Global existence and convergence of a flow to Kazdan–Warner equation with non-negative prescribed function

Abstract

We consider an evolution problem associated to the Kazdan–Warner equation on a closed Riemann surface $(\Sigma ,g)$

$$\begin{aligned} -\Delta _{g}u=8\pi \left( \dfrac{he^{u}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-\dfrac{1}{\int _{\Sigma }\mathop {}\mathrm {d}\mu _{g}}\right) \end{aligned}$$

where the prescribed function $h\ge 0$ and $\max _{\Sigma }h>0$. We prove the global existence and convergence under additional assumptions such as

$$\begin{aligned} \Delta _{g}\ln h(p_0)+8\pi -2K(p_0)>0 \end{aligned}$$

for any maximum point $p_0$ of the sum of $2\ln h$ and the regular part of the Green function, where K is the Gaussian curvature of $\Sigma$. In particular, this gives a new proof of the existence result by Yang and Zhu (Pro Am Math Soc 145:3953–3959, 2017) which generalizes existence result of Ding et al. (Asian J Math 1:230–248, 1997) to the non-negative prescribed function case.

1 Introduction

Let $\Sigma$ be a closed Riemann surface with a fixed conformal structure. Choose a conformal metric g in the conformal class such that the area of $\Sigma _{g}:=\left( \Sigma ,g\right)$ is one. Let h be a non-negative but nonzero smooth function on $\Sigma$. We consider the following Kazdan–Warner equation

$$\begin{aligned} -\Delta _{g}u=8\pi \left( \dfrac{he^{u}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-1\right) . \end{aligned}$$

(1.1)

Here $\Delta _{g}$ is the Laplace–Beltrami operator. The solutions to (1.1) are the critical points of the following functional:

$$\begin{aligned} J(u):=\int _{\Sigma }\left( \dfrac{1}{2}\left|\nabla _{g}u\right|_{g}^2+8\pi u\right) \mathop {}\mathrm {d}\mu _{g}-8\pi \ln \left( \int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}\right) . \end{aligned}$$

Many mathematicians have contributed to the study of Kazdan–Warner equation. Forty years ago, Kazdan and Warner [22] considered the solvability of the equation

$$\begin{aligned} -\Delta _{g}u=he^u-\rho , \end{aligned}$$

where $\rho$ is a constant and h is some smooth prescribed function. When $\rho >0$, the equation above is equivalent to

$$\begin{aligned} -\Delta _{g}u=\rho (he^u-1). \end{aligned}$$

In particular, when $\Sigma _{g}$ is the standard sphere $\mathbb {S}^2$, it is called the Nirenberg problem, which comes from the conformal geometry. It has been studied by Moser [28], Kazdan and Warner [22], Chen and Ding [10], Chang and Yang [7] and others.

The Kazdan–Warner equation can be also viewed as a special case of the following mean field equation:

$$\begin{aligned} -\Delta _{g}u=\rho \left( \dfrac{fe^{u}}{\int _{\Sigma }fe^{u}\mathop {}\mathrm {d}\mu _{g}}-1\right) , \end{aligned}$$

(1.2)

where f is a smooth function on $\Sigma$. The mean field Eq. (1.2) appears in various context such as the abelian Chern–Simons–Higgs models (see for example [3, 32, 33]). When $f>0$, the equation (1.2) is equivalent to the following equation:

$$\begin{aligned} -\Delta _{g}u=\rho \dfrac{e^{u}}{\int _{\Sigma }e^{u}\mathop {}\mathrm {d}\mu _{g}}-Q, \end{aligned}$$

(1.3)

where $Q\in C^\infty (\Sigma )$ is a given function such that $\int _\Sigma Q\mathop {}\mathrm {d}\mu _{g}=\rho$. The existence of solutions of (1.3) has been widely studied in recent decades. Many partial existence results have been obtained for noncritical cases according to the Euler characteristic of $\Sigma$ (see for example Brezis and Merle [2], Chen and Lin [9], Ding, Jost, Li and Wang [13], Lin [26], Malchiodi [27] and the references therein). Djadli [14] established the existence of solutions for all surfaces $\Sigma$ when $\rho \ne 8k\pi$ by studying the topology of sublevels to achieve a min-max scheme which already introduced by Djadli and Malchiodi in [15]. At this point we want to mention another generalization in [35], where the author considered the mean field equations on a closed Riemannian surface with the action of an isometric group.

The following evolution problem associated to (1.3) was also well studied by Castéras for noncritical cases.

$$\begin{aligned} \dfrac{\partial e^{u}}{\partial t}=\Delta _{g}u+\rho \dfrac{e^u}{\int _\Sigma e^u\mathop {}\mathrm {d}\mu _{g}}-Q,\quad u(\cdot ,0)=u_0 \end{aligned}$$

(1.4)

where $u_0\in C^{2+\alpha }(\Sigma )$. This flow possesses a structure that is very similar to the Calabi and Ricci-Hamilton flows. When Q is a constant equal to the scalar curvature of $\Sigma$ with respect to the metric g, the flow (1.4) has been studied by Struwe [30]. A flow approaching to Nirenberg’s problem was studied by Struwe in [31]. The global existence and convergence of (1.4) were proved by Castéras in [4]. However, the convergence result there does not include the critical cases, i.e. $\rho =8k\pi$ for $k\in \mathbb {N}$. Recently, when $\rho =8\pi$, a sufficient condition for convergence was given by Li and Zhu in [23]. This gives a new proof of the result of Ding, Jost, Li and Wang in [12] which was extended by Lin and Chen to general critical cases [8] and recently generalized by Yang and Zhu to non-negative prescribed function cases in [34, 36].

Motivated by these results, we consider the following evolution problem for (1.1) with non-negative prescribed function:

$$\begin{aligned} \dfrac{\partial e^{u}}{\partial t}=\Delta _{g}u+8\pi \left( \dfrac{he^{u}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-1\right) ,\quad u(\cdot ,0)=u_0 \end{aligned}$$

(1.5)

where $u_0\in H^{2}(\Sigma )$ and h is a non-negative but nonzero smooth function on $\Sigma$. Since the prescribed function h may be zero on some nonempty subset of $\Sigma$, the global existence and convergence of this flow are subtle. Precisely, we can not use the lower bound of h to do a priori estimates. Therefore, Castéras’s proof of global existence for positive prescribed function does not apply to our situation. In addition, the condition (ii) of (1.6) in Castéras’s compactness result [5] actually assumes

$$\begin{aligned} -\frac{\partial e^{u_n}}{\partial t}+\rho e^{u_n}\ge -C, \quad \forall x\in \Sigma , \forall n\ge 1, \end{aligned}$$

for a sequence of time-slices $u_n:=u(\cdot ,t_n)$. This condition was proved in Proposition 2.1 [4]. However, the proof also need the prescribed function h to be positive. Thus, our a priori estimates in the proof of global existence and blow-up analysis used in the proof of global convergence are both new.

First, we prove the global existence of the flow (1.5).

Theorem 1.1

(Global existence) For $u_0\in H^2(\Sigma )$, there is a unique global solution $u\in C^{\infty }\left( \Sigma \times (0,\infty )\right)$ to (1.5) with

$$\begin{aligned} u\in \cap _{0<T<\infty }\left( L^{\infty }\left( 0,T;H^2\left( \Sigma \right) \right) \cap H^1\left( 0,T;H^1\left( \Sigma \right) \right) \cap H^2\left( 0,T;H^{-1}\left( \Sigma \right) \right) \right) . \end{aligned}$$

Moreover, for every $0<T<\infty$, there is a positive constant $C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right)$ depending only on T, the upper bound of $\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }$ and $\Sigma _{g}$,

$$\begin{aligned}&ess\,sup_{0\le t\le T}\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }+\left( \int _0^T\left( \left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{H^1\left( \Sigma \right) }^2+\left\Vert \dfrac{\partial ^2 u(t)}{\partial t^2}\right\Vert _{H^{-1}\left( \Sigma \right) }^2\right) \mathop {}\mathrm {d}t\right) ^{1/2}\nonumber \\&\quad \le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right) , \end{aligned}$$

(1.6)

where $u(t):=u(\cdot ,t)$. In particular, if $u_0$ is smooth, then u is smooth. Here the Sobolev spaces $H^{k}(0,T;X):=W^{k,2}(0,T;X)$ and $W^{k,p}(0,T;X)$ consists of all functions $u\in L^p(X\times [0,T])$ such that $\frac{\partial u}{\partial t}, \ldots c,\frac{\partial ^k}{\partial t^k}$ exists in the weak sense and belongs to $L^p(X\times [0,T])$ and

$$\begin{aligned} \left\Vert u\right\Vert _{W^{k,p}(0,T;X)}:={\left\{ \begin{array}{ll} \left( \int _0^T\left( \left\Vert u(t)\right\Vert _X^p+\sum _{i=1}^k\left\Vert \frac{\partial ^iu(t)}{\partial t^i}\right\Vert _{X}^p\right) \mathop {}\mathrm {d}t\right) ^{1/p},&{} 1\le p<\infty ,\\ {\mathop {ess\,sup}\limits _{0\le t\le T}}\left( \left\Vert u(t)\right\Vert _X+\sum _{i=1}^k\left\Vert \frac{\partial ^iu(t)}{\partial t^i}\right\Vert _{X}\right) ,&{} p=\infty . \end{array}\right. } \end{aligned}$$

Then it is interesting to consider the convergence of the flow. To do so, we begin with the monotonicity formula. It gives us that a sequence of positive numbers $t_n\rightarrow \infty$ as $n\rightarrow \infty$ with

$$\begin{aligned} \int _{\Sigma }e^{u_n}\left| \frac{\partial u_n}{\partial t}\right| ^2\mathop {}\mathrm {d}\mu _{g}\rightarrow 0, \quad \text {as} \quad n\rightarrow \infty , \end{aligned}$$

where $u_n:=u(t_n)$. If $\Vert u_n\Vert _{H^2(\Sigma )}$ is uniformly bounded, then $u_n$ subsequentially converges to a smooth solution of (1.1). Otherwise, we can get the following lower bound of the functional J along the flow (1.5).

Theorem 1.2

If the flow (1.5) develops a singularity at the infinity, then we have

$$\begin{aligned} J(u(t))\ge C_0=-4\pi \max _{x\in \Sigma }(A(x)+2\ln h(x))-8\pi \ln \pi -8\pi ,\quad \forall t\ge 0, \end{aligned}$$

where A is the regular part of the Green function G which has the following expansion in the normal coordinate system:

$$\begin{aligned} G(x,p)=-4\ln r+A(p)+b_1x_1+b_2x_2+c_1x_1^2+2c_2x_1x_2+c_3x_2^2+O(r^3), \end{aligned}$$

where $r(x)=\mathrm {dist}_g(x,p)$.

Last, by imposing certain geometric condition, we get functions whose value under J is strictly less than $C_0$. Consequently, when the flow starts with these functions, the previous $u_n$ will converges in $H^2(\Sigma )$. Moreover, it follows from the Łojasiewicz-Simon gradient inequality that the convergence of the flow is actually global in time.

Theorem 1.3

(Global convergence) There exists an initial data $u_0\in C^{\infty }(\Sigma )$ such that u(t) converges in $H^2(\Sigma )$ to a smooth solution of (1.1) provided that

$$\begin{aligned} \begin{aligned}&\Delta _{g} h\left( p_0\right) +2\left( b_1(p_0)k_1\left( p_0\right) +b_2\left( p_0\right) k_2\left( p_0\right) \right) \\&\quad >-\left( 8\pi +b_1^2\left( p_0\right) +b_2^2\left( p_0\right) -2K\left( p_0\right) \right) h\left( p_0\right) , \end{aligned} \end{aligned}$$

(1.7)

where K is the Gaussian curvature of $\Sigma$, $\nabla _{g} h(p_0)=(k_1(p_0),k_2(p_0))$ in the normal coordinate system, $p_0$ is the maximum point of the function $q\mapsto A(q)+2\ln h(q)$.

