On the shape of Meissner solutions to the 2-dimensional Ginzburg–Landau system

Abstract

This paper concerns the asymptotic behavior of the stable solution $(f_\lambda ,{\mathbf {Q}}_\lambda )$ of the full Meissner state equation for a two-dimensional superconductor with penetration depth $\lambda$ and Ginzburg–Landau parameter $\kappa$, and subjected to an applied magnetic field ${\mathcal {H}}^e$. It is known that the solution is stable if the minimum value of $|f_\lambda (x)|^2-|{\mathbf {Q}}_\lambda (x)|^2$ is larger than 1/3, and the solution loses its stability when the minimum value reached 1/3. It has been conjectured that the location of the minimum points of $|f_\lambda (x)|^2-|{\mathbf {Q}}_\lambda (x)|^2$ has connection with the location of vortex nucleation of the superconductor. In this paper, we prove that if the penetration depth $\lambda$ is small, the solution $(f_\lambda , {\mathbf {Q}}_\lambda )$ exhibits boundary layer behavior, and $(1-f_\lambda , {\mathbf {Q}}_\lambda )$ exponentially decays in the normal direction away from the boundary. Moreover, the minimum points of $|f_\lambda (x)|^2-|{\mathbf {Q}}_\lambda (x)|^2$ locate near the set $S({\mathcal {H}}^e)$, which is determined by the applied magnetic field ${\mathcal {H}}^e$ and the geometry of the domain. In the special case where the applied magnetic field ${\mathcal {H}}^e$ is constant, the minimum points of $|f_\lambda (x)|^2-|{\mathbf {Q}}_\lambda (x)|^2$ locate near the maximum points of the curvature of the domain boundary.

Appendices

Appendix A: Uniqueness of the solution to system (2.11)

Lemma A.1

If (2.11) has a solution $(f_{0}, {\mathbf {Q}}_{0})\in C^2({\mathbb {R}}_{+}^2, {\mathbb {R}}^3)$ satisfying (2.12), then it is unique.

Proof

The uniqueness has been proved in Proposition 2.2, where we used the fact that the functional ${\mathcal {E}}$ is strictly convex. Here we give a direct proof. The idea of the proof goes back to Lemma 4.2 in [26] where the case of the bounded domains was treated.

Let $(f_{1}, {\mathbf {Q}}_1)$ and $(f_{2}, {\mathbf {Q}}_2)$ be two solutions, both satisfying

$$\begin{aligned} |f_{0}|^2-|{\mathbf {Q}}_{0}|^2>\frac{1}{3}+\delta ^2. \end{aligned}$$

Let $h\in H^1(\mathbb {R}^2)$ and ${\mathbf {B}}\in H^1(\text {curl}, \mathbb {R}^2)$, both with compact support. We have

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}_{+}^2}&\Big \{\frac{1}{\kappa ^2} \nabla (f_1-f_2)\cdot \nabla h- \left[ (1-|f_1|^2-|{\mathbf {Q}}_1|^2)f_1-(1-|f_2|^2-|{\mathbf {Q}}_2|^2)f_2\right] h\\&\qquad +(|f_1|^2 {\mathbf {Q}}_1-|f_2|^2 {\mathbf {Q}}_2)\cdot {\mathbf {B}}+\text {curl}({\mathbf {Q}}_1-{\mathbf {Q}}_2)\cdot \text {curl}{\mathbf {B}}\Big \}dz=0. \end{aligned} \end{aligned}$$

Take $h=\eta ^2(f_1-f_2)$ and ${\mathbf {B}}=\eta ^2({\mathbf {Q}}_1-{\mathbf {Q}}_2)$, where $\eta$ is a smooth function with compact support in $\mathbb {R}^2.$ Then we have

$$\begin{aligned}&\int _{\mathbb {R}_{+}^2}\big \{\frac{1}{\kappa ^2} |\nabla (\eta (f_1-f_2))|^2+|\text {curl}(\eta ({\mathbf {Q}}_1-{\mathbf {Q}}_2))|^2dx \nonumber \\&\quad +\int _{\mathbb {R}_{+}^2}\int _0^1 \{| f_t({\mathbf {Q}}_1-{\mathbf {Q}}_2)+2(f_1-f_2){\mathbf {Q}}_t|^2 +(3f_t^2-3|{\mathbf {Q}}_t|^2-1)|f_1-f_2|^2\big \}\eta ^2 dtdz \nonumber \\&\quad =\int _{\mathbb {R}_{+}^2}\big \{\frac{1}{\kappa ^2} |(f_1-f_2)\nabla \eta |^2+|({\mathbf {Q}}_1-{\mathbf {Q}}_2)\times \nabla \eta |^2dx, \end{aligned}$$

(A.1)

where $f_t=f_1+t(f_1-f_2)$ and ${\mathbf {Q}}_t={\mathbf {Q}}_1+t({\mathbf {Q}}_1-{\mathbf {Q}}_2)$. Note that

$$\begin{aligned} \begin{aligned}&| f_t({\mathbf {Q}}_1-{\mathbf {Q}}_2)+2(f_1-f_2){\mathbf {Q}}_t|^2+(3f_t^2-3|{\mathbf {Q}}_t|^2-1)|f_1-f_2|^2 \ge \frac{\delta ^2}{9}|{\mathbf {Q}}_1-{\mathbf {Q}}_2|^2,\\&3f_t^2-3|{\mathbf {Q}}_t|^2-1\ge \delta ^2. \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned}&| f_t({\mathbf {Q}}_1-{\mathbf {Q}}_2)+2(f_1-f_2){\mathbf {Q}}_t|^2+(3f_t^2-3|{\mathbf {Q}}_t|^2-1)|f_1-f_2|^2\\&\quad \ge \frac{\delta ^2}{18}|{\mathbf {Q}}_1-{\mathbf {Q}}_2|^2+\frac{\delta ^2}{2}|f_1-f_2|^2. \end{aligned} \end{aligned}$$

Taking $\eta =e^{-\sigma r} \xi (r)$, where $\xi (r)$ is a smooth cut-off function such that $\xi (r)=1$ for $r<R$, $\xi (r)=0$ for $r > R + 1,$ and $\xi '(r)\le 2.$ Then we have

$$\begin{aligned} \begin{aligned}&\int _{B_R^{+}}\left( \frac{\delta ^2}{18}|{\mathbf {Q}}_1-{\mathbf {Q}}_2|^2 +\frac{\delta ^2}{2}|f_1-f_2|^2\right) e^{-2\sigma r} dz\\&\quad \le \frac{\sigma ^2}{\kappa ^2}\int _{B_{R+1}^{+}}|f_1-f_2|^2e^{-2\sigma r} dz+ \sigma ^2\int _{B_{R+1}^{+}}|{\mathbf {Q}}_1-{\mathbf {Q}}_2|^2e^{-2\sigma r} dz\\&\qquad + 4 e^{-2\sigma R}\int _{B_{R+1}^{+}\backslash B_{R}^{+}}|{\mathbf {Q}}_1-{\mathbf {Q}}_2|^2dz +\frac{4}{\kappa ^2}e^{-2\sigma R}\int _{B_{R+1}^{+}\backslash B_{R}^{+}}|f_1-f_2|^2dz. \end{aligned}\nonumber \\ \end{aligned}$$

Letting $R\rightarrow \infty$ first and then letting $\sigma \rightarrow 0$ in the above inequality , we obtain that $f_1=f_2$ and ${\mathbf {Q}}_1={\mathbf {Q}}_2.$ $\square$

Appendix B: Exponential decay for some ODEs

Consider the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} u''=a_{11}(z_2) u+a_{12}(z_2)v+b_1(z_2) &{}\quad \text {in }\mathbb {R}_{+},\\ v''=a_{21}(z_2) u+a_{22}(z_2)v+b_2(z_2) &{}\quad \text {in }\mathbb {R}_{+},\\ u'(0)=u_0,\quad v'(0)=v_0,\\ u(\infty )=0,\quad v(\infty )=0.&{} \end{array}\right. } \end{aligned}$$

(B.1)

Definition B.1

We say that the coefficient matrix $A(z_2)=(a_{ij}(z_2))_{2\times 2}$ is elliptic if there exist positive constants $\lambda$ and M such that

$$\begin{aligned} \lambda |\xi |^2\le \sum ^{2}_{i,j=1} a_{ij}(z_2)\xi _i\xi _j\le M |\xi |^2. \end{aligned}$$

(B.2)

for all $\xi \in {\mathbb {R}}^2$ and almost every $z_2 \in {\mathbb {R}}_{+}$.

