Splash Singularities for the Free Boundary Navier-Stokes Equations

Abstract

In this paper, we prove the existence of smooth initial data for the 2D free boundary incompressible Navier-Stokes equations, for which the smoothness of the interface breaks down in finite time into a splash singularity.

Appendices

A Results on the Linear System

This appendix is devoted to prove Theorem 4.1. The proof is an adaptation of [5, Theorem 4.3] to the tilde domain.

A.1 Case $g=h=v_0=0$

We would like to solve the following system:

$$\begin{aligned} v_t - Q^{2} \Delta v + A^{*} \nabla q&= f \nonumber \\ Tr(\nabla v A)&= 0 \nonumber \\ (qI - ((\nabla v A) +(\nabla v A)^{*})B_1 n&= 0 \nonumber \\ \left. v\right| _{t=0}&= 0 , \end{aligned}$$

(53)

where

$$\begin{aligned} B_1 = -JA^{-1}J=\frac{1}{Q^2}A^* \end{aligned}$$

and

$$\begin{aligned} A_{ij}=\partial _{j}P^i\circ P^{-1}. \end{aligned}$$

Our first purpose will be to obtain a weak formulation of the time independent part of this system. In order to do it we will use the following identities

$$\begin{aligned} AA^{*}&= Q^{2} I\\ \Delta v&= \text {div}\left( AA^{*}\frac{1}{Q^{2}}\nabla v\right) \end{aligned}$$

and also

$$\begin{aligned} Q^{2} \partial _{l}\left( \frac{1}{Q^{2}}A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})\right) = Q^{2} \Delta v \end{aligned}$$

Using this relation it is easy to arrive to the following identity:

$$\begin{aligned}&\int _{\Omega } \partial _{l}\left( \frac{1}{Q^{2}}A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})\right) \phi ^{i} dx\\&\quad = -\int _{\Omega }\frac{1}{Q^{2}}A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})\partial _{l}\phi ^{i} dx\\&\qquad + \int _{\partial \Omega } n^{l} A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j}) \phi ^{i}\frac{1}{Q^{2}}d\sigma \\&\quad = -\int _{\Omega }\frac{1}{Q^{2}}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})A_{lj}\partial _{l}\phi ^{i} dx\\&\qquad + \int _{\partial \Omega } n^{l} A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j}) \phi ^{i} \frac{1}{Q^2}d\sigma \\&\quad = -\frac{1}{2}\int _{\Omega }\frac{1}{Q^{2}}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})(A_{lj}\partial _{l}\phi ^{i}+A_{li}\partial _{l} \phi ^{j}) dx\\&\qquad + \int _{\partial \Omega } n^{l} A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j}) \phi ^{i}\frac{1}{Q^2} d\sigma \end{aligned}$$

for $\phi ^{i} \in C^{\infty }(\overline{\Omega })$.

We also have that

$$\begin{aligned} \int _{\Omega } A_{ki} \partial _{k} q \phi ^{i} \frac{1}{Q^{2}} dx = - \int _{\Omega } q \partial _{k} \left( A_{ki} \phi ^{i} \frac{1}{Q^{2}}\right) dx + \int _{\partial \Omega } q A_{ki} n^{k} \phi ^{i}\frac{d\sigma }{Q^2} \end{aligned}$$

The following identities hold:

$$\begin{aligned} \partial _{k}\left( A_{ki} \phi ^{i} \frac{1}{Q^{2}}\right) = A_{ki} \partial _{k} \phi ^{i} \frac{1}{Q^{2}}, \quad \partial _{k} \left( \frac{A_{ki}}{Q^2}\right) = 0 \end{aligned}$$

The energy identity for the time independent version of (53) reads:

$$\begin{aligned}&\frac{1}{2}\int _{\Omega }\frac{1}{Q^{2}}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})(A_{lj}\partial _{l}\phi ^{i}+A_{li}\partial _{l} \phi ^{j}) dx - \int _{\Omega } q \partial _{k} \left( A_{ki} \frac{1}{Q^{2}} \phi ^{i}\right) dx \\&\quad - \int _{\partial \Omega } n^{l} A_{lj}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j}) \phi ^{i} \frac{1}{Q^{2}} d\sigma + \int _{\partial \Omega } \phi ^{i} q A_{ki} n^{k} \frac{d\sigma }{Q^{2}} = \int _{\Omega } f \cdot \phi \frac{1}{Q^{2}} dx \end{aligned}$$

and therefore

$$\begin{aligned}&\frac{1}{2}\int _{\Omega }\frac{1}{Q^{2}}(A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v^{j})(A_{lj}\partial _{l}\phi ^{i}+A_{li}\partial _{l} \phi ^{j}) dx - \int _{\Omega } q \partial _{k} \left( A_{ki} \frac{1}{Q^{2}} \phi ^{i}\right) dx \\&\quad = -\int _{\partial \Omega } (q \delta ^{ij} - (A_{kj} \partial _{k} v^{i} + A_{ki} \partial _{k} v ^{j}))A_{lj} n^{l} \phi ^{i} \frac{d\sigma }{Q^{2}} + \int _{\Omega } f \cdot \phi \frac{1}{Q^{2}} dx, \end{aligned}$$

where

$$\begin{aligned} \partial _{k} \left( A_{ki} \frac{1}{Q^{2}} \phi ^{i}\right) = \frac{A_{ki}}{Q^{2}} \partial _{k} \phi ^{i} =\frac{1}{Q^2} Tr(\nabla \phi A), \end{aligned}$$

so that we finally write as follows

$$\begin{aligned}&\frac{1}{2}\int _{\Omega }Tr\left( \left( \nabla v A+A^*\nabla v^*\right) \left( \nabla \phi A +A^*\nabla \phi ^*\right) \right) \frac{1}{Q^2}dx-\int _{\Omega }q \,Tr(\nabla \phi A) \frac{1}{Q^2} dx\nonumber \\&\quad =\int _{\Omega } f\cdot \phi \frac{dx}{Q^2} -\int _{\partial \Omega } \left( q\mathbb {I}-\left( \nabla v A+ A^*\nabla v^*\right) \right) A^* n \cdot \phi \frac{1}{Q^2}d\sigma . \end{aligned}$$

(54)

Therefore (54) is the time independent weak formulation of our system.

Next we will define a kind of Leray projector. Let $H^{0}_{d}$ the subspace of $H^{0}$ formed by vectors $A^{*} \nabla \phi$ such that $\phi \in H^{1}_{0}$. Let $H^{0}_{\sigma }$ the orthogonal complement of $H^{0}_{d}$ with the following vector valued $H^{0}$ scalar product:

$$\begin{aligned} ( f,g )_{H^{0}} = \int _{\Omega } f \cdot g \frac{1}{Q^{2}} dx. \end{aligned}$$

Then it is easy to check that if $v\in H^1\cap H^{0}_\sigma$, then v must satisfy

$$\begin{aligned} Tr(\nabla v A) = 0. \end{aligned}$$

For $v\in L^2$ we define Rv to be the orthogonal projection of v onto $H^{0}_{\sigma }$ . For $v\in H^1$ we have that

$$\begin{aligned} Rv = v - A^{*} \nabla \psi , \end{aligned}$$

(55)

where

$$\begin{aligned} Q^{2} \Delta \psi = Tr(\nabla v A)\quad&\text {in }\Omega _0 \nonumber \\ \psi = 0\quad&\text {on }\partial \Omega _0, \end{aligned}$$

(56)

The next lemma will deal with some properties about this projector R. Note that we have defined R for functions of x. For functions of (x, t) we apply R for every t.

Lemma A.1

Let $0 \le s \le 4$. We have:

• i) R is a bounded operator on $H^{s}$.

• ii) R is a bounded operator on $H_{(0)}^{ht,s}$, with norm bounded uniformly if T is bounded above.

• iii) If $\phi \in H^{s+1}$, then $R(A^{*} \nabla \phi ) = A^{*} \nabla \phi _1$, with $Q^{2} \Delta \phi _1 = 0$ in $\Omega _0$, $\phi _1 = \phi$ on $\partial \Omega _0$.

Proof

• i) For $v \in H^{s}, 0 \le s \le 4$, it is easy to see that $Tr(\nabla v A) \in H^{s-1}$. Therefore the solution of the system (56) satisfies:

  $$\begin{aligned} \Vert \psi \Vert _{H^{s+1}} \le C \left\| \frac{Tr(\nabla v A)}{Q^{2}}\right\| _{H^{s-1}} \le \Vert v\Vert _{H^{s}}, \end{aligned}$$

  by elliptic theory, since both A and $Q^{2}$ are regular. The identity (55) provides

  $$\begin{aligned} \Vert R v \Vert _{H^{s}} \le \Vert v\Vert _{H^{s}} + \Vert \nabla \psi \Vert _{H^{s}} \le C \Vert v\Vert _{H^{s}}. \end{aligned}$$

• ii) It is easy to check that $\partial _{t}$ commutes with R, since

  $$\begin{aligned} \partial _{t}^{k} Rv = \partial _{t}^{k} v - A^{*} \nabla \partial _{t}^{k} \psi = R \partial _{t}^{k} v. \end{aligned}$$

  This proves the result for an integer number of derivatives. By interpolation we get the result for fractional derivatives.

• iii) For $\phi \in H^{s+1}$, if $v = A^{*} \nabla \phi$, then

  $$\begin{aligned} R(A^{*} \nabla \phi ) = A^{*} \nabla \phi - A^{*} \nabla \psi = A^{*} \nabla (\phi - \psi ). \end{aligned}$$

  Thus, we have that

  $$\begin{aligned} 0 = Tr(\nabla (R(A^{*} \nabla \phi )) A ) = Q^{2} \Delta (\phi - \psi ), \quad \left. \psi \right| _{\partial \Omega _{0}} = 0, \end{aligned}$$

  which implies that $\left. \phi - \psi \right| _{\partial \Omega _{0}} = \phi$ and we can take $\phi _1 = \phi - \psi$.

$\square$

Once we have obtained the energy identity (54) and Lemma A.1 we pass to announce the main theorem of this section:

Theorem A.2

Let $f \in H_{(0)}^{ht,s-1}$, let v, q solve (53) and $2<s\le 3$. Then

$$\begin{aligned} \Vert v\Vert _{H_{(0)}^{ht,s+1}} + \Vert \nabla q\Vert _{H_{(0)}^{ht,s-1}} + |q|_{H_{(0)}^{ht,s-\frac{1}{2}}} \le C \Vert f\Vert _{H_{(0)}^{ht,s-1}}. \end{aligned}$$

The constant is independent of T.

The rest of this subsection is devoted to prove this theorem.

First of all we will modify the equation by considering the new variables

$$\begin{aligned} u = e^{-t} v; \quad p = e^{-t} q; \quad \overline{f} = e^{-t} f \end{aligned}$$

We should remark that $f \in H_{(0)}^{ht,s-1} \Leftrightarrow \overline{f} \in H_{(0)}^{ht,s-1}$. Then, the equation reads

$$\begin{aligned} \partial _t u&= -e^{-t} v + e^{-t} v_t = -u + e^{-t}(Q^{2} \Delta v - A^{*} \nabla q + f) \\&= -u + Q^{2} \Delta u - A^{*} \nabla p + \overline{f}. \end{aligned}$$

We will solve therefore

$$\begin{aligned} \partial _t u +u - Q^{2} \Delta u + A^{*} \nabla p = \overline{f}. \end{aligned}$$

Let’s start projecting onto $H^0_\sigma$ to obtain

$$\begin{aligned} \partial _t u +u-Q^2 \Delta u +A^*\nabla q_1 = R\overline{f}, \end{aligned}$$

since $Tr(\nabla (Q^{2} \Delta \tilde{v}) A) = 0$ and therefore $R(Q^{2} \Delta u) = Q^{2} \Delta u$, where $A^*\nabla q_1=RA^*\nabla p$.

We now introduce the operator

$$\begin{aligned} S_A: V^{s+1}(\Omega ) \rightarrow RH^{s-1} \end{aligned}$$

defined via:

$$\begin{aligned} S_Au = -Q^{2} \Delta u + u + A^{*} \nabla q_1, \quad A^{*} \nabla q_1 \equiv RA^{*} \nabla p, \end{aligned}$$

where $V^{r}(\Omega ) = \{u \in RH^{r}, A^*t_0\left( \nabla u A+A^*\nabla u^*\right) A^* n_0=0 \text { on } \partial \Omega \}$ and $RH^{s-1} = \{Rf, f \in H^{s-1}\}$.

The following lemma deals with the invertibility of $S_A$.

Lemma A.3

$S_A$ has a bounded inverse for $1\le s \le 3$, and

$$\begin{aligned} \Vert S_{A}^{-1} f\Vert _{H^{s+1}} \le C \Vert f\Vert _{H^{s-1}}. \end{aligned}$$

Proof

Let $f \in RH^{s-1}$. We will show that there exists $u \in V^{s+1}$ such that $S_Au = f$, i.e. $u - Q^{2} \Delta u + A^{*} \nabla q_1 = Rf$ (we will keep in the notation Rf instead of f, although $Rf=f$, to keep in mind this fact). Using the energy identity (54), we observe that

$$\begin{aligned} (u,\phi ) + \langle u,\phi \rangle = - \int _{\partial \Omega }h\cdot \phi \frac{d\sigma }{Q^{2}} + \int _{\Omega } R\overline{f} \cdot \phi \frac{dx}{Q^{2}} + \int _{\Omega } q_1 Tr(\nabla \phi A) \frac{d\tilde{x}}{Q^{2}} \end{aligned}$$

is the weak formulation of

$$\begin{aligned} u - Q^{2} \Delta u + A^{*} \nabla q_1&= R \overline{f} \\ \left( q_1-\left( \nabla u A+A^*\nabla u^*\right) \right) A^* n_0&= h, \end{aligned}$$

where

$$\begin{aligned} (u,\phi )&= \int u \phi \frac{dx}{Q^{2}} \end{aligned}$$

(57)

$$\begin{aligned} \langle u, \phi \rangle&= \int _{\Omega } Tr((\nabla u A + A^{*} \nabla u^{*})(\nabla \phi A + A^{*} \nabla \phi ^{*})) \frac{dx}{Q^{2}} \end{aligned}$$

(58)

If we look for $h = 0$ we then have to solve

$$\begin{aligned} (u,\phi ) + \langle u, \phi \rangle = \int _{\Omega } R \overline{f} \cdot \phi \frac{dx}{Q^{2}} + \int _{\Omega } q_1 Tr(\nabla \phi A) \frac{d\tilde{x}}{Q^{2}}. \end{aligned}$$

(59)

We will find a solution to this equation in $RH^1$. By this we mean that there exists $u\in RH^1$ such that (59) holds for all $\phi \in RH^1$. We notice that the last term vanishes since $\phi \in RH^{1}$, and henceforth, in $RH^1$, the equation (59) is equivalent to

$$\begin{aligned} (u,\phi ) + \langle u, \phi \rangle = \int _{\Omega } R \overline{f} \phi \frac{dx}{Q^{2}}. \end{aligned}$$

(60)

for all $\phi \in RH^1$.

Via [29, Corollary 4.7],

$$\begin{aligned} (u,u) + \langle u, u \rangle \ge C ||u||_{H^1}^2 \end{aligned}$$

and therefore it is a coercive bilinear form. That implies Lax-Milgram’s Theorem can be applied in $RH^{1}$ to obtain a solution $u \in RH^{1}$ of (60).

The next step is to show that there exists $p \in L^{2}$ such that

$$\begin{aligned} (u,\phi ) + \langle u, \phi \rangle = \int _{\Omega } Rf \phi + \int _{\Omega } p\, Tr(\nabla \phi A) \quad \forall \phi \in H^{1}. \end{aligned}$$

To do this, we decompose $\phi = R\phi + A^{*} \nabla \pi _{\phi }$, and then

$$\begin{aligned} (u,\phi ) + \langle u, \phi \rangle&= (u,R\phi ) + \langle u, R\phi \rangle + \langle u, A^{*} \nabla \pi _{\phi } \rangle + \underbrace{(u, A^{*} \nabla \pi _{\phi })}_{0} \\&= (Rf, R\phi ) + \langle u, A^{*} \nabla \pi _{\phi } \rangle = (Rf,\phi ) + \langle u, A^{*} \nabla \pi _{\phi } \rangle . \end{aligned}$$

Therefore we have to show that there exists $p\in L^2$ such that

$$\begin{aligned} (p,Tr(\nabla \phi A)) = \langle u, A^{*} \nabla \pi _{\phi } \rangle \quad \forall \phi \in H^{1}. \end{aligned}$$

Let us assume that u is smooth and suppose we take p satisfying:

$$\begin{aligned} Q^{2} \Delta p&= 0 \nonumber \\ \left. p \right| _{\partial \Omega }&= (A^{*}n, ((\nabla u A) + (\nabla u A)^{*}) A^{*}n) \end{aligned}$$

(61)

$$\begin{aligned}&= Q^{2}(n (A \nabla u + \nabla u^{*}A^* ) n). \end{aligned}$$

(62)

Then, on the one hand

$$\begin{aligned} \int p Tr(\nabla \phi A) \frac{dx}{Q^{2}}&= \int p \Delta \pi _{\phi } dx = \int p \text {div}\left( \frac{AA^{*}}{Q^{2}}\nabla \pi _{\phi }\right) dx \nonumber \\&= -\int _{\Omega } \nabla p \cdot AA^{*} \nabla \pi _{\phi } \frac{dx}{Q^{2}} + \int _{\partial \Omega } p \cdot n AA^{*} \nabla \pi _{\phi } \frac{dx}{Q^{2}} \nonumber \\&= - \underbrace{\int _{\Omega } \nabla p Q^{2} \nabla \pi _{\phi } \frac{dx}{Q^{2}}}_{0} + \int _{\partial \Omega } p (A^{*}n) \cdot A^{*} \nabla \pi _{\phi } \frac{dx}{Q^{2}}. \end{aligned}$$

(63)

On the other hand

$$\begin{aligned} \langle u, A^{*} \nabla \pi _{\phi }\rangle = \frac{1}{2} \int Tr((\nabla u A + A^{*} \nabla u^{*})(\nabla (A^{*} \nabla \pi _{\phi }) A + A^{*} \nabla (A^{*} \nabla \pi _{\phi }))^{*}) \frac{dx}{Q^{2}}, \end{aligned}$$

where

$$\begin{aligned} (\nabla (A^{*} \nabla \pi _{\phi })A)_{ij}&= \partial _{l}(A^{*} \nabla \pi _{\phi })^{i} A_{lj} = \partial _{l}(A_{ki} \partial _{k} \pi _{\phi }) A_{lj} \\&= A_{lj} \partial _{l} A_{ki} \partial _{k} \pi _\phi + A_{ki} A_{lj} \partial _{lk}^{2} \pi _\phi \\ \langle u, A^{*} \nabla \pi _{\phi }\rangle&= \int (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li}) A_{mj} \partial _{m}(\partial _{k} \pi _\phi A_{ki}) \frac{dx}{Q^{2}} \\&= -\int \partial _{m}\left( \left( A_{mj} \partial _{l} u^{i} A_{lj} + A_{mj} \partial _{l} u^{j} A_{li}\right) \frac{1}{Q^{2}}\right) \partial _{k} \pi _\phi A_{ki} dx \\&\quad +\int _{\partial \Omega } (n_{0})_m\left( A_{mj} \partial _{l} u^{i} A_{lj} + A_{mj} \partial _{l} u^{j} A_{li}\right) \partial _{k} \pi _\phi A_{ki} \frac{dx}{Q^{2}} \\&= -\int \partial _{m}\left( \left( \delta ^{ml} \partial _{l} u^{i} + A_{mj} \partial _{l} u^{j} A_{li}\right) \frac{1}{Q^{2}}\right) \partial _{k} \pi _\phi A_{ki} dx \\&\quad +\int _{\partial \Omega } A^{*}n_0 \cdot (\nabla u A + A^{*} \nabla u) A^{*} \nabla \pi _\phi \frac{dx}{Q^{2}}. \end{aligned}$$

The first term is zero because of the orthogonality and because of the condition $Tr(\nabla u A)=0$. We do the calculations for the second term:

$$\begin{aligned} \int _{\partial \Omega } (A^{*} n_0)\cdot (\nabla u A + A^{*} \nabla u^{*}) A^{*} \nabla \pi _\phi \frac{d\sigma }{Q^{2}}&= \int n_0(A \nabla u + \nabla u^{*} A^{*}) \nabla \pi _\phi \\&= \int n_0(A \nabla u + \nabla u^{*} A^{*}) n_0 \partial _{n} \pi _\phi , \end{aligned}$$

since $t_0\cdot \nabla \pi _\phi = 0$. Comparing with (62) we have that

$$\begin{aligned} (p,Tr(\nabla \phi A)) = \langle u, A^{*} \nabla \pi _\phi \rangle , \quad \forall \phi \in H^{1}. \end{aligned}$$

for smooth u. Let $u \in H^{1}$ and $\{u_m\}_{m=1}^{\infty }$ such that $u_m \in H^{\infty }$, and $tr(\nabla u_m A) = 0$ for every m, and $u_m\rightarrow u$ strongly in $H^1$ (for example we extend u by zero to ${\mathbb {R}}^2$, we make the convolution with a mollifier $\rho _{\frac{1}{m}}$ and finally we project onto $H_\sigma$). Let $p_m$ be given by

$$\begin{aligned} \Delta p_m&= 0\\ \left. p_m\right| _{\partial \Omega }&=A^{-1} n_0 (\nabla u_{m} A + A^{*} \nabla (u_{m})^{*})A^{-1} n_0. \end{aligned}$$

Then, we have that

$$\begin{aligned} (p_m, Tr(\nabla \phi A)) = \langle u_m, A^{*} \nabla \pi _\phi \rangle \quad \forall \phi \in H^{1}. \end{aligned}$$

In particular, we take $\phi _{m}$ such that $Tr(\nabla \phi _{m}A) = p_m$. This implies that $Q^{2} \Delta \pi _{\phi _{m}} = p_{m}, \left. \pi _{\phi _m}\right| _{\partial \Omega } = 0$. Showing the existence of such $\phi _{m}$ is trivial since one can choose $\phi _{m} = A^{*} \nabla \psi$, with $Q^{2} \Delta \psi = p_m$. Then, we can bound the $L^{2}$ norm of $p_m$ in the following way:

$$\begin{aligned} \Vert p_m\Vert _{L^{2}}^{2} \le C \langle u_{m}, A^{*} \nabla \pi _{\phi _m}\rangle \le C \Vert u_{m}\Vert _{H^{1}} \Vert \pi _{\phi _m}\Vert _{H^{2}} \le C \Vert u\Vert _{H^{1}} \Vert p_m\Vert _{L^{2}}, \end{aligned}$$

which shows that $p_m$ is bounded in $L^{2}$. Therefore there exists a subsequence $p_{m_i}$ which converges weakly to a function p in $L^{2}$ and

$$\begin{aligned} (p, Tr(\nabla \phi A)) = \langle u, \nabla \pi _\phi \rangle \quad \forall \phi \in H^{1}. \end{aligned}$$

We have shown that there exist $(u,p) \in H^{1} \times L^{2}$ such that

$$\begin{aligned} (u,\phi ) + \langle u,\phi \rangle = (Rf,\phi ) + \int p Tr(\nabla \phi A) \frac{dx}{Q^{2}} \quad \forall \phi \in H^{1}. \end{aligned}$$

Indeed, $u \in H^{2}, p \in H^{1}$. We now show that improvement on the regularity.

For every $\Omega ^{\flat } \Subset \Omega$, it is easy to obtain the interior regularity estimate $\Vert u\Vert _{H^{2}(\Omega ^{\flat })} \le C, \Vert p\Vert _{H^{1}(\Omega ^{\flat })} \le C$. We focus here on the boundary estimates. We perform the following change of coordinates in $\Omega '$, where $\Omega '$ is a tubular neighborhood of $\partial \Omega$:

$$\begin{aligned} x(s,\lambda ) = z(s) + \lambda z_s^{\perp }(s) \end{aligned}$$

We would like to check that it is indeed a diffeomorphism. We have following:

$$\begin{aligned} |z_s|^{2}&= 1 \\ x_s(s)&= z_{s}(s) + \lambda z_{ss}^{\perp }(s) \\ z_{ss}^{\perp }&= \langle z_{ss}^{\perp }, z_s(s) \rangle z_s(s) = -\kappa (s) z_s(s) \\ x_s(s)&= (1 - \lambda \kappa (s))z_s(s) \\ x_{\lambda }&= z_s^{\perp }(s) \end{aligned}$$

Computing more,

$$\begin{aligned} \text {det} \left( \begin{array}{cc} x^{1}_{s} &{} x^{2}_{s} \\ x^{1}_{\lambda } &{} x^{2}_{\lambda } \end{array} \right) = x_{s}^{1} x_{\lambda }^{2} - x_{s}^{2} x_{\lambda }^{2} = x_{s}^{\perp } x_{\lambda } = 1 - \lambda \kappa (s) \end{aligned}$$

This fixes the width of the tubular neighborhood to be $\lambda _0 < \sup _{s} \frac{1}{\kappa (s)}$. Under these assumptions $x(s,\lambda )$ is a diffeomorphism. Fix $x_{0} \in \partial \Omega$, and consider the following cutoff function $\psi$, defined by

$$\begin{aligned} \psi (x) = \left\{ \begin{array}{cc} 1 &{} B\left( x_0,\frac{\lambda _0}{2}\right) \cap \Omega ' \\ 0 &{} \left( B\left( x_0,\frac{3\lambda _0}{4}\right) \cap \Omega '\right) ^{c} \end{array} \right\} , \quad 0 \le \psi (x) \le 1, \quad \psi (x) \in C^{\infty }. \end{aligned}$$

We extend $\psi$ to $\Omega$ by zero. We define the set C as $C = x^{-1}(B(x_0,\lambda _0) \cap \Omega ')$, and the set $C/2 = x^{-1}(B(x_0,\frac{\lambda _0}{2}) \cap \Omega ')$.

