Determination of the spectrum of frequencies and vibrations of a rectangular plate, mobily employed around the edge, in different environments

Мұқаба

Дәйексөз келтіру

Толық мәтін

Аннотация

The spectrum of frequencies and shapes of bending vibrations of a rectangular plate in contact with a liquid or gas are determined. A derivation of the expression for the distributed transverse load on a plate movably embedded along the contour is given. The surfaces of the plate are in contact with media of different densities and pressures. The medium can be compressible during surface deformation and incompressible. The influence on the bending of the interaction of average pressure and changes in the curvature of the middle surface, as well as the added mass of the gaseous medium, is determined.

Толық мәтін

1. Введение. В работах [1–17] исследуется спектр частот пластин и оболочек, контактирующих с жидкостью и газом, обзор которых приводится в [18]. В последней работе определяется низшая частота изгибных колебаний пластины, контактирующей с жидкостью или газом, в предположении ее цилиндрического изгиба. Поверхности пластины контактируют со средой одинаковой плотности и давления. Среда может быть сжимаемой в процессе деформации поверхности и несжимаемой. Определяется влияние на изгиб взаимодействия среднего давления и изменения кривизны срединной поверхности, а также присоединенной массы газовой среды. Исследовано влияние давления окружающей среды на низшую частоту колебаний пластины с учетом взаимодействия среднего избыточного давления на ее поверхности и кривизны срединной поверхности, а также действие присоединенной массы газовой среды с удаленными границами.

На основе использования дискретно структурной модели деформирования многослойных пластин при малых перемещениях, деформациях и учете внутреннего трения материалов слоев по модели Кельвина–Фойгта в работе [19] рассмотрены две задачи о прохождении моногармонической звуковой волны сквозь тонкую композитную прямоугольную пластину, шарнирно закрепленную в проеме абсолютно жесткой перегородки. При постановке первой задачи предполагается, что пластина находится между двумя полубесконечными пространствами и на нее падает плоская звуковая волна с заданным амплитудным значением давления звуковой волны. При постановке второй задачи считается, что пластина находится между двумя абсолютно жесткими преградами, одна из них за счет гармонических колебаний с заданной амплитудой перемещений формирует падающую на пластину звуковую волну, а другая неподвижна и имеет деформируемое энергопоглощающее покрытие.

В работе [20] исследовались собственные колебания прямоугольных металлических пластин. Для определения частот собственных колебаний применялись расчетные методы, в частности аналитический расчет и расчет методом конечных элементов. За основу аналитического расчета было принято уравнение движения тонкой прямоугольной пластины. Затем применялся асимптотический метод, учитывающий динамический краевой эффект. В результате были определены частоты собственных колебаний пластины.

В работах [21–24] изучены колебания прямоугольной пластины с различными граничными условиями на краях. Установлены энергетические неравенства, из которых следует единственность решения поставленных начально-граничных задач. Решения построены в виде суммы рядов с обоснованием сходимости в классах классических и обобщенных решений. Установлена устойчивость решений от начальных данных.

В данной работе определяется спектр частот и формы изгибных колебаний прямоугольной пластины, подвижно заделанной по контуру, которая помещена в жидкость или газ. Изучен вопрос о взаимном влиянии эффекта среднего давления и известного из литературы эффекта присоединенной массы жидкости на деформацию пластины. Получены формулы для вычисления частот и формы изгибных колебаний прямоугольной пластины, находящейся в несжимаемой и сжимаемой жидкости.

Для описания колебаний тонкой прямоугольной пластины рассмотрим дифференциальное уравнение четвертого порядка [25, с. 99]:

  D 4 w x 4 +2 4 w x 2 y 2 + 4 w y 4 +ρh 2 w t 2 =q,D= E h 3 12 1 ν 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGebWaaeWaaeaadaWcaaqaaiabgkGi2oaaCa aaleqabaGaaGinaaaakiaadEhaaeaacqGHciITcaWG4bWaaWbaaSqa beaacaaI0aaaaaaakiabgUcaRiaaikdadaWcaaqaaiabgkGi2oaaCa aaleqabaGaaGinaaaakiaadEhaaeaacqGHciITcaWG4bWaaWbaaSqa beaacaaIYaaaaOGaeyOaIyRaamyEamaaCaaaleqabaGaaGOmaaaaaa GccqGHRaWkdaWcaaqaaiabgkGi2oaaCaaaleqabaGaaGinaaaakiaa dEhaaeaacqGHciITcaWG5bWaaWbaaSqabeaacaaI0aaaaaaaaOGaay jkaiaawMcaaiabgUcaRiabeg8aYjaadIgadaWcaaqaaiabgkGi2oaa CaaaleqabaGaaGOmaaaakiaadEhaaeaacqGHciITcaWG0bWaaWbaaS qabeaacaaIYaaaaaaakiabg2da9iaadghacaGGSaGaaGPaVlaaykW7 caaMc8Uaamiraiabg2da9maalaaabaGaamyraiaadIgadaahaaWcbe qaaiaaiodaaaaakeaacaaIXaGaaGOmamaabmaabaGaaGymaiabgkHi Tiabe27aUnaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaaaaaa a@6ADF@ , (1.1)

где E, ν, ρ – модуль упругости, коэффициент Пуассона, плотность материала, h – толщина пластины, w(x,y,t) – прогиб, х, y, t – координаты, время, q – поперечная распределенная нагрузка.

На нижнюю и верхнюю поверхность пластины действуют давления р0 + р1 и р0 + р2 жидкостей с плотностями ρ1 и ρ2 (рис. 1). Здесь р0 – давление сборки, в частности атмосферное давление, действующее на все поверхности, р1, р2 – избыточные давления. При определении нагрузки q будем предполагать, что ρ1, ρ2 и р1, р2 являются постоянными и, вообще говоря, они могут быть равными или неравными соответственно.

 

Рис. 1. Элементы dx и dy срединной поверхности изогнутой пластины.

 

2. Несжимаемая среда. Предполагаем, что области, занятые жидкостями, простираются неограниченно, опоры не препятствуют свободному перетеканию жидкости вдоль пластины в направлении осей x и y. Возникающие в результате движения пластины давления на нижнюю и верхнюю поверхность обозначим через p ¯ 1 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaaceWGWbGbaebadaWgaaWcbaGaaGymaaqabaaaaa@32FE@  и p ¯ 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaaceWGWbGbaebadaWgaaWcbaGaaGOmaaqabaaaaa@32FF@ . Уравнения динамики несжимаемой жидкости в прямоугольных координатах x, y, z относительно потенциала скорости φi(x, y, z, t) имеют вид [1–3]:

  2 φ i x 2 + 2 φ i y 2 + 2 φ i z 2 =0, p ¯ i = ρ i φ i t ,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeduuDJXwAKbYu51MyVXgarqqr1ngBPrgifHhD YfgasaacHOWxh9vrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vspGe9FjuP0=fs0xXdbba9pGe9xq=Jbba9suk9fr=xfr=x frpeWZqaaeaabiGaaiaadaqabeaabeqacqaaaOqaaabaaaaaaaaape WaaSaaa8aabaWdbiabgkGi2+aadaahaaWcbeqaa8qacaaIYaaaaOGa eqOXdO2damaaBaaaleaapeGaamyAaaWdaeqaaaGcbaWdbiabgkGi2k aadIhapaWaaWbaaSqabeaapeGaaGOmaaaaaaGccqGHRaWkdaWcaaWd aeaapeGaeyOaIy7damaaCaaaleqabaWdbiaaikdaaaGccqaHgpGApa WaaSbaaSqaa8qacaWGPbaapaqabaaakeaapeGaeyOaIyRaamyEa8aa daahaaWcbeqaa8qacaaIYaaaaaaakiabgUcaRmaalaaapaqaa8qacq GHciITpaWaaWbaaSqabeaapeGaaGOmaaaakiabeA8aQ9aadaWgaaWc baWdbiaadMgaa8aabeaaaOqaa8qacqGHciITcaWG6bWdamaaCaaale qabaWdbiaaikdaaaaaaOGaeyypa0JaaeimaiaacYcapaGaaGjcVlaa yIW7caaMi8UaaGjcV=qaceWGWbGbaebapaWaaSbaaSqaa8qacaWGPb aapaqabaGcpeGaeyypa0JaeyOeI0IaeqyWdi3damaaBaaaleaapeGa amyAaaWdaeqaaOWdbmaalaaapaqaa8qacqGHciITcqaHgpGApaWaaS baaSqaa8qacaWGPbaapaqabaaakeaapeGaeyOaIyRaamiDaaaacaGG SaWdaiaayIW7caaMc8UaaGjcV=qacaWGPbGaeyypa0JaaeymaiaabY cacaqGYaaaaa@7708@ . (2.1)

Задаются условия на поверхностях контакта со средой:

  φ 1 z = w t ,z= h 2 , φ 2 z = w t ,z= h 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeduuDJXwAKbYu51MyVXgarqqr1ngBPrgifHhD YfgasaacHOWxh9vrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vspGe9FjuP0=fs0xXdbba9pGe9xq=Jbba9suk9fr=xfr=x frpeWZqaaeaabiGaaiaadaqabeaabeqacqaaaOqaaabaaaaaaaaape WaaSaaa8aabaWdbiabgkGi2kabeA8aQ9aadaWgaaWcbaWdbiaaigda a8aabeaaaOqaa8qacqGHciITcaWG6baaaiabg2da9maalaaapaqaa8 qacqGHciITcaWG3baapaqaa8qacqGHciITcaWG0baaaiaabYcapaGa aGjcVlaayIW7peGaamOEaiabg2da9iabgkHiTmaalaaapaqaa8qaca WGObaapaqaa8qacaaIYaaaaiaacYcapaGaaGjcVlaayIW7caaMi8Ua aGjcVlaayIW7peWaaSaaa8aabaWdbiabgkGi2kabeA8aQ9aadaWgaa WcbaWdbiaaikdaa8aabeaaaOqaa8qacqGHciITcaWG6baaaiabg2da 9maalaaapaqaa8qacqGHciITcaWG3baapaqaa8qacqGHciITcaWG0b aaaiaabYcapaGaaGjcVlaayIW7peGaamOEaiabg2da9maalaaapaqa a8qacaWGObaapaqaa8qacaaIYaaaaaaa@6C05@ . (2.2)

На большом удалении от поверхности возмущения среды от пластины исчезают:

  φ 1 z 0,z, φ 2 z 0,z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeduuDJXwAKbYu51MyVXgarqqr1ngBPrgifHhD YfgasaacHOWxh9vrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vspGe9FjuP0=fs0xXdbba9pGe9xq=Jbba9suk9fr=xfr=x frpeWZqaaeaabiGaaiaadaqabeaabeqacqaaaOqaaabaaaaaaaaape WaaSaaa8aabaWdbiabgkGi2kabeA8aQ9aadaWgaaWcbaWdbiaaigda a8aabeaaaOqaa8qacqGHciITcaWG6baaaiabgkziUkaaicdacaGGSa WdaiaayIW7caaMi8+dbiaadQhacqGHsgIRcqGHsislcqGHEisPcaGG SaWdaiaayIW7caaMi8UaaGjcVlaayIW7caaMi8+dbmaalaaapaqaa8 qacqGHciITcqaHgpGApaWaaSbaaSqaa8qacaaIYaaapaqabaaakeaa peGaeyOaIyRaamOEaaaacqGHsgIRcaaIWaGaaiila8aacaaMi8UaaG jcV=qacaWG6bGaeyOKH4QaeyOhIukaaa@65ED@ (2.3)

Элементарные длины dx1 и dx2 нижней и верхней поверхностей, выраженные через длину dx срединной поверхности пластины, определяются по формулам (рис. 1a)

  d x 1 = 1+ ε x h 2 dx,d x 2 = 1+ ε x h 2 dx MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGKbGaamiEamaaBaaaleaacaaIXaaabeaaki abg2da9maabmaabaGaaGymaiabgUcaRiabew7aLnaaBaaaleaacaWG 4baabeaakmaabmaabaGaeyOeI0YaaSaaaeaacaWGObaabaGaaGOmaa aaaiaawIcacaGLPaaaaiaawIcacaGLPaaacaWGKbGaamiEaiaacYca caaMc8UaaGPaVlaadsgacaWG4bWaaSbaaSqaaiaaikdaaeqaaOGaey ypa0ZaaeWaaeaacaaIXaGaey4kaSIaeqyTdu2aaSbaaSqaaiaadIha aeqaaOWaaeWaaeaadaWcaaqaaiaadIgaaeaacaaIYaaaaaGaayjkai aawMcaaaGaayjkaiaawMcaaiaadsgacaWG4baaaa@53C8@ , (2.4)

где деформации в соответствии с гипотезами Кирхгоффа [18] равны

  ε x h 2 = h 2 2 w x 2 , ε x h 2 = h 2 2 w x 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH1oqzdaWgaaWcbaGaamiEaaqabaGcdaqada qaaiabgkHiTmaalaaabaGaamiAaaqaaiaaikdaaaaacaGLOaGaayzk aaGaeyypa0ZaaSaaaeaacaWGObaabaGaaGOmaaaadaWcaaqaaiabgk Gi2oaaCaaaleqabaGaaGOmaaaakiaadEhaaeaacqGHciITcaWG4bWa aWbaaSqabeaacaaIYaaaaaaakiaacYcacaaMc8UaaGPaVlaaykW7ca aMc8UaeqyTdu2aaSbaaSqaaiaadIhaaeqaaOWaaeWaaeaadaWcaaqa aiaadIgaaeaacaaIYaaaaaGaayjkaiaawMcaaiabg2da9iabgkHiTm aalaaabaGaamiAaaqaaiaaikdaaaWaaSaaaeaacqGHciITdaahaaWc beqaaiaaikdaaaGccaWG3baabaGaeyOaIyRaamiEamaaCaaaleqaba GaaGOmaaaaaaaaaa@58E2@ . (2.5)

Аналогично определяются элементарные длины dy1 и dy2 нижней и верхней поверхностей по оси y, выраженные через длину dy срединной поверхности пластины (рис. 1б)

d y 1 = 1+ ε y h 2 dy,d y 2 = 1+ ε y h 2 dy MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGKbGaamyEamaaBaaaleaacaaIXaaabeaaki abg2da9maabmaabaGaaGymaiabgUcaRiabew7aLnaaBaaaleaacaWG 5baabeaakmaabmaabaGaeyOeI0YaaSaaaeaacaWGObaabaGaaGOmaa aaaiaawIcacaGLPaaaaiaawIcacaGLPaaacaWGKbGaamyEaiaacYca caaMc8UaaGPaVlaadsgacaWG5bWaaSbaaSqaaiaaikdaaeqaaOGaey ypa0ZaaeWaaeaacaaIXaGaey4kaSIaeqyTdu2aaSbaaSqaaiaadMha aeqaaOWaaeWaaeaadaWcaaqaaiaadIgaaeaacaaIYaaaaaGaayjkai aawMcaaaGaayjkaiaawMcaaiaadsgacaWG5baaaa@53CE@

и деформации в соответствии с гипотезами Кирхгоффа:

ε y h 2 = h 2 2 w y 2 , ε y h 2 = h 2 2 w y 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH1oqzdaWgaaWcbaGaamyEaaqabaGcdaqada qaaiabgkHiTmaalaaabaGaamiAaaqaaiaaikdaaaaacaGLOaGaayzk aaGaeyypa0ZaaSaaaeaacaWGObaabaGaaGOmaaaadaWcaaqaaiabgk Gi2oaaCaaaleqabaGaaGOmaaaakiaadEhaaeaacqGHciITcaWG5bWa aWbaaSqabeaacaaIYaaaaaaakiaacYcacaaMc8UaaGPaVlaaykW7ca aMc8UaeqyTdu2aaSbaaSqaaiaadMhaaeqaaOWaaeWaaeaadaWcaaqa aiaadIgaaeaacaaIYaaaaaGaayjkaiaawMcaaiabg2da9iabgkHiTm aalaaabaGaamiAaaqaaiaaikdaaaWaaSaaaeaacqGHciITdaahaaWc beqaaiaaikdaaaGccaWG3baabaGaeyOaIyRaamyEamaaCaaaleqaba GaaGOmaaaaaaaaaa@58E6@ .

Распределенная сила q определяется аналогично работам [15, 16]:

qdxdy= p 0 + p 1 + p ¯ 1 d x 1 d y 1 p 0 + p 2 + p ¯ 2 d x 2 d y 2 = = p 0 + p 1 + p ¯ 1 1+ h 2 2 w x 2 + h 2 2 w y 2 + h 2 4 2 w x 2 2 w y 2 dxdy p 0 + p 2 + p ¯ 2 1 h 2 2 w x 2 h 2 2 w y 2 + h 2 4 2 w x 2 2 w y 2 dxdy, MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiaadghacaWGKbGaamiEaiaadsgacaWG5b Gaeyypa0ZaaeWaaeaacaWGWbWaaSbaaSqaaiaaicdaaeqaaOGaey4k aSIaamiCamaaBaaaleaacaaIXaaabeaakiabgUcaRiqadchagaqeam aaBaaaleaacaaIXaaabeaaaOGaayjkaiaawMcaaiaadsgacaWG4bWa aSbaaSqaaiaaigdaaeqaaOGaamizaiaadMhadaWgaaWcbaGaaGymaa qabaGccqGHsisldaqadaqaaiaadchadaWgaaWcbaGaaGimaaqabaGc cqGHRaWkcaWGWbWaaSbaaSqaaiaaikdaaeqaaOGaey4kaSIabmiCay aaraWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaGaamizaiaa dIhadaWgaaWcbaGaaGOmaaqabaGccaWGKbGaamyEamaaBaaaleaaca aIYaaabeaakiabg2da9aqaaiabg2da9maabmaabaGaamiCamaaBaaa leaacaaIWaaabeaakiabgUcaRiaadchadaWgaaWcbaGaaGymaaqaba GccqGHRaWkceWGWbGbaebadaWgaaWcbaGaaGymaaqabaaakiaawIca caGLPaaadaqadaqaaiaaigdacqGHRaWkdaWcaaqaaiaadIgaaeaaca aIYaaaamaalaaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4D aaqaaiabgkGi2kaadIhadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaS YaaSaaaeaacaWGObaabaGaaGOmaaaadaWcaaqaaiabgkGi2oaaCaaa leqabaGaaGOmaaaakiaadEhaaeaacqGHciITcaWG5bWaaWbaaSqabe aacaaIYaaaaaaakiabgUcaRmaalaaabaGaamiAamaaCaaaleqabaGa aGOmaaaaaOqaaiaaisdaaaWaaSaaaeaacqGHciITdaahaaWcbeqaai aaikdaaaGccaWG3baabaGaeyOaIyRaamiEamaaCaaaleqabaGaaGOm aaaaaaGcdaWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOmaaaakiaadE haaeaacqGHciITcaWG5bWaaWbaaSqabeaacaaIYaaaaaaaaOGaayjk aiaawMcaaiaadsgacaWG4bGaamizaiaadMhacqGHsislaeaacqGHsi sldaqadaqaaiaadchadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWG WbWaaSbaaSqaaiaaikdaaeqaaOGaey4kaSIabmiCayaaraWaaSbaaS qaaiaaikdaaeqaaaGccaGLOaGaayzkaaWaaeWaaeaacaaIXaGaeyOe I0YaaSaaaeaacaWGObaabaGaaGOmaaaadaWcaaqaaiabgkGi2oaaCa aaleqabaGaaGOmaaaakiaadEhaaeaacqGHciITcaWG4bWaaWbaaSqa beaacaaIYaaaaaaakiabgkHiTmaalaaabaGaamiAaaqaaiaaikdaaa WaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWG3baabaGa eyOaIyRaamyEamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaWcaa qaaiaadIgadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaamaalaaa baGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4DaaqaaiabgkGi2k aadIhadaahaaWcbeqaaiaaikdaaaaaaOWaaSaaaeaacqGHciITdaah aaWcbeqaaiaaikdaaaGccaWG3baabaGaeyOaIyRaamyEamaaCaaale qabaGaaGOmaaaaaaaakiaawIcacaGLPaaacaWGKbGaamiEaiaadsga caWG5bGaaiilaaaaaa@BFBA@

откуда следует:

  q= p 1 p 2 + 2 p 0 + p 1 + p 2 h 2 2 w x 2 + 2 w y 2 + p ¯ 1 p ¯ 2 + p ¯ 1 + p ¯ 2 h 2 2 w x 2 + 2 w y 2 + + p 1 p 2 h 2 4 2 w x 2 2 w y 2 + p ¯ 1 p ¯ 2 h 2 4 2 w x 2 2 w y 2 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaaceqaaiaadghacqGH9aqpcaWGWbWaaSbaaSqaai aaigdaaeqaaOGaeyOeI0IaamiCamaaBaaaleaacaaIYaaabeaakiab gUcaRmaalaaabaWaaeWaaeaacaaIYaGaamiCamaaBaaaleaacaaIWa aabeaakiabgUcaRiaadchadaWgaaWcbaGaaGymaaqabaGccqGHRaWk caWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaGaamiAaa qaaiaaikdaaaGaeyyXIC9aaeWaaeaadaWcaaqaaiabgkGi2oaaCaaa leqabaGaaGOmaaaakiaadEhaaeaacqGHciITcaWG4bWaaWbaaSqabe aacaaIYaaaaaaakiabgUcaRmaalaaabaGaeyOaIy7aaWbaaSqabeaa caaIYaaaaOGaam4DaaqaaiabgkGi2kaadMhadaahaaWcbeqaaiaaik daaaaaaaGccaGLOaGaayzkaaGaey4kaSIabmiCayaaraWaaSbaaSqa aiaaigdaaeqaaOGaeyOeI0cabaGaeyOeI0IaaGPaVlqadchagaqeam aaBaaaleaacaaIYaaabeaakiabgUcaRmaalaaabaWaaeWaaeaaceWG WbGbaebadaWgaaWcbaGaaGymaaqabaGccqGHRaWkceWGWbGbaebada WgaaWcbaGaaGOmaaqabaaakiaawIcacaGLPaaacaWGObaabaGaaGOm aaaacqGHflY1daqadaqaamaalaaabaGaeyOaIy7aaWbaaSqabeaaca aIYaaaaOGaam4DaaqaaiabgkGi2kaadIhadaahaaWcbeqaaiaaikda aaaaaOGaey4kaSYaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaa GccaWG3baabaGaeyOaIyRaamyEamaaCaaaleqabaGaaGOmaaaaaaaa kiaawIcacaGLPaaacqGHRaWkaeaacqGHRaWkdaWcaaqaamaabmaaba GaamiCamaaBaaaleaacaaIXaaabeaakiabgkHiTiaadchadaWgaaWc baGaaGOmaaqabaaakiaawIcacaGLPaaacaWGObWaaWbaaSqabeaaca aIYaaaaaGcbaGaaGinaaaadaWcaaqaaiabgkGi2oaaCaaaleqabaGa aGOmaaaakiaadEhaaeaacqGHciITcaWG4bWaaWbaaSqabeaacaaIYa aaaaaakmaalaaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4D aaqaaiabgkGi2kaadMhadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaS YaaSaaaeaadaqadaqaaiqadchagaqeamaaBaaaleaacaaIXaaabeaa kiabgkHiTiqadchagaqeamaaBaaaleaacaaIYaaabeaaaOGaayjkai aawMcaaiaadIgadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaamaa laaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4Daaqaaiabgk Gi2kaadIhadaahaaWcbeqaaiaaikdaaaaaaOWaaSaaaeaacqGHciIT daahaaWcbeqaaiaaikdaaaGccaWG3baabaGaeyOaIyRaamyEamaaCa aaleqabaGaaGOmaaaaaaGccaGGUaaaaaa@AA16@

