Theorem 4.1
Under the same assumptions as in Theorem
3.1, there exist
\(u^{\star}=(u_{1}^{\star},u_{2}^{\star })\in V_{z}\)
and
\(p^{\star}\in L_{0}^{2}(\Omega)\), such that
$$\begin{aligned} &\hat{u}_{i}^{\varepsilon} \rightharpoonup u_{i}^{\star} \quad( 1\leq i\leq2 ) \textit{ weakly in} V_{z}, \end{aligned}$$
(4.1)
$$\begin{aligned} &\varepsilon\frac{\partial\hat{u}_{i}^{\varepsilon}}{\partial x_{j}} \rightharpoonup0 \quad( 1\leq i,j\leq2 ) \textit{ weakly in }L^{2} ( \Omega ) , \end{aligned}$$
(4.2)
$$\begin{aligned} &\varepsilon\frac{\partial\hat{u}_{3}^{\varepsilon}}{\partial z} \rightharpoonup0 \quad\textit{weakly in }L^{2} ( \Omega ) , \end{aligned}$$
(4.3)
$$\begin{aligned} &\varepsilon^{2}\frac{\partial\hat{u}_{3}^{\varepsilon}}{\partial x_{i}} \rightharpoonup0\quad ( 1\leq i\leq2 ) \textit{ weakly in }L^{2} ( \Omega ) , \end{aligned}$$
(4.4)
$$\begin{aligned} &\varepsilon\hat{u}_{3}^{\varepsilon} \rightharpoonup 0\quad \textit{weakly in }L^{2} ( \Omega ) , \end{aligned}$$
(4.5)
$$\begin{aligned} &\widehat{p}^{\varepsilon} \rightharpoonup p^{\star} \quad\textit{weakly in }L^{2} ( \Omega ) , \textit{ depend only of }x^{\prime }. \end{aligned}$$
(4.6)
For the proof of this theorem, we follow the same steps as in [6].
Theorem 4.2
With the same assumptions of Theorem
3.1, the solution
\((u^{\star},p^{\star})\)
satisfies the following relations:
$$\begin{aligned} &\sum_{i=1} ^{{2} } \int _{\Omega}\frac{\partial u_{i}^{\star}}{\partial z}\frac{\partial}{\partial z} \bigl(\hat{ \phi}_{i}-u_{i}^{\star} \bigr) \,dx^{\prime}\,dz- \int _{\Omega}p^{\star} \bigl( x^{\prime } \bigr) \biggl( \frac{\partial\hat{\phi}_{1}}{\partial x_{1}}+\frac{\partial\hat{\phi}_{2}}{\partial x_{2}} \biggr) \,dx^{\prime }\,dz \\ &\qquad{}- \int _{\omega}p^{\star} \bigl( x^{\prime} \bigr) \biggl[ \hat{\phi }_{1} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \frac{\partial h ( x^{\prime} ) }{\partial x_{1}}+\hat{\phi}_{2} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) \frac{\partial h ( x^{\prime} ) }{\partial x_{2}} \biggr] \,dx^{\prime} \\ &\qquad{}+ \sum_{i=1}^{{2}}\widehat{l} \int _{\omega}u_{i}^{\star } \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigl[ \hat{ \phi}_{i} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) -u_{i}^{\star} \bigl( x^{\prime },h \bigl(x^{\prime } \bigr) \bigr) \bigr] \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl( \vert \hat{\phi} \vert - \bigl\vert u^{\star} \bigr\vert \bigr) \,dx^{\prime}+\widehat{ \alpha}\int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2}} \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\qquad{}-\widehat{\alpha} \int _{\Omega} \Biggl(\frac{1}{2}\sum _{i=1} ^{{2}} \biggl( \frac{\partial u_{i}^{\star}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\quad \geq\sum _{i=1}^{{2}} \bigl(\hat{f}_{i},\hat{ \phi}_{i}-u_{i}^{\star} \bigr), \quad\forall\hat { \phi}\in \Pi(V). \end{aligned}$$