Remark 1.4

As pointed by Ding, Li, Jost and Wang in [12, Remark 1.1], the inequality (1.7) is implied by the following one:

$$\begin{aligned} \Delta _{g}\ln {h}(p_0)+8\pi -2K(p_0)>0 \end{aligned}$$

where $p_0$ is the maximum point of the function $q\mapsto A(q)+2\ln h(q)$.

Remark 1.5

For $\rho \in (0,8\pi )$ and any initial data $u_0\in C^{\infty }\left( \Sigma \right)$, by using a similarly argument, the

$$\begin{aligned} \dfrac{\partial e^{u}}{\partial t}=\Delta _{g}u+\rho \left( \dfrac{he^{u}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-1\right) ,\quad u(\cdot ,0)=u_0 \end{aligned}$$

admits a unique global smooth solution which converges to a solution to

$$\begin{aligned} \Delta _{g}u_{\infty }+\rho \left( \dfrac{he^{u_{\infty }}}{\int _{\Sigma }he^{u_{\infty }}\mathop {}\mathrm {d}\mu _{g}}-1\right) =0. \end{aligned}$$

The remaining part of this paper will be organized as follows. In Sect. 2, we prove the global existence of the flow (1.5). In Sect. 3, we prove the number of the singularities is at most one. In Sect. 4, we show the lower bound of J along the flow if the singularity occurs. In the last Sect. 5, we prove the global convergence of the flow.

2 Global existence

The aim of this section is to prove the global existence of the mean field flow (1.5), i.e. Theorem 1.1.

Proof of Theorem 1.1

First, we assume $u_0\in C^{\infty }\left( \Sigma \right)$. Since the flow is parabolic, the short time existence of (1.5) follows from the standard method (e.g. [19]). Thus, there exists $T>0$ such that $u\in C^{\infty }(\Sigma \times [0,T])$ is a solution of (1.5).

Along the flow (1.5), it is easy to see

$$\begin{aligned} \dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }e^{u(t)}\mathop {}\mathrm {d}\mu _{g}=0 \end{aligned}$$

(2.1)

and

$$\begin{aligned} \dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}J(u(t))=-\int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

(2.2)

According to (2.2) and (2.1), we get

$$\begin{aligned} \int _{\Sigma }\left( \dfrac{1}{2}\left|\nabla _{g}u(t)\right|_{g}^2+8\pi u(t)\right) \mathop {}\mathrm {d}\mu _{g}\le J(u_0)+8\pi \ln \max _{\Sigma }h+8\pi \ln \int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

(2.3)

Recall the Trudinger-Moser inequality (cf. [18, Theorem 1.7])

(2.4)

where c is a constant depending only on the Riemann surface $\left( \Sigma ,g\right)$. As an immediately consequence of (2.4),

$$\begin{aligned} J(u(t))\ge 8\pi \ln \int _{\Sigma }e^{u(t)}\mathop {}\mathrm {d}\mu _{g}-8\pi c-8\pi \ln \int _{\Sigma }he^{u(t)}\mathop {}\mathrm {d}\mu _{g} \end{aligned}$$

(2.5)

and (2.1) imply that

$$\begin{aligned} 0<C^{-1}\exp \left( -C\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }^2\right) \le \int _{\Sigma }he^{u(t)}\mathop {}\mathrm {d}\mu _{g}\le C\exp \left( C\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }^2\right) . \end{aligned}$$

(2.6)

Together with

$$\begin{aligned} \begin{aligned} \dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }e^{pu(t)}\mathop {}\mathrm {d}\mu _{g}&=p\int _{\Sigma }e^{(p-1)u(t)}\left( \Delta _{g}u(t)+8\pi \dfrac{he^u(t)}{\int _{\Sigma }he^u(t)}-8\pi \right) \mathop {}\mathrm {d}\mu _{g}\\&=-p(p-1)\int _{\Sigma }e^{(p-1)u(t)}\left|\nabla _g{u(t)}\right|_g^2\mathop {}\mathrm {d}\mu _{g}\\&\quad +8p\pi \left( \dfrac{\int _{\Sigma }he^{pu(t)}\mathop {}\mathrm {d}\mu _{g}}{\int _{\Sigma }he^{u(t)}\mathop {}\mathrm {d}\mu _{g}}-\int _{\Sigma }e^{(p-1)u(t)}\mathop {}\mathrm {d}\mu _{g}\right) , \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} \dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }e^{pu(t)}\mathop {}\mathrm {d}\mu _{g}\le Cp\exp \left( C\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }^2\right) \int _{\Sigma }e^{pu(t)}\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

Thus,

$$\begin{aligned} \int _{\Sigma }e^{pu(t)}\mathop {}\mathrm {d}\mu _{g}\le \exp \left[ Cp\exp \left( C\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }^2\right) t\right] \int _{\Sigma }e^{pu_0}\mathop {}\mathrm {d}\mu _{g}, \quad \forall p\ge 1. \end{aligned}$$

(2.7)

In order to get the global existence of solution when $u_0\in H^1(\Sigma )$, it is necessary to derive several a priori estimates (1.6). To do this, we split three steps.

• Step 1  

  $\left\Vert u(t)\right\Vert _{H^1\left( \Sigma \right) }\le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right)$ for any $t\in [0,T]$.

  Set

  $$\begin{aligned} A(t)=\left\{ x\in \Sigma : e^{u(x,t)}\ge \dfrac{1}{2}\int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}.\right\} \end{aligned}$$

  According to (2.1) and (2.7), we have

  $$\begin{aligned} \int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}=&\int _{\Sigma }e^{u(t)}\mathop {}\mathrm {d}\mu _{g}=\int _{\Sigma {\setminus } A(t)}e^{u(t)}\mathop {}\mathrm {d}\mu _{g}+\int _{A(t)}e^{u(t)}\mathop {}\mathrm {d}\mu _{g}\\ \le&\dfrac{1}{2}\int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}+C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left|A(t)\right|_{g}^{1/2}, \end{aligned}$$

  where $\left|A(t)\right|_{g}$ stands for the area of A(t). This gives

  $$\begin{aligned} \left|A(t)\right|_{g}\ge C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) ^{-1}>0,\quad \left|\int _{A(t)}u(t) \mathop {}\mathrm {d}\mu _{g}\right|\le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \end{aligned}$$

  and

  $$\begin{aligned} \left|\bar{u}(t)\right|:=&\left|\int _{\Sigma }u(t) \mathop {}\mathrm {d}\mu _{g}\right|\\ \le&\left|\int _{\Sigma {\setminus } A(t)}u(t) \mathop {}\mathrm {d}\mu _{g}\right|+\left|\int _{A(t)}u(t) \mathop {}\mathrm {d}\mu _{g}\right|\\ \le&\left|\Sigma {\setminus } A(t)\right|^{1/2}\left( \int _{\Sigma {\setminus } A(t)}u(t)^2 \mathop {}\mathrm {d}\mu _{g}\right) ^{1/2}+C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \\ \le&\sqrt{1-C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) ^{-1}}\left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }+C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) . \end{aligned}$$

  Then, by Poincaré inequality, we get

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }\le&c\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }+\left|\bar{u}(t)\right|\\ \le&c\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }+\sqrt{1-C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) ^{-1}}\left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }\\&\quad +C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) , \end{aligned}$$

  which implies the following $L^2$-estimate

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }\le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left( 1+\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }\right) . \end{aligned}$$

  (2.8)

  Now, applying Young’s inequality to (2.3), we obtain

  $$\begin{aligned} C\ge \int _{\Sigma }\left|\nabla _{g}u(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}-\varepsilon \int _{\Sigma }u(t)^2\mathop {}\mathrm {d}\mu _{g}-C_{\varepsilon }. \end{aligned}$$

  Choosing small $\varepsilon$, together with (2.8), we can conclude that

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{H^1\left( \Sigma \right) }\le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) . \end{aligned}$$

  (2.9)

• Step 2  

  $\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }+\left( \int _0^T\left\Vert \frac{\partial u(t)}{\partial t}\right\Vert _{H^1\left( \Sigma \right) }^2\mathop {}\mathrm {d}t\right) ^{1/2}\le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right)$ for any $t\in [0,T]$.

  Set $w(t)=e^{\frac{u(t)}{2}}\frac{\partial u(t)}{\partial t}$. Then

  $$\begin{aligned}&\dfrac{1}{2}\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }\left|\Delta _{g}u(t)\right|^2\mathop {}\mathrm {d}\mu _{g}\\&\quad =\int _{\Sigma }\Delta _{g}u(t)\Delta _{g}\dfrac{\partial u(t)}{\partial t}\mathop {}\mathrm {d}\mu _{g}\\&\quad =\int _{\Sigma }\left( e^{\frac{u(t)}{2}}w(t)-8\pi \left( \dfrac{he^{u(t)}}{\int _{\Sigma }he^{u(t)\mathop {}\mathrm {d}\mu _{g}}}-1\right) \right) \Delta _{g}\left( e^{-\frac{u(t)}{2}}w(t)\right) \mathop {}\mathrm {d}\mu _{g}\\&\quad =-\int _{\Sigma }\left|\nabla _{g}w(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}+\dfrac{1}{4}\int _{\Sigma }w(t)^2\left|\nabla _{g}u(t)\right|^2\mathop {}\mathrm {d}\mu _{g}\\&\qquad +\dfrac{8\pi }{\int _{\Sigma }he^{u(t)}\mathop {}\mathrm {d}\mu _{g}}\int _{\Sigma }\left\langle e^{\frac{u(t)}{2}}\left( \nabla _{g}h+h\nabla _{g}u(t)\right) ,\nabla _{g}w(t)-\dfrac{1}{2}w(t)\nabla _{g}u(t)\right\rangle _{g}\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

  According to (2.6) and (2.9), we know that

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{H^1\left( \Sigma \right) }+\dfrac{1}{\int _{\Sigma }he^{u(t)}\mathop {}\mathrm {d}\mu _{g}}\le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) . \end{aligned}$$

  Therefore, Young’s inequality implies that

  $$\begin{aligned}&\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }\left|\Delta _{g}u(t)\right|^2\mathop {}\mathrm {d}\mu _{g}\le -\int _{\Sigma }\left|\nabla _{g}w(t)\right|^2_{g}\mathop {}\mathrm {d}\mu _{g}\\&\quad +\int _{\Sigma }w(t)^2\left|\nabla _{g}u(t)\right|^2\mathop {}\mathrm {d}\mu _{g}+C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left( 1+\left\Vert \nabla _{g}u(t)\right\Vert _{L^4\left( \Sigma \right) }^2\right) . \end{aligned}$$

  Since for all $f\in H^1\left( \Sigma \right)$, we have the following interpolation inequality

  $$\begin{aligned} \left\Vert f\right\Vert ^2_{L^4\left( \Sigma \right) }\le c\left\Vert f\right\Vert _{L^2\left( \Sigma \right) }\left\Vert f\right\Vert _{H^1\left( \Sigma \right) }. \end{aligned}$$

  (2.10)

  We estimate

  $$\begin{aligned} \int _{\Sigma }w(t)^2\left|\nabla _{g}u(t)\right|^2\mathop {}\mathrm {d}\mu _{g}\le&c\left\Vert w(t)\right\Vert ^2_{L^4\left( \Sigma \right) }\left\Vert \nabla _{g}u(t)\right\Vert _{L^4\left( \Sigma \right) }^2\\ \le&c\left\Vert w(t)\right\Vert _{L^2\left( \Sigma \right) }\left\Vert w(t)\right\Vert _{H^1\left( \Sigma \right) }\left\Vert u(t)\right\Vert _{H^1\left( \Sigma \right) }\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }\\ \le&C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left\Vert w(t)\right\Vert _{L^2\left( \Sigma \right) }\left\Vert w(t)\right\Vert _{H^1\left( \Sigma \right) }\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) } \end{aligned}$$

  and

  $$\begin{aligned} \left\Vert \nabla _{g}u(t)\right\Vert _{L^4\left( \Sigma \right) }^2\le c\left\Vert u\right\Vert _{H^1\left( \Sigma \right) }\left\Vert u\right\Vert _{H^2\left( \Sigma \right) }\le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left\Vert u\right\Vert _{H^2\left( \Sigma \right) }. \end{aligned}$$

