Gipergeometrik tenglama, uning echimlari va gipergeometrik funksiyalar haqida
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hol. π(π§) funksiya bitta ildizga ega π(π§) = (π§ β π) va πβ²(π) β 0 boβlsa, u holda π§ = π + ππ chiziqli almashtirish yordamida (2) tenglama
π π2π’ π(π + ππ ) ππ’ π ππ 2 + π koβrinishga keltiriladi. + ππ’ = 0 ππ πΎ = π(π), πΌ = βππ belgilashlar kiritib va π = β 1 ππ(π) deb tanlasak, tenglama quyidagicha yoziladi: π π’β²β² + (πΎ β π )π’β² β πΌπ’ = 0. (3) bu tenglama aynigan (vΡrojdennaya) gipergeometrik tenglama deb aytiladi. hol. agar π(π§) β‘ ππππ π‘ boβlsa, umumiylikka zid keltirmasdan π(π§) β‘ 1 deb hisoblash mumkin. agar, πβ²(π ) = 0 boβlsa, (1) tenglama chiziqli bir jinsli oβzgarmas koeffitsientli tenglama boβladi. bu holni qaramaymiz. πβ²(π ) β 0 holni qaraymiz. π§ = π + ππ chiziqli almashtirish yordamida (1) tenglama 1 π2π’ π(π + ππ ) ππ’ π ππ 2 + π koβrinishga keltiriladi. + ππ’ = 0 ππ π β soni π(π) = 0 tenglamani ildizi boβlsin. π2 = β 2 ππ(π ) deb tanlaymiz va π2π ni 2π deb belgilaymiz. natijada (1) tenglama π’β²β² β 2π π’β² + 2ππ’ = 0 (4) koβrinishga keladi. (4) tenglama ermit tenglamasi deb ataladi. π ning manfiy boβlmagan butun qiymatlarida koβphadlar uchun ermit tenglamasi bilan bir xil boβladi. endi (1) tenglamaning echimini oβrganish bilan shugβullanamiz. agar π β 0, β1, β2, β¦ boβlsa, unda β (π) (π) π§π π’1 = β π π β‘ πΉ(π, π; π; π§) (π)ππ! π=0 (1) gipergeometrik tenglamaning echimi boβladi va π§ = 0 nuqtada regulyar boβladi [1]. ikkinchi tomondan π π π > π π π > 0 da 1 Ξ(π) πΉ(π, π; π; π§) = Ξ(π)Ξ(π β π) β« 0 π‘πβ1(1 β π‘)πβπβ1 (1 β π‘π§)π ππ‘, (5) tenglik oβrinli boβlib, bu eyler formulasi deyiladi. (5) ning oβng tomoni |arg(1 β π§)| < π sohada π§ ga bogβliq bir qiymatli analitik funsiya, ya'ni πΉ(π, π; π; π§) funksiyaning analitik davomi boβladi. (5) formulani oβrinli ekanligini isbotlaymiz. (5) tenglikda |π§| < 1 da (1 β π‘π§)βπ ni binomial qatorga yoyamiz va hosil boβlgan qatorni hadma-had integrallaymiz. shunda biz beta integrallarga ega boβlamiz. ular eylerning gamma va beta funksiyalari orqali hisoblanadi. quyidagi ayniyatdan π2 π π‘πβ1(1 β π‘)πβπβ1 {π§(1 β π§) ππ§2 + [π β (π + π + 1)π§] ππ§ β ππ} (1 β π‘π§)π = = βπ π [π‘π(1 β π‘)πβπ(1 β π‘π§)βπβ1] (2.1.28) ππ‘ kelib chiqadiki, ushbu tenglikning oβng tomonidagi ifoda (1) differensial tenglamaning echimi boβladi. π = βπ‘ deb olinsa, π π π > 0, π π (π + 1 β π) > 0 va |arg π§| < π da β β« π πβ1(1 + π )πβπβ1(1 + π π§)βπππ 0 integral (1) gipergeometrik differensial tenglamaning echimi boβladi. π = π 1βπ oβrniga qoβyish orqali quyidagi koβrinishdagi integralga kelamiz: 1 β« ππβ1(1 β π)πβπ[1 β π(1 β π§)]βπππ‘. 