(4.7)
Proof
As \(\int_{\omega}\widehat{l}\varepsilon ^{2}\hat {u}_{3}^{\varepsilon}(x^{\prime},h(x^{\prime})\hat{u}_{3}^{\varepsilon }(x^{\prime},h(x^{\prime})))\sqrt{1+ \vert \nabla h(x^{\prime}) \vert ^{2}}\,dx^{\prime }\geq0\), then applying \(\lim_{\varepsilon\rightarrow0}\inf\) in the left part of (3.2) and the \(\lim_{\varepsilon \rightarrow0}\) in the right part of (3.2) and from the convergence results of Theorem 4.1, we obtain
$$\begin{aligned} &\sum_{i=1}^{{2}} \int _{\Omega}\frac{\partial u_{i}^{\star}}{\partial z}\frac{\partial u_{i}^{\star}}{\partial z}\,dx^{\prime}\,dz+ \sum_{i=1}^{2} \widehat{l} \int _{\omega }u_{i}^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) u_{i}^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl\vert u^{\star} \bigr\vert \,dx^{\prime }+\widehat{\alpha} \int _{\Omega} \Biggl(\frac{1}{2}\sum _{i=1}^{{2}} \biggl( \frac{\partial u_{i}^{\star}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\quad \leq\sum _{i=1}^{ {2} } \int _{\Omega}\frac{\partial u_{i}^{\star}}{\partial z}\frac {\partial\hat{\phi}_{i}}{\partial z} \,dx^{\prime}\,dz \\ &\qquad{}+\sum_{i=1} ^{{2} }\widehat{l} \int _{\omega}u_{i}^{\star } \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \hat{\phi}_{i} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) \,dx^{\prime}+ \int _{\Omega}p^{\star} \bigl( x^{\prime} \bigr) \frac{\partial\hat{\phi}_{3}}{\partial z}\,dx^{\prime }\,dz \\ &\qquad{}+ \int _{\omega}\hat{k} \vert \hat{\phi} \vert \,dx^{\prime }+ \widehat{\alpha} \int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2}} \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\qquad{}+ \int _{\Omega}p^{\star} \bigl( x^{\prime} \bigr) \biggl( \frac {\partial\hat{\phi}_{1}}{\partial x_{1}}+\frac{\partial\hat{\phi}_{2} }{\partial x_{2}} \biggr) \,dx^{\prime}\,dz+ \sum _{i=1}^{{2}} \int _{\Omega}\hat{f}_{i} \bigl( \hat{ \phi}_{i}-u_{i}^{\star} \bigr) \,dx^{\prime}\,dz \end{aligned}$$
(4.8)
as \(\int _{\Omega}p^{\star} ( x^{\prime} ) \frac {\partial\hat{\phi}_{3}}{\partial z}\,dx^{\prime}\,dz=\int _{\omega}p^{\star} ( x^{\prime} ) \hat{\phi}_{3} ( x^{\prime },h(x^{\prime}) ) \,dx^{\prime}\) and \(\hat{\phi}_{3}=\hat{\phi}_{1}\frac{\partial h}{\partial x_{1}}+\hat{\phi}_{2}\frac{\partial h}{\partial x_{2}}\), we have
$$ \int _{\Omega}p^{\star} \bigl( x^{\prime} \bigr) \frac{\partial \hat{\phi}_{3}}{\partial z}\,dx^{\prime}\,dz= \int _{\omega}p^{\star} \bigl( x^{\prime} \bigr) \biggl( \hat{\phi}_{1}\frac{\partial h}{\partial x_{1}}+\hat{\phi}_{2} \frac{\partial h}{\partial x_{2}} \biggr) \,dx^{\prime } $$
(4.9)
