1 Introduction

Let \((M,{\bar{g}})\) be a two-dimensional, smooth, closed, connected, oriented Riemann manifold endowed with a smooth background metric \({\bar{g}}\). A classical problem raised by Kazdan and Warner in [11] and [10] is the question which smooth functions \(f :M \rightarrow \mathbb {R}\) arise as the Gauss curvature \(K_g\) of a conformal metric \(g(x)=\textrm{e}^{2u(x)}{\bar{g}}(x)\) on M and to characterise the set of all such metrics.

For a constant function f, this prescribed Gauss curvature problem is exactly the statement of the Uniformisation Theorem (see e.g. [12, 16]):

There exists a metric g which is pointwise conformal to \({\bar{g}}\) and has constant Gauss curvature \(K_{g}\equiv {\bar{K}}\in \mathbb {R}\).

We now use this statement to assume in the following without loss of generality that the background metric \({\bar{g}}\) itself has constant Gauss curvature \(K_{{\bar{g}}}\equiv {\bar{K}}\in \mathbb {R}\). Furthermore we can normalise the volume of \((M,{\bar{g}})\) to one. We recall that the Gauss curvature of a conformal metric \(g(x)=\textrm{e}^{2u(x)}{\bar{g}}(x)\) on M is given by the Gauss equation

$$\begin{aligned} K_{g}(x) = \textrm{e}^{-2u(x)}(-\Delta _{{\bar{g}}}u(x) + {\bar{K}}). \end{aligned}$$
(1.1)

Therefore the problem reduces to the question for which functions f there exists a conformal factor u solving the equation

$$\begin{aligned} -\Delta _{{\bar{g}}}u(x) + {\bar{K}} = f(x) \textrm{e}^{2u(x)} \qquad \text {in }M. \end{aligned}$$
(1.2)

Given a solution u, we may integrate (1.2) with respect to the measure \(\mu _{{\bar{g}}}\) on M induced by the Riemannian volume form. Using the Gauss–Bonnet Theorem, we then obtain the identity

$$\begin{aligned} \int _M f(x) d\mu _g(x) = \int _M {\bar{K}} d\mu _{{\bar{g}}}(x) = {\bar{K}} {\text {vol}}_{{\bar{g}}}={\bar{K}}= 2\pi \chi (M), \end{aligned}$$
(1.3)

where \(d\mu _g(x) = \textrm{e}^{2u(x)}d\mu _{{\bar{g}}}(x)\) is the element of area in the metric \(g(x)=\textrm{e}^{2u(x)}{\bar{g}}(x)\). We note that (1.3) immediately yields necessary conditions on f for the solvability of the prescribed Gauss curvature problem. In particular, if \(\pm \chi (M)>0\), then \(\pm f\) must be positive somewhere. Moreover, if \(\chi (M)=0\), then f must change sign or must be identically zero.

In the present paper we focus on the case \(\chi (M)<0\), so M is a surface of genus greater than one and \({\bar{K}} < 0\). The complementary cases \(\chi (M) \ge 0\)—i.e., the cases where \(M = S^2\) or \(M=T\), the 2-torus—will be discussed briefly at the end of this introduction, and we also refer the reader to [2, 8, 18, 19] and the references therein. Multiplying Eq. (1.2) with the factor \(\textrm{e}^{-2u}\) and integrating over M with respect to the measure \(\mu _{{\bar{g}}}\), we get the following necessary condition—already mentioned by Kazdan and Warner in [11]—for the average \({\bar{f}}:=\frac{1}{{\text {vol}}_{{\bar{g}}}}\int _M f(x)d\mu _{{\bar{g}}}(x)\), with \({\text {vol}}_{{\bar{g}}}:=\int _Md\mu _{{\bar{g}}}(x)\):

$$\begin{aligned} \begin{aligned} {\bar{f}}&=\frac{1}{{\text {vol}}_{{\bar{g}}}}\int _M f(x)d\mu _{{\bar{g}}}(x)=\int _M(-\Delta _{{\bar{g}}}u(x)+{\bar{K}})\textrm{e}^{-2u(x)}d\mu _{{\bar{g}}}(x)\\&=\int _M(-2|\nabla _{{\bar{g}}}u(x)|^2_{{\bar{g}}}+{\bar{K}})\textrm{e}^{-2u(x)}d\mu _{{\bar{g}}}(x)<0. \end{aligned} \end{aligned}$$
(1.4)

This condition is not sufficient. Indeed, it has already been pointed out in [11, Theorem 10.5] that in the case \(\chi (M)<0\) there always exist functions \(f \in C^\infty (M)\) with \({\bar{f}}< 0\) and the property that (1.2) has no solution.

We recall that solutions of (1.2) can be characterised as critical points of the functional

$$\begin{aligned}{} & {} E_f:H^1(M,{\bar{g}})\rightarrow \mathbb {R};\quad \nonumber \\{} & {} E_f(u) := \frac{1}{2}\int _M \left( |\nabla _{{\bar{g}}} u(x)|^2_{{\bar{g}}} + 2{\bar{K}} u(x) - f(x)\textrm{e}^{2u(x)}\right) d\mu _{{\bar{g}}}(x). \end{aligned}$$
(1.5)

Under the assumption \(\chi (M)<0\), i.e., \({\bar{K}} < 0\), the functional \(E_f\) is strictly convex and coercive on \(H^1(M,{\bar{g}})\) if \(f\le 0\) and f does not vanish identically. Hence, as noted in [7], the functional \(E_f\) admits a unique critical point \(u_f \in H^1(M,{\bar{g}})\) in this case, which is a strict absolute minimiser of \(E_f\) and a (weak) solution of (1.2). The situation is more delicate in the case where \(f_\lambda =f_0+\lambda \), where \(f_0\le 0\) is a smooth, nonconstant function on M with \(\max _{x\in M}f_0(x)=0\), and \(\lambda >0\). In the case where \(\lambda >0\) sufficiently small (depending on \(f_0\)), it was shown in [7] and [1] that the corresponding functional \(E_{f_\lambda }\) admits a local minimiser \(u_\lambda \) and a further critical point \(u^\lambda \ne u_\lambda \) of mountain pass type.

These results motivate our present work, where we suggest a new flow approach to the prescribed Gausss curvature problem in the case \(\chi (M)<0\). It is important to note here that there is an intrinsic motivation to formulate the static problem in a flow context. Typically, elliptic theories are regarded as the static case of the corresponding parabolic problem; in that sense, many times the better-understood elliptic theory has been a source of intuition to generalise the corresponding results in the parabolic case. Examples of this feedback are minimal surfaces/mean curvature flow, harmonic maps/solutions of the heat equation, and the Uniformisation Theorem/the two-dimensional normalised Ricci flow.

In this spirit, a flow approach to (1.2), the so-called prescribed Gauss curvature flow, was first introduced by Struwe in [19] (and [2]) for the case \(M=S^2\) with the standard background metric and a positive function \(f \in C^2(M)\). More precisely, he considers a family of metrics \((g(t,\cdot ))_{t\ge 0}\) which fulfils the initial value problem

$$\begin{aligned} \partial _tg(t,x)&=2(\alpha (t)f(x)-K_{g(t,\cdot )}(x))g(t,x)\quad \text {in }(0,T)\times M; \end{aligned}$$
(1.6)
$$\begin{aligned} g(0,x)&= g_0(x)\quad \text {on }\{0\}\times M, \end{aligned}$$
(1.7)

with

$$\begin{aligned} \alpha (t)=\frac{\int _MK_{g(t,\cdot )}(x)d\mu _{g(t,\cdot )}(x)}{\int _M f(x)d\mu _{g(t,\cdot )}(x)}=\frac{2\pi \chi (M)}{\int _Mf(x)d\mu _{g(t,\cdot )}(x)}. \end{aligned}$$
(1.8)

This choice of \(\alpha (t)\) ensures that the volume of \((M,g(t,\cdot ))\) remains constant throughout the deformation, i.e.,

$$\begin{aligned} \int _M d\mu _{g(t,\cdot )}(x)=\int _M\textrm{e}^{2u(t,x)}d\mu _{{\bar{g}}}(x)\equiv {\text {vol}}_{g_0}\quad \text {for all }t\ge 0, \end{aligned}$$

where \(g_0\) denotes the initial metric on M. Equivalently one may consider the evolution equation for the associated conformal factor u given by \(g(t,x)=\textrm{e}^{2u(t,x)}{\bar{g}}(x)\):

$$\begin{aligned} \partial _tu(t,x)&=\alpha (t) f(x)-K_{g(t,\cdot )}(x)\quad \text {in }(0,T)\times M; \end{aligned}$$
(1.9)
$$\begin{aligned} u(0,x)&=u_0(x)\quad \text {on }\{0\}\times M. \end{aligned}$$
(1.10)

Here the initial value \(u_0\) is given by \(g_0(x)=\textrm{e}^{2u_0(x)}{\bar{g}}(x)\). The flow associated to this parabolic equation is usually called the prescribed Gauss curvature flow. With the help of this flow, Struwe [19] provided a new proof of a result by Chang and Yang [6] on sufficient criteria for a function f to be the Gauss curvature of a metric \(g(x)=\textrm{e}^{2u(x)}g_{S^2}(x)\) on \(S^2\). He also proved the sharpness of these criteria.

In the case of surfaces with genus greater than one, i.e., with negative Euler characteristic, the prescribed Gauss curvature flow was used by Ho in [9] to prove that any smooth, strictly negative function on a surface with negative Euler characteristic can be realised as the Gaussian curvature of some metric. More precisely, assuming that \(\chi (M)< 0\) and that \(f \in C^\infty (M)\) is a strictly negative function, he proves that Eq. (1.9) has a solution which is defined for all times and converges to a metric \(g_\infty \) with Gaussian curvature \(K_{g_\infty }\) satisfying

$$\begin{aligned} K_{g_\infty }(x)=\alpha _\infty f(x) \end{aligned}$$

for some constant \(\alpha _\infty \).

While the prescribed Gauss curvature flow is a higly useful tool in the cases where f is of fixed sign, it cannot be used in the case where f is sign-changing. Indeed, in this case we may have \(\int _Mf(x)d\mu _{g(t,\cdot )}(x)=0\) along the flow and then the normalising factor \(\alpha (t)\) is not well-defined by (1.8). As a consequence, a long-time solution of (1.9) might not exist. In particular, the static existence results of [7] and [1] can not be recovered and reinterpreted with the standard prescribed Gauss curvature flow.

In this paper we develop a new flow approach to (1.2) in the case \(\chi (M)<0\) for general \(f \in C^\infty (M)\), which sheds new light on the results in [1, 7] and [9]. The main idea is to replace the multiplicative normalisation in (1.9) by an additive normalisation, as will be described in details in the next chapter.

At this point, it should be noted that the normalisation factor \(\alpha (t)\) in the prescribed Gauss curvature flow given by (1.8) is also not the appropriate choice in the case of the torus, where, as noted before, f has to change sign or be identically zero in order to arise as the Gauss curvature of a conformal metric. The case of the torus was considered by Struwe in [18], where, in particular, he used to a flow approach to reprove and partially improve a result by Galimberti [8] on the static problem. In this approach, the normalisation in (1.8) is replaced by

$$\begin{aligned} \alpha (t)=\frac{\int _M f(x)K_{g(t,\cdot )}(x)d\mu _{g(t,\cdot )}(x)}{\int _Mf^2(x)d\mu _{g(t,\cdot )}(x)}. \end{aligned}$$
(1.11)

With this choice, Struwe shows that for any smooth

$$\begin{aligned} u_0\in C^*:=\left\{ u\in H^1(M,{\bar{g}})\mid \int _Mf(x)\textrm{e}^{2u(x)}d\mu _{{\bar{g}}}(x)=0,\,\int _M\textrm{e}^{2u(x)}d\mu _{{\bar{g}}}(x)=1\right\} \end{aligned}$$

there exists a unique, global smooth solution u of (1.9) satisfying \(u(t,\cdot )\in C^*\) for all \(t>0\). Moreover, \(u(t,\cdot )\rightarrow u_\infty (\cdot )\) in \(H^2(M,{\bar{g}})\) (and smoothly) as \(t\rightarrow \infty \) suitably, where \(u_\infty +c_\infty \) is a smooth solution of (1.2) for some \(c_\infty \in \mathbb {R}\).

In principle, the normalisation (1.11) could also be considered in the case \(\chi (M)<0\), but then the flow is not volume-preserving anymore, which results in a failure of uniform estimates for solutions of (1.9). Consequently, we were not able to make use of the associated flow in this case.

The paper is organised as follows. In Sect. 2 we set up the framework for the new variant of the prescribed Gauss curvature flow with additive normalisation, and we collect basic properties of it. In Sect. 3, we then present our main result on the long-time existence and convergence of the flow (for suitable times \(t_k \rightarrow \infty \)) to solutions of the corresponding static problem. In particular, our results show how sign changing functions of the form \(f_\lambda = f_0 + \lambda \) arise depending on various assumptions on the shape of \(f_0\) and on the fixed volume A of M with respect to the metric g(t). Before proving our results on the time-dependent problem, we first derive, in Sect. 4, some results on the static problem with volume constraint. Most of these results will then be used in Sect. 5, where the parabolic problem is studied in detail and the main results of the paper are proved. In the appendix, we provide some regularity estimates and a variant of a maximum principle for a class of linear evolution problems with Hölder continuous coefficients.

In the remainder of the paper, we will use the short form f, g(t), u(t), \(K_{g(t)}\), \({\text {vol}}_{g(t)}:=\int _M d\mu _{g(t)}=\int _M\textrm{e}^{2u(t)}d\mu _{{\bar{g}}}\), and so on instead of f(x), g(tx), u(tx), \(K_{g(t,\cdot )}(x)\), \(\int _M d\mu _{g(t,\cdot )}(x)=\int _M\textrm{e}^{2u(t,x)}d\mu _{{\bar{g}}}(x)\), et cetera.

2 A new flow approach and some of its properties

Before introducing the additively rescaled prescribed Gauss curvature flow, we recall an important and highly useful estimate. The following lemma (see e.g. [5, Corollary 1.7]) is a consequence of the Trudinger’s inequality [20] which was improved by Moser in [15] (for more details see e.g. [18, Theorem 2.1 and Theorem 2.2]):

Lemma 2.1

For a two-dimensional, closed Riemannian manifold \((M,{\bar{g}})\) there are constants \(\eta >0\) and \(C_{\text {MT}}>0\) such that

$$\begin{aligned} \int _M\textrm{e}^{(u-{\bar{u}})}d\mu _{{\bar{g}}}\le C_{\text {MT}}\exp \left( \eta \Vert \nabla _{{\bar{g}}}u\Vert ^2_{L^2(M,{\bar{g}})}\right) \end{aligned}$$
(2.1)

for all \(u\in H^1(M,{\bar{g}})\) where

$$\begin{aligned} {\bar{u}}:=\frac{1}{{\text {vol}}_{{\bar{g}}}}\int _Mu\,d\mu _{{\bar{g}}}=\int _Mu\,d\mu _{{\bar{g}}}, \end{aligned}$$

in view of our assumption that \({\text {vol}}_{{\bar{g}}}=1\).

