Xo'sh, texnik jihatdan bu erda qilishimiz kerak bo'lgan birinchi narsa o'zgaruvchilarni ajratishni qo'llashdir. Oldingida buni qilgan bo'lsak ham, keling, nima qilganimizni takrorlaymiz.

Birinchidan, biz yechim shaklni oladi deb taxmin qilamiz,

\[u\left( {x,t} \right) = \varphi \left( x \right)G\left( t \right)\]

va biz buni qisman differentsial tenglama va chegara shartlariga qo'shamiz. Biz tenglamani bir tomondan faqat \(t\) funktsiyasini va boshqa tomonda faqat \(x\) funktsiyasini olish uchun ajratamiz va keyin ajratish konstantasini kiritamiz. Bu bizga ikkita oddiy differentsial tenglamani qoldiradi.

Biz bularning barchasini ichkarida qildik 1-misol oldingi bo'lim va ikkita oddiy differentsial tenglamalar,

\[\begin{align*}\frac{{dG}}{{dt}} = - k\lambda G\hspace{0.25in} & \frac{{{d^2}\varphi }}{{d{x^2}}} + \lambda \varphi = 0\\ & \varphi \left( 0 \right) = 0\hspace{0.25in}\varphi \left( L \right) = 0\end{align*}\]

Vaqtga bog'liq tenglamani har qanday vaqtda hal qilish mumkin, ammo biz \(\lambda \) nima ekanligini bilmasligimiz uchun, keling, buni to'xtatib turaylik. Shuni ham yodda tutingki, ko'p masalalarda faqat chegaraviy masala bu nuqtada echilishi mumkin, shuning uchun har doim ikkalasini ham hal qila olishni kutmang.

Fazoviy tenglama chegaraviy masala bo'lib, biz oldingi bobdagi ishimizdan bilamizki, u faqat \(\lambda \) ning ma'lum qiymatlari uchun ahamiyatsiz bo'lmagan echimlarga ega bo'ladi (buni biz xohlaymiz), biz eslaymizki, ular xususiy qiymatlar deb ataladi. Ularga ega bo'lgach, har bir \(\lambda \) uchun ahamiyatsiz bo'lmagan echimlarni aniqlashimiz mumkin, i.e. xos funktsiyalar.

Endi biz fazoviy muammoni hal qildik,

\[\begin{align*}& \frac{{{d^2}\varphi }}{{d{x^2}}} + \lambda \varphi = 0\\ & \varphi \left( 0 \right) = 0\hspace{0.25in}\varphi \left( L \right) = 0\end{align*}\]

in 1-misol \(L = 2\pi \) uchun oldingi bobning Xususiy qiymatlar va xususiy funktsiyalar bo'limi. Shunday qilib, biz buni ma'lum bir \(L\) uchun bir marta hal qilganimiz va ish umumiy \(L\) uchun unchalik farq qilmaganligi sababli, biz bu erda ko'p tushuntirishlarni kiritmaymiz va agar sizga biror narsa qanday ishlashi yoki nima uchun nima qilganimiz haqida eslatma kerak bo'lsa, Eslatma uchun Xususiy qiymatlar va xos funktsiyalar bo'limidagi 1-misolga qayting.

Bizda uchta ishni hal qilishimiz kerak, shuning uchun ketaylik.

\(\underline {\lambda > 0} \)
Bu holda biz bilamizki, differentsial tenglamaning yechimi:

\[\varphi \left( x \right) = {c_1}\cos \left( {\sqrt \lambda \,x} \right) + {c_2}\sin \left( {\sqrt \lambda \,x} \right)\]

Birinchi chegara shartini qo'llash,

\[0 = \varphi \left( 0 \right) = {c_1}\]

Endi ikkinchi chegara shartini qo'llash va yuqoridagi natijadan foydalanish, albatta, beradi:

\[0 = \varphi \left( L \right) = {c_2}\sin \left( {L\sqrt \lambda } \right)\]

Endi biz ahamiyatsiz bo'lmagan echimlarga intilamiz va bu shuni anglatadiki, bizda shunday bo'lishi kerak:

\[\sin \left( {L\sqrt \lambda } \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}L\sqrt \lambda = n\pi \hspace{0.25in}n = 1,2,3, \ldots \]

Bu chegaraviy masalaning musbat xos qiymatlari va ularga mos keladigan xos funktsiyalari quyidagicha:

\[{\lambda _{\,n}} = {\left( {\frac{{n\pi }}{L}} \right)^2}\hspace{0.25in}{\varphi _n}\left( x \right) = \sin \left( {\frac{{n\,\pi \,x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \]

E'tibor bering, bizga \({c_2}\) xos funktsiyada kerak emas, chunki u keyinchalik biz oladigan boshqa konstantaga singib ketadi.

