Numerical approximation of random periodic solutions of stochastic differential equations
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(2) 119. Page 2 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. both periodic and random nature, e.g. daily temperature, energy consumption, airline passenger volumes, CO2 concentration. The concept and its study are relevant to modelling random periodicity in the real world. In the literature, there have been a number of recent works such as [3] on random attractors of the stochastic TJ model in climate dynamics; [2] on stochastic lattice systems; [4] on stochastic resonance; [7] for SDEs with multiplicative linear noise; and [25] on bifurcations of stochastic reaction diffusion equations. All these results are theoretical on the existence of random periodic paths. In general, neither stationary solutions nor random periodic solutions can be constructed explicitly, so numerical approximation is another indispensable tool to study stochastic dynamics, especially to physically relevant problems. It is worth mentioning here that this is a numerical approximation of an infinite time horizon problem. There are numerous works on numerical analysis of SDEs on a finite horizon, and a number of excellent monographs [14,17]. However, there are only a few works on infinite horizon problems. A numerical analysis of approximation to the stationary solutions and invariant measures of SDEs through discretising the pull-back was given in [16,22–24,26]. Numerical approximations to stable zero solutions of SDEs were given in [10,14]. In this paper, we study stochastic differential equations, which possess random periodic solutions, and approximate them by Euler–Maruyama and Milstein schemes. As far as we know, this is the first paper addressing analysis of numerical approximations of random periodic solutions. Consider the following m-dimensional SDE dXtt0 = [AXtt0 + f (t, Xtt0 )]dt + g(t, Xtt0 )dWt. (1.3). = ξ, where f : R × R → R , g : R × R → R , A is a symmetric and negative-definite with d m × m matrix, Wt is a two-sided Wiener process in R on a probability space (Ω, F, P). The filtration t = s≤t Fst , the random variable is defined as follows: Fst = σ{Wu − Wv : s ≤ v ≤ u ≤ t}, F t = F−∞ ξ is F t0 -measurable. We assume that the functions f and g are τ -periodic in time. By the variation of constant formula, the solution of (1.3) is given Xtt00. Xtt0 (ξ). m. m. A(t−t0 ). At. =e. m. t ξ+e. −As. e. m×d. t f (s, Xst0 )ds. t0. At. +e. e−As g(s, Xst0 )dWs .. (1.4). t0. Denote the standard P -preserving ergodic Wiener shift by θ : R × Ω → Ω, θt (ω)(s) := W (t + s) − W (t), t, s ∈ R. The solution X of the non-autonomous SDE does not satisfy the cocycle property, but u(t, t0 ) : Ω × Rm → Rm given by u(t, t0 )ξ = Xtt0 (ξ) satisfies the semi-flow property (1.1) and periodicity (1.2). Denote by Xr−kτ (ξ, ω) the solution starting from time −kτ . We will show that when k → ∞, the pull-back Xr−kτ (ξ) has a limit Xr∗ in L2 (Ω) and Xr∗ is the random periodic solution of SDE (1.3). It satisfies the infinite horizon stochastic integral equation (IHSIE) r r ∗ A(r−s) ∗ Xr = e f (s, Xs )ds + eA(r−s) g(s, Xs∗ )dWs . −∞. −∞. We separate the linear term AX from the nonlinear term in (1.3) to enable us to represent the random periodic solution by IHSIE [5,7]. This is helpful to formulate the scheme for SPDEs for which random periodic solutions were considered in [6]. Numerical analysis for random periodic solutions was not considered in previous work. The infinite horizon stochastic integral equation (IHSIE) method can deal with anticipated cases [5–7]. But it is still not clear how to numerically approximate two-sided IHSIE and anticipating random periodic solutions. The pull-back method used in this paper is a popular way to study random attractors. Here we use this to deal with stable adapted random periodic solutions of dissipative systems for the first time. The pull-back method has some advantages. First, stability can be obtained immediately. Secondly, it can deal with some dissipative equations that cannot be dealt with by the IHSIE; especially, the current IHSIE.
(3) ZAMP. Numerical approximation of random periodic solutions. Page 3 of 32. 119. technique requires equations to have multiplicative linear noise or additive noise and f being bounded. Thirdly in this paper, we study numerical approximations of random periodic solutions of dissipative SDEs and with the pull-back idea, a random periodic solution of the discretised system can be obtained as well. We will first study the Euler–Maruyama numerical scheme in infinite horizon and obtain an approxi√ ∗ . We will prove that the latter converges to the exact r.p.s. in L2 (Ω) at the rate of Δt mating r.p.s. X r √ when the time mesh Δt tends to zero. This √ result will be numerically verified. Despite its lower order of the approximation only at the rate of Δt, the advantage of this scheme is its simplicity and it is relatively easy to implement in actual computations. It works well for the SDE we consider in this paper. We also consider more advanced numerical schemes, e.g. √ Milstein scheme [13,14,24], for high-order convergence. We improve the rate of approximation from Δt in Euler scheme to Δt. We will also do some numerical simulations to sample paths of the r.p.s. (Fig. 1). However, simulation of one pathwise trajectory is not a reliable way to tell whether or not it is random periodic though it looks very much like to be. Here we provide two reliable methods for this from numerical simulations. One method is to simulate {Xt∗ (ω), t ∈ R} and {Xt∗ (θ−τ ω), t ∈ R} for the same ω. These two trajectories should be repeating each other, but with a shift of one period of time. See Fig. 1 as an example. The other way is to simulate {Xt∗ (θ−t ω), t ∈ R}, which is periodic if and only if Xt∗ (ω) is random periodic. As an example, see Fig. 2. These two approaches would apply to any other stochastic differential equations should they have a random periodic solution. It was known from the recent work [8] that the law of the random periodic solution is the periodic measure of the corresponding Markov semi-group. Thus, we will consider the convergence of transition probabilities generated by (1.3) and its numerical scheme along the integral multiples of period to the periodic measure and discretised periodic measure, respectively, and error estimate of the two periodic measures in the weak topology.. 2. Assumptions and preliminary results p 1/p. First we fix some notation. Let p ≥ 1 and denote the Lp -norm of a random variable ξ by ξ p = (E |ξ| ) d1 d2 2 12 and the Frobenius norm of any d1 × d2 matrix B by |B| = ( i=1 j=1 Bij ) .. ,. 2.1. Conditions for the SDE We assume the following conditions. Condition (A). The eigenvalues of the symmetric matrix A, {λj , j = 1, 2, . . . , m}, satisfy 0 > λ1 ≥ λ2 ≥ . . . ≥ λm . Condition (1). Assume there exists a constant τ > 0 such that for any t ∈ R, x ∈ Rm , f (t + τ, x) = β2 f (t, x), g(t + τ, x) = g(t, x) and there exist constants C0 , β1 , β2 > 0 with β1 + 22 < |λ1 | such that for any s, t ∈ R and x, y ∈ Rm , |f (s, x) − f (t, y)| ≤ C0 |s − t|. 1/2. + β1 |x − y| ,. 1/2. + β2 |x − y| .. |g(s, x) − g(t, y)| ≤ C0 |s − t|. Condition (2). There exists a constant K ∗ > 0 such that ξ 2 ≤ K ∗ . From Condition (1) it follows that for any x ∈ Rm , the linear growth condition also holds: |f (t, x)| ≤ β1 |x| + C1 , |g(t, x)| ≤ β2 |x| + C2 , where the constants C1 , C2 > 0 are constants. It is easy to see that β2 there exists a constant α such that β1 + 22 < α < |λ1 |. In the following, we always assume that α satisfies this condition in all the following proofs. Set ρ := |λm |..