The energy identity can be written down as:

$$\begin{aligned}&\frac{1}{2} \int _{\Omega } (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li}) (\partial _{k} \phi ^{i} A_{kj} + \partial _{k} \phi ^{j} A_{ki}) \frac{dx}{Q^{2}} + \int _{\Omega } u^{i} \phi ^{i} \frac{dx}{Q^{2}} \\&\quad = \int _{\Omega } f^{i} \phi ^{i} \frac{dx}{Q^{2}} + \int _{\Omega } p \partial _{l} \phi ^{i} A_{li} \frac{dx}{Q^{2}} \end{aligned}$$

We choose as test function $\phi = \varphi \psi$. Then:

$$\begin{aligned}&\int _{\Omega } (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li}) (\partial _{k} (\varphi ^{i} \psi ) A_{kj} + \partial _{k} (\varphi ^{j} \psi )A_{ki}) \frac{dx}{Q^{2}} + \int _{\Omega } u^{i} \phi ^{i} \frac{dx}{Q^{2}} \\&\quad = \int _{\Omega } f^{i} \phi ^{i} \frac{dx}{Q^{2}} + \int _{\Omega } p \partial _{l} \phi ^{i} A_{li} \frac{dx}{Q^{2}} \\&M_1 + M_2 = M_3 + M_4 \end{aligned}$$

We start developing each of the terms one by one

$$\begin{aligned} M_1&= \int _{\Omega '} (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li}) (\partial _{k} (\varphi ^{i}) A_{kj} + \partial _{k} (\varphi ^{j})A_{ki}) \psi \frac{dx}{Q^{2}} \\&\quad + \int _{\Omega '} (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li}) (\varphi ^{i} \partial _{k} \psi A_{kj} + \varphi ^{j} \partial _{k} \psi A_{ki}) \frac{dx}{Q^{2}} \\&= \int _{\Omega '} (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li}) \psi (\partial _{k} (\varphi ^{i}) A_{kj} + \partial _{k} (\varphi ^{j})A_{ki}) \frac{dx}{Q^{2}} \\&\quad + \int _{\Omega '} (\partial _{l} u^{i} A_{lj} + \partial _{l} u^{j} A_{li})\partial _{k} \psi (\varphi ^{i} A_{kj} + \varphi ^{j} A_{ki}) \frac{dx}{Q^{2}} \end{aligned}$$

We now do the change of variables

$$\begin{aligned} \underline{dx} = \left( \frac{dx}{dsd\lambda }\right) ds d\lambda = \underbrace{(1+\kappa (s) \lambda )}_{J(s,\lambda )} ds d\lambda , \end{aligned}$$

and define

$$\begin{aligned}&\overline{A}(s,\lambda ) = A \circ x(s,\lambda ),\quad \overline{Q}(s,\lambda ) = Q \circ x(s,\lambda ),\quad \overline{u}(s,\lambda ) = u \circ x(s,\lambda ),\\&\quad \overline{p}(s,\lambda )=p\circ x(s,\lambda ),\\&\overline{\varphi }(s,\lambda ) = \varphi \circ x(s,\lambda ),\quad \overline{\psi }(s,\lambda ) = \psi \circ x(s,\lambda ),\quad \overline{f}(s,\lambda ) = f\circ x(s,\lambda ). \end{aligned}$$

We can compute the derivatives in the bar-coordinates

$$\begin{aligned} u(x)&= \overline{u} \circ s(x), \quad \partial _{l} u^{i} = \partial _{k} \overline{u}^{i} \circ s(x) \partial _{l} s^{k} \\ \partial _{l} u^{i} \circ x(s,\lambda )&= \partial _{k} \overline{u}^{i}(s,\lambda ) \partial _{l} s^{k} \circ x(s,\lambda ) \end{aligned}$$

Setting $x = x(s,\lambda ), \quad x^{j} = x^{j}(s,\lambda )$, we obtain:

$$\begin{aligned} \partial _{l} x^{j}&= \partial _{l} x^{j}(s,\lambda ) \frac{\partial s^{l}}{\partial x^{i}} \\ \delta ^{ij}&= \partial _{l} x^{j}(s,\lambda ) \frac{\partial s^{l}}{\partial x^{i}} \\ \delta ^{ij}&= \partial _{l} x^{j} \circ s(x) \frac{\partial s^{l}}{\partial x^{i}}(x) \\ \delta ^{ij}&= (\nabla x)_{jl} (\nabla s)_{li} \circ x(s,\lambda ) = (\nabla x \nabla s)_{ji}\\ I&= \nabla x \nabla s \circ x(s,\lambda ), \quad \nabla s \circ x(s,\lambda ) = (\nabla x)^{-1} \equiv \Sigma \end{aligned}$$

Plugging this result into the equation for the derivatives of u, we get

$$\begin{aligned} \partial _{l} u^{i} \circ x(s,\lambda )&= \Sigma _{kl} \partial _{k} \overline{u}^{i} \\ \Sigma&= \frac{1}{1 - \kappa (s)\lambda } \left( \begin{array}{cc} x^{2}_{\lambda } &{} -x^{1}_{\lambda } \\ -x^{2}_{s} &{} x^{1}_{s} \end{array} \right) \end{aligned}$$

Plugging this expression into $M_1$, and letting $\Sigma \overline{A} = B$, we obtain

$$\begin{aligned} 2 M_1&= \int _{C} (\partial _{l} u^{i} \circ x(s,\lambda ) \overline{A}_{lj} + \partial _{l} u^{j} \circ x(s,\lambda ) \overline{A}_{li}) \\&\qquad \left( \psi (\partial _{k} \overline{\varphi }^{i} \circ x(s,\lambda ) \overline{A}_{kj} + \partial _{k} \overline{\varphi }^{j} \circ x(s,\lambda ) \overline{A}_{ki}\right) \frac{J}{\overline{Q}^{2}} ds d\lambda \\&\quad + \int _{C} \partial _{a} \overline{\psi } (\partial _{m} \overline{u}^{i} B_{mj} + \partial _{m} \overline{u}^{j} B_{mi}) (\overline{\varphi }^{i} B_{aj} + \overline{\varphi }^{j} B_{ai}) \frac{J}{\overline{Q}^{2}}dsd\lambda \\&= \int _{C} (\partial _{m} \overline{u}^{i} \Sigma _{ml} \overline{A}_{lj} + \partial _{m} \overline{u}^{j} \Sigma _{ml} \overline{A}_{li}) \overline{\psi } (\partial _{a} \overline{\varphi }^{i} \Sigma _{ak} \overline{A}_{kj} + \partial _{a} \overline{\varphi }^{j} \Sigma _{ak} \overline{A}_{ki}) \frac{J}{\overline{Q}^{2}} ds d\lambda \\&\quad + \int _{C} \partial _{a} \overline{\psi } (\partial _{m} \overline{u}^{i} B_{mj} + \partial _{m} \overline{u}^{j} B_{mi}) (\overline{\varphi }^{i} B_{aj} + \overline{\varphi }^{j} B_{ai}) \frac{J}{\overline{Q}^{2}}dsd\lambda \\&= \int _{C} \overline{\psi } (\partial _{m} \overline{u}^{i} B_{mj} + \partial _{m} \overline{u}^{j} B_{mi}) (\partial _{a} \overline{\varphi }^{i} B_{aj} + \partial _{a} \overline{\varphi }^{j} B_{ai}) \frac{J}{\overline{Q}^{2}} ds d\lambda \\&\quad + \int _{C} \partial _{a} \overline{\psi } (\partial _{m} \overline{u}^{i} B_{mj} + \partial _{m} \overline{u}^{j} B_{mi}) (\overline{\varphi }^{i} B_{aj} + \overline{\varphi }^{j} B_{ai}) \frac{J}{\overline{Q}^{2}}dsd\lambda \end{aligned}$$

Moreover,

$$\begin{aligned} M_2&= \int _{\Omega '} u^{i} \varphi ^{i} \psi \frac{dx}{Q^{2}} = \int _{C} \overline{u}^{i} \overline{\varphi }^{i} \overline{\psi } \frac{J}{\overline{Q}^{2}} ds d\lambda \\ M_3&= \int _{C} \overline{f}^{i} \overline{\varphi }^{i} \overline{\psi } \frac{J}{\overline{Q}^{2}} ds d\lambda \\ M_4&= \int _{C} \overline{p} \partial _{a}(\overline{\varphi }^{i} \overline{\psi }) B_{ai} \frac{J}{\overline{Q}^{2}} ds d\lambda , \end{aligned}$$

with

$$\begin{aligned} M_1 + M_2 = M_3 + M_4 \Rightarrow M_1 = -M_2 + M_3 + M_4. \end{aligned}$$

We start computing $M_1$. We have that

$$\begin{aligned} 2M_1&= \int _{C} (\partial _{l} \overline{u}^{i} B_{lj} + \partial _{l} \overline{u}_{j} B_{li})(\partial _{k} \overline{\varphi }^{i} B_{kj} + \partial _{k} \overline{\varphi }^{j} B_{ki}) \overline{\psi } \frac{J}{\overline{Q}^{2}} \\&\quad + \int _{C} (\partial _{l} \overline{u}^{i} B_{lj} + \partial _{l} \overline{u}_{j} B_{li})( \overline{\varphi }^{i} B_{kj} + \overline{\varphi }^{j} B_{ki}) \partial _{k} \overline{\psi } \frac{J}{\overline{Q}^{2}} \\&= m + l_1. \end{aligned}$$

We will use the convention that the m terms will denote bad terms from now on. We further split m into four terms:

$$\begin{aligned} m&= \int _{C} \partial _{l} \overline{u}^{i} B_{lj} \partial _k \overline{\varphi }^{i}B_{kj} \overline{\psi }\frac{J}{\overline{Q}^{2}} + \int _{C} \partial _{l} \overline{u}^{j} B_{li} \partial _k \overline{\varphi }^{i}B_{kj} \overline{\psi }\frac{J}{\overline{Q}^{2}} \\&\quad + \int _{C} \partial _{l} \overline{u}^{i} B_{lj} \partial _k \overline{\varphi }^{j}B_{ki} \overline{\psi }\frac{J}{\overline{Q}^{2}} + \int _{C} \partial _{l} \overline{u}^{j} B_{li} \partial _k \overline{\varphi }^{j}B_{ki} \overline{\psi }\frac{J}{\overline{Q}^{2}} \\&= m_{1} + m_{2} + m_{3} + m_{4}. \end{aligned}$$

If we are able to estimate any one of them we can estimate all of them.

We will use the following formulas related to finite differences:

$$\begin{aligned} D_{s}^{h}(fg)&= \frac{f(s+h)g(s+h) - f(s)g(s)}{h} = \frac{(f(s+h)-f(s))g(s+h)}{h} \\&\quad + f(s)\frac{g(s+h)-g(s)}{h} \\&= g(s+h)D_{s}^{h}f + f(s)D_{s}^{h}g, \\ D_{s}^{-h}D_{s}^{h}(fg)&= D_{s}^{-h}(D_{s}^{h}f g(s)) + D_{s}^{-h}(f(s)D_{s}^{h}g) \\&= (D_{s}^{-h}D_{s}^{h}f) g(s) + D_{s}^{h}fD_{s}^{-h}(g(s+h)) + D_{s}^{-h}f D_{s}^{h}g(s-h) \\&\quad + f(s) D_{s}^{-h} D_{s}^{h} g, \\ D_{s}^{-h}(g(s+h))&= \frac{g(s)-g(s+h)}{-h} = D_{s}^{h} g, \\ D_{s}^{h} (g(s-h))&= \frac{g(s)-g(s-h)}{h} = D_{s}^{-h} g, \\ \Rightarrow D_{s}^{-h} D_{s}^{h} (fg)&= (D_{s}^{-h} D_{s}^{h} f) g(s) + D_{s}^{h} f D_{s}^{h}g + D_{s}^{-h} f D_{s}^{-h} g + f(s) D_{s}^{-h} D_{s}^{h} g. \end{aligned}$$

We take $\overline{\varphi } = D_{s}^{-h} D_{s}^{h} \overline{u}$. Since divergence free condition, we have that

$$\begin{aligned} D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{l} \overline{u}^{j})&= 0 = (D_{s}^{-h} D_{s}^{h} B_{lj}) \partial _{l} \overline{u}^{j}(s) + D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l} \overline{u}^{j} + D_{s}^{-h} B_{lj} D_{s}^{-h} \partial _{l} \overline{u}^{j} \\&\quad + B_{lj} \partial _{l} D_{s}^{-h} D_{s}^{h} \overline{u}^{j}. \end{aligned}$$

Therefore

$$\begin{aligned} B_{lj} \partial _{l} D_{s}^{-h} D_{s}^{h} \overline{u}^{j} =&- (D_{s}^{-h} D_{s}^{h} B_{lj}) \partial _{l} \overline{u}^{j}(s) - D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l} \overline{u}^{j} - D_{s}^{-h} B_{lj} D_{s}^{-h} \partial _{l} \overline{u}^{j} \\ B_{lj} \partial _{l} D_{s}^{-h} D_{s}^{h}(\overline{u}^{j} \overline{\psi }) =&D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{l}(\overline{u}^{j} \overline{\psi })) - D_{s}^{-h} D_{s}^{h} B_{lj} \partial _{l}(\overline{u}^{j}(s)\overline{\psi }(s)) \\&- D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l} (\overline{u}^{j} \overline{\psi }) - D_{s}^{-h} B_{lj} D_{s}^{-h}(\partial _{l}(\overline{u}^{j} \overline{\psi })), \end{aligned}$$

and

$$\begin{aligned} B_{lj} \partial _{l}(\overline{\psi } \overline{u}^{j}) = B_{lj} \partial _{l} \overline{\psi } \overline{u}^{j}. \end{aligned}$$

Expanding the calculations, we obtain

$$\begin{aligned} B_{lj} \partial _{l} D_{s}^{-h} D_{s}^{h}(\overline{u}^{j} \overline{\psi }) =&D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{j} \overline{\psi } \overline{u}^{j}) - D_{s}^{-h} D_{s}^{h} B_{lj} \partial _{l}(\overline{u}^{j}(s) \overline{\psi }(s)) \\&- D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l}(\overline{u}^{j} \overline{\psi }) - D_{s}^{-h} B_{lj} D_{s}^{-h}(\partial _{l} \overline{u}^{j} \overline{\psi })) \\ \overline{\psi } B_{lj} \partial _{l} D_{s}^{-h} D_{s}^{h} (\overline{u}^{j} \overline{\psi }) =&\overline{\psi } D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{l} \overline{\psi } \overline{u}^{j}) - \overline{\psi } D_{s}^{-h} D_{s}^{h} B_{lj} \partial _{l} (\overline{u}^{j} \overline{\psi }) \\&- \overline{\psi } D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l}(\overline{u}^{j} \overline{\psi }) - \overline{\psi } D_{s}^{-h} B_{lj} D_{s}^{-h}(\partial _{l}(\overline{u}^{j} \overline{\psi })). \end{aligned}$$

Finally, we get to

$$\begin{aligned}&\overline{\psi } D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{l} \overline{\psi } \overline{u}^{j}) = D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{l} \overline{\psi } \overline{u}^{j}\overline{\psi }) \\&\quad - D_{s}^{-h}(B_{lj}\partial _l \overline{\psi }\overline{u}^j)D_{s}^{-h}\overline{\psi }-D_{s}^{h}(B_{lj}\partial _l \overline{\psi }\overline{u}^j)D_{s}^{h}\overline{\psi }-B_{lj}\partial _l\overline{\psi }\overline{u}^jD_s^{-h}D_s^{h}\overline{\psi }. \end{aligned}$$

Therefore

$$\begin{aligned} \overline{\psi } B_{lj} \partial _{l} D_{s}^{-h} D_{s}^{h}(\overline{u}^{j} \overline{\psi })&= D_{s}^{-h} D_{s}^{h}(B_{lj} \partial _{l} \overline{\psi } \overline{u}^{j}\overline{\psi }) - \overline{\psi } D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l}(\overline{u}^{j} \overline{\psi }) \\&\quad - \overline{\psi } D_{s}^{h} B_{lj} D_{s}^{h} \partial _{l}(\overline{u}^{j} \overline{\psi }) + \text {LOW} \\&= B_{lj} \partial _{l} \overline{\psi } D_{s}^{-h} D_{s}^{h} (\overline{u}^{j} \overline{\psi }) - \overline{\psi } D_{s}^{h} B_{lj} D_{s}^{-h} \partial _{l}(\overline{u}^{j} \overline{\psi }) \\&\quad - \overline{\psi } D_{s}^{-h} B_{lj} D_{s}^{h} \partial _{l}(\overline{u}^{j} \overline{\psi }) + \text {LOW}, \end{aligned}$$

where we say that a term T is LOW is $\Vert T\Vert _{L^{2}} \le C\Vert \overline{u}\Vert _{H^{1}}$. We also say that a term T is SAFE if, for any $\delta > 0$,

$$\begin{aligned} \Vert T\Vert _{L^{2}} \le C_{\delta } + \delta (\Vert \nabla D_{s}^{-h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2} + \Vert \nabla D_{s}^{h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2}), \end{aligned}$$

where $C_{\delta }$ may depend on $\Vert \overline{u}\Vert _{H^{1}}, \Vert \overline{p}\Vert _{L^{2}}, \Vert \overline{f}\Vert _{L^{2}}$ and $\overline{\psi }, B, \overline{Q}$ or J. We now look at what the terms $M_1, \ldots , M_4$ look like. We have that

$$\begin{aligned} M_2&= \int _{C} \overline{\psi } \overline{u}^{i} D_{s}^{-h} D_{s}^{h} \overline{u}^{i} \frac{J}{\overline{Q}^{2}} = \int _{C} D_{s}^{h}(\overline{\psi } \overline{u}^{i}) D_{s}^{h} \overline{u}^{i} \frac{J}{Q^{2}} + \text {LOW}, \\ M_3&= \int _{C} \overline{f}^{i} D_{s}^{-h} D_{s}^{h} \overline{u}^{i} \frac{J}{\overline{Q}^{2}} \le C_{\delta }\Vert \overline{f}^{i}\Vert _{L^{2}}^{2} + \delta \Vert D_{s}^{-h}D_{s}^{h} (\overline{u} \overline{\psi })\Vert _{L^{2}}^{2}+\text {LOW}, \\ M_4&= \int _{C} \overline{p} B_{lj} \partial _{j} \overline{\psi } D_{s}^{-h} D_{s}^{h} \overline{u}^{i} \frac{J}{Q^{2}}\le C_{\delta }\Vert \overline{p}\Vert _{L^{2}}^{2} + \delta \Vert D_{s}^{-h} D_{s}^{h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2}, \\ l_1&= \int _{C} \partial _{l} \overline{u}^{i} B_{lj} \phi ^{j} \partial _{k} \overline{\psi } B_{kj} \frac{J}{\overline{Q}^{2}} \le C_{\delta } \Vert \overline{u}\Vert _{H^{1}}^{2} + \delta \Vert D_{s}^{-h} D_{s}^{h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2}, \\ M_1&= \int _{C} \partial _{l} \overline{u}^{i} B_{lj} B_{kj} \partial _{k} D_{s}^{-h} D_{s}^{h}(\overline{u}^{i} \overline{\psi }) \overline{\psi } \frac{J}{\overline{Q}^{2}}\\&= \int _{C} D_{s}^{h}\left( \overline{\psi } \partial _{l} \overline{u}^{i} B_{lj} B_{kj}\frac{J}{\overline{Q}^{2}}\right) D_{s}^{h}(\partial _{k}(\overline{u}^{i} \overline{\psi })) \\&= \int _{C} B_{lj} B_{kj} D_{s}^{h}(\overline{\psi } \partial _{l} \overline{u}^{i}) D_{s}^{h} \partial _{k}(\overline{u}^{i} \overline{\psi }) \frac{J}{\overline{Q}^{2}} + \text {SAFE} \\&= \int _{C} B_{lj} B_{kj} D_{s}^{h}(\partial _{l} \overline{\psi } \overline{u}^{i})D_{s}^{h} \partial _{k}(\overline{u}^{i} \overline{\psi }) \frac{J}{\overline{Q}^{2}} + \text {SAFE}. \end{aligned}$$

We can then bound

$$\begin{aligned} \int _{C} |\nabla D_{s}^{h}(\overline{\psi } \overline{u})|^2&\le M_1 \le C_{\delta } + \delta \Vert \nabla D_{s}^{-h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2} + \delta \Vert \nabla D_{s}^{h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2}, \\ \int _{C} |\nabla D_{s}^{-h}(\overline{\psi } \overline{u})|^2&\le M_1 \le C_{\delta } + \delta \Vert \nabla D_{s}^{-h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2} + \delta \Vert \nabla D_{s}^{h}(\overline{u} \overline{\psi })\Vert _{L^{2}}^{2}. \end{aligned}$$

Similar computations show that we can control $\Vert \nabla \partial _{s} u \Vert _{L^{2}}$. This implies that

$$\begin{aligned} \Vert \partial _{s} u \Vert _{H^{1}(C/2)} \le C. \end{aligned}$$

We proceed to control the pressure terms. From the energy identity:

$$\begin{aligned}&\frac{1}{2}\int _{C}(\partial _{l} \overline{u}^{i} B_{lj} + \partial _{l} \overline{u}^{j} B_{li})(\partial _{k} \overline{\varphi }^{i} B_{kj} + \partial _{k} \overline{\varphi }^{j} B_{ki}) \frac{J}{\overline{Q}^{2}} \\&\qquad + \int _{C}(\partial _{l} \overline{u}^{i} B_{lj} + \partial _{l} \overline{u}^{j} B_{li})(\overline{\varphi }^{i} B_{kj} + \overline{\varphi }^{j} B_{ki}) \partial _{k} \overline{\psi } \frac{J}{\overline{Q}^{2}} + \int _{C} \overline{u}^{i} \overline{\partial }^{i} \psi \frac{J}{\overline{Q}^{2}} \\&\quad = \int _{C} \overline{f}^{i} \overline{\varphi }^{i} \overline{\psi } \frac{J}{\overline{Q}^{2}} + \int _{C} \overline{p} Tr(\nabla \overline{\varphi } B) \overline{\psi } \frac{J}{\overline{Q}^{2}} + \overline{p} B \nabla \overline{\psi } \overline{\varphi } \frac{J}{\overline{Q}^{2}} \\&P_{1} + P_{2} = P_{3} + P_{4} + P_{5} \end{aligned}$$

We choose $\overline{\varphi } = B^{*} \nabla \overline{\Pi }$, with

$$\begin{aligned} \frac{J}{\overline{Q}^{2}} D_{s}^{-h} D_{s}^{h}(\overline{p} \overline{\psi }) = \text {div}(J\Sigma \Sigma ^{*} \nabla \overline{\Pi }). \end{aligned}$$

Then:

$$\begin{aligned} P_{4}&= \int _{C} \overline{p} Tr(\nabla \overline{\varphi } B) \overline{\psi } \frac{J}{\overline{Q}^{2}} = \int _{C} \overline{\psi } \overline{p} D_{s}^{-h} D_{s}^{h}(\overline{p} \overline{\psi }) \frac{J}{\overline{Q}^{2}} \\&= -\int _{C}(D_{s}^{h}(\overline{\psi } \overline{p}))^{2} \frac{J}{\overline{Q}^{2}} + \int _{C}\left( D_{s}^{h}\left( \frac{J}{\overline{Q}^{2}} \overline{\psi } \overline{p}\right) - \frac{J}{\overline{Q}^{2}} D_{s}^{h}(\overline{\psi } \overline{p})\right) \\&= m_{5} + l_{1}, \end{aligned}$$

where

$$\begin{aligned} |l_{1}| \le c \Vert p\Vert _{L^{2}}. \end{aligned}$$

So, if we control $m_{5}$ we can control $\Vert \partial _{s} \overline{p}\Vert _{L^{2}\left( C/2\right) }$. It is not hard to see that

$$\begin{aligned} |P_1| + |P_2| + |P_3| + |P_5|&\le C_{\delta }(\Vert \overline{u}\Vert _{L^{2}}^{2} + \Vert \nabla \partial _{s} \overline{u}\Vert _{L^{2}}^{2} + \Vert \nabla \overline{u}\Vert _{L^{2}}^{2} + \Vert \overline{p}\Vert _{L^{2}}^{2}) \\&\quad + \delta \Vert D_{s}^{h}(\overline{p} \overline{\psi })\Vert _{L^{2}}^{2} \end{aligned}$$

which implies that $\Vert D_{s}^{h}(\overline{p} \overline{\psi })\Vert _{L^2} \le C$ independent of h. Because of the interior regularity estimate we have that the solution is strong in the interior of $\Omega$ and we can write

$$\begin{aligned} \frac{\overline{Q}^{2}}{J}\text {div}\left( \frac{1}{J}\left( \begin{array}{cc}1 &{} 0 \\ 0 &{} J^{2}\end{array}\right) \nabla \overline{u}^{i}\right) + \overline{u}^{i} + (B^{*} \nabla \overline{p})^{i} = \overline{f}^{i}, \quad Tr(\nabla \overline{u} B) = 0 \end{aligned}$$

where we recall that $J = (1 + \lambda \kappa (s))$ and that we have used

$$\begin{aligned} \Delta u \circ x(s,\lambda )&= \frac{1}{J(s,\lambda )} \text {div}\left( J(s,\lambda )\Sigma \Sigma ^{*} \nabla \overline{u}\right) (s,\lambda )\\ \Sigma \Sigma ^{*}&= \frac{1}{(1+\kappa (s)\lambda )^{2}}\left( \begin{array}{cc}x_{\lambda }^{2} &{} -x_{\lambda }^{1} \\ -x_{s}^{2} &{} x_{s}^{1} \end{array}\right) \left( \begin{array}{cc}x_{\lambda }^{2} &{} -x_{s}^{2} \\ -x_{\lambda }^{1} &{} x_{s}^{1} \end{array}\right) \\&= \frac{1}{(1+\kappa (s)\lambda )^{2}}\left( \begin{array}{cc}|x_{\lambda }|^{2} &{} -x_{\lambda }\cdot x_{s} \\ -x_{\lambda } \cdot x_{s} &{} |x_{s}|^{2} \end{array}\right) \end{aligned}$$