\(\underline {\lambda = 0} \)
Bu holda differensial tenglamaning yechimi:

\[\varphi \left( x \right) = {c_1} + {c_2}x\]

Chegaraviy shartlarni qo'llash,

\[0 = \varphi \left( 0 \right) = {c_1}\hspace{0.25in}0 = \varphi \left( L \right) = {c_2}L\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_2} = 0\]

Demak, bu holda yagona yechim arzimas yechimdir va shuning uchun \(\lambda = 0\) bu chegaraviy masala uchun xos qiymat emas.

\(\underline {\lambda < 0} \)
Bu erda differensial tenglamaning yechimi:

\[\varphi \left( x \right) = {c_1}\cosh \left( {\sqrt { - \lambda } \,x} \right) + {c_2}\sinh \left( {\sqrt { - \lambda } \,x} \right)\]

Birinchi chegara shartini qo'llash,

\[0 = \varphi \left( 0 \right) = {c_1}\]

va ikkinchisini qo'llash,

\[0 = \varphi \left( L \right) = {c_2}\sinh \left( {L\sqrt { - \lambda } } \right)\]

Shunday qilib, biz \(\lambda < 0\) va shunga o'xshash \(L\sqrt { - \lambda } \ne 0\) deb taxmin qilamiz va bu \(\sinh \left( {L\sqrt { - \lambda } } \right) \ne 0\) degan ma'noni anglatadi. Shuning uchun bizda \({c_2} = 0\) bo'lishi kerak va shuning uchun biz bu holatda faqat ahamiyatsiz yechimni olishimiz mumkin.

Shuning uchun bu chegaraviy masala uchun salbiy xos qiymatlar bo'lmaydi.

Ushbu muammo uchun xos qiymatlar va o'z funktsiyalarining to'liq ro'yxati quyidagicha:

\[{\lambda _{\,n}} = {\left( {\frac{{n\pi }}{L}} \right)^2}\hspace{0.25in}{\varphi _n}\left( x \right) = \sin \left( {\frac{{n\,\pi \,x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \]

Endi vaqt differensial tenglamasini yechamiz,

\[\frac{{dG}}{{dt}} = - k{\lambda _n}G\]

Shuni yodda tutingki, biz \(\lambda \)-ni hozir bilsak ham, tartibsizlikni minimallashtirish uchun uni hali ulab bo'lmaydi. Ammo endi biz \({\lambda _n}\) dan foydalanib, bu erda cheksiz ko'p qiymatlar mavjudligini eslatib o'tamiz.

Bu oddiy chiziqli (va bu masala uchun ajratilishi mumkin) 1^st differensial tenglamani buyurtma qiling va shuning uchun biz sizga yechim ekanligini tekshirishga ruxsat beramiz,

\[G\left( t \right) = c{{\bf{e}}^{ - k{\lambda _{\,n}}t}} = c{{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\]

Xo'sh, endi biz oddiy differentsial tenglamalarning ikkalasini ham echganimizdan so'ng, biz nihoyat yechimni yozishimiz mumkin. Shunga qaramay, biz cheksiz ko'p echimlarni topdik, chunki cheksiz ko'p echimlar mavjud (i.e. xos funktsiyalar) fazoviy muammoga.

Bizning mahsulot yechimimiz shundan iboratki,

\[{u_n}\left( {x,t} \right) = {B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\hspace{0.25in}n = 1,2,3, \ldots \]

Biz \({u_n}\) mahsulot yechimini \(n\) ning har bir qiymati boshqacha yechim berishiga e’tirof etish uchun belgiladik. Shuni ham yodda tutingki, biz \(c\) ni vaqt masalasini hal qilishda \({B_n}\) ga o'zgartirdik, chunki u \(n\) ning har bir qiymati uchun ham har xil bo'lishi mumkin va agar biz \({c_2}\) ni o'z funksiyasi bilan saqlaganimizda edi, biz uni \(c\) ga singdirgan bo'lardik.

