Bir o‘lchovli Dirak tenglamasi uchun teskari masalani yechishda tezkor hisoblash algoritmini qurish masalalari

Аннотация

Berilgan spectral ma’limotlar asosida bir o‘lchovli Dirak operatori parametrlarini tiklash masalasi qaralqai. Masalani yechisha qulay almashtirishlar hisoboga ixcham hisoblash algoritmi qurilgan. Kalit so‘zlar: Bir o’lchovli Dirak tenglamasi, xos qiymat, xos funksiya, Volterra tenglamasi. Aniq va asimtotik echim, hisoblash xatoligi. Uchbu maqolada bir o‘lchovli Dirak tenlamasi uchun teskari masalani echishda qulay almashtirishlar hisobiga ixcham hisoblash algoritmi qurish masalasi muhokama 1 qilinadi. Bunda qaralayotgan differensial tengamaning echimini 𝑂( ) aniqlikda 𝜆3 elemetar funksiyalar orqali ifodalash, olingan yechimni qaralayotgan oraliq chegarasidagi berilgan 𝑦 (0)𝑠𝑖𝑛𝛽 + 𝐻𝑦 (𝛽) = 0 1 2 chegaraviy shartga qo’yib, oldindan berilgan 𝜆 ning har xil qiymatlari yordamida noma’lum 𝑝(𝑥), 𝑞(𝑥) larga nisbatan tenglamalar sistemasi hosil qilinadi. Bu bilan hisoblash amallari ketma-ketligining keskin qisqarishiga va o’z navbatida kutilayotgan echimdan chetlanish xatoligining kamayishiga erishiladi [1], [2]. Quyidagi teorema qo’yilgan masala mazmunini o’zida to’liq ifodalaydi. Teorema. Bir o’lchovli 0 1 𝑝(𝑥) 0 𝑦 (𝑥 ) ( )𝑌′ + ( )𝑌 = 𝜆𝑌, 𝑌 = ( 1 ) (1) −1 0 