Proposition B.2

Assume that the matrix $A(z_2)=(a_{ij}(z_2))_{2\times 2}$ is elliptic, and suppose there exist positive constants $\alpha _5$, $\beta _5$, $M_1$ and $M_2$ such that

$$\begin{aligned} |b_1(z_2)|\le M_1 e^{-\alpha _5 z_2}, \quad |b_2(z_2)|\le M_2 e^{-\beta _5 z_2},\quad z_2\ge 0. \end{aligned}$$

(B.3)

Then system (B.1) has a unique solution $(u, v)\in C^2(\mathbb {R}_+)\cap H^1(\mathbb {R}_+)$. Moreover, for any real number $\mu$ satisfying $0<\mu <\min \{\sqrt{\lambda },\alpha _5,\beta _5\}$ we have

$$\begin{aligned} |u(z_2)|\le C e^{-\mu z_2}, \quad |v(z_2)|\le C e^{-\mu z_2}, \quad z_2\ge 0, \end{aligned}$$

(B.4)

where the constant C depends on the constants in (B.2) and (B.3).

Proof

Replacing u by $u-u_0 e^{-\lambda z_2}$ and v by $v-v_0 e^{-\lambda z_2}$, we see that there is no loss of generality in assuming $u_0=v_0=0.$ Let us fix a constant $\mu$ with $0<\mu <\min \{\sqrt{\lambda },\alpha _5,\beta _5\}$, and take a function $\eta \in C^2(\mathbb {R}_+)$ satisfying

$$\begin{aligned} \eta (z_2)=1\quad \text {for } z_2\in [0,1],\quad e^{-\mu x}\eta (z_2)<2\quad \text {and}\quad |\eta '(z_2)|\le \mu \eta (z_2)\quad \text {for all } z_2\ge 0.\nonumber \\ \end{aligned}$$

(B.5)

Define a space

$$\begin{aligned} {\mathscr {Y}}=\left\{ (u, v):\; (\eta u)\in H^1(\mathbb {R}_+),\; (\eta v)\in H^1(\mathbb {R}_+),\; u'(0)=0, \; v'(0)=0\right\} . \end{aligned}$$

Equipped with the norm

$$\begin{aligned} \Vert (u,v)\Vert _{{\mathscr {Y}}}=\left( \Vert \eta u\Vert _{H^1(\mathbb {R}_{+})}^2+\Vert \eta v\Vert _{H^1(\mathbb {R}_{+})}^2\right) ^{1/2} \end{aligned}$$

and the inner product

$$\begin{aligned} \langle (u_1, v_1),(u_2, v_2)\rangle =\int _{\mathbb {R}_+}\{\eta ^2 (u_1 u_2+v_1 v_2)+ (\eta u_1)'(\eta u_2)'+ (\eta v_1)'(\eta v_2)'\}dz_2, \end{aligned}$$

${\mathscr {Y}}$ is a Hilbert space.

Define a bilinear form ${\mathcal {B}}[(\cdot ,\cdot ),(\cdot ,\cdot )]$ on ${\mathscr {Y}}$ by

$$\begin{aligned} \begin{aligned}&{\mathcal {B}}[(u,v),(u^*, v^*)]\\&\quad =\int _{\mathbb {R}^+} \Big \{(\eta u)' (\eta u^*)'+(\eta v)' (\eta v^*)'-\frac{\eta '^2}{\eta ^2}(\eta u)(\eta u^*) -\frac{\eta '^2}{\eta ^2}(\eta v)(\eta v^*)\\&\qquad -\frac{\eta '}{\eta }[(\eta u^*)'(\eta u)-(\eta u^*)(\eta u)'] -\frac{\eta '}{\eta }[(\eta v^*)'(\eta v)-(\eta v^*)(\eta v)']\\&\qquad +(a_{11}(z_2) \eta u+a_{12}(z_2)\eta v)(\eta u^*)+(a_{21}(z_2) \eta u+a_{22}(z_2)\eta v)(\eta v^*) \Big \} dz_2. \end{aligned} \end{aligned}$$

Using the condition (B.2) on the coefficient matrix A and the assumption (B.5) on the function $\eta$, and by the Cauchy’s inequality, there exists a constant K depending only on the constants in (B.2) and $\mu$, such that for all (u, v) and $(u^*,v^*)$ in ${\mathscr {Y}}$ we have

$$\begin{aligned} \begin{aligned} {\mathcal {B}}[(u,v),(u^*, v^*)]&\le K \Vert (u,v)\Vert _{{\mathscr {Y}}}\Vert (u^*,v^*)\Vert _{{\mathscr {Y}}},\\ {\mathcal {B}}[(u,v),(u, v)]&\ge \min \{1, \lambda -\mu ^2\}\Vert (u,v)\Vert _{{\mathscr {Y}}}^2. \end{aligned} \end{aligned}$$

Therefore, ${\mathcal {B}}$ is bounded and coercive on ${\mathscr {Y}}$. Then the existence and uniqueness of the solution to (B.1) in ${\mathscr {Y}}$ follows from the Lax-Milgram lemma.

Set ${\check{u}}=\eta u$ and ${\check{v}}=\eta v$. Then $({\check{u}}, {\check{v}})$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} {\check{u}}''=\frac{2\eta '}{\eta }{\check{u}}'+\frac{\eta ''}{\eta }{\check{u}} -\frac{2\eta '^2}{\eta ^2}{\check{u}}+a_{11}(z_2) {\check{u}}+a_{12}(z_2){\check{v} }+\eta b_1(z_2) &{}\quad \text {in }\mathbb {R}_{+},\\ {\check{v}}''=\frac{2\eta '}{\eta }{\check{v}}'+\frac{\eta ''}{\eta }{\check{v}} -\frac{2\eta '^2}{\eta ^2}{\check{v}}+a_{21}(z_2) {\check{u}}+a_{22}(z_2){\check{v}} +\eta b_2(z_2) &{}\quad \text {in }\mathbb {R}_{+},\\ {\check{u}}'(0)=0,\quad {\check{v}}'(0)=0,\\ {\check{u}}(\infty )=0,\quad {\check{v}}(\infty )=0.&{} \end{array}\right. } \end{aligned}$$

(B.6)

Also, we have

$$\begin{aligned}&\min \{1, \lambda -\mu ^2\}\left( \Vert {\check{u}}\Vert _{H^1(\mathbb {R}_{+})}^2+\Vert {\check{v}}\Vert _{H^1(\mathbb {R}_{+})}^2\right) \le {\mathcal {B}}[(u,v),(u, v)]\\&\quad =\int _{\mathbb {R}^+}\left( \eta b_1(z_2) {\check{u}}+\eta b_2(z_2){\check{v}}\right) dz_2. \end{aligned}$$

Then by the Cauchy’s inequality we get

$$\begin{aligned} \Vert {\check{u}}\Vert _{H^1(\mathbb {R}_{+})}+\Vert {\check{v}}\Vert _{H^1(\mathbb {R}_{+})}\le C. \end{aligned}$$

Since $H^1(\mathbb {R}_{+})$ is continuously embedded into $C^0(\mathbb {R}_{+})$, then we have

$$\begin{aligned} \Vert {\check{u}}\Vert _{C^0(\mathbb {R}_{+})}+\Vert {\check{v}}\Vert _{C^0(\mathbb {R}_{+})}\le C. \end{aligned}$$

This proves (B.4). $\square$

Proof of Proposition 2.5

From Proposition 2.2, we obtain the decay estimate for $|{\hat{f}}_{0}(y_1, z_2)|$ and $|{\hat{{\mathbf {Q}}}}_0(y_1,z_2)|$ at $y_1=0$. Next we derive the estimates for $\partial _{y_1}{\hat{f}}_{0}(y_1, z_2)$ and $\partial _{y_1}{\hat{{\mathbf {Q}}}}_0(y_1,z_2)$ at $y_1=0.$ Recall that

$$\begin{aligned} p(z_2):=\partial _{y_1} {\hat{f}}_{0}(0,z_2),\quad (q (z_2),0):=\partial _{y_1} {\hat{{\mathbf {Q}}}}_{0}(0,z_2). \end{aligned}$$