This implies that

$$\begin{aligned}&\frac{1}{J(s,\lambda )} \text {div}\left( \frac{1}{(1+\kappa (s)\lambda )^{2}}\left( \begin{array}{cc}|x_{\lambda }|^{2} &{} -x_{\lambda }\cdot x_{s} \\ -x_{\lambda } \cdot x_{s} &{} |x_{s}|^{2} \end{array}\right) \nabla \overline{u}^{i}\right) \\&\quad = \frac{1}{J(s,\lambda )} \text {div}\left( \frac{1}{(1+\kappa (s)\lambda )^{2}}\left( \begin{array}{cc}1 &{} 0 \\ 0 &{} (1+\kappa (s)\lambda )^{2} \end{array}\right) \nabla \overline{u}^{i}\right) . \end{aligned}$$

Let us define

$$\begin{aligned} \beta&= \left( \begin{array}{cc}1/J &{} 0 \\ 0 &{} J\end{array}\right) \\ \text {div}(\beta \nabla \overline{u}^{i})&= \partial _{k}(\beta _{kl} \partial _{l} \overline{u}^{i}) = \underbrace{\partial _{k} B_{kl}}_{\Gamma _{l}} \partial _{l} \overline{u}^{i} + \beta _{kl} \partial _{k} \partial _{l} \overline{u}^{i} \\&= \Gamma \nabla \overline{u}^{i} + \frac{1}{J} \partial _{s}^{2} \overline{u}^{i} + J \partial _{\lambda }^{2} \overline{u}^{i}. \end{aligned}$$

And then:

$$\begin{aligned}&\frac{\overline{Q}^{2}}{J} \partial _{s}^{2} \overline{u}^{i} + \overline{Q}^{2} \partial _{\lambda }^{2} \overline{u}^{i} + \Gamma \nabla \overline{u}^{i} + B_{li} \partial _{l} \overline{p} = \overline{f}^{i} \\&\quad \Rightarrow \overline{Q}^{2} \partial _{\lambda }^{2} \overline{u}^{i} + B_{li} \partial _{l} \overline{p} = -\frac{\overline{Q}^{2}}{J} \partial _{s}^{2} \overline{u}^{i} - \Gamma \nabla \overline{u}^{i} + \overline{f}^{i} = g^{i}, \end{aligned}$$

and we know that $g^{i} \in L^{2}\left( C/2\right)$. We also have that

$$\begin{aligned} (\partial _{l} \overline{u}^{i} B_{li})&= 0 \\ \partial _{s} \overline{u}^{1} B_{11} + \partial _{\lambda } \overline{u}^{i} B_{21} + \partial _{s} \overline{u}^{2} B_{12} + \partial _{\lambda } \overline{u}^{2} B_{22}&= 0 \end{aligned}$$

Thus

$$\begin{aligned} \partial _{\lambda }^{2} \overline{u}^{1} B_{21} + \partial _{\lambda }^{2} \overline{u}^{2} B_{22} = g^{3}, \text { where } g^{3} \in L^{2} \end{aligned}$$

Plus

$$\begin{aligned} \overline{Q}^{2} \partial _{\lambda }^{2} \overline{u}^{1} + B_{11} \partial _{s} \overline{p} + B_{21} \partial _{\lambda } \overline{p}&= g^{1} \\ \overline{Q}^{2} \partial _{\lambda }^{2} \overline{u}^{2} + B_{12} \partial _{s} \overline{p} + B_{22} \partial _{\lambda } \overline{p}&= g^{2} \\ \overline{Q}^{2} B_{21} \partial _{\lambda }^{2} \overline{u}^{1} + B_{12} B_{11} \partial _{s} \overline{p} + B_{12}^{2} \partial _{\lambda } \overline{p}&= B_{12} g^{1} \\ \overline{Q}^{2} B_{22} \partial _{\lambda }^{2} \overline{u}^{2} + B_{22} B_{12} \partial _{s} \overline{p} + B_{22}^{2} \partial _{\lambda } \overline{p}&= B_{22} g^{2} \\ \end{aligned}$$

This implies

$$\begin{aligned} (B_{21}^{2} + B_{22}^{2}) \partial _{\lambda } \overline{p} = g^{4}, \text { where } g^{4} \in L^{2}\left( C/2\right) , \end{aligned}$$

and since $(B_{21}^{2} + B_{22}^{2}) > 0$ for a small enough $\lambda _0$, and we get that $\partial _{\lambda } \overline{p} \in L^{2}\left( C/2\right)$. Finally, we use this to get that

$$\begin{aligned} \overline{Q}^{2} \partial _{\lambda }^{2} \overline{u}^{1}&= g^{5}, \text { where } g^{5} \in L^{2}\left( C/2\right) . \\ \overline{Q}^{2} \partial _{\lambda }^{2} \overline{u}^{2}&= g^{6}, \text { where } g^{6} \in L^{2}\left( C/2\right) . \end{aligned}$$

This completes the regularity proof, since we can cover a $\frac{\lambda _0}{2}$ neighborhood by a finite number of sets of type C / 2.

We are only left to show that we can apply [2, Theorem 10.5, p.78] to our system in order to obtain higher regularity. Indeed we can apply this theorem to show:

$$\begin{aligned} \Vert u\Vert _{H^{s+1}} + \Vert p\Vert _{H^{s}} \le \Vert f\Vert _{H^{s-1}}. \end{aligned}$$

(64)

What follows is a confirmation that our problem fulfills the elliptical conditions of [2]. To adapt our notation to the one in [2] we will write $(u^{1}, u^{2}, u^{3}) = (u^{1}, u^{2}, p)$.

In [2, pp.38, 42] the system is written like

$$\begin{aligned} l_{ij}(x,\partial )u^j=F_i, \end{aligned}$$

with the boundary condition

$$\begin{aligned} B_{h,j}(x,\partial )u_j= \phi _h. \end{aligned}$$

In our case we have the correspondence

$$\begin{aligned} l_{11}&= 1- Q^{2} \Delta , \quad l_{12} = 0, \quad l_{13} = A_{k1} \partial _{l}, \quad F_{1} = f_1 \\ l_{21}&= 0, \quad l_{22} =1 - Q^{2} \Delta , \quad l_{23} = A_{k2} \partial _{l}, \quad F_{2} = f_2 \\ l_{31}&= A_{k1} \partial _{k}, \quad l_{32} = A_{k2} \partial _{k}, \quad l_{33} = 0, \quad F_{3} = 0 \end{aligned}$$

And the indices $t_j$ and $s_j$ can be taken as

$$\begin{aligned} t_1&= 2, \quad t_2 = 2, \quad t_3 = 1 \\ s_1&= 0, \quad s_2 = 0, \quad s_3 = -1 \end{aligned}$$

It can be checked that with this choice $l'_{ij}$ is given by

$$\begin{aligned} l'_{11}&= - Q^{2} \Delta , \quad l'_{12} = 0, \quad l'_{13} = A_{k1} \partial _{l}, \\ l'_{21}&= 0, \quad l'_{22} = - Q^{2} \Delta , \quad l'_{23} = A_{k2} \partial _{l}, \\ l'_{31}&= A_{k1} \partial _{k}, \quad l'_{32} = A_{k2} \partial _{k}, \quad l'_{33} = 0, \end{aligned}$$

where $l'_{ij}$ is defined in pages 38-39 of [2]. Also

$$\begin{aligned} B_{2l}&= (t^{i} A_{il} n_{j} + t^{j} A_{il} n_i) \partial _{j}, \quad l = 1,2 \\ B_{23}&= 0 \end{aligned}$$

and the indices $r_1=-1$ and $r_2=-1$. With this choice $B'_{ij}=B_{ij}$ (see [2, p.42]).

We can write

$$\begin{aligned} (l'(x,\xi ))_{ij} = \left( \begin{array}{ccc} -Q^{2}(\xi _{1}^{2} + \xi _{2}^{2}) &{} 0 &{} A_{k1} \xi _{k} \\ 0 &{} -Q^{2}(\xi _{1}^{2} + \xi _{2}^{2}) &{} A_{k2} \xi _{k} \\ A_{k1} \xi _{k} &{} A_{k2} \xi _{k} &{} 0\\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} B(x,\xi )_{ij} = \left( \begin{array}{ccc} -2n^{i} A_{i1}(n \cdot \xi ) &{} - 2n^{i} A_{i2}(n\cdot \xi ) &{} 1 \\ t^{i} A_{i1}(\xi \cdot n) + n^{i} A_{i1}(t\cdot \xi )&{} t^{i} A_{i2} (\xi \cdot n) + n^{i} A_{i2}(t \cdot \xi ) &{} 0 \end{array} \right) \end{aligned}$$

Let $L= \det (l'_{ij}(x,\xi ))$, i.e.

$$\begin{aligned} L(x,\xi )&= \det (l'(x,\xi )) = Q^{2}(\xi _{1}^{2} + \xi _{2}^{2})(A_{k1}\xi _{k})^{2} + Q^{2}(\xi _{1}^{2} + \xi _{2}^{2})(A_{k2}\xi _{k})^{2} \\ (A_{11} \xi _{1} + A_{21} \xi _{2})^{2}&= A_{11}^{2} \xi _{1}^{2} + A_{21}^{2} \xi _{2}^{2} + 2A_{11} A_{21} \xi _{1} \xi _{2} \\ (A_{12} \xi _{1} + A_{22} \xi _{2})^{2}&= A_{12}^{2} \xi _{1}^{2} + A_{22}^{2} \xi _{2}^{2} + 2A_{12} A_{22} \xi _{1} \xi _{2} \\ \Rightarrow L(P,\xi )&= Q^{2}(\xi _{1}^{2} + \xi _{2}^{2})Q^{2}(\xi _{1}^{2} + \xi _{2}^{2}) = Q^{4}(\xi _{1}^{2} + \xi _{2}^{2})^{2} = Q^{4}|\xi |^{4} \end{aligned}$$

To know whether the system is uniformly elliptic, we are left to check that it satisfies the Supplementary condition (see [2, p.39]). The degree is 4 and therefore $m = 2$. We need to compute the solutions of $L(P,\xi +\tau \xi ') = 0$, which are the solutions to $|\xi + \tau \xi '|^{2} = 0$.

$$\begin{aligned} |\xi + \tau \xi '|^{2} = |\xi |^{2} + \tau ^{2}|\xi '|^{2} + 2 \tau \xi \cdot \xi ' = 0 \end{aligned}$$

Solving in $\tau$ yields

$$\begin{aligned} \tau = \frac{-2 \xi \cdot \xi ' + \sqrt{4(\xi \cdot \xi ')^{2} - 4|\xi |^{2}|\xi '|^{2}}}{2|\xi '|^{2}}. \end{aligned}$$

If $\xi$ and $\xi '$ are linearly independent, then the discriminant is strictly negative, which implies that there is a complex root with positive imaginary part. Since the roots have multiplicity 2, the Supplementary condition is satisfied. In addition uniform ellipticity is easy to obtain. Next we check that the Complementing Boundary Condition is satisfied ([2, p. 42]).

Let $t_0$ be tangential vector and $n_0$ the normal one.

Since i is a double root of $L(x,t_0+\tau n)$ and $M^+(x,\xi ,\tau )$ ([2, p. 42]) is given by

$$\begin{aligned} M^+(x,\xi ,\tau ) = (\tau - i)^{2}. \end{aligned}$$

In addition

$$\begin{aligned} (l'(x,t_0+\tau n))_{ij} = \left( \begin{array}{ccc} -Q^{2}(1+\tau ^{2}) &{} 0 &{} A_{k1}(t_0^{k} + \tau n^{k}) \\ 0 &{} -Q^{2}(1+\tau ^{2}) &{} A_{k2}(t_0^{k} + \tau n^{k}) \\ A_{k1}(t_0^{k} + \tau n^{k}) &{} A_{k2}(t_0^{k} + \tau n^{k}) &{} 0 \end{array} \right) \end{aligned}$$

We define $L_{ij}$ as in [2, p.42], and we have that. $L(x,t_0+\tau n) = \overline{l'(x,t+\tau n)}$. Also, it can be computed that

$$\begin{aligned}&B(x,t+\tau n) = \left( \begin{array}{ccc} -2n^{i} A_{i1} \tau &{} -2n^{i} A_{i2} \tau &{} 1 \\ (n+\tau t)^{i} A_{i1} &{} (n+\tau t)^{i} A_{i2} &{} 0 \end{array} \right) \\&L_{ij}(x,t_0+\tau n) = \left( \begin{array}{ccc} -Q^{2}(1+\overline{\tau }^{2}) &{} 0 &{} A_{k1}(t+\overline{\tau }n)^{k} \\ 0 &{} -Q^{2}(1+\overline{\tau }^{2}) &{} A_{k2}(t+\overline{\tau }n)^{k} \\ A_{k1}(t+\overline{\tau }n)^{k} &{} A_{k2}(t+\overline{\tau }n)^{k} &{} 0 \end{array} \right) \\&{\scriptstyle (B(x,t+\tau n) L(x,t+\tau n))_{ij}} \\&\quad {\scriptstyle = \left( \begin{array}{ccc} 2n_{k} A_{k1} \tau Q^{2}(1 + \overline{\tau }^{2}) + A_{k1}(t+\overline{\tau }n)^{k} &{} 2n_{k} A_{k2} \tau Q^{2}(1+\overline{\tau }^{2}) + A_{k2} (t+\overline{\tau }n)^{k} &{} n^{i}A_{i1} \tau A_{k1}(t + \overline{\tau } n)^{k} + \tau n^{i} A_{i2}A_{k2}(t+\overline{\tau }n)^{k} \\ -(n+\tau t)^{k}A_{k1} Q^{2}(1+\overline{\tau }^{2}) &{} -(n+\tau t)^{i} A_{i2} Q^{2} (1+ \overline{\tau }^{2}) &{} (n + \tau t)^{i} A_{i1} A_{k1} (t+\overline{\tau }n)^{k} + (n+\tau t)^{i} A_{i2} A_{k2}(t+\overline{\tau }n)^{k} \end{array} \right) } \end{aligned}$$

If the rows of $(BL)_{ij}$ are linearly independent modulo $(\tau -i)^2$, that means that the condition

$$\begin{aligned} C_h B_{hj}L_{jk}= 0 \quad \mod \,\ M^+ \end{aligned}$$

implies that all $C_h=0$ ([2, p.43]). If this condition is satisfied in particular

$$\begin{aligned} C_h B_{hj}L_{jk}|_{\tau =i}= 0. \end{aligned}$$

Then $c_3$ must be zero and the following system of equations must be satisfied:

$$\begin{aligned} c_1 A_{k1} t^{k} + c_2 A_{k2} t^{k}&= 0 \\ c_1 A_{k1} n^{k} + c_2 A_{k2} n^{k}&= 0 \end{aligned}$$

In matrix form:

$$\begin{aligned} \left( \begin{array}{cc} A_{k1} t^{k} &{} A_{k2} t^{k}\\ A_{k1} n^{k} &{} A_{k2} n^{k} \end{array} \right) \left( \begin{array}{c} c_1 \\ c_2 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right) \end{aligned}$$

But the determinant of this matrix satisfies

$$\begin{aligned}&A_{k1} t^{k} A_{k2} n^{k} - A_{k2} t^{k} A_{k1} n^{k} \\&(A^{*}t)^{1}(A^{*}n)^{2} - (A^{*}t)^{2}(A^{*}n)^{1} \\&(A^{*} t, JA^{*}n) = (t,AJA^{*}n) \ne 0, \end{aligned}$$

and therefore $c_1=c_2=0$. Thus we have checked that the complementing Boundary Condition is satisfied.

Finally we notice that our system can be written as in [2, p. 71], where the coefficients $a_{ij,P}$ are smooth. In our case the index $l_1$ in [2, p. 77] is $l_1=0$. The index l in [2] coincides with $s-1$. That means $l=0,1,2$. The regularity we ask for the coefficient $b_{hj,\sigma }$ is $C^{l-r_h}$ (cf. [2, p. 77]). The most we need is therefore $b_{hj,\sigma }\in C^3$. Since these coefficients are one derivative less regular that the boundary, a $C^4$ boundary is enough for our purpose. This fact finishes the proof of the inequality (64) if s is an integer. For the rest of the values we proceed by interpolation.

This concludes the proof of Lemma A.3. $\square$

Once we have studied the operator $S_A$ we will solve the time evolution. First we will show the following lemma.

Lemma A.4

Let $1 \le s \le 3$, $\lambda \in \mathbb {C}$, $\mathfrak {R}(\lambda ) \ge 0$. Then the operator $\lambda + S_A: V^{s+1} \rightarrow RH^{s-1}$ is invertible. The inverse satisfies:

$$\begin{aligned} \Vert (\lambda + S_A)^{-1}Rf\Vert _{H^{s+1}} \le C(\Vert Rf\Vert _{H^{s-1}} + |\lambda |^{\frac{s-1}{2}}\Vert Rf\Vert _{L^{2}}) \end{aligned}$$

(65)

Proof

As before, we look for a weak solution of $(\lambda + S_A)v = Rf, \quad f \in L^{2}$. Therefore,

$$\begin{aligned} (1+\lambda )(v,w) + \langle v,w \rangle = \int _{\Omega } Rf\bar{w} \frac{dx}{Q^{2}} \quad \forall w \in RH^{1}. \end{aligned}$$

The solution is given by Lax-Milgram’s Theorem. For $w=v$, by virtue of Korn’s inequality and $\mathfrak {R}(\lambda ) \ge 0$, one obtains

$$\begin{aligned} |(1+\lambda )(v,v) + \langle v,v \rangle | \ge C((1+|\lambda |)\Vert v\Vert _{L^{2}}^{2} + \Vert \nabla v \Vert _{L^{2}}^{2}), \end{aligned}$$

with C independent of $\lambda$. Easily, the following bound is obtained:

$$\begin{aligned} \Vert v\Vert _{L^{2}} \le \frac{1}{1+|\lambda |} \Vert Rf\Vert _{L^{2}}. \end{aligned}$$

(66)

If $f \in H^{s-1}$, then Lemma A.3 gives $v = S_{A}^{-1}(Rf - \lambda v) \in H^{s+1}$, and we can get the following bounds:

$$\begin{aligned} \Vert v\Vert _{H^{s+1}}&\le C\Vert S_A v\Vert _{H^{s-1}} \le C(\Vert (S_A + \lambda ) v\Vert _{H^{s-1}} + |\lambda |\Vert v\Vert _{H^{s-1}}) \\&\le C\Vert Rf\Vert _{H^{s-1}} + |\lambda | \Vert v\Vert _{H^{s+1}}^{\frac{s-1}{s+1}}\Vert v\Vert _{L^{2}}^{\frac{2}{s+1}}. \end{aligned}$$

The case $s = 1$ is already solved using (66). Young’s inequality provides

$$\begin{aligned} \Vert v\Vert _{H^{s+1}} \le C\Vert Rf\Vert _{H^{s-1}} + \frac{1}{2}\Vert v\Vert _{H^{s+1}} + |\lambda |^{\frac{s+1}{2}}\Vert v\Vert _{L^{2}}. \end{aligned}$$

and we can get (65) using (66). $\square$

In order to find the solution of

$$\begin{aligned} v_t + S_A(v) = Rf \quad \text { in } V^{s+1}, \end{aligned}$$

we take Fourier transforms in time. Since $f \in H_{(0)}^{ht,s-1}, Rf(0) = 0$, we can extend Rf to a function $\overline{Rf}$ defined in $H^{ht,s-1}(\Omega \times \mathbb {R})$, with $\overline{Rf}(t) = 0$ for all $t < 0$. Since $\frac{s-1}{2} \le \frac{3}{4} < 1$, using Lemma 3.1 (ii) we get

$$\begin{aligned} \Vert \overline{Rf}\Vert _{H_{(0)}^{ht,s-1}(\mathbb {R}\times \Omega _0)} \le C\Vert Rf\Vert _{H_{(0)}^{ht,s-1}([0,T]\times \Omega _0)}\le C\Vert f\Vert _{H_{(0)}^{ht,s-1}([0,T]\times \Omega _0)}, \end{aligned}$$

with C independent of T. We look for a solution of

$$\begin{aligned} v_t + S_A(v) = \overline{Rf}, \quad \forall t \in \mathbb {R}, \quad v(0) = 0. \end{aligned}$$

By Fourier, $i \tau \hat{v}(\tau ) + S_A(\hat{v})(\tau ) = \hat{\overline{Rf}}$, and therefore the solution is given by $\hat{v}(\tau ) = (i \tau + S_A)^{-1}\hat{\overline{Rf}}$. Using (65) and (66) we can bound

$$\begin{aligned} \Vert v\Vert ^2_{H^{ht,s+1}(\mathbb {R}\times \Omega _0)}&= \int _{{\mathbb {R}}}(\Vert \hat{v}\Vert _{H^{s+1}}^{2}(\tau ) + |\tau |^{s+1}\Vert \hat{v}\Vert _{L^{2}}^{2}(\tau )) d\tau \\&\le C\int _{{\mathbb {R}}}(\Vert \overline{Rf}\Vert _{H^{s-1}}^{2}(\tau ) + |\tau |^{s-1}\Vert \overline{Rf}\Vert _{L^{2}}^{2}) d\tau \\&\le C\Vert \overline{Rf}\Vert ^2_{H^{ht,s-1}} \le C\Vert f\Vert ^2_{H_{(0)}^{ht,s-1}}. \end{aligned}$$

Since $\overline{Rf}(t) = 0$ for every $t < 0$, $\hat{\overline{Rf}}(\tau )$ has an analytic extension in $\tau$ to $\mathfrak {I}(\tau ) < 0$. Using Lemma A.4, $\hat{v}(\tau )$ also has that extension. Moreover, (66) gives $\Vert \hat{v}(\tau )\Vert _{L^{2}} \le C\Vert \hat{\overline{Rf}(\tau )}\Vert _{L^{2}}$. Thus, Paley-Wiener provides $v(t) = 0 \; \forall t < 0$. Since $v \in H^{\frac{s+1}{2}}({\mathbb {R}}; L^{2})$ and $\frac{3}{2}<\frac{s+1}{2}$ we have continuity in time and hence $v(0) = 0$. Since $\hat{v} \in L^{2}({\mathbb {R}}; V^{s+1}(\Omega ))$ we have that $v \in L^{2}([0,T]; V^{s+1}(\Omega ))$ and therefore $Tr(\nabla v A) = 0$ and $(q + \nabla v A + (\nabla v A)^{*})A^{-1}n = 0$. We have already solved

$$\begin{aligned} v_t - Q^{2}\Delta v+ A^{*} \nabla q_1 = Rf, \end{aligned}$$

where $q_1$ satisfies

$$\begin{aligned} Q^{2} \Delta q_1&= 0, \text { in } \Omega _0\times [0,T] \\ q_1&= A^{-1}n(\nabla v A + (\nabla v A)^{*}) A^{-1}n, \text { on } \partial \Omega \times [0,T]. \end{aligned}$$

We have that $\partial _tv(0)=0$ and then $||v||_{H_{(0)}^{ht,s+1}}\le C||f||_{H_{(0)}^{ht,s-1}}$. Definition

$$\begin{aligned} A^{*} \nabla q =(I-R)f + A^{*} \nabla q_1, \end{aligned}$$

gives us the solution that we were looking for

$$\begin{aligned} v_t - Q^{2} \Delta v + A^{*} \nabla q = f. \end{aligned}$$

The properties of R allow us to obtain

$$\begin{aligned} \Vert \nabla q\Vert _{H_{(0)}^{ht,s-1}}&\le C\Vert A^{*} \nabla q\Vert _{H_{(0)}^{ht,s-1}} \le \Vert (I-R)f\Vert _{H_{(0)}^{ht,s-1}} + \Vert A^{*} \nabla q_1\Vert _{H_{(0)}^{ht,s-1}}\\&\le C( \Vert f\Vert _{H_{(0)}^{ht,s-1}} + \Vert A^{*} \nabla q_1\Vert _{H_{(0)}^{ht,s-1}}). \end{aligned}$$