0 𝑞(𝑥) 𝑦 (𝑥 ) 2 Dirak tenglamasining 𝑦 (0) = 𝑐𝑜𝑠𝛼, 𝑦 (0) = −𝑠𝑖𝑛𝛼 (2) 1 2 boshlang’ich shartlarni qanoatlantiruvchi yechimini 𝐹 𝐹 𝑐𝑜𝑠𝛼 𝑌(𝑥,𝜆,𝛼) = 𝐻(𝜆𝑥)( 11 12 )( ) (3) 𝐹 𝐹 −𝑠𝑖𝑛𝛼 21 22 ko’rinishda yozish mumkin. Bu yerda 𝐹 = 𝑐ℎ(𝜑)+ 𝑐𝑜𝑠(𝜂) − 2(1 − 𝜓) + 𝑒−𝜓 + 𝐼 + 𝑓 11 11 11 𝐹 = −𝑠ℎ(𝜑)+ 𝑠𝑖𝑛(𝜂) + 2(𝜑 − 𝜂) + 𝐼 12 12 𝐹 = −𝑠ℎ(𝜑)− 𝑠𝑖𝑛(𝜂) + 2(𝜑 + 𝜂) + 𝐼 12 12 𝐹 = 𝑐ℎ(𝜑)+ 𝑐𝑜𝑠(𝜂) − 2(1 + 𝜓) + 𝑒𝜓 + 𝐼 + 𝑓 22 22 22 𝑥 𝑐𝑜𝑠𝜆𝑥 −𝑠𝑖𝑛𝜆𝑥 𝐼 = ∫(𝜑′𝜂 − 𝜂′𝜑)𝑑𝑡 𝐼 = − 𝐼 𝐻(𝜆𝑥) = ( ) 11 22 11 𝑠𝑖𝑛𝜆𝑥 𝑐𝑜𝑠𝜆𝑥 0 𝑥 𝑥 𝐼 = ∫(𝜓′(𝜑 − 𝜂) − (𝜑′ − 𝜂′)𝜓)𝑑𝑡 𝐼 = ∫(𝜓′(𝜑 − 𝜂) + (𝜑′ − 𝜂′)𝜓)𝑑𝑡 12 21 0 0 186 𝑥 𝜓 𝑓 (𝑥,𝜆) ≈ exp (−𝜑2 − 𝜂2)∫[ 𝜓′(𝜑2 − 𝜂2) ]𝑑𝑡 1 𝜆3 0 𝑥 𝜑 𝑓 (𝑥,𝜆) ≈ exp (−𝜓2 − 𝜂2)∫[ 𝜑′(𝜓2 − 𝜂2) ]𝑑𝑡 2 𝜆3 0 𝑥 𝑥 𝑥 𝑝 − 𝑞 𝑝 − 𝑞 𝑝 + 𝑞 𝜑 = ∫ 𝑐𝑜𝑠2𝜆𝑡𝑑𝑡, 𝜓 = ∫ 𝑠𝑖𝑛2𝜆𝑡𝑑𝑡, 𝜂 = ∫ 𝑑𝑡 2 2 2 0 0 0 𝑝 = 𝑝(𝑥), 𝑞 = 𝑞(𝑥) ∈ 𝐶1 Isbot. Laplas akslantirishini qo’llab, berilgan (2) boshlang’ish shartga ko’ra (1) tenglamaning yechimini 𝑥 𝑐𝑜𝑠𝛼 𝑌(𝑥,𝜆,𝛼) = 𝐻(𝜆𝑥)[( )− ∫𝑅(𝑡)𝐻(𝜆𝑡)𝑌(𝑡,𝜆,𝛼)𝑑𝑡] (4) −𝑠𝑖𝑛𝛼 0 ko’rinishda yozish mumkin. Bu yerda 𝑝 + 𝑞 𝑝 − 𝑞 𝑝 + 𝑞 𝑠𝑖𝑛2𝜆𝑥 − 𝑐𝑜𝑠2𝜆𝑥 2 2 2 𝑅(𝜆𝑥) = ( ), 𝑝 − 𝑞 𝑝 + 𝑞 𝑝 + 𝑞 + 𝑐𝑜𝑠2𝜆𝑥 𝑠𝑖𝑛2𝜆𝑥 2 2 2 Olingan (4) tenglama Volterra tenglamasi bo’lib uni ketma ket yaqinlashish usulida echish mumkin. Dastlabki yaqinlashish sifatida 𝑐𝑜𝑠𝛼 𝑌̃ = 𝐻(𝜆𝑥)( ) 0 −𝑠𝑖𝑛𝛼 ifodani tanlaymiz. Uni (4) ga qo’yib 𝑌̃ uchun quyidagi ifodani olamiz. 