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Convergence rate of the truncated Milstein method of stochastic differential delay equations
Applied Mathematics and Computation 357 (2019) 263–281
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Convergence rate of the truncated Milstein method of stochastic differential delay equationsR Wei Zhang a, Xunbo Yin b, M.H. Song b,∗, M.Z. Liu b a b
School of Mathematical Sciences, Heilongjiang University, Harbin, Heilongjiang, China Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
a r t i c l e
i n f o
Keywords: Stochastic differential delay equation Truncated Milstein method Local Lipschitz condition Khasminskii-type condition Strong convergence
a b s t r a c t This paper is concerned with the strong convergence of highly nonlinear stochastic differential delay equations (SDDEs) without the linear growth condition. On the one hand, these nonlinear SDDEs do not have explicit solutions, therefore implementable numerical methods for such SDDEs are required. On the other hand, the implicit Euler methods are known to converge strongly to the exact solution of such SDDEs. However, they require additional computational efforts. In this article, we propose the truncated Milstein method which is an explicit method under the local Lipschitz condition plus Khasminskii-type condition, study its pth monent boundedness (p is a parameter in Khasminskii-type condition) and show that its rate of strong convergence is close to one. © 2019 Elsevier Inc. All rights reserved.
1. Introduction Stochastic differential delay equations (SDDEs) have come to play an important role in many branches of science and industry. Hence the theory of SDDEs received increasing attentions and obtained significant results. Many SDDEs have no explicit solutions and numerical approximations are required. The numerical solutions of stochastic differential equations (SDEs) have also been studied extensively by many authors (see e.g., [1,4,6–9,17]). To the best of our knowledge, the coefficients of SDDEs satisfying the global Lipschitz condition or the local Lipschitz condition and the linear growth condition (see e.g., [2,10,11]) are required by the classical strong convergence theorem of SDDEs. However, there are many SDDEs which do not satisfy the linear growth condition. In 2005, Mao [12] established a Khasminskii-type theorem for SDDEs where the linear growth condition was no longer required. In 2011, Mao [13] considered numerical solutions of SDDEs under the generalized Khasminskii-type condition. In 2012, [3] proposed an explicit strongly convergent numerical scheme, called the tamed Euler method for SDEs. Motivated by this work, [18] introduce a tamed version of the Milstein scheme for SDEs with commutative noise. In 2015 and 2016, Mao [14,15] showed the truncated Euler-Maruyama method for SDEs and gave its rate. In 2018, Zhang investigated strong convergence of the partially truncated Euler-Maruyama method for SDDEs and showed its rate was close to 1/2 under the polynomial growth condition plus Khasminskii-type condition in [19]. So far there is little numerical theory on SDDEs under Khasminskii-type conditions.
R ∗
This work is supported by the NSF of P.R. China (No.11671113). Corresponding author. Tel.:+86045186403055. E-mail addresses: [email protected] (W. Zhang), [email protected] (X. Yin), [email protected] (M.H. Song), [email protected] (M.Z. Liu).
https://doi.org/10.1016/j.amc.2019.04.001 0 096-30 03/© 2019 Elsevier Inc. All rights reserved.
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In this paper we consider the following 1-dimensional SDDE
dx(t ) = f (x(t ), x(t − τ ))dt + g(x(t ), x(t − τ ))dw(t )
(1)
with initial data x0 = ξ ∈ C ([−τ , 0], R ). Motivated by the techniques discussed in [14,15,18,19], we propose a truncated Milstein method which converge to SDDEs (1) with super-linear drift and diffusion coefficients. For the convenience, we will, in Section 2, present the truncated Milstein method. We will study the pth moment boundedness of the truncated Milstein method and its convergence and give its rate in Section 3. We will conclude our paper in Section 4. 2. The truncated Milstein method Throughout this paper, let (, F, {F t }t≥0 , P ) denote a complete probability space with a filtration {F t }t≥0 satisfying the usual conditions (i.e, it is right continuous and increasing while F0 contains all P-null sets), and let E be the expectation corresponding to P. Let w(t ) be a 1-dimensional Brownian motion defined on the probability space. Let | · | denote both Euclidean norm in R and the trace norm in R2 × 2 . Denote by C ([−τ , 0], R ) the family of continuous functions from [−τ , 0] to R with the norm η = sup−τ ≤u≤0 |η (u )|. For a, b ∈ R, we use a∨b and a∧b for max {a, b} and min {a, b}, respectively. Consider a 1-dimensional SDDE
dx(t ) = f (x(t ), x(t − τ ))dt + g(x(t ), x(t − τ ))dw(t )
(2)
on t ∈ [0, ∞) with initial data {ξ (θ ) : −τ ≤ θ ≤ 0} = ξ ∈ C ([−τ , 0]; R ), where f: R × R → R and g: R × R → R. Throughout this paper, we impose the following five assumptions as standing hypotheses: (A1) For every constant Z ≥ 1, there exists a positive constant LZ > 0, such that
f (x, y ) − f (xˆ, yˆ ) ∨ g(x, y ) − g(xˆ, yˆ ) ≤ LZ |x − xˆ|2 + |y − yˆ|2
(3)
for all x, y, xˆ, yˆ ∈ R with |x| ∨ |y| ∨ |xˆ| ∨ |yˆ| ≤ Z. (A2) We also assume that the coefficients f and g satisfy the Khasminskii-type condition: there is a pair of constants p > 2 and K > 0 such that
xT f (x, y ) + ( p − 1 )|g(x, y )|2 ≤ K 1 + |x|2 + |y|2
(4)
for all x, y ∈ R. (A3) We also assume that the coefficients f and g satisfy the following condition: there are constants q > 2 and K1 > 0 such that
(x − x¯ )T ( f (x, y ) − f (x¯, y¯ )) + (q − 1 )|g(x, y ) − g(x¯, y¯ )|2 ≤ K1 |x − x¯|2 + |y − y¯|2
(5)
for all x, y, x¯ y¯ ∈ R. (A4) There is a constant K0 > 0 such that
|ξ (t ) − ξ (s )| ≤ K0 |t − s|
(6)
for all s, t ∈ [−τ , 0]. (A5) There are constants r > 0 and K2 > 0 such that
| fi (x, y )| ∨ |gi (x, y )| ∨ | fi j (x, y )| ∨ |gi j (x, y )|∨ ≤ K2 1 + |x|r+1 + |y|r+1
(7)
for all x, y ∈ R, i, j = 1, 2, where
∂ f ( x1 , x2 ) ∂ g( x 1 , x 2 ) , gi ( x1 , x2 ) = , ∂ xi ∂ xi ∂ 2 f ( x1 , x2 ) ∂ 2 g( x 1 , x 2 ) f i j ( x1 , x2 ) = , gi j ( x1 , x2 ) = . ∂ xi x j ∂ xi x j f i ( x1 , x2 ) =
We state a known result (see, e.g., [16]) as a lemma for the use of this paper. Lemma 2.1. Under (A1) and (A2), the SDDE (2) has a unique global solution y(t) and, moreover,
sup E |x(t )| p ≤ C,
t∈[−τ ,T ]
∀T > 0.
(8)
where C stands for a generic positive real constant (but independent of and Z later) and its value may change between occurrences. Motivated by Mao [14], we first define the truncated functions. Choose a strictly increasing continuous function φ : R+ → R+ such that φ (s) → ∞ as s → ∞ and for i = 1, 2,
sup (| f (x, y )| ∨ |g(x, y )| ∨ |gi (x, y )| ) ≤ φ (s ),
|x|∨|y|≤s
∀s ≥ 0.
(9)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
265
Let φ −1 be the inverse function of φ and we see that φ −1 is a strictly increasing continuous function from [φ (0), ∞) to R+ . We also choose a number ∗ ∈ (0, 1] and a strictly decreasing function ψ : (0, ∗ ] → (1, ∞) such that
ψ (∗ ) ≥ φ (1 ),
(10)
ψ () = ∞
(11)
lim
→ 0
and
1/4 ψ 3/2 () ≤ 1,
∀ ∈ ( 0 , ∗ ] .
(12)
For a given step size ∈ (0, 1), let us define the truncated functions
x y , |y| ∧ φ −1 (ψ ()) , |x| |y| x y gˆ(x, y ) = g |x| ∧ φ −1 (ψ ()) , |y| ∧ φ −1 (ψ ()) , |x| |y| x y gˆi (x, y ) = gi |x| ∧ φ −1 (ψ ()) , |y| ∧ φ −1 (ψ ()) |x| |y| fˆ(x, y ) = f
|x| ∧ φ −1 (ψ ())
(13)
for all x, y ∈ R, i = 1, 2, where we set x/|x| = 0 when x = 0. It is easy to see that
ˆ f (x, y ) ∨ gˆ(x, y ) ∨ gˆi (x, y ) ≤ φ φ −1 (ψ ()) = ψ ()
(14)
for all x, y ∈ R, i = 1, 2, the truncated functions fˆ, gˆ and gˆi are bounded although f, g and gi may not. Moreover, we can show that truncated functions fˆ and gˆ preserve the Khasminskii-type condition. Lemma 2.2. Let (A2) hold. Then, for all ∈ (0, ∗ ], we have
2 xT fˆ(x, y ) + ( p − 1 )gˆ(x, y ) ≤ 3K 1 + |x|2 + |y|2
(15)
for all x, y ∈ R. Proof. The proof is similar to that of Lemma 2.2 in [19].
Without loss of any generality, we may assume that T and τ are rational numbers. Moreover, T = M and τ = N. We can now form the discrete-time truncated Milstein numerical solutions Y(tk ) ≈ y(tk ) for tk = k by setting Y (0 ) = ξ (0 ) and computing
⎧ Y (t ) = ξ (tk ), k = −N, −N + 1, . . . , 0, ⎪ ⎨ k Y (tk+1 ) = Y (tk ) + fˆ(Y (tk ), Y (tk−N ) ) + gˆ(Y (tk ), Y (tk−N ) )wk + gˆ1 (Y (tk ), Y (tk−N ) )gˆ(Y (tk ), Y (tk−N ) )I1 ⎪ ⎩ + gˆ2 (Y (tk ), Y (tk−N ) )gˆ(Y (tk−N ), Y (tk−2N ) )I2 , k = 0, 1, . . . , M,
where
I1 =
tk
I2 =
tk+1
s
tk
tk+1
tk
s
tk
dw(t )dw(s ) =
(wk )2 − 2
,
dw(t − τ )dw(s ).
wk = w(tk ) and wk = wk+1 − wk . Let us now establish two continuous-time truncated Milstein solutions. The first one is defined by
y¯ (t ) =
M−1
Y (tk )1[tk ,tk+1 ) (t ),
(16)
k=0
with 1G denoting the indicator function for the set G. This is a simple step process, so its sample paths are not continuous. The other continuous-time step-process truncated Milstein solution is defined by
⎧ ⎪ ⎪ y(t ) = ξ (t ), t ∈ [−τ , 0], ⎪ t ⎪ t ⎪ ⎪ ⎪ y(t ) = ξ (0 ) + fˆ(y¯ (s ), y¯ (s − τ ))ds + gˆ(y¯ (s ), y¯ (s − τ ))dw(s ) ⎪ ⎨ 0 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
+
0
+
t
0
t
gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s )
(17)
gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s ), t ∈ [0, T ],
where w¯ (t ) = M k=−N (w (t ) − w (tk ))1[tk ,tk+1 ) (t ). (17) is referred as the continuous-time continuous-sample truncated Milstein solution. In particular, this shows that Y (tk ) = y¯ (tk ) = y(tk ) for all k = 0, 1, . . . , M. Moreover, for all t ∈ [0, T], we have
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E |y¯ (t )| ≤ sup E |y¯ (t )| = sup E |Y (tk )| 0≤t≤T
0≤tk ≤T
= sup E |y(tk )| ≤ sup E |y(s )|, 0≤tk ≤T
(18)
0≤t≤T
3. Convergence of the truncated Milstein method 3.1. Moment boundedness of the truncated Milstein method We first show that y(t) and y¯ (t ) are close to each other in L pˆ where pˆ > 0. Lemma 3.1. For any ∈ (0, ∗ ] and pˆ > 0, we have
E |y(t ) − y¯ (t )| pˆ ≤ C pˆ/2 (ψ ())2 pˆ ,
∀t > 0.
(19)
Consequently
lim E |y(t ) − y¯ (t )| pˆ = 0,
∀t > 0.
→ 0
(20)
Proof. Fix any ∈ (0, ∗ ] and t ≥ 0. There is a unique integer k ≥ 0 such that tk ≤ t < tk+1 . We divided the proof into two cases. Case (I): If pˆ ≥ 2, applying Hölder inequality, Itô isometry and (14), we obtain
E |y(t ) − y¯ (t )| pˆ = E |y(t ) − y(tk )| pˆ pˆ t pˆ t ˆ ¯ ¯ ˆ ¯ ¯ ≤ C E f (y(s ), y(s − τ ))ds + E g(y(s ), y(s − τ ))dw(s ) tk
tk
t pˆ + E gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s ) tk t pˆ + E gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s ) tk t t pˆ ˆ ¯ ( pˆ−2 )/2 gˆ(y¯ (s ), y¯ (s − τ ))pˆ ds ¯ ≤ C pˆ−1 E f ( ( s ) , ( s − τ )) ds + E y y tk
+ ( pˆ−2)/2 E + ( pˆ−2)/2 E ≤ C
pˆ/2
tk
t
tk
t
tk
gˆ1 (y¯ (s ), y¯ (s − τ ) )gˆ(y¯ (s ), y¯ (s − τ ) )w¯ (s ) pˆ ds gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )pˆ ds
(ψ () )2 pˆ ,
∀t > 0.
which is (19). Observe from (12) that pˆ/2 (ψ ())2 pˆ ≤ pˆ/2 , it is easy to get (20) from (19). Case (II): For any pˆ ∈ (0, 2 ), let l ≥ 2, using Hölder inequality, we have
E |y(t ) − y¯ (t )| pˆ ≤ E |y(t ) − y¯ (t )|l
pˆ/l
≤ C l/2 (ψ () )2l
pˆ/l
≤ C pˆ/2 (ψ () )2 pˆ . Then (20) holds.
Lemma 3.2. Let (A1) and (A2) hold. Then, for all t ∈ [0, T], we have
sup
sup E |y(t )| p ≤ C,
0<≤∗ −τ ≤t≤T
∀T > 0.
