A representation formula of the viscosity solution of the contact Hamilton-Jacobi equation and its applications
aa r X i v : . [ m a t h . A P ] J a n A REPRESENTATION FORMULA OF THE VISCOSITY SOLUTIONOF THE CONTACT-TYPE HAMILTON-JACOBI EQUATION ANDITS APPLICATIONS
PANRUI NI, LIN WANG, AND JUN YAN
Abstract.
Assume M is a closed, connected and smooth Riemannian manifold. Weconsider the following two forms of Hamilton-Jacobi equations ( ∂ t u ( x, t ) + H ( x, u ( x, t ) , ∂ x u ( x, t )) = 0 , ( x, t ) ∈ M × (0 , + ∞ ) .u ( x,
0) = ϕ ( x ) , x ∈ M, ϕ ∈ C ( M, R ) . and H ( x, u ( x ) , ∂ x u ( x )) = 0 , where H ( x, u, p ) is continuous, convex and coercive in p , uniformly Lipschitz in u . Byintroducing a solution semigroup, we provide a representation formula of the viscositysolution of the evolutionary equation. As its applications, we obtain a necessaryand sufficient condition for the existence of the viscosity solutions of the stationaryequations. Moreover, we extend partial results established in [28, 29] to more generalcases. Besides, we prove a new comparison result with a necessary neighborhood for aspecial class of Hamilton-Jacobi equations that does not satisfy the “proper” conditionintroduced in the seminal paper [12]. Contents
1. Introduction and main results 12. Preliminaries 83. Representation of the evolutionary solution 104. Existence of stationary solutions 175. Existence of forward weak KAM solutions 216. Strictly increasing case 237. Strictly decreasing case 258. More about regularity 30Appendix A. 32References 391.
Introduction and main results
The study of the theory of viscosity solutions of the following two forms of Hamilton-Jacobi equations ∂ t u ( x, t ) + H ( x, u ( x, t ) , ∂ x u ( x, t )) = 0 , (1.1)and H ( x, u ( x ) , ∂ x u ( x )) = 0 (1.2) Date : January 12, 2021.2020
Mathematics Subject Classification.
Key words and phrases. first order PDEs, viscosity solutions, solution semigroup. as a long history. There are numerous results on the existence, uniqueness, stabilityand large-time behavior problems for the viscosity solutions of the above first-ordernonlinear partial differential equations. Especially for the cases with the Hamiltonianindependent of the argument u , whose characteristic equations are classical Hamiltonequations.For the Hamilton-Jacobi equations depending on the unknown functions, the corre-sponding characteristic equations are called the contact Hamilton equations. In [26],the authors introduced an implicit variational principle for the contact Hamilton equa-tions. Based on the implicit variational principle, [27] gave the representation formulafor the unique viscosity solution of the evolutionary equation by introducing a solutionsemigroup. The existence of the solutions for the ergodic problem was also proved. [28]developed the Aubry-Mather theory for contact Hamiltonian systems with moderateincreasing in u . In [29], the authors further studied the strictly decreasing case, anddiscussed large time behavior of the solution of the evolutionary case.It is worth mentioning that an alternative remarkable variational formulation wasprovided in [9, 10] in light of G. Herglotz’s work, which is of explicit form with non-holonomic constraints. By using the Herglotz variational principle, various kinds ofrepresentation formulas for the viscosity solutions of (1.1) were also obtained in [17].In order to get the C -regularity of the minimizers, [26] assumes H ( x, u, p ) to be atleast C . Thus, the results in [27–29] based on the implicit variational principle, allrequire the contact Hamiltonian to be at least C . This paper is devoted to reducingthe dynamical assumptions on the Hamiltonian: C , strictly convex and superlinear tocontinuous, convex and coercive. In this general case, the contact Hamiltonian equationscan not be defined.Throughout this paper, we assume M is a closed, connected and smooth Riemannianmanifold and H : T ∗ M × R → R satisfies (C): H ( x, u, p ) is continuous; (CON): H ( x, u, p ) is convex in p , for any ( x, u ) ∈ M × R ; (CER): H ( x, u, p ) is coercive in p , i.e. lim k p k→ + ∞ (inf x ∈ M H ( x, , p )) = + ∞ ; (LIP): H ( x, u, p ) is Lipschitz in u , uniformly with respect to ( x, p ), i.e., there exists λ > | H ( x, u, p ) − H ( x, v, p ) | ≤ λ | u − v | , for all ( x, p ) ∈ T ∗ M and all u, v ∈ R .Using [14] Proposition 2.7, one can prove the corresponding Lagrangian L ( x, u, ˙ x ) := sup p ∈ T ∗ x M ( h ˙ x, p i − H ( x, u, p ))satisfies the following properties (LSC): L ( x, u, ˙ x ) is lower semicontinuous in ˙ x ; (CON): L ( x, u, ˙ x ) is convex in ˙ x , for any ( x, u ) ∈ M × R ; (LIP): L ( x, u, ˙ x ) is Lipschitz in u Lipschitz, uniformly with respect to ( x, ˙ x ), i.e., thereexists λ > | L ( x, u, ˙ x ) − L ( x, v, ˙ x ) | ≤ λ | u − v | , for all ( x, ˙ x ) ∈ T M and all u, v ∈ R .We list notations which will be used later in the present paper. (cid:5) diam( M ) denotes the diameter of M . (cid:5) d ( x, y ) denotes the distance between x and y induced by the Riemannian metric g on M . (cid:5) k · k denotes the norms induced by g on both tangent and cotangent spaces of M . C ( M ) stands for the space of continuous functions on M . Lip ( M ) stands forthe space of Lipschitz continuous functions on M . (cid:5) k·k ∞ stands for the supremum norm of the vector valued functions on its domain.1.1. The representation of the solution of the evolutionary equation.
Considerthe viscosity solution of the Cauchy problem ( ∂ t u ( x, t ) + H ( x, u ( x, t ) , ∂ x u ( x, t )) = 0 , ( x, t ) ∈ M × (0 , + ∞ ) .u ( x,
0) = ϕ ( x ) , x ∈ M. ( CP H )We have the following results. Main Result 1.
Assume H : T ∗ M × R → R satisfies (C)(CON)(CER)(LIP). thefollowing backward solution semigroup T − t ϕ ( x ) = inf γ ( t )= x (cid:26) ϕ ( γ (0)) + Z t L ( γ ( τ ) , T − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) d τ (cid:27) (T-) is well-defined. The infimum is taken among absolutely continuous curves γ : [0 , t ] → M with γ ( t ) = x . If the initial condition ϕ is continuous, then u ( x, t ) := T − t ϕ ( x ) representsthe unique continuous viscosity solution of ( CP H ). If ϕ is Lipschitz continuous, then u ( x, t ) := T − t ϕ ( x ) is also Lipschitz continuous on M × [0 , + ∞ ) . For the uniqueness of the solution of ( CP H ) with the continuous initial condition,one can see [21] Theorem 1.1. Definition 1.1.
Define F ( x, u, p ) := H ( x, − u, − p ) , according to Main Result 1, thereexists a backward solution semigroup ¯ T − t corresponding to F . Then the forward solutionsemigroup of H can be defined by T + t ϕ := − ¯ T − t ( − ϕ ) . Remark 1.2.
The forward solution semigroup satisfies T + t ϕ ( x ) = sup γ (0)= x (cid:26) ϕ ( γ ( t )) − Z t L ( γ ( τ ) , T + t − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ (cid:27) . (T+) Proposition 1.3.
The solution semigroups defined by (T-) and (T+) have the followingproperties (1)
For ϕ and ϕ ∈ C ( M ) , if ϕ ( x ) ≤ ϕ ( x ) for all x ∈ M , we have T − t ϕ ( x ) ≤ T − t ϕ ( x ) and T + t ϕ ( x ) ≤ T + t ϕ ( x ) for all ( x, t ) ∈ M × (0 , + ∞ ) . (2) Given any ϕ and ψ ∈ C ( M ) , we have k T − t ϕ − T − t ψ k ∞ ≤ e λt k ϕ − ψ k ∞ and k T + t ϕ − T + t ψ k ∞ ≤ e λt k ϕ − ψ k ∞ for all t > . The proofs of Remark 1.2 and Proposition 1.3 are provided in the appendix.1.2.
Weak KAM solutions, fixed points and viscosity solutions.
Following Fathi[15], one can extend the definition of weak KAM solutions of equation (1.2) by usingLipschitz calibrated curves instead of C curves. Definition 1.4.
A function u − ∈ C ( M ) is called a backward weak KAM solution of( E H ) if (1) For each absolutely curve γ : [ t ′ , t ] → M , we have u − ( γ ( t )) − u − ( γ ( t ′ )) ≤ Z tt ′ L ( γ ( s ) , u − ( γ ( s )) , ˙ γ ( s )) ds. The above condition reads that u − is dominated by L and denoted by u − ≺ L . For each x ∈ M , there exists a Lipschitz curve γ − : ( −∞ , → M with γ − (0) = x such that u − ( γ − ( t )) − u − ( x ) = Z t L ( γ − ( s ) , u − ( γ − ( s )) , ˙ γ − ( s )) ds, ∀ t < . The curves satisfying the above equality are called ( u − , L, -calibrated curves. Definition 1.5.
Similar to the definition above, a function v + ∈ C ( M ) is called aforward weak KAM solution of ( E H ) if (1) For each absolutely curve γ : [ t ′ , t ] → M , we have v + ( γ ( t )) − v + ( γ ( t ′ )) ≤ Z tt ′ L ( γ ( s ) , v + ( γ ( s )) , ˙ γ ( s )) ds. The above condition reads that v + is dominated by L and denoted by v + ≺ L . (2) For each x ∈ M , there exists a Lipschitz curve γ + : [0 , + ∞ ) → M with γ + (0) = x such that v + ( γ + ( t )) − v + ( x ) = Z t L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) ds, ∀ t > . The curves satisfying the above equality are called ( v + , L, -calibrated curves. We denote by S − and S + the set of all backward weak KAM solutions and the set ofall forward weak KAM solutions respectively. Remark 1.6.
The backward weak KAM solutions defined in Definition 1.4 are the fixedpoints of T − t . Similarly, the forward weak KAM solutions defined in Definition 1.5 arethe fixed points of T + t . The proof of Remark 1.6 is given in the appendix. By Main Result 1, the fixed pointsof T − t , which are denoted by u − , are viscosity solutions of H ( x, u ( x ) , ∂ x u ( x )) = 0 . ( E H )1.3. The existence of the solutions of the stationary equation.Remark 1.7.
Assume H : T ∗ M × R → R satisfies (C)(CER)(LIP), according to theclassical Perron method [18] (see also [20, Theorem 1.1]), if ( E H ) has a subsolution ψ and a supersolution Ψ , both are Lipschitz continuous and satisfy ψ ≤ Ψ , then the aboveequation admits a Lipschitz viscosity solution. Besides (LSC), (CON), (LIP), we assume the Lagrangian L ( x, u, ˙ x ) satisfies (LB): For any
R >
0, the Lagrangian L ( x, u, ˙ x ) is bounded on M × [ − R, R ] × B (0 , R ),and the bound depends on R . Here B (0 , R ) denotes the ball contained in T x M centered at the origin with radius R .Compared to [18], we introduce another necessary and sufficient condition for ( E H )to admit solutions. Main Result 2.
Assume H : T ∗ M × R → R satisfies (C)(CON)(CER)(LIP), and thecorresponding Lagrangian L ( x, u, ˙ x ) satisfies (LB), then ( E H ) admits viscosity solutionsif and only if one of the following conditions holds. (1) There exist two continuous functions ϕ and ϕ such that T − t ϕ has a lowerbound independent of t , and T − t ϕ has a upper bound independent of t ; (2) There exist two continuous functions ϕ and ψ , and two constants t a and t b > such that T − t a ϕ ≥ ϕ and T − t b ψ ≤ ψ . f ( E H ) admits a solution u − , we can take u − as the initial function, the condition(1) and (2) holds obviously. Thus, we only need to show the opposite direction.In the discussion below, we need to introduce the following assumption (S): The set S − is nonempty, i.e. equation ( E H ) admits a viscosity solution u − . Main Result 3.
Assume H : T ∗ M × R → R satisying (C)(CON)(CER)(LIP) andthe condition (S), and the corresponding Lagrangian L ( x, u, ˙ x ) satisfies (LB), then lim t → + ∞ T + t u − converges uniformly to u + , where u + ∈ S + and − u + is a viscositysolution of F ( x, u ( x ) , ∂ x u ( x )) = 0 . ( E F ) Moreover, the existence of the viscosity solutions of ( E H ) is equivalent to the existenceof the viscosity solutions of ( E F ). Definition 1.8.
Define the projected Ma˜n´e set I ( u − ,u + ) := { x ∈ M : u − ( x ) = u + ( x ) } . It can be shown (see Corollary 5.6 below) that for each t >
0, there exists x t ∈ M such that T + t u − ( x t ) = u − ( x t ). At the same time, we have u + = lim t → + ∞ T + t u − . Since M is compact, taking the accumulation point x ∗ of the sequence { x t n } n ∈ N , we have u + ( x ∗ ) = u − ( x ∗ ). Therefore, the set defined above is nonempty. Remark 1.9. If H ( x, u, p ) satisfies (C) (CON) (CER), and nondecreasing in u , thenthe condition (S) holds if and only if there exists a constant a ∈ R such that c ( H a ) := inf u ∈ C ∞ ( M ) max x ∈ M H ( x, a, ∂ x u ) = 0 . The proof of Remark 1.9 is given in the appendix.1.4.
Strictly increasing case.
We assume H : T ∗ M × R → R satisfies (C), (CON),(CER), (LIP) and (STI): H ( x, u, p ) is strictly increasing in u .The corresponding Lagrangian satisfies (LSC)(CON)(LIP)(LB) and (STD): L ( x, u, ˙ x ) is strictly decreasing in u . Main Result 4.
Assume H : T ∗ M × R → R satisfies (C)(CON)(CER)(STI) and thecondition (S), and the corresponding Lagrangian L ( x, u, ˙ x ) satisfies (LB), then u − isthe unique viscosity solution of ( E H ), and (1) For the limit function u + = lim t → + ∞ T + t u − , we have u − = lim t → + ∞ T − t u + . Wecall ( u − , u + ) a conjugate pair; (2) u + is the maximal element in S + ; (3) For v + ∈ S + , define I v + := { x ∈ M : u − ( x ) = v + ( x ) } , then I v + is nonempty, and I v + ⊆ I ( u − ,u + ) . Strictly decreasing case.
In this part, we are concerned with viscosity solutionstheory for a special class of Hamilton-Jacobi equations that does not satisfy the “proper”condition introduced in the seminal paper [12], H ( x, r, p ) H ( x, s, p ) whenever r s. We consider the viscosity solutions of ( E F ), or equivalently, the set of the forward weakKAM solutions corresponding to H . If H ( x, u, p ) satisfies (STI), then F ( x, u, p ) = H ( x, − u, − p ) satisfies STD): F ( x, u, p ) is strictly decreasing in u .We denote by VS ( F ) the set of the viscosity solutions of ( E F ). We also define apartial order (cid:22) on this set, i.e. v (cid:22) v if and only if v ( x ) ≤ v ( x ) for all x ∈ M .Under the assumption of Main Result 3, the existence of u − ∈ S − guarantees theexistence of u + ∈ S + . Note that − u + ∈ VS ( F ). It means VS ( F ) is nonempty. In viewof [28, Example 1.1] (see also Remark 1.10 below), this set may not be a singleton. Main Result 5.
Assume F : T ∗ M × R → R satisfies (C)(CON)(CER)(STD). Thecorresponding Lagrangian L F := sup p ∈ T ∗ x M ( h ˙ x, p i − F ( x, u, p )) satisfies (LB). Then (1) The set of VS ( F ) is compact in the topology induced by the W , ∞ -norm; (2) Let v − := min v ∈VS ( F ) v , then v − ∈ VS ( F ) . The maximal elements in VS ( F ) exist in the sense of the partial order; (3) Denote by ¯ u + the unique fixed point of the forward semigroup ¯ T + t correspondingto F . For v ∈ VS ( F ) , define I v := { x ∈ M : v ( x ) = ¯ u + ( x ) } . Let v and v ∈ VS ( F ) , then (i) If v ≤ v , then ∅ 6 = I v ⊆ I v ⊆ I v − , where v − := min v ∈VS ( F ) v ; (ii) If there exists a neighbourhood O of I v such that v | O ≤ v | O , then v ≤ v everywhere; (iii) If I v = I v and v | O = v | O , then v = v everywhere. Remark 1.10.
