In this paper, unless otherwise stated, we shall assume that
$$\begin{aligned}& N\geq1,\qquad \frac{2N}{N+1}< p< +\infty,\qquad 1 \leq r_{i} \leq \min\bigl\{ p,p'\bigr\} ,\quad i = 1,2, \\& \frac{1}{p}+\frac{1}{p'}=1,\qquad \frac{1}{r_{1}}+ \frac{1}{r_{1}'} = 1, \qquad\frac{1}{r_{2}}+\frac{1}{r_{2}'} = 1. \end{aligned}$$
In (1.5), Ω is a bounded conical domain of a Euclidean space \(\mathbb {R}^{N}\) with its boundary \(\Gamma\in C^{1}\) [6], T is a positive constant, \(\lambda_{1}\), \(\lambda_{2}\) and ε are non-negative constants, and ϑ denotes the exterior normal derivative of Γ. We shall assume that Green’s formula is available.
Suppose that \(g:\Omega\times \mathbb {R}^{N+2} \rightarrow \mathbb {R}\) is a given function satisfying the following conditions:
-
(a)
Carathéodory’s conditions
$$\begin{aligned}& x \rightarrow g(x,r)\mbox{ is measurable on } \Omega, \quad\mbox{for all } r \in \mathbb {R}^{N+2}; \\& r \rightarrow g(x,r)\mbox{ is continuous on } \mathbb {R}^{N+2}, \quad\mbox{for almost all }x\in\Omega. \end{aligned}$$
-
(b)
Growth condition
$$g(x,s_{1},\ldots,s_{N+2})\leq h(x)+ k_{3} |s_{1}|^{\min \{p/p', 1 \}}, $$
where \((s_{1}, s_{2}, \ldots, s_{N+2})\in \mathbb {R}^{N+2} \), \(h(x)\in L^{2}(\Omega) \cap L^{p'}(\Omega)\) and \(k_{3} \) is a positive constant.
-
(c)
Monotone condition
g is monotone with respect to \(r_{1}\), i.e.,
$$\bigl(g(x,s_{1},\ldots,s_{N+2})-g(x,t_{1}, \ldots,t_{N+2})\bigr) (s_{1} - t_{1}) \geq0 $$
for all \(x \in\Omega\) and \((s_{1},\ldots,s_{N+2}), (t_{1},\ldots,t_{N+2})\in \mathbb {R}^{N+2}\).
-
(d)
Coercive condition
$$g(x,s_{1},\ldots,s_{N+2})s_{1} \geq k_{4} s_{1}^{2}, $$
where \(k_{4}\) is a fixed positive constant.
Now, we present our discussion in the sequel.
Lemma 2.1
([18])
Let
\(X_{0}\)
denote the closed subspace of all constant functions in
\(W^{1,p}(\Omega)\). Let
X
be the quotient space
\(W^{1,p}(\Omega)/X_{0}\). For
\(u\in W^{1,p}(\Omega)\), define the mapping
\(P:W^{1,p}(\Omega) \rightarrow X_{0}\)
by
$$Pu = \frac{1}{\operatorname{meas}(\Omega)} \int_{\Omega}u \,dx. $$
Then there is a constant
\(k_{5}>0\)
such that for all
\(u\in W^{1,p}(\Omega)\),
$$\|u-Pu\|_{p}\leq k_{5}\|\nabla u\|_{(L^{p}(\Omega))^{N}}. $$
Lemma 2.2
Define the mapping
\(B: L^{p}(0,T; W^{1,p}(\Omega))\rightarrow L^{p'}(0,T;(W^{1,p}(\Omega))^{*})\)
by
$$\begin{aligned} (w,Bu) =& \int_{0}^{T}\int_{\Omega} \biggl\langle \int_{\Omega}\alpha\bigl(|\nabla u|^{p} \bigr)|\nabla u|^{p-2}\nabla u, \nabla w \biggr\rangle \,dx\,dt+ \lambda_{1}\int_{0}^{T} \int _{\Omega}|u|^{r_{1}-2}uw\,dx\,dt \\ &{} + \lambda_{2} \int_{0}^{T}\int _{\Omega}|u|^{r_{2}-2}uw\,dx\,dt \end{aligned}$$
for any
\(u,w\in L^{p}(0,T; W^{1,p}(\Omega))\). Then
B
is strictly monotone, pseudo-monotone, and coercive.
(Here, \(\langle\cdot,\cdot\rangle\)
and
\(|\cdot|\)
denote the Euclidean inner-product and Euclidean norm in
\(\mathbb {R}^{N}\), respectively.)
Proof
Step 1. B is everywhere defined.
For \(u,w \in L^{p}(0,T; W^{1,p}(\Omega))\), we find
$$\begin{aligned} \bigl|(w,Bu)\bigr| \leq& \int_{0}^{T}\int _{\Omega}k_{1}|\nabla u|^{p-1}|\nabla w|\,dx\,dt + \lambda_{1} \int_{0}^{T}\int _{\Omega}|u|^{r_{1}-1}|w|\,dx\,dt\\ &{}+\lambda_{2} \int _{0}^{T}\int_{\Omega}|u|^{r_{2}-1}|w|\,dx\,dt \\ \leq& k_{1}\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}^{p/p'}\|w\|_{L^{p}(0,T; W^{1,p}(\Omega))}+ \lambda_{1}\|w\|_{L^{r_{1}}(0,T; L^{r_{1}}(\Omega))} \|u\|_{L^{r_{1}}(0,T; L^{r_{1}}(\Omega))}^{r_{1}/r_{1}'} \\ &{}+ \lambda_{2}\|w\|_{L^{r_{2}}(0,T; L^{r_{2}}(\Omega))} \|u\|_{L^{r_{2}}(0,T; L^{r_{2}}(\Omega))}^{r_{2}/r'_{2}}. \end{aligned}$$
Since \(W^{1,p}(\Omega)\hookrightarrow L^{p}(\Omega)\hookrightarrow L^{r_{1}}(\Omega)\) and \(W^{1,p}(\Omega)\hookrightarrow L^{p}(\Omega)\hookrightarrow L^{r_{2}}(\Omega)\), for \(v \in W^{1,p}(\Omega)\), we have \(\|v\|_{L^{r_{1}}(\Omega)} \leq k_{6} \|v\|_{W^{1,p}(\Omega)}\), \(\|v\|_{L^{r_{2}}(\Omega)} \leq k_{7} \|v\|_{W^{1,p}(\Omega)}\), where \(k_{6}\) and \(k_{7}\) are positive constants. Hence,
$$\begin{aligned} \bigl|(w,Bu)\bigr| \leq& k_{1}\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}^{p/p'}\|w \|_{L^{p}(0,T; W^{1,p}(\Omega))}\\ &{}+\lambda_{1} k_{6} \|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}^{r_{1}/r_{1}'} \|w\|_{L^{p}(0,T; W^{1,p}(\Omega))}\\ &{} + \lambda_{2} k_{7} \|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}^{r_{2}/r_{2}'} \|w\|_{L^{p}(0,T; W^{1,p}(\Omega))}, \end{aligned}$$
which implies that B is everywhere defined.
