The following result about the existence and uniqueness of solutions of equation (1.1) can be obtained by the standard Faedo-Galerkin methods, here we only formulate the result.
Theorem 3.1
Let (1.2), (1.3) be satisfied and
\(g(x,t)\in L^{2}_{\mathrm{loc}}(\mathbb {R};L ^{2}(\Omega ))\). For any
\(\varepsilon \geqslant 0\), \(I=[\tau ,T]\)
and
\(u_{\tau } \in H_{0}^{1}(\Omega )\), equation (1.1) admits a unique solution
u
satisfying
$$\begin{aligned} u\in C\bigl(I;H^{1}_{0}(\Omega )\bigr),\quad \partial _{t}u\in L^{2}\bigl(I;H^{1}_{0}(\Omega ) \bigr). \end{aligned}$$
Moreover, the solution continuously depends on the initial data in
\(H^{1}_{0}(\Omega )\).
By Theorem 3.1, for each \(\varepsilon \geqslant 0\) and \(g\in L ^{2}_{b}(\mathbb {R};L^{2}(\Omega ))\), we define a process as follows:
$$ U^{\varepsilon }_{g}(t,\tau )u_{\tau }=u(t)\quad \mbox{for all }t\geqslant \tau \mbox{ and }u_{\tau }\in H_{0}^{1}( \Omega ), $$
and the mapping \(U^{\varepsilon }_{g}(t,\tau ):H_{0}^{1}(\Omega )\to H_{0}^{1}(\Omega )\) is continuous.
Lemma 3.1
Let (1.2), (1.3) be satisfied. Assume
\(g_{0}\in L^{2}_{b}(\mathbb {R};L ^{2}(\Omega ))\)
and Σ is the hull of
\(g_{0}\)
in
\(L^{2}_{w,\mathrm{loc}}(\mathbb {R};L ^{2}(\Omega ))\). There exists
\(\delta >0\)
such that for any
\(\tau \in \mathbb {R}\)
and any initial data
\(u_{\tau }\in H_{0}^{1}(\Omega )\), the solutions of equation (1.1) satisfy: for all
\(\varepsilon \geqslant 0\), \(t\geqslant \tau \)
and
\(g\in \Sigma \),
$$\begin{aligned} &\bigl\Vert \nabla u(t) \bigr\Vert ^{2}\leqslant e^{-\delta (t-\tau )}Q \bigl(\Vert \nabla u_{\tau } \Vert \bigr)+M_{0}, \end{aligned}$$
(3.1)
$$\begin{aligned} & \int_{t}^{t+1}\bigl(\bigl\Vert \partial _{t}u(s) \bigr\Vert ^{2}+\varepsilon \bigl\Vert \nabla \partial _{t}u(s) \bigr\Vert ^{2}\bigr)\,ds \leqslant Q\bigl(\Vert \nabla u_{\tau } \Vert \bigr)+ M_{0}, \end{aligned}$$
(3.2)
where
\(M_{0}>0\)
depends on
\(\Vert g_{0} \Vert _{L^{2}_{b}}\), but is independent of
ε.
Proof
The proof is classical (see, e.g., [8–10]), we only sketch the main steps of the reasoning. Multiplying the first equation of (1.1) by \(\partial _{t}u+\delta u\) and integrating over Ω, we have
$$\begin{aligned} &\frac{d}{dt} \bigl( \delta \Vert u \Vert ^{2}+(1+\varepsilon \delta )\Vert \nabla u \Vert ^{2}+2\bigl\langle F(u),1\bigr\rangle \bigr) +2\Vert \partial _{t}u \Vert ^{2}+2 \varepsilon \Vert \nabla \partial _{t}u \Vert ^{2} \\ & \quad{} +2\delta \Vert \nabla u \Vert ^{2}+2\delta \bigl\langle f(u),u \bigr\rangle =2\bigl\langle g(x,t),\partial _{t}u+\delta u\bigr\rangle , \end{aligned}$$
(3.3)
where \(F(u)=\int_{0}^{u}f(s)\,ds\) and \(\delta >0\) is sufficiently small which will be given precisely later.
