A new conformal heat flow of harmonic maps

Woongbae Park 522 Thackeray Hall, Department of Mathematics, University of Pittsburgh, Pittsburgh, Pennsylvania, 15260 [email protected]

Abstract.

We introduce and study a conformal heat flow of harmonic maps defined by an evolution equation for a pair consisting of a map and a conformal factor of metric on the two-dimensional domain. This flow is designed to postpone finite time singularity but does not get rid of possibility of bubble forming. We show that Struwe type global weak solution exists, which is smooth except at most finitely many points.

Key words and phrases:

harmonic maps, conformal heat flow, short time existence, global weak solution

2020 Mathematics Subject Classification:

Primary 58E20, 53E99, 53C43; Secondary 35K58

1. Introduction

Consider a map $f_{0}:\Sigma\times[0,T)\rightarrow N$ from a compact Riemann surface $(\Sigma,g_{0})$ with metric $g_{0}$ to a Riemannian manifold $(N,h)$ . Under the usual harmonic map heat flow, $f_{0}$ evolves to a map $f(t)$ according to the evolution equation $f_{t}=\tau_{g_{0}}(f)$ , where $\tau_{g}(f)=\operatorname{tr}_{g}(\nabla^{g}df)$ is the tension field with respect to the metric $g$ . In this paper we consider the generalization in which both the map and the metric evolve with $(f(t),g(t))$ satisfying the equations

(1a)			$\displaystyle f_{t}\ =\tau_{g}(f)$
(1b)			$\displaystyle g_{t}\ =(2b\|df\|_{g}^{2}-2a)g$

where $a,b>0$ are constants and $|df|_{g}^{2}=g^{ij}h_{\alpha\beta}f_{i}^{\alpha}f_{j}^{\beta}$ is the energy density. We assume that the initial map $f(0)=f_{0}$ and metric $g(0)=g_{0}$ are smooth.

The first of these equations is the harmonic map heat flow, with varying metric $g$ . The second equation is designed to attenuate energy concentration. If the energy density become large in some region $\Omega\subset\Sigma$ , then under the flow (1b), the metric is conformally enlarged; this increases the area of $\Omega$ and decreases the energy density. This suggests that the system (1b) may be better behaved than the harmonic map heat flow, where energy concentration at points is an impediment to convergence.

Writing the metric $g(t)=e^{2u}g_{0}$ for a real-valued function $u(t)$ , equations (1b) are equivalent to the following equations for the pair $(f(t),u(t))$ :

(2a)			$\displaystyle f_{t}\ =e^{-2u}\tau(f)$
(2b)			$\displaystyle u_{t}\ =be^{-2u}\|df\|^{2}-a$

where $\tau$ and $|df|^{2}$ are with respect to the fixed metric $g_{0}$ , and where the initial conditions are $f(0)=f_{0}$ , $u(0)=0$ . In this form, the flow is more easily analyzed.

The main Theorem of this paper is the following.

Theorem 1.1.

(Existence of global weak solution) For any $f_{0}\in W^{3,2}(\Sigma,N)$ , a global weak solution $(f,u)$ of (2b) exists on $\Sigma\times[0,\infty)$ which is smooth on $\Sigma\times(0,\infty)$ except at most finitely many points.

There is a long history of harmonic maps and related fields. We could not list all such literatures but only few, including [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12] and therein. In terms of heat flow of harmonic maps, see for example [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23], [24], [25] and therein. Note that usual heat flow can have finite time singularity, see Chang-Ding-Ye [26], Raphael-Schweyer [27], or more recently Dávila-Del Pino-Wei [28].

There are several directions to allow metric change along harmonic map heat flow. The most well-known direction is Teichmüller flow, where metric lies in Teichmüller space of constant curvature. Teichmüller flow is the $L^{2}$ gradient flow of the energy and hence reduce the energy in the fastest sense. For relevant literature, see for example Rupflin-Topping [29], Huxol-Rupflin-Topping [30] or Rupflin-Topping [31] and therein. Another direction is Ricci-harmonic map flow. This is a combination of harmonic map heat flow and Ricci flow of the metric. Surprisingly, this flow is more regular than both harmonic map heat flow and Ricci flow. See for example, Muller [32], Williams [33] or Buzano-Rupflin [34] among others. Recently in Huang-Tam [35], harmonic map heat flow together with evolution equation of metric is considered under time-dependent curvature restriction and smooth short time existence is obtained. Because we do not assume a priori curvature bounds of the domain, the result cannot be applied into our case.

The paper is organized as follows. In Section 2 we look at some preliminaries, including volume formula and its asymptotic limit if the map $f$ is steady solution, that is, harmonic. Next, in Section 3 we define Hilbert spaces $X,Y,Z$ and their closed subsets $B,B^{\prime}$ . So, from Section 3 we consider $f\in B$ and $u\in B^{\prime}$ . Then Section 4 defines the operator $S_{1},S_{2}$ and shows their properties. Briefly, we can show that $S_{1}:B\times B^{\prime}\to B$ and $S_{2}:B\times B^{\prime}\to B^{\prime}$ and they satisfy twisted partial contraction properties, see Lemmas 4.3, 4.4, 4.7, and 4.8. In the last section 5 we define the operator $\mathcal{S}$ on $B\times B^{\prime}$ mapping into itself defined by $\mathcal{S}=(S_{1},S_{2})$ . For $T$ small enough, $\mathcal{S}$ is a contraction and hence we can prove short time existence.

Next we are working on types of singularity. Ultimately we will show that the solution is singular only when energy concentrates, similar with Struwe’s result. In Section 6 we show local estimate and obtain bounds for $\iint e^{2u}|f_{t}|^{4}$ . This is used in Section 7 to show $W^{2,2}$ and higher estimate, which implies boundedness of $|df|$ . Finally in Section 8 we prove the main theorem 1.1.

1.1. Notation

Even though our equation is heat-type equation for varying metric, we use initial metric $g_{0}$ as default. So, all terms using metric use $g_{0}$ unless we specify the metric. For example, $|df|^{2}$ is calculated in terms of $g_{0}$ and $|df|_{g}^{2}$ is calculated in terms of $g$ . If the volume form is calculated in terms of metric $g$ , we denote it as $dvol_{g}$ . We also omit $dvol_{g_{0}}$ and $dt$ if there is no confusion. We also use the simplifications $\|\cdot\|_{W^{k,p}}=\|\cdot\|_{W^{k,p}(\Sigma\times[0,T])}$ , $\|\cdot\|_{C^{0}}=\|\cdot\|_{C^{0}(\Sigma\times[0,T])}$ and $\|\cdot\|_{L^{p}}=\|\cdot\|_{L^{p}(\Sigma\times[0,T])}$ . Also, the constant $c$ is universal and changed line by line.

2. Preliminaries

Before we show the main result, we record a few facts about solutions to the flow equations (2b).

2.1. Energy and Volume

First note that the 2-form $|df|^{2}\,dvol_{g}$ is conformally invariant, and that the energy

(3)

E(t)=\tfrac{1}{2}\int|df|^{2}\,dvol_{g}

satisfies

(4)

E^{\prime}(t)=\int\ \langle df,df_{t}\rangle\ =\ -\int\langle\nabla df,e^{-2u}\tau(f)\rangle\ =\ -\int e^{-2u}|\tau(f)|^{2}\ \leq\ 0.

Thus $E(t)\leq E_{0}$ for all $t$ .

Lemma 2.1.

The volume satisfies $V(t)\leq e^{-2at}V(0)+\tfrac{2b}{a}E_{0}$ , and hence is finite for all $t$ .

Proof.

The second equation (2b) can be explicitly solved, yielding

(5)

e^{2u}=e^{-2at}\left(1+2b\int_{0}^{t}e^{2as}|df|^{2}(s)ds\right).

Consequently, the volume

(6)

V(t)=\int_{\Sigma}dvol_{g(t)}=\int_{\Sigma}e^{2u}dvol_{g_{0}}

can be written as

(7)

V(t)=e^{-2at}\left(V(0)+4b\int_{0}^{t}e^{2as}E(s)ds\right).

The lemma follows by noting that $E(s)\leq E_{0}$ and integrating. ∎

2.2. Asymptotic behavior of steady solution

Now we consider steady solution.

Lemma 2.2.

Let $(f,u)$ be a solution of (2b) and $f(0)$ a harmonic with energy $E$ . Then $f(t)$ is harmonic for all $t$ and as $t\rightarrow\infty$ ,

e^{2u}\rightarrow\tfrac{b}{a}|df|^{2}

and hence by (6) the volume $V(t)$ converges to

V(\infty)=\tfrac{2b}{a}E.

Proof.

If $f(0)$ is harmonic, then $f_{t}=0$ and hence $f$ and $|df|^{2}$ are independent of $t$ . Integrating (5) then shows that, as $t\rightarrow\infty$ ,

	$\displaystyle e^{2u}$	$\displaystyle=e^{-2at}\left(1+2b\|df\|^{2}\frac{e^{2at}-1}{2a}\right)$
		$\displaystyle=e^{-2at}+\tfrac{b}{a}\|df\|^{2}(1-e^{-2at})\rightarrow\tfrac{b}{a}\|df\|^{2}.$

∎

This means that, for solutions as in Lemma 2.2, the energy density $|df|_{g}^{2}=|df|^{2}e^{-2u}$ converges as $t\to\infty$ to the constant $\frac{a}{b}$ . Hence the conformal heat flow forces the conformal factor and the energy density be distributed evenly. Remark that, because the image $f(\Sigma)$ does not change, this flow modifies the domain toward the space which is similar to the image with the similarity ratio $\frac{a}{b}$ .

3. Construction of Hilbert spaces

In this section we build Hilbert spaces $X_{T},Y_{T},Z_{T}$ and their closed subsets $B,B^{\prime}$ . For parabolic theory used here, see Mantegazza-Martinazzi [36], Evans [37] or Lieberman [38]. From now on, we consider the target manifold being isometrically embedded, $N\hookrightarrow\mathbb{R}^{L}$ .

3.1. Spaces $X$ , $Y$ and $Z$

The set

Y_{T}\ =L^{2}([0,T],W^{4,2}(\Sigma,\mathbb{R}^{L}))\cap W^{1,2}([0,T],W^{2,2}(\Sigma,\mathbb{R}^{L}))\cap W^{2,2}([0,T],L^{2}(\Sigma,\mathbb{R}^{L}))

is a Hilbert space with norm

\|f\|^{2}_{Y}\ =\ \int_{0}^{T}\int_{\Sigma}\ |\nabla^{4}f|^{2}+|f|^{2}+|\nabla^{2}f_{t}|^{2}+|f_{t}|^{2}+|f_{tt}|^{2}\ dvol_{g_{0}}\ dt.

As in Proposition 4.1 in [36],

Y_{T}\hookrightarrow C^{0}([0,T],C^{1}(\Sigma,\mathbb{R}^{L}))\cap L^{4}([0,T],W^{3,4}(\Sigma,\mathbb{R}^{L}))\cap W^{1,4}([0,T],W^{1,4}(\Sigma,\mathbb{R}^{L}))

and there is a constant ${c}$ such that

(8)

\displaystyle\|f\|_{C^{0}}+\|\nabla f\|_{C^{0}}+\|\nabla^{3}f\|_{L^{4}}+\|\nabla f_{t}\|_{L^{4}}\leq{c}\|f\|_{Y}.

Also, by standard parabolic theory (See, for example, [37]), $f\in Y_{T}$ implies $f\in C^{0}([0,T],W^{3,2}(\Sigma,\mathbb{R}^{L}))$ , $f_{t}\in C^{0}([0,T],W^{1,2}(\Sigma,\mathbb{R}^{L}))$ and

(9)

\displaystyle\max_{0\leq t\leq T}\|f(t)\|_{W^{3,2}(\Sigma)},\max_{0\leq t\leq T}\|f_{t}(t)\|_{W^{1,2}(\Sigma)}\leq{c}\|f\|_{Y}.

This also implies that

(10)

\displaystyle\max_{0\leq t\leq T}\|f(t)\|_{W^{2,8}(\Sigma)}\leq{c}\|f\|_{Y}.

Next, denote

X_{T}=L^{2}([0,T],W^{2,2}(\Sigma,\mathbb{R}^{L}))\cap W^{1,2}([0,T],L^{2}(\Sigma,\mathbb{R}^{L}))

be another Hilbert space with norm

\|f\|^{2}_{X}=\int_{0}^{T}\int_{\Sigma}|f|^{2}+|\nabla^{2}f|^{2}+|f_{t}|^{2}\ dvol_{g_{0}}\ dt.

Note that in the notation of [36], $Y=P^{2}$ and $X=P^{1}$ .

Now we define spaces for $u$ . The set

Z_{T}\ =L^{2}([0,T],W^{3,2}(\Sigma))\cap W^{1,2}([0,T],W^{1,2}(\Sigma))

is a Hilbert space with norm

\|u\|_{Z}^{2}\ =\ \int_{0}^{T}\int_{\Sigma}\ |\nabla^{3}u|^{2}+|u|^{2}+|\nabla u_{t}|^{2}+|u_{t}|^{2}\ dvol_{g_{0}}\ dt.

Similar to above, there is a constant ${c}$ such that

(11)

\|\nabla^{2}u\|_{L^{4}}+\|u_{t}\|_{L^{4}}\leq{c}\|u\|_{Z}

and

(12)

\max_{0\leq t\leq T}\|u(t)\|_{W^{2,2}(\Sigma)}+\max_{0\leq t\leq T}\|u_{t}(t)\|_{L^{2}(\Sigma)}\leq{c}\|u\|_{Z}.

Also, by Sobolev embedding, we have

(13)

\max_{0\leq t\leq T}\|u(t)\|_{W^{1,8}(\Sigma)}\leq{c}\|u\|_{Z}.

Moreover, $u$ is continuous and there is a constant $C_{2}$ such that for all $u\in Z_{T}$ ,

(14)

\|u\|_{C^{0}}\leq C_{2}\|u\|_{Z}.

3.2. The ball $B$ and $B^{\prime}$

Now we fix $f_{0}\in W^{3,2}(\Sigma)$ throughout the section and thereafter. Consider the operator $\partial_{t}-e^{-2u}\Delta$ . If $\|u\|_{C^{0}}\leq 1$ , this operator is uniformly elliptic. So, Proposition 2.3 of [36] then says that the map $f\mapsto\big{(}f_{0},(\partial_{t}-e^{-2u}\Delta)f\big{)}$ is a linear isomorphism

Y_{T}\to W^{3,2}(\Sigma)\times X_{T}.

