This section is devoted to the proof of the existence of solutions to problem (1). We prove the following theorem.
Theorem 3
Let
\(p \ge2\)
and
\(\gamma \in(0,1)\). Under the assumption (2), problem (1) admits a weak solution
u
with
\(\frac{{\partial{u^{\mu}}}}{{\partial t}} \in {L^{2}}({\Omega_{T}})\)
where
$$ \mu = \frac{{\gamma p}}{{2(p - 1)}} + \frac{1}{2}. $$
(8)
To prove Theorem 3, let us consider the approximation problem
$$ \left \{ \textstyle\begin{array}{l} { ( {{u_{\varepsilon}}} )_{t}} - {u_{\varepsilon}}\operatorname {div} ( {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}}} ) - \gamma{ \vert {\nabla {u_{\varepsilon}}} \vert ^{p}} - {\beta_{\varepsilon}}({u_{0\varepsilon }} - {u_{\varepsilon}}) = 0,\quad \mbox{in } {\Omega_{T}}, \\ {u_{\varepsilon}}(x,0) = {u_{0\varepsilon}}(x) = {u_{0}}(x) + \varepsilon , \quad \mbox{on }\Omega, \\ {u_{\varepsilon}}(x,t) = 0,\quad \mbox{on }\partial\Omega \times(0,T), \end{array}\displaystyle \right . $$
(9)
where \({\beta_{\varepsilon}}( \cdot)\) is the penalty function satisfying
$$ \begin{aligned} &0 < \varepsilon \le1, \qquad {\beta_{\varepsilon}}(x) \in{C^{2}}(R),\qquad {\beta _{\varepsilon}}(x) \le0, \qquad { \beta_{\varepsilon}}(0) = - 1, \\ &{\beta'_{\varepsilon}}(x) \ge0,\qquad {\beta''_{\varepsilon}}(x) \le0,\qquad \lim_{\varepsilon \to0} {\beta_{\varepsilon}}(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} 0,& x > 0, \\ - \infty, &x < 0. \end{array}\displaystyle \displaystyle \displaystyle \right . \end{aligned} $$
(10)
Definition 4
A nonnegative function \({u_{\varepsilon}}\) is called a weak solution of problem (9), if
-
(a)
\({u_{\varepsilon}} \in{L^{\infty}}({\Omega_{T}}) \cap {L^{p}}(0,T;W_{0}^{1,p}(\Omega))\),
-
(b)
\(\int_{\Omega_{T}} { ( {{u_{\varepsilon}}{\varphi_{t}} + {u_{\varepsilon}}{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}}\nabla\varphi + (1 - \gamma){{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\varphi - {\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}})\varphi} )}{\,\mathrm{d}x\,\mathrm{d}t}= 0\) hold, for any \(\forall\varphi \in C_{0}^{\infty}({\Omega_{T}})\),
-
(c)
\(\lim_{t \to\infty} \int_{\Omega}{ \vert {u_{\varepsilon}^{\mu}(x,t) - u_{0\varepsilon}^{\mu}(x)} \vert \,\mathrm{d}x} = 0\) holds for some \(\mu > 0\).
According to the standard theory for parabolic equations [7], problem (9) admits a weak solution
$$ {u_{\varepsilon}} \in{L^{\infty}}({\Omega_{T}}) \cap {L^{p}}\bigl(0,T;W_{0}^{1,p}({\Omega_{T}}) \bigr) $$
(11)
in the sense of Definition 4, which satisfies
$$ { \bigl( {u_{\varepsilon}^{\mu}} \bigr)_{t}} \in{L_{2}}({\Omega_{T}}),\qquad \nabla {u_{\varepsilon}} \in{L_{p}}({\Omega_{T}}). $$
(12)
Further, it follows by the comparison principle and the maximum principle [8, 9] that
$$ \varepsilon \le{u_{\varepsilon}} \le{u_{0\varepsilon}} \le{ \vert {{u_{0}}} \vert _{\infty}} + \varepsilon,\qquad {u_{{\varepsilon_{1}}}} \le{u_{{\varepsilon_{2}}}} \quad \mbox{for } {\varepsilon_{1}} \le{\varepsilon_{2}}. $$
(13)
Moreover, from (12) and (13), we assert that there exists a subsequence ε (still denoted by ε) such that
$$\begin{aligned}& {u_{\varepsilon}} \to u \in{L^{p}}\bigl(0,T;W_{0}^{1,p}({ \Omega_{T}})\bigr) \quad \mbox{as } \varepsilon \to0, \end{aligned}$$
(14)
$$\begin{aligned}& {u_{\varepsilon}} \ge u \ge0\quad \mbox{for any } \varepsilon > 0. \end{aligned}$$
(15)
Lemma 5
Assume
\(Q_{c}^{\varepsilon}= \{ {(x,t) \in{\Omega _{T}};{u_{\varepsilon}} \ge c,c > 0} \}\), \(Q_{c} = \{ {(x,t) \in{\Omega_{T}};u \ge c,c > 0} \}\)
such that, as
\(\varepsilon \to0\),
$$ \int_{Q_{c}^{\varepsilon}} {{{\vert {\nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \to0,\qquad \int _{{Q_{c}}} {{{\vert {\nabla {u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \to0. $$
(16)
Proof
Choosing \(\varphi = u_{\varepsilon}^{p - 2}(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}})\) as the test function in