(4) 119. Page 4 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. For the SDE case, the quantity ρ is certainly finite, and for simplicity, we choose numerical schemes to treat the linear part explicitly, which simplify the proof of the pull-back convergence to the random periodic solutions for the discretised systems. However, in a case of SPDEs, this technical assumption is no longer true, but can be removed by employing exponential Euler–Maruyama method and Milstein scheme [1,12]. This will be studied in future work. 2.2. Existence and uniqueness of random periodic solution We first consider the boundedness of the solution in L2 (Ω). Lemma 2.1. Assume Conditions (A), (1) and (2). Then there exists a constant C > 0 such that for any 2 k ∈ N, r ≥ −kτ , we have E Xr−kτ ≤ C. 2 Proof. First, using Itˆ o’s formula to e2αr Xr−kτ , we have r r T 2 2αr −kτ 2 −2αkτ 2αs −kτ 2 Xr Xs =e |ξ| + 2α e ds + 2 e2αs Xs−kτ AXs−kτ ds e r +2 −kτ r. +2. −kτ. −kτ. T e2αs Xs−kτ f (s, Xs−kτ )ds +. r. 2 e2αs g(s, Xs−kτ ) ds. −kτ. T e2αs Xs−kτ g(s, Xs−kτ )dWs .. (2.1). −kτ. Firstly, note the sum of the second and third terms of the right-hand side is non-positive as the matrix αI + A is non-positive definite. Take the expectation of both sides of (2.1), apply the above inequality and use linear growth conditions to obtain r 2 2 2αr −kτ 2 −2αkτ 2 ≤e ξ 2 + (2β1 + β2 ) e2αs E Xs−kτ ds e E Xr r +2(C1 + β2 C2 ). −kτ. e2αs E Xs−kτ ds + (2α)−1 C22 e2αr − e−2αkτ .. −kτ. Also, there exits ε > 0, such that β1 +. β22 2. (1 + ε) < α < |λ1 | . By Young’s inequality. (C1 + β2 C2 )2 2 2(C1 + β2 C2 ) Xs−kτ ≤ + ε(2β1 + β22 ) Xs−kτ . 2 ε(2β1 + β2 ) Then we have 2αr. e. 2 E Xr−kτ ≤K1 + K2 e2αr + K3. r. 2 e2αs Xs−kτ 2 ds,. −kτ. where.
(5). C22 (C1 + β2 C2 )2 + − K1 = e e−2αkτ , 2α 2αε(2β1 + β22 ) (C1 + β2 C2 )2 C2 K2 = 2 + , K3 = (2β1 + β22 )(1 + ε) < 2α. 2α 2αε(2β1 + β22 ) −2αkτ. 2 ξ 2.
(6) ZAMP. Numerical approximation of random periodic solutions. Page 5 of 32. 119. Now applying Gronwall’s inequality, we have 2αr. e. 2 E Xr−kτ ≤ K1 + K2 e2αr +. r. . 2αs. K1 + K2 e. . r. K3 es. K3 dr. ds. −kτ. ≤ (K1 e2αkτ + K2 )e2αr +. K2 K3 2αr e . 2α − K3. 2. Here we notice that K1 e2αkτ + K2 = ξ 2 . Therefore, by Condition (2) 2 2αK2 2αK2 2 E Xr−kτ ≤ ξ 2 + ≤ K∗ + , 2α − K3 2α − K3 −kτ. −kτ In the next lemma, we will also obtain a bound on the norm Xt1 − Xt2 2 for any fixed time t1 , t2 . This will be essential for us to estimate the error of the numerical approximation in Sect. 4. Lemma 2.2. Assume Conditions (A), (1) and (2). Then there exist constants C3 > 0, C4 > 0, such that ≤ for any positive k ∈ N and any t1 , t2 ≥ 0, t1 ≥ t2 , the solution of (1.3) satisfies Xt−kτ − Xt−kτ 1 2 2 √ C3 (t1 − t2 ) + C4 t1 − t2 . Proof. From (1.4), we see that −kτ. Xt ≤ e2Akτ ξ eAt1 − eAt2 − Xt−kτ 2 1 2 2. t1 t2 At. −As −kτ At2 −As −kτ 1. + e e f (s, Xs )ds − e e f (s, Xs )ds. −kτ −kτ 2. t1 t2 At. −As −kτ At2 −As −kτ 1. e + e g(s, X )dW − e e g(s, X )dW s s . s s. −kτ. −kτ. (2.2). 2. We evaluate each term on the right-hand side of (2.2). First, we consider the first term. By Lemma 1 in o’s [26], eAt1 − eAt2 ≤ |A| (t1 − t2 ). Now we estimate the third term with the Minkowski inequality, Itˆ isometry and the linear growth property:. t1 t2. At −As −kτ At2 −As −kτ e 1 e g(s, Xs )dWs − e e g(s, Xs )dWs . −kτ −kτ 2 t t. 2 1. At1 At2 −As −kτ −A(s−t1 ) −kτ. e e ≤ − e g(s, X )dW + e g(s, X )dW s s s s. t2 −kτ 2 2 t2 2 2 ≤ |(eAt1 − eAt2 ) e−As | E β2 Xs−kτ + C2 ds −kτ. t1 2 2 + e−A(s−t1 ) E β2 Xs−kτ + C2 ds t2. t2 . 2 2 |(eAt1 − eAt2 ) e−As | 2β22 E Xs−kτ + 2C22 ds ≤ −kτ.
(7) 119. Page 6 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. t1 . 2 2 + e−A(s−t1 ) 2β22 E Xs−kτ + 2C22 ds t2. t1 t2 2 2 ≤ K4 |(eAt1 − eAt2 ) e−As | ds + K4 e−A(s−t1 ) ds. −kτ. t2. . √ T r(−A) (t1 − t2 ) + K4 t1 − t2 . 2 2 Here we take some constant K4 because E Xs−kτ is bounded above according to Lemma 2.1. Lastly, we consider the second term of (2.2) with Minkowski inequality. t1 t2. At −As −kτ At2 −As −kτ e 1 e f (s, Xs )ds − e e f (s, Xs )ds. −kτ −kτ 2 t. t. 2. 1. At1 At2 −As −kτ −A(s−t1 ) −kτ. ≤ (e − e )e f (s, X )ds + e f (s, X )ds s s. ≤ K4. −kτ. t2 ≤. 2. At (e 1. ≤ −kτ. 2. 2. −kτ. t2. t2. t1 . At2 −As −kτ − e )e f (s, Xs ) 2 ds + e−A(s−t1 ) f (s, Xs−kτ ) ds t2. At (e 1 − eAt2 )e−As f (s, Xs−kτ ) ds + 2 ⎛. ≤ K5 ⎝. t1 . −A(s−t1 ) e f (s, Xs−kτ ) 2 ds t2. t2. At e 1. ⎞ t1 − eAt2 e−As ds + e−A(s−t1 ) ds⎠. −kτ. t2. ≤ 2K5 (t1 − t2 ), for a constant K5 > 0. Combining the above estimates we obtain the lemma with the constants C3 , C4 being independent of k and t1 , t2 . Now we continue to consider the difference of the solutions under various initial values. For simplicity, we here study two different initial values ξ and η. Lemma 2.3. Denote by Xr−kτ and Yr−kτ two solutions of (1.3) with different initial values ξ and η, respec tively. Assume Conditions (A), (1) and Condition (2) for both initial values. Then Xr−kτ − Yr−kτ 2 ≤
(8). e. β1 +. 2 β2 2. −α (r+kτ ). ξ − η 2 .. Proof. According to (1.4) we have Xr−kτ. −. Yr−kτ. A(r+kτ ). =e. r (ξ − η) + e. r + eAr −kτ. Ar. e−As f (s, Xs−kτ ) − f (s, Ys−kτ ) ds. −kτ. e−As g(s, Xs−kτ ) − g(s, Ys−kτ ) dWs ..
