Appendix A. Formally Matched Expansions
We use the matched asymptotic expansions to obtain (3.3a) and (3.3), and thus give a complete proof of Proposition 3.1, and of (1.9). The construction will employ the solvability of an ODE, see [2, Lemma 4.1].
Lemma A.1
Let \(\ell ,m, n\in \{0, 1, 2\}\) and \(\widetilde{A}(z,x,t):\mathbb {R}\times \Gamma ^0(\delta )\rightarrow \mathbb {R}\) be a function satisfying
$$\begin{aligned} \partial _z^\ell \partial _x^m\partial _t^n \widetilde{A}(z,x,t)=O(e^{-C|z|})~\text {as}~z\rightarrow \pm \infty ,~\text {uniformly in}~(x,t)\in \Gamma ^0(\delta ) \end{aligned}$$
(7.30)
and the following compatibility condition:
$$\begin{aligned} \int _{\mathbb {R}}\widetilde{A}(z,x,t)\theta '(z)\mathrm{dz}=0,~(x,t)\in \Gamma ^0(\delta ). \end{aligned}$$
(A.1)
Then the equation \(\mathscr {L} \widetilde{U}=\widetilde{A}\) has a bounded solution so that
$$\begin{aligned} \partial _z^\ell \partial _x^m\partial _t^n \widetilde{U}(z,x,t)=O(e^{-C|z|})~\text {as}~z\rightarrow \pm \infty ,~\text {uniformly in}~(x,t)\in \Gamma ^0(\delta ). \end{aligned}$$
(A.2)
Moreover, there exists a smooth function U(x, t) such that
$$\begin{aligned} {\widetilde{U}(z,x,t)=U(x,t)\theta '(z)+\left[ \int _{0}^{z}(\theta '(\zeta ))^{-2}\left( \int _{\zeta }^{+\infty }\widetilde{A}(\tau ,x,t)\theta '(\tau )d\tau \right) d\zeta \,\right] \theta '(z).} \end{aligned}$$
(A.3)
The unique solution satisfying \(\widetilde{U}(0,x,t)=0\) corresponds to \(U\equiv 0\).
Recall from Section 2.1 that \(d^{(0)}\) is the signed-distance to \(\Gamma ^0\). We need the following lemma whose proof can be found in [5].
Lemma A.2
The interface \(\Gamma ^0\) evolves under the Willmore flow (1.5) if and only if \(d^{(0)}\) fulfills
$$\begin{aligned} \partial _td^{(0)}+\Delta ^2 d^{(0)}=\Delta d^{(0)}{D^{(0)}} +\nabla d^{(0)}\cdot \nabla {D^{(0)}}\ \ \ \text {on}\ \ \Gamma ^0. \end{aligned}$$
(A.4)
Following [2], we employ the stretched variable \(z=\frac{d_\varepsilon }{\varepsilon }\in \mathbb {R}\) (see (2.1) for the definition of \(d_\varepsilon \)) and set the following Ansatz in \(\Gamma ^0(3\delta )\) for the inner expansion
$$\begin{aligned} \widetilde{\phi }^\varepsilon (z,x,t)=\sum _{\ell \ge 0}\varepsilon ^\ell \widetilde{\phi }^{(\ell )}(z,x,t),\qquad \widetilde{\mu }^\varepsilon (z,x,t)=\sum _{\ell \ge 0}\varepsilon ^\ell \widetilde{\mu }^{(\ell )}(z,x,t), \end{aligned}$$
(A.5)
which should fulfill the following matching conditions in \((x,t)\in \Gamma ^0(3\delta )\):
$$\begin{aligned}&D_z^\gamma D_x^\alpha D_t^\beta \Big (\widetilde{\phi }^{(i)}(z,x,t)-\phi ^{(i)}_{\pm }(x,t)\Big )=O(e^{-\nu |z|}), \end{aligned}$$
(A.6)
$$\begin{aligned}&D_z^\gamma D_x^\alpha D_t^\beta \Big (\widetilde{\mu }^{(i)}(z,x,t) -\mu ^{(i)}_{\pm }(x,t)\Big )=O(e^{-\nu |z|}). \end{aligned}$$
(A.7)
Here \(\nu \) is a positive fixed constant and \(0\le \alpha ,\beta ,\gamma \le 2\). It follows from the Taylor expansion and (A.6) that
$$\begin{aligned} f'(\widetilde{\phi }^\varepsilon )&=f'(\widetilde{\phi }^{(0)})+ f''(\widetilde{\phi }^{(0)})\sum _{i\ge 1}\varepsilon ^i\widetilde{\phi }^{(i)}+\sum _{i\ge 1} \varepsilon ^{i} g_{i-1}\big (\widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(i-1)}\big ), \end{aligned}$$
(A.8)
$$\begin{aligned} f''(\widetilde{\phi }^\varepsilon )&=f''(\widetilde{\phi }^{(0)})+ f'''(\widetilde{\phi }^{(0)})\sum _{i\ge 1}\varepsilon ^i\widetilde{\phi }^{(i)}+ \sum _{i\ge 1} \varepsilon ^i g_{i-1}^*\big (\widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(i-1)}\big ), \end{aligned}$$
(A.9a)
where \(g_i,g_i^*\) enjoy the following property:
Lemma A.3
For \(i\ge 1\), \(g_i(x_0,\cdots ,x_i)\) and \(g^*_i(x_0,\cdots ,x_i)\) are polynomials of \(i+1\) variables. Moreover, they vanish when \(x_1=\cdots =x_{i-1}=0\), and \(g_0=g^*_0=0\).
