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SIAM J. CONTROL OPTIM. c 2005 Society for Industrial and Applied MathematicsVol. 44, No. 4, pp. 12591288

HYBRID CONTROL SYSTEMS AND VISCOSITY SOLUTIONS

SHEETAL DHARMATTI AND MYTHILY RAMASWAMY

Abstract. We investigate a model of hybrid control system in which both discrete and contin-uous controls are involved. In this general model, discrete controls act on the system at a given setinterface. The state of the system is changed discontinuously when the trajectory hits predefinedsets, namely, an autonomous jump set A or a controlled jump set C where the controller can chooseto jump or not. At each jump, the trajectory can move to a different Euclidean space. We provethe continuity of the associated value function V with respect to the initial point. Using the dy-namic programming principle satisfied by V, we derive a quasi-variational inequality satisfied by Vin the viscosity sense. We characterize the value function V as the unique viscosity solution of thequasi-variational inequality by the comparison principle method.

Key words. dynamic programming principle, viscosity solution, quasi-variational inequality,hybrid control

AMS subject classifications. 34H05, 34K35, 49L20, 49L25

DOI. 10.1137/040618072

1. Introduction. Many complicated control systems, like flight control andtransportation, perform computer coded checks and issue logical as well as contin-uous control commands. The interaction of these different types of dynamics andinformation leads to hybrid control problems. Thus hybrid control systems are thosehaving continuous and discrete dynamics and continuous and discrete controls. Manycontrol systems, which involve both logical decision making and continuous evolution,are of this type. Typical examples of such systems are constrained robotic systems [1]and automated highway systems [8]. See [5], [6], and the references therein for moreexamples of such systems.

In [5], Branicky, Borkar, and Mitter presented a model for the most general hybridcontrol system in which continuous controls are present and, in addition, discrete

controls act at a given set interface, which corresponds to the logical decision makingprocess as in the above examples. The state of the system is changed discontinuouslywhen the trajectory hits these predefined sets, namely, an autonomous jump set A or acontrolled jump set Cwhere the controller can choose to jump or not. They prove rightcontinuity of the value function corresponding to this hybrid control problem. Usingthe dynamic programming principle they arrive at the partial differential equationsatisfied by the value function, which turns out to be the quasi-variational inequality,referred hereafter as QVI.

In [4], Bensoussan and Menaldi study a similar system and prove that the valuefunction u is close to a certain u which they mention to be continuous indicating theuse of the basic ordinary differential equation estimate for continuous trajectories andthe continuity of the first hitting time (see [4, Theorem 2.5 and Remark 3.5]). They

Received by the editors November 1, 2004; accepted for publication (in revised form) March 24,2005; published electronically October 7, 2005. This work was partially supported by DRDO 508and ISRO 050 grants to Nonlinear Studies Group, IISc.

http://www.siam.org/journals/sicon/44-4/61807.htmlDepartment of Mathematics, Indian Institute of Science, Bangalore 560012, India (sheetal@

math.iisc.ernet.in). This author is a UGC Research Fellow and the financial support from UGC isgratefully acknowledged.

IISc-TIFR Mathematics Program, TIFR Center, P.O. Box 1234, Bangalore 560012, India([email protected]).

1259

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1260 SHEETAL DHARMATTI AND MYTHILY RAMASWAMY

prove its uniqueness as a viscosity solution of the QVI in a certain special case wherethe autonomous jump set is empty and the controlled jump set is the whole space.

In our work, we study this problem in a more general case in which the au-tonomous jump set is nonempty and the controlled jump set can be arbitrary. Our

model is based on that of [5]. Our main aim is to prove uniqueness in the most gen-eral case when the sets A and C are nonempty and also to obtain precise estimatesto improve the earlier continuity results. Our motivation comes from the fact that inall the real-life models mentioned above, logical decision making is always involvedas well as the continuous control. This will correspond to a nonempty autonomous

jump set A.Here we prove the local Holder continuity of the value function under a transver-

sality condition, the same as the one assumed in [5] and [4] (see (2.36) in [4]). Forthis we need to follow the trajectories starting from two neighboring points, throughtheir continuous evolution, and through their discrete jumps since the autonomous

jump set is nonempty. This involves careful estimation of the distance between thetrajectories in various time intervals and summing up these terms to show that thedistance remains small for initial points sufficiently close enough. Although the basic

estimates used are similar to those available in the literature (e.g., [3], [4]), the crucialpoint in our proof is the convergence of the above summation. This also allows us toget the precise Holder exponent for the continuity of the value function.

As in [5] and [4], using the dynamic programming principle, we arrive at the QVIsatisfied by the value function. Then we show that the value function is the uniqueviscosity solution of the QVI. Our proof is very different from [4]. Their approachusing a fixed point method does not seem to be suitable, as it is for the general case ofa nonempty autonomous jump set. Our approach is based on the comparison principlein the class of bounded continuous functions. It is inspired by earlier work on impulseand switching control and game theoretic problems in the literature, namely, [2], [7],[9], particularly the idea of defining a sequence of new auxiliary functions. But thepresence of the autonomous and controlled jump sets leads to different equations onthese sets, and hence some new ideas are needed to arrive at the conclusion.

2. Notation and assumptions. In a hybrid control system, as in [5], the statevector during continuous evolution is given by the solution of the following problem:

X(t) = f(X(t), u(t)),(2.1)

X(0) = x,(2.2)

where X(t) := i i {i}, with each i a closed connected subset of Rdi , i,di Z+; x ; and f : U . Actually, f = fi with the understanding thatX(t) = fi(X(t), u(t)) whenever x i. U is the continuous control set

U= {u : [ 0, ) U | u measurable, U compact metric space} .The trajectory also undergoes discrete jumps when it hits predefined sets A, the

autonomous jump set, and C, the controlled jump set. A predefined set D is thedestination set for both autonomous jumps as well as controlled jumps:

A =i

Ai {i}, Ai i Rdi ,C =

i

Ci {i}, Ci i Rdi ,D =

i

Di {i}, Di i Rdi .