(9) ZAMP. Numerical approximation of random periodic solutions. Page 7 of 32. 119. For simplicity, denote ζr−kτ = Xr−kτ − Yr−kτ . Then according to the method used in Lemma 2.1, and the global Lipschitz condition, we have 2αr. e. −kτ 2 ζr ≤ e−2αkτ ξ − η 2 + 2 2 2. r. −kτ r. − f (s, Ys−kτ )) ds +. e2αs E (ζs−kτ )T (f (s, Xs−kτ ) 2 e2αs E g(s, Xs−kτ ) − g(s, Ys−kτ ) ds.. −kτ −2αkτ. ≤e. ξ −. 2 η 2. . + 2β1 +. β22. . r. 2 e2αs ζs−kτ 2 ds.. −kτ. . Then the result follows from the Gronwall inequality. Now we can prove the following theorem.. Theorem 2.4. Assume Conditions (A), (1). Then there exists a unique random periodic solution X ∗ (r, ·) ∈ L2 (Ω), r ≥ 0 such that for any initial value ξ satisfying Condition (2), the solution of (1.3) satisfies limk→∞ Xr−kτ (ξ) − X ∗ (r) 2 = 0. Proof. Condition (2) implies that the initial value ξ belongs to L2 (Ω). According to Lemma 2.1, Xr−kτ (·) maps L2 (Ω) to itself. Now we use the semi-flow property to get that for any r, k, p ≥ 0, Xr−kτ −pτ (ξ) = −(k+p)τ ∗ Xr−kτ (ω) ◦ X−kτ (ω, ξ). Thus, we can apply Lemma 2.3 to have for any ε > 0 there exists k > 0. −(k+p)τ such that for any k ≥ k ∗ , Xr−kτ (ξ) − Xr (ξ) < ε. This means that there exists N > 0 such 2 −lτ that for any l, m ≥ N , we have Xr (ξ) − Xr−mτ (ξ) 2 < ε, i.e. {Xr−kτ (ξ)}k∈N is a Cauchy sequence, so converges to some X ∗ (r, ω) in L2 (Ω), when k → ∞. Set u(t, r)(ξ) = Xtr (ξ), then u(t, r) : Ω × Rm → Rm defines a semi-flow of homeomorphism (Kunita k→∞ [15]). By the continuity of Xtr (ω) : L2 (Ω, Rm ) → L2 (Ω, Rm ), t ≥ r, then u(t, r, ω) Xr−kτ (ξ, ω) −−2−−→ L (Ω). u(t, r, ω) ◦ (X ∗ (r, ω)) . But k→∞ u(t, r, ω) Xr−kτ (ξ, ω) = Xt−kτ (ξ, ω) −−2−−→ X ∗ (t, ω). L (Ω). So u(t, r, ω) (X ∗ (r, ω)) = X ∗ (t, ω), P − a.s. Taking some other initial value η satisfying Condition (2), we have. ∗ Xr − Xr−kτ (η) ≤ Xr∗ − Xr−kτ (ξ) + Xr−kτ (ξ) − Xr−kτ (η) . 2 2 2 Applying Lemma 2.3 again, we can make the right-hand side small enough when k → ∞. Therefore, the convergence is independent of the initial value. Now we need to prove the random periodicity of the X ∗ (r, ω). Note by the continuity of f and g, −(k−1)τ (ξ) Xr+τ. A(r+kτ ). =e. r ξ+. −(k−1)τ −(k−1)τ s , eA(r−s) f (s, Xs+τ (ξ))ds + g(s, Xs+τ (ξ))dW. −kτ. s := (θτ ω)(s) = Ws+τ − Wτ . On the other hand, where W r (ξ) θτ X−kτ. A(r+kτ ). =e. r θτ ξ + −kτ. s . eA(r−s) f (s, θτ Xs−kτ )ds + g(s, θτ Xs−kτ )dW.
(10) 119. Page 8 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. By pathwise uniqueness of the solution of (1.3), we have −(k−1)τ. Xr−kτ (θτ ω, ξ(θτ ω)) = θτ Xr−kτ (ξ) = Xr+τ. (ω, ξ(ω)).. (2.3). From the proof of convergence we have −(k−1)τ. Xr+τ. (ω, ξ) −−2−−→ X ∗ (r + τ, ω), Xr−kτ (θτ ω, ξ(θτ ω)) −−2−−→ X ∗ (r, θτ ω). k→∞. k→∞. L (Ω). L (Ω). Therefore, X ∗ (r + τ, ω) = X ∗ (r, θτ ω), P − a.s.. . 3. Numerical approximation for random periodic solution 3.1. Euler–Maruyama scheme In this section, we will introduce the basic Euler–Maruyama method to approximate the solution on infinite horizon. Take Δt = τ /n, which will be taken to be sufficiently small such that Δt ≤ ρ1 , for some n ∈ N, in the remaining part of the paper. Let N = kn. The time domain from time −kτ to time 0 is divided into N intervals of length Δt such that N Δt = kτ . The scheme starts from an F −kτ -measurable −kτ random variable ξ at a time −kτ . At each of the points iΔt we set the value X −kτ +iΔt with the iteration formula. −kτ −kτ −kτ −kτ X −kτ +(i+1)Δt = X−kτ +iΔt + AX−kτ +iΔt Δt + f iΔt, X−kτ +iΔt Δt. −kτ +g iΔt, X W−kτ +(i+1)Δt − W−kτ +iΔt , (3.1) −kτ +iΔt −kτ where i = 0, 1, 2, . . . , and X −kτ +0Δt = ξ. It is easy to see that for any M ≥ 0, M −kτ X −kτ +M Δt = (I + AΔt) ξ + Δt. M −1 . −kτ (I + AΔt)M −i−1 f iΔt, X −kτ +iΔt. i=0. +. M −1 . . −kτ W−kτ +(i+1)Δt − W−kτ +iΔt . (I + AΔt)M −i−1 g iΔt, X −kτ +iΔt. (3.2). i=0. ˆ jΔt (ξ), i ≥ j, i, j ∈ {−kn, −kn + Moreover, we can set up a discrete semi-flow given by u ˆi,j (ξ) = X iΔt ˆ Then it is easy to see that u satisfies the semi-flow property u ˆi,j (ω) ◦ 1, · · · }, θˆ = θΔt , θˆn = θˆθˆ · · · θ. u ˆj,l (ω) = u ˆi,l (ω), for i ≥ j ≥ l, and the periodic property u ˆi+n,j+n (ω) = u ˆi,j (θˆn ω) for i ≥ j. In order to prove the convergence of the discretised semi-flow to a random periodic solution, we first derive some similar estimates as in Lemmas 2.1 and 2.3. Then a discrete analogue of Theorem 2.4 will give us the result. > 0 such that for any Lemma 3.1. Assume Conditions (A), (1) and (2). Then there exists a constant C −kτ natural numbers k ≥ 0, M ≥ 0, and sufficiently small Δt, the numerical solution X −kτ +M Δt defined by 2 −kτ (3.2) satisfies E X −kτ +M Δt ≤ C. 2. Proof. We still choose α such that β1 + β2 < α < |λ1 | . Then for any M ≥ 0, 2 −2M −kτ (1 − αΔt) X−kτ +M Δt ⎛ ⎞ 2 −kτ M −1 X 2 −kτ +(i+1)Δt −kτ ⎟ 2 −2i ⎜ = |ξ| + (1 − αΔt) − X ⎝ −kτ +iΔt ⎠ . 2 (1 − αΔt) i=0. (3.3).
(11) ZAMP. Numerical approximation of random periodic solutions. Page 9 of 32. 119. This is not hard to verify by expanding the sum and noting cancellations. Notice that 2 −kτ X−kτ +(i+1)Δt 2. (1 − αΔt)
(12) −kτ X =. 2 −kτ − X −kτ +iΔt .
(13) I + AΔt. Δt T −kτ f (iΔt, X −kτ +iΔt ) 1 − αΔt T −kτ )T W−kτ +(i+1)Δt − W−kτ +iΔt g(iΔt, X −kτ +iΔt. +. 1 − αΔt. −I. +. −kτ +iΔt. 1 − αΔt. I + AΔt Δt −kτ −kτ +I X f (iΔt, X × −kτ +iΔt + −kτ +iΔt ) 1 − αΔt 1 − αΔt −kτ g(iΔt, X −kτ +iΔt ) W−kτ +(i+1)Δt − W−kτ +iΔt . + 1 − αΔt
(14)
(15). (3.4). is non-positive definite, where Δt satisfies 0 < Δt ≤ ρ1 as −kτ −kτ defined before, and for each i, f (iΔt, X −kτ +iΔt ) and g(iΔt, X−kτ +iΔt ) are both independent of W−kτ +(i+1)Δt − W−kτ +iΔt . Take expectation on both sides of (3.3), consider (3.4), apply the linear growth property and Young’s inequality to have Note. I+AΔt 1−αΔt. −I. I+AΔt 1−αΔt. +I. 2 −kτ E X −kτ +M Δt
(16) M −1 2 −2i (1 − αΔt) ≤ ξ 2 +. (1 − αΔt). −2M. i=0. +. M −1 . (1 − αΔt). −2i. (1 − αΔt). −2i. +. Δt. 2Δt (1 − αΔt). i=0. 1 + (1 − αΔt) ≤K. Δt 1 − αΔt. (1 − αΔt). i=0 M −1 . (3.5). −2M. 2 + K 3 K. 2. 2 −kτ E f (iΔt, X ) −kτ +iΔt . 2 −kτ g(iΔt, X E ) −kτ +iΔt 2 2E. M −1 . −kτ X −kτ +iΔt. (1 − αΔt). T. −2i. ! −kτ (I + AΔt) f (iΔt, X−kτ +iΔt ). 2 −kτ E X −kτ +iΔt ,. i=0. where 1 = ξ 2 , K 3 = K 2 2 = K. C12. Δt (1 − αΔt). 2. (Δt) + C22 Δt. 2αΔt − α2 (Δt). 2. +. 2. (1 + ε) 2β1 + β22 + Δt β12 + 2β1 |A| , 2. (C1 + β2 C2 + ΔtC1 (β1 + |A|)) . 2 ε 2 2 2 2αΔt − α (Δt) (2β1 + β2 + Δt (β1 + 2β1 |A|)) Δt. Here Δt and ε need to be chosen small enough such that (1 + ε) 2β1 + β22 + Δt β12 + 2β1 |A| + α2 Δt < 2α..