We have the following bounds:

$$\begin{aligned} \Vert A^{*}\nabla q_1\Vert _{L^{2}H^{s-1}}&\le \Vert \nabla q_1\Vert _{L^{2}H^{s-1}} \le |A^{-1}n ((\nabla v A)+(\nabla v A)^{*})A^{-1}n|_{L^{2}H^{s-\frac{1}{2}}} \\&\le \Vert (\nabla v A) + (\nabla v A)^{*}\Vert _{L^{2}H^{s}}\le \Vert \nabla v\Vert _{L^{2}H^{s}} \le \Vert v\Vert _{L^{2}([0,T]; H^{s+1})},\\ \Vert A^{*}\nabla q_1\Vert _{H_{(0)}^{\frac{s-1}{2}}L^{2}}&\le |A^{-1}n ((\nabla v A)+(\nabla v A)^{*})A^{-1}n|_{H_{(0)}^{\frac{s-1}{2}}H^{\frac{1}{2}}}. \end{aligned}$$

Decomposing $\nabla v_i$ into the tangential and normal components: $\nabla v_i = (\nabla v_i \cdot n_0) n_0 + (\nabla v_i \cdot t_0) t_0$, we can bound each of them by

$$\begin{aligned} |\nabla v_i \cdot t_0|_{H_{(0)}^{\frac{s-1}{2}}H^{\frac{1}{2}}}&\le C|v_i|_{H_{(0)}^{\frac{s-1}{2}} H^{\frac{3}{2}}} \le C\Vert v_i\Vert _{H_{(0)}^{\frac{s-1}{2}}H^{2}} \le C\Vert v_i\Vert _{H_{(0)}^{ht,s+1}}, \\ |\nabla v_i \cdot n_0|_{H_{(0)}^{\frac{s-1}{2}}H^{\frac{1}{2}}}&\le |\nabla v_i \cdot n_0|_{H_{(0)}^{ht,s-\frac{1}{2}}} \le C\Vert v_i\Vert _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

Above we have used Lemma 3.2 (i). This yields

$$\begin{aligned} \Vert A^{*} \nabla q_1\Vert _{H_{(0)}^{\frac{s-1}{2}}L^{2}} \le C\Vert v\Vert _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

and therefore

$$\begin{aligned} \Vert A^{*} \nabla q\Vert _{H_{(0)}^{ht,s-1}} \le C\Vert f\Vert _{H_{(0)}^{ht,s-1}}. \end{aligned}$$

Since $(I-R) f |_{\partial \Omega _0} = 0$, we have that $q = q_1$ on $\partial \Omega _0$. This implies

$$\begin{aligned} |q|_{H_{(0)}^{ht,s-\frac{1}{2}}}&= |q_1|_{H_{(0)}^{ht,s-\frac{1}{2}}} = |A^{-1}n((\nabla v A) + (\nabla v A)^{*} )A^{-1}n|_{H_{(0)}^{ht,s-\frac{1}{2}}} \\&\le |A^{-1}n((\nabla v A) + (\nabla v A)^{*} )A^{-1}n|_{L^{2}H^{s-\frac{1}{2}}} \\&\quad + |A^{-1}n((\nabla v A) + (\nabla v A)^{*} )A^{-1}n|_{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^{2}}. \end{aligned}$$

Decomposing into the tangential and normal components as before, we find finally

$$\begin{aligned} |q|_{H_{(0)}^{ht,s-\frac{1}{2}}}&\le \Vert \nabla v\Vert _{L^{2}H^{s}} + |\nabla v_i \cdot t_0|_{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^{2}} + |\nabla v_i \cdot n_0|_{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^{2}} \\&\le \Vert v\Vert _{L^{2}H^{s+1}} + |v_i|_{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{1}} + |\nabla v_i \cdot n_0|_{H_{(0)}^{ht,s-\frac{1}{2}}}\\&\le \Vert v\Vert _{H_{(0)}^{ht,s+1}} + \Vert v\Vert _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{1+\frac{1}{2}}} + \Vert v\Vert _{H_{(0)}^{ht,s+1}} \le \Vert v\Vert _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

A.2 Reduction for Arbitrary g and h

In this section we want to extend the result with $g = 0$ and $h = 0$ to the case:

$$\begin{aligned} v_t - \nu Q^{2} \Delta v + A^{*} \nabla q&= f&\text { in } \Omega _0 \times [0,T] \nonumber \\ Tr(\nabla v A)&= g&\text { in } \Omega _0 \times [0,T] \nonumber \\ (q + (\nabla v A) + (\nabla v A)^{*})A^{-1} n&= h&\text { on } \partial \Omega _0 \times [0,T] \nonumber \\ v(x,0)&= 0&\text { in } \Omega _0 \end{aligned}$$

(67)

To get the space where g belongs we proceed as follows. For $\phi \in H^{1}_{0}$ it is easy to find

$$\begin{aligned} \int \partial _t^{j}(Tr(\nabla v A)) \phi (x) \frac{dx}{Q^{2}(x)} = \int Tr(\nabla \partial _{t}^{j} v A) \phi (x) \frac{dx}{Q^{2}(x)} = \int \partial _{t}^{j} v A^* \nabla \phi \frac{dx}{Q^{2}(x)}. \end{aligned}$$

Then

$$\begin{aligned} \left| \int \partial _t^{j}(Tr(\nabla v A)) \phi (x) \frac{dx}{Q^{2}(x)}\right| \le \Vert \partial _{t}^{j} v\Vert _{L^{2}}(t)\Vert \nabla \phi \Vert _{H^{1}}, \end{aligned}$$

and duality provides

$$\begin{aligned} \Vert \partial _{t}^{j}(Tr(\nabla v A))\Vert _{H^{-1}} \le \Vert \partial _{t}^{j} v\Vert _{L^{2}}(t), \text { with } H^{-1} = (H^{1}_{0})^*. \end{aligned}$$

For $j = \frac{s+1}{2}$, integration in time yields

$$\begin{aligned} \Vert Tr(\nabla v A)\Vert _{H^{\frac{s+1}{2}}H^{-1}} \le \Vert v\Vert _{H^{ht,s+1}}. \end{aligned}$$

Here we remark that we use (28) for the norm of fractional derivatives on time. Also, $Tr(\nabla v A) \in L^{2}([0,T]; H^{s})$, which implies:

$$\begin{aligned} Tr(\nabla v A) \in L^{2}([0,T]; H^{s}) \cap H^{\frac{s+1}{2}}([0,T]; H^{-1}), \end{aligned}$$

To prove this fact, one can proceed for an integer number of derivatives, then interpolate for fractional ones (see [27]).

We check next the compatibility conditions of the initial data:

$$\begin{aligned} Tr(\nabla v_0 A)&= g(0)&\text { in } \Omega _0 \end{aligned}$$

(68)

$$\begin{aligned} (A^{-1} n)^{\perp } (\nabla v_0 A + (\nabla v_0 A)^{*}) A^{-1} n&= h(0) (A^{-1} n)^{\perp }&\text { on } \partial \Omega _0. \end{aligned}$$

(69)

Defining the following spaces:

$$\begin{aligned} X_0&:= \{(v,q) : v \in H_{(0)}^{ht,s+1}, q \in H^{ht,s}_{pr,\,(0)}\} \\ Y_0&:= \{(f,g,h): f \in H_{(0)}^{ht,s-1}, g \in \overline{H}_{(0)}^{ht,s}, h \in H_{(0)}^{ht,s-\frac{1}{2}}(\partial \Omega \times [0,T]), \} \end{aligned}$$

(here we remark that in $X_0$ and $Y_0$: $v(0)=\partial _tv(0)=q(0)=f(0)=g(0)=\partial _tg(0)=h(0)=0$), then, for $(f,g,h)\in Y_0$, (33) is equivalent to solve:

$$\begin{aligned} L(v,q) = (f,g,h,0); \quad L: X_0 \rightarrow Y_0, \quad 2< s < \frac{5}{2}. \end{aligned}$$

Theorem A.5

$L: X_0 \rightarrow Y_0$ is invertible for $2<s< \frac{5}{2}$. Moreover, $\Vert L^{-1}\Vert$ is bounded uniformly if T is bounded above.

Proof of Theorem 4.1

We structure the proof in 3 steps, in order to get to the previous case ($g = h = v_0 = 0$). Let $(f,g,h) \in Y_0$.

• Step 1: A -divergence adjustment. We want to find $(v^{1},q^{1}) \in X$ such that $L(v^{1},q^{1}) = (f^{1},g^{1},h^{1},0)$ with

  $$\begin{aligned} f^{1}(0) = 0; \quad g^{1}(t) = g(t); \; \forall t \in [0,T] \quad h^{1}(0) = 0. \end{aligned}$$

  (70)

  We define $\phi$ by solving the following elliptic problem for every $t \in (-\infty ,\infty )$, after extending g to the whole real line:

  $$\begin{aligned} Q^{2} \Delta \phi&= g(t) \qquad \text { in } \Omega _0\times {\mathbb {R}}\\ \phi&= 0 \qquad \text { on } \partial \Omega _0\times {\mathbb {R}}\\ \end{aligned}$$

  This system satisfies $\Vert \phi \Vert _{H^{s+2}}(t) \le C\Vert g\Vert _{H^{s}}(t)$. In particular

  $$\begin{aligned} \Vert \nabla \phi \Vert _{H^{s+1}}(t) \le C\Vert g\Vert _{H^{s}}(t). \end{aligned}$$

  (71)

  Taking the Fourier Transform in time, one has, for every $\tau$:

  $$\begin{aligned} Q^{2} \Delta \hat{\phi }(\tau )&= \hat{g}(\tau ) \qquad \text { in } \Omega _0\times {\mathbb {R}}, \\ \hat{\phi }(\tau )&= 0 \qquad \text { on } \partial \Omega _0\times {\mathbb {R}}. \end{aligned}$$

  For $\lambda \in H^{1}_{0}$ it is possible to find

  $$\begin{aligned} \int \hat{\overline{g}}(\tau ) \lambda \frac{dx}{Q^{2}}&=\int \Delta \hat{\phi }(\tau ) \lambda dx = \int \nabla (\nabla \hat{\phi } A)A \lambda \frac{dx}{Q^{2}} = - \int \nabla \hat{\phi } A \underbrace{\nabla \lambda A}_{A^{*} \nabla \lambda } \frac{dx}{Q^{2}} \\&= - \int \nabla \hat{\phi } \nabla \lambda dx \end{aligned}$$

  Therefore:

  $$\begin{aligned} \int \nabla \hat{\phi } \nabla \lambda dx = -\int \hat{g}(\tau ) \lambda \frac{dx}{Q^{2}} \quad \forall \lambda \in H^{1}_{0}. \end{aligned}$$

  This implies:

  $$\begin{aligned} \Vert \nabla \hat{\phi }\Vert _{L^{2}}(\tau )&\le C\Vert \hat{g}\Vert _{H^{-1}}(\tau ) \\ \Rightarrow \int |\tau |^{\frac{s+1}{2}}\Vert \nabla \hat{\phi }\Vert _{L^{2}}(\tau ) d\tau&\le C \int |\tau |^{\frac{s+1}{2}} \Vert \hat{g}\Vert _{H^{-1}}(\tau ) d\tau . \end{aligned}$$

  We can conclude that $\nabla \phi \in H_{(0)}^{ht,s+1}$. We now define

  $$\begin{aligned} v^{1}&= A^{*} \nabla \phi \in H_{(0)}^{ht,s+1}, \quad q^{1} = q. \end{aligned}$$

  It is easy to check that

  $$\begin{aligned} f^{1}(0)&= A^{*} \nabla \phi _t(0)=0. \end{aligned}$$

  By construction, it is obvious that $g^{1}(t) = g(t)$ for every $t \in [0,T]$. Since $\phi (0) = 0$, $h^{1}(0) = 0$. This shows that (70) is satisfied.

• Step 2: Adjusting the boundary conditions in the tangential direction without modifying the A-divergence.

  We want to find $(v^{2},q^{2}) \in X_0$ such that $L(v^{2},q^{2}) = (f^{2},g^{2},h^{2}, 0)$ with

  $$\begin{aligned}&f^{2}(0) = 0; \quad g^{2}(t) = g(t); \; \forall t \in [0,T] \nonumber \\&h^{2}(t) \cdot (A^{-1} n)^{\perp } = h(t) \cdot (A^{-1} n)^{\perp }; \; \forall t \in [0,T], \quad h^{2}(0) = 0. \end{aligned}$$

  (72)

  We will use the following Lemma:

Lemma A.6

Let $\eta \in H_{(0)}^{ht,s-\frac{1}{2}}(\partial \Omega _0 \times [0,T])$, $2< s < \frac{5}{2}$, $\eta (0) = 0$. Then there exists $w \in H_{(0)}^{ht,s+1}$ such that $\Vert w\Vert _{H_{(0)}^{ht,s+1}} \le C|\eta |_{H_{(0)}^{ht,s-\frac{1}{2}}}$, $w(0) = w_t(0) = Tr(\nabla w A) = 0$ and

$$\begin{aligned} (A^{-1}n)^{\perp }(\nabla w A + (\nabla w A)^{*})A^{-1}n = \eta \text { on } \partial \Omega _0 \times [0,T]. \end{aligned}$$

Proof

Let $\psi \in H_{(0)}^{ht,s+2}$, with $\psi (0) = \psi _t(0) = 0, \psi (x) = \partial _{n} \psi (x) = 0, \; \forall x \in \partial \Omega _0$, and moreover $\partial _{n}^{2} \psi (x) = \eta (x) \; \forall x \in \partial \Omega _0\times [0,T]$. This choice is possible because of the parabolic trace. All that is left is to check that the compatibility conditions from Lemma 3.2. Defining

$$\begin{aligned} w = \nabla _{A}^{\perp } \psi = (- \partial _2 P_1 \circ P^{-1} \partial _1 \psi - \partial _2 P_2 \circ P^{-1} \partial _2 \psi , \partial _1 P_1 \circ P^{-1} \partial _1 \psi + \partial _1 P_2 \circ P^{-1} \partial _2 \psi ), \end{aligned}$$

it is immediate that $w(0) = w_t(0) = 0$. A straightforward, but long calculation gives:

$$\begin{aligned} Tr(\nabla (\nabla _{A}^{\perp } \psi ) A) = 0 = Tr(\nabla w A). \end{aligned}$$

We will now show that $(A^{-1}n)^{\perp }(\nabla w A + (\nabla w A)^{*})(A^{-1}n) = \eta$. Let $x_0 \in \partial \Omega _0$. We perform an euclidean change of coordinates in a way that $x_0 = 0, n_0 = (0,1)$. Thus, $\psi (0,0) = 0$, as well as $\partial _1 \psi (0,0)$ and $\partial _1^{2} \psi (0,0)$. The condition $\partial _n \psi (0,0) = 0$ implies $\partial _2 \psi (0,0) = 0, \partial _1 \partial _2 \psi (0,0) = 0$. That gives:

$$\begin{aligned} \nabla (\nabla _A^{\perp } \psi ) = \left( \begin{array}{cc} 0 &{} - \partial _2 P_2 \circ P^{-1} \partial _2^{2} \psi (0,0) \\ 0 &{} \partial _1 P_2 \circ P^{-1} \partial _2^{2} \psi (0,0) \end{array}\right) . \end{aligned}$$

Computing further:

$$\begin{aligned} \nabla (\nabla _A^{\perp } \psi )A = \partial _{2}^{2} \psi (0,0) \left( \begin{array}{cc} -\partial _2 P_2 \partial _1 P_2 &{} -(\partial _2 P_2)^2 \\ (\partial _1 P_2)^2 &{} \partial _1 P_2 \partial _2 P_2 \end{array}\right) \circ P^{-1} \end{aligned}$$

Thus

$$\begin{aligned}&( (\nabla (\nabla _A^{\perp } \psi )A) + ( \nabla (\nabla _A^{\perp } \psi )A)^{*}) \\&\quad = \partial _{2}^{2} \psi (0,0) \left( \begin{array}{cc} -2\partial _2 P_2 \partial _1 P_2 &{} (\partial _1 P_2)^2 -(\partial _2 P_2)^2 \\ (\partial _1 P_2)^2 -(\partial _2 P_2)^2 &{} 2\partial _1 P_2 \partial _2 P_2 \end{array}\right) \circ P^{-1}. \end{aligned}$$

We also have that

$$\begin{aligned} A^{-1}n_0 = \frac{1}{Q^{2}} \left( \begin{array}{c}-\partial _2 P_1 \\ \partial _1 P_1\end{array}\right) \circ P^{-1}; \quad (A^{-1}n_0)^{\perp } = \frac{1}{Q^{2}} \left( \begin{array}{c}-\partial _1 P_1 \\ -\partial _2 P_1\end{array}\right) \circ P^{-1}; \end{aligned}$$

Combining everything, we get

$$\begin{aligned}&( (\nabla (\nabla _A^{\perp } \psi )A) + ( \nabla (\nabla _A^{\perp } \psi )A)^{*})A^{-1} n_0 \\&\quad = \frac{\partial _2 \psi (0,0)}{Q^{2}}(-\partial _1 P_1\circ P^{-1} Q^{2}, \partial _1 P_2\circ P^{-1} Q^{2}) \\&\qquad \Rightarrow (A^{-1}n)^{\perp }( (\nabla (\nabla _A^{\perp } \psi )A) + ( \nabla (\nabla _A^{\perp } \psi )A)^{*})A^{-1} n_0 = \partial _2^{2} \psi (0,0) \end{aligned}$$

Since $\partial _n^{2} \psi (0,0) = \sum _{i,j}(n_0)_i \partial _i \partial _j \psi (n_0)_j = \partial _2^{2} \psi (0,0)$, we are done. $\square$

We now apply Lemma A.6 to $\eta = h(t)(A^{-1}n)^{\perp } - h^{1}(t) (A^{-1}n)^{\perp }$. Equation (70) shows that $\eta (0) = 0$. Then, we can take:

$$\begin{aligned} v^{2} = v^{1} + w, \quad q^{2} = q^{1}. \end{aligned}$$

Since $w(0) = w_t(0) = 0$, we have that $f^2(0)=0$. Also, since $Tr(\nabla w A) = 0$, by (70), $g^{2}(t) = g(t) \; \forall t \in [0,T]$. By construction, one has that $h^{2}(t) (A^{-1}n)^{\perp } = h(t)(A^{-1}n)^{\perp }$ and since $\eta (0)=0$ we have that $h^2(0)=0$.

• Step 3: $h(t) = h^{3}(t)\, \forall t$ without modifying the rest. We want to have $(v^{3},q^{3}) \in X_0$. We have that $L(v^{3},q^{3}) = (f^{3},g^{3},h^{3},0)$ with

  $$\begin{aligned}&f^{3}(0) = 0; \quad g^{3}(t) = g(t); \; \forall t \in [0,T] \nonumber \\&\quad h^{3}(t) = h(t); \; \forall t \in [0,T] . \end{aligned}$$

  (73)

  We first take $v^{3} = v^{2}$ and define $\overline{q}$ by

  $$\begin{aligned} \overline{q}&= h(t) \cdot \frac{A^{-1}n_0}{|A^{-1}n_0|^{2}} - \underbrace{(q^{2}I + (\nabla v^{2}A) + (\nabla v^{2}A)^{*})A^{-1}n_0}_{h^{2}(t)} \\&\quad \cdot \frac{A^{-1}n_0}{|A^{-1}n_0|^{2}} \text { on } \partial \Omega _0 \times [0,T],\\ \overline{q}(x,0)&=0,\quad \text { in } \Omega _0. \end{aligned}$$

  Using once again the parabolic trace, the compatibility condition is satisfied since $\overline{q}(x,0) = 0$ in $\partial \Omega \times [0,T]$ by means of (72). Therefore, we take $q^{3} = q^{2} + \overline{q}$.

  Since $\nabla \overline{q}(x,0) = 0$ we have as before $f^{3}(0) = 0$, and $Tr(\nabla v^{3} A) = 0$ because the velocity was not modified. At the boundary we find:

  $$\begin{aligned}&((q^{3}I + (\nabla v^{3}A) + (\nabla v^{3}A)^{*})A^{-1}n\\&\quad = \overline{q}A^{-1}n + (q^{2}I + (\nabla v^{2}A) + (\nabla v^{2}A)^{*}) A^{-1}n\\ = \overline{q}A^{-1}n + h^{2}(t)&= h(t) \cdot \frac{A^{-1}n}{|A^{-1}n|}\frac{A^{-1}n}{|A^{-1}n|} - h^{2}(t) \cdot \frac{A^{-1}n}{|A^{-1}n|}\frac{A^{-1}n}{|A^{-1}n|} + h^{2}(t)\\&\quad = h(t) \cdot \frac{A^{-1}n}{|A^{-1}n|}\frac{A^{-1}n}{|A^{-1}n|} + h^{2}(t) \cdot \frac{(A^{-1}n)^{\perp }}{|A^{-1}n|}\frac{(A^{-1}n)^{\perp }}{|A^{-1}n|}\\&\quad {\mathop {=}\limits ^{72}} h(t) \cdot \frac{A^{-1}n}{|A^{-1}n|}\frac{A^{-1}n}{|A^{-1}n|} + h(t) \cdot \frac{(A^{-1}n)^{\perp }}{|A^{-1}n|}\frac{(A^{-1}n)^{\perp }}{|A^{-1}n|} = h(t). \end{aligned}$$

  We find $v^{3}(0) = 0$ trivially. By construction, we conclude that

  $$\begin{aligned} \Vert (v^{3},q^{3})\Vert _{X_0} \le C \Vert (f,g,h)\Vert _{Y_0}. \end{aligned}$$

  If we consider the variables $v_{\tilde{f}} = v - v^{3}, q_{\tilde{f}} = q - q^{3}$, we obtain the following problem:

  $$\begin{aligned} \partial _t v_{\tilde{f}} - Q^{2} \Delta v_{\tilde{f}} + A^{*} \nabla q_{\tilde{f}}&= \tilde{f}&\text { in } \Omega \times [0,T] \nonumber \\ Tr(\nabla v_{\tilde{f}} A)&= 0&\text { in } \Omega \times [0,T] \nonumber \\ (q_{\tilde{f}} + (\nabla v_{\tilde{f}} A) + (\nabla v_{\tilde{f}} A)^{*})A^{-1} n&= 0&\text { in } \partial \Omega \times [0,T] \nonumber \\ v_{\tilde{f}}(x,0)&= 0&\text { in } \Omega , \end{aligned}$$

  (74)

  with $\tilde{f}(0) = 0$. Using Theorem A.2, the Theorem is proved.

$\square$

B Proofs of Structural Stability Theorems

B.1 Proof of Proposition 5.3

Part 1:

First we notice that $||\int _{0}^t\tau \psi ||_{F^{s+1}}=\frac{1}{2}\left| \left| t^2\psi \right| \right| _{F^{s+1}}$ thus we need to control $\left| \left| t^2\psi \right| \right| _{L^\infty _{1/4}H^{s+1}}\le C[v_0]T^{1.75}$ and $\left| \left| t^2\psi \right| \right| _{H^{2}_{(0)}H^{\gamma }}\le C[v_0]||t^2||_{H_{(0)}^2}\le C[v_0]$.

In addition, we have that

$$\begin{aligned} f_\phi \equiv -\partial _t \phi +Q^2\Delta \phi -A^*\nabla q_{\phi }=tQ^2\Delta \left( Q^2\Delta v_0-A^*\nabla q_{\phi }\right) \equiv tQ^2\Delta \psi . \end{aligned}$$

Therefore $f_\phi |_{t=0}=0$. By definition (36) of $q_{\phi }$ we have that $A^{-1} n_0\cdot h_{\phi }|_{t=0}=0$ and the tangential component of $h_{\phi }|_{t=0}=0$ by the choice of the initial data. Then $h_\phi |_{t=0}=0$ and in fact $h_\phi =O(t)$ when $t\rightarrow 0$ . Also $g_\phi |=O(t^2)$, when $t\rightarrow 0$ by the choice of the initial data (incompressibility condition). Then we can apply theorem 4.1 to obtain that $\left| \left| L^{-1}(f_\phi ,\overline{g}_\phi ,h_\phi )\right| \right| _{H_{(0)}^{ht,s+1}\times H^{ht,s}_{pr \, (0)}}$ is bounded independently of T for T small if the norms $||f_\phi ||_{H_{(0)}^{ht,s-1}}$, $||g||_{\overline{H}_{(0)}^{ht,s}}$ and $|h|_{H_{(0)}^{ht,s-\frac{1}{2}}}$ are bounded independently of T for T small. Since $||t^n||_{H_{(0)}^{\frac{s-1}{2}}}\le C[n]$, for $n=1,2,...$, we see that $f_\phi$ is under control. Since $||t^n||_{H_{(0)}^{\frac{s+1}{2}}}\le C[n]$, for $n=2,...$, $g_\phi$ is under control. Finally $h_\phi$ is under control because $||t^n||_{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}}\le C[n]$, for $n=1,2,...$.

