Interior regularity criterion for incompressible Ericksen-Leslie system

Abstract

An interior regularity criterion of suitable weak solutions is formulated for the Ericksen-Leslie system of liquid crystals. Such a criterion is point-wise, with respect to some appropriate norm of velocity u and the gradient of d, and it can be viewed as a sort of simply sufficient condition on the local regularity of suitable weak solutions.

1 Introduction and main results

In this paper, we investigate the local regularity of weak solutions to the following 3D incompressible Ericksen-Leslie liquid crystal system:

$$\begin{aligned} &\partial_{t}u+(u\cdot\nabla)u-\Delta u+\nabla P=-\nabla\cdot(\nabla d\odot\nabla d), \end{aligned}$$

(1.1a)

$$\begin{aligned} & \nabla\cdot u=0, \end{aligned}$$

(1.1b)

$$\begin{aligned} &\partial_{t}d+(u\cdot\nabla) d =\Delta d-f(d), \end{aligned}$$

(1.1c)

with the initial boundary conditions

$$\begin{aligned} \begin{aligned} &(u, d) (x,t)|_{t=0}= \bigl(u_{0}(x), d_{0}(x) \bigr),\qquad \nabla\cdot u_{0}=0, \quad x\in\Omega, \\ &(u,d) (x,t)|_{x\in\partial\Omega}= \bigl(0, d_{0}(x) \bigr),\qquad u_{0}(x)\in H_{0}^{1}(\Omega ),\qquad d_{0}(x)\in H_{0}^{2}(\Omega), \end{aligned} \end{aligned}$$

(1.2)

where $u, d, P$ denote the velocity of the fluid, the uniaxial molecular direction, and the pressure, respectively, the $i,j$th element of $\nabla d\odot\nabla d$ is $\partial_{i}d^{k}\partial_{j}d^{k}$, $d_{0}(x)$ is a unit vector, $\Omega\subset \mathbb {R}^{3}$ is a smooth domain. Additionally, $f(d)=\nabla F(d)$, and $F(d)=\frac{1}{\zeta^{2}}( \vert d \vert ^{2}-1)^{2}, \zeta$ is a small number, formally speaking, as $\zeta\to0, d$ tends to a unit vector.

The dynamic flows of liquid crystals have been successfully described by the Ericksen-Leslie theory [1–4]. System (1.1a)-(1.1c) is a coupled system of the Navier-Stokes equations with a parabolic system. It is Leray [5] and Hopf [6] that established the global existence of weak solutions to the 3D Navier-Stokes; however, the regularity of the weak solutions is still an open problem. Since the regularity of weak solutions to the 3D Navier-Stokes equations is hard to get, some related conditions or criteria for the regularity of the weak solutions are considered, such as the well-known Serrin type criterion [7] and the Beale-Kato-Majda type criterion [8]. Furthermore, based on the suitable weak solutions, some point-wise sufficient regularity criteria were imposed in [9–12].

The global existence of suitable weak solutions to system (1.1a)-(1.1c) was established in [13, 14] by Lin and Liu; however, noticing that system (1.1a)-(1.1c) contains the 3D Navier-Stokes equations as a subsystem, the uniqueness and regularity of these weak solutions are not known. In this paper, we would extend some point-wise sufficient conditions, which guarantee the local regularity of weak solutions for 3D Navier-Stokes equations, to the Ericksen-Leslie system (1.1a)-(1.1c). We would like to mention that when $f(d)$ in system (1.1a)-(1.1c) is replaced by $- \vert \nabla d \vert ^{2}d$, the global existence of weak solutions to the resulting system in three dimensions has only been known under the additional assumption that $d_{3}\geq0$ or small initial data (see [15, 16]). Without these conditions, the general existence of weak solutions is still open. However, the Serrin type criterion and the Beale-Kato-Majda type criterion still hold true even for a weak solution (if it exists) (see [17, 18]).

The suitable weak solution established in [14] can be stated as below.

