Opuscula Math. 38, no. 3 (2018), 307–326 https://doi.org/10.7494/OpMath.2018.38.3.307
Opuscula Mathematica
BACKWARD STOCHASTIC VARIATIONAL INEQUALITIES DRIVEN BY MULTIDIMENSIONAL FRACTIONAL BROWNIAN MOTION Dariusz Borkowski and Katarzyna Jańczak-Borkowska Communicated by Tomasz Zastawniak Abstract. We study the existence and uniqueness of the backward stochastic variational inequalities driven by m-dimensional fractional Brownian motion with Hurst parameters Hk (k = 1, . . . m) greater than 1/2. The stochastic integral used throughout the paper is the divergence type integral. Keywords: backward stochastic differential equation, fractional Brownian motion, backward stochastic variational inequalities, subdifferential operator. Mathematics Subject Classification: 60H05, 60H07, 60H22.
1. INTRODUCTION A centred Gaussian process B H = {BtH , t ≥ 0} is called a fractional Brownian motion (fBm for short) with Hurst parameter H ∈ (0, 1) if it has the covariance function 1 2H RH (s, t) = E BsH BtH = t + s2H − |t − s|2H . 2
H This process for any H is a self-similar, i.e. Bat has the same law as aH BtH for any a > 0, it has homogeneous increments. For H = 1/2 we obtain a standard Wiener process, but for H = 6 1/2, the process B H is not a semimartingale. If H > 1/2, B H H H H has P∞a long range dependence, that is if we put r(n) = cov(B1 , Bn+1 − Bn ), then n=1 r(n) = +∞. These properties make this process a useful tool in models related to network traffic analysis, mathematical finance, physics, signal processing and many other fields. Unfortunately we cannot use the classical theory of stochastic calculus to define the fractional stochastic integral because generally B H is not a semimartingale (it happens only if H = 1/2). Nevertheless an efficient stochastic calculus of B H
c Wydawnictwa AGH, Krakow 2018
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Dariusz Borkowski and Katarzyna Jańczak-Borkowska
has been developed. Firstly definitions of integrals of Stratonovich type with respect to fBm were introduced (see [7, 14]), but they did not satisfy the natural property Rt E o fs dBsH = 0. Next, a new type of stochastic integral with respect to fBm was defined to satisfy the mentioned property. The definition introduced in [8] is the divergence operator (Skorokhod integral), defined as the adjoint of the derivative operator in the framework of the Malliavin calculus and the equivalent for H > 1/2 definition introduced in [9] is based on the Wick product as the limit of Riemann sums. Pardoux and Peng were first who proved the existence and uniqueness of solution of the backward stochastic differential equations (BSDEs) with respect to Wiener process (see [19]). Since then many papers have been devoted to the study of BSDEs, mainly due to their applications. The main purpose of studying these equations was to give a probabilistic interpretation for viscosity solutions of semilinear partial differential equations. BSDEs driven by a fBm with Hurst parameter H > 1/2 were first considered in [3] and with Hurst parameter H ∈ (0, 1) in [2]. Next, Hu and Peng in [12] considered the nonlinear BSDEs with respect to a fBm with Hurst parameter H > 1/2. However, the existence and uniqueness of the solution of the BSDE driven by a fBm was obtained there with some restrictive assumption. Maticiuc and Nie, [15] improved their result and omitted this assumption. Moreover, they developed a theory of backward stochastic variational inequalities, that is they proved the existence and uniqueness of the solution of the reflected BSDEs driven by a fBm. The existence and uniqueness of the generalized BSDEs driven by a fBm with Hurst parameter H greater than 1/2 was shown in [13] and in [4] the existence and uniqueness of the generalized backward stochastic variational inequalities driven by a fBm with Hurst parameter H greater than 1/2 was proven. The existence and uniqueness of BSDE driven by multidimensional fBm was shown with a very restrictive assumption by Miao and Yang in [17]. In [5] it was shown that this assumption is redundant. In the current paper we show the existence and uniqueness of the backward stochastic variational inequalities (BSVI for short) driven by multidimensional fBm, i.e. we consider the reflected BSDE. In our approach we use as stochastic integral the divergence operator. The paper is organized as follows. In Section 2 we recall some definitions and results about a fractional stochastic integral, which will be needed throughout the paper. Section 3 contains assumptions, the definition of BSDE driven by multidimensional fBm and formulation of the main theorem of the paper. Section 4 is devoted for some a priori estimates. Finally, in Section 5 we prove the main theorem using a penalization method. 2. FRACTIONAL CALCULUS – MULTIDIMENSIONAL FBM Assume that B H1 , B H2 , . . . , B Hm are independent fractional Brownian motions with Hurst parameters H1 , H2 , . . . , Hm respectively, where Hk ∈ 12 , 1 , k = 1, 2, . . . , m.