Using \(|\nabla d_\varepsilon |=1\) and the chain-rule, we have the following expansions
$$\begin{aligned} \partial _t \widetilde{\phi }^\varepsilon (\tfrac{d_\varepsilon }{\varepsilon },x,t)&=\partial _t\widetilde{\phi }^\varepsilon +\varepsilon ^{-1}\partial _z\widetilde{\phi }^\varepsilon \partial _td_\varepsilon ,\\ \Delta \widetilde{\mu }^\varepsilon (\tfrac{d_\varepsilon }{\varepsilon },x,t)&=\varepsilon ^{-2}\partial _z^2\widetilde{\mu }^\varepsilon +2\varepsilon ^{-1}\nabla _x\partial _z\widetilde{\mu }^\varepsilon \cdot \nabla d_\varepsilon +\varepsilon ^{-1}\partial _z\widetilde{\mu }^\varepsilon \Delta d_\varepsilon +\Delta _x\widetilde{\mu }^\varepsilon ,\\ \Delta \widetilde{\phi }^\varepsilon (\tfrac{d_\varepsilon }{\varepsilon },x,t)&=\varepsilon ^{-2}\partial _z^2\widetilde{\phi }^\varepsilon +2\varepsilon ^{-1}\nabla _x\partial _z\widetilde{\phi }^\varepsilon \cdot \nabla d_\varepsilon +\varepsilon ^{-1}\partial _z\widetilde{\phi }^\varepsilon \Delta d_\varepsilon +\Delta _x\widetilde{\phi }^\varepsilon . \end{aligned}$$
We expect \((\widetilde{\phi }^\varepsilon (z,x,t),\widetilde{\mu }^\varepsilon (z,x,t))|_{z= d_\varepsilon /\varepsilon }\) to satisfy (1.1) up to a high order term in \(\varepsilon \) and thus determine the terms in (A.6):
$$\begin{aligned} \partial _z^2\widetilde{\mu }^\varepsilon -f''(\widetilde{\phi }^\varepsilon )\widetilde{\mu }^\varepsilon +2\varepsilon \nabla _x\partial _z\widetilde{\mu }^\varepsilon \cdot \nabla d_\varepsilon +\varepsilon \partial _z\widetilde{\mu }^\varepsilon \Delta d_\varepsilon -\varepsilon ^2\partial _z\widetilde{\phi }^\varepsilon \partial _td_\varepsilon \nonumber \\+\varepsilon ^2\Delta _x\widetilde{\mu }^\varepsilon -\varepsilon ^3\partial _t\widetilde{\phi }^\varepsilon = O(\varepsilon ^{k}), \end{aligned}$$
(A.9b)
$$\begin{aligned} -\partial _z^2\widetilde{\phi }^\varepsilon +f'(\widetilde{\phi }^\varepsilon )-2\varepsilon \nabla _x\partial _z\widetilde{\phi }^\varepsilon \cdot \nabla d_\varepsilon -\varepsilon \partial _z\widetilde{\phi }^\varepsilon \Delta d_\varepsilon -\varepsilon \widetilde{\mu }^\varepsilon -\varepsilon ^2\Delta _x\widetilde{\phi }^\varepsilon = O(\varepsilon ^{k}). \end{aligned}$$
(A.10)
Since \(z=\frac{d_\varepsilon }{\varepsilon }\), we need the above two equations to hold merely on
$$S^\varepsilon \triangleq \{(z,x,t)\in \mathbb {R}\times \Gamma ^0(3\delta ):z= d_\varepsilon /\varepsilon \}.$$
So we can add in (A.10) terms which are multiplied by \(d_\varepsilon -\varepsilon z\). These terms will give more degrees of freedom to construct and to solve the equations for \(d^{(\ell )}\). See Remark A.1 below and [2]. So we modify (A.10) as follows
$$\begin{aligned} \partial _z^2\widetilde{\mu }^\varepsilon -f''(\widetilde{\phi }^\varepsilon )\widetilde{\mu }^\varepsilon&+2\varepsilon \nabla _x\partial _z\widetilde{\mu }^\varepsilon \cdot \nabla d_\varepsilon +\varepsilon \partial _z\widetilde{\mu }^\varepsilon \Delta d_\varepsilon -\varepsilon ^2\partial _z\widetilde{\phi }^\varepsilon \partial _td_\varepsilon \nonumber \\&+\varepsilon ^2\Delta _x\widetilde{\mu }^\varepsilon -\varepsilon ^3\partial _t\widetilde{\phi }^\varepsilon +\varepsilon ^2\chi ^\varepsilon (d_\varepsilon -\varepsilon z)\eta '= O(\varepsilon ^{k}), \end{aligned}$$
(A.11)
where \(\eta (z)\) is a smooth non-decreasing function satisfying
$$\begin{aligned} \eta (z)=0\ \text {if}\ z\le -1;\ \eta (z)=1\ \text {if}\ z\ge 1;\ \ \eta '(z)\ \text {is\ even}, \end{aligned}$$
(A.12)
and \(\chi ^{\varepsilon }(x, t)=\sum _{i=0}^{\infty }\varepsilon ^{i}\chi ^{(i)}(x, t)\) with \(\chi ^{(i)}\) being determined later on.
Definition A.4
We shall use \(\varepsilon ^\ell \)-scale to denote the terms of form \(\varepsilon ^\ell g(z,x,t)\). The \(\ell \)-order will refer to those indexed by \(\ell \) if \(\ell \ge 0\), and by 0 if \(\ell <0\). Moreover, \(\widetilde{\Psi }^{(\ell )}(z,x,t)\) and \(\Xi ^{(\ell )}(x,t)\) will denote generic terms which might change from line to line, and will depend on terms of order at most \(\ell \).
1.1
\(\varepsilon ^1\)-Scale
Collecting all terms of \(\varepsilon ^0\)-scale in (A.11)–(A.12), we have \(\partial _z^2\widetilde{\phi }^{(0)}=f'(\widetilde{\phi })\) and \(\partial _z^2\widetilde{\mu }^{(0)}=f''(\widetilde{\phi }^{(0)})\widetilde{\mu }^{(0)}\). Together with the matching condition (A.7) and (A.8), we obtain
$$\begin{aligned} \widetilde{\phi }^{(0)}=\theta (z),\quad \widetilde{\mu }^{(0)}(z,x,t)=\mu _0(x,t)\theta '(z), \end{aligned}$$
(A.13)
for some function \(\mu _0\) which will be determined later on. To proceed we recall the operator \(\mathscr {L}\) defined at (1.16), which enjoys
$$\begin{aligned} \mathscr {L}~\text {is self-adjoint},~ \partial _z\mathscr {L}-\mathscr {L}\partial _z=f'''(\theta )\theta '\mathcal {I}. \end{aligned}$$
(A.14)
Collecting all terms of \(\varepsilon ^1\)-scale in (A.11)–(A.12), and using \(\widetilde{\phi }^{(0)}=\theta (z)\), we have
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(1)}&=-f'''(\theta )\widetilde{\phi }^{(1)}\widetilde{\mu }^{(0)} +2\nabla _x\partial _z\widetilde{\mu }^{(0)}\cdot \nabla d^{(0)}+\partial _z\widetilde{\mu }^{(0)}\Delta d^{(0)}, \end{aligned}$$
(A.15)
$$\begin{aligned} \mathscr {L}\widetilde{\phi }^{(1)}&=\widetilde{\mu }^{(0)}+\partial _z\theta \Delta d^{(0)}. \end{aligned}$$
(A.16a)
Here \(\mu _0\) is chosen such that (A.16b) fulfills (A.2), i.e. \(\mu _0=-\Delta d^{(0)}\). Thus
$$\begin{aligned} \widetilde{\mu }^{(0)}(z,x,t)=-\Delta d^{(0)}\theta ',~\widetilde{\phi }^{(1)}(z,x,t)=0. \end{aligned}$$
(A.16b)
This together with Lemma A.3 implies
$$\begin{aligned} g_{1}=g_{2}=g_{1}^*=g_{2}^*=0. \end{aligned}$$
(A.17)
Substituting this into (A.16a) yields \({\mathscr {L}\widetilde{\mu }^{(1)}}=-{2D^{(0)}}\theta ''\) where \(D^{(0)}\) is given by
$$\begin{aligned} D^{(0)}(x,t)=\nabla \Delta d^{(0)}\cdot \nabla d^{(0)}+\tfrac{1}{2}(\Delta d^{(0)})^2. \end{aligned}$$
(A.18)
Using (A.4) we deduce
$$\begin{aligned} \widetilde{\mu }^{(1)}(z,x,t)&=D^{(0)}z\theta '(z)+\mu _1(x,t)\theta '(z) \end{aligned}$$
(A.19)
for some \(\mu _1(x,t)\) which will be determined later on.