Observe that
$$\begin{aligned} 2\bigl\langle g(x,t),\partial _{t}u+\delta u\bigr\rangle \leqslant 2\bigl\Vert g(x,t) \bigr\Vert ^{2}+\bigl\Vert \partial _{t}u\bigl\Vert ^{2}+\delta ^{2} \bigr\Vert u \bigr\Vert ^{2}. \end{aligned}$$
(3.4)
Thus
$$\begin{aligned} \frac{d}{dt}E(t)+\delta E(t)+\Vert \partial _{t}u \Vert ^{2}+2\varepsilon \Vert \nabla \partial _{t}u \Vert ^{2}+\Pi (t)\leqslant 2\bigl\Vert g(x,t) \bigr\Vert ^{2}, \end{aligned}$$
(3.5)
where
$$\begin{aligned} &E(t)=\delta \Vert u \Vert ^{2}+(1+\varepsilon \delta )\Vert \nabla u \Vert ^{2}+2\bigl\langle F(u),1 \bigr\rangle +2C_{\rho } \end{aligned}$$
and
$$\begin{aligned} &\Pi (t)=2\delta \Vert \nabla u \Vert ^{2}+2\delta \bigl\langle f(u),u \bigr\rangle -\delta ^{2} \Vert u \Vert ^{2}-\delta E(t). \end{aligned}$$
Thanks to (1.2) and (1.3), we estimate that
$$\begin{aligned} &\bigl\langle f(u),u\bigr\rangle \geqslant -\rho \Vert u \Vert ^{2}-C_{\rho }, \end{aligned}$$
(3.6)
$$\begin{aligned} &\bigl\langle F(u),1\bigr\rangle \geqslant -\frac{1}{2}\rho \Vert u \Vert ^{2}-C_{ \rho }, \end{aligned}$$
(3.7)
$$\begin{aligned} &\bigl\langle f(u),u\bigr\rangle -\bigl\langle F(u),1\bigr\rangle \geqslant - \frac{1}{2} \rho \Vert u \Vert ^{2}-C_{\rho }, \end{aligned}$$
(3.8)
$$\begin{aligned} &\bigl\langle F(u),1\bigr\rangle \leqslant C\bigl(\Vert \nabla u \Vert ^{\frac{2N}{N-2}}+1\bigr) \end{aligned}$$
(3.9)
for some positive constants \(\rho <\lambda_{1}\) and \(C_{\rho }\).
Let \(0<\delta \leqslant \frac{1-\rho \lambda _{1}^{-1}}{1+2\lambda _{1}^{-1}}\), then
$$\begin{aligned} E(t)\geqslant 0\quad \mbox{and}\quad \Pi (t)\geqslant -C_{0}, \end{aligned}$$
(3.10)
where \(C_{0}=4\delta C_{\rho }\). Moreover, from (3.7), (3.9), there exist positive constants \(C_{1}\), \(C_{2}\) such that
$$\begin{aligned} C_{1}\Vert \nabla u \Vert ^{2}\leqslant E(t)\leqslant C_{2}\bigl( \Vert \nabla u \Vert ^{ \frac{2N}{N-2}}+1 \bigr). \end{aligned}$$
(3.11)
Hence, by (3.5) and (3.10), we get
$$\begin{aligned} \frac{d}{dt}E(t)+\delta E(t)+\Vert \partial _{t}u \Vert ^{2}+2\varepsilon \Vert \nabla \partial _{t}u \Vert ^{2}\leqslant 2\bigl\Vert g(x,t) \bigr\Vert ^{2}+C_{0}. \end{aligned}$$
(3.12)
Note that [14], Proposition V.4.2, implies that \(\Vert g \Vert _{L^{2} _{b}}\leqslant \Vert g_{0} \Vert _{L^{2}_{b}}, \forall g\in \Sigma \). Applying Gronwall’s inequality to (3.12), we obtain
$$\begin{aligned} E(t)\leqslant e^{-\delta (t-\tau )}E(\tau )+C\bigl(\Vert g_{0} \Vert ^{2}_{L^{2}_{b}}+1\bigr), \quad \forall g\in \Sigma , \end{aligned}$$
and this together with (3.11) implies (3.1).