$$ \int_{{\Omega_{T}}} { \bigl( {{u_{\varepsilon}} { \varphi_{t}} + {u_{\varepsilon}} {{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}}\nabla\varphi + (1 - \gamma){{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\varphi - { \beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}})\varphi} \bigr)\, \mathrm{d}x\,\mathrm{d}t} = 0. $$
(17)
Then it is easy to see that
$$\begin{aligned}& \int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} \\& \quad = - \int_{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}}\nabla \bigl\{ {u_{\varepsilon}^{p - 1}\bigl(u_{\varepsilon}^{2} - { \varepsilon^{2}} - {u^{2}}\bigr)} \bigr\} \,\mathrm{d}x\, \mathrm{d}t} \\& \qquad {} + \gamma\int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2}{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\bigl(u_{\varepsilon}^{2} - {\varepsilon ^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\, \mathrm{d}t} \\& \qquad {} + \int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr){\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}})\,\mathrm{d}x\,\mathrm{d}t} \\& \quad = - \int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 1}{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}} \nabla \bigl(u_{\varepsilon}^{2} - {u^{2}}\bigr)\,\mathrm{d}x \,\mathrm{d}t} \\& \qquad {} + \int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr){\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}})\,\mathrm{d}x\,\mathrm{d}t} \\& \qquad {} + (\gamma - p + 1)\int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2}{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\, \mathrm{d}t} . \end{aligned}$$
(18)
From (15) and the definition of \({\beta_{\varepsilon}}\), we derive
$$\begin{aligned}& \int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2}{{\vert {\nabla {u_{\varepsilon}}} \vert }^{p}}\bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\, \mathrm{d}t} \ge - {\varepsilon^{2}}\int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2}{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t}, \end{aligned}$$
(19)
$$\begin{aligned}& \int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr){\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}})\,\mathrm{d}x\,\mathrm{d}t} \\& \quad \le{ \bigl( {{{\vert {{u_{0}}} \vert }_{\infty}} + \varepsilon} \bigr)^{p - 2}}\biggl\vert {\int_{{\Omega_{T}}} { \bigl(u_{\varepsilon}^{2} - {\varepsilon ^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} } \biggr\vert . \end{aligned}$$
(20)
Observing that\(\gamma - p + 1 < 0\), and combing (18), (19), and (20), we have
$$\begin{aligned}& \int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} \\& \quad \le - {2^{1 - p}}\int_{{\Omega_{T}}} {{{\bigl\vert { \nabla u_{\varepsilon}^{2}} \bigr\vert }^{p - 2}}\nabla u_{\varepsilon}^{2}\nabla\bigl(u_{\varepsilon}^{2} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} + p{\varepsilon^{2}} \int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2}{{\vert { \nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \\& \qquad {}+ { \bigl( {{{\vert {{u_{0}}} \vert }_{\infty}} + \varepsilon} \bigr)^{p - 2}}\biggl\vert {\int_{{\Omega_{T}}} { \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} } \biggr\vert . \end{aligned}$$
(21)
Note that \(\varepsilon \le{u_{\varepsilon}} \le{ \vert {{u_{0}}} \vert _{\infty}} + \varepsilon\). Thus, it follows by the trigonometrical inequality and the Hölder inequality that