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HYBRID CONTROL SYSTEMS AND VISCOSITY SOLUTIONS 1261

The trajectory starting from x i, on hitting A, that is the respective Ai i, jumps to the destination set D according to the given transition map g. g usesdiscrete controls from the discrete control set V1 and can move the trajectory fromAi to Dj j Rdj . The trajectory then will continue its evolution under fj tillit again hits A or C, in particular Aj or Cj . On hitting C the controller can chooseeither to jump or not to jump. If the controller chooses to jump, then the trajectoryis moved to a new point in D. In this case the controller can also move from i toany of the j .

This gives rise to a sequence of hitting times of A, which we denote by i, and asequence of hitting times of C, where the controller chooses to make a jump which isdenoted by i. Thus i and i are the times when continuous and discrete dynamicsinteract. Hence the trajectory of this problem is composed of continuous evolutiongiven by (2.1) between two hitting times and discrete jumps at the hitting times. Wedenote (X(i ), u()) by xi and g(X(i ), v) by xi and the destination of X(+i , u())by X(i)

. In general we take the trajectory to be left continuous so that Xx(i)

means Xx(i ) and Xx(i) means Xx(

i ), whereas Xx(

+i ) will be denoted by x

i and

Xx(+i ) will be denoted by Xx(i)

.

We give the inductive limit topology on , namely,

(xn, in) converges to (x, i) if for some N large and n N,

in = i, x, xn i, i Rdi for some i, and xn xRdi < .

With the understanding of the above topology we suppress the second variable i from. We follow the same for A, C, and D. We make the following basic assumptionson the sets A,C,D, and on functions f and g.

(A1): Each i is the closure of a connected, open subset ofRdi .

(A2): Ai, Ci, Di are closed, Ai,Ci are C2. For all i and for all x Di, |x| < R,

and Ai i for all i.(A3): g : A V1 D is a bounded, uniformly Lipschitz continuous map, withLipschitz constant G with the understanding that g = {gi} and gi : Ai V Dj .(A4): Vector field f is Lipschitz continuous with Lipschitz constant L in the state

variable x and uniformly continuous in control variable u. Also,

|f(x, u)| F x and u U.(2.3)

(A5): We assume Ai is compact for all i, and for some 0 > 0, following trans-versality condition holds

f(x0, u) (x0) 20 x0 Ai u U,(2.4)

where (x0) is the unit outward normal to Ai at x0. We assume a similar transver-sality condition on Ci.

(A6):

infi

d(Ai, Ci) and inf i

d(Ai, Di) > 0,(2.5)

where d is the appropriate Euclidean distance. Note that the above rules out infinitelymany jumps in finite time.

(A7): We assume the control sets U and V1 to be compact metric spaces.

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Now (u(), v , i, X(i)) is the control, and the total discounted cost is given by

J(x, u(), v , i, X(i)) =0

K(Xx(t), u(t))etdt +

i=0Ca(X(i), v)e

i(2.6)

+

Cc(X(i), X(i))ei ,

where is the discount factor, K : U R+ is the running cost, Ca : A V1 R+ is the autonomous jump cost, and Cc : C D R+ is the controlled jump cost.The value function V is then defined as

V(x) = inf (UV1[0,)D)

J(x, u(), v , i, X(i)).(2.7)

We assume the following conditions on the cost functionals.(C1): K is Lipschitz continuous in the x variable with Lipschitz constant K1 and

is uniformly continuous in the u variable. Moreover, K is bounded by K0.(C2): Ca and Cc are uniformly continuous in both variables and bounded below

by C > 0. Moreover, Ca is Lipschitz continuous in the x variable with Lipschitzconstant C1 and is bounded above by C0. Also we assume

Cc(x, y) < Cc(x, z) + Cc(z, y) x Ci, z D Cj , y D.

We now give two simple examples of hybrid control systems. For more examples,see [5].

Example 2.1 (collisions). Consider the ball of mass mwhich is moving in verticaland horizontal directions in a room under gravity with gravitational constant g. Thedynamics can be given as

x = vx, vx = 0,

y = vy, vy =

mg.

On hitting the boundaries of the room A1 = {(x, y)|y = 0, or y = R1} we instantlyset vy to vy for some [0, 1], the coefficient of restitution. Similarly we reset vxto vx on hitting the boundary A2{(x, y)|x = 0 or x = R2}. Thus in this case thesets A1 and A2 are autonomous jump sets. We can generalize the above system byallowing dynamics to occur in different Rd after hitting.

The next example illustrates the importance of the transversality condition, inthe absence of which the optimal trajectory and hence the optimal control may failto exist.

Example 2.2. Consider the dynamical system in R2 given by

x1(t) = 1, x1(0) = 0,

x2

(t) = u, x2

(0) = 0,

where u [0, 1], and when the trajectory hits the set A given by A = {(x1, x2)|(x1 1)2 + (x2 + 1)2 = 1} it jumps to (1010, 1010). The cost is given by

0

et min{|x1(t) +x2(t)|, 21010}.

Here the vector field (u, 1) is not transversal to the boundary at (1, 0) for u =0. Hence optimal trajectory does not exist and, moreover, the value function isdiscontinuous at (1, 0).

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In the following sections we are interested in exploring the value function of thehybrid control problem defined in (2.7). In section 2 we show that the value function isbounded and locally Holder continuous with respect to the initial point. In section 3,we use viscosity solution techniques and the dynamic programming principle to derive

a partial differential equation satisfied by V in the viscosity sense, which turns out tobe the HamiltonJacobiBellman QVI. Section 4 deals with uniqueness of the solutionof the QVI. We give a comparison principle proof characterizing the value function asunique viscosity solution of the QVI.

3. Continuity of the value function. Let the trajectory given by the solutionof (2.1) and starting from the point x be denoted by Xx(t, u()). Since x , itbelongs in particular to some i. Then we have from the theory of ordinary differentialequations

|Xx(t, u()) Xz(t, u())| eLt|x z|,(3.1)|Xx(t, u()) Xx(t, u())| F|t t|,(3.2)

where F and L are as in (A4).Define the first hitting time of the trajectory as

T(x) = infu

{t > 0 | Xx(t, u) A} .

Notice that this T(x) is in particular with respect to Ai as x i. By assuming asuitable transversality condition on Ai and Ci we prove the continuity of T in thetopology ofRdi . This is equivalent to proving the continuity of T on with respectto the inductive limit topology on . Hereafter by convention we assume the topologyto be of that i, in which the respective points belong.

