Navier–Stokes Equations in a Curved Thin Domain, Part II: Global Existence of a Strong Solution

Abstract

We consider the Navier–Stokes equations in a three-dimensional curved thin domain around a given closed surface under Navier’s slip boundary conditions. When the thickness of the thin domain is sufficiently small, we establish the global existence of a strong solution for large data. We also show several estimates for the strong solution with constants explicitly depending on the thickness of the thin domain. The proofs of these results are based on a standard energy method and a good product estimate for the convection and viscous terms following from a detailed study of average operators in the thin direction. We use the average operators to decompose a three-dimensional vector field on the thin domain into the almost two-dimensional average part and the residual part, and derive good estimates for them which play an important role in the proof of the product estimate.

Appendices

Appendix A: Notations on Vectors and Matrices

In this appendix we fix notations on vectors and matrices. For $m\in {\mathbb {N}}$ we consider a vector $a\in {\mathbb {R}}^m$ as a column vector

$$\begin{aligned} a = \begin{pmatrix} a_1 \\ \vdots \\ a_m \end{pmatrix} = (a_1, \ldots , a_m)^T \end{aligned}$$

and denote the i-th component of a by $a_i$ or sometimes by $a^i$ or $[a]_i$ for $i=1,\dots ,m$. A matrix $A\in {\mathbb {R}}^{l\times m}$ with $l,m\in {\mathbb {N}}$ is expressed as

$$\begin{aligned} A = (A_{ij})_{i,j} = \begin{pmatrix} A_{11} &{} \quad \cdots &{}\quad A_{1m} \\ \vdots &{}\quad &{}\quad \vdots \\ A_{l1} &{}\quad \cdots &{} \quad A_{lm} \end{pmatrix} \end{aligned}$$

and the (i, j)-entry of A is denoted by $A_{ij}$ or sometimes by $[A]_{ij}$ for $i=1,\dots ,l$ and $j=1,\dots m$. We denote the transpose of A by $A^T$ and, when $l=m$, the symmetric part of A by $A_S:=(A+A^T)/2$. Also, we write $I_m$ for the $m\times m$ identity matrix. The tensor product of $a\in {\mathbb {R}}^l$ and $b\in {\mathbb {R}}^m$ is defined as

$$\begin{aligned} a\otimes b := (a_ib_j)_{i,j} = \begin{pmatrix} a_1b_1 &{}\quad \cdots &{}\quad a_1b_m \\ \vdots &{}\quad &{}\quad \vdots \\ a_lb_1 &{}\quad \cdots &{}\quad a_lb_m \end{pmatrix}, \quad a = \begin{pmatrix} a_1 \\ \vdots \\ a_l \end{pmatrix}, \quad b = \begin{pmatrix} b_1 \\ \vdots \\ b_m \end{pmatrix}. \end{aligned}$$

For three-dimensional vector fields $u=(u_1,u_2,u_3)^T$ and $\varphi$ on an open set in ${\mathbb {R}}^3$ let

$$\begin{aligned} \nabla u&:= \begin{pmatrix} \partial _1u_1 &{}\quad \partial _1u_2 &{}\quad \partial _1u_3 \\ \partial _2u_1 &{}\quad \partial _2u_2 &{}\quad \partial _2u_3 \\ \partial _3u_1 &{}\quad \partial _3u_2 &{}\quad \partial _3u_3 \end{pmatrix}, \quad |\nabla ^2u|^2 := \sum _{i,j,k=1}^3|\partial _i\partial _ju_k|^2 \quad \left( \partial _i := \frac{\partial }{\partial x_i}\right) , \\ (\varphi \cdot \nabla )u&:= \begin{pmatrix} \varphi \cdot \nabla u_1 \\ \varphi \cdot \nabla u_2 \\ \varphi \cdot \nabla u_3 \end{pmatrix} = (\nabla u)^T\varphi . \end{aligned}$$

Also, we define the inner product of $3\times 3$ matrices A and B and the norm of A by

$$\begin{aligned} A: B := \mathrm {tr}[A^TB] = \sum _{i=1}^3AE_i\cdot BE_i, \quad |A| := \sqrt{A:A}, \end{aligned}$$

where $\{E_1,E_2,E_3\}$ is an orthonormal basis of ${\mathbb {R}}^3$. Note that A : B does not depend on a choice of $\{E_1,E_2,E_3\}$. In particular, taking the standard basis of ${\mathbb {R}}^3$ we get

$$\begin{aligned} A:B = \sum _{i,j=1}^3A_{ij}B_{ij} = B:A = A^T:B^T, \quad AB:C = A:CB^T = B:A^TC \end{aligned}$$

for $A,B,C\in {\mathbb {R}}^{3\times 3}$. Also, for $a,b\in {\mathbb {R}}^3$ we have $|a\otimes b|=|a||b|$.

Appendix B: Proofs of Auxiliary Lemmas

The purpose of this appendix is to present the proofs of Lemmas 4.1, 4.3, and 7.4. First we prove Lemma 4.1 after giving two auxiliary statements. Recall that $\Gamma$ is a two-dimensional closed surface in ${\mathbb {R}}^3$ of class $C^5$.

Lemma B.1

[34, Lemma B.4] Let U be an open set in ${\mathbb {R}}^2$, $\mu :U\rightarrow \Gamma$ a $C^5$ local parametrization of $\Gamma$, and ${\mathcal {K}}$ a compact subset of U. For $p\in [1,\infty ]$ if $\eta \in L^p(\Gamma )$ is supported in $\mu ({\mathcal {K}})$, then $\eta ^\flat :=\eta \circ \mu \in L^p(U)$ and

$$\begin{aligned} c^{-1}\Vert \eta ^\flat \Vert _{L^p(U)} \le \Vert \eta \Vert _{L^p(\Gamma )} \le c\Vert \eta ^\flat \Vert _{L^p(U)}. \end{aligned}$$

(B.1)

If in addition $\eta \in W^{1,p}(\Gamma )$, then $\eta ^\flat \in W^{1,p}(U)$ and

$$\begin{aligned} c^{-1}\Vert \nabla _s\eta ^\flat \Vert _{L^p(U)} \le \Vert \nabla _\Gamma \eta \Vert _{L^p(\Gamma )} \le c\Vert \nabla _s\eta ^\flat \Vert _{L^p(U)}, \end{aligned}$$

(B.2)

where $\nabla _s\eta ^\flat =(\partial _{s_1}\eta ^\flat ,\partial _{s_2}\eta ^\flat )^T$ is the gradient of $\eta ^\flat$ in $s\in {\mathbb {R}}^2$.

We omit the proof of Lemma B.1 since it is given in our first paper [34].

Lemma B.2

Let U be an open set in ${\mathbb {R}}^2$. Then

$$\begin{aligned} \Vert \varphi \Vert _{L^4(U)} \le \sqrt{2}\Vert \varphi \Vert _{L^2(U)}^{1/2}\Vert \nabla _s\varphi \Vert _{L^2(U)}^{1/2} \end{aligned}$$

(B.3)

for all $\varphi \in H_0^1(U)$.

The inequality (B.3) is the well-known Ladyzhenskaya inequality on ${\mathbb {R}}^2$ (see [25, Chapter 1, Section 1.1, Lemma 1]). We give its proof for the readers’ convenience.

