Issiqlik tarqalish tenglamasi uchun boshlang‘ich-chegaraviy aralash masalalarni Furye usulida yechish
![boshlang’ichshartlarniqanoatlantiruvchi u(x, t )yechiminitoping.
Y echish: (1 .1.11),(1 .1.12),(1 .1.13)masalaniFuryeusulidayechib,uningyechimini
u(x, t ) =∞ a
n e − ( anπ l ) 2 t · cos πnx l ko’rinishidaifodalay
miz,buyerda
P
n =0
l
a
0 =1 l
Z
xdx = l 2, (1 .1.14)
0
l
a
n =2 l
Z
x cos πnx l dx, n ∈N (1 .1.
15)
0
(1 .1.15)nisoddalashtiribquyidagi(1 .1.16)nihosilqilamiz:
a
n =2 l
| x = l x =0 −l
πn
Z l
sin πnx l
(πn )
2 cos
πnx
| x = l x =0 = lπnx ·
sin πnx
dx =2 lπn
sin πn +
0
=
l
2( πn )
2[cos
πn −cos 0] =
l
2( πn ) 2[(−1) n −1] =
0,aga rn = 2k
4l −(πn )
2,agar n =
k= 0 , 1 , 2 ...
(1 .1.14) va (1 .1.16) ga asosan (1 .1.11), (1 .1.12), (1 .1.13) masalaning yechimini
quyidagicha ifodalaymiz:
l 4l
∞ 1
−
(2 k +1) aπ 2 · (2 k + 1) πx XX
l
t
2 (πn )
2
k =0 (2 k + 1) 2 l
2-misol.
u
t = a 2 u xx , 0 < x < l, t > 0
tenglamaning
u(0 , t ) = u
x (l, t ) + hu (l,
t) = 0 ,
t > 0 (1 .1.17)
chegaraviyva u(x, 0) = 1boshlang’ichshartlarniqanoatlantiruvchi u(x, t )yechiminito-
ping.
Y echish: (1 .1.11)va(1 .1.17)ni(1 .1.4)gaqo’yibquyidagilarni:
T
′(t) + a 2 λ 2 T(t) = 0 (1 .1.18)
X
′′ (x) + λ 2 X(x) = 0 (1 .1.19)
X(0) = 0 ,X
′ (l) + hX (l) = 0 (1 .1.20)
hosilqilamiz.Endi(1 .1.19)va(1 .1.20)masalaniyechamiz.
(1 .1.19) −oddiydifferensialtenglamaningumumiyyechimi
X(x) = c
1 cos λx + c 2 sin λx (1 .1.21)
6](https://docx.uz/documents/8fd85d81-468c-4319-ba11-a38b548910a9/page_8.png?v=1)
![X(0) = c1 = 0 , c 2 [λ cos λl + h sin λl ] = 0 .
c
2 ̸= 0 ,λ cos λl = −h sin λl ⇒ h tan λl = −λ ⇒ tan µ n = −µ n hl , µ n =λ n
l
(1 .1.22)
(1 .1.22) tenglamaning musbat yechimlarini λ
n = µ n l deb belgilasak,unga xos
funksiya
X
n (x) = sin λ n x, n ∈N (1 .1.23)
ko’rinishdabo’ladi.
(1 .1.18)tenglamaning λ
n gamosyechimlari
T
n (t) = a n e − λ 2 n a 2 t (1 .1.24)
ko’rinishdabo’ladi,buyerda a
n −ixtiyoriyo’zgarmaslar.
Demak,(1 .1.11),(1 .1.17)va u(x, 0)= ϕ(x)=1masalaningyechimi(1 .1.23)va
(1 .1.24)gako’raquyidagigateng:
∞
u(x, t ) = X a
n e − λ 2 n a 2 t sin λ n x (1 .1.25)
n =1
buyerda
l
a= 1 Z ϕ(x) sin λxdx,
n
∥sin λ
n x∥ n
0
∥sin λx ∥= Z
l sin
2 λxdx = l(h 2 +
λ
2 n ) + h
.n
0 n
ϕ(x) ≡1tengliknie’tiborgaolib a
n koefitsiyentnihisoblaymiz:
2( h
2 + λ 2
n
)
Z l
sin λ
n xdx =
2( h 2 +
λ
2 n )
= −
a
n = l(h
2 + λ 2 n )
+ h
0
λ n [l(h 2 +
λ
2 n ) + h ]
(1 .1.26)
2( h
2 + λ 2 n )
λ
n [l(h 2 + λ 2 n ) + h ][cos λ n l −cos 0] =
2( h 2 + λ 2 n )
λ
n [l(h 2 + λ 2 n ) + h ][1
(1 .1.26)ni(1 .1.25)gaqo’yib,qo’yilganmasalaningyechiminitopamiz:
u(x, t ) = 2
∞ λ
n [l(h 2 + λ 2 n ) + h ][1 −cos
µ
n ]e − λ 2 (h 2 + λ 2 n )
Xn =1
7](https://docx.uz/documents/8fd85d81-468c-4319-ba11-a38b548910a9/page_10.png?v=1)
![1.2.Birj inslibo’l maganissiqlik o’t k azuv chanl ik t englamasi.
