# Very weak solutions of the Dirichlet problem for 2-Hessian equation

Tongtong Li  Institute of Mathematics, Academy of Mathematics and Systems
Science, Chinese Academy of Sciences, No.55 Zhongguancun East Road, 100190,
Beijing, China<EMAIL_ADDRESS>and  Guohuan Qiu  Institute of
Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of
Sciences, No.55 Zhongguancun East Road, 100190, Beijing, China
<EMAIL_ADDRESS>
###### Abstract.

For any $\alpha$ small, we construct infinitely many $C^{1,\alpha}$ very weak
solutions to the 2-Hessian equation with prescribed boundary value. The proof
relies on the convex integration method and cut-off technique.

###### Key words and phrases:

2-Hessian equation; Dirichlet problem; very weak solutions; Nash-Kuiper
construction

###### 2020 Mathematics Subject Classification:

Primary 35M10; Secondary 76B03

##  1\. Introduction

Given a smooth function $v:\mathbb{R}^{n}\to\mathbb{R}$, the $2$-Hessian
equation of $v$ is defined by111Our definition of $\sigma_{2}$ is different
from the standard one by a factor $2$.

$\sigma_{2}(\nabla^{2}v):=\sum^{n}_{i,j=1}[\partial_{ii}v\partial_{jj}v-(\partial_{ij}v)^{2}]=f.$
(1.1)

In the seminal work [5], Caffarelli, Nirenberg and Spruck established the
existence and uniqueness of a classical solution to the Dirichlet problem:

$\begin{cases}\sigma_{2}(\nabla^{2}v)=f>0&\text{in
}\Omega\subset\mathbb{R}^{n},\ n\geq 2,\\\ v=g&\text{on
}\partial\Omega,\end{cases}$ (1.2)

Furthermore, existence and uniqueness of the Dirichlet problem (1.2) also hold
for 2-convex weak solutions in the viscosity sense, see [43, 39].

The 2-Hessian equation (1.1) is a simplified model of the scalar curvature
equation for hypersurfaces. Motivated by the Weyl isometric embedding problem,
E. Heinz [20] established the interior estimates when $n=2$. This is a fully
nonlinear partial differential equation for which the regularity is a highly
nontrivial problem. Pogorelov [33] constructed non-$C^{2}$ convex solutions to
the equation $\det D^{2}u=1$ in $B_{r}\subseteq\mathbb{R}^{n}$ when $n\geq 3$.
Warren and Yuan [44] established the interior estimate for
$\sigma_{2}(D^{2}u)=1$ when $n=3$. For smooth $f>0$ and $n=3$, the second
named author [34, 35] proved the interior estimates for 2-Hessian equations
and scalar curvature equations. Recently, Shankar and Yuan [38] derived a
priori interior Hessian estimate and interior regularity for
$\sigma_{2}(D^{2}u)=1$ in dimension four. Combining the solvability and
uniqueness of the Dirichlet problem (1.2) for viscosity weak solutions, the
above a priori estimates imply that viscosity solutions become smooth when
$n=2,3,4$ (at least for $f=1$ when $n=4$). In higher dimensions, additional
convexity conditions are assumed to obtain the interior estimate and
regularity of the 2-Hessian equation, as seen in papers like [18, 27, 36, 28,
37]. The question of whether viscosity solutions to (1.1) become smooth in
dimensions $n\geq 5$ remains a famous open problem in the field. The most
crucial ingredient for obtaining the interior estimate is the validity of the
Jacobi inequality when $n=3$ and $4$, as demonstrated in [44, 34, 38]. But the
crucial Jacobi inequality does not hold when $n\geq 5$, see [38]. Thus one may
inquire whether it is possible to construct $C^{1,\alpha}$ viscosity weak
solutions by combining the convex integration method with other approaches.
The motivation behind our study of _very weak solutions_ , as defined in 1.1,
is to explore possible ways to construct singular $C^{1,\alpha}$ viscosity
weak solutions when $n\geq 5$. Although _very weak solutions_ and viscosity
solutions are entirely different types of weak solutions, it is still worth
exploring possible connections between them.

In this paper, we focus on the non-uniqueness of _very weak solutions_ ,
defined as follows.

######  Definition 1.1.

Let $\Omega$ be a domain in $\mathbb{R}^{n}$ and $v\in W^{1,2}_{loc}(\Omega)$.
We say $v$ solves $\sigma_{2}(\nabla^{2}v)=f$ in very weak sense if

$-\sum^{n}_{i,j=1}\int_{\Omega}\sigma_{2}^{ij}(\nabla^{2}\phi)\partial_{i}v\partial_{j}v=\int_{\Omega}f\phi,\quad\forall\phi\in
C^{\infty}_{c}(\Omega),$

where

$\sigma_{2}^{ij}(\nabla^{2}\phi)=\Delta\phi\delta_{ij}-\partial_{ij}\phi.$

And $v$ is a very weak solution to (1.2) if further $v\in C(\bar{\Omega})$ and
$v=g$ on $\partial\Omega$.

This definition is motivated by the double divergence structure of
$\sigma_{2}(\nabla^{2}v)$ (see Section 2). For the Laplace equation $\Delta
u=f$, the well-known Weyl lemma guarantees the smoothness of locally
integrable weak solutions in the following sense:

$\int_{\Omega}u\Delta\phi=\int_{\Omega}f\phi,\quad\forall\phi\in
C^{\infty}_{c}(\Omega).$

One interesting point compared with the Laplace equation is that after
integrating by parts twice, there are quadratic gradient terms for 2-Hessian
equation. For the Monge-Ampere equation, one well-known concept of weak
solutions was introduced by Alexandrov [1], now commonly referred to as
Alexandrov solutions, which are equivalent to viscosity solutions, as shown in
[19]. After Trudinger and Wang’s series of challenging works [40, 41, 42],
Alexandrov solutions have been successfully extended to weak solutions defined
by the $2$-Hessian measure for the $\sigma_{2}$ operator. It is important to
note that the concept of the _very weak_ solution is distinct from the weak
solution defined in [40, 41, 42].

In dimension two, (1.1) reduces to Monge-Ampère equation. In this case, the
concept of distributional $\sigma_{2}$ was introduced by Iwaniec [21] in the
name of _weak Hessian_. Lewicka and Pakzad [26] (see also [13, 10]) noticed
that (1.1) is closely related to isometric immersion of surfaces into
$\mathbb{R}^{3}$. Using the convex integration method there, they showed that
$C^{1,\alpha}(\bar{\Omega})$ very weak solutions to (1.1) are dense in
$C^{0}(\bar{\Omega})$ for any $\alpha<1/7$ and $f\in L^{7/6}(\Omega)$. This
was improved by Cao and Székelyhidi Jr. [8] to $\alpha<1/7$ and recently by
Cao, Hirsch, and Inauen [9] to $\alpha<1/3$. For the corresponding Dirichlet
problem (1.2), Cao [6] proved Theorem 1.2 and 1.3 with $\alpha<1/5$.

In general dimensions $n\geq 2$, Lewicka [24] considered the following Monge-
Ampère system

$\mathfrak{Det}\nabla^{2}v=F\quad\text{in }\Omega\subset\mathbb{R}^{n},$

where $F=(F_{ij,st})_{1\leq i,j,s,t\leq n}:\Omega\to\mathbb{R}^{n^{4}}$ is
given, $v:\Omega\to\mathbb{R}^{k}$ is unknown and

$(\mathfrak{Det}\nabla^{2}v)_{ij,st}:=\langle\partial_{is}v,\partial_{jt}v\rangle-\langle\partial_{it}v,\partial_{js}v\rangle.$

She derived the denseness result for $\alpha<\frac{1}{1+(n^{2}+n)/k}$ (cf.
[25]). When $k=1$, this covers the counterpart of Theorem 1.2 without boundary
conditions, since
$\sigma_{2}(\nabla^{2}v)=\sum_{i,j}(\mathfrak{Det}\nabla^{2}v)_{ij,ij}$.

In this paper, we prove the following theorems, generalizing the result of Cao
[6] to higher dimensions.

######  Theorem 1.2.

Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with $C^{2,\kappa}$
boundary $\partial\Omega$, $f\in C^{0,\kappa}(\bar{\Omega})$,
$\kappa\in(0,1)$. Then there exists some $g\in C^{2,\kappa}(\partial\Omega)$,
such that: for any

$\alpha<\frac{1}{1+n+n^{2}},$

there exist infinitely many very weak solutions $v$ to the Dirichlet problem
(1.2) such that $v\in C^{1,\alpha}(\bar{\Omega})$.

######  Theorem 1.3.

Let $\Omega,f,\kappa$ be as in Theorem (1.2). Assume further $f>0$. Then for
any $g\in C^{0,\kappa}(\partial\Omega)$ and any

$\alpha<\frac{1}{1+n+n^{2}},$

there exist infinitely many very weak solutions $v$ to the Dirichlet problem
(1.2) such that $v\in C^{1,\alpha}(\bar{\Omega})$.

The proof is based on an iteration scheme called convex integration, which was
originally introduced by Nash-Kuiper [29, 22] to construct $C^{1}$ isometric
embedding of manifolds, and later generalized (and named) by Gromov [15, 16]
to more general setting to prove the abundance of (weak) solutions (Gromov’s
$h$-principle). After Borisov’s earlier work [2, 4], Conti, De Lellis, and
Székelyhidi [12, 11] and Cao and Székelyhidi [7] showed that any $C^{1}$
differentiable immersion (resp. embedding) of a smooth closed $n$-manifold
into $\mathbb{R}^{n+1}$ can be deformed to an $C^{1,\alpha}$ isometric
immersion (resp. embedding) if $\alpha<1/(1+n+n^{2})$. For surfaces this was
improved to $\alpha<1/5$ in [14, 7].

These results asserting the _abundance_ of (weak) solutions are often referred
to as _flexibility_ results. On the other hand, one obtains _rigidity_ ,
namely _uniqueness_ of solutions, if higher regularity is required. For
instance, Borisov [3] showed that any $C^{1,\alpha},\alpha>2/3$ isometric
_embedding_ of a smooth closed surface with positive Guass curvature into
$\mathbb{R}^{3}$ is unique up to rigid motion. A modern proof can be found in
[12, 11]. This result is based on the work of Pogorelov [31, 32]. Gromov [17,
Section 3.5.5.C, Open Problem 34-36] conjectures the regularity threshold of
rigidity vs. flexibility to be $\alpha=1/2$. See [23, 30] and references
therein for recent developments. In [30], Pakzad also proved the rigidity of
$C^{1,\alpha}$ very weak solutions to (1.2) for $n=2,\alpha>2/3$ and
$\Omega=\mathbb{R}^{2}$. One interesting question regarding the rigidity
aspect of the 2-Hessian equation is whether a $C^{1,\alpha}$ (for
$\alpha\approx 1$) very weak solution of (1.2) become the viscosity weak
solution.

This article is organized as follows. Section 2 collects some basic facts, and
in Remark 2.2 we explain the idea of the proof. In Section 3 we omit the
boundary condition $v=g$ and construct very weak solutions by convex
integration. The main ingredient is the one stage induction Proposition 3.5,
relying on the decomposition Lemma 3.4. In Section 4 we apply the cut-off
technique in [6] such that the boundary condition is satisfied, thus finishing
the proof of Theorem 1.2 and 1.3.

##  2\. Preliminaries.

###  2.1. Distributional $\sigma_{2}$

We observe that $\sigma_{2}(\nabla^{2}v)$ has the following divergence
structure

$\displaystyle\sigma_{2}(\nabla^{2}v)$
$\displaystyle=\sum_{i,j}\partial_{i}(\partial_{i}v\partial_{jj}v)-\partial_{j}(\partial_{i}v\partial_{ij}v)$
(2.1)
$\displaystyle=\sum_{i,j}\partial_{ij}(\partial_{i}v\partial_{j}v)-\partial_{i}(\partial_{ij}v\partial_{j}v)-\partial_{j}(\partial_{i}v\partial_{ij}v)$
(2.2)
$\displaystyle=\sum_{i,j}\partial_{ij}(\partial_{i}v\partial_{j}v)-\frac{1}{2}\partial_{ii}(\partial_{j}v)^{2}-\frac{1}{2}\partial_{jj}(\partial_{i}v)^{2}.$
(2.3)

By (2.3), there holds

$\displaystyle\int\sigma_{2}(\nabla^{2}v)\phi$
$\displaystyle=\sum_{i,j}\int\partial_{ij}\phi\partial_{i}v\partial_{j}v-\frac{1}{2}\partial_{ii}\phi(\partial_{j}v)^{2}-\frac{1}{2}\partial_{jj}\phi(\partial_{i}v)^{2}$
$\displaystyle=-\sum_{i,j}\int\sigma_{2}^{ij}(\nabla^{2}\phi)\partial_{i}v\partial_{j}v,\quad\forall\phi\in
C^{\infty}_{c}(\mathbb{R}^{n}),$

where

$\sigma_{2}^{ij}(\nabla^{2}\phi)=\Delta\phi\delta_{ij}-\partial_{ij}\phi.$

This allows us to define $\sigma_{2}(\nabla^{2}v)$ in distributional sense for
any $v\in W^{1,2}_{loc}(\mathbb{R}^{n})$. Accordingly, we can define _very
weak solutions_ to $\sigma_{2}$ equation as in Definition 1.1.

