5.9-bo'lim: Takroriy xos qiymatlar
Bu biz ko'rib chiqishimiz kerak bo'lgan oxirgi holat. Ushbu bo'limda biz tizimning echimlarini ko'rib chiqamiz,
\[\vec x' = A\vec x\]bu yerda xos qiymatlar takrorlanuvchi xos qiymatlardir. Biz \(A\) \(2 \times 2\) matritsasi bo'lgan tizimlar bilan ishlayotganimiz uchun biz boshidanoq bu taxminni qilamiz. Shunday qilib, tizim \(\lambda \) qo'sh xususiy qiymatga ega bo'ladi.
Bu bizga muammo tug'diradi. Biz umumiy yechimni shakllantirishimiz uchun ikkita chiziqli mustaqil echimlarni xohlaymiz. Biroq, er-xotin xos qiymat bilan bizda faqat bitta bo'ladi,
\[{\vec x_1} = \vec \eta {{\bf{e}}^{\lambda t}}\]Shunday qilib, biz ikkinchi yechimni topishimiz kerak. Eslatib o'tamiz, biz qaraganimizda qo'sh ildiz Ikkinchi tartibli differensial tenglamalar bilan biz shunga o'xshash muammoga duch keldik. Ushbu bo'limda biz yechimga shunchaki \(t\) qo'shdik va ikkinchi yechimni olishga muvaffaq bo'ldik. Keling, xuddi shu narsa bu holatda ham ishlaydimi yoki yo'qligini ko'rib chiqaylik. Ko'ramiz, agar
\[\vec x = t\,{{\bf{e}}^{\lambda t}}\vec \eta \]ham yechim bo‘ladi.
Tekshirish uchun biz qilishimiz kerak bo'lgan yagona narsa tizimga ulanish. Farq qilganingizda, taklif qilingan yechimni mahsulot bilan tartibga solishni unutmang!
\[\vec \eta {{\bf{e}}^{\lambda t}} + \lambda \vec \eta t{{\bf{e}}^{\lambda t}} = A\vec \eta t{{\bf{e}}^{\lambda t}}\]Endi biz chap tomonda ikkita funktsiyani oldik, o'z-o'zidan eksponensial va eksponensial marta \(t\). Shunday qilib, bizning taxminimiz yechim bo'lishi uchun biz talab qilishimiz kerak:
\[\begin{align*}A\vec \eta & = \lambda \vec \eta \hspace{0.25in} \Rightarrow \hspace{0.25in}\left( {A - \lambda I} \right)\vec \eta = \vec 0\\ \vec \eta & = \vec 0\end{align*}\]Birinchi talab muammo emas, chunki bu faqat \(\lambda \) xos qiymat va uning xos vektori \(\vec \eta \) ekanligini aytadi. Biz buni allaqachon bilardik, shuning uchun u erda hech qanday yangilik yo'q. Biroq, ikkinchisi muammo. \(\vec \eta \) xos vektor bo'lgani uchun biz u nolga teng bo'lmasligini bilamiz, lekin ikkinchi shartni qondirish uchun u bo'lishi kerak.
Shunday qilib, bizning taxminimiz noto'g'ri edi. Muammo shundaki, unda faqat eksponentsi bo'lgan yolg'iz atama bor, shuning uchun buni tuzatish uchun taxminlarimizni tuzata olmasligimizni bilib olaylik. Keling, quyidagi taxminni sinab ko'raylik.
\[\vec x = t\,{{\bf{e}}^{\lambda t}}\vec \eta + {{\bf{e}}^{\lambda t}}\vec \rho \]bu erda \(\vec \rho \) biz aniqlashimiz kerak bo'lgan noma'lum vektor.
Birinchi taxminda bo'lgani kabi, keling, buni tizimga ulab, nima olishimizni ko'raylik.
