5.10-bo'lim: Bir hil bo'lmagan tizimlar
Endi biz bir hil bo'lmagan tizimlarga qisqacha to'xtalib o'tishimiz kerak. Biz ikkinchi tartibli differensial tenglamalar bobida ko'rib chiqqan ikkala usuldan ham bu erda foydalanish mumkin. Ko'rib turganimizdek Aniqlanmagan koeffitsientlar tizimlarda foydalanilganda deyarli bir xil bo'ladi Parametrlarning o'zgarishi yangi formula olingan bo'lishi kerak bo'ladi, lekin bo'ladi aslida tizimlarga qo'llanilganda biroz osonroq bo'ladi.
Aniqlanmagan koeffitsientlar
Tizimlar uchun Aniqlanmagan koeffitsientlar usuli ikkinchi tartibli differensial tenglama holati bilan deyarli bir xil. Faqatgina farq shundaki, koeffitsientlar endi vektor bo'lishi kerak.
Keling, bir misolni tez ko'rib chiqaylik.
Yechim
Bizda qo'shimcha yechim mavjud, chunki biz bu qismni haqiqiy xos qiymatga qaytardik bo'limi. It is,
\[{\vec x_c}\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) + {c_2}{{\bf{e}}^{4t}}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right)\]Muayyan yechimning shaklini taxmin qilish biz ushbu usulni birinchi marta ko'rib chiqqanimizda xuddi shunday ishlaydi. Bizda chiziqli polinom bor va shuning uchun bizning taxminimiz chiziqli polinom bo'lishi kerak. Yagona farq shundaki, "koeffitsientlar" doimiylar o'rniga vektor bo'lishi kerak. Maxsus yechim shaklga ega bo'ladi,
\[{\vec x_P} = t\vec a + \vec b = t\left( {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right)\]Shunday qilib, biz taxminni farqlashimiz kerak
\[{\vec x'_P} = \vec a = \left( {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right)\]
Tizimga ulanishdan oldin ishimizga yordam berish uchun yozuvni biroz soddalashtiraylik. Biz tizimni shunday yozamiz:
\[\vec x' = \left( {\begin{array}{*{20}{c}}1&2\\3&2\end{array}} \right)\vec x + t\left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right) = A\vec x + t\vec g\]Bu keyingi ishni biroz osonlashtiradi. Endi narsalarni tizimga ulaymiz.
\[\begin{align*}\vec a& = A\left( {t\vec a + \vec b} \right) + t\vec g\\ \vec a & = tA\vec a + A\vec b + t\vec g\\ \vec 0 & = t\left( {A\vec a + \vec g} \right) + \left( {A\vec b - \vec a} \right)\end{align*}\]Endi biz koeffitsientlarni tenglashtirishimiz kerak. Buni qilish,
\[\begin{align*}& {t^1}: & A\vec a + \vec g & = \vec 0 & \hspace{0.25in}A\vec a & = - \vec g\\ & {t^0}: & A\vec b - \vec a & = \vec 0 & \hspace{0.25in}A\vec b & = \vec a\end{align*}\]Endi birinchi tenglamada faqat \(\vec a\) noma'lum, shuning uchun tizimni yechish uchun Gauss eliminatsiyasidan foydalanishimiz mumkin. Bu ishni tekshirishni sizga qoldiramiz.
\[\left( {\begin{array}{*{20}{c}}1&2\\3&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\end{array}} \right) = - \left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}\vec a = \left( {\begin{array}{*{20}{c}}3\\{ - \frac{5}{2}}\end{array}} \right)\]Endi biz \(\vec a\) ni bilganimizdan keyin \(\vec b\) uchun ikkinchi tenglamani yechamiz.
