A New Conformal Heat Flow of Harmonic Maps

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.

1 Introduction

Consider a map $f_0 : \varSigma \times [0,T) \rightarrow N$ from a compact Riemann surface $(\varSigma ,g_0)$ 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 $\tau _g(f) = {{\,\textrm{tr}\,}}_g (\nabla ^g \textrm{d}f)$ 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

where $a,b>0$ are constants and $|\textrm{d}f|_g^2 = g^{ij}h_{\alpha \beta } f_i^{\alpha }f_j^{\beta }$ 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 $\varOmega \subset \varSigma$, then under the flow (1b), the metric is conformally enlarged; this increases the area of $\varOmega$ and decreases the energy density. This suggests that the system (1) may be better behaved than the harmonic map heat flow, where energy concentration at points is an impediment to convergence.

Writing the metric $g(t) = \textrm{e}^{2u} g_0$ for a real-valued function u(t), equations (1) are equivalent to the following equations for the pair (f(t), u(t)):

where $\tau$ and $|\textrm{d}f|^2$ 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

(Existence of global weak solution) For any $f_0 \in W^{3,2}(\varSigma ,N),$ a global weak solution (f, u) of (2) exists on $\varSigma \times [0,\infty )$ which is smooth on $\varSigma \times (0,\infty )$ 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. A pioneering work in this direction was the result of Ding–Li–Liu [29] in the torus case, and later in higher genus case by Rupflin [30], Rupflin–Topping [31], and Rupflin–Topping–Zhu [32]. For further references, see for example Rupflin–Topping [33], Huxol–Rupflin–Topping [34] or Rupflin–Topping [35] 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 [36], Williams [37] or Buzano–Rupflin [38] among others. Recently in Huang–Tam [39], 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 Sect. 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 Sect. 3 we define Hilbert spaces X, Y, Z and their closed subsets $B,B'$. So, from Sect. 3 we consider $f \in B$ and $u \in B'$. Then Sect. 4 defines the operator $S_1, S_2$ and shows their properties. Briefly, we can show that $S_1 : B \times B' \rightarrow B$ and $S_2 : B \times B' \rightarrow B'$ and they satisfy twisted partial contraction properties, see Lemmas 6, 7, 10, and 11. In Section 5 we define the operator ${\mathcal {S}}$ on $B \times B'$ 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 solution for harmonic map heat flow. In Sect. 6 we show local estimate and obtain bounds for $\iint \textrm{e}^{2u}|f_t|^4$. This is used in Sect. 7 to show $W^{2,2}$ and higher estimate, which implies boundedness of $|\textrm{d}f|$. Finally in Sect. 8 we prove the main theorem 1 and in Sect. 9 some remarks about finite time singularity are provided.

1.1 Notation

Even though our equation is heat-type equation for varying metric, we use initial metric $g_0$ as default. So, all computations use the metric $g_0$ unless we specify the metric. For example, $|\textrm{d}f|^2$ is calculated in terms of $g_0$ and $|\textrm{d}f|_g^2$ is calculated in terms of g. If the volume form is calculated in terms of metric g, we denote it as $\textrm{dvol}_g$. We also omit $\textrm{dvol}_{g_0}$ and $\textrm{d}t$ if there is no confusion. We also use the simplifications $\Vert \cdot \Vert _{W^{k,p}} = \Vert \cdot \Vert _{W^{k,p}(\varSigma \times [0,T])}$, $\Vert \cdot \Vert _{C^0} = \Vert \cdot \Vert _{C^0(\varSigma \times [0,T])}$ and $\Vert \cdot \Vert _{L^p} = \Vert \cdot \Vert _{L^p(\varSigma \times [0,T])}$. 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 (2).

2.1 Energy and Volume

First note that the 2-form $|\textrm{d}f|^2 \, \textrm{dvol}_g$ is conformally invariant, and that the energy

$$\begin{aligned} E(t) = \tfrac{1}{2} \int |\textrm{d}f|^2 \, \textrm{dvol}_g \end{aligned}$$

(3)

satisfies

$$\begin{aligned} E'(t) = \int \ \langle \textrm{d}f, \textrm{d}f_t\rangle \ =\ - \int \langle \nabla \textrm{d}f, \textrm{e}^{-2u}\tau (f)\rangle \ =\ - \int \textrm{e}^{-2u} |\tau (f)|^2\ \le \ 0. \end{aligned}$$

(4)

Thus $E(t)\le E_0$ for all t.

Lemma 2

The volume satisfies $V(t)\le \textrm{e}^{-2at} V(0) + \tfrac{2b}{a} E_0,$ and hence is finite for all t.

Proof

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

$$\begin{aligned} \textrm{e}^{2u} = \textrm{e}^{-2at} \left( 1 + 2b \int _{0}^{t}\textrm{e}^{2as}|\textrm{d}f|^2 (s)\textrm{d}s \right) . \end{aligned}$$

(5)

Consequently, the volume

$$\begin{aligned} V(t) = \int _{\varSigma } \textrm{dvol}_{g(t)} = \int _{\varSigma } \textrm{e}^{2u} \textrm{dvol}_{g_0} \end{aligned}$$

(6)

can be written as

$$\begin{aligned} V(t) = \textrm{e}^{-2at} \left( V(0) + 4b \int _{0}^{t}\textrm{e}^{2as} E(s){ \mathrm d}s \right) . \end{aligned}$$

(7)

The lemma follows by noting that $E(s)\le E_0$ and integrating. $\square$

2.2 Asymptotic Behavior of Steady Solution

In this subsection we consider the steady solution.

Lemma 3

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

$$\begin{aligned} \textrm{e}^{2u} \rightarrow \tfrac{b}{a}|\textrm{d}f|^2 \end{aligned}$$

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

$$\begin{aligned} V(\infty ) = \tfrac{2b}{a} E. \end{aligned}$$

Proof

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

$$\begin{aligned} \textrm{e}^{2u}&= \textrm{e}^{-2at} \left( 1 + 2b |\textrm{d}f|^2 \frac{\textrm{e}^{2at}-1}{2a} \right) \\&= \textrm{e}^{-2at} + \tfrac{b}{a} |\textrm{d}f|^2 (1-\textrm{e}^{-2at}) \rightarrow \tfrac{b}{a} |\textrm{d}f|^2. \end{aligned}$$

$\square$

This means that, for solutions as in Lemma 3, the energy density $|\textrm{d}f|_g^2 = |\textrm{d}f|^2 \textrm{e}^{-2u}$ converges as $t \rightarrow \infty$ 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 $f(\varSigma )$ 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 $B,B'$. For parabolic theory used here, see Mantegazza–Martinazzi [40], Evans [41] or Lieberman [42]. 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

$$\begin{aligned} Y_T\ = L^2([0,T],W^{4,2}(\varSigma ,{\mathbb {R}}^L)) \cap W^{1,2}([0,T],W^{2,2}(\varSigma ,{\mathbb {R}}^L)) \cap W^{2,2}([0,T],L^2(\varSigma ,{\mathbb {R}}^L)) \end{aligned}$$

is a Hilbert space with norm

$$\begin{aligned} \Vert f\Vert ^2_Y\ =\ \int _0^T\int _\varSigma \ |\nabla ^4f|^2 + |f|^2+|\nabla ^2 f_t|^2 + |f_t|^2+|f_{tt}|^2 \ \textrm{dvol}_{g_0} \ \textrm{d}t. \end{aligned}$$

As in Proposition 4.1 in [40],

$$\begin{aligned} Y_T \hookrightarrow C^0([0,T],C^1(\varSigma ,{\mathbb {R}}^L)) \cap L^4([0,T],W^{3,4}(\varSigma ,{\mathbb {R}}^L)) \cap W^{1,4}([0,T],W^{1,4}(\varSigma ,{\mathbb {R}}^L)) \end{aligned}$$

and there is a constant c such that

$$\begin{aligned} \Vert f\Vert _{C^0} + \Vert \nabla f\Vert _{C^0} + \Vert \nabla ^3 f\Vert _{L^4} + \Vert \nabla f_t\Vert _{L^4} \le {c}\Vert f\Vert _Y. \end{aligned}$$

(8)

Also, by standard parabolic theory (see, for example, [41]), $f \in Y_T$ implies $f \in C^0([0,T],W^{3,2}(\varSigma ,{\mathbb {R}}^L))$, $f_t \in C^0([0,T],W^{1,2}(\varSigma ,{\mathbb {R}}^L))$ and

$$\begin{aligned} \max _{0 \le t \le T} \Vert f(t)\Vert _{W^{3,2}(\varSigma )}, \max _{0 \le t \le T} \Vert f_t(t)\Vert _{W^{1,2}(\varSigma )} \le {c} \Vert f\Vert _Y. \end{aligned}$$

(9)

This also implies that

$$\begin{aligned} \max _{0 \le t \le T} \Vert f(t)\Vert _{W^{2,8}(\varSigma )} \le {c} \Vert f\Vert _Y. \end{aligned}$$

(10)

Next, denote

$$\begin{aligned} X_T = L^2([0,T],W^{2,2}(\varSigma ,{\mathbb {R}}^L)) \cap W^{1,2}([0,T],L^2(\varSigma ,{\mathbb {R}}^L)) \end{aligned}$$

be another Hilbert space with norm

$$\begin{aligned} \Vert f\Vert ^2_{X} = \int _{0}^{T} \int _{\varSigma } |f|^2 + |\nabla ^2 f|^2 + |f_t|^2 \ \textrm{dvol}_{g_0} \ \textrm{d}t. \end{aligned}$$

Note that in the notation of [40], $Y = P^2$ and $X = P^1$.

Now we define spaces for u. The set

$$\begin{aligned} Z_T\ = L^2([0,T],W^{3,2}(\varSigma )) \cap W^{1,2}([0,T],W^{1,2}(\varSigma )) \end{aligned}$$

is a Hilbert space with norm

$$\begin{aligned} \Vert u\Vert _{Z}^2 \ = \ \int _0^T\int _\varSigma \ |\nabla ^3 u|^2 + |u|^2+|\nabla u_t|^2 + |u_t|^2 \ \textrm{dvol}_{g_0} \ \textrm{d}t. \end{aligned}$$

Similar to above, there is a constant c such that

$$\begin{aligned} \Vert \nabla ^2 u\Vert _{L^4} + \Vert u_t\Vert _{L^4} \le {c} \Vert u\Vert _Z \end{aligned}$$

(11)

and

$$\begin{aligned} \max _{0 \le t \le T} \Vert u(t)\Vert _{W^{2,2}(\varSigma )} + \max _{0 \le t \le T} \Vert u_t(t)\Vert _{L^2(\varSigma )} \le {c} \Vert u\Vert _{Z}. \end{aligned}$$

(12)

Also, by Sobolev embedding, we have

$$\begin{aligned} \max _{0 \le t \le T} \Vert u(t)\Vert _{W^{1,8}(\varSigma )} \le {c} \Vert u\Vert _Z. \end{aligned}$$

(13)

Moreover, u is continuous and there is a constant $C_2$ such that for all $u \in Z_T$,

$$\begin{aligned} \Vert u\Vert _{C^0} \le C_2\Vert u\Vert _Z. \end{aligned}$$

(14)

3.2 The Ball B and $B'$

Now we fix $f_0 \in W^{3,2}(\varSigma )$ throughout the section and thereafter. Consider the operator $\partial _t - \textrm{e}^{-2u}\varDelta$. If $\Vert u\Vert _{C^0} \le 1$, this operator is uniformly elliptic. So, Proposition 2.3 of [40] then says that the map $f\mapsto \big (f_0, (\partial _t-\textrm{e}^{-2u}\Delta )f \big )$ is a linear isomorphism

$$\begin{aligned} Y_T \rightarrow W^{3,2}(\varSigma ) \times X_T. \end{aligned}$$

Hence there is a constant $C_1$ such that for each $f_0 \in W^{3,2}(\varSigma )$ and $g \in X_T$, there is a unique solution $h(t,x) \in Y_T$ of the initial value problem

$$\begin{aligned} (\partial _t-\textrm{e}^{-2u}\Delta )h=g \quad h(0)=f_0 \end{aligned}$$

(15)

with

$$\begin{aligned} \Vert h\Vert _Y\le C_1 \big (\Vert f_0\Vert _{3,2} + \Vert g\Vert _{X}\big ). \end{aligned}$$

(16)

Let $h_0(t,x)$ be the unique solution of

$$\begin{aligned} (\partial _t-\Delta )h=0 \quad h(0)=f_0. \end{aligned}$$

(17)

By (16) there is a constant $C_0$, depending on $C_1$ and $\Vert f_0\Vert _{3,2}$ such that

$$\begin{aligned} \Vert h_0\Vert _{Y}\ \le \ C_0. \end{aligned}$$

(18)

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

$$\begin{aligned} B = B_\delta \ =\ \big \{f\in Y_{T}\, \big |\, f(0)=f_0 \ \text{ and }\ \Vert f-h_0\Vert _Y\le \delta \big \} \end{aligned}$$

(19)

is a closed subset of $Y_T$. Note that each $f\in B_\delta$ satisfies

$$\begin{aligned} \Vert f\Vert _Y\ \le \ \Vert f-h_0\Vert _Y+\Vert h_0\Vert _Y\ \le \ \delta +C_0 . \end{aligned}$$

(20)

Also let the ball

$$\begin{aligned} B' = B_{\delta '}' = \{u \in Z_T \, \big | \, u(0) = 0 \ \text{ and } \ \Vert u\Vert _{Z} \le \delta '\} \end{aligned}$$

be a closed subset of $Z_T$. Obviously $h_0 \in B_\delta$ and $0 \in B_{\delta '}'$. For simplicity, we denote $B= B_\delta$ and $B' = B_{\delta '}'$.

