7.1-bo'lim: Asosiy tushunchalar
Biz ushbu bobni ko'pgina darsliklar ushbu bobda yoritiladigan materiallardan boshlaymiz. Biz materialni dan olamiz Ikkinchi tartib bo'lim va uni \(n^{\text{th}}\) tartibli chiziqli differentsial tenglamalarga kengaytiring. Biz deyarli barcha 2 ni ko'rib chiqamiznd Buyurtma materiali tabiiy ravishda \(n^{\text{th}}\) buyurtmasiga ozgina yangi material bilan uzaytiriladi.
Shunday qilib, keling, \(n^{\text{th}}\) tartibli chiziqli differensial tenglamalar uchun ba'zi asosiy tushunchalardan boshlaylik. Eng umumiy \(n^{\text{th}}\) tartibli chiziqli differentsial tenglama:
\[\begin{equation}{P_n}\left( t \right){y^{\left( n \right)}} + {P_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {P_1}\left( t \right)y' + {P_0}\left( t \right)y = G\left( t \right)\label{eq:eq1}\end{equation}\]qaerda, umid qilamanki, buni eslaysiz,
\[{y^{\left( m \right)}} = \frac{{{d^m}y}}{{d{x^m}}}\]Ushbu material uchun ko'plab teoremalar va g'oyalar \({y^{\left( n \right)}}\) ning 1 koeffitsientiga ega bo'lishini talab qiladi va shuning uchun biz \({P_n}\left( t \right)\) ga bo'linsak, biz quyidagilarni olamiz:
\[\begin{equation}{y^{\left( n \right)}} + {p_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {p_1}\left( t \right)y' + {p_0}\left( t \right)y = g\left( t \right)\label{eq:eq2}\end{equation}\]\(n^{\text{th}}\) tartibli differensial tenglama uchun IVP dan shubhalanishimiz mumkin bo'lganidek, quyidagi \(n\) boshlang'ich shartlarini talab qiladi.
\[\begin{equation}y\left( {{t_0}} \right) = {\overline{y}_0},\hspace{0.25in}y'\left( {{t_0}} \right) = {\overline{y}_1},\hspace{0.25in} \cdots ,\hspace{0.25in}{y^{\left( {n - 1} \right)}}\left( {{t_0}} \right) = {\overline{y}_{n - 1}}\label{eq:eq3}\end{equation}\]Quyidagi teorema bizga \(\eqref{eq:eq2}\) va \(\eqref{eq:eq3}\) tomonidan berilgan IVP uchun yagona yechim bo'lishini kutishimiz mumkinligini aytadi.
Yechimlarning o'ziga xosligi
Faraz qilaylik, \({p_0},{p_1}, \ldots ,{p_{n - 1}}\) va \(g\left( t \right)\) funksiyalari \({t_0}\) ni o'z ichiga olgan ba'zi ochiq intervalda \(I\) uzluksiz bo'lsa, \(\eqref{eq:eq2}\) va \(\eqref{eq:eq3}\) tomonidan berilgan IVP uchun yagona yechim mavjud va yechim \(I\) da barcha \(t\) uchun mavjud bo'ladi.
Bu teorema biz shunga o'xshash teoremaning juda tabiiy davomidir ko'rgan 1 dast materialga buyurtma berish.
Keyinchalik \(n^{\text{th}}\) tartibli chiziqli bir hil differensial tenglamani muhokama qilishga o'tishimiz kerak,
\[\begin{equation}{y^{\left( n \right)}} + {p_{n - 1}}\left( t \right){y^{\left( {n - 1} \right)}} + \cdots + {p_1}\left( t \right)y' + {p_0}\left( t \right)y = 0\label{eq:eq4}\end{equation}\]Faraz qilaylik, biz \({y_1}\left( t \right),{y_2}\left( t \right), \ldots ,{y_n}\left( t \right)\) barcha \(\eqref{eq:eq4}\) yechimlari ekanligini bilamiz, keyin kengaytma orqali. Superpozitsiya printsipi buni bilamiz
\[y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)\]\(\eqref{eq:eq4}\) ga ham yechim bo'ladi. Bu erda haqiqiy savol, bu \(\eqref{eq:eq4}\) uchun umumiy echim bo'ladimi yoki yo'qmi.
Bu umumiy yechim bo'lishi uchun biz \({t_0}\) ning har qanday tanlovi (1 teoremadan \(I\) oralig'ida) va \({\overline{y}_1},{\overline{y}_2}, \ldots ,{\overline{y}_n}\) ning istalgan tanlovi uchun \({c_1},{c_2}, \ldots ,{c_n}\) konstantalarini topishimiz kerak bo'ladi. Yoki, boshqacha qilib aytganda, biz hal qiladigan \({c_1},{c_2}, \ldots ,{c_n}\) ni topishimiz kerak,
\[\begin{align*}{c_1}{y_1}\left( {{t_0}} \right) + {c_2}{y_2}\left( {{t_0}} \right) + \cdots + {c_n}{y_n}\left( {{t_0}} \right) & = {{\overline{y}}_0}\\ {c_1}{{y'}_1}\left( {{t_0}} \right) + {c_2}{{y'}_2}\left( {{t_0}} \right) + \cdots + {c_n}{{y'}_n}\left( {{t_0}} \right) & = {{\overline{y}}_1}\\ & \vdots \\ {c_1}y_1^{\left( {n - 1} \right)}\left( {{t_0}} \right) + {c_2}y_2^{\left( {n - 1} \right)}\left( {{t_0}} \right) + \cdots + {c_n}y_n^{\left( {n - 1} \right)}\left( {{t_0}} \right) & = {{\overline{y}}_{n - 1}}\end{align*}\]Xuddi biz kabi qildi 2 uchunnd Differensial tenglamalarni tartiblash uchun biz Kramer qoidasidan foydalanishimiz mumkin va har bir javobning maxraji \(n\) x \(n\) matritsasining quyidagi determinanti bo'ladi.
