The Lax-Oleinik semigroup on graphs

1.   Introduction.

There has been an increasing interest in the study of dynamical systems and differential equation on networks. In the same vein, the study of control problems on networks has interesting applications in various fields. A typical optimal control problem is the minimum time problem, which consists of finding the shortest path between an initial position and a given target set. If the cost changes in a continuous way along the edges and the dynamics is continuous in time, the minimum time problem can be seen as a continuous-state continuous-time control problem where the admissible trajectories of the system are constrained to remain on the network. Control problems with state constrained in closures of open sets have been intensively studied but the there is much fewer literature on problems on networks.

Networks are the simplest examples of ramified spaces. Ramified spaces are the natural settings for problems of interaction between different media which are described by differential equations on the branches and transition conditions on the ramifications. Those interaction problems have different applications in physics, chemistry, and biology. Concerning the analysis of these models, the possibility of the application of several mathematical methods does not only depend on the structure of the differential equations on the branches, but also they depend strongly on the properties of the transition conditions. In fact, their effect on existence, uniqueness, and regularity of solutions is quite considerable. Many well-known results for elliptic and parabolic problems in the classical non-ramified situation have been extended to ramified spaces, providing results for boundary or initial value problems, now stablished for collections of media with transition conditions. In many cases, the core of the difficulties found on ramified spaces are already present in the simple setting of networks.

Viscosity solutions of the Hamilton Jacobi equation on networks have been studied for several authors, but the problem of the long time behaviour of the solutions to the Cauchy problem for the Hamilton Jacobi equation has not been considered. Upon addressing that problem, we found more convenient to follow the variational approach, considering the viscosity solution given by the Lax Oleinik semigroup.

In the first part of this article we study the Lax-Oleinik semi-group $\mathcal{L}_t$ defined by a Tonelli Lagrangian on a graph and prove that for any continuous function $u$, $\mathcal{L}_tu+c t$ converges as $t\to\infty$ where $c$ is the critical value of the Lagrangian.

For Lagrangians on compact manifolds, Fathi [6, 7] proved the convergence using the Euler-Lagrange flow and conservation of energy. In our case we do not have those tools but we can follow ideas of Roquejoffre [11] and Davini-Siconolfi [5]. See also [10] for the convergence when the manifold is the whole euclidian space.

Camilli and collaborators [1, 4, 3] have studied viscosity solutions of the Hamilton-Jacobi equation, and given sufficient conditions for a set to be a uniqueness set and a representation formula. See also the recent papers [2, 9].

In the second part of this article we prove that, under the assumption that the Lagrangian is symmetric at the vertices, the sets of weak KAM and viscosity solutions of the Hamilton-Jacobi equation coincide.

We consider a graph $G$ without boundary consisting of finite sets of unoriented edges $\mathcal{I}=\{I_j\}$ and vertices $\mathcal{V}=\{e_l\}$. The interior of $I_j$ is $I_j-\mathcal{V}$. Parametrizing each edge by arc length $\sigma_j:I_j\to[0, s_j]$ we can write its tangent bundle as $TI_j=I_j\times\mathbb{R}$ and

where $(i, x, v)\sim(j, y, w)\iff (i, x, v)=(j, y, w)$ or $x=y\in I_i\cap I_j, v=w=0$. Thus, a function $L: TG\to\mathbb{R}$ is given by a collection of functions $L_j: TI_j\to\mathbb{R}$ such that $L_i (e_l, 0) = L_j(e_l, 0)$ for $e_l\in I_i\cap I_j$. A Lagrangian in $G$ is a function $L: TG\to\mathbb{R}$ such that each $L_j$ is $C^k$, $k\ge 2$, and $L_j(x, \cdot)$ is strictly convex and super-linear for any $x\in I_j$. We will say that a Lagrangian is it symmetric at the vertices if at each vertex $e_l$ there is a function $\lambda_l:\{u\in\mathbb{R}:u\ge 0\}\to\mathbb{R}$ such that $L_j(e_l, z)=\lambda_l(|z|)$ if $e_l\in I_j$. As an example consider the mechanical Lagrangian given by $L_j(x, v)=\frac 12v^2-U_j(x)$, with $U_j(e_l)=a_l$ if $e_l\in I_j$. For $x\in I_j\backslash \mathcal{V}$, we say that $(x, v)$ points towards $\sigma^{-1}(s_j)$ if $v>0$ and points towards $\sigma^{-1}(0)$ if $v<0$.

We say that $(\sigma_j^{-1}(0), v)$ is an $I_j$-incoming or outgoing vector according to whether $v>0$ or $v<0$, and we say that $(\sigma_j^{-1}(s_j), v)$ is an $I_j$-incoming or outgoing vector according to wether $v<0$ or $v>0$. We let $T^+_{e_l}I_j$ ($T^-_{e_l}I_j$) to be the set of $I_j$-outgoing (incoming or zero) vectors in $T_{e_l}I_j$.

2.   Basic properties of the action.

2.1.   A distance on a graph.

We start defining a distance in the most natural way. We say a continuous path $\alpha :[a, b]\to G$ is a unit speed geodesic (u.s.g.) if there is a partition $a=t_0<\ldots<t_m=b$ such that for each $1\le i\le m$ there is $j(i)$ such that

$\sigma_{j(i)}\circ\alpha |_{[t_{i-1}, t_i]}$ is differentiable and either $(\sigma_{j(i)}\circ\alpha )'\equiv 1$ or $(\sigma_{j(i)}\circ\alpha )'\equiv-1$. We set the length of a u.s.g. to be

and define a distance on $G$ by

2.2.   Absolute continuity.

We say a path $\gamma :[a, b]\to G$ is absolutely continuous if for any $\varepsilon >0$ there is $\delta >0$ such that for any finite collection of disjoint intervals $\{[c_i, d_i]\}$ with $\sum_i(d_i-c_i)<\delta$ we have $\sum_id(\gamma (c_i), \gamma (d_i))<\varepsilon$.

If $\gamma :[a, b]\to G$, $\gamma (t)\in\mathcal{V}$, we define ${\dot \gamma }(t)=0$ if for any $\varepsilon >0$ there is $\delta >0$ such that $d(\gamma (s), \gamma (t))<\varepsilon |s-t|$ when $|s-t|<\delta$.

