We will perform the last limit \(\varepsilon\rightarrow0\) in this section and assume that the initial function \(u_{0\varepsilon}\) converges to \(u_{0}\) strongly in \(L^{2}(\varOmega )\).
By letting \(\delta=0\) in the definition of \(\varPhi _{\delta}(\cdot)\), we can define \(\varPhi _{0}(\cdot)\) as
$$ \varPhi _{0}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{(1-n)(2-n)}x^{2-n}-\frac{1}{1-n}x+\frac{1}{2-n} & \mbox{if } n\in[0,2), n\neq1; \\ x\ln x-x+1 & \mbox{if } n=1. \end{array}\displaystyle \right . $$
Lemma 7
In the sense of
\(\mathcal{D}'(0, T)\), there exists a constant
\(C_{0}>0\)
such that
$$ \frac{\mathrm{d}}{\mathrm{d}t} \int_{\varOmega }\varPhi _{0}(\overline{u})\, \mathrm{d}x+C_{0} \int_{\varOmega }|\varDelta \overline{u}|^{2}\,\mathrm{d}x +\nu \int_{\varOmega }|\overline{u}_{t}|^{2}\, \mathrm{d}x \leq A \int_{\varOmega }\frac{|\nabla\overline{u}|^{2}}{(\overline {u}+\varepsilon)^{n+\alpha}(1+\varepsilon|\nabla\overline{u}|^{2})}\, \mathrm{d}x. $$
Proof
From the idea of (23) and the \(L^{p}\)-estimate, we get
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t} \int_{\varOmega }\varPhi \bigl(u(x,t)\bigr)\,\mathrm {d}x+C_{0} \int_{\varOmega }|\varDelta u|^{2}\,\mathrm{d}x+\nu \int_{\varOmega }|u_{t}|^{2}\, \mathrm{d}x \\& \quad \leq \frac{\mathrm{d}}{\mathrm{d}t} \int_{\varOmega }\varPhi \bigl(u(x,t)\bigr)\, \mathrm{d}x+ \int_{\varOmega }|w|^{2}\,\mathrm{d}x+\nu \int_{\varOmega }|u_{t}|^{2}\, \mathrm{d}x \\& \quad = A \int_{\varOmega }\frac{|\nabla u|^{2}}{(u_{+}+\varepsilon)^{n+\alpha }(1+\varepsilon|\nabla u|^{2})}\,\mathrm{d}x. \end{aligned}$$
(40)
Since \(u\rightarrow\overline{u}\) in \(C(\overline{Q}_{T})\) as \(\delta \rightarrow0\), we have
$$ - \int_{0}^{T}\phi'(t) \int_{\varOmega }\varPhi (u)\,\mathrm{d}x\,\mathrm{d}t \rightarrow- \int_{0}^{T}\phi'(t) \int_{\varOmega }\varPhi _{0}(\overline{u})\,\mathrm {d}x\, \mathrm{d}t $$
(41)
for any nonnegative function \(\phi\in\mathcal{D}'(0, T)\). By applying the limit \(\varDelta u\rightharpoonup \varDelta \overline{u}\) in \(L^{2}(Q_{T})\) as \(\delta\rightarrow0\), one has
$$ \liminf_{\delta\rightarrow0} \int_{0}^{T}\phi(t) \int_{\varOmega }|\varDelta u|^{2}\, \mathrm{d}x\,\mathrm{d}t \geq \int_{0}^{T}\phi(t) \int_{\varOmega }|\varDelta \overline{u}|^{2}\,\mathrm{d}x\, \mathrm{d}t. $$
(42)
Finally, it is easy to check that
$$\begin{aligned}& A \iint_{Q_{T}}\frac{|\nabla u|^{2}\phi(t)}{(u_{+}+\varepsilon)^{n+\alpha }(1+\varepsilon|\nabla u|^{2})}\,\mathrm{d}x\,\mathrm{d}t \\& \quad \rightarrow A \iint_{Q_{T}}\frac{|\nabla\overline{u}|^{2}\phi (t)}{(\overline{u}+\varepsilon)^{n+\alpha}(1+\varepsilon|\nabla \overline{u}|^{2})}\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(43)
Equations (40)-(43) give the result of this lemma. □
Lemma 8
If one of the following conditions holds:
-
(I)
\(\int_{\varOmega }\varPhi _{0}(w_{0})\,\mathrm{d}x<\infty\), \(A\leq 0\), and
-
(II)
\(\int_{\varOmega }\varPhi _{0}(w_{0})\,\mathrm{d}x<\infty\), \(\alpha \leq1-n\), \(n<1\), one has
\(\overline{u}\in L^{2}(0,T;H_{\mathrm{per}}^{2}(\varOmega ))\), \(\overline{w}, \overline{u}_{t}\in L^{2}(Q_{T})\)
independent of
ε.
