O'zbekiston respublikasi oliy va o'rta maxsus talim vazirligi samarqand davlat universiteti haydarov Akram matematik fizika va analizning zamonaviy usullari va nokorrekt masalalari
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u(x,y, 0) = g(x,y) = cos 2x, uz(x,y, 0) = x — 2 siny.
Koshi masalasini yeching: utt = Au tenglamaning u(x, y, z, 0) = xyz, ut(x, у, z, 0) = x — 2 sin 2y. Koshi masalasini yeching: utt = Au u(x, y, z, 0) = x2 + у2 + z2, ut(x, y, z, 0) = xy. Koshi masalasini yeching: utt = Au u(x,y,z, 0) = 6X cos у, ut(x,y,z, 0) = x2 -y2 Koshi masalasini yeching: utt = Au u(x, y, z, 0) = x2 + y2, ut(x,y,z, 0) = 1 Koshi masalasini yeching: utt = Au u(x, y, z, 0) = ex, ut(x, y, z, 0) = Koshi masalasini yeching: utt = Au 1 u(x,y,z, 0) = —, ut(x,y,z, 0) = 0. ОС Koshi masalasini yeching: utt = Au u(x,y,z, 0) = —, ut(x,y,z, 0) = 0. У Koshi masalasini yeching: utt = Au 1 u(x,y,z, 0) = —, ut(x,y,z, 0) = 0. z Koshi masalasini yeching: utt = Au + x + t u(x, у, z, 0) = xyz, ut(x, y, z, 0) = xy + z. Koshi masalasini yeching: u(x, 0) = e~x, ut(x, 0) = 1. Koshi masalasini yeching: — ^xx "l" (uctf 0) — sinx Koshi masalasini yeching: utt = uxx + a sin bt u(x, 0) = cos x, ut(x, 0) = sin x Koshi masalasini yeching: u(x, 0) = sin x, ut(x, 0) = cos x. Koshi masalasini yeching: u(x, 0) = e~x, ut(x, 0) = 3 Garmonik o'lchov yordamida garmonik funksiyani toping u(e*) = f г> 0<в<71 ^ } l-l, 7Г < в < 2n Garmonik o'lchov yordamida garmonik funksiyani toping 0, u{ei(P) = 1, 7Г <6 <71 3n л <6 < — 3n —<в<2п A 21 22 7Г o 7Г -2, (.ei(P) = и 0 < в <- 4 7Г -1, - < в < 2л 4 24) Garmonik o'lchov yordamida garmonik funksiyani toping 7Г 0 — < в и 2, 1, (.ei(P) = (Д 7Г < в < 2n 25) Garmonik o'lchov yordamida garmonik funksiyani toping ( ТТ О, О < в < - u(ei(P) = { , 71 ^ п ^ v } 1, — < 6 <л 4 Д 7Г < в < 2л 2. Mustaqil o'zlashtirish uchun savollar ut = uxx tenglama uchun minimum qiymat prinsipini isbotlang. ut = uxx tenglama yechimi uchun maksimum qiymati prinsipini isbotlang. ut = uxx tenglama uchun aralash masalasi yechimining yagonaligini isbotlang. ut = uxx tenglama uchun aralash masala yechimining turg'unligini isbotlang. ut = 4uxx tenglama uchun aralash masala yechimining yagonaligini isbotlang. ut = 4uxx tenglama uchun aralash masala yechimining turg'unligini isbotlang. utt = a2uxx to'lqin tenglama uchun Koshi masalasi mavjudligini isbotlang. 8 )utt = uxx to'lqin tenglama uchun Koshi masalasi yechimining mavjudligini isbotlang. utt = uxx to'lqin tenglamasi uchun Koshi masalasi yechimining turg'unligini isbotlang. utt = uxx tenglama uchun aralash masala yechimi mavjudligini isbotlang. utt = uxx tenglama uchun aralash masala yechimining turg'unligini isbotlang. utt = 4uxx tenglama uchun Koshi masalasi yechimining turg'unligini isbotlang. utt = uxx tenglamada to'lqin tarqalishi tenglamasini nimaga teng. utt = 4uxx,u(x,0) = (p{x),ut{x, 0) = ip(x) masala yechimini toping. utt = uXX)u(x, 0) = (p{x),ut{x, 0) = ip(x) Koshi masalasining yechimi qanday cp va ip funksiyalar uchun regulyar bo'ladi? Agar utt = uxx tenglama yechimi u(x, t) bo'lsa, xux + tut funksiya ham shu tenglama yechimi bo'lishini isbotlang. 17) Agar u(x,t) funksiya utt = uxx tenglama yechimi bo'lsa, u2 + u2 funksiya ham shu tenglama yechimi bo'lishini isbotlang. 18 21 22 23 Tenglama umumiy yechimini toping uyy - ил, = 4e: Tenglama umumiy yechimini toping uxx Tenglama umumiy yechimini toping uxy + yuy — и = 0. 3. Mustaqil o'zlashtirish uchun savollar 1 )u(x,y) = xy funksiyaning x2+y2< 1 doirada ekstremumlarini Zaremba - jiro prinsipi orqali aniqlang. u(x,y) = xy funksiyaning D = {(x,y); x2 + y2 < 1} doirada minimumini toping. u(x,y) = xy funksiyaning D = {(x,y); x2 + y2 < 1} doirada maksimumini toping. с x2 y2 \ u(x,y) = x2 - y2 funksiyaning D = Ux,y); — + — < lj ellipsda ekstremumlarini Zaremba - jiro prinsipi orqali aniqlang. u(x,y) = x2 - y2 funksiyaning D = j(x,y); + < lj ellipsda minimumini toping. u(x,y) = x2 + y2 funksiyaning D = j(x,y); + < lj ellipsda maksimumini toping. Agar w(x) funksiya D sohada ikkinchi tartibli hosilalari bilan uzluksiz bo'lsa, u holda Aw < 0 bo'lganda w(x) nisbiy manfiy minimumga erishmasligini isbotlang. Agar w(x) fnuksiya D sohada ikkinchi tartibligacha uzluksiz hosilaga ega bo'lib, Aw >0 bo'lsa, u holda D sohada musbat nisbiy maksimumga ega boTmasligini isbotlang. u(x,y) = xy funksiyaning D = {(x,y); x2 + y2 < 1} soha chegarasida ^ — normal hosilasini toping. и = x2 — у2 funksiya uchun D = j(x,y); + < lj soha дъi chegarasida — - normal hosilasini toping. Agar и funksiya D sohada garmonik bo'lib, D ning chegarasi S gacha uzluksiz hosilaga ega bo'lsa, u holda x0 £ S nuqtada funksiya maksimumga ega du bo'lsa, bu nuqtada — > 0 bo'lishini isbotlang. Agar и funksiya D sohada garmonik bo'lib, soha chegarasi S gacha uzluksiz hosilaga ega bo'lsa, u holda x0 £ S minimum nuqta bo'lganda, bu du nuqtada — < 0 bo'lishini isbotlang. Agar /(z) analitik funksiyaning haqiqiy qismi u(x,y) bo'lsa, u holda quyidagi Gursa formulasi o'rinli ekanligini isbotlang. f(z) = 2u{^'Yi)~ u(°>°) + ci e D> z = x + iy u(x,y) = xy3 — yx3 bo'lsa, /(z) ni Gursa formulasi orqali toping. 6>6> Download 391.68 Kb. Do'stlaringiz bilan baham: |
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