Proximal Calculus and Universal Feedback Strategies in Two Person Non-Zero Sum Differential Games

0.13 MB·English

by  Yurii Averboukh

Proximal Calculus and Universal Feedback Strategies in
Two Person Non-Zero Sum Differential Games∗
Yurii Averboukh
Institute of Mathematics and Mechanics UrB RAS
S. Kovalevskayastreet 16
620219,GSP-384, Ekaterinburg Russia
1
[email protected], [email protected]
1
0
2

n
a
J Abstract

5
2 In this paper we introduce the discontinuous universal feedback for the problem of

Nash equilibrium in two person non-zero sum differential game. We assume that there
]
C exist functions satisfying some conditions analogous to the infinitesimal conditions on value
O function in zero sum differential games. Under this assumption we prove the existence of
. universal feedback Nash equilibrium.
h
t Mathematics Subject Classification (2010): 49N70; 49N35.
a
m Keywords: Nash equilibrium, differential games, optimal feedback syntheses.
[

1 1 Introduction
v
5
2 In this paper we introduce the discontinuous universal feedback for the problem of
9
Nash equilibrium in two person non-zero sum differential game. This approach is close to
4
. the extremal shift rule suggested by N.N. Krasovskii and A.I. Subbotin for the zero-sum
1
0 differential games. The extremal shift rule was suggested in [1] to prove the existence of
1 value function. The value function of zero sum differential game is the viscosity solution
1
of corresponding Hamilton-Jacobi equation. The universal feedback synthesis is based on
:
v the properties of value function and the definition of viscosity solutions. N.N. Krasovskii
i
X designed the universal feedback ε-strategies [2]. His scheme use the minimization of the
r value function of the game in ε-neighborhood. A.I. Subbotin introduced the universal
a
feedback using the notion of quasigradient [3], [4]. In the paper of F. H. Clarke, Yu. S.
Ledyaev, and A. I. Subbotin [5] the aiming in the direction of proximal subgradients of the
value function was studied. Constructed strategy is universal also.
The problem of Nash equilibrium in the differential game is connected with the Cauchy
problem for the system of Hamilton-Jacobi equations [6]. In the general case the Hamilto-
nians are discontinuous. If the smooth solution of Cauchy problem exists and the controls
of players are continuous then there exists universal feedback Nash equilibrium. If the so-
lutions doesn’t exist or it isn’t smooth then the situation becomes more complicated. A
particular case when the universal feedback Nash equilibrium exists were considered by P.
∗
TheworkissupportedbyRFBR(grantNo09-01-00436-a),GrantofPresidentofRussianFederation(project
MK-7320.2010.1),RAS Presidium Programof Fundamental Researches “Mathematical Theory of Control”.
1
Cardaliaguet and S. Plaskacz [7]. The approach based on the solution of the system of con-
servation laws is developed by A. Bressan and W. Shen[8], [9]. They studyonedimensional
games under condition of hyperbolicity and design the universal feedback strategies.
In this paper we consider the general case of finite horizon non-zero sum differential
game. We assume that there exist functions satisfying some conditions analogous to the
infinitesimal conditions on value function in zero sum differential games [4], [5]. Under this
assumption we prove the existence of universal feedback Nash equilibrium.
2 Definitions and Designations
We consider the doubly controlled system
x˙ = f(t,x,u,v), t ∈ [t ,ϑ ], x ∈Rn, u∈ P, v ∈ Q. (1)
0 0
Here u and v are controls of the Player I and the Player II respectively. The purposes of
the Players are nonantagonistic. The Player I wants to maximize the functional
ϑ0
Λ (x(·),u(·),v(·)) = σ (x(ϑ ))+ g (ξ,x(ξ),u(ξ),v(ξ))dξ,
1 1 0 1
Zt
the Player II wants to maximize the functional
ϑ0
Λ (x(·),u(·),v(·)) = σ (x(ϑ ))+ g (ξ,x(ξ),u(ξ),v(ξ))dξ.
2 2 0 2
Zt
WeassumethatthesetsP andQarecompacts,thefunctionf,σ andσ arecontinuous,
1 2
moreoverf,g andg areLipschitzcontinuouswithrespecttothephasevariable,andsatisfy
1 2
the sublinear growth condition with respect to x.
We usethe discontinuous feedback control scheme firstsuggested by N.N. Krasovskii for
zero-sum differential games. We consider the two cases:
• the Players choose feedback strategies and make consistent the corrective moments;
• one of the Players deviates.