Theorem 3.1. Assume (A1)(A7). Let X(t) be the trajectory given by thesolution of (2.1). Let the first hitting time T(x) be finite. Then it is locally Lipschitzcontinuous, i.e., there exists a 1 > 0 depending on f, 0, and the distance function

from Ai such that for all y, y in B(x, 1), a 1 neighborhood of x in

|T(y) T(y)| < C|y y|, where C depends on 0.

Proof. Step 1. Estimates for points near A. First we show that there exist > 0 and C > 0 such that

T(x) < C d(x) x B(Ai, )\A,

where B(Ai, ) is a neighborhood of Ai and d(x) is a signed distance of x from Aigiven by

d(x) =

dist(x,Ai) if x

Ai,

0 if x Ai,dist(x,Ai) if x Aci .

For simplicity of notation we drop the suffix i from now on, remembering that thedistances are in Rdi . It is possible to choose R > 0 such that in a small neighborhoodof A, say B(A,R), the above signed distance function d is C1, thanks to ourassumption (A2).

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Now for x0 A choose u0 in U such that u0(t) = u0 for all t and r0 < R suchthat

f(x, u0) Dd(x) < 0 x B(x0, r0).(3.3)

Observe that we can choose r0 independent of x0 by using compactness of A. Nowconsider the trajectory starting from x, given by

X(t) = f(X(t), u0),

X(0) = x,

where x B(x0, r0). Then

d(X(s)) d(x) =s0

Dd(x) f(x, u0) d+s0

Dd(X()) Dd(x) f(X(), u0) d

+

s0

Dd(x) (f(X(), u0) f(x, u0)) d.

By using (3.3) and (2.3),

d(X(s)) d(x) s0

0 d + Fs0

(Dd(X()) Dd(x)) d

+

s0

Dd(x) (f(X(), u0) f(x, u0)) d.

Let c be the bound on Dd on B(A,r0). Restricting s to be small so that X() is inthe r0 neighborhood of A, we are assured that Dd is continuous. So is f. Thus

d(X(s)) d(x) 0s + o(F s) + o(cLs) 0 | X(t) A, X(t) = f(X(t), u), X(0) = x}.

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For given 0 < < 1, and x by the definition of T(x), we can choose u Usuchthat

t = t(x, u) < T(x) + .(3.4)

Using estimate (3.1),

|Xx

t, u Xxt, u| |x x|eLt |x x|eL(T(x)+).(3.5)

Define 1 = eL(T(x)+1), where is as in Step 1. Let us choose x such that |x x| 0, choose u1 Usuch that

T

Xx(t, u) tXx(t, u), u1 1.

Define a new control u2 by

u2(s) =

u(s) ifs t,u1(s t) ifs > t.

Then

T(x) t(x, u2) t + t (Xx(t, u), u1) t + T (Xx(t, u)) + 1.

Since 1 is arbitrary, this proves (3.7). Using (3.4) and (3.7) for x B(x, 1) we getT(x) T(x) + TXx(t, u)+

T(x) + C dXx(t, u)+ by (3.6).Notice that d(Xx(t, u)) |Xx(t, u) Xx(t, u)|. So by (3.5)

T(x) T(x) + C |x x| eL(T(x)+ ) + .Interchanging the roles of x and x we get

|T(x) T(x)| C |x x| eL(T(x)T(x))(3.8)as tends to 0, where T(x) T(x) = max{T(x), T(x)}. Also observe that

T(x) T(x) + C |x x| eL(T(x)+) + T(x) + C + T(x) + C + 1 T(x) + 2.

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Hence for all x belonging to B(x, 1), T is bounded. Let this bound be T0. Then wehave

|T(x) T(x)| < C|x x|eLT0 .Hence we conclude that the first hitting time of trajectory is locally Lipschitz contin-uous with respect to the initial point.

Now we take up the issue of continuity of the value function. For this proof weneed some estimates on hitting times of trajectories starting from two nearby points.We prove these estimates in the following lemmas. We fix the controls u and v andsuppress them in the following calculations.

Lemma 3.2. Let1 and 1 be the first hitting times of trajectories evolving withfixed controls u and v according to (2.1) starting from x and z, respectively. Let x1and z1 be points where these trajectories hit A for the first time:

x1 = Xx(1), z1 = Xz(1), x1, z1 A.If

|x

z

|< 1, where 1 is as in Theorem 3.1, then

|x1 z1| (1 + F C)eL(11)|x z|.(3.9)Proof. Note here that by Theorem 3.1 we have the estimate on |1 1| given

by (3.8),

|1 1| < C eL(11) |x z|.(3.10)Using this we estimate |x1 z1|. Without loss of generality we assume that 1 > 1,

|x1 z1| = |Xx(1) Xz(1)| |Xx(1) Xz(1)| + |Xz(1) Xz(1)|.

Using (3.1) we get

|Xx(1) Xz(1)| < eL1 |x z|,while (3.2) and (3.10) lead to

|Xz(1) Xz(1)| F|1 1| F CeL1 |x z|.Combining these estimates, we get

|x1 z1| eL1 |x z|(1 + F C) for z B(x, 1).Observe that the destination points of x1 and z1, which are denoted by x1 =

g(x1, v) and z1 = g(z1, v), may belong to j Rdj . Without loss of generality weassume that x1, z1

2

R

d2 , and the evolution of trajectories takes place in 2till the next hitting time. Let 2 and 2 be the next hitting times of the trajectorieswhen they hit A once again. The next lemma deals with the estimate of |2 2|.

Lemma 3.3. Let the first hitting time of trajectories starting from x and z, andevolving with fixed control u, be 1 and 1, and the second hitting times are 2 and2. Then there exists 2 such that for |x z| < 2,

|2 2| Ce(22)(F C+ G(F C+ 1))|x z|(3.11)

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and if we denote

x2 = Xx1(2 1), x2 = g(x2),z2 = Xz1(2 1), z2 = g(z2),

then

|x2 z2| (F C+ 1)eL(22)(F C+ G(F C+ 1))|x z|.(3.12)

Proof. Without loss of generality let 1 < 1. Observe that 2 and 2 are thefirst hitting times of trajectories starting from points Xx1(1 1) and z1 at timet = 1. Then

T(z1) = (2 1) and T(Xx1(1 1)) = 2 1.