Proof
By Lemma 7 and the condition (I), we can prove the result easily.
If the condition (II) holds, Lemma 1 and Lemma 7 give
$$\begin{aligned}& \int_{\varOmega }\varPhi _{0}(\overline{u})\, \mathrm{d}x+C_{0} \int_{\varOmega }|\varDelta \overline{u}|^{2}\,\mathrm{d}x +\nu \iint_{Q_{T}}|\overline{u}_{t}|^{2}\, \mathrm{d}x\,\mathrm{d}t \\& \quad \leq \int_{\varOmega }\varPhi _{0}(\overline{u}_{0})\, \mathrm{d}x+|A| \iint _{Q_{T}}\frac{|\nabla\overline{u}|^{2}}{(\overline{u}+\varepsilon )(1+\varepsilon|\nabla\overline{u}|^{2})}\,\mathrm{d}x\,\mathrm {d}t \\& \quad \leq \int_{\varOmega }\varPhi _{0}(\overline{u}_{0})\, \mathrm{d}x+|A| \iint _{Q_{T}}\frac{|\nabla\overline{u}|^{2}}{(\overline{u}+\varepsilon)^{\alpha +n}}\,\mathrm{d}x\,\mathrm{d}t \\& \quad \leq \int_{\varOmega }\varPhi _{0}(\overline{u}_{0})\, \mathrm{d}x +|A| \biggl( \iint_{Q_{T}}\frac{|\nabla\overline{u}|^{4}}{(\overline {u}+\varepsilon)^{2}}\,\mathrm{d}x\,\mathrm{d}t \biggr)^{\frac{1}{2}} \biggl( \iint_{Q_{T}}(\overline{u}+\varepsilon)^{2(1-(\alpha+n))}\,\mathrm {d}x\,\mathrm{d}t \biggr)^{\frac{1}{2}} \\& \quad \leq \int_{\varOmega }\varPhi _{0}(\overline{u}_{0})\, \mathrm{d}x +C \biggl( \int_{0}^{T}\|\overline{u}\|_{H^{2}(\varOmega )}\, \mathrm{d}t \biggr)^{\frac{1}{2}} \biggl( \iint_{Q_{T}}(\overline{u}+\varepsilon)\,\mathrm{d}x\,\mathrm {d}t \biggr)^{1-(\alpha+n)} \\& \quad \leq \frac{C_{0}}{2} \iint_{Q_{T}}|\varDelta \overline{u}|^{2}\,\mathrm{d}x\, \mathrm{d}t+C, \end{aligned}$$
(44)
which yields \(\varDelta \overline{u}\in L^{2}(Q_{T})\). Applying the second equation of Proposition 1, we get \(\overline{w}\in L^{2}(Q_{T})\). □
Now we are in the position to prove Theorem 1.