1 𝑥 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛼 𝑌̃ = 𝐻(𝜆𝑥)[( ) − ∫𝑅(𝑡)𝐻(𝜆𝑡)𝐻(𝜆𝑡)( )𝑑𝑡] = 1 −𝑠𝑖𝑛𝛼 −𝑠𝑖𝑛𝛼 0 𝑥 𝑐𝑜𝑠𝛼 = 𝐻(𝜆𝑥)[𝐼 − ∫𝑅(𝑡)𝐻(2𝜆𝑡)𝑑𝑡]( ) −𝑠𝑖𝑛𝛼 0 Xuddi shu usulda davom etib keying yaqinlashishlar uchun quyidagi ifodalarga ega bo’lamiz 𝑥 𝑥 𝑐𝑜𝑠𝛼 𝑌̃ = 𝐻(𝜆𝑥)[( )− ∫𝑅(𝑡)𝐻(𝜆𝑡)𝑌̃𝑑𝑡] = 𝐻(𝜆𝑥)[𝐼 − ∫𝑅(𝑡)𝐻(2𝜆𝑡)𝑑𝑡 + 2 −𝑠𝑖𝑛𝛼 1 0 0 𝑥 𝑡 𝑐𝑜𝑠𝛼 +∫𝑅(𝑡)𝐻(2𝜆𝑡)(∫𝑅(𝜏)𝐻(2𝜆𝜏)𝑑𝜏)𝑑𝑡]( ) = −𝑠𝑖𝑛𝛼 0 0 𝑐𝑜𝑠𝛼 = 𝐻(𝜆𝑥){𝐼 − 𝐴 + 𝐴 }( ); ⋯ 1 2 −𝑠𝑖𝑛𝛼 𝑐𝑜𝑠𝛼 𝑌̃ = 𝐻(𝜆𝑥){𝐼 − 𝐴 + 𝐴 − 𝐴 + ⋯+ (−1)𝑛𝐴 }( ) (5) 𝑛 1 2 3 𝑛 −𝑠𝑖𝑛𝛼 Agar 187 𝑅(𝑡)𝐻(2𝜆𝑡) = 𝑝 + 𝑞 𝑝 − 𝑞 𝑝 + 𝑞 𝑠𝑖𝑛2𝜆𝑡 − 𝑐𝑜𝑠2𝜆𝑡 2 2 2 𝑐𝑜𝑠2𝜆𝑡 −𝑠𝑖𝑛2𝜆𝑡 = ( )( ) = 𝑝 − 𝑞 𝑝 + 𝑞 𝑝 + 𝑞 𝑠𝑖𝑛2𝜆𝑡 𝑐𝑜𝑠2𝜆𝑡 + 𝑐𝑜𝑠2𝜆𝑡 𝑠𝑖𝑛2𝜆𝑡 2 2 2 𝑝 − 𝑞 𝑝 − 𝑞 𝑝 + 𝑞 𝑠𝑖𝑛2𝜆𝑡 𝑐𝑜𝑠2𝜆𝑡 − 2 2 2 𝜓′ 𝜑′ − 𝜂′ = ( ) = ( ) 𝑝 − 𝑞 𝑝 + 𝑞 𝑝 − 𝑞 𝜑′ + 𝜂′ −𝜓′ 𝑐𝑜𝑠2𝜆𝑡 + − 𝑠𝑖𝑛2𝜆𝑡 2 2 2 ekanligi hisobga olinsa 𝐴 lar uchun quyidagi hisoblash formulasiga ega bo’lamiz 𝑛 𝑥 𝑥 𝜓′ 𝜑′ − 𝜂′ 𝜓 𝜑 − 𝜂 𝐴 = 𝐴 = ∫𝑅(𝑡)𝐻(2𝜆𝑡)𝑑𝑡 =∫( )𝑑𝑡 = ( ), 1 𝜑′ + 𝜂′ −𝜓′ 𝜑 + 𝜂 −𝜓 0 0 𝑥 𝑎 𝑏 𝐴 = ( 𝑛 𝑛 ) = ∫𝑅(𝑡)𝐻(2𝜆𝑡)𝐴 (𝑡)𝑑𝑡 = 𝑛 𝑐 𝑑 𝑛−1 𝑛 𝑛 0 𝑥 𝑥 𝜓′ 𝜑′ − 𝜂′ 𝑎 𝑏 = ∫𝐴′(𝑡)𝐴 (𝑡)𝑑𝑡 =∫( )( 𝑛−1 𝑛−1 )𝑑𝑡 = 𝑛−1 𝜑′ + 𝜂′ −𝜓′ 𝑐 𝑑 𝑛−1 𝑛−1 0 0 𝑥 𝜓′𝑎 + (𝜑′ − 𝜂′)𝑐 𝜓′𝑏 + (𝜑′ − 𝜂′)𝑑 𝑛−1 𝑛−1 𝑛−1 𝑛−1 = ∫( )𝑑𝑡 𝜑′ + 𝜂′)𝑎 − 𝜓′𝑐 (𝜑′ + 𝜂′)𝑏 − 𝜓′𝑑 𝑛−1 𝑛−1 𝑛−1 𝑛−1 0 𝑎 = 𝑑 = 1 , 𝑏 = с = 0, 0 0 0 0 Qulaylik uchun hisoblashlarni indekslarning juft va toq holatlari uchun alohida alohida amalga oshiramiz, 𝑛 > 1 bo’ganda 𝑥 𝜓2𝑛−1 1 𝑎 = + ∫[ 𝜓′(𝜑2(𝑛−1) − (−1)𝑛𝜂2(𝑛−1)) + 𝑎 ]𝑑𝑡 2𝑛−1 2(𝑛−1) (2𝑛 − 1)! (2𝑛 − 2)! 