The vanishing discount problem for monotone systems of Hamilton-Jacobi equations. Part 1: linear coupling
Hitoshi Ishii

TL;DR
This paper proves a convergence theorem for the vanishing discount problem in weakly coupled Hamilton-Jacobi systems, introducing new measure concepts to handle linear coupling.
Contribution
It introduces viscosity Mather and Green-Poisson measures for systems, advancing the analysis of the vanishing discount problem with linear coupling.
Findings
Established convergence theorem for linear coupling systems
Introduced viscosity Mather and Green-Poisson measures
Utilized convex duality in the analysis
Abstract
We establish a convergence theorem for the vanishing discount problem for a weakly coupled system of Hamilton-Jacobi equations. The crucial step is the introduction of Mather measures and their relatives for the system, which we call respectively viscosity Mather and Green-Poisson measures. This is done by the convex duality and the duality between the space of continuous functions on a compact set and the space of Borel measures on it. This is part 1 of our study of the vanishing discount problem for systems, which focuses on the linear coupling, while part 2 will be concerned with nonlinear coupling.
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Taxonomy
TopicsGeometric Analysis and Curvature Flows · Point processes and geometric inequalities
The vanishing discount problem for
monotone systems of Hamilton-Jacobi equations.
Part 1: linear coupling
Hitoshi Ishii
Institute for Mathematics and Computer Science
Tsuda University
2-1-1 Tsuda-machi, Kodaira-shi, Tokyo, 187-8577 Japan.
Dedicated to Italo Capuzzo Dolcetta with respect, admiration, and friendship on the occasion of his retirement.
Abstract.
We establish a convergence theorem for the vanishing discount problem for a weakly coupled system of Hamilton-Jacobi equations. The crucial step is the introduction of Mather measures and their relatives for the system, which we call respectively viscosity Mather and Green-Poisson measures. This is done by the convex duality and the duality between the space of continuous functions on a compact set and the space of Borel measures on it. This is part 1 of our study of the vanishing discount problem for systems, which focuses on the linear coupling, while part 2 will be concerned with nonlinear coupling.
Key words and phrases:
systems of Hamilton-Jacobi equations, Mather measures, vanishing discount
2010 Mathematics Subject Classification:
35B40, 35F50, 49L25
1. Introduction
We consider the weakly coupled -system of Hamilton-Jacobi equations
[TABLE]
where , is a nonnegative constant, called the discount factor in terms of optimal control. Here denotes the -dimensional flat torus, is a family of Hamiltonians given by
[TABLE]
where , is a given compact metric space, and . The unknown in (1) is an -valued function on , is a linear map represented by a matrix , that is,
[TABLE]
We use the abbreviated notation to denote . The system is called weakly coupled since the -th equation depends on only through but not on , with . Problem (1) can be stated in the component-wise style as
[TABLE]
We are mainly concerned with the asymptotic behavior of the solution of (1) as . Asymptotic problems in this class are called the vanishing discount problem, in view that the constant in (1) appears as a discount factor in the dynamic programming PDE in optimal control.
Recently, there has been a keen interest in the vanishing discount problem concerned with Hamilton-Jacobi equations and, furthermore, fully nonlinear degenerate elliptic PDEs. We refer to [DFIZ, AAIY, IJ, MT, IMT1, IMT2, GMT, CCIZ, DZ2, IS] for relevant work. The asymptotic analysis in these papers relies heavily on Mather measures or their generalizations and, thus, it is considered part of the Aubry-Mather and weak KAM theories. For the development of these theories we refer to [Fa1, Fa2, Ev1] and the references therein.
We are here interested in the case of systems of Hamilton-Jacobi equations and, indeed, Davini and Zavidovique in [DZ2] have established a general convergence result for the vanishing discount problem for (1). We establish a result (Theorem 9 below) similar to the main result of [DZ2]. In establishing our convergence result, we adapt the argument in [IMT1] (see also [Go]) to the case of systems, especially, to construct generalized Mather measures for (1). Regarding the recent developments of the weak KAM theory and asymptotic analysis in its influence for systems, we refer to [CGT, CLLN, MSTY, MT1, MT2, MT3, Ter].