309
Backward stochastic variational inequalities. . .
It is well known that EBtHk = 0 and 1 |t|2Hk + |s|2Hk − |t − s|2Hk δkj , E BtHk · BsHj = 2
1 ≤ k, j ≤ m, where δkj = 1 for k = j and δkj = 0 for k 6= j. Now, fix a Hurst index Hk ∈ ( 12 , 1) and define φk (x) = Hk (2Hk − 1)|x|2Hk −2 ,
Let ξ, η : [0, T ] → R be two continuous function. Define hξ, ηik,t =
Zt Zt 0
0
x ∈ R.
φk (u − v)ξ(u)η(v)dudv,
0≤t≤T
and for ξ = η, |ξ|2k,T
= hξ, ξik,t =
ZT ZT 0
0
φk (u − v)ξ(u)ξ(v)dudv < ∞.
The Malliavin derivative operator DHk of an element T Z ZT F (ω) = f ξ1 (t)dBtHk , . . . , ξl (t)dBtHk , 0
0
where f is a polynomial function of l variables, is defined as follows: T Z ZT l X ∂f ξ1 (t)dBtHk , . . . , ξl (t)dBtHk ξi (s), s ∈ [0, T ]. DsHk F = ∂x i i=1 0
0
Now, introduce another derivative k DH t F
=
ZT 0
φk (t − s)DsHk F ds.
Theorem 2.1. Let F : (Ω, F, P) → H be a stochastic process such that ZT ZT 2 k < ∞. E kF k2k,T + |DH s Ft | dsdt 0
0
RT Then, the Itô-type stochastic integral denoted by 0 Fs dBsHk exists in L2 (Ω, F). Moreover, T Z E Fs dBsHk = 0 0
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Dariusz Borkowski and Katarzyna Jańczak-Borkowska
T 2 Z ZT ZT Hk k DH E Fs dBsHk = E kF k2k,T + s Ft Dt Fs dsdt . 0
0
0
Theorem 2.2 (multidimensional fractional Itô formula). Let σik ∈ L2 ([0, T ]), i = 1, . . . d, k = 1, . . . , m, be deterministic functions. Suppose that hσik , σjk ik,t are continuously differentiable as functions of t ∈ [0, T ]. Set Xt = (Xt1 , . . . , Xtd ) where Xti = X0i +
Zt
bi (s)ds +
m Z X
t
σik (s)dBsHk ,
t ∈ [0, T ],
k=1 0
0
i = 1, . . . , d,
X0 = (X01 , . . . , X0d ) is a constant vector and bi are deterministic continuous functions RT with 0 |bi (s)|ds < ∞, i = 1, . . . , d. Let F be continuously differentiable with respect to t and twice continuously differentiable with respect to x. Then F (t, Xt ) = F (0, X0 ) +
Zt 0
+
d X
Zt
i,j=1 0
2
Zt 0
+
d X m Zt X i=1 k=1 0 d Zt
1 X 2 i,j=1
t
0
∂ F (s, Xs ) ∂xi xj
= F (0, X0 ) +
+
d Z X ∂F ∂F (s, Xs )ds + (s, Xs )dXsi ∂s ∂x i i=1
0
m X
j k σik (s)DH s (Xs )ds
k=1
d Z X ∂F ∂F (s, Xs )ds + (s, Xs )bi (s)ds ∂s ∂x i i=1 t
0
∂F (s, Xs )σik (s)dBsHk ∂xi X d ∂2F (hσik , σjk ik,s ) ds, (s, Xs ) ∂xi xj ds m
k=1
t ∈ [0, T ].