(17) 119. Page 10 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. 2 3 < 1. By the discrete Gronwall inequality, This guarantees that (1 − αΔt) 1 + K 2 −kτ −2M E X (1 − αΔt) −kτ +M Δt. 1 + K 2 (1 − αΔt) ≤K. −2M. +. M −1 . M −i−1 1 + K 3 1 + K 3 2 (1 − αΔt)−2i K K .. i=0. It turns out that 2 . M −kτ 3 (1 − αΔt)2 2 + K 1 1 + K E X ≤ K −kτ +M Δt
(18). M 3 (1 − αΔt)2 1 − 1 + K 2K 3 (1 − αΔt)2 K . + ≤ C. 3 (1 − αΔt)2 1− 1+K is independent of k and the lemma holds for sufficiently small time Note the choice of the constant C step Δt and constant ε. The following lemma is a discrete analogue of Lemma 2.3. −kτ −kτ Lemma 3.2. Denote by X −kτ +M Δt and Y−kτ +M Δt solutions of the Euler scheme with initial values ξ and η, respectively. Assume Conditions (A), (1) and Condition (2) for both initial values. Let Δt = τ /n, n ∈ Z+ , be sufficiently small such that 0 < Δt ≤ ρ1 . Then for any ε > 0, there exists an integer M ∗ > 0. −kτ −kτ such that for any M ≥ M ∗ , we have X −kτ +M Δt − Y−kτ +M Δt < ε. 2. Proof. According to scheme (3.2) we have M −kτ −kτ X (ξ − η) + Δt −kτ +M Δt − Y−kτ +M Δt = (I + AΔt). M −1 . (I + AΔt). M −i−1. Fi. i=0. +. M −1 . (I + AΔt). M −i−1. i W−kτ +(i+1)Δt − W−kτ +iΔt . G. i=0. −kτ −kτ −kτ Here Fi = f (iΔt, X −kτ +iΔt ) − f (iΔt, Y−kτ +iΔt ), Gi = g(iΔt, X−kτ +iΔt ) − g(iΔt, −kτ i = X −kτ −kτ ≤ β ζ and F Y−kτ ). Denote ζ − Y . Then by Condition (1), we have Gi ≤ i 1 i +iΔt −kτ +iΔt −kτ +iΔt β2 ζi . According to the method used in Lemma 3.1, we get the following result similar to inequality (3.5) (1 − αΔt). −2M.
(19) M −1 2 2 −2i E ζM ≤ ξ − η 2 + (1 − αΔt) i=0. +. M −1 . (1 − αΔt). −2i. (1 − αΔt). −2i. (1 − αΔt). i=0. +. M −1 i=0. 4 ≤ ξ − η 2 + K 2. Δt. M −1 i=0. 2Δt (1 − αΔt). (1 − αΔt). Δt 1 − αΔt. 2. 2 E Fi . 2 E 2 Gi 2E. −2i. ! T ζi (I + AΔt) Fi. 2 E ζi ,.
(20) ZAMP. Numerical approximation of random periodic solutions. Page 11 of 32. 119. Δt 4 = 2β1 + β22 + Δt β12 + 2β1 |A| . We choose Δt small enough such that 2β1 + β22 + where K (1−αΔt)2. 2 4 < 1. Again the discrete Gronwall Δt β12 + 2β1 |A| + α2 Δt < 2α. Then, we have (1 − αΔt) 1 + K inequality implies M −1 2. M " −2M 2 4 = ξ − η 2 1 + K 4 1+K E ζM ≤ ξ − η 2 . (1 − αΔt) 2 i=0. 2. M 2 2 4 < ε with sufficiently large M . Finally, E ζM ≤ ξ − η 2 (1 − αΔt) 1 + K. . In the numerical scheme we consider the process as two parts, [−kτ, 0) and [0, r]. Define −kτ := X(r, 0, ω) ◦ X −kτ , X 0. r. (3.6). 0, ω), r ≥ 0, is finite time Euler approximation of the solution of stochastic differential where X(r, equation with time step size Δt, till N Δt ≤ r, where N is the unique number such that N Δt ≤ r and (N + 1)Δt > r. If N Δt < r, define 0, ω) = X(N Δt, 0, ω) + f (N Δt, X(N Δt, 0, ω))(r − N Δt) X(r, Δt, 0, ω))(Wr − WN Δt ). +g(N Δt, X(N. (3.7). #r0 and Y#r0 Lemma 3.3. (Continuity of the discrete semi-flow with respect to the initial value) Denote by X # the solutions of the finite time Euler scheme with the initial values ξ and η# at time 0. Assume Conditions (A), (1) and Condition (2) for both initial Let Δt be sufficiently small, p ≥ 1. Then for any ε > 0,. values.. # there exists a δ > 0 such that for any ξ − η# < δ, we have p. #0 # − Y# 0 (ω, η#) (3.8) < ε. Xr (ω, ξ) r p. # 0 and Y# 0 satisfy analogues of (3.2), with initial value ξ# and η# at time 0 instead Proof. Note that X N Δt N Δt of −kτ . Apply the Euler scheme on the finite time r = N Δt to obtain p #0 # − Y# 0 (ω, η#) Xr (ω, ξ) r p N −1 p p−1 pN # p−1 p pN −i−1 # Fi ≤3 (I + AΔt) (I + AΔt) ξ − η# + 3 (Δt) (I + AΔt) i=0 p −1 N p−1 pN −i−1 # Gi W(i+1)Δt − WiΔt , +3 (I + AΔt) (3.9) (I + AΔt) i=0 # 0 ) − f (iΔt, Y# 0 ), G # i := g(iΔt, X # 0 ) − g(iΔt, Y# 0 ). Denote ζ#i := X # 0 − Y# 0 . where F#i := f (iΔt, X iΔt iΔt iΔt iΔt iΔt iΔt p−1 . Taking expectation on both sides of (3.9), For convenience, we denote Cp = 3p−1 , Cp,N = 3p−1 N and noting the Lipschitz condition of function f and g, we have −1 N p p. p . # −pN # p (1 − αΔt)−(i+1)p β1p ζ#i (1 − αΔt) ζN ≤ Cp ξ − η# + Cp,N (Δt) p. p. + Cp,N (Δt)p/2. p. i=0 . −1 N . p (1 − αΔt)−(i+1)p β2p ζ#i p. i=0 N −1 p. p . # = Cp ξ# − η# + K (1 − αΔt)−ip ζ#i , . p. i=0. p.
