In this section, we put our emphasis on the proofs of Theorem 2.1 and Theorem 2.2. They contain two parts:
-
1.
Recover the system in \(\Omega_{1}\).
-
2.
Investigate the optimal boundary conditions on \(\partial\Omega_{1}\).
Proof of the first part in Theorem 2.1 and Theorem 2.2
By Lemma 2.1 and the standard compactness theorem, we have \(u^{\delta}\rightharpoonup u\) weakly in \(L^{2}(0,T;H^{1}(\Omega_{1}))\) and \(p^{\delta}\rightharpoonup p\) weakly in \(L^{2}(\Omega_{1}\times(0,T))\) as \(\delta\rightarrow0\). Next, we choose the test function
$$ \eta(r,s,t)=\left \{ \textstyle\begin{array}{l@{\quad}l} g & \text{in } \Omega_{1}, \\ \sqrt{\mu}\tau\varphi_{1}+\frac{\sqrt{\sigma}\nu}{1+s\kappa}\varphi _{2} & \text{in } \Omega_{2}, \end{array}\displaystyle \right . $$
(23)
where \(\varphi=(\varphi_{1},\varphi_{2})^{\tau}\) is the solution of the boundary layer problem. Note that \(\operatorname{div}_{x_{1},x_{2}}\eta(r, s,t)=0 \) in \(\Omega_{1}\cup\Omega_{2}\) and \(\eta(r,s,t)|_{\partial\Omega}=0\). Multiplying (1) by \(\eta(r,s,t)\) and integrating over \(\Omega \times(0,T)\), we have
$$ \int_{0}^{T} \int_{\Omega}f\cdot\eta= \int_{0}^{T} \int_{\Omega}\eta\cdot \bigl[u_{t}^{\delta}+ \bigl(u^{\delta}\cdot\nabla\bigr)u^{\delta}-{\operatorname{div}}\bigl( \mathcal{A}\nabla u^{\delta}\bigr)+\nabla p^{\delta}\bigr]. $$
(24)
To avoid the influence on \(\partial\Omega_{1}\), we set \(g=0\) on \(\partial \Omega_{1}\), then (15) admits a unique solution \(\varphi=0\). Then we have
$$ \int_{0}^{T} \int_{\Omega_{1}}f\cdot g= \int_{0}^{T} \int_{\Omega_{1}}g\cdot \bigl[u_{t}^{\delta}+ \bigl(u^{\delta}\cdot\nabla\bigr)u^{\delta}-\operatorname{div}\bigl( \mathcal{A}\nabla u^{\delta}\bigr)+\nabla p^{\delta}\bigr]. $$
Passing to the limit \(\delta\rightarrow0\) and using the uniform bounds of those quantities, we recover the system and the initial condition
$$ u_{t}+(u\cdot\nabla)u-\triangle u+\nabla p=f \quad \text{in } \Omega _{1}, \qquad u(x,0)=u_{0}. $$
(25)
At last, by using the same test function, we have
$$ 0= \int_{\Omega_{1}}\operatorname{div} u^{\delta}g=- \int_{\Omega_{1}} u^{\delta}\cdot \nabla g\rightarrow- \int_{\Omega_{1}} u\cdot\nabla g, $$
which implies
$$ \operatorname{div} u=0. $$
(26)
The first part is then proved. □
Note that (25) and (26) are not determined because of lacking of boundary conditions. In the following, we investigate the boundary conditions on \(\partial\Omega_{1}\) satisfied by u.
