2: = x,, y: = y,, u: = u,. The first iteration step is somewhat different from the subsequent ones, therefore we describe it in detail and then we formulate the common rule.
According t o Proposition 1 there exists a measurable function y ; ( t ) E I i ( t , x z ( t ) ) such t h a t
Since K ( t , x )
c
l i o ( t , x ) we havefor some measurable selection
d ( t > E U ( t , x Z ( t ) ) . Thus for some measurable & ( t ) E F 2 ( t , z : ( t ) , y ; ( t ) )
Let X be the Lipschitz constant of F ( t ,
.,
.) corresponding t o the compact set G = D x ( K ( [ O , T ] , D )+
v B ) , where v is an upper bound of y,(.). According t o Proposition 1 the values of ( x : ( t ) , y z ( t ) ) and ( x z ( t ) , y ; ( t ) ) belong t o this set. T h e same will concern t h e values of the other pairs ( x f ( t ) , yf+'(t)) defined iteratively below as it will become clear in the end.There is a measurable selection v E ( t ) E F 2 ( t , x z ( t ) , y:(t)) such t h a t
Then the projection
< E ( ~ ) = p ~ 2 ( t , ~ : ( t ) , y f ( t ) ) 7 1 ~ ( ~ )
is measurable and
( J E ( t ) - 71,(t)l
F
X y E ( t ) .Subtracting ( 6 1 ) from ( 6 2 ) we obtain
Condition i) in Supposition A6 implies that the projection map PLins v ( t , x ) ( . ) is continuous in ( t , x ) , therefore there exists a continuous n x n-matrix C ( t , x ) such t h a t PLins u ( t , x ) ( E ) = C ( t , x)l for each 1 E Rn. Multiplying (64) by C ( t , x:(t)) and then by B ( t , x:(t)) and taking into account that uz(t) - @:(t) E Lins U ( t , x:(t)) we obtain that
= - B C ( t , x:(t))(J,(t) - & ( t ) ) - B C ( t , x,(t))(l;l,(t) - J , ( t ) )
+
B C ( ( t , x:(t))~3/:(t).Since J,(t) - & ( t ) E Lins Fz(t, x:(t), yZ(t)) Supposition A6 ii) implies t h a t t h e first term in the right-hand side is zero, hence
where
W E ( . ) = ([,(a) - rlE(.))
+
&Y:(t).Having in mind also (63) we obtain that x: satisfies the differential inclusion x:(t> E ~ ~ ( t , x:(t), ~ , 0 ( t ) )
+
~ ( t , ~ : ( t ) ) u : ( t )+
B C ( ~ , x : ( t ) ) ~ , ( t )c
~ l ( t , x : ( t ) , ~ , l ( t ) )+
~ ( t , x:(t))u;(t)+
B C ( ~ , x:(t))w,(t)+
~ y , ( t ) ~ , x , ( o ) 0 = x O . (65) We shall apply Lemma 3 t o the differential inclusion ( 6 5 ) forHaving in mind (63) and the properties of y, given in Proposition 1 we can represent
where a,(t) + 0 uniformly. In order t o estimate the first term in the right-hand side we take one row d,(t) of B C ( s , x,(t)) and keep in mind that x,(.) are uniformly bounded and equi-Lipschitz, hence d,(.) are uniformly bounded (by a constant M) and equi-continuous.
We know also that y, and E Y , are uniformly bounded (by constants M and
I?,
respectively).Take an arbitrary positive number 6 and t E ( 0 , TI. Denote tj = jtlp, j = 1,.
.
.,
p , where p is fixed so large that for each two numbers t', t" E [0, t ] for which It' - t"J<
T l p it holdsWithout any restriction we may assume t h a t E
<
6 / ( 4 M M p ) , since all the constants in t h e right-hand side are independent of E . We haveThus we obtain that p(q,)(-) (we use the notation from Lemma 3) converges uniformly t o zero. Then Lemma 3 implies the existence of a solution x:(.) of
for which
P E = ( I x ~ ( . ) - X ~ ( . ) I I C + 0.
3. D e s c r i p t i o n o f t h e ( k t 1)-st i t e r a t i o n s t e p . Now we begin with the description of the k-th iteration step, supposing that absolutely continuous x i ( . ) and measurable y:(-), u 3 . ) are already defined for i = 0 , . .