Feedback strategyofthePlayerIisafunctionU(t,x)withvaluesinP,feedbackstrategy
of the Player II is a function V(t,x) with the values in Q.
Let us consider the first case. We assume that the Player I chooses the strategy U,
the Player II chooses the strategy V. Let (t ,x ) be an initial position. Suppose that the
∗ ∗
Players choose the partion of time segment [t ,ϑ ] ∆ = {τ }m . Further let d(∆) denote
∗ 0 k k=1
the fineness of partition ∆. Let xc[·,t ,x ,U,V,∆] be an unique solution of the problem
∗ ∗
t
x[t]= x[τ ]+ f(ξ,x[ξ],u(τ ,x[τ ]),v(τ ,x[τ ]))dξ, t ∈ [τ ,τ ), i= 0,m−1, x[t ] = x .
i i i i i i i+1 ∗ ∗
Zτi
The value of cost functional of i-th Player in this case is equal to
Λc(t ,x ,U,V,∆) = σ (xc[ϑ ,t ,x ,U,V,∆])+
i ∗ ∗ i 0 ∗ ∗
m−1 τj+1
g (t,xc[t,t ,x ,U,V,∆],u(τ ,xc[τ ,t ,x ,U,V,∆]),v(τ ,xc[τ ,t ,x ,U,V,∆])dt.
i ∗ ∗ i j ∗ ∗ i j ∗ ∗
j=0 Zτj
X
Now we suppose that the Player II chooses a measurable control v[·]. Let
x1[·,t ,x ,U,∆,v[·]] be an unique solution of the problem
∗ ∗
t
x[t] = x[τ ]+ f(ξ,x[ξ],u(τ ,x[τ ]),v[ξ])dξ, t ∈ [τ ,τ ), i= 0,m−1, x[t ] = x .
i i i i i+1 ∗ ∗
Zτi
2
The value of cost functional of Player II in this case is equal to
Λd(t ,x ,U,∆,v[·]) = σ (x1[ϑ ,t ,x ,U,∆,v[·]])+
2 ∗ ∗ i 0 ∗ ∗
m−1 τj+1
g (t,x1[t,t ,x ,U,∆,v[·]],u(τ ,xc[τ ,t ,x ,U,∆,v[·]]),v(t)dt.
i ∗ ∗ i j ∗ ∗
j=0 Zτj
X
In thesame way the case when the Player IIdeviates is considered. Denote the motion gen-
eratedbythestrategyofthePlayerIIandthecontrolofthePlayerIbyx2[·,t ,x ,V,∆,u[·]].
∗ ∗
The value of cost functional of Player I in this case is equal to
Λd(t ,x ,V,∆,u[·]) = σ (x2[ϑ ,t ,x ,V,∆,u[·]])+
1 ∗ ∗ i 0 ∗ ∗
m−1 τj+1
g (t,x2[t,t ,x ,V,∆,u[·]],u[t],x2[τ ,x2[τ ,t ,x ,V,∆,u[·]])dt.
i ∗ ∗ i j ∗ ∗
j=0 Zτj
X
Let us introduce the following values:
Υ (t ,x ,U,V) = liminfΛc(t ,x ,U,V,∆),
i ∗ ∗ i ∗ ∗
d(∆)↓0
Γ (t ,x ,V)= limsupsupΛd(t ,x ,V,∆,u[·]),
1 ∗ ∗ 1 ∗ ∗
d(∆)↓0 u[·]
Γ (t ,x ,U) = limsupsupΛd(t ,x ,U,∆,v[·]).
2 ∗ ∗ 2 ∗ ∗
d(∆)↓0 v[·]
We say that the family of strategies is universal feedback Nash equilibrium on compact
D ⊂ [t ,ϑ ]×Rn, if there exists nonnegative functions η (α), i = 1,2, η (α) → 0 α → 0,
0 0 0 i i
such that for sufficiently small α the following inequalities are fulfilled for all (t ,x ) ∈ D
∗ ∗ 0
Γ (t ,x ,Vα) ≤ Υ (t ,x ,Uα,Vα)+η (α),
1 ∗ ∗ 1 ∗ ∗ 1
Γ (t ,x ,Uα)≤ Υ (t ,x ,Uα,Vα)+η (α).
2 ∗ ∗ 2 ∗ ∗ 2
In this paper we develop the approach based on system of Hamilton-Jacobi equations.
The definition of Hamiltonians involves the Nash equilibrium in a static game. This game
is an analog of a small game used in the theory of zero-sum games [1].
