Báo cáo nghiên cứu khoa học: "Đánh giá tính ổn định cho ph­ương trình dạng Burgers ng­ược thời gian" pps

     u t = (a(x, t)u x ) x − uu x , (x, t) ∈ (0, 1) × (0, 1), u(0, t) = u(1, t) = 0, 0  t  1, u(x, 1) = ϕ(x), 0  x  1. 1. (1.1)      u t = (a(x, t)u x ) x − uu x , (x, t) ∈ (0, 1) × (0, 1), u(0, t) = u(1, t) = 0, 0  t  1, u(x, 1) = ϕ(x), 0  x  1. ϕ(x) C 2,1 1 D = {(x, t) : 0 < x < 1, 0 < t < 1}. a(x, t) C 2,1 (D) C 2,1 (D) x t D D β > 0 a(x, t)  β > 0, (x, t) ∈ D 1 u(x, t) V u(x, t) C 2,1 (D)  M, M V u xx a(t) u xx 2. u 1 (x, t) u 2 (x, t) ϕ 1 (x) ϕ 2 (x) z(x, t) = u 1 (x, t) − u 2 (x, t) f(t) =  1 0 z 2 (x, t)dx. 1 u i (x, t) ∈ V, i = 1, 2, c 1 c 2 u i (x, t) f(t)  e (c 2 /c 1 )[t−α] [f(0)] (1−α) [f(1)] α , ∀t ∈ [0, 1],(2.1) α = exp{c 1 t} − 1 exp{c 1 } − 1 .(2.2) u i (x, t), i = 1, 2, z t = (az x ) x − u 2 z x − u 1x z.(2.3) f(t) =  1 0 z 2 dx f  (t) = 2  1 0 zz t dx = 2  1 0 z{(az x ) x − u 2 z x − u 1x z}dx = 2  1 0 z(az x ) x dx − 2  1 0 u 2 zz x dx − 2  1 0 u 1x z 2 dx = 2az x z    1 0 − 2  1 0 a(z x ) 2 dx − u 2 z 2    1 0 +  1 0 {u 2x − 2u 1x }z 2 dx = −2  1 0 a(z x ) 2 dx −  1 0 u 1x z 2 dx −  1 0 z 2 z x dx = −2  1 0 a(z x ) 2 dx −  1 0 u 1x z 2 dx f  (t) = −2  1 0 a t (z x ) 2 dx − 4  1 0 az x z xt dx −  1 0 u 1xt z 2 dx − 2  1 0 u 1x zz t dx = −2  1 0 a t (z x ) 2 dx − 4az x z t    1 0 + 4  1 0 z t (az x ) x dx −  1 0 u 1xt z 2 dx − 2  1 0 u 1x zz t dx = −2  1 0 a t (z x ) 2 dx + 4  1 0 z t {z t + u 2 z x + u 1x z}dx −  1 0 u 1xt z 2 dx − 2  1 0 u 1x zz t dx = 4  1 0 z 2 t dx − 2  1 0 a t (z x ) 2 dx + 4  1 0 u 2 z x z t dx + 2  1 0 u 1x zz t dx −  1 0 u 1xt z 2 dx = 4  1 0 z 2 t dx − 2  1 0 a t (z x ) 2 dx + 2  1 0 (2u 2 z x + u 1x z)z t dx +  1 0 2u 1t zz x dx = 4  1 0 z 2 t dx − 2  1 0 a t (z x ) 2 dx + 2  1 0 (2u 2 z x + u 1x z){(az x ) x − u 2 z x − u 1x z}dx +  1 0 2u 1t zz x dx = 4  1 0 z 2 t dx +  1 0 {4u 2 (a x − u 2 ) − 2(au 1x ) − 1 2 (au 2 ) x − 2a t }(z x ) 2 dx +2  1 0 (u 1x a x − (au 1x ) x − 3u 1x u 2 + 2u 1t )(zz x )dx − 2  1 0 (u 1x ) 2 z 2 dx. u ∈ V a ∈ C 2,1 ( ¯ D) c 3 u 1 , u 2 |u 1x a x − (au 1x ) x − 3u 1x u 2 + 2u 1t |  c 3 , ∀(x, t) ∈ D. |bc|  εb 2 2 + c 2 2ε ε > 0 b, c 2  1 0 (u 1x a x − (au 1x ) x − 3u 1x u 2 )(zz x )dx  −c 3  ε  1 0 (z x ) 2 dx + 1 ε  1 0 z 2 dx  . f  (t)  4  1 0 z 2 t dx +  1 0 {4u 2 (a x − u 2 ) − 2(au 1x ) − 1 2 (au 2 ) x − 2a t − c 3 ε}(z x ) 2 dx −  1 0 ( c 3  + 2u 2 1x )z 2 dx = = 4  1 0 z 2 t dx − 2  1 0 {−2u 2 (a x − u 2 ) + (au 1x ) + 1 4 (au 2 ) x + a t + 1 2 c 3 ε}(z x ) 2 dx −  1 0 ( c 3  + 2u 2 1x )z 2 dx = = 4  1 0 z 2 t dx − 2  1 0 {−2u 2 (a x − u 2 ) + (au 1x ) + 1 4 (au 2 ) x + a t + 1 2 c 3 ε} a .a(z x ) 2 dx −  1 0 ( c 3  + 2u 2 1x )z 2 dx. c 1 u 1 , u 2 {−2u 2 (a x − u 2 ) + (au 1x ) + 1 4 (au 2 ) x + a t + 1 2 c 3 ε} a  c 1 , ∀(x, t) ∈ D. f  (t)  4  1 0 z 2 t dx − 2c 1  1 0 a(z x ) 2 dx −  1 0 ( c 3  + 2u 2 1x )z 2 dx = 4  1 0 z 2 t dx + c 1 {f  (t) +  1 0 u 1x z 2 dx} −  1 0 ( c 3  + 2u 2 1x )z 2 dx = 4  1 0 z 2 t dx + c 1 f  (t) +  1 0 {c 1 u 1x − ( c 3  + 2u 2 1x )}z 2 dx. c 2 u 1 , u 2 |c 1 u 1x − ( c 3  + 2u 2 1x )|  c 2 , ∀(x, t) ∈ D. f  (t)  4  1 0 z 2 t dx + c 1 f  (t) − c 2  1 0 z 2 dx = 4  1 0 z 2 t dx + c 1 f  (t) − c 2 f(t).