Theorem 2.3 (fractional Itô chain rule). Let T ∈ (0, ∞) and let b1 (s), b2 (s), σ1k (s), σ2k (s) be in D1,2 and T Z E (|bi (s)| + |σij (s)|) ds < ∞, i = 1, 2, j = 1, . . . , m. 0
Assume that Dt j σij (s) are continuously differentiable with respect to (s, t) ∈ [0, T ] × [0, T ] for almost all ω ∈ Ω. Suppose that H
E
ZT ZT 0
0
|Dt j σij (s)|2 dsdt < ∞, H
i = 1, 2, j = 1, . . . , m.
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Backward stochastic variational inequalities. . .
Denote F (t) = F (0) +
Zt
b1 (s)ds +
m Z X
0
k=1 0
Zt
m Z X
t
σ1k (s)dBsHk ,
t ∈ [0, T ]
σ2k (s)dBsHk ,
t ∈ [0, T ].
and G(t) = G(0) +
b2 (s)ds +
t
k=1 0
0
Then F (t)G(t) = F (0)G(0) +
+
m X
k=1
Zt
F (s)dG(s) +
0
Zt 0
F (s)b2 (s)ds +
G(s)b1 (s)ds +
0
+
m X
k=1
Zt 0
Zt
0
k DH s G(s)σ1k (s)ds
m Z X
m Zt X
t
F (s)σ2k (s)dBsHk
G(s)σ1k (s)dBsHk
k=1 0
Zt
k=1 0
0
+
G(s)dF (s)
0
k DH s F (s)σ2k (s)ds +
= F (0)G(0) + Zt
Zt
k DH s F (s)σ2k (s)ds +
Zt 0
k DH s G(s)σ1k (s)ds .
The above theorems can be found in [9, 10, 12, 15, 17] and for a deeper discussion we refer the reader to [10, 18]. In solving backward stochastic differential equation with respect to a fractional Brownian motion the major problem is the absence of a martingale representation type theorem. Considering such equations an important role plays the quasi conditional expectation, which was introduced in [11]. For a while, in what follows to simplify the notations we will drop the superscripts k in the Hurst index H and in a function φ. For any natural n define the set H⊗n of all real symmetric Borel functions f of n variable such that kf k2H⊗n ,T
=
Z
[0,T ]2n
n Y
i=1
φ(si − ri )f (s1 , . . . , sn )f (r1 , . . . , rn )ds1 . . . dsn dr1 . . . drn < ∞.
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Then we can define the iterated integral in the sense of Itô-Skorokhod (see [9]) Z InH (f ) = f (t1 , . . . , tn )dBtH1 . . . dBtHn [0,T ]n
= n!
Z
0≤t1 0 such that for all t ∈ [0, T ], s2Hk −1 σ ˆk (s) ≤ ≤ Mk s2Hk −1 . Mk σk (s)
(4.2)
Denote M = max1≤k≤m Mk . By the above, integrating (4.1) we have m Z 2 X s2Hk −1 |Zsk |2 ds E|Yt | + M T
2
k=1 t
m Z X σ ˆk (s) k 2 ≤ E|Yt | + 2 |Z | ds σk (s) s T
2
(4.3)
k=1 t
2
= E|ξ| + 2E
ZT t
Ys f (s, ηs , Ys , Zs )ds − 2E
ZT t
Ys Us ds.