\[\left| {\begin{array}{*{20}{c}}{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{{{y'}_1}}&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{y_1^{\left( {n - 1} \right)}}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right|\] Biz 2 bilan qaytganimizdeknd materialga buyurtma bering, biz buni aniqlaymiz Vronskiy va uni bilan belgilang, \[W\left( {{y_1},{y_2}, \ldots {y_n}} \right)\left( t \right) = \left| {\begin{array}{*{20}{c}}{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{{{y'}_1}}&{{{y'}_2}}& \cdots &{{{y'}_n}}\\ \vdots & \vdots & \ddots & \vdots \\{y_1^{\left( {n - 1} \right)}}&{y_2^{\left( {n - 1} \right)}}& \cdots &{y_n^{\left( {n - 1} \right)}}\end{array}} \right|\]Endi bizda Wronskianning ta'rifi yo'q bo'lib, biz savolga qaytishimiz kerak. Wronskian \({c_i}\) ning har bir yechimida maxraj bo'lgani uchun, biz Vronskianni baholash uchun tanlagan \(t = {t_0}\) qiymati uchun nolga teng bo'lmasa, bizda yechim borligini ko'rishimiz mumkin. Quyidagi teorema bularning barchasini umumlashtiradi.
Yechimlarning asosiy to'plami
Faraz qilaylik, \({p_0},{p_1}, \ldots ,{p_{n - 1}}\) funksiyalarining barchasi \(I\) ochiq intervalida uzluksiz va bundan keyin \({y_1}\left( t \right),{y_2}\left( t \right), \ldots {y_n}\left( t \right)\) \(\eqref{eq:eq4}\) ning barcha yechimlari deb faraz qilaylik. Agar \(I\) da har bir \(t\) uchun \(W\left( {{y_1},{y_2}, \ldots {y_n}} \right)\left( t \right) \ne 0\), \({y_1}\left( t \right),{y_2}\left( t \right), \ldots {y_n}\left( t \right)\) hosil qiladi Yechimlarning asosiy to'plami va \(\eqref{eq:eq4}\) ning umumiy yechimi:
\[y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)\]Eslatib o'tamiz, agar echimlar to'plami asosiy echimlar to'plamini tashkil qilsa, ular ham to'plam bo'ladi chiziqli mustaqil funktsiyalar.
Biz ushbu bo'limni \(\eqref{eq:eq2}\) bir jinsli bo'lmagan differensial tenglamaning yechimlarini qanday topishimiz haqida tezkor eslatma bilan yopamiz. Bizga birinchi navbatda teoremaning \(n^{\text{th}}\) tartibli versiyasi kerak ko'rgan qaytib 2nd materialga buyurtma berish.
\(\eqref{eq:eq2}\) va \(\eqref{eq:eq4}\) yechimlari o'rtasidagi aloqa
Faraz qilaylik, \({Y_1}\left( t \right)\) va \({Y_2}\left( t \right)\) \(\eqref{eq:eq2}\) ning ikkita yechimi va \({y_1}\left( t \right),{y_2}\left( t \right), \ldots {y_n}\left( t \right)\) bir jinsli \(\eqref{eq:eq4}\) differensial tenglamasining asosiy yechimlari to‘plamidir. keyin,
\[{Y_1}\left( t \right) - {Y_2}\left( t \right)\]\(\eqref{eq:eq4}\) ning yechimidir va uni shunday yozish mumkin
\[{Y_1}\left( t \right) - {Y_2}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)\]Endi, xuddi 2 bilan qilganimiz kabind agar biz \(Y\left( t \right)\) ni \(\eqref{eq:eq2}\) ning umumiy yechimi bo‘lsin va agar \({Y_P}\left( t \right)\) \(\eqref{eq:eq2}\) ning har qanday yechimi bo‘lsin, agar materialga buyurtma bersak, bu teorema natijasidan foydalanib, biz shunday bo‘lishi kerakligini ko‘ramiz:
\[Y\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right) + {Y_P}\left( t \right) = {y_c}\left( t \right) + {Y_P}\left( t \right)\]bu erda \({y_c}\left( t \right) = {c_1}{y_1}\left( t \right) + {c_2}{y_2}\left( t \right) + \cdots + {c_n}{y_n}\left( t \right)\) deyiladi qo'shimcha yechim va \({Y_P}\left( t \right)\) a deb ataladi maxsus yechim.
Keyingi bir nechta bo'limlar davomida biz \(n^{\text{th}}\) tartibli differensial tenglamalar uchun bir-birini to'ldiruvchi va maxsus echimlarni topishdagi farqlarni muhokama qilamiz.nd tartibli differentsial tenglamalar. Ko'ramiz, ko'p hollarda usullar bir xil. Biroq, jalb qilingan ish miqdori ko'pincha sezilarli darajada ko'proq bo'ladi.