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t} } \\ &\ge{2^{ - {p^{+} }}} \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t)}} \,\mathrm{d}x\,\mathrm{d}t} } \ge0. \end{aligned} $$
(11)
Noting that \(\frac{{p(x,t)}}{{p(x,t) - 1}} \ge\frac{{{p^{+} }}}{{{p^{+} } - 1}} \ge\frac{\sigma}{2} = \alpha > 1\) and applying Young’s inequality, we can estimate the integrand of \({J_{3}}\) in the following way:
$$ \begin{aligned}[b] & \bigl\vert {\bigl({u^{\sigma}} - {v^{\sigma}}\bigr){w^{ - \alpha}}} { \vert {\nabla w} \vert ^{p(x,t) - 2}} {\nabla u\nabla w} \bigr\vert \\ &\quad= \biggl\vert {\sigma w \int_{0}^{1} {{{\bigl(\theta u + (1 - \theta )v \bigr)}^{\sigma - 1}}\,d\theta{w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t) - 2}}\nabla u\nabla w} } \biggr\vert \\ &\quad\le\frac{C}{{{w^{\alpha}}}}\biggl[\frac{{{v^{\sigma}} + {d_{0}}}}{C}\biggr]{ \vert {\nabla w} \vert ^{p(x,t)}} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}} \bigr){ \vert w \vert ^{p'(x,t)}} { \vert {\nabla u} \vert ^{p(x,t)}}] \\ &\quad= \frac{{({v^{\sigma}} + {d_{0}})}}{{{2^{{p^{+} } + 1}}{w^{\alpha}}}}{ \vert {\nabla w} \vert ^{p(x,t)}} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}}\bigr){ \vert w \vert ^{p'(x,t) - \alpha }} { \vert {\nabla u} \vert ^{p(x,t)}} \\ &\quad\le\frac{{({v^{\sigma}} + {d_{0}})}}{{{2^{{p^{+} } + 1}}{w^{\alpha}}}}{ \vert {\nabla w} \vert ^{p(x,t)}} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}}\bigr){ \vert {\nabla u} \vert ^{p(x,t)}}. \end{aligned} $$
(12)
Substituting (12) into \(J_{3}\), we get
$$ {J_{3}} \le\frac{1}{2}{J_{2}} + C{ \int{ \int_{{Q_{\varepsilon,\tau}}} { \vert {\nabla u} \vert } } ^{p(x,t)}} \,\mathrm{d}x\,\mathrm {d}t. $$
(13)
Second, we consider the case \(1 < {p^{-} } \le p(x,t) < 2\), \({p^{+} } \ge2\). According to the second inequality of Lemma 3.1, it is easily seen that the following inequalities hold:
$$ \begin{aligned}[b] {J_{2}} = \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}}\bigl({{ \vert {\nabla w} \vert }^{p(x,t) - 2}} \nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla w \,\mathrm{d}x\,\mathrm{d}t} } \\ \ge\bigl({p^{-} } - 1\bigr) \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}} {{\bigl( \vert {\nabla w} \vert + \vert {\nabla v} \vert \bigr)}^{p(x,t) - 2}} {{ \vert {\nabla w} \vert }^{2}} \,\mathrm{d}x\,\mathrm{d}t} }. \end{aligned} $$
(14)
Using the conditions \(1 < \alpha \le\frac{{{p^{+} }}}{{{p^{+} } - 1}} \le2\) and Young’s inequality, we can evaluate the integrand of \(J_{3}\) as follows:
$$ \begin{aligned}[b] & \bigl\vert {\bigl({u^{\sigma}} - {v^{\sigma}}\bigr){w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t) - 2}}\nabla u\nabla w} \bigr\vert \\ &\quad = \biggl\vert {\sigma w \int_{0}^{1} {{{\bigl(\theta u + (1 - \theta )v \bigr)}^{\sigma - 1}}\,d\theta{w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t) - 2}}\nabla u\nabla w} } \biggr\vert \\ &\quad \le\frac{{({v^{\sigma}} + {d_{0}})({p^{-} } - 1)}}{{2{w^{\alpha}}}}{\bigl( \vert {\nabla w} \vert + \vert {\nabla v} \vert \bigr)^{p(x,t) - 2}} { \vert {\nabla w} \vert ^{2}} \\ &\qquad{} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}} \bigr){ \vert w \vert ^{2 - \alpha}} \vert {\nabla w} \vert + \vert { \nabla v} \vert {)^{p(x,t)}}. \end{aligned} $$
(15)
Plugging (15) into \(J_{3}\), we get
$$ {J_{3}} \le\frac{1}{2}{J_{2}} + C \int{ \int_{{Q_{\varepsilon,\tau}}} { \vert {\nabla w} \vert + \vert {\nabla v} \vert {)^{p(x,t)}}} } \,\mathrm{d}x\,\mathrm{d}t. $$
(16)
Plugging estimates (10), (11), (13) and (10), (14), (16) into (9) and dropping the nonnegative terms, we arrive at the inequality
$$ (\delta - 2\varepsilon) \bigl(1 - {2^{1 - \alpha}}\bigr){\varepsilon^{1 - \alpha}} \mu({\Omega_{\delta}}) \le\tilde{C} $$
(17)
with a constant C̃ independent of ε.