  Hence

  $$\begin{aligned}&\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }\left|\Delta _{g}u(t)\right|^2\mathop {}\mathrm {d}\mu _{g}\\&\quad \le -\dfrac{1}{2}\int _{\Sigma }\left|\nabla _{g}w(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}+\dfrac{1}{2}\int _{\Sigma }w(t)^2\mathop {}\mathrm {d}\mu _{g}\\&\quad +C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left\Vert w(t)\right\Vert _{L^2\left( \Sigma \right) }^2\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }^2\\&\qquad +C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left( 1+\left\Vert u\right\Vert _{H^2\left( \Sigma \right) }\right) \\&\quad \le -\dfrac{1}{4}\int _{\Sigma }\left|\nabla _{g}w(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}+C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \\&\qquad \times \left( 1+\left\Vert w(t)\right\Vert ^2_{L^2\left( \Sigma \right) }\right) \left( 1+\left\Vert \Delta _{g}u(t)\right\Vert ^2_{L^2\left( \Sigma \right) }\right) . \end{aligned}$$

  Thus

  $$\begin{aligned}&\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\ln \left( 1+\left\Vert \Delta _{g}u(t)\right\Vert ^2_{L^2\left( \Sigma \right) } +\int _0^t\int _{\Sigma }\left|\nabla _{g}w(\tau )\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}\mathop {}\mathrm {d}\tau \right) \\&\quad \le C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \left( 1+\left\Vert w(t)\right\Vert ^2_{L^2\left( \Sigma \right) }\right) . \end{aligned}$$

  Together with $u_0\in H^2(\Sigma )$, we obtain

  $$\begin{aligned}&\ln \left( 1+\left\Vert \Delta _{g}u(t)\right\Vert ^2_{L^2\left( \Sigma \right) }+\int _0^T\int _{\Sigma }\left|\nabla _{g}w(\tau )\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}\mathop {}\mathrm {d}\tau \right) \\&\quad \le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right) +C\left( T,\left\Vert u_0\right\Vert _{H^1\left( \Sigma \right) }\right) \int _0^T\left( 1+\left\Vert w(t)\right\Vert ^2_{L^2\left( \Sigma \right) }\right) \mathop {}\mathrm {d}t. \end{aligned}$$

  By (2.2), we know that

  $$\begin{aligned} \int _0^T\left\Vert w(t)\right\Vert ^2_{L^2\left( \Sigma \right) }\mathop {}\mathrm {d}t=\int _{0}^T\int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}\mathop {}\mathrm {d}t=J\left( u(0)-J\left( u(T)\right) \right) \le C. \end{aligned}$$

  Consequently, by using Sobolev embedding, we conclude

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }+\left( \int _0^T\left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{H^1\left( \Sigma \right) }^2\mathop {}\mathrm {d}t\right) ^{1/2}\le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right) . \end{aligned}$$

• Step 3  

  $\left( \int _{0}^T\left\Vert \frac{\partial ^2u}{\partial t^2}\right\Vert _{H^{-1}\left( \Sigma \right) }^2\mathop {}\mathrm {d}t\right) ^{1/2}\le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right)$ for any $t\in [0,T]$.

  Differential the Eq. (1.5) with respect to t, we get

  $$\begin{aligned} e^u\ddot{u}+e^u\dot{u}^2=\Delta \dot{u}+8\pi \left( \frac{he^u\dot{u}}{\int _{\Sigma }he^u\mathop {}\mathrm {d}\mu _{g}}-\frac{he^u\int _{\Sigma }he^u\dot{u}\mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^u\mathop {}\mathrm {d}\mu _{g}\right) ^2}\right) \end{aligned}$$

  where $\ddot{u}=\frac{\partial ^2u}{\partial t^2}$ and $\dot{u}=\frac{\partial u}{\partial t}$. Then for all $\psi \in H^1(\Sigma )$ with $\Vert \psi \Vert _{H^1(\Sigma )}\le 1$, we have

  $$\begin{aligned} \int _\Sigma \ddot{u}\psi \mathop {}\mathrm {d}\mu _{g} =&\int _\Sigma e^{-u}\left( \Delta \dot{u}+8\pi \left( \frac{he^u\dot{u}}{\int _{\Sigma }he^u\mathop {}\mathrm {d}\mu _{g}}-\frac{he^u\int _{\Sigma }he^u\dot{u}\mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^u\mathop {}\mathrm {d}\mu _{g}\right) ^2}\right) \right) \psi \mathop {}\mathrm {d}\mu _{g}\\&\quad -\int _\Sigma |\dot{u}|^2\psi \mu _{g}\\ =&-\int _{\Sigma }\left\langle \nabla _{g}\dot{u},-\nabla _{g} u\psi +\nabla _{g}\psi \right\rangle _{g}e^{-u}\mathop {}\mathrm {d}\mu _{g}\\&+\int _\Sigma 8\pi \left( \dfrac{h\dot{u}}{\int _{\Sigma }he^u\mathop {}\mathrm {d}\mu _{g}}-\dfrac{h\int _{\Sigma }he^u\dot{u}\mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^u\mathop {}\mathrm {d}\mu _{g}\right) ^2}\right) \psi \mathop {}\mathrm {d}\mu _{g}-\int _\Sigma |\dot{u}|^2\psi \mu _{g}\\ \le&C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right) \Vert \dot{u}\Vert _{H^1(\Sigma )}. \end{aligned}$$

  Thus $\left\Vert \ddot{u}(t)\right\Vert _{H^{-1}\left( \Sigma \right) }\le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right) \left\Vert \dot{u(t)}\right\Vert _{H^1(\Sigma )}$ which implies the desired estimate.

Since we have the following embedding (cf. [16, page 304, Theorem 2] and [16, page 305, Theorem 3])

$$\begin{aligned}&C\left( 0,T;H^1\left( \Sigma \right) \right) \subset H^1\left( 0,T;H^1\left( \Sigma \right) \right) ,\quad C\left( 0,T;L^2\left( \Sigma \right) \right) \\&\quad \subset L^2\left( 0,T;H^1\left( \Sigma \right) \right) \cap H^1\left( 0,T;H^{-1}\left( \Sigma \right) \right) , \end{aligned}$$

we get

$$\begin{aligned} u\in&\cap _{0<T<\infty }\left( L^{\infty }\left( 0,T;H^2\left( \Sigma \right) \right) \cap H^1\left( 0,T;H^1\left( \Sigma \right) \right) \cap H^2\left( 0,T;H^{-1}\left( \Sigma \right) \right) \right. \\&\left. \cap C\left( 0,T;H^{1}\left( \Sigma \right) \right) \cap C^1\left( 0,T;L^{2}\left( \Sigma \right) \right) \right) . \end{aligned}$$

By using the parabolic Sobolev embedding theorems (cf. [6, pages 368–369]) together with the interpolation inequality (2.10), we get

$$\begin{aligned} u\in \cap _{0< T<\infty } W^{2,1}_4\left( \Sigma \times [0,T]\right) \subset \cap _{0< T<\infty }C^{\alpha ,\alpha /2}\left( \Sigma \times [0,T]\right) ,\quad \forall 0<\alpha <1. \end{aligned}$$

Here $W^{2,1}_p\left( \Sigma \times [0,T]\right) =L^p\left( 0;T;W^{2,p}\left( \Sigma \right) \right) \cap W^{1,p}\left( 0,T; L^p\left( \Sigma \right) \right)$ stands for the usual parabolic Sobolev space.

Then the standard regularity theory for parabolic equation gives

$$\begin{aligned} \left\Vert u(t)\right\Vert _{C^{2+k+\alpha ,(2+k+\alpha )/2}\left( \Sigma \times [0,T]\right) }\le C\left( T,k,\left\Vert u_0\right\Vert _{C^{2+k+\alpha }(\Sigma )}\right) \end{aligned}$$

for all integer number $k\ge 0$. In particular, we can extend this flow to infinity and u is smooth in $\Sigma \times (0,\infty )$.

Now assume $u_0\in H^2\left( \Sigma \right)$ and choose a sequence of smooth functions $u_{0,\varepsilon }$ on $\Sigma$ such that $u_{0,\varepsilon }$ converges to $u_0$ in $H^2\left( \Sigma \right)$ as $\varepsilon \rightarrow 0$. Let $u_{\varepsilon }$ be the unique smooth solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{\partial e^{u_{\varepsilon }}}{\partial t}=\Delta _{g}u_{\varepsilon }+8\pi \left( \dfrac{he^{u_{\varepsilon }}}{\int _{\Sigma }he^{u_{\varepsilon }}\mathop {}\mathrm {d}\mu _{g}}-1\right) ,&{}\Sigma \times (0,\infty ),\\ u_{\varepsilon }(\cdot ,0)=u_{0,\varepsilon },&{}\Sigma . \end{array}\right. } \end{aligned}$$

The a prior estimates (1.6) gives the following estimates

$$\begin{aligned}&ess\,sup_{0\le t\le T}\left\Vert u_{\varepsilon }(t)\right\Vert _{H^2\left( \Sigma \right) }+\left( \int _0^T\left( \left\Vert \dfrac{\partial u_{\varepsilon }(t)}{\partial t}\right\Vert _{H^1\left( \Sigma \right) }^2+\left\Vert \dfrac{\partial ^2 u_{\varepsilon }(t)}{\partial t^2}\right\Vert _{H^{-1}\left( \Sigma \right) }^2\right) \mathop {}\mathrm {d}t\right) ^{1/2}\\&\quad \le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }\right) ,\quad \forall 0<T<\infty . \end{aligned}$$

Thus we obtain a solution $u\in C^{\infty }\left( \Sigma \times (0,\infty )\right)$ to (1.5) with

$$\begin{aligned}&u\in \cap _{0<T<\infty }\left( L^{\infty }\left( 0,T;H^2\left( \Sigma \right) \right) \cap H^1\left( 0,T;H^1\left( \Sigma \right) \right) \right. \\&\left. \cap H^2\left( 0,T;H^{-1}\left( \Sigma \right) \right) \right) \end{aligned}$$

and the desired a priori estimates (1.6).

To prove the uniqueness of the solution, we assume that u and v are two solutions to (1.5) with initial data $u_0$ and $v_0$ respectively. Denote $w=u-v$. By direct computations, we have

$$\begin{aligned} a\dfrac{\partial w}{\partial t}=\Delta _{g}w+f-bw \end{aligned}$$

(2.11)

where

$$\begin{aligned} a=&\int _0^1e^{su+(1-s)v}\mathop {}\mathrm {d}s,\quad b=\int _0^1e^{su+(1-s)v}\left( s\dfrac{\partial u}{\partial t}+(1-s)\dfrac{\partial v}{\partial t}\right) \mathop {}\mathrm {d}s=\dfrac{\partial a}{\partial t},\\ f=&8\pi \int _0^1\dfrac{he^{su+(1-s)v}}{\int _{\Sigma }he^{su+(1-s)v}\mathop {}\mathrm {d}\mu _{g}}w\mathop {}\mathrm {d}s -8\pi \int _0^1\dfrac{he^{su+(1-s)v}\int _{\Sigma }he^{su+(1-s)v}w\mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^{su+(1-s)v\mathop {}\mathrm {d}\mu _{g}}\right) ^2}\mathop {}\mathrm {d}s. \end{aligned}$$

One can check that there is a constant C depends only on $T, \left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }$ and $\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }$ such that for all $0\le t\le T$

$$\begin{aligned} C^{-1}\le&a(t)\le C, \quad \left|b(t)\right|\le C\left( \left|\dfrac{\partial u(t)}{\partial t}\right|+\left|\dfrac{\partial v(t)}{\partial t}\right|\right) ,\\ \left|f(t)\right|\le&C\left( \left|w(t)\right|+\left\Vert w(t)\right\Vert _{L^2\left( \Sigma \right) }\right) ,\quad \int _{\Sigma }f(t)w(t)\le C \int _{\Sigma }w(t)^2. \end{aligned}$$

