Citation: Piermarco Cannarsa, Rossana Capuani, Pierre Cardaliaguet. C1;1-smoothness of constrained solutions in the calculus of variations with application to mean field games[J]. Mathematics in Engineering, 2019, 1(1): 174-203. doi: 10.3934/Mine.2018.1.174
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The centrality of necessary conditions in optimal control is well-known and has originated an immense literature in the fields of optimization and nonsmooth analysis, see, e.g., [3,16,17,29,33,35].
In control theory, the celebrated Pontryagin Maximum Principle plays the role of the classical Euler-Lagrange equations in the calculus of variations. In the case of unrestricted state space, such conditions provide Lagrange multipliers---the so-called co-states---in the form of solutions to a suitable adjoint system satisfying a certain transversality condition. Among various applications of necessary optimality conditions is the deduction of further regularity properties for minimizers which, a priori, would just be absolutely continuous.
When state constraints are present, a large body of results provide adaptations of the Pontryagin Principle by introducing appropriate corrections in the adjoint system. The price to pay for such extensions usually consists of reduced regularity for optimal trajectories which, due to constraint reactions, turn out to be just Lipschitz continuous while the associated co-states are of bounded variation, see [20].
The maximum principle under state constraints was first established by Dubovitskii and Milyutin [17] (see also the monograph [35] for different forms of such a result). It may happen that the maximum principle is degenerate and does not yield much information (abnormal maximum principle). As explained in [8,10,18,19] in various contexts, the so-called "inward pointing condition" generally ensures the normality of the maximum principle under state constraints. In our setting (calculus of variation problem, with constraints on positions but not on velocities), this will never be an issue. The maximum principle under state constraints generally involves an adjoint state which is the sum of a
Let
Γ={γ∈AC(0,T;Rn): γ(t)∈¯Ω, ∀t∈[0,T]}, |
with the uniform metric. For any
Γ[x]={γ∈Γ:γ(0)=x}. |
We consider the problem of minimizing the classical functional of the calculus of variations
J[γ]=∫T0f(t,γ(t),˙γ(t))dt+g(γ(T)). |
Let
infγ∈Γ[x]J[γ], where J[γ]={∫T0f(t,γ(t),˙γ(t))dt+g(γ(T))}, | (1.1) |
where
{˙γ⋆(t)=−DpH(t,γ⋆(t),p(t)) for all t∈[0,T]˙p(t)=DxH(t,γ⋆(t),p(t))−Λ(t,γ⋆,p)1∂Ω(γ⋆)DbΩ(γ⋆(t))for a. e. t∈[0,T] | (1.2) |
where
infγ∈AC(0,T;Rn)γ(0)=x{∫T0[f(t,γ(t),˙γ(t))+1ϵ dΩ(γ(t))]dt+1δ dΩ(γ(T))+g(γ(T))}. |
Then, we show that all solutions of the penalized problem remain in
A direct consequence of our necessary conditions is the Lipschitz regularity of the value function associated to (1.1) (Proposition 4.1).
Our interest is also motivated by application to mean field games, as we explain below. Mean field games (MFG) theory has been developed simultaneously by Lasry and Lions ([25,26,27]) and by Huang, Malhamé and Caines ([23,24]) in order to study differential games with an infinite number of rational players in competition. The simplest MFG model leads to systems of partial differential equations involving two unknown functions: the value function
This paper is organised as follows. In Section 2, we introduce the notation and recall preliminary results. In Section 3, we derive necessary conditions for the constrained problem. Moreover, we prove the
Throughout this paper we denote by
||A||=maxx∈Rn,|x|=1||Ax|| . |
For any subset
1S(x)={1 x∈S,0x∈Sc. |
We write
Let
‖ϕ‖k,∞:=supx∈U|α|≤k|Dαϕ(x)|<∞ |
Let
The distance function from
dΩ(x):=infy∈¯Ω|x−y| (x∈Rn). |
We define the oriented boundary distance from
bΩ(x)=dΩ(x)−dΩc(x) (x∈Rn). |
We recall that, since the boundary of
bΩ(⋅)∈C2b on Σρ0={y∈B(x,ρ0):x∈∂Ω}. | (2.1) |
Throughout the paper, we suppose that
Take a continuous function
p⋅(y−x)≤f(y)−f(x)+C|y−x|2 for all y that satisfy |y−x|≤ϵ. |
The set of all proximal subetaadients of
The set of all limiting subetaadients of
Lemma 2.1. Let
∂pdΩ(x)=∂dΩ(x)={DbΩ(x) 0<bΩ(x)<ρ0,DbΩ(x)[0,1]x∈∂Ω,0x∈Ω, |
where
The proof is given in the Appendix.