Notice that \({\lim_{\varepsilon \to0}}(\delta - 2\varepsilon)(1 - {2^{1 - \alpha}}){\varepsilon^{1 - \alpha}}\mu({\Omega_{\delta}}) = + \infty\), a contradiction. This means that \(\mu({\Omega_{\delta}}) = 0\) and \(w \le0\) a.e. in \({Q_{\tau}}\). □
Lemma 3.3
Let
\({u_{\varepsilon}}\)
be weak solutions of (5). Then
$$\begin{aligned}& {u_{0\varepsilon}} \le{u_{\varepsilon}} \le{ \vert {{u_{0}}} \vert _{\infty}} + \varepsilon, \end{aligned}$$
(18)
$$\begin{aligned}& {u_{{\varepsilon_{1}}}} \le{u_{{\varepsilon_{2}}}} \quad \textit{for } { \varepsilon_{1}} \le{\varepsilon_{2}}. \end{aligned}$$
(19)
Proof
First, we prove \({u_{\varepsilon}} \ge{u_{0\varepsilon}}\) by contradiction. Assume that \({u_{\varepsilon}} \le{u_{0\varepsilon}}\) in \(Q_{T}^{0}\), \(Q_{T}^{0} \subset {Q_{T}}\). Noting that \({u_{\varepsilon}} \ge{u_{0\varepsilon}}\) on \(\partial{Q_{T}}\), we may assume that \({u_{\varepsilon}} = {u_{0\varepsilon}}\) on \(\partial Q_{T}^{0}\). With (5) and letting \(t = 0\), we deduce that
$$\begin{aligned}& L{u_{0\varepsilon}} = {\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{0\varepsilon}}) = - 1, \end{aligned}$$
(20)
$$\begin{aligned}& L{u_{\varepsilon}} = {\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0\varepsilon}}) \le - 1. \end{aligned}$$
(21)
From Lemma 3.2 we conclude that
$$ u(x,t) \le{u_{0\varepsilon}}(x) \quad \text{for any } (x,t) \in { \Omega_{T}}, $$
(22)
obtaining a contradiction.
Second, we pay attention to \({u_{\varepsilon}}(t,x) \le{| {{u_{0}}} |_{\infty}} + \varepsilon\). Applying the definition of \({\beta_{\varepsilon}}( \cdot )\), we have
$$ L\bigl( {{{ \vert {{u_{0}}} \vert }_{\infty}} + \varepsilon} \bigr) = 0, \quad L{u_{\varepsilon}} \le0. $$
(23)
From (5) it is easy to prove that \({u_{\varepsilon}}(x,t) \ge \varepsilon\) on \(\partial\Omega \times(0,T)\) and \({u_{0\varepsilon}}(x) \ge\varepsilon\) in Ω. Thus, combining (21) and (23) and repeating Lemma 3.3, we have
$$ {u_{\varepsilon}}(x,t) \ge\varepsilon \quad\text{in } \Omega \times (0,T). $$
(24)
Third, we aim to prove (19). Since
$$\begin{aligned}& L{u_{{\varepsilon_{1}}}} - {\beta_{{\varepsilon_{1}}}}({u_{0{\varepsilon _{1}}}} - {u_{{\varepsilon_{1}}}}) = 0, \end{aligned}$$
(25)
$$\begin{aligned}& L{u_{{\varepsilon_{2}}}} - {\beta_{{\varepsilon_{2}}}}({u_{0{\varepsilon _{2}}}} - {u_{{\varepsilon_{2}}}}) = 0. \end{aligned}$$
(26)
It follows by \({\varepsilon_{1}} \le{\varepsilon_{2}}\) and the definition of \({\beta_{\varepsilon}}( \cdot )\) that
$$ \begin{aligned}[b] &{\partial_{t}} {u_{{\varepsilon_{2}}}} - L{u_{{\varepsilon_{2}}}} - {\beta _{{\varepsilon_{1}}}}({u_{0{\varepsilon_{1}}}} - {u_{{\varepsilon_{2}}}}) \\ &\quad={\beta_{{\varepsilon_{2}}}}({u_{0{\varepsilon_{2}}}} - {u_{{\varepsilon_{2}}}}) - { \beta_{{\varepsilon_{1}}}}({u_{0{\varepsilon _{1}}}} - {u_{{\varepsilon_{2}}}}) \ge{ \beta_{{\varepsilon _{2}}}}({u_{0{\varepsilon_{2}}}} - {u_{{\varepsilon_{2}}}}) - {\beta _{{\varepsilon_{1}}}}({u_{0{\varepsilon_{2}}}} - {u_{{\varepsilon_{2}}}}) \ge0. \end{aligned} $$