(21)
Proof. Fix any ∈ (0, ∗ ]. Applying the Itô formula and Lemma 2.2, we derive from (17) that, for 0 ≤ t ≤ T,
E |y(t )| p ≤
|ξ ( 0 )| p + E
t 0
p|y(s )| p−2 yT (s ) fˆ(y¯ (s ), y¯ (s − τ ))
p−1 + |gˆ(y¯ (s ), y¯ (s − τ )) + gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
|ξ ( 0 )| p + E
≤
t
267
p|y(s )| p−2 y¯ T (s ) fˆ(y¯ (s ), y¯ (s − τ )) + ( p − 1 )|gˆ(y¯ (s ), y¯ (s − τ ))|2 ds
0
|y(s )| p−2 gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 0 2 + gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds + p( p − 1 ) E
+E
t
0
t
p|y(s )| p−2 (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds
|ξ (0 )| p + 3KE
≤
+ p( p − 1 ) E
0 t
0
t
p|y(s )| p−2 1 + |y¯ (s )|2 + |y¯ (s − τ )|2 ds
|y(s )| p−2 |gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds t +E p|y(s )| p−2 (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds. 0
Using the Young inequality and (14), it is easy to obtain
t
pE 0
t
pE 0
|y(s )| p−2 |y¯ (s )|2 ds ≤ ( p − 2 ) ≤ p
t
pE 0
|y(s )| p−2 ds ≤ ( p − 2 )E
t
t
0 t
|y(s )| p ds + 2T ,
E |y(s )| p ds + 2
0
t
0
E |y¯ (s )| p ds
sup E |y(u )| p ds,
0 0≤u≤s
|y(s )| p−2 |y¯ (s − τ )|2 ds ≤ ( p − 2 )
0
t
E |y(s )| p ds + 2
0
t
E |y¯ (s − τ )| p ds
t
sup E |y(u )| p ds + 2τ ξ , 0 0 ≤u≤s t pE |y(s )| p−2 gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 0 2 + gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds t t ≤ ( p − 2 )E |y(s )| p ds + 2E gˆ1 (y¯ (s ), y¯ (s − τ ) )gˆ(y¯ (s ), y¯ (s − τ ) )w¯ (s ) 0 0 p + gˆ2 (y¯ (s ), y¯ (s − τ ) )gˆ(y¯ (s − τ ), y¯ (s − 2τ ) )w¯ (s − τ ) ds ≤ p
≤ ( p − 2 )E ≤ ( p − 2) and
t
E 0
0
t
|y(s )| p ds + C p/2 (ψ () )2 p
0 t
E |y(s )| p ds + C
p|y(s )| p−2 (y(s ) − y¯ (s ) )T fˆ(y¯ (s ), y¯ (s − τ ))ds
≤ ( p − 2 )E
t
0
|y(s )| p ds + 2E
t 0
p/2 |y(s ) − y¯ (s )| p/2 fˆ(y¯ (s ), y¯ (s − τ )) ds.
By Lemma 3.1 and inequalities (14) and (12), we obtain
T
E 0
p/2 |y(s ) − y¯ (s )| p/2 fˆ(y¯ (s ), y¯ (s − τ )) ds
≤ (ψ () ) p/2 E
T
0 3 p/2
≤ CT (ψ () ) ≤ C.
|y(s ) − y¯ (s )| p/2 ds
p/4 (22)
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Applying the well-known Gronwall inequality, we get
sup E |y(u )| p ≤ C.
0≤u≤t
It is easy to show
sup E |y(u )| p ≤ C.
−τ ≤u≤t
As this holds for any ∈ (0, ∗ ], we have the required (21). The proof is complete.
Corollary 3.3. Assume that (A1) and (A2) hold. Then we have
sup
sup E|y¯ (t )| p ≤ C
(23)
0<≤∗ −τ ≤t≤T
for all T > 0. 3.2. Strong convergence To show that both truncated Milstein solutions y(T) and y¯ (T ) will converge to the exact solution x(T) in L2 for any T > 0, we need to present several lemmas. In the following of this section, we fix T > 0 arbitrarily. Lemma 3.4. Let (A1) and (A2) hold. For any real number Z > ξ , define the stopping time
σZ = inf {t ≥ 0 : |x(t )| ≥ Z }, where throughout this paper we set inf = ∞ (and as usual denotes the empty set). Then
P ( σZ ≤ T ) ≤
C . Z2
(24)
Proof. Using the Itô formula and (A2), we derive that
E |x(t ∧ σZ )|2 ≤ ≤
|ξ ( 0 )|2 + E
t∧σZ 0
|ξ (0 )|2 + 2KE
t∧σZ 0
2xT (s ) f (x(s ), x(s − τ )) + |g(x(s ), x(s − τ ))|2 ds
1 + |x(s )|2 + |x(s − τ )|2 ds
≤ (1 + 2K τ )ξ + 2KT + 4K
2
t
0
E |x(s ∧ σZ )| ds 2
for any 0 ≤ t ≤ T. The Gronwall inequality yields that
E |y(t ∧ σZ )|2 ≤ C. This implies
Z 2 P ( σZ ≤ T ) ≤ C and the assertion follows.
Lemma 3.5. Let (A1) and (A2) hold. For any real number Z > ξ and ∈ (0, ∗ ], define the stopping time
ρ,Z = inf{t ≥ 0 : |y(t )| ≥ Z }. Then
P (ρ,Z ≤ T ) ≤
C . Z2
(25)
Proof. We simply write ρ,Z = ρ . Using the Itô formula, we have that for 0 ≤ t ≤ T,
E |y(t ∧ ρ )|2 ≤ |ξ ( 0 )|2 + E
t∧ρ
0
2y(s )T fˆ(y¯ (s ), y¯ (s − τ )) + |gˆ(y¯ (s ), y¯ (s − τ ))
+ gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds t∧ρ ≤ |ξ ( 0 )|2 + 2E y¯ (s )T fˆ(y¯ (s ), y¯ (s − τ )) + |gˆ(y¯ (s ), y¯ (s − τ ))|2 ds + 2E
0
t∧ρ
0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
269
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds t∧ρ + 2E (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds. 0
Applying Lemma 2.2, we then get that,
E |y(t ∧ ρ )|2 ≤
|ξ ( 0 )|2 + E +E
t∧ρ
0
+ 2E
t∧ρ 0
6K 1 + |y¯ (s )|2 + |y¯ (s − τ )|2 ds
2|y(s ) − y¯ (s )|| fˆ(y¯ (s ), y¯ (s − τ ))|ds
t∧ρ 0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w(s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w(s − τ )|2 )ds t∧ρ ≤ |ξ (0 )|2 + 6KT + 24K E |y(s )|2 ds + 24K
0 T
+ 2E + 2E
0
0
T
E |y(s ) − y¯ (s )|2 ds + 12K τ ξ 2
|y(s ) − y¯ (s ))|| fˆ(y¯ (s ), y¯ (s − τ ))|ds
t∧ρ
0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w(s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w(s − τ )|2 )ds. By (14), we have
t∧ρ
E 0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds ≤ 2T (ψ ())4 ≤ 2T . Using Lemma 3.1, we get
T 0
E |y(s ) − y¯ (s )|2 ds ≤ CT (ψ ())4 ≤ C.
While by Lemma 3.1, (12) and (14), we derive that
T
E 0
|y(s ) − y¯ (s ))|| fˆ(s, y¯ (s )|ds ≤ CT 1/2 (ψ ())3 ≤ C.
The Gronwall inequality yields that
E |y(T ∧ ρ )|2 ≤ C. This implies the assertion (25) easily.
Theorem 3.6. Let (A1)–(A5) hold. Then,
lim E |x(T ) − y(T )|2 = 0
→ 0
and
lim E |x(T ) − y¯ (T )|2 = 0.
→ 0
(26)
Proof. Let σ Z and ρ ,Z be the same as before. Define
θ :=σZ ∧ ρ,Z ,
e (T ) :=
x ( T ) − y ( T ).
Using the Young inequality, we obtain that for any > 0,
E |e (T )|2 = E |e (T )|2 1{θ >T } + E |e (T )|2 1{θ ≤T } 2 p−2 ≤ E |e (T )|2 1{θ >T } + E |e ( T )| p + P ( θ ≤ T ). p p 2/( p−2) Applying Lemmas 2.1 and 3.2, we get
E |e (T )| ≤ C. p
(27)
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W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
While using Lemmas 3.4 and 3.5,
P (θ ≤ T ) ≤ P (σZ ≤ T ) + P (ρ,Z ≤ T ) ≤ We hence obtain
E |e (T )| ≤ E |e (T )|2 1{θ >T } + 2
C . Z2
2C ( p − 2 )C + . p p 2/( p−2) Z 2
Let ε > 0 be arbitrary. Choose sufficiently small such that ( p−2 )C
≤ 3ε . We then derive from (28) that
p 2/( p−2 ) Z 2
E |e (T )| ≤ E |e (T )|2 1{θ >T } + 2
2ε . 3
(28) 2C p
≤ ε3 and then choose Z sufficiently large such that
(29)
If we can show that for any sufficiently small ,
E |e (T )|2 1{θ >T } ≤
ε
3
,
(30)
we then have
lim E |e (T )| = 0 2
→ 0
and then using Lemma 3.1, we also obtain
lim E |x(T ) − y¯ (T )|2 = 0.
→ 0
Namely, in order to complete our proof, what we need is to show (30). For this purpose, we define the truncated functions
FZ (x, y ) = f (|x| ∧ Z ) and
x y , ( |y| ∧ Z ) |x| |y|
GZ (x, y ) = g (|x| ∧ Z )
x y , ( |y| ∧ Z ) |x| |y|
for any x, y ∈ R, without loss of generality, we may assume that ∗ is already sufficiently small for φ −1 (ψ (∗ )) ≥ Z. Hence, for all ∈ (0, ∗ ], we get that
FZ (x, y ) = fˆ(x, y ) and GZ (x, y )
= gˆ(x, y )
for any x, y ∈ R with |x|∨|y| ≤ Z. Consider the SDDE
dX (t ) = FZ (X (t ), X (t − τ ))dt + GZ (X (t ), X (t − τ ))dw(t )
(31)
on t ≥ 0 with initial data η (t ) ∈ C ([−τ , 0]; R ). By (A1), it is easy to see that both FZ (X (t ), X (t − τ )) and GZ (X (t ), X (t − τ )) are globally Lipschitz continuous with the Lipschitz constant LZ . So the SDDE (31) has a unique global solution X(t) on t ≥ 0. It is easy to see that
x(t ∧ σZ ) = X (t ∧ σZ )
a.s. f or all t ≥ 0.
(32)
∗ ],
On the other hand, for each step size ∈ (0, we use the truncated Milstein method to the SDDE (31) and let X¯ (t ) denote the continuous-time continuous-sample truncated Milstein solution. It is obvious that
y(t ∧ ρ,Z ) = X¯ (t ∧ ρ,Z )
a.s. f or all t ≥ 0.
(33)
However, it is known that (see Theorem 6.1 in [5])
sup |X (t ) − X¯ (t )|2
E
0≤t≤T
≤ O(2 ).
Therefore,
E
2
sup X (t ∧ θ ) − X¯ (t ∧ θ )
0≤t≤T
Using (32) and (33), we then obtain
sup |x(t ∧ θ ) − y(t ∧ θ )|2
E
0≤t≤T
which implies
≤ O(2 ).
≤ O(2 ).
E |x(T ∧ θ ) − y(T ∧ θ )|2 ≤ O(2 ).
(34)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
Hence
E |e (T )|2 1{θ >T } = E |e (T ∧ θ )|2 1{θ >T }
271
≤ E |x ( T ∧ θ ) − y ( T ∧ θ |2
≤ O(2 ). This implies (30) holds. The proof is therefore complete.
3.3. Convergence rate In the previous section, we showed that
lim E |x(T ) − y(T )|2 = 0
lim E |x(T ) − y¯ (T )|2 = 0
and
→ 0
→ 0
for any T > 0. This is sufficient for some applications. However, we sometimes need a stronger convergence result like
sup |x(t ) − y(t )|
lim E
→ 0
2
=0
0≤t≤T
(35)
We impose an additional assumption. (A6) Assume that there is a pair of constants γ > 0 and K3 > 0 such that
| f (x, y ) − f (xˆ, yˆ )| ∨ |g(x, y ) − g(xˆ, yˆ )| ≤ K3 1 + |x|γ + |y|γ + |xˆ|γ + |yˆ|γ · |x − xˆ| + |y − yˆ|
(36)
for all x, y, xˆ, yˆ ∈ R. See from (A6), we get
| f (x, y )| ∨ |g(x, y )| ≤ K4 1 + |x|γ +1 + |y|γ +1
(37)
for all x, y ∈ R, where K4 independent of K3 , |f(0, 0)|, |g(0, 0)| and ρ . Lemma 3.7. Under (A1), (A2), (A5) and (A6), assume that p > 2γ + 2, set p¯ = p − 2γ , then
E
sup
t∈[−τ ,T ]
|x(t )|
p¯
≤ C.
(38)
Proof. By the Itô formula, we can show that
|x(t )| p¯ ≤|ξ (0 )| p¯ + +
t 0
t 0
p¯ |x(s )| p¯ −2 xT (s )g(x(s ), x(s − τ ))dw(s )
p¯ |x(s )| p¯ −2 xT (s ) f (x(s ), x(s − τ )) +
p¯ − 1 |g(x(s ), x(s − τ ))|2 ds. 2
Using (A2), we get
E ( sup |x(t )| p¯ ) ≤|ξ (0 )| p¯ + E 0≤t≤T
+E
T 0
t
sup 0≤t≤T
0
2K p¯ |x(s )| p¯ −2 1 + |x(s )|2 + |x(s − τ )|2 ds
p¯ |x(s )|
x (s )g(x(s ), x(s − τ ))dw(s )
p¯ −2 T
Applying the Burkholder-Davis-Gundy inequality, Young inequality and (A6), we then have
E
sup |x(t )| p¯
0≤t≤T
≤
|ξ (0 )| p¯ + 2K ( p¯ − 2 )E
T 0
|x(s )| p¯ ds + 2K p¯ E
T
T 0
|x(s )| p¯ ds
|x(s − τ )| p¯ ds T √ + 4 2 p¯ E sup |x(s )| p¯ |x(s )| p¯−2 |g(x(s ), x(s − τ ))|2 ds 1/2 + 2K T + 2K E
0
0≤t≤T
0
272
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
≤
|ξ (0 )| + CE
T
p¯
0
|x(s )| ds + 2K T + 2K τ E ξ p¯
1 + E 2
p¯
sup |x(t )|
p¯
0≤t≤T
|x(s )| p¯−2 1 + |x(s )|2γ +2 + |x(s − τ )|2γ +2 ds 1 ≤ C + E sup |x(t )| p¯ + 16K¯ p¯ 2 E
2
T
0
0≤t≤T
+ 16K¯ p¯ 2 E
|x(s )| p¯−2 1 + |x(s )|2γ +2 + |x(s − τ )|2γ +2 ds.
T
0
Noting that p¯ + 2ρ = p, by the Young inequality and Lemma 2.1, we obtain
T
p¯ E
0
0
= p¯ E ≤ p¯ E
T
T 0
|x(s )| p¯−2 1 + |x(s )|2γ +2 + |x(s − τ )|2γ +2 ds |x(s )| p¯−2 ds + p¯ E
T 0
|x(s )| p¯ ds + 2T + CE
|x(s )| p ds + p¯ E
T 0
T 0
|x(s )| p ds + CE
|x(s )| p¯−2 |x(s − τ )|2γ +2 ds
T
0
|x(s − τ )| p ds
≤ C. This implies
sup |x(t )|
E
p¯
≤C.