and from (4.8)-(4.9), we deduce (4.7). □
Remark 4.1
If ϕ̂ satisfies the condition \(( D^{\prime} ) \), the inequality (4.7) is reduced as follows:
$$\begin{aligned} &\sum_{i=1}^{2} \int _{\Omega}\frac{\partial u_{i}^{\star}}{\partial z} \biggl( \frac{\partial\hat{\phi }_{i}}{\partial z}- \frac{\partial u_{i}^{\star}}{\partial z} \biggr) \,dx^{\prime}\,dz+\sum _{i=1}^{{2} }\widehat{l} \int _{\omega}u_{i}^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigl( \hat{ \phi}_{i} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) -u_{i}^{\star} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) \bigr) \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl( \vert \hat{\phi} \vert - \bigl\vert u^{\star} \bigr\vert \bigr) \,dx^{\prime}+\widehat{\alpha} \int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\qquad -\widehat{\alpha} \int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2}} \biggl( \frac{\partial u_{i}^{\star}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime }\,dz \\ &\quad{}\geq\sum^{2}_{i=1} \int _{\Omega}\hat{f}_{i} \bigl( \hat{\phi }_{i}-u_{i}^{\star} \bigr) \,dx^{\prime}\,dz,\quad \forall\hat{\phi }\in \sum ( V ) . \end{aligned}$$
(4.10)
Theorem 4.3
The limit functions
\(u^{\star}\)
and
\(p^{\star}\)
satisfy
$$\begin{aligned} &p^{\star} ( x_{1},x_{2},z ) =p^{\star} ( x_{1},x_{2} ) \quad\textit{in }\Omega\quad\textit{and}\quad p^{\star }\in H^{1} ( \omega ) , \end{aligned}$$
(4.11)
$$\begin{aligned} &{-}\mu\frac{\partial^{2}u_{i}^{\star}}{\partial z^{2}}+\frac{\partial p^{\star}}{\partial x_{i}} =\hat{f}_{i},\quad i=1,2\textit{ in }L^{2} ( \Omega ) . \end{aligned}$$
(4.12)
Proof
Choosing in (3.2) \(\hat{\phi}_{i}=\hat{u}_{i}^{\varepsilon}\) for \(i=1,2\), \(\hat{\phi}_{3}=\hat {u}_{3}^{\varepsilon}\pm\psi\) with \(\psi\in H_{0}^{1} ( \Omega ) \), we obtain
$$\begin{aligned} &\sum_{j=1} ^{{2} } \int _{\Omega}\mu\varepsilon ^{2} \biggl( \varepsilon^{2}\frac{\partial\hat{u}_{3}^{\varepsilon}}{\partial x_{j}}+\frac{\partial\hat{u}_{j}^{\varepsilon}}{\partial z} \biggr) \frac {\partial\psi}{\partial x_{j}}\,dx^{\prime}\,dz+ \int _{\Omega } \biggl( 2\mu\varepsilon^{2} \frac{\partial\hat{u}_{3}^{\varepsilon }}{\partial z}-\widehat{p}^{\varepsilon} \biggr) \frac{\partial\psi}{\partial z} \,dx^{\prime}\,dz \\ &\qquad{}+\overset{2}{\sum_{i=1}} \int _{\Omega}\varepsilon^{4}\hat{u}_{i}^{\varepsilon} \frac{\partial\hat{u}_{3}^{\varepsilon}}{\partial x_{i}}\psi \,dx^{\prime}\,dz+ \int _{\Omega}\varepsilon^{4}\hat{u}_{3}^{\varepsilon} \frac{\partial\hat{u}_{3}^{\varepsilon}}{\partial z}\psi \,dx^{\prime}\,dz\\ &\quad = \int_{\Omega}\varepsilon\hat{f}_{3}\psi \,dx^{\prime}\,dz. \end{aligned}$$
By the results of convergence (4.1)-(4.7), we deduce