The new argument, which is different from that of [DZ2], makes it fairly easy to build a generalized Mather measure for systems in a wide class. One advantage of our argument is that it allows us to treat the case where the coupling matrix in (1) depends on the space variable . As in [IJ, IMT1], our approach is applicable to the system with nonlinear coupling of fully nonlinear second-order elliptic PDEs, but we restrict ourselves in this paper to the case of the linearly coupled system of first-order Hamilton-Jacobi equations. Another possible approach for constructing generalized Mather measures is the so-called adjoint method (see [Ev2, MT, GMT, CGT, Ter]).
This paper is part 1 of our study of the vanishing discount problem for weakly coupled systems of Hamilton-Jacobi equations and deals only with the linear coupling and with compact control sets . These restrictions make the presentation of our results clear and transparent. In part 2 [IJ], we remove these restrictions and establish a general convergence result extending Theorem 9 below. Sections 5 and 6 are devoted to the study of ergodic problems of the form , where is an unknown as well. Also, thanks to the linearity of the coupling, our results on the ergodic problems are applied to extend the scope of Theorem 9. On the other hand, the role of the ergodic problem, with general right-hand side , is not clear at least for the author in the vanishing discount problem for the systems with the nonlinear coupling.
In this paper, we adopt the notion of viscosity solution to (1), for which the reader may consult [BaCa, BarB, CIL, PLL].
To proceed, we give our main assumptions on the system (1).
We assume that is coercive, that is, for any ,
[TABLE]
This is a convenient assumption, under which any upper semicontinuous subsolution of (1) is Lipschitz continuous on .
We assume that is a monotone matrix for every , that is, it satisfies
[TABLE]
This is a natural assumption that (1) should possess the comparison principle between a subsolution and a supersolution.
In what follows we set, for ,
[TABLE]
and (1) can be written as
[TABLE]
We use the symbol (resp., ) for -vectors to indicate (resp., ) for all .
The following theorem is well-known: see [EL, IK] for instance.
Theorem 1**.**
Assume (C) and (M). Let . Then the exists a unique solution of (1). Also, if are, respectively, upper and lower semicontinuous on and a subsolution and a supersolution of (1), then on .
Henceforth, let denote the vector .
Outline of proof.
We follow the line of the arguments in [IK]. Although [IK] is concerned with the case when the domain is an open subset of a Euclidean space, the results in [IK] is valid in the case when the domain is .
Choose a large constant so that the constant functions are a supersolution and a subsolution of (1), respectively. (See also (6) below.) According to [IK, Theorems 3.3, Lemma 4.8], there is a function such that the upper and lower semicontinuous envelopes and are a subsolution and a supersolution of (1), respectively. By the coercivity assumption (C), we find (see [CL, Theorem I.14], [IPerron, Example 1]) that the functions are Lipschitz continuous on . Let be a Lipschitz bound of the functions . To take into account the Lipschitz property of , we modify the Hamiltonian . Fix any so that
[TABLE]
and choose constants and so that
[TABLE]
and, in view of (1),
[TABLE]
Define by
[TABLE]
By the choice of , it is easy to see that is a subsolution of
[TABLE]
Also, since , is a supersolution of (2). Observe furthermore that, if , then
[TABLE]
the functions are uniformly continuous on , and hence, for some continuous function on , with ,
[TABLE]
The last inequality above shows that satisfies [IK, (A.2)], which allows us to apply [IK, Theorem 4.7], to conclude that on and, moreover, that . Similarly, we deduce that the comparison assertion is valid. Thus, is a unique solution of (1). ∎
Regarding the coercivity (C), the following proposition is well-knwon.
Proposition 2**.**
The function given by (H) satisfies (C) if and only if there exists such that
[TABLE]
where designates “convex hull” and denotes the open ball with origin at the origin and radius .
Outline of proof.
Set Assume that (3) is valid for some and observe that
[TABLE]
which shows that holds.