Слагаемыми, содержащими квадрат h, можно пренебречь. В линейной задаче также пренебрегаем слагаемым, содержащим произведение среднего динамического давления на сумму вторых производных от прогиба по координатам х, y. Тогда получим:

  q= p 1 p 2 + 2 p 0 + p 1 + p 2 h 2 2 w x 2 + 2 w y 2 + p ¯ 1 p ¯ 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGXbGaeyypa0JaamiCamaaBaaaleaacaaIXa aabeaakiabgkHiTiaadchadaWgaaWcbaGaaGOmaaqabaGccqGHRaWk daWcaaqaamaabmaabaGaaGOmaiaadchadaWgaaWcbaGaaGimaaqaba GccqGHRaWkcaWGWbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiC amaaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaaiaadIgaaeaaca aIYaaaaiabgwSixpaabmaabaWaaSaaaeaacqGHciITdaahaaWcbeqa aiaaikdaaaGccaWG3baabaGaeyOaIyRaamiEamaaCaaaleqabaGaaG OmaaaaaaGccqGHRaWkdaWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOm aaaakiaadEhaaeaacqGHciITcaWG5bWaaWbaaSqabeaacaaIYaaaaa aaaOGaayjkaiaawMcaaiabgUcaRiqadchagaqeamaaBaaaleaacaaI XaaabeaakiabgkHiTiqadchagaqeamaaBaaaleaacaaIYaaabeaaaa a@5C04@ . (2.6)

По условию прямоугольная пластина по осям x и y подвижно заделана на опоры, расположенные на равных расстояниях a и b. Это означает, что

  w x,y,t x = 3 w x,y,t x 3 =0,|x|=0,a,2a,... w x,y,t y = 3 w x,y,t y 3 =0,|y|=0,b,2b,... MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaamaalaaabaGaeyOaIyRaam4Damaabmaaba GaamiEaiaacYcacaWG5bGaaiilaiaadshaaiaawIcacaGLPaaaaeaa cqGHciITcaWG4baaaiabg2da9maalaaabaGaeyOaIy7aaWbaaSqabe aacaaIZaaaaOGaam4DamaabmaabaGaamiEaiaacYcacaWG5bGaaiil aiaadshaaiaawIcacaGLPaaaaeaacqGHciITcaWG4bWaaWbaaSqabe aacaaIZaaaaaaakiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caGG8bGaamiEaiaacYhacqGH9aqpcaaIWaGaai ilaiaadggacaGGSaGaaeOmaiaadggacaGGSaGaaiOlaiaac6cacaGG UaaabaWaaSaaaeaacqGHciITcaWG3bWaaeWaaeaacaWG4bGaaiilai aadMhacaGGSaGaamiDaaGaayjkaiaawMcaaaqaaiabgkGi2kaadMha aaGaeyypa0ZaaSaaaeaacqGHciITdaahaaWcbeqaaiaaiodaaaGcca WG3bWaaeWaaeaacaWG4bGaaiilaiaadMhacaGGSaGaamiDaaGaayjk aiaawMcaaaqaaiabgkGi2kaadMhadaahaaWcbeqaaiaaiodaaaaaaO Gaeyypa0JaaGimaiaacYcacaaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaacYhacaWG5bGaaiiFaiabg2da9iaaicdacaGGSaGaamOyaiaacY cacaqGYaGaamOyaiaacYcacaGGUaGaaiOlaiaac6cacaaMc8oaaaa@8E5B@  (2.7)

Разделяя переменные w x,y,t =v x,y f t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG3bWaaeWaaeaacaWG4bGaaiilaiaadMhaca GGSaGaamiDaaGaayjkaiaawMcaaiabg2da9iaadAhadaqadaqaaiaa dIhacaGGSaGaamyEaaGaayjkaiaawMcaaiaadAgadaqadaqaaiaads haaiaawIcacaGLPaaaaaa@4185@  в уравнении (1.1) при q = 0, относительно функции v x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG2bWaaeWaaeaacaWG4bGaaiilaiaadMhaai aawIcacaGLPaaaaaa@3639@ , получим спектральную задачу:

Δ Δv λ 2 v=0 v 0,y x = 3 v 0,y x 3 = v a,y x = 3 v a,y x 3 =0,0yb  v x,0 y = 3 v x,0 y 3 = v x,b y = 3 v x,b y 3 =0,0xa.  MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiabfs5aenaabmaabaGaeuiLdqKaamODaa GaayjkaiaawMcaaiabgkHiTiabeU7aSnaaCaaaleqabaGaaGOmaaaa kiaadAhacqGH9aqpcaaIWaaabaWaaSaaaeaacqGHciITcaWG2bWaae WaaeaacaaIWaGaaiilaiaadMhaaiaawIcacaGLPaaaaeaacqGHciIT caWG4baaaiabg2da9maalaaabaGaeyOaIy7aaWbaaSqabeaacaaIZa aaaOGaamODamaabmaabaGaaGimaiaacYcacaWG5baacaGLOaGaayzk aaaabaGaeyOaIyRaamiEamaaCaaaleqabaGaaG4maaaaaaGccqGH9a qpdaWcaaqaaiabgkGi2kaadAhadaqadaqaaiaadggacaGGSaGaamyE aaGaayjkaiaawMcaaaqaaiabgkGi2kaadIhaaaGaeyypa0ZaaSaaae aacqGHciITdaahaaWcbeqaaiaaiodaaaGccaWG2bWaaeWaaeaacaWG HbGaaiilaiaadMhaaiaawIcacaGLPaaaaeaacqGHciITcaWG4bWaaW baaSqabeaacaaIZaaaaaaakiabg2da9iaaicdacaGGSaGaaGPaVlaa ykW7caaMc8UaaGPaVlaaicdacaaMc8UaeyizImQaamyEaiabgsMiJk aadkgacaqGGaaabaWaaSaaaeaacqGHciITcaWG2bWaaeWaaeaacaWG 4bGaaiilaiaaicdaaiaawIcacaGLPaaaaeaacqGHciITcaWG5baaai abg2da9maalaaabaGaeyOaIy7aaWbaaSqabeaacaaIZaaaaOGaamOD amaabmaabaGaamiEaiaacYcacaaIWaaacaGLOaGaayzkaaaabaGaey OaIyRaamyEamaaCaaaleqabaGaaG4maaaaaaGccqGH9aqpdaWcaaqa aiabgkGi2kaadAhadaqadaqaaiaadIhacaGGSaGaamOyaaGaayjkai aawMcaaaqaaiabgkGi2kaadMhaaaGaeyypa0ZaaSaaaeaacqGHciIT daahaaWcbeqaaiaaiodaaaGccaWG2bWaaeWaaeaacaWG4bGaaiilai aadkgaaiaawIcacaGLPaaaaeaacqGHciITcaWG5bWaaWbaaSqabeaa caaIZaaaaaaakiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7caaMc8 UaaGPaVlaaicdacaaMc8UaeyizImQaamiEaiabgsMiJkaadggacaGG UaGaaeiiaaaaaa@B3A5@

Собственные функции этой задачи определяются по формуле [23]

v 00 x,y = v 00 = 1 ab , v mn x,y = 2 ab cos πmx a cos πny b ,m,n=0,1,2,... MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeduuDJXwAKbYu51MyVXgarqqr1ngBPrgifHhD YfgasaacHOWxh9vrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vspGe9FjuP0=fs0xXdbba9pGe9xq=Jbba9suk9fr=xfr=x frpeWZqaaeaabiGaaiaadaqabeaabeqacqaaaOqaaabaaaaaaaaape GaamODamaaBaaaleaacaqGWaGaaeimaaqabaGcdaqadaWdaeaapeGa amiEaiaacYcacaWG5baacaGLOaGaayzkaaGaeyypa0JaamODa8aada WgaaWcbaWdbiaaicdacaaIWaaapaqabaGcpeGaeyypa0ZaaSaaa8aa baWdbiaaigdaa8aabaWdbmaakaaapaqaa8qacaWGHbGaamOyaaWcbe aaaaGccaGGSaGaamODa8aadaWgaaWcbaWdbiaad2gacaWGUbaapaqa baGcpeWaaeWaa8aabaWdbiaadIhacaGGSaGaamyEaaGaayjkaiaawM caaiabg2da9maalaaapaqaa8qacaaIYaaapaqaa8qadaGcaaWdaeaa peGaamyyaiaadkgaaSqabaaaaOGaae4yaiaab+gacaqGZbWaaSaaa8 aabaWdbiabec8aWjaad2gacaWG4baapaqaa8qacaWGHbaaaiaaboga caqGVbGaae4Camaalaaapaqaa8qacqaHapaCcaWGUbGaamyEaaWdae aapeGaamOyaaaacaGGSaWdaiaayIW7caaMi8UaaGjcV=qacaWGTbGa aiilaiaad6gacqGH9aqpcaaIWaGaaiilaiaaigdacaGGSaGaaGOmai aacYcacaGGUaGaaiOlaiaac6capaGaaGjcVdaa@7483@ , (2.8)

которые соответствуют собственным значениям:

  λ mn = π 2 m 2 a 2 + π 2 n 2 b 2 ,m,n=0,1,2,... MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeduuDJXwAKbYu51MyVXgarqqr1ngBPrgifHhD YfgasaacHOWxh9vrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vspGe9FjuP0=fs0xXdbba9pGe9xq=Jbba9suk9fr=xfr=x frpeWZqaaeaabiGaaiaadaqabeaabeqacqaaaOqaaabaaaaaaaaape Gaeq4UdW2damaaBaaaleaapeGaamyBaiaad6gaa8aabeaak8qacqGH 9aqpdaWcaaWdaeaapeGaeqiWda3damaaCaaaleqabaWdbiaaikdaaa GccaWGTbWdamaaCaaaleqabaWdbiaaikdaaaaak8aabaWdbiaadgga paWaaWbaaSqabeaapeGaaGOmaaaaaaGccqGHRaWkdaWcaaWdaeaape GaeqiWda3damaaCaaaleqabaWdbiaaikdaaaGccaWGUbWdamaaCaaa leqabaWdbiaaikdaaaaak8aabaWdbiaadkgapaWaaWbaaSqabeaape GaaGOmaaaaaaGccaGGSaGaamyBaiaacYcacaWGUbGaeyypa0JaaGim aiaacYcacaaIXaGaaiilaiaaikdacaGGSaGaaiOlaiaac6cacaGGUa aaaa@5770@ (2.9)

Отметим, что система собственных функций (2.8) является полной и образует ортонормированный базис в пространстве L2(G), где G – область переменных (x, y): 0 < x < a, 0 < y < b.

Тогда изгибные колебания пластины будем искать по формуле:

w x,y,t = W 00 t v 00 + m,n=0 m+n0 N W mn t v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG3bWaaeWaaeaacaWG4bGaaiilaiaadMhaca GGSaGaamiDaaGaayjkaiaawMcaaiabg2da9iaadEfadaWgaaWcbaGa aGimaiaaicdaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaGaam ODamaaBaaaleaacaaIWaGaaGimaaqabaGccqGHRaWkdaaeWbqaaiaa dEfadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG0baaca GLOaGaayzkaaGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqa daqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaaaWceaqabeaaca WGTbGaaiilaiaad6gacqGH9aqpcaaIWaaabaGaamyBaiabgUcaRiaa d6gacqGHGjsUcaaIWaaaaeaacaWGobaaniabggHiLdaaaa@5AF4@ . (2.10)

Функции φ i x,y,z,t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeduuDJXwAKbYu51MyVXgarqqr1ngBPrgifHhD YfgasaacHOWxh9vrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFf ea0dXdd9vspGe9FjuP0=fs0xXdbba9pGe9xq=Jbba9suk9fr=xfr=x frpeWZqaaeaabiGaaiaadaqabeaabeqacqaaaOqaaiabeA8aQnaaBa aaleaacaWGPbaabeaakmaabmaabaGaamiEaiaacYcacaWG5bGaaiil aiaadQhacaGGSaGaamiDaaGaayjkaiaawMcaaaaa@42F2@  будем искать исходя из условий (2.1), (2.3), (2.2) и (2.7) в виде:

  φ i x,y,z,t = Φ i00 z v 00 g 00 t + m,n=0 m+n>1 N Φ imn z v mn x,y g mn t ,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHgpGAdaWgaaWcbaGaamyAaaqabaGcdaqada qaaiaadIhacaGGSaGaamyEaiaacYcacaWG6bGaaiilaiaadshaaiaa wIcacaGLPaaacqGH9aqpcqqHMoGrdaWgaaWcbaGaamyAaiaaicdaca aIWaaabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiaadAhadaWg aaWcbaGaaGimaiaaicdaaeqaaOGaam4zamaaBaaaleaacaaIWaGaaG imaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaacqGHRaWkdaae WbqaaiabfA6agnaaBaaaleaacaWGPbGaamyBaiaad6gaaeqaaOWaae WaaeaacaWG6baacaGLOaGaayzkaaGaamODamaaBaaaleaacaWGTbGa amOBaaqabaGcdaqadaqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawM caaiaadEgadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG 0baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2 da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaigdaaaqa aiaad6eaa0GaeyyeIuoakiaacYcacaaMc8UaaGPaVlaaykW7caWGPb Gaeyypa0JaaGymaiaacYcacaaIYaaaaa@7505@ , (2.11)

где Φ imn z , g mn t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaamyAaiaad2gacaWGUb aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiaacYcacaWGNbWa aSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiDaaGaayjkai aawMcaaaaa@3E4E@  – пока неизвестные функции.

Подставим (2.10) в уравнение Лапласа:

  Δ φ i x,y,z,t = g 00 t v 00 x,y d 2 Φ i00 z d z 2 + + m,n=0 m+n>1 N g mn t v mn x,y d 2 Φ imn z d z 2 π 2 m 2 a 2 + π 2 n 2 b 2 Φ imn z =0 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiabgs5aejabeA8aQnaaBaaaleaacaWGPb aabeaakmaabmaabaGaamiEaiaacYcacaWG5bGaaiilaiaadQhacaGG SaGaamiDaaGaayjkaiaawMcaaiabg2da9iaadEgadaWgaaWcbaGaaG imaiaaicdaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaGaamOD amaaBaaaleaacaaIWaGaaGimaaqabaGcdaqadaqaaiaadIhacaGGSa GaamyEaaGaayjkaiaawMcaamaalaaabaGaamizamaaCaaaleqabaGa aGOmaaaakiabfA6agnaaBaaaleaacaWGPbGaaGimaiaaicdaaeqaaO WaaeWaaeaacaWG6baacaGLOaGaayzkaaaabaGaamizaiaadQhadaah aaWcbeqaaiaaikdaaaaaaOGaey4kaScabaGaey4kaSYaaabCaeaaca WGNbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiDaaGa ayjkaiaawMcaaiaadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaae WaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaadaWadaqaamaa laaabaGaamizamaaCaaaleqabaGaaGOmaaaakiabfA6agnaaBaaale aacaWGPbGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG6baacaGLOaGa ayzkaaaabaGaamizaiaadQhadaahaaWcbeqaaiaaikdaaaaaaOGaey OeI0YaaeWaaeaadaWcaaqaaiabec8aWnaaCaaaleqabaGaaGOmaaaa kiaad2gadaahaaWcbeqaaiaaikdaaaaakeaacaWGHbWaaWbaaSqabe aacaaIYaaaaaaakiabgUcaRmaalaaabaGaeqiWda3aaWbaaSqabeaa caaIYaaaaOGaamOBamaaCaaaleqabaGaaGOmaaaaaOqaaiaadkgada ahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzkaaGaeuOPdy0aaSba aSqaaiaadMgacaWGTbGaamOBaaqabaGcdaqadaqaaiaadQhaaiaawI cacaGLPaaaaiaawUfacaGLDbaacqGH9aqpcaaIWaaalqaabeqaaiaa d2gacaGGSaGaamOBaiabg2da9iaaicdaaeaacaWGTbGaey4kaSIaam OBaiabg6da+iaaigdaaaqaaiaad6eaa0GaeyyeIuoakiaac6caaaaa @9770@

Отсюда получим дифференциальные уравнения относительно неизвестных функций Φ imn z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaamyAaiaad2gacaWGUb aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaaaa@3815@ :

  d 2 Φ i00 z d z 2 =0, d 2 Φ imn z d z 2 λ mn 2 Φ imn z =0,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccqqHMoGrdaWgaaWcbaGaamyAaiaaicdacaaIWaaabeaakmaabmaa baGaamOEaaGaayjkaiaawMcaaaqaaiaadsgacaWG6bWaaWbaaSqabe aacaaIYaaaaaaakiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7daWc aaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccqqHMoGrdaWgaaWcba GaamyAaiaad2gacaWGUbaabeaakmaabmaabaGaamOEaaGaayjkaiaa wMcaaaqaaiaadsgacaWG6bWaaWbaaSqabeaacaaIYaaaaaaakiabgk HiTiabeU7aSnaaDaaaleaacaWGTbGaamOBaaqaaiaaikdaaaGccqqH MoGrdaWgaaWcbaGaamyAaiaad2gacaWGUbaabeaakmaabmaabaGaam OEaaGaayjkaiaawMcaaiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7 caWGPbGaeyypa0JaaGymaiaacYcacaaIYaaaaa@63E8@ . (2.12)

Дифференциальное уравнение (2.12) при m + n > 0 имеет общие решения:

Φ 1mn z = C 11mn exp λ mn z + C 12mn exp λ mn z Φ 2mn z = C 21mn exp λ mn z + C 22mn exp λ mn z , MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiabfA6agnaaBaaaleaacaaIXaGaamyBai aad6gaaeqaaOWaaeWaaeaacaWG6baacaGLOaGaayzkaaGaeyypa0Ja am4qamaaBaaaleaacaaIXaGaaGymaiaad2gacaWGUbaabeaakiGacw gacaGG4bGaaiiCamaabmaabaGaeq4UdW2aaSbaaSqaaiaad2gacaWG UbaabeaakiaadQhaaiaawIcacaGLPaaacqGHRaWkcaWGdbWaaSbaaS qaaiaaigdacaaIYaGaamyBaiaad6gaaeqaaOGaciyzaiaacIhacaGG WbWaaeWaaeaacqGHsislcqaH7oaBdaWgaaWcbaGaamyBaiaad6gaae qaaOGaamOEaaGaayjkaiaawMcaaaqaaiabfA6agnaaBaaaleaacaaI YaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG6baacaGLOaGaayzkaa Gaeyypa0Jaam4qamaaBaaaleaacaaIYaGaaGymaiaad2gacaWGUbaa beaakiGacwgacaGG4bGaaiiCamaabmaabaGaeq4UdW2aaSbaaSqaai aad2gacaWGUbaabeaakiaadQhaaiaawIcacaGLPaaacqGHRaWkcaWG dbWaaSbaaSqaaiaaikdacaaIYaGaamyBaiaad6gaaeqaaOGaciyzai aacIhacaGGWbWaaeWaaeaacqGHsislcqaH7oaBdaWgaaWcbaGaamyB aiaad6gaaeqaaOGaamOEaaGaayjkaiaawMcaaiaacYcaaaaa@7B4C@

а при m = n = 0

Φ 100 z = C 1100 z+ C 1200 , Φ 200 z = C 2100 z+ C 2200 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaaicdacaaIWa aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadoea daWgaaWcbaGaaGymaiaaigdacaaIWaGaaGimaaqabaGccaWG6bGaey 4kaSIaam4qamaaBaaaleaacaaIXaGaaGOmaiaaicdacaaIWaaabeaa kiaacYcacaaMc8UaeuOPdy0aaSbaaSqaaiaaikdacaaIWaGaaGimaa qabaGcdaqadaqaaiaadQhaaiaawIcacaGLPaaacqGH9aqpcaWGdbWa aSbaaSqaaiaaikdacaaIXaGaaGimaiaaicdaaeqaaOGaamOEaiabgU caRiaadoeadaWgaaWcbaGaaGOmaiaaikdacaaIWaGaaGimaaqabaaa aa@557C@ ,

где C ijmn i,j=1,2, MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGdbWaaSbaaSqaaiaadMgacaWGQbGaamyBai aad6gaaeqaaOGaaGPaVlaaykW7caWGPbGaaiilaiaadQgacqGH9aqp caaIXaGaaiilaiaaikdacaGGSaaaaa@3F4A@  – произвольные постоянные. В силу условий (2.3) при m + n > 0 находим:

Φ 1mn z = A 1mn exp λ mn z 0,  z, Φ 2mn z = A 2mn exp λ mn z 0,z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaad2gacaWGUb aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadgea daWgaaWcbaGaaGymaiaad2gacaWGUbaabeaakiGacwgacaGG4bGaai iCamaabmaabaGaeq4UdW2aaSbaaSqaaiaad2gacaWGUbaabeaakiaa dQhaaiaawIcacaGLPaaacqGHsgIRcaaIWaGaaeilaiaabccacaqGGa GaamOEaiabgkziUkabgkHiTiabg6HiLkaacYcacaaMc8UaeuOPdy0a aSbaaSqaaiaaikdacaWGTbGaamOBaaqabaGcdaqadaqaaiaadQhaai aawIcacaGLPaaacqGH9aqpcaWGbbWaaSbaaSqaaiaaikdacaWGTbGa amOBaaqabaGcciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTiabeU 7aSnaaBaaaleaacaWGTbGaamOBaaqabaGccaWG6baacaGLOaGaayzk aaGaeyOKH4QaaGimaiaacYcacaaMc8UaaGPaVlaaykW7caWG6bGaey OKH4QaeyOhIukaaa@73C7@   Φ 1mn z = A 1mn exp λ mn z 0,  z, Φ 2mn z = A 2mn exp λ mn z 0,z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaad2gacaWGUb aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadgea daWgaaWcbaGaaGymaiaad2gacaWGUbaabeaakiGacwgacaGG4bGaai iCamaabmaabaGaeq4UdW2aaSbaaSqaaiaad2gacaWGUbaabeaakiaa dQhaaiaawIcacaGLPaaacqGHsgIRcaaIWaGaaeilaiaabccacaqGGa GaamOEaiabgkziUkabgkHiTiabg6HiLkaacYcacaaMc8UaeuOPdy0a aSbaaSqaaiaaikdacaWGTbGaamOBaaqabaGcdaqadaqaaiaadQhaai aawIcacaGLPaaacqGH9aqpcaWGbbWaaSbaaSqaaiaaikdacaWGTbGa amOBaaqabaGcciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTiabeU 7aSnaaBaaaleaacaWGTbGaamOBaaqabaGccaWG6baacaGLOaGaayzk aaGaeyOKH4QaaGimaiaacYcacaaMc8UaaGPaVlaaykW7caWG6bGaey OKH4QaeyOhIukaaa@73C7@ ,

а при m = n = 0

Φ 100 z = A 100 , Φ 200 z = A 200 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaaicdacaaIWa aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadgea daWgaaWcbaGaaGymaiaaicdacaaIWaaabeaakiaacYcacaaMc8Uaeu OPdy0aaSbaaSqaaiaaikdacaaIWaGaaGimaaqabaGcdaqadaqaaiaa dQhaaiaawIcacaGLPaaacqGH9aqpcaWGbbWaaSbaaSqaaiaaikdaca aIWaGaaGimaaqabaaaaa@486D@ .