0 shuningdek, πΉ(π, π; π + π + 1 β π; 1 β π§) = β = Ξ(π + π + 1 β π) β« π πβ1(1 + π )πβπβ1(1 + π π§)βπππ Ξ(π)Ξ(π + 1 β π) 0 ham gipergeometrik tenglamaning echimi boβladi. bundan tashqari, quyidagi koβrinishdagi ixtiyoriy integral β« π‘πβ1(1 β π‘)πβπβ1(1 β π‘π§)βπππ‘ πΆ (1) gipergeometrik tenglamaning echimi boβladi, agarda πΆ integralosti funksiyaning riman sirtida yopiq kontur yoki kontur chekkalari π‘π(1 β π‘)πβπ(1 β π‘π§)βπβ1 funksiyaning nollari boβlsa. (1 β π‘π§)βπ ni binomial qatorga yoyib va beta funksiya uchun kontur integrallarini qoβllab, quyidagilarni topamiz: πΞ(π) exp[ππ(π β π)] (1+) π‘πβ1(1 β π‘)πβπβ1 πΉ(π, π; π; π§) = Ξ(π)Ξ(π β π)2 sin[π(π β π)] β« 0 (1 β π‘π§)π ππ‘, Re π > 0, |arg(1 β π§)| < π, π β π β 1,2,3, β¦ ; (1+) βπΞ(π) exp[πππ] πΉ(π, π; π; π§) = β« Ξ(π)Ξ(π β π)2 sin[ππ] 0 π‘πβ1(1 β π‘)πβπβ1 (1 β π‘π§)π ππ‘, Re π > Re π, |arg(βπ§)| < π, π β 1,2,3, β¦ ; πΉ(π, π; π; π§) = (1+,0+,1β,0β) βΞ(π) exp[πππ] = Ξ(π)Ξ(π β π)4 sin ππ sin[π(π β π)] β« π‘πβ1(1 β π‘)πβπβ1 (1 β π‘π§)π ππ‘, |arg(βπ§)| < π, π, 1 β π, π β π β 1,2,3, β¦ . faraz qilamiz, barcha hollarda integrallash yoβli π‘πβ1(1 β π‘)πβπβ1(1 β π‘π§)βπ uchun riman sirtidagi nuqtadan boshlanadi, π‘ haqiqiy, 0 β€ π‘ β€ 1 va π‘π, (1 β π‘)πβπ lar funksiyaning bosh qiymatlari, (1 β π‘π§)βπ aniqlangan va π§ β 0 da (1 β π‘π§)βπ β 1 boβladi. agar π§ = 1 ni qoβysak, (5) ning oβng tomoni beta integral boβladi va quyidagi kelib chiqadi: Ξ(π)Ξ(π β π β π) πΉ(π, π; π; 1) = ( ) ( , Ξ π β π Ξ π β π) bunda Re π > Re π > 0, Re(π β π β π) > 0. ushbu tenglikni parametrlarga kuchsiz shartlar qoβyilganda ham oβrinli boβlishini koβrsatish mumkin, xususan π β 0, β1, β2, β¦ va Re(π β π β π) > 0 boβlishi formulaning oβrinli boβlishi uchun etarli. ilmiy izlanishlarda qulaylik tugβdirish uchun bir nechta adabiyotlardan gipergeometrik funksiya haqida toβliqroq ma'lumotlar toβplashga harakat qilindi. xususan, uning bir nechta xossalarini keltiramiz: 1. πΉ(π, π, π; π§) = πΉ(π, π, π, π§) (gipergeometrik funsiya birinchi va ikkinchi argumentlari boβyicha simmetrik); 2. πΉ(π, π, π; π§) = (1 β π§)βπ; 3. πΉ(π, π, π; 1) = g(π)g(πβπβπ) , π π(π β π β π) > 0; g(πβπ)g(πβπ) 4. πΉ(π, π, π; 0) = πΉ(0, π, π, π§) = 1; 5. πΉ(π, π, π; π§) = (1 β π§)βππΉ (π, π β π, π; π§ ) = π§β1 = (1 β π§)βππΉ (π β π, π, π; π§ π§ β 1 ) , |arg(1 β π§)| < π; 6. πΉ(π, π, π; π§) = g(π)g(πβπ) (1 β π§)βππΉ (π, π β π, π β π + 1; 1 ) + g(πβπ)g(π) + g(π)g(π β π) (1 β π§)βππΉ (π β π, π, π β π + 1; 1 ), π§β1 g(π β π)g(π) π§ β 1 π β π β 0, Β±1, Β±2, β¦ , |arg(βπ§)| < π, |arg(1 β π§)| < π; 7. πΉ(π, π, π, π§) = g(π)g(πβπβπ) π§βππΉ (π, π β π + 1, π + π + 1 β π; 1 β