Define for (t,x) ∈ [t ,ϑ ]×Rn, p,q ∈ Rn, u∈ P, v ∈ Q two criterions:
0 0
χ (t,x,p,q,u,v) , hp,f(t,x,u,v)i+g (t,x,u,v),
1 1
χ (t,x,p,q,u,v) , hq,f(t,x,u,v)i+g (t,x,u,v).
2 2
Fix (t,x) ∈ [t ,ϑ ]×Rn, p,q ∈ Rn and consider the static games
0 0
χ (t,x,p,q,u,v) → max ,
1 u∈P
(2)
( χ2(t,x,p,q,u,v) → maxv∈Q.
Denote the set of Nash equilibriums of game (2) by NE(t,x,p,q). Further we assume
that for all (t,x) ∈ [t ,ϑ ]×Rn, p,q ∈ Rn the set NE(t,x,p,q) is nonempty.
0 0
Remark 1. If the sets
{χ (t,x,p,q,u,v ) :u ∈P}, {χ (t,x,p,q,u ,v) :v ∈ Q}
1 ∗ 2 ∗
are convex for all (t,x) ∈ [t ,ϑ ] × Rn, p,q ∈ Rn, u ∈ P, v ∈ Q then NE(t,x,p,q) is
0 0 ∗ ∗
nonempty.
The pair of values (χ (t,x,p,q,u∗,v∗),χ (t,x,p,q,u∗,v∗)) for (u∗,v∗) ∈ NE(t,x,p,q) is
1 2
an analog of Hamiltonian.
3
3 Elements of Proximal Calculus
Inthissectionwefollowthedefinitionsfrom[5]. Letφ :[t ,ϑ ]×Rn → Rbeacontinuous
0 0
function. The vector (ζ−,ζ−) ∈ R1+n is said to be a proximal subgradient at the position
t x
(t,x), if there exists a constant σ− > 0 such that for (t′,x′) sufficiently close to (t,x) the
following inequality holds
φ(t′,x′) ≥ φ(t,x)+ζ−(t′−t)+hζ−,x′−xi−σ−k(t′−t,x′−x)k2.
t x
Analogously, the vector (ζ+,ζ+) ∈ R1+n is said to be a proximal supergradient at the
t x
position (t,x), if there exists a constant σ+ > 0 such that for (t′,x′) sufficiently close to
(t,x) the following inequality holds
φ(t′,x′) ≤ φ(t,x)+ζ+(t′−t)+hζ+,x′−xi+σ+k(t′−t,x′−x)k2.
t x
Denote the set of all proximal subgradients by ∂−φ(t,x), the set of all proximal supergra-
P
dients by ∂+φ(t,x).
P
Let D ⊂ [t ,ϑ ]×Rn be a compact. Denote by D the reachable set from D . Let
0 0 0 0
D , {(t,x) ∈ [t ,ϑ ]×Rn : kτ −t,x−yk ≤ 1, (t,x) ∈D}. (3)
1 0 0
Let us introduce the following transformation of φ:
1
φα(t,x) = max φ(τ,y)− kt−τ,x−yk2 . (4)
(τ,y)∈D1 2α2
h i
Let (τα,yα) maximize the right hand of (4) for the fixed position (t,x) ∈ [t ,ϑ ] × Rn.
0 0
Denote
t−τα x−yα
ζα[φ] , , ζα[φ] , . (5)
t α2 x α2
In the same way we define the following transformation of the function φ
1
φ = min φ(τ,y)+ kt−τ,x−yk2 . (6)
α (τ,y)∈D1 2α2
h i
Let (τ ,y ) minimize the right hand of (6). Denote
α α
t−τ x−y
ζ [φ] , α, ζ [φ] , α. (7)
α t α2 α x α2
Lemmas 1–5 formulated below are analogs of lemmas 3.1–3.5 of [5]. Therefore lemmas
1–5 are not proved here. In lemmas 1–5 we assume that the position (t,x) ∈ D is fixed,
(τα,yα) maximizes the right hand of (4), and (τ ,y ) minimizes the right hand of (4).
α α
Lemma 1. Let (τ ,y ) be an inner point of D , then
α α 1
(ζ [φ] ,ζ [φ] )∈ ∂−φ(τα,xα).
α t α x P
Analogously, if (τα,yα) is an inner point of D , then
1
(ζα[φ] ,ζα[φ] ) ∈ ∂+φ(τα,xα).
t x P
Let
k , min{φ(t,x) : (t,x) ∈ D }, k , max{φ(t,x) : (t,x) ∈ D },
1 1 2 1
C , 2(k −k ). (8)
2 2 1
p
4
Lemma 2. The following estimates are valid
k(t−τ ,x−y )k ≤ C α, k(t−τα,x−yα)k ≤C α.