Then we obtain

$$\begin{aligned} \dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }a(t)w(t)^2\mathop {}\mathrm {d}\mu _{g}=&\int _{\Sigma }b(t)w(t)^2\mathop {}\mathrm {d}\mu _{g}+2\int _{\Sigma }a(t)w(t)\dfrac{\partial w(t)}{\partial t}\mathop {}\mathrm {d}\mu _{g}\\ =&-2\int _{\Sigma }\left|\nabla _{g}w(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}+2\int _{\Sigma }f(t)w(t)\mathop {}\mathrm {d}\mu _{g}-\int _{\Sigma }b(t)w(t)^2\mathop {}\mathrm {d}\mu _{g}\\ \le&-\int _{\Sigma }\left|\nabla _{g}w(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}+C\int _{\Sigma }a(t)w(t)^2\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

Gronwall’s inequality implies

$$\begin{aligned} \int _{\Sigma }w(t)^2\mathop {}\mathrm {d}\mu _{g}\le&C\int _{\Sigma }a(t)w(t)^2\mathop {}\mathrm {d}\mu _{g}\le Ce^t\int _{\Sigma }a(0)w(0)^2\mathop {}\mathrm {d}\mu _{g}\nonumber \\ =&Ce^t\left\Vert u_0-v_0\right\Vert _{L^2\left( \Sigma \right) }^2,\quad \forall 0<t<T. \end{aligned}$$

(2.12)

The uniqueness then follows from the above inequality and we finish the proof. $\square$

Remark 2.1

One check that the difference of two solutions u and v satisfies

$$\begin{aligned} \left\Vert u-v\right\Vert _{W^{2,1}_2\left( \Sigma \times [0,T]\right) }\le C\left( T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) },\left\Vert v_0\right\Vert _{H^2\left( \Sigma \right) }\right) \left\Vert u_0-v_0\right\Vert _{H^2\left( \Sigma \right) }. \end{aligned}$$

The proof is standard. Roughly speaking, (2.11) implies

$$\begin{aligned} \left|a^{1/2}\dfrac{\partial w}{\partial t}-a^{-1/2}\Delta _{g}\right|^2=a\left|\dfrac{\partial w}{\partial t}\right|^2+a^{-1}\left|\Delta _{g}w\right|^2-2\left\langle \dfrac{\partial w}{\partial t},\Delta _{g}w\right\rangle _{g}. \end{aligned}$$

Integration by parts,

$$\begin{aligned}&\int _{\Sigma } \left( a\left|\dfrac{\partial w}{\partial t}w\right|^2+a^{-1}\left|\Delta _{g}w\right|^2\right) \mathop {}\mathrm {d}\mu _{g}+\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\int _{\Sigma }\left|\nabla _{g}w\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}\\&\quad \le C\int _{\Sigma }\left|w\right|^2\mathop {}\mathrm {d}\mu _{g}+C\left( \int _{\Sigma }\left|b\right|^4\mathop {}\mathrm {d}\mu _{g}\right) ^{1/2}\left( \int _{\Sigma }\left|w\right|^4\mathop {}\mathrm {d}\mu _{g}\right) ^{1/2}. \end{aligned}$$

Applying the interpolation inequality (2.10) and the $L^2$-estimate (2.12) of the w, we have

$$\begin{aligned}&\int _0^T\int _{\Sigma } \left( a\left|\dfrac{\partial w}{\partial t}w\right|^2+a^{-1}\left|\Delta _{g}w\right|^2\right) \mathop {}\mathrm {d}\mu _{g}\mathop {}\mathrm {d}t+\max _{0\le t\le T}\int _{\Sigma }\left|\nabla _{g}w(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}\\&\quad \le C\int _{\Sigma }\left|u_0-v_0\right|^2\mathop {}\mathrm {d}\mu _{g} \end{aligned}$$

where the constant C depends only on $T,\left\Vert u_0\right\Vert _{H^2\left( \Sigma \right) }$ and $\left\Vert v_0\right\Vert _{H^2\left( \Sigma \right) }$.

3 Blowup analysis

In this section, we prove an estimate of a Dirac measure at the blowup points. Consequently, we show the fact that the flow develops at most one blowup point when the time goes to infinity.

According to (2.2) and (2.5), we know that

$$\begin{aligned} \int _{0}^{\infty }\int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}\mathop {}\mathrm {d}t\le C. \end{aligned}$$

There is a sequence of positive numbers $\left\{ t_n\right\}$ such that $n\le t_n\le n+1$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{\Sigma }e^{u(t_n)}\left|\dfrac{\partial u(t_n)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}=0. \end{aligned}$$

Set

$$\begin{aligned} u_n=u(t_n),\quad V_n=\dfrac{8\pi h}{\int _{\Sigma }he^{u_n}\mathop {}\mathrm {d}\mu _{g}},\quad \rho =8\pi ,\quad f_n=e^{u_n/2}\dfrac{\partial u(t_n)}{\partial t}, \end{aligned}$$

(3.1)

then

$$\begin{aligned} -\Delta _{g}u_n=V_ne^{u_n}-\rho -f_ne^{u_n/2},\quad \text {in}\ \Sigma , \end{aligned}$$

(3.2)

and $u_n,V_n,\rho ,f_n$ are smooth functions on $\Sigma$ satisfying

$$\begin{aligned} \rho >0,\quad 0\le V_n\le C,\quad \quad \lim _{n\rightarrow \infty }\left\Vert f_n\right\Vert _{L^2\left( \Sigma \right) }=0. \end{aligned}$$

(3.3)

One can check that

$$\begin{aligned} \int _{\Sigma }e^{u_n}\mathop {}\mathrm {d}\mu _{g}\le C. \end{aligned}$$

(3.4)

We say that a sequence $\left\{ u_n\right\}$ which satisfies (3.2) and (3.3) is a blowup sequence if $\limsup \limits _{n\rightarrow \infty }\max \limits _{\Sigma }u_n=+\infty$.

Lemma 3.1

If $\left\{ u_n\right\}$ is not a blowup sequence, then $\left\{ u_n\right\}$ is bounded in $H^{2}\left( \Sigma \right)$.

Proof

By definition, $\left\{ u_n^{+}\right\}$ is bounded in $L^{\infty }\left( \Sigma \right)$. By the standard elliptic estimates and the normalization $\int _{\Sigma }\mathop {}\mathrm {d}\mu _{g}=1$, we conclude that $\left\{ u_n-\bar{u}_n\right\}$ is bounded in $H^{2}\left( \Sigma \right)$, where $\bar{u}_n:=\bar{u}(t_n)=\int _{\Sigma }u(t_n)\mathop {}\mathrm {d}\mu _{g}$. By Jensen’s inequality, according to (3.4), we have $\bar{u}_n\le C$. It suffices to prove that $\bar{u}_n\ge -C$. Otherwise, there is a subsequence $\left\{ u_{n_k}\right\}$ such that $\lim _{n_k\rightarrow \infty }\bar{u}_{n_k}=-\infty$. Notice that

$$\begin{aligned} -\Delta _{g}\left( u_{n_k}-\bar{u}_{n_k}\right) =V_{n_k} e^{\bar{u}_{n_k}}e^{u_{n_k}-\bar{u}_{n_k}}-\rho -f_{n_k}e^{\bar{u}_{n_k}/2}e^{\left( u_{n_k}-\bar{u}_{n_k}\right) /2},\quad \text {in}\ \Sigma . \end{aligned}$$

We may assume $u_{n_k}-\bar{u}_{n_k}$ converges weakly to $\hat{u}$ in $H^2\left( \Sigma \right)$ and strongly in $L^1\left( \Sigma \right)$. Then $\left\{ e^{p\left( u_{n_k}-\bar{u}_{n_k}\right) }\right\}$ converges strongly to $e^{p\hat{u}}$ in $L^1\left( \Sigma \right)$ for each $p>0$. Thus $\hat{u}$ is a weak solution to

$$\begin{aligned} -\Delta _{g}\hat{u}=-\rho . \end{aligned}$$

It is well know that $\hat{u}\in C^{\infty }\left( \Sigma \right)$ and $\rho =0$ which is a contradiction. Therefore, $\left\{ u_n\right\}$ is bounded in $H^2\left( \Sigma \right)$. $\square$

From now on, we assume $\left\{ u_n\right\}$ is a blowup sequence. Since $\left\{ V_n e^{u_n}\right\}$ is bounded in $L^1\left( \Sigma \right)$, we may assume $\left\{ V_n e^{u_{n}}\mathop {}\mathrm {d}\mu _{g}\right\}$ converges to a nonzero Radon measure $\mu$ on $\Sigma$ in the sense of measures. By using the method of potential estimates (cf. [20, Lemma 7.12]), we get

$$\begin{aligned} \left\Vert u_n-\bar{u}_n\right\Vert _{W^{1,p}\left( \Sigma \right) }\le C_p,\quad \forall 1\le p<2. \end{aligned}$$

We may assume $\left\{ u_n-\bar{u}_n\right\}$ converges weakly to G in $W^{1,p}\left( \Sigma \right)$ and strongly in $L^p\left( \Sigma \right)$ for every $1<p<2$. Hence G satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{g}G=\mu -\rho ,\quad \text {in}\ \Sigma ,\\ \int _{\Sigma }G\mathop {}\mathrm {d}\mu _{g}=0, \end{array}\right. } \end{aligned}$$

in the sense of distribution. Define the singular set S of the sequence $\left\{ u_n\right\}$ as follows

$$\begin{aligned} S=\left\{ x\in \Sigma :\mu \left( \left\{ x\right\} \right) \ge 4\pi .\right\} \end{aligned}$$

It is easy to check that S is a finite nonempty subset of $\Sigma$.

Recall Brezis-Merle’s estimate ([2, Theorem 1]).

Lemma 3.2

(cf. [12]) Let $\Omega \subset \Sigma$ be a smooth domain. Assume u is a solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{g}u=f,&{}\text {in}\ \Omega ,\\ u=0,&{}\text {on}\ \partial \Omega , \end{array}\right. } \end{aligned}$$

where $f\in L^1\left( \Omega \right)$. For every $0<\delta <4\pi$, there is a constant C depending only on $\delta$ and $\Omega$ such that

$$\begin{aligned} \int _{\Omega }\exp \left( \dfrac{\left( 4\pi -\delta \right) \left|u\right|}{\left\Vert f\right\Vert _{L^1\left( \Omega \right) }}\right) \mathop {}\mathrm {d}\mu _{g}\le C. \end{aligned}$$

As a consequence, we have the following Lemma (cf. [12, Lemma 2.8]).

