Let Ω, \(\Gamma_{0}\), \(\Gamma_{1}\) be given as in Section 1. For any \(1\leq s<\infty\), define
$$\Vert u \Vert _{s}^{s}= \int_{\Omega}\bigl\vert u(x) \bigr\vert ^{s}\, \mathrm{d}x, \quad\quad \Vert u \Vert _{\Gamma_{1},s}^{s}= \int_{\Gamma_{1}} \bigl\vert u(x) \bigr\vert ^{s}\, \mathrm{d}\sigma, $$
and the Hilbert space
$$H\triangleq H_{\Gamma_{0}}^{1}(\Omega)=\bigl\{ u\in H^{1}( \Omega): u=0 \text{ on } \Gamma_{0}\bigr\} . $$
Since \(\Gamma_{0}\) has positive \(N-1\) dimensional Lebesgue measure, we can equip H with the norm \(\Vert u \Vert _{H}= \Vert \nabla u \Vert _{2}\) that is equivalent to the standard one (see [21]).
Throughout this paper, the relaxation function g and the parameter p are supposed to satisfy
$$ g(s)\geq0,\qquad g'(s)\leq0,\qquad1- \int_{0}^{\infty}g(s)\,\mathrm{d}s=l>0, $$
(7)
and
$$ 2< p\leq\frac{2N-3}{N-2},\quad\text{if } N>3; \qquad p>2,\quad \text{if } N=2. $$
(8)
Before going further, we present the definition of strong solutions to Problem (1), which was given in [1, 22]. Local existence of such a solution was proved in [23] (the first three steps in the proof of Theorem 6) for a little more general problems which contain Problem (1) as a special case, by applying Galerkin’s method and the contraction mapping principle. The lengthy proof will not be repeated here.
Definition 2.1
We call \(u(x,t)\) a strong solution to Problem (1) if \(u\in C([0,T);H)\cap C^{1}([0,T);L^{2}(\Omega))\) and satisfies
$$ \int_{0}^{t} \int_{\Omega}\biggl(u_{t}\phi+\nabla u\nabla\phi- \int_{0}^{s}g(s-\tau )\nabla u(\tau)\nabla\phi(s)\, \mathrm{d}\tau \biggr)\,\mathrm{d}x\,\mathrm{d}s= \int_{0}^{t} \int_{\Gamma_{1}} \vert u \vert ^{p-2}u\phi\,\mathrm{d} \sigma\,\mathrm{d}s $$
for all \(t\in[0,T)\) and all \(\phi\in C([0,T);H^{1}(\Omega))\).
Assumption (7) is necessary to ensure that the equation in (1) is of parabolic type, and assumption (8) implies \(\vert u \vert ^{p-2}u\in L^{2}(\Gamma_{1})\) by the Sobolev trace embedding theorem (Theorem 5.36 in [24]) and hence \(\int_{\Gamma_{1}} \vert u \vert ^{p-2}u\phi\,\mathrm{d}\sigma\) makes sense.
Let \(u(x,t)\) be a strong solution to Problem (1) and define
$$ E\bigl(u(t)\bigr)=\frac{1}{2}(g\diamond\nabla u) (t)+ \frac{1}{2} \biggl(1- \int _{0}^{t}g(s)\,\mathrm{d}s \biggr) \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}-\frac {1}{p} \bigl\Vert u(t) \bigr\Vert _{\Gamma_{1},p}^{p}, $$
(9)
where \((g\diamond u)(t)\) is given in (5). Then we have the following lemma.