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Dariusz Borkowski and Katarzyna Jańczak-Borkowska
By Remark 3.1, the last component of the right hand side is non-positive. By the Lipschitz continuity of f and the simple inequality 2ab ≤ a/ε + εb, we have 2yf (s, η, y, z) ≤ 2L|y| (|y| + kzk) + 2|y||f (s, η, 0, 0)| v um uX ≤ (2L + 1) |y|2 + |f (s, η, 0, 0)|2 + 2Lt |y|2 |z k |2 k=1
≤ (2L + 1) |y|2 + |f (s, η, 0, 0)|2 + m X M L2 2L + 1 + s2Hk −1
≤
k=1
and therefore
!
m X
k=1
2L|y||z k |
|y|2 + |f (s, η, 0, 0)|2 +
m 1 X 2Hk −1 k 2 s |z | M k=1
ZT m ZT 2 X 2Hk −1 k 2 2 2 s |Zs | ds ≤ E |ξ| + |f (s, ηs , 0, 0)| ds E|Yt | + M 2
k=1 t
+E
ZT t
m X M L2 2L + 1 + s2Hk −1 k=1
Denoting
!
t
Z m 1 X |Ys | ds + E s2Hk −1 |Zsk |2 ds. M
Θ(t, T ) = E |ξ|2 +
we can write
T
2
k=1
ZT t
t
|f (s, ηs , 0, 0)|2
m ZT X 1 E |Yt |2 + s2Hk −1 |Zsk |2 ds ≤ Θ(t, T ) M k=1 t
+E
ZT t
m X M L2 2L + 1 + s2Hk −1 k=1
!
(4.4) |Ys |2 ds.
By the Gronwall inequality, (
m X T 2−2Hk − t2−2Hk E|Yt | ≤ Θ(t, T ) exp (2L + 1)(T − t) + M L 2 − 2Hk 2
2
k=1
and by (4.4) also
E
m Z X
k=1 t
T
s2Hk −1 |Zsk |2 ds ≤ CΘ(t, T ).
)
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Backward stochastic variational inequalities. . .
˜ U ˜ ) be two solutions Proposition 4.2. Assume (H1 )–(H4 ) and let (Y, Z, U ) and (Y˜ , Z, ˜ ˜ of (3.2) with data (ξ, f ) and (ξ, f ), respectively. Then
m Z X
T
E |Yt − Y˜t |2 +
2Hk −1
s
k=1 t
˜2 + ≤ CE |ξ − ξ|
ZT t
|Zsk
k 2 ˜ − Zs | ds
2 ˜ |f (s, ηs , Ys , Zs ) − f (s, ηs , Ys , Zs )| ds .
Proof. By the Itô formula, computing similarly as in the previous theorem m Z 2 X 2 ˜ s2Hk −1 |Zsk − Z˜sk |2 ds |Yt − Yt | + M T
k=1 t
˜2 + 2 ≤ |ξ − ξ| −2
ZT X m t
k=1
ZT t
(Ys − Y˜s ) f (s, ηs , Ys , Zs ) − f˜(s, ηs , Y˜s , Z˜s ) ds
(Ys −
Y˜s )(Zsk
−
Z˜sk )dBsHk
−2
ZT t
˜s )ds. (Ys − Y˜s )(Us − U
From assumptions we get 2(y − y˜)(f (s, η, y, z) − f˜(s, η, y˜, z˜)) ≤ 2(y − y˜)(f (s, η, y, z) − f˜(s, η, y, z)) ! m m X X L2 M s2Hk −1 k 2 + 2L + |y − y ˜ | + |z − z˜k |2 2H −1 s k M k=1
k=1
≤ |f (s, η, y, z) − f˜(s, η, y, z)|2 ! m m X 1 X 2Hk −1 k L2 M 2 + 2L + 1 + |y − y ˜ | + s |z − z˜k |2 . s2Hk −1 M k=1
k=1
˜t ∈ ∂ϕ(Y˜t ), Since Ut ∈ ∂ϕ(Yt ) and U ˜t )(Yt − Y˜t ) = Ut (Yt − Y˜t ) + U ˜t (Y˜t − Yt ) (Ut − U ≥ ϕ(Yt ) − ϕ(Y˜t ) + ϕ(Y˜t ) − ϕ(Yt ) = 0.