Hence there is a constant $C_{1}$ such that for each $f_{0}\in W^{3,2}(\Sigma)$ and $g\in X_{T}$ , there is a unique solution $h(t,x)\in Y_{T}$ of the initial value problem

(15)

\displaystyle(\partial_{t}-e^{-2u}\Delta)h=g\qquad h(0)=f_{0}

with

(16)

\displaystyle\|h\|_{Y}\leq C_{1}\big{(}\|f_{0}\|_{3,2}+\|g\|_{X}\big{)}.

Let $h_{0}(t,x)$ be the unique solution of

(17)

\displaystyle(\partial_{t}-\Delta)h=0\qquad h(0)=f_{0}.

By (16) there is a constant $C_{0}$ , depending on $C_{1}$ and $\|f_{0}\|_{3,2}$ such that

(18)

\|h_{0}\|_{Y}\ \leq\ C_{0}.

Because of (8), $\big{\{}f\in Y_{T}\,\big{|}\,f(0)=f_{0}\big{\}}$ is a closed affine subspace of $Y_{T}$ . Hence the ball

(19)

\displaystyle B=B_{\delta}\ =\ \big{\{}f\in Y_{T}\,\big{|}\,f(0)=f_{0}\ \mbox{and}\ \|f-h_{0}\|_{Y}\leq\delta\big{\}}

is a closed subset of $Y_{T}$ . Note that each $f\in B_{\delta}$ satisfies

(20)

\displaystyle\|f\|_{Y}\ \leq\ \|f-h_{0}\|_{Y}+\|h_{0}\|_{Y}\ \leq\ \delta+C_{0}.

Also let the ball

B^{\prime}=B_{\delta^{\prime}}^{\prime}=\{u\in Z_{T}\,\big{|}\,u(0)=0\ \mbox{and}\ \|u\|_{Z}\leq\delta^{\prime}\}

be a closed subset of $Z_{T}$ . Obviously $h_{0}\in B_{\delta}$ and $0\in B_{\delta^{\prime}}^{\prime}$ . For simplicity, we denote $B=B_{\delta}$ and $B^{\prime}=B_{\delta^{\prime}}^{\prime}$ .

Now fix $\delta>0$ and define

(21)

C_{3}:=1600C_{0}C_{1}C_{2}.

Choose $\delta^{\prime}$ small enough so that $C_{2}\delta^{\prime}<1$ which implies $\|u\|_{C^{0}}\leq 1$ . Also we assume $\delta^{\prime}\leq\frac{\delta}{C_{3}}$ .

4. Construction of operators

In this section we will construct operators $S_{1}:Y_{T}\times Z_{T}\to Y_{T}$ and $S_{2}:Y_{T}\times Z_{T}\to Z_{T}$ . First fix $f\in Y_{T}$ and $u\in Z_{T}$ . $f$ and $u$ are considered to be fixed throughout this section and after unless we mention any choice of them.

First we show a lemma that is needed in several places.

Lemma 4.1.

Fix $f_{0}\in W^{3,2}(\Sigma)$ . Then there is an $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ , for each $h\in B$ and $u_{1},u_{2}\in B^{\prime}$ ,

(22)

\|(e^{2u_{2}-2u_{1}}-1)\partial_{t}h\|_{X}\leq\frac{C_{3}}{2C_{1}}\|u_{1}-u_{2}\|_{Z}.

Proof.

Denote

g:=(e^{2u_{2}-2u_{1}}-1)\partial_{t}h.

Recall that

\left|e^{2u_{1}-2u_{2}}-1\right|\leq e^{2|u_{1}-u_{2}|}\left|1-e^{-2|u_{1}-u_{2}|}\right|\leq 2e^{4}|u_{1}-u_{2}|

if $\|u_{1}-u_{2}\|_{C^{0}}\leq 2$ , which comes from $u_{1},u_{2}\leq B^{\prime}$ . Using $2e^{4}\leq 200$ and by (9) and (20),

	$\displaystyle\\|g\\|_{L^{2}}^{2}$	$\displaystyle\leq 200^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}(t)\\|_{L^{2}}^{2}\,T$
		$\displaystyle\leq\frac{C_{3}^{2}}{16C_{1}^{2}}\\|u_{1}-u_{2}\\|_{Z}^{2}$

if we choose $T$ small enough.

Next, consider $\left|\nabla^{2}g\right|^{2}$ .

	$\displaystyle\left\|\nabla^{2}g\right\|$	$\displaystyle=\left\|\nabla^{2}\left((e^{2u_{2}-2u_{1}}-1)\partial_{t}h_{2}\right)\right\|$
		$\displaystyle\leq 800\|u_{1}-u_{2}\|\|\nabla(u_{1}-u_{2})\|^{2}\|\partial_{t}h_{2}\|+400\|u_{1}-u_{2}\|\|\nabla^{2}(u_{1}-u_{2})\|\|\partial_{t}h_{2}\|$
		$\displaystyle\quad+400\|u_{1}-u_{2}\|\|\nabla(u_{1}-u_{2})\|\|\nabla\partial_{t}h_{2}\|+200\|u_{1}-u_{2}\|\|\nabla^{2}\partial_{t}h_{2}\|.$

Hence, by integrating, we have

	$\displaystyle\\|\nabla^{2}g\\|_{L^{2}}^{2}$	$\displaystyle\leq 1600^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla(u_{1}-u_{2})\\|_{L^{8}}^{4}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}(t)\\|_{L^{4}(\Sigma)}^{2}\,T^{1/2}$
		$\displaystyle\quad+800^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}(u_{1}-u_{2})\\|_{L^{4}}^{2}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}(t)\\|_{L^{8}(\Sigma)}^{4}\,T^{1/2}$
		$\displaystyle\quad+800^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla(u_{1}(t)-u_{2}(t))\\|_{L^{4}(\Sigma)}^{2}\\|\nabla\partial_{t}h_{2}\\|_{L^{4}}^{2}\,T^{1/2}$
		$\displaystyle\quad+400^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}\partial_{t}h_{2}\\|_{L^{2}}^{2}$
		$\displaystyle\leq 400^{2}C_{2}^{2}\\|u_{1}-u_{2}\\|_{Z}^{2}2C_{0}^{2}$
		$\displaystyle=\frac{C_{3}^{2}}{8C_{1}^{2}}\\|u_{1}-u_{2}\\|_{Z}^{2}$

if we choose $T$ small enough.

Finally, we will compute $\|g_{t}\|_{L^{2}}^{2}$ .

\displaystyle|g_{t}|

\displaystyle\leq 400|u_{1}-u_{2}||\partial_{t}h_{2}||(u_{1}-u_{2})_{t}|+200|u_{1}-u_{2}||\partial_{tt}h_{2}|.

Hence,

	$\displaystyle\\|g_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq 2(400)^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|(u_{1}-u_{2})_{t}\\|_{L^{4}}^{2}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}\\|_{L^{4}(\Sigma)}^{2}\,T^{1/2}$
		$\displaystyle\quad+2(200)^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\partial_{tt}h_{2}\\|_{L^{2}}^{2}$
		$\displaystyle\leq 2(200)^{2}C_{2}^{2}\\|u_{1}-u_{2}\\|_{Z}^{2}2C_{0}^{2}$
		$\displaystyle=\frac{C_{3}^{2}}{16C_{1}^{2}}\\|u_{1}-u_{2}\\|_{Z}^{2}$

if we choose $T$ small enough.

Combining all the estimates above, we get

\|(e^{2u_{2}-2u_{1}}-1)\partial_{t}h\|_{X}\leq\frac{C_{3}}{2C_{1}}\|u_{1}-u_{2}\|_{Z}

which proves the lemma. ∎

4.1. The construction $S_{1}$

Define an operator

S_{1}:Y_{T}\times Z_{T}\to Y_{T}

by $S_{1}(f,u)=h$ where $h\in Y_{T}$ is the unique solution of

(23)

\displaystyle(\partial_{t}-e^{-2u}\Delta)h=e^{-2u}A_{f}(df,df)\qquad h(0)=f_{0}.

Lemma 4.2.

Fix $f_{0}\in W^{3,2}(\Sigma)$ . Then there is $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ , $S_{1}$ restricts to an operator $S_{1}:B\times B^{\prime}\to B$ .

Proof.

We also can assume $\|A\|,\|DA\|,\|D^{2}A\|,\|D^{3}A\|\leq{c}$ where ${c}$ depends only on the geometry of $N$ . Then the vector-valued function $A_{f}(df,df)$ satisfies the pointwise bound $|A_{f}(df,df)|^{2}\leq{c}|df|^{4}$ . Fix $f\in B$ and $u\in B^{\prime}$ .

Now we estimate $X$ norm of

g=e^{-2u(t)}A_{f}(df,df).

First, $|g|^{2}\leq{c}|df|^{4}$ , so $\|g\|_{L^{2}}^{2}\leq{c}\|f\|_{Y}^{4}|\Sigma|T$ . Hence if we choose $T$ small enough, we have $\|g\|_{L^{2}}^{2}\leq\frac{\delta^{2}}{6C_{1}^{2}}$ . Next, compute $|\nabla^{2}g|^{2}$ .

	$\displaystyle\|\nabla^{2}g\|$	$\displaystyle\leq{c}\|df\|^{2}\|\nabla^{2}u\|+{c}\|df\|^{2}\|\nabla u\|^{2}+{c}\|df\|^{3}\|\nabla u\|+{c}\|\nabla df\|\|df\|\|\nabla u\|$
		$\displaystyle\quad+{c}\|df\|^{4}+{c}\|df\|^{2}\|\nabla df\|+{c}\|\nabla^{3}f\|\|df\|+{c}\|\nabla df\|^{2}.$

So, using Young’s inequality

{c}\|df\|_{C^{0}}^{2}\iint|\nabla df|^{2}|\nabla u|^{2}\leq{c}\|df\|_{C^{0}}^{4}\iint|\nabla u|^{4}+{c}\iint|\nabla df|^{4},

we get, by (8), (9), (10), (12) and (20),

	$\displaystyle\\|\nabla^{2}g\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{4}\iint(\|\nabla^{2}u\|^{2}+\|\nabla u\|^{4})+{c}\\|df\\|_{C^{0}}^{6}\iint\|\nabla u\|^{2}$
		$\displaystyle\quad+{c}\\|df\\|_{C^{0}}^{2}\iint\|\nabla df\|^{2}\|\nabla u\|^{2}+{c}\\|df\\|_{C^{0}}^{8}\|\Sigma\|T+{c}\\|df\\|_{C^{0}}^{4}\iint\|\nabla df\|^{2}$
		$\displaystyle\quad+{c}\\|df\\|_{C^{0}}^{2}\iint\|\nabla^{3}f\|^{2}+{c}\iint\|\nabla df\|^{4}$
		$\displaystyle\leq{c}\\|f\\|_{Y}^{4}\left(\max_{0\leq t\leq T}\\|\nabla^{2}u(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{4}(\Sigma)}^{4}\right)T$
		$\displaystyle\quad+{c}\\|f\\|_{Y}^{6}\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{2}(\Sigma)}^{2}T+{c}\\|f\\|_{Y}^{8}\|\Sigma\|T$
		$\displaystyle\quad+{c}\\|f\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla df(t)\\|_{L^{2}(\Sigma)}^{2}T+{c}\\|f\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla^{3}f(t)\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\quad+{c}\max_{0\leq t\leq T}\\|\nabla df\\|_{L^{4}(\Sigma)}^{4}T$
		$\displaystyle\leq\frac{\delta^{2}}{6C_{1}^{2}}$

if we choose $T$ small enough. Finally,

\displaystyle|g_{t}|

\displaystyle\leq{c}|df|^{2}|u_{t}|+{c}|df|^{2}|f_{t}|+{c}|df_{t}||df|

and

	$\displaystyle\\|g_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{4}\iint\|u_{t}\|^{2}+\|f_{t}\|^{2}+{c}\\|df\\|_{C^{0}}^{2}\iint\|df_{t}\|^{2}$
		$\displaystyle\leq{c}\\|f\\|_{Y}^{4}\left(\max_{0\leq t\leq T}\\|u_{t}(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|f_{t}(t)\\|_{L^{2}(\Sigma)}^{2}\right)T$
		$\displaystyle\quad+{c}\\|f\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla f_{t}\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\leq\frac{\delta^{2}}{6C_{1}^{2}}$

if we choose $T$ small enough.

Therefore, if we choose $T$ small enough, we have $\|g\|_{X}^{2}\leq\frac{\delta^{2}}{2C_{1}^{2}}$ . Noting that $S(f)-h_{0}=h-h_{0}$ satisfies

(\partial_{t}-e^{-2u}\Delta)(h-h_{0})=g+(e^{-2u}-1)\Delta h_{0}\qquad(h-h_{0})(0)=0.

The bounds (16) give

	$\displaystyle\\|S(f)-h_{0}\\|_{Y}^{2}\$	$\displaystyle\leq\ C_{1}^{2}\left(\\|g\\|_{X}^{2}+\\|(e^{-2u}-1)\Delta h_{0}\\|_{X}^{2}\right)$
		$\displaystyle=C_{1}^{2}\left(\\|g\\|_{X}^{2}+\\|(e^{-2u}-1)\partial_{t}h_{0}\\|_{X}^{2}\right)$

because $h_{0}$ satisfies (17).

Now by Lemma 4.1 with $h=h_{0}$ , $u_{1}=u$ , $u_{2}=0$ ,

\|(e^{-2u}-1)\partial_{t}h_{0}\|_{X}\leq\frac{C_{3}}{2C_{1}}\|u\|_{Z}\leq\frac{\delta}{2C_{1}}.

This implies

\|S(f)-h_{0}\|_{Y}^{2}\ \leq\ C_{1}^{2}\left(\frac{\delta^{2}}{2C_{1}^{2}}+\frac{\delta^{2}}{4C_{1}^{2}}\right)\leq\delta^{2}.

Therefore $S(f)\in B$ . ∎

Lemma 4.3.

Fix $f_{0}\in W^{3,2}(\Sigma)$ and $u\in B^{\prime}$ . Then there is an $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ and for each $f_{1},f_{2}\in B$ ,

(24)

\displaystyle\|S_{1}(f_{1},u)-S_{1}(f_{2},u)\|_{Y}\leq\frac{1}{3}\|f_{1}-f_{2}\|_{Y}.

Proof.

Set $h_{i}=S_{1}(f_{i},u)$ and $g_{i}=e^{-2u}A_{f_{i}}(df_{i},df_{i})$ for $i=1,2$ and subtracting, the function $h_{1}-h_{2}$ satisfies

(\partial_{t}-e^{-2u}\Delta)(h_{1}-h_{2})=g_{1}-g_{2}\qquad(h_{1}-h_{2})(0)=0.