Definition 1.1

Suitable weak solutions in $\Omega\times(0, T)\subset\mathbb {R}^{3}\times(0,\infty)$

A pair $(u, d)$ is called a suitable weak solution to system (1.1a)-(1.1c) and (1.2) in an open set $\mathcal {O}\subset\mathbb {R}^{3}\times(0,\infty)$ (we set $\mathcal {O}_{t}=\mathcal {O}\cap(\mathbb {R}^{3}\times\{t\} )$), if it satisfies the following properties:

• $(u, d)$ is a weak solution in the sense of distribution;

• $u\in L^{\infty}(0,T;L^{2}(\Omega ))\cap L^{2}(0,T;H^{1}(\Omega)), d\in L^{\infty}(0,T;H^{1}(\Omega))\cap L^{2}(0,T;H^{2}(\Omega))$, or generally, there exist constants $E_{1}, E_{2}$, such that

  $$\begin{aligned} &\int_{\mathcal {O}_{t}} \bigl[ \vert u \vert ^{2}+ \vert \nabla d \vert ^{2}+F(d) \bigr]\,\mathrm{ d}x< E_{1},\\ &\int\!\int_{\mathcal {O}} \bigl[ \vert \nabla u \vert ^{2}+ \bigl\vert \Delta d-f(d) \bigr\vert ^{2}+F(d) \bigr]\,\mathrm{ d}x\,\mathrm{ d}t< E_{2}; \end{aligned}$$

• for any $\varphi\in C_{c}^{\infty}(\mathcal {O})$, more specifically, for any $\varphi\in C_{c}^{\infty}(B(x_{0}, R)\times(t_{0}-R^{2}, t_{0}))$, the following generalized energy inequality holds

  $$\begin{aligned} &\int_{B(x_{0}, R)} \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr) \varphi\,\mathrm{ d}x+2 \int _{t_{0}-R^{2}}^{t} \int_{B(x_{0}, R)} \bigl( \vert \nabla u \vert ^{2}+ \bigl\vert \nabla^{2} d \bigr\vert ^{2} \bigr)\varphi\,\mathrm{ d}x \,\mathrm{ d}\tau \\ &\quad\leq \int_{t_{0}-R^{2}}^{t} \int_{B(x_{0}, R)} \bigl\{ \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr) (\varphi _{t}+\Delta \varphi)+ \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2}+2P \bigr)u\cdot\nabla\varphi \bigr\} \,\mathrm{ d}x\,\mathrm{ d}\tau \\ &\qquad{}+2 \int_{t_{0}-R^{2}}^{t} \int_{B(x_{0}, R)} \bigl((u\cdot\nabla) d\nabla d\nabla \varphi-\nabla f(d):\nabla d\varphi \bigr)\,\mathrm{ d}x\,\mathrm{ d}\tau. \end{aligned}$$

  (1.3)

In the following, we can take $Q((x_{0}, t_{0}), R)\equiv B(x_{0}, R)\times(t_{0}-R^{2}, t_{0})$, $B(x_{0}, R)\equiv\{y\in\mathbb {R}^{3}| \vert y-x_{0} \vert < R\}, z_{0}\equiv(x_{0}, t_{0})$ for simplicity.

We now state our main result of this paper.

Theorem 1.2

Let $(u, d)$ be a suitable weak solution to liquid crystal system (1.1a)-(1.1c) in $Q(z_{0}, R)$. The real numbers $l\geq1$ and $s\geq1$ satisfy

$$ \frac{1}{2}\geq\frac{3}{s}+\frac{2}{l}-\frac{3}{2}>\max \biggl\{ \frac{1}{2l}, \frac{1}{2}-\frac{1}{s}, \frac{1}{s}- \frac{1}{6} \biggr\} . $$

Then there is a positive number $\varepsilon=\varepsilon(s,l)$, such that if

$$ M^{s, l}(z_{0}, R)=\frac{1}{R^{\kappa}} \int_{t_{0}-R^{2}}^{t_{0}} \biggl( \int_{B(x_{0}, R)} \vert u \vert ^{s}+ \vert \nabla d \vert ^{s}\,\mathrm{ d}x \biggr)^{\frac{l}{s}}\,\mathrm{ d}t< \varepsilon,\quad \kappa=\frac{3l}{s}+2-l, $$

then $z_{0}$ is a regular point of $(u,\nabla d)$, i.e. $(u,\nabla d)$ is Hölder continuous in $Q(z_{0},r)$, for some $r\in(0, R]$.