######  Remark 2.1.

In fact, we can provide different distributional definitions of $\sigma_{2}$
based on (2.1) (2.2) (2.3). Refer to [21] for a discussion in dimension two.

###  2.2. Notations

Basically we follow the notations in [8, Section 2]. The set of $n\times n$
symmetric matrices is denoted by ${\mathcal{S}_{n}}$. We use usual Hölder
norms $\|\cdot\|_{k},\|\cdot\|_{k,\beta}$. Also we write
$\|(v,w)\|_{k}=\|v\|_{k}+\|w\|_{k}$.

We fix a non-negative radial function $\phi\in C^{\infty}_{c}(B_{1}(0))$, such
that

$\int_{\mathbb{R}^{n}}\phi(x)dx=1.$

This $\phi$ will serve as a mollifier. Namely, let
$\phi_{l}(x)=l^{-n}\phi(\frac{x}{l})$. For any function $h$ on
$\mathbb{R}^{n}$, the mollification of $h$ at length $l$ is

$h*\phi_{l}(x)=\int_{\mathbb{R}^{n}}h(x-y)\phi_{l}(y)dy.$

A constant is called universal if it depends only on $n,\Omega,f,g$. $C>1$
denotes a positive universal constant that may vary from line to line.
Besides, we write $a\lesssim b$ if $a\leq Cb$, and $a\sim b$ if
$\frac{b}{C}\leq a\leq Cb$, for some universal constant $C$.

###  2.3. Corrugation functions

Let

$\Gamma_{1}(s,t)=\frac{s}{\pi}\sin(2\pi
t),\quad\Gamma_{2}(s,t)=-\frac{s^{2}}{4\pi}\sin(4\pi t)$

which satisfy the important relation

$\partial_{t}\Gamma_{2}(s,t)+\frac{1}{2}|\partial_{t}\Gamma_{1}(s,t)|^{2}=s^{2}.$
(2.4)

It’s easy to get following bounds: for $0\leq k\leq 3$,

$\displaystyle|\partial_{t}^{k}\Gamma_{1}(s,t)|\leq
Cs,\quad|\partial_{s}\partial_{t}^{k}\Gamma_{1}(s,t)|\leq C,$ (2.5)
$\displaystyle|\partial_{t}^{k}\Gamma_{2}(s,t)|\leq
Cs^{2},\quad|\partial_{s}\partial_{t}^{k}\Gamma_{2}(s,t)|\leq
Cs,\quad|\partial_{s}^{2}\partial_{t}^{k}\Gamma_{2}(s,t)|\leq C.$

######  Remark 2.2.

Such periodic functions $\Gamma_{1},\Gamma_{2}$ satisfying (2.4) are crucial
in convex integration, whose existence is guaranteed by the quadratic
structure in distributional $\sigma_{2}$. We give some explanations below.

It turns out that (1.2) is equivalent to (3.3) (3.4) with $v=g$ on
$\partial\Omega$. (3.3) is easy to solve since the elements of $A$ are
independent. The idea of convex integration is, starting from arbitrary
$(v^{b},w^{b})$, to construct inductively $\\{(V_{q},W_{q})\\}_{q=1}^{\infty}$
that finally converges in $C^{1,\alpha}$ sense to some solution $(v,w)$ of
(3.4).

By linearization of (3.4), the derivative
$(\nabla(v-v^{b}),\nabla(w-w^{b})+\text{sym}\nabla(v-v^{b})\otimes\nabla
v^{b}))$ has to satisfy a relation similar to $y+\frac{1}{2}x^{2}=1$, cf.
(3.17). Since the origin lies in the convex hull of the curve
$y+\frac{1}{2}x^{2}=1$, we let the derivative move along the curve, such that
its integration is small. Then $(v,w)$ oscillates around $(v^{b},w^{b})$,
leading to the $C^{0}$ closeness $\|(v^{b}-v,w^{b}-w)\|_{0}\leq\epsilon$. This
is where the name ”convex integration” comes from.

This is achieved by adding higher and higher frequency oscillatory functions
(i.e. corrugation) produced by $\Gamma_{1},\Gamma_{2}$ at each stage
$(V_{q},W_{q})\to(V_{q+1},W_{q+1})$. Since (3.4) is actually a matrix
equation, each stage has to be decomposed into $N_{*}=n(n+1)/2$ steps, which
determines $(v,w)\in C^{1,\alpha}$ with $\alpha<1/(1+2N_{*})$, cf. [8, 9].

###  2.4. Mollification estimates.

Recall that for any $r,s\geq 0$

$\displaystyle[h*\phi_{l}]_{r}\leq[h]_{r},$ (2.6)
$\displaystyle[h-h*\phi_{l}]_{r}\leq
C[h]_{r+s}l^{s},\quad[h*\phi_{l}]_{r+s}\leq C[h]_{r}l^{-s},$ (2.7)
$\displaystyle[(h_{1}h_{2})*\phi_{l}-(h_{1}*\phi_{l})(h_{2}*\phi_{l})]_{r}\leq
C\|h_{1}\|_{1}\|h_{2}\|_{1}l^{2-r},$ (2.8)

where $[h]_{r}=\|\nabla^{r}h\|_{0}$. See [12, Lemma 1] for the proof.
Mollification is important to control higher order derivatives, and the
commutator inequality (2.8) is used to estimate deficit matrices.

##  3\. Convex integration.

In this section we temporarily omit the boundary condition and construct very
weak solutions to

$\sigma_{2}(\nabla^{2}v)=f\quad\text{in }\Omega.$ (3.1)

We first introduce some notations. Let $L$ be a linear differential operator
acting on ${\mathcal{S}_{n}}$-valued functions defined by

$L(A):=\sum_{i,j}\partial_{ii}A_{jj}+\partial_{jj}A_{ii}-2\partial_{ij}A_{ij},\quad
A=(A_{ij})_{1\leq i,j\leq n}:\Omega\to{\mathcal{S}_{n}}.$ (3.2)

Then equation (1.2) can be written as

$-\frac{1}{2}L(\nabla v\otimes\nabla v)=f\quad\text{in}\ \Omega.$

The operator $L$ generalizes the curl curl operator in dimension two. In
particular, for $\forall$ $w=(w_{1},...,w_{n}):\Omega\to\mathbb{R}^{n}$, let

$\text{sym}\nabla w:=\frac{1}{2}(\nabla w+\nabla
w^{T})=(\frac{\partial_{i}w_{j}+\partial_{j}w_{i}}{2})_{1\leq i,j\leq n},$

then by simple computation, we can check that

$L(\text{sym}\nabla w)=0$

in distributional sense. In fact, due to
$\partial_{i}\sigma_{2}^{ij}(\nabla^{2}\phi)=\partial_{j}\sigma_{2}^{ij}(\nabla^{2}\phi)=0$,
there holds

$\displaystyle\sum_{i,j}\int\sigma_{2}^{ij}(\nabla^{2}\phi)(\frac{\partial_{i}w_{j}+\partial_{j}w_{i}}{2})=-\frac{1}{2}\sum_{i,j}\int\partial_{i}\sigma_{2}^{ij}(\nabla^{2}\phi)w_{j}+\partial_{j}\sigma_{2}^{ij}(\nabla^{2}\phi)w_{i}=0$

for any $\phi\in C_{c}^{\infty}(\Omega)$.

So to solve (3.1) it suffices to find $(A,v,w)\in
C^{0}(\bar{\Omega},{\mathcal{S}_{n}})\times C^{1}(\bar{\Omega})\times
C^{1}(\bar{\Omega},\mathbb{R}^{n})$ such that

$\displaystyle-L(A)=f\quad\text{in}\ \Omega,$ (3.3) $\displaystyle
A=\frac{1}{2}\nabla v\otimes\nabla v+\text{sym}\nabla w\quad\text{in}\
\Omega.$ (3.4)

Part one: solve (3.3). Let $A=(u+\tau)Id$, where $\tau>0$ is a constant and
$u$ solves the Dirichlet problem

$-(2n-2)\Delta u=f\ \text{in}\ \Omega,\quad u=0\ \text{on}\ \partial\Omega.$

Then

$-L(A)=-(2n-2)\Delta u=f$

and $A\in C^{2,\kappa}(\bar{\Omega},{\mathcal{S}_{n}})$ by classical elliptic
PDE theory.

Part two: solve (3.4). Replacing $v^{b}$ by its mollification, we may assume
$v^{b}\in C^{\infty}(\bar{\Omega})$, and let $w^{b}=0$. Denote the initial
deficit matrix by

$D^{b}:=A-\frac{1}{2}\nabla v^{b}\otimes\nabla v^{b}-\text{sym}\nabla w^{b}.$
(3.5)

Recall $A=(u+\tau)Id$. By letting $\tau$ large enough, the problem boils down
to the following:

######  Theorem 3.1 ([24], Theorem 1.1, $(d,k)=(n,1)$).

Let $\Omega\subset\mathbb{R}^{n}$ be an open, bounded domain. Given a function
$v^{b}\in C^{1}(\bar{\Omega})$ a vector field $w^{b}\in
C^{1}(\bar{\Omega},\mathbb{R}^{n})$ and a matrix field $A\in
C^{0,\beta}(\bar{\Omega},{\mathcal{S}_{n}})$, assume that:

$D^{b}:=A-\frac{1}{2}\nabla v^{b}\otimes\nabla v^{b}-\text{sym}\nabla
w^{b}\quad\text{satisfies}\quad D^{b}>\bar{\tau}Id\ on\ \bar{\Omega},$ (3.6)

for some $\bar{\tau}>0$, in the sense of matrix inequalities. Fix $\epsilon>0$
and let:

$0<\alpha<\min\\{\frac{\beta}{2},\frac{1}{1+n+n^{2}}\\}.$

Then, there exists $v\in C^{1,\alpha}(\bar{\Omega})$ and $w\in
C^{1,\alpha}(\bar{\Omega},\mathbb{R}^{n})$ such that the following holds:

$\displaystyle\|v-v^{b}\|_{0}\leq\epsilon,\quad\|w-w^{b}\|_{0}\leq\epsilon,$
(3.7) $\displaystyle A=\frac{1}{2}\nabla v\otimes\nabla v+\text{sym}\nabla
w\quad\text{in }\Omega.$ (3.8)

For our purpose we only prove for $A\in
C^{2,\kappa}(\bar{\Omega},{\mathcal{S}_{n}})$, and w.l.o.g we assume
$\bar{\tau}>1$. We divide its proof into four steps following [8]. The key is
the one stage induction Proposition 3.5 in Step 3.

######  Remark 3.2.

Step 1-2 are preparations in nature, both consisting of inductive
constructions that are very similar to (and in fact easier than) Step 3.
Besides, after finishing this note, we learned a trick from [9, Section 4]
that helps to bypass Step 1-2. So the readers may directly skip to Step 3. The
trick goes as follows. Recall $A=(u+\tau)Id$. Fix

$\tau=(K+\sigma^{-1})(\|u\|_{0}+\|v^{b}\|_{2}^{2}+100),$

and let

$\bar{A}=\delta_{1}\tau^{-1}A,\quad
V_{0}=\delta_{1}^{\frac{1}{2}}\tau^{-1/2}v^{b},\quad W_{0}=0.$

Then, replacing $A$ by $\bar{A}$, the assumptions (3.36)-(3.38) hold at $q=0$.
By induction we get a solution $(v,w)$ to (3.1) with $A$ replaced by
$\bar{A}$. Then $(\delta_{1}^{-\frac{1}{2}}\tau^{1/2}v,\delta_{1}^{-1}\tau w)$
is a solution to (3.1).

Step 1 (Preparation): $(v^{b},w^{b})\to(\bar{v}_{N},\bar{w}_{N})$. In this
step we aim to reduce $D^{b}$ to the neighborhood of identity matrix $Id$,
i.e. to construct $(\bar{v}_{i},\bar{w}_{i}),1\leq i\leq N$ inductively such
that

$\displaystyle\|(\bar{v}_{N},\bar{w}_{N})-(v^{b},w^{b})\|_{0}\leq\frac{\epsilon}{3},$
(3.9) $\displaystyle\|\bar{D}\|_{0}\leq\frac{\sigma_{*}}{3},$ (3.10)

where $\sigma_{*}>0$ is given as in Lemma 3.4 and

$\bar{D}:=A-\frac{1}{2}\nabla\bar{v}_{N}\otimes\nabla\bar{v}_{N}-\text{sym}\nabla\bar{w}_{N}-Id.$
(3.11)

Note that by (3.5)

$\bar{D}=(D^{b}-Id)+(\frac{1}{2}\nabla v^{b}\otimes\nabla
v^{b}+\text{sym}\nabla
w^{b})-(\frac{1}{2}\nabla\bar{v}_{N}\otimes\nabla\bar{v}_{N}+\text{sym}\nabla\bar{w}_{N})$
(3.12)

The construction relies on the following lemma 3.3.

