# A nonlinear semigroup approach to Hamilton-Jacobi equations–revisited

Panrui Ni  Lin Wang

###### Abstract

We consider the Hamilton-Jacobi equation

${H}(x,Du)+\lambda(x)u=c,\quad x\in M,$

where $M$ is a connected, closed and smooth Riemannian manifold. The functions
${H}(x,p)$ and $\lambda(x)$ are continuous. ${H}(x,p)$ is convex, coercive
with respect to $p$, and $\lambda(x)$ changes the signs. The first
breakthrough to this model was achieved by Jin-Yan-Zhao [12] under the Tonelli
conditions. In this paper, we consider more detailed structure of the
viscosity solution set and large time behavior of the viscosity solution on
the Cauchy problem.

Keywords. Hamilton-Jacobi equations, viscosity solutions, weak KAM theory

††Panrui Ni: Shanghai Center for Mathematical Sciences, Fudan University,
Shanghai 200433, China; e-mail<EMAIL_ADDRESS>
Lin Wang: School of Mathematics and Statistics, Beijing Institute of
Technology, Beijing 100081, China; e-mail<EMAIL_ADDRESS>†† _Mathematics
Subject Classification (2020):_ 37J50; 35F21; 35D40

###### Contents

  1. 1 Introduction and main results
  2. 2 Preliminaries
  3. 3 a priori estimates on subsolutions
  4. 4 Proof of Theorem 1
  5. 5 Proof of Theorem 2
  6. 6 Proof of Theorem 3
  7. 7 An example with the unique solution
  8. A Proof of Proposition 2.5
  9. B Generalization of JYZ’s results
  10. C The locally Lipschitz estimate of $T^{-}_{t}\varphi(x)$
  11. References

##  1 Introduction and main results

Inspired by Aubry-Mather theory and weak KAM theory for classical Hamiltonian
systems, an action minimizing method for contact Hamiltonian systems was
developed in a series of papers [14, 15, 16, 17, 18]. Let $H(x,p,u)$ be a
contact Hamiltonian. It turns out that the dependence of $H$ on $u$ plays a
crucial role in exploiting the dynamics generated by $H$. By using previous
dynamical approaches, some progress on viscosity solutions of Hamilton-Jacobi
(HJ) equations have been achieved [14, 16, 18, 19]. It was found that the
properties of viscosity solutions are closely related to the dependence of $H$
on $u$. For the cases with nondecreasing dependence of $H$ on $u$, the large
time behavior of the solution on the Cauchy problem was shown in [14]. In
[16], the existence of the solutions for the ergodic problem was proved for
the HJ equations with Lipschitz dependence on $u$. In [18] (resp. [19]), the
authors focused on the strictly decreasing case (resp. Lipschitz depending
case), and discussed large time behavior of the solution of the evolutionary
equation. For the HJ equations depending Lipschitz continuously on $u$, it was
also studied in the framework of the semicontinuous viscosity solutions due to
Barron-Jensen [11].

More recently, for a class of HJ equations with non-monotone dependence on
$u$, the first breakthrough was achieved by L. Jin, J. Yan and K. Zhao [12]
under the Tonelli conditions. In that work, they provided a description of the
solution set of the stationary equation (formulated as (E0) below) and
revealed a bifurcation phenomenon with respect to the value $c$ in the right
hand side, which opened a way to exploit further properties of viscosity
solutions beyond well-posedness for HJ equations with non-monotone dependence
on $u$. The main results in this paper are highly motivated by [12].

Let us consider the stationary equation:

${H}(x,Du)+\lambda(x)u=c,\quad x\in M.$ (E0)

Throughout this paper, we assume $M$ is a closed, connected and smooth
Riemannian manifold. Denote by $TM$ and $T^{*}M$ the tangent bundle and
cotangent bundle of $M$ respectively (see Section 2.1 below for more
notations). Let $H:T^{*}M\rightarrow\mathbb{R}$ satisfy

  * (C):

$H(x,p)$ is continuous;

  * (CON):

$H(x,p)$ is convex in $p$, for any $x\in M$;

  * (CER):

$H(x,p)$ is coercive in $p$, i.e.
$\lim_{\|p\|\rightarrow+\infty}H(x,p)=+\infty$.

Correspondingly, one has the Lagrangian associated to $H$:

$L(x,\dot{x}):=\sup_{p\in
T^{*}_{x}M}\\{\langle\dot{x},p\rangle_{x}-H(x,p)\\},$

where $\langle\cdot,\cdot\rangle_{x}$ represents the canonical pairing between
$T_{x}M$ and $T^{*}_{x}M$. The Lagrangian $L(x,\dot{x})$ satisfies the
following properties:

  * (LSC):

$L(x,\dot{x})$ is lower semicontinous in $\dot{x}$, and continuous on the
interior of its domain $\textrm{dom}(L):=\\{(x,\dot{x})\in TM:\
L(x,\dot{x})<+\infty\\}$;

  * (CON):

$L(x,\dot{x})$ is convex in $\dot{x}$, for any $x\in M$.

We also assume

  * ($\pm$):

there exist $x_{1},x_{2}\in M$ such that $\lambda(x_{1})>0$ and
$\lambda(x_{2})<0$.

From an economical point of view, the assumption ($\pm$) may be corresponding
to a model with fluctuating interest rates. To be more precise, the rates
depend on the space variable and admit values above and below zero at some
places. From a theoretical point of view, it is a toy model of the HJ
equations with non-monotone dependence on $u$, by which certain differences
from the cases with monotone dependence on $u$ can be revealed.

In [13], the well-posedness of the Lax-Oleinik semigroup was verified for
contact HJ equations under mild conditions. By virtue of that, we generalize
the results in [12] to the cases from the Tonelli conditions to the
assumptions (C), (CON) and (CER). Henceforth, for simplicity of notation, we
omit the word “viscosity”.

######  Generalization of [12].

Let

$c_{0}:=\inf_{u\in C^{\infty}(M)}\sup_{x\in
M}\bigg{\\{}H(x,Du)+\lambda(x)u\bigg{\\}}.$

Then $c_{0}$ is finite. Given $c\geq c_{0}$, the
$\|\cdot\|_{W^{1,\infty}}$-norm of all subsolutions of (E0) is bounded.
Moreover,

  * (1)

(E0) has a solution if and only if $c\geq c_{0}$;

  * (2)

if $c>c_{0}$, then (E0) has at least two solutions.

Under the assumptions (C), (CON) and (CER), some new ingredients are needed
for a priori estimates of subsolutions, which will be provided in Section 3.
The other parts are similar to the proof in [12], which is postponed to
Appendix B. With more efforts, we obtain the more detailed structure of the
solution set.

######  Theorem 1.

Let $c\geq c_{0}$. There exist the maximal element $u_{\max}$ and the minimal
element $u_{\min}$ in the set of solutions of (E0).

We will discuss an example with the unique solution for the case with
$c=c_{0}$ in Section 7 below, which together with [12, Example 3.1] shows
various possibilities about the solution set of (E0) in this case. In
particular, different from the case with $c>c_{0}$, both the phenomena with
the unique solution and many solutions may happen. Hence, $c=c_{0}$ is called
the critical value.

In the second part, let us consider the evolutionary equation:

$\left\\{\begin{aligned}
&\partial_{t}u(x,t)+H(x,\partial_{x}u(x,t))+\lambda(x)u(x,t)=c,\quad(x,t)\in
M\times(0,+\infty).\\\ &u(x,0)=\varphi(x),\quad x\in M,\\\
\end{aligned}\right.$ (CP)

where $\varphi\in C(M)$. It is well known that the viscosity solution of (CP)
is unique (see [11, Corollary 3.2] for instance). By [13, Theorem 1], this
solution can be represented by $u(x,t):=T_{t}^{-}\varphi(x)$, where
$T^{-}_{t}:C(M)\rightarrow C(M)$ is defined by

$T^{-}_{t}\varphi(x)=\inf_{\gamma(t)=x}\left\\{\varphi(\gamma(0))+\int_{0}^{t}\bigl{[}L(\gamma(\tau),\dot{\gamma}(\tau))-\lambda(x)T^{-}_{\tau}\varphi(\gamma(\tau))+c\bigl{]}\textrm{d}\tau\right\\}.$
(T-)

Due to the non-monotone issue, it is not easy to obtain the locally Lipschitz
property of $(x,t)\mapsto T_{t}^{-}\varphi(x)$ on $M\times(0,+\infty)$ under
the assumptions (C), (CON) and (CER). From a dynamical point of view, this
property can be verified by a standard procedure once we have the Lipschitz
regularity of minimizers of $T_{t}^{-}\varphi(x)$. Since $H$ is only supposed
to be continuous, one can not use contact Hamilton’s equations to improve
regularity of minimizers of $T_{t}^{-}\varphi(x)$. Following [2], we will deal
with this issue by establishing the Erdmann condition for a non-smooth “energy
function”, see Lemma C.2. The Hamiltonian considered in Appendix C has the
form $H(x,p)+f(x,u)$, which includes the case $H(x,p)+\lambda(x)u$ considered
here. Hence, Appendix C is of its own interest. In order to ensure the locally
Lipschitz property of $T^{-}_{t}\varphi(x)$, we have to require the following
extra assumption:

  * (*)

$H(x,p)$ is strictly convex in $p$ for any $x\in M$, and there are two
superlinear functions $\theta_{1}$, $\theta_{2}$ $:[0,+\infty)\to[0,+\infty)$
such that $\theta_{1}(\|p\|)\leq H(x,p)\leq\theta_{2}(\|p\|)$.

######  Theorem 2.

Assume (*) holds. Let $u(x,t)$ be the solution of (CP) with $c\geq c_{0}$.
Then $u(x,t)$ is locally Lipschitz on $M\times(0,+\infty)$. Moreover,

  * (1)

if $\varphi\geq u_{\max}$, then $u(x,t)$ converges to $u_{\max}$ uniformly as
$t\to+\infty$;

  * (2)

if $\varphi<u_{\min}$, then $\min_{x\in M}u(x,t)$ tends to $-\infty$ as
$t\to+\infty$.

If $c>c_{0}$ and the initial data is located between the maximal solution
$u_{\max}$ and the minimal solution $u_{\min}$, then we also obtain large time
behavior of the solution of (CP).

######  Theorem 3.

Let $u(x,t)$ be the solution of (CP) with $c>c_{0}$. If $u_{\min}<u_{\max}$,
then for all initial functions $\varphi\in C(M)$ satisfying
$u_{\min}<\varphi\leq u_{\max}$, the solution $u(x,t)$ converges to $u_{\max}$
uniformly as $t\to+\infty$.

######  Remark 1.1.

It is worthy mentioning that Theorem 3 is proved without using (*). An example
for the case $u_{\min}<u_{\max}$ is given in [12, Section 3.2] with $c$ large
enough.

The rest of this paper is organized as follows. Section 2 gives some
preliminaries on viscosity solutions of HJ equations with Lipschitz dependence
on $u$. In Section 3, a priori estimates on subsolutions of (E0) are
established. Theorem 1 is proved in Section 4, Theorem 2 is proved in Section
5, and Theorem 3 is proved in Section 6. In Section 7, An example with the
unique solution is verified to show the variety of the solution set in the
critical case. For the sake of completeness, some supporting materials are
provided in Appendix A, B, and C.

##  2 Preliminaries

###  2.1 Notations

  * $\centerdot$

$\textrm{diam}(M)$ denotes the diameter of $M$.

  * $\centerdot$

$d(x,y)$ denotes the distance between $x$ and $y$ induced by the Riemannian
metric $g$ on $M$.

  * $\centerdot$

$\|\cdot\|$ denotes the norms induced by $g$ on both tangent and cotangent
spaces of $M$.

  * $\centerdot$

$D$ denotes the spacial gradient with respect to $x$ for $x\in M$.

  * $\centerdot$

$C(M)$ stands for the space of continuous functions on $M$. Lip$(M)$ stands
for the space of Lipschitz continuous functions on $M$.

  * $\centerdot$

$\|\cdot\|_{\infty}$ stands for the supremum norm of the vector-valued
functions on their domains.

###  2.2 Some facts on $T_{t}^{\pm}$, weak KAM solutions and Aubry sets

Consider the evolutionary equation:

$\left\\{\begin{aligned}
&\partial_{t}u(x,t)+H(x,u(x,t),\partial_{x}u(x,t))=0,\quad(x,t)\in
M\times(0,+\infty).\\\ &u(x,0)=\varphi(x),\quad x\in M.\\\
\end{aligned}\right.$ (A0)

and the stationary equation:

$H(x,u(x),Du(x))=0$ (B0)

We collect some facts in [13] used in this paper. These facts hold under more
general assumptions on the dependence of $u$. Let
$H:T^{*}M\times\mathbb{R}\rightarrow\mathbb{R}$ be a continuous Hamiltonian
satisfying

  * (CON):

$H(x,u,p)$ is convex in $p$, for any $(x,u)\in M\times\mathbb{R}$;

  * (CER):

$H(x,u,p)$ is coercive in $p$, i.e. $\lim_{\|p\|\rightarrow+\infty}(\inf_{x\in
M}H(x,0,p))=+\infty$;

  * (LIP):

$H(x,u,p)$ is Lipschitz in $u$, uniformly with respect to $(x,p)$, i.e., there
exists $\Theta>0$ such that $|H(x,u,p)-H(x,v,p)|\leq\Theta|u-v|$, for all
$(x,p)\in\ T^{*}M$ and all $u,v\in\mathbb{R}$.

Correspondingly, one has the Lagrangian associated to $H$:

$L(x,u,\dot{x}):=\sup_{p\in
T^{*}_{x}M}\\{\langle\dot{x},p\rangle-H(x,u,p)\\}.$

Due to the absence of superlinearity of $H$, the corresponding Lagrangian $L$
may take the value $+\infty$. Define

$\text{dom}(L):=\\{(x,\dot{x},u)\in TM\times\mathbb{R}\ |\
L(x,u,\dot{x})<+\infty\\}.$

Then $L(x,u,\dot{x})$ satisfies the following properties:

  * (LSC):

$L(x,u,\dot{x})$ is lower semicontinuous, and continuous on the interior of
$\textrm{dom}(L)\times\mathbb{R}$;

  * (CON):

$L(x,u,\dot{x})$ is convex in $\dot{x}$, for any $(x,u)\in M\times\mathbb{R}$;

  * (LIP):

$L(x,u,\dot{x})$ is Lipschitz in $u$, uniformly with respect to $(x,\dot{x})$,
i.e., there exists $\Theta>0$ such that
$|L(x,u,\dot{x})-L(x,v,\dot{x})|\leq\Theta|u-v|$, for all $(x,\dot{x})\in\
\textrm{dom}(L)$ and all $u,v\in\mathbb{R}$.

See [13, Remark 1.1] for more illustrations for the assumption (CER) and
dom($L$).