\[\left( {\begin{array}{*{20}{c}}1&2\\3&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3\\{ - \frac{5}{2}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}\vec b = \left( {\begin{array}{*{20}{c}}{ - \frac{{11}}{4}}\\{\frac{{23}}{8}}\end{array}} \right)\]Shunday qilib, biz ikkala tenglamani ham hal qila olganimiz sababli, maxsus yechim shunday bo'ladi:
\[{\vec x_P} = t\left( {\begin{array}{*{20}{c}}3\\{ - \frac{5}{2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - \frac{{11}}{4}}\\{\frac{{23}}{8}}\end{array}} \right)\]Umumiy yechim shunday bo'ladi:
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) + {c_2}{{\bf{e}}^{4t}}\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}3\\{ - \frac{5}{2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - \frac{{11}}{4}}\\{\frac{{23}}{8}}\end{array}} \right)\]Ko'rib turganingizdek, aniqlanmagan koeffitsientlar biz birinchi marta ko'rganimiz bilan deyarli bir xil. Biroq, "doimiy" ni echishdagi ish biroz chalkashroq.
Parametrlarning o'zgarishi
Bunday holda, tizimlar uchun parametrlarni o'zgartirish uchun yangi formulani olishimiz kerak bo'ladi. Bu safar chiqarish biz birinchi bo'lganimizga qaraganda ancha sodda bo'ladi ko'rgan parametrlarning o'zgarishi.
Avval \(X(t)\) matritsa bo'lsin, uning \(i^{\text{th}}\) ustuni \(i^{\text{th}}\) tizimning chiziqli mustaqil yechimi bo'lsin,
\[\vec x' = A\vec x\]Endi \(X(t)\) quyidagi differentsial tenglamaning yechimi bo'lishini ko'rsatish mumkin.
\[\begin{equation}X' = AX\label{eq:eq1}\end{equation}\]Bu asl vektor o'rniga matritsaga ega bo'lgan asl tizimdan boshqa narsa emas.
Biz aniq bir yechim topishga harakat qilamiz
\[\vec x' = A\vec x + \vec g\left( t \right)\]Biz shaklning yechimini topishimiz mumkin deb taxmin qilamiz,
\[{\vec x_P} = X\left( t \right)\,\vec v\left( t \right)\]bu erda \(\vec v\left( t \right)\) vektorini aniqlashimiz kerak bo'ladi. Buning uchun biz uni bir hil bo'lmagan tizimga ulashimiz kerak. Tahminni tizimga ulashda mahsulotning muayyan echimini boshqarishni unutmang.
\[X'\,\vec v + X\,\vec v' = A\,X\,\vec v + \vec g\]E'tibor bering, biz eslatmani biroz soddalashtirish uchun narsalarning \(\left( t \right)\) qismini tashladik. Endi \(\eqref{eq:eq1}\) yordamida biz buni biroz qayta yozishimiz mumkin.
\[\begin{align*}X'\,\vec v + X\,\vec v' & = X'\,\vec v + \vec g\\ X\,\vec v' & = \vec g\end{align*}\]Biz \(X\)-ni chiziqli mustaqil echimlar yordamida yaratganimiz uchun biz \(\det(X)\) noldan farqli bo'lishi kerakligini bilamiz va bu o'z navbatida biz topishimiz mumkinligini anglatadi. teskari \(X\). Shunday qilib, ikkala tomonni \(X\) ning teskarisi bilan ko'paytiring.
\[\,\vec v' = {X^{ - 1}}\vec g\]Endi biz qilishimiz kerak bo'lgan narsa - \(\vec v\left( t \right)\) ni olish uchun ikkala tomonni birlashtirish.
\[\,\vec v\left( t \right) = \int{{{X^{ - 1}}\vec g\,dt}}\]Ikkinchi tartibli differensial tenglamada bo'lgani kabi, biz integratsiyaning har qanday konstantalarini e'tiborsiz qoldiramiz. Maxsus yechim shunday bo'ladi:
\[\begin{equation}{\vec x_P} = \,X\int{{{X^{ - 1}}\vec g\,dt}}\label{eq:eq2}\end{equation}\]Buni ishlatib, tezkor misol bilan ishlaylik.
