1 Introduction
This article concerned with initial-boundary problem whose model is
with
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 \({Q_{T}}\).
$$ \textstyle\begin{cases} \min\{ Lu,u(x,0) - {u_{0}}(x)\} = 0,& (x,t) \in{Q_{T}},\\ u(x,t) = 0,& (x,t) \in{\Gamma_{T}},\\ u(x,0) = {u_{0}}(x),& x \in\Omega, \end{cases} $$
(1)
$$Lu = {u_{t}} - \operatorname{div}\bigl( {a(u){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u} \bigr) - f(x,t),\quad a(u) = {u^{\sigma}} + {d_{o}}, $$
This type of variational inequality was studied initially by Chen and Yi [1], who proposed the equation
for modeling the American option. When r and σ are positive constant, the existence and uniqueness of solutions to problem (4) were also studied in [2‐4].
$$ \textstyle\begin{cases} \frac{\partial}{{\partial\tau}}V - \frac{1}{2}{\sigma^{2}}\frac {{{\partial^{2}}}}{{\partial{x^{2}}}}V - ( {r - \frac{1}{2}{\sigma^{2}}} )\frac{\partial}{{\partial x}}V + rV \ge0 &\text{in } {\Omega_{T}},\\ V \ge g(x), &\text{in } {\Omega_{T}},\\ ( {\frac{\partial}{{\partial\tau}}V - \frac{1}{2}{\sigma^{2}}\frac {{{\partial^{2}}}}{{\partial{x^{2}}}}V - ( {r - \frac{1}{2}{\sigma^{2}}} )\frac{\partial}{{\partial x}}V + rV} )(V - g(x)) = 0 &\text{in }{\Omega_{T}},\\ V(t,x) = 0 &\text{on } \partial{\Omega_{T}},\\ V(x,0) = g(x) &\text{in } \Omega \end{cases} $$
(2)
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In 2014, the authors in [5] discussed the problem
with second-order elliptic operator
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.
$$\textstyle\begin{cases} {u_{t}} - Lu - F(x,t,u,\nabla u) \ge0&\text{in }{Q_{T}},\\ u(x,t) \ge{u_{0}}(x)& \text{in } {Q_{T}},\\ ( {{u_{t}} - Lu - F(x,t,u,\nabla u)} )(u - {u_{0}}(x)) = 0 &\text{in } {Q_{T}},\\ u(x,0) = {u_{0}}(x) &\text{on } \Omega,\\ u(x,t) = g(x) &\text{on } \partial\Omega \times(0,T) \end{cases} $$
$$L(x,t) = \sum_{i,j = 1}^{d} { \frac{\partial}{{\partial{x_{j}}}}} \biggl( {{a^{ij}}(x,t)\frac{\partial}{{\partial{x_{i}}}}} \biggr) - \sum_{i,j = 1}^{d} {{b^{i}}(x,t) \frac{\partial}{{\partial{x_{i}}}}} - c(x,t). $$
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:
where
$$\begin{aligned}& 1 < {p^{-} } = \inf_{(x,t) \in Q} p(x,t) \le p(x,t) \le{p^{+} } = \sup _{(x,t) \in Q} p(x,t) < \infty, \end{aligned}$$
(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)
$$\mathop{\lim\sup}\limits _{\tau \to{0^{+} }} \omega(\tau)\ln\frac {1}{\tau} = C < + \infty. $$
The outline of this paper is as follows. In Section 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. Section 3 is devoted to the proof of the existence and uniqueness of the solution obtained in Section 2.
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2 Basic spaces and the main results
To study our problems, let us introduce the Banach spaces:
and denote by \(W'({Q_{T}})\) the dual of \(W({Q_{T}})\) with respect to the inner product in \({L^{2}}({Q_{T}})\).
$$\begin{aligned}& {L^{p(x,t)}}({Q_{T}}) = \biggl\{ u(x,t) \Big\vert u \text{ is measurable in } {Q_{T}},{A_{p(\cdot)}}(u) = \int {\int_{{Q_{T}}} {{{\left| u \right|}^{p(x,t)}}\, {\mathrm{d}}x\, {\mathrm{d}}t < \infty } } \biggr\} , \\& { \Vert u \Vert _{p(\cdot)}} = \inf\bigl\{ \lambda > 0,{A_{p(\cdot)}}(u/\lambda) \le1\bigr\} , \\& {V_{t}}(\Omega) = \bigl\{ u \vert {u \in{L^{2}}(\Omega) \cap W_{0}^{1,1}(\Omega)}, \vert {\nabla u} \vert \in {L^{p(x,t)}}(\Omega)\bigr\} ,\qquad { \Vert u \Vert _{{V_{t}}(\Omega)}} = { \Vert u \Vert _{2,\Omega}} + { \vert {\nabla u} \vert _{p(\cdot )\Omega}}, \\& W({Q_{t}}) = \bigl\{ u:[0,T] \mapsto{V_{t}}(\Omega) \vert {u \in {L^{2}}({Q_{t}})}, \vert {\nabla u} \vert \in {L^{p(x,t)}}({Q_{T}}),u = 0\text{ on } {\Gamma_{T}} \bigr\} , \\& { \Vert u \Vert _{W({Q_{t}})}} = { \Vert u \Vert _{2,{Q_{T}}}} + { \vert {\nabla u} \vert _{p(x,t),{Q_{T}}}} \end{aligned}$$
In spirit of [3] and [4], we introduce the following maximal monotone graph:
In addition, we define the following function class for the solution:
$$G(\lambda) = \textstyle\begin{cases} 0,& \lambda > 0,\\ [0,+\infty),& \lambda = 0. \end{cases} $$
$$B = \bigl\{ {u \in W({Q_{T}}) \cap{L^{\infty}} \bigl(0,T;{L^{\infty}}(\Omega)\bigr)} \bigr\} . $$
Definition 2.1
A pair \((u,\xi) \in B \times{L^{\infty}}({\Omega_{T}})\) is called a weak solution of problem (1) if (a) \(u(x,t) \le{u_{0}}(x)\), (b) \(u(x,0) = {u_{0}}(x)\), (c) \(\xi \in G(u - {u_{0}})\), (d) 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)–(4). Suppose also that the following conditions hold:
Then problem (1) has at least one weak solution in the sense of Definition 2.1.