######  Proposition 2.1.

[13, Theorem 1] Both backward Lax-Oleinik semigroup

$T^{-}_{t}\varphi(x)=\inf_{\gamma(t)=x}\left\\{\varphi(\gamma(0))+\int_{0}^{t}L(\gamma(\tau),T^{-}_{\tau}\varphi(\gamma(\tau)),\dot{\gamma}(\tau))\textrm{d}\tau\right\\}$
(2.1)

and forward Lax-Oleinik semigroup

$T^{+}_{t}\varphi(x)=\sup_{\gamma(0)=x}\left\\{\varphi(\gamma(t))-\int_{0}^{t}L(\gamma(\tau),T^{+}_{t-\tau}\varphi(\gamma(\tau)),\dot{\gamma}(\tau))d\tau\right\\}.$
(2.2)

are well-defined for $\varphi\in C(M)$. The infimum (resp. supremum) is taken
among absolutely continuous curves $\gamma:[0,t]\rightarrow M$ with
$\gamma(t)=x$ (resp. $\gamma(0)=x$). If $\varphi$ is continuous, then
$u(x,t):=T^{-}_{t}\varphi(x)$ represents the unique continuous viscosity
solution of (A0). If $\varphi$ is Lipschitz continuous, then
$u(x,t):=T^{-}_{t}\varphi(x)$ is also Lipschitz continuous on
$M\times[0,+\infty)$.

######  Proposition 2.2.

[13, Proposition 3.1] The Lax-Oleinik semigroups have the following properties

  * (1)

For $\varphi_{1}$ and $\varphi_{2}\in C(M)$, if
$\varphi_{1}(x)<\varphi_{2}(x)$ for all $x\in M$, we have
$T^{-}_{t}\varphi_{1}(x)<T^{-}_{t}\varphi_{2}(x)$ and
$T^{+}_{t}\varphi_{1}(x)<T^{+}_{t}\varphi_{2}(x)$ for all $(x,t)\in
M\times(0,+\infty)$.

  * (2)

Given any $\varphi$ and $\psi\in C(M)$, we have
$\|T^{-}_{t}\varphi-T^{-}_{t}\psi\|_{\infty}\leq e^{\Theta
t}\|\varphi-\psi\|_{\infty}$ and
$\|T^{+}_{t}\varphi-T^{+}_{t}\psi\|_{\infty}\leq e^{\Theta
t}\|\varphi-\psi\|_{\infty}$ for all $t>0$.

Similar to [17, Proposition 2.8], one has

######  Proposition 2.3.

Let $\varphi\in C(M)$. Then

$\displaystyle-T^{+}_{t}(-\varphi)=\bar{T}^{-}_{t}\varphi,\quad-T^{-}_{t}(-\varphi)=\bar{T}^{+}_{t}\varphi,\quad\forall
t\geq 0.$ (2.3)

where $\bar{T}_{t}^{\pm}$ denote the Lax-Oleinik semigroups associated to
$H(x,-u,-Du)=0$.

Following Fathi [7], one can extend the definitions of weak KAM solutions of
equation (B0) by using absolutely continuous calibrated curves instead of
$C^{1}$ curves (see [13, Definition D.1 and D.2]).

######  Proposition 2.4.

[13, Proposition D.4] Let $u_{-}\in C(M)$. The following statements are
equivalent:

  * (1)

$u_{-}$ is a fixed point of $T^{-}_{t}$;

  * (2)

$u_{-}$ is a backward weak KAM solution of (B0);

  * (3)

$u_{-}$ is a viscosity solution of (B0).

Similarly, let $v_{+}\in C(M)$. The following statements are equivalent:

  * (1’)

$v_{+}$ is a fixed point of $T^{+}_{t}$;

  * (2’)

$v_{+}$ is a forward weak KAM solution of (B0);

  * (3)’

$-v_{+}$ is a viscosity solution of $H(x,-u(x),-Du(x))=0$.

The following result is well known for Hamilton-Jacobi equations independent
of $u$. It is also true in contact cases. We will give a proof of this fact in
Appendix A.

######  Proposition 2.5.

Let $\varphi\in Lip(M)$. The following conditions are equivalent:

  * (1)

$\varphi$ is a Lipschitz subsolution of (B0);

  * (2)

$\varphi\prec L$;

  * (3)

for each $t\geq 0$,

$T_{t}^{-}\varphi\geq\varphi\geq T_{t}^{+}\varphi.$

######  Proposition 2.6.

[13, Theorem 3 and Remark 3.5] Let $\varphi\in C(M)$.

  * (1)

If $T^{-}_{t}\varphi(x)$ has a bound independent of $t$, then the lower half
limit

$\check{\varphi}(x)=\lim_{r\rightarrow 0+}\inf\\{T^{-}_{t}\varphi(y):\
d(x,y)<r,\ t>1/r\\}$

is a Lipschitz continuous viscosity solution of (B0).

  * (2)

If $T^{+}_{t}\varphi(x)$ has a bound independent of $t$, then the upper half
limit

$\hat{\varphi}(x)=\lim_{r\rightarrow 0+}\sup\\{T^{+}_{t}\varphi(y):\
d(x,y)<r,\ t>1/r\\},$

is a Lipschitz continuous forward weak KAM solution of (B0).

  * (3)

Let $u_{-}$ be a viscosity solution of (B0). Then $T_{t}^{+}u_{-}\leq u_{-}$,
the limit $u_{+}:=\lim_{t\rightarrow+\infty}T_{t}^{+}u_{-}$ exists, and
$u_{+}$ is a forward weak KAM solution of (B0).

  * (4)

Let $v_{+}$ be a forward weak KAM solution of (B0). Then $T_{t}^{-}v_{+}\geq
v_{+}$. The limit $v_{-}:=\lim_{t\rightarrow+\infty}T_{t}^{-}v_{+}$ exists,
and $v_{-}$ is a viscosity solution of (B0).

######  Proposition 2.7.

[13, Theorem 3] Let $u_{-}$ and $u_{+}$ be viscosity solution and forward weak
KAM solution of (B0) respectively. We define the projected Aubry set with
respect to $u_{-}$ by

$\mathcal{I}_{u_{-}}:=\\{x\in M:\
u_{-}(x)=\lim_{t\rightarrow+\infty}T^{+}_{t}u_{-}(x)\\}.$

Correspondingly, we define the projected Aubry set with respect to $u_{+}$ by

$\mathcal{I}_{u_{+}}:=\\{x\in M:\
u_{+}(x)=\lim_{t\rightarrow+\infty}T^{-}_{t}u_{+}(x)\\}.$

Both $\mathcal{I}_{u_{-}}$ and $\mathcal{I}_{u_{+}}$ are nonempty. In
particular, if $u_{+}(x)=\lim_{t\rightarrow+\infty}T^{+}_{t}u_{-}(x)$ and
$u_{-}(x)=\lim_{t\rightarrow+\infty}T^{-}_{t}u_{+}(x)$, then

$\mathcal{I}_{u_{-}}=\mathcal{I}_{u_{+}},$

which is denoted by $\mathcal{I}_{(u_{-},u_{+})}$, following the notation
introduced by Fathi [7].

##  3 a priori estimates on subsolutions

In this section, we assume the existence of subsolutions of (E0) and prove
some a priori estimates on subsolutions. The existence of subsolutions will be
verified for $c\geq c_{0}$ in Proposition B.3 below.

Let $L(x,\dot{x})$ be the Legendre transformation of $H(x,p)$. Let
$T_{t}^{\pm}$ be the Lax-Oleinik semigroups associated to

$L(x,\dot{x})-\lambda(x)u(x)+c.$

Similar to [9, Lemma 2.3], one can prove the local boundedness of
$L(x,\dot{x})$ in a neighborhood of the zero section of $TM$.

######  Lemma 3.1.

For $H(x,p)$ satisfying (C)(CON)(CER), there exist constants $\delta>0$ and
$C_{L}>0$ such that the Lagrangian $L(x,\dot{x})$ corresponding to $H(x,p)$
satisfies

$L(x,\xi)\leq C_{L},\quad\forall(x,\xi)\in M\times\bar{B}(0,\delta).$ (3.1)

Throughout this paper, we define $\mu:=\textrm{diam}(M)/\delta$.

######  Lemma 3.2.

Let $\varphi\in C(M)$.

  * (1)

$T^{-}_{t}\varphi$ has an upper bound independent of $t$;

  * (2)

$T^{+}_{t}\varphi$ has a lower bound independent of $t$.

###### Proof.

Take $x_{1}\in M$ with $\lambda(x_{1})>0$. We first show

$T^{-}_{t}\varphi(x_{1})\leq\max\bigg{\\{}\varphi(x_{1}),\frac{L(x_{1},0)+c}{\lambda(x_{1})}\bigg{\\}},\quad\forall
t\geq 0.$

Otherwise, there is $t>0$ such that

$T^{-}_{t}\varphi(x_{1})>\max\bigg{\\{}\varphi(x_{1}),\frac{L(x_{1},0)+c}{\lambda(x_{1})}\bigg{\\}}\geq\frac{L(x_{1},0)+c}{\lambda(x_{1})}.$

There are two cases:

(i) For all $s\in[0,t]$ we have

$T^{-}_{s}\varphi(x_{1})>\frac{L(x_{1},0)+c}{\lambda(x_{1})}.$

Take the constant curve $\gamma\equiv x_{1}$, we have

$T^{-}_{t}\varphi(x_{1})\leq\varphi(x_{1})+\int_{0}^{t}\bigg{[}L(x_{1},0)+c-\lambda(x_{1})T^{-}_{s}\varphi(x_{1})\bigg{]}ds<\varphi(x_{1}),$

which leads to a contradiction.

(ii) There is $t_{0}\geq 0$ such that

$T^{-}_{t_{0}}\varphi(x_{1})=\frac{L(x_{1},0)+c}{\lambda(x_{1})},$

and

$T^{-}_{s}\varphi(x_{1})>\frac{L(x_{1},0)+c}{\lambda(x_{1})},\quad\forall
s\in(t_{0},t].$

Take the constant curve $\gamma\equiv x_{1}$, we have

$T^{-}_{t}\varphi(x_{1})\leq
T^{-}_{t_{0}}\varphi(x_{1})+\int_{0}^{t}\bigg{[}L(x_{1},0)+c-\lambda(x_{1})T^{-}_{s}\varphi(x_{1})\bigg{]}ds<\frac{L(x_{1},0)+c}{\lambda(x_{1})},$

which leads to a contradiction.

We then prove that for all $x\in M$ and all $t>0$, $T_{t}^{-}\varphi(x)$ is
bounded from above. It suffices to prove that for all $x\in M$,
$T_{t+\mu}^{-}\varphi(x)$ is bounded from above, for all $t>0$. Let
$\alpha:[0,\mu]\rightarrow M$ be a geodesic connecting $x_{1}$ and $x$ with
constant speed, then $\|\dot{\alpha}\|\leq\delta$. Let

$K_{0}:=\max\bigg{\\{}\varphi(x_{1}),\frac{L(x_{1},0)+c}{\lambda(x_{1})}\bigg{\\}}.$

Given $x\neq x_{1}$. We assume $T^{-}_{t+\mu}\varphi(x)>K_{0}$. Otherwise the
proof is completed. Since $T^{-}_{t}\varphi(x_{1})\leq K_{0}$, there exists
$\sigma\in[0,\mu)$ such that $T^{-}_{{t}+\sigma}\varphi(\alpha(\sigma))=K_{0}$
and $T^{-}_{{t}+s}\varphi(\alpha(s))>K_{0}$ for all $s\in(\sigma,\mu]$. By
definition

$\displaystyle T^{-}_{{t}+s}\varphi(\alpha(s))$ $\displaystyle\leq
T^{-}_{{t}+\sigma}\varphi(\alpha(\sigma))+\int_{\sigma}^{s}\bigg{[}L(\alpha(\tau),\dot{\alpha}(\tau))-\lambda(\alpha(\tau))\cdot
T^{-}_{{t}+\tau}\varphi(\alpha(\tau))+c\bigg{]}d\tau$
$\displaystyle=K_{0}+\int_{\sigma}^{s}\bigg{[}L(\alpha(\tau),\dot{\alpha}(\tau))-\lambda(\alpha(\tau))\cdot
T^{-}_{{t}+\tau}\varphi(\alpha(\tau))+c\bigg{]}d\tau,$

which implies

$\displaystyle
T^{-}_{{t}+s}\varphi(\alpha(s))-K_{0}\leq\int_{\sigma}^{s}\bigg{[}L(\alpha(\tau),\dot{\alpha}(\tau))-\lambda(\alpha(\tau))\cdot
T^{-}_{{t}+\tau}\varphi(\alpha(\tau))+c\bigg{]}d\tau$
$\displaystyle\leq\int_{\sigma}^{s}\bigg{[}L(\alpha(\tau),\dot{\alpha}(\tau))-\lambda(\alpha(\tau))\cdot
K_{0}+c\bigg{]}d\tau+\lambda_{0}\int_{\sigma}^{s}\bigg{[}T^{-}_{{t}+\tau}\varphi(\alpha(\tau))-K_{0}\bigg{]}d\tau$
$\displaystyle\leq
L_{0}\mu+\lambda_{0}\int_{\sigma}^{s}\bigg{[}T^{-}_{{t}+\tau}\varphi(\alpha(\tau))-K_{0}\bigg{]}d\tau,$

where $\lambda_{0}$ is given by (B.1) and

$L_{0}:=C_{L}+\lambda_{0}K_{0}+c.$

where $C_{L}$ is given by (3.1). By the Gronwall inequality, we have

$T^{-}_{{t}+s}\varphi(\alpha(s))-K_{0}\leq L_{0}\mu
e^{\lambda_{0}(s-\sigma)}\leq L_{0}\mu e^{\lambda_{0}\mu},\quad\forall
s\in(\sigma,\mu].$

Take $s=\mu$ we have $T^{-}_{{t}+\mu}\varphi(x)\leq K_{0}+L_{0}\mu
e^{\lambda_{0}\mu}$.

Similar to the argument above, by choosing constant curve $\gamma(\tau)\equiv
x_{2}$ with $\tau\in[0,t]$ and replacing $T_{t+\mu}^{-}\varphi$ by
$T_{t-\mu}^{+}\varphi$, one has

$T_{t}^{+}\varphi(x)\geq\min\bigg{\\{}\varphi(x_{2}),\frac{L(x_{2},0)+c}{\lambda(x_{2})}\bigg{\\}}-L_{0}\mu
e^{\lambda_{0}\mu}.$ (3.2)

This completes the proof. ∎

######  Corollary 3.3.

Let $u_{0}$ be a Lipschitz subsolution of (E0). Then $T_{t}^{-}u_{0}$ (resp.
$T_{t}^{+}u_{0}$) has an upper (resp. lower) bound independent of $t$ and
$u_{0}$.