Throughout this paper, we use c to denote a generic positive constant which can be different from line to line.

2 Preliminaries

As the preparation for proving Theorem 1.2, we first give two auxiliary lemmas.

Lemma 2.1

We have

$$ D(z_{0}, r; p)\leq c \biggl[\frac{r}{\rho}D(z_{0}, \rho; p)+ \biggl(\frac{\rho}{r} \biggr)^{2}C(z_{0}, \rho; u, \nabla d) \biggr], $$

(2.1)

where

$$ C(z_{0},r;u,\nabla d)=\frac{1}{r^{2}} \int_{Q(z_{0},r)} \bigl( \vert u \vert ^{3}+ \vert \nabla d \vert ^{3} \bigr) \,\mathrm{ d}z, \qquad D(z_{0},r;p)= \frac{1}{r^{2}} \int_{Q(z_{0},r)} \vert p \vert ^{\frac{3}{2}}\,\mathrm{ d}z. $$

Proof

Step 1. For (1.1a), we choose the test function $w=\chi\nabla q$, for any $\chi\in C_{c}^{\infty}((t_{0}-\rho^{2}, t_{0})), q\in C_{c}^{\infty}(B(x_{0}, \rho))$, then it yields

$$ \int_{Q(z_{0}, \rho)}-u\cdot\partial_{t}\chi\nabla q-(u\otimes u+ \nabla d\odot\nabla d):\chi\nabla^{2} q-u\cdot\chi\nabla\Delta q\,\mathrm{ d}z= \int _{Q(z_{0}, \rho)}p\chi\Delta q \,\mathrm{d}z. $$

It follows from $\nabla\cdot u=0$ that

$$\begin{aligned} - \int_{Q(z_{0}, \rho)}p\chi\Delta q \,\mathrm{d}z= \int_{Q(z_{0}, \rho)}\chi (u\otimes u+\nabla d\odot\nabla d): \nabla^{2} q\,\mathrm{ d}z. \end{aligned}$$

Therefore, for a.e. $t\in(t_{0}-\rho^{2}, t_{0})$, we have

$$ - \int_{B(x_{0}, \rho)}p\Delta q \,\mathrm{d}x= \int_{B(x_{0}, \rho)}(u\otimes u+\nabla d\odot\nabla d): \nabla^{2} q\,\mathrm{ d}x,\quad \forall q\in C_{c}^{\infty}\bigl(B(x_{0}, \rho) \bigr). $$

(2.2)

Step 2. Approximate p with $p_{1}$ by confining q in $W^{2, 3}(B(x_{0}, \rho))$.

Set $p_{1}\in L^{\frac{3}{2}}(Q(z_{0}, \rho))$ such that, for a.e. $t\in(t_{0}-\rho ^{2}, t_{0})$,

$$ - \int_{B(x_{0}, \rho)}p_{1}\Delta q \,\mathrm{d}x= \int_{B(x_{0}, \rho )}(u\otimes u+\nabla d\odot\nabla d): \nabla^{2} q\,\mathrm{ d}x, $$

(2.3)

for any $q(\cdot, t)\in W^{2, 3}(B(x_{0}, \rho))$, and $q(\cdot, t)=0 \text{ on } \partial B(x_{0}, \rho)$. The existence of $p_{1}$ is established due to the Lax-Milgram theorem with appropriate approximating process on u and d (see [11]).