$$ \int _{\Omega}p^{\star}\frac{\partial\psi}{\partial z}\,dx^{\prime } \,dz=0, \quad\forall\psi\in H_{0}^{1} ( \Omega ) , $$
(4.13)
then
$$ \frac{\partial p^{\star}}{\partial z}=0\quad\text{in }H^{-1} ( \Omega ) . $$
(4.14)
Choosing now \(\hat{\phi}_{i}=\hat{u}_{i}^{\varepsilon}\pm\psi_{i}\) for \(i=1,2\), with \(\psi_{i}\in H_{0}^{1} ( \Omega ) \) and \(\hat {\phi}_{3}=\hat{u}_{3}^{\varepsilon}\) in (3.2) we obtain
$$\begin{aligned} &\sum_{1\leq i,j\leq2} \int _{\Omega} \biggl[ \mu\varepsilon ^{2} \biggl( \frac{\partial\hat{u}_{i}^{\varepsilon}}{\partial x_{j}}+\frac{\partial\hat{u}_{j}^{\varepsilon}}{\partial x_{i}} \biggr) -\delta_{ij} \widehat{p}^{\varepsilon} \biggr] \frac{\partial\psi_{i}}{\partial x_{j}}\,dx^{\prime}\,dz \\ &\qquad{}+\mu\sum_{i=1}^{{2}} \int _{\Omega} \biggl( \frac {\partial\hat{u}_{i}^{\varepsilon}}{\partial z}+\varepsilon^{2} \frac {\partial\hat{u}_{3}^{\varepsilon}}{\partial x_{i}} \biggr) \frac {\partial\psi_{i}}{\partial z}\,dx^{\prime}\,dz+\sum _{1 \leq i,j\leq 2} \int _{\Omega}\varepsilon^{2}\hat{u}_{i}^{\varepsilon} \frac {\partial\hat{u}_{j}^{\varepsilon}}{\partial x_{i}}\psi_{j}\,dx^{\prime }\,dz \\ &\qquad{}+\overset{2}{\sum_{i=1}} \int _{\Omega}\varepsilon^{2}\hat{u}_{3}^{\varepsilon} \frac{\partial\hat{u}_{i}^{\varepsilon}}{\partial z}\psi_{i}\,dx^{\prime}\,dz\\ &\quad = \sum _{i=1}^{{2} } \int_{\Omega }\hat {f}_{i}\psi_{i} \,dx^{\prime}\,dz. \end{aligned}$$
Using (4.1)-(4.7), and choosing \(\psi_{1}=0\) and \(\psi_{2}\in H_{0}^{1} ( \Omega ) \), then \(\psi_{2}=0\) and \(\psi_{1}\in H_{0}^{1} ( \Omega ) \), we obtain
$$ \sum_{i=1}^{-{2} } \int _{\Omega}p^{\star}\frac {\partial\psi_{i}}{\partial x_{i}}\,dx^{\prime} \,dz+ \mu\sum_{i=1}^{{2}} \int _{\Omega}\frac{\partial u_{i}^{\star}}{\partial z}\frac{\partial\psi_{i}}{\partial z} \,dx^{\prime}\,dz= \sum_{i=1}^{{2} } \int_{\Omega}\hat{f}_{i}\psi_{i} \,dx^{\prime}\,dz. $$
(4.15)
By utilizing the Green formula, we obtain (4.12) but in \(H^{-1} ( \Omega ) \).
To prove \(p^{\star}\in H^{1} ( \omega ) \), we see from (4.14) that \(p^{\star}\) does not depend on z, then following [23], we choose \(\psi_{i}\) in (4.15) such that \(\psi_{i} ( x^{\prime},z ) =z ( z-h(x^{\prime}) ) \theta ( x^{\prime} ) \) with \(\theta\in H_{0}^{1} ( \omega ) \), and using the Green formula, we obtain
$$\frac{1}{6} \int _{\omega}p^{\star}\frac{\partial(h^{3}\theta )}{\partial x_{i}} \,dx^{\prime}-2 \int _{\omega}h\tilde{u}_{i}^{\star}\theta \,dx^{\prime}+ \int _{\omega}h \bigl(x^{\prime} \bigr) \bigl[ u_{i}^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) +u_{i}^{\star} \bigl( x^{\prime},0 \bigr) \bigr] \theta \,dx^{\prime}= \int _{\omega}\widetilde{f_{i}}\theta \,dx^{\prime}, $$
where