Step 2. B is strictly monotone.
For \(u, v \in L^{p}(0,T;W^{1,p}(\Omega))\), we have
$$\begin{aligned} (u-v,Bu-Bv) \geq& \int_{0}^{T}\int _{\Omega} \bigl[\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-1}- \alpha\bigl(|\nabla v|^{p}\bigr)|\nabla v|^{p-1} \bigr] \bigl(|\nabla u|-|\nabla v|\bigr)\,dx\,dt \\ &{} + \lambda_{1} \int_{0}^{T}\int _{\Omega} \bigl(|u|^{r_{1}-1}-|v|^{r_{1}-1} \bigr) \bigl(|u|-|v|\bigr)\,dx\,dt \\ &{} + \lambda_{2} \int_{0}^{T}\int _{\Omega} \bigl(|u|^{r_{2}-1}-|v|^{r_{2}-1} \bigr) \bigl(|u|-|v|\bigr)\,dx\,dt. \end{aligned}$$
If we set \(f(s) = s^{1-\frac{1}{p}}\alpha(s)\), \(s > 0\), then in view of the assumption of α, we have
$$f'(s) = \biggl[ \biggl(1-\frac{1}{p} \biggr)\alpha(s) + s \alpha'(s) \biggr]s^{-\frac{1}{p}} > 0, $$
which implies that f is strictly monotone. Hence, B is strictly monotone.
Step 3. B is hemi-continuous.
It suffices to show that for any \(u,v,w\in L^{p}(0,T;W^{1,p}(\Omega))\) and \(t\in[0,1]\), \((w,B(u+tv)-Bu) \rightarrow0\) as \(t\rightarrow 0\). Since
$$\begin{aligned} &\bigl|\bigl(w,B(u+tv)-Bu\bigr)\bigr| \\ &\quad\leq\int_{0}^{T}\int_{\Omega} \bigl|\alpha\bigl(|\nabla u+t\nabla v|^{p}\bigr)|\nabla u+ t \nabla v|^{p-2}(\nabla u + t \nabla v)\\ &\qquad{}-\alpha\bigl(|\nabla u|^{p} \bigr)|\nabla u|^{p-2}\nabla u \bigr||\nabla w| \,dx\,dt \\ &\qquad{}+ \lambda_{1} \int_{0}^{T} \int_{\Omega} \bigl||u+tv|^{r_{1}-2}(u+tv)-|u|^{r_{1}-2}u \bigr||w|\,dx\,dt \\ &\qquad{}+ \lambda_{2} \int_{0}^{T}\int _{\Omega} \bigl||u+tv|^{r_{2}-2}(u+tv)-|u|^{r_{2}-2}u \bigr||w|\,dx\,dt, \end{aligned}$$
by Lebesque’s dominated convergence theorem and noting that α is continuous, we find
$$\lim_{t\rightarrow0}\bigl(w,B(u+tv)-Bu\bigr)=0. $$
Hence, B is hemi-continuous.
Step 4. B is coercive.
We shall first show that for \(u\in L^{p}(0,T; W^{1,p}(\Omega))\),
$$ \|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}\leq k_{8} \biggl(\int_{0}^{T} \int_{\Omega}|\nabla u|^{p} \,dx\,dt \biggr)^{\frac {1}{p}}+ k_{9}, $$
(2.1)
where \(k_{8}\) and \(k_{9}\) are positive constants.
