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2Vaqt bo'yicha kasr tartibli uzulishli koeffitsiyentli diffuziya tenglamasi uchun aralash masala
Shahnoza Farhod qizi Jumayeva Osiyo Xalqaro Univeristeti
Annotatsiya: Bu maqolada diffuziya koeffitsiyenti uzulishga ega bo'lgan kasr tartibli diffuziya tenglamasi uchun aralash masala qaralgan.
Kalit so'zlar: kasr tartibli diffuziya tenglamasi, anomal diffuziya, 1-tur uzulish, Kaputo hosilasi
A mixed problem for the diffusion equation with a fractional-order discontinuity coefficient in time
Shahnoza Farhod kizi Jumayeva Asian International University
Abstract: In this article, mixed problem for the time-fractional diffusion equation with discontinuous diffusion coefficient is studied.
Keywords: fractional diffusion equation, anomalous diffusion, discontinuity, Caputo derivative
So'nggi yillarda klassik diffuziyadan farqli bo'lgan anomal diffuziya hodisalar ko'p kuzatilmoqda.Shu jumladan, uzulishga ega bo'lgan holat. Diffuziya koeffitsiyenti uzulishga ega bo'lgan diffuziya tenglamasi Hald[6], Suzuki va Murayama[7], Pierce[8] maqolalarida uchraydi. Kasr tartibli koeffitsiyentlari uzluksiz bo'lgan anomal diffuziya tenglamasi uchun teskari masala Cheng[1] maqolasida batafsil ko'ril chiqilgan. Ushbu maqolada diffuziya koeffitsiyenti ma'lum bir nuqtada uzulishga ega bo'lgan kasr tartibli anomal diffuziya tenglamasi uchun aralash masalani ko'rib chiqamiz.
Biz quyidagi kasr tartibli differensial tenglamani ko'rib chiqamiz:
« d d tu ( x, t ) = —
dx
du )
P ( x ) — ( x,t )
dx
0 < x < l, 0 < t < T
(1)
u(x,0) = S(x) Ç2)
d u d u
— (0, t) = — (l, t) = 0, 0 < t < T d x d x (3)
Bu yerda T > 0,l > 0 tayinlangan x^ Dirakning delta funksiyasi va ö'u(x'') -u (x'' ) funksiyaning Kaputo ma'nosidagi kasr tartibli hosilasi:
a 1 ' — a ÖU (X, 5)
ö tu(x,t) =-I (t — s) -ds
' r(1 — a) 0 ös (4)
Bizda 0 < a < 1 va p (x) funksiya x0 nuqta uzulishga ega, ya'ni
, W» ^„„ü^i w» 0
2
f a2, 0 < x < x„ , p (x) = \ 2 0
[ b , x0 < x < l, a ^ b ^^
Biz (2) boshlang'ich shart Dirakning delta funksiyasi bo'lganligi sababli bu masala uchun kuchsiz yechim qaraymiz.Buning uchun biz (1)-(3) ga kuchsiz yechim uchun tegishli ta'rifni kiritishimiz va yechimni kuchsiz yechim ekanligini tekshirishimiz kerak. Kuchsiz yechim ta'rifi kiritilishi uchun zarur bo'lgan funksional fazolarni kiritamiz.
Birinchi biz L (0'1) da Ap operatorni
f a V "(x), 0 < x < x0 APV (x) = \ 2
[b v"(xX x0 < x < l (6)
ko'rinishida va uning aniqlanish sohasi D (Ap) ni quyidagicha
Ve C[0,l],
D (Ap) = \We C '[0, x C'[ x 0, l ]
V e C 2(0, x0),v e C 2( x 0, l)
kiritamiz.
V(x0 - 0) = V(x0 + 0)
aV(x0 - 0) = bV(x0 + 0) \ V (0) = V (l) = 0
Bundan ma'lumki, Ap operatorning xos sonlari haqiqiy va oddiy
sonlar bo'ladi va quyidagi shartni 0 = A, < A, < ..., lim A = .
1 2 7 n in\
n(7)
qanoatlantiradi.
Ap operatorning aniqlash sohasi L2(0l^ fazoning qism fazosini bo'ladi, ya'ni
D ( Ap ) c L2(0, l)
Lemma 1. Ap operator uchun quyidagilar o'rinli:
1) (Apy, z) = (y, ApZ), Vy, z e D (Ap)
p 112
2) (APy'y) ^ CH '
Isboti.
