Stochastic differential equations with time-dependent coefficients driven by fractional Brownian motion

Stochastic differential equations with time-dependent coefficients driven by fractional Brownian motion

Physica A 530 (2019) 121565 Contents lists available at ScienceDirect Physica A journal homepage: www.elsevier.com/locate/physa Stochastic differen...

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Physica A 530 (2019) 121565

Contents lists available at ScienceDirect

Physica A journal homepage: www.elsevier.com/locate/physa

Stochastic differential equations with time-dependent coefficients driven by fractional Brownian motion✩ ∗

Zhi Li , Wentao Zhan, Liping Xu School of Information and Mathematics, Yangtze University, Jingzhou, Hubei 434023, PR China

highlights • We prove the existence and uniqueness of a strong solutions for (5) when b1 is linear growth but is discontinuous in x. • Our results improve the results of the Theorem 3.2 and 3.4 in [1]. • Using the unique strong solution for (5), we obtain existence of the weak solution for (1) only assuming that satisfies (3) without any continuity assumption.

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a b s t r a c t In this paper, we study a class of stochastic differential equations with a time-dependent diffusion driven by a fractional Brownian motion with Hurst parameter 1/2 < H < 1. By using a transformation formula for fractional Brownian motion, we prove the existence of weak solutions to this kind of equations under the linear growth condition, but the drift can be discontinuous. Some known results are improved. © 2019 Elsevier B.V. All rights reserved.

Article history: Received 26 April 2019 Available online 29 May 2019 MSC: 60H15 60G15 60H05 Keywords: Stochastic differential equations Fractional Brownian motion Girsanov theorem

1. Introduction In this paper, we are interested in the following stochastic differential equation (SDE):

∫ xt = x0 +

t

(b1 (s, xs ) + b2 (s, xs ))ds +

0

t



σ (s)dBHs ,

(1.1)

0

where b1 , b2 , σ : [0, T ]× R are Borel functions, BH = {BH t , t ∈ [0, T ]} is a fractional Brownian motion with Hurst parameter H ∈ (0, 1). In the case H = 1/2, the process BH is the standard Brownian motion and Zvonkin [2] and Veretennikov [1] studied the existence of a weak (strong) solution to the following stochastic differential equation xt = x0 + BH (t) +

t



b1 (s, xs )ds,

(1.2)

0

✩ This research is partially supported by the NNSF of China (No.11571071), and Natural Science Foundation of Hubei Province, PR China (No.2016CFB479). ∗ Corresponding author. E-mail address: [email protected] (Z. Li). https://doi.org/10.1016/j.physa.2019.121565 0378-4371/© 2019 Elsevier B.V. All rights reserved.

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Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

assuming only that the coefficient b1 (t , x) satisfies the following linear growth in x uniformly in time t

|b1 (t , x)| ≤ C (1 + |x|).

(1.3)

In the singular case 0 < H < 1/2, Nualart and Ouknine [3] established the existence of a weak solution to (1.2) by applying the Girsanov theorem, also assuming only that the coefficient b(t , x) has linear growth in x uniformly in time t. But, in the regular case 1/2 < H < 1, since the application of the Girsanov theorem in the fractional Brownian motion with Hurst parameter 1/2 < H < 1 requires Hölder continuity conditions of the coefficient b1 , Nualart and Ouknine [3] proved that (1.2) has a weak solution if the coefficient b1 satisfies the following Hölder continuity condition

|b1 (t , x) − b1 (s, y)| ≤ C (|x − y|α + |t − s|γ ),

(1.4)

where 1 − 1/2H < α < 1 and γ > H − 1/2. By taking advantage of the properties of the sign function, Mishura and Nualart [4] established the existence of a weak solution for (1.2) when√ the function b1 was Hölder continuous of order 1 − 1/2H < α < 1 in a finite number of intervals and 1/2 < H < (1 + 5)/4. Denis et al. [5] investigated the existence of a weak solution for (1.1) with 1/2 < H < 1 when σ ≡ 1, b1 is only dependent on time t and b2 satisfies the condition (1.4). Boufoussi and Ouknine [6] found a weak solution for (1.1) with σ ≡ 1, where the function b2 satisfies the condition (1.4) and b1 (s, ·) is left-continuous and nondecreasing or b1 (s, ·) is linear growth, nonincreasing and continuous function for each s. The main aim of our work is to consider the existence of a weak solution for (1.1) with 1/2 < H < 1 under some less restrictive conditions. As far as we know, in the case of time-dependent diffusion σ , the proofs involve heavy calculations which have to be carefully modified if σ ≡ 1 is not satisfied. On the other hand, in the regular case H ∈ (1/2, 1) a less limited integration theory is available, and in the singular case H ∈ (0, 1/2) the Girsanov transform is applicable under less restrictive conditions. To this end, we first consider the existence of the strong solution for the following stochastic differential equation (SDE): t



b1 (s, xs )ds +

xt = x0 +

t



σ (s)dBHs .