Next, assume that (3) does not hold for any . Then there exists for each such that
[TABLE]
For each select and so that is the point of closest to . (Notice that is a compact convex set.) Setting , we find that
[TABLE]
Sending along an appropriate subsequence, say , we find that there are a unit vector of , and such that
[TABLE]
If , then we have , since , and the inequality above reads
[TABLE]
These observations imply that for ,
[TABLE]
which shows that (C) does not hold. This completes the proof. ∎
The rest of this paper is organized as follows. In Section 2, we recall some basic facts concerning monotone matrices. In Section 3, we study viscosity Green-Poisson measures for our system, which are crucial in our asymptotic analysis. We establish the main result for the vanishing discount problem in Section 4. We study the ergodic problem (P0) in the cases when is irreducible, and is a constant matrix, respectively, in Sections 5 and 6, and combine the results with the analysis on the vanishing discount problem of Section 4.
2. Monotone matrices
Here we are concerned with real matrix .
Let denote the vector , with and if .
Lemma 3**.**
Let be a real matrix. It is monotone if and only if
[TABLE]
We remark that if satisfies (4), then
[TABLE]
Proof.
We assume first that is monotone. Since
[TABLE]
By the monotonicity of , we have
[TABLE]
Similarly, if and , then we have and hence,
[TABLE]
from which we find by sending that
[TABLE]
Hence, (4) is satisfied.
Next, we assume that (4) holds. Let satisfy
[TABLE]
Then we observe that, since for all ,
[TABLE]
Thus, is monotone. ∎
Lemma 4**.**
Let and be a constant. Let be an real monotone matrix. Then we have
[TABLE]
Proof.
Using Lemma 3, we see that
[TABLE]
which states that . It is then obvious to compute that
[TABLE]
and therefore,
[TABLE]
3. Viscosity Green-Poisson measures
For we write for the set of all such that is a subsolution of
[TABLE]
where and
[TABLE]
In the above, since is bounded on , if satisfies (C), then satisfies (C).
Lemma 5**.**
The set is a convex cone in with vertex at the origin.
Proof.
Recall [Bar, Remark 2.5] that for any , is a subsolution of
[TABLE]
if and only if for any ,
[TABLE]
and by the coercivity (C) that for any , we have .
Fix and . Fix and observe that
[TABLE]
which imply that there is a set of Lebesgue measure zero such that
[TABLE]
Multiplying the first and second by and , respectively, adding the resulting inequalities and setting , we obtain
[TABLE]
which readily implies that . ∎
We refer the reader to [IMT1, Lemma 2.2] for another proof of the above lemma.
We establish a representation formula for the solution of (1), with , by modifying the argument in [IMT1] (see also [Go]).
For any nonnegative Borel measure on and , we write
[TABLE]
Similarly, for any collection of nonnegative Borel measures on and , we write
[TABLE]
Note that any collection of nonnegative Borel measures on is regarded as a nonnegative Borel measure on and vice versa.
We set
[TABLE]
Note that
[TABLE]
By assumption (M) and Lemma 3, we have on for all .
Given a constant , let denote the set of of nonnegative Borel measures on such that
[TABLE]
The last condition reads
[TABLE]
where denotes the total mass of on . Note as well that can be identified with the space of Borel probability measures on by the correspondence between and , where indicates the product of two measures and denotes the Dirac measure at . If we set and consider as a collection of measures on , then . We denote simply by the space of Borel probability measures on .
For and we set
[TABLE]
and
[TABLE]
Theorem 6**.**
Assume (H), (C) and (M). Let and . Let be the unique solution of (1). Then there exists a such that
[TABLE]
We remark that for any we have and, accordingly, in the theorem above, the measures has the minimizing property:
[TABLE]
We call any minimizing family of the optimization problem above a viscosity Green-Poisson measure for (1).
Proof.
Note first that and hence, for any ,
[TABLE]
Next, we show that
[TABLE]
Note that for ,
[TABLE]
Hence, in order to prove (11), we only need to show that
[TABLE]
We postpone the proof of (12) and, assuming temporarily that (11) is valid, we prove that there exists such that
[TABLE]
which, together with (10), completes the proof.