Let
supp(η):={x∈X:η(V)>0 for each neighborhood V of x}. |
We say that a sequence
limi→∞∫Xf(x)dηi(x)=∫Xf(x)dη ∀f∈Cb(X). |
We denote by
d1(m,m′)=sup{∫Xf(x)dm(x)−∫Xf(x)dm′(x) | f:X→R is 1-Lipschitz}, | (2.2) |
for all
Let
d1(m(t),m(s))≤C|t−s|, ∀t,s∈[0,T], | (2.3) |
for some constant
Let
Γ={γ∈AC(0,T;Rn): γ(t)∈¯Ω, ∀t∈[0,T]}. |
For any
Γ[x]={γ∈Γ:γ(0)=x}. |
Let
infγ∈Γ[x]J[γ], where J[γ]={∫T0f(t,γ(t),˙γ(t))dt+g(γ(T))}. | (3.1) |
We denote by
X[x]={γ⋆∈Γ[x]:J[γ⋆]=infΓ[x]J[γ]}. |
We assume that
(g1)
(f0)
|f(t,x,0)|+|Dxf(t,x,0)|+|Dvf(t,x,0)|≤M ∀ (t,x)∈[0,T]×U. | (3.2) |
(f1) For all
Iμ≤D2vvf(t,x,v)≤Iμ, | (3.3) |
||D2vxf(t,x,v)||≤μ(1+|v|), | (3.4) |
for all
(f2) For all
|f(t,x,v)−f(s,x,v)|≤κ(1+|v|2)|t−s| | (3.5) |
|Dvf(t,x,v)−Dvf(s,x,v)|≤κ(1+|v|)|t−s| | (3.6) |
for all
Remark 3.1. By classical results in the calculus of variation (see, e.g., [15, Theorem 11.1i]), there exists at least one minimizer of (3.1) in
In the next lemma we show that (f0)-(f2) imply the useful growth conditions for
Lemma 3.1. Suppose that (f0)-(f2) hold. Then, there exists a positive constant
|Dvf(t,x,v)|≤C(μ,M)(1+|v|), | (3.7) |
|Dxf(t,x,v)|≤C(μ,M)(1+|v|2), | (3.8) |
14μ|v|2−C(μ,M)≤f(t,x,v)≤4μ|v|2+C(μ,M), | (3.9) |
for all
Proof. By (3.2), and by (3.3) one has that
|Dvf(t,x,v)|≤|Dvf(t,x,v)−Dvf(t,x,0)|+|Dvf(t,x,0)|≤∫10|D2vvf(t,x,τv)||v|dτ+|Dvf(t,x,0)|≤μ|v|+M≤C(μ,M)(1+|v|) |
and so (3.7) holds. Furthermore, by (3.2), and by (3.4) we have that
|Dxf(t,x,v)|≤|Dxf(t,x,v)−Dxf(t,x,0)|+|Dxf(t,x,0)|≤∫10|D2xvf(t,x,τv)||v|dτ+M≤μ(1+|v|)|v|+M≤C(μ,M)(1+|v|2). |
Therefore, (3.8) holds. Moreover, fixed
f(t,x,v)=f(t,x,0)+⟨Dvf(t,x,0),v⟩+12⟨D2vvf(t,x,ξ)v,v⟩. |
By (3.2), (3.3), and by (3.7) we have that
−C(μ,M)+14μ|v|2≤−M−C(μ,M)|v|+12μ|v|2≤f(t,x,v)≤M+C(μ,M)|v|+μ2|v|2≤C(μ,M)+4μ|v|2, |
and so (3.9) holds. This completes the proof.
In the next result we show a special property of the minimizers of (3.1).
Lemma 3.2. For any
∫T014μ|˙γ⋆(t)|2dt≤K, |
where
K:=T(C(μ,M)+M)+2maxU|g(x)|. | (3.10) |
Proof. Let
∫T0f(t,γ⋆(t),˙γ⋆(t))dt+g(γ⋆(T))≤∫T0f(t,x,0)dt+g(x)≤Tmax[0,T]×U|f(t,x,0)|+maxU|g(x)|. | (3.11) |
Using (3.2) and (3.9) in (3.11), one has that
∫T014μ|˙γ⋆(t)|2dt≤K, |
where
K:=T(C(μ,M)+M)+2maxU|g(x)|. |
We denote by
H(t,x,p)=supv∈Rn{−⟨p,v⟩−f(t,x,v)},∀ (t,x,p)∈[0,T]×U×Rn. |
Our assumptions on
(H0)
|H(t,x,0)|+|DxH(t,x,0)|+|DpH(t,x,0)|≤M′ ∀ (t,x)∈[0,T]×U. | (3.12) |
(H1) For all
Iμ≤D2ppH(t,x,p)≤Iμ, | (3.13) |
||D2pxH(t,x,p)||≤C(μ,M′)(1+|p|), | (3.14) |
for all
(H2) For all
|H(t,x,p)−H(s,x,p)|≤κC(μ,M′)(1+|p|2)|t−s|, | (3.15) |
|DpH(t,x,p)−DpH(s,x,p)|≤κC(μ,M′)(1+|p|)|t−s|, | (3.16) |
for all
Remark 3.2. Arguing as in Lemma 3.1 we deduce that
|DpH(t,x,p)|≤C(μ,M′)(1+|p|), | (3.17) |
|DxH(t,x,p)|≤C(μ,M′)(1+|p|2), | (3.18) |
14μ|p|2−C(μ,M′)≤H(t,x,p)≤4μ|p|2+C(μ,M′), | (3.19) |
for all
Under the above assumptions on
Theorem 3.1. For any
(i)
(ii) There exist:
(a) a Lipschitz continuous arc
(b) a constant
0≤ν≤max{1,2μ supx∈U|DpH(T,x,Dg(x))|}, |
which satisfy the adjoint system
{˙γ⋆=−DpH(t,γ⋆,p) for all t∈[0,T],˙p=DxH(t,γ⋆,p)−Λ(t,γ⋆,p)1∂Ω(γ⋆)DbΩ(γ⋆)for a.e. t∈[0,T], | (3.20) |
and the transversality condition
p(T)=Dg(γ⋆(T))+νDbΩ(γ⋆(T))1∂Ω(γ⋆(T)), |
where
Moreover,
(iii) the following estimate holds
||˙γ⋆||∞≤L⋆, ∀γ⋆∈X[x], | (3.21) |
where
The (feedback) function
In this section, we prove Theorem 3.1 in the special case of
The proof is based on [12, Theorem 2.1] where the Maximum Principle under state constraints is obtained for a Mayer problem. The reasoning requires several intermediate steps.