Then from the equation (5.18) in section 5, we see that $(p(z_2), q (z_2))$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{\kappa ^2} p''(z_2)=(3|f_0|^2+|{Q}_0^1|^2-1)p+2Q_0^1 f_0 q &{}\quad \text {in }\mathbb {R}_{+},\\ q''(z_2)=2f_0 {Q}_0^1 p+{f}_0^2 q &{}\quad \text {in }{\mathbb {R}}_{+},\\ p'(0)=0,\quad q' (0)=-\hat{{\mathcal {H}}}_{y_1}^e(0),\\ p(\infty )=0,\quad q(\infty )=0.&{} \end{array}\right. } \end{aligned}$$

(B.7)

Let $\lambda (z_2)$ be the minimum eigenvalue of the matrix

$$\begin{aligned} \left( \begin{matrix} 3|f_0|^2+|{Q}_0^1|^2-1 &{} \quad 2f_0 Q_0^1 \\ 2f_0 Q_0^1 &{}\quad |f_0|^2 \end{matrix}\right) . \end{aligned}$$

Then $\lambda (z_2)\rightarrow 1$ as $z_2\rightarrow \infty .$ Now we can apply Proposition B.2 to conclude that, for any real number $\beta _1$ satisfying $0<\beta _1<1$ we have

$$\begin{aligned} |p(z_2)|+|q (z_2)|\le C(\kappa , \beta _1, \Omega , {\mathcal {H}}^e) e^{-\beta _1 z_2}. \end{aligned}$$

Applying Proposition B.2 again for the first equation in (B.7) and noting that $|Q_0^1|\le C e^{-\beta _1 z_2},$ for any real number $\alpha _1$ satisfying $0<\alpha _1<\min \{2, \sqrt{2}\kappa \}$, we have

$$\begin{aligned} |p(z_2)|\le C(\kappa , \alpha _1, \beta _1, \Omega , {\mathcal {H}}^e) e^{-\alpha _1 z_2}. \end{aligned}$$

We derive the higher derivative estimates of ${\hat{f}}_{0}(y_1, z_2)$ and ${\hat{{\mathbf {Q}}}}_0(y_1,z_2)$ at $y_1=0.$ From the equation (3.5), we see that

$$\begin{aligned} u(z_2):=\partial _{y_1^{i}}^{i} {\hat{f}}_{0}(0,z_2),\quad v(z_2):=\partial _{y_1^{i}}^{i} {\hat{Q}}_{0}^1(0,z_2) \quad \text {for }i=2,3 \end{aligned}$$

satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{\kappa ^2} u''(z_2)=(3|f_0|^2+|{Q}_0^1|^2-1)u+2Q_0^1 f_0 v+F_{i}(z_2) &{}\quad \text {in }{\mathbb {R}}_{+},\\ v''(z_2)=2f_0 {Q}_0^1 u+{f}_0^2 v +G_{i}(z_2) &{}\quad \text {in }{\mathbb {R}}_{+},\\ u'(0)=0,\quad v' (0)=-\hat{{\mathcal {H}}}_{y_1^{i}}^e(0),\\ u(\infty )=0,\quad v(\infty )=0,&{} \end{array}\right. } \end{aligned}$$

(B.8)

where

$$\begin{aligned} \begin{aligned} |F_{i}(z_2)|&\le C(\kappa , \alpha _1, \beta _1, \Omega , {\mathcal {H}}^e) e^{-\min \{3\alpha _1, 2\beta _1\} z_2},\\ |G_{i}(z_2)|&\le C(\kappa , \alpha _1, \beta _1, \Omega , {\mathcal {H}}^e) e^{-(\alpha _1+\beta _1) z_2} \end{aligned} \end{aligned}$$

for $i=2,3.$ As the proof of the estimates of $p(z_2)$ and $q(z_2)$, by applying Proposition B.2 we can obtain the decay estimates of $\partial _{y_1^i}^i {\hat{f}}_{0}(0,z_2)$ and $\partial _{y_1^i}^i {\hat{Q}}_{0}^1(0,z_2)$ for $i=2,3.$

Applying the above argument to the equations of $\partial _{y_1^i}^i {\hat{f}}_{0}(0,z_2)$ and $\partial _{y_1^i}^i {\hat{Q}}_{0}^1(0,z_2)$ respectively, we immediately obtain the decay estimates of $|\partial _{y_1^i z_2^2}^{i+2}{\hat{f}}_{0}(y_1, z_2)|$ and $|\partial _{y_1^i z_2^2}^{i+2}{\hat{{\mathbf {Q}}}}_0(y_1,z_2)|$ for $i=0,\cdots ,3$.

Integrating from $z_2$ to $\infty$ on both sides of the equations of $\partial _{y_1^i}^i {\hat{f}}_{0}(0,z_2)$ and $\partial _{y_1^i}^i {\hat{Q}}_{0}^1(0,z_2)$ respectively, we can obtain the decay estimates of $|\partial _{y_1^i z_2}^{i+1}{\hat{f}}_{0}(y_1, z_2)|$ and $|\partial _{y_1^i z_2}^{i+1}{\hat{{\mathbf {Q}}}}_0(y_1,z_2)|$ for $i=0,\ldots ,3$.

Now we have proved Proposition 2.5 for $y_1=0.$ Replacing $\hat{{\mathcal {H}}}^e(0)$ by $\hat{{\mathcal {H}}}^e(y_1)$ in (B.7) and in (B.8), then noting that $\hat{{\mathcal {H}}}^e\in C^3(\partial \Omega ),$ we see that Proposition 2.5 also holds for $y_1\ne 0.$ Now we have completed the proof. $\square$

Appendix C: Derivation of system (2.20)

To derive equation (2.20) we need the local coordinate expansions introduced in [24, section 3]. Here we keep the notations in section 2. We use ${\mathfrak {R}}_i(|y_1^3|)$, $i=1,2,\cdots$, to denote a function of $y_1$ and $z_2$ which is of order $(|y_1^3|)$ uniformly for $z_2$, and use ${\mathfrak {R}}_{i}(\lambda ^k)$, $k>0$, $i=1,2,\cdots$, to denote a function of $y_1$ and $z_2$ which is of order $(\lambda ^k)$.

For the function g defined in (2.2) we have, for $\lambda >0$ small,

$$\begin{aligned} \begin{aligned} g(z)&=1-\lambda k(0)z_2-\lambda ^2k'(0)z_1 z_2+O(\lambda ^3),\\ {1\over g(z)}&=1+\lambda k(0)z_2+\lambda ^2\left( k^2(0) z_2^2+k'(0)z_1 z_2\right) +O(\lambda ^3), \end{aligned} \end{aligned}$$

(C.1)

where $k'(0)={\mathrm {d} k\over \mathrm {d} s}(0)={\mathrm {d} k\over \mathrm {d} y_1}(0).$ For any fixed $z_2\ge 0$, we have the formal asymptotic expansions for ${\hat{f}}_{0}(y_1, z_2)$ and ${\hat{{\mathbf {Q}}}}_{0}(y_1, z_2)$ with respect to the variable $y_1$ at the point $(0, z_2)$:

$$\begin{aligned} \begin{aligned} {\hat{f}}_{0}(y_1, z_2)&=f_0+y_1 \partial _{y_1} {\hat{f}}_{0}(0,z_2)+\frac{1}{2}y_1^2 \partial _{y_1^2} {\hat{f}}_{0}(0,z_2) +{\mathfrak {R}}_1(|y_1^3|),\\ {\hat{{\mathbf {Q}}}}_{0}(y_1, z_2)&={\mathbf {Q}}_0+y_1 \partial _{y_1} {\hat{{\mathbf {Q}}}}_{0}(0,z_2)+\frac{1}{2}y_1^2 \partial _{y_1^2} {\hat{{\mathbf {Q}}}}_{0}(0,z_2) +{\mathfrak {R}}_2(|y_1^3|), \end{aligned} \end{aligned}$$

(C.2)

where

$$\begin{aligned} (f_0, {\mathbf {Q}}_0)=({\hat{f}}_{0}(0,z_2), {\hat{{\mathbf {Q}}}}_{0}(0,z_2))=({\hat{f}}_{0}(0,z_2), ({\hat{Q}}_0^1(0,z_2), {\hat{Q}}_0^2(0,z_2))) \end{aligned}$$

is the solution of (2.14), $({\hat{f}}_0(y_1, z_2), \hat{{{\textbf {Q}}}}_0(y_1, z_2))$ is the solution of (5.18).