Proof

By a density argument, it is sufficient to prove (B.3) for all $\varphi \in C_c^\infty (U)$. We extend $\varphi$ to ${\mathbb {R}}^2$ by setting it to be zero outside U. Then

$$\begin{aligned} |\varphi (s_1,s_2)|^2 = \int _{-\infty }^{s_1}\frac{\partial }{\partial t_1}\bigl (|\varphi (t_1,s_2)|^2\bigr )\,dt_1 = 2\int _{-\infty }^{s_1}\varphi (t_1,s_2)\partial _{t_1}\varphi (t_1,s_2)\,dt_1 \end{aligned}$$

for each $s=(s_1,s_2)\in {\mathbb {R}}^2$. Thus Hölder’s inequality implies

$$\begin{aligned} |\varphi (s_1,s_2)|^2 \le 2\left( \int _{-\infty }^\infty |\varphi (t_1,s_2)|^2\,dt_1\right) ^{1/2}\left( \int _{-\infty }^\infty |\partial _{t_1}\varphi (t_1,s_2)|^2\,dt_1\right) ^{1/2}. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} |\varphi (s_1,s_2)|^2 \le 2\left( \int _{-\infty }^\infty |\varphi (s_1,t_2)|^2\,dt_2\right) ^{1/2}\left( \int _{-\infty }^\infty |\partial _{t_2}\varphi (s_1,t_2)|^2\,dt_2\right) ^{1/2}. \end{aligned}$$

From the above two inequalities we deduce that

$$\begin{aligned}&\int _{{\mathbb {R}}^2}|\varphi (s_1,s_2)|^4\,ds_1\,ds_2 \nonumber \\&\quad \le 4\int _{-\infty }^\infty \left( \int _{-\infty }^\infty |\varphi (t_1,s_2)|^2\,dt_1\right) ^{1/2}\left( \int _{-\infty }^\infty |\partial _{t_1}\varphi (t_1,s_2)|^2\,dt_1\right) ^{1/2}ds_2 \nonumber \\&\qquad \times \int _{-\infty }^\infty \left( \int _{-\infty }^\infty |\varphi (s_1,t_2)|^2\,dt_2\right) ^{1/2}\left( \int _{-\infty }^\infty |\partial _{t_2}\varphi (s_1,t_2)|^2\,dt_2\right) ^{1/2}ds_1. \end{aligned}$$

(B.4)

We again use Hölder’s inequality to get

$$\begin{aligned}&\int _{-\infty }^\infty \left( \int _{-\infty }^\infty |\varphi (t_1,s_2)|^2\,dt_1\right) ^{1/2}\left( \int _{-\infty }^\infty |\partial _{t_1}\varphi (t_1,s_2)|^2\,dt_1\right) ^{1/2}ds_2 \\&\quad \le \left( \int _{-\infty }^\infty \int _{-\infty }^\infty |\varphi (t_1,s_2)|^2\,dt_1ds_2\right) ^{1/2}\left( \int _{-\infty }^\infty \int _{-\infty }^\infty |\partial _{t_1}\varphi (t_1,s_2)|^2\,dt_1ds_2\right) ^{1/2} \\&\quad \le \Vert \varphi \Vert _{L^2({\mathbb {R}}^2)}\Vert \nabla _s\varphi \Vert _{L^2({\mathbb {R}}^2)} \end{aligned}$$

and a similar inequality for the last line of (B.4). By these inequalities and (B.4),

$$\begin{aligned} \Vert \varphi \Vert _{L^4({\mathbb {R}}^2)}^4 \le 4\Vert \varphi \Vert _{L^2({\mathbb {R}}^2)}^2\Vert \nabla _s\varphi \Vert _{L^2({\mathbb {R}}^2)}^2. \end{aligned}$$

Since $\varphi \in C_c^\infty (U)$ is compactly supported in U, this inequality implies (B.3). $\square$

Proof of Lemma 4.1

Since $\Gamma$ is compact and without boundary, we can take a finite number of bounded open sets in $\mathbb {R}^2$ and local parametrizations of $\Gamma$

$$\begin{aligned} U_k\subset \mathbb {R}^2,\quad \mu _k : \text {U}_k\rightarrow \Gamma , \quad k=1,\ldots , k_0 \end{aligned}$$

such that $\{\mu _k(U_k)\}_{k=1}^{k_0}$ is an open covering of $\Gamma$. Let $\{\xi _k\}_{k=1}^{k_0}$ be a partition of unity on $\Gamma$ subordinate to $\{\mu _k(U_k)\}_{k=1}^{k_0}$ consisting of $C^1$ functions on $\Gamma$. We may assume that $\xi _k$ is supported in $\mu _k({\mathcal {K}}_k)$ with some compact subset ${\mathcal {K}}_k$ of $U_k$ for $k = 1,\ldots , k_0$. For $\eta \in H^1(\Gamma )$ and $k = 1,\ldots , k_0$ let $(\xi _k\eta )^{\flat }:= (\xi _k\eta )\circ \mu _k$ on $U_k$. Then $(\xi _k\eta )^{\flat }\in H_0^1(U_k)$ by Lemma (B.1), since $\xi _k\eta$ belongs to $H^1(\Gamma )$ and is supported in $\mu _k(\mathcal {K}_k)$. Hence

$$\begin{aligned} \Vert (\xi _k\eta )^{\flat }\Vert _{L^4(U_k)} \le \sqrt{2}\Vert (\xi _k\eta )^{\flat }\Vert _{L^2(U_k)}^{1/2}\Vert \nabla _s(\xi _k\eta )^{\flat }\Vert _{L^2(U_k)}^{1/2} \end{aligned}$$

by (B.3). We apply (B.1) and (B.2) to this inequality to get

$$\begin{aligned} \Vert \xi _k\eta \Vert _{L^4(\Gamma )} \le c \Vert \xi _k\eta \Vert _{L^2(\Gamma )}^{1/2}\Vert \nabla _{\Gamma }(\xi _k\eta )\Vert _{L^2(\Gamma )}^{1/2} \le c\Vert \eta \Vert _{L^2(\Gamma )}^{1/2}\Vert \eta \Vert _{H^1(\Gamma )}^{1/2}. \end{aligned}$$

Since this inequality holds for all $k = 1,\ldots , k_0$, we obtain (4.1). $\square$

Next we present the proof of Lemma 4.3.

Proof of Lemma 4.3

To prove (4.6) we use the anisotropic Agmon inequality

$$\begin{aligned} \Vert \Phi \Vert _{L^\infty (V)} \le c\Vert \Phi \Vert _{L^2(V)}^{1/4}\prod _{i=1}^3\left( \Vert \Phi \Vert _{L^2(V)}+\Vert \partial _i\Phi \Vert _{L^2(V)}+\Vert \partial _i^2\Phi \Vert _{L^2(V)}\right) ^{1/4} \end{aligned}$$

(B.5)

for $V=(0,1)^3$ and $\Phi \in H^2(V)$ (see [54, Proposition 2.2]). For this purpose, we use a partition of unity on $\Gamma$ to localize a function on $\Omega _\varepsilon$.

Since $\Gamma$ is compact and without boundary, we can take a finite number of bounded open sets in ${\mathbb {R}}^2$ and local parametrizations of $\Gamma$

$$\begin{aligned} U_k \subset {\mathbb {R}}^2, \quad \mu _k:U_k\rightarrow \Gamma , \quad k=1,\dots ,k_0 \end{aligned}$$

such that $\{\mu _k(U_k)\}_{k=1}^{k_0}$ is an open covering of $\Gamma$. Let $\{\eta _k\}_{k=1}^{k_0}$ be a partition of unity on $\Gamma$ subordinate to $\{\mu _k(U_k)\}_{k=1}^{k_0}$. We may assume that $\eta _k$ is supported in $\mu _k({\mathcal {K}}_k)$ with some compact subset ${\mathcal {K}}_k$ of $U_k$ for each $k=1,\dots ,k_0$. Let ${\bar{\eta }}_k:=\eta _k\circ \pi$ be the constant extension of $\eta _k$ and

$$\begin{aligned} \zeta _k(s)&:= \mu _k(s')+\varepsilon \{(1-s_3)g_0(\mu _k(s'))+s_3g_1(\mu _k(s'))\}n(\mu _k(s')), \\ s&= (s',s_3) \in V_k := U_k\times (0,1),\, k=1,\dots ,k_0. \end{aligned}$$