\[\begin{align*}\vec \eta {{\bf{e}}^{\lambda t}} + \lambda \vec \eta t{{\bf{e}}^{\lambda t}} + \lambda \vec \rho {{\bf{e}}^{\lambda t}} & = A\left( {\vec \eta t{{\bf{e}}^{\lambda t}} + \vec \rho {{\bf{e}}^{\lambda t}}} \right)\\ \left( {\vec \eta + \lambda \vec \rho } \right){{\bf{e}}^{\lambda t}} + \lambda \vec \eta t{{\bf{e}}^{\lambda t}} & = A\vec \eta t{{\bf{e}}^{\lambda t}} + A\vec \rho {{\bf{e}}^{\lambda t}}\end{align*}\]Endi koeffitsientlarni yana tenglashtiring,
\[\begin{align*}\lambda \vec \eta & = A\vec \eta & \Rightarrow \hspace{0.25in}\left( {A - \lambda I} \right)\vec \eta & = \vec 0\\ \vec \eta + \lambda \vec \rho & = A\vec \rho & \Rightarrow \hspace{0.25in}\left( {A - \lambda I} \right)\vec \rho & = \vec \eta \end{align*}\]Bizning birinchi taxminimiz kabi, birinchi tenglama bizga biz bilmagan hech narsani aytmaydi. Bu safar ikkinchi tenglama muammo emas. Ikkinchi tenglamaning barchasi \(\vec \rho \) bu tenglamaning yechimi bo'lishi kerakligini aytadi.
Bizning ikkinchi taxminimiz ishlaganga o'xshaydi. Shuning uchun,
\[{\vec x_2} = t\,{{\bf{e}}^{\lambda t}}\vec \eta + {{\bf{e}}^{\lambda t}}\vec \rho \]\(\vec \rho \) bilan ta'minlangan tizimga yechim bo'ladi
\[\left( {A - \lambda I} \right)\vec \rho = \vec \eta \]Bundan tashqari, bu yechim va birinchi yechim chiziqli mustaqildir va shuning uchun ular asosiy echimlar to'plamini tashkil qiladi va shuning uchun qo'sh xususiy qiymat holatida umumiy yechim:
\[\vec x = {c_1}\,{{\bf{e}}^{\lambda t}}\vec \eta + {c_2}\left( {t\,{{\bf{e}}^{\lambda t}}\vec \eta + {{\bf{e}}^{\lambda t}}\vec \rho } \right)\]Keling, misol bilan ishlaylik.
Yechim
Avval tizim uchun xos qiymatlarni toping.
\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{7 - \lambda }&1\\{ - 4}&{3 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} - 10\lambda + 25\\ & = {\left( {\lambda - 5} \right)^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}{\lambda _{1,2}} = 5\end{align*}\]Shunday qilib, biz ikki tomonlama xos qiymatga ega bo'ldik. Albatta, biz bo'lgan bo'limni hisobga olsak, bu juda ajablanarli emas. Keling, ushbu xos qiymat uchun xos vektorni topaylik.
\[\left( {\begin{array}{*{20}{c}}2&1\\{ - 4}&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\hspace{0.25in} \Rightarrow \,\hspace{0.25in}2{\eta _1} + {\eta _2} = 0\hspace{0.25in}{\eta _2} = - 2{\eta _1}\]Xususiy vektor u holda,
\[\begin{align*}\vec \eta & = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{ - 2{\eta _1}}\end{array}} \right) & \hspace{0.25in}{\eta _1} & \ne 0\\ {{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) & \hspace{0.25in}{\eta _1} & = 1\end{align*}\]Keyingi qadam \(\vec \rho \) ni topishdir. Buning uchun biz hal qilishimiz kerak,
\[\left( {\begin{array}{*{20}{c}}2&1\\{ - 4}&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rho _1}}\\{{\rho _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right)\hspace{0.25in} \Rightarrow \,\hspace{0.25in}2{\rho _1} + {\rho _2} = 1\hspace{0.25in}{\rho _2} = 1 - 2{\rho _1}\]E'tibor bering, bu biz xos qiymatni topish uchun hal qiladigan tizim bilan deyarli bir xil. Farqi faqat o'ng tomonda. Eng keng tarqalgan \(\vec \rho \)
\[\vec \rho = \left( {\begin{array}{*{20}{c}}{{\rho _1}}\\{1 - 2{\rho _1}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}\vec \rho = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\hspace{0.25in}{\mbox{if }}{\rho _1} = 0\]Bunday holda, o'z vektor tizimidan farqli o'laroq, biz doimiyni xohlagan narsamiz bo'lishi uchun tanlashimiz mumkin, shuning uchun hayotimizni osonlashtirish uchun uni tanlashimiz mumkin. Bu odatda uni nol qilib tanlashni bildiradi.
Endi biz tizimning umumiy yechimini yozishimiz mumkin.