Write

$$\begin{aligned} p(z_2):=\partial _{y_1} {\hat{f}}_{0}(0,z_2),\quad (q (z_2),0):=\partial _{y_1} {\hat{{\mathbf {Q}}}}_{0}(0,z_2). \end{aligned}$$

(C.3)

Then we take the expansions for ${\tilde{f}}$ and ${\tilde{{\mathbf {Q}}}}$ in (2.6) with respect to $\lambda$, and have

$$\begin{aligned} {\tilde{f}}\!=\!f_0+\lambda (p z_1+f_1)+{\mathfrak {R}}_3(\lambda ^2),\quad {\tilde{{\mathbf {Q}}}}\!=\!{\mathbf {Q}}_0\!+\!\lambda ((q z_1,0)+{\mathbf {Q}}_1)+{\mathfrak {R}}_4(\lambda ^2),\nonumber \\ \end{aligned}$$

(C.4)

where $f_1={\hat{f}}_{1}(0, z_2)$ and ${\mathbf {Q}}_1={\hat{{\mathbf {Q}}}}_{1}(0, z_2)=(Q_1^1, Q_1^2)$ are to be determined.

Firstly, we have

$$\begin{aligned} \begin{aligned} \lambda \text {curl}{\mathbf {Q}}(x)&=\frac{1}{g}\left[ \partial _{z_1} {\tilde{Q}}_2 -\partial _{z_2} (g {\tilde{Q}}_1)\right] \\&=-(Q_0^1)'+\lambda (k(0)Q_0^1-(Q_1^1)'-z_1 q'(z_2)) +{\mathfrak {R}}_5(\lambda ^2). \end{aligned} \end{aligned}$$

(C.5)

Then,

$$\begin{aligned} \begin{aligned}&\frac{1}{g}\left( \partial _{z_1} \left( \frac{1}{g}\partial _{z_1} {\tilde{f}}\right) +\partial _{z_2}\left( g \partial _{z_2} {\tilde{f}}\right) \right) \\&\quad =(f_0)''+\lambda \left( (f_1)''-k(0) (f_0)'+p''(z_2)z_1\right) +{\mathfrak {R}}_6(\lambda ^2) \end{aligned} \end{aligned}$$

(C.6)

and

$$\begin{aligned} \begin{aligned} (1-|{\tilde{f}}|^2-|{\tilde{{\mathbf {Q}}}}|^2){\tilde{f}}&=(1-|f_0|^2-|{\mathbf {Q}}_0|^2)f_0+\lambda \big ((1-|f_0|^2-|{\mathbf {Q}}_0|^2)(p z_1+f_1)\\&\quad -2f_0(f_0( p z_1+f_1)+{\mathbf {Q}}_0\cdot (( q z_1,0)+{\mathbf {Q}}_1))\big )+{\mathfrak {R}}_7(\lambda ^2), \end{aligned}\nonumber \\ \end{aligned}$$

(C.7)

where $p=p(z_2)$ and $q=q(z_2)$ are defined in (C.2). Using (C.5), for $\mathcal {M}_1(\lambda z)$ and $\mathcal {M}_2(\lambda z)$ defined by (2.3) we have

$$\begin{aligned} \begin{aligned} \mathcal {M}_1(\lambda z)&=-(Q_0^1)''-\lambda \big [(Q_1^1)''+q''(z_2) z_1-k(0) (Q_0^1)'\big ]+{\mathfrak {R}}_8(\lambda ^2),\\ \mathcal {M}_2(\lambda z)&=\lambda q'(z_2) +{\mathfrak {R}}_9(\lambda ^2). \end{aligned} \end{aligned}$$

(C.8)

Also, we have

$$\begin{aligned} |{\tilde{f}}|^2 {\tilde{{\mathbf {Q}}}}=|f_0|^2 {\mathbf {Q}}_0+\lambda \left[ 2f_0(p z_1+f_1){\mathbf {Q}}_0+|f_0|^2 ((q z_1,0)+{\mathbf {Q}}_1)\right] +{\mathfrak {R}}_{10}(\lambda ^2).\nonumber \\ \end{aligned}$$

(C.9)

We now consider the equations at the point $(0,z_2)$. We have

$$\begin{aligned} \frac{1}{g}\left( \partial _{z_1} \left( \frac{1}{g}\partial _{z_1} {\tilde{f}}\right) +\partial _{z_2}\left( g \partial _{z_2} {\tilde{f}}\right) \right) =(f_0)''+\lambda \left( (f_1)''-k(0) \partial _2 f_0\right) +{\mathfrak {R}}_{11}(\lambda ^2)\nonumber \\ \end{aligned}$$

(C.10)

and

$$\begin{aligned} \begin{aligned}&(1-|{\tilde{f}}|^2-|{\tilde{{\mathbf {Q}}}}|^2){\tilde{f}} =(1-|f_0|^2-|Q_0^1|^2)f_0\\&\quad +\lambda \big ((1-|f_0|^2-|Q_0^1|^2)f_1 -2f_0(f_0 f_1+Q_0^1 Q_1^1)\big )+{\mathfrak {R}}_{12}(\lambda ^2). \end{aligned} \end{aligned}$$

(C.11)

For $\mathcal {M}_1(\lambda z)$ we have

$$\begin{aligned} \mathcal {M}_1(\lambda z)=-(Q_0^1)''+\lambda \big [-(Q_1^1)''+k(0) (Q_0^1)' \big ]+{\mathfrak {R}}_{13}(\lambda ^2). \end{aligned}$$

(C.12)

Also, we have

$$\begin{aligned} |{\tilde{f}}|^2 {\tilde{{\mathbf {Q}}}}=|f_0|^2 {\mathbf {Q}}_0+\lambda \left[ 2f_0 f_1 {\mathbf {Q}}_0+|f_0|^2 {\mathbf {Q}}_1\right] +{\mathfrak {R}}_{14}(\lambda ^2). \end{aligned}$$

(C.13)

Comparing with the coefficients of $\lambda$, we obtain the equations (2.20) for the first order terms.

Appendix D: Derivation of system (2.22) and proof of (3.5)

We follow the notations used in section 2 and in appendix C. Let ${\mathfrak {R}}_{i}(\lambda ^2)$ be the terms appear in appendix C, and it has been proved in section 3 that these terms have the order $O(\lambda ^2)$ uniformly for $y_1$ and $z_2$. In this section we shall expand these terms in the form

$$\begin{aligned} {\mathfrak {R}}_{i}(\lambda ^2)=\lambda ^2 R_i+{\mathfrak {R}}_{i}(\lambda ^3)\quad \quad \text {for }i=3,\cdots , 14, \end{aligned}$$

where $R_i$ denotes a functions of $y_1$ and $z_2$ which is independent of $\lambda$, and ${\mathfrak {R}}_{i}(\lambda ^3)$ denotes a function of $y_1$ and $z_2$ which is of the order $O(\lambda ^3)$.

From the inner expansion (2.9) and the expansion (C.2), we have the expansions for the function ${\mathfrak {R}}_3$ and the vector field ${\mathfrak {R}}_4$ in (C.4):

$$\begin{aligned} {\mathfrak {R}}_3(\lambda ^2)=\lambda ^2\left( \frac{1}{2} z_1^2 \partial _{y_1^2 } {\hat{f}}_{0}(0,z_2) + z_1 \partial _{y_1} {\hat{f}}_{1}(0,z_2)+f_2\right) +{\mathfrak {R}}_{16}(\lambda ^3), \end{aligned}$$

(D.1)

and

$$\begin{aligned} \begin{aligned} {\mathfrak {R}}_4(\lambda ^2)&=\lambda ^2\Big ( \frac{1}{2} z_1^2 \partial _{y_1^2} {\hat{Q}}_{0}^1(0,z_2) + z_1 \partial _{y_1} {\hat{Q}}_{1}^1(0,z_2)+Q_{2}^1(0,z_2), \\&\qquad z_1 \partial _{y_1} {\hat{Q}}_{1}^2(0,z_2)+Q_{2}^2(0,z_2)\Big )+{\mathfrak {R}}_{17}(\lambda ^3), \end{aligned} \end{aligned}$$

(D.2)

where

$$\begin{aligned} ({\hat{f}}_{0}(0,z_2), {\hat{{\mathbf {Q}}}}_{0}(0,z_2))=({f}_{0}, (Q_0^1, 0)) \end{aligned}$$

is the solution of (2.14),

$$\begin{aligned} ({\hat{f}}_{1}(0,z_2), {\hat{{\mathbf {Q}}}}_{1}(0,z_2))=({f}_{1}, (Q_1^1, Q_1^2)) \end{aligned}$$

is the solution of (2.20), and

$$\begin{aligned} ({\hat{f}}_{2}(0,z_2), {\hat{{\mathbf {Q}}}}_{2}(0,z_2))=({f}_{2}, (Q_2^1, Q_2^2)) \end{aligned}$$

is to be determined now.