Finally, integrating (3.12) over \([t,t+1]\) with \(\delta =0\) leads to (3.2). □
By Lemma 3.1, we can construct a uniformly (w.r.t. \(g\in \Sigma \), \(\varepsilon \in [0,\infty )\)) absorbing set, which is independent of ε, for the family of processes \(\{U^{\varepsilon }_{g}(t, \tau )\mid g\in \Sigma , \varepsilon \in [0,\infty )\}\).
Corollary 3.1
Under the assumptions of Lemma
3.1, there exists a bounded uniformly (w.r.t. \(g\in \Sigma\)
and
\(\varepsilon \in [0,\infty )\)) absorbing set
\(\mathcal {B}\)
of
\(H_{0}^{1}(\Omega )\)
for the family of processes
\(\{U^{\varepsilon }_{g}(t,\tau )\mid g\in \Sigma , \varepsilon \in [0,\infty )\}\)
associated with equation (1.1), that is, for any
\(\tau \in \mathbb {R}\)
and any bounded subset
\(B\subset H_{0}^{1}(\Omega )\), there exists
\(T=T(B)\geqslant \tau \)
such that
\(\bigcup_{\varepsilon \geqslant 0} \bigcup_{g\in \Sigma }U^{\varepsilon }_{g}(t,\tau )B\subset \mathcal {B}\)
for all
\(t\geqslant T\).
Lemma 3.2
Let (1.2), (1.3) be satisfied. Assume
\(g_{0},\partial _{t}g_{0} \in L^{2}_{b}(\mathbb {R};L^{2}(\Omega ))\)
and Σ is the hull of
\(g_{0}\)
in
\(L^{2}_{w,\mathrm{loc}}(\mathbb {R};L^{2}(\Omega ))\). For any
\(\tau \in \mathbb {R}\), \(T>\tau \), any initial data
\(u_{\tau }\in H_{0}^{1}(\Omega )\)
and any
\(g\in \Sigma \), the solutions of equation (1.1) satisfy the following estimate:
$$\begin{aligned} \bigl\Vert \partial _{t}u(t) \bigr\Vert ^{2}+ \varepsilon \bigl\Vert \nabla \partial _{t}u(t) \bigr\Vert ^{2} \leqslant \frac{Q}{(t- \tau )^{2}},\quad \forall t\in (\tau ,T], \forall \varepsilon \geqslant 0, \end{aligned}$$
(3.13)
where
Q
depends on
τ, T, \(\Vert \nabla u_{\tau } \Vert \), \(\Vert g_{0} \Vert _{L^{2}_{b}}\)
and
\(\Vert \partial _{t}g_{0} \Vert _{L^{2}_{b}}\), but is independent of
ε.
Proof
From Lemma 3.1, we observe that
$$\begin{aligned} \int_{\tau }^{T}\bigl(\bigl\Vert \partial _{t}u(t) \bigr\Vert ^{2}+\varepsilon \bigl\Vert \nabla \partial _{t}u(t) \bigr\Vert ^{2}\bigr)\,dt \leqslant M, \end{aligned}$$
(3.14)
where M depends on τ, T, \(\Vert \nabla u_{\tau } \Vert \) and \(\Vert g_{0} \Vert _{L^{2}_{b}}\).
