Fizika-matematika fakulteti

Yechish. Tenglikning ikkala tomonini ikki marta differensiallab, quyidagilarni hosil qilamiz:
x
^{u(x) cosx}^{cos(xt)u(t)dt,} 0

(1.17)

ketma-ket

x
^{u(x) sin xu(x)}^{sin( xt)u(t)dt.} 0
x
^{Berilgan differensial tenglama va oxirgi tenglikdan} ^{sin( xt)u(t)dt ni}
0
qisqartirib, izlanayotgan funksiyaga nisbatan ushbu u^(x) 0 oddiy differensial tenglamaga kelamiz. Berilgan differensial tenglama va birinchi tenglikdan ushbu u(0) = 0 va u(0) = 1 boshlang„ich shartlarni topamiz. Natijada y(x) = x yechimga kelamiz.

7-Misol. Ushbu
x
^{u(x) x2sin x1}^{(xt)u(t)dt (1.17)} 0

16

integral tenglamani yeching.
Yechish. Tenglikning ikkala tomonini ikki marta ketma-ket differensiallab, quyidagilarni hosil qilamiz:
x
^{u(x) 12cosx}^{u(t)dt ,}
0
u^(x) 2sin xu(x).
yoki uni standart shalda yozsak, quyidagi differensial tenglamaga kelamiz: u^(x)u(x) 2sin x
Berilgan tenglama va birinchi tenglikdan x = 0 da ushbu u(0) = –1 va u(0)=3 boshlang„ich shartlarni topamiz. Hosil bo„lgan Koshi masalasining yechimi quyidagi natijani beradi: y(x) = 2sinx +(x - 1)cosx.


1.3. Chiziqli chegaraviy masalalarning Fredgolm integral tenglamasi bilan bog‘liqligi





  
Ushbu L upx u^p^x u^qx u f x _(1.21) ikkinchi tartibli o„z-o„ziga qo„shma differensial tenglama uchun quyidagi bir jinsli chegaraviy masalani qaraymiz:



à uuau^a0,_
Ã_b u₀ub₁u^b0,_{ (1.22)} 0, 0 ,
_bunda px , p^x , qx , f xa,b_{da aniql}^a_{ngan uzluksiz fu}ⁿ_{ksiyalar hamda} px0, xa,b .
Ta’rif. Agar Gx,tfunksiya quyidagi 1)-4) shartlarni qanoatlantirsa,
u holda unga (1.21)-(1.22) chegaraviy masalaning Grin funksiyasi
deyiladi:
₁₎ Gx,ta,b² _{da aniqlangan uzluksiz fu}ⁿ_{ksiya; 2)} L G x,t 0 _agar x t _bo„lsa;
3) Ã_a Gx,t0, Ã_b Gb,t0;

1

 
₄₎ G_xt 0,tG_xt 0,t_pt^_.
Agar Gx,tGrin funksiyasi topilgan (ma‟lum) bo„lsa, (1.21)-(1.22) chegaraviy masalaning yechimi ushbu

17


b
^{ux}^{Gx,tf tdt (1.23)} a


formula bilan ifodalanishini tekshirib ko„rish qiyin emas. Bundan tashqari




G x,t Grin funksiya simmetrik shartni qanoatlantiradi, ya‟ni Gx,tG t,x , x,ta,b.


Endi Shturm-Liuvill chegaraviy masalasini qaraymiz:



.


px u^qxxu 0, _(1.24) Ã_a uÃ_b u0 px0, x0(1.25)
_Ushbu x u _{hadni erkin had deb hisobl}^ab_{, (1.24) ni integrallasak, (1.23)}
formulaga ko„ra quyidagi bir jinsli Fredgolm integral tenglamasini hosil qilamiz


b


^{ux}^{Kx,t utdt, (1.26)} a

t .
bunda Kx,tGx,t _(1.27) Ko„rinib turibdiki, Shturm-Liuvill masalasining xos qiymati (1.26) integral tenglamaning ham xos qiymati bo„lar ekan.
(1.26) teglamani simmetrik yadroli integral tenglamaga keltirish
mumkin. Haqiqatdan,
zxuxx
deb belgilasak, (1.26) va (1.27) formulaga ko„ra


b
^{zx}^K*^{x,tztdt,} a
bunda yadro
K^*x,tGx,txt
simmetrikligi ko„rinib turibdi. Bu yerdan esa, yadroning simmetrikligiga ko„ra, barcha xos qiymatlar haqiqiy bo„lishi kelib chiqadi.


Mashqlar

1) Boshlangich shartlari bilan birga berilgan quyidagi differensial tenglamalarga mos bo„lgan integral tenglamalar tuzilsin.

1. y–y = 0, x = 0 bo„lganda y = 1.

18


x
^{Javobi: u(x) 1}^u(t)dt 0
2. y–5y+6y = 0, x = 0 bo„lganda y = 0, y= -1.
x
^{Javobi: u(x) 56x}^{(6x6t 5)u(t)dt.} 0
3. yy = 0, x = 0 bo„lganda y = 0, y= -1.


x
Javobi: ^{u(x) x}^{(xt)u(t)dt.} 0
4. y–xy = 0, x = 0 bo„lganda y = 0, y= -1.


x
^{Javobi: xu(x) x}^(x2 ^{xt 2)u(t)dt.} 0
5. y–ycosx = x, x = 0 bo„lganda y = 0, y= 0.
x
^{Javobi: u(x) xcosx}^{(xt)cosxu(t)dt.} 0
6. y–ysinx+y = 1, x = 0 bo„lganda y = 0, y= 1.


x
^{Javobi: u(x) xsin x}^{(sin xxt)u(t)dt.} 0
7. y+n²y = hsinpx ( ^{p n}), x = 0 bo„lganda y = a, y= d.
x
^{Javobi: u(x) hsin px6n}2^xan2 ^n2 ^{(xt)u(t)dt.} 0
8. y+2hy+n²y = 0, x = 0 bo„lganda y = a, y= d.

 
x
^{Javobi: u(x) (dn}2^x2dhan2^)ⁿ2^{(xt)2hu(t)dt.} 0


3
9. y^₂ xy^2 , y(0) = –3, y(0) = 1, y(0) = –1 . 10. y^xxy² y^2 , y(0)= 1, y(0) = y(0)= 0 .
_{11. y}_{xy e}^x _{, y(0) = 1, y(0) = y(0)= 0} ^.
12. y^IV y^y 0 , y(0) = y(0) = y(0)= 0, y(0) = 1.

2) Quyidagi integral tenglamalarni oddiy differensial tenglamalarga keltirib yeching:

19
x
^{1. u(x) e}x ^^u(t)dt. 0
x
^{2. u(x) 1}^tu(t)dt. 0

1


x
3. u(x) _1x2 0sin( x t)u(t)dt.


x
^{4. u(x) e}x ^cosx^cosxe(xt)^u(t)dt. 0
x
^{5. u(x) 4e}x ^3x4^{(xt)u(t)dt.} 0