From (C.5), we have

$$\begin{aligned} \begin{aligned} \mathfrak {R}_5(\lambda ^2)=&\,\lambda ^2 \Big (\, \partial _{y_1} \hat{Q}_1^2+k'(0)z_1 Q_0^1+k'(0)z_1 z_2 (Q_0^1)' +k(0)\left( z_1 \partial _{y_1} \hat{Q}_0^1\big |_{y_1=0}+Q_1^1\right) \\&\quad - (Q_2^1)' +k(0)z_2\left[ z_1 \partial _{y_1 z_2} \hat{Q}_0^1\big |_{y_1=0} +(Q_1^1)'\right] -\frac{1}{2}z_1^2\partial _{y_1^2 z_2}\hat{Q}_0^1\big |_{y_1=0}\\&\quad -z_1 \partial _{y_1 z_2}\hat{Q}_1^1\big |_{y_1=0}+k(0)z_2 \left[ k(0)Q_0^1-(Q_1^1)'-z_1\partial _{y_1 z_2}\hat{Q}_0^1\big |_{y_1=0}\right] \\&\quad -(Q_0^1)'\left[ k^2(0)z_2^2 +k'(0)z_1 z_2\right] \Big ) +\mathfrak {R}_{18}(\lambda ^3).\end{aligned} \end{aligned}$$

(D.3)

Then we have the expansions for ${\mathfrak {R}}_6$ in (C.6) and for ${\mathfrak {R}}_7$ in (C.7):

$$\begin{aligned} \begin{aligned} \mathfrak {R}_6(\lambda ^2)=&\lambda ^2\Big (\, \partial _{y_1^2}\hat{f}_0\big |_{y_1=0} +\frac{1}{2}z_1^2 \partial _{z_2^2 y_1^2} \hat{f}_0\big |_{y_1=0} +z_1 \partial _{z_2^2 y_1} \hat{f}_1\big |_{y_1=0}-k(0) \left[ p' z_1+(f_1)'\right] \\&\quad +(f_2)''-k(0) z_2\left[ p'' z_1+(f_1)''\right] -(f_0)' k'(0) z_1-(f_0)'' k'(0) z_1 z_2\\&\quad +k(0) z_2 \left[ (f_1)''+p'' z_1 -k(0) (f_0)' -k(0) z_2 (f_0)''\right] \\&\quad +(f_0)''\left[ k^2(0) z_2^2+k'(0) z_1 z_2\right] \Big )+\mathfrak {R}_{19}(\lambda ^3), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} {\mathfrak {R}}_7(\lambda ^2)&=\lambda ^2\big (\left[ \frac{1}{2}z_1^2 \partial _{ y_1^2} {\hat{f}}_0\big |_{y_1=0}+z_1 \partial _{y_1} {\hat{f}}_1\big |_{y_1=0}+f_2\right] (1-f_0^2-|{\mathbf {Q}}_0|^2)\\&\qquad +(pz_1+f_1)[-2f_0(pz_1+f_1)-2Q_0^1(q z_1+Q_1^1)-2Q_0^2 Q_1^2]\\&\qquad +f_0\left[ -(pz_1+f_1)^2-2f_0\left( \frac{1}{2}z_1^2 \partial _{ y_1^2} {\hat{f}}_0\big |_{y_1=0}+z_1 \partial _{y_1} {\hat{f}}_1\big |_{y_1=0}+f_2\right) \right. \\&\qquad -(q z_1+Q_1^1)^2 -2Q_0^1\left( \frac{1}{2}z_1^2 \partial _{ y_1^2} {\hat{Q}}_0^1\big |_{y_1=0} +z_1 \partial _{y_1} {\hat{Q}}_1^1\big |_{y_1=0}+Q_2^1\right) \\&\qquad \left. -(Q_1^2)^2-2Q_0^2(z_1 \partial _{y_1} {\hat{Q}}_1^2\big |_{y_1=0}+Q_2^2)\right] \big )+{\mathfrak {R}}_{20}(\lambda ^3). \end{aligned} \end{aligned}$$

From (D.1), (D.2) and (C.8), we see that

$$\begin{aligned} \begin{aligned} \mathfrak {R}_8(\lambda ^2) =&\lambda ^2\Big (\, \partial _{z_2 y_1}\hat{Q}_1^2\big |_{y_1=0} -\frac{1}{2}z_1^2 \partial _{z_2^2 y_1^2} \hat{Q}_2^1\big |_{y_1=0}-z_1 \partial _{z_2^2 y_1} \hat{Q}_1^1\big |_{y_1=0} +k'(0) z_1 z_2(Q_0^1)''\\&\quad -(Q_2^1)''+2k(0) \left[ q' z_1+(Q_1^1)'\right] +k(0) z_2\left[ q'' z_1+(Q_1^1)''\right] +k'(0) z_1(Q_0^1)'\\&\quad -k(0) \left[ q'z_1+(Q_1^1)'-k(0) Q_0^1\right] -(Q_0^1)''\left[ k^2(0) z_2^2+k'(0) z_1 z_2\right] \\&\quad -k(0) z_2\left[ q'' z_1+(Q_1^1)''-k(0) (Q_0^1)' -k(0) z_2 (Q_0^1)''\right] \Big ) +\mathfrak {R}_{21}(\lambda ^3) \end{aligned} \end{aligned}$$

(D.4)

and

$$\begin{aligned} \begin{aligned} \mathfrak {R}_9(\lambda ^2) =&\lambda ^2 \Big (\, z_1\partial _{z_2 y_1^2}\hat{Q}_0^1\big |_{y_1=0}+\partial _{z_2 y_1}\hat{Q}_1^1\big |_{y_1=0}-k(0) q -k'(0) Q_0^1\\&\quad +k(0) z_2 q'\Big )+\mathfrak {R}_{22}(\lambda ^3). \end{aligned} \end{aligned}$$

(D.5)

From (D.1), (D.2) and (C.9), we have

$$\begin{aligned} \begin{aligned} \mathfrak {R}_{10}(\lambda ^2) =&\lambda ^2 \Big (\left[ (p z_1+f_1)^2+2f_0\left( \frac{1}{2}z_1^2 \partial _{y_1^2} \hat{f}_0\big |_{y_1=0}+z_1 \partial _{y_1} \hat{f}_1\big |_{y_1=0}+f_2\right) \right] {\mathbf {Q}}_0\\&\quad +f_0^2\left( \frac{1}{2}z_1^2 \partial _{y_1 y_1} \hat{Q}_0^1\big |_{y_1=0}+z_1\partial _{y_1} \hat{Q}_1^1\big |_{y_1=0}+Q_2^1, z_1\partial _{y_1} \hat{Q}_1^2\big |_{y_1=0}+Q_2^2\right) \\&\quad +2f_0 (p z_1+f_1)\left[ (q z_1,0)+{\mathbf {Q}}_1\right] \Big )+\mathfrak {R}_{23}(\lambda ^3). \end{aligned}\nonumber \\ \end{aligned}$$

(D.6)

We now consider the equations at the point $(0, z_2).$ From the expression of ${\mathfrak {R}}_{6}(\lambda ^3)$, it follows that,

$$\begin{aligned} {\mathfrak {R}}_{11}(\lambda ^2) =\lambda ^2 \left( \partial _{y_1^2} {\hat{f}}_0\big |_{y_1=0}+ f_2^{''}-k(0) (f_1)'-k^2(0) z_2 (f_0)'\right) +{\mathfrak {R}}_{24}(\lambda ^3).\qquad \end{aligned}$$

