(1.3). We derive a general formula for the commutators of the material
derivative of high order $\mathcal{D}_{t}^{k}$ with spacial derivatives. We
apply this to calculate $[\mathcal{D}_{t}^{k},\nabla]v$ and
$[\mathcal{D}_{t}^{k},\nabla]p$, which will produce two types of error terms,
(4.13) and (4.14), defined in the fluid domain $\Omega_{t}$. We will also
calculate the formula for $\mathcal{D}_{t}^{k}p$ on the moving boundary
$\Sigma_{t}$ in Lemma 4.7, which includes third type of error term defined in
(4.26). The precise structures of these error terms are complicated and we
only need to trace the order of the derivatives that appear. Therefore we
effectively use the notation from [31]

$\nabla^{k}f\star\nabla^{l}g$

to denote a contraction of some indexes of tensors $\nabla^{i}f$ and
$\nabla^{j}g$ for $i\leq k$ and $j\leq l$. Note that we include the lower
order derivatives.

We begin by recalling the following formulas from [49]

(4.1) $[\mathcal{D}_{t},\nabla]f=\mathcal{D}_{t}\nabla
f-\nabla\mathcal{D}_{t}f=-\nabla v^{T}\nabla f,$ (4.2)
$[\mathcal{D}_{t},\nabla_{\tau}]f=-(\nabla_{\tau}v)^{T}\nabla_{\tau}f$

and

(4.3) $[\mathcal{D}_{t},\Delta_{\Sigma}]f=\nabla_{\tau}^{2}f\star\nabla
v-\nabla_{\tau}f\cdot\Delta_{\Sigma}v+B\star\nabla v\star\nabla_{\tau}f.$

Let us also recall the material derivative of the normal field. We use the
shorthand notation $\nu=\nu_{\Sigma_{t}}$, $B=B_{\Sigma_{t}}$ and
$v_{n}=v\cdot\nu$. We have by [49]

(4.4) $\mathcal{D}_{t}\nu=-(\nabla_{\tau}v)^{T}\nu.$

Since $\nabla_{\tau}v_{n}=\nabla_{\tau}v^{T}\nu+Bv_{\tau}$, we may write (4.4)
as

(4.5) $\mathcal{D}_{t}\nu=-\nabla_{\tau}v_{n}+Bv_{\tau}.$

We need higher order versions of the commutation formula (4.1), i.e., for

$[\mathcal{D}_{t}^{l},\nabla^{k}]f=\mathcal{D}_{t}^{l}\nabla^{k}f-\nabla^{k}\mathcal{D}_{t}^{l}f.$

Recall the definition of the norm of an index vector
$\alpha=(\alpha)_{i=1}^{k}\in\mathbb{N}^{k}$

$|\alpha|=\sum_{i=1}^{k}\alpha_{i}$

and note that we include zero in the set of natural numbers $\mathbb{N}$.

######  Lemma 4.1.

For $l,k\in\mathbb{N}$ with $l,k\geq 1$ it holds

$[\mathcal{D}_{t}^{l},\nabla^{k}]f=\sum_{\begin{subarray}{c}|\alpha|\leq
k-1\\\ |\beta|\leq
l-1\end{subarray}}\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l}}\mathcal{D}_{t}^{\beta_{l}}v\star\nabla^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}f.$

###### Proof.

Let us first assume $l=1$ and prove

(4.6)
$\mathcal{D}_{t}\nabla^{k}f-\nabla^{k}\mathcal{D}_{t}f=\sum_{|\alpha|\leq
k-1}\nabla^{1+\alpha_{1}}v\star\nabla^{1+\alpha_{2}}f.$

We argue by induction over $k$ and observe immediately that the case $k=1$
follows from (4.1). Assume that (4.6) holds for $k-1$ and note that by (4.1)
we have

$\mathcal{D}_{t}\nabla^{k}f=\mathcal{D}_{t}\,\nabla(\nabla^{k-1}f)=\nabla\,\mathcal{D}_{t}(\nabla^{k-1}f)+\nabla
v\star(\nabla^{k}f).$

By induction assumption we have

$\begin{split}\nabla\,\mathcal{D}_{t}(\nabla^{k-1}f)&=\nabla\big{(}\nabla^{k-1}\mathcal{D}_{t}f+\sum_{|\alpha|\leq
k-2}\nabla^{1+\alpha_{1}}v\star\nabla^{1+\alpha_{2}}f\big{)}\\\
&=\nabla^{k}\mathcal{D}_{t}f+\sum_{|\alpha|\leq
k-1}\nabla^{1+\alpha_{1}}v\star\nabla^{1+\alpha_{2}}f.\end{split}$

This yields the claim for $l=1$.

The proof for $l\geq 1$ follows from a similar induction argument. Assume that
the claim holds for $l-1$ and note that

$\mathcal{D}_{t}^{l}\nabla^{k}f=\nabla^{k}\mathcal{D}_{t}^{l}f+\mathcal{D}_{t}([\mathcal{D}_{t}^{l-1},\nabla^{k}]f)+[\mathcal{D}_{t},\nabla^{k}](\mathcal{D}_{t}^{l-1}f).$

By the first claim we have

$[\mathcal{D}_{t},\nabla^{k}](\mathcal{D}_{t}^{l-1}f)=\sum_{|\alpha|\leq
k-1}\nabla^{1+\alpha_{1}}v\star\nabla^{1+\alpha_{2}}\mathcal{D}_{t}^{l-1}f.$

On the other hand, by the induction assumption we have

$\begin{split}\mathcal{D}_{t}[\mathcal{D}_{t}^{l-1},\nabla^{k}]f=\mathcal{D}_{t}\sum_{\begin{subarray}{c}|\alpha|\leq
k-1\\\ |\beta|\leq
l-2\end{subarray}}\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l-1}}\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla^{1+\alpha_{l}}\mathcal{D}_{t}^{\beta_{l}}f.\end{split}$

We use the Leibniz rule and the first claim to deduce that

$\begin{split}\mathcal{D}_{t}\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v=\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{1+\beta_{1}}v+\sum_{|\tilde{\alpha}|\leq\alpha_{1}}\nabla^{1+\tilde{\alpha}_{1}}v\star\nabla^{1+\tilde{\alpha}_{2}}\mathcal{D}_{t}^{\beta_{1}}v.\end{split}$

Similar formula holds also for
$\mathcal{D}_{t}\nabla^{1+\alpha_{l}}\mathcal{D}_{t}^{\beta_{l}}f$. Hence, we
obtain the claim. ∎

Let us next prove higher order commutation formulas for (4.2) and a formula
for $\mathcal{D}_{t}^{l}\nu$ and $\mathcal{D}_{t}B$. Below $a_{\beta}(\nu)$
and $a_{\alpha,\beta}(\nu,B)$ denote bounded coefficients which depend on
$\nu$ and on $\nu$ and $B$ respectively.

######  Lemma 4.2.

For $l\geq 1$ it holds

(4.7) $[\mathcal{D}_{t}^{l},\nabla_{\tau}]f=\sum_{|\beta|\leq
l-1}a_{\beta}(\nu)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v\star\nabla_{\tau}\mathcal{D}_{t}^{\beta_{l+1}}f$

and

(4.8)
$[\mathcal{D}_{t}^{l},\nabla^{2}_{\tau}]f=\sum_{\begin{subarray}{c}|\alpha|\leq
1\\\ |\beta|\leq
l-1\end{subarray}}a_{\alpha,\beta}(\nu,B)\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l}}\mathcal{D}_{t}^{\beta_{l}}v\star\nabla_{\tau}^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}f.$

Moreover we have

(4.9) $\mathcal{D}_{t}^{l}\nu=\sum_{|\beta|\leq
l-1}a_{\beta}(\nu)\,\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v$

and

(4.10) $\mathcal{D}_{t}^{l}B=\sum_{\begin{subarray}{c}|\alpha|\leq 1\\\
|\beta|\leq
l-1\end{subarray}}a_{\alpha,\beta}(\nu,B)\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{i}}v\star\cdots\star\nabla^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}v.$

###### Proof.

Let us first prove (4.9). First, the claim holds for $l=1$ by (4.4). Let us
assume that (4.9) holds for $l-1$. Then

$\mathcal{D}_{t}^{l}\nu=\mathcal{D}_{t}\sum_{|\beta|\leq
l-2}a_{\beta}(\nu)\,\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l-1}}v.$

By (4.4) it holds $\mathcal{D}_{t}a_{\beta}(\nu)=\tilde{a}_{\beta}(\nu)\nabla
v$ and by (4.1) we have

$\mathcal{D}_{t}\nabla\mathcal{D}_{t}^{\beta_{i}}v=\nabla\mathcal{D}_{t}^{\beta_{i}+1}v+\nabla
v\star\nabla\mathcal{D}_{t}^{\beta_{i}}v.$

Thus we deduce

$\mathcal{D}_{t}\sum_{|\beta|\leq
l-2}a_{\beta}(\nu)\,\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l-1}}v=\sum_{|\beta|\leq
l-1}\tilde{a}_{\beta}(\nu)\,\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v$

which implies (4.9).

Let us next prove (4.7). By (4.2) the claim holds for $l=1$. Let us assume
that the claim holds for $l-1$. Then

$\mathcal{D}_{t}^{l}\nabla_{\tau}f=\mathcal{D}_{t}\nabla_{\tau}\mathcal{D}_{t}^{l-1}f+\mathcal{D}_{t}\sum_{|\beta|\leq
l-2}a_{\beta}(\nu)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla_{\tau}\mathcal{D}_{t}^{\beta_{l}}f.$

As before we have by (4.4)
$\mathcal{D}_{t}a_{\beta}(\nu)=\tilde{a}_{\beta}(\nu)\nabla v$ and by (4.2)

$\mathcal{D}_{t}\nabla_{\tau}\mathcal{D}_{t}^{\beta_{i}}f=\nabla_{\tau}\mathcal{D}^{\beta_{i}+1}_{t}f+a(\nu)\nabla
v\star\nabla_{\tau}\mathcal{D}_{t}^{\beta_{i}}f.$

Therefore we obtain by Leibniz rule

$\mathcal{D}_{t}^{l}\nabla_{\tau}f=\nabla_{\tau}\mathcal{D}^{l}_{t}f+\sum_{|\beta|\leq
l-1}\tilde{a}_{\beta}(\nu)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v\star\nabla_{\tau}\mathcal{D}_{t}^{\beta_{l}}f$

and (4.7) follows.

We notice next that (4.10) follows from $B=\nabla_{\tau}\nu$ and by combining
(4.7) with (4.9). Finally we obtain (4.8) by first applying (4.7) as

$\begin{split}\mathcal{D}_{t}^{l}\nabla_{\tau}^{2}f&=\nabla_{\tau}(\mathcal{D}_{t}^{l}\nabla_{\tau}f)+\sum_{|\beta|\leq
l-1}a_{\beta}(\nu)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v\star\nabla_{\tau}\mathcal{D}_{t}^{\beta_{l+1}}\nabla_{\tau}f.\end{split}$

The claim then follows by differentiating (4.7). ∎

######  Remark 4.3.

By Lemma 4.2 we have in particular that

$\mathcal{D}_{t}^{l}v_{n}=\sum_{|\beta|\leq
l}a_{\beta}(\nu)\,\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v\star
D_{t}^{\beta_{l+1}}v,$

where $\beta_{i}\leq l-1$ for $i\leq l$. Moreover, since we may write the
Laplace-Beltrami operator as $\Delta_{\Sigma}f=\text{Tr}(\nabla_{\tau}^{2}f)$
then Lemma 4.2 yields

$\mathcal{D}_{t}^{l}\Delta_{\Sigma}f=\Delta_{\Sigma}\mathcal{D}_{t}^{l}f+\sum_{\begin{subarray}{c}|\alpha|\leq
1\\\ |\beta|\leq
l-1\end{subarray}}a_{\alpha,\beta}(\nu,B)\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l}}\mathcal{D}_{t}^{\beta_{l}}v\star\nabla_{\tau}^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}f.$

Let us next derive formulas for the divergence and the curl of the vector
field $\mathcal{D}_{t}^{l}v$. Recall that by (1.3) we have
$\operatorname{div}v=0$ which then implies

(4.11) $-\Delta p=\operatorname{div}(\mathcal{D}_{t}v)=\text{Tr}((\nabla
v)^{2})=\operatorname{div}\operatorname{div}(v\otimes v).$

For the curl we have $\operatorname{curl}(D_{t}v)=0$ and
$\omega=\operatorname{curl}v=\nabla v-\nabla v^{T}$ satisfies (see e.g. [49])

(4.12) $\mathcal{D}_{t}\omega=-\nabla v^{T}\omega-\omega\nabla v.$

We will derive formulas for $\operatorname{div}\mathcal{D}_{t}^{l}v$ and
$\operatorname{curl}\mathcal{D}_{t}^{l}v$ below by using (4.11), (4.12) and
the commutation formula in Lemma 4.1. To this aim we introduce two type of
error functions. The first type we denote by $R_{\operatorname{div}}^{l}$,
which stands for any function which can be written in the form

(4.13) $R^{l}_{\operatorname{div}}=\sum_{|\beta|\leq l}a_{\beta}(\nabla
v)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v,$

for $l\geq 0$ We also use the convention that the indexes are ordered as
$\beta_{1}\geq\beta_{2}\geq\dots\geq\beta_{l}$. The second type of error
function is slightly more general and it can be written in the form

(4.14) $R_{bulk}^{l}=\sum_{|\alpha|\leq 1,|\beta|\leq
l}a_{\alpha,\beta}(\nabla
v)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l}}v\star\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{l+1}}v,$

for $l\geq 0$. Note that $R_{bulk}^{l}$ has one higher order term compared to
$R^{l}_{\operatorname{div}}$. In particular, all functions of type
$R^{l}_{\operatorname{div}}$ are contained in $R_{bulk}^{l}$. The reason for
these two notations is that we need to estimate them in different norms. We
will do this in the next section. Note that using Lemma 4.1 and $-\nabla
p=\mathcal{D}_{t}v$ we deduce that

(4.15) $[\mathcal{D}_{t}^{l+1},\nabla]p=R_{bulk}^{l}.$

######  Lemma 4.4.

Let $l\geq 1$ and denote $\omega=\operatorname{curl}v$. Then it holds

$\mathcal{D}_{t}\nabla^{l}\omega=\nabla
v\star\nabla^{l}\omega+\sum_{|\alpha|\leq
l}\nabla^{1+\alpha_{1}}v\star\nabla^{1+\alpha_{2}}v.$

Moreover, $\operatorname{curl}\mathcal{D}_{t}^{l}v$ and
$\operatorname{div}\mathcal{D}_{t}^{l}v$ can be written in the form

$\operatorname{curl}\mathcal{D}_{t}^{l}v=R_{\operatorname{div}}^{l-1}\qquad\text{and}\qquad\operatorname{div}\mathcal{D}_{t}^{l}v=R_{\operatorname{div}}^{l-1}.$

We may also write the divergence of $\mathcal{D}_{t}^{l+1}v$ as

$\operatorname{div}\mathcal{D}_{t}^{l+1}v=\operatorname{div}\operatorname{div}(v\otimes\mathcal{D}_{t}^{l}v)+\operatorname{div}R^{l-1}_{bulk}.$

###### Proof.

The first claim is an immediate consequence of Lemma 4.1 and (4.12). The
second claim follows from Lemma 4.1 and from
$\operatorname{curl}(\mathcal{D}_{t}v)=0$. Similarly the third claim follows
from Lemma 4.1 and $\operatorname{div}v=0$.

Let us then prove the last claim. We begin by proving two useful identities.
First, we claim that for a vector field $F$ it holds

(4.16) $[\mathcal{D}_{t},\operatorname{div}]F=-\operatorname{div}(\nabla vF).$

Indeed, since $\operatorname{div}v=0$ we have

$\mathcal{D}_{t}\operatorname{div}F-\operatorname{div}\mathcal{D}_{t}F=\sum_{i,j=1}^{3}v_{i}\partial_{i}(\partial_{j}F_{j})-\partial_{i}(\partial_{j}F_{i}v_{j})=-\sum_{i,j=1}^{3}\partial_{i}v_{j}\partial_{j}F_{i}=-\operatorname{div}(\nabla
vF).$

The second identity follows also from $\operatorname{div}v=0$ and we may write
it

(4.17)
$\operatorname{div}\operatorname{div}(v\otimes\mathcal{D}_{t}^{l}v)=\operatorname{div}(\nabla\mathcal{D}_{t}^{l}v\,v).$

Let us prove the last claim in the case $l=1$. We use (4.1), (4.11), (4.16),
(4.17) and the definition of $R_{bulk}$ in (4.14) to deduce

$\begin{split}\operatorname{div}\mathcal{D}_{t}^{2}v&=\mathcal{D}_{t}\operatorname{div}\mathcal{D}_{t}v-[\mathcal{D}_{t},\operatorname{div}]\mathcal{D}_{t}v\\\
&=\mathcal{D}_{t}\operatorname{div}(\nabla v\,v)+\operatorname{div}(\nabla
v\,\mathcal{D}_{t}v)\\\ &=\operatorname{div}(\mathcal{D}_{t}(\nabla
v\,v))-\operatorname{div}(\nabla v\,\nabla
v\,v)+\operatorname{div}(R_{bulk}^{0})\\\
&=\operatorname{div}(\nabla\mathcal{D}_{t}v\,v))+\operatorname{div}(R_{bulk}^{0})\\\
&=\operatorname{div}\operatorname{div}(v\otimes\mathcal{D}_{t}v))+\operatorname{div}(R_{bulk}^{0}).\end{split}$

Let us assume that the claim holds for $l-1$. We argue as before and obtain by
(4.16), (4.17) and by the induction assumption

$\begin{split}\operatorname{div}\mathcal{D}_{t}^{l+1}v&=\mathcal{D}_{t}\operatorname{div}\mathcal{D}_{t}^{l}v-[\mathcal{D}_{t},\operatorname{div}]\mathcal{D}_{t}^{l}v\\\
&=\mathcal{D}_{t}\operatorname{div}\left(\nabla\mathcal{D}_{t}^{l-1}v\,v+R_{\operatorname{div}}^{l-2}\right)+\operatorname{div}(\nabla
v\,\mathcal{D}_{t}^{l}v).\end{split}$

We use (4.16), (4.1) and the definition of $R_{bulk}^{l-1}$ in (4.14) and
obtain

$\begin{split}\mathcal{D}_{t}\operatorname{div}&\left(\nabla\mathcal{D}_{t}^{l-1}v\,v\right)=\operatorname{div}\left(\mathcal{D}_{t}(\nabla\mathcal{D}_{t}^{l-1}v\,v)\right)+[\mathcal{D}_{t},\operatorname{div}](\nabla\mathcal{D}_{t}^{l-1}v\,v)\\\
&=\operatorname{div}(\nabla\mathcal{D}_{t}^{l}v\,v)+\operatorname{div}\left(\nabla\mathcal{D}_{t}^{l-1}v\star\nabla
v\star v+\nabla\mathcal{D}_{t}^{l-1}v\star\mathcal{D}_{t}v\right)\\\
&=\operatorname{div}(\nabla\mathcal{D}_{t}^{l}v\,v)+\operatorname{div}R_{bulk}^{l-1}\\\
&=\operatorname{div}\operatorname{div}(v\otimes\mathcal{D}_{t}^{l}v)+\operatorname{div}R_{bulk}^{l-1}.\end{split}$

Similarly we have

$\mathcal{D}_{t}\operatorname{div}R_{\operatorname{div}}^{l-2}=\operatorname{div}R_{\operatorname{div}}^{l-1}\qquad\text{and}\qquad\operatorname{div}(\nabla
v\,\mathcal{D}_{t}^{l}v)=\operatorname{div}R_{\operatorname{div}}^{l-1}$

and the claim follows. ∎

Let us then turn our focus on the pressure. By (4.11) and (1.3) we have that
$p$ is a solution of

$\begin{cases}-\Delta p=\operatorname{div}\operatorname{div}(v\otimes
v),\quad\text{in }\,\Omega_{t},\\\ p=H-\frac{Q(t)}{2}|\nabla
U|^{2},\quad\text{on }\,\Sigma_{t},\end{cases}$

where $Q(t)$ is a real valued function of time defined in (2.1) as

$Q(t)=\frac{Q}{(\text{Cap}(\Omega_{t}))^{2}},$

$U=U_{\Omega_{t}}$ is the capacitary potential and $H=H_{\Sigma_{t}}$ is the
mean curvature.

We need to derive the equation for $\mathcal{D}_{t}^{l}p$. We obtain the
equation for $\mathcal{D}_{t}^{l}p$ in the bulk from Lemma 4.4.

######  Remark 4.5.

By Lemma 4.4 and by (4.15) the function $-\Delta\mathcal{D}_{t}^{l+1}p$ can be
written as

$\begin{split}-\Delta\mathcal{D}_{t}^{l+1}p&=-\operatorname{div}\mathcal{D}_{t}^{l+1}\nabla
p+\operatorname{div}[\mathcal{D}_{t}^{l+1},\nabla]p=\operatorname{div}\mathcal{D}_{t}^{l+2}v+\operatorname{div}R_{bulk}^{l}\\\
&=\operatorname{div}\operatorname{div}(v\otimes\mathcal{D}_{t}^{l+1}v)+\operatorname{div}(R_{bulk}^{l}).\end{split}$

To find the formula for $\mathcal{D}_{t}^{l}p$ on the boundary $\Sigma_{t}$ is
more challenging. To that aim we first need to study the capacitary potential
$U$. We introduce an error term which appears when we deal with the capacity
term on the boundary, i.e., for $l\geq 0$ we denote by $R_{U}^{l}$ as
functions on $\Sigma_{t}$, which can be written in the form

(4.18) $R_{U}^{l}=\sum_{\begin{subarray}{c}|\alpha|+|\beta|\leq l+1\\\
|\beta|\leq
l\end{subarray}}a_{\alpha,\beta}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l}}v\star\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U.$

We note that $v$ is defined in $\Omega_{t}$ while $U$ is defined in
$\Omega_{t}^{c}$, but they are both well-defined on the boundary $\Sigma_{t}$.
We have the following formulas for $U$ on $\Sigma_{t}$.