Proof of Theorem 2.1
As in [1], suppressing the t dependence, we write (24) in the following way:
$$ I_{1}^{\delta}- \int_{\Omega}f\cdot\eta=I_{2}^{\delta}, $$
where
$$\begin{aligned}& I_{1}^{\delta}=- \int_{\Omega_{1}}\eta_{t}\cdot u^{\delta}+ \int_{\Omega_{1}}\nabla \eta:\nabla u^{\delta}- \int_{\Omega_{1}}\eta_{0}\cdot u^{\delta}_{0}+ \int_{\Omega_{1}}\nabla p^{\delta}\cdot\eta+ \int_{\Omega_{1}}\bigl(u^{\delta}\cdot\nabla\bigr)u^{\delta}\cdot\eta, \\& I_{2}^{\delta}= \int_{\Omega_{2}}\eta_{t}\cdot u^{\delta}- \int_{\Omega_{2}}\nabla \eta:\mathcal{A}\nabla u^{\delta}+ \int_{\Omega_{2}}\eta_{0}\cdot u^{\delta}_{0}- \int_{\Omega_{2}}\nabla p^{\delta}\cdot\eta- \int_{\Omega_{2}}\bigl(u^{\delta}\cdot\nabla\bigr)u^{\delta}\cdot\eta. \end{aligned}$$
Denote
$$\begin{aligned} B_{1}^{\delta} =& - \int_{\Omega_{2}}\nabla\eta:\mathcal{A}\nabla u^{\delta} \\ =& - \int_{0}^{l} \int_{0}^{\delta}(1+s\kappa)\biggl[\sigma \eta_{s} u_{s}^{\delta}+\frac{\mu }{(1+s\kappa)^{2}} \eta_{r} u_{r}^{\delta}\biggr] \\ =& B_{11}^{\delta}+B_{12}^{\delta}, \end{aligned}$$
(27)
where
$$ B_{11}^{\delta}=- \int_{0}^{l} \int_{0}^{\delta}\bigl(\sigma\eta_{s} u_{s}^{\delta}+\mu\eta_{r} u_{r}^{\delta}\bigr) ,\qquad B_{12}^{\delta}= \int_{0}^{l} \int_{0}^{\delta}s\kappa\biggl(\frac{\mu}{1+s\kappa}\eta _{r} u_{r}^{\delta}-\sigma\eta_{s} u_{s}^{\delta}\biggr). $$
(28)
Also
$$\begin{aligned} B_{2}^{\delta} =&- \int_{0}^{l} \int_{0}^{\delta}\bigl[\tau p_{r}+\nu(1+sk) p_{s}\bigr]\cdot\biggl(\sqrt{\mu}\tau \varphi_{1}+ \frac{\sqrt{\sigma}\nu}{1+s\kappa}\varphi_{2}\biggr)\,dr\,ds \\ =&- \int_{0}^{l} \int_{0}^{\delta}(\sqrt{\mu} p_{r} \varphi_{1}+\sqrt{\sigma} p_{s}\varphi_{2})\,dr\,ds \\ =&\sqrt{\sigma} \int_{0}^{l} p(r,0,t)\varphi_{2}(r,0,t)\, dr. \end{aligned}$$
(29)
Also
$$ \begin{aligned} &B_{3}^{\delta}= - \int_{0}^{l} \int_{0}^{\delta}\biggl[(1+s\kappa)\sqrt{\mu}\varphi _{1}\bigl(u_{2}^{\delta}u_{1s}^{\delta}-u_{1}^{\delta}u_{2s}^{\delta}\bigr)+\frac{\sqrt{\sigma }\varphi_{2}}{1+s\kappa} \bigl(u_{1}^{\delta}u_{2r}^{\delta}-u_{2}^{\delta}u_{1r}^{\delta}\bigr)\biggr]\,dr\,ds \\ &\hphantom{B_{3}^{\delta}}= B_{31}^{\delta}+B_{32}^{\delta}, \\ &B_{31}^{\delta}=- \int_{0}^{l} \int_{0}^{\delta}\bigl[\sqrt{\mu}\varphi_{1} \bigl(u_{2}^{\delta}u_{1s}^{\delta}-u_{1}^{\delta}u_{2s}^{\delta}\bigr)+\sqrt{\sigma}\varphi _{2} \bigl(u_{1}^{\delta}u_{2r}^{\delta}-u_{2}^{\delta}u_{1r}^{\delta}\bigr)\bigr]\,dr\,ds \\ &\hphantom{B_{31}^{\delta}}=B_{311}^{\delta}+B_{312}^{\delta}. \end{aligned} $$