Let $\gamma :[a, b]\to G$ be absolutely continuous and consider the closed set $V=\gamma ^{-1}(\mathcal{V})$ so that $(a, b)\backslash V=\bigcup_i(a_i, b_i)$ where the intervals $(a_i, b_i)$ are disjoint and $\gamma ([a_i, b_i])\subset I_{j(i)}$. It is clear that each $\sigma_{j(i)}\circ\gamma :[a_i, b_i]\to[0, s_{j(i)}]$ is absolutely continuous. We set ${\dot \gamma }(t)=(\sigma_{j(i)}\circ\gamma )'(t)$ whenever is defined.

Next Proposition will allow us to define the action of an absolutely continuous curve.

Proposition 1. Let $\gamma :[a, b]\to G$ be absolutely continuous and $V=\gamma ^{-1}(\mathcal{V})$

(a)${\dot \gamma }=0$ Lebesgue almost everywhere in $V$.

(b)${\dot \gamma }$ is integrable and for any $[c, d]\subset[a, b]$ we have $d(\gamma (c), \gamma (d))\le\int\limits_c^d|{\dot \gamma }|$.

proof. Write $(a, b)\backslash V=\bigcup_i(a_i, b_i)$ as above with the intervals $(a_i, b_i)$ disjoint. Since ${\dot \gamma }=0$ on the interior of $V$ and $\cup\{a_n, b_n\}$ is numerable to stablish item (a) it remains to prove that ${\dot \gamma }=0$ Lebesgue almost everywhere in $\partial V\backslash \cup\{a_n, b_n\}$.

Let $\bar s=\min_j s_j$ and take $\delta >0$ such that $d(\gamma (t_1), \gamma (t_2))<\bar s$ if $|t_1-t_2|<\delta$. There is $N$ such that $b_i-a_i<\delta$ for $i>N$. Since $\gamma (a_i), \gamma (b_i)\in\mathcal{V}$ we have that $\gamma (a_i)=\gamma (b_i)$ for $i>N$. We change the labeling of the first $N$ terms to have $a_1<b_1\le a_2<\cdots<b_m$ and $\gamma (a_i)=\gamma (b_i)$ for $i>m$. Letting $J_0=[a, a_1]$, $J_i=[b_i, a_{i+1}]$, $1\le i<m$, $J_m=[b_m, b]$, and $V_i=V\cap J_i$ we have $\gamma (V_i)=e_{l_i}$, $0\le i\le m$. We can forget about the cases $b_i=a_{i+1}$.

Define the function $f_i:J_i\to\mathbb{R}$ by $f_i(t)=d(e_{l_i}, \gamma (t))$. For $t, s\in J_i$ we have $|f_i(t)-f_i(s)|\le d(\gamma (t), \gamma (s))$, so $f_i$ is absolutely continuous and then $f_i'$ exists Lebesgue almost everywhere in $J_i$. Let $t\in\partial V_i\backslash \bigcup_n\{a_n, b_n\}$ be a point where $f_i'$ exists. There is a sequence $n_k\to\infty$ such that $a_{n_k}\to t$ and $\gamma (a_{n_k})=e_{l_i}$. Thus $f_i'(t)=0$, which means that ${\dot \gamma }(t)=0$.

If $(a_j, b_j)\subset J_i$ then $|{\dot \gamma }|=|f_i'|$ Lebesgue almost everywhere in $(a_j, b_j)$, so that

and then

It is also easy to see that for $t, s\in J_i$ we have $d(\gamma (t), \gamma (s))\le\int\limits_s^t|f_i'|$, and using the partition $a\le a_1<b_1\le a_2<\cdots<b_m\le b$ to make a partition of $[c, d]$ we get $d(\gamma (c), \gamma (d))\le\int\limits_c^d|{\dot \gamma }|$.

2.3.   Lower semicontinuity and apriori bounds.

In this crucial part of the paper we prove that in the framework of graphs we have the lower semicontinuity of the action and apriori bounds for the Lipschitz norm of minimizers. The proofs have the same spirit as in euclidean space, paying attention to what happens at the vertices. Denote by $\mathcal{C}^{ac}([a, b])$ the set of absolutely continuous functions $\gamma :[a, b]\to G$ provided with the topology of uniform convergence. We define the action of $\gamma \in \mathcal{C}^{ac}([a, b])$ as

A minimizer is a $\gamma \in \mathcal{C}^{ac}([a, b])$ such that for any $\alpha \in\mathcal{C}^{ac}([a, b])$ with $\alpha (a)=\gamma (a)$, $\alpha (b)=\gamma (b)$ we have

The following two properties of the Lagrangian are important to achieve our goal and follow from its strict convexity and super-linearity.

Proposition 2. If $C\ge 0$, $\varepsilon >0$, there is $\eta>0$ such that for $x, y\in I_j$, $d(x, y)<\eta$ and $v, w\in\mathbb{R}$, $|v|\le C$, we have

Proposition 3. If $L_{vv}\ge\theta >0$, $C\ge 0$, $\varepsilon >0$, there is $\eta>0$ such that for $x, y\in I_j$, $d(x, y)<\eta$ and $v, w\in\mathbb{R}$, $|v|\le C$, we have

Lemma 2.1. Let $L$ be a Lagrangian on $G$. If a sequence $\gamma _n\in\mathcal{C}^{ac}([a, b])$ converges uniformly to the curve $\gamma :[a, b]\to G$ and

then the curve $\gamma$ is absolutely continuous and

proof. By the super-linearity of $L$ we may assume that $L\ge 0$. Let $c= \liminf\limits_{n\to\infty} A(\gamma _n)$. Passing to a subsequence we can assume that

Fix $\varepsilon >0$ and take $B> 2(c+1)/\varepsilon$. Again by super-linearity there is a positive number $C(B)$ such that

From Proposition 1 and $L\ge 0$, for $E\subset[a, b]$ measurable we have

Thus

Choosing $0<\delta <\dfrac{\varepsilon B}{2C(B)}$ we have that

Since the sequence ${\dot \gamma }_n$ is uniformly integrable, we have that $\gamma$ is absolutely continuous and ${\dot \gamma }_n$ converges to ${\dot \gamma }$ in the $\sigma(L^1, L^\infty)$ weak topology.