g(πβπ)g(πβπ) 1) + π§ g(π)g(π + π β π) π§πβπ(1 β π§)πβπβππΉ (π β π, 1 β π, π + 1 β π β π; 1 β 1), g(π)g(π) π§ π β π β π β 0, Β±1, Β±2, β¦ , |arg(βπ§)| < π, |arg(1 β π§)| < π; 8. πΉ(π, 1 β π, π; π§) = (1 β π§)πβ1πΉ (πβπ , π+πβ1 , π; 4π§(1 β π§)) ; 2 2 9. πΉ(π, 1 β π, π, βπ§) = (1 + π§)πβ1(β1 + π§ + βπ§)2β2πβ2π β 1 β2 πΉ (π + π β 1, π β 2 , 2π β 1; 4βπ§(1 + π§)(β1 + π§ + βπ§) ) ; 10. πΉ(π, π, π; π§) = π πΉ(π, π + 1, π + 1; π§) + πβπ πΉ(π, π, π + 1; π§); π π 11. ππ ππ§π 12. ππ ππ§π πΉ(π, π, π, π§) = (π)π(π)π πΉ(π + π, π + π, π + π; π§); (π)π [π§πβ1(1 β π§)πβπ+ππΉ(π, π, π; π§)] = (π β π)ππ§πβ1βπ(1 β π§)πβππΉ(π β π, π, π β π; π§). gipergeometrik funksiya turli xil xossalarga ega. uning universal funksiya ekanligi ham shundaki, parametrlarining turli qiymatlarida u bir qator elementar va maxsus qiymatlarni ifodalaydi. jumladan, parametrlarning xususiy qiymatlarida gipergeometrik funksiyaning elementar va maxsus funsiyalarni ifodalashiga doir jadvalni keltiramiz: π 1
π 2 1 ; 1; π2) = β« ππ = πΎ(π); 2 2 2 β1 β π2π ππ2π 0 π 2) π πΉ (β 1 , 1 ; 1; π2) = β«2 β1 β π2π ππ2πππ = πΈ(π); 2 2 2 0 1 β π₯ 3) πΉ (π + 1, βπ; 1; 2 ) = ππ(π₯), 2 π g(π + π + 1) 4) ππ,π(π₯) = (1 β π₯ ) 2 2πg(π β π + 1)g(π + 1) β 1 β π₯ 5) πΉ (π + π + 1, π β π; π + 1; ) ; 2 π₯π£ 2 6) π½ (π₯) = lim [ 2 πΉ (π, π; π + 1; βπ₯ )] ; π£ π,πββ g(π + 1) 4ππ 7) πΉ(1, π½; π½; π₯) = 1 + π₯ + π₯2 + β― + π₯π + β― = 1 ; 1 β π₯ 8) πΉ(βπ, π½; π½; π₯) = (1 + π₯)π; πΉ(πΌ, π½; π½; βπ₯) β 1) 9) | πΌ πΌ=0 = log(1 + π₯) ; 10) πΉ(π, π½; π½; π₯) = (1 + π₯)π; 11) πΉ(πΌ, π½; π½; π₯) = (1 β π₯)βπΌ; 1 12) π₯πΉ(1,1; 2; π₯) = ln ; 1 β π₯ 1 13) 2π₯πΉ ( 2 3 , 1; 2 ; π₯ 2) = ln 1 + π₯ ; 1 β π₯ 14) lim πΉ(1, π; 1; π₯βπ) = π π₯; πβ0 15) πΉ (π + 1, βπ; 1; 1βπ₯) = π (π₯), bu erda π (π₯) β lejandr koβphadi; lim π,πβ0 πΉ (π, π; π + 1; β 2 π₯2 4ππ π π ) = π½π£(π₯), π½π£(π₯) β besel funksiyasi; 1 1 17) π₯πΉ ( , 2 2 3 ; ; π₯ 2 2) = ππππ πππ₯; 1 18) π₯πΉ ( 2 3 , 1; 2 ; βπ₯ 2) = ππππ‘ππ₯; π π 19) πΉ ( , β 2 2 1 ; ; π₯ 2 2) = cos(πππππ πππ₯) ; 1 + π 1 β π 20) ππ₯πΉ ( , 3 ; ; π₯ 2) = π ππ(πππππ πππ₯) ; 2 2 2 1 1 21) πΉ ( 2 , 1; 1; π₯) = (1 β π₯)β2; 22) πΉ(βπ, 1; 1; 1 β π₯) = π₯π; πΌ πΌ(πΌ + 1) ( ) 2 23) πΉ πΌ, β2; πΎ; π₯ = 1 β 2 π₯ + π₯ ; πΎ πΎ(πΎ + 1) 24) πΉ (πΌ, πΌ + 1 ; 2πΌ + 1; π₯) = [ 2 1 + (1 β π₯)1β2 2 β2πΌ ] ; 25) πΉ (πΌ, πΌ + 1 ; 2πΌ; π₯) = (1 β π₯)β1β2 [ 2 1 + (1 β π₯)1β2 2 1β2πΌ ] ; 1 3 26) πΉ ( , 2 1 1 ; ; ) = . 