Здесь постоянные A1mn и A2mn – неизвестные, для определения которых воспользуемся условиями (2.2). Для этого воспользуемся формулой Тейлора для разложения функции ∂φ1/∂z в окрестности точки z = 0:

φ 1 z = φ 1 x,y,0,t z + z 1! 2 φ 1 x,y,0,t z 2 + z 2 2! 3 φ 1 x,y,θ,t z 3 ,0 <θ<z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiabgkGi2kabeA8aQnaaBaaaleaaca aIXaaabeaaaOqaaiabgkGi2kaadQhaaaGaeyypa0ZaaSaaaeaacqGH ciITcqaHgpGAdaWgaaWcbaGaaGymaaqabaGcdaqadaqaaiaadIhaca GGSaGaamyEaiaacYcacaaIWaGaaiilaiaadshaaiaawIcacaGLPaaa aeaacqGHciITcaWG6baaaiabgUcaRmaalaaabaGaamOEaaqaaiaaig dacaGGHaaaaiaaykW7daWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOm aaaakiabeA8aQnaaBaaaleaacaaIXaaabeaakmaabmaabaGaamiEai aacYcacaWG5bGaaiilaiaaicdacaGGSaGaamiDaaGaayjkaiaawMca aaqaaiabgkGi2kaadQhadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaS YaaSaaaeaacaWG6bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaiaa cgcaaaGaaGPaVpaalaaabaGaeyOaIy7aaWbaaSqabeaacaaIZaaaaO GaeqOXdO2aaSbaaSqaaiaaigdaaeqaaOWaaeWaaeaacaWG4bGaaiil aiaadMhacaGGSaGaeqiUdeNaaiilaiaadshaaiaawIcacaGLPaaaae aacqGHciITcaWG6bWaaWbaaSqabeaacaaIZaaaaaaakiaacYcacaaM c8UaaGPaVlaaykW7caaMc8UaaGimaiaabccacqGH8aapcqaH4oqCcq GH8aapcaWG6baaaa@7E75@ . (2.13)

Пренебрегая последним слагаемым с учетом малости h2 и первого условия из (2.2), имеем:

  φ 1 z = φ 1 x,y,0,t z h 2 2 φ 1 x,y,0,t z 2 = w t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiabgkGi2kabeA8aQnaaBaaaleaaca aIXaaabeaaaOqaaiabgkGi2kaadQhaaaGaeyypa0ZaaSaaaeaacqGH ciITcqaHgpGAdaWgaaWcbaGaaGymaaqabaGcdaqadaqaaiaadIhaca GGSaGaamyEaiaacYcacaaIWaGaaiilaiaadshaaiaawIcacaGLPaaa aeaacqGHciITcaWG6baaaiabgkHiTmaalaaabaGaamiAaaqaaiaaik daaaGaaGPaVpaalaaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGa eqOXdO2aaSbaaSqaaiaaigdaaeqaaOWaaeWaaeaacaWG4bGaaiilai aadMhacaGGSaGaaGimaiaacYcacaWG0baacaGLOaGaayzkaaaabaGa eyOaIyRaamOEamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpdaWcaa qaaiabgkGi2kaadEhaaeaacqGHciITcaWG0baaaaaa@6027@ . (2.14)

Подставляя в (2.14) функции (2.11), (2.10), получим:

m,n=0 m+n>0 N λ mn A 1mn v mn x,y g mn t h 2 A 1mn λ mn 2 v mn x,y g mn t = d W 00 t dt v 00 + m,n=0 m+n>0 N d W mn t dt v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaaeWbqaaiabeU7aSnaaBaaaleaacaWGTbGaam OBaaqabaGccaWGbbWaaSbaaSqaaiaaigdacaWGTbGaamOBaaqabaGc caWG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEai aacYcacaWG5baacaGLOaGaayzkaaGaam4zamaaBaaaleaacaWGTbGa amOBaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaacqGHsislda WcaaqaaiaadIgaaeaacaaIYaaaaaWceaqabeaacaWGTbGaaiilaiaa d6gacqGH9aqpcaaIWaaabaGaamyBaiabgUcaRiaad6gacqGH+aGpca aIWaaaaeaacaWGobaaniabggHiLdGccaWGbbWaaSbaaSqaaiaaigda caWGTbGaamOBaaqabaGccqaH7oaBdaqhaaWcbaGaamyBaiaad6gaae aacaaIYaaaaOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqa daqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaaiaadEgadaWgaa WcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzk aaGaeyypa0ZaaSaaaeaacaWGKbGaam4vamaaBaaaleaacaaIWaGaaG imaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaaaeaacaWGKbGa amiDaaaacaWG2bWaaSbaaSqaaiaaicdacaaIWaaabeaakiabgUcaRm aaqahabaWaaSaaaeaacaWGKbGaam4vamaaBaaaleaacaWGTbGaamOB aaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaaaeaacaWGKbGaam iDaaaacaWG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGa amiEaiaacYcacaWG5baacaGLOaGaayzkaaaalqaabeqaaiaad2gaca GGSaGaamOBaiabg2da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiab g6da+iaaicdaaaqaaiaad6eaa0GaeyyeIuoakiaaykW7caaMc8oaaa@9415@

или

  m,n=0 m+n>0 N A 1mn λ mn 1 λ mn h 2 v mn x,y g mn t = d W 00 t dt v 00 + m,n=0 m+n>0 N d W mn t dt v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaaeWbqaaiaadgeadaWgaaWcbaGaaGymaiaad2 gacaWGUbaabeaakiabeU7aSnaaBaaaleaacaWGTbGaamOBaaqabaGc daqadaqaaiaaigdacqGHsisldaWcaaqaaiabeU7aSnaaBaaaleaaca WGTbGaamOBaaqabaGccaWGObaabaGaaGOmaaaaaiaawIcacaGLPaaa caWG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEai aacYcacaWG5baacaGLOaGaayzkaaGaam4zamaaBaaaleaacaWGTbGa amOBaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaaaSabaeqaba GaamyBaiaacYcacaWGUbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWk caWGUbGaeyOpa4JaaGimaaaabaGaamOtaaqdcqGHris5aOGaeyypa0 ZaaSaaaeaacaWGKbGaam4vamaaBaaaleaacaaIWaGaaGimaaqabaGc daqadaqaaiaadshaaiaawIcacaGLPaaaaeaacaWGKbGaamiDaaaaca WG2bWaaSbaaSqaaiaaicdacaaIWaaabeaakiabgUcaRmaaqahabaWa aSaaaeaacaWGKbGaam4vamaaBaaaleaacaWGTbGaamOBaaqabaGcda qadaqaaiaadshaaiaawIcacaGLPaaaaeaacaWGKbGaamiDaaaacaWG 2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacY cacaWG5baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGaamOB aiabg2da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaic daaaqaaiaad6eaa0GaeyyeIuoaaaa@820D@ . (2.15)

При выполнении условий

  d W 00 t dt =0, d W mn t dt = ω ¯ mn g mn t ,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaadsgacaWGxbWaaSbaaSqaaiaaic dacaaIWaaabeaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaa dsgacaWG0baaaiabg2da9iaaicdacaGGSaWaaSaaaeaacaWGKbGaam 4vamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadshaaiaa wIcacaGLPaaaaeaacaWGKbGaamiDaaaacqGH9aqpcuaHjpWDgaqeam aaBaaaleaacaWGTbGaamOBaaqabaGccaWGNbWaaSbaaSqaaiaad2ga caWGUbaabeaakmaabmaabaGaamiDaaGaayjkaiaawMcaaiaacYcaca WGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaa@5379@ (2.16)

из равенства (2.15) найдем:

  A 1mn = ω ¯ mn λ mn 1 λ mn h 2 1 ,m,n=0,1,2,...,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGbbWaaSbaaSqaaiaaigdacaWGTbGaamOBaa qabaGccqGH9aqpdaWcaaqaaiqbeM8a3zaaraWaaSbaaSqaaiaad2ga caWGUbaabeaaaOqaaiabeU7aSnaaBaaaleaacaWGTbGaamOBaaqaba aaaOWaaeWaaeaacaaIXaGaeyOeI0YaaSaaaeaacqaH7oaBdaWgaaWc baGaamyBaiaad6gaaeqaaOGaamiAaaqaaiaaikdaaaaacaGLOaGaay zkaaWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaaiilaiaad2gacaGG SaGaamOBaiabg2da9iaaicdacaGGSaGaaGymaiaacYcacaaIYaGaai ilaiaac6cacaGGUaGaaiOlaiaacYcacaWGTbGaey4kaSIaamOBaiab g6da+iaaicdaaaa@57FE@ . (2.17)

Аналогично находим постоянные:

  A 2mn = ω ¯ mn λ mn 1 λ mn h 2 1 ,m,n=0,1,2,...,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGbbWaaSbaaSqaaiaaikdacaWGTbGaamOBaa qabaGccqGH9aqpcqGHsisldaWcaaqaaiqbeM8a3zaaraWaaSbaaSqa aiaad2gacaWGUbaabeaaaOqaaiabeU7aSnaaBaaaleaacaWGTbGaam OBaaqabaaaaOWaaeWaaeaacaaIXaGaeyOeI0YaaSaaaeaacqaH7oaB daWgaaWcbaGaamyBaiaad6gaaeqaaOGaamiAaaqaaiaaikdaaaaaca GLOaGaayzkaaWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaaiilaiaa d2gacaGGSaGaamOBaiabg2da9iaaicdacaGGSaGaaGymaiaacYcaca aIYaGaaiilaiaac6cacaGGUaGaaiOlaiaacYcacaWGTbGaey4kaSIa amOBaiabg6da+iaaicdaaaa@58EC@ . (2.18)

Таким образом, у функций φ i x,y,z,t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHgpGAdaWgaaWcbaGaamyAaaqabaGcdaqada qaaiaadIhacaGGSaGaamyEaiaacYcacaWG6bGaaiilaiaadshaaiaa wIcacaGLPaaaaaa@3B77@  найдены постоянные A1mn и A2mn при m + n > 0, которые определяются по формулам (2.17) и (2.18), а A100 и A200 остаются произвольными постоянными.

Также, используя формулы (2.13), (2.16), (2.17) и (2.18), определим динамические давления:

  p ¯ 1 = ρ 1 φ 1 x,y,0,t t h 2 2 φ 1 x,y,0,t tz = = ρ 1 A 100 v 00 d g 00 t dt + + m,n=0 m+n>0 N A 1mn v mn x,y d g mn t dt h 2 m,n=0 m+n>0 N A 1mn λ mn v mn x,y d g mn t dt = = ρ 1 A 100 v 00 d g 00 t dt + m,n=0 m+n>0 N A 1mn v mn x,y d g mn t dt 1 λ mn h 2 = = ρ 1 A 100 v 00 d g 00 t dt + m,n=0 m+n>0 N ω ¯ mn λ mn v mn x,y d g mn t dt = = ρ 1 A 100 v 00 d g 00 t dt + m,n=0 m+n>0 N 1 λ mn v mn x,y d 2 W mn t d t 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaaceqaaiqadchagaqeamaaBaaaleaacaaIXaaabe aakiabg2da9iabgkHiTiabeg8aYnaaBaaaleaacaaIXaaabeaakmaa dmaabaWaaSaaaeaacqGHciITcqaHgpGAdaWgaaWcbaGaaGymaaqaba GcdaqadaqaaiaadIhacaGGSaGaamyEaiaacYcacaaIWaGaaiilaiaa dshaaiaawIcacaGLPaaaaeaacqGHciITcaWG0baaaiabgkHiTmaala aabaGaamiAaaqaaiaaikdaaaGaaGPaVpaalaaabaGaeyOaIy7aaWba aSqabeaacaaIYaaaaOGaeqOXdO2aaSbaaSqaaiaaigdaaeqaaOWaae WaaeaacaWG4bGaaiilaiaadMhacaGGSaGaaGimaiaacYcacaWG0baa caGLOaGaayzkaaaabaGaeyOaIyRaamiDaiabgkGi2kaadQhaaaaaca GLBbGaayzxaaGaeyypa0dabaGaeyypa0JaeyOeI0IaeqyWdi3aaSba aSqaaiaaigdaaeqaaOWaaiqaaeaacaWGbbWaaSbaaSqaaiaaigdaca aIWaGaaGimaaqabaGccaWG2bWaaSbaaSqaaiaaicdacaaIWaaabeaa kmaalaaabaGaamizaiaadEgadaWgaaWcbaGaaGimaiaaicdaaeqaaO WaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizaiaadshaaaGa ey4kaScacaGL7baaaeaacqGHRaWkdaWadaqaamaaqahabaGaamyqam aaBaaaleaacaaIXaGaamyBaiaad6gaaeqaaOGaamODamaaBaaaleaa caWGTbGaamOBaaqabaGcdaqadaqaaiaadIhacaGGSaGaamyEaaGaay jkaiaawMcaamaalaaabaGaamizaiaadEgadaWgaaWcbaGaamyBaiaa d6gaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizai aadshaaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2da9iaaicda aeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaqaaiaad6eaa0 GaeyyeIuoakiabgkHiTmaalaaabaGaamiAaaqaaiaaikdaaaWaaabC aeaacaWGbbWaaSbaaSqaaiaaigdacaWGTbGaamOBaaqabaGccqaH7o aBdaWgaaWcbaGaamyBaiaad6gaaeqaaOGaamODamaaBaaaleaacaWG TbGaamOBaaqabaGcdaqadaqaaiaadIhacaGGSaGaamyEaaGaayjkai aawMcaamaalaaabaGaamizaiaadEgadaWgaaWcbaGaamyBaiaad6ga aeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizaiaads haaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2da9iaaicdaaeaa caWGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaqaaiaad6eaa0Gaey yeIuoaaOGaay5waiaaw2faaiabg2da9aqaaiabg2da9iabgkHiTiab eg8aYnaaBaaaleaacaaIXaaabeaakmaadmaabaGaamyqamaaBaaale aacaaIXaGaaGimaiaaicdaaeqaaOGaamODamaaBaaaleaacaaIWaGa aGimaaqabaGcdaWcaaqaaiaadsgacaWGNbWaaSbaaSqaaiaaicdaca aIWaaabeaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaadsga caWG0baaaiabgUcaRmaaqahabaGaamyqamaaBaaaleaacaaIXaGaam yBaiaad6gaaeqaaOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGc daqadaqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaamaalaaaba GaamizaiaadEgadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaa caWG0baacaGLOaGaayzkaaaabaGaamizaiaadshaaaaalqaabeqaai aad2gacaGGSaGaamOBaiabg2da9iaaicdaaeaacaWGTbGaey4kaSIa amOBaiabg6da+iaaicdaaaqaaiaad6eaa0GaeyyeIuoakmaabmaaba GaaGymaiabgkHiTmaalaaabaGaeq4UdW2aaSbaaSqaaiaad2gacaWG UbaabeaakiaadIgaaeaacaaIYaaaaaGaayjkaiaawMcaaaGaay5wai aaw2faaiabg2da9aqaaiabg2da9iabgkHiTiabeg8aYnaaBaaaleaa caaIXaaabeaakmaadmaabaGaamyqamaaBaaaleaacaaIXaGaaGimai aaicdaaeqaaOGaamODamaaBaaaleaacaaIWaGaaGimaaqabaGcdaWc aaqaaiaadsgacaWGNbWaaSbaaSqaaiaaicdacaaIWaaabeaakmaabm aabaGaamiDaaGaayjkaiaawMcaaaqaaiaadsgacaWG0baaaiabgUca RmaaqahabaWaaSaaaeaacuaHjpWDgaqeamaaBaaaleaacaWGTbGaam OBaaqabaaakeaacqaH7oaBdaWgaaWcbaGaamyBaiaad6gaaeqaaaaa kiaadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG4b GaaiilaiaadMhaaiaawIcacaGLPaaadaWcaaqaaiaadsgacaWGNbWa aSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiDaaGaayjkai aawMcaaaqaaiaadsgacaWG0baaaaWceaqabeaacaWGTbGaaiilaiaa d6gacqGH9aqpcaaIWaaabaGaamyBaiabgUcaRiaad6gacqGH+aGpca aIWaaaaeaacaWGobaaniabggHiLdaakiaawUfacaGLDbaacqGH9aqp aeaacqGH9aqpcqGHsislcqaHbpGCdaWgaaWcbaGaaGymaaqabaGcda WadaqaaiaadgeadaWgaaWcbaGaaGymaiaaicdacaaIWaaabeaakiaa dAhadaWgaaWcbaGaaGimaiaaicdaaeqaaOWaaSaaaeaacaWGKbGaam 4zamaaBaaaleaacaaIWaGaaGimaaqabaGcdaqadaqaaiaadshaaiaa wIcacaGLPaaaaeaacaWGKbGaamiDaaaacqGHRaWkdaaeWbqaamaala aabaGaaGymaaqaaiabeU7aSnaaBaaaleaacaWGTbGaamOBaaqabaaa aOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadI hacaGGSaGaamyEaaGaayjkaiaawMcaamaalaaabaGaamizamaaCaaa leqabaGaaGOmaaaakiaadEfadaWgaaWcbaGaamyBaiaad6gaaeqaaO WaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizaiaadshadaah aaWcbeqaaiaaikdaaaaaaaabaeqabaGaamyBaiaacYcacaWGUbGaey ypa0JaaGimaaqaaiaad2gacqGHRaWkcaWGUbGaeyOpa4JaaGimaaaa baGaamOtaaqdcqGHris5aaGccaGLBbGaayzxaaaaaaa@6A7D@  (2.19)

p¯2=ρ2φ2x,y,0,tt+h22φ2x,y,0,ttz==ρ2A200v00dg00tdt++m,n=0m+n>0NA2mnvmnx,ydgmntdth2m,n=0m+n>0NA2mnλmnvmnx,ydgmntdt==ρ2A200v00dg00tdt+m,n=0m+n>0NA2mnvmnx,ydgmntdt1λmnh2=

=ρ2A200v00dg00tdtm,n=0m+n>0Nω¯mnλmnvmnx,ydgmntdt==ρ2A200v00dg00tdtm,n=0m+n>0N1λmnvmnx,yd2Wmntdt2. (2.20)

Теперь на основании формулы (2.6) с учетом (2.19), (2.20) найдем:

  q= p 1 p 2 + ρ 2 A 200 ρ 1 A 100 v 00 d g 00 t dt + + 2 p 0 + p 1 + p 2 h 2 2 w x 2 + 2 w y 2 ρ 1 + ρ 2 m,n=0 m+n>0 N 1 λ mn v mn x,y d 2 W mn t d t 2 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiaadghacqGH9aqpcaWGWbWaaSbaaSqaai aaigdaaeqaaOGaeyOeI0IaamiCamaaBaaaleaacaaIYaaabeaakiab gUcaRmaabmaabaGaeqyWdi3aaSbaaSqaaiaaikdaaeqaaOGaamyqam aaBaaaleaacaaIYaGaaGimaiaaicdaaeqaaOGaeyOeI0IaeqyWdi3a aSbaaSqaaiaaigdaaeqaaOGaamyqamaaBaaaleaacaaIXaGaaGimai aaicdaaeqaaaGccaGLOaGaayzkaaGaamODamaaBaaaleaacaaIWaGa aGimaaqabaGcdaWcaaqaaiaadsgacaWGNbWaaSbaaSqaaiaaicdaca aIWaaabeaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaadsga caWG0baaaiabgUcaRaqaaiabgUcaRiaaykW7daWcaaqaamaabmaaba GaaGOmaiaadchadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWGWbWa aSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiCamaaBaaaleaacaaIYa aabeaaaOGaayjkaiaawMcaaiaadIgaaeaacaaIYaaaaiabgwSixpaa bmaabaWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWG3b aabaGaeyOaIyRaamiEamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWk daWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOmaaaakiaadEhaaeaacq GHciITcaWG5bWaaWbaaSqabeaacaaIYaaaaaaaaOGaayjkaiaawMca aiabgkHiTaqaaiabgkHiTmaabmaabaGaeqyWdi3aaSbaaSqaaiaaig daaeqaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaGccaGL OaGaayzkaaWaaabCaeaadaWcaaqaaiaaigdaaeaacqaH7oaBdaWgaa WcbaGaamyBaiaad6gaaeqaaaaakiaadAhadaWgaaWcbaGaamyBaiaa d6gaaeqaaOWaaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPa aadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWGxbWaaSba aSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiDaaGaayjkaiaawM caaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaaqaabeqa aiaad2gacaGGSaGaamOBaiabg2da9iaaicdaaeaacaWGTbGaey4kaS IaamOBaiabg6da+iaaicdaaaqaaiaad6eaa0GaeyyeIuoakiaac6ca aaaa@9EC8@ (2.21)

