9.3-bo'lim: Terminologiya
Bizda qisman differensial tenglamalarni echishni boshlashdan oldin e'tiborga olishimiz kerak bo'lgan yana bir bo'lim bor. Bu juda qisqa bo'lim bo'lib, biz o'zgaruvchilarni ajratish usulini joriy qilganimizda keyingi bo'limda bizga kerak bo'ladigan ba'zi asosiy atamalarni qamrab oladi.
Keling, operator g'oyasidan boshlaylik. Operator aslida raqamlar o'rniga funktsiyani argument sifatida qabul qiladigan funktsiyadir, chunki biz funktsiyalarda ishlashga odatlanganmiz. Siz allaqachon bir nechta operatorlarni bilasiz, hatto ular operator ekanligini bilmagan bo'lsangiz ham. Quyida operatorlarning ba'zi misollari keltirilgan.
\[L = \frac{d}{{dx}}\hspace{0.25in}L = \int{{\,dx}}\hspace{0.25in}L = \int_{a}^{b}{{\,dx}}\hspace{0.25in}L = \frac{\partial }{{\partial t}}\]Yoki, agar biz funktsiyani qo'shsak, aytaylik \(u\left( x \right)\), ularning har biriga biz erishamiz,
\[L\left( u \right) = \frac{{du}}{{dx}}\hspace{0.25in}L\left( u \right) = \int{{u\left( x \right)\,dx}}\hspace{0.25in}L\left( u \right) = \int_{a}^{b}{{u\left( x \right)\,dx}}\hspace{0.25in}L\left( u \right) = \frac{{\partial u}}{{\partial t}}\]Bularning barchasi operatorlarning juda oddiy misollari, lekin hosila va integral operatorlardir. Keyinchalik murakkab operator bo'ladi issiqlik operatori. Biz issiqlik operatorini biroz qayta yozishdan olamiz issiqlik tenglamasi manbalarsiz. Issiqlik operatori,
\[L = \frac{\partial }{{\partial t}} - k\frac{{{\partial ^2}}}{{\partial {x^2}}}\]Endi biz bu erda aniqlamoqchi bo'lgan narsa operator emas, balki a chiziqli operator. Chiziqli operator - bu qanoatlantiradigan har qanday operator
\[L\left( {{c_1}{u_1} + {c_2}{u_2}} \right) = {c_1}L\left( {{u_1}} \right) + {c_2}L\left( {{u_2}} \right)\]Issiqlik operatori chiziqli operatorning namunasidir va buni qisman hosilaning asosiy xususiyatlaridan foydalangan holda ko'rsatish juda oson, shuning uchun buni qilaylik.
\[\begin{align*}L\left( {{c_1}{u_1} + {c_2}{u_2}} \right) & = \frac{\partial }{{\partial t}}\left( {{c_1}{u_1} + {c_2}{u_2}} \right) - k\frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {{c_1}{u_1} + {c_2}{u_2}} \right)\\ & = \frac{\partial }{{\partial t}}\left( {{c_1}{u_1}} \right) + \frac{\partial }{{\partial t}}\left( {{c_2}{u_2}} \right) - k\left[ {\frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {{c_1}{u_1}} \right) + \frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {{c_2}{u_2}} \right)} \right]\\ & = {c_1}\frac{{\partial {u_1}}}{{\partial t}} + {c_2}\frac{{\partial {u_2}\,}}{{\partial t}} - k\left[ {{c_1}\frac{{{\partial ^2}{u_1}}}{{\partial {x^2}}} + {c_2}\frac{{{\partial ^2}{u_2}}}{{\partial {x^2}}}} \right]\\ & = {c_1}\frac{{\partial {u_1}}}{{\partial t}} - k{c_1}\frac{{{\partial ^2}{u_1}}}{{\partial {x^2}}} + {c_2}\frac{{\partial {u_2}\,}}{{\partial t}} - k{c_2}\frac{{{\partial ^2}{u_2}}}{{\partial {x^2}}}\\ & = {c_1}\left[ {\frac{{\partial {u_1}}}{{\partial t}} - k\frac{{{\partial ^2}{u_1}}}{{\partial {x^2}}}} \right] + {c_2}\left[ {\frac{{\partial {u_2}\,}}{{\partial t}} - k\frac{{{\partial ^2}{u_2}}}{{\partial {x^2}}}} \right]\\ & = {c_1}L\left( {{u_1}} \right) + {c_2}L\left( {{u_2}} \right)\end{align*}\]Yuqorida biz bergan lotin va integral operatorlar ham chiziqli operatorlar ekanligini o'zingiz tekshirib ko'rishingiz mumkin. Aslida, issiqlik operatorining chiziqli operator ekanligini ko'rsatish jarayonida biz birinchi va ikkinchi tartibli qisman hosila operatorlari ham chiziqli ekanligini ko'rsatdik.
Biz belgilashimiz kerak bo'lgan keyingi atama a chiziqli tenglama. Chiziqli tenglama - bu ko'rinishdagi tenglama,
\[\begin{equation}L\left( u \right) = f\label{eq:eq1}\end{equation}\]bu erda \(L\) chiziqli operator va \(f\) ma'lum funksiya.