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Dariusz Borkowski and Katarzyna Jańczak-Borkowska
Therefore, we obtain
1 E |Yt − Y˜t |2 + M
ZT X m t
2Hk −1
s
k=1
|Zsk
k 2 ˜ − Zs | ds
ZT 2 ˜ + f (s, ηs , Ys , Zs )− f˜(s, ηs , Ys , Zs ) ds ≤ E |ξ − ξ| t
+E
ZT
2
2L + 1 + L M
m X
k=1
t
1
s2Hk −1
!
|Ys − Y˜s |2 ds.
Using the Gronwall Lemma we get the required inequality. 5. PENALIZATION SCHEME We will approximate the function ϕ by a sequence of convex, C 1 class functions ϕε , ε > 0, defined by 1 1 ϕε (y) = inf |y − v|2 + ϕ(v) : v ∈ R = |y − Jε (y)|2 + ϕ(Jε (y)), (5.1) 2ε 2ε where Jε (y) = y − ε∇ϕε (y). Here are some properties of ϕε (see [1] or [6]):
y − Jε (y) ∈ ∂ϕ(Jε (y)), ε |Jε (y) − Jε (v)| ≤ |y − v| and lim Jε (y) = πDomϕ (y),
∇ϕε (y) =
ε&0
0 ≤ ϕε (y) ≤ y∇ϕε (y),
(5.2) (5.3) (5.4)
where by πDomϕ (y) we denote the projection of y on the closure of the set Domϕ. Note that if for some a ≤ 0 we define convex indicator function ( 0, y ≥ a, ϕ(y) = ∞, y < a, then ∇ϕε (y) = − 1ε (y − a)− , where x− = max(−x, 0). Consider a sequence of BSDEs Ytε
=ξ+
ZT t
m Z X
T
f (s, ηs , Ysε , Zsε )ds
−
k=1 t
Zsk,ε dBsHk
−
ZT t
∇ϕε (Ysε )ds,
t ∈ [0, T ]. (5.5)
Since ∇ϕε is Lipschitz continuous function, then by [5] (5.5) has a unique solution (Y ε , Z ε ).
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Backward stochastic variational inequalities. . .
Proposition 5.1. Let assumptions (H1 )–(H4 ) hold. Then E
m Z X
T
|Ytε |2
+
2Hk −1
s
k=1 t
|Zsk,ε |2 ds
+
ZT t
Ysε ∇ϕε (Ysε )ds
ZT 2 2 ≤ CE |ξ| + |f (s, ηs , 0, 0)| ds = CΘ(t, T ). t
Proof. Similarly as in the proof of Theorem 4.1, we have m ZT 1 X 2Hk −1 k,ε 2 ε 2 s |Zs | ds E |Yt | + M k=1 t
ZT 2 2 ≤ E |ξ| + |f (s, ηs , 0, 0)| ds t
+E
ZT t
m X M L2 2L + 1 + s2Hk −1 k=1
!
|Ysε |2 ds
− 2E
ZT t
Ysε ∇ϕε (Ysε )ds.
By (5.4), the last component of the right hand side is non-positive, so we obtain ZT m ZT 1 X s2Hk −1 |Zsk,ε |2 ds + 2 Ysε ∇ϕε (Ysε )ds E |Ytε |2 + M k=1 t
≤ Θ(t, T ) + E
ZT
t
2L + 1 +
m X M L2 s2Hk −1
k=1
t
!
|Ysε |2 ds.
Now using similar arguments as in the proof of Theorem 4.1 we finish the proof. Proposition 5.2. Under assumptions (H1 )–(H4 ) there exists a positive constant C such that for any t ∈ [0, T ] m Z X
T
(a) E (b) E (c) E
k=1 t m X 2Hk −1
t
k=1 m X
k=1
(d) E
s2Hk −1 |∇ϕε (Ysε )|2 ds ≤ CΘ2 (t, T ), 2
|Ytε − Jε (Ytε )| ≤ ε · CΘ2 (t, T ),
t2Hk −1 ϕ (Jε (Ytε )) ≤ CΘ2 (t, T ),
m Z X
k=1 t
T 2
s2Hk −1 |Ysε − Jε (Ysε )| ds ≤ ε2 CΘ2 (t, T ),
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Dariusz Borkowski and Katarzyna Jańczak-Borkowska
where m X
Θ2 (t, T ) = E
T
2Hk −1
ϕ(ξ) +
k=1
ZT t
s2Hk −1 |Ysε |2 + |Zsk,ε |2 + |f (s, ηs , 0, 0)|2
.