4 4 3 3 ββ 4 + 3β4 + 4 β β2 β 3β4 β 2 β2 β 3β4 yuqorida keltirilganlardan xulosa qilib, shuni aytishimiz mumkinki, gipergeometrik funksiyalarning eng koβp qoβllaniladigan sohalaridan biri matematikaning differensial tenglamalar sohasi hisoblanadi. hozirgi vaqtda differensial tenglamalar nazariyasi bilan bir qatorda, xususiy hosilali differensial tenglamalar nazariyasining muhim yoβnalishlaridan biri - qaralayotgan sohada buzilish chizigβiga ega boβlgan tenglamalarni oβrganish ham jadal rivojlanib bormoqda. ikkinchi tomondan buzilish chizigβiga ega boβlgan tenglamalarning echimlari mexanika, fizika va texnika masalalarida keng koβlamli tarzda amaliyotda tadbiq etilishi katta qiziqish uygβotadi. buzilish chizigβiga ega tenglamalar deb qaralayotgan sohaning ichida elliptik, soha chegarasida parabolik (yoki soha ichida giperbolik, soha chegarasida parabolik) tipga tegishli boβlgan tenglamalarga aytiladi. bu tipdagi tenglamalar uchun dirixle va neyman (elliptik tip uchun) hamda koshi - gursa (giperbolik tip uchun) masalalarining echimlari gipergeometrik funksiyalar orqali ifodalanadi. jumladan, f.frankl yassi devorli idishdan tovush tezligidan yuqori tezlikda suyuqlik yoki gazning oqib chiqish (idish ichida tezlik tovush tezligidan past) masalasi a.s.chaplΡginning πΎ(π¦)ππ₯π₯ + ππ¦π¦ = 0 (πΎ(0) = 0, πΎβ²(u) > 0) tenglamasi uchun chegaraviy masalaga kelishi koβrsatilgan. ushbu tenglamalar tipiga kiruvchi quyidagi tenglamani qaraylik: β(βπ¦)πππ₯π₯ + π₯πππ¦π¦ = 0. ushbu tenglama buzilish chizigβiga ega boβlgan tenglamalarga kiradi. tenglamani xarakteristik koordinatalarga oβtkazsak π½ ππ¦ β π β π (ππ¦ β π) = 0 boβladi, bunda βπ = π₯π β (βπ¦)π, βπ = π₯π + (βπ¦)π, π = π + 2. agar tenglamada π‘ = πβπ almashtirish bajarsak, tegishli parametrlar orqali ifodalangan (1) tenglama β gipergeometrik tenglamaga kelamiz. ushbu xususiy hosilali differensial tenglama uchun koshi masalasining echimi riman funksiyasi (πβ² β πβ²)2π½ π(πβ², πβ²; π, π) = (πβ πβ²)π½(πβ² β π)π½ (π β πβ²)(πβ² β π) πΉ(π½, π½, 1, π§), π§ = (πβ² β π)(π β πβ²) orqali yoziladi [2], bu erda 2π½ = πβ(π + 2). bu esa gipergeometrik tenglama, uning echimlari va gipergeometrik funksiyalarning keng amaliy ahamiyatga egaligini koβrsatadi. [2-23] ilmiy izlanishlarda gipergeometrik tenglamalar va gipergeometrik funksiyalarning amaliy ahamiyati hamda oddiy differensial tenglamalar ishtirok etgan bogβliq masalalar ishlangan va ular haqida kengroq ma'lumotlar berilgan. shu oβrinda aytish lozimki, ilmiy ishlarni oβrganish va tahlil qilish talabalar uchun bir qator qiyinchiliklar tugβdiradi. shu sababli ushbu maqolada talabalarning ilmiy maqolalarni oβrganishlarini osonlashtirish uchun matematikani fanini oβqitishga bagβishlangan ilgβor pedagogik texnologiyalarning [24-30] ayrim elementlari ham qoβllanildi. Download 96.53 Kb. Do'stlaringiz bilan baham: |
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