Lemma 2.1
\(E(u(t))\)
defined in (9) is nonincreasing in
t
and satisfies
$$ E\bigl(u(t)\bigr)+ \int_{0}^{t} \int_{\Omega}u_{t}^{2}\,\mathrm{d}x\, \mathrm{d}t\leq E(u_{0}). $$
(10)
Proof
Assume first that \(u(x,t)\) is sufficiently smooth. Multiplying the equation in (1) by \(u_{t}\) and integrating by parts over Ω yield
$$\begin{aligned} \int_{\Omega}u_{t}^{2}\,\mathrm{d}x =& \int_{\Omega}\Delta u u_{t}\,\mathrm{d}x- \int _{\Omega}u_{t} \biggl( \int_{0}^{t}g(t-s)\Delta u(x,s)\,\mathrm{d}s \biggr) \,\mathrm{d}x \\ =& \int_{\Gamma_{1}}u_{t} \biggl(\frac{\partial u}{\partial\nu}- \int _{0}^{t}g(t-s)\frac{\partial u}{\partial\nu}\,\mathrm{d}s \biggr)\,\mathrm{d}\sigma- \int_{\Omega}\nabla u\nabla u_{t}\,\mathrm{d}x \\ &{}+ \int_{\Omega}\biggl( \int_{0}^{t}g(t-s)\nabla u(s)\nabla u_{t}(t)\,\mathrm{d}s \biggr)\,\mathrm{d}x \\ =&-\frac{\mathrm{d}}{\mathrm{d}t} \biggl(\frac{1}{2} \int_{\Omega} \vert \nabla u \vert ^{2}\,\mathrm{d}x \biggr)+\frac{\mathrm{d}}{\mathrm {d}t} \biggl(\frac{1}{p} \int_{\Gamma_{1}} \vert u \vert ^{p}\,\mathrm{d}\sigma \biggr) \\ &{}+ \int_{0}^{t}g(t-s) \int_{\Omega}\nabla u(s)\nabla u_{t}(t)\,\mathrm{d}x\, \mathrm{d}s. \end{aligned}$$
(11)
Rewrite the last term of the right-hand side of the above equality as follows:
$$\begin{aligned}& \int_{0}^{t}g(t-s) \int_{\Omega}\nabla u(s)\nabla u_{t}(t)\,\mathrm{d}x\, \mathrm{d}s \\& \quad = -\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int_{0}^{t}g(t-s) \int _{\Omega}\bigl\vert \nabla u(s)-\nabla u(t) \bigr\vert ^{2}\,\mathrm{d}x\,\mathrm{d}s \biggr)+\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \biggl( \int _{0}^{t}g(s)\,\mathrm{d}s \int_{\Omega}\bigl\vert \nabla u(t) \bigr\vert ^{2} \,\mathrm{d}x \biggr) \\& \quad\quad{} +\frac{1}{2} \int_{0}^{t}g'(t-s) \int_{\Omega}\bigl\vert \nabla u(s)-\nabla u(t) \bigr\vert ^{2}\,\mathrm{d}x\,\mathrm{d}s-\frac{1}{2}g(t) \int_{\Omega}\bigl\vert \nabla u(t) \bigr\vert ^{2} \,\mathrm{d}x. \end{aligned}$$
(12)
Substituting (12) into (11) and recalling (7) and (9), we see that
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}E\bigl(u(t)\bigr) =&- \int_{\Omega}u_{t}^{2}\,\mathrm{d}x- \frac{1}{2}g(t) \int_{\Omega}\bigl\vert \nabla u(t) \bigr\vert ^{2} \,\mathrm{d}x \\ &{}+\frac{1}{2} \int_{0}^{t}g'(t-s) \int_{\Omega}\bigl\vert \nabla u(s)-\nabla u(t) \bigr\vert ^{2}\,\mathrm{d}x\,\mathrm{d}s \\ \leq&- \int_{\Omega}u_{t}^{2}\,\mathrm{d}x\leq0. \end{aligned}$$
Therefore, the conclusion of this lemma is true for smooth solution \(u(x,t)\). By a standard density argument, it is known that the same result can be established for strong solutions and for almost every t. The proof is complete. □
Next, let l be the constant given in (7). For any \(u\in H\), set
$$\begin{aligned}& J(u)=\frac{l}{2} \Vert \nabla u \Vert _{2}^{2}- \frac{1}{p} \Vert u \Vert _{\Gamma_{1},p}^{p}, \\& I(u)=l \Vert \nabla u \Vert _{2}^{2}- \Vert u \Vert _{\Gamma_{1},p}^{p}, \\& \mathcal{N}=\bigl\{ u\in H: I(u)=0, \Vert \nabla u \Vert _{2}\neq 0\bigr\} . \end{aligned}$$
We then define the potential well and its corresponding set as follows:
$$\begin{aligned}& W=\bigl\{ u\in H: I(u)>0, J(u)< d\bigr\} \cup\{0\}, \\& V=\bigl\{ u\in H: I(u)< 0, J(u)< d\bigr\} , \end{aligned}$$
where the depth d of the potential well W is characterized as
$$ d=\inf_{u\in H\backslash\{0\}} \Bigl\{ \sup_{\lambda >0}J( \lambda u) \Bigr\} =\inf_{u\in\mathcal{N}}J(u). $$
(13)
The positivity of d is given in the next lemma.
