Yechish. Tenglikning ikkala tomonini ikki marta differensiallab, quyidagilarni hosil qilamiz:
x
u(x) cosx cos(xt)u(t)dt, 0
(1.17)
ketma-ket
x
u(x) sin xu(x) sin( xt)u(t)dt. 0
x
Berilgan differensial tenglama va oxirgi tenglikdan sin( xt)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) x2sin x1 (xt)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) 12cosxu(t)dt ,
0
u(x) 2sin xu(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 upx upx uqx u f x (1.21) ikkinchi tartibli o„z-o„ziga qo„shma differensial tenglama uchun quyidagi bir jinsli chegaraviy masalani qaraymiz:
à uuaua0,
Ãb u0ub1ub0, (1.22) 0, 0 ,
bunda px , px , qx , f xa,bda aniqlangan uzluksiz funksiyalar hamda px0, xa,b .
Ta’rif. Agar Gx,tfunksiya quyidagi 1)-4) shartlarni qanoatlantirsa,
u holda unga (1.21)-(1.22) chegaraviy masalaning Grin funksiyasi
deyiladi:
1) Gx,ta,b2 da aniqlangan uzluksiz funksiya; 2) L G x,t 0 agar x t bo„lsa;
3) Ãa Gx,t0, Ãb Gb,t0;
1
4) Gxt 0,tGxt 0,tpt.
Agar Gx,tGrin funksiyasi topilgan (ma‟lum) bo„lsa, (1.21)-(1.22) chegaraviy masalaning yechimi ushbu
17
b
uxGx,tf tdt (1.23) a
formula bilan ifodalanishini tekshirib ko„rish qiyin emas. Bundan tashqari
G x,t Grin funksiya simmetrik shartni qanoatlantiradi, ya‟ni Gx,tG t,x , x,ta,b.
Endi Shturm-Liuvill chegaraviy masalasini qaraymiz:
.
px uqxxu 0, (1.24) Ãa uÃb u0 px0, x0(1.25)
Ushbu x u hadni erkin had deb hisoblab, (1.24) ni integrallasak, (1.23)
formulaga ko„ra quyidagi bir jinsli Fredgolm integral tenglamasini hosil qilamiz
b
uxKx,t utdt, (1.26) a
t .
bunda Kx,tGx,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,
zxuxx
deb belgilasak, (1.26) va (1.27) formulaga ko„ra
b
zxK*x,tztdt, a
bunda yadro
K*x,tGx,txt
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) 1u(t)dt 0
2. y–5y+6y = 0, x = 0 bo„lganda y = 0, y= -1.
x
Javobi: u(x) 56x(6x6t 5)u(t)dt. 0
3. yy = 0, x = 0 bo„lganda y = 0, y= -1.
x
Javobi: u(x) x(xt)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) xcosx(xt)cosxu(t)dt. 0
6. y–ysinx+y = 1, x = 0 bo„lganda y = 0, y= 1.
x
Javobi: u(x) xsin x(sin xxt)u(t)dt. 0
7. y+n2y = hsinpx ( p n), x = 0 bo„lganda y = a, y= d.
x
Javobi: u(x) hsin px6n2xan2 n2 (xt)u(t)dt. 0
8. y+2hy+n2y = 0, x = 0 bo„lganda y = a, y= d.
x
Javobi: u(x) (dn2x2dhan2)n2(xt)2hu(t)dt. 0
3
9. y2 xy2 , y(0) = –3, y(0) = 1, y(0) = –1 . 10. yxxy2 y2 , y(0)= 1, y(0) = y(0)= 0 .
11. yxy ex , y(0) = 1, y(0) = y(0)= 0 .
12. yIV yy 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) ex u(t)dt. 0
x
2. u(x) 1tu(t)dt. 0
1
x
3. u(x) 1x2 0sin( x t)u(t)dt.
x
4. u(x) ex cosxcosxe(xt)u(t)dt. 0
x
5. u(x) 4ex 3x4(xt)u(t)dt. 0
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