Then $\{\zeta _k(V_k)\}_{k=1}^{k_0}$ is an open covering of $\Omega _\varepsilon$ and $\{{\bar{\eta }}_k\}_{k=1}^{k_0}$ is a partition of unity on $\Omega _\varepsilon$ subordinate to $\{\zeta _k(V_k)\}_{k=1}^{k_0}$. For $\varphi \in H^2(\Omega _\varepsilon )$ we define

$$\begin{aligned} \varphi _k := {\bar{\eta }}_k\varphi \quad \text {on}\quad \Omega _\varepsilon ,\, \, k=1,\dots ,k_0. \end{aligned}$$

Then $\varphi _k$ is supported in $\zeta _k({\mathcal {K}}_k\times (0,1))\subset \zeta _k(V_k)$ and

$$\begin{aligned} \partial _n\varphi _k = {\bar{\eta }}_k\partial _n\varphi , \quad \partial _n^2\varphi _k = {\bar{\eta }}_k\partial _n^2\varphi \quad \text {in}\quad \Omega _\varepsilon ,\, \, k=1,\dots ,k_0 \end{aligned}$$

by (4.3). Therefore, if we prove

$$\begin{aligned} \Vert \varphi _k\Vert _{L^\infty (\zeta _k(V_k))}&\le c\varepsilon ^{-1/2}\Vert \varphi _k\Vert _{L^2(\zeta _k(V_k))}^{1/4}\Vert \varphi _k\Vert _{H^2(\zeta _k(V_k))}^{1/2} \nonumber \\&\quad \times \left( \Vert \varphi _k\Vert _{L^2(\zeta _k(V_k))}+\varepsilon \Vert \partial _n\varphi _k\Vert _{L^2(\zeta _k(V_k))}+\varepsilon ^2\Vert \partial _n^2\varphi _k\Vert _{L^2(\zeta _k(V_k))}\right) ^{1/4} \end{aligned}$$

(B.6)

for all $k=1,\dots ,k_0$, then we obtain (4.6) for $\varphi$.

Let us show (B.6). In what follows, we fix and suppress the index k. Hence we assume that $\varphi \in H^2(\Omega _\varepsilon )$ is supported in $\zeta ({\mathcal {K}}\times (0,1))\subset \zeta (V)$ with some compact subset ${\mathcal {K}}$ of U. Taking U small and scaling it, we may further assume that

$$\begin{aligned} U = (0,1)^2, \quad V = U\times (0,1) = (0,1)^3. \end{aligned}$$

The local parametrization $\zeta$ of $\Omega _\varepsilon$ is of the form

$$\begin{aligned} \zeta (s) = \mu (s')+h_\varepsilon (s)n(\mu (s')), \quad s = (s',s_3) \in V = U\times (0,1), \end{aligned}$$

(B.7)

where $\mu :U\rightarrow \Gamma$ is a $C^5$ local parametrization of $\Gamma$ and

$$\begin{aligned} h_\varepsilon (s) := \varepsilon \{(1-s_3)g_0(\mu (s'))+s_3g_1(\mu (s'))\}. \end{aligned}$$

(B.8)

Since $g_0$, $g_1$, and n are of class $C^4$ on $\Gamma$, ${\mathcal {K}}$ is compact in U, and $h_\varepsilon$ is an affine function of $s_3$, there exists a constant $c>0$ independent of $\varepsilon$ such that

$$\begin{aligned} |\partial _{s_i}\zeta (s)| \le c, \quad |\partial _{s_i}\partial _{s_j}\zeta (s)| \le c, \quad s\in {\mathcal {K}}\times (0,1),\,\,i,j=1,2,3. \end{aligned}$$

(B.9)

Let $\nabla _s\zeta$ be the gradient matrix of $\zeta$ in $s\in {\mathbb {R}}^3$, J the function given by (3.30), and $\theta$ the Riemannian metric of $\Gamma$ given by

$$\begin{aligned} \begin{gathered} \theta (s') := \nabla _{s'}\mu (s')\{\nabla _{s'}\mu (s')\}^T, \quad s'\in U, \\ \nabla _{s'}\mu := \begin{pmatrix} \partial _{s_1}\mu _1 &{}\quad \partial _{s_1}\mu _2 &{}\quad \partial _{s_1}\mu _3 \\ \partial _{s_2}\mu _1 &{}\quad \partial _{s_2}\mu _2 &{}\quad \partial _{s_2}\mu _3 \end{pmatrix}. \end{gathered} \end{aligned}$$

(B.10)

Then

$$\begin{aligned} \det \nabla _s\zeta (s) = \varepsilon g(\mu (s'))J(\mu (s'),h_\varepsilon (s))\sqrt{\det \theta (s')}, \quad s=(s',s_3)\in V, \end{aligned}$$

(B.11)

which we prove at the end of the proof. Moreover, since $\det \theta$ is continuous and strictly positive on U and ${\mathcal {K}}$ is compact in U, we have

$$\begin{aligned} \det \theta (s') \ge c, \quad s'\in {\mathcal {K}}. \end{aligned}$$

Applying this inequality, (2.1), and (3.31) to (B.11) we obtain

$$\begin{aligned} \det \nabla _s\zeta (s) \ge c\varepsilon , \quad s\in {\mathcal {K}}\times (0,1). \end{aligned}$$

(B.12)

Now let $\Phi :=\varphi \circ \zeta$ on V. Then

$$\begin{aligned} \Vert \Phi \Vert _{L^\infty (V)} = \Vert \varphi \Vert _{L^\infty (\zeta (V))} \end{aligned}$$

(B.13)

and $\Phi$ is supported in ${\mathcal {K}}\times (0,1)$ since $\varphi$ is supported in $\zeta ({\mathcal {K}}\times (0,1))$. Also, since

$$\begin{aligned} \int _{\zeta (V)}\varphi (x)\,dx = \int _V\Phi (s)\det \nabla _s\zeta (s)\,ds, \end{aligned}$$

(B.14)

we observe by (B.12) that

$$\begin{aligned} \Vert \Phi \Vert _{L^2(V)} \le c\varepsilon ^{-1/2}\Vert \varphi \Vert _{L^2(\zeta (V))}. \end{aligned}$$

(B.15)

We differentiate $\Phi (s)=\varphi (\zeta (s))$ in $s\in V$. Then

$$\begin{aligned} \partial _{s_i}\Phi (s)&= \partial _{s_i}\zeta (s)\cdot \nabla \varphi (\zeta (s)), \\ \partial _{s_i}^2\Phi (s)&= \partial _{s_i}^2\zeta (s)\cdot \nabla \varphi (\zeta (s))+\partial _{s_i}\zeta (s)\cdot \nabla ^2\varphi (\zeta (s))\partial _{s_i}\zeta (s) \end{aligned}$$

for $s\in V$ and $i=1,2$, and

$$\begin{aligned} \partial _{s_3}\Phi (s) = \varepsilon g(\mu (s'))\partial _n\varphi (\zeta (s)), \quad \partial _{s_3}^2\Phi (s) = \varepsilon ^2g(\mu (s'))^2\partial _n^2\varphi (\zeta (s)) \end{aligned}$$

for $s=(s',s_3)\in V$. Hence (B.9) and the boundedness of g on $\Gamma$ imply that

$$\begin{aligned} |\partial _{s_i}\Phi (s)|&\le c|\nabla \varphi (\zeta (s))|, \\ |\partial _{s_i}^2\Phi (s)|&\le c(|\nabla \varphi (\zeta (s))|+|\nabla ^2\varphi (\zeta (s))|), \\ |\partial _{s_3}^k\Phi (s)|&\le c\varepsilon ^k|\partial _n^k\varphi (\zeta (s))|, \end{aligned}$$

for $i,k=1,2$ and $s\in {\mathcal {K}}\times (0,1)$. Since $\Phi$ is supported in ${\mathcal {K}}\times (0,1)$, we deduce from these inequalities and (B.14) that

$$\begin{aligned} \begin{aligned} \Vert \partial _{s_i}^k\Phi \Vert _{L^2(V)}&\le c\varepsilon ^{-1/2}\Vert \varphi \Vert _{H^k(\zeta (V))}, \\ \Vert \partial _{s_3}^k\Phi \Vert _{L^2(V)}&\le c\varepsilon ^{k-1/2}\Vert \partial _n^k\varphi \Vert _{L^2(\zeta (V))} \end{aligned} \end{aligned}$$

(B.16)

for $i,k=1,2$. Similarly, for $i,j=1,2,3$ with $i\ne j$ we have

$$\begin{aligned} \Vert \partial _{s_i}\partial _{s_j}\Phi \Vert _{L^2(V)}\le c\varepsilon ^{-1/2}\Vert \varphi \Vert _{H^2(\zeta (V))} \end{aligned}$$

(B.17)

and thus $\Phi \in H^2(V)$. Hence we can apply (B.5) to $\Phi =\varphi \circ \zeta$ and use (B.13), (B.15), and (B.16) to obtain (B.6).