1. L2(0l^ da skalyar ko'paytmani qo'llaymiz.
l x0 l (Apy, z) = J Apyzdx = -J a2y"(x)z(x)dx - J b2y"(x)z(x)dx =
0 0 x0
x0 1 ' = -a2y'(x)z(x)|o° + J a2y'(x)z' (x)dx -b2y'(x)z(x)| + J b2y'(x)z'(x)dx =
0 0 x 0
x0
= -a2y'(x0 - 0)z(x0 - 0) + a2y'(0)z(0) + a2y(x)z' (x)|o° - Ja2y(x)z" (x)dx -
0
/
-b2y'(x0 + 0)z(x0 + 0) + b2y'(l)z(l) + b2y(x)z' (x)[ - J b2y (x)z" (x)dx
x0
x„
Endi D (Ap) operatorning aniqlash sohasidagi Xo nuqtadagi y (0) z (l ) 0 shartdan foydalansak,
(Apy, z) = - a2 y' (x o - 0) z (Xo - 0) + a2 y (Xo - 0) z' (Xo - 0) + b2 y' (Xo + 0) z (Xo + 0) -
X0 l -b2y(x0 + 0)z'(x0 + 0) - J a2y(X)z"(z)dX - J b2y(x)z"(z)dX =
= a 2 [ y ( X o - 0) y' ( X o - 0) - y' ( X o - 0) y ( Xo - 0) ] + + b2 [y(Xo + 0)y'(Xo + 0) - y'(Xo + 0)y(Xo + 0)] + (y, Apz)
hosil bo'ladi va Xo nuqtadagi ulash shartlani foydalanib,
( Apy, z ) = ( y, Apz )
ega bo'lamiz.
2. L2(0l) da skalyar ko'paytmani qo'llaymiz.
l X0 l (Apy, y) = J Apy • ydX = -J a2 y"( x ) y ( x ) dX - J b2 y" ( x ) y ( x ) dX =
0 0 x0
Xo l l
y' (x) y (x) + J a2 y' (x) y' (x) dx - b2 y' (x) y (x) [ + J b2 y' (x) y' (x) dx
0 x0
x0
= - a2 y'(x - 0) y (x0 - 0) + 0 - 0 + b2 y'(^ + 0) y (x0 + 0) + a2 J y " (x)dx + b2 J y " (x)dx =
0
X0 l = a2 J y,2(x)dx + b2 J y'2(x)dx
2
= - a ,
0
1 2,2,
-min{a ,b }
Endi 2 tanlaymiz , quyidagi o'rinli bo'ladi:
xo
(Apy, y) = a2 J y'2(x)dx + b2 J y ,2(x)dx > — min{a2, b2} Jy ,2(x)dx = — min{a2, b2}||y'|
0 x0 2 0 2
Quyidagi baholashlarni bajaramiz:
i i i x fx \2 fx \2 fl \2 fx ^ y2(x) = 2Jy(x)y'(x)dx < 2 Jy'2(x)dx Jy2(x)dx < 2 Jy,2(x)dx Jy2(x)dx
lL2(0,l)
v o
v o
v 0
v 0
o
x
0
x
0
l f l \ 2 l f x \ 2 ........
J y 2( x ) dx < 2 J y'2( x ) dx J J y 2( x ) dx dx < 2\\y IIL (0,lJ|y|L2 (0,l) l
0 V 0 J 0 V 0 J
llyIL2 (0,l) < 2 l lly IL2 (0,l ) II yIL2 (0,l)
412
■Jy2(x)dx < jy'2(x)dx
Bundan quyidagi baholash kelib chiqadi:
min{a2 b2) , (Apy,y) >-}Jy'2(x)dx >
min{a2,b2} l
812
J y 2(x)dx = C ||y|
Il2 (0,l)
C =
min{a2, b2} 8?
mavjudki, L 2(0'1) da quyidagi tengsizlik o'rinli:
Ya' ni shunday
(Apy, y) > C||y||2
Lemma isbotlandi.
A
Biz p operatorning spektri sof diskret bo'lgani va faqat xos qiymatlaridan
A
tuzilgan o'z-o'ziga qo'shma kengaytirib, bu kengaytmani p bilan belgilaymiz.
A = A
p p
Lemma 2. A p operator x0 nuqtadagi ulash shartlar va
(/ - x0 )a _ " _ D xQb k
Shart orqali quyidagi xos sonlarga ega bo'ladi:
2 7 2
-k k
0 a 2 2 A =-k n =
n 4x02 4(l - x0)2 (8) An xos songa mos xos funksiyani Vn bilan belgilaymiz va u v(0) =1 ni
qanoatlantiradi.
Bizda V«(x) xos funksiya
cos
JA JA
y n / 1 \ \ n
-(l - x )cos-.