Set $V=\gamma ^{-1}(\mathcal{V})$. Let $\varepsilon >0$ and $E_k=\bigl\{t:|{\dot \gamma }(t)|\le k, d(t, V)\ge\dfrac 1k\bigr\}$.

By Propositions 2, 1, for $n$ large,

Letting $n\to+\infty$ we have that

Since $E_k\uparrow [a, b]\backslash V$ when $k\to+\infty$ and $L\ge 0$, we have

Now let $\varepsilon \to 0$.

Lemma 2.1 implies

Theorem 2.2. Let $L$ be a Lagrangian on $G$. The action $A:\mathcal{C}^{ac}([a, b])\to\mathbb{R}\cup\{\infty\}$ is lower semicontinuous.

Proposition 4. Let $L$ be a Lagrangian on $G$.

The set $\{\gamma \in\mathcal{C}^{ab}([a, b]): A(\gamma )\le K\}$ is compact with the topology of uniform convergence.

Let $C_t=\sup\{L(x, v):x\in G, |v|\le\frac{\text{diam}(G)}t\}$, then for any minimizer $\gamma :[a, b]\to G$ with $b-a\ge t$ we have

Proposition 5. Suppose $\gamma _n\in\mathcal{C}^{ab}([a, b])$ converge uniformly to $\gamma :[a, b]\to G$ and $A(\gamma _n)$ converges to $A(\gamma )$, then $\dot\gamma _n$ converges to $\dot\gamma$ in $L^1[a, b]$

proof. Let $F\subset[a, b]$ be a finite union of intervals. From Lemma 2.1 we have

Since

we have

If $V=\gamma ^{-1}(\mathcal{V})\subset F$ then

Thus

As in Lemma 2.1, the ${\dot \gamma }_n$ are uniformly integrable, so they converge to ${\dot \gamma }$ in the $\sigma(L_1, L_\infty)$ weak topology and then, for any Borel set $B$ where ${\dot \gamma }$ is bounded,

Given $\varepsilon >0$, from Proposition 3 we have that for $n$ large enough

Which together with Proposition 1 and equations (2), (3) give $\limsup\limits_n\int\limits_{ V}|{\dot \gamma }_n|^2\le\varepsilon$ for any $\varepsilon >0$. So

For $k>0$ let $D_k:=\{t\in[a, b]:|{\dot \gamma }(t)|>k\}$, $B_k:=\{t\in[a, b]: d(t, V)>\frac 1k\}$. Then $\lim\limits_{k\to\infty}$Leb$(D_k)=\lim\limits_{k\to\infty}$Leb$([a, b]\backslash V\backslash B_k)=0$. Let $F_k$ be a finite union of intervals such that $D_k\cap B_k\subset F_k\subset B_k$ and Leb$(F_k\backslash (D_k\cap B_k))<\frac 1k.$ Then $\lim\limits_{k\to\infty}$ Leb$(F_k)=0$.

Given $\varepsilon >0$, from Proposition 3 we have that for $n$ large enough

From (1), (3) we get that $\limsup\limits_n\int\limits_{B_k\backslash F_k}|{\dot \gamma }_n-{\dot \gamma }|^2\le\varepsilon$ for any $\varepsilon >0$. So

Since $\{{\dot \gamma }_n\}$ is uniformly integrable, given $\varepsilon >0$, for $k$ sufficiently large we have

From (4), (5), (6) and Cauchy-Schwartz inequality, we have that for any $\varepsilon >0$

Lemma 2.3. Let $L$ be a Lagrangian in $G$. For $\varepsilon >0$ there exists $K_{\varepsilon }$ that is a Lipschitz constant for any minimizer $\gamma :[a, b]\to G$ with $b-a\ge \varepsilon$.

proof. Note that if $\gamma$ is a minimizer and $\gamma (c, d)\subset I_j$ then $\gamma |_{(c, d)}$ is a solution of the Euler Lagrange equation for $L_j$.

Suppose the Lemma is not true, then by Proposition 1, for any $i\in\mathbb{N}$ there are a minimizer $\gamma _i: [s_i, t_i]\to G$ with $t_i-s_i\ge\varepsilon$ and a set $E_i\subset [s_i, t_i]\backslash \gamma _i^{-1}(\mathcal{V})$ with Leb$(E_i)>0$ such that $|{\dot \gamma }|>i$ on $E_i$. Let $c_i\in E_i$. Translating $[s_i, t_i]$ we can assume that $c_i=c$ for all $i$ and taking a subsequence that there is $a\in\mathbb{R}$ such that $\gamma _i$ is defined in $[a, a+\frac{\varepsilon }2], c\in[a, a+\frac{\varepsilon }2]$. As $A(\gamma _i|[a, a+\frac{\varepsilon }2])$ is bounded, by Proposition 4 there is subsequence $\gamma _i|[a, a+\frac{\varepsilon }2]$ which converges uniformly to $\gamma :[a, a+\frac{\varepsilon }2]\to G$. Since $\gamma$ is limit of minimizers, it is a minimizer and $A(\gamma )\le\liminf A(\gamma _i|[a, a+\frac{\varepsilon }2])$. We can not have that $A(\gamma )<\text{lim sup} A(\gamma _i|[a, a+\frac{\varepsilon }2])$ because that would contradict that the $\gamma _i$ are minimizers. Thus $A(\gamma )=\lim A(\gamma _i|[a, a+\frac{\varepsilon }2])$.

If $\gamma (c)\in I_j\backslash \mathcal{V}$, there is $\delta >0$ such that $\gamma ([c-\delta, c+\delta])\subset I_j$.

If $\gamma (c)=e_l$ we have 2 possibilities (not mutually exclusive)

a) There is an edge $I_j$ with $e_l\in I_j$ and infinitely many $i$'s such that $\gamma _i(c)\in I_j$ and $\dot\gamma _i(c)$ points towards $e_l$.

b) There is an edge $I_j$ with $e_l\in I_j$ and infinitely many $i$'s such that $\gamma _i(c)\in I_j$ and $\dot\gamma _i(c)$ points towards the other vertex.

In case a) there is $\delta >0$ such that $\gamma ([c-\delta, c])\subset I_j$.