(27)
Thus, Lemma 3.3 can be proved by combining initial and boundary conditions in (5). □
Moreover, with (18), we assert that there exists a subsequence ε (still denoted by ε) such that
$$\begin{aligned}& {u_{\varepsilon}} \to u \in{L^{p}}\bigl(0,T;W_{0}^{1,p}({ \Omega_{T}})\bigr) \quad \text{as } \varepsilon \to0, \end{aligned}$$
(28)
$$\begin{aligned}& {u_{\varepsilon}} \ge u \ge0 \quad \text{for any } \varepsilon > 0. \end{aligned}$$
(29)
Lemma 3.4
Let
\({u_{\varepsilon}}\)
be a solution of problem (5). For any
\(\varepsilon > 0\), we have
$$ { \Vert {{u_{\varepsilon}}} \Vert _{\infty,{Q_{T}}}} \le{ \Vert {{u_{0}}} \Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \bigl\Vert {f(x,t)} \bigr\Vert }_{\infty,\Omega}}\,\mathrm{d}t} = K(T) < \infty. $$
(30)
Proof
Define
$${u_{\varepsilon M}} = \textstyle\begin{cases} M & \text{if } {u_{\varepsilon}} > M,\\ {u_{\varepsilon}} & \text{if } \vert {{u_{\varepsilon}}} \vert \le M,\\ - M &\text{if } {u_{\varepsilon}} < - M. \end{cases} $$
Choosing \(u_{\varepsilon M}^{2k - 1}\) as a test-function in (8) and letting \({t_{1}} = t\) and \({t_{2}} = t + h\), we conclude that
$$ \begin{aligned}[b] &\frac{1}{{2k}} \int_{t}^{t + h} {\frac{\mathrm{d}}{{\mathrm{d}t}}\biggl( { \int _{\Omega}{u_{\varepsilon M}^{2k}\,dx} } \biggr)} \,\mathrm{d}t + \int_{t}^{t + h} { \int_{\Omega}{(2k - 1){a_{\varepsilon,M}}({u_{\varepsilon M}})} } u_{\varepsilon M}^{2(k - 1)}{ \vert {\nabla{u_{\varepsilon M}}} \vert ^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t \\ &\quad = \int_{t}^{t + h} { \int_{\Omega}{fu_{\varepsilon M}^{2k - 1}\,\mathrm{d}x \,\mathrm{d}t} - \int_{t}^{t + h} { \int_{\Omega}{{\beta _{\varepsilon}}u_{\varepsilon M}^{2k - 1} \,\mathrm{d}x} } }. \end{aligned} $$
(31)
Letting \(h \to0\) and applying Lebesgue’s dominated convergence theorem, we have that, for all \(t \in(0,T)\),
$$ \begin{aligned}[b] &\frac{1}{{2k}}\frac{\mathrm{d}}{{\mathrm{d}t}} \int_{\Omega}{u_{\varepsilon M}^{2k}\,\mathrm{d}x} + \int_{\Omega}{(2k - 1){a_{\varepsilon,M}}({u_{\varepsilon M}})} u_{\varepsilon M}^{2(k - 1)}{ \vert {\nabla{u_{\varepsilon M}}} \vert ^{p(x,t)}}\,\mathrm{d}x \\ &\quad = \int_{\Omega}{fu_{\varepsilon M}^{2k - 1}\,\mathrm{d}x} - \int _{\Omega}{{\beta_{\varepsilon}}u_{\varepsilon M}^{2k - 1} \,\mathrm{d}x}. \end{aligned} $$
(32)
Using Holder’s inequality, we obtain
$$\begin{aligned}& \biggl\vert { \int_{\Omega}{fu_{\varepsilon M}^{2k - 1}\,dx} } \biggr\vert \le \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{2k,\Omega}^{2k - 1} \cdot{ \bigl\Vert {f(\cdot,t)} \bigr\Vert _{2k,\Omega}},\quad k = 1,2,\dots, \\& \biggl\vert { \int_{\Omega}{{\beta_{\varepsilon}}u_{\varepsilon M}^{2k - 1} \,\mathrm{d}x} } \biggr\vert \le \int_{\Omega}{u_{\varepsilon M}^{2k - 1}\,\mathrm{d}x} \le \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{2k,\Omega}^{2k - 1}, \end{aligned}$$
whence