$$\begin{aligned}& \int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} \\& \qquad {} + \int_{{\Omega_{T}}} { \bigl[ {{{\bigl\vert {\nabla u_{\varepsilon}^{2}} \bigr\vert }^{p - 2}}\nabla u_{\varepsilon}^{2} - {{\bigl\vert {\nabla u^{2}} \bigr\vert }^{p - 2}}\nabla u^{2}} \bigr]\nabla \bigl(u_{\varepsilon}^{2} - {u^{2}}\bigr)\,\mathrm{d}x\, \mathrm{d}t} \\& \quad \le - {2^{1 - p}}\int_{{\Omega_{T}}} {{{\bigl\vert { \nabla u^{2}} \bigr\vert }^{p - 2}}\nabla u^{2}\nabla \bigl(u_{\varepsilon}^{2} - {u^{2}}\bigr)\,\mathrm{d}x\, \mathrm{d}t} + p{\varepsilon^{2}}\int_{{\Omega_{T}}} {u_{\varepsilon}^{p - 2}{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \\& \qquad {} + { \bigl( {{{\vert {{u_{0}}} \vert }_{\infty}} + \varepsilon } \bigr)^{p - 2}}\biggl\vert {\int_{{\Omega_{T}}} {\bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} } \biggr\vert \\& \quad \le{2^{1 - p}} { \biggl( {\int_{{\Omega_{T}}} {{{\bigl\vert {\nabla u^{2}} \bigr\vert }^{p}}\,\mathrm{d}x\, \mathrm{d}t} } \biggr)^{\frac{{p - 1}}{p}}} { \biggl( {\int_{{\Omega_{T}}} {{{\bigl\vert {\nabla\bigl(u_{\varepsilon}^{2} - {u^{2}} \bigr)} \bigr\vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{1}{p}}} \\& \qquad {} + {\bigl(\vert {{u_{0}}} \vert + \varepsilon \bigr)^{p - 2}}\mu {\varepsilon^{2}}\int_{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\, \mathrm{d}t} \\& \qquad {} + { \bigl( {{{\vert {{u_{0}}} \vert }_{\infty}} + \varepsilon } \bigr)^{p - 2}}\biggl\vert {\int_{{\Omega_{T}}} {\bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} } \biggr\vert . \end{aligned}$$
(22)
Since \(\frac{\partial}{{\partial t}}u_{\varepsilon}^{\mu}\in {L^{2}}({\Omega_{T}})\), using the Hölder inequality, one derives
$$\begin{aligned}& \int_{{\Omega_{T}}} {\biggl\vert {\frac{{\partial u_{\varepsilon}^{p - 1}}}{{\partial t}}} \biggr\vert \,\mathrm{d}x\,\mathrm{d}t} \\& \quad = (p - 1)\int_{{\Omega _{T}}} {u_{\varepsilon}^{p - 2} \biggl\vert {\frac{{\partial u_{\varepsilon}}}{{\partial t}}} \biggr\vert \,\mathrm{d}x\,\mathrm{d}t} \\& \quad \le(p - 1){\bigl({\vert {{u_{0}}} \vert _{\infty}} + \varepsilon\bigr)^{p - 2}}\int_{{\Omega_{T}}} {\biggl\vert { \frac{{\partial u_{\varepsilon}}}{{\partial t}}} \biggr\vert \,\mathrm{d}x\,\mathrm{d}t} \\& \quad \le(p - 1){\bigl({\vert {{u_{0}}} \vert _{\infty}} + \varepsilon\bigr)^{p - 2}}\sqrt{\vert {{\Omega_{T}}} \vert } \sqrt{\int_{{\Omega_{T}}} {{{\biggl\vert {\frac{{\partial u_{\varepsilon}}}{{\partial t}}} \biggr\vert }^{2}}\,\mathrm{d}x\,\mathrm{d}t} } < \infty. \end{aligned}$$
(23)
From the above equation and (13), we may conclude that, as \(\varepsilon \to0\),
$$\begin{aligned}& \biggl\vert {\int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{p - 2} \bigl(u_{\varepsilon}^{2} - {\varepsilon ^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} } \biggr\vert \\& \quad \le\frac{1}{{p - 1}}\int_{{\Omega_{T}}} {\biggl\vert { \frac{{\partial u_{\varepsilon}^{p - 1}}}{{\partial t}}\bigl(u_{\varepsilon}^{2} - {\varepsilon ^{2}} - {u^{2}}\bigr)} \biggr\vert \,\mathrm{d}x\, \mathrm{d}t} \\& \quad \le\frac{1}{{p - 1}}\sqrt{\int_{{\Omega_{T}}} {{{ \biggl( {\frac {{\partial u_{\varepsilon}^{p - 1}}}{{\partial t}}} \biggr)}^{2}}\,\mathrm{d}x\,\mathrm{d}t} } \cdot\sqrt{\int_{{\Omega_{T}}} {{{\bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)}^{2}}\, \mathrm{d}x\,\mathrm{d}t} } \\& \quad \le\frac{1}{{p - 1}}{\varepsilon^{2}}\sqrt{\vert {{ \Omega_{T}}} \vert } \sqrt{\int_{{\Omega_{T}}} {{{ \biggl( {\frac{{\partial u_{\varepsilon}^{p - 1}}}{{\partial t}}} \biggr)}^{2}}\,\mathrm{d}x\,\mathrm{d}t} } \to0, \end{aligned}$$
(24)
$$\begin{aligned}& { \bigl( {{{\vert {{u_{0}}} \vert }_{\infty}} + \varepsilon} \bigr)^{p - 2}}\biggl\vert {\int_{{\Omega_{T}}} { \bigl(u_{\varepsilon}^{2} - {\varepsilon^{2}} - {u^{2}}\bigr)\,\mathrm{d}x\,\mathrm{d}t} } \biggr\vert \to0. \end{aligned}$$
(25)
Substituting (24) and (25) into (22) yields
$$ \limsup_{\varepsilon \to0} \int_{{\Omega_{T}}} { \bigl[ {{{\bigl\vert {\nabla u_{\varepsilon}^{2}} \bigr\vert }^{p - 2}} \nabla u_{\varepsilon}^{2} - {{\bigl\vert {\nabla u^{2}} \bigr\vert }^{p - 2}}\nabla u^{2}} \bigr]\nabla \bigl(u_{\varepsilon}^{2} - {u^{2}}\bigr)\,\mathrm{d}x\, \mathrm{d}t} \le0. $$
(26)
This and Lemma 2 lead to
