5.4-bo'lim: Differensial tenglamalar sistemalari
Ushbu bo'limning kirish qismida biz o'lja va yirtqichlarning populyatsiyasini hisobga oladigan populyatsiya muammosidan qanday qilib differentsial tenglamalar sistemasi paydo bo'lishi mumkinligini qisqacha muhokama qildik. Mavjud o'lja soni mavjud yirtqichlarning soniga ta'sir qilishi mantiqan. Xuddi shunday, mavjud yirtqichlarning soni ham mavjud bo'lgan o'ljalar soniga ta'sir qiladi. Shuning uchun o'lja yoki yirtqichning populyatsiyasini boshqaradigan differentsial tenglama qaysidir ma'noda ikkinchisining populyatsiyasiga bog'liq bo'lishi kerak. Bu o'lja va yirtqichning populyatsiyasini aniqlash uchun bir vaqtning o'zida echilishi kerak bo'lgan ikkita differentsial tenglamaga olib keladi.
Buning mohiyati shundaki, differensial tenglamalar sistemasi tabiiy vaziyatlardan osongina paydo bo'lishi mumkin. Differensial tenglamalarning samarali yirtqich-o'lja tizimini ishlab chiqish ushbu bobning mavzusi emas. Biroq, tizimlar \(n^{\text{th}}\) tartibli chiziqli differentsial tenglamalardan ham paydo bo'lishi mumkin. Biroq, bunga kirishdan oldin, keling, tizimni yozamiz va ba'zi atamalarni yo'ldan chiqaramiz.
Biz differensial tenglamalarning birinchi tartibli chiziqli sistemalarini ko'rib chiqamiz. Bu atamalar shu paytgacha ma'nosini bildirgan ma'noni anglatadi. Tizimning istalgan joyidagi eng katta hosila birinchi hosila bo'ladi va barcha noma'lum funktsiyalar va ularning hosilalari faqat birinchi darajaga to'g'ri keladi va boshqa noma'lum funktsiyalarga ko'paytirilmaydi. Birinchi tartibli sistemaga, chiziqli differensial tenglamalarga misol keltiramiz.
\[\begin{align*}{{x'}_1} & = {x_1} + 2{x_2}\\ {{x'}_2} & = 3{x_1} + 2{x_2}\end{align*}\]Biz bunday tizimni a deb ataymiz bog'langan tizim, chunki \(x_{1}\) ni topish uchun \(x_{2}\) bilimi talab qilinadi va \(x_{2}\) ni topish uchun ham \(x_{1}\) bilimi talab qilinadi. Biz ularni qanday hal qilish haqida o'ylaymiz keyinroq. Ayni paytda biz faqat tizimlarning ba'zi asoslari bilan tanishishdan manfaatdormiz.
Endi yuqorida aytib o'tilganidek, tizim sifatida \(n^{\text{th}}\) tartibli chiziqli differentsial tenglamani yozishimiz mumkin. Keling, buni qanday qilish mumkinligini ko'rib chiqaylik.
Yechim
Biz yuqori tartibli differensial tenglamalarni o'zgaruvchining juda oddiy o'zgarishiga ega tizim sifatida yozishimiz mumkin. Biz quyidagi ikkita yangi funktsiyani belgilashdan boshlaymiz.
\[\begin{align*}{x_1}\left( t \right) & = y\left( t \right)\\ {x_2}\left( t \right) & = y'\left( t \right)\end{align*}\]Endi e'tibor bering, agar biz ikkala tomonni farq qilsak,
\[\begin{align*}{{x'}_1} & = y' = {x_2}\\ {{x'}_2} & = y'' = - \frac{1}{2}y + \frac{5}{2}y' = - \frac{1}{2}{x_1} + \frac{5}{2}{x_2}\end{align*}\]Ikkinchi tenglamada differentsial tenglamadan foydalanishga e'tibor bering. Biz boshlang'ich shartlarni yangi funksiyalarga ham o'zgartirishimiz mumkin.
\[\begin{align*}{x_1}\left( 3 \right) & = y\left( 3 \right) = 6\\ {x_2}\left( 3 \right) & = y'\left( 3 \right) = - 1\end{align*}\]Bularning barchasini jamlash quyidagi differensial tenglamalar sistemasini beradi.
\[\begin{align*}{{x'}_1} & = {x_2} & \hspace{0.25in}{x_1}\left( 3 \right) & = 6\\ {{x'}_2} & = - \frac{1}{2}{x_1} + \frac{5}{2}{x_2} & \hspace{0.25in}{x_2}\left( 3 \right) & = - 1\end{align*}\]Yuqoridagi misoldagi tizimni a deb nomlaymiz Boshlang'ich qiymat muammosi xuddi boshlang'ich shartli differensial tenglamalar uchun qilganimizdek.
Keling, yana bir misolni ko'rib chiqaylik.