Lemma 3.3

If $x\notin S$, then there is a geodesic ball $B^{g}_r(x)\subset \Sigma {\setminus } S$ and a positive constant $C=C_x$ such that

$$\begin{aligned} \left\Vert u_n-\bar{u}_n\right\Vert _{L^{\infty }\left( B^{g}_r(x)\right) }\le C. \end{aligned}$$

Proof

There exist $\delta =\delta _x\in \left( 0,2\pi \right) , r=r_x\in \left( 0,\mathrm {inj}\left( \Sigma \right) /4\right) , N=N_x\in \mathbb {N}$ such that

$$\begin{aligned} \int _{B^{g}_{4r}(x)}\left|V_n e^{u_n}-f_ne^{u_n/2}\right|\mathop {}\mathrm {d}\mu _{g}\le 4\pi -2\delta ,\quad \forall n\ge N. \end{aligned}$$

Solve

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{g}y_n=-\rho ,&{}\text {in}\ B^{g}_{4r}(x),\\ y_n=0,&{}\text {on}\ \partial B^{g}_{4r}(x). \end{array}\right. } \end{aligned}$$

It is well know that $\left\{ y_n\right\}$ is bounded in $L^{\infty }\left( B^{g}_{4r}(x)\right)$. Solve

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{g}w_n=V_n e^{u_n}-f_ne^{u_n/2},&{}\text {in}\ B^{g}_{4r}(x),\\ w_n=0,&{}\text {on}\ \partial B^{g}_{4r}(x). \end{array}\right. } \end{aligned}$$

(3.5)

According to Lemma 3.2, we have

$$\begin{aligned} \left\Vert e^{\left|w_n\right|}\right\Vert _{L^{p}\left( B^{g}_{4r}(x)\right) }\le C,\quad p=\dfrac{4\pi -\delta }{4\pi -2\delta }>1. \end{aligned}$$

In particular, $\left\{ w_n\right\}$ is bounded in $L^1\left( B^{g}_{4r}(x)\right)$. Since $h_n:=u_n-\bar{u}_n-y_n-w_n$ is harmonic in $B^{g}_{4r}(x)$, we have

$$\begin{aligned} \left\Vert h_n\right\Vert _{L^{\infty }\left( B^{g}_{2r}(x)\right) }\le&C\left\Vert h_n\right\Vert _{L^{1}\left( B^{g}_{4r}(x)\right) }\\ \le&C\left( \left\Vert u_n-\bar{u}_n\right\Vert _{L^{1}\left( B^{g}_{4r}(x)\right) }+\left\Vert y_n\right\Vert _{L^{1}\left( B^{g}_{4r}(x)\right) }+\left\Vert w_n\right\Vert _{L^{1}\left( B^{g}_{4r}(x)\right) }\right) \\ \le&C\left( \left\Vert u_n-\bar{u}_n\right\Vert _{L^{1}\left( \Sigma \right) }+\left\Vert w_n\right\Vert _{L^{1}\left( B^{g}_{4r}(x)\right) }+\left\Vert y_n\right\Vert _{L^{1}\left( B^{g}_{4r}(x)\right) }\right) \\ \le&C. \end{aligned}$$

Thus

$$\begin{aligned} \left\Vert e^{u_n}\right\Vert _{L^p\left( B_{2r}^{g}(x)\right) }\le C. \end{aligned}$$

Applying the standard elliptic estimates for (3.5), we get

$$\begin{aligned} \left\Vert w_n\right\Vert _{L^{\infty }\left( B^g_{r}(x)\right) }\le C. \end{aligned}$$

Hence

$$\begin{aligned} \left\Vert u_n-\bar{u}_n\right\Vert _{L^{\infty }\left( B^g_{r}(x)\right) }\le C. \end{aligned}$$

$\square$

Theorem 3.4

If $\left\{ u_n\right\}$ is a blowup sequence, then S is nonempty and

$$\begin{aligned} S=\left\{ x\in \Sigma : \exists \ \left\{ x_n\right\} \subset \Sigma ,\ \lim _{n\rightarrow \infty }x_n=x,\ \lim _{n\rightarrow \infty }u_n\left( x_n\right) =+\infty .\right\} \end{aligned}$$

Moreover $\lim _{n\rightarrow \infty }\bar{u}_n=-\infty$. Thus $\left\{ u_n\right\}$ converges to $-\infty$ uniformly on compact subsets of $\Sigma {\setminus } S$ and $\mu =\sum _{x\in S}\mu \left( \left\{ x\right\} \right) \delta _{x}$ is a Dirac measure.

Proof

According to Lemma 3.3, we know that $\left\{ u_n-\bar{u}_n\right\}$ is bounded in $L^{\infty }_{loc}\left( \Sigma {\setminus } S\right)$.

If $S=\emptyset$, then $\left\{ u_n-\bar{u}_n\right\}$ is bounded in $L^{\infty }\left( \Sigma \right)$ which implies that $\left\{ u_n^{+}\right\}$ is bounded in $L^{\infty }\left( \Sigma \right)$ which is a contradiction.

We claim that $\lim _{n\rightarrow \infty }\bar{u}_n=-\infty$. Otherwise, there is a subsequence of $\left\{ u_n\right\}$ which also denoted by $\left\{ u_n\right\}$ such that

$$\begin{aligned} \bar{u}_n\ge -C. \end{aligned}$$

For $x\in S$, choose $r>0$ such that $B^{g}_{2r}(x)\cap S=\left\{ x\right\}$. According to Lemma 3.3, $\left\{ u_n\right\}$ is bounded in $L^{\infty }_{loc}\left( B^{g}_{2r}(x){\setminus }\left\{ x\right\} \right)$. In particular, $M:=\sup _{n}\left\Vert u_n\right\Vert _{L^{\infty }\left( \partial B^{g}_{r}\left( x\right) \right) }<\infty$. Solve

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{g} z_n=V_ne^{u_n}-\rho -f_ne^{u_n/2},&{}\text {in}\ B^{g}_{r}(x),\\ z_n=-M,&{}\text {on}\ B^{g}_{r}(x). \end{array}\right. } \end{aligned}$$

By potential estimates, we know that $z_n$ is bounded in $W^{1,p}\left( B^{g}_{r}(x)\right)$ for all $1<p<2$. Thus, up to a subsequence, $z_n$ converges weakly to $z\in W^{1,p}\left( B^{g}_{r}(x)\right)$ for all $1<p<2$ and strongly in $L^q\left( B^{g}_{r}(x)\right)$ for all $1<q<\infty$. Then z is a weak solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{g} z=\mu \left( \left\{ x\right\} \right) \delta _{x}-\rho ,&{}\text {in}\ B^{g}_{r}(x),\\ z=-M,&{}\text {on}\ B^{g}_{r}(x). \end{array}\right. } \end{aligned}$$

Thus

$$\begin{aligned} z(\cdot )\ge -\dfrac{\mu \left( \left\{ x\right\} \right) }{2\pi }\ln \mathrm {dist}_{g}(\cdot ,x)-C. \end{aligned}$$

Since $\mu \left( \left\{ x\right\} \right) \ge 4\pi$, we get

$$\begin{aligned} \int _{B^{g}_{r}(x)}e^{z}\mathop {}\mathrm {d}\mu _{g}=\infty . \end{aligned}$$

On the other hand, the maximum principle implies that $z_n\le u_n$. By Fatou’s Lemma,

$$\begin{aligned} \infty =\int _{B^{g}_{r}(x)}e^{z}\mathop {}\mathrm {d}\mu _{g}\le \liminf _{n\rightarrow \infty }\int _{B^{g}_{r}(x)}e^{z_n}\mathop {}\mathrm {d}\mu _{g}\le \liminf _{n\rightarrow \infty }\int _{B^{g}_{r}(x)}e^{u_n}\mathop {}\mathrm {d}\mu _{g}\le C, \end{aligned}$$

which is a contradiction.

Hence $\left\{ u_n\right\}$ converges to $-\infty$ uniformly on compact subsets of $\Sigma {\setminus } S$. Thus for every domain $\Omega \subset \Sigma$

$$\begin{aligned} \mu \left( \Omega \right) =&\lim _{n\rightarrow \infty }\int _{\Omega }V_n e^{u_n}\mathop {}\mathrm {d}\mu _{g}\\ =&\sum _{x\in S}\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{\Omega {\setminus } B_{r}^{g}(x)}V_n e^{u_n}\mathop {}\mathrm {d}\mu _{ g}+\sum _{x\in S}\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{\Omega \cap B_{r}^{g}(x)}V_n e^{u_n}\mathop {}\mathrm {d}\mu _{g}\\ =&\sum _{x\in S\cap \Omega }\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{ B_{r}^{g}(x)}V_n e^{u_n}\mathop {}\mathrm {d}\mu _{g}\\ =&\sum _{x\in S\cap \Omega }\mu \left( \left\{ x\right\} \right) \\ =&\mu \left( \Omega \cap S\right) . \end{aligned}$$

In other words, $\mu =\sum _{x\in S}\mu \left( \left\{ x\right\} \right) \delta _{x}$ is a Dirac measure.

According to Lemma 3.3, we obtain

$$\begin{aligned} \left\{ x\in \Sigma : \exists \ \left\{ x_n\right\} \subset \Sigma ,\ \lim _{n\rightarrow \infty }x_n=x,\ \lim _{n\rightarrow \infty }u_n\left( x_n\right) =+\infty .\right\} \subset S. \end{aligned}$$

For $x_0\in S$, choose a geodesic ball $B_{2r}^{g}\left( x_0\right)$ such that $B_{2r}^{g}\left( x_0\right) \cap S=\left\{ x_0\right\}$. Choose $x_n\in \overline{B_{r}^{g}\left( x_0\right) }$ such that

$$\begin{aligned} \lambda _n:=\max _{\overline{B_{r}^{\Sigma }\left( x_0\right) }}u_n=u_n\left( x_n\right) . \end{aligned}$$

• Fact 1. $\lim _{n\rightarrow \infty }\lambda _n=\infty$.

  Otherwise, up to a subsequence, $\left\{ u_n^{+}\right\}$ is bounded in $L^{\infty }\left( B^{g}_r\left( x_0\right) \right)$. Thus $\left\{ e^{u_n}\right\}$ is bounded in $L^{\infty }\left( B^{g}_r\left( x_0\right) \right)$ which is a contradiction.

• Fact 2. $\lim _{n\rightarrow \infty }x_n=x_0$.

  Otherwise, up to a subsequence, $\lim _{n\rightarrow \infty }x_n=\tilde{x}\in \overline{B_{r}\left( x_0\right) }{\setminus }\left\{ x_0\right\}$. Thus $\tilde{x}$ is not a singular point which is impossible according Lemma 3.3 and the above claim.

Consequently,

$$\begin{aligned} S\subset \left\{ x\in \Sigma : \exists \ \left\{ x_n\right\} \subset \Sigma ,\ \lim _{n\rightarrow \infty }x_n=x,\ \lim _{n\rightarrow \infty }u_n\left( x_n\right) =+\infty .\right\} \end{aligned}$$

$\square$

Now we want to prove that $\mu \left( \left\{ x_0\right\} \right) \ge 8\pi$. We assume additionally that $V_n$ converges to V in $C^0\left( \Sigma \right)$.

Lemma 3.5

For each $x_0\in S$, we have $V(x_0)>0$ and $\mu \left( \left\{ x_0\right\} \right) \ge 8\pi$.

Proof

The proof is similar to [25, Lemma 1]. Assume $B^{g}_{2r}(x_0)\cap S=\left\{ x_0\right\}$. Choose $x_n\in B^{g}_{2r}(x_0)$ such that

$$\begin{aligned} \lambda _n:=\max _{\overline{B^{g}_{r}(x_0)}}u_n=u_{n}\left( x_n\right) . \end{aligned}$$

It is easy to check that

$$\begin{aligned} \lim _{n\rightarrow \infty }\lambda _n=+\infty ,\quad \lim _{n\rightarrow \infty }x_n=x_0. \end{aligned}$$

Now choose a conformal coordinate $\left\{ x\right\}$ centered at $x_0$. We have $g=e^{\phi (x)}\left|\mathop {}\mathrm {d}x\right|^2$ and

$$\begin{aligned} -\Delta _{\mathbb {R}^2}u_n=V_n e^{\phi } e^{u_n}-e^{\phi }\rho -f_ne^{\phi }e^{u_n/2},\quad \left|x\right|<2\tilde{r}. \end{aligned}$$

Consider

$$\begin{aligned} \tilde{u}_n(x)=u_n\left( x_n+e^{-\lambda _n/2}x\right) -\lambda _n, \end{aligned}$$

then for $\left|x\right|<e^{\lambda _n/2}\tilde{r}$,

$$\begin{aligned} -\Delta _{\mathbb {R}^2}\tilde{u}_n(x)=&V_n\left( x_n+e^{-\lambda _n/2}x\right) e^{\phi \left( x_n+e^{-\lambda _n/2}x\right) }e^{\tilde{u}_n(x)}\\&-e^{\phi \left( x_n+e^{-\lambda _n/2}x\right) }\rho -f_n\left( x_n+e^{-\lambda _n/2}x\right) e^{\phi \left( x_n+e^{-\lambda _n/2}x\right) -\lambda _n/2}e^{\tilde{u}_n(x)/2}. \end{aligned}$$