(\(H_{1}\))
\(\max\{ {1,\frac{{2N}}{{N + 2}}} \} < {p^{-} } < N\), \(2 \le\sigma < \frac {{2{p^{+} }}}{{{p^{+} } - 1}}\),
(\(H_{2}\))
\({u_{0}} \ge0\), \(f \ge0\), \({ \Vert {{u_{0}}} \Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \Vert {f(x,t)} \Vert}_{\infty,\Omega}}} \,\mathrm{d}t = K(T) < \infty\).
3 Proof of the main results
In this section, we consider the family of auxiliary parabolic problems
Here, M is a positive parameter to be chosen later. Moreover,
and \({\beta_{\varepsilon}}( \cdot)\) is the penalty function satisfying
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:
and
$$ \textstyle\begin{cases} {L_{\varepsilon}}{u_{\varepsilon}} + {\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) = 0, & (x,t) \in{Q_{T}},\\ u(x,t) = \varepsilon, & (x,t) \in{\Gamma_{T}},\\ u(x,0) = {u_{0}}(x) + \varepsilon, & x \in\Omega. \end{cases} $$
(5)
$$\begin{aligned}& {L_{\varepsilon}} {u_{\varepsilon}} = {\partial_{t}} {u_{\varepsilon}} - \operatorname{div}\bigl( {{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}}} \bigr) - f(x,t), \\& 0 < {d_{0}} \le{a_{\varepsilon,M}}(u) = {\bigl(\min\bigl({ \vert u \vert ^{2}},{M^{2}}\bigr) + {\varepsilon^{2}} \bigr)^{\frac{\sigma}{2}}} + {d_{0}} \le \bigl({M^{2}} + 1\bigr) + {d_{0}},\quad 0 < \varepsilon < 1, \end{aligned}$$
$$ \begin{gathered} 0 < \varepsilon \le1,\qquad {\beta_{\varepsilon}}(x) \in{C^{2}}(R),\qquad {\beta _{\varepsilon}}(x) \le0,\qquad {\beta_{\varepsilon}}(0) = - 1, \\ {\beta'_{\varepsilon}}(x) \ge0,\qquad {\beta''_{\varepsilon}}(x) \le0,\qquad \lim_{\varepsilon \to0} {\beta_{\varepsilon}}(x) = \textstyle\begin{cases} 0, & x > 0,\\ - \infty, & x < 0. \end{cases}\displaystyle \end{gathered} $$
(6)
$$ \begin{aligned}[b] & \int_{{t_{1}}}^{{t_{2}}} { \int_{\Omega}{\bigl[{u_{\varepsilon}} {\varphi_{t}} - {a_{\varepsilon}},M({u_{\varepsilon}}){{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla\varphi + f(x,t)\varphi \bigr]\,\mathrm{d}x\,\mathrm{d}t} } \\ &\quad = \int_{{t_{1}}}^{{t_{2}}} { \int_{\Omega}{{\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}})\varphi\,\mathrm{d}x\,\mathrm{d}t} } + \int _{\Omega}{{u_{\varepsilon}}\varphi\,\mathrm{d}x} \bigg\vert _{{t_{1}}}^{{t_{2}}} \end{aligned} $$
(7)
$$ \int_{{t_{1}}}^{{t_{2}}} \int_{\Omega}\bigl[({\partial_{t}} {u_{\varepsilon}})\varphi + {a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \varphi - f(x,t)\varphi + { \beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}})\varphi\bigr] \,\mathrm{d}x\,\mathrm{d}t = 0. $$
(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
where \(\delta > 2\varepsilon > 0\) and \(\alpha = \frac{\sigma}{2}\). Let a test-function \(\xi = {F_{\varepsilon}}(w) \in Z\) in (8). Then
where \({Q_{\varepsilon,\tau}} = {Q_{\tau}} \cap\{ (x,t) \in{Q_{\tau}}| {w > \varepsilon} \} \),
Now, let \({t_{0}} = \inf\{ t \in(0,\tau]:w > \varepsilon\} \). Then we estimate \({J_{1}}\), \({J_{2}}\), and \({J_{3}}\) as follows:
Let us first consider the case \({p^{-} } \ge2\). By the first inequality of Lemma 3.1 we get
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:
Substituting (12) into \(J_{3}\), we get
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:
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:
Plugging (15) into \(J_{3}\), we get
Plugging estimates (10), (11), (13) and (10), (14), (16) into (9) and dropping the nonnegative terms, we arrive at the inequality
with a constant C̃ independent of ε.