Подставляя выражение (2.21) в уравнение (1.1), получим:

D 4 w x 4 +2 4 w x 2 y 2 + 4 w y 4 +ρh 2 w t 2 2 p 0 + p 1 + p 2 h 2 2 w x 2 + 2 w y 2 + + ρ 1 + ρ 2 m,n=0 m+n>0 N 1 λ mn v mn x,y d 2 W mn t d t 2 ρ 2 A 200 ρ 1 A 100 v 00 d g 00 t dt = p 1 p 2 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaaceqaaiaadseadaqadaqaamaalaaabaGaeyOaIy 7aaWbaaSqabeaacaaI0aaaaOGaam4DaaqaaiabgkGi2kaadIhadaah aaWcbeqaaiaaisdaaaaaaOGaey4kaSIaaGOmamaalaaabaGaeyOaIy 7aaWbaaSqabeaacaaI0aaaaOGaam4DaaqaaiabgkGi2kaadIhadaah aaWcbeqaaiaaikdaaaGccqGHciITcaWG5bWaaWbaaSqabeaacaaIYa aaaaaakiabgUcaRmaalaaabaGaeyOaIy7aaWbaaSqabeaacaaI0aaa aOGaam4DaaqaaiabgkGi2kaadMhadaahaaWcbeqaaiaaisdaaaaaaa GccaGLOaGaayzkaaGaey4kaSIaeqyWdiNaamiAamaalaaabaGaeyOa Iy7aaWbaaSqabeaacaaIYaaaaOGaam4DaaqaaiabgkGi2kaadshada ahaaWcbeqaaiaaikdaaaaaaOGaeyOeI0YaaeWaaeaacaaIYaGaamiC amaaBaaaleaacaaIWaaabeaakiabgUcaRiaadchadaWgaaWcbaGaaG ymaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGL OaGaayzkaaWaaSaaaeaacaWGObaabaGaaGOmaaaacqGHflY1daqada qaamaalaaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4Daaqa aiabgkGi2kaadIhadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaS aaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWG3baabaGaeyOa IyRaamyEamaaCaaaleqabaGaaGOmaaaaaaaakiaawIcacaGLPaaacq GHRaWkaeaacqGHRaWkdaqadaqaaiabeg8aYnaaBaaaleaacaaIXaaa beaakiabgUcaRiabeg8aYnaaBaaaleaacaaIYaaabeaaaOGaayjkai aawMcaamaaqahabaWaaSaaaeaacaaIXaaabaGaeq4UdW2aaSbaaSqa aiaad2gacaWGUbaabeaaaaGccaWG2bWaaSbaaSqaaiaad2gacaWGUb aabeaakmaabmaabaGaamiEaiaacYcacaWG5baacaGLOaGaayzkaaWa aSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam4vamaaBaaale aacaWGTbGaamOBaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaa aeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaaaaeaqabeaaca WGTbGaaiilaiaad6gacqGH9aqpcaaIWaaabaGaamyBaiabgUcaRiaa d6gacqGH+aGpcaaIWaaaaeaacaWGobaaniabggHiLdGccqGHsislae aacqGHsislcaaMc8+aaeWaaeaacqaHbpGCdaWgaaWcbaGaaGOmaaqa baGccaWGbbWaaSbaaSqaaiaaikdacaaIWaGaaGimaaqabaGccqGHsi slcqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaWGbbWaaSbaaSqaaiaa igdacaaIWaGaaGimaaqabaaakiaawIcacaGLPaaacaWG2bWaaSbaaS qaaiaaicdacaaIWaaabeaakmaalaaabaGaamizaiaadEgadaWgaaWc baGaaGimaiaaicdaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaa aabaGaamizaiaadshaaaGaeyypa0JaamiCamaaBaaaleaacaaIXaaa beaakiabgkHiTiaadchadaWgaaWcbaGaaGOmaaqabaGccaGGUaaaaa a@C489@  (2.22)

Подставляя в (2.22) функции (2.7) при p1p2 = p, будем иметь:

D m,n=0 m+n>0 N π 4 m 4 a 4 +2 π 4 m 2 n 2 a 2 b 2 + π 4 n 4 b 4 v mn x,y W mn t + +ρh m,n=1 N v mn x,y d 2 W mn t d t 2 + + 2 p 0 + p 1 + p 2 h 2 m,n=0 m+n>0 N λ mn 2 v mn x,y W mn t + + ρ 1 + ρ 2 m,n=0 m+n>0 N 1 λ mn v mn x,y d 2 W mn t d t 2 ρ 2 A 200 ρ 1 A 100 v 00 d g 00 t dt = = m,n=0 N p mn v mn x,y = p 00 v 00 + m,n=0 m+n>0 N p mn v mn x,y , MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaaceqaaiaadseadaWadaqaamaaqahabaWaaeWaae aadaWcaaqaaiabec8aWnaaCaaaleqabaGaaGinaaaakiaad2gadaah aaWcbeqaaiaaisdaaaaakeaacaWGHbWaaWbaaSqabeaacaaI0aaaaa aakiabgUcaRiaaikdadaWcaaqaaiabec8aWnaaCaaaleqabaGaaGin aaaakiaad2gadaahaaWcbeqaaiaaikdaaaGccaWGUbWaaWbaaSqabe aacaaIYaaaaaGcbaGaamyyamaaCaaaleqabaGaaGOmaaaakiaadkga daahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacqaHapaCda ahaaWcbeqaaiaaisdaaaGccaWGUbWaaWbaaSqabeaacaaI0aaaaaGc baGaamOyamaaCaaaleqabaGaaGinaaaaaaaakiaawIcacaGLPaaaca WG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaa cYcacaWG5baacaGLOaGaayzkaaGaam4vamaaBaaaleaacaWGTbGaam OBaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaaaSabaeqabaGa amyBaiaacYcacaWGUbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWkca WGUbGaeyOpa4JaaGimaaaabaGaamOtaaqdcqGHris5aaGccaGLBbGa ayzxaaGaey4kaScabaGaey4kaSIaaGPaVlabeg8aYjaadIgadaaeWb qaaiaadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG 4bGaaiilaiaadMhaaiaawIcacaGLPaaadaWcaaqaaiaadsgadaahaa WcbeqaaiaaikdaaaGccaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaa kmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaadsgacaWG0bWaaW baaSqabeaacaaIYaaaaaaaaeaacaWGTbGaaiilaiaad6gacqGH9aqp caaIXaaabaGaamOtaaqdcqGHris5aOGaey4kaScabaGaey4kaSYaae WaaeaacaaIYaGaamiCamaaBaaaleaacaaIWaaabeaakiabgUcaRiaa dchadaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaai aaikdaaeqaaaGccaGLOaGaayzkaaWaaSaaaeaacaWGObaabaGaaGOm aaaacqGHflY1daaeWbqaaiabeU7aSnaaDaaaleaacaWGTbGaamOBaa qaaiaaikdaaaGccaWG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaa bmaabaGaamiEaiaacYcacaWG5baacaGLOaGaayzkaaGaam4vamaaBa aaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadshaaiaawIcacaGL PaaaaSabaeqabaGaamyBaiaacYcacaWGUbGaeyypa0JaaGimaaqaai aad2gacqGHRaWkcaWGUbGaeyOpa4JaaGimaaaabaGaamOtaaqdcqGH ris5aOGaey4kaScabaGaey4kaSIaaGPaVpaabmaabaGaeqyWdi3aaS baaSqaaiaaigdaaeqaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaikda aeqaaaGccaGLOaGaayzkaaWaaabCaeaadaWcaaqaaiaaigdaaeaacq aH7oaBdaWgaaWcbaGaamyBaiaad6gaaeqaaaaakiaadAhadaWgaaWc baGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG4bGaaiilaiaadMhaai aawIcacaGLPaaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGc caWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiDaa GaayjkaiaawMcaaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYaaa aaaaaqaabeqaaiaad2gacaGGSaGaamOBaiabg2da9iaaicdaaeaaca WGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaqaaiaad6eaa0Gaeyye IuoakiabgkHiTaqaaiabgkHiTmaabmaabaGaeqyWdi3aaSbaaSqaai aaikdaaeqaaOGaamyqamaaBaaaleaacaaIYaGaaGimaiaaicdaaeqa aOGaeyOeI0IaeqyWdi3aaSbaaSqaaiaaigdaaeqaaOGaamyqamaaBa aaleaacaaIXaGaaGimaiaaicdaaeqaaaGccaGLOaGaayzkaaGaamOD amaaBaaaleaacaaIWaGaaGimaaqabaGcdaWcaaqaaiaadsgacaWGNb WaaSbaaSqaaiaaicdacaaIWaaabeaakmaabmaabaGaamiDaaGaayjk aiaawMcaaaqaaiaadsgacaWG0baaaiabg2da9aqaaiabg2da9maaqa habaGaamiCamaaBaaaleaacaWGTbGaamOBaaqabaGccaWG2bWaaSba aSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG5b aacaGLOaGaayzkaaaaleaacaWGTbGaaiilaiaad6gacqGH9aqpcaaI WaaabaGaamOtaaqdcqGHris5aOGaeyypa0JaamiCamaaBaaaleaaca aIWaGaaGimaaqabaGccaWG2bWaaSbaaSqaaiaaicdacaaIWaaabeaa kiabgUcaRmaaqahabaGaamiCamaaBaaaleaacaWGTbGaamOBaaqaba GccaWG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiE aiaacYcacaWG5baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSa GaamOBaiabg2da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da +iaaicdaaaqaaiaad6eaa0GaeyyeIuoakiaacYcaaaaa@2D80@

где

p mn = G p v mn x,y dxdy = p ab ,m=n=0 0,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGWbWaaSbaaSqaaiaad2gacaWGUbaabeaaki abg2da9maapifabaGaamiCaiaadAhadaWgaaWcbaGaamyBaiaad6ga aeqaaOWaaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaaca WGKbGaamiEaiaadsgacaWG5baaleaacaWGhbaabeqdcqGHRiI8cqGH RiI8aOGaeyypa0ZaaiqaaeaafaqabeGabaaabaGaamiCamaakaaaba GaamyyaiaadkgaaSqabaGccaGGSaGaaGPaVlaaykW7caWGTbGaeyyp a0JaamOBaiabg2da9iaaicdaaeaacaaIWaGaaiilaiaaykW7caaMc8 UaamyBaiabgUcaRiaad6gacqGH+aGpcaaIWaaaaaGaay5Eaaaaaa@5CA6@ .

Отсюда в силу полноты системы функций (2.8) в L2(G) получим дифференциальные уравнения:

ρ 1 A 100 ρ 2 A 200 d g 00 t dt = p 00 ,m=n=0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaqadaqaaiabeg8aYnaaBaaaleaacaaIXaaabe aakiaadgeadaWgaaWcbaGaaGymaiaaicdacaaIWaaabeaakiabgkHi Tiabeg8aYnaaBaaaleaacaaIYaaabeaakiaadgeadaWgaaWcbaGaaG OmaiaaicdacaaIWaaabeaaaOGaayjkaiaawMcaamaalaaabaGaamiz aiaadEgadaWgaaWcbaGaaGimaiaaicdaaeqaaOWaaeWaaeaacaWG0b aacaGLOaGaayzkaaaabaGaamizaiaadshaaaGaeyypa0JaamiCamaa BaaaleaacaaIWaGaaGimaaqabaGccaGGSaGaaGPaVlaaykW7caWGTb Gaeyypa0JaamOBaiabg2da9iaaicdaaaa@5343@

D λ mn 4 + 2 p 0 + p 1 + p 2 h 2 λ mn 2 W mn t + ρh+ ρ 1 + ρ 2 λ mn d 2 W mn t d t 2 =0,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWadaqaaiaadseacqaH7oaBdaqhaaWcbaGaam yBaiaad6gaaeaacaaI0aaaaOGaey4kaSYaaeWaaeaacaaIYaGaamiC amaaBaaaleaacaaIWaaabeaakiabgUcaRiaadchadaWgaaWcbaGaaG ymaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGL OaGaayzkaaWaaSaaaeaacaWGObaabaGaaGOmaaaacqaH7oaBdaqhaa WcbaGaamyBaiaad6gaaeaacaaIYaaaaaGccaGLBbGaayzxaaGaam4v amaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadshaaiaawI cacaGLPaaacqGHRaWkdaqadaqaaiabeg8aYjaadIgacqGHRaWkdaWc aaqaaiabeg8aYnaaBaaaleaacaaIXaaabeaakiabgUcaRiabeg8aYn aaBaaaleaacaaIYaaabeaaaOqaaiabeU7aSnaaBaaaleaacaWGTbGa amOBaaqabaaaaaGccaGLOaGaayzkaaWaaSaaaeaacaWGKbWaaWbaaS qabeaacaaIYaaaaOGaam4vamaaBaaaleaacaWGTbGaamOBaaqabaGc daqadaqaaiaadshaaiaawIcacaGLPaaaaeaacaWGKbGaamiDamaaCa aaleqabaGaaGOmaaaaaaGccqGH9aqpcaaIWaGaaiilaiaaykW7caaM c8UaaGPaVlaad2gacqGHRaWkcaWGUbGaeyOpa4JaaGimaaaa@74A0@ .

Из первого уравнения при условии ρ 1 A 100 ρ 2 A 200 0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaWGbb WaaSbaaSqaaiaaigdacaaIWaGaaGimaaqabaGccqGHsislcqaHbpGC daWgaaWcbaGaaGOmaaqabaGccaWGbbWaaSbaaSqaaiaaikdacaaIWa GaaGimaaqabaGccqGHGjsUcaaIWaaaaa@4032@  находим:

g 00 t = p 00 t ρ 1 A 100 ρ 2 A 200 + C 00 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGNbWaaSbaaSqaaiaaicdacaaIWaaabeaakm aabmaabaGaamiDaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiC amaaBaaaleaacaaIWaGaaGimaaqabaGccaWG0baabaGaeqyWdi3aaS baaSqaaiaaigdaaeqaaOGaamyqamaaBaaaleaacaaIXaGaaGimaiaa icdaaeqaaOGaeyOeI0IaeqyWdi3aaSbaaSqaaiaaikdaaeqaaOGaam yqamaaBaaaleaacaaIYaGaaGimaiaaicdaaeqaaaaakiabgUcaRiaa doeadaWgaaWcbaGaaGimaiaaicdaaeqaaaaa@4AC1@ ,

где C 00 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGdbWaaSbaaSqaaiaaicdacaaIWaaabeaaaa a@3372@  – произвольная постоянная. Второе дифференциальное уравнение перепишем в виде:

  d 2 W mn t d t 2 + ω mn 2 W mn t =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiD aaGaayjkaiaawMcaaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYa aaaaaakiabgUcaRiabeM8a3naaDaaaleaacaWGTbGaamOBaaqaaiaa ikdaaaGccaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaaba GaamiDaaGaayjkaiaawMcaaiabg2da9iaaicdaaaa@4804@ , (2.23)

где частота ωmn колебаний определяется по формуле:

  ω mn 2 = D λ mn 4 + 2 p 0 + p 1 + p 2 h 2 λ mn 2 ρh+ ρ 1 + ρ 2 λ mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaOGaeyypa0ZaaSaaaeaacaWGebGaeq4UdW2aa0baaSqaaiaa d2gacaWGUbaabaGaaGinaaaakiabgUcaRmaabmaabaGaaGOmaiaadc hadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaa igdaaeqaaOGaey4kaSIaamiCamaaBaaaleaacaaIYaaabeaaaOGaay jkaiaawMcaamaalaaabaGaamiAaaqaaiaaikdaaaGaeq4UdW2aa0ba aSqaaiaad2gacaWGUbaabaGaaGOmaaaaaOqaaiabeg8aYjaadIgacq GHRaWkdaWcaaqaaiabeg8aYnaaBaaaleaacaaIXaaabeaakiabgUca Riabeg8aYnaaBaaaleaacaaIYaaabeaaaOqaaiabeU7aSnaaBaaale aacaWGTbGaamOBaaqabaaaaaaaaaa@5AA7@ . (2.24)

Формулу (2.24) перепишем в следующей форме:

  ω mn 2 = ω 0mn 2 1+ α mn 1+ μ mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaOGaeyypa0JaeqyYdC3aa0baaSqaaiaaicdacaWGTbGaamOB aaqaaiaaikdaaaGcdaWcaaqaaiaaigdacqGHRaWkcqaHXoqydaWgaa WcbaGaamyBaiaad6gaaeqaaaGcbaGaaGymaiabgUcaRiabeY7aTnaa BaaaleaacaWGTbGaamOBaaqabaaaaaaa@46DF@ , (2.25)

здесь

ω 0mn 2 = D λ mn 4 ρh , α mn = p 0 + p 1 + p 2 /2 h D λ mn 2 , μ mn = ρ 1 + ρ 2 ρh λ mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaaGimaiaad2gacaWGUb aabaGaaGOmaaaakiabg2da9maalaaabaGaamiraiabeU7aSnaaDaaa leaacaWGTbGaamOBaaqaaiaaisdaaaaakeaacqaHbpGCcaWGObaaai aacYcacaaMc8UaaGPaVlaaykW7caaMc8UaeqySde2aaSbaaSqaaiaa d2gacaWGUbaabeaakiabg2da9maalaaabaWaaeWaaeaacaWGWbWaaS baaSqaaiaaicdaaeqaaOGaey4kaSYaaSGbaeaadaqadaqaaiaadcha daWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaaik daaeqaaaGccaGLOaGaayzkaaaabaGaaGOmaaaaaiaawIcacaGLPaaa caWGObaabaGaamiraiabeU7aSnaaDaaaleaacaWGTbGaamOBaaqaai aaikdaaaaaaOGaaiilaiaaykW7caaMc8UaaGPaVlaaykW7cqaH8oqB daWgaaWcbaGaamyBaiaad6gaaeqaaOGaeyypa0ZaaSaaaeaacqaHbp GCdaWgaaWcbaGaaGymaaqabaGccqGHRaWkcqaHbpGCdaWgaaWcbaGa aGOmaaqabaaakeaacqaHbpGCcaWGObGaeq4UdW2aaSbaaSqaaiaad2 gacaWGUbaabeaaaaaaaa@7545@ .

Здесь ω0mn – частота пластины, не контактирующей с жидкостью. Параметры αmn и µmn определяют влияние давления и плотности окружающей среды. Таким образом, давление повышает, плотность понижает собственную частоту пластины. При αmn << 1, µmn << 1 их влияние исчезает. Через исходные данные параметры αmn, µmn принимают вид:

α mn = 12 1 ν 2 p 0 + p 1 + p 2 /2 a 2 b 2 π 2 E h 2 m 2 b 2 + n 2 a 2 , μ mn = ρ 1 + ρ 2 ab πρh m 2 b 2 + n 2 a 2 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHXoqydaWgaaWcbaGaamyBaiaad6gaaeqaaO Gaeyypa0ZaaSaaaeaacaaIXaGaaGOmamaabmaabaGaaGymaiabgkHi Tiabe27aUnaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaamaabm aabaGaamiCamaaBaaaleaacaaIWaaabeaakiabgUcaRmaalyaabaWa aeWaaeaacaWGWbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiCam aaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaaaqaaiaaikdaaaaa caGLOaGaayzkaaGaamyyamaaCaaaleqabaGaaGOmaaaakiaadkgada ahaaWcbeqaaiaaikdaaaaakeaacqaHapaCdaahaaWcbeqaaiaaikda aaGccaWGfbGaamiAamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaam yBamaaCaaaleqabaGaaGOmaaaakiaadkgadaahaaWcbeqaaiaaikda aaGccqGHRaWkcaWGUbWaaWbaaSqabeaacaaIYaaaaOGaamyyamaaCa aaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaaaacaGGSaGaaGPaVlaa ykW7caaMc8UaaGPaVlabeY7aTnaaBaaaleaacaWGTbGaamOBaaqaba GccqGH9aqpdaWcaaqaamaabmaabaGaeqyWdi3aaSbaaSqaaiaaigda aeqaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaGccaGLOa GaayzkaaGaamyyaiaadkgaaeaacqaHapaCcqaHbpGCcaWGObWaaOaa aeaacaWGTbWaaWbaaSqabeaacaaIYaaaaOGaamOyamaaCaaaleqaba GaaGOmaaaakiabgUcaRiaad6gadaahaaWcbeqaaiaaikdaaaGccaWG HbWaaWbaaSqabeaacaaIYaaaaaqabaaaaOGaaiOlaaaa@7E67@

При E = 2·105МПа, ν = 0.3, ρ = 7.8·103 кг/м3, ρ1 = ρ2 = 103 кг/м3, p0 = 0, p1 = 1 МПа, p2 = 2 МПа, a = 0.20 м, b = 0.20 м, h = 0.001м, m = 1, n = 1, α11 = 0.16, µ11 = 11.5, m = 2, n = 1, α21 = 0.06, µ21 = 7.3, α12 = α21, µ12 = µ21, m = 2, n = 2, α22 = 0.04, µ22 = 5.77. Следовательно, влияние давления незначительное, имеется значительное снижение собственной частоты за счет присоединенной массы. По модели несжимаемой жидкости в случае воды имеется только снижение собственной частоты. Это известный результат [1–3], однако учет влияния давления вносит некоторое изменение частоты.