Vn(x) = \
cos
[ a
x cos
JA
\ n
(l - x),
0 < x < x„
x„ < x < l
ko'rinishda bo'ladi. (p" xos funksiyalar L da to'la orthogonal bo'ladi.
- L (0, / } dagi skalyar ko'paytinani aniqlaydi va uning ko'rinishi quyidagicha
l
( f, g ) = J f ( x ) g ( x ) dx
1
0
bo'ladi. Bu skalyar ko'paytma orqali L (0, ( ) da normani kiritamiz
i
\\ <p H ^ = \\ <p H =
Quyidagi belgilashni kiritamiz: Pn . U holda xos funksiyalar kengaytmasi
to
¥ = ZPn (¥ ,Vn )Vn
n \ I ^ / n ' ' n n = 1
Bundan tashqari,p« 'n e asimptotik xarakteriga ega. Lemma 3. Shunday c° > 0 mavjudki,
pn = c0 + O(1)
2 a
Endi biz ixtiyoriy o'zgarmas M > 0 ni tanlab L (0'l) da p■M operatorni
d ( d \
dx
( ap m¥ )( x ) = "— p ( x )— ¥ ( x ) + m ¥, 0 < x <
y dx j
i
i
jd{apm) = \¥ e h\o,0;—(0) = —(0 = o 1.
[ [ dx dx J
(m ) 2 Tiyf []
aniqlaymiz. Bu operatorning xos sonlarning to'plami » = A„ + M » e bo'ladi
2<a/ ' n d D i A1 ^
va » > 'n e . Endi > 0 uchun quyidagi - pM ' funksional fazoni
D (a;m) = ty e l\ojy±pn i ar n (v.pj l'< «}
d(a;m )
kiritamiz. Bu ^ pM ' fazo quyidagi norma bilan Banax fazosi bo'ladi:
i
"¥l I. ( ^ ,= {]CPn^nM 'I2;|(¥ (P } 2
3
O- 0 ";<7 u D ( A; ) = H2; (0, l) u tl •
Biz agar 4 bo lsa, v p-M ' J ga ega bo lamiz.
D {Apm )<= L (°> iigidan; o'z-o'ziga qo'shma L ni aniqlashdan,
D( )cL2p (0, l) D( A; )) , tl . D ( A";) = ( D( A; )) ,, ( p- ) ( , ) ( ( p- )) ga ega bolamiz. Biz ( p-) ( ( p-)) deb
f e( D ( )) ¥g( D ( A \ )) f
belgilash kiritamiz. - va - uchun ning ¥ ga ta'sirini
{ro
Z p |TM)|
L-i ~ n \ n I n =1
(M ) |-2; j2 I n I ' ' n
norma
< f ¥ > 1 1 1 -, • d (a-;,) _ d(apm)=
-*< f ¥>k bilan belgilaymiz. ( p-M ) fazo
1
0 < t < —
bilan Banax fazosi bo'ladi. Endi 2 ni tayinlaymiz. Sobolevning joylashtirish
f _1 - T ^ r _1_ T A ™
5 = Y p v
p, M ' n ' n
teoremasiga ko'ra, ^ ' va ^ ^ da n 1 ga ega bo'lamiz. Biz
5 e D I ApM I D
' _H ' belgilash kiritib olamiz. Brezis[4] ga ko'ra agar f e L (0'1 ^ va
W G D (Ap'M ) bo;lsa, u holda < f >= (f bo'rinli.
Ta'rif. Agar u funksiya quyidagi shartlarni bajarsa, (2.1.1)-(2.1.3) masalaning kuchsiz yechimi deyiladi:
u(-,t)G L2(0j), 0<t<T,
1
u e C
f 1_ T
[0'T];DI ApM
! 5 ( _ 1_T ^
I- u' ' Ap ' e C ^ (0, T ]; D (Ap ' )j,
lim II u (■' t) _ 5 II , , = 0
t ^ 0 [ _ — T !
D | Vj
<ö> (■' t )'^> + (u (■' t)' Ap^ ) = 0' t e (0' T ],W e D (Af ).
r r _-OA
u(x't) e C | [0'T];D i Ap4_ I |
Teorema. ^ ^ ^ funksiya (1)-(3) masalaning kuchsiz
yechim bo'ladi va uning ko'rinishi quyidagicha bo'ladi:
u (x' t) = Y PnEaA _aJa )vn (x)-
n=1
Bu yerda « > 0 va ^ e D uchun E a-ß ^ Mittag-Leffler funksiyasidir.
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