Differentiate the first equation of (1.1) with respect to t and let \(v=\partial _{t}u\), then v satisfies the following equality:
$$\begin{aligned} \partial _{t}v-\varepsilon \Delta \partial _{t}v-\Delta v+f'(u)v=\partial _{t}g(x,t). \end{aligned}$$
(3.15)
Multiplying (3.15) by v and integrating over Ω, using (1.3), after standard transformations, we obtain
$$\begin{aligned} \frac{d}{dt}G_{\varepsilon }(t)+\Vert \nabla v \Vert ^{2}\leqslant 2lG_{\varepsilon }(t)+ \lambda _{1}^{-1} \bigl\Vert \partial _{t}g(x,t) \bigr\Vert ^{2} \end{aligned}$$
(3.16)
for some \(l\geqslant \lambda _{1}\), where \(G_{\varepsilon }(t)=\Vert v(t) \Vert ^{2}+\varepsilon \Vert \nabla v(t) \Vert ^{2}\).
Multiplying (3.16) by \((t-\tau )^{2}\), we obtain
$$\begin{aligned} &\frac{d}{dt}(t-\tau )^{2}G_{\varepsilon }(t) \\ &\quad \leqslant 2l(t-\tau )^{2}G_{\varepsilon }(t)+2(t-\tau )G_{\varepsilon }(t)+\lambda _{1} ^{-1}(t-\tau )^{2} \bigl\Vert \partial _{t}g(x,t) \bigr\Vert ^{2} \\ &\quad \leqslant (2l+1) (t-\tau )^{2}G_{\varepsilon }(t)+G_{\varepsilon }(t)+ \lambda _{1}^{-1}(t- \tau )^{2}\bigl\Vert \partial _{t}g(x,t) \bigr\Vert ^{2}. \end{aligned}$$
Then, by Gronwall’s inequality, [14], Proposition V.4.2, (3.14) and noting that
$$ \int_{\tau }^{T}\bigl\Vert \partial _{t}g(x,s) \bigr\Vert ^{2}\,ds \leqslant \bigl([T-\tau ]+1\bigr)\Vert \partial _{t}g\Vert ^{2}_{L^{2}_{b}}\leqslant \bigl([T-\tau ]+1\bigr) \Vert \partial _{t}g_{0} \Vert ^{2} _{L^{2}_{b}}, $$
we obtain (3.13) immediately. □
Lemma 3.3
Under the assumptions of Lemma
3.2, for any
\(\tau \in \mathbb {R}\), any bounded subsets
\(B\subset H_{0}^{1}(\Omega )\)
and
\(I\subset [0, \infty )\), the following estimate holds true:
$$\begin{aligned} &\sup_{ g\in \Sigma }\bigl\Vert \nabla \bigl(U^{\varepsilon _{1}}_{g}(t, \tau )u_{1}-U^{\varepsilon _{2}}_{g}(t,\tau )u_{2} \bigr)\bigl\Vert ^{2}\leqslant Q\bigl( \bigr\Vert \nabla (u_{1}-u_{2}) \bigr\Vert ^{2}+\vert \varepsilon _{1}-\varepsilon _{2} \vert \bigr), \\ & \quad \forall t\geqslant \tau , \forall u_{1},u_{2}\in B, \forall \varepsilon _{1},\varepsilon _{2}\in I, \end{aligned}$$
where
Q
depends on
t, τ, \(\Vert g_{0} \Vert _{L^{2}_{b}}\), \(\Vert \partial _{t}g_{0} \Vert _{L^{2}_{b}}\), \(\vert I \vert \)
and
\(H_{0}^{1}\)-bounds of
B.
Proof
Let \(u_{i}(t)=U_{g}^{\varepsilon _{i}}(t,\tau )u_{i}\) be the solution of problem (1.1) with \(\varepsilon =\varepsilon _{i}\) and the initial data \(u_{i}(\tau )=u _{i}\) (\(i=1,2\)).
Set \(w(t)=u_{1}(t)- u_{2}(t)\), then the following equality holds true:
$$\begin{aligned} \partial _{t}w-\Delta w-\varepsilon _{2}\Delta \partial _{t}w-(\varepsilon _{1}-\varepsilon _{2})\Delta \partial _{t} u_{1}+f(u_{1})-f(u_{2})=0, \end{aligned}$$
(3.17)
with the initial data \(w(\tau )=u_{1}-u_{2}\).