1.2
\(\varepsilon ^2\)-Scale
Substituting (A.6) into (A.10)–(A.11), and collecting all terms of \(\varepsilon ^2\)-scale, we obtain
$$\begin{aligned} 0&=\partial _z^2\widetilde{\mu }^{(2)}-f''(\theta )\widetilde{\mu }^{(2)}-f'''(\theta )\widetilde{\phi }^{(1)}\widetilde{\mu }^{(1)} -\big (f'''(\theta )\widetilde{\phi }^{(2)}+g_1^*\big (\widetilde{\phi }^{(0)},\widetilde{\phi }^{(1)}\big )\big )\widetilde{\mu }^{(0)} \nonumber \\&\quad +2\nabla _x\partial _z\widetilde{\mu }^{(1)}\cdot \nabla d^{(0)} +2\nabla _x\partial _z\widetilde{\mu }^{(0)}\cdot \nabla d^{(1)} +\partial _z\widetilde{\mu }^{(1)}\Delta d^{(0)}+\partial _z\widetilde{\mu }^{(0)}\Delta d^{(1)} \nonumber \\&\quad -\partial _z\theta \partial _td^{(0)}+ \Delta _x\widetilde{\mu }^{(0)}+\chi ^{(0)} d^{(0)}\eta ',\\ 0&=-\partial _z^2\widetilde{\phi }^{(2)}+f''(\theta )\widetilde{\phi }^{(2)} +g_1\big (\widetilde{\phi }^{(0)},\widetilde{\phi }^{(1)}\big ) -2\nabla _x\partial _z\widetilde{\phi }^{(1)}\cdot \nabla d^{(0)}-2\nabla _x\partial _z\theta \cdot \nabla d^{(1)} \nonumber \\&\quad -\partial _z\widetilde{\phi }^{(1)}\Delta d^{(0)}-\partial _z\theta \Delta d^{(1)}-\widetilde{\mu }^{(1)}-\Delta _x\widetilde{\phi }^{(0)}. \end{aligned}$$
In view of (A.18), the above two equations can be simplified as
$$\begin{aligned} {\mathscr {L}\widetilde{\mu }^{(2)}}&= -f'''(\theta )\widetilde{\phi }^{(2)}\widetilde{\mu }^{(0)} +2\nabla _x\partial _z\widetilde{\mu }^{(1)}\cdot \nabla d^{(0)} +2\nabla _x\partial _z\widetilde{\mu }^{(0)}\cdot \nabla d^{(1)} \nonumber \\&\quad +\partial _z\widetilde{\mu }^{(1)}\Delta d^{(0)}+\partial _z\widetilde{\mu }^{(0)}\Delta d^{(1)}-\partial _z\theta \partial _td^{(0)}+ \Delta _x\widetilde{\mu }^{(0)}+\chi ^{(0)} d^{(0)}\eta ', \end{aligned}$$
(A.20)
$$\begin{aligned} {\mathscr {L}\widetilde{\phi }^{(2)}}&=\partial _z\theta \Delta d^{(1)}+\widetilde{\mu }^{(1)} =\big (\Delta d^{(1)}+\mu _1\big )\theta '+{D^{(0)}}z\theta '(z). \end{aligned}$$
(A.21a)
Recall (A.20) that \(\mu _1\) shall be determined such that (A.21b) fulfills (A.2), i.e. \(\big (\Delta d^{(1)}+\mu _1\big )\sigma =0\), where \(\sigma \triangleq \int _{\mathbb {R}}(\theta ')^2\mathrm{dz}\). This leads to the formula for \(\mu _1\) and completes formula (A.20):
$$\begin{aligned} \mu _1(x,t)=-\Delta d^{(1)},\qquad \widetilde{\mu }^{(1)}(z,x,t)&=D^{(0)}z\theta '(z)-\Delta d^{(1)}\theta '(z). \end{aligned}$$
(A.21b)
As a result, (A.21b) is simplified to
$$\begin{aligned} {\mathscr {L}\widetilde{\phi }^{(2)}}={D^{(0)}}z\theta '(z). \end{aligned}$$
(A.22)
Using \(\int _{\mathbb {R}}z(\theta ')^2\mathrm{dz}=0\) and formula (A.4), we can solve (A.23):
$$\begin{aligned} \widetilde{\phi }^{(2)}(z,x,t)={D^{(0)}}\theta '(z) \alpha (z),~\text {with}~\alpha (z)\triangleq \int _{0}^{z}(\theta '(\zeta ))^{-2}\int _{\zeta }^{+\infty }\tau (\theta '(\tau ))^2d\tau d\zeta \end{aligned}$$
(A.23)
being an odd function. This implies that \(\widetilde{\phi }^{(2)}\) is odd with respect to z. On the other hand, \(\chi ^{(0)}\) is determined so that the right hand side of (A.21a) fulfills (A.2), e.g.
$$\begin{aligned} \chi ^{(0)}d^{(0)}\sigma ^{-1}\overline{\sigma }=\mathscr {G}_0d^{(0)},~\text {with}~\overline{\sigma }=\int _{\mathbb {R}}\eta '\theta ' \mathrm{dz},\end{aligned}$$
(A.24)
$$\begin{aligned} \mathscr {G}_0d^{(0)}\triangleq \partial _td^{(0)}+\Delta ^2 d^{(0)}-\Delta d^{(0)}{D^{(0)}} -\nabla d^{(0)}\cdot \nabla {D^{(0)}}. \end{aligned}$$
(A.25)
Note that we used the following formula which is due to (A.15) and (A.23):
$$\begin{aligned} \int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(2)}(\theta ')^2\mathrm{dz}=\int _{\mathbb {R}}\partial _z\big ({\mathscr {L}\widetilde{\phi }^{(2)}}\big )\theta '\mathrm{dz}-\int _{\mathbb {R}} \mathscr {L}\left( \partial _z\widetilde{\phi }^{(2)}\right) \theta ' \mathrm{dz}=\tfrac{\sigma }{2} D^{(0)}. \end{aligned}$$
(A.26)
Combining (A.25) and (A.5) leads to the choice of \(\chi ^{(0)}\):
$$\begin{aligned} \chi ^{(0)}\triangleq \left\{ \begin{array}{llr} \sigma (\overline{\sigma })^{-1}\left( \mathscr {G}_0d^{(0)}\right) /d^{(0)}, \ &{}\forall (x,t)\in \Gamma ^0(3\delta )\backslash \Gamma ^0,\\ \sigma (\overline{\sigma })^{-1}\nabla \left( \mathscr {G}_0d^{(0)}\right) \cdot \nabla d^{(0)}, \ &{}\forall (x,t)\in \Gamma ^0. \end{array} \right. \end{aligned}$$
(A.27)
Remark A.1
If we do not modify the equation (A.10) into (A.12), then we would require the equation (A.5) to hold in \(\Gamma ^0(3\delta )\), which is not compatible with \(|\nabla d^{(0)}|=1\) in general.