###### Proof.

We only prove $T_{t}^{-}u_{0}$ has an upper bound independent of $t$ and
$u_{0}$. The case with $T_{t}^{+}u_{0}$ is similar. Let

$\mathbf{e}_{0}:=\min_{(x,p)\in T^{*}M}H(x,p).$ (3.3)

By (CER), $\mathbf{e}_{0}>-\infty$. By the definition of the subsolution,
$H(x_{1},p)+\lambda(x_{1})u_{0}(x_{1})\leqslant 0$ for any $p\in
D^{*}u_{0}(x_{1})$, where $D^{*}$ denotes the reachable gradients. It implies

$\lambda(x_{1})u_{0}(x_{1})\leqslant-\min_{(x,p)\in
T^{*}M}H(x,p)=-\mathbf{e}_{0}.$

Hence, for each subsolution $u_{0}$, we have

$\displaystyle u_{0}(x_{1})\leq-\frac{\mathbf{e}_{0}}{\lambda(x_{1})}.$

Let

$K_{0}:=\max\bigg{\\{}-\frac{\mathbf{e}_{0}}{\lambda(x_{1})},\frac{L(x_{1},0)+c}{\lambda(x_{1})}\bigg{\\}},\quad
L_{0}:=C_{L}+\lambda_{0}K_{0}+c,$

where $\lambda_{0}$ is given by (B.1). By Lemma 3.2, we have

$T_{t}^{-}u_{0}(x)\leqslant K_{0}+L_{0}\mu e^{\lambda_{0}\mu}.$ (3.4)

This completes the proof. ∎

######  Proposition 3.4.

There exists a constant $C>0$ such that for any subsolution $u:M\to\mathbb{R}$
of (E0), there holds

$\|u\|_{W^{1,\infty}}\leqslant C.$

###### Proof.

By Proposition 2.5, for each $t\geq 0$,

$T_{t}^{+}u\leq u\leq T_{t}^{-}u.$

By Corollary 3.3, there exist $C_{1}$, $C_{2}$ independent of $u$ such that

$C_{2}\leq u\leq C_{1}.$

For each $x,y\in M$, let $\alpha:[0,d(x,y)/\delta]\rightarrow M$ be a geodesic
of length $d(x,y)$ with constant speed $\|\dot{\alpha}\|=\delta$ and
connecting $x$ and $y$. Then

$L(\alpha(s),\dot{\alpha}(s))\leq C_{L},\quad\forall s\in[0,d(x,y/\delta)].$

By Proposition 2.5,

$\displaystyle u(y)-u(x)$
$\displaystyle\leq\int_{0}^{d(x,y)/\delta}\bigg{[}L(\alpha(s),\dot{\alpha}(s))-\lambda(\alpha(s))u(\alpha(s))+c\bigg{]}ds$
$\displaystyle\leq\frac{1}{\delta}\bigg{(}C_{L}+\lambda_{0}\max\\{|C_{1}|,|C_{2}|\\}+c\bigg{)}d(x,y)=:\kappa
d(x,y).$

Note that $\kappa$ is independent of the choice of the subsolution $u$. We get
the equi-Lipschitz continuity of $u$ by exchanging the role of $x$ and $y$. ∎

######  Proposition 3.5.

Let $u_{0}$ be a Lipschitz subsolution of (E0). Then

$u_{-}:=\lim_{t\to+\infty}{T_{t}^{-}}u_{0}(x),\quad
u_{+}:=\lim_{t\to+\infty}{T_{t}^{+}}u_{0}(x)$

exist, and the limit procedure is uniform in $x$. Moreover, $u_{-}$ is a
viscosity solution of (E0), and $u_{+}$ is a forward weak KAM solution of
(E0). In particular, (E0) has a solution $u_{-}$ for $c\geq c_{0}$.

###### Proof.

We only prove that $u_{-}:=\lim_{t\to+\infty}{T_{t}^{-}}u_{0}(x)$ exits, and
is a viscosity solution of (E0). The existence of $u_{+}$ is similar . By
Proposition 2.6

$\check{u}_{-}(x):=\lim_{r\rightarrow 0+}\inf\\{T^{-}_{t}u_{0}(y):\ d(x,y)<r,\
t>1/r\\}$

is a backward weak KAM solution of (E0). Corollary 3.3 implies that the
pointwise limit $\lim_{t\rightarrow+\infty}T^{-}_{t}u_{0}(x)$ exists and
satisfies $\check{u}_{-}(x)\leq\lim_{t\rightarrow+\infty}T^{-}_{t}u_{0}(x)$.
Since $T^{-}_{t}u_{0}$ is increasing in $t$ by Proposition 2.5, for all $t>0$,
we have

$\displaystyle T^{-}_{t}u_{0}(x)$ $\displaystyle=\lim_{r\rightarrow
0+}\inf\\{T^{-}_{t}u_{0}(y):\ d(x,y)<r\\}$
$\displaystyle\leq\lim_{r\rightarrow 0+}\inf\\{T^{+}_{t+s}u_{0}(y):\
d(x,y)<r,\ t+s>1/r\\}=\check{u}_{-}(x).$

Then $\lim_{t\rightarrow+\infty}T^{-}_{t}u_{0}=\check{u}_{-}$, which is a
viscosity solution of (E0). By the Dini theorem, the family $T^{-}_{t}u_{0}$
uniformly converges to $\check{u}_{-}$. ∎

##  4 Proof of Theorem 1

###  4.1 The maximal solution

We first prove the existence of the maximal solution. Since each solution is a
subsolution, by Proposition 3.4, there are $C_{1}$ and $C_{2}$ such that
$C_{2}\leq u_{-}\leq C_{1}$ for all $u_{-}\in\mathcal{S}_{-}$. Note that
solutions of (E0) are fixed points of $T^{-}_{t}$. Take a Lipschitz continuous
function $\varphi>C_{1}$ as the initial function. By Proposition 2.2 (1),
$T^{-}_{t}\varphi$ is larger than every solution of (E0). By Lemma 3.2(1),
$T^{-}_{t}\varphi$ has an upper bound independent of $t$. By Proposition 2.6
(1), the lower half limit

$\check{\varphi}(x)=\lim_{r\rightarrow 0+}\inf\\{T^{-}_{t}\varphi(y):\
d(x,y)<r,\ t>1/r\\}$

is a Lipschitz continuous viscosity solution of (E0). Since $T^{-}_{t}\varphi$
is larger than every solution of (E0), we have

$\displaystyle\check{\varphi}(x)$ $\displaystyle=\lim_{r\rightarrow
0+}\inf\\{T^{-}_{t}\varphi(y):\ d(x,y)<r,\ t>1/r\\}$
$\displaystyle\geq\lim_{r\rightarrow 0+}\inf\\{v_{-}(y):\
d(x,y)<r\\}=v_{-}(x),$

for all $v_{-}\in\mathcal{S}_{-}$. Thus, $\check{\varphi}(x)$ is the maximal
solution of (E0).

###  4.2 The minimal solution

Since each forward weak KAM solution is dominated by
$L(x,\dot{x})-\lambda(x)u+c$, by Proposition 2.4, it is a subsolution of (E0).
By Proposition 3.4, there are $C_{1}$ and $C_{2}$ such that $C_{2}\leq
u_{+}\leq C_{1}$ for all $u_{+}\in\mathcal{S}_{+}$. Take a Lipschitz
continuous function $\varphi<C_{2}$ as the initial function. By Proposition
2.2 (1), $T^{+}_{t}\varphi$ is smaller than every forward weak KAM solution of
(E0). By Lemma 3.2(2), $T^{+}_{t}\varphi$ has a lower bound independent of
$t$. By Proposition 2.6 (2), the upper half limit

$\hat{\varphi}(x)=\lim_{r\rightarrow 0+}\sup\\{T^{+}_{t}\varphi(y):\
d(x,y)<r,\ t>1/r\\}$

is a forward weak KAM solution of (E0). Since $T^{+}_{t}\varphi$ is smaller
than every forward weak KAM solutions of (E0), we have

$\displaystyle\hat{\varphi}(x)$ $\displaystyle=\lim_{r\rightarrow
0+}\sup\\{T^{+}_{t}\varphi(y):\ d(x,y)<r,\ t>1/r\\}$
$\displaystyle\leq\lim_{r\rightarrow 0+}\sup\\{v_{+}(y):\
d(x,y)<r\\}=v_{+}(x).$

for all $v_{+}\in\mathcal{S}_{+}$. Thus, $\hat{\varphi}(x)$ is the minimal
forward weak KAM solution of (E0). By Proposition 2.6 (4),
$\hat{\varphi}_{\infty}:=\lim_{t\rightarrow+\infty}T^{-}_{t}\hat{\varphi}$
exists, and it is a solution of (E0).

######  Lemma 4.1.

$\hat{\varphi}_{\infty}$ is the minimal solution of (E0).

###### Proof.

Define

$\mathcal{P}_{-}:=\\{u_{-}\in\mathcal{S}_{-}:\ \exists
u_{+}\in\mathcal{S}_{+}\ \textrm{such\ that}\
u_{-}=\lim_{t\rightarrow+\infty}T^{-}_{t}u_{+}\\}.$

We first prove that for each $v_{-}\in\mathcal{P}_{-}$, there holds
$v_{-}\geq\hat{\varphi}_{\infty}$. In fact, by definition of
$\mathcal{P}_{-}$, there is $u_{+}\in\mathcal{S}_{+}$ such that
$v_{-}=\lim_{t\rightarrow+\infty}T^{-}_{t}u_{+}$. Since $\hat{\varphi}$ is the
minimal forward weak KAM solution, we have $u_{+}\geq\hat{\varphi}$. Acting
$T^{-}_{t}$ and letting $t\rightarrow+\infty$, we have
$v_{-}\geq\hat{\varphi}_{\infty}$.

We then prove that for each
$v_{-}\in\mathcal{S}_{-}\backslash\mathcal{P}_{-}$,
$v_{-}\geq\hat{\varphi}_{\infty}$ still holds. Let
$v_{+}:=\lim_{t\rightarrow+\infty}T^{+}_{t}v_{-}$ and
$u_{-}:=\lim_{t\rightarrow+\infty}T^{-}_{t}v_{+}$. Then
$u_{-}\in\mathcal{P}_{-}$, which implies $u_{-}\geq\hat{\varphi}_{\infty}$. By
Proposition 2.6 (3), $v_{+}\leq v_{-}$. Then we have $T^{-}_{t}v_{+}\leq
T^{-}_{t}v_{-}=v_{-}$. Taking $t\rightarrow+\infty$ we get $u_{-}\leq v_{-}$.
Therefore, $v_{-}\geq u_{-}\geq\hat{\varphi}_{\infty}$. ∎

##  5 Proof of Theorem 2

###  5.1 Above the maximal solution

Let $\varphi\geq u_{\max}$. Then $T_{t}^{-}\varphi\geq u_{\max}$. Combining
with Lemma 3.2(1), $T^{-}_{t}\varphi(x)$ has a bound independent of $t$. Then
the pointwise limit

$\bar{u}(x):=\limsup_{t\rightarrow+\infty}T^{-}_{t}\varphi(x)$

exists.

Assume (*) holds. By Corollary C.4, the family
$\\{T^{-}_{t}\varphi(x)\\}_{t\geq 1}$ is equi-Lipschitz in $x$. We denote by
$\kappa$ the Lipschitz constant of $T^{-}_{t}\varphi(x)$ in $x$. Since

$|\sup_{s\geq t}T^{-}_{s}\varphi(x)-\sup_{s\geq
t}T^{-}_{s}\varphi(y)|\leq\sup_{s\geq
t}|T^{-}_{s}\varphi(x)-T^{-}_{s}\varphi(y)|\leq\kappa d(x,y),$

the limiting procedure

$\bar{u}(x)=\lim_{t\rightarrow+\infty}\sup_{s\geq t}T^{-}_{s}\varphi(x)$

is uniform in $x$. Thus, the function $\bar{u}(x)$ is Lipschitz continuous. We
assert that $\bar{u}$ is a subsolution. If the assertion is true. By
Proposition 3.5, $\lim_{t\rightarrow+\infty}T^{-}_{t}\bar{u}(x)$ exits, and it
is a solution. Sine $T^{-}_{t}\varphi\geq u_{\max}$, we have $\bar{u}\geq
u_{\max}$. Thus, $\lim_{t\rightarrow+\infty}T^{-}_{t}\bar{u}=u_{\max}$. Based
on Section 4.1, the lower half limit $\check{\varphi}=u_{\max}$. By the
definition of $\check{\varphi}$, we have

$\liminf_{t\rightarrow+\infty}T^{-}_{t}\varphi(x)\geq\check{\varphi}(x)=u_{\max}.$

On the other hand,

$\limsup_{t\rightarrow+\infty}T^{-}_{t}\varphi(x)=\bar{u}(x)\leq\lim_{t\rightarrow+\infty}T^{-}_{t}\bar{u}(x)=u_{\max}(x).$

It follows that $\lim_{t\rightarrow+\infty}T^{-}_{t}\varphi=u_{\max}$
uniformly as $t\to+\infty$.

It remains to prove $\bar{u}$ is a subsolution. By Proposition 2.5(1), we only
need to show $T^{-}_{t}\bar{u}$ is non-decreasing in $t$.

We claim that for every $\varepsilon>0$, there exists a constant $s_{0}>0$
independent of $x$ such that for any $s\geq s_{0}$,

$T^{-}_{s}\varphi(x)\leq\bar{u}(x)+\varepsilon.$

Fixing $x\in M$, by definition of $\limsup$, for every $\varepsilon>0$, there
is $s_{0}(x)>0$ such that for any $s\geq s_{0}(x)$,

$T^{-}_{s}\varphi(x)\leq\bar{u}(x)+\frac{\varepsilon}{3}.$

Take $r:=\frac{\varepsilon}{3\kappa}$. For $s\geq s_{0}(x)$, we have

$\displaystyle T^{-}_{s}\varphi(y)$ $\displaystyle\leq
T^{-}_{s}\varphi(x)+\kappa d(x,y)\leq\bar{u}(x)+\frac{\varepsilon}{3}+\kappa
d(x,y)$ $\displaystyle\leq\bar{u}(y)+\frac{\varepsilon}{3}+2\kappa
d(x,y)\leq\bar{u}(y)+\varepsilon,\quad\forall y\in B_{r}(x).$

Since $M$ is compact, there are finite points $x_{i}\in M$ such that for each
$y\in M$, there is a point $x_{i}$ such that $y\in B_{r}(x_{i})$. Let
$s_{0}:=\max_{i}s_{0}(x_{i})$ and the claim is proved.

By Proposition 2.2, for each $t>0$ we have

$T^{-}_{t}(T^{-}_{s}\varphi(x))\leq T^{-}_{t}(\bar{u}(x)+\varepsilon)\leq
T^{-}_{t}\bar{u}(x)+\varepsilon e^{\lambda_{0}t},$

where $\lambda_{0}:=\|\lambda(x)\|_{\infty}>0$. Taking the limit
$s\rightarrow+\infty$, we have

$\bar{u}(x)=\limsup_{s\rightarrow+\infty}T^{-}_{t}(T^{-}_{s}\varphi(x))\leq
T^{-}_{t}\bar{u}(x)+\varepsilon e^{\lambda_{0}t}.$

Letting $\varepsilon\rightarrow 0+$ we get $\bar{u}(x)\leq
T^{-}_{t}\bar{u}(x)$, which means $T^{-}_{t}\bar{u}(x)$ is non-decreasing in
$t$.

###  5.2 Below the minimal solution

We have proved that for each $\varphi\geq u_{\max}$,
$\lim_{t\rightarrow+\infty}T^{-}_{t}\varphi=u_{\max}$ uniformly as
$t\to+\infty$. Combining with Proposition 2.4 and 2.3, one has

######  Lemma 5.1.

Let $\varphi\in C(M)$. If $\varphi\leq u^{+}_{\min}$, then
$\lim_{t\rightarrow+\infty}T^{+}_{t}\varphi=u^{+}_{\min}$ uniformly as
$t\to+\infty$.