0≤t≤T
Hence the required assertion (39) holds. The proof is complete.
Lemma 3.8. Under (A1), (A2), (A5) and (A6), assume that p > 2γ + 2, set p¯ = p − 2γ , then
sup E
|y(t )|
sup
−τ ≤t≤T
0<≤∗
p¯
≤ C.
(39)
Proof. Fix any ∈ (0, ∗ ]. By the Itô formula and (A2), we can show in the same way as the proof of Lemma 3.2 that
E
sup |y(s )|
0≤s≤t
≤
|ξ (0 )| p¯ + E
p¯
t 0
p¯ |y(s )| p¯ −2 yT (s ) fˆ(y¯ (s ), y¯ (s − τ ))
p¯ − 1 + |gˆ(y¯ (s ), y¯ (s − τ )) + gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
+E
s
p¯ |y(r )| p¯ −2 yT (r ) gˆ(y¯ (r ), y¯ (r − τ ))
sup 0≤s≤t
0
+ gˆ1 (y¯ (r ), y¯ (r − τ ))gˆ(y¯ (r ), y¯ (r − τ ))w¯ (r )
+ gˆ2 (y¯ (r ), y¯ (r − τ ))gˆ(y¯ (r − τ ), y¯ (r − 2τ ))w¯ (r − τ ) dw(r ) ≤
|ξ (0 )| p¯ + E +E +E
t 0 t 0
t 0
p¯ |y(s )| p¯ −2 y¯ T (s ) fˆ(y¯ (s ), y¯ (s − τ )) + ( p¯ − 1 )|gˆ(y¯ (s ), y¯ (s − τ ))|2 ds
p¯ |y(s )| p¯ −2 (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds
p¯ ( p¯ − 1 )|y(s )| p¯ −2 gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds √ + 4 2E
0
t
p¯ |y(s )|2 p¯ −2 gˆ(y¯ (s ), y¯ (s − τ ))
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
+ gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) + gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) Similarly to Lemma 3.2, we have
E sup |y(s )| p¯ 0≤s≤t
≤ C +C + CE
t 0 t
0
2
ds
1/2
.
sup E |y(u )| p¯ ds
0≤u≤s
|y(s )| p¯−2 gˆ(y¯ (s ), y¯ (s − τ ))
+ gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds. Using Lemma 3.1 and (A6), we get
t
E 0
|y(s )| p¯−2 |gˆ(y¯ (s ), y¯ (s − τ ))|2 ds
t
≤ K¯ E
p¯ |y(s )| p¯ −2 1 + |y¯ (s )|2γ +2 + |y¯ (s − τ )|2γ +2 ds
0
t
t
|y(s )| p¯ ds + K¯ T + K¯ E |y¯ (s )| p ds + K¯ E 0 0 t ≤C E sup |y(u )| p¯ ds + 1 , ≤C E
t 0
t 0
|y¯ (s − τ )| p ds
0≤u≤s
0
E
p¯ |y(s )| p¯ −2 |gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )|2 ds
≤ CE
t 0
|y(s )| p¯ ds + C
t
t 0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )| p¯ ds
t
|y(s )| p¯ ds + C p¯/2 (ψ ())4 p¯ ds 0 0 t p¯ ≤C E sup |y(u )| ds + 1 . ≤ CE
0≤u≤s
0
Similarly, we get
t
E 0
p¯ |y(s )| p¯ −2 |gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
≤C
t
sup |y(u )|
E 0
0≤u≤s
p¯
ds + 1 .
Hence the required assertion (39) follows. The proof is complete.
Due to [20], we give the following Taylor formula. If a function h: R × R → R is twice differentiable, define
∂ h(x, y ) , ∂x ∂ h(x, y ) h2 (x, y ) := , ∂y h1 (x, y ) :=
then we have the following Taylor formula
h(x, y ) − h(xˆ, yˆ ) = h1 (xˆ, yˆ )(x − xˆ) + h2 (xˆ, yˆ )(y − yˆ ) + R(h(ζ1 , ζ2 ) ), where ζ1 = x + λ(x − xˆ), ζ2 = y + λ(y − yˆ ) with λ ∈ (0, 1) and
x − xˆ 1 R(h(xˆ, yˆ )) = (x − xˆ, y − yˆ )J (h(ζ1 , ζ2 )) , 2 y − yˆ
J (h(x, y )) =
h11 (x, y ) h12 (x, y )
h21 (x, y ) h22 (x, y )
273
274
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
for x, y, xˆ, yˆ ∈ R, where
∂ 2 h(x, y ) , ∂ x2 ∂ 2 h(x, y ) h22 (x, y ) := , ∂ y2
∂ 2 h(x, y ) , ∂ x∂ y ∂ 2 h(x, y ) h21 (x, y ) := . ∂ y∂ x
h11 (x, y ) :=
h12 (x, y ) :=
Hence, we have
h(y(t ), y(t − τ )) − h(y¯ (t ), y¯ (t − τ )) = h1 (y¯ (t ), y¯ (t − τ ))(y(t ) − y¯ (t )) + h2 (y¯ (t ), y¯ (t − τ ))(y(t − τ ) − y¯ (t − τ )) + R(h ) t = h1 (y¯ (t ), y¯ (t − τ )) gˆ(y¯ (s ), y¯ (s − τ ))dw(s ) + Rˆ(h(y¯ (t ), y¯ (t − τ ))) + h2 (y¯ (t ), y¯ (t − τ ))
tk
t−τ
t k −τ
gˆ(y¯ (s − τ ), y¯ (s − 2τ ))dw(s ),
where
Rˆ(h(y¯ (t ), y¯ (t − τ ))) = h1 (y¯ (t ), y¯ (t − τ )) +
t
tk t
+
tk
gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s )
t−τ
t k −τ
+
fˆ(y¯ (s ), y¯ (s − τ ))ds
tk
gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s )
+ h2 (y¯ (t ), y¯ (t − τ )) +
t
(40)
t−τ
t k −τ
t−τ
t k −τ
fˆ(y¯ (s ), y¯ (s − τ ))ds
gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s )
gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s )
+ R(h(y¯ (t ), y¯ (t − τ ))).
(41)
By Lemma 2.1, Lemma 3.2 and Corollary 3.3, we have the following two lemmas. Lemma 3.9. Let (A1), (A2), (A5) and (A6) hold. Assume that q(γ + 1 ) ∨ q(r + 1 ) < p, then for any i, j = 1, 2, we have
sup E | f (x(t ), x(t − τ ))| ∨ E |g(x(t ), x(t − τ ))| q
q
0≤t≤T
∨ E | fi (x(t ), x(t − τ ) )| ∨ E |gi (x(t ), x(t − τ ))| q
q
q q ∨ E fi j (x(t ), x(t − τ ) ) ∨ E gi j (x(t ), x(t − τ )) ≤ C, q q sup sup E | f (y(t ), y(t − τ ))| ∨ E |g(y(t ), y(t − τ ))| 0<<∗ 0≤t≤T
∨ E | fi (y(t ), y(t − τ ) )| ∨ E |gi (y(t ), y(t − τ ))| q
q
q q ∨ E fi j (y(t ), y(t − τ ) ) ∨ E gi j (y(t ), y(t − τ )) ≤ C and
sup
sup
0<<∗ 0≤t≤T
E | f (y¯ (t ), y¯ (t − τ ))| ∨ E |g(y¯ (t ), y¯ (t − τ ))| q
q
∨ E | fi (y¯ (t ), y¯ (t − τ ) )| ∨ E |gi (y¯ (t ), y¯ (t − τ ))| q
q
q q ∨ E fi j (y¯ (t ), y¯ (t − τ ) ) ∨ E gi j (y¯ (t ), y¯ (t − τ )) ≤ C Lemma 3.10. Let (A1), (A2), (A5) and (A6) hold. Assume that 2q(r + 1 ) < p, then we have
sup
sup E |Rˆ( f (y(t ), y(t − τ )))|q ∨ E |Rˆ(g(y(t ), y(t − τ )))|q
0<<∗ 0
∨ [E |R( f (y(t ), y(t − τ )))|q ] ∨ [E |R(g(y(t ), y(t − τ )))|q ] ≤ C q 1 + (ψ ())2q .
(42)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
and
sup
sup E |Rˆ( f (y¯ (t ), y¯ (t − τ )))|q ∨ E |Rˆ(g(y¯ (t ), y¯ (t − τ )))|q
0<<∗ 0
275
∨ [E |R( f (y¯ (t ), y¯ (t − τ )))|q ] ∨ [E |R(g(y¯ (t ), y¯ (t − τ )))|q ] ≤ C q 1 + (ψ () )2q .
(43)
Proof. By the definition of R( f (y(t ), y(t − τ ))), for any t ∈ [tk , tk+1 ], we have
E |R( f (y(t ), y(t − τ )) )|
q
≤ C E |y(t ) − y¯ (t )|
2q
+ |y(t − τ ) − y¯ (t − τ )|2q
× E (| f11 (y(t ), y(t − τ ))|
2q
1/2
+ 2| f12 (y(t ), y(t − τ ))|
2q
+ | f22 (y(t ), y(t − τ ))| ) 2q
1/2
≤ C q 1 + (ψ () )2q .
(44)
Hence
E |Rˆ( f (y(t ), y(t − τ )))|q
q
q
≤ C q E f1 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ )) + E f2 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ ) )
q
+ E f1 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ ))((w¯ (t ))2 − ) q t + E f1 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) w¯ (s − τ )dw(s ) tk
q + E f2 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ )) (w¯ (t ))2 − q t + E f2 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) w¯ (s − τ )dw(s ) tk q + |R( f (y(t ), y(t − τ ))| .
(45)
Applying elementary inequality and Hölder inequality, we get
q q t 2 E (w¯ (t )) − ∨ w¯ (s − τ )dw(s ) ≤ C q , tk q E f1 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ ))) ≤ C (ψ ())q , q E f2 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ ))) ≤ C (ψ ())q , q E f1 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ )) ≤ C (ψ ())2q , q E f1 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) ≤ C (ψ ())2q , q E f2 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ )) ≤ C (ψ ())2q
and
q
E f2 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) ≤ C (ψ ())2q . Consequently, we obtain
q
E Rˆ( f (y(t ), y(t − τ ))) ≤ C q 1 + (ψ () )2q . Similarly, we can show
E |R(g(y(t ), y(t − τ )))|2 ∨ E |Rˆ(g(y(t ), y(t − τ )))|q ≤ C q 1 + ψ () )2q
and (48).
Lemma 3.11. Let (A2) – (A6) hold. Let σ Z and ρ ,Z be the same as before and 2(1 + γ ) ∨ 4(1 + r ) < p. Define
θ := σZ ∧ ρ,Z and e (t ) := x(t ) − y(t ). ¯ ∈ (0, ∗ ] be sufficient small such that ψ () ≥ φ (Z). Then, Let
E
sup
s∈[0,T ∧θ ]
|e (s )|2 ≤ C 2 1 ∨ (ψ ())4 , ∀ T > 0.
(46)
276
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
Proof. Using Itô formula, we can show that for all t ∈ [0, T],
E
sup
s∈[0,t∧θ ]
≤E
|e ( s )|2
T ∧θ
2eT (s ) f (x(s ), x(s − τ )) − fˆ(y¯ (s ), y¯ (s − τ ))
0
+ |g(x(s ), x(s − τ )) − gˆ(y¯ (s ), y¯ (s − τ )) − gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
− gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
+E
r
sup
r∈[0,t∧θ ] 0
2eT (s )[g(x(s ), x(s − τ )) − gˆ(y¯ (s ), y¯ (s − τ ))
− gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
− gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )]dw(s ) .
(47)
Since s ∈ [0, t∧θ ], |y¯ (s )| ≤ Z and ψ () ≥ φ (Z), we have |y¯ (s )| ≤ φ −1 (ψ ()) and fˆ(y¯ (s ), y¯ (s − τ )) = f (y¯ (s ), y¯ (s − τ )), gˆ(y¯ (s ), y¯ (s − τ )) = g(y¯ (s ), y¯ (s − τ )) and gˆ(y¯ (s − τ ), y¯ (s − 2τ )) = g(y¯ (s − τ ), y¯ (s − 2τ )). Consequently, we have
E
sup
s∈[0,t∧θ ]
≤E
|e ( s )|
2
t∧θ
2eT (s )[ f (x(s ), x(s − τ )) − f (y¯ (s ), y¯ (s − τ ))]
0
+ |g(x(s ), x(s − τ )) − g(y¯ (s ), y¯ (s − τ )) − g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s )
− g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
√ +4 2 E
t∧θ
0
2eT (s ) g(x(s ), x(s − τ )) − g(y¯ (s ), y¯ (s − τ ))
− g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s ) − g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )
2
ds
1/2
≤ H1 + H2 + H3 + H4 , where
H1 = 2E
H3 ≤ E
t∧θ
2eT (s )[ f (y(s ), y(s − τ ) − f (y¯ (s ), y¯ (s − τ ))]ds,
0 t∧θ 0
√ H4 = 4 2E
eT (s )[ f (x(s ), x(s − τ )) − f (y(s ), y(s − τ ))] + |g(x(s ), x(s − τ )) − g(y(s ), y(s − τ ))|2 ds,
0
H2 = E
t∧θ
(48)
2
2Rˆ(g(y¯ (s ), y¯ (s − τ )) ds,
t∧θ 0
2eT (s ) g(x(s ), x(s − τ )) − g(y¯ (s ), y¯ (s − τ ))
− g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s ) − g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) By (A3) and (A4), we get
H1 ≤ 2K1 E ≤C
t 0
t∧θ 0
|x(s ) − y(s )|2 + |x(s − τ ) − y(s − τ )|2 ds
E |e (s ∧ θ )| ds. 2
2
ds
1/2
.