α α 2 2
Lemma 3. The following inequalities hold
kx−yαk2 kx−yαk2
≤ ω (C α), ≤ ω (C α).
2α2 φ 2 2α2 φ 2
Here ω is a modulus of continuity of the function φ on D .
φ 1
Following [5] we consider the sets
Fα[φ] = {(t,x) ∈D :τα = ϑ }, (9)
0
F [φ] = {(t,x) ∈ D : τ = ϑ }. (10)
α α 0
Lemma 4. Let α ∈ (0,1/C ). Then one of the following statements are fulfilled:
2
• (ζ [φ] ,ζ [φ] ) ∈ ∂−φ(τ ,y );
α t α x P α α
• (t,x) ∈ F and |ϑ −τα| ≤ C α.
α 0 2
Analogously, one of the following statements is fulfilled:
• (ζα[φ] ,ζα[φ] )∈ ∂+φ(τα,yα);
t x p
• (t,x) ∈ Fα and |ϑ −τα| ≤C α.
0 2
Lemma 5. For any δ > 0 and f ∈ Rn the following inequalities hold
δ2
φ (t+δ,x+δf) ≤ φ (t,x)+δ(ζ [φ] +hζ [φ] ,fi)+ (1+kfk2),
α α α t α x 2α2
δ2
φα(t+δ,x+δf) ≥ φα(t,x)+δ(ζα[φ] +hζα[φ] ,fi)− (1+kfk2).
t x 2α2
4 Main result
Consider the functions (t,x,p,q) 7→ u∗(t,x,p,q), (t,x,p,q) 7→ v∗(t,x,p,q) such that
(u∗(t,x,p,q),v∗(t,x,p,q)) ∈ NE(t,x,p,q). Put
H (t,x,p,q) , χ (t,x,p,q,u∗(t,x,p,q),v∗(t,x,p,q)), i = 1,2.
i 1
Theorem. Suppose that there exist functions ϕ(t,x), ψ(t,x), and ω(t,x,d) with properties:
1. ω ≥ 0, ω(t,x,d) → 0 as d → 0 uniformly on any compact D, (t,x) ∈ D.
2. ϕ(ϑ ,·) = σ (·);
0 1
3. ψ(ϑ ,·) = σ (·);
0 2
4. a+ +H (t,x,p+,q+) ≥ −ω(t,x,d) for all (a+,p+) ∈ ∂+ϕ(θ,z), (b+,q+) ∈ ∂+ψ(τ,y),
1 P P
such that k(τ −t,y−x)k,k(θ−t,z−x)k ≤ d;
5. b+ +H (t,x,p+,q+) ≥ −ω(t,x,d) for all (a+,p+) ∈ ∂+ϕ(θ,z), (b+,q+) ∈ ∂+ψ(τ,y),
2 P P
such that k(τ −t,y−x)k,k(θ−t,z−x)k ≤ d;
6. a− +H (t,x,p−,q+) ≤ ω(t,x,d) for all (a+,p+) ∈ ∂+ϕ(θ,z), (b+,q+) ∈ ∂+ψ(τ,y),
1 P P
such that k(τ −t,y−x)k,k(θ−t,z−x)k ≤ d;
5
7. b− +H (τ,y,p+,q−) ≤ ω(t,x,τ,y) for all (a+,p+) ∈ ∂+ϕ(θ,z), (b+,q+) ∈ ∂+ψ(τ,y),
2 P P
such that k(τ −t,y−x)k,k(θ−t,z−x)k ≤ d;
Then the pair of strategies Uα(t,x) = u∗(t,x,ζα[ϕ] ,ζα[ψ] ), Vα(t,x) =
x x
v∗(t,x,ζα[ϕ] ,ζα[ψ] ) is Nash equilibrium on each compacts D . Here (ζ[ϕ] ,ζ[ϕ] ),
x x 0 t x
(ζ[ψ] ,ζ[ψ] ) are defined by formula (5) at the position (t,x) for the functions ϕ and ψ
t x
respectively. Moreover
liminfΥ (t,x,Uα,Vα) =ϕ(t,x),
1
α→0
liminfΥ (t,x,Uα,Vα) = ψ(t,x).