. ,
k in such a way thatwhere Co = X(L t M 1 ( l
+
L ) ) , L is the Lipschitz constant of K ( t , -) in D and MI is a bound of the norm of BC on [0, TIx
D. From (66), (59), (60) and (67) we see t h a t (68)-(71) are fulfilled for k = 1. Only (72) is not fulfilled because of the specificity of the first step (for k = 1 we have (58), where y,(-) is only point-wise convergent). But, in fact, we shall not use the validity of (72) for k when proving that it holds for k+
1.In the end it will become clear that all (x,k(t), y;(t)), k
>
1 belong t o Dx
1i([O, TI, D ) for all sufficiently small E , therefore we skip checking this a t every step.From (69) and (71) we obtain
Then there exists a measurable y,k+'(-) such that
T h e second relation is just ( 7 2 ) for k
+
1. T h e first of t h e above relations means t h a t there is a measurable u,k+'(t) E ~ ( t , x , k ( t ) ) , such t h a tT h u s ( 7 0 ) a n d ( 6 9 ) a r e also fulfilled for k
+
1.O n t h e other hand from ( 6 9 ) , ( 7 1 ) a n d t h e second relation in ( 7 3 ) we obtain
c
~ z ( ~ ,~ , k ( ~ ) ,
Y $ + ' ( ~ ) )+
u,k(t)+
Aa,k(t)(l+ L ) B . Thanks t o Supposition A6 ii) we obtain exactly a s in the first stepwhere
l ~ E ( . ) l
5
Aa,k(t)(l+
L ) .Using the above two relations a n d the second estimation in ( 7 3 ) we get
F1(t, x,k(t), yEk+'(t))
+
B ( t , x,k(t))u,k+'(t)+
A ( ~ ' a , k ( t ) ( l + L )+
~ a : ( t ) ) B= ~ 1 ( t , x,k(t), y$+'(t))
+
~ ( t , x,k(t))u,k+'(t)+
c o a , k ( t ) B . According t o t h e (classical) Filippov theorem there exists a solution x:+'(.) ofi,k+l E Fl ( t , x,k+'(t), yt+'(t))
+
~ ( t , x,k+' (t))u,k+' ( t ) , x,k+l ( 0 ) = x O , such t h a tt
x k + l ( t ) - x : ( t ) l < e m 1 c 0 a : ( s ) d s =
t S k - l
( e A T c o ) ( e m c o ) k - l
P, /
d s = ( e " ~ 0 ) k%P,
t k = .k+l,
0 ( k - l ) !
T h e above two relations shaw t h a t ( 6 8 ) a n d ( 7 1 ) a r e fulfilled also for k
+
1. This completes t h e ( k+
1)-st iteration step.4) Passing to a limit. Notice t h a t sum of all a,k(t) is finite and can be estimated by
where C 1 is independent of E a n d t . According t o Proposition 1 x:(t)
+
K Bc
int D for someK
>
0. Sincep,
+ 0 with E one may chose EO>
0 so small t h a t C I P ,<
K for E E (O,EO].T h e n x:(t) E D for all E
<
E O which legitimates the use of t h e constants A , L a n d M' a s they where defined above.The convergence of the sum in (74) together with (71) and (72) implies also uniform con- vergence of x,k(.) and ?/!(.) t o respective xE(.) and y€(.). Since u,k(t) € U([O,T], D) which is bounded, one can take a n L2-weakly convergent subsequent with a limit iiE(.) (we shall use the same indexes).
5 . End of the proof. Thanks t o the uniform convergence of x,k(.), the convexity and the continuity of U(t, x ) , and the Mazur theorem we obtain in a standard way
By the same argument we have also
Choosing a L2-weakly convergent subsequence of
~ , k ( . )
and employing for third time the Mazur theorem we obtain alsoT h e last three inclusions mean that (xE(.), yE(.)) E Zo. Moreover,
Because of the properties of the function yE(.) from Proposition 1 we have
This completes the proof of the theorem.
Appendix B
In the proof of the lower semicontinuity of the trajectory set we shall use the following lemma, which we prove for completeness.
Lemma 4 Let ( t , u)
-
M ( t , u ) be a continuous mapping from [O,T] x U to the convex compact subsets of Rn. Then, given measurable functions uo(-) and yo(-) satisfyingand a positive number a, there exist continuous Go(.) and Lipschitz continuous ilo(.) such that
~ o ( t ) E M ( t , Co(t)>, E U and
Iluo(.> - fiO(.)llLl
I
a, IlYo(.) - YO(.>IIL,I
a .Proof. T h e proof consists of three steps. First we approximate uo(-) by a n appropriate Lipschitz selection ilo(-) of U. Then we approximate yo(-) by a continuous function that is not necessary a selection of M ( t , ilo(t)), finaly we "project" it on M ( t , il,-,(t)) preserving the continuity.