In case b) there is $\delta >0$ such that $\gamma ([c, c+\delta])\subset I_j$.

We have that $\gamma$ is a solution of the Euler-Lagrange equation for $L_j$ either on $[c-\delta, c]$ or on $[c, c+\delta]$ and then $|\dot\gamma (t)|\le K$ on $[c-\delta, c]$ or $[c, c+\delta]$. For some $0<\delta _1<\delta$ we have that $\gamma _i$ are solutions of the Euler-Lagrange equation for $L_j$ on $[c-\delta _1, c]$ or on $[c, c+\delta _1]$. For $i$ suficiently large, we have that $|\dot\gamma _i|>2 K$ either on $[c-\delta _1, c]$ or on $[c, c+\delta _1]$. This would contradict Proposition 5.

3.   Weak KAM theory on graphs.

The content of this section is similar to that for Lagrangians on compact manifolds. We only give the proofs that are different from those in the compact manifold case, which can be found in [7], as well as extensions in [8].

3.1.   The Peierls barrier.

Given $x, y \in G$ let $\mathcal{C}^{ac}(x, y, t)$ be the set of curves $\alpha\in\mathcal{C}^{ac}([0, t])$ such that $\alpha (0)=x$ and $\alpha (t) = y$. For a given real number $k$ define

and

Lemma 3.1. For $\varepsilon >0$ the function $F:[\varepsilon, \infty)\times G\times G\to\mathbb{R}$ defined by $F(t, x, y)= h_t(x, y)$ is Lipschitz.

Lemma 3.2. There exists a real $c$ independent of $x$ and $y$ such that

1. For all $k>c$ we have $h^k(x, y) =\infty$.

2. For all $k<c$ we have $h^k(x, y) =-\infty$

3.$h^c(x, y)$ is finite. The function $h:=h^c$ is called the Peierls barrier.

Lemma 3.3. The value $c$ is the infimum of $k$ such that $\int\limits_\gamma L+k\ge 0$ for all closed curves $\gamma$.

Definition 3.4. The Mañé potencial $\Phi:G\times G\to\mathbb{R}$ is defined by

Clearly we have $\Phi(x, y)\le h(x, y)$ for any $x, y\in G$.

Proposition 6. Functions $h$ and $\Phi$ have the following properties.

1. $\Phi(x, z)\le \Phi(x, y)+\Phi(y, z)$.

2. $h(x, z)\le h(x, y)+\Phi(y, z)$, $h(x, z)\le\Phi(x, y)+h(y, z)$.

3. $h$ and $\Phi$ are Lipschitz

4. If $\gamma _n:[0, t_n]\to G$ is a sequence of absolutely continuous curves with $t_n\to\infty$ and $\gamma _n(0)\to x$, $\gamma _n(t_n)\to y$, then

Definition 3.5. A curve $\gamma :J\to G$ defined on an interval $J$ is called

$\bullet$ semi-static if

for any $t, s\in J$, $t\le s$.

$\bullet$ static if

for any $t, s\in J$, $t\le s$.

$\bullet$ The Aubry set $\mathcal{A}$ is the set of points $x\in G$ such that $h(x, x)=0$.

Notice that by item (2) in Proposition 6, $h(x, z)=\Phi(x, z)$ if $x\in\mathcal{A}$ or $z\in\mathcal{A}$.

Proposition 7. If $\eta:\mathbb{R}\to G$ is static then $\eta(s)\in\mathcal{A}$ for any $s\in\mathbb{R}$.

We do not have a Lagrangian flow and therefore we can not speak of conservation of energy. Nevertheless the following Proposition says that semi-static curves have energy $c(L)$.

Proposition 8. Let $\eta:J\to G$ be semi-static. For almost every $t\in J$

proof. For $\lambda >0$, let $\eta_\lambda(t):=\eta(\lambda t)$ so that $\dot\eta_\lambda(t)=\lambda\dot\eta(\lambda t)$ almost everywhere.

For $r, s\in J$ let

Since $\eta$ is a free-time minimizer, differentiating $\mathcal{A}_{rs}(\lambda)$ at $\lambda=1$, we have that

Since this holds for any $r, s\in J$ we have

for almost every $t\in J$.

3.2.   Weak KAM solutions.

Following Fathi [7], we define weak KAM solutions and give some of their properties

Definition 3.6. Let $c$ be given by Lemma 3.2.

$\bullet$ A function $u:G\to\mathbb{R}$ is dominated if for any $x, y\in G$, we have

or equivalently

$\bullet$ $\gamma :I\to G$ calibrates a dominated function $u:G\to\mathbb{R}$ if

$\bullet$ A continuous function $u:G\to\mathbb{R}$ is a backward (forward) weak KAM solution if it is dominated and for any $x\in G$ there is $\gamma :(-\infty, 0]\to G$ ($\gamma :[0, \infty)\to G$) that calibrates $u$ and $\gamma (0)=x$

Corollary 1. Any static curve $\gamma :J\to G$ calibrates any dominated function $u:G\to\mathbb{R}$

Proposition 9. For any $x\in G$, $h(x, \cdot)$ is a backward weak KAM solution and $-h(\cdot, x)$ is a forward weak KAM solution.

proof. By item (2) of Proposition 6, $h(x, \cdot)$ is dominated.

The standard construction of calibrating curves for compact manifolds involves the Euler Lagange flow that we do not have, so we use a diagonal trick. Let $\gamma _n:[-t_n, 0]\to G$ be a sequence of minimizing curves connecting $x$ to $y$ such that

By Lemma 2.3, $\{\gamma _n\}$ is uniformly Lipschitz and then equicontinuous. It follows from the Arzela Ascoli Theorem that there is a sequence $n^1_j\to\infty$ such that $\gamma _{n^1_j}$ converges uniformly on $[-1, 0]$. Again, by the Arzela Ascoli Theorem, there is a subsequence $(n^2_j)_j$ of the sequence $(n^1_j)_j$ such that $\gamma _{n^2_j}$ converges uniformly on $[-2, 0]$. By induction, this procedure gives for each $k\in\mathbb{N}$ a sequence $(n^k_j)_j$ that is a subsequence of the sequence $(n^{k-1}_j)_j$ and such that $\gamma _{n^k_j}$ converges uniformly on $[-k, 0]$ as $j\to\infty$. Letting $m_k=n^k_k$, the sequence $\gamma _{m_k}$ converges uniformly on each $[-l, 0]$. For $s<0$ define $\gamma (s)=\lim\limits_{k\to\infty}\gamma _{m_k}(s)$. Fix $t<0$, for $k$ large $t+t_{m_k}\ge 0$ and

Since $\gamma _{m_k}$ converges to $\gamma$ uniformly on $[t, 0]$, we have

From item (4) of Proposition 6 we have

Taking $\liminf\limits_{k\to\infty}$ in (8) we get

So $\gamma$ calibrates $h(x, \cdot)$.