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
By (40), we have that
H2 ≤ E
t∧θ 0
2e (s ) f1 (y¯ (s ), y¯ (s − τ ))
+ f2 (y¯ (s ), y¯ (s − τ )) ≤ CE where
t∧θ
0
H5 ≤E
s −τ
s−τ
s
g(y¯ (r ), y¯ (r − τ ))dw(r )
g(y¯ (r ), y¯ (r − τ ))dw(r ) + Rˆ( f (y¯ (s ), y¯ (s − τ ))) ds
eT (s ) f1 (y¯ (s ), y¯ (s − τ ))
0
s
2 |e (s )|2 + Rˆ( f (y¯ (s ), y¯ (s − τ ))) ds + 2H5 ,
+ f2 (y¯ (s ), y¯ (s − τ )) Define
t∧θ
T
f1 (y¯ (s ), y¯ (s − τ ))
+ f2 (y¯ (s ), y¯ (s − τ ))
s−τ
s
s
s −τ
s
s
g(y¯ (r ), y¯ (r − τ ))dw(r )
g(y¯ (r ), y¯ (r − τ ))dw(r ) ds.
g(y¯ (r ), y¯ (r − τ ))dw(r )
s −τ
g(y¯ (r − τ ), y¯ (r − 2τ ))dw(r )
s−τ
:= v(y¯ (s )). Noting that
x (s ) − y (s ) = x ( s ) − y ( s ) + + − −
s s
s
s
[ f (x(r ), x(r − τ )) − f (y¯ (r ), y¯ (r − τ ))]dr
[g(x(r ), x(r − τ )) − g(y¯ (r ), y¯ (r − τ ))]dw(r )
s s s s
g1 (y(r ), y(r − τ ))g(y¯ (r ), y¯ (r − τ ))w(r )dw(r ) g2 (y(r ), y(r − τ ))g(y¯ (r − τ ), y¯ (r − 2τ ))w(r − τ )dw(r ),
we have
H5 ≤J1 + J2 + J3 + J4 , where
J1 = E
t∧θ 0
J2 = E
t∧θ
0
J3 = E
s
[ f (x(r ), x(r − τ )) − f (y¯ (r ), y¯ (r − τ ))]dr
s t∧θ
0
J4 = E
[x(s ) − y(s )]T v(y¯ (s ))ds,
s
s t∧θ
0
s
s
T
v(y¯ (s ))ds,
[g(x(r ), x(r − τ )) − g(y(r ), y(r − τ ))]dw(r )
T
v(y¯ (s ))ds,
[Rˆ(g(y(r ), y(r − τ )))dw(r )
T
v(y¯ (s ))ds.
Computing that
J1 = E
t∧θ −1 t k+1 k=0
tk
[x(tk ) − y(tk )]T f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ds + E = J11 + J12 ,
t∧θ
t∧θ
[x(s ) − y(s )]T v(y¯ (s ))ds
277
278
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
where t∧θ −1
J11 ≤ E
tk+1
tk
k=0
[x(tk ) − y(tk )]T f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ds t∧θ −1
≤E
tk+1
tk
k=0
t∧θ −1
sup
s∈[0,t∧θ ]
k=0
1/2
[x(tk ) − y(tk )]T f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ≤E
2
|x ( s ) − y ( s )|2
tk+1
tk
2
1/2
ds
f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
2 1/2
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ds 1 ≤ E 16
|e ( s )|
sup
s∈[0,t∧θ ]
+ C E
2
t∧θ −1 k=0
tk+1
tk
| f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s )| ds 2
1 ≤ E 16
|e ( s )|
sup
s∈[0,t∧θ ]
2
+ C 2 ,
t∧θ
|x ( s ) − y ( s )| |v(y¯ (s ))|ds t∧θ s∈[0,t∧θ ] 1 ≤ E sup |e (s )|2 + C 2 .
J12 ≤ E
sup
16
s∈[0,t∧θ ]
Similarly, we get
J2 = E
t∧θ
0
s
s
[| f1 (y¯ (r ), y¯ (r − τ ))||x(r ) − y¯ (r )| + |R( f (y¯ (r ), y¯ (r − τ )))|
+ | f2 (y¯ (r ), y¯ (r − τ ))||x(r − τ ) − y¯ (r − τ )|]dr ≤ CE
t∧θ
s
0
s
T
| f1 (y¯ (r ), y¯ (r − τ ))||x(r ) − y(r )|
+ | f2 (y¯ (r ), y¯ (r − τ ))||x(r − τ ) − y(r − τ )| dr + CE
t∧θ 0
s
s
T
v(y¯ (s ))ds
[| f1 (y¯ (r ), y¯ (r − τ ))||y(r ) − y¯ (r )|
+ | f2 (y¯ (r ), y¯ (r − τ ))||y(r − τ ) − y¯ (r − τ )|]dr
T
v(y¯ (s ))ds
|R( f (y¯ (r ), y¯ (r − τ )))|dr T v(y¯ (s ))ds 0 s t∧θ s ≤ CE |x(r ) − y(r )|dr T f1T (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))ds 0 s t∧θ s + CE |x(r ) − y(r )|dr T f2T (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))ds
+E
t∧θ
0
v(y¯ (s ))ds
s
s
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
t∧θ
s
[| f1 (y¯ (r ), y¯ (r − τ ))||y(r ) − y¯ (r )|dr
v(y¯ (s ))ds t∧θ s + CE [| f2 (y¯ (r ), y¯ (r − τ ))||y(r − τ ) − y¯ (r − τ )|]dr T v(y¯ (s ))ds 0 s t∧θ s +E |R( f (y¯ (r ), y¯ (r − τ )))|dr T v(y¯ (s ))ds + CE
0
s
0
T
s
s 2 t∧θ C ds + C E ≤ E | x ( r ) − y ( r ) | dr | f1 (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))|2 ds 0 s 0 2 t∧θ s t∧θ C |x(r ) − y(r )|dr ds + C E + E | f2 (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))|2 ds 0 s 0 2 t∧θ s t∧θ C ds + C E ¯ ¯ ¯ + E [ | f ( ( r ) , ( r − τ )) || y ( r ) − y ( r ) | dr y y |v(y¯ (s ))|2 ds 1 0 s 0 2 t∧θ s C [| f2 (y¯ (r ), y¯ (r − τ ))||y(r − τ ) − y¯ (r − τ )|]dr ds + E 0 s
t∧θ
t∧θ
|v(y¯ (s ))|2 ds 2 t∧θ s t∧θ C |R( f (y¯ (r ), y¯ (r − τ )))|dr ds + C E + E |v(y¯ (s ))|2 ds 0 s 0 t 2 ≤C E sup |e (u )| ds + C 2 (ψ ())4 + C 2 , + C E
0
u∈[0,t∧θ ]
0
Similarly, we get
J3 ≤ C
t
E
sup
u∈[0,t∧θ ]
0
|e (u )|2 ds + C 2 (ψ ())4 + C 2 ,
s 2 t∧θ Rˆ(g(y(r ), y(r − τ )))dw(r ) ds + C E |v(y¯ (s ))|2 ds 0 s 0 ≤ C 2 1 + (ψ ())4 .
J4 ≤
C
E
t∧θ
Using the elementary inequality, we obtain
√ H4 ≤ 4 2E
sup
s∈[0,t∧θ ]
|x ( s ) − y ( s )|2
t∧θ
0
4|g(x(s ), x(s − τ ))
− g(y¯ (s ), y¯ (s − τ )) − g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s )
− g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds ≤
1 E 16
sup
s∈[0,t∧θ ]
|e (s )|2 + CE
t∧θ 0
1/2
g(x(s ), x(s − τ ))
− g(y¯ (s ), y¯ (s − τ )) − g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s )
2
− g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds 1 ≤ E 16
sup
s∈[0,t∧θ ]
|e ( s )|
2
+ CE
t∧θ 0
t∧θ
Rˆ(g(y¯ (s ), y¯ (s − τ )))2 ds
|g(x(s ), x(s − τ )) − g(y(s ), y(s − τ ))|2 ds t∧θ 1 Rˆ(g(y¯ (s ), y¯ (s − τ )))2 ds ≤ E sup |e (s )|2 + CE + CE
16
0
s∈[0,t∧θ ]
0
279
280
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
+ CE
t∧θ
0
|g1 (y(s ), y(s − τ ))|2 |x(s ) − y(s )|2
+ |g2 (y(s ), y(s − τ ))|2 |x(s − τ ) − y(s − τ )|2 + R(g(y(s ), y(s − τ ))) ds 1 ≤ E 16
|e ( s )|
sup
s∈[0,t∧θ ]
2
2
+C
t
0
4
E sup |e (r ∧ θ )|
2
ds
r∈[0,s]
+ C 1 + (ψ ()) . Substituting these into (48), we get
E
sup
s∈[0,t∧θ ]
|e ( s )|
2
≤C
t
E
sup
r∈[0,s∧θ ]
0
|e (r )| ds + C 2 1 ∨ (ψ ())4 . 2
The Gronwall inequality yields the assertion (46).
Theorem 3.12. Let (A1), (A2), (A4) – (A6) hold and assume that (A3) is satisfied for any p > 2γ + 2. Let p¯ = p − 2γ and 2(γ + 1 ) ∨ 4(r + 1 ) < p and ε ∈ (0, 1/2). Define
φ (u ) = H¯ u1+γ , u ≥ 0 and
ψ () = −ε/2 , ∈ (0, 1], where H¯ is a positive constant. for all sufficient small ∈ (0, ∗ ), then for every such small ,
sup |x(t ) − y(t )|
E
≤ O(2(1−ε ) ),
2
0≤t≤T
Consequently,
lim E
→ 0
(49)
sup |x(t ) − y(t )|
2
0≤t≤T
= 0,
Proof. The notation is same as in the proof of Theorem 3.11. For our proof, assume that γ > 0, p > 2γ + 2, p¯ = p − 2γ , let p¯ > 2 sufficiently large such that
ε 2
>
2 (1 + γ ) . ( p¯ − 2 )
(50)
Applying the Young inequality, we can see that for any ∈ (0, ∗ ), > 0 and Z > ξ , we have
sup |e (t )|2
E
0≤t≤T
sup |e (t )|2 1{θ >T }
=E
0≤t≤T
+
sup |e (t )|2 1{θ >T }
≤E
0≤t≤T
p¯ − 2
p¯ 2/( p¯ −2)
+E +
sup |e (t )|2 1{θ ≤T }
0≤t≤T
2 E p¯
sup |e (t )| p¯
0≤t≤T
P ( θ ≤ T ).
(51)
Using Lemmas 3.7 and 3.8, we have
sup |e (t )|
E
p¯
0≤t≤T
≤C.
(52)
By Lemmas 3.4 and 3.5, we obtain
P (θ ≤ T ) ≤
C . Z p¯
Hence we get
(53)
sup |e (t )|2
E
≤E
0≤t≤T
sup |e (t )|2 1{θ >T }
+
0≤t≤T
2 C ( p¯ − 2 )C + . p¯ p¯ 2/( p¯ −2) Z p¯
(54)
Choosing = 2(1−ε ) and Z = −2(1−ε )/( p¯ −2 ) , we have that
E
sup |e (t )|2
0≤t≤T
≤E
sup |e (t )|2 1{θ >T }
0≤t≤T
+ C 2 ( 1 − ε ) .
(55)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
281
According to (50), we see that for sufficient small ,
φ −1 (ψ ()) ≥ −2(1−ε )/( p¯−2) = Z. In view of Lemma 3.11, we hence obtain from (55) that
E
sup |e (t )|2
0≤t≤T
≤ C 2 ( 1 − ε ) .
The proof is therefore complete.
4. Conclusion For most highly nonlinear SDDEs, there are no explicit solutions, it is important to develop some approximate numerical methods. Due to the advantages of small amount of calculation and short time of explicit method, our aim in this paper is to present an explicit method named truncated Milstein method of the highly nonlinear SDDEs. We study strong convergence of the truncated Milstein method under the local Lipschitz condition plus Khasminskii-type condition and shows its convergence rate is close to 1 under an additional condition. References [1] D.J. Higham, X. Mao, A.M. Stuart, Strong convergence of euler-type methods for nonlinear stochastic differential equations, SIAM J. Numer. Anal 40 (3) (2002) 1041–1063. [2] D.J. Higham, Exponential mean-square stability of numerical solutions to stochastic differential equations, LMS J. Comput. Math. 6 (6) (2003) 275–294. [3] M. Hutzenthaler, A. Jentzen, P.E. Kloeden, Strong convergence of an explicit numerical method for SDEs with nonglobally lipschitz continuous coefficients, Ann. Appl. Probab. 22 (4) (2012) 1611–1641. [4] P.E. Kloeden, E. Platen, Numerical Solution of Stochastic Differential Equations, Springer, Berlin, 1992. [5] P.E. Kloeden, T. Shardlow, The Milstein scheme for stochastic delay differential equations without using anticipative calculus, J. Sto. Anal. Appl. 30 (2) (2012) 181–202. [6] W. Liu, X. Mao, Almost sure stability of the Euler-Maruyama method with random variable stepsize for stochastic differential equations, Numer. Algorithms (2016) 1–20. [7] Q. Luo, X. Mao, Y. Shen, Generalised theory on asymptotic stability and boundedness of stochastic functional differential equations, Automatica 47 (9) (2011) 2075–2081. [8] X. Mao, Numerical solutions of stochastic functional differential equations, LMS J. Comput. Math 6 (4) (2003) 1821–1841. [9] X. Mao, C. Yuan, Stochastic Differential Equations with Markovian Switching, Imperial College Press, 2006. [10] X. Mao, Stochastic Differential Equations and Applications, Horwood Pub Limited, 1997. [11] X. Mao, Approximate solutions for a class of stochastic evolution equations with variable delays, Numer. Funct. Anal. Optim. 12 (1) (1991) 525–533. [12] X. Mao, M.J. Rassias, Khasminskii-type theorems for stochastic differential delay equations, Stoch. Anal. Appl. 23 (2005) 1045–1069. [13] X. Mao, Numerical solutions of stochastic differential delay equations under the generalized Khasminskii-type conditions, Appl. Math. Comput. 217 (12) (2011) 5512–5524. [14] X. Mao, The truncated Euler-Maruyama method for stochastic differential equations, J. Comput. Appl. Math. (2015) 370–384. [15] X. Mao, Convergence rates of the truncated Euler-Maruyama method for stochastic differential equations, J. Comput. Appl. Math 296 (2016) 362–375. [16] M. Song, L. Hu, X. Mao, L. Zhang, Khasminskii-type theorems for stochastic functional differential equations, Discrete Contin. Dyn. Syst. Ser. B 6 (6) (2013) 1697–1714. [17] Z. Wang, C. Zhang, An analysis of stability of Milstein method for stochastic differential equations with delay, Comput. Math. Appl. 51 (9) (2006) 1445–1452. [18] X. Wang, S. Gan, The tamed Milstein method for commutative stochastic differential equations with non-globally Lipschitz continuous coefficients, J. Differ. Equ. Appl. 19 (3) (2013) 466–490. [19] W. Zhang, M.H. Song, M.Z. Liu, Strong convergence of the partially truncated Euler-Maruyama method for a class of stochastic differential delay equations, J. Comput. Appl. Math. 335 (2018) 114–128. [20] V.A. Zorich, Mathematical Analysis I, second, Springer, 2004.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Convergence rate of the truncated Milstein method of stochastic differential delay equationsR Wei Zhang a, Xunbo Yin b, M.H. Song b,∗, M.Z. Liu b a b
School of Mathematical Sciences, Heilongjiang University, Harbin, Heilongjiang, China Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
a r t i c l e
i n f o
Keywords: Stochastic differential delay equation Truncated Milstein method Local Lipschitz condition Khasminskii-type condition Strong convergence
a b s t r a c t This paper is concerned with the strong convergence of highly nonlinear stochastic differential delay equations (SDDEs) without the linear growth condition. On the one hand, these nonlinear SDDEs do not have explicit solutions, therefore implementable numerical methods for such SDDEs are required. On the other hand, the implicit Euler methods are known to converge strongly to the exact solution of such SDDEs. However, they require additional computational efforts. In this article, we propose the truncated Milstein method which is an explicit method under the local Lipschitz condition plus Khasminskii-type condition, study its pth monent boundedness (p is a parameter in Khasminskii-type condition) and show that its rate of strong convergence is close to one. © 2019 Elsevier Inc. All rights reserved.