1. First we shall define an appropriate Go(.). Obviously uO(.) can be approximated in L1 with any accuracy l / k by a Lipschitz continuous selection u k ( . ) of U, since U is convex.
Denote pk(t) = Iuo(t)-uk(t)J. Since pk + 0 in L1, there is a subsequence (we keep the same indexes) such t h a t p k ( t ) + 0 almost everywhere. Moreover, pk(.) is uniformly bounded, since U is compact. The modulus of continuity wM(.) of M is a monotone increasing function and limh,o wM(h) = wM(0) = 0. Therefore the function wM(pk(.)) is measurable and uniformly bounded, hence integrable. By the dominated convergence theorem
thus
I I ~ M ( P ~ ( . ) : I I I L ,
5
a for ko 2 a-' for which l / k<
a . We define ilo(.) = uko(.).2. Now we take an arbitrary continuous approximation Go(.) of yo(.) such t h a t
We have
3. Finally we approximate Go(.) by a selection of M. Namely, we consider the mapping M ( t , ho(t))
n
[Go (t)+
2dist(Go(t), M ( t , ilo(t)))B].Lemma 9.4.2 in [3] claims that for every two convex compact sets M1 and M2 in Rn and every t,wo points yl and y2 in Rn it holds
This implies t h a t the above mapping is continuous (and convex valued), therefore has a continuous selection yo(.). We have
T h e proof is complete. Q.E.D.
Proof of Theorem 3. We split the proof in a number of steps.
1. First we introduce some notations. By a compactness argument Supposition C3 is fulfilled also for some smaller compact set D' such t h a t D'
+
p Bc
D for some p>
0.Let G'
c
Rn be a compact set such t h a tGo U k o ( [ 0 , T I , D, U) C G'.
(see suppositions C2 a n d C5). Denote G = G'
+
B . Let L be a Lipschitz constant of Fl and F2 on D x G x U (uniformly in t E [0, TI) a n d let M be a bound ofI
FlI
and JF21 in t h e same set. T h e modulus of continuity of F2 with respect t o t (uniformly in ( x , y , u ) E Dx
G x U) will be denoted by w t ( . ) . Similarly,i
will denote a Lipschitz constant ofko
with respect t o x E D (uniformly in (t, u)) and G ( . ) will denote t h e modulus of continuityko
on [O,T] x D x U.Consider
S E F l ( t , x , G , U), x(0) = x 0 .
Since x0 E D', there is
Po >
0 such t h a t all solutions remain in D'+
0.5pB on [O,Po].2. Suppose t h a t the claim of the theorem is not true. Then there exist zo = ( x O ( - ) , yo(.)) E Zo, a sequence ~k -+ 0 and a
>
0 such t h a t for each zk = (xk(.), yk(.)) E Z,, eitherLet uo(.) be t h e selection of U corresponding t o zo(.).
3. We continue with some preparatory work.
3.1. According t o Lemma 9.4.2 [3] (see also (75)) the mapping
is locally Lipschitz with respect t o ( x , y, u) (uniformly in t ) and measurable in t. Obviously it is also convex and compact valued. Then its Steiner point El(t,x, y , u ) has the same properties (Theorem 9.4.1 in [3]). We keep the notation I, also for its Lipschitz constant in the set D
x
Gx
U (uniformly in t E [0, TI).3.2. According t o Lemma 4 applied t o M ( t , u) = k o ( t , s o ( t ) , u) for every cr
>
0 there are continuous Go(-) and Go(-) such t h a tlluo(.> - Go(.)IIL,
5
f f , IIYo(.) - YO(.)IIL~<
f f .Denote by w,(.) the modulus of continuity of Go(.) and by L ( a ) - the Lipschitz constant of Go(.).
T h e value of cr will be considered as fixed, but the way we fix it will be specified later on.
3.3 Denote by yO[x, t] the Steiner point of the mapping
As above, we can prove t h a t yo is continuous. Its modulus of continuity in [0, TI x D will be denoted by wz(.).
4. In the next step we derive some implications of (77),(78), which will lead t o a contra- diction in the end.
4.1. Consider the system
Each solution (obviously locally such exist) will be considered on a maximal interval of extendibility [0,Ok(zk)] C [0, TI in the set D