From Proposition 9 we have

Corollary 2. If $x\in\mathcal{A}$ there exists a curve $\gamma :\mathbb{R}\to G$ such that $\gamma (0)=x$ and for all $t\ge 0$

In particular the curve $\gamma$ is static and calibrates any dominated function $u:G\to\mathbb{R}$.

Theorem 3.7. The function $\Phi(x, \cdot)$ is a backward weak KAM solution if and only if $x\in\mathcal{A}$.

Corollary 3. Let $C\subset G$ and $w_0:C\to\mathbb{R}$ be bounded from below. Let

1. $w$ is the maximal dominated function not exceeding $w_0$ on $C$.

2. If $C\subset\mathcal{A}$, $w$ is a backward weak KAM solution.

3. If for all $x, y\in C$

then $w$ coincides with $w_0$ on $C$.

For $u:G\to\mathbb{R}$ let $I(u)$ be the set of points $x\in G$ for which exists $\gamma :\mathbb{R}\to G$ such that $\gamma (0)=x$ and $\gamma$ calibrates $u$.

Corollary 4.

Proposition 10. For each $x, y\in G$ with $x\ne y$ we can find $\varepsilon >0$ and a curve $\gamma :[-\varepsilon, 0]\to G$ such that $\gamma (0)=y$ and for all $t\in[0, \varepsilon]$

In particular, for each $x\in G$ the function $G\backslash \{x\}\to\mathbb{R}$; $y\mapsto\Phi(x, y)$ is a backward weak KAM solution.

Theorem 3.8. $\mathcal{A}$ is nonempty and if $u:G\to\mathbb{R}$ is a backward weak KAM solution then

Corollary 5.

4.   The Lax semigroup and its convergence.

4.1.   The Lax semigroup.

Let $\mathcal{F}$ be the set of real functions on $G$, bounded from below.

The backward Lax semigroup $\mathcal{L}_t:\mathcal{F}\to\mathcal{F}$, $t>0$ is defined by

It is clear that $f\in \mathcal{F}$ is dominated if and only if $f\le\mathcal{L}_tf+ct$ for any $t>0$.

It follows at once that $\mathcal{L}_t\circ\mathcal{L}_s=\mathcal{L}_{t+s}$ and

The proof of the following Lemma is the same as in the compact manifold case.

Lemma 4.1. Given $\varepsilon >0$ there is $K_\varepsilon >0$ such that for each $u:G\to\mathbb{R}$ continuous, $t\ge\varepsilon$, we have $\mathcal{L}_tu:G\to\mathbb{R}$ is a Lipschitz with constant $K_\varepsilon$.

Theorem 4.2. A continuous function $u:G\to\mathbb{R}$ is a fixed point of the semigroup $\mathcal{L}_t+ct$ if and only if it is a backward weak KAM solution

proof. Suppose $u:G\to\mathbb{R}$ is a fixed point of the semigroup $\mathcal{L}_t+ct$. For each $T\ge 2$ there is a curve $\alpha _T: [-T, 0]\to G$ such that $\alpha _T(0)=x$ and

By Lemma 2.3 $\{\alpha _T\}$ is uniformly Lipschitz. As in Propostion 9 one obtains a sequence $t_k\to\infty$ and $\gamma :(-\infty, 0]\to G$ such that $\alpha _{t_k}$ converges to $\gamma$, uniformly on each $[-n, 0]$.

By Lemma 2.2

Suppose now that $u:G\to\mathbb{R}$ is a backward weak KAM solution. Since $u$ is dominated, $u\le\mathcal{L}_t u+ct$. For $x\in G$ let $\gamma :(-\infty, 0]\to G$ be such that $\gamma (0)=x$ and for all $t>0$

Thus

From Proposition 9 and Theorem 4.2 one obtains

Corollary 6. The semigroup $\mathcal{L}_t+ct$ has fixed points.

4.2.   Convergence of the Lax semigroup

Without loss of generality assume $c=0$. For $u\in C(G)$ define

Proposition 11. Let $\psi=\lim\limits_{n\to\infty}\mathcal{L}_{t_n}u$ for some $t_n\to\infty$, then

proof. For $x\in G$ let $\gamma _n:[0, t_n]\to G$ be such that $\gamma _n(t_n)=x$ and

Passing to a subsequence if necessary we may assume that $\gamma _n(0)$ converges to $y\in G$. Taking $\liminf$ in (13), we have from item (4) of Proposition 6

Proposition 12. If $\mathcal{L}_tu$ converges as $t\to\infty$, then the limit is function $v$ defined in.

proof. For $x\in G$ let $z\in G$ be such that $v(z)=u(z)+h(z, x)$. Since $\mathcal{L}_t u(x)\le u(z)+h_t(z, x)$, we have

which together with Proposition (11) gives $\lim\limits_{t\to\infty}\mathcal{L}_t u=v$.

Thus, given $u\in C(G)$ our goal is to prove that $\mathcal{L}_tu$ converges to $v$ defined in (11).

Remark 1. Using Corollary 5 we can write (11) as

Item (1) of Corollary 3 states that $w$ is the maximal dominated function not exceeding $u$. Items (2), (3) of the same Corollary imply that $v$ is the unique backward weak KAM solution that coincides with $w$ on $\mathcal{A}$.