Hence (15) gives a bound

(25)

\displaystyle\|h_{1}-h_{2}\|_{Y}\leq C_{1}\|g_{1}-g_{2}\|_{X}.

Next, we have

	$\displaystyle g_{1}-g_{2}\$	$\displaystyle=\ e^{-2u}(A_{f_{1}}-A_{f_{2}})(df_{1},df_{1})+e^{-2u}A_{f_{2}}(df_{1}+df_{2},df_{1}-df_{2})$
		$\displaystyle=I+II.$

So, there is a constant ${c}$ with

|g_{1}-g_{2}|^{2}\ \leq\ {c}|{f_{1}}-f_{2}|^{2}|df_{1}|^{4}+{c}|df_{1}-df_{2}|^{2}\left(|df_{1}|^{2}+|df_{2}|^{2}\right).

Integrating and applying Holder’s inequality, (8), and (20) gives

	$\displaystyle\\|g_{1}-g_{2}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|{f_{1}}-f_{2}\\|_{C^{0}}^{2}\iint\|df_{1}\|^{4}\ +\ {c}\\|df_{1}-df_{2}\\|_{L^{4}}^{2}\left(\\|df_{1}\\|_{L^{4}}^{2}+\\|df_{2}\\|_{L^{4}}^{2}\right)$
		$\displaystyle\leq{c}\\|{f_{1}}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{4}\|\Sigma\|T\ +\ {c}\\|{f_{1}}-f_{2}\\|_{Y}^{2}(\\|f_{1}\\|_{Y}^{2}+\\|f_{2}\\|_{Y}^{2})\|\Sigma\|^{1/2}T^{1/2}$
		$\displaystyle\leq\frac{1}{27C_{1}^{2}}\,\\|{f_{1}}-f_{2}\\|_{Y}^{2}$

if we choose $T$ small enough.

For $\nabla^{2}(g_{1}-g_{2})$ , first note that

	$\displaystyle\nabla(A_{f_{1}}-A_{f_{2}})$	$\displaystyle=DA_{f_{1}}df_{1}-DA_{f_{2}}df_{2}=(DA_{f_{1}}-DA_{f_{2}})df_{1}+DA_{f_{2}}(df_{1}-df_{2})$
	$\displaystyle\nabla(DA_{f_{1}}-DA_{f_{2}})$	$\displaystyle=(D^{2}A_{f_{1}}-D^{2}A_{f_{2}})df_{1}+D^{2}A_{f_{2}}(df_{1}-df_{2}).$

So, we get

	$\displaystyle\|\nabla^{2}I\|$	$\displaystyle\leq{c}\|f_{1}-f_{2}\|\|df_{1}\|^{2}\|\nabla^{2}u\|+{c}\|f_{1}-f_{2}\|\|df_{1}\|^{2}\|\nabla u\|^{2}+{c}\|f_{1}-f_{2}\|\|df_{1}\|^{3}\|\nabla u\|$
		$\displaystyle\qquad+{c}\|df_{1}-df_{2}\|\|df_{1}\|^{2}\|\nabla u\|+{c}\|f_{1}-f_{2}\|\|\nabla df_{1}\|\|df_{1}\|\|\nabla u\|$
		$\displaystyle\quad+{c}\|f_{1}-f_{2}\|\|df_{1}\|^{4}+{c}\|f_{1}-f_{2}\|\|\nabla df_{1}\|\|df_{1}\|^{2}$
		$\displaystyle\qquad+{c}\|df_{1}-df_{2}\|\|df_{2}\|\|df_{1}\|^{2}+{c}\|\nabla df_{1}-\nabla df_{2}\|\|df_{1}\|^{2}$
		$\displaystyle\qquad+{c}\|df_{1}-df_{2}\|\|\nabla df_{1}\|\|df_{1}\|$
		$\displaystyle\quad+{c}\|f_{1}-f_{2}\|\|\nabla^{3}f_{1}\|\|df_{1}\|+{c}\|f_{1}-f_{2}\|\|\nabla df_{1}\|^{2}.$

Using (8), (9), (10), (12), (13), and using Young’s inequality, we can estimate it term by term.

	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|df_{1}\|^{4}\|\nabla^{2}u\|^{2}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla^{2}u(t)\\|_{L^{2}(\Sigma)}^{2}T$
	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|df_{1}\|^{4}\|\nabla u\|^{4}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{4}(\Sigma)}^{4}T$
	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|df_{1}\|^{6}\|\nabla u\|^{2}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{6}\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{2}(\Sigma)}^{2}T$
	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|\nabla df_{1}\|^{2}\|df_{1}\|^{2}\|\nabla u\|^{2}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{4}(\Sigma)}^{4}T$
		$\displaystyle\quad+\\|f_{1}-f_{2}\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla df_{1}(t)\\|_{L^{4}(\Sigma)}^{4}T$

	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|df_{1}\|^{8}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{8}\|\Sigma\|T$
	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|\nabla df_{1}\|^{2}\|df_{1}\|^{4}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla df_{1}(t)\\|_{L^{2}(\Sigma)}^{2}T$
	$\displaystyle\iint\|df_{1}-df_{2}\|^{2}\|df_{2}\|^{2}\|df_{1}\|^{4}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{4}\|\Sigma\|T$
	$\displaystyle\iint\|\nabla df_{1}-\nabla df_{2}\|^{2}\|df_{1}\|^{4}$	$\displaystyle\leq\\|f_{1}\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla df_{1}(t)-\nabla df_{2}(t)\\|_{L^{2}(\Sigma)}^{2}T$
	$\displaystyle\iint\|df_{1}-df_{2}\|^{2}\|\nabla df_{1}\|^{2}\|df_{1}\|^{2}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla df_{1}(t)\\|_{L^{2}(\Sigma)}^{2}T$
	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|\nabla^{3}f_{1}\|^{2}\|df_{1}\|^{2}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\\|f_{1}\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla^{3}f_{1}(t)\\|_{L^{2}(\Sigma)}^{2}T$
	$\displaystyle\iint\|f_{1}-f_{2}\|^{2}\|\nabla df_{1}\|^{4}$	$\displaystyle\leq\\|f_{1}-f_{2}\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla df_{1}(t)\\|_{L^{4}(\Sigma)}^{4}T.$

Hence, using (20), if we choose $T$ small enough, we get

\displaystyle\|\nabla^{2}I\|_{L^{2}}^{2}

\displaystyle\leq\frac{1}{54C_{1}^{2}}\|f_{1}-f_{2}\|_{Y}^{2}.

We obtain similar result for $II$ if we choose $T$ small enough:

\displaystyle\|\nabla^{2}II\|_{L^{2}}^{2}

\displaystyle\leq\frac{1}{54C_{1}^{2}}\|f_{1}-f_{2}\|_{Y}^{2}.

Hence, we obtain that $\|\nabla^{2}(g_{1}-g_{2})\|_{L^{2}}^{2}\leq\frac{1}{27C_{1}^{2}}\|f_{1}-f_{2}\|_{Y}^{2}$ .

Finally, compute $\partial_{t}(g_{1}-g_{2})$ . As above, note that

\partial_{t}(A_{f_{1}}-A_{f_{2}})=DA_{f_{1}}\partial_{t}f_{1}-DA_{f_{2}}\partial_{t}f_{2}=(DA_{f_{1}}-DA_{f_{2}})\partial_{t}f_{1}+DA_{f_{2}}(\partial_{t}(f_{1}-f_{2})).

So,

	$\displaystyle\|\partial_{t}(g_{1}-g_{2})\|$	$\displaystyle\leq{c}\|f_{1}-f_{2}\|\|df_{1}\|^{2}\|u_{t}\|+{c}\|f_{1}-f_{2}\|\|\partial_{t}f_{1}\|\|df_{1}\|^{2}+{c}\|\partial_{t}(f_{1}-f_{2})\|\|df_{1}\|^{2}$
		$\displaystyle\quad+{c}\|f_{1}-f_{2}\|\|\partial_{t}df_{1}\|\|df_{1}\|$
		$\displaystyle\quad+{c}\|df_{1}+df_{2}\|\|df_{1}-df_{2}\|\|u_{t}\|+{c}\|\partial_{t}f_{2}\|\|df_{1}+df_{2}\|\|df_{1}-df_{2}\|$
		$\displaystyle\quad+{c}\|\partial_{t}(df_{1}+df_{2})\|\|df_{1}-df_{2}\|+{c}\|df_{1}+df_{2}\|\|\partial_{t}(df_{1}-df_{2})\|.$

Similar with above, by (8), (9), (10), (12), (13) and (20),

	$\displaystyle\\|\partial_{t}$	$\displaystyle(g_{1}-g_{2})\\|_{L^{2}}^{2}\leq{c}\\|f_{1}\\|_{Y}^{4}\\|f_{1}-f_{2}\\|_{Y}^{2}\left(\max_{0\leq t\leq T}\\|u_{t}(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|\partial_{t}f_{1}(t)\\|_{L^{2}(\Sigma)}^{2}\right)T$
		$\displaystyle\,\,+{c}\\|f_{1}\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\partial_{t}(f_{1}(t)-f_{2}(t))\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\,\,+{c}\\|f_{1}\\|_{Y}^{2}\\|f_{1}-f_{2}\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\partial_{t}df_{1}(t)\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\,\,+{c}(\\|f_{1}\\|_{Y}^{2}+\\|f_{2}\\|_{Y}^{2})\\|f_{1}-f_{2}\\|_{Y}^{2}\left(\max_{0\leq t\leq T}\\|u_{t}(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|\partial_{t}f_{2}(t)\\|_{L^{2}(\Sigma)}^{2}\right)T$
		$\displaystyle\,\,+{c}\\|f_{1}-f_{2}\\|_{Y}^{2}\left(\max_{0\leq t\leq T}\\|\partial_{t}df_{1}(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|\partial_{t}df_{2}(t)\\|_{L^{2}(\Sigma)}^{2}\right)T$
		$\displaystyle\,\,+{c}(\\|f_{1}\\|_{Y}^{2}+\\|f_{2}\\|_{Y}^{2})\max_{0\leq t\leq T}\\|\partial_{t}(df_{1}(t)-df_{2}(t))\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\leq\frac{1}{27C_{1}^{2}}\\|f_{1}-f_{2}\\|_{Y}^{2}$

if we choose $T$ small enough.

Combine all of them,

\|h_{1}-h_{2}\|_{Y}\leq C_{1}\|g_{1}-g_{2}\|_{X}\leq\frac{1}{3}\|f_{1}-f_{2}\|_{Y}

which proves the lemma. ∎

Lemma 4.4.

Fix $f_{0}\in W^{3,2}(\Sigma)$ and $f\in B$ . Then there is an $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ and for each $u_{1},u_{2}\in B^{\prime}$ ,

(26)

\displaystyle\|S_{1}(f,u_{1})-S_{1}(f,u_{2})\|_{Y}\leq\frac{C_{3}}{2}\|u_{1}-u_{2}\|_{Z}.

Proof.

Set $h_{i}=S_{1}(f,u_{i})$ . Multiplying $e^{2u_{i}}$ to the equation for $h_{i}$ respectively and subtracting them gives

	$\displaystyle e^{2u_{1}}\partial_{t}(h_{1}-h_{2})-\Delta(h_{1}-h_{2})$	$\displaystyle=-(e^{2u_{1}}-e^{2u_{2}})\partial_{t}h_{2}$
	$\displaystyle(\partial_{t}-e^{-2u_{1}}\Delta)(h_{1}-h_{2})$	$\displaystyle=(e^{2u_{2}-2u_{1}}-1)\partial_{t}h_{2}.$

So, $h_{1}-h_{2}$ satisfies the estimate from (16), and by Lemma 4.1,

\|h_{1}-h_{2}\|_{Y}\leq C_{1}\|(e^{2u_{2}-2u_{1}}-1)\partial_{t}h_{2}\|_{X}\leq\frac{C_{3}}{2}\|u_{1}-u_{2}\|_{Z}

if we choose $T$ small enough.

∎

4.2. The construction $S_{2}$

Define an operator

S_{2}:Y_{T}\times Z_{T}\to Z_{T}

by $S_{2}(f,u)=v$ where $v\in Z_{T}$ is the unique solution of

(27)

\displaystyle\partial_{t}v=b|df|^{2}e^{-2u}-a\qquad v(0)=0.

Lemma 4.5.

In the above definition, $v\in Z_{T}$ .

Proof.

From (27), we directly get

(28)

v(t)=\int_{0}^{t}(b|df|^{2}e^{-2u}-a).

So, $\|v\|_{L^{2}}$ and $\|v_{t}\|_{L^{2}}$ is trivially bounded if $f\in Y_{T}$ and $u\in Z_{T}$ . (Because $u\in Z_{T}$ , we have $e^{-2u}$ is pointwise uniformly bounded by $e^{C\|u\|_{Z}}$ .) Applying Cauchy-Schwarz, we obtain the pointwise bound

	$\displaystyle\|\nabla^{3}v\|^{2}$	$\displaystyle=\left\|b\int_{0}^{T}\nabla^{2}\left(\langle\nabla df,df\rangle e^{-2u}-2\|df\|^{2}e^{-2u}\nabla u\right)\right\|^{2}$
		$\displaystyle\leq{c}T\int_{0}^{T}\Big{(}\|\nabla^{4}f\|^{2}\|df\|^{2}+\|\nabla^{3}f\|^{2}\|\nabla df\|^{2}+\|\nabla^{3}f\|^{2}\|df\|^{2}\|\nabla u\|^{2}+\|\nabla^{2}f\|^{4}\|\nabla u\|^{2}$
		$\displaystyle\qquad+\|\nabla^{2}f\|^{2}\|df\|^{2}(\|\nabla u\|^{4}+\|\nabla^{2}u\|^{2})+\|df\|^{4}(\|\nabla u\|^{6}+\|\nabla^{2}u\|^{2}\|\nabla u\|^{2}+\|\nabla^{3}u\|^{2})\Big{)}$

	$\displaystyle\\|\nabla^{3}v\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}T^{2}\\|df\\|_{C^{0}}^{2}\\|\nabla^{4}f\\|_{L^{2}}^{2}+{c}T^{2}\\|\nabla^{3}f\\|_{L^{4}}^{2}\\|\nabla^{2}f\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{2}\\|\nabla^{3}f\\|_{L^{4}}^{2}\\|\nabla u\\|_{L^{4}}^{2}+{c}T^{2}\\|\nabla^{2}f\\|_{L^{8}}^{4}\\|\nabla u\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{2}\\|\nabla^{2}f\\|_{L^{4}}^{2}(\\|\nabla u\\|_{L^{8}}^{4}+\\|\nabla^{2}u\\|_{L^{4}}^{2})$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{4}(\\|\nabla u\\|_{L^{6}}^{6}+\\|\nabla^{2}u\\|_{L^{4}}^{2}\\|\nabla u\\|_{L^{4}}^{2}+\\|\nabla^{3}u\\|_{L^{2}}^{2})$

which is bounded if $f\in Y_{T}$ and $u\in Z_{T}$ .