It remains to show the formula (B.11). In what follows, we use the notation

$$\begin{aligned} \eta ^\flat (s') := \eta (\mu (s')), \quad s'\in U \end{aligned}$$

for a function $\eta$ on $\Gamma$. Note that, since $\eta (\mu (s'))={\bar{\eta }}(\mu (s'))$ by $\mu (s')\in \Gamma$,

$$\begin{aligned} \partial _{s_i}\eta ^\flat (s') = \partial _{s_i}\mu (s')\cdot \nabla {\bar{\eta }}(\mu (s')) = \partial _{s_i}\mu (s')\cdot \nabla _\Gamma \eta (\mu (s')) \end{aligned}$$

(B.18)

for $i=1,2$ by (3.5). By (B.7) and (B.8) we have

$$\begin{aligned} \zeta (s) = \mu (s')+\varepsilon \{(1-s_3)g_0^\flat (s')+s_3g_1^\flat (s')\}n^\flat (s'), \quad s = (s',s_3) \in V = U\times (0,1). \end{aligned}$$

We differentiate $\zeta (s)$ and apply (B.18) and $-\nabla _\Gamma n=W=W^T$ on $\Gamma$ to get

$$\begin{aligned} \begin{aligned} \partial _{s_i}\zeta (s)&= \{I_3-h_\varepsilon (s)W^\flat (s')\}\partial _{s_i}\mu (s')+\eta _\varepsilon ^i(s)n^\flat (s'), \quad i=1,2, \\ \partial _{s_3}\zeta (s)&= \varepsilon g^\flat (s')n^\flat (s') \end{aligned} \end{aligned}$$

(B.19)

for $s=(s',s_3)\in V$, where $h_\varepsilon (s)$ is given by (B.8) and

$$\begin{aligned} \eta _\varepsilon ^i(s) := \varepsilon \partial _{s_i}\mu (s')\cdot \{(1-s_3)(\nabla _\Gamma g_0)^\flat (s')+s_3(\nabla _\Gamma g_1)^\flat (s')\}, \quad i=1,2. \end{aligned}$$

From now on, we fix and suppress the arguments $s'$ and s. By (B.19) we have

$$\begin{aligned} \nabla _s\zeta = \begin{pmatrix} \partial _{s_1}\zeta _1 &{}\quad \partial _{s_1}\zeta _2 &{}\quad \partial _{s_1}\zeta _3 \\ \partial _{s_2}\zeta _1 &{} \quad \partial _{s_2}\zeta _2 &{}\quad \partial _{s_2}\zeta _3 \\ \partial _{s_3}\zeta _1 &{}\quad \partial _{s_3}\zeta _2 &{}\quad \partial _{s_3}\zeta _3 \end{pmatrix} = \begin{pmatrix} \nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^T+\eta _\varepsilon \otimes n^\flat \\ \varepsilon g^\flat (n^\flat )^T \end{pmatrix}. \end{aligned}$$

Here we consider $n^\flat \in {\mathbb {R}}^3$ and $\eta _\varepsilon :=(\eta _\varepsilon ^1,\eta _\varepsilon ^2)^T\in {\mathbb {R}}^2$ as column vectors. Since $\partial _{s_1}\mu$ and $\partial _{s_2}\mu$ are tangent to $\Gamma$ at $\mu (s')$ we have $(\nabla _{s'}\mu )n^\flat =0$. Moreover,

$$\begin{aligned} W^\flat n^\flat =0, \quad (\eta _\varepsilon \otimes n^\flat )n^\flat =|n^\flat |^2\eta _\varepsilon =\eta _\varepsilon , \quad (\eta _\varepsilon \otimes n^\flat )(n^\flat \otimes \eta _\varepsilon ) = \eta _\varepsilon \otimes \eta _\varepsilon . \end{aligned}$$

From these equalities and the symmetry of the matrix $W^\flat$ it follows that

$$\begin{aligned} \nabla _s\zeta (\nabla _s\zeta )^T = \begin{pmatrix} \nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^2(\nabla _{s'}\mu )^T+\eta _\varepsilon \otimes \eta _\varepsilon &{}\quad \varepsilon g^\flat \eta _\varepsilon \\ \varepsilon g^\flat \eta _\varepsilon ^T &{}\quad \varepsilon ^2(g^\flat )^2 \end{pmatrix}. \end{aligned}$$

Hence by elementary row operations we have

$$\begin{aligned} \det [\nabla _s\zeta (\nabla _s\zeta )^T]&= \det \begin{pmatrix} \nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^2(\nabla _{s'}\mu )^T+\eta _\varepsilon \otimes \eta _\varepsilon &{}\quad \varepsilon g^\flat \eta _\varepsilon \\ \varepsilon g^\flat \eta _\varepsilon ^T &{}\quad \varepsilon ^2(g^\flat )^2 \end{pmatrix} \\&= \det \begin{pmatrix} \nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^2(\nabla _{s'}\mu )^T &{}\quad 0 \\ \varepsilon g^\flat \eta _\varepsilon ^T &{}\quad \varepsilon ^2 (g^\flat )^2 \end{pmatrix} \\&= \varepsilon ^2 (g^\flat )^2\det [\nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^2(\nabla _{s'}\mu )^T]. \end{aligned}$$

Since $\det [\nabla _s\zeta (\nabla _s\zeta )^T]=(\det \nabla _s\zeta )^2$, the above equality implies that

$$\begin{aligned} (\det \nabla _s\zeta )^2 = \varepsilon ^2(g^\flat )^2\det [\nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^2(\nabla _{s'}\mu )^T]. \end{aligned}$$

(B.20)

To compute the right-hand side we define $3\times 3$ matrices

$$\begin{aligned} A := \begin{pmatrix} \nabla _{s'}\mu \\ (n^\flat )^T \end{pmatrix}, \quad A_h := \begin{pmatrix} \nabla _{s'}\mu (I_3-h_\varepsilon W^\flat ) \\ (n^\flat )^T \end{pmatrix}. \end{aligned}$$

Then by $(\nabla _{s'}\mu )n^\flat =0$, $W^\flat n^\flat =0$, the symmetry of $W^\flat$, and (B.10) we have

$$\begin{aligned} A_h&= A(I_3-h_\varepsilon W^\flat ), \\ AA^T&= \begin{pmatrix} \theta &{} \quad 0 \\ 0 &{}\quad 1 \end{pmatrix}, \quad A_hA_h^T = \begin{pmatrix} \nabla _{s'}\mu (I_3-h^\flat W^\flat )^2(\nabla _{s'}\mu )^T &{}\quad 0 \\ 0 &{}\quad 1 \end{pmatrix}. \end{aligned}$$

Noting that A and $I_3-h_\varepsilon W^\flat$ are $3\times 3$ matrices, we use these equalities to get

$$\begin{aligned} \begin{aligned} \det [\nabla _{s'}\mu (I_3-h_\varepsilon W^\flat )^2(\nabla _{s'}\mu )^T]&= \det [A_hA_h^T] = \det [A(I_3-h_\varepsilon W^\flat )^2A^T] \\&= \det [(I_3-h_\varepsilon W^\flat )^2]\det [AA^T] \\&= J(\mu ,h_\varepsilon )^2\det \theta , \end{aligned} \end{aligned}$$

where the last equality follows from $\det (I_3-h_\varepsilon W^\flat )=J(\mu ,h_\varepsilon )$. From this equality and (B.20) we deduce that

$$\begin{aligned} (\det \nabla _s\zeta )^2 = \varepsilon ^2(g^\flat )^2J(\mu ,h_\varepsilon )^2\det \theta . \end{aligned}$$

This equality yields (B.11) since $g^\flat$ and $J(\mu ,h_\varepsilon )$ are positive by (2.1) and (3.31). $\square$

Finally, let us prove Lemma 7.4. To this end, we give an auxiliary result.