To prove (13), we observe that and, by Lemma 5, are convex,
[TABLE]
is convex and continuous, in the topology of weak convergence of measures, for any and
[TABLE]
is concave and continuous for any . Hence, noting moreover that is a compact set, we apply the minimax theorem ([Te, Si]), to find from (11) that
[TABLE]
Observe by using the cone property of that
[TABLE]
This and (14) yield
[TABLE]
which proves (13).
It remains to show (12). For this, we argue by contradiction and thus suppose that (12) does not hold. Accordingly, we have
[TABLE]
for some . We may select so that
[TABLE]
That is, for any , we have
[TABLE]
Plugging , with any , into the above, we find that
[TABLE]
Hence, we have
[TABLE]
This ensures that is a subsolution of
[TABLE]
which implies that is a subsolution of (1). By comparison (Theorem 1), we get
[TABLE]
The -th component of the above, evaluated at , yields an obvious contradiction. Thus we conclude that (12) holds. ∎
We have the following characterization of .
Proposition 7**.**
Assume (H), (C) and (M) hold. Let and . Then we have if and only if
[TABLE]
Proof.
Assume first that . Fix any and define by
[TABLE]
Observe that satisfy, respectively,
[TABLE]
and, hence,
[TABLE]
Since , we have
[TABLE]
respectively, which shows that (15) is valid.
Now, assume that (15) is satisfied. Fix any . As noted in the proof of Theorem 1, we have . By the standard mollification technique, given a positive constant , we can approximate by a smooth function so that
[TABLE]
The last inequality reads
[TABLE]
Integrating the above by , summing up in and using (15), we get
[TABLE]
Sending shows that . ∎
It is convenient to restate the theorem above as follows. For and , consider given by
[TABLE]
(Notice by the above definition that .) Observe that for ,
[TABLE]
and that for any , we have if and only if
[TABLE]
The condition above is stated in the spirit of Proposition 7 as
[TABLE]
We define
[TABLE]
The following proposition is an immediate consequence of Theorem 6.
Corollary 8**.**
Assume (H), (C) and (M). Let and . Let be the unique solution of (1). Then there exists a such that
[TABLE]
4. A convergence result for the vanishing discount problem
We study the asymptotic behavior of the solution of (1), with , as .
We make a convenient assumption on the system (P0):
[TABLE]
If for all , then Theorem 1 assures that there exists a unique solution of (4). In this situation, it is not difficult to show that the uniform convergence, as , of to the unique solution on . In general, existence and uniqueness of a solution of (P0) may fail. In fact, one can prove at least in the case when the are constants (see Theorem 18) that there exists such that
[TABLE]
has a solution and possibly multiple solutions. If such a exists, then the introduction of a new family of Hamiltonians,
[TABLE]
allows us to view (17) as in the form of (P0). The link between two vanishing discount problems for the original (1) and for (1), with in place of , is discussed in Sections 5 and 6.
Theorem 9**.**
Assume (H), (C), (M) and (4). Let be the unique solution of (1) for . Then there exists a solution of (P0) such that the functions converge to uniformly on as for all .
Lemma 10**.**
Under the hypotheses of Theorem 9, there exists a constant such that for any ,
[TABLE]
Proof.
Let be the solution of (P0). Choose a constant so that
[TABLE]
and observe by the monotonicity of (Lemma 4) that and are a supersolution and a subsolution of (P0), respectively. Noting that and , we deduce that and are a supersolution and a subsolution of (1) for any , respectively. By comaprison (Theorem 1), we see that, for any , on and, moreover, on . Thus, (18) holds with . ∎
Lemma 11**.**
Under the hypotheses of Theorem 9, the family is equi-Lipschitz continuous on .
Indeed, the family is equi-Lipschitz continuous on , which we do not need here.
Proof.
By Lemma 10, there is a constant such that
[TABLE]
Hence, as is a solution of (1), we deduce by (C) that there exists a constant such that the are subsolutions of in . It is a standard fact that the are Lipschitz continuous on with as their Lipschitz bound. ∎
In the proof of Theorem 9, Corollary 8 has a crucial role. We need also results for similar to the corollary.