In order to explain the necessity of the neighbourhood condition above,we consider the following example − λu ( x ) + 12 | ∂ x u ( x ) | + V ( x ) = 0 , x ∈ S ≃ ( − , , (1.3) where V ( x ) is the restriction of x / on S . Then F ( x, u, p ) = − λu + | p | / V ( x ) defined on T ∗ S × R is Lipschitz continuous and smooth for x ∈ ( − , . Assume λ > ,then two viscosity solutions of the above equation are u ( x ) = λ + √ λ − V ( x ) , u ( x ) = λ − √ λ − V ( x ) . One can prove that the unique forward weak KAM solution ¯ u + satisfies ¯ u + (0) = u (0) = u (0) and ¯ u + ( x ) < u ( x ) on (0 , . The proof of Remark 1.10 is given in the appendix.Following [29], we then consider the large-time behavior of the viscosity solution ofthe following Chauchy problem ( ∂ t u ( x, t ) + F ( x, u ( x, t ) , ∂ x u ( x, t )) = 0 , ( x, t ) ∈ M × (0 , + ∞ ) .u ( x,
0) = ϕ ( x ) , x ∈ M. ( CP F ) Main Result 6.
Assume F : T ∗ M × R → R satisfies (C)(CON)(CER)(STD), andthe corresponding Lagrangian L F satisfies (LB). We also assume that ( E F ) admitssolutions. Denote by ¯ u + the unique forward weak KAM solution corresponding to F .For given ϕ ∈ Lip ( M ) , let w ( x, t ) be the unique Lipschitz viscosity solution of ( CP F ),then (1) The uniform boundedness of w ( x, t ) : (1-1) If ϕ satisfies the following condition ( ∗ ) ϕ ≥ ¯ u + and there exists a point x such that ϕ ( x ) = ¯ u + ( x ) .Then w ( x, t ) is bounded by a constant independent of t ; (1-2) If the condition ( ∗ ) does not hold, then we have the following: a) if there exists a point x such that ϕ ( x ) < ¯ u + ( x ) , then w ( x, t ) tendsto −∞ uniformly on x as t tends to + ∞ ; (b) if ϕ > ¯ u + , then w ( x, t ) tends to + ∞ uniformly on x as t tends to + ∞ . (2) The convergence of w ( x, t ) : (2-1) If ¯ u + ≤ ϕ ≤ v − , where (¯ u + , v − ) is a conjugate pair (defined in Main Result4), then lim t → + ∞ ¯ T − t ϕ ( x ) = v − ( x ) uniformly on x ∈ M ; (2-2) If the solution of ( E F ) is unique and ϕ satisfies ( ∗ ), then ϕ ∞ ( x ) = lim t → + ∞ ¯ T − t ϕ ( x );(2-3) Let v ∗ be a maximal element of VS ( F ) . If ϕ satisfies ( ∗ ) and ϕ ≥ v ∗ , then v ∗ = lim t → + ∞ w ( x, t ) . As a corollary, we have
Corollary 1.11.
There is a uniform bound
K > such that for each ϕ satisfying thecondition ( ∗ ), there is a constant T ( ϕ ) > depending on ϕ , such that | w ( x, t ) | ≤ K ,for all ( x, t ) ∈ M × [ T ( ϕ ) , + ∞ ) . More about regularity.
In the final part of this paper, we require the Hamil-tonian H : T ∗ M × R → R to have higher regularity. We assume H ( x, u, p ) satisfies(C)(CER)(LIP) and (STC): H ( x, u, p ) is strictly convex in p , for any ( x, u ) ∈ M × R ; (LL): ( x, p ) H ( x, u, p ) is locally Lipschitz, for all u ∈ R .Here we note that the condition (STC) is necessary. On one hand, the locally semicon-cavity of the viscosity solutions of ( E H ) and ( E F ) requires (STC), on the other hand,the proof of the differentiability of u ± on the calibrated curves also depends on thiscondition. Main Result 7. If H : T ∗ M × R → R satisfies (LL)(STC)(CER)(LIP) and (S), thecorresponding Lagrangian L ( x, u, ˙ x ) satisfies (LB), then for a conjugate pair ( u − , u + ) ,we have (1) The backward weak KAM solutions u − and forward weak KAM solutions u + aredifferentiable on I ( u − ,u + ) with the same derivative. We then define ˜ I ( u − ,u + ) := { ( x, u, p ) : x ∈ I ( u − ,u + ) , u = u − ( x ) , p = ∂ x u − ( x ) } . (2) The backward weak KAM solutions u − and forward weak KAM solutions u + are of class C on I ( u − ,u + ) , or equivalently, the lift from I ( u − ,u + ) to ˜ I ( u − ,u + ) iscontinuous.We then require higher regularity on H ( x, u, p ) . (1) If H ( x, u, p ) is of class C , then the contact Hamilton equations can be defined.For x ∈ I ( u − ,u + ) , there exists a C curve γ : ( −∞ , ∞ ) → M with γ (0) = x suchthat u − ( γ ( t )) = u + ( γ ( t )) for all t ∈ R , and x ( t ) := γ ( t ) , u ( t ) := u ± ( γ ( t )) , p ( t ) := ∂ γ ( t ) u ± ( γ ( t )) satisfies the contact Hamilton equations ˙ x = ∂ p H ( x, u, p ) , ˙ p = − ∂ x H ( x, u, p ) − ∂ u H ( x, u, p ) p, a.e. ˙ u = ∂ p H ( x, u, p ) · p. (1.4) If H ( x, u, p ) is in addition of class C , , then the backward weak KAM solution u − and the forward weak KAM solution u + are of class C , on I ( u − ,u + ) . Equiv-alently, the projection π : T ∗ M × R → M induces a bi-Lipschitz map between I ( u − ,u + ) and ˜ I ( u − ,u + ) . The rest of this paper is organized as follows. Section 2 gives the basic results on theexistence and regularity of the minimizers of the one dimensional variational problem.In Section 3.1, we prove Main result 1 under the superlinear condition and the Lipschitzinitial condition. In Section 3.2, we complete the proof of Main Result 1. Main Result2 is proved in Section 4. Main Result 3 is proved in Section 5. In Section 6 and 7,we consider the monotone cases. Main Result 4 is proved in Section 6. Main Result 5and 6 are proved in Section 7. The regularity of weak KAM solutions is considered inSection 8, and Main Result 7 is proved.2.
Preliminaries
The following results are important in the proof of the existence and regularity of theminimizers in (T-), which all come from [6] and [24]. The results in the present sectionwere all proved for the case in the Euclidean space R n . One can easily generalise themfor the case in the Riemannian manifold M .2.1. The direct method in the calculus of variations.Lemma 2.1.
Let J be a bounded interval. Assume that F ( t, x, ˙ x ) is lower semicontin-uous, convex in ˙ x , and has a lower bound. Then the integral functional F ( γ ) = Z J F ( s, γ ( s ) , ˙ γ ( s )) ds is sequentially weakly lower semicontinuous in W , ( J, M ) . Proposition 2.2.
Let M be a compact connected smooth manifold. Denote by I =( a, b ) ⊂ R a bounded interval, and let F ( t, x, ˙ x ) be a Lagrangian defined on I × T M .Assume F satisfies (i) F ( t, x, ˙ x ) is measurable in t for all ( x, ˙ x ) , and continuous in ( x, ˙ x ) for almostevery t ; (ii) F ( t, x, ˙ x ) is convex in ˙ x ; (iii) F ( t, x, ˙ x ) is superlinear in ˙ x .Then for any given boundary condition x and x ∈ M , there exists a minimizer of R I F ( t, x, ˙ x ) dt in { x ( t ) ∈ W , ([ a, b ] , M ) : x ( a ) = x , x ( b ) = x } . -convergence.Definition 2.3. Let X be a topological space. Given a sequence F n : X → [ −∞ , + ∞ ] ,then we define (Γ − lim inf n → + ∞ F n )( x ) = sup U ∈N ( x ) lim inf n → + ∞ inf y ∈ U F n ( y ) , (Γ − lim sup n → + ∞ F n )( x ) = sup U ∈N ( x ) lim sup n → + ∞ inf y ∈ U F n ( y ) . Here the neighbourhoods N ( x ) can be replaced by the topological basis. When the superiorlimit equals to the inferior limit, we can define the Γ -limit. efinition 2.4. Let X be a topological space. For every function F : X → [ −∞ , + ∞ ] ,the lower semicontinuous envelope sc − F of F is defined for every x ∈ X by ( sc − F )( x ) = sup G ∈G ( F ) G ( x ) , where G ( F ) is the set of all lower semicontinuous functions G on X such that G ( y ) ≤ F ( y ) for every y ∈ X . Lemma 2.5. If F n is an increasing sequence, then Γ − lim n → + ∞ F n = lim n → + ∞ sc − F n = sup n ∈ N sc − F n . Remark 2.6. If F n is an increasing sequence of lower semicontinuous functions whichconverges pointwise to a function F , then F is lower semicontinuous and F n has a Γ -convergence to F by Lemma 2.5. Lemma 2.7.
If the sequence F n has a Γ -convergence in X to F , and there is a compactset K ⊂ X such that inf x ∈ X F n ( x ) = inf x ∈ K F n ( x ) , then F takes its minimum in X , and min x ∈ X F ( x ) = lim n → + ∞ inf x ∈ X F n ( x ) . (2.1)2.3. Regularity of minimizers in t -dependent cases. The following results con-sider the regularity of minimizers, which comes form [5]. Consider the following one-dimensional variational problem I ( γ ) := Z ba F ( t, γ ( t ) , ˙ γ ( t )) dt + Ψ( γ ( a ) , γ ( b )) , (P)where γ is taken in the class of absolutely continuous curves, Ψ takes its value in R ∪ { + ∞} and stands for the constraints on the two ends of the curves γ . In thefollowing, we focus on a minimizer of the above integral functional, which is denoted by γ ∗ ∈ W , ([ a, b ] , M ). (Lt): F takes its value in R , there exist a constant ε > k : [ a, b ] × (0 , + ∞ ) → R such that k ( t, ∈ L [ a, b ], and, fora.e. t ∈ [ a, b ], for all σ > | F ( t , γ ∗ ( t ) , σ ˙ γ ∗ ( t )) − F ( t , γ ∗ ( t ) , σ ˙ γ ∗ ( t )) | ≤ k ( t, σ ) | t − t | , where t , ∈ [ t − ε, t + ε ] ∩ [ a, b ]. Lemma 2.8.
Let γ ∗ be a minimizer of (P). If F satisfies (Lt), then there exists anabsolutely continuous function p ∈ W , ([ a, b ] , R ) such that for a.e. t ∈ [ a, b ] , we have F (cid:18) t, γ ∗ ( t ) , ˙ γ ∗ ( t ) v (cid:19) v − F ( t, γ ∗ ( t ) , ˙ γ ∗ ( t )) ≥ p ( t )( v − , ∀ v > , (W) and | p ′ ( t ) | ≤ k ( t, for a.e. t ∈ [ a, b ] . Lemma 2.9.
Let γ ∗ be a minimizer of (P). Assume F is a Borel measurable function.If F satisfies (Lt) and (1) Superlinearity: There exists a function
Θ : R → R satisfying lim r → + ∞ Θ( r ) r = + ∞ , and F ( t, γ ∗ ( t ) , ξ ) ≥ Θ( k ξ k ) for all ξ ∈ T γ ∗ ( t ) M. Local boundedness: There exists ρ > and M ≥ such that for a.e. t ∈ [ a, b ] ,we have F ( t, γ ∗ ( t ) , ξ ) ≤ M for all ξ ∈ T γ ∗ ( t ) M with k ξ k = ρ .Then the minimizer γ ∗ is Lipschitz continuous. Moreover, if k ˙ γ ∗ ( t ) k > ρ , we take v = k ˙ γ ∗ ( t ) k /ρ > in (W), then F (cid:18) t, γ ∗ ( t ) , ρ ˙ γ ∗ ( t ) k ˙ γ ∗ ( t ) k (cid:19) ≥ ρ Θ( k ˙ γ ∗ ( t ) k ) k ˙ γ ∗ ( t ) k − k p k ∞ . Therefore k ˙ γ ∗ ( t ) k ≤ max { ρ, R } where R := inf { s : ρ Θ( s ) s > M + k p k ∞ } . Representation of the evolutionary solution
The proof of Main Result 1 is organized as follows. In Section 3.1, we prove the MainResult when the initial condition ϕ is Lipschitz continuous and the the Hamiltonian H ( x, u, p ) is superlinear in p . In Section 3.2, we first prove the Main Result when ϕ isLipschitz continuous and the the Hamiltonian H ( x, u, p ) satisfies (CER). We then provethe Main Result when ϕ is a continuous function.3.1. Superlinear condition.
In this section, we assume the Hamiltonian H : T ∗ M × R → R satisfies (C)(CON)(LIP) and (SL): For every ( x, u ) ∈ M × R , H ( x, u, p ) is superlinear in p , i.e. there exists afunction Θ : R → R satisfyinglim r → + ∞ Θ( r ) r = + ∞ , and H ( x, u, p ) ≥ Θ( k p k ) for every ( x, u, p ) ∈ T ∗ M × R . One can prove the corresponding Lagrangian satisfies (CON)(LIP) and (C): L ( x, u, ˙ x ) is continuous; (SL): For every ( x, u ) ∈ M × R , L ( x, u, ˙ x ) is superlinear in ˙ x , i.e. there exists afunction Θ : R → R satisyinglim r → + ∞ Θ( r ) r = + ∞ , and L ( x, u, ˙ x ) ≥ Θ( k ˙ x k ) for every ( x, u, ˙ x ) ∈ T M × R . Lemma 3.1.
For given
T > and ϕ ∈ C ( M ) , if v ( x, t ) is a Lipschitz continuousfunction on M × [0 , T ] , then for any ( x, t ) ∈ M × [0 , T ] , the minimizers of u ( x, t ) = inf γ ( t )= x (cid:26) ϕ ( γ (0)) + Z t L ( γ ( τ ) , v ( γ ( τ ) , τ ) , ˙ γ ( τ )) d τ (cid:27) (3.1) are Lipschitz continuous. For any r > , if d ( x, x ′ ) ≤ r and | t − t ′ | ≤ r/ , where t ≥ r > , then the Lipschitz constant of the minimizers of u ( x ′ , t ′ ) is independent of ( x ′ , t ′ ) , and only depends on ( x, t ) and r .Proof. According to (LIP) and the Lipschitz continuity of v ( x, t ) on M × [0 , T ], foreach τ ∈ [0 , t ], the map s L ( γ ( τ ) , v ( γ ( τ ) , s ) , ˙ γ ( τ )) satisfies the condition (Lt), where k ≡ λ k ∂ t v ( x, t ) k ∞ . By Lemma 2.9, for every ( x, t ) ∈ M × [0 , T ], the minimizers of u ( x, t ) are Lipschitz continuous. However, the Lipschitz constant depends on the endpoint ( x, t ). We are now going to show for ( x ′ , t ′ ) sufficiently close to ( x, t ), the Lipschgitzconstant of the minimizers of u ( x ′ , t ′ ) is independent of ( x ′ , t ′ ). or any r >
0, if d ( x, x ′ ) ≤ r and | t − t ′ | ≤ r/
2, where t ≥ r >
0, we denote by γ ( s ; x, t ) and γ ( s ; x ′ , t ′ ) the minimizers of u ( x, t ) and u ( x ′ , t ′ ) respectively, then we have u ( x ′ , t ′ ) = ϕ ( γ (0; x ′ , t ′ )) + Z t ′ L ( γ ( s ; x ′ , t ′ ) , v ( γ ( s ; x ′ , t ′ ) , s ) , ˙ γ ( s ; x ′ , t ′ )) ds ≤ ϕ ( γ (0; x, t )) + Z t − r L ( γ ( s ; x, t ) , v ( γ ( s ; x, t ) , s ) , ˙ γ ( s ; x, t )) ds + Z t ′ t − r L ( α ( s ) , v ( α ( s ) , s ) , ˙ α ( s )) ds, where α : [ t − r, t ′ ] → M is a geodesic connecting γ ( t − r ; x, t ) and x ′ with constantspeed. Noticing k ˙ α k ≤ t ′ − ( t − r ) (cid:0) d ( γ ( t − r ; x, t ) , x ) + d ( x, x ′ ) (cid:1) ≤ (cid:18) r Z tt − r k ˙ γ ( s ; x, t ) k ds + 1 (cid:19) , we obtain that Z t ′ L ( γ ( s ; x ′ , t ′ ) , v ( γ ( s ; x ′ , t ′ ) , s ) , ˙ γ ( s ; x ′ , t ′ )) ds has a bound depending only on ( x, t ) and r . By (SL), there exists a constant M ( x, t, r ) > Z t ′ k ˙ γ ( s ; x ′ , t ′ ) k ds ≤ M ( x, t, r ) , where t ′ ≥ t − r/ >
0. It means k ˙ γ ( s ; x ′ , t ′ ) k are equi-integrable. Therefore, for ( x ′ , t ′ )sufficiently close to ( x, t ), there exists a constant R ( x, t, r ) > s ∈ [0 , t ′ ] suchthat k ˙ γ ( s ; x ′ , t ′ ) k ≤ R ( x, t, r ). By Lemma 2.8, there exists an absolutely continuousfunction p ( t ; x ′ , t ′ ) satisfying | p ′ ( t ; x ′ , t ′ ) | ≤ λ k ∂ t v ( x, t ) k ∞ such that L ( γ ( s ; x ′ , t ′ ) , v ( γ ( s ; x ′ , t ′ ) , s ) , ˙ γ ( s ; x ′ , t ′ ) θ ) θ − L ( γ ( s ; x ′ , t ′ ) , v ( γ ( s ; x ′ , t ′ ) , s ) , ˙ γ ( s ; x ′ , t ′ )) ≥ p ( s ; x ′ , t ′ )( θ − , ∀ θ > . One can take θ = 2 and t = s to obtain the upper bound of p ( s ). One can take θ = 1 / t = s to obtain the lower bound of p ( s ). Note that p ′ ( t ) is bounded,we finally obtain the bound of k p ( t ) k ∞ which is independent of ( x ′ , t ′ ). Since L ( x, u, ˙ x )satisfies (SL), according to Lemma 2.9 and taking ρ = 1, we have L ( γ ( s ; x ′ , t ′ ) , v ( γ ( s ; x ′ , t ′ ) , s ) , ˙ γ ( s ; x ′ , t ′ ) k ˙ γ ( s ; x ′ , t ′ ) k ) ≥ Θ( k ˙ γ ( s ; x ′ , t ′ ) k ) k ˙ γ ( s ; x ′ , t ′ ) k − k p ( s ; x ′ , t ′ ) k ∞ . Therefore, for ( x ′ , t ′ ) sufficiently close to ( x, t ), the minimizers γ ( s ; x ′ , t ′ ) have a Lipschitzconstant independent of ( x ′ , t ′ ). (cid:3) Lemma 3.2.