Lemma B.3

Let $E_1$, $E_2$, and $E_3$ be vector fields on an open subset U of ${\mathbb {R}}^3$ such that $\{E_1(x),E_2(x),E_3(x)\}$ is an orthonormal basis of ${\mathbb {R}}^3$ for each $x\in U$ and

$$\begin{aligned} E_1\times E_2 = E_3, \quad E_2\times E_3 = E_1, \quad E_3\times E_1 = E_2 \quad \text {in}\quad U. \end{aligned}$$

Then for $u\in C^1(U)^3$ we have

$$\begin{aligned} \mathrm {curl}\,u&= \{(E_2\cdot \nabla )u\cdot E_3-(E_3\cdot \nabla )u\cdot E_2\}E_1 \nonumber \\&\quad +\{(E_3\cdot \nabla )u\cdot E_1-(E_1\cdot \nabla )u\cdot E_3\}E_2 \nonumber \\&\quad +\{(E_1\cdot \nabla )u\cdot E_2-(E_2\cdot \nabla )u\cdot E_1\}E_3 \quad \text {in}\quad U. \end{aligned}$$

(B.21)

Proof

By the assumption, $\mathrm {curl}\,u=\sum _{i=1}^3(\mathrm {curl}\,u\cdot E_i)E_i$. Since $E_1=E_2\times E_3$,

$$\begin{aligned} \mathrm {curl}\,u\cdot E_1&= \mathrm {curl}\,u\cdot (E_2\times E_3) = E_2\cdot (E_3\times \mathrm {curl}\,u) \\&= E_2\cdot \{(\nabla u)E_3-(\nabla u)^TE_3\} \\&= (\nabla u)^TE_2\cdot E_3-(\nabla u)^TE_3\cdot E_2 \\&= (E_2\cdot \nabla )u\cdot E_3-(E_3\cdot \nabla )u\cdot E_2. \end{aligned}$$

Calculating $\mathrm {curl}\,u\cdot E_i$, $i=2,3$ in the same way we obtain (B.21). $\square$

Proof of Lemma 7.4

Let $u\in C^1(\Omega _\varepsilon )^3$ and $u^a$ be given by (6.47). Since the surface $\Gamma$ is compact and without boundary, we can take a finite number of relatively open subsets $O_k$ of $\Gamma$ and pairs of tangential vector fields $\{\tau _1^k,\tau _2^k\}$ on $O_k$, $k=1,\dots ,k_0$ such that $\Gamma =\bigcup _{k=1}^{k_0}O_k$, the triplet $\{\tau _1^k,\tau _2^k,n\}$ forms an orthonormal basis of ${\mathbb {R}}^3$ on $O_k$, and

$$\begin{aligned} \tau _1^k\times \tau _2^k = n, \quad \tau _2^k\times n = \tau _1^k, \quad n\times \tau _1^k = \tau _2^k \quad \text {on}\quad O_k \end{aligned}$$

for each $k=1,\dots ,k_0$. Then since $\Omega _\varepsilon =\bigcup _{k=1}^{k_0}U_k$ with

$$\begin{aligned} U_k:=\{y+rn(y)\mid y\in O_k,\,r\in (\varepsilon g_0(y),\varepsilon g_1(y))\}, \quad k=1,\dots ,k_0, \end{aligned}$$

it is sufficient to show (7.5) in $U_k$ for each $k=1,\dots ,k_0$.

From now on, we fix and suppress the index k and carry out calculations in U unless otherwise stated. We apply (B.21) to $u^a$ with

$$\begin{aligned} E_1 := {\bar{\tau }}_1, \quad E_2 := {\bar{\tau }}_2, \quad E_3 := {\bar{n}} \end{aligned}$$

and use $P\tau _i=\tau _i$ for $i=1,2$ and $Pn=0$ on O to get

$$\begin{aligned} {\overline{P}}\,\mathrm {curl}\,u^a = \{({\bar{\tau }}_2\cdot \nabla )u^a\cdot {\bar{n}}-({\bar{n}}\cdot \nabla )u^a\cdot {\bar{\tau }}_2\}{\bar{\tau }}_1+\{({\bar{n}}\cdot \nabla )u^a\cdot {\bar{\tau }}_1-({\bar{\tau }}_1\cdot \nabla )u^a\cdot {\bar{n}}\}{\bar{\tau }}_2. \end{aligned}$$

By this equality, $({\bar{n}}\cdot \nabla )u^a=\partial _nu^a$, and $|{\bar{\tau }}_1|=|{\bar{\tau }}_2|=|{\bar{n}}|=1$ we get

$$\begin{aligned} \left| {\overline{P}}\,\mathrm {curl}\,u^a\right| \le c\left( |\partial _nu^a|+|({\bar{\tau }}_1\cdot \nabla )u^a\cdot {\bar{n}}|+|({\bar{\tau }}_2\cdot \nabla )u^a\cdot {\bar{n}}|\right) . \end{aligned}$$

(B.22)

Let us estimate each term on the right-hand side. By (4.3) and (6.47) we have

$$\begin{aligned} \partial _nu^a = \partial _n\Bigl [\overline{M_\tau u}+\Bigl (\overline{M_\tau u}\cdot \Psi _\varepsilon \Bigr ){\bar{n}}\Bigr ] = \Bigl (\overline{M_\tau u}\cdot \partial _n\Psi _\varepsilon \Bigr ){\bar{n}}. \end{aligned}$$

Hence it follows from (4.19) that

$$\begin{aligned} |\partial _nu^a| = \left| \overline{M_\tau u}\cdot \partial _n\Psi _\varepsilon \right| \le c\left| \overline{M_\tau u}\right| = c\left| {\overline{PMu}}\right| \le c\left| {\overline{Mu}}\right| . \end{aligned}$$

(B.23)

To estimate the other terms we set

$$\begin{aligned} u_\tau ^a := {\overline{P}}u^a = \overline{M_\tau u}, \quad u_n^a := (u^a\cdot {\bar{n}}){\bar{n}} = \Bigl (\overline{M_\tau u}\cdot \Psi _\varepsilon \Bigr ){\bar{n}} \end{aligned}$$

(B.24)

so that $u^a=u_\tau ^a+u_n^a$. Let $i=1,2$. Since $u_\tau ^a\cdot {\bar{n}}=0$ in U, we have

$$\begin{aligned} ({\bar{\tau }}_i\cdot \nabla )u_\tau ^a\cdot {\bar{n}} = ({\bar{\tau }}_i\cdot \nabla )(u_\tau ^a\cdot {\bar{n}})-u_\tau ^a\cdot ({\bar{\tau }}_i\cdot \nabla ){\bar{n}} = -u_\tau ^a\cdot ({\bar{\tau }}_i\cdot \nabla ){\bar{n}}. \end{aligned}$$

