Bizga birinchi navbatda matritsa uchun xos qiymatlar va xos vektorlar kerak.

\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{3 - \lambda }&9\\{ - 4}&{ - 3 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} + 27\hspace{0.25in}{\lambda _{1,2}} = \pm 3\sqrt 3 \,i\end{align*}\]

Shunday qilib, bizda xos qiymatlar bor ekan, biz faqat xos qiymatlardan biri uchun xos vektorni olishimiz kerakligini eslaymiz, chunki biz birinchi xos vektordan ikkinchi xos vektorni bepul olishimiz mumkin.

\({\lambda _1} = 3\sqrt 3 \,i\):
Biz quyidagi tizimni hal qilishimiz kerak.

\[\left( {\begin{array}{*{20}{c}}{3 - 3\sqrt 3 \,i}&9\\{ - 4}&{ - 3 - 3\sqrt 3 \,i}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\]

Biz olgan birinchi tenglamadan foydalanib,

\[\begin{align*}\left( {3 - 3\sqrt 3 \,i} \right){\eta _1} + 9{\eta _2} & = 0\\ {\eta _2} & = - \frac{1}{3}\left( {1 - \sqrt 3 \,i} \right){\eta _1}\end{align*}\]

Demak, birinchi xos vektor,

\[\begin{align*}\vec \eta & = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{ - \frac{1}{3}\left( {1 - \sqrt 3 \,i} \right){\eta _1}}\end{array}} \right)\\ {{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}3\\{ - 1 + \sqrt 3 \,i}\end{array}} \right)\hspace{0.25in}{\eta _1} = 3\end{align*}\]

Bunday hollarda xos vektorlarni topayotganda, kompleks son har qanday kasrning suratida ko'rinishiga ishonch hosil qiling, chunki u keyinchalik hisoblagichda kerak bo'ladi. Shuningdek, konstantani to'g'ri tanlash orqali har qanday kasrlarni tozalashga harakat qiling. Bu bizning hayotimizni yo'lda osonlashtiradi.

Endi ikkinchi xos vektor,

\[{\vec \eta ^{\left( 2 \right)}} = \left( {\begin{array}{*{20}{c}}3\\{ - 1 - \sqrt 3 \,i}\end{array}} \right)\]

Biroq, ko'rib turganimizdek, bizga bu xos vektor kerak emas.

Birinchi xos qiymat va xos vektordan olingan yechim:

\[{\vec x_1}\left( t \right) = {{\bf{e}}^{3\sqrt 3 \,i\,t}}\left( {\begin{array}{*{20}{c}}3\\{-1 + \sqrt 3 \,i}\end{array}} \right)\]

Shunday qilib, biz ko'rib turganimizdek, eksponensial va vektorda kompleks sonlar mavjud bo'lib, biz buni yechim sifatida ishlatishimiz uchun ulardan xalos bo'lishimiz kerak. Biz foydalanishimiz mumkin bo'lgan ikkinchi tartibli differentsial tenglamalar bo'limining murakkab ildizlar bo'limidan eslaylik Eyler formulasi kompleks sonni eksponensialdan chiqarish uchun. Bu bizga beradi,

\[{\vec x_1}\left( t \right) = \left( {\cos \left( {3\sqrt 3 t} \right) + i\sin \left( {3\sqrt 3 t} \right)} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 1 + \sqrt 3 \,i}\end{array}} \right)\]

Keyingi qadam kosinuslar va sinuslarni vektorga ko'paytirishdir.

\[{\vec x_1}\left( t \right) = \left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right) + 3i\sin \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - i\sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,i\cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right)\]

Endi shartlarni ulardagi “\(i\)” bilan birlashtiring va ushbu shartlarni “\(i\)” ni o‘z ichiga olmagan atamalardan ajrating. Shuningdek, ushbu vektordan "\(i\)" ni hisobga oling.

\[\begin{align*}{{\vec x}_1}\left( t \right) & = \left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right) + i\left( {\begin{array}{*{20}{c}}{3\sin \left( {3\sqrt 3 t} \right)}\\{ - \sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,\cos \left( {3\sqrt 3 t} \right)}\end{array}} \right)\\ & = \vec u\left( t \right) + i\,\vec v\left( t \right)\end{align*}\]

Endi \(\vec u\left( t \right)\) va \(\vec v\left( t \right)\) differensial tenglamalar sistemasining ikkita chiziqli mustaqil yechimlari ekanligini ko'rsatish mumkin (tafsilotlarni sizga qoldiramiz). Bu shuni anglatadiki, biz ulardan umumiy yechim yaratish uchun foydalanishimiz mumkin va ular ikkalasi ham haqiqiy echimlardir.

