Differensial tenglamalar kafedrasi

^t) = f / , ^exp^{

Аж a it


}

4a t


h===^exP{-^{(X S }(}-^s))ds + j-j=¹

-r^J4жа t ^{4a t} о д/4ж


4жа t


exp{- ——^S^ }


4a²t

ntegral ostida juft va toq funksiyalarning ko ’paytmasi turibdi , shuning uchun u nolga teng . Chegaravi shart bajarilayapdi. endi yechim uchun to ’liq formulani olamiz:

+f I / \ 2 +r i / \ 2

■ f . ^exP^{- ll }(fr{s)^ds + f ^exP^{{- (x}—^s)-}^)ds

o
+r ₁

= f ¹

о у/4ж a²/


exp {-^(x-At} - exp {- ^}


4a t


4a t


(f(s)ds.

^4жa²t ^{4a t} o^4жa²t ^{4a t}

D
+r 2

u( x, t) = f-j= 0 v4ж^


^exp^{{- (x}^^sr^{} - ex}p^{{- (X}+^{1 }}


4a t


4a t


о л/4ж a t

buyarim to ’g’ri chiziqda birinchi chegaraviy masalaningyechimi bo’ladi.


(2.4.3)

emak,

2. 
   Yarim to’g’ri chiziqda ikkinchi chegaraviy masala
   

Yarim to ’gri chiziqda ikkinchi chegaraviy masala qo ’ydagi ko ’rinishga ega:

^utt = ^{a u}xx^{, x} > 0>t > 0;

u
(2.4.5)

_x(0,t) = 0, t > 0;

u(x,0) = ф(x), x > 0.

Yana yechimni topish uchun boshlang ’ich shartni beruvchi funksiyani endi juft qilib davom ettiramiz:

f ф(x), x > 0;

Ф(х) = \ ^{ф( ), ;}

[ ф(-x), x < 0.

Boshlang’ich shartni o ’zgartirib , quyidagi koshi masalasini olamiz:

U(0,t) = 0, t > 0;

U(х,0) = ф(х), -да< x < +да.

Xuddi shunday uningyechimi

(
(x - 2 )²
\ +f 1 _f (x - 5)

^{U t)}= J I — ^eXP ^{— ~j~T7

-^/4^a²t ^{4a t}

funksiya bo ’ladi.

Aytaylik (x, t)g(r⁺ x R⁺ ) da u(x, t) = U(x, t) bo’lsin. Yana

u_t = a u_xx , x > 0, t > 0;

u(x,0) = ф(x), x > 0.

ekanligi aniq.

C
+да -j

⁽x^t) = ^Ux^{(x, t)}= J i

—»y 4л a¹


C (x — s)1 (x — s)²

'exp{— у— }Ф(5)^5 ^


t


u


+^ -j

(0, t )=Ux (0, t )=h=

—X 4xa"


V 2a ²t

f


4a²1


t


2 I 2

— Iexp{ — }Ф(5)^5 Vt > 0.

V2a t J 4a t

hegaraviy masalaning bajarilishini tekshiramiz:

Hosil bo ’lgan integral ostida 2 ta juft va bitta toq funksiyaning ko ’paytmas turibdi, demak u nolga Doiradi. Chegaraviy shart bajarilmoqda. (2.4.4) ning yechimi uchun qo ’ydagi formilani xosil qilamiz:

+
^{(x — s)2}

u
f 1 (x + s)² 0 1 (x — s)<sup>2

(</sup>x^t) = J , ₉ ^exP^{{— (} . ₂^{2 }(}s)ds + J , ^exP^{{— ( 2 }ф(—}s)ds =

o
+w -j

hr


exp^{{— (x} ₂^{S) }} + exp^{{— (x} + ₂^{5) }}


ф(5^5.


о т]4лa"t

Bu yarim to’g’ri chiziqda 2-chegaraviy masalaning yechimidir.


4a²1


4a ²t

-\]4xa²t ^{4a t} —_му4лa²t ^{4a t}

5. 
   §. Maple paketi orqali yarim chegaralangan sohada issiqlik
   o’tkazuvchanlik tenglamasini Fur’ye usuli (o’zgaruvchilarni ajratish usuli)
   

yordamida yechish

Y
_S

dt


u( t, x ) = a²


fd²


\

J

arim chegaralangan sterjenda issiqlik tarqalish jarayonini qaraymiz [6-8]. Buning uchun quyidagi

bir jinsli tenglamani

u(0, x) = f( x)

boshlang’ich va quyidagi:

^{k u(}t⁰⁾ 1 = ^{hl (u(}t^{0) - T1)}

^{-k f}^ ^u(t L⁾1 = ^h2 ^(u(t L⁾ - ^T2 ⁾

chegaraviy shartlar bilan yechamiz (k - issiqlik o’tkazuvchanlik koeffisiyenti, hl va h2 - sterjenning chetlarida issiqlik almashinish koeffisiyentlari, Tl va T2 - sterjen chetlaridagi (chegaralaridagi) temperaturalar).

• 
  restart;
  

Bir jinsli tenglama va uning yechimini o’zgaruvchilami almashtirish usuli bilan yechamiz:

• 
  P
  PDE
  
  
  d_
  
  dt
  
  
  u( t, x )
  
  
  = a
  
  
  2
  
  
  f d^_ _Kdx²
  
  
  \
  
  u( t, x )
  
  J
  
  
  struc := (u(t, x) = T(t) X(x))&where
  
  
  d
  
  ^{dt,^T('⁾
  
  
  d<sup>2
  
  C</sup>-^T('⁾’ dx2^{X( x)}
  
  
  C ^{X( x)}
  
  
  a