######  Lemma 4.6.

Let $l\geq 1$. Then on $\Sigma_{t}$ it holds

$\mathcal{D}_{t}^{l}\nabla U=R_{U}^{l-1}$

and

$\begin{split}\mathcal{D}_{t}^{l+1}\nabla
U=&\nabla\partial_{t}^{l+1}U+\nabla^{2}U\,\mathcal{D}_{t}^{l}v\\\
&+\sum_{\begin{subarray}{c}\alpha+|\beta|+\gamma\leq l+1\\\ |\beta|\leq
l-1,\gamma\leq
l\end{subarray}}a_{\alpha,\beta,\gamma}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla^{1+\alpha}\partial_{t}^{\gamma}U.\end{split}$

Moreover we have the following formula for $\partial_{t}^{l+1}U$

$\partial_{t}^{l+1}U=-\partial_{\nu}U\,(\mathcal{D}_{t}^{l}v\cdot\nu)+R_{U}^{l-1}\qquad\text{on
}\,\Sigma_{t}.$

###### Proof.

The proof of the first statement is straightforward. Note that

$\mathcal{D}_{t}\nabla U=\nabla\partial_{t}U+\nabla^{2}Uv$

and

$\mathcal{D}_{t}^{2}\nabla
U=\nabla\partial_{t}^{2}U+\nabla^{2}U\mathcal{D}_{t}v+\nabla^{3}U\star v\star
v+\nabla^{2}\partial_{t}U\star v.$

Thus the first claim holds for $l=1,2$ and the second for $l=1$. The general
case $l\geq 1$ follows by an induction argument.

For the third claim we recall that the potential satisfies $U=1$ on
$\Sigma_{t}$. Therefore it holds $\mathcal{D}_{t}U=0$ on $\Sigma_{t}$ which we
write as

$\partial_{t}U=-(\nabla U\cdot v).$

Differentiating this yields

$\mathcal{D}_{t}^{l}\partial_{t}U=-(\nabla
U\cdot\mathcal{D}_{t}^{l}v)+\sum_{\begin{subarray}{c}i+j=l\\\ i\leq
l-1\end{subarray}}\mathcal{D}_{t}^{i}v\star\mathcal{D}_{t}^{j}\nabla U.$

By the first claim we have $\mathcal{D}_{t}^{j}\nabla U=R_{U}^{j-1}$ and thus
by the definition of $R_{U}^{j-1}$ in (4.18) we may write

$\mathcal{D}_{t}^{l}\partial_{t}U=-(\nabla
U\cdot\mathcal{D}_{t}^{l}v)+\sum_{\begin{subarray}{c}|\alpha|+|\beta|\leq l\\\
|\beta|\leq
l-1\end{subarray}}a_{\alpha,\beta}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U.$

It also holds

$\mathcal{D}_{t}^{l}\partial_{t}U=\partial_{t}^{l+1}U+\sum_{\begin{subarray}{c}|\alpha|+|\beta|\leq
l\\\ |\beta|\leq
l-1\end{subarray}}a_{\alpha,\beta}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U.$

Since $U$ is constant on $\Sigma_{t}$ it holds $\nabla
U=\partial_{\nu}U\,\nu$. This implies the third claim. ∎

We conclude this section by deriving a formula for $\mathcal{D}_{t}^{l+1}p$.
Recall that

(4.19) $p=H-\frac{Q(t)}{2}|\nabla U|^{2}\qquad\text{on }\,\Sigma.$

It is well known that (e.g. [15])

(4.20)
$\mathcal{D}_{t}H=-\Delta_{\Sigma}v_{n}-|B|^{2}v_{n}+\nabla_{\tau}H\cdot v$

where $v_{n}=v\cdot\nu$. Using the geometric fact

(4.21) $\Delta_{\Sigma}\nu=-|B|^{2}\nu+\nabla_{\tau}H$

and (4.19) we obtain the formula

(4.22)
$\mathcal{D}_{t}p=-\Delta_{\Sigma}v\cdot\nu-2B:\nabla_{\tau}v-Q(t)(\mathcal{D}_{t}\nabla
U\cdot\nabla U)-\frac{Q^{\prime}(t)}{2}|\nabla U|^{2}.$

We may write (4.22) in a different form. Indeed we use $\nabla
U=\partial_{\nu}U\,\nu=-|\nabla U|\nu$ and obtain

$\begin{split}-\mathcal{D}_{t}\nabla U\cdot\nabla
U&=-(\nabla\partial_{t}U\cdot\nabla U)-(\nabla^{2}U\nu\cdot
v)\,\partial_{\nu}U\\\ &=-(\nabla\partial_{t}U\cdot\nabla
U)+(\nabla^{2}U\nu\cdot\nu)\,v_{n}\,|\nabla U|-(\nabla^{2}U\nabla U\cdot
v_{\tau}).\end{split}$

We notice that

(4.23) $(\nabla^{2}U\nabla U\cdot v_{\tau})=\frac{1}{2}(\nabla_{\tau}|\nabla
U|^{2}\cdot v).$

Moreover, we recall that $U$ is harmonic in $\Omega_{t}^{c}$ and constant on
$\Sigma_{t}$. Therefore it holds by (3.4)

(4.24) $0=\Delta
U=\overbrace{\Delta_{\tau}U}^{=0}+(\nabla^{2}U\nu\cdot\nu)+H\partial_{\nu}U=(\nabla^{2}U\nu\cdot\nu)-H|\nabla
U|.$

Thus we have by (4.19), (4.20), (4.23), (4.24) that

(4.25)
$\mathcal{D}_{t}p=-\Delta_{\Sigma}v_{n}-|B|^{2}v_{n}-Q(t)\,\big{(}\partial_{\nu}U\,(\partial_{\nu}\partial_{t}U)-H|\nabla
U|^{2}v_{n}\big{)}+\langle\nabla_{\tau}p,v\rangle-Q^{\prime}(t)\frac{|\nabla
U|^{2}}{2}.$

In the next lemma we find a suitable expression for $\mathcal{D}_{t}^{l}p$ for
$l\geq 1$. Again we will have an error term which in this case is defined on
the boundary $\Sigma_{t}$ and is more complicated than the previous ones. We
define the error term $R_{p}^{l}$ for $l\geq 1$ as

(4.26) $R_{p}^{l}=R_{I}^{l}+R_{II}^{l}+R_{III}^{l}.$

where

(4.27) $\begin{split}&R_{I}^{l}=-(|B|^{2}-Q(t)H\,|\nabla
U|^{2})(\mathcal{D}_{t}^{l}v\cdot\nu)+(\nabla_{\tau}p\cdot\mathcal{D}_{t}^{l}v),\\\
&R_{II}^{l}=\sum_{|\alpha|\leq 1,\,|\beta|\leq
l-1}a_{\alpha,\beta,\gamma}(B)\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}v\qquad\text{and}\\\
&R_{III}^{l}=\sum_{\begin{subarray}{c}|\alpha|+|\beta|+|\gamma|\leq l+1\\\
|\beta|\leq l-1,\gamma_{i}\leq
l\end{subarray}}a_{\alpha,\beta,\gamma,Q}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{2}}U,\end{split}$

where the coefficients $a_{\alpha,\beta,\gamma,Q}(v)$ depend on $v$ and on the
derivatives $Q^{(k)}(t)$ for $k\leq l+1$. Above $a_{\alpha,\beta,\gamma}(B)$
means that the coefficient depends on the second fundamental form. For $l=1$
we need to quantify this dependence in which case $R_{II}^{1}$ reads as

(4.28) $R_{II}^{1}=a_{1}(\nu,\nabla v)\star B+a_{2}(\nu,\nabla
v)\star\nabla^{2}v.$

The reason why $R_{p}^{l}$ has three terms is that $R_{II}^{l}$ contains the
error terms arising from the surface tension and $R_{III}^{l}$ from the
capacity. The first term $R_{I}^{l}$ is separate merely from notational
reasons as it contains the highest order material derivatives.

######  Lemma 4.7.

For $l\geq 2$ It holds

$\mathcal{D}_{t}^{l}p=-\Delta_{\Sigma}(\mathcal{D}_{t}^{l-1}v\cdot\nu)-Q(t)\,\partial_{\nu}U\,(\partial_{\nu}\partial_{t}^{l}U)+R_{p}^{l-1}$

on $\Sigma_{t}$.

###### Proof.

We first claim that it holds

(4.29)
$\begin{split}\mathcal{D}_{t}^{l}p&=-(\Delta_{\Sigma}\mathcal{D}_{t}^{l-1}v)\cdot\nu-2B:\nabla_{\tau}(\mathcal{D}_{t}^{l-1}v)\\\
&-Q(t)(\nabla\partial_{t}^{l}U\cdot\nabla U)-Q(t)(\nabla^{2}U\nabla
U\cdot\mathcal{D}_{t}^{l-1}v)+R_{II}^{l-1}+R_{III}^{l-1}.\end{split}$

To obtain the claim (4.29) for $l=2$ we first recall that by (4.3) we have

$[\mathcal{D}_{t},\Delta_{\Sigma}]v=a_{1}(\nu,\nabla v)\star
B+a_{2}(\nu,\nabla v)\star\nabla^{2}v,$

and that (4.4) implies $\mathcal{D}_{t}\nu=-(\nabla_{\tau}v)^{T}\nu$ and (4.2)
implies $[\mathcal{D}_{t},\nabla_{\tau}]v=a(\nu)\nabla v\star\nabla v$. We use
(4.2) and (4.4) to obtain

$\begin{split}\mathcal{D}_{t}B=\mathcal{D}_{t}(\nabla_{\tau}\nu)&=\nabla_{\Sigma}(\mathcal{D}_{t}\nu)+[\mathcal{D}_{t},\nabla_{\tau}]\nu\\\
&=-\nabla_{\tau}((\nabla_{\tau}v)^{T}\nu)+a_{1}(\nu,\nabla v)\star B\\\
&=a_{1}(\nu,\nabla v)\star B+a_{2}(\nu,\nabla v)\star\nabla^{2}v.\end{split}$

We differentiate (4.22) and use the above identities and have

$\begin{split}\mathcal{D}_{t}^{2}p=-(\Delta_{\Sigma}\mathcal{D}_{t}v)\cdot\nu-2B:\nabla_{\tau}(\mathcal{D}_{t}v)-\mathcal{D}_{t}\Big{(}Q(t)(\mathcal{D}_{t}\nabla
U\cdot\nabla U)+\frac{Q^{\prime}(t)}{2}|\nabla
U|^{2}\Big{)}+R_{II}^{1}.\end{split}$

Lemma 4.6 yields

$\begin{split}\mathcal{D}_{t}(\mathcal{D}_{t}\nabla U\cdot\nabla
U)&=(\mathcal{D}_{t}^{2}\nabla U\cdot\nabla U)+(\mathcal{D}_{t}\nabla
U\cdot\mathcal{D}_{t}\nabla U)\\\ &=(\nabla\partial_{t}^{2}U\cdot\nabla
U)+(\nabla^{2}U\nabla U\cdot\,\mathcal{D}_{t}v)\\\
&\,\,\,\,\,\,\,\,\,\,+\sum_{\begin{subarray}{c}|\alpha|+|\beta|\leq 2,\\\
\beta_{i}\leq
1\end{subarray}}a_{\alpha,\beta}(v)\nabla^{1+\alpha_{1}}\partial_{t}^{\beta_{1}}U\star\nabla^{1+\alpha_{2}}\partial_{t}^{\beta_{2}}U.\end{split}$

We may embed the rest of the terms to $R_{III}^{1}$. This implies (4.29) for
$l=2$.

To obtain the claim (4.29) for $l\geq 2$ we differentiate (4.22)
$(l-1)$-times. Since the argument is similar to the case $l=2$, we only
highlight the most subtle steps. To identify the error terms we recall that by
Lemma 4.2 we have for $i\leq l-1$

$\mathcal{D}_{t}^{i}\nu=\sum_{|\beta|\leq
i-1}a_{\beta}(\nu)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{i}}v,$
$[\mathcal{D}_{t}^{i},\nabla_{\Sigma}]v=\sum_{|\beta|\leq
i-1}a_{\beta}(\nu)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{i+1}}v,$

and by Remark 4.3

$[\mathcal{D}_{t}^{i},\Delta_{\Sigma}]v=\sum_{\begin{subarray}{c}|\alpha|\leq
1\\\ |\beta|\leq
i-1\end{subarray}}a_{\alpha,\beta}(B)\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{i}}\mathcal{D}_{t}^{\beta_{i}}v\star\nabla_{\tau}^{1+\alpha_{i+1}}\mathcal{D}_{t}^{\beta_{i+1}}v.$

In order to treat the capacitary terms we first observe that

$\mathcal{D}_{t}^{l-1}\langle\mathcal{D}_{t}\nabla U,\nabla
U\rangle=\langle\mathcal{D}_{t}^{l}\nabla U,\nabla
U\rangle+\sum_{\begin{subarray}{c}i+j\leq l\\\ i,j\leq
l-1\end{subarray}}\mathcal{D}_{t}^{i}\nabla U\star\mathcal{D}_{t}^{i}\nabla U$

and then use Lemma 4.6 to deduce

$\begin{split}\mathcal{D}_{t}^{l-1}&(\mathcal{D}_{t}\nabla U\cdot\nabla
U)=(\nabla\partial_{t}^{l}U\cdot\nabla U)+(\nabla^{2}U\nabla
U\cdot\mathcal{D}_{t}^{l-1}v)\\\
&+\sum_{\begin{subarray}{c}|\alpha|+|\beta|+|\gamma|\leq l,\\\ |\beta|\leq
l-2,\,\gamma_{i}\leq
l-1\end{subarray}}a_{\alpha,\beta,\gamma}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l-1}}\star\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla^{1+\alpha_{2}}\partial_{t}^{\gamma_{2}}U.\end{split}$

This implies (4.29).

We proceed by calculating and by using (4.21)

$\begin{split}\Delta_{\Sigma}(\mathcal{D}_{t}^{l-1}v\cdot\nu)&=(\Delta_{\Sigma}\mathcal{D}_{t}^{l-1}v)\cdot\nu+2B:\nabla_{\tau}(\mathcal{D}_{t}^{l-1}v)+(\Delta_{\Sigma}\nu)\cdot(\mathcal{D}_{t}^{l-1}v)\\\
&=(\Delta_{\Sigma}\mathcal{D}_{t}^{l-1}v)\cdot\nu+2B:\nabla_{\tau}(\mathcal{D}_{t}^{l-1}v)-|B|^{2}(\mathcal{D}_{t}^{l-1}v\cdot\nu)+(\nabla_{\tau}H\cdot\mathcal{D}_{t}^{l-1}v).\end{split}$

Moreover, we recall that $\nabla U=\partial_{\nu}U\nu$ and that (4.24) implies
$(\nabla^{2}U\nu\cdot\nu)=H\partial_{\nu}U$. Therefore we have

$\begin{split}\langle\nabla^{2}U\nabla
U,\mathcal{D}_{t}^{l-1}v\rangle&=\langle\nabla^{2}U\nu,\nu\rangle\,\partial_{\nu}U\,(\mathcal{D}_{t}^{l-1}v\cdot\nu)+\langle\nabla^{2}U\nabla
U,(\mathcal{D}_{t}^{l-1}v)_{\tau}\rangle\\\ &=H|\nabla
U|^{2}(\mathcal{D}_{t}^{l-1}v\cdot\nu)+(\nabla_{\tau}\frac{|\nabla
U|^{2}}{2}\cdot\mathcal{D}_{t}^{l-1}v).\end{split}$

By combining the previous identities with (4.29) implies

$\begin{split}\mathcal{D}_{t}^{l}p&=-\Delta_{\Sigma}(\mathcal{D}_{t}^{l-1}v\cdot\nu)-Q(t)\,\partial_{\nu}U\,(\partial_{\nu}\partial_{t}^{l}U)\\\
&-(|B|^{2}-Q(t)\,H|\nabla
U|^{2})(\mathcal{D}_{t}^{l-1}v\cdot\nu)+\big{(}\nabla_{\tau}(H-Q(t)\frac{|\nabla
U|^{2}}{2})\cdot\mathcal{D}_{t}^{l-1}v\big{)}+R_{p}^{l-1}.\end{split}$

Hence, the claim follows from (4.19). ∎

##  5\. Estimation of the error terms

In the previous section we introduced four error terms
$R_{\operatorname{div}}^{l},R_{bulk}^{l},R_{U}^{l}$ and $R_{p}^{l}$, which
will appear later in the proof of the Main Theorem. These are nonlinear and
characterized by their order $l\geq 0$ and their precise forms can be found in
(4.13), (4.14) (4.18) and (4.26) respectively. The first two terms
$R_{\operatorname{div}}^{l}$ and $R_{bulk}^{l}$ are defined in the fluid
domain and appear already in the case when the shape of the drop does not
change. The term $R_{U}^{l}$ is due to the nonlinearity of the capacitary
term. The term $R_{p}^{l}$ is due to the nonlinear behavior of the pressure on
the moving boundary and it is by far the most difficult to treat.

In this section our goal is to estimate these error terms by the energy
quantity of order $l\in\ \mathbb{N}$ which we define as

(5.1)
$E_{l}(t):=\sum_{k=0}^{l}\|\mathcal{D}_{t}^{l+1-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}+\|v\|_{H^{\lfloor\frac{3}{2}(1+1)\rfloor}(\Omega_{t})}^{2}+\|\mathcal{D}_{t}^{l}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2}+1.$

The most difficult is to estimate the lowest order terms $R_{div}^{1},\dots$,
i.e., the case $l=1$, and we treat it separately. The difficulty of the case
$l=1$ makes the arguments in this section long and cumbersome.

As we explained in the introduction, the proof of the Main Theorem is by
induction argument, where we assume that we have the bound $E_{l-1}(t)\leq C$
and then use this to bound $E_{l}(t)$. We begin this section by proving that
the bound $E_{l-1}(t)\leq C$ for $l\geq 2$ implies

$\|B\|_{H^{\frac{3}{2}l-1}(\Sigma_{t})}\leq C.$

This will guarantee that every step improves the regularity of the flow.
Perhaps the most challenging part is to start the argument and we show in
Section 6 that the a priori bounds (1.7) imply the following estimate on the
pressure

$\|p\|_{H^{1}(\Omega_{t})}\leq C$

for all $t\leq T$. We will show that this implies the following curvature
bound

$\|B\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq C,$

which in particular implies the bound $\|B\|_{L^{4}(\Sigma_{t})}\leq C$.

We notice that the above curvature bounds ensure that $\Sigma_{t}$ satisfies
the condition (Hm) for $m=\lfloor\frac{3}{2}l\rfloor+1$ when $l\geq 2$ and
$m=2$ for $l=1$. This means that the results from Section 2 such as
Proposition 2.1, Proposition 2.7, Corollary 2.9, Proposition 2.10, Proposition
2.11 and Proposition 2.12 hold for all $k\leq m$. We take this for granted in
the calculations throughout this section without further mention in order to
make the presentation less heavy.

We begin by estimating the capacitary potential $U$ by the pressure via the
identity (4.19). We note that in the next lemma the a priori
$C^{1,\alpha}$-bound for the boundary $\Sigma_{t}=\partial\Omega_{t}$ is
crucial.

######  Lemma 5.1.

Let $l\geq 1$ and assume that $\Sigma_{t}$ is uniformly
$C^{1,\alpha}(\Gamma)$-regular and satisfies the condition
$\|B\|_{L^{4}(\Sigma_{t})}\leq C$ when $l=1$ and
$\|B\|_{H^{\frac{3}{2}l-1}(\Sigma_{t})}\leq C$ when $l\geq 2$. Let $U$ be the
capacitary potential defined in (1.2). Then it holds

$\|\nabla^{1+k}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{k}(\Sigma_{t})})\qquad\text{on }\,\Sigma_{t}$

for all integers $k\leq\frac{3}{2}l+\frac{1}{2}$.

###### Proof.

Let us note that the assumptions on the curvature imply that $\Sigma_{t}$
satisfies the condition (Hm) for $m=\lfloor\frac{3}{2}l\rfloor+1$. In
particular, the condition $k\leq\frac{3}{2}l+\frac{1}{2}$ implies $k\leq m$.

Let us prove the claim by induction over $k$ and consider first the case
$k=0$. This follows immediately from Theorem 3.9 and from
$\|B\|_{L^{4}(\Sigma_{t})}\leq C$ as

$\|\nabla U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|B\|_{L^{2}(\Sigma_{t})})\leq C.$

Let us then fix $k$ and assume that the claim holds for $k-1$. Since $U$ is
constant on $\Sigma_{t}$, Theorem 3.9 implies

$\|\nabla^{1+k}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|B\|_{H^{k}(\Sigma_{t})}).$

By Proposition 2.12 we have

$\|B\|_{H^{k}(\Sigma_{t})}\leq C(1+\|H\|_{H^{k}(\Sigma_{t})})\leq
C(1+\|p\|_{H^{k}(\Sigma_{t})}+\||\nabla U|^{2}\|_{H^{k}(\Sigma_{t})}).$

Proposition 2.10 yields

$\||\nabla U|^{2}\|_{H^{k}(\Sigma_{t})}\leq C\|\nabla
U\|_{L^{\infty}(\Sigma_{t})}\|\nabla U\|_{H^{k}(\Sigma_{t})}.$

Since $\Omega_{t}$ is uniformly $C^{1,\alpha}$-regular we have by Schauder
estimates $\|\nabla U\|_{C^{1,\alpha}(\Sigma_{t})}\leq C$. By combining the
previous inequalities we obtain

(5.2) $\|\nabla^{1+k}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{k}(\Sigma_{t})}+\|\nabla U\|_{H^{k}(\Sigma_{t})})$

We claim next that under the assumptions of the lemma, it holds for every
smooth function $u:\Omega_{t}\to\mathbb{R}$ and for all $k\leq m$

(5.3) $\|\nabla u\|_{H^{k}(\Sigma_{t})}\leq
C(\|u\|_{L^{2}(\Sigma_{t})}+\|\nabla^{1+k}u\|_{L^{2}(\Sigma_{t})}).$

Indeed, for $k=1$ we have $\|\nabla
u\|_{H^{1}(\Sigma_{t})}\leq\|u\|_{L^{2}(\Sigma_{t})}+\|\nabla^{2}u\|_{L^{2}(\Sigma_{t})}$,
while for $k=2$ the assumption $\|B\|_{L^{4}}\leq C$ and the Sobolev embedding
imply

$\begin{split}\|\nabla
u\|_{H^{2}(\Sigma_{t})}&\leq\|u\|_{L^{2}(\Sigma_{t})}+\|\nabla^{3}u\|_{L^{2}(\Sigma_{t})}+\|B\star\nabla^{2}u\|_{L^{2}(\Sigma_{t})}\\\
&\leq\|u\|_{L^{2}(\Sigma_{t})}+\|\nabla^{3}u\|_{L^{2}(\Sigma_{t})}+\|B\|_{L^{4}(\Sigma_{t})}\|\nabla^{2}u\|_{L^{4}(\Sigma_{t})}\\\
&\leq
C(\|u\|_{L^{2}(\Sigma_{t})}+\|\nabla^{3}u\|_{L^{2}(\Sigma_{t})}).\end{split}$

The case $k\geq 3$ follows from the same argument. We will take (5.3) for
granted from now on.