Proof of Theorem 1
By Lemma 8, we can show the existence of two functions \(u\geq 0\) and w such that, as \(\varepsilon\rightarrow0\),
$$\begin{aligned}& \overline{u}\rightharpoonup u\quad \mbox{in } L^{2}\bigl(0, T; H_{\mathrm {per}}^{2}(\varOmega )\bigr); \end{aligned}$$
(45)
$$\begin{aligned}& \overline{u}_{t}\rightharpoonup u_{t}\quad \mbox{in } L^{2}(Q_{T}); \end{aligned}$$
(46)
$$\begin{aligned}& \overline{w}\rightharpoonup w\quad \mbox{in } L^{2}(Q_{T}); \end{aligned}$$
(47)
$$\begin{aligned}& \overline{u}\rightarrow u\quad \mbox{in } C\bigl([0, T]; H_{\mathrm {per}}^{1}( \varOmega )\bigr); \end{aligned}$$
(48)
$$\begin{aligned}& \overline{u}\rightarrow u\quad \mbox{in } L^{2}\bigl(0, T; H_{\mathrm {per}}^{1}(\varOmega )\bigr); \end{aligned}$$
(49)
$$\begin{aligned}& \overline{u}\rightarrow u,\qquad \nabla\overline{u}\rightarrow\nabla u\quad \mbox{a.e. in } Q_{T}. \end{aligned}$$
(50)
Furthermore, Lemma 3 yields
$$\begin{aligned}& \|\overline{u}\|_{C([0, T]; H_{\mathrm{per}}^{s}(\varOmega )) }\leq C; \end{aligned}$$
(51)
$$\begin{aligned}& \|u\|_{C([0, T]; H_{\mathrm{per}}^{s}(\varOmega )) }\leq C \end{aligned}$$
(52)
for \(\frac{3}{2}< s<2\). By the Sobolev embedding theorem with the case \(N\leq3\), we have \(\|\overline{u}\|_{L^{\infty}(Q_{T})}\leq C\) and \(\|u\|_{L^{\infty }(Q_{T})}\leq C\).
Step 1. By using (51)-(52) and Vitali’s theorem, we get \(\overline{u}^{n}\rightarrow u^{n}\) in \(L^{q}(Q_{T})\) for any \(q>0\) and thus one has
$$ \iint_{Q_{T}}\overline{u}^{n}\overline{w}\varDelta \phi\, \mathrm{d}x\,\mathrm {d}t\rightarrow \iint_{Q_{T}}u^{n}w\varDelta \phi\,\mathrm{d}x\,\mathrm{d}t $$
(53)
as \(\varepsilon\rightarrow0\) for any test function \(\phi\in C^{\infty }([0, T]; C_{\mathrm{per}}^{2}(\overline{\varOmega }))\).
Step 2. In this step, we will prove the limit \(\overline {u}^{n-1}\nabla\overline{u}\rightarrow u^{n-1}\nabla u\) in \(L^{2}(Q_{T})\).
First of all, the Bernis inequality yields \(\iint_{Q_{T}}\vert \frac {\nabla\overline{u}}{\sqrt{\overline{u}}}\vert ^{4}\,\mathrm{d}x\, \mathrm{d}t\leq C\) and then we have
$$\begin{aligned} \iint_{\varDelta _{0}}\overline{u}^{n-1}|\nabla \overline{u}|^{2}\,\mathrm {d}x\,\mathrm{d}t =& \iint_{\varDelta _{0}}\overline{u}^{2n-1}\frac{|\nabla\overline{u}|^{2}}{\sqrt {\overline{u}}}\, \mathrm{d}x\,\mathrm{d}t \\ \leq& C \biggl( \iint_{\varDelta _{0}}\overline{u}^{4n-2}\,\mathrm{d}x\,\mathrm {d}t \biggr)^{\frac{1}{2}} \rightarrow0 \end{aligned}$$
(54)
as \(\varepsilon\rightarrow0\) with \(\varDelta _{0}=\{(x, t)\in Q_{T}|u(x, t)=0\} \). On the other hand, it is easy to get
$$\frac{\nabla\overline{u}}{\sqrt{\overline{u}}}\rightarrow\frac{\nabla u}{\sqrt{u}}\quad \mbox{a.e. in } Q_{T}\backslash \varDelta _{0} $$
as \(\varepsilon\rightarrow0\). By Vitali’s theorem, we have
$$ \overline{u}^{n-1}\nabla\overline{u}\rightarrow u^{n-1}\nabla u \quad \mbox{in } L^{2}(Q_{T} \backslash \varDelta _{0}). $$
(55)
Hence, we have
$$ \overline{u}^{n-1}\nabla\overline{u}\rightarrow u^{n-1}\nabla u \quad \mbox{in } L^{2}(Q_{T}), $$
(56)
where we define \(u^{n-1}\nabla u=0\) on \(\varDelta _{0}\).
Step 3. In this step, we prove the limit \(F_{\varepsilon}=\frac{\overline{u}^{n}\nabla\overline{u}}{(\overline {u}+\varepsilon)^{n+\alpha}(1+\varepsilon|\nabla\overline {u}|^{2})}\rightarrow u^{-\alpha}\nabla u\) in \(L^{2}(Q_{T})\).