Hence by (3.8)

|2

2

| CeL(21)

|Xx1(1

1)

z1

|whenever |Xx1(1 1) z1| 1. Now

|Xx1(1 1) z1| |Xx1(1 1) x1| + |x1 z1|.

Hence by using estimate (3.2) and (3.10) for the first term we have

|Xx1(1 1) x1| F|1 1| F CeL1 |x z|,

whereas using Lipschitz continuity of g and (3.9) for the second term we get

|x1 z1| G|x1 z1| G(F C+ 1)eL1 |x z| for z B(x, 1).

Combining the above two estimates we have

|Xx1(1 1) z1| eL1(F C+ G(F C+ 1))|x z|(3.13)

and by our choice of 2 = min{1, 1eL1FC+G(FC+1)}, |Xx1(1 1) z1| < 1. Using(3.13) in the estimate of |2 2| for z B(x, 2) we have

|2 2| CeL2(F C+ G(F C+ 1))|x z|.(3.14)

Now we estimate |x2 z2|:

|x2 z2| = |Xx1(2 1) Xz1(2 1)| |Xx1(2 1) Xz1(2 1)| + |Xz1(2 1) Xz1(2 1)|.

Observe that by the semigroup property

Xx1(2 1) = XXx1(11)(2 1).

Hence

|Xx1(2 1) Xz1(2 1)| = |XXx1 (11)(2 1) Xz1(2 1)|

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and by (3.1)

|Xx1(2 1) Xz1(2 1)| eL(21)|Xx1(1 1) z1|.(3.15)From (3.2) and (3.14) we get

|Xz1(2 1) Xz1(2 1)| F|2 1 (2 1)| F CeL2(F C+ G(F C+ 1))|x z|.(3.16)

Together these estimates yield, for z B(x, 2),|x2 z2| eL2(F C+ 1)(F C+ G(F C+ 1))|x z|.

Let i and i be the ith hitting times of trajectories starting from x and z,respectively. With the above notation we assume that xi

, zi i+1 Rdi+1 . We

apply Theorem 3.1 and the above lemmas recursively to find estimates on successivehitting times and points where trajectories hit A. We generalize the above estimatesfor the ith hitting times of trajectories when they hit A. For simplicity of calculations

we denote F C+ G(F C+ 1) by P hereafter.Remark 3.4. Let the control u be fixed. Let i and i be the ith consecutive

hitting time of the trajectory starting from x and z, respectively, when they hit A, andlet xi, zi be the points on A where trajectories hit A. Then proceeding along linessimilar to those of Lemmas 3.2 and3.3 we get the estimates for |ii| and |xizi|which are given by

|i i| CeLiPi1|x z|,|xi zi| eLi(F C+ 1)Pi1|x z|

whenever |x z| < i, where i := min{1, 2, . . . , 1eLiPi1 }.Theorem 3.5 (continuity of the value function). Under the assumptions of

Theorem 3.1, value function V of hybrid control problem defined by (2.7) is bounded

and locally Holder continuous with respect to the initial point.Proof. First we show that the value function is bounded. For any u U and

v V1,

V(x) 0


i=0

Ca(X(i), v)ei .

By our assumptions (C1) and (C2),

V(x) K0+0

et dt +

+i=1

C0ei K0

+ C0

+i=1

ei .

From (A5), recalling that = inf d(Ai, Di),

i+1 i + sup |f(x, u)| i + /F.(3.17)

Hence we get

i=1

ei e1i=1

e/F

i e1 11 e/F ,(3.18)

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leading to

V(x) K

+ C0e1

1

1 e/F .

This proves V(x) is bounded.We now show that V defined in (2.7) is locally Holder continuous with respect

to the initial point. Let x, z . Regarding V(x) as in (2.7), we assume that thecontroller chooses not to make any controlled jumps. Note that the controller hasthis choice because in the interior of C he can always choose not to jump. On theboundary ofC that is C by the transversality condition, vector field is nonzero andhence he can continue the evolution without jumping. Thus in any case he can choosenot to jump. Then given > 0, we can choose the controls u, v depending on suchthat

V(z) 0

K(Xz(t), u(t))etdt +

i=1Ca(Xz(i), v)e

i .

Also

V(x) 0


i=1

Ca(Xx(i), v)ei .

Hence

V(x) V(z) 0

|K(Xx(t), u(t)) K(Xz(t), u(t))|etdt

+i=1

|Ca(Xx(i), v) Ca

Xz(i), v|e(ii) + ,

where i i = max{i, i}. Now for T large to be chosen precisely later on we splitthe integral and summation as follows:

V(x) V(z) T0

|K(Xx(t), u(t)) K(Xz(t), u(t))|etdt(3.19)

+Ni=1

|Ca(Xx(i), v) Ca(Xz(i), v)|e(ii)

+

T

|K(Xx(t), u(t)) K(Xz(t), u(t))|etdt

+

i=N+1|Ca(Xx(i), v) Ca(Xz(i), v)|e(ii) + ,

where T will be chosen so that the tail end of the integral and summation becomesmall and T is in between the Nth and (N + 1)th hitting times of the trajectories.By using the bound K0 on K given by (C1) we get

T

|K(Xx(t), u(t)) K(Xz(t), u(t))|etdt 2K0

eT(3.20)

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and by using bound C0 on Ca given by (C2) and doing calculations along lines similarto those of (3.18) we get the estimate

i=N+1

|Ca(Xx(i), v) Ca(Xz(i), v)|e(ii) 2C0

e/FN 1

1 e/F .(3.21)

Now we calculateT0

|K(Xx(t), u(t)) K(Xz(t), u(t))|etdt. We will show thatthere exists > 0 such that if |x z| < , then the sequence of i and i can be, forexample,

0 1 1 2 2 n n T(3.22)

or 0 1 1 n n T.

That is, every A hitting time of trajectory starting from x is followed by A hittingtime of trajectory starting from z.

Without loss of generality let us assume 1 < 1. If 1 < 1, the followingcalculations go through with appropriate changes and hence we split this integral,assuming (3.22) as follows:

T0

Ietdt 10

Ietdt +

11

Ietdt +

21

Ietdt + (3.23)

+

nn

Ietdt +

n+1n

Ietdt,

where I = |K(Xx(t), u(t)) K(Xz(t), u(t))|. In this there are two types of integrals:1. i

iIetdt;

2.i+1i

Ietdt.