(30)
It is obvious that
$$\begin{aligned}& \bigl\vert B_{311}^{\delta}\bigr\vert \leq C \sqrt{\mu}, \qquad \bigl\vert B_{312}^{\delta}\bigr\vert \leq C \sqrt {\sigma}, \\& B_{32}^{\delta}=- \int_{0}^{l} \int_{0}^{\delta}s\kappa\biggl[\sqrt{\mu}\varphi _{1}\bigl(u_{2}^{\delta}u_{1s}^{\delta}-u_{1}^{\delta}u_{2s}^{\delta}\bigr)-\frac{\sqrt{\sigma }\varphi_{2}}{1+s\kappa}\bigl(u_{1}^{\delta}u_{2r}^{\delta}-u_{2}^{\delta}u_{1r}^{\delta}\bigr)\biggr]\,dr\,ds \\& \hphantom{B_{32}^{\delta}}=B_{321}^{\delta}+B_{322}^{\delta}. \end{aligned}$$
(31)
We also have
$$ \bigl\vert B_{321}^{\delta}\bigr\vert \leq C\delta\sqrt{\mu}, \qquad \bigl\vert B_{322}^{\delta}\bigr\vert \leq C\delta \sqrt{\sigma}. $$
(32)
In \(\Omega_{2}\), η satisfies
$$\begin{aligned} -\mu\eta_{rr}-\sigma\eta_{ss} =&-\mu\sqrt{\mu}\tau \pi_{r}-\frac{\sigma\sqrt{\mu}\nu}{1+s\kappa}\pi_{s}-\mu \sqrt{\mu}( \tau_{r}\varphi_{1})_{r}-\mu\sqrt{\mu} \tau_{r}\varphi_{1r} \\ &{}-\biggl[\biggl(\frac{\mu\sqrt{\sigma}\nu}{1+s\kappa}\biggr)_{r}\varphi_{2} \biggr]_{r}-\biggl(\frac{\mu \sqrt{\sigma}\nu}{1+s\kappa}\biggr)_{r} \varphi_{2r} \\ &{}-\biggl[\biggl(\frac{\sigma\sqrt{\sigma}\nu}{1+s\kappa}\biggr)_{s} \varphi_{2}\biggr]_{s}-\biggl(\frac {\sigma\sqrt{\sigma}\nu}{1+s\kappa} \biggr)_{s}\varphi_{2s}. \end{aligned}$$
(33)
Multiplying (33) by η and integrating over \([0,l)\times (0,\delta)\) and elaborated computation based on (33), and by integration by parts, we have
$$\begin{aligned}& \int_{0}^{l} \int_{0}^{\delta}\mu \vert \eta_{r}\vert ^{2}+\sigma \vert \eta_{s}\vert ^{2} \\& \quad = -\sigma\mu \int_{0}^{l}\varphi_{1}(r,0,t) \varphi_{1s}(r,0,t)-\sigma^{2} \int _{0}^{l}\varphi_{2}(r,0,t) \varphi_{2s}(r,0,t) \\& \qquad {} +\sigma\sqrt{\mu\sigma} \int_{0}^{l}\pi(r,0,t)\varphi_{2}(r,0,t)+2 \mu\sigma \int_{0}^{l} \int_{0}^{\delta}\kappa\bigl\vert \varphi_{2}(r,0,t)\bigr\vert ^{2} \\& \qquad {} + \int_{0}^{l} \int_{0}^{\delta}\biggl[\mu^{2}- \frac{\mu\sigma}{(1+s\kappa)^{2}}\biggr]\pi\varphi_{1r} -\sigma\sqrt{\mu\sigma} \int_{0}^{l} \int_{0}^{\delta}\frac{2\kappa}{(1+s\kappa )^{3}}\pi \varphi_{2} \\& \qquad {} +\mu^{2} \int_{0}^{l} \int_{0}^{\delta}\kappa^{2}\vert \varphi_{1}\vert ^{2} -\mu\sigma \int_{0} ^{l} \int_{0}^{\delta}\biggl(\frac{\kappa}{1+s\kappa} \biggr)^{2}\vert \varphi _{2}\vert ^{2} \\& \qquad {} +\mu\sigma \int_{0}^{l} \int_{0}^{\delta}\biggl\vert \biggl(\frac{1}{1+s\kappa} \biggr)_{r}\varphi_{2}\biggr\vert ^{2} + \sigma^{2} \int_{0}^{l} \int_{0}^{\delta}\biggl\vert \biggl(\frac{1}{1+s\kappa} \biggr)_{s}\varphi_{2}\biggr\vert ^{2}. \end{aligned}$$