We obtain by (5.2) and (5.3) that

$\|\nabla^{1+k}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{k}(\Sigma_{t})}+\|\nabla^{1+k}U\|_{L^{2}(\Sigma_{t})}).$

We deduce by Lemma 3.3, by interpolation and by the induction assumption (that
the claim holds for $k-1$)

$\begin{split}\|\nabla^{1+k}U\|_{L^{2}(\Sigma_{t})}&\leq
C(1+\|\nabla^{k}U\|_{H^{1}(\Sigma_{t})})\leq\varepsilon\|\nabla^{k}U\|_{H^{\frac{3}{2}}(\Sigma_{t})}+C_{\varepsilon}(1+\|\nabla^{k}U\|_{L^{2}(\Sigma_{t})})\\\
&\leq\varepsilon\|\nabla^{1+k}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}(1+\|p\|_{H^{k-1}(\Sigma_{t})}).\end{split}$

Thus by choosing $\varepsilon$ small enough we obtain the claim. ∎

From Lemma 5.1 we deduce that an estimate on the pressure implies bound on the
curvature. The statement follows from the proof of Lemma 5.1 and we leave the
proof for the reader.

######  Lemma 5.2.

Let $l\geq 1$ and assume that $\Sigma_{t}$ is uniformly
$C^{1,\alpha}(\Gamma)$-regular and for $l=1$ it holds
$\|p\|_{H^{1}(\Omega_{t})}\leq C$ and for $l\geq 2$ it holds $E_{l-1}(t)\leq
C$. In the case $l=1$ we have

$\|B\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq C$

and $\|B\|_{L^{4}(\Sigma_{t})}\leq C$. In the case $l\geq 2$ we have

$\|B\|_{H^{\frac{3}{2}l-1}(\Sigma_{t})}\leq C.$

Moreover for $l\geq 1$ we have

$\|B\|_{H^{k}(\Sigma_{t})}\leq C(1+\|p\|_{H^{k}(\Sigma_{t})})$

for integers $k\leq\frac{3}{2}l+\frac{1}{2}$.

From now on we will assume that, in addition to (1.7), we have for $l=1$ the
estimate $\|p\|_{H^{1}(\Omega_{t})}\leq C$ and for $l\geq 2$ $E_{l-1}(t)\leq
C$. By Lemma 5.2 these imply curvature bounds that we mentioned at the
beginning of the section.

We begin to estimate the error terms and we begin with
$R_{\operatorname{div}}^{l}$ defined in (4.13).

######  Lemma 5.3.

Consider $R_{\operatorname{div}}^{l}$ defined in (4.13). Assume that (1.7)
holds and $\|p\|_{H^{1}(\Omega_{t})}\leq C$. Then we have

$\|R_{\operatorname{div}}^{1}\|_{H^{\frac{1}{2}}(\Omega_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

Let $l\geq 2$ and assume $E_{l-1}(t)\leq C$. Then it holds

(5.4) $\|R_{\operatorname{div}}^{l}\|_{H^{\frac{1}{2}}(\Omega_{t})}^{2}\leq
CE_{l}(t)$

and for integers $1\leq k\leq l$ and every $\varepsilon>0$ it holds

(5.5)
$\|R_{\operatorname{div}}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Omega_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

###### Proof.

For $l=1$ we have by the definition of $R_{\operatorname{div}}^{1}$ (4.13)
that

$R_{\operatorname{div}}^{1}=a(\nabla v)\star\nabla\mathcal{D}_{t}v,$

where $a$ is smooth. Note that in this case $E_{1}(t)$, defined in (5.1),
reads as

$E_{1}(t)=\|\mathcal{D}_{t}^{2}v\|_{L^{2}(\Omega_{t})}^{2}+\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}+\|v\|_{H^{3}(\Omega_{t})}^{2}+\|\mathcal{D}_{t}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2}+1.$

Since $\|B\|_{L^{4}}\leq C$, we may extend $\nabla v$ and
$\nabla\mathcal{D}_{t}v$ to $\mathbb{R}^{3}$, denote the extensions $F_{v}$
and $G_{v_{t}}$ respectively, such that the extensions satisfy

$\|F_{v}\|_{L^{\infty}(\mathbb{R}^{3})}\leq C\|\nabla
v\|_{L^{\infty}(\Omega_{t})},\qquad\|F_{v}\|_{H^{2}(\mathbb{R}^{3})}\leq
C\|\nabla v\|_{H^{2}(\Omega_{t})}$

and

$\|G_{v_{t}}\|_{L^{2}(\mathbb{R}^{3})}\leq
C\|\nabla\mathcal{D}_{t}v\|_{L^{2}(\Omega_{t})},\qquad\|G_{v_{t}}\|_{H^{\frac{1}{2}}(\mathbb{R}^{3})}\leq
C\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}}(\Omega_{t})}.$

Moreover, since $\Omega_{t}$ is bounded we may assume that $F_{v},G_{v_{t}}\in
C_{0}^{\infty}(B_{R})$.

We use the Kato-Ponce inequality (2.15) in $\mathbb{R}^{3}$ with
$p_{1}=2,q_{1}=\infty$, $p_{2}=\frac{12}{5}$ and $q_{2}=12$ to deduce

$\begin{split}\|R_{\operatorname{div}}^{1}\|_{H^{\frac{1}{2}}(\Omega_{t})}&\leq
C\|\nabla\mathcal{D}_{t}v\star a(\nabla v)\|_{H^{\frac{1}{2}}(\Omega_{t})}\leq
C\|G_{v_{t}}\star a(F_{v})\|_{H^{\frac{1}{2}}(\mathbb{R}^{3})}\\\ &\leq
C\|G_{v_{t}}\|_{H^{\frac{1}{2}}(\mathbb{R}^{3})}\|F_{v}\|_{L^{\infty}(\mathbb{R}^{3})}+C\|G_{v_{t}}\|_{L^{\frac{12}{5}}(\mathbb{R}^{3})}\|F_{v}\|_{W^{\frac{1}{2},12}(\mathbb{R}^{3})}.\end{split}$

Since $\|F_{v}\|_{L^{\infty}}\leq C$ we have

$\|G_{v_{t}}\|_{H^{\frac{1}{2}}(\mathbb{R}^{3})}\|F_{v}\|_{L^{\infty}(\mathbb{R}^{3})}\leq
C\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}}(\Omega_{t})}\leq
CE_{1}(t)^{\frac{1}{2}}.$

We have by using the Sobolev embedding $\|u\|_{L^{p}(B_{R})}\leq
C\|u\|_{H^{s}(B_{R})}=C\|u\|_{W^{s,2}(B_{R})}$, for $p=\frac{6}{3-2s}$ and
$s=\frac{1}{4}$, and by the general Gagliardo-Nirenberg inequality (2.13) that

$\|G_{v_{t}}\|_{L^{\frac{12}{5}}(\mathbb{R}^{3})}\leq
C\|G_{v_{t}}\|_{H^{\frac{1}{4}}(\mathbb{R}^{3})}\leq
C\|G_{v_{t}}\|_{H^{\frac{1}{2}}(\mathbb{R}^{3})}^{\frac{1}{2}}\|G_{v_{t}}\|_{L^{2}(\mathbb{R}^{3})}^{\frac{1}{2}}.$

By (2.13) we also have

$\|F_{v}\|_{W^{\frac{1}{2},12}(\mathbb{R}^{3})}\leq
C\|F_{v}\|_{W^{1,6}(\mathbb{R}^{3})}^{\frac{1}{2}}\|F_{v}\|_{L^{\infty}(\mathbb{R}^{3})}^{\frac{1}{2}}\leq
C\|F_{v}\|_{H^{2}(\mathbb{R}^{3})}^{\frac{1}{2}}\|F_{v}\|_{L^{\infty}(\mathbb{R}^{3})}^{\frac{1}{2}}.$

Therefore by $\|F_{v}\|_{L^{\infty}(\mathbb{R}^{3})}\leq C$,
$\|F_{v}\|_{H^{2}(\mathbb{R}^{3})}\leq C\|\nabla v\|_{H^{2}(\Omega_{t})}\leq
C\|v\|_{H^{3}(\Omega_{t})}$ and

$\|G_{v_{t}}\|_{L^{2}(\mathbb{R}^{3})}\leq
C\|\nabla\mathcal{D}_{t}v\|_{L^{2}(\Omega_{t})}=C\|p\|_{H^{2}(\Omega_{t})}$

we have

$\|G_{v_{t}}\|_{L^{\frac{12}{5}}(\mathbb{R}^{3})}\|F_{v}\|_{W^{\frac{1}{2},12}(\mathbb{R}^{3})}\leq
C\|p\|_{H^{2}(\Omega_{t})}^{\frac{1}{2}}\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{\frac{1}{2}}\|F_{v}\|_{H^{2}(\mathbb{R}^{3})}^{\frac{1}{2}}\leq
C\|p\|_{H^{2}(\Omega_{t})}^{\frac{1}{2}}E_{1}(t)^{\frac{1}{2}}.$

This implies the first inequality.

Let $l\geq 2$. In order to estimate the product (4.13) we use Proposition 2.10
to deduce

$\|R^{l}_{\operatorname{div}}\|_{H^{\frac{1}{2}}(\Omega_{t})}\leq\sum_{|\beta|\leq
l}\|\nabla\mathcal{D}^{\beta_{1}}_{t}v\|_{H^{\frac{1}{2}}(\Omega_{t})}\prod_{i=2}^{l}\|\nabla\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}(\Omega_{t})},$

where we use the convention that
$\beta_{1}\geq\beta_{2}\geq\dots\geq\beta_{l}$. By Recalling the definition of
$E_{l-1}(t)$ in (5.1), by the assumption $E_{l-1}(t)\leq C$ and by Sobolev
embedding it holds for $\beta_{i}\leq l-2$

$\|\nabla\mathcal{D}^{\beta_{i}}_{t}v\|_{L^{\infty}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\beta_{i}}v\|_{H^{3}(\Omega_{t})}^{2}\leq\sum_{k=0}^{l-1}\|\mathcal{D}_{t}^{l-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

For future purpose we also note that by the same argument we have

(5.6)
$\|\nabla^{1+\alpha}\mathcal{D}^{\beta}_{t}v\|_{L^{\infty}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\quad\text{for }\,\alpha+\beta\leq l-2.$

Moreover, by the same argument it holds

$\|\nabla\mathcal{D}_{t}^{l-1}v\|^{2}_{L^{\infty}(\Omega_{t})}\leq CE_{l}(t).$

We also have

$\|\nabla\mathcal{D}_{t}^{l-1}v\|_{H^{\frac{1}{2}}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l-1}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

Recall that by the definition of $R_{\operatorname{div}}^{l}$ above, the norm
of the index is $|\beta|\leq l$. Therefore since $l\geq 2$ it holds
$\beta_{i}\leq l-1$ for $i\geq 2$ and $\beta_{i}\leq l-2$ for $i\geq 3$. Thus
we conclude by the above estimates that

$\|R^{l}_{\operatorname{div}}\|_{H^{\frac{1}{2}}(\Omega_{t})}\leq
C(1+\|\nabla\mathcal{D}^{l}_{t}v\|_{H^{\frac{1}{2}}(\Omega_{t})}+\|\nabla\mathcal{D}_{t}^{l-1}v\|_{H^{\frac{1}{2}}(\Omega_{t})}\|\mathcal{D}_{t}^{l-1}v\|_{L^{\infty}(\Omega_{t})})\leq
CE_{l}(t)^{\frac{1}{2}},$

which implies (5.4).

The proof of (5.5) follows from similar argument and we merely sketch it. For
$k=l$ the statement is trivial. For $1\leq k\leq l-1$ we recall that

$R^{l-k}_{\operatorname{div}}=\sum_{|\beta|\leq l-k}a_{\beta}(\nabla
v)\nabla\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla\mathcal{D}_{t}^{\beta_{l-k}}v.$

First, if $k=1$ then by applying the previous estimate for $l-1$ we obtain

$\|R^{l-1}_{\operatorname{div}}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{l-1}(t).$

But now the condition $E_{l-1}(t)\leq C$ yields

$\|p\|_{H^{2}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}v\|_{H^{1}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

This implies the inequality for $k=1$.

Assume $2\leq k\leq l-1$. We apply Proposition 2.10 to bound

$\|R^{l-k}_{\operatorname{div}}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq\sum_{|\beta|\leq
l-k}\|\nabla\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{\infty}(\Sigma_{t})}\cdots\|\nabla\mathcal{D}_{t}^{\beta_{l-k-1}}v\|_{L^{\infty}(\Sigma_{t})}\|\nabla\mathcal{D}_{t}^{\beta_{l-k}}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}.$

Since $k\geq 2$ then $\beta_{i}\leq l-2$ for all $i$. Therefore by (5.6) we
have $\|\nabla\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}(\Omega_{t})}\leq C$
for all $i$. Moreover since $\beta_{i}\leq l-k$ it holds by the Trace Theorem
and by interpolation

$\begin{split}\|\nabla\mathcal{D}_{t}^{\beta_{i}}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}&\leq
C\|\mathcal{D}_{t}^{\beta_{i}}v\|_{H^{\frac{3}{2}k+1}(\Omega_{t})}^{2}\\\
&\leq\varepsilon\|\mathcal{D}_{t}^{\beta_{i}}v\|_{H^{\frac{3}{2}(k+1)}(\Omega_{t})}^{2}+C_{\varepsilon}\|\mathcal{D}_{t}^{\beta_{i}}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\\\
&\leq\varepsilon E_{l}(t)+C_{\varepsilon}E_{l-1}(t)\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.\end{split}$

Hence, we have (5.5). ∎

We proceed to bound the $L^{2}$-norm of $R_{bulk}^{l}$, which is defined in
(4.14), in the fluid domain $\Omega_{t}$. Formally $R_{bulk}^{l}$ is of order
$1/2$ higher than $R^{l}_{\operatorname{div}}$, and therefore this bound is of
the same order than the previous lemma.

######  Lemma 5.4.

Consider $R_{bulk}^{l}$ defined in (4.14). Assume that (1.7) holds and
$\|p\|_{H^{1}(\Omega_{t})}\leq C$. Then we have

$\|R_{bulk}^{1}\|_{L^{2}(\Omega_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

Let $l\geq 2$ and assume $E_{l-1}(t)\leq C$. Then we have

$\|R_{bulk}^{l}\|_{L^{2}(\Omega_{t})}^{2}\leq CE_{l}(t)$

and for integers $1\leq k\leq l-1$ and for $\varepsilon>0$ it holds

$\|R_{bulk}^{l-k}\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

###### Proof.

By (4.14) and the uniform bound on $\nabla v$ given by (1.7) we have a
pointwise estimate

$|R_{bulk}^{l}|\leq C\sum_{|\alpha|\leq 1,|\beta|\leq
l}|\nabla\mathcal{D}_{t}^{\beta_{1}}v|\cdots|\nabla\mathcal{D}_{t}^{\beta_{l}}v|\,|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{l+1}}v|.$

Let us first consider the case $l=1$. Then we have by the above inequality, by
$\mathcal{D}_{t}v=-\nabla p$ and by ignoring the terms of the form $|\nabla
v|$, as they are uniformly bounded, and obtain a pointwise bound

$|R_{bulk}^{1}|\leq
C\left(1+|\mathcal{D}_{t}^{2}v|+|\nabla\mathcal{D}_{t}v||\nabla p|\right).$

Therefore we have by Hölder’s inequality and by the Sobolev embedding

$\begin{split}\|R_{bulk}^{1}\|_{L^{2}(\Omega_{t})}^{2}&\leq
C(\|\mathcal{D}_{t}^{2}v\|_{L^{2}(\Omega_{t})}^{2}+\|\nabla
p\|_{L^{6}(\Omega_{t})}^{2}\|\nabla\mathcal{D}_{t}v\|_{L^{3}(\Omega_{t})}^{2})\\\
&\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})(1+\|\mathcal{D}_{t}^{2}v\|_{L^{2}(\Omega_{t})}^{2}+\|\mathcal{D}_{t}v\|_{H^{3/2}(\Omega_{t})}^{2})\\\
&\leq C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})(1+E_{1}(t)).\end{split}$

This implies the claim for $l=1$.

Let us then treat the case $l\geq 2$. Let us assume that the first $l$ indexes
are ordered as $\beta_{1}\geq\beta_{2}\geq\dots\geq\beta_{l}$. As before we
ignore all the terms in above which are uniformly bounded by the a priori
assumption and by the assumption $E_{l-1}(t)\leq C$. Recall first that by
(5.6) it holds

$\|\nabla\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}}\leq C\quad\text{when
}\quad\beta_{i}\leq l-2.$

Recall that it holds $|\beta|\leq l$. Therefore, if $\beta_{l+1}\geq l-1$ then
$\beta_{1}\leq 1$ and $\beta_{i}=0$ for $i\geq 2$. When $\beta_{l+1}=l-2$ then
the only possible other none-zero indexes are when $\beta_{1}=2$ or when
$\beta_{1}=\beta_{2}=1$. Finally when $l\geq 3$ and $\beta_{l+1}\leq l-3$ then
$\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{l+1}}v$ is itself
uniformly bounded and the only nontrivial terms are given by the indexes
$\beta_{1}=l-1$ and $\beta_{2}=1$. Hence, we have a pointwise bound which we
write by relabelling the indexes as

$\begin{split}&|R_{bulk}^{l}|\leq C\sum_{|\alpha|\leq
1}|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+l}v|+C\sum_{|\alpha|\leq
1,\beta_{i}\leq
l-1}|\nabla\mathcal{D}_{t}^{\beta_{1}}v||\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v|\\\
&+C\sum_{|\alpha|\leq
1}(|\nabla\mathcal{D}_{t}^{2}v|+|\nabla\mathcal{D}_{t}v|^{2})|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+l-2}v|+C(|\nabla\mathcal{D}_{t}^{l}v|+|\nabla\mathcal{D}_{t}^{l-1}v||\nabla\mathcal{D}_{t}v|)|\nabla\mathcal{D}_{t}v|\end{split}$

Then by Hölder’s inequality we deduce

(5.7) $\begin{split}\|R_{bulk}^{l}\|_{L^{2}(\Omega_{t})}^{2}&\leq
C\sum_{|\alpha|\leq
1}\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+l}v\|_{L^{2}}^{2}+C\sum_{|\alpha|\leq
1,\beta_{i}\leq
l-1}\|\nabla\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{3}}^{2}\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{L^{6}}^{2}\\\
&+C\sum_{|\alpha|\leq
1}(\|\nabla\mathcal{D}_{t}^{2}v\|_{L^{3}}^{2}+\|\nabla\mathcal{D}_{t}v\|_{L^{6}}^{4})\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+l-2}v\|_{L^{6}}^{2}\\\
&+C(\|\nabla\mathcal{D}_{t}^{l}v\|_{L^{3}}^{2}+\|\nabla\mathcal{D}_{t}^{l-1}v\|_{L^{6}}^{2}\|\nabla\mathcal{D}_{t}v\|_{L^{6}}^{2})\|\mathcal{D}_{t}v\|_{L^{6}}^{2}.\end{split}$

We bound the first term on RHS of (5.7) for $\alpha_{1}+\alpha_{2}\leq 1$

$\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+l}v\|_{L^{2}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}^{l+1}v\|_{L^{2}(\Omega_{t})}^{2}+\|\mathcal{D}_{t}^{l}v\|_{H^{3/2}(\Omega_{t})}^{2}\leq
E_{l}(t).$

We claim that the next term with the $L^{3}$-norm,
$\|\nabla\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{3}}$, is bounded. Indeed, we use
the Sobolev embedding, the induction assumption and the fact that
$\beta_{1}\leq l-1$ and have

$\|\nabla\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{3}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}\leq\sum_{k=0}^{l}\|\mathcal{D}_{t}^{l-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\leq
E_{l-1}(t)\leq C.$

We bound the coupling term with $\alpha_{1}+\alpha_{2}\leq 1$ and
$\beta_{2}\leq l-1$ by

$\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{L^{6}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{H^{1+\alpha_{1}}(\Omega_{t})}^{2}\leq
C\sum_{k=0}^{l}\|\mathcal{D}_{t}^{l+1-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\leq
CE_{l}(t).$

We proceed to the next row in (5.7) and for $\alpha_{1}+\alpha_{2}\leq 1$ we
have

$\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+l-2}v\|_{L^{6}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}^{\alpha_{2}+l-2}v\|_{H^{1+\alpha_{1}}(\Omega_{t})}^{2}\leq
C\sum_{k=0}^{l}\|\mathcal{D}_{t}^{l-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

We also have

$\|\nabla\mathcal{D}_{t}^{2}v\|_{L^{3}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}^{2}v\|_{H^{3/2}(\Omega_{t})}^{2}\leq
CE_{2}(t)\leq CE_{l}(t)$

since $l\geq 2$. Moreover, by interpolation it holds

$\|\nabla\mathcal{D}_{t}v\|_{L^{6}}\leq
C\|\nabla\mathcal{D}_{t}v\|_{H^{2}}^{\frac{1}{2}}\|\nabla\mathcal{D}_{t}v\|_{L^{2}}^{\frac{1}{2}}\leq
CE_{2}(t)^{\frac{1}{4}}E_{1}(t)^{\frac{1}{4}}\leq CE_{l}(t)^{\frac{1}{4}},$

when $l\geq 2$. Hence, the second row in (5.7) is bounded by $E_{l}(t)$.