If |x z| < N, where N = min{1, 2, . . . , 1eLNPN1 }, we can estimate the aboveintegrals using Lemmas 3.2 and 3.3 and Remark 3.4. We use the bound on K toevaluate the first integral.

ii

Ietdt 2K0

ei ei 2K0

|i i|.

Using Remark 3.4,

i

i

Ietdt 2K0CPi1eLi .(3.24)

To evaluate the second integral we use the Lipschitz continuity of K.i+1i

Ietdt =

i+1i

|KXxi(t i) KXzi(t i)|etdt(3.25) K1

i+1i

|Xxi(t i) Xzi(t i)|etdt.

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By the semigroup property,

|Xxi(t i) Xzi(t i)| = |XXxi(ii)(t i) Xzi(t i)|

eL(ti)

|Xxi(i

i)

zi

|by (3.1).

Now by generalizing the estimate in (3.13) we get

|Xxi(i i) zi| PieLi |x z|.(3.26)Hence substituting the above estimates in (3.25), we geti+1

i

Ietdt K1eLiPieLi |x z|i+1i

e(L)tdt.

For L = , i+1i

Ietdt K1Pi |x z| e(L)(i+1) e(L)i

L (3.27)

K1Pi|x z|e(L)T

1

L and for L = , i+1

i

Ietdt K1eLiPieLi |x z|i+1i

dt(3.28)

K1Pi|x z| |i+1 i| K1Pi |x z| 2T.

For L = , by using (3.24), (3.27), T0

Ietdt becomes

T

0

Ietdt N

i=1

2K0CPi1eLT|x z| +

N

i=1

K1L

Pie(L)T 1|x z|.

Hence T0

Ietdt 2K0CPN1

P1

|x z|

+ K1

PN1P1

e(L)T1

L |x z|

for L = (3.29)

and for L = , using (3.24) and (3.28),

T0

Ietdt Ni=1

2K0|i i| +Ni=1

K1T Pi|x z|

Ni=1

2K0CPi1

|x z| +Ni=1

K1T Pi

|x z|.Thus T

0

Ietdt 2K0C

PN1P1

|x z|

+2K1T

PN1P1

|x z|

for L = .(3.30)

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Furthermore, by using (C2) and Remark 3.4 we get

N

i=1|Ca(xi, v) Ca(zi, v)|e(ii)

N

i=12C1|xi zi|e(ii)

2C1Ni=1

(F C+ 1)eLTPi1|x z|,

Ni=1

|Ca(xi, v) Ca(zi, v)|e(ii) 2C1(F C+ 1)eLT|x z|

PN1 1P 1

.

(3.31)

Since P is a constant, without loss of generality we can assume

PN

P

1

< 2PN.(3.32)

Also observe that i i+1 /F implies that T N+1 1 N/F and hence

N < T F/.(3.33)

Using (3.20), (3.21), (3.29), (3.31), (3.32), (3.33) in (3.19) for L = we have

V(x) V(z) 4K0CeLTPTF/ |x z| + 2K1PTF/ e(L)T 1

L |x z|+

2K0

eT + 2C1eLTPTF/ |x z|

+ 2C0

e/FTF/ 1

1 e/F .

Now we further restrict |x z| < (1)1

1 for some such that 0 < < 1. Thenchoose T such that

PTF/eLT = |x z|.

This gives

T = log |x z| + F log P /

.(3.34)

This together with the choice of |x z| implies

N =1

eLN PN1>

1eLTPTF/

= 1|x z| > |x z|.(3.35)

Thus |x z| < N and hence the above estimate holds true for our choice of T. Thensubstituting the value of T in the above estimate, for L = , we get

V(x) V(z) 4K0C|x z|1 + K1L |x z|

1 + C1|x z|1

+2K0

|x z| (F log P/)+L + 2C0|x z|

(F logP/)+L .

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Here we have used the fact that e(L)T 1 < eLT. Thus we have proved that inthe

11

1 ball around x,

V(x) V(z) < C1|x z|1 for some constant C1,where

1 = min

1 ,

(F log P/) + L

for 0 < < 1.

For L = , using (3.20), (3.21), (3.30), (3.31), (3.32), and (3.34) in (3.19), wehave

V(x) V(z) 4K0C|x z|1 + 2 K1(F log P/) + L

log(|x z|)|x z|1

+ 2C1(F C+ 1)|x z|1 + 2K0L

|x z| L(F log P/)+L

+ 2C0|x z|L

(F logP/)+L .

Since |x z|1 goes to 0 faster than log(|x z|) goes to as |x z| 0, allterms on the right-hand side (RHS) go to 0. The modulus of continuity ofV is the

same as that of log(r)r1. This suggests that in the 1

1

1 ball around x,

V(x) V(z) < C1|x z|1 for some constant C1and for all 1 such that

1 < min

1 , L

(F log P/) + L

for 0 < < 1.

Thus in any case we have shown that (for 1 chosen depending on L = or L = )V(x)

V(z)

C1

|x

z

|1 for some constant C1.

Interchanging the roles of x and z we will get

V(z) V(x) C2|x z|1 for some constant C2.Together these will give

|V(x) V(z)| C|x z|1 for some constant C.This proves the Holder continuity of V.

Now we want to justify our claim in (3.22), i.e., if 1 < 1, we can choose|x z| small enough such that (3.22) holds. If we restrict |x z| such that |x z| min( 4FC , (

4CF )

11 ), then by Remark 3.4,

|i i| CeLT

(F C+ G(F C+ 1))TF/

|x z|.By our choice of T,

|i i| C|x z|1 14

F 0,

V(x) = inf u,v,i,X(i)

T0


i

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Now taking the infimum over discrete controls v belonging to Vin the last two termswe get

V(x)

1

0

K(X(t), u(t))etdt + MV(Xx(1)).

Further taking the infimum over continuous controls u in U we have the one-wayinequality in (DPPA). For the reverse inequality, let > 0 be given. Choose the control = (u, v, i, X(i)

) such that

V(x) + 10

K(X(t), u(t))etdt + Ca(X(1), v)e

1

+ e1

1

K(X(t), u(t))etdt +

i=2

Ca(X(i), v)ei

+

i=1Cc

X(i), X(i)

ei

with calculations similar to those earlier, we can conclude that

V(x) + 10

K(X(t), u(t))etdt + MV(Xx(1))

infu10

K(X(t), u(t))etdt + MV(Xx(1)).