Therefore N is a number that does not depend on T for T small. Indeed $N\le C[v_0]$.

The definition of $F^{s+1}$ and $v^{(n)}$ imply

$$\begin{aligned}&\left| \left| X^{(n+1)}-\alpha -\int _{0}^tA\phi \right| \right| _{F^{s+1}}\le \left| \left| \int _{0}^tA\circ X^{(n)}w^{(n)}\right| \right| _{F^{s+1}}\\&\qquad +\left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) \phi d\tau \right| \right| _{F^{s+1}}\\&\quad \le \left| \left| \int _{0}^tA\circ X^{(n)}w^{(n)}\right| \right| _{F^{s+1}}+\left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) v_0 d\tau \right| \right| _{F^{s+1}}\\&\qquad +\left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) \tau \psi d\tau \right| \right| _{F^{s+1}}\\&\quad \le \left| \left| \int _{0}^tA\circ X^{(n)}w^{(n)}\right| \right| _{L^\infty _{1/4}H^{s+1}}+\left| \left| \int _{0}^tA\circ X^{(n)}w^{(n)}\right| \right| _{H^{2}_{(0)}H^{\gamma }}\\&\qquad + \left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) v_0 d\tau \right| \right| _{L^\infty _{1/4}H^{s+1}}+\left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) v_0 d\tau \right| \right| _{H^{2}_{(0)}H^{\gamma }}\\&\qquad + \left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) \tau \psi d\tau \right| \right| _{L^\infty _{1/4}H^{s+1}}+\left| \left| \int _{0}^t\left( A\circ X^{(n)}- A\right) \tau \psi d\tau \right| \right| _{H^{2}_{(0)}H^{\gamma }}\\&\qquad \equiv I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_6. \end{aligned}$$

To estimate $I_1$ we proceed as follows. Applying Hölder and Minkowski inequalities yields

$$\begin{aligned} \left| \left| \int _{0}^t A\circ X^{(n)}w^{(n)}d\tau \right| \right| _{L^{\infty }_{1/4}H^{s+1}}&\le T^\frac{1}{4} \left| \left| A\circ X^{(n)}w^{(n)}\right| \right| _{L^2H^{s+1}}\\&\le T^\frac{1}{4}\left| \left| A\circ X^{(n)}\right| \right| _{L^\infty H^{s+1}}\left| \left| w^{(n)}\right| \right| _{H^{0}_{(0)}H^{s+1}}, \end{aligned}$$

In order to bound $\left| \left| A\circ X^{(n)}\right| \right| _{L^\infty H^{s+1}}$ we will use Lemma 3.10. Therefore

$$\begin{aligned} I_1\le T^\frac{1}{4}C[v_0]\left| \left| w^{(n)}\right| \right| _{L^2H^{s+1}}\le T^\frac{1}{4}C[v_0]||w^{(n)}||_{H_{(0)}^{ht,s+1}}\le T^\frac{1}{4}C[v_0]. \end{aligned}$$

For $I_2$ we have that

$$\begin{aligned} I_2&\le \left| \left| \int _{0}^t A\circ X^{(n)}w^{(n)}\right| \right| _{H^{2}_{(0)}H^{\gamma }}\le C \left| \left| A\circ X^{(n)}w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\le C\left| \left| \left( A\circ X^{(n)}-A\right) w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }} +\left| \left| Aw^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\le C \left( \left| \left| A\circ X^{(n)}-A\right| \right| _{H^{1}_{(0)}H^{\gamma }}\left| \left| w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}+\left| \left| Aw^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\right) \\ \end{aligned}$$

By lemma 3.11, we have that $I_2\le C[v_0]\left| \left| w\right| \right| _{H^{1}_{(0)}H^{\gamma }}$. In addition

$$\begin{aligned} \left| \left| w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}&\le \left| \left| \int _{0}^t \partial _t w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\le C\left| \left| \int _{0}^t \partial _t w^{(n)}\right| \right| _{H^{1+\delta -\varepsilon }_{(0)}H^{\gamma }} \\&\le CT^\varepsilon \left| \left| w^{(n)}\right| \right| _{H^{1+\delta }_{(0)}H^{\gamma }}\le CT^\varepsilon C[v_0]. \end{aligned}$$

Here $\varepsilon >0$, $\varepsilon <\delta$, $\delta < \frac{s-1-\gamma }{2}$ and we have used Lemmas 3.3 and 3.4.

For $I_3$ we have that

$$\begin{aligned}&\left| \left| \int _{0}^t \left( A\circ X-A\right) v_0d\tau \right| \right| _{L^\infty _{1/4}H^{1+s}}= \sup _{t\in [0,T]}\frac{1}{t^\frac{1}{4}}\left| \left| \int _{0}^t\left( A\circ X^{(n)}-A\right) v_0d\tau \right| \right| _{H^{s+1}}\\&\quad \le C[v_0]T^\frac{3}{4}\left| \left| A\circ X-A\right| \right| _{L^\infty H^{1+s}}\le C[v_0]T^\frac{3}{4} \end{aligned}$$

by lemma 3.10.

For $I_4$ by applying the second part of Lemma 3.4 with $\varepsilon =0$ we have that

$$\begin{aligned} I_4\le \left| \left| (A\circ X^{(n)}-A)v_0 \right| \right| _{H^{1}_{(0)}H^{\gamma }}\le C[v_0] \left| \left| (A\circ X^{(n)}-A) \right| \right| _{H^{1}_{(0)}H^{\gamma }} \end{aligned}$$

Now we apply Lemma 3.11 and we obtain that, for small enough T, $I_4\le C[v_0]||X^{(n)}-\alpha ||_{H_{(0)}^1H^\gamma }$. In addition

$$\begin{aligned} \left| \left| X^{(n)}-\alpha \right| \right| _{H^{1}_{(0)}H^{\gamma }}\le \left| \left| X^{(n)}-\alpha -Av_0t\right| \right| _{H^{1}_{(0)}H^{\gamma }}+\left| \left| Av_0t\right| \right| _{H^{1}_{(0)}H^{\gamma }} \end{aligned}$$

with $\left| \left| Av_0t\right| \right| _{H^{1}_{(0)}H^{\gamma }}\le C[v_0]||t||_{H_{(0)}^1}\le C[v_0]T^\frac{1}{2}$. Also

$$\begin{aligned}&\left| \left| X^{(n)}-\alpha -Av_0t\right| \right| _{H^{1}_{(0)}H^{\gamma }}=\left| \left| \int _{0}^t \partial _t(X^{(n)}-\alpha -Av_0t)\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\quad \le \left| \left| \int _{0}^t \partial _t(X^{(n)}-\alpha -Av_0t)\right| \right| _{H^{1+\delta -\varepsilon }_{(0)}H^{\gamma }}\\&\quad \le C T^\varepsilon \left| \left| X^{(n)}-\alpha -Av_0t\right| \right| _{H^{1+\delta }_{(0)}H^{\gamma }}\le C T^\varepsilon \left| \left| X^{(n)}-\alpha -Av_0t\right| \right| _{H^{1+\delta }_{(0)}H^{\gamma }}\le CT^\varepsilon C[v_0]. \end{aligned}$$

where we have applied Lemma 3.4, for $\varepsilon >0$ and $\varepsilon<\delta <\frac{1}{2}$. This concludes the estimate for $I_4$. The estimates for $I_5$ and $I_6$ follow in a similar way. Therefore

$$\begin{aligned} \left| \left| X^{(n+1)}-\alpha -\int _{0}^tA\phi \right| \right| _{F^{s+1}}\le C[v_0]T^\varepsilon \end{aligned}$$

This concludes the proof of part 1 of proposition 5.3.

Part 2:

From (26) we have that

$$\begin{aligned} \left| \left| X^{(n+1)}-X^{(n)}\right| \right| _{F^{s+1}}&=\left| \left| \int _0^t\left( A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}\right) d\tau \right| \right| _{F^{s+1}}\\&\le \left| \left| \int _0^t\left( A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}\right) d\tau \right| \right| _{L^\infty _{1/4}H^{s+1}}\\&\quad + \left| \left| \int _0^t\left( A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}\right) d\tau \right| \right| _{H^{2}_{(0)}H^{\gamma }}\\&\equiv I_1 +I_2 \end{aligned}$$

In order to bound $I_1$ we notice that

$$\begin{aligned}&\left| \left| \int _0^t\left( A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}\right) d\tau \right| \right| _{H^{s+1}}\\&\quad \le t^\frac{1}{2}\left| \left| A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}d\tau \right| \right| _{L^2H^{s+1}} \end{aligned}$$

Therefore

$$\begin{aligned} I_1&\le T^\frac{1}{4}\left| \left| \left( A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}\right) \right| \right| _{L^2H^{s+1}}\\&\le T^\frac{1}{4}\left| \left| \left( A\circ X^{(n)}-A\circ X^{(n-1)}\right) v^{(n)}\right| \right| _{L^2H^{s+1}}\\&\quad + T^\frac{1}{4}\left| \left| \left( v^{(n)}-v^{(n-1)}\right) A\circ X^{(n-1)}\right| \right| _{L^2H^{s+1}} \\&\equiv I_{11} + I_{12} \end{aligned}$$

In addition

$$\begin{aligned} I_{11}\le T^\frac{1}{4}\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{L^\infty H^{s+1}}\left| \left| v^{(n)}\right| \right| _{L^2H^{s+1}}. \end{aligned}$$

Here we notice that $\left| \left| v^{(n)}\right| \right| _{L^2H^{s+1}}\le \left| \left| w^{(n)}\right| \right| _{L^2H^{s+1}}+\left| \left| v_0\right| \right| _{L^2H^{s+1}}+\left| \left| t\psi \right| \right| _{L^2H^{s+1}}\le C[v_0]$. Then by applying Lemma 3.12 we have that $I_{11}\le C[v_0]T^\frac{1}{2}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}$

Also we have, by applying Lemma 3.10, that

$$\begin{aligned} I_{12}\le C[v_0]T^\frac{1}{4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{L^2H^{s+1}} \end{aligned}$$

Thus

$$\begin{aligned} I_1\le C[v_0]T^\frac{1}{4}\left( \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty _{1/4}H^{s+1}}+\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\right) . \end{aligned}$$

It remains to bound $I_2$.

$$\begin{aligned} I_2=\left| \left| \int _{0}^t (A\circ X^{(n)}v^{(n)}- A\circ X^{(n-1)}v^{(n-1)})d\tau \right| \right| _{H^{2}_{(0)}H^{\gamma }} \end{aligned}$$

and applying Lemma 3.4 with $\varepsilon =0$ we have that

$$\begin{aligned} I_2\le \left| \left| A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}. \end{aligned}$$

We will decompose that term in the following way

$$\begin{aligned} A\circ X^{(n)}v^{(n)}-A\circ X^{(n-1)}v^{(n-1)}=&(A\circ X^{(n)}-A\circ X^{(n-1)})w^{(n)}\\&+ (A\circ X^{(n)}-A\circ X^{(n-1)})\phi \\&+ A\circ X^{(n-1)}(w^{(n)}-w^{(n-1)}). \end{aligned}$$

Thus

$$\begin{aligned} I_2\le I_{21}+I_{22}+I_{23} \end{aligned}$$

with

$$\begin{aligned} I_{21}=&\left| \left| (A\circ X^{(n)}-A\circ X^{(n-1)})w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\ I_{22}=&\left| \left| (A\circ X^{(n)}-A\circ X^{(n-1)})\phi \right| \right| _{H^{1}_{(0)}H^{\gamma }}\\ I_{23}=&\left| \left| A\circ X^{(n-1)}(w^{(n)}-w^{(n-1)})\right| \right| _{H^{1}_{(0)}H^{\gamma }}. \end{aligned}$$

For $I_{21}$, by applying Lemma 3.7 ($1<\gamma <s-1$), we have that

$$\begin{aligned}&\left| \left| (A\circ X^{(n)}-A\circ X^{(n-1)})w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\quad \le C\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\left| \left| w^{(n)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\quad \le C[v_0]\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}, \end{aligned}$$

because of Lemma 3.12. Then

$$\begin{aligned} I_{21}\le&C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\le \left| \left| \int _0^t(\partial _t\left( X^{(n)}-X^{(n-1)}\right) d\tau )\right| \right| _{H^{1+\delta -\varepsilon }_{(0)}H^{\gamma }}\\&\le C[v_0]T^\varepsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{1+\delta }_{(0)}H^{\gamma }}, \end{aligned}$$

where we have applied Lemma 3.4 with $\varepsilon >0$, $\delta >\varepsilon$ and $\delta <\frac{1}{2}$.

The term $I_{22}$ is quite similar to $I_{21}$. The only difference is that we can not take $\left| \left| \phi \right| \right| _{H^{1}_{(0)}H^{\gamma }}$. Instead of that,

$$\begin{aligned}&\left| \left| (A\circ X^{(n)}- A\circ X^{(n-1)})\phi \right| \right| _{H^{1}_{(0)}H^{\gamma }}\le C[v_0]||v_0||_{H^\gamma }\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\quad +C[v_0]\left| \left| t\psi \right| \right| _{H^{1}_{(0)}H^{\gamma }}\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}. \end{aligned}$$

Finally

$$\begin{aligned} I_{23}&\le C\left( \left| \left| A\circ X^{(n-1)}-A\right| \right| _{H^{1}_{(0)}H^{\gamma }}+1\right) \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}\\&\le C[v_0]\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H^{1}_{(0)}H^{\gamma }}, \end{aligned}$$

by applying Lemma 3.10. Also we can compute

$$\begin{aligned} I_{23}\le&C[v_0] \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H^{1+\delta -\varepsilon }_{(0)}H^{\gamma }}\le C[v_0]T^\varepsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H^{1+\delta }_{(0)}H^{\gamma }}\\ \le&C[v_0]T^\varepsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

for $\varepsilon >0$, $\delta >\varepsilon$ and $\delta <\frac{s-1-\gamma }{2}$.

This concludes the proof of part 2 of proposition 5.3. Therefore proposition 5.3 is shown.

B.2 Proof of Proposition 5.4

Proof of Part 1:

We split the proof in three parts, corresponding with the functions $f^{(n)}$, $\overline{g}^{(n)}$ and $h^{(n)}$:

P1.1. Estimate for $f^{(n)}$:

In this section we have to deal with $f^{(n)}$ to estimate

$$\begin{aligned} \left| \left| f^{(n)}\right| \right| _{H_{(0)}^{ht,s-1}}=\left| \left| f^{(n)}\right| \right| _{L^2H^{s-1}}+\left| \left| f^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}. \end{aligned}$$

We will gather terms by writing $f^{(n)}=f^{(n)}_w+f^{(n)}_\phi +f^{(n)}_{q}$ as follows,

$$\begin{aligned} f^{(n)}_w= & {} Q^2\circ X^{(n)}\zeta ^{(n)}\partial \left( \zeta ^{(n)}\partial w^{(n)}\right) -Q^2\Delta w^{(n)},\nonumber \\ f^{(n)}_\phi= & {} Q^2\circ X^{(n)}\zeta ^{(n)}\partial \left( \zeta ^{(n)}\partial \phi \right) -Q^2\Delta \phi , \end{aligned}$$

(75)

and

$$\begin{aligned} f^{(n)}_{q}=-A\circ X^{(n)}\zeta ^{(n)}\partial q^{(n)}+A\partial q^{(n)}. \end{aligned}$$

Above we removed subscripts to alleviate notation. Next we also ignore the superscripts for the same reason. We firstly bound $\left| \left| \cdot \right| \right| _{L^2H^{s-1}}$:

$$\begin{aligned} \left| \left| f_w\right| \right| _{L^2H^{s-1}}\le&\left| \left| (Q^2\circ X-Q^2)\zeta \partial (\zeta \partial w)\right| \right| _{L^2H^{s-1}}+\left| \left| Q^2(\zeta -\mathbb {I})\partial (\zeta \partial w)\right| \right| _{L^2H^{s-1}}\\&+\left| \left| Q^2\partial ((\zeta -\mathbb {I})\partial w)\right| \right| _{L^2H^{s-1}}\equiv I_{1}+I_{2}+I_{3}. \end{aligned}$$

We can deal with these three terms as follows. For $I_1$ one can get

$$\begin{aligned} I_{1}\le&\left| \left| (Q^2\circ X -Q^2)\zeta \partial \zeta \partial w\right| \right| _{L^2H^{s-1}}+\left| \left| (Q^2\circ X-Q^2)\zeta \zeta \partial ^2 w\right| \right| _{L^2H^{s-1}}\\ \le&\left| \left| Q^2\circ X-Q^2\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta \right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta \right| \right| _{L^\infty H^{s-1}}\left| \left| \partial w\right| \right| _{L^2H^{s-1}}\\&+ \left| \left| Q^2\circ X-Q^2\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta \right| \right| _{L^\infty H^{s-1}}^2\left| \left| \partial ^2w\right| \right| _{L^2H^{s-1}}, \end{aligned}$$

By applying Lemma 3.16 we have that $\left| \left| Q^2\circ X-Q^2\right| \right| _{L^\infty H^{s-1}}\le C[v_0]\left| \left| X-\alpha \right| \right| _{L^\infty H^{s-1}}$ and by applying Lemma 3.13, $\left| \left| \zeta \right| \right| _{L^\infty H^{s-1}}$ and $\left| \left| \zeta \right| \right| _{L^\infty H^{s-1}}\le C[v_0]$.

$$\begin{aligned} I_{1}&\le C\left| \left| Q^2\circ X-Q^2\right| \right| _{L^\infty H^{s-1}}\left| \left| X\right| \right| _{L^\infty H^{s+1}}^2\left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}\\&\le C[v_0]\left| \left| X-\alpha \right| \right| _{L^\infty H^{s-1}}\left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}} \le C[v_0]T^\frac{1}{4}. \end{aligned}$$

Also, by Lemma 3.13,

$$\begin{aligned} I_{2}\le&\left| \left| Q^2(\zeta -\mathbb {I})\partial \zeta \partial w\right| \right| _{L^2H^{s-1}}+\left| \left| Q^2(\zeta -\mathbb {I})\zeta \partial ^2 w\right| \right| _{L^2H^{s-1}}\\ \le&\,C\left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s-1}}(\left| \left| \partial \zeta \right| \right| _{L^\infty H^{s-1}}\left| \left| \partial w\right| \right| _{L^2H^{s-1}}+\left| \left| \zeta \right| \right| _{L^\infty H^{s-1}}\left| \left| \partial ^2 w\right| \right| _{L^2H^{s-1}})\\ \le \,&C[v_0] \left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s-1}}\left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

The identity $\zeta -\mathbb {I}=\zeta \left( \mathbb {I}-\nabla X\right) =\zeta \nabla \left( \alpha -X\right)$, together with Lemma 3.13, allows us to get

$$\begin{aligned} \left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s}}&\le \left| \left| \zeta \right| \right| _{L^\infty H^{s}}\left| \left| X-\alpha \right| \right| _{L^\infty H^{s+1}}\nonumber \\&\le C[v_0]\left( \left| \left| X-\alpha -Av_0t\right| \right| _{L^\infty H^{s+1}}+T||Av_0||_{s+1}\right) \le C[v_0]T^\frac{1}{4}, \end{aligned}$$

(76)

Thus

$$\begin{aligned} I_{2}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

It remains $I_{3}$ for which we compute

$$\begin{aligned} I_{3}=&\left| \left| Q^2\partial ((\zeta -\mathbb {I})\partial w)\right| \right| _{L^2H^{s-1}}\le \left| \left| Q^2\partial \zeta \partial w\right| \right| _{L^2H^{s-1}}+\left| \left| Q^2(\zeta -\mathbb {I})\partial ^2 w\right| \right| _{L^2H^{s-1}}\\ \le \,&C[v_0](\left| \left| \partial \zeta \right| \right| _{L^\infty H^{s-1}}\left| \left| w\right| \right| _{L^2H^{s}}+\left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s-1}}\left| \left| w\right| \right| _{L^2H^{s+1}})\\ \le \,&C[v_0]\left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

We are done with $\left| \left| f_w\right| \right| _{L^2H^{s-1}}$. At this point it is easy to check that an analogous procedure yields

$$\begin{aligned} \left| \left| f_\phi \right| \right| _{L^2H^{s-1}}\le C[v_0]T^\frac{1}{4}\left| \left| \phi \right| \right| _{L^2H^{s+1}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

We notice that in the bound of $\left| \left| f_w\right| \right| _{L^2H^{s-1}}$ the only bound we really need on w is for $\left| \left| w\right| \right| _{L^2H^{s+1}}$. In addition $\left| \left| \phi \right| \right| _{L^2H^{s+1}}\le C[v_0]$.

Next we deal with $\left| \left| f_{q}\right| \right| _{L^2H^{s-1}}$. We separate into two terms, $\left| \left| f_{q_w}+f_{q_\phi }\right| \right| _{L^2H^{s-1}}\le \left| \left| f_{q_w}\right| \right| _{L^2H^{s-1}}+\left| \left| f_{q_\phi }\right| \right| _{L^2H^{s-1}}$Indeed, by lemmas 3.10, 3.13 and expression (76) he have that

$$\begin{aligned} \left| \left| f_{q_w}\right| \right| _{L^2H^{s-1}}\le&\left| \left| A\circ X(\zeta -\mathbb {I})\partial q_w\right| \right| _{L^2H^{s-1}}+\left| \left| (A-A\circ X)\partial q_w\right| \right| _{L^2H^{s-1}}\\ \le&(\left| \left| A\circ X\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s-1}}\\&+\left| \left| A\circ X-A\right| \right| _{L^\infty H^{s-1}})\left| \left| \partial q_w\right| \right| _{L^2H^{s-1}}\\ \le&C[v_0]T^\frac{1}{4}||q_w||_{H^{ht,{s}}_{pr\, (0)}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

Here we notice that in the previous inequality the only bound we really need on $q_w$ is for the norm $\left| \left| \partial q_w\right| \right| _{L^2H^{s-1}}$. Therefore we have that

$$\begin{aligned} \left| \left| f_{q_\phi }\right| \right| _{L^2H^{s-1}}\le C[v_0]T^\frac{1}{4}\left| \left| \partial q_\phi \right| \right| _{L^2H^{s-1}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

Thus we are done with $\left| \left| f\right| \right| _{L^2H^{s-1}}$.

Using the same splitting we now deal with $\left| \left| f\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}$:

$$\begin{aligned} \left| \left| f_w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le&\left| \left| (Q^2\circ X-Q^2)\zeta \partial (\zeta \partial w)\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}+\left| \left| Q^2(\zeta -\mathbb {I})\partial (\zeta \partial w)\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&+\left| \left| Q^2\partial ((\zeta -\mathbb {I})\partial w)\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\equiv I_{4}+I_{5}+I_{6}. \end{aligned}$$

We need to split

$$\begin{aligned}&(Q^2\circ X-Q^2)\zeta \partial \left( \zeta \partial w\right) \\&\quad =(Q^2\circ X-Q^2)(\zeta -\mathbb {I})\partial \left( \zeta \partial w\right) \\&\qquad + (Q^2\circ X-Q^2)\partial \left( (\zeta -\mathbb {I})\partial w\right) +(Q^2\circ X-Q^2)\partial ^2 w\\&\quad =(Q^2\circ X-Q^2)(\zeta -\mathbb {I})\partial \left( (\zeta -\mathbb {I})\partial w\right) +(Q^2\circ X-Q^2)(\zeta -\mathbb {I})\partial ^2 w\\&\qquad + (Q^2\circ X-Q^2)\partial \left( (\zeta -\mathbb {I})\partial w\right) \\&\qquad +(Q^2\circ X-Q^2)\partial ^2 w. \end{aligned}$$

We bound $I_{4}$ by using lemma 3.7 as follows,

$$\begin{aligned} I_4&\le \left| \left| Q^2\circ X-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left( 1+\left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\right) \\&\quad \left( \left| \left| \partial \left( (\zeta -\mathbb {I})\partial w\right) \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}+\left| \left| \partial ^2 w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\right) . \end{aligned}$$

Using lemma 3.6, with $\frac{1}{q}=\varepsilon$, yields

$$\begin{aligned}&\left| \left| \partial (\zeta -\mathbb {I})\partial w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le \left| \left| \partial (\zeta -\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\varepsilon }}\left| \left| \partial w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\varepsilon }}\\&\quad \le C[v_0]\left| \left| X-\alpha \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\varepsilon }}\left| \left| w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2-\varepsilon }}\\&\quad \le C[v_0]\left( \left| \left| X-\alpha -Av_0t\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\varepsilon }}+\left| \left| tv_0\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\varepsilon }}\right) \le C[v_0], \end{aligned}$$

for $\varepsilon >0$ and small enough. Also

$$\begin{aligned} \left| \left| (\zeta -\mathbb {I})\partial ^2 w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le \left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| \partial ^2w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le C[v_0]. \end{aligned}$$

Finally,

$$\begin{aligned} \left| \left| Q^2\circ X-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\le C[v_0] \left( \left| \left| X-\alpha -Av_0t\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}+\left| \left| tAv_0\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\right) \end{aligned}$$

(77)

and we have that $||t||_{H_{(0)}^\frac{s-1}{2}}\le C\sqrt{T}$. Also,

$$\begin{aligned}&\left| \left| \partial _t\!\!\! \int _{0}^t\!\!(X-\alpha -Av_0t)\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\&\quad \le C[v_0]\left| \left| \int _{0}^t (X-\alpha -Av_0t)\right| \right| _{H^{\frac{s-1}{2}+\varepsilon +1-\varepsilon }_{(0)}H^{1+\delta }} \le C[v_0]T^\varepsilon \left| \left| X-\alpha -Av_0t\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }}\\&\quad \le C[v_0]T^\varepsilon \left| \left| X-\alpha -Av_0t\right| \right| _{F^{s+1}}\le C[v_0]T^\varepsilon , \end{aligned}$$

where we have used lemma 3.4, with $\frac{1}{2}<\frac{s-1}{2}+\varepsilon <1$ and Lemma 3.8 . With this last inequality we conclude the estimate for $I_{4}$.