As a consequence of Lemma 2.1, we have

$$\begin{aligned} \int _M \textrm{e}^{p u}d\mu _{{\bar{g}}} = \textrm{e}^{p {\bar{u}}} \int _M\textrm{e}^{(pu-\bar{pu})}d\mu _{{\bar{g}}} \le \textrm{e}^{p {\bar{u}}} C_{\text {MT}}\exp \left( \eta \Vert \nabla _{{\bar{g}}}(pu)\Vert ^2_{L^2(M,{\bar{g}})}\right) < \infty \end{aligned}$$
(2.2)

for every \(u \in H^1(M,{\bar{g}})\) and \(p>0\). Therefore, for a given \(A>0\), the set

$$\begin{aligned} {\mathcal {C}}_{A}:= \left\{ u\in H^1(M,{\bar{g}})\mid \int _M \textrm{e}^{2u}d\mu _{{\bar{g}}}=A\right\} \end{aligned}$$
(2.3)

is well defined. We also note that

$$\begin{aligned} {\bar{u}}\le \frac{1}{2}\log (A) \qquad \text {for }u\in {\mathcal {C}}_{A}, \end{aligned}$$
(2.4)

since by Jensen’s inequality and our assumption that \({\text {vol}}_{{\bar{g}}}=1\) we have

(2.5)

Next, we let \(f\in C^\infty (M)\) be a fixed smooth function. As a consequence of (2.2), the energy functional \(E_f\) given in (1.5) is then well defined and of class \(C^1\) on \(H^1(M,{\bar{g}})\). Moreover, we have

$$\begin{aligned} E_f(u) \le \frac{1}{2} \Vert \nabla u\Vert _{L^2(M,{\bar{g}})}^2 + |{\bar{K}}| \Vert u\Vert _{L^1(M,{\bar{g}})} + \frac{A}{2} \Vert f\Vert _{L^\infty (M,{\bar{g}})} \quad \text {for }u \in {\mathcal {C}}_{A} \end{aligned}$$
(2.6)

We now consider the additively rescaled prescribed Gauss curvature flow given by the evolution equation

$$\begin{aligned} \partial _tu(t)=f-K_{g(t)}-\alpha (t) = f+\textrm{e}^{-2u(t)}( \Delta _{{\bar{g}}}u(t)-{\bar{K}})-\alpha (t) \quad \text {in }(0,T)\times M, \end{aligned}$$
(2.7)

where \(\alpha (t)\) is chosen such that the volume \({\text {vol}}_{g(t)}\) of M with respect to the metric \(g(t)=\textrm{e}^{2u(t)}{\bar{g}}\) remains constant along the flow. The latter condition requires that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}{\text {vol}}_{g(t)}= & {} \int _M\partial _t u(t) d\mu _{g(t)}=\int _M(f-K_{g(t)}-\alpha (t))d\mu _{g(t)}\nonumber \\= & {} \int _Mf d\mu _{g(t)} -\alpha (t){\text {vol}}_{g(t)} - {\bar{K}} \end{aligned}$$
(2.8)

vanishes for \(t >0\) and therefore suggest the definition of \(\alpha (t)\) given in (2.11) below. We first note the following observations.

Proposition 2.2

Let \(T>0\), \(f \in C^\infty (M)\), \(A>0\), let \(u_0 \in {\mathcal {C}}_{A}\), and let \(u \in C([0,T), H^1(M,{\bar{g}}))\cap C^1((0,T),H^2(M,{\bar{g}}))\) be a solution of the initial value problem

$$\begin{aligned} \partial _tu(t)&=f-K_{g(t)}-\alpha (t)\quad \text {in }(0,T)\times M; \end{aligned}$$
(2.9)
$$\begin{aligned} u(0)&=u_0 \quad \text {on }\{0\}\times M, \end{aligned}$$
(2.10)

where

$$\begin{aligned} \alpha (t)=\frac{1}{A}\left( \int _Mf d \mu _{g(t)}-{\bar{K}}\right) =\frac{1}{A}\left( \int _Mf \textrm{e}^{2u(t)}d \mu _{{\bar{g}}}-{\bar{K}}\right) \end{aligned}$$
(2.11)

Then

  1. 1.

    the volume \({\text {vol}}_{g(t)}\) of (Mg(t)) is preserved along the flow, i.e., \({\text {vol}}_{g(t)}\equiv {\text {vol}}_{g_0}=A\) and therefore \(u(t) \in {\mathcal {C}}_{A}\) for \(t \in [0,T)\);

  2. 2.

    along this trajectory, we have a uniform bound for \(\alpha \) given by

    $$\begin{aligned} |\alpha (t)| \le \alpha _0\quad \text {for }t \in [0,T)\;\text { with}\quad \alpha _0:=\Vert f\Vert _{L^\infty (M,{\bar{g}})} +\frac{|{\bar{K}}|}{A}; \end{aligned}$$
    (2.12)
  3. 3.

    the Eq. (2.9) remains invariant under adding a constant \(c \in \mathbb {R}\) to the function f;

  4. 4.

    the function \(t \mapsto E_f(u(t))\) is decreasing on [0, T), so in particular \(E_f(u(t))\le E_f(u_0)\) for \(t \in [0,T)\);

  5. 5.

    there exist constants \(c_0=c_0(u_0)>0\), \(c_1=c_1(u_0)>0\) depending only on \(u_0\) with the property that

    $$\begin{aligned} \Vert \nabla _{{\bar{g}}}u(t)\Vert ^2_{L^2(M,{\bar{g}})}\le c_0 + c_1 \Vert f\Vert _{L^\infty (M,{\bar{g}})} \qquad \text {for }t \in [0,T); \end{aligned}$$
    (2.13)
  6. 6.

    there exist constants \(m_0=m_0(u_0) \in \mathbb {R}\), \(m_1=m_1(u_0)>0\) depending only on \(u_0\) with the property that

    $$\begin{aligned} m_0- m_1 \Vert f\Vert _{L^\infty (M,{\bar{g}})} \le {\bar{u}}(t)\le \frac{1}{2}\log (A) \qquad \text {for }t \in [0,T); \end{aligned}$$
    (2.14)
  7. 7.

    for every \(p \in \mathbb {R}\) there exist constants \(\nu _0= \nu _0(u_0,p),\,\nu _1 = \nu _1(u_0,p)>0\) with

    $$\begin{aligned} \int _M \textrm{e}^{2pu(t)}d\mu _{{\bar{g}}}\le \nu _0 \textrm{e}^{\nu _1 \Vert f\Vert _{L^\infty (M,{\bar{g}})}} \qquad \text {for }t \in [0,T). \end{aligned}$$
    (2.15)

Proof

1. Let \(h(t)=\frac{1}{2}\bigl ({\text {vol}}_{g(t)}-A\bigr )\). Then by (2.8) we have

$$\begin{aligned} \dot{h}(t)&= \frac{1}{2}\frac{d}{dt}{\text {vol}}_{g(t)} = \int _Mf d\mu _{g(t)} -\alpha (t){\text {vol}}_{g(t)} - {\bar{K}}\\&= \left( \int _Mf d\mu _{g(t)} - {\bar{K}}\right) \left( 1- \frac{{\text {vol}}_{g(t)}}{A}\right) \\&= \frac{2}{A}\left( \int _Mf d\mu _{g(t)} - {\bar{K}}\right) h(t) \qquad \text {for }t \in (0,T). \end{aligned}$$

Since h is continuous in 0 and \(h(0)=0\), Gronwall’s inequality (see e.g. [3]) implies that \(h(t)=0\) and therefore \({\text {vol}}_{g(t)} = A\) for \(t \in [0,T)\).

2. follows directly from (2.11).

To show 3., we note that replacing f by \(f+ c\) in (2.9) gives

$$\begin{aligned} f+ c-K_{g(t)}-\frac{1}{A}\left( \int _M(f+ c)d\mu _{g(t)}-{\bar{K}}\right)= & {} f-K_{g(t)}-\frac{1}{A}\left( \int _Mfd\mu _{g(t)}-{\bar{K}}\right) \\= & {} \partial _tu(t), \end{aligned}$$

so the equation remains unchanged.

To see 4., we use (2.8) and get

$$\begin{aligned} \begin{aligned} \frac{d}{dt}E_{f}(u(t))&=\int _M(-\Delta _{{\bar{g}}}u(t)+{\bar{K}}-f\textrm{e}^{2u(t)})\partial _tu(t)d\mu _{{\bar{g}}}\\&=\int _M((-\Delta _{{\bar{g}}}u(t)+{\bar{K}})\textrm{e}^{-2u(t)}-f)\textrm{e}^{2u(t)}\partial _tu(t)d\mu _{{\bar{g}}}\\&=\int _M((-\Delta _{{\bar{g}}}u(t)+{\bar{K}})\textrm{e}^{-2u(t)}-f)\partial _tu(t)d\mu _{g(t)}\\&=\int _M(K_{g(t)}-f)\partial _tu(t)d\mu _{g(t)}=\int _M(K_{g(t)}-f+\alpha (t))\partial _tu(t)d\mu _{g(t)}\\&=-\int _M|\partial _tu(t)|^2d\mu _{g(t)}\le 0. \end{aligned}\nonumber \\ \end{aligned}$$
(2.16)

Therefore, we have

$$\begin{aligned} E_{f}(u(\tau ))+\int _0^\tau \int _M|\partial _tu(t)|^2d\mu _{g(t)}dt= E_{f}(u(0)) \qquad \text {for }0< \tau < T. \end{aligned}$$
(2.17)

5. Since \(u(t)\in {\mathcal {C}}_A\) for \(t \in [0,T)\) by 1., we may use 4., (2.5) and (2.6) to observe that

$$\begin{aligned} \begin{aligned} \Vert \nabla _{{\bar{g}}}u(t)\Vert ^2_{L^2(M,{\bar{g}})}&=2E_{f}(u(t))-\int _M(2{\bar{K}}u(t)-f\textrm{e}^{2u(t)})d\mu _{{\bar{g}}}\\&=2E_{f}(u(t))+\int _M(2|{\bar{K}}|u(t)+f\textrm{e}^{2u(t)})d\mu _{{\bar{g}}}\\&\le 2 E_{f}(u_0)+|{\bar{K}}|\log (A)+ A \Vert f\Vert _{L^\infty (M,{\bar{g}})}\\&\le \Vert \nabla u_0 \Vert _{L^2(M,{\bar{g}})}^2 + |{\bar{K}}|\Bigl (\log (A)+ 2\Vert u_0\Vert _{L^1(M,{\bar{g}})}\Bigr ) + 2 A \Vert f\Vert _{L^\infty (M,{\bar{g}})}\\&\le c_0 + c_1 \Vert f\Vert _{L^\infty (M,{\bar{g}})} \qquad \text {for }t \in [0,T). \end{aligned}\nonumber \\ \end{aligned}$$
(2.18)

with constants \(c_0,c_1>0\) depending only on \(u_0\) (recall here that \(A = \int _M \textrm{e}^{2u_0(t)}d\mu _{{\bar{g}}}\)).

6. With (2.13) and Lemma 2.1 we can estimate

$$\begin{aligned} A&=\int _M\textrm{e}^{2u(t)}d\mu _{{\bar{g}}}=\textrm{e}^{2{\bar{u}}(t)}\int _M\textrm{e}^{2(u(t)-{\bar{u}}(t))}d\mu _{{\bar{g}}} \le \textrm{e}^{2{\bar{u}}(t)}C_{\text {MT}}\exp (\eta _1\Vert \nabla _{{\bar{g}}}(2u(t))\Vert ^2_{L^2(M,{\bar{g}})})\\&\le \textrm{e}^{2{\bar{u}}(t)}C_{\text {MT}}\exp \bigl (\eta _1(c_1 + c_2 \Vert f\Vert _{L^\infty (M,{\bar{g}})})\bigr ) \end{aligned}$$

and therefore

$$\begin{aligned} {\bar{u}}(t)\ge \frac{1}{2}\log \left( \frac{A}{C_{\text {MT}}}\right) - \frac{1}{2}\eta _1(c_1 + c_2 \Vert f\Vert _{L^\infty (M,{\bar{g}})})= m_0 - m_1 \Vert f\Vert _{L^\infty (M,{\bar{g}})} \end{aligned}$$

with constants \(m_0 \in \mathbb {R}\), \(m_1 >0\) depending only on \(u_0\). Combining this lower bound with the upper bound given by (2.4), we obtain (2.14).

7. With Lemma 2.1, (2.5), and (2.18) we directly get for any \(p\in \mathbb {R}\) that

$$\begin{aligned} \int _M\textrm{e}^{2pu(t)}d\mu _{{\bar{g}}}&=\textrm{e}^{2p{\bar{u}}(t)}\int _M\textrm{e}^{2p(u(t)-{\bar{u}}(t))}d\mu _{{\bar{g}}}\\&\le \textrm{e}^{ p \log (A)}C_{\text {MT}}\exp (4\eta _2p^2\Vert \nabla _{{\bar{g}}}u(t)\Vert ^2_{L^2(M,{\bar{g}})})\\&\le A^p C_{\text {MT}}\exp \Bigl ((4\eta _2p^2\bigl (c_1 + c_2 \Vert f\Vert _{L^\infty (M,{\bar{g}})}\bigr ) \Bigr )\\&\le C_{\text {MT}} A^{p}\textrm{e}^{4\eta _2p^2 c_1} \exp \bigl (4\eta _2p^2 c_2 \Vert f\Vert _{L^\infty (M,{\bar{g}})}\bigr )\\&= \nu _0 \textrm{e}^{\nu _1 \Vert f\Vert _{L^\infty (M,{\bar{g}})}} \end{aligned}$$

with constants \(\nu _i=\nu _i(u_0,p)>0\), \(i\in \{0,1\}\). \(\square \)

3 Main results

In the following, we put

$$\begin{aligned} {\mathcal {C}}_{p,A}:= & {} W^{2,p}(M,{\bar{g}}) \cap {\mathcal {C}}_{p,A} \nonumber \\= & {} \left\{ v\in W^{2,p}(M,{\bar{g}})\mid \int _M\textrm{e}^{2v}d\mu _{{\bar{g}}}=A\right\} \qquad \text {for }p>2, A>0. \end{aligned}$$
(3.1)

The following is our first main result.

Theorem 3.1

Let \(f\in C^\infty (M)\), \(p>2\), and \(u_0\in {\mathcal {C}}_{p,A}\) for a given \(A>0\).

Then the initial value problem (2.9), (2.10) admits a unique global solution

$$\begin{aligned} u \in C([0,\infty )\times M)\cap C([0,\infty );H^1(M,{\bar{g}}))\cap C^\infty ((0,\infty )\times M) \end{aligned}$$

satisfying the energy bound \(E_{f}(u(t))\le E_{f}(u_0)\) for all \(t \ge 0\).

Moreover, u is uniformly bounded in the sense that

$$\begin{aligned} \sup _{t>0}\Vert u(t)\Vert _{L^\infty (M,{\bar{g}})}<\infty . \end{aligned}$$

Furthermore, if \((t_l)_l \subset (0,\infty )\) is a sequence with \(t_l \rightarrow \infty \) as \(l \rightarrow \infty \), then, after passing to a subsequence, \(u(t_l)\) converges in \(H^2(M,{\bar{g}})\) to a function \(u_\infty \in H^2(M,{\bar{g}}) \cap {\mathcal {C}}_{A}\) solving the equation

$$\begin{aligned} -\Delta _{{\bar{g}}}u_\infty + {\bar{K}} = f_\lambda \textrm{e}^{2u_\infty } \qquad \text {in }M, \end{aligned}$$
(3.2)

where \(f_\lambda := f +\lambda \) with

$$\begin{aligned} \lambda =\frac{1}{A}\left( {\bar{K}}- \int _M f \textrm{e}^{2u_\infty }d\mu _{{\bar{g}}}\right) . \end{aligned}$$
(3.3)

In other words, \(u_\infty \) induces a metric \(g_\infty \) with \({\text {vol}}_{g_\infty }= A\) and Gauss curvature \(K_{g_\infty }\) satisfying

$$\begin{aligned} K_{g_\infty }(x)= f_{\lambda }(x)=f(x)+\lambda \quad \text {for}\quad x\in M. \end{aligned}$$
(3.4)

Some remarks are in order.

Remark 3.2

It follows in a standard way that, under the assumptions of Theorem 3.1, the \(\omega \)-limit set

$$\begin{aligned} \omega (u_0):= \bigcap _{T>0} \overline{ \{u(t)\,:\, T \le t < \infty \}} \end{aligned}$$

is a compact connected subset of \(H^2(M,{\bar{g}}) \cap {\mathcal {C}}_{A}\) (with respect to the \(H^2\)-topology) consisting of solutions of (3.2), (3.3), which are precisely the critical points of the restriction of the energy functional \(E_f\) to \({\mathcal {C}}_{A}\).

In particular, the connectedness implies that, if \(u_\infty \) in Theorem 3.1 is an isolated critical point in \({\mathcal {C}}_{A}\), then \(\omega (u_0)= \{u_\infty \}\) and therefore we have the full convergence of the flow line

$$\begin{aligned} u(t) \rightarrow u_\infty \quad \text {in }H^2(M,{\bar{g}})\quad \text {as }t \rightarrow \infty . \end{aligned}$$
(3.5)

In particular, (3.5) holds if \(u_\infty \) is a strict local minimum of the restriction of \(E_f\) to \({\mathcal {C}}_{A}\).

Remark 3.3

For functions \(f<0\), the convergence of the flow (1.9) is shown in [9]. For the additively rescaled flow (2.9) with initial data (2.10) we get convergence for arbitrary functions \(f\in C^\infty (M)\). In general we do not have any information about \(\lambda \) and therefore no information about the sign of \(f_{\lambda }\) in Theorem 3.1. On the other hand, more information can be derived for certain functions \(f \in C^\infty (M)\) and certain values of \(A>0\).