Yechim
Oxirgi misolda qilganimizdek, biz ba'zi yangi funktsiyalarni aniqlashimiz kerak. Bu safar bizga 4 ta yangi funksiya kerak bo'ladi.
\[\begin{align*}{x_1} & = y & \Rightarrow \hspace{0.25in}{{x'}_1} & = y' = {x_2}\\ {x_2} & = y' & \Rightarrow \hspace{0.25in}{{x'}_2} & = y'' = {x_3}\\ {x_3} & = y'' & \Rightarrow \hspace{0.25in}{{x'}_3} & = y''' = {x_4}\\ {x_4} & = y''' & \Rightarrow \hspace{0.25in}{{x'}_4}& = {y^{\left( 4 \right)}} = - 8y + \sin \left( t \right)y' - 3y'' + {t^2} = - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2}\end{align*}\]Tizim dastlabki shartlar bilan birgalikda,
\[\begin{align*}{{x'}_1} & = {x_2} & \hspace{0.25in}{x_1}\left( 0 \right) & = 1\\ {{x'}_2} & = {x_3} & \hspace{0.25in}{x_2}\left( 0 \right) & = 2\\ {{x'}_3} & = {x_4} & \hspace{0.25in}{x_3}\left( 0 \right) & = 3\\ {{x'}_4} & = - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2} & \hspace{0.25in}{x_4}\left( 0 \right) & = 4\end{align*}\]Endi, biz nihoyat ularni hal qilishga kirishganimizda, biz odatda tizimlarni ushbu bo'limda ularga bergan shaklda hal qilmasligimizni ko'ramiz. Differensial tenglamalar sistemalariga aylantirilishi mumkin matritsa shakli va bu biz odatda tizimlarni hal qilishda foydalanadigan shakldir.
Yechim
Avval tizimni shunday yozingki, har bir tomon vektor bo'lsin.
\[\left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{4{x_1} + 7{x_2}}\\{ - 2{x_1} - 5{x_2}}\end{array}} \right)\]Endi o'ng tomonni matritsani ko'paytirish sifatida yozish mumkin,
\[\left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&7\\{ - 2}&{ - 5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\]Endi aniqlasak,
\[\vec x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\]keyin,
\[\vec x' = \left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\end{array}} \right)\]Keyin tizim matritsa shaklida yozilishi mumkin,
\[\vec x' = \left( {\begin{array}{*{20}{c}}4&7\\{ - 2}&{ - 5}\end{array}} \right)\vec x\]Yechim
Biz 1-misoldan tizimdan boshlaymiz.
\[\begin{align*}{{x'}_1} & = {x_2}\hspace{0.25in}{x_1}\left( 3 \right) = 6\\ {{x'}_2} & = - \frac{1}{2}{x_1} + \frac{5}{2}{x_2}\hspace{0.25in}{x_2}\left( 3 \right) = - 1\end{align*}\]Avval belgilang,
\[\vec x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\]Keyin tizim,
\[\vec x' = \left( {\begin{array}{*{20}{c}}0&1\\{ - \frac{1}{2}}&{\frac{5}{2}}\end{array}} \right)\vec x\hspace{0.25in}\vec x\left( 3 \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( 3 \right)}\\{{x_2}\left( 3 \right)}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 1}\end{array}} \right)\]Endi 2-misoldan tizimni bajaramiz.
\[\begin{align*}{{x'}_1} & = {x_2}\hspace{0.25in}{x_1}\left( 0 \right) = 1\\ {{x'}_2} & = {x_3}\hspace{0.25in}{x_2}\left( 0 \right) = 2\\ {{x'}_3} & = {x_4}\hspace{0.25in}{x_3}\left( 0 \right) = 3\\ {{x'}_4} & = - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2}\hspace{0.25in}{x_4}\left( 0 \right) = 4\end{align*}\]Bunday holda, biz ehtiyot bo'lishimiz kerak t2 oxirgi tenglamada. Biz tizimni yana vektor sifatida yozishni boshlaymiz va keyin uni ikkita vektorga ajratamiz, biri noma'lum funktsiyalarni, ikkinchisi esa ma'lum funktsiyalarni o'z ichiga oladi.
\[\left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\\{{{x'}_3}}\\{{{x'}_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_3}}\\{{x_4}}\\{ - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_3}}\\{{x_4}}\\{ - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0\\0\\0\\{{t^2}}\end{array}} \right)\]Endi birinchi vektorni matritsani ko'paytirish sifatida yozish mumkin va biz ikkinchi vektorni yolg'iz qoldiramiz.
\[\vec x' = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 8}&{\sin \left( t \right)}&{ - 3}&0\end{array}} \right)\vec x + \left( {\begin{array}{*{20}{c}}0\\0\\0\\{{t^2}}\end{array}} \right)\hspace{0.25in}\vec x\left( 0 \right) = \left( {\begin{array}{*{20}{c}}1\\2\\3\\4\end{array}} \right)\]qayerda,
\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\\{{x_3}\left( t \right)}\\{{x_4}\left( t \right)}\end{array}} \right)\]E'tibor bering, vaqti-vaqti bilan bu kabi "katta" tizimlar uchun biz bir qadam oldinga boramiz va tizimni quyidagicha yozamiz:
\[\vec x' = A\vec x + \vec g\left( t \right)\]Ushbu bo'limda qilishimiz kerak bo'lgan oxirgi narsa - terminologiyadan biroz voz kechish. bilan boshlab
\[\vec x' = A\vec x + \vec g\left( t \right)\]tizim deb aytamiz bir hil agar \(\vec g\left( t \right) = \vec 0\) va biz tizim deymiz bir hil bo'lmagan if \(\vec g\left( t \right) \ne \vec 0\).