For given
T > and ϕ ∈ C ( M ) , if v ( x, t ) is a Lipschitz continuousfunction on M × [0 , T ] , then for any ( x, t ) ∈ M × (0 , T ] , the value function u ( x, t ) defined in (3.1) is locally Lipschitz.Proof. We first show that u ( x, t ) is locally Lipschitz in x . For any δ >
0, given ( x , t ) ∈ M × [ δ, T ] and x , x ′ ∈ B ( x , δ/ d = d ( x, x ′ ) ≤ δ the Riemannian distance etween x and x ′ , then u ( x ′ , t ) − u ( x, t ) ≤ Z tt − d L ( α ( s ) , v ( α ( s ) , s ) , ˙ α ( s )) ds − Z tt − d L ( γ ( s ; x, t ) , v ( γ ( s ; x, t ) , s ) , ˙ γ ( s ; x, t )) ds, where γ ( s ; x, t ) is a minimizer of u ( x, t ) and α : [ t − d , t ] → M is a geodesic connecting γ ( t − d ; x, t ) and x ′ with constant speed. By Lemma 3.1, if x ∈ B ( x , δ/ k ˙ γ ( s ; x, t ) k depends only on x and δ . Noticing that k ˙ α ( s ) k ≤ d ( γ ( t − d ; x, t ) , x ′ ) d ≤ d ( γ ( t − d ; x, t ) , x ) d + 1 , and that d ( γ ( t − d ; x, t ) , x ) ≤ R tt − d k ˙ γ ( s ; x, t ) k ds , the bound of k ˙ α ( s ) k also depends onlyon x and δ . Exchanging the role of ( x, t ) and ( x ′ , t ), one obtain that | u ( x, t ) − u ( x ′ , t ) | ≤ J d ( x, x ′ ), where J depends only on x and δ . We conclude that for t ∈ (0 , T ], thevalue function u ( · , t ) is (locally) Lipschitz on M .We are now going to show the locally Lipschitz continuity of u ( x, t ) in t . For given t ≥ δ/ t and t ′ ∈ [ t − δ/ , t + δ/ t ′ > t ,then u ( x, t ′ ) − u ( x, t ) ≤ u ( γ ( t ; x, t ′ ) , t ) − u ( x, t )+ Z t ′ t L ( γ ( s ; x, t ′ ) , v ( γ ( s ; x, t ′ ) , s ) , ˙ γ ( s ; x, t ′ )) ds, here the bound of k ˙ γ ( s ; x, t ′ ) k depends only on t and δ . We have shown that for t ≥ δ ,the following holds u ( γ ( t ; x, t ′ ) , t ) − u ( x, t ) ≤ J d ( γ ( t ; x, t ′ ) , x ) ≤ J Z t ′ t k ˙ γ ( s ; x, t ′ ) k ds ≤ J ( t ′ − t ) . Thus, u ( x, t ′ ) − u ( x, t ) ≤ J ( t ′ − t ), where J depends only on t and δ . The condition t ′ < t is similar. We conclude the locally Lipschitz continuity of u ( x, · ) on (0 , T ]. (cid:3) Lemma 3.3.
For given
T > and ϕ ∈ C ( M ) , if v ( x, t ) is a Lipschitz continuousfunction on M × [0 , T ] , then the value function defined by (3.1) is a continuous viscositysolution of ( ∂ t u ( x, t ) + H ( x, v ( x, t ) , ∂ x u ( x, t )) = 0 ,u ( x,
0) = ϕ ( x ) . (3.2) on M × [0 , T ] . The proof of Lemma 3.3 is given in the appendix.
Lemma 3.4.
For given
T > and ϕ ∈ Lip ( M ) , if v ( x, t ) is a Lipschitz continuousfunction on M × [0 , T ] , then the value function u ( x, t ) defined by (3.1) is Lipschitzcontinuous on M × [0 , T ] , and the Lipschitz constant depends only on k v ( x, t ) k ∞ and k ∂ x ϕ ( x ) k ∞ .Proof. Define M v := sup {| H ( x, u, p ) | : x ∈ M, | u | ≤ k v ( x, t ) k ∞ , k p k ≤ k ∂ x ϕ ( x ) k ∞ } , hen the Lipschitz function w ( x, t ) = ϕ ( x ) − M v t satisfies ∂ t w + H ( x, v ( x, t ) , ∂ x w ) ≤ h >
0, define a continuous subsolution¯ w ( x, t ) = ( ϕ ( x ) − M v t, t ≤ h.u ( x, t − h ) − M v h, t > h. By [3, Theorem 5.1], since ϕ ( x ) − M v t is Lipschitz in x , we have the comparison result¯ w ( x, h ) = ϕ ( x ) − M v h ≤ u ( x, h ). By Lemma 3.2, since for any given h > u ( x, t ) isLipschitz on M × [ h, T ], we have the comparison result u ( x, t ) − M v h = ¯ w ( x, t + h ) ≤ u ( x, t + h ) , ∀ t ≥ , h > . Therefore ∂ t u ( x, t ) ≥ − M v . Plugging into (3.2), one obtain H ( x, v ( x, t ) , ∂ x u ( x, t )) ≤ M v , so H ( x, , ∂ x u ( x, t )) ≤ M v + λ k v ( x, t ) k ∞ . Thus k ∂ x u ( x, t ) k ∞ is bounded on M × [0 , T ]. Plugging into (3.2) again, one obtain that k ∂ t u ( x, t ) k ∞ is bounded on M × [0 , T ]. We finally prove that u ( x, t ) is Lipschitz on M × [0 , T ], and the Lipschitz constant depends only on k v ( x, t ) k ∞ and k ∂ x ϕ ( x ) k ∞ . (cid:3) We are now going to prove Main Result 1 under the assumption ϕ ∈ Lip ( M ). Let u = ϕ ∈ Lip ( M ). For k ∈ N + , we define the following iteration procedure u k ( x, t ) = inf γ ( t )= x (cid:26) ϕ ( γ (0)) + Z t L ( γ ( τ ) , u k − ( γ ( τ ) , τ ) , ˙ γ ( τ ))d τ (cid:27) . (3.3)By Lemma 3.4, u ( x, t ) is Lipschitz on M × [0 , T ]. Using Lemma 3.4, one can obtainthat u k ( x, t ) is the unique Lipshcitz continuous viscosity solution of ( ∂ t u ( x, t ) + H ( x, u k − ( x, t ) , ∂ x u ( x, t )) = 0 ,u ( x,
0) = ϕ ( x ) . (3.4)on M × [0 , T ] by induction.Given T >
0, there exists a constant k large enough such that ( λT ) k / ( k )! < α := ( λT ) k / ( k )!, according to [27, Lemma 4.1] and (LIP), we have k u ( k +1) k − u kk k ∞ ≤ α k u kk − u ( k − k k ∞ ≤ · · · ≤ α k k u k − ϕ k ∞ , ∀ k ∈ N . Thus for k > k , we have k u k k ( x, t ) − u k k ( x, t ) k ∞ ≤ k − X k = k α k k u k ( x, t ) − ϕ ( x ) k ∞ ≤ α k − α k − α k u k ( x, t ) − ϕ ( x ) k ∞ . Therefore, the sequence { u kk ( x, t ) } k ∈ N is a Cauchy sequence in ( C ( M × [0 , T ]) , k · k ∞ )and uniformly converges to a continuous function u ( x, t ). This function is the uniquefixed point of the operator A k ϕ , where we define the operator A ϕ [ u ]( x, t ) = inf γ ( t )= x (cid:26) ϕ ( γ (0)) + Z t L ( γ ( τ ) , u ( γ ( τ ) , τ ) , ˙ γ ( τ ))d τ (cid:27) . Since A k +1 ϕ [ u ]( x, t ) = A ϕ [ u ]( x, t ), one can figure out that A ϕ [ u ]( x, t ) is also a fixedpoint of A k ϕ . Therefore u ( x, t ) = A ϕ [ u ]( x, t ), i.e. it is the unique fixed point of A ϕ ,equivalently it satisfies (T-). We have shown that u k ( x, t ) has a subsequence uniformlyconverges to the semigroup (T-). We still denote this subsequence by u k ( x, t ). t remains to show that the Lipschitz constant of u k ( x, t ) is uniform with respectto k on M × [0 , T ]. Thus the uniform limit u ( x, t ) is Lipschitz continuous. Because { u k ( x, t ) } k ∈ N uniformly converges, we can define a constant l := sup k ∈ N k u k ( x, t ) k ∞ .Define M l := sup {| H ( x, u, p ) | : x ∈ M, | u | ≤ l, k p k ≤ k ∂ x ϕ ( x ) k ∞ } , then the Lipschitz function w ( x, t ) = ϕ ( x ) − M l t is a subsolution of (3.4), for all k ∈ N + . Similar to the proof in Lemma 3.4, one can prove that ∂ t u k ( x, t ) ≥ − M l , so H ( x, u k − ( x, t ) , ∂ x u k ( x, t )) ≤ M l , which implies H ( x, , ∂ x u k ( x, t )) ≤ M l + λl. Therefore k ∂ x u k ( x, t ) k ∞ has a bound independent of k . We finally prove that u ( x, t ) isLipschitz continuous. According to the Comparison Theorem, since H ( x, u k ( x, t ) , p ) := H k ( t, x, p ) converges uniformly on compact subsets of R × T ∗ M , and u k ( x, t ) convergesuniformly on compact subsets of M × (0 , + ∞ ), the backward semigroup u ( x, t ) = T − t ϕ ( x ) is the unique Lipschitz viscosity solution of ( CP H ) by the stability of viscositysolutions. Remark 3.5. If u ( x, t ) has a bound independent of t , using the discussion above, forany t > , there exists a constant K ( t ) > such that for k ≥ K ( t ) , we have k u k ( x, t ) − u ( x, t ) k ∞ ≤ . Define l := k u ( x, t ) k ∞ + 1 and M l := sup {| H ( x, u, p ) | : x ∈ M, | u | ≤ k u ( x, t ) k ∞ + 1 , k p k ≤ k ∂ x ϕ ( x ) k ∞ } . Similar to the proof in Lemma 3.4, one can obtain that H ( x, , ∂ x u k ) ≤ M l + λ ( k u ( x, t ) k ∞ + 1) for all k ≥ K ( t ) . Thus, the uniform limit u ( x, t ) admits a Lipschitz constant in x , whichis independent of t . Similarly, if T + t ϕ ( x ) has a bound independent of t , then it has aLipschitz constant in x independent of t . Coercive condition.
In this section, we assume the Hamiltonian H : T ∗ M × R → R satisfies (C)(CON)(CER)(LIP). Since the Lagrangian is not locally bounded, wecan not use the results in [5] to obtain the Lispchitz regularity of the minimizers of T − t ϕ ( x ). In fact, for arbitrary two points, the minimizers between them may not exist.A counterexample can be given via the Weierstrass case (see [6, Section 3.2]). We makea modification as in [14, Appendix A] H n ( x, u, p ) := H ( x, u, p ) + max {k p k − n , } , n ∈ N . Obviously H n is superlinear in p . The sequence H n is decreasing, and converges uni-formly to H on compact subsets of T ∗ M × R . The sequence of the correspondingLagrangians L n is increasing, and converges to L pointwisely. Denote by u n,k ( x, t ) theviscosity solution of (3.4) with H replaced by H n . Lemma 3.6.
For given x ∈ M , T > and ϕ ∈ C ( M, R ) , take any v ∈ C ( M × [0 , T ] , R ) and fix t ∈ [0 , T ] , the functional L t ( γ ) := ϕ ( γ (0)) + Z t L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s )) ds reaches its infimum in the class of curves X t ( x ) = { γ ∈ W , ([0 , t ] , M ) : γ ( t ) = x } . roof. The proof is similar to [14, Proposition A.6]. Define L tn ( γ ) = ϕ ( γ (0)) + Z t L n ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s )) ds. It is well-known that the above functionals admit minimizers in the class of absolutelycontinuous curves with γ ( t ) = x . To prove the existence of the minimizers of L t ( γ ), wedefine Θ( r ) := inf x ∈ M (cid:18) inf k ˙ x k≥ r L ( x, , ˙ x ) (cid:19) , ∀ r ≥ . Because L is obtained by modifying outside k p k >
1, the function Θ( r ) is superlinear,and Θ( k ˙ x k ) ≤ L n ( x, , ˙ x ) ≤ L n ( x, u, ˙ x ) + λ | u |≤ L ( x, u, ˙ x ) + λ | u | , ∀ n ∈ N , ∀ ( x, u, ˙ x ) ∈ T M × R . For any sequence γ n on X t ( x ) with lim n L t ( γ n ) < + ∞ , we have sup n R t Θ( k ˙ γ n k ) ds < + ∞ , so γ n admits a weakly sequentially converging subsequence. By Lemma 2.1, thefunctionals L t and L tn are sequentially weakly lower semicontinuous on X t ( x ). Since X t ( x ) is a metric space, the functionals L t and L tn are also lower semicontinuous. Since L tn is an increasing sequence, converges pointwisely to L t on X t ( x ), and both L t and L tn ( γ ) are lower semicontinuous, by Lemma 2.5, we conclude Γ − lim n → + ∞ L tn = L t on X t ( x ).If the minimizers γ n of L tn are contained in a compact subset of X t ( x ), then by Lemma2.7 one can obtain that L t admits a minimum on X t ( x ). It remains to show there existsa compact set in X t ( x ) such that all minimizers γ n are contained in this set. Considerthe set K t ( x ) := (cid:26) γ ∈ X t ( x ) : Z t Θ( k ˙ γ k ) ds ≤ k φ k ∞ + K t + 2 λKt (cid:27) , where K := sup x ∈ M L ( x, , K := k v ( x, t ) k ∞ . The set K t ( x ) is weakly sequentiallycompact in W , ([0 , t ] , M ). According to [6, Theorem 2.13], K t ( x ) is compact in X t ( x ).For constant curve γ x ≡ x , we have Z t Θ( k ˙ γ x k ) ds ≤ L tn ( γ x ) + λKt ≤ L t ( γ x ) + λKt ≤ k φ k ∞ + K t + 2 λKt, therefore γ x is contained in K t ( x ). Similarly, for minimizers γ n , we have Z t Θ( k ˙ γ n k ) ds ≤ L tn ( γ n ) + λKt ≤ L tn ( γ x ) + λKt ≤ L t ( γ x ) + λKt ≤ k φ k ∞ + K t + 2 λKt, thus γ n are all contained in K t ( x ). (cid:3) Lemma 3.7.