(34)
Multiplying (33) by \(u^{\delta}\) and integrating over \([0,l)\times (0,\delta)\), we have
$$\begin{aligned}& \int_{0}^{l} \int_{0}^{\delta}(\mu\eta_{rr}+\sigma \eta_{ss})\cdot u^{\delta} \\& \quad = 2\sigma\sqrt{\sigma} \int_{0}^{l}\kappa\nu\varphi_{2}(r,0,t) \cdot u^{\delta}(r,0,t) \\& \qquad {}+\sigma\sqrt{\mu} \int_{0}^{l}\nu\pi(r,0,t)u^{\delta}(r,0,t) +\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\tau_{r}\pi\cdot u^{\delta} \\& \qquad {}+\mu\sqrt{\mu } \int_{0}^{l} \int_{0}^{\delta}\tau\pi\cdot u^{\delta}_{r}+\sigma\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\frac{\nu}{1+s\kappa}\pi\cdot u^{\delta}_{s} \\& \qquad {} -\sigma\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\frac{\nu\kappa}{(1+s\kappa )^{2}}\pi\cdot u^{\delta}-2\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\kappa\nu \varphi_{1}\cdot u^{\delta}_{r} \\& \qquad {} +2\mu\sqrt{\sigma} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa } \biggr)_{r}\varphi_{2}\cdot u^{\delta}_{r}+2 \sigma\sqrt{\sigma} \int_{0}^{l} \int _{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa} \biggr)_{s}\varphi_{2}\cdot u^{\delta}_{s} \\& \qquad {} -\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\kappa'\nu \varphi_{1}\cdot u^{\delta}-\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\kappa^{2}\tau \varphi_{1}\cdot u^{\delta} \\& \qquad {} +\mu\sqrt{\sigma} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa } \biggr)_{rr}\varphi_{2}\cdot u^{\delta}+\sigma\sqrt{ \sigma} \int_{0}^{l} \int _{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa} \biggr)_{ss}\varphi_{2}\cdot u^{\delta}. \end{aligned}$$
(35)
Integrating by parts, combining with (22), (23), and (35), we have
$$\begin{aligned} B_{11}^{\delta} =& \int_{0}^{l} \int_{0}^{\delta}(\mu\eta_{rr}+\sigma\eta _{ss})\cdot u^{\delta}+\sigma \int_{0}^{l}\eta_{s}(r,0,t)\cdot u^{\delta}(r,0,t)\,dr \\ =& -\frac{\sigma}{\delta}\sqrt{\mu} \int_{0}^{l}\tau g_{1}(r,t)u^{\delta}(r,0,t)\,dr-\frac{\sigma}{\delta}\sqrt{\sigma} \int_{0}^{l}\nu g_{2}(r,t)u^{\delta}(r,0,t)\,dr \\ &{} +\sigma\sqrt{\sigma} \int_{0}^{l}\nu\kappa\varphi_{2}(r,0,t)u^{\delta}(r,0,t)\,dr-\sigma\sqrt{\mu} \int_{0}^{l}\nu\pi(r,0,t)u^{\delta}(r,0,t)\,dr \\ &{} -\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\tau_{rr} \varphi_{1}\cdot u^{\delta}-2\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\tau_{r} \varphi_{1}\cdot u^{\delta}_{r} \\ &{} -\mu\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\tau_{r}\pi\cdot u^{\delta}-\mu\sqrt {\mu} \int_{0}^{l} \int_{0}^{\delta}\tau\pi\cdot u^{\delta}_{r} \\ &{} -\mu\sqrt{\sigma} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa } \biggr)_{rr}\varphi_{2}\cdot u^{\delta}-2\mu\sqrt{ \sigma} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa} \biggr)_{r}\varphi_{2}\cdot u^{\delta}_{r} \\ &{}-\sigma\sqrt{\sigma} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa } \biggr)_{ss}\varphi_{2}\cdot u^{\delta}-2\sigma\sqrt{ \sigma} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa} \biggr)_{s}\varphi_{2}\cdot u^{\delta}_{s} \\ &{} -\sigma\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\biggl(\frac{\nu}{1+s\kappa} \biggr)_{s}\pi \cdot u^{\delta}-\sigma\sqrt{\mu} \int_{0}^{l} \int_{0}^{\delta}\frac{\nu}{1+s\kappa }\pi\cdot u^{\delta}_{s}+\frac{\sigma}{\delta}\sqrt{\mu}o \bigl(h^{2}\bigr). \end{aligned}$$
(36)
By using the Hölder inequality and (34), we have
$$ \bigl\vert B_{12}^{\delta}\bigr\vert \leq C\delta\biggl( \int_{0}^{l} \int_{0}^{\delta}\mu \vert \eta_{r}\vert ^{2}+\sigma \vert \eta_{s}\vert ^{2} \biggr)^{\frac{1}{2}} \biggl( \int_{0}^{l} \int_{0}^{\delta}\mu\bigl\vert u^{\delta}_{r} \bigr\vert ^{2}+\sigma\bigl\vert u^{\delta}_{s} \bigr\vert ^{2}\biggr)^{\frac {1}{2}}. $$
(37)
Now we consider the limit to (29), (30), (36), and (37) as \(\delta\rightarrow0\). We study three cases:
1. If \(\lim_{\delta\rightarrow0}\frac{\sigma^{\frac {3}{2}}}{\delta}=0\), in this case, \(\lim_{\delta\rightarrow 0}\sigma=\lim_{\delta\rightarrow0}\mu=0\), we have
$$ B_{1}^{\delta}=B_{11}^{\delta}+B_{12}^{\delta}\rightarrow0,\qquad B_{2}^{\delta}\rightarrow0, \qquad B_{3}^{\delta}\rightarrow0. $$
We obtain the boundary condition
$$ \frac{\partial u}{\partial\nu}=(u\otimes u+pI)\cdot\nu \quad \text{on } \partial \Omega_{1}. $$
2. If \(\lim_{\delta\rightarrow0}\frac{\sigma^{\frac {3}{2}}}{\delta}=\alpha\in(0,+\infty)\), in this case, \(\lim_{\delta\rightarrow0}\sigma=\lim_{\delta\rightarrow 0}\mu=0\), we consider two different subcases.