Proof. In the proof below we will use similar arguments as in the proof of Proposition 2.2 in [20] and in the proof of Proposition 11 in [16]. Since ∇ϕε (Yrε ) ∈ ∂ϕ(Jε (Yrε )), ∇ϕε (Yrε ) · (Ysε − Yrε ) ≤ ϕε (Ysε ) − ϕε (Yrε )
and
ϕε (Ysε ) ≥ ∇ϕε (Yrε ) · (Ysε − Yrε ) + ϕε (Yrε ).
Now for any k ∈ {1, 2, . . . , m} and s > r ≥ 0
s2Hk −1 ϕε (Ysε ) ≥ s2Hk −1 ϕε (Yrε ) + s2Hk −1 ∇ϕε (Yrε ) · (Ysε − Yrε )
≥ r2Hk −1 ϕε (Yrε ) + s2Hk −1 ∇ϕε (Yrε ) · (Ysε − Yrε ) .
Take s = ti+1 ∧ T , r = ti ∧ T , where 0 = t0 < t1 < . . . < t ∧ T and ti+1 − ti = 1/n. Summing up over i and passing to the limit as n → ∞, we deduce T 2Hk −1 ϕε (YTε ) ≥ t2Hk −1 ϕε (Ytε ) +
ZT t
s2Hk −1 ∇ϕε (Ysε )dYsε .
Therefore, t
2Hk −1
ϕε (Ytε )
2Hk −1
≤T
2Hk −1
=T
ϕε (ξ) −
ZT
s2Hk −1 ∇ϕε (Ysε )dYsε
ϕε (ξ) +
ZT
s2Hk −1 ∇ϕε (Ysε )f (s, ηs , Ysε , Zsε )ds
t
t
−
ZT
2Hk −1
s
t
∇ϕε (Ysε )
and t
2Hk −1
ϕε (Ytε )
+
ZT t
≤T −
2Hk −1
ϕε (ξ) +
t
Zsj,ε dBsHj
j=1
−
ZT t
2
s2Hk −1 |∇ϕε (Ysε )| ds
2
s2Hk −1 |∇ϕε (Ysε )| ds ZT t
ZT
m X
s2Hk −1 ∇ϕε (Ysε )f (s, ηs , Ysε , Zsε )ds
s2Hk −1 ∇ϕε (Ysε )
m X j=1
Zsj,ε dBsHj .
(5.6)
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Backward stochastic variational inequalities. . .
Note that ∇ϕε (y)f (s, η, y, z) ≤ |∇ϕε (y)| (L|y| + Lkzk + |f (s, η, 0, 0)|) 1 3 2 2 L |y| + L2 kzk2 + |f (s, η, 0, 0)|2 . ≤ |∇ϕε (y)|2 + 2 2
Using the fact that ϕε (ξ) ≤ ϕ(ξ) and integrating the inequality (5.6) we get 2Hk −1
Et
ϕε (Ytε )
1 + E 2
ZT t
2
s2Hk −1 |∇ϕε (Ysε )| ds
ZT 3 2Hk −1 ≤ ET ϕ(ξ) + E s2Hk −1 L2 |Ysε |2 + L2 kZsε k2 + |f (s, ηs , 0, 0)|2 ds. 2
(5.7)
t
Since Hk was arbitrary, we can write similar inequalities for every k = 1, . . . , m. Now, summing up (5.7) over k we have
E
m X
m Z X
T
t
2Hk −1
ϕε (Ytε )
+E
k=1
k=1 t
2
s2Hk −1 |∇ϕε (Ysε )| ds
ZT m X 2Hk −1 ≤C E T ϕ(ξ) + s2Hk −1 L2 |Ysε |2 + L2 kZsε k2 + |f (s, ηs , 0, 0)|2 ds k=1
t
= CΘ2 (t, T ).