Given ϕ ∈ Lip ( M ) , for each k ∈ N + , the function u k ( x, t ) defined by(3.3) is the unique Lipschitz continuous viscosity solution of (3.4).Proof. We first prove by induction that u n,k ( x, t ) converges uniformly to u k ( x, t ) on M × [0 , T ]. When k = 1, by [7, Theorem 4.10], for given T >
0, the sequence u n, ( x, t )converges uniformly to u ( x, t ) on M × [0 , T ]. Assume u n,k ( x, t ) converges uniformly to u k ( x, t ) on M × [0 , T ]. Then u k ( x, t ) is continuous, and l k := sup n ∈ N + k u n,k ( x, t ) k ∞ is nite. Plugging the continuous function u k ( x, t ) into the definition formula of u k +1 ( x, t )and by Lemma 3.6, the minimizers of u k +1 ( x, t ) = ϕ ( γ (0)) + Z t L ( γ ( τ ) , u k ( γ ( τ ) , τ ) , ˙ γ ( τ ))d τ exist in the class of absolutely continuous curves. Define M k +1 := sup {| H ( x, u, p ) | : x ∈ M, | u | ≤ l k , k p k ≤ k ∂ x ϕ ( x ) k ∞ } . By definition, for n ≥ k ∂ x ϕ ( x ) k , we have M k +1 = sup {| H ( x, u, p ) | : x ∈ M, | u | ≤ l k , k p k ≤ k ∂ x ϕ ( x ) k ∞ } = sup {| H n ( x, u, p ) | : x ∈ M, | u | ≤ l k , k p k ≤ k ∂ x ϕ ( x ) k ∞ } . Similar to the proof in Lemma 3.4, one can prove that H ( x, , ∂ x u n,k +1 ( x, t )) ≤ H n ( x, , ∂ x u n,k +1 ( x, t )) ≤ M k +1 + λl k , ∀ n ≥ k ∂ x ϕ ( x ) k . Therefore, the Lispchitz constant of u n,k +1 ( x, t ) is uniform with respect to n . Note that u n,k +1 ( x,
0) = ϕ ( x ), the sequence u n,k +1 ( x, t ) is uniformly bounded, so it has a converg-ing subsequence. According to (2.1), u n,k +1 ( x, t ) converges to u k +1 ( x, t ) pointwisely, sothe uniform limit of u n,k +1 ( x, t ) is u k +1 ( x, t ). According to the uniform convergence of u n,k ( x, t ) with respect to n , we obtain the Lipschitz continuity of each u k ( x, t ). Notethat the Lipschitz constant may depend on k . Since that H n uniformly converges to H on compact subsets of T ∗ M × R , and u n,k ( x, t ) uniformly converges to u k ( x, t ) on M × [0 , T ], by the stability of the viscosity solutions, we conclude that u k ( x, t ) is theunique Lipschitz continuous viscosity solution of (3.4). (cid:3) Using a similar argument in Section 3.1, one can prove that u k ( x, t ) has a subse-quence converging to u ( x, t ). We still denote this subsequence by u k ( x, t ). The Lips-chitz constants of the sequence u k ( x, t ) are uniform with respect to k . Therefore thelimit function u ( x, t ) = T − t ϕ ( x ) of the sequence { u k ( x, t ) } k ∈ N + is the unique Lipschitzcontinuous viscosity solution of ( CP H ). The Main Result 1 has been proved when ϕ isLipschitz continuous.Now we begin to consider the initial function ϕ to be continuous. We first prove thatthe functions u k ( x, t ) defined in (3.3) is continuous. Lemma 3.8.
Given
T > and ϕ ∈ C ( M ) , The value function u k ( x, t ) defined in (3.3)is a continuous function on M × [0 , T ] .Proof. For any ϕ ∈ C ( M ), there exists a sequence of Lipschitz functions ϕ m uniformlyconverging to ϕ . We have already proven in Lemma 3.7 that, for initial functions ϕ m ,the solutions of (3.4), which are denoted by u mk ( x, t ), are Lipschitz continuous. We thenprove by induction. It is obvious that u m uniformly converges to u . Assume u mk − uniformly converges to u k − , then u k − is continuous. By Lemma 3.6, u k ( x, t ) admitsa minimizer γ . By definition u mk ( x, t ) − u k ( x, t ) ≤ ϕ m ( γ (0)) − ϕ ( γ (0)) + λ k u mk − ( x, t ) − u k − ( x, t ) k ∞ T. Exchanging the role of u mk ( x, t ) and u k ( x, t ), we obtain | u mk ( x, t ) − u k ( x, t ) | → (cid:3) By a similar argument in Section 3.1, one can obtain that u k ( x, t ) has a subsequenceconverges to u ( x, t ), and the limit function satisfies (T-). We have proven that for ϕ ∈ Lip ( M ), T − t ϕ ( x ) is the unique Lipschitz continuous viscosity solution of ( CP H ).Using Proposition 1.3 (2), for t ∈ [0 , T ], T − t ϕ m uniformly converges to T − t ϕ . According o the stability of viscosity solutions, we conclude that T − t ϕ is the unique continuousviscosity solution of ( CP H ) under the initial condition u ( x,
0) = ϕ ( x ).4. Existence of stationary solutions
Lemma 4.1.
For any given ϕ ∈ C ( M ) and σ > , fixing t > , we have inf s ≥ σ T − t + s ϕ ( x ) = T − t ( inf s ≥ σ T − s ϕ ( x )) , sup s ≥ σ T − t + s ϕ ( x ) = T − t (sup s ≥ σ T − s ϕ ( x )) . By a similar argument, one can show that the forward semigroup T + t has the sameproperty.Proof. We first prove that T − t commutes with inf. On one hand, by definition we haveinf s ≥ σ T − s ϕ ( x ) ≤ T − s ϕ ( x ) , ∀ x ∈ M. By Proposition 1.3 (1) we get T − t ( inf s ≥ σ T − s ϕ ( x )) ≤ T − t ◦ T − s ϕ ( x ) , ∀ x ∈ M. Therefore T − t ( inf s ≥ σ T − s ϕ ( x )) ≤ inf s ≥ σ T − t + s ϕ ( x ) , ∀ x ∈ M. On the other hand, for any given x ∈ M and any ε >
0, there exists s ≥ σ such that T − s ϕ ( x ) ≤ inf s ≥ σ T − s ϕ ( x ) + ε. By Proposition 1.3 (1) we get T − t ( T − s ϕ ( x )) ≤ T − t ( inf s ≥ σ T − s ϕ ( x ) + ε ) . By Proposition 1.3 (2) we getinf s ≥ σ T − t + s ϕ ( x ) ≤ T − t + s ϕ ( x )) ≤ T − t ( inf s ≥ σ T − s ϕ ( x )) + εe λt . Let ε → + , we concludeinf s ≥ σ T − t + s ϕ ( x ) ≤ T − t ( inf s ≥ σ T − s ϕ ( x )) , ∀ x ∈ M. We then prove that T − t commutes with sup. On one hand, by definition we have T − s ϕ ( x ) ≤ sup s ≥ σ T − s ϕ ( x ) , ∀ x ∈ M. By Proposition 1.3 (1) we get T − t ◦ T − s ϕ ( x ) ≤ T − t (sup s ≥ σ T − s ϕ ( x )) , ∀ x ∈ M. Therefore sup s ≥ σ T − t + s ϕ ( x ) ≤ T − t (sup s ≥ σ T − s ϕ ( x )) , ∀ x ∈ M. On the other hand, for any given x ∈ M and any ε >
0, there exists s ≥ σ such that T − s ϕ ( x ) ≥ sup s ≥ σ T − s ϕ ( x ) − ε. By Proposition 1.3 (1) we get T − t ( T − s ϕ ( x )) ≥ T − t (sup s ≥ σ T − s ϕ ( x ) − ε ) . y Proposition 1.3 (2) we getsup s ≥ σ T − t + s ϕ ( x ) ≥ T − t + s ϕ ( x )) ≥ T − t (sup s ≥ σ T − s ϕ ( x )) − εe λt . Let ε → + we concludesup s ≥ σ T − t + s ϕ ( x ) ≥ T − t (sup s ≥ σ T − s ϕ ( x )) , ∀ x ∈ M. The proof is now completed. (cid:3)
Lemma 4.2.
Given ϕ ∈ Lip ( M ) , if T − t ϕ ( x ) has a bound independent of t , then both ˇ ϕ = lim inf t → + ∞ T − t ϕ ( x ) , ˆ ϕ ( x ) = lim sup t → + ∞ T − t ϕ ( x ) exist, and are both backward weak KAM solutions. By a similar argument, one canprove that if T + t ϕ ( x ) has a bound independent of t , then both ˇ ϕ = lim inf t → + ∞ T + t ϕ ( x ) , ˆ ϕ ( x ) = lim sup t → + ∞ T + t ϕ ( x ) exist, and are both forward weak KAM solutions.Proof. According to Remark 3.5, the function T − t ϕ ( x ) has a Lipschitz constant in x independent of t . We denote it by κ . Therefore ˇ ϕ ( x ) and ˆ ϕ ( x ) are well-defined. Wethen prove that ˇ ϕ ( x ) is a fixed point of T − t . Similarly, one can prove that ˆ ϕ ( x ) is alsoa fixed point of T − t . Note that | inf s ≥ t T − s ϕ ( x ) − inf s ≥ t T − s ϕ ( y ) | ≤ sup s ≥ t | T − s ϕ ( x ) − T − s ϕ ( y ) | ≤ κd ( x, y ) , so the limit procedure ˇ ϕ ( x ) := lim t → + ∞ inf s ≥ t T − s ϕ ( x )is uniform in x .For given t >
0, combing Proposition 1.3 (2) with Lemma 4.1, we have k inf s ≥ σ T − t + s ϕ ( x ) − T − t ˇ ϕ ( x ) k ∞ ≤ e λt k inf s ≥ σ T − s ϕ ( x ) − ˇ ϕ ( x ) k ∞ for all σ >
0. Let σ → + ∞ , the right hand side tends to zero, and inf s ≥ σ T − t + s ϕ tendsto ˇ ϕ , therefore T − t ˇ ϕ = ˇ ϕ . (cid:3) Lemma 4.3.
Given ϕ ∈ C ( M ) , then (1) If T − t ϕ ( x ) does not have a upper bound as t tends to infinity, then for any c ∈ R ,there exists t c > such that T − t c ϕ ( x ) > ϕ ( x ) + c for all x ∈ M ; (2) If T − t ϕ ( x ) does not have a lower bound as t tends to infinity, then for any c ∈ R ,there exists t c > such that T − t c ϕ ( x ) < ϕ ( x ) + c for all x ∈ M .Proof. (1) We argue by contradiction. Assume that there exists c ∈ R such that forany t >
0, we have a point x t ∈ M satisfying T − t c ϕ ( x t ) ≤ ϕ ( x t ) + c . Let α : [0 , → M be a geodesic connecting x t and x with constant speed, then k ˙ α k ≤ diam( M ). If T − t +1 ϕ ( x ) > ϕ ( x t ) + c , since T − t ϕ ( x t ) ≤ ϕ ( x t ) + c , there exists σ ∈ [0 ,
1) such that T − t + σ ϕ ( α ( σ )) = ϕ ( x t ) + c and T − t + s ϕ ( α ( s )) > ϕ ( x t ) + c for all s ∈ ( σ, T − t + s ϕ ( α ( s )) ≤ T − t + σ ϕ ( α ( σ )) + Z sσ L ( α ( τ ) , T − t + τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ = ϕ ( x t ) + c + Z sσ L ( α ( τ ) , T − t + τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ, hich implies T − t + s ϕ ( α ( s )) − ( ϕ ( x t ) + c ) ≤ Z sσ L ( α ( τ ) , T − t + τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ ≤ Z sσ L ( α ( τ ) , ϕ ( x t ) + c , ˙ α ( τ )) dτ + λ Z sσ ( T − t + τ ϕ ( α ( τ )) − ( ϕ ( x t ) + c )) dτ ≤ L + λ Z sσ ( T − t + τ ϕ ( α ( τ )) − ( ϕ ( x t ) + c )) dτ, where L := sup x ∈ M, | u |≤k ϕ k ∞ + c , k ˙ x k≤ diam( M ) | L ( x, u, ˙ x ) | . By the Gronwall inequality, we have T − t + s ϕ ( α ( s )) − ( ϕ ( x t ) + c ) ≤ L e λ ( s − σ ) ≤ L e λ , ∀ s ∈ ( σ, . Take s = 1 we have T − t +1 ϕ ( x ) ≤ ϕ ( x t ) + c + L e λ . We conclude that T − t +1 ϕ ( x ) has aupper bound independent of t , which contradicts the assumption.(2) We argue by contradiction. Assume that there exists c ∈ R such that for any s >
0, we have a point x s ∈ M satisfying T − s ϕ ( x s ) ≥ ϕ ( x s ) + c . Take s = t − T − t +1 ϕ ( x t +1 ) ≥ ϕ ( x t +1 ) + c . Let α : [0 , → M be a geodesic connecting x and x t +1 with constant speed, then k ˙ α k ≤ diam( M ). If T − t ϕ ( x ) < ϕ ( x t +1 ) + c , since T − t +1 ϕ ( x t +1 ) ≥ ϕ ( x t +1 ) + c , there exists σ ∈ (0 ,
1] such that T − t + σ ϕ ( α ( σ )) = ϕ ( x t +1 ) + c and T − t + s ϕ ( α ( s )) < ϕ ( x t +1 ) + c for all s ∈ [0 , σ ). By definition ϕ ( x t +1 ) + c = T − t + σ ϕ ( α ( σ )) ≤ T − t + s ϕ ( α ( s )) + Z σs L ( α ( τ ) , T − t + τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ, which implies ϕ ( x t +1 ) + c − T − t + s ϕ ( α ( s )) ≤ Z σs L ( α ( τ ) , T − t + τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ ≤ Z σs L ( α ( τ ) , ϕ ( x t +1 ) + c , ˙ α ( τ )) dτ + λ Z σs ( ϕ ( x t +1 ) + c − T − t + τ ϕ ( α ( τ ))) dτ ≤ L + λ Z σs ( ϕ ( x t +1 ) + c − T − t + τ ϕ ( α ( τ ))) dτ, where L := sup x ∈ M, | u |≤k ϕ k ∞ + c , k ˙ x k≤ diam( M ) | L ( x, u, ˙ x ) | . Let G ( σ − s ) = ϕ ( x t +1 ) + c − T − t + s ϕ ( α ( s )), then G ( σ − s ) ≤ L + λ Z σ − s G ( τ ) dτ. By the Gronwall inequality, we have ϕ ( x t +1 ) + c − T − t + s ϕ ( α ( s )) ≤ L e λ ( s − σ ) ≤ L e λ , ∀ s ∈ [0 , σ ) . Take s = 0 we have T − t ϕ ( x ) ≥ ϕ ( x t +1 ) + c − L e λ . We conclude that T − t ϕ ( x ) has alower bound independent of t , which contradicts the assumption. (cid:3) Lemma 4.4.
