5.
Айталық болсын. Ол облысында шенелген. Онда
яғни , әрі айнымалылары мейлінше аз болғанда оң таңбалы. Мына теңсіздік
орында алатын нүктелер бар болғандықтан жүйенің нөлдік шешімі орнықсыз. Бірақ Ляпуновтың қосымша теоремасы бойынша нөлдік шешімнің толығынан орнықсыз болатыны туралы тұжырымдауға болмайды. Егер де үшін функциясын алып Ляпуновтың үшінші теоремасын қолдансақ,
болғандықтан жүйенің нөлдік шешімі толығынан орнықсыз.
6.
Жүйенің оң жағы -дан айқын түрде тәуелді емес. Егер функциясын -дан тәуелсіз етіп алсақ, оның туындысы
тек қана нүктесінде ғана емес, өсінің өн бойында нөлге тең болады. Яғни анықталған таңбалы бола алмайды. Басқа сөзбен айтқанда -дан тәуелсіз анықталған таңбалы болмайды. Сондықтан Ляпуновтың үшінші теоремасын қолдану ештеңе бермейді. Егер де деп алып, соңғы теореманы пайдаланатын болсақ,
болғандықтан жүйенің нөлдік шешімінің орнықсыз екендігін анықтаймыз.
7. Мына жүйенің
нөлдік шешімінің орнықты не орнықсыз болатындығын тексеру керек.
Четаев теоремасындағы функциясы үшін
функциясын алайық. Ол үзіліссіз дифференциалданады. Оң таңбалы мән қабылдайтын облысы
не
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17-сурет
Бұл облыстарының мына
облысының ішінде жатқан шекараларында
Біз облысын қарастырамыз. Ол облыста функциясы мына шарттарды қанағаттандырады:
облыстарында шенелген,
яғни
Мұнда -дәрежесі 5-тен кем емес мүшелердің қосындысы, ол мейлінше аз болғанда
облысынан бір ішкі облыс алайық:
Онда
Сонымен Четаев теоремасының барлық шарты қанағаттандырылады. Олай болса жүйенің шешімі орнықсыз.
Достарыңызбен бөлісу: |