2(i). If \(\lim_{\delta\rightarrow0}\frac{\mu}{\sigma}=0\), we have
$$ B_{1}^{\delta}=B_{11}^{\delta}+B_{12}^{\delta}\rightarrow-\alpha \int_{0}^{l}\nu g_{2}(r,t)u(r,0,t)\, dr, \qquad B_{2}^{\delta}\rightarrow0, \qquad B_{3}^{\delta}\rightarrow0. $$
Then the boundary condition is
$$ \frac{\partial u}{\partial\nu}-(u\otimes u+pI)\cdot\nu=\alpha\nu \cdot u(r,0,t)e_{2} \quad \text{on } \partial\Omega_{1}. $$
2(ii). If \(\lim_{\delta\rightarrow0}\frac{\mu}{\sigma}=\beta\in (0,1]\), we also have \(B_{2}^{\delta}\rightarrow0\) and \(B_{3}^{\delta}\rightarrow0\),
$$ B_{1}^{\delta}\rightarrow-\alpha\sqrt{\beta} \int_{0}^{l}\tau g_{1}(r,t)u(r,0,t)\, dr- \alpha \int_{0}^{l}\nu g_{2}(r,t)u(r,0,t)\, dr, $$
which implies
$$ \frac{\partial u}{\partial\nu}-(u\otimes u+pI)\cdot\nu=\alpha\sqrt {\beta}\tau\cdot u e_{1}+\alpha\nu\cdot u e_{2}\quad \text{on } \partial \Omega_{1}. $$
3. If \(\lim_{\delta\rightarrow0}\frac{\sigma^{\frac {3}{2}}}{\delta}=+\infty\), in this case, we also consider two different subcases.
3(i). If \(\lim_{\delta\rightarrow0}\frac{\mu}{\sigma}=0\), we have
$$ \frac{\delta}{\sigma\sqrt{\sigma}}B_{1}^{\delta}=\frac{\delta}{\sigma\sqrt {\sigma}}B_{11}^{\delta}+ \frac{\delta}{\sigma\sqrt{\sigma}}B_{12}^{\delta}. $$
Moreover, if \(\sigma\geq M>0\),
$$ \frac{\delta}{\sigma\sqrt{\sigma}}\bigl\vert B_{2}^{\delta}\bigr\vert \leq \frac{\delta}{M\sqrt {M}} \int_{0}^{l}\bigl\vert p^{\delta}(r,0,t) \varphi_{2}(r,0,t)\bigr\vert \, dr\rightarrow0. $$
If \(0<\sigma< M\), we have
$$ \frac{\delta}{\sigma\sqrt{\sigma}}\bigl\vert B_{2}^{\delta}\bigr\vert \leq \frac{\delta}{\sigma \sqrt{\sigma}}\sqrt{M} \int_{0}^{l}\bigl\vert p^{\delta}(r,0,t) \varphi _{2}(r,0,t)\bigr\vert \, dr\rightarrow0. $$
The limit of \(B_{3}^{\delta}\) can be derived as \(B_{2}^{\delta}\) and we have \(\frac{\delta}{\sigma\sqrt{\sigma}}|B_{3}^{\delta}|\rightarrow0\).
Then we obtain
$$ \int_{0}^{l}\nu g_{2}(r,t)u^{\delta}(r,0,t) \, dr\rightarrow \int_{0}^{l}\nu g_{2}(r,t)u(r,0,t)\, dr=0, $$
which implies
$$ u(r,0,t)\cdot\nu=0 \quad \text{on } \partial\Omega_{1}. $$
3(ii). If \(\lim_{\delta\rightarrow0}\frac{\mu}{\sigma}=\beta\in (0,1]\), we have
$$\begin{aligned} \frac{\delta}{\sigma\sqrt{\sigma}}B^{\delta} =&\frac{\delta}{\sigma\sqrt {\sigma}}B_{11}^{\delta}+ \frac{\delta}{\sigma\sqrt{\sigma}}B_{12}^{\delta}\\ \rightarrow&-\sqrt{\beta} \int_{0}^{l}\tau g_{1}(r,t)u(r,0,t)\, dr- \int_{0}^{l}\nu g_{2}(r,t)u(r,0,t)\, dr=0. \end{aligned}$$
Following the steps in 3(i), we also have
$$ \frac{\delta}{\sigma\sqrt{\sigma}}B_{2}^{\delta}\rightarrow0, \qquad \frac {\delta}{\sigma\sqrt{\sigma}}B_{3}^{\delta}\rightarrow0. $$
Then we obtain the boundary condition on u,
$$ u(r,0,t)\cdot\tau|_{\partial\Omega_{1}}=0, \qquad u(r,0,t)\cdot\nu |_{\partial\Omega_{1}}=0, $$