(1.5)

0

0

By a new approximation argument, we prove the existence and uniqueness of a strong solutions for (1.5) when |b1 (s, x)| ≤ m(s)(1 +|x|), m(s) ∈ L1 ([0, T ]) and b1 is a left (or right) continuous in x, which improves the results of the Theorem 3.2 and 3.4 in [6]. Subsequently, motivated mainly by [7], we will transform the fractional Brownian motion with Hurst parameter 1/2 < H < 1 into some integral with respect to the fractional Brownian motion with Hurst parameter 0 < H < 1/2 by using of the transformation formula for fractional Brownian motion. Then, we can transform the application of the Girsanov theorem in the fractional Brownian motion with Hurst parameter 1/2 < H < 1 into the application of the Girsanov theorem in the fractional Brownian motion with Hurst parameter 0 < H < 1/2. Lastly, by using the unique strong solution for (1.5), we will obtain existence of the weak solution for (1.1) with Hurst parameter 1/2 < H < 1 only assuming that σ −1 b2 satisfies (1.3) without any continuity assumption on σ −1 b2 . The rest of this paper is organized as follows. In Section 2, we give some preliminaries on fractional Brownian motion. In Section 3, we consider the existence and uniqueness of a strong solutions for (1.5) by a new approximation argument. In Section 4, we consider the existence of a weak solution for (1.1) with H ∈ (1/2, 1) under some less restrictive conditions. 2. Preliminaries We recall that a stochastic process BH = {BH t , t ≥ 0}, defined on a filtered probability space (Ω , F , P, F = {Ft }t ≥0 ), is called an F-fractional Brownian motion with Hurst parameter H ∈ (0, 1) if (i) BH is a Gauss process with continuous paths and BH 0 = 0; H (ii) for each t ≥ 0, BH is F -measurable and E B = 0, for each t ≥ 0; t t t (iii) for all t , s ≥ 0, it holds that H RH (t , s) = E(BH t Bs ) =

1 2

(t 2H + s2H − |t − s|2H ).

In particular, if H = 21 , BH is a standard Brownian motion. It is well known that if H ̸ = 12 , BH does not have independent increments and has α -order Hölder continuous path for all α ∈ (0, H). For more details of fBm and proofs we refer the readers, for instance, to [8]. On the other hand, from [8], we know the covariance kernel RH (t , s) can be written as RH (t , s) =

t ∧s



KH (t , r)KH (s, r)dr , 0

where KH is a square integrable kernel given by KH (t , s) = Γ (H +

1 2

1

)−1 (t − s)H − 2 F (H −

1 1

,

2 2

1

t

2

s

− H , H + , 1 − ),

Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

3

in which F (·, ·, ·, ·) is the Gauss hypergeometric function and Γ (·) denotes the Euler Gamma function. Moreover, according to [9], fractional Brownian motion has the following integral representation with respect to the standard Brownian motion B = (Bt )t ≥0 BH t =

t



KH (t , s)dBs . 0 H+ 1

By [8], the operator KH : L2 ([0, T ]; R) → I0+ 2 (L2 [0, T ]; R) associated with the square integrable kernel KH (·, ·) is defined as follows t



KH (t , s)f (s)ds, f ∈ L2 ([0, T ]; R).

(KH f )(t) := 0 H+ 1

where I0+ 2 is the H + 12 -order left fractional Riemann–Liouville operator on [0, T ], one can see [10]. It is an isomorphism and for each f ∈ L2 ([0, T ]; R), 1

1

−H

1

−H 2 2 (KH f )(s) = I02H I0 + s H − 2 f , H ≤ +s 1 H− 1

1

(KH f )(s) = I01+ sH − 2 I0+ 2 s 2 −H f , H ≥

1 2 1 2

, .

H+ 1

As a consequence, for every h ∈ I0+ 2 (L2 [0, T ]; R), the inverse operator KH−1 is of the following form H− 1

1

1

(KH−1 h)(s) = sH − 2 D0+ 2 s 2 −H h′ , H > 1

1

−H

1 2

1

,

(KH−1 h)(s) = s 2 −H D02+ sH − 2 D2H 0+ h, H < H− 1

1

(2.1)

1 2

,

(2.2)

−H

where D0+ 2 (D02+ ) is H − 12 ( 12 − H)-order left-sided Riemann–Liouville derivative, one can see [10]. In particular, if h is absolutely continuous, we have 1

1

−H

1

(KH−1 h)(s) = sH − 2 I02+ s 2 −H h′ , H <

1 2

.