Now fix $\delta >0$ and define

$$\begin{aligned} C_3 := 1600 C_0 C_1 C_2. \end{aligned}$$

(21)

Choose $\delta '$ small enough so that $C_2 \delta ' < 1$ which implies $\Vert u\Vert _{C^0} \le 1$. Also we assume $\delta ' \le \frac{\delta }{C_3}$.

4 Construction of Operators

In this section we will construct operators $S_1 : Y_T \times Z_T \rightarrow Y_T$ and $S_2 : Y_T \times Z_T \rightarrow Z_T$. 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

Fix $f_0 \in W^{3,2}(\varSigma )$. Then there is an $T_0 = T_0(C_0,\delta ,\delta ')>0$ such that for all $T \le T_0,$ for each $h \in B$ and $u_1,u_2 \in B',$

$$\begin{aligned} \Vert (\textrm{e}^{2u_2-2u_1}-1) \partial _t h\Vert _X \le \frac{C_3}{2C_1}\Vert u_1-u_2\Vert _Z. \end{aligned}$$

(22)

Proof

Denote

$$\begin{aligned} g := (\textrm{e}^{2u_2-2u_1}-1) \partial _t h. \end{aligned}$$

Recall that

$$\begin{aligned} \left| \textrm{e}^{2u_1-2u_2}-1 \right| \le \textrm{e}^{2|u_1-u_2|} \left| 1-\textrm{e}^{-2|u_1-u_2|} \right| \le 2\textrm{e}^4 |u_1-u_2| \end{aligned}$$

if $\Vert u_1-u_2\Vert _{C^0} \le 2$, which comes from $u_1,u_2 \le B'$. Using $2\textrm{e}^4 \le 200$ and by (9) and (20),

$$\begin{aligned} \Vert g\Vert _{L^2}^2&\le 200^2 \Vert u_1-u_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \partial _t h_2(t)\Vert _{L^2}^2 \, T \\&\le \frac{C_3^2}{16C_1^2}\Vert u_1-u_2\Vert _{Z}^2 \end{aligned}$$

if we choose T small enough.

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

$$\begin{aligned} \left| \nabla ^2 g \right|&= \left| \nabla ^2 \left( (\textrm{e}^{2u_2-2u_1}-1) \partial _t h_2 \right) \right| \\&\le 800 |u_1-u_2| |\nabla (u_1-u_2)|^2 |\partial _t h_2| + 400 |u_1-u_2| |\nabla ^2 (u_1-u_2)| |\partial _t h_2|\\&\quad + 400 |u_1-u_2| |\nabla (u_1-u_2)| |\nabla \partial _t h_2| + 200 |u_1-u_2| |\nabla ^2 \partial _t h_2|. \end{aligned}$$

Hence, by integrating, we have

$$\begin{aligned} \Vert \nabla ^2 g\Vert _{L^2}^2&\le 1600^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla (u_1-u_2)\Vert _{L^8}^4 \max _{0 \le t \le T} \Vert \partial _t h_2(t)\Vert _{L^4(\varSigma )}^2 \, T^{1/2}\\&\quad + 800^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^2 (u_1-u_2)\Vert _{L^4}^2 \max _{0 \le t \le T} \Vert \partial _t h_2(t)\Vert _{L^8(\varSigma )}^4 \, T^{1/2}\\&\quad + 800^2 \Vert u_1-u_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla (u_1(t)-u_2(t))\Vert _{L^4(\varSigma )}^2 \Vert \nabla \partial _t h_2\Vert _{L^4}^2 \, T^{1/2}\\&\quad + 400^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^2 \partial _t h_2\Vert _{L^2}^2\\&\le 400^2 C_2^2 \Vert u_1-u_2\Vert _Z^2 2 C_0^2\\&= \frac{C_3^2}{8C_1^2} \Vert u_1-u_2\Vert _Z^2 \end{aligned}$$

if we choose T small enough.

Finally, we will compute $\Vert g_t\Vert _{L^2}^2$.

$$\begin{aligned} |g_t|&\le 400 |u_1-u_2||\partial _t h_2| |(u_1-u_2)_t| + 200 |u_1-u_2| |\partial _{tt} h_2|. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert g_t\Vert _{L^2}^2&\le 2(400)^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert (u_1-u_2)_t\Vert _{L^4}^2 \max _{0 \le t \le T} \Vert \partial _t h_2\Vert _{L^4(\varSigma )}^2 \, T^{1/2}\\&\quad + 2(200)^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \partial _{tt} h_2\Vert _{L^2}^2 \\&\le 2(200)^2 C_2^2 \Vert u_1-u_2\Vert _Z^2 2 C_0^2\\&= \frac{C_3^2}{16C_1^2} \Vert u_1-u_2\Vert _Z^2 \end{aligned}$$

if we choose T small enough.

Combining all the estimates above, we get

$$\begin{aligned} \Vert (\textrm{e}^{2u_2-2u_1}-1) \partial _t h\Vert _X \le \frac{C_3}{2C_1} \Vert u_1-u_2\Vert _Z \end{aligned}$$

which proves the lemma. $\square$

4.1 The Construction $S_1$

Define an operator

$$\begin{aligned} S_1: Y_T \times Z_T \rightarrow Y_T \end{aligned}$$

by $S_1(f,u)=h$ where $h\in Y_{T}$ is the unique solution of

$$\begin{aligned} (\partial _t-\textrm{e}^{-2u}\Delta )h=\textrm{e}^{-2u} A_f(\textrm{d}f,\textrm{d}f)\quad h(0)=f_0. \end{aligned}$$

(23)

Lemma 5

Fix $f_0\in W^{3,2}(\varSigma )$. Then there is $T_0 = T_0(C_0,\delta ,\delta ') >0$ such that for all $T \le T_0$, $S_1$ restricts to an operator $S_1 : B \times B' \rightarrow B$.

Proof

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

Now we estimate X norm of

$$\begin{aligned} g=\textrm{e}^{-2u(t)} A_f(\textrm{d}f,\textrm{d}f). \end{aligned}$$

First, $|g|^2 \le {c}|\textrm{d}f|^4$, so $\Vert g\Vert _{L^2}^2 \le {c}\Vert f\Vert _Y^4 |\varSigma | T$. Hence if we choose T small enough, we have $\Vert g\Vert _{L^2}^2 \le \frac{\delta ^2}{6C_1^2}$. Next, compute $|\nabla ^2 g|^2$.

$$\begin{aligned} |\nabla ^2 g|&\le {c} |\textrm{d}f|^2 |\nabla ^2 u| + {c} |\textrm{d}f|^2 |\nabla u|^2 + {c} |\textrm{d}f|^3 |\nabla u| + {c}|\nabla \textrm{d}f| |\textrm{d}f| |\nabla u|\\&\quad + {c}|\textrm{d}f|^4 + {c}|\textrm{d}f|^2 |\nabla \textrm{d}f| + {c} |\nabla ^3 f| |\textrm{d}f| + {c} |\nabla \textrm{d}f|^2. \end{aligned}$$

So, using Young’s inequality

$$\begin{aligned} {c}\Vert \textrm{d}f\Vert _{C^0}^2 \iint |\nabla \textrm{d}f|^2 |\nabla u|^2 \le {c}\Vert \textrm{d}f\Vert _{C^0}^4 \iint |\nabla u|^4 + {c} \iint |\nabla \textrm{d}f|^4, \end{aligned}$$

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

$$\begin{aligned} \Vert \nabla ^2 g\Vert _{L^2}^2&\le {c} \Vert \textrm{d}f\Vert _{C^0}^4 \iint (|\nabla ^2 u|^2 + |\nabla u|^4) + {c} \Vert \textrm{d}f\Vert _{C^0}^6 \iint |\nabla u|^2\\&\quad + {c} \Vert \textrm{d}f\Vert _{C^0}^2 \iint |\nabla \textrm{d}f|^2 |\nabla u|^2 + {c} \Vert \textrm{d}f\Vert _{C^0}^8 |\varSigma | T + {c} \Vert \textrm{d}f\Vert _{C^0}^4 \iint |\nabla \textrm{d}f|^2 \\&\quad + {c} \Vert \textrm{d}f\Vert _{C^0}^2 \iint |\nabla ^3 f|^2 + {c} \iint |\nabla \textrm{d}f|^4\\&\le {c} \Vert f\Vert _Y^4 \left( \max _{0 \le t \le T} \Vert \nabla ^2 u(t)\Vert _{L^2(\varSigma )}^2 + \max _{0 \le t \le T} \Vert \nabla u(t)\Vert _{L^4(\varSigma )}^4 \right) T\\&\quad + {c}\Vert f\Vert _Y^6 \max _{0 \le t \le T} \Vert \nabla u(t)\Vert _{L^2(\varSigma )}^2 T + {c}\Vert f\Vert _Y^8 |\varSigma | T \\&\quad + {c} \Vert f\Vert _Y^4 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f (t)\Vert _{L^2(\varSigma )}^2 T + {c} \Vert f\Vert _Y^2 \max _{0 \le t \le T}\Vert \nabla ^3 f(t)\Vert _{L^2(\varSigma )}^2 T \\&\quad + {c} \max _{0 \le t \le T} \Vert \nabla \textrm{d}f\Vert _{L^4(\varSigma )}^4 T\\&\le \frac{\delta ^2}{6C_1^2} \end{aligned}$$

if we choose T small enough. Finally,

$$\begin{aligned} |g_t|&\le {c} |\textrm{d}f|^2 |u_t| + {c} |\textrm{d}f|^2 |f_t| + {c}|\textrm{d}f_t| |\textrm{d}f| \end{aligned}$$

and

$$\begin{aligned} \Vert g_t\Vert _{L^2}^2&\le {c}\Vert \textrm{d}f\Vert _{C^0}^4 \iint |u_t|^2 + |f_t|^2 + {c} \Vert \textrm{d}f\Vert _{C^0}^2 \iint |\textrm{d}f_t|^2\\&\le {c}\Vert f\Vert _{Y}^4 \left( \max _{0 \le t \le T} \Vert u_t(t) \Vert _{L^2(\varSigma )}^2 + \max _{0 \le t \le T} \Vert f_t(t)\Vert _{L^2(\varSigma )}^2 \right) T \\&\quad + {c} \Vert f\Vert _Y^2 \max _{0 \le t \le T} \Vert \nabla f_t\Vert _{L^2(\varSigma )}^2 T\\&\le \frac{\delta ^2}{6C_1^2} \end{aligned}$$

if we choose T small enough.

Therefore, if we choose T small enough, we have $\Vert g\Vert _X^2 \le \frac{\delta ^2}{2C_1^2}$. Noting that $S(f)-h_0=h-h_0$ satisfies

$$\begin{aligned} (\partial _t-\textrm{e}^{-2u}\Delta )(h-h_0)=g + (\textrm{e}^{-2u}-1) \Delta h_0 \quad (h-h_0)(0)=0. \end{aligned}$$

The bounds (16) give

$$\begin{aligned} \Vert S(f)-h_0\Vert _Y^2\ {}&\le \ C_1^2 \left( \Vert g\Vert _{X}^2 + \Vert (\textrm{e}^{-2u}-1) \Delta h_0\Vert _{X}^2 \right) \\&= C_1^2 \left( \Vert g\Vert _X^2 + \Vert (\textrm{e}^{-2u}-1) \partial _t h_0\Vert _{X}^2 \right) \end{aligned}$$

because $h_0$ satisfies (17).