Bu aslida ko'rinadiganidan osonroq. Buning uchun yuqoridagi mahsulot yechimida \(n = 1\) va \({B_1} = 6\) ni tanlashimiz kifoya,

\[u\left( {x,t} \right) = 6\sin \left( {\frac{{\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{\pi }{L}} \right)}^2}\,t}}\]

va bizda kerakli yechim bor. Bu birinchi misol uchun mahsulot yechimidir va shuning uchun qisman differentsial tenglama va chegara shartlarini qondiradi va boshlang'ich shartni qondiradi, chunki \(t = 0\) ni ulash eksponensialni yo'qotadi.

Bu deyarli birinchi qism kabi oddiy. dan eslang Superpozitsiya printsipi Agar chiziqli bir hil differensial tenglamaning ikkita yechimiga ega bo'lsak (bu erda bizda mavjud), ularning yig'indisi ham yechimdir. Shunday qilib, biz qilishimiz kerak bo'lgan yagona narsa, dastlabki shartning har bir qismini qanoatlantiradigan yechimni olish va keyin ularni qo'shish uchun birinchi qismda qilganimiz kabi \(n\) va \({B_n}\) ni tanlashdir. Buni qilish,

\[u\left( {x,t} \right) = 12\sin \left( {\frac{{9\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{9\pi }}{L}} \right)}^2}\,t}} - 7\sin \left( {\frac{{4\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{4\pi }}{L}} \right)}^2}\,t}}\]

Bu aslida dastlabki shart va chegara shartlariga mos kelishini tekshirish uchun buni sizga qoldiramiz.

Bu erda unchalik ko'p narsa qilishning hojati yo'q, chunki biz yuqoridagi misollar va muhokamalarda ko'p narsani qildik.

Birinchidan, yechim,

\[u\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \]

Koeffitsientlar quyidagicha ifodalanadi:

\[{B_n} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{20\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\left( {\frac{{20L\left( {1 - \cos \left( {n\pi } \right)} \right)}}{{n\pi }}} \right) = \frac{{40\left( {1 - {{\left( { - 1} \right)}^n}} \right)}}{{n\pi }}\]

Agar biz ularni ulab olsak, biz yechimga erishamiz,

\[u\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {\frac{{40\left( {1 - \,{{\left( { - 1} \right)}^n}} \right)}}{{n\,\pi }}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \]

Biz ushbu muammoga o'zgaruvchilarni ajratishni qo'lladik 2-misol oldingi bo'limdan. Shunday qilib, bizning yechimimiz shaklda deb hisoblagandan so'ng,

\[u\left( {x,t} \right) = \varphi \left( x \right)G\left( t \right)\]

va o'zgaruvchilarni ajratishni qo'llash orqali biz hal qilishimiz kerak bo'lgan quyidagi ikkita oddiy differensial tenglamani olamiz.

\[\begin{align*}\frac{{dG}}{{dt}} = - k\lambda G\hspace{0.25in} & \frac{{{d^2}\varphi }}{{d{x^2}}} + \lambda \varphi = 0\\ & \frac{{d\varphi }}{{dx}}\left( 0 \right) = 0\hspace{0.25in}\frac{{d\varphi }}{{dx}}\left( L \right) = 0\end{align*}\]

Biz chegaraviy qiymat muammosini hal qildik 2-misol \(L = 2\pi \) uchun oldingi bobning Xususiy qiymatlar va xususiy funktsiyalar bo'limida, shuning uchun ushbu bo'limdagi birinchi misolda biz bu erda ish haqida ko'p tushuntirish bermoqchi emasmiz. Agar sizga bu qanday ishlashi haqida eslatma kerak bo'lsa, avvalgi bobga qayting va biz u erda ishlagan misolni ko'rib chiqing. Keling, ushbu muammoni hal qilish uchun ishlashimiz kerak bo'lgan uchta holatni ko'rib chiqaylik.