  1. (i)

    In the case where \(A \le - \frac{{\bar{K}}}{\Vert f\Vert _{L^\infty (M,{\bar{g}})}}\), it follows that

    $$\begin{aligned} \lambda= & {} \frac{1}{A}\left( {\bar{K}}- \int _M f \textrm{e}^{2u}d\mu _{{\bar{g}}}\right) \le \frac{{\bar{K}}}{A} + \frac{\Vert f\Vert _{L^\infty (M,{\bar{g}})}}{A} \int _M \textrm{e}^{2u}d\mu _{{\bar{g}}} \\= & {} \frac{{\bar{K}}}{A} + \Vert f\Vert _{L^\infty (M,{\bar{g}})} \le 0 \end{aligned}$$

    for every solution \(u \in {\mathcal {C}}_{2,A}:=\left\{ v\in H^2(M,{\bar{g}})\mid \int _M\textrm{e}^{2v}d\mu _{{\bar{g}}}=0\right\} \) of the static problem (3.2), and therefore this also applies to \(\lambda \) in Theorem 3.1 in this case.

  2. (ii)

    The following theorems show that \(f_{\lambda }\) in Theorem 3.1 may change sign if \(A > - \frac{{\bar{K}}}{\Vert f\Vert _{L^\infty (M,{\bar{g}})}}\), so in this case we get a solution of the static problem (1.2) for sign-changing functions \(f\in C^\infty (M)\) by using the additively rescaled prescribed Gauss curvature flow (2.9).

Theorem 3.4

Let \(p>2\). For every \(A>0\) and \(c> - \frac{{\bar{K}}}{A}\) there exists \(\varepsilon = \varepsilon (c,A,{\bar{K}})>0\) with the following property.

If \(u_0 \equiv \frac{1}{2}\log (A) \in {\mathcal {C}}_{p,A}\) and \(f\in C^\infty (M)\) with \(-c \le f \le 0\) and \(\Vert f+c\Vert _{L^1(M,{\bar{g}})} < \varepsilon \) are chosen in Theorem 3.1, then the value \(\lambda \) defined in (3.3) is positive.

In particular, if f has zeros on M, then \(f_\lambda \) in (3.4) is sign changing.

Under fairly general assumptions on f, we can prove that \(\lambda >0\) if A is sufficiently large and \(u_0 \in {\mathcal {C}}_{p,A}\) is chosen suitably.

Theorem 3.5

Let \(f \in C^\infty (M)\) be nonconstant with \(\max _{x\in M} f(x) = 0\). Then there exists \(\kappa >0\) with the property that for every \(A \ge \kappa \) there exists \(u_0 \in {\mathcal {C}}_{p,A}\) such that the value \(\lambda \) defined in (3.3) is positive.

In fact we have even more information on the associated limit \(u_\infty \) in this case, see Corollary 4.7 below.

It remains open how large \(\lambda \) can be depending on A and f. The only upper bound we have is

$$\begin{aligned} \lambda <- \int _M f d\mu _{{\bar{g}}}, \end{aligned}$$
(3.6)

since we must have

$$\begin{aligned} {\bar{f}}_\lambda =\frac{1}{{\text {vol}}_{{\bar{g}}}}\int _M f_\lambda d\mu _{{\bar{g}}}=\int _M f d\mu _{{\bar{g}}}+\lambda \overset{!}{<}0, \end{aligned}$$

so that \(f_\lambda \) fulfills the necessary condition (1.4) provided by Kazdan and Warner in [11].

4 The static minimisation problem with volume constraint

To obtain additional information on the limiting function \(u_\infty \) and the value \(\lambda \in \mathbb {R}\) associated to it by (3.3) and (3.4), we need to consider the associated static setting for the prescribed Gauss curvature problem with the additional condition of prescribed volume. In this setting, we wish to find, for given \(f \in C^\infty (M)\) and \(A>0\), critical points of the restriction of the functional \(E_{f}\) defined in (1.5) to the set \({\mathcal {C}}_{A}\) defined in (2.3). A critical point \(u \in {\mathcal {C}}_{A}\) of this restriction is a solution of (3.2) for some \(\lambda \in \mathbb {R}\), where, here and in the following, we put again \(f_\lambda := f + \lambda \in C^\infty (M)\). In other words, such a critical point induces, similarly as the limit \(u_\infty \) in Theorem 3.1, a metric \(g^u\) with Gauss curvature \(K_{g^u}\) satisfying \(K_{g^u}(x)= f_{\lambda }(x)=f(x)+\lambda .\) The unknown \(\lambda \in \mathbb {R}\) arises in this context as a Lagrange multiplier and is a posteriori characterised again by

$$\begin{aligned} \lambda =\frac{1}{A}\left( {\bar{K}}- \int _M f \textrm{e}^{2u}d\mu _{{\bar{g}}}\right) . \end{aligned}$$

In the study of critical points of the restriction of \(E_{f}\) to \({\mathcal {C}}_{A}\), it is natural to consider the minimisation problem first. For this we set

$$\begin{aligned} m_{f,A} = \inf _{u \in {\mathcal {C}}_{A}}E_f(u). \end{aligned}$$

We have the following estimates for \(m_{f,A}\):

Lemma 4.1

Let \(f \in C^\infty (M)\), \(A>0\). Then we have

$$\begin{aligned} m_{f,A} \le \frac{1}{2} \left( {\bar{K}} \log (A) - A\int _{M} f d \mu _{{\bar{g}}}\right) . \end{aligned}$$
(4.1)

Moreover, if \(\max f \ge 0\), then we have

$$\begin{aligned} \limsup _{A \rightarrow \infty } \frac{m_{f,A}}{A} \le 0. \end{aligned}$$
(4.2)

Proof

Let \(u_0(A)\equiv \frac{1}{2}\log (A)\), so that \(\int _M\textrm{e}^{2u_0(A)}d\mu _{{\bar{g}}}=A\). Hence \(u_0(A)\) is the (unique) constant function in \({\mathcal {C}}_{A}\), and

$$\begin{aligned} m_{f,A}&\le E_{f}(u_0(A))=\frac{1}{2}\int _M(|\nabla _{{\bar{g}}}u_0(A)|^2_{{\bar{g}}}+2{\bar{K}}u_0(A)-f\textrm{e}^{2u_0(A)})d\mu _{{\bar{g}}}\\&=\frac{1}{2}\int _M({\bar{K}}\log (A)-f A)d\mu _{{\bar{g}}} = \frac{1}{2} \left( {\bar{K}} \log (A) - A\int _{M} f d \mu _{{\bar{g}}}\right) . \end{aligned}$$

This shows (4.1). To show (4.2), we let \(\varepsilon >0\). Since \(f \in C^\infty (M)\) and \(\max f \ge 0\) by assumption, there exists an open set \(\Omega \subset M\) with \(f \ge -\varepsilon \) on \(\Omega \). Next, let \(\psi \in C^\infty (M)\), \(\psi \ge 0\), be a function supported in \(\Omega \) and with \(\Vert \psi \Vert _{L^\infty (M,{\bar{g}})} = 2\). Consequently, the set \(\Omega ':= \{x \in M\mid \psi >1\}\) is a nonempty open subset of \(\Omega \), and therefore \(\mu _{{\bar{g}}}(\Omega ')>0\).

Next we consider the continuous function

$$\begin{aligned} h: [0,\infty ) \rightarrow [0,\infty );\quad h(\tau )= \int _M\textrm{e}^{2 \tau \psi }d\mu _{{\bar{g}}} \end{aligned}$$

and we note that \(h(0)= \int _M d\mu _{{\bar{g}}}=1\), and that

$$\begin{aligned} h(\tau ) \ge \int _{\Omega '} \textrm{e}^{2 \tau \psi }d\mu _{{\bar{g}}} \ge \textrm{e}^{2\tau } \mu _{{\bar{g}}}(\Omega ') \qquad \text {for }\tau \ge 0. \end{aligned}$$

Hence for every \(A \ge 1\) there exists

$$\begin{aligned} 0 \le \tau _A \le \frac{1}{2}\Bigl (\log (A) -\log (\mu _{{\bar{g}}}(\Omega '))\Bigr ) \end{aligned}$$
(4.3)

with \(h(\tau _A)=A\) and therefore \(\tau _A \psi \in {\mathcal {C}}_{A}\). Consequently,

$$\begin{aligned} m_{f,A}&\le E_f(\tau _A \psi ) = \frac{1}{2} \int _M(|\nabla _{{\bar{g}}}\tau _A \psi |^2_{{\bar{g}}}+2{\bar{K}} \tau _A \psi -f\textrm{e}^{2\tau _A \psi })d\mu _{{\bar{g}}}\\&= \tau _A^2 c_1 - \tau _A c_2 -c_3 - \frac{1}{2} \int _{\Omega } f\textrm{e}^{2\tau _A \psi }d\mu _{{\bar{g}}} \end{aligned}$$

with

$$\begin{aligned} c_1 = \frac{1}{2}\int _M |\nabla _{{\bar{g}}}\psi |^2_{{\bar{g}}}d\mu _{{\bar{g}}},\quad c_2 = - {\bar{K}} \int _{M} \psi d\mu _{{\bar{g}}} \quad \text {and}\quad c_3 = \frac{1}{2} \int _{M \setminus \Omega } f d\mu _{{\bar{g}}}. \end{aligned}$$

Since \(f \ge -\varepsilon \) on \(\Omega \), we thus deduce that

$$\begin{aligned} m_{f,A} \le \tau _A^2 c_1 -2 \tau _A c_2 +c_3 + \frac{\varepsilon }{2} \int _{\Omega } \textrm{e}^{2\tau _A \psi }d\mu _{{\bar{g}}} \le \tau _A^2 c_1 -2 \tau _A c_2 +c_3 +\frac{\varepsilon A}{2}. \end{aligned}$$

Since \(\frac{\tau _A}{A} \rightarrow 0\) as \(A \rightarrow \infty \) by (4.3), we conclude that

$$\begin{aligned} \limsup _{A \rightarrow \infty } \frac{m_{f,A}}{A} \le \frac{\varepsilon }{2}. \end{aligned}$$

Since \(\varepsilon >0\) was chosen arbitrarily, (4.2) follows. \(\square \)

Lemma 4.2

Let \(f \in C^\infty (M)\) nonconstant with \(\max _{x\in M} f(x) = 0\). For every \(\varepsilon >0\) there exists \(\kappa _0>0\) with the following property. If \(A \ge \kappa _0\) and \(u \in {\mathcal {C}}_{A}\) is a solution of

$$\begin{aligned} -\Delta _{{\bar{g}}}u + {\bar{K}} = (f+\lambda ) \textrm{e}^{2u} \end{aligned}$$
(4.4)

for some \(\lambda \in \mathbb {R}\) with \(E_f(u)< \frac{\varepsilon A}{2}\), then we have \(\lambda <\varepsilon \).

Proof

For given \(\varepsilon >0\), we may choose \(\kappa _0>0\) sufficiently large so that \(\frac{|{\bar{K}}|}{2} \frac{\log (A)}{|A|}< \frac{\varepsilon }{2}\) for \(A \ge \kappa _0\).

Now, let \(A \ge \kappa _0\), and let \(u \in {\mathcal {C}}_{A}\) be a solution of (4.4) satisfying \(E_f(u)< \frac{\varepsilon A}{2}\). Integrating (4.4) over M with respect to \(\mu _{{\bar{g}}}\) and using that \({\text {vol}}_{{\bar{g}}}(M)=1\) and \(\int _{M} \textrm{e}^{2u}d\mu _{{\bar{g}}}=A\), we obtain

$$\begin{aligned} \lambda&= \frac{1}{A}\left( {\bar{K}} - \int _{M} f \textrm{e}^{2u}d\mu _{{\bar{g}}}\right) \le - \frac{1}{A}\int _{M} f \textrm{e}^{2u}d\mu _{{\bar{g}}} \\&= \frac{1}{A} \left( E_{f}(u)-\frac{1}{2}\int _M(|\nabla _{{\bar{g}}}u|^2_{{\bar{g}}}+2{\bar{K}} u)d\mu _{{\bar{g}}} \right) \le \frac{1}{A}\left( E_f(u) +|{\bar{K}}| {\bar{u}}\right) \\&\le \frac{\varepsilon }{2} + \frac{|{\bar{K}}|}{2} \frac{\log (A)}{A} <\varepsilon , \end{aligned}$$

as claimed. Here we used (2.4) to estimate \({\bar{u}}\). \(\square \)

Proposition 4.3

Let \(f \in C^\infty (M)\) be a nonconstant function with \(\max _{x\in M} f(x) =0\). Moreover, let \(\lambda _n \rightarrow 0^+\) for \(n\rightarrow \infty \), and let \((u_n)_{n\in \mathbb {N}}\) be a sequence of solutions of

$$\begin{aligned} -\Delta _{{\bar{g}}} u_n + {\bar{K}} = (f +\lambda _n)\textrm{e}^{2u_n} \quad \text {in }M \end{aligned}$$
(4.5)

which are weakly stable in the sense that

$$\begin{aligned} \int _{M}(|\nabla _{{\bar{g}}} h|_{{\bar{g}}}^2 -2 (f+\lambda _n) \textrm{e}^{2u_n} h^2) d \mu _{{\bar{g}}} \ge 0 \quad \text {for all }h \in H^1(M). \end{aligned}$$
(4.6)

Then \(u_n \rightarrow u_0\) in \(C^2(M)\), where \(u_0\) is the unique solution of

$$\begin{aligned} -\Delta _{{\bar{g}}} u_0 + {\bar{K}} = f \textrm{e}^{2 u_0} \qquad \text {in }M. \end{aligned}$$
(4.7)

Proof

We only need to show that

$$\begin{aligned} (u_n)_{n\in \mathbb {N}}\text { is bounded in }C^{2,\alpha }(M)\text { for some }\alpha >0. \end{aligned}$$
(4.8)

Indeed, assuming this for the moment, we may complete the argument as follows. Suppose by contradiction that there exists \(\varepsilon >0\) and a subsequence, also denoted by \((u_n)_{n\in \mathbb {N}}\), with the property that

$$\begin{aligned} \Vert u_n-u_0\Vert _{C^2(M)} \ge \varepsilon \qquad \text {for all }n \in \mathbb {N}. \end{aligned}$$
(4.9)

By (4.8) and the compactness of the embedding \(C^{2,\alpha }(M) \hookrightarrow C^2(M)\), we may then pass to a subsequence, still denoted by \((u_n)_{n\in \mathbb {N}}\), with \(u_n \rightarrow u_*\) in \(C^2(M)\) for some \(u_* \in C^2(M)\). Passing to the limit in (4.5), we then see that \(u_*\) is a solution of (4.7), which by uniqueness implies that \(u_* = u_0\). This contradicts (4.9), and thus the claim follows.