2
α→0
Corollary. Suppose that one can choose the functions u∗(t,x,p,q) and v∗(t,x,p,q) such
that its dependence on (t,x) is continuous. Moreover we assume that there exists a con-
tinuous function ϕ(t,x), ψ(t,x), and ω(t,x,d) such that the conditions 1–3 of the Theorem
hold and the following conditions are fulfilled:
4′. a+ +H (t,x,p+,q+) ≥ −ω(t,x,d) for all (a+,p+) ∈ ∂+ϕ(t,x), (b+,q+) ∈ ∂+ψ(τ,y),
1 P P
such that k(τ −t,y−x)k ≤ d;
5′. b+ +H (t,x,p+,q+) ≥ −ω(t,x,d) for all (a+,p+) ∈ ∂+ϕ(t,x), (b+,q+) ∈ ∂+ψ(τ,y),
2 P P
such that k(τ −t,y−x)k ≤ d;
6′. a− +H (t,x,p−,q+) ≤ ω(t,x,d) for all (a+,p+) ∈ ∂+ϕ(t,x), (b+,q+) ∈ ∂+ψ(τ,y),
1 P P
such that k(τ −t,y−x)k ≤ d;
7′. b− +H (τ,y,p+,q−) ≤ ω(t,x,τ,y) for all (a+,p+) ∈ ∂+ϕ(t,x), (b+,q+) ∈ ∂+ψ(τ,y),
2 P P
such that k(τ −t,y−x)k ≤ d;
Then the conclusion of the Theorem holds.
Proof of the Theorem. Let us consider the compact D . Denote by D the reachable set
0
from D . Also we assume that D is defined by (3).
0 1
Put
F = {(ϑ ,x) :x ∈ Rn}∪Fα[ψ]∪F [ϕ].
∗ 0 α
Let(t ,x ) ∈D beaninitialposition. LetΞ = {ξ }r beapartitionoftheinterval[t ,ϑ ].
∗ ∗ 0 j j=0 ∗ 0
Further we will consider a motion of the system x[·], either x[·] = x[·,t ,x ,V,Ξ,u[·]], or
∗ ∗
x[·] = x[·,t ,x ,U,V,Ξ]. For simplification we put x = x[ξ ].
∗ ∗ j j
Let a constant C be define by the rule
3
C , max{kf(t,x,u,v)k : (t,x) ∈ D ,u∈ P,v ∈ Q}.
3 1
Also define
C , 1+C2,
∗ 3
q
C , {|g (t,x,u,v)| :(t,x) ∈D ,u ∈ P,v ∈Q}.
4 1 1
Consider the following moduli of continuity on D :
1
ω (δ) , {kf(t′,x′,u,v)−f(t′′,x′′,u,v)k :
f
(t′,x′),(t′′,x′′) ∈ D , u∈ P,v ∈Q, |t′−t′′| ≤δ,kx′ −x′′k ≤ C δ},
1 3
ω (δ) , {kg (t′,x′,u,v)−g (t′′,x′′,u,v)k :
g1 1 1
(t′,x′),(t′′,x′′)∈ D u∈ P,v ∈Q, |t′−t′′| ≤δ,kx′ −x′′k ≤ C δ}.
1 3
6
First we consider the case then the Player I deviates. We prove that for any control of
the Player I u[·], the following inequality is valid
Λd(t ,x ,V,Ξ,u[·]) ≤ ϕ(t ,x )+η(α,d(Ξ)), (11)
1 ∗ ∗ ∗ ∗
where
lim limη(α,δ) = 0.
α→0δ→0
In this case we have x[·] = x[·,t ,x ,Vα,Ξ,u(·)]. Let l be a minimal number such that
∗ ∗
(ξ ,x[ξ ]) ∈ F .Thenbylemma4wehavethat|ξ −θ |≤ C α. Hence, kx[ϑ ]−x[ξ ]k ≤ C α.
l l ∗ l 0 2 0 l 3
Then
r−1 ξk+1
σ (x[ϑ ]) = ϕ(ϑ ,x[ϑ ]) ≤ϕ(ξ ,x[ξ ])− g (ξ,x[ξ],u[ξ],v )dξ +ω (C α)+C α.