$$ \begin{aligned}[b] & \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega}^{2k - 1}\frac{\mathrm{d}}{{\mathrm{d}t}}\bigl({ \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega}}\bigr) + (2k - 1) \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon M}})} u_{\varepsilon M}^{2(k - 1)}{ \vert {\nabla{u_{\varepsilon,M}}} \vert ^{p(x,t)}}\,\mathrm{d}x \\ &\quad \le \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega }^{2k - 1} \cdot{ \bigl\Vert {f( \cdot,t)} \bigr\Vert _{2k,\Omega}} + C(T) \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega}^{2k - 1},\quad k = 1,2,\dots. \end{aligned} $$
(33)
By integration over \((0, t)\), for all t, we have
$${ \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{2k,\Omega}} \le { \bigl\Vert {{u_{\varepsilon M}}(\cdot,0)} \bigr\Vert _{2k,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{2k,\Omega}} \,\mathrm{d}t} + C(T),\quad \forall k \in\mathbb{N}. $$
Then, as \(k \to\infty\),
$$\begin{aligned}[b] { \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{\infty,\Omega }} &\le{ \bigl\Vert {{u_{\varepsilon M}}(\cdot,0)} \bigr\Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{\infty,\Omega}} \,\mathrm{d}t} \\ &\le{ \Vert {{u_{0}}} \Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{\infty,\Omega}} \,\mathrm{d}t} + C(T) = K(T). \end{aligned} $$
□
If we chose \(M > K(T)\) then \({u_{\varepsilon M}}(\cdot,t) \le\sup| {{u_{\varepsilon M}}(\cdot,t)} | \le K(T) < M\), and therefore \({u_{\varepsilon M}}(\cdot,t) = {u_{\varepsilon}}(\cdot,t)\).
Corollary 3.1
Choosing
M
large enough, we have
$$\min\bigl\{ u_{\varepsilon}^{2},{M^{2}}\bigr\} = u_{\varepsilon}^{2} \quad \textit{and}\quad {a_{\varepsilon,M}} \bigl(u_{\varepsilon M}\bigr) = {a_{\varepsilon,M}}({a_{\varepsilon}}) = \bigl({\varepsilon^{2}} + u_{\varepsilon}^{2} \bigr)^{\sigma/ 2} + {d_{0}}. $$
Corollary 3.2
If
\({u_{0}} \ge0\)
and
\(f \ge0\), then the solution
\({u_{\varepsilon}}(x,t)\)
is nonnegative in
\({Q_{T}}\).
Proof
Set \(u_{\varepsilon}^{-} = \min\{ {u_{\varepsilon}},0\} \). Then \(u_{\varepsilon}^{-} (x,0) = 0\), \(u_{\varepsilon}^{-} | {{\Gamma_{T}}} = 0\), and
$$\frac{1}{2}\frac{\mathrm{d}}{{\mathrm{d}t}}\bigl( \bigl\Vert {u_{\varepsilon}^{-} (x,t)} \bigr\Vert _{2,\Omega}^{2}\bigr) + \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}})} { \bigl\vert {\nabla u_{\varepsilon}^{-} } \bigr\vert ^{p(x,t)}}\,\mathrm{d}x \le0. $$
It follows that, for every \(t > 0\),
$$\bigl\Vert {u_{\varepsilon}^{-} (x,t)} \bigr\Vert _{2,\Omega} \le \bigl\Vert {u_{\varepsilon}^{-} (\cdot,0)} \bigr\Vert _{2,\Omega} = 0, $$
and the required assertion follows. □
Lemma 3.5
The solution of problem (5) satisfies the estimates
$$\begin{aligned}& { \int{ \int_{{Q_{T}}} {u_{\varepsilon}^{\sigma}{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t \le K(T) \vert \Omega \vert } } ^{\frac{1}{2}}}, \end{aligned}$$
(34)
$$\begin{aligned}& {\varepsilon^{\sigma}} { \int{ \int_{{Q_{T}}} {{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t \le K(T) \vert \Omega \vert } } ^{\frac{1}{2}}}, \end{aligned}$$
