9.6-bo'lim: Nol bo'lmagan harorat chegaralari bilan issiqlik tenglamasi
Ushbu bo'limda biz oldingi qismdagi holatlardan birini biroz kengaytirmoqchimiz. Oldingida biz quyidagi issiqlik muammosini ko'rib chiqamiz.
\[\begin{align*}& \frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ & u\left( {x,0} \right) = f\left( x \right)\hspace{0.25in}u\left( {0,t} \right) = 0\hspace{0.25in}\,\,\,\,\,u\left( {L,t} \right) = 0\end{align*}\]Endi, bu muammoda hech qanday noto'g'ri narsa yo'q, lekin biz har ikki tomondagi haroratni nolga o'rnatayotganimiz biroz haqiqatga to'g'ri kelmaydi. Biz ko'rib chiqqan boshqa ikkita muammo, izolyatsiyalangan chegaralar va ingichka halqa, biroz realroq muammolar, ammo bu unchalik real emas, shuning uchun biz uni biroz kengaytirmoqchimiz.
Ushbu bo'limda biz nima qilishni xohlaymiz, buning o'rniga quyidagi muammoni ko'rib chiqing.
\[\begin{equation}\begin{aligned}& \frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ & u\left( {x,0} \right) = f\left( x \right)\hspace{0.25in}u\left( {0,t} \right) = {T_1}\hspace{0.25in}\,\,\,\,\,u\left( {L,t} \right) = {T_2}\end{aligned}\label{eq:eq1}\end{equation}\]Bu holda biz chegaralar har qanday belgilangan harorat bo'lishiga ruxsat beramiz, \({T_1}\) yoki \({T_2}\). Muammo shundaki, o'zgaruvchilarni ajratish endi bu muammoda ishlamaydi, chunki chegara shartlari endi bir hil emas. Esda tutingki, o'zgaruvchilarni ajratish faqat qisman differentsial tenglama va chegara shartlari chiziqli va bir hil bo'lsa ishlaydi. Shunday qilib, biz buni hal qilishdan oldin chegara shartlari bilan qandaydir tarzda shug'ullanishimiz kerak.
Yaxshiyamki, biz uchun ular bilan kurashishning oson yo'li bor. Keling, bu muammoni bir oz ko'rib chiqaylik. Barning biron bir joyida issiqlik energiyasini qo'shish / ayirish uchun manbalar yo'q. Bundan tashqari, bizning chegara shartlarimiz belgilangan haroratdir va shuning uchun vaqt o'tishi bilan o'zgarmasdir va biz issiqlik energiyasini doimiy ravishda qo'shish/ayirish uchun chegaralarda issiqlik oqimini belgilamaymiz. Demak, bularning barchasi shuni anglatadiki, issiqlik energiyasi hech qachon barga kirish yoki tashqariga chiqmaydi va shuning uchun issiqlik energiyasining bir qismi tabiiy ravishda so'nggi nuqtalarda bardan tashqariga oqib chiqishi mumkin, chunki harorat o'zgarib, bardagi harorat taqsimoti barqarorlashishi va endi vaqtga bog'liq bo'lmasligi kerak.