If \(\alpha\leq\frac{1}{2}\), we have
$$\begin{aligned} \iint_{\varDelta _{0}}|F_{\varepsilon}|^{2}\,\mathrm{d}x\, \mathrm{d}t \leq& \iint_{\varDelta _{0}}\overline{u}^{1-2\alpha}\frac{|\nabla\overline {u}|^{2}}{\overline{u}}\, \mathrm{d}x\,\mathrm{d}t \\ \leq& C \biggl( \iint_{\varDelta _{0}}\overline{u}^{2-4\alpha}\,\mathrm{d}x\, \mathrm{d}t \biggr)^{\frac{1}{2}} \rightarrow0 \end{aligned}$$
(57)
as \(\varepsilon\rightarrow0\). Beside, it is easy to show \(F_{\varepsilon}\rightarrow u^{-\alpha }\nabla u\) a.e. in \(Q_{T}\backslash \varDelta _{0}\) and Vitali’s theorem yields
$$ \iint_{\varDelta _{0}}\bigl\vert F_{\varepsilon}-u^{-\alpha} \nabla u\bigr\vert ^{2}\,\mathrm{d}x\, \mathrm{d}t\rightarrow0 $$
(58)
as \(\varepsilon\rightarrow0\). By (57)-(58), we have
$$ F_{\varepsilon}\rightarrow u^{-\alpha}\nabla u \quad \mbox{in } L^{2}(Q_{T}), $$
(59)
where we define \(u^{-\alpha}\nabla u=0\) on \(\varDelta _{0}\).
As \(\varepsilon\rightarrow0\), the convergence (56) and (46)-(47) give \(\iint_{Q_{T}}\overline{u}_{t}\phi\,\mathrm{d}x\,\mathrm{d}t\rightarrow \iint_{Q_{T}}u_{t}\phi\,\mathrm{d}x\,\mathrm{d}t\) and \(\iint_{Q_{T}}\overline{u}^{n-1}\nabla\overline{u}\overline{w}\nabla\phi \,\mathrm{d}x\,\mathrm{d}t \rightarrow\iint_{Q_{T}}u^{n-1}\nabla u w\nabla\phi\,\mathrm{d}x\, \mathrm{d}t\). Step 3 yields
$$\iint_{Q_{T}}\frac{\overline{u}^{n}\nabla\overline{u}\nabla\phi }{(\overline{u}+\varepsilon)^{n+\alpha}(1+\varepsilon|\nabla\overline {u}|^{2})}\,\mathrm{d}x\,\mathrm{d}t \rightarrow \iint_{Q_{T}}u^{-\alpha}\nabla u\nabla\phi\,\mathrm{d}x\, \mathrm{d}t. $$
Now we can take the limit \(\varepsilon\rightarrow0\) in the equality
$$\begin{aligned}& \iint_{Q_{T}}\overline{u}_{t}\phi\,\mathrm{d}x\, \mathrm{d}t+ \iint _{Q_{T}}\overline{u}^{n} \overline{w}\varDelta \phi\, \mathrm{d}x\,\mathrm{d}t \\& \quad {} +n \iint_{Q_{T}}\overline{u}^{n-1}\nabla\overline{u} \overline{w}\nabla \phi\,\mathrm{d}x\,\mathrm{d}t -A \iint_{Q_{T}}\frac{\overline{u}^{n}\nabla\overline{u}\nabla\phi }{(\overline{u}+\varepsilon)^{n+\alpha}(1+\varepsilon|\nabla\overline {u}|^{2})}\,\mathrm{d}x\,\mathrm{d}t=0, \\& \iint_{Q_{T}}\overline{w}\phi\,\mathrm{d}x\,\mathrm{d}t=- \iint _{Q_{T}}\varDelta \overline{u}\phi\,\mathrm{d}x\,\mathrm{d}t + \nu \iint_{Q_{T}}\overline{u}_{t}\phi\,\mathrm{d}x\, \mathrm{d}t \end{aligned}$$
for any test function \(\phi\in C([0, T]; C_{\mathrm{per}}^{2}(\overline {\varOmega }))\). For the initial value, this holds in the sense of \(u\in C([0, T]; H_{\mathrm{per}}^{1}(\varOmega ))\). □