$$ \lim_{\varepsilon \to0} \int_{{\Omega_{T}}} {{{\bigl\vert { \nabla u_{\varepsilon}^{2} - \nabla u^{2}} \bigr\vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} = 0. $$
(27)
Hence, in view of
$$\begin{aligned}& \nabla u_{\varepsilon}^{2} - \nabla{u^{2}} = 2{u_{\varepsilon}}\nabla {u_{\varepsilon}} - 2u\nabla u = 2{u_{\varepsilon}}( \nabla{u_{\varepsilon}} - \nabla u) + 2\nabla u ( {{u_{\varepsilon}} - u} ), \end{aligned}$$
(28)
$$\begin{aligned}& \nabla u_{\varepsilon}^{2} - \nabla{u^{2}} = 2{u_{\varepsilon}}\nabla {u_{\varepsilon}} - 2u\nabla u = 2u( \nabla{u_{\varepsilon}} - \nabla u) + 2\nabla{u_{\varepsilon}}({u_{\varepsilon}} - u), \end{aligned}$$
(29)
we derive
$$\begin{aligned}& {2^{p}}\int_{{\Omega_{T}}} {u_{\varepsilon}^{p}{{ \vert {\nabla {u_{\varepsilon}} - \nabla u} \vert }^{p}}\, \mathrm{d}x\,\mathrm{d}t} \\& \quad \le{2^{p}}\int_{{\Omega_{T}}} {{{\bigl\vert {\nabla u_{\varepsilon}^{2} - \nabla{u^{2}}} \bigr\vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} + {4^{p}}\int _{{\Omega_{T}}} {{{\vert {\nabla u} \vert }^{p}} {{\vert {{u_{\varepsilon}} - u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \to0 \quad (\varepsilon \to0), \end{aligned}$$
(30)
$$\begin{aligned}& {2^{p}}\int_{{\Omega_{T}}} {{u^{p}} {{\vert { \nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\, \mathrm{d}t} \\& \quad \le{2^{p}}\int_{{\Omega_{T}}} {{{\bigl\vert {\nabla u_{\varepsilon}^{2} - \nabla{u^{2}}} \bigr\vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} + {4^{p}}\int _{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} {{\vert {{u_{\varepsilon}} - u} \vert }^{p}}\, \mathrm{d}x\,\mathrm{d}t} \to0\quad (\varepsilon \to0). \end{aligned}$$
(31)
Note that \(Q_{c}^{\varepsilon}\subset{\Omega_{T}}\), \(Q_{c} \subset {\Omega_{T}}\). Then it is easy to see that as \(\varepsilon \to0\),
$$\begin{aligned}& {c^{p}}\int_{Q_{c}^{\varepsilon}} {{{\vert {\nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \le\int _{{\Omega_{T}}} {u_{\varepsilon}^{p}{{\vert { \nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\, \mathrm{d}t} \to0, \end{aligned}$$
(32)
$$\begin{aligned}& {c^{p}}\int_{Q_{c}} {{{\vert {\nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \le\int _{{\Omega_{T}}} {{u^{p}} {{\vert {\nabla {u_{\varepsilon}} - \nabla u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \to0. \end{aligned}$$
(33)
Thus, the lemma is proved. □
Lemma 6
The solution of (9) satisfies
$$ \int_{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}u_{\varepsilon}^{ - \alpha}\,\mathrm{d}x\,\mathrm{d}t} \le C, $$
(34)
where
C
is independent of
ε, \(\alpha \in[0,1 - \gamma)\).
Proof
Multiply (9) by \(u_{\varepsilon}^{ - \alpha}\) and integrate both sides of the equation over \({\Omega_{T}}\). After integrating by parts, we obtain
$$\begin{aligned}& \int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{ - \alpha} \,\mathrm{d}x\,\mathrm{d}t} \\& \quad = \int_{{\Omega_{T}}} {u_{\varepsilon}^{1 - \alpha} \operatorname {div} \bigl\{ {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}}} \bigr\} +\gamma u_{\varepsilon}^{ - \alpha}{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p}} + u_{\varepsilon}^{ - \alpha}{\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}})\, \mathrm{d}x\,\mathrm{d}t} \\& \quad = \int_{0}^{T} {\mathrm{d}t\int _{\partial\Omega} { \biggl\{ {u_{\varepsilon}^{1 - \alpha}{{ \vert { \nabla{u_{\varepsilon}}} \vert }^{p - 2}}\frac {{\partial{u_{\varepsilon}}}}{{\partial\nu}}} \biggr\} \, \mathrm{d}x} } - (1 - \alpha - \gamma)\int_{{\Omega_{T}}} {u_{\varepsilon}^{ - \alpha }{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \\& \qquad {}+ \int_{{\Omega_{T}}} {u_{\varepsilon}^{ - \alpha}{ \beta_{\varepsilon}} ( {{u_{0\varepsilon}} - {u_{\varepsilon}}} )\,\mathrm{d}x \,\mathrm{d}t} , \end{aligned}$$
(35)
where ν denotes the outward normal to \(\partial\Omega \times(0,T)\).