######  Lemma 5.2.

For each $\varphi\in C(M)$, we have $T^{+}_{t}\circ
T^{-}_{t}\varphi\leq\varphi\leq T^{-}_{t}\circ T^{+}_{t}\varphi$ for all
$t\geq 0$.

###### Proof.

We only prove $\varphi\leq T^{-}_{t}\circ T^{+}_{t}\varphi$, the other side is
similar. We argue by a contradiction. Assume that there is $x\in M$ and $t>0$
such that

$T^{-}_{t}\circ T^{+}_{t}\varphi(x)<\varphi(x).$

Let $\gamma:[0,t]\rightarrow M$ with $\gamma(t)=x$ be a minimizer of
$T^{-}_{t}\circ T^{+}_{t}\varphi(x)$, and define

$F(s):=T^{+}_{t-s}\varphi(\gamma(s))-T^{-}_{s}\circ
T^{+}_{t}\varphi(\gamma(s)).$

Then $F(0)=0$ and $F(t)>0$. By continuity, there is $\sigma\in[0,t)$ such that
$F(\sigma)=0$ and $F(\tau)>0$ for all $\tau\in(\sigma,t]$. By definition, for
$s\in(\sigma,t]$ we have

$\displaystyle T^{-}_{s}\circ T^{+}_{t}\varphi(\gamma(s))$
$\displaystyle=T^{-}_{\sigma}\circ
T^{+}_{t}\varphi(\gamma(\sigma))+\int_{\sigma}^{s}\bigg{[}L(\gamma(\tau),\dot{\gamma}(\tau))-\lambda(\gamma(\tau))T^{-}_{\tau}\circ
T^{+}_{t}\varphi(\gamma(\tau))+c\bigg{]}d\tau$
$\displaystyle=T^{+}_{t-\sigma}\varphi(\gamma(\sigma))+\int_{\sigma}^{s}\bigg{[}L(\gamma(\tau),\dot{\gamma}(\tau))-\lambda(\gamma(\tau))T^{-}_{\tau}\circ
T^{+}_{t}\varphi(\gamma(\tau))+c\bigg{]}d\tau$ $\displaystyle\geq
T^{+}_{t-s}\varphi(\gamma(s))-\int_{\sigma}^{s}\bigg{[}L(\gamma(\tau),\dot{\gamma}(\tau))-\lambda(\gamma(\tau))T^{+}_{t-\tau}\varphi(\gamma(\tau))+c\bigg{]}d\tau$
$\displaystyle\quad\quad\quad+\int_{\sigma}^{s}\bigg{[}L(\gamma(\tau),\dot{\gamma}(\tau))-\lambda(\gamma(\tau))T^{-}_{\tau}\circ
T^{+}_{t}\varphi(\gamma(\tau))+c\bigg{]}d\tau$ $\displaystyle\geq
T^{+}_{t-s}\varphi(\gamma(s))-\lambda_{0}\int_{\sigma}^{s}F(\tau)d\tau,$

which implies

$F(s)\leq\lambda_{0}\int_{\sigma}^{s}F(\tau)d\tau.$

By the Gronwall inequality, we have $F(s)\equiv 0$ for $s\in[\sigma,t]$, which
contradicts $F(t)>0$. ∎

######  Lemma 5.3.

Let $\varphi\in C(M)$ and there is a point $x_{0}\in M$ such that
$\varphi(x_{0})<u_{\min}^{+}(x_{0})$, then $\min_{x\in M}T^{-}_{t}\varphi(x)$
tends to $-\infty$ as $t\rightarrow+\infty$.

###### Proof.

We argue by contradiction. Assume there is a constant $K_{1}$ and a sequence
$\\{t_{n}\\}_{n\in\mathbb{N}}$ such that $T^{-}_{t_{n}}\varphi\geq K_{1}$. By
Lemma 3.2, $T^{-}_{t_{n}}\varphi$ also has a upper bound independent of $t$.
Thus, the function $v_{n}(x):=T^{-}_{t_{n}}\varphi(x)$ is bounded continuous
for each $n$. By Lemma 5.2, we have $\varphi(x_{0})\geq
T^{+}_{t_{n}}v_{n}(x_{0})$. By Proposition 3.4, all of subsolutions are
uniformly bounded. Denote by $K_{2}$ their lower bound. Let
$K^{\prime}:=\min\\{K_{1},K_{2}\\}$, then $T^{+}_{t_{n}}v_{n}\geq
T^{+}_{t_{n}}K^{\prime}$. By Lemma 3.2(2), $T^{+}_{t}K^{\prime}$ has a lower
bound independent of $t$. Since $K^{\prime}\leq K_{2}$, $T^{+}_{t}K^{\prime}$
is smaller than every forward weak KAM solution of (E0). By Lemma 5.1,
$\lim_{t\rightarrow+\infty}T^{+}_{t}K^{\prime}$ exists and it equals to
$u^{+}_{\min}$. We conclude

$u^{+}_{\min}(x_{0})\leq\limsup_{n}T^{+}_{t_{n}}v_{n}(x_{0})\leq\varphi(x_{0})<u^{+}_{\min}(x_{0}),$

which leads to a contradiction. ∎

Based on Section 4.2, the set $\mathcal{S}_{+}$ contains the minimal element.
Denote by $u_{\min}^{+}$ the minimal element. By Lemma 4.1,

$u_{\min}=\lim_{t\to+\infty}T_{t}^{-}u_{\min}^{+}.$ (5.1)

By Proposition 2.7, the set $\mathcal{I}_{u^{+}_{\min}}$ is not empty. Thus,
if $\varphi<u_{\min}$, there is a point $x_{0}$ such that
$\varphi(x_{0})<u_{\min}(x_{0})=u_{\min}^{+}(x_{0})$. By Lemma 5.3,
$\min_{x\in M}T^{-}_{t}\varphi(x)$ tends to $-\infty$ as $t\to+\infty$.

##  6 Proof of Theorem 3

According to Proposition B.3, for $c\geq c_{0}$, (E0) has a Lipschitz
subsolution. Let $u_{0}$ be a subsolution of (E0) with $c=c_{0}$. For
$c>c_{0}$, there holds

$T_{t}^{+}u_{0}<u_{0}<T_{t}^{-}u_{0}.$

One can construct two different solutions $u_{-}$ and $v_{-}$ of (E0) from
$u_{0}$ by Proposition B.4. Precisely, we have

$u_{-}=\lim_{t\rightarrow+\infty}T^{-}_{t}u_{0},\
u_{+}=\lim_{t\rightarrow+\infty}T^{+}_{t}u_{0},\ \
v_{-}=\lim_{t\rightarrow+\infty}T^{-}_{t}u_{+}.$ (6.1)

It follows that $u_{+}<u_{0}<u_{-}$.

######  Lemma 6.1.

Let $c>c_{0}$. For each $\alpha\in(0,1]$ and each solution $w_{-}$ of (E0),
the convex combination

$u_{\alpha}:=\alpha u_{0}+(1-\alpha)w_{-}$

is a strict subsolution of (E0). In particular, we have
$T^{+}_{t}u_{\alpha}<u_{\alpha}<T^{-}_{t}u_{\alpha}$.

###### Proof.

Since $u_{0}$ is a Lipschitz subsolution of (E0) with $c=c_{0}$, we have

$H(x,Du_{0}(x))+\lambda(x)u_{0}(x)+(c-c_{0})\leq c,\quad a.e\ x\in M.$

Since $w_{-}$ is a solution of (E0), we have

$H(x,Dw_{-}(x))+\lambda(x)w_{-}(x)=c,\quad a.e.\ x\in M.$

Therefore

$\displaystyle\alpha H(x,Du_{0}(x))$ $\displaystyle+(1-\alpha)H(x,Dw_{-}(x))$
$\displaystyle+\lambda(x)\bigg{(}\alpha
u_{0}(x)+(1-\alpha)w_{-}(x)\bigg{)}+\alpha(c-c_{0})\leq c,\quad a.e.\ x\in M.$

By the convexity of $H(x,p)$ with respect to $p$, the Jensen’s inequality
gives

$H(x,Du_{\alpha}(x))+\lambda(x)u_{\alpha}(x)\leq(1-\alpha)c+\alpha c_{0},\quad
a.e.\ x\in M.$

Let $\epsilon_{0}:=\alpha(c-c_{0})>0$. Then

$H(x,Du_{\alpha}(x))+\lambda(x)u_{\alpha}(x)+\epsilon_{0}\leq c,\quad a.e.\
x\in M.$

By Lemma A.2, $T^{+}_{t}u_{\alpha}<u_{\alpha}<T^{-}_{t}u_{\alpha}$. ∎

######  Lemma 6.2.

Let $c>c_{0}$. Define $u_{-}$ and $v_{-}$ as in (6.1). Then $u_{-}$ is the
maximal solution of (E0), and $v_{-}$ is the minimal solution of (E0).

###### Proof.

We first prove that there is no solution $w_{-}$ different from $u_{-}$ such
that $w_{-}\geq u_{-}$. Assume that there is such a solution $w_{-}$. Since
$u_{0}<u_{-}\leq w_{-}$, there is $\alpha\in(0,1)$ such that
$u_{\alpha}=\alpha u_{0}+(1-\alpha)w_{-}$ satisfies

$\min_{x\in M}(u_{-}(x)-u_{\alpha}(x))=0.$

Let $x_{0}\in M$ be the point at which the above minimum is attained. Then

$T^{-}_{t}u_{\alpha}\leq T^{-}_{t}u_{-}.$

By Lemma 6.1, we have
$T^{-}_{t}u_{\alpha}(x_{0})>u_{\alpha}(x_{0})=u_{-}(x_{0})=T^{-}_{t}u_{-}(x_{0})$,
which leads to a contradiction.

We then prove that $w_{-}\leq u_{-}$ for all solutions $w_{-}$. Assume that
there is a solution $w_{-}$ such that

$\max_{x\in M}(w_{-}(x)-u_{-}(x))>0.$

Let $y_{0}\in M$ be the point at which the above maximum is attained. Then the
function $\bar{u}(x):=\max\\{u_{-}(x),w_{-}(x)\\}$ is a subsolution. By
Proposition 3.5, we get a solution
$\bar{w}_{-}:=\lim_{t\rightarrow+\infty}T^{-}_{t}\bar{u}$. We also have

$\bar{w}_{-}(y_{0})\geq\bar{u}(y_{0})=w_{-}(y_{0})>u_{-}(y_{0}),$

which leads to a contradiction by the first step above.

Similar to the argument above, we conclude that $u_{+}$ is the minimal forward
weak KAM solution of (E0). By Lemma 4.1, $v_{-}$ is the minimal solution of
(E0). ∎

Let us recall $u_{0}$ is a subsolution of (E0) with $c=c_{0}$. For $c>c_{0}$,
there holds

$T_{t}^{+}u_{0}<u_{0}<T_{t}^{-}u_{0}.$

By Proposition 2.2(1) and Lemma 5.2, we have

$T^{-}_{t+s}u_{0}\geq T^{-}_{t+s}\circ
T^{+}_{t}u_{0}=T^{-}_{s}\circ(T^{-}_{t}\circ T^{+}_{t}u_{0})\geq
T^{-}_{s}u_{0}$

for all $t,s\geq 0$. Letting $s\rightarrow+\infty$ we have

$\lim_{s\rightarrow+\infty}T^{-}_{t+s}\circ T^{+}_{t}u_{0}=u_{\max},$ (6.2)

for each $t>0$. Let $\varphi\in C(M)$ satisfy $u_{\min}<\varphi\leq u_{\max}$.
Since $u_{+}=\lim_{t\rightarrow+\infty}T^{+}_{t}u_{0}\leq u_{\min}$, there is
$t_{0}>0$ such that $T^{+}_{t_{0}}u_{0}\leq\varphi$ on $M$. Then we have

$T^{-}_{t_{0}+s}\circ T^{+}_{t_{0}}u_{0}\leq T^{-}_{t_{0}+s}\varphi\leq
u_{\max}.$

Letting $s\rightarrow+\infty$ and by (6.2), we have

$\lim_{t\rightarrow+\infty}T^{-}_{t}\varphi=u_{\max}.$

The proof is now complete.

##  7 An example with the unique solution

In [12], Jin, Yan and Zhao provided a detailed analysis for the following
example:

######  Example 7.1.

$\frac{1}{2}|u^{\prime}(x)|^{2}+\sin x\cdot u(x)=c,\quad
x\in\mathbb{S}^{1}\simeq[0,2\pi).$ (7.1)

It was shown that the critical value $c_{0}=0$ and there are uncountably many
solutions of (7.1) in the critical case. As a complement, we consider

######  Example 7.2.

$\frac{1}{2}|u^{\prime}(x)|^{2}+\sin x\cdot u(x)+\cos 2x-1=c,\quad
x\in\mathbb{S}^{1}\simeq[0,2\pi).$ (7.2)

We will prove that the critical value is also $c_{0}=0$, but (7.2) admits a
unique solution in the critical case. The Hamiltonian of (7.2) is formulated
as

$H(x,u,p)=\frac{p^{2}}{2}+\sin x\cdot u+\cos 2x-1.$ (7.3)

A rough picture of the solution is given by Figure 1.

Figure 1: The unique solution of (7.2)

###### Proof.

We first show $c_{0}=0$. Assume (7.2) admits a smooth subsolution $u_{0}$ when
$c<0$, then we have $|u^{\prime}_{0}(0)|^{2}\leq 2c<0$, which is impossible.
When $c=0$, the constant function $\varphi\equiv 0$ is a subsolution of (7.2).
Therefore $c_{0}=0$. By Proposition 3.5, there is a solution $u_{-}$ of (7.2)
given by

$u_{-}:=\lim_{t\to+\infty}T_{t}^{-}\varphi.$

Since $T_{t}^{-}\varphi\geq\varphi$, then $u_{-}\geq 0$.

We then divide the proof into the following steps:

  * •

In Step 1, we discuss the dynamical behavior of the contact Hamiltonian flow
$\Phi^{H}_{t}$ generated by $H(x,u,p)$, which is restricted on a two
dimensional energy shell $M^{0}$.

$\centerdot$ In Step 1.1, we show that the non-wandering set of $\Phi^{H}_{t}$
consists of four fixed points;

$\centerdot$ In Step 1.2, we classify these fixed points by linearization;

$\centerdot$ In Step 1.3, we show that for each solution $v_{-}$ of (7.2), the
$\alpha$-limit set of any $(v_{-},L,0)$-calibrated curve
$\gamma:(-\infty,0]\rightarrow\mathbb{S}^{1}$ with $\gamma(0)\neq\pi/2$ and
$3\pi/2$ can only be $0$ or $\pi$. We only focus on the projected
$\alpha$-limit set defined on $M$. For simplicity, we define

$\alpha(\gamma):=\\{x\in M:\ \textrm{there\ exists\ a\ sequence}\
t_{n}\rightarrow-\infty\ \textrm{such\ that}\ \text{d}(\gamma(t_{n}),x)\to
0\\},$

where $\gamma:(-\infty,0]\rightarrow M$ is a $(u_{-},L,0)$-calibrated curve.
Moreover, we check the constant curves $\gamma(t)\equiv 0,\pi$ are calibrated
curves, which implies $v_{-}(0)=v_{-}(\pi)=0$,
$v^{\prime}_{-}(0)=v^{\prime}_{-}(\pi)=0$.