Proposition 13. Suppose that $u$ is dominated, then $\mathcal{L}_tu$ converges uniformly as $t\to\infty$ to the function $v$ given by (11).

proof. Since $u$ is dominated, the function $t\mapsto\mathcal{L}_tu$ is nondecreasing. As well, in this case, $w$ given by (15) coincides with $u$. Items (1) and (3) of Corollary 3 imply that $v$ is the maximal dominated function that coincides with $u$ on $\mathcal{A}$ and then $u\le v$ on $G$.

Since the semigroup $\mathcal{L}_t$ is monotone and $v$ is a backward weak KAM solution

Thus the uniform limit $\lim\limits_{t\to\infty}u$ exists.

We now address the convergence of $\mathcal{L}_t$ following the lines in [5] and [11].

For $u\in C(G)$ let

From these and Proposition 11

Proposition 14. Let $u\in C(G)$, $v$ be the function given by (11), $\underline{u}, \overline{u}$ defined in (16) and (17). Then

Proposition 15. For $u\in C(G)$, function $\underline{u}$ given by (16) is dominated.

proof. Let $x, y\in G$. Given $\varepsilon >0$ there is $\psi=\lim\limits_{n\to\infty}\mathcal{L}_{t_n}u$ such that $\underline{u}(x)-\varepsilon <\psi(x).$ For $n>N(\varepsilon )$ and $a>0$

Choose a divergent sequence $n_j$ such that $(\mathcal{L}_{t_{n_j}-a}u)_j$ converges uniformly. For $j>\bar N(\varepsilon )$, $\mathcal{L}_{t_{n_j}-a}u(y)<\underline{u}(y)+\varepsilon$, and then

Denote by $\mathcal{K}$ the family of static curves $\eta:\mathbb{R} \to G$, and for $y\in\mathcal{A}$ denote by $\mathcal{K}(y)$ the set of curves $\eta\in\mathcal{K}$ with $\eta(0)=y$.

Proposition 16. $\mathcal{K}$ is a compact metric space with respect to the uniform convergence on compact intervals.

proof. Let $\{\eta_n\}$ be a sequence in $\mathcal{K}$. By Lemma 2.3, $\{\eta_n\}$ is uniformly Lipschitz. As in Proposition 9 we obtain a sequence $n_k\to\infty$ such that $\eta_{n_k}$ converges to $\eta:\mathbb{R}\to G$ uniformly on each $[a, b]$ and then $\eta$ is static.

Proposition 17. Two dominated functions that coincide on $\mathcal{M}=\bigcup\limits_{\eta\in\mathcal{K}}\omega (\eta)$ also coincide on $\mathcal{A}$.

proof. Let $\varphi_1$, $\varphi_2$ be two dominated functions coinciding on $\mathcal{M}$. Let $y\in\mathcal{A}$ and $\eta\in\mathcal{K}(y)$. Let $(t_n)_n$ be a diverging sequence such that $\lim_n\eta(t_n)= x\in\mathcal{M}$. By Corollary 1

for every $n\in N, i=1, 2$. Sending $n$ to $\infty$, we get

Proposition 18. Let $\eta\in\mathcal{K}$, $\psi\in C(G)$ and $\varphi$ be a dominated function. Then the function $t\mapsto(\mathcal{L}_t\psi)(\eta(t))-\varphi(\eta(t))$ is nonincreasing on $\mathbb{R}_+$.

proof. From Corollary 1, for $t<s$ we have

Lemma 4.3. There is a $M>0$ such that, if $\eta$ is any curve in $\mathcal{K}$ and $\lambda$ is sufficiently close to 1, we have

for any $t_2>t_1$, where $\eta_\lambda(t)=\eta(\lambda t)$.

proof. Let $K>0$ be a Lipschitz constant for any minimizer $\gamma :[a, b]\to G$ with $b-a>1$, $2R=\sup\{|L_{vv}(x, v)|:|v|\le K\}$. For $\lambda\in(1-\delta, 1+\delta )$ fixed, using Proposition 8

Proposition 19. Let $\eta\in\mathcal{K}$, $\psi\in C(G)$ and $\varphi$ be a dominated function. Assume that $D^+((\psi-\varphi)\circ\eta)(0)\backslash \{0\}\ne\emptyset$ where $D^+$ denote the super-differential. Then for all $t>0$ we have

proof. Fix $t>0$. By Corollary 1 it is enough to prove (20) for $\varphi=-\Phi(\cdot, \eta(t))$. Since $\mathcal{L}_t(\psi+a)=\mathcal{L}_t\psi+a$ we can assume that $\psi(\eta(0))=\varphi(\eta(0))$.

thus, by Lemma 4.3

If $m\in D^+((\psi-\varphi)\circ\eta)(0)\backslash \{0\}$, we have

where $\lim\limits_{\lambda\to 1}\dfrac{o((1-\lambda)t)}{1-\lambda}=0$. Choosing appropriately $\lambda$ close to $1$, we get

Proposition 20. Suppose $\varphi$ is dominated and $\psi\in\omega _\mathcal{L}(u)$. For any $y\in\mathcal{M}$ there exists $\gamma \in\mathcal{K}(y)$ such that the function $t\mapsto\psi(\gamma (t))-\varphi(\gamma (t))$ is constant.

proof. Let $(s_k)_k$ and $(t_k)_k$ be diverging sequences, $\eta$ be a curve in $\mathcal{K}$ such that $y=\lim\limits_k \eta(s_k)$, and $\psi$ is the uniform limit of $\mathcal{L}_{t_k}u$. As in Proposition 9, we can assume that the sequence of functions $t\mapsto\eta(s_k+t )$ converges uniformly on compact intervals to $\gamma :\mathbb{R}\to G$, and so $\gamma \in\mathcal{K}$. We may assume moreover that $t_k-s_k\to\infty$, as $k\to\infty$, and that $\mathcal{L}_{t_k-s_k}u$ converges uniformly to $\psi_1\in\omega _\mathcal{L}(u)$. By the semi-group property and (10)

which implies that $\mathcal{L}_{s_k}\psi_1$ converges uniformly to $\psi$. From Proposition 18, we have that for any $\tau\in\mathbb{R}$ $s\mapsto(\mathcal{L}_s\psi_1)(\eta(\tau+s))-\varphi(\eta(\tau+s))$ is a nonincreasing function in $\mathbb{R}^+$, and hence it has a limit $l(\tau)$ as $s\to\infty$, which is finite since $l(\tau)\ge-\|\overline u-\varphi\|_\infty$. Given $t>0$, we have

The function $t\mapsto(\mathcal{L}_t\psi)(\gamma (\tau+t))-\varphi(\gamma (\tau+t))$ is therefore constant on $\mathbb{R}^+$. Applying Proposition 19 to the curve $\gamma (\tau +\cdot)\in\mathcal{K}$, we have $D^+((\psi-\varphi)\circ\gamma )(\tau)\backslash \{0\} =\emptyset$ for any $\tau\in\mathbb{R}$. This implies that $\psi-\varphi$ is constant on $\gamma$.