If there exist two continuous functions ϕ and ϕ on M such that T − t ϕ has a lower bound independent of t , and T − t ϕ has a upper bound independent of t , thenthere is a constant A such that T − t A ( x ) is uniformly bounded. roof. Define A := k ϕ k ∞ and A := −k ϕ k ∞ , then A ≤ A and T − t A ( x ) ≥ T − t ϕ ( x ), T − t A ( x ) ≤ T − t ϕ ( x ) for all x ∈ M . If T − t A ( x ) has a upper bound in-dependent of t , then A is the constant A we are looking for. If T − t A ( x ) does not havea upper bound independent of t , we define the following A ∗ := inf { A : ∃ t A > T − t A A ( x ) ≥ A, ∀ x ∈ M } . By Lemma 4.3 (1), we take c = 0, it is obvious that if T − t A ( x ) is unbounded formabove, then A ∗ ≤ A < + ∞ . The discussion is divided into two situations.Case (1): A ∗ > −∞ , then we can prove that A ∗ is the constant A we are looking for.We first show that T − t A ∗ ( x ) has a upper bound independent of t , then A ∗ < A . Weargue by contradiction. If T − t A ∗ ( x ) does not have a upper bound, by Lemma 4.3 (1),for c = 1, there is t > T − t A ∗ ( x ) > A ∗ + 1 for all x ∈ M . By Proposition 1.3(2), for any ε >
0, we have T − t ( A ∗ − ε )( x ) ≥ T − t A ∗ ( x ) − e λt ε > A ∗ + 1 − e λt ε . For every0 < ε < ( e λt − − , we have T − t ( A ∗ − ε )( x ) > A ∗ − ε . It means that we have found asmaller constant A ∗ − ε such that if we take t A ∗ − ε = t , then T − t A ∗− ε ( A ∗ − ε )( x ) > A ∗ − ε ,which contradicts the definition of A ∗ .We then prove that T − t A ∗ has a lower bound independent of t . We argue by con-tradiction. If T − t A ∗ ( x ) does not have a lower bound, by Lemma 4.3 (2), for c = − t > T − t A ∗ ( x ) < A ∗ − x ∈ M . By Proposition 1.3 (2) and A ∗ < A , there is a constant δ > A ∗ + δ < A and T − t ( A ∗ + δ )( x ) < A ∗ −
12 + δ < A ∗ + δ, (4.1)for all δ ∈ [0 , δ ). By the definition of A ∗ , there is ¯ A ∈ [ A ∗ , A ∗ + δ ) and t := t ¯ A > T − t ¯ A ( x ) ≥ ¯ A. (4.2)Define B ∗ := ¯ A − . According to the continuity of T − t ϕ ( x ) at t = 0 (see Lemma 3.3),there exists ε > ≤ σ < ε , we have T − σ B ∗ ( x ) ≤ ¯ A − . (4.3)For t and t >
0, there exist n and n ∈ N , and ε ∈ [0 , ε ) such that n t + ε = n t .By Proposition 1.3 (1) and (4.1), we have T − n t ¯ A ( x ) ≤ T − t ¯ A ( x ) < B ∗ . (4.4)Take σ = ε in (4.3), by Proposition 1.3 (1) and the second inequality in (4.4), we get T − ε ◦ T − n t ¯ A ( x ) ≤ T − ε B ∗ ( x ) ≤ ¯ A − . (4.5)By (4.2) one can easily figure out that T − n t ¯ A ( x ) ≥ ¯ A , thus T − ε ◦ T − n t ¯ A ( x ) = T − n t ¯ A ( x ) ≥ ¯ A, (4.6)which contradicts (4.5).Case (2): A ∗ = −∞ , then we can prove that for any A < A , the function T − t A ( x )is uniformly bounded. Since T − t A ( x ) ≤ T − t A ( x ), T − t A ∗ ( x ) has a upper bound. Theproof of the existence of the lower bound of T − t A ( x ) is similar to Case (1). (cid:3) If there is ϕ ∈ C ( M ) and t a > T − t a ϕ ≥ ϕ , for any t >
0, there is n ∈ N and r ∈ [0 , t a ) such that t = nt a + r . By Proposition 1.3 (1) we have T − t ϕ ≥ T − r ϕ , i.e. T − t ϕ has a lower bound independent of t . Similarly, if there is ψ ∈ C ( M ) and t b > uch that T − t b ψ ≤ ψ , T − t ψ has a upper bound independent of t . By Lemma 4.4, thereexists a constant A such that T − t A is uniformly bounded. By Lemma 4.2, the condition(S) holds. 5. Existence of forward weak KAM solutions
Proposition 5.1.
Given ϕ ∈ C ( M ) , if ϕ satisfies the following condition ( ∗ ’) ϕ ≤ u − and there exists a point x such that ϕ ( x ) = u − ( x ) .then T + t ϕ ( x ) has a bound independent of t . We divide the proof into three parts, that is, Lemmas 5.2, 5.5 and 5.7.
Lemma 5.2.
Suppose ϕ satisfies the condition ( ∗ ’), then T + t ϕ ( x ) ≤ u − ( x ) for all t > .Proof. We argue by contradiction. Assume there exists ( x, t ) such that T + t ϕ ( x ) > u − ( x ).Let γ : [0 , t ] → M with γ (0) = x be a minimizer of T + t ϕ ( x ). Define F ( s ) = T + t − s ϕ ( γ ( s )) − u − ( γ ( s )) , s ∈ [0 , t ] . Then F ( s ) is continuous and F ( t ) = ϕ ( γ ( t )) − u − ( γ ( t )) ≤
0. By assumption F (0) > τ ∈ (0 , t ] such that F ( τ ) = 0 and F ( τ ) > s ∈ [0 , τ ). For each τ ∈ [0 , τ ], we have T + t − τ ϕ ( γ ( τ )) = T + t − τ ϕ ( γ ( τ )) − Z τ τ L ( γ ( s ) , T + t − s ϕ ( γ ( s )) , ˙ γ ( s )) ds. Since u − = T − t u − for all t >
0, we have u − ( γ ( τ )) ≤ u − ( γ ( τ )) + Z τ τ L ( γ ( s ) , u − ( γ ( s )) , ˙ γ ( s )) ds. Thus F ( τ ) ≤ F ( τ ) + λ R τ τ F ( s ) ds , where F ( τ ) = 0. Define F ( s ) = G ( τ − s ), we get G ( τ − τ ) ≤ λ Z τ − τ G ( σ ) dσ. By the Gronwall inequality, we conclude F ( τ ) = G ( τ − τ ) ≡ τ ∈ [0 , τ ], whichcontradicts F (0) > (cid:3) Corollary 5.3.
Let u − ∈ S − , then T + t u − ≤ u − for each t > . Combining Corollary 5.3 with Proposition 1.3 (1), one can easily obtain that T + t u − = T + s ◦ T + t − s u − ≤ T + s u − for all t > s , then we have Corollary 5.4. T + t u − is decreasing in t . Lemma 5.5.
Suppose ϕ satisfies the condition ( ∗ ’), then for each t > , there exists apoint x t ∈ M such that T + t ϕ ( x t ) = u − ( x t ) .Proof. Since u − ( x ) = T − t u − ( x ), let γ : [0 , t ] → M with γ ( t ) = x be a minimizerof T − t u − ( x )). From the above discussion, for each s ∈ [0 , t ], we have u − ( γ ( s )) ≥ T + t − s ϕ ( γ ( s )). Define F ( s ) = u − ( γ ( s )) − T + t − s ϕ ( γ ( s )) , then F ( s ) ≥ F ( t ) = 0. If F (0) >
0, then there is s ∈ (0 , t ] such that F ( s ) = 0and F ( s ) > s ∈ [0 , s ). By definition, for s ∈ [0 , s ), we have u − ( γ ( s )) = u − ( γ ( s )) + Z s s L ( γ ( s ) , u − ( γ ( s )) , ˙ γ ( s )) ds, nd T + t − s ϕ ( γ ( s )) ≥ T + t − s ϕ ( γ ( s )) − Z s s L ( γ ( s ) , T + t − s ϕ ( γ ( s )) , ˙ γ ( s )) ds, which implies F ( s ) ≤ F ( s ) + λ Z s s F ( s ) ds. By the Gronwall inequality, we conclude F ( s ) = G ( s − s ) ≡ s ∈ [0 , s ], whichcontracts F (0) >
0. Therefore T + t ϕ ( γ (0)) = u − ( γ (0)). (cid:3) Corollary 5.6.
For each t > , there exists x t ∈ M such that T + t u − ( x t ) = u − ( x t ) . Lemma 5.7.
Suppose ϕ satisfies the condition ( ∗ ’), then T + t ϕ ( x ) has a lower boundindependent of t .Proof. Let α : [0 , → M be a geodesic connecting x and x t − with constant speed,then k ˙ α k ≤ diam( M ). If T + t ϕ ( x ) ≥ u − ( x t − ), then the proof is finished. If T + t ϕ ( x )
1] such that T + t − σ ϕ ( α ( σ )) = u − ( x t − ) and T + t − s ϕ ( α ( s )) < u − ( x t − ) for all s ∈ [0 , σ ). By definition T + t − s ϕ ( α ( s )) ≥ T + t − σ ϕ ( α ( σ )) − Z σs L ( α ( τ ) , T + t − τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ = u − ( x t − ) − Z σs L ( α ( τ ) , T + t − τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ, which implies u − ( x t − ) − T + t − s ϕ ( α ( s )) ≤ Z σs L ( α ( τ ) , T + t − τ ϕ ( α ( τ )) , ˙ α ( τ )) dτ ≤ Z σs L ( α ( τ ) , u − ( x t − ) , ˙ α ( τ )) dτ + λ Z σs ( u − ( x t − ) − T + t − τ ϕ ( α ( τ ))) dτ ≤ L + λ Z σs ( u − ( x t − ) − T + t − τ ϕ ( α ( τ ))) dτ, where L := sup x ∈ M, | u |≤k u − k ∞ , k ˙ x k≤ diam( M ) | L ( x, u, ˙ x ) | . Let G ( σ − s ) = u − ( x t − ) − T + t − s ϕ ( α ( s )), then G ( σ − s ) ≤ L + λ Z σ − s G ( τ ) dτ. By the Gronwall inequality, we have u − ( x t − ) − T + t − s ϕ ( α ( s )) = G ( σ − s ) ≤ L e λ ( σ − s ) ≤ L e λ , ∀ s ∈ [0 , σ ) . Thus T + t ϕ ( x ) ≥ u − ( x t − ) − L e λ . We finally get a lower bound of T + t ϕ ( x ) independentof t . (cid:3) Corollary 5.8. T + t u − has a lower bound independent of t . Lemma 5.9. If u ≺ L , then u is a Lipschitz continuous function defined on M . roof. For each x, y ∈ M , let α : [0 , d ( x, y )] → M be a geodesic of length d ( x, y ),parameterized by arclength and connecting x and y . Since M is compact, L ( x, u, ˙ x )satisfies (LB), and u is continuous, let κ := sup { L ( x, u, ˙ x ) : x ∈ M, | u | ≤ k u k ∞ , k ˙ x k = 1 } . Then by u ≺ L we have u ( y ) − u ( x ) ≤ Z d ( x,y )0 L ( α ( s ) , u ( α ( s )) , ˙ α ( s )) ds ≤ κd ( x, y ) . Exchanging the role of x and y , we get the Lipschitz continuity of u . (cid:3) Note that u − ≺ L , it is Lipschitz. By Remark 3.5, the bound and the Lipschitzconstant in x of T + t u − is independent of t . By the Arzela-Ascoli theorem, T + t u − has aconverging subsequence as t → + ∞ . Since T + t u − is decreasing in t , the limit functionlim t → + ∞ T + t u − = u + exists. By Lemma 4.2, u + is a fixed point of T + t . Thus, − u + is aviscosity solution of ( E F ). 6. Strictly increasing case
Remark 6.1.
According to [13, Theorem II.2], if the set of fixed points of T − t isnonempty, it must be a singleton. Using Main Result 3 and exchanging the role of( E H ) and ( E F ), one obtain that T − t v + converges to the unique viscosity of ( E F ) de-noted by u − , where v + is a fixed point of T + t . Therefore, u − and u + = lim t → + ∞ T + t u − conjugate to each other. Proposition 6.2. u + is the maximal element in S + .Proof. For each v + ∈ S + , we first prove that v + ≤ T − t v + for all t >
0. We argue bycontradiction. Assume there exist x ∈ M and t > v + ( x ) > T − t v + ( x ). Let γ : [0 , t ] → M with γ ( t ) = x be a minimizer of T − t v + ( x ). Let F ( s ) := v + ( γ ( s )) − T − s v + ( γ ( s )), then F (0) = 0. By assumption F ( t ) >
0. Then there exists s ∈ [0 , t ) suchthat F ( s ) = 0 and F ( s ) > s ∈ ( s , t ]. By definition, we have T − s v + ( γ ( s )) = T − s v + ( γ ( s )) + Z ss L ( γ ( s ) , T − s v + ( γ ( s )) , ˙ γ ( s )) ds, and v + ( γ ( s )) ≥ v + ( γ ( s )) − Z ss L ( γ ( s ) , v + ( γ ( s )) , ˙ γ ( s )) ds. Thus F ( s ) ≤ F ( s ) + λ Z ss F ( σ ) dσ, we get F ( s ) ≡ s ∈ [ s , t ], which contradicts F ( t ) > v + ≤ T − t v + , by a limit procedure we get v + ≤ u − , applying T + t we get v + ≤ T + t u − , by a limit procedure again we finally conclude that v + ≤ u + for all v + ∈ S + . (cid:3) Proposition 6.3.
For each v + ∈ S + , the corresponding projected Ma˜n´e set I v + isnonempty and I v + ⊆ I ( u − ,u + ) .Proof. For given v + ∈ S + , we define the barrier function B v + ( x ) := u − ( x ) − v + ( x ) . ccording to the discussion above, we have B v + ( x ) ≥
0. We first show B v + ( γ + ( t )) isnonnegative nonincreasing along a ( v + , L, γ + . By definition v + ( γ + ( t ′ )) − v + ( γ + ( t )) = Z t ′ t L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) ds, and u − ( γ + ( t ′ )) − u − ( γ + ( t )) ≤ Z t ′ t L ( γ + ( s ) , u − ( γ + ( s )) , ˙ γ + ( s )) ds. Since L satisfies (STD) and u − ≥ v + , we have u − ( γ + ( t ′ )) − u − ( γ + ( t )) ≤ v + ( γ + ( t ′ )) − v + ( γ + ( t )) , which implies B v + ( γ + ( t ′ )) ≤ B v + ( γ + ( t )) , for t ′ > t ≥ . Therefore the limit lim t → + ∞ B v + ( γ + ( t )) = δ ≥ δ = 0, whichimplies that I v + is nonempty. If not, assume δ > v + ∈ S + , since L ( x, u, ˙ x ) satisfies (LB), by Proposition 7.1, it is Lipschitzcontinuous. Take R ≥ max {k ∂ x u − k ∞ , k ∂ x v + k ∞ } , we can apply a modification on H by H R ( x, u, p ) := H ( x, u, p ) + max {k p k − R , } , then u − and v + are also the viscosity solutions of H R ( x, u ( x ) , ∂ x u ( x )) = 0 , F R ( x, u ( x ) , ∂ x u ( x )) = 0respectively. It is obvious that H R satisfies (SL). Therefore, without any loss of gener-ality, we can assume that H ( x, u, p ) satisfies (SL).By definition of the calibrated curves, for any s > s ≥
0, we have v + ( γ + ( s )) − v + ( γ + ( s )) = Z s s L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) ds. (6.1)By Lemma 5.9, the forward weak KAM solution v + is Lipschitz continuous, we have v + ( γ + ( s )) − v + ( γ + ( s )) γ + ( s ) − γ + ( s ) · γ + ( s ) − γ + ( s ) s − s ≤ k ∂ x v + k ∞ γ + ( s ) − γ + ( s ) s − s . (6.2)By continuity, there is T > s − s ≤ T we have L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) ≥ L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) − , s ∈ [ s , s ] . The Jensen inequality implies1 s − s Z s s L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) ds ≥ s − s Z s s L ( γ + ( s ) , v + ( γ + ( s )) , ˙ γ + ( s )) ds − ≥ L (cid:18) γ + ( s ) , v + ( γ + ( s )) , s − s Z s s ˙ γ + ( s ) ds (cid:19) − L (cid:18) γ + ( s ) , v + ( γ + ( s )) , γ + ( s ) − γ + ( s ) s − s (cid:19) − ≥ Θ (cid:18) γ + ( s ) − γ + ( s ) s − s (cid:19) − . (6.3)From (6.1)(6.2)(6.3), we conclude that γ + ( t ) has a Lipschitz constant independent of t .Since M is compact, we can take a converging subsequence γ + ( t n ) → ¯ x with t n → + ∞ .For any T >
0, the sequence of curves γ n ( s ) := γ + ( t n + s ) is uniformly bounded and qui-Lipschitz on s ∈ [0 , T ]. Therefore it admits a uniformly converging subsequence,which will be still denoted by γ n . We denote the uniform limit by ¯ γ .We then prove that ¯ γ is a ( v + , L, , T ]. By definition2 k v + k ∞ ≥ v + ( γ n ( s )) − v + ( γ n (0))= Z s L ( γ n ( τ ) , v + ( γ n ( τ )) , ˙ γ n ( τ )) dτ ≥ Z s Θ( k ˙ γ n ( τ ) k ) dτ, ∀ s ∈ [0 , T ] . Therefore the sequence γ n is weakly compact in W , ([0 , T ] , M ). Since γ n is a uniformlyconverging sequence v + (¯ γ ( s )) − v + (¯ x ) = lim n → + ∞ ( v + ( γ n ( s )) − v + ( γ n (0)))= lim n → + ∞ Z s L ( γ n ( τ ) , v + ( γ n ( s )) , ˙ γ n ( τ )) dτ ≥ Z s L (¯ γ ( τ ) , v + (¯ γ ( τ )) , ˙¯ γ ( τ )) dτ, where the last inequality comes from Lemma 2.1. Therefore ¯ γ is a ( v + , L, s ∈ [0 , T ], by definition we have B v + (¯ γ ( s )) = lim n → + ∞ B v + ( γ n ( s )) = lim n → + ∞ B v + ( γ + ( t n + s )) ≡ δ. Since the Lagrangian satisfies (STD) and by the assumption δ > v + (¯ γ ( T )) − v + (¯ γ (0)) = Z T L (¯ γ ( s ) , v + (¯ γ ( s )) , ˙¯ γ ( s )) ds> Z T L (¯ γ ( s ) , u − (¯ γ ( s )) , ˙¯ γ ( s )) ds ≥ u − (¯ γ ( T )) − u − (¯ γ (0)) . We conclude B v + (¯ γ (0)) > B v + (¯ γ ( T )), which gives a contradiction.We then prove I v + ⊆ I ( u − ,u + ) . It is obvious that for x ∈ I v + , we have u − ( x ) = v + ( x ) ≤ u + ( x ) ≤ u − ( x ), then u − ( x ) = v + ( x ) = u + ( x ), that is, x ∈ I ( u − ,u + ) . (cid:3) Strictly decreasing case
Proposition 7.1.