Finally, compute $\nabla v_{t}$ from (27).

	$\displaystyle\|\nabla v_{t}\|^{2}$	$\displaystyle\leq{c}\|\nabla df\|\|df\|+\|df\|^{2}\|\nabla u\|$
	$\displaystyle\\|\nabla v_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{2}\\|\nabla df\\|_{L^{2}}^{2}+{c}\\|df\\|_{C^{0}}^{4}\\|\nabla u\\|_{L^{2}}^{2}$

which is bounded if $f\in Y_{T}$ and $u\in Z_{T}$ . ∎

In fact, we can show further.

Lemma 4.6.

Fix $f_{0}\in W^{3,2}(\Sigma)$ . Then there is $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ , $S_{2}$ restricts to an operator $S_{2}:B\times B^{\prime}\to B^{\prime}$ .

Proof.

From previous calculation, we have

\|\nabla^{3}v\|_{L^{2}}^{2}\leq{c}T^{2}\|f\|_{Y}^{4}(1+\|u\|_{Z}^{2}+\|u\|_{Z}^{4}+\|u\|_{Z}^{6}).

So, if we choose $T$ small enough, we get $\|\nabla^{3}v\|_{L^{2}}^{2}\leq\frac{{\delta^{\prime}}^{2}}{4}$ . Also, because $|v_{t}|\leq{c}(\|df\|_{C^{0}}+1)$ and $|v|\leq{c}T(\|df\|_{C^{0}}+1)$ , we can make $\|v_{t}\|_{L^{2}}^{2},\|v\|_{L^{2}}^{2}\leq\frac{{\delta^{\prime}}^{2}}{4}$ if we choose $T$ small. Finally,

\|\nabla v_{t}\|_{L^{2}}^{2}\leq{c}\|df\|_{C^{0}}^{2}\max_{0\leq t\leq T}\|\nabla df(t)\|_{L^{2}(\Sigma)}^{2}\,T+{c}\|df\|_{C^{0}}^{4}\max_{0\leq t\leq T}\|\nabla u(t)\|_{L^{2}(\Sigma)}^{2}\,T

so if we choose $T$ small enough, we get $\|\nabla v_{t}\|_{L^{2}}^{2}\leq\frac{{\delta^{\prime}}^{2}}{4}$ . This proves the lemma. ∎

Lemma 4.7.

Fix $f_{0}\in W^{3,2}(\Sigma)$ and $u\in B^{\prime}$ . Then there is an $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ and for each $f_{1},f_{2}\in B$ ,

(29)

\displaystyle\|S_{2}(f_{1},u)-S_{2}(f_{2},u)\|_{Z}\leq T^{1/4}\|f_{1}-f_{2}\|_{Y}.

Proof.

Set $v_{i}=S_{2}(f_{i},u)$ . Then from (27), subtracting them gives

	$\displaystyle(v_{1}-v_{2})_{t}$	$\displaystyle=b(\|df_{1}\|^{2}-\|df_{2}\|^{2})e^{-2u}=be^{-2u}\langle df_{1}+df_{2},df_{1}-df_{2}\rangle$
	$\displaystyle v_{1}-v_{2}$	$\displaystyle=b\int_{0}^{t}e^{-2u}\langle df_{1}+df_{2},df_{1}-df_{2}\rangle.$

So,

	$\displaystyle\\|v_{1}-v_{2}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}T^{2}(\\|df_{1}\\|_{C^{0}}^{2}+\\|df_{2}\\|_{C^{0}}^{2})\\|df_{1}-df_{2}\\|_{L^{2}}^{2}$
		$\displaystyle\leq\frac{\sqrt{T}}{4}\\|f_{1}-f_{2}\\|_{Y}^{2}$

if we choose $T$ small enough. Also,

	$\displaystyle\\|(v_{1}-v_{2})_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}(\\|f_{1}\\|_{Y}^{2}+\\|f_{2}\\|_{Y}^{2})\max_{0\leq t\leq T}\\|df_{1}(t)-df_{2}(t)\\|_{L^{2}(\Sigma)}^{2}\,T$
		$\displaystyle\leq\frac{\sqrt{T}}{4}\\|f_{1}-f_{2}\\|_{Y}^{2}$

if we choose $T$ small enough.

Next, compute $\nabla^{3}(v_{1}-v_{2})$ .

	$\displaystyle\nabla(v_{1}-v_{2})$	$\displaystyle=b\int_{0}^{t}e^{-2u}\Big{(}\langle\nabla(df_{1}+df_{2}),df_{1}-df_{2}\rangle+\langle df_{1}+df_{2},\nabla(df_{1}-df_{2})\rangle$
		$\displaystyle\qquad\qquad-\langle df_{1}+df_{2},df_{1}-df_{2}\rangle 2\nabla u\Big{)}.$

So,

	$\displaystyle\|\nabla^{3}(v_{1}-v_{2})\|$	$\displaystyle\leq{c}\int_{0}^{T}e^{-2u}\Big{(}\|\nabla^{3}(df_{1}+df_{2})\|\|df_{1}-df_{2}\|+\|\nabla^{2}(df_{1}+df_{2})\|\|\nabla(df_{1}-df_{2})\|$
		$\displaystyle\qquad+\|\nabla(df_{1}+df_{2})\|\|\nabla^{2}(df_{1}-df_{2})\|+\|df_{1}+df_{2}\|\|\nabla^{3}(df_{1}-df_{2})\|$
		$\displaystyle\quad+\|\nabla^{2}(df_{1}+df_{2})\|\|df_{1}-df_{2}\|\|\nabla u\|+\|\nabla(df_{1}+df_{2})\|\|\nabla(df_{1}-df_{2})\|\|\nabla u\|$
		$\displaystyle\qquad+\|df_{1}+df_{2}\|\|\nabla^{2}(df_{1}-df_{2})\|\|\nabla u\|$
		$\displaystyle\quad+\|\nabla(df_{1}+df_{2})\|\|df_{1}-df_{2}\|(\|\nabla u\|^{2}+\|\nabla^{2}u\|)$
		$\displaystyle\qquad+\|df_{1}+df_{2}\|\|\nabla(df_{1}-df_{2})\|(\|\nabla u\|^{2}+\|\nabla^{2}u\|)$
		$\displaystyle\quad+\|df_{1}+df_{2}\|\|df_{1}-df_{2}\|(\|\nabla u\|^{3}+\|\nabla^{2}u\|\|\nabla u\|+\|\nabla^{3}u\|)\Big{)}.$

Integrating over $\Sigma\times[0,T]$ gives

	$\displaystyle\\|\nabla^{3}$	$\displaystyle(v_{1}-v_{2})\\|_{L^{2}}^{2}$
		$\displaystyle\leq{c}T^{2}\\|df_{1}-df_{2}\\|_{C^{0}}^{2}\\|\nabla^{4}(f_{1}+f_{2})\\|_{L^{2}}^{2}+{c}T^{2}\\|\nabla^{3}(f_{1}+f_{2})\\|_{L^{4}}^{2}\\|\nabla^{2}(f_{1}-f_{2})\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}\\|\nabla^{2}(f_{1}+f_{2})\\|_{L^{4}}^{2}\\|\nabla^{3}(f_{1}-f_{2})\\|_{L^{4}}^{2}+{c}T^{2}\\|df_{1}+df_{2}\\|_{C^{0}}^{2}\\|\nabla^{4}(f_{1}-f_{2})\\|_{L^{2}}^{2}$
		$\displaystyle+{c}T^{2}\\|df_{1}-df_{2}\\|_{C^{0}}^{2}\\|\nabla^{3}(f_{1}+f_{2})\\|_{L^{4}}^{2}\\|\nabla u\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}\\|\nabla^{2}(f_{1}-f_{2})\\|_{L^{4}}^{2}\\|\nabla^{2}(f_{1}+f_{2})\\|_{L^{8}}^{4}\\|\nabla u\\|_{L^{8}}^{4}$
		$\displaystyle\quad+{c}T^{2}\\|df_{1}+df_{2}\\|_{C^{0}}^{2}\\|\nabla^{3}(f_{1}-f_{2})\\|_{L^{4}}^{2}\\|\nabla u\\|_{L^{4}}^{2}$
		$\displaystyle+{c}T^{2}\\|df_{1}-df_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}(f_{1}+f_{2})\\|_{L^{4}}^{2}(\\|\nabla u\\|_{L^{8}}^{4}+\\|\nabla^{2}u\\|_{L^{4}}^{2})$
		$\displaystyle\quad+{c}T^{2}\\|df_{1}+df_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}(f_{1}-f_{2})\\|_{L^{4}}^{2}(\\|\nabla u\\|_{L^{8}}^{4}+\\|\nabla^{2}u\\|_{L^{4}}^{2})$
		$\displaystyle+{c}T^{2}\\|df_{1}-df_{2}\\|_{C^{0}}^{2}\\|df_{1}+df_{2}\\|_{C^{0}}^{2}(\\|\nabla u\\|_{L^{6}}^{6}+\\|\nabla^{2}u\\|_{L^{4}}^{2}\\|\nabla u\\|_{L^{4}}^{2}+\\|\nabla^{3}u\\|_{L^{2}}^{2})$
		$\displaystyle\leq\frac{\sqrt{T}}{4}\\|f_{1}-f_{2}\\|_{Y}^{2}$

if we choose $T$ small enough.

Finally consider $\nabla(v_{1}-v_{2})_{t}$ .

	$\displaystyle\nabla(v_{1}-v_{2})_{t}$	$\displaystyle=be^{-2u}\left(\langle\nabla(df_{1}+df_{2}),df_{1}-df_{2}\rangle+\langle df_{1}+df_{2},\nabla(df_{1}-df_{2})\rangle\right.$
		$\displaystyle\qquad\qquad\left.-\langle df_{1}+df_{2},df_{1}-df_{2}\rangle 2\nabla u\right).$

So,

	$\displaystyle\\|\nabla(v_{1}-v_{2})_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df_{1}-df_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla^{2}(f_{1}(t)+f_{2}(t))\\|_{L^{2}(\Sigma)}^{2}\,T$
		$\displaystyle\quad+{c}\\|df_{1}+df_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla^{2}(f_{1}(t)-f_{2}(t))\\|_{L^{2}(\Sigma)}^{2}\,T$
		$\displaystyle\quad+{c}\\|df_{1}+df_{2}\\|_{C^{0}}^{2}\\|df_{1}-df_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{2}(\Sigma)}^{2}\,T$
		$\displaystyle\leq\frac{\sqrt{T}}{4}\\|f_{1}-f_{2}\\|_{Y}^{2}$

if we choose $T$ small enough.

In summary, we get

\|v_{1}-v_{2}\|_{Z}^{2}\leq\sqrt{T}\|f_{1}-f_{2}\|_{Y}^{2}

which proves the lemma.

∎

Lemma 4.8.

Fix $f_{0}\in W^{3,2}(\Sigma)$ and $f\in B$ . Then there is an $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ and for each $u_{1},u_{2}\in B^{\prime}$ ,

(30)

\displaystyle\|S_{2}(f,u_{1})-S_{2}(f,u_{2})\|_{Z}\leq\frac{1}{3}\|u_{1}-u_{2}\|_{Z}.

Proof.

Set $v_{i}=S_{2}(f,u_{i})$ . Subtracting them gives

	$\displaystyle(v_{1}-v_{2})_{t}$	$\displaystyle=b\|df\|^{2}(e^{-2u_{1}}-e^{-2u_{2}})$
	$\displaystyle v_{1}-v_{2}$	$\displaystyle=b\int_{0}^{t}\|df\|^{2}(e^{-2u_{1}}-e^{-2u_{2}}).$

Using $|e^{-2u_{1}}-e^{-2u_{2}}|\leq{c}|u_{1}-u_{2}|$ , we have

	$\displaystyle\\|v_{1}-v_{2}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}T^{2}\\|df\\|_{C^{0}}^{4}\\|u_{1}-u_{2}\\|_{L^{2}}^{2}$
	$\displaystyle\\|(v_{1}-v_{2})_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{4}\max_{0\leq t\leq T}\\|u_{1}(t)-u_{2}(t)\\|_{L^{2}(\Sigma)}^{2}\,T$

so if we choose $T$ small enough, we have that $\|v_{1}-v_{2}\|_{L^{2}}^{2},\|(v_{1}-v_{2})_{t}\|_{L^{2}}^{2}\leq\frac{1}{36}\|u_{1}-u_{2}\|_{Z}^{2}$ .

Next, compute $\nabla^{3}(v_{1}-v_{2})$ .

\displaystyle\nabla(v_{1}-v_{2})

\displaystyle=b\int_{0}^{t}\Big{(}2\langle\nabla df,df\rangle(e^{-2u_{1}}-e^{-2u_{2}})-|df|^{2}(e^{-2u_{1}}-e^{-2u_{2}})2(\nabla u_{1}-\nabla u_{2})\Big{)}.

So,

	$\displaystyle\|\nabla^{3}$	$\displaystyle(v_{1}-v_{2})\|\leq{c}\int_{0}^{T}\Big{(}\|\nabla^{4}f\|\|df\|\|u_{1}-u_{2}\|+\|\nabla^{3}f\|\|\nabla^{2}f\|\|u_{1}-u_{2}\|$
		$\displaystyle\quad+(\|\nabla^{3}f\|\|df\|+\|\nabla^{2}f\|^{2})\|u_{1}-u_{2}\|\|\nabla(u_{1}-u_{2})\|$
		$\displaystyle\quad+\|\nabla^{2}f\|\|df\|\|u_{1}-u_{2}\|(\|\nabla(u_{1}-u_{2})\|^{2}+\|\nabla^{2}(u_{1}-u_{2})\|)$
		$\displaystyle\quad+\|df\|^{2}\|u_{1}-u_{2}\|(\|\nabla(u_{1}-u_{2})\|^{3}+\|\nabla^{2}(u_{1}-u_{2})\|\|\nabla(u_{1}-u_{2})\|+\|\nabla^{3}(u_{1}-u_{2})\|)\Big{)}.$

Now we integrate over $\Sigma\times[0,T]$ .