Fix
infγ∈AC(0,T;Rn)γ(0)=x{∫T0[f(t,γ(t),˙γ(t))+1ϵ dΩ(γ(t))]dt+1δ dΩ(γ(T))+g(γ(T))}. | (3.22) |
Then, we will show that, for
Observe that the Hamiltonian associated with the penalized problem is given by
Hϵ(t,x,p)=supv∈Rn{−⟨p,v⟩−f(t,x,v)−1ϵ dΩ(x)}=H(t,x,p)−1ϵ dΩ(x), | (3.23) |
for all
By classical results in the calculus of variation (see, e.g., [15, Section 11.2]), there exists at least one mimimizer of (3.22) in
Remark 3.3. Arguing as in Lemma 3.2 we have that, for any
∫T0[14μ|˙γ(t)|2+1ϵ dΩ(γ(t))]dt≤K, | (3.24) |
where
The first step of the proof consists in showing that the solutions of the penalized problem remain in a neighborhood of
Lemma 3.3. Let
∀ x∈¯Ω, γ∈Xϵ,δ[x] ⟹ supt∈[0,T]dΩ(γ(t))≤ρ. | (3.25) |
Proof. We argue by contradiction. Assume that, for some
ϵk↓0, δk>0, tk∈[0,T], xk∈¯Ω, γk∈Xϵk,δk[xk] and dΩ(γk(tk))>ρ, for all k≥1. |
By Remark 3.3, one has that for all
∫T0[14μ|˙γk(t)|2+1ϵk dΩ(γk(t))]dt≤K, |
where
dΩ(γk(tk))−dΩ(γk(s))≤(4μK)1/2|tk−s|1/2, s∈[0,T]. |
Since
dΩ(γk(s))>ρ−(4μK)1/2|tk−s|1/2. |
Hence,
K≥1ϵk∫T0dΩ(γk(t))dt≥1ϵk∫JdΩ(γk(t))dt≥1ϵkρ332μK. |
But the above inequality contradicts the fact that
In the next lemma, we show the necessary conditions for the minimizers of the penalized problem.
Lemma 3.4. Let
(i)
(ii) there exists an arc
{˙γ(t)=−DpH(t,γ(t),p(t)), for all t∈[0,T],˙p(t)=DxH(t,γ(t),p(t))−λ(t)ϵ DbΩ(γ(t)),for a.e. t∈[0,T],p(T)=Dg(γ(T))+βδ DbΩ(γ(T)), | (3.26) |
where
λ(t)∈{{0}ifγ(t)∈Ω,{1}if0<dΩ(γ(t))<ρ,[0,1]ifγ(t)∈∂Ω, | (3.27) |
and
β∈{{0}ifγ(T)∈Ω,{1}if0<dΩ(γ(T))<ρ,[0,1]ifγ(T)∈∂Ω. | (3.28) |
Moreover,
(iii) the function
r(t):=H(t,γ(t),p(t))−1ϵ dΩ(γ(t)), ∀t∈[0,T] |
belongs to
∫T0|˙r(t)|dt≤κ(T+4μK), |
where
(iv) the following estimate holds
|p(t)|2≤4μ[1ϵdΩ(γ(t))+C1δ2], ∀t∈[0,T], | (3.29) |
where
Proof. In order to use the Maximum Principle in the version of [35, Theorem 8.7.1], we rewrite (3.22) as a Mayer problem in a higher dimensional state space. Define
X(t)=(γ(t)z(t)), |
where
{˙X(t)=(˙γ(t)˙z(t))=Fϵ(t,X(t),u(t)),X(0)=(x00). |
where
Fϵ(t,X,u)=(uLϵ(t,x,u)) |
and
min{Φ(Xu(T)):u∈L1}, | (3.30) |
where
Hϵ(t,X,P,u)=−⟨P,Fϵ(t,X,u)⟩,∀(t,X,P,u)∈[0,T]×Rn+1×Rn+1×Rn. |
We observe that, as
(ⅰ)
(ⅱ)
(ⅲ)
(ⅳ)
(ⅴ)
where
(p,b,λ0)≢(0,0,0), | (3.31) |
(˙r(t),˙p(t))∈−b(t) co ∂t,xLϵ(t,γ(t),˙γ(t)), | (3.32) |
˙b(t)=0, | (3.33) |
p(T)∈λ0 ∂(g+1δ dΩ)(γ(T)), | (3.34) |
b(T)=λ0, | (3.35) |
r(t)=Hϵ(t,γ(t),p(t)), | (3.36) |
where
Note that the Weierstrass Condition (ⅳ) becomes