Yoki, boshqacha qilib aytganda, \(t \to \infty \) bizning harorat taqsimotimiz sifatida \(u\left( {x,t} \right)\) shunday harakat qilishini kutishimiz mantiqiydir:
\[\mathop {\lim }\limits_{t \to \infty } u\left( {x,t} \right) = {u_E}\left( x \right)\]bu erda \({u_E}\left( x \right)\) deyiladi muvozanat harorati. Shuni ham yodda tutingki, u hali ham issiqlik tenglamasi va chegara shartlarini qondirishi kerak. Biroq, u boshlang'ich shartni qoniqtirmaydi, chunki u \(t \to \infty \) kabi harorat taqsimoti, dastlabki holat esa \(t = 0\) da. Shunday qilib, muvozanat harorat taqsimoti qondirishi kerak,
\[\begin{equation}0 = \frac{{{d^2}{u_E}}}{{d{x^2}}}\hspace{0.25in}{u_E}\left( 0 \right) = {T_1}\hspace{0.25in}\,\,\,\,\,{u_E}\left( L \right) = {T_2}\label{eq:eq2}\end{equation}\]Bu juda oson 2nd oddiy differensial tenglamani yechish uchun tartib bering. Agar biz ikki marta integratsiya qilsak, biz olamiz,
\[{u_E}\left( x \right) = {c_1}x + {c_2}\]va chegara shartlarini qo'llash (buni tekshirishni sizga qoldiramiz) bizga,
\[{u_E}\left( x \right) = {T_1} + \frac{{{T_2} - {T_1}}}{L}x\]Xo'sh, buning yuqoridagi \(\eqref{eq:eq1}\) tomonidan berilgan muammoni hal qilish bilan qanday aloqasi bor? Funktsiyani aniqlaymiz,
\[\begin{equation}v\left( {x,t} \right) = u\left( {x,t} \right) - {u_E}\left( x \right)\label{eq:eq3}\end{equation}\]bu yerda \(u\left( {x,t} \right)\) \(\eqref{eq:eq1}\) yechimi va \({u_E}\left( x \right)\) \(\eqref{eq:eq1}\) uchun muvozanat harorati.
Endi buni quyidagicha qayta yozamiz,
\[u\left( {x,t} \right) = v\left( {x,t} \right) + {u_E}\left( x \right)\]va ba'zi hosilalarni olaylik.
\[\frac{{\partial u}}{{\partial t}} = \frac{{\partial v}}{{\partial t}} + \frac{{\partial {u_E}}}{{\partial t}} = \frac{{\partial v}}{{\partial t}}\hspace{0.25in}\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}} + \frac{{{\partial ^2}{u_E}}}{{\partial {x^2}}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}\]Bu hosilalarning ikkalasida biz \({u_E}\left( x \right)\) muvozanat harorati ekanligini va shuning uchun \(t\) vaqtga bog'liq emasligi va \(\eqref{eq:eq2}\) dagi differentsial tenglamani qondirishi kerakligidan foydalandik.
Bu bizga \(u\left( {x,t} \right)\) va \(v\left( {x,t} \right)\) bir xil qisman differentsial tenglamani qondirishi kerakligini aytadi. Keling, \(v\left( {x,t} \right)\) uchun dastlabki shartlar va chegara shartlari qanday bo'lishi kerakligini ko'rib chiqaylik.
\[\begin{align*}v\left( {x,0} \right) & = u\left( {x,0} \right) - {u_E}\left( x \right) = f\left( x \right) - {u_E}\left( x \right)\\ v\left( {0,t} \right) & = u\left( {0,t} \right) - {u_E}\left( 0 \right) = {T_1} - {T_1} = 0\\ v\left( {L,t} \right) & = u\left( {L,t} \right) - {u_E}\left( L \right) = {T_2} - {T_2} = 0\end{align*}\]Shunday qilib, boshlang'ich holat potentsial ravishda chigallashadi, lekin chegara shartlari endi bir hil! \(v\left( {x,t} \right)\) qanoatlantirishi kerak bo'lgan qisman differentsial tenglama:
\[\begin{align*}\frac{{\partial v}}{{\partial t}} & = k\frac{{{\partial ^2}v}}{{\partial {x^2}}}\\ v\left( {x,0} \right) & = f\left( x \right) - {u_E}\left( x \right)\hspace{0.25in}v\left( {0,t} \right) = 0\hspace{0.25in}\,\,\,\,\,v\left( {L,t} \right) = 0\end{align*}\]We ko'rgan Oldingida buni qanday hal qilish kerak va shuning uchun biz yechim,
\[v\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \]bu erda koeffitsientlar berilgan,
\[{B_n} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{\left( {f\left( x \right) - {u_E}\left( x \right)} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,\hspace{0.25in}n = 1,2,3, \ldots \]\(\eqref{eq:eq1}\) ning yechimi shundan iboratki,
\[\begin{align*}u\left( {x,t} \right) & = {u_E}\left( x \right) + v\left( {x,t} \right)\\ & = {T_1} + \frac{{{T_2} - {T_1}}}{L}x + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \end{align*}\]va koeffitsientlar yuqorida keltirilgan.