  * •

In Step 2, we prove the uniqueness of the solution of (7.2).

$\centerdot$ In Step 2.1, we first prove that $v_{-}$ is unique near $0$ and
$\pi$ by the the local existence and uniqueness of the solution of ODEs. We
then prove that $v_{-}$ is unique on $[0,\pi]$ by the the comparison principle
for the Dirichlet problem;

$\centerdot$ In Step 2.2, we first prove that $v_{-}$ is unique near $\pi$ and
$2\pi$ by the the local existence and uniqueness of the solution of ODEs. We
then prove that $v_{-}$ is unique on $[\pi,2\pi]$ by the comparison along
calibrated curves via the Gronwall inequality.

Step 1. The dynamical behavior of the contact Hamiltonian flow.

For each solution $v_{-}$ of (7.2), let
$\gamma:(-\infty,0]\rightarrow\mathbb{S}^{1}$ be a $(v_{-},L,0)$-calibrated
curve. Similar to the analysis at the beginning of [12, Section 3.2], the
derivative $v^{\prime}_{-}(\gamma(t))$ exists for each $t\in(-\infty,0)$ and
the orbit $(\gamma(t),v_{-}(\gamma(t)),v^{\prime}_{-}(\gamma(t)))$ satisfies
the contact Hamilton equations generated by the Hamiltonian $H(x,u,p)$ defined
in (7.3). Then the proof of the uniqueness of the solution of (7.2) is related
to the contact Hamiltonian flow $\Phi^{H}_{t}$ generated by $H(x,u,p)$.

Since $c_{0}=0$ and
$H(\gamma(t),v_{-}(\gamma(t)),v^{\prime}_{-}(\gamma(t)))=0$ for
$t\in(-\infty,0)$, we discuss the flow on the two dimensional energy shell

$M^{0}:=\\{(x,u,p)\in T^{*}\mathbb{S}^{1}\times\mathbb{R}:\ H(x,u,p)=0\\}.$

Note that along the contact Hamiltonian flow, we have $dH/dt=-H\partial
H/\partial u$, which equals to zero on the set $M^{0}$. Thus, $M^{0}$ is an
invariant set under the action of $\Phi^{H}_{t}$. Since we are interested in
the orbit $(\gamma(t),v_{-}(\gamma(t)),v^{\prime}_{-}(\gamma(t)))$, we then
consider the flow $\Phi^{H}_{t}$ restrict on $M^{0}$. The contact Hamilton
equations then reduce to

$\left\\{\begin{aligned} &\dot{x}=p,\\\ &\dot{p}=-(\cos x\cdot u-2\sin
2x)-\sin x\cdot p,\\\ &\dot{u}=p^{2}.\\\ \end{aligned}\right.$ (7.4)

Step 1.1. The non-wandering set. We first consider the non-wandering set
$\Omega$ of $\Phi^{H}_{t}|_{M^{0}}$. Suppose there is an orbit
$(x(t),u(t),p(t))$ belongs to $\Omega$. Since $\dot{u}=p^{2}\geq 0$, $u(t)$
equals to a constant $c_{u}$ and $p(t)\equiv 0$. By $\dot{x}(t)=p(t)=0$,
$x(t)$ also equals to a constant $c_{x}$. By $H(x,u,p)=0$ and $p=0$, we have

$\sin x\cdot u+\cos 2x-1=0.$

By $p=0$ and $\dot{p}=0$ we have

$\cos x\cdot u-2\sin 2x=0.$

A direct calculation shows that the only non-wandering points are

$P_{1}=(0,0,0),\quad P_{2}=(\pi,0,0),\quad P_{3}=(\frac{\pi}{2},2,0),\quad
P_{4}=(\frac{3\pi}{2},-2,0).$

Step 1.2. The classification of fixed points. We then consider the dynamical
behavior of $\Phi^{H}_{t}|_{M^{0}}$ near the fixed points. Near the points
$P_{1}$ and $P_{2}$, the linearised equation of (7.4) is

$\dot{x}=p,\quad\dot{p}=4x,\quad\dot{u}=0.$

Thus, $P_{1}$ and $P_{2}$ are hyperbolic fixed points for the dynamical system
$\Phi^{H}_{t}|_{M^{0}}$. Near the points $P_{3}$ and $P_{4}$, the linearised
equations of (7.4) are

$\dot{x}=p,\quad\dot{p}=-2x-p,\quad\dot{u}=0$

and

$\dot{x}=p,\quad\dot{p}=-2x+p,\quad\dot{u}=0$

respectively. Thus, $P_{3}$ is a stable focus, and $P_{4}$ is an unstable
focus.

Step 1.3. The $\alpha$-limit set of calibrated curves. The $\alpha$-limit set
of a $(v_{-},L,0)$-calibrated curve $\gamma$ is contained in the projection of
$\Omega$. If $\gamma$ itself is not a fixed point, and the $\alpha$-limit of
$\gamma$ is a focus, then there are two constants $t_{1}<t_{2}<0$ with
$\gamma(t_{1})=\gamma(t_{2})$ such that $v^{\prime}_{-}(\gamma(t_{1}))\neq
v^{\prime}_{-}(\gamma(t_{2}))$, which is impossible. In other words, the obits
near a focus can not form a 1-graph. Thus, the $\alpha$-limit of
$\gamma:(-\infty,0]\rightarrow\mathbb{S}^{1}$ with $\gamma(0)\neq\pi/2,3\pi/2$
can only be either $0$ or $\pi$. For constant curve
$\gamma:(-\infty,0]\rightarrow\mathbb{S}^{1}$ with $\gamma(t)\equiv x_{0}$ and
$x_{0}$ equals to either $0$ or $\pi$, we have

$v_{-}(x_{0})-v_{-}(x_{0})=0=\int_{0}^{t}L(x_{0},v_{-}(x_{0}),0)ds,$

where

$L(x,u,\dot{x})=\frac{\dot{x}^{2}}{2}-\sin x\cdot u-\cos 2x+1$

is the Lagrangian corresponding to $H(x,u,p)$. Then the constant curve
$\gamma$ is a $(v_{-},L,0)$-calibrated curve. We then have

$\lim_{t\rightarrow-\infty}v_{-}(\gamma(t))=v_{-}(0)=v_{-}(\pi)=c_{u}=0,$

and

$\lim_{t\rightarrow-\infty}v^{\prime}_{-}(\gamma(t))=v^{\prime}_{-}(0)=v^{\prime}_{-}(\pi)=0.$

Step 2. The uniqueness of the solution $v_{-}$ of (7.2).

Step 2.1. We first consider the uniqueness on $[0,\pi]$. For
$x\in[0,\pi]\backslash\\{\pi/2\\}$, let
$\gamma:(-\infty,0]\rightarrow\mathbb{S}^{1}$ with $\gamma(0)=x$ be a
$(v_{-},L,0)$-calibrated curve. Since
$\dot{\gamma}(t)=v^{\prime}_{-}(\gamma(t))$ by the first equality of (7.4),
$\gamma$ can not backtrack. We claim that there is a constant $\delta>0$ such
that for $x\in[0,\delta]$, the $\alpha$-limit of the calibrated curve $\gamma$
is $0$. If not, $v_{-}$ is decreasing on $(0,\pi]$, since $v_{-}$ is
nondecreasing along $\gamma$ by the last equality of (7.4). By Step 1.3,
$v_{-}(0)=v_{-}(\pi)=0$, we get $v_{-}\equiv 0$ on $[0,\pi]$, which is
impossible. Thus, for $x\in[0,\delta]$, $v_{-}(x)$ is increasing in $x$. We
then have

$v^{\prime}_{-}=\sqrt{2(1-\cos 2x-\sin x\cdot v_{-})}.$

For $x\in[0,\delta]$, we define

$F_{1}(x,v):=\left\\{\begin{aligned} &\sqrt{2(1-\cos 2x-v\sin x)},\quad 0\leq
v\leq x,\\\ &\sqrt{2(1-\cos 2x-x\sin x)},\quad x\leq v\leq\delta.\\\
\end{aligned}\right.$

Since $v_{-}(0)=v^{\prime}_{-}(0)=0$, we have $\lim_{x\rightarrow
0+}v_{-}(x)/x=0$. By continuity, $v_{-}(x)\leq x$ for $x>0$ small enough.
Thus, for $\delta$ small enough, $v_{-}(x)$ is a solution of
$v^{\prime}=F_{1}(x,v)$ with $v(0)=0$. For $0\leq v\leq x$, the derivative of
$F_{1}(x,v)$ with respect to $v$ is

$\partial_{v}F_{1}(x,v)=\frac{-\sin x}{2\sqrt{2(1-\cos 2x-v\sin x)}},$

and satisfies

$\frac{-\sin x}{2\sqrt{2(1-\cos 2x-x\sin x)}}\leq\partial_{v}F_{1}(x,v)\leq
0.$

Since

$\lim_{x\rightarrow 0+}\frac{-\sin x}{2\sqrt{2(1-\cos 2x-x\sin
x)}}=-\frac{1}{2\sqrt{2}},$

the derivative $\partial_{v}F_{1}(x,v)$ is bounded for $x$ near $0$. By the
local existence and uniqueness of the solutions of ODEs, and shrink $\delta$
if necessary, $v_{-}$ is unique on $[0,\delta]$. Similarly, $v_{-}$ is unique
on $[\pi-\delta,\pi]$. Since $\sin x\geq\sin\delta>0$ on
$[\delta,\pi-\delta]$, by the uniqueness of the solution of the Dirichlet
problem (cf. [4, Theorem 3.3]), $v_{-}$ is unique on $[0,\pi]$.

Step 2.2. It remains to consider the uniqueness on $[\pi,2\pi)$. For
$x\in[\pi,2\pi)\backslash\\{3\pi/2\\}$, let
$\gamma:(-\infty,0]\rightarrow\mathbb{S}^{1}$ with $\gamma(0)=x$ be a
$(v_{-},L,0)$-calibrated curve. Similar to Step 2.1, one can prove that there
is a constant $\delta>0$ such that for $x\in[\pi,\pi+\delta]$, the
$\alpha$-limit of the calibrated curve $\gamma$ is $\pi$. Thus, for
$x\in[\pi,\pi+\delta]$, $v_{-}(x)$ is increasing in $x$. We then have

$v^{\prime}_{-}=\sqrt{2(1-\cos 2x-\sin x\cdot v_{-})}.$

For $x\in[\pi,\pi+\delta]$, we define

$F_{2}(x,v):=\sqrt{2(1-\cos 2x-v\sin x)},\quad 0\leq v\leq\delta.$

Then $v_{-}(x)$ is a solution of $Dv=F_{2}(x,v)$ with $v(\pi)=0$. The
derivative of $F_{2}(x,v)$ with respect to $v$ is

$\partial_{v}F_{2}(x,v)=\frac{-\sin x}{2\sqrt{2(1-\cos 2x-v\sin x)}},$

and satisfies

$0\leq\partial_{v}F_{2}(x,v)\leq\frac{-\sin x}{2\sqrt{2(1-\cos 2x)}}.$

Since

$\lim_{x\rightarrow\pi+}\frac{-\sin x}{2\sqrt{2(1-\cos 2x)}}=\frac{1}{4},$

the derivative $\partial_{v}F_{2}(x,v)$ is bounded for $x$ near $\pi$. By the
local existence and uniqueness of the solutions of ODEs, and shrink $\delta$
if necessary, $v_{-}$ is unique on $[\pi,\pi+\delta]$. Similarly, $v_{-}$ is
unique on $[2\pi-\delta,2\pi]$.

Assume that there are two solutions $u_{-}$ and $v_{-}$ satisfying
$u_{-}(x)>v_{-}(x)$ at some point $x\in(\pi+\delta,3\pi/2)$. Let $\gamma$ be a
$(v_{-},L,0)$-calibrated curve with $\gamma(0)=x$. Without any loss of
generality, we assume the $\alpha$-limit of $\gamma$ is $\pi$. Define
$t_{0}<0$ such that $\gamma(t_{0})=\pi+\delta$, and let

$G(s):=u_{-}(\gamma(s))-v_{-}(\gamma(s)),\quad s\in[t_{0},0].$

Then $G(t_{0})=0$ and $G(0)>0$. By continuity, there is
$\sigma_{0}\in[t_{0},0)$ such that $G(\sigma_{0})=0$ and $G(\sigma)>0$ for all
$\sigma\in(\sigma_{0},0]$. By definition we have

$u_{-}(\gamma(\sigma))-u_{-}(\gamma(\sigma_{0}))\leq\int_{\sigma_{0}}^{\sigma}L(\gamma(s),u_{-}(\gamma(s)),\dot{\gamma}(s))ds,$

and

$v_{-}(\gamma(\sigma))-v_{-}(\gamma(\sigma_{0}))=\int_{\sigma_{0}}^{\sigma}L(\gamma(s),v_{-}(\gamma(s)),\dot{\gamma}(s))ds,$

which implies

$G(\sigma)\leq\int_{\sigma_{0}}^{\sigma}G(s)ds.$

By the Gronwall inequality, we have $G(\sigma)\equiv 0$ for all
$\sigma\in(\sigma_{0},0]$, which contradicts $u_{-}(x)>v_{-}(x)$. The case
$x\in(3\pi/2,2\pi-\delta)$ is similar. By the continuity of $v_{-}$ at
$3\pi/2$, we finally conclude that the solution is unique on $[\pi,2\pi)$. ∎

Acknowledgements: The authors would like to thank Professor J. Yan for many
helpful discussions. Lin Wang is supported by NSFC Grant No. 12122109,
11790273.

##  Appendix A Proof of Proposition 2.5

######  Lemma A.1.

If $\varphi$ is a Lipschitz subsolution of (B0), then $\varphi\prec L$.

###### Proof.

Without loss of generality, we assume $M$ is an open set of $\mathbb{R}^{n}$.
In fact, for each absolutely continuous curve $\gamma:[0,t]\rightarrow M$, we
use a covering of it by local coordinate charts. Clearly, there exists
$N\in\mathbb{N}$ such that $[0,t]=\cup_{i=0}^{N-1}[t_{i},t_{i+1}]$ with
$t_{0}=0$, $t_{N}=t$, such that $\gamma|_{[t_{i},t_{i+1}]}$ is contained in an
open subset of $\mathbb{R}^{n}$.