Further, putting together (10) and (13) implies
$$ \int_{{\Omega_{T}}} {u_{\varepsilon}^{ - \alpha}{ \beta_{\varepsilon}} ( {{u_{0\varepsilon}} - {u_{\varepsilon}}} )\,\mathrm{d}x \,\mathrm{d}t} \le0. $$
(36)
Since \({u_{\varepsilon}} \ge\varepsilon\), we have
$$ \frac{{\partial{u_{\varepsilon}}}}{{\partial\nu}} \le0,\quad \mbox{on } \partial\Omega \times(0,T). $$
(37)
This leads to
$$ \int_{0}^{T} {\int_{\partial\Omega} { \biggl\{ {u_{\varepsilon}^{1 - \alpha }{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\frac{{\partial {u_{\varepsilon}}}}{{\partial\nu}}} \biggr\} \,\mathrm{d}x\,\mathrm{d}t} } \le 0. $$
(38)
Now, we drop the non-positive terms (36) and (38) in (35) to get
$$ \int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{ - \alpha} \,\mathrm{d}x\,\mathrm{d}t} \le - (1 - \alpha - \gamma)\int_{{\Omega_{T}}} {u_{\varepsilon}^{ - \alpha}{{ \vert {\nabla {u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} . $$
(39)
Clearly, using integration by parts, we derive
$$ \int_{{\Omega_{T}}} {\frac{{\partial{u_{\varepsilon}}}}{{\partial t}}u_{\varepsilon}^{ - \alpha} \,\mathrm{d}x\,\mathrm{d}t} = \frac{1}{{1 - \alpha}}\int_{\Omega}{u_{\varepsilon}^{1 - \alpha}(x,T) - u_{\varepsilon}^{1 - \alpha}(x,0)\, \mathrm{d}x} . $$
(40)
This and (39) lead to
$$ \int_{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}u_{\varepsilon}^{ - \alpha}\,\mathrm{d}x\,\mathrm{d}t} \le \frac{1}{{(1 - \alpha - \gamma)(1 - \alpha)}}\int_{\Omega}{u_{\varepsilon}^{ - \alpha}(x,0) \,\mathrm{d}x} \le{C_{1}}, $$
(41)
where \({C_{1}} > 0\) depending on α, γ, Ω, and \(\vert {{u_{0}}} \vert \). Hence, the proof is completed. □
Lemma 7
As
\(\varepsilon \to0\), we have
$$\begin{aligned}& \int_{{\Omega_{T}}} {\bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} - {{\vert {\nabla u} \vert }^{p}}} \bigr\vert \,\mathrm{d}x\,\mathrm{d}t} \to 0, \end{aligned}$$
(42)
$$\begin{aligned}& \int_{{\Omega_{T}}} {\bigl\vert {{u_{\varepsilon}} {{\vert { \nabla {u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}} - u{{ \vert {\nabla u} \vert }^{p}}\nabla u} \bigr\vert \,\mathrm{d}x\, \mathrm{d}t} \to0, \end{aligned}$$
(43)
$$\begin{aligned}& {\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}}) \to\xi \in G({u_{0}} - u). \end{aligned}$$
(44)
Proof
Let \({\chi_{\eta}}\) and \(\chi_{\eta}^{(\varepsilon)}\) be the characteristic functions of \(\{ {(x,t) \in{\Omega_{T}};u(x,t) < \eta} \}\) and \(\{ (x,t) \in{\Omega_{T}};{u_{\varepsilon}}(x,t) < \eta \}\), respectively. Since \({u_{\varepsilon}} \to u\), as \(\varepsilon \to0\), \({\chi_{\eta}} \le\chi_{\eta}^{(\varepsilon)}\), we have