−⟨p(t),˙γ(t)⟩−f(t,γ(t),˙γ(t))=supu∈Rn{−⟨p(t),u⟩−f(t,γ(t),u)}. | (3.37) |
Therefore
˙γ(t)=−DpH(t,γ(t),p(t)),a.e.t∈[0,T]. | (3.38) |
By Lemma 2.1, by the definition of
∂t,xLϵ(t,x,u)⊂{[−κ(1+|u|2),κ(1+|u|2)]×Dxf(t,x,u)ifx∈Ω,[−κ(1+|u|2),κ(1+|u|2)]×(Dxf(t,x,u)+1ϵ DbΩ(x))if0<bΩ(x)<ρ,[−κ(1+|u|2),κ(1+|u|2)]×(Dxf(t,x,u)+1ϵ[0,1] DbΩ(x))ifx∈∂Ω. |
Thus (3.32) implies that there exists
|˙r(t)|≤κ(1+|˙γ(t)|2), ∀t∈[0,T], | (3.39) |
˙p(t)=−Dxf(t,γ(t),˙γ(t))−λ(t)ϵ DbΩ(γ(t)), a.e. t∈[0,T]. | (3.40) |
Hence, by (3.39), and by Remark 3.3 we conclude that
∫T0|˙r(t)|dt≤κ∫T0(1+|˙γ(t)|2)dt≤κ(T+4μK). | (3.41) |
Moreover, by Lemma 2.1, and by assumption on
∂(g+1δ dΩ)(x)⊂{Dg(x)ifx∈Ω,Dg(x)+1δ DbΩ(x)if0<bΩ(x)<ρ,Dg(x)+1δ[0,1] DbΩ(x)ifx∈∂Ω. |
So, by (3.34), there exists
p(T)=Dg(x)+βδ DbΩ(x). | (3.42) |
Finally, by well-known properties of the Legendre transform one has that
DxH(t,x,p)=−Dxf(t,x,−DpH(t,x,p)). |
So, recalling (3.38), (3.40) can be rewritten as
˙p(t)=DxH(t,γ(t),p(t))−λ(t)ϵ DbΩ(γ(t)),a.e. t∈[0,T]. |
We have to prove estimate (3.29). Recalling (3.23) and (3.19), we have that
Hϵ(t,γ(t),p(t))=H(t,γ(t),p(t))−1ϵ dΩ(γ(t))≥14μ|p(t)|2−C(μ,M′)−1ϵ dΩ(γ(t)). |
So, using (3.41) one has that
|Hϵ(T,γ(T),p(T))−Hϵ(t,γ(t),p(t))|=|r(T)−r(t)|≤∫Tt|˙r(s)|ds≤κ(T+4μK). |
Moreover, (3.42) implies that
14μ|p(t)|2−C(μ,M′)−1ϵ dΩ(γ(t))≤Hϵ(t,γ(t),p(t))≤Hϵ(T,γ(T),p(T))+κ(T+4μK)≤4μ|p(T)|2+C(μ,M′)+κ(T+4μK)≤8μ[1δ2+||Dg||2∞]+C(μ,M′)+κ(T+4μK). |
Hence,
|p(t)|2≤4μ[1ϵdΩ(γ(t))+C1δ2], |
where
Finally, by the regularity of
Lemma 3.5. Let
Proof. Let
˙γ(t)=−DpH(t,γ(t),p(t)),˙p(t)=DxH(t,γ(t),p(t))−1ϵDbΩ(γ(t)), |
for
In the next two lemmas, we show that, for
ϵ0=ϵ(ρ0), where ρ0 is such that (2.1) holds and ϵ(⋅) is given by Lemma 3.3. |
Lemma 3.6. Let
δ=12μN∧1, | (3.43) |
where
N=supx∈Rn|DpH(T,x,Dg(x))|. |
Fix any
⟨˙γ(T),DbΩ(γ(T))⟩≤0. |
Proof. As
⟨DpH(T,γ(T),p(T)),DbΩ(γ(T))⟩=⟨DpH(T,γ(T),Dg(γ(T))),DbΩ(γ(T))⟩+⟨DpH(T,γ(T),Dg(γ(T))+1δ DbΩ(γ(T)))−DpH(T,γ(T),Dg(γ(T))),DbΩ(γ(T))⟩. |
Recalling that
⟨DpH(T,γ(T),Dg(γ(T))+1δ DbΩ(γ(T)))−DpH(T,γ(T),Dg(γ(T))),1δ DbΩ(γ(T))⟩≥12μ1δ2 |DbΩ(γ(T))|2=12δ2μ. |
So,
⟨DpH(T,γ(T),p(T)),DbΩ(γ(T))⟩≥12δμ−|DpH(T,γ(T),Dg(γ(T)))|. |
Therefore, we obtain
⟨˙γ(T),DbΩ(γ(T))⟩=−⟨DpH(T,γ(T),p(T)),DbΩ(γ(T))⟩≤−12δμ+|DpH(T,γ(T),Dg(γ(T)))|. |
Thus, choosing
Lemma 3.7. Fix
∀x∈¯Ω, γ∈Xϵ,δ[x] ⟹ γ(t)∈¯Ω ∀t∈[0,T]. |
Proof. We argue by contradiction. Assume that there exist sequences
