# Study of weak solutions for parabolic variational inequalities with nonstandard growth conditions

## Abstract

In this paper, we study the degenerate parabolic variational inequality problem in a bounded domain. First, the weak solutions of the variational inequality are defined. Second, the existence and uniqueness of the solutions in the weak sense are proved by using the penalty method and the reduction method.

## 1 Introduction

$$\textstyle\begin{cases} \min\{ Lu,u(x,0) - {u_{0}}(x)\} = 0,& (x,t) \in{Q_{T}},\\ u(x,t) = 0,& (x,t) \in{\Gamma_{T}},\\ u(x,0) = {u_{0}}(x),& x \in\Omega, \end{cases}$$
(1)

with

$$Lu = {u_{t}} - \operatorname{div}\bigl( {a(u){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u} \bigr) - f(x,t),\quad a(u) = {u^{\sigma}} + {d_{o}},$$

where $$\Omega \subset{\mathbb{R}^{+} }$$ is a bounded simply connected domain, $${Q_{T}} = \Omega \times(0,T]$$, and $${\Gamma_{T}}$$ denotes the lateral boundary of the cylinder $${Q_{T}}$$.

This type of variational inequality was studied initially by Chen and Yi [1], who proposed the equation

$$\textstyle\begin{cases} \frac{\partial}{{\partial\tau}}V - \frac{1}{2}{\sigma^{2}}\frac {{{\partial^{2}}}}{{\partial{x^{2}}}}V - ( {r - \frac{1}{2}{\sigma^{2}}} )\frac{\partial}{{\partial x}}V + rV \ge0 &\text{in } {\Omega_{T}},\\ V \ge g(x), &\text{in } {\Omega_{T}},\\ ( {\frac{\partial}{{\partial\tau}}V - \frac{1}{2}{\sigma^{2}}\frac {{{\partial^{2}}}}{{\partial{x^{2}}}}V - ( {r - \frac{1}{2}{\sigma^{2}}} )\frac{\partial}{{\partial x}}V + rV} )(V - g(x)) = 0 &\text{in }{\Omega_{T}},\\ V(t,x) = 0 &\text{on } \partial{\Omega_{T}},\\ V(x,0) = g(x) &\text{in } \Omega \end{cases}$$
(2)

for modeling the American option. When r and Ïƒ are positive constant, the existence and uniqueness of solutions to problem (4) were also studied in [2â€“4].

In 2014, the authors in [5] discussed the problem

$$\textstyle\begin{cases} {u_{t}} - Lu - F(x,t,u,\nabla u) \ge0&\text{in }{Q_{T}},\\ u(x,t) \ge{u_{0}}(x)& \text{in } {Q_{T}},\\ ( {{u_{t}} - Lu - F(x,t,u,\nabla u)} )(u - {u_{0}}(x)) = 0 &\text{in } {Q_{T}},\\ u(x,0) = {u_{0}}(x) &\text{on } \Omega,\\ u(x,t) = g(x) &\text{on } \partial\Omega \times(0,T) \end{cases}$$

with second-order elliptic operator

$$L(x,t) = \sum_{i,j = 1}^{d} { \frac{\partial}{{\partial{x_{j}}}}} \biggl( {{a^{ij}}(x,t)\frac{\partial}{{\partial{x_{i}}}}} \biggr) - \sum_{i,j = 1}^{d} {{b^{i}}(x,t) \frac{\partial}{{\partial{x_{i}}}}} - c(x,t).$$

They proved the existence and uniqueness of a solution to this problem with some conditions on $${u_{0}}$$, F, and L. Later, the authors in [6, 7] extended the relative conclusions with the assumption that $$a(u)$$ and $$p(x)$$ are two positive constants. The author discussed the existence and uniqueness of a solution by the penalty method.

The existence and uniqueness of such a problem with the assumption that $$p(x)$$ and $$a(u)$$ are variables were less studied.

