Báo cáo nghiên cứu khoa học: "Đánh giá tính ổn định nghiệm cuả bài toán Cauchy cho phương trình truyền nhiệt ngược thời gian" pot

¨o    ∂u ∂t = a(t) ∂ 2 u ∂x 2 , (x, t) ∈ (−∞; +∞) × (0; 1), u(x, 1) = ϕ(x). 1. ¨o (1.1)    ∂u ∂t = a(t) ∂ 2 u ∂x 2 , (x, t) ∈ (−∞; +∞) × (0; 1), u(x, 1) = ϕ(x). ϕ(x) E u(·, 0) :=   +∞ −∞ u 2 (x, 0)dx  1 2  E. 1 1 u(x, t) U u(·, 0)  E, E U a(t) 2. 1 ( ¨o p > 1 q > 1 1 p + 1 q = 1 f ∈ L p (R), g ∈ L q (R) fg ∈ L 1 (R) fg 1  f p g q . 2 ( L 1 (R) f ∈ L 1 (R) f  f(ξ) := 1 √ 2π  +∞ −∞ e −ix.ξ f(x)dx (y ∈ R). 2 ( f ∈ L 1 (R) ∩L 2 (R)  f ∈ L 2 (R) f =   f  ·  L 2 (R) 3 ( L 2 (R)  f f ∈ L 2 (R) {f k } ∞ k=1 ⊂ L 1 (R) ∩ L 2 (R) f k → f L 2 (R)   f k −  f j  =   f k − f j  = f k − f j  {  f k } ∞ k=1 L 2 (R)  f k →  f L 2 (R)  f f L 2 (R)  f {  f k } ∞ k=1 3 ( f, g ∈ L 2 (R) +∞  −∞ f ¯gdx = +∞  −∞ ˆ f ˆgdξ  D α f = (iξ) α  f α D α f ∈ L 2 (R) 3. 1 ( u(x, t) (3.1)        ∂u ∂t = a(t) ∂ 2 u ∂x 2 , (x, t) ∈ (−∞; +∞) × (0; 1), u(·, 1)  , u(·, 0)  E, (0 <  < E), a(t)  B > 0, t ∈ [0; 1] u(·, t)   µ(t) E 1−µ(t) , t ∈ [0; 1] u(·, t) =   +∞ −∞ u 2 (x, t)dx  1 2 , t ∈ [0, 1] µ(t) = A(t) A(1) , ∀t ∈ [0, 1], A(t) =  t 0 a(τ)dτ, ∀t ∈ [0, 1]. u(x, t) ∂u ∂t = a(t) ∂ 2 u ∂x 2 ∂u ∂t (ξ, t) = −ξ 2 a(t)u(ξ, t).(3.2) u(ξ, t) = e A(1)(1−µ(t))ξ 2 u(ξ, 1), (ξ, t) ∈ R × [0, 1].(3.3) ˆu(·, t) ∈ L 2 (R), t ∈ [0, 1] |u(ξ, t)| µ(t) = e A(1)µ(t)(1−µ(t))ξ 2 |u(ξ, 1)| µ(t) , (ξ, t) ∈ R ×[0, 1].(3.4) t = 0 u(ξ, 0) = e A(1)ξ 2 u(ξ, 1), ξ ∈ R,(3.5) u(ξ, 1) = e −A(1)ξ 2 u(ξ, 0), ξ ∈ R.(3.6) u(ξ, t) = e −A(1)µ(t)ξ 2 u(ξ, 0), (ξ, t) ∈ R × [0, 1].(3.7) |u(ξ, t)| (1−µ(t)) = e −A(1)µ(t)(1−µ(t))ξ 2 |u(ξ, 0)| (1−µ(t)) , (ξ, t) ∈ R ×[0, 1].(3.8) |u(ξ, t)| = | u(ξ, 1)| µ(t) |u(ξ, 0)| (1−µ(t)) , ξ ∈ R. |u(·, t)| = |u(·, 1)| µ(t) |u(·, 0)| (1−µ(t)) . t = 0 t = 1 t ∈ (0, 1) f = |u(·, 1)| 2µ(t) , g = |u(·, 0)| 2(1−µ(t)) , p = 1 µ(t) , q = 1 1 − µ(t) u(·, t) 2 = u(·, t) 2 =  +∞ −∞ |u(ξ, 1)| 2µ(t) |u(ξ, 0)| 2(1−µ(t)) dξ =  +∞ −∞ f(ξ)g(ξ)dξ =  +∞ −∞ |f(ξ)g(ξ)|dξ = f g 1  f p g q =   +∞ −∞ |f(ξ)| p dξ  µ(t) .   +∞ −∞ |g(ξ)| q dξ  (1−µ(t)) =   +∞ −∞ u 2 (ξ, 1)dξ  µ(t) .   +∞ −∞ u 2 (ξ, 0)dξ  (1−µ(t)) = u(·, 1) 2µ(t) .u(·, 0) 2(1−µ(t)) = u(·, 1) 2µ(t) .u(·, 0) 2(1−µ(t))   2µ(t) E 2(1−µ(t)) . t ∈ (0, 1)  1 a(t) a(t)  0 a(t) = 0 a(t) ∈ L 1 (0, 1) [1] K. A. Ames and B. Straughan, Aca- demic Press, San Diego, 1997. [2] Dinh Nho Hao, Journal of Mathematical Analysis and Applications, 1996, 873-909. ¨o    ∂u ∂t = a(t) ∂ 2 u ∂x 2 , (x, t) ∈ (−∞; +∞) × (0; 1), u(x, 1) = ϕ(x).