We are left with the last row in (5.7). We bound the first term by

$\|\nabla\mathcal{D}_{t}^{l}v\|_{L^{3}(\Omega_{t})}^{2}\leq\|\nabla\mathcal{D}_{t}^{l}v\|_{H^{\frac{1}{2}}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}^{l}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}\leq
E_{l}(t)$

and the last by
$\|\mathcal{D}_{t}v\|_{L^{6}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}v\|_{H^{1}(\Omega_{t})}^{2}\leq
CE_{1}(t)\leq C$. Finally we treat the two remaining terms by the same
argument. Indeed for $\beta\leq l-1$ we have by interpolation as before

$\|\nabla\mathcal{D}_{t}^{\beta}v\|_{L^{6}(\Omega_{t})}\leq
C\|\nabla\mathcal{D}_{t}^{\beta}v\|_{H^{2}(\Omega_{t})}^{\frac{1}{2}}\|\nabla\mathcal{D}_{t}^{\beta}v\|_{L^{2}(\Omega_{t})}^{\frac{1}{2}}\leq
CE_{l}(t)^{\frac{1}{4}}E_{l-1}(t)^{\frac{1}{4}}\leq CE_{l}(t)^{\frac{1}{4}}.$

Hence, we have

$\|R_{bulk}^{l}\|_{L^{2}(\Omega_{t})}^{2}\leq CE_{l}(t).$

We are left with the last inequality. For $k=1$ the claim follows by applying
the previous inequality with $l-1$. Let us then assume $l\geq 3$ and $2\leq
k\leq l-1$. By definition of $R_{bulk}^{l-k}$ in (4.14) it holds $|\beta|\leq
l-k\leq l-2$. Therefore (5.6) implies

$\|\nabla\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}(\Omega_{t})}\leq\|\mathcal{D}_{t}^{\beta_{i}}v\|_{H^{3}(\Omega_{t})}\leq
C$

for all $i$. Therefore by Proposition 2.10 and by relabelling the indexes we
have

$\begin{split}\|R_{bulk}^{l-k}\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}\leq
C\sum_{\begin{subarray}{c}{|\beta|\leq l-k}\\\ {|\alpha|\leq
1}\end{subarray}}&\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}\\\
&+\|\nabla\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{L^{\infty}(\Omega_{t})}.\end{split}$

Since $\beta_{2}\leq l-k$, we may estimate the first term on RHS as

$\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}^{2}\leq
C\sum_{i=0}^{l-1}\|\mathcal{D}_{t}^{l-i}v\|_{H^{\frac{3}{2}i}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

We estimate the second similarly by using $\beta_{1}\leq l-k$

$\|\nabla\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}^{2}\leq\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

Finally we bound the last term by the Sobolev embedding, by $\beta_{2}\leq
l-k$, $\alpha_{1}+\alpha_{2}\leq 1$ and by interpolation

$\begin{split}\|\nabla^{\alpha_{1}}\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{L^{\infty}(\Omega_{t})}^{2}&\leq
C\|\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{H^{2+\alpha_{1}}(\Omega_{t})}^{2}\\\
&\leq\varepsilon\|\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{H^{\frac{3}{2}(2+\alpha_{1})}(\Omega_{t})}^{2}+C_{\varepsilon}\|\mathcal{D}_{t}^{\alpha_{2}+\beta_{2}}v\|_{L^{2}(\Omega_{t})}^{2}\\\
&\leq\varepsilon E_{l}(t)+C_{\varepsilon}E_{l-1}(t)\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.\end{split}$

Thus we have

$\|R_{bulk}^{l-k}\|_{H^{\frac{3}{2}(k-1)}(\Omega_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

∎

The two previous error bounds in Lemma 5.3 and Lemma 5.4 are similar in the
sense that they only involve the material derivatives of the velocity field.
We proceed to the error terms which involve the time derivatives of the
capacity potential. Note that $\partial_{t}^{k}U$ for all $k$ is a harmonic
function in $\Omega_{t}^{c}$ but not constant on $\Sigma_{t}$. We will use
again Theorem 3.9 together with Lemma 4.6 which gives the formula for
$\partial_{t}^{k}U$ on the boundary $\Sigma_{t}$.

We first prove a generic bound which will be useful when we bound the
pressure.

######  Lemma 5.5.

Let $l\geq 2$ and assume that (1.7) and the condition $E_{l-1}(t)\leq C$ hold.
Let $\alpha,\beta\geq 0$ be integers. When $\alpha+\beta\leq l$, it holds

(5.8)
$\|\nabla^{1+\alpha}\partial_{t}^{\beta}U\|_{H^{\frac{\alpha}{2}+\frac{1}{2}}(\Sigma_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

On the other hand, when $\alpha+\beta\leq l+1$ and $\beta\leq l$ then

(5.9)
$\|\nabla^{1+\alpha}\partial_{t}^{\beta}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
CE_{l}(t).$

###### Proof.

Instead of (5.8) we prove in fact a slightly stronger result, namely

(5.10)
$\begin{split}&\|\nabla^{1+\alpha}\partial_{t}^{\beta}U\|_{H^{\frac{\alpha}{2}+\frac{1}{2}}(\Sigma_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}\quad\text{when }\,\alpha\,\text{is even and }\\\
&\|\nabla^{1+\alpha}\partial_{t}^{\beta}U\|_{H^{\frac{\alpha}{2}+1}(\Sigma_{t})}^{2}\leq
CE_{l}(t)\quad\text{when }\,\alpha\,\text{ is odd}.\end{split}$

The inequality (5.8) then follows from (5.10) by interpolation.

We prove (5.10) by induction over $\beta$ and consider first the case
$\beta=0$. Note that then $\alpha\leq l$. Let us first consider the case when
$\alpha$ is even. Then by Lemma 5.1 and by interpolation we have

$\begin{split}\|\nabla^{1+\alpha}U\|_{H^{\frac{\alpha}{2}+\frac{1}{2}}(\Sigma_{t})}&\leq
C(1+\|p\|_{H^{\frac{3}{2}\alpha}(\Sigma_{t})})\leq\varepsilon\|p\|_{H^{\frac{3}{2}\alpha+\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}\|p\|_{L^{2}(\Sigma_{t})}\\\
&\leq\varepsilon\|p\|_{H^{\frac{3}{2}\alpha+\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}.\end{split}$

We use Lemma 3.7, $-\nabla p=\mathcal{D}_{t}v$, $\alpha\leq l$ and the
definition of $E_{l}(t)$ to estimate

$\|p\|_{H^{\frac{3}{2}\alpha+\frac{1}{2}}(\Sigma_{t})}^{2}\leq
C\|p\|_{H^{\frac{3}{2}\alpha+1}(\Omega_{t})}^{2}\leq
C(1+\|\mathcal{D}_{t}v\|_{H^{\frac{3l}{2}}(\Omega_{t})}^{2})\leq CE_{l}(t).$

This implies (5.10) when $\alpha$ is even. When $l$ is odd we have again by
Lemma 5.1, Lemma 3.7 and by the definition of $E_{l}$ that

$\|\nabla^{1+\alpha}U\|_{H^{\frac{\alpha}{2}+1}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{\frac{3\alpha}{2}+\frac{1}{2}}(\Sigma_{t})}^{2})\leq
C(1+\|p\|_{H^{\frac{3l}{2}+1}(\Omega_{t})}^{2})\leq CE_{l}(t).$

Let us assume that (5.10) holds for $\beta\leq k-1$ for $2\leq k\leq l$. In
particular, this implies that (5.8) holds for $\beta\leq k-1$. Let us consider
only the case when $\alpha$ is even, since the argument for $\alpha$ odd is
similar. We first observe that since $\alpha\leq l-k\leq l-1$ then by the
Trace Theorem

(5.11) $\|p\|_{H^{\frac{3}{2}\alpha}(\Sigma_{t})}^{2}\leq C(1+\|\nabla
p\|_{H^{\frac{3}{2}\alpha}(\Omega_{t})}^{2})\leq
C(1+\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}(l-1)}(\Omega_{t})}^{2})\leq
CE_{l-1}(t)\leq C.$

We have then by Theorem 3.9, Lemma 5.2 and by (5.11) that

$\begin{split}\|\nabla^{1+\alpha}\partial_{t}^{k}U\|_{H^{\frac{\alpha}{2}+\frac{1}{2}}(\Sigma_{t})}&\leq
C(1+\|p\|_{H^{\frac{3}{2}\alpha}(\Sigma_{t})}+\|\partial_{t}^{k}U\|_{H^{\frac{3}{2}\alpha+\frac{3}{2}}(\Sigma_{t})})\\\
&\leq
C(1+\|\partial_{t}^{k}U\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}).\end{split}$

We use the expression of $\partial_{t}^{k}U$ from Lemma 4.6 and Proposition
2.10 to estimate the last term in above as follows

$\begin{split}\|\partial_{t}^{k}U\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}\leq&\|\nabla
U\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}\|\mathcal{D}_{t}^{k-1}v\|_{L^{\infty}(\Sigma_{t})}\\\
&+\|\nabla
U\|_{L^{\infty}(\Sigma_{t})}\|\mathcal{D}_{t}^{k-1}v\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}+\|R^{k-2}_{U}\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}.\end{split}$

To estimate the first term on RHS let us first assume that $k\leq l-1$. Then
by (5.6) $\|\mathcal{D}_{t}^{k-1}v\|_{L^{\infty}(\Sigma_{t})}\leq C$ and since
$\alpha\leq l-1$ we have by (5.8) for $\beta=0$ that

$\|\nabla U\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

On the other hand, when $k=l$ then $\alpha=0$. Thus again by (5.8) it holds

$\|\nabla U\|_{H^{\frac{3}{2}}(\Sigma_{t})}^{2}\leq CE_{1}(t)\leq
CE_{l-1}(t)\leq C.$

By the Sobolev embedding, by interpolation, by the Trace Theorem and by
$\alpha+k\leq l$ we have

$\begin{split}\|\mathcal{D}_{t}^{k-1}v\|_{L^{\infty}(\Sigma_{t})}^{2}&\leq
C\|\mathcal{D}_{t}^{k-1}v\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}^{2}\\\
&\leq\varepsilon\|\mathcal{D}_{t}^{k-1}v\|_{H^{\frac{3}{2}\alpha+2}(\Sigma_{t})}^{2}+C_{\varepsilon}\|\mathcal{D}_{t}^{k-1}v\|_{L^{2}(\Sigma_{t})}^{2}\\\
&\leq
C\varepsilon\|\mathcal{D}_{t}^{k-1}v\|_{H^{\frac{3}{2}(\alpha+2)}(\Omega_{t})}^{2}+C_{\varepsilon}E_{l-1}(t)\leq
C\varepsilon E_{l}(t)+C_{\varepsilon}.\end{split}$

Hence, we need yet to prove

(5.12)
$\|R^{k-2}_{U}\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon},$

where $k\leq l$.

By the definition in (4.18) and by the Kato-Ponce inequality (Proposition
2.10) we may estimate

(5.13)
$\begin{split}\|R^{k-2}_{U}\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}\leq&C\sum_{\begin{subarray}{c}|\gamma|+|\beta|\leq
k-1\\\ |\beta|\leq
k-2\end{subarray}}\Big{(}\|\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{\infty}}\cdots\|\mathcal{D}_{t}^{\beta_{k-2}}v\|_{L^{\infty}}\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}\\\
&\,\,\,\,\,\,\,+\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}\cdots\|\mathcal{D}_{t}^{\beta_{k-2}}v\|_{L^{\infty}}\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{L^{\infty}(\Sigma_{t})}\Big{)}.\end{split}$

Since $|\beta|\leq k-2\leq l-2$, (5.6) implies
$\|\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{\infty}}\leq C$ for all $i$. Note that
$\alpha+k\leq l$ and $\beta_{1}\leq k-2$ implies $\alpha+\beta_{1}\leq l-2$.
Therefore we have by the Trace Theorem and by the definition of $E_{l-1}(t)$

$\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{3}{2}(\alpha+2)}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

We bound the both capacitary terms by the Sobolev embedding and by the
induction assumption, which states that (5.8) holds for $\beta\leq k-1$.
Indeed we have by $\gamma_{1}+\alpha\leq|\gamma|+\alpha\leq k-1+\alpha\leq
l-1$ and $\gamma_{2}\leq k-1$ that

(5.14)
$\begin{split}\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{L^{\infty}(\Sigma_{t})}^{2}&\leq
C\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{3}{2}(1+\alpha)}(\Sigma_{t})}^{2}\\\
&\leq
C(1+\|\nabla^{1+(1+\alpha+\gamma_{1})}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{1}{2}(1+\alpha+\gamma_{1})}(\Sigma_{t})}^{2})\\\
&\leq\varepsilon E_{l}(t)+C_{\varepsilon}.\end{split}$

Hence, we have (5.12) when $\alpha$ is even.

Let us then prove (5.9). We notice that Theorem 3.9 and (5.8) imply (5.9) when
$2\leq\alpha\leq l$. On the other Lemma 5.1 implies (5.9) when $\alpha=l+1$.
(Note that the assumption $\alpha\leq\frac{3}{2}l$ is satisfied for all
$\alpha\leq l+1$ since $l\geq 2$.) We need thus to consider the case
$\alpha=1$ and $\beta=l$. For this the argument is similar than before and we
only sketch it.

By Theorem 3.9 and by (5.11) we have

$\begin{split}\|\nabla^{2}\partial_{t}^{l}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}&\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}+\|\partial_{t}^{l}U\|_{H^{\frac{5}{2}}(\Sigma_{t})})\\\
&\leq C(1+\|\partial_{t}^{l}U\|_{H^{\frac{5}{2}}(\Sigma_{t})}).\end{split}$

We use the expression of $\partial_{t}^{l}U$ from Lemma 4.6 and Proposition
2.10 to estimate the last term in above as follows

$\begin{split}\|\partial_{t}^{l}U\|_{H^{\frac{5}{2}}(\Sigma_{t})}&\leq\|\nabla
U\|_{H^{\frac{5}{2}}(\Sigma_{t})}\|\mathcal{D}_{t}^{l-1}v\|_{L^{\infty}(\Sigma_{t})}\\\
&+\|\nabla
U\|_{L^{\infty}(\Sigma_{t})}\|\mathcal{D}_{t}^{l-1}v\|_{H^{\frac{5}{2}}(\Sigma_{t})}+\|R^{l-2}_{U}\|_{H^{\frac{5}{2}}(\Sigma_{t})}.\end{split}$

By Lemma 5.1 and Lemma 3.7 we have

$\begin{split}\|\nabla U\|_{H^{\frac{5}{2}}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Sigma_{t})}^{2})\leq C(1+\|\nabla
p\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2})\leq CE_{1}(t)\leq C.\end{split}$

Recall also that $\|\nabla U\|_{L^{\infty}}\leq C$. Sobolev embedding and
Trace Theorem yield

$\|\mathcal{D}_{t}^{l-1}v\|_{L^{\infty}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l-1}v\|_{H^{2}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l-1}v\|_{H^{3}(\Omega_{t})}^{2}\leq CE_{l}(t).$

Therefore we need yet to estimate
$\|R^{l-2}_{U}\|_{H^{\frac{5}{2}}(\Sigma_{t})}$.

Arguing as in (5.13) we obtain

$\begin{split}\|R^{l-2}_{U}\|_{H^{\frac{5}{2}}(\Sigma_{t})}\leq&\sum_{\begin{subarray}{c}|\gamma|+|\beta|\leq
l-1\\\ |\beta|\leq
l-2\end{subarray}}\Big{(}\|\mathcal{D}_{t}^{\beta_{1}}v\|_{L^{\infty}}\cdots\|\mathcal{D}_{t}^{\beta_{l-2}}v\|_{L^{\infty}}\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{5}{2}}(\Sigma_{t})}\\\
&\,\,\,\,\,\,\,+\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{5}{2}}(\Sigma_{t})}\cdots\|\mathcal{D}_{t}^{\beta_{l-2}}v\|_{L^{\infty}}\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{L^{\infty}(\Sigma_{t})}\Big{)}.\end{split}$

Arguing as before we deduce $\|\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}}\leq
C$ for all $i$ and

$\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{\frac{5}{2}}(\Sigma_{t})}^{2}\leq\|\mathcal{D}_{t}^{\beta_{1}}v\|_{H^{3}(\Omega_{t})}^{2}\leq
E_{l-1}(t)\leq C.$

Finally we use the fact that (5.9) holds for $\beta\leq l-1$ and
$\gamma_{1}+\gamma_{2}\leq l-1$ to conclude

$\begin{split}\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{L^{\infty}(\Sigma_{t})}^{2}&\leq
C\|\nabla^{1+\gamma_{1}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{5}{2}}(\Sigma_{t})}^{2}\\\
&\leq
C(1+\|\nabla^{1+(2+\gamma_{1})}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2})\leq
CE_{l}(t).\end{split}$

This concludes the proof. ∎

We need also the following bound on the capacitary potential and for the error
term $R_{U}^{l}$, defined in (4.18), associated with it. In the first
statement of the following lemma we need to relax the usual assumption on the
quantity (1.5) being bounded to assume that the set $\Omega_{t}$ is uniformly
$C^{1,\alpha}(\Gamma)$-regular and that the velocity satisfies
$\|v\|_{W^{1,4}(\Sigma_{t})}\leq C$. The point is that we need the following
estimate when we do not have the Lipschitz bound on $v$. This does not
complicate the proof and will be useful later.

######  Lemma 5.6.

Consider $R_{U}^{l}$ defined in (4.18). Assume that $\Sigma_{t}$ is
$C^{1,\alpha}(\Gamma)$-regular and satisfies

$\|p\|_{H^{1}(\Omega_{t})}+\|v\|_{W^{1,4}(\Sigma_{t})}\leq C.$

Then it holds

(5.15)
$\|\nabla\partial_{t}^{2}U\|_{L^{2}(\Sigma_{t})}^{2}+\|R_{U}^{1}\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

Let $l\geq 2$ and assume that $E_{l-1}(t)\leq C$. Then it holds

(5.16)
$\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Sigma_{t})}^{2}+\|R_{U}^{l}\|_{L^{2}(\Sigma_{t})}^{2}\leq
CE_{l}(t).$

###### Proof.

This time we only prove (5.15) since (5.16) follows from similar argument.
Note also that the $C^{1,\alpha}$-bound on $\Sigma_{t}$ implies
$C^{1,\alpha}$-bound on $U$. Recall also that $\|p\|_{H^{1}(\Omega_{t})}\leq
C$ implies
$\|B\|_{L^{4}(\Sigma_{t})}\leq\|B\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq C$ by
Lemma 5.2. We begin by noticing that, since $\partial_{t}^{2}U$ is harmonic,
it holds by Lemma 3.3

$\|\nabla\partial_{t}^{2}U\|_{L^{2}(\Sigma_{t})}\leq
C(1+\|\partial_{t}^{2}U\|_{H^{1}(\Sigma_{t})}).$

Then by Lemma 4.6 we have

$\|\partial_{t}^{2}U\|_{H^{1}(\Sigma_{t})}\leq
C(1+\|\partial_{\nu}U(\mathcal{D}_{t}v\cdot\nu)\|_{H^{1}(\Sigma_{t})}+\|R_{U}^{0}\|_{H^{1}(\Sigma_{t})}),$

where

$R_{U}^{0}=\sum_{|\alpha|\leq
1}a_{\alpha}(v)\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U.$

Since $\|\nabla U\|_{L^{\infty}(\Sigma_{t})}\leq C$, we may estimate by
Proposition 2.10, $\|B\|_{L^{4}}\leq C$ and by the Sobolev embedding

(5.17) $\begin{split}&\|\partial_{t}^{2}U\|_{H^{1}(\Sigma_{t})}\\\ &\leq
C+C\|\mathcal{D}_{t}v\cdot\nu\|_{H^{1}(\Sigma_{t})}+C(\|\nabla^{2}U\|_{L^{4}(\Sigma_{t})}+\|B\|_{L^{4}(\Sigma_{t})})\|\mathcal{D}_{t}v\cdot\nu\|_{L^{4}(\Sigma_{t})}+\|R_{U}^{0}\|_{H^{1}(\Sigma_{t})}\\\
&\leq
C+C(1+\|\nabla^{2}U\|_{L^{4}(\Sigma_{t})})\|\mathcal{D}_{t}v\cdot\nu\|_{H^{1}(\Sigma_{t})}+\|R_{U}^{0}\|_{H^{1}(\Sigma_{t})}.\end{split}$

We have by the Sobolev embedding and by Lemma 5.1

(5.18)
$\|\nabla^{2}U\|_{L^{4}(\Sigma_{t})}\leq\|\nabla^{2}U\|_{H^{1/2}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}).$

Since $\|\mathcal{D}_{t}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2}\leq E_{1}(t)$, we
need yet to show that $\|R_{U}^{0}\|_{H^{1}(\Sigma_{t})}^{2}\leq CE_{1}(t)$.