(D.7)

From the expression of ${\mathfrak {R}}_{7}(\lambda ^3)$, we have

$$\begin{aligned} \begin{aligned} \mathfrak {R}_{12}(\lambda ^2) =&\lambda ^2\Big (\, f_2(1-3f_0^2-|{\mathbf {Q}}_0|^2)+f_0\left[ -f_1^2-(Q_1^1)^2-(Q_1^2)^2\right] \\&\quad -2f_0 Q_0^1Q_2^1+f_1\left[ -2f_0f_1-2Q_0^1Q_1^1\right] \Big )+\mathfrak {R}_{25}(\lambda ^3). \end{aligned}\nonumber \\ \end{aligned}$$

(D.8)

From the expressions of ${\mathfrak {R}}_{8}(\lambda ^3)$ and ${\mathfrak {R}}_{9}(\lambda ^3)$, we have

$$\begin{aligned} \begin{aligned} \mathfrak {R}_{13}(\lambda ^2) =&\lambda ^2\Big (\partial _{z_2 y_1}\hat{Q}_1^2\big |_{y_1=0}-(Q_2^1)''+k(0) (Q_1^1)' +k(0) Q_0^1\\&\quad +k^2(0) z_2(Q_0^1)' \Big )+\mathfrak {R}_{26}(\lambda ^3) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \mathfrak {R}_{14}(\lambda ^2) =\lambda ^2 \left( \partial _{z_2 y_1}\hat{Q}_1^1\big |_{y_1=0}-k(0) q -k'(0) Q_0^1+k(0) z_2q'\right) +\mathfrak {R}_{27}(\lambda ^3). \end{aligned}$$

From ${\mathfrak {R}}_{10}(\lambda ^3)$, it follows that

$$\begin{aligned} \begin{aligned} {\mathfrak {R}}_{15}(\lambda ^2) =\lambda ^2 ((f_1^2+2f_0 f_2){\mathbf {Q}}_0+2f_0 f_1{\mathbf {Q}}_1+f_0^2{\mathbf {Q}}_2)+{\mathfrak {R}}_{28}(\lambda ^3). \end{aligned} \end{aligned}$$

Comparing with the coefficients of $\lambda ^2$, we obtain the equations (2.22) for the second order terms.

Proof of Lemma 3.1

Step 1. From (3.2), we can see that

$$\begin{aligned} {\mathbf {b}}(x,\lambda )=(0,{{\textbf {0}}})\quad \text {for all }x\in \Omega \backslash \sigma _2, \end{aligned}$$

where $\sigma _n$ is defined by (3.1). Then (3.5) holds for $x\in \Omega \backslash \sigma _2$.

Step 2. We show the estimate (3.5) when $x\in \sigma _4.$ We consider this problem in a neighborhood ${\mathcal {U}}$ of $X_0\in \partial \Omega$. We follow the notation used in section 2 and in appendix C.

To obtain ${\tilde{{\mathbf {b}}}}$, we replace the expressions of ${\tilde{f}}$ and ${\tilde{{\mathbf {Q}}}}$ by

$$\begin{aligned} {\hat{f}}_{0}(y_1, z_2) +\lambda {\hat{f}}_{1}(y_1, z_2)+\lambda ^2 {\hat{f}}_{2}(y_1, z_2)\quad \text {and}\quad {\hat{{\mathbf {Q}}}}_{0}(y_1, z_2) +\lambda {\hat{{\mathbf {Q}}}}_{1}(y_1, z_2)+\lambda ^2 {\hat{{\mathbf {Q}}}}_{2}(y_1, z_2) \end{aligned}$$

under the $z-$ coordinate system in (C.4) respectively.

We first estimate ${\tilde{b}}_1(z_1, z_2,\lambda ),$ where ${\tilde{b}}_1(z_1, z_2, \lambda )$ is the representation of $b_1(x,\lambda )$ under the z-coordinate system. At the point $(0, z_2),$ from (D.7) and (D.8) we have

$$\begin{aligned} \begin{aligned} \mathfrak {R}_{24}(\lambda ^3)=&\lambda ^3 \kappa ^{-2}\Big (\, g_0^{-3}k'(0) z_2 \left[ \partial _{y_1} \hat{f}_0\big |_{y_1=0}+\lambda \partial _{y_1} \hat{f}_1\big |_{y_1=0}+\lambda ^2 \partial _{y_1} \hat{f}_2\big |_{y_1=0}\right] \\&\qquad \quad +g_0^{-2}\left[ (1+g_0)k(0) z_2 \partial _{y_1^2} \hat{f}_0\big |_{y_1=0}+\partial _{y_1^2} \hat{f}_1\big |_{y_1=0}+ \lambda \partial _{y_1^2} \hat{f}_2\big |_{y_1=0}\right] \\ {}&\qquad \quad -g_0^{-1}\left[ k(0)(f_0)' -\lambda k^3(0)z_2^2(f_1)'-\lambda k^2(0)z_2(f_2)'\right] \\&\qquad \quad -k^3(0) z_2^2(f_0)'-k^2(0) z_2(f_1)'-k(0) (f_2)'\Big )\end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \mathfrak {R}_{25}(\lambda ^3)=&-\lambda ^3 \Big (\, (f_1 +\lambda f_2)\Big [2 Q_0^1Q_2^1+2 f_0 f_2+f_1^2+(Q_1^1)^2+(Q_1^2)^2\Big ]\\&\qquad \quad +2f_2(f_0 f_1+Q_0^1 Q_1^1)+(f_0+\lambda f_1 +\lambda ^2 f_2)\Big [2 Q_1^1Q_2^1+2 f_1 f_2\\&\qquad \quad +2 Q_1^2 Q_2^2+\lambda \left( f_2^2+(Q_1^1)^2+(Q_1^2)^2\right) \Big ]\Big ), \end{aligned} \end{aligned}$$

where $k_0=k(0)$ is the curvature of $\partial \Omega$ at the point $X_0$, $k_0'={\mathrm {d} k\over \mathrm {d} s}(0)= {\mathrm {d} k\over \mathrm {d} y_1}(0)$,

$$\begin{aligned} g_0=g(0, \lambda z_2)=1-\lambda k(0) z_2. \end{aligned}$$

We see that ${\mathfrak {R}}_{24}(\lambda ^3)$ and ${\mathfrak {R}}_{25}(\lambda ^3)$ are the polynomials of ${\hat{f}}_{0}, {\hat{f}}_{1},{\hat{f}}_{2}, {\hat{{\mathbf {Q}}}}_{0}, {\hat{{\mathbf {Q}}}}_{1}, {\hat{{\mathbf {Q}}}}_{2}$ and their derivatives up to the order 2 at $(0, z_2)$. From Proposition 2.5, Proposition 2.7 and Proposition 2.9, it follows that

$$\begin{aligned} |{\mathfrak {R}}_{24}(\lambda ^3)|+|{\mathfrak {R}}_{25}(\lambda ^3)|\le C(\Omega ,{\mathcal {H}}^e, \kappa )\lambda ^3. \end{aligned}$$

For any $x\in \sigma _4$, let $\psi$ be defined by (2.1), $x=\psi (y_1,y_2)$, $z_1=y_1/\lambda , z_2=y_2/\lambda .$ Then ${\tilde{b}}_1(0, z_2, \lambda )={\mathfrak {R}}_{24}(\lambda ^3)+{\mathfrak {R}}_{25}(\lambda ^3).$ Note that ${\hat{f}}_{0}, {\hat{f}}_{1},{\hat{f}}_{2}, {\hat{{\mathbf {Q}}}}_{0}, {\hat{{\mathbf {Q}}}}_{1}, {\hat{{\mathbf {Q}}}}_{2}$ and their derivatives up to the order 2 are continuously differentiable with respect to the parameter $y_1$ (by applying the continuous differentiability of solutions with respect to parameters in the theory of ODEs). Therefore, for each $z_1\ne 0,$ we also have

$$\begin{aligned} |{\tilde{b}}_1(z_1, z_2,\lambda )|\le C(\Omega ,{\mathcal {H}}^e, \kappa )\lambda ^3. \end{aligned}$$