$${F_{\varepsilon}}(\xi) = \textstyle\begin{cases} \frac{1}{{\alpha - 1}}{\varepsilon^{1 - \alpha}} - \frac{1}{{\alpha - 1}}{\xi^{1 - \alpha}}& \text{if }\xi > \varepsilon,\\ 0& \text{if }\xi \le\varepsilon, \end{cases} $$
$$ \begin{aligned}[b] 0 &\ge \int{ \int_{{Q_{T}}} {\bigl[{w_{t}} {F_{\varepsilon}}(w) + \bigl({v^{\sigma}} + {d_{0}}\bigr) \bigl({{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla {F_{\varepsilon}}(w)\bigr]\,\mathrm{d}x \,\mathrm{d}t} } \\ &\quad{}+ \int{ \int_{{Q_{T}}} {\bigl({u^{\sigma}} - {v^{\sigma}} \bigr){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u\nabla {F_{\varepsilon}}(w)\,\mathrm{d}x\,\mathrm{d}t} } = {J_{1}} + {J_{2}} + {J_{3}}, \end{aligned} $$
(9)
$$\begin{aligned}& {J_{1}} = \int{ \int_{{Q_{T}}} {{w_{t}} {F_{\varepsilon}}(w)\,\mathrm{d}x \,\mathrm {d}t} }, \qquad{J_{2}} = \int{ \int_{{Q_{T}}} {\bigl({u^{\sigma}} - {v^{\sigma}} \bigr){w^{ - \alpha}} {{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u \nabla w\,\mathrm{d}x\,\mathrm{d}t} }, \\& {J_{3}} = \int{ \int_{{Q_{T}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}}\bigl({{ \vert {\nabla u} \vert }^{p(x,t) - 2}} \nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla w \,\mathrm {d}x\,\mathrm{d}t} }. \end{aligned}$$
$$ \begin{aligned}[b] {J_{1}} &= \int{ \int_{{Q_{\varepsilon,\tau}}} {{w_{t}} {F_{\varepsilon}}(w)\,\mathrm{d}x \,\mathrm{d}t} = \int_{\Omega}{\biggl( \int_{0}^{{t_{0}}} {{w_{t}} {F_{\varepsilon}}(w) \,\mathrm{d}t} + \int_{{t_{0}}}^{\tau}{{w_{t}} {F_{\varepsilon}}(w)\,\mathrm{d}t} \biggr)} } \,\mathrm{d}x \\ &\ge \int_{\Omega}{ \int_{\varepsilon}^{w(x,\tau)} {{F_{\varepsilon}}(s)\,\mathrm{d}s \,\mathrm{d}x} \ge} \int_{{\Omega_{\delta}}} { \int _{\varepsilon}^{w(x,\tau)} {{F_{\varepsilon}}(s)\,\mathrm{d}s \,\mathrm{d}x} } \\ &\ge \int_{{\Omega_{\delta}}} {(w - 2\varepsilon)} {F_{\varepsilon}}(2 \varepsilon)\,\mathrm{d}x \ge(\delta - 2\varepsilon){F_{\varepsilon}}(2 \varepsilon)\mu({\Omega_{\delta}}). \end{aligned} $$
(10)
$$ \begin{aligned}[b] {J_{2}} &= \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}}\bigl({{ \vert {\nabla u} \vert }^{p(x,t) - 2}} \nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla w \,\mathrm{d}x\,\mathrm{d}t} } \\ &\ge \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}} {2^{ - p(x,t)}} {{ \vert {\nabla w} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t} } \\ &\ge{2^{ - {p^{+} }}} \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t)}} \,\mathrm{d}x\,\mathrm{d}t} } \ge0. \end{aligned} $$
(11)
$$ \begin{aligned}[b] & \bigl\vert {\bigl({u^{\sigma}} - {v^{\sigma}}\bigr){w^{ - \alpha}}} { \vert {\nabla w} \vert ^{p(x,t) - 2}} {\nabla u\nabla w} \bigr\vert \\ &\quad= \biggl\vert {\sigma w \int_{0}^{1} {{{\bigl(\theta u + (1 - \theta )v \bigr)}^{\sigma - 1}}\,d\theta{w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t) - 2}}\nabla u\nabla w} } \biggr\vert \\ &\quad\le\frac{C}{{{w^{\alpha}}}}\biggl[\frac{{{v^{\sigma}} + {d_{0}}}}{C}\biggr]{ \vert {\nabla w} \vert ^{p(x,t)}} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}} \bigr){ \vert w \vert ^{p'(x,t)}} { \vert {\nabla u} \vert ^{p(x,t)}}] \\ &\quad= \frac{{({v^{\sigma}} + {d_{0}})}}{{{2^{{p^{+} } + 1}}{w^{\alpha}}}}{ \vert {\nabla w} \vert ^{p(x,t)}} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}}\bigr){ \vert w \vert ^{p'(x,t) - \alpha }} { \vert {\nabla u} \vert ^{p(x,t)}} \\ &\quad\le\frac{{({v^{\sigma}} + {d_{0}})}}{{{2^{{p^{+} } + 1}}{w^{\alpha}}}}{ \vert {\nabla w} \vert ^{p(x,t)}} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}}\bigr){ \vert {\nabla u} \vert ^{p(x,t)}}. \end{aligned} $$
(12)
$$ {J_{3}} \le\frac{1}{2}{J_{2}} + C{ \int{ \int_{{Q_{\varepsilon,\tau}}} { \vert {\nabla u} \vert } } ^{p(x,t)}} \,\mathrm{d}x\,\mathrm {d}t. $$