\(\underline {\lambda > 0} \)
Differensial tenglamaning yechimi:

\[\varphi \left( x \right) = {c_1}\cos \left( {\sqrt \lambda \,x} \right) + {c_2}\sin \left( {\sqrt \lambda \,x} \right)\]

Birinchi chegara shartini qo'llash,

\[0 = \frac{{d\varphi }}{{dx}}\left( 0 \right) = \sqrt \lambda \,{c_2}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_2} = 0\]

Ikkinchi chegara sharti quyidagilarni beradi:

\[0 = \frac{{d\varphi }}{{dx}}\left( L \right) = - \sqrt \lambda \,{c_1}\sin \left( {L\sqrt \lambda } \right)\]

Eslatib o'tamiz, \(\lambda > 0\) va shuning uchun biz buni talab qilsak, faqat ahamiyatsiz bo'lmagan echimlarni olamiz,

\[\sin \left( {L\sqrt \lambda } \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}L\sqrt \lambda = n\pi \hspace{0.25in}n = 1,2,3, \ldots \]

Bu chegaraviy masalaning musbat xos qiymatlari va ularga mos keladigan xos funktsiyalari quyidagicha:

\[{\lambda _{\,n}} = {\left( {\frac{{n\pi }}{L}} \right)^2}\hspace{0.25in}{\varphi _n}\left( x \right) = \cos \left( {\frac{{n\,\pi x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \]

\(\underline {\lambda = 0} \)
Umumiy yechim:

\[\varphi \left( x \right) = {c_1} + {c_2}x\]

Birinchi chegara shartini qo'llash,

\[0 = \frac{{d\varphi }}{{dx}}\left( 0 \right) = {c_2}\]

Buning yordamida umumiy yechim shunday bo'ladi:

\[\varphi \left( x \right) = {c_1}\]

va bu ikkinchi chegara shartini arzimas darajada qondirishini unutmang. Shuning uchun \(\lambda = 0\) bu BVP uchun xos qiymat va bu xos qiymatga mos keladigan xos funksiyalar:

\[\varphi \left( x \right) = 1\]

\(\underline {\lambda < 0} \)
Bu erda umumiy yechim:

\[\varphi \left( x \right) = {c_1}\cosh \left( {\sqrt { - \lambda } \,x} \right) + {c_2}\sinh \left( {\sqrt { - \lambda } \,x} \right)\]

Birinchi chegara shartini qo'llash,

\[0 = \frac{{d\varphi }}{{dx}}\left( 0 \right) = \sqrt { - \lambda } \,{c_2}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_2} = 0\]

Ikkinchi chegara sharti quyidagilarni beradi:

\[0 = \frac{{d\varphi }}{{dx}}\left( L \right) = \sqrt { - \lambda } \,{c_1}\sinh \left( {L\sqrt { - \lambda } } \right)\]

Biz \(L\sqrt { - \lambda } \ne 0\) va shunga o'xshash \(\sinh \left( {L\sqrt { - \lambda } } \right) \ne 0\) ni bilamiz. Shuning uchun bizda \({c_1} = 0\) bo'lishi kerak va shuning uchun bu chegaraviy masalada salbiy xos qiymatlar bo'lmaydi.

Shunday qilib, ushbu muammo uchun xos qiymatlar va o'z funktsiyalarining to'liq ro'yxati quyidagicha:

\[\begin{align*}{\lambda _{\,n}} & = {\left( {\frac{{n\pi }}{L}} \right)^2} & \hspace{0.25in}{\varphi _n}\left( x \right) & = \cos \left( {\frac{{n\,\pi x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \\ {\lambda _{\,0}} & = 0 & \hspace{0.25in}{\varphi _0}\left( x \right) & = 1\end{align*}\]

E'tibor bering, agar biz \(n = 0\) birinchi to'plamda ruxsat etilsa, biz \({\lambda _{\,0}} = 0\) o'z qiymatini va uning o'z funktsiyasini olamiz va shuning uchun biz quyidagini o'z qiymatlari va o'z funktsiyalari to'plami sifatida ishlatamiz.

\[{\lambda _{\,n}} = {\left( {\frac{{n\pi }}{L}} \right)^2}\hspace{0.25in}{\varphi _n}\left( x \right) = \cos \left( {\frac{{n\,\pi x}}{L}} \right)\hspace{0.25in}n = 0,1,2,3, \ldots \]