######  Lemma 3.3 ([29], Lemma 1).

Given two constants $\Lambda>\lambda>0$, there exists constants
$\Lambda_{*}>\lambda_{*}\geq 0$, a positive integer $N$, constant unit vectors
$\bar{\xi}_{1},...,\bar{\xi}_{N}$ in $\mathbb{R}^{n}$ and $C^{\infty}$
functions $\bar{d}_{1},...\bar{d}_{N}$ from $S_{n}$ to $\mathbb{R}$, depending
only on $n,\lambda,\Lambda$, such that for any $D\in{\mathcal{S}_{n}}$ and

$\lambda Id<D<\Lambda Id$

we have

$D=\sum_{i=1}^{N}\bar{d}_{i}^{2}\bar{\xi}_{i}\otimes\bar{\xi}_{i},\
\bar{d}_{i}=\bar{d}_{i}(D)\in[\lambda_{*},\Lambda_{*}]\ \forall i.$

###### Proof of (3.9)-(3.10)..

Because of (3.6) and $\bar{\tau}>1$, we can apply Lemma 3.3 to $D^{b}-Id$,
which yields

$D^{b}-Id=\sum_{i=1}^{N}\bar{a}_{i}^{2}\bar{\xi}_{i}\otimes\bar{\xi}_{i},\
\bar{a}_{i}\in[\lambda_{*},\Lambda_{*}].$ (3.13)

Now we define, for $1\leq i\leq N$,

$\displaystyle(\bar{v}_{0},\bar{w}_{0})=(v^{b},w^{b}),$ (3.14)
$\displaystyle\bar{v}_{i}=\bar{v}_{i-1}+\frac{1}{\bar{\mu}_{i}}\Gamma_{1}(\bar{a}_{i},\bar{\mu}_{i}x\cdot\bar{\xi}_{i}),$
(3.15)
$\displaystyle\bar{w}_{i}=\bar{w}_{i-1}-\frac{1}{\bar{\mu}_{i}}\Gamma_{1}(\bar{a}_{i},\bar{\mu}_{i}x\cdot\bar{\xi}_{i})\nabla\bar{v}_{i-1}+\frac{1}{\bar{\mu}_{i}}\Gamma_{2}(\bar{a}_{i},\bar{\mu}_{i}x\cdot\bar{\xi}_{i})\bar{\xi}_{i},$
(3.16)

where the frequencies $\bar{\mu}_{i}>1$ are constants determined later. It
will be clear later that the $i$-th step
$(\bar{v}_{i-1},\bar{w}_{i-1})\to(\bar{v}_{i},\bar{w}_{i})$ cancels the term
$\bar{a}_{i}^{2}\bar{\xi}_{i}\otimes\bar{\xi}_{i}$. Denote the $i$-th step
error by

$\bar{\mathcal{E}}_{i}:=\bar{a}_{i}^{2}\bar{\xi}_{i}\otimes\bar{\xi}_{i}+(\frac{1}{2}\nabla\bar{v}_{i-1}\otimes\nabla\bar{v}_{i-1}+\text{sym}\nabla\bar{w}_{i-1})-(\frac{1}{2}\nabla\bar{v}_{i}\otimes\nabla\bar{v}_{i}+\text{sym}\nabla\bar{w}_{i}).$

Then, combining (3.12) and (3.13)

$\bar{D}=\sum_{i=1}^{N}\bar{\mathcal{E}}_{i}.$

We now estimate $\bar{\mathcal{E}}_{i}$. Using (3.15) and (3.16), we get

$\displaystyle\nabla\bar{v}_{i}$
$\displaystyle=\nabla\bar{v}_{i-1}+\frac{1}{\bar{\mu}_{i}}\nabla\Gamma_{1},$
$\displaystyle\nabla\bar{w}_{i}$
$\displaystyle=\nabla\bar{w}_{i-1}-\frac{1}{\bar{\mu}_{i}}\nabla\Gamma_{1}\otimes\nabla\bar{v}_{i-1}-\frac{1}{\bar{\mu}_{i}}\Gamma_{1}\nabla^{2}\bar{v}_{i-1}+\frac{1}{\bar{\mu}_{i}}\nabla\Gamma_{2}\otimes\bar{\xi}_{i},$

Here we write
$\Gamma_{1}=\Gamma_{1}(\bar{a}_{i},\bar{\mu}_{i}x\cdot\bar{\xi}_{i}),\Gamma_{2}=\Gamma_{2}(\bar{a}_{i},\bar{\mu}_{i}x\cdot\bar{\xi}_{i})$,
hence

$\displaystyle\nabla\Gamma_{1}=(\partial_{s}\Gamma_{1})\nabla\bar{a}_{i}+\bar{\mu}_{i}(\partial_{t}\Gamma_{1})\bar{\xi}_{i},\quad\nabla\Gamma_{2}=(\partial_{s}\Gamma_{2})\nabla\bar{a}_{i}+\bar{\mu}_{i}(\partial_{t}\Gamma_{2})\bar{\xi}_{i}.$

It follows that

$\displaystyle(\frac{1}{2}\nabla\bar{v}_{i}\otimes\nabla\bar{v}_{i}+\text{sym}\nabla\bar{w}_{i})-(\frac{1}{2}\nabla\bar{v}_{i-1}\otimes\nabla\bar{v}_{i-1}+\text{sym}\nabla\bar{w}_{i-1})$
$\displaystyle=-\frac{1}{\bar{\mu}_{i}}\Gamma_{1}\nabla^{2}\bar{v}_{i-1}+\frac{1}{\bar{\mu}_{i}}\text{sym}(\nabla\Gamma_{2}\otimes\bar{\xi}_{i})+\frac{1}{2\bar{\mu}_{i}^{2}}\nabla\Gamma_{1}\otimes\nabla\Gamma_{1}$
$\displaystyle=-\frac{1}{\bar{\mu}_{i}}\Gamma_{1}\nabla^{2}\bar{v}_{i-1}+\frac{1}{\bar{\mu}_{i}}[\partial_{s}\Gamma_{2}+(\partial_{s}\Gamma_{1})(\partial_{t}\Gamma_{1})]\text{sym}(\nabla\bar{a}_{i}\otimes\bar{\xi}_{i})$
$\displaystyle\quad+(\partial_{t}\Gamma_{2}+\frac{1}{2}|\partial_{t}\Gamma_{1}|^{2})\bar{\xi}_{i}\otimes\bar{\xi}_{i}+\frac{1}{2\bar{\mu}_{i}^{2}}|\partial_{s}\Gamma_{1}|^{2}\nabla\bar{a}_{i}\otimes\nabla\bar{a}_{i}$

Recalling (2.4), we have

$\partial_{t}\Gamma_{2}+\frac{1}{2}|\partial_{t}\Gamma_{1}|^{2}=\bar{a}_{i}^{2},$
(3.17)

therefore

$\displaystyle\bar{\mathcal{E}}_{i}$
$\displaystyle=\frac{1}{\bar{\mu}_{i}}\Gamma_{1}\nabla^{2}\bar{v}_{i-1}-\frac{1}{\bar{\mu}_{i}}[\partial_{s}\Gamma_{2}+(\partial_{s}\Gamma_{1})(\partial_{t}\Gamma_{1})]\text{sym}(\nabla\bar{a}_{i}\otimes\bar{\xi}_{i})$
(3.18)
$\displaystyle\quad-\frac{1}{2\bar{\mu}_{i}^{2}}|\partial_{s}\Gamma_{1}|^{2}\nabla\bar{a}_{i}\otimes\nabla\bar{a}_{i}$

Using (2.5), it is easy to see

$\displaystyle\|\bar{\mathcal{E}}_{i}\|_{0}\leq\frac{C(\bar{v}_{i-1},\bar{a}_{i})}{\bar{\mu}_{i}},$
$\displaystyle\|(\bar{v}_{i}-\bar{v}_{i-1},\bar{w}_{i}-\bar{w}_{i-1})\|_{0}\leq\frac{C(\bar{v}_{i-1},\bar{a}_{i})}{\bar{\mu}_{i}}.$

Choosing $\bar{\mu}_{i}$ large enough inductively, we can obtain

$\|\bar{\mathcal{E}}_{i}\|_{0}\leq\frac{\sigma_{*}}{3N},\quad\|(\bar{v}_{i}-\bar{v}_{i-1},\bar{w}_{i}-\bar{w}_{i-1})\|_{0}\leq\frac{\epsilon}{3N}$

where $\sigma_{*}>0$ is given in Lemma 3.4 below. As a result

$\|\bar{D}\|_{0}\leq\frac{\sigma_{*}}{3},\quad\|(\bar{v}_{N}-v^{b},\bar{w}_{N}-w^{b})\|_{0}\leq\frac{\epsilon}{3}.$
(3.19)

∎

Step 2 ($0$th approximate): $(\bar{v}_{N},\bar{w}_{N})\to(V_{0},W_{0})$. In
this step we construct $0$th stage approximate solution $(V_{0},W_{0})$ such
that the induction assumptions in Proposition 3.5 holds at $q=0$. This relies
on the following decomposition. The proof can be found in [12, Lemma 5.2].

######  Lemma 3.4.

Let $N_{*}=\frac{1}{2}n(n+1)$ be the dimension of ${\mathcal{S}_{n}}$. Then
there exist constants $\sigma_{*}\in(0,1/2),C_{*}>c_{*}>0$, unit vectors
$\xi_{1},...,\xi_{N_{*}}$ in $\mathbb{R}^{n}$ and $C^{\infty}$ functions
$d_{1},...,d_{N_{*}}$ from ${\mathcal{S}_{n}}$ to $\mathbb{R}$, depending only
on $n$, such that for any $D\in{\mathcal{S}_{n}}$ and

$\|D-Id\|_{0}\leq\sigma_{*},$

we have

$D=\sum_{i=1}^{N_{*}}d_{i}^{2}\xi_{i}\otimes\xi_{i},\
d_{i}=d_{i}(D)\in[c_{*},C_{*}],\ \forall i.$

We construct $(V_{0},W_{0})$ as follows. Let

$\delta_{1}\leq\frac{\sigma_{*}}{3}$ (3.20)

be a constant given by $\delta_{1}=a^{-b}$ as in Step 3. Then by (3.19)

$\|\bar{D}-\delta_{1}Id\|_{0}\leq\sigma_{*},$

thus by Lemma 3.4

$\bar{D}-\delta_{1}Id=\sum_{i=1}^{N_{*}}\hat{a}_{i}^{2}\xi_{i}\otimes\xi_{i},$

where

$\|\hat{a}_{i}\|_{k}\lesssim 1,\ \forall 0\leq k\leq 3.$ (3.21)

with implicit constants depending on $v^{b},w^{b}$. Similar to Step 1, we
define

$\displaystyle(\hat{v}_{0},\hat{w}_{0})=(\bar{v}_{N},\bar{w}_{N}),$ (3.22)
$\displaystyle\hat{v}_{i}=\hat{v}_{i-1}+\frac{1}{\hat{\mu}_{i}}\Gamma_{1}(\hat{a}_{i},\hat{\mu}_{i}x\cdot\xi_{i}),$
(3.23)
$\displaystyle\hat{w}_{i}=\hat{w}_{i-1}-\frac{1}{\hat{\mu}_{i}}\Gamma_{1}(\hat{a}_{i},\hat{\mu}_{i}x\cdot\xi_{i})\nabla\hat{v}_{i-1}+\frac{1}{\hat{\mu}_{i}}\Gamma_{2}(\hat{a}_{i},\hat{\mu}_{i}x\cdot\xi_{i})\xi_{i},$
(3.24) $\displaystyle(V_{0},W_{0})=(\hat{v}_{N_{*}},\hat{w}_{N_{*}}),$ (3.25)

with constants $1\ll\hat{\mu}_{1}<...<\hat{\mu}_{N_{*}}$ determined later. The
$i$-th step error is

$\displaystyle\hat{\mathcal{E}}_{i}$
$\displaystyle:=\hat{a}_{i}^{2}\xi_{i}\otimes\xi_{i}+(\frac{1}{2}\nabla\hat{v}_{i-1}\otimes\nabla\hat{v}_{i-1}+\text{sym}\nabla\hat{w}_{i-1})-(\frac{1}{2}\nabla\hat{v}_{i}\otimes\nabla\hat{v}_{i}+\text{sym}\nabla\hat{w}_{i})$
$\displaystyle=\frac{1}{\hat{\mu}_{i}}\Gamma_{1}\nabla^{2}\hat{v}_{i-1}-\frac{1}{\hat{\mu}_{i}}[\partial_{s}\Gamma_{2}+(\partial_{s}\Gamma_{1})(\partial_{t}\Gamma_{1})]\text{sym}(\nabla\hat{a}_{i}\otimes\xi_{i})$
$\displaystyle-\frac{1}{2\hat{\mu}_{i}^{2}}|\partial_{s}\Gamma_{1}|^{2}\nabla\hat{a}_{i}\otimes\nabla\hat{a}_{i},$

and by (3.35)