Hence by (3.17) and $|{\bar{\tau }}_i|=1$ we get

$$\begin{aligned} |({\bar{\tau }}_i\cdot \nabla )u_\tau ^a\cdot {\bar{n}}| \le c|u_\tau ^a| \le c\left| {\overline{Mu}}\right| , \quad i=1,2. \end{aligned}$$

(B.25)

Next we deal with $({\bar{\tau }}_i\cdot \nabla )u_n^a\cdot {\bar{n}}$. Since $\tau _i=P\tau _i$, $P=P^T$, and $|\tau _i|=1$ on O,

$$\begin{aligned} |({\bar{\tau }}_i\cdot \nabla )u_n^a| = \left| (\nabla u_n^a)^T{\overline{P}}{\bar{\tau }}_i\right| = \left| \Bigl ({\overline{P}}\nabla u_n^a\Bigr )^T{\bar{\tau }}_i\right| \le \left| {\overline{P}}\nabla u_n^a\right| . \end{aligned}$$

(B.26)

Moreover, by the definition (B.24) of $u_n^a$ we have

$$\begin{aligned} {\overline{P}}\nabla u_n^a = \left[ \left\{ {\overline{P}}\nabla \Bigl (\overline{M_\tau u}\Bigr )\right\} \Psi _\varepsilon +\Bigl ({\overline{P}}\nabla \Psi _\varepsilon \Bigr )\overline{M_\tau u}\right] \otimes {\bar{n}}+\Bigl (\overline{M_\tau u}\cdot \Psi _\varepsilon \Bigr ){\overline{P}}\nabla {\bar{n}} \end{aligned}$$

and thus we deduce from (3.13), (3.17), (4.19), and (4.20) that

$$\begin{aligned} \left| {\overline{P}}\nabla u_n^a\right| \le c\varepsilon \left( \left| \overline{M_\tau u}\right| +\left| \overline{\nabla _\Gamma M_\tau u}\right| \right) \le c\varepsilon \left( \left| {\overline{Mu}}\right| +\left| \overline{\nabla _\Gamma Mu}\right| \right) . \end{aligned}$$

(B.27)

In the last inequality we also used $M_\tau u=PMu$ on $\Gamma$ and the $C^4$-regularity of P on $\Gamma$. By (B.26) and (B.27) we observe that

$$\begin{aligned} |({\bar{\tau }}_i\cdot \nabla )u_n^a\cdot {\bar{n}}| \le |({\bar{\tau }}_i\cdot \nabla )u_n^a| \le c\varepsilon \left( \left| {\overline{Mu}}\right| +\left| \overline{\nabla _\Gamma Mu}\right| \right) , \quad i=1,2. \end{aligned}$$

(B.28)

Noting that $u^a=u_\tau ^a+u_n^a$, we conclude by (B.22), (B.23), (B.25), and (B.28) that the inequality (7.5) holds in U. $\square$

Appendix C: Weaker Assumption on the Friction Coefficients

In this appendix we discuss the global existence of a strong solution to (1.2) under an assumption on $\gamma _\varepsilon ^0$ and $\gamma _\varepsilon ^1$ weaker than Assumption 2.1.

For a fixed $\delta \in [0,1]$ we assume that there exists a constant $c>0$ such that

$$\begin{aligned} \gamma _\varepsilon ^0 \le c\varepsilon ^\delta , \quad \gamma _\varepsilon ^1 \le c\varepsilon ^\delta \end{aligned}$$

(C.1)

for all $\varepsilon \in (0,1]$ and that either of the condition (A2) of Assumption 2.2 or the following condition is satisfied:

• (A4) There exists a constant $c>0$ such that

  $$\begin{aligned} \gamma _\varepsilon ^0 \ge c\varepsilon ^\delta \quad \text {for all}\quad \varepsilon \in (0,1] \quad \text {or}\quad \gamma _\varepsilon ^1 \ge c\varepsilon ^\delta \quad \text {for all}\quad \varepsilon \in (0,1]. \end{aligned}$$

Then we can show the global existence of a strong solution to (1.2) under an additional condition on $\delta$ as in the case of flat thin domains [12, 14]. To see this, let us explain which inequalities in the first part [34] and this paper are modified. In what follows, we omit the proofs of modified inequalities if they are obtained just by replacing the estimates for $\gamma _\varepsilon ^0$ and $\gamma _\varepsilon ^1$ in the proofs of the original ones.

As in Sect. 2, let ${\mathcal {H}}_\varepsilon :=L_\sigma ^2(\Omega _\varepsilon )$ and ${\mathcal {V}}_\varepsilon :={\mathcal {H}}_\varepsilon \cap H^1(\Omega _\varepsilon )^3$. First we note that we can take the same $\varepsilon _0\in (0,1]$ as in [34, Theorem 2.4] to get the boundedness and coerciveness of the bilinear form $a_\varepsilon$ on ${\mathcal {V}}_\varepsilon$. However, they are not uniform in $\varepsilon$ and the basic inequalities for the Stokes operator $A_\varepsilon$ given in [34, Lemma 2.5] change as follows.

Lemma C.1

For all $\varepsilon \in (0,\varepsilon _0]$ and $u\in {\mathcal {V}}_\varepsilon$ we have

$$\begin{aligned} \begin{aligned} \Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}&\le c\left( \Vert \nabla u\Vert _{L^2(\Omega _\varepsilon )}+\varepsilon ^{(\delta -1)/2}\Vert u\Vert _{L^2(\Omega _\varepsilon )}\right) , \\ \Vert \nabla u\Vert _{L^2(\Omega _\varepsilon )}&\le c\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )} \end{aligned} \end{aligned}$$

(C.2)

and

$$\begin{aligned} \Vert u\Vert _{L^2(\Omega _\varepsilon )} \le {\left\{ \begin{array}{ll} c\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )} &{}\text {if the condition (A2) is imposed}, \\ c\varepsilon ^{(1-\delta )/2}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )} &{}\text {if the condition (A4) is imposed}. \end{array}\right. } \end{aligned}$$

(C.3)

Moreover, if $u\in D(A_\varepsilon )$, then

$$\begin{aligned} \Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )} \le {\left\{ \begin{array}{ll} c\Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )} &{}\text {if the condition (A2) is imposed}, \\ c\varepsilon ^{(1-\delta )/2}\Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )} &{}\text {if the condition (A4) is imposed}. \end{array}\right. } \end{aligned}$$

For a vector field u on $\Omega _\varepsilon$, the auxiliary vector field G(u) given by (7.2) satisfies

$$\begin{aligned} |\nabla G(u)| \le c(|\nabla u|+\varepsilon ^{\delta -1}|u|) \quad \text {in}\quad \Omega _\varepsilon \end{aligned}$$

(C.4)

instead of the second inequality of (7.3) (see [34, Lemma 7.2]). Using this inequality we can show the following estimate as in the proof of [34, Theorem 2.6].

Lemma C.2

For all $\varepsilon \in (0,\varepsilon _0]$ and $u\in D(A_\varepsilon )$ we have

$$\begin{aligned} \Vert A_\varepsilon u+\nu \Delta u\Vert _{L^2(\Omega _\varepsilon )} \le c\left( \Vert \nabla u\Vert _{L^2(\Omega _\varepsilon )}+\varepsilon ^{\delta -1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}\right) . \end{aligned}$$

(C.5)

We also observe that the uniform a priori estimate for the vector Laplace operator shown in [34, Lemma 6.1] gets slightly worse.