Ushbu
u(x, t ) = a2 u xx + f(x, t ) (1 .2.1)
tenglamaning
u|
x =0 = ψ 1 (t), u| x = l = ψ 2 (t), t ≥ 0 (1 .2.2)
chegaraviyva
u|
t =0 = ϕ (x) (1 .2.3)
boshlang’ichshartlarniqanoatlantiruvchiyechiminitopi
ng. (1 .2.1),(1 .2.2),(1 .2.3)masalaningyechimini
u(x, t ) = ϑ (x, t ) + U (x, t ) (1 .2.4)
ko’rinishdaizlaymiz,buyerda
U(x, t ) = ψ
1 (t) + [ ψ 2 (t) −ψ 1 (t)]x l , (1 .2.5)
ϑ(x, t )−funksiya ϑ
t = a 2 ϑ xx +F(x, t ) tenglamaning bir jinsli chegaraviy shartlarni,
ya’ni ϑ|
x =0 =ϑ| x = l =0 ,t> 0hamda ϑ| t =0 =ϕ(x)boshlang’ichshartniqanoatlantiruvchi
yechimidir.
Xuddi shunday (1 .1.1) tenglama uchun Furye usuli yordamida II , II I aralash
masalalarni yechishmumkin.
1.3.To’g’riburchak liplast ink adaissiqlik t arqalishihaqidagi masala.
Ushbu
u
t = a 2 (u xx + u yy ) (1 .3.1)
tenglamaning T={(x, y, t ):0 <x<p, 0 <y<s, t> 0}sohadaaniqlangan,uzluksiz va
u|
x =0 = u | x = p = 0 , 0 ≤ y≤ s, t ≥ 0; u| y =0 = u | y = s = 0 , 0 ≤ x ≤ p, t ≥ 0 (1 .3.2)
chegaraviyva
u|
t =0 = ϕ (x, y ),0 ≤ x ≤ p, 0 ≤ y≤ s (1 .3.3)
boshlang’ichshartlarniqanoatlantiruvchiyechiminitopi
ng. (1 .3.1),(1 .3.2),(1 .3.3)masalaningyechimini
u(x, y, t ) = X (x)Y (y)T(t) (1 .3.4)
ko’rinishdaizlatymiz,buyerda X(x),Y (y),T(t)funksiyalarmosravishdaquyidagi
X
′′ (x) + λ 2 X(x) = 0 ,Y ′′ (y) + µ 2 Y (y) = 0 ,T ′ (t) + a 2 (λ 2 + µ 2 )T(t) = 0
8](https://docx.uz/documents/8fd85d81-468c-4319-ba11-a38b548910a9/page_11.png?v=1)
![Xk (x) = sin λ k x = ⇒ λ k cos λ k x + h sin λ k x = 0 = ⇒ λ k
h= −tg λ
k p
.
(∗)
Y
n ( y) = co
, µ n =(2 n −1) π2s
T
kn (t) = A kn e − a 2 ( λ 2 k + µ 2 n ) t , k, n ∈N .
u(x, y, 0) = f(x, y )boshlang’ichshartdanquyidaginihosilqilamiz:
∞
A
kn sin λ k · cos (2 n −1) πy 2s
= f(x, y ) X
k,n =1
p
|| X
k (x)|| 2 =
Z
sin 2 λ k xdx = p
2−sin 2 pλ
k
=p 2− tg pλ k
2(1 +
tg
2 pλ k )= [( ∗)]0
=p 2+ 2( h
2 +
λ
2
hλ
k
.
Uholda A
kn koeffitsiyentlarquyidagichaaniqlanadi:
4( h
2 + λ 2 k ) p Z
s · (2 n −1) y Z
A
kn = (p(h
2 + λ 2 k )
+ hλ
k )s 0 0
f(x, y ) sin λ k x
cos
dxdy.
Demak,berilganmasalaningyechiminiquyidagichabo’ladi:
u(x, y, t ) =
∞ A
kn e − a 2 ( λ 2 k + µ 2 n ) t ·
sin λ
k x · cos (2 n
. ▲ Xk,n =1
2.3.4-masala. T={(x, y, t ):0 <x<p, 0 <y<s, t> 0}sohadaquyidagiaralash
masalaningyechiminitoping.
u
t = a 2 (u xx + u yy ) + f(x, y, t )
u(0 , y, t ) = u
x (p, y, t ) = 0 ,
u(x, 0 , t ) = u (x, s, t ) = 0 ,](https://docx.uz/documents/8fd85d81-468c-4319-ba11-a38b548910a9/page_42.png?v=1)
MUNDARIJA
KIRISH
ASOSIY QISM
I bob. Bir o’lchovli issiqlik o’tkazuvchanlik tenglamasi uchun aralash masalalar
1.1. Bir jinsli issiqlik o’tkazuvchanlik tenglamasi
1.2. Bir jinsli bo’lmagan issiqlik o’tkazuvchanlik tenglamasi
1.3. To’g’ri burchakli plastinkada issiqlik tarqalishi haqidagi masala
II bob. Furye usuli yordamida aralash masalalarni yechish
2.1. Bir jinsli masalalarni yechish
2.2. Bir jinsli bo’lmagan masalalarni yechish
2.3. To’g’ri burchakli plastinkada issiqlik tarqalishi masalalarini yechish
XULOSA