$D_{0}=\sum_{i=1}^{N_{*}}\hat{\mathcal{E}}_{i}.$

Unlike before, we estimate $\hat{\mathcal{E}}_{i}$ more carefully. One can see
that

$\displaystyle\|\nabla^{k}\hat{v}_{i}-\nabla^{k}\hat{v}_{i-1}\|_{0}\lesssim\hat{\mu}_{i}^{k-1},\
0\leq k\leq 3,$
$\displaystyle\|\nabla^{k}\hat{w}_{i}-\nabla^{k}\hat{w}_{i-1}\|_{0}\lesssim\hat{\mu}_{i}^{k-1},\
0\leq k\leq 2.$

Besides $\|(\hat{v}_{0},\hat{w}_{0})\|_{3}\leq C$ for some $C$ depending on
$v^{b},w^{b}$. Thus

$\displaystyle\|(\hat{v}_{i},\hat{w}_{i})\|_{1}\lesssim 1,$
$\displaystyle\|(\hat{v}_{i},\hat{w}_{i})\|_{2}\lesssim
1+\hat{\mu}_{1}+...+\hat{\mu}_{i}\lesssim\hat{\mu}_{i}.$

Hence we get

$\|\hat{\mathcal{E}}_{i}\|_{0}\lesssim\frac{1}{\hat{\mu}_{i}}\|\nabla^{2}\hat{v}_{i-1}\|_{0}+\frac{1}{\hat{\mu}_{i}}+\frac{1}{\hat{\mu}_{i}^{2}}\lesssim\frac{\hat{\mu}_{i-1}}{\hat{\mu}_{i}}$

where we write $\hat{\mu}_{0}=1$. It follows that

$\displaystyle\|D_{0}\|_{0}\leq\sum_{i=1}^{N_{*}}\|\hat{\mathcal{E}}_{i}\|_{0}\lesssim\frac{1}{\hat{\mu}_{1}}+\frac{\hat{\mu}_{1}}{\hat{\mu}_{2}}+...+\frac{\hat{\mu}_{N_{*}-1}}{\hat{\mu}_{N_{*}}},$
$\displaystyle\|(V_{0},W_{0})\|_{2}\lesssim\hat{\mu}_{N_{*}}.$

We are ready to determine $\hat{\mu}_{i}$. Let

$\hat{\mu}_{i}:=(\frac{C}{\sigma\delta_{1}})^{i}.$

By choosing $C>1$ large enough, and then $K>1$ large enough, we get

$\displaystyle\|D_{0}\|_{0}\leq{\sigma}\delta_{1},$ (3.26)
$\displaystyle\|(V_{0},W_{0})\|_{2}\leq C(\sigma\delta_{1})^{-N_{*}}\leq
K\delta_{1}^{-N_{*}}.$ (3.27)

where $\sigma>0$ and $K>1$ are universal constants as in Proposition 3.5. We
now require

$\delta_{1}^{-N_{*}}\leq\delta_{0}^{\frac{1}{2}}\lambda_{0},$ (3.28)

hence

$\displaystyle\|D_{0}\|_{0}\leq{\sigma}\delta_{1},$ (3.29)
$\displaystyle\|(V_{0},W_{0})\|_{2}\leq K\delta_{0}^{\frac{1}{2}}\lambda_{0}.$
(3.30)

Besides, if $K>1$ is large enough

$\displaystyle\|(V_{0}-\bar{v}_{N},W_{0}-\bar{w}_{N})\|_{0}\lesssim\frac{1}{\hat{\mu}_{1}}+...+\frac{1}{\hat{\mu}_{N}}\leq\sigma\delta_{1},$
(3.31)
$\displaystyle\|(V_{0},W_{0})\|_{1}=\|(\hat{v}_{N_{*}},\hat{w}_{N_{*}})\|_{1}\lesssim
1\leq\frac{\sqrt{K}}{2}.$ (3.32)

Step 3 (Induction): $(V_{q},W_{q})\to(V_{q+1},W_{q+1})$. We are ready to do
induction. Set $a>1,1<b<2,c>0$ as parameters. Take a sequence of (small)
amplitudes

$\delta_{q}:=a^{-b^{q}}$ (3.33)

and a sequence of (large) frequencies

$\lambda_{q}:=a^{cb^{q}}$ (3.34)

Denote the $q$-th stage deficit matrix by

$D_{q}:=A-\frac{1}{2}\nabla V_{q}\otimes\nabla V_{q}-\text{sym}\nabla
W_{q}-\delta_{q+1}Id.$ (3.35)

We need the following one stage induction:

######  Proposition 3.5.

There exist some positive universal constants $\sigma\in(0,\sigma_{*}/3)$ and
$K>1$, such that: if $(V_{q},W_{q})\in C^{2}(\bar{\Omega})\times
C^{2}(\bar{\Omega},\mathbb{R}^{n})$ satisfy

$\displaystyle\|(V_{q},W_{q})\|_{1}\leq\sqrt{K},$ (3.36)
$\displaystyle\|(V_{q},W_{q})\|_{2}\leq K\delta_{q}^{\frac{1}{2}}\lambda_{q},$
(3.37) $\displaystyle\|D_{q}\|_{0}\leq{\sigma}\delta_{q+1}.$ (3.38)

we can construct $(V_{q+1},W_{q+1})\in C^{2}(\bar{\Omega})\times
C^{2}(\bar{\Omega},\mathbb{R}^{n})$ with

$\displaystyle\|(V_{q+1}-V_{q},W_{q+1}-W_{q})\|_{1}\leq
K\delta_{q+1}^{\frac{1}{2}},$ (3.39)
$\displaystyle\|(V_{q+1},W_{q+1})\|_{2}\leq
K\delta_{q+1}^{\frac{1}{2}}\lambda_{q+1},$ (3.40)
$\displaystyle\|D_{q+1}\|_{0}\leq{\sigma}\delta_{q+2}.$ (3.41)

###### Proof.

In the proof all implicit (universal) constants are independent of $\sigma,K$.
For convenience we write

$\mu_{0}:=K\delta_{q+1}^{-\frac{1}{2}}\delta_{q}^{\frac{1}{2}}\lambda_{q},$
(3.42)

thus

$\|(V_{q},W_{q})\|_{2}\leq\delta_{q+1}^{\frac{1}{2}}\mu_{0}.$ (3.43)

Let $\tilde{A}=A*\phi_{l},\tilde{v}=V_{q}*\phi_{l},\tilde{w}=W_{q}*\phi_{l}$
be the mollification of $A,V_{q},W_{q}$, respectively, at length $l$.

######  Remark 3.6.

Unlike before, here we need mollification to avoid the dependence on
uncontrolled higher derivatives of $(V_{q},W_{q})$, see Remark 3.7.
Mollification is usually done in the entire space $\mathbb{R}^{n}$. To this
end we actually first extend $v^{b},f$ to $C_{c}(\tilde{\Omega})$ for some
$\tilde{\Omega}\Supset\Omega$. and then solve (3.1) in $\tilde{\Omega}$. This
does not affect the arguments in the next section, since we cut-off everything
there inside $\Omega$.

Denote

$\tilde{D}:=\tilde{A}-\frac{1}{2}\nabla\tilde{v}\otimes\nabla\tilde{v}-\text{sym}\nabla\tilde{w}-\delta_{q+2}Id.$

Then by (3.35)

$\tilde{D}-\delta_{q+1}Id=D_{q}*\phi_{l}+\frac{1}{2}[(\nabla
V_{q}\otimes\nabla
V_{q})*\phi_{l}-\nabla\tilde{v}\otimes\nabla\tilde{v}]-\delta_{q+2}Id.$

Assume

$\delta_{q+2}\leq{\sigma}\delta_{q+1},$ (3.44)

and injecting (2.6) (2.8) and assumptions (3.36) (3.43) (3.38), we see

$\displaystyle\|\tilde{D}-\delta_{q+1}Id\|_{0}$
$\displaystyle\leq\|D_{q}*\phi_{l}\|_{0}+\frac{1}{2}\|(\nabla
V_{q}\otimes\nabla
V_{q})*\phi_{l}-\nabla\tilde{v}\otimes\nabla\tilde{v}\|_{0}+\delta_{q+2}$
$\displaystyle\leq\|D_{q}\|_{0}+Cl^{2}\|V_{q}\|_{2}^{2}+\delta_{q+2}$
$\displaystyle\leq 2{\sigma}\delta_{q+1}+C(\mu_{0}l)^{2}\delta_{q+1}$

Now we take

$l:=\frac{\sigma}{C\mu_{0}}=\frac{\sigma}{CK}\delta_{q+1}^{\frac{1}{2}}\delta_{q}^{-\frac{1}{2}}\lambda_{q}^{-1}$
(3.45)

for some constant $C>1$ large enough, independent of $\sigma$ and $K$, such
that

$\|\tilde{D}-\delta_{q+1}Id\|_{0}\leq
3\sigma\delta_{q+1},\quad\text{i.e.}\quad\|\frac{\tilde{D}}{\delta_{q+1}}-Id\|_{0}\leq
3\sigma\leq\sigma_{*}.$

Similarly

$\|\tilde{D}-\delta_{q+1}Id\|_{k}\leq C\sigma\delta_{q+1}l^{-k},\quad\forall
1\leq k\leq 3.$

Now Lemma 3.4 applies to $\frac{\tilde{D}}{\delta_{q+1}}$, yielding

$\tilde{D}=\sum_{i=1}^{N_{*}}a_{i}^{2}\xi_{i}\otimes\xi_{i},\quad\text{i.e.}\quad\tilde{A}-\frac{1}{2}\nabla\tilde{v}\otimes\nabla\tilde{v}-\text{sym}\nabla\tilde{w}-\delta_{q+2}Id=\sum_{i=1}^{N_{*}}a_{i}^{2}\xi_{i}\otimes\xi_{i},$
(3.46)

where

$\displaystyle\|a_{i}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}},$ (3.47)
$\displaystyle\|\nabla^{k}a_{i}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}\|\frac{\tilde{D}}{\delta_{q+1}}-Id\|_{k}\lesssim\sigma\delta_{q+1}^{\frac{1}{2}}l^{-k},\
\forall 1\leq k\leq 3.$ (3.48)

Now we define, for $1\leq i\leq N_{*}$

$\displaystyle(v_{0},w_{0})=(\tilde{v},\tilde{w})$ (3.49) $\displaystyle
v_{i}=v_{i-1}+\frac{1}{\mu_{i}}\Gamma_{1}(a_{i},\mu_{i}x\cdot\xi_{i}),$ (3.50)
$\displaystyle
w_{i}=w_{i-1}-\frac{1}{\mu_{i}}\Gamma_{1}(a_{i},\mu_{i}x\cdot\xi_{i})\nabla
v_{i-1}+\frac{1}{\mu_{i}}\Gamma_{2}(a_{i},\mu_{i}x\cdot\xi_{i})\xi_{i},$
(3.51) $\displaystyle(V_{q+1},W_{q+1})=(v_{N_{*}},w_{N_{*}}),$ (3.52)

with constants

$\mu_{0}\leq\mu_{1}\leq...\leq\mu_{N_{*}}=\lambda_{q+1}$

determined later. Using (2.7) (2.8) and (3.43), we see

$\displaystyle\|(v_{0}-V_{q},w_{0}-W_{q})\|_{1}\lesssim\|(V_{q}.W_{q})\|_{2}l\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{0}l\lesssim\sigma\delta_{q+1}^{\frac{1}{2}},$
(3.53)
$\displaystyle\|v_{0}\|_{2+k}\lesssim\|V_{q}\|_{2}l^{-k}\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{0}l^{-k},\
k=0,1.$ (3.54)

We first estimate $v_{i}$ as follows: by (2.5) and (3.47) (3.48), for $0\leq
k\leq 3$ there holds

$\displaystyle\|v_{i}-v_{i-1}\|_{k}$
$\displaystyle\lesssim\frac{1}{\mu_{i}}(\mu_{i}^{k}\|a_{i}\|_{0}+\mu_{i}^{k-1}\|a_{i}\|_{1}+...+\|a_{i}\|_{k})$
(3.55)
$\displaystyle\lesssim\frac{1}{\mu_{i}}(\mu_{i}^{k}\delta_{q+1}^{\frac{1}{2}}+\mu_{i}^{k-1}\cdot\sigma\delta_{q+1}^{\frac{1}{2}}l^{-1}+...+\sigma\delta_{q+1}^{\frac{1}{2}}l^{-k})$
$\displaystyle\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{i}^{k-1}$

where we assumed $l^{-1}\leq\mu_{1}$, namely

$\frac{C\mu_{0}}{\sigma}\leq\mu_{1}\leq...\leq\mu_{N_{*}}=\lambda_{q+1}.$
(3.56)

As a byproduct

$\displaystyle\|\nabla v_{i}\|_{0}\leq\|\nabla
V_{q}\|_{0}+C\delta_{q+1}^{\frac{1}{2}}\leq\sqrt{K}+C\lesssim\sqrt{K},$
$\displaystyle\|\nabla^{2}v_{i}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}(\mu_{0}+\mu_{1}+...+\mu_{i})\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{i},$
$\displaystyle\|\nabla^{3}v_{0}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{0}l^{-1},$
$\displaystyle\|\nabla^{3}v_{i}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{0}l^{-1}+\delta_{q+1}^{\frac{1}{2}}(\mu_{1}^{2}+...+\mu_{i}^{2})\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{i}^{2}.$

These will be used to estimate $w_{i}$ and $\mathcal{E}_{i}$.