Lemma C.3

For all $\varepsilon \in (0,1]$ and $u\in H^2(\Omega _\varepsilon )^3$ satisfying (4.8) we have

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega _\varepsilon )} \le c\left( \Vert \Delta u\Vert _{L^2(\Omega _\varepsilon )}+\varepsilon ^{(\delta -1)/2}\Vert u\Vert _{H^1(\Omega _\varepsilon )}\right) . \end{aligned}$$

Using Lemmas C.1–C.3 and noting that $A_\varepsilon =-\nu {\mathbb {P}}_\varepsilon \Delta$ and ${\mathbb {P}}_\varepsilon$ is the orthogonal projection from $L^2(\Omega _\varepsilon )^3$ onto ${\mathcal {H}}_\varepsilon$, we obtain the following norm equivalence for $A_\varepsilon$ instead of [34, Theorem 2.7] (also note that $1\le \varepsilon ^{(\delta -1)/2}\le \varepsilon ^{\delta -1}$ by $\delta -1\le 0$).

Lemma C.4

For all $\varepsilon \in (0,\varepsilon _0]$ and $u\in D(A_\varepsilon )$ we have

$$\begin{aligned} \Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )} \le c\Vert u\Vert _{H^2(\Omega _\varepsilon )}. \end{aligned}$$

Moreover, if the condition (A2) is imposed, then

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega _\varepsilon )} \le c\left( \Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )}+\varepsilon ^{(\delta -1)/2}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}+\varepsilon ^{\delta -1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}\right) \end{aligned}$$

and, if the condition (A4) is imposed,

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega _\varepsilon )} \le c\Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )}. \end{aligned}$$

In particular, under both of the conditions (A2) and (A4) we have

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega _\varepsilon )} \le c\left( \Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )}+\varepsilon ^{\delta -1}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}\right) . \end{aligned}$$

(C.6)

Now let us move to the results of this paper. When the inequalities (C.1) are valid, for $p\in [1,\infty )$ and $u\in W^{2,p}(\Omega _\varepsilon )^3$ satisfying (4.8) on $\Gamma _\varepsilon ^0$ or on $\Gamma _\varepsilon ^1$ we have

$$\begin{aligned} \left\| {\overline{P}}\partial _nu+{\overline{W}}u\right\| _{L^p(\Omega _\varepsilon )} \le c\left( \varepsilon \Vert u\Vert _{W^{2,p}(\Omega _\varepsilon )}+\varepsilon ^\delta \Vert u\Vert _{L^p(\Omega _\varepsilon )}\right) \end{aligned}$$

instead of (4.16). This modified inequality, however, does not affect the other parts of this paper since (4.16) was used only in the proof of (6.61) and we get the same result even if we replace (4.16) by the above inequality due to the fact that the Weingarten map W of $\Gamma$ does not vanish in general. We also note that most of the estimates in [34] and this paper are worse than the corresponding estimates for flat thin domains given in [12, 14] because of the nonzero curvatures of the limit set $\Gamma$.

Next we see that the estimate (7.1) for the trilinear term changes to the one in which an additional term appears due to the modified inequalities (C.4) and (C.5).

Lemma C.5

For any $\alpha >0$ there exist constants $c_\alpha ^1,c_\alpha ^2,c_\alpha ^3>0$ such that

$$\begin{aligned} \left| \bigl ((u\cdot \nabla )u,A_\varepsilon u\bigr )_{L^2(\Omega _\varepsilon )}\right|&\le \left( \alpha +c_\alpha ^1\varepsilon ^{1/2}\Vert u\Vert _{H^1(\Omega _\varepsilon )}\right) \Vert u\Vert _{H^2(\Omega _\varepsilon )}^2 \nonumber \\&\quad +c_\alpha ^2\left( \Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^4+\varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^2\right) \nonumber \\&\quad +c_\alpha ^3\varepsilon ^{-2+4\delta /3}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^{4/3} \end{aligned}$$

(C.7)

for all $\varepsilon \in (0,\varepsilon _0]$ and $u\in D(A_\varepsilon )$. (In fact, $c_\alpha ^1$ does not depend on $\alpha$.)

Proof

The proof is the same as that of Lemma 7.1 except for the estimates of

$$\begin{aligned} J_2 = (\omega \times u^a,A_\varepsilon u+\nu \Delta u)_{L^2(\Omega _\varepsilon )}, \quad J_3^1 = -\nu (\mathrm {curl}\,G(u),\Phi )_{L^2(\Omega _\varepsilon )} \end{aligned}$$

with $\omega =\mathrm {curl}\,u$ and $\Phi =\omega \times u^a$. For $J_2$ we apply (7.7) and (C.5) instead of (5.6). Then we get

$$\begin{aligned} |J_2| \le \Vert \omega \times u^a\Vert _{L^2(\Omega _\varepsilon )}\Vert A_\varepsilon u+\nu \Delta u\Vert _{L^2(\Omega _\varepsilon )} \le c(K_1^\prime +K_2^\prime ), \end{aligned}$$

where

$$\begin{aligned} K_1^\prime&:= \varepsilon ^{-1/2}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^{1/2}\Vert u\Vert _{H^1(\Omega _\varepsilon )}^2\Vert u\Vert _{H^2(\Omega _\varepsilon )}^{1/2}, \\ K_2^\prime&:= \varepsilon ^{-3/2+\delta }\Vert u\Vert _{L^2(\Omega _\varepsilon )}^{3/2}\Vert u\Vert _{H^1(\Omega _\varepsilon )}\Vert u\Vert _{H^2(\Omega _\varepsilon )}^{1/2}. \end{aligned}$$

We observe by (5.8) and Young’s inequality $ab\le \alpha a^2+c_\alpha b^2$ that

$$\begin{aligned} K_1^\prime&\le \varepsilon ^{-1/2}\Vert u\Vert _{L^2(\Omega _\varepsilon )}\Vert u\Vert _{H^1(\Omega _\varepsilon )}\Vert u\Vert _{H^2(\Omega _\varepsilon )} \\&\le \alpha \Vert u\Vert _{H^2(\Omega _\varepsilon )}^2+c_\alpha \varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^2. \end{aligned}$$

Also, applying Young’s inequality $ab\le \alpha a^4+c_\alpha b^{4/3}$ to $K_2^\prime$ we have

$$\begin{aligned} K_2^\prime \le \alpha \Vert u\Vert _{H^2(\Omega _\varepsilon )}^2+c_\alpha \varepsilon ^{-2+4\delta /3}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^{4/3}. \end{aligned}$$

Combining the above estimates we obtain (after replacing $c\alpha$ by $\alpha$)

$$\begin{aligned}&|J_2| \le \alpha \Vert u\Vert _{H^2(\Omega _\varepsilon )}^2 \nonumber \\&+c_\alpha \left( \varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^2+\varepsilon ^{-2+4\delta /3}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert u\Vert _{H^1(\Omega _\varepsilon )}^{4/3}\right) . \end{aligned}$$

(C.8)

For $J_3^1$ we deduce from (7.7) and (C.4) instead of (7.3) that

$$\begin{aligned} |J_3^1| \le c\Vert \nabla G(u)\Vert _{L^2(\Omega _\varepsilon )}\Vert \Phi \Vert _{L^2(\Omega _\varepsilon )} \le c(K_1^\prime +K_2^\prime ) \end{aligned}$$

with the same $K_1^\prime$ and $K_2^\prime$ as above. Thus (C.8) holds for $J_3^1$ and (C.7) follows. $\square$

Now we make an additional assumption on $\delta$, which is the same as those in [12, 14], to derive a good estimate for the trilinear term in terms of $A_\varepsilon$.

Lemma C.6

Suppose that

$$\begin{aligned} \delta _0 \le \delta \le 1, \quad \text {where}\quad \delta _0 := {\left\{ \begin{array}{ll} 3/4 &{}\text {if the condition (A2) is imposed}, \\ 2/3 &{}\text {if the condition (A4) is imposed}. \end{array}\right. } \end{aligned}$$

(C.9)

Then there exist constants $d_1,d_2,d_3>0$ such that

$$\begin{aligned}&\left| \bigl ((u\cdot \nabla )u,A_\varepsilon u\bigr )_{L^2(\Omega _\varepsilon )}\right| \nonumber \\&\quad \le \left( \frac{1}{4}+d_1\varepsilon ^{1/2}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}\right) \left( \Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )}^2+\varepsilon ^{-1}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2\right) \nonumber \\&\qquad +d_2\left( \Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^4+\varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2\right) \nonumber \\&\qquad +d_3\varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^{4/3}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2 \end{aligned}$$

(C.10)

for all $\varepsilon \in (0,\varepsilon _0]$ and $u\in D(A_\varepsilon )$.