Multiplying equation (3.17) by w and integrating over Ω, gives
$$\begin{aligned} &\frac{d}{dt}\bigl(\Vert w \Vert ^{2}+\varepsilon _{2}\Vert \nabla w \Vert ^{2}\bigr)+2\Vert \nabla w \Vert ^{2} \\ &\quad {}+2(\varepsilon _{1}-\varepsilon _{2})\langle -\Delta \partial _{t}u_{1},w\rangle +2\bigl\langle f(u _{1})-f(u_{2}),w \bigr\rangle =0. \end{aligned}$$
(3.18)
Observing that
$$\begin{aligned} \bigl\vert (\varepsilon _{1}-\varepsilon _{2})\langle -\Delta \partial _{t}u_{1},w\rangle \bigr\vert \leqslant \frac{(\varepsilon _{1}-\varepsilon _{2})^{2}}{2} \Vert \nabla \partial _{t}u_{1} \Vert ^{2}+ \frac{1}{2}\Vert \nabla w \Vert ^{2} \end{aligned}$$
(3.19)
and by (1.3), we have
$$\begin{aligned} \bigl\langle f(u_{1})-f(u_{2}),w\bigr\rangle \geqslant -l\Vert w \Vert ^{2} \end{aligned}$$
(3.20)
for some \(l\geqslant \lambda _{1}\).
Collecting (3.18)-(3.20), we arrive at
$$\begin{aligned} \frac{d}{dt}\bigl(\Vert w \Vert ^{2}+\varepsilon _{2} \Vert \nabla w \Vert ^{2}\bigr)\leqslant 2l\bigl(\Vert w \Vert ^{2}+\varepsilon _{2}\Vert \nabla w \Vert ^{2}\bigr)+( \varepsilon _{1}-\varepsilon _{2})^{2}\Vert \nabla \partial _{t}u_{1} \Vert ^{2}. \end{aligned}$$
Thus, by Gronwall’s inequality and noting that \(\lambda _{1}\Vert u \Vert ^{2} \leqslant \Vert \nabla u \Vert ^{2}\), we have
$$\begin{aligned} \bigl\Vert w(t) \bigr\Vert ^{2}+\varepsilon _{2} \bigl\Vert \nabla w(t) \bigr\Vert ^{2} \leqslant C \bigl( \Vert \nabla (u_{1}-u_{2})\bigr\Vert ^{2} +(\varepsilon _{1}-\varepsilon _{2})^{2}\int_{\tau }^{t} \bigl\Vert \nabla \partial _{t}u_{1}(s) \bigr\Vert ^{2}\,ds \biggr) , \end{aligned}$$
(3.21)
where C depends on t, τ and \(\vert I \vert \).
Now we divide the argument into two cases.
Case 1: \(\varepsilon _{1}\varepsilon _{2}\neq 0\). Without loss of generality, let \(\varepsilon _{1}\geqslant \varepsilon _{2}>0\), from Lemma 3.1 and (3.21), we readily get
$$\begin{aligned} &\sup_{ g\in \Sigma } \bigl\Vert \nabla \bigl(U^{\varepsilon _{1}}_{g}(t,\tau )u_{1}-U^{\varepsilon _{2}}_{g}(t, \tau )u_{2}\bigr) \bigr\Vert ^{2} \\ &\quad \leqslant C \biggl( \bigl\Vert \nabla (u_{1}-u_{2}) \bigr\Vert ^{2}+ 2(\varepsilon _{1}-\varepsilon _{2}) \cdot \varepsilon _{1} \int_{\tau }^{t}\bigl\Vert \nabla \partial _{t}u_{1}(s) \bigr\Vert ^{2}\,ds \biggr) \\ &\quad \leqslant Q\bigl(\bigl\Vert \nabla (u_{1}-u_{2}) \bigr\Vert ^{2}+(\varepsilon _{1}-\varepsilon _{2})\bigr), \end{aligned}$$
(3.22)
where Q depends on t, τ, \(\Vert g \Vert _{L^{2}_{b}}\), \(\Vert \partial _{t}g \Vert _{L^{2}_{b}}\) and \(H_{0}^{1}\)-bounds of B.