Since $\|v\|_{L^{\infty}(\Sigma_{t})}\leq C\|v\|_{W^{1,4}(\Sigma_{t})}\leq C$
we have by the Sobolev embedding

$\begin{split}\|R_{U}^{0}\|_{H^{1}(\Sigma_{t})}&\leq C\sum_{|\alpha|\leq
1}\|v\|_{L^{\infty}}\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{H^{1}(\Sigma_{t})}+C\sum_{|\alpha|\leq
1}\|v\|_{W^{1,4}}\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{L^{4}(\Sigma_{t})}\\\
&\leq
C(1+\|\nabla^{2}U\|_{H^{1}(\Sigma_{t})}+\|\nabla\partial_{t}U\|_{H^{1}(\Sigma_{t})}).\end{split}$

Lemma 5.1 and Lemma 3.7 yield

(5.19) $\begin{split}\|\nabla^{3}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Sigma_{t})}^{2})\leq C(1+\|\nabla
p\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2})\leq CE_{1}(t).\end{split}$

We bound $\|\nabla^{2}\partial_{t}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}$ with a
similar argument and thus we only sketch it. First, we recall that it holds
$\partial_{t}U=-\nabla U\cdot v$ on $\Sigma_{t}$. We use Theorem 3.9, Lemma
5.2, Proposition 2.10 and (5.19) to deduce

(5.20)
$\begin{split}\|\nabla^{2}\partial_{t}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}&\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|\nabla U\cdot
v\|_{H^{\frac{5}{2}}(\Sigma_{t})}^{2})\\\ &\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|\nabla^{3}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}+\|v\|_{H^{3}(\Omega_{t})}^{2})\\\
&\leq CE_{1}(t).\end{split}$

We thus deduce by (5.17), (5.18), (5.19) and (5.20) that

(5.21) $\|\nabla\partial_{t}^{2}U\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

We are left with

(5.22) $\|R_{U}^{1}\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

To this aim we recall the definition of $R_{U}^{1}$ in (4.18)

$R_{U}^{1}=\sum_{|\alpha|+\beta\leq 2,\,\beta\leq
1}a_{\alpha,\beta}\mathcal{D}_{t}^{\beta}v\star\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U.$

We use Hölder’s inequality as

$\|R_{U}^{1}\|_{L^{2}(\Sigma_{t})}\leq C\big{(}\sum_{|\alpha|\leq
2}\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{L^{2}(\Sigma_{t})}+\|\mathcal{D}_{t}v\|_{L^{4}(\Sigma_{t})}\sum_{|\alpha|\leq
1}\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{L^{4}(\Sigma_{t})}\big{)}.$

We have by (5.19), (5.20) and (5.21)

$\sum_{|\alpha|\leq
2}\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

By the Sobolev embedding and $-\nabla p=\mathcal{D}_{t}v$ it holds

$\|\mathcal{D}_{t}v\|_{L^{4}(\Sigma_{t})}\leq\|\mathcal{D}_{t}v\|_{H^{1}(\Omega_{t})}\leq
C\|p\|_{H^{2}(\Omega_{t})}.$

Moreover (5.19) and (5.20) imply for $\alpha_{1}+\alpha_{2}\leq 1$

$\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{L^{4}(\Sigma_{t})}^{2}\leq\|\nabla^{1+\alpha_{1}}\partial_{t}^{\alpha_{2}}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
CE_{1}(t).$

Hence we have (5.22). ∎

Finally we need to bound the error term $R_{p}^{l}$, defined in (4.26), which
is associated with the pressure. This term is the most difficult to treat and
it turns out that the lower order case $l=1$ is the most challenging to deal
with. Therefore we state it as an own lemma.

######  Lemma 5.7.

Let $R_{p}^{l}$ be as defined in (4.26). Assume that (1.7) holds and
$\|p\|_{H^{1}(\Omega_{t})}\leq C$. Then it holds

$\|R_{p}^{1}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

###### Proof.

Let us begin by recalling that by the definition of $R_{p}^{1}$ in (4.26),
(4.27) (4.28) we may write

(5.23) $\begin{split}R_{p}^{1}=&-|\nabla_{\tau}p|^{2}-(|B|^{2}-Q(t)H|\nabla
U|^{2})\,\partial_{\nu}p+a_{1}(\nu,\nabla v)\star B+a_{2}(\nu,\nabla
v)\star\nabla^{2}v\\\ &+\sum_{|\alpha|+|\gamma|\leq 2,\,\gamma_{i}\leq
1}a_{\alpha,\gamma,Q}(v)\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla^{1+\alpha_{2}}\partial_{t}^{\gamma_{2}}U.\end{split}$

We first recall it holds $\|U\|_{C^{1,\alpha}(\Sigma_{t})}\leq C$. We may
bound the curvature by Sobolev embedding and interpolation as

$\|B\|_{C^{\alpha}(\Sigma_{t})}\leq C(1+\|p\|_{C^{\alpha}(\Sigma_{t})})\leq
C(1+\|\bar{\nabla}p\|_{L^{\frac{11}{5}}(\Sigma_{t})})\leq
C(1+\|p\|_{H^{2}(\Sigma_{t})}^{\theta}\|p\|_{L^{4}(\Sigma_{t})}^{1-\theta}),$

for $\theta<\frac{4}{9}$. Recall that $\|p\|_{H^{1}(\Omega_{t})}\leq C$
implies $\|p\|_{L^{4}(\Sigma_{t})},\|B\|_{L^{4}}\leq C$. By the Sobolev
embedding and by Lemma 3.7 we have

(5.24) $\begin{split}\|p\|_{H^{2}(\Sigma_{t})}^{2}\leq C\|\nabla
p\|_{H^{1}(\Sigma_{t})}^{2}\leq C\|\nabla
p\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}\leq CE_{1}(t).\end{split}$

Therefore we obtain

(5.25) $\|B\|_{C^{\alpha}(\Sigma_{t})}^{2}\leq
CE_{1}(t)^{\theta}\qquad\text{for }\,\theta<\frac{4}{9}.$

We may also bound the curvature simply as

(5.26) $\|B\|_{L^{\infty}(\Sigma_{t})}\leq\|B\|_{C^{\alpha}(\Sigma_{t})}\leq
C(1+\|p\|_{C^{\alpha}(\Sigma_{t})})\leq C(1+\|p\|_{H^{2}(\Omega_{t})}).$

Let us bound the first term in (5.23) which is of the highest order. We
observe that by interpolation it holds

$\|\nabla p\|_{L^{8}(\Sigma_{t})}\leq C\|\nabla
p\|_{H^{1}(\Sigma_{t})}^{\frac{1}{2}}\|\nabla
p\|_{L^{4}(\Sigma_{t})}^{\frac{1}{2}}.$

By using this, the Sobolev embedding and (2.17) we have

$\begin{split}\||\nabla_{\tau}p|^{2}\|_{H^{1/2}(\Sigma_{t})}&\leq C\||\nabla
p|^{2}\|_{H^{\frac{1}{2}}(\Sigma_{t})}+C\|\nu\|_{W^{1,4}(\Sigma_{t})}\||\nabla
p|^{2}\|_{L^{4}(\Sigma_{t})}\\\ &\leq C\||\nabla
p|^{2}\|_{H^{1}(\Omega_{t})}+C\|B\|_{L^{4}(\Sigma_{t})}\|\nabla
p\|_{L^{8}(\Sigma_{t})}^{2}\\\ &\leq C\|\nabla
p\|_{L^{6}(\Omega_{t})}\|\nabla^{2}p\|_{L^{3}(\Omega_{t})}+C\|\nabla
p\|_{L^{4}(\Sigma_{t})}\|\nabla p\|_{H^{1}(\Sigma_{t})}\\\ &\leq
C\|p\|_{H^{2}(\Omega_{t})}\|\nabla p\|_{H^{\frac{3}{2}}(\Omega_{t})}\\\ &\leq
C(1+\|p\|_{H^{2}(\Omega_{t})})E_{1}(t)^{\frac{1}{2}}.\end{split}$

This gives bound for the first term.

In order to bound the next term in (5.23) we let $\tilde{\nu}$ and $\tilde{B}$
be the harmonic extensions of the normal $\nu$ and of the second fundamental
form $B$ to $\Omega_{t}$. By maximum principle
$\|\tilde{B}\|_{L^{\infty}(\Omega_{t})}\leq\|B\|_{L^{\infty}(\Sigma_{t})}$,
while by standard results from harmonic analysis [16] it holds

$\|\tilde{B}\|_{W^{1,3}(\Omega_{t})}\leq C\|B\|_{W^{1,3}(\Sigma_{t})}$

and by (3.12) $\|\tilde{\nu}\|_{W^{1,4}(\Omega_{t})}\leq C$. Then we have

(5.27) $\|\nabla p\cdot\tilde{\nu}\|_{H^{1}(\Omega_{t})}\leq
C\|p\|_{H^{2}(\Omega_{t})}+C\|\tilde{\nu}\|_{W^{1,4}(\Omega_{t})}\|p\|_{W^{1,4}(\Omega_{t})}\leq
C\|p\|_{H^{2}(\Omega_{t})}.$

We by (5.25), (5.27) and by the Sobolev embedding

$\begin{split}\||B|^{2}\,&\partial_{\nu}p\|_{H^{1/2}(\Sigma_{t})}^{2}\leq
C(1+\|\nabla(|\tilde{B}|^{2}\,(\nabla
p\cdot\tilde{\nu}))\|_{L^{2}(\Omega_{t})}^{2}\\\ &\leq
C(1+\|B\|_{L^{\infty}(\Sigma_{t})}^{4}\|\nabla
p\cdot\tilde{\nu}\|_{H^{1}(\Omega_{t})}^{2}+\|B\|_{L^{\infty}(\Sigma_{t})}^{2}\|\nabla\tilde{B}\|_{L^{3}(\Omega_{t})}^{2}\|\nabla
p\|_{L^{6}(\Omega_{t})}^{2})\\\ &\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t)^{2\theta}+E_{1}(t)^{\theta}\|B\|_{W^{1,3}(\Sigma_{t})}^{2}\|p\|_{H^{2}(\Omega_{t})}^{2}),\end{split}$

for $\theta<\frac{4}{9}$. By interpolation in Proposition 2.8, by Lemma 5.2
and (5.24) we have

(5.28) $\|B\|_{W^{1,3}(\Sigma_{t})}\leq
C\|B\|_{H^{2}(\Sigma_{t})}^{\frac{5}{9}}\|B\|_{L^{4}(\Sigma_{t})}^{\frac{4}{9}}\leq
C(1+\|p\|_{H^{2}(\Sigma_{t})}^{\frac{5}{9}})\leq
CE_{1}(t)^{\frac{1}{2}\cdot\frac{5}{9}}.$

Therefore, since $\theta<\frac{4}{9}$, we may bound the second term as

$\||B|^{2}\,\partial_{\nu}p\|_{H^{1/2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

By the same argument we also have

(5.29) $\|H|\nabla U|^{2}\partial_{\nu}p\|_{H^{1/2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

Indeed, the same calculations as above lead to

$\|H|\nabla U|^{2}\partial_{\nu}p\|_{H^{1/2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t)^{\frac{4}{9}}(\|B\|_{W^{1,3}(\Sigma_{t})}^{2}+\||\nabla
U|^{2}\|_{W^{1,3}(\Sigma_{t})}^{2}).$

Recall that (5.28) yields $\|B\|_{W^{1,3}(\Sigma_{t})}^{2}\leq
CE_{1}(t)^{\frac{5}{9}}$. By Lemma 5.1 we deduce

$\begin{split}\||\nabla U|^{2}\|_{H^{2}(\Sigma_{t})}&\leq C\|\nabla
U\|_{L^{\infty}}\|\nabla U\|_{H^{2}(\Sigma_{t})}\\\ &\leq
C(1+\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})})\leq
C(1+\|p\|_{H^{2}(\Sigma_{t})})\leq CE_{1}(t)^{\frac{1}{2}}.\end{split}$

Hence, by interpolation

$\||\nabla U|^{2}\|_{W^{1,3}(\Sigma_{t})}\leq C\||\nabla
U|^{2}\|_{H^{2}(\Sigma_{t})}^{\frac{1}{3}}\||\nabla
U|^{2}\|_{L^{\infty}(\Sigma_{t})}^{\frac{2}{3}}\leq
CE_{1}(t)^{\frac{1}{2}\cdot\frac{1}{3}}$

and (5.29) follows.

The term $a_{1}(\nu,\nabla v)\star B$ is easy to bound and leave the details
for the reader. Also the term $a_{2}(\nu,\nabla v)\,\nabla^{2}v$ is not
difficult and we merely point out that by interpolation

$\|\nabla^{2}v\|_{L^{4}(\Omega_{t})}\leq
C\|v\|_{H^{3}(\Omega_{t})}^{1/2}\|\nabla
v\|_{L^{\infty}(\Omega_{t})}^{1/2}\leq C\|v\|_{H^{3}(\Omega_{t})}^{1/2}.$

Thus we have by (5.26)

$\begin{split}\|a_{1}(\nu,\nabla
v)\,\nabla^{2}v\|_{H^{\frac{1}{2}}(\Sigma_{t})}&\leq\|a_{1}(\nu,\nabla
v)\,\nabla^{2}v\|_{H^{1}(\Omega_{t})}\\\ &\leq
C\|\nabla^{3}v\|_{H^{3}(\Omega_{t})}+C\|\nabla^{2}v\|_{L^{4}(\Omega_{t})}^{2}+C\|B\|_{C^{\alpha}(\Sigma_{t})}\|\nabla^{2}v\|_{L^{2}(\Omega_{t})}\\\
&\leq C(1+\|p\|_{H^{2}(\Omega_{t})})\|v\|_{H^{3}(\Omega_{t})}.\end{split}$

Before we treat the last term in (5.23) we need to show that the coefficients
$a_{\alpha,\gamma,Q}$ are bounded. To this aim we need to show that
$Q^{(1)},Q^{(2)}$ are bounded since $a_{\alpha,\gamma,Q}$ depend smoothly on
them. It is clear that it is enough to show that the first and second
derivative of $\text{Cap}(\Omega_{t})$ are bounded. It is easy to see, and in
fact we already calculated, that

$\frac{d}{dt}\text{Cap}(\Omega_{t})=-\frac{1}{2}\int_{\Sigma_{t}}|\nabla
U|^{2}v_{n}d\mathcal{H}^{2}.$

This is clearly bounded. We calculate further and obtain

(5.30)
$\frac{d^{2}}{dt^{2}}\text{Cap}(\Omega_{t})=-\frac{1}{2}\int_{\Sigma_{t}}H_{\Sigma_{t}}|\nabla
U|^{2}v_{n}d\mathcal{H}^{2}-\int_{\Sigma_{t}}(\nabla\partial_{t}U\cdot\nabla
U)v_{n}+|\nabla U|^{2}\partial_{t}(v_{n})\,d\mathcal{H}^{2}.$

The first term on RHS is clearly bounded. For the second term on RHS in (5.30)
we note that since $U$ is constant on $\Sigma_{t}$ we have $\nabla U=-|\nabla
U|\nu$. Therefore

$|\int_{\Sigma_{t}}\nabla\partial_{t}U\cdot\nabla
Uv_{n}d\mathcal{H}^{2}|=|\int_{\Sigma_{t}}|\nabla
U|v_{n}\nabla\partial_{t}U\cdot\nu d\mathcal{H}^{2}|\leq\|\nabla
Uv_{n}\|_{H^{\frac{1}{2}}(\Sigma_{t})}\|\nabla\partial_{t}U\cdot\nu\|_{H^{-\frac{1}{2}}(\Sigma)}.$

We note that we may use the Kato Ponce inequality (Proposition 2.10) and Lemma
5.1 to deduce

$\|\nabla Uv_{n}\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq\|\nabla
U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\|v_{n}\|_{L^{\infty}(\Sigma_{t})}+\|v_{n}\|_{H^{\frac{1}{2}}(\Sigma_{t})}\|\nabla
U\|_{L^{\infty}(\Sigma_{t})}\leq C.$

Next we let $\tilde{U}_{t}$ the harmonic extension of $\partial_{t}U$ in
$\Omega_{t}$ and note that for any $\phi\in H^{\frac{1}{2}}(\Sigma_{t})$ it
holds

$\begin{split}\int_{\Sigma_{t}}\phi\nabla\partial_{t}U\cdot\nu\,d\mathcal{H}^{2}=\int_{\Omega_{t}}\operatorname{div}(\phi\nabla\tilde{U}_{t})\,dx&\leq\|\nabla\phi\|_{L^{2}(\Omega_{t})}\|\nabla\partial_{t}\tilde{U}\|_{L^{2}(\Omega_{t})}\\\
&\leq\|\phi\|_{H^{\frac{1}{2}}(\Sigma_{t})}\|\partial_{t}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}.\end{split}$

This and $\partial_{t}U=-\nabla U\cdot v$ imply

$\|\nabla\partial_{t}U\cdot\nu\|_{H^{-\frac{1}{2}}(\Sigma_{t})}\leq\|\partial_{t}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}=\|\partial_{\nu}Uv_{n}\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C.$

Let us estimate the last term in (5.30). We note that

$\partial_{t}(v_{n})=\mathcal{D}_{t}v\cdot\nu+a(\nabla v)\star
B_{\Sigma_{t}}.$

By (4.11) it holds $\operatorname{div}\mathcal{D}_{t}v=-\text{Tr}((\nabla
v)^{2})$. Therefore we estimate

$\begin{split}|\int_{\Sigma_{t}}|\nabla
U|^{2}\partial_{t}v_{n}\,d\mathcal{H}^{2}|&\leq C+\int_{\Sigma_{t}}|\nabla
U|^{2}\mathcal{D}_{t}v\cdot\nu d\mathcal{H}^{2}\\\ &\leq
C(1+\|\operatorname{div}\mathcal{D}_{t}v\|_{L^{2}(\Omega_{t})}+\|\nabla
U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\|\mathcal{D}_{t}v\|_{L^{2}(\Omega_{t})})\leq
C.\end{split}$

Thus we have $|Q^{(2)}(t)|\leq C$ and the coefficients $a_{\alpha,\gamma,Q}$
are bounded.

Let us treat the last term in (5.23). We may assume that
$\alpha_{1}+\gamma_{1}\geq\alpha_{2}+\gamma_{2}$ and assume first that
$\alpha_{1}+\gamma_{1}=2$ (in which case $\alpha_{2}=\gamma_{2}=0$). This
means that either $\alpha_{1}=2,\gamma_{1}=0$ or $\alpha_{1}=1,\gamma_{1}=1$.
Therefore we have by (5.19) and (5.20)

(5.31)
$\|\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
CE_{1}(t).$

We extend $v$ to the complement $\Omega_{t}^{c}$ such that it remains
Lipschitz. Recall that $U$ (and $\partial_{t}U$) is harmonic in
$\Omega_{t}^{c}$. Since $\Omega_{t}$ is bounded we may choose a large ball
such that $\Omega_{t}\subset B_{R/2}$ and
$\|U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\simeq\|\nabla
U\|_{L^{2}(B_{R}\setminus\Omega_{t})}$. Then by (5.18), (5.31) and by the
Sobolev embedding it holds

$\begin{split}\|a_{\alpha,\gamma}(v)&\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla
U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C\|a_{\alpha,\gamma}(v)\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla
U\|_{H^{1}(B_{R}\setminus\Omega_{t})}\\\ &\leq
C(1+\|\nabla^{2+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\|_{L^{2}(B_{R}\setminus\Omega_{t})}+\|\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\|_{L^{4}(B_{R}\setminus\Omega_{t})}\|\nabla^{2}U\|_{L^{4}(B_{R}\setminus\Omega_{t})})\\\
&\leq
C(1+\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})})(1+\|\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\|_{H^{\frac{1}{2}}(\Sigma_{t})})\\\
&\leq(1+\|p\|_{H^{1}(\Sigma_{t})})E_{1}(t)^{\frac{1}{2}}.\end{split}$

We are left with the last term (5.23) in the case when
$\alpha_{i}+\gamma_{i}\leq 1$ for $i=1,2$. We estimate this by the Kato-Ponce
inequality (Proposition 2.10) and the Sobolev embedding as

$\begin{split}\sum_{\alpha_{i}+\gamma_{i}\leq
1,\,i=1,2}&\|a_{\alpha,\gamma}(v)\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla^{1+\alpha_{2}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\\\
&\leq C\sum_{\alpha+\gamma\leq
1}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{L^{\infty}(\Sigma_{t})}\sum_{\alpha+\gamma\leq
1}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\\\
&\leq C\sum_{\alpha+\gamma\leq
1}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{3}{2}}(\Sigma_{t})}\sum_{\alpha+\gamma\leq
1}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}.\end{split}$

We bound the first term in the last row by (5.31)

$\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{3}{2}}(\Sigma_{t})}^{2}\leq
CE_{1}(t).$

We bound the last term in the last row when $\alpha=1$ and $\gamma=0$ by
(5.18) as before $\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})})$. We need yet to prove

(5.32) $\|\nabla\partial_{t}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})})$

to conclude the proof. We use the fact that $\partial_{t}U$ is harmonic and
$\partial_{t}U=\partial_{\nu}U\,v_{n}$ and therefore by Theorem 3.9 we deduce

$\|\nabla\partial_{t}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|\partial_{\nu}U\,v_{n}\|_{H^{3/2}(\Sigma_{t})}).$

We recall that $\|B\|_{L^{4}(\Sigma_{t})}\leq C$ and
$\|B\|_{H^{1}(\Sigma_{t})}\leq C(1+\|p\|_{H^{1}(\Sigma_{t})})$. Therefore we
deduce by by Proposition 2.10, the apriori bound
$\|v_{n}\|_{H^{2}(\Sigma_{t})}\leq C$ in (1.7) and by (5.18) that

$\begin{split}\|\partial_{\nu}U\,v_{n}\|_{H^{3/2}(\Sigma_{t})}&\leq
C(\|v_{n}\|_{H^{\frac{3}{2}}(\Sigma_{t})}+\|\nabla
U\|_{H^{\frac{3}{2}}(\Sigma_{t})}+\|\nu\|_{H^{\frac{3}{2}}(\Sigma_{t})})\\\
&\leq
C(1+\|v_{n}\|_{H^{2}(\Sigma_{t})}+\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}+\|B\|_{H^{1}(\Sigma_{t})})\\\
&\leq C(1+\|p\|_{H^{1}(\Sigma_{t})}).\end{split}$

Hence, we have (5.32) and the claim follows. ∎

We conclude this section with the higher order version of Lemma 5.7.