Hence as 0 we have other way inequality. Thus (DPPA) is proved. Now weproceed to prove (DPPC). Let t1 be the first hitting time of C where the controllerchooses to jump. In this case we write t1 = 1. Then

V(x)

1

0

K(X(t), u(t))etdt + CcX(1), X(1))e1+

1

K(X(t), u(t))etdt +

i=1

Ca(X(i), v)ei

+i=2

Cc

X(i), X(i)

ei

.

Doing the change of variables t = t1 in the square brackets and taking the infimumover the control variables, it is the value function of trajectory starting from ( Xx(1))

.Hence,

V(x) 10

K(X(t), u(t))etdt + e1Cc(X(1), X(1)) + e1V(Xx(1)

).

Now taking the infimum over (Xx(1)) D in the last two terms we get

V(x) 10

K(X(t), u(t))etdt + NV(Xx(1)),

and taking the infimum over u in Uon the RHS we will get the one-way inequality of(DPPC).

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For the reverse inequality, given > 0 choose = (u, v, i, X(i)) such that

V(x) + 10

K(X(t), u(t))etdt + N V(Xx(1))

infu 10

K(X(t), u(t))etdt + N V(Xx(1)).

As 0 we will get

V(x) = infu

10

K(X(t), u(t))etdt + N V(Xx(1))

,

which proves (DPPC). The proof of (DPP) for any T > 0 follows similarly, which weskip here.

Theorem 4.2 (quasi-variational inequality). Under the assumptions (A1)(A7)and(C1), (C2), the value function V described in (2.7) satisfies the following the QVIin the viscosity sense:

V(x) =

MV(x) x A,min {N V(x), H(x,DV(x))} x C,H(x,DV(x)) x \ A C,

(QVI)

where H is the Hamiltonian given by

H(x, p) = supuU

K(x, u) f(x, u) p

.

Proof. Let x A. In this case we have to show that V(x) = MV(x). Sincex A, the first hitting time of trajectory is 1 = 0. Hence, by (DPPA) we getV(x) = MV(x).

Now we consider the case x \ A C. In this case we want to show that Vsatisfies the HamiltonJacobiBellman (HJB) equation in the viscosity sense. For weneed to show the following: for all C1() and x local maximum ofV

V(x) + H(x,D(x)) 0and for all C1() and x local minimum ofV

V(x) + H(x,D(x)) 0.Let r = min {d(x,A), d(x,C)}. Choose R < r. Then in the ball B(x, R) noimpulses are applied. Now V is continuous at x, and assume that V has localmaximum at x. Choose small enough such that Xx() B(x, R). By our choice ofR and , is less than the first hitting time. Then, since x is the local maximum ofV ,

(x) (Xx()) V(x) V(Xx())0

K(Xx(t), u(t))etdt + (e 1)V(Xx()),

where the second inequality follows by (DPP), since < 1 and < 1. Dividing by and taking the limit as 0 we get

D(x) f(x) K(x, u(0)) V(x),

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which implies

V(x) +K(x, u(0) D(x) f(x)

0.

Taking the supremum over all u Uwe will getV(x) + H(x,D(x)) 0.

Hence V is a viscosity subsolution of HJB equation.To show that V is a viscosity supersolution, let V have local minimum at x.

Then for such that Xx() B(x, R),(Xx()) (x) V(Xx()) V(x)

(1 e)V(Xx()) 0

K(Xx(t), u(t))etdt by (DPP).

Dividing by and taking the limit as 0 we getV(x)

K(x, u(0))

D(x)

f(x)

0,

V(x) +K(x, u(0)) D(x) f(x)

0.

Taking the supremum over all u we will get

V(x) + H(x,D(x)) 0.Hence V is a viscosity supersolution of the HJB equation. Thus we have shown thatin the case x \ A C, V satisfies the HJB equation in the viscosity sense.

Now consider the case x C. We observe that if x C, and the controllerchooses to jump, then by (DPPC), V should satisfy N V(x). Whereas if the controllerdecides not to jump, then the system undergoes some continuous evolution and we

can analyze as before to conclude that V satisfies the HJB equation in the viscositysense. In this case we have to show that V satisfies the following equation in theviscosity sense:

min{V(x) N V(x), V(x) + H(x,DV(x))} = 0.For this we need to show that, for all C1(), x local minimum ofV

min{V(x) N V(x), V(x) + H(x,DV(x))} 0,and for all C1(), x local maximum ofV ,

min{V(x) N V(x), V(x) + H(x,DV(x))} 0.Now if V(x) = N V(x), there is nothing to prove.

Suppose V(x) < N V(x); then we need to show that V satisfies the HJB equationin the viscosity sense. We show that whenever V(x) < N V(x) there exists r > 0 anda ball B(x, r) around x such that it is not optimal to apply any impulses on B(x, r).Then we can do the analysis in this ball to conclude as in the case of x \ A C.For we claim that there exists > 0 such that

V(x) = inf u,v,i,X(i)

t10

K(Xx(t), u(t))etdt + N V(Xx(t1)) | t1 >

.

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Suppose not; then = 0, which implies 1 = 0, which by (DPPC) implies V(x) =N V(x); this is a contradiction of our hypothesis V(x) < N V(x). Hence > 0. Chooser < min{d(x, Xx()), d(A, C)}. Then in the ball B(x, r), no impulses are applied. Sowe can do the analysis in this ball around x and conclude as in the earlier case. This

proves the QVI for the case x C.5. Uniqueness. We take up the issue of uniqueness of the viscosity solutions of

(QVI) in this section. Inspired by the earlier work on impulse control problem (see[2], [9]), we prove the comparison between any two solutions of the QVI.

Theorem 5.1. Assume (A1)(A7) and(C1), (C2). Letu1, u2 BC(), boundedcontinuous functions on, be two viscosity solutions of the QVI given by (QVI). Thenu1 = u2.

Proof. The idea of the proof is to show that u1(x) u2(x) for all x . Wedefine the following auxiliary function on

i=1 (i i) that is i on each ii

by

i(x, y) = u1(x) u2(y) 1|x y|2 |x|

2 + |y|2,(5.1)where and are small positive parameters to be chosen suitably later on. Observethat for each i, i attains its supremum over i i, thanks to the last two terms,which become large negative as |x|, |y| goes to 0. We prove the theorem in two steps.In the first step of the proof we show that sup i supii

i(x, y) 0. In the nextstep we prove the uniqueness using Step 1.