(13)
$$ \begin{aligned}[b] {J_{2}} = \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}}\bigl({{ \vert {\nabla w} \vert }^{p(x,t) - 2}} \nabla u - {{ \vert {\nabla v} \vert }^{p(x,t) - 2}}\nabla v\bigr)\nabla w \,\mathrm{d}x\,\mathrm{d}t} } \\ \ge\bigl({p^{-} } - 1\bigr) \int{ \int_{{Q_{\varepsilon,\tau}}} {\bigl({v^{\sigma}} + {d_{0}} \bigr){w^{ - \alpha}} {{\bigl( \vert {\nabla w} \vert + \vert {\nabla v} \vert \bigr)}^{p(x,t) - 2}} {{ \vert {\nabla w} \vert }^{2}} \,\mathrm{d}x\,\mathrm{d}t} }. \end{aligned} $$
(14)
$$ \begin{aligned}[b] & \bigl\vert {\bigl({u^{\sigma}} - {v^{\sigma}}\bigr){w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t) - 2}}\nabla u\nabla w} \bigr\vert \\ &\quad = \biggl\vert {\sigma w \int_{0}^{1} {{{\bigl(\theta u + (1 - \theta )v \bigr)}^{\sigma - 1}}\,d\theta{w^{ - \alpha}} {{ \vert {\nabla w} \vert }^{p(x,t) - 2}}\nabla u\nabla w} } \biggr\vert \\ &\quad \le\frac{{({v^{\sigma}} + {d_{0}})({p^{-} } - 1)}}{{2{w^{\alpha}}}}{\bigl( \vert {\nabla w} \vert + \vert {\nabla v} \vert \bigr)^{p(x,t) - 2}} { \vert {\nabla w} \vert ^{2}} \\ &\qquad{} + {C_{1}}\bigl(\sigma,{d_{0}},K(T),{p^{\pm}} \bigr){ \vert w \vert ^{2 - \alpha}} \vert {\nabla w} \vert + \vert { \nabla v} \vert {)^{p(x,t)}}. \end{aligned} $$
(15)
$$ {J_{3}} \le\frac{1}{2}{J_{2}} + C \int{ \int_{{Q_{\varepsilon,\tau}}} { \vert {\nabla w} \vert + \vert {\nabla v} \vert {)^{p(x,t)}}} } \,\mathrm{d}x\,\mathrm{d}t. $$
(16)
$$ (\delta - 2\varepsilon) \bigl(1 - {2^{1 - \alpha}}\bigr){\varepsilon^{1 - \alpha}} \mu({\Omega_{\delta}}) \le\tilde{C} $$
(17)
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
From Lemma 3.2 we conclude that
obtaining a contradiction.
$$\begin{aligned}& L{u_{0\varepsilon}} = {\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{0\varepsilon}}) = - 1, \end{aligned}$$
(20)
$$\begin{aligned}& L{u_{\varepsilon}} = {\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0\varepsilon}}) \le - 1. \end{aligned}$$
(21)
$$ u(x,t) \le{u_{0\varepsilon}}(x) \quad \text{for any } (x,t) \in { \Omega_{T}}, $$
(22)
Second, we pay attention to \({u_{\varepsilon}}(t,x) \le{| {{u_{0}}} |_{\infty}} + \varepsilon\). Applying the definition of \({\beta_{\varepsilon}}( \cdot )\), we have
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
$$ L\bigl( {{{ \vert {{u_{0}}} \vert }_{\infty}} + \varepsilon} \bigr) = 0, \quad L{u_{\varepsilon}} \le0. $$
(23)
$$ {u_{\varepsilon}}(x,t) \ge\varepsilon \quad\text{in } \Omega \times (0,T). $$
(24)
Third, we aim to prove (19). Since
It follows by \({\varepsilon_{1}} \le{\varepsilon_{2}}\) and the definition of \({\beta_{\varepsilon}}( \cdot )\) that
Thus, Lemma 3.3 can be proved by combining initial and boundary conditions in (5). □
$$\begin{aligned}& L{u_{{\varepsilon_{1}}}} - {\beta_{{\varepsilon_{1}}}}({u_{0{\varepsilon _{1}}}} - {u_{{\varepsilon_{1}}}}) = 0, \end{aligned}$$
(25)
$$\begin{aligned}& L{u_{{\varepsilon_{2}}}} - {\beta_{{\varepsilon_{2}}}}({u_{0{\varepsilon _{2}}}} - {u_{{\varepsilon_{2}}}}) = 0. \end{aligned}$$
(26)
$$ \begin{aligned}[b] &{\partial_{t}} {u_{{\varepsilon_{2}}}} - L{u_{{\varepsilon_{2}}}} - {\beta _{{\varepsilon_{1}}}}({u_{0{\varepsilon_{1}}}} - {u_{{\varepsilon_{2}}}}) \\ &\quad={\beta_{{\varepsilon_{2}}}}({u_{0{\varepsilon_{2}}}} - {u_{{\varepsilon_{2}}}}) - { \beta_{{\varepsilon_{1}}}}({u_{0{\varepsilon _{1}}}} - {u_{{\varepsilon_{2}}}}) \ge{ \beta_{{\varepsilon _{2}}}}({u_{0{\varepsilon_{2}}}} - {u_{{\varepsilon_{2}}}}) - {\beta _{{\varepsilon_{1}}}}({u_{0{\varepsilon_{2}}}} - {u_{{\varepsilon_{2}}}}) \ge0. \end{aligned} $$
(27)
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
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