Shunday qilib, murakkab ildizlarga ega bo'lgan tizimning umumiy yechimi

\[\vec x\left( t \right) = {c_1}\vec u\left( t \right) + {c_2}\vec v\left( t \right)\]

Bu yerda \(\vec u\left( t \right)\) va \(\vec v\left( t \right)\) birinchi yechimni quyidagicha yozish orqali topiladi

\[\vec x\left( t \right) = \vec u\left( t \right) + i\,\vec v\left( t \right)\]

Bizning tizimimiz uchun umumiy yechim:

\[\vec x\left( t \right) = {c_1}\left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{3\sin \left( {3\sqrt 3 t} \right)}\\{ - \sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,\cos \left( {3\sqrt 3 t} \right)}\end{array}} \right)\]

Endi konstantalarni topish uchun boshlang'ich shartni qo'llashimiz kerak.

\[\left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}0\\{\sqrt 3 }\end{array}} \right)\]

Bu quyidagi tenglamalar sistemasini echishga olib keladi:

\[\left. {\begin{array}{*{20}{r}}{3{c_1} = 2}\\{ - {c_1} + \sqrt 3 {c_2} = - 4}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = \frac{2}{3},\,\,\,{c_2} = \frac{{ - 10}}{{3\sqrt 3 }}\]

Haqiqiy yechim shundan iboratki,

\[\vec x\left( t \right) = \frac{2}{3}\left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right) - \frac{{10}}{{3\sqrt 3 }}\left( {\begin{array}{*{20}{c}}{3\sin \left( {3\sqrt 3 t} \right)}\\{ - \sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,\cos \left( {3\sqrt 3 t} \right)}\end{array}} \right)\]

\(A\) matritsasining xos qiymatlari sof murakkab bo'lsa, ular bu holatda bo'lgani kabi, yechimlarning traektoriyalari koordinata markazida joylashgan doiralar yoki ellipslar bo'ladi. Bu erda biz tashvishlanishimiz kerak bo'lgan yagona narsa - ular soat yo'nalishi bo'yicha yoki soat miliga teskari yo'nalishda aylanadimi.

Buni qilish yetarlicha oson. Biz ushbu bosqichli portretlarni birinchi marta ko'rib chiqqanimizni eslang bo'limlar oldin Agar biz \(\vec x\left( t \right)\) qiymatini tanlab, uni tizimimizga ulab olsak, biz o'sha nuqtadagi traektoriyaga tegib turadigan va traektoriya harakatlanayotgan yo'nalishga ishora qiluvchi vektorni olamiz. Shunday qilib, keling, quyidagi nuqtani tanlaymiz va biz nimani olganimizni ko'rib chiqamiz.

\[\vec x = \left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}\vec x' = \left( {\begin{array}{*{20}{c}}3&9\\{ - 4}&{ - 3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right)\]

Shunday qilib, fazalar tekisligidagi \(\left( {1,0} \right)\) nuqtasida traektoriya pastga yo'naltirilgan bo'ladi. Buning yagona yo'li, agar traektoriyalar soat yo'nalishi bo'yicha harakatlansa.

Mana bu muammoning ba'zi traektoriyalarining eskizi.

Ishdagi muvozanat yechimi a deyiladi markaz va barqaror.

Keling, matritsa uchun xos qiymatlar va xos vektorlarni olaylik.

\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 13}\\5&{1 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} - 4\lambda + 68\hspace{0.25in}{\lambda _{1,2}} = 2 \pm 8\,i\end{align*}\]

Endi birinchi xos qiymat uchun xos vektorni oling.

\({\lambda _1} = 2 + 8i\):
Biz quyidagi tizimni hal qilishimiz kerak.

\[\left( {\begin{array}{*{20}{c}}{1 - 8i}&{ - 13}\\5&{ - 1 - 8i}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\]

Ikkinchi tenglamadan foydalanib, biz olamiz,

\[\begin{align*}5{\eta _1} + \left( { - 1 - 8\,i} \right){\eta _2} & = 0\\ {\eta _1} & = \frac{1}{5}\left( {1 + 8\,i} \right){\eta _2}\end{align*}\]

Demak, birinchi xos vektor,

\[\begin{align*}\vec \eta & = \left( {\begin{array}{*{20}{c}}{\frac{1}{5}\left( {1 + 8\,i} \right){\eta _2}}\\{{\eta _2}}\end{array}} \right)\\ {{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}{1 + 8\,i}\\5\end{array}} \right)\hspace{0.25in}{\eta _2} = 5\end{align*}\]