The formula (A.25) reduces (A.21a) to
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(2)}&= \big (f'''(\theta )(\theta ')^2\alpha +z\theta ''\big )\Delta d^{(0)}{D^{(0)}} +\big (\theta '+2z\theta ''\big )\nabla d^{(0)}\cdot \nabla {D^{(0)}}\nonumber \\&-2\theta ''{D^{(1)}} +\chi ^{(0)}d^{(0)}\eta '- \sigma ^{-1}\overline{\sigma }\chi ^{(0)}d^{(0)}\theta ', \end{aligned}$$
(A.28)
$$\begin{aligned} D^{(1)}&=\nabla \Delta d^{(1)}\cdot \nabla d^{(0)}+\nabla \Delta d^{(0)}\cdot \nabla d^{(1)}+\Delta d^{(0)}\Delta d^{(1)}. \end{aligned}$$
(A.29)
Note that \(D^{(1)}\) is consistent with (3.4). We can solve (A.29) by employing (A.4),
$$\begin{aligned} \widetilde{\mu }^{(2)}(z,x,t)=&\Delta d^{(0)}{D^{(0)}}\theta '(z)\gamma _1(z) +\nabla d^{(0)}\cdot \nabla {D^{(0)}}\theta '(z)\gamma _2(z) \nonumber \\&+{D^{(1)}}z\theta ' +\mu _2(x,t)\theta ' +\chi ^{(0)}d^{(0)}\theta '(z)\gamma _3(z), \end{aligned}$$
(A.30)
where \(\mu _2(x,t)\) is a smooth function which will be determined by the \(\varepsilon ^3\)-scale below, and \(\gamma _1(z)\) and \(\gamma _2(z)\) and \(\gamma _3(z)\) are three even functions defined by
$$\begin{aligned} \begin{aligned}&\gamma _1(z)=\int _{0}^{z}(\theta '(\zeta ))^{-2}\int _{\zeta }^{+\infty }\theta '(\tau )\big (f'''(\theta )(\theta ')^2\alpha +\tau \theta ''\big )(\tau )d\tau d\zeta , \quad \gamma _2(z)=-z^2/2, \\&\gamma _3(z)=\int _{0}^{z}(\theta '(\zeta ))^{-2}\int _{\zeta }^{+\infty }\theta '(\tau )\big (\eta '(\tau )-\sigma ^{-1}\overline{\sigma }\theta '(\tau )\big )d\tau d\zeta . \end{aligned} \end{aligned}$$
(A.31)
1.3
\(\varepsilon ^3\)-Scale
We substitute (A.6) into (A.11)–(A.12), then use (A.18) and collect all the terms of \(\varepsilon ^3\)-scale:
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(3)}&=-f'''(\theta )\widetilde{\phi }^{(2)}\widetilde{\mu }^{(1)} -f'''(\theta )\widetilde{\phi }^{(3)}\widetilde{\mu }^{(0)}+\big (\chi ^{(0)}d^{(1)}+\chi ^{(1)}d^{(0)}\big ) \eta '-\chi ^{(0)}z\eta ' \nonumber \\&\quad +2\nabla _x\partial _z\widetilde{\mu }^{(2)}\cdot \nabla d^{(0)} +2\nabla _x\partial _z\widetilde{\mu }^{(0)}\cdot \nabla d^{(2)}+2\nabla _x\partial _z\widetilde{\mu }^{(1)}\cdot \nabla d^{(1)} \nonumber \\&\quad +\partial _z\widetilde{\mu }^{(2)}\Delta d^{(0)}+\partial _z\widetilde{\mu }^{(0)}\Delta d^{(2)}+\partial _z\widetilde{\mu }^{(1)}\Delta d^{(1)} -\partial _z\theta \partial _td^{(1)}+\Delta _x\widetilde{\mu }^{(1)} , \end{aligned}$$
(A.32)
$$\begin{aligned} {\mathscr {L}\widetilde{\phi }^{(3)}}&= 2\nabla _x\partial _z\widetilde{\phi }^{(2)}\cdot \nabla d^{(0)} +\partial _z\widetilde{\phi }^{(2)}\Delta d^{(0)}+\partial _z\theta \Delta d^{(2)}+\widetilde{\mu }^{(2)}. \end{aligned}$$
(A.33a)
We determine \(\mu _2(x,t)\) in (A.31) so that (A.33b) satisfies (A.2), i.e.
$$\begin{aligned} \big (\Delta d^{(2)}+\mu _2\big )\sigma&=-\left( \nabla d^{(0)}\cdot \nabla {D^{(0)}} +\tfrac{1}{2}\Delta d^{(0)}{D^{(0)}}\right) \int _{\mathbb {R}}\int _{z}^{+\infty }\tau (\theta '(\tau ))^2d\tau \mathrm{dz} \nonumber \\&\quad -\Delta d^{(0)}{D^{(0)}}\int _{\mathbb {R}}(\theta ')^2\gamma _1(z)\mathrm{dz} -\nabla d^{(0)}\cdot \nabla {D^{(0)}}\int _{\mathbb {R}}(\theta ')^2\gamma _2(z)\mathrm{dz} \nonumber \\&\quad -\chi ^{(0)}d^{(0)}\int _{\mathbb {R}} (\theta '(z))^2\gamma _3(z)\mathrm{dz}. \end{aligned}$$
(A.33b)
To prove (A.34), it follows from (A.24) and integration by parts that
$$\begin{aligned}&\int _{\mathbb {R}}\left( 2\nabla _x\partial _z\widetilde{\phi }^{(2)}\cdot \nabla d^{(0)} +\partial _z\widetilde{\phi }^{(2)}\Delta d^{(0)}+\partial _z\theta \Delta d^{(2)}\right) \theta '\mathrm{dz} \nonumber \\&=-\big (2\nabla d^{(0)}\cdot \nabla D^{(0)}+D^{(0)}\Delta d^{(0)}\big )\int _{\mathbb {R}}\alpha \theta '\theta ''\mathrm{dz}+\Delta d^{(2)}\sigma \nonumber \\&=\left( \nabla d^{(0)}\cdot \nabla {D^{(0)}} +\tfrac{1}{2}\Delta d^{(0)}{D^{(0)}}\right) \int _{\mathbb {R}}\int _{z}^{+\infty }\tau (\theta '(\tau ))^2d\tau \mathrm{dz}+\Delta d^{(2)}\sigma . \end{aligned}$$
(A.34)
This together with (A.31) leads to (A.34). So we can use (A.34) to rewrite (A.31) as
$$\begin{aligned} \widetilde{\mu }^{(2)}(z,x,t)&=-\Delta d^{(2)}(x,t)\theta '(z)+D^{(1)}(x,t)z\theta '(z)+\widetilde{\Psi }^{(0)}(z,x,t), \end{aligned}$$
(A.35)
where \(\widetilde{\Psi }^{(0)}\) only depends on 0-order terms:
$$\begin{aligned} \widetilde{\Psi }^{(0)}&=\Delta d^{(0)}D^{(0)}\theta '\gamma _1+\nabla d^{(0)}\cdot \nabla D^{(0)}\theta '\gamma _2 \nonumber \\&\quad - (2\sigma )^{-1}\left( \int _{\mathbb {R}}\int _{z}^{+\infty }\tau (\theta '(\tau ))^2d\tau \mathrm{dz}\right) \bigg (\Delta d^{(0)}D^{(0)}+2\nabla d^{(0)}\cdot \nabla D^{(0)}\bigg )\theta ' \nonumber \\&\quad -\sigma ^{-1}\bigg (\Delta d^{(0)}{D^{(0)}}\int _{\mathbb {R}}(\theta ')^2\gamma _1(z)\mathrm{dz} +\nabla d^{(0)}\cdot \nabla {D^{(0)}}\int _{\mathbb {R}}(\theta ')^2\gamma _2(z)\mathrm{dz}\bigg )\theta ' \nonumber \\&\quad -\sigma ^{-1}\bigg (\int _{\mathbb {R}}(\theta '(z))^2 \gamma _3(z)\mathrm{dz}\bigg )\chi ^{(0)}d^{(0)}\theta '. \end{aligned}$$
(A.36)
Finally, applying Lemma A.1 to (A.33b) yields a solution \(\widetilde{\phi }^{(3)}\):
Lemma A.5
\(\widetilde{\Psi }^{(0)}\) satisfies (A.1) and the equation (A.33b) has a unique smooth solution \(\widetilde{\phi }^{(3)}\) depending up to 1-order terms, satisfying \(\widetilde{\phi }^{(3)}|_{z=0}=0\) and (A.3).