Общая оценка рассматриваемых эффектов состоит в том, что при αmn > µmn преобладает повышающее частоту влияние давления среды, а при αmn < µmn – понижающее влияние плотности или присоединенной массы. Через входные параметры эти неравенства имеют вид:

12 1 ν 2 p 0 + p 1 + p 2 /2 ρab πEh ρ 1 + ρ 2 m 2 a 2 + n 2 b 2 >1, 12 1 ν 2 p 0 + p 1 + p 2 /2 ρab πEh ρ 1 + ρ 2 m 2 a 2 + n 2 b 2 <1 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaaigdacaaIYaWaaeWaaeaacaaIXa GaeyOeI0IaeqyVd42aaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzk aaWaaeWaaeaacaWGWbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaS GbaeaadaqadaqaaiaadchadaWgaaWcbaGaaGymaaqabaGccqGHRaWk caWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaaabaGaaG OmaaaaaiaawIcacaGLPaaacqaHbpGCcaWGHbGaamOyaaqaaiabec8a WjaadweacaWGObWaaeWaaeaacqaHbpGCdaWgaaWcbaGaaGymaaqaba GccqGHRaWkcqaHbpGCdaWgaaWcbaGaaGOmaaqabaaakiaawIcacaGL PaaadaGcaaqaaiaad2gadaahaaWcbeqaaiaaikdaaaGccaWGHbWaaW baaSqabeaacaaIYaaaaOGaey4kaSIaamOBamaaCaaaleqabaGaaGOm aaaakiaadkgadaahaaWcbeqaaiaaikdaaaaabeaaaaGccqGH+aGpca aIXaGaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8+aaSaaaeaacaaIXaGaaGOmamaabmaabaGaaGymaiabgk HiTiabe27aUnaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaamaa bmaabaGaamiCamaaBaaaleaacaaIWaaabeaakiabgUcaRmaalyaaba WaaeWaaeaacaWGWbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiC amaaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaaaqaaiaaikdaaa aacaGLOaGaayzkaaGaeqyWdiNaamyyaiaadkgaaeaacqaHapaCcaWG fbGaamiAamaabmaabaGaeqyWdi3aaSbaaSqaaiaaigdaaeqaaOGaey 4kaSIaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaWa aOaaaeaacaWGTbWaaWbaaSqabeaacaaIYaaaaOGaamyyamaaCaaale qabaGaaGOmaaaakiabgUcaRiaad6gadaahaaWcbeqaaiaaikdaaaGc caWGIbWaaWbaaSqabeaacaaIYaaaaaqabaaaaOGaeyipaWJaaGymaa aa@958C@ .

Первый случай реализуется для весьма тонких пластин из материала с малым модулем упругости и при предельно высоком давлении в контактирующей среде. Второй случай всегда реализуется при невысоких давлениях в плотной среде.

Далее найдем формулу для определения колебаний пластины с учетом найденных частот ωmn. Построим общее решение дифференциального уравнения (2.23):

W mn t = C 1mn cos ω mn t+ C 2mn sin ω mn t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakm aabmaabaGaamiDaaGaayjkaiaawMcaaiabg2da9iaadoeadaWgaaWc baGaaGymaiaad2gacaWGUbaabeaakiGacogacaGGVbGaai4CaiabeM 8a3naaBaaaleaacaWGTbGaamOBaaqabaGccaWG0bGaey4kaSIaam4q amaaBaaaleaacaaIYaGaamyBaiaad6gaaeqaaOGaci4CaiaacMgaca GGUbGaeqyYdC3aaSbaaSqaaiaad2gacaWGUbaabeaakiaadshaaaa@4F15@ ,

где C 1mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGdbWaaSbaaSqaaiaaigdacaWGTbGaamOBaa qabaaaaa@349E@  и C 2mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGdbWaaSbaaSqaaiaaikdacaWGTbGaamOBaa qabaaaaa@349F@  – произвольные постоянные. Тогда функция (2.10) принимает вид:

  w x,y,t = W 00 v 00 + m,n=0 m+n>0 N W mn t v mn x,y = = W 00 v 00 + m,n=0 m+n>0 N C 1mn cos ω mn t+ C 2mn sin ω mn t v mn x,y . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaaceqaaiaadEhadaqadaqaaiaadIhacaGGSaGaam yEaiaacYcacaWG0baacaGLOaGaayzkaaGaeyypa0Jaam4vamaaBaaa leaacaaIWaGaaGimaaqabaGccaWG2bWaaSbaaSqaaiaaicdacaaIWa aabeaakiabgUcaRmaaqahabaGaam4vamaaBaaaleaacaWGTbGaamOB aaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaacaWG2bWaaSbaaS qaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG5baa caGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2da9i aaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaqaaiaa d6eaa0GaeyyeIuoakiabg2da9aqaaiabg2da9iaadEfadaWgaaWcba GaaGimaiaaicdaaeqaaOGaamODamaaBaaaleaacaaIWaGaaGimaaqa baGccqGHRaWkdaaeWbqaamaabmaabaGaam4qamaaBaaaleaacaaIXa GaamyBaiaad6gaaeqaaOGaci4yaiaac+gacaGGZbGaeqyYdC3aaSba aSqaaiaad2gacaWGUbaabeaakiaadshacqGHRaWkcaWGdbWaaSbaaS qaaiaaikdacaWGTbGaamOBaaqabaGcciGGZbGaaiyAaiaac6gacqaH jpWDdaWgaaWcbaGaamyBaiaad6gaaeqaaOGaamiDaaGaayjkaiaawM caaiaadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG 4bGaaiilaiaadMhaaiaawIcacaGLPaaaaSabaeqabaGaamyBaiaacY cacaWGUbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWkcaWGUbGaeyOp a4JaaGimaaaabaGaamOtaaqdcqGHris5aOGaaiOlaaaaaa@8CF3@  (2.26)

Чтобы найти в формуле (2.26) постоянные C1mn, C2mn, нужно задать начальные условия:

  w x,y,0 =τ x,y , w x,y,0 t =ψ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG3bWaaeWaaeaacaWG4bGaaiilaiaadMhaca GGSaGaaGimaaGaayjkaiaawMcaaiabg2da9iabes8a0naabmaabaGa amiEaiaacYcacaWG5baacaGLOaGaayzkaaGaaiilaiaaykW7caaMc8 UaaGPaVlaaykW7caaMc8+aaSaaaeaacqGHciITcaWG3bWaaeWaaeaa caWG4bGaaiilaiaadMhacaGGSaGaaGimaaGaayjkaiaawMcaaaqaai abgkGi2kaadshaaaGaeyypa0JaeqiYdK3aaeWaaeaacaWG4bGaaiil aiaadMhaaiaawIcacaGLPaaaaaa@5881@ , (2.27)

Удовлетворим функцию (2.26) условиям (2.27):

  W 00 v 00 + m,n=0 m+n>0 N C 1mn v mn x,y =τ x,y = τ 00 v 00 + m,n=0 m+n>0 N τ mn v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGxbWaaSbaaSqaaiaaicdacaaIWaaabeaaki aadAhadaWgaaWcbaGaaGimaiaaicdaaeqaaOGaey4kaSYaaabCaeaa caWGdbWaaSbaaSqaaiaaigdacaWGTbGaamOBaaqabaGccaWG2bWaaS baaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG 5baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2 da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaqa aiaad6eaa0GaeyyeIuoakiabg2da9iabes8a0naabmaabaGaamiEai aacYcacaWG5baacaGLOaGaayzkaaGaeyypa0JaeqiXdq3aaSbaaSqa aiaaicdacaaIWaaabeaakiaadAhadaWgaaWcbaGaaGimaiaaicdaae qaaOGaey4kaSYaaabCaeaacqaHepaDdaWgaaWcbaGaamyBaiaad6ga aeqaaOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaai aadIhacaGGSaGaamyEaaGaayjkaiaawMcaaaWceaqabeaacaWGTbGa aiilaiaad6gacqGH9aqpcaaIWaaabaGaamyBaiabgUcaRiaad6gacq GH+aGpcaaIWaaaaeaacaWGobaaniabggHiLdaaaa@7422@ , (2.28)

  m,n=0 m+n>0 N C 2mn ω mn v mn x,y =ψ x,y = ψ 00 v 00 + m,n=0 m+n>0 N ψ mn v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaaeWbqaaiaadoeadaWgaaWcbaGaaGOmaiaad2 gacaWGUbaabeaakiabeM8a3naaBaaaleaacaWGTbGaamOBaaqabaGc caWG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEai aacYcacaWG5baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGa amOBaiabg2da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+i aaicdaaaqaaiaad6eaa0GaeyyeIuoakiabg2da9iabeI8a5naabmaa baGaamiEaiaacYcacaWG5baacaGLOaGaayzkaaGaeyypa0JaeqiYdK 3aaSbaaSqaaiaaicdacaaIWaaabeaakiaadAhadaWgaaWcbaGaaGim aiaaicdaaeqaaOGaey4kaSYaaabCaeaacqaHipqEdaWgaaWcbaGaam yBaiaad6gaaeqaaOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGc daqadaqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaaaWceaqabe aacaWGTbGaaiilaiaad6gacqGH9aqpcaaIWaaabaGaamyBaiabgUca Riaad6gacqGH+aGpcaaIWaaaaeaacaWGobaaniabggHiLdaaaa@7219@  (2.29)

где коэффициенты tmn и ymn разложения функций τ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHepaDdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@3703@  и ψ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@370C@  в ряд по системе функций (2.8) определяются по формулам:

  τ 00 = G τ x,y v 00 dxdy, τ mn = G τ x,y v mn x,y dxdy, m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHepaDdaWgaaWcbaGaaGimaiaaicdaaeqaaO Gaeyypa0Zaa8GuaeaacqaHepaDdaqadaqaaiaadIhacaGGSaGaamyE aaGaayjkaiaawMcaaiaadAhadaWgaaWcbaGaaGimaiaaicdaaeqaaO GaamizaiaadIhacaWGKbGaamyEaiaacYcaaSqaaiaadEeaaeqaniab gUIiYlabgUIiYdGccaaMc8UaaGPaVlaaykW7caaMc8UaeqiXdq3aaS baaSqaaiaad2gacaWGUbaabeaakiabg2da9maapifabaGaeqiXdq3a aeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaacaWG2bWaaS baaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG 5baacaGLOaGaayzkaaGaamizaiaadIhacaWGKbGaamyEaiaacYcaaS qaaiaadEeaaeqaniabgUIiYlabgUIiYdGccaaMc8UaaGPaVlaaykW7 caWGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaa@726E@ , (2.30)

ψ 00 = G ψ x,y v 00 dxdy, ψ mn = 1 ω mn G ψ x,y v mn x,y dxdy, m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaWgaaWcbaGaaGimaiaaicdaaeqaaO Gaeyypa0Zaa8GuaeaacqaHipqEdaqadaqaaiaadIhacaGGSaGaamyE aaGaayjkaiaawMcaaiaadAhadaWgaaWcbaGaaGimaiaaicdaaeqaaO GaamizaiaadIhacaWGKbGaamyEaiaacYcaaSqaaiaadEeaaeqaniab gUIiYlabgUIiYdGccaaMc8UaaGPaVlaaykW7cqaHipqEdaWgaaWcba GaamyBaiaad6gaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaeqyY dC3aaSbaaSqaaiaad2gacaWGUbaabeaaaaGcdaWdsbqaaiabeI8a5n aabmaabaGaamiEaiaacYcacaWG5baacaGLOaGaayzkaaGaamODamaa BaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadIhacaGGSaGaam yEaaGaayjkaiaawMcaaiaadsgacaWG4bGaamizaiaadMhacaGGSaaa leaacaWGhbaabeqdcqGHRiI8cqGHRiI8aOGaaGPaVlaaykW7caWGTb Gaey4kaSIaamOBaiabg6da+iaaicdaaaa@742F@ . (2.31)

Тогда из равенств (2.28) и (2.29) в силу полноты и ортонормированности системы (2.8) в пространстве L2(G) находим:

W 00 = τ 00 , ψ 00 =0; C 1mn = τ mn , C 2mn = ψ mn / ω mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGxbWaaSbaaSqaaiaaicdacaaIWaaabeaaki abg2da9iabes8a0naaBaaaleaacaaIWaGaaGimaaqabaGccaGGSaGa aGPaVlaaykW7caaMc8UaaGPaVlabeI8a5naaBaaaleaacaaIWaGaaG imaaqabaGccqGH9aqpcaaIWaGaai4oaiaaykW7caaMc8UaaGPaVlaa doeadaWgaaWcbaGaaGymaiaad2gacaWGUbaabeaakiabg2da9iabes 8a0naaBaaaleaacaWGTbGaamOBaaqabaGccaGGSaGaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caWGdbWaaSbaaSqaaiaaikdacaWGTbGaam OBaaqabaGccqGH9aqpdaWcgaqaaiabeI8a5naaBaaaleaacaWGTbGa amOBaaqabaaakeaacqaHjpWDdaWgaaWcbaGaamyBaiaad6gaaeqaaa aaaaa@66E6@ .

Замечание. Из разложений (2.28) и (2.29) видно, что функции τ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHepaDdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@3703@  и ψ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@370C@  должны удовлетворять условиям (2.7) и обладать гладкостью функций (2.8).

Таким образом, нами установлены следующие утверждения.

Утверждение 1. Если параметры αmn и µmn, определяющие соответственно влияние давления и плотности окружающей среды, то при

а) αmn << 1, µmn << 1 или αmn = µmn их влияние исчезает;

б) αmn > µmn преобладает повышающее частоту ωmn колебаний влияние давления среды;

в) αmn < µmn преобладает понижающее частоту ωmn колебаний влияние плотности среды.

Утверждение 2. Если начальные функции τ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHepaDdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@3703@  и ψ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@370C@  удовлетворяют условиям замечания и ψ 00 =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaWgaaWcbaGaaGimaiaaicdaaeqaaO Gaeyypa0JaaGimaaaa@3642@ , то колебания прямоугольной однородной пластины в указанной среде при избыточных давлениях p1, p2 и плотностях ρ 1 , ρ 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaGGSa GaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaaa@3713@ , удовлетворяющих условию ρ 1 A 100 ρ 2 A 200 0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaWGbb WaaSbaaSqaaiaaigdacaaIWaGaaGimaaqabaGccqGHsislcqaHbpGC daWgaaWcbaGaaGOmaaqabaGccaWGbbWaaSbaaSqaaiaaikdacaaIWa GaaGimaaqabaGccqGHGjsUcaaIWaaaaa@4032@ , определяется по формуле

w x,y,t = τ 00 v 00 + m,n=0 m+n>0 N τ mn cos ω mn t+ 1 ω mn ψ mn sin ω mn t v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG3bWaaeWaaeaacaWG4bGaaiilaiaadMhaca GGSaGaamiDaaGaayjkaiaawMcaaiabg2da9iabes8a0naaBaaaleaa caaIWaGaaGimaaqabaGccaWG2bWaaSbaaSqaaiaaicdacaaIWaaabe aakiabgUcaRmaaqahabaWaaeWaaeaacqaHepaDdaWgaaWcbaGaamyB aiaad6gaaeqaaOGaci4yaiaac+gacaGGZbGaeqyYdC3aaSbaaSqaai aad2gacaWGUbaabeaakiaadshacqGHRaWkdaWcaaqaaiaaigdaaeaa cqaHjpWDdaWgaaWcbaGaamyBaiaad6gaaeqaaaaakiabeI8a5naaBa aaleaacaWGTbGaamOBaaqabaGcciGGZbGaaiyAaiaac6gacqaHjpWD daWgaaWcbaGaamyBaiaad6gaaeqaaOGaamiDaaGaayjkaiaawMcaai aadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG4bGa aiilaiaadMhaaiaawIcacaGLPaaaaSabaeqabaGaamyBaiaacYcaca WGUbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWkcaWGUbGaeyOpa4Ja aGimaaaabaGaamOtaaqdcqGHris5aaaa@7177@ . (2.32)

Собственные колебания пластины находятся по формуле:

  w mn x,y,t = τ mn cos ω mn t+ 1 ω mn ψ mn sin ω mn t v mn x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG3bWaaSbaaSqaaiaad2gacaWGUbaabeaakm aabmaabaGaamiEaiaacYcacaWG5bGaaiilaiaadshaaiaawIcacaGL PaaacqGH9aqpdaqadaqaaiabes8a0naaBaaaleaacaWGTbGaamOBaa qabaGcciGGJbGaai4BaiaacohacqaHjpWDdaWgaaWcbaGaamyBaiaa d6gaaeqaaOGaamiDaiabgUcaRmaalaaabaGaaGymaaqaaiabeM8a3n aaBaaaleaacaWGTbGaamOBaaqabaaaaOGaeqiYdK3aaSbaaSqaaiaa d2gacaWGUbaabeaakiGacohacaGGPbGaaiOBaiabeM8a3naaBaaale aacaWGTbGaamOBaaqabaGccaWG0baacaGLOaGaayzkaaGaamODamaa BaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadIhacaGGSaGaam yEaaGaayjkaiaawMcaaaaa@60A2@ , (2.33)

а собственные частоты wmn по формуле (2.25) при условиях (2.7) и (2.27), где коэффициенты tmn, ymn определяются соответственно по формулам (2.30), (2.31).

3. Сжимаемая среда. В случае сжимаемой среды вместо уравнений (2.1) имеем трехмерные волновые уравнения [1–3]

  2 φ i x 2 + 2 φ i y 2 + 2 φ i z 2 1 c i 2 2 φ i t 2 =0,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOmaa aakiabeA8aQnaaBaaaleaacaWGPbaabeaaaOqaaiabgkGi2kaadIha daahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaacqGHciITda ahaaWcbeqaaiaaikdaaaGccqaHgpGAdaWgaaWcbaGaamyAaaqabaaa keaacqGHciITcaWG5bWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRm aalaaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaeqOXdO2aaSba aSqaaiaadMgaaeqaaaGcbaGaeyOaIyRaamOEamaaCaaaleqabaGaaG OmaaaaaaGccqGHsisldaWcaaqaaiaaigdaaeaacaWGJbWaa0baaSqa aiaadMgaaeaacaaIYaaaaaaakiabgwSixpaalaaabaGaeyOaIy7aaW baaSqabeaacaaIYaaaaOGaeqOXdO2aaSbaaSqaaiaadMgaaeqaaaGc baGaeyOaIyRaamiDamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpca aIWaGaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyAaiab g2da9iaaigdacaGGSaGaaGOmaaaa@6A5A@  (3.1)

  p ¯ i = ρ i φ i t , c i 2 = κ i p i ρ i ,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaaceWGWbGbaebadaWgaaWcbaGaamyAaaqabaGccq GH9aqpcqGHsislcqaHbpGCdaWgaaWcbaGaamyAaaqabaGcdaWcaaqa aiabgkGi2kabeA8aQnaaBaaaleaacaWGPbaabeaaaOqaaiabgkGi2k aadshaaaGaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaadogadaqhaaWcbaGaamyAaaqaaiaaikdaaaGccqGH9aqpcqaH6o WAdaWgaaWcbaGaamyAaaqabaGcdaWcaaqaaiaadchadaWgaaWcbaGa amyAaaqabaaakeaacqaHbpGCdaWgaaWcbaGaamyAaaqabaaaaOGaai ilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyAaiabg2da9iaa igdacaGGSaGaaGOmaaaa@60EE@ (3.2)

где ci – скорость звука, κ i MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH6oWAdaWgaaWcbaGaamyAaaqabaaaaa@33D6@  – коэффициент адиабаты. В отличие от случая несжимаемой жидкости здесь давление и плотность не являются независимыми, а связаны изотермическим законом.

На основании функции (2.8) аналогично (2.11) функции φ i x,y,z,t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHgpGAdaWgaaWcbaGaamyAaaqabaGcdaqada qaaiaadIhacaGGSaGaamyEaiaacYcacaWG6bGaaiilaiaadshaaiaa wIcacaGLPaaaaaa@3B77@  будем искать в виде:

φ i x,y,z,t = Φ i00 z v 00 g 00 t + m,n=0 m+n>1 N Φ imn z v mn x,y g mn t ,i=1,2,... MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHgpGAdaWgaaWcbaGaamyAaaqabaGcdaqada qaaiaadIhacaGGSaGaamyEaiaacYcacaWG6bGaaiilaiaadshaaiaa wIcacaGLPaaacqGH9aqpcqqHMoGrdaWgaaWcbaGaamyAaiaaicdaca aIWaaabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiaadAhadaWg aaWcbaGaaGimaiaaicdaaeqaaOGaam4zamaaBaaaleaacaaIWaGaaG imaaqabaGcdaqadaqaaiaadshaaiaawIcacaGLPaaacqGHRaWkdaae WbqaaiabfA6agnaaBaaaleaacaWGPbGaamyBaiaad6gaaeqaaOWaae WaaeaacaWG6baacaGLOaGaayzkaaGaamODamaaBaaaleaacaWGTbGa amOBaaqabaGcdaqadaqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawM caaiaadEgadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG 0baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2 da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaigdaaaqa aiaad6eaa0GaeyyeIuoakiaacYcacaaMc8UaaGPaVlaaykW7caWGPb Gaeyypa0JaaGymaiaacYcacaaIYaGaaiilaiaac6cacaGGUaGaaiOl aaaa@77CB@ (3.3)

Подставляя (3.3) в волновое уравнение (3.1), получим:

d 2 Φ i00 z d z 2 v 00 g 00 t + + m,n=0 m+n>0 N Φ imn z v mn x,y g mn t d 2 Φ imn z d z 2 1 Φ imn z λ mn 2 1 c i 2 d 2 g mn t d t 2 1 g mn t =0, i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaalaaabaGaamizam aaCaaaleqabaGaaGOmaaaakiabfA6agnaaBaaaleaacaWGPbGaaGim aiaaicdaaeqaaOWaaeWaaeaacaWG6baacaGLOaGaayzkaaaabaGaam izaiaadQhadaahaaWcbeqaaiaaikdaaaaaaOGaamODamaaBaaaleaa caaIWaGaaGimaaqabaGccaWGNbWaaSbaaSqaaiaaicdacaaIWaaabe aakmaabmaabaGaamiDaaGaayjkaiaawMcaaiabgUcaRaqaaiabgUca RmaaqahabaGaeuOPdy0aaSbaaSqaaiaadMgacaWGTbGaamOBaaqaba GcdaqadaqaaiaadQhaaiaawIcacaGLPaaacaWG2bWaaSbaaSqaaiaa d2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG5baacaGLOa GaayzkaaGaam4zamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqa aiaadshaaiaawIcacaGLPaaadaWadaabaeqabaWaaSaaaeaacaWGKb WaaWbaaSqabeaacaaIYaaaaOGaeuOPdy0aaSbaaSqaaiaadMgacaWG TbGaamOBaaqabaGcdaqadaqaaiaadQhaaiaawIcacaGLPaaaaeaaca WGKbGaamOEamaaCaaaleqabaGaaGOmaaaaaaGcdaWcaaqaaiaaigda aeaacqqHMoGrdaWgaaWcbaGaamyAaiaad2gacaWGUbaabeaakmaabm aabaGaamOEaaGaayjkaiaawMcaaaaacqGHsislaeaacqGHsislcaaM c8Uaeq4UdW2aa0baaSqaaiaad2gacaWGUbaabaGaaGOmaaaakiabgk HiTmaalaaabaGaaGymaaqaaiaadogadaqhaaWcbaGaamyAaaqaaiaa ikdaaaaaaOWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam 4zamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadshaaiaa wIcacaGLPaaaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaa GcdaWcaaqaaiaaigdaaeaacaWGNbWaaSbaaSqaaiaad2gacaWGUbaa beaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaaaaaGaay5waiaaw2 faaiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7aSabaeqabaGaamyB aiaacYcacaWGUbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWkcaWGUb GaeyOpa4JaaGimaaaabaGaamOtaaqdcqGHris5aOGaamyAaiabg2da 9iaaigdacaGGSaGaaGOmaaaaaa@F0D8@ . (3.4)

В силу (2.15) и (2.22): d 2 g mn t d t 2 = ω mn 2 g mn t MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccaWGNbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiD aaGaayjkaiaawMcaaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYa aaaaaakiabg2da9iabgkHiTiabeM8a3naaDaaaleaacaWGTbGaamOB aaqaaiaaikdaaaGccaWGNbWaaSbaaSqaaiaad2gacaWGUbaabeaakm aabmaabaGaamiDaaGaayjkaiaawMcaaaaa@4775@ , тогда из (3.4) имеем:

  d 2 Φ i00 z d z 2 =0, d 2 Φ imn z d z 2 κ mn 2 Φ imn z =0,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccqqHMoGrdaWgaaWcbaGaamyAaiaaicdacaaIWaaabeaakmaabmaa baGaamOEaaGaayjkaiaawMcaaaqaaiaadsgacaWG6bWaaWbaaSqabe aacaaIYaaaaaaakiabg2da9iaaicdacaGGSaGaaGPaVlaaykW7caaM c8UaaGPaVpaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiabfA 6agnaaBaaaleaacaWGPbGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG 6baacaGLOaGaayzkaaaabaGaamizaiaadQhadaahaaWcbeqaaiaaik daaaaaaOGaeyOeI0IaeqOUdS2aa0baaSqaaiaad2gacaWGUbaabaGa aGOmaaaakiabfA6agnaaBaaaleaacaWGPbGaamyBaiaad6gaaeqaaO WaaeWaaeaacaWG6baacaGLOaGaayzkaaGaeyypa0JaaGimaiaacYca caaMc8UaaGPaVlaaykW7caWGPbGaeyypa0JaaGymaiaacYcacaaIYa aaaa@6887@ , (3.5)

где

κ mn 2 = λ mn 2 ω mn 2 c i 2 ,i=1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH6oWAdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaOGaeyypa0Jaeq4UdW2aa0baaSqaaiaad2gacaWGUbaabaGa aGOmaaaakiabgkHiTmaalaaabaGaeqyYdC3aa0baaSqaaiaad2gaca WGUbaabaGaaGOmaaaaaOqaaiaadogadaqhaaWcbaGaamyAaaqaaiaa ikdaaaaaaOGaaiilaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaam yAaiabg2da9iaaigdacaGGSaGaaGOmaaaa@5013@ .