(35)
$$\begin{aligned}& {d_{0}} { \int{ \int_{{Q_{T}}} {{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t \le K(T) \vert \Omega \vert } } ^{\frac{1}{2}}}. \end{aligned}$$
(36)
Proof
Similarly as in Lemma 3.4, we take \(k = 1\) in (32) to get
$$\frac{\mathrm{d}}{{\mathrm{d}t}}{ \bigl\Vert {{u_{\varepsilon}}(\cdot,t)} \bigr\Vert _{2,\Omega}} + \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x} \le{ \Vert f \Vert _{2,\Omega}} + C(T),\quad \forall t \in(0,T). $$
Clearly, integrating over \((0,t)\), we have
$$ { \bigl\Vert {{u_{\varepsilon}}(\cdot,t)} \bigr\Vert _{2,\Omega}} + \int _{0}^{t} { \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x} \,\mathrm{d}t \le} { \bigl\Vert {{u_{\varepsilon}}(\cdot,t)} \bigr\Vert _{2,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{2,\Omega }} \,\mathrm{d}t} + C(T). $$
(37)
Note that the first term on the left-hand side is nonnegative. We drop the nonpositive term in (37) to get
$$\int_{0}^{t} { \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x} \,\mathrm{d}t \le} K(T){ \vert \Omega \vert ^{\frac{1}{2}}}. $$
If \({a_{\varepsilon,M}}({u_{\varepsilon}}) \ge{d_{0}}\), then we have inequality (36), and if \({a_{\varepsilon,M}}({u_{\varepsilon}}) \ge{\varepsilon^{\sigma}}\), then we have inequality (35) such that \(M > K(T)\), and we have \({a_{\varepsilon,M}}({u_{\varepsilon}}) \ge u_{\varepsilon}^{\sigma}\). Furthermore, we get inequality (34). □
Lemma 3.6
The solution of problem (5) satisfies the estimate
$$ { \Vert {{u_{\varepsilon t}}} \Vert _{W'({Q_{T}})}} \le C\bigl( { \sigma,{p^{\pm}},K(T), \vert \Omega \vert } \bigr). $$
(38)
Proof
From identity (7) we get
$$\begin{aligned}[b] \int{ \int_{{Q_{T}}} {{u_{\varepsilon t}}\xi\,\mathrm{d}x\,\mathrm{d}t} } &= - \int{ \int_{{Q_{T}}} {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ { \varepsilon ^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}} } \nabla{u_{\varepsilon}}\nabla\xi\,\mathrm {d}x\,\mathrm{d}t \\ &\quad{}+ \int{ \int_{{Q_{T}}} {f(x,t)\xi(x,t)\,\mathrm{d}x\,\mathrm{d}t} } - \int{ \int_{{Q_{T}}} {{\beta_{\varepsilon}}(x,t)\xi(x,t)\,\mathrm {d}x \,\mathrm{d}t} }. \end{aligned} $$
Applying the fact that \({\beta_{\varepsilon}}(x,t) \in(0,1)\), we get
$$\begin{aligned} \int{ \int_{{Q_{T}}} {{u_{\varepsilon t}}\xi\,\mathrm{d}x\,\mathrm{d}t} } &\le \int_{0}^{T} { \int_{\Omega}{\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ { \varepsilon ^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} } \nabla{u_{\varepsilon}}\nabla\xi\,\mathrm {d}x\,\mathrm{d}t\\ &\quad {} + \int_{0}^{T} { \int_{\Omega}{ \vert {f + 1} \vert \cdot \vert \xi \vert } } \,\mathrm{d}x\,\mathrm{d}t. \end{aligned} $$
Using the Hölder inequality repeatedly, we have that