The proof of (4.8) follows by similar arguments as in [7, p. 1063 f.]. Since the framework is slightly different, we sketch the main steps here for the convenience of the reader. We first note that, by the same argument as in [7, p. 1063 f.], there exists a constant \(C_0>0\) with

$$\begin{aligned} u_n \ge -C_0 \qquad \text {for all }n. \end{aligned}$$
(4.10)

Since \(\{f < 0\}\) is a nonempty open subset of M by assumption, we may fix a nonempty open subdomain \(\Omega \subset \subset \{f < 0\}\). By [1, Appendix], there exists a constant \(C_1>0\) with

$$\begin{aligned} \Vert u_n^+\Vert _{H^1(\Omega ,{\bar{g}})} \le C_1 \qquad \text {for all }n \end{aligned}$$

and therefore

$$\begin{aligned} \int _{\Omega }\textrm{e}^{2u_n}d\mu _{{\bar{g}}} \le \int _{\Omega }\textrm{e}^{2u_n^+}d\mu _{{\bar{g}}} \le C_2 \qquad \text {for all }n \end{aligned}$$
(4.11)

for some \(C_2>0\) by the Moser–Trudinger inequality. Next, we consider a nontrivial, nonpositive function \(h \in C^\infty _c(\Omega ) \subset C^\infty (M)\) and the unique solution \(w \in C^\infty (M)\) of the equation

$$\begin{aligned} -\Delta _{{\bar{g}}} w + {\bar{K}} = h\textrm{e}^{2w} \quad \text {in }M. \end{aligned}$$

Moreover, we let \(w_n:= u_n - w\), and we note that \(w_n\) satisfies

$$\begin{aligned} -\Delta _{{\bar{g}}} w_n +h\textrm{e}^{2w} = (f+{\lambda _n})\textrm{e}^{2u_n} \quad \text {in }M. \end{aligned}$$

Multiplying this equation by \(\textrm{e}^{2w_n}\) and integrating by parts, we obtain

$$\begin{aligned} \int _{M}(f+\lambda _n)\textrm{e}^{2(u_n+w_n)} d\mu _{{\bar{g}}}&= \int _{M}\Bigl (-\Delta _{{\bar{g}}} w_n + h \textrm{e}^{2w} \Bigr )\textrm{e}^{2w_n} d\mu _{{\bar{g}}}\nonumber \\&= \int _{M} \Bigl (2\textrm{e}^{2w_n} |\nabla _{{\bar{g}}}w_n|_{{\bar{g}}}^2 + h \textrm{e}^{2(w+w_n)} \Bigr ) d\mu _{{\bar{g}}} \nonumber \\&= 2 \int _{M}|\nabla _{{\bar{g}}} \textrm{e}^{w_n}|_{{\bar{g}}}^2 d \mu _{{\bar{g}}} + \int _{\Omega } h \textrm{e}^{2u_n} d\mu _{{\bar{g}}}. \end{aligned}$$
(4.12)

Moreover, applying (4.6) to \(h= \textrm{e}^{w_n}\) gives

$$\begin{aligned} \int _{M}(f+\lambda _n)\textrm{e}^{2(u_n+w_n)} d\mu _{{\bar{g}}} \le \frac{1}{2} \int _{M}|\nabla _{{\bar{g}}} \textrm{e}^{w_n}|_{{\bar{g}}}^2 d \mu _{{\bar{g}}}. \end{aligned}$$
(4.13)

Combining (4.11), (4.12) and (4.13) yields

$$\begin{aligned} \Vert \nabla _{{\bar{g}}} \textrm{e}^{w_n}\Vert _{L^2(M,{\bar{g}})}^2 \le -\frac{2}{3} \int _{\Omega } h \textrm{e}^{2u_n} d\mu _{{\bar{g}}} \le \frac{2}{3}\Vert h\Vert _{L^\infty (M,{\bar{g}})}C_2 \quad \text {for all }n. \end{aligned}$$
(4.14)

Next we claim that also \(\Vert \textrm{e}^{w_n}\Vert _{L^2(M,{\bar{g}})}\) remains uniformly bounded. Suppose by contradiction that

$$\begin{aligned} \Vert \textrm{e}^{w_n}\Vert _{L^2(M,{\bar{g}})} \rightarrow \infty \qquad \text {as }n \rightarrow \infty . \end{aligned}$$
(4.15)

We then set \(v_n:= \frac{\textrm{e}^{w_n}}{\Vert \textrm{e}^{w_n}\Vert _{L^2(M,{\bar{g}})}}\), and we note that

$$\begin{aligned} \Vert v_n\Vert _{L^2(M,{\bar{g}})} = 1 \quad \text {for all }n\quad \text {and}\quad \Vert \nabla _{{\bar{g}}} v_n\Vert _{L^2(M,{\bar{g}})}^2 \rightarrow 0 \quad \text {as }n \rightarrow \infty \end{aligned}$$
(4.16)

by (4.14). Consequently, we may pass to a subsequence satisfying \(v_n \rightharpoonup v\) in \(H^1(M,{\bar{g}})\), where v is a constant function with

$$\begin{aligned} \Vert v\Vert _{L^2(M,{\bar{g}})}=1. \end{aligned}$$
(4.17)

However, since

$$\begin{aligned} \Vert \textrm{e}^{w_n}\Vert _{L^2(\Omega ,{\bar{g}})} \le \Vert \textrm{e}^{u_n}\Vert _{L^2(\Omega ,{\bar{g}})} \Vert \textrm{e}^{-w}\Vert _{L^\infty (\Omega ,{\bar{g}})} \le \sqrt{C_2} \Vert \textrm{e}^{-w}\Vert _{L^\infty (\Omega ,{\bar{g}})}\quad \text {for all }n \in \mathbb {N}\end{aligned}$$

by (4.11) and therefore

$$\begin{aligned} \Vert v\Vert _{L^2(\Omega ,{\bar{g}})} = \lim _{n \rightarrow \infty }\Vert v_n\Vert _{L^2(\Omega ,{\bar{g}})}= \lim _{n \rightarrow \infty }\frac{\Vert \textrm{e}^{w_n}\Vert _{L^2(\Omega ,{\bar{g}})}}{\Vert \textrm{e}^{w_n}\Vert _{L^2(M,{\bar{g}})}} = 0 \end{aligned}$$

by (4.15), we conclude that the constant function v must vanish identically, contradicting (4.17).

Consequently, \(\Vert \textrm{e}^{w_n}\Vert _{L^2(M,{\bar{g}})}\) remains uniformly bounded, which by (4.14) implies that \(\textrm{e}^{w_n}\) remains bounded in \(H^1(M,{\bar{g}})\) and therefore in \(L^p(M,{\bar{g}})\) for any \(p < \infty \). Since \(\textrm{e}^{u_n} \le \Vert \textrm{e}^{w}\Vert _{L^\infty (M,{\bar{g}})} \textrm{e}^{w_n}\) on M for all \(n \in \mathbb {N}\), it thus follows that also \(\textrm{e}^{u_n}\) remains bounded in \(L^p(M,{\bar{g}})\) for any \(p< \infty \). Moreover, by (4.10), the same applies to the sequence \(u_n\) itself. Therefore, applying successively elliptic \(L^p\) and Schauder estimates to (4.5), we deduce (4.8), as required. \(\square \)

In the proof of the next proposition, we need the following classical interpolation inequality, see e.g. [4].

Lemma 4.4

(Gagliardo–Nirenberg–Ladyžhenskaya inequality) For every \(r>2\), there exists a constant \(C_{\text {GNL}}=C_{\text {GNL}}(r)>0\) with

$$\begin{aligned} \Vert \zeta \Vert ^{r}_{L^r(M,{\bar{g}})} \le C_{\text {GNL}}\Vert \zeta \Vert ^{2}_{L^2(M,{\bar{g}})}\Vert \zeta \Vert ^{r-2}_{H^1(M,{\bar{g}})} \qquad \text {for every }\zeta \in H^1(M,{\bar{g}}). \end{aligned}$$

Proposition 4.5

Let \(f \in C^\infty (M)\) be a nonconstant function with \(\max _{x\in M} f(x) =0\). Then there exists \(\lambda _\sharp \) and a \(C^1\)-curve \((-\infty ,\lambda _{\sharp }] \rightarrow C^2(M); \quad \lambda \mapsto u_\lambda \) with the following properties.

  1. (i)

    If \(\lambda \le 0\), then \(u_\lambda \) is the unique solution of

    $$\begin{aligned} -\Delta _{{\bar{g}}} u + {\bar{K}} = f_\lambda \textrm{e}^{2 u} \quad \text {in }M \end{aligned}$$
    (4.18)

    and a global minimum of \(E_{f_\lambda }\).

  2. (ii)

    If \(\lambda \in (0,\lambda _\sharp ]\), then \(u_\lambda \) is the unique weakly stable solution of (4.18) in the sense of (4.6), and it is a local minimum of \(E_{f_\lambda }\).

  3. (iii)

    The curve of functions \(\lambda \mapsto u_\lambda \) is pointwisely strictly increasing on M, and so the volume function

    $$\begin{aligned} (-\infty ,\lambda _\sharp ] \rightarrow [0,\infty );\quad \lambda \mapsto V(\lambda ):= \int _{M}\textrm{e}^{2u_\lambda } d\mu _{{\bar{g}}} \end{aligned}$$
    (4.19)

    is continuous and strictly increasing.

Proof

We already know that, for \(\lambda \le 0\), the energy \(E_{f_\lambda }\) admits a strict global minimiser \(u_\lambda \) which depends smoothly on \(\lambda \). Moreover, by [1, Proposition 2.4], the curve \(\lambda \mapsto u_\lambda \) can be extended as a \(C^1\)-curve to an interval \((-\infty ,\lambda _{\sharp }]\) for some \(\lambda _{\sharp }>0\). We also know from [1, Proposition 2.4] that, for \(\lambda \in (-\infty ,\lambda _{\sharp }]\), the solution \(u_\lambda \) is strongly stable in the sense that

$$\begin{aligned} C_\lambda := \inf _{h \in H^1(M,{\bar{g}})} \frac{1}{\Vert h\Vert _{H^1(M,{\bar{g}})}^2}\int _{M}\Bigl (|\nabla _{{\bar{g}}} h|_{{\bar{g}}}^2 -2 f_\lambda \textrm{e}^{2u_\lambda } h^2\Bigr ) d \mu _{{\bar{g}}} >0. \end{aligned}$$
(4.20)

Here we note that the function \(\lambda \mapsto C_\lambda \) is continuous since \(u_\lambda \) depends continuously on \(\lambda \) with respect to the \(C^2\)-norm. Next we prove that, after making \(\lambda _\sharp >0\) smaller if necessary, the function \(u_\lambda \) is the unique weakly stable solution of (4.18) for \(\lambda \in (0,\lambda _{\sharp }]\). Arguing by contradiction, we assume that there exists a sequence \(\lambda _n \rightarrow 0^+\) and corresponding weakly stable solutions \((u_n)_{n\in \mathbb {N}}\) of

$$\begin{aligned} -\Delta _{{\bar{g}}} u_n + {\bar{K}} = (f + \lambda _n)\textrm{e}^{2u_n} \quad \text {in }M \end{aligned}$$
(4.21)

with the property that \(u_n \not = u_{\lambda _n}\) for every \(n \in \mathbb {N}\). By Proposition 4.3, we know that \(u_n \rightarrow u_0\) in \(C^2(M)\). Consequently, \(v_n:= u_n - u_{\lambda _n} \rightarrow 0\) in \(C^2(M)\) as \(n \rightarrow \infty \), whereas the functions \(v_n\) solve

$$\begin{aligned} -\Delta _{{\bar{g}}} v_n= & {} (f + \lambda _n)\bigl (\textrm{e}^{2u_n}-\textrm{e}^{2u_{\lambda _n}}\bigr )\nonumber \\= & {} (f + \lambda _n)\textrm{e}^{2u_{\lambda _n}} \bigl (\textrm{e}^{2v_n}-1\bigr ) \quad \text {in }M \quad \text {for every }n \in \mathbb {N}. \end{aligned}$$
(4.22)

Combining this fact with (4.20), we deduce that

$$\begin{aligned} \Vert v_n\Vert _{H^1(M,{\bar{g}})}^2&\le \frac{1}{C_\lambda } \int _{M}\Bigl (|\nabla _{{\bar{g}}} v_n|_{{\bar{g}}}^2 -2 (f+\lambda _n) \textrm{e}^{2u_{\lambda _n}} v_n^2\Bigr )d\mu _{{\bar{g}}}\\&= \frac{1}{C_\lambda } \int _{M}(f+\lambda _n) \textrm{e}^{2u_{\lambda _n}} \bigl (\textrm{e}^{2v_n}-1- 2v_n\Bigr )v_nd\mu _{{\bar{g}}}. \end{aligned}$$

Since \(v_n \rightarrow 0\) in \(C^2(M)\), there exists a constant \(C>0\) with \(|(\textrm{e}^{2v_n}-1- 2v_n)v_n| \le C |v_n|^3\) on M for all \(n \in \mathbb {N}\), which then implies with Hölder’s inequality and Lemma 4.4 that

$$\begin{aligned} \Vert v_n\Vert _{H^1(M,{\bar{g}})}^2&\le C \Vert (f+\lambda _n) \textrm{e}^{2u_{\lambda _n}}\Vert _{L^\infty (M,{\bar{g}})}\Vert v_n\Vert _{L^3(M,{\bar{g}})}^3\\&\le C\left( \int _M|v_n|^{3\cdot \frac{4}{3}}d\mu _{{\bar{g}}}\right) ^{\frac{3}{4}}= C \Vert v_n\Vert _{L^4(M,{\bar{g}})}^3\le C \Vert v_n\Vert _{H^1(M,{\bar{g}})}^3 \end{aligned}$$

with a constant \(C>0\) independent on M. This contradicts the fact that \(v_n \rightarrow 0\) in \(H^1(M)\) as \(n \rightarrow \infty \). The claim thus follows.

It remains to prove that the curve of functions \(\lambda \mapsto u_\lambda \) is pointwisely strictly increasing on M. This is a consequence of the uniqueness of weakly stable solutions stated in (ii) and the fact that, as noted in [7], if \(u_{\lambda _0}\) is a solution for some \(\lambda _0 \in (-\infty ,\lambda _{\sharp }]\), it is possible to construct, via the method of sub- and supersolutions, for every \(\lambda < \lambda _0\), a weakly stable solution \(u_\lambda \) with \(u_\lambda < u_{\lambda _0}\) everywhere in M. \(\square \)

Corollary 4.6

Let \(f \in C^\infty (M)\) be nonconstant with \(\max _{x\in M} f (x)= 0\), and let \(\lambda _\sharp >0\) be given as in Proposition 4.5. Then there exists \(\kappa _1>0\) with the following property.

If \(A \ge \kappa _1\) and \(u \in {\mathcal {C}}_{A}\) is a solution of

$$\begin{aligned} -\Delta _{{\bar{g}}}u + {\bar{K}} = (f+\lambda ) \textrm{e}^{2u} \end{aligned}$$
(4.23)

for some \(\lambda \in \mathbb {R}\) with \(E_f(u)< \frac{\lambda _\sharp A}{2}\), then \(0< \lambda < \lambda _\sharp \), and u is not a weakly stable solution of (4.23), so \(u \not = u_\lambda \).

Proof

Let \(\kappa _0>0\) be given as in Lemma 4.2 for \(\varepsilon = \lambda _\sharp >0\). Moreover, let

$$\begin{aligned} \kappa _1:= \max \left\{ \kappa _0, V(u_{\lambda _\sharp })\right\} \end{aligned}$$

with V defined in (4.19). Next, let \(u \in {\mathcal {C}}_{A}\) be a solution of (4.23) for some \(\lambda \in \mathbb {R}\) with \(E_f(u)< \frac{\lambda _\sharp A}{2}\). From Lemma 4.2, we then deduce that \(0< \lambda < \lambda _\sharp \), and by Proposition 4.5 (iii) we have \(u \not = u_\lambda \). Since \(u_\lambda \) is the unique weakly stable solution of (4.23), it follows that u is not weakly stable. \(\square \)

Corollary 4.7

Let \(p>2\), \(f \in C^\infty (M)\) be nonconstant with \(\max _{x\in M} f (x)= 0\), and let \(\lambda _\sharp >0\) be given as in Proposition 4.5. Then there exists \(\kappa >0\) with the property that for every \(A \ge \kappa \) the set

$$\begin{aligned} \tilde{{\mathcal {C}}}:=\left\{ u_0\in {\mathcal {C}}_{p,A} \mid E_f(u_0)<\frac{\lambda _\sharp A}{2}\right\} \end{aligned}$$

is nonempty, and for every \(u_0\in \tilde{{\mathcal {C}}}\) the global solution \(u \in C([0,\infty )\times M)\cap C([0,\infty );H^1(M,{\bar{g}}))\cap C^\infty ((0,\infty )\times M)\) of the initial value problem (2.9), (2.10) converges, as \(t \rightarrow \infty \) suitably, to a solution \(u_\infty \) of the static problem (4.23) for some \(\lambda \in (0,\lambda _\sharp )\) which is not weakly stable and hence no local minimiser of \(E_{f_\lambda }\).

Proof

Let \(\kappa _1>0\) be given by Corollary 4.6. By (4.2), there exists \(\kappa \ge \kappa _1>0\) with \(m_{f,A}<\frac{\lambda _\sharp A}{4}\) for fixed \(A>\kappa \). Consequently, there exists \(u_0 \in {\mathcal {C}}_{A}\cap W^{2,p}(M,{\bar{g}})\) with \(E_f(u_0)< \frac{\lambda _\sharp A}{2}\). By Theorem 3.1, the global solution \(u \in C([0,\infty )\times M)\cap C([0,\infty );H^1(M,{\bar{g}}))\cap C^\infty ((0,\infty )\times M)\) of the initial value problem (2.9), (2.10) converges, as \(t \rightarrow \infty \) suitably, to a solution \(u_\infty \in {\mathcal {C}}_{A}\) of the static problem (4.23) for some \(\lambda \in \mathbb {R}\), whereas \(E_f(u_\infty ) \le E_f(u_0) < \frac{\lambda _\sharp A}{2}\). Consequently, \(\lambda \in (0,\lambda _\sharp )\) by Corollary 4.6, and \(u_\infty \) is not weakly stable. \(\square \)

5 Proof of the main results

5.1 Preliminaries

In the following, we consider, for fixed \(T>0\), the spaces

$$\begin{aligned} L^p_tL^r_x:=L^p([0,T];L^r(M,{\bar{g}}))\quad \text {and}\quad L^p_tH^q_x:=L^p([0,T];H^q(M,{\bar{g}})). \end{aligned}$$

We stress that, although these spaces depend on T, we prefer to use a T-independent notation. We also note that, since \(T<\infty \) and \({\text {vol}}_{{\bar{g}}}=1\), we have \(L^q_tL^r_x\subset L^s_tL^p_x\) for \(p,q,r,s\in [1,\infty ]\) with \(q\ge s\), \(r\ge p\).