Step 1. Let

supi

supii

i(x, y) = C > 0.

Fix > 0 such that < min{C2 , C

2 }. If the above supremum is achieved at some(x0, y0), the following proof gets simplified. If not, corresponding to this we canchoose (x, y) in some i

i, say, 1

1, such that

1(x, y) > C > C2

.(5.2)

Let 1 attain its supremum at some finite point, say, at (x0, y0) in 1 1. Then

sup11

1(x, y) = 1(x0, y0) > C > C2

.(5.3)

Since x0 and y0 can lie in different sets in 1, u1(x0) and u2(y0) will satisfydifferent equations from the QVI. We list below the different cases which arise:

1. (x0, y0) A C or C A.2. (x0, y0) \ (A C) \ (A C).3. x0, y0 /

A and one of x0 or y0

C. This takes care of (x0, y0)

C

\(A C), (x0, y0) \ (A C) C, (x0, y0) C C.4. x0, y0 / C and one of the x0 or y0 A, i.e., (x0, y0) A A or (x0, y0)

A \ (A C), (x0, y0) \ (A C) A.Our idea is to show that in any of these cases, u1(x) u2(x) is arbitrarily small for and small. For this we will estimate u1(x0) u2(y0) at the maximum point (x0, y0)of 1 or u1(xn) u2(yn) at the maximum point (xn, yn) of n, a suitably definedauxiliary function. The crucial point in our proof is that after at most finitely many

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steps, say n0, at the maximum point ofn0 both u1 and u2 satisfy the HJB equation.Then we can use the usual comparison principle available in the literature. We firstlist some standard estimates needed later in the proof.

Lemma 5.2. Let and (x0, y0) be as above. Then

(i) |x0y0|2

C for some C independent of and ;(ii)

|x0|,

|y0| C for some C independent of and ;

(iii) |x0y0|2

1(

C), where 1 is the local modulus of continuity of both

u1 and u2 in the ball of radius R, dependent on but independent of ,R = R() = C/

in 1.

Proof. By our assumption

21(x0, y0) 1(x0, x0) + 1(y0, y0).(5.4)

Hence

2

|x0 y0|2 u1(x0) u1(y0) + u2(x0) u2(y0).(5.5)

Since u1 and u2 are bounded,

|x0 y0|2

C,

which proves (i). This also implies

|x0 y0|

C.

To prove (ii), fix some z 1 such that |z| = 1; then 1(x0, y0) 1(z, z), whichimplies

|x0|2 + |y0|2 u1(x0) u1(z) u2(y0) + u2(z)

1

|x0 y0|2 + 2|z|2

C+ 2 C+ 2.Hence

|x0| C, where C is independent of and . Similarly,

|y0| C. This

proves (ii). Hence x0 and y0 lie in some ball BR of radius R = R().Now using the estimate in (i) and the modulus of continuity of u1 and u2 in the

compact set BR() in 1, we get

|x0 y0|2

1(

C).

This proves (iii).Now we consider the different cases listed earlier.Case 1. (x0, y0) A C or C A.Claim. This case does not occur.Without loss of generality let (x0, y0) A C. Since d(A, C) > ,

|x0 y0| > .

On the other hand by Lemma 5.2(i),

|x0 y0| u(x) + cc(x, x)

or u(x) = Mu(x) = u(g(x, v)) + ca(x, v),

then there exists an 1 > 0 depending only on the uniform continuity of u onD1 1but independent of and such that

|x x| > 1(5.13)

or |x g(x, v)| > 1,(5.14)

depending on which equation u(x) satisfies.Proof. We claim that there exists 1 > 0 such that |x x| > 1. Suppose the

contrary. That is, there exists sequence xn, xn 1 such that

u(xn) > u

xn

+ cc

xn, xn and |xn xn| 0.

Then by continuity of u, |u(xn) u(xn)| 0. But

|u(xn) u

xn| = ccxn, xn > C > C2 > 0,

which is a contradiction. Hence given C

4 choose the corresponding 1 given by uniform

continuity of u on D1 1 such that |y z| < 1 |u(y) u(z)| < C4 . Then

|x x| > 1.

This proves (5.13).

To prove that |x g(x, v

)| > 1, we proceed with arguments similar to thoseabove and choose 1 corresponding to the

C

4 in the definition of uniform continuityof u on D1.

In the next lemma we estimate the difference 1(x0, y0) and 2(y0, y

0), which we

are going to use to define new auxiliary function 1.Lemma 5.4. Let be as defined in (5.1) and let (x0, y0) 11 be as in(5.3),

the point where 1 attains supremum. Let y0 D2 be such that

u2(y0) = N u2(y0) > u2

y0

+ cc

y0, y0

.(5.15)Then

1(x0, y0) 2y0, y

0 K

for some constant K > 1 depending only on the constants of the problem and inde-pendent of and .

Proof.

1(x0, y0) 2

y0, y0

= u1(x0) u2(y0) 1

|x0 y0|2

|x0|2 + |y0|2 u1

y0

+ u2

y0

+ 2|y0|2.

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Using (5.15) we get

1(x0, y0) 2

y0, y

0

< u1(x0) cc

y0, y

0

1

|x0 y0|2

|x0|2 + |y0|2

u1y0+ 2|y0|2 + .Also u1(y0) N u1(y0) u1(y0) + cc(y0, y0). Hence,

1(x0, y0) 2

y0, y0

u1(x0) u1(y0) 1|x0 y0|2

|x0|2 + |y0|2+ 2|y0|2+ u1(x0) u1(y0) + 2|y0|2 +

u1(x0) u1(y0) + 2R2 + 1

C

+ 2R2 + .

Using the modulus of continuity of u1, on BR in 1 for a given > 0 choose > 0such that

1

C

< 1

(x0, y0) 2

y

0, y

0 K.

This proves the lemma.We use the above difference to define another auxiliary function 1. We further

restrict 2 given by Lemma 5.3, if necessary, so that 2 1,i.e., 1 has support in the 1 ball around (y0, y

0) 2 2, having maximum at

(y0, y0) and it vanishes on all i i other than i = 2.