(2.4) f(t)  0 f  (t).f(t)  4  1 0 z 2 t dx.  1 0 z 2 dx + c 1 f  (t)f(t) − c 2 f 2 (t)   2  1 0 zz t dx  2 + c 1 f  (t)f(t) − c 2 f 2 (t)  (f  (t)) 2 + c 1 f  (t)f(t) − c 2 f 2 (t), ∀t ∈ (0, 1).(2.5) f  (t).f(t) − (f  (t)) 2  c 1 f  (t)f(t) − c 2 f 2 (t), ∀t ∈ (0, 1).(2.6) t 0 ∈ (0, 1) f(t 0 ) = 0 f(t) 0 f(t) > 0, t ∈ (0, 1) f 2 (t) f  (t).f(t) − (f  (t)) 2 f 2 (t)  c 1 f  (t) f(t) − c 2 , ∀t ∈ (0, 1).(2.7)  f  (t) f(t)    c 1 f  (t) f(t) − c 2 , ∀t ∈ (0, 1) ⇔ ( f  (t) f(t) )  − c 1 f  (t) f(t)  −c 2 , ∀t ∈ (0, 1) ⇔ e −c 1 t {( f  (t) f(t) )  − c 1 f  (t) f(t) }  −c 2 e −c 1 t , ∀t ∈ (0, 1) ⇔ {e −c 1 t f  (t) f(t) }   −c 2 e −c 1 t , ∀t ∈ (0, 1).(2.8) s = e c 1 t d ds  1 f df ds   − c 2 c 2 1 1 s 2 . ψ 1 (s) = ln(f s − c 2 c 2 1 ) (ψ  1 (s)  0) ln(exp{− c 2 c 1 t}f(t))  exp{c 1 } − exp{c 1 t} exp{c 1 } − 1 ln(exp{− c 2 c 1 0}f(0)) + exp{c 1 t} − 1 exp{c 1 } − 1 ln(exp{− c 2 c 1 1}f(1)).(2.9)  1 u i (x, t) ∈ V, i = 1, 2,      u t = u xx − uu x , (x, t) ∈ (0, 1) × (0, 1), u(0, t) = u(1, t) = 0, 0  t  1, u(x, 1) = ϕ i (x), 0  x  1, (i = 1, 2) ϕ 1 − ϕ 2  = {  1 0 (ϕ 1 (x) − ϕ 2 (x)) 2 dx} 1 2  ε u 1 (·, t) − u 2 (·, t)  e (c 2 /c 1 )[t−α] [2M] (1−α) ε α , ∀t ∈ [0, 1],(2.10)  ·  L 2 (0, 1) α = exp{c 1 t} − 1 exp{c 1 } − 1 ; c 1 = 3M 2 + 9 4 M c 2 = 3M 3 + 35 4 M 2 + 9 2 M. 2 u i (x, t) ∈ V, i = 1, 2,      u t = ((x + t + 1)u x ) x − uu x , (x, t) ∈ (0, 1) × (0, 1), u(0, t) = u(1, t) = 0, 0  t  1, u(x, 1) = ϕ i (x), 0  x  1, (i = 1, 2) ϕ 1 − ϕ 2  = {  1 0 (ϕ 1 (x) − ϕ 2 (x)) 2 dx} 1 2  ε u 1 (·, t) − u 2 (·, t)  e (c 2 /c 1 )[t−α] [2M] (1−α) ε α , ∀t ∈ [0, 1],(2.11)  ·  L 2 (0, 1) α = exp{c 1 t} − 1 exp{c 1 } − 1 ; c 1 = 3M 2 + 21 2 M + 1 c 2 = 3M 3 + 14M 2 + 7 2 M. a(x, t) = 1, (x, t) ∈ D c 3 = 3M 2 + 3M  = 2 3 c 1 = 3M 2 + 9 4 M c 2 = 3M 3 + 35 4 M 2 + 9 2 M a(x, t) = x + t + 1, (x, t) ∈ D c 3 = 3M 2 + 5M  = 2 c 1 = 3M 2 + 21 2 M + 1 c 2 = 3M 3 + 14M 2 + 7 2 M. ε L 2 (0, 1) L 2 (0, 1) t ∈ (0, 1] V a(x, t) c 1 c 2 M L 2 [1] K. A. Ames and B. Straughan, Aca- demic Press, San Diego, 1997. [2] A. Carasso,  Journal of Mathematical Analysis and Applications, 59 (1977), 169-209. [3] L. E. Payne and B. Straughan, , Int. J. Nonlinear Mech., 24 (1989), 209–214. [4] S. M. Ponomarev, Soviet Math. Dokl., 33 (1986), 621-624. [5] A. Friedman, Prentice-Hall, Engle- wood Cliffs, N. J., 1964.      u t = (a(x, t)u x ) x − uu x , (x, t) ∈ (0, 1) × (0, 1), u(0, t) = u(1, t) = 0, 0  t  1, u(x, 1) = ϕ(x), 0  x  1.