For $I_{5}$ we have to make the following splitting $Q^2(\zeta -\mathbb {I})\partial (\zeta \partial w)=Q^2(\zeta -\mathbb {I})\partial ((\zeta -\mathbb {I})\partial w)+Q^2(\zeta -\mathbb {I})\partial ^2 w$ and then

$$\begin{aligned} I_{5}\le ||Q^2||_{H^{1+\delta }}\left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left( \left| \left| \partial \left( (\zeta -\mathbb {I})\partial w\right) \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}+\left| \left| \partial ^2w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\right) . \end{aligned}$$

The terms inside of the parenthesis in the previous expression have already been estimated in $I_{4}$. Using lemmas 3.8 and 3.4 we find that

$$\begin{aligned} \left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}&\le C[v_0]T^\varepsilon \left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }}\le C(M)T^\varepsilon \left| \left| X-\alpha \right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{2+\delta }}, \end{aligned}$$

by Lemma 3.13. Proceeding as for $I_4$ we have that $I_5\le C[v_0]T^\varepsilon$. We are done with $I_5$. For $I_{6}$ we have that

$$\begin{aligned} I_{6}\le \left| \left| Q^2\partial (\zeta -\mathbb {I})\partial w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2} +\left| \left| Q^2(\zeta -\mathbb {I})\partial ^2 w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}. \end{aligned}$$

Both terms can be handled as before. In fact

$$\begin{aligned} \left| \left| Q^2(\zeta -\mathbb {I})\partial ^2 w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le&||Q^2||_{H^{1+\delta }}\left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| \partial ^2w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\ \le&C[v_0] T^\varepsilon \left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}\le C[v_0] T^\varepsilon , \end{aligned}$$

and

$$\begin{aligned} \left| \left| Q^2\partial (\zeta -\mathbb {I})\partial w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le ||Q^2||_{H^{1+\delta }}\left| \left| \partial (\zeta -\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\varepsilon }}\left| \left| \partial w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\varepsilon }}\\&\le C \left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2-\varepsilon }}\le C[v_0] T^\varepsilon \left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}\\&\le C[v_0] T^\varepsilon , \end{aligned}$$

for $\varepsilon >0$ and small enough. We are done with $I_6$ and therefore with $\left| \left| f_w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}$. A similar procedure allows us to get

$$\begin{aligned} \left| \left| f_\phi \right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le C[v_0]T^\epsilon . \end{aligned}$$

Here we remark the main differences to get it. We need split $\phi =v_0+t\psi$. For the terms coming from $v_0$ we do not find any problem because $v_0$ does not depend on time, and for the terms coming from $t\psi$ we can use that $\psi$ does not depend on t and $||t||_{H_{(0)}^\frac{s-1}{2}}\le C\sqrt{T}$.

The estimate of $\left| \left| f_{q}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}$ is obtained as follows. First we split $f_q=f_{q_w}+f_{q_\phi }$, we can write

$$\begin{aligned} f_{q_w}=\left( -A\circ X+A\right) \left( \zeta -\mathbb {I}\right) \partial q_{w}+(-A\circ X+A)\partial q_w-A(\zeta -\mathbb {I})\partial q_w. \end{aligned}$$

Then we can estimate

$$\begin{aligned}&\left| \left| f_{q_w}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\quad \le \left| \left| (A\circ X-A)(\zeta -\mathbb {I})\partial q_w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}+\left| \left| (A\circ X-A)\partial q_w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\\&\qquad +\left| \left| A(\zeta -\mathbb {I})\partial q_w\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\\&\quad \le C[v_0]\left( \left| \left| (A\circ X-A)\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\left| \left| (\zeta -\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}+\left| \left| (A\circ X-A)\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}+\left| \left| (\zeta -\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\right) \\&\quad \le C[v_0]T^\varepsilon . \end{aligned}$$

For $f_{q_\phi }$ we have that

$$\begin{aligned} f_{q_\phi }=\left( -A\circ X+A\right) \left( \zeta -\mathbb {I}\right) \partial q_{\phi }+(-A\circ X+A)\partial q_\phi -A(\zeta -\mathbb {I})\partial q_\phi . \end{aligned}$$

Here we can not take $\left| \left| \nabla q_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}$ because $\nabla q_\phi |_{t=0}\ne 0$. Fortunately we do not need it since $q_\phi$ does not depend on t. Similarly to $\left| \left| f_{q_w}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}$ we have that $\left| \left| f_{q_\phi }\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\le C[v_0]T^\varepsilon$. This finishes the bounds for $\left| \left| f\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}$ and therefore we are done with $\left| \left| f\right| \right| _{H_{(0)}^{ht,s-1}}$.

P1.2. Estimate for $\overline{g}^{(n)}$:

We recall that

$$\begin{aligned} \left| \left| \overline{g}^{(n)}\right| \right| _{\overline{H}_{(0)}^{ht,s}}=\left| \left| \overline{g}^{(n)}\right| \right| _{L^2H^{s}}+\left| \left| \overline{g}^{(n)}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}. \end{aligned}$$

We will first estimate the $H^0H^s$-norm and after that the $H^{\frac{s+1}{2}}H^{-1}$-norm. We will split $\overline{g}^{(n)}$ in the following terms,

$$\begin{aligned} \overline{g}^{(n)}&=-Tr\left( \nabla v^{(n)}\zeta ^{(n)}A\circ X^{(n)}\right) +Tr\left( \nabla v^{(n)}A\right) \\&\quad +Tr\left( \nabla \phi \zeta _\phi A_\phi \right) -Tr\left( \nabla \phi A\right) \\&=-Tr\left( \nabla w^{(n)}\zeta ^{(n)}A\circ X^{(n)}\right) -Tr\left( \nabla \phi (\zeta ^{(n)}A\circ X^{(n)}-\zeta _\phi A_\phi )\right) \\&\quad + Tr\left( \nabla w^{(n)}A\right) +Tr\left( \nabla \phi \zeta _\phi A_\phi \right) \\&=-Tr\left( \nabla w^{(n)}\left( \zeta ^{(n)}-\mathbb {I}\right) A\circ X^{(n)}\right) -Tr\left( \nabla \phi \left( \zeta ^{(n)}-\zeta _\phi \right) A\circ X^{(n)}\right) \\&\quad + Tr\left( \nabla w^{(n)}\left( A-A\circ X^{(n)}\right) \right) +Tr\left( \nabla \phi \zeta _\phi \left( A_\phi -A\circ X^{(n)}\right) \right) , \end{aligned}$$

From the partition of $\overline{g}^{(n)}$ we have that

$$\begin{aligned} \left| \left| \overline{g}^{(n)}\right| \right| _{L^2H^{s}}&=\left| \left| \nabla w^{(n)}\left( \zeta ^{(n)}-\mathbb {I}\right) A\circ X^{(n)}\right| \right| _{L^2H^{s}}+\left| \left| \nabla \phi \left( \zeta ^{(n)}-\zeta _\phi \right) A\circ X^{(n)}\right| \right| _{L^2H^{s}}\\&\quad +\left| \left| \nabla w^{(n)}\left( A-A\circ X^{(n)}\right) \right| \right| _{L^2H^{s}}+\left| \left| \nabla \phi \left( A_\phi -A\circ X^{(n)}\right) \right| \right| _{L^2H^{s}}\\&\equiv I_1+I_2+I_3+I_4. \end{aligned}$$

Since $H^s$ is an algebra for $s>1$ as stated in Lemma 3.5 we have that

$$\begin{aligned} I_1\le \left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| A\circ X^{(n)}\right| \right| _{L^\infty H^{s}}\left| \left| \nabla w^{(n)}\right| \right| _{L^2H^{s}}. \end{aligned}$$

Therefore, applying lemma 3.10 yields

$$\begin{aligned} I_1\le C[v_0]\left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| w^{(n)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

and using (76) we get $I_1\le C[v_0]\left| \left| w^{(n)}\right| \right| _{H_{(0)}^{ht,s+1}}T^\frac{1}{4}$. Similarly we obtain that $I_2\le C[v_0]T^\frac{1}{4}$. In addition, by using lemma 3.10 it can be checked that $I_3\le C[v_0]T^\frac{1}{4}$, since $\left| \left| \nabla \phi \right| \right| _{L^2H^{s}}\le C[v_0]$. And by lemma 3.10 $I_4\le C[v_0]T^\frac{1}{4}$.

Then we have proved that $\left| \left| \overline{g}^{(n)}\right| \right| _{L^2H^{s}}\le C[v_0] T^\frac{1}{4}$.

To estimate the $H^{\frac{s+1}{2}}H^{-1}$-norm of $\overline{g}^{(n)}$, we split in a different way

$$\begin{aligned} \overline{g}^{(n)}&= -Tr\left( \nabla w^{(n)}\zeta ^{(n)}A\circ X^{(n)}\right) + Tr\left( \nabla w^{(n)}A\right) \nonumber \\&\quad -Tr\left( \nabla \phi (\zeta ^{(n)}-\zeta _\phi ) A\circ X^{(n)}\right) +Tr\left( \nabla \phi \zeta _\phi (A_\phi -A\circ X^{(n)})\right) \nonumber \\&\equiv -\overline{g}_w^{(n)}-\overline{g}_\phi ^{(n)}, \end{aligned}$$

(78)

where

$$\begin{aligned} \overline{g}^{(n)}_w =\,&Tr\left( \nabla w^{(n)}\zeta ^{(n)}A\circ X^{(n)}\right) - Tr\left( \nabla w^{(n)}A\right) \\&= Tr\left( \nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi )\right) + Tr\left( \nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi ) A_\phi \right) \\&\quad +Tr\left( \nabla w^{(n)}\zeta _\phi (A\circ X^{(n)}-A_\phi )\right) + Tr\left( \nabla w^{(n)}(\zeta _\phi -\mathbb {I}) A_\phi \right) \\&\quad +Tr\left( \nabla w^{(n)}(A_\phi -A)\right) \end{aligned}$$

and

$$\begin{aligned} \overline{g}_\phi ^{(n)}&=Tr\left( \nabla \phi (\zeta ^{(n)}-\zeta _\phi ) A\circ X^{(n)}\right) -Tr\left( \nabla \phi \zeta _\phi (A_\phi -A\circ X^{(n)})\right) \\&=Tr\left( \nabla \phi (\zeta ^{(n)}-\zeta _\phi ) (A\circ X^{(n)}-A_\phi )\right) -Tr\left( \nabla \phi \zeta _\phi (A_\phi -A\circ X^{(n)})\right) \\&\quad +Tr\left( \nabla \phi \zeta _\phi A_\phi \right) . \end{aligned}$$

First we will bound $\overline{g}_{w}^{(n)}$ and then we will do the same with $\overline{g}_\phi ^{(n)}$. Taking $H^{\frac{s+1}{2}}H^{-1}$-norms yields,

$$\begin{aligned}&\left| \left| \overline{g}^{(n)}_w\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\le \left| \left| \nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )\left( A\circ X^{(n)}-A_\phi \right) \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\nonumber \\&\qquad +\left| \left| \nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )A_\phi \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\nonumber \\&\qquad +\left| \left| \nabla w^{(n)}\zeta _\phi (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\nonumber \\&\qquad +\left| \left| \nabla w^{(n)}(A_\phi -A)+\nabla w^{(n)}(\zeta _\phi -\mathbb {I}) A_\phi \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\nonumber \\&\quad =I_1+I_2+I_3+I_4. \end{aligned}$$

(79)

For $I_1$ we have that

$$\begin{aligned} I_1&=\left| \left| \nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&=\left| \left| \int _{0}^t\partial _t\left( \nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi )\right) \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\le \left| \left| \int _{0}^t\partial _t\nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi ) \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\quad +\left| \left| \int _{0}^t\nabla w^{(n)}\partial _t(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\quad +\left| \left| \int _{0}^t\nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )\partial _t (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\equiv I_{11}+I_{12}+I_{13}. \end{aligned}$$

And we bound $I_{11}$, $I_{12}$ and $I_{13}$ as follows

$$\begin{aligned} I_{11}\le \left| \left| \partial _t\nabla w^{(n)}(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}, \end{aligned}$$

since $0<\frac{s-1}{2}<1$ and we can apply Lemma 3.4 with $\varepsilon =0$. Moreover, applying Lemma 3.7 we obtain that

$$\begin{aligned} I_{11}\le C[v_0]\left| \left| \nabla \partial _tw^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}. \end{aligned}$$

In addition

$$\begin{aligned}&\left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\le \left| \left| \partial _t\int _{0}^t\zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\&\quad \le C\left| \left| \int _{0}^t\zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}+\varepsilon +1-\varepsilon }_{(0)}H^{1+\delta }}\le C T^\varepsilon \left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }} \end{aligned}$$

by Lemma 3.4, for $0<\frac{s-1}{2}+\varepsilon <1$. In addition, by Lemma 3.14,

$$\begin{aligned} \left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }}\le C[v_0]\left| \left| X^{(n)}-\alpha -Av_0t\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{2+\delta }}. \end{aligned}$$

Then Lemmas 3.17 and 3.8 close the estimate for $I_{11}$.

For $I_{12}$ we have that, applying lemma 3.5 and lemma 3.7,

$$\begin{aligned} I_{12}\le C[v_0] \left| \left| \nabla w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| \partial _t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}. \end{aligned}$$

By Lemmas 3.3 and 3.17 we obtain

$$\begin{aligned} I_{12}\le C[v_0]\left| \left| \partial _t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}. \end{aligned}$$

Again we can apply Lemma 3.4 to get

$$\begin{aligned} I_{12}\le C[v_0] T^\varepsilon \left| \left| \partial _t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{0}}\le C[v_0] T^\varepsilon \left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s+1}{2}+\varepsilon }_{(0)}H^{0}} \end{aligned}$$

for $0<\frac{s-1}{2}+\varepsilon <1$. Therefore lemmas 3.14 and 3.8 yield a suitable estimate for $I_{12}$. The term $I_{13}$ is bounded in a similar way by using lemmas 3.17, 3.8 and 3.3.

Next we bound the second term in (79). In order to do it we split $\zeta _\phi = \mathbb {I}-t\nabla (Av_0)$. The terms coming from the identity can be bounded as $I_{13}$. For the terms containing the factor $t\nabla (A v_0)$ we just notice that we can proceed as for $I_{13}$ but putting the factor t together with $w^{(n)}$ (here we remark that we can not take $||t||_{H_{(0)}^\frac{s+1}{2}}$ since $\frac{s+1}{2}>1.5$). Indeed

$$\begin{aligned} I_2&= \left| \left| t\nabla w^{(n)}\nabla (Av_0) (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\le C[v_0]\left| \left| t\nabla w^{(n)}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{1+\delta }}, \end{aligned}$$

where

$$\begin{aligned} \left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{1+\delta }}&=\left| \left| \int _0^t\partial _t (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}+1}_{(0)}H^{1+\delta }}\\&\le CT^\varepsilon \left| \left| \partial _t(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }}, \end{aligned}$$

for $0<\frac{s-1}{2}+\varepsilon <1$. Finally we can apply lemma 3.17, 3.8 and 3.9.

For $I_3$ we can proceed in a similar way that for $I_2$. Finally for $I_4$ we just need to use Lemma 3.9.

The estimate of $\left| \left| \overline{g}_\phi \right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}$ follows similar steps. We just notice the need to split $\phi =v_0+t\psi$ and use Lemma 3.9 and the fact that $Tr\left( \nabla \phi \zeta _\phi A_\phi \right) =O(t^2)$.

P1.3. Estimate for $h^{(n)}$:

We will show the appropriate estimate for $h^{(n)}$ decomposing $h^{(n)}=h^{(n)}_v+h^{(n)}_{v*}+h^{(n)}_q$ given by

$$\begin{aligned} h^{(n)}_v= & {} (\nabla v^{(n)}\zeta ^{(n)}\nabla _{J}X^{(n)}-\nabla v^{(n)})n_0,\nonumber \\ h^{(n)}_{v*}= & {} ((\nabla v^{(n)}\zeta ^{(n)}A\circ X^{(n)})^*A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}-(\nabla v^{(n)}A)^*A^{-1})n_0,\nonumber \\ h^{(n)}_q= & {} (-q^{(n)}A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}+q^{(n)}A^{-1})n_0. \end{aligned}$$

(80)

As before, we ignore the superscripts for simplicity. We deal first with the $\left| \,\cdot \,\right| _{L^2H^{s-\frac{1}{2}}}$ norm. Then

$$\begin{aligned} \left| h_v\right| _{L^2H^{s-\frac{1}{2}}}\le \left| \nabla v (\zeta -\mathbb {I})\nabla _{J}X \right| _{L^2H^{s-\frac{1}{2}}}+\left| \nabla v (\nabla _{J}X-\mathbb {I})\right| _{L^2H^{s-\frac{1}{2}}}\equiv I_1+I_2. \end{aligned}$$

For $I_1$ we find

$$\begin{aligned} I_1&\le C \left| \nabla v\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta -\mathbb {I}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \nabla X\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C \left| \left| v\right| \right| _{L^2H^{s+1}} \left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| X\right| \right| _{L^\infty H^{s+1}}\\&\le C[v_0]\left| \left| X-\alpha \right| \right| _{L^\infty H^{s+1}}\le C[v_0]T^\frac{1}{4} \left| \left| X-\alpha \right| \right| _{L^\infty _{1/4}H^{s+1}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

For $I_2$ the computation is analogous:

$$\begin{aligned} I_2&\le C \left| \nabla v\right| _{L^2H^{s-\frac{1}{2}}} \left| \nabla X-\mathbb {I}\right| _{L^\infty H^{s-\frac{1}{2}}}\le C[v_0]\left| \left| X-\alpha \right| \right| _{L^\infty H^{s+1}}\\&\le C[v_0]T^\frac{1}{4} \left| \left| X-\alpha \right| \right| _{L^\infty _{1/4}H^{s+1}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

We are done with $\left| h_v\right| _{L^2H^{s-\frac{1}{2}}}$. Next we deal with $\left| h_{v*}\right| _{L^2H^{s-\frac{1}{2}}}$. Indeed

$$\begin{aligned} \left| h_{v*}\right| _{L^2H^{s-\frac{1}{2}}}&\le \left| (\nabla v(\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{L^2H^{s-\frac{1}{2}}}\\&\quad +\!\left| (\nabla v (A\!\circ \! X-A))^* A^{-1}\!\!\circ \! X \nabla _{J}X\right| _{L^2H^{s-\frac{1}{2}}}\\&\quad +\left| (\nabla v A )^* (A^{-1}\!\!\circ \!X-A^{-1})\nabla _{J}X\right| _{L^2H^{s-\frac{1}{2}}}\\&\quad +\left| (\nabla v A)^*A^{-1}(\nabla _{J}X-\mathbb {I})\right| _{L^2H^{s-\frac{1}{2}}}\equiv I_3+I_4+I_5+I_6. \end{aligned}$$

It is possible to obtain

$$\begin{aligned} I_3&\le C\left| \nabla v\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta -\mathbb {I}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| X\right| _{L^\infty H^{s-\frac{1}{2}}}^2\left| \nabla X\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C \left| \left| v\right| \right| _{L^2H^{s+1}} \left| \left| \zeta -\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| X\right| \right| _{L^\infty H^{s+1}}^3\le C[v_0]\left| \left| X-\alpha \right| \right| _{L^\infty H^{s+1}}\\&\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

Similarly, using Lemma 3.10

$$\begin{aligned} I_4+I_5+I_6&\le C[v_0]\left| \nabla v\right| _{L^2H^{s-\frac{1}{2}}} \left| \left| X-\alpha \right| \right| _{L^\infty H^{s+1}}(\left| \left| X\right| \right| _{L^\infty H^{s+1}}^2+\left| \left| X\right| \right| _{L^\infty H^{s+1}}+1)\\&\le C[v_0]T^\frac{1}{4} \left| \left| X-\alpha \right| \right| _{L^\infty _{1/4}H^{s+1}}\le C[v_0]T^\frac{1}{4}. \end{aligned}$$

It remains to control $\left| h_{q}\right| _{L^2H^{s-\frac{1}{2}}}$. We proceed as follows:

$$\begin{aligned} \left| h_{q}\right| _{L^2H^{s-\frac{1}{2}}}\le&\left| q (A^{-1}\circ X-A^{-1}) \nabla _{J}X\right| _{L^2H^{s-\frac{1}{2}}}+\left| q A^{-1} (\nabla _{J}X-\mathbb {I})\right| _{L^2H^{s-\frac{1}{2}}}\\ \le&C[v_0]\left| q\right| _{L^2H^{s-\frac{1}{2}}}\left| \left| X-\alpha \right| \right| _{L^\infty H^{s+1}}(\left| \left| \nabla X\right| \right| _{L^\infty H^{s+1}}+1)\\ \le&C[v_0]T^{1/4}, \end{aligned}$$

using again Lemma 3.10 to end with the bounds for $\left| h\right| _{L^2H^{s-\frac{1}{2}}}$.

The next step is to deal with $\left| h\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}$.

$$\begin{aligned} \left| h_v\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le \left| \nabla v (\zeta -\mathbb {I})\nabla _{J}X \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| \nabla v (\nabla _{J}X-\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2} \equiv K_1+K_2. \end{aligned}$$

This splitting provides

$$\begin{aligned} K_1&\le \left| \nabla w (\zeta -\mathbb {I})\nabla _{J}X \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| \nabla v_0 (\zeta -\mathbb {I})\nabla _{J}X \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&\quad +\left| t\nabla \psi (\zeta -\mathbb {I})\nabla _{J}X \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\equiv K_{11}+K_{12}+K_{13}. \end{aligned}$$

and therefore

$$\begin{aligned} K_{11}&\le \left| \nabla w (\zeta -\mathbb {I})(\nabla _{J}X-\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| \nabla w (\zeta -\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&\le C\left| \nabla w\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| \zeta -\mathbb {I}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }} \left( \left| \nabla _{J}X-\mathbb {I}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}+1\right) . \end{aligned}$$

We remark that the constant above is independent of time due to Lemma 3.7. Then we use that $\nabla w_i=(\nabla w_i\cdot n_0) n_0+(\nabla w_i\cdot t_0) t_0$ for $i=1,2$ and

$$\begin{aligned} \left| (\nabla w_i\cdot t_0)t_0\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C\left| \nabla w_i\cdot t_0\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C\left| w_i\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{1}}\\&\le C\left| \left| w_i\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{\frac{3}{2}}}\le C\left| \left| w_i\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

together with

$$\begin{aligned} \left| (\nabla w_i\cdot n_0)n_0\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C\left| \nabla w_i\cdot n_0\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le \left| \nabla w_i\cdot n_0\right| _{H_{(0)}^{ht,s-\frac{1}{2}}}\le C\left| \left| w_i\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

These two yield

$$\begin{aligned} \left| \nabla w\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C\left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

(81)

and therefore

$$\begin{aligned} K_{11}&\le C \left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }} \left( \left| \left| X-\alpha \right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{2+\delta }}+1\right) \\&\le C[v_0]\left( \left| \left| X-\alpha -Av_0t\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{2+\delta }}+T^{\frac{1}{2}}\right) \\&\le C[v_0]\left( T^\epsilon \left| \left| X-\alpha -Av_0t\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}+\epsilon }_{(0)}H^{2+\delta }}+T^{\frac{1}{2}}\right) \\&\le C[v_0]\left( T^\epsilon \left| \left| X-\alpha -Av_0t\right| \right| _{F^{s+1}}+T^{\frac{1}{2}}\right) \le C[v_0]T^\epsilon . \end{aligned}$$

For $K_{12}$ it is easy to find

$$\begin{aligned} K_{12}\le C\Vert v_0\Vert _{H^{\frac{3}{2}}}\left| (\zeta -\mathbb {I})\nabla _{J}X \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\le C[v_0]T^\epsilon , \end{aligned}$$

proceeding as before. For the last term $K_{13}$ we obtain

$$\begin{aligned} K_{13}&\le C\left| t\nabla \psi \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| (\zeta -\mathbb {I})\nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\le CT^{\frac{1}{2}}\Vert \psi \Vert _{H^{\frac{3}{2}}} \left| (\zeta -\mathbb {I})\nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\le C[v_0]T^\epsilon , \end{aligned}$$

as before. We are then done with $K_{1}$.

Using that $v=w+v_0+t\psi$, it is possible to estimate $K_2$ similarly as $K_1$. Therefore the appropriate estimate for $\left| h_v\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}$ follows.