In fact, using Lemma 2.1, we know that, for \(u \in L^{p}(0,T; W^{1,p}(\Omega))\),
$$\biggl\Vert u - \frac{1}{\operatorname{meas}(\Omega)} \int_{\Omega}u \,dx\biggr\Vert _{L^{p}(\Omega)} \leq k_{5} \biggl(\int_{\Omega}| \nabla u|^{p} \,dx \biggr)^{\frac{1}{p}}. $$
Thus,
$$\begin{aligned} &\biggl\Vert u - \frac{1}{\operatorname{meas}(\Omega)} \int_{\Omega}u \,dx\biggr\Vert ^{p}_{W^{1,p}(\Omega)} \\ &\quad= \biggl\Vert u - \frac{1}{\operatorname{meas}(\Omega)}\int_{\Omega}u \,dx \biggr\Vert ^{p}_{L^{p}(\Omega)} + \biggl\Vert \nabla \biggl(u - \frac{1}{\operatorname{meas}(\Omega)}\int_{\Omega}u \,dx \biggr)\biggr\Vert ^{p}_{(L^{p}(\Omega))^{N}} \\ &\quad\leq \bigl(k_{5}^{p} +1 \bigr) \int _{\Omega}|\nabla u|^{p} \,dx. \end{aligned}$$
Since
$$\biggl\Vert u - \frac{1}{\operatorname{meas}(\Omega)}\int_{\Omega}u \,dx\biggr\Vert _{W^{1,p}(\Omega)} \geq \|u\|_{W^{1,p}(\Omega)} - \biggl\Vert \frac{1}{\operatorname{meas}(\Omega)}\int_{\Omega}u \,dx\biggr\Vert _{W^{1,p}(\Omega)}, $$
we have
$$\|u\|_{W^{1,p}(\Omega)}\leq\biggl\Vert u - \frac{1}{\operatorname{meas}(\Omega)}\int _{\Omega}u \,dx\biggr\Vert _{W^{1,p}(\Omega)}+\mbox{Const}. $$
Therefore,
$$\begin{aligned} \|u\|_{L^{p}(0,T; W^{1,p}(\Omega))} \leq& \biggl\Vert u - \frac{1}{\operatorname{meas}(\Omega)}\int _{\Omega}u \,dx\biggr\Vert _{L^{p}(0,T; W^{1,p}(\Omega))}+ k_{9} \\ \leq& \bigl(k_{5}^{p} +1 \bigr)^{\frac{1}{p}} \biggl( \int_{0}^{T}\int_{\Omega}|\nabla u|^{p} \,dx\,dt \biggr)^{\frac{1}{p}}+ k_{9}. \end{aligned}$$
If we set \(k_{8} = (k_{5}^{p} + 1 )^{\frac{1}{p}}\), then (2.1) is true.
Since \(\lim_{t \rightarrow+\infty}\alpha(t) = k_{2} > 0\), there exists sufficiently large \(K > 0\) such that \(\alpha(t)> \frac{l}{2}\) whenever \(t > K\). Now, for \(u\in L^{p}(0,T; W^{1,p}(\Omega))\), let \(\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}\rightarrow+\infty\). Using (2.1), we find
$$\begin{aligned} &\frac{(u, Bu)}{\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}} \\ &\quad= \frac{\int_{0}^{T}\int_{\Omega}\alpha (|\nabla u|^{p})|\nabla u|^{p} \,dx\,dt}{\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}}+ \lambda_{1} \frac{\int_{0}^{T}\int_{\Omega}|u|^{r_{1}} \,dx\,dt}{\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}}+ \lambda_{2} \frac{\int_{0}^{T}\int_{\Omega}|u|^{r_{2}} \,dx\,dt}{\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}} \\ &\quad> \frac{1}{\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}} \biggl[\frac{l}{2}\int_{0}^{T} \int_{\Omega}|\nabla u|^{p} \,dx\,dt + \lambda_{1} \int_{0}^{T}\int_{\Omega}|u|^{r_{1}} \,dx\,dt \\ &\qquad{}+ \lambda_{2}\int_{0}^{T}\int _{\Omega}|u|^{r_{2}} \,dx\,dt \biggr] \\ &\quad> \frac{1}{\|u\|_{L^{p}(0,T; W^{1,p}(\Omega))}} \frac{l}{2}\int_{0}^{T} \int_{\Omega}|\nabla u|^{p} \,dx\,dt \rightarrow+\infty. \end{aligned}$$
This completes the proof. □
Lemma 2.3
The mapping
\(\Phi: L^{p}(0,T; W^{1,p}(\Omega))\rightarrow \mathbb {R}\)
defined by
$$\Phi(u) = \int_{0}^{T} \int_{\Gamma}\varphi_{x}\bigl(u|_{\Gamma}(x,t)\bigr)\,d\Gamma(x)\,dt, $$
for any
\(u \in L^{p}(0,T; W^{1,p}(\Omega))\), is proper, convex, and lower-semicontinuous on
\(L^{p}(0,T; W^{1,p}(\Omega))\). Moreover, the subdifferential
∂Φ of Φ is maximal monotone in view of Lemma
1.2.
Proof
The proof is similar to that of Lemma 3.1 in [27]. □
Lemma 2.4
([19])
Define
\(S: D(S) \rightarrow L^{p'}(0,T; (W^{1,p}(\Omega))^{*})\)
by
$$Su(x,t) = \frac{\partial u}{\partial t}, $$
where
$$D(S) = \biggl\{ u\in L^{p}\bigl(0,T; W^{1,p}(\Omega)\bigr) \Big| \frac{\partial u}{\partial t}\in L^{p'}\bigl(0,T; \bigl(W^{1,p}( \Omega)\bigr)^{*}\bigr), u(x,0) = u(x,T) \biggr\} . $$
The mapping
S
is linear maximal monotone.
Definition 2.1
Define a mapping \(A :L^{2}(0,T; L^{2}( \Omega))\rightarrow2^{L^{2}(0,T; L^{2}( \Omega))}\) by
$$Au = \bigl\{ w(x)\in L^{2}\bigl(0,T; L^{2}( \Omega) \bigr) | w(x)\in Bu+\partial\Phi(u)+Su \bigr\} $$
for \(u\in D(A)= \{u\in L^{2}(0,T; L^{2}( \Omega)) \vert \mbox{ there exists a }w(x)\in L^{2}(0,T; L^{2}( \Omega))\mbox{ such that } w(x) \in Bu+\partial\Phi(u)+Su\}\).
Lemma 2.5
Define the mapping
\(F : L^{p}(0,T; W^{1,p}(\Omega)) \rightarrow L^{p'}(0,T;(W^{1,p}(\Omega))^{*})\)
by
$$(v, Fu) = \int_{0}^{T}\int_{\Omega}g \biggl(x,u,\frac{\partial u}{\partial t},\varepsilon\nabla u \biggr)v(x,t) \,dx\,dt $$
for
\(u(x,t), v(x,t) \in L^{p}(0,T; W^{1,p}(\Omega))\). Then
F
is everywhere defined.
Proof
Step 1. For \(u(x,t) \in L^{p}(0,T; W^{1,p}(\Omega))\), \(x \rightarrow g (x, u, \frac{\partial u}{\partial t}, \varepsilon\nabla u )\) is measurable on Ω.