Lemma 2.2
The depth
d
of the potential well
W
is positive.
Proof
Since p satisfies (8), H can be embedded into \(L^{p}(\Gamma _{1})\) continuously. Let \(S>0\) be the best embedding constant, i.e., \(\Vert u \Vert _{\Gamma_{1},p}\leq S \Vert \nabla u \Vert _{2}\), \(\forall u\in H\).
For any \(u\in\mathcal{N}\), we have
$$l \Vert \nabla u \Vert _{2}^{2}= \Vert u \Vert ^{p}_{\Gamma _{1},p}\leq S^{p} \Vert \nabla u \Vert ^{p}_{2}, $$
which implies \(\Vert \nabla u \Vert _{2}\geq (\frac {l}{S^{p}} )^{\frac{1}{p-2}}\). Therefore, for any \(u\in\mathcal{N}\),
$$ J(u)=\frac{l}{2} \Vert \nabla u \Vert _{2}^{2}-\frac{1}{p} \Vert u \Vert _{\Gamma_{1},p}^{p}=\biggl(\frac{1}{2}-\frac{1}{p}\biggr)l \Vert \nabla u \Vert _{2}^{2}\geq\frac{(p-2)l}{2p} \biggl(\frac {l}{S^{p}} \biggr)^{\frac{2}{p-2}}>0. $$
(14)
By the definition of d, it is seen that \(d>0\). The proof is complete. □
The next lemma describes the invariance of V with respect to the semiflow of (1) under some additional conditions.
Lemma 2.3
Let (7) and (8) hold and
\(u(x,t)\)
be a local solution to Problem (1). If there exists
\(t_{0}\in[0,T)\)
such that
\(u(\cdot, t_{0})\in V\)
and
\(E(u(t_{0}))< d\), then
\(u(x,t)\)
remains inside
V
for any
\(t\in[t_{0},T)\), where
T
is the maximal existence time of
\(u(x,t)\).
Proof
Suppose on the contrary that there exists \(t_{1}\in[0,T)\) such that \(u(x,t)\in V\) for \(t\in[t_{0},t_{1})\) and \(u(x,t_{1})\notin V\). By the definition of V and the continuity of \(J(u(x,t))\) and \(I(u(x,t))\) with respect to t, we have either (i) \(J(u(x,t_{1}))=d\) or (ii) \(I(u(x,t_{1}))=0\).
By Lemma 2.1 we know that \(E(u(t_{1}))\leq E(u(t_{0}))\), which implies
$$ J\bigl(u(x,t_{1})\bigr)\leq E\bigl(u(t_{1})\bigr)\leq E \bigl(u(t_{0})\bigr)< d. $$
So case (i) is impossible.
To prove that case (ii) is also impossible, we first show that there is a positive constant \(c^{*}\) such that for any \(u\in V\), \(\Vert \nabla u \Vert _{2}\geq c^{*}\). Indeed, for any \(u\in V\), it holds that
$$ l \Vert \nabla u \Vert _{2}^{2}< \Vert u \Vert ^{p}_{\Gamma _{1},p}\leq S^{p} \Vert \nabla u \Vert ^{p}_{2}, $$
which shows \(\Vert \nabla u \Vert _{2}> (\frac{l}{S^{p}} )^{\frac{1}{p-2}}\triangleq c^{*}\). Here \(S>0\) is the embedding constant given in Lemma 2.2.
Since \(u(x,t)\in V\) for \(t\in[t_{0},t_{1})\), we have \(\Vert \nabla u(\cdot,t) \Vert _{2}\geq c^{*}\) for all \(t\in[t_{0},t_{1})\). By continuity, it also holds that \(\Vert \nabla u(\cdot,t_{1}) \Vert _{2}\geq c^{*}\). This together with (ii) implies that \(u(x,t_{1})\in\mathcal{N}\). By the definition of d, we have \(J(u(x,t_{1}))\geq d\), a contradiction. The proof is complete. □