(2.3)

Now we aim at introducing the Wiener integral with respect to the fBm BH . Let T > 0 and denote by Λ the linear space of R-valued step functions on [0, T ], that is, ϕ ∈ Λ if

ϕ (t) =

n−1 ∑

xk I[tk ,tk+1 ) ,

k=0

for some n ∈ N, t ∈ [0, T ], 0 = t0 < t1 < · · · < tN = T , xk ∈ R, IA denoting an indicator function of A. Then we define the integral of a function ϕ with respect to fBm as T



ϕ (s)dBHs := 0

n−1 ∑

H xk (BH tk+1 − Btk ).

k=0

Let H be the Hilbert space defined as the closure of Λ with respect to the scalar product

⟨I[0,t ] , I[0,s] ⟩H = RH (t , s). Then the mapping

ϕ=

n−1 ∑

T



ϕ (s)dBHs

xk I[tk ,tk+1 ) → 0

k=0

is an isometry between Λ and the linear space span {BH t , t ∈ [0, T ]}, which can be extended to an isometry between H L2 (Ω )

and the first Wiener chaos of the fBm span {BHt , t ∈ [0, T ]}. The image of an element ϕ ∈ H by this isometry is called H the Wiener integral of ϕ with respect to B . Consider the linear operator KH∗ : Λ → L2 ([0, T ]; R) given by (KH∗ ϕ )(s) =

t



ϕ (t) s

∂ KH (t , s)dt . ∂t

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Then (KH∗ I[0,t ] )(s) = KH (t , s)I[0,t ] (s) and KH∗ is an isometry between I and L2 ([0, T ]; R) that can be extended to H. Then for all ϕ, ψ ∈ Λ we have T

⟨∫

ϕ (s)dBHs ,

E

T



ψ (s)dBHs

R

0

0



= ⟨KH∗ (ϕ ), KH∗ (ψ )⟩L2 ([0,T ];R) =: ⟨ϕ, ψ⟩H

(2.4)

Considering W = {Wt , t ∈ [0, T ]} defined by Wt = BH ((KH∗ )−1 I[0,t ] ), it turns out that W is a Wiener process and BH has the following Wiener integral representation: BH t =

t



KH (t , s)dWs . 0

In addition, for any ϕ ∈ H, T



ϕ (s)dBHs = 0

T



(KH∗ ϕ )(t)dWt 0

if and only if KH∗ ϕ ∈ L2 ([0, T ]; R). Remark 2.1. Denoting L2H ([0, T ]; R) = {ϕ ∈ H, KH∗ ϕ ∈ L2 ([0, T ]; R)}. In the case H ∈ (0, 1/2), the inclusion C r ([0, T ]; R) ⊂ L2H ([0, T ]; R) holds for any 1/2 − H < r < 1 (see [11]), where C r ([0, T ]; R) denotes the space of all Hölder continuous functions of order r from the interval [0, T ] to R. In the case H ∈ (1/2, 1), we have L1/H ([0, T ]; R) ⊂ L2H ([0, T ]; R) (see [12]). Proposition 2.1. Let H ∈ (1/2, 1) and σ ∈ L2 ([0, T ]; R). Then there exists a version of

{∫

t 0

} σ (s)dBHs , t ∈ [0, T ] with Hölder

H

continuous trajectories of order γ ∈ (0, H − 1/2), which is an FtB -adapted centered Gaussian process. Proof. Since H ∈ (1/2, 1), L2 ([0, T ]; R) ⊂ L1/H ([0, T ]; R) and the integral

∫t 0

σ (s)dBHs is well defined. Let

t



σ (s)dBHs ,

Zt =

t ∈ [0, T ].

0

Then, by using the Lemma 2 of [13], we have for any 0 ≤ s < t ≤ T

E|Zt − Zs |2 ≤ cH (2H − 1)(t − s)2H −1 ∥σ (·)∥2L2 ([0,T ];R) . Since Zt − Zs is Gaussian with zero mean for each k ∈ N, there exists a constant C (k) > 0 such that

E|Zt − Zs |2k ≤ C (k)(t − s)(2H −1)k holds, hence by the Kolmogorov–Chentsov Theorem the process {Zt , t ∈ [0, T ]} has a Hölder continuous version of order γ < ((2H − 1)k − 1)/2k for all k ∈ N satisfying (2H − 1)k > 1. Taking k → ∞ completes the proof. □ {∫ } t H Remark 2.2. For fixed 0 ≤ γ < H − 1/2 we identify the process σ (s)dB , t ∈ [ 0 , T ] with its version having the s 0 Hölder continuous trajectories for order γ for σ satisfying the assumptions of Proposition 2.1. 3. Existence and uniqueness of the strong solution for (1.5) In this section, we consider the existence and uniqueness of a strong solution for (1.5). To this end, we make the following assumptions on the coefficients: (H1) |b1 (s, x)| ≤ m(s)(1 + |x|) with m(s) ∈ L1 ([0, T ]; R). (H2) For all s ∈ [0, T ], b1 (s, ·) is left-continuous and lower semi-continuous, i.e., for each x0 ∈ R and all s ∈ [0, T ], lim b1 (s, x) = b1 (s, x0 ), and lim inf b1 (s, x) ≥ b1 (s, x0 ).

x→x− 0

x→x+ 0

Remark 3.1. Obviously, if (H2) holds true for b1 , then, for each x0 ∈ R, we have lim inf b1 (t , x) ≥ b1 (t , x0 ). x→x0

Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

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Theorem 3.1. Assume that there exists a function c : [0, T ] → R+ in L1 ([0, T ]; R) such that for any t ∈ [0, T ] and x, y ∈ R

|b1 (t , x)| ≤ c(t)(1 + |x|),

|b1 (t , x) − b1 (t , y)| ≤ c(t)|x − y|.