Now turn to $w_{i}$. By (2.5) (3.47) (3.48) and above bounds for $v_{i-1}$, we
have

$\displaystyle\|w_{i}-w_{i-1}\|_{1}$
$\displaystyle\lesssim\frac{1}{\mu_{i}}(\|a_{i}\|_{0}\|\nabla
v_{i-1}\|_{0}\mu_{i}+\|a_{i}\|_{1}\|\nabla v_{i-1}\|_{0}+\|a_{i}\|_{0}\|\nabla
v_{i-1}\|_{1})$ (3.57)
$\displaystyle\quad+\frac{1}{\mu_{i}}(\|a_{i}^{2}\|_{0}\mu_{i}+\|a_{i}^{2}\|_{1})$
$\displaystyle\lesssim\frac{1}{\mu_{i}}(\delta_{q+1}^{\frac{1}{2}}\cdot\sqrt{K}\mu_{i}+\delta_{q+1}^{\frac{1}{2}}l^{-1}\cdot\sqrt{K}+\delta_{q+1}^{\frac{1}{2}}\cdot\delta_{q+1}^{\frac{1}{2}}\mu_{i-1})$
$\displaystyle\quad+\frac{1}{\mu_{i}}(\delta_{q+1}\mu_{i}+\delta_{q+1}l^{-1})$
$\displaystyle\lesssim\sqrt{K}\delta_{q+1}^{\frac{1}{2}}.$

Similarly

$\|w_{i}-w_{i-1}\|_{2}\lesssim\sqrt{K}\delta_{q+1}^{\frac{1}{2}}\mu_{i}.$
(3.58)

######  Remark 3.7.

Although to solve (3.4) only requires $W_{q}$ converges in $C^{1}$ sense,
(3.58) is necessary: note that to estimate $\|\nabla^{2}a_{i}\|_{0}$ used in
(3.55), we need to mollify $W_{q}$, which produces an error
$\|w_{0}-W_{q}\|_{1}$, hence the bound of $\|W_{q}\|_{2}$, in particular
(3.58), is unavoidable.

By (3.53) (3.55) (3.57)

$\displaystyle\|(V_{q+1}-V_{q},W_{q+1}-W_{q})\|_{1}$ (3.59)
$\displaystyle\lesssim\|(v_{0},w_{0})-(V_{q},W_{q})\|_{1}+\sum_{i=1}^{N_{*}}(\|v_{i}-v_{i-1}\|_{1}+\|w_{i}-w_{i-1}\|_{1})$
$\displaystyle\lesssim\sqrt{K}\delta_{q+1}^{\frac{1}{2}}\leq
K\delta_{q+1}^{\frac{1}{2}},$

and similarly by (3.54) (3.55) (3.58)

$\|(V_{q+1}-V_{q},W_{q+1}-W_{q})\|_{2}\lesssim\sqrt{K}\delta_{q+1}^{\frac{1}{2}}(\mu_{0}+...+\mu_{N_{*}})\leq
K\delta_{q+1}^{\frac{1}{2}}\lambda_{q+1},$ (3.60)

as long as $K>1$ is large enough. This verifies (3.39) (3.40).

It remains to prove (3.41). The $i$-th step error is

$\displaystyle{\mathcal{E}}_{i}$
$\displaystyle={a}_{i}^{2}\xi_{i}\otimes\xi_{i}+(\frac{1}{2}\nabla{v}_{i-1}\otimes\nabla{v}_{i-1}+\text{sym}\nabla{w}_{i-1})-(\frac{1}{2}\nabla{v}_{i}\otimes\nabla{v}_{i}+\text{sym}\nabla{w}_{i})$
$\displaystyle=\frac{1}{{\mu}_{i}}\Gamma_{1}\nabla^{2}{v}_{i-1}-\frac{1}{{\mu}_{i}}(\partial_{s}\Gamma_{2}+(\partial_{s}\Gamma_{1})(\partial_{t}\Gamma_{1}))\text{sym}(\nabla{a}_{i}\otimes\xi_{i})$
$\displaystyle-\frac{1}{2{\mu}_{i}^{2}}|\partial_{s}\Gamma_{1}|^{2}\nabla{a}_{i}\otimes\nabla{a}_{i}.$

Then by (3.35) (3.46)

$D_{q+1}=A-\tilde{A}+\sum_{i=1}^{N_{*}}\mathcal{E}_{i}.$ (3.61)

Recalling (2.5), one obtains

$\displaystyle\|\mathcal{E}_{i}\|_{0}$
$\displaystyle\lesssim\frac{1}{\mu_{i}}\|\Gamma_{1}\|_{0}\|\nabla^{2}v_{i}\|_{0}+\frac{1}{\mu_{i}}(\|\partial_{s}\Gamma_{2}\|_{0}+\|\partial_{s}\Gamma_{1}\|_{0}\|\partial_{t}\Gamma_{1}\|_{0})\|\nabla
a_{i}\|_{0}$ (3.62)
$\displaystyle+\frac{1}{\mu_{i}^{2}}\|\partial_{s}\Gamma_{1}\|_{0}^{2}\|\nabla
a_{i}\|_{0}^{2}$
$\displaystyle\lesssim\frac{1}{\mu_{i}}\delta_{q+1}^{\frac{1}{2}}\cdot\delta_{q+1}^{\frac{1}{2}}\mu_{i-1}+\frac{1}{\mu_{i}}\cdot\delta_{q+1}^{\frac{1}{2}}\cdot\sigma\delta_{q+1}^{\frac{1}{2}}l^{-1}+\frac{1}{\mu_{i}^{2}}(\sigma\delta_{q+1}^{\frac{1}{2}}l^{-1})^{2}$
$\displaystyle\lesssim\delta_{q+1}\frac{\mu_{i-1}}{\mu_{i}}$

Besides, since $A\in C^{2,\kappa}(\bar{\Omega},{\mathcal{S}_{n}})$,

$\|A-\tilde{A}\|_{0}\lesssim\|A\|_{1}l\lesssim l.$

So we conclude

$\|D_{q+1}\|_{0}\lesssim
l+\delta_{q+1}(\frac{\mu_{0}}{\mu_{1}}+\frac{\mu_{1}}{\mu_{2}}+...+\frac{\mu_{N_{*}-1}}{\mu_{N_{*}}}).$

We now take frequencies $\mu_{i}$ as

$\mu_{i}:=\mu_{0}^{1-\frac{i}{N_{*}}}\mu_{N_{*}}^{\frac{i}{N_{*}}}$

and require

$l\leq\frac{\sigma\delta_{q+2}}{C},\quad\frac{\mu_{0}}{\mu_{N_{*}}}\leq(\frac{\sigma\delta_{q+2}}{C\delta_{q+1}})^{N_{*}}$
(3.63)

for some universal constant $C>1$ large enough. In particular (3.56) holds.
Now we get

$\|D_{q+1}\|_{0}\lesssim
l+\delta_{q+1}(\frac{\mu_{0}}{\mu_{N_{*}}})^{\frac{1}{N_{*}}}\leq{\sigma}\delta_{q+2}$

which is exactly (3.41). This concludes the proof of Proposition 3.5. ∎

Step 4 (Conclusion): $(V_{q},W_{q})\to(v,w)$. Now it is easy to prove Theorem
3.1. First take $\sigma=\sigma_{*}/3$ and take $K$ large enough, such that
(3.27) (3.32) (3.59) (3.60) hold.

Next we determine the parameters $b,c$ depending on $N_{*},\alpha$, and $a>1$
depending on $K,\sigma,\epsilon$ and $b,c$. The inequalities we require are
(3.28) (3.63) (recall (3.42) (3.45))

$\displaystyle\delta_{1}^{-N_{*}}\leq\delta_{0}^{\frac{1}{2}}\lambda_{0},$
$\displaystyle\frac{\sigma\delta_{q+1}^{\frac{1}{2}}}{K\delta_{q}^{\frac{1}{2}}\lambda_{q}}\leq\frac{\sigma\delta_{q+2}}{C},$
$\displaystyle
K\frac{\delta_{q}^{\frac{1}{2}}\lambda_{q}}{\delta_{q+1}^{\frac{1}{2}}\lambda_{q+1}}\leq(\frac{\sigma\delta_{q+2}}{C\delta_{q+1}})^{N_{*}}$

and (3.20) (3.44) (recall $\sigma=\sigma_{*}/3$)

$\delta_{1}\leq{\sigma},\quad\delta_{q+2}\leq{\sigma}\delta_{q+1}.$ (3.64)

Recall $\delta_{q}=a^{-b^{q}},\lambda_{q}=a^{cb^{q}}$. By direct computation,
the first three are equivalent to

$c\geq
N_{*}b+\frac{1}{2}+\frac{1}{(b-1)b^{q}}\log_{a}(\frac{CK}{\sigma^{N_{*}}})$
(3.65)

On the other hand, by (3.39) (3.40) and interpolation

$\|V_{q+1}-V_{q}\|_{1+\alpha}\leq\|V_{q+1}-V_{q}\|_{1}^{1-\alpha}\|V_{q+1}-V_{q}\|_{2}^{\alpha}\leq
K\delta_{q+1}^{\frac{1}{2}}\lambda_{q+1}^{\alpha}.$ (3.66)

So convergence in the $C^{1,\alpha}$ norm is equivalent to

$-\frac{1}{2}+c\alpha<0.$ (3.67)

As long as

$\alpha<\frac{1}{1+2N_{*}}=\frac{1}{1+n+n^{2}},$

(3.64) (3.65) and (3.67) can be achieved by taking $b>1$ close to $1$,
$c>N_{*}+\frac{1}{2}$ close to $N_{*}+\frac{1}{2}$ and $a>1$ sufficiently
large. We fix such $b,c$ and below take $a$ even larger.

(3.29) (3.30) (3.32) in Step 2 yield the induction assumptions (3.36) (3.37)
and (3.38) of Proposition 3.5 at $q=0$. Moreover, as

$\|(V_{0},W_{0})\|_{1}\leq\frac{\sqrt{K}}{2},\quad\|(V_{q+1}-V_{q},W_{q+1}-W_{q})\|_{1}\leq
K\delta_{q+1}^{\frac{1}{2}},$

by taking $a>1$ large enough, we may assume

$\|(V_{q},W_{q})\|_{1}\leq\sqrt{K},\ \forall\ q\geq 0.$

Now successive use of Proposition 3.5 produces a series of $(V_{q},W_{q})$
satisfying (3.39)-(3.41) and hence the interpolation bound (3.66), so

$\displaystyle(V_{q},W_{q})\to(v,w)\ \text{in}\
C^{1,\alpha}(\bar{\Omega})\times C^{1,\alpha}(\bar{\Omega},\mathbb{R}^{n}),$
$\displaystyle D_{q}\to 0\ \text{in}\ C^{0}(\bar{\Omega},{\mathcal{S}_{n}}).$

Recall

$D_{q}=A-\frac{1}{2}\nabla V_{q}\otimes\nabla V_{q}-\text{sym}\nabla
W_{q}-\delta_{q+1}Id.$

Taking limit, we get

$A-\frac{1}{2}\nabla v\otimes\nabla v-\text{sym}\nabla w=0.$

Moreover by (3.19) (3.31) and (3.39),

$\displaystyle\|v-v^{b}\|_{0}$
$\displaystyle\leq\|\bar{v}_{N}-v^{b}\|_{0}+\|V_{0}-\bar{v}_{N}\|_{0}+\sum_{q=0}^{+\infty}\|V_{q+1}-V_{q}\|_{0}$
$\displaystyle\leq\frac{\epsilon}{3}+\sigma\delta_{1}+\sum_{q=0}^{+\infty}K\delta_{q+1}^{\frac{1}{2}}$
$\displaystyle\leq\epsilon$

and similarly $\|w-w^{b}\|_{0}\leq\epsilon$, as long as $a$ is large enough.
So (3.7) and (3.8) are justified. This concludes the proof of Theorem 3.1.