\(d^{(1)}\) is determined so that the right hand side of (A.33a) fulfills (A.2):
Lemma A.6
There exists \(\Xi ^{(0)}(x,t)\) depending on 0-order terms such that
$$\begin{aligned}&\mathscr {G}_1d^{(1)}=\tfrac{\overline{\sigma }}{\sigma }\big (\chi ^{(0)}d^{(1)} +\chi ^{(1)}d^{(0)}\big )+\Xi ^{(0)}~\text {in}~\Gamma ^0(3\delta ), \end{aligned}$$
(A.37)
$$\begin{aligned}&\text {where}\quad \mathscr {G}_1d^{(1)}\triangleq \partial _td^{(1)}+\Delta ^2d^{(1)}-\sum \limits _{i=0,1}\left( \nabla D^{(i)}\cdot \nabla d^{(1-i)}+ D^{(i)}\Delta d^{(1-i)}\right) . \end{aligned}$$
(A.38)
Proof
We note that \(f'''(\theta )(\theta '(z))^3\alpha (z) z\) is an odd function, so it follows from (A.22), (A.24) and (A.27) that
$$\begin{aligned} -\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(2)}\widetilde{\mu }^{(1)}\theta '\mathrm{dz}&=\Delta d^{(1)}\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(2)}(\theta ')^2 \mathrm{dz}=\frac{\sigma }{2}D^{(0)}\Delta d^{(1)}. \end{aligned}$$
(A.39)
Using (A.17), (A.33b) and (A.24), we can proceed in the same way as we obtain (A.27) and yield
$$\begin{aligned}&-\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(3)}\widetilde{\mu }^{(0)}\theta '\mathrm{dz}=\Delta d^{(0)}\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(3)}(\theta ')^2\mathrm{dz} \nonumber \\&=\Delta d^{(0)}\int _{\mathbb {R}}\partial _z\left( 2\nabla _x\partial _z\widetilde{\phi }^{(2)}\cdot \nabla d^{(0)} +\partial _z\widetilde{\phi }^{(2)}\Delta d^{(0)}+\partial _z\theta \Delta d^{(2)}+\widetilde{\mu }^{(2)}\right) \theta '\mathrm{dz} \nonumber \\&=-\Delta d^{(0)}\int _{\mathbb {R}}\widetilde{\mu }^{(2)}\theta ''\mathrm{dz}. \end{aligned}$$
This combined with (A.36) leads to
$$\begin{aligned}&-\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(3)}\widetilde{\mu }^{(0)}\theta '\mathrm{dz} \nonumber \\&=-\Delta d^{(0)}\int _{\mathbb {R}}\big (-\Delta d^{(2)}\theta '(z)+D^{(1)}z\theta '(z)+\widetilde{\Psi }^{(0)}(z,x,t)\big )\theta ''\mathrm{dz} \nonumber \\&=\frac{\sigma }{2}D^{(1)}\Delta d^{(0)}-\Delta d^{(0)}\int _{\mathbb {R}}\widetilde{\Psi }^{(0)}(z,x,t)\theta ''\mathrm{dz}. \end{aligned}$$
(A.40)
We continue treating the terms on the right hand side of (A.33a). It follows from (A.17), (A.22), and (A.36) that
$$\begin{aligned}&\int _{\mathbb {R}}\left( 2\nabla _x\partial _z\widetilde{\mu }^{(2)}\cdot \nabla d^{(0)} +2\nabla _x\partial _z\widetilde{\mu }^{(0)}\cdot \nabla d^{(2)}+2\nabla _x\partial _z\widetilde{\mu }^{(1)}\cdot \nabla d^{(1)}\right) \theta '\mathrm{dz} \nonumber \\&=\sigma \big (\nabla D^{(1)}\cdot \nabla d^{(0)}+\nabla D^{(0)}\cdot \nabla d^{(1)}\big ) -2\nabla d^{(0)}\cdot \int _{\mathbb {R}}\nabla _x\widetilde{\Psi }^{(0)}(z,x,t)\theta ''\mathrm{dz}. \end{aligned}$$
Moreover, we have the following two identities:
$$\begin{aligned}&\int _{\mathbb {R}}\left( \partial _z\widetilde{\mu }^{(2)}\Delta d^{(0)}+\partial _z\widetilde{\mu }^{(0)}\Delta d^{(2)}+\partial _z\widetilde{\mu }^{(1)}\Delta d^{(1)}\right) \theta '\mathrm{dz} \nonumber \\&\qquad =\frac{\sigma }{2}\big (D^{(1)}\Delta d^{(0)}+D^{(0)}\Delta d^{(1)}\big ) -\Delta d^{(0)}\int _{\mathbb {R}}\widetilde{\Psi }^{(0)}(z,x,t)\theta ''\mathrm{dz},\\&\int _{\mathbb {R}} \big (-\partial _z\theta \partial _td^{(1)}+\Delta _x\widetilde{\mu }^{(1)}\big )\theta '\mathrm{dz}=-\sigma \big (\partial _td^{(1)}+\Delta ^2d^{(1)}\big ). \end{aligned}$$
(A.41)
Therefore, using the notation (A.39) , we deduce that \(d^{(1)}\) satisfies (A.38) and
$$\begin{aligned} \Xi ^{(0)}=-\frac{2}{\sigma }\nabla d^{(0)}\cdot \int _{\mathbb {R}}\nabla _x\widetilde{\Psi }^{(0)}(z,x,t)\theta ''\mathrm{dz}-\frac{2}{\sigma }\Delta d^{(0)}\int _{\mathbb {R}}\widetilde{\Psi }^{(0)}(z,x,t)\theta ''\mathrm{dz}. \end{aligned}$$
To determine \(d^{(1)}\) and \(\chi ^{(1)}\) so that (A.38) holds, we need the following result:
Corollary A.7
The following equation of \(d^{(1)}\) has a local in time classical solution:
$$\begin{aligned} \mathscr {G}_1d^{(1)}=\sigma ^{-1}\overline{\sigma }\chi ^{(0)}d^{(1)}+\Xi ^{(0)}~\text {on}~\Gamma ^0. \end{aligned}$$
(A.42)
Moreover, if we define \(\chi ^{(1)}\) by
$$\begin{aligned} \chi ^{(1)}\triangleq \left\{ \begin{array}{ll} \sigma (\overline{\sigma })^{-1}\left( \mathscr {G}_1d^{(1)}-\sigma ^{-1}\overline{\sigma }\chi ^{(0)}d^{(1)}-\Xi ^{(0)}\right) /d^{(0)}&{} \ \text {in}~\Gamma ^0(3\delta )\backslash \Gamma ^0,\\ \sigma \big (\overline{\sigma }\big )^{-1}\nabla \big ( \mathscr {G}_1d^{(1)}-\sigma ^{-1}\overline{\sigma }\chi ^{(0)}d^{(1)}-\Xi ^{(0)}\big )\cdot \nabla d^{(0)}&{} \ \text {on}~\Gamma ^0, \end{array} \right. \end{aligned}$$
(A.43)
then (A.38) holds in \(\Gamma ^0(3\delta )\).