1. Introduction Stochastic differential delay equations (SDDEs) have come to play an important role in many branches of science and industry. Hence the theory of SDDEs received increasing attentions and obtained significant results. Many SDDEs have no explicit solutions and numerical approximations are required. The numerical solutions of stochastic differential equations (SDEs) have also been studied extensively by many authors (see e.g., [1,4,6–9,17]). To the best of our knowledge, the coefficients of SDDEs satisfying the global Lipschitz condition or the local Lipschitz condition and the linear growth condition (see e.g., [2,10,11]) are required by the classical strong convergence theorem of SDDEs. However, there are many SDDEs which do not satisfy the linear growth condition. In 2005, Mao [12] established a Khasminskii-type theorem for SDDEs where the linear growth condition was no longer required. In 2011, Mao [13] considered numerical solutions of SDDEs under the generalized Khasminskii-type condition. In 2012, [3] proposed an explicit strongly convergent numerical scheme, called the tamed Euler method for SDEs. Motivated by this work, [18] introduce a tamed version of the Milstein scheme for SDEs with commutative noise. In 2015 and 2016, Mao [14,15] showed the truncated Euler-Maruyama method for SDEs and gave its rate. In 2018, Zhang investigated strong convergence of the partially truncated Euler-Maruyama method for SDDEs and showed its rate was close to 1/2 under the polynomial growth condition plus Khasminskii-type condition in [19]. So far there is little numerical theory on SDDEs under Khasminskii-type conditions.
R ∗
This work is supported by the NSF of P.R. China (No.11671113). Corresponding author. Tel.:+86045186403055. E-mail addresses: [email protected] (W. Zhang), [email protected] (X. Yin), [email protected] (M.H. Song), [email protected] (M.Z. Liu).
https://doi.org/10.1016/j.amc.2019.04.001 0 096-30 03/© 2019 Elsevier Inc. All rights reserved.
264
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In this paper we consider the following 1-dimensional SDDE
dx(t ) = f (x(t ), x(t − τ ))dt + g(x(t ), x(t − τ ))dw(t )
(1)
with initial data x0 = ξ ∈ C ([−τ , 0], R ). Motivated by the techniques discussed in [14,15,18,19], we propose a truncated Milstein method which converge to SDDEs (1) with super-linear drift and diffusion coefficients. For the convenience, we will, in Section 2, present the truncated Milstein method. We will study the pth moment boundedness of the truncated Milstein method and its convergence and give its rate in Section 3. We will conclude our paper in Section 4. 2. The truncated Milstein method Throughout this paper, let (, F, {F t }t≥0 , P ) denote a complete probability space with a filtration {F t }t≥0 satisfying the usual conditions (i.e, it is right continuous and increasing while F0 contains all P-null sets), and let E be the expectation corresponding to P. Let w(t ) be a 1-dimensional Brownian motion defined on the probability space. Let | · | denote both Euclidean norm in R and the trace norm in R2 × 2 . Denote by C ([−τ , 0], R ) the family of continuous functions from [−τ , 0] to R with the norm η = sup−τ ≤u≤0 |η (u )|. For a, b ∈ R, we use a∨b and a∧b for max {a, b} and min {a, b}, respectively. Consider a 1-dimensional SDDE
dx(t ) = f (x(t ), x(t − τ ))dt + g(x(t ), x(t − τ ))dw(t )
(2)
on t ∈ [0, ∞) with initial data {ξ (θ ) : −τ ≤ θ ≤ 0} = ξ ∈ C ([−τ , 0]; R ), where f: R × R → R and g: R × R → R. Throughout this paper, we impose the following five assumptions as standing hypotheses: (A1) For every constant Z ≥ 1, there exists a positive constant LZ > 0, such that
f (x, y ) − f (xˆ, yˆ ) ∨ g(x, y ) − g(xˆ, yˆ ) ≤ LZ |x − xˆ|2 + |y − yˆ|2
(3)
for all x, y, xˆ, yˆ ∈ R with |x| ∨ |y| ∨ |xˆ| ∨ |yˆ| ≤ Z. (A2) We also assume that the coefficients f and g satisfy the Khasminskii-type condition: there is a pair of constants p > 2 and K > 0 such that
xT f (x, y ) + ( p − 1 )|g(x, y )|2 ≤ K 1 + |x|2 + |y|2
(4)
for all x, y ∈ R. (A3) We also assume that the coefficients f and g satisfy the following condition: there are constants q > 2 and K1 > 0 such that
(x − x¯ )T ( f (x, y ) − f (x¯, y¯ )) + (q − 1 )|g(x, y ) − g(x¯, y¯ )|2 ≤ K1 |x − x¯|2 + |y − y¯|2
(5)
for all x, y, x¯ y¯ ∈ R. (A4) There is a constant K0 > 0 such that
|ξ (t ) − ξ (s )| ≤ K0 |t − s|
(6)
for all s, t ∈ [−τ , 0]. (A5) There are constants r > 0 and K2 > 0 such that
| fi (x, y )| ∨ |gi (x, y )| ∨ | fi j (x, y )| ∨ |gi j (x, y )|∨ ≤ K2 1 + |x|r+1 + |y|r+1
(7)
for all x, y ∈ R, i, j = 1, 2, where
∂ f ( x1 , x2 ) ∂ g( x 1 , x 2 ) , gi ( x1 , x2 ) = , ∂ xi ∂ xi ∂ 2 f ( x1 , x2 ) ∂ 2 g( x 1 , x 2 ) f i j ( x1 , x2 ) = , gi j ( x1 , x2 ) = . ∂ xi x j ∂ xi x j f i ( x1 , x2 ) =
We state a known result (see, e.g., [16]) as a lemma for the use of this paper. Lemma 2.1. Under (A1) and (A2), the SDDE (2) has a unique global solution y(t) and, moreover,
sup E |x(t )| p ≤ C,
t∈[−τ ,T ]
∀T > 0.
(8)
where C stands for a generic positive real constant (but independent of and Z later) and its value may change between occurrences. Motivated by Mao [14], we first define the truncated functions. Choose a strictly increasing continuous function φ : R+ → R+ such that φ (s) → ∞ as s → ∞ and for i = 1, 2,
sup (| f (x, y )| ∨ |g(x, y )| ∨ |gi (x, y )| ) ≤ φ (s ),
|x|∨|y|≤s
∀s ≥ 0.
(9)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
265
Let φ −1 be the inverse function of φ and we see that φ −1 is a strictly increasing continuous function from [φ (0), ∞) to R+ . We also choose a number ∗ ∈ (0, 1] and a strictly decreasing function ψ : (0, ∗ ] → (1, ∞) such that
ψ (∗ ) ≥ φ (1 ),
(10)
ψ () = ∞
(11)
lim
→ 0
and
1/4 ψ 3/2 () ≤ 1,
∀ ∈ ( 0 , ∗ ] .
(12)
For a given step size ∈ (0, 1), let us define the truncated functions
x y , |y| ∧ φ −1 (ψ ()) , |x| |y| x y gˆ(x, y ) = g |x| ∧ φ −1 (ψ ()) , |y| ∧ φ −1 (ψ ()) , |x| |y| x y gˆi (x, y ) = gi |x| ∧ φ −1 (ψ ()) , |y| ∧ φ −1 (ψ ()) |x| |y| fˆ(x, y ) = f
|x| ∧ φ −1 (ψ ())
(13)
for all x, y ∈ R, i = 1, 2, where we set x/|x| = 0 when x = 0. It is easy to see that
ˆ f (x, y ) ∨ gˆ(x, y ) ∨ gˆi (x, y ) ≤ φ φ −1 (ψ ()) = ψ ()
(14)
for all x, y ∈ R, i = 1, 2, the truncated functions fˆ, gˆ and gˆi are bounded although f, g and gi may not. Moreover, we can show that truncated functions fˆ and gˆ preserve the Khasminskii-type condition. Lemma 2.2. Let (A2) hold. Then, for all ∈ (0, ∗ ], we have
2 xT fˆ(x, y ) + ( p − 1 )gˆ(x, y ) ≤ 3K 1 + |x|2 + |y|2
(15)
for all x, y ∈ R. Proof. The proof is similar to that of Lemma 2.2 in [19].
Without loss of any generality, we may assume that T and τ are rational numbers. Moreover, T = M and τ = N. We can now form the discrete-time truncated Milstein numerical solutions Y(tk ) ≈ y(tk ) for tk = k by setting Y (0 ) = ξ (0 ) and computing
⎧ Y (t ) = ξ (tk ), k = −N, −N + 1, . . . , 0, ⎪ ⎨ k Y (tk+1 ) = Y (tk ) + fˆ(Y (tk ), Y (tk−N ) ) + gˆ(Y (tk ), Y (tk−N ) )wk + gˆ1 (Y (tk ), Y (tk−N ) )gˆ(Y (tk ), Y (tk−N ) )I1 ⎪ ⎩ + gˆ2 (Y (tk ), Y (tk−N ) )gˆ(Y (tk−N ), Y (tk−2N ) )I2 , k = 0, 1, . . . , M,
where
I1 =
tk
I2 =
tk+1
s
tk
tk+1
tk
s
tk
dw(t )dw(s ) =
(wk )2 − 2
,
dw(t − τ )dw(s ).
wk = w(tk ) and wk = wk+1 − wk . Let us now establish two continuous-time truncated Milstein solutions. The first one is defined by
y¯ (t ) =
M−1
Y (tk )1[tk ,tk+1 ) (t ),
(16)
k=0
with 1G denoting the indicator function for the set G. This is a simple step process, so its sample paths are not continuous. The other continuous-time step-process truncated Milstein solution is defined by
⎧ ⎪ ⎪ y(t ) = ξ (t ), t ∈ [−τ , 0], ⎪ t ⎪ t ⎪ ⎪ ⎪ y(t ) = ξ (0 ) + fˆ(y¯ (s ), y¯ (s − τ ))ds + gˆ(y¯ (s ), y¯ (s − τ ))dw(s ) ⎪ ⎨ 0 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
+
0
+
t
0
t
gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s )
(17)
gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s ), t ∈ [0, T ],
where w¯ (t ) = M k=−N (w (t ) − w (tk ))1[tk ,tk+1 ) (t ). (17) is referred as the continuous-time continuous-sample truncated Milstein solution. In particular, this shows that Y (tk ) = y¯ (tk ) = y(tk ) for all k = 0, 1, . . . , M. Moreover, for all t ∈ [0, T], we have
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E |y¯ (t )| ≤ sup E |y¯ (t )| = sup E |Y (tk )| 0≤t≤T
0≤tk ≤T
= sup E |y(tk )| ≤ sup E |y(s )|, 0≤tk ≤T
(18)
0≤t≤T
3. Convergence of the truncated Milstein method 3.1. Moment boundedness of the truncated Milstein method We first show that y(t) and y¯ (t ) are close to each other in L pˆ where pˆ > 0. Lemma 3.1. For any ∈ (0, ∗ ] and pˆ > 0, we have
E |y(t ) − y¯ (t )| pˆ ≤ C pˆ/2 (ψ ())2 pˆ ,
∀t > 0.
(19)
Consequently
lim E |y(t ) − y¯ (t )| pˆ = 0,
∀t > 0.
→ 0
(20)
Proof. Fix any ∈ (0, ∗ ] and t ≥ 0. There is a unique integer k ≥ 0 such that tk ≤ t < tk+1 . We divided the proof into two cases. Case (I): If pˆ ≥ 2, applying Hölder inequality, Itô isometry and (14), we obtain
E |y(t ) − y¯ (t )| pˆ = E |y(t ) − y(tk )| pˆ pˆ t pˆ t ˆ ¯ ¯ ˆ ¯ ¯ ≤ C E f (y(s ), y(s − τ ))ds + E g(y(s ), y(s − τ ))dw(s ) tk
tk
t pˆ + E gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s ) tk t pˆ + E gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s ) tk t t pˆ ˆ ¯ ( pˆ−2 )/2 gˆ(y¯ (s ), y¯ (s − τ ))pˆ ds ¯ ≤ C pˆ−1 E f ( ( s ) , ( s − τ )) ds + E y y tk
+ ( pˆ−2)/2 E + ( pˆ−2)/2 E ≤ C
pˆ/2
tk
t
tk
t
tk
gˆ1 (y¯ (s ), y¯ (s − τ ) )gˆ(y¯ (s ), y¯ (s − τ ) )w¯ (s ) pˆ ds gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )pˆ ds
(ψ () )2 pˆ ,
∀t > 0.
which is (19). Observe from (12) that pˆ/2 (ψ ())2 pˆ ≤ pˆ/2 , it is easy to get (20) from (19). Case (II): For any pˆ ∈ (0, 2 ), let l ≥ 2, using Hölder inequality, we have
E |y(t ) − y¯ (t )| pˆ ≤ E |y(t ) − y¯ (t )|l
pˆ/l
≤ C l/2 (ψ () )2l
pˆ/l
≤ C pˆ/2 (ψ () )2 pˆ . Then (20) holds.
Lemma 3.2. Let (A1) and (A2) hold. Then, for all t ∈ [0, T], we have
sup
sup E |y(t )| p ≤ C,
0<≤∗ −τ ≤t≤T
∀T > 0.