Now by Lemma 4 with $h=h_0$, $u_1=u$, $u_2=0$,

$$\begin{aligned} \Vert (\textrm{e}^{-2u}-1) \partial _t h_0\Vert _{X} \le \frac{C_3}{2C_1}\Vert u\Vert _Z \le \frac{\delta }{2C_1}. \end{aligned}$$

This implies

$$\begin{aligned} \Vert S(f)-h_0\Vert _Y^2\ \le \ C_1^2 \left( \frac{\delta ^2}{2C_1^2} +\frac{\delta ^2}{4C_1^2} \right) \le \delta ^2. \end{aligned}$$

Therefore $S(f)\in B$. $\square$

Lemma 6

Fix $f_0 \in W^{3,2}(\varSigma )$ and $u \in B'$. Then there is an $T_0 = T_0(C_0,\delta ,\delta ')>0$ such that for all $T \le T_0$ and for each $f_1,f_2\in B,$

$$\begin{aligned} \Vert S_1(f_1,u)-S_1(f_2,u)\Vert _Y \le \frac{1}{3} \Vert f_1-f_2\Vert _Y. \end{aligned}$$

(24)

Proof

Set $h_i=S_1(f_i,u)$ and $g_i=\textrm{e}^{-2u} A_{f_i}(\textrm{d}f_i,\textrm{d}f_i)$ for $i=1,2$ and subtracting, the function $h_1-h_2$ satisfies

$$\begin{aligned} (\partial _t-\textrm{e}^{-2u}\Delta )(h_1-h_2)=g_1-g_2 \quad (h_1-h_2)(0)= 0. \end{aligned}$$

Hence (15) gives a bound

$$\begin{aligned} \Vert h_1-h_2\Vert _Y \le C_1 \Vert g_1-g_2\Vert _{X}. \end{aligned}$$

(25)

Next, we have

$$\begin{aligned} g_1-g_2\ {}&= \ \textrm{e}^{-2u} (A_{f_1}-A_{f_2})(\textrm{d}f_1,\textrm{d}f_1) + \textrm{e}^{-2u} A_{f_2}(\textrm{d}f_1+\textrm{d}f_2,\textrm{d}f_1-\textrm{d}f_2) \\&= \textrm{I} + \textrm{II}. \end{aligned}$$

So, there is a constant c with

$$\begin{aligned} |g_1-g_2|^2\ \le \ {c} |{f_1}-f_2|^2 |\textrm{d}f_1|^4 + {c} |\textrm{d}f_1-\textrm{d}f_2|^2 \left( |\textrm{d}f_1|^2+|\textrm{d}f_2|^2\right) . \end{aligned}$$

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

$$\begin{aligned} \Vert g_1-g_2\Vert _{L^2}^2&\le {c} \Vert {f_1}-f_2\Vert _{C^0}^2 \iint |\textrm{d}f_1|^4 \ +\ {c} \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{L^4}^2 \left( \Vert \textrm{d}f_1\Vert _{L^4}^2 +\Vert \textrm{d}f_2\Vert _{L^4}^2 \right) \\&\le {c} \Vert {f_1}-f_2\Vert _Y^2 \Vert f_1\Vert _Y^4 |\varSigma | T \ +\ {c} \Vert {f_1}-f_2\Vert _Y^2 (\Vert f_1\Vert _Y^2 + \Vert f_2\Vert _Y^2) |\varSigma |^{1/2} T^{1/2}\\&\le \frac{1}{27C_1^2}\, \Vert {f_1}-f_2\Vert _Y^2 \end{aligned}$$

if we choose T small enough.

For $\nabla ^2 (g_1-g_2)$, first note that

$$\begin{aligned} \nabla (A_{f_1}-A_{f_2})&= DA_{f_1} \textrm{d}f_1 - DA_{f_2} \textrm{d}f_2 = (DA_{f_1} - DA_{f_2}) \textrm{d}f_1 + DA_{f_2} (\textrm{d}f_1-\textrm{d}f_2)\\ \nabla (DA_{f_1} - DA_{f_2})&= (D^2 A_{f_1} - D^2 A_{f_2})\textrm{d}f_1 + D^2 A_{f_2}(\textrm{d}f_1-\textrm{d}f_2). \end{aligned}$$

So, we get

$$\begin{aligned} |\nabla ^2 I|&\le {c}|f_1-f_2| |\textrm{d}f_1|^2 |\nabla ^2 u| + {c} |f_1-f_2| |\textrm{d}f_1|^2 |\nabla u|^2 + {c} |f_1-f_2| |\textrm{d}f_1|^3 |\nabla u| \\&\quad + {c} |\textrm{d}f_1-\textrm{d}f_2| |\textrm{d}f_1|^2 |\nabla u| + {c} |f_1-f_2| |\nabla \textrm{d}f_1| |\textrm{d}f_1| |\nabla u|\\&\quad + {c} |f_1-f_2||\textrm{d}f_1|^4 + {c}|f_1-f_2| |\nabla \textrm{d}f_1| |\textrm{d}f_1|^2\\&\quad + {c}|\textrm{d}f_1-\textrm{d}f_2||\textrm{d}f_2||\textrm{d}f_1|^2 + {c} |\nabla \textrm{d}f_1-\nabla \textrm{d}f_2| |\textrm{d}f_1|^2\\&\quad + {c}|\textrm{d}f_1-\textrm{d}f_2| |\nabla \textrm{d}f_1| |\textrm{d}f_1|\\&\quad + {c} |f_1-f_2| |\nabla ^3 f_1| |\textrm{d}f_1| + {c} |f_1-f_2| |\nabla \textrm{d}f_1|^2. \end{aligned}$$

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

$$\begin{aligned} \iint |f_1-f_2|^2 |\textrm{d}f_1|^4 |\nabla ^2 u|^2&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^4 \max _{0 \le t \le T} \Vert \nabla ^2 u(t)\Vert _{L^2(\varSigma )}^2 T\\ \iint |f_1-f_2|^2 |\textrm{d}f_1|^4 |\nabla u|^4&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^4 \max _{0 \le t \le T} \Vert \nabla u(t)\Vert _{L^4(\varSigma )}^4 T\\ \iint |f_1-f_2|^2 |\textrm{d}f_1|^6 |\nabla u|^2&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^6 \max _{0 \le t \le T} \Vert \nabla u(t)\Vert _{L^2(\varSigma )}^2 T\\ \iint |f_1-f_2|^2 |\nabla \textrm{d}f_1|^2 |\textrm{d}f_1|^2 |\nabla u|^2&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^4 \max _{0 \le t \le T} \Vert \nabla u(t)\Vert _{L^4(\varSigma )}^4 T\\&\quad + \Vert f_1-f_2\Vert _Y^2 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f_1(t)\Vert _{L^4(\varSigma )}^4 T\\ \iint |f_1-f_2|^2 |\textrm{d}f_1|^8&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^8 |\varSigma | T\\ \iint |f_1-f_2|^2 |\nabla \textrm{d}f_1|^2 |\textrm{d}f_1|^4&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^4 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f_1(t)\Vert _{L^2(\varSigma )}^2 T\\ \iint |\textrm{d}f_1-\textrm{d}f_2|^2 |\textrm{d}f_2|^2 |\textrm{d}f_1|^4&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_2\Vert _Y^2 \Vert f_1\Vert _Y^4 |\varSigma | T\\ \iint |\nabla \textrm{d}f_1 - \nabla \textrm{d}f_2|^2 |\textrm{d}f_1|^4&\le \Vert f_1\Vert _Y^4 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f_1(t) - \nabla \textrm{d}f_2(t)\Vert _{L^2(\varSigma )}^2 T\\ \iint |\textrm{d}f_1-\textrm{d}f_2|^2 |\nabla \textrm{d}f_1|^2 |\textrm{d}f_1|^2&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^2 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f_1(t)\Vert _{L^2(\varSigma )}^2 T\\ \iint |f_1-f_2|^2 |\nabla ^3 f_1|^2 |\textrm{d}f_1|^2&\le \Vert f_1-f_2\Vert _Y^2 \Vert f_1\Vert _Y^2 \max _{0 \le t \le T} \Vert \nabla ^3 f_1(t)\Vert _{L^2(\varSigma )}^2 T\\ \iint |f_1-f_2|^2 |\nabla \textrm{d}f_1|^4&\le \Vert f_1-f_2\Vert _Y^2 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f_1(t)\Vert _{L^4(\varSigma )}^4 T. \end{aligned}$$

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

$$\begin{aligned} \Vert \nabla ^2 I\Vert _{L^2}^2&\le \frac{1}{54C_1^2} \Vert f_1-f_2\Vert _Y^2. \end{aligned}$$

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

$$\begin{aligned} \Vert \nabla ^2 II\Vert _{L^2}^2&\le \frac{1}{54C_1^2} \Vert f_1-f_2\Vert _Y^2. \end{aligned}$$

Hence, we obtain that $\Vert \nabla ^2 (g_1-g_2)\Vert _{L^2}^2 \le \frac{1}{27C_1^2} \Vert f_1-f_2\Vert _Y^2$.

Finally, compute $\partial _t (g_1-g_2)$. As above, note that

$$\begin{aligned} \partial _t (A_{f_1}-A_{f_2}) = DA_{f_1} \partial _t f_1 - DA_{f_2} \partial _t f_2 = (DA_{f_1}-DA_{f_2}) \partial _t f_1 + DA_{f_2} (\partial _t (f_1-f_2)). \end{aligned}$$

So,

$$\begin{aligned} |\partial _t (g_1-g_2)|&\le {c}|f_1-f_2| |\textrm{d}f_1|^2 |u_t| + {c}|f_1-f_2| |\partial _t f_1| |\textrm{d}f_1|^2 + {c}|\partial _t (f_1-f_2)| |\textrm{d}f_1|^2\\&\quad + {c}|f_1-f_2| |\partial _t \textrm{d}f_1| |\textrm{d}f_1| \\&\quad + {c} |\textrm{d}f_1+\textrm{d}f_2| |\textrm{d}f_1-\textrm{d}f_2| |u_t| +{c} |\partial _t f_2| |\textrm{d}f_1+\textrm{d}f_2| |\textrm{d}f_1-\textrm{d}f_2| \\&\quad + {c}|\partial _t (\textrm{d}f_1+\textrm{d}f_2)| |\textrm{d}f_1-\textrm{d}f_2| + {c} |\textrm{d}f_1+\textrm{d}f_2| |\partial _t (\textrm{d}f_1-\textrm{d}f_2)|. \end{aligned}$$

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

$$\begin{aligned}&\Vert \partial _t (g_1-g_2)\Vert _{L^2}^2 \le {c} \Vert f_1\Vert _Y^4 \Vert f_1-f_2\Vert _Y^2 \left( \max _{0 \le t \le T} \Vert u_t(t)\Vert _{L^2(\varSigma )}^2 + \max _{0 \le t \le T} \Vert \partial _t f_1(t)\Vert _{L^2(\varSigma )}^2 \right) T \\&\qquad + {c} \Vert f_1\Vert _Y^4 \max _{0 \le t \le T} \Vert \partial _t (f_1(t)-f_2(t))\Vert _{L^2(\varSigma )}^2 T\\&\qquad + {c} \Vert f_1\Vert _Y^2 \Vert f_1-f_2\Vert _Y^2 \max _{0 \le t \le T} \Vert \partial _t \textrm{d}f_1(t) \Vert _{L^2(\varSigma )}^2 T\\&\qquad + {c} (\Vert f_1\Vert _Y^2 + \Vert f_2\Vert _Y^2) \Vert f_1-f_2\Vert _Y^2 \left( \max _{0 \le t \le T} \Vert u_t(t)\Vert _{L^2(\varSigma )}^2 + \max _{0 \le t \le T} \Vert \partial _t f_2(t)\Vert _{L^2(\varSigma )}^2 \right) T\\&\qquad + {c} \Vert f_1-f_2\Vert _Y^2 \left( \max _{0 \le t \le T} \Vert \partial _t \textrm{d}f_1(t)\Vert _{L^2(\varSigma )}^2 + \max _{0 \le t \le T} \Vert \partial _t \textrm{d}f_2(t)\Vert _{L^2(\varSigma )}^2 \right) T\\&\qquad + {c} (\Vert f_1\Vert _Y^2 + \Vert f_2\Vert _Y^2) \max _{0 \le t \le T} \Vert \partial _t (\textrm{d}f_1(t)-\textrm{d}f_2(t))\Vert _{L^2(\varSigma )}^2 T\\&\quad \le \frac{1}{27C_1^2} \Vert f_1-f_2\Vert _Y^2 \end{aligned}$$

if we choose T small enough.

Combine all of them,

$$\begin{aligned} \Vert h_1-h_2\Vert _Y \le C_1 \Vert g_1-g_2\Vert _X \le \frac{1}{3} \Vert f_1-f_2\Vert _Y \end{aligned}$$

which proves the lemma. $\square$

Lemma 7

Fix $f_0 \in W^{3,2}(\varSigma )$ and $f \in B$. Then there is an $T_0 = T_0(C_0,\delta ,\delta ')>0$ such that for all $T \le T_0$ and for each $u_1,u_2\in B',$

$$\begin{aligned} \Vert S_1(f,u_1)-S_1(f,u_2)\Vert _Y \le \frac{C_3}{2} \Vert u_1-u_2\Vert _{Z}. \end{aligned}$$

(26)

Proof

Set $h_i=S_1(f,u_i)$. Multiplying $\textrm{e}^{2u_i}$ to the equation for $h_i$ respectively and subtracting them gives

$$\begin{aligned} \textrm{e}^{2u_1} \partial _t (h_1 - h_2) - \Delta (h_1-h_2)&= -(\textrm{e}^{2u_1}-\textrm{e}^{2u_2}) \partial _t h_2,\\ (\partial _t - \textrm{e}^{-2u_1} \Delta ) (h_1-h_2)&= (\textrm{e}^{2u_2-2u_1}-1) \partial _t h_2. \end{aligned}$$

So, $h_1-h_2$ satisfies the estimate from (16), and by Lemma 4,

$$\begin{aligned} \Vert h_1-h_2\Vert _Y \le C_1 \Vert (\textrm{e}^{2u_2-2u_1}-1) \partial _t h_2\Vert _X \le \frac{C_3}{2} \Vert u_1-u_2\Vert _Z \end{aligned}$$

if we choose T small enough.