При условии kmn > 0 дифференциальное уравнение (3.5) при m + n > 0 имеет общее решение:

Φ imn z = C i1mn exp κ mn z + C i2mn exp κ mn z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaamyAaiaad2gacaWGUb aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadoea daWgaaWcbaGaamyAaiaaigdacaWGTbGaamOBaaqabaGcciGGLbGaai iEaiaacchadaqadaqaaiabeQ7aRnaaBaaaleaacaWGTbGaamOBaaqa baGccaWG6baacaGLOaGaayzkaaGaey4kaSIaam4qamaaBaaaleaaca WGPbGaaGOmaiaad2gacaWGUbaabeaakiGacwgacaGG4bGaaiiCamaa bmaabaGaeyOeI0IaeqOUdS2aaSbaaSqaaiaad2gacaWGUbaabeaaki aadQhaaiaawIcacaGLPaaaaaa@5663@ ,

а при m = n = 0

Φ 100 z = C 1100 z+ C 1200 , Φ 200 z = C 2100 z+ C 2200 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaaicdacaaIWa aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadoea daWgaaWcbaGaaGymaiaaigdacaaIWaGaaGimaaqabaGccaWG6bGaey 4kaSIaam4qamaaBaaaleaacaaIXaGaaGOmaiaaicdacaaIWaaabeaa kiaacYcacaaMc8UaeuOPdy0aaSbaaSqaaiaaikdacaaIWaGaaGimaa qabaGcdaqadaqaaiaadQhaaiaawIcacaGLPaaacqGH9aqpcaWGdbWa aSbaaSqaaiaaikdacaaIXaGaaGimaiaaicdaaeqaaOGaamOEaiabgU caRiaadoeadaWgaaWcbaGaaGOmaiaaikdacaaIWaGaaGimaaqabaaa aa@557C@ ,

где C ijmn i,j=1,2, MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGdbWaaSbaaSqaaiaadMgacaWGQbGaamyBai aad6gaaeqaaOGaaGPaVlaadMgacaGGSaGaamOAaiabg2da9iaaigda caGGSaGaaGOmaiaacYcaaaa@3DBF@  – произвольные постоянные. В силу условий (2.3) при m + n > 0 найдем:

Φ 1mn z = B 1mn exp κ mn z 0,z, Φ 2mn z = B 2mn exp κ mn z 0,z MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaad2gacaWGUb aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadkea daWgaaWcbaGaaGymaiaad2gacaWGUbaabeaakiGacwgacaGG4bGaai iCamaabmaabaGaeqOUdS2aaSbaaSqaaiaad2gacaWGUbaabeaakiaa dQhaaiaawIcacaGLPaaacqGHsgIRcaaIWaGaaiilaiaaykW7caWG6b GaeyOKH4QaeyOeI0IaeyOhIuQaaiilaiaaykW7cqqHMoGrdaWgaaWc baGaaGOmaiaad2gacaWGUbaabeaakmaabmaabaGaamOEaaGaayjkai aawMcaaiabg2da9iaadkeadaWgaaWcbaGaaGOmaiaad2gacaWGUbaa beaakiGacwgacaGG4bGaaiiCamaabmaabaGaeyOeI0IaeqOUdS2aaS baaSqaaiaad2gacaWGUbaabeaakiaadQhaaiaawIcacaGLPaaacqGH sgIRcaaIWaGaaeilaiaaykW7caWG6bGaeyOKH4QaeyOhIukaaa@70F4@ ,

а при m = n = 0

Φ 100 z = B 100 , Φ 200 z = B 200 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqqHMoGrdaWgaaWcbaGaaGymaiaaicdacaaIWa aabeaakmaabmaabaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadkea daWgaaWcbaGaaGymaiaaicdacaaIWaaabeaakiaacYcacaaMc8Uaeu OPdy0aaSbaaSqaaiaaikdacaaIWaGaaGimaaqabaGcdaqadaqaaiaa dQhaaiaawIcacaGLPaaacqGH9aqpcaWGcbWaaSbaaSqaaiaaikdaca aIWaGaaGimaaqabaaaaa@486F@ ,

где постоянные B1mn и B2mn найдем из условий (2.2) аналогично вышеизложенному:

B 1mn = ω ¯ mn k 1 κ mn 1 κ mn h 2 1 , B 2mn = ω ¯ mn k 2 κ mn 1 κ mn h 2 1 ,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGcbWaaSbaaSqaaiaaigdacaWGTbGaamOBaa qabaGccqGH9aqpdaWcaaqaaiqbeM8a3zaaraWaaSbaaSqaaiaad2ga caWGUbaabeaakiaadUgadaWgaaWcbaGaaGymaaqabaaakeaacqaH6o WAdaWgaaWcbaGaamyBaiaad6gaaeqaaaaakmaabmaabaGaaGymaiab gkHiTmaalaaabaGaeqOUdS2aaSbaaSqaaiaad2gacaWGUbaabeaaki aadIgaaeaacaaIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaeyOe I0IaaGymaaaakiaacYcacaaMc8UaaGPaVlaadkeadaWgaaWcbaGaaG Omaiaad2gacaWGUbaabeaakiabg2da9iabgkHiTmaalaaabaGafqyY dCNbaebadaWgaaWcbaGaamyBaiaad6gaaeqaaOGaam4AamaaBaaale aacaaIYaaabeaaaOqaaiabeQ7aRnaaBaaaleaacaWGTbGaamOBaaqa baaaaOWaaeWaaeaacaaIXaGaeyOeI0YaaSaaaeaacqaH6oWAdaWgaa WcbaGaamyBaiaad6gaaeqaaOGaamiAaaqaaiaaikdaaaaacaGLOaGa ayzkaaWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaaiilaiaaykW7ca aMc8UaaGPaVlaad2gacqGHRaWkcaWGUbGaeyOpa4JaaGimaaaa@7187@ . (3.6)

Тем самым функции (3.3) построены, где B1mn и B2mn при m + n > 0 находятся по формулам (3.6), а постоянные B100 и B200 остаются произвольными постоянными.

Далее аналогично пункту 2 найдем:

  p ¯ 1 = ρ 1 φ 1 t = ρ 1 B 100 v 00 d g 00 t dt + m,n=0 m+n>0 N 1 κ mn v mn x,y d 2 W mn t d t 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaaceWGWbGbaebadaWgaaWcbaGaaGymaaqabaGccq GH9aqpcqGHsislcqaHbpGCdaWgaaWcbaGaaGymaaqabaGcdaWcaaqa aiabgkGi2kabeA8aQnaaBaaaleaacaaIXaaabeaaaOqaaiabgkGi2k aadshaaaGaeyypa0JaeyOeI0IaeqyWdi3aaSbaaSqaaiaaigdaaeqa aOWaamWaaeaacaWGcbWaaSbaaSqaaiaaigdacaaIWaGaaGimaaqaba GccaWG2bWaaSbaaSqaaiaaicdacaaIWaaabeaakmaalaaabaGaamiz aiaadEgadaWgaaWcbaGaaGimaiaaicdaaeqaaOWaaeWaaeaacaWG0b aacaGLOaGaayzkaaaabaGaamizaiaadshaaaGaey4kaSYaaabCaeaa daWcaaqaaiaaigdaaeaacqaH6oWAdaWgaaWcbaGaamyBaiaad6gaae qaaaaakiaadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWaaeWaaeaa caWG4bGaaiilaiaadMhaaiaawIcacaGLPaaadaWcaaqaaiaadsgada ahaaWcbeqaaiaaikdaaaGccaWGxbWaaSbaaSqaaiaad2gacaWGUbaa beaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaadsgacaWG0b WaaWbaaSqabeaacaaIYaaaaaaaaqaabeqaaiaad2gacaGGSaGaamOB aiabg2da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaic daaaqaaiaad6eaa0GaeyyeIuoaaOGaay5waiaaw2faaaaa@7576@

  p ¯ 2 = ρ 2 φ 2 t = ρ 2 B 200 v 00 d g 00 t dt m,n=0 m+n>0 N 1 κ mn v mn x,y d 2 W mn t d t 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeGabaaOAiqadchagaqeamaaBaaaleaacaaIYaaabe aakiabg2da9iabgkHiTiabeg8aYnaaBaaaleaacaaIYaaabeaakmaa laaabaGaeyOaIyRaeqOXdO2aaSbaaSqaaiaaikdaaeqaaaGcbaGaey OaIyRaamiDaaaacqGH9aqpcqGHsislcqaHbpGCdaWgaaWcbaGaaGOm aaqabaGcdaWadaqaaiaadkeadaWgaaWcbaGaaGOmaiaaicdacaaIWa aabeaakiaadAhadaWgaaWcbaGaaGimaiaaicdaaeqaaOWaaSaaaeaa caWGKbGaam4zamaaBaaaleaacaaIWaGaaGimaaqabaGcdaqadaqaai aadshaaiaawIcacaGLPaaaaeaacaWGKbGaamiDaaaacqGHsisldaae WbqaamaalaaabaGaaGymaaqaaiabeQ7aRnaaBaaaleaacaWGTbGaam OBaaqabaaaaOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqa daqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaamaalaaabaGaam izamaaCaaaleqabaGaaGOmaaaakiaadEfadaWgaaWcbaGaamyBaiaa d6gaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizai aadshadaahaaWcbeqaaiaaikdaaaaaaaabaeqabaGaamyBaiaacYca caWGUbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWkcaWGUbGaeyOpa4 JaaGimaaaabaGaamOtaaqdcqGHris5aaGccaGLBbGaayzxaaaaaa@7643@

и на основании формулы (2.6) вычислим q:

  q= p 1 p 2 + ρ 2 B 200 ρ 1 B 100 v 00 d g 00 t dt + 2 p 0 + p 1 + p 2 h 2 × × 2 w x 2 + 2 w y 2 ρ 1 + ρ 2 m,n=0 m+n>0 N 1 κ mn v mn x,y d 2 W mn t d t 2 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiaadghacqGH9aqpcaWGWbWaaSbaaSqaai aaigdaaeqaaOGaeyOeI0IaamiCamaaBaaaleaacaaIYaaabeaakiab gUcaRmaabmaabaGaeqyWdi3aaSbaaSqaaiaaikdaaeqaaOGaamOqam aaBaaaleaacaaIYaGaaGimaiaaicdaaeqaaOGaeyOeI0IaeqyWdi3a aSbaaSqaaiaaigdaaeqaaOGaamOqamaaBaaaleaacaaIXaGaaGimai aaicdaaeqaaaGccaGLOaGaayzkaaGaamODamaaBaaaleaacaaIWaGa aGimaaqabaGcdaWcaaqaaiaadsgacaWGNbWaaSbaaSqaaiaaicdaca aIWaaabeaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaadsga caWG0baaaiabgUcaRmaalaaabaWaaeWaaeaacaaIYaGaamiCamaaBa aaleaacaaIWaaabeaakiabgUcaRiaadchadaWgaaWcbaGaaGymaaqa baGccqGHRaWkcaWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaay zkaaGaamiAaaqaaiaaikdaaaGaey41aqlabaGaey41aq7aaeWaaeaa daWcaaqaaiabgkGi2oaaCaaaleqabaGaaGOmaaaakiaadEhaaeaacq GHciITcaWG4bWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRmaalaaa baGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4DaaqaaiabgkGi2k aadMhadaahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzkaaGaeyOe I0YaaeWaaeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccqGHRaWkcq aHbpGCdaWgaaWcbaGaaGOmaaqabaaakiaawIcacaGLPaaadaaeWbqa amaalaaabaGaaGymaaqaaiabeQ7aRnaaBaaaleaacaWGTbGaamOBaa qabaaaaOGaamODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqa aiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaamaalaaabaGaamizam aaCaaaleqabaGaaGOmaaaakiaadEfadaWgaaWcbaGaamyBaiaad6ga aeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizaiaads hadaahaaWcbeqaaiaaikdaaaaaaaabaeqabaGaamyBaiaacYcacaWG UbGaeyypa0JaaGimaaqaaiaad2gacqGHRaWkcaWGUbGaeyOpa4JaaG imaaaabaGaamOtaaqdcqGHris5aOGaaiOlaaaaaa@9D51@  (3.7)

Подставляя выражение (3.7) в уравнение (1.1), получим:

D 4 w x 4 +2 4 w x 2 y 2 + 4 w y 4 +ρh 2 w t 2 2 p 0 + p 1 + p 2 h 2 2 w x 2 + 2 w y 2 + + v 00 d g 00 t dt ρ 2 B 200 ρ 1 B 100 + ρ 1 + ρ 2 m,n=0 m+n>0 N 1 κ mn v mn x,y d 2 W mn t d t 2 ρ 2 B 200 ρ 1 B 100 v 00 d g 00 t dt = p 1 p 2 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaaiaadseadaqadaqaamaalaaabaGaeyOaIy 7aaWbaaSqabeaacaaI0aaaaOGaam4DaaqaaiabgkGi2kaadIhadaah aaWcbeqaaiaaisdaaaaaaOGaey4kaSIaaGOmamaalaaabaGaeyOaIy 7aaWbaaSqabeaacaaI0aaaaOGaam4DaaqaaiabgkGi2kaadIhadaah aaWcbeqaaiaaikdaaaGccqGHciITcaWG5bWaaWbaaSqabeaacaaIYa aaaaaakiabgUcaRmaalaaabaGaeyOaIy7aaWbaaSqabeaacaaI0aaa aOGaam4DaaqaaiabgkGi2kaadMhadaahaaWcbeqaaiaaisdaaaaaaa GccaGLOaGaayzkaaGaey4kaSIaeqyWdiNaamiAamaalaaabaGaeyOa Iy7aaWbaaSqabeaacaaIYaaaaOGaam4DaaqaaiabgkGi2kaadshada ahaaWcbeqaaiaaikdaaaaaaOGaeyOeI0YaaSaaaeaadaqadaqaaiaa ikdacaWGWbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaamiCamaaBa aaleaacaaIXaaabeaakiabgUcaRiaadchadaWgaaWcbaGaaGOmaaqa baaakiaawIcacaGLPaaacaWGObaabaGaaGOmaaaacqGHflY1daqada qaamaalaaabaGaeyOaIy7aaWbaaSqabeaacaaIYaaaaOGaam4Daaqa aiabgkGi2kaadIhadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaS aaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWG3baabaGaeyOa IyRaamyEamaaCaaaleqabaGaaGOmaaaaaaaakiaawIcacaGLPaaacq GHRaWkaeaacqGHRaWkcaWG2bWaaSbaaSqaaiaaicdacaaIWaaabeaa kmaalaaabaGaamizaiaadEgadaWgaaWcbaGaaGimaiaaicdaaeqaaO WaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGaamizaiaadshaaaWa aeWaaeaacqaHbpGCdaWgaaWcbaGaaGOmaaqabaGccaWGcbWaaSbaaS qaaiaaikdacaaIWaGaaGimaaqabaGccqGHsislcqaHbpGCdaWgaaWc baGaaGymaaqabaGccaWGcbWaaSbaaSqaaiaaigdacaaIWaGaaGimaa qabaaakiaawIcacaGLPaaacqGHRaWkdaqadaqaaiabeg8aYnaaBaaa leaacaaIXaaabeaakiabgUcaRiabeg8aYnaaBaaaleaacaaIYaaabe aaaOGaayjkaiaawMcaamaaqahabaWaaSaaaeaacaaIXaaabaGaeqOU dS2aaSbaaSqaaiaad2gacaWGUbaabeaaaaGccaWG2bWaaSbaaSqaai aad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG5baacaGL OaGaayzkaaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaam 4vamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadshaaiaa wIcacaGLPaaaaeaacaWGKbGaamiDamaaCaaaleqabaGaaGOmaaaaaa aaeaqabeaacaWGTbGaaiilaiaad6gacqGH9aqpcaaIWaaabaGaamyB aiabgUcaRiaad6gacqGH+aGpcaaIWaaaaeaacaWGobaaniabggHiLd GccqGHsislaeaacqGHsisldaqadaqaaiabeg8aYnaaBaaaleaacaaI YaaabeaakiaadkeadaWgaaWcbaGaaGOmaiaaicdacaaIWaaabeaaki abgkHiTiabeg8aYnaaBaaaleaacaaIXaaabeaakiaadkeadaWgaaWc baGaaGymaiaaicdacaaIWaaabeaaaOGaayjkaiaawMcaaiaadAhada WgaaWcbaGaaGimaiaaicdaaeqaaOWaaSaaaeaacaWGKbGaam4zamaa BaaaleaacaaIWaGaaGimaaqabaGcdaqadaqaaiaadshaaiaawIcaca GLPaaaaeaacaWGKbGaamiDaaaacqGH9aqpcaWGWbWaaSbaaSqaaiaa igdaaeqaaOGaeyOeI0IaamiCamaaBaaaleaacaaIYaaabeaakiaac6 caaaaa@DCA9@  (3.8)

Теперь подставим функцию (2.10) в уравнение (3.8). Отсюда при условии p1p2 = p имеем:

m,n=0 m+n>0 N D λ mn 4 v mn x,y W mn t +ρh v mn x,y d 2 W mn t d t 2 + + p 0 + p 1 + p 2 2 h λ mn 2 v mn x,y W mn t + ρ 1 + ρ 2 κ mn v mn x,y d 2 W mn t d t 2 ρ 2 B 200 ρ 1 B 100 v 00 d g 00 t dt = p 00 v 00 + m,n=0 m+n>0 N p mn v mn x,y . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakqaabeqaamaaqahabaWaamqaaeaacaWGebGaeq4UdW 2aa0baaSqaaiaad2gacaWGUbaabaGaaGinaaaakiaadAhadaWgaaWc baGaamyBaiaad6gaaeqaaOWaaeWaaeaacaWG4bGaaiilaiaadMhaai aawIcacaGLPaaacaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaa bmaabaGaamiDaaGaayjkaiaawMcaaiabgUcaRiabeg8aYjaadIgaca WG2bWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaa cYcacaWG5baacaGLOaGaayzkaaWaaSaaaeaacaWGKbWaaWbaaSqabe aacaaIYaaaaOGaam4vamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqa daqaaiaadshaaiaawIcacaGLPaaaaeaacaWGKbGaamiDamaaCaaale qabaGaaGOmaaaaaaGccqGHRaWkaiaawUfaaaWceaqabeaacaWGTbGa aiilaiaad6gacqGH9aqpcaaIWaaabaGaamyBaiabgUcaRiaad6gacq GH+aGpcaaIWaaaaeaacaWGobaaniabggHiLdaakeaadaWacaqaaiab gUcaRmaabmaabaGaamiCamaaBaaaleaacaaIWaaabeaakiabgUcaRm aalaaabaGaamiCamaaBaaaleaacaaIXaaabeaakiabgUcaRiaadcha daWgaaWcbaGaaGOmaaqabaaakeaacaaIYaaaaaGaayjkaiaawMcaai aadIgacqaH7oaBdaqhaaWcbaGaamyBaiaad6gaaeaacaaIYaaaaOGa amODamaaBaaaleaacaWGTbGaamOBaaqabaGcdaqadaqaaiaadIhaca GGSaGaamyEaaGaayjkaiaawMcaaiaadEfadaWgaaWcbaGaamyBaiaa d6gaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaGaey4kaSYaaS aaaeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccqGHRaWkcqaHbpGC daWgaaWcbaGaaGOmaaqabaaakeaacqaH6oWAdaWgaaWcbaGaamyBai aad6gaaeqaaaaakiaadAhadaWgaaWcbaGaamyBaiaad6gaaeqaaOWa aeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaadaWcaaqaai aadsgadaahaaWcbeqaaiaaikdaaaGccaWGxbWaaSbaaSqaaiaad2ga caWGUbaabeaakmaabmaabaGaamiDaaGaayjkaiaawMcaaaqaaiaads gacaWG0bWaaWbaaSqabeaacaaIYaaaaaaaaOGaayzxaaGaeyOeI0ca baGaeyOeI0YaaeWaaeaacqaHbpGCdaWgaaWcbaGaaGOmaaqabaGcca WGcbWaaSbaaSqaaiaaikdacaaIWaGaaGimaaqabaGccqGHsislcqaH bpGCdaWgaaWcbaGaaGymaaqabaGccaWGcbWaaSbaaSqaaiaaigdaca aIWaGaaGimaaqabaaakiaawIcacaGLPaaacaWG2bWaaSbaaSqaaiaa icdacaaIWaaabeaakmaalaaabaGaamizaiaadEgadaWgaaWcbaGaaG imaiaaicdaaeqaaOWaaeWaaeaacaWG0baacaGLOaGaayzkaaaabaGa amizaiaadshaaaGaeyypa0JaamiCamaaBaaaleaacaaIWaGaaGimaa qabaGccaWG2bWaaSbaaSqaaiaaicdacaaIWaaabeaakiabgUcaRmaa qahabaGaamiCamaaBaaaleaacaWGTbGaamOBaaqabaGccaWG2bWaaS baaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiEaiaacYcacaWG 5baacaGLOaGaayzkaaaalqaabeqaaiaad2gacaGGSaGaamOBaiabg2 da9iaaicdaaeaacaWGTbGaey4kaSIaamOBaiabg6da+iaaicdaaaqa aiaad6eaa0GaeyyeIuoakiaac6caaaaa@D96F@