1 0 0 0 l l 1 k ϕ ∗ 4
k=lZξk
X
We obtain the inequality
r−1 ξk+1
σ (x[ϑ ])+ g (ξ,x[ξ],u[ξ],v )dξ ≤ ϕ(ξ ,x[ξ ])+ω (C α)+C α. (12)
1 0 1 k l l ϕ ∗ 4
k=lZξk
X
Now let j < l. Denote
1 ξj+1
fˆ, f(ξ,x[ξ],u[ξ],v ).
j
ξ −ξ
j+1 j Zξj
Put u , Uα(ξ ,x[ξ ]), v , Vα(ξ ,x[ξ ]) . Denote the position (τ ,y ) for the function
j j j j j j α α
ϕ and (ξ ,x[ξ ]) by (τ−,y−). Denote the corresponding pair (ζ [ϕ] ,ζ [ϕ] ) by (a−,p−).
j j 1 1 α t α x
Analogously denote by (τ+,y+) the position (τα,yα) for ϕ and (ξ ,x[ξ ]). Also denote the
1 1 j j
correspondingpair(ζα[ϕ] ,ζα[ϕ] )by(a+,p+). Lettheposition(τα,yα)forψ and(ξ ,x[ξ ])
t x j j
be denoted by (τ+,y+), the corresponding pair (ζα[ϕ] ,ζα[ϕ] ) be denoted (b+,q+).
2 2 t x
By lemma 3 we have inequalities
kp−k,kp+k ≤ ω (C α).
ϕ 2
By lemma 5 we obtain that
(ξ −ξ )2
ϕ (ξ ,x[ξ ])≤ ϕ (ξ ,x[ξ ])+(ξ −ξ )(a− +hp−,fˆi)+ j+1 j (1+kfˆk2)=
α j+1 j+1 α j j j+1 j α2
ξj+1 (ξ −ξ )2
ϕ (ξ ,x[ξ ])+(ξ −ξ )a− + hp−,f(ξ,x[ξ],u[ξ],v )idξ + j+1 j (1+kfˆk2) =
α j j j+1 j j α2
Zξj
ξj+1
ϕ (ξ ,x[ξ ])+(ξ −ξ )a− + hp−,f(ξ ,x[ξ ],u[ξ],v )idξ+
α j j j+1 j j j j
Zξj
(ξ −ξ )2
(ξ −ξ )ω (C α)ω (ξ −ξ )+ j+1 j (1+kfˆk2) ≤
j+1 j ϕ 2 f j+1 j α2
ϕ (ξ ,x[ξ ])+(ξ −ξ )a−+(ξ −ξ )max hp−,f(ξ ,x[ξ ],u,v )i+g (ξ ,x[ξ ],u,v ) −
α j j j+1 j j+1 j j j j 1 j j j
u∈P
ξj h i
g (ξ,x[ξ],u[ξ],v )dξ +(ξ −ξ )ω (ξ −ξ )+
1 j j+1 j g1 j+1 j
Zξj
(ξ −ξ )2
(ξ −ξ )ω (C α)ω (ξ −ξ )+ j+1 j (1+kfˆk2). (13)
j+1 j ϕ 2 f j+1 j α2
7
By the choice of v we have that
j
a− +max hp−,f(ξ ,x[ξ ],u,v )i+g (ξ ,x[ξ ],u,v ) = a−+H (ξ ,x[ξ ],p−,q+). (14)
j j j 1 j j j 1 j j
u∈P
h i
It follows from the properties (a−,p−)∈ ∂−ϕ(τ−,y−), (a−,p−) ∈ ∂+ϕ(τ+,y+),
P 1 1 P 2 2
k(τ− −ξ ,y−−x[ξ ])k,k(τ+ −ξ ,y+−x[ξ ])k ≤ C α
1 j 1 j 2 j 2 j 2
and condition 6 of the Theorem that
a− +H (t,x,p−,q+)≤ ω(τ−,y−,C α). (15)
1 1 1 2
Put
γ(α) ,sup{ω(t′,x′,C α) :(t′,x′) ∈ D}.
2
By the condition 1 of the Theorem we have that γ(α) → 0, as α→ 0. From the inequalities
(13), (14) and (15) it follows that
ξj+1
ϕ (ξ ,x[ξ ])+ g (ξ,x[ξ],u[ξ],v )dξ ≤
α j+1 j+1 1 j
Zξj
(ξ −ξ )
ϕ (ξ ,x[ξ ])+(ξ −ξ ) γ(α)+ω (C α)ω (ξ −ξ )+ω (ξ −ξ )+ j+1 j C2 .
α j j j+1 j ϕ 2 f j+1 j g1 j+1 j α2 ∗
h (i16)
Using the estimate (16) for j = 0,l−1 we conclude that
l−1 ξk+1
ϕ (ξ ,x[ξ ])+ g (ξ,x[ξ],u[ξ],v )dξ ≤
α l l 1 k
k=0Zξk
X
δ
ϕ (t ,x )+(ξ −t ) γ(α)+ω (C α)ω (δ)+ω (δ)+ C2 .