Case 2: \(\varepsilon _{1}\varepsilon _{2}=0\). Without loss of generality, let \(\varepsilon _{2}=0\), then (3.21) can be simplified as
$$\begin{aligned} \bigl\Vert w(t) \bigr\Vert ^{2}\leqslant C \biggl( \bigl\Vert \nabla (u_{1}-u_{2}) \bigr\Vert ^{2}+ \varepsilon ^{2} _{1} \int_{\tau }^{t}\bigl\Vert \nabla \partial _{t}u_{1}(s) \bigr\Vert ^{2}\,ds \biggr) . \end{aligned}$$
(3.23)
Multiplying (3.17) by w and using (3.20), we obtain
$$\begin{aligned} \bigl\Vert \nabla w(t) \bigr\Vert ^{2}\leqslant C \bigl( \bigl\Vert \partial _{t}w(t) \bigr\Vert \bigl\Vert w(t) \bigr\Vert +\varepsilon _{1} ^{2}\bigl\Vert \nabla \partial _{t}u_{1}(t) \bigr\Vert ^{2}+\bigl\Vert w(t) \bigr\Vert ^{2} \bigr) . \end{aligned}$$
(3.24)
Hence, on account of Lemmas 3.1, 3.2 and (3.23), similar to (3.22), we find
$$\begin{aligned} \sup_{ g\in \Sigma } \bigl\Vert \nabla \bigl(U^{\varepsilon _{1}}_{g}(t,\tau )u_{1}-U^{\varepsilon _{2}}_{g}(t, \tau )u_{2}\bigr)\bigl\Vert ^{2}\leqslant Q\bigl( \bigr\Vert \nabla (u_{1}-u_{2}) \bigr\Vert ^{2}+\varepsilon _{1}\bigr). \end{aligned}$$
(3.25)
Combining (3.22) and (3.25), we can get the expected result. □
Proof of Theorem 1.2
If (1.9) is not correct, we can find \(\delta >0\), \(\varepsilon _{0} \geqslant 0\) and \(\{\varepsilon _{n}\}_{n\in \mathbb{N}}\subset [0,\infty )\) with \(\varepsilon _{n}\to \varepsilon _{0}\) such that
$$\begin{aligned} \operatorname{dist}_{H_{0}^{1}}\bigl(\mathcal {A}^{\varepsilon _{n}}_{\Sigma }, \mathcal {A}^{\varepsilon _{0}}_{\Sigma }\bigr)\geqslant \delta ,\quad \forall n\in \mathbb{N}. \end{aligned}$$
Hence, there exists \(\{y_{n}\}_{n\in \mathbb{N}}\subset \mathcal {A}^{\varepsilon _{n}}_{\Sigma }\) such that
$$\begin{aligned} \operatorname{dist}_{H_{0}^{1}}\bigl(y_{n},\mathcal {A}^{\varepsilon _{0}}_{\Sigma }\bigr)\geqslant \delta ,\quad \forall n\in \mathbb{N}. \end{aligned}$$
(3.26)
Let \(\mathcal {B}\) be the uniformly (w.r.t. \(\sigma \in \Sigma \), \(\varepsilon \in [0, \infty )\)) absorbing set given by Corollary 3.1. Then we can choose \(m>0\) sufficiently large to guarantee that
$$\begin{aligned} \bigcup_{\varepsilon \geqslant 0}\bigcup _{g\in \Sigma }U_{g}^{\varepsilon }(t,0)\mathcal {B} \subset \mathcal {B}, \quad \forall t\geqslant m, \end{aligned}$$
(3.27)
and
$$\begin{aligned} \sup_{g\in \Sigma }\operatorname{dist}\bigl(U^{\varepsilon _{0}}_{g}(m,0) \mathcal {B},\mathcal {A} ^{\varepsilon _{0}}_{\Sigma }\bigr)\leqslant \frac{\delta }{4}. \end{aligned}$$