Отсюда получаем дифференциальные уравнения:

  d 2 W mn t d t 2 + ω mn 2 W mn t =0,m+n>0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaa GccaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaabaGaamiD aaGaayjkaiaawMcaaaqaaiaadsgacaWG0bWaaWbaaSqabeaacaaIYa aaaaaakiabgUcaRiabeM8a3naaDaaaleaacaWGTbGaamOBaaqaaiaa ikdaaaGccaWGxbWaaSbaaSqaaiaad2gacaWGUbaabeaakmaabmaaba GaamiDaaGaayjkaiaawMcaaiabg2da9iaaicdacaGGSaGaaGPaVlaa ykW7caaMc8UaamyBaiabgUcaRiaad6gacqGH+aGpcaaIWaaaaa@51DE@ , (3.9)

где

  ω mn 2 = D λ mn 4 + 2 p 0 + p 1 + p 2 h 2 λ mn 2 ρh+ ρ 1 + ρ 2 / κ mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaOGaeyypa0ZaaSaaaeaacaWGebGaeq4UdW2aa0baaSqaaiaa d2gacaWGUbaabaGaaGinaaaakiabgUcaRmaabmaabaGaaGOmaiaadc hadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaa igdaaeqaaOGaey4kaSIaamiCamaaBaaaleaacaaIYaaabeaaaOGaay jkaiaawMcaamaalaaabaGaamiAaaqaaiaaikdaaaGaeq4UdW2aa0ba aSqaaiaad2gacaWGUbaabaGaaGOmaaaaaOqaaiabeg8aYjaadIgacq GHRaWkdaWcgaqaamaabmaabaGaeqyWdi3aaSbaaSqaaiaaigdaaeqa aOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaay zkaaaabaGaeqOUdS2aaSbaaSqaaiaad2gacaWGUbaabeaaaaaaaaaa @5C34@ (3.10)

ρ 1 B 100 ρ 2 B 200 d g 00 t dt = p 00 ,m=n=0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaqadaqaaiabeg8aYnaaBaaaleaacaaIXaaabe aakiaadkeadaWgaaWcbaGaaGymaiaaicdacaaIWaaabeaakiabgkHi Tiabeg8aYnaaBaaaleaacaaIYaaabeaakiaadkeadaWgaaWcbaGaaG OmaiaaicdacaaIWaaabeaaaOGaayjkaiaawMcaamaalaaabaGaamiz aiaadEgadaWgaaWcbaGaaGimaiaaicdaaeqaaOWaaeWaaeaacaWG0b aacaGLOaGaayzkaaaabaGaamizaiaadshaaaGaeyypa0JaamiCamaa BaaaleaacaaIWaGaaGimaaqabaGccaGGSaGaaGPaVlaaykW7caaMc8 UaaGPaVlaad2gacqGH9aqpcaWGUbGaeyypa0JaaGimaaaa@565B@ .

Решение последнего уравнения при условии ρ 1 B 100 ρ 2 B 200 0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaWGcb WaaSbaaSqaaiaaigdacaaIWaGaaGimaaqabaGccqGHsislcqaHbpGC daWgaaWcbaGaaGOmaaqabaGccaWGcbWaaSbaaSqaaiaaikdacaaIWa GaaGimaaqabaGccqGHGjsUcaaIWaaaaa@4034@  определяется по формуле:

g 00 t = p 00 t ρ 1 B 100 ρ 2 B 200 + d 00 , d 00 =const MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGNbWaaSbaaSqaaiaaicdacaaIWaaabeaakm aabmaabaGaamiDaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiC amaaBaaaleaacaaIWaGaaGimaaqabaGccaWG0baabaGaeqyWdi3aaS baaSqaaiaaigdaaeqaaOGaamOqamaaBaaaleaacaaIXaGaaGimaiaa icdaaeqaaOGaeyOeI0IaeqyWdi3aaSbaaSqaaiaaikdaaeqaaOGaam OqamaaBaaaleaacaaIYaGaaGimaiaaicdaaeqaaaaakiabgUcaRiaa dsgadaWgaaWcbaGaaGimaiaaicdaaeqaaOGaaiilaiaaykW7caaMc8 UaaGPaVlaaykW7caWGKbWaaSbaaSqaaiaaicdacaaIWaaabeaakiab g2da9iaabogacaqGVbGaaeOBaiaabohacaqG0baaaa@5A19@ .

Поскольку kmn зависит от wmn, то равенство (3.10) перепишем в виде:

D λ mn 4 + 2 p 0 + p 1 + p 2 h 2 λ mn 2 ρh+ ρ 1 + ρ 2 / κ mn ω mn 2 =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGebGaeq4UdW2aa0baaSqaaiaad2gacaWGUb aabaGaaGinaaaakiabgUcaRmaabmaabaGaaGOmaiaadchadaWgaaWc baGaaGimaaqabaGccqGHRaWkcaWGWbWaaSbaaSqaaiaaigdaaeqaaO Gaey4kaSIaamiCamaaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMca amaalaaabaGaamiAaaqaaiaaikdaaaGaeq4UdW2aa0baaSqaaiaad2 gacaWGUbaabaGaaGOmaaaakiabgkHiTmaadmaabaWaaSGbaeaacqaH bpGCcaWGObGaey4kaSYaaeWaaeaacqaHbpGCdaWgaaWcbaGaaGymaa qabaGccqGHRaWkcqaHbpGCdaWgaaWcbaGaaGOmaaqabaaakiaawIca caGLPaaaaeaacqaH6oWAdaWgaaWcbaGaamyBaiaad6gaaeqaaaaaaO Gaay5waiaaw2faaiabeM8a3naaDaaaleaacaWGTbGaamOBaaqaaiaa ikdaaaGccqGH9aqpcaaIWaaaaa@5FC7@ .

Из данного уравнения с заменами

ω 0mn 2 = D λ mn 4 ρh , α mn = 2 p 0 + p 1 + p 2 λ mn 2 2ρ ω 0mn 2 , μ mn = ρ 1 + ρ 2 ρh λ mn , η mn = ω 0mn 2 c i 2 λ mn 2 , x mn = ω mn 2 ω 0mn 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaaGimaiaad2gacaWGUb aabaGaaGOmaaaakiabg2da9maalaaabaGaamiraiabeU7aSnaaDaaa leaacaWGTbGaamOBaaqaaiaaisdaaaaakeaacqaHbpGCcaWGObaaai aacYcacaaMc8UaeqySde2aaSbaaSqaaiaad2gacaWGUbaabeaakiab g2da9maalaaabaWaaeWaaeaacaaIYaGaamiCamaaBaaaleaacaaIWa aabeaakiabgUcaRiaadchadaWgaaWcbaGaaGymaaqabaGccqGHRaWk caWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaGaeq4UdW 2aa0baaSqaaiaad2gacaWGUbaabaGaaGOmaaaaaOqaaiaaikdacqaH bpGCcqaHjpWDdaqhaaWcbaGaaGimaiaad2gacaWGUbaabaGaaGOmaa aaaaGccaGGSaGaaGPaVlaaykW7caaMc8UaeqiVd02aaSbaaSqaaiaa d2gacaWGUbaabeaakiabg2da9maalaaabaGaeqyWdi3aaSbaaSqaai aaigdaaeqaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaGc baGaeqyWdiNaamiAaiabeU7aSnaaBaaaleaacaWGTbGaamOBaaqaba aaaOGaaiilaiaaykW7caaMc8Uaeq4TdG2aaSbaaSqaaiaad2gacaWG Ubaabeaakiabg2da9maalaaabaGaeqyYdC3aa0baaSqaaiaaicdaca WGTbGaamOBaaqaaiaaikdaaaaakeaacaWGJbWaa0baaSqaaiaadMga aeaacaaIYaaaaOGaeq4UdW2aa0baaSqaaiaad2gacaWGUbaabaGaaG OmaaaaaaGccaGGSaGaaGPaVlaaykW7caWG4bWaaSbaaSqaaiaad2ga caWGUbaabeaakiabg2da9maalaaabaGaeqyYdC3aa0baaSqaaiaad2 gacaWGUbaabaGaaGOmaaaaaOqaaiabeM8a3naaDaaaleaacaaIWaGa amyBaiaad6gaaeaacaaIYaaaaaaaaaa@9AEE@

  ω 0mn 2 = D λ mn 4 ρh , α mn = 2 p 0 + p 1 + p 2 λ mn 2 2ρ ω 0mn 2 , μ mn = ρ 1 + ρ 2 ρh λ mn , η mn = ω 0mn 2 c i 2 λ mn 2 , x mn = ω mn 2 ω 0mn 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaaGimaiaad2gacaWGUb aabaGaaGOmaaaakiabg2da9maalaaabaGaamiraiabeU7aSnaaDaaa leaacaWGTbGaamOBaaqaaiaaisdaaaaakeaacqaHbpGCcaWGObaaai aacYcacaaMc8UaeqySde2aaSbaaSqaaiaad2gacaWGUbaabeaakiab g2da9maalaaabaWaaeWaaeaacaaIYaGaamiCamaaBaaaleaacaaIWa aabeaakiabgUcaRiaadchadaWgaaWcbaGaaGymaaqabaGccqGHRaWk caWGWbWaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaGaeq4UdW 2aa0baaSqaaiaad2gacaWGUbaabaGaaGOmaaaaaOqaaiaaikdacqaH bpGCcqaHjpWDdaqhaaWcbaGaaGimaiaad2gacaWGUbaabaGaaGOmaa aaaaGccaGGSaGaaGPaVlaaykW7caaMc8UaeqiVd02aaSbaaSqaaiaa d2gacaWGUbaabeaakiabg2da9maalaaabaGaeqyWdi3aaSbaaSqaai aaigdaaeqaaOGaey4kaSIaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaGc baGaeqyWdiNaamiAaiabeU7aSnaaBaaaleaacaWGTbGaamOBaaqaba aaaOGaaiilaiaaykW7caaMc8Uaeq4TdG2aaSbaaSqaaiaad2gacaWG Ubaabeaakiabg2da9maalaaabaGaeqyYdC3aa0baaSqaaiaaicdaca WGTbGaamOBaaqaaiaaikdaaaaakeaacaWGJbWaa0baaSqaaiaadMga aeaacaaIYaaaaOGaeq4UdW2aa0baaSqaaiaad2gacaWGUbaabaGaaG OmaaaaaaGccaGGSaGaaGPaVlaaykW7caWG4bWaaSbaaSqaaiaad2ga caWGUbaabeaakiabg2da9maalaaabaGaeqyYdC3aa0baaSqaaiaad2 gacaWGUbaabaGaaGOmaaaaaOqaaiabeM8a3naaDaaaleaacaaIWaGa amyBaiaad6gaaeaacaaIYaaaaaaaaaa@9AEE@ (3.11)

получим алгебраическое уравнение относительно xmn:

  1 x mn + α mn μ mn x mn 1 η mn x mn =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaaIXaGaeyOeI0IaamiEamaaBaaaleaacaWGTb GaamOBaaqabaGccqGHRaWkcqaHXoqydaWgaaWcbaGaamyBaiaad6ga aeqaaOGaeyOeI0YaaSaaaeaacqaH8oqBdaWgaaWcbaGaamyBaiaad6 gaaeqaaOGaamiEamaaBaaaleaacaWGTbGaamOBaaqabaaakeaadaGc aaqaaiaaigdacqGHsislcqaH3oaAdaWgaaWcbaGaamyBaiaad6gaae qaaOGaamiEamaaBaaaleaacaWGTbGaamOBaaqabaaabeaaaaGccqGH 9aqpcaaIWaaaaa@4CA3@ . (3.12)

При условиях

1 η mn x mn >0,1 x mn + α mn 0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaaIXaGaeyOeI0Iaeq4TdG2aaSbaaSqaaiaad2 gacaWGUbaabeaakiaadIhadaWgaaWcbaGaamyBaiaad6gaaeqaaOGa aeOpaiaabcdacaqGSaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca qGXaGaeyOeI0IaamiEamaaBaaaleaacaWGTbGaamOBaaqabaGccqGH RaWkcqaHXoqydaWgaaWcbaGaamyBaiaad6gaaeqaaOGaeyyzImRaaG imaaaa@4F40@  

уравнение (3.12) принимает вид:

x mn 3 + a 1 x mn 2 + a 2 x mn a 3 =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG4bWaa0baaSqaaiaad2gacaWGUbaabaGaaG 4maaaakiabgUcaRiaadggadaWgaaWcbaGaaGymaaqabaGccaWG4bWa a0baaSqaaiaad2gacaWGUbaabaGaaGOmaaaakiabgUcaRiaadggada WgaaWcbaGaaGOmaaqabaGccaWG4bWaaSbaaSqaaiaad2gacaWGUbaa beaakiabgkHiTiaadggadaWgaaWcbaGaaG4maaqabaGccqGH9aqpca aIWaaaaa@45C6@ , (3.13)

где

a 1 = μ mn 2 1 η mn 2 1+ α mn , a 2 = 1+ α mn 2 + 2 1+ α mn η mn , a 3 = 1+ α mn 2 η mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGHbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0 ZaaSaaaeaacqaH8oqBdaqhaaWcbaGaamyBaiaad6gaaeaacaaIYaaa aOGaeyOeI0IaaGymaaqaaiabeE7aOnaaBaaaleaacaWGTbGaamOBaa qabaaaaOGaeyOeI0IaaGOmamaabmaabaGaaGymaiabgUcaRiabeg7a HnaaBaaaleaacaWGTbGaamOBaaqabaaakiaawIcacaGLPaaacaGGSa GaaGPaVlaaykW7caaMc8UaamyyamaaBaaaleaacaaIYaaabeaakiab g2da9maabmaabaGaaGymaiabgUcaRiabeg7aHnaaBaaaleaacaWGTb GaamOBaaqabaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGc cqGHRaWkdaWcaaqaaiaaikdadaqadaqaaiaaigdacqGHRaWkcqaHXo qydaWgaaWcbaGaamyBaiaad6gaaeqaaaGccaGLOaGaayzkaaaabaGa eq4TdG2aaSbaaSqaaiaad2gacaWGUbaabeaaaaGccaGGSaGaaGPaVl aaykW7caWGHbWaaSbaaSqaaiaaiodaaeqaaOGaeyypa0ZaaSaaaeaa daqadaqaaiaaigdacqGHRaWkcqaHXoqydaWgaaWcbaGaamyBaiaad6 gaaeqaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGcbaGa eq4TdG2aaSbaaSqaaiaad2gacaWGUbaabeaaaaaaaa@7519@ .

Кубическое уравнение (3.13) на числовой прямой имеет хотя бы один вещественный корень. Пусть xmn = x0mn = x0 такой корень. Тогда

x 3 + a 1 x 2 + a 2 x a 3 = x x 0 x 2 + a 1 + x 0 x 2 + a 2 +x a 1 + x 0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaS IaamyyamaaBaaaleaacaaIXaaabeaakiaadIhadaahaaWcbeqaaiaa ikdaaaGccqGHRaWkcaWGHbWaaSbaaSqaaiaaikdaaeqaaOGaamiEai abgkHiTiaadggadaWgaaWcbaGaaG4maaqabaGccqGH9aqpdaqadaqa aiaadIhacqGHsislcaWG4bWaaSbaaSqaaiaaicdaaeqaaaGccaGLOa GaayzkaaWaamWaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4k aSYaaeWaaeaacaWGHbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaam iEamaaBaaaleaacaaIWaaabeaaaOGaayjkaiaawMcaaiaadIhadaah aaWcbeqaaiaaikdaaaGccqGHRaWkcaWGHbWaaSbaaSqaaiaaikdaae qaaOGaey4kaSIaamiEamaabmaabaGaamyyamaaBaaaleaacaaIXaaa beaakiabgUcaRiaadIhadaWgaaWcbaGaaGimaaqabaaakiaawIcaca GLPaaaaiaawUfacaGLDbaaaaa@5C32@ .

Обозначим через f x MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaa aaaa@347B@  левую часть уравнения (3.13), где xmn = x. Функция f x = x 3 + a 1 x 2 + a 2 x a 3 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaa Gaeyypa0JaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaadgga daWgaaWcbaGaaGymaaqabaGccaWG4bWaaWbaaSqabeaacaaIYaaaaO Gaey4kaSIaamyyamaaBaaaleaacaaIYaaabeaakiaadIhacqGHsisl caWGHbWaaSbaaSqaaiaaiodaaeqaaaaa@428E@  – это многочлен, по крайней мере непрерывная на всей числовой прямой функция. В точке x = 0: f 0 = a 3 <0. MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaaIWaaacaGLOaGaayzkaa Gaeyypa0JaeyOeI0IaamyyamaaBaaaleaacaaIZaaabeaakiabgYda 8iaaicdacaGGUaaaaa@3A74@  Выясним знак функции f x MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaa aaaa@347B@  в точке x = 1:

f 1 =1+ a 1 + a 2 a 3 = μ mn 2 α mn 2 η mn + α mn 2 = μ mn 2 η mn + α mn 2 η mn 1 η mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaaIXaaacaGLOaGaayzkaa Gaeyypa0JaaGymaiabgUcaRiaadggadaWgaaWcbaGaaGymaaqabaGc cqGHRaWkcaWGHbWaaSbaaSqaaiaaikdaaeqaaOGaeyOeI0Iaamyyam aaBaaaleaacaaIZaaabeaakiabg2da9maalaaabaGaeqiVd02aa0ba aSqaaiaad2gacaWGUbaabaGaaGOmaaaakiabgkHiTiabeg7aHnaaDa aaleaacaWGTbGaamOBaaqaaiaaikdaaaaakeaacqaH3oaAdaWgaaWc baGaamyBaiaad6gaaeqaaaaakiabgUcaRiabeg7aHnaaDaaaleaaca WGTbGaamOBaaqaaiaaikdaaaGccqGH9aqpdaWcaaqaaiabeY7aTnaa DaaaleaacaWGTbGaamOBaaqaaiaaikdaaaaakeaacqaH3oaAdaWgaa WcbaGaamyBaiaad6gaaeqaaaaakiabgUcaRiabeg7aHnaaDaaaleaa caWGTbGaamOBaaqaaiaaikdaaaGcdaWcaaqaaiabeE7aOnaaBaaale aacaWGTbGaamOBaaqabaGccqGHsislcaaIXaaabaGaeq4TdG2aaSba aSqaaiaad2gacaWGUbaabeaaaaaaaa@6A5B@ .