Yechim
Biz ushbu tizimning to'ldiruvchi yechimini haqiqiy xos qiymatda topdik bo'limi. It is,
\[{\vec x_c}\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) + {c_2}{{\bf{e}}^{ - 6t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\]Endi \(X\) matritsasi,
\[X = \left( {\begin{array}{*{20}{c}}{{{\bf{e}}^{ - t}}}&{ - {{\bf{e}}^{ - 6t}}}\\{4{{\bf{e}}^{ - t}}}&{{{\bf{e}}^{ - 6t}}}\end{array}} \right)\]Endi biz ushbu matritsaning teskarisini topishimiz kerak. Biz ikkinchi chiziqli algebra sharhida matritsalarning teskarisini qanday topishni ko'rdik bo'limi va bizda doimiy yozuvlar bo'lmasa ham, jarayon bu erda bir xil. Tafsilotlarni tekshirish uchun sizga qoldiramiz.
\[{X^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{5}{{\bf{e}}^t}}&{\frac{1}{5}{{\bf{e}}^t}}\\{ - \frac{4}{5}{{\bf{e}}^{6t}}}&{\frac{1}{5}{{\bf{e}}^{6t}}}\end{array}} \right)\]Endi integralda ko'paytirishni bajaring.
\[{X^{ - 1}}\vec g = \left( {\begin{array}{*{20}{c}}{\frac{1}{5}{{\bf{e}}^t}}&{\frac{1}{5}{{\bf{e}}^t}}\\{ - \frac{4}{5}{{\bf{e}}^{6t}}}&{\frac{1}{5}{{\bf{e}}^{6t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{6{{\bf{e}}^{2t}}}\\{ - {{\bf{e}}^{2t}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{{\bf{e}}^{3t}}}\\{ - 5{{\bf{e}}^{8t}}}\end{array}} \right)\]Now do the integral.
\[\int{{{X^{ - 1}}\vec g\,dt}} = \int{{\left( {\begin{array}{*{20}{c}}{{{\bf{e}}^{3t}}}\\{ - 5{{\bf{e}}^{8t}}}\end{array}} \right)\,dt}} = \left( {\begin{array}{*{20}{c}}{\int{{{{\bf{e}}^{3t}}\,dt}}}\\{\int{{ - 5{{\bf{e}}^{8t}}\,dt}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{1}{3}{{\bf{e}}^{3t}}}\\{ - \frac{5}{8}{{\bf{e}}^{8t}}}\end{array}} \right)\]Esda tutingki, matritsa yoki vektorni birlashtirish uchun siz faqat alohida yozuvlarni birlashtirasiz.
Endi biz maxsus yechimni olishimiz mumkin.
\[\begin{align*}{{\vec x}_P} & = \,X\int{{{X^{ - 1}}\vec g\,dt}}\\ & = \left( {\begin{array}{*{20}{c}}{{{\bf{e}}^{ - t}}}&{ - {{\bf{e}}^{ - 6t}}}\\{4{{\bf{e}}^{ - t}}}&{{{\bf{e}}^{ - 6t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{3}{{\bf{e}}^{3t}}}\\{ - \frac{5}{8}{{\bf{e}}^{8t}}}\end{array}} \right)\\ & = \left( {\begin{array}{*{20}{c}}{\frac{{23}}{{24}}{{\bf{e}}^{2t}}}\\{\frac{{17}}{{24}}{{\bf{e}}^{2t}}}\end{array}} \right)\\ & = {{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{\frac{{23}}{{24}}}\\{\frac{{17}}{{24}}}\end{array}} \right)\end{align*}\]Umumiy yechim shunday bo'ladi:
\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - t}}\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) + {c_2}{{\bf{e}}^{ - 6t}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) + {{\bf{e}}^{2t}}\left( {\begin{array}{*{20}{c}}{\frac{{23}}{{24}}}\\{\frac{{17}}{{24}}}\end{array}} \right)\]Shunday qilib, ba'zi ishlar biroz chalkash bo'lishi mumkin, lekin umuman yomon emas.
Biz bu erda bir hil bo'lmagan differentsial tenglamalarni echishning ikkita usulini ko'rib chiqdik va ish biroz chalkash bo'lishi mumkin bo'lsa-da, ular unchalik yomon emas. Albatta, biz bu erda bir hil bo'lmagan qismni ham juda sodda qilib qo'ydik. Murakkabroq muammolar katta hajmdagi ishlarni o'z ichiga oladi.