We now estimate ${\tilde{{\mathbf {b}}}}_2=({\tilde{b}}_2^1, {\tilde{b}}_2^2).$ At the point $(y_1, z_2)$ with $y_1=\lambda z_2$, from (D.4) we have

$$\begin{aligned} \begin{aligned} \tilde{\mathfrak {R}}_{21}(\lambda ^3)=&\lambda ^3\Big [s_3+\lambda k(y_1)z_2 (s_2+\lambda s_3) +\lambda ^2 k^2(y_1) z_2^2(s_1+\lambda s_2+\lambda ^2 s_3) \\&\quad +\lambda ^3 k^3(y_1) z_2^3 g^{-1}(-\partial _{z_2^2} \hat{Q}_0^1 +\lambda s_1+\lambda ^2 s_2+\lambda ^3 s_3)\Big ]\\&+\lambda ^3\Big [k(y_1)s_5 +2\lambda ^2 k^2(y_1) z_2 (s_4+\lambda s_5)\\&\qquad +\lambda ^3 k^3(y_1) z_2^2 g^{-2}(2g+1)(-\partial _{z_2}\hat{Q}_0^1 +\lambda s_4+\lambda ^2 s_5)\Big ], \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} s_1&=k(y_1){\hat{Q}}_0^1+k(y_1) z_2 \partial _{z_2^2} {\hat{Q}}_0^1 +k(y_1)\partial _{z_2}{\hat{Q}}_0^1-\partial _{z_2^2}{\hat{Q}}_1^1, \\ s_2&=\partial _{y_1 z_2}{\hat{Q}}_1^2+k(y_1){\hat{Q}}_1^1+k(y_1) z_2 \partial _{z_2^2} {\hat{Q}}_1^1-\partial _{z_2^2}{\hat{Q}}_2^1+k(y_1)\partial _{z_2} {\hat{Q}}_1^1,\\ s_3&=\partial _{y_1 z_2}{\hat{Q}}_2^2+k(y_1){\hat{Q}}_2^1+k(y_1) z_2 \partial _{z_2^2} {\hat{Q}}_2^1+k(y_1)\partial _{z_2} {\hat{Q}}_2^1 \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} s_4&=k(y_1)z_2 \partial _{z_2} \hat{Q}_0^1+k(y_1)\hat{Q}_0^1-\partial _{z_2}\hat{Q}_1^1,\\ s_5&=\partial _{y_1}\hat{Q}_1^2+k(y_1)\hat{Q}_1^1+k(y_1) z_2 \partial _{z_2} \hat{Q}_1^1-\partial _{z_2}\hat{Q}_2^1 \lambda \Big (\partial _{y_1}\hat{Q}_2^2+k(y_1)\hat{Q}_2^1+k(y_1) z_2 \partial _{z_2} \hat{Q}_2^1\Big ). \end{aligned} \end{aligned}$$

From (D.5) we have

$$\begin{aligned} \begin{aligned} \tilde{\mathfrak {R}}_{22}(\lambda ^3)=&-\lambda ^3\Big [s_7+2 k(y_1)z_2(s_6+\lambda s_7) + k^2(y_1) z_2^2 g^{-2}(2g+1)(- \partial _{y_1 z_2}\hat{Q}_0^1+\lambda s_6+\lambda ^2 s_7)\Big ]\\&-\lambda ^3\Big [ k'(y_1) z_2 s_8+ k(y_1)k'(y_1) z_2^2 g^{-3}(1+g+g^2)(-\partial _{z_2} \hat{Q}_0^1+\lambda s_8)\Big ],\\ \end{aligned} \end{aligned}$$

where g is defined by (2.2),

$$\begin{aligned} \begin{aligned} s_6=&k(y_1)\partial _{y_1} \hat{Q}_0^1+k'(y_1) \hat{Q}_0^1+\partial _{y_1 z_2}\hat{Q}_0^1 -\partial _{y_1 z_2}\hat{Q}_1^1+k'(y_1) z_2\partial _{z_2} \hat{Q}_0^1,\\ s_7=&\partial _{y_1^2}\hat{Q}_1^2+\lambda \partial _{y_1^2}\hat{Q}_2^2+k(y_1) \partial _{y_1}\hat{Q}_1^1 +\lambda k(y_1)\partial _{y_1}\hat{Q}_2^1+k'(y_1)\hat{Q}_1^1+\lambda k'(y_1)\hat{Q}_2^1\\&+k(y_1) z_2 \partial _{y_1 z_2}\hat{Q}_1^1-\partial _{y_1 z_2}\hat{Q}_2^1+\lambda k(y_1)z_2 \partial _{y_1 z_2}\hat{Q}_2^1+k'(y_1)z_2 (\partial _{z_2} \hat{Q}_1^1+\lambda \partial _{z_2}\hat{Q}_2^1),\end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} s_8=&k(y_1)z_2 \partial _{z_2} \hat{Q}_0^1+k(y_1)\hat{Q}_0^1-\partial _{z_2}Q_1^1\\&+\lambda \Big (\partial _{y_1}\hat{Q}_1^2+k(y_1)\hat{Q}_1^1+k(y_1) z_2 \partial _{z_2} \hat{Q}_1^1-\partial _{z_2}\hat{Q}_2^1\Big )\\&+\lambda ^2\Big (\partial _{y_1}\hat{Q}_2^2+k(y_1)\hat{Q}_2^1+k(y_1) z_2 \partial _{z_2} \hat{Q}_2^1\Big ).\end{aligned} \end{aligned}$$

From (D.6) we have

$$\begin{aligned} \begin{aligned} \tilde{\mathfrak {R}}_{23}^1(\lambda ^3)&=\lambda ^3 \Big [(\hat{Q}_0^1+\lambda \hat{Q}_1^1+\lambda ^2 \hat{Q}_2^1)(2 \hat{f}_1 \hat{f}_2+\lambda \hat{f}_2^2) +(\hat{Q}_1^1+\lambda \hat{Q}_2^1)(\hat{f}_1^2 +2\hat{f}_0 \hat{f}_2)+2\hat{f}_0 \hat{f}_1 \hat{Q}_2^1\Big ], \\ \tilde{\mathfrak {R}}_{23}^2(\lambda ^3)&=\lambda ^3\Big [(\lambda \hat{Q}_1^2+\lambda ^2 \hat{Q}_2^2)(2 \hat{f}_1 \hat{f}_2+\lambda \hat{f}_2^2)+(\hat{Q}_1^2+\lambda \hat{Q}_2^2)(\hat{f}_1^2 +2\hat{f}_0 \hat{f}_2)+2\hat{f}_0 \hat{f}_1 \hat{Q}_2^2\Big ].\end{aligned} \end{aligned}$$

Then

$$\begin{aligned} {\tilde{b}}_2^1=-\tilde{{\mathfrak {R}}}_{21}-\tilde{{\mathfrak {R}}}_{23}^1,\quad {\tilde{b}}_2^2=-\tilde{{\mathfrak {R}}}_{22}-\tilde{{\mathfrak {R}}}_{23}^2. \end{aligned}$$

We see that ${\tilde{b}}_2^1$ and ${\tilde{b}}_2^2$ are the polynomials of ${\hat{f}}_{0}, {\hat{f}}_{1},{\hat{f}}_{2}, {\hat{{\mathbf {Q}}}}_{0}, {\hat{{\mathbf {Q}}}}_{1}, {\hat{{\mathbf {Q}}}}_{2}$ and their derivatives up to the order 2. Using Proposition 2.5, Proposition 2.7 and Proposition 2.9 again, we have

$$\begin{aligned} \Vert {\tilde{{\mathbf {b}}}}_2\Vert _{C^0(\psi ^{-1}(\sigma _4))} +\Vert {{\,\mathrm{div}\,}}_z{\tilde{{\mathbf {b}}}}_2\Vert _{C^1(\psi ^{-1}(\sigma _4))} \le C(\Omega ,{\mathcal {H}}^e, \kappa )\lambda ^3. \end{aligned}$$

where $\sigma _4$ is defined in (3.1), $\psi$ is defined by (2.1). Thus, we have (3.5) for $x\in \sigma _4$ by the scaling argument.