Next, choose $q_{0}(\cdot, t)\in W^{2, 3}(B(x_{0}, \rho))$, such that, for a.e. $t\in(t_{0}-\rho^{2}, t_{0})$,

$$\begin{aligned} \Delta q_{0}(\cdot, t)=- \bigl\vert p_{1}(\cdot, t) \bigr\vert ^{\frac{1}{2}} \operatorname{sgn}p_{1}(\cdot, t),\quad \text{in } B(x_{0}, \rho),\qquad q_{0}(\cdot, t)=0, \quad\text{on } \partial B(x_{0}, \rho). \end{aligned}$$

Then, by the Calderon-Zygmund inequality, it yields

$$\begin{aligned} \biggl( \int_{B(x_{0}, \rho)} \bigl\vert \nabla^{2}q_{0}( \cdot, t) \bigr\vert ^{3}\,\mathrm{ d}x \biggr)^{\frac{1}{3}}\leq c \biggl( \int_{B(x_{0}, \rho)} \bigl\vert p_{1}(\cdot, t) \bigr\vert ^{\frac{3}{2}}\,\mathrm{ d}x \biggr)^{\frac{1}{3}}, \quad\mbox{a.e. }t\in \bigl(t_{0}- \rho^{2}, t_{0} \bigr). \end{aligned}$$

Therefore, it follows from (2.3) and the Hölder inequality that

$$\begin{aligned} \int_{B(x_{0}, \rho)} \bigl\vert p_{1}(\cdot, t) \bigr\vert ^{\frac{3}{2}}\,\mathrm{ d}x &\leq c \biggl( \int_{B(x_{0}, \rho)} \vert u \vert ^{3}+ \vert \nabla d \vert ^{3}\,\mathrm{ d}x \biggr)^{\frac{2}{3}} \biggl( \int_{B(x_{0}, \rho)} \bigl\vert \nabla^{2}q \bigr\vert ^{3}\,\mathrm{ d}x \biggr)^{\frac{1}{3}} \\ &\leq c \biggl( \int_{B(x_{0}, \rho)} \vert u \vert ^{3}+ \vert \nabla d \vert ^{3}\,\mathrm{ d}x \biggr)^{\frac{2}{3}} \biggl( \int_{B(x_{0}, \rho)} \vert p_{1} \vert ^{\frac{3}{2}} \,\mathrm{ d}x \biggr)^{\frac{1}{3}}, \end{aligned}$$

which yields $\int_{Q(z_{0}, \rho)} \vert p_{1}(\cdot, t) \vert ^{\frac{3}{2}}\,\mathrm{ d}z\leq c\rho^{2}C(z_{0}, \rho; u, \nabla d)$.

Step 3. Estimates for the remainder $p-p_{1}$.

For a.e. $t\in(t_{0}-\rho^{2}, t_{0})$, let $p_{2}=p-p_{1}$, then from (2.2)-(2.3) one infers that

$$\begin{aligned} \Delta p_{2}(\cdot, t)=0,\quad \text{in } B(x_{0}, \rho). \end{aligned}$$

By the harmonic property, one can get

$$\begin{aligned} \frac{1}{r^{3}} \int_{Q(z_{0}, r)} \vert p_{2} \vert ^{\frac{3}{2}} \,\mathrm{ d}z\leq \frac{c}{\rho ^{3}} \int_{Q(z_{0}, \rho)} \vert p_{2} \vert ^{\frac{3}{2}} \,\mathrm{ d}z,\quad \forall r< \rho, \end{aligned}$$

while

$$ \int_{Q(z_{0}, \rho)} \vert p_{2} \vert ^{\frac{3}{2}} \,\mathrm{ d}z \leq \int_{Q(z_{0}, \rho)} \bigl( \vert p \vert ^{\frac{3}{2}}+ \vert p_{1} \vert ^{\frac{3}{2}} \bigr)\,\mathrm{ d}z\leq c\rho ^{2} \bigl(D(z_{0}, \rho; p)+C(z_{0}, \rho; u, \nabla d) \bigr). $$

Step 4. Estimates for p.