Lemma 5.1

(Sobolev inequality) There exists a constant \(C_S>0\) such that for every \(T \le 1\) and every \(\rho \in L^\infty _t H^1_x\) we have

$$\begin{aligned} \Vert \rho \Vert ^2_{L^4_tL^4_x}\le C_S(\Vert \rho \Vert ^2_{L^\infty _tL^2_x}+\Vert \nabla _{{\bar{g}}}\rho \Vert ^2_{L^2_tL^2_x})<\infty . \end{aligned}$$
(5.1)

Proof

By Lemma 4.4, applied with \(r=4\), there exists a constant \(C_{\text {GNL}}=C_{\text {GNL}}(4)>0\) with the property that, for all \(T\le 1\),

$$\begin{aligned} \Vert \rho \Vert ^4_{L^4_tL^4_x}&=\int _0^T\Vert \rho (t)\Vert ^4_{L^4(M,{\bar{g}})}dt \le C_{\text {GNL}}\int _0^T\Vert \rho (t)\Vert ^2_{L^2(M,{\bar{g}})}\Vert \rho (t)\Vert ^2_{H^1(M,{\bar{g}})}dt\\&\le C_{\text {GNL}}\Vert \rho \Vert ^2_{L^\infty _tL^2_x}\int _0^T(\Vert \rho (t)\Vert _{L^2(M,{\bar{g}})}^2+\Vert \nabla _{{\bar{g}}}\rho (t)\Vert ^2_{L^2(M,{\bar{g}})})dt\\&\le C_{\text {GNL}}\cdot T\,\Vert \rho \Vert ^4_{L^\infty _tL^2_x}+C_{\text {GNL}}\Vert \rho \Vert ^2_{L^\infty _tL^2_x}\Vert \nabla _{{\bar{g}}}\rho \Vert ^2_{L^2_tL^2_x}\\&\le C_{\text {GNL}}\left( \Vert \rho \Vert ^4_{L^\infty _tL^2_x}+\Vert \rho \Vert ^2_{L^\infty _tL^2_x}\Vert \nabla _{{\bar{g}}}\rho \Vert ^2_{L^2_tL^2_x}\right) \\&\le C_{\text {GNL}}\left( \frac{3}{2}\Vert \rho \Vert ^4_{L^\infty _tL^2_x}+\frac{1}{2}\Vert \nabla _{{\bar{g}}}\rho \Vert ^4_{L^2_tL^2_x}\right) \\&\le \frac{3 C_{\text {GNL}}}{2}\left( \Vert \rho \Vert ^2_{L^\infty _tL^2_x}+\Vert \nabla _{{\bar{g}}}\rho \Vert ^2_{L^2_tL^2_x}\right) ^2. \end{aligned}$$

Hence the first inequality in (5.1) holds with \(C_S = \Bigl (\frac{3 C_{\text {GNL}}}{2}\Bigr )^{\frac{1}{2}}\). Moreover, since T is finite, \(\rho \in L^\infty _tH^1_x\) implies that \(\rho \in L^p_tH^1_x\) for all \(p\in [1,\infty ]\) which shows that the RHS in (5.1) is finite. \(\square \)

Now we can turn to the proofs of the main results.

5.2 Short-time existence

Let \(A>0\) and \(p>2\) be fixed. We are looking for a short-time solution of (2.9), (2.10) with initial value \(u_0 \in {\mathcal {C}}_{p,A}\), where \({\mathcal {C}}_{p,A}\) is defined in (3.1). Using the Gauss Eq. (1.1) we can rewrite (2.9), (2.10) in the following way:

$$\begin{aligned} \partial _tu(t)&=f-K_{g(t)}-\alpha (t)\nonumber \\&=\textrm{e}^{-2u(t)}\Delta _{{\bar{g}}}u(t)- \textrm{e}^{-2u(t)}{\bar{K}} + f -\alpha (t)\end{aligned}$$
(5.2)
$$\begin{aligned}&=\textrm{e}^{-2u(t)}\Delta _{{\bar{g}}}u(t) + {\bar{K}} \left( \frac{1}{A}-\textrm{e}^{-2u(t)}\right) +f-\frac{1}{A}\int _Mf\textrm{e}^{2u(t)} d\mu _{{\bar{g}}};\nonumber \\ u(0)&=u_0\in {\mathcal {C}}_{p,A}, \end{aligned}$$
(5.3)

where

$$\begin{aligned} \alpha (t)=\frac{1}{A}\left( \int _M fd\mu _{g(t)}-{\bar{K}}\right) . \end{aligned}$$

To find a solution of (5.2), (5.3) on a short time interval, we consider the linear equation

$$\begin{aligned} \partial _tu(t)&=\textrm{e}^{-2v(t)}\Delta _{{\bar{g}}}u(t)+{\bar{K}}\left( \frac{1}{A}-\textrm{e}^{-2v(t)}\right) +f-\frac{1}{A}\int _Mf \textrm{e}^{2v(t)}d\mu _{{\bar{g}}}; \end{aligned}$$
(5.4)
$$\begin{aligned} u(0)&=u_0\in {\mathcal {C}}_{p,A}, \end{aligned}$$
(5.5)

and use a fixed point argument in the Banach space

$$\begin{aligned} (X,\Vert \cdot \Vert _X):=(C([0,T]\times M),\Vert \cdot \Vert _{L^\infty ([0,T]\times M)}). \end{aligned}$$
(5.6)

For this we first observe that Eq. (5.4) is strongly parabolic for \(v \in X\). Furthermore, since \(p>2\) and M is compact, we have \(u_0\in {\mathcal {C}}_{p,A} \subset H^2(M,{\bar{g}})\), and therefore \(u_0\in C(M)\).

For the fixed point argument we fix \(u_0\in {\mathcal {C}}_{p,A}\) and set

$$\begin{aligned} R=R(u_0):=\Vert u_0\Vert _{L^\infty (M,{\bar{g}})}+1. \end{aligned}$$

For fixed \(T>0\) and \(v\in X\), we then get, by Proposition 6.2 in the appendix, a unique solution \(u_v\in W^{2,1}_p((0,T)\times M)\) of (5.4) which satisfies (5.5) in the initial trace sense. Here \(W^{2,1}_p((0,T)\times M)\) denotes the space of functions \(u \in L^p((0,T)\times M)\) which have weak derivatives \(Du, D^2u\) and \(\partial _t u\) in \(L^p((0,T)\times M)\), so this space is compactly embedded in C(X) by Lemma 6.1 in the appendix. On \(X_R=\{ U\in X\mid \Vert U\Vert _X\le R\}\), we now define the function \(\Phi \) as follows: for \(v\in X_R\), let \(\Phi (v)=:u_v\) be the unique solution of (5.4), (5.5). First, we show that \(\Phi :X_R\rightarrow X_R\) if \(T>0\) is chosen small enough.

Lemma 5.2

If \(T>0\) is fixed with

$$\begin{aligned} T \le \left( |{\bar{K}}|\textrm{e}^{2(\Vert u_0\Vert _{L^\infty (M,{\bar{g}})}+1)}+\Vert f\Vert _{L^\infty (M,{\bar{g}})}\left( 1+ \frac{\textrm{e}^{2(\Vert u_0\Vert _{L^\infty (M,{\bar{g}})}+1)}}{A}\right) \right) ^{-1} \end{aligned}$$
(5.7)

and \(v\in X_R\), then \(\Phi (v)\in X_R\).

Proof

With Proposition 6.4 (ii) we directly get

$$\begin{aligned} \Vert \Phi (v)\Vert _{X} = \Vert u_v\Vert _{X} \le \Vert u_0^+\Vert _{L^\infty (M,{\bar{g}})}+T d_T \end{aligned}$$
(5.8)

where

$$\begin{aligned} d_T\le & {} |{\bar{K}}|\textrm{e}^{2\Vert v\Vert _{X}}+\Vert f\Vert _{L^\infty (M,{\bar{g}})}+\frac{\Vert f\Vert _{L^\infty (M,{\bar{g}})}\textrm{e}^{2\Vert v\Vert _{X}}}{A}\\\le & {} |{\bar{K}}|\textrm{e}^{2R}+\Vert f\Vert _{L^\infty (M,{\bar{g}})}\Bigl (1+ \frac{\textrm{e}^{2R}}{A}\Bigr ), \end{aligned}$$

hence

$$\begin{aligned} \Vert \Phi (v)\Vert _{X}&\le T\left( |{\bar{K}}|\textrm{e}^{2R}+\Vert f\Vert _{L^\infty (M,{\bar{g}})}\left( 1+ \frac{\textrm{e}^{2R}}{A}\right) \right) +\Vert u^+_0\Vert _{L^\infty (M,{\bar{g}})}\\&\le 1+ \Vert u_0\Vert _{L^\infty (M,{\bar{g}})} = R, \end{aligned}$$

by (5.7) and since \(R=\Vert u_0\Vert _{L^\infty (M,{\bar{g}})}+1\), which shows the claim. \(\square \)

We now use Schauder’s fixed point Theorem [17] to show the following proposition.

Proposition 5.3

If \(u_0 \in {\mathcal {C}}_{p,A}\subset W^{2,p}(M,{\bar{g}})\) and \(T>0\) is fixed with (5.7), then there exists a short-time solution \(u \in X \cap C^{\infty }((0,T)\times M)\) of (5.2), (5.3).

Moreover, any such solution satisfies \(u \in C([0,T), H^1(M,{\bar{g}}))\).

Proof

Step 1: First we recall Schauder’s Theorem: If H is a nonempty, convex, and closed subset of a Banach space B and F is a continuous mapping of H into itself such that F(H) is a relatively compact subset of H, then F has a fixed point.

In our case, \(B\hat{=}X=C([0,T] \times M)\), \(H\hat{=}X_R=\{u\in X\mid \Vert u\Vert _X=\Vert u\Vert _{C_tC_x}\le R\}\), and \(F\hat{=}\Phi \). So to show the existence of a fixed point of \(\Phi \) in \(X_R\), it remains to show that

  1. 1.

    \(\Phi : X_R\rightarrow X_R\) is continuous and

  2. 2.

    \(\Phi (X_R)\subset X_R\) is relatively compact.

First, we show that \(\Phi :X_R\rightarrow X_R\) is continuous. For this, let \(v\in X_R\), and let \((v_n)_{n}\subset X_R\) be a sequence with \(\Vert v_n-v\Vert _X\rightarrow 0\). Moreover, let \(u= \Phi (v)\) and \(u_n= \Phi (v_n)\) for \(n \in \mathbb {N}\). By Proposition 6.2, we know that

$$\begin{aligned}{} & {} \Vert u_n\Vert _{W^{2,1}_p}\le C(\Vert u_0\Vert _{W^{2,p}(M,{\bar{g}})}+\Vert d_n\Vert _{L^p_tL^p_x})\\{} & {} \qquad \text {and}\qquad \Vert u\Vert _{W^{2,1}_p}\le C(\Vert u_0\Vert _{W^{2,p}(M,{\bar{g}})}+\Vert d\Vert _{L^p_tL^p_x})\end{aligned}$$

for \(n \in \mathbb {N}\) with

$$\begin{aligned} d_n(t):= & {} {\bar{K}}\left( \frac{1}{A}-\textrm{e}^{-2v_n(t)}\right) +f-\frac{1}{A}\int _Mf\textrm{e}^{2v_n(t)}d\mu _{{\bar{g}}}\qquad \text {and}\qquad \\ d(t):= & {} {\bar{K}}\left( \frac{1}{A}-\textrm{e}^{-2v(t)}\right) +f-\frac{1}{A}\int _Mf\textrm{e}^{2v(t)}d\mu _{{\bar{g}}}. \end{aligned}$$

Since \(v_n\rightarrow v\) in X, we have \(\textrm{e}^{\pm 2v_n}\rightarrow \textrm{e}^{\pm 2v}\) and therefore also \(d_n\rightarrow d\) in X, which also implies that \(d_n\rightarrow d\) in \(L^p_tL^p_x\) for all p. Moreover, the difference \(u_n-u= \Phi (v_n)-\Phi (v)\) fulfils the equation

$$\begin{aligned} \partial _t(u_n-u)(t)&=\textrm{e}^{-2v_n(t)}\Delta _{{\bar{g}}}u_n(t)+d_n(t)-\textrm{e}^{-2v(t)}\Delta _{{\bar{g}}}u(t)- d(t)\\&=\textrm{e}^{-2v_n(t)}\Delta _{{\bar{g}}}(u_n-u)(t)+(\textrm{e}^{-2v_n(t)}-\textrm{e}^{-2v(t)})\Delta _{{\bar{g}}}u(t)+d_n(t)-d(t). \end{aligned}$$

Since also \([u_n-u](0)=0\), we have, again by Proposition 6.2,

$$\begin{aligned} \Vert u_n-u\Vert _{W^{2,1}_p}&\le C\Vert (\textrm{e}^{-2v_n}-\textrm{e}^{-2v})\Delta _{{\bar{g}}}u+d_n-d\Vert _{L^p_tL^p_x}\\&\le C\left( \Vert \textrm{e}^{-2v_n}-\textrm{e}^{-2v}\Vert _{X}\Vert \Delta _{{\bar{g}}}u\Vert _{L^p_tL^p_x}+\Vert d_n-d\Vert _{L^p_tL^p_x}\right) \end{aligned}$$

Since \(\Vert \Delta _{{\bar{g}}}u\Vert _{L^p_tL^p_x}\) is finite, it thus follows that \(\Phi (v_n)-\Phi (v)= u_n-u \rightarrow 0\) in \(W^{2,1}_p\) and therefore also \(\Phi (v_n)-\Phi (v) \rightarrow 0\) in X, since \(W^{2,1}_p\) is embedded in X by Lemma 6.1. Together with 5.2, this shows the continuity of \(\Phi :X_R\rightarrow X_R\).

Next, we show that \(\Phi (X_R)\) is relatively compact. For this let \((u_n)_{n\in \mathbb {N}}\subset \Phi (X_R)\) be an arbitrary sequence in \(\Phi (X_R)\), and let \(v_n\in X_R\) with \(\Phi (v_n)=u_n\) for \(n \in \mathbb {N}\). So, by definition of \(\Phi \) and by Proposition 6.2, we see that

$$\begin{aligned} \Vert u_n\Vert _{W^{2,1}_p}&\le C \left( \Vert u_0\Vert _{W^{2,p}(M,{\bar{g}})}+\frac{T|{\bar{K}}|}{A}+\Vert {\bar{K}}\textrm{e}^{-2v_n}\Vert _{L^p_tL^p_x}+\Vert f\Vert _{L^p_tL^p_x}+\left\| \frac{1}{A}\int _Mf\textrm{e}^{2v_n}d\mu _{{\bar{g}}}\right\| _{L^p_tL^p_x}\right) \\&\le C\left( \Vert u_0\Vert _{W^{2,p}(M,{\bar{g}})}+\frac{T|{\bar{K}}|}{A}+|{\bar{K}}|\textrm{e}^{2R}+T\Vert f\Vert _{L^\infty (M,{\bar{g}})}+\frac{T}{A}\Vert f\Vert _{L^\infty (M,{\bar{g}})}\textrm{e}^{2R}\right) \end{aligned}$$

for \(n \in \mathbb {N}\). Hence \((u_n)_{n\in \mathbb {N}}\) is uniformly bounded in \(W^{2,1}_p((0,T)\times M)\). Using now that \(W^{2,1}_p((0,T)\times M)\) is compactly embedded in X by Lemma 6.1, we conclude the claim.

We have thus proved that \(\Phi \) has a fixed point u in \(X_R\), which then is a (strong) solution \(u \in W^{2,1}_p((0,T)\times M)\) of (5.2), (5.3).