The set S + is compact in the topology induced by the Sobolev W , ∞ -norm.Proof. Since v + ≤ u − , we only need to show that v + ∈ S + has a uniform lower bound.Take y ∈ I v + , then v + ( y ) = u − ( y ). Let α : [0 , → M be a geodesic connecting x and y with constant speed, then k ˙ α k ≤ diam( M ). If v + ( x ) ≥ u − ( y ), then the proof isfinished. If T +1 v + ( x ) = v + ( x ) < u − ( y ), since v + ( y ) = u − ( y ), there is σ ∈ (0 ,
1] suchthat v + ( α ( σ )) = u − ( y ) and v + ( α ( s )) < u − ( y ) for all s ∈ [0 , σ ). By definition we have v + ( α ( s )) ≥ v + ( α ( σ )) − Z σs L ( α ( τ ) , v + ( α ( τ )) , ˙ α ( τ )) dτ = u − ( y ) − Z σs L ( α ( τ ) , v + ( α ( τ )) , ˙ α ( τ )) dτ, hich implies u − ( y ) − v + ( α ( s )) ≤ Z σs L ( α ( τ ) , v + ( α ( τ )) , ˙ α ( τ )) dτ ≤ Z σs L ( α ( τ ) , u − ( y ) , ˙ α ( τ )) dτ + λ Z σs ( u − ( y ) − v + ( α ( τ ))) dτ ≤ L + λ Z σs ( u − ( y ) − v + ( α ( τ ))) dτ, where L := sup x ∈ M, | u |≤k u − k ∞ , k ˙ x k≤ diam( M ) | L ( x, u, ˙ x ) | . Let G ( σ − s ) = u − ( y ) − v + ( α ( s )), then G ( σ − s ) ≤ L + λ Z σ − s G ( τ ) dτ. By the Gronwall inequality we get u − ( y ) − v + ( α ( s )) = G ( σ − s ) ≤ L e λ ( σ − s ) ≤ L e λ , ∀ s ∈ [0 , σ ) . Therefore v + ( x ) ≥ u − ( y ) − L e λ , i.e. v + is uniformly bounded from below. Then thereexists a constant K > k v + k ∞ ≤ K .We then prove that v + are equi-Lipschitz continuous. For each x, y ∈ M , let α : [0 , d ( x, y )] → M be a geodesic of length d ( x, y ) , parameterized by arclength andconnecting x and y . Since M is compact and u is continuous, let κ := sup { L ( x, u, ˙ x ) : x ∈ M, | u | ≤ K, k ˙ x k = 1 } . Since v + ≺ L , we have v + ( y ) − v + ( x ) ≤ Z d ( x,y )0 L ( α ( s ) , v + ( α ( s )) , ˙ α ( s )) ds ≤ κd ( x, y ) . Exchanging the role of x and y , we finally prove the uniform boundedness of k v + k W , ∞ . (cid:3) Proposition 7.2.
Let A be a totally ordered subset in S + . Since v + ∈ S + is boundedform below, the function ¯ u := inf u ∈ A u ( x ) is well-defined for each x ∈ M . Then ¯ u ∈ S + .Proof. If A is a finite set, the proof is finished. We then consider A being a infinite set.By Proposition 7.1, the pointwise convergence of a sequence u n in S + implies the uniformconvergence. We first proof the limit function is contained in S + . By Proposition 1.3(2), we have k T + t u n − T + t ¯ u k ∞ ≤ e λt k u n − ¯ u k ∞ , the right hand side tends to zero uniformly, so T + t ¯ u = lim n → + ∞ T + t u n = lim n → + ∞ u n = ¯ u. We then proof the existence of a pointwisely converging sequence in A with the limitfunction ¯ u . Then by the discussion above we have ¯ u ∈ S + . According to Proposition7.1, every u ∈ S + are κ -Lipschitz, we have¯ u ( x ) − ¯ u ( y ) ≤ sup u ∈S + | u ( x ) − u ( y ) | ≤ κ | x − y | . (7.1)Since M is compact, it is also separable. Namely on can find a countable dense subsetdenoted by U := { x , x , . . . , x n , . . . } . ssertion. There exists a sequence { u n } n ∈ N ⊂ A such that for a given n ∈ N and each i ∈ { , , . . . , n } , we have 0 ≤ u n ( x i ) − ¯ u ( x i ) < n . (7.2)Using the assertion above, one can prove that u n has a pointwise convergence to ¯ u . INfact, for each x ∈ M , there exists a subsequence V := { x m } m ∈ N ⊂ U such that | x m − x | < m . Given x ∈ M and n ∈ N , for each x i ∈ { x , x , . . . , x n } ∩ V , up to arearrangement of i , using (7.1) we have | u n ( x ) − ¯ u ( x ) | ≤ | u n ( x ) − u n ( x i ) | + | u n ( x i ) − ¯ u ( x i ) | + | ¯ u ( x ) − ¯ u ( x i ) |≤ κ | x i − x | + 1 n ≤ κi + 1 n . Let n and i tend to infinity, we get the pointwise convergence of u n to ¯ u .It remains to prove the assertion above. By the definition of ¯ u , we have u n ≥ ¯ u .In the following, we construct a sequence { u n } n ∈ N ⊂ A such that for a given n ∈ N and each i ∈ { , , . . . , n } , (7.2) holds. First of all, for x ∈ U , we take v ∈ A suchthat v ( x ) − ¯ u ( x ) < /n . Let x j ∈ U with j ≤ n satisfying v ( x j ) − ¯ u ( x j ) ≥ /n and v ( x i ) − ¯ u ( x i ) < /n for each i ≤ j −
1. For x j , we take v ∈ A satisfying v ( x j ) − ¯ u ( x j ) < /n . Then v ( x j ) < ¯ u ( x j ) + 1 n ≤ v ( x j ) . Note that A is totally ordered, it yields v ≤ v . Therefore, for each i ≤ j − v ( x i ) − ¯ u ( x i ) ≤ v ( x i ) − ¯ u ( x i ) < n . Thus, we have found v ∈ A satisfying that for all i ∈ { , , . . . , j } , we have v ( x i ) − ¯ u ( x i ) < /n . Repeating the process above, we finally obtain v k ∈ A with k ≤ n satisfying for each i ∈ { , , . . . , n } , the following property holds v k ( x i ) − ¯ u ( x i ) < n . Take u n = v k , then the proof is completed. (cid:3) Proposition 7.3.
Let v + and v ′ + be two forward weak KAM solutions of ( E H ), thefollowing holds: (1) If v + ≤ v ′ + , then ∅ 6 = I v + ⊆ I v ′ + ⊆ I u + ; (2) If there is a neighborhood O of I v + such that v ′ + | O ≥ v + | O , then v ′ + ≥ v + everywhere; (3) If I v ′ + = I v + and v ′ + | O = v + | O , then v ′ + = v + everywhere.Proof. The result (1) comes from Proposition 6.3, the result (3) comes from (2). Itremains to prove the result (2). By Proposition 6.3, for a ( v + , L, γ + : [0 , + ∞ ) → M with γ + (0) = x , we have lim t → + ∞ B v + ( γ + ( t )) = 0. Then there is a t large enough, such that γ + ( t ) ∈ O . Define F ( s ) = v + ( γ + ( s )) − v ′ + ( γ + ( s )) , s ∈ [0 , t ] . f v + > v ′ + , then F (0) = v + ( x ) − v ′ + ( x ) > F ( t ) = v + ( γ + ( t )) − v ′ + ( γ + ( t )) ≤ σ ∈ (0 , t ] such that F ( σ ) = 0 and F ( s ) > s ∈ [0 , σ ). By definitionwe have v + ( γ + ( σ )) − v + ( γ + ( s )) = Z σs L ( γ + ( τ ) , v + ( γ + ( τ )) , ˙ γ + ( τ )) dτ, and v ′ + ( γ + ( σ )) − v ′ + ( γ + ( s )) ≤ Z σs L ( γ + ( τ ) , v ′ + ( γ + ( τ )) , ˙ γ + ( τ )) dτ, which implies F ( s ) ≤ F ( σ ) + λ Z σs F ( τ ) dτ. By the Gronwall inequality we conclude F ( s ) = G ( σ − s ) ≡ s ∈ [0 , σ ], whichcontradicts F (0) >
0. We finally have v ′ + ≥ v + . (cid:3) Proposition 7.4.
Given ϕ ∈ C ( M ) , we have the following results. (1) If ϕ satisfies the condition ( ∗ ’) stated in Proposition 5.1, then T + t ϕ ( x ) has abound independent of t ; (2) If ( ∗ ’) does not hold, then there are two possible cases: (a) There is x such that ϕ ( x ) > u − ( x ) , then T + t ϕ ( x ) tends to + ∞ uniformlyas t tends to infinity; (b) ϕ < u − , then T + t ϕ ( x ) tends to −∞ uniformly as t tends to infinity.Proof. The proof of (1) was given in Lemma 5.1. It remains to prove (2). We only show(a) here, the proof of (b) is similar. We first show that T − t ϕ is bounded from below.Define ϕ ( x ) := ϕ ( x ) − ( ϕ ( x ) − u − ( x )), then ϕ ( x ) satisfies the condition ( ∗ ’), thus T + t ϕ is uniformly bounded. By Proposition 1.3 (1), we have T + t ϕ ≥ T + t ϕ .We argue by contradiction. Assume there is a sequence t n → + ∞ such that T + t n ϕ ( x ) isbounded by C . For any given t n , the function v n ( x ) := T + t n ϕ ( x ) is a bounded continuousfunction defined on M . We are going to show ϕ ( x ) ≤ T − t n v n ( x ). If not, then ϕ ( x ) >T − t n v n ( x ) holds. Let γ : [0 , t n ] → M with γ ( t n ) = x be a minimizer of T − t n v n ( x ).Define F ( s ) := T + t n − s ϕ ( γ ( s )) − T − s v n ( γ ( s )) , s ∈ [0 , t n ] . By assumption F ( t n ) >
0. There are two possible cases:Case (1): There is σ ∈ [0 , t n ] such that F ( σ ) = 0 and F ( s ) > s ∈ ( σ, t n ]. Bydefinition T − s v n ( γ ( s )) = T − σ v n ( γ ( σ )) + Z sσ L ( γ ( τ ) , T − τ v n ( γ ( τ )) , ˙ γ ( τ )) dτ, and T + t n − σ ϕ ( γ ( σ )) ≥ T + t n − s ϕ ( γ ( s )) − Z sσ L ( γ ( τ ) , T + t n − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ, which implies F ( s ) ≤ F ( σ ) + λ Z sσ F ( τ ) dτ. By the Gronwall inequality we conclude F ( s ) ≡ s ∈ [ σ, t n ], which contradicts F ( t n ) > ase (2): For each s ∈ [0 , t n ], we have F ( s ) >
0. By definition T − s v n ( γ ( s )) = v n ( γ (0)) + Z s L ( γ ( τ ) , T − τ v n ( γ ( τ )) , ˙ γ ( τ )) dτ = T + t n ϕ ( γ (0)) + Z s L ( γ ( τ ) , T − τ v n ( γ ( τ )) , ˙ γ ( τ )) dτ ≥ T + t n − s ϕ ( γ ( s )) − Z s L ( γ ( τ ) , T + t n − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ + Z s L ( γ ( τ ) , T − τ v n ( γ ( τ )) , ˙ γ ( τ )) dτ, which implies that F ( s ) ≤ λ Z s F ( τ ) dτ, thus F ( s ) ≡
0, which contradicts F ( t n ) > ϕ ( x ) ≤ T − t n v n ( x ). By Proposition 1.3 (1) we have T − t n ( − C )( x ) ≤ T − t n v n ( x ) ≤ T − t n C ( x ) , ∀ x ∈ M. We assert that lim n → + ∞ T − t n v n ( x ) = lim t → + ∞ T − t ( ± C )( x ) = u − ( x ). In fact, since H ( x, u, p ) satisfies (STI), u − ( x ) + k u − − C k ∞ and u − ( x ) − k u − − C k ∞ are viscos-ity supersolution and subsolution of ( E H ) respectively. By the comparison theorem wehave u − ( x ) − k u − − C k ∞ ≤ T − t C ( x ) ≤ u − ( x ) + k u − − C k ∞ for all x ∈ M . Thus T − t C ( x )has a bound independent of t . Similarly, T − t ( − C )( x ) has a bound independent of t . ByRemark 3.5, T − t ( ± C ) has a Lipschitz constant in x independent of t . By Lemma 4.2one can see that lim inf t → + ∞ T − t ( ± C )( x ) and lim sup t → + ∞ T − t ( ± C )( x ) are both fixedpoints of T − t . By Remark 6.1, the fixed point of T − t is unique, then the limit functionlim t → + ∞ T − t ( ± C )( x ) does exist, and equals to u − .Therefore u − ( x ) = lim n → + ∞ T − t n v n ( x ) ≥ ϕ ( x ) > u − ( x ) , which gives a contradiction. (cid:3) Proposition 7.5.
Given ϕ ∈ Lip ( M ) , if (1) u + ≤ ϕ ≤ u − , then lim t → + ∞ T + t ϕ ( x ) = u + ( x ) uniformly on x ∈ M ; (2) S + = { ϕ ∞ } is a singleton and ϕ satisfying the condition ( ∗ ’), then ϕ ∞ ( x ) =lim t → + ∞ T + t ϕ ( x ) ; (3) Let v ∗ + be a minimal element in S + . If the condition ( ∗ ’) holds and ϕ ≤ v ∗ + ,then v ∗ + = lim t → + ∞ T + t ϕ ( x ) .Proof. Proof of (1): The proof is straightforward. By Proposition 1.3 (1) we have T + t u + ≤ T + t ϕ ≤ T + t u − . Meanwhile we have u + = T + t u + and u + = lim t → + ∞ T + t u − .One can easily check the result by taking a limit with respect to t .Proof of (2): Since ϕ satisfies the condition ( ∗ ’), T + t ϕ ( x ) is uniformly bounded. ByRemark 3.5 the Lipschitz constant of T + t ϕ ( x ) in x , which is denoted by κ , is independentof t . By the Arzela-Ascoli theorem, the following two functions are well-definedˆ ϕ ( x ) := lim sup t → + ∞ T + t ϕ ( x ) , ˇ ϕ ( x ) := lim inf t → + ∞ T + t ϕ ( x ) . By Lemma 4.2, both ˆ ϕ ( x ) and ˇ ϕ ( x ) are fixed points of T + t . Since S + is a singleton, thelimit function lim t → + ∞ T + t ϕ ( x ) does exist, and equals to ϕ ∞ . roof of (3): Since the condition ( ∗ ’) is satisfied, the functions ˆ ϕ and ˇ ϕ given aboveare well-defined. Both of them are fixed points of T + t . Since ϕ ≤ v ∗ + , we have T + t ϕ ≤ T + t v ∗ + = v ∗ + . Therefore both ˆ ϕ and ˇ ϕ are no larger than v ∗ + . Since v ∗ + is the minimalelement in S + , we get ˆ ϕ = ˇ ϕ = v ∗ + . Therefore the limit function lim t → + ∞ T + t ϕ doesexist, and equals to v ∗ + . (cid:3) Corollary 7.6.