ϵk↓0, tk∈[0,T], xk∈¯Ω, γk∈Xϵk,δ[xk] and γk(tk)∉¯Ω, for all k≥1. | (3.44) |
Then, for each
{dΩ(γk(ak))=0,dΩ(γk(t))>0 t∈(ak,bk),dΩ(γk(bk))=0 or else bk=T. |
Let
dΩ(γk(¯tk))=maxt∈[ak,bk]dΩ(γk(t)). |
We note that, by Lemma 3.5,
Step 1
We claim that
d2dt2dΩ(γk(t))|t=¯tk≤0. | (3.45) |
Indeed, (3.45) is trivial if
ddtdΩ(γk(t))|t=¯tk≥0. |
Moreover, Lemma 3.6 yields
ddtdΩ(γk(t))|t=¯tk≤0. |
So,
ddtdΩ(γk(t))|t=¯tk=0, |
and we have that (3.45) holds true at
Step 2
Now, we prove that
1μϵk≤C(μ,M′,κ)[1+4μC1δ2+4μϵk dΩ(γk(¯tk))], ∀k≥1, | (3.46) |
where
¨γ(¯tk)=−D2ptH(˜tk,γ(˜tk),p(˜tk))−⟨D2pxH(˜tk,γ(˜tk),p(˜tk)),˙γ(˜tk)⟩−⟨D2ppH(˜tk,γ(˜tk),p(˜tk)),˙p(˜tk)⟩. | (3.47) |
Developing the second order derivative of
0≥⟨D2dΩ(γ(˜tk))˙γ(˜tk),˙γ(˜tk)⟩+⟨DdΩ(γ(˜tk)),¨γ(˜tk)⟩=⟨D2dΩ(γ(˜tk))DpH(˜tk,γ(˜tk),p(˜tk)),DpH(˜tk,γ(˜tk),p(˜tk))⟩−⟨DdΩ(γ(˜tk)),D2ptH(˜tk,γ(˜tk),p(˜tk))⟩+⟨DdΩ(γ(˜tk)),D2pxH(˜tk,γ(˜tk),p(˜tk))DpH(˜tk,γ(˜tk),p(˜tk))⟩−⟨DdΩ(γ(˜tk)),D2ppH(˜tk,γ(˜tk),p(˜tk))DxH(˜tk,γ(˜tk),p(˜tk))⟩+1ϵ⟨DdΩ(γ(˜tk)),D2ppH(˜tk,γ(˜tk),p(˜tk))DdΩ(γ(˜tk))⟩. |
We now use the growth properties of
1μϵk≤C(μ,M′)(1+|p(˜tk)|)2+κC(μ,M′)(1+|p(˜tk)|)≤C(μ,M′,κ)(1+|p(˜tk)|2), |
where the constant
1μϵk≤C(μ,M′,κ)[1+4μC1δ2+4μϵkdΩ(γ(˜tk))], ∀ k≥1, |
where
Conclusion
Let
supt∈[0,T]dΩ(γ(t))≤ρ, ∀γ∈Xϵ,δ[x]. |
Hence, using (3.46), we deduce that
12μϵk≤4C(μ,M′,κ)[1+4μC1δ2]. |
Since the above inequality fails for
An obvious consequence of Lemma 3.7 is the following:
Corollary 3.1. Fix
We are now ready to complete the proof of Theorem 3.1.
Proof of Theorem 3.1. Let
p(T)=Dg(γ⋆(T))+ν DbΩ(γ⋆(T)). | (3.48) |
By Lemma 3.4
˙γ⋆(t)=−DpH(t,γ⋆(t),p(t)), ∀ t∈[0,T]. | (3.49) |
Moreover,
|p(t)|≤2√μC1δ, ∀t∈[0,T], |
where
||˙γ⋆||∞=supt∈[0,T]|DpH(t,γ⋆(t),p(t))|≤C(μ,M′)(supt∈[0,T]|p(t)|+1)≤C(μ,M′)(2√μC1δ+1))=L⋆, |
where
Finally, we want to find an explicit expression for
D={t∈[0,T]:γ⋆(t)∈∂Ω}andDρ0={t∈[0,T]:|bΩ(γ⋆(t))|<ρ0}, |
where
˙ψ(t)=⟨DbΩ(γ⋆(t)),˙γ⋆(t)⟩=⟨DbΩ(γ⋆(t)),−DpH(t,γ⋆(t),p(t))⟩. |
Since
¨ψ(t)=−⟨D2bΩ(γ⋆(t))˙γ⋆(t),DpH(t,γ⋆(t),p(t))⟩−⟨DbΩ(γ⋆(t)),D2ptH(t,γ⋆(t),p(t))⟩−⟨DbΩ(γ⋆(t)),D2pxH(t,γ⋆(t),p(t))˙γ⋆(t)⟩−⟨DbΩ(γ⋆(t)),D2ppH(t,γ⋆(t),p(t))˙p(t)⟩=⟨D2bΩ(γ⋆(t))DpH(t,γ⋆(t),p(t)),DpH(t,γ⋆(t),p(t))⟩−⟨DbΩ(γ⋆(t)),D2ptH(t,γ⋆(t),p(t))⟩+⟨DbΩ(γ⋆(t)),D2pxH(t,γ⋆(t),p(t))DpH(t,γ⋆(t),p(t))⟩−⟨DbΩ(γ⋆(t)),D2ppH(t,γ⋆(t),p(t))DxH(t,γ⋆(t),p(t))⟩+λ(t)ϵ ⟨DbΩ(γ⋆(t)),D2ppH(t,γ⋆(t),p(t))DbΩ(γ⋆(t))⟩. |
Let
⟨DbΩ(γ⋆(t)),D2ppH(t,γ⋆(t),p(t))DbΩ(γ⋆(t))⟩>0,a.e.t∈Dρ0. |
So, for a.e.