Proof
Note that \(d^{(1)}\) might not fulfill (A.44) in \(\Gamma ^0(3\delta )\). Since \(\partial _r d^{(1)}=0\) (see (2.2)), it suffices to determine \(d^{(1)}\) on \(\Gamma ^0\) and then extends constantly in the normal direction. Using (6.9) we can convert mixed derivatives of \(d^{(1)}\) into tangential ones. This combined with (A.30) and (A.44) yields
$$\begin{aligned} \partial _td^{(1)}+\Delta ^2d^{(1)}-(\nabla d^{(0)}\otimes \nabla d^{(0)}):\nabla ^2 \Delta d^{(1)} =\mathfrak {T}(d^{(1)}), \end{aligned}$$
(A.44)
where \(\mathfrak {T}\) is a generic term that includes at most third-order (tangential) derivatives of \(d^{(1)}\). Using (6.9), (2.9) and (2.11) yields
$$\begin{aligned} \begin{aligned}&\Delta ^2d^{(1)}-(\nabla d^{(0)}\otimes \nabla d^{(0)}):\nabla ^2 \Delta d^{(1)}\\ =&{\text {div}}_{\Gamma _0} (\nabla \Delta _{\Gamma _0} d^{(1)})=\Delta ^2_{\Gamma _0} d^{(1)}+\left( {\text {div}}_{\Gamma _0}\mathbf {n}\right) \partial _r (\Delta _{\Gamma _0}d^{(1)}). \end{aligned} \end{aligned}$$
(A.45)
So we can write (A.46) as (see [2] for similar arguments)
$$\begin{aligned} \partial _td^{(1)}+ \Delta _{\Gamma _0}^2 d^{(1)} =\mathfrak {T}(d^{(1)}). \end{aligned}$$
(A.46)
This is a surface evolutionary equation and has a local in time smooth solution.
1.4
\(\varepsilon ^{K}\)-scale
With Definition A.4, we set the following statements indexed by K:
$$\begin{aligned} A_K&:\widetilde{\phi }^{(i)}~\text {depends on terms of order up to}~(i-2);~2\le i\le K+1, \end{aligned}$$
(A.47)
$$\begin{aligned} B_K&:\widetilde{\mu }^{(i)}=-\Delta d^{(i)}\theta '+{D^{(i-1)}z\theta '}+\widetilde{\Psi }^{(i-2)}~\text {for}~2\le i\le K,\end{aligned}$$
(A.48)
$$\begin{aligned} C_K&:\widetilde{\mu }^{(K+1)}=\mu _{K+1}(x,t)\theta '+{D^{(K)}z\theta '} +\widetilde{\Psi }^{(K-1)},\end{aligned}$$
(A.49a)
$$\begin{aligned} D_K&: (d^{(K)}, \chi ^{(K)})~\text {depend on terms up to order}\, (K-1)\,\text {through}\nonumber \\&\qquad \mathscr {G}_K d^{(K)}=\sigma ^{-1}\overline{\sigma }\big (\chi ^{(0)}d^{(K)}+\chi ^{(K)}d^{(0)}\big )+\Xi ^{(K-1)}, \end{aligned}$$
(A.49b)
where \(D^{(i)}\) is defined by (3.4), \(\widetilde{\Psi }^{(K)}\) satisfies the decay property (A.3), and
$$\begin{aligned}&\mathscr {G}_K d^{(K)}\triangleq \partial _td^{(K)}+\Delta ^2d^{(K)}-\sum _{\ell =0,K}\left( \nabla D^{(\ell )}\cdot \nabla d^{(K-\ell )}+D^{(\ell )}\Delta d^{(K-\ell )}\right) , \end{aligned}$$
(A.49c)
$$\begin{aligned}&\chi ^{(K)}\triangleq \left\{ \begin{array}{ll} \sigma \overline{\sigma }^{-1}\left( \mathscr {G}_K d^{(K)}-\sigma ^{-1}\overline{\sigma }\chi ^{(0)}d^{(K)}-\Xi ^{(K-1)}\right) /d^{(0)}&{} \ \text {in}~\Gamma ^0(3\delta )\backslash \Gamma ^0, \\ \sigma \overline{\sigma }^{-1}\nabla \big (\mathscr {G}_K d^{(K)}-\sigma ^{-1}\overline{\sigma }\chi ^{(0)}d^{(K)}-\Xi ^{(K-1)}\big )\cdot \nabla d^{(0)}&{} \ \text {in}~\Gamma ^0. \end{array} \right. \end{aligned}$$
(A.49d)
Lemma A.8
The statements \((A_1,B_1,C_1)\) and \((A_2,B_2,C_2,D_1)\) are valid.
Proof
Recall the results in previous subsections. Using \(d^{(1)}\) we can determine \(\widetilde{\mu }^{(1)}\) through (A.22). Using \(d^{(2)}\) determined by (A.50) with \(K=2\), we obtain \(\widetilde{\mu }^{(2)}\) by (A.36) and \(\widetilde{\phi }^{(3)}\) by solving (A.33b). Finally we can rewrite (A.33a) as
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(3)}&=-2\theta ''D^{(2)}+\widetilde{\Psi }^{(1)},~\text {where}~D^{(2)}\nonumber \\&=\sum \limits _{0\le \ell \le 2}\left( \nabla \Delta d^{(\ell )}\cdot \nabla d^{(2-\ell )}+\tfrac{1}{2}\Delta d^{(\ell )}\Delta d^{(2-\ell )}\right) , \end{aligned}$$
(A.50)
and \(\widetilde{\Psi }^{(1)}\) satisfies (A.1). Applying (A.4) yields
$$\begin{aligned} \widetilde{\mu }^{(3)}(z,x,t)=\mu _3(x,t)\theta '(z)+D^{(2)}(x,t)z\theta '(z)+\widetilde{\Psi }^{(1)}(z,x,t). \end{aligned}$$
(A.51)
where \(\widetilde{\Psi }^{(1)}\) satisfies (A.1), and \(\mu _3(x,t)\) shall be determined by the \(\varepsilon ^4\)-scale.