If σ (·) ∈ L2 ([0, T ]; R), then (1.5) admits a unique strong solution. Proof. We just have to use the standard iteration of Picard, more precisely: we define x0t = x0 for all t ∈ [0, T ] and then for all n ∈ N and t ∈ [0, T ]: xnt +1

t



σ

= x0 +

(s)dBH s



t

+

b1 (s, xns )ds,

0

0

which is well defined and so for all t ∈ [0, T ] xnt +1

|

∫ t ∫ t ⏐∫ t ⏐ ⏐ H⏐ n | ≤ |x0 | + ⏐ c(s)ds + c(s)ds. σ (s)dBs ⏐ + sup |xs | t ∈[0,T ]

0

0

0

Then, by Proposition 2.1 we have

|xnt +1 | ≤ |x0 | + BT H −1/2 + sup |xns | t ∈[0,T ]

t



t



c(s)ds,

c(s)ds + 0

0

where B is a constant which depends only on σ . On the other hand, we clearly have for all n ∈ N and t ∈ [0, T ]:

⏐ ∫ t ⏐∫ t ⏐ ⏐ c(s)|xns − xns −1 |ds |xnt +1 − xnt | =⏐ (b1 (s, xns ) − b1 (s, xns −1 ))ds⏐ ≤ 0 0 (∫ )n ≤

t 0

c(s)ds

sup |x1s − x0s |.

n!

s∈[0,T ]

It is now standard to conclude. □ Lemma 3.1. Assume that b(t , x) satisfies the condition (H1) and (H2). Then the sequence of functions bn (t , x) = inf {b(t , y) + (n + m(t))|x − y|},

(3.1)

y∈R

is well defined for n ∈ N and it satisfies (i) (ii) (iii) (iv)

n → bn (t , x) is nonincreasing for all x ∈ R and t ∈ [0, T ]; |bn (t , x)| ≤ m(t)(1 + |x|) for all n ∈ N, t ∈ [0, T ] and x ∈ R; |bn (t , x) − bn (t , y)| ≤ m(t)|x − y| for all n ∈ N, t ∈ [0, T ] and x, y ∈ R; For any sequence xn converging to x0 ∈ R and t ∈ [0, T ], we have lim bn (t , xn ) = b(t , x0 ).

n→∞

Proof. It is clear that (i) holds true from (3.1). In view of the inequality |x| ≥ |x − y| − |y|, it follows from the assumption (H1) that for each y ∈ R b(t , y) + (n + m(t))|x − y| ≥ −m(t)(1 + |y|) + (n + m(t))|x − y| ≥ −m(t)(1 + |x|). Thus, for each n ∈ N and each x ∈ R bn (t , x) = inf {b(t , y) + (n + m(t))|x − y|} ≥ −m(t)(1 + |x|). y∈R

On the other hand, it follows from the definition of bn and the assumption (H1) that for each n ∈ N and x ∈ R, bn (t , x) ≤ b(t , x) ≤ m(t)(1 + |x|).

(3.2)

So, we have proven (ii). Now, we can prove (iii). It follows from the definition of bn that for any x, y ∈ R,

|bn (t , x) − bn (t , y)| =| inf {b(t , u) + (n + m(t))|x − u|} − inf {b(t , u) + (n + m(t))|y − u|}| u∈R

u∈R

≤ sup |(b(t , u) + (n + m(t))|x − u|) − (b(t , u) + (n + m(t))|y − u|)| u∈R

≤(n + m(t))|x − y|.

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Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

Finally, we prove (iv). Assume that xn → x− as n → ∞. By the definition of bn and the assumption (H1) we can take a sequence {vn }n≥0 such that bn (t , xn ) ≥ b(t , vn ) + (n + m(t))|xn − vn | −

≥ − m(t)(1 + |vn |) + (n + m(t))|xn − vn | − ≥ − m(t) − m(t)|xn | −

1 n

1 n

1

(3.3)

n

+ (n + m(t) − 1)|xn − vn |

which means that (n + m(t) − 1)|xn − vn | ≤ 2m(t)(1 + |xn |) +

1 n

and then lim sup(n + m(t) − 1)|xn − vn | < ∞. n→∞

Therefore limn→∞ vn = x0 . Then, it follows from (3.2) and the assumption (H2) that, in view of Remark 3.1, lim inf bn (t , xn ) ≥ lim inf b(t , vn ) ≥ lim inf b(t , x) ≥ b(t , x0 ). n→∞

n→∞

x→x0

On the other hand, from (H2) and (3.1) we can also deduce that lim sup bn (t , xn ) ≤ lim sup b(t , xn ) = lim = b(t , x0 ). n→∞

x→x− 0

n→∞

Hence, (iv) holds. The proof is complete. □ Theorem 3.2. Suppose that b1 (s, x) satisfies the assumption (H1) and (H2). If σ (·) ∈ L2 ([0, T ]; R), then (1.5) has a strong solution x and there exists a version of the process {xt , t ∈ [0, T ]} with the Hölder continuous trajectories of order 0 < γ < H − 1/2. Moreover, for any 0 < γ < H − 1/2 the estimate