$$\begin{aligned}[b] \int{ \int_{{Q_{T}}} {{u_{\varepsilon t}}\xi\,\mathrm{d}x\,\mathrm{d}t} } & \le2{ \bigl\Vert {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ {\varepsilon ^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\Vert _{p'(x,t)}} { \Vert {\nabla\xi} \Vert _{p(x,t)}}\\ &\quad {} + 2{ \Vert {f + 1} \Vert _{p'(x,t)}} \cdot{ \Vert \xi \Vert _{p(x,t)}} \\ &\le2\max\{ {F_{1}},{F_{2}}\} { \Vert {\nabla\xi} \Vert _{p(x,t)}} + 2\max\{ {{F_{3}},{F_{4}}} \}{ \Vert \xi \Vert _{p(x,t)}} \\ & \le\bigl(2{\bigl({\bigl({K^{2}}(T) + 1\bigr)^{{\sigma / 2}}} + {d_{0}}\bigr)^{\frac{1}{{{p^{\pm}} - 1}}}}K(T) \vert \Omega \vert + 2{ \vert {f + 1} \vert _{\infty}} \vert T \vert \bigr){ \Vert \xi \Vert _{W({Q_{T}})}}, \end{aligned} $$
where
$$\begin{aligned}& {F_{1}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{\bigl\{ {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ {\varepsilon^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\} }^{\frac {{p(x,t)}}{{p(x,t) - 1}}}}\,\mathrm{d}x\,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' + }}}}, \\& {F_{2}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{\bigl\{ {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ {\varepsilon^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\} }^{\frac {{p(x,t)}}{{p(x,t) - 1}}}}\,\mathrm{d}x\,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' - }}}}, \\& {F_{3}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{ \vert f \vert }^{p'(x,t)}}\,\mathrm{d}x \,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' + }}}}, \qquad {F_{4}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{ \vert {f + 1} \vert }^{p'(x,t)}} \,\mathrm{d}x\,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' - }}}}. \end{aligned}$$
Then (38) follows from Lemma 3.5. □
From [6] we can get the following inclusions:
$$\begin{aligned}& {u_{\varepsilon}} \in W({Q_{T}}) \subseteq{L^{{p^{-} }}} \bigl(0,T;W_{0}^{1,{p^{-} }}(\Omega)\bigr), \qquad {u_{\varepsilon t}} \in W'({Q_{T}}) \subseteq {L^{\frac{{{p^{+} }}}{{{p^{+} } - 1}}}}\bigl(0,T;{V_{+} }(\Omega)\bigr), \\& W_{0}^{1,{p^{-} }}(\Omega) \subset{L^{2}}(\Omega) \subset{V_{+} }^{\prime}(\Omega) \quad \text{with } {V_{+} }(\Omega) = \bigl\{ u(x)\vert {u \in{L^{2}}(\Omega) \cap W_{0}^{1,1}( \Omega)}, \vert {\nabla u} \vert \in{L^{{p^{+} }}}\bigr\} . \end{aligned}$$
These conclusions, together with the uniform estimates in ε, allow us to extract from the sequence \(\{ {u_{\varepsilon}}\} \) a subsequence (for simplicity, we assume that it merely coincides with the whole sequence) such that
$$ \textstyle\begin{cases} {u_{\varepsilon}} \to u & \mbox{a.e. in }{Q_{T}},\\ \nabla{u_{\varepsilon}} \to\nabla u &\text{weakly in }{L^{p(x,t)}}({Q_{T}}),\\ u_{\varepsilon}^{\sigma}{ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}{D_{i}}{u_{\varepsilon}} \to{A_{i}}(x,t) & \text{weakly in } {L^{p'(x,t)}}({Q_{T}}),\\ { \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}{D_{i}}{u_{\varepsilon}} \to{W_{i}}(x,t) & \text{weakly in }{L^{p'(x,t)}}({Q_{T}}) \end{cases} $$
(39)
for some functions \(u \in W({Q_{T}})\), \({A_{i}}(x,t) \in {L^{p'(x,t)}}({Q_{T}})\), and \({W_{i}}(x,t) \in{L^{p'(x,t)}}({Q_{T}})\).