	$\displaystyle\\|\nabla^{3}$	$\displaystyle(v_{1}-v_{2})\\|_{L^{2}}^{2}$
		$\displaystyle\leq{c}T^{2}\\|df\\|_{C^{0}}^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{4}f\\|_{L^{2}}^{2}+{c}T^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{3}f\\|_{L^{4}}^{2}\\|\nabla^{2}f\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}(\\|df\\|_{C^{0}}^{2}\\|\nabla^{3}f\\|_{L^{4}}^{2}+\\|\nabla^{2}f\\|_{L^{8}}^{4})\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla(u_{1}-u_{2})\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}f\\|_{L^{4}}^{2}(\\|\nabla(u_{1}-u_{2})\\|_{L^{8}}^{4}+\\|\nabla^{2}(u_{1}-u_{2})\\|_{L^{4}}^{2})$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{4}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla(u_{1}-u_{2})\\|_{L^{6}}^{6}$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{4}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}(u_{1}-u_{2})\\|_{L^{4}}^{2}\\|\nabla(u_{1}-u_{2})\\|_{L^{4}}^{2}$
		$\displaystyle\quad+{c}T^{2}\\|df\\|_{C^{0}}^{4}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{3}(u_{1}-u_{2})\\|_{L^{2}}^{2}$
		$\displaystyle\leq\frac{1}{36}\\|u_{1}-u_{2}\\|_{Z}^{2}$

if we choose $T$ small enough.

Finally,

\displaystyle|\nabla(v_{1}-v_{2})_{t}|

\displaystyle\leq{c}|\nabla df||df||u_{1}-u_{2}|+|df|^{2}|u_{1}-u_{2}||\nabla(u_{1}-u_{2})|

	$\displaystyle\\|\nabla(v_{1}-v_{2})_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla^{2}f(t)\\|_{L^{2}(\Sigma)}^{2}\,T$
		$\displaystyle\quad+{c}\\|df\\|_{C^{0}}^{4}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla(u_{1}(t)-u_{2}(t))\\|_{L^{2}(\Sigma)}^{2}\,T$
		$\displaystyle\leq\frac{1}{36}\\|u_{1}-u_{2}\\|_{Z}^{2}$

if we choose $T$ small enough.

In summary, we get

\|v_{1}-v_{2}\|_{Z}^{2}\leq\frac{1}{9}\|u_{1}-u_{2}\|_{Z}^{2}

which proves the lemma.

∎

5. Existence of fixed point

Because $Y_{T}$ and $Z_{T}$ are Hilbert space, $Y_{T}\times Z_{T}$ is also a Hilbert space and we can equip the norm

(31)

\|(f,u)\|_{Y\times Z}=(C_{3})^{-1}\|f\|_{Y}+\|u\|_{Z}.

Define an operator $\mathcal{S}:Y_{T}\times Z_{T}\rightarrow Y_{T}\times Z_{T}$ by

(32)

\mathcal{S}(f,u)=(S_{1}(f,u),S_{2}(f,u)).

Proposition 5.1.

Fix $f_{0}\in W^{3,2}(\Sigma)$ . Then there is an $T_{0}=T_{0}(C_{0},\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ ,

(a)

$\mathcal{S}$ restricts to an operator $\mathcal{S}:B\times B^{\prime}\to B\times B^{\prime}$ .

(b)

For each $f_{1},f_{2}\in B$ and $u_{1},u_{2}\in B^{\prime}$ ,

(33)

\displaystyle\|\mathcal{S}(f_{1},u_{1})-\mathcal{S}(f_{2},u_{2})\|_{Y\times Z}\leq\frac{5}{6}\|(f_{1},u_{1})-(f_{2}-u_{2})\|_{Y\times Z}.

Proof.

By Lemma 4.2 and Lemma 4.6, (a) is proved. For (b), using Lemmas 4.3, 4.4, 4.7, 4.8, there is $T_{0}=T_{0}(\delta,\delta^{\prime})>0$ such that for all $T\leq T_{0}$ ,

	$\displaystyle\\|\mathcal{S}(f_{1},u_{1})$	$\displaystyle-\mathcal{S}(f_{2},u_{2})\\|_{Y\times Z}$
		$\displaystyle=(C_{3})^{-1}\\|S_{1}(f_{1},u_{1})-S_{1}(f_{2},u_{2})\\|_{Y}+\\|S_{2}(f_{1},u_{1})-S_{2}(f_{2},u_{2})\\|_{Z}$
		$\displaystyle\leq(C_{3})^{-1}\\|S_{1}(f_{1},u_{1})-S_{1}(f_{2},u_{1})\\|_{Y}+(C_{3})^{-1}\\|S_{1}(f_{2},u_{1})-S_{1}(f_{2},u_{2})\\|_{Y}$
		$\displaystyle\quad+\\|S_{2}(f_{1},u_{1})-S_{2}(f_{2},u_{1})\\|_{Z}+\\|S_{2}(f_{2},u_{1})-S_{2}(f_{2},u_{2})\\|_{Z}$
		$\displaystyle\leq\frac{1}{3}(C_{3})^{-1}\\|f_{1}-f_{2}\\|_{Y}+\frac{1}{2}\\|u_{1}-u_{2}\\|_{Z}$
		$\displaystyle\quad+T^{1/4}\\|f_{1}-f_{2}\\|_{Y}+\frac{1}{3}\\|u_{1}-u_{2}\\|_{Z}$
		$\displaystyle\leq\frac{5}{6}\left((C_{3})^{-1}\\|f_{1}-f_{2}\\|_{Y}+\\|u_{1}-u_{2}\\|_{Z}\right)$
		$\displaystyle=\frac{5}{6}\\|(f_{1},u_{1})-(f_{2},u_{2})\\|_{Y\times Z}$

if $T^{1/4}\leq\frac{1}{2}(C_{3})^{-1}$ . ∎

Theorem 5.2.

(Short time existence for strong solution) There is $T_{0}>0$ such that there exists a smooth solution $(f,u)\in B\times B^{\prime}$ of (2b) on $\Sigma\times[0,T_{0}]$ .

Proof.

The existence of solution $f,u$ comes from 5.1. The fact $f(\Sigma\times[0,T_{0}])\subset N$ can be easily shown using nearest point projection, see for example [24]. Moreover, the operator $\partial_{t}-e^{-2u}\Delta$ is uniformly parabolic, so $|(\partial_{t}-e^{-2u}\Delta)f|\in L^{p}(\Sigma\times[0,T_{0}])$ for any $1\leq p<\infty$ , by standard parabolic theory. This implies

\nabla^{2}f,\partial_{t}f\in L^{p}(\Sigma\times[0,T_{0}])

for any $1\leq p<\infty$ .

Next, by direct computation from (2b), we have

e^{2u}=e^{-2at}\left(1+2b\int_{0}^{t}e^{2as}|df|^{2}\right)

hence

	$\displaystyle\nabla u$	$\displaystyle=e^{-2u-2at}2b\int_{0}^{t}e^{2as}\langle\nabla df,df\rangle$
	$\displaystyle\int\|\nabla u\|^{p}$	$\displaystyle\leq(4b)^{p}\int\left(\int_{0}^{t}\|\nabla df\|\|df\|\right)^{p}$
		$\displaystyle\leq(4b)^{p}t^{p-1}\int\int_{0}^{t}\|\nabla df\|^{p}\|df\|^{p}$

which implies $\nabla u\in L^{p}(\Sigma\times[0,T_{0}])$ for any $1\leq p<\infty$ . Now taking $\nabla$ in the equation (2a) to get

|(\partial_{t}-e^{-2u}\Delta)\nabla f|\leq C\left(|\nabla u||\Delta f|+|\nabla u||df|^{2}+|\nabla df||df|+|df|^{3}\right)\in L^{p}(\Sigma\times[0,T_{0}])

which implies

\nabla^{3}f,\partial_{t}(\nabla f)\in L^{p}(\Sigma\times[0,T_{0}])

for any $1\leq p<\infty$ .

Finally, from Sobolev embedding, we have $f,df\in C^{\alpha}(\Sigma\times[0,T_{0}])$ for some $\alpha>0$ . This implies $(\partial_{t}-e^{-2u}\Delta)f\in C^{\alpha,\alpha/2}(\Sigma\times[0,T_{0}])$ where $C^{\alpha,\alpha/2}$ is parabolic Hölder space of exponent $\alpha$ . Now by Schauder estimate and standard bootstrapping argument, we conclude that $f$ is smooth, so $u$ is. ∎

6. Local estimate

To get global weak solution, we will follow Struwe’s idea: Run the flow until singularity occurs. Then take weak limit as new initial condition, run the flow again. Keep going this process and we will have only finitely many singularities due to finiteness of the energy. Because our flow is coupled, we need to re-establish the whole process with $f$ and $u$ . And this requires some condition on $b$ , which can be interpreted as the sensitiveness of the conformal evolution of the metric with respect to high energy density. Let $C_{N}>0$ be a constant only depending on the embedding $N\hookrightarrow\mathbb{R}^{L}$ such that $\|R^{N}\|,\|A\|,\|DA\|\leq C_{N}$ where $R^{N}$ is the Riemannian curvature tensor of $N$ . And from now on, assume $b\geq C_{N}^{2}$ .

6.1. Energy estimate

Now we establish local energy estimate. Fix $B_{2r}$ and let $\varphi$ be a cut-off function supported on $B_{2r}$ such that $\varphi\equiv 1$ on $B_{r}$ , $0\leq\varphi\leq 1$ and $|\nabla\varphi|\leq\frac{4}{r}$ .

Proposition 6.1.

For solutions $(f,u)$ of (2b), we have

(34)		$\displaystyle\int_{t_{1}}^{t_{2}}\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}+$	$\displaystyle\int_{B_{2r}}\|df\|^{2}\varphi^{2}(t_{2})-\int_{B_{2r}}\|df\|^{2}\varphi^{2}(t_{1})$
(34)			$\displaystyle\leq\frac{4^{2}}{ar^{2}}(e^{2at_{2}}-e^{2at_{1}})E_{0}.$

Especially, we have

(35)

E(B_{r},t_{2})-E(B_{2r},t_{1})\leq\frac{4^{2}}{2ar^{2}}(e^{2at_{2}}-e^{2at_{1}})E_{0}.

Proof.

From the equation (2a), multiplying $e^{2u}f_{t}\varphi^{2}$ gives

	$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}$	$\displaystyle=\int_{B_{2r}}\langle f_{t},\tau(f)\varphi^{2}\rangle$
		$\displaystyle=-\int_{B_{2r}}\langle df_{t},df\varphi^{2}\rangle-2\int_{B_{2r}}\langle f_{t},f_{i}\rangle\varphi\nabla_{i}\varphi$
		$\displaystyle\leq-\frac{1}{2}\frac{d}{dt}\int_{B_{2r}}\|df\|^{2}\varphi^{2}+\frac{1}{2}\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}+2\int_{B_{2r}}e^{-2u}\|df\|^{2}\|\nabla\varphi\|^{2}.$

So, we have

	$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}+\frac{d}{dt}\int_{B_{2r}}\|df\|^{2}\varphi^{2}$	$\displaystyle\leq 4\int_{B_{2r}}e^{-2u}\|df\|^{2}\|\nabla\varphi\|^{2}$
		$\displaystyle\leq 4\frac{4^{2}}{r^{2}}e^{2at}\int_{B_{2r}}\|df\|^{2}$
		$\displaystyle\leq 4\frac{4^{2}}{r^{2}}e^{2at}2E_{0}.$

Integrating from $t_{1}$ to $t_{2}$ gives the result. ∎

Lemma 6.2.

Furthermore, assume

\sup_{t_{1}\leq t\leq t_{2}}E(B_{2r},t)<\varepsilon_{1}.

Then we have

(36)		$\displaystyle\int_{t_{1}}^{t_{2}}\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}$	$\displaystyle\leq 4^{2}\varepsilon_{1}\left(1+\frac{e^{2at_{2}}-e^{2at_{1}}}{2ar^{2}}\right)$
(37)		$\displaystyle\int_{t_{1}}^{t_{2}}\int_{B_{2r}}\|f_{t}\|^{2}\varphi^{2}$	$\displaystyle\leq e^{2at_{2}}4^{2}\varepsilon_{1}\left(1+\frac{e^{2at_{2}}-e^{2at_{1}}}{2ar^{2}}\right).$

Proof.

The first equation directly comes from (34), by changing $E_{0}$ to $\varepsilon_{1}$ . Also, it is easy to see that

	$\displaystyle\int_{t_{1}}^{t_{2}}\int_{B_{2r}}\|f_{t}\|^{2}\varphi^{2}$	$\displaystyle=\int_{t_{1}}^{t_{2}}\int_{B_{2r}}e^{-2u}e^{2u}\|f_{t}\|^{2}\varphi^{2}\leq e^{2at_{2}}\int_{t_{1}}^{t_{2}}\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}$
		$\displaystyle\leq e^{2at_{2}}4^{2}\varepsilon_{1}\left(1+\frac{e^{2at_{2}}-e^{2at_{1}}}{2ar^{2}}\right).$

∎

6.2. Estimate for $\int|f_{t}|^{2}$

The next step is to get estimate for derivative of $\int_{B_{2r}}|f_{t}|^{2}\varphi^{2}$ , which will lead to the control of itself. For the future purpose, here we introduce more general version of it. For now, we need $p=0$ .

Proposition 6.3.

Let $(f,u)$ are solutions of (2b). For $p\geq 0$ , we have

(38)	$\displaystyle\frac{d}{dt}\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}$	$\displaystyle\leq 2a(p+1)\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}+4(p+2)\int_{B_{2r}}\|f_{t}\|^{p+2}\|\nabla\varphi\|^{2}$
		$\displaystyle-\frac{p+2}{4}\int_{B_{2r}}\|\nabla f_{t}\|^{2}\|f_{t}\|^{p}\varphi^{2}$
		$\displaystyle+\left((p+2)C_{N}+\frac{(p+2)C_{N}^{2}}{2}-2b(p+1)\right)\int_{B_{2r}}\|df\|^{2}\|f_{t}\|^{p+2}\varphi^{2}.$

Especially, we have

(39)		$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}(t)$	$\displaystyle\leq e^{2a(p+1)(t-t_{0})}$
(39)			$\displaystyle\quad\cdot\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}(t_{0})+4(p+2)\int_{t_{0}}^{t}\int_{B_{2r}}\|f_{t}\|^{p+2}\|\nabla\varphi\|^{2}\right).$

Proof.

By taking time-derivative to (2a), we have

(e^{2u}f_{t})_{t}=\Delta f_{t}+A(df,df)_{t}.