$\square$

4.2 The Construction $S_2$

Define an operator

$$\begin{aligned} S_2: Y_T \times Z_T \rightarrow Z_T \end{aligned}$$

by $S_2(f,u)=v$ where $v\in Z_{T}$ is the unique solution of

$$\begin{aligned} \partial _t v =b|\textrm{d}f|^2 \textrm{e}^{-2u} - a \quad v(0)=0. \end{aligned}$$

(27)

Lemma 8

In the above definition, $v \in Z_T$.

Proof

From (27), we directly get

$$\begin{aligned} v(t) = \int _{0}^{t} (b|\textrm{d}f|^2 \textrm{e}^{-2u} - a). \end{aligned}$$

(28)

So, $\Vert v\Vert _{L^2}$ and $\Vert v_t\Vert _{L^2}$ is trivially bounded if $f \in Y_T$ and $u \in Z_T$. (Because $u \in Z_T$, we have $\textrm{e}^{-2u}$ is pointwise uniformly bounded by $\textrm{e}^{C\Vert u\Vert _Z}$.) Applying Cauchy–Schwarz, we obtain the pointwise bound

$$\begin{aligned} |\nabla ^3 v|^2&= \left| b \int _{0}^{T} \nabla ^2 \left( \langle \nabla \textrm{d}f, \textrm{d}f \rangle \textrm{e}^{-2u} -2 |\textrm{d}f|^2 \textrm{e}^{-2u} \nabla u \right) \right| ^2\\&\le {c} T \int _{0}^{T} \Big ( |\nabla ^4 f|^2 |\textrm{d}f|^2 + |\nabla ^3 f|^2 |\nabla \textrm{d}f|^2 + |\nabla ^3 f|^2 |\textrm{d}f|^2 |\nabla u|^2 + |\nabla ^2 f|^4 |\nabla u|^2 \\&\quad + |\nabla ^2 f|^2 |\textrm{d}f|^2 (|\nabla u|^4 + |\nabla ^2 u|^2) + |\textrm{d}f|^4 (|\nabla u|^6 + |\nabla ^2 u|^2 |\nabla u|^2 + |\nabla ^3 u|^2) \Big ) \end{aligned}$$

so

$$\begin{aligned} \Vert \nabla ^3 v\Vert _{L^2}^2&\le {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^2 \Vert \nabla ^4 f\Vert _{L^2}^2 + {c} T^2 \Vert \nabla ^3 f\Vert _{L^4}^2 \Vert \nabla ^2 f\Vert _{L^4}^2 \\&\quad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^2 \Vert \nabla ^3 f\Vert _{L^4}^2 \Vert \nabla u\Vert _{L^4}^2 + {c} T^2 \Vert \nabla ^2 f\Vert _{L^8}^4 \Vert \nabla u\Vert _{L^4}^2\\&\quad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^2 \Vert \nabla ^2 f\Vert _{L^4}^2 \left( \Vert \nabla u\Vert _{L^8}^4 + \Vert \nabla ^2 u\Vert _{L^4}^2\right) \\&\quad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^4 \left( \Vert \nabla u\Vert _{L^6}^6 + \Vert \nabla ^2 u\Vert _{L^4}^2 \Vert \nabla u\Vert _{L^4}^2 + \Vert \nabla ^3 u\Vert _{L^2}^2\right) \end{aligned}$$

which is bounded if $f \in Y_T$ and $u \in Z_T$.

Finally, compute $\nabla v_t$ from (27).

$$\begin{aligned} |\nabla v_t|^2&\le {c} |\nabla \textrm{d}f| |\textrm{d}f| + |\textrm{d}f|^2 |\nabla u|,\\ \Vert \nabla v_t\Vert _{L^2}^2&\le {c} \Vert \textrm{d}f\Vert _{C^0}^2 \Vert \nabla \textrm{d}f\Vert _{L^2}^2 + {c} \Vert \textrm{d}f\Vert _{C^0}^4 \Vert \nabla u\Vert _{L^2}^2 \end{aligned}$$

which is bounded if $f \in Y_T$ and $u \in Z_T$. $\square$

In fact, we can show further.

Lemma 9

Fix $f_0\in W^{3,2}(\varSigma )$. Then there is $T_0 = T_0(C_0,\delta ,\delta ') >0$ such that for all $T \le T_0,$ $S_2$ restricts to an operator $S_2 : B \times B' \rightarrow B'$.

Proof

From previous calculation, we have

$$\begin{aligned} \Vert \nabla ^3 v\Vert _{L^2}^2 \le {c} T^2 \Vert f\Vert _Y^4 (1 + \Vert u\Vert _Z^2 + \Vert u\Vert _Z^4 + \Vert u\Vert _Z^6). \end{aligned}$$

So, if we choose T small enough, we get $\Vert \nabla ^3 v\Vert _{L^2}^2 \le \frac{{\delta '}^2}{4}$. Also, because $|v_t| \le {c}(\Vert \textrm{d}f\Vert _{C^0} + 1)$ and $|v| \le {c} T (\Vert \textrm{d}f\Vert _{C^0} + 1)$, we can make $\Vert v_t\Vert _{L^2}^2, \Vert v\Vert _{L^2}^2 \le \frac{{\delta '}^2}{4}$ if we choose T small. Finally,

$$\begin{aligned} \Vert \nabla v_t\Vert _{L^2}^2 \le {c}\Vert \textrm{d}f\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla \textrm{d}f(t)\Vert _{L^2(\varSigma )}^2 \, T + {c} \Vert \textrm{d}f\Vert _{C^0}^4 \max _{0 \le t \le T} \Vert \nabla u(t)\Vert _{L^2(\varSigma )}^2 \, T \end{aligned}$$

so if we choose T small enough, we get $\Vert \nabla v_t\Vert _{L^2}^2 \le \frac{{\delta '}^2}{4}$. This proves the lemma. $\square$

Lemma 10

Fix $f_0 \in W^{3,2}(\varSigma )$ and $u \in B'$. Then there is an $T_0 = T_0(C_0,\delta ,\delta ')>0$ such that for all $T \le T_0$ and for each $f_1,f_2\in B,$

$$\begin{aligned} \Vert S_2(f_1,u)-S_2(f_2,u)\Vert _Z \le T^{1/4} \Vert f_1-f_2\Vert _Y. \end{aligned}$$

(29)

Proof

Set $v_i = S_2(f_i,u)$. Then from (27), subtracting them gives

$$\begin{aligned} (v_1-v_2)_t&= b(|\textrm{d}f_1|^2 - |\textrm{d}f_2|^2) \textrm{e}^{-2u} = b \textrm{e}^{-2u} \langle \textrm{d}f_1+\textrm{d}f_2, \textrm{d}f_1-\textrm{d}f_2 \rangle ,\\ v_1-v_2&= b \int _{0}^{t} \textrm{e}^{-2u} \langle \textrm{d}f_1+\textrm{d}f_2, \textrm{d}f_1-\textrm{d}f_2 \rangle . \end{aligned}$$

So,

$$\begin{aligned} \Vert v_1-v_2\Vert _{L^2}^2&\le {c} T^2 (\Vert \textrm{d}f_1\Vert _{C^0}^2 + \Vert \textrm{d}f_2\Vert _{C^0}^2) \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{L^2}^2\\&\le \frac{\sqrt{T}}{4} \Vert f_1-f_2\Vert _Y^2 \end{aligned}$$

if we choose T small enough. Also,

$$\begin{aligned} \Vert (v_1-v_2)_t\Vert _{L^2}^2&\le {c} (\Vert f_1\Vert _Y^2 + \Vert f_2\Vert _Y^2) \max _{0 \le t \le T} \Vert \textrm{d}f_1(t)-\textrm{d}f_2(t)\Vert _{L^2(\varSigma )}^2 \, T\\&\le \frac{\sqrt{T}}{4} \Vert f_1-f_2\Vert _Y^2 \end{aligned}$$

if we choose T small enough.

Next, compute $\nabla ^3 (v_1-v_2)$.

$$\begin{aligned} \nabla (v_1-v_2)&= b \int _{0}^{t} \textrm{e}^{-2u} \Big ( \langle \nabla (\textrm{d}f_1+\textrm{d}f_2), \textrm{d}f_1-\textrm{d}f_2 \rangle + \langle \textrm{d}f_1+\textrm{d}f_2, \nabla (\textrm{d}f_1-\textrm{d}f_2) \rangle \\&\quad - \langle \textrm{d}f_1+\textrm{d}f_2, \textrm{d}f_1-\textrm{d}f_2 \rangle 2\nabla u \Big ). \end{aligned}$$

So,

$$\begin{aligned} |\nabla ^3 (v_1-v_2)|&\le {c} \int _{0}^{T} \textrm{e}^{-2u} \Big ( |\nabla ^3 (\textrm{d}f_1+\textrm{d}f_2)| |\textrm{d}f_1-\textrm{d}f_2| + |\nabla ^2 (\textrm{d}f_1+\textrm{d}f_2)| |\nabla (\textrm{d}f_1-\textrm{d}f_2)| \\&\quad + |\nabla (\textrm{d}f_1+\textrm{d}f_2)| |\nabla ^2 (\textrm{d}f_1-\textrm{d}f_2)| + |\textrm{d}f_1+\textrm{d}f_2| |\nabla ^3 (\textrm{d}f_1-\textrm{d}f_2)| \\&\quad + |\nabla ^2 (\textrm{d}f_1+\textrm{d}f_2)| |\textrm{d}f_1-\textrm{d}f_2 | |\nabla u| + |\nabla (\textrm{d}f_1+\textrm{d}f_2)| |\nabla (\textrm{d}f_1-\textrm{d}f_2)| |\nabla u|\\&\quad + |\textrm{d}f_1+\textrm{d}f_2| |\nabla ^2 (\textrm{d}f_1-\textrm{d}f_2)| |\nabla u|\\&\quad + |\nabla (\textrm{d}f_1+\textrm{d}f_2)| |\textrm{d}f_1-\textrm{d}f_2| (|\nabla u|^2 + |\nabla ^2 u|)\\&\quad + |\textrm{d}f_1+\textrm{d}f_2| |\nabla (\textrm{d}f_1-\textrm{d}f_2)| (|\nabla u|^2 + |\nabla ^2 u|)\\&\quad + |\textrm{d}f_1+\textrm{d}f_2| |\textrm{d}f_1-\textrm{d}f_2| (|\nabla u|^3 + |\nabla ^2 u| |\nabla u| + |\nabla ^3 u|) \Big ). \end{aligned}$$

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

$$\begin{aligned}&\Vert \nabla ^3 (v_1-v_2)\Vert _{L^2}^2\\&\le {c} T^2 \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{C^0}^2 \Vert \nabla ^4 (f_1+ f_2)\Vert _{L^2}^2 + {c} T^2 \Vert \nabla ^3 ( f_1 + f_2)\Vert _{L^4}^2 \Vert \nabla ^2 (f_1-f_2)\Vert _{L^4}^2\\&\qquad + {c} T^2 \Vert \nabla ^2 (f_1+f_2)\Vert _{L^4}^2 \Vert \nabla ^3 (f_1-f_2)\Vert _{L^4}^2 + {c} T^2 \Vert \textrm{d}f_1+\textrm{d}f_2\Vert _{C^0}^2 \Vert \nabla ^4 (f_1-f_2)\Vert _{L^2}^2\\&\qquad + {c} T^2 \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{C^0}^2 \Vert \nabla ^3 (f_1+f_2)\Vert _{L^4}^2 \Vert \nabla u\Vert _{L^4}^2\\&\qquad + {c} T^2 \Vert \nabla ^2 (f_1-f_2) \Vert _{L^4}^2 \Vert \nabla ^2 (f_1+f_2)\Vert _{L^8}^4 \Vert \nabla u\Vert _{L^8}^4\\&\qquad + {c} T^2 \Vert \textrm{d}f_1+\textrm{d}f_2\Vert _{C^0}^2 \Vert \nabla ^3 (f_1-f_2)\Vert _{L^4}^2 \Vert \nabla u\Vert _{L^4}^2\\&\qquad + {c} T^2 \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{C^0}^2 \Vert \nabla ^2 (f_1+f_2)\Vert _{L^4}^2 (\Vert \nabla u\Vert _{L^8}^4 + \Vert \nabla ^2 u\Vert _{L^4}^2)\\&\qquad + {c} T^2 \Vert \textrm{d}f_1+\textrm{d}f_2\Vert _{C^0}^2 \Vert \nabla ^2 (f_1-f_2)\Vert _{L^4}^2 (\Vert \nabla u\Vert _{L^8}^4 + \Vert \nabla ^2 u\Vert _{L^4}^2)\\&\qquad + {c} T^2 \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{C^0}^2 \Vert \textrm{d}f_1+\textrm{d}f_2\Vert _{C^0}^2 ( \Vert \nabla u\Vert _{L^6}^6 + \Vert \nabla ^2 u \Vert _{L^4}^2 \Vert \nabla u\Vert _{L^4}^2 + \Vert \nabla ^3 u\Vert _{L^2}^2)\\&\quad \le \frac{\sqrt{T}}{4} \Vert f_1-f_2\Vert _Y^2 \end{aligned}$$

if we choose T small enough.