Letting \(h \to0\) and applying Lebesgue’s dominated convergence theorem, we have that, for all \(t \in(0,T)\),
Using Holder’s inequality, we obtain
whence
By integration over \((0, t)\), for all t, we have
Then, as \(k \to\infty\),
□
$${u_{\varepsilon M}} = \textstyle\begin{cases} M & \text{if } {u_{\varepsilon}} > M,\\ {u_{\varepsilon}} & \text{if } \vert {{u_{\varepsilon}}} \vert \le M,\\ - M &\text{if } {u_{\varepsilon}} < - M. \end{cases} $$
$$ \begin{aligned}[b] &\frac{1}{{2k}} \int_{t}^{t + h} {\frac{\mathrm{d}}{{\mathrm{d}t}}\biggl( { \int _{\Omega}{u_{\varepsilon M}^{2k}\,dx} } \biggr)} \,\mathrm{d}t + \int_{t}^{t + h} { \int_{\Omega}{(2k - 1){a_{\varepsilon,M}}({u_{\varepsilon M}})} } u_{\varepsilon M}^{2(k - 1)}{ \vert {\nabla{u_{\varepsilon M}}} \vert ^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t \\ &\quad = \int_{t}^{t + h} { \int_{\Omega}{fu_{\varepsilon M}^{2k - 1}\,\mathrm{d}x \,\mathrm{d}t} - \int_{t}^{t + h} { \int_{\Omega}{{\beta _{\varepsilon}}u_{\varepsilon M}^{2k - 1} \,\mathrm{d}x} } }. \end{aligned} $$
(31)
$$ \begin{aligned}[b] &\frac{1}{{2k}}\frac{\mathrm{d}}{{\mathrm{d}t}} \int_{\Omega}{u_{\varepsilon M}^{2k}\,\mathrm{d}x} + \int_{\Omega}{(2k - 1){a_{\varepsilon,M}}({u_{\varepsilon M}})} u_{\varepsilon M}^{2(k - 1)}{ \vert {\nabla{u_{\varepsilon M}}} \vert ^{p(x,t)}}\,\mathrm{d}x \\ &\quad = \int_{\Omega}{fu_{\varepsilon M}^{2k - 1}\,\mathrm{d}x} - \int _{\Omega}{{\beta_{\varepsilon}}u_{\varepsilon M}^{2k - 1} \,\mathrm{d}x}. \end{aligned} $$
(32)
$$\begin{aligned}& \biggl\vert { \int_{\Omega}{fu_{\varepsilon M}^{2k - 1}\,dx} } \biggr\vert \le \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{2k,\Omega}^{2k - 1} \cdot{ \bigl\Vert {f(\cdot,t)} \bigr\Vert _{2k,\Omega}},\quad k = 1,2,\dots, \\& \biggl\vert { \int_{\Omega}{{\beta_{\varepsilon}}u_{\varepsilon M}^{2k - 1} \,\mathrm{d}x} } \biggr\vert \le \int_{\Omega}{u_{\varepsilon M}^{2k - 1}\,\mathrm{d}x} \le \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{2k,\Omega}^{2k - 1}, \end{aligned}$$
$$ \begin{aligned}[b] & \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega}^{2k - 1}\frac{\mathrm{d}}{{\mathrm{d}t}}\bigl({ \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega}}\bigr) + (2k - 1) \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon M}})} u_{\varepsilon M}^{2(k - 1)}{ \vert {\nabla{u_{\varepsilon,M}}} \vert ^{p(x,t)}}\,\mathrm{d}x \\ &\quad \le \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega }^{2k - 1} \cdot{ \bigl\Vert {f( \cdot,t)} \bigr\Vert _{2k,\Omega}} + C(T) \Vert {{u_{\varepsilon M}}} \Vert _{2k,\Omega}^{2k - 1},\quad k = 1,2,\dots. \end{aligned} $$
(33)
$${ \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{2k,\Omega}} \le { \bigl\Vert {{u_{\varepsilon M}}(\cdot,0)} \bigr\Vert _{2k,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{2k,\Omega}} \,\mathrm{d}t} + C(T),\quad \forall k \in\mathbb{N}. $$
$$\begin{aligned}[b] { \bigl\Vert {{u_{\varepsilon M}}(\cdot,t)} \bigr\Vert _{\infty,\Omega }} &\le{ \bigl\Vert {{u_{\varepsilon M}}(\cdot,0)} \bigr\Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{\infty,\Omega}} \,\mathrm{d}t} \\ &\le{ \Vert {{u_{0}}} \Vert _{\infty,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{\infty,\Omega}} \,\mathrm{d}t} + C(T) = K(T). \end{aligned} $$
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
It follows that, for every \(t > 0\),
and the required assertion follows. □
$$\frac{1}{2}\frac{\mathrm{d}}{{\mathrm{d}t}}\bigl( \bigl\Vert {u_{\varepsilon}^{-} (x,t)} \bigr\Vert _{2,\Omega}^{2}\bigr) + \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}})} { \bigl\vert {\nabla u_{\varepsilon}^{-} } \bigr\vert ^{p(x,t)}}\,\mathrm{d}x \le0. $$
$$\bigl\Vert {u_{\varepsilon}^{-} (x,t)} \bigr\Vert _{2,\Omega} \le \bigl\Vert {u_{\varepsilon}^{-} (\cdot,0)} \bigr\Vert _{2,\Omega} = 0, $$