Proof

First we observe by (C.2), (C.3), and $\varepsilon ^{(1-\delta )/2}\le 1$ that

$$\begin{aligned} \Vert u\Vert _{H^1(\Omega _\varepsilon )}\le c\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}. \end{aligned}$$

We apply this inequality and (C.6) to (C.7) to get

$$\begin{aligned}&\left| \bigl ((u\cdot \nabla )u,A_\varepsilon u\bigr )_{L^2(\Omega _\varepsilon )}\right| \nonumber \\&\quad \le \left( c\alpha +d_\alpha ^1\varepsilon ^{1/2}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}\right) \left( \Vert A_\varepsilon u\Vert _{L^2(\Omega _\varepsilon )}^2+\varepsilon ^{2\delta -2}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2\right) \nonumber \\&\qquad +d_\alpha ^2\left( \Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^4+\varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2\right) \nonumber \\&\qquad +d_\alpha ^3\varepsilon ^{-2+4\delta /3}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^{4/3} \end{aligned}$$

(C.11)

with positive constants c, $d_\alpha ^1$, $d_\alpha ^2$, and $d_\alpha ^3$ independent of $\varepsilon$. Moreover, since

$$\begin{aligned}&\varepsilon ^{-2+4\delta /3}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^{4/3} \le c\varepsilon ^{-1+\sigma }\Vert u\Vert _{L^2(\Omega _\varepsilon )}^{4/3}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2, \\&\sigma := {\left\{ \begin{array}{ll} -1+4\delta /3 &{}\text {if the condition (A2) is imposed}, \\ -2/3+\delta &{}\text {if the condition (A4) is imposed} \end{array}\right. } \end{aligned}$$

by (C.3), under the assumption (C.9) we have $\sigma \ge 0$ and thus (note that $\varepsilon \le 1$)

$$\begin{aligned} \varepsilon ^{-2+4\delta /3}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^2\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^{4/3} \le c\varepsilon ^{-1}\Vert u\Vert _{L^2(\Omega _\varepsilon )}^{4/3}\Vert A_\varepsilon ^{1/2}u\Vert _{L^2(\Omega _\varepsilon )}^2. \end{aligned}$$

We also observe that $\varepsilon ^{2\delta -2}=\varepsilon ^{-1+(2\delta -1)}\le \varepsilon ^{-1}$ since $2\delta -1\ge 0$ by (C.9). Applying these inequalities to (C.11) and taking $\alpha =1/4c$ we obtain (C.10). $\square$

Finally, we observe that the global existence of a strong solution to (1.2) can be established as in Theorem 2.6.

Theorem C.7

Suppose that the inequalities (C.1) are valid and that either of the conditions (A2) and (A4) is satisfied. Then under the condition (C.9) there exists a constant $c_0\in (0,1]$ such that the following statement holds: for each $\varepsilon \in (0,\varepsilon _0]$ suppose that the given data

$$\begin{aligned} u_0^\varepsilon \in {\mathcal {V}}_\varepsilon , \quad f^\varepsilon \in L^\infty (0,\infty ;L^2(\Omega _\varepsilon )^3) \end{aligned}$$

satisfy

$$\begin{aligned}&\varepsilon ^{\delta -1}\Vert u_0^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2+\Vert \nabla u_0^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2+\Vert {\mathbb {P}}_\varepsilon f^\varepsilon \Vert _{L^\infty (0,\infty ;L^2(\Omega _\varepsilon ))}^2 \nonumber \\&\quad +\Vert M_\tau u_0^\varepsilon \Vert _{L^2(\Gamma )}^2+\Vert M_\tau {\mathbb {P}}_\varepsilon f^\varepsilon \Vert _{L^\infty (0,T;H^{-1}(\Gamma ,T\Gamma ))}^2 \le c_0\varepsilon ^{-1}. \end{aligned}$$

(C.12)

Then there exists a global-in-time strong solution

$$\begin{aligned} u^\varepsilon \in C([0,\infty );{\mathcal {V}}_\varepsilon )\cap L_{loc}^2([0,\infty );D(A_\varepsilon ))\cap H_{loc}^1([0,\infty );{\mathcal {H}}_\varepsilon ) \end{aligned}$$

to the Navier–Stokes equations (1.2).

Note that (C.12) implies $\Vert u_0^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2\le c_0\varepsilon ^{-\delta }$ with $\delta \le 1$, which is slightly worse than the condition on the $L^2(\Omega _\varepsilon )$-norm of $u_0^\varepsilon$ in (2.11).

Proof

We only show the points modified from the proof of Theorem 2.6. First we deduce from (C.2) and (C.12) that

$$\begin{aligned} \Vert A_\varepsilon ^{1/2}u_0^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2 \le C_2\left( \varepsilon ^{\delta -1}\Vert u_0^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2+\Vert \nabla u_0^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2\right) \le C_2c_0\varepsilon ^{-1} \end{aligned}$$

as in (8.6). Also, (8.8) holds by (8.7), (C.12), and $\delta \le 1$. Thus we can show (8.15) for the strong solution $u^\varepsilon$ on $[0,T_{\max })$ as in the proof of Theorem 2.6.

Next we prove (8.16) by contradiction. Assume to the contrary that there exists $T\in (0,T_{\max })$ such that (8.17) and (8.18) hold. Then as in (8.20) we have

$$\begin{aligned} \frac{d}{dt}\Vert A_\varepsilon ^{1/2}u^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2+\frac{1}{2}\Vert A_\varepsilon u^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2 \le \xi \Vert A_\varepsilon ^{1/2}u^\varepsilon \Vert _{L^2(\Omega _\varepsilon )}^2+\zeta +\psi \quad \text {on}\quad (0,T] \end{aligned}$$

by (C.10) and Young’s inequality, where $\xi$ and $\zeta$ are given by (8.21) and

$$\begin{aligned} \psi (t) := \left( 1+2d_3\Vert u^\varepsilon (t)\Vert _{L^2(\Omega _\varepsilon )}^{4/3}\right) \varepsilon ^{-1}\Vert A_\varepsilon ^{1/2}u^\varepsilon (t)\Vert _{L^2(\Omega _\varepsilon )}^2, \quad t\in (0,T]. \end{aligned}$$

For this additional function $\psi$, we observe by (8.15) and $c_0\le 1$ that

$$\begin{aligned} \psi (t) \le c(1+c_0^{2/3})\varepsilon ^{-1}\Vert A_\varepsilon ^{1/2}u^\varepsilon (t)\Vert _{L^2(\Omega _\varepsilon )}^2 \le c\varepsilon ^{-1}\Vert A_\varepsilon ^{1/2}u^\varepsilon (t)\Vert _{L^2(\Omega _\varepsilon )}^2 \end{aligned}$$

for all $t\in (0,T]$ and, if $T\ge 1$ and $t\in [1,T]$,

$$\begin{aligned} \int _{t-1}^t\psi (s)\,ds \le c\varepsilon ^{-1}\int _{t-1}^t\Vert A_\varepsilon ^{1/2}u^\varepsilon (s)\Vert _{L^2(\Omega _\varepsilon )}^2\,ds \le cc_0\varepsilon ^{-1}. \end{aligned}$$

Therefore, proceeding as in the proof of Theorem 2.6 with $\zeta$ replaced by $\zeta +\psi$, we can get a contradiction and conclude that (8.16) is valid, which yields $T_{\max }=\infty$. $\square$