From the above inequality (a) is clear. Condition (c) follows additionally from inequality ϕ(Jε (y)) ≤ ϕε (y). From |y − Jε (y)|2 ≤ 2εϕε (y) follows (b). Finally (d) we get from y − Jε (y) = ε∇ϕε (y). Proposition 5.3. Let assumptions (H1 )–(H4 ) be satisfied. Then (Y ε , Z ε ) is a Cauchy sequence, i.e. for ε, δ > 0 ZT m X 2Hk −1 ε δ 2 E t |Yt − Yt | + s2Hk −2 |Ysε − Ysδ |2 ds
k=1
t
+
ZT t
2Hk −1
s
m X j=1
≤ C · (ε + δ) · Θ2 (t, T ).
2Hj −1
s
|Zsj,ε
−
Zsj,δ |2 ds
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Dariusz Borkowski and Katarzyna Jańczak-Borkowska
Proof. Put Yˇ = Y ε − Y δ and Zˇ = Z ε − Z δ . For any k ∈ {1, 2, . . . , m}, we have 2Hk −1
t
|Yˇt | + 2
ZT t
(2Hk − 1)s2Hk −2 |Yˇs |2 ds
ZT ZT m X σ ˆj (s) ˇ j 2 = T 2Hk −1 |YˇT |2 − 2 s2Hk −1 Yˇs dYˇs − 2 s2Hk −1 |Z | ds. σ (s) s j=1 j t
t
Therefore, ZT ZT m X 2 2Hk −1 ˇ 2 2Hk −2 ˇ 2 E t |Yt | + (2Hk − 1)s s2Hk −1 s2Hj −1 |Zˇsj |2 ds |Ys | ds + M j=1 t
t
≤ 2E
ZT t
− 2E
s2Hk −1 Yˇs f (s, ηs , Ysε , Zsε ) − f (s, ηs , Ysδ , Zsδ ) ds
ZT t
s2Hk −1 Yˇs ∇ϕε (Ysε ) − ∇ϕδ (Ysδ ) ds.
Note that m 2 X L M |ˇ y |2 2s2Hk −1 yˇ · f (s, η, y ε , z ε ) − f (s, η, y δ , z δ ) ≤ s2Hk −1 2L + 2Hj −1 s j=1 m 1 2Hk −1 X 2Hj −1 j 2 s s |ˇ z | . + M j=1
Moreover, by the definition of ϕε we get 0 ≤ ∇ϕε (Ysε ) − ∇ϕδ (Ysδ ) · Jε (Ysε ) − Jδ (Ysδ ) = ∇ϕε (Ysε ) − ∇ϕδ (Ysδ ) · Ysε − Ysδ − ε∇ϕε (Ysε ) + δ∇ϕδ (Ysδ ) = ∇ϕε (Ysε ) − ∇ϕδ (Ysδ ) · Ysε − Ysδ − ε|∇ϕε (Ysε )|2 − δ|∇ϕδ (Ysδ )|2 + (ε + δ)∇ϕε (Ysε ) · ∇ϕδ (Ysδ )
and then ∇ϕε (Ysε ) − ∇ϕδ (Ysδ ) · Ysε − Ysδ ≥ −(ε + δ)∇ϕε (Ysε ) · ∇ϕδ (Ysδ ).
323
Backward stochastic variational inequalities. . .
Therefore, ZT m ZT 1 X 2Hk −2 ˇ 2 2Hk −1 ˇ 2 |Ys | ds + E t |Yt | + (2Hk − 1)s s2Hk −1 s2Hj −1 |Zˇsj |2 ds M j=1 t t ZT m X L2 M s2Hk −1 |Yˇs |2 ds (5.8) ≤ E 2L + s2Hj −1 j=1 t
+ 2(ε + δ)E
ZT t
s2Hk −1 ∇ϕε (Ysε ) ∇ϕδ (Ysδ )ds.
By the Gronwall Lemma and by the simple inequality ab ≤ a2 /2 + b2 /2, Et2Hk −1 |Yˇt |2 ≤ C(ε + δ)E
ZT
≤ C(ε + δ)E
ZT
t
t
s2Hk −1 ∇ϕε (Ysε ) ∇ϕδ (Ysδ )ds s2Hk −1 |∇ϕε (Ysε )|2 + |∇ϕδ (Ysδ )|2 ds.