Lemma 3.7
For almost all
\((x,t) \in{Q_{T}}\),
$$\lim_{\varepsilon \to{0^{+} }} \int{ \int_{{Q_{T}}} {\bigl({{\bigl(u_{\varepsilon}^{2} + { \varepsilon^{2}}\bigr)}^{\frac{\sigma}{2}}} - u_{\varepsilon}^{\sigma}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} \nabla{u_{\varepsilon}}\nabla\xi\,\mathrm{d}x\,\mathrm{d}t = 0,\quad \forall\xi = W({Q_{T}})} }. $$
Proof
We first compute
$$\begin{aligned}[b] I &\stackrel{\Delta}{=} \int{ \int_{{Q_{T}}} {\bigl({{\bigl(u_{\varepsilon}^{2} + { \varepsilon^{2}}\bigr)}^{\frac{\sigma}{2}}} - u_{\varepsilon}^{\sigma}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}}\nabla\xi\,\mathrm{d}x\,\mathrm{d}t} } \\ & = \frac{\sigma}{2}{\varepsilon^{2}} \int{ \int_{{Q_{T}}} {\biggl( { \int_{0}^{1} {{{\bigl(u_{\varepsilon}^{2} + s{\varepsilon^{2}}\bigr)}^{\frac{{\sigma - 2}}{2}}}\,ds} } \biggr)} } { \vert { \nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \xi\,\mathrm{d}x\,\mathrm{d}t \\ & \le\sigma{\varepsilon^{2}} {\bigl({K^{2}}(T) + 1 \bigr)^{{\frac{{\sigma - 2}}{2}}}} { \bigl\Vert {{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\Vert _{p'(x,t)}} { \Vert {\nabla\xi } \Vert _{p(x,t)}} \\ & \le C{\varepsilon^{2}}\max\biggl\{ {{{\biggl( \int{ \int_{{Q_{T}}} {{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)}^{\frac{{{p^{+} } - 1}}{{{p^{+} }}}}},{{\biggl( \int{ \int_{{Q_{T}}} {{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)}^{\frac{{{p^{-} } - 1}}{{{p^{-} }}}}}} \biggr\} { \Vert {\nabla\xi} \Vert _{p(x,t)}}. \end{aligned} $$
By (35) we get
$$I \le C{\varepsilon^{2 - \frac{{\sigma({p^{+} } - 1)}}{{{p^{+} }}}}} { \Vert {\nabla\xi} \Vert _{p(x,t)}}. $$
Passing to the limit as \(\varepsilon \to0\), we obtain Lemma 3.7. □
Lemma 3.8
For almost all
\((x,t) \in{Q_{T}}\), we have
$${A_{i}}(x,t) = {u^{\sigma}} {W_{i}}(x,t),\quad i = 1,2, \dots,N. $$
Proof
In (39), letting \(\varepsilon \to0\), we have
$$\begin{aligned}& { \int \int_{{Q_{T}}} {u_{\varepsilon}^{\sigma} \vert {\nabla {u_{\varepsilon}}} \vert } ^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \xi\,\mathrm{d}x\,\mathrm{d}t \to\sum{ \int{ \int_{{Q_{T}}} {{A_{i}}(x,t){D_{i}}\xi \,\mathrm{d}x\,\mathrm{d}t} } }, \end{aligned}$$
(40)
$$\begin{aligned}& { \int \int_{{Q_{T}}} {u_{\varepsilon}^{\sigma} \vert {\nabla {u_{\varepsilon}}} \vert } ^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \xi\,\mathrm{d}x\,\mathrm{d}t \to\sum{ \int{ \int_{{Q_{T}}} {{W_{i}}(x,t){D_{i}}\xi \,\mathrm{d}x\,\mathrm{d}t} } }. \end{aligned}$$
(41)
By Lebesgue’s dominated convergence theorem we have
$$ \lim_{\varepsilon \to0} \sum_{i = 1}^{N} { \int{ \int_{{Q_{T}}} {\bigl(u_{\varepsilon}^{\sigma}- {u^{\sigma}}\bigr){A_{i}}(x,t){D_{i}}\xi\,\mathrm {d}x \,\mathrm{d}t = 0} }.} $$
(42)
So
$$\begin{aligned}[b] &\lim_{\varepsilon \to0} \sum{ \int{ \int_{{Q_{T}}} {\bigl(u_{\varepsilon}^{\sigma}{{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} {D_{i}} {u_{\varepsilon}} - {u^{\sigma}} {W_{i}}(x,t) \bigr){D_{i}}\xi\,\mathrm {d}x\,\mathrm{d}t} } } \\ &\quad= \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\bigl(u_{\varepsilon}^{\sigma}- {u^{\sigma}}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} {D_{i}} {u_{\varepsilon}} + {u^{\sigma}} \bigl({{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} {D_{i}} {u_{\varepsilon}} - {W_{i}}(x,t) \bigr){D_{i}}\xi\,\mathrm{d}x\,\mathrm{d}t} }\\ &\quad = 0. \end{aligned} $$
By (40)–(42) and the previous inequalities, we complete the proof of Lemma 3.8. □
Lemma 3.9
For almost all
\((x,t) \in{Q_{T}}\), we have
$${W_{i}}(x,t) = { \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}} {D_{i}}u,\quad i = 1,2,\dots,N. $$
Proof
In (8), choosing \(\xi = ({u_{\varepsilon}} - u)\Phi\) with \(\Phi \in W({Q_{T}})\), \(\Phi \ge0\), we have