By [10] Proposition 2.4, there is a function $q\in
L^{\infty}([0,t],\mathbb{R}^{n})$ such that for almost all $s\in[0,t]$, we
have

$\frac{d}{ds}\varphi(\gamma(s))=q(s)\cdot\dot{\gamma}(s),$

and the vector $q(s)$ belongs to $\partial_{c}\varphi(\gamma(s))$. Here we
recall the definition of the Clarke’s generalized gradient

$\partial_{c}\varphi(x):=\bigcap_{r>0}\overline{\textrm{co}}\\{D\varphi(y):\
y\in B(x,r),\ \textrm{and}\ \varphi\ \textrm{is\ differentiable\ at}\ y\\},$

where $\overline{\textrm{co}}$ stands for the closure of the convex
combination. Since $\varphi$ is a Lipschitz subsolution of (B0), if $\varphi$
is differentiable at $y$, we have

$H(y,\varphi(y),D\varphi(y))\leq 0.$

By the convexity of $H$ with respect to $p$, and the definition of
$\partial_{c}\varphi(x)$, we have

$H(x,\varphi(x),q)\leq 0,\quad\forall q\in\partial_{c}\varphi(x).$

We conclude that

$\displaystyle\varphi(\gamma(t))-\varphi(\gamma(0))$
$\displaystyle=\int_{0}^{t}\frac{d}{ds}\varphi(\gamma(s))ds=\int_{0}^{t}q(s)\cdot\dot{\gamma}(s)ds$
$\displaystyle\leq\int_{0}^{t}\bigg{[}L(\gamma(s),\varphi(\gamma(s)),\dot{\gamma}(s))+H(\gamma(s),\varphi(\gamma(s)),q(s))\bigg{]}ds$
$\displaystyle\leq\int_{0}^{t}L(\gamma(s),\varphi(\gamma(s)),\dot{\gamma}(s))ds,$

which implies $\varphi\prec L$. ∎

######  Lemma A.2.

If $\varphi\prec L$, then for each $t\geq 0$, we have
$T_{t}^{-}\varphi\geq\varphi\geq T_{t}^{+}\varphi$. Moreover, if there exists
$\epsilon_{0}>0$ such that for a.e. $x\in M$,

$H(x,u,Du)+\epsilon_{0}\leq 0.$

then

$T_{t}^{+}\varphi<\varphi<T_{t}^{-}\varphi.$

###### Proof.

In the following, we only prove $T_{t}^{-}\varphi\geq\varphi$ for each $t\geq
0$, since the proof of $T_{t}^{+}\varphi\leq\varphi$ is similar. By
contradiction, we assume there exists $x_{0}\in M$ such that
$\varphi(x_{0})>T_{t}^{-}\varphi(x_{0})$. Let $\gamma:[0,t]\rightarrow M$ be a
minimizer of $T_{t}^{-}\varphi$ with $\gamma(t)=x_{0}$, i.e.

$T_{t}^{-}\varphi(x)=\varphi(\gamma(0))+\int_{0}^{t}L(\gamma(\tau),T_{\tau}^{-}\varphi(\gamma(\tau)),\dot{\gamma}(\tau))d\tau.$
(A.1)

Let $F(\tau):=\varphi(\gamma(\tau))-T_{\tau}^{-}\varphi(\gamma(\tau))$. Since
$F(t)>0$ and $F(0)=0$, then one can find $s_{0}\in[0,t)$ such that
$F(s_{0})=0$ and $F(s)>0$ for $s\in(s_{0},t]$. A direct calculation shows

$F(s)\leq\Theta\int_{s_{0}}^{s}F(\tau)d\tau,$

which implies $F(s)\leq 0$ for $s\in(s_{0},t]$ from the Gronwall inequality.
It contradicts $F(t)>0$.

Next, we assume there exists $\epsilon_{0}>0$ such that for a.e. $x\in M$,

$H(x,u,Du)+\epsilon_{0}\leq 0.$

Let us denote

$\tilde{L}(x,u,\dot{x}):=L(x,u,\dot{x})-\epsilon_{0},$

and let $\tilde{T}_{t}^{-}$ be the Lax-Oleinik semigroup associated to
$\tilde{L}$. By a similar argument above, we have
$\tilde{T}_{t}^{-}\varphi\geq\varphi$ and
$\tilde{T}_{t}^{+}\varphi\leq\varphi$. Note that $\tilde{L}<L$. Using a
similar argument as [16, Proposition 3.1],
$\tilde{T}_{t}^{-}\varphi<{T}_{t}^{-}\varphi$ and
$\tilde{T}_{t}^{+}\varphi>{T}_{t}^{+}\varphi$ for each $t>0$. Therefore,
$T_{t}^{-}\varphi>\varphi$ and $T_{t}^{+}\varphi<\varphi$ for each $t>0$. This
completes the proof. ∎

######  Lemma A.3.

If for each $t>0$, $T^{-}_{t}\varphi\geq\varphi$, then $\varphi$ is a
Lipschitz subsolution of (E0).

###### Proof.

Fix $T>0$, by assumption we have $T^{-}_{t}\varphi\geq\varphi$ for each
$t\in[0,T]$. By [13], there is a constant $R>0$ depending on $T$ and
$\|D\varphi\|_{\infty}$, such that
$\|\partial_{x}T^{-}_{t}\varphi(x)\|_{\infty}\leq R$. We make a modification

$H_{R}(x,u,p):=H(x,u,p)+\max\\{\|p\|^{2}-R^{2},0\\}.$

Then $T^{-}_{t}\varphi$ is also the solution of (A0) with $H$ replaced by
$H_{R}$. One can prove that the Lagrangian $L_{R}$ corresponding to $H_{R}$ is
continuous. By the uniqueness of the solution of (A0), we have
$T^{-}_{t}\varphi=T^{R}_{t}\varphi$, where $T^{R}_{t}\varphi$ is defined by
(2.1) with $L$ replaced by $L_{R}$.

Let $\varphi$ be differentiable at $x\in M$. For each $v\in T_{x}M$, there is
a $C^{1}$ curve $\gamma:[0,T]\rightarrow M$ with $\gamma(0)=x$ and
$\dot{\gamma}(0)=v$. By assumption for each $t\in[0,T]$, we have

$\varphi(\gamma(t))\leq
T^{-}_{t}\varphi(\gamma(t))=T^{R}_{t}\varphi(\gamma(t))\leq\varphi(x)+\int_{0}^{t}L_{R}(\gamma(s),T^{R}_{s}\varphi(\gamma(s)),\dot{\gamma}(s))ds.$

Dividing by $t$ and let $t$ tend to zero, using the continuity of $\gamma$,
$L_{R}$ and $T^{R}_{t}\varphi(x)$. We get

$D\varphi(x)\cdot v\leq L_{R}(x,\varphi(x),v).$

Since $v$ is arbitrary, we have

$H_{R}(x,\varphi(x),D\varphi(x))=\sup_{v\in T_{x}M}\bigg{[}D\varphi(x)\cdot
v-L_{R}(x,\varphi(x),v)\bigg{]}\leq 0.$

Therefore, $\varphi$ is a Lipschitz subsolution of

$H_{R}(x,u(x),Du(x))=0.$

Since $H_{R}$ converge to $H$ uniformly on compact subsets of
$T^{*}M\times\mathbb{R}$, by the stability of subsolutions, $\varphi$ is also
a Lipschitz subsolution of (B0). ∎

##  Appendix B Generalization of JYZ’s results

###  B.1 $c_{0}$ and subsolutions

Inspired by [5], we denote

$c_{0}:=\inf_{u\in C^{\infty}(M)}\sup_{x\in
M}\bigg{\\{}H(x,Du)+\lambda(x)u\bigg{\\}}.$

######  Proposition B.1.

$c_{0}$ is finite.

###### Proof.

Choose $u(x)\equiv 0$, then by definition,

$c_{0}\leq\sup_{x\in M}H(x,0)<+\infty.$

Let us recall

$\mathbf{e}_{0}:=\min_{(x,p)\in T^{*}M}H(x,p)>-\infty.$

By the assumption ($\pm$), there exists $x_{0}\in M$ such that
$\lambda(x_{0})=0$. Thus for each $u\in C^{\infty}(M)$,

$\displaystyle c_{0}=$ $\displaystyle\inf_{u\in C^{\infty}(M)}\sup_{x\in
M}\,\,\bigg{\\{}H(x,Du(x))+\lambda(x)u(x)\bigg{\\}}$ $\displaystyle\geqslant$
$\displaystyle\,\inf_{u\in
C^{\infty}(M)}\,\,\bigg{\\{}H(x_{0},Du(x_{0}))+\lambda(x_{0})u(x_{0})\bigg{\\}}$
$\displaystyle=$ $\displaystyle\,\inf_{u\in
C^{\infty}(M)}H(x_{0},Du(x_{0}))\geqslant\mathbf{e}_{0}.$

This means $c_{0}$ is finite. ∎

######  Proposition B.2.

For $c<c_{0}$, (E0) has no continuous subsolutions.

###### Proof.

By contradiction, we assume for $c<c_{0}$, (E0) admits a continuous
subsolution $u:M\rightarrow\mathbb{R}$. Let

$\lambda_{0}:=\|\lambda(x)\|_{\infty}>0.$ (B.1)

By the definition of the subsolution, for any $p\in D^{+}u(x)$,

$\displaystyle H(x,p)\leq c-\lambda(x)u(x)\leq c+\lambda_{0}\|u\|_{\infty}.$

Combining (CER), one can conclude that $u$ is Lipschitz continuous (see [8,
Proposition 1.14] for more details). By [6, Lemma 2.2], for all
$\varepsilon>0$, there exists $u_{\varepsilon}\in C^{\infty}(M)$ such that
$\|u-u_{\varepsilon}\|_{\infty}<\varepsilon$ and for all $x\in M$,

$H(x,Du_{\varepsilon}(x))+\lambda(x)u(x)\leq c+\varepsilon.$

We choose $\varepsilon=\frac{1}{2(1+\lambda_{0})}(c_{0}-c)>0$, then

$\displaystyle H(x,Du_{\varepsilon}(x))+\lambda(x)u_{\varepsilon}(x)$
$\displaystyle\leqslant$
$\displaystyle\,H(x,Du_{\varepsilon}(x))+\lambda(x)u(x)+\lambda_{0}\|u-u_{\varepsilon}\|_{\infty}$
$\displaystyle\leqslant$ $\displaystyle\,c+(1+\lambda_{0})\varepsilon<c_{0},$

this contradicts the definition of $c_{0}$. ∎

###  B.2 Existence of subsolutions and solutions

Let us recall that $T_{t}^{\pm}$ denote the Lax-Oleinik semigroups associated
to

$L(x,\dot{x})-\lambda(x)u(x)+c.$

######  Proposition B.3.

For $c\geq c_{0}$, (E0) has a Lipschitz subsolution. Let $u_{0}$ be a
subsolution of (E0) with $c=c_{0}$. For $c>c_{0}$, there holds

$T_{t}^{+}u_{0}<u_{0}<T_{t}^{-}u_{0}.$

###### Proof.

By the definition of $c_{0}$, there exists $u_{n}\in C^{\infty}(M)$ such that
for all $x\in M$,

$H(x,Du_{n}(x))+\lambda(x)u_{n}(x)\leqslant c_{0}+\frac{1}{n}.$ (B.2)

Namely, $u_{n}$ is a subsolution of

$H(x,Du)+\lambda(x)u=c_{0}+\frac{1}{n},$

By (3.4) and Proposition 3.4, $\\{u_{n}\\}_{n\geq 1}$ is equi-bounded and
equi-Lipschitz continuous. Then by the Ascoli-Arzelà theorem, it contains a
subsequence $\\{u_{n_{k}}\\}_{k\in\mathbb{N}}$ uniformly converging on $M$ to
some $u_{0}\in$ Lip$(M)$. By the stability of subsolutions (see [3, Theorem
5.2.5]), $u_{0}$ is a subsolution of

$H(x,Du)+\lambda(x)u=c_{0}.$

Moreover, for $c>c_{0}$ and a.e. $x\in M$, we have

$H(x,Du_{0})+\lambda(x)u_{0}+(c-c_{0})\leq c.$

By Lemma A.2,

$T_{t}^{+}u_{0}<u_{0}<T_{t}^{-}u_{0}.$

This completes the proof. ∎

Combining Propositions B.2, B.3 and 3.5, we conclude that (E0) has a solution
if and only if $c\geq c_{0}$. It remains to prove the following result.

######  Proposition B.4.

(E0) has at least two solutions for $c>c_{0}$.

###### Proof.

By Proposition B.3, if $c>c_{0}$, there exists a strict Lipschitz subsolution
$u_{0}$ of (E0). Based on Proposition 2.5, for $t>0$,

$T_{t}^{-}u_{0}(x)>u_{0}(x),\quad T_{t}^{+}u_{0}(x)<u_{0}(x).$ (B.3)

Denote

$u_{-}:=\lim_{t\to+\infty}{T_{t}^{-}}u_{0}(x),\quad
u_{+}:=\lim_{t\to+\infty}{T_{t}^{+}}u_{0}(x).$ (B.4)

and

$v_{-}:=\lim_{t\to+\infty}T_{t}^{-}u_{+}(x).$ (B.5)

By Proposition 3.5, $u_{-}$ and $v_{-}$ are solutions of (E0).

It remains to verify $u_{-}\neq v_{-}$. By contradiction, we assume
$u_{-}\equiv v_{-}$ on $M$. In view of (B.5), we have

$u_{-}=\lim_{t\to+\infty}T_{t}^{-}u_{+}(x).$ (B.6)

Based on (B.6), it follows from Proposition 2.7 that

$\mathcal{I}_{u_{+}}:=\\{x\in M:\ u_{-}(x)=u_{+}(x)\\}\neq\emptyset.$ (B.7)

On the other hand, from (B.3) and (B.4), it follows that for any $x\in M$,

$u_{+}(x)<u_{0}(x)<u_{-}(x),$ (B.8)

which implies

$\mathcal{I}_{u_{+}}=\emptyset.$

This contradicts (B.7). ∎

##  Appendix C The locally Lipschitz estimate of $T^{-}_{t}\varphi(x)$

In this section, we prove the locally Lipschitz property of
$T^{-}_{t}\varphi(x)$ generated by

$\left\\{\begin{aligned}
&\partial_{t}u(x,t)+H(x,\partial_{x}u(x,t))+f(x,u(x,t))=0,\quad(x,t)\in
M\times(0,+\infty).\\\ &u(x,0)=\varphi(x),\quad x\in M,\\\
\end{aligned}\right.$

where $\varphi\in C(M)$. In the following, we assume that $f(x,u)$ satisfies

  * (Cf):

$f(x,u)$ and $\frac{\partial f}{\partial u}(x,u)$ are continuous;

  * (LIP):

there is a constant $\lambda>0$ such that $|\frac{\partial f}{\partial
u}(x,u)|\leq\lambda$.

We also assume that $H(x,p)$ satisfies

  * (CH):

$H(x,p)$ is continuous;

  * (STC):

$H(x,p)$ is strictly convex in $p$, for any $x\in M$;

  * (SL):

there are two superlinear functions $\theta_{1}(r)$ and $\theta_{2}(r)$ such
that $\theta_{1}(\|p\|)\leq H(x,p)\leq\theta_{2}(\|p\|)$.