##  4\. Cut-off technique.

In this section we take into consideration the boundary value $v=g$. As in
[6], we adapt the arguments in the last section by further introducing a cut-
off procedure. More precisely, we will cut-off at the level set of a function
$\psi$, which represents the size of $D^{b}$, and in each stage
$(V_{q},W_{q})\to(V_{q+1},W_{q+1})$, we only change the data in the set

$\\{\psi>\delta_{q+2}\\}\Subset\Omega.$

Let’s now explain the details. First we take $v^{b}\in
C^{2,\kappa}(\bar{\Omega})$ to solve

$\Delta v^{b}=0\ \text{in}\ \Omega,\quad v^{b}=g\ \text{on}\ \partial\Omega.$
(4.1)

With this initial data $v^{b}$, we will show that

######  Theorem 4.1.

Under the conditions of Theorem 1.2 or 1.3, for any $\epsilon>0$, there exists
a very weak solution $v\in C^{1,\alpha}(\bar{\Omega})$ to (1.2) such that

$\|v-v^{b}\|_{0}\leq\epsilon,\quad v=v^{b}=g\ \text{ on }\ \partial\Omega.$

This clearly yields Theorem 1.2 and 1.3. The proof is in the same line with
Section 3 and essentially follows [6]. Remark 4.3 explains in more detail the
construction of $\tilde{D}$, corresponding to the quantity $H_{q+1}$ in [6,
page 19].

Part one: solve (3.3). Assuming that first $f\in C^{0,\kappa}(\bar{\Omega})$
and $f>0$ aligns with the assumption in Theorem 1.3. Given $g\in
C^{2,\kappa}(\partial\Omega)$, we let $v^{b}$ solves (4.1) and take $\psi$
such that

$\displaystyle-(2n-2)\Delta\psi$
$\displaystyle=f-\sigma_{2}(\nabla^{2}v^{b})\quad\text{in}\ \Omega,$
$\displaystyle\psi$ $\displaystyle=0\quad\text{on}\ \partial\Omega.$

Then by Newton inequality
$\sigma_{2}(\nabla^{2}v^{b})\leq\frac{n-1}{2n}(\Delta v^{b})^{2}=0$, so

$f-\sigma_{2}(\nabla^{2}v^{b})>0.$

For general $f\in C^{0,\kappa}(\bar{\Omega})$ as in Theorem 1.2, we pick
$v^{b}\in C^{3}(\bar{\Omega})$ such that

$f-\sigma_{2}(\nabla^{2}v^{b})>0,$

e.g. $v^{b}(x)=C(x_{1}^{2}-x_{2}^{2})$ for $C>1$ sufficiently large. Then we
take $g=v^{b}\mid_{\partial\Omega}$ and $\psi$ as above. In both cases,
$\psi\in C^{2,\kappa}(\bar{\Omega})$ and by strong maximum principle

$\psi>0\ \text{in}\ \Omega,\quad\psi=0\ \text{on}\ \partial\Omega.$

Let

$A=\psi Id+\frac{1}{2}\nabla v^{b}\otimes\nabla v^{b},$

Then

$-L(A)=-(2n-2)\Delta\psi+\sigma_{2}(\nabla^{2}v^{b})=f.$

Part two: solve (3.4). Again, we will construct approximate solutions
$(V_{q},W_{q})$ inductively.

Step 1 (Preparation). Set $w^{b}=0$. Define the initial deficit matrix by

$D^{b}:=A-\frac{1}{2}\nabla v^{b}\otimes\nabla v^{b}-\text{sym}\nabla
w^{b}=\psi Id.$ (4.2)

Unlike before, here we do not need to construct $(\bar{v}_{N},\bar{w}_{N})$ to
reduce $D^{b}$, since $D^{b}$ is merely a multiple of $Id$. In particular,
Lemma 3.3 is never used. The fact that $D^{b}$ is merely a (non-negative)
multiple of $Id$ is crucial to the cut-off technique: we will cut-off at the
level set of $\psi$.

Let’s introduce some notations. We decompose

$Id=\sum_{i=1}^{N_{*}}d_{i}^{*2}\xi_{i}\otimes\xi_{i}.$

for some constants $d_{i}^{*}\in[c_{*},C_{*}]$, by applying Lemma 3.4 to $Id$.

Let

$\displaystyle\Omega_{q}:=\\{x\in\Omega:\psi(x)>2\delta_{q}\\},$
$\displaystyle\tilde{\Omega}_{q}:=\\{x\in\Omega:\psi(x)>\frac{3}{2}\delta_{q}\\}.$

Clearly $\Omega_{q}\Subset\tilde{\Omega}_{q}\Subset\Omega_{q+1}\Subset\Omega$
and

$dist(\tilde{\Omega}_{q},\partial\Omega)\geq
dist(\Omega_{q},\partial\tilde{\Omega}_{q})\gtrsim\delta_{q}.$

Here and below the implicit constants may depend on $\|\psi\|_{1}$.

Let $\eta_{q}$ be a smooth cut-off function subordinate to
$\Omega_{q}\subset\tilde{\Omega}_{q}$, i.e.

$\displaystyle\eta_{q}=1\quad\text{in}\ \Omega_{q},$
$\displaystyle\eta_{q}=0\quad\text{in}\ \Omega-\tilde{\Omega}_{q},$
$\displaystyle\|\nabla^{k}\eta_{q}\|_{0}\lesssim\delta_{q}^{-k},\ \forall
0\leq k\leq 3.$

Let $\psi_{q}$ be a smooth truncation of $\psi$ at $\delta_{q}$, defined by

$\psi_{q}:=\eta_{q}^{2}\delta_{q}+(1-\eta_{q}^{2})\psi.$ (4.3)

Note that

$\|\nabla\psi_{q}\|_{0}\lesssim\delta_{q}\|\nabla\eta_{q}\|_{0}+\|\nabla\eta_{q}\|_{0}\|\psi\|_{C^{0}(\Omega-\Omega_{q})}+\|\nabla\psi\|_{0}\lesssim
1.$

Step 2 ($0$th approximate): $(v^{b},w^{b})\to(V_{0},W_{0})$. By mollification
in $\tilde{\Omega}_{0}$, we may assume $v^{b}\in
C^{\infty}(\tilde{\Omega}_{0})$. Let

$\hat{a}_{i}:=\eta_{1}(\psi-\delta_{1})^{\frac{1}{2}}d_{i}^{*},$

in other words (recall $D^{b}=\psi Id$)

$\eta_{1}^{2}(D^{b}-\delta_{1}Id)=\sum_{i=1}^{N_{*}}\hat{a}_{i}^{2}\xi_{i}\otimes\xi_{i}.$
(4.4)

Then

$\displaystyle\|\nabla^{k}\hat{a}_{i}\|_{0}\lesssim\delta_{1}^{-k},\ \forall
0\leq k\leq 2,$ (4.5) $\displaystyle\hat{a}_{i}=0\quad\text{in
}\Omega-\tilde{\Omega}_{1}.$ (4.6)

As in Step 2 in Section 3, we define

$\displaystyle(\hat{v}_{0},\hat{w}_{0})=(v^{b},w^{b}),$ (4.7)
$\displaystyle\hat{v}_{i}=\hat{v}_{i-1}+\frac{1}{\hat{\mu}_{i}}\Gamma_{1}(\hat{a}_{i},\hat{\mu}_{i}x\cdot\xi_{i}),$
(4.8)
$\displaystyle\hat{w}_{i}=\hat{w}_{i-1}-\frac{1}{\hat{\mu}_{i}}\Gamma_{1}(\hat{a}_{i},\hat{\mu}_{i}x\cdot\xi_{i})\nabla\hat{v}_{i-1}+\frac{1}{\hat{\mu}_{i}}\Gamma_{2}(\hat{a}_{i},\hat{\mu}_{i}x\cdot\xi_{i})\xi_{i},$
(4.9) $\displaystyle(V_{0},W_{0})=(\hat{v}_{N_{*}},\hat{w}_{N_{*}}),$ (4.10)

with constants $1\ll\hat{\mu}_{1}<...<\hat{\mu}_{N_{*}}$ determined later.
Denote the $0$th stage deficit matrix by

$D_{0}:=A-\frac{1}{2}\nabla V_{0}\otimes\nabla V_{0}-\text{sym}\nabla
W_{0}-\psi_{1}Id.$ (4.11)

Note that by (4.6)

$(V_{0},W_{0},D_{0})=(v^{b},w^{b},0)\quad\text{in}\
\Omega-\tilde{\Omega}_{1}.$ (4.12)

The $i$-th step error is

$\displaystyle\hat{\mathcal{E}}_{i}$
$\displaystyle=\hat{a}_{i}^{2}\xi_{i}\otimes\xi_{i}+(\frac{1}{2}\nabla\hat{v}_{i-1}\otimes\nabla\hat{v}_{i-1}+\text{sym}\nabla\hat{w}_{i-1})-(\frac{1}{2}\nabla\hat{v}_{i}\otimes\nabla\hat{v}_{i}+\text{sym}\nabla\hat{w}_{i})$
$\displaystyle=\frac{1}{\hat{\mu}_{i}}\Gamma_{1}\nabla^{2}\hat{v}_{i-1}-\frac{1}{\hat{\mu}_{i}}[\partial_{s}\Gamma_{2}+(\partial_{s}\Gamma_{1})(\partial_{t}\Gamma_{1})]\text{sym}(\nabla\hat{a}_{i}\otimes\xi_{i})$
$\displaystyle-\frac{1}{2\hat{\mu}_{i}^{2}}|\partial_{s}\Gamma_{1}|^{2}\nabla\hat{a}_{i}\otimes\nabla\hat{a}_{i}$

Injecting (4.2) and (4.4) into (4.11)

$\displaystyle D_{0}$ $\displaystyle=A-\frac{1}{2}\nabla V_{0}\otimes\nabla
V_{0}-\text{sym}\nabla W_{0}-\eta_{1}^{2}\delta_{1}Id-(1-\eta_{1}^{2})\psi Id$
$\displaystyle=A-D^{b}-\frac{1}{2}\nabla V_{0}\otimes\nabla
V_{0}-\text{sym}\nabla W_{0}+\eta_{1}^{2}(D^{b}-\delta_{1}Id)$
$\displaystyle=\frac{1}{2}\nabla v^{b}\otimes\nabla v^{b}+\text{sym}\nabla
w^{b}-\frac{1}{2}\nabla V_{0}\otimes\nabla V_{0}-\text{sym}\nabla
W_{0}+\sum_{i=1}^{N_{*}}\hat{a}_{i}^{2}\xi_{i}\otimes\xi_{i}$
$\displaystyle=\sum_{i=1}^{N_{*}}\hat{\mathcal{E}}_{i}$

We turn to required estimates. Using (2.5) (4.5)

$\displaystyle\|\nabla^{2}\hat{v}_{i}-\nabla^{2}\hat{v}_{i-1}\|_{0}$
$\displaystyle\lesssim\frac{1}{\hat{\mu}_{i}}(\|\nabla^{2}\hat{a}_{i}\|_{0}+\hat{\mu}_{i}\|\nabla\hat{a}_{i}\|_{0}+\hat{\mu}_{i}^{2}\|\hat{a}_{i}\|_{0})$
$\displaystyle\lesssim\frac{1}{\hat{\mu}_{i}}(\delta_{1}^{-2}+\hat{\mu}_{i}\delta_{1}^{-1}+\hat{\mu}_{i}^{2})$
$\displaystyle\lesssim\hat{\mu}_{i}$

where we assumed

$\hat{\mu}_{i}\geq\delta_{1}^{-1}.$ (4.13)

Using this and (2.5) (4.5), we get

$\displaystyle\|\hat{\mathcal{E}}_{i}\|_{0}\lesssim\frac{1}{\hat{\mu}_{i}}\|\hat{a}_{i}\|_{0}\|\nabla^{2}\hat{v}_{i-1}\|_{0}+\frac{1}{\hat{\mu}_{i}}\|\hat{a}_{i}\|_{0}\|\nabla\hat{a}_{i}\|_{0}+\frac{1}{\mu_{i}^{2}}\|\nabla\hat{a}_{i}\|_{0}^{2}\lesssim\frac{\hat{\mu}_{i-1}}{\hat{\mu}_{i}}$

where we write $\hat{\mu}_{0}=1$. This is the same as in Step 2 of Section 3.
Similarly we deduce

$\displaystyle\|D_{0}\|_{0}\leq\sum_{i=1}^{N_{*}}\|\hat{\mathcal{E}}_{i}\|_{0}\lesssim\frac{1}{\hat{\mu}_{1}}+\frac{\hat{\mu}_{1}}{\hat{\mu}_{2}}+...+\frac{\hat{\mu}_{N_{*}-1}}{\hat{\mu}_{N_{*}}},$
$\displaystyle\|(V_{0},W_{0})\|_{2}\lesssim\hat{\mu}_{N_{*}}.$

Hence we take

$\hat{\mu}_{i}:=(\frac{C}{\sigma\delta_{1}})^{i}$

in particular (4.13) holds. Then, assuming

$\delta_{1}^{-N_{*}}\leq\delta_{0}^{\frac{1}{2}}\lambda_{0}$

and $K>1$ large enough, we conclude

$\displaystyle\|D_{0}\|_{0}\leq{\sigma}\delta_{1},\quad\|(V_{0},W_{0})\|_{2}\leq
K\delta_{0}^{\frac{1}{2}}\lambda_{0},$
$\displaystyle\|(V_{0}-v^{b},W_{0}-w^{b})\|_{0}\leq\sigma\delta_{1},\quad\|(V_{0},W_{0})\|_{1}\leq\frac{\sqrt{K}}{2}.$

Together with (4.12), these ensure the assumptions (4.15)-(4.18) below at
$q=0$.