which gives
$$ u(r,0,t)=0 \quad \text{on } \partial\Omega_{1}. $$
Theorem 2.1 is proved. □
Proof of Theorem 2.2
Under the conditions in Theorem 2.2, (29), (33), and (34) are still satisfied. To obtain the boundary condition as \(\delta \rightarrow0\), we consider three cases:
1. If \(\lim_{\delta\rightarrow0}\frac{\mu^{\frac{3}{2}}}{\delta }=0\), then \(\lim_{\delta\rightarrow0}\frac{\sigma\sqrt{\mu }}{\delta} =\lim_{\delta\rightarrow0}\frac{\sigma\sqrt{\sigma}}{\delta }=0\), \(\lim_{\delta\rightarrow0}\sigma=\lim_{\delta \rightarrow0}\mu\)=0, we have
$$ B_{1}^{\delta}=B_{11}^{\delta}+B_{12}^{\delta}\rightarrow0, \qquad B_{2}^{\delta}\rightarrow0, \qquad B_{3}^{\delta}\rightarrow0. $$
Then we obtain the optimal boundary condition on u,
$$ \frac{\partial u}{\partial\nu}-(u\otimes u+pI)\cdot\nu=0\quad \text{on } \partial \Omega_{1}. $$
2. If \(\lim_{\delta\rightarrow0}\frac{\mu^{\frac{3}{2}}}{\delta }=\alpha\in(0,+\infty)\), then \(\lim_{\delta\rightarrow0}\sigma =\lim_{\delta\rightarrow0}\mu=0\), and we consider two subcases.
2(i). If \(\lim_{\delta\rightarrow0}\frac{\sigma}{\mu}=0\), then we have \(\lim_{\delta\rightarrow0}\frac{\sigma\sqrt{\sigma }}{\delta}\leq \lim_{\delta\rightarrow0}\frac{\sigma\sqrt{\mu}}{\delta} =\lim_{\delta\rightarrow0}(\frac{\mu\sqrt{\mu}}{\delta}\times \frac{\sigma}{\mu})=0\), passing to the limit, we have
$$ B_{1}^{\delta}=B_{11}^{\delta}+B_{12}^{\delta}\rightarrow0,\qquad B_{2}^{\delta}\rightarrow0, \qquad B_{2}^{\delta}\rightarrow0. $$
Then we obtain the optimal boundary condition on u,
$$ \frac{\partial u}{\partial\nu}-(u\otimes u+pI)\cdot\nu=0\quad \text{on } \partial \Omega_{1}. $$
2(ii). If \(\lim_{\delta\rightarrow0}\frac{\sigma}{\mu}=\beta\in (0,1]\), then we have \(\lim_{\delta\rightarrow0}\frac{\sigma\sqrt {\sigma}}{\delta}= \lim_{\delta\rightarrow0}(\frac{\mu\sqrt{\mu}}{\delta}\times \frac{\sigma\sqrt{\sigma}}{\mu\sqrt{\mu}})=\alpha\beta^{\frac{3}{2}}\), \(\lim_{\delta\rightarrow0}\frac{\sigma\sqrt{\mu}}{\delta} =\lim_{\delta\rightarrow0}(\frac{\mu\sqrt{\mu}}{\delta}\times \frac{\sigma}{\mu})=\alpha\beta\), passing to the limit, we have
$$ B_{1}^{\delta}\rightarrow\alpha\beta \int_{0}^{l}(\tau\cdot ue_{1})\cdot g\, dr+\alpha\beta ^{\frac{3}{2}} \int_{0}^{l}(\nu\cdot ue_{2})\cdot g\, dr, \qquad B_{2}^{\delta}\rightarrow 0, \qquad B_{3}^{\delta}\rightarrow0. $$
Then we obtain the optimal boundary condition on u,
$$ \frac{\partial u}{\partial\nu}-(u\otimes u+pI)\cdot\nu=\alpha\beta \tau\cdot u e_{1}+\alpha\beta^{\frac{3}{2}}\nu\cdot u e_{2}\quad \text{on } \partial\Omega_{1}. $$
3. If \(\lim_{\delta\rightarrow0}\frac{\mu^{\frac{3}{2}}}{\delta }=+\infty\), then \(\lim_{\delta\rightarrow0}\frac{\sigma\sqrt{\mu }}{\delta}=+\infty\). We consider two subcases.