Step 2: We now show that \(u \in C^\infty ((0,T)\times M)\). To see this, we first note the trivial fact that \(u \in W^{2,1}_p((0,T)\times M)\) is a strong solution of (5.4), (5.5) with \(v = u\). Since then \(v \in W^{2,1}_p((0,T)\times M) \subset C^\alpha ([0,T]\times M)\), [14, Theorems 5.9 and 5.10] imply the existence of a classical solution \({{\tilde{u}}} \in X \cap C^{2+\alpha ',1+\alpha '}_{loc}((0,T)\times M)\) of (5.4), (5.5) with \(v = u\) for some \(\alpha '>0\). Here \(C^{2+\alpha ',1+\alpha '}_{loc}((0,T)\times M)\) denotes the space of functions \(f \in C^{2,1}((0,T)\times M)\) with the property that \(\partial _t f\) and all derivatives up to second order of f with respect to \(x \in M\) are locally \(\alpha '\)-Hölder continuous. In particular, \({{\tilde{u}}} \in W^{2,1}_p((\varepsilon ,T-\varepsilon )\times M)\) for \(\varepsilon \in (0,T)\). The function \(w:= u- {{\tilde{u}}} \in W^{2,1}_p((\varepsilon ,T-\varepsilon )\times M)\) is then a strong solution of the initial value problem

$$\begin{aligned} \partial _t w(t)=\textrm{e}^{-2v(t)}\Delta _{{\bar{g}}}w(t) \quad \text {for }t \in (\varepsilon ,T-\varepsilon ), \qquad w(\varepsilon )= u(\varepsilon ,\cdot )-{{\tilde{u}}}(\varepsilon ,\cdot ). \end{aligned}$$

By Proposition 6.4 (ii) we then have \(|w| \le \Vert u(\varepsilon ,\cdot )-{{\tilde{u}}}(\varepsilon ,\cdot )\Vert _{L^\infty (M,{\bar{g}})}\) on \((\varepsilon ,T-\varepsilon )\times M\), whereas \(\Vert u(\varepsilon ,\cdot )-{{\tilde{u}}}(\varepsilon ,\cdot )\Vert _{L^\infty (M,{\bar{g}})} \rightarrow 0\) as \(\varepsilon \rightarrow 0\) by the continuity of u and \({{\tilde{u}}}\). It thus follows that \(u \equiv {{\tilde{u}}}\) on \((0,T)\times M)\), and therefore \(u \in C^{2+\alpha ',1+\alpha '}_{loc}((0,T)\times M)\). Since u solves (5.4), (5.5) with \(v = u \in C^{2+\alpha ',1+\alpha '}_{loc}((0,T)\times M)\), we can apply [14, Theorems 5.9] and the above argument again to get \(u \in C^{4+\alpha '',2+\alpha ''}_{loc}((0,T)\times M)\) for some \(\alpha ''>0\). Repeating this argument inductively, we get \(u \in C^{k,\frac{k}{2}}_{loc}((0,T)\times M)\) for every \(k>0\), and hence \(u \in C^\infty ((0,T)\times M)\).

Step 3: It remains to show that any solution \(u \in X \cap C^{\infty }((0,T)\times M)\) of (5.2), (5.3) also satisfies \(u \in C([0,T), H^1(M,{\bar{g}}))\). Since \(u \in C^{\infty }((0,T)\times M)\), only the continuity in \(t=0\) needs to be proved. Setting \(\phi (t)= \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2\) for \(t \in (0,T)\), we see that

$$\begin{aligned} \frac{1}{2}(\phi (t_2)-\phi (t_1))&= \frac{1}{2} \int _{t_1}^{t_2}\partial _t \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2\,dt \\ {}&=\int _{t_1}^{t_2}\int _{M} \Bigl (u(t) \partial _t u(t) + \nabla u(t) \nabla \partial _t u(t)\Bigr )d\mu _{{\bar{g}}}dt \\&=\int _{t_1}^{t_2}\int _{M} \Bigl (u(t) \partial _t u(t) - [\Delta u(t)] \partial _t u(t)\Bigr )d\mu _{{\bar{g}}}dt \end{aligned}$$

and therefore, by Hölder’s inequality,

$$\begin{aligned} \frac{1}{2}|\phi (t_2)-\phi (t_1)|&\le \int _{t_1}^{t_2} \int _{M}\bigl (|u||\partial _t u|+ |\Delta u||\partial _t u|\bigr )d\mu _{{\bar{g}}}dt\\&\le C \Vert \partial _t u\Vert _{L^p((0,T) \times M)} \bigl (\Vert u\Vert _{L^p((0,T) \times M)} + \Vert \Delta u\Vert _{L^p((0,T) \times M)} \bigr ) (t_2-t_1)^\beta \\&\le C \Vert u\Vert _{W^{1,2}_p((0,T) \times M)}(t_2-t_1)^\beta , \end{aligned}$$

for \(0<t_1<t_2<T\) with some \(\beta >0\) depending on \(p>2\), which implies that the function \(\phi \) is uniformly continuous and therefore bounded on (0, T).

We now assume by contradiction that u is not continuous at \(t=0\) with respect to the \(H^1(M,{\bar{g}})\)-norm. Then there exists a sequence \((t_n)_{n\in \mathbb {N}}\) in (0, T) and \(\varepsilon >0\) with \(t_n \rightarrow 0^+\) as \(n \rightarrow \infty \) and

$$\begin{aligned} \Vert u(t_n)-u_0\Vert _{{H^1(M,{\bar{g}})}} \ge \varepsilon \qquad \text {for all }n \in \mathbb {N}. \end{aligned}$$
(5.9)

Since \(\Vert u(t_n)\Vert _{H^1(M,{\bar{g}})}^2 = \phi (t_n)\) remains bounded as \(n \rightarrow \infty \), we conclude that, passing to a subsequence, the sequence \(u(t_n)\) converges weakly in \(H^1(M,{\bar{g}})\) and therefore strongly in \(L^2(M, {\bar{g}})\). Since the strong \(L^2\)-limit of \(u(t_n)\) must be \(u_0=u(0)\) as a consequence of the fact that \(u \in X\), we deduce that \(u(t_n) \rightharpoonup u_0\) weakly in \({H^1(M,{\bar{g}})}\) as \(n \rightarrow \infty \). Combining this information with Proposition 6.2 from the appendix, we deduce that

$$\begin{aligned} \limsup _{n \rightarrow \infty }\Vert u(t_n)\Vert _{H^1(M,{\bar{g}})}^2 \le \Vert u_0\Vert _{H^1(M,{\bar{g}})}^2 \le \liminf _{n \rightarrow \infty }\Vert u(t_n)\Vert _{H^1(M,{\bar{g}})}^2 \end{aligned}$$
(5.10)

and therefore \(\Vert u(t_n)\Vert _{H^1(M,{\bar{g}})} \rightarrow \Vert u_0\Vert _{H^1(M,{\bar{g}})}\). Note here that this part of Proposition 6.2 applies since u solves (5.4), (5.5) with \(v = u \in W^{2,1}_p((0,T)\times M) \subset C^\alpha ([0,T]\times M)\) for some \(\alpha >0\). From (5.10) and the uniform convexity of the Hilbert space \({H^1(M,{\bar{g}})}\), we conclude that \(u(t_n) \rightarrow u_0\) strongly in \(H^1(M,{\bar{g}})\), contrary to (5.9). \(\square \)

5.3 Uniqueness

We now show that the solution from Proposition 5.3 is unique.

Lemma 5.4

Let \(u_0 \in W^{2,p}(M,{\bar{g}})\), \(p>2\), and \(T>0\) be fixed with (5.7). Then the short-time solution of \(u \in X \cap C^{\infty }((0,T)\times M)\) of (5.2), (5.3) given by Proposition 5.3 is unique.

Proof

Let \(u_1, u_2 \in X \cap C^{\infty }((0,T)\times M)\) be two solutions of (5.2), (5.3). The difference \(u:= u_1-u_2 \in X \cap C^{\infty }((0,T)\times M)\) then fulfils

$$\begin{aligned} \begin{aligned} \partial _t u(t)&=\textrm{e}^{-2u_1(t)}\Delta _{{\bar{g}}}u_1(t)-\textrm{e}^{-2u_2(t)}\Delta _{{\bar{g}}}u_2(t)\\&\quad -{\bar{K}}(\textrm{e}^{-2u_1(t)}-\textrm{e}^{-2u_2(t)})-\frac{1}{A}\int _Mf(\textrm{e}^{2u_1(t)}-\textrm{e}^{2u_2(t)})d\mu _{{\bar{g}}} \\&= \textrm{e}^{-2u_1(t)}\Delta _{{\bar{g}}}u(t) + \Delta _{{\bar{g}}}u_2(t)\bigl ( \textrm{e}^{-2u_1(t)} -\textrm{e}^{-2u_2(t)}\bigr )\\&\quad -{\bar{K}}(\textrm{e}^{-2u_1(t)}-\textrm{e}^{-2u_2(t)})-\frac{1}{A}\int _Mf(\textrm{e}^{2u_1(t)}-\textrm{e}^{2u_2(t)})d\mu _{{\bar{g}}}\quad \text {for }t \in (0,T). \end{aligned}\nonumber \\ \end{aligned}$$
(5.11)

In the following, the letter C denotes different positive constants. Multiplying (5.11) with 2u and integrating over M gives

$$\begin{aligned}&\frac{d}{dt} \Vert u(t)\Vert _{L^2(M,{\bar{g}})}^2 \nonumber \\&= 2 \int _{M} u(t) \partial _t u(t) d\mu _{{\bar{g}}} \nonumber \\&=2\int _M\textrm{e}^{-2u_1(t)}u(t)\Delta _{{\bar{g}}}u(t)d\mu _{{\bar{g}}} \nonumber \\&\qquad + 2\int _Mu(t)\Delta _{{\bar{g}}}u_2(t)\bigl ( \textrm{e}^{-2u_1(t)} -\textrm{e}^{-2u_2(t)}\bigr )d\mu _{{\bar{g}}} \end{aligned}$$
(5.12)
$$\begin{aligned}&\qquad -2\int _M{\bar{K}}u(t)(\textrm{e}^{-2u_1(t)}-\textrm{e}^{-2u_2(t)})d\mu _{{\bar{g}}}-\frac{2}{A}\int _Mf(\textrm{e}^{2u_1(t)}-\textrm{e}^{2u_2(t)})d\mu _{{\bar{g}}}\int _Mu(t)d\mu _{{\bar{g}}}\nonumber \\&\quad \le 2 \int _M \textrm{e}^{-2u_1(t)} u(t) \Delta _{{\bar{g}}}u(t) + 2\int _M V(t,x)u^2(t) + 2\rho (t) \Vert u(t)\Vert _{L^2(M,{\bar{g}})} \int _M |u(t)|d\mu _{{\bar{g}}} \nonumber \\&\quad \le 2 \left( -\int _M \textrm{e}^{-2u_1(t)} |\nabla _{{\bar{g}}} u(t)|_{{\bar{g}}}^2 + 2 \int _M \textrm{e}^{-2u_1(t)} u(t) \langle \nabla _{{\bar{g}}} u_1(t),\nabla _{{\bar{g}}}u(t)\rangle _{{\bar{g}}}d\mu _{{\bar{g}}}\right) \nonumber \\&\qquad + 2\Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})} \Vert u(t)\Vert _{L^{2p'}(M,{\bar{g}})}^2 + C \Vert u(t)\Vert _{L^2(M,{\bar{g}})}^2 \nonumber \\&\quad \le C \Vert \nabla _{{\bar{g}}}u_1(t)\Vert _{L^4(M,{\bar{g}})} \Vert u(t)\Vert _{L^4(M,{\bar{g}})} \Vert \nabla _{{\bar{g}}}u(t)\Vert _{L^2(M,{\bar{g}})}\nonumber \\&\qquad + 2\Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})} \Vert u(t)\Vert _{L^{2p'}(M,{\bar{g}})}^2 +C \Vert u(t)\Vert _{L^2(M,{\bar{g}})}^2 \nonumber \\&\quad \le C \Bigl (\Vert u_1(t)\Vert _{H^2(M,{\bar{g}})} \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2 + 2\Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})} \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2 + \Vert u(t)\Vert _{L^2(M,{\bar{g}})}^2\Bigr ) \nonumber \\&\quad \le C \Bigl (\Vert u_1(t)\Vert _{H^2(M,{\bar{g}})} + 2\Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})}+1\Bigr ) \Vert u\Vert _{H^1(M,{\bar{g}})}^2, \end{aligned}$$
(5.13)

with functions \(V \in L^p((0,T) \times M) \cap C^\infty ((0,T) \times M)\) and \(\rho \in L^\infty (0,T)\). Here we used the Sobolev embeddings \(H^1(M,{\bar{g}}) \hookrightarrow L^\rho (M)\) for \(\rho \in [1,\infty )\). Multiplying (5.11) with \(-2\Delta u\) and integrating over M yields

$$\begin{aligned} \frac{d}{dt} \Vert \nabla _g u(t)\Vert _{L^2(M,{\bar{g}})}^2&= 2 \int _M \nabla u(t) \nabla \partial _t u(t)d\mu _{{\bar{g}}}= -2 \int _M \Delta _g u(t) \partial _t u(t)d\mu _{{\bar{g}}} \nonumber \\&\le - 2 \int _M \textrm{e}^{-2u_1(t)} |\Delta _{{\bar{g}}}u(t)|^2 d\mu _{{\bar{g}}} +2 \int _M V(t,x)|u(t)||\Delta u(t)|d\mu _{{\bar{g}}} \nonumber \\&\le -\kappa \Vert \Delta _{{\bar{g}}}u(t)\Vert _{L^2(M,{\bar{g}})}^2 + 2\Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})} \Vert u\Vert _{L^\alpha (M,{\bar{g}})} \Vert \Delta _g u\Vert _{L^2(M,{\bar{g}})} \nonumber \\&\le -\kappa \Vert \Delta _{{\bar{g}}}u(t)\Vert _{L^2(M,{\bar{g}})}^2 + \frac{1}{\kappa } \Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})}^2 \Vert u\Vert _{L^\alpha (M,{\bar{g}})}^2 + \kappa \Vert \Delta _g u\Vert _{L^2(M,{\bar{g}})}^2 \nonumber \\&= \frac{1}{\kappa } \Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})}^2 \Vert u\Vert _{L^\alpha (M,{\bar{g}})}^2 \le C \Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})}^2 \Vert u\Vert _{H^1(M,{\bar{g}})}^2, \end{aligned}$$
(5.14)

where we used first Hölder’s inequality with \(\alpha = \frac{2p}{p-2}\), then Young’s inequality and finally Sobolev embeddings again. Here we note that, by making \(C>0\) larger if necessary, we may assume that the constants are the same in (5.13) and (5.14). Combining these estimates gives

$$\begin{aligned} \frac{d}{dt} \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2 \le g(t) \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2 \qquad \text {for }t \in (0,T) \end{aligned}$$
(5.15)

with the function \(g \in L^1(0,T)\) given by \(g(t)= C \Bigl (\Vert u_1(t)\Vert _{H^2(M,{\bar{g}})} + 3\Vert V(t,\cdot )\Vert _{L^p(M,{\bar{g}})}+1\Bigr )\). Integrating and using the fact that \(u \in C([0,T), H^1(M,{\bar{g}}))\) by Proposition 5.3 with \(u(0)=u_1(0)-u_2(0)=0\), we see that

$$\begin{aligned} \Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2 \le \int _0^t g(s)\Vert u(s)\Vert _{H^1(M,{\bar{g}})}^2\,ds \qquad \text {for }t \in [0,T). \end{aligned}$$

It then follows from Gronwall’s inequality [3] that \(\Vert u(t)\Vert _{H^1(M,{\bar{g}})}^2 \equiv 0\) on [0, T), hence \(u_1 \equiv u_2\). \(\square \)

5.4 Global existence

Let \(f \in C^\infty (M)\), \(A>0\), \(p>2\) and \(u_0 \in {\mathcal {C}}_{p,A}\). In this section, we wish to show that the (unique) local solution

$$\begin{aligned} u\in C([0,T]\times M)\cap C([0,T],H^1(M,{\bar{g}}))\cap C^\infty ((0,T)\times M) \end{aligned}$$

of the initial value problem (5.2), (5.3) for small \(T>0\) can be extended to a global und uniformly bounded solution defined for all positive times.

We first need the following local boundedness property on open time intervals.

Lemma 5.5

Let, for some \(T>0\), \(u\in C([0,T)\times M)\cap C([0,T),H^1(M,{\bar{g}}))\cap C^\infty ((0,T)\times M)\) be a solution of (5.2), (5.3) on [0, T). Then we have

$$\begin{aligned} \sup _{t\in [0,T)}\Vert u(t)\Vert _{L^\infty (M,{\bar{g}})} \le {\mathcal {M}} \end{aligned}$$
(5.16)

with some \({\mathcal {M}}={\mathcal {M}}(\Vert u_0\Vert _{L^\infty (M,{\bar{g}})},\Vert f\Vert _{L^\infty (M,{\bar{g}})},T)>0\) which is increasing in all of its variables.