$$\tilde{u}_{i}^{\star}=\frac{1}{h ( x^{\prime} ) }{ \int _{0}^{h ( x^{\prime} ) }} u_{i}^{\star} \bigl( x^{\prime},z \bigr) \,dz\quad\text{and}\quad\widetilde {f}_{i}={ \int _{0}^{h ( x^{\prime} ) }} z \bigl( z-h \bigl(x^{\prime} \bigr) \bigr) \hat{f}_{i} \bigl( x^{\prime},z \bigr) \,dz; $$
so
$$ h \bigl(x^{\prime} \bigr) \bigl[ u_{i}^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) +u_{i}^{\star} \bigl( x^{\prime},0 \bigr) \bigr] -2h\tilde {u}_{i}^{\star }- \frac{1}{6}h^{3}\frac{\partial p^{\star}}{\partial x_{i}}=\widetilde {f}_{i},\quad i=1,2\text{ in }H^{-1} ( \omega ) $$
(4.16)
as \(\hat{f}_{i}\in L^{2} ( \Omega ) \), \(u_{i}^{\star}\in V_{z}\) in particular in \(L^{2} ( \omega ) \), therefore \(\tilde {u}_{i}^{\star }\) and \(\widetilde{f}_{i}\) are in \(L^{2} ( \omega ) \). Then from (4.16) we get (4.11). As \(\hat{f}_{i}\in L^{2} ( \omega ) \), from (4.16) we have \(\frac{\partial^{2}u_{i}^{\star}}{\partial z^{2}}\) in \(L^{2} ( \Omega ) \); then (4.12) holds. □
Theorem 4.4
Under the assumptions of the preceding theorems, the traces
\(s^{\star}\), \(\tau^{\star}\)
satisfy the following equality:
$$ \int _{\omega} \biggl( \frac{h^{3}}{12\mu}\nabla p^{\star}- \frac{h}{2}s^{\star}-\frac{h}{2}s_{h}^{\star}+ \widetilde{F} \biggr) \nabla\hat {\phi }\,dx^{\prime}+ \int _{\partial\omega}h\widetilde{u}^{\star}\phi \cdot n\,d\sigma=0, \quad \forall\hat{\phi}\in H^{1} ( \omega ) , $$
(4.17)
where
$$\begin{aligned} &s^{\star} \bigl( x^{\prime} \bigr) =u^{\star} \bigl( x^{\prime},0 \bigr) ,\qquad s_{h}^{\star} \bigl( x^{\prime} \bigr) =u^{\star} \bigl( x^{\prime },h \bigl( x^{\prime} \bigr) \bigr) ,\\ & \tau^{\star}=\frac {\partial u^{\star}}{\partial z} \bigl( x^{\prime},0 \bigr) ,\qquad \tau _{h}^{\star}= \frac{\partial u^{\star}}{\partial z} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr), \\ &\widetilde{g} \bigl( x^{\prime} \bigr) ={ \int _{0}^{h ( x^{\prime} ) }} \widehat{g} \bigl( x^{\prime},z \bigr) \,dz, \quad\forall x^{\prime}\in \partial \omega, \\ &F_{i} \bigl( x^{\prime},z \bigr) ={ \int _{0}^{z}} { \int _{0}^{\zeta}} \hat{f}_{i} \bigl( x^{\prime},z \bigr) \,dt\,d\zeta,\qquad \widetilde{F}_{i} \bigl( x^{\prime} \bigr) =\frac{1}{\mu}{ \int _{0}^{h ( x^{\prime} ) }} F_{i} \bigl( x^{\prime},z \bigr) \,dz-\frac{h}{2\mu}F \bigl(x^{\prime },h \bigl(x^{\prime } \bigr) \bigr). \end{aligned}$$
Proof
Integrating twice (4.12) between 0 and z, we obtain
$$ \mu u^{\star} \bigl( x^{\prime},z \bigr) =\mu s^{\star} \bigl( x^{\prime } \bigr) +\frac{z^{2}}{2}\nabla p^{\star}+\mu z \tau^{\star}-F \bigl( x^{\prime},z \bigr) . $$
(4.18)
Replacing z by h, we obtain
$$\mu u^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) = \mu s^{\star} \bigl( x^{\prime} \bigr) +\frac{h^{2}}{2}\nabla p^{\star}+\mu h\tau^{\star }-F \bigl( x^{\prime},h \bigr) $$
so