From Theorem 1.1 we know \(\mathcal {A}^{\varepsilon _{n}}_{\Sigma }=\omega ^{\varepsilon _{n}}_{0,\Sigma }(\mathcal {B})\) (\(n\in \mathbb {N}\)). Therefore, there exist sequences \(\{g_{n}\}_{n\in \mathbb{N}}\subset \Sigma \), \(\{x_{n}\} _{n\in \mathbb{N}}\subset \mathcal {B}\) and \(\{t_{n}\}_{n\in \mathbb{N}} \subset \mathbb {R}^{+}\) with \(t_{n}\to \infty \), without loss of generality, we let \(t_{n}\geqslant 2m\) satisfy
$$\begin{aligned} \bigl\Vert U^{\varepsilon _{n}}_{g_{n}}(t_{n},0)x_{n}-y_{n} \bigr\Vert _{H_{0}^{1}}\leqslant \frac{\delta }{4},\quad \forall n\in \mathbb{N}. \end{aligned}$$
Let \(\widetilde{x}_{n}=U_{g_{n}}^{\varepsilon _{n}}(t_{n}-m,0)x_{n}\) and \(g'_{n}=T(t _{n}-m)g_{n}\), by (1.6), (1.7), (3.27) and noticing that \(t_{n}\geqslant 2m\), we have
$$\begin{aligned} \{\widetilde{x}_{n}\}_{n\in \mathbb{N}}\subset \mathcal {B} \end{aligned}$$
and
$$\begin{aligned} U_{g_{n}}^{\varepsilon _{n}}(t_{n},0)x_{n} &=U_{g_{n}}^{\varepsilon _{n}}(t_{n},t_{n}-m)\widetilde{x} _{n}=U_{g_{n}'}^{\varepsilon _{n}}(m,0)\widetilde{x}_{n}. \end{aligned}$$
On the other hand, due to Lemma 3.3, we can choose \(N\in \mathbb{N}\) large enough such that
$$\begin{aligned} \bigl\Vert U^{\varepsilon _{N}}_{g_{N}'}(m,0)\widetilde{x}_{N}-U^{\varepsilon _{0}}_{g_{N}'}(m,0) \widetilde{x}_{N} \bigr\Vert \leqslant \frac{\delta }{4}. \end{aligned}$$
Therefore, from the above analysis we find
$$\begin{aligned} &\operatorname{dist}_{H_{0}^{1}}\bigl(y_{N}, \mathcal {A}^{\varepsilon _{0}}_{\Sigma }\bigr) \\ &\quad \leqslant \operatorname{dist}_{H_{0}^{1}} \bigl( y_{N}, U_{g_{N}'}^{\varepsilon _{N}}(m,0)\widetilde{x}_{N} \bigr) \\ &\quad \quad {} +\operatorname{dist}_{H_{0}^{1}} \bigl( U^{\varepsilon _{N}}_{g_{N}'}(m,0) \widetilde{x} _{N}, U^{\varepsilon _{0}}_{g_{N}'}(m,0)\widetilde{x}_{N} \bigr) \\ &\quad \quad {} +\operatorname{dist}_{H_{0}^{1}} \bigl( U^{\varepsilon _{0}}_{g_{N}'}(m,0) \widetilde{x} _{N}, U^{\varepsilon _{0}}_{g_{N}'}(m,0)\mathcal {B} \bigr) \\ &\quad \quad {} +\operatorname{dist}_{H_{0}^{1}} \bigl( U^{\varepsilon _{0}}_{g_{N}'}(m,0) \mathcal {B}, \mathcal {A}_{\Sigma }^{\varepsilon _{0}} \bigr) \\ &\quad \leqslant \frac{\delta }{4}+\frac{\delta }{4}+0+\frac{\delta }{4}= \frac{3\delta }{4}, \end{aligned}$$
which contradicts (3.26).