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) + {c_2}\left( {{{\bf{e}}^{5t}}t\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) + {{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)} \right)\]Konstantalarni topish uchun dastlabki shartni qo'llash bizga,
\[\left( {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\] \[\left. {\begin{array}{*{20}{r}}{{c_1} = 2}\\{ - 2{c_1} + {c_2} = - 5}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = 2,\,\,\,{c_2} = - 1\]Haqiqiy yechim shundan iboratki,
\[\begin{align*}\vec x\left( t \right) & = 2{{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) - \left( {t{{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) + {{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)} \right)\\ & = {{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right) - {{\bf{e}}^{5t}}t\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) - {{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\\ & = {{\bf{e}}^{5t}}\left( {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right) - {{\bf{e}}^{5t}}t\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right)\end{align*}\]E'tibor bering, biz bu erda yechimni biroz soddalashtirish uchun ozgina birlashtirdik.
Shunday qilib, keyingi misol ushbu tizim uchun fazali portretni chizish bo'ladi.
Yechim
Ular xuddi haqiqiy, aniq o'ziga xos fazali portretlar boshlanganidek boshlanadi. Biz birinchi navbatda xos vektorga parallel bo'lgan traektoriyani chizamiz va xos qiymat ijobiy bo'lganligi sababli traektoriya boshlang'ichdan uzoqlashishini ta'kidlaymiz.
Endi, agar oldimizda bittasi bo'lsa, fazali portretning qolgan qismini tushuntirish osonroq bo'ladi. Shunday qilib, bu erda yana bir nechta traektoriyalar chizilgan to'liq bosqichli portret.
Bunday hollarda traektoriyalar har doim boshlang'ichdan xos vektorga parallel bo'lgan yo'nalishda paydo bo'ladi (yoki harakat qiladi). Xuddi shunday, ular orqaga burilib, boshqa tomonga o'tishdan oldin bir yo'nalishda boshlanadi. Biz xos vektorga mos keladigan traektoriyaning qaysi tomonida ekanligimizga qarab, ular harakatlanadigan yo'nalishlar qarama-qarshidir. Bundan tashqari, traektoriyalar boshlang'ichdan uzoqlashganda, u xos vektorga mos keladigan traektoriyaga parallel bo'lishni boshlashi kerak.
Xo'sh, biz yo'nalishni qanday aniqlaymiz? Murakkab holatda qilgan ishni qilishimiz mumkin. Biz \(\left( {1,0} \right)\) ni tizimga ulaymiz va shu nuqtada traektoriyalar qaysi yo‘nalishda harakatlanayotganini ko‘ramiz. Bu nuqta to'g'ridan-to'g'ri boshlang'ichning o'ng tomonida bo'lganligi sababli, bu nuqtadagi traektoriya allaqachon aylangan bo'lishi kerak va shuning uchun u aylangandan keyin harakatlanadigan yo'nalishni beradi.
Fazali portretimizni tekshirish uchun ushbu muammoni bajarish,
\[\left( {\begin{array}{*{20}{c}}7&1\\{ - 4}&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}7\\{ - 4}\end{array}} \right)\]Bu vektor to'rtinchi kvadrantga pastga ishora qiladi va shuning uchun traektoriya ham to'rtinchi kvadrantga o'tishi kerak. Bu bizning fazali portretimizga mos keladi.
Bunday hollarda muvozanat a deyiladi tugun va bu holatda beqaror. E'tibor bering, ba'zida siz chaqirilgan takroriy o'z qiymat holati uchun tugunlarni eshitasiz degenerativ tugunlar or noto'g'ri tugunlar.
Keling, yana bir misol bilan ishlaylik.
Yechim
Avval tizim uchun xos qiymat.
\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{ - 1 - \lambda }&{\frac{3}{2}}\\{ - \frac{1}{6}}&{ - 2 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} + 3\lambda + \frac{9}{4}\\ & = {\left( {\lambda + \frac{3}{2}} \right)^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}{\lambda _{1,2}} = - \frac{3}{2}\end{align*}\]Endi xos vektorni olaylik.