Next we deal with $\left| h_{v*}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}$. Indeed

$$\begin{aligned} \left| h_{v*}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le&\left| (\nabla v(\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&+\left| (\nabla v (A\circ X-A))^* A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&+\left| (\nabla v A )^* (A^{-1}\circ X-A^{-1})\nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&+\left| (\nabla v A)^*A^{-1}(\nabla _{J}X-\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&\quad \equiv K_3+K_4+K_5+K_6. \end{aligned}$$

Taking $v=w+v_0+t\psi$, we find

$$\begin{aligned} K_3\le&\left| (\nabla w(\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&+\left| (\nabla v_0(\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&+\left| t(\nabla \psi (\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\equiv K_{31}+K_{32}+K_{33}. \end{aligned}$$

Then using (81) we find

$$\begin{aligned} K_{31}\le&C\left| \nabla w\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| ((\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\ \le&C\left| \left| w\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| ((\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}, \end{aligned}$$

where the constant is independent of time due to Lemma 3.7. The splitting

$$\begin{aligned}&((\zeta -\mathbb {I})A\circ X)^*A^{-1}\circ X \nabla _{J}X=((\zeta -\mathbb {I}) (A\circ X-A))^*(A^{-1}\circ X-A^{-1})(\nabla _{J}X-\mathbb {I})\nonumber \\&\quad +((\zeta -\mathbb {I}) (A\circ X-A))^*(A^{-1}\circ X-A^{-1})+((\zeta -\mathbb {I}) (A\circ X-A))^*A^{-1} (\nabla _{J}X-\mathbb {I})\nonumber \\&\quad +((\zeta -\mathbb {I}) (A\circ X-A))^*A^{-1}+((\zeta -\mathbb {I})A)^*(A^{-1}\circ X-A^{-1})(\nabla _{J}X-\mathbb {I})\nonumber \\&\quad +((\zeta -\mathbb {I})A)^*(A^{-1}\circ X-A^{-1})+((\zeta -\mathbb {I})A)^*A^{-1}(\nabla _{J}X-\mathbb {I})+((\zeta -\mathbb {I})A)^*A^{-1} \end{aligned}$$

(82)

allows us to bound $K_{31}$ with bounds independent of time due to Lemma 3.11 to find

$$\begin{aligned} K_{31}\le C[v_0]\left| \left| \zeta -\mathbb {I}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\le C[v_0](\left| \left| X-\alpha -Av_0t\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{2+\delta }}+T^{\frac{1}{2}})\le C[v_0]T^{\epsilon }. \end{aligned}$$

We proceed for $K_{32}$ as for $K_{12}$ getting

$$\begin{aligned} K_{32}\le C\Vert v_0\Vert _{H^{3/2}}\left| ((\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}, \end{aligned}$$

so that splitting (82) allows us to bound as before $K_{32}\le C[v_0]T^\epsilon$. We continue by using previous estimates to obtain

$$\begin{aligned} K_{33}\le C\left| t\nabla \psi \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| ((\zeta -\mathbb {I}) A\circ X)^*A^{-1}\circ X \nabla _{J}X\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\le C[v_0]T^{\frac{1}{2}}. \end{aligned}$$

We are done with $K_3$. Taking $v=w+v_0+t\psi$ it is possible to control $K_4$, $K_5$ and $K_6$ analogously

$$\begin{aligned} K_4+K_5+K_6 \le C[v_0]T^\epsilon . \end{aligned}$$

The term $\left| h_{v^*}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}$ is then controlled and it remains to handle $\left| h_{q}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}$. We proceed as follows:

$$\begin{aligned} \left| h_{q}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le&\left| q (A^{-1}\circ X-A^{-1})(\nabla _{J}X-\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| q (A^{-1}\circ X-A^{-1})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&+\left| q A^{-1} (\nabla _{J}X-\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\equiv L_1+L_2+L_3. \end{aligned}$$

In $L_1$ we split $q=q_{w}+q_{\phi }$ to find

$$\begin{aligned} L_1\le&C (\left| q_{w}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+|q_\phi |_{L^2})\left| (A^{-1}\circ X-A^{-1})(\nabla _{J}X-\mathbb {I})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\ \le&C(||q_{w}||_{H^{ht,s}_{pr \, (0)}}+C[v_0])\left| \left| (A^{-1}\circ X-A^{-1})(\nabla _{J}X-\mathbb {I})\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\le C[v_0]T^{\epsilon }, \end{aligned}$$

where the time $T^{\epsilon }$ is found as before using Lemmas 3.7, 3.4, 3.16 and 3.13. The terms $L_2$ and $L_3$ are estimated analogously to get $L_2+L_3\le C[v_0]T^{\epsilon }$ and finally

$$\begin{aligned} \left| h_{q}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C[v_0]T^{\epsilon }, \end{aligned}$$

to end with the bounds for $\left| h\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}$. We are done with h.

Proof of Part 2:

It will be enough to show that

$$\begin{aligned} \left| \left| f^{(n)}-f^{(n-1)}\right| \right| _{H_{(0)}^{ht,s-1}}\le \,&C[v_0]\left( \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}+\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\right. \\&\left. +||q^{(n)}-q^{(n-1)}||_{H^{ht,s}_{pr \, (0)}}\right) ,\\ \left| \left| g^{(n)}-g^{(n-1)}\right| \right| _{\overline{H}_{(0)}^{ht,s}}\le \,&C[v_0]\left( \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}+\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\right) ,\\ \left| h^{(n)}-h^{(n-1)}\right| _{H_{(0)}^{ht,s-\frac{1}{2}}}\le \,&C[v_0]\left( \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}+\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\right. \\&\left. +||q^{(n)}-q^{(n-1)}||_{H^{ht,s}_{pr \, (0)}}\right) . \end{aligned}$$

Again we split the proof in three parts:

P2.1. Estimate for $f^{(n)}-f^{(n-1)}$

We split as in (75): $f^{(j)}=f^{(j)}_w+f^{(j)}_\phi +f^{(j)}_q$. In $f^{(n)}_w-f^{(n-1)}_w$ we split further $f^{(n)}_w-f^{(n-1)}_w=d_1+...+d_6$ with the following differences

$$\begin{aligned} d_1= & {} (Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)})\zeta ^{(n)}\partial \left( \zeta ^{(n)}\partial w^{(n)}\right) ,\nonumber \\ d_2= & {} Q^2\circ X^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})\partial \left( \zeta ^{(n)}\partial w^{(n)}\right) ,\nonumber \\ d_3= & {} Q^2\circ X^{(n-1)}\zeta ^{(n-1)}\partial \left( (\zeta ^{(n)}-\zeta ^{(n-1)})\partial w^{(n)}\right) ,\nonumber \\ d_4= & {} (Q^2\circ X^{(n-1)}-Q^2)\zeta ^{(n-1)}\partial \left( \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)})\right) ,\nonumber \\ d_5= & {} Q^2(\zeta ^{(n-1)}-\mathbb {I})\partial \left( \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)})\right) ,\nonumber \\ d_6= & {} Q^2\partial \left( (\zeta ^{(n-1)}-\mathbb {I})\partial (w^{(n)}-w^{(n-1)})\right) . \end{aligned}$$

(83)

Above we do not distinguish from coordinates and partial derivatives, as all the cases can be handled in the same manner. Next we estimate $d_1$. In order to do that we split further $d_1=d_{11}+d_{12}$ with

$$\begin{aligned} d_{11}= & {} (Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)})\zeta ^{(n)}\partial \zeta ^{(n)}\partial w^{(n)},\\ d_{12}= & {} (Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)})\zeta ^{(n)}\zeta ^{(n)}\partial ^2 w^{(n)}. \end{aligned}$$

We take

$$\begin{aligned}&\left| \left| d_{11}\right| \right| _{L^2H^{s-1}}\\&\quad \le C \left| \left| Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta ^{(n)}\partial w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0] \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta ^{(n)}\right| \right| _{L^\infty H^{s-1}} \left| \left| \partial w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and, since

$$\begin{aligned}&d_{11}= \left( Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)}\right) \left( \zeta ^{(n)}-\mathbb {I}\right) \partial \left( \zeta ^{(n)}-\mathbb {I}\right) \partial w^{(n)}\\&\qquad +\left( Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)}\right) \partial (\zeta ^{(n)}-\mathbb {I})\partial w^{(n)},\\&\left| \left| d_{11}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le C \left| \left| Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\qquad \times \left( \left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\right) \left| \left| \partial (\zeta ^{(n)}-\mathbb {I})\partial w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\quad \le C[v_0] \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }} \left| \left| \partial (\zeta ^{(n)}-\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\epsilon }} \left| \left| \partial w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\epsilon }}. \end{aligned}$$

We deal with above terms as before to get

$$\begin{aligned} \left| \left| d_{11}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

For $d_{12}$ we find

$$\begin{aligned} \left| \left| d_{12}\right| \right| _{L^2H^{s-1}}&\le C \left| \left| Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n)}\right| \right| _{L^\infty H^{s-1}}^2\left| \left| \partial ^2 w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\le C[v_0] \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}} \end{aligned}$$

and

$$\begin{aligned} \left| \left| d_{12}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\quad \times (\left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}^2+1)\left| \left| \partial ^2 w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

In order to continue we decompose the next term, $d_2=d_{21}+d_{22}$ where

$$\begin{aligned} d_{21}= & {} Q^2\circ X^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})\partial \zeta ^{(n)}\partial w^{(n)},\\ d_{22}= & {} Q^2\circ X^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})\zeta ^{(n)}\partial ^2 w^{(n)}. \end{aligned}$$

We take

$$\begin{aligned} \left| \left| d_{21}\right| \right| _{L^2H^{s-1}}&\le C \left| \left| Q^2\circ X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta ^{(n)}\partial w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\le C[v_0] \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s}}\left| \left| \partial \zeta ^{(n)}\right| \right| _{L^\infty H^{s-1}} \left| \left| \partial w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}.\\ \left| \left| d_{21}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C (\left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1)\\&\quad \times \left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| \partial \zeta ^{(n)}\partial w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\epsilon }} \left| \left| \partial (\zeta ^{(n)}-\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\epsilon }} \left| \left| \partial w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\epsilon }}\\&\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

For $d_{22}$ we find

$$\begin{aligned} \left| \left| d_{22}\right| \right| _{L^2H^{s-1}}&\le C[v_0] \left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial ^2 w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\le C[v_0] \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s}}\le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d_{22}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C[v_0]\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| \partial ^2 w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\epsilon }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

In $d_3$ we split as follows $d_3=d_{31}+d_{32}$ with

$$\begin{aligned} d_{31}=Q^2\circ X^{(n-1)}\zeta ^{(n-1)}\partial (\zeta ^{(n)}-\zeta ^{(n-1)})\partial w^{(n)}, \end{aligned}$$

and

$$\begin{aligned} d_{32}=Q^2\circ X^{(n-1)}\zeta ^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})\partial ^2 w^{(n)}. \end{aligned}$$

For $d_{31}$ it is possible to get

$$\begin{aligned}&\left| \left| d_{31}\right| \right| _{L^2H^{s-1}}\\&\quad \le C \left| \left| Q^2\circ X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial (\zeta ^{(n)}-\zeta ^{(n-1)})\partial w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]\left| \left| \partial (\zeta ^{(n)}-\zeta ^{(n-1)})\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d_{31}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C (\left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1)\\&\quad \times \left( \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\right) \left| \left| \partial (\zeta ^{(n)}-\zeta ^{(n-1)})\partial w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| \partial (\zeta ^{(n)}-\zeta ^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\delta }}\left| \left| \partial w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\delta }}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\delta }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

Here, in order to bound $\left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}$ we can do the same than in (77).

Next, for $d_{32}$ we obtain

$$\begin{aligned}&\left| \left| d_{32}\right| \right| _{L^2H^{s-1}}\\&\quad \le C \left| \left| Q^2\circ X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| (\zeta ^{(n)}-\zeta ^{(n-1)})\partial ^2 w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial ^2 w^{(n)}\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s}} \le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

The usual approach also gives

$$\begin{aligned} \left| \left| d_{32}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \Big (\left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\Big )\\&\quad \times \Big (\left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\Big )\left| \left| (\zeta ^{(n)}-\zeta ^{(n-1)})\partial ^2 w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| \partial ^2 w^{(n)}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{2+\delta }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

In $d_{4}$ we proceed by considering $d_{4}=d_{41}+d_{42}$ with

$$\begin{aligned} d_{41}=(Q^2\circ X^{(n-1)}-Q^2)\zeta ^{(n-1)}\partial \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)}), \end{aligned}$$

and

$$\begin{aligned} d_{42}=(Q^2\circ X^{(n-1)}-Q^2)\zeta ^{(n-1)}\zeta ^{(n-1)}\partial ^2(w^{(n)}-w^{(n-1)}). \end{aligned}$$

For $d_{41}$ it is possible to get

$$\begin{aligned}&\left| \left| d_{41}\right| \right| _{L^2H^{s-1}}\\&\quad \le C \left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{L^\infty H^{s-1}}\\&\qquad \times \left| \left| \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)}) \right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]\left| \left| X^{(n-1)}-\alpha \right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial (w^{(n)}-w^{(n-1)})\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]T^{1/4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

and

$$\begin{aligned}&\left| \left| d_{42}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\quad \le C \left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\qquad \times \Big (\left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\Big )\left| \left| \partial \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)})\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\quad \le C[v_0]\left| \left| X^{(n-1)}-\alpha \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| \partial (\zeta ^{(n-1)}-\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\delta }}\\&\qquad \left| \left| \partial (w^{(n)}-w^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\delta }}\\&\quad \le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

For $d_{42}$ it is possible to get

$$\begin{aligned}&\left| \left| d_{42}\right| \right| _{L^2H^{s-1}}\\&\quad \le C \left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}^2\left| \left| \partial ^2(w^{(n)}-w^{(n-1)})\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]\left| \left| X^{(n)}-\alpha \right| \right| _{L^\infty H^{s-1}}\left| \left| \partial ^2 (w^{(n)}-w^{(n-1)})\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]T^{1/4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d_{42}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| Q^2\circ X^{(n-1)}-Q^2\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\quad \times \Big (\left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}^2+1\Big )\left| \left| \partial ^2(w^{(n)}-w^{(n-1)})\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]\left| \left| X^{(n)}-\alpha \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| \partial ^2(w^{(n)}-w^{(n-1)})\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]T^{\epsilon }\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

Analogously, one could take $d_5=d_{51}+d_{52}$ with

$$\begin{aligned} d_{51}=Q^2(\zeta ^{(n-1)}-\mathbb {I})\partial \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)}), \end{aligned}$$

and

$$\begin{aligned} d_{52}=Q^2(\zeta ^{(n-1)}-\mathbb {I})\zeta ^{(n-1)}\partial ^2(w^{(n)}-w^{(n-1)}). \end{aligned}$$

The splitting yields

$$\begin{aligned} \left| \left| d_{51}\right| \right| _{L^2H^{s-1}}&\le C \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial (w^{(n)}-w^{(n-1)}) \right| \right| _{L^2H^{s-1}}\\&\le C[v_0]T^{1/4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

and

$$\begin{aligned}&\left| \left| d_{51}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\quad \le C \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| \partial (\zeta ^{(n-1)}-\mathbb {I})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\delta }} \left| \left| \partial (w^{(n)}-w^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\delta }}\\&\quad \le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

For $d_{52}$ we proceed as follows

$$\begin{aligned} \left| \left| d_{52}\right| \right| _{L^2H^{s-1}}&\le C \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{L^\infty H^{s-1}}\left| \left| \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial ^2 (w^{(n)}-w^{(n-1)}) \right| \right| _{L^2H^{s-1}}\\&\le C[v_0]T^{1/4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}},\\ \left| \left| d_{52}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\quad \times \Big (\left| \left| \zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\Big )\left| \left| \partial ^2 (w^{(n)}-w^{(n-1)})\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

Finally, writing $d_6=d_{61}+d_{62}$ where

$$\begin{aligned} d_{61}=Q^2\partial \zeta ^{(n-1)}\partial (w^{(n)}-w^{(n-1)}), \text{ and } d_{62}=Q^2(\zeta ^{(n-1)}-\mathbb {I})\partial ^2(w^{(n)}-w^{(n-1)}), \end{aligned}$$

it is possible to find

$$\begin{aligned} \left| \left| d_{61}\right| \right| _{L^2H^{s-1}}&\le C\left| \left| \partial \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \partial (w^{(n)}-w^{(n-1)}) \right| \right| _{L^2H^{s-1}}\\&\le C[v_0]T^{1/4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d_{61}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| \partial \zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{\delta }} \left| \left| \partial (w^{(n)}-w^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1-\delta }}\\&\le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

For $d_{62}$ we proceed as for $d_{52}$ to get

$$\begin{aligned} \left| \left| d_{62}\right| \right| _{L^2H^{s-1}}&\le C[v_0] T^{1/4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}},\\ \left| \left| d_{62}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

We are done with $f^{(n)}_w-f^{(n-1)}_w$. To continue with $f^{(n)}_\phi -f^{(n-1)}_\phi$ we decompose as before but this time using $f^{(n)}_\phi -f^{(n-1)}_\phi =d_7+d_8+d_9$ where

$$\begin{aligned} d_7= & {} (Q^2\circ X^{(n)}-Q^2\circ X^{(n-1)})\zeta ^{(n)}\partial \left( \zeta ^{(n)}\partial \phi \right) ,\\ d_8= & {} Q^2\circ X^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})\partial \left( \zeta ^{(n)}\partial \phi \right) ,\\ d_9= & {} Q^2\circ X^{(n-1)}\zeta ^{(n-1)}\partial \left( (\zeta ^{(n)}-\zeta ^{(n-1)})\partial \phi \right) . \end{aligned}$$

Here we need to split $\phi =v_0+t\psi$. After that we proceed as for $d_1$, $d_2$ and $d_3$ to find

$$\begin{aligned} \left| \left| d_7+d_8+d_9\right| \right| _{H_{(0)}^{ht,s-1}}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

We move to the f term involving q. We split $f^{(n)}_{q}=f^{(n)}_{q_w}+f^{(n)}_{q_\phi }$. The splitting $f^{(n)}_{q_w}-f^{(n-1)}_{q_w}=d^q_1+d^q_2+d^q_3+d^q_4$ with

$$\begin{aligned} d^q_1= & {} -(A\circ X^{(n)}(\zeta ^{(n)}-\zeta ^{(n-1)}))^*\nabla q^{(n)}_w,\nonumber \\ d^q_2= & {} -((A\circ X^{(n)}-A\circ X^{(n-1)})\zeta ^{(n-1)})^*\nabla q^{(n)}_w,\nonumber \\ d^q_3= & {} -(A\circ X^{(n-1)}(\zeta ^{(n-1)}-\mathbb {I}))^*\nabla (q^{(n)}_w-q^{(n-1)}_w),\nonumber \\ d^q_4= & {} -(A\circ X^{(n-1)}-A)^*\nabla (q^{(n)}_w-q^{(n-1)}_w), \end{aligned}$$

(84)

allows us to do the work. In fact

$$\begin{aligned} \left| \left| d^q_1\right| \right| _{L^2H^{s-1}}&\le C \left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| A\circ X^{(n)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \nabla q^{(n)}_w\right| \right| _{L^2H^{s-1}}\\&\le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}},\\ \left| \left| d^q_{1}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\quad \times \Big (\left| \left| A\circ X^{(n)}-A\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\Big )\left| \left| \nabla q^{(n)}_w\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

Sharing norms in the same manner gives

$$\begin{aligned} \left| \left| d^q_2\right| \right| _{L^2H^{s-1}}\le C[v_0]T^{1/4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d^q_2\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

In a similar way

$$\begin{aligned}&\left| \left| d^q_3\right| \right| _{L^2H^{s-1}}\\&\quad \le C \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{L^\infty H^{s-1}}\left| \left| A\circ X^{(n-1)}\right| \right| _{L^\infty H^{s-1}}\left| \left| \nabla (q^{(n)}_w-q^{(n-1)}_w)\right| \right| _{L^2H^{s-1}}\\&\quad \le C[v_0]T^{1/4}\left| \left| \nabla (q^{(n)}_w-q^{(n-1)}_w)\right| \right| _{L^2H^{s-1}}\le C[v_0]T^{1/4}\Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d^q_{3}\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\\&\quad \times \Big (\left| \left| A\circ X^{(n-1)}-A\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}+1\Big )\left| \left| \nabla (q^{(n)}_w-q^{(n-1)}_w)\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]T^\epsilon \left| \left| \nabla (q^{(n)}_w-q^{(n-1)}_w)\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]T^{\epsilon }\Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}. \end{aligned}$$

Finally

$$\begin{aligned} \left| \left| d^q_4\right| \right| _{L^2H^{s-1}}&\le C \left| \left| A\circ X^{(n-1)}-A\right| \right| _{L^\infty H^{s-1}}\left| \left| \nabla (q^{(n)}_w-q^{(n-1)}_w)\right| \right| _{L^2H^{s-1}}\\&\le C[v_0]T^{1/4}\Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}, \end{aligned}$$

and

$$\begin{aligned} \left| \left| d^q_4\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}&\le C \left| \left| A\circ X^{(n-1)}-A\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\epsilon }}\left| \left| \nabla (q^{(n)}_w-q^{(n-1)}_w)\right| \right| _{H_{(0)}^{\frac{s-1}{2}}L^2}\\&\le C[v_0]T^{\epsilon }\Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}. \end{aligned}$$

The estimation for $f^{(n)}_{q_\phi }-f^{(n-1)}_{q_\phi }$ follows similar steps. We only need to take into account that $q_\phi$ does not depend on time.

P2.2. Estimate for $\overline{g}^{(n)}-\overline{g}^{(n-1)}$:

We are concerned with the estimates of the norms

$$\begin{aligned} \left| \left| \overline{g}^{(n)}-\overline{g}^{(n-1)}\right| \right| _{\overline{H}_{(0)}^{ht,s}}\le \left| \left| \overline{g}^{(n)}-\overline{g}^{(n-1)}\right| \right| _{L^2H^{s}}+ \left| \left| \overline{g}^{(n)}-\overline{g}^{(n-1)}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}, \end{aligned}$$

We will split $\overline{g}^{(n)}= -\overline{g}^{(n)}_w-\overline{g}^{(n)}_\phi$ (see (78)). First we will estimate the $\left| \left| \cdot \right| \right| _{L^2H^{s}}$-norm. For $\overline{g}^{(n)}_w-\overline{g}^{(n-1)}_w$ we have that

$$\begin{aligned}&\overline{g}^{(n)}_w-\overline{g}^{(n-1)}_w= Tr\left( \left( \nabla w^{(n)}-\nabla w^{(n-1)}\right) \zeta ^{(n)}A\circ X^{(n)}\right) \\&\qquad +Tr\left( \nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})A\circ X^{(n)}\right) \\&Tr\left( \nabla w^{(n-1)}\zeta ^{(n-1)}\left( A\circ X^{(n)}-A\circ X^{(n-1)}\right) \right) -Tr\left( \left( \nabla w^{(n)}-\nabla w^{(n-1)}\right) A\right) \\&\quad = Tr\left( \left( \nabla w^{(n)}-\nabla w^{(n-1)}\right) (\zeta ^{(n)}-\mathbb {I}) A\circ X^{(n)}\right) \\&\qquad +Tr\left( \nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})A\circ X^{(n)}\right) \\&Tr\left( \nabla w^{(n-1)}\zeta ^{(n-1)}\left( A\circ X^{(n)}-A\circ X^{(n-1)}\right) \right) \\&\qquad +Tr\left( \left( \nabla w^{(n)}-\nabla w^{(n-1)}\right) (A\circ X^{(n)}-A\right) \\&\quad =d_1+d_2+d_3+d_4. \end{aligned}$$

With $d_1$ we proceed as follows

$$\begin{aligned} \left| \left| d_1\right| \right| _{L^2H^{s}}=&\left| \left| (\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\mathbb {I})A\circ X^{(n)}\right| \right| _{L^2H^{s}}\nonumber \\ \le&\left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| A\circ X^{(n)}\right| \right| _{L^\infty H^{s}}\left| \left| \nabla w^{(n)}-\nabla w^{(n-1)}\right| \right| _{L^2H^{s}}. \end{aligned}$$

(85)

Then lemma 3.13 and (76) implies

$$\begin{aligned}&\left| \left| d_1\right| \right| _{L^2H^{s}}\le C[v_0]\left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\\&\quad \le C[v_0]T^\frac{1}{4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

For $d_2$ we have that

$$\begin{aligned} \left| \left| d_2\right| \right| _{L^2H^{s}}\le \left| \left| \nabla w^{(n-1)}\right| \right| _{L^2H^{s}}\left| \left| A\circ X^{(n)}\right| \right| _{L^\infty H^{s}}\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s}}. \end{aligned}$$

Thus, using Lemma 3.15 yields,

$$\begin{aligned} \left| \left| d_2\right| \right| _{L^2H^{s}}\le C[v_0]T^\frac{1}{4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty _{1/4}H^{s+1}}. \end{aligned}$$

The estimate for $d_3$ follows the next steps

$$\begin{aligned} \left| \left| d_3\right| \right| _{L^2H^{s}}&\le \left| \left| \nabla w^{(n-1)}\right| \right| _{L^2H^{s}}\left| \left| \zeta ^{(n-1)}\right| \right| _{L^\infty H^{s}}\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{L^\infty H^{s}}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s}}\le C[v_0]T^\frac{1}{4}\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty _{1/4}H^{s}}. \end{aligned}$$

And for $d_4$ we have that, using 3.10 and proceeding as in (77)

$$\begin{aligned} \left| \left| d_4\right| \right| _{L^2H^{s}}&\le \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| A\circ X^{(n)}-A\right| \right| _{L^\infty H^{s}}\\&\le C[v_0]\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| X^{(n)}-\alpha \right| \right| _{L^\infty H^{s}}\\&\le C[v_0]T^\frac{1}{4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

It remains to estimate the $\left| \left| \,\cdot \,\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}$-norm.