Bu xos qiymat va xos vektorga mos keladigan yechim

\[\begin{align*}{{\vec x}_1}\left( t \right) & = {{\bf{e}}^{\left( {2 + 8i} \right)t}}\left( {\begin{array}{*{20}{c}}{1 + 8\,i}\\5\end{array}} \right)\\ & = {{\bf{e}}^{2t}}{{\bf{e}}^{8i\,t}}\left( {\begin{array}{*{20}{c}}{1 + 8\,i}\\5\end{array}} \right)\\ & = {{\bf{e}}^{2t}}\left( {\cos \left( {8t} \right) + i\sin \left( {8t} \right)} \right)\left( {\begin{array}{*{20}{c}}{1 + 8\,i}\\5\end{array}} \right)\end{align*}\]

Birinchi misolda bo'lgani kabi, kosinuslar va sinuslarni vektorga ko'paytiring va uni ajrating. Bu safar yechimdagi eksponensial haqida unutmang.

\[\begin{align*}{{\vec x}_1}\left( t \right) & = {{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{\cos \left( {8t} \right) - 8\sin \left( {8t} \right)}\\{5\cos \left( {8t} \right)}\end{array}} \right) + i{{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{8\cos \left( {8t} \right) + \sin \left( {8t} \right)}\\{5\sin \left( {8t} \right)}\end{array}} \right)\\ & = \vec u\left( t \right) + i\,\vec v\left( t \right)\end{align*}\]

Bu tizimning umumiy yechimi,

\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{\cos \left( {8t} \right) - 8\sin \left( {8t} \right)}\\{5\cos \left( {8t} \right)}\end{array}} \right) + {c_2}{{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{8\cos \left( {8t} \right) + \sin \left( {8t} \right)}\\{5\sin \left( {8t} \right)}\end{array}} \right)\]

Endi boshlang'ich shartni qo'llang va doimiylarni toping.

\[\left( {\begin{array}{*{20}{c}}3\\{ - 10}\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}8\\0\end{array}} \right)\] \[\left. {\begin{array}{*{20}{c}}{{c_1} + 8{c_2} = 3}\\{5{c_1} = - 10}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = - 2,\,\,\,{c_2} = \frac{5}{8}\]

Haqiqiy yechim shundan iboratki,

\[\vec x\left( t \right) = - 2{{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{\cos \left( {8t} \right) - 8\sin \left( {8t} \right)}\\{5\cos \left( {8t} \right)}\end{array}} \right) + \frac{5}{8}{{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{8\cos \left( {8t} \right) + \sin \left( {8t} \right)}\\{5\sin \left( {8t} \right)}\end{array}} \right)\]

Tizimning o'ziga xos qiymatlari haqiqiy qism bilan murakkab bo'lsa, traektoriyalar boshlang'ichga yoki undan tashqariga aylanadi. Haqiqiy qismga qarab, qaysi biri bo'lishini aniqlashimiz mumkin. Haqiqiy qism eksponensial funktsiyaning ko'rsatkichi bo'lishi uchun (bu tizimning yechimida ko'rganimizdek), agar haqiqiy qism ijobiy bo'lsa, \(t\) o'sishi bilan yechim juda katta bo'ladi. Xuddi shunday, agar haqiqiy qism manfiy bo'lsa, \(t\) ortishi bilan yechim o'ladi.

Shunday qilib, agar haqiqiy qism ijobiy bo'lsa, traektoriyalar boshlang'ichdan tashqariga chiqadi va haqiqiy qism salbiy bo'lsa, ular boshlang'ichga aylanadi. Aylanish yo'nalishini (soat yo'nalishi bo'yicha va teskari yo'nalishda) xuddi markaz uchun qilganimizdek aniqlaymiz.

Bizning holatimizda traektoriyalar boshlang'ichdan tashqariga chiqadi, chunki haqiqiy qism ijobiy va

\[\vec x' = \left( {\begin{array}{*{20}{c}}3&{ - 13}\\5&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3\\5\end{array}} \right)\]

oxirgi misol kabi soat miliga teskari yo'nalishda aylanadi.

Mana, ushbu tizim uchun ba'zi traektoriyalarning eskizi.

Bu yerda muvozanat yechimini a deb ataymiz spiral (g'alati ...) va bu holda u beqaror, chunki traektoriyalar kelib chiqish joyidan uzoqlashadi.