We argue by induction on K. Assuming \(\left( A_K,B_K,C_K,D_{K-1}\right) \). We substitute (A.6) into (A.11)–(A.12) and use () and (A.18) to sort all terms of \(\varepsilon ^{K+2}\)-scale:
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(K+2)} =&-\sum \limits _{2\le i\le K+2}\left( f'''(\theta )\widetilde{\phi }^{(i)}+g_{i-1}^*\big (\widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(i-1)}\big )\right) \widetilde{\mu }^{(K+2-i)} \nonumber \\&+2\sum \limits _{0\le i\le K+1}\nabla _x\partial _z\widetilde{\mu }^{(i)}\cdot \nabla d^{(K+1-i)} +\sum \limits _{0\le i\le K+1}\partial _z\widetilde{\mu }^{(i)}\Delta d^{(K+1-i)} \nonumber \\&-\sum \limits _{1\le i\le K}\partial _z\widetilde{\phi }^{(i)}\partial _td^{(K-i)}-\partial _z\theta \partial _td^{(K)} +\Delta _x\widetilde{\mu }^{(K)}-\partial _t\widetilde{\phi }^{(K-1)} \nonumber \\&+\Big (\chi ^{(0)}d^{(K)}+\sum \limits _{1\le i\le K-1}\chi ^{(i)} d^{(K-i)}+\chi ^{(K)}d^{(0)}\Big )\eta '-\chi ^{(K-1)}z\eta ', \end{aligned}$$
(A.52)
$$\begin{aligned} \mathscr {L}\widetilde{\phi }^{(K+2)}=&-g_{K+1}^*\left( \widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(K+1)}\right) +2\sum \limits _{2\le i\le K+1}\nabla _x\partial _z\widetilde{\phi }^{(i)}\cdot \nabla d^{(K+1-i)} \nonumber \\&\quad +\theta '\Delta d^{(K+1)}+\sum \limits _{2\le i\le K+1} \partial _z\widetilde{\phi }^{(i)}\cdot \Delta d^{(K+1-i)}+\widetilde{\mu }^{(K+1)}+\Delta _x\widetilde{\phi }^{(K)}. \end{aligned}$$
(A.53)
Using \(A_K\) and \(C_K\), we can write (A.54b) as
$$\begin{aligned} {\mathscr {L}\widetilde{\phi }^{(K+2)}=\theta '\Delta d^{(K+1)}+\mu _{K+1}(x,t)\theta '+D^{(K)} z\theta '+\widetilde{\Psi }^{(K-1)}.} \end{aligned}$$
(A.54a)
To fulfill the compatibility condition (A.2), we choose \(\mu _{K+1}=-\Delta d^{(K+1)}+{\Xi ^{(K-1)}}\). This together with \(C_K\) implies \(B_{K+1}\), and reduces (A.55) to the following equation, which leads to \(A_{K+1}\):
$$\begin{aligned} {\mathscr {L}\widetilde{\phi }^{(K+2)}=D^{(K)} z\theta '+\widetilde{\Psi }^{(K-1)}.} \end{aligned}$$
(A.54b)
Proposition A.9
The equation (A.54a) can be written as
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(K+2)}={-2\sum _{\ell =0,K+1} \nabla \Delta d^{(\ell )}\cdot \nabla d^{(K+1-\ell )}\theta ''}-2\Delta d^{(0)}\Delta d^{(K+1)}\theta ''+\widetilde{\Psi }^{(K)}, \end{aligned}$$
(A.55)
and its compatibility condition is guaranteed by \(D_K\).
Proof
We consider the right hand side of (A.54a). Using (A.18), (A.49a) and (A.49b),
$$\begin{aligned}&-\sum \limits _{2\le i\le K+2}\left( f'''(\theta )\widetilde{\phi }^{(i)}+ g_{i-1}^*\big (\widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(i-1)}\big )\right) \widetilde{\mu }^{(K+2-i)} \nonumber \\&=-f'''(\theta )\widetilde{\phi }^{(2)}\widetilde{\mu } ^{(K)}-f'''(\theta )\widetilde{\phi }^{(K+2)}\widetilde{\mu }^{(0)} -g_{K+1}^*\big (\widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(K+1)}\big ) \nonumber \\&\qquad -\sum \limits _{3\le i\le K+1} \left( f'''(\theta )\widetilde{\phi }^{(i)}+g_{i-1}^*\big (\widetilde{\phi }^{(0)},\cdots , \widetilde{\phi }^{(i-1)}\big )\right) \widetilde{\mu }^{(K+2-i)} \nonumber \\&=\Delta d^{(K)}f'''(\theta )\widetilde{\phi }^{(2)}\theta ' -f'''(\theta ) \widetilde{\phi }^{(K+2)}\widetilde{\mu }^{(0)}+\widetilde{\Psi }^{(K-1)} \end{aligned}$$
In a similar way,
$$\begin{aligned}&2\sum \limits _{0\le i\le K+1}\nabla _x\partial _z\widetilde{\mu }^{(i)}\cdot \nabla d^{(K+1-i)}\nonumber \\&=2\nabla _x\partial _z\widetilde{\mu }^{(0)}\cdot \nabla d^{(K+1)}+2\nabla _x\partial _z\widetilde{\mu }^{(1)}\cdot \nabla d^{(K)}+2\nabla _x\partial _z\widetilde{\mu }^{(K)}\cdot \nabla d^{(1)} \nonumber \\&\quad +2\nabla _x\partial _z\widetilde{\mu }^{(K+1)}\cdot \nabla d^{(0)}+2\sum \limits _{2\le i\le K-1}\nabla _x\partial _z\widetilde{\mu }^{(i)}\cdot \nabla d^{(K+1-i)} \nonumber \\&{=-2\sum _{\ell =0,1,K,K+1} \nabla \Delta d^{(\ell )}\cdot \nabla d^{(K+1-\ell )}\theta ''} \nonumber \\&\quad +2\left( \nabla D^{(K)}\cdot \nabla d^{(0)}+\nabla D^{(0)}\cdot \nabla d^{(K)}\right) (z\theta ')'+\widetilde{\Psi }^{(K-1)},\\&\quad \sum \limits _{0\le i\le K+1}\partial _z\widetilde{\mu }^{(i)}\Delta d^{(K+1-i)} \nonumber \\&=\partial _z\widetilde{\mu }^{(0)}\Delta d^{(K+1)}+\partial _z\widetilde{\mu }^{(1)}\Delta d^{(K)}+\sum \limits _{2\le i\le K-1}\partial _z\widetilde{\mu }^{(i)}\Delta d^{(K+1-i)}\\&\quad +\partial _z\widetilde{\mu }^{(K)}\Delta d^{(1)}+\partial _z\widetilde{\mu }^{(K+1)}\Delta d^{(0)} \nonumber \\&=-2(\Delta d^{(0)}\Delta d^{(K+1)}+\Delta d^{(1)}\Delta d^{(K)})\theta ''+(D^{(0)}\Delta d^{(K)}\\&\quad +D^{(K)}\Delta d^{(0)})(z\theta ')'+\widetilde{\Psi }^{(K-1)}. \end{aligned}$$
Finally,
$$\begin{aligned}&-\sum \limits _{1\le i\le K}\partial _z\widetilde{\phi }^{(i)}\partial _td^{(K-i)}-\partial _z\theta \partial _td^{(K)} +\Delta _x\widetilde{\mu }^{(K)}-\partial _t\widetilde{\phi }^{(K-1)}\\&\qquad +\big (\chi ^{(0)}d^{(K)}+\sum \limits _{1\le i\le K-1}\chi ^{(i)}d^{(K-i)} +\chi ^{(K)}d^{(0)}\big )\eta '-\chi ^{(K-1)}z\eta ' \\&=-\big (\partial _td^{(K)}+\Delta ^2d^{(K)}\big )\theta '+\eta ' \big (\chi ^{(0)}d^{(K)}+\chi ^{(K)}d^{(0)}\big )+\widetilde{\Psi }^{(K-1)}. \end{aligned}$$