We have $\tilde{u}_n\le 0, \tilde{u}_n(0)=0$ and

$$\begin{aligned}&\int _{B_{e^{\lambda _n/2}\tilde{r}}}e^{\tilde{u}_n}\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}\le \int _{B^{g}_{2\tilde{r}}(x_0)}e^{u_n}\mathop {}\mathrm {d}\mu _{g}\le C,\\&\quad \int _{B_{e^{\lambda _n/2}\tilde{r}}}\left|f_n\left( x_n+e^{-\lambda _n/2}x\right) e^{\phi \left( x_n+e^{-\lambda _n/2}x\right) -\lambda _n/2}\right|^2\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}\le \int _{B^{g}_{2\tilde{r}}(x_0)}f_n^2\mathop {}\mathrm {d}\mu _{g}\rightarrow 0. \end{aligned}$$

Thus up to a subsequence, $\left\{ \tilde{u}_n\right\}$ converges weakly to $\tilde{u}_{\infty }$ in $H^{2}_{loc}\left( \mathbb {R}^2\right)$ and strongly in $H^{1}_{loc}\left( \mathbb {R}^2\right)$. In particular, $\tilde{u}_{\infty }$ is a weak solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _{\mathbb {R}^2}\tilde{u}_{\infty }=V(x_0) e^{\phi (0)}e^{\tilde{u}_{\infty }},&{}\text {in}\ \mathbb {R}^2,\\ \int _{\mathbb {R}^2}e^{\tilde{u}_{\infty }}\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}<\infty . \end{array}\right. } \end{aligned}$$

By a classification theorem of Chen-Li [11], we know that

$$\begin{aligned} \int _{\mathbb {R}^2}V(x_0) e^{\phi (0)}e^{\tilde{u}_{\infty }}\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}=8\pi . \end{aligned}$$

In particular $V(x_0)>0$. By Fatou’s Lemma, we have

$$\begin{aligned} \int _{\mathbb {R}^2}V(x_0) e^{\phi (0)}e^{\tilde{u}_{\infty }}\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}=&\lim _{R\rightarrow \infty }\int _{B_{R}}V(x_0) e^{\phi (0)}e^{\tilde{u}_{\infty }}\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}\\ \le&\lim _{R\rightarrow \infty }\liminf _{n\rightarrow \infty }\int _{B_{R}}V_n\left( x_n+e^{-\lambda _n/2}x\right) e^{\phi \left( x_n+e^{-\lambda _n/2}x\right) }e^{\tilde{u}_n(x)}\mathop {}\mathrm {d}\mu _{\mathbb {R}^2}\\ =&\lim _{R\rightarrow \infty }\liminf _{n\rightarrow \infty }\int _{B^{g}_{e^{-\lambda _n/2}R}\left( x_n\right) }V_n\left( x_n+e^{-\lambda _n/2}x\right) e^{u_n}\mathop {}\mathrm {d}\mu _{g}\\ \le&\lim _{r\rightarrow 0}\liminf _{n\rightarrow \infty }\int _{B^{g}_{r}\left( x_0\right) }V e^{u_n}\mathop {}\mathrm {d}\mu _{g}\\ =&\mu \left( \left\{ x_0\right\} \right) . \end{aligned}$$

Thus

$$\begin{aligned} \mu \left( \left\{ x_0\right\} \right) \ge 8\pi . \end{aligned}$$

In our initial model (3.1), we must have $\mu \left( \left\{ x_0\right\} \right) =8\pi$ and $\#S=1$. Moreover,

$$\begin{aligned} V=\lim _{n}V_n=\lim _{n\rightarrow \infty }\dfrac{8\pi h}{\int _{\Sigma }he^{u_n}\mathop {}\mathrm {d}\mu _{g}}=\dfrac{8\pi h}{h(x_0)\int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}} \end{aligned}$$

and $h(x_0)>0$.

4 Lower bound for the functional

In this section, we give a lower bound for J(u(t)) along the flow, i.e. we give the proof of Theorem 1.2.

Proof of Theorem 1.2

Suppose our flow develops a singularity as time goes to infinity, we will analyse the asymptotic behavior of the flow near and away from the blow-up point and derive a lower bound of J(u). From the previous compactness argument, there is only one blow-up point when $\rho =8\pi$, denoted by $x_0$. Then there is a sequence of points $\{x_n\}$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }x_n=x_0, \ \lambda _n=u_n(x_n)=\max _{\Sigma }u_n=\max _{\Sigma }u(t_n)=+\infty , \end{aligned}$$

where $t_n\rightarrow \infty$ as $n\rightarrow \infty$. In an isothermal coordinate system $\{x\}$ around $x_0$, we still denote $u_n$ and $x_n$ in this coordinate by $u_n$ and $x_n$, respectively. Set $r_n=e^{-\lambda _n/2}$ and

$$\begin{aligned} \tilde{u}_n:=u_n\left( x_n+r_n x\right) -\lambda _n. \end{aligned}$$

Then we have $\tilde{u}_n$ weakly converges to $\tilde{u}_{\infty }$ satisfying

$$\begin{aligned} \tilde{u}_{\infty }=-2\ln \left( 1+a|x|^2\right) , \ a=\frac{\pi e^{\phi (0)}}{\int _\Sigma e^{u_0}\mathop {}\mathrm {d}\mu _{g}} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{2}\int _{B^g_{r_nR}(x_n)}|du_n|_g^2d\mu _g&=\lim _{n\rightarrow \infty }\frac{1}{2}\int _{B_R(x_n)}|d\tilde{u}_n|^2d\mu _{\mathbb {R}^2}\\&=\pi \int _{0}^R\left| \frac{4ar}{1+ar^2}\right| ^2rdr\\&=8\pi \int _{0}^{aR^2}\frac{s}{(1+s)^2}ds\\&=8\pi \int _{0}^{aR^2}\left( \frac{1}{1+s}-\frac{1}{(1+s)^2}\right) ds\\&=8\pi \left( \ln (1+aR^2)+\frac{1}{1+aR^2}-1\right) \\&=8\pi \ln (aR^2)-8\pi +o_R(1). \end{aligned} \end{aligned}$$

(4.1)

Here and in the following, we use $o_{R}(1),o_{n}(1), o_{\delta }(1)$ to denote those functions which converges to zero as $R\rightarrow +\infty , n\rightarrow \infty , \delta \rightarrow 0$ respectively.

Since $u_n-\bar{u}_n$ converges to G weakly in $W^{1,p}(\Sigma )$ for $1<p<2$ and strongly in $H^2_{loc}(\Sigma {\setminus }\{x_0\})$ (see Proposition 3.5 in [23]) and G satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _gG=8\pi (\delta _{x_0}-1),&{}\Sigma ,\\ \int _\Sigma Gd\mu _g=0. \end{array}\right. } \end{aligned}$$

we get

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{2}\int _{\Sigma {\setminus } B^g_\delta (x_n)}|du_n|^2d\mu _g&=\lim _{n\rightarrow \infty }\frac{1}{2}\int _{\Sigma {\setminus } B^g_\delta (x_0)}|du_n|^2d\mu _g\\&=\frac{1}{2}\int _{\Sigma {\setminus } B^g_\delta (x_0)}|dG|^2d\mu _g\\&=-\frac{1}{2}\int _{\Sigma {\setminus } B^g_\delta (x_0)}G\Delta _gGd\mu _g-\frac{1}{2}\int _{\partial B^g_\delta (x_0)}G\frac{\partial }{\partial \nu }Gd\mu _g\\&=4\pi \int _{B^g_\delta (x_0)}G\mathop {}\mathrm {d}\mu _{g}-\frac{1}{2}\int _{\partial B^g_\delta (x_0)}G\frac{\partial }{\partial \nu }Gd\mu _g, \end{aligned} \end{aligned}$$

where $\nu$ is the normal vector field on $\partial B^g_\delta (x_0)$ pointing to the complement of $B^g_\delta (x_0)$.

In normal coordinate, G has the following expansion

$$\begin{aligned} G(x)=-4\ln |x-x_0|+A\left( x_0\right) +\left( b,x-x_0\right) +\left( x-x_0\right) ^Tc\left( x-x_0\right) +O\left( |x-x_0|^3\right) . \end{aligned}$$

Then

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{2}\int _{\Sigma {\setminus } B^g_\delta \left( x_n\right) }|du_n|^2d\mu _g=-16\pi \ln \delta +4\pi A\left( x_0\right) +o_\delta (1). \end{aligned}$$

(4.2)

Define

$$\begin{aligned} u_n^*(r)=\frac{1}{2\pi }\int _0^{2\pi }u_n\left( x_n+re^{i\theta }\right) d\theta . \end{aligned}$$

Then for $r_nR\le r<s\le \delta$, we have (cf. [24, equation (3.4)])

$$\begin{aligned} \int _{B_s{\setminus } B_r}|du_n^*|^2dx\le \int _{B_s{\setminus } B_r}\left| \frac{\partial u_n}{\partial r}\right| ^2dx. \end{aligned}$$

Notice that

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }\left( u_n^*\left( r_nR\right) +2\ln r_n\right) =&-2\ln \left( 1+aR^2\right) =-2\ln \left( aR^2\right) +o_R(1),\\ \lim _{n\rightarrow \infty }\left( u_n^*(\delta )-\bar{u}_n\right) =&\frac{1}{2\pi }\int _0^{2\pi }G\left( \delta e^{i\theta }\right) d\theta =-4\ln \delta +A\left( x_0\right) +o_\delta (1). \end{aligned} \end{aligned}$$

(4.3)

Let $w_n$ be the harmonic functions in the neck domains $B^g_\delta (x_n){\setminus } B^g_{r_nR}(x_n)$ such that

$$\begin{aligned} w_n|_{\partial B^g_\delta \left( x_n\right) }=u_n^*(\delta ), \ w_n|_{\partial B^g_{r_nR}(x_n)}=u_n^*\left( r_nR\right) . \end{aligned}$$

Then we have

$$\begin{aligned} \begin{aligned} \frac{1}{2}\int _{B^g_\delta (x_n){\setminus } B^g_{r_nR}(x_n)}|du_n|^2d\mu _g&\ge \frac{1}{2}\int _{B^g_\delta (x_n){\setminus } B^g_{r_nR}(x_n)}|du_n^*|^2d\mu _g\\&\ge \frac{1}{2}\int _{B^g_\delta (x_n){\setminus } B^g_{r_nR}(x_n)}|dw_n|^2d\mu _g\\&\ge \frac{\pi (u_n^*(\delta )-u_n^*(r_nR))^2}{\ln \delta -\ln (r_nR)}. \end{aligned} \end{aligned}$$

Set $\tau _n:=u_n^*(\delta )-u_n^*(r_nR)-\bar{u}_n-2\ln r_n$. It follows from (4.3) that

$$\begin{aligned} \lim _{n\rightarrow \infty }\tau _n=-4\ln \delta +A\left( x_0\right) +2\ln \left( aR^2\right) +o_R(1)+o_\delta (1). \end{aligned}$$

Then we get

$$\begin{aligned} \begin{aligned} \frac{1}{2}\int _{B^g_\delta \left( x_n\right) {\setminus } B^g_{r_nR}\left( x_n\right) }|du_n|^2d\mu _g&\ge \frac{\pi \left( \tau _n+\bar{u}_n+2\ln r_n\right) ^2}{-\ln r_n}\left( 1-\frac{\ln \left( R/\delta \right) }{-\ln r_n}\right) ^{-1}\\&\ge \frac{\pi \left( \tau _n+\bar{u}_n-2\ln r_n\right) ^2}{-\ln r_n}-8\pi \bar{u}_n+32\pi \ln \delta -8\pi A(x_0)\\&\quad -16\ln (aR^2)\\&\quad +\pi \left( 2+\frac{\tau _n}{\ln r_n}+\frac{\bar{u}_n}{\ln r_n}\right) ^2\ln (R/\delta )+o_R(1)+o_\delta (1) \end{aligned} \end{aligned}$$

(4.4)

for large n.