(21) 119. Page 12 of 32. # = where K. C. Feng, Y. Liu and H. Zhao. ZAMP. Cp,N ((Δt)p β1p +(Δt)p/2 β2p ) , (1−αΔt)p. p. which is bounded for any 1 ≤ p < +∞. Then by the Gronwall p . N . # # . Note (1 + K)(1 − αΔt)p ≤ (1 − inequality, we have ζ#N ≤ Cp ξ# − η# (1 + K)(1 − αΔt)p p p αΔt)p + Cp,N (Δt)p β1p + (Δt)p/2 β2p ≤ 1 + Cp,N . Result (3.8) at r = N Δt follows by taking δ = ε Cp. (1 + Cp,N ). −N . . Finally, (3.8) at time r follows from (3.7) and the estimate at r = N Δt.. . Theorem 3.4. Assume that Condition (1) and Δt is fixed and small enough. The time domain is divided r∗ ∈ L2 (Ω) such that for any initial values ξ satisfying Condition (2), as τ = nΔt. Then there exists X the solution of the Euler–Maruyama scheme satisfies. −kτ ∗ lim X (ξ) − X (3.10) r r = 0, 2. k→∞. r∗ satisfies the random periodicity property. and X −kτ can be made similarly as Proof. Firstly, we note that the proof of the convergence of the process X 0 2 −kτ that of Theorem 2.4. According to Lemma 3.1 we know that for any M , we have X −kτ +M Δt ∈ L (Ω). We use a similar construction of a Cauchy sequence as in Theorem 2.4. As we assume that τ = nΔt and kτ = knΔt =: N Δt, we have the following result by using semi-flow property, for any m ≥ 1, −(N +mn)Δt = X −N Δt ◦ X −(N +mn)Δt . −(k+m)τ = X X 0 0 0 −N Δt −N Δt with a different initial value. By Lemma 3.2 we have that for any ε > 0 It is a same process as X 0 ∗ there exists N such that for any N ≥ N ∗ , Δt > 0, we have. −kτ −N Δt −(k+m)τ −(N +mn)Δt −X X0 − X = X < ε. 0 0 0 2. 2. i = X −iτ , which converges to some X ∗ in L2 (Ω). We now use Then we construct the Cauchy sequence X 0 the same method to prove the convergence is independent of the initial point. Note for fixed Δt,. N →∞ ∗ ∗ −kτ −kτ (η) −kτ (ξ) −kτ (η) X X ≤ − X + (ξ) − X X − X. −−−−→ 0, 0 0 0 0 2. 2. 2. where N → ∞ is equivalent to k → ∞. 0, ω) ◦ X ∗ , r ≥ 0. According to Lemma 3.3, we have ∗ (r, ω) := X(r, Define X 0, ω) ◦ X ∗ (ω) = X −kτ (ω) = X(r, 0, ω) ◦ X −kτ (ω) −k→∞ ∗ (r, ω), X −2−−→ X(r, r 0 L (Ω). so (3.10) holds. On the other hand, similar to the proof of (2.3), we obtain τ (ω, ξ(ω)) = X 0 (ω, ξ(ω)). 0 (θτ ω, ξ(θτ ω)) = θτ X X r+τ r r Therefore, 0, θτ ω) ◦ X −kτ (θτ ω) = X(r, 0, θτ ω) ◦ X −kτ (θτ ω) −k→∞ ∗ (θτ ω) = X ∗ (r, θτ ω). X −2−−→ X(r, r 0 L (Ω). But, −kτ +τ (ω) −k→∞ ∗ (r + τ, ω), and X −kτ +τ (ω) = X r−kτ (θτ ω), P − a.s; X −2−−→ X r+τ r+τ L (Ω). ∗ (r + τ, ω) = X ∗ (r, θτ ω), P − a.s. thus, we have X. . Example 3.5. Consider a specific SDE dXtt0 = −πXtt0 dt + sin(πt)dt + Xtt0 dWt .. (3.11). According to Theorem 2.4, (3.11) has a random periodic solution. By Theorem 3.4, its Euler–Maruyama dissertation also has a random periodic path. To see the “periodicity” numerically, we provided two.
(22) ZAMP. Numerical approximation of random periodic solutions. Page 13 of 32. 119. ˆ ∗ (ω), −5 ≤ t ≤ 0} and {X ˆ ∗ (θ−2 ω), −5 ≤ t ≤ 2} Fig. 1. Simulations of the processes {X t t. ˆ ∗ (ω) = X t−6 (ω, 0.5), −5 ≤ t ≤ 0, and X ˆ ∗ (θ−2 ω) = methods. One approach is to simulate the processes X t t t−6 (θ−2 ω, 0.5), −5 ≤ t ≤ 2, with the same ω and step size Δt = 0.01 (Fig. 1). One can see that these two X ˆ t∗ (θ−2 ω) trajectories exactly repeat each with a time shift of one period (only comparing the graph of X ∗ ˆ t (θ−t ω), 0 ≤ t ≤ 6} for the same realisation ω for −3 ≤ t ≤ 2). The second method is the simulation of {X and step size as before (Fig. 2). One can easily see that Fig. 2 is a perfect periodic curve. This agrees with ˆ ∗ (ω) is a random periodic path iff X ˆ ∗ (θ−t ω) is periodic, i.e. X ˆ ∗ (θ−(t+τ ) ω) = X ˆ ∗ (θ−t ω). the fact that if X t t t+τ t ˆ t∗ = X ˆ t−∞ , but we take pull-back time −6 as this is already enough to generate a good Note in theory X ˆ t∗ (·) for t ≥ −5 by the solution starting at −6 from 0.5 for convergence to the random periodic paths X both cases. The choice of the initial position does not affect random periodic paths, but the time to take for the convergence.. 3.2. Modified Milstein scheme We will consider the Milstein scheme which will increase the convergence order for the infinite horizon problem. Condition (1 ). Assume there exists a constant τ > 0 such that for any t ∈ R, x ∈ Rm , f (t + τ, x) = β2 f (t, x), g(t + τ, x) = g(t, x), and there exist constants C0 , β1 , β2 > 0 with β1 + 22 < |λ1 | such that for any s, t ∈ R and x ∈ Rm , |f (s, x) − f (t, y)| ≤ C0 |s − t| + β1 |x − y| , |g(s, x) − g(t, y)| ≤ C0 |s − t| + β2 |x − y| . Meanwhile, we assume the boundedness of first-order partial derivative of function f and g with respect to x..
(23) 119. Page 14 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. ˆ ∗ (θ−t ω), 0 ≤ t ≤ 6} Fig. 2. Simulation of the process {X t. The iteration formula for the modified SRK scheme is −kτ −kτ −kτ −kτ X −kτ +(i+1)Δt = X−kτ +iΔt + AX−kτ +iΔt Δt + f (iΔt, X−kτ +iΔt )Δt −kτ +g(iΔt, X −kτ +iΔt ) (ΔW i ). . ΔZ i −kτ −kτ + (X + √ f iΔt, Υ −kτ +iΔt )) − f iΔt, Υ− (X−kτ +iΔt ) 2 Δt. (ΔW i )2 − Δt + (X −kτ −kτ √ g iΔt, Υ + , −kτ +iΔt )) − g iΔt, Υ− (X−kτ +iΔt ) 4 Δt. (3.12). with −kτ −kτ −kτ −kτ ± (X Υ −kτ +iΔt ) = X−kτ +iΔt + AX−kτ +iΔt Δt + f (iΔt, X−kτ +iΔt )Δt √ −kτ ± g(iΔt, X −kτ +iΔt ) Δt and −kτ +(i+1)Δt . dWs = W−kτ +(i+1)Δt − W−kτ +iΔt ,. ΔW i = −kτ +iΔt. −kτ +(i+1)Δt . s. ΔZ i =. dWu ds, −kτ +iΔt. −kτ X −kτ +0Δt. where i = 0, 1, 2, . . . , and Kloeden and Platen in [14].. −kτ +iΔt. = ξ. Here we used the approximation of ΔZ i by the method of. Theorem 3.6. Assume that Conditions (A), (1 ) hold and Δt is fixed and small enough. The time domain ∗ ∈ L2 (Ω) such that for any initial values ξ satisfying Condition is divided as τ = nΔt. Then there exists X r (2), the solution of the Milstein scheme satisfies. −kτ ∗ lim X (ξ) − X (3.13) r r = 0, k→∞. and. r∗ X. 2. satisfies the random periodicity property.. Proof. The proof is by a similar argument as Theorem 3.4. As it is tedious and there is no special difficulty, so omitted here. .
(24) ZAMP. Numerical approximation of random periodic solutions. Page 15 of 32. 119. 1, K 2, K 3, K 4 in the proof of Euler– Remark 3.7. For the Milstein scheme, the existence of constants as K Maruyama scheme is guaranteed by the boundedness of partial derivatives of functions f and g. Then we still have the convergence of the discrete processes. The for different initial. values and the boundedness. −kτ −kτ ΔZ i addition term 2√Δt [f iΔt, Υ+ (X−kτ +iΔt )) − f iΔt, Υ− (X−kτ +iΔt ) ] in the scheme does not influence the result of the convergence. However, when we analyse the error between approximation and the exact solution of random periodic solutions, this term is necessary for infinite horizon case to satisfy the order of error.. 4. The error estimate 4.1. Euler–Maruyama method In the last two sections, we proved the existence of random periodic solutions of SDE (1.3) and its discretisations as the limits of semi-flows when the starting times were pushed to −∞. The next step is to estimate the error between these two limits. Now we need to consider the difference between the discrete approximate solution and the exact solution. The exact solution at time −kτ + M Δt is as follows −kτ AM Δt ξ + eA(M Δt−kτ ) X−kτ +M Δt (ω, ξ) = e. M Δt−kτ . e−As f (s, Xs−kτ )ds. −kτ. +eA(M Δt−kτ ). M Δt−kτ . e−As g(s, Xs−kτ )dWs .. (4.1). −kτ. Lemma 4.1. Assume Conditions (A), (1) and (2). Choose Δt = τ /n for some n ∈ N and N = kn. Then there exists a constant K > 0 such that for any sufficiently small fixed Δt and N ∈ N, , we have. √ −kτ −kτ lim sup XN Δt, Δt − XN Δt ≤ K 2. k→∞. −kτ −kτ where XN Δt and XN Δt are the exact and the numerical solutions given by (4.1) and (3.2), respectively, K is independent of N and Δt.. ˆ · the constant derived from the underlining compuProof. In the following proof, we always denote by K tation unless otherwise stated. For any M ∈ N, we have −kτ X−kτ +M Δt. M AM Δt −kτ ξ + eA(M Δt−kτ ) −X − (I + AΔt) −kτ +M Δt = e. M Δt−kτ . e−As f (s, Xs−kτ )ds. −kτ. −. M −1 . (I + AΔt). M −i−1. A(M Δt−kτ ) −kτ f iΔt, X −kτ +iΔt Δt + e. i=0. −. M −1 . M Δt−kτ . e−As g(s, Xs−kτ )dWs. −kτ. (I + AΔt). M −i−1. −kτ g iΔt, X −kτ +iΔt. . W−kτ +(i+1)Δt − W−kτ +iΔt .. i=0. Similar to the method of Lemma 3.1, firstly consider.