From the fact that \(u(x,t), \frac{\partial u}{\partial x_{i}} \in L^{p}(\Omega)\), \(i = 1 ,2, \ldots, N\), we see that \(x \rightarrow (u, \frac{\partial u}{\partial x_{1}}, \ldots, \frac{\partial u}{\partial x_{N}} )\) is measurable on Ω. Combining with the fact that g satisfies Carathéodory’s conditions, we know that \(x \rightarrow g (x, u, \frac{\partial u}{\partial t}, \varepsilon \nabla u )\) is measurable on Ω.
Step 2. F is everywhere defined.
For \(u,v\in L^{p}(0,T; W^{1,p}(\Omega))\), we have
$$\begin{aligned} \bigl|(v, Fu)\bigr| \leq& \int_{0}^{T}\int _{\Omega }\bigl|h(x)\bigr|\bigl|v(x,t)\bigr|\,dx\,dt + k_{3}\int _{0}^{T}\int_{\Omega}\bigl|u(x,t)\bigr|^{p/p'}\bigl|v(x,t)\bigr|\,dx\,dt \\ \leq& \bigl(T^{\frac{1}{p'}}\bigl\| h(x)\bigr\| _{L^{p'}(\Omega )}+ k_{3} \|u \|^{p/p'}_{L^{p}(0,T; W^{1,p}(\Omega))} \bigr)\|v\|_{L^{p}(0,T; W^{1,p}(\Omega))}, \end{aligned}$$
which implies that F is everywhere defined.
This completes the proof. □
Definition 2.2
Define the mapping \(H : L^{2}(0,T; L^{2}( \Omega))\rightarrow L^{2}(0,T; L^{2}( \Omega))\) by
$$H u(x) = \bigl\{ v(x)\in L^{2}\bigl(0,T; L^{2}( \Omega) \bigr) | v(x) = Fu(x)\bigr\} $$
for \(u \in D(H) = \{u(x) \in L^{2}(0,T; L^{2}( \Omega)) | \mbox{ there exists }v(x)\in L^{2}(0,T; L^{2}( \Omega)) \mbox{ such that }v(x) = Fu(x)\}\), where F is the same as in Lemma 2.5.
Lemma 2.6
The mapping
\(H : L^{2}(0,T; L^{2}( \Omega))\rightarrow L^{2}(0,T; L^{2}( \Omega))\)
defined in Definition
2.2
is bounded, coercive, hemi-continuous, and monotone.
Proof
Step 1. H is bounded.
From condition (b) of g, we know that
$$\begin{aligned} \|Hu\|_{L^{2}(0,T; L^{2}( \Omega))}^{2} &= \int_{0}^{T} \int_{\Omega}\biggl\vert g \biggl(x,u,\frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr)\biggr\vert ^{2} \,dx\,dt \\ &\leq k_{10}\|u\|_{L^{2}(0,T; L^{2}( \Omega))}^{2}+k_{11}\bigl\| h(x) \bigr\| _{L^{2}(\Omega)}^{2}, \end{aligned}$$
where \(k_{10}\) and \(k_{11}\) are positive constants. This implies that H is bounded.
Step 2. H is coercive.
From condition (d) of g, we know that
$$\begin{aligned} (u,Hu) &= \int_{0}^{T}\int_{\Omega}g \biggl(x, u, \frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr)u \,dx\,dt \geq k_{4} \int_{0}^{T}\int _{\Omega}|u|^{2} \,dx\,dt \\ &= k_{4}\|u \|^{2}_{L^{2}(0,T; L^{2}( \Omega))} \rightarrow +\infty, \end{aligned}$$
as \(\|u\|_{L^{2}(0,T; L^{2}( \Omega))} \rightarrow +\infty\). Hence, H is coercive.
Step 3. H is hemi-continuous.
Since g satisfies condition (a), we have, for any \(w(x,t) \in L^{2}(0,T; L^{2}( \Omega))\),
$$\begin{aligned} &\bigl(w, H(u+tv)-Hu\bigr) \\ &\quad= \int\int_{\Omega}\biggl[g \biggl(x, u+tv, \frac {\partial u}{\partial t}+t\frac{\partial v}{\partial t}, \varepsilon(\nabla u+ t \nabla v) \biggr)-g \biggl(x,u,\frac{\partial u}{\partial t},\varepsilon \nabla u \biggr) \biggr]w \,dx\,dt \rightarrow0, \end{aligned}$$
as \(t \rightarrow0\), which implies that H is hemi-continuous.
Step 4. H is monotone.
In view of condition (c) of g, we have
$$\begin{aligned} &(u - v, Hu - Hv)\\ &\quad= \int_{0}^{T}\int _{\Omega}\biggl[g \biggl(x, u, \frac {\partial u}{\partial t}, \varepsilon \nabla u \biggr)-g \biggl(x,v,\frac {\partial v}{\partial t}, \varepsilon\nabla v \biggr) \biggr]\bigl(u(x,t)-v(x,t)\bigr) \,dx\,dt \geq 0, \end{aligned}$$
which implies that H is monotone.
This completes the proof. □
Lemma 2.7
For all
\(u, v \in L^{p}(0,T; W^{1,p}(\Omega))\), we have
$$\bigl(v,\partial\Phi(u)\bigr) = \int_{0}^{T}\int _{\Omega}\beta_{x}\bigl(u|_{\Gamma}(x,t)\bigr) v |_{\Gamma}(x,t)\,d\Gamma(x)\,dt. $$
Moreover, \(0 \in\partial\Phi(0)\).
Proof
The idea of the proof mainly comes from Proposition 3.2(ii) in [27]. For completeness, we give the outline of the proof as follows.