We will split $\overline{g}_w^{(n)}-\overline{g}_w^{(n-1)}$ in the following terms

$$\begin{aligned}&\overline{g}_w^{(n)}-\overline{g}_w^{(n-1)}\\&\quad =Tr\left( (\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\zeta _\phi )(A\circ X^{(n)}-A_\phi )\right) \\&\qquad +Tr\left( (\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\zeta _\phi )A_\phi \right) \\&\qquad +Tr\left( (\nabla w^{(n)}-\nabla w^{(n-1)})\zeta _\phi (A\circ X^{(n)}-A_\phi )\right) \\&\qquad +Tr\left( \nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})(A\circ X^{(n)}-A_\phi )\right) \\&\qquad +Tr\left( \nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})A_\phi \right) \\&\qquad +Tr\left( \nabla w^{(n-1)}(\zeta ^{(n-1)}-\zeta _\phi )(A\circ X^{(n)}-A\circ X^{(n-1)})\right) \\&\qquad +Tr\left( \nabla w^{(n-1)}\zeta _\phi \left( A\circ X^{(n)}-A\circ X^{(n-1)}\right) \right) \\&\qquad +Tr\left( (\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta _\phi -\mathbb {I})A_\phi +(\nabla w^{(n)}-\nabla w^{(n-1)})(A_\phi -A)\right) \\&\qquad =D_1+D_2+D_3+D_4+D_5+D_6+D_7+D_8. \end{aligned}$$

For $D_1$ we have that

$$\begin{aligned} D_1= & {} -\int _{0}^t Tr(\partial _t((\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\zeta _\phi ) (A\circ X^{(n)}-A_\phi ))) d\tau \\= & {} D_{11}+D_{12}+D_{13} \end{aligned}$$

where

$$\begin{aligned} D_{11}=&-\int _{0}^t Tr(\partial _t(\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\zeta _\phi ) (A\circ X^{(n)}-A_\phi ))d\tau ,\nonumber \\ D_{12}=&-\int _{0}^t Tr((\nabla w^{(n)}-\nabla w^{(n-1)})\partial _t(\zeta ^{(n)}-\zeta _\phi ) (A\circ X^{(n)}-A_\phi ))d\tau ,\nonumber \\ D_{13}=&-\int _{0}^t Tr((\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\zeta _\phi )\partial _t (A\circ X^{(n)}-A_\phi ))d\tau . \end{aligned}$$

(86)

By applying Lemma 3.4 with $\varepsilon =0$ we obtain

$$\begin{aligned} \left| \left| D_{11}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\le \left| \left| \partial _t(\nabla w^{(n)}-\nabla w^{(n-1)})(\zeta ^{(n)}-\zeta _\phi ) (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}. \end{aligned}$$

Now we use Lemma 3.5 to yield

$$\begin{aligned} \left| \left| D_{11}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}&\le \left| \left| \partial _t(\nabla w^{(n)}-\nabla w^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\\&\quad \left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }} \end{aligned}$$

In addition Lemmas 3.17, 3.8 and 3.3 and proceeding as in (77) imply

$$\begin{aligned} \left| \left| D_{11}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\le C[v_0]\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}. \end{aligned}$$

And we can apply Lemma 3.4 to get

$$\begin{aligned} \left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}&\le \left| \left| \partial _t\int _{0}^t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| \int _{0}^t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}+\varepsilon +1-\varepsilon }_{(0)}H^{1+\delta }}\\&\le CT^\varepsilon \left| \left| \zeta ^{(n)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }}, \end{aligned}$$

for $0<\frac{s-1}{2}+\varepsilon <1$. Thus, by lemmas 3.14 and 3.8,

$$\begin{aligned} \left| \left| D_{11}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}} \le C[v_0]T^\varepsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

Next we bound $D_{12}$. Indeed

$$\begin{aligned}&\left| \left| D_{12}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\quad =\left| \left| (\nabla w^{(n)}-\nabla w^{(n-1)})\partial _t(\zeta ^{(n)}-\zeta _\phi ) (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\\&\quad \le C \left| \left| (\nabla w^{(n)}-\nabla w^{(n-1)})\partial _t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\&\quad \le C[v_0]\left| \left| \nabla w^{(n)}-\nabla w^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\left| \left| \partial _t(\zeta ^{(n)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}. \end{aligned}$$

Since, $\left| \left| \partial _t\zeta ^{(n)}-\partial _t\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\le C[v_0] T^\varepsilon$, and $\left| \left| \nabla w^{(n)}-\nabla w^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\le C[v_0]T^\varepsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}$ we obtain a suitable estimate for $D_{12}$.

The term $D_{13}$ does not cause any difficulty and it can be handled as before.

For $D_2$ we just split $(A_\phi )_{ij}=A_{ij}+t \partial _k A_{ij}A_{kl}v_{0}^l$. For the terms coming from A we just notice that A does not depend on t. For the term coming from $t\partial _k A_{ij}A_{kl}v_{0}^l$ we use Lemma 3.9 and the fact that $\partial _k A_{ij}A_{kl}v_{0}^l$ does not depend on t. For $D_3$ a similar argument holds after splitting $\zeta _\phi =t-\nabla (Av_0)$.

The estimate for $D_4$ follows the following steps,

$$\begin{aligned} \left| \left| D_{4}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\le&\left| \left| \nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\nonumber \\ \le&\left| \left| \int _{0}^t\partial _t (\nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})(A\circ X^{(n)}-A_\phi ))\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}, \end{aligned}$$

(87)

and therefore

$$\begin{aligned} \left| \left| D_{4}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\le&\left| \left| \partial _t\nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\\&+ \left| \left| \nabla w^{(n-1)}\partial _t(\zeta ^{(n)}-\zeta ^{(n-1)})(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\\&+\left| \left| \nabla w^{(n-1)}(\zeta ^{(n)}-\zeta ^{(n-1)})\partial _t(A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}. \end{aligned}$$

Each addend is estimated in a different manner as follows

$$\begin{aligned}&\left| \left| D_{4}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\quad \le C[v_0]\left| \left| \nabla \partial _tw^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }} \\&\qquad +C[v_0] \left| \left| \nabla w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\left| \left| \partial _t (\zeta ^{(n)}-\zeta ^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\left| \left| A\circ X^{(n)}-A_\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\&\qquad + C[v_0]\left| \left| \nabla w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\left| \left| \partial _t (A\circ X^{(n)}-A_\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}, \end{aligned}$$

and we have already bounded every term in here.

For $D_5$ we proceed as for $D_2$ and $D_3$.

We will give full detail for $D_6$.

$$\begin{aligned} D_6=-\int _{0}^t Tr\partial _t(\nabla w^{(n-1)}(\zeta ^{(n-1)}-\zeta _\phi ) (A\circ X^{(n)}-A\circ X^{(n-1)}))d\tau =d_{31}+d_{32}+d_{33} \end{aligned}$$

where

$$\begin{aligned} D_{61}=&-\int _0^t Tr(\partial _t\nabla w^{(n-1)}(\zeta ^{(n-1)}-\zeta _\phi )(A\circ X^{(n)}-A\circ X^{(n-1)}))d\tau \nonumber \\ D_{62}=&-\int _0^t Tr(\nabla w^{(n-1)}\partial _t(\zeta ^{(n-1)}-\zeta _\phi ) (A\circ \dot{}X^{(n)}-A\circ X^{(n-1)})) d\tau \nonumber \\ D_{63}=&-\int _0^t Tr(\nabla w^{(n-1)}(\zeta ^{(n-1)}-\zeta _\phi ) \partial _t(A\circ X^{(n)}-A\circ X^{(n-1)}))d\tau . \end{aligned}$$

(88)

The estimates of these three terms follow similar steps as those one for $D_{1}$ and $D_4$. First we apply lemma 3.4 with $\varepsilon =0$ since $0<\frac{s-1}{2}<1$ to find

$$\begin{aligned}&\left| \left| D_{61}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\\&\quad \le \left| \left| \partial _t\nabla w^{(n-1)}(\zeta ^{(n-1)}-\zeta _\phi )(A\circ X^{(n)}-A\circ X^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\\&\quad \le C[v_0]\left| \left| \nabla \partial _tv^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}}\left| \left| \zeta ^{(n-1)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }})\\&\qquad \left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\&\quad \le C[v_0]\left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}. \end{aligned}$$

Above lemma 3.5, lemma 3.7, lemma 3.14 and lemma 3.17. In addition

$$\begin{aligned} \left| \left| D_{61}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}\le&C[v_0]\left| \left| \partial _t\int _{0}^tX^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\ \le&\left| \left| \int _{0}^tX^{(n)}- X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}+\varepsilon +1-\varepsilon }_{(0)}H^{1+\delta }}\\ \le&CT^\varepsilon \left| \left| X^{(n)}- X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{1+\delta }}\le CT^\varepsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

for $0<\frac{s-1}{2}+\varepsilon <1$. The estimate for $D_{62}$ is quite similar. Indeed, lemma 3.4 with $\varepsilon =0$ to obtain

$$\begin{aligned} \left| \left| D_{62}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}&\le C\left| \left| \nabla w^{(n-1)}\partial _t(\zeta ^{(n-1)}-\zeta _\phi )(A\circ X^{(n)}-A\circ X^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{-1}} \\&\le C[v_0]\left| \left| \nabla w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\left| \left| \partial _t(\zeta ^{(n-1)}-\zeta _\phi )\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\\&\quad \left| \left| A\circ X^{(n)}-A\circ X^{(n-1)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}. \end{aligned}$$

Above we apply lemma 3.5 and the rest of the estimate follows the same steps. Next we can make use of 3.4 with $\varepsilon =0$ since $0<\frac{s-1}{2}<1$ to obtain

$$\begin{aligned} \left| \left| D_{63}\right| \right| _{H^{\frac{s+1}{2}}_{(0)}H^{-1}}&\le C[v_0]\left| \left| \nabla w^{(n)}\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1}}\left| \left| \zeta ^{(n-1)}-\zeta _\phi \right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{1+\delta }}\\&\quad \left| \left| \partial _t (A\circ X^{(n)}-A\circ X^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}, \end{aligned}$$

and therefore, it is enough to estimate $\left| \left| \partial _t (X^{(n)}-X^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}$. As we did before we obtain

$$\begin{aligned} \left| \left| \partial _t (X^{(n)}-X^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}}_{(0)}H^{0}}\le&CT^\varepsilon \left| \left| \partial _t(X^{(n)}-X^{(n-1)})\right| \right| _{H^{\frac{s-1}{2}+\varepsilon }_{(0)}H^{0}} \\\le&CT^\varepsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

To estimate $D_7$ we just split $\zeta _\phi =\mathbb {I}-t\nabla (Av_0)$ and use lemma 3.9. Finally we estimate $D_8$ by using lemma 3.9.

P2.3. Estimate for $h^{(n)}-h^{(n-1)}$:

We split using (80) and decompose further $h^{(n)}_v-h^{(n-1)}_v=d_1+d_2+d_3+d_4$ where the differences are given by

$$\begin{aligned}&d_1=\nabla v^{(n)}(\zeta ^{(n)}-\zeta ^{(n-1)})\nabla _{J}X^{(n)}n_0, \quad d_2=\nabla v^{(n)}\zeta ^{(n-1)}(\nabla _{J}X^{(n)}-\nabla _{J}X^{(n-1)}) n_0,\nonumber \\&d_3=\nabla (v^{(n)}-v^{(n-1)})(\zeta ^{(n-1)}-\mathbb {I})\nabla _{J}X^{(n-1)}n_0, \nonumber \\&d_4=\nabla (v^{(n)}-v^{(n-1)})(\nabla _{J}X^{(n-1)}-\mathbb {I}) n_0. \end{aligned}$$

(89)

We estimate as follows. For $d_1$ we find:

$$\begin{aligned} \left| d_1\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| \nabla v^{(n)}\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \nabla X^{(n)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C \left| \left| v^{(n)}\right| \right| _{L^2H^{s+1}} \left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s}}\left| \left| X^{(n)}\right| \right| _{L^\infty H^{s+1}}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s+1}}\le C[v_0]T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

The time norm needs the splitting $v^{(n)}=w^{(n)}+v_0+t\psi$ which gives

$$\begin{aligned} \left| d_1\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C \Big (\left| \nabla w^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\Vert v_0\Vert _{H^{3/2}}+\left| t\nabla \psi \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\Big )\\&\quad \left| (\zeta ^{(n)}-\zeta ^{(n-1)})\nabla _{J}X^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\le C \Big (\left| \left| w^{(n)}\right| \right| _{H_{(0)}^{ht,s+1}}+C[v_0]\Big )\left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\\&\quad \Big ( \left| \left| \nabla _{J}X^{(n)}-\mathbb {I}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}+1\Big )\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{2+\delta }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

Above we have used (81) together with Lemmas 3.4, 3.7 and 3.15. Similarly

$$\begin{aligned} \left| d_2\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| \nabla v^{(n)}\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta ^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \nabla X^{(n)}-\nabla X^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C[v_0]T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| d_2\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C \Big (\left| \nabla w^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\Vert v_0\Vert _{H^{3/2}}+\left| t\nabla \psi \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\Big )\\&\quad \left| \zeta ^{(n-1)}(\nabla _{J}X^{(n)}-\nabla _{J}X^{(n-1)})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\le C[v_0]\Big (\left| \left| \zeta ^{(n)}-\mathbb {I}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}+1\Big ) \left| \left| \nabla _{J}X^{(n)}-\nabla _{J}X^{(n-1)}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\\&\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

For $d_3$ we find:

$$\begin{aligned} \left| d_3\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| \nabla (v^{(n)}-v^{(n-1)})\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta ^{(n-1)}-\mathbb {I}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \nabla X^{(n)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{L^2H^{s+1}} \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{L^\infty H^{s}}\left| \left| X^{(n)}\right| \right| _{L^\infty H^{s+1}}\\&\le C[v_0]T^\frac{1}{4} \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| d_3\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C \left| \nabla (v^{(n)}-v^{(n-1)})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| \zeta ^{(n-1)}-\mathbb {I}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\quad \Big (\left| \nabla _{J}X^{(n)}-\mathbb {I}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}+1\Big )\\&\le C \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\\&\quad \Big (\left| \left| X^{(n)}-\alpha \right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{2+\delta }}+1\Big )\\&\le C[v_0]T^\epsilon \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

Above we have used (81) together with Lemmas 3.4, 3.8 and 3.13. Similarly

$$\begin{aligned}&\left| d_4\right| _{L^2H^{s-\frac{1}{2}}}\le C[v_0]T^\frac{1}{4} \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}},\,\text{ and } \\&\quad \left| d_4\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C[v_0]T^\epsilon \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

We are done with $h_{v}^{(n)}-h_{v}^{(n-1)}$. In order to deal with $h_{v*}^{(n)}-h_{v*}^{(n-1)}=d_1^*+...+d_8^*$ we use the splitting where

$$\begin{aligned} d_1^*= & {} (\nabla v^{(n)}(\zeta ^{(n)}-\zeta ^{(n-1)}) A\circ X^{(n)})^* A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}n_0,\nonumber \\ d_2^*= & {} (\nabla v^{(n)}\zeta ^{(n-1)}(A\circ X^{(n)}-A\circ X^{(n-1)}))^* A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}n_0,\nonumber \\ d_3^*= & {} (\nabla v^{(n)}\zeta ^{(n-1)}A\circ X^{(n-1)})^* (A^{-1}\circ X^{(n)}-A^{-1}\circ X^{(n-1)})) \nabla _{J}X^{(n)}n_0,\nonumber \\ d_4^*= & {} (\nabla v^{(n)}\zeta ^{(n-1)}A\circ X^{(n-1)})^*A^{-1}\circ X^{(n-1)}(\nabla _{J}X^{(n)}-\nabla _{J}X^{(n-1)}) n_0,\nonumber \\ d_5^*= & {} (\nabla (v^{(n)}-v^{(n-1)})(\zeta ^{(n-1)}-\mathbb {I}) A\circ X^{(n-1)})^* A^{-1}\circ X^{(n-1)}\nabla _{J}X^{(n-1)}n_0,\nonumber \\ d_6^*= & {} (\nabla (v^{(n)}-v^{(n-1)})(A\circ X^{(n-1)}-A))^* A^{-1}\circ X^{(n-1)}\nabla _{J}X^{(n-1)}n_0,\nonumber \\ d_7^*= & {} (\nabla (v^{(n)}-v^{(n-1)}) A)^* (A^{-1}\circ X^{(n-1)}-A^{-1})\nabla _{J}X^{(n-1)}n_0,\nonumber \\ d_8^*= & {} (\nabla (v^{(n)}-v^{(n-1)}) A)^* A^{-1}(\nabla _{J}X^{(n)}-\mathbb {I}) n_0. \end{aligned}$$

(90)

Then

$$\begin{aligned} \left| d_1^*\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| \nabla v^{(n)}\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| X^{(n)}\right| _{L^\infty H^{s-\frac{1}{2}}}^2\left| \nabla X^{(n)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C \left| \left| v^{(n)}\right| \right| _{L^2H^{s+1}} \left| \left| \zeta ^{(n)}-\zeta ^{(n-1)}\right| \right| _{L^\infty H^{s}}\left| \left| X^{(n)}\right| \right| _{L^\infty H^{s+1}}^3\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{L^\infty H^{s+1}}\le C[v_0]T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

For the norm in time we split as follows

$$\begin{aligned} \left| d_1^*\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C \Big (\left| \nabla w^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\Vert v_0\Vert _{H^{3/2}}+\left| t\nabla \psi \right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\Big )\\&\quad \times \left| ((\zeta ^{(n)}-\zeta ^{(n-1)}) A\circ X^{(n)})^* A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\le C[v_0]\left| \left| ((\zeta ^{(n)}-\zeta ^{(n-1)}) A\circ X^{(n)})^* A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}. \end{aligned}$$

With a decomposition similar to (82), it is possible to find

$$\begin{aligned} \left| d_1^*\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{2+\delta }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

with the use of Lemmas 3.4 and 3.8. In an analogous manner

$$\begin{aligned}&\left| d_2^*\right| _{L^2H^{s-\frac{1}{2}}}\\&\quad \le C[v_0] \left| \nabla v^{(n)}\right| _{L^2H^{s-\frac{1}{2}}} \left| \zeta ^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| X^{(n)}-X^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \left| X^{(n)}\right| \right| _{L^\infty H^{s+1}}^2\\&\quad \le C[v_0]T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned} \left| d_2^*\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C[v_0] \left| ((\zeta ^{(n-1)}) (A\circ X^{(n)}-A\circ X^{(n-1)})^* A^{-1}\circ X^{(n)}\nabla _{J}X^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\le C[v_0]\left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

We proceed for $d^*_3$ and $d^*_4$ as for $d^*_2$ to find

$$\begin{aligned} \left| d^*_3\right| _{L^2H^{s-\frac{1}{2}}}+\left| d^*_4\right| _{L^2H^{s-\frac{1}{2}}}&\le C[v_0]T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}},\\ \left| d^*_3\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| d^*_4\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

A similar approach is used to find

$$\begin{aligned} \left| d^*_5\right| _{L^2H^{s-\frac{1}{2}}}&\le C[v_0] \left| \nabla (v^{(n)}-v^{(n-1)})\right| _{L^2H^{s-\frac{1}{2}}}\left| \zeta ^{(n-1)}-\mathbb {I}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\quad \Big (\left| \left| X^{(n-1)}\right| \right| _{L^\infty H^{s+1}}^3+1\Big )\\&\le C[v_0] \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{L^2H^{s+1}} \left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{L^\infty H^{s}}\\&\le C[v_0]T^\frac{1}{4}\left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}, \end{aligned}$$

and

$$\begin{aligned}&\left| d^*_5\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&\quad \le C \left| \nabla (v^{(n)}-v^{(n-1)})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| (\zeta ^{(n-1)}-\mathbb {I}) A\circ X^{(n-1)})^* A^{-1}\circ X^{(n-1)}\nabla _{J}X^{(n-1)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\quad \le C[v_0] \left| \left| v^{(n)}-v^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}\left| \left| \zeta ^{(n-1)}-\mathbb {I}\right| \right| _{H^{\frac{s}{2}-\frac{1}{4}}_{(0)}H^{1+\delta }}\\&\quad \le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

It is possible to distribute the norms in a similar manner as in $d_5^*$ to conclude that

$$\begin{aligned} \left| d^*_6\right| _{L^2H^{s-\frac{1}{2}}}+\left| d^*_7\right| _{L^2H^{s-\frac{1}{2}}}+\left| d^*_8\right| _{L^2H^{s-\frac{1}{2}}}&\le C[v_0]T^\frac{1}{4} \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}},\\ \left| d^*_6\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| d^*_7\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+\left| d^*_8\right| _{L^2H^{s-\frac{1}{2}}}&\le C[v_0]T^\epsilon \left| \left| w^{(n)}-w^{(n-1)}\right| \right| _{H_{(0)}^{ht,s+1}}. \end{aligned}$$

Next, we deal with $h^{(n)}_q-h^{(n-1)}_q=d_1^q+d_2^q+d_3^q+d_4^q$ where

$$\begin{aligned} d_1^q= & {} q^{(n-1)}(A^{-1}\circ X^{(n-1)}-A^{-1}\circ X^{(n)})\nabla _{J}X^{(n-1)}n_0,\nonumber \\ d_2^q= & {} q^{(n-1)}A^{-1}\circ X^{(n)}(\nabla _{J}X^{(n-1)}-\nabla _{J}X^{(n)}) n_0,\nonumber \\ d_3^q= & {} (q^{(n-1)}_w-q^{(n)}_w) (A^{-1}\circ X^{(n)}-A^{-1}) \nabla _{J}X^{(n)}n_0,\nonumber \\ d_4^q= & {} (q^{(n-1)}_w-q^{(n)}_w) A^{-1} (\nabla _{J}X^{(n)}-\mathbb {I}) n_0. \end{aligned}$$

(91)

We start as follows

$$\begin{aligned} \left| d_1^q\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| q^{(n-1)}\right| _{L^2H^{s-\frac{1}{2}}}\left| X^{(n)}-X^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \nabla X^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C[v_0] T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned}&\left| d_1^q\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&\quad \le C \Big (\left| q^{(n-1)}_w\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+|q_{\phi }|_{L^2}\Big )\left| (A^{-1}\circ X^{(n-1)}-A^{-1}\circ X^{(n)})\nabla _{J}X^{(n-1)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\quad \le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

following before estimates. Likewise, it is possible to find next

$$\begin{aligned} \left| d_2^q\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| q^{(n-1)}\right| _{L^2H^{s-\frac{1}{2}}}\left| X^{(n)}\right| _{L^\infty H^{s-\frac{1}{2}}}\left| \nabla X^{(n)}-\nabla X^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C[v_0]T^\frac{1}{4} \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}, \end{aligned}$$

and

$$\begin{aligned}&\left| d_2^q\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\\&\quad \le C \Big (\left| q^{(n-1)}_w\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}+|q_{\phi }|_{L^2}\Big )\left| A^{-1}\circ X^{(n)}(\nabla _{J}X^{(n-1)}-\nabla _{J}X^{(n)})\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\quad \le C[v_0]T^\epsilon \left| \left| X^{(n)}-X^{(n-1)}\right| \right| _{F^{s+1}}. \end{aligned}$$

The next term provides

$$\begin{aligned} \left| d_3^q\right| _{L^2H^{s-\frac{1}{2}}}&\le C \left| q^{(n)}_w-q^{(n-1)}_w\right| _{L^2H^{s-\frac{1}{2}}}\left| X^{(n)}-\alpha \right| _{L^\infty H^{s+\frac{1}{2}}}\left| \nabla X^{(n-1)}\right| _{L^\infty H^{s-\frac{1}{2}}}\\&\le C[v_0]T^\frac{1}{4} \Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}, \end{aligned}$$

and

$$\begin{aligned} \left| d_3^q\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}&\le C \left| q^{(n)}_w-q^{(n-1)}_w\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\left| (A^{-1}\circ X^{(n)}-A^{-1}) \nabla _{J}X^{(n)}\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}H^{\frac{1}{2}+\delta }}\\&\le C[v_0]T^\epsilon \Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}. \end{aligned}$$

As for $d_3^q$ we find

$$\begin{aligned}&\left| d_4^q\right| _{L^2H^{s-\frac{1}{2}}}\le C[v_0]T^\frac{1}{4}\Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}},\\&\quad \left| d_4^q\right| _{H_{(0)}^{\frac{s}{2}-\frac{1}{4}}L^2}\le C[v_0]T^\epsilon \Vert q^{(n)}_w-q^{(n-1)}_w\Vert _{H_{pr \, (0)}^{ht,s}}. \end{aligned}$$