Then the corresponding Lagrangian $L(x,\dot{x})$ satisfies

  * (CL):

$L(x,\dot{x})$ and $\frac{\partial L}{\partial\dot{x}}(x,\dot{x})$ are
continuous;

  * (CON):

$L(x,\dot{x})$ is convex in $\dot{x}$, for any $x\in M$;

  * (SL):

there are two superlinear functions $\theta^{*}_{1}(r)$ and
$\theta^{*}_{2}(r)$ such that $\theta^{*}_{2}(\|\dot{x}\|)\leq
L(x,\dot{x})\leq\theta^{*}_{1}(\|\dot{x}\|)$.

######  Lemma C.1.

Let

$f(\varepsilon):=\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon})\cdot\frac{\dot{x}}{1+\varepsilon}-L(x,\frac{\dot{x}}{1+\varepsilon}),$

then the following statements hold

  * (1)

The function $f(\varepsilon)$ is decreasing for $\varepsilon>-1$. In
particular, we have $f(\varepsilon)\geq
f(+\infty)=-L(x,0)\geq-\theta^{*}_{1}(0)$.

  * (2)

If $\varepsilon_{1},\varepsilon_{2}>-1$, then

$L(x,\frac{\dot{x}}{1+\varepsilon_{2}})\leq(\kappa+1)^{-1}L(x,\frac{\dot{x}}{1+\varepsilon_{1}})+\kappa(\kappa+1)^{-1}\theta^{*}_{1}(0)$

and

$f(\varepsilon_{2})\leq\kappa^{-1}L(x,\frac{\dot{x}}{1+\varepsilon_{1}})-(\kappa^{-1}+1)L(x,\frac{\dot{x}}{1+\varepsilon_{2}}),$

where $\kappa:=(\varepsilon_{2}-\varepsilon_{1})/(1+\varepsilon_{1})>0$.

###### Proof.

The proof is quite similar to [1, Lemma 5].

(1) Let $\varepsilon_{2}>\varepsilon_{1}>-1$, by (CON) we have

$L(x,\frac{\dot{x}}{1+\varepsilon_{2}})\geq
L(x,\frac{\dot{x}}{1+\varepsilon_{1}})+\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{1}})\cdot(\frac{\dot{x}}{1+\varepsilon_{2}}-\frac{\dot{x}}{1+\varepsilon_{1}}),$

where $\dot{x}/(1+\varepsilon_{1})$ and $\dot{x}/(1+\varepsilon_{2})$ belong
to the linear space $T_{x}M$. It follows that

$\displaystyle f(\varepsilon_{1})-f(\varepsilon_{2})$
$\displaystyle\geq\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{1}})\cdot\frac{\dot{x}}{1+\varepsilon_{1}}-\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}})\cdot\frac{\dot{x}}{1+\varepsilon_{2}}$
$\displaystyle\quad+\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{1}})\cdot(\frac{\dot{x}}{1+\varepsilon_{2}}-\frac{\dot{x}}{1+\varepsilon_{1}})$
$\displaystyle=\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{1}})\cdot\frac{\dot{x}}{1+\varepsilon_{2}}-\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}})\cdot\frac{\dot{x}}{1+\varepsilon_{2}}$
$\displaystyle=(\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{1}})-\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}}))(\frac{\dot{x}}{1+\varepsilon_{1}}-\frac{\dot{x}}{1+\varepsilon_{2}})(\frac{1}{1+\varepsilon_{1}}-\frac{1}{1+\varepsilon_{2}})^{-1}\frac{1}{1+\varepsilon_{1}}$
$\displaystyle\geq 0.$

(2) By (CON) we have

$\displaystyle L(x,\frac{\dot{x}}{1+\varepsilon_{1}})$ $\displaystyle\geq
L(x,\frac{\dot{x}}{1+\varepsilon_{2}})+\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}})\cdot(\frac{\dot{x}}{1+\varepsilon_{1}}-\frac{\dot{x}}{1+\varepsilon_{2}})$
$\displaystyle=L(x,\frac{\dot{x}}{1+\varepsilon_{2}})+\kappa\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}})\cdot\frac{\dot{x}}{1+\varepsilon_{2}}.$

By (1) we have

$\displaystyle
L(x,\frac{\dot{x}}{1+\varepsilon_{1}})-(\kappa+1)L(x,\frac{\dot{x}}{1+\varepsilon_{2}})$
$\displaystyle\geq\kappa(-L(x,\frac{\dot{x}}{1+\varepsilon_{2}})+\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}})\cdot\frac{\dot{x}}{1+\varepsilon_{2}})\geq-\kappa\theta^{*}_{1}(0),$

which implies the first assertion. Moreover we have

$\frac{\partial
L}{\partial\dot{x}}(x,\frac{\dot{x}}{1+\varepsilon_{2}})\cdot\frac{\dot{x}}{1+\varepsilon_{2}})\leq\kappa^{-1}(L(x,\frac{\dot{x}}{1+\varepsilon_{1}})-L(x,\frac{\dot{x}}{1+\varepsilon_{2}})),$

which leads the second assertion. ∎

######  Lemma C.2.

(Erdmann condition). For each $(x,t)\in M\times(0,+\infty)$, let
$\gamma:[0,t]\rightarrow M$ be a minimizer of $T^{-}_{t}\varphi(x)$. Set
$T^{-}_{s}\varphi(\gamma(s)):=u_{1}(s)$ with $s\in[0,t]$, and

$E_{0}(s):=\frac{\partial
L}{\partial\dot{x}}(\gamma(s),\dot{\gamma}(s))\cdot\dot{\gamma}(s)-L(\gamma(s),\dot{\gamma}(s)),$

then

$E(s):=e^{\int_{0}^{t}\partial f/\partial
u(\gamma(r),u_{1}(r))dr}[E_{0}(s)+f(\gamma(s),u_{1}(s))]$

satisfies $\dot{E}(s)=0$ a.e on $[0,t]$.

###### Proof.

The proof is quite similar to [2, Theorem 2.2], one can also refer to [1,
Lemma 6] for the autonomous case.

Step 1. Reparameterization. Let $\alpha:[0,t]\rightarrow[1/2,3/2]$ be a
measurable function satisfying $\int_{0}^{t}\alpha(s)ds=t$. We denote the set
of such functions by $\Omega$. Define $\tau(s):=\int_{0}^{s}\alpha(r)dr$. Then
$\tau(s)$ is a bi-Lipschitz map from $[0,t]$ to itself. The inverse of $\tau$
satisfies $s^{\prime}(\tau)=(\alpha(s(\tau)))^{-1}$ for a.e. $\tau\in[0,t]$.
We define an absolutely continuous curve $\eta:[0,t]\rightarrow M$ by
$\eta(\tau):=\gamma(s(\tau))$, then
$\dot{\eta}(\tau)=\dot{\gamma}(s(\tau))/\alpha(s(\tau))$. It is clear that
$L(\gamma(s),\dot{\gamma}(s))\in L^{1}[0,t]$.

Step 1.1. We are now going to show that there is a measurable function
$\delta(s)$ such that for almost all $s\in[0,t]$, if
$|\alpha(s)-1|\leq\delta(s)$, then
$L(\gamma(s),\dot{\gamma}(s)/\alpha(s))\alpha(s)\in L^{1}[0,t]$. We denote by
$\Omega_{0}$ the set of all such functions $\alpha(s)$. The proof is quite
similar to Step II of the proof of [2, Theorem 2.2]. For $\alpha\in[1/2,3/2]$,
we define

$\Phi_{1}(s,\alpha):=L(\gamma(s),\dot{\gamma}(s)/\alpha)\alpha-L(\gamma(s),\dot{\gamma}(s)).$

For almost all $s$, by continuity of $L$ with respect to $\dot{x}$, there
exists $\delta_{1}(s)\in(0,1/2]$ such that

$-1\leq\Phi_{1}(s,\alpha)-\Phi_{1}(s,1)\leq
1,\quad\forall\alpha\in[1-\delta_{1}(s),1+\delta_{1}(s)].$

We define a set-valued map $G:[0,t]\rightrightarrows\mathbb{R}$ by

$[0,t]\ni s\mapsto G(s)=\\{\delta>0:\
\Phi_{1}(s,[1-\delta,1+\delta])\subset\Phi_{1}(s,1)+[-1,1]\\},$

where the value set is nonempty by the existence of $\delta_{1}(s)$. For each
$k\in\mathbb{N}$, we define a set-valued map
$G_{k}:[0,t]\rightrightarrows\mathbb{R}$ by

$\textrm{dom}(G_{k})\ni s\mapsto G_{k}(s)=\\{\delta>1/k:\
\Phi_{1}(s,[1-\delta,1+\delta])\subset\Phi_{1}(s,1)+[-1,1]\\}.$

By a standard measurable selection theorem, for each $k$, there exists a
measurable selection $g_{k}:\textrm{dom}(G_{k})\rightarrow\mathbb{R}$ such
that $g_{k}(s)\in G_{k}(s)$ for all $s\in[0,t]\cap\textrm{dom}(G_{k})$. Notice
that we can assume that $g_{k}$ is nondecreasing and converges to a measurable
selection $g$ of $G$ as $k\rightarrow+\infty$. Thus, we can assume
$\delta(\cdot)$ is measurable and $\delta(s)>0$ for almost all $s\in[0,t]$.

Step 1.2. For $\alpha(s)\in\Omega_{0}$, since
$L(\gamma(s),\dot{\gamma}(s)/\alpha(s))\alpha(s)\in L^{1}[0,t]$ by Step 1.1,
the following Cauchy problem

$\left\\{\begin{aligned}
&\dot{u}_{\alpha}(s)=[L(\gamma(s),\dot{\gamma}(s)/\alpha(s))-f(\gamma(s),u_{\alpha}(s))]\alpha(s),\quad
a.e.\ s\in[0,t],\\\ &u_{\alpha}(0)=\varphi(\gamma(0)).\end{aligned}\right.$

admits an absolutely continuous solution. By (LIP) and [2, Proposition A.1],
the solution is unique. It is direct to see that
$u_{\eta}(\tau)=u_{\alpha}(s)$, where $u_{\eta}(\tau)$ satisfies

$\left\\{\begin{aligned}
&\dot{u}_{\eta}(\tau)=L(\eta(\tau),\dot{\eta}(\tau))-f(\eta(\tau),u_{\eta}(\tau)),\quad
a.e.\ \tau\in[0,t],\\\ &u_{\eta}(0)=\varphi(\eta(0)).\end{aligned}\right.$

We also note that $T^{-}_{s}\varphi(\gamma(s))=u_{1}(s)$.

Step 1.3. Now we show that $u_{1}(t)\leq u_{\alpha}(t)$ for all
$\alpha\in\Omega_{0}$. We argue by contradiction. Assume

$T^{-}_{t}\varphi(x)=u_{1}(t)>u_{\alpha}(t)=u_{\eta}(t).$

Define

$F(\tau):=T^{-}_{\tau}\varphi(\eta(\tau))-u_{\eta}(\tau),\quad\tau\in(0,t].$

Then $F(0)=0$ and $F(t)>0$. By continuity, there is $s_{0}\in[0,t]$ such that
$F(s_{0})=0$ and $F(\tau)>0$ for all $\tau\in(s_{0},t]$. By definition we have

$T^{-}_{\sigma}\varphi(\eta(\sigma))\leq
T^{-}_{s_{0}}\varphi(\eta(s_{0}))+\int_{s_{0}}^{\sigma}\bigg{[}L(\eta(\tau),\dot{\eta}(\tau))-f(\eta(\tau),T^{-}_{\tau}\varphi(\eta(\tau)))\bigg{]}d\tau,$

and

$u_{\eta}(\sigma)=u_{\eta}(s_{0})+\int_{s_{0}}^{\sigma}\bigg{[}L(\eta(\tau),\dot{\eta}(\tau))-f(\eta(\tau),u_{\eta}(\tau))\bigg{]}d\tau.$

By the Gronwall inequality we have

$F(\sigma)\leq\lambda\int_{s_{0}}^{\sigma}F(\tau)d\tau,$

which implies that $F(t)=0$. This contradicts $F(t)>0$.

Step 2. Calculation of variation. Define the functional
$\Lambda:\Omega_{0}\rightarrow\mathbb{R}$ by $\Lambda(\alpha):=u_{\alpha}(t)$,
then $\Lambda$ attains its minimum at $\alpha=1$ by Step 1.3. For
$0\neq\beta\in L^{\infty}[0,t]$ such that $1+\beta\in\Omega_{0}$, we have
$1+\varepsilon\beta\in\Omega_{0}$ for $|\varepsilon|\leq 1$. We are now going
to show that the derivative
$\frac{d}{d\varepsilon}|_{\varepsilon=0}\Lambda(1+\varepsilon\beta)$ exists,
which naturally equals to zero.

Step 2.1. For all $\alpha(s)\in\Omega_{0}$ and almost all $s\in[0,t]$, we have

$\displaystyle\dot{u}_{\alpha}-\dot{u}_{1}$
$\displaystyle=[L(\gamma,\dot{\gamma}/\alpha)-f(\gamma,u_{\alpha})]\alpha-[L(\gamma,\dot{\gamma})-f(\gamma,u_{1})]$
(C.1)
$\displaystyle=[L(\gamma,\dot{\gamma}/\alpha)-f(\gamma,u_{\alpha})]\alpha-[L(\gamma,\dot{\gamma}/\alpha)-f(\gamma,u_{1})]\alpha$
$\displaystyle\quad+[L(\gamma,\dot{\gamma}/\alpha)-f(\gamma,u_{1})]\alpha-[L(\gamma,\dot{\gamma})-f(\gamma,u_{1})]$
$\displaystyle=\widehat{L^{\alpha}_{u}}(u_{\alpha}-u_{1})+[L(\gamma,\dot{\gamma}/\alpha)-f(\gamma,u_{1})]\alpha-[L(\gamma,\dot{\gamma})-f(\gamma,u_{1})],$

where

$\widehat{L^{\alpha}_{u}}(s):=\int_{0}^{1}-\frac{\partial f}{\partial
u}\bigg{(}\gamma(s),u_{1}(s)+\mu(u_{\alpha}(s)-u_{1}(s))\bigg{)}\alpha(s)d\mu.$

By solving (C.1) we conclude that

$u_{\alpha}(s)-u_{1}(s)=\int_{0}^{s}e^{\int_{\sigma}^{s}}\widehat{L^{\alpha}_{u}}(r)dr\bigg{\\{}[L(\gamma,\dot{\gamma}/\alpha)-f(\gamma,u_{1})]\alpha-[L(\gamma,\dot{\gamma})-f(\gamma,u_{1})]\bigg{\\}}d\sigma.$

It follows that for $\varepsilon>0$, we have

$0\leq\frac{\Lambda(1+\varepsilon\beta)-\Lambda(1)}{\varepsilon}=\int_{0}^{t}e^{\int_{s}^{t}\widehat{L^{\varepsilon}_{u}}(r)dr}\lambda_{\varepsilon}(s)ds,$

where
$\widehat{L^{\varepsilon}_{u}}(s):=\widehat{L^{1+\varepsilon\beta}_{u}}(s)$
and