(21)
Proof. Fix any ∈ (0, ∗ ]. Applying the Itô formula and Lemma 2.2, we derive from (17) that, for 0 ≤ t ≤ T,
E |y(t )| p ≤
|ξ ( 0 )| p + E
t 0
p|y(s )| p−2 yT (s ) fˆ(y¯ (s ), y¯ (s − τ ))
p−1 + |gˆ(y¯ (s ), y¯ (s − τ )) + gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
|ξ ( 0 )| p + E
≤
t
267
p|y(s )| p−2 y¯ T (s ) fˆ(y¯ (s ), y¯ (s − τ )) + ( p − 1 )|gˆ(y¯ (s ), y¯ (s − τ ))|2 ds
0
|y(s )| p−2 gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 0 2 + gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds + p( p − 1 ) E
+E
t
0
t
p|y(s )| p−2 (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds
|ξ (0 )| p + 3KE
≤
+ p( p − 1 ) E
0 t
0
t
p|y(s )| p−2 1 + |y¯ (s )|2 + |y¯ (s − τ )|2 ds
|y(s )| p−2 |gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds t +E p|y(s )| p−2 (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds. 0
Using the Young inequality and (14), it is easy to obtain
t
pE 0
t
pE 0
|y(s )| p−2 |y¯ (s )|2 ds ≤ ( p − 2 ) ≤ p
t
pE 0
|y(s )| p−2 ds ≤ ( p − 2 )E
t
t
0 t
|y(s )| p ds + 2T ,
E |y(s )| p ds + 2
0
t
0
E |y¯ (s )| p ds
sup E |y(u )| p ds,
0 0≤u≤s
|y(s )| p−2 |y¯ (s − τ )|2 ds ≤ ( p − 2 )
0
t
E |y(s )| p ds + 2
0
t
E |y¯ (s − τ )| p ds
t
sup E |y(u )| p ds + 2τ ξ , 0 0 ≤u≤s t pE |y(s )| p−2 gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 0 2 + gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds t t ≤ ( p − 2 )E |y(s )| p ds + 2E gˆ1 (y¯ (s ), y¯ (s − τ ) )gˆ(y¯ (s ), y¯ (s − τ ) )w¯ (s ) 0 0 p + gˆ2 (y¯ (s ), y¯ (s − τ ) )gˆ(y¯ (s − τ ), y¯ (s − 2τ ) )w¯ (s − τ ) ds ≤ p
≤ ( p − 2 )E ≤ ( p − 2) and
t
E 0
0
t
|y(s )| p ds + C p/2 (ψ () )2 p
0 t
E |y(s )| p ds + C
p|y(s )| p−2 (y(s ) − y¯ (s ) )T fˆ(y¯ (s ), y¯ (s − τ ))ds
≤ ( p − 2 )E
t
0
|y(s )| p ds + 2E
t 0
p/2 |y(s ) − y¯ (s )| p/2 fˆ(y¯ (s ), y¯ (s − τ )) ds.
By Lemma 3.1 and inequalities (14) and (12), we obtain
T
E 0
p/2 |y(s ) − y¯ (s )| p/2 fˆ(y¯ (s ), y¯ (s − τ )) ds
≤ (ψ () ) p/2 E
T
0 3 p/2
≤ CT (ψ () ) ≤ C.
|y(s ) − y¯ (s )| p/2 ds
p/4 (22)
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W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
Applying the well-known Gronwall inequality, we get
sup E |y(u )| p ≤ C.
0≤u≤t
It is easy to show
sup E |y(u )| p ≤ C.
−τ ≤u≤t
As this holds for any ∈ (0, ∗ ], we have the required (21). The proof is complete.
Corollary 3.3. Assume that (A1) and (A2) hold. Then we have
sup
sup E|y¯ (t )| p ≤ C
(23)
0<≤∗ −τ ≤t≤T
for all T > 0. 3.2. Strong convergence To show that both truncated Milstein solutions y(T) and y¯ (T ) will converge to the exact solution x(T) in L2 for any T > 0, we need to present several lemmas. In the following of this section, we fix T > 0 arbitrarily. Lemma 3.4. Let (A1) and (A2) hold. For any real number Z > ξ , define the stopping time
σZ = inf {t ≥ 0 : |x(t )| ≥ Z }, where throughout this paper we set inf = ∞ (and as usual denotes the empty set). Then
P ( σZ ≤ T ) ≤
C . Z2
(24)
Proof. Using the Itô formula and (A2), we derive that
E |x(t ∧ σZ )|2 ≤ ≤
|ξ ( 0 )|2 + E
t∧σZ 0
|ξ (0 )|2 + 2KE
t∧σZ 0
2xT (s ) f (x(s ), x(s − τ )) + |g(x(s ), x(s − τ ))|2 ds
1 + |x(s )|2 + |x(s − τ )|2 ds
≤ (1 + 2K τ )ξ + 2KT + 4K
2
t
0
E |x(s ∧ σZ )| ds 2
for any 0 ≤ t ≤ T. The Gronwall inequality yields that
E |y(t ∧ σZ )|2 ≤ C. This implies
Z 2 P ( σZ ≤ T ) ≤ C and the assertion follows.
Lemma 3.5. Let (A1) and (A2) hold. For any real number Z > ξ and ∈ (0, ∗ ], define the stopping time
ρ,Z = inf{t ≥ 0 : |y(t )| ≥ Z }. Then
P (ρ,Z ≤ T ) ≤
C . Z2
(25)
Proof. We simply write ρ,Z = ρ . Using the Itô formula, we have that for 0 ≤ t ≤ T,
E |y(t ∧ ρ )|2 ≤ |ξ ( 0 )|2 + E
t∧ρ
0
2y(s )T fˆ(y¯ (s ), y¯ (s − τ )) + |gˆ(y¯ (s ), y¯ (s − τ ))
+ gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds t∧ρ ≤ |ξ ( 0 )|2 + 2E y¯ (s )T fˆ(y¯ (s ), y¯ (s − τ )) + |gˆ(y¯ (s ), y¯ (s − τ ))|2 ds + 2E
0
t∧ρ
0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
269
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds t∧ρ + 2E (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds. 0
Applying Lemma 2.2, we then get that,
E |y(t ∧ ρ )|2 ≤
|ξ ( 0 )|2 + E +E
t∧ρ
0
+ 2E
t∧ρ 0
6K 1 + |y¯ (s )|2 + |y¯ (s − τ )|2 ds
2|y(s ) − y¯ (s )|| fˆ(y¯ (s ), y¯ (s − τ ))|ds
t∧ρ 0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w(s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w(s − τ )|2 )ds t∧ρ ≤ |ξ (0 )|2 + 6KT + 24K E |y(s )|2 ds + 24K
0 T
+ 2E + 2E
0
0
T
E |y(s ) − y¯ (s )|2 ds + 12K τ ξ 2
|y(s ) − y¯ (s ))|| fˆ(y¯ (s ), y¯ (s − τ ))|ds
t∧ρ
0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w(s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w(s − τ )|2 )ds. By (14), we have
t∧ρ
E 0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds ≤ 2T (ψ ())4 ≤ 2T . Using Lemma 3.1, we get
T 0
E |y(s ) − y¯ (s )|2 ds ≤ CT (ψ ())4 ≤ C.
While by Lemma 3.1, (12) and (14), we derive that
T
E 0
|y(s ) − y¯ (s ))|| fˆ(s, y¯ (s )|ds ≤ CT 1/2 (ψ ())3 ≤ C.
The Gronwall inequality yields that
E |y(T ∧ ρ )|2 ≤ C. This implies the assertion (25) easily.
Theorem 3.6. Let (A1)–(A5) hold. Then,
lim E |x(T ) − y(T )|2 = 0
→ 0
and
lim E |x(T ) − y¯ (T )|2 = 0.
→ 0
(26)
Proof. Let σ Z and ρ ,Z be the same as before. Define
θ :=σZ ∧ ρ,Z ,
e (T ) :=
x ( T ) − y ( T ).
Using the Young inequality, we obtain that for any > 0,
E |e (T )|2 = E |e (T )|2 1{θ >T } + E |e (T )|2 1{θ ≤T } 2 p−2 ≤ E |e (T )|2 1{θ >T } + E |e ( T )| p + P ( θ ≤ T ). p p 2/( p−2) Applying Lemmas 2.1 and 3.2, we get
E |e (T )| ≤ C. p
(27)
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W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
While using Lemmas 3.4 and 3.5,
P (θ ≤ T ) ≤ P (σZ ≤ T ) + P (ρ,Z ≤ T ) ≤ We hence obtain
E |e (T )| ≤ E |e (T )|2 1{θ >T } + 2
C . Z2
2C ( p − 2 )C + . p p 2/( p−2) Z 2
Let ε > 0 be arbitrary. Choose sufficiently small such that ( p−2 )C
≤ 3ε . We then derive from (28) that
p 2/( p−2 ) Z 2
E |e (T )| ≤ E |e (T )|2 1{θ >T } + 2
2ε . 3
(28) 2C p
≤ ε3 and then choose Z sufficiently large such that
(29)
If we can show that for any sufficiently small ,
E |e (T )|2 1{θ >T } ≤
ε
3
,
(30)
we then have
lim E |e (T )| = 0 2
→ 0
and then using Lemma 3.1, we also obtain
lim E |x(T ) − y¯ (T )|2 = 0.
→ 0
Namely, in order to complete our proof, what we need is to show (30). For this purpose, we define the truncated functions
FZ (x, y ) = f (|x| ∧ Z ) and
x y , ( |y| ∧ Z ) |x| |y|
GZ (x, y ) = g (|x| ∧ Z )
x y , ( |y| ∧ Z ) |x| |y|
for any x, y ∈ R, without loss of generality, we may assume that ∗ is already sufficiently small for φ −1 (ψ (∗ )) ≥ Z. Hence, for all ∈ (0, ∗ ], we get that
FZ (x, y ) = fˆ(x, y ) and GZ (x, y )
= gˆ(x, y )
for any x, y ∈ R with |x|∨|y| ≤ Z. Consider the SDDE
dX (t ) = FZ (X (t ), X (t − τ ))dt + GZ (X (t ), X (t − τ ))dw(t )
(31)
on t ≥ 0 with initial data η (t ) ∈ C ([−τ , 0]; R ). By (A1), it is easy to see that both FZ (X (t ), X (t − τ )) and GZ (X (t ), X (t − τ )) are globally Lipschitz continuous with the Lipschitz constant LZ . So the SDDE (31) has a unique global solution X(t) on t ≥ 0. It is easy to see that
x(t ∧ σZ ) = X (t ∧ σZ )
a.s. f or all t ≥ 0.
(32)
∗ ],
On the other hand, for each step size ∈ (0, we use the truncated Milstein method to the SDDE (31) and let X¯ (t ) denote the continuous-time continuous-sample truncated Milstein solution. It is obvious that
y(t ∧ ρ,Z ) = X¯ (t ∧ ρ,Z )
a.s. f or all t ≥ 0.
(33)
However, it is known that (see Theorem 6.1 in [5])
sup |X (t ) − X¯ (t )|2
E
0≤t≤T
≤ O(2 ).
Therefore,
E
2
sup X (t ∧ θ ) − X¯ (t ∧ θ )
0≤t≤T
Using (32) and (33), we then obtain
sup |x(t ∧ θ ) − y(t ∧ θ )|2
E
0≤t≤T
which implies
≤ O(2 ).
≤ O(2 ).
E |x(T ∧ θ ) − y(T ∧ θ )|2 ≤ O(2 ).
(34)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
Hence
E |e (T )|2 1{θ >T } = E |e (T ∧ θ )|2 1{θ >T }
271
≤ E |x ( T ∧ θ ) − y ( T ∧ θ |2
≤ O(2 ). This implies (30) holds. The proof is therefore complete.
3.3. Convergence rate In the previous section, we showed that
lim E |x(T ) − y(T )|2 = 0
lim E |x(T ) − y¯ (T )|2 = 0
and
→ 0
→ 0
for any T > 0. This is sufficient for some applications. However, we sometimes need a stronger convergence result like
sup |x(t ) − y(t )|
lim E
→ 0
2
=0
0≤t≤T
(35)
We impose an additional assumption. (A6) Assume that there is a pair of constants γ > 0 and K3 > 0 such that
| f (x, y ) − f (xˆ, yˆ )| ∨ |g(x, y ) − g(xˆ, yˆ )| ≤ K3 1 + |x|γ + |y|γ + |xˆ|γ + |yˆ|γ · |x − xˆ| + |y − yˆ|
(36)
for all x, y, xˆ, yˆ ∈ R. See from (A6), we get
| f (x, y )| ∨ |g(x, y )| ≤ K4 1 + |x|γ +1 + |y|γ +1
(37)
for all x, y ∈ R, where K4 independent of K3 , |f(0, 0)|, |g(0, 0)| and ρ . Lemma 3.7. Under (A1), (A2), (A5) and (A6), assume that p > 2γ + 2, set p¯ = p − 2γ , then
E
sup
t∈[−τ ,T ]
|x(t )|
p¯
≤ C.
(38)
Proof. By the Itô formula, we can show that
|x(t )| p¯ ≤|ξ (0 )| p¯ + +
t 0
t 0
p¯ |x(s )| p¯ −2 xT (s )g(x(s ), x(s − τ ))dw(s )
p¯ |x(s )| p¯ −2 xT (s ) f (x(s ), x(s − τ )) +
p¯ − 1 |g(x(s ), x(s − τ ))|2 ds. 2
Using (A2), we get
E ( sup |x(t )| p¯ ) ≤|ξ (0 )| p¯ + E 0≤t≤T
+E
T 0
t
sup 0≤t≤T
0
2K p¯ |x(s )| p¯ −2 1 + |x(s )|2 + |x(s − τ )|2 ds
p¯ |x(s )|
x (s )g(x(s ), x(s − τ ))dw(s )
p¯ −2 T
Applying the Burkholder-Davis-Gundy inequality, Young inequality and (A6), we then have
E
sup |x(t )| p¯
0≤t≤T
≤
|ξ (0 )| p¯ + 2K ( p¯ − 2 )E
T 0
|x(s )| p¯ ds + 2K p¯ E
T
T 0
|x(s )| p¯ ds
|x(s − τ )| p¯ ds T √ + 4 2 p¯ E sup |x(s )| p¯ |x(s )| p¯−2 |g(x(s ), x(s − τ ))|2 ds 1/2 + 2K T + 2K E
0
0≤t≤T
0
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≤
|ξ (0 )| + CE
T
p¯
0
|x(s )| ds + 2K T + 2K τ E ξ p¯
1 + E 2
p¯
sup |x(t )|
p¯
0≤t≤T
|x(s )| p¯−2 1 + |x(s )|2γ +2 + |x(s − τ )|2γ +2 ds 1 ≤ C + E sup |x(t )| p¯ + 16K¯ p¯ 2 E
2
T
0
0≤t≤T
+ 16K¯ p¯ 2 E
|x(s )| p¯−2 1 + |x(s )|2γ +2 + |x(s − τ )|2γ +2 ds.
T
0
Noting that p¯ + 2ρ = p, by the Young inequality and Lemma 2.1, we obtain
T
p¯ E
0
0
= p¯ E ≤ p¯ E
T
T 0
|x(s )| p¯−2 1 + |x(s )|2γ +2 + |x(s − τ )|2γ +2 ds |x(s )| p¯−2 ds + p¯ E
T 0
|x(s )| p¯ ds + 2T + CE
|x(s )| p ds + p¯ E
T 0
T 0
|x(s )| p ds + CE
|x(s )| p¯−2 |x(s − τ )|2γ +2 ds
T
0
|x(s − τ )| p ds
≤ C. This implies
sup |x(t )|
E
p¯
≤C.
0≤t≤T
Hence the required assertion (39) holds. The proof is complete.
Lemma 3.8. Under (A1), (A2), (A5) and (A6), assume that p > 2γ + 2, set p¯ = p − 2γ , then
sup E
|y(t )|
sup
−τ ≤t≤T
0<≤∗
p¯
≤ C.