λ(t)ϵ=1⟨DbΩ(γ⋆(t)),D2ppH(t,γ⋆(t),p(t))DbΩ(γ⋆(t))⟩ [⟨DbΩ(γ⋆(t)),D2ptH(t,γ⋆(t),p(t))⟩−⟨D2bΩ(γ⋆(t))DpH(t,γ⋆(t),p(t)),DpH(t,γ⋆(t),p(t))⟩−⟨DbΩ(γ⋆(t)),D2pxH(t,γ⋆(t),p(t))DpH(t,γ⋆(t),p(t))⟩+⟨DbΩ(γ⋆(t)),D2ppH(t,γ⋆(t),p(t))DxH(t,γ⋆(t),p(t))⟩]. |
Since
Remark 3.4. The above proof gives a representation of
Λ(t,x,p)=1θ(t,x,p) [−⟨D2bΩ(x)DpH(t,x,p),DpH(t,x,p)⟩−⟨DbΩ(x),D2ptH(t,x,p)⟩−⟨DbΩ(x),D2pxH(t,x,p)DpH(t,x,p)⟩+⟨DbΩ(x),D2ppH(t,x,p)DxH(t,x,p)⟩], |
where
We now want to remove the extra assumption
{ξ(x)=0 if x∈(−∞,13],0<ξ(x)<1if x∈(13,23),ξ=1if x∈[23,+∞). | (3.50) |
Lemma 3.8. Let
σ0=dist(¯Ω,Rn∖U)>0. |
Suppose that
˜f(t,x,v)=ξ(bΩ(x)σ)|v|22+(1−ξ(bΩ(x)σ))f(t,x,v), ∀ (t,x,v)∈[0,T]×Rn×Rn, |
that satisfies conditions (f0)-(f2) with
˜g(x)=(1−ξ(bΩ(x)σ))g(x), ∀x∈Rn, |
that satisfies condition (g1) with
Note that, since
Proof. By construction we note that
Dv˜f(t,x,v)=ξ(bΩ(x)σ)v+(1−ξ(bΩ(x)σ))Dvf(t,x,v), |
and
D2vv˜f(t,x,v)=ξ(bΩ(x)σ)I+(1−ξ(bΩ(x)σ))D2vvf(t,x,v). |
Hence, by the definition of
(1∧1μ)I≤D2vv˜f(t,x,v)≤(1∨μ)I, ∀ (t,x,v)∈[0,T]×Rn×Rn. |
Since
Moreover, since
Dx(Dv˜f(t,x,v))=˙ξ(bΩ(x)σ)v⊗DbΩ(x)σ+(1−ξ(bΩ(x)σ))D2vxf(t,x,v)−˙ξ(bΩ(x)σ)Dvf(t,x,v)⊗DbΩ(x)σ, |
and by (3.4) we obtain that
||D2vx˜f(t,x,v)||≤C(μ,M)(1+|v|) ∀(t,x,v)∈[0,T]×Rn×Rn. |
For all
|˜f(t,x,v)−˜f(s,x,v)|=|(1−ξ(bΩ(x)σ))[f(t,x,v)−f(s,x,v)]|≤κ(1+|v|2)|t−s| |
for all
|Dv˜f(t,x,v))−Dv˜f(s,x,v))|≤|(1−ξ(bΩ(x)σ))[Dvf(t,x,v)−Dvf(s,x,v))]|≤κ(1+|v|)|t−s|, |
for all
Finally, by the regularity of
Suppose that
u(t,x)=infγ∈Γγ(t)=x∫Ttf(s,γ(s),˙γ(s))ds+g(γ(T)). | (4.1) |
Proposition 4.1. Let
Proof. First, we shall prove that
{ˉγ(t)=y,˙ˉγ(s)=˙γ(s)+x−yτ if s∈[t,t+τ] a.e.,˙ˉγ(s)=˙γ(s) otherwise, |
where
(a)
(b)
(c)
(d)
Indeed, by the definition of
ˉγ(t+τ)−ˉγ(t)=ˉγ(t+τ)−y=∫t+τt(˙γ(s)+x−yτ)ds=γ(t+τ)−y, |
and this gives (a). Moreover, by (a), and by the definition of
|ˉγ(s)−γ(s)|≤|y−x+∫st(˙ˉγ(σ)−˙γ(σ))dσ|=|y−x+∫stx−yτdσ|≤|y−x| |
and so (c) holds. Since
|ˉγ(s)−x0|≤|ˉγ(s)−y|+|y−x0|≤|∫st˙ˉγ(σ)dσ|+r≤∫st |˙γ(σ)+x−yτ|dσ+r≤∫st[|˙γ(σ)|+|x−y|τ]dσ+r≤L⋆(s−t)+|x−y|τ(s−t)+r≤L⋆τ+|x−y|+r. |
Recalling that
|ˉγ(s)−x0|≤|x−y|2+|x−y|+r≤4r. |
Therefore,
Now, owing to the dynamic programming principle, by (a) one has that
u(t,y)≤∫t+τtf(s,ˉγ(s),˙ˉγ(s))ds+u(t+τ,γ(t+τ)). | (4.2) |
Since
u(t,y)≤u(t,x)+∫t+τt[f(s,ˉγ(s),˙ˉγ(s))−f(s,γ(s),˙γ(s))]ds. |
By (3.7), (3.8), and the definition of
|f(s,ˉγ(s),˙ˉγ(s))−f(s,γ(s),˙γ(s))|≤|f(s,ˉγ(s),˙ˉγ(s))−f(s,ˉγ(s),˙γ(s))|+|f(s,ˉγ(s),˙γ(s))−f(s,γ(s),˙γ(s))|≤∫10|⟨Dvf(s,ˉγ(s),λ˙ˉγ(s)+(1−λ)˙γ(s)),˙ˉγ(s)−˙γ(s)⟩|dλ+∫10|Dxf(s,λˉγ(s)+(1−λ)γ(s),˙γ(s)),ˉγ(s)−γ(s)⟩|dλ≤C(μ,M)|˙ˉγ(s)−˙γ(s)|∫10(1+|λ˙ˉγ(s)+(1−λ)˙γ(s)|)dλ+C(μ,M)|ˉγ(s)−γ(s)|∫10(1+|˙γ(s)|2)dλ. |
By Theorem 3.1 one has that
∫10(1+|λ˙ˉγ(s)+(1−λ)˙γ(s)|)dλ≤1+4L⋆, | (4.3) |
∫10(1+|˙γ(s)|2)dλ≤1+(L⋆)2. | (4.4) |
Using (4.3), (4.4), and (c), by the definition of
|f(s,ˉγ(s),˙ˉγ(s))−f(s,γ(s),˙γ(s))|≤C(μ,M)(1+4L⋆)|x−y|τ+C(μ,M)(1+(L⋆)2)|x−y|, | (4.5) |
for a.e.