The aim of this paper is to study the existence and uniqueness of solutions for a degenerate parabolic variational inequality problem. Throughout the paper, we assume that the exponent $$p(x,t)$$ is continuous in $$Q = \overline{{Q_{T}}}$$ with logarithmic module of continuity:

\begin{aligned}& 1 < {p^{-} } = \inf_{(x,t) \in Q} p(x,t) \le p(x,t) \le{p^{+} } = \sup _{(x,t) \in Q} p(x,t) < \infty, \end{aligned}
(3)
\begin{aligned}& \forall z = (x,t) \in Q,\qquad \xi = (y,s) \in{Q_{T}},\qquad \vert {z - \xi} \vert < 1,\qquad \bigl\vert {p(z) - p(\xi)} \bigr\vert \le\omega \bigl( \vert {z - \xi} \vert \bigr), \end{aligned}
(4)

where

$$\mathop{\lim\sup}\limits _{\tau \to{0^{+} }} \omega(\tau)\ln\frac {1}{\tau} = C < + \infty.$$

The outline of this paper is as follows. In SectionÂ 2, we introduce the function spaces of Orlicz-Sobolev type, give the definition of a weak solution to the problem, and prove the existence and uniqueness. SectionÂ 3 is devoted to the proof of the existence and uniqueness of the solution obtained in SectionÂ 2.

## 2 Basic spaces and the main results

To study our problems, let us introduce the Banach spaces:

\begin{aligned}& {L^{p(x,t)}}({Q_{T}}) = \biggl\{ u(x,t) \Big\vert u \text{ is measurable in } {Q_{T}},{A_{p(\cdot)}}(u) = \int {\int_{{Q_{T}}} {{{\left| u \right|}^{p(x,t)}}\, {\mathrm{d}}x\, {\mathrm{d}}t < \infty } } \biggr\} , \\& { \Vert u \Vert _{p(\cdot)}} = \inf\bigl\{ \lambda > 0,{A_{p(\cdot)}}(u/\lambda) \le1\bigr\} , \\& {V_{t}}(\Omega) = \bigl\{ u \vert {u \in{L^{2}}(\Omega) \cap W_{0}^{1,1}(\Omega)}, \vert {\nabla u} \vert \in {L^{p(x,t)}}(\Omega)\bigr\} ,\qquad { \Vert u \Vert _{{V_{t}}(\Omega)}} = { \Vert u \Vert _{2,\Omega}} + { \vert {\nabla u} \vert _{p(\cdot )\Omega}}, \\& W({Q_{t}}) = \bigl\{ u:[0,T] \mapsto{V_{t}}(\Omega) \vert {u \in {L^{2}}({Q_{t}})}, \vert {\nabla u} \vert \in {L^{p(x,t)}}({Q_{T}}),u = 0\text{ on } {\Gamma_{T}} \bigr\} , \\& { \Vert u \Vert _{W({Q_{t}})}} = { \Vert u \Vert _{2,{Q_{T}}}} + { \vert {\nabla u} \vert _{p(x,t),{Q_{T}}}} \end{aligned}

and denote by $$W'({Q_{T}})$$ the dual of $$W({Q_{T}})$$ with respect to the inner product in $${L^{2}}({Q_{T}})$$.

In spirit of [3] and [4], we introduce the following maximal monotone graph:

$$G(\lambda) = \textstyle\begin{cases} 0,& \lambda > 0,\\ [0,+\infty),& \lambda = 0. \end{cases}$$

In addition, we define the following function class for the solution:

$$B = \bigl\{ {u \in W({Q_{T}}) \cap{L^{\infty}} \bigl(0,T;{L^{\infty}}(\Omega)\bigr)} \bigr\} .$$