α ∗ ∗ l ∗ ϕ 2 f g1 α2 ∗
h i
Here δ = d(Ξ).
Since
ϕ (ξ ,x[ξ ]) ≥ ϕ(ξ ,x[ξ ])−ω (α),
α l l l l ϕ
using the inequality (12) we have that
r−1 ξk+1
σ(x[ϑ ])+ g (ξ,x[ξ],u[ξ],v )dξ ≤
0 1 k
k=0Zξk
X
δ
ϕ (t ,x )+(ϑ −t ) γ(α)+ω (C α)ω (δ)+ω (δ)+ C2 +ω (α)+C α+ω (α).
α ∗ ∗ 0 ∗ ϕ 2 f g1 α2 ∗ ϕ 4 ϕ
h i
Denoting
δ
η(α,δ) , (ϑ −t ) γ(α)+ω (C α)ω (δ)+ω (δ)+ C2 +ω (C α)+C α+ω (α),
0 0 ϕ 2 f g1 α2 ∗ ϕ ∗ 4 ϕ
h i
we obtain the estimate (11).
Now let us consider the case when the Player I and Player II use the strategies Uα and
Vα respectively. Let x[·] denote the motion xc[·,t ,x ,U,V,Ξ].
∗ ∗
Put
F∗ , {(ϑ ,x) :x ∈ Rn}∪Fα[ϕ]∪Fα[ψ].
0
8
Let l be a minimal number such that (ξ ,x[ξ ]) ∈ F∗. By lemma 4, |ξ − θ | ≤ C α.
l l l 0 2
Therefore, kx[ϑ ]−x[ξ ]k ≤ C α. Estimating the variation of the function ϕ we have that
0 l 3
ϑ0
σ (x[ϑ ]) = ϕ(ϑ ,x[ϑ ]) ≥ ϕ(ξ ,x[ξ ])− g (ξ,x[ξ],uc(ξ,Ξ),vc[ξ,Ξ])dξ−ω (C α)−C α.
1 0 0 0 l l 1 ϕ ∗ 4
Zξl
Hence we conclude
r−1 ξk+1
σ (x[ϑ ])+ g (ξ,x[ξ],u ,v )dξ ≥ ϕ(ξ ,x[ξ ])−ω (C α)−C α. (17)
1 0 1 k k l l ϕ ∗ 4
k=lZξk
X
Now let j < l. Denote
1 ξj+1
f˜= f(ξ,x[ξ],u ,v )dξ.
j j
ξ −ξ
j+1 j Zξj
As to the rest we use the notations introduced above. By lemma 5 we obtain that
(ξ −ξ )2
ϕ(ξ ,x[ξ ]) ≥ ϕ(ξ ,x[ξ ])+(ξ −ξ )(a+ +hp+,f˜i)− j+1 j (1+kf˜k2)=
j+1 j+1 j j j+1 j 2α2
ξj+1 (ξ −ξ )2
ϕ(ξ ,x[ξ ])+(ξ −ξ )a+ + hp+,f(ξ,x[ξ],u ,v )dξi)dξ − j+1 j (1+kf˜k2)≥
j j j+1 j j j 2α2
Zξj
ϕ(ξ ,x[ξ ])+(ξ −ξ )[a++hp+,f(ξ ,x[ξ ],u ,v )i+g (ξ ,x[ξ ],u ,v )]−
j j j+1 j j j j j 1 j j j j
ξj+1
− g (ξ,x[ξ],u ,v )dξ −(ξ −ξ )(ω (C α)ω (ξ −ξ )+ω (ξ −ξ ))−
1 j j j+1 j ϕ 2 f j+1 j g1 j+1 j
Zξj
(ξ −ξ )2
j+1 j C2. (18)
2α2 ∗
It follows from the definition of the strategies Uα, Vα and the elements u , v that
j j
a++hp+,f(ξ ,x[ξ ],u ,v )i+g (ξ ,x[ξ ],u ,v ) = a++H (ξ ,x[ξ ],p+,q+).
j j j j 1 j j j j 1 j j
By the lemma 2 we have that
kτ+ −ξ ,y+ −x[ξ ]k,kτ+ −ξ ,y+ −x[ξ ]k ≤ C α
1 j 1 j 2 j 2 j 2
The condition 4 of the Theorem yields that
a++H(t,x,p+,q+)≥ −ω(t,x,C α).
2
Using the function γ(·) we have
a+ +H (ξ ,x[ξ ],p+,q+) ≥ −γ(α).