For each continuous function ϕ satisfying the condition ( ∗ ’), there isa uniform bound K > and T ( ϕ ) > depending on ϕ , such that | T + t ϕ | ≤ K , for all ( x, t ) ∈ M × [ T ( ϕ ) , + ∞ ) .Proof. By Proposition 7.2, the following constant is well-defined K := k inf v + ∈S + v + k ∞ + k u + k ∞ + 1 . Since the condition ( ∗ ’) is satisfied, the functions ˆ ϕ and ˇ ϕ can be defined, and both ofthem are fixed points of T + t . Thereforeinf v + ∈S + v + ( x ) − ≤ ˇ ϕ ( x ) − ≤ T + t ϕ ( x ) ≤ ˆ ϕ ( x ) + 1 ≤ u + ( x ) + 1 , for all t ≥ T ( ϕ ) and all x ∈ M . (cid:3) More about regularity
For a given conjugate pair ( u − , u + ), since L ( x, u, ˙ x ) satisfies (LB), by Proposition5.9, both of them are Lipschitz continuous. Define R ≥ max {k ∂ x u − k ∞ , k ∂ x u + k ∞ } , we can apply a modification on H via H R ( x, u, p ) := H ( x, u, p ) + max {k p k − R , } , then u − and u + are the viscosity solutions of H R ( x, u ( x ) , ∂ x u ( x )) = 0 , F R ( x, u ( x ) , ∂ x u ( x )) = 0respectively. It is obvious that H R satisfies (SL). Thus, without any loss of general-ity, we assume H ( x, u, p ) satisfies (SL). By [14, Proposition 2.7], if H ( x, u, p ) satisfies(C)(STC)(SL)(LIP), then the corresponding Lagrangian satisfies (CON)(SL)(LIP) and (CD): L ( x, u, ˙ x ) is continuous and differentiable in ˙ x . The map ( x, u, ˙ x ) ∂ ˙ x L ( x, u, ˙ x )is continuous.In addition, if H ( x, u, p ) satisfies (LL), by [16, Proposition 7.2 (iv)], the correspondingLagrangian satisfies (CON)(SL)(LIP) and (LLD): the map ( x, ˙ x ) L ( x, u, ˙ x ) is locally Lipschitz continuous for any u . L ( x, u, ˙ x )is differentiable in ˙ x , and the map ( x, u, ˙ x ) ∂ ˙ x L ( x, u, ˙ x ) is continuous. Lemma 8.1.
Given a > . If u ≺ L , let γ : [ − a, a ] → M be a ( u, L, -calibrated curve,then γ is of class C and u is differentiable at γ (0) . The proof of Lemma 8.1 is similar to [28, Lemma 4.3]. We give it in the appendix forthe reader’s convenience.
Lemma 8.2.
Given a conjugate pair ( u − , u + ) , for x ∈ I ( u − ,u + ) , there exists a C curve γ : ( −∞ , ∞ ) → M with γ (0) = x such that u − ( γ ( t )) = u + ( γ ( t )) , and u ± ( γ ( t ′ )) − u ± ( γ ( t )) = Z t ′ t L ( γ ( s ) , u ± ( γ ( s )) , ˙ γ ( s )) ds, ∀ t ≤ t ′ ∈ R . (8.1) In addition, u ± are differentiable at x with the derivative. roof. For x ∈ I ( u − ,u + ) , there is a ( u − , L, γ − : ( −∞ , → M with γ − (0) = x and a ( u + , L, γ + : [0 , + ∞ ) → M with γ + (0) = x , con-necting these two curves, we get a curve γ : ( −∞ , ∞ ) → M with γ (0) = x .We then prove u + ( γ + ( s )) = u − ( γ + ( s )), the proof of u + ( γ − ( s )) = u − ( γ − ( s )) is similar.We argue by contradiction. Assume there exists s ∈ (0 , + ∞ ) such that u + ( γ + ( s ))
0. Then there is τ ∈ [0 , s )such that F ( τ ) = 0 and F ( τ ) > τ ∈ ( τ , s ]. By definition we have u + ( γ + ( τ )) − u + ( γ + ( τ )) = Z ττ L ( γ + ( s ) , u + ( γ + ( s )) , ˙ γ + ( s )) ds,u − ( γ + ( τ )) − u − ( γ + ( τ )) ≤ Z ττ L ( γ + ( s ) , u + ( γ + ( s )) , ˙ γ + ( s )) ds. Therefore F ( τ ) ≤ λ R ττ F ( s ) ds . By the Gronwall inequality we get F ( τ ) ≡ τ ∈ [ τ , s ], which contradicts F ( s ) >
0. Therefore γ is a ( u ± , L, R , i.e. (8.1) holds. By Lemma 8.1, γ is a C curve and ∂ x u ± ( x ) = ∂ ˙ x L ( x, u ± ( x ) , ˙ γ (0)). (cid:3) Lemma 8.3.
The conjugate pair u − and u + are both of class C on I ( u − ,u + ) .Proof. By Lemma 5.9, if L ( x, u, ˙ x ) satisfies (LB), then u − is Lipschitz continuous. By [8,Theorem 5.3.7], if H ( x, u, p ) satisfies (LL)(STC), u − is locally semiconcave. Similarly,since − u + is a viscosity solution of ( E F ), it is also locally semiconcave. Equivalently u + is locally semiconvex. Then by [8, Theorem 3.3.7], for x ∈ I ( u − ,u + ) , the conjugatepair u − and u + are both of class C . (cid:3) Lemma 8.4.
Assume H : T ∗ M × R → R satisfies (STC)(SL)(LIP), then for a givenconjugate pair ( u − , u + ) we have (1) If H ( x, u, p ) is of class C , for x ∈ I ( u − ,u + ) , there exists a C curve γ :( −∞ , ∞ ) → M with γ (0) = x such that u − ( γ ( t )) = u + ( γ ( t )) for all t ∈ R , and ( γ ( t ) , u ± ( γ ( t ) , ∂ γ ( t ) u ± ( γ ( t ))) satisfies the contact Hamiltonian equations (1.4)on R . (2) In addition, if H ( x, u, p ) is of class C , , the conjugate pair u − and u + are ofclass C , on I ( u − ,u + ) . Equivalently, the projection π : T ∗ M × R → M inducesa bi-Lipschitz map between I ( u − ,u + ) and ˜ I ( u − ,u + ) .Proof. (1) By Lemma 8.2, the C curve γ exists and γ ( t ) ∈ I ( u − ,u + ) for all t . Since u − is a viscosity solution of ( E H ) and differentiable on γ ( t ), we have H ( γ ( t ) , u − ( γ ( t )) , ∂ γ − ( t ) u − ( γ − ( t ))) = 0 . Denote u ( γ ( t )) = u ± ( γ ( t )), be definition we have u ( γ ( t ′ )) − u ( γ ( t )) = Z t ′ t L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s )) ds. Dividing by t ′ − t on both sides and letting t ′ → t + , we have ∂ γ ( t ) u ( γ ( t )) · ˙ γ ( t ) = L ( γ ( t ) , u ( γ ( t )) , ˙ γ ( t )) . From the definition of the Legendre transformation, one easily get˙ γ ( t ) = ∂ p H ( γ ( t ) , u ( γ ( t )) , ∂ γ ( t ) u ( γ ( t ))) , ∂ γ ( t ) u ( γ ( t )) = ∂ ˙ x L ( γ ( t ) , u ( γ ( t )) , ˙ γ ( t )) . y a direct calculation, we get˙ u ( γ ( t )) = ∂ γ ( t ) u ( γ ( t )) · ˙ γ ( t ) = ∂ γ ( t ) u ( γ ( t )) · ∂ p H ( γ ( t ) , u ( γ ( t )) , ∂ γ ( t ) u ( γ ( t ))) . By Lemma 8.3, u is of class C on γ ( t ). Denote ¯ L ( x, v ) = L ( x, u ( x ) , v ), then it is of class C along γ . By [11, Theorem 2.1 (i)], for almost all t , the calibrated curve γ satisfies ddt ( ∂ ˙ x ¯ L ( γ ( t ) , ˙ γ ( t ))) = ∂ x ¯ L ( γ ( t ) , ˙ γ ( t )) , which implies ddt ( ∂ γ ( t ) u ( γ ( t ))) = − ∂ x H ( γ ( t ) , u ( γ ( t )) , ∂ γ ( t ) u ( γ ( t ))) − ∂ u H ( γ ( t ) , u ( γ ( t )) , ∂ γ ( t ) u ( γ ( t ))) ∂ γ ( t ) u ( γ ( t )) , a.e. (2) If H is of class C , , it satisfies the assumptions in [8, Theorem 5.3.6], so u − isa semiconcave function with a linear modulus. Similarly − u + is also a semiconcavefunction with a linear modulus. Equivalently, u + is a semiconvex function with a linearmodulus. By [8, Theorem 3.3.7], the conjugate pair u − and u + are both of class C , on I ( u − ,u + ) . (cid:3) Acknowledgements:
The authors would like to thank Professor A. Davini, H. Ishiiand A. Siconolfi for suggesting us to consider the previous results in [28, 29] undermore general assumptions. The authors are also grateful to Professor Wei Cheng andKaizhi Wang for many useful discussions. Lin Wang is supported by NSFC Grant No.11790273, 11631006. Jun Yan is supported by NSFC Grant No. 11631006, 11790273.
Appendix A. Proof of Remark 1.2.
According to the Legendre transformation, it is obvious that¯ L ( x, u, ˙ x ) = L ( x, − u, − ˙ x ). By Lemma 3.6, the minimizers of ¯ T − t ( − ϕ )( x ) exist, and wedenote one of them by ¯ γ , then¯ T − t ( − ϕ )( x ) = − ϕ (¯ γ (0)) + Z t ¯ L (¯ γ ( s ) , ¯ T − s ( − ϕ )(¯ γ ( s )) , ˙¯ γ ( s )) ds = − ϕ (¯ γ (0)) + Z t L (¯ γ ( s ) , − ¯ T − s ( − ϕ )(¯ γ ( s )) , − ˙¯ γ ( s )) ds = − ϕ ( γ ( t )) + Z t L ( γ ( t − s ) , − ¯ T − s ( − ϕ )( γ ( t − s )) , ˙ γ ( t − s )) ds = − ϕ ( γ ( t )) + Z t L ( γ ( τ ) , − ¯ T − t − τ ( − ϕ )( γ ( τ )) , ˙ γ ( τ )) dτ, where ¯ γ ( s ) = γ ( t − s ) and τ = t − s . By the definition of minimizers, it is obvious that − ¯ T − t ( − ϕ )( x ) satisfies (T+). (cid:3) Proof of Proposition 1.3. (1) We argue by contradiction. Assume there exists ( x, t )such that T − t ϕ ( x ) > T − t ϕ ( x ). Let γ : [0 , t ] → M be a minimizer of T − t ϕ ( x ). Define F ( s ) = T − s ϕ ( γ ( s )) − T − s ϕ ( γ ( s )) , s ∈ [0 , t ] . Then F ( s ) is a continuous function defined on [0 , t ], and F (0) ≤
0. By assumption F ( t ) >
0. Then there is s ∈ [0 , t ) such that F ( s ) = 0 and F ( s ) > s ∈ ( s , t ]. ince γ is a minimizer of T − t ϕ ( x ), we have T − s ϕ ( γ ( s )) = T − s ϕ ( γ ( s )) + Z ss L ( γ ( τ ) , T − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ, and T − s ϕ ( γ ( s )) ≤ T − s ϕ ( γ ( s )) + Z ss L ( γ ( τ ) , T − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ, which implies F ( s ) ≤ F ( s ) + λ R ss F ( τ ) dτ . Here F ( s ) = 0, thus F ( s ) ≤ λ Z ss F ( τ ) dτ. By the Gronwall inequality, we conclude F ( s ) ≡ s ∈ [ s , t ]., which contradicts F ( t ) > x ∈ M and t >
0, if T − t ϕ ( x ) = T − t ψ ( x ), then the proof is finished. Wethen consider T − t ϕ ( x ) > T − t ψ ( x ). Let γ be a minimizer of T − t ψ ( x ), define F ( s ) := T − s ϕ ( γ ( s )) − T − s ψ ( γ ( s )) , ∀ s ∈ [0 , t ] . By assumption F ( t ) >
0. If there is σ ∈ [0 , t ) such that F ( σ ) = 0 and F ( s ) > s ∈ ( σ, t ], by definition T − s ϕ ( γ ( s )) ≤ T − t ϕ ( γ ( σ )) + Z sσ L ( γ ( τ ) , T − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ, and T − s ψ ( γ ( s )) = T − t ψ ( γ ( σ )) + Z sσ L ( γ ( τ ) , T − τ ψ ( γ ( τ )) , ˙ γ ( τ )) dτ, which implies F ( s ) ≤ F ( σ ) + λ Z sσ F ( τ ) dτ, where F ( σ ) = 0. By the Gronwall inequality we conclude F ( s ) ≡ s ∈ [ σ, t ],which contradicts F ( t ) > σ ∈ [0 , t ], we have F ( σ ) >
0. Here 0 < F (0) ≤ k ϕ − ψ k ∞ . Bydefinition T − s ϕ ( γ ( σ )) ≤ T − t ϕ ( γ (0)) + Z σ L ( γ ( τ ) , T − τ ϕ ( γ ( τ )) , ˙ γ ( τ )) dτ, and T − s ψ ( γ ( σ )) = T − t ψ ( γ (0)) + Z σ L ( γ ( τ ) , T − τ ψ ( γ ( τ )) , ˙ γ ( τ )) dτ, which implies F ( σ ) ≤ F (0) + λ Z σ F ( τ ) dτ. By the Gronwall inequality we get F ( σ ) ≤ k ϕ − ψ k ∞ e λσ , which implies T − t ϕ ( x ) − T − t ψ ( x ) ≤ k ϕ − ψ k ∞ e λt by taking σ = t . Exchanging the role of ϕ and ψ , we finallyobtain that | T − t ϕ ( x ) − T − t ψ ( x ) | ≤ k ϕ − ψ k ∞ e λt .By definition 1.1, one can show the corresponding properties of T + easily. (cid:3) Proof of Remark 1.6.
According to the definition of the backward weak KAM solu-tions, for u − ∈ S − we have u − ( x ) = inf γ ( t )= x (cid:26) u − ( γ (0)) + Z t L ( γ ( τ ) , u − ( γ ( τ )) , ˙ γ ( τ )) dτ (cid:27) , here the infimum is taken in the class of absolutely continuous curves. We show u − ( x ) ≤ T − t u − ( x ), the opposite direction is similar. We argue by contradiction. Assume u − ( x ) > T − t u − ( x ). Let γ : [0 , t ] → M with γ ( t ) = x be a minimizer of T − t u − ( x ). Define F ( τ ) := u − ( γ ( τ )) − T − τ u − ( γ ( τ )) . Since F ( t ) > F (0) = 0, there is s ∈ [0 , t ) such that F ( s ) = 0 and F ( s ) > s ∈ ( s , t ]. By definition T − s u − ( γ ( s )) = T − s u − ( γ ( s )) + Z ss L ( γ ( τ ) , T − τ u − ( γ ( τ )) , ˙ γ ( τ )) dτ, and u − ( γ ( s )) ≤ u − ( γ ( s )) + Z ss L ( γ ( τ ) , u − ( γ ( τ )) , ˙ γ ( τ )) dτ, which implies F ( s ) ≤ λ Z ss F ( τ ) dτ. By the Gronwall inequality, we conclude F ( s ) ≡ s ∈ [ s , t ], which contradicts F ( t ) > v + ∈ S + we have v + ( x ) = sup γ (0)= x (cid:26) v + ( γ ( t )) − Z t L ( γ ( τ ) , v + ( γ ( τ )) , ˙ γ ( τ )) dτ (cid:27) , where the infimum is taken in the class of absolutely continuous curves. We show v + ( x ) ≤ T + t v + ( x ), the opposite direction is similar. We argue by contradiction. Assume v + ( x ) > T + t v + ( x ). Let γ + : [0 , t ] → M with γ + (0) = x be a ( v + , L, F ( τ ) := v + ( γ + ( τ )) − T + t − τ v + ( γ + ( τ )) . Since F ( t ) = 0 and F (0) >
0, there is s ∈ (0 , t ] such that F ( s ) = 0 and F ( s ) > s ∈ [0 , s ). By definition T + t − s v + ( γ + ( s )) ≥ T + t − s v + ( γ + ( s )) − Z s s L ( γ + ( τ ) , T + t − τ v + ( γ + ( τ )) , ˙ γ + ( τ )) dτ, and v + ( γ + ( s )) = v + ( γ + ( s )) + Z s s L ( γ + ( τ ) , v + ( γ + ( τ )) , ˙ γ + ( τ )) dτ, which implies F ( s ) ≤ λ Z s s F ( τ ) dτ. By the Gronwall inequality, we get F ( s ) ≡ s ∈ [0 , s ], which contradicts F (0) > (cid:3) Proof of Remark 1.9.