u(t,y)≤u(t,x)+C(μ,M)(1+4L⋆)∫t+τt|x−y|τds+C(μ,M)(1+(L⋆)2)∫t+τt|x−y|ds≤u(t,x)+C(μ,M)(1+4L⋆)|x−y|+τC(μ,M)(1+(L⋆)2)|x−y|≤u(t,x)+CL⋆|x−y| |
where
In order to prove Lipschitz continuity in time, let
|u(t2,x)−u(t1,x)|≤|u(t2,x)−u(t2,γ(t2))|+|u(t2,γ(t2))−u(t1,x)|. | (4.6) |
The first term on the right-side of (4.6) can be estimated using the Lipschitz continuity in space of
|u(t2,x)−u(t2,γ(t2))|≤CL⋆|x−γ(t2)|≤CL⋆∫t2t1|˙γ(s)|ds≤L⋆CL⋆(t2−t1). | (4.7) |
We only have to estimate the second term on the right-side of (4.6). By the dynamic programming principle, (3.9), and the assumptions on
|u(t2,γ(t2))−u(t1,x)|=|∫t2t1f(s,γ(s),˙γ(s))ds|≤∫t2t1|f(s,γ(s),˙γ(s))|ds≤∫t2t1[C(μ,M)+4μ|˙γ(s)|2]ds≤[C(μ,M)+4μL⋆](t2−t1) | (4.8) |
Using (4.7) and (4.8) to bound the right-hand side of (4.6), we obtain that
In this section we want to apply Theorem 3.1 to a mean field game (MFG) problem with state constraints. Such a problem was studied in [11], where the existence and uniqueness of constrained equilibria was obtained under fairly general assumptions on the data. Here, we will apply our necessary conditions to deduce the existence of more regular equilibria than those constructed in [11], assuming the data
Assumptions
Let
(D1) For all
|F(x,m1)−F(x,m2)|+|G(x,m1)−G(x,m2)|≤κd1(m1,m2), | (4.9) |
for any
(D2) For all
|DxF(x,m)|+|DxG(x,m)|≤κ, ∀ x∈U, ∀ m∈P(¯Ω). |
Let
(L0)
|L(x,0)|+|DxL(x,0)|+|DvL(x,0)|≤M, ∀ x∈U. | (4.10) |
(L1)
Iμ≤D2vvL(x,v)≤Iμ, | (4.11) |
||D2vxL(x,v)||≤μ(1+|v|), | (4.12) |
for all
Remark 4.1. (ⅰ)
(ⅱ) Arguing as Lemma 3.1 we deduce that there exists a positive constant
|DxL(x,v)|≤C(μ,M)(1+|v|2), | (4.13) |
|DvL(x,v)|≤C(μ,M)(1+|v|), | (4.14) |
|v|24μ−C(μ,M)≤L(x,v)≤4μ|v|2+C(μ,M), | (4.15) |
for all
Let
H(t,x,p)=HL(x,p)−F(x,m(t)), ∀ (t,x,p)∈[0,T]×U×Rn, |
where
HL(x,p)=supv∈Rn{−⟨p,v⟩−L(x,v)}, ∀ (x,p)∈U×Rn. |
The assumptions on
1.
|HL(x,0)|+|DxHL(x,0)|+|DpHL(x,0)|≤M′, ∀x∈U. | (4.16) |
2.