The above four results imply (A.57). They also imply the compatibility condition
$$\begin{aligned}&\sigma \big (\partial _td^{(K)}+\Delta ^2d^{(K)}\big )\nonumber \\&=\Delta d^{(K)}\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(2)} (\theta ')^2\mathrm{dz}-\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi } ^{(K+2)}\widetilde{\mu }^{(0)}\theta '\mathrm{dz}+\int _{\mathbb {R}}\widetilde{\Psi }^{(K-1)}\theta ' \mathrm{dz} \nonumber \\&\quad +\sigma \left( \nabla D^{(K)}\cdot \nabla d^{(0)}+\nabla D^{(0)}\cdot \nabla d^{(K)}+\tfrac{1}{2}\big (D^{(0)}\Delta d^{(K)}+D^{(K)}\Delta d^{(0)}\big )\right) . \end{aligned}$$
(A.56)
It remains to calculate the first two terms on the right hand side of (A.58). Using (A.27) yields
$$\begin{aligned} \Delta d^{(K)}\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(2)}(\theta ')^2\mathrm{dz} =\frac{\sigma }{2}D^{(0)}\Delta d^{(K)}. \end{aligned}$$
(A.57)
With the aid of (A.17) and (A.15) we have
$$\begin{aligned} -\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(K+2)}\widetilde{\mu }^{(0)}\theta '\mathrm{dz}&=\Delta d^{(0)}\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(K+2)}(\theta ')^2\mathrm{dz} \nonumber \\&=\Delta d^{(0)}\int _{\mathbb {R}}\left( \partial _z\big (\mathscr {L}\widetilde{\phi }^{(K+2)})\theta '-\big (\mathscr {L}\partial _z\widetilde{\phi }^{(K+2)}\big )\theta '\right) \mathrm{dz} \nonumber \\&=-\Delta d^{(0)}\int _{\mathbb {R}}\mathscr {L}\widetilde{\phi }^{(K+2)}\theta ''\mathrm{dz}. \end{aligned}$$
In view of (A.56), the above two formulas together lead to
$$\begin{aligned} -\int _{\mathbb {R}}f'''(\theta )\widetilde{\phi }^{(K+2)} \widetilde{\mu }^{(0)}\theta '\mathrm{dz}&=-\Delta d^{(0)}\int _{\mathbb {R}}\left( D^{(K)}z\theta '+\widetilde{\Psi }^{(K-1)}\right) \theta ''\mathrm{dz}\\&=\frac{\sigma }{2}D^{(K)}\Delta d^{(0)}+\Xi ^{(K-1)}. \end{aligned}$$
Substituting (A.59) and the above formula into (A.58) leads to \(D_K\).
Using (3.4), we can write (A.57) by
$$\begin{aligned} \mathscr {L}\widetilde{\mu }^{(K+2)}=-2D^{(K+1)}\theta ''+\widetilde{\Psi }^{(K)}(z,x,t). \end{aligned}$$
(A.58)
Applying Lemma A.1 to the above equation implies \(C_{K+1}\). To conclude \(D_K\), we need:
Corollary A.10
The following equation of \(d^{(K)}\) has a local in time classical solution
$$\begin{aligned} \mathscr {G}_K d^{(K)}=\sigma ^{-1} \overline{\sigma }\chi ^{(0)}d^{(K)}+\Xi ^{(K-1)}~\text {on}~\Gamma ^0. \end{aligned}$$
(A.59)
Moreover, if we define \(\chi ^{(K)}\) by (A.51), then \(D_K\) holds in \(\Gamma ^0(3\delta )\).
Note that the local in time solution follows from the same argument for Corollary A.7. So we have shown \((A_{K+1}, B_{K+1},C_{K+1}, D_K)\) and the induction for (A.49) is completed.
Proposition A.11
Assume (1.5) has a smooth solution \(\Gamma ^0\) within [0, T], starting from a smooth closed hypersurface \(\Gamma ^0_0\subset \mathbb {R}^N\), and let \(d^{(0)}\) be the signed-distance, defined in \(\Gamma ^0(3\delta )\). Then we can construct the inner expansion with Ansatz (A.6) so that for \(i\ge 1\)
$$\begin{aligned} D_x^\alpha D_t^\beta D_z^\gamma \widetilde{\phi }^{(i)}(z,x,t)=O(e^{-C|z|}),D_x^\alpha D_t^\beta D_z^\gamma \widetilde{\mu }^{(i)}(z,x,t)=O(e^{-C|z|}), \end{aligned}$$
(A.60)
as \(z\rightarrow \pm \infty \) for \((x,t)\in \Gamma ^0(3\delta )\) and \(0\le \alpha ,\beta ,\gamma \le 2\). Moreover,
$$\begin{aligned} \hat{\phi }^I_a(x,t)=\sum _{0\le i\le k}\varepsilon ^i\widetilde{\phi }^{(i)}(z,x,t)\big |_{z= \frac{ d^{[k]}(x,t) }{\varepsilon }},~\hat{\mu }^I_a(x,t)=\sum _{0\le i\le k}\varepsilon ^i\widetilde{\mu }^{(i)}(z,x,t)\big |_{z=\frac{d^{[k]}(x,t)}{\varepsilon }}, \end{aligned}$$
(A.61)
satisfies for \((x,t)\in \Gamma ^0(3\delta )\)
$$\begin{aligned} \varepsilon ^3\partial _t\hat{\phi }_a^I&= \varepsilon ^2\Delta \hat{\mu }_a^I-f''(\hat{\phi }_a^I)\hat{\mu }_a^I +O(\varepsilon ^{k}), \end{aligned}$$
(A.62)
$$\begin{aligned} \varepsilon \hat{\mu }_a^I&=-\varepsilon ^2\Delta \hat{\phi }_a^I+f'(\hat{\phi }_a^I) +O(\varepsilon ^{k}). \end{aligned}$$
(A.63)
Proof
It follows from (A.6) and chain-rule that
$$\begin{aligned}&-\left( \varepsilon ^3\partial _t\hat{\phi }^I_a-\varepsilon ^2\Delta \hat{\mu }^I_a+f''(\hat{\phi }^I_a)\hat{\mu }^I_a\right) \nonumber \\ =&\partial _z^2\widetilde{\mu }^\varepsilon \big |\nabla d^{[k]}\big |^2-f''(\widetilde{\phi }^\varepsilon )\widetilde{\mu }^\varepsilon +2\varepsilon \nabla \partial _z\widetilde{\mu }^\varepsilon \cdot \nabla d^{[k]}\nonumber \\&+\varepsilon \partial _z\widetilde{\mu }^\varepsilon \Delta d^{[k]}- \varepsilon ^2\partial _z\widetilde{\phi }^\varepsilon \partial _td^{[k]}+\varepsilon ^2 \Delta _x\widetilde{\mu }^\varepsilon -\varepsilon ^3 \partial _t \widetilde{\phi }^\varepsilon ,\qquad \text {with}~z=d^{[k]}(x,t)/\varepsilon . \end{aligned}$$
If we replace \(d_\varepsilon \) by \(d^{[k]}\) in (A.10), and compare it with the above formula, then we arrive at (A.64). In a similar way we can show (A.65) and the details are omitted.