Finally consider $\nabla (v_1-v_2)_t$.

$$\begin{aligned} \nabla (v_1-v_2)_t&= b \textrm{e}^{-2u} \left( \langle \nabla (\textrm{d}f_1+\textrm{d}f_2), \textrm{d}f_1-\textrm{d}f_2 \rangle + \langle \textrm{d}f_1+\textrm{d}f_2, \nabla (\textrm{d}f_1-\textrm{d}f_2) \rangle \right. \\&\quad \left. - \langle \textrm{d}f_1+\textrm{d}f_2, \textrm{d}f_1-\textrm{d}f_2 \rangle 2\nabla u \right) . \end{aligned}$$

So,

$$\begin{aligned} \Vert \nabla (v_1-v_2)_t\Vert _{L^2}^2&\le {c} \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla ^2 (f_1(t)+f_2(t))\Vert _{L^2(\varSigma )}^2 \, T \\&\quad + {c} \Vert \textrm{d}f_1+\textrm{d}f_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla ^2 (f_1(t)-f_2(t))\Vert _{L^2(\varSigma )}^2 \, T \\&\quad + {c} \Vert \textrm{d}f_1+\textrm{d}f_2\Vert _{C^0}^2 \Vert \textrm{d}f_1-\textrm{d}f_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla u(t) \Vert _{L^2(\varSigma )}^2 \, T\\&\le \frac{\sqrt{T}}{4} \Vert f_1-f_2\Vert _Y^2 \end{aligned}$$

if we choose T small enough.

In summary, we get

$$\begin{aligned} \Vert v_1-v_2\Vert _Z^2 \le \sqrt{T} \Vert f_1-f_2\Vert _Y^2 \end{aligned}$$

which proves the lemma.

$\square$

Lemma 11

Fix $f_0 \in W^{3,2}(\varSigma )$ and $f \in B$. Then there is an $T_0 = T_0(C_0,\delta ,\delta ')>0$ such that for all $T \le T_0$ and for each $u_1,u_2\in B',$

$$\begin{aligned} \Vert S_2(f,u_1)-S_2(f,u_2)\Vert _Z \le \frac{1}{3} \Vert u_1-u_2\Vert _{Z}. \end{aligned}$$

(30)

Proof

Set $v_i=S_2(f,u_i)$. Subtracting them gives

$$\begin{aligned} (v_1-v_2)_t&= b |\textrm{d}f|^2 (\textrm{e}^{-2u_1}-\textrm{e}^{-2u_2}),\\ v_1-v_2&= b \int _{0}^{t} |\textrm{d}f|^2 (\textrm{e}^{-2u_1}-\textrm{e}^{-2u_2}). \end{aligned}$$

Using $|\textrm{e}^{-2u_1}-\textrm{e}^{-2u_2}| \le {c}|u_1-u_2|$, we have

$$\begin{aligned} \Vert v_1-v_2\Vert _{L^2}^2&\le {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^4 \Vert u_1-u_2\Vert _{L^2}^2,\\ \Vert (v_1-v_2)_t\Vert _{L^2}^2&\le {c} \Vert \textrm{d}f\Vert _{C^0}^4 \max _{0 \le t \le T} \Vert u_1(t)-u_2(t)\Vert _{L^2(\varSigma )}^2 \, T, \end{aligned}$$

so if we choose T small enough, we have that $\Vert v_1-v_2\Vert _{L^2}^2, \Vert (v_1-v_2)_t\Vert _{L^2}^2 \le \frac{1}{36} \Vert u_1-u_2\Vert _Z^2$.

Next, compute $\nabla ^3 (v_1-v_2)$.

$$\begin{aligned} \nabla (v_1-v_2)&= b \int _{0}^{t} \Big ( 2\langle \nabla \textrm{d}f, \textrm{d}f \rangle (\textrm{e}^{-2u_1}-\textrm{e}^{-2u_2}) - |\textrm{d}f|^2 (\textrm{e}^{-2u_1}-\textrm{e}^{-2u_2}) 2 (\nabla u_1 - \nabla u_2) \Big ). \end{aligned}$$

So,

$$\begin{aligned}&|\nabla ^3 (v_1-v_2)| \le {c} \int _{0}^{T} \Big ( |\nabla ^4 f| |\textrm{d}f| |u_1-u_2| + |\nabla ^3 f| |\nabla ^2 f| |u_1-u_2|\\&\quad + (|\nabla ^3 f| |\textrm{d}f| + |\nabla ^2 f|^2) |u_1-u_2| |\nabla (u_1-u_2)|\\&\quad + |\nabla ^2 f| |\textrm{d}f| |u_1-u_2| (|\nabla (u_1-u_2)|^2 + |\nabla ^2 (u_1-u_2)|)\\&\quad + |\textrm{d}f|^2 |u_1-u_2| (|\nabla (u_1-u_2)|^3 + |\nabla ^2 (u_1-u_2)| |\nabla (u_1-u_2)| + |\nabla ^3 (u_1-u_2)|) \Big ). \end{aligned}$$

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

$$\begin{aligned}&\Vert \nabla ^3 (v_1-v_2)\Vert _{L^2}^2\\&\quad \le {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^4 f\Vert _{L^2}^2 + {c} T^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^3 f\Vert _{L^4}^2 \Vert \nabla ^2 f\Vert _{L^4}^2\\&\qquad + {c} T^2 (\Vert \textrm{d}f\Vert _{C^0}^2 \Vert \nabla ^3 f\Vert _{L^4}^2 + \Vert \nabla ^2 f\Vert _{L^8}^4) \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla (u_1-u_2)\Vert _{L^4}^2\\&\qquad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^2 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^2 f\Vert _{L^4}^2 (\Vert \nabla (u_1-u_2)\Vert _{L^8}^4 + \Vert \nabla ^2 (u_1-u_2)\Vert _{L^4}^2)\\&\qquad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^4 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla (u_1-u_2)\Vert _{L^6}^6\\&\qquad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^4 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^2 (u_1-u_2)\Vert _{L^4}^2 \Vert \nabla (u_1-u_2)\Vert _{L^4}^2\\&\qquad + {c} T^2 \Vert \textrm{d}f\Vert _{C^0}^4 \Vert u_1-u_2\Vert _{C^0}^2 \Vert \nabla ^3 (u_1-u_2)\Vert _{L^2}^2\\&\quad \le \frac{1}{36} \Vert u_1-u_2\Vert _Z^2 \end{aligned}$$

if we choose T small enough.

Finally,

$$\begin{aligned} |\nabla (v_1-v_2)_t|&\le {c} |\nabla \textrm{d}f| |\textrm{d}f| |u_1-u_2| + |\textrm{d}f|^2 |u_1-u_2| |\nabla (u_1 - u_2)| \end{aligned}$$

so

$$\begin{aligned} \Vert \nabla (v_1-v_2)_t\Vert _{L^2}^2&\le {c} \Vert \textrm{d}f\Vert _{C^0}^2 \Vert u_1-u_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla ^2 f(t) \Vert _{L^2(\varSigma )}^2 \, T\\&\quad + {c} \Vert \textrm{d}f\Vert _{C^0}^4 \Vert u_1-u_2\Vert _{C^0}^2 \max _{0 \le t \le T} \Vert \nabla (u_1(t)-u_2(t))\Vert _{L^2(\varSigma )}^2 \, T\\&\le \frac{1}{36} \Vert u_1-u_2\Vert _Z^2 \end{aligned}$$

if we choose T small enough.

In summary, we get

$$\begin{aligned} \Vert v_1-v_2\Vert _Z^2 \le \frac{1}{9} \Vert u_1-u_2\Vert _Z^2 \end{aligned}$$

which proves the lemma.

$\square$

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

$$\begin{aligned} \Vert (f,u)\Vert _{Y \times Z} = (C_3)^{-1} \Vert f\Vert _Y + \Vert u\Vert _Z. \end{aligned}$$

(31)

Define an operator ${\mathcal {S}} : Y_T \times Z_T \rightarrow Y_T \times Z_T$ by

$$\begin{aligned} {\mathcal {S}} (f,u) = (S_1(f,u),S_2(f,u)). \end{aligned}$$

(32)

Proposition 12

Fix $f_0 \in W^{3,2}(\varSigma )$. Then there is an $T_0 = T_0(C_0,\delta ,\delta ')>0$ such that for all $T \le T_0$,

• (a) ${\mathcal {S}}$ restricts to an operator ${\mathcal {S}}: B \times B' \rightarrow B \times B'$.

• (b) For each $f_1,f_2 \in B$ and $u_1,u_2 \in B',$

  $$\begin{aligned} \Vert {\mathcal {S}}(f_1,u_1)-{\mathcal {S}}(f_2,u_2)\Vert _{Y \times Z} \le \frac{5}{6} \Vert (f_1,u_1)-(f_2-u_2)\Vert _{Y \times Z}. \end{aligned}$$

  (33)

Proof

By Lemmas 5 and 9, (a) is proved. For (b), using Lemmas 6, 7, 10, 11, there is $T_0 = T_0(\delta ,\delta ')>0$ such that for all $T \le T_0$,

$$\begin{aligned}&\Vert {\mathcal {S}}(f_1,u_1)-{\mathcal {S}}(f_2,u_2)\Vert _{Y \times Z}\\&\quad = (C_3)^{-1}\Vert S_1(f_1,u_1)- S_1(f_2,u_2) \Vert _Y + \Vert S_2(f_1,u_1) - S_2(f_2,u_2)\Vert _{Z}\\&\quad \le (C_3)^{-1} \Vert S_1(f_1,u_1)-S_1(f_2,u_1)\Vert _Y + (C_3)^{-1} \Vert S_1(f_2,u_1)-S_1(f_2,u_2)\Vert _Y\\&\qquad + \Vert S_2(f_1,u_1) - S_2(f_2,u_1)\Vert _Z + \Vert S_2(f_2,u_1)-S_2(f_2,u_2)\Vert _Z\\&\quad \le \frac{1}{3} (C_3)^{-1} \Vert f_1-f_2\Vert _Y + \frac{1}{2} \Vert u_1-u_2\Vert _Z\\&\qquad + T^{1/4} \Vert f_1-f_2\Vert _Y + \frac{1}{3} \Vert u_1-u_2\Vert _Z\\&\quad \le \frac{5}{6} \left( (C_3)^{-1} \Vert f_1-f_2\Vert _Y + \Vert u_1-u_2\Vert _Z \right) \\&\quad = \frac{5}{6}\Vert (f_1,u_1)-(f_2,u_2)\Vert _{Y \times Z} \end{aligned}$$

if $T^{1/4} \le \frac{1}{2} (C_3)^{-1}$. $\square$

Theorem 13

(Short time existence for strong solution) There is $T_0>0$ such that there exists a smooth solution $(f,u) \in B \times B'$ of (2) on $\varSigma \times [0,T_0]$.

Proof

The existence of solution f, u comes from Proposition 12. The fact $f(\varSigma \times [0,T_0]) \subset N$ can be easily shown using nearest point projection, see for example [24]. Moreover, the operator $\partial _t - \textrm{e}^{-2u}\Delta$ is uniformly parabolic, so $|(\partial _t - \textrm{e}^{-2u}\Delta )f| \in L^p(\varSigma \times [0,T_0])$ for any $1 \le p < \infty$, by standard parabolic theory. This implies

$$\begin{aligned} \nabla ^2 f, \partial _t f \in L^p(\varSigma \times [0,T_0]) \end{aligned}$$

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

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

$$\begin{aligned} \textrm{e}^{2u} = \textrm{e}^{-2at} \left( 1 + 2b \int _{0}^{t} \textrm{e}^{2as}|\textrm{d}f|^2 \right) \end{aligned}$$

hence

$$\begin{aligned} \nabla u&= \textrm{e}^{-2u-2at} 2b \int _{0}^{t} \textrm{e}^{2as} \langle \nabla \textrm{d}f,\textrm{d}f \rangle \\ \int |\nabla u|^p&\le (4b)^p \int \left( \int _{0}^{t} |\nabla \textrm{d}f| |\textrm{d}f| \right) ^p\\&\le (4b)^p t^{p-1} \int \int _{0}^{t} |\nabla \textrm{d}f|^p |\textrm{d}f|^p \end{aligned}$$

which implies $\nabla u \in L^{p}(\varSigma \times [0,T_0])$ for any $1 \le p < \infty$. Now taking $\nabla$ in the Eq. (2a) to get

$$\begin{aligned} |(\partial _t - \textrm{e}^{-2u} \Delta ) \nabla f|{} & {} \le C \left( |\nabla u| |\Delta f| + |\nabla u| |\textrm{d}f|^2 + |\nabla \textrm{d}f| |\textrm{d}f| + |\textrm{d}f|^3 \right) \\{} & {} \quad \in L^p(\varSigma \times [0,T_0]) \end{aligned}$$

which implies

$$\begin{aligned} \nabla ^3 f, \partial _t (\nabla f) \in L^p(\varSigma \times [0,T_0]) \end{aligned}$$

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

Finally, from Sobolev embedding, we have $f,\textrm{d}f \in C^{\alpha }(\varSigma \times [0,T_0])$ for some $\alpha >0$. This implies $(\partial _t - \textrm{e}^{-2u}\Delta )f \in C^{\alpha ,\alpha /2}(\varSigma \times [0,T_0])$ 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. $\square$

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 $\Vert R^N\Vert ,\Vert A\Vert ,\Vert DA\Vert \le C_N$ where $R^N$ is the Riemannian curvature tensor of N. And from now on, assume $b \ge C_N^2$.