Iμ≤DppHL(x,p)≤Iμ, ∀ (x,p)∈U×Rn, | (4.17) |
||D2pxHL(x,p)||≤C(μ,M′)(1+|p|), ∀ (x,p)∈U×Rn, | (4.18) |
where
For any
et(γ)=γ(t), ∀γ∈Γ. |
For any
mη(t)=et♯η ∀t∈[0,T]. |
Remark 4.2. We observe that for any
(ⅰ)
(ⅱ) Let
For any fixed
Jη[γ]=∫T0[L(γ(t),˙γ(t))+F(γ(t),mη(t))] dt+G(γ(T),mη(T)), ∀γ∈Γ. |
For all
Γη[x]={γ∈Γ[x]:Jη[γ]=minΓ[x]Jη}. |
It is shown in [11] that, for every
Definition 4.1. Let
supp(η)⊆⋃x∈¯ΩΓη[x]. |
Let
Definition 4.2. Let
PLipm0(Γ′)={η∈Pm0(Γ′):mη∈Lip(0,T;P(¯Ω))}. |
Remark 4.3. We note that
j(x)(t)=x ∀t∈[0,T]. |
Then,
η:=j♯m0 |
is a Borel probability measure on
In order to show that
e0♯η(B)=η(e−10(B))=2∑i=1λiηi(e−10(B))=2∑i=1λie0♯ηi(B)=2∑i=1λim0(B)=m0(B). |
So,
In the next result, we apply Theorem 3.1 to prove a useful property of minimizers of
Proposition 4.2. Let
||˙γ||∞≤L0, ∀γ∈Γη[x], | (4.19) |
where
Proof. Let
||˙γ||∞≤L0, ∀γ∈Γη[x], |
where
We denote by
ΓL0={γ∈Γ:||˙γ||∞≤L0}. | (4.20) |
Lemma 4.1. Let
Proof. Arguing as in Remark 4.3, we obtain that
Let
d1(mη(t2),mη(t1))=sup{∫¯Ωϕ(x)(mη(t2,dx)−mη(t1,dx)) | ϕ:¯Ω→R is 1-Lipschitz}. |
Since
∫¯Ωϕ(x)(mη(t2,dx)−mη(t1,dx))=∫Γ[ϕ(et2(γ))−ϕ(et1(γ))]dη(γ)=∫Γ[ϕ(γ(t2))−ϕ(γ(t1))]dη(γ)≤∫Γ|γ(t2)−γ(t1)|dη(γ). |
Since
∫Γ|γ(t2)−γ(t1)|dη(γ)≤L0∫Γ|t2−t1|dη(γ)=L0|t2−t1| |
and so
In the next result, we deduce the existence of more regular equilibria than those constructed in [11].
Theorem 4.1. Let
Proof. First of all, we recall that for any
*We say that
{η(dγ)=∫¯Ωηx(dγ)m0(dx),supp(ηx)⊂Γ[x] m0−a.e. x∈¯Ω | (4.21) |
(see, e.g., [2, Theorem 5.3.1]). Proceeding as in [11], we introduce the set-valued map
E:Pm0(Γ)⇉Pm0(Γ), |
by defining, for any
E(η)={ˆη∈Pm0(Γ):supp(ˆηx)⊆Γη[x] m0−a.e. x∈¯Ω}. | (4.22) |
We recall that, by [11, Lemma 3.6], the map
Now, we consider the restriction
E0:PLipm0(ΓL0)⇉Pm0(Γ), E0(η)=E(η) ∀η∈PLipm0(ΓL0). |
We will show that the set-valued map
E0(PLipm0(ΓL0))⊆PLipm0(ΓL0). | (4.23) |
Indeed, let
Γη[x]⊂ΓL0 ∀x∈¯Ω, |
and by definition of
supp(ˆη)⊂ΓL0. |
So,
Since
We recall the definition of a mild solution of the constrained MFG problem, given in [11].
Definition 4.3. We say that
(i)
(ii) u is given by
u(t,x)=infγ∈Γγ(t)=x{∫Tt[L(γ(s),˙γ(s))+F(γ(s),m(s))] ds+G(γ(T),m(T))}, | (4.24) |
for
Theorem 4.2. Let
(i)
(ii)
The question of the Lipschitz continuity up to the boundary of the value function under state constraints was addressed in [28] and [34], for stationary problems, and in a very large literature that has been published since. We refer to the survey paper [20] for references.
Proof. Let
Remark 4.4. Recall that
∫¯Ω(F(x,m1)−F(x,m2))d(m1−m2)(x) ≥ 0, | (4.25) |
for any
Suppose that
In this Appendix we prove Lemma 2.1. The only case which needs to be analyzed is when
dΩ(y)−dΩ(x)−⟨p,y−x⟩≥C|y−x|2, for any y such that |y−x|≤ϵ, | (5.1) |
for some constant
dΩ(y)−dΩ(x)−⟨DbΩ(x),y−x⟩≥bΩ(y)−bΩ(x)−⟨DbΩ(x),y−x⟩≥C|y−x|2. |
This shows that
dΩ(y)−dΩ(x)−⟨λDbΩ(x),y−x⟩≥λ(dΩ(y)−dΩ(x)−⟨DbΩ(x),y−x⟩) ∀λ∈[0,1], |
we further obtain the inclusion
DbΩ(x)[0,1]⊂∂dΩ(x). |
Next, in order to show the reverse inclusion, let
bΩ(y)−bΩ(x)−⟨p,y−x⟩≥C|y−x|2, |y−x|≤ϵ. | (5.2) |
Since
bΩ(y)−bΩ(x)≤⟨DbΩ(x),y−x⟩+C|y−x|2 | (5.3) |
for some constant
⟨DbΩ(x)−p,y−x|y−x|⟩≥C|y−x|. |
Hence, passing to the limit for
⟨DbΩ(x)−p,v⟩≥0, ∀v∈TΩc(x), |
where
DbΩ(x)−p=λv(x), |
where
p=(1−λ)DbΩ(x). |
Now, we prove that
0=dΩ(y)≥(1−λ)⟨DbΩ(x),y−x⟩+C|y−x|2. |
Hence,
(1−λ)⟨DbΩ(x),y−x|y−x|⟩≤−C|y−x|. |
Passing to the limit for
(1−λ)⟨DbΩ(x),w⟩≤0, ∀w∈T¯Ω(x), |
where
This work was partly supported by the University of Rome Tor Vergata (Consolidate the Foundations 2015) and by the Istituto Nazionale di Alta Matematica "F. Severi" (GNAMPA 2016 Research Projects). The authors acknowledge the MIUR Excellence Department Project awarded to the Department of Mathematics, University of Rome Tor Vergata, CUP E83C18000100006. The second author is grateful to the Universitá Italo Francese (Vinci Project 2015).
The authors declare no conflict of interest.
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