Taking inner product with $f_{t}|f_{t}|^{p}\varphi^{2}$ and integrating gives

	$\displaystyle\int\langle(e^{2u}f_{t})_{t},f_{t}\|f_{t}\|^{p}\varphi^{2}\rangle$	$\displaystyle=\int\langle\Delta f_{t},f_{t}\|f_{t}\|^{p}\varphi^{2}\rangle+\int\langle A(df,df)_{t},f_{t}\|f_{t}\|^{p}\varphi^{2}\rangle$
		$\displaystyle=-\int\|\nabla f_{t}\|^{2}\|f_{t}\|^{p}\varphi^{2}-\int\langle\nabla f_{t},f_{t}\rangle p\|f_{t}\|^{p-2}\varphi^{2}\langle\nabla f_{t},f_{t}\rangle$
		$\displaystyle\qquad-2\int\langle\nabla f_{t},f_{t}\rangle\|f_{t}\|^{p}\varphi\nabla\varphi+\int\langle DA(df,df)\cdot f_{t},f_{t}\|f_{t}\|^{p}\varphi^{2}\rangle$
		$\displaystyle\qquad+\int\langle A(df_{t},df),f_{t}\|f_{t}\|^{p}\varphi^{2}\rangle$
		$\displaystyle=-\int\|\nabla f_{t}\|^{2}\|f_{t}\|^{p}\varphi^{2}-p\int\|\langle\nabla f_{t},f_{t}\rangle\|^{2}\|f_{t}\|^{p-2}\varphi^{2}$
		$\displaystyle\qquad+III+IV+V.$

Now we have

	$\displaystyle III$	$\displaystyle\leq\frac{1}{4}\int\|\nabla f_{t}\|^{2}\varphi^{2}\|f_{t}\|^{p}+4\int\|f_{t}\|^{p+2}\|\nabla\varphi\|^{2}$
	$\displaystyle IV$	$\displaystyle\leq C_{N}\int\|df\|^{2}\|f_{t}\|^{p+2}\varphi^{2}$
	$\displaystyle V$	$\displaystyle\leq C_{N}\int\|\nabla f_{t}\|\|df\|\|f_{t}\|^{p+1}\varphi^{2}$
		$\displaystyle\leq\frac{1}{2}\int\|\nabla f_{t}\|^{2}\varphi^{2}\|f_{t}\|^{p}+\frac{C_{N}^{2}}{2}\int\|df\|^{2}\|f_{t}\|^{p+2}\varphi^{2}.$

On the other hand, LHS becomes

	$\displaystyle\int\langle(e^{2u}f_{t})_{t},f_{t}\|f_{t}\|^{p}\varphi^{2}\rangle$	$\displaystyle=\frac{1}{p+2}\frac{d}{dt}\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}+2\frac{p+1}{p+2}\int e^{2u}\|f_{t}\|^{p+2}u_{t}\varphi^{2}$
		$\displaystyle=\frac{1}{p+2}\frac{d}{dt}\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}+2b\frac{p+1}{p+2}\int\|df\|^{2}\|f_{t}\|^{p+2}\varphi^{2}$
		$\displaystyle\qquad-2a\frac{p+1}{p+2}\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}.$

All together, we have

	$\displaystyle\frac{d}{dt}\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}$	$\displaystyle\leq 2a(p+1)\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}+4(p+2)\int\|f_{t}\|^{p+2}\|\nabla\varphi\|^{2}$
		$\displaystyle\qquad-\frac{p+2}{4}\int\|\nabla f_{t}\|^{2}\|f_{t}\|^{p}\varphi^{2}$
		$\displaystyle\qquad+\left((p+2)C_{N}+\frac{(p+2)C_{N}^{2}}{2}-2b(p+1)\right)\int\|df\|^{2}\|f_{t}\|^{p+2}\varphi^{2}.$

By the choice of $b$ , the last term is negative for all $p\geq 0$ . Hence,

	$\displaystyle\frac{d}{dt}\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}$	$\displaystyle\leq 2a(p+1)\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}+4(p+2)\int\|f_{t}\|^{p+2}\|\nabla\varphi\|^{2}$
	$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}(t)$	$\displaystyle\leq e^{2a(p+1)(t-t_{0})}$
		$\displaystyle\quad\cdot\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}(t_{0})+4(p+2)\int_{t_{0}}^{t}\int_{B_{2r}}\|f_{t}\|^{p+2}\|\nabla\varphi\|^{2}\right)$

by Gronwall’s inequality. ∎

Lemma 6.4.

Let $(f,u)$ are solutions of (2b). Assume that

\sup_{T-2\delta r^{2}\leq t\leq T}E(B_{2r},t)<\varepsilon_{1}.

Then for $t\in[T-\delta r^{2},T]$ , we have

(40)

\int_{B_{2r}}e^{2u}|f_{t}|^{2}\varphi^{2}(t)\leq C_{1}(r,\delta,t)C_{2}(r,\delta,t)\varepsilon_{1}

where

(41)		$\displaystyle C_{1}(r,\delta,t)$	$\displaystyle=4^{2}\left(1+e^{2at}\frac{1-e^{-2a\delta r^{2}}}{2ar^{2}}\right)$
(42)		$\displaystyle C_{2}(r,\delta,t)$	$\displaystyle=e^{6a\delta r^{2}}\left(\frac{1}{\delta r^{2}}+\frac{16(4)^{2}}{r^{2}}e^{2at}\right).$

Proof.

Suppose $\varphi$ be a cut-off function supported on $B_{3r/2}$ and $\varphi\equiv 1$ on $B_{r}$ and $|\nabla\varphi|\leq\frac{4}{r}$ . Also, let $\psi$ be a cut-off function supported on $B_{2r}$ and $\psi\equiv 1$ on $B_{3r/2}$ and $|\nabla\psi|\leq\frac{4}{r}$ . From (39) for $p=0$ and using (37), we have

	$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}(t)$	$\displaystyle\leq e^{2a(t-t_{0})}\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}(t_{0})+8\int_{t_{0}}^{t}\int_{B_{3r/2}}\|f_{t}\|^{2}\|\nabla\varphi\|^{2}\right)$
		$\displaystyle\leq e^{2a(t-t_{0})}\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}(t_{0})+\frac{8(4)^{2}}{r^{2}}\int_{t_{0}}^{t}\int_{B_{2r}}\|f_{t}\|^{2}\psi^{2}\right)$
		$\displaystyle\leq e^{2a(t-t_{0})}\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}(t_{0})$
		$\displaystyle\quad+e^{2a(t-t_{0})}\frac{8(4)^{2}}{r^{2}}e^{2at}4^{2}\varepsilon_{1}\left(1+\frac{e^{2at}-e^{2at_{0}}}{2ar^{2}}\right).$

Now take $t_{0}\in[t-\delta r^{2},t]$ such that

\int_{B_{2r}}e^{2u}|f_{t}|^{2}\varphi^{2}(t_{0})=\min_{t-\delta r^{2}\leq s\leq t}\int_{B_{2r}}e^{2u}|f_{t}|^{2}\varphi^{2}(s).

Then by (36),

\int_{B_{2r}}e^{2u}|f_{t}|^{2}\varphi^{2}(t_{0})\leq\frac{1}{\delta r^{2}}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}|f_{t}|^{2}\varphi^{2}\leq\frac{1}{\delta r^{2}}4^{2}\varepsilon_{1}\left(1+\frac{e^{2at}-e^{2a(t-\delta r^{2})}}{2ar^{2}}\right).

Therefore,

\displaystyle\int_{B_{2r}}e^{2u}|f_{t}|^{2}\varphi^{2}(t)

\displaystyle\leq 4^{2}\varepsilon_{1}\left(1+\frac{e^{2at}-e^{2at-2a\delta r^{2}}}{2ar^{2}}\right)\left(\frac{1}{\delta r^{2}}+\frac{8(4)^{2}}{r^{2}}e^{2at}\right)e^{2a\delta r^{2}}.

This completes the proof. ∎

Corollary 6.5.

Under the same assumption as above, we also have

(43)		$\displaystyle\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|\nabla f_{t}\|^{2}\varphi^{2}$	$\displaystyle\leq CC_{1}(r,\delta,t)C_{2}(r,\delta,t)\varepsilon_{1}$
(44)		$\displaystyle\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|df\|^{2}\|f_{t}\|^{2}\varphi^{2}$	$\displaystyle\leq CC_{1}(r,\delta,t)C_{2}(r,\delta,t)\varepsilon_{1}.$

Proof.

From the equation (38) with $p=0$ , we can integrate from $t-\delta r^{2}$ to $t$ .

	$\displaystyle\left.\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}\right\|^{t}_{t-\delta r^{2}}$	$\displaystyle\leq 2a\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}\|f_{t}\|^{2}\varphi^{2}+8\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|f_{t}\|^{2}\|\nabla\varphi\|^{2}$
		$\displaystyle-\frac{1}{2}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|\nabla f_{t}\|^{2}\varphi^{2}$
		$\displaystyle+\left(2C_{N}+C_{N}^{2}-2b\right)\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|df\|^{2}\|f_{t}\|^{2}\varphi^{2}.$

Hence, we have

	$\displaystyle\frac{1}{2}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|\nabla f_{t}\|^{2}\varphi^{2}$	$\displaystyle\leq 2C_{1}(r,\delta,t)C_{2}(r,\delta,t)\varepsilon_{1}+2aC_{1}(r,\delta,t)\varepsilon_{1}+8\frac{4^{2}}{r^{2}}e^{2at}C_{1}(r,\delta,t)\varepsilon_{1}$
		$\displaystyle\leq CC_{1}(r,\delta,t)C_{2}(r,\delta,t)\varepsilon_{1}.$

The other inequality is similar. ∎

6.3. Higher estimate for time derivatives

In this subsection we will get estimate for $e^{2u}|f_{t}|^{4}$ . We first build up $(p+2)$ -version of Equation 34.

Proposition 6.6.

For solutions $(f,u)$ of (2b) and for $p\geq 1$ , we have

(45)		$\displaystyle\int_{t_{1}}^{t_{2}}\int_{B_{2r}}e^{2u}\|f_{t}\|^{p+2}\varphi^{2}$	$\displaystyle\leq C\int_{t_{1}}^{t_{2}}\int_{B_{2r}}\|f_{ti}\|^{2}\|f_{t}\|^{p-1}\varphi^{2}+C\int_{t_{1}}^{t_{2}}\int_{B_{2r}}\|f_{t}\|^{p+1}\|\nabla\varphi\|^{2}$
(45)			$\displaystyle\quad+C\int_{t_{1}}^{t_{2}}\int_{B_{2r}}\|df\|^{2}\|f_{t}\|^{p+1}\varphi^{2}.$

Proof.

First note that for any $p\geq 1$ , $\nabla_{i}|f_{t}|^{p}=p|f_{t}|^{p-2}\langle f_{ti},f_{t}\rangle$ . Also, for simplicity, denote $\int\int=\int_{t_{1}}^{t_{2}}\int_{B_{2r}}$ . Multiplying $\tau(f)$ to (2a) gives

2e^{2u}|f_{t}|^{2}=-2\langle f_{ti},f_{i}\rangle+\nabla_{i}(2\langle f_{t},f_{i}\rangle).

Multiplying $|f_{t}|^{p}\varphi^{2}$ for $p\geq 1$ and integrating gives

	$\displaystyle 2\int\int e^{2u}\|f_{t}\|^{p+2}\varphi^{2}$	$\displaystyle=-2\int\int\langle f_{ti},f_{i}\rangle\|f_{t}\|^{p}\varphi^{2}-4\int\int\langle f_{t},f_{i}\rangle\|f_{t}\|^{p}\varphi\nabla_{i}\varphi$
		$\displaystyle\qquad-2p\int\int\langle f_{t},f_{i}\rangle\varphi^{2}\|f_{t}\|^{p-2}\langle f_{ti},f_{t}\rangle$
		$\displaystyle=I+II+III.$

Now

	$\displaystyle I$	$\displaystyle\leq C\int\int\|f_{ti}\|^{2}\|f_{t}\|^{p-1}\varphi^{2}+C\int\int\|df\|^{2}\|f_{t}\|^{p+1}\varphi^{2}$
	$\displaystyle II$	$\displaystyle\leq C\int\int\|f_{t}\|^{p+1}\|\nabla\varphi\|^{2}+C\int\int\|df\|^{2}\|f_{t}\|^{p+1}\varphi^{2}$
	$\displaystyle III$	$\displaystyle\leq C\int\int\|f_{ti}\|^{2}\|f_{t}\|^{p-1}\varphi^{2}+C\int\int\|df\|^{2}\|f_{t}\|^{p+1}\varphi^{2}.$

This completes the proof. ∎

Now we will show the desired estimate.

Proposition 6.7.

Let $(f,u)$ are solutions of (2b). Assume that

\sup_{T-2\delta r^{2}\leq t\leq T}E(B_{2r},t)<\varepsilon_{1}.

Then for $t\in[T-\delta r^{2},T]$ , we have

(46)

\int_{B_{2r}}e^{2u}|f_{t}|^{4}\varphi^{2}(t)\leq C_{3}

where

(47)

C_{3}=CC_{1}(r,\delta,t)C_{2}(r,\delta,t)^{3}\varepsilon_{1}.

Note that $C_{3}$ depends on $r,t,\delta$ .

Proof.

For simplicity, denote $C_{1}=C_{1}(r,\delta,t)$ , $C_{2}=C_{2}(r,\delta,t)$ . Also, denote $C$ for any number appeared in computations. Suppose $\varphi$ be a cut-off function supported on $B_{3r/2}$ and $\varphi\equiv 1$ on $B_{r}$ and $|\nabla\varphi|\leq\frac{4}{r}$ . Also, let $\psi$ be a cut-off function supported on $B_{2r}$ and $\psi\equiv 1$ on $B_{3r/2}$ and $|\nabla\psi|\leq\frac{4}{r}$ . Let $t_{1}=t-\delta r^{2}$ and $t_{2}=t$ .

The proof consists of several steps, increasing power of $|f_{t}|$ .

Step 1. Estimate for $\int\int e^{2u}|f_{t}|^{3}\varphi^{2}$ .

From (45) with $p=1$ and using (37), (43) and (44), we have

(48)

\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}|f_{t}|^{3}\varphi^{2}\leq CC_{1}C_{2}\varepsilon_{1}

and

(49)

\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}|f_{t}|^{3}\varphi^{2}\leq e^{2at}CC_{1}C_{2}\varepsilon_{1}.

Step 2. Estimate for $\int e^{2u}|f_{t}|^{3}\varphi^{2}$ .

Now let $t_{0}\in[t-\delta r^{2},t]$ be such that

\int_{B_{2r}}e^{2u}|f_{t}|^{3}\varphi^{2}(t_{0})=\min_{t-\delta r^{2}\leq s\leq t}\int_{B_{2r}}e^{2u}|f_{t}|^{3}\varphi^{2}(s).