Define the mapping \(G:L^{p}(0,T;L^{p}(\Gamma))\rightarrow L^{p'}(0,T; L^{p'}(\Gamma))\) by \(Gu = \beta_{x}(u) \), for any \(u\in L^{p}(0,T; L^{p}(\Gamma))\). Also, define the mapping \(K:L^{p}(0,T; W^{1,p}(\Omega))\rightarrow L^{p}(0,T; L^{p}(\Gamma)) \) by \(K(v)= v|_{\Gamma}\), for any \(v\in L^{p}(0,T; W^{1,p}(\Omega))\). Then \(K^{*}GK =\partial\Phi\), where Φ is the same as in Lemma 2.3.
In fact, it is obvious that G is continuous. For \(u(x,t), v(x,t) \in L^{p}(0,T; L^{p}(\Gamma))\), we have \((u - v, Gu - Gv)= \int_{0}^{T}\int_{\Gamma}(\beta_{x}(u)- \beta_{x}(v)) (u- v)\,d\Gamma(x)\,dt \geq 0\), since \(\beta_{x}\) is monotone. Thus, G is monotone. In view of Lemma 1.1, \(G : L^{p}(0,T; L^{p}(\Gamma))\rightarrow L^{p'}(0,T; L^{p'}(\Gamma))\) is maximal monotone.
Define \(\Psi: L^{p}(0,T; L^{p}(\Gamma)) \rightarrow \mathbb {R}\) by \(\Psi(u)= \int_{0}^{T} \int_{\Gamma}\varphi_{x}(u)\,d\Gamma(x)\,dt\). It is easy to see that Ψ is a proper, convex, and lower-semicontinuous function on \(L^{p}(0,T; L^{p}(\Gamma))\), which implies that \(\partial\Psi: L^{p}(0,T; L^{p}(\Gamma))\rightarrow L^{p'}(0,T; L^{p'}(\Gamma))\) is maximal monotone in view of Lemma 1.2. Since
$$\begin{aligned} \Psi(u)-\Psi(v)&= \int_{0}^{T} \int _{\Gamma} \bigl[\varphi_{x}(u)-\varphi_{x}(v) \bigr]\,d\Gamma (x)\,dt \\ &\geq \int_{0}^{T}\int _{\Gamma}\beta_{x}(v) (u-v)\,d\Gamma(x)\,dt = (Gv,u-v) \end{aligned}$$
for all \(u(x,t), v(x,t) \in L^{p}(0,T; L^{p}(\Gamma))\), we have \(Gv \in\partial\Psi(v)\). So \(G = \partial\Psi\).
Now, it is clear that \(K^{*}GK: L^{p}(0,T; W^{1,p}(\Omega))\rightarrow L^{p'}(0,T; (W^{1,p}(\Omega))^{*}) \) is maximal monotone since both K and G are continuous. Finally, for any \(u,v\in L^{p}(0,T; W^{1,p}(\Omega))\), we have
$$\begin{aligned} \Phi(v) - \Phi(u) =& \Psi(Kv)-\Psi(Ku) \\ = & \int_{0}^{T}\int_{\Gamma} \bigl[\varphi_{x}\bigl(v|_{\Gamma }(x,t)\bigr)-\varphi _{x}\bigl(u|_{\Gamma}(x,t)\bigr) \bigr]\,d\Gamma(x)\,dt \\ \geq& \int_{0}^{T}\int_{\Gamma} \beta_{x} \bigl(u|_{\Gamma }(x,t) \bigr) \bigl(v|_{\Gamma}(x,t)-u|_{\Gamma}(x,t) \bigr)\,d\Gamma(x)\,dt \\ = & (GKu,Kv-Ku) = \bigl(K^{*}GKu,v-u\bigr). \end{aligned}$$
Hence, we get \(K^{*}GK \subset\partial\Phi\) and so \(K^{*}GK = \partial\Phi\).
It now follows that for all \(u, v \in L^{p}(0,T; W^{1,p}(\Omega))\),
$$\bigl(v, \partial\Phi(u)\bigr) = \int_{0}^{T}\int _{\Omega}\beta_{x}\bigl(u|_{\Gamma}(x,t)\bigr) v |_{\Gamma}(x,t)\,d\Gamma(x)\,dt. $$
Moreover, \(0 \in\partial\Phi(0)\) since \(0 \in\beta_{x}(0)\). This completes the proof. □
Lemma 2.8
The mapping
\(A : L^{2}(0,T; L^{2}(\Omega)) \rightarrow L^{2}(0,T; L^{2}(\Omega))\)
defined in Definition
2.1
is maximal monotone.
Proof
Noting Lemmas 2.2-2.4, we can easily get the result that A is monotone.
Next, we shall show that \(R(I+A) = L^{2}(0,T; L^{2}(\Omega))\), which ensures that A is maximal monotone.
Case 1. \(p \geq2\). We define \(\overline{F}:L^{p}(0,T; W^{1,p}(\Omega)) \rightarrow L^{p'}(0,T; (W^{1,p}(\Omega))^{*})\) by
$$\overline{F}u = u, (v,\overline{F}u)_{L^{p}(0,T; W^{1,p}(\Omega ))\times L^{p'}(0,T; (W^{1,p}(\Omega))^{*})} = (v,u)_{L^{2}(0,T; L^{2}(\Omega))}, $$
where \((\cdot,\cdot)_{L^{2}(0,T; L^{2}(\Omega))}\) denotes the inner-product of \(L^{2}(0,T; L^{2}(\Omega))\). Then \(\overline{F}\) is everywhere defined, monotone and hemi-continuous, which implies that \(\overline{F}\) is maximal monotone in view of Lemma 1.1. Combining with the facts of Lemmas 1.3, 2.2-2.4, we have \(R(B+\partial\Phi+ S + \overline{F}) = L^{p'}(0,T; (W^{1,p}(\Omega))^{*})\).
For \(f \in L^{2}(0,T; L^{2}(\Omega)) \subset L^{p'}(0,T; (W^{1,p}(\Omega))^{*})\), there exists \(u \in L^{p}(0,T; W^{1,p}(\Omega ))\subset L^{2}(0,T; L^{2}(\Omega))\) such that
$$f = Bu + \partial\Phi(u) + Su+\overline{F}u = Au + u, $$
which implies that \(R(I+A)= L^{2}(0,T; L^{2}(\Omega))\).