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
Clearly, integrating over \((0,t)\), we have
Note that the first term on the left-hand side is nonnegative. We drop the nonpositive term in (37) to get
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). □
$$\frac{\mathrm{d}}{{\mathrm{d}t}}{ \bigl\Vert {{u_{\varepsilon}}(\cdot,t)} \bigr\Vert _{2,\Omega}} + \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x} \le{ \Vert f \Vert _{2,\Omega}} + C(T),\quad \forall t \in(0,T). $$
$$ { \bigl\Vert {{u_{\varepsilon}}(\cdot,t)} \bigr\Vert _{2,\Omega}} + \int _{0}^{t} { \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x} \,\mathrm{d}t \le} { \bigl\Vert {{u_{\varepsilon}}(\cdot,t)} \bigr\Vert _{2,\Omega}} + \int_{0}^{T} {{{ \Vert f \Vert }_{2,\Omega }} \,\mathrm{d}t} + C(T). $$
(37)
$$\int_{0}^{t} { \int_{\Omega}{{a_{\varepsilon,M}}({u_{\varepsilon}}){{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x} \,\mathrm{d}t \le} K(T){ \vert \Omega \vert ^{\frac{1}{2}}}. $$
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
Applying the fact that \({\beta_{\varepsilon}}(x,t) \in(0,1)\), we get
Using the Hölder inequality repeatedly, we have that
where
Then (38) follows from Lemma 3.5. □
$$\begin{aligned}[b] \int{ \int_{{Q_{T}}} {{u_{\varepsilon t}}\xi\,\mathrm{d}x\,\mathrm{d}t} } &= - \int{ \int_{{Q_{T}}} {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ { \varepsilon ^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}} } \nabla{u_{\varepsilon}}\nabla\xi\,\mathrm {d}x\,\mathrm{d}t \\ &\quad{}+ \int{ \int_{{Q_{T}}} {f(x,t)\xi(x,t)\,\mathrm{d}x\,\mathrm{d}t} } - \int{ \int_{{Q_{T}}} {{\beta_{\varepsilon}}(x,t)\xi(x,t)\,\mathrm {d}x \,\mathrm{d}t} }. \end{aligned} $$
$$\begin{aligned} \int{ \int_{{Q_{T}}} {{u_{\varepsilon t}}\xi\,\mathrm{d}x\,\mathrm{d}t} } &\le \int_{0}^{T} { \int_{\Omega}{\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ { \varepsilon ^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} } \nabla{u_{\varepsilon}}\nabla\xi\,\mathrm {d}x\,\mathrm{d}t\\ &\quad {} + \int_{0}^{T} { \int_{\Omega}{ \vert {f + 1} \vert \cdot \vert \xi \vert } } \,\mathrm{d}x\,\mathrm{d}t. \end{aligned} $$
$$\begin{aligned}[b] \int{ \int_{{Q_{T}}} {{u_{\varepsilon t}}\xi\,\mathrm{d}x\,\mathrm{d}t} } & \le2{ \bigl\Vert {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ {\varepsilon ^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\Vert _{p'(x,t)}} { \Vert {\nabla\xi} \Vert _{p(x,t)}}\\ &\quad {} + 2{ \Vert {f + 1} \Vert _{p'(x,t)}} \cdot{ \Vert \xi \Vert _{p(x,t)}} \\ &\le2\max\{ {F_{1}},{F_{2}}\} { \Vert {\nabla\xi} \Vert _{p(x,t)}} + 2\max\{ {{F_{3}},{F_{4}}} \}{ \Vert \xi \Vert _{p(x,t)}} \\ & \le\bigl(2{\bigl({\bigl({K^{2}}(T) + 1\bigr)^{{\sigma / 2}}} + {d_{0}}\bigr)^{\frac{1}{{{p^{\pm}} - 1}}}}K(T) \vert \Omega \vert + 2{ \vert {f + 1} \vert _{\infty}} \vert T \vert \bigr){ \Vert \xi \Vert _{W({Q_{T}})}}, \end{aligned} $$
$$\begin{aligned}& {F_{1}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{\bigl\{ {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ {\varepsilon^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\} }^{\frac {{p(x,t)}}{{p(x,t) - 1}}}}\,\mathrm{d}x\,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' + }}}}, \\& {F_{2}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{\bigl\{ {\bigl[{{\bigl(u_{\varepsilon}^{\sigma}+ {\varepsilon^{2}}\bigr)}^{{\sigma / 2}}} + {d_{0}}\bigr]{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\} }^{\frac {{p(x,t)}}{{p(x,t) - 1}}}}\,\mathrm{d}x\,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' - }}}}, \\& {F_{3}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{ \vert f \vert }^{p'(x,t)}}\,\mathrm{d}x \,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' + }}}}, \qquad {F_{4}} = {\biggl( { \int_{0}^{T} { \int_{\Omega}{{{ \vert {f + 1} \vert }^{p'(x,t)}} \,\mathrm{d}x\,\mathrm{d}t} } } \biggr)^{\frac{1}{{p' - }}}}. \end{aligned}$$