$$ h\tau^{\star}=\frac{1}{\mu}F \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) +s_{h}^{\star} \bigl( x^{\prime} \bigr) -s^{\star} \bigl( x^{\prime } \bigr) - \frac{h^{2}}{2\mu}\nabla p^{\star}; $$
(4.19)
integrating (4.18) with respect to z, in the interval \((0,h(x^{\prime}))\), we obtain
$$ h\widetilde{u}^{\star} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) =hs^{\star } \bigl( x^{\prime} \bigr) + \frac{h^{3}}{6\mu} \nabla p^{\star}+\frac {h^{2}}{2}\tau^{\star}- \frac{1}{\mu}{ \int _{0}^{z}} F \bigl( x^{\prime},z \bigr) \,dz $$
(4.20)
we set for any function φ
$$\widetilde{\varphi} \bigl( x^{\prime} \bigr) =\frac{1}{h(x^{\prime})}{ \int _{0}^{h ( x^{\prime} ) }} \varphi \bigl( x^{\prime},z \bigr) \,dz, \quad\forall x^{\prime} \in \omega, $$
on the other hand, \(\forall\phi\in H^{1}(\omega)\),
$$\begin{aligned} { \int _{\Omega}} \phi\operatorname{div} \bigl( \hat{u}^{\varepsilon} \bigr)\,dx^{\prime}\,dz & =0={ \int _{\omega}} \phi \bigl( x^{\prime} \bigr) { \int _{0}^{h(x^{\prime})}} { \sum_{i=1}^{2}} \biggl( \frac{\partial\hat{u}_{i}^{\varepsilon}}{\partial x_{i}}+\frac{\partial\hat{u}_{3}^{\varepsilon}}{\partial z} \biggr) \,dz\,dx^{\prime} \\ & ={ \int _{\omega}} \phi \bigl( x^{\prime} \bigr) \Biggl[ { \sum_{i=1}^{2}} \frac{\partial ( h\widetilde{\hat{u}}_{i}^{\varepsilon} ) }{\partial x_{i}}+\hat{u}_{3}^{\varepsilon} \bigl( x^{\prime },h \bigl(x^{\prime } \bigr) \bigr) -\hat{u}_{3}^{\varepsilon} \bigl(x^{\prime},0 \bigr) \Biggr] \,dx^{\prime}, \end{aligned}$$
we have \(\hat{u}_{3}^{\varepsilon}(x^{\prime},0)=0\) in ∂Ω. Then
$${ \int _{\omega}} \phi \bigl( x^{\prime} \bigr) { \sum_{i=1}^{2}} \frac{\partial ( h\widetilde{\hat{u}}_{i}^{\varepsilon} ) }{\partial x_{i}}\,dx^{\prime}=0.$$
Using the Green formula, we get
$${ \sum_{i=1}^{2}} { \int _{\omega}} \frac{\partial\phi}{\partial x_{i}}\widetilde{ \hat{u}}_{i}^{\varepsilon }h\,dx^{\prime}={ \sum_{i=1}^{2}} { \int _{\partial\omega}} h\widetilde{\hat{u}}_{i}^{\varepsilon} \phi\cdot n_{i}\,d\sigma.$$
As \(\hat{u}_{i}^{\varepsilon}\rightharpoonup u^{\star}\) in \(V_{z}\) and consequently in \(L^{2} ( \omega ) \), therefore \(\widetilde {\hat {u}_{i}^{\varepsilon}}\rightharpoonup\widetilde{u}_{i}^{\star}\) in \(L^{2} ( \omega ) \), we deduce
$$ { \sum_{i=1}^{2}} { \int _{\omega}} \frac{\partial\phi}{\partial x_{i}}\widetilde{u}_{i}^{\star }h \,dx^{\prime}={ \sum _{i=1}^{2}} { \int _{\partial\omega}} h\widetilde{u}_{i}^{\star} \phi\cdot n_{i}\,d\sigma,\quad \forall\phi \in H^{1}( \omega), $$
(4.21)
using (4.19) to eliminate the term containing \(\tau^{\star}\) from (4.20). Multiplying now (4.20) by ∇ϕ, then integrating it in ω and using (4.21), we obtain (4.16). □
Theorem 4.5
The solution
\((u^{\star },p^{\star})\)
of our limit problem is unique.