x
G. T h e set of all (maximal in t h e above sense) solutions will be denoted byzk.
For fixed a , k and zk(.) = ( x k ( . ) , yk(.)) E
zk
we denote for brevity y;(t) = yO[xk(t), t].Obviously this is a continuous function with modulus w;(h) = w z ( ( M
+
1 ) h ) andMoreover,
IY;(~> - Go(t)l
5
2dist(Go(t), k o ( t , xk(t), Go(t)))4.2. Denote
d e f
Pk = '[o,~k(Zk)](~k(.), YE('))
(here r is the Tikhonov metric defined in Section 2 , taken on the interval [0, d k ( z k ) ] ) . By the definition of r ( . , .) we have p k
5
dk(zk). Since y k ( t ) , yo(t) E G , we haveIyk(t) - yo(t)
1 5
diam(G) for t<
p k . ( 8 1 )By the definition of the metric r for every two bounded functions y' a n d y" it holds lyl(t) -
yl'(t)l
<
r ( y l , y") for t>
r ( y l , y"). Then for t E ( p k , d k ( z k ) ] (if it has happened t h a t d k ( z k )>
p k ) we haveTaking into account the differential equations t h a t x k ( . ) a n d s o ( . ) satisfy a n d using (81) a n d ( 8 2 ) we obtain
t
5
L1
1xk(s) - X O ( S ) ~ ds+
La+
Ldiam(G)pk+
LTpk+
2 L iJot
l x k ( s ) - x o ( s ) ds+
La.T h e same inequality obviously holds also for t E [0, pk]. Using t h e Grunwall inequality we estimate
Ixk(t) -
X O ( ~ ) I i
Clpk+
C Z ~ , (83)where C1 a n d Cz d o not depend on a a n d k. Then (82) implies
In particular
Since C 2 , C4 a n d C6 are independent of a a n d k , we may suppose t h a t cr is fixed in advance so small t h a t
P 1 u
C2a
< -,
C4a< -,
c s a< -.
2 2 2
T h e n for all t
>
p k (in case of dk(zk)>
p k ) we haveLet
d e f 1 p l a a
p = min{-,Po, - - -
-1.
2 4C1
'
4c3'
4C1'
4C5Assume t h a t pk
I
2 p for some k and in the same time B k ( z k )>
2 p . Then ( 8 5 ) implies t h a t z k ( t ) E int ( D x G ) for t 2 pk, which means t h a t B k ( z k ) = T. Since t h e right-hand sides of ( 8 3 ) and ( 8 4 ) are not bigger than o we obtain a contradiction with ( 7 7 ) , ( 7 8 ) . Thus we conclude t h a t if ( 7 7 ) , ( 7 8 ) are fulfilled, then for every k and for every zk Ezk
eitheror pk
>
2 p , which implies that5 . In t h e next step we prove t h a t the set
zk
contains a n element zk = ( x k ( . ) , y k ( . ) ) for which l y k ( t ) - y O [ x k ( t ) , t]l "quickly" decreases towards zero right from t = 0 .5.1. Denote
6 = 6 ( p / 2 ) , v = 612, TO = r 0 ( v ) ,
where 6 ( . ) and T O ( . ) are defined in suppositions C 4 and C 5 , respectively.
Let
ik
be t h e set of those zk Ezk
for whichI x i ( t ) - xo(t)l
5
P5
for all t E [ O , & k ~ o ] , ( 8 8 )5.2. We shall prove t h a t
ik
is non-empty for all sufficiently large k. According t o Suppo- sition C 5 and in view of the inclusion y O [ x k ( 0 ) , 0 ] Eko(O,
x O , io ( 0 ) ) the systemhas a solution X ( T )
=
x O , y ( - ) , for which y ( r ) E G o for each T 2 0 andConsider on [ 0 , TO] the system
We shall apply the Filippov theorem t o the last system with the reference trajectory x ( . ) , y(.). For this reason we estimate
T h e number a being fixed, the quantity in the right-hand side is arbitrarily small for sufficiently large k . According t o the Filippov theorem there is a solution ( Z k ( . ) , j j k ( . ) ) of ( 9 1 ) , ( 9 2 ) on [ O , TO] such t h a t
IW
- Y ( T ) II
d 7 ( ~ , E ~ ) .Hence, x k ( t ) = Z k ( t / & k ) , y k ( t ) = g k ( t / ~ k ) satisfy ( 7 9 ) , ( 8 0 ) On [ O , E ~ T O ] and
Moreover, x o ( t ) E D', y!(t) E G' and y ( r ) E G o
c
G'. Thus, the following inequalities ensure (88)-(90):d 7 ( a , r x )
+
M ~ o E ~<
P2,
d 7 ( a , & k )
<
1,T h e number a being fixed, the above inequalities are obviously fulfilled for all sufficiently large k , hence
zk # 8.