$$\begin{aligned}& \int_{{\Omega_{T}}} {\bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} - {{\vert {\nabla u} \vert }^{p}}} \bigr\vert \,\mathrm{d}x\,\mathrm{d}t} \\& \quad \le\int_{{\Omega_{T}}} {\bigl\vert {{{\vert {\nabla u} \vert }^{p}} {\chi _{\eta}} - {{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\chi_{\eta}^{(\varepsilon)}} \bigr\vert \, \mathrm{d}x\,\mathrm{d}t} + \int_{{\Omega_{T}}} {\bigl\vert {{{\vert {\nabla u} \vert }^{p}}(1 - {\chi_{\eta}}) - {{\vert { \nabla{u_{\varepsilon}}} \vert }^{p}}\bigl(1 - \chi_{\eta}^{(\varepsilon )} \bigr)} \bigr\vert \,\mathrm{d}x\,\mathrm{d}t} \\& \quad \le\int_{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\chi_{\eta}^{(\varepsilon)}\,\mathrm{d}x\, \mathrm{d}t} + \int_{{\Omega _{T}}} {{{\vert {\nabla u} \vert }^{p}} {\chi_{\eta}}\,\mathrm{d}x\,\mathrm{d}t} \\& \qquad {}+ \int_{{\Omega_{T}}} {{{\vert {\nabla u} \vert }^{p}}\bigl(\chi_{\eta}^{(\varepsilon)} - {\chi _{\eta}} \bigr)\,\mathrm{d}x\,\mathrm{d}t} + \int_{{\Omega_{T}}} {\bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} - {{\vert { \nabla u} \vert }^{p}}} \bigr\vert \bigl(1 - \chi_{\eta}^{(\varepsilon)} \bigr)\,\mathrm{d}x\,\mathrm{d}t} \\& \quad = {H_{1}} + {H_{2}} + {H_{3}} + {H_{4}}. \end{aligned}$$
(45)
Taking \(\alpha = (1 - \gamma)/2\) in Lemma 6 one obtains
$$\begin{aligned} \begin{aligned}[b] {H_{1}} &= \int_{{\Omega_{T}}} {{{\vert { \nabla{u_{\varepsilon}}} \vert }^{p}}\frac{{u_{\varepsilon}^{\alpha}}}{{u_{\varepsilon}^{\alpha}}}\chi _{\eta}^{(\varepsilon)}\,\mathrm{d}x\,\mathrm{d}t} \\ &\le{\eta^{\alpha}}\int_{{\Omega_{T}}} {{{\vert {\nabla {u_{\varepsilon}}} \vert }^{p}}u_{\varepsilon}^{ - \alpha} \chi_{\eta}^{(\varepsilon)}\,\mathrm{d}x\,\mathrm{d}t} \\ &\le{\eta^{\alpha}}\int_{{\Omega_{T}}} {{{\vert {\nabla {u_{\varepsilon}}} \vert }^{p}}u_{\varepsilon}^{ - \alpha}\, \mathrm{d}x\,\mathrm{d}t} \le C{\eta^{\alpha}} \to0 \quad (\eta \to0). \end{aligned} \end{aligned}$$
(46)
Applying Lemma 5, (46), and the fact that \({\chi_{\eta}} \le\chi_{\eta}^{(\varepsilon)}\) implies
$$\begin{aligned}& {H_{2}} \le\int_{{\Omega_{T}}} {\chi_{\eta}^{(\varepsilon)}{{ \vert {\nabla u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} \\& \hphantom{{H_{2}}} \le\int_{{\Omega_{T}}} {\chi_{\eta}^{(\varepsilon)}{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\, \mathrm{d}t} + \int_{{\Omega_{T}}} {\chi_{\eta}^{(\varepsilon)}{{ \vert {\nabla {u_{\varepsilon}} - \nabla u} \vert }^{p}}\, \mathrm{d}x\,\mathrm{d}t} \to0\quad (\eta \to0), \end{aligned}$$
(47)
$$\begin{aligned}& {H_{4}} \to0 \quad (\eta \to0). \end{aligned}$$
(48)
For fixed \(\eta > 0\), \(\chi_{\eta}^{(\varepsilon)} \to{\chi_{\eta}}\) (\(\varepsilon \to0\)) a.e. in \({\Omega_{T}}\), so
$$ {H_{3}} \to0\quad (\eta \to0). $$
(49)
Putting together (46), (47), (48), and (49), we have
$$ \int_{{\Omega_{T}}} {\bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} - {{\vert {\nabla u} \vert }^{p}}} \bigr\vert \,\mathrm{d}x\,\mathrm{d}t} \to 0 \quad (\eta \to0). $$
(50)
Thus (42) holds.