Case 2. \(\frac{2N}{N+1} < p<1\), then \(p' \geq 2\). Similar to Lemma 2.2, we define \(\widehat{B}: L^{p'}(0,T; W^{1,p}(\Omega))\rightarrow L^{p}(0,T;(W^{1,p}(\Omega))^{*})\) by
$$\begin{aligned} (w,\widehat{B}u) =& \int_{0}^{T}\int _{\Omega} \biggl\langle \int_{\Omega}\alpha\bigl(| \nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u, \nabla w \biggr\rangle \,dx\,dt+\lambda_{1}\int_{0}^{T} \int_{\Omega}|u|^{r_{1}-2}uw\,dx\,dt \\ &{} + \lambda_{2} \int_{0}^{T}\int _{\Omega}|u|^{r_{2}-2}uw\,dx\,dt \end{aligned}$$
for any \(u,w\in L^{p'}(0,T; W^{1,p}(\Omega))\). Then \(\widehat{B}\) is maximal monotone and coercive. Similar to Lemma 2.3, define the mapping \(\widehat{\Phi} : L^{p'}(0,T; W^{1,p}(\Omega))\rightarrow \mathbb {R}\) by
$$\widehat{\Phi}(u) = \int_{0}^{T} \int _{\Gamma}\varphi_{x}\bigl(u|_{\Gamma}(x,t) \bigr)\,d\Gamma(x)\,dt, $$
for any \(u \in L^{p'}(0,T; W^{1,p}(\Omega))\), then \(\partial\widehat{\Phi}\) is maximal monotone. Similar to Lemma 2.4, define \(\widehat{S}: D(\widehat{S})= \{ u\in L^{p'}(0,T; W^{1,p}(\Omega)) | \frac{\partial u}{\partial t}\in L^{p}(0,T; (W^{1,p}(\Omega))^{*}), u(x,0) = u(x,T) \} \rightarrow L^{p}(0,T; (W^{1,p}(\Omega ))^{*})\) by
$$\widehat{S}u(x,t) = \frac{\partial u}{\partial t}. $$
Then \(\widehat{S}\) is linear maximal monotone. Similar to Case 1, define \(\overline{\overline{F}}:L^{p'}(0,T; W^{1,p}(\Omega)) \rightarrow L^{p}(0,T; (W^{1,p}(\Omega))^{*})\) by
$$\overline{\overline{F}}u = u,\quad (v,\overline{\overline {F}}u)_{L^{p'}(0,T; W^{1,p}(\Omega))\times L^{p}(0,T; (W^{1,p}(\Omega ))^{*})} = (v,u)_{L^{2}(0,T; L^{2}(\Omega))}, $$
then we have \(R(\widehat{B}+\partial\widehat{\Phi}+\widehat{S}+\overline {\overline{F}}) = L^{p}(0,T; (W^{1,p}(\Omega))^{*})\). So, for \(f \in L^{2}(0,T; L^{2}(\Omega)) \subset L^{p}(0,T; (W^{1,p}(\Omega))^{*})\), there exists \(u \in L^{p'}(0,T; W^{1,p}(\Omega ))\subset L^{2}(0,T; L^{2}(\Omega))\) such that
$$f = \widehat{B}u + \partial\widehat{\Phi}(u) + \widehat{S}u+\overline{ \overline{F}}u = Au + u, $$
which implies that \(R(I+A)= L^{2}(0,T; L^{2}(\Omega))\). □
Theorem 2.1
For
\(f(x,t) \in L^{2}(0,T; L^{2}(\Omega))\), the nonlinear parabolic equation (1.5) has a unique solution
\(u(x,t) \)
in
\(L^{2}(0,T; L^{2}(\Omega))\), i.e.,
-
(a)
\(\frac{\partial u}{\partial t}-\operatorname{div} [\alpha(|\nabla u|^{p})|\nabla u|^{p-2}\nabla u ]+\lambda_{1} |u|^{r_{1}-2}u + \lambda_{2} |u|^{r_{2}-2}u + g (x,u,\frac{\partial u}{\partial t}, \varepsilon \nabla u ) = f(x,t)\), a.e. \((x,t)\in\Omega\times(0,T)\);
-
(b)
\(- \langle\vartheta,\alpha(|\nabla u|^{p})|\nabla u|^{p-2}\nabla u \rangle\in\beta_{x}(u(x,t))\), a.e. \(x\in \Gamma \times(0,T)\);
-
(c)
\(u(x,0) = u(x,T)\), \(x \in\Omega\).
Proof
We split our proof into two steps.
Step 1. There exists a unique \(u(x,t)\) which satisfies \(Hu+\lambda Au = f\), where \(f(x,t) \in L^{2}(0,T; L^{2}(\Omega))\) is a given function.
From Theorem 1.5, Lemmas 2.6 and 2.8, we know that A is H-monotone. Thus, \(R(H+\lambda A)= L^{2}(0,T; L^{2}(\Omega))\). Then, for \(f(x,t) \in L^{2}(0,T; L^{2}(\Omega))\) in (1.5), there exists \(u(x,t) \in L^{2}(0,T; L^{2}(\Omega))\) such that \(Hu(x,t)+\lambda Au(x,t) = f(x,t)\). Next, we shall prove that \(u(x,t)\) is unique.
Suppose that \(u(x,t)\) and \(v(x,t)\) satisfy \(Hu+\lambda Au = f\) and \(Hv + \lambda Av = f\), respectively. Then \(0 \leq\lambda(u - v, Au - Av) = - (u - v, Hu - Hv)\leq0\), which ensures that
$$0 = (u - v, Au - Av)= (u - v, Bu - Bv) + \bigl(u - v, \partial\Phi(u)- \partial \Phi(v) \bigr)+(u - v, Su - Sv). $$
Using Lemmas 2.2, 2.3, and 2.4, we have \((u - v, Bu - Bv)= 0\), which implies that \(u(x,t) = v(x,t)\), since B is strictly monotone.