6. T h e next step will be t o prove that for all sufficiently large k and for each zk E
zk
thereexists t
2
E ~ T O for whichI ~ k ( t ) - yO[xk(t),t]I
>
P. ( 9 3 )Take an arbitrary zk E
zk
and denote for brevity y!(t) = y O [ x k ( t ) , t ] . Suppose first that O k ( z k )>
2p. Then according t o the alternative ( 8 6 ) , ( 8 7 ) we have ( 9 3 ) for some t2
p , hence for some t>
E ~ T O , provided t h a t k is sufficiently large.Now we consider the second possibility: O k ( z k )
<
p. If it occurs that y k ( O k ( z k ) ) E a G , then ( 9 3 ) is also fulfilled at t = O k ( z k )2
E ~ T O , since y O [ x k ( t ) , t ] E G' and p<
1.If y k ( O k ( z k ) ) E int G , then x k ( O k ( z k ) ) must belong t o the boundary of D , according t o the definition of O k ( z k ) . But this contradicts the inequality p
< Po,
since x k ( - ) satisfies the inclusion ( 7 6 ) on [ 0 , 8 ( z k ) ] , thus x k ( O k ( z k ) ) E D'+
rho12
c
int D .7. In the next step we prove existence of an "extremal" (in a specific sense) element 2k E z k .
7.1 According t o Step 6 for each zk E
ik
(k is sufficiently large) there is a maximal t = tk(zk) E [ E ~ T O , ~ ~ ( Z ~ ) ) such thatSince according t o the definition of
ik
there is a last moment t = t i ( z k ) E [ E ~ T ~ , tk(zk)) a t which still
Recapitulating, we have
I~k(~Ok(~k>) - ~O[~k(tOk(zk)),t;(zk)]l = 6,
for numbers t
>
t k ( z k ) arbitrarily close t o tk(zk).7 . 2 We shall prove that there exists Tk E i k such t h a t
For a fixed k (large enough so t h a t the consideration in Step 7.1 is valid) we take a sequence z i E i k such t h a t
t",c.)
-
t,. 0Each z i is defined a t least on [O,tk(zi)]. Denote
and consider z i on [0, fk] (extending i t , if necessary, t o the right by z i ( t k ( z i ) ) ) . By a standard compactness argument there is a convergent subsequence of {zi); with limit Tk(.) and zk(.) is a solution of (79),(80) on [0, fk]. Since (88)-(90) are fulfilled for z i and yo[-, .]
is continuous, the same relations are fulfilled also for zk. Thus Tk E
ik.
Since
Ivi(t) - Y0[~'k(t),t1I = 6
for t = t i ( z i ) and yo[.,
-1
is continuous, the same equality is fulfilled also for ,i?k and t = t i . Since obviously E ~ T O5
t i5
t k ( z k ) we have ti(Tk)1
t i . Thus the supremum of t i ( . ) onik
is attained a t Tk.
8. In the last step we shall modify the maximal element 2 k = ( i t k ( . ) , y k ( . ) ) in a way that contradicts its maximality. This contradiction completes t h e proof, since it is caused by our assumption t h a t t h e claim of the theorem is not true.
For the maximal element 2k denote
ik
= t k ( 2 k ) , YE(.) = y O [ x k ( . ) ,-1.
Denoter
= r o ( b , v), where r o ( . , .) is defined in Supposition C4.We have
-0 0
Y k ( t k ) E k o ( t : , ~ k ( t j r ) , c o p : ) )
-0 0
l ~ k ( ~ : ) - ? j k ( t k ) l = 6 , thus, according t o Supposition C4 the system
has a solution x ( r )
=
i t k ( $ ) , y ( - ) for which-0 0 P
I Y ( ~ ) - ~ k ( t k ) l
<
- for every r >_ 0 , 2Moreover,
x ( r ) = i t k ( t i ) E D , y ( r ) E G'
+
p B C G .Then repeating t h e same Filippov's argument as in Step 5.2 we prove existence of a solution ( Z k ( . ) , ? k ( . ) ) of
such t h a t
where
- L7 -
? ( a , & ) = E M
+
w t ( & r )+
L L ( o I ) E T , d = e r .Now we extend ~k on
[$,ti +
E ~ T ] as Z k ( t ) = ? k ( ( t - t i ) / ~ k ) , y k ( t ) = f i k ( ( t -Obviously it satisfies ( 7 9 ) , ( 8 0 ) on [ t i ,
ti + Ekr].
Moreover, for all sufficiently large k5 %
2 &'(a, EX)+
u:((M+
].)&IT)5
p.Similarly, for t = tg
+
& k TThe inequality (94) implies t h a t tg+&kT
5
tk(Zk) and then the last inequality gives t!(Zk) 2 t! + & k T . This is a contradiction with the maximality oft!. T h e proof is complete. Q.E.D.References
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