(25) 119. Page 16 of 32. C. Feng, Y. Liu and H. Zhao. 2 −kτ −kτ X−kτ +M Δt − X −kτ +M Δt ⎛ ⎞ 2 −kτ −kτ M −1 X − X −kτ +(i+1)Δt −kτ +(i+1)Δt 2 ⎟ −kτ −2i ⎜ −kτ (1 − αΔt) − X−kτ = ⎝ +iΔt − X−kτ +iΔt ⎠ . 2 (1 − αΔt) i=0. (1 − αΔt). −2M. ZAMP. (4.2). For simplicity we denote 1 B1 = 1 − αΔt. B2 =. 1 1 − αΔt. (i+1)Δt−kτ . −kτ ) ds, e−A(s+kτ −(i+1)Δt) f (s, Xs−kτ ) − f (iΔt, X −kτ +iΔt. −kτ e−A(s+kτ −(i+1)Δt) g(s, Xs−kτ ) − g(iΔt, X ) dWs . −kτ +iΔt. iΔt−kτ (i+1)Δt−kτ . iΔt−kτ. Therefore, −kτ −kτ X−kτ +(i+1)Δt − X−kτ +(i+1)Δt −kτ −kτ = eAΔt X−kτ +iΔt − (I + AΔt) X−kτ +iΔt + (1 − αΔt) (B1 + B2 ) .. Now we consider 2 −kτ −kτ X−kτ +(i+1)Δt − X −kτ +(i+1)Δt (1 − αΔt). 2. 2 −kτ −kτ − X−kτ − X +iΔt −kτ +iΔt .
(26) AΔt . T
(27) eAΔt. e −kτ −kτ −kτ −kτ X−kτ − I + I = X−kτ − X +iΔt −kτ +iΔt +iΔt − X−kτ +iΔt 1 − αΔt 1 − αΔt. T
(28) eAΔt − I − AΔt 2 . T T −kτ −kτ X + X −kτ +iΔt −kτ +iΔt + B1 B1 + B2 B2 1 − αΔt. T
(29) eAΔt
(30) eAΔt − I − AΔt . −kτ −kτ −kτ X +2 X−kτ − X +iΔt −kτ +iΔt −kτ +iΔt 1 − αΔt 1 − αΔt
(31)
(32) AΔt
(33) . T I + AΔt −kτ T e −kτ +2 X−kτ − X B1 +iΔt −kτ +iΔt 1 − αΔt 1 − αΔt
(34)
(35) AΔt . T
(36) I + AΔt −kτ T e −kτ +2 X−kτ (4.3) − X B2 + 2B1T B2 . +iΔt −kτ +iΔt 1 − αΔt 1 − αΔt. AΔt. AΔt e e −I + I can be non-positive definite when we choose the Δt We note that the matrix 1−αΔt 1−αΔt small enough. Now we consider each term in (4.3). First, $. %. T
(37) eAΔt − I − AΔt 2 −kτ X E −kτ +iΔt 1 − αΔt 1 A2 (Δt)2 2 . −kτ 2. −kτ 5 (Δt)4 . ≤ X−kτ +iΔt X−kτ +iΔt ≤ K 2 1 − αΔt 2. −kτ X −kτ +iΔt.
(38) ZAMP. Numerical approximation of random periodic solutions. Page 17 of 32. 119. Next, ⎛ 2 E B1T B1 = E |B1 | ≤. 2(1 + μ) μ (1 − αΔt) ⎛. +. 2(1 + μ) 2. μ (1 − αΔt). ⎜ ⎝. ⎛ +. 1+μ (1 − αΔt). 2. ⎜ ⎝. 2. ⎜ ⎝. (i+1)Δt−kτ . iΔt−kτ. (i+1)Δt−kτ . iΔt−kτ (i+1)Δt−kτ . ⎞2 . −A(s+kτ −(i+1)Δt) ⎟ − I f (s, Xs−kτ ) 2 ds⎠ e ⎞2. ⎟. f (s, Xs−kτ ) − f (iΔt, X −kτ −kτ +iΔt ) 2 ds⎠. ⎞2. ⎟ −kτ −kτ f (iΔt, X−kτ +iΔt ) − f (iΔt, X−kτ +iΔt ) ds⎠ , 2. (4.4). iΔt−kτ. where μ is a small number from Young’s inequality, which will be fixed later. By linear growth property. of f and Lemma 2.1, we know that f (s, Xs−kτ ) 2 is bounded. So for the first term in (4.4) we only need to estimate (i+1)Δt−kτ . (Δt)2 −A(s+kτ −(i+1)Δt) T r (−A) . − I ds ≤ e 2. iΔt−kτ. By Condition (1) and Lemma 2.2, the second term in (4.4) becomes (i+1)Δt−kτ . f (s, Xs−kτ ) − f (iΔt, X −kτ −kτ +iΔt ) 2 ds. iΔt−kτ (i+1)Δt−kτ . ( f (s, Xs−kτ ) − f (iΔt, Xs−kτ ) 2. ≤ iΔt−kτ. −kτ. + f (iΔt, Xs−kτ ) − f (iΔt, X−kτ +iΔt ) 2 )ds (i+1)Δt−kτ . ≤. C0 |s − iΔt + kτ | iΔt−kτ. 1/2. (i+1)Δt−kτ . ds +. −kτ. β1 Xs−kτ − X−kτ +iΔt 2 ds. iΔt−kτ. (i+1)Δt−kτ . √ (C0 + β1 C4 ) s − iΔt + kτ ds. ≤ iΔt−kτ 3. 6 (Δt) 2 . ≤K Applying the global Lipschitz condition, the third term of (4.4) becomes (i+1)Δt−kτ . −kτ −kτ f (iΔt, X−kτ +iΔt ) − f (iΔt, X−kτ +iΔt ) ds 2. iΔt−kτ. −kτ. −kτ ≤ β1 Δt X−kτ +iΔt − X−kτ +iΔt . 2.
(39) 119. Page 18 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. We summarise the above inequalities to have 2 2 7 (Δt)3 + (1 + μ)β1 (Δt) E B1T B1 ≤ K 2 (1 − αΔt). 2 −kτ. −kτ X−kτ +iΔt − X −kτ +iΔt . 2. (4.5). 2 −kτ. −kτ This term is of the third order of Δt and second order of Δt with X−kτ − X +iΔt −kτ +iΔt . 2 T Similar to the E B1 B1 , the following term can be estimated as 2 E B2T B2 = E |B2 | ≤. (i+1)Δt−kτ . 2(1 + μ) μ (1 − αΔt) +. +. 2 iΔt−kτ. (i+1)Δt−kτ . 2(1 + μ) 2. μ (1 − αΔt). 2. g(s, Xs−kτ ) − g(iΔt, X −kτ −kτ +iΔt ) 2 ds. iΔt−kτ (i+1)Δt−kτ . 1+μ (1 − αΔt). 2 2 −A(s+kτ −(i+1)Δt) − I g(s, Xs−kτ ) 2 ds e. 2. 2. −kτ −kτ g(iΔt, X−kτ +iΔt ) − g(iΔt, X−kτ +iΔt ) ds, 2. (4.6). iΔt−kτ. where μ is a small number from Young’s inequality, which. 2 will be fixed later. By the linear growth property of g and Lemma 2.1, we know that g(s, Xs−kτ ) 2 is bounded. So we only need to estimate (i+1)Δt−kτ . 2 2 −A(s+kτ −(i+1)Δt) 3 − I ds ≤ (Δt) T r A2 . e 3. iΔt−kτ. By Condition (1) and Lemma 2.2, the second term in (4.6) becomes (i+1)Δt−kτ . 2. g(s, Xs−kτ ) − g(iΔt, X −kτ −kτ +iΔt ) 2 ds. iΔt−kτ (i+1)Δt−kτ . ≤. 8 (Δt)2 . 2(C02 + β22 C42 ) |s − iΔt + kτ | ds ≤ K. iΔt−kτ. The third term follows from the global Lipschitz condition (i+1)Δt−kτ . 2. −kτ −kτ ) − g(iΔt, X ) g(iΔt, X−kτ +iΔt −kτ +iΔt ds 2. iΔt−kτ. 2. −kτ −kτ ≤ β22 Δt X−kτ +iΔt − X−kτ +iΔt . 2. Conclude the above results to obtain. 2 2 . −kτ 9 (Δt)2 + (1 + μ)β2 Δt −kτ E B2T B2 ≤ K 2 X−kτ +iΔt − X−kτ +iΔt . 2 (1 − αΔt). (4.7).