0 𝜑2𝑛−1 + (−1)𝑛𝜂2𝑛−1 𝑏 = + 2𝑛−1 (2𝑛 − 1)! 𝑥 1 + ∫[𝜑′(𝜓2(𝑛−1) − (−1)𝑛𝜂2(𝑛−1)) − 𝜂′(𝜓2(𝑛−1) + 𝜑2(𝑛−1)) + 𝑏 ]𝑑𝑡 (2𝑛 − 2)! 2(𝑛−1) 0 𝜑2𝑛−1 − (−1)𝑛𝜂2𝑛−1 𝑐 = + 2𝑛−1 (2𝑛 − 1)! 𝑥 1 + ∫[𝜑′(𝜓2(𝑛−1) − (−1)𝑛𝜂2(𝑛−1)) + 𝜂′(𝜓2(𝑛−1) + 𝜑2(𝑛−1)) + 𝑐 ]𝑑𝑡 (2𝑛 − 2)! 2(𝑛−1) 0 𝑥 𝜓2𝑛−1 1 𝑑 = − − ∫[ 𝜓′(𝜑2(𝑛−1) − (−1)𝑛𝜂2(𝑛−1)) + 𝑑 ]𝑑𝑡 2𝑛−1 2(𝑛−1) (2𝑛 − 1)! (2𝑛 − 2)! 0 188 𝜓2𝑛 + 𝜑2𝑛 + (−1)𝑛𝜂2𝑛 𝑎 = 2𝑛 (2𝑛)! 𝑥 1 − ∫[(−1)𝑛𝜑′𝜂2𝑛−1 + 𝜂′𝜑2𝑛−1 + 𝑎 ]𝑑𝑡 (2𝑛 − 1)! 2𝑛−1 0 𝑥 1 𝑏 = ∫{𝜓′[𝜑2𝑛−1 + (−1)𝑛𝜂2𝑛−1] − (𝜑′ − 𝜂′)𝜓2𝑛−1+𝑏 }𝑑𝑡 + 2𝑛 (2𝑛 − 1)! 2𝑛−1 0 𝑥 1 𝑐 = ∫{(𝜑′ + 𝜂′)𝜓2𝑛−1 − 𝜓′[𝜑2𝑛−1 − (−1)𝑛𝜂2𝑛−1]+𝑐 }𝑑𝑡 + 2𝑛 (2𝑛 − 1)! 2𝑛−1 0 𝑥 𝜓2𝑛 + 𝜑2𝑛 + (−1)𝑛𝜂2𝑛 1 𝑑 = + ∫[(−1)𝑛𝜑′𝜂2𝑛−1 + 𝜂′𝜑2𝑛−1+𝑑 ]𝑑𝑡 2𝑛 (2𝑛)! (2𝑛 − 1)! 2𝑛−1 0 Topilgan ushbu matrisa elementlarini (5) ga qo’yib, ba’zi elementar almashtirishlardan so’ng (3) tenglikka kelamiz [3],[4]. Formuladagi 𝑓 ва 𝑓 1 2 funksiyalar ∞ ∞ 𝑥 𝜓 𝑓 = ∑(𝑎 − 𝑑 )∑(𝑎 − 𝑑 ) ≈ exp (−𝜑2 − 𝜂2)∫[ 𝜓′(𝜑2 − 𝜂2) ]𝑑𝑡 1 2𝑖 2𝑖 2𝑖+1 2𝑖+1 𝜆3 𝑖=3 𝑖=3 0 ∞ ∞ 𝑥 𝜑 𝑓 = ∑(𝑏 − 𝑐 )∑(𝑏 − 𝑐 ) ≈ exp (−𝜓2 − 𝜂2)∫[ 𝜑′(𝜓2 − 𝜂2) ]𝑑𝑡 2 2𝑖 2𝑖 2𝑖+1 2𝑖+1 𝜆3 𝑖=3 𝑖= 0 munosabatlardan kelib chiqadi. Foydalanilgan adabiyotlar ro‘yxati: 1. Rabimkul Abdunazarov. (2023). Problems of constructing a