######  Lemma 5.8.

Let $l\geq 2$ and consider $R_{p}^{l}$ defined in (4.26). Assume that (1.7)
holds and $E_{l-1}(t)\leq C$. Then

(5.33) $\|R_{p}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq CE_{l}(t)$

and for integers $1\leq k\leq l-1$ and $\varepsilon>0$ it holds

(5.34) $\|R_{p}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

###### Proof.

Let us first prove (5.33). We begin by showing

(5.35) $\|R_{I}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq CE_{l}(t),$

where

$R_{I}^{l}=-(|B|^{2}+Q(t)\,H\,|\nabla
U|^{2})(\mathcal{D}_{t}^{l}v\cdot\nu)+\langle\nabla_{\tau}p,\mathcal{D}_{t}^{l}v\rangle.$

Let us first recall that $E_{1}(t)\leq C$ implies
$\|B\|_{H^{2}(\Sigma_{t})}\leq C$ and $\|p\|_{H^{2}(\Sigma_{t})}\leq C$. By
Lemma 5.1 this implies $\|\nabla^{3}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq C$.
In particular, $\|B\|_{L^{\infty}}\leq C$ and
$\|\nabla^{2}U\|_{L^{\infty}}\leq C$. Therefore we may bound by Sobolev
embedding and by Proposition 2.10

$\begin{split}\|(|B|^{2}+Q(t)\,H\,|\nabla
U|^{2})&(\mathcal{D}_{t}^{l}v\cdot\nu)\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq\|(|B|^{2}+Q(t)\,H\,|\nabla
U|^{2})(\mathcal{D}_{t}^{l}v\cdot\nu)\|_{H^{1}(\Sigma_{t})}^{2}\\\ &\leq
C(1+\|B\|_{W^{1,4}(\Sigma_{t})}^{2}\|\mathcal{D}_{t}^{l}v\cdot\nu\|_{L^{4}(\Sigma_{t})}^{2}+\|\mathcal{D}_{t}^{l}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2})\\\
&\leq
C(1+\|B\|_{H^{2}(\Sigma_{t})}^{2})\|\mathcal{D}_{t}^{l}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2}\\\
&\leq CE_{l}(t).\end{split}$

In order bound
$\|\nabla_{\tau}p\cdot\mathcal{D}_{t}^{l}v\|_{H^{\frac{1}{2}}(\Sigma_{t})}$ we
observe that by the curvature bound $\|B\|_{L^{\infty}}\leq C$, by $-\nabla
p=\mathcal{D}_{t}v$ and by the Sobolev embeddings
$\|u\|_{L^{3}(\Omega_{t})}\leq C\|u\|_{H^{\frac{1}{2}}(\Omega_{t})}$ and
$\|u\|_{L^{6}(\Omega_{t})}\leq C\|u\|_{H^{1}(\Omega_{t})}$ we have

$\begin{split}\|\nabla_{\tau}p\cdot\mathcal{D}_{t}^{l}v\|_{H^{\frac{1}{2}}(\Sigma_{t})}&\leq
C\|\mathcal{D}_{t}v\cdot\mathcal{D}_{t}^{l}v\|_{H^{1}(\Omega_{t})}\\\ &\leq
C(1+\|\nabla\mathcal{D}_{t}v\cdot\mathcal{D}_{t}^{l}v\|_{L^{2}(\Omega_{t})}+\|\mathcal{D}_{t}v\cdot\nabla\mathcal{D}_{t}^{l}v\|_{L^{2}(\Omega_{t})})\\\
&\leq
C(1+\|\nabla\mathcal{D}_{t}v\|_{L^{3}(\Omega_{t})}\|\mathcal{D}_{t}^{l}v\|_{L^{6}(\Omega_{t})}+\|\mathcal{D}_{t}v\|_{L^{6}(\Omega_{t})}\|\nabla\mathcal{D}_{t}^{l}v\|_{L^{3}(\Omega_{t})})\\\
&\leq
C(1+\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}}(\Omega_{t})}\|\mathcal{D}_{t}^{l}v\|_{H^{1}(\Omega_{t})}+\|\mathcal{D}_{t}v\|_{H^{1}(\Omega_{t})}\|\mathcal{D}_{t}^{l}v\|_{H^{\frac{3}{2}}(\Omega_{t})}).\end{split}$

By definition of $E_{l}(t)$ it holds

$\|\mathcal{D}_{t}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}\leq E_{1}(t)\leq
E_{l-1}(t)\leq
C\quad\text{and}\quad\|\mathcal{D}_{t}^{l}v\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}\leq
E_{l}(t).$

Therefore we have
$\|\nabla_{\tau}p\cdot\mathcal{D}_{t}^{l}v\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
CE_{l}(t)$ and (5.35) follows.

Let us next show

(5.36) $\|R_{II}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq CE_{l}(t),$

where

$R_{II}^{l}=\sum_{|\alpha|\leq 1,\,|\beta|\leq
l-1}a_{\alpha,\beta}(B)\overbrace{\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}v}^{=:R_{\alpha,\beta}(v)}.$

We first observe that we may ignore the coefficients $a_{\alpha,\beta}(B)$.
Indeed, we may extend $B$ to $\Omega_{t}$, call the extension $\tilde{B}$,
such that $\|\tilde{B}\|_{H^{2}(\Omega_{t})}\leq C$. Then by the above
notation

(5.37) $\begin{split}\|R_{II}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}&\leq
C\|R_{II}^{l}\|_{H^{1}(\Omega_{t})}\leq C(1+\sum_{\alpha,\beta}\|\nabla(\nabla
a_{\alpha,\beta}(\tilde{B})\star R_{\alpha,\beta}(v))\|_{L^{2}(\Omega_{t})}\\\
&\leq C(1+\sum_{\alpha,\beta}\|\nabla\tilde{B}\star
R_{\alpha,\beta}(v)\|_{L^{2}(\Omega_{t})}+\|\nabla
R_{\alpha,\beta}(v)\|_{L^{2}(\Omega_{t})})\\\ &\leq
C(1+\sum_{\alpha,\beta}\|\nabla\tilde{B}\|_{L^{4}(\Omega_{t})}\|R_{\alpha,\beta}(v)\|_{L^{4}(\Omega_{t})}+\|\nabla
R_{\alpha,\beta}(v)\|_{L^{2}(\Omega_{t})})\\\ &\leq
C(1+\sum_{\alpha,\beta}\|\nabla R_{\alpha,\beta}(v)\|_{L^{2}(\Omega_{t})})\\\
&\leq C\big{(}1+\sum_{|\alpha|\leq 2,\,|\beta|\leq
l-1}\|\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}v\|_{L^{2}(\Omega_{t})}\big{)}.\end{split}$

By the assumption $E_{l-1}(t)\leq C$ and by (5.6) we deduce
$\|\nabla^{1+\alpha_{i}}\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}}\leq C$ for
$\alpha_{i}+\beta_{i}\leq l-2$. We note also that by $|\alpha|\leq 2$ and
$|\beta|\leq l-1$ it follows that $|\alpha|+|\beta|\leq l+1$. We ignore all
the terms in the last row of (5.37) which indexes satisfy
$\alpha_{i}+\beta_{i}\leq l-2$ as these are uniformly bounded. For the rest of
the terms we use Hölder’s inequality and relabel the indexes (note that below
we assume $\alpha\leq 2$ and $\beta\leq l-1$)

(5.38) $\begin{split}\sum_{|\alpha|\leq 2,\,|\beta|\leq
l-1}&\|\nabla^{1+\alpha_{1}}\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\nabla^{1+\alpha_{l+1}}\mathcal{D}_{t}^{\beta_{l+1}}v\|_{L^{2}(\Omega_{t})}\\\
&\leq C\sum_{\alpha\leq 2,\beta\leq
l-1}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{2}}^{2}+\sum_{\alpha+\beta=l}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{6}}^{2}\cdot\sum_{\alpha+\beta=1}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{3}}^{2}\\\
&\,\,\,\,\,\,\,\,+\sum_{\alpha+\beta=l-1}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{3}}^{2}\cdot\sum_{\alpha+\beta=2}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{6}}^{2}+\sum_{\alpha+\beta\leq
l-1}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{6}}^{6}.\end{split}$

To bound the first term on the RHS of (5.38) we simply note that for
$\beta\leq l-1$ and $\alpha\leq 2$ it holds

$\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{2}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\beta}v\|_{H^{3}(\Omega_{t})}^{2}\leq CE_{l}(t).$

For the second and the third terms we have first for $\alpha+\beta\leq l$ and
$\beta\leq l-1$ (which include the case $\alpha+\beta=2$ as $l\geq 2$) that

$\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{6}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\beta}v\|_{H^{l-\beta+2}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l+1-(l+1+\beta)}v\|_{H^{\frac{3}{2}(l+1+\beta)}(\Omega_{t})}^{2}\leq
CE_{l}(t).$

For $\alpha+\beta\leq l-1$ (which includes the case $\alpha+\beta=1$) we
deduce

(5.39)
$\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{3}(\Omega_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{1}{2}+(l-\beta)}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

For the last term we interpolate in the fluid domain
$\Omega_{t}\subset\mathbb{R}^{3}$ for $\alpha+\beta\leq l-1$ as

$\begin{split}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{6}(\Omega_{t})}&\leq\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{H^{2}(\Omega_{t})}^{1/3}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{3}(\Omega_{t})}^{2/3}\leq
CE_{l}(t)^{1/6}\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{3}}^{2/3}.\end{split}$

By (5.39) we have $\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{3}}\leq
C$ and thus

$\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{L^{6}}^{6}\leq CE_{l}(t).$

By combing the previous estimates with (5.37) and (5.38) we obtain

$\|R_{II}^{l}\|_{H^{1/2}(\Sigma_{t})}^{2}\leq C(1+\|\nabla
R_{II}^{l}\|_{L^{2}(\Omega_{t})}^{2})\leq CE_{l}(t),$

and (5.36) follows.

We are left to prove

(5.40) $\|R_{III}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq CE_{l}(t),$

where

$R_{III}^{l}=\sum_{\begin{subarray}{c}|\alpha|+|\beta|+|\gamma|\leq l+1\\\
|\beta|\leq l-1,\gamma_{i}\leq
l\end{subarray}}a_{\alpha,\beta,\gamma,Q}(v)\mathcal{D}_{t}^{\beta_{1}}v\star\cdots\star\mathcal{D}_{t}^{\beta_{l-1}}v\star\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\star\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{2}}U$

and the coefficients $a_{\alpha,\beta,\gamma,Q}$ depend on the time
derivatives of $Q(t)$ up to order $l+1$.

This time we will not give the argument which proves the boundedness of
$Q(t)^{(l+1)}=\frac{d^{l+1}}{dt^{l+1}}Q(t)$ as it simpler than the rest of the
proof and is similar to the argument in (5.30).

Recall that by (5.6) it holds
$\|\mathcal{D}_{t}^{\beta_{i}}v\|_{L^{\infty}}\leq C$ for $\beta_{i}\leq l-2$.
On the other hand for $\alpha+\gamma\leq l-1$ we have by Lemma 5.5

(5.41)
$\begin{split}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{L^{\infty}(\Sigma_{t})}^{2}\leq
C\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{3}{2}}(\Sigma_{t})}^{2}\leq
C(1+\|\nabla^{2+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2})\leq
CE_{l-1}(t)\leq C.\end{split}$

Similarly we have

(5.42) $\|\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C\quad\text{for }\,\beta\leq
l-1\quad\text{and}\quad\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C\quad\text{for}\,\alpha+\gamma\leq l.$

Therefore we may ignore all the terms with indexes which satisfy
$\beta_{i}\leq l-2$ and $\alpha_{i}+\gamma_{i}\leq l-1$. Recall that we assume
$\beta_{1}\geq\dots\geq\beta_{l-1}$. Therefore we may estimate by Proposition
2.10, (5.41) and by (5.42)

$\begin{split}&\|R_{III}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}\\\
&\leq\sum_{\alpha+\gamma\leq
l}C\big{(}\|\mathcal{D}_{t}^{l-1}v\|_{L^{\infty}(\Sigma_{t})}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}+\|\mathcal{D}_{t}^{l-1}v\|_{H^{\frac{1}{2}}(\Sigma_{t})}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{L^{\infty}(\Sigma_{t})}\big{)}\\\
&\,\,\,\,\,\,\,\,\,\,\,+C\sum_{\alpha+\gamma\leq l+1,\gamma\leq
l}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\\\
&\leq
C\big{(}\|\mathcal{D}_{t}^{l-1}v\|_{L^{\infty}(\Sigma_{t})}+\sum_{\alpha+\gamma\leq
l}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{L^{\infty}(\Sigma_{t})}+\sum_{\begin{subarray}{c}\alpha+\gamma\leq
l+1\\\ \gamma\leq
l\end{subarray}}\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\big{)}.\end{split}$

We estimate the first term in the last row by Sobolev embedding

$\|\mathcal{D}_{t}^{l-1}v\|_{L^{\infty}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l-1}v\|_{H^{3}(\Omega_{t})}^{2}\leq CE_{l}(t).$

For the second we use Sobolev embedding and Lemma 5.5 and obtain for
$\alpha+\gamma\leq l$

$\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{L^{\infty}(\Sigma_{t})}^{2}\leq
C(1+\|\nabla^{2+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2})\leq
CE_{l}(t).$

The same argument also implies

$\|\nabla^{1+\alpha}\partial_{t}^{\gamma}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\leq
CE_{l}(t)$

for $\alpha+\gamma\leq l+1$ and $\gamma\leq l$. Hence we obtain (5.40) and
this concludes the proof of (5.33).

Let us then prove (5.34). Let us first treat the first term in the definition
of $R_{p}^{l-k}$ and bound $\|R_{I}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}$.
We first observe that the case $k=1$ follows from (5.33). Let us then assume
$k\geq 2$. By the Sobolev embedding it holds $\|u\|_{L^{\infty}(\Sigma)}\leq
C\|u\|_{L^{\frac{3}{2}k-1}(\Sigma)}$. We use this and the Kato-Ponce
inequality in Proposition 2.10 to deduce that

$\begin{split}&\|R_{I}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\\\ &\leq
C(1+\|B\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}+\|\nabla
U\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}+\|\mathcal{D}_{t}^{l-k}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}+\|\nabla
p\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}).\end{split}$

Let us show that all the terms on RHS are bounded.

To this aim we first recall that the bound $E_{l-1}(t)\leq C$ implies
$\|B\|_{H^{\frac{3}{2}l-1}(\Sigma_{t})}\leq C$. Since $k\leq l-1$ we deduce
$\|B\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C$. Lemma 5.1 implies $\|\nabla
U\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq CE_{l-1}(t)\leq C$. The condition
$k\leq l-1$ and the Trace Theorem also yields

$\|\mathcal{D}_{t}^{l-k}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

Similarly we deduce by $-\nabla p=\mathcal{D}_{t}v$ that $\|\nabla
p\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C$. Hence, we have

$\|R_{I}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C.$

Let us then bound $\|R_{II}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}$. As
before, Proposition 2.10 and the Sobolev embedding yield

$\|R_{II}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq\sum_{\alpha\leq
1,\beta\leq
l-k-1}C(1+\|a_{\alpha,\beta}(B)\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}+\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{q})$

for $q\geq 1$. Recall that $\|B\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C$ and
$k\geq 2$. Therefore $\|B\|_{L^{\infty}}\leq C$ and thus
$\|a_{\alpha,\beta}(B)\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C$. On the other
hand by $\alpha\leq 1$, $\beta\leq l-(k+1)$ and by Lemma 3.7 we deduce

(5.43)
$\|\nabla^{1+\alpha}\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{3}{2}(k+1)}(\Omega_{t})}^{2}\leq
C\sum_{i=0}^{l-1}\|\mathcal{D}_{t}^{l-i}v\|_{H^{\frac{3}{2}i}(\Omega_{t})}^{2}\leq
CE_{l-1}(t)\leq C.$

Hence, we have $\|R_{II}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C$.

Let us finally treat $R_{III}^{l-k}$. We note that it holds
$\|v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq CE_{l-1}(t)\leq C$ and
therefore we may ignore the coefficients $a_{\alpha,\beta,\gamma,Q}(v)$. Then
by Proposition 2.10 and by the Sobolev embedding we have

(5.44) $\begin{split}\|R_{III}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}&\leq
C(1+\sum_{\beta\leq
l-(k+1)}\|\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{q}\cdot\\\
&\cdot\sum_{\begin{subarray}{c}|\alpha|+|\gamma|\leq l-k+1,\\\ |\gamma|\leq
l-k\end{subarray}}\|\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\|\nabla^{1+\alpha_{2}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})})\end{split}$

for $q\geq 1$. By (5.43) we have
$\|\mathcal{D}_{t}^{\beta}v\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}\leq C$ for all
$\beta\leq l-(k+1)$. To bound the last term we may assume that
$\alpha_{1}+\gamma_{1}\geq\alpha_{1}+\gamma_{1}$. If
$\alpha_{1}+\gamma_{1}=l-k+1$ then necessarily $\alpha_{2}+\gamma_{2}\leq
l-k$. Therefore by Lemma 5.5

$\|\nabla^{1+\alpha_{1}}\partial_{t}^{\gamma_{1}}U\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq
C(1+\|\nabla^{1+(\alpha_{1}+k-1)}\partial_{t}^{\gamma_{1}}U\|_{H^{\frac{k-1}{2}+\frac{1}{2}}(\Sigma_{t})}^{2})\leq\varepsilon
E_{l}(t)+C_{\varepsilon}$

and

$\|\nabla^{1+\alpha_{2}}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq
C(1+\|\nabla^{1+(\alpha_{2}+k-1)}\partial_{t}^{\gamma_{2}}U\|_{H^{\frac{k-1}{2}+\frac{1}{2}}(\Sigma_{t})}^{2})\leq
CE_{l-1}(t)\leq C.$

Therefore we deduce by (5.44) that

$\|R_{III}^{l-k}\|_{H^{\frac{3}{2}k-1}(\Sigma_{t})}^{2}\leq\varepsilon
E_{l}(t)+C_{\varepsilon}.$

This concludes the proof of (5.34). ∎

##  6\. First regularity estimates

In this section we prove our first regularity estimates for the solution of
(1.3). We assume that the solution satisfies the a priori estimates (1.7),
i.e., $\Lambda_{T}<\infty$ and $\sigma_{T}>0$, where $\Lambda_{T}$ and
$\sigma_{T}$ are defined in (1.5) and (1.4). We recall that

$\Lambda_{T}:=\sup_{t\in(0,T]}\left(\|h(\cdot,t)\|_{C^{1,\alpha}(\Sigma_{t})}+\|\nabla
v(\cdot,t)\|_{L^{\infty}(\Omega_{t})}+\|v_{n}(\cdot,t)\|_{H^{2}(\Sigma_{t})}+\right)<\infty.$

In particular, we do not have curvature, and thus we need to be careful as we
may not use e.g. the interpolation results from Proposition 2.8. Our goal in
this section is to show that the a priori estimates (1.7) imply the following
bounds for the pressure

$\sup_{t\leq T}\|p\|_{H^{1}(\Omega_{t})}\leq
C\qquad\text{and}\qquad\int_{0}^{T}\|p\|_{H^{2}(\Omega_{t})}^{2}\,dt\leq C.$

The first bound above is important as it implies
$\|B_{\Sigma}\|_{L^{4}(\Sigma_{t})}\leq C$, which is crucial e.g. for the
interpolation inequality in Proposition 2.8 to hold. The second estimate is
important for the first order energy estimate which we prove in the next
section in Proposition 7.1.

Let us begin by stating regularity estimates that we have by the a priori
estimate. First, recall once again that by the $C^{1,\alpha}(\Gamma)$-bound we
have for the capacitary potential $U$ that
$\|U\|_{C^{1,\alpha}(\bar{\Omega}_{t}^{c})}\leq C$. Let us prove the following
estimates for the second fundamental form and for the capcacitary potential.

######  Lemma 6.1.

Assume that (1.7) holds for $T>0$. Then for all $t<T$ we have

$\|B\|_{L^{4}(\Sigma_{t})}^{4}\leq\varepsilon\|p\|_{H^{1}(\Sigma_{t})}^{2}+C_{\varepsilon}$

and

$\|B\|_{H^{1}(\Sigma_{t})}+\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}).$

###### Proof.

We denote the height-function by $h=h(\cdot,t)$. Then by standard calculations
(see e.g. [22, 42]) we may write the second fundamental form on $\Sigma$ as
$B=a(h,\bar{\nabla}h)\bar{\nabla}^{2}h$. Therefore we may bound

$\|B\|_{L^{4}(\Sigma_{t})}^{4}\leq
C\|\bar{\nabla}^{2}h\|_{L^{4}(\Gamma)}^{4}\quad\text{and}\quad\|\bar{\nabla}^{3}h\|_{L^{2}(\Gamma)}^{2}\leq
C\|\bar{\nabla}B\|_{L^{2}(\Sigma_{t})}^{2}+C\|\bar{\nabla}^{2}h\|_{L^{4}(\Gamma)}^{4}.$

We use interpolation on $\Gamma$ as

$\|\bar{\nabla}^{2}h\|_{L^{4}(\Gamma)}\leq
C\|\bar{\nabla}^{3}h\|_{L^{2}(\Gamma)}^{\theta}\|h\|_{C^{1,\alpha}(\Gamma)}^{1-\theta}\leq
C\|\bar{\nabla}^{3}h\|_{L^{2}(\Gamma)}^{\theta}$

for $\theta<1/2$. This implies by Young’s inequality
$\|\bar{\nabla}^{2}h\|_{L^{4}(\Gamma)}^{4}\leq\varepsilon\|\bar{\nabla}^{3}h\|_{L^{2}(\Gamma)}^{2}+C_{\varepsilon}$.
Thus by choosing $\varepsilon$ small we obtain

$\|\bar{\nabla}^{3}h\|_{L^{2}(\Gamma)}^{2}\leq
C(1+\|\bar{\nabla}B\|_{L^{2}(\Sigma_{t})}^{2})$

and

$\|B\|_{L^{4}(\Sigma_{t})}^{4}\leq\varepsilon\|\bar{\nabla}^{3}h\|_{L^{2}(\Gamma)}^{2}+C_{\varepsilon}.$

By the Simon’s identity (2.11) we deduce

$\|\bar{\nabla}B\|_{L^{2}(\Sigma_{t})}^{2}\leq\|\bar{\nabla}H\|_{L^{2}(\Sigma_{t})}^{2}+C\|B\|_{L^{4}(\Sigma_{t})}^{4}.$

Therefore we have

(6.1)
$\|B\|_{L^{4}(\Sigma_{t})}^{4}\leq\varepsilon\|H\|_{H^{1}(\Sigma_{t})}^{2}+C_{\varepsilon}$

and

(6.2) $\|B\|_{H^{1}(\Sigma_{t})}\leq C(1+\|H\|_{H^{1}(\Sigma_{t})}).$

Let us consider the capacitary potential $U$. Let us show that even though we
may not use Proposition 2.8, the $C^{1,\alpha}(\Gamma)$-regularity still
implies the following weak interpolation inequality

(6.3) $\|\nabla
U\|_{H^{1}(\Sigma_{t})}\leq\varepsilon\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}\|\nabla
U\|_{L^{2}(\Sigma_{t})}\leq\varepsilon\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}.$

In order to prove (6.3) we first observe that the
$C^{1,\alpha}(\Gamma)$-regularity of $\Sigma_{t}$ implies the following
inequalities for $p\in(1,2)$ and $u,v\in C^{\infty}(\Sigma_{t})$

$\|u\|_{L^{2}(\Sigma_{t})}\leq\varepsilon\|u\|_{H^{\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}\|u\|_{L^{p}(\Sigma_{t})}\quad\text{and}\quad\|\nabla_{\tau}v\|_{L^{p}(\Sigma_{t})}\leq\delta\|v\|_{H^{1}(\Sigma_{t})}+C_{\delta}\|v\|_{L^{2}(\Sigma_{t})}.$

We apply these for $u=\nabla^{2}U$ and $v=\nabla U$ and have

$\begin{split}&\|\nabla^{2}U\|_{L^{2}(\Sigma_{t})}\leq\varepsilon\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}+C_{\varepsilon}\|\nabla^{2}U\|_{L^{p}(\Sigma_{t})}\quad\text{and}\\\
&\|\nabla_{\tau}\nabla U\|_{L^{p}(\Sigma_{t})}\leq\delta\|\nabla
U\|_{H^{1}(\Sigma_{t})}+C_{\delta}\|\nabla
U\|_{L^{2}(\Sigma_{t})}.\end{split}$

Since $\partial_{x_{i}}U$ is harmonic and $\Sigma_{t}$ is
$C^{1,\alpha}(\Gamma)$-regular we have by [19] that

$\|\nabla^{2}U\|_{L^{p}(\Sigma_{t})}\leq C(\|\nabla_{\tau}\nabla
U\|_{L^{p}(\Sigma_{t})}+\|\nabla U\|_{L^{2}(\Omega_{t})}).$

Therefore by first choosing $\varepsilon$ small and then $\delta$ even smaller
we obtain (6.3).