Proof

Since \({\bar{K}} <0\), we have

$$\begin{aligned} \partial _tu(t)= & {} \textrm{e}^{-2u(t)}\Delta _{{\bar{g}}}u(t)-\textrm{e}^{-2u(t)}{\bar{K}}+f-\alpha (t)\\= & {} \textrm{e}^{-2u(t)}\Delta _{{\bar{g}}}u(t)+\textrm{e}^{-2u(t)}|{\bar{K}}|+ f-\alpha (t) \qquad \text {for }t \in [0,T) \end{aligned}$$

by (5.2), where

$$\begin{aligned} |\alpha (t)| \le \alpha _0:=\Vert f\Vert _{L^\infty (M,{\bar{g}})}+\frac{|{\bar{K}}|}{A} \qquad \text {for }t \in [0,T) \end{aligned}$$

by (2.12). Hence the function \(v=-u\) satisfies

$$\begin{aligned} \partial _t v(t)= \textrm{e}^{2 v(t)}\Delta _{{\bar{g}}}v(t)-\textrm{e}^{2v(t)}|{\bar{K}}| -f+\alpha (t) \le \textrm{e}^{2 v(t)}\Delta _{{\bar{g}}}v(t)+ c \qquad \text {for }t \in (0,T) \end{aligned}$$

with \(c = \Vert f\Vert _{L^\infty (M,{\bar{g}})}+ \alpha _0\). Next, let \((T_k)_k \subset (0,T)\) be a sequence with \(T_k\rightarrow T\) for \(k\rightarrow \infty \). For fixed \(k \in \mathbb {N}\) the continuous function \(\textrm{e}^{2v}\) is then bounded from below by a positive constant on the compact set \([0,T_k]\times M\). Therefore Proposition 6.4 (ii) from the appendix implies that

$$\begin{aligned} v(t,x) \le \Vert u_0\Vert _{L^\infty (M,{\bar{g}})} + T_k c \qquad \text {for }(t,x) \in [0,T_k]\times M. \end{aligned}$$

Letting \(k \rightarrow \infty \), we deduce that

$$\begin{aligned} u(t,x) =- v(t,x) \ge - \Vert u_0\Vert _{L^\infty (M,{\bar{g}})} - T c \qquad \text {for }(t,x) \in [0,T)\times M. \end{aligned}$$
(5.17)

In order to derive an upper bound for u, we now observe that

$$\begin{aligned} \partial _tu(t)=\textrm{e}^{-2u(t)}\Delta _{{\bar{g}}}u(t)+\textrm{e}^{-2u(t)}|{\bar{K}}|+ f-\alpha (t)\\ \le \textrm{e}^{-2u(t)} \Delta _{{\bar{g}}}u(t) + \textrm{e}^{2(\Vert u_0\Vert _{L^\infty (M,{\bar{g}})} + T c)}+ c \end{aligned}$$

on M for \(t \in [0,T)\). Applying Proposition 6.4 (ii) in the same way as above therefore gives

$$\begin{aligned} u(t,x) \le \Vert u_0\Vert _{L^\infty (M,{\bar{g}})} + T \Bigl (\textrm{e}^{2(\Vert u_0\Vert _{L^\infty (M,{\bar{g}})} + T c)}+ c\Bigr ). \end{aligned}$$
(5.18)

Combining (5.17) and (5.18) yields

$$\begin{aligned} \begin{aligned}&\sup _{\begin{array}{c} t\in [0,T)\\ x\in M \end{array}}|u(t,x)| \le {\mathcal {M}}\quad \text {with}\\&\quad {\mathcal {M}}={\mathcal {M}}(\Vert u_0\Vert _{L^\infty }(M,{\bar{g}}),\Vert f\Vert _{L^\infty (M,{\bar{g}})},T):= \Vert u_0\Vert _{L^\infty } + T \Bigl (\textrm{e}^{2(\Vert u_0\Vert _{L^\infty (M,{\bar{g}})} + T c)}+ c\Bigr ), \end{aligned} \end{aligned}$$
(5.19)

as claimed in (5.18). \(\square \)

Corollary 5.6

The initial value problem (5.2), (5.3) admits a unique global solution \(u\in C([0,\infty )\times M)\cap C([0,\infty ),H^1(M,{\bar{g}}))\cap C^\infty ((0,\infty )\times M)\).

Proof

This follows from Proposition 5.3, Lemma 5.4 and Lemma 5.5 by a standard continuation argument using condition  (5.7). \(\square \)

In the next lemma, with the help of (2.17), we turn (5.16) into a uniform estimate for all time.

Lemma 5.7

Let u be the global, smooth solution of the initial value problem (5.2), (5.3). Then we have

$$\begin{aligned} \sup _{t>0}\Vert u(t)\Vert _{L^\infty (M,{\bar{g}})}\le {\mathcal {N}} \end{aligned}$$

with some \({\mathcal {N}}={\mathcal {N}}(u_0,\Vert f\Vert _{L^\infty (M,{\bar{g}})})>0\) which is increasing in its second variable.

Proof

We argue similarly as in the proof of [18, Lemma 2.5].

By using the fact that u(t) is a volume preserving solution of (5.2) with \(u(0)=u_0\in {\mathcal {C}}_{p,A}\) and therefore \(\int _M\textrm{e}^{2u(t)}d\mu _{{\bar{g}}}\equiv A\), we get with (2.4) and the fact that \({\bar{K}}<0\) that

$$\begin{aligned} \begin{aligned} E_{f}(u(t))&=\frac{1}{2}\Vert \nabla _{{\bar{g}}}u(t)\Vert ^2_{L^2(M, {\bar{g}})}+\int _M{\bar{K}}u(t)d\mu _{{\bar{g}}}-\frac{1}{2}\int _Mf\textrm{e}^{2u(t)}d\mu _{{\bar{g}}}\\&\ge \frac{{\bar{K}}}{2}\int _M2u(t)d\mu _{{\bar{g}}}-\frac{1}{2}\int _Mf\textrm{e}^{2u(t)}d\mu _{{\bar{g}}} \ge \frac{{\bar{K}}}{2}\log (A)-\frac{A}{2}\Vert f\Vert _{L^\infty (M,{\bar{g}})}>-\infty . \end{aligned} \end{aligned}$$
(5.20)

For the function

$$\begin{aligned} t \mapsto F(t):=\int _M|\partial _tu(t)|^2d\mu _{g(t)}=\int _M|\partial _tu(t)|^2\textrm{e}^{2u(t)}d\mu _{{\bar{g}}}, \end{aligned}$$
(5.21)

we then obtain, by combining (5.20) with (2.17), the estimate

$$\begin{aligned} \int _0^\infty F(t)dt= & {} \lim _{T \rightarrow \infty } \int _0^T\int _M|\partial _tu(t)|^2d\mu _{g(t)}dt\nonumber \\\le & {} E_{f}(u_0)+\frac{|{\bar{K}}|}{2}|\log (A)|+\frac{A}{2}\Vert f\Vert _{L^\infty (M,{\bar{g}})}. \end{aligned}$$
(5.22)

Hence, for any \(T>0\) we find \(t_T\in [T,T+1]\) such that

$$\begin{aligned} F(t_T)&=\inf _{t\in (T,T+1)}F(t)\le E_{f}(u_0)+\frac{|{\bar{K}}|}{2}|\log (A)|+\frac{A}{2}\Vert f\Vert _{L^\infty (M,{\bar{g}})} \nonumber \\&\le \frac{1}{2} \Vert \nabla u_0\Vert _{L^2(M,{\bar{g}})}^2 + |{\bar{K}}| \bigl (\frac{1}{2}|\log (A)|+ \Vert u\Vert _{L^1(M,{\bar{g}})}\bigr ) + A \Vert f\Vert _{L^\infty (M,{\bar{g}})} \nonumber \\&= d_1 + d_2 \Vert f\Vert _{L^\infty (M,{\bar{g}})} \end{aligned}$$
(5.23)

with constants \(d_i = d_i(u_0)>0\). Here we used (2.6).

So, at time \(t_T\) we get with (2.7), Hölders inequality, Young’s inequality, (2.15), and (5.23) that

$$\begin{aligned} \begin{aligned}&\Vert \Delta _{{\bar{g}}}u(t_T)\Vert _{L^{\frac{3}{2}}(M,{\bar{g}})}\\&\quad \le \Vert \textrm{e}^{2u(t_T)}\partial _tu(t_T)\Vert _{L^{\frac{3}{2}}(M,{\bar{g}})}+\Vert {\bar{K}}\Vert _{L^{\frac{3}{2}}(M,{\bar{g}})}+\Vert \textrm{e}^{2u(t_T)}f\Vert _{L^{\frac{3}{2}}(M,{\bar{g}})}+\Vert \textrm{e}^{2u(t_T)}\alpha (t_T)\Vert _{L^{\frac{3}{2}}(M,{\bar{g}})}\\&\quad \le \Vert \textrm{e}^{u(t_T)}\Vert _{L^6(M,{\bar{g}})}F(t_T)^{\frac{1}{2}}+|{\bar{K}}|+ \Vert f\Vert _{L^\infty (M,{\bar{g}})} \Bigl (\int _M\textrm{e}^{3u(t_T)}d\mu _{{\bar{g}}}\Bigr )^{\frac{2}{3}}+ |\alpha (t_T)| \Bigl (\int _M\textrm{e}^{3u(t_T)} d\mu _{{\bar{g}}}\Bigr )^{\frac{2}{3}}\\&\quad \le \frac{1}{2} \Vert \textrm{e}^{u(t_T)}\Vert _{L^6(M,{\bar{g}})}^2 + \frac{1}{2} F(t_T) +|{\bar{K}}|+ \frac{1}{3}\Bigl ( \Vert f\Vert _{L^\infty (M,{\bar{g}})}^3 + |\alpha (t_T)|^3\Bigr )+\frac{4}{3} \int _M\textrm{e}^{3u(t_T)} d\mu _{{\bar{g}}}\\&\quad \le \frac{1}{2} \Bigl (\nu _0(u_0,6) \textrm{e}^{\nu _1(u_0,6) \Vert f\Vert _{L^\infty (M,{\bar{g}})}}\Bigr )^{1/3} + \frac{1}{2} \Bigl (d_1 + d_2 \Vert f\Vert _{L^\infty (M,{\bar{g}})}\Bigr ) +|{\bar{K}}|\\&\qquad + \frac{1}{3}\Bigl ( \Vert f\Vert _{L^\infty (M,{\bar{g}})}^3 + |\alpha (t_T)|^3\Bigr )+\frac{4}{3} \nu _0(u_0,3) \textrm{e}^{\nu _1(u_0,3) \Vert f\Vert _{L^\infty (M,{\bar{g}})}}\\&\quad \le d_3 \textrm{e}^{d_4 \Vert f\Vert _{L^\infty (M,{\bar{g}})}} \end{aligned} \end{aligned}$$
(5.24)

with constants \(d_i=d_i(u_0)\), \(i\in \{3,4\}\). Here the constants \(\nu _i(u_0,3)\), \(i\in \{0,1\}\) are given in (2.15).

Furthermore, with Sobolev’s embedding theorem we have \(W^{2,\frac{3}{2}}(M)\subset C^{0,\frac{2}{3}} \subset L^\infty (M,{\bar{g}})\). Therefore we get with Poincaré’s inequality, the Calderón–Zygmund inequality for closed surfaces, and with (5.24) that

$$\begin{aligned} \begin{aligned} \Vert u(t_T)-{\bar{u}}(t_T)\Vert ^{\frac{3}{2}}_{L^\infty (M,{\bar{g}})}&\le d_5\Vert u(t_T)-{\bar{u}}(t_T)\Vert ^{\frac{3}{2}}_{W^{2,\frac{3}{2}}(M,{\bar{g}})}\le d_6 \Vert \nabla _{{\bar{g}}}^2u(t_T)\Vert ^{\frac{3}{2}}_{L^{\frac{3}{2}}(M,{\bar{g}})}\\&\le d_7 \Vert \Delta _{{\bar{g}}}u(t_T)\Vert ^{\frac{3}{2}}_{L^{\frac{3}{2}}(M,{\bar{g}})}\le d_8 \textrm{e}^{d_9 \Vert f\Vert _{L^\infty (M,{\bar{g}})}}. \end{aligned} \end{aligned}$$
(5.25)

with constants \(d_i>0\), \(i\in \{5,6,7\}\) and \(d_i=d_i(u_0)>0\), \(i\in \{8,9\}\). With (2.14) we therefore obtain the uniform bound

$$\begin{aligned} \Vert u(t_T)\Vert _{L^\infty (M,{\bar{g}})}\le d_8 \textrm{e}^{d_9 \Vert f\Vert _{L^\infty (M,{\bar{g}})}} +\max \left\{ |m_0|+ m_1 \Vert f\Vert _{L^{\infty }(M)},\frac{1}{2}|\log (A)|\right\} . \end{aligned}$$
(5.26)

Upon shifting time by \(t_T\), we therefore get from Lemma 5.5

$$\begin{aligned} \begin{aligned}&\sup _{s\in [T+1,T+2]}\Vert u(s)\Vert _{L^\infty (M,{\bar{g}})} \\&\quad \le \sup _{s\in [t_T,t_T+3)}\Vert u(s)\Vert _{L^\infty (M,{\bar{g}})} \le {\mathcal {M}}(\Vert u(t_T)\Vert _{L^\infty (M,{\bar{g}})},\Vert f\Vert _{L^\infty (M,{\bar{g}})},3)\\&\quad \le {\mathcal {M}}\Bigl (d_8 \textrm{e}^{d_9 \Vert f\Vert _{L^\infty (M,{\bar{g}})}} +\max \left\{ |m_0|+ m_1 \Vert f\Vert _{L^{\infty }(M,{\bar{g}})},\frac{1}{2}|\log (A)|\right\} ,\Vert f\Vert _{L^\infty (M,{\bar{g}})},3\Bigr )\\&\quad =:{\mathcal {N}}(u_0,\Vert f\Vert _{L^\infty (M,{\bar{g}})}). \end{aligned}\nonumber \\ \end{aligned}$$
(5.27)

Since \({\mathcal {M}}\) is increasing in its first and second variables by Lemma 5.5, we see that \({\mathcal {N}}\) is increasing in \(\Vert f\Vert _{L^\infty (M,{\bar{g}})}\), as claimed. Since \(T>0\) was arbitrary, the claim follows. \(\square \)

5.5 Convergence of the flow

Let \(f \in C^\infty (M)\), \(A>0\), \(p>2\) and \(u_0 \in {\mathcal {C}}_{p,A}\) as before, and let u denote the global, smooth solution of the initial value problem (5.2), (5.3). In this section we shall show that for a suitable sequence \(t_l\rightarrow \infty \), \(l\rightarrow \infty \), the associated sequence of metrics \(g(t_l)\) tends to a limit metric \(g_\infty =\textrm{e}^{2u_\infty }{\bar{g}}\) with Gauss curvature \(K_{g_\infty }\), which then implies that \(K_{g_\infty }=f-\alpha ^\infty \) with a constant \(\alpha ^\infty \). Afterwards, we shall have a closer look at this constant \(\alpha ^\infty \).

By (5.22), we know that, for a suitable sequence \(t_l\rightarrow \infty \), \(l\rightarrow \infty \) we have

$$\begin{aligned} \int _M|\partial _tu(t_l)|^2d\mu _{g(t_l)}=\int _M|f-K_{g_l}-\alpha (t_l)|^2d\mu _{g(t_l)}\rightarrow 0\quad \text {for }l\rightarrow \infty . \end{aligned}$$
(5.28)

We can strengthen this observation as follows.

Lemma 5.8

For \(F(t)=\int _M|\partial _tu(t)|^2d\mu _{g(t)}\) as above, we have \(F(t)\rightarrow 0\) for \(t\rightarrow \infty \).