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 $B_{r}$, $0 \le \varphi \le 1$ and $|\nabla \varphi | \le \frac{4}{r}$.

Proposition 14

For solutions (f, u) of (2), we have

$$\begin{aligned} \begin{aligned}&\int _{t_1}^{t_2} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^2 \varphi ^2 + \int _{B_{2r}}|\textrm{d}f|^2 \varphi ^2 (t_2) - \int _{B_{2r}}|\textrm{d}f|^2 \varphi ^2 (t_1)\\&\le \frac{4^2}{ar^2} (\textrm{e}^{2at_2}-\textrm{e}^{2at_1}) E_0. \end{aligned} \end{aligned}$$

(34)

Especially, we have

$$\begin{aligned} E(B_{r},t_2) - E(B_{2r},t_1) \le \frac{4^2}{2ar^2}(\textrm{e}^{2at_2}-\textrm{e}^{2at_1})E_0. \end{aligned}$$

(35)

Proof

From the Eq. (2a), multiplying $\textrm{e}^{2u} f_t \varphi ^2$ gives

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^2 \varphi ^2&= \int _{B_{2r}}\langle f_t, \tau (f) \varphi ^2 \rangle \\&= -\int _{B_{2r}} \langle \textrm{d}f_t, \textrm{d}f \varphi ^2 \rangle - 2 \int _{B_{2r}} \langle f_t, f_i \rangle \varphi \nabla _i \varphi \\&\le -\frac{1}{2} \frac{d}{\textrm{d}t} \int _{B_{2r}} |\textrm{d}f|^2 \varphi ^2 + \frac{1}{2}\int _{B_{2r}} \textrm{e}^{2u} |f_t|^2 \varphi ^2 +2 \int _{B_{2r}} \textrm{e}^{-2u} |\textrm{d}f|^2 |\nabla \varphi |^2. \end{aligned}$$

So, we have

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u} |f_t|^2 \varphi ^2 + \frac{d}{\textrm{d}t} \int _{B_{2r}} |\textrm{d}f|^2 \varphi ^2&\le 4\int _{B_{2r}} \textrm{e}^{-2u}|\textrm{d}f|^2 |\nabla \varphi |^2\\&\le 4\frac{4^2}{r^2}\textrm{e}^{2at} \int _{B_{2r}} |\textrm{d}f|^2\\&\le 4\frac{4^2}{r^2}\textrm{e}^{2at} 2E_0. \end{aligned}$$

Integrating from $t_1$ to $t_2$ gives the result. $\square$

Lemma 15

Furthermore, assume

$$\begin{aligned} \sup _{t_1 \le t \le t_2}E(B_{2r},t) < \varepsilon _1. \end{aligned}$$

Then we have

$$\begin{aligned} \int _{t_1}^{t_2}\int _{B_{2r}}\textrm{e}^{2u}|f_t|^2 \varphi ^2&\le 4^2 \varepsilon _1 \left( 1 +\frac{\textrm{e}^{2at_2}-\textrm{e}^{2at_1}}{2ar^2}\right) , \end{aligned}$$

(36)

$$\begin{aligned} \int _{t_1}^{t_2}\int _{B_{2r}} |f_t|^2 \varphi ^2&\le \textrm{e}^{2at_2} 4^2 \varepsilon _1 \left( 1 + \frac{\textrm{e}^{2at_2}-\textrm{e}^{2at_1}}{2ar^2} \right) . \end{aligned}$$

(37)

Proof

The first equation directly comes from (34), by changing $E_0$ to $\varepsilon _1$. Also, it is easy to see that

$$\begin{aligned} \int _{t_1}^{t_2}\int _{B_{2r}} |f_t|^2 \varphi ^2&=\int _{t_1}^{t_2}\int _{B_{2r}} \textrm{e}^{-2u} \textrm{e}^{2u}|f_t|^2 \varphi ^2 \le \textrm{e}^{2at_2} \int _{t_1}^{t_2}\int _{B_{2r}} \textrm{e}^{2u} |f_t|^2 \varphi ^2\\&\le \textrm{e}^{2at_2} 4^2 \varepsilon _1 \left( 1 + \frac{\textrm{e}^{2at_2}-\textrm{e}^{2at_1}}{2ar^2} \right) . \end{aligned}$$

$\square$

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 16

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

$$\begin{aligned} \begin{aligned} \frac{d}{\textrm{d}t}\int _{B_{2r}} \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2&\le 2a(p+1)\int _{B_{2r}}\textrm{e}^{2u}|f_t|^{p+2}\varphi ^2 + 4(p+2) \int _{B_{2r}}|f_t|^{p+2}|\nabla \varphi |^2\\&\quad - \frac{p+2}{4} \int _{B_{2r}}|\nabla f_t|^{2}|f_t|^{p} \varphi ^2\\&\quad + \left( (p+2)C_N + \frac{(p+2)C_N^2}{2} - 2b(p+1) \right) \int _{B_{2r}}|\textrm{d}f|^2 |f_t|^{p+2}\varphi ^2. \end{aligned} \end{aligned}$$

(38)

Especially, we have

$$\begin{aligned} \begin{aligned}&\int _{B_{2r}}\textrm{e}^{2u}|f_t|^{p+2}\varphi ^2 (t) \le \textrm{e}^{2a(p+1)(t-t_0)}\\&\quad \cdot \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^{p+2}\varphi ^2(t_0) + 4(p+2) \int _{t_0}^{t} \int _{B_{2r}}|f_t|^{p+2}|\nabla \varphi |^2 \right) . \end{aligned} \end{aligned}$$

(39)

Proof

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

$$\begin{aligned} (\textrm{e}^{2u}f_t)_t = \Delta f_t + A(\textrm{d}f,\textrm{d}f)_t. \end{aligned}$$

Taking inner product with $f_t |f_t|^p \varphi ^2$ and integrating gives

$$\begin{aligned} \int \langle (\textrm{e}^{2u}f_t)_t,f_t |f_t|^p \varphi ^2 \rangle&= \int \langle \Delta f_t, f_t |f_t|^p \varphi ^2 \rangle + \int \langle A(\textrm{d}f,\textrm{d}f)_t,f_t |f_t|^p \varphi ^2 \rangle \\&= - \int |\nabla f_t|^2 |f_t|^p \varphi ^2 - \int \langle \nabla f_t,f_t \rangle p |f_t|^{p-2} \varphi ^2 \langle \nabla f_t,f_t \rangle \\&\quad - 2 \int \langle \nabla f_t,f_t \rangle |f_t|^p \varphi \nabla \varphi + \int \langle DA(\textrm{d}f,\textrm{d}f) \cdot f_t,f_t |f_t|^p \varphi ^2 \rangle \\&\quad + \int \langle A(\textrm{d}f_t,\textrm{d}f),f_t |f_t|^p \varphi ^2 \rangle \\&= -\int |\nabla f_t|^2 |f_t|^p \varphi ^2 - p\int |\langle \nabla f_t,f_t \rangle |^2 |f_t|^{p-2}\varphi ^2\\&\quad + \textrm{III} + \textrm{IV} + \textrm{V}. \end{aligned}$$

Now we have

$$\begin{aligned} \textrm{III}&\le \frac{1}{4} \int |\nabla f_t|^2 \varphi ^2 |f_t|^{p} + 4 \int |f_t|^{p+2} |\nabla \varphi |^2,\\ \textrm{IV}&\le C_N \int |\textrm{d}f|^2 |f_t|^{p+2} \varphi ^2,\\ \textrm{V}&\le C_N \int |\nabla f_t| |\textrm{d}f| |f_t|^{p+1} \varphi ^2,\\&\le \frac{1}{2} \int |\nabla f_t|^2 \varphi ^2 |f_t|^p + \frac{C_N^2}{2} \int |\textrm{d}f|^2 |f_t|^{p+2} \varphi ^2. \end{aligned}$$

On the other hand, LHS becomes

$$\begin{aligned} \int \langle (\textrm{e}^{2u}f_t)_t,f_t |f_t|^p \varphi ^2 \rangle&= \frac{1}{p+2}\frac{d}{\textrm{d}t}\int \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2 + 2\frac{p+1}{p+2} \int \textrm{e}^{2u}|f_t|^{p+2} u_t \varphi ^2\\&= \frac{1}{p+2}\frac{d}{\textrm{d}t}\int \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2 + 2b\frac{p+1}{p+2} \int |\textrm{d}f|^2|f_t|^{p+2} \varphi ^2\\&\quad - 2a \frac{p+1}{p+2} \int \textrm{e}^{2u}|f_t|^{p+2}\varphi ^2. \end{aligned}$$

All together, we have

$$\begin{aligned} \frac{d}{\textrm{d}t}\int \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2&\le 2a(p+1) \int \textrm{e}^{2u}|f_t|^{p+2}\varphi ^2 + 4(p+2) \int |f_t|^{p+2}|\nabla \varphi |^2\\&\quad - \frac{p+2}{4} \int |\nabla f_t|^2 |f_t|^p \varphi ^2\\&\quad + \left( (p+2)C_N + \frac{(p+2)C_N^2}{2} - 2b(p+1) \right) \int |\textrm{d}f|^2 |f_t|^{p+2}\varphi ^2. \end{aligned}$$

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

$$\begin{aligned}&\frac{d}{\textrm{d}t}\int \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2 \le 2a(p+1) \int \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2 + 4(p+2) \int |f_t|^{p+2} |\nabla \varphi |^2\\&\int _{B_{2r}} \textrm{e}^{2u}|f_t|^{p+2} \varphi ^2(t) \le \textrm{e}^{2a(p+1)(t-t_0)}\\&\quad \cdot \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^{p+2} \varphi ^2 (t_0) + 4(p+2) \int _{t_0}^{t} \int _{B_{2r}}|f_t|^{p+2} |\nabla \varphi |^2 \right) \end{aligned}$$

by Gronwall’s inequality. $\square$

Lemma 17

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

$$\begin{aligned} \sup _{T-2\delta r^2 \le t \le T}E(B_{2r},t) < \varepsilon _1. \end{aligned}$$

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

$$\begin{aligned} \int _{B_{2r}} \textrm{e}^{2u} |f_t|^2 \varphi ^2 (t) \le C_1(r,\delta ,t)C_2(r,\delta ,t) \varepsilon _1 \end{aligned}$$

(40)

where

$$\begin{aligned} C_1(r,\delta ,t)&= 4^2 \left( 1 + \textrm{e}^{2at} \frac{1-\textrm{e}^{-2a\delta r^2}}{2ar^2} \right) , \end{aligned}$$

(41)

$$\begin{aligned} C_2(r,\delta ,t)&= \textrm{e}^{6a\delta r^2} \left( \frac{1}{\delta r^2} + \frac{16(4)^2}{r^2}\textrm{e}^{2at} \right) . \end{aligned}$$

(42)

Proof

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

$$\begin{aligned} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^{2} \varphi ^2(t)&\le \textrm{e}^{2a(t-t_0)} \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^{2} \varphi ^2 (t_0) + 8 \int _{t_0}^{t} \int _{B_{3r/2}}|f_t|^{2} |\nabla \varphi |^2 \right) \\&\le \textrm{e}^{2a(t-t_0)} \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^{2} \varphi ^2 (t_0) + \frac{8(4)^2}{r^2} \int _{t_0}^{t} \int _{B_{2r}}|f_t|^{2} \psi ^2 \right) \\&\le \textrm{e}^{2a(t-t_0)}\int _{B_{2r}}\textrm{e}^{2u}|f_t|^{2} \varphi ^2 (t_0)\\&\quad + \textrm{e}^{2a(t-t_0)} \frac{8(4)^2}{r^2}\textrm{e}^{2at} 4^2 \varepsilon _1 \left( 1 + \frac{\textrm{e}^{2at}-\textrm{e}^{2at_0}}{2ar^2} \right) . \end{aligned}$$

Now take $t_0 \in [t-\delta r^2,t]$ such that

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^2 \varphi ^2 (t_0) = \min _{t-\delta r^2 \le s \le t} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^2 \varphi ^2 (s). \end{aligned}$$

Then by (36),

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^{2} \varphi ^2 (t_0) \le \frac{1}{\delta r^2}\int _{t-\delta r^2}^{t} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^2 \varphi ^2 \le \frac{1}{\delta r^2} 4^2 \varepsilon _1 \left( 1 + \frac{\textrm{e}^{2at}-\textrm{e}^{2a(t-\delta r^2)}}{2ar^2} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^{2} \varphi ^2(t)&\le 4^2 \varepsilon _1 \left( 1 + \frac{\textrm{e}^{2at}-\textrm{e}^{2at-2a\delta r^2}}{2ar^2} \right) \left( \frac{1}{\delta r^2} + \frac{8 (4)^2}{r^2} \textrm{e}^{2at} \right) \textrm{e}^{2a\delta r^2}. \end{aligned}$$