We first prove that (A) implies (S). By Lemma 4.2, we onlyneed to show that T − t a ( x ) has a bound independent of t . We first show that T − t a ( x )is bounded from below. If for any ( x, t ) ∈ M × (0 , + ∞ ), there holds T − t a ( x ) ≥ a ,then the proof is finished. If not, there exists ( x , t ) such that T − t a ( x ) < a . Let γ : [0 , t ] → M with γ ( t ) = x be a minimizer of T − t a ( x ), then there is s ∈ [0 , t ) uch that T − s a ( γ ( s )) = a and T − s a ( γ ( s )) < a for all s ∈ ( s , t ]. By the nondecreasingproperty of H ( x, u, p ) in u , we have T − t a ( x ) = T − s a ( γ ( s )) + Z t s L ( γ ( s ) , T − s a ( γ ( s )) , ˙ γ ( s )) ds ≥ a + Z t s L ( γ ( s ) , a, ˙ γ ( s )) ds. Since c ( H a ) = 0, the following equation H ( x, a, ∂ x u ( x )) = 0 (A.1)admits viscosity solutions. We denote it by u a . Then u a ( γ ( t )) − u a ( γ ( s )) ≤ Z t s L ( γ ( s ) , a, ˙ γ ( s )) ds, which implies T − t a ( x ) has a lower bound independent of t .We then show that T − t a ( x ) is bounded from above. If for any ( x, t ) ∈ M × (0 , + ∞ ),we have T − t a ( x ) ≤ a , then the proof is finished. If not, there is ( x , t ) such that T − t a ( x ) > a . Let T − a,t be the backward Lax-Oleinik semigroup defined by L ( x, a, ˙ x ).Since u a is a viscosity solution of (A.1), we have T − a,t u a ( x ) = u a ( x ). Let γ : [0 , t ] → M with γ ( t ) = x be a minimizer of T − a,t u a ( x ), then there is s ∈ [0 , t ) such that T − s a ( γ ( s )) = a and T − s a ( γ ( s )) > a for all s ∈ ( s , t ]. By the nondecreasing propertyof H ( x, u, p ) in u , we have T − t a ( x ) ≤ T − s a ( γ ( s )) + Z t s L ( γ ( s ) , T − s a ( γ ( s )) , ˙ γ ( s )) ds ≤ a + Z t s L ( γ ( s ) , a, ˙ γ ( s )) ds. By definition u a ( γ ( t )) − u a ( γ ( s )) = Z t s L ( γ ( s ) , a, ˙ γ ( s )) ds, which implies that T − t a ( x ) has a upper bound independent of t .It remains to prove that (S) implies (A). The proof is similar to Step 3 in the proofof [28, Theorem B.1]. Let u − be the unique viscosity solution of ( E H ), define a = sup x ∈ M u − ( x ) , a = inf x ∈ M u − ( x ) . Let T − a i ,t be the backward Lax-Oleinik semigroup defined by L ( x, a i , ˙ x ), i = 1 ,
2. By thenondecreasing property of H ( x, u, p ) in u and the definition of a i one can easily obtian T − a ,t u − ( x ) ≤ u − ( x ) , T − a ,t u − ( x ) ≥ u − ( x ) , for all t > x ∈ M . Therefore( T − a ,t u − ( x ) + c ( H a ) t ) − c ( H a ) t ≤ u − ( x ) , and ( T − a ,t u − ( x ) + c ( H a ) t ) − c ( H a ) t ≥ u − ( x ) , for all t > x ∈ M . From the convergence of the Lax-Oleinik semigroup forautonomous systems, it is easy to see that c ( H a ) ≥ c ( H a ) ≤
0. By the continuityof c ( H a ) in a , it has a zero point between a and a . (cid:3) Proof of Remark 1.10.
We have already known that ¯ u + ≤ u − for all u − ∈ S − . Itis sufficient to show ¯ u + ( x ) < u ( x ) for x ∈ ( − , \{ } . By symmetry we only consider x ∈ (0 , u + is a semiconvex function and satisfies ¯ u + ( x ) ≤ u ( x ), ¯ u + cannot equal to u at x = 1. We then assume that there exists x ∈ (0 ,
1) such that u + ( x ) = u ( x ). Since we assume λ >
2, we have λu ( x ) > V ( x ) for all x ∈ (0 , z > V ( x ), we set f ( x, z ) = λ p z − V ( x )) , then f ( x, z ) is of class C on (0 , × { z ∈ R : z > V ( x ) } . By the classical theory ofordinary differential equations, for x ∈ (0 , λu ( x ) is the unique solution of dzdx = f ( x, z ) , z ( x ) = u ( x ) . (A.2)If ¯ u + is differentiable on (0 , u + satisfies (1.3) in the classical sense. Since¯ u + ≤ u and ¯ u + ( x ) = u ( x ), λ ¯ u + is the unique solution of (A.2). That is, ¯ u + = u on(0 , u + . Therefore, for all x ∈ (0 , u + < u .It remains to show that ¯ u + is differentiable on (0 , y ∈ (0 , u + is not differentiable at y . Since ¯ u + is the unique forward weak KAMsolution of (1.3), there is l > ± l ∈ D ∗ ¯ u + ( y ), where D ∗ stands for the set ofall reachable gradients. By the semiconvexity of ¯ u + , it is decreasing on the left side of y .Since ¯ u + (0) = 0 and ¯ u + ( y ) ≥
0, there is z ∈ (0 , y ) which is a local maximum of ¯ u + .By the semiconvexity of ¯ u + , it is differentiable at z , then ¯ u ′ + ( z ) = 0. By (1.3) we have − λ ¯ u + ( z ) + V ( z ) = 0. Since ¯ u ′ + ( x ) exists for almost all x , there is z ∈ ( z , y ) suchthat ¯ u ′ + ( z ) exists, | ¯ u ′ + ( z ) | > u + ( z ) ≥ ¯ u + ( z ) ≥
0. We also have V ( z ) > V ( z ).Therefore − λ ¯ u + ( z ) + 12 | ¯ u ′ + ( z ) | + V ( z ) > − λ ¯ u + ( z ) + V ( z ) = 0 , which contradicts that ¯ u + satisfies (1.3) at z in the classical sense. (cid:3) Proof of Lemma 3.3.
We first prove that u ( x, t ) is continuous at t = 0. We take theinitial functions in (3.1) as ϕ and ϕ , and denote by u ( x, t ) and u ( x, t ) the corre-sponding value functions respectively. Since v ( x, t ) is fixed, by the non-expansiveness ofthe Lax-Oleinik semigroup, we have k u ( x, t ) − u ( x, t ) k ∞ ≤ k ϕ − ϕ k ∞ . Thus, withoutany loss of generality, we assume the initial function to be Lipschitz continuous in thefollowing discussion. Take a constant curve α ( t ) ≡ x and let γ be a minimizer of u ( x, t ),it is obvious that u ( x, t ) = ϕ ( γ (0)) + Z t L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s )) ds ≤ ϕ ( x ) + Z t L ( x, v ( x, s ) , ds, so lim sup t → + u ( x, t ) ≤ ϕ ( x ). By (SL), there exists a constant C > Z t L ( γ ( τ ) , v ( γ ( τ ) , τ ) , ˙ γ ( τ )) dτ ≥ Z t k ∂ x ϕ k ∞ k ˙ γ ( τ ) k dτ + Ct ≥ k ∂ x ϕ k ∞ d ( γ (0) , γ ( t )) + Ct, which implies that Z t L ( γ ( τ ) , v ( γ ( τ ) , τ ) , ˙ γ ( τ )) dτ + ϕ ( γ (0)) ≥ ϕ ( x ) + Ct.
Therefore lim inf t → + u ( x, t ) ≥ ϕ ( x ). Combining with Lemma 3.2, the conclusion that u ( x, t ) is continuous on M × [0 , T ] is then proved.We are now going to show that the value function u ( x, t ) is a continuous viscositysolution of (3.2). We first show that u ( x, t ) is a viscosity subsolution. Let V be an opensubset of M and φ : V × [0 , T ] → R be a C test function such that u ( x, t ) − φ ( x, t ) takesits maximum at ( x , t ). Equivalently we have φ ( x , t ) − φ ( x, t ) ≤ u ( x , t ) − u ( x, t ) for ll ( x, t ) ∈ V × [0 , T ]. Given a constant δ >
0, we take a C curve γ : [ t − δ, t + δ ] → M taking its value in V , satisfying γ ( t ) = x and ˙ γ ( t ) = ξ . For t ∈ [ t − δ, t ], we have φ ( x , t ) − φ ( γ ( t ) , t ) ≤ u ( x , t ) − u ( γ ( t ) , t ) ≤ Z t t L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s ))d s. Dividing by t − t on both side of the above inequality, we have φ ( x , t ) − φ ( γ ( t ) , t ) t − t ≤ t − t Z t t L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s ))d s. Let t → t − , we have φ t ( x , t ) + φ x ( x , t ) · ξ ≤ L ( x , v ( x , t ) , v ). By definition of theLagrangian via Legendre transformation, we have φ t ( x , t ) + H ( x , v ( x , t ) , φ x ( x , t )) ≤ . Then we show that u ( x, t ) is a supersolution. Let ψ : V × [0 , T ] → R be a C testfunction such that u ( x, t ) − ψ ( x, t ) takes its minimum at ( x , t ). Equivalently we have ψ ( x , t ) − ψ ( x, t ) ≥ u ( x , t ) − u ( γ ( t ) , t ) for all ( x, t ) ∈ V × [0 , T ]. Let γ be a minimmizerof u ( x , t ), for t ∈ [ t − δ, t ] with γ ( t − δ ) ∈ V , we have ψ ( x , t ) − ψ ( γ ( t ) , t ) ≥ u ( x , t ) − u ( γ ( t ) , t ) = Z t t L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s ))d s. (A.3)Let t → t − . When t is close enough to t , the curve γ ( s ) is contained in a coordinateneighbourhood of x . In the local coordinate, we can assume M equals to an opensubset of R n . Since v ( x, t ) is Lipschitz continuous on M × [0 , T ], the minimizer γ is aLipschitz curve. Therefore k x − γ ( t ) k / | t − t | is bounded. One can take a sequence t n → t − such that ( x − γ ( t n )) / ( t − t n ) converges to some ξ ′ ∈ R n . By the continuityof L ( x, u, ˙ x ), v ( x, t ) and γ , for any ε >
0, there exists a large enough n ∈ N such that L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s )) ≥ L ( x , v ( x , t ) , ˙ γ ( s )) − ε, ∀ s ∈ [ t n , t ] . Since L ( x, u, · ) is convex, the Jensen inequality implies that1 t − t n Z t t n L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s )) ds ≥ L (cid:18) x , v ( x , t ) , t − t n Z t t n ˙ γ ( s ) ds (cid:19) − ε = L (cid:18) x , v ( x , t ) , x − γ ( t n ) t − t n (cid:19) − ε. When n is large enough, ε can be arbitrary small. Dividing by t − t n on both side of(A.3), we havelim n → + ∞ ψ ( x , t ) − ψ ( γ ( t n ) , t n ) t − t n = ψ t ( x , t ) + ψ x ( x , t ) · ξ ′ ≥ lim sup n → + ∞ t − t n Z t t n L ( γ ( s ) , v ( γ ( s ) , s ) , ˙ γ ( s )) ds ≥ L ( x , v ( x , t ) , ξ ′ ) . Therefore ψ t ( x , t ) + H ( x , v ( x , t ) , ψ x ( x , t )) ≥ ψ t ( x , t ) + ψ x ( x , t ) · ξ ′ − L ( x , v ( x , t ) , ξ ′ ) ≥ . Finally, we have proven that u ( x, t ) is a continuous viscosity solution of (3.2) on M × [0 , T ]. (cid:3) roof of Lemma 8.1. By Lemma 5.9, u is Lipschitz continuous. By the condition(LLD), for k v k and k v k less than R , there exists a constant K ( R, k u ( x ) k ∞ ) > | L ( x , u ( x ) , v ) − L ( x , u ( x ) , v ) |≤ | L ( x , u ( x ) , v ) − L ( x , u ( x ) , v ) | + | L ( x , u ( x ) , v ) − L ( x , u ( x ) , v ) |≤ λ k ∂ x u ( x ) k ∞ d ( x , x ) + K ( R, k u ( x ) k ∞ )( d ( x , x ) + k v − v k ) . Therefore ( x, v ) L ( x, u ( x ) , v ) is locally Lipschitz continuous. By [11, Theorem 2.1(ii)], the minimizer γ is a C curve.Since we are arguing locally near the point x := γ (0), it suffices to prove the lemmafor the case when M is an open subset U of R n . We are going to show for each y ∈ U ,there holdslim sup η → + u ( x + ηy ) − u ( x ) η ≤ ∂L∂ ˙ x ( x, u ( x ) , ˙ γ (0)) · y ≤ lim inf η → + u ( x + ηy ) − u ( x ) η . (A.4)For η > < ε ≤ a , define γ η : [ − ε, → U by γ η ( s ) = γ ( s ) + s + εε ηy , then γ η (0) = x + ηy and γ η ( − ε ) = γ ( − ε ). Since γ is a ( u, L, u ( x + ηy ) − u ( γ ( − ε )) ≤ Z − ε L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ η ( s )) ds,u ( x ) − u ( γ ( − ε )) = Z − ε L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s )) ds. It follows that u ( x + ηy ) − u ( x ) η ≤ η Z − ε ( L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ η ( s )) − L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s ))) ds. By the locally Lipschitz continuity of the map ( x, v ) L ( x, u ( x ) , v ), there exists K ′ ( k ˙ γ ( s ) k ) such thatlim sup η → + u ( x + ηy ) − u ( x ) η ≤ lim sup η → + η Z − ε ( L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ η ( s )) − L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ ( s )))+ ( L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ ( s )) − L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s ))) ds ≤ Z − ε ( 1 ε ∂L∂ ˙ x ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s )) + K ′ ( k ˙ γ ( s ) k ) s + εε k y k ) ds. Let ε → + , we get the first equality in (A.4).Similarly, define γ η : [0 , ε ] → U by γ η ( s ) = γ ( s ) + ε − sε ηy , then γ η (0) = x + ηy and γ η ( ε ) = γ ( ε ). Since γ is a ( u, L, u ( γ ( ε )) − u ( x + ηy ) ≤ Z ε L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ η ( s )) ds,u ( γ ( ε )) − u ( x ) = Z ε L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s )) ds. It follows that u ( x + ηy ) − u ( x ) η ≥ η Z ε ( L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s )) − L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ η ( s ))) ds. y the locally Lipschitz continuity of the map ( x, v ) L ( x, u ( x ) , v ), there exists K ′ ( k ˙ γ ( s ) k ) such thatlim inf η → + u ( x + ηy ) − u ( x ) η ≤ lim sup η → + η Z − ε ( L ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s )) − L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ ( s )))+ ( L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ ( s )) − L ( γ η ( s ) , u ( γ η ( s )) , ˙ γ η ( s ))) ds ≤ Z − ε ( K ′ ( k ˙ γ ( s ) k ) s − εε k y k + 1 ε ∂L∂ ˙ x ( γ ( s ) , u ( γ ( s )) , ˙ γ ( s ))) ds. Let ε → + , we get the second equality in (A.4). (cid:3) References [1] V. I. Arnold.
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Shanghai Center for Mathematical Sciences, Fudan University, Shanghai 200433, China
Email address : [email protected] Yau Mathematical Sciences Center, Tsinghua University, Beijing 100084, China
Email address : [email protected] School of Mathematical Sciences, Fudan University, Shanghai 200433, China
Email address : [email protected]@fudan.edu.cn