$$\begin{aligned}[b] & \int{ \int_{{Q_{T}}} {\bigl[{u_{\varepsilon t}}({u_{\varepsilon}} - u) \Phi + \Phi\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 2}} \nabla{u_{\varepsilon}}\nabla({u_{\varepsilon}} - u)\bigr]\,\mathrm{d}x \,\mathrm{d}t} } \\ &\quad{} + \int{ \int_{{Q_{T}}} {\bigl[({u_{\varepsilon}} - u) \bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}} \nabla\Phi - f(x,t) ({u_{\varepsilon}} - u)\Phi\bigr]\,\mathrm{d}x\,\mathrm{d}t} } \\ &\quad{}+ \int{ \int_{{Q_{T}}} {\bigl({{\bigl(u_{\varepsilon}^{\sigma}- { \varepsilon^{2}}\bigr)}^{\frac{\sigma}{2}}} - u_{\varepsilon}^{\sigma}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}}\nabla\xi\,\mathrm{d}x\,\mathrm{d}t} } = 0. \end{aligned} $$
It follows that
$$ \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla({u_{\varepsilon}} - u) \,\mathrm{d}x\,\mathrm{d}t = 0.} } $$
(43)
On the other hand, from \({u_{\varepsilon}},u \in{L^{\infty}}({Q_{T}})\) and \(| {\nabla u} | \in {L^{p(x,t)}}({Q_{T}})\) we get
$$\begin{aligned}& \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl(u^{\sigma}+ {d_{0}}\bigr){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u \nabla({u_{\varepsilon}} - u)\,\mathrm{d}x\,\mathrm{d}t = 0} }, \end{aligned}$$
(44)
$$\begin{aligned}& \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl(u_{\varepsilon}^{\sigma}+ {u^{\sigma}}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla({u_{\varepsilon}} - u) \,\mathrm{d}x\,\mathrm{d}t = 0.} } \end{aligned}$$
(45)
Note that
$$ \begin{aligned}[b] 0 &\le\bigl({ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla{u_{\varepsilon}} - { \vert {\nabla u} \vert ^{p(x,t) - 2}}\nabla u\bigr)\nabla({u_{\varepsilon}} - u) \\ &\le\frac{1}{{{d_{0}}}}\bigl[\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla {u_{\varepsilon}} - \bigl(u_{\varepsilon}^{\sigma}- u^{\sigma}\bigr){ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla u\bigr]\nabla ({u_{\varepsilon}} - u) \\ &\quad{} - \frac{1}{{{d_{0}}}}\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla u\nabla({u_{\varepsilon}} - u). \end{aligned} $$
(46)
By (44)–(46) we have
$$ \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl({{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}} - } } { \vert { \nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla u\bigr) \nabla({u_{\varepsilon}} - u)\,\mathrm {d}x\,\mathrm{d}t = 0. $$
(47)
□
Lemma 3.10
As
\(\varepsilon \to0\),we have
$$ {\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) \to\xi \in G(u - {u_{0}}). $$
(48)
Proof
Using (7) and the definition of \({\beta_{\varepsilon}}\), we have
$${\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) \to\xi\quad \text{as } \varepsilon \to0. $$
Now we prove that \(\xi \in G(u - {u_{0}})\). According to the definition of \(G( \cdot)\), we only need to prove that if \(u({x_{0}},{t_{0}}) > {u_{0}}({x_{0}})\), then \(\xi({x_{0}},{t_{0}}) = 0\). In fact, if \(u({x_{0}},{t_{0}}) > {u_{0}}(x)\), there exist a constant \(\lambda > 0\) and a δ neighborhood \({B_{\delta}}({x_{0}},{t_{0}})\) such that if ε is small enough, we have
$${u_{\varepsilon}}(x,t) \ge{u_{0}}(x) + \lambda, \quad \forall(x,t) \in {B_{\delta}}({x_{0}},{t_{0}}). $$
Thus, if ε is small enough, then we have
$$0 \ge{\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) \ge{\beta _{\varepsilon}}(\lambda) = 0, \quad \forall(x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
Furthermore, it follows by \(\varepsilon \to0\) that
$$\xi(x,t) = 0, \quad \forall(x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
Hence, (48) holds, and the proof of Lemma 3.10 is completed. □
Applying (28), (29), and Lemma 3.10, it is clear that
$$u(x,t) \le{u_{0}}(x) \quad\text{in } {\Omega_{T}}, \qquad u(x,0) = {u_{0}}(x) \quad\text{in } \Omega, \xi \in G(u - {u_{0}}), $$
and thus (a), (b), and (c) hold. The remaining arguments of the existence part are the same as those of Theorem 2.1 in [8], and we omit the details. Moreover, the uniqueness of solutions can be proved by repeating Lemma 3.1.