Next, we prove (1.10). If it is not correct, we can find \(\delta >0\), \(t_{0}\in \mathbb {R}\), \(g\in \Sigma \), \(\varepsilon _{0}\geqslant 0\) and \(\{\varepsilon _{n}\}_{n\in \mathbb{N}}\subset [0,\infty )\) with \(\varepsilon _{n} \to \varepsilon _{0}\) such that
$$\begin{aligned} \operatorname{dist}_{H_{0}^{1}}\bigl(\mathcal {K}^{\varepsilon _{n}}_{g}(t_{0}), \mathcal {K}^{\varepsilon _{0}} _{g}(t_{0})\bigr)\geqslant \delta ,\quad \forall n\in \mathbb{N}. \end{aligned}$$
(3.28)
Let \(\mathcal {K}^{\varepsilon }_{g}\) be the kernel of the process \(U^{\varepsilon }_{g}(t, \tau )\). By (3.28), for every \(n\in \mathbb {N}\), there exists a complete trajectory \(u_{n}(\cdot )\in \mathcal {K}_{g}^{\varepsilon _{n}}\) satisfying
$$\begin{aligned} \operatorname{dist}_{H_{0}^{1}}\bigl(u_{n}(t_{0}), \mathcal {K}^{\varepsilon _{0}}_{g}(t_{0})\bigr) \geqslant \delta ,\quad \forall n\in \mathbb{N}, \end{aligned}$$
(3.29)
and
$$\begin{aligned} u_{n}(t)=U^{\varepsilon _{n}}_{g}(t,\tau )u_{n}(\tau ),\quad \forall t \geqslant \tau , \forall \tau \in \mathbb {R}. \end{aligned}$$
(3.30)
For every \(s\in \mathbb {R}\), since \(u_{n}(s)\in \mathcal {K}^{\varepsilon _{n}}_{g}(s) \subset \mathcal {A}^{\varepsilon _{n}}_{\Sigma }\), by (1.8), (1.9) and the compactness of \(\mathcal {A}^{\varepsilon _{0}}_{\Sigma }\), there exists \(u(s)\in \mathcal {A}^{\varepsilon _{0}}_{\Sigma }\) such that
$$\begin{aligned} \bigl\Vert u_{n}(s)-u(s) \bigr\Vert _{H_{0}^{1}} \stackrel{n\to \infty }{\longrightarrow }0. \end{aligned}$$
(3.31)
Consequently, Lemma 3.3 yields
$$\begin{aligned} \bigl\Vert U^{\varepsilon _{n}}_{g}(t,\tau )u_{n}(\tau )-U^{\varepsilon _{0}}_{g}(t,\tau )u(\tau ) \bigr\Vert _{H_{0}^{1}}\stackrel{n\to \infty }{\longrightarrow }0,\quad \forall t \geqslant \tau , \forall \tau \in \mathbb {R}. \end{aligned}$$
(3.32)
Combining (3.31) and (3.32), taking \(n\to \infty \) in (3.30), we find
$$\begin{aligned} u(t)=U^{\varepsilon _{0}}_{g}(t,\tau )u(\tau ),\quad \forall t\geqslant \tau , \forall \tau \in \mathbb {R}, \end{aligned}$$
(3.33)
that is, \(u(\cdot )\in \mathcal {K}^{\varepsilon _{0}}_{g}\) and \(u(t)\in \mathcal {K}^{\varepsilon _{0}}_{g}(t)\) (\(\forall t\in \mathbb {R}\) ).
Hence
$$\begin{aligned} \operatorname{dist}_{H_{0}^{1}}\bigl(u_{n}(t_{0}), \mathcal {K}_{g}^{\varepsilon _{0}}(t_{0})\bigr) \leqslant \operatorname{dist}_{H_{0}^{1}}\bigl(u_{n}(t_{0}),u(t_{0}) \bigr)\stackrel{n \to \infty }{\longrightarrow }0, \end{aligned}$$
(3.34)
and this contradicts (3.29). The proof is completed. □