1 j j
From this and the inequality (18) it follows that
ξj+1
ϕ(ξ ,x[ξ ])+ g (ξ,x[ξ],u ,v )dξ ≥
j+1 j+1 1 j j
Zξj
(ξ −ξ )
ϕ(ξ ,x[ξ ])−(ξ −ξ ) γ(α)+ω (C α)ω (ξ −ξ )+ω (ξ −ξ )+ j+1 j C2 .
j j j+1 j ϕ 2 f j+1 j g1 j+1 j 2α2 ∗
h (i19)
9
Using the inequalities (19) for j =0,l−1 we obtain that
l−1 ξk+1
ϕ(ξ ,x[ξ ])+ g (ξ,x[ξ],u ,v )dξ ≥
l l 1 k k
k=0Zξk
X
δ
ϕ(t ,x )−(ξ −t ) γ(α)+ω (C α)ω (δ)+ω (δ)+ C2 . (20)
∗ ∗ l ∗ ϕ 2 f g1 2α2 ∗
h i
Here δ = d(Ξ). Using the estimate (17) and the property
ϕα(ξ ,x[ξ ]) ≤ ϕ(ξ ,x[ξ ])+ω (α),
l l l l ϕ
we conclude from (20) that
r−1 ξk+1
Λc(t ,x ,Uα,Vα,∆) = σ(x[ϑ ])+ g (ξ,x[ξ],u ,v )dξ ≥
i ∗ ∗ 0 1 k k
k=0Zξk
X
δ
ϕα(t ,x )−(ϑ −t ) γ(α)+ω (C α)ω (δ)+ω (δ)+ C2 −ω (α)−C α−ω (α).
∗ ∗ 0 ∗ ϕ 2 f g1 α2 ∗ ϕ 4 ϕ
h i (21)
Letting δ to 0 in (11) and using the definition of η we obtain that
Γ (t ,x ,Vα) ≤ ϕ(t ,x )+(ϑ −t )γ(α)+ω (C α)+C α+ω (α). (22)
1 ∗ ∗ ∗ ∗ 0 0 ϕ ∗ 4 ϕ
In the same manner we obtain from (21) that
Υ[t ,x ,Uα,Vα]≥ ϕ(t ,x )−(ϑ −t )γ(α)−ω (α)−C α−ω (α). (23)
∗ ∗ ∗ ∗ 0 ∗ ϕ 4 ϕ
Therefore
Γ[t ,x ,Vα]−κ(α) ≤ϕ(t ,x ) ≤ Υ[t ,x ,Uα,Vα]+κ(α). (24)
∗ ∗ ∗ ∗ ∗ ∗
Here
κ(α) ,(ϑ −t )γ(α)+ω (α)+C α+ω (α).
0 ∗ ϕ 4 ϕ
The analog of the inequality (24) is fulfilled for the function of the Player II. Therefore the
pairofstrategies Uα,Vα istheuniversalNashfeedback. Moreover thepayoffs ofthePlayers
I and the Player II at the position (t ,x ) are equal to ϕ(t ,x ) and ψ(t ,x ) respectively.
∗ ∗ ∗ ∗ ∗ ∗
Proof of the Corollary. Inthis casetheinequalities (22)and(23)arestill valid ifwe replace
the function γ(·) with the function γ˜(·)
γ˜(α) , γ(α)+ω (C α)ω∗(C α)+ω∗ (C α),
ϕ 2 f 2 g1 2
where ω∗ is the following modulus of continuity
f
ω∗(δ) , sup{kf(t′,x′,u∗(t′,x′,p,q),v∗(t′,x′,p,q))−f(t′′,x′′,u∗(t′′,x′′,p,q),v∗(t′′,x′′,p,q)) :
f
(t′,x′),(t′′,x′′) ∈ D ,k(t′ −t′′,x′−x′′)k ≤ δ,kpk ≤ ω (δ),kqk ≤ ω (δ)}.
1 ϕ ψ
The function ω∗ (·) is defined in the same way. Now let us consider the analog of the
g1
inequality (14). We use the designations introduced in the proof of the Theorem. We have
that
a−+max hp−,f(ξ ,x[ξ ],u,v )i+g (ξ ,x[ξ ],u,v ) =
j j j 1 j j j
u∈P
h a−+hp−,f(ξ ,x ,u∗(ξ ,x ,p−,q+),vi∗(ξ ,x ,p−,q+))i+
j j j j j j
g (ξ ,x ,u∗(ξ ,x ,p−,q+),v∗(ξ ,x ,p−,q+)
1 j j j j j j
10