3(i). If \(\lim_{\delta\rightarrow0}\frac{\sigma}{\mu}=0\), then we have
$$ \frac{\delta}{\sigma\sqrt{\mu}} B_{1}^{\delta}=\frac{\delta}{\sigma\sqrt {\mu}}B_{11}^{\delta}+ \frac{\delta}{\sigma\sqrt{\mu}}B_{12}^{\delta}\rightarrow- \int_{0}^{l}(\tau\cdot ue_{1})\cdot g\, dr. $$
Moreover, if \(\mu\geq a>0\),
$$ \frac{\delta}{\sigma\sqrt{\mu}}\bigl\vert B_{2}^{\delta}\bigr\vert \leq \frac{\delta}{a}\sqrt {M} \int_{0}^{l}\bigl\vert p^{\delta}(r,0,t) \varphi_{2}(r,0,t)\bigr\vert \, dr\rightarrow0. $$
If \(0<\mu< M\), we have
$$ \frac{\delta}{\sigma\sqrt{\mu}}\bigl\vert B_{2}^{\delta}\bigr\vert \leq \frac{\delta}{\sigma \sqrt{\mu}}\sqrt{M} \int_{0}^{l}\bigl\vert p^{\delta}(r,0,t) \varphi _{2}(r,0,t)\bigr\vert \, dr\rightarrow0. $$
The limit of \(B_{3}^{\delta}\) can be derived as \(B_{2}^{\delta}\) and we have \(\frac{\delta}{\sigma\sqrt{\sigma}}|B_{3}^{\delta}|\rightarrow0\). Then we obtain the optimal boundary condition on u,
$$ u\cdot\tau=0\quad \text{on } \partial\Omega_{1}. $$
3(ii). If \(\lim_{\delta\rightarrow0}\frac{\sigma}{\mu}=\beta\in (0,1]\), then we have
$$\begin{aligned} \frac{\delta}{\sigma\sqrt{\mu}} B_{1}^{\delta} =&\frac{\delta}{\sigma\sqrt {\mu}}B_{11}^{\delta}+ \frac{\delta}{\sigma\sqrt{\mu}}B_{12}^{\delta}\\ \rightarrow&- \int_{0}^{l}(\tau\cdot ue_{1})\cdot g \, dr- \int_{0}^{l}(\sqrt{\beta }\nu\cdot ue_{2}) \cdot g\, dr. \end{aligned}$$
Following the steps in 3(i), we have
$$ \frac{\delta}{\sigma\sqrt{\mu}}\bigl\vert B_{2}^{\delta}\bigr\vert \rightarrow0. $$
Then we obtain the optimal boundary condition on u,
$$ \tau\cdot ue_{1}+\sqrt{\beta}\nu\cdot ue_{2}=0 \quad \text{on } \partial \Omega_{1}, $$
which implies
$$ u=0\quad \text{on } \partial\Omega_{1}, $$
Theorem 2.2 is then proved. □
Remark 4.1
As stated in the introduction, to protect \(\partial\Omega_{1}\) from eroding, it is desirable that the eigenvector corresponding to the smallest eigenvalue is orthogonal to the boundary of the body in order to directly confront the ambient effect. In our problem, it requires \(\delta\leq\sigma\leq\mu\). The optimal boundary conditions to our problem are given in Theorem 2.2. We must acknowledge that we need the condition \(\delta\leq\sigma\leq\mu\leq M\sigma\) for some constants \(M>0\) so that we can ensure \(0< h\leq1\) as δ is small enough. If this condition does not hold, the estimate (22) will not be satisfied. The case that \(h>1\) we will consider in the future.