We have

$$\begin{aligned} D(z_{0}, r; p)&\leq c \biggl(\frac{1}{r^{2}} \int_{Q(z_{0}, r)} \vert p_{1} \vert ^{\frac{3}{2}} \,\mathrm{ d}z+ \frac{r}{\rho^{3}} \int_{Q(z_{0}, \rho)} \vert p_{2} \vert ^{\frac{3}{2}} \,\mathrm{ d}z \biggr) \\ &\leq c \biggl(\frac{\rho^{2}}{r^{2}}\frac{1}{\rho^{2}} \int_{Q(z_{0}, r)} \vert p_{1} \vert ^{\frac{3}{2}} \,\mathrm{ d}z+ \frac{r}{\rho}\frac{1}{\rho^{2}} \int_{Q(z_{0}, \rho)} \vert p_{2} \vert ^{\frac{3}{2}} \,\mathrm{ d}z \biggr) \\ &\leq c \biggl[\frac{\rho^{2}}{r^{2}}C(z_{0}, \rho; u, \nabla d)+ \frac{r}{\rho } \bigl(D(z_{0}, \rho; p)+C(z_{0}, \rho; u, \nabla d) \bigr) \biggr] \\ &\leq c \biggl[\frac{r}{\rho}D(z_{0}, \rho; p)+ \biggl( \frac{\rho}{r} \biggr)^{2}C(z_{0}, \rho; u, \nabla d) \biggr]. \end{aligned}$$

 □

We denote

$$\begin{aligned} &A(\rho)=\operatorname{ess}\sup_{{t_{0}}-\rho ^{2}< t< {t_{0}}}\frac{1}{\rho}\int_{B(x_{0},\rho )} \bigl( \bigl\vert u(t) \bigr\vert ^{2}+ \bigl\vert \nabla d(t) \bigr\vert ^{2} \bigr)\,\mathrm{ d}x, \\ & E(\rho)=\frac{1}{\rho}\int_{Q(z_{0},\rho)} \bigl( \vert \nabla u \vert ^{2}+ \bigl\vert \nabla^{2} d \bigr\vert ^{2} \bigr)\,\mathrm{ d}z,\qquad H( \rho)=\frac{1}{\rho^{3}} \int_{Q(z_{0},\rho )} \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr) \,\mathrm{ d}z. \end{aligned}$$

Lemma 2.2

Under the assumptions of Theorem 1.2, we have

$$C(\rho)\leq c\epsilon^{\frac{1}{q}} \bigl(E(\rho)+A(\rho)+1 \bigr), $$

where $q=2l(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})$, and $q'=\frac{q}{q-1}$.

Proof

With the help of the Hölder and Sobolev embedding inequalities, one gets

$$\begin{aligned} \int_{B(x_{0}, \rho)} \vert v \vert ^{3}\,\mathrm{ d}x={}& \int_{B(x_{0}, \rho)} \vert v \vert ^{\lambda s+2\mu+6\gamma}\,\mathrm{ d}x \\ \leq {}&\biggl( \int_{B(x_{0}, \rho)} \vert v \vert ^{2}\,\mathrm{ d}x \biggr)^{\mu}\biggl( \int _{B(x_{0}, \rho)} \vert v \vert ^{s}\,\mathrm{ d}x \biggr)^{\lambda}\biggl( \int_{B(x_{0}, \rho)} \vert v \vert ^{6}\,\mathrm{ d}x \biggr)^{\gamma}\\ \leq{}& \frac{c}{2}\rho^{\mu}\biggl(\operatorname{ess}\sup_{t_{0}-\rho^{2}< t< t_{0}} \frac{1}{\rho}\int _{B(x_{0},\rho)} \vert v \vert ^{2}\,\mathrm{ d}x \biggr)^{\mu}\biggl( \int_{B(x_{0}, \rho)} \vert v \vert ^{s}\,\mathrm{ d}x \biggr)^{\lambda}\\ &{}\times \biggl( \int_{B(x_{0}, \rho)} \vert \nabla v \vert ^{2}+ \frac{1}{\rho ^{2}} \vert v \vert ^{2} \,\mathrm{ d}x \biggr)^{3\gamma}, \end{aligned}$$