From the above and from (5.8) ZT ZT m X E t2Hk −1 |Yˇt |2 + s2Hk −2 |Yˇs |2 ds + s2Hk −1 s2Hj −1 |Zˇsj |2 ds t
≤ C(ε + δ)E
ZT t
j=1
t
s2Hk −1 |∇ϕε (Ysε )|2 + |∇ϕδ (Ysδ )|2 ds.
Summing up over k, k = 1, 2, . . . , m and using Proposition 5.2 a) we get the result. Now we can give a proof of Theorem 3.3. Proof of Theorem 3.3. First, we show the uniqueness. From the proof of Proposition 4.2 it follows that for (Y, Z, U ) and (Y 0 , Z 0 , U 0 ) being two solutions of (3.2), we have m ZT X 0 2 2Hk −1 k 0k 2 E |Yt − Yt | + s |Zs − Zs | ds = 0, k=1 t
and E
ZT t
(Ys −
Ys0 )(Us
which means that the solution is unique.
−
Us0 )ds
≤ 0,
324
Dariusz Borkowski and Katarzyna Jańczak-Borkowska
Now we will show that the limit of (Y ε , Z ε , ∇ϕε (Y ε )) converges to a solution of (3.2). Since by Proposition 5.3 (Y ε , Z ε ) is a Cauchy sequence, there exists its limit, i.e. there exists a pair of processes (Y, Z) such that Y ∈ V˜TH1 ∩ V˜TH2 . . . ∩ V˜THm , Z = (Z 1 , . . . , Z k ), where H +H −1/2 H +H −1/2 H +H −1/2 Z k ∈ V˜T 1 k ∩ V˜T 2 k . . . ∩ V˜T m k
and lim E
ε&0
X m ZT k=1 t
2Hk −1
s
|Ysε
2
− Ys | ds +
m ZT X
k,j=1 t
s2(Hk +Hj −1/2)−1 |Zsj,ε − Zsj |2 ds = 0.
From Proposition 5.2 c), lim E
ε&0
and we have
m X
k=1
2
t2Hk −1 |Ytε − Jε (Ytε )| = 0,
lim Jε (Y ε ) = Y in V˜TH1 ∩ V˜TH2 . . . ∩ V˜THm .
ε&0
Denoting U ε = ∇ϕε (Y ε ) from Proposition 5.2 a) we obtain m Z X
T
E
k=1 0
s2Hk −1 |Usε |2 ds ≤ C.
Hence there exist a subsequence εn & 0 and process U such that U εn → U weakly in V˜TH1 ∩ V˜TH2 . . . ∩ V˜THm
and from the Fatou Lemma m Z X
T
E
k=1 0
s2Hk −1 |Us |2 ds ≤ C.
Passing now with ε to 0 in (5.5) we obtain (3.2). Moreover, since Utε ∈ ∂ϕ(Jε (Ytε )), for all u ∈ V˜TH1 ∩ V˜TH2 . . . ∩ V˜THm we have Utε · (ut − Jε (Ytε )) + ϕ(Jε (Ytε )) ≤ ϕ(ut ). Therefore we can deduce (passing to limes infimum) that Ut · (ut − Yt ) + ϕ(Yt ) ≤ ϕ(ut ), which means that (Yt , Ut ) ∈ ∂ϕ, t ∈ [0, T ]. This completes the proof.
Backward stochastic variational inequalities. . .
325
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Dariusz Borkowski
[email protected] Nicolaus Copernicus University Faculty of Mathematics and Computer Science ul. Chopina 12/18, 87-100 Toruń, Poland
Katarzyna Jańczak-Borkowska
[email protected] University of Science and Technology Institute of Mathematics and Physics al. prof. S. Kaliskiego 7, 85-796 Bydgoszcz, Poland
Received: March 7, 2017. Revised: October 22, 2017. Accepted: November 17, 2017.