Step 3. From Proposition 2.5, Proposition 2.7 and Proposition 2.9, we see that, each component of ${\tilde{{\mathbf {b}}}}(z,\lambda )$ is a linear combinations of exponentially decaying terms with respect to $z_2$. Therefore, for $x\in \sigma _2\backslash \sigma _4$ we also have (3.5) for small $\lambda$.

We now finish the proof of (3.5). $\square$

Appendix E: Derivation of the boundary condition (3.8)

To derive the boundary condition of $R_{f}$, we use the boundary condition of ${\hat{f}}_{0}$ in (5.18), that of ${\hat{f}}_{1}$ in (5.19), and that of ${\hat{f}}_{2}$ (see (2.22) when $y_1=0$), and find

$$\begin{aligned} \frac{\partial R_{f}}{\partial {\mathbf {n}}}=\frac{\partial f}{\partial {\mathbf {n}}}+\frac{\partial }{\partial z_2}\Big (\hat{f}_{0}(y_1, z_2) +\lambda \hat{f}_{1}(y_1, z_2)+\lambda ^2 \hat{f}_{2}(y_1, z_2)\Big )\Big |_{z_2=0}=0. \end{aligned}$$

To derive the boundary condition for ${\mathbf {R}}_{{\mathbf {Q}}}$, we first consider the value of $\lambda \,\text {curl}{\mathbf {R}}_{{\mathbf {Q}}}$ at $X_0\in \partial \Omega$. We keep the notation used in section 2. We use equality (C.5) in appendix C with ${\mathbf {Q}}(x)$ replaced by ${\mathbf {Q}}_{\mathrm ap}(x)$, and use equality (D.3) in appendix D. Then we have

$$\begin{aligned} \begin{aligned} \lambda \,\text {curl}\, {\mathbf {Q}}_{\mathrm ap}(X_0)&=\Big (-\partial _2 Q_0^1+\lambda (k_0 Q_0^1-\partial _2 Q_1^1) +\lambda ^2 (\partial _{y_1}\hat{Q}_1^2\big |_{y_1=0}+k_0 Q_1^1-\partial _2Q_2^1)\\&+\lambda ^3 (k_0 Q_2^1+\partial _1 Q_2^2)\Big )\Big |_{z_2=0}, \end{aligned} \end{aligned}$$

where $k_0$ is the curvature of $\partial \Omega$ at $X_0$. Then we use the boundary condition of $Q_0^1$ in (2.14), the boundary condition of $(Q_1^1, Q_1^2)$ in (2.20), and that of $(Q_2^1, Q_2^2)$ in (2.22) to get

$$\begin{aligned} \lambda \text {curl}{\mathbf {Q}}_{\mathrm ap}(X_0) ={\mathcal {H}}^e(X_0)+\lambda ^3 (k_0 Q_2^1+\partial _1 Q_2^2)\big )\Big |_{z_2=0}. \end{aligned}$$

Similarly, for any $x\in \partial \Omega ,$ we also have

$$\begin{aligned} \lambda \text {curl}{\mathbf {Q}}_{\mathrm ap}(x)={\mathcal {H}}^e(x)-{\mathcal {B}}_3(x)\quad \text {on }\partial \Omega , \end{aligned}$$

(E.1)

where

$$\begin{aligned} \mathcal {B}_3(x)=-\lambda ^3 \Big (k(x) \hat{Q}_2^1(y_1,z_2)+\partial _1 \hat{Q}_2^2(y_1,z_2)\Big )\Big |_{z_2=0},\end{aligned}$$

(E.2)

k(x) is the curvature of $\partial \Omega$ at x, $z_2=y_2/\lambda ,$ $x=\psi (y_1,y_2)$ and $\psi$ is defined by (2.1). From Proposition 2.9, we have

$$\begin{aligned} \Vert {\mathcal {B}}_3\Vert _{C^2(\partial \Omega )}\le C\left( \Omega , {\mathcal {H}}^e\right) \lambda ^3. \end{aligned}$$

(E.3)

Combining (E.1) with the boundary condition $\lambda \text {curl}{\mathbf {Q}}={\mathcal {H}}^e$ on $\partial \Omega$, we immediately obtain that

$$\begin{aligned} \lambda \,\text {curl}{\mathbf {R}}_{{\mathbf {Q}}}={\mathcal {B}}_3\quad \text {on }\partial \Omega . \end{aligned}$$

Now we compute the value of ${\mathbf {n}}\cdot {\mathbf {R}}_{{\mathbf {Q}}}$ on $\partial \Omega$. We first calculate the value of $f_{\mathrm ap}^2 {\mathbf {n}}\cdot {\mathbf {Q}}_{\mathrm ap}(x)$ at $X_0\in \partial \Omega$. From (2.21) and (2.23), we have

$$\begin{aligned} |f_0|^2 Q_1^2\big |_{z_2=0}=\partial _{y_1 z_2} {\hat{Q}}_{0}^1\big |_{y_1=0, z_2=0}=\hat{{\mathcal {H}}}^e_{y_1}(0), \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} (|f_0|^2 Q_2^2+2 f_0 f_1 Q_1^2)\big |_{z_2=0}&=-(\partial _{z_2 y_1}{\hat{Q}}_1^1 -\partial _{y_1}(k {\hat{Q}}_{0}^1))\big |_{y_1=0, z_2=0}\\&\quad =-\partial _{y_1}(\partial _{z_2 }{\hat{Q}}_1^1 -(k {\hat{Q}}_{0}^1))\big |_{y_1=0, z_2=0}=0. \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} |f_{\mathrm ap}|^2 {\mathbf {n}}\cdot {\mathbf {Q}}_{\mathrm ap}(X_0)&=-(f_0+\lambda f_1+\lambda ^2 f_2)^2(\lambda Q_1^2+\lambda ^2 Q_2^2)\\&=-\lambda \hat{{\mathcal {H}}}^e_{y_1}(0) +\lambda ^3 \tilde{{\mathfrak {R}}}_2(y_1, z_2)\Big |_{y_1=0, z_2=0}, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \tilde{\mathfrak {R}}_2(y_1, z_2)=&(f_1^2+2f_0 f_2)Q_1^2+2f_0 f_1 Q_2^2 +\lambda \Big (2f_1 f_2 Q_1^2+(f_1^2+2f_0 f_2)Q_2^2\Big )\\&+\lambda ^2 \Big (f_2^2 Q_1^2+2f_1 f_2 Q_2^2\Big )+\lambda ^3 f_2^2 Q_2^2.\end{aligned} \end{aligned}$$

Similarly, for any $x\in \partial \Omega ,$ we also have

$$\begin{aligned} f_{\mathrm ap}^2 {\mathbf {n}}\cdot {\mathbf {Q}}_{\mathrm ap}(x)=-\lambda \nabla _{\mathrm {tan}} ({\mathcal {H}}^e)(x)-{\mathcal {B}}_4(x), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {B}}_4(x)=\lambda ^3 \tilde{{\mathfrak {R}}}_2(y_1, z_2)\Big |_{z_2=0}, \end{aligned}$$

(E.4)

$z_2=y_2/\lambda ,$ $x=\psi (y_1,y_2)$, and $\psi$ is defined by (2.1). From Proposition 2.5, Proposition 2.7 and Proposition 2.9, we have

$$\begin{aligned} \Vert {\mathcal {B}}_4\Vert _{C^2(\partial \Omega )}\le C\left( \Omega , {\mathcal {H}}^e\right) \lambda ^3. \end{aligned}$$

(E.5)

From the second and the third equations in (1.1), it follows that

$$\begin{aligned} f^2 {\mathbf {n}}\cdot {\mathbf {Q}}=-\lambda ^2{\mathbf {n}}\cdot \text {curl}^2 {\mathbf {Q}}=-\lambda \nabla _{\mathrm {tan}} (\lambda \text {curl}{\mathbf {Q}}) =-\lambda \nabla _{\mathrm {tan}}({\mathcal {H}}^e). \end{aligned}$$

This gives that

$$\begin{aligned} \nu \cdot {\mathbf {R}}_{{\mathbf {Q}}} =f_{\mathrm ap}^{-2}\left[ \mathcal {B}_4 +\lambda |f|^{-2}(|f|^2-|f_{\mathrm ap}|^2)\nabla _{\mathrm {tan}}(\mathcal H^e)\right] :=\mathcal {B}_5. \end{aligned}$$

(E.6)

Summarizing, we obtain the boundary conditions (3.8) for system (3.7).