Определение спектра частот и колебаний прямоугольной пластинки, подвижно заделаннной по краю, в разных средах

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Аннотация

Определяется спектр частот и формы изгибных колебаний прямоугольной пластины, контактирующей с жидкостью или газом. Дается вывод выражения распределенной поперечной нагрузки на пластину, подвижно заделанной по контуру. Поверхности пластины контактируют со средой разной 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.

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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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Об авторах

К. Б. Сабитов

Институт механики им. Р.Р. Мавлютова УФИЦ РАН

Автор, ответственный за переписку.
Email: sabitov_fmf@mail.ru
Россия, Уфа

А. Г. Хакимов

Институт механики им. Р.Р. Мавлютова УФИЦ РАН

Email: hakimov@anrb.ru
Россия, Уфа

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

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3. Рис. 2. Зависимость первой частоты изгибных колебаний пластинки f11 (Hz) от давления p2 (MPa) для давления p1 = 0.5 МПа: (а) для разных газов: = 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 = 1.9768 кг/м3 (сплошная, пунктирная линии соответственно).

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4. Рис. 3. Зависимость второй частоты изгибных колебаний пластинки f22 (Hz) от давления p2 (MPa) для давления p1 = 0.5 МПа: (а) для разных газов: = 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 = 1.9768 кг/м3 (сплошная, пунктирная линии соответственно).

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