Proposition 21. Let $\eta\in\mathcal{K}$, $\psi\in\omega _\mathcal{L}(u)$ and $v$ be defined by (11). For any $\varepsilon >0$ there exists $\tau\in\mathbb{R}$ such that

proof. Since the curve $\eta$ is contained in $\mathcal{A}$, we have

and hence $v(\eta(0)) = u(z_0) + \Phi(z_0, \eta(0))$, for some $z_0\in G$. Take a curve $\gamma :[0, T]\to G$ such that

Choosing a divergent sequence $(t_n)_n$ such that $\mathcal{L}_{t_n}u$ converges uniformly to $\psi$ we have for $n$ sufficiently large

Take $\tau=t_n-T$

From Propositions 20 and 21 we obtain

Theorem 4.4. Let $\psi\in\omega _\mathcal{L}(u)$ and $v$ be defined by (11). Then $\psi=v$ on $\mathcal{M}$.

Theorem 4.5. Let $u\in C(G)$, then $\mathcal{L}_tu$ converges uniformly as $t\to\infty$ to $v$ given by (11).

proof. The function $\underline{u}$ is dominated and coincides with $v$ on $\mathcal{M}$ by Theorem 4.4. Proposition 17 implies that $\underline{u}$ coincide with $v$ on $\mathcal{A}$ and so does with $w$. By item (1) of Corollary 3 we have $\underline{u}\le v$.

5.   Viscosity solutions of the Hamilton -Jacobi equation

In this section we compare weak KAM and viscosity solutions.

Definition 5.1.   $\bullet$ A continuous real function $\varphi$ defined on the neighborhood of $e_l$ is $C^1$ if for every $j$ with $e_l\in I_j$, $\varphi|I_j$ is $C^1$.

$\bullet$ A continuous real function $\varphi$ defined on the neighborhood of $(e_l, t)$ is $C^1$ if for every $j$ with $e_l\in I_j$, $\varphi|I_j\times(t-\delta, t+\delta )$ is $C^1$.

Note that if $\alpha : [0, \delta]\to I_j$ is differentiable and $\alpha (0)=e_l$, then $\alpha _+'(0)\in T^-_{e_l}I_j$ and we have

We consider the Hamiltonian consisting in functions $H_j:I_j\times\mathbb{R}\to\mathbb{R}$ given by

and the Hamilton Jacobi equations

Note that if $L$ is symmetric at the vertices, then for any vertex $e_l$ there is a function $h_a$ such that $H_j(e_l, p)=h_a(|p|)$ for any $j$ with $e_l\in I_j$. This kind of Hamiltonians are called of eikonal type [3].

The following definition appeared in [3] and [4].

Definition 5.2. A function $u:G\to\mathbb{R}$ is a

$\bullet$ viscosity subsolution of (21) if satisfies the usual definition in $G\backslash \mathcal{V}$ and for any $C^1$ function $\varphi$ on the neighborhood of any $e_l$ s.t. $u-\varphi$ has a maximum at $e_l$ we have

$\bullet$ viscosity supersolution of (21) if satisfies the usual definition in $G\backslash \mathcal{V}$ and for any $C^1$ function $\varphi$ on the neighborhood of any $e_l$ s.t. $u-\varphi$ has a minimum at $e_l$ we have

$\bullet$ viscosity solution if it is both, a subsolution and a supersolution.

A function $u:G\times[0, \infty)\to\mathbb{R}$ is a$\bullet$ viscosity subsolution of (22) if satisfies the usual definition in $G\backslash \mathcal{V}\times[0, \infty)$ and for any $C^1$ function $\varphi$ on the neighborhood of any $(e_l, t)$ s.t. $u-\varphi$ has a maximum at $(e_l, t)$ we have

$\bullet$ viscosity supersolution of (21) if satisfies the usual definition in $G\backslash \mathcal{V}\times[0, \infty)$ and for any $C^1$ function $\varphi$ on the neighborhood of any $(e_l, t)$ s.t. $u-\varphi$ has a minimum at $(e_l, t)$ we have

$\bullet$ viscosity solution if it is both, a subsolution and a supersolution.

Proposition 22. If $u:G\to\mathbb{R}$ is dominated then then it is a viscosity subsolution of (21). If $u$ is a backward weak KAM solution then it is a viscosity solution.

proof. Suppose $u:G\to\mathbb{R}$ is dominated. Let $\varphi$ be a $C^1$ function on the neighborhood of $e_l$ s.t. $u-\varphi$ has a maximum at $e_l$, $j$ s.t. $e_l\in I_j$, $\alpha :[0, \delta]\to I_j$ differentiable with $\alpha (0)=e_l$, $z=\alpha '(0)$. Define $\gamma :[-\delta, 0]\to I_j$ by $\gamma (s)=\alpha (-s)$.

So $u$ is a subsolution.

Let $\varphi$ be a $C^1$ function on the neighborhood of $e_l$ s.t. $u-\varphi$ has a minimum at $e_l$. Let $\gamma :(-\infty, 0]\to G$ be such that $\gamma (0)=e_l$ and for $t<0$

Let $\delta >0, j$ be such that $\gamma ([-\delta, 0])\subset I_j$.

Define $\alpha :[0, \delta]\to I_j$ by $\alpha (t)=\gamma (-t)$, $z=\alpha '(0)$,

So $u$ is a supersolution.

Our approach to get a converse to Proposition 22 is to prove uniqueness of solutions to the Cauchy problem for (22), using a comparison principle. For that purpose the symmetry Lagrangian is a sufficient condition. It may be possible that other assumptions imply the required uniqueness or that a different approach gives a converse to Proposition 22.