We consider the condition for ,
[TABLE]
We denote by the subset of consisting of those which satisfy (19).
Theorem 12**.**
Assume (H), (C), (M) and (4). Assume that on for every . Then there exists a such that
[TABLE]
Proof.
We fix a . By Corollary 8, for each there exists such that
[TABLE]
Since is a family of Borel probability measures on a compact space , there exists a sequence converging to zero such that the sequence converges weakly in the sense of measures to a Borel probability measure on . It follows from (21) and Lemma 10 that
[TABLE]
Observe that if , then, for any , is a subsolution of
[TABLE]
and hence, , with . Hence, the inclusion yields
[TABLE]
Multiplying the above by and sending , in view of Lemma 10, we get
[TABLE]
This shows that . These observations together with (19) for guarantee that
[TABLE]
We state a characterization of in the next, similar to Proposition 7, which we leave to the reader to verify.
Proposition 13**.**
Assume (H), (C) and (M). Let . We have if and only if
[TABLE]
We call any minimizer of the optimization problem (20) a viscosity Mather measure.
We denote by the set of all Borel nonnegative measures on . We set
[TABLE]
Theorem 14**.**
Let . Assume (H), (C), (M) and (4). For any , let be the unique solution of (1) and be a minimizer of (17). Then there exists a subsequence of , which is denoted again by the same symbol, such that, as ,
[TABLE]
weakly in the sense of measures for some , and satisfies
[TABLE]
In particular,
[TABLE]
Notice that the minimization problem (23) is trivial since is a minimizer.
Proof.
The proof is similar to that of Theorem 12.
We fix a . For each , we have
[TABLE]
Observe that
[TABLE]
Accordingly, since is a compact metric space, the families have a common subsequence, along which all the families converge to some Borel nonnegative measures weakly in the sense of measures. We may assume by replacing the original sequence by its subsequence that
[TABLE]
weakly in the sense of measures. Combine this with (24) yields
[TABLE]
It is obvious to see that .
Let . As before, we have , with and moreover
[TABLE]
Multiplying the above by and sending , we get
[TABLE]
This shows that . ∎
Proof of Theorem 9.
Let denote the set of accumulation points in the space of as . In view of the Ascoli-Arzela theorem, Lemmas 10 and 11 guarantee that the family is relatively compact in . In particular, the set is nonempty. Note by the stability of the viscosity property under uniform convergence that any is a solution of (P0).
If is a singleton, then it is obvious that the whole family converges to the unique element of in as .
We need only to show that is a singleton. It is enough to show that for any and , the inequality holds.
Fix any and . Select sequences and converging to zero so that
[TABLE]
By Corollary 8, there exists a sequence such that
[TABLE]
In view of Theorem 14, we may assume by passing to a subsequence if necessary that, as ,
[TABLE]
for all and for some and, moreover,
[TABLE]
Since and , in view of (26), we have
[TABLE]
which yields after dividing by and then sending along
[TABLE]
Now, note that is a solution of
[TABLE]
and thus, and infer by (25) that
[TABLE]
Sending now yields
[TABLE]
This together with (27) shows that , which completes the proof. ∎
5. The ergodic problem for irreducible matrix
We consider the problem of finding and such that is a solution of
[TABLE]
The pair of such and is also called a solution of (28). This problem is called the ergodic problem in this paper although the term, ergodic problem, should be used only when the condition that holds for some .
Henceforth, denotes the diagonal matrix
[TABLE]
where, as before, .
Throughout this section, we treat the case when
[TABLE]
The irreducibility of is stated as follows: for any nonempty subset of , which is not identical to , there exists a pair of and such that .
The following result has been established in Davini-Zavidovique [DZ1, Theorem 2.10] (see also [CLLN, MT3]).
Proposition 15**.**
Assume (H), (C), (M), (29), and that
[TABLE]
Then there exist and such that the pair is a solution of (28).
We remark that (30) is satisfied if and only if for all , which holds if and only if for all .
The next theorem states the central result of this section.