where $\lambda s+2\mu+6\gamma=3, \lambda+\mu+\gamma=1$. Substituting v by u and ∇d, respectively, then one can get the summation

$$\begin{aligned} \int_{B(x_{0}, \rho)} \vert u \vert ^{3}+ \vert \nabla d \vert ^{3}\,\mathrm{ d}x \leq{}&c\rho^{\mu}A^{\mu}( \rho) \biggl( \int_{B(x_{0}, \rho)} \bigl( \vert u \vert ^{s}+ \vert \nabla d \vert ^{s} \bigr) \,\mathrm{ d}x \biggr)^{\lambda}\\ &{}\times \biggl( \int_{B(x_{0}, \rho)} \bigl( \vert \nabla u \vert ^{2}+ \bigl\vert \nabla^{2} d \bigr\vert ^{2} \bigr)+ \frac{1}{\rho^{2}} \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr)\,\mathrm{ d}x \biggr)^{3\gamma}. \end{aligned}$$

Therefore, by choosing appropriate parameters $\lambda=\frac{1}{2s(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})}$, $\mu=\frac {\frac{3}{s}+\frac{3}{l}-2}{2(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})}$, $\gamma= \frac{\frac{2}{s}+\frac{1}{l}-1}{2(\frac{3}{s}+\frac{2}{l}-\frac{3}{2})}$, and integrating from $t_{0}-\rho^{2}$ to $t_{0}$ with the variable t, it follows from the Hölder and Young inequalities that

$$\begin{aligned} C(\rho)\leq{}&c\rho^{\mu-2} A^{\mu}(\rho){ \biggl( \int_{Q(z_{0}, \rho )} \bigl( \vert \nabla u \vert ^{2}+ \bigl\vert \nabla^{2} d \bigr\vert ^{2} \bigr)+ \frac{1}{\rho^{2}} \bigl( \vert u \vert ^{2}+ \vert \nabla d \vert ^{2} \bigr)\,\mathrm{ d}z \biggr)}^{\frac{1}{q'}} \\ &{}\times \biggl[ \int_{t_{0}-\rho^{2}}^{t_{0}} \biggl( \int_{B(x_{0}, \rho )} \bigl( \vert u \vert ^{s}+ \vert \nabla d \vert ^{s} \bigr) \,\mathrm{ d}x \biggr)^{\frac{l}{s}} \,\mathrm{ d}t \biggr]^{\frac{1}{q}} \\ \leq{}&c\rho^{\mu-2} A^{\mu}(\rho)\rho^{\frac{1}{q'}} \bigl(E(\rho)+H( \rho) \bigr)^{\frac{1}{q'}} \bigl(\rho^{\kappa}M^{s,l}(\rho) \bigr)^{\frac{1}{q}} \\ \leq{}&c A^{\mu}(\rho) \bigl(E(\rho)+H(\rho) \bigr)^{\frac{1}{q'}} \bigl(M^{s,l}(\rho) \bigr)^{\frac{1}{q}} \\ \leq{}&c \epsilon^{\frac{1}{q}} A^{\mu}(\rho) \bigl(E(\rho)+H(\rho) \bigr)^{\frac{1}{q'}} \\ \leq{}&c \epsilon^{\frac{1}{q}} \bigl(A^{\mu q}(\rho)+E(\rho)+H(\rho) \bigr) \\ \leq{}&c \epsilon^{\frac{1}{q}} \bigl(E(\rho)+A(\rho)+1 \bigr), \end{aligned}$$

where $\kappa=\frac{3l}{s}+2-l$ as in Theorem 1.2, and in the last step, we used the fact that $\mu q\leq1, H(\rho)\leq A(\rho)$. □

3 Proof of Theorem 1.2

Due to the induction argument as Proposition 2.6 in [10] or Lemma 2.2 in [19] (the parabolic version of the Campanato criterion), to get the desired consequence, it suffices to prove $C(\theta^{k})+D(\theta^{k})< \epsilon_{0}$ for some small $\epsilon_{0}$. Here θ is a small number, which will be chosen later.