(40) ZAMP. Numerical approximation of random periodic solutions. Page 19 of 32. 119. The fifth term of (4.3) can be estimate as follows. T
(41) eAΔt
(42) eAΔt − I − AΔt . ! −kτ X−kτ +iΔt E 2 1 − αΔt 1 − αΔt 1 A2 (Δt)2 . 2 −kτ. −kτ −kτ X − X ≤ 2 X−kτ. +iΔt −kτ +iΔt −kτ +iΔt 2 (1 − αΔt)2 2. 2 −kτ −kτ 10 (Δt) X ≤K −kτ +iΔt − X−kτ +iΔt .. −kτ X−kτ +iΔt. −kτ −X −kτ +iΔt. 2. To estimate the sixth term of (4.3),
(43). . T
(44) I + AΔt ! eAΔt −kτ E 2 − X−kτ +iΔt B1 1 − αΔt 1 − αΔt
(45) AΔt ! T −kτ e I + AΔt − = E 2 X−kτ B1 +iΔt 1 − αΔt 1 − αΔt. T
(46) I + AΔt !. −kτ −kτ + E 2 X−kτ − X B1 . +iΔt −kτ +iΔt 1 − αΔt . T −kτ X−kτ +iΔt.
(47). (4.8). Now we discuss these two terms separately, 1 2 2 A (Δt) 2 −kτ T −kτ e I + AΔt. X − B1 2 E 2 X−kτ ≤ 2 B 1 +iΔt −kτ +iΔt 2 1 − αΔt 1 − αΔt 1 − αΔt √. 3 . −kτ 12 (Δt)7/2 + 1 + μβ1 K11 (Δt) −kτ ≤K − X X −kτ +iΔt −kτ +iΔt . (1 − αΔt)2 2
(48). AΔt. !. And,. T
(49) I + AΔt ! −kτ −kτ E 2 X−kτ − X B1 +iΔt −kτ +iΔt 1 − αΔt &. 7 (Δt)3/2 2 K −kτ. −kτ ≤ X−kτ +iΔt − X −kτ +iΔt (1 + Δt |A|) 1 − αΔt 2 √ 2 2 1 + μβ1 Δt −kτ. −kτ + − X X (1 + Δt |A|) . −kτ +iΔt −kτ +iΔt (1 − αΔt)2 2. (4.9). We use the conditional expectation to eliminate the seventh term
(50)
(51) AΔt . T
(52) I + AΔt ! −kτ T e −kτ X−kτ +iΔt E − X−kτ +iΔt B2 1 − αΔt 1 − αΔt
(53)
(54) AΔt
(55) !. T I + AΔt −kτ T e −kτ iΔt−kτ X−kτ +iΔt =E − X−kτ +iΔt E B2 |F 1 − αΔt 1 − αΔt = 0. For the last term, E 2B1T B2 ≤ 2 B1T 2 · B2 2. 2. −kτ 13 (Δt)5/2 + K 14 (Δt)3/2 −kτ ≤K X−kτ +iΔt − X−kτ +iΔt . 2.
(56) 119. Page 20 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. Combining all the estimation above, we have 2 −kτ −kτ 2 X−kτ +(i+1)Δt − X −kτ +(i+1)Δt −kτ −kτ − X−kτ +iΔt − X−kτ +iΔt 2 (1 − αΔt) ' ( 2. (1 + μ)β22 Δt 2 (1 + μ)β1 Δt. −kτ 16 (Δt)3/2 −kτ X ≤ + K − X. −kτ +iΔt −kτ +iΔt 2 + 2 (1 − αΔt) 2 (1 − αΔt) ⎞ ⎛ &. 7 (Δt)3/2 2 K. −kτ 17 (Δt)2 ⎠ 15 (Δt)2 + ⎝ −kτ +K +K − X X−kτ +iΔt −kτ +iΔt . 1 − αΔt 2 2. −kτ −kτ − X Now we notice that the term X−kτ +iΔt −kτ +iΔt has coefficients, the largest of which contains a 2. 2 constant multiplied by Δt. The largest free term contains a constant multiplied by (Δt) . Choosing μ −kτ. −kτ and Δt small enough and applying Young’s inequality for the term (Δt)3/2 X−kτ +iΔt − X−kτ +iΔt , 2. and from (4.2) we get. 2 −kτ. −kτ X−kτ +M Δt − X −kτ +M Δt 2
(57). M −1 2. . −kτ −2i −kτ 2 K20 Δt X−kτ +iΔt − X−kτ +iΔt + K18 (Δt) ≤ (1 − αΔt) −2M. (1 − αΔt). 2. i=0. 20 (Δt) 19 (Δt)(1 − αΔt)−2M + K ≤K. M −1 . (1 − αΔt). −2i. 2 −kτ. −kτ X−kτ +iΔt − X −kτ +iΔt ,. i=0. 2. (4.10). where 2 2 2 20 = (1 + μ)(2β1 + β2 + ε) . 19 = K18 (1 − αΔt) (Δt) = K18 (1 − αΔt) , K K 2 2α − α2 (Δt) (1 − αΔt)2 2αΔt − α2 (Δt). 20 Δt + 1 (1 − αΔt)2 < 1. Now Here μ, ε and the time step Δt are chosen small enough such that K. using the discrete time Gronwall inequality, from (4.10), we have. 2 −kτ. −kτ X−kτ +M Δt − X −kτ +M Δt 2. M . 20 Δt (1 − αΔt)2 1− 1+K 20 (Δt)2 19 Δt + K 19 K 21 Δt.. ≤K ≤K 20 Δt (1 − αΔt)2 1− 1+K 21 which is independent of M and Δt. Finally, we take M = N + N , where We can find a constant K N Δt = kτ , N ∈ Z, then. −kτ −kτ. −kτ −kτ lim sup XN Δt − XN Δt = lim sup X−kτ +(N +N )Δt − X−kτ +(N +N )Δt 2 2 N →∞ k→∞ & √ 21 Δt. ≤ K So we get the result.. . We have proved the estimation of error from −kτ to N Δt as k → ∞ can be controlled under the 1/2 order of the time step. And the upper bound is uniform in time. The following theorem will give us a −kτ , r > 0 be given by (3.6). more general result, which is from −kτ to time r. Let X r.