Theorem 16**.**
Assume (H), (C), (M), (29), and (30). Let be the unique solution of (1) for . Then there exists a constant and a function such that the functions converge to uniformly on as . Moreover, the pair is a solution of (28).
Proof.
Thanks to Proposition 15, there exists a solution of (28). We set , and note that, since for all , the function satisfies, in the viscosity sense,
[TABLE]
By Theorem 9, there exists a solution of in such that, as , in . Noting that is a solution of (28), we finish the proof. ∎
The condition (30) in Proposition 15 can be removed and the following theorem is valid.
Theorem 17**.**
Assume (H), (C), (M), and (29). Then there exist and such that the pair is a solution of (28).
Proof.
For , we set
[TABLE]
and note that is irreducible and (30) holds with replaced by . Note also that for all .
Thanks to Proposition 15, there exist and which solve
[TABLE]
We choose a constant so that and set , respectively. Observe that, since and for all , the functions and are a supersolution and subsolution of
[TABLE]
that is, , respectively. In view of the Perron method, the function given by
[TABLE]
is a solution of (28), with . ∎
Even without the assumption (30), it is immediate from Theorem 9 that, under the hypotheses of Theorem 17, if , then the convergence holds for the whole family of the solutions of (1), with . A typical case when is realized is given by [CLLN, Theorem 4.2](see also [DZ1, MT1]).
6. The ergodic problem for constant matrix
Throughout this section we assume that is a constant matrix, that is, independent of .
The main results in this section are as follows.
Theorem 18**.**
Assume (H), (C), (M), and that is a constant matrix. Then (28) has a solution .
Theorem 19**.**
Under the same hypotheses of Theorem 18, let be a solution of (28) and let be the unique solution of (1) for . Then there exists a function such that the functions converge to uniformly on as . Moreover, the pair is a solution of (28).
Proof.
It is well-known (and easily checked) that due to the monotonicity of , is invertible for any . We set for and also for . Observe that, in the viscosity sense,
[TABLE]
It is clear that satisfies (H) and (C) and that is a solution of in . By Theorem 9, we conclude that there exists a solution of in such that in as . Noting that is a solution of (28), we finish the proof. ∎
For the proof of Theorem 18, we begin with a preliminary remark on the permutations.
For a given permutation , we define the matrix by
[TABLE]
where if and otherwise. Note that and that for any ,
[TABLE]
The system of Hamilton-Jacobi equations
[TABLE]
can be written component-wise as
[TABLE]
By the use of , the system above is expressed as
[TABLE]
and furthermore, if ,
[TABLE]
Set and observe that if is monotone, then
[TABLE]
and
[TABLE]
Consequently, if is monotone, then is monotone as well, and the system (32), by using the permutation matrix , is converted to (33).
Proof of Theorem 18.
It is well-known (see for instance [Var, Section 2.3]) that, given a monotone matrix , one can find a permutation such that
[TABLE]
where, is given by (31), is a diagonal matrix of order and, for , are irreducible matrices of order . In view of the preliminary remark before this proof, to seek for a solution of (28), we may and do assume henceforth has the normal form of the right hand side of (34).
Set
[TABLE]
Notice that . If , then we first show that there exist an -vector and a function such that is a solution of
[TABLE]
where . The system is, in fact, a collection of single equations
[TABLE]
and thus the existence of a solution of (35) is a classical result. Indeed, for each , if , then (36) has a (unique) solution for any choice of . If , then (36) has a solution (see [LPV]). If , then we are done.
Next, assume that (and equivalently, ) and we show that there exist a vector and a function such that is a solution of the system
[TABLE]
where
[TABLE]
According to Proposition 15, there exist and which satisfy (37). This way (by induction), we find and such that
[TABLE]
and satisfies
[TABLE]
where
[TABLE]
We define and by setting
[TABLE]
and observe that
[TABLE]
This completes the proof. ∎
Acknowledgements
The author would like to thank the anonymous referee for useful and critical comments on the original version of this paper, which have helped significantly to improve the presentation. This work was partially supported by the KAKENHI #16H03948, #18H00833, #20H03688, JSPS.
References