Если μ mn α mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH8oqBdaWgaaWcbaGaamyBaiaad6gaaeqaaO GaeyyzImRaeqySde2aaSbaaSqaaiaad2gacaWGUbaabeaaaaa@3A51@  или η mn 1 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH3oaAdaWgaaWcbaGaamyBaiaad6gaaeqaaO GaeyyzImRaaGymaaaa@3752@ , то f 1 >0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaaIXaaacaGLOaGaayzkaa GaeyOpa4JaaGimaaaa@35FB@  и график функции f x MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaa aaaa@347B@  пересекает ось Ox между точками x = 0 и x = 1, т.е. существует точка x = x0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqGHiiIZaaa@328E@  (0, 1), такая, что f x 0 =0. MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaWG4bWaaSbaaSqaaiaaic daaeqaaaGccaGLOaGaayzkaaGaeyypa0JaaGimaiaac6caaaa@37DD@  

Рассмотрим параметр η mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH3oaAdaWgaaWcbaGaamyBaiaad6gaaeqaaa aa@34C7@  и оценим его снизу:

  η mn = ω 0mn 2 c i 2 λ mn 2 = D λ mn 4 ρh c i 2 λ mn 2 = D λ mn 2 ρh c i 2 D λ 11 2 ρh c i 2 1 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH3oaAdaWgaaWcbaGaamyBaiaad6gaaeqaaO Gaeyypa0ZaaSaaaeaacqaHjpWDdaqhaaWcbaGaaGimaiaad2gacaWG UbaabaGaaGOmaaaaaOqaaiaadogadaqhaaWcbaGaamyAaaqaaiaaik daaaGccqaH7oaBdaqhaaWcbaGaamyBaiaad6gaaeaacaaIYaaaaaaa kiabg2da9maalaaabaGaamiraiabeU7aSnaaDaaaleaacaWGTbGaam OBaaqaaiaaisdaaaaakeaacqaHbpGCcaWGObGaam4yamaaDaaaleaa caWGPbaabaGaaGOmaaaakiabeU7aSnaaDaaaleaacaWGTbGaamOBaa qaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacaWGebGaeq4UdW2aa0ba aSqaaiaad2gacaWGUbaabaGaaGOmaaaaaOqaaiabeg8aYjaadIgaca WGJbWaa0baaSqaaiaadMgaaeaacaaIYaaaaaaakiabgwMiZoaalaaa baGaamiraiabeU7aSnaaDaaaleaacaaIXaGaaGymaaqaaiaaikdaaa aakeaacqaHbpGCcaWGObGaam4yamaaDaaaleaacaWGPbaabaGaaGOm aaaaaaGccqGHLjYScaaIXaaaaa@6D9E@ . (3.14)

Отсюда видно, что при выборе данных D, ρ, h, a, b, ci всегда можно добиться выполнения неравенства (3.14). Если же η mn <1 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH3oaAdaWgaaWcbaGaamyBaiaad6gaaeqaaO GaeyipaWJaaGymaaaa@3690@  и по условию 1 η mn x mn >0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaaIXaGaeyOeI0Iaeq4TdG2aaSbaaSqaaiaad2 gacaWGUbaabeaakiaadIhadaWgaaWcbaGaamyBaiaad6gaaeqaaOGa eyOpa4JaaGimaaaa@3B53@ , то тогда имеем:

  1 η mn x mn 1 2 1 1 2 η mn x mn 1 8 η mn x mn 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaqadaqaaiaaigdacqGHsislcqaH3oaAdaWgaa WcbaGaamyBaiaad6gaaeqaaOGaamiEamaaBaaaleaacaWGTbGaamOB aaqabaaakiaawIcacaGLPaaadaahaaWcbeqaamaalaaabaGaaGymaa qaaiaaikdaaaaaaOGaeyisISRaaGymaiabgkHiTmaalaaabaGaaGym aaqaaiaaikdaaaGaeq4TdG2aaSbaaSqaaiaad2gacaWGUbaabeaaki aadIhadaWgaaWcbaGaamyBaiaad6gaaeqaaOGaeyOeI0YaaSaaaeaa caaIXaaabaGaaGioaaaadaqadaqaaiabeE7aOnaaBaaaleaacaWGTb GaamOBaaqabaGccaWG4bWaaSbaaSqaaiaad2gacaWGUbaabeaaaOGa ayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaa@5462@ . (3.15)

Тогда уравнение (3.12) с учетом (3.15) примет вид:

1 x mn + α mn 1 η mn x mn 2 1 8 η mn x mn 2 μ mn x mn =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaqadaqaaiaaigdacqGHsislcaWG4bWaaSbaaS qaaiaad2gacaWGUbaabeaakiabgUcaRiabeg7aHnaaBaaaleaacaWG TbGaamOBaaqabaaakiaawIcacaGLPaaacqGHflY1daWadaqaaiaaig dacqGHsisldaWcaaqaaiabeE7aOnaaBaaaleaacaWGTbGaamOBaaqa baGccaWG4bWaaSbaaSqaaiaad2gacaWGUbaabeaaaOqaaiaaikdaaa GaeyOeI0YaaSaaaeaacaaIXaaabaGaaGioaaaadaqadaqaaiabeE7a OnaaBaaaleaacaWGTbGaamOBaaqabaGccaWG4bWaaSbaaSqaaiaad2 gacaWGUbaabeaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaa aOGaay5waiaaw2faaiabgkHiTiabeY7aTnaaBaaaleaacaWGTbGaam OBaaqabaGccaWG4bWaaSbaaSqaaiaad2gacaWGUbaabeaakiabg2da 9iaaicdaaaa@5EE9@

или

  x mn 3 + b 1 x mn 2 b 2 x mn + b 3 =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG4bWaa0baaSqaaiaad2gacaWGUbaabaGaaG 4maaaakiabgUcaRiaadkgadaWgaaWcbaGaaGymaaqabaGccaWG4bWa a0baaSqaaiaad2gacaWGUbaabaGaaGOmaaaakiabgkHiTiaadkgada WgaaWcbaGaaGOmaaqabaGccaWG4bWaaSbaaSqaaiaad2gacaWGUbaa beaakiabgUcaRiaadkgadaWgaaWcbaGaaG4maaqabaGccqGH9aqpca aIWaaaaa@45C9@ , (3.16)

где

  b 1 = 4 η mn 1 α mn , b 2 = 4 1+ α mn η mn + 8 μ mn +1 η mn 2 , b 3 = 8 1+ α mn η mn 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGIbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0 ZaaSaaaeaacaaI0aaabaGaeq4TdG2aaSbaaSqaaiaad2gacaWGUbaa beaaaaGccqGHsislcaaIXaGaeyOeI0IaeqySde2aaSbaaSqaaiaad2 gacaWGUbaabeaakiaacYcacaaMc8UaaGPaVlaaykW7caWGIbWaaSba aSqaaiaaikdaaeqaaOGaeyypa0ZaaSaaaeaacaaI0aWaaeWaaeaaca aIXaGaey4kaSIaeqySde2aaSbaaSqaaiaad2gacaWGUbaabeaaaOGa ayjkaiaawMcaaaqaaiabeE7aOnaaBaaaleaacaWGTbGaamOBaaqaba aaaOGaey4kaSYaaSaaaeaacaaI4aWaaeWaaeaacqaH8oqBdaWgaaWc baGaamyBaiaad6gaaeqaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaa qaaiabeE7aOnaaDaaaleaacaWGTbGaamOBaaqaaiaaikdaaaaaaOGa aiilaiaaykW7caaMc8UaamOyamaaBaaaleaacaaIZaaabeaakiabg2 da9maalaaabaGaaGioamaabmaabaGaaGymaiabgUcaRiabeg7aHnaa BaaaleaacaWGTbGaamOBaaqabaaakiaawIcacaGLPaaaaeaacqaH3o aAdaqhaaWcbaGaamyBaiaad6gaaeaacaaIYaaaaaaaaaa@7272@ .

Обозначим левую часть уравнения (3.16) через

g x = x 3 + b 1 x 2 b 2 x+ b 3 =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGNbWaaeWaaeaacaWG4baacaGLOaGaayzkaa Gaeyypa0JaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaadkga daWgaaWcbaGaaGymaaqabaGccaWG4bWaaWbaaSqabeaacaaIYaaaaO GaeyOeI0IaamOyamaaBaaaleaacaaIYaaabeaakiaadIhacqGHRaWk caWGIbWaaSbaaSqaaiaaiodaaeqaaOGaeyypa0JaaGimaaaa@445C@ ,

где x = xmn, аналогично функции f x MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaa aaaa@347B@  убеждаемся в существовании корня уравнения (3.16). Действительно, вычислим g 0 = b 3 >0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGNbWaaeWaaeaacaaIWaaacaGLOaGaayzkaa Gaeyypa0JaamOyamaaBaaaleaacaaIZaaabeaakiabg6da+iaaicda aaa@38DB@  и

  g 1 =1+ b 1 b 2 + b 3 = α mn 4 α mn η mn 8 η mn 2 μ mn α mn <0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGNbWaaeWaaeaacaaIXaaacaGLOaGaayzkaa Gaeyypa0JaaGymaiabgUcaRiaadkgadaWgaaWcbaGaaGymaaqabaGc cqGHsislcaWGIbWaaSbaaSqaaiaaikdaaeqaaOGaey4kaSIaamOyam aaBaaaleaacaaIZaaabeaakiabg2da9iabgkHiTiabeg7aHnaaBaaa leaacaWGTbGaamOBaaqabaGccqGHsislcaaI0aWaaSaaaeaacqaHXo qydaWgaaWcbaGaamyBaiaad6gaaeqaaaGcbaGaeq4TdG2aaSbaaSqa aiaad2gacaWGUbaabeaaaaGccqGHsisldaWcaaqaaiaaiIdaaeaacq aH3oaAdaqhaaWcbaGaamyBaiaad6gaaeaacaaIYaaaaaaakmaabmaa baGaeqiVd02aaSbaaSqaaiaad2gacaWGUbaabeaakiabgkHiTiabeg 7aHnaaBaaaleaacaWGTbGaamOBaaqabaaakiaawIcacaGLPaaacqGH 8aapcaaIWaaaaa@5F22@

при условии μ mn α mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH8oqBdaWgaaWcbaGaamyBaiaad6gaaeqaaO GaeyyzImRaeqySde2aaSbaaSqaaiaad2gacaWGUbaabeaaaaa@3A51@ . Последнее неравенство всегда имеет место, если

ρ 1 + ρ 2 ρ 6 1 ν 2 2 p 0 + p 1 + p 2 ab πE a 2 + b 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaadaWcaaqaaiabeg8aYnaaBaaaleaacaaIXaaabe aakiabgUcaRiabeg8aYnaaBaaaleaacaaIYaaabeaaaOqaaiabeg8a YbaacqGHLjYSdaWcaaqaaiaaiAdadaqadaqaaiaaigdacqGHsislcq aH9oGBdaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaadaqadaqa aiaaikdacaWGWbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaamiCam aaBaaaleaacaaIXaaabeaakiabgUcaRiaadchadaWgaaWcbaGaaGOm aaqabaaakiaawIcacaGLPaaacaWGHbGaamOyaaqaaiabec8aWjaadw eadaGcaaqaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG IbWaaWbaaSqabeaacaaIYaaaaaqabaaaaaaa@543B@ .

Поскольку показано существование решений уравнений (3.13) и (3.16), то частоты колебаний в случае сжимаемой среды определяются по формуле

  ω mn = ω 0mn x mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaWgaaWcbaGaamyBaiaad6gaaeqaaO Gaeyypa0JaeqyYdC3aaSbaaSqaaiaaicdacaWGTbGaamOBaaqabaGc daGcaaqaaiaadIhadaWgaaWcbaGaamyBaiaad6gaaeqaaaqabaaaaa@3DB8@ . (3.17)

Покажем, что частоты, определенные по формуле (3.12), меньше, чем соответствующие частоты в несжимаемой среде.

Действительно, из уравнения (3.12) имеем:

1+ α mn = 1+ μ mn 1 η mn x mn x mn > 1+ μ mn x mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaqGXaGaey4kaSIaeqySde2aaSbaaSqaaiaad2 gacaWGUbaabeaakiabg2da9maabmaabaGaaGymaiabgUcaRmaalaaa baGaeqiVd02aaSbaaSqaaiaad2gacaWGUbaabeaaaOqaamaakaaaba GaaGymaiabgkHiTiabeE7aOnaaBaaaleaacaWGTbGaamOBaaqabaGc caWG4bWaaSbaaSqaaiaad2gacaWGUbaabeaaaeqaaaaaaOGaayjkai aawMcaaiaadIhadaWgaaWcbaGaamyBaiaad6gaaeqaaOGaeyOpa4Za aeWaaeaacaaIXaGaey4kaSIaeqiVd02aaSbaaSqaaiaad2gacaWGUb aabeaaaOGaayjkaiaawMcaaiaadIhadaWgaaWcbaGaamyBaiaad6ga aeqaaaaa@5523@ .

Отсюда следует:

x mn < 1+ α mn 1+ μ mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWG4bWaaSbaaSqaaiaad2gacaWGUbaabeaaki abgYda8maalaaabaGaaeymaiabgUcaRiabeg7aHnaaBaaaleaacaWG TbGaamOBaaqabaaakeaacaaIXaGaey4kaSIaeqiVd02aaSbaaSqaai aad2gacaWGUbaabeaaaaaaaa@3FEA@  или ω mn 2 < ω 0mn 2 1+ α mn 1+ μ mn MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaOGaaeipaiabeM8a3naaDaaaleaacaaIWaGaamyBaiaad6ga aeaacaaIYaaaaOWaaSaaaeaacaaIXaGaey4kaSIaeqySde2aaSbaaS qaaiaad2gacaWGUbaabeaaaOqaaiaaigdacqGHRaWkcqaH8oqBdaWg aaWcbaGaamyBaiaad6gaaeqaaaaaaaa@4698@ .

Правая часть полученной оценки представляет формулу (2.25), по которой определяются частоты колебаний в несжимаемой среде.

Обозначим f mn = ω mn / 2π, MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaSbaaSqaaiaad2gacaWGUbaabeaaki abg2da9maalyaabaGaeqyYdC3aaSbaaSqaaiaad2gacaWGUbaabeaa aOqaaiaaikdacqaHapaCcaGGSaGaaGPaVdaaaaa@3DC8@  тогда первая полная собственная частота колебаний равна f 11 = ω 11 / 2π, ω 11 = ω 011 x 11 . MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacaWGMbWaaSbaaSqaaiaaigdacaaIXaaabeaaki abg2da9maalyaabaGaeqyYdC3aaSbaaSqaaiaaigdacaaIXaaabeaa aOqaaiaaikdacqaHapaCcaGGSaGaaGPaVlaaykW7aaGaeqyYdC3aaS baaSqaaiaaigdacaaIXaaabeaakiabg2da9iabeM8a3naaBaaaleaa caaIWaGaaGymaiaaigdaaeqaaOWaaOaaaeaacaWG4bWaaSbaaSqaai aaigdacaaIXaaabeaaaeqaaOGaaiOlaaaa@4A92@

При m = 1, n = 1, E = 2·105 МПа, ν = 0.3, ρ = 7800 кг/м3, h = 0.001 м, a = 0.2 м, b = 0.2 м, κ 1,2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH6oWAdaWgaaWcbaGaaGymaiaacYcacaaIYa aabeaaaaa@350F@  = 1.4, атмосферном давлении pa = 0.1 МПа, плотности воздуха при атмосферном давлении ρ1a = 1.2928 кг/м3, p1 = p2 = p = 2 МПа численное решение уравнения (3.10) дает корень: x11 = 0.93948. Соответствующая частота равна f11 = 116.6 Гц.

Для проверки выполнения условия κ mn 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH6oWAdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaaaa@358A@  > 0 учтем полученное выражение ω 11 = ω 011 x 11 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHjpWDdaWgaaWcbaGaaGymaiaaigdaaeqaaO Gaeyypa0JaeqyYdC3aaSbaaSqaaiaaicdacaaIXaGaaGymaaqabaGc daGcaaqaaiaadIhadaWgaaWcbaGaaGymaiaaigdaaeqaaaqabaaaaa@3C6B@ , где ω011 определяется из первой формулы из (3.11). В рассмотренном примере x11 ≈ 0.94, т.е. имеет место преобладание влияния давления воздуха над его плотностью. Условие κ mn 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaH6oWAdaqhaaWcbaGaamyBaiaad6gaaeaaca aIYaaaaaaa@358A@  > 0 для случая стальной пластины и воды выполняется всегда, а в случае газов − при больших значениях a/h (например, a/h > 15).

В таблице приводятся частоты изгибных колебаний прямоугольной. Из таблицы следует, что частоты, вычисленные по формулам для несжимаемой и сжимаемой сред, отличаются незначительно, причем частоты по формуле для несжимаемой среды больше, чем частоты по формуле для сжимаемой среды.

 

Таблица. 1 Частоты изгибных колебаний прямоугольной пластины для разных mn по формулам (2.25) и (3.17) соответственно для несжимаемой и сжимаемой сред

m, n

fmn, Hz, формула (2.25)

fmn, Hz, формула (3.17)

0, 1; 1, 0

59.91

59.89

1, 1

117.55

117.50

2, 2

469.1

468.8

3, 3

1061.1

1059.8

 

На рис. 2а приводится зависимость первой частоты изгибных колебаний пластинки от давления для разных газов. Из рис. 2а видно, что с ростом давления собственная частота колебаний убывает. А с увеличением плотности газа происходит уменьшение собственной частоты изгибных колебаний. На рис. 2б приводится зависимость первой частоты изгибных колебаний пластинки от давления по формулам для несжимаемой и сжимаемой жидкостей для двуокиси углерода. Из рис. 2б видно, что частоты по модели несжимаемой жидкости выше частот по модели для сжимаемой жидкости, причем с ростом давления разность частот колебаний возрастает.

 

Рис. 2. Зависимость первой частоты изгибных колебаний пластинки f11 (Hz) от давления p2 (MPa) для давления p1 = 0.5 МПа: (а) для разных газов: ρ1a=ρ2a= 0.1785 (гелий), 1.2928 (воздух), 1.9768 (двуокись углерода) кг/м3 (пунктирная, штриховая, сплошная линии соответственно); (b) по формулам для несжимаемой (2.25) и сжимаемой (3.17) жидкостей для двуокиси углерода ρ1a=ρ2a= 1.9768 кг/м3 (сплошная, пунктирная линии соответственно).

 

На рис. 3,а приводится зависимость второй частоты изгибных колебаний пластинки от давления для разных газов для m = n = 2. Из рис. 3,а видно, что с ростом давления собственная частота колебаний убывает. А с увеличением плотности газа происходит уменьшение собственной частоты изгибных колебаний. На рис. 3,b приводится зависимость второй частоты изгибных колебаний пластинки от давления по формулам для несжимаемой и сжимаемой жидкостей для двуокиси углерода. Из рис. 3,b видно, что частоты по модели несжимаемой жидкости выше частот по модели для сжимаемой жидкости, причем с ростом давления разность частот колебаний возрастает.

 

Рис. 3. Зависимость второй частоты изгибных колебаний пластинки f22 (Hz) от давления p2 (MPa) для давления p1 = 0.5 МПа: (а) для разных газов: ρ1a=ρ2a= 0.1785 (гелий), 1.2928 (воздух), 1.9768 (двуокись углерода) кг/м3 (пунктирная, штриховая, сплошная линии соответственно); (b) по формулам для несжимаемой (2.25) и сжимаемой (3.17) жидкостей для двуокиси углерода ρ1a=ρ2a= 1.9768 кг/м3 (сплошная, пунктирная линии соответственно).

 

Аналогично п. 2 находится общее решение дифференциального уравнения (3.9) и строится формула (2.32) для определения формы колебаний пластины с учетом найденных частот wmn по формуле (3.17).

Таким образом, в случае сжимаемой среды имеют место следующие утверждения.

Утверждение 3. Частоты в случае сжимаемой среды меньше, чем соответствующие частоты в несжимаемой среде.

Утверждение 4. С ростом давления собственная частота колебаний возрастает для гелия и убывает для воздуха и углекислого газа. А с увеличением плотности газа происходит уменьшение собственной частоты изгибных колебаний.

Утверждение 5. Если начальные функции τ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHepaDdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@3703@  и ψ x,y MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaqadaqaaiaadIhacaGGSaGaamyEaa GaayjkaiaawMcaaaaa@370C@  удовлетворяют условиям замечания из пункта 2 и ψ 00 =0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHipqEdaWgaaWcbaGaaGimaiaaicdaaeqaaO Gaeyypa0JaaGimaaaa@3642@ , то колебания прямоугольной однородной пластины в сжимаемой среде при избыточных давлениях p1, p2 и плотностях ρ 1 , ρ 2 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaGGSa GaeqyWdi3aaSbaaSqaaiaaikdaaeqaaaaa@3713@ , удовлетворяющих условию ρ 1 B 100 ρ 2 B 200 0 MathType@MTEF@5@5@+= feaahGart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbov2D09 MBdbqedmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqik8fkY=xipgYl h9vqqj=hEeeu0xXdi9arFj0xirFj0dXdbba91qpK0=yr0RYxfr=Jbb f9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciaacaGaaeqa baqabeGadaaakeaacqaHbpGCdaWgaaWcbaGaaGymaaqabaGccaWGcb WaaSbaaSqaaiaaigdacaaIWaGaaGimaaqabaGccqGHsislcqaHbpGC daWgaaWcbaGaaGOmaaqabaGccaWGcbWaaSbaaSqaaiaaikdacaaIWa GaaGimaaqabaGccqGHGjsUcaaIWaaaaa@4034@ , определяется по формуле (2.32), собственные колебания пластины находятся по формуле (2.33), а собственные частоты по формуле (3.17) при условиях (2.7) и (2.27), где коэффициенты τmn и ψmn определяются соответственно по формулам (2.30), (2.31).

4. Заключение. Хорошо известно из литературы (например, [1–3], что собственные частоты изгибных колебаний пластины при ее контакте с жидкостью значительно снижаются. Это объясняется влиянием присоединенной массы жидкости. Установлено [15, 16], что учет разности площадей противоположных поверхностей пластины, образующейся при ее изгибе, может оказывать повышающее влияние на собственные частоты. Учет этого эффекта приводит к появлению распределенной поперечной силы, равной произведению кривизны срединной поверхности и среднего давления на поверхности пластины.

Одновременное влияние указанных факторов на частоты колебаний в случае несжимаемой жидкости зависит от отношения среднего давления к модулю упругости материала, плотностей материала и жидкости и отношения длины пластины к ее толщине. Для реальных параметров характерно превалирующее влияние плотности среды над давлением в ней. Однако давление может оказывать заметное влияние на результат.

Для сжимаемой жидкости влияние носит более сложный характер, так как присоединенная масса зависит от скорости звука и от самой частоты колебаний. Кроме того, давление и плотность газовой среды не являются независимыми.

Влияние контактирующей среды на частоты колебаний является значительным для весьма тонких пластин и пленок с низким модулем упругости. Учет его необходим особенно в случае элементов микро- и наноразмерных толщин.

С ростом давления собственная частота колебаний возрастает. В случае легких газов (водород, гелий) влияние давления может превалировать над их плотностью. Эти результаты могут быть использованы при моделировании колебаний пластинок, контактирующих с жидкостью и газом, в том числе микро- и наноразмеров.

Работа проведена в порядке выполнения госзадания (FMRS-2023-0015).

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Авторлар туралы

K. Sabitov

Mavlyutov Institute of Mechanics

Хат алмасуға жауапты Автор.
Email: sabitov_fmf@mail.ru
Ресей, Ufa

A. Khakimov

Mavlyutov Institute of Mechanics

Email: hakimov@anrb.ru
Ресей, Ufa

Әдебиет тізімі

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Қосымша файлдар

Қосымша файлдар
Әрекет
1. JATS XML
2. Fig. 1. Elements dx and dy of the middle surface of a curved plate.

Жүктеу (13KB)
3. Fig. 2. Dependence of the first frequency of bending vibrations of the plate f11 (Hz) on the pressure p2 (MPa) for the pressure p1 = 0.5 MPa: (a) for different gases: = 0.1785 (helium), 1.2928 (air), 1.9768 (carbon dioxide) kg/m3 (dotted, dashed, solid lines, respectively); (b) according to the formulas for incompressible (2.25) and compressible (3.17) liquids for carbon dioxide = 1.9768 kg/m3 (solid, dotted lines, respectively).

Жүктеу (20KB)
4. Fig. 3. Dependence of the second frequency of bending vibrations of the plate f22 (Hz) on the pressure p2 (MPa) for the pressure p1 = 0.5 MPa: (a) for different gases: = 0.1785 (helium), 1.2928 (air), 1.9768 (carbon dioxide) kg/m3 (dotted, dashed, solid lines, respectively); (b) according to the formulas for incompressible (2.25) and compressible (3.17) liquids for carbon dioxide = 1.9768 kg/m3 (solid, dotted lines, respectively).

Жүктеу (22KB)

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