Observe that by the definition of i1,

21

y0, y0

= 2

y0, y

0

+ 2K

1(x0, y0) K + 2K supi supii

i

(x, y) + K 21(x, y) 2K 1(x, y) + (K 1).

As 1 is 0 for all (x, y) i i, i = 2, and for (x, y) outside the 1 ball around(y0, y

0) in 2 2, we have for all such (x, y)

21

y0, y0

> 21(x, y).

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Hence 21 has the supremum over 2 2 in the 1 ball around (y0, y0). Let (x1, y1)be such that

sup22

21 = 21(x1, y1).

Then

21(x1, y1) 11(x0, y0) = 1(x0, y0) > C .(5.16)

Since 1 u2

y1

+ cc

y1, y1

.We define 2 on

i i, that is, i2 on i i, by

i2(x, y) = i(x, y) + 2K

2j=1

j(x, y),

where 2(y1, y1) = 1 and 2 has support in the 3 ball around (y

1, y

1) in 3 3

with the properties 2 C0 ( ), 0 2 1, |D2|

23 , 2(x, y) < 1 if

(x, y) = (y1, y1). Hence as before we can show that the supremum of2 is attained inthe 3 ball around (y1, y

1). Also we can show that

32 satisfies the inequality similar

to (5.16), namely,

22(x1, y1) 21(x1, y1) = 1(x0, y0) > C .Thus we can proceed to define 3, 4, . . . , n and so on, in case u2(yi) = N u2(yi).We now claim that this process has to terminate in finitely many steps, which is thecontent of the following lemma.

Lemma 5.5. Suppose (xn, yn) n+1 n+1, yn Dn+2 are sequences suchthat

u2(yn) = N u2(yn) > u2(yn) + cc(yn, y

n)

, yn

B(yn1, n+1);

n(x, y) = n1(x, y) + 2K n(x, y); n(xn, yn) = supn+1n+1

n(x, y);

where n is such that n C0 ( ); actually n has support in the n+1 ballaround (yn, y

n) n+2 n+2. 0 n 1; |Dn| < 2n+1 ; n(yn1, yn1) = 1,

n = 1, 2, . . . . Then n < n0 = [8CC ], where C is a bound on u1 and u2 and C

is thelower bound on cc.

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Proof. Observe that yi, yi+1 Di+2. By uniform continuity of u2 on Di+2 i+2, for all i,

|yi+1 yi| < i+1 |u2(yi+1) u2yi| u2

y0

+ cc

y0, y0

> u2

y0

+ C ; because cc C > 0

> u2(y1) C

4+ C = u2(y1) + 3

4C

> u2

y1

+ cc

y1, y1

+

3

4C 2 > u2

y1

+ C +3

4C 2

> u2(y2) C

4+ C +

3

4C 2 = u2(y2) + 6

4C 2.

Therefore, at the nth stage we will get

C u2(y0) > u2(yn) + 34

nC n.

By using < C

2 , ifn > n0 = [8CC ], then u2(y0) > C, which is a contradiction, because

|u2| < C.Thus we have only a finite sequence of {yn} such that u2(yn) = N u2(yn) . So,

for n > n0 = [8CC ] necessarily u2(yn) < N u2(yn) and hence

u2(yn) + H(yn, Du2(yn)) 0.Hence both u1 and u2 satisfy the HJB at the supremum point of auxiliary functionn. Now we proceed as in Case 2 taking care of the extra terms.

In this case we define test functions 1 and 2 by

1(x) = u2(yn) + 1|x yn|2 +

|x|2 + |yn|2 2K nj=1

j(x, yn),(5.17)

2(y) = u1(xn) 1|xn y|2

|xn|2 + |y|2+ 2K nj=1

j(xn, y).(5.18)

Then by the definition of (xn, yn), u1 1 has maximum at xn and u2 2 has mini-mum at yn. Using u1 as the viscosity subsolution and u2 as the viscosity supersolution,we get

u1(xn) u2(yn) H(yn, D2(yn)) H(xn, D1(xn)).Let = min{1, . . . , n+1}. Also, whenever (xn, yn) j+1 j+1 we can write

D1(xn) = 2

(xn yn) + 2xn 2K nj=1

Dj(xn, yn),(5.19)

D2(yn) =2

(xn yn) 2yn + 2K

nj=1

Di(xn, yn),(5.20)

|D1(yn)| 2

(xn yn) + 2|yn| + 4nK

.(5.21)

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Hence by structural condition on H given by (5.11),

u1(xn) u2(yn) L|D2(yn)| |xn yn| + K1|xn yn| + F|D2(yn) D2(xn)|.(5.22)

By using (5.19), (5.20), (5.21) in the above we get

u1(xn) u2(yn) 2L

|xn yn|2 + 2L|yn| |xn yn| +

4Kn

|xn yn|(5.23)

+ K1|xn yn| + 4F (|xn| + |yn|) + 8Kn

.

Now by using the technique of Lemma 5.2 for n, we can prove that

|xn yn| 1 depending only on the constants of the problem and inde-pendent of and .

Proof.

1(x0, y0) 2(g(y0, v0),(g(y0, v0)) = u1(x0) u2(y0) 1|x0 y0|2 (|x0|2 + |y0|2)

u1(g(y0, v0)) + u2(g(y0, v0)) + 2|g(y0, v0)|2

= u1(x0) ca(y0, v0) 1|x0 y0|2

(|x0|2 + |y0|2)u1(g(y0, v0))+2|g(y0, v0)|2.We add and subtract u1(y0) in the above, and observing that u1(y0) Mu1(y0) u1(g(y0, v0)) + ca(y0, v0), we get

1(x0, y0) 2(g(y0, v0), g(y0, v0)) u1(x0) u1(y0) ca(y0, v0) u1(g(y0, v0)) + u1(y0) + 2|g(y0, v0)|2

u1(x0) u1(y0) + 2|g(y0, v0)|2 1(|x0 y0|) + 2R2.

We can choose such that 1(

C) < . Then by the Lemma 5.2,

1

(|x0 y0|) 1

C

< 1(x0, y0) 2(g(y0, v0), g(y0, v0)) K,where K depends on the modulus of continuity of u1 and R. This proves thelemma.

To proceed, if necessary, we restrict 2

Sic on 61807

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