$\displaystyle\lambda_{\varepsilon}(s)$
$\displaystyle:=\frac{1}{\varepsilon}\bigg{\\{}[L(\gamma,\dot{\gamma}/(1+\varepsilon\beta))-f(\gamma,u_{1})](1+\varepsilon\beta)-[L(\gamma,\dot{\gamma})-f(\gamma,u_{1})]\bigg{\\}}$
$\displaystyle=[L(\gamma,\dot{\gamma}/(1+\varepsilon\beta))-f(\gamma,u_{1})]\beta$
$\displaystyle\quad+\frac{1}{\varepsilon}\bigg{\\{}[L(\gamma,\dot{\gamma}/(1+\varepsilon\beta))-f(\gamma,u_{1})]-[L(\gamma,\dot{\gamma})-f(\gamma,u_{1})]\bigg{\\}}.$

Step 2.2. We are now going to show that

$\displaystyle
0=\frac{d}{d\varepsilon}\bigg{|}_{\varepsilon=0}\Lambda(1+\varepsilon\beta)=-\int_{0}^{t}e^{\int_{s}^{t}-\partial
f/\partial u(\gamma(r),u_{1}(r))dr}[E_{0}(s)+f(\gamma(s),u_{1}(s))]\beta ds.$
(C.2)

Set

$l_{\varepsilon}(s):=\frac{\partial
L}{\partial\dot{x}}(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})\cdot\frac{\dot{\gamma}}{1+\varepsilon\beta}-L(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})+f(\gamma,u_{1}).$

By Lemma C.1 (1),
$l_{\varepsilon}\geq-\theta^{*}_{1}(0)-\|f(x,0)\|_{\infty}-\lambda\|T^{-}_{t}\varphi\|_{\infty}$.
By (CON) we have

$\displaystyle
L(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})-L(\gamma,\dot{\gamma})\leq\frac{\partial
L}{\partial\dot{x}}(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})\cdot(\frac{\dot{\gamma}}{1+\varepsilon\beta}-\dot{\gamma})=-\varepsilon\beta\frac{\partial
L}{\partial\dot{x}}(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})\cdot\frac{\dot{\gamma}}{1+\varepsilon\beta}.$

It follows that

$\lambda_{\varepsilon}(s)\leq-\beta(s)\left(\frac{\partial
L}{\partial\dot{x}}(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})\cdot\frac{\dot{\gamma}}{1+\varepsilon\beta}-L(\gamma,\frac{\dot{\gamma}}{1+\varepsilon\beta})+f(\gamma,u_{1})\right)\leq-\beta(s)l_{\varepsilon}(s).$
(C.3)

Let $\varepsilon\in[0,1]$, we rewrite $\lambda_{\varepsilon}$,
$l_{\varepsilon}$ as $\lambda^{\beta}_{\varepsilon}$,
$l^{\beta}_{\varepsilon}$ respectively. Let $\beta^{+}$ and $\beta^{-}$ be the
positive part and negative part of $\beta$ respectively, then
$\beta=\beta^{+}-\beta^{-}$ and $\beta^{\pm}\geq 0$. By (C.3) we have

$\lambda_{\varepsilon}(s)+\beta^{+}(s)l^{\beta}_{\varepsilon}(s)\leq\beta^{-}(s)l^{\beta}_{\varepsilon}(s).$

Observe that
$\beta^{+}(s)l^{\beta}_{\varepsilon}(s)=\beta^{+}(s)l^{\beta^{+}}_{\varepsilon}(s)$
and
$\beta^{-}(s)l^{\beta}_{\varepsilon}(s)=\beta^{-}(s)l^{\beta^{-}}_{\varepsilon}(s)$,
then

$\lambda_{\varepsilon}(s)+\beta^{+}(s)l^{\beta^{+}}_{\varepsilon}(s)\leq\beta^{-}(s)l^{\beta^{-}}_{\varepsilon}(s).$
(C.4)

By Lemma C.1 (1)

$\beta^{-}(s)l^{\beta^{-}}_{-\varepsilon}(s)\leq\beta^{-}(s)l^{\beta^{-}}_{-1}(s),\quad\forall\varepsilon\in(0,1).$

By Lemma C.1 (2)

$\displaystyle\beta^{-}l^{\beta^{-}}_{-\varepsilon}$
$\displaystyle=\beta^{-}\left(\frac{\partial
L}{\partial\dot{x}}(\gamma,\frac{\dot{\gamma}}{1-\varepsilon\beta^{-}})\cdot\frac{\dot{\gamma}}{1-\varepsilon\beta^{-}}-L(\gamma,\frac{\dot{\gamma}}{1-\varepsilon\beta^{-}})+f(\gamma,u_{1})\right)$
$\displaystyle\leq(\kappa^{\beta^{-}}_{\varepsilon})^{-1}L(\gamma,\frac{\dot{\gamma}}{1-\beta^{-}})-((\kappa^{\beta^{-}}_{\varepsilon})^{-1}+\beta^{-})L(\gamma,\frac{\dot{\gamma}}{1-\varepsilon\beta^{-}})+\beta^{-}f(\gamma,u_{1}),$

where $(\kappa^{\beta^{-}}_{\varepsilon})^{-1}=(1-\beta^{-})/(1-\varepsilon)$.
Since $1-\varepsilon\beta\in\Omega_{0}$ for all $\varepsilon\in[0,1]$, and
$l^{\beta^{-}}_{-\varepsilon}$ has a lower bound, we conclude that
$\beta^{-}l^{\beta^{-}}_{-\varepsilon}\in L^{1}[0,t]$ for all
$\varepsilon\in(0,1]$ uniformly. Thus, integrating (C.4) and by the Lebesgue
theorem, we obtain that

$0\leq\int_{0}^{t}e^{\int_{s}^{t}-\partial f/\partial
u(\gamma,u_{1})dr}l_{0}(s)\beta^{+}(s)ds\leq\int_{0}^{t}e^{\int_{s}^{t}-\partial
f/\partial u(\gamma,u_{1})dr}l_{0}(s)\beta^{-}(s)ds,$

which implies that

$\int_{0}^{t}e^{\int_{s}^{t}-\partial f/\partial
u(\gamma,u_{1})dr}l_{0}(s)\beta(s)ds=0,$

since $\beta$ is arbitrary. Then (C.2) follows.

Step 3. Erdmann condition. Now observe that the primitive
$\mu(s):=\int_{0}^{s}\beta(r)dr$ gives a one-to-one corresponding between
$\Omega_{0}$ and the set

$\Omega_{1}:=\\{\mu:[0,t]\rightarrow\mathbb{R}:\ \mu\ \textrm{is\ Lischitz\
continuous\ with}\ \mu(0)=\mu(t)=0,\ \mu^{\prime}\in\Omega_{0}\\}.$

Thus, by (C.2) we have

$0=e^{\int_{0}^{t}-\partial f/\partial
u(\gamma,u_{1})dr}\int_{0}^{t}E(s)\mu^{\prime}(s)ds,\quad\forall\mu\in\Omega_{1}.$

Recalling that $E(s)=e^{\int_{0}^{s}\partial f/\partial
u(\gamma,u_{1})dr}l_{0}(s)\in L^{1}[0,t]$, we have $\dot{E}(s)=0$ for almost
all $s\in[0,t]$ by the du Bois-Reymond lemma [2, Theorem 2.1]. ∎

######  Theorem C.3.

Given two positive constants $\delta$ and $T$ with $\delta<T$. For each
$\varphi\in C(M)$ and $t\in[\delta,T]$, the Lipschitz constant of
$T^{-}_{t}\varphi(x)$ depends only on $\|\varphi\|_{\infty}$, $\delta$ and
$T$.

###### Proof.

Step 1. Lipschitz estimate of minimizers. Given $(x,t)\in M\times[\delta,T]$.
In the following, we denote by $\gamma:[0,t]\rightarrow M$ a minimizer of
$T^{-}_{t}\varphi(x)$. We focus on the Lipschitz regularity of the curve
$\gamma$. The proof is quite similar to [2, Theorem 2.4], one can also refer
to [1, Proposition 2] for the autonomous case. Note that
$T^{-}_{t}(-\|\varphi\|_{\infty})\leq T^{-}_{t}\varphi\leq
T^{-}_{t}\|\varphi\|_{\infty}$, $T^{-}_{t}\varphi$ is bounded by a constant
$K$ depends only on $\|\varphi\|_{\infty}$ and $T$. We then have

$\displaystyle K$ $\displaystyle\geq
T^{-}_{t}\varphi(x)=\varphi(\gamma(0))+\int_{0}^{t}\bigg{[}L(\gamma(s),\dot{\gamma}(s))-f(\gamma(s),T^{-}_{s}\varphi(\gamma(s)))\bigg{]}ds$
$\displaystyle\geq-\|\varphi\|_{\infty}-(\|f(x,0)\|_{\infty}+\lambda
K)T+\int_{0}^{t}L(\gamma(s),\dot{\gamma}(s))ds.$

By (SL), there is a constant $D$ such that
$L(\gamma(s),\dot{\gamma}(s))\geq\|\dot{\gamma}(s)\|+D$, then we have

$K+(\|f(x,0)\|_{\infty}+\lambda
K+|D|)T+\|\varphi\|_{\infty}\geq\int_{0}^{t}\|\dot{\gamma}(s)\|ds.$

Thus, there is $s_{0}\in[0,t]$ such that $\|\dot{\gamma}(s_{0})\|$ is bounded
by a constant depends only on $\|\varphi\|_{\infty}$, $\delta$ and $T$. Since
$\dot{E}(s)=0$, we have

$E_{0}(s)\leq e^{\lambda T}(|E_{0}(s_{0})|+\|f(x,0)\|_{\infty}+\lambda
K)+\|f(x,0)\|_{\infty}+\lambda K:=F_{1}.$

By (CON) we have

$\displaystyle
L(\gamma(s),\frac{\dot{\gamma}(s)}{1+\|\dot{\gamma}(s)\|})-L(\gamma(s),\dot{\gamma}(s))$
$\displaystyle\geq(\frac{1}{1+\|\dot{\gamma}(s)\|}-1)\frac{\partial
L}{\partial\dot{x}}(\gamma(s),\dot{\gamma}(s))\cdot\dot{\gamma}(s)$
$\displaystyle\geq(\frac{1}{1+\|\dot{\gamma}(s)\|}-1)(F_{1}+L(\gamma(s),\dot{\gamma}(s))).$

We denote by $K_{3}$ the bound of $L(x,\dot{x})$ for $\|\dot{x}\|\leq 1$. Then
we have

$L(\gamma(s),\dot{\gamma}(s))\leq K_{3}+(F_{1}+K_{3})\|\dot{\gamma}(s)\|.$

By (SL), $\|\dot{\gamma}(s)\|$ is bounded by a constant depends only on
$\|\varphi\|_{\infty}$, $\delta$ and $T$.

Step 2. Lipschitz estimate of $(x,t)\mapsto T^{-}_{t}\varphi(x)$. We first
show that $u(x,t):=T^{-}_{t}\varphi(x)$ is locally Lipschitz in $x$. For any
$r>0$ with $2r<\delta$, given $(x_{0},t)\in M\times[\delta,T]$ and $x$,
$x^{\prime}\in B(x_{0},r)$, denoted by $d_{0}:=d(x,x^{\prime})\leq 2r<\delta$
the Riemannian distance between $x$ and $x^{\prime}$, then

$\displaystyle u(x^{\prime},t)-u(x,t)\leq$
$\displaystyle\int_{t-d_{0}}^{t}\bigg{[}L(\alpha(s),\dot{\alpha}(s))-f(\alpha(s),u(\alpha(s),s))\bigg{]}ds$
$\displaystyle-\int_{t-d_{0}}^{t}\bigg{[}L(\gamma(s),\dot{\gamma}(s))-f(\gamma(s),u(\gamma(s),s))\bigg{]}ds,$

where $\gamma(s)$ is a minimizer of $u(x,t)$ and
$\alpha:[t-d_{0},t]\rightarrow M$ is a geodesic satisfying
$\alpha(t-d_{0})=\gamma(t-d_{0})$ and $\alpha(t)=x^{\prime}$ with constant
speed. By Step 1, the bound of $\|\dot{\gamma}(s)\|$ depends only on
$\|\varphi\|_{\infty}$, $\delta$ and $T$. Since

$\|\dot{\alpha}(s)\|\leq\frac{d(\gamma(t-d_{0}),x^{\prime})}{d_{0}}\leq\frac{d(\gamma(t-d_{0}),x)}{d_{0}}+1,$

and $d(\gamma(t-d_{0}),x)\leq\int_{t-d_{0}}^{t}\|\dot{\gamma}(s)\|ds$, the
bound of $\|\dot{\alpha}(s)\|$ also depends only on $\|\varphi\|_{\infty}$,
$\delta$ and $T$. Exchanging the role of $(x,t)$ and $(x^{\prime},t)$, one
obtain that $|u(x,t)-u(x^{\prime},t)|\leq J_{1}d(x,x^{\prime})$, where $J_{1}$
depends only on $\|\varphi\|_{\infty}$, $\delta$ and $T$. By the compactness
of $M$, we conclude that for $t\in[\delta,T]$, the value function $u(\cdot,t)$
is Lipschitz on $M$.

We are now going to show the locally Lipschitz continuity of $u(x,t)$ in $t$.
Given $t$ and $t^{\prime}$ with $\delta\leq t<t^{\prime}\leq T$. Let
$\gamma:[0,t^{\prime}]\rightarrow M$ be a minimizer of $u(x,t^{\prime})$, then

$\displaystyle
u(x,t^{\prime})-u(x,t)=u(\gamma(t),t)-u(x,t)+\int_{t}^{t^{\prime}}\bigg{[}L(\gamma(s),\dot{\gamma}(s))-f(\gamma(s),u(\gamma(s),s))\bigg{]}ds,$

where the bound of $\|\dot{\gamma}(s)\|$ depends only on
$\|\varphi\|_{\infty}$, $\delta$ and $T$. We have shown that for
$t\geq\delta$, the following holds

$u(\gamma(t),t)-u(x,t)\leq J_{1}d(\gamma(t),x)\leq
J_{1}\int_{t}^{t^{\prime}}\|\dot{\gamma}(s)\|ds\leq J_{2}(t^{\prime}-t).$

Thus, $u(x,t^{\prime})-u(x,t)\leq J_{3}(t^{\prime}-t)$, where $J_{3}$ depends
only on $\|\varphi\|_{\infty}$, $\delta$ and $T$. The condition $t^{\prime}<t$
is similar. We conclude the Lipschitz continuity of $u(x,\cdot)$ on
$[\delta,T]$. ∎

######  Corollary C.4.

If $T^{-}_{t}\varphi(x)$ has a bound independent of $t$, then the family
$\\{T^{-}_{t}\varphi(x)\\}_{t\geq 1}$ is equi-Lipschitz.

###### Proof.

Let $\|T^{-}_{t}\varphi(x)\|_{\infty}\leq K$ for all $t\geq 0$, with the bound
$K$ independent of $t$. Note that $T^{-}_{t}\varphi(x)=T^{-}_{1}\circ
T^{-}_{t-1}\varphi(x)$. Take $\delta=1/2$ and $T=1$ in Theorem C.3, the
Lipschitz constant of $T^{-}_{1}\circ T^{-}_{t-1}\varphi(x)$ depends only on
$K$, which is independent of $t$. ∎

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