∥x∥C γ ([0,T ];R) ≤ A(1 + ∥Z ∥C γ ([0,T ];R) ) P − a.s. is valid for a constant A ≡ A(T , x0 , H , γ ). Furthermore, if for all s ∈ [0, T ], b1 (s, ·) is a nonincreasing function, there exists a unique strong solution for (1.5). Proof. For any n ≥ 1, let bn1 be as in the Lemma 3.1. Then, Theorem 3.1 implies that there exists a unique strong solution xn for the equation xnt

t



σ

= x0 +

(s)dBH s

t



bn1 (s, xns )ds.

+ 0

0

Since {bn1 }n≥1 is nonincreasing, comparison theorem entails that {xn }n≥1 is nonincreasing. By the Gronwall’s lemma we may deduce that xn converges a.s. to x, which is clearly a strong solution to the SDE (1.5). The proof of the second assertion on the Hölder continuous trajectories and estimate is similar to the proof of Theorem 3.6 of [14]. So we omit it. Moreover, if x1 and x2 are two solutions of (1.5), using the fact that b1 (s, ·) is nonincreasing, we get by applying the Tanaka’s formula to x1 − x2 , (x1 − x2 )+ =

t



sign(x1s − x2s )(b1 (s, x1s ) − b1 (s, x2s ))ds ≤ 0. 0

Then we have the pathwise uniqueness of the solution.



Remark 3.2. In [6], the authors also obtained the existence and uniqueness of strong solution for (1.5) when σ ≡ 1 under the following assumptions on b1 : (H1′ ) sups∈[0,T ] supx∈R |b1 (s, x)| ≤ M(1 + |x|); (H2′ ) for all s ∈ [0, T ], b1 (s, ·) is a nonincreasing and continuous function. Notice that b1 (s, ·) is a left-continuous and lower semi-continuous function if b1 (s, ·) is a continuous function. But, b1 (s, ·) is not necessarily continuous function if b1 (s, ·) is a left-continuous and lower semi-continuous function. On the other hand, comparing with (H1′ ), m(t) in (H1) can be some generalized integrable functions but is not necessary to be some constants as the assumption (H1′ ). Thus, our Theorem 3.2 improved and generalized Theorem 3.4 of [6].

Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

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Remark 3.3. By the same way as in Theorem 3.2, by replacing (3.1) with bn (t , x) = sup{b(t , y) − (n + m(t))|x − y|} y∈R

we can prove the following existence and uniqueness result on the strong solution for (1.5). Theorem 3.3. Suppose that b1 (s, x) satisfies the assumption (H1) and the following (H2′′ ): (H2′′ ) for all s ∈ [0, T ], b1 (s, ·) is right-continuous and upper semi-continuous, i.e., for each x0 ∈ R and all s ∈ [0, T ], lim b1 (s, x) = b1 (s, x0 ), and lim sup b1 (s, x) ≤ b1 (s, x0 ).

x→x+ 0

x→x− 0

If σ (·) ∈ L2 ([0, T ]; R), then (1.5) has a strong solution x and there exists a version of the process {xt , t ∈ [0, T ]} with the Hölder continuous trajectories of order 0 < γ < H − 1/2. Moreover, for any 0 < γ < H − 1/2 the estimate

∥x∥C γ ([0,T ];R) ≤ A(1 + ∥Z ∥C γ ([0,T ];R) ) P − a.s. is valid for a constant A ≡ A(T , x0 , H , γ ). Furthermore, if for all s ∈ [0, T ], b1 (s, ·) is a nonincreasing function, there exists a unique strong solution for (1.5). 4. Existence of a weak solution for (1.1) In this section, we consider the existence of a weak solution for (1.1) with H ∈ (1/2, 1). By a weak solution to (1.1), we mean a couple of adapted continuous processes (BH , x) on a filtered probability space (Ω , F , P, {Ft , t ∈ [0, T ]}), such that BH is an Ft -fractional Brownian motion and x and BH satisfy (1.1). To construct the weak solution for (1.1), we need the following lemma. Lemma 4.1. Suppose that trajectories of the process {u(t), t ∈ [0, T ]} are P-a.s. in L2 ([0, T ]; R). Then for any σ (·) ∈ L2 ([0, T ]; R) t



σ (s)dBHs =

t



0

σ (s)d˜ BH s + 0

H where {˜ BH t = Bt −

∫t 0



t

0

σ (s)u(s)ds, ˜ P − a.s., t ∈ [0, T ],

(4.1)

H

u(s)ds, t ∈ [0, T ]} is the FtB -fBm on (Ω , F , ˜ P) and the measures P and ˜ P are equivalent.