From (39) with $p=1$ and using (48) and (49), we have

	$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{3}\varphi^{2}(t)$	$\displaystyle\leq e^{4a\delta r^{2}}\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{3}\varphi^{2}(t_{0})+12\int_{t_{0}}^{t}\int_{B_{3r/2}}\|f_{t}\|^{3}\|\nabla\varphi\|^{2}\right)$
		$\displaystyle\leq e^{4a\delta r^{2}}\left(\frac{1}{\delta r^{2}}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}\|f_{t}\|^{3}\varphi^{2}+12\frac{4^{2}}{r^{2}}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|f_{t}\|^{3}\psi^{2}\right)$
		$\displaystyle\leq e^{4a\delta r^{2}}\left(\frac{1}{\delta r^{2}}CC_{1}C_{2}\varepsilon_{1}+12\frac{4^{2}}{r^{2}}e^{2at}CC_{1}C_{2}\varepsilon_{1}\right)$
		$\displaystyle=CC_{1}C_{2}\varepsilon_{1}\left(\frac{1}{\delta r^{2}}+\frac{12(4)^{2}}{r^{2}}e^{2at}\right)e^{4a\delta r^{2}}.$

So, simply,

(50)

\int_{B_{2r}}e^{2u}|f_{t}|^{3}\varphi^{2}(t)\leq CC_{1}C_{2}^{2}\varepsilon_{1}.

Step 3. Estimate for $\int\int|\nabla f_{t}|^{2}|f_{t}|\varphi^{2}$ and $\int\int|df|^{2}|f_{t}|^{3}\varphi^{2}$ .

From (38) with $p=1$ , we can integrate from $t-\delta r^{2}$ to $t$ .

	$\displaystyle\left.\int_{B_{2r}}e^{2u}\|f_{t}\|^{3}\varphi^{2}\right\|^{t}_{t-\delta r^{2}}$	$\displaystyle\leq 4a\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}\|f_{t}\|^{3}\varphi^{2}+12\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|f_{t}\|^{3}\|\nabla\varphi\|^{2}$
		$\displaystyle-\frac{3}{4}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|\nabla f_{t}\|^{2}\|f_{t}\|\varphi^{2}$
		$\displaystyle+\left(3C_{N}+\frac{3C_{N}^{2}}{2}-4b\right)\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|df\|^{2}\|f_{t}\|^{3}\varphi^{2}.$

Note that $3C_{N}+\frac{3C_{N}^{2}}{2}-4b<0$ . Now, from (48), (49), and (50), we have

	$\displaystyle\frac{3}{4}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|\nabla f_{t}\|^{2}\|f_{t}\|\varphi^{2}$	$\displaystyle\leq 2CC_{1}C_{2}^{2}\varepsilon_{1}+4aCC_{1}C_{2}\varepsilon_{1}+12\frac{4^{2}}{r^{2}}e^{2at}CC_{1}C_{2}\varepsilon_{1}$
		$\displaystyle\leq CC_{1}C_{2}^{2}\varepsilon_{1}.$

So, we have

(51)

\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}|\nabla f_{t}|^{2}|f_{t}|\varphi^{2}\leq CC_{1}C_{2}^{2}\varepsilon_{1}.

Similarly,

(52)

\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}|df|^{2}|f_{t}|^{3}\varphi^{2}\leq CC_{1}C_{2}^{2}\varepsilon_{1}.

Step 4. Estimate for $\int\int e^{2u}|f_{t}|^{4}\varphi^{2}$ .

From (45) with $p=2$ and using (49), (51) and (52), we have

(53)

\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}|f_{t}|^{4}\varphi^{2}\leq CC_{1}C_{2}^{2}\varepsilon_{1}

and

(54)

\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}|f_{t}|^{4}\varphi^{2}\leq e^{2at}CC_{1}C_{2}^{2}\varepsilon_{1}.

Step 5. Estimate for $\int e^{2u}|f_{t}|^{4}\varphi^{2}$ .

Now let $t_{0}\in[t-\delta r^{2},t]$ be such that

\int_{B_{2r}}e^{2u}|f_{t}|^{4}\varphi^{2}(t_{0})=\min_{t-\delta r^{2}\leq s\leq t}\int_{B_{2r}}e^{2u}|f_{t}|^{4}\varphi^{2}(s).

From (39) with $p=2$ and using (53) and (54), we have

	$\displaystyle\int_{B_{2r}}e^{2u}\|f_{t}\|^{4}\varphi^{2}(t)$	$\displaystyle\leq e^{6a\delta r^{2}}\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{4}\varphi^{2}(t_{0})+16\int_{t_{0}}^{t}\int_{B_{3r/2}}\|f_{t}\|^{4}\|\nabla\varphi\|^{2}\right)$
		$\displaystyle\leq e^{6a\delta r^{2}}\left(\frac{1}{\delta r^{2}}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}e^{2u}\|f_{t}\|^{4}\varphi^{2}+16\frac{4^{2}}{r^{2}}\int_{t-\delta r^{2}}^{t}\int_{B_{2r}}\|f_{t}\|^{3}\psi^{2}\right)$
		$\displaystyle\leq e^{6a\delta r^{2}}\left(\frac{1}{\delta r^{2}}CC_{1}C_{2}^{2}\varepsilon_{1}+16\frac{4^{2}}{r^{2}}e^{2at}CC_{1}C_{2}^{2}\varepsilon_{1}\right)$
		$\displaystyle=CC_{1}C_{2}^{2}\varepsilon_{1}\left(\frac{1}{\delta r^{2}}+\frac{16(4)^{2}}{r^{2}}e^{2at}\right)e^{6a\delta r^{2}}.$

So, simply,

\int_{B_{2r}}e^{2u}|f_{t}|^{4}\varphi^{2}(t)\leq CC_{1}C_{2}^{3}\varepsilon_{1}.

∎

Remark 6.8.

We can keep going on to get bounds for $\int_{B_{2r}}e^{2u}|f_{t}|^{n}\varphi^{2}(t)\leq C_{3}(n)$ for any $n$ . However, these bounds blow up to infinity as $n\rightarrow\infty$ .

7. $W^{2,2}$ and gradient estimate

In this section we will get $W^{2,2}$ estimate and gradient estimate for the solution $f$ of (2a). For simplicity, denote $\|\cdot\|_{k,p}=\|\cdot\|_{W^{k,p}(B_{2r})}$ and $\|\cdot\|_{p}=\|\cdot\|_{0,p}$ . First observe the following.

Lemma 7.1.

Let $u$ be a solution of (2b). For $p>2$ and for any $r>0$ ,

(55)

\int_{B_{2r}}e^{pu}\varphi^{r}(t)\leq\int_{B_{2r}}e^{pu}\varphi^{r}(t_{0})+\frac{2b^{2}(p-2)}{pa}\int_{t_{0}}^{t}\int_{B_{2r}}|df|^{p}\varphi^{r}.

Proof.

Note that

\partial_{t}(e^{pu})=pe^{pu}u_{t}=pbe^{(p-2)u}|df|^{2}-pae^{pu}.

So, multiplying $\varphi^{r}$ and integrating over $B_{2r}$ gives

	$\displaystyle\frac{d}{dt}\int_{B_{2r}}e^{pu}\varphi^{r}$	$\displaystyle=pb\int_{B_{2r}}e^{(p-2)u}\|df\|^{2}\varphi^{r}-pa\int_{B_{2r}}e^{pu}\varphi^{r}$
		$\displaystyle\leq b\lambda(p-2)\int_{B_{2r}}e^{pu}\varphi^{r}+2b\lambda^{-1}\int_{B_{2r}}\|df\|^{p}\varphi^{r}-pa\int_{B_{2r}}e^{pu}\varphi^{r}$
		$\displaystyle=\frac{2b^{2}(p-2)}{pa}\int_{B_{2r}}\|df\|^{p}\varphi^{r}$

by Young’s inequality with weight $\lambda=\frac{pa}{b(p-2)}$ . Hence, by integrating, we obtain the result. ∎

Lemma 7.2.

Let $f$ be any smooth function and let $\varphi\in C^{\infty}_{0}(B_{2r})$ be a cut-off function. Then for any $r>1$ and $p\geq 2$ , we have

(56)

\|\,|df|^{r}\varphi\|_{p}\leq C\|\,|df|^{r-1}\|_{p}\|f\varphi\|_{2,2}.

Proof.

Let $1\leq s<2$ be such that $p=2s(2-s)$ . By Sobolev embedding,

	$\displaystyle\\|\,\|df\|^{r}\varphi\\|_{p}$	$\displaystyle\leq C\\|\nabla(\|df\|^{r}\varphi)\\|_{s}$
		$\displaystyle\leq C\\|\|df\|^{r-1}\nabla(\|df\|\varphi)\\|_{s}$
		$\displaystyle\leq C\\|\,\|df\|^{r-1}\\|_{p}\\|f\varphi\\|_{2,2}.$

∎

Next, we will show $W^{2,2}$ estimate.

Proposition 7.3.

Let $(f,u)$ are solutions of (2b). Then there exists $\varepsilon_{1}>0$ such that the following holds:

Assume that

\sup_{T-2\delta r^{2}\leq t\leq T}E(B_{2r},t)\leq\varepsilon_{1},\qquad\int_{B_{2r}\times\{T-2\delta r^{2}\}}e^{6u}\leq\varepsilon_{1}.

Then for $t\in[T-\delta r^{2},T]$ , we have

(57)

\|f\varphi\|_{2,2}\leq C_{4}=C_{4}(r,\delta,T,\varepsilon_{1},C_{N})

where

C_{4}=\left(CC_{3}\varepsilon_{1}+C\varepsilon_{1}^{4}\right)^{1/4}\left(1+C_{3}\varepsilon_{1}\delta r^{2}\exp(C_{3}\varepsilon_{1}\delta r^{2})\right)^{1/4}.

Proof.

Without loss of generality, assume $\int_{\Omega}f=0$ . Then we have, by Poincare,

\|f\|_{p}\leq C_{p}\|df\|_{p}.

From the equation $\Delta f+A(df,df)=e^{2u}f_{t}$ , multiplying $\varphi$ and arranging terms gives

	$\displaystyle\|\Delta(f\varphi)\|$	$\displaystyle\leq\|A(df,df)\varphi\|+\|e^{2u}f_{t}\varphi\|+k(\varphi)(\|f\|+\|df\|)$
		$\displaystyle\leq C_{N}\|\,\|df\|^{2}\varphi\|+\|e^{2u}f_{t}\varphi\|+k(\varphi)(\|f\|+\|df\|).$

By the $L^{p}$ estimate, we have

(58)

\|f\varphi\|_{2,p}\leq C\left(C_{N}\|\,|df|^{2}\varphi\|_{p}+\|e^{2u}|f_{t}|\varphi\|_{p}+\|df\|_{p}\right)

where the constant $C$ only depends on $p$ and $r$ .

Now let $p=2$ . Note that, by (46) and (55),

	$\displaystyle\\|e^{2u}\|f_{t}\|\varphi\\|_{2}^{4}$	$\displaystyle=\left(\int_{B_{2r}}e^{4u}\|f_{t}\|^{2}\varphi^{2}\right)^{2}$
		$\displaystyle\leq\left(\int_{B_{3r/2}}e^{2u}\|f_{t}\|^{4}\right)\left(\int_{B_{2r}}e^{6u}\varphi^{4}\right)$
		$\displaystyle\leq\left(\int_{B_{2r}}e^{2u}\|f_{t}\|^{4}\psi^{2}\right)\left(\int_{B_{2r}}e^{6u}\varphi^{4}\right)$
		$\displaystyle\leq C_{3}\left(\int_{B_{2r}}e^{6u}\varphi^{4}(t_{0})+\frac{8b^{2}}{6a}\int_{t_{0}}^{t}\int_{B_{2r}}\|df\|^{6}\varphi^{4}\right)$
		$\displaystyle\leq C_{3}\varepsilon_{1}+C_{3}\int_{t_{0}}^{t}\int_{B_{2r}}\|df\|^{6}\varphi^{4}.$

Now applying Lemma 7.2 with $r=3/2$ , $q=4$ gives

	$\displaystyle\left(\int_{B_{2r}}\|df\|^{6}\varphi^{4}\right)^{1/4}$	$\displaystyle=\\|\,\|df\|^{3/2}\varphi\\|_{4}\leq C\\|\,\|df\|^{1/2}\\|_{4}\\|f\varphi\\|_{2,2}$
		$\displaystyle\leq C\varepsilon_{1}^{1/4}\\|f\varphi\\|_{2,2}.$

On the other hand, applying Lemma 7.2 with $r=2$ , $q=2$ gives

\|\,|df|^{2}\varphi\|_{2}\leq C\|df\|_{2}\|f\varphi\|_{2,2}\leq C\varepsilon_{1}^{1/2}\|f\varphi\|_{2,2}.

All together, we have

\|f\varphi\|_{2,2}^{4}\leq CC_{N}^{4}\varepsilon_{1}^{2}\|f\varphi\|_{2,2}^{4}+CC_{3}\varepsilon_{1}+CC_{3}\varepsilon_{1}\int_{t_{0}}^{t}\|f\varphi\|_{2,2}+C\varepsilon_{1}^{4}

Let $X=\|f\varphi\|_{2,2}^{4}$ . Then the above equation becomes

(1-CC_{N}^{4}\varepsilon_{1}^{2})X\leq CC_{3}\varepsilon_{1}+C\varepsilon_{1}^{4}+C_{3}\varepsilon_{1}\int_{t_{0}}^{t}X.

So, if $\varepsilon_{1}$ is small enough so that $1-CC_{N}^{4}\varepsilon_{1}^{2}>1/2$ , then by Gronwall’s inequality, we have

	$\displaystyle\\|f\varphi\\|_{2,2}^{4}$	$\displaystyle\leq\left(CC_{3}\varepsilon_{1}+C\varepsilon_{1}^{4}\right)\left(1+C_{3}\varepsilon_{1}(t-t_{0})\exp(C_{3}\varepsilon_{1}(t-t_{0}))\right)$
		$\displaystyle\leq\left(CC_{3}\varepsilon_{1}+C\varepsilon_{1}^{4}\right)\left(1+C_{3}\varepsilon_{1}\delta r^{2}\exp(C_{3}\varepsilon_{1}\delta r^{2})\right).$

This completes the proof. ∎

From Sobolev embedding, we now have, for $t\in[T-\delta r^{2},T]$ ,

(59)

\|f\varphi\|_{1,p}\leq C_{4}

for any $p>1$ .

Now we will show gradient estimate. This can be achieved by obtaining better estimate than $W^{2,2}$ , say $W^{2,3}$ .

Proposition 7.4.

Assume the same as in 7.3. In addition, we assume that

\int_{B_{2r}\times\{T-2\delta r^{2}\}}e^{18u}\leq\varepsilon_{1}.