By $p=H-\frac{Q(t)}{2}|\nabla U|^{2}$ we may estimate

$\|H\|_{H^{1}(\Sigma_{t})}\leq C(\|p\|_{H^{1}(\Sigma_{t})}+\|\nabla
U\|_{H^{1}(\Sigma_{t})}).$

Then by (6.3) we have

$\|H\|_{H^{1}(\Sigma_{t})}\leq
C_{\varepsilon}(1+\|p\|_{H^{1}(\Sigma_{t})})+\varepsilon\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}.$

We use Theorem 3.9 and (6.2) and have

$\|\nabla^{2}U\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq
C(1+\|B\|_{H^{1}(\Sigma_{t})})\leq C(1+\|H\|_{H^{1}(\Sigma_{t})}).$

Therefore we deduce from the two above inequalities that

$\|H\|_{H^{1}(\Sigma_{t})}\leq C(1+\|p\|_{H^{1}(\Sigma_{t})}).$

The claim follows from this together with (6.1) and (6.2). ∎

Let us proceed to the following regularity estimate.

######  Lemma 6.2.

Assume that the a priori estimates (1.7) hold for $T>0$. Then

$\sup_{t\in[0,T]}\|p\|_{L^{2}(\Sigma_{t})}^{2}+\int_{0}^{T}\|p\|_{H^{1}(\Sigma_{t})}^{2}\,dt\leq
C(1+T),$

where the constant $C$ depends on $\Lambda_{T}$ and $\sigma_{T}$.

###### Proof.

The idea is to consider the following function

$\Psi(t):=\int_{\Sigma_{t}}p\,(\nabla v\,\nu\cdot\nu)+\varepsilon
p^{2}\,d\mathcal{H}^{2},$

where the choice of $\varepsilon$ will be clear later. First, we observe that
under the a priori estimates (1.7) $v$ is uniformly Lipschitz and therefore
$\Psi$ is bounded from below by

(6.4)
$\Psi(t)\geq-C\|p(\cdot,t)\|_{L^{1}(\Sigma_{t})}+\varepsilon\|p(\cdot,t)\|_{L^{2}(\Sigma_{t})}^{2}\geq-
C_{\varepsilon}+\frac{\varepsilon}{2}\|p(\cdot,t)\|_{L^{2}(\Sigma_{t})}^{2}.$

By differentiating and using the a priori estimates (1.7) we obtain

(6.5) $\begin{split}\frac{d}{dt}&\int_{\Sigma_{t}}p\,(\nabla
v\,\nu\cdot\nu)\,d\mathcal{H}^{2}\\\ &=\int_{\Sigma_{t}}p\,(\nabla
v\,\nu\cdot\nu)\operatorname{div}_{\tau}v\,d\mathcal{H}^{2}+\int_{\Sigma_{t}}\mathcal{D}_{t}p\,(\nabla
v\,\nu\cdot\nu)\,d\mathcal{H}^{2}+\int_{\Sigma_{t}}p\,\mathcal{D}_{t}(\nabla
v\,\nu\cdot\nu)\,d\mathcal{H}^{2}\\\ &\leq
C_{\varepsilon}+\varepsilon\int_{\Sigma_{t}}|\mathcal{D}_{t}p|^{2}\,d\mathcal{H}^{2}+\int_{\Sigma_{t}}p\,\mathcal{D}_{t}(\nabla
v\,\nu\cdot\nu)\,d\mathcal{H}^{2}.\end{split}$

We estimate the second last term in (6.5) by (4.25) and (1.7) and have

$\int_{\Sigma_{t}}|\mathcal{D}_{t}p|^{2}\,d\mathcal{H}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|B\|_{L^{4}(\Sigma_{t})}^{4}+\|\nabla\partial_{t}U\|_{L^{2}(\Sigma_{t})}^{2}).$

Lemma 6.1 yields $\|B\|_{L^{4}(\Sigma_{t})}^{4}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2})$. On the other hand, by Lemma 3.3 it holds
$\|\nabla\partial_{t}U\|_{\Sigma_{t}}^{2}\leq
C\|\partial_{t}U\|_{H^{1}(\Sigma_{t})}^{2}$. Since $\partial_{t}U=-\nabla
U\cdot v$, we may estimate by Theorem 3.9 and by Lemma 6.1

$\begin{split}\|\nabla\partial_{t}U\|_{L^{2}(\Sigma_{t})}^{2}&\leq
C(1+\|\bar{\nabla}\partial_{t}U\|_{L^{2}(\Sigma_{t})}^{2})\leq
C(1+\|\bar{\nabla}(\nabla U\cdot v)\|_{L^{2}(\Sigma_{t})}^{2})\leq
C(1+\|\nabla^{2}U\|_{L^{2}(\Sigma_{t})}^{2})\\\ &\leq
C(1+\|\nabla^{2}U\|_{H^{1/2}(\Sigma_{t})}^{2})\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).\end{split}$

Therefore we may bound

(6.6) $\int_{\Sigma_{t}}|\mathcal{D}_{t}p|^{2}\,d\mathcal{H}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).$

Let us treat the last in term in (6.5). First, we have by (4.1), (4.4) and
(1.7) that

$\begin{split}\int_{\Sigma_{t}}p\,\mathcal{D}_{t}(\nabla
v\,\nu\cdot\nu)\,d\mathcal{H}^{2}&\leq\int_{\Sigma_{t}}p\,(\nabla\mathcal{D}_{t}v\,\nu\cdot\nu)\,d\mathcal{H}^{2}+\varepsilon\|p\|_{L^{2}(\Sigma_{t})}^{2}+C_{\varepsilon}\\\
&=-\int_{\Sigma_{t}}p\,(\nabla^{2}p\,\nu\cdot\nu)\,d\mathcal{H}^{2}+\varepsilon\|p\|_{L^{2}(\Sigma_{t})}^{2}+C_{\varepsilon}.\end{split}$

We use (4.11) to estimate $|\Delta p|\leq C\|\nabla v\|_{L^{\infty}}^{2}\leq
C$ and recall that by (3.4) it holds $(\nabla^{2}p\,\nu\cdot\nu)=\Delta
p-\Delta_{\Sigma}p-H\partial_{\nu}p$ to deduce

$\begin{split}-\int_{\Sigma_{t}}p\,(\nabla^{2}p\,\nu\cdot\nu)\,d\mathcal{H}^{2}&\leq
C+\int_{\Sigma_{t}}p\,\Delta_{\Sigma}p\,d\mathcal{H}^{2}+\int_{\Sigma_{t}}Hp\,\partial_{\nu}p\,d\mathcal{H}^{2}\\\
&\leq
C-\int_{\Sigma_{t}}|\bar{\nabla}p|^{2}\,d\mathcal{H}^{2}+\int_{\Sigma_{t}}(\varepsilon|\partial_{\nu}p|^{2}+C_{\varepsilon}(1+|H|^{4}))\,d\mathcal{H}^{2},\end{split}$

where in the last inequality we have used $p=H-\frac{Q(t)}{2}|\nabla U|^{2}$.
Lemma 3.3 yields $\|\partial_{\nu}p\|_{L^{2}(\Sigma_{t})}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})})$ while Lemma 6.1 implies
$\|H\|_{L^{4}}^{4}\leq\delta\|p\|_{H^{1}(\Sigma_{t})}^{2}+C_{\delta}$.
Therefore by first choosing $\varepsilon$ small and then $\delta$ even
smaller, we obtain

$-\int_{\Sigma_{t}}p\,(\nabla^{2}p\,\nu\cdot\nu)\,d\mathcal{H}^{2}\leq-\|\bar{\nabla}p\|_{L^{2}(\Sigma_{t})}^{2}+\varepsilon\|p\|_{H^{1}(\Sigma_{t})}^{2}+C_{\varepsilon}.$

By direct calculation and by using (6.6) we have

$\frac{d}{dt}\int_{\Sigma_{t}}p^{2}\,d\mathcal{H}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).$

Combining the two above inequalities with (6.5) and (6.6) we conclude

(6.7)
$\frac{d}{dt}\Psi(t)\leq-\frac{1}{2}\|\bar{\nabla}p\|_{L^{2}(\Sigma_{t})}^{2}+\varepsilon\|p\|_{L^{2}(\Sigma_{t})}^{2}+C_{\varepsilon},$

when $\varepsilon$ is small. Finally we use Lemma 6.1 to estimate

$\|p\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|H\|_{L^{2}(\Sigma_{t})}^{2})\leq\|B\|_{L^{4}(\Sigma_{t})}^{4}+C\leq\varepsilon\|p\|_{H^{1}(\Sigma_{t})}^{2}+C_{\varepsilon}.$

This yields $\|p\|_{L^{2}(\Sigma_{t})}^{2}\leq
2\|\bar{\nabla}p\|_{L^{2}(\Sigma_{t})}^{2}+C_{\varepsilon}$ when $\varepsilon$
is small. Thus we may estimate (6.7)

$\frac{d}{dt}\Psi(t)\leq-\frac{1}{4}\|\bar{\nabla}p\|_{L^{2}(\Sigma_{t})}^{2}+C_{\varepsilon},$

when $\varepsilon$ is small. The conclusion follows by integrating the above
over $[0,T]$ and using (6.4). ∎

We proceed to higher order regularity estimate which is uniform in time.

######  Proposition 6.3.

Assume that the a priori estimates (1.7) hold for $T>0$. Then

$\sup_{t\in(0,T]}\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}\leq
e^{C(1+T)}(1+\|\nabla p\|_{L^{2}(\Omega_{0})}^{2}),$

where the constant $C$ depends on $\Lambda_{T}$ and $\sigma_{T}$.

###### Proof.

We differentiate

$\begin{split}\frac{d}{dt}\frac{1}{2}\int_{\Omega_{t}}|\nabla
p|^{2}\,dx&=\frac{1}{2}\int_{\Omega_{t}}|\nabla
p|^{2}\overbrace{\operatorname{div}(v)}^{=0}\,dx+\int_{\Omega_{t}}(\mathcal{D}_{t}\nabla
p\cdot\nabla p)\,dx\\\ &=\int_{\Omega_{t}}(\nabla\mathcal{D}_{t}p\cdot\nabla
p)\,dx+\int_{\Omega_{t}}([\mathcal{D}_{t},\nabla]p\cdot\nabla
p)\,dx.\end{split}$

By (4.1) and $\|\nabla v\|_{L^{\infty}}\leq C$ we have a pointwise estimate
$|[\mathcal{D}_{t},\nabla]p|\leq C|\nabla v||\nabla p|\leq C|\nabla p|$.
Therefore we deduce

(6.8) $\begin{split}\frac{d}{dt}\frac{1}{2}\int_{\Omega_{t}}|\nabla
p|^{2}\,dx&\leq\int_{\Omega_{t}}(\nabla\mathcal{D}_{t}p\cdot\nabla
p)\,dx+C\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}\\\
&=\int_{\Omega_{t}}\operatorname{div}(\mathcal{D}_{t}p\,\nabla
p)\,dx-\int_{\Omega_{t}}\mathcal{D}_{t}p\,\Delta p\,dx+C\|\nabla
p\|_{L^{2}(\Omega_{t})}^{2}\\\
&=\int_{\Sigma_{t}}\mathcal{D}_{t}p\,\partial_{\nu}p\,dx-\int_{\Omega_{t}}\mathcal{D}_{t}p\,\Delta
p\,dx+C\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}\\\
&\leq\|\mathcal{D}_{t}p\|_{L^{2}(\Sigma_{t})}^{2}+\|\partial_{\nu}p\|_{L^{2}(\Sigma_{t})}^{2}-\int_{\Omega_{t}}\mathcal{D}_{t}p\,\Delta
p\,dx+C\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}.\end{split}$

We have by (6.6) that $\|\mathcal{D}_{t}p\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2})$ and by Lemma 3.3

$\|\partial_{\nu}p\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|\Delta p\|_{L^{2}(\Omega_{t})}^{2})\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).$

We are left with the second last term in (6.8).

To that aim let $u:\Omega_{t}\to\mathbb{R}$ be the solution of

$\begin{cases}-\Delta u=\Delta p\quad\text{in }\,\Omega_{t}\\\
u=0\quad\text{on }\,\Sigma_{t}.\end{cases}$

Then it holds

$-\int_{\Omega_{t}}\mathcal{D}_{t}p\,\Delta
p\,dx=\int_{\Omega_{t}}\mathcal{D}_{t}p\,\Delta
u\,dx=\int_{\Omega_{t}}\Delta\mathcal{D}_{t}p\,u\,dx+\int_{\Sigma_{t}}\mathcal{D}_{t}p\,\partial_{\nu}u\,d\mathcal{H}^{2}.$

Since $|\Delta u|=|\Delta p|\leq C$ and $u=0$ on $\Sigma_{t}$ it holds
$\|u\|_{H^{1}(\Omega_{t})}\leq C$ and by Lemma 3.3 we deduce $\|\nabla
u\|_{L^{2}(\Sigma_{t})}\leq C$. We may bound the last term on the RHS by (6.6)

$\int_{\Sigma_{t}}\mathcal{D}_{t}p\,\partial_{\nu}u\,d\mathcal{H}^{2}\leq\|\mathcal{D}_{t}p\|_{L^{2}(\Sigma_{t})}^{2}+\|\partial_{\nu}u\|_{L^{2}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).$

We are thus left with the second last term.

We have by (3.2), by Lemma 6.1, by Sobolev embedding and by interpolation that

$\begin{split}\|\nabla^{2}u\|_{L^{2}(\Omega_{t})}^{2}&\leq
C+\int_{\Sigma_{t}}|H_{\Sigma_{t}}||\nabla u|^{2}\,d\mathcal{H}^{2}\leq
C+C_{\varepsilon}\|H_{\Sigma_{t}}\|_{L^{3}(\Sigma_{t})}^{3}+\varepsilon\|\nabla
u\|_{L^{3}(\Sigma_{t})}^{3}\\\ &\leq
C_{\varepsilon}(1+\|p\|_{L^{2}(\Sigma_{t})}^{2})+\varepsilon\|\nabla
u\|_{L^{4}(\Sigma_{t})}^{2}\|\nabla u\|_{L^{2}(\Sigma_{t})}\\\ &\leq
C_{\varepsilon}(1+\|p\|_{L^{2}(\Sigma_{t})}^{2})+C\varepsilon\|\nabla^{2}u\|_{L^{2}(\Omega_{t})}^{2}.\end{split}$

Therefore it holds

$\|u\|_{H^{2}(\Omega_{t})}^{2}\leq C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).$

By Remark 4.5 we have

$\Delta\mathcal{D}_{t}p=\operatorname{div}\operatorname{div}(v\otimes\nabla
p)+\operatorname{div}(R_{bulk}^{0}).$

Therefore by integrating by parts

$\begin{split}\int_{\Omega_{t}}\Delta\mathcal{D}_{t}p\,u\,dx&=\int_{\Omega_{t}}(v\otimes\nabla
p):\nabla^{2}u\,dx+\int_{\Omega_{t}}R_{bulk}^{0}\star\nabla
u\,dx-\int_{\Sigma_{t}}(\nabla p\cdot\nu)(\nabla u\cdot
v)\,d\mathcal{H}^{2}\\\ &\leq C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|\nabla
p\|_{L^{2}(\Omega_{t})}^{2}).\end{split}$

We deduce by (6.8) and by the above estimates that

$\frac{d}{dt}\frac{1}{2}\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}).$

This implies

$\frac{d}{dt}\log(1+\|\nabla p\|_{L^{2}(\Omega_{t})}^{2})\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2})$

and the claim follows from Lemma 6.2. ∎

An important consequence of Proposition 6.3 is that by Lemma 5.2 we have the
following bound for the curvature

(6.9) $\|B\|_{L^{4}(\Sigma_{t})}+\|B\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq C.$

This means that from now on we may use the general interpolation inequality
from Proposition 2.8.

At the end of this section we improve Lemma 6.2. We recall the definition of
the energy quantity $E_{1}(t)$ from (5.1)

$E_{1}(t)=\|\mathcal{D}_{t}^{2}v\|_{L^{2}(\Omega_{t})}^{2}+\|\nabla
p\|_{H^{\frac{3}{2}}(\Omega_{t})}^{2}+\|v\|_{H^{3}(\Omega_{t})}^{2}+\|\partial_{\nu}p\|_{H^{1}(\Sigma_{t})}^{2}+1.$

In particular, $E_{1}(0)$ denotes the above quantity at time $t=0$. It is
clear that

$\|p\|_{H^{1}(\Omega_{t})}^{2}\leq E_{1}(t).$

######  Lemma 6.4.

Assume that the a priori estimates (1.7) hold for $T>0$. Then

$\int_{0}^{T}\|p\|_{H^{2}(\Omega_{t})}^{2}\,dt\leq C,$

where the constant $C$ depends on $T,\Lambda_{T},\sigma_{T}$ and on
$E_{1}(0)$.

###### Proof.

The proof is similar to Lemma 6.2. This time we differentiate

$\Phi(t):=-\int_{\Sigma_{t}}p\Delta_{\Sigma_{t}}v_{n}\,d\mathcal{H}^{2}.$

Note that by the a priori estimates (1.7) and by Proposition 6.3, $\Phi$ is
uniformly bounded on $[0,T]$. Note also that by Proposition 6.3 it holds

$\sup_{t<T}\|p\|_{H^{1}(\Omega_{t})}^{2}\leq C,$

where the constant depends on $T,\Lambda_{T}$ and on $E_{1}(0)$.

We calculate as in (6.5) by using (6.6) and $\|v_{n}\|_{H^{2}(\Sigma_{t})}\leq
C$ that

$\begin{split}\frac{d}{dt}\Phi(t)&\leq
C+\|\mathcal{D}_{t}p\|_{L^{2}(\Sigma_{t})}^{2}-\int_{\Sigma_{t}}p\,(\mathcal{D}_{t}\Delta_{\Sigma_{t}}v_{n})\,d\mathcal{H}^{2}\\\
&\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2})-\int_{\Sigma_{t}}p\,(\mathcal{D}_{t}\Delta_{\Sigma_{t}}v_{n})\,d\mathcal{H}^{2}.\end{split}$

To bound the last term we recall that by (4.3) it holds

$(\mathcal{D}_{t}\Delta_{\Sigma_{t}}v_{n})=\Delta_{\Sigma_{t}}(\mathcal{D}_{t}v_{n})+\nabla_{\tau}^{2}v_{n}\star\nabla
v-\nabla_{\tau}v_{n}\cdot\Delta_{\Sigma_{t}}v+B\star\nabla
v\star\nabla_{\tau}v_{n}.$

Therefore by $\|\nabla v\|_{L^{\infty}}+\|v_{n}\|_{H^{2}(\Sigma_{t})}\leq C$,
by
$\|\nabla_{\tau}v_{n}\|_{L^{4}(\Sigma_{t})}\leq\|v_{n}\|_{H^{2}(\Sigma_{t})}$
and by (6.9) it holds

(6.10)
$\begin{split}&-\int_{\Sigma_{t}}p\,(\mathcal{D}_{t}\Delta_{\Sigma_{t}}v_{n})\,d\mathcal{H}^{2}\leq-\int_{\Sigma_{t}}p\,\Delta_{\Sigma_{t}}(\mathcal{D}_{t}v_{n})\,d\mathcal{H}^{2}+\int_{\Sigma_{t}}p\,\nabla_{\tau}v_{n}\cdot\Delta_{\Sigma_{t}}v\,d\mathcal{H}^{2}\\\
&\,\,\,\,\,\,\,\,+C(1+\|p\|_{L^{2}(\Sigma_{t})}\|B\|_{L^{4}(\Sigma_{t})}\|\bar{\nabla}v_{n}\|_{L^{4}(\Sigma_{t})})\\\
&\leq-\int_{\Sigma_{t}}p\,\Delta_{\Sigma_{t}}(\mathcal{D}_{t}v\cdot\nu)\,d\mathcal{H}^{2}-\int_{\Sigma_{t}}p\,\Delta_{\Sigma_{t}}(\mathcal{D}_{t}\nu\cdot
v)\,d\mathcal{H}^{2}-\int_{\Sigma_{t}}(\nabla_{\tau}(p\,\nabla_{\tau}v_{n})\cdot\nabla_{\tau}v)\,d\mathcal{H}^{2}+C.\end{split}$

We may write the first term on RHS in (6.10) by the formula (3.2),
$\mathcal{D}_{t}v=-\nabla p$ and by the estimate (6.9)

$\begin{split}-\int_{\Sigma_{t}}p\,\Delta_{\Sigma_{t}}(\mathcal{D}_{t}v\cdot\nu)\,d\mathcal{H}^{2}&=\int_{\Sigma_{t}}\Delta_{\Sigma_{t}}p\,\partial_{\nu}p\,d\mathcal{H}^{2}\\\
&\leq-\frac{1}{2}\int_{\Omega_{t}}(|\nabla^{2}p|^{2}-|\Delta
p|^{2})\,dx+C\int_{\Sigma_{t}}|B||\nabla p|^{2}\,d\mathcal{H}^{2}\\\
&\leq-\frac{1}{2}\int_{\Omega_{t}}|\nabla^{2}p|^{2}\,dx+C+C_{\varepsilon}\|B\|_{L^{4}}+\varepsilon\|\nabla
p\|_{L^{\frac{8}{3}}(\Sigma_{t})}^{2}\\\
&\leq-\frac{1}{2}\int_{\Omega_{t}}|\nabla^{2}p|^{2}\,dx+C_{\varepsilon}+\varepsilon\|\nabla
p\|_{L^{\frac{8}{3}}(\Sigma_{t})}^{2}.\end{split}$

By the Sobolev embedding it holds $\|\nabla
p\|_{L^{\frac{8}{3}}(\Sigma_{t})}^{2}\leq C\|\nabla
p\|_{H^{1/2}(\Sigma_{t})}^{2}\leq
C(1+\|\nabla^{2}p\|^{2}_{L^{2}(\Omega_{t})}).$ Therefore by choosing
$\varepsilon>0$ small we deduce

$-\int_{\Sigma_{t}}p\,\Delta_{\Sigma_{t}}(\mathcal{D}_{t}v\cdot\nu)\,d\mathcal{H}^{2}\leq-\frac{1}{3}\int_{\Omega_{t}}|\nabla^{2}p|^{2}\,dx+C_{\varepsilon}.$

We bound the third term on RHS in (6.10) simply by $\|\nabla
v\|_{L^{\infty}}+\|v_{n}\|_{H^{2}(\Sigma_{t})}\leq C$ as

$-\int_{\Sigma_{t}}\nabla_{\tau}(p\,\nabla_{\tau}v_{n})\cdot\nabla_{\tau}v\,d\mathcal{H}^{2}\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).$

For the remaining term in (6.10) we recall (4.5) which states
$\mathcal{D}_{t}\nu=-\nabla_{\tau}v_{n}+Bv_{\tau}$. Therefore we obtain by
Lemma 6.1 and by $\|\nabla v\|_{L^{\infty}}+\|v_{n}\|_{H^{2}(\Sigma_{t})}\leq
C$ that

$\begin{split}-\int_{\Sigma_{t}}p\,\Delta_{\Sigma_{t}}(\mathcal{D}_{t}\nu\cdot
v)\,d\mathcal{H}^{2}&=\int_{\Sigma_{t}}\langle\bar{\nabla}p,\bar{\nabla}(\mathcal{D}_{t}\nu\cdot
v)\rangle\,d\mathcal{H}^{2}\\\ &\leq
C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}+\|B\|_{H^{1}(\Sigma_{t})}^{2}+\|B\|_{L^{4}(\Sigma_{t})}^{4})\\\
&\leq C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2}).\end{split}$

Hence, we have

$\frac{d}{dt}\Phi(t)\leq-\frac{1}{3}\int_{\Omega_{t}}|\nabla^{2}p|^{2}\,dx+C(1+\|p\|_{H^{1}(\Sigma_{t})}^{2})$

and the claim follows from Lemma 6.2. ∎

##  7\. Energy estimates

As we mentioned before, the fundamental property of the solution of (1.3) is
the conservation of the energy (1.1). In this section we define high order
energy functions and show that their derivatives are controlled by the
quantity (5.1) of the same order. This will be the first step in proving that
the high order energy quantities remain bounded along the flow.