Proof. For any t ∈ [0, T ], 0 ≤ a < b ≤ t, if σ = I[a,b) then the left-hand side of (4.1) is equal to t



σ (s)dBHs = BHb − BHa .

(4.2)

0

Further, with this choice of σ , we have t



σ (s)d˜ BH s + 0

t

∫ 0

∫ b H ˜ σ (s)u(s)ds =˜ BH − B + u(s)ds b a a ∫ b ∫ ( ) ( = ˜ BH u(s)ds − ˜ BH b + a + 0

a

)

u(s)ds 0

=BHb − BHa . So (3.1) has been shown for σ = I[a,b) , 0 ≤ a < b ≤ t. By linearity we can extend (3.1) to all σ ∈ Λ. Noticing that for any σ ∈ L2 ([0, T ]; R), then there exists the sequence σn ∈ Λ, n ∈ N such that

∥σn − σ ∥L2 ([0,T ];R) → 0,

n → +∞.

By Theorem 1.1 of [15], we know that (L1/H ([0, T ]; R), ∥ · ∥L1/H ([0,T ];R) ) ↪→ (H, ∥ · ∥H ). Thus, for any σ ∈ L2 ([0, T ]; R) ⊂ L1/H ([0, T ]; R), there exist two constants bH and bH ,T such that

∥σ ∥H ≤ bH ∥σ ∥L1/H ([0,T ];R) ≤ bH ,T ∥σ ∥L2 ([0,T ];R) , which implies that

∥σn − σ ∥H → 0,

n → +∞.

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Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

Using the isometry (2.4), we have

∫ t ∫ ∫ t   2 H ˜ ∗ ˜ − σ (s)d B σn (s)d˜ BH  s dP =∥KH (σn − σ )(s)∥L2 ([0,t ];R) s Ω

0

0

≤∥KH∗ (σn − σ )(s)∥2L2 ([0,T ];R) =∥σn − σ ∥2H → 0, n → +∞. It means that t



σn (s)d˜ BH s →

t



2 ˜ σ (s)d˜ BH s , n → +∞ in L (Ω , P; R).

0

0

Thus, there exists a subsequence of (σn ) (denoted again by (σn )) such that t



σn (s)d˜ BH s →

t



0

˜ σ (s)d˜ BH s , n → +∞ in P − a.s.

0

(4.3)

Repeating the same procedure for {BH t , t ∈ [0, T ]} we get t



σn (s)dBHs →

t



σ (s)dBHs , n → +∞ in P − a.s. 0

0

and since the measures P and ˜ P are equivalent we also have t



σn (s)dBHs →

t



0

0

σ (s)dBHs , n → +∞ in ˜ P − a.s.

(4.4)

On the other hand, we have by using the Hölder’s inequality

∫ t ⏐ ⏐ ⏐∫ t ⏐∫ t ⏐ ⏐ ⏐ ⏐ σ (s)u(s)ds⏐ =⏐ (σn (s) − σ (s))u(s)ds⏐ σn (s)u(s)ds − ⏐ 0

0

0

(4.5)

≤∥u∥L2 ([0,T ];R) ∥σn − σ ∥L2 ([0,T ];R) → 0 n → +∞ ˜ P − a.s. Our statement follows from t



σn (s)dBHs = 0

t



σn (s)d˜ BH s +

t



0

0

σn (s)u(s)ds ˜ P − a.s., t ∈ [0, T ],

by letting n → +∞ and using (4.3)–(4.5). □ Theorem 4.1. Assume that 1/2 < H < 1, σ ∈ L2 ([0, T ]; R), b1 satisfies the assumption (H1) and (H2). Furthermore, if σ −1 (t)b2 (t , x) ∈ L2 ([0, T ]; R) and for some C > 0, it holds that for any t ∈ [0, T ], x ∈ R

|σ −1 (t)b2 (t , x)| ≤ C (1 + |x|).

(4.6)

Then for any T > 0, the SDE (1.1) has a weak solution on [0, T ]. Proof. Let BH be a fractional Brownian motion with Hurst parameter H ∈ (1/2, 1). Since, σ ∈ L2 ([0, T ]; R), b1 satisfies the assumption (H1) and (H2), (1.5) has a unique strong x1 . Let u(s) = −σ −1 (s)b(s, x1 ). For any 0 ≤ t ≤ T , we set H ˜ BH t = Bt −

t



σ −1 (s)b(s, x1 )ds. 0

Note that by means of the integral representation of fractional Brownian motion, the definition of the operator KH and transformation formulas for fractional Brownian motion (see the Corollary 5.2 of [7]), we get that for the any 0 ≤ t ≤ T ,

˜ BH t =

t



u(v )dv + BH t 0 t



u(v )dv + c˜H

= 0

t

∫ 0

(t − v )2H −1 dB1v−H

t

∫ = 0

c˜H (t − v )2H −1 [˜cH−1 (t − v )1−2H u(v )dv + dB1v−H ].