Step 2. If \(u(x,t)\in L^{2}(0,T; L^{2}(\Omega))\) satisfies \(f = H u + Au\), then \(u(x,t)\) is the solution of (1.5).
Since \(\Phi(u+ \varphi) = \Phi(u)\) for any \(\varphi\in C^{\infty}_{0}(\Omega\times(0,T))\), we have \((\varphi, \partial\Phi(u)) = 0\). Then, for \(\varphi\in C_{0}^{\infty}(\Omega\times(0,T))\), we have
$$(\varphi, f-Hu)= (\varphi, Bu)+\bigl(\varphi, \partial\Phi(u)\bigr)+(\varphi, Su) = (\varphi, Bu)+(\varphi, Su). $$
So
$$\begin{aligned} &\int_{0}^{T}\int_{\Omega}\biggl(f - g \biggl(x,u,\frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr) \biggr) \varphi \,dx\,dt \\ &\quad= \int_{0}^{T}\int_{\Omega}\bigl\langle \alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2} \nabla u, \nabla\varphi \bigr\rangle \,dx\,dt + \lambda_{1} \int _{0}^{T}\int_{\Omega}|u|^{r_{1}-2}u \varphi \,dx\,dt \\ &\qquad{}+ \lambda_{2} \int_{0}^{T} \int_{\Omega}|u|^{r_{2}-2}u\varphi \,dx\,dt + \int_{0}^{T}\int_{\Omega} \frac{\partial u}{\partial t} \varphi \,dx\,dt \\ &\quad= - \int_{0}^{T}\int_{\Omega}\operatorname{div} \bigl[\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr] \varphi \,dx\,dt + \lambda_{1} \int _{0}^{T}\int_{\Omega}|u|^{r_{1}-2}u \varphi \,dx\,dt \\ &\qquad{}+ \lambda_{2}\int_{0}^{T} \int_{\Omega}|u|^{r_{2}-2}u\varphi \,dx\,dt + \int_{0}^{T}\int_{\Omega} \frac{\partial u}{\partial t} \varphi \,dx\,dt, \end{aligned}$$
which implies that the equation
$$\begin{aligned} &\frac{\partial u}{\partial t }-\operatorname{div} \bigl[\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr] + \lambda_{1} |u|^{r_{1}-2}u \\ &\quad{}+ \lambda_{2} |u|^{r_{2}-2}u +g \biggl(x,u, \frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr)= f(x,t), \quad \mbox{a.e. }x\in \Omega\times(0,T), \end{aligned}$$
(2.2)
is true.
By using (2.2) and Green’s formula, we have
$$\begin{aligned} &\int_{0}^{T}\int _{\Gamma} \bigl\langle \vartheta,\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr\rangle v|_{\Gamma}\,d\Gamma(x)\,dt \\ &\quad= \int_{0}^{T}\int_{\Omega} \operatorname{div} \bigl[\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr] v \,dx\,dt + \int_{0}^{T} \int_{\Omega} \bigl\langle \alpha\bigl(|\nabla u|^{p} \bigr)|\nabla u|^{p-2}\nabla u, \nabla v \bigr\rangle \,dx\,dt \\ &\quad= \biggl(v, \frac{\partial u}{\partial t}+ \lambda_{1}|u|^{r_{1}-2}u+ \lambda_{2}|u|^{r_{2}-2}u+ g \biggl(x,u, \frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr) - f \biggr) \\ &\qquad{}+ \bigl(v, Bu - \lambda_{1} |u|^{r_{1}-2}u- \lambda_{2} |u|^{r_{2}-2}u \bigr) = (v, Su+Bu + Hu - f) = \bigl(v, - \partial\Phi(u)\bigr) \\ &\quad= - \int _{0}^{T}\int_{\Gamma}\beta_{x}\bigl(u|_{\Gamma}(x)\bigr)v|_{\Gamma}(x)\,d\Gamma(x)\,dt. \end{aligned}$$
(2.3)
Then
$$ - \bigl\langle \vartheta, \alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr\rangle \in\beta_{x}\bigl(u(x,t)\bigr), \quad \mbox{a.e. on } \Gamma\times(0,T). $$
(2.4)
From the definition of S, we can easily obtain \(u(x,0) = u(x,T)\) for all \(x \in\Omega\). Combining with (2.2) and (2.4) we see that u is the unique solution of (1.5).
This completes the proof. □
Lemma 2.9
Define
\(\widetilde{B} : L^{p}(0,T; W^{1,p}(\Omega)) \rightarrow L^{p'}(0,T; (W^{1,p}(\Omega))^{*})\)
by
\(\widetilde{B}u \equiv Bu - f(x,t)\), for
\(u \in L^{p}(0,T; W^{1,p}(\Omega))\). Then
\(\widetilde{B}\)
is maximal monotone.
Proof
Similar to the proof of Lemma 2.2, we know that \(\widetilde{B}\) is everywhere defined, monotone, and hemi-continuous. It follows that \(\widetilde{B}\) is maximal monotone. □
Definition 2.3
Define a mapping \(\widetilde{A} :L^{2}(0,T; L^{2}( \Omega))\rightarrow2^{L^{2}(0,T; L^{2}( \Omega))}\) by
$$\widetilde{A}u = \bigl\{ w(x)\in L^{2}\bigl(0,T; L^{2}( \Omega)\bigr) | w(x)\in\widetilde{B}u+\partial\Phi(u)+Su \bigr\} $$
for \(u\in D(\widetilde{A})= \{u\in L^{2}(0,T; L^{2}( \Omega)) | \mbox{there exists }w(x) \in L^{2}(0,T; L^{2}( \Omega ))\mbox{ such that }w(x) \in\widetilde{B}u +\partial\Phi(u)+Su\}\).