We define the energy of order $l\geq 1$ as

(7.1)
$\begin{split}\mathcal{E}_{l}(t)=\frac{1}{2}\int_{\Omega_{t}}&|\mathcal{D}_{t}^{l+1}v|^{2}\,dx+\frac{1}{2}\int_{\Sigma_{t}}|\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu)|^{2}\,d\mathcal{H}^{2}\\\
&-\frac{Q(t)}{2}\int_{\Omega_{t}^{c}}|\nabla(\partial_{t}^{l+1}U)|^{2}\,dx+\int_{\Omega_{t}}|\nabla^{\lfloor\frac{1}{2}(3l+1)\rfloor}\omega|^{2}\,dx,\end{split}$

where $\lfloor\frac{1}{2}(3l+1)\rfloor$ is the integer part of
$\frac{1}{2}(3l+1)$, $\omega$ is the curl of $v$ defined as

$\omega=\operatorname{curl}v=\nabla v-\nabla v^{T}.$

In this section we calculate $\frac{d}{dt}\mathcal{E}_{l}$ and estimate it in
terms of the $E_{l}(t)$, which we recall is defined in (5.1) as

$E_{l}(t)=\sum_{k=0}^{l}\|\mathcal{D}_{t}^{l+1-k}v\|_{H^{\frac{3}{2}k}(\Omega_{t})}^{2}+\|v\|_{H^{\lfloor\frac{3}{2}(1+1)\rfloor}(\Omega_{t})}^{2}+\|\mathcal{D}_{t}^{l}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2}+1.$

In particular, it holds $\mathcal{E}_{l}(t)\leq CE_{l}(t)$. We state the main
results of this section below and prove them later.

######  Proposition 7.1.

Assume that the a priori estimates (1.7) hold for $T>0$. Then for all $t<T$ it
holds

$\frac{d}{dt}\mathcal{E}_{1}(t)\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t),$

where the constant depends on $T,\Lambda_{T},\sigma_{T}$ and $E_{1}(0)$, i.e.,
$E_{1}(t)$ at time $t=0$.

######  Proposition 7.2.

Let $l\geq 2$ and assume that (1.7) and $E_{l-1}(t)\leq\tilde{C}$ hold for all
$t\in[0,T)$. Then all $t<T$ it holds

$\frac{d}{dt}\mathcal{E}_{l}(t)\leq CE_{l}(t)$

where the constant depends on $l,T,\Lambda_{T},\sigma_{T}$ and on
$\sup_{t<T}E_{l-1}(t)$.

The proof of the both above energy estimates is based on the calculations of
the differential of $\mathcal{E}_{l}(t)$, which we state first for all $l\geq
1$. In the proof of Proposition 7.1 and Proposition 7.2 we then need to
estimate the remainder terms by the quantity $E_{l}(t)$.

We begin by differentiating the first term of $\mathcal{E}_{l}(t)$ in (7.1)
and obtain by $\operatorname{div}v=0$, by (4.15) and by the definition of
$E_{l}(t)$ that

(7.2)
$\begin{split}&\frac{d}{dt}\frac{1}{2}\int_{\Omega_{t}}|\mathcal{D}_{t}^{l+1}v|^{2}\,dx=\int_{\Omega_{t}}(\mathcal{D}_{t}^{l+2}v\cdot\mathcal{D}_{t}^{l+1}v)\,dx\\\
&=-\int_{\Omega_{t}}(\nabla\mathcal{D}_{t}^{l+1}p\cdot\mathcal{D}_{t}^{l+1}v)\,dx-\int_{\Omega_{t}}([\mathcal{D}_{t}^{l+1},\nabla]p\cdot\mathcal{D}_{t}^{l+1}v)\,dx\\\
&\leq-\int_{\Omega_{t}}(\nabla\mathcal{D}_{t}^{l+1}p\cdot\mathcal{D}_{t}^{l+1}v)\,dx+\|\mathcal{D}_{t}^{l+1}v\|_{L^{2}(\Omega_{t})}^{2}+\|[\mathcal{D}_{t}^{l+1},\nabla]p\|_{L^{2}(\Omega_{t})}^{2}\\\
&=-\int_{\Omega_{t}}\operatorname{div}(\mathcal{D}_{t}^{l+1}p\,\mathcal{D}_{t}^{l+1}v)\,dx+\int_{\Omega_{t}}\mathcal{D}_{t}^{l+1}p\,\operatorname{div}(\mathcal{D}_{t}^{l+1}v)\,dx+E_{l}(t)+\|R_{bulk}^{l}\|_{L^{2}(\Omega_{t})}^{2}\\\
&=-\int_{\Sigma_{t}}\mathcal{D}_{t}^{l+1}p\,(\mathcal{D}_{t}^{l+1}v\cdot\nu)\,d\mathcal{H}^{2}+E_{l}(t)+\|R_{bulk}^{l}\|_{L^{2}(\Omega_{t})}^{2}+\int_{\Omega_{t}}\mathcal{D}_{t}^{l+1}p\,\operatorname{div}(\mathcal{D}_{t}^{l+1}v)\,dx.\end{split}$

Next we differentiate the second term in the energy and obtain by $\|\nabla
v\|_{L^{\infty}}\leq C$ and (4.2)

(7.3)
$\begin{split}&\frac{d}{dt}\frac{1}{2}\int_{\Sigma_{t}}|\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu)|^{2}\,d\mathcal{H}^{2}\\\
&=\int_{\Sigma_{t}}(\mathcal{D}_{t}\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu)\cdot\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu))\,d\mathcal{H}^{2}+\frac{1}{2}\int_{\Sigma_{t}}|\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu)|^{2}(\operatorname{div}_{\tau}v)\,d\mathcal{H}^{2}\\\
&\leq\int_{\Sigma_{t}}(\nabla_{\tau}\mathcal{D}_{t}(\mathcal{D}_{t}^{l}v\cdot\nu)\cdot\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu))\,d\mathcal{H}^{2}+C\|\mathcal{D}_{t}^{l}v\cdot\nu\|_{H^{1}(\Sigma_{t})}^{2}\\\
&=-\int_{\Sigma_{t}}(\Delta_{\Sigma_{t}}(\mathcal{D}_{t}^{l}v\cdot\nu))(\mathcal{D}_{t}(\mathcal{D}_{t}^{l}v\cdot\nu))\,d\mathcal{H}^{2}+CE_{l}(t)\\\
&=-\int_{\Sigma_{t}}(\Delta_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu))(\mathcal{D}_{t}^{l+1}v\cdot\nu)\,d\mathcal{H}^{2}\\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\int_{\Sigma_{t}}(\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\nu)\cdot\nabla_{\tau}(\mathcal{D}_{t}^{l}v\cdot\mathcal{D}_{t}\nu))\rangle\,d\mathcal{H}^{2}+CE_{l}(t)\\\
&\leq-\int_{\Sigma_{t}}(\Delta_{\Sigma_{t}}(\mathcal{D}_{t}^{l}v\cdot\nu))(\mathcal{D}_{t}^{l+1}v\cdot\nu)\,d\mathcal{H}^{2}+CE_{l}(t)+\|\mathcal{D}_{t}^{l}v\cdot\mathcal{D}_{t}\nu\|_{H^{1}(\Sigma_{t})}^{2}.\end{split}$

We differentiate the third term, use the fact that $\partial_{t}^{l+1}U$ is
harmonic and Lemma 4.6 and have (recall that it holds
$\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Omega_{t}^{c})}\leq
C\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Sigma_{t})}$)

(7.4)
$\begin{split}\frac{d}{dt}&-\frac{Q(t)}{2}\int_{\Omega_{t}^{c}}|\nabla\partial_{t}^{l+1}U|^{2}\,dx=-Q(t)\int_{\Omega_{t}^{c}}\langle\nabla\partial_{t}^{l+2}U,\nabla\partial_{t}^{l+1}U\rangle\,dx\\\
&\,\,\,\,\,\,\,\,+\frac{Q(t)}{2}\int_{\Sigma_{t}}|\nabla\partial_{t}^{l+1}U|^{2}v_{n}\,d\mathcal{H}^{2}-\frac{Q^{\prime}(t)}{2}\int_{\Omega_{t}^{c}}|\nabla\partial_{t}^{l+1}U|^{2}\,dx\\\
&\leq
Q(t)\int_{\Sigma_{t}}\partial_{t}^{l+2}U\,(\partial_{\nu}\partial_{t}^{l+1}U)\,d\mathcal{H}^{2}+C\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Sigma_{t})}^{2}\\\
&=-Q(t)\int_{\Sigma_{t}}(\partial_{\nu}U(\mathcal{D}_{t}^{l+1}v\cdot\nu)+R_{U}^{l})\,(\partial_{\nu}\partial_{t}^{l+1}U)\,d\mathcal{H}^{2}+C\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Sigma_{t})}^{2}\\\
&\leq-Q(t)\int_{\Sigma_{t}}(\partial_{\nu}U\,\partial_{\nu}\partial_{t}^{l+1}U)(\mathcal{D}_{t}^{l+1}v\cdot\nu)\,d\mathcal{H}^{2}+\|R_{U}^{l}\|_{L^{2}(\Sigma_{t})}^{2}+C\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Sigma_{t})}^{2},\end{split}$

where $R_{U}^{l}$ in the remainder term defined in (4.18).

Finally, we differentiate the fourth term involving the curl. To that aim we
denote $\lambda_{l}:=\lfloor\frac{1}{2}(3l+1)\rfloor$. We have by Lemma 4.4
and by (2.16)

(7.5)
$\begin{split}\frac{d}{dt}\int_{\Omega_{t}}|\nabla^{\lambda_{l}}\omega|^{2}\,dx&=\int_{\Omega_{t}}\nabla
v\star\nabla^{\lambda_{l}}\omega\star\nabla^{\lambda_{l}}\omega\,dx+C\|\nabla^{\lambda_{l}}\omega\|_{L^{2}(\Omega_{t})}^{2}\\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\sum_{|\alpha|\leq\lambda_{l}}\|\nabla^{1+\alpha_{1}}v\star\nabla^{1+\alpha_{2}}v\|_{L^{2}(\Omega_{t})}^{2}\\\
&\leq C\|\nabla^{\lambda_{l}}\omega\|_{L^{2}(\Omega_{t})}^{2}+C\|\nabla
v\|_{L^{\infty}(\Omega_{t})}^{2}\|\nabla
v\|_{H^{\lambda_{l}}(\Omega_{t})}^{2}\\\ &\leq
C\|\nabla^{\lambda_{l}}\omega\|_{L^{2}(\Omega_{t})}^{2}+C\|v\|_{H^{\lfloor\frac{3}{2}(l+1)\rfloor}(\Omega_{t})}^{2}\leq
CE_{l}(t).\end{split}$

Let us next show that the highest order terms in (7.2), (7.3) and (7.4) cancel
out. Indeed, this follows from Lemma 4.7 which states that

$\mathcal{D}_{t}^{l+1}p=-\Delta_{\Sigma_{t}}(\mathcal{D}_{t}^{l}v\cdot\nu)-Q(t)\,\partial_{\nu}U\,(\partial_{\nu}\partial_{t}^{l+1}U)+R_{p}^{l},$

where $R_{p}^{l}$ denotes the error term defined in (4.26). Note that we may
estimate

$\big{|}\int_{\Sigma_{t}}R_{p}^{l}(\mathcal{D}_{t}^{l+1}v\cdot\nu)\,d\mathcal{H}^{2}\big{|}\leq\|R_{p}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}+\|\mathcal{D}_{t}^{l+1}v\cdot\nu\|_{H^{-\frac{1}{2}}(\Sigma_{t})}^{2}.$

By divergence theorem and by Lemma 4.4 we have

$\|\mathcal{D}_{t}^{l+1}v\cdot\nu\|_{H^{-\frac{1}{2}}(\Sigma_{t})}^{2}\leq
C\|\mathcal{D}_{t}^{l+1}v\|_{L^{2}(\Omega_{t})}^{2}+\|R_{\operatorname{div}}^{l}\|_{L^{2}(\Omega_{t})}^{2}\leq
CE_{l}(t)+\|R_{\operatorname{div}}^{l}\|_{L^{2}(\Omega_{t})}^{2}.$

Therefore we have by (7.2), (7.3), (7.4) and (7.5) that

(7.6) $\begin{split}\frac{d}{dt}\mathcal{E}_{l}(t)&\leq
CE_{l}(t)+\|R_{bulk}^{l}\|_{L^{2}(\Omega_{t})}^{2}+\|R_{U}^{l}\|_{L^{2}(\Sigma_{t})}^{2}+\|R_{\operatorname{div}}^{l}\|_{L^{2}(\Omega_{t})}^{2}+\|R_{p}^{l}\|_{H^{\frac{1}{2}}(\Sigma_{t})}^{2}\\\
&\,\,\,\,+C\|\nabla\partial_{t}^{l+1}U\|_{L^{2}(\Sigma_{t})}^{2}+\|\mathcal{D}_{t}^{l}v\cdot\mathcal{D}_{t}\nu\|_{H^{1}(\Sigma_{t})}^{2}+\int_{\Omega_{t}}\mathcal{D}_{t}^{l+1}p\,\operatorname{div}(\mathcal{D}_{t}^{l+1}v)\,dx.\end{split}$

We need thus to bound the remainder terms in (7.6). As we mentioned before, we
prove the energy bounds by induction such that we bound $E_{l}(t)$ when we
know that $E_{l-1}(t)$ is already bounded. The difficulty is to bound the
first order quantity $E_{1}(t)$ and we do this separately in Proposition 7.1.

###### Proof of Proposition 7.1.

First we recall that Proposition 6.3 implies

$\sup_{t\in(0,T)}\|\nabla p\|_{L^{2}(\Omega_{t})}^{2}\leq e^{CT}(1+\|\nabla
p\|_{L^{2}(\Omega_{0})}^{2})\leq C,$

where the constant $C$ on the RHS depends on $T,\Lambda_{T}$ defined in (1.7)
and on $E_{1}(0)$, which is the energy quantity $E_{1}(t)$ at time $t=0$. Then
we have the bound (6.9) which we recall is

$\|B\|_{L^{4}(\Sigma_{t})}+\|B\|_{H^{\frac{1}{2}}(\Sigma_{t})}\leq C$

for all $t<T$.

We recall the estimate (7.6) for $l=1$ and use Lemma 5.3, Lemma 5.4, estimate
(5.15) from Lemma 5.6 and Lemma 5.7 to deduce

$\frac{d}{dt}\mathcal{E}_{1}(t)\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t)+\|\mathcal{D}_{t}v\cdot\mathcal{D}_{t}\nu\|_{H^{1}(\Sigma_{t})}^{2}+\int_{\Omega_{t}}\mathcal{D}_{t}^{2}p\,\operatorname{div}(\mathcal{D}_{t}^{2}v)\,dx.$

Hence, we need to bound the two last terms. The second last term is easy to
treat and we merely claim that it holds

$\|\mathcal{D}_{t}v\cdot\mathcal{D}_{t}\nu\|_{H^{1}(\Sigma_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

Indeed, this follows from $\mathcal{D}_{t}v=-\nabla p$,
$\mathcal{D}_{t}\nu=-(\nabla_{\tau}v)^{T}\nu$ from (4.4) and from Proposition
2.10. We leave the details for the reader. The last term is challenging and we
will prove that

(7.7)
$\int_{\Omega_{t}}\mathcal{D}_{t}^{2}p\,\operatorname{div}(\mathcal{D}_{t}^{2}v)\,dx\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

The argument is similar than in the proof of Proposition 6.3. The idea is to
use the fact that the term $\operatorname{div}(\mathcal{D}_{t}^{2}v)$ is lower
order due to the fact that $\operatorname{div}v=0$. Indeed, we have by Lemma
4.4 that
$\operatorname{div}(\mathcal{D}_{t}^{2}v)=R^{1}_{\operatorname{div}}$, where
$R^{1}_{\operatorname{div}}$ is defined in (4.13) and is lower order than
$\mathcal{D}_{t}^{2}v$. To this aim let $u$ be a solution of

$\begin{cases}&-\Delta
u=\operatorname{div}(\mathcal{D}_{t}^{2}v),\quad\text{in }\,\Omega_{t}\\\
&\,\,\,\,\,\,\,u=0\qquad\text{on }\,\Sigma_{t}.\end{cases}$

We have by integration by parts

$\int_{\Omega_{t}}\mathcal{D}_{t}^{2}p\,\operatorname{div}(\mathcal{D}_{t}^{2}v)\,dx=-\int_{\Omega_{t}}\mathcal{D}_{t}^{2}p\,\Delta
u\,dx=-\int_{\Omega_{t}}\Delta\mathcal{D}_{t}^{2}p\,u\,dx-\int_{\Sigma_{t}}\mathcal{D}_{t}^{2}p\,\partial_{\nu}u\,d\mathcal{H}^{2}.$

We use Remark 4.5 and write

$-\Delta\mathcal{D}_{t}^{2}p=\operatorname{div}\operatorname{div}(v\otimes\mathcal{D}_{t}^{2}v)+\operatorname{div}(R_{bulk}^{1}).$

By integration by parts we deduce

(7.8)
$\begin{split}-\int_{\Omega_{t}}\Delta\mathcal{D}_{t}^{2}p\,u\,dx=&\int_{\Omega_{t}}(v\otimes\mathcal{D}_{t}^{2}v):\nabla^{2}u\,dx+\int_{\Omega_{t}}R_{bulk}^{1}\star\nabla
u\,dx\\\ &-\int_{\Sigma_{t}}(\mathcal{D}_{t}^{2}v\cdot\nu)(\nabla u\cdot
v)\,d\mathcal{H}^{2}.\end{split}$

We use divergence theorem, the definition of $E_{1}(t)$,
$\operatorname{div}(\mathcal{D}_{t}^{2}v)=R^{1}_{\operatorname{div}}$ and
Lemma 5.3 for the last term

$\begin{split}-\int_{\Sigma_{t}}(\mathcal{D}_{t}^{2}v\cdot\nu)(\nabla u\cdot
v)\,d\mathcal{H}^{2}&=-\int_{\Omega_{t}}\operatorname{div}\big{(}(\nabla
u\cdot v)\mathcal{D}_{t}^{2}v\big{)}\\\ &\leq
CE_{1}(t)+\|u\|_{H^{2}(\Omega_{t})}^{2}+C\|R^{1}_{\operatorname{div}}\|_{L^{2}(\Omega_{t})}^{2}\\\
&\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t)+\|u\|_{H^{2}(\Omega_{t})}^{2}.\end{split}$

Since $-\Delta u=\operatorname{div}(D_{t}^{2}v)$, we may use the inequality
(3.19) and Lemma 5.3 to obtain

$\|u\|_{H^{2}(\Omega_{t})}^{2}\leq
C\|R^{1}_{\operatorname{div}}\|_{L^{2}(\Omega_{t})}^{2}\leq
C(1+\|p\|_{H^{2}(\Omega_{t})}^{2})E_{1}(t).$

Therefore we have by (7.8), by the definition of $E_{1}(t)$ and by Lemma 5.4
that

(7.9)