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
for some functions \(u \in W({Q_{T}})\), \({A_{i}}(x,t) \in {L^{p'(x,t)}}({Q_{T}})\), and \({W_{i}}(x,t) \in{L^{p'(x,t)}}({Q_{T}})\).
$$ \textstyle\begin{cases} {u_{\varepsilon}} \to u & \mbox{a.e. in }{Q_{T}},\\ \nabla{u_{\varepsilon}} \to\nabla u &\text{weakly in }{L^{p(x,t)}}({Q_{T}}),\\ u_{\varepsilon}^{\sigma}{ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}{D_{i}}{u_{\varepsilon}} \to{A_{i}}(x,t) & \text{weakly in } {L^{p'(x,t)}}({Q_{T}}),\\ { \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}{D_{i}}{u_{\varepsilon}} \to{W_{i}}(x,t) & \text{weakly in }{L^{p'(x,t)}}({Q_{T}}) \end{cases} $$
(39)
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
By (35) we get
Passing to the limit as \(\varepsilon \to0\), we obtain Lemma 3.7. □
$$\begin{aligned}[b] I &\stackrel{\Delta}{=} \int{ \int_{{Q_{T}}} {\bigl({{\bigl(u_{\varepsilon}^{2} + { \varepsilon^{2}}\bigr)}^{\frac{\sigma}{2}}} - u_{\varepsilon}^{\sigma}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}}\nabla\xi\,\mathrm{d}x\,\mathrm{d}t} } \\ & = \frac{\sigma}{2}{\varepsilon^{2}} \int{ \int_{{Q_{T}}} {\biggl( { \int_{0}^{1} {{{\bigl(u_{\varepsilon}^{2} + s{\varepsilon^{2}}\bigr)}^{\frac{{\sigma - 2}}{2}}}\,ds} } \biggr)} } { \vert { \nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \xi\,\mathrm{d}x\,\mathrm{d}t \\ & \le\sigma{\varepsilon^{2}} {\bigl({K^{2}}(T) + 1 \bigr)^{{\frac{{\sigma - 2}}{2}}}} { \bigl\Vert {{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 1}}} \bigr\Vert _{p'(x,t)}} { \Vert {\nabla\xi } \Vert _{p(x,t)}} \\ & \le C{\varepsilon^{2}}\max\biggl\{ {{{\biggl( \int{ \int_{{Q_{T}}} {{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)}^{\frac{{{p^{+} } - 1}}{{{p^{+} }}}}},{{\biggl( \int{ \int_{{Q_{T}}} {{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t)}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)}^{\frac{{{p^{-} } - 1}}{{{p^{-} }}}}}} \biggr\} { \Vert {\nabla\xi} \Vert _{p(x,t)}}. \end{aligned} $$
$$I \le C{\varepsilon^{2 - \frac{{\sigma({p^{+} } - 1)}}{{{p^{+} }}}}} { \Vert {\nabla\xi} \Vert _{p(x,t)}}. $$
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
By Lebesgue’s dominated convergence theorem we have
So
By (40)–(42) and the previous inequalities, we complete the proof of Lemma 3.8. □
$$\begin{aligned}& { \int \int_{{Q_{T}}} {u_{\varepsilon}^{\sigma} \vert {\nabla {u_{\varepsilon}}} \vert } ^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla \xi\,\mathrm{d}x\,\mathrm{d}t \to\sum{ \int{ \int_{{Q_{T}}} {{A_{i}}(x,t){D_{i}}\xi \,\mathrm{d}x\,\mathrm{d}t} } }, \end{aligned}$$
(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)
$$ \lim_{\varepsilon \to0} \sum_{i = 1}^{N} { \int{ \int_{{Q_{T}}} {\bigl(u_{\varepsilon}^{\sigma}- {u^{\sigma}}\bigr){A_{i}}(x,t){D_{i}}\xi\,\mathrm {d}x \,\mathrm{d}t = 0} }.} $$
(42)
$$\begin{aligned}[b] &\lim_{\varepsilon \to0} \sum{ \int{ \int_{{Q_{T}}} {\bigl(u_{\varepsilon}^{\sigma}{{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} {D_{i}} {u_{\varepsilon}} - {u^{\sigma}} {W_{i}}(x,t) \bigr){D_{i}}\xi\,\mathrm {d}x\,\mathrm{d}t} } } \\ &\quad= \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\bigl(u_{\varepsilon}^{\sigma}- {u^{\sigma}}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} {D_{i}} {u_{\varepsilon}} + {u^{\sigma}} \bigl({{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}} {D_{i}} {u_{\varepsilon}} - {W_{i}}(x,t) \bigr){D_{i}}\xi\,\mathrm{d}x\,\mathrm{d}t} }\\ &\quad = 0. \end{aligned} $$