(58) ZAMP. Numerical approximation of random periodic solutions. Page 21 of 32. 119. Theorem 4.2. Assume Conditions (A), (1) and (2). We choose Δt = τ /n for some n ∈ N, N = kn. For # > 0 such that for any sufficiently small fixed Δt, any r ≥ 0, there exists a constant K. √. −kτ # Δt, lim sup Xr−kτ − X ≤K r 2. k→∞. −kτ is the numerical solution and K # is independent of Δt and where Xr−kτ is the exact solution, while X r r. Proof. Assume for any r ≥ 0, N is the unique integer such that N Δt ≤ r, (N + 1)Δt > r. According to the semi-flow property, we have −kτ (ω) =X N Δt (ω) ◦ X −kτ N Δt (ω) ◦ X −kτ Xr−kτ (ω) − X r r N Δt (ω) − Xr N Δt (ω), rN Δt is finite time Euler approximation of solution of (1.3) from N Δt to r and X −kτ where X N Δt is defined as before. So,. −kτ r−kτ (4.11). Xr − X 2. . N Δt −kτ −kτ N Δt −kτ rN Δt ◦ X −kτ ≤ XrN Δt ◦ XN ◦X ◦ XN Δt − X Δt − Xr N Δt + Xr N Δt . 2 2. √ −kτ −kτ For the first term on the right-hand side, by Lemma 4.1, we have XN Δt − XN Δt ≤ K Δt. By the . continuity of XrN Δt (·) with respect to initial values in L2 (Ω) [15], then. √ N Δt. −kτ −kτ N Δt −kτ −kτ X ◦ XN ◦X ≤ C − X Xr. Δt − Xr N Δt N Δt N Δt ≤ C5 Δt, 2. 2. where C5 is independent of Δt. For the second term on the right-hand side of (4.11), it is finite time Euler approximation with same initial value. By Theorem 10.2.2 in Kloeden and Platen [14], there exists a constant C6 > 0 such that for sufficiently Δt > 0,. √. N Δt −kτ N Δt ◦ X −kτ ◦ XN Δt − X Xr r N Δt ≤ C6 Δt, 2. # = C5 + C6 . where the choice of C6 is independent of Δt. The result follows by taking K. . Corollary 4.3. For any r ≥ 0, the exact and numerical approximating random periodic solutions of ∗ , given in Theorems 2.4 and 3.4, respectively, satisfy Eq. (1.3), Xr∗ and X r. √ ∗ r∗ # Δt. Xr − X ≤K 2. Proof. The result follows from. . . ∗ −kτ −kτ ∗ −kτ ∗ Xr∗ − Xr−kτ + X − X + − X Xr − X. . X. r ≤ lim sup r r r r 2 2. k→∞. 2. 2. 4.2. Modified Milstein method For Milstein method, we can use the similar calculation as Euler–Maruyama scheme to get an improved error estimate between discrete approximate solution and the exact solution. Theorem 4.4. Assume Conditions (A), (1 ) and (2). Then there exists a constant K ∗ > 0 such that for any sufficiently small fixed Δt, the error between the exact solution Xr−kτ and the numerical solution r−kτ r−kτ given by Milstein scheme (3.12) is lim supk→∞ X Xr−kτ − X ≤ K ∗ Δt, for all r ≥ 0, where K ∗ 2 is independent of Δt..
(59) 119. Page 22 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. ˆ · the constant derived from the underlining compuProof. In the following proof, we always denote by K tation unless otherwise stated. We consider the error in the similar way as Lemma 4.1. 2 −2M −kτ −kτ (1 − αΔt) X−kτ +M Δt − X −kτ +M Δt ⎛ ⎞ 2 −kτ −kτ M −1 − X X 2 −kτ +(i+1)Δt −kτ +(i+1)Δt −kτ ⎟ −2i ⎜ −kτ = (1 − αΔt) − X−kτ (4.12) ⎝ +iΔt − X−kτ +iΔt ⎠ . 2 (1 − αΔt) i=0 For simplicity we denote ˜1 = B. 1 1 − αΔt s − iΔt−kτ. 1 ˜2 = B 1 − αΔt s −. $ (i+1)Δt−kτ . −kτ e−A(s+kτ −(i+1)Δt) f (s, Xs−kτ ) − f (iΔt, X −kτ +iΔt ). iΔt−kτ. ⎤. −kτ ⎦ Fi (X −kτ +iΔt )dWυ ds. (1). $ (i+1)Δt−kτ . −kτ e−A(s+kτ −(i+1)Δt) g(s, Xs−kτ ) − g(iΔt, X −kτ +iΔt ). iΔt−kτ. ⎤. (1) −kτ ⎦ Gi (X −kτ +iΔt )dWυ dWs ,. iΔt−kτ. with. 1 (1) + (x) − f iΔt, Υ − (x) , Fi (x) = √ f iΔt, Υ 2 Δt. 1 (1) + (x) − g iΔt, Υ − (x) . g iΔt, Υ Gi (x) = √ 2 Δt. Therefore, −kτ −kτ X−kτ +(i+1)Δt − X−kτ +(i+1)Δt. −kτ −kτ ˜ ˜ = eAΔt X−kτ +iΔt − (I + AΔt) X−kτ +iΔt + (1 − αΔt) B1 + B2 .. Now we consider 2 −kτ −kτ X−kτ +(i+1)Δt − X −kτ +(i+1)Δt . =. (1 − αΔt) −kτ X−kτ +iΔt. 2. −kτ −X −kτ +iΔt. 2 −kτ −kτ − X−kτ +iΔt − X−kτ +iΔt .
(60) AΔt. T
(61) eAΔt e −I +I 1 − αΔt 1 − αΔt. −kτ −kτ × X−kτ +iΔt − X−kτ +iΔt. T
(62) eAΔt − I − AΔt 2 . −kτ −kτ ˜T ˜ ˜T ˜ X + X −kτ +iΔt −kτ +iΔt + B1 B1 + B2 B2 1 − αΔt. T
(63) eAΔt
(64) eAΔt − I − AΔt . −kτ −kτ −kτ X +2 X−kτ − X +iΔt −kτ +iΔt −kτ +iΔt 1 − αΔt 1 − αΔt.
(65) ZAMP. Numerical approximation of random periodic solutions.
(66). T
(67) I + AΔt ˜1 B +2 1 − αΔt
(68)
(69) AΔt . T
(70) I + AΔt T −kτ e −kτ ˜ ˜2 + 2B ˜T B B +2 X−kτ +iΔt − X−kτ +iΔt 1 2. 1 − αΔt 1 − αΔt . T −kτ X−kτ +iΔt.
(71). Page 23 of 32. eAΔt 1 − αΔt. 119. −kτ − X −kτ +iΔt. (4.13). By the similar analysis as (4.5) and (4.7), we have 2 2 2. −kτ 22 (Δt)4 + (1 + μ)β1 (Δt) ˜1 ≤ K ˜T B −kτ E B − X X 1 −kτ +iΔt −kτ +iΔt 2 2 (1 − αΔt). 2. −kτ 2 (Δt)3 −kτ +K X−kτ 23 +iΔt − X−kτ +iΔt , 2. and. 2 (1 + μ)β22 Δt −kτ. 3 ˜ ˜T B −kτ E B − X X 2 2 ≤ K24 (Δt) + −kτ +iΔt −kτ +iΔt 2 2 (1 − αΔt). 2. −kτ 25 (Δt)2 −kτ +K X−kτ +iΔt − X−kτ +iΔt . 2. The crossing product terms in (4.13) are estimated similar as (4.8) as follows:
(72)
(73) AΔt . T
(74) I + AΔt ! −kτ T e ˜1 −kτ B E 2 X−kτ − X +iΔt −kτ +iΔt 1 − αΔt 1 − αΔt √. 27 (Δt)3 1 + μβ1 K −kτ. 4 −kτ X − X ≤ K26 (Δt) +. −kτ +iΔt −kτ +iΔt 2 (1 − αΔt) 2 &. 22 (Δt)2 2 K −kτ. −kτ + X−kτ +iΔt − X −kτ +iΔt (1 + Δt |A|) 1 − αΔt 2 √. 2 1 + μβ1 Δt −kτ 2 −kτ + X−kτ +iΔt − X −kτ +iΔt (1 + Δt |A|) 2 (1 − αΔt) 2. 2. −kτ −kτ 23 (Δt)3/2 + 2K X−kτ +iΔt − X−kτ +iΔt (1 + Δt |A|) .. (4.14). 2. The seventh term remains 0 under conditional expectation.
(75) AΔt
(76). T
(77) I + AΔt ! −kτ T e ˜2 = 0. −kτ X−kτ +iΔt B E − X −kτ +iΔt 1 − αΔt 1 − αΔt For the last term,. ˜ ˜T ˜2 ≤ 2 ˜T B E 2B B 1 1 · B2 2. 2. 28 (Δt) ≤K. 7/2. 2. −kτ 29 (Δt)3/2 −kτ +K − X X−kτ +iΔt −kτ +iΔt . 2. Combining all the estimation above, we have 2 −kτ −kτ 2 X−kτ +(i+1)Δt − X −kτ +(i+1)Δt −kτ −kτ − X−kτ +iΔt − X−kτ +iΔt 2 (1 − αΔt) ' (. 2 (1 + μ)β22 Δt 2 (1 + μ)β1 Δt. −kτ 40 (Δt)3/2 −kτ ≤ + K − X X. −kτ +iΔt −kτ +iΔt 2 + 2 (1 − αΔt) 2 (1 − αΔt). −kτ 41 (Δt)3 + K 42 (Δt)2 −kτ +K X−kτ +iΔt − X−kτ +iΔt . 2. (4.15).
(78) 119. Page 24 of 32. C. Feng, Y. Liu and H. Zhao. ZAMP. Choosing μ and Δt small enough and applying Young’s inequality to the term. −kτ 2 −kτ (Δt) X−kτ +iΔt − X−kτ +iΔt , and from (4.12) we get 2. 2. −2M −kτ −kτ (1 − αΔt) X−kτ +M Δt − X −kτ +M Δt 2
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