(39)
Proof. Fix any ∈ (0, ∗ ]. By the Itô formula and (A2), we can show in the same way as the proof of Lemma 3.2 that
E
sup |y(s )|
0≤s≤t
≤
|ξ (0 )| p¯ + E
p¯
t 0
p¯ |y(s )| p¯ −2 yT (s ) fˆ(y¯ (s ), y¯ (s − τ ))
p¯ − 1 + |gˆ(y¯ (s ), y¯ (s − τ )) + gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) 2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
+E
s
p¯ |y(r )| p¯ −2 yT (r ) gˆ(y¯ (r ), y¯ (r − τ ))
sup 0≤s≤t
0
+ gˆ1 (y¯ (r ), y¯ (r − τ ))gˆ(y¯ (r ), y¯ (r − τ ))w¯ (r )
+ gˆ2 (y¯ (r ), y¯ (r − τ ))gˆ(y¯ (r − τ ), y¯ (r − 2τ ))w¯ (r − τ ) dw(r ) ≤
|ξ (0 )| p¯ + E +E +E
t 0 t 0
t 0
p¯ |y(s )| p¯ −2 y¯ T (s ) fˆ(y¯ (s ), y¯ (s − τ )) + ( p¯ − 1 )|gˆ(y¯ (s ), y¯ (s − τ ))|2 ds
p¯ |y(s )| p¯ −2 (y(s ) − y¯ (s ))T fˆ(y¯ (s ), y¯ (s − τ ))ds
p¯ ( p¯ − 1 )|y(s )| p¯ −2 gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds √ + 4 2E
0
t
p¯ |y(s )|2 p¯ −2 gˆ(y¯ (s ), y¯ (s − τ ))
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
+ gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s ) + gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) Similarly to Lemma 3.2, we have
E sup |y(s )| p¯ 0≤s≤t
≤ C +C + CE
t 0 t
0
2
ds
1/2
.
sup E |y(u )| p¯ ds
0≤u≤s
|y(s )| p¯−2 gˆ(y¯ (s ), y¯ (s − τ ))
+ gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
2
+ gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds. Using Lemma 3.1 and (A6), we get
t
E 0
|y(s )| p¯−2 |gˆ(y¯ (s ), y¯ (s − τ ))|2 ds
t
≤ K¯ E
p¯ |y(s )| p¯ −2 1 + |y¯ (s )|2γ +2 + |y¯ (s − τ )|2γ +2 ds
0
t
t
|y(s )| p¯ ds + K¯ T + K¯ E |y¯ (s )| p ds + K¯ E 0 0 t ≤C E sup |y(u )| p¯ ds + 1 , ≤C E
t 0
t 0
|y¯ (s − τ )| p ds
0≤u≤s
0
E
p¯ |y(s )| p¯ −2 |gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )|2 ds
≤ CE
t 0
|y(s )| p¯ ds + C
t
t 0
|gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )| p¯ ds
t
|y(s )| p¯ ds + C p¯/2 (ψ ())4 p¯ ds 0 0 t p¯ ≤C E sup |y(u )| ds + 1 . ≤ CE
0≤u≤s
0
Similarly, we get
t
E 0
p¯ |y(s )| p¯ −2 |gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
≤C
t
sup |y(u )|
E 0
0≤u≤s
p¯
ds + 1 .
Hence the required assertion (39) follows. The proof is complete.
Due to [20], we give the following Taylor formula. If a function h: R × R → R is twice differentiable, define
∂ h(x, y ) , ∂x ∂ h(x, y ) h2 (x, y ) := , ∂y h1 (x, y ) :=
then we have the following Taylor formula
h(x, y ) − h(xˆ, yˆ ) = h1 (xˆ, yˆ )(x − xˆ) + h2 (xˆ, yˆ )(y − yˆ ) + R(h(ζ1 , ζ2 ) ), where ζ1 = x + λ(x − xˆ), ζ2 = y + λ(y − yˆ ) with λ ∈ (0, 1) and
x − xˆ 1 R(h(xˆ, yˆ )) = (x − xˆ, y − yˆ )J (h(ζ1 , ζ2 )) , 2 y − yˆ
J (h(x, y )) =
h11 (x, y ) h12 (x, y )
h21 (x, y ) h22 (x, y )
273
274
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
for x, y, xˆ, yˆ ∈ R, where
∂ 2 h(x, y ) , ∂ x2 ∂ 2 h(x, y ) h22 (x, y ) := , ∂ y2
∂ 2 h(x, y ) , ∂ x∂ y ∂ 2 h(x, y ) h21 (x, y ) := . ∂ y∂ x
h11 (x, y ) :=
h12 (x, y ) :=
Hence, we have
h(y(t ), y(t − τ )) − h(y¯ (t ), y¯ (t − τ )) = h1 (y¯ (t ), y¯ (t − τ ))(y(t ) − y¯ (t )) + h2 (y¯ (t ), y¯ (t − τ ))(y(t − τ ) − y¯ (t − τ )) + R(h ) t = h1 (y¯ (t ), y¯ (t − τ )) gˆ(y¯ (s ), y¯ (s − τ ))dw(s ) + Rˆ(h(y¯ (t ), y¯ (t − τ ))) + h2 (y¯ (t ), y¯ (t − τ ))
tk
t−τ
t k −τ
gˆ(y¯ (s − τ ), y¯ (s − 2τ ))dw(s ),
where
Rˆ(h(y¯ (t ), y¯ (t − τ ))) = h1 (y¯ (t ), y¯ (t − τ )) +
t
tk t
+
tk
gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s )
t−τ
t k −τ
+
fˆ(y¯ (s ), y¯ (s − τ ))ds
tk
gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s )
+ h2 (y¯ (t ), y¯ (t − τ )) +
t
(40)
t−τ
t k −τ
t−τ
t k −τ
fˆ(y¯ (s ), y¯ (s − τ ))ds
gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )dw(s )
gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )dw(s )
+ R(h(y¯ (t ), y¯ (t − τ ))).
(41)
By Lemma 2.1, Lemma 3.2 and Corollary 3.3, we have the following two lemmas. Lemma 3.9. Let (A1), (A2), (A5) and (A6) hold. Assume that q(γ + 1 ) ∨ q(r + 1 ) < p, then for any i, j = 1, 2, we have
sup E | f (x(t ), x(t − τ ))| ∨ E |g(x(t ), x(t − τ ))| q
q
0≤t≤T
∨ E | fi (x(t ), x(t − τ ) )| ∨ E |gi (x(t ), x(t − τ ))| q
q
q q ∨ E fi j (x(t ), x(t − τ ) ) ∨ E gi j (x(t ), x(t − τ )) ≤ C, q q sup sup E | f (y(t ), y(t − τ ))| ∨ E |g(y(t ), y(t − τ ))| 0<<∗ 0≤t≤T
∨ E | fi (y(t ), y(t − τ ) )| ∨ E |gi (y(t ), y(t − τ ))| q
q
q q ∨ E fi j (y(t ), y(t − τ ) ) ∨ E gi j (y(t ), y(t − τ )) ≤ C and
sup
sup
0<<∗ 0≤t≤T
E | f (y¯ (t ), y¯ (t − τ ))| ∨ E |g(y¯ (t ), y¯ (t − τ ))| q
q
∨ E | fi (y¯ (t ), y¯ (t − τ ) )| ∨ E |gi (y¯ (t ), y¯ (t − τ ))| q
q
q q ∨ E fi j (y¯ (t ), y¯ (t − τ ) ) ∨ E gi j (y¯ (t ), y¯ (t − τ )) ≤ C Lemma 3.10. Let (A1), (A2), (A5) and (A6) hold. Assume that 2q(r + 1 ) < p, then we have
sup
sup E |Rˆ( f (y(t ), y(t − τ )))|q ∨ E |Rˆ(g(y(t ), y(t − τ )))|q
0<<∗ 0
∨ [E |R( f (y(t ), y(t − τ )))|q ] ∨ [E |R(g(y(t ), y(t − τ )))|q ] ≤ C q 1 + (ψ ())2q .
(42)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
and
sup
sup E |Rˆ( f (y¯ (t ), y¯ (t − τ )))|q ∨ E |Rˆ(g(y¯ (t ), y¯ (t − τ )))|q
0<<∗ 0
275
∨ [E |R( f (y¯ (t ), y¯ (t − τ )))|q ] ∨ [E |R(g(y¯ (t ), y¯ (t − τ )))|q ] ≤ C q 1 + (ψ () )2q .
(43)
Proof. By the definition of R( f (y(t ), y(t − τ ))), for any t ∈ [tk , tk+1 ], we have
E |R( f (y(t ), y(t − τ )) )|
q
≤ C E |y(t ) − y¯ (t )|
2q
+ |y(t − τ ) − y¯ (t − τ )|2q
× E (| f11 (y(t ), y(t − τ ))|
2q
1/2
+ 2| f12 (y(t ), y(t − τ ))|
2q
+ | f22 (y(t ), y(t − τ ))| ) 2q
1/2
≤ C q 1 + (ψ () )2q .
(44)
Hence
E |Rˆ( f (y(t ), y(t − τ )))|q
q
q
≤ C q E f1 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ )) + E f2 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ ) )
q
+ E f1 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ ))((w¯ (t ))2 − ) q t + E f1 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) w¯ (s − τ )dw(s ) tk
q + E f2 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ )) (w¯ (t ))2 − q t + E f2 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) w¯ (s − τ )dw(s ) tk q + |R( f (y(t ), y(t − τ ))| .
(45)
Applying elementary inequality and Hölder inequality, we get
q q t 2 E (w¯ (t )) − ∨ w¯ (s − τ )dw(s ) ≤ C q , tk q E f1 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ ))) ≤ C (ψ ())q , q E f2 (y(t ), y(t − τ )) fˆ(y(t ), y(t − τ ))) ≤ C (ψ ())q , q E f1 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ )) ≤ C (ψ ())2q , q E f1 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) ≤ C (ψ ())2q , q E f2 (y(t ), y(t − τ ))gˆ1 (y(t ), y(t − τ ))gˆ(y(t ), y(t − τ )) ≤ C (ψ ())2q
and
q
E f2 (y(t ), y(t − τ ))gˆ2 (y(t ), y(t − τ ))gˆ(y(t − τ ), y(t − 2τ )) ≤ C (ψ ())2q . Consequently, we obtain
q
E Rˆ( f (y(t ), y(t − τ ))) ≤ C q 1 + (ψ () )2q . Similarly, we can show
E |R(g(y(t ), y(t − τ )))|2 ∨ E |Rˆ(g(y(t ), y(t − τ )))|q ≤ C q 1 + ψ () )2q
and (48).
Lemma 3.11. Let (A2) – (A6) hold. Let σ Z and ρ ,Z be the same as before and 2(1 + γ ) ∨ 4(1 + r ) < p. Define
θ := σZ ∧ ρ,Z and e (t ) := x(t ) − y(t ). ¯ ∈ (0, ∗ ] be sufficient small such that ψ () ≥ φ (Z). Then, Let
E
sup
s∈[0,T ∧θ ]
|e (s )|2 ≤ C 2 1 ∨ (ψ ())4 , ∀ T > 0.
(46)
276
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
Proof. Using Itô formula, we can show that for all t ∈ [0, T],
E
sup
s∈[0,t∧θ ]
≤E
|e ( s )|2
T ∧θ
2eT (s ) f (x(s ), x(s − τ )) − fˆ(y¯ (s ), y¯ (s − τ ))
0
+ |g(x(s ), x(s − τ )) − gˆ(y¯ (s ), y¯ (s − τ )) − gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
− gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
+E
r
sup
r∈[0,t∧θ ] 0
2eT (s )[g(x(s ), x(s − τ )) − gˆ(y¯ (s ), y¯ (s − τ ))
− gˆ1 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s ), y¯ (s − τ ))w¯ (s )
− gˆ2 (y¯ (s ), y¯ (s − τ ))gˆ(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )]dw(s ) .
(47)
Since s ∈ [0, t∧θ ], |y¯ (s )| ≤ Z and ψ () ≥ φ (Z), we have |y¯ (s )| ≤ φ −1 (ψ ()) and fˆ(y¯ (s ), y¯ (s − τ )) = f (y¯ (s ), y¯ (s − τ )), gˆ(y¯ (s ), y¯ (s − τ )) = g(y¯ (s ), y¯ (s − τ )) and gˆ(y¯ (s − τ ), y¯ (s − 2τ )) = g(y¯ (s − τ ), y¯ (s − 2τ )). Consequently, we have
E
sup
s∈[0,t∧θ ]
≤E
|e ( s )|
2
t∧θ
2eT (s )[ f (x(s ), x(s − τ )) − f (y¯ (s ), y¯ (s − τ ))]
0
+ |g(x(s ), x(s − τ )) − g(y¯ (s ), y¯ (s − τ )) − g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s )
− g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds
√ +4 2 E
t∧θ
0
2eT (s ) g(x(s ), x(s − τ )) − g(y¯ (s ), y¯ (s − τ ))
− g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s ) − g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )
2
ds
1/2
≤ H1 + H2 + H3 + H4 , where
H1 = 2E
H3 ≤ E
t∧θ
2eT (s )[ f (y(s ), y(s − τ ) − f (y¯ (s ), y¯ (s − τ ))]ds,
0 t∧θ 0
√ H4 = 4 2E
eT (s )[ f (x(s ), x(s − τ )) − f (y(s ), y(s − τ ))] + |g(x(s ), x(s − τ )) − g(y(s ), y(s − τ ))|2 ds,
0
H2 = E
t∧θ
(48)
2
2Rˆ(g(y¯ (s ), y¯ (s − τ )) ds,
t∧θ 0
2eT (s ) g(x(s ), x(s − τ )) − g(y¯ (s ), y¯ (s − τ ))
− g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s ) − g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) By (A3) and (A4), we get
H1 ≤ 2K1 E ≤C
t 0
t∧θ 0
|x(s ) − y(s )|2 + |x(s − τ ) − y(s − τ )|2 ds
E |e (s ∧ θ )| ds. 2
2
ds
1/2
.