Proposition 23. Let $f:G\to\mathbb{R}$ be continuous and define $u:G\times[0, \infty)\to\mathbb{R}$ by $u(x, t)=\mathcal{L}_tf(x)$, then $u$ is a viscosity solution of (22)

proof. Since $\mathcal{L}_tf=\mathcal{L}_{t-s}(\mathcal{L}_sf)$ if $0\le s<t$, for any $\gamma :[s, t]\to G$

and for any $x\in G$ there is $\gamma :[s, t]\to G$ with $\gamma (t)=x$ such that equality in (23) holds.

Let $\varphi$ be a $C^1$ function on the neighborhood of $(e_l, t)$ s.t. $u-\varphi$ has a maximum at $(e_l, t)$, $j$ s.t. $e_l\in I_j$, $\alpha :[0, \delta]\to I_j$ differentiable with $\alpha (0)=e_l$, $z=\alpha '(0)$. Define $\gamma :[t-\delta, t]\to I_j$ by $\gamma (s)=\alpha (t-s)$.

So $u$ is subsolution.

Let $\varphi$ be a $C^1$ function on the neighborhood of $(e_l, t)$ s.t. $u-\varphi$ has a minimum at $(e_l, t)$. Let $\gamma :[t-1, t]\to G$ be such that $\gamma (t)=e_l$ and

Let $\delta >0, j$ be such that $\gamma ([t-\delta, t])\subset I_j$. For $s\in[t-\delta, t]$

Define $\alpha :[0, \delta]\to I_j$ by $\alpha (s)=\gamma (t-s)$, $z=\alpha '(0)$,

So $u$ is supersolution.

Proposition 24. Suppose the Lagrangian is symmetric at the vertices. Let $u, v:G\times[0, T]\to\mathbb{R}$ be respectively a Lipschitz viscosity sub, supersolution of (22) such that $u(x, 0)\le v(x, 0)$, for any $x\in G$. Then $u\le v$.

proof. Suppose that there are $x^*, t^*$ such that $\delta =u(x^*, t^*)- v(x^*, t^*)>0$. Let $0<\rho\le\dfrac\delta {4t^*}$ and define $\Phi:G^2\times[0, T]^2$ by

From the previous definitions we have

It follows from $\Phi(x_\varepsilon, x_\varepsilon, t_\varepsilon, t_\varepsilon )+\Phi(y_\varepsilon, y_\varepsilon, s_\varepsilon, s_\varepsilon )\le 2\Phi(x_\varepsilon, y_\varepsilon, t_\varepsilon, s_\varepsilon )$ that

Thus, there is a sequence $\varepsilon \to 0$ such that $x_\varepsilon, y_\varepsilon$ converge to $\overline x \in G$ and $t_\varepsilon, s_\varepsilon$ converge to $\overline t \in[0, T]$ and (24) gives

and so $\overline t \ne 0$.

Define the test functions

Since $u-\varphi$ has maximum at $(x_\varepsilon, t_\varepsilon )$, $v-\psi$ has minimum at $(y_\varepsilon, s_\varepsilon )$, $u$ is subsolution and $v$ is supersolution,

Since $\rho>0$ we can not have $x_\varepsilon =y_\varepsilon$.

If $\overline x$ is not a vertex, $\overline x \in I_j$, for $\varepsilon >0$ small we have

If we denote by $a(x_\varepsilon, y_\varepsilon )$ this common value, then (25) becomes

with $a(x_\varepsilon, y_\varepsilon )$ bounded as $\varepsilon \to 0$, giving a contradiction.

Suppose now that $\overline x =e_l$. For $\varepsilon >0$ small we distinguish the following cases

1. Neither $x_\varepsilon$ nor $y_\varepsilon$ is a vertex. If $x_\varepsilon, y_\varepsilon \in I_j$, $d(x_\varepsilon, y_\varepsilon )=|\sigma_j(x_\varepsilon )-\sigma_j(y_\varepsilon )|$. If $x_\varepsilon \in I_i$, $y_\varepsilon \in I_j$, and $e_l\in I_i\cap I_j$, then $d(x_\varepsilon, y_\varepsilon )=d(x_\varepsilon, e_l)+d(e_l, y_\varepsilon )$. In both subcases

Then (25) becomes

2. Suppose $x_\varepsilon =e_l$, $y_\varepsilon \in I_j\backslash \mathcal{V}$.

Since

we have that (25) becomes

3. If $y_\varepsilon =e_l$, $x_\varepsilon \in I_j\backslash \mathcal{V}$ we get in the same way that (25) becomes

Since $\dfrac{d(x_\varepsilon, y_\varepsilon )}{\varepsilon }$ remains bounded as $\varepsilon \to 0$, we get a contradiction.

Corollary 7. Suppose the Lagrangian is symmetric at the vertices. Let $u, v:G\times[0, T]\to\mathbb{R}$ be viscosity solutions of (22) such that $u(x, 0)= v(x, 0)$ for any $x\in G$. Then $u=v$.

Corollary 8. Suppose the Lagrangian is symmetric at the vertices. Let $f:G\to\mathbb{R}$ be a viscosity solution of (21), then $f$ is a fixed point of the Lax semigroup $\mathcal{L}_t+ct$.

proof. We next show that $u(x, t)=f(x)-ct$ is a viscosity solution of (21). Proposition 23 and Corollary 7 then imply that $f-ct=\mathcal{L}_tf$.

Let $\varphi$ be a $C^1$ function on the neighborhood of $(e_l, t)$ s.t. $u-\varphi$ has a maximum at $(e_l, t)$. Then $s\to -cs-\varphi(e_l, s)$ has a maximum at $t$ and so $\varphi_t(e_l, t)=-c$. Since $f-\varphi(\cdot, t)$ has a maximum at $e_l$ we have

so $u$ is a subsolution of (22). Similarly $u$ is a supersolution of (22).

Corollary 9. Suppose the Lagrangian is symmetric at the vertices. Let $u:G\to\mathbb{R}$ be a viscosity solution of (21) then the representation formula (9) holds.

proof. By Proposition 22 and Corollary 9, $u$ is a backward weak KAM solution and by Theorem (3.8), formula (9) holds.