Next we prove (43). It follows by the trigonometrical inequality that
$$\begin{aligned}& \int_{{\Omega_{T}}} {\bigl\vert {{u_{\varepsilon}} {{\vert { \nabla {u_{\varepsilon}}} \vert }^{p - 2}}\nabla{u_{\varepsilon}} - u{{ \vert {\nabla u} \vert }^{p - 2}}\nabla u} \bigr\vert \,\mathrm{d}x\, \mathrm{d}t} \\& \quad \le\int_{{\Omega_{T}}} {\vert {{u_{\varepsilon}} - u} \vert {{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 1}}\,\mathrm{d}x\, \mathrm{d}t} + \int_{{\Omega_{T}}} {u{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}}\vert {\nabla{u_{\varepsilon}} - \nabla u} \vert \, \mathrm{d}x\,\mathrm{d}t} \\& \qquad {}+ \int_{{\Omega_{T}}} {u\vert {\nabla u} \vert \cdot\bigl\vert {{{ \vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}} - {{ \vert {\nabla u} \vert }^{p - 2}}} \bigr\vert \,\mathrm{d}x\, \mathrm{d}t} \\& \quad = {H_{5}} + {H_{6}} + {H_{7}}. \end{aligned}$$
(51)
Using the Hölder inequality and Lemma 6, we obtain
$$ {H_{5}} \le C{ \biggl( {\int_{{\Omega_{T}}} {{{\vert {{u_{\varepsilon}} - u} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{1}{p}}} \to0\quad \mbox{as } \varepsilon \to0. $$
(52)
With the inequality \(\vert {{a^{r}} - {b^{r}}} \vert \le{ \vert {a - b} \vert ^{r}}\) (\(r \in[0,1]\), \(a,b \ge0\)), the Hölder inequality, and (42), we have
$$\begin{aligned} {H_{7}} =& \int_{{\Omega_{T}}} {u\vert {\nabla u} \vert \cdot\bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}} - {{\vert {\nabla u} \vert }^{p - 2}}} \bigr\vert \,\mathrm{d}x\, \mathrm{d}t} \\ =& \int_{{\Omega_{T}}} {u\vert {\nabla u} \vert \cdot \bigl( {{{ \bigl({{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} \bigr)}^{\frac{{p - 2}}{p}}} - {{\bigl({{\vert {\nabla u} \vert }^{p}} \bigr)}^{\frac{{p - 2}}{p}}}} \bigr)\,\mathrm{d}x\,\mathrm{d}t} \\ \le& C\int_{{\Omega_{T}}} {\vert {\nabla u} \vert \cdot {{\bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} - {{\vert {\nabla u} \vert }^{p}}} \bigr\vert }^{\frac{{p - 2}}{p}}}\,\mathrm{d}x \,\mathrm{d}t} \\ \le& C{ \biggl( {\int_{{\Omega_{T}}} {\vert {\nabla u} \vert \cdot \bigl\vert {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}} - {{ \vert {\nabla u} \vert }^{p}}} \bigr\vert \,\mathrm{d}x\, \mathrm{d}t} } \biggr)^{\frac{{p - 2}}{p}}} \\ &{}\times { \biggl( {\int_{{\Omega_{T}}} {{{\vert {\nabla u} \vert }^{\frac{p}{2}}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{2}{p}}} \to0\quad (\varepsilon \to0). \end{aligned}$$
(53)
Finally, we estimate \({H_{6}}\). Again by using trigonometrical inequality, we arrive at
$$\begin{aligned} {H_{6}} =& \int_{{\Omega_{T}}} {u{{\vert { \nabla{u_{\varepsilon}}} \vert }^{p - 2}} \cdot \vert { \nabla{u_{\varepsilon}} - \nabla u} \vert {\chi_{\eta}}\,\mathrm{d}x\, \mathrm{d}t} \\ &{}+ \int_{{\Omega_{T}}} {u{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}} \cdot \vert {\nabla {u_{\varepsilon}} - \nabla u} \vert (1 - {\chi_{\eta}})\, \mathrm{d}x\,\mathrm{d}t} \\ \le&\eta\int_{{\Omega_{T}}} {{{\vert {\nabla {u_{\varepsilon}}} \vert }^{p - 2}} \cdot \vert {\nabla{u_{\varepsilon}} - \nabla u} \vert {\chi_{\rho}}\,\mathrm{d}x\,\mathrm{d}t} \\ &{} + C \biggl( {\int_{{\Omega_{T}}} {{{\vert {\nabla {u_{\varepsilon}}} \vert }^{(p - 2)p/(p - 1)}}}\,\mathrm{d}x\,\mathrm{d}t} \biggr)^{p/(p-1)} \biggl( {\int_{{\Omega_{T}}} {{{\vert { \nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}(1 - {\chi _{\rho}})\, \mathrm{d}x\,\mathrm{d}t} } \biggr)^{1/p} \\ =& \eta\int_{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p - 2}} \cdot \vert {\nabla{u_{\varepsilon}} - \nabla u} \vert {\chi_{\rho}}\,\mathrm{d}x\,\mathrm{d}t} \\ &{}+ C \biggl( {\int _{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}(1 - {\chi _{\rho}})\, \mathrm{d}x\,\mathrm{d}t} } \biggr)^{1/p} \\ =& {H_{8}} + {H_{9}}. \end{aligned}$$
(54)
For all \(\delta > 0\) and \(\varepsilon \in(0,1)\), let η be small enough and use Lemma 5 such that, as \(\varepsilon \to0\),
$$\begin{aligned} {H_{8}} &\le{ \biggl( {\int_{{\Omega_{T}}} {{{\vert { \nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{{p - 2}}{p}}} { \biggl( {\int_{{\Omega_{T}}} {{{\vert { \nabla{u_{\varepsilon}} - \nabla u} \vert }^{\frac{p}{2}}} \chi_{\rho}^{\frac{2}{p}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{2}{p}}} \\ &\le{ \vert {{\Omega_{T}}} \vert ^{\frac{1}{p}}} { \biggl( {\int _{{\Omega_{T}}} {{{\vert {\nabla{u_{\varepsilon}}} \vert }^{p}}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{{p - 2}}{p}}} { \biggl( { \int_{{\Omega _{T}}} {{{\vert {\nabla{u_{\varepsilon}} - \nabla u} \vert }^{p}}\chi _{\rho}\,\mathrm{d}x\,\mathrm{d}t} } \biggr)^{\frac{1}{p}}} \to0. \end{aligned}$$
(55)
Clearly, for fixed \(\eta > 0\), using Lemma 5, we have
$$ {H_{9}} \to0\quad \mbox{as } \varepsilon \to0. $$
(56)
Substituting (55) and (56) into (54), we obtain
$$ {H_{6}} \to0\quad \mbox{as } \varepsilon \to0 . $$
(57)
Hence, (43) is proved by putting together (52), (53), and (57).