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
By (40), we have that
H2 ≤ E
t∧θ 0
2e (s ) f1 (y¯ (s ), y¯ (s − τ ))
+ f2 (y¯ (s ), y¯ (s − τ )) ≤ CE where
t∧θ
0
H5 ≤E
s −τ
s−τ
s
g(y¯ (r ), y¯ (r − τ ))dw(r )
g(y¯ (r ), y¯ (r − τ ))dw(r ) + Rˆ( f (y¯ (s ), y¯ (s − τ ))) ds
eT (s ) f1 (y¯ (s ), y¯ (s − τ ))
0
s
2 |e (s )|2 + Rˆ( f (y¯ (s ), y¯ (s − τ ))) ds + 2H5 ,
+ f2 (y¯ (s ), y¯ (s − τ )) Define
t∧θ
T
f1 (y¯ (s ), y¯ (s − τ ))
+ f2 (y¯ (s ), y¯ (s − τ ))
s−τ
s
s
s −τ
s
s
g(y¯ (r ), y¯ (r − τ ))dw(r )
g(y¯ (r ), y¯ (r − τ ))dw(r ) ds.
g(y¯ (r ), y¯ (r − τ ))dw(r )
s −τ
g(y¯ (r − τ ), y¯ (r − 2τ ))dw(r )
s−τ
:= v(y¯ (s )). Noting that
x (s ) − y (s ) = x ( s ) − y ( s ) + + − −
s s
s
s
[ f (x(r ), x(r − τ )) − f (y¯ (r ), y¯ (r − τ ))]dr
[g(x(r ), x(r − τ )) − g(y¯ (r ), y¯ (r − τ ))]dw(r )
s s s s
g1 (y(r ), y(r − τ ))g(y¯ (r ), y¯ (r − τ ))w(r )dw(r ) g2 (y(r ), y(r − τ ))g(y¯ (r − τ ), y¯ (r − 2τ ))w(r − τ )dw(r ),
we have
H5 ≤J1 + J2 + J3 + J4 , where
J1 = E
t∧θ 0
J2 = E
t∧θ
0
J3 = E
s
[ f (x(r ), x(r − τ )) − f (y¯ (r ), y¯ (r − τ ))]dr
s t∧θ
0
J4 = E
[x(s ) − y(s )]T v(y¯ (s ))ds,
s
s t∧θ
0
s
s
T
v(y¯ (s ))ds,
[g(x(r ), x(r − τ )) − g(y(r ), y(r − τ ))]dw(r )
T
v(y¯ (s ))ds,
[Rˆ(g(y(r ), y(r − τ )))dw(r )
T
v(y¯ (s ))ds.
Computing that
J1 = E
t∧θ −1 t k+1 k=0
tk
[x(tk ) − y(tk )]T f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ds + E = J11 + J12 ,
t∧θ
t∧θ
[x(s ) − y(s )]T v(y¯ (s ))ds
277
278
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
where t∧θ −1
J11 ≤ E
tk+1
tk
k=0
[x(tk ) − y(tk )]T f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ds t∧θ −1
≤E
tk+1
tk
k=0
t∧θ −1
sup
s∈[0,t∧θ ]
k=0
1/2
[x(tk ) − y(tk )]T f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ≤E
2
|x ( s ) − y ( s )|2
tk+1
tk
2
1/2
ds
f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
2 1/2
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s ) ds 1 ≤ E 16
|e ( s )|
sup
s∈[0,t∧θ ]
+ C E
2
t∧θ −1 k=0
tk+1
tk
| f1 (yk , yk−N )g(yk , yk−N )w¯ (s )
+ f2 (yk , yk−N )g(yk−N , yk−2N )w¯ (s )| ds 2
1 ≤ E 16
|e ( s )|
sup
s∈[0,t∧θ ]
2
+ C 2 ,
t∧θ
|x ( s ) − y ( s )| |v(y¯ (s ))|ds t∧θ s∈[0,t∧θ ] 1 ≤ E sup |e (s )|2 + C 2 .
J12 ≤ E
sup
16
s∈[0,t∧θ ]
Similarly, we get
J2 = E
t∧θ
0
s
s
[| f1 (y¯ (r ), y¯ (r − τ ))||x(r ) − y¯ (r )| + |R( f (y¯ (r ), y¯ (r − τ )))|
+ | f2 (y¯ (r ), y¯ (r − τ ))||x(r − τ ) − y¯ (r − τ )|]dr ≤ CE
t∧θ
s
0
s
T
| f1 (y¯ (r ), y¯ (r − τ ))||x(r ) − y(r )|
+ | f2 (y¯ (r ), y¯ (r − τ ))||x(r − τ ) − y(r − τ )| dr + CE
t∧θ 0
s
s
T
v(y¯ (s ))ds
[| f1 (y¯ (r ), y¯ (r − τ ))||y(r ) − y¯ (r )|
+ | f2 (y¯ (r ), y¯ (r − τ ))||y(r − τ ) − y¯ (r − τ )|]dr
T
v(y¯ (s ))ds
|R( f (y¯ (r ), y¯ (r − τ )))|dr T v(y¯ (s ))ds 0 s t∧θ s ≤ CE |x(r ) − y(r )|dr T f1T (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))ds 0 s t∧θ s + CE |x(r ) − y(r )|dr T f2T (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))ds
+E
t∧θ
0
v(y¯ (s ))ds
s
s
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
t∧θ
s
[| f1 (y¯ (r ), y¯ (r − τ ))||y(r ) − y¯ (r )|dr
v(y¯ (s ))ds t∧θ s + CE [| f2 (y¯ (r ), y¯ (r − τ ))||y(r − τ ) − y¯ (r − τ )|]dr T v(y¯ (s ))ds 0 s t∧θ s +E |R( f (y¯ (r ), y¯ (r − τ )))|dr T v(y¯ (s ))ds + CE
0
s
0
T
s
s 2 t∧θ C ds + C E ≤ E | x ( r ) − y ( r ) | dr | f1 (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))|2 ds 0 s 0 2 t∧θ s t∧θ C |x(r ) − y(r )|dr ds + C E + E | f2 (y¯ (s ), y¯ (s − τ ))v(y¯ (s ))|2 ds 0 s 0 2 t∧θ s t∧θ C ds + C E ¯ ¯ ¯ + E [ | f ( ( r ) , ( r − τ )) || y ( r ) − y ( r ) | dr y y |v(y¯ (s ))|2 ds 1 0 s 0 2 t∧θ s C [| f2 (y¯ (r ), y¯ (r − τ ))||y(r − τ ) − y¯ (r − τ )|]dr ds + E 0 s
t∧θ
t∧θ
|v(y¯ (s ))|2 ds 2 t∧θ s t∧θ C |R( f (y¯ (r ), y¯ (r − τ )))|dr ds + C E + E |v(y¯ (s ))|2 ds 0 s 0 t 2 ≤C E sup |e (u )| ds + C 2 (ψ ())4 + C 2 , + C E
0
u∈[0,t∧θ ]
0
Similarly, we get
J3 ≤ C
t
E
sup
u∈[0,t∧θ ]
0
|e (u )|2 ds + C 2 (ψ ())4 + C 2 ,
s 2 t∧θ Rˆ(g(y(r ), y(r − τ )))dw(r ) ds + C E |v(y¯ (s ))|2 ds 0 s 0 ≤ C 2 1 + (ψ ())4 .
J4 ≤
C
E
t∧θ
Using the elementary inequality, we obtain
√ H4 ≤ 4 2E
sup
s∈[0,t∧θ ]
|x ( s ) − y ( s )|2
t∧θ
0
4|g(x(s ), x(s − τ ))
− g(y¯ (s ), y¯ (s − τ )) − g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s )
− g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ )|2 ds ≤
1 E 16
sup
s∈[0,t∧θ ]
|e (s )|2 + CE
t∧θ 0
1/2
g(x(s ), x(s − τ ))
− g(y¯ (s ), y¯ (s − τ )) − g1 (y¯ (s ), y¯ (s − τ ))g(y¯ (s ), y¯ (s − τ ))w¯ (s )
2
− g2 (y¯ (s ), y¯ (s − τ ))g(y¯ (s − τ ), y¯ (s − 2τ ))w¯ (s − τ ) ds 1 ≤ E 16
sup
s∈[0,t∧θ ]
|e ( s )|
2
+ CE
t∧θ 0
t∧θ
Rˆ(g(y¯ (s ), y¯ (s − τ )))2 ds
|g(x(s ), x(s − τ )) − g(y(s ), y(s − τ ))|2 ds t∧θ 1 Rˆ(g(y¯ (s ), y¯ (s − τ )))2 ds ≤ E sup |e (s )|2 + CE + CE
16
0
s∈[0,t∧θ ]
0
279
280
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
+ CE
t∧θ
0
|g1 (y(s ), y(s − τ ))|2 |x(s ) − y(s )|2
+ |g2 (y(s ), y(s − τ ))|2 |x(s − τ ) − y(s − τ )|2 + R(g(y(s ), y(s − τ ))) ds 1 ≤ E 16
|e ( s )|
sup
s∈[0,t∧θ ]
2
2
+C
t
0
4
E sup |e (r ∧ θ )|
2
ds
r∈[0,s]
+ C 1 + (ψ ()) . Substituting these into (48), we get
E
sup
s∈[0,t∧θ ]
|e ( s )|
2
≤C
t
E
sup
r∈[0,s∧θ ]
0
|e (r )| ds + C 2 1 ∨ (ψ ())4 . 2
The Gronwall inequality yields the assertion (46).
Theorem 3.12. Let (A1), (A2), (A4) – (A6) hold and assume that (A3) is satisfied for any p > 2γ + 2. Let p¯ = p − 2γ and 2(γ + 1 ) ∨ 4(r + 1 ) < p and ε ∈ (0, 1/2). Define
φ (u ) = H¯ u1+γ , u ≥ 0 and
ψ () = −ε/2 , ∈ (0, 1], where H¯ is a positive constant. for all sufficient small ∈ (0, ∗ ), then for every such small ,
sup |x(t ) − y(t )|
E
≤ O(2(1−ε ) ),
2
0≤t≤T
Consequently,
lim E
→ 0
(49)
sup |x(t ) − y(t )|
2
0≤t≤T
= 0,
Proof. The notation is same as in the proof of Theorem 3.11. For our proof, assume that γ > 0, p > 2γ + 2, p¯ = p − 2γ , let p¯ > 2 sufficiently large such that
ε 2
>
2 (1 + γ ) . ( p¯ − 2 )
(50)
Applying the Young inequality, we can see that for any ∈ (0, ∗ ), > 0 and Z > ξ , we have
sup |e (t )|2
E
0≤t≤T
sup |e (t )|2 1{θ >T }
=E
0≤t≤T
+
sup |e (t )|2 1{θ >T }
≤E
0≤t≤T
p¯ − 2
p¯ 2/( p¯ −2)
+E +
sup |e (t )|2 1{θ ≤T }
0≤t≤T
2 E p¯
sup |e (t )| p¯
0≤t≤T
P ( θ ≤ T ).
(51)
Using Lemmas 3.7 and 3.8, we have
sup |e (t )|
E
p¯
0≤t≤T
≤C.
(52)
By Lemmas 3.4 and 3.5, we obtain
P (θ ≤ T ) ≤
C . Z p¯
Hence we get
(53)
sup |e (t )|2
E
≤E
0≤t≤T
sup |e (t )|2 1{θ >T }
+
0≤t≤T
2 C ( p¯ − 2 )C + . p¯ p¯ 2/( p¯ −2) Z p¯
(54)
Choosing = 2(1−ε ) and Z = −2(1−ε )/( p¯ −2 ) , we have that
E
sup |e (t )|2
0≤t≤T
≤E
sup |e (t )|2 1{θ >T }
0≤t≤T
+ C 2 ( 1 − ε ) .
(55)
W. Zhang, X. Yin and M.H. Song et al. / Applied Mathematics and Computation 357 (2019) 263–281
281
According to (50), we see that for sufficient small ,
φ −1 (ψ ()) ≥ −2(1−ε )/( p¯−2) = Z. In view of Lemma 3.11, we hence obtain from (55) that
E
sup |e (t )|2
0≤t≤T
≤ C 2 ( 1 − ε ) .
The proof is therefore complete.
4. Conclusion For most highly nonlinear SDDEs, there are no explicit solutions, it is important to develop some approximate numerical methods. Due to the advantages of small amount of calculation and short time of explicit method, our aim in this paper is to present an explicit method named truncated Milstein method of the highly nonlinear SDDEs. We study strong convergence of the truncated Milstein method under the local Lipschitz condition plus Khasminskii-type condition and shows its convergence rate is close to 1 under an additional condition. References [1] D.J. Higham, X. Mao, A.M. Stuart, Strong convergence of euler-type methods for nonlinear stochastic differential equations, SIAM J. Numer. Anal 40 (3) (2002) 1041–1063. [2] D.J. Higham, Exponential mean-square stability of numerical solutions to stochastic differential equations, LMS J. Comput. Math. 6 (6) (2003) 275–294. [3] M. Hutzenthaler, A. Jentzen, P.E. Kloeden, Strong convergence of an explicit numerical method for SDEs with nonglobally lipschitz continuous coefficients, Ann. Appl. Probab. 22 (4) (2012) 1611–1641. [4] P.E. Kloeden, E. Platen, Numerical Solution of Stochastic Differential Equations, Springer, Berlin, 1992. [5] P.E. Kloeden, T. Shardlow, The Milstein scheme for stochastic delay differential equations without using anticipative calculus, J. Sto. Anal. Appl. 30 (2) (2012) 181–202. [6] W. Liu, X. Mao, Almost sure stability of the Euler-Maruyama method with random variable stepsize for stochastic differential equations, Numer. Algorithms (2016) 1–20. [7] Q. Luo, X. Mao, Y. Shen, Generalised theory on asymptotic stability and boundedness of stochastic functional differential equations, Automatica 47 (9) (2011) 2075–2081. [8] X. Mao, Numerical solutions of stochastic functional differential equations, LMS J. Comput. Math 6 (4) (2003) 1821–1841. [9] X. Mao, C. Yuan, Stochastic Differential Equations with Markovian Switching, Imperial College Press, 2006. [10] X. Mao, Stochastic Differential Equations and Applications, Horwood Pub Limited, 1997. [11] X. Mao, Approximate solutions for a class of stochastic evolution equations with variable delays, Numer. Funct. Anal. Optim. 12 (1) (1991) 525–533. [12] X. Mao, M.J. Rassias, Khasminskii-type theorems for stochastic differential delay equations, Stoch. Anal. Appl. 23 (2005) 1045–1069. [13] X. Mao, Numerical solutions of stochastic differential delay equations under the generalized Khasminskii-type conditions, Appl. Math. Comput. 217 (12) (2011) 5512–5524. [14] X. Mao, The truncated Euler-Maruyama method for stochastic differential equations, J. Comput. Appl. Math. (2015) 370–384. [15] X. Mao, Convergence rates of the truncated Euler-Maruyama method for stochastic differential equations, J. Comput. Appl. Math 296 (2016) 362–375. [16] M. Song, L. Hu, X. Mao, L. Zhang, Khasminskii-type theorems for stochastic functional differential equations, Discrete Contin. Dyn. Syst. Ser. B 6 (6) (2013) 1697–1714. [17] Z. Wang, C. Zhang, An analysis of stability of Milstein method for stochastic differential equations with delay, Comput. Math. Appl. 51 (9) (2006) 1445–1452. [18] X. Wang, S. Gan, The tamed Milstein method for commutative stochastic differential equations with non-globally Lipschitz continuous coefficients, J. Differ. Equ. Appl. 19 (3) (2013) 466–490. [19] W. Zhang, M.H. Song, M.Z. Liu, Strong convergence of the partially truncated Euler-Maruyama method for a class of stochastic differential delay equations, J. Comput. Appl. Math. 335 (2018) 114–128. [20] V.A. Zorich, Mathematical Analysis I, second, Springer, 2004.