This completes the proof. $\square$

Corollary 18

Under the same assumption as above, we also have

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}} |\nabla f_t|^2 \varphi ^2&\le C C_1(r,\delta ,t) C_2(r,\delta ,t) \varepsilon _1, \end{aligned}$$

(43)

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}} |\textrm{d}f|^2 |f_t|^2 \varphi ^2&\le C C_1(r,\delta ,t) C_2(r,\delta ,t)\varepsilon _1. \end{aligned}$$

(44)

Proof

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

$$\begin{aligned} \left. \int _{B_{2r}} \textrm{e}^{2u}|f_t|^{2} \varphi ^2 \right| ^t_{t-\delta r^2}&\le 2a \int _{t-\delta r^2}^{t}\int _{B_{2r}}\textrm{e}^{2u}|f_t|^{2}\varphi ^2 + 8 \int _{t-\delta r^2}^{t} \int _{B_{2r}}|f_t|^{2}|\nabla \varphi |^2\\&\quad - \frac{1}{2}\int _{t-\delta r^2}^{t} \int _{B_{2r}}|\nabla f_t|^{2} \varphi ^2\\&\quad + \left( 2C_N + C_N^2 - 2b \right) \int _{t-\delta r^2}^{t} \int _{B_{2r}}|\textrm{d}f|^2 |f_t|^{2}\varphi ^2. \end{aligned}$$

Hence, we have

$$\begin{aligned} \frac{1}{2}\int _{t-\delta r^2}^{t} \int _{B_{2r}}|\nabla f_t|^2 \varphi ^2&\le 2C_1(r,\delta ,t) C_2(r,\delta ,t) \varepsilon _1 + 2a C_1(r,\delta ,t) \varepsilon _1\\&\quad + 8 \frac{4^2}{r^2} \textrm{e}^{2at} C_1(r,\delta ,t) \varepsilon _1 \le C C_1(r,\delta ,t) C_2(r,\delta ,t)\varepsilon _1. \end{aligned}$$

The other inequality is similar. $\square$

6.3 Higher Estimate for Time Derivatives

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

Proposition 19

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

$$\begin{aligned} \begin{aligned} \int _{t_1}^{t_2} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^{p+2}\varphi ^2&\le C \int _{t_1}^{t_2} \int _{B_{2r}} |f_{ti}|^2 |f_t|^{p-1} \varphi ^2 + C \int _{t_1}^{t_2} \int _{B_{2r}}|f_t|^{p+1}|\nabla \varphi |^2\\&\quad + C \int _{t_1}^{t_2}\int _{B_{2r}}|\textrm{d}f|^2 |f_t|^{p+1}\varphi ^2. \end{aligned} \end{aligned}$$

(45)

Proof

First note that for any $p\ge 1$, $\nabla _i |f_t|^p = p |f_t|^{p-2}\langle f_{ti},f_t \rangle$. Also, for simplicity, denote $\int \int = \int _{t_1}^{t_2} \int _{B_{2r}}$. Multiplying $\tau (f)$ to (2a) gives

$$\begin{aligned} 2\textrm{e}^{2u}|f_t|^2 = -2 \langle f_{ti},f_i \rangle + \nabla _i ( 2 \langle f_t,f_i \rangle ). \end{aligned}$$

Multiplying $|f_t|^p \varphi ^2$ for $p \ge 1$ and integrating gives

$$\begin{aligned} 2 \int \int \textrm{e}^{2u}|f_t|^{p+2}\varphi ^2&= -2 \int \int \langle f_{ti},f_i \rangle |f_t|^p \varphi ^2 - 4 \int \int \langle f_t,f_i \rangle |f_t|^p \varphi \nabla _i \varphi \\&\quad - 2p \int \int \langle f_t,f_i \rangle \varphi ^2 |f_t|^{p-2} \langle f_{ti},f_t \rangle \\&= \textrm{I} + \textrm{II} + \textrm{III}. \end{aligned}$$

Now

$$\begin{aligned} \textrm{I}&\le C\int \int |f_{ti}|^2 |f_t|^{p-1} \varphi ^2 + C\int \int |\textrm{d}f|^2 |f_t|^{p+1} \varphi ^2,\\ \textrm{II}&\le C \int \int |f_t|^{p+1} |\nabla \varphi |^2 + C \int \int |\textrm{d}f|^2 |f_t|^{p+1} \varphi ^2,\\ \textrm{III}&\le C \int \int |f_{ti}|^2 |f_t|^{p-1} \varphi ^2 + C \int \int |\textrm{d}f|^2 |f_t|^{p+1} \varphi ^2. \end{aligned}$$

This completes the proof. $\square$

Now we will show the desired estimate.

Proposition 20

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

$$\begin{aligned} \sup _{T-2\delta r^2 \le t \le T}E(B_{2r},t) < \varepsilon _1. \end{aligned}$$

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

$$\begin{aligned} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^4 \varphi ^2(t) \le C_3, \end{aligned}$$

(46)

where

$$\begin{aligned} C_3 = C C_1(r,\delta ,t) C_2(r,\delta ,t)^3 \varepsilon _1. \end{aligned}$$

(47)

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 $|\nabla \varphi | \le \frac{4}{r}$. Also, let $\psi$ be a cut-off function supported on $B_{2r}$ and $\psi \equiv 1$ on $B_{3r/2}$ and $|\nabla \psi | \le \frac{4}{r}$. 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 $\int \int \textrm{e}^{2u}|f_t|^3 \varphi ^2$.

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

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^3 \varphi ^2 \le CC_1C_2 \varepsilon _1 \end{aligned}$$

(48)

and

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}} |f_t|^3 \varphi ^2 \le \textrm{e}^{2at} CC_1C_2 \varepsilon _1. \end{aligned}$$

(49)

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

Now let $t_0 \in [t-\delta r^2,t]$ be such that

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^3 \varphi ^2 (t_0) = \min _{t-\delta r^2 \le s \le t} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^3 \varphi ^2 (s). \end{aligned}$$

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

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^3 \varphi ^2(t)&\le \textrm{e}^{4a\delta r^2} \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^3 \varphi ^2(t_0) + 12 \int _{t_0}^{t} \int _{B_{3r/2}}|f_t|^3 |\nabla \varphi |^2 \right) \\&\le \textrm{e}^{4a\delta r^2} \left( \frac{1}{\delta r^2} \int _{t-\delta r^2}^{t} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^3 \varphi ^2 + 12 \frac{4^2}{r^2} \int _{t-\delta r^2}^{t}\int _{B_{2r}} |f_t|^3 \psi ^2 \right) \\&\le \textrm{e}^{4a\delta r^2} \left( \frac{1}{\delta r^2}C C_1C_2 \varepsilon _1 + 12 \frac{4^2}{r^2} \textrm{e}^{2at} C C_1C_2 \varepsilon _1 \right) \\&= C C_1 C_2 \varepsilon _1 \left( \frac{1}{\delta r^2} + \frac{12(4)^2}{r^2} \textrm{e}^{2at}\right) \textrm{e}^{4a\delta r^2}. \end{aligned}$$

So, simply,

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^3 \varphi ^2(t) \le C C_1 C_2^2 \varepsilon _1. \end{aligned}$$

(50)

Step 3. Estimate for $\int \int |\nabla f_t|^2 |f_t| \varphi ^2$ and $\int \int |\textrm{d}f|^2 |f_t|^3 \varphi ^2$.

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

$$\begin{aligned} \left. \int _{B_{2r}} \textrm{e}^{2u}|f_t|^{3} \varphi ^2 \right| ^t_{t-\delta r^2}&\le 4a \int _{t-\delta r^2}^{t}\int _{B_{2r}}\textrm{e}^{2u}|f_t|^{3}\varphi ^2 + 12 \int _{t-\delta r^2}^{t} \int _{B_{2r}}|f_t|^{3}|\nabla \varphi |^2\\&\quad - \frac{3}{4}\int _{t-\delta r^2}^{t} \int _{B_{2r}}|\nabla f_t|^{2} |f_t| \varphi ^2\\&\quad + \left( 3C_N + \frac{3 C_N^2}{2} - 4b \right) \int _{t-\delta r^2}^{t} \int _{B_{2r}}|\textrm{d}f|^2 |f_t|^{3}\varphi ^2. \end{aligned}$$

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

$$\begin{aligned} \frac{3}{4}\int _{t-\delta r^2}^{t} \int _{B_{2r}}|\nabla f_t|^2 |f_t| \varphi ^2&\le 2C C_1C_2^2 \varepsilon _1 + 4a C C_1 C_2 \varepsilon _1 + 12 \frac{4^2}{r^2} \textrm{e}^{2at} CC_1 C_2 \varepsilon _1\\&\le C C_1 C_2^2 \varepsilon _1. \end{aligned}$$

So, we have

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}}|\nabla f_t|^2 |f_t| \varphi ^2 \le C C_1 C_2^2 \varepsilon _1. \end{aligned}$$

(51)

Similarly,

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}}|\textrm{d}f|^2 |f_t|^{3}\varphi ^2 \le C C_1 C_2^2 \varepsilon _1. \end{aligned}$$

(52)

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

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

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^4 \varphi ^2 \le CC_1C_2^2 \varepsilon _1 \end{aligned}$$

(53)

and

$$\begin{aligned} \int _{t-\delta r^2}^{t} \int _{B_{2r}} |f_t|^4 \varphi ^2 \le \textrm{e}^{2at}C C_1C_2^2 \varepsilon _1. \end{aligned}$$

(54)

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

Now let $t_0 \in [t-\delta r^2,t]$ be such that

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^4 \varphi ^2 (t_0) = \min _{t-\delta r^2 \le s \le t} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^4 \varphi ^2 (s). \end{aligned}$$

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

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^4 \varphi ^2(t)&\le \textrm{e}^{6a \delta r^2} \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^4 \varphi ^2(t_0) + 16 \int _{t_0}^{t} \int _{B_{3r/2}}|f_t|^4 |\nabla \varphi |^2 \right) \\&\le \textrm{e}^{6a \delta r^2} \left( \frac{1}{\delta r^2} \int _{t-\delta r^2}^{t} \int _{B_{2r}} \textrm{e}^{2u}|f_t|^4 \varphi ^2 + 16 \frac{4^2}{r^2} \int _{t-\delta r^2}^{t}\int _{B_{2r}} |f_t|^3 \psi ^2 \right) \\&\le \textrm{e}^{6a \delta r^2} \left( \frac{1}{\delta r^2} CC_1C_2^2 \varepsilon _1 + 16 \frac{4^2}{r^2} \textrm{e}^{2at} CC_1C_2^2 \varepsilon _1 \right) \\&= C C_1 C_2^2 \varepsilon _1 \left( \frac{1}{\delta r^2} + \frac{16(4)^2}{r^2} \textrm{e}^{2at}\right) \textrm{e}^{6a \delta r^2}. \end{aligned}$$

So, simply,

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{2u}|f_t|^4 \varphi ^2(t) \le C C_1 C_2^3 \varepsilon _1. \end{aligned}$$

$\square$

Remark 1

We can keep going on to get bounds for $\int _{B_{2r}}\textrm{e}^{2u}|f_t|^n \varphi ^2(t) \le C_3(n)$ 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 $\Vert \cdot \Vert _{k,p} = \Vert \cdot \Vert _{W^{k,p}(B_{2r})}$ and $\Vert \cdot \Vert _p = \Vert \cdot \Vert _{0,p}$. First observe the following.