Step 3 (Induction): $(V_{q},W_{q})\to(V_{q+1},W_{q+1})$. Denote the $q$-th
stage deficit matrix by

$D_{q}:=A-\frac{1}{2}\nabla V_{q}\otimes\nabla V_{q}-\text{sym}\nabla
W_{q}-\psi_{q+1}Id.$ (4.14)

Similar to Proposition 3.5, we will prove

######  Proposition 4.2.

There exist some positive universal constants $\sigma\in(0,\sigma_{*}/3)$ and
$K>1$, such that: if $(V_{q},W_{q})\in C^{2}(\bar{\Omega})\times
C^{2}(\bar{\Omega},\mathbb{R}^{n})$ satisfy

$\displaystyle\|(V_{q},W_{q})\|_{1}\leq\sqrt{K},$ (4.15)
$\displaystyle\|(V_{q},W_{q})\|_{2}\leq K\delta_{q}^{\frac{1}{2}}\lambda_{q},$
(4.16) $\displaystyle\|D_{q}\|_{0}\leq{\sigma}\delta_{q+1}.$ (4.17)

and

$(V_{q},W_{q},D_{q})=(v^{b},w^{b},0)\quad\text{in}\
\Omega-\tilde{\Omega}_{q+1},$ (4.18)

we can construct $(V_{q+1},W_{q+1})\in C^{2}(\bar{\Omega})\times
C^{2}(\bar{\Omega},\mathbb{R}^{n})$ with

$\displaystyle\|(V_{q+1}-V_{q},W_{q+1}-W_{q})\|_{1}\leq
K\delta_{q+1}^{\frac{1}{2}}$ (4.19)
$\displaystyle\|(V_{q+1},W_{q+1})\|_{2}\leq
K\delta_{q+1}^{\frac{1}{2}}\lambda_{q+1},$ (4.20)
$\displaystyle\|D_{q+1}\|_{0}\leq{\sigma}\delta_{q+2}.$ (4.21)

and

$(V_{q+1},W_{q+1},D_{q+1})=(v^{b},w^{b},0)\quad\text{in}\
\Omega-\tilde{\Omega}_{q+2},\\\ $ (4.22)

###### Proof.

Let $\tilde{A},\tilde{v},\tilde{w},\tilde{\psi}_{q+1}$ be the mollification of
$A,V_{q},W_{q},\psi_{q+1}$, respectively, at length $l$ in
$\tilde{\Omega}_{q+2}$, where as in (3.42) (3.45)

$\mu_{0}:=K\delta_{q+1}^{-\frac{1}{2}}\delta_{q}^{\frac{1}{2}}\lambda_{q},\quad
l:=\frac{\sigma}{C\mu_{0}}.$ (4.23)

In particular (4.16) is

$\|(V_{q},W_{q})\|_{2}\leq\delta_{q+1}^{\frac{1}{2}}\mu_{0}.$ (4.24)

Let $\eta$ be a smooth cut-off function such that

$\displaystyle\eta=1\quad\text{if }\psi\geq\frac{5}{4}\delta_{q+1},$
$\displaystyle\eta=0\quad\text{if }\psi\leq\delta_{q+1},$

and

$\|\nabla^{k}\eta\|_{0}\lesssim\delta_{q+1}^{-k},\ \forall 0\leq k\leq 3.$
(4.25)

We assume

$l\leq\frac{\delta_{q+2}}{4\|\psi\|_{1}+1}.$ (4.26)

Then

  1. ($l_{1}$)

$l\leq dist(\tilde{\Omega}_{q+2},\partial\Omega)$ and the mollification is
well-defined;

  2. ($l_{2}$)

$\eta=1$ in the $l$-neighborhood of $\tilde{\Omega}_{q+1}$.

Define

$\displaystyle\tilde{D}$
$\displaystyle:=\eta^{2}(\tilde{A}-\frac{1}{2}\nabla\tilde{v}\otimes\nabla\tilde{v}-\text{sym}\nabla\tilde{w}-\delta_{q+2}Id)$
(4.27)
$\displaystyle\quad+(1-\eta^{2})\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id.$

Compared with (4.14), the first line is

$\displaystyle\eta^{2}(\tilde{A}-\frac{1}{2}\nabla\tilde{v}\otimes\nabla\tilde{v}-\text{sym}\nabla\tilde{w}-\delta_{q+2}Id)$
$\displaystyle=\eta^{2}D_{q}*\phi_{l}+\eta^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id.$

Since $\eta^{2}\eta_{q+2}^{2}=\eta^{2}$, the second line is

$(1-\eta^{2})\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id=\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id-\eta^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id.$

Therefore

$\displaystyle\tilde{D}$
$\displaystyle=\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id$ (4.28)
$\displaystyle\quad+\eta^{2}D_{q}*\phi_{l}+\frac{1}{2}\eta^{2}[(\nabla
V_{q}\otimes\nabla V_{q})*\phi_{l}-\nabla\tilde{v}\otimes\nabla\tilde{v}].$

Now we construct $a_{i}$ in two cases separately.

Case 1: $\psi\geq\delta_{q+1}$. Then by (2.6) (4.17)

$\|\eta^{2}D_{q}*\phi_{l}\|_{0}\leq\|D_{q}\|_{0}\leq\sigma\delta_{q+1},$

and by (4.23) (4.24) and (2.8)

$\|\frac{1}{2}\eta^{2}[(\nabla V_{q}\otimes\nabla
V_{q})*\phi_{l}-\nabla\tilde{v}\otimes\nabla\tilde{v}]\|_{0}\lesssim\|V_{q}\|_{2}^{2}l^{2}\leq\sigma\delta_{q+1},$

so

$\|\tilde{D}-\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id\|_{0}\leq
2\sigma\delta_{q+1}.$

Let us assume $\sigma\leq\frac{\sigma_{*}}{3}\leq\frac{1}{6}$, such that

$\delta_{q+2}\leq\sigma\delta_{q+1}\leq\frac{1}{6}\delta_{q+1}.$ (4.29)

So from $\psi\geq\delta_{q+1}$, we have $\eta_{q+2}=1$. And by (4.26) and the
definition of

$\psi_{q+1}=\eta^{2}_{q+1}\delta_{q+1}+(1-\eta^{2}_{q+1})\psi,$

we obtain

$\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})\geq\delta_{q+1}-\frac{5}{4}\delta_{q+2}\geq\frac{2}{3}\delta_{q+1}.$

Therefore

$\|\frac{\tilde{D}}{\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})}-Id\|_{0}\leq
3\sigma\leq\sigma_{*}.$

Note that since

$\|\frac{\tilde{D}}{\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})}-Id\|_{k}\lesssim
l^{-k},\forall 0\leq k\leq 3$

where we omit the universal constant $\sigma_{\ast}$. Applying Lemma 3.4, we
get

$\tilde{D}=\sum_{i=1}^{N_{*}}a_{i,1}^{2}\xi_{i}\otimes\xi_{i},$ (4.30)

and

$\|\nabla^{k}a_{i,1}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}l^{-k},\ \forall
0\leq k\leq 3.$

Case 2: $\psi\leq\delta_{q+1}$. Then $\eta=0$ and

$\tilde{D}=\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id.$

Recall $Id=\sum_{i=1}^{N_{*}}d_{i}^{*2}\xi_{i}\otimes\xi_{i}$. By (4.26) and
the definition of $\tilde{\Omega}_{q+2}$

$\tilde{\psi}_{q+1}-\delta_{q+2}\geq\frac{3}{2}\delta_{q+2}-\frac{1}{4}\delta_{q+2}-\delta_{q+2}=\frac{1}{4}\delta_{q+2}\quad\text{in}\
\tilde{\Omega}_{q+2}.$ (4.31)

Let

$a_{i,2}=\eta_{q+2}(\tilde{\psi}_{q+1}-\delta_{q+2})^{\frac{1}{2}}d_{i}^{*}$

Note that $a_{i,2}=0$ in $\Omega-\tilde{\Omega}_{q+2}$. Then

$\tilde{D}=\sum_{i=1}^{N_{*}}a_{i,2}^{2}\xi_{i}\otimes\xi_{i},$

and

$\|\nabla^{k}a_{i,2}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}l^{-k},\ \forall
0\leq k\leq 3.$

Now we define $a_{i}$ as

$a_{i}=\begin{cases}a_{i,1},&\text{if }\psi\geq\delta_{q+1},\\\
a_{i,2},&\text{if }\psi\leq\delta_{q+1}.\end{cases}$ (4.32)

Then

$\tilde{D}=\sum_{i=1}^{N_{*}}a_{i}^{2}\xi_{i}\otimes\xi_{i}.$ (4.33)

Note that $a_{i,1}$ and $a_{i,2}$ are both obtained by the decomposition of
$\tilde{D}$, and $\tilde{D}$ is smooth, so $a_{i,1}$ and $a_{i,2}$ patches
smoothly. Thus $a_{i}$ is smooth with

$\|\nabla^{k}a_{i}\|_{0}\lesssim\delta_{q+1}^{\frac{1}{2}}l^{-k},\ \forall
0\leq k\leq 3$

and

$a_{i}=0\quad\text{in}\ \Omega-\tilde{\Omega}_{q+2}.$ (4.34)

######  Remark 4.3.

The above arguments aim to find such $a_{i}$ smoothly compactly supported in
$\tilde{\Omega}_{q+2}$, which are used as usual to reduce $D_{q}$ to near
$\delta_{q+2}Id$. To this end, we have expected to construct $a_{i}$ in two
domains separately and then glue them together. In $\tilde{\Omega}_{q+1}$ we
resort to Lemma 3.4. While in $\Omega-\Omega_{q+1}$ we have
$D_{q}=\psi_{q+1}Id$, so we simply proceed as in Step 2. But this gluing
procedure brings extra error which should be cancelled by $a_{i}$. So instead
we first glue the deficit matrix to obtain $\tilde{D}$, so that $a_{i}$
automatically glue together in a smooth way. The exact expression of
$\tilde{D}$ is determined by (4.43).

Let

$\displaystyle(v_{0},w_{0})=\eta^{2}(\tilde{v},\tilde{w})+(1-\eta^{2})(V_{q},W_{q}),$
(4.35) $\displaystyle
v_{i}=v_{i-1}+\frac{1}{\mu_{i}}\Gamma_{1}(a_{i},\mu_{i}x\cdot\xi_{i}),$ (4.36)
$\displaystyle
w_{i}=w_{i-1}-\frac{1}{\mu_{i}}\Gamma_{1}(a_{i},\mu_{i}x\cdot\xi_{i})\nabla
v_{i-1}+\frac{1}{\mu_{i}}\Gamma_{2}(a_{i},\mu_{i}x\cdot\xi_{i})\xi_{i},$
(4.37) $\displaystyle(V_{q+1},W_{q+1})=(v_{N_{*}},w_{N_{*}}).$ (4.38)

From the definition of $\eta$ and (4.29) (4.34), it is clear that (4.22)
holds, i.e.