Proof
Let \((U^{1},p^{1})\), \((U^{2},p^{2})\) be two solutions of the limit problem (4.10), then
$$\begin{aligned} &\sum_{i=1} ^{{2} } \int _{\Omega}\frac{\partial U_{i}^{1}}{\partial z} \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z}- \frac{\partial U_{i}^{1}}{\partial z} \biggr) \,dx^{\prime}\,dz \\ &\qquad{}+\sum _{i=1}^{{2} }\widehat{l} \int _{\omega}U_{i}^{1} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigl( \hat{ \phi}_{i} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) -U_{i}^{1} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigr) \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl( \vert \hat{\phi} \vert - \bigl\vert U^{1} \bigr\vert \bigr) \,dx^{\prime}+\widehat{ \alpha}\int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1} ^{{2} } \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\qquad{}-\widehat{\alpha} \int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac{\partial U_{i}^{1}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\quad\geq\sum_{i=1} ^{{2} } \int _{\Omega}\hat{f}_{i} \bigl( \hat{ \phi}_{i}-U_{i}^{1} \bigr) \,dx^{\prime}\,dz, \quad \forall\hat{\phi}\in\sum ( V ) , \end{aligned}$$
(4.22)
and
$$\begin{aligned} &\sum_{i=1}^{{2} } \int _{\Omega}\frac{\partial U_{i}^{2}}{\partial z} \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z}- \frac{\partial U_{i}^{2}}{\partial z} \biggr) \,dx^{\prime}\,dz \\ &\qquad{}+\sum _{i=1}^{{2}}\widehat{l} \int _{\omega}U_{i}^{2} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigl( \hat{ \phi}_{i} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) -U_{i}^{2} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigr) \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl( \vert \hat{\phi} \vert - \bigl\vert U^{2} \bigr\vert \bigr) \,dx^{\prime}+\widehat{ \alpha}\int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac{\partial\hat{\phi}_{i}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\qquad{}-\widehat{\alpha} \int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac{\partial U_{i}^{2}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\quad\geq\sum_{i=1}^{{2}} \int _{\Omega}\hat{f}_{i} \bigl( \hat{ \phi}_{i}-U_{i}^{2} \bigr) \,dx^{\prime}\,dz,\quad \forall\hat{\phi}\in\sum ( V ) . \end{aligned}$$
(4.23)
Taking \(\hat{\phi}=U^{2}\) in (4.22) and \(\hat{\phi}=U^{1}\) in (4.23), we obtain
$$\begin{aligned} &\sum_{i=1}^{{2}} \int _{\Omega}\frac{\partial U_{i}^{1}}{\partial z} \biggl( \frac{\partial U_{i}^{2}}{\partial z}- \frac {\partial U_{i}^{1}}{\partial z} \biggr) \,dx^{\prime}\,dz \\ &\qquad{}+\sum _{i=1}^{{2} }\widehat{l} \int _{\omega}U_{i}^{1} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) \bigl( U_{i}^{2} \bigl( x^{\prime},h \bigl(x^{\prime } \bigr) \bigr) -U_{i}^{1} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigr) \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl( \bigl\vert U^{2} \bigr\vert - \bigl\vert U^{1} \bigr\vert \bigr) \,dx^{\prime}+ \widehat{\alpha} \int _{\Omega } \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac {\partial U_{i}^{2}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime }\,dz \\ &\qquad{}-\widehat{\alpha} \int _{\Omega} \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac{\partial U_{i}^{1}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\quad\geq\sum_{i=1}^{{2} } \int _{\Omega}\hat{f}_{i} \bigl( U_{i}^{2}-U_{i}^{1} \bigr) \,dx^{\prime}\,dz , \end{aligned}$$