Proof

First we consider the evolution equation of the curvature \(K_{g(t)}\) and of \(\alpha (t)\). By the Gauss Eq. (1.1) and (5.2) we have

$$\begin{aligned} \partial _tK_{g(t)}= & {} \partial _t(-\textrm{e}^{-2u(t)}\Delta _{{\bar{g}}}u(t)+\textrm{e}^{-2u(t)}{\bar{K}})\nonumber \\= & {} -2\partial _tu(t)K_{g(t)}-\Delta _{g(t)}\partial _tu(t) \nonumber \\= & {} 2K_{g(t)}(K_{g(t)}-f+\alpha (t))+\Delta _{g(t)}(K_{g(t)}-f+\alpha (t)) \nonumber \\= & {} 2(K_{g(t)}-f+\alpha (t))^2+2(f-\alpha (t))(K_{g(t)}-f \nonumber \\{} & {} +\alpha (t))+\Delta _{g(t)}(K_{g(t)}-f+\alpha (t)) \nonumber \\= & {} 2(\partial _t u(t))^2-2(f-\alpha (t))\partial _t u(t)- \Delta _{g(t)}\partial _t u(t)\nonumber \\ \end{aligned}$$
(5.29)

for \(t>0\). Moreover, by (2.11) we have

$$\begin{aligned} \frac{d}{dt}\alpha (t)=\frac{2}{A}\int _Mf\textrm{e}^{2u(t)}\partial _tu(t)d\mu _{{\bar{g}}}=\frac{2}{A}\int _Mf \partial _tu(t)d\mu _{g(t)}. \end{aligned}$$
(5.30)

Combining (5.2), (5.29) and (5.30), we arrive at

$$\begin{aligned} \begin{aligned} \partial _{tt}u(t)&= \partial _t \bigl (f - K_{g(t)}-\alpha (t)\bigr )\\&=-2(\partial _t u(t))^2+2(f-\alpha (t))\partial _t u(t)+ \Delta _{g(t)}\partial _t u(t)+\frac{2}{A}\int _Mf \partial _tu(t)d\mu _{g(t)}. \end{aligned} \end{aligned}$$
(5.31)

We therefore get, using (2.12), that

$$\begin{aligned} \begin{aligned} \frac{1}{2}\frac{d}{dt}F(t)&= \frac{1}{2} \frac{d}{dt}\int _{M}|\partial _t u(t)|^2 \textrm{e}^{2u(t)}d \mu _{{\bar{g}}}= \int _{M}\bigl (\partial _t u(t) \partial _{tt}u(t) + |\partial _t u(t)|^2\partial _t u(t) \bigr )d \mu _{g(t)} \\&=\int _{M}\Bigl (- (\partial _t u(t))^3 + 2\bigl (f-\alpha (t)\bigr )(\partial _t u(t))^2 + \partial _t u(t) \Delta _{g(t)}\partial _t u(t)\Bigr ) d \mu _{g(t)} \\&\le - \int _{M} (\partial _t u(t))^3d \mu _{g(t)} + 2\bigl (\Vert f\Vert _{L^\infty (M,{\bar{g}})}+\alpha _0\bigr )F(t)-G(t) \end{aligned} \end{aligned}$$
(5.32)

with

$$\begin{aligned} G(t):=\int _M |\nabla _{g(t)}\partial _tu(t)|^2_{g(t)}d\mu _{g(t)} \qquad \text {for }t >0. \end{aligned}$$

With Lemma 4.4, applied with \(r=3\), \(C_{\text {GNL}}= C_{\text {GNL}}(3)>0\), (5.2) and Lemma 5.7 we can furthermore estimate

$$\begin{aligned} \begin{aligned}&- \int _{M} (\partial _t u(t))^3d \mu _{g(t)} \\&\quad \le \int _{M} |\partial _tu(t)|^3 \textrm{e}^{2u(t)} d\mu _{{\bar{g}}} \le \textrm{e}^{2{\mathcal {N}}} \Vert \partial _tu(t)\Vert ^3_{L^3(M,{\bar{g}})}\\&\quad \le \textrm{e}^{2{\mathcal {N}}} C_{\text {GNL}} \Vert \partial _tu(t)\Vert _{L^2(M,{\bar{g}})}^2 \Vert \partial _tu(t)\Vert _{H^1(M,{\bar{g}})}\\&\quad = \textrm{e}^{2{\mathcal {N}}} C_{\text {GNL}} \int _M|\partial _tu(t)|^2\textrm{e}^{-2u(t)}d\mu _{g(t)} \Bigg (\int _M|\partial _tu(t)|^2\textrm{e}^{-2u(t)}d\mu _{g(t)} \\&\qquad + \int _M|\nabla _{g(t)} \partial _tu(t)|^2d\mu _{g(t)} \Bigg )^{\frac{1}{2}}\\&\quad \le \textrm{e}^{6{\mathcal {N}}} C_{\text {GNL}} \int _M|\partial _tu(t)|^2 d\mu _{g(t)} \left( \int _M|\partial _tu(t)|^2d\mu _{g(t)} + \int _M|\nabla _{g(t)} \partial _tu(t)|^2d\mu _{g(t)} \right) ^{\frac{1}{2}}\\&\quad = \textrm{e}^{6{\mathcal {N}}} C_{\text {GNL}} F(t) \Bigl (F(t) + G(t) \Bigr )^{\frac{1}{2}} \le \frac{\bigl (\textrm{e}^{6{\mathcal {N}}} C_{\text {GNL}}\bigr )^2}{2}F^2(t) + \frac{1}{2} \Bigl (F(t) + G(t) \Bigr ), \end{aligned} \end{aligned}$$
(5.33)

where we used Young’s inequality and the fact that

$$\begin{aligned} G(t) =\int _M |\nabla _{g(t)}\partial _tu(t)|^2_{g(t)}d\mu _{g(t)}= \int _M|\nabla _{{\bar{g}}}\partial _tu(t)|^2_{{\bar{g}}}d\mu _{{\bar{g}}} \qquad \text {for }t >0. \nonumber \\ \end{aligned}$$

Combining (5.32) and (5.33) and using that \(G(t) \ge 0\) gives

$$\begin{aligned} \begin{aligned} \frac{d}{dt}F(t) \le \frac{d}{dt}F(t)+G(t)&\le \bigl (\textrm{e}^{6{\mathcal {N}}} C_{\text {GNL}}\bigr )^2 F^2(t)+ \bigl (4(\Vert f\Vert _{L^\infty (M,{\bar{g}})}+\alpha _0)+1\bigr ) F(t)\\&=:{{\tilde{C}}}_1F(t)+{{\tilde{C}}}_2 F^2(t). \end{aligned} \nonumber \\ \end{aligned}$$
(5.34)

By integrating (5.34) over \((t_l,t)\subset (t_l,T)\) and taking the supremum over \(t \in (t_l,T)\) we get

$$\begin{aligned} \sup _{t\in (t_l,T)}F(t)&\le F(t_l)+{{\tilde{C}}}_1\int _{t_l}^TF(t)dt+{{\tilde{C}}}_2\int _{t_l}^TF^2(t)dt\\&\le F(t_l)+{{\tilde{C}}}_1\int _{t_l}^\infty F(t)dt+ {{\tilde{C}}}_2\sup _{t\in (t_l,T)}F(t)\int _{t_l}^\infty F(t)dt. \end{aligned}$$

With (5.22) we also have \(\int _{t_l}^\infty F(t)dt\rightarrow 0\) for \(l\rightarrow \infty \) and thus \(1-{{\tilde{C}}}_2\int _{t_l}^\infty F(t) dt>0\) for l sufficiently large. For these l and \(T > t_l\) we thus have

$$\begin{aligned} \sup _{t\in (t_l,T)}F(t)\le \frac{1}{\left( 1-{{\tilde{C}}}_2\int _{t_l}^\infty F(t) dt\right) } \left( F(t_l)+{{\tilde{C}}}_1 \int _{t_l}^\infty F(t) dt\right) . \end{aligned}$$

Letting \(T\rightarrow \infty \) yields

$$\begin{aligned} \sup _{t\in (t_l,\infty )}F(t)\le \frac{1}{\left( 1-{{\tilde{C}}}_2\int _{t_l}^\infty F(t) dt\right) } \left( F(t_l)+{{\tilde{C}}}_1\int _{t_l}^\infty F(t) dt\right) \rightarrow 0\quad \text {as }l\rightarrow \infty \end{aligned}$$

which shows the claim. \(\square \)

To prove now the convergence of the flow, we first note u(t) is uniformly (in \(t \in (0,\infty )\)) bounded in \(H^1(M,{\bar{g}})\) by Proposition 2.25. and Lemma 5.8. We now consider a sequence \(t_l\rightarrow \infty \), \(l\rightarrow \infty \) and the associated sequence of functions \(u_l:=u(t_l)\). This sequence is bounded in \(H^1(M,{\bar{g}})\), hence there exists a subsequence, again denoted by \((u_l)_l\), with \(u_l\rightarrow u_\infty \) weakly in \(H^1(M,{\bar{g}})\) and therefore strongly in \(L^2(M,{\bar{g}})\). Furthermore with (2.12) we know that \(\alpha _l:=\alpha (t_l)\rightarrow \alpha _\infty \) as \(l\rightarrow \infty \) after passing again to a subsequence. Moreover we claim that \(\textrm{e}^{\pm u_l}\rightarrow \textrm{e}^{\pm u_\infty }\) (as \(l\rightarrow \infty \)) in \(L^p(M,{\bar{g}})\) for any \(2\le p<\infty \). Indeed, using Lemma 5.7 and the elementary estimate

$$\begin{aligned} |1-\textrm{e}^x| \le |x| \textrm{e}^{|x|} \qquad \text {for }x \in \mathbb {R}, \end{aligned}$$
(5.35)

we find that

$$\begin{aligned} \Vert \textrm{e}^{u_l}-\textrm{e}^{u_\infty }\Vert ^p_{L^p(M,{\bar{g}})}&=\int _M\textrm{e}^{pu_l}|1-\textrm{e}^{u_\infty -u_l}|^pd\mu _{{\bar{g}}}\\&\le \textrm{e}^{p{\mathcal {N}}}\int _M|1-\textrm{e}^{u_\infty -u_l}|^pd\mu _{{\bar{g}}}\\&\le \textrm{e}^{p{\mathcal {N}}}\int _M|u_\infty -u_l|^p\textrm{e}^{p|u_\infty -u_l|}|d\mu _{{\bar{g}}}\\&\le \textrm{e}^{p{\mathcal {N}}}\textrm{e}^{2p{\mathcal {N}}}\int _M|u_\infty -u_l|^{p-2}|u_\infty -u_l|^2d\mu _{{\bar{g}}}\\&\le \textrm{e}^{3p{\mathcal {N}}}(2{\mathcal {N}})^{p-2}\Vert u_\infty -u_l\Vert ^2_{L^2(M,{\bar{g}})}\rightarrow 0\quad \text {as }l\rightarrow \infty . \end{aligned}$$

Replacing \(u_l\) by \(-u_l\) we get also \(\textrm{e}^{-u_l}\rightarrow \textrm{e}^{-u_\infty }\) in \(L^p(M,{\bar{g}})\) as \(l\rightarrow \infty \) for any \(p<\infty \). Furthermore, we have

$$\begin{aligned} \Vert \textrm{e}^{2u_l}\alpha _l-\textrm{e}^{2u_\infty }\alpha _\infty \Vert _{L^2(M,{\bar{g}})}&\le \Vert \textrm{e}^{2u_l}(\alpha _l-\alpha _\infty )\Vert _{L^2(M,{\bar{g}})}+\Vert \alpha _\infty (\textrm{e}^{2u_l}-\textrm{e}^{2u_\infty })\Vert _{L^2(M,{\bar{g}})}\\&\le \Vert \textrm{e}^{2u_l}\Vert _{L^\infty (M,{\bar{g}})}|\alpha _l-\alpha _\infty |A^{\frac{1}{2}}+|\alpha _\infty |\Vert \textrm{e}^{2u_l}-\textrm{e}^{2u_\infty }\Vert _{L^2(M,{\bar{g}})}\\&\rightarrow 0\quad \text {for }l\rightarrow \infty . \end{aligned}$$

Since moreover \(\textrm{e}^{2u_l}\partial _tu_l\rightarrow 0\) in \(L^2(M,{\bar{g}})\) as \(l\rightarrow \infty \) with Lemma 5.7 and Lemma 5.8, the evolution Eq. (5.2) yields

$$\begin{aligned} \Delta _{{\bar{g}}}u_l=\textrm{e}^{2u_l}\partial _tu_l+{\bar{K}}-\textrm{e}^{2u_l}f+\textrm{e}^{2u_l}\alpha _l \;\rightarrow \; {\bar{K}}-\textrm{e}^{2u_\infty }f+\textrm{e}^{2u_\infty }\alpha _\infty \qquad \text {in }L^2(M,{\bar{g}}). \end{aligned}$$

Since the Laplace operator \(\Delta _{{\bar{g}}}\) is closed in \(L^2(M,{\bar{g}})\) with domain \(H^2(M,{\bar{g}})\), we deduce that \(u_\infty \) in \(H^2(M,{\bar{g}})\) with

$$\begin{aligned} \Delta _{{\bar{g}}}u_\infty = {\bar{K}}-\textrm{e}^{2u_\infty }f+\textrm{e}^{2u_\infty }\alpha _\infty \end{aligned}$$
(5.36)

and thus

$$\begin{aligned} \Vert \Delta _{{\bar{g}}}(u_l-u_\infty )\Vert _{L^2(M,{\bar{g}})}\rightarrow 0\quad \text {as}\quad l\rightarrow \infty . \end{aligned}$$

So, we even have strong convergence \(u_l\rightarrow u_\infty \) in \(H^2(M,{\bar{g}})\) and uniformly, which implies that \(u_\infty \in {\mathcal {C}}_A\) and therefore

$$\begin{aligned} \alpha _\infty = \frac{1}{A}\left( \int _M f d\mu _{g_\infty } -{\bar{K}}\right) \end{aligned}$$

by integrating (5.36) over M. Consequently, for the Gauss curvature \(K_{g_\infty }\) of the limit metric \(g_\infty =\textrm{e}^{2u_\infty }{\bar{g}}\) we get from (1.1) and (5.36) that

$$\begin{aligned} K_{g_\infty }=\textrm{e}^{-2u_\infty }\bigl (-\Delta _{{\bar{g}}}u_\infty + {\bar{K}}\bigr )= f-\alpha _\infty =f+ \frac{1}{A}\left( {\bar{K}}-\int _Mfd\mu _{g_\infty }\right) \end{aligned}$$

which shows the convergence of the flow.

5.6 The Sign of the Constant \(\alpha _\infty \)

In this subsection we complete the proofs of Theorem 3.4 and Theorem 3.5. For this we show, under certain assumptions, that the expression

$$\begin{aligned} \lambda = \frac{1}{A}\left( {\bar{K}}-\int _Mfd\mu _{g_\infty } \right) \end{aligned}$$

is positive. The proof of Theorem 3.5 is already completed by the statement of Corollary 4.7. So we can turn to Theorem 3.4.

Proof of Theorem 3.4 (completed)

We have seen in Lemma 5.7 that in the case where \(u_0 \equiv \frac{1}{2}\log (A) \in {\mathcal {C}}_{p,A}\), the uniform \(L^\infty \)-bound on the global solution of the initial value problem (5.2), (5.3) only depends on A and an upper bound on \(\Vert f\Vert _{L^\infty (M,{\bar{g}})}\). In other words, if \(A>0\) and \(c>0\) are fixed, then there exists \(\tau >0\) with the property that

$$\begin{aligned} \sup _{t>0}\Vert u(t)\Vert _{L^\infty (M,{\bar{g}})}\le \tau \end{aligned}$$

for every \(f \in C^\infty (M)\) with \(\Vert f\Vert _{L^\infty (M,{\bar{g}})} \le c\) and the corresponding solution u of the initial value problem (5.2), (5.3) with \(u_0 \equiv \frac{1}{2}\log (A) \in {\mathcal {C}}_{p,A}\). Consequently, we also have \(\Vert u_\infty \Vert _{L^\infty (M,{\bar{g}})} \le \tau \) under the current assumptions on f, which implies that

$$\begin{aligned} \lambda&=\frac{1}{A}\left( {\bar{K}}- \int _M f \textrm{e}^{2u_\infty }d\mu _{{\bar{g}}}\right) = \frac{1}{A}\left( {\bar{K}} +cA - \int _M (f+c) \textrm{e}^{2u_\infty }d\mu _{{\bar{g}}}\right) \\&\ge c + \frac{{\bar{K}}}{A} - \Vert f+c\Vert _{L^1(M,{\bar{g}})}\Vert \textrm{e}^{2u_\infty }\Vert _{L^\infty (M,{\bar{g}})} \ge c + \frac{{\bar{K}}}{A} - \Vert f+c\Vert _{L^1(M,{\bar{g}})}\textrm{e}^{2\tau }. \end{aligned}$$

Hence, if \(\Vert f+c\Vert _{L^1(M,{\bar{g}})} <\varepsilon :=\frac{c + \frac{{\bar{K}}}{A}}{\textrm{e}^{2\tau }}\), we have \(\lambda >0\). \(\square \)