$(V_{q+1},W_{q+1},D_{q+1})=(V_{q},W_{q},D_{q})=(v^{b},w^{b},0)\quad\text{in}\
\Omega-\tilde{\Omega}_{q+2}.$

We derive necessary estimates. We have

$(v_{0},w_{0})-(V_{q},W_{q})=\eta^{2}(\tilde{v}-V_{q},\tilde{w}-W_{q}).$

Using (2.7) (4.23) (4.24) (4.26) (4.29) and
$\|\nabla\eta\|_{0}\lesssim\delta_{q+1}^{-1}$, we get

$\displaystyle\|(v_{0},w_{0})-(V_{q},W_{q})\|_{1}$
$\displaystyle\lesssim\|\eta^{2}\|_{0}\|(\tilde{v}-V_{q},\tilde{w}-W_{q})\|_{1}+\|\nabla\eta^{2}\|_{0}\|(\tilde{v}-V_{q},\tilde{w}-W_{q})\|_{0}$
$\displaystyle\lesssim\|(V_{q},W_{q})\|_{2}(l+l^{2}\|\nabla\eta\|_{0})$
$\displaystyle\lesssim\sigma\delta_{q+1}^{\frac{1}{2}}$

and

$\|(v_{0},w_{0})-(V_{q},W_{q})\|_{2+k}\lesssim\|(V_{q},W_{q})\|_{2}l^{-k}\lesssim\delta_{q+1}^{\frac{1}{2}}\mu_{0}l^{-k},\
k=0,1.$

As usual, the $i$-th step error is

$\mathcal{E}_{i}:={a}_{i}^{2}\xi_{i}\otimes\xi_{i}+(\frac{1}{2}\nabla{v}_{i-1}\otimes\nabla{v}_{i-1}+\text{sym}\nabla{w}_{i-1})-(\frac{1}{2}\nabla{v}_{i}\otimes\nabla{v}_{i}+\text{sym}\nabla{w}_{i}).$
(4.39)

Using the above estimates, we can proceed as in Step 3 of Section 2 to get
(4.19) (4.20) and

$\|\mathcal{E}_{i}\|_{0}\lesssim\delta_{q+1}\frac{\mu_{i-1}}{\mu_{i}}.$ (4.40)

It remains to derive (4.21). By the definition of $D_{q+1}$

$\displaystyle D_{q+1}$ $\displaystyle=A-\frac{1}{2}\nabla
V_{q+1}\otimes\nabla V_{q+1}-\text{sym}\nabla W_{q+1}-\psi_{q+2}Id$
$\displaystyle=A-\frac{1}{2}\nabla v_{0}\otimes\nabla v_{0}-\text{sym}\nabla
w_{0}-\psi_{q+2}Id$ (4.41)
$\displaystyle\quad+\sum_{i=1}^{N_{*}}(\frac{1}{2}\nabla{v}_{i-1}\otimes\nabla{v}_{i-1}+\text{sym}\nabla{w}_{i-1})-(\frac{1}{2}\nabla{v}_{i}\otimes\nabla{v}_{i}+\text{sym}\nabla{w}_{i}).$
(4.42)

We reformulate it. First, by the construction

$\displaystyle\nabla v_{0}$
$\displaystyle=\eta^{2}\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}+(\nabla\eta^{2})(\tilde{v}-V_{q}),$ $\displaystyle\nabla w_{0}$
$\displaystyle=\eta^{2}\nabla\tilde{w}+(1-\eta^{2})\nabla
W_{q}+(\nabla\eta^{2})\otimes(\tilde{w}-W_{q}),$

so

$\displaystyle\nabla v_{0}\otimes\nabla v_{0}$
$\displaystyle=\eta^{4}\nabla\tilde{v}\otimes\nabla\tilde{v}+(1-\eta^{2})^{2}\nabla
V_{q}\otimes\nabla
V_{q}+2\eta^{2}(1-\eta^{2})\text{sym}\nabla\tilde{v}\otimes\nabla V_{q}$
$\displaystyle\quad+2\text{sym}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\eta^{2}\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}]$
$\displaystyle\quad+[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\nabla\eta^{2})(\tilde{v}-V_{q})]$
$\displaystyle=\eta^{2}\nabla\tilde{v}\otimes\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}\otimes\nabla
V_{q}-\eta^{2}(1-\eta^{2})\nabla(\tilde{v}-V_{q})\otimes\nabla(\tilde{v}-V_{q})$
$\displaystyle\quad+2\text{sym}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\eta^{2}\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}]$
$\displaystyle\quad+[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\nabla\eta^{2})(\tilde{v}-V_{q})].$

Since $V_{q}=0$ in $\Omega-\tilde{\Omega}_{q+1}$ and $\eta=1$ in the
$l$-neighbourhood of $\tilde{\Omega}_{q+1}$, there holds

$\eta^{2}(1-\eta^{2})\nabla(\tilde{v}-V_{q})\otimes\nabla(\tilde{v}-V_{q})=0.$

Next, by the definition of $\psi_{q},\eta,\eta_{q}$, we find

$\displaystyle\psi_{q+2}$ $\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2}_{q+2})\psi$
$\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2}_{q+1})(1-\eta^{2}_{q+2})\psi$
$\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2}_{q+2})[(1-\eta^{2}_{q+1})\psi+\eta^{2}_{q+1}\delta_{q+1}-\eta^{2}_{q+1}\delta_{q+1}]$
$\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2}_{q+2})[\psi_{q+1}-\eta^{2}_{q+1}\delta_{q+1}]$
$\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2})(1-\eta^{2}_{q+2})[\psi_{q+1}-\eta^{2}_{q+1}\delta_{q+1}]$
$\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2})(1-\eta^{2}_{q+2})\psi_{q+1}$
$\displaystyle=$
$\displaystyle\eta^{2}_{q+2}\delta_{q+2}+(1-\eta^{2})(1-\eta^{2}_{q+2})\psi_{q+1}+\eta^{2}\delta_{q+2}(1-\eta^{2}_{q+2})$
$\displaystyle=$
$\displaystyle\eta^{2}\delta_{q+2}+(1-\eta^{2})\psi_{q+1}-(1-\eta^{2})\eta_{q+2}^{2}(\psi_{q+1}-\delta_{q+2}).$

That is

$\displaystyle\psi_{q+2}$
$\displaystyle=\eta^{2}\delta_{q+2}+(1-\eta^{2})\psi_{q+1}-(1-\eta^{2})\eta_{q+2}^{2}(\psi_{q+1}-\delta_{q+2})$
$\displaystyle=\eta^{2}\delta_{q+2}+(1-\eta^{2})\psi_{q+1}-(1-\eta^{2})\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})$
$\displaystyle\quad-(1-\eta^{2})\eta_{q+2}^{2}(\psi_{q+1}-\tilde{\psi}_{q+2}).$

Finally we write

$A=\eta^{2}\tilde{A}+(1-\eta^{2})A+\eta^{2}(A-\tilde{A}).$

Putting these together, we reformulate (4.41) as

$\displaystyle A-\frac{1}{2}\nabla v_{0}\otimes\nabla v_{0}-\text{sym}\nabla
w_{0}-\psi_{q+2}Id$ $\displaystyle=\left\\{\begin{aligned}
&\eta^{2}(\tilde{A}-\frac{1}{2}\nabla\tilde{v}\otimes\nabla\tilde{v}-\text{sym}\nabla\tilde{w}-\delta_{q+2}Id)\\\
&+(1-\eta^{2})(A-\frac{1}{2}\nabla V_{q}\otimes\nabla V_{q}-\text{sym}\nabla
W_{q}-\psi_{q+1}Id)\\\
&+(1-\eta^{2})\eta_{q+2}^{2}(\tilde{\psi}_{q+1}-\delta_{q+2})Id\end{aligned}\right\\}(*)$
$\displaystyle\quad+\eta^{2}(A-\tilde{A})+(1-\eta^{2})\eta_{q+2}^{2}(\psi_{q+1}-\tilde{\psi}_{q+1})Id$
$\displaystyle\quad-\text{sym}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\eta^{2}\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}]$
$\displaystyle\quad-\frac{1}{2}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\nabla\eta^{2})(\tilde{v}-V_{q})]$
$\displaystyle\quad-\text{sym}(\nabla\eta^{2})\otimes(\tilde{w}-W_{q})$

Compared with the definitions of $\tilde{D},D_{q}$ in (4.27) (4.14), we see

$(*)=\tilde{D}+(1-\eta^{2})D_{q}.$ (4.43)

As for (4.42), recall the definition of $\mathcal{E}_{i}$, then we see

$\tilde{D}+\sum_{i=1}^{N_{*}}(\frac{1}{2}\nabla{v}_{i-1}\otimes\nabla{v}_{i-1}+\text{sym}\nabla{w}_{i-1})-(\frac{1}{2}\nabla{v}_{i}\otimes\nabla{v}_{i}+\text{sym}\nabla{w}_{i})=\sum_{i=1}^{N_{*}}\mathcal{E}_{i}.$

Therefore we conclude

$\displaystyle D_{q+1}$
$\displaystyle=\sum_{i=1}^{N_{*}}\mathcal{E}_{i}+(1-\eta^{2})D_{q}$
$\displaystyle\quad+\eta^{2}(A-\tilde{A})+(1-\eta^{2})\eta_{q+2}^{2}(\psi_{q+1}-\tilde{\psi}_{q+1})Id$
$\displaystyle\quad-\text{sym}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\eta^{2}\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}]$
$\displaystyle\quad-\frac{1}{2}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\nabla\eta^{2})(\tilde{v}-V_{q})]$
$\displaystyle\quad-\text{sym}(\nabla\eta^{2})\otimes(\tilde{w}-W_{q}).$

We estimate term by term. First, because of (4.40), there holds

$\|\sum_{i=1}^{N_{*}}\mathcal{E}_{i}\|_{0}\lesssim\delta_{q+1}(\frac{\mu_{0}}{\mu_{1}}+\frac{\mu_{1}}{\mu_{2}}+...+\frac{\mu_{N_{*}-1}}{\mu_{N_{*}}}).$

And by assumption (4.18)

$(1-\eta^{2})D_{q}=0.$

Next, recall $A\in C^{2,\kappa}(\bar{\Omega},{\mathcal{S}_{n}})$, hence

$\|\eta^{2}(A-\tilde{A})\|_{0}\lesssim\|A\|_{1}l\lesssim l.$

Also recall $\|\psi_{q+1}\|_{1}\lesssim 1$, so

$\|(1-\eta^{2})\eta_{q+2}^{2}(\psi_{q+1}-\tilde{\psi}_{q+1})Id\|_{0}\lesssim\|\psi_{q+1}\|_{1}l\lesssim
l.$

Similarly, using (2.7), (4.15)-(4.17) and
$\|\nabla\eta\|_{0}\lesssim\delta_{q+1}^{-1}$ , we obtain

$\displaystyle\|-\text{sym}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\eta^{2}\nabla\tilde{v}+(1-\eta^{2})\nabla
V_{q}]\|_{0}$
$\displaystyle\lesssim\|\eta\|_{1}\|V_{q}\|_{2}l^{2}\cdot\|V_{q}\|_{1}\lesssim\sqrt{K}\delta_{q+1}^{-\frac{1}{2}}l,$
$\displaystyle\|-\frac{1}{2}[(\nabla\eta^{2})(\tilde{v}-V_{q})]\otimes[(\nabla\eta^{2})(\tilde{v}-V_{q})]\|_{0}$
$\displaystyle\lesssim(\|\eta\|_{1}\|V_{q}\|_{2}l^{2})^{2}\lesssim\delta_{q+1}^{-1}l^{2},$
$\displaystyle\|-\text{sym}(\nabla\eta^{2})\otimes(\tilde{w}-W_{q})\|\lesssim\|\eta\|_{1}\|W_{q}\|_{2}l^{2}\lesssim\delta_{q+1}^{-\frac{1}{2}}l.$

As a result

$\|D_{q+1}\|_{0}\lesssim\delta_{q+1}^{-1}l^{2}+\sqrt{K}\delta_{q+1}^{-\frac{1}{2}}l+\delta_{q+1}(\frac{\mu_{0}}{\mu_{1}}+\frac{\mu_{1}}{\mu_{2}}+...+\frac{\mu_{N_{*}-1}}{\mu_{N_{*}}}).$

We take

$\mu_{i}:=\mu_{0}^{1-\frac{i}{N_{*}}}\mu_{N_{*}}^{\frac{i}{N_{*}}}$

and require

$\sqrt{K}\delta_{q+1}^{-\frac{1}{2}}l\leq\frac{\sigma\delta_{q+2}}{C},\quad\frac{\mu_{0}}{\mu_{N_{*}}}\leq(\frac{\sigma\delta_{q+2}}{C\delta_{q+1}})^{N_{*}}$
(4.44)

so that (4.21) follows. This concludes the proof of Proposition 4.2. ∎

Step 4 (Conclusion): $(V_{q},W_{q})\to(v,w)$. The new inequalities we need are
(4.26)

$l=\frac{\sigma\delta_{q+1}^{\frac{1}{2}}}{CK\delta_{q}^{\frac{1}{2}}\lambda_{q}}\leq\frac{\delta_{q+2}}{C_{1}},$

and the first inequality in (4.44)

$l=\frac{\sigma\delta_{q+1}^{\frac{1}{2}}}{CK\delta_{q}^{\frac{1}{2}}\lambda_{q}}\leq\frac{\sigma}{C_{2}\sqrt{K}}\delta_{q+1}^{\frac{1}{2}}\delta_{q+2}.$

These amount to

$\frac{1}{2}-c\leq-b^{2}+\frac{1}{b^{q}}\log_{a}C(\sigma,K).$

Recall at the end of Section 3, we take $a>1$ sufficiently large, $b>1$ close
to $1$ and $c>N_{*}+\frac{1}{2}\geq\frac{3}{2}$, so this can be achieved
without breaking the other inequalities we need. The rest is the same as in
Section 3. This finishes the whole proof of Theorem 4.1, which yields Theorem
1.2 and Theorem 1.3.

## Acknowledgments

The second named author is partially supported by CAS Project for Young
Scientists in Basic Research, Grant No.YSBR-031.

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