\[\left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&{\frac{3}{2}}\\{ - \frac{1}{6}}&{ - \frac{1}{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\hspace{0.25in} \Rightarrow \,\hspace{0.25in}\frac{1}{2}{\eta _1} + \frac{3}{2}{\eta _2} = 0\,\,\,\,\,\,\,\,\,{\eta _1} = - 3{\eta _2}\] \[\begin{align*}\vec \eta & = \left( {\begin{array}{*{20}{c}}{ - 3{\eta _2}}\\{{\eta _2}}\end{array}} \right) & \hspace{0.25in}{\eta _2} & \ne 0\\ {{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) & \hspace{0.25in}{\eta _2} & = 1\end{align*}\]Endi \(\vec \rho \) toping,
\[\left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&{\frac{3}{2}}\\{ - \frac{1}{6}}&{ - \frac{1}{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rho _1}}\\{{\rho _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right)\hspace{0.25in} \Rightarrow \,\hspace{0.25in}\frac{1}{2}{\rho _1} + \frac{3}{2}{\rho _2} = - 3\,\,\,\,\,\,\,\,\,{\rho _1} = - 6 - 3{\rho _2}\] \[\vec \rho = \left( {\begin{array}{*{20}{c}}{ - 6 - 3{\rho _2}}\\{{\rho _2}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}\vec \rho = \left( {\begin{array}{*{20}{c}}{ - 6}\\0\end{array}} \right)\hspace{0.25in}{\mbox{if }}{\rho _2} = 0\]Tizim uchun umumiy yechim shunday bo'ladi:
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - \frac{{3t}}{2}}}\left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) + {c_2}\left( {t{{\bf{e}}^{ - \frac{{3t}}{2}}}\left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) + {{\bf{e}}^{ - \frac{{3t}}{2}}}\left( {\begin{array}{*{20}{c}}{ - 6}\\0\end{array}} \right)} \right)\]Dastlabki shartni qo'llash,
\[\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right) = \vec x\left( 2 \right) = {c_1}{{\bf{e}}^{ - 3}}\left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) + {c_2}\left( {2{{\bf{e}}^{ - 3}}\left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) + {{\bf{e}}^{ - 3}}\left( {\begin{array}{*{20}{c}}{ - 6}\\0\end{array}} \right)} \right)\]E'tibor bering, biz bu safar \(t=0\) dan foydalanmadik! Endi biz quyidagi tizimni hal qilishimiz kerak,
\[\left. {\begin{array}{*{20}{r}}{ - 3{{\bf{e}}^{ - 3}}{c_1} - 12{{\bf{e}}^{ - 3}}{c_2} = 1}\\{{{\bf{e}}^{ - 3}}{c_1} + 2{{\bf{e}}^{ - 3}}{c_2} = 0}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = \frac{{{{\bf{e}}^3}}}{3},\,\,\,{c_2} = - \frac{{{{\bf{e}}^3}}}{6}\]Haqiqiy yechim shundan iboratki,
\[\begin{align*}\vec x\left( t \right) &= \frac{{{{\bf{e}}^3}}}{3}{{\bf{e}}^{ - \frac{{3t}}{2}}}\left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) - \frac{{{{\bf{e}}^3}}}{6}\left( {t{{\bf{e}}^{ - \frac{{3t}}{2}}}\left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right) + {{\bf{e}}^{ - \frac{{3t}}{2}}}\left( {\begin{array}{*{20}{c}}{ - 6}\\0\end{array}} \right)} \right)\\ & = {{\bf{e}}^{ - \frac{{3t}}{2} + 3}}\left( {\begin{array}{*{20}{c}}0\\{\frac{1}{3}}\end{array}} \right) + t{{\bf{e}}^{ - \frac{{3t}}{2} + 3}}\left( {\begin{array}{*{20}{c}}{\frac{1}{2}}\\{ - \frac{1}{6}}\end{array}} \right)\end{align*}\]Va biz qilgan barcha boshqa muammolarga mos kelish uchun fazali portretni chizamiz.
Yechim
Avval shuni ta'kidlaylikki, bu holda xos qiymat manfiy bo'lgani uchun traektoriyalarning barchasi boshlang'ich tomon harakatlanishi kerak. Keling, \(\left( {1,0} \right)\) da traektoriyalar yo'nalishini tekshiramiz.
\[\left( {\begin{array}{*{20}{c}}{ - 1}&{\frac{3}{2}}\\{ - \frac{1}{6}}&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - \frac{1}{6}}\end{array}} \right)\]Shunday qilib, traektoriyalar \(\left( {1,0} \right)\) da uchinchi kvadrantga ishora qilganga o'xshaydi. Bu quyidagi bosqich portretini beradi.