(4.7)

Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

9

For t given in (4.7) and 0 ≤ s ≤ t, let

˜ B1s −H =

s



c˜H−1 (t − v )1−2H u(v )dv + B1s −H

0 s



c˜H−1 (t − v )1−2H u(v )dv +

=

∫0 s =

s

K1−H (s, v )dWv

0

[(

K1−H (s, v )

·



K1−−1H

0

)

c˜H−1 (t − z)1−2H u(z)dz (v )dv + dWv

]

0

(

where c˜H =



) 12

2H

.

Γ (2H)Γ (3−2H)

Now, let Rε (t) = exp

[

∫ t(

K1−−1H



1



2

K1−−1H

0

−1

c˜H (t − z)

1−2H

) u(z)dz

(v )dWv

0

0

∫ t(

·



·



c˜H−1 (t − z)1−2H u(z)dz

)2

(v )dv

]

.

0 H

B Using Corollary 5.2 of [7], we immediately know that (˜ BH t )0≤t ≤T is an Ft -fractional Brownian motion with Hurst 1−H parameter H ∈ ( 12 , 1) under the new probability ˜ P(dω) = Rε (T )P(dω) if (˜ B1t −H )0≤t ≤T is an FtB -fractional Brownian

motion with Hurst parameter 1 − H under the new probability ˜ P(dω) = Rε (T )P(dω). 1−H Next we want to show (˜ B1s −H )0≤s≤t is an FsB -fractional Brownian motion with Hurst parameter H ∈ ( 21 , 1) under

the new probability ˜ P(dω) = Rε (t)P(dω). Due to [3], it suffices to show that EP Rε (t) = 1. Since absolutely continuous, we have by (2.3) that

(

K1−−1H

·



)

1

H− 1

1

c˜H−1 (t − z)1−2H u(z)dz (v ) = v 2 −H I0+ 2 v H − 2 c˜H−1 (t − v )1−2H u(v ),

∫· 0

c˜H−1 (t − z)1−2H u(z)dz is

v ∈ [0, t ].

0

Hence, by using the assumption (4.6) we have further that for any v ∈ [0, t ],

∫ · ⏐( ) ⏐ ⏐ −1 ⏐ c˜H−1 (t − z)1−2H u(z)dz (v )⏐ ⏐ K1−H 0 ∫ v ⏐ ⏐ 1 ⏐ ⏐ −1 21 −H H − 12 1−2H −1 1 − 23 +H ˜ v =⏐ c z (t − z) σ (z)b(z , x )( v − z) dz ⏐ H Γ (H − 21 ) 0 ∫ v C 1 3 −1 12 −H ˜ v z H − 2 (t − z)1−2H (v − z)− 2 +H (1 + |x1z |)dz . c ≤ H 1 Γ (H − 2 ) 0 Then, by Theorem 3.2 we can obtain for any γ ∈ (0, H − 1/2)

∫ · ⏐( ) ⏐ ⏐ −1 ⏐ K c˜H−1 (t − z)1−2H u(z)dz (v )⏐ ⏐ 1−H 0 ∫ v C 3 1 −1 21 −H ≤ c˜H v z H − 2 (t − z)1−2H (v − z)− 2 +H (1 + |x0 | + z γ )dz 1 Γ (H − 2 ) 0 ∫ C (1 + |x0 |) −1 1 −H v H − 1 3 = c˜H v 2 z 2 (t − z)1−2H (v − z)− 2 +H dz Γ (H − 12 ) 0 ∫ v C 1 3 −1 21 −H ˜ + c v z H − 2 +γ (t − z)1−2H (v − z)− 2 +H dz H Γ (H − 12 ) 0 =:I1 + I2 . For I1 , by using the Hölder’s inequality, we have for any 1 < p < I1 ≤



C (1 + |x0 |)

Γ (H − 12 ) C (1 + |x0 |)

Γ (H − 12 )

c˜H v −1

1 −H 2

v

(∫

(z

H − 12

(v − z)

1

2 3−2H

) 1p (∫ v ) p−p 1 p(1−2H) ) dz · (t − z) p−1 dz

− 32 +H p

0 3

(4.8)

0 1

1

3

2

2

H− + c˜H−1 v 2 p B p (pH − p + 1, Hp − p + 1)

t

2−2H − 1p

where B(·, ·) and Γ (·) are the standard Beta and Gamma functions.

−1 | 2p−p2pH | −1

(4.9)

1

+ (t − v )2−2H − p p−1 p

,

10

Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

For I2 , by using again the Hölder’s inequality, we have for any 1 < p < I2 ≤



C

1

Γ (H − 12 ) C

Γ (H − 12 )

c˜H−1 v 2 −H

v

(∫

1

3

(z H − 2 +γ (v − z)− 2 +H )p dz

) p−p 1 ) 1p (∫ v p(1−2H) · (t − z) p−1 dz 0

0

c˜H−1 v

2 3−2H

H − 23 +γ + 1p

1

B p (pH −

1 2

3

p + γ p + 1, Hp −

2

p + 1)

t

2−2H − 1p

(4.10)

1

+ (t − v )2−2H − p

−1 | | 2p−p2pH −1

p−1 p

.