From the generalized energy inequality, it is easy to check that, for $\rho\in(0, R]$,

$$ A \biggl(\frac{\rho}{2} \biggr)+E \biggl(\frac{\rho}{2} \biggr)\leq c \bigl[C^{\frac{2}{3}}(\rho)+C(\rho)+D(\rho) \bigr]. $$

Denoting $G(\rho)=A(\rho)+E(\rho)+D(\rho)$, due to Lemmas 2.1-2.2, and the fact that $C(2\theta\rho)\leq\frac{1}{4\theta^{2}}C(\rho)$, we can get

$$\begin{aligned} G(\theta\rho)\leq{}& c \biggl[C^{\frac{2}{3}}(2\theta\rho)+C(2\theta \rho)+D(2 \theta \rho)+\theta D(\rho)+\frac{1}{\theta^{2}}C(\rho) \biggr] \\ \leq{}& c \biggl[\frac{1}{\theta^{\frac{4}{3}}}C^{\frac{2}{3}}(\rho)+ \frac{1}{\theta^{2}}C( \rho )+\theta D(\rho) \biggr] \\ \leq{}& c \biggl[\frac{\epsilon^{\frac{2}{3q}}}{\theta^{\frac{4}{3}}} \bigl(G(\rho)+1 \bigr)^{\frac{2}{3}}+ \frac{\epsilon^{\frac{1}{q}}}{\theta^{2}} \bigl(G(\rho)+1 \bigr)+\theta G(\rho) \biggr] \\ \leq{}& c \biggl[ \biggl(\theta+\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}} \biggr)G(\rho)+ \frac {\epsilon^{\frac{1}{2q}}}{\theta^{2}} \biggr], \end{aligned}$$

where in the last step we have used $\frac{\epsilon^{\frac{2}{3q}}}{\theta^{\frac{4}{3}}}(G(\rho)+1)^{\frac{2}{3}}\leq c[\epsilon^{\frac{1}{q}}+\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}(G(\rho)+1)]$. Now choosing θ and ϵ such that $c\theta<\frac{1}{4}$ and $c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}<\frac{1}{4}$, then it yields $G(\theta\rho)\leq\frac{1}{2}G(\rho)+c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}$. Iterating the above process, we obtain $G(\theta^{k} \rho)\leq\frac{1}{2^{k}}G(\rho)+c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}$, therefore,

$$ D \bigl(\theta^{k}\rho \bigr)\leq\frac{1}{2^{k}}G( \rho)+c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}. $$

(3.1)

For $C(\theta^{k}\rho)$, by Lemma 2.2, we have

$$\begin{aligned} C \bigl(\theta^{k} \rho \bigr)\leq c \epsilon^{\frac{1}{q}} \bigl[ G \bigl(\theta^{k} \rho \bigr)+1 \bigr]\leq c \epsilon^{\frac{1}{q}} \biggl[\frac{1}{2^{k}}G(\rho)+ \frac{\epsilon^{\frac{1}{2q}}}{\theta ^{2}}+1 \biggr]\leq c \biggl[\frac{1}{2^{k}}G(\rho)+ \frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}} \biggr], \end{aligned}$$

(3.2)

where in the last step we use the fact that $\epsilon^{\frac{1}{q}}\leq \frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}$ for ϵ small enough. With these inequalities in hand, for fixed ρ and $\epsilon_{0}$, we can choose $k_{0}$ large enough such that $c\frac{1}{2^{k_{0}}}G(\rho)<\frac{\epsilon_{0}}{4}$, and choose ϵ small enough, such that $c\frac{\epsilon^{\frac{1}{2q}}}{\theta^{2}}<\frac{\epsilon_{0}}{4}$. With these prerequisites and (3.1)-(3.2), it follows that $D(\theta^{k}\rho)+C(\theta^{k}\rho)<\epsilon_{0}$.