Thus, (4.1), (4.2) and (4.4) give us

$$\begin{aligned} J(u_n)&\ge 8\pi \ln \left( aR^2\right) -8\pi -16\pi \ln \delta +4\pi A(x_0)+8\pi \left( \bar{u}_n-\ln h(x_0)-\ln \int _\Sigma e^{u_0}\mathop {}\mathrm {d}\mu _{g}\right) \\&\quad +\frac{\pi \left( \tau _n+\bar{u}_n-2\ln r_n\right) ^2}{-\ln r_n}-8\pi \bar{u}_n+32\pi \ln \delta -8\pi A\left( x_0\right) -16\ln \left( aR^2\right) \\&\quad +\pi \left( 2+\frac{\tau _n}{\ln r_n}+\frac{\bar{u}_n}{\ln r_n}\right) ^2\ln (R/\delta )+o_R(1)+o_\delta (1)+o_n(1)\\&=-4\pi (A(x_0)+2\ln h(x_0))-8\pi \ln \pi -8\pi +o_R(1)+o_\delta (1)+o_n(1)\\&\quad +\frac{\pi \left( \tau _n+\bar{u}_n-2\ln r_n\right) ^2}{-\ln r_n}+\pi \left[ \left( 2+\frac{\tau _n}{\ln r_n}+\frac{\bar{u}_n}{\ln r_n}\right) ^2-16\right] \ln (R/\delta ). \end{aligned}$$

Since $J(u_n)\le J(u_0)$, we have $\lim _{n\rightarrow \infty }\bar{u}_n\rightarrow 2\ln r_n$. Hence,

$$\begin{aligned} \lim _{n\rightarrow \infty }J(u_n)\ge -4\pi \max _{x\in \Sigma }(A(x)+2\ln h(x))-8\pi \ln \pi -8\pi . \end{aligned}$$

By the monotonicity formula (2.2), we conclude that

$$\begin{aligned} J(u(t))\ge C_0=-4\pi \max _{x\in \Sigma }(A(x)+2\ln h(x))-8\pi \ln \pi -8\pi ,\quad \forall t\ge 0. \end{aligned}$$

$\square$

5 Global convergence

Proof of Theorem 1.3

Notice that there is a sequence of positive numbers $\left\{ t_n\right\}$ such that $n\le t_n\le n+1$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{\Sigma }e^{u(t_n)}\left|\dfrac{\partial u(t_n)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}=0. \end{aligned}$$

By the lower bound of J along the flow stated in Theorem 1.2, the existence of mean field Eq. (1.1) is reduced to construct a function whose value under J is strictly less than $C_0$. In fact, such kind of functions were constructed in [12] provided that

$$\begin{aligned} \begin{aligned}&\Delta _{g} h\left( p_0\right) +2\left( b_1\left( p_0\right) k_1\left( p_0\right) +b_2\left( p_0\right) k_2\left( p_0\right) \right) \\&\quad >-\left( 8\pi +b_1^2(p_0)+b_2^2\left( p_0\right) -2K\left( p_0\right) \right) h\left( p_0\right) , \end{aligned} \end{aligned}$$

where K(x) is the Gaussian curvature of $\Sigma$, $\nabla h(p_0)=(k_1(p_0),k_2(p_0))$ in the normal coordinate system, $p_0$ is the maximum point of $A(q)+2\ln h(q)$ and $b_1(p_0)$, $b_2(p_0)$ are the constants in the following expression of Green function G:

$$\begin{aligned} G\left( x,p_0\right)&=-4\ln r+A\left( p_0\right) +b_1\left( p_0\right) x_1+b_2\left( p_0\right) x_2\nonumber \\&\quad +c_1x_1^2+2c_2x_1x_2+c_3x_2^2+O\left( r^3\right) , \end{aligned}$$

where $r(x)=\mathrm {dist}_g(x,p_0)$. The sequence $\left\{ u_n\right\}$ can not blowup by our assumption. By Lemma 3.1, $\left\{ u_n\right\}$ is bounded in $H^2(\Sigma )$ and there is a function $u_\infty \in H^2(\Sigma )$ and a subsequence $\left\{ u_{n_k}\right\}$ of $\left\{ u_{n}\right\}$ such that

$$\begin{aligned} u_{n_k}\rightarrow u_\infty \quad \text {weakly in} \quad H^2(\Sigma ) \end{aligned}$$

and

$$\begin{aligned} u_{n_k}\rightarrow u_\infty \quad \text {in} \quad C^\alpha (\Sigma ) \end{aligned}$$

for $\alpha \in (0,1)$ as $n_k\rightarrow \infty$. It is easy to see that $u_\infty$ is a smooth solution to

$$\begin{aligned} -\Delta _{g} u_{\infty }+8\pi =8\pi \frac{he^{u_{\infty }}}{\int _\Sigma he^{u_{\infty }}\mathop {}\mathrm {d}\mu _{g}}. \end{aligned}$$

To obtain the strong convergence for $\left\{ u_{n_k}\right\}$, please notice that

$$\begin{aligned} \begin{aligned}&\int _{\Sigma }\left( \Delta _{g}u_{n_k}-\Delta _{g}u_{\infty }\right) ^2\\&\quad =\int _{\Sigma }\left( \frac{\partial {e^{u_{n_k}}}}{\partial {t}}+8\pi \left( \dfrac{he^{u_\infty }}{\int _{\Sigma }he^{u_\infty }\mathop {}\mathrm {d}\mu _{g}}-\dfrac{he^{u_{n_k}}}{\int _{\Sigma }he^{u_{n_k}}\mathop {}\mathrm {d}\mu _{g}}\right) \right) ^2\mathop {}\mathrm {d}\mu _{g}\\&\quad \le C\int _{\Sigma }\left( e^{u_\infty }-e^{u_{n_k}}\right) ^2\mathop {}\mathrm {d}\mu _{g}+C\int _{\Sigma }\bigg |\frac{\partial {u_{n_k}}}{\partial {t}}\bigg |^2e^{u_{n_k}}\mathop {}\mathrm {d}\mu _{g}\rightarrow 0 \end{aligned} \end{aligned}$$

as $n_k\rightarrow +\infty$.

We use Łojasiewicz-Simon gradient inequality to get the global convergence of the flow. When $h>0$, one can refer to [4] for non-critical cases, i.e. $\rho \ne 8k\pi$ and [23] for $\rho =8\pi$. In both papers, the authors just provided the paper by Simon [29] and no more details were given. In this section, we give a detailed proof and some references. We divide the proof of the global convergence to several steps.

• Step 1  

  $\left\Vert u(t)^+\right\Vert _{L^{\infty }\left( \Sigma \right) }\le C.$

  Since

  $$\begin{aligned} \dfrac{\partial u}{\partial t}\le e^{-u}\Delta _{g}u+C. \end{aligned}$$

  Applying the maximum principle, we have $\frac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\left( \max _{\Sigma }u(t)-Ct\right) \le 0$. By using the fact $\left\{ u_{n}\right\}$ is bounded in $L^{\infty }\left( \Sigma \right)$ and $n\le t_n\le n+1$, we conclude that $u(t)^{+}$ is bounded in $L^{\infty }\left( \Sigma \right)$.

• Step 2  

  $\left\Vert u(t)\right\Vert _{H^1\left( \Sigma \right) }\le C$.

  Denote

  $$\begin{aligned} A(t)=\left\{ x\in \Sigma : e^{u(t)}\ge \dfrac{1}{2}\int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}\right\} . \end{aligned}$$

  Then

  $$\begin{aligned} \int _{A(t)}u(t)\mathop {}\mathrm {d}\mu _{g}\ge \ln \left( \dfrac{1}{2}\int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}\right) \left|A(t)\right|_{g}\ge -C, \end{aligned}$$

  and

  $$\begin{aligned} \int _{A(t)}u(t)\mathop {}\mathrm {d}\mu _{g}\le \int _{A(t)}e^{u(t)}\le \int _{\Sigma }e^{u(t)}=\int _{\Sigma }e^{u_0}\le C. \end{aligned}$$

  Thus

  $$\begin{aligned} \left|\int _{A(t)}u(t)\mathop {}\mathrm {d}\mu _{g}\right|\le C. \end{aligned}$$

  Notice that

  $$\begin{aligned} \int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}=&\int _{\Sigma }e^{u(t)}\mathop {}\mathrm {d}\mu _{g}\\ =&\int _{\Sigma {\setminus } A(t)}e^{u(t)}\mathop {}\mathrm {d}\mu _{g}+\int _{A(t)}e^{u(t)}\mathop {}\mathrm {d}\mu _{g}\\ \le&\dfrac{1}{2}\int _{\Sigma }e^{u_0}\mathop {}\mathrm {d}\mu _{g}+C\left|A(t)\right|_{g}\\ =&\dfrac{1}{2}\int _{\Sigma }e^{u(t)}\mathop {}\mathrm {d}\mu _{g}+C\left|A(t)\right|_{g}, \end{aligned}$$

  we get

  $$\begin{aligned} \left|A(t)\right|_{g}\ge C^{-1}. \end{aligned}$$

  By Poincaré inequality,

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }\le&C\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }+\left|\bar{u}(t)\right|\\ \le&C\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }+\left|\int _{A(t)}u(t)\mathop {}\mathrm {d}\mu _{g}\right|+\left|\int _{\Sigma {\setminus } A(t)}u(t)\mathop {}\mathrm {d}\mu _{g}\right|\\ \le&C\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }+C+\sqrt{\left|\Sigma {\setminus } A(t)\right|_{g}}\left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }. \end{aligned}$$

  Hence

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{L^2\left( \Sigma \right) }\le&C\left\Vert \nabla _{g}u(t)\right\Vert _{L^2\left( \Sigma \right) }+C. \end{aligned}$$

  Notice that

  $$\begin{aligned} \dfrac{1}{2}\int _{\Sigma }\left|\nabla _{g}u(t)\right|_{g}^2\mathop {}\mathrm {d}\mu _{g}= J(u(t))-8\pi \bar{u}(t)+8\pi \ln \int _{\Sigma }he^{u(t)}\mathop {}\mathrm {d}\mu _{g}\le C+C\bar{u}(t). \end{aligned}$$

  By Young’s inequality, we conclude that

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{H^1\left( \Sigma \right) }\le C. \end{aligned}$$

• Step 3  

  $\lim _{t\rightarrow \infty }\int _{\Sigma }e^{u(t)}\left|\frac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}=0$.