Lemma 21

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

$$\begin{aligned} \int _{B_{2r}}\textrm{e}^{pu}\varphi ^r(t) \le \int _{B_{2r}}\textrm{e}^{pu}\varphi ^r(t_0) + \frac{2b^2 (p-2)}{pa} \int _{t_0}^{t} \int _{B_{2r}} |\textrm{d}f|^p\varphi ^r. \end{aligned}$$

(55)

Proof

Note that

$$\begin{aligned} \partial _t (\textrm{e}^{pu}) = p \textrm{e}^{pu}u_t = pb \textrm{e}^{(p-2)u}|\textrm{d}f|^2 - pa\textrm{e}^{pu}. \end{aligned}$$

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

$$\begin{aligned} \frac{d}{\textrm{d}t}\int _{B_{2r}} \textrm{e}^{pu} \varphi ^r&= pb \int _{B_{2r}}\textrm{e}^{(p-2)u}|\textrm{d}f|^2\varphi ^r - pa \int _{B_{2r}} \textrm{e}^{pu}\varphi ^r\\&\le b \lambda (p-2) \int _{B_{2r}}\textrm{e}^{pu}\varphi ^r + 2b \lambda ^{-1} \int _{B_{2r}}|\textrm{d}f|^p\varphi ^r - pa \int _{B_{2r}}\textrm{e}^{pu}\varphi ^r\\&= \frac{2b^2 (p-2)}{pa} \int _{B_{2r}}|\textrm{d}f|^p\varphi ^r \end{aligned}$$

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

Lemma 22

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 $p \ge 2,$ we have

$$\begin{aligned} \Vert \,|\textrm{d}f|^r \varphi \Vert _{p} \le C \Vert \,|\textrm{d}f|^{r-1}\Vert _{p} \Vert f \varphi \Vert _{2,2}. \end{aligned}$$

(56)

Proof

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

$$\begin{aligned} \Vert \,|\textrm{d}f|^r \varphi \Vert _{p}&\le C \Vert \nabla (|\textrm{d}f|^r \varphi ) \Vert _{s}\\&\le C \Vert |\textrm{d}f|^{r-1} \nabla (|\textrm{d}f| \varphi ) \Vert _{s}\\&\le C \Vert \,|\textrm{d}f|^{r-1}\Vert _{p} \Vert f \varphi \Vert _{2,2}. \end{aligned}$$

$\square$

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

Proposition 23

Let (f, u) are solutions of (2). Then there exists $\varepsilon _1>0$ such that the following holds:

Assume that

$$\begin{aligned} \sup _{T-2\delta r^2 \le t \le T}E(B_{2r},t) \le \varepsilon _1 , \quad \int _{B_{2r} \times \{T-2\delta r^2\}}\textrm{e}^{6u} \le \varepsilon _1. \end{aligned}$$

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

$$\begin{aligned} \Vert f\varphi \Vert _{2,2} \le C_4 = C_4 (r,\delta ,T,\varepsilon _1,C_N), \end{aligned}$$

(57)

where

$$\begin{aligned} C_4 = \left( C C_3 \varepsilon _1 + C \varepsilon _1^4\right) ^{1/4} \left( 1 + C_3 \varepsilon _1 \delta r^2 \exp (C_3 \varepsilon _1 \delta r^2 ) \right) ^{1/4}. \end{aligned}$$

Proof

Suppose $\varphi$ be a cut-off function supported on $B_{3r/2}$ and $\varphi \equiv 1$ on $B_{r}$ and $|\nabla \varphi | \le \frac{4}{r}$. Also, let $\psi$ be a cut-off function supported on $B_{2r}$ and $\psi \equiv 1$ on $B_{3r/2}$ and $|\nabla \psi | \le \frac{4}{r}$. Let $t_0 = T-2\delta r^2$.

Without loss of generality, assume $\int _{\varOmega } f = 0$. Then we have, by Poincare,

$$\begin{aligned} \Vert f\Vert _{p} \le C_p \Vert \textrm{d}f\Vert _{p}. \end{aligned}$$

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

$$\begin{aligned} |\Delta (f\varphi )|&\le |A(\textrm{d}f,\textrm{d}f) \varphi | + |\textrm{e}^{2u}f_t \varphi | + k(\varphi ) (|f| + |\textrm{d}f|)\\&\le C_N | \, |\textrm{d}f|^2 \varphi | + |\textrm{e}^{2u}f_t \varphi | + k(\varphi ) (|f| + |\textrm{d}f|). \end{aligned}$$

By the $L^p$ estimate, we have

$$\begin{aligned} \Vert f\varphi \Vert _{2,p} \le C \left( C_N \Vert \, |\textrm{d}f|^2 \varphi \Vert _{p} + \Vert \textrm{e}^{2u} |f_t| \varphi \Vert _{p} + \Vert \textrm{d}f\Vert _{p} \right) , \end{aligned}$$

(58)

where the constant C only depends on p and r.

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

$$\begin{aligned} \Vert \textrm{e}^{2u} |f_t| \varphi \Vert _{2}^4&= \left( \int _{B_{2r}}\textrm{e}^{4u}|f_t|^2 \varphi ^2 \right) ^2\\&\le \left( \int _{B_{3r/2}}\textrm{e}^{2u}|f_t|^4 \right) \left( \int _{B_{2r}} \textrm{e}^{6u}\varphi ^4\right) \\&\le \left( \int _{B_{2r}}\textrm{e}^{2u}|f_t|^4 \psi ^2\right) \left( \int _{B_{2r}} \textrm{e}^{6u}\varphi ^4\right) \\&\le C_3 \left( \int _{B_{2r}} \textrm{e}^{6u}\varphi ^4 (t_0) + \frac{8b^2}{6a} \int _{t_0}^{t} \int _{B_{2r}}|\textrm{d}f|^6 \varphi ^4\right) \\&\le C_3 \varepsilon _1 + C_3 \int _{t_0}^{t}\int _{B_{2r}}|\textrm{d}f|^6 \varphi ^4. \end{aligned}$$

Now applying Lemma 22 with $r=3/2$,$q=4$ gives

$$\begin{aligned} \left( \int _{B_{2r}}|\textrm{d}f|^6 \varphi ^4 \right) ^{1/4}&= \Vert \,|\textrm{d}f|^{3/2} \varphi \Vert _{4} \le C \Vert \,|\textrm{d}f|^{1/2}\Vert _{4} \Vert f \varphi \Vert _{2,2}\\&\le C \varepsilon _1^{1/4} \Vert f \varphi \Vert _{2,2}. \end{aligned}$$

On the other hand, applying Lemma 22 with $r=2$, $q=2$ gives

$$\begin{aligned} \Vert \,|\textrm{d}f|^2 \varphi \Vert _{2} \le C \Vert \textrm{d}f \Vert _{2} \Vert f \varphi \Vert _{2,2} \le C \varepsilon _1^{1/2} \Vert f \varphi \Vert _{2,2}. \end{aligned}$$

All together, we have

$$\begin{aligned} \Vert f\varphi \Vert _{2,2}^4 \le C C_N^4 \varepsilon _1^2 \Vert f \varphi \Vert _{2,2}^4 + C C_3 \varepsilon _1 + C C_3 \varepsilon _1 \int _{t_0}^{t} \Vert f \varphi \Vert _{2,2} + C \varepsilon _1^4. \end{aligned}$$

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

$$\begin{aligned} (1- C C_N^4 \varepsilon _1^2) X \le C C_3 \varepsilon _1 + C \varepsilon _1^4 +C_3 \varepsilon _1 \int _{t_0}^{t} X. \end{aligned}$$

So, if $\varepsilon _1$ is small enough so that $1- C C_N^4 \varepsilon _1^2 >1/2$, then by Gronwall’s inequality, we have

$$\begin{aligned} \Vert f\varphi \Vert _{2,2}^4&\le \left( C C_3 \varepsilon _1 + C \varepsilon _1^4\right) \left( 1 + C_3 \varepsilon _1 (t-t_0) \exp (C_3 \varepsilon _1 (t-t_0)) \right) \\&\le \left( C C_3 \varepsilon _1 + C \varepsilon _1^4\right) \left( 1 + C_3 \varepsilon _1 \delta r^2 \exp (C_3 \varepsilon _1 \delta r^2) \right) . \end{aligned}$$

This completes the proof. $\square$

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

$$\begin{aligned} \Vert f\varphi \Vert _{1,p} \le C_4 \end{aligned}$$

(59)

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 24

Assume the same as in Proposition 23. In addition, we assume that

$$\begin{aligned} \int _{B_{2r} \times \{T-2\delta r^2\}}\textrm{e}^{18u} \le \varepsilon _1. \end{aligned}$$

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

$$\begin{aligned} \Vert f\varphi \Vert _{2,3} \le C_5 = C_5 (r,\delta ,T,\varepsilon _1,C_N) \end{aligned}$$

(60)

where

$$\begin{aligned} C_5 = C \left( C_NC_4 ^2 + C_3^{1/12} \varepsilon _1^{1/12} + C_3^{1/12} \delta ^{1/12} r^{1/6} C_4 ^{3/2} + C_4 \right) . \end{aligned}$$

In particular,

$$\begin{aligned} \sup _{B_r} |\textrm{d}f| \le C_5 . \end{aligned}$$

(61)

Proof

By (59), we have uniform bound for $|\textrm{d}f|^p$ for any p. Now from Eq. (58), we have

$$\begin{aligned} \Vert f\varphi \Vert _{2,p} \le C \left( C_N \Vert \, \textrm{d}f \Vert _{2p}^2 + \Vert \textrm{e}^{2u} |f_t| \varphi \Vert _{p} + \Vert \textrm{d}f\Vert _{p} \right) . \end{aligned}$$

Now let $p=3$ and $t_0 = T-2\delta r^2$. Then we have, using (55) and (59),

$$\begin{aligned} \Vert \textrm{e}^{2u} |f_t| \varphi \Vert _{3}^{12}&\le \Vert \textrm{e}^{u/2}|f_t| \varphi \Vert _{4}^{12} \Vert \textrm{e}^{3u/2}\Vert _{12}^{12}\\&\le C_3 \int _{B_{2r}}\textrm{e}^{18u}\\&\le C_3 \left( \int _{B_{2r}} \textrm{e}^{18u} (t_0) + C \int _{t_0}^{t} \int _{B_{2r}}|\textrm{d}f|^{18} \right) \\&\le C_3 \varepsilon _1 + C C_3 \delta r^2 C_4 ^{18}. \end{aligned}$$

Applying (59) completes the proof. $\square$

8 Global Weak Solution

In this section, we will prove the main Theorem 1.

Lemma 25

There exists $\varepsilon _1>0$ such that if (f, u) be a smooth solution of (2) on $B_{2r} \times [T-2\delta r^2,T]$ and

$$\begin{aligned} \sup _{T-2\delta r^2 \le t \le T}E(B_{2r},t) \le \varepsilon _1 \quad \textrm{and } \quad \int _{B_{2r} \times \{T-2\delta r^2\}}\textrm{e}^{18u} \le \varepsilon _1, \end{aligned}$$

(62)

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 $|\textrm{d}f|$, we have $\textrm{e}^{-2u(f)} \le \textrm{e}^{2aT}$ and

$$\begin{aligned} \textrm{e}^{-2u(f)}&= \frac{\textrm{e}^{2at}}{1+2b \int _{0}^{t}\textrm{e}^{2as}|\textrm{d}f|^2(x,s){ \mathrm d}s} \ge \frac{\textrm{e}^{2at}}{1+2b {M'} ^2 \frac{\textrm{e}^{2at}-1}{2a}}\\&\ge \frac{1}{1 + \frac{b}{a} {M'} ^2}. \end{aligned}$$

Hence the operator $\partial _t - \textrm{e}^{-2u}\Delta$ is uniformly parabolic on $[0,T_0)$.

Similar in proof of Theorem 13, we conclude the desired estimate. $\square$

Proof

(Proof of Theorem 1) First consider $f_0$ is smooth. By Theorem 13, there exists a smooth solution in $(\varSigma \times [0,T))$ 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 $\limsup _{t \nearrow T_1} E(B_{2r}(x),t) \le \varepsilon _1$ for any $x \in \varSigma$ 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 $x \in \varSigma$ such that

$$\begin{aligned} \limsup _{t \nearrow T_1} E(B_{2r}(x),t) > \varepsilon _1. \end{aligned}$$

Since the total energy is finite, there are at most finitely many such points $\{x_1, \ldots , x_{k_1}\}$. Then by above lemma, we get smooth solution $(f_1,u_1)$ on $\varSigma \times [0,T] \setminus \{(x_i^1,T_1)\}_{i=1, \ldots , k_1}$. 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 $W^{1,2}_\textrm{loc}(\varSigma \setminus \{x_i^1\})$.

Next, denote $g_1 = \textrm{e}^{2u_1(x,T_1)}g_0$ and consider the flow (2) with initial map $f_1$ and initial metric $g_1$. As above, there is a smooth solution $(f_2,u_2)$ on $\varSigma \times [0,T_2] \setminus \{(x_i^2,T_2)\}_{i=1, \ldots , k_2}$. From these we can set up a smooth solution (f, u) on $\varSigma \times [0,T_1+T_2]$ 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.

$\square$

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 26

([23]) There exist a compact target manifold N,  a smooth map $f_0 : D \rightarrow N$ and $\varepsilon >0$ such that every smooth map $f : D \rightarrow N$ homotopic to $v_0$ fails to be harmonic. If furthermore $E(f) \le E(f_0),$ then

$$\begin{aligned} \int _{D} |\tau (f)|^2 \ge \varepsilon . \end{aligned}$$

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

$$\begin{aligned} E(0)-E(t) = \int _{0}^{t} \int _{D} \textrm{e}^{-2u}|\tau (f(t))|^2. \end{aligned}$$

So, if u is large, $\int _{D} \textrm{e}^{-2u}|\tau (f(t))|^2$ 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 $\Vert \tau \Vert _{L^2} \rightarrow 0$. And the assumption $\Vert \tau \Vert _{L^2} \rightarrow 0$ is essential in the no-neck property as there is a counter example of Parker [7] where $\Vert \tau (f_i)\Vert _{L^1}$ is uniformly bounded. In fact, the energy identity and no-neck property of approximate harmonic map with $\Vert \tau (f_i)\Vert _{L^p}$ for some $p>1$ uniformly bounded was proved in Wang–Wei–Zhang [43]. 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.