Chiziqli qisman differensial tenglamalarga misollar keltiramiz.
\[\begin{align*}\frac{{\partial u}}{{\partial t}} & = k\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{Q\left( {x,t} \right)}}{{c\rho }}\\ \frac{{{\partial ^2}u}}{{\partial {t^2}}} & = {c^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} & = {\nabla ^2}u = 0\\ \frac{{\partial u}}{{\partial t}} - 4\frac{{{\partial ^2}u}}{{\partial {t^2}}} & = \frac{{{\partial ^3}u}}{{\partial {x^3}}} + 8u - g\left( {x,t} \right)\end{align*}\]Ushbu ro'yxatdagi dastlabki ikkitasi, albatta, issiqlik tenglamasi va to'lqin tenglamasidir. Uchinchisi dan foydalanadi Laplasiyalik va odatda deyiladi Laplas tenglamasi. Biz aslida Laplas tenglamasining 2 o'lchamli versiyasini bir necha bo'limda echamiz. To'rtinchi tenglama shunchaki murakkabroq misol keltirish uchun tuzilgan.
Issiqlik tenglamasi va yuqoridagi to'rtinchi misol bilan bir qatorda \(Q\left( {x,t} \right)\) va \(g\left( {x,t} \right)\) ning mavjudligi ularning chiziqli tenglamalar bo'lishiga to'sqinlik qilmasligiga e'tibor bering. Bularning chiziqli tenglamalar bo‘lishiga imkon beruvchi asosiy masala, har biridagi operatorning chiziqli bo‘lishidir.
Endi to'liq bo'lish uchun bu erda chiziqli bo'lmagan qisman differentsial tenglamalarga bir nechta misollar.
\[\begin{align*}\frac{{\partial u}}{{\partial t}} & = k\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {u^2}\\ \frac{{{\partial ^2}u}}{{\partial {t^2}}} - \frac{{\partial u}}{{\partial x}}\frac{{\partial u}}{{\partial t}} & = u + f\left( {x,t} \right)\end{align*}\]Ularning har biridagi operatorlar chiziqli emasligini tekshirishni sizga qoldiramiz, biroq birinchisida muammo atamasi \({u^2}\), ikkinchisida esa ikkita hosilaning mahsuloti muammoli atamadir.
Endi, agar biz \(\eqref{eq:eq1}\) ga qaytsak va \(f = 0\) deb faraz qilsak, biz unga erishamiz,
\[\begin{equation}L\left( u \right) = 0\label{eq:eq2}\end{equation}\]Biz buni a deb ataymiz chiziqli bir jinsli tenglama (esda tuting, \(L\) chiziqli operatordir).
E'tibor bering, \(u = 0\) har doim chiziqli bir hil tenglamaning yechimi bo'ladi (chiziqli bo'lish nimani anglatishiga qayting va \({c_1} = {c_2} = 0\) ni istalgan ikkita yechim bilan ishlating va buni tekshirish oson). Biz \(u = 0\) deb ataymiz ahamiyatsiz yechim. Aslida, bu tenglamaning bir hilligini aniqlashning juda yaxshi usuli. Agar \(L\) chiziqli operator bo'lsa va biz \(u = 0\) ni tenglamaga ulab, \(L\left( u \right) = 0\) ni olsak, operator bir hil ekanligini bilib olamiz.
Biz, shuningdek, chiziqlilik va bir jinslilik g'oyalarini chegara shartlariga kengaytirishimiz mumkin. Agar biz issiqlik tenglamasi uchun muhokama qilgan turli chegara shartlariga qaytadigan bo'lsak, biz ularni chiziqli va/yoki bir hil deb tasniflashimiz mumkin.
Belgilangan harorat chegarasi shartlari,
\[u\left( {0,t} \right) = {g_1}\left( t \right)\hspace{0.25in}u\left( {L,t} \right) = {g_2}\left( t \right)\]ular chiziqli va faqat \({g_1}\left( t \right) = 0\) va \({g_2}\left( t \right) = 0\) bo'lsa, bir hil bo'ladi.
Belgilangan issiqlik oqimi chegara shartlari,
\[ - {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = {\varphi _1}\left( t \right)\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = {\varphi _2}\left( t \right)\]ular chiziqli va faqat \({\varphi _1}\left( t \right) = 0\) va \({\varphi _2}\left( t \right) = 0\) bo'lsa, yana bir hil bo'ladi.
Keyinchalik, Nyutonning sovutish qonunidan chegara shartlari,
\[ - {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = - H\left[ {u\left( {0,t} \right) - {g_1}\left( t \right)} \right]\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = H\left[ {u\left( {L,t} \right) - {g_2}\left( t \right)} \right]\]ular yana chiziqli va faqat \({g_1}\left( t \right) = 0\) va \({g_2}\left( t \right) = 0\) bo'lsa, bir hil bo'ladi.
Biz ko'rib chiqqan yakuniy chegara shartlari davriy chegara shartlari edi,
\[u\left( { - L,t} \right) = u\left( {L,t} \right)\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( { - L,t} \right) = \frac{{\partial u}}{{\partial x}}\left( {L,t} \right)\]va bular ham chiziqli, ham bir jinsli.
Ushbu bo'limdagi yakuniy mavzu aslida terminologiya emas, balki biz ushbu eslatmalarda bir necha bor ko'rgan haqiqatni qayta ko'rib chiqishdir.
Superpozitsiya printsipi
Agar \({u_1}\) va \({u_2}\) chiziqli bir hil tenglamaning yechimlari bo'lsa, \({c_1}{u_1} + {c_2}{u_2}\) \({c_1}\) va \({c_2}\) ning har qanday qiymatlari uchun ham shunday bo'ladi.
Yuqorida aytib o'tganimizdek, biz buni bu semestrda bir necha bor ko'rganmiz, lekin biz u bilan ko'p ish qilmadik. Biroq, biz qisman differensial tenglamalarni echishga kirishganimizda, bu asosiy g'oya bo'ladi. Bu faktsiz biz qisman differensial tenglamalarning eng asosiylaridan tashqari hammasini yecha olmaymiz.