  We will follow the argument of Brendle [1] (see also [4]). For every $\varepsilon >0$, there exist $k_0$ such that for all $k\ge k_0$

  $$\begin{aligned} \int _{\Sigma }e^{u\left( t_{n_k}\right) }\left|\dfrac{\partial u\left( t_{n_k}\right) }{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}<\varepsilon . \end{aligned}$$

  Assume for all $k\ge k_0$,

  $$\begin{aligned} m_k=\inf \left\{ t>t_{n_{k}}: \int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}\ge 2\varepsilon \right\} <\infty . \end{aligned}$$

  For $t_{n_k}\le t\le m_k$, we have

  $$\begin{aligned} \int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}\le 2\varepsilon . \end{aligned}$$

  Since u(t) is bounded in $H^1\left( \Sigma \right)$ and $u(t)^{+}$ is bounded in $L^{\infty }\left( \Sigma \right)$, we conclude that

  $$\begin{aligned} \left|\Delta u(t)\right|\le C\varepsilon +C,\quad \forall t_{n_k}\le t\le t_{m_k}. \end{aligned}$$

  Thus

  $$\begin{aligned} \left\Vert u(t)\right\Vert _{L^{\infty }\left( \Sigma \right) }\le C\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }\le C_{\varepsilon },\quad \forall t_{n_k}\le t\le m_k. \end{aligned}$$

  Set

  $$\begin{aligned} y(t)=\int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

  Denote by $\dot{u} =\frac{\partial u}{\partial t},\ \ddot{u}=\frac{\partial ^2u}{\partial t^2}$. Notice that

  $$\begin{aligned} \dot{u}=e^{-u}\left( \Delta _{g}u-8\pi \right) +\dfrac{8\pi h}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}. \end{aligned}$$

  We get

  $$\begin{aligned} \ddot{u}=&e^{-u}\Delta _{g}\dot{u}-\dot{u}e^{-u}\left( \Delta _{g}u-8\pi \right) -\dfrac{8\pi h\int _{\Sigma }he^{u}\dot{u}\mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}\right) ^2}\\ =&e^{-u}\Delta _{g}\dot{u}-\dot{u}^2+\dfrac{8\pi h\dot{u}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-\dfrac{8\pi h\int _{\Sigma }he^{u}\dot{u}\mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}\right) ^2}. \end{aligned}$$

  Hence

  $$\begin{aligned} \dot{y}=&\int _{\Sigma }\left( e^{u}\dot{u}^3+2e^{u}\dot{u}\ddot{u}\right) \mathop {}\mathrm {d}\mu _{g}\\ =&-2\int _{\Sigma }\left|\nabla _{g}\dot{u}\right|^2\mathop {}\mathrm {d}\mu _{g}-\int _{\Sigma }e^{u}\dot{u}^3\mathop {}\mathrm {d}\mu _{g}+16\pi \left[ \dfrac{\int _{\Sigma } he^{u}\dot{u}^2\mathop {}\mathrm {d}\mu _{g}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-\left( \dfrac{\int _{\Sigma }he^{u}\dot{u}\mathop {}\mathrm {d}\mu _{g}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}\right) ^2\right] \\ \le&-2\int _{\Sigma }\left|\nabla _{g}\dot{u}\right|^2\mathop {}\mathrm {d}\mu _{g}-\int _{\Sigma }e^{u}\dot{u}^3\mathop {}\mathrm {d}\mu _{g}+Cy. \end{aligned}$$

  We estimate the second term in the RHS of the above inequality as follows: for all $t_{n_k}\le t\le m_k$,

  $$\begin{aligned} -\int _{\Sigma }e^{u}\dot{u}^3\mathop {}\mathrm {d}\mu _{g}\le&C\int _{\Sigma }\left|\dot{u}\right|^3\mathop {}\mathrm {d}\mu _{g}\\ \le&C\left\Vert \dot{u}\right\Vert _{L^2\left( \Sigma \right) }^2\left\Vert \dot{u}\right\Vert _{H^1\left( \Sigma \right) }\\ \le&C_{\varepsilon }y\left( y+\int _{\Sigma }\left|\nabla _{g}\dot{u}\right|^2_{g}\mathop {}\mathrm {d}\mu _{g}\right) ^{1/2}. \end{aligned}$$

  Since $\int _{\Sigma }e^{u}\dot{u}\mathop {}\mathrm {d}\mu _{g}=0$, applying the Poincaré inequality to obtain

  $$\begin{aligned}&\int _{\Sigma }e^{u}\dot{u}^2\mathop {}\mathrm {d}\mu _{g}\le \dfrac{1}{\lambda _{1,e^{u}g}}\int _{\Sigma }\left|\nabla _{e^{u}g}\dot{u}\right|^2_{e^{u}g}\mathop {}\mathrm {d}\mu _{e^{u}g}=\dfrac{1}{\lambda _{1,e^{u}g}}\int _{\Sigma }\left|\nabla _{g}\dot{u}\right|^2_{g}\mathop {}\mathrm {d}\mu _{g}\\&\quad \le C\int _{\Sigma }\left|\nabla _{g}\dot{u}\right|^2_{g}\mathop {}\mathrm {d}\mu _{g}. \end{aligned}$$

  Thus for all $t_{n_k}\le t\le m_k$,

  $$\begin{aligned} -\int _{\Sigma }e^{u}\dot{u}^3\mathop {}\mathrm {d}\mu _{g}\le C_{\varepsilon }y^{1/2}\left( \int _{\Sigma }\left|\nabla _{g}\dot{u}\right|^2_{g}\mathop {}\mathrm {d}\mu _{g}\right) ^{1/2}, \end{aligned}$$

  which implies

  $$\begin{aligned} \dot{y}\le C_{\varepsilon }y. \end{aligned}$$

  Hence

  $$\begin{aligned} y\left( t_{m_k}\right) \le y\left( t_{n_k}\right) +C_{\varepsilon }\int _{t_{n_k}}^{\infty }y(t)\mathop {}\mathrm {d}t. \end{aligned}$$

  Thus

  $$\begin{aligned} \varepsilon \le C_{\varepsilon }\int _{t_{n_k}}^{\infty }y(t)\mathop {}\mathrm {d}t\rightarrow 0,\quad \text {as}\quad t_{n_k}\rightarrow \infty \end{aligned}$$

  which is a contradiction. Therefore

  $$\begin{aligned} \lim _{t\rightarrow \infty }\int _{\Sigma }e^{u(t)}\left|\dfrac{\partial u(t)}{\partial t}\right|^2\mathop {}\mathrm {d}\mu _{g}=0. \end{aligned}$$

• Step 4  

  $\left\Vert u(t)\right\Vert _{H^2\left( \Sigma \right) }\le C$ which implies that $\left\Vert u(t)\right\Vert _{C^{\gamma }\left( \Sigma \right) }\le C_{\gamma }$ for every $0<\gamma <1$.

  This is a direct consequence of the standard elliptic estimates and Sobolev inequalities.

• Step 5  

  $\lim _{t\rightarrow \infty }\left\Vert u(t)-u_\infty \right\Vert _{L^2\left( \Sigma \right) }=0$ implies $\lim _{t\rightarrow \infty }\left\Vert u(t)-u_{\infty }\right\Vert _{H^2\left( \Sigma \right) }=0$.

  Since

  $$\begin{aligned} \dfrac{\partial e^{u}}{\partial t}=\Delta _{g}\left( u-u_{\infty }\right) +8\pi \left( \dfrac{he^{u}}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-\dfrac{he^{u_{\infty }}}{\int _{\Sigma }he^{u_{\infty }}\mathop {}\mathrm {d}\mu _{g}}\right) , \end{aligned}$$

  we get

  $$\begin{aligned} \left|\Delta _{g}\left( u(t)-u_{\infty }\right) \right|\le C\left( \left|\dfrac{\partial u(t)}{\partial t}\right|+\left|u(t)-u_{\infty }\right|+\left\Vert u(t)-u_{\infty }\right\Vert _{L^1\left( \Sigma \right) }\right) \end{aligned}$$

  which implies

  $$\begin{aligned} \left\Vert u(t)-u_{\infty }\right\Vert _{H^2\left( \Sigma \right) }\le C\left( \left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }+\left\Vert u(t)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }\right) . \end{aligned}$$

  The claim follows by letting $t\rightarrow +\infty$.

• Step 6  

  There are positive constants $\sigma$ and $\theta \in (1/2,1)$ such that

  $$\begin{aligned} \forall u\in H^2\left( \Sigma \right) ,\ \left\Vert u-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }<\sigma \quad \Longrightarrow \quad \left|J(u)-J(u_{\infty })\right|^{\theta }\le \left\Vert \mathcal {M}(u)\right\Vert _{L^{2}\left( \Sigma \right) }. \end{aligned}$$

  Notice that the functional $J:H^1\left( \Sigma \right) \longrightarrow \mathbb {R}$ is analytic and the gradient map $\mathcal {M}:H^1\left( \Sigma \right) \longrightarrow H^{-1}\left( \Sigma \right)$ is given by

  $$\begin{aligned} u\mapsto \mathcal {M}(u)=-\Delta _{g}u-8\pi \left( \dfrac{he^u}{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-1\right) . \end{aligned}$$

  The Jacobi operator $\mathcal {L}:H^1\left( \Sigma \right) \longrightarrow H^{-1}\left( \Sigma \right)$ of J at a critical point $u\in C^{\infty }\left( \Sigma \right)$ of J is given by

  $$\begin{aligned} \xi \mapsto \mathcal {L}(\xi )=-\Delta _{g}\xi -8\pi \left( \dfrac{he^{u}\xi }{\int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}}-\dfrac{he^{u}\int _{\Sigma }he^{u}\xi \mathop {}\mathrm {d}\mu _{g}}{\left( \int _{\Sigma }he^{u}\mathop {}\mathrm {d}\mu _{g}\right) ^2}\right) \end{aligned}$$

  is a Fredohom operator with index zero. Since $\mathcal {M}\left( H^2\left( \Sigma \right) \right) \subset L^2\left( \Sigma \right)$, applying the Łojasiewicz-Simon gradient inequality (cf. [21, Proposition 1.3] or [17, Theorem 2]), there are positive constants $\tilde{\sigma }$ and $\theta \in (1/2,1)$ such that

  $$\begin{aligned} \forall u\in H^2\left( \Sigma \right) ,\ \left\Vert u-u_{\infty }\right\Vert _{H^2\left( \Sigma \right) }<\tilde{\sigma }\quad \Longrightarrow \quad \left|J(u)-J(u_{\infty })\right|^{\theta }\le \left\Vert \mathcal {M}(u)\right\Vert _{L^{2}\left( \Sigma \right) }. \end{aligned}$$

  Hence we obtain this claim by choosing $\sigma$ small.

• Step 7  

  $\lim _{t\rightarrow \infty }\left\Vert u(t)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }=0$ which gives the global convergence.

  We will follow the approach of Jendoubi [21]. For every $0<\varepsilon<<\sigma$, there exist $k_1$ such that for all $k\ge k_1$,

  $$\begin{aligned} \left\Vert u\left( t_{k}\right) -u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }<\varepsilon . \end{aligned}$$

  Assume for all $k\ge k_1$,

  $$\begin{aligned} s_k=\inf \left\{ t>t_{n_k}: \left\Vert u(t)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }\ge \sigma \right\} <\infty . \end{aligned}$$

  Then for all $n_k\le t<s_k$,

  $$\begin{aligned} \left\Vert u(t)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }<\sigma =\left\Vert u(s_k)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }. \end{aligned}$$

  Without loss of generality, assume $J(u(t))>J(u_{\infty })$ for all $t>0$. For $t_{n_k}\le t<s_k$, we have

  $$\begin{aligned} -\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\left( J(u(t))-J(u_{\infty })\right) ^{1-\theta }=&-\left( 1-\theta \right) \left( J(u(t))-J(u_{\infty })\right) ^{-\theta }\dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}J(u(t))\\ =&\left( 1-\theta \right) \left( J(u(t))-J(u_{\infty })\right) ^{-\theta }\left\Vert e^{u(t)/2}\dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }^2\\ \ge&\dfrac{1-\theta }{C}\left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }. \end{aligned}$$

  Thus

  $$\begin{aligned} \int _{t_{n_k}}^{s_k}\left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }\mathop {}\mathrm {d}t\le \dfrac{C}{1-\theta }\left( J(u(t_{n_k}))-J(u_{\infty })\right) ^{1-\theta }. \end{aligned}$$

  Since

  $$\begin{aligned} \dfrac{\mathop {}\mathrm {d}}{\mathop {}\mathrm {d}t}\left\Vert u(t)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }\le \left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }, \end{aligned}$$

  we get

  $$\begin{aligned} \sigma =&\left\Vert u(s_k)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }\\ \le&\left\Vert u(t_{n_k})-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }+\int _{t_{n_k}}^{s_k}\left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }\mathop {}\mathrm {d}t\\ \le&\left\Vert u(t_{n_k})-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }+\dfrac{C}{1-\theta }\left( J(u(t_{n_k}))-J(u_{\infty })\right) ^{1-\theta } \end{aligned}$$

  which is a contradiction when $n_{k}\rightarrow +\infty$. Hence we have $s_{k_2}=+\infty$ for some $k_2$. We conclude that

  $$\begin{aligned} \int _{0}^{\infty }\left\Vert \dfrac{\partial u(t)}{\partial t}\right\Vert _{L^2\left( \Sigma \right) }\mathop {}\mathrm {d}t<+\infty \end{aligned}$$

  which gives

  $$\begin{aligned} \lim _{t\rightarrow \infty }\left\Vert u(t)-u_{\infty }\right\Vert _{L^2\left( \Sigma \right) }=0. \end{aligned}$$

  $\square$