Finally we prove (44). Using (13) and the definition of \({\beta _{\varepsilon}}\), we have
$$ {\beta_{\varepsilon}}({u_{\varepsilon}} - {u_{0\varepsilon}}) \to\xi\quad \mbox{as } \varepsilon \to0 . $$
(58)
Now we prove \(\xi \in G({u_{0}} - u)\). According to the definition of \(G( \cdot)\), we only need to prove that if \(u({x_{0}},{t_{0}}) < {u_{0}}({x_{0}})\), \(\xi({x_{0}},{t_{0}}) = 0\). 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
$$ {u_{\varepsilon}}(x,t) \le{u_{0\varepsilon}}(x) - \lambda,\quad \forall (x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
(59)
Thus, if ε is small enough, we have
$$ 0 \ge{\beta_{\varepsilon}}({u_{0\varepsilon}} - {u_{\varepsilon}}) \ge { \beta_{\varepsilon}}(\lambda) = 0, \quad \forall(x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
(60)
Furthermore, it follows by \(\varepsilon \to0\) that
$$ \xi(x,t) = 0,\quad \forall(x,t) \in{B_{\delta}}({x_{0}},{t_{0}}). $$
(61)
Hence, (44) holds, and the proof of Lemma 7 is completed. □
With Lemma 7 and (14), it is easy to check that u satisfies (c) and (d) in Definition 1. Moreover, applying (13), it is clear that
$$ u(x,t) \le{u_{0}}(x), \quad \mbox{in } {\Omega_{T}}, \qquad u(x,0) = {u_{0}}(x), \quad \mbox{in } \Omega. $$
(62)
Thus (a) and (b) hold.
To show the existence of problem (1), we only need to prove that (e) holds. Define
$$ I = \int_{\Omega}{\bigl\vert {u_{\varepsilon}^{\mu}- u_{0\varepsilon}^{\mu}} \bigr\vert \,\mathrm{d}x}. $$
(63)
Applying the Hölder inequality twice, we obtain
$$\begin{aligned} I &= \int_{\Omega}{\bigl\vert {u_{\varepsilon}^{\mu}(x,t) - u_{0\varepsilon }^{\mu}(x)} \bigr\vert \,\mathrm{d}x} = \int _{\Omega}{\biggl\vert {\int_{0}^{t} {\frac {\partial}{{\partial s}}} u_{\varepsilon}^{\mu}\,\mathrm{d}x} \biggr\vert \,\mathrm{d}x} \\ &\le\sqrt{t} \int_{\Omega}{\biggl\vert {\sqrt{\int _{0}^{t} {{{ \biggl( {\frac{{\partial u_{\varepsilon}^{\mu}}}{{\partial s}}} \biggr)}^{2}}} \,\mathrm{d}t} } \biggr\vert \,\mathrm{d}x} \le{ \vert \Omega \vert ^{\frac {1}{2}}}\sqrt{t} \int_{\Omega}{\int _{0}^{t} {{{ \biggl( {\frac{{\partial u_{\varepsilon}^{\mu}}}{{\partial s}}} \biggr)}^{2}}} \,\mathrm{d}t\,\mathrm{d}x} \\ &\le\sqrt{t} {\vert \Omega \vert ^{\frac{1}{2}}}\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(64)
It follows by (12) that
$$ \int_{\Omega}{\bigl\vert {u_{\varepsilon}^{\mu}(x,t) - u_{0\varepsilon}^{\mu}(x)} \bigr\vert \,\mathrm{d}x} \le C\sqrt{t}, $$
(65)
where C is independent of ε. Using (14) and letting \(\varepsilon \to0\) yields
$$ \int_{\Omega}{\bigl\vert {u^{\mu}(x,t) - u_{0}^{\mu}(x)} \bigr\vert \,\mathrm{d}x} \le C\sqrt{t}. $$
(66)
So
$$ \int_{\Omega}{\bigl\vert {u^{\mu}(x,t) - u_{0}^{\mu}(x)} \bigr\vert \,\mathrm{d}x} \to0 \quad \mbox{as } t \to0. $$
(67)
Thus the proof of existence is completed.