Definition 2.4
Let \(\mathcal{H}\) be a Hilbert space and A be a H-monotone operator. The resolvent operator of A, \(R_{A, \lambda}^{H} : \mathcal{H} \rightarrow\mathcal{H}\), is defined by
$$R_{A, \lambda}^{H} (u) = (H+\lambda A)^{-1}u, \quad \forall u \in \mathcal{H}. $$
Theorem 2.2
\(u(x,t) = R^{H}_{\widetilde{A},1}(0)\)
if and only if
\(u(x,t) \in L^{2}(0,T; L^{2}( \Omega))\)
is the solution of (1.5).
Proof
Let \(u(x,t)\) be the solution of (1.5). Then, using Green’s formula and Lemma 2.7, we have
$$\begin{aligned} & \bigl(v, (H+\widetilde{A} )u \bigr) \\ &\quad= \int_{0}^{T}\int_{\Omega} \bigl\langle \alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2} \nabla u,\nabla v \bigr\rangle \,dx\,dt +\lambda_{1} \int _{0}^{T}\int_{\Omega}|u|^{r_{1}-1}uv\,dx\,dt \\ &\qquad{}+ \lambda_{2}\int_{0}^{T}\int _{\Omega}|u|^{r_{2}-1}uv \,dx\,dt- \int_{0}^{T} \int_{\Omega}f(x,t)v(x,t) \,dx\,dt \\ &\qquad{}+ \int_{0}^{T}\int_{\Omega}g \biggl(x,u,\frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr)v(x,t)\,dx \,dt +\bigl(v, \partial\Phi(u)\bigr)+ \int_{0}^{T}\int _{\Omega}\frac{\partial u}{\partial t}v \,dx\,dt \\ &\quad= - \int_{0}^{T}\int_{\Omega} \operatorname{div} \bigl[\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr] v \,dx\,dt \\ &\qquad{}+ \int_{0}^{T} \int_{\Gamma} \bigl\langle \vartheta,\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr\rangle v|_{\Gamma}\,d\Gamma(x)\,dt \\ &\qquad{}+ \lambda_{1} \int_{0}^{T} \int_{\Omega}|u|^{r_{1}-1}uv\,dx\,dt + \lambda_{2} \int _{0}^{T}\int_{\Omega}|u|^{r_{2}-1}uv \,dx\,dt \\ &\qquad{}- \int_{0}^{T}\int_{\Omega}f(x,t)v(x,t) \,dx\,dt +\int_{0}^{T}\int_{\Omega}g \biggl(x,u,\frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr)v(x,t) \,dx\,dt \\ &\qquad{}+ \int _{0}^{T}\int_{\Gamma} \beta_{x}(u|_{\Gamma})v|_{\Gamma}\,d\Gamma(x)\,dt + \int_{0}^{T}\int_{\Omega} \frac{\partial u}{\partial t}v \,dx\,dt \\ &\quad= \int_{0}^{T} \int_{\Gamma} \bigl\langle \vartheta,\alpha\bigl(|\nabla u|^{p}\bigr)|\nabla u|^{p-2}\nabla u \bigr\rangle v|_{\Gamma}\,d\Gamma(x)\,dt+ \int _{0}^{T}\int_{\Gamma} \beta_{x}(u|_{\Gamma})v|_{\Gamma}\,d\Gamma(x)\,dt \\ &\quad= - \int_{0}^{T}\int_{\Gamma} \beta_{x}(u|_{\Gamma})v|_{\Gamma}\,d\Gamma(x)\,dt+ \int _{0}^{T}\int_{\Gamma} \beta_{x}(u|_{\Gamma})v|_{\Gamma}\,d\Gamma (x)\,dt = 0. \end{aligned}$$
Thus, \(u(x,t) = R^{H}_{\widetilde{A},1}(0)\).
If \(u(x,t) = R^{H}_{\widetilde{A},1}(0)\), then noting Lemma 2.7, we have for \(\varphi\in C^{\infty}_{0}(\Omega\times(0,T))\),
$$\begin{aligned} 0 =& \int_{0}^{T}\int_{\Omega} \frac{\partial u}{\partial t} \varphi \,dx\,dt + \int_{0}^{T}\int _{\Omega}\bigl\langle \alpha\bigl(|\nabla u|^{p}\bigr)| \nabla u|^{p-2}\nabla u, \nabla\varphi \bigr\rangle \,dx\,dt \\ &{} + \lambda_{1} \int_{0}^{T}\int _{\Omega} |u|^{r_{1}-2}u \,dx\,dt + \lambda_{2} \int _{0}^{T} \int_{\Omega} |u|^{r_{2}-2}u \,dx\,dt \\ &{} - \int_{0}^{T}\int_{\Omega}f \varphi \,dx\,dt + \int_{0}^{T}\int _{\Omega}g \biggl(x,u, \frac{\partial u}{\partial t}, \varepsilon\nabla u \biggr)\varphi \,dx\,dt, \end{aligned}$$
which implies that the equation
$$\begin{aligned} &\frac{\partial u}{\partial t} -\operatorname{div} \bigl[\alpha\bigl(|\nabla u|^{p} \bigr)|\nabla u|^{p-2}\nabla u \bigr] + \lambda_{1} |u|^{r_{1}-2}u \\ &\quad{}+ \lambda_{2} |u|^{r_{2}-2}u+ g \biggl(x,u, \frac{\partial u }{\partial t},\varepsilon\nabla u \biggr) = f(x,t), \quad \mbox{a.e. } (x,t) \in\Omega\times(0,T), \end{aligned}$$
is true.
Similar to the last part of Theorem 2.1, we know that \(- \langle\vartheta, \alpha(|\nabla u|^{p})|\nabla u|^{p-2}\nabla u \rangle\in\beta_{x}(u(x,t))\). From the definition of S, we know that \(u(x,0) = u(x,T)\) for all \(x\in\Omega\), which implies that \(u(x,t)\) is the solution of (1.5). This completes the proof. □