Báo cáo nghiên cứu khoa học: "Về một điều kiện đủ cho tính ổn định mũ của một lớp phương trình vi phân ngẫu nhiên có trễ" pot

dx(t) = f(x(t), x(t − τ), t)dt + σ(t, x(t))dw(t). dx(t) = f(x(t), x(t − τ), t)dt. dx(t) = f(x(t), x(t−τ), t)dt+σ(t)dw(t). (1.1) dx(t) = f(x(t), x(t−τ), t)dt+σ(t, x(t))dw(t). (1.2) (Ω, , { t } t0 , P ) { t } t0 |x| x ∈ R n A A = sup{|Ax| : |x| = 1} B T 1 A = (a ij ) Trace(A) =  a ii τ C([−τ, 0]; R d ) R d − [−τ, 0] L 2  t ([−τ, 0]; R d )  t − C([−τ, 0]; R d ) ξ = {ξ(u) : −τ  u  0} ξ 2 E = sup −τ u0 E|ξ(u)| 2 < ∞. dx(t) = f(x(t), x(t−τ), t)dt+σ(t, x(t))dw(t); on t ≥ 0 (2.1) x(t) = ξ(u) −τ  u  0 f : R d × R d × R + → R d , σ : R d ×R + → R d×m w ξ ∈ L 2  0 ([−τ, 0]; R d ) x(t, ξ) (2.1) δ K ξ ∈ L 2  0 ([−τ, 0]; R d ) E|x(t, ξ)|  Kξ 2 E e −δt , ∀t ≥ 0. (2.2) δ K u(t) v(t) N 0 t ≥ s u(t)  N 0 +  t s u(t 1 )v(t 1 )dt 1 . t ≥ s u(t)  N 0 exp{  t s v(t 1 )dt 1 }. (2.3) c 1 − c 3 2x T f(x, y, t)  −c 1 |x| 2 + c 2 |y| 2 , T race(σ(t, x)σ T (t, x))  c 3 |x| 2 , c 2 e c 1 τ + c 3 < c 1 , x, y ∈ R d ; t  0. (2.1) Proof. For all ξ ∈ L 2 F 0 ([−τ, 0]; R d ) Fix ξ arbitrarily and write x(t, ξ) = x(t) simple. By Ito’s formula and assumption, e c 1 t |x(t)| 2 = |x(0)| 2 + M(t) + N(t) for all t ≥ 0, where M(t) = 2  t 0 e c 1 s x T (s)σ(s, x(s))dw(s). N(t) =  t 0 e c 1 s (c 1 |x(s)| 2 + 2x(s) T f(x(s), x(s − τ), s) + trace(σ(s, x(s))σ T (s, x(s)))ds. By (i), (ii) we have e c 1 t |x(t)| 2  |x(0)| 2 +M(t)+  t 0 e c 1 s (c 2 |x(s−τ)| 2 +c 3 |x(s)| 2 )ds. (4.4) But  t 0 e c 1 s |x(s − τ)| 2 ds   τ 0 e c 1 s |x(s − τ)| 2 ds +  max {τ,t} τ e c 1 s |x(s − τ)| 2 ds   τ 0 e c 1 s |x(s−τ)| 2 ds+e c 1 τ  t 0 e c 1 s |x(s)| 2 ds. (4.5). From inequalities (4.5) and (4.4) follow that e c 1 t |x(t)| 2  |x(0)| 2 + M(t) +  τ 0 e c 1 s c 2 |x(s − τ)| 2 ds +  t 0 (c 3 + c 2 e c 1 τ )e c 1 s |x(s)| 2 ds because EM(t) = 0, moreover we have  τ 0 e c 1 s c 2 E|x(s − τ )| 2 ds  c 2 c 1 (e c 1 τ − 1)ξ 2 E ; E|x(0)| 2  ξ 2 E so that e c 1 t E|x(t)| 2  E|x(0)| 2 +  τ 0 e c 1 s c 2 E|x(s − τ )| 2 ds +  t 0 (c 3 + c 2 e c 1 τ )e c 1 s E|x(s)| 2 ds  (1+ c 2 c 1 (e c 1 τ −1))ξ 2 E +  t 0 (c 3 +c 2 e c 1 τ )e c 1 s E|x(s)| 2 ds. (4.6). From (4.6) and applying lemma 2.2 with u(t) = e c 1 t E|x(t)| 2 ; v(t) = c 3 + c 2 e c 1 τ ; N 0 = (1 + c 2 c 1 (e c 1 τ − 1))ξ 2 E we have e c 1 t E|x(t)| 2  (1 + c 2 c 1 (e c 1 τ − 1))ξ 2 E e (c 3 +c 2 e c 1 τ )t . Hence we obtain E|x(t)| 2  (1 + c 2 c 1 (e c 1 τ − 1))ξ 2 E e (c 3 +c 2 e c 1 τ −c 1 )t By assumptions (iii) we can rewrite E|x(t)| 2  Kξ 2 E e −δt , where K = 1 + c 2 c 1 (e c 1 τ − 1) > 0 and δ = c 1 − c 3 − c 2 e c 1 τ > 0. In other words,the stochastic differential equations (2.1) is e xponential stability in mean square. The proof is completed.  Acknowledgement. dx(t) = f(x(t), x(t − τ), t)dt + σ(t, x(t))dw(t). dx(t) = f(x(t), x(t − τ), t)dt.