Then for $t\in[T-\delta r^{2},T]$ , we have

(60)

\|f\varphi\|_{2,3}\leq C_{5}=C_{5}(r,\delta,T,\varepsilon_{1},C_{N})

where

C_{5}=C\left(C_{N}C_{4}^{2}+C_{3}^{1/12}\varepsilon_{1}^{1/12}+C_{3}^{1/12}\delta^{1/12}r^{1/6}C_{4}^{3/2}+C_{4}\right).

In particular,

(61)

\sup_{B_{r}}|df|\leq C_{5}.

Proof.

By Equation 59, we have uniform bound for $|df|^{p}$ for any $p$ . Now from equation (58), we have

\|f\varphi\|_{2,p}\leq C\left(C_{N}\|\,df\|_{2p}^{2}+\|e^{2u}|f_{t}|\varphi\|_{p}+\|df\|_{p}\right).

Now let $p=3$ and $t_{0}=T-2\delta r^{2}$ . Then we have, using (55) and Equation 59,

	$\displaystyle\\|e^{2u}\|f_{t}\|\varphi\\|_{3}^{12}$	$\displaystyle\leq\\|e^{u/2}\|f_{t}\|\varphi\\|_{4}^{12}\\|e^{3u/2}\\|_{12}^{12}$
		$\displaystyle\leq C_{3}\int_{B_{2r}}e^{18u}$
		$\displaystyle\leq C_{3}\left(\int_{B_{2r}}e^{18u}(t_{0})+C\int_{t_{0}}^{t}\int_{B_{2r}}\|df\|^{18}\right)$
		$\displaystyle\leq C_{3}\varepsilon_{1}+CC_{3}\delta r^{2}C_{4}^{18}.$

Applying Equation 59 completes the proof. ∎

8. Global weak solution

In this section, we will prove the main theorem 1.1.

Lemma 8.1.

There exists $\varepsilon_{1}>0$ such that if $(f,u)$ be a smooth solution of (2b) on $B_{2r}\times[T-2\delta r^{2},T]$ and

(62)

\sup_{T-2\delta r^{2}\leq t\leq T}E(B_{2r},t)\leq\varepsilon_{1}\quad\textrm{ and }\quad\int_{B_{2r}\times\{T-2\delta r^{2}\}}e^{18u}\leq\varepsilon_{1},

then Hölder norms of $f,u$ and their derivatives are all bounded by constants only depending on $T,r,\delta,\varepsilon_{1},C_{N}$ .

Proof.

By the sup bound of $|df|$ , we have $e^{-2u(f)}\leq e^{2aT}$ and

	$\displaystyle e^{-2u(f)}$	$\displaystyle=\frac{e^{2at}}{1+2b\int_{0}^{t}e^{2as}\|df\|^{2}(x,s)ds}\geq\frac{e^{2at}}{1+2b{M^{\prime}}^{2}\frac{e^{2at}-1}{2a}}$
		$\displaystyle\geq\frac{1}{1+\frac{b}{a}{M^{\prime}}^{2}}.$

Hence the operator $\partial_{t}-e^{-2u}\Delta$ is uniformly parabolic on $[0,T_{0})$ .

Similar in proof of Theorem 5.2, we conclude the desired estimate. ∎

Proof.

(Proof of Theorem 1.1) First consider $f_{0}$ is smooth. By Theorem 5.2, there exists a smooth solution in $(\Sigma\times[0,T))$ for some $T>0$ . Let $T_{1}$ be the maximal existence time. If $T_{1}=\infty$ then we obtain global solution which is smooth everywhere. So suppose $T_{1}<\infty$ .

If we have $\limsup_{t\nearrow T_{1}}E(B_{2r}(x),t)\leq\varepsilon_{1}$ for any $x\in\Sigma$ and $r>0$ , then by above lemma Hölder norms of $f,u$ and their derivatives are all bounded, hence $f,u$ can be extended beyond the time $T_{1}$ . This contradicts with maximality of $T_{1}$ . So there should be a point $x\in\Sigma$ such that

\limsup_{t\nearrow T_{1}}E(B_{2r}(x),t)>\varepsilon_{1}.

Since the total energy is finite, there are at most finitely many such points $\{x_{1},\cdots,x_{k_{1}}\}$ . Then by above lemma, we get smooth solution $(f_{1},u_{1})$ on $\Sigma\times[0,T]\setminus\{(x_{i}^{1},T_{1})\}_{i=1,\cdots,k_{1}}$ . If we denote $f(x,T_{1})$ and $u(x,T_{1})$ as the weak limit of $f(x,t)$ and $u(x,t)$ as $t\nearrow T_{1}$ , then $f(t),u(t)$ converges to $f(T_{1}),u(T_{1})$ strongly in $W^{1,2}_{loc}(\Sigma\setminus\{x_{i}^{1}\})$ .

Next, denote $g_{1}=e^{2u_{1}(x,T_{1})}g_{0}$ and consider the flow (2b) with initial map $f_{1}$ and initial metric $g_{1}$ . As above, there is a smooth solution $(f_{2},u_{2})$ on $\Sigma\times[0,T_{2}]\setminus\{(x_{i}^{2},T_{2})\}_{i=1,\cdots,k_{2}}$ . From these we can set up a smooth solution $(f,u)$ on $\Sigma\times[0,T_{1}+T_{2}]$ which is smooth except $\{(x_{i}^{1},T_{1})\}\cup\{(x_{i}^{2},T_{2})\}$ . Iterate this process to obtain global solution with exception points, which are at most finitely many because the total energy is finite.

∎

9. Finite time singularity

As the conformal heat flow is developed to postpone the finite time singularity, it is expected to have no finite time singularity. In this section we will discuss few remarks about finite time singularity.

Recall the following

Lemma 9.1.

([23]) There exist a compact target manifold $N$ , a smooth map $f_{0}:D\to N$ and $\varepsilon>0$ such that every smooth map $f:D\to N$ homotopic to $v_{0}$ fails to be harmonic. If furthermore $E(f)\leq E(f_{0})$ , then

\int_{D}|\tau(f)|^{2}\geq\varepsilon.

Together with energy decreasing property of harmonic map heat flow $f(t)$ , the above lemma implies that no heat flow starting with initial map $f_{1}$ homotopic to $f_{0}$ above can be smooth after the time $t=\frac{E(f_{1})}{\varepsilon}$ .

This argument can be avoided in conformal heat flow. From (4), we have

E(0)-E(t)=\int_{0}^{t}\int_{D}e^{-2u}|\tau(f(t))|^{2}.

So, if $u$ is large, $\int_{D}e^{-2u}|\tau(f(t))|^{2}$ can be smaller than $\varepsilon$ even if $\int_{D}|\tau(f(t))|^{2}>\varepsilon$ .

The proof of the above lemma relies on no-neck property of approximate harmonic map with $\|\tau\|_{L^{2}}\to 0$ . And the assumption $\|\tau\|_{L^{2}}\to 0$ is essential in the no-neck property as there is a counter example of Parker without the assumption. The conformal heat flow makes the tension field converge to zero with different scale. Hence the information about the converging scale of the tension field will play an important role in the property of the flow.

Acknowledgement

The author would like to thanks Armin Schikorra and Thomas Parker for valuable comments and advice.

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	$\displaystyle\left\|\nabla^{2}g\right\|$	$\displaystyle=\left\|\nabla^{2}\left((e^{2u_{2}-2u_{1}}-1)\partial_{t}h_{2}\right)\right\|$
		$\displaystyle\leq 800\|u_{1}-u_{2}\|\|\nabla(u_{1}-u_{2})\|^{2}\|\partial_{t}h_{2}\|+400\|u_{1}-u_{2}\|\|\nabla^{2}(u_{1}-u_{2})\|\|\partial_{t}h_{2}\|$
		$\displaystyle\quad+400\|u_{1}-u_{2}\|\|\nabla(u_{1}-u_{2})\|\|\nabla\partial_{t}h_{2}\|+200\|u_{1}-u_{2}\|\|\nabla^{2}\partial_{t}h_{2}\|.$

	$\displaystyle\\|\nabla^{2}g\\|_{L^{2}}^{2}$	$\displaystyle\leq 1600^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla(u_{1}-u_{2})\\|_{L^{8}}^{4}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}(t)\\|_{L^{4}(\Sigma)}^{2}\,T^{1/2}$
		$\displaystyle\quad+800^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}(u_{1}-u_{2})\\|_{L^{4}}^{2}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}(t)\\|_{L^{8}(\Sigma)}^{4}\,T^{1/2}$
		$\displaystyle\quad+800^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\max_{0\leq t\leq T}\\|\nabla(u_{1}(t)-u_{2}(t))\\|_{L^{4}(\Sigma)}^{2}\\|\nabla\partial_{t}h_{2}\\|_{L^{4}}^{2}\,T^{1/2}$
		$\displaystyle\quad+400^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\nabla^{2}\partial_{t}h_{2}\\|_{L^{2}}^{2}$
		$\displaystyle\leq 400^{2}C_{2}^{2}\\|u_{1}-u_{2}\\|_{Z}^{2}2C_{0}^{2}$
		$\displaystyle=\frac{C_{3}^{2}}{8C_{1}^{2}}\\|u_{1}-u_{2}\\|_{Z}^{2}$

	$\displaystyle\\|g_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq 2(400)^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|(u_{1}-u_{2})_{t}\\|_{L^{4}}^{2}\max_{0\leq t\leq T}\\|\partial_{t}h_{2}\\|_{L^{4}(\Sigma)}^{2}\,T^{1/2}$
		$\displaystyle\quad+2(200)^{2}\\|u_{1}-u_{2}\\|_{C^{0}}^{2}\\|\partial_{tt}h_{2}\\|_{L^{2}}^{2}$
		$\displaystyle\leq 2(200)^{2}C_{2}^{2}\\|u_{1}-u_{2}\\|_{Z}^{2}2C_{0}^{2}$
		$\displaystyle=\frac{C_{3}^{2}}{16C_{1}^{2}}\\|u_{1}-u_{2}\\|_{Z}^{2}$

	$\displaystyle\\|\nabla^{2}g\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{4}\iint(\|\nabla^{2}u\|^{2}+\|\nabla u\|^{4})+{c}\\|df\\|_{C^{0}}^{6}\iint\|\nabla u\|^{2}$
		$\displaystyle\quad+{c}\\|df\\|_{C^{0}}^{2}\iint\|\nabla df\|^{2}\|\nabla u\|^{2}+{c}\\|df\\|_{C^{0}}^{8}\|\Sigma\|T+{c}\\|df\\|_{C^{0}}^{4}\iint\|\nabla df\|^{2}$
		$\displaystyle\quad+{c}\\|df\\|_{C^{0}}^{2}\iint\|\nabla^{3}f\|^{2}+{c}\iint\|\nabla df\|^{4}$
		$\displaystyle\leq{c}\\|f\\|_{Y}^{4}\left(\max_{0\leq t\leq T}\\|\nabla^{2}u(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{4}(\Sigma)}^{4}\right)T$
		$\displaystyle\quad+{c}\\|f\\|_{Y}^{6}\max_{0\leq t\leq T}\\|\nabla u(t)\\|_{L^{2}(\Sigma)}^{2}T+{c}\\|f\\|_{Y}^{8}\|\Sigma\|T$
		$\displaystyle\quad+{c}\\|f\\|_{Y}^{4}\max_{0\leq t\leq T}\\|\nabla df(t)\\|_{L^{2}(\Sigma)}^{2}T+{c}\\|f\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla^{3}f(t)\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\quad+{c}\max_{0\leq t\leq T}\\|\nabla df\\|_{L^{4}(\Sigma)}^{4}T$
		$\displaystyle\leq\frac{\delta^{2}}{6C_{1}^{2}}$

	$\displaystyle\\|g_{t}\\|_{L^{2}}^{2}$	$\displaystyle\leq{c}\\|df\\|_{C^{0}}^{4}\iint\|u_{t}\|^{2}+\|f_{t}\|^{2}+{c}\\|df\\|_{C^{0}}^{2}\iint\|df_{t}\|^{2}$
		$\displaystyle\leq{c}\\|f\\|_{Y}^{4}\left(\max_{0\leq t\leq T}\\|u_{t}(t)\\|_{L^{2}(\Sigma)}^{2}+\max_{0\leq t\leq T}\\|f_{t}(t)\\|_{L^{2}(\Sigma)}^{2}\right)T$
		$\displaystyle\quad+{c}\\|f\\|_{Y}^{2}\max_{0\leq t\leq T}\\|\nabla f_{t}\\|_{L^{2}(\Sigma)}^{2}T$
		$\displaystyle\leq\frac{\delta^{2}}{6C_{1}^{2}}$

A new conformal heat flow of harmonic maps

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

Theorem 1.1.

1.1. Notation

2. Preliminaries

2.1. Energy and Volume

Lemma 2.1.

Proof.

2.2. Asymptotic behavior of steady solution

Lemma 2.2.

Proof.

3. Construction of Hilbert spaces

3.1. Spaces XX, YY and ZZ

3.2. The ball BB and B′B^{\prime}

4. Construction of operators

Lemma 4.1.

Proof.

4.1. The construction S1S_{1}

Lemma 4.2.

Proof.

Lemma 4.3.

Proof.

Lemma 4.4.

Proof.

4.2. The construction S2S_{2}

Lemma 4.5.

Proof.

Lemma 4.6.

Proof.

Lemma 4.7.

Proof.

Lemma 4.8.

Proof.

5. Existence of fixed point

Proposition 5.1.

Proof.

Theorem 5.2.

Proof.

6. Local estimate

6.1. Energy estimate

Proposition 6.1.

Proof.

Lemma 6.2.

Proof.

6.2. Estimate for ∫|ft|2\int|f_{t}|^{2}

Proposition 6.3.

Proof.

Lemma 6.4.

Proof.

Corollary 6.5.

Proof.

6.3. Higher estimate for time derivatives

Proposition 6.6.

Proof.

Proposition 6.7.

Proof.

Remark 6.8.

7. W2,2W^{2,2} and gradient estimate

Lemma 7.1.

Proof.

Lemma 7.2.

Proof.

Proposition 7.3.

Proof.

Proposition 7.4.

Proof.

8. Global weak solution

Lemma 8.1.

Proof.

Proof.

9. Finite time singularity

Lemma 9.1.

Acknowledgement

References

3.1. Spaces $X$ , $Y$ and $Z$

3.2. The ball $B$ and $B^{\prime}$

4.1. The construction $S_{1}$

4.2. The construction $S_{2}$

6.2. Estimate for $\int|f_{t}|^{2}$

7. $W^{2,2}$ and gradient estimate