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
It follows that
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
Note that
By (44)–(46) we have
□
$$\begin{aligned}[b] & \int{ \int_{{Q_{T}}} {\bigl[{u_{\varepsilon t}}({u_{\varepsilon}} - u) \Phi + \Phi\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p(x,t) - 2}} \nabla{u_{\varepsilon}}\nabla({u_{\varepsilon}} - u)\bigr]\,\mathrm{d}x \,\mathrm{d}t} } \\ &\quad{} + \int{ \int_{{Q_{T}}} {\bigl[({u_{\varepsilon}} - u) \bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}} \nabla\Phi - f(x,t) ({u_{\varepsilon}} - u)\Phi\bigr]\,\mathrm{d}x\,\mathrm{d}t} } \\ &\quad{}+ \int{ \int_{{Q_{T}}} {\bigl({{\bigl(u_{\varepsilon}^{\sigma}- { \varepsilon^{2}}\bigr)}^{\frac{\sigma}{2}}} - u_{\varepsilon}^{\sigma}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}}\nabla\xi\,\mathrm{d}x\,\mathrm{d}t} } = 0. \end{aligned} $$
$$ \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla{u_{\varepsilon}}\nabla({u_{\varepsilon}} - u) \,\mathrm{d}x\,\mathrm{d}t = 0.} } $$
(43)
$$\begin{aligned}& \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl(u^{\sigma}+ {d_{0}}\bigr){{ \vert {\nabla u} \vert }^{p(x,t) - 2}}\nabla u \nabla({u_{\varepsilon}} - u)\,\mathrm{d}x\,\mathrm{d}t = 0} }, \end{aligned}$$
(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)
$$ \begin{aligned}[b] 0 &\le\bigl({ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla{u_{\varepsilon}} - { \vert {\nabla u} \vert ^{p(x,t) - 2}}\nabla u\bigr)\nabla({u_{\varepsilon}} - u) \\ &\le\frac{1}{{{d_{0}}}}\bigl[\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla {u_{\varepsilon}} - \bigl(u_{\varepsilon}^{\sigma}- u^{\sigma}\bigr){ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla u\bigr]\nabla ({u_{\varepsilon}} - u) \\ &\quad{} - \frac{1}{{{d_{0}}}}\bigl(u_{\varepsilon}^{\sigma}+ {d_{0}}\bigr){ \vert {\nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla u\nabla({u_{\varepsilon}} - u). \end{aligned} $$
(46)
$$ \lim_{\varepsilon \to0} \int{ \int_{{Q_{T}}} {\Phi\bigl({{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p(x,t) - 2}}\nabla {u_{\varepsilon}} - } } { \vert { \nabla{u_{\varepsilon}}} \vert ^{p(x,t) - 2}}\nabla u\bigr) \nabla({u_{\varepsilon}} - u)\,\mathrm {d}x\,\mathrm{d}t = 0. $$
(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
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
Thus, if ε is small enough, then we have
Furthermore, it follows by \(\varepsilon \to0\) that
Hence, (48) holds, and the proof of Lemma 3.10 is completed. □
$${\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) \to\xi\quad \text{as } \varepsilon \to0. $$
$${u_{\varepsilon}}(x,t) \ge{u_{0}}(x) + \lambda, \quad \forall(x,t) \in {B_{\delta}}({x_{0}},{t_{0}}). $$
$$0 \ge{\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0}}) \ge{\beta _{\varepsilon}}(\lambda) = 0, \quad \forall(x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
$$\xi(x,t) = 0, \quad \forall(x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
Applying (28), (29), and Lemma 3.10, it is clear that
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.
$$u(x,t) \le{u_{0}}(x) \quad\text{in } {\Omega_{T}}, \qquad u(x,0) = {u_{0}}(x) \quad\text{in } \Omega, \xi \in G(u - {u_{0}}), $$
Acknowledgements
This work was supported by Scientific research Foundation of Shaanxi Railway Institute (No. KY2016-04) and Natural Science Foundation of Shaanxi Provincial Department of Education (No. 17JK0170). The author is sincerely grateful to the reference and the Associate Editor handling the paper for their valuable comments.
Competing interests
The author declares that there is no conflict of interests regarding the publication of this paper.
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