Thus, combining (4.8)–(4.10) we can further obtain that for any 0 ≤ t ≤ T ,

∫ t ⏐( ∫ · ) ⏐2 ⏐ −1 ⏐ c˜H−1 (t − z)1−2H u(z)dz (v )⏐ dv ⏐ K1−H 0 0 ∫ t 2 2 2 2H −3+ 2p ≤C1 (v + v 2H −3+2γ + p )[t 4−4H − p + (t − v )4−4H − p ]dv, 0

where C1 = 4

]2 (

−1 C (1+|x0 |)c˜H

[

p−1 −1 p Γ (H − 21 )| 2p−p2pH | −1

)

2

2

B p (pH − 12 p + 1, Hp − 32 p + 1) + B p (pH − 21 p + γ p + 1, Hp − 23 p + 1) .

Furthermore, we have

∫ t ⏐( ∫ · ) ⏐2 ⏐ −1 ⏐ K c˜H−1 (t − z)1−2H u(z)dz (v )⏐ dv ⏐ 1−H 0 0 ∫ t 2 2 2H −3+ 2p 4−4H − 2p 2−2H 2−2H +2γ (v ≤C2 (t +t ) + C1 (t − v ) + v 2H −3+2γ + p (t − v )4−4H − p )dv,

(4.11)

0

where C2 = C1 ·

1 2H −2+ 2p

. If

1 2

< H ≤ 43 , then 5 − 4H −

2 p

≥2−

2 p

> 0. Thus,

∫ t ⏐( ∫ · ) ⏐2 ⏐ ⏐ −1 c˜H−1 (t − z)1−2H u(z)dz (v )⏐ dv ⏐ K1−H 0

≤ C2 (t

0

2−2H

2

2

p 2

p

+ t 2−2H +2γ ) + C1 t 2−2H B(2H − 2 + , 5 − 4H − )

+ C1 t 2−2H +2γ B(2H − 2 + If < H < 1, then 1 < that for any 0 ≤ t ≤ T , 3 4

2 5−4H

<

2 p

(4.12)

+ 2γ , 5 − 4H − ). p

2 . 3−2H

Thus, we have 5 − 4H −

2 p

> 0 for

2 5−4H


2 3−2H

and we can further obtain

∫ t ⏐( ∫ · ) ⏐2 ⏐ −1 ⏐ c˜H−1 (t − z)1−2H u(z)dz (v )⏐ dv ⏐ K1−H 0

≤ C2 (t

0

2−2H

2

2

p 2

p

+ t 2−2H +2γ ) + C1 t 2−2H B(2H − 2 + , 5 − 4H − )

+ C1 t 2−2H +2γ B(2H − 2 +

2 p

(4.13)

+ 2γ , 5 − 4H − ). p

As a consequence, using the well-known Novikov criterion, one can have EP Rε (t) = 1. So, we can claim ˜ BH is an fractional H 1 ˜ ˜ Brownian motion under the probability measure P. So it is sufficient to show that the process (B , x ) satisfies Eq. (1.1) for all t ∈ [0, T ], ˜ P-a.s. Using Lemma 4.1 and the fact that x1 is a solution for (1.5) we have x1t =x0 +

t



b1 (s, x1s )ds +

t



0

σ (s)dBHs 0

t

∫ t σ (s)d˜ BH + σ (s)σ −1 (s)b2 (s, x1s )ds s 0 ∫0 t ∫ t0 1 1 ˜ =x0 + (b1 (s, xs ) + b2 (s, xs ))ds + σ (s)d˜ BH s , P − a.s., ∫

b1 (s, x1s )ds +

= x0 +

0



t

0

therefore (˜ BH , x1 ) satisfies Eq. (1.1) for all t ∈ [0, T ], ˜ P-a.s. It means that (˜ BH , x1 ) is a weak solution for (1.1). The proof is complete. □

Remark 4.1. In [14], the author proved the existence of a weak solution for (1.1) with H ∈ (1/2, 1) when σ ∈ L∞ ([0, T ]; R) and σ −1 b2 satisfies the condition (1.4). However, in our Theorem 4.1 we need only that σ −1 b2 has the linear growth but not any continuity on σ −1 b2 . So, our Theorem 4.1 improves the results of [14].

Z. Li, W. Zhan and L. Xu / Physica A 530 (2019) 121565

11

Remark 4.2. In [6], Boufoussi and Ouknine showed the existence of a weak solution for (1.1) with H ∈ (1/2, 1) when σ ≡ 1, the function b2 satisfies the condition (1.4) and is b1 (s, ·) is linear growth and continuous function for each s. Obviously, our conditions on b1 and b2 in our Theorem 4.1 are much weaker than those conditions on b1 and b2 in [6]. Thus, our Theorem 4.1 improves and generalizes the results of [14]. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]

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