### Definition 2.1

A pair $$(u,\xi) \in B \times{L^{\infty}}({\Omega_{T}})$$ is called a weak solution of problem (1) if (a)Â $$u(x,t) \le{u_{0}}(x)$$, (b)Â $$u(x,0) = {u_{0}}(x)$$, (c)Â $$\xi \in G(u - {u_{0}})$$, (d)Â for all $${t_{1}},{t_{2}} \in[0,T]$$, the following identity holds:

$$\int_{{t_{1}}}^{{t_{2}}} \int_{\Omega}\bigl[u{\varphi_{t}} - \bigl({u^{\sigma}} + {d_{0}}\bigr){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u\nabla\varphi + f(x,t)\varphi - \xi\varphi\bigr] \,\mathrm{d}x\,\mathrm{d}t = \int_{\Omega}{u\varphi\,\mathrm{d}x \bigg\vert _{{t_{1}}}^{{t_{2}}}}.$$

The main theorem in this section is the following:

### Theorem 2.1

Let $$p(x,t)$$ satisfy conditions (3)â€“(4). Suppose also that the following conditions hold:

($$H_{1}$$):

$$\max\{ {1,\frac{{2N}}{{N + 2}}} \} < {p^{-} } < N$$, $$2 \le\sigma < \frac {{2{p^{+} }}}{{{p^{+} } - 1}}$$,

($$H_{2}$$):

$${u_{0}} \ge0$$, $$f \ge0$$, $${ \Vert {{u_{0}}} \Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \Vert {f(x,t)} \Vert}_{\infty,\Omega}}} \,\mathrm{d}t = K(T) < \infty$$.

Then problem (1) has at least one weak solution in the sense of DefinitionÂ 2.1.

### Theorem 2.2

Suppose that the conditions in TheoremÂ 2.1 are fulfilled and $${p^{+} } \ge2$$. Then problem (1) admits a unique solution in the sense of DefinitionÂ 2.1.

## 3 Proof of the main results

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 LemmaÂ 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 Youngâ€™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 LemmaÂ 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 Youngâ€™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 CÌƒ independent of Îµ.

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 $${Q_{\tau}}$$.â€ƒâ–¡

### 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 LemmaÂ 3.2 we conclude that

$$u(x,t) \le{u_{0\varepsilon}}(x) \quad \text{for any } (x,t) \in { \Omega_{T}},$$
(22)

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 Î©. Thus, combining (21) and (23) and repeating LemmaÂ 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, LemmaÂ 3.3 can be proved by combining initial and boundary conditions in (5).â€ƒâ–¡

Moreover, with (18), we assert that there exists a subsequence Îµ (still denoted by Îµ) 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 Lebesgueâ€™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 Holderâ€™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}

â€ƒâ–¡

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.â€ƒâ–¡

### 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 LemmaÂ 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).â€ƒâ–¡

### 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 HÃ¶lder 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 LemmaÂ 3.5.â€ƒâ–¡

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 Îµ, 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 LemmaÂ 3.7.â€ƒâ–¡

### 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 Lebesgueâ€™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)â€“(42) and the previous inequalities, we complete the proof of LemmaÂ 3.8.â€ƒâ–¡

### 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)â€“(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)

â€ƒâ–¡

### 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 Î´ neighborhood $${B_{\delta}}({x_{0}},{t_{0}})$$ such that if Îµ 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 Îµ 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 LemmaÂ 3.10 is completed.â€ƒâ–¡

Applying (28), (29), and LemmaÂ 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 TheoremÂ 2.1 in [8], and we omit the details. Moreover, the uniqueness of solutions can be proved by repeating LemmaÂ 3.1.

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## Acknowledgements

This work was supported by Scientific research Foundation of Shaanxi Railway Institute (No.Â KY2016-04) and Natural Science Foundation of Shaanxi Provincial Department of Education (No.Â 17JK0170). The author is sincerely grateful to the reference and the Associate Editor handling the paper for their valuable comments.

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The author read and approved the final manuscript.

### Corresponding author

Correspondence to Yan Dong.

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Dong, Y. Study of weak solutions for parabolic variational inequalities with nonstandard growth conditions. J Inequal Appl 2018, 35 (2018). https://doi.org/10.1186/s13660-018-1623-0