(4.24)
$$\begin{aligned} &\sum_{i=1}^{{2} } \int _{\Omega}\frac{\partial U_{i}^{2}}{\partial z} \biggl( \frac{\partial U_{i}^{1}}{\partial z}- \frac {\partial U_{i}^{2}}{\partial z} \biggr) \,dx^{\prime}\,dz \\ &\qquad{}+\sum _{i=1}^{{2} }\widehat{l} \int _{\omega}U_{i}^{2} \bigl( x^{\prime },h \bigl(x^{\prime} \bigr) \bigr) \bigl( U_{i}^{1} \bigl( x^{\prime},h \bigl(x^{\prime } \bigr) \bigr) -U_{i}^{2} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigr) \,dx^{\prime} \\ &\qquad{}+ \int _{\omega}\hat{k} \bigl( \bigl\vert U^{1} \bigr\vert - \bigl\vert U^{2} \bigr\vert \bigr) \,dx^{\prime}+ \widehat{\alpha} \int _{\Omega } \Biggl( \frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac {\partial U_{i}^{1}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime }\,dz \\ &\qquad{}-\widehat{\alpha} \int _{\Omega} \Biggl(\frac{1}{2}\sum _{i=1}^{{2} } \biggl( \frac{\partial U_{i}^{2}}{\partial z} \biggr) ^{2} \Biggr) ^{\frac{1}{2}}\,dx^{\prime}\,dz \\ &\quad\geq\sum_{i=1} ^{{2} } \int _{\Omega}\hat{f}_{i} \bigl( U_{i}^{1}-U_{i}^{2} \bigr) \,dx^{\prime}\,dz. \end{aligned}$$
(4.25)
By adding the two inequalities (4.24), (4.25), it follows that
$$\begin{aligned} & \sum_{i=1}^{{2} } \int _{\Omega}\frac{\partial U_{i}^{1}}{\partial z} \biggl( \frac{\partial U_{i}^{2}}{\partial z}- \frac{\partial U_{i}^{1}}{\partial z} \biggr) \,dx^{\prime}\,dz+\sum _{i=1} ^{{2} } \int _{\Omega}\frac{\partial U_{i}^{2}}{\partial z} \biggl( \frac{\partial U_{i}^{1}}{\partial z}- \frac {\partial U_{i}^{2}}{\partial z} \biggr) \,dx^{\prime}\,dz \\ & \quad\geq\sum_{i=1}^{2}\widehat{l} \int _{\omega } \bigl\vert U_{i}^{2} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) -U_{i}^{1} \bigl( x^{\prime},h \bigl(x^{\prime} \bigr) \bigr) \bigr\vert ^{2}\,dx^{\prime }, \end{aligned}$$
then \(\sum_{i=1}^{{2}}\int _{\Omega } \vert \frac{\partial}{\partial z} ( U_{i}^{2}-U_{i}^{1} ) \vert ^{2}\,dx^{\prime}\,dz\leq0\).
Using the Poincaré inequality, we deduce that
$$ \bigl\Vert U_{i}^{2}-U_{i}^{1} \bigr\Vert _{V_{z}}=0, $$
(4.26)
so \(u^{\star}\) is unique.
The uniqueness of \(p^{\star}\) in \(L_{0}^{2}(\omega )\cap H^{1}(\omega)\) follows then from the specific weak Reynolds equation (4.17), indeed we have
$$\begin{aligned} & \int _{\omega} \biggl( \frac{h^{3}}{12}\nabla p^{1}- \frac{h}{2}U^{1} \bigl(x^{\prime},0 \bigr)- \frac{h}{2}U^{1} \bigl(x^{\prime},h \bigl(x^{\prime } \bigr) \bigr)+\widetilde {F} \biggr) \nabla\hat{\phi}\,dx^{\prime}+ \int _{\partial\omega }h\widetilde{U^{1}}\hat{\phi}\cdot n\,d \sigma =0 \\ &\quad\forall\hat{\phi}\in H^{1} ( \omega ) , \end{aligned}$$
(4.27)
$$\begin{aligned} & \int _{\omega} \biggl( \frac{h^{3}}{12}\nabla p^{2}- \frac{h}{2}U^{2} \bigl(x^{\prime},0 \bigr)- \frac{h}{2}U^{2} \bigl(x^{\prime},h \bigl(x^{\prime } \bigr) \bigr)+\widetilde {F} \biggr) \nabla\hat{\phi}\,dx^{\prime}+ \int _{\partial\omega }h\widetilde{U^{2}}\hat{\phi}\cdot n\,d \sigma =0 \\ &\quad\quad\forall\hat{\phi}\in H^{1} ( \omega ). \end{aligned}$$
(4.28)
Subtracting (4.28) and (4.27) and using (4.26), we obtain
$$\int _{\omega}\frac{h^{3}}{12} \bigl( \nabla p^{1}- \nabla p^{2} \bigr) \nabla\hat{\phi}\,dx^{\prime}=0, $$
taking \(\hat{\phi}=p^{1}-p^{2}\) and by Poincaré’s inequality, we get \(\Vert p^{1}-p^{2} \Vert _{L^{2} ( \omega ) }=0\).
This ends the proof of the uniqueness. □