Bu erda vaqt muammosi biz ko'rib chiqqan birinchi muammo bilan bir xil,

\[G\left( t \right) = c{{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\]

Bizning mahsulot yechimlarimiz shunday bo'ladi,

\[{u_n}\left( {x,t} \right) = {A_n}\cos \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\hspace{0.25in}n = 0,1,2,3, \ldots \]

va bu qisman differentsial tenglamaning yechimi:

\[u\left( {x,t} \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \]

Agar biz boshlang'ich shartni qo'llasak, natijaga erishamiz.

\[u\left( {x,0} \right) = f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]

va biz bu boshqa narsa emasligini ko'rishimiz mumkin Furye kosinuslar qatori \(f\left( x \right)\) on \(0 \le x \le L\) va shunga o'xshash koeffitsientlarni olish uchun kosinuslarning ortogonalligidan foydalanishimiz mumkin yoki biz buni oldingi bobda qilganimizni eslashimiz mumkin va koeffitsientlar tomonidan berilganligini bilishimiz mumkin:

\[{A_n} = \left\{ {\begin{array}{*{20}{l}}{\frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}}}&{\,\,\,\,\,n = 0}\\{\frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}}&{\,\,\,\,\,n \ne 0}\end{array}} \right.\]

Biz ushbu muammoga o'zgaruvchilarni ajratishni qo'lladik 3-misol oldingi bo'limdan. Demak, yechimni shaklda deb hisoblasak,

\[u\left( {x,t} \right) = \varphi \left( x \right)G\left( t \right)\]

yechishimiz kerak bo'lgan quyidagi ikkita oddiy differensial tenglamani olamiz.

\[\begin{align*}\frac{{dG}}{{dt}} = - k\lambda G\hspace{0.25in} & \frac{{{d^2}\varphi }}{{d{x^2}}} + \lambda \varphi = 0\\ & \varphi \left( { - L} \right) = \varphi \left( L \right)\hspace{0.25in}\frac{{d\varphi }}{{dx}}\left( { - L} \right) = \frac{{d\varphi }}{{dx}}\left( L \right)\end{align*}\]

Oldingi ikkita muammoda ko'rganimizdek, biz oldingi bobning Xususiy qiymatlar va xususiy funktsiyalar bo'limida shunga o'xshash chegaraviy muammoni hal qildik, 3-misol aniq \(L = \pi \) bilan. Shunday qilib, agar sizga nima bo'layotganini biroz ko'proq tushuntirish kerak bo'lsa, bu misolga qayting va biroz ko'proq tushuntirishni ko'rishingiz mumkin.

Bizda yana uchta holat bor.

\(\underline {\lambda > 0} \)
Differensial tenglamaning umumiy yechimi:

\[\varphi \left( x \right) = {c_1}\cos \left( {\sqrt \lambda \,x} \right) + {c_2}\sin \left( {\sqrt \lambda \,x} \right)\]

Birinchi chegaraviy shartni qo‘llash va kosinus juft funktsiya, sinus esa toq funksiya ekanligini esga olish bizga shunday beradi:

\[\begin{align*}{c_1}\cos \left( { - L\sqrt \lambda } \right) + {c_2}\sin \left( { - L\sqrt \lambda } \right)& = {c_1}\cos \left( {L\sqrt \lambda } \right) + {c_2}\sin \left( {L\sqrt \lambda } \right)\\ - {c_2}\sin \left( {L\sqrt \lambda } \right) & = {c_2}\sin \left( {L\sqrt \lambda } \right)\\ 0 & = 2{c_2}\sin \left( {L\sqrt \lambda } \right)\end{align*}\]

Ushbu bosqichda biz hech narsa deya olmaymiz, chunki \({c_2}\) yoki sinus nolga teng bo'lishi mumkin. Shunday qilib, keling, ikkinchi chegara shartini qo'llaymiz va biz nimani qo'lga kiritamiz.

\[\begin{align*} - \sqrt \lambda \,{c_1}\sin \left( { - L\sqrt \lambda } \right) + \sqrt \lambda \,{c_2}\cos \left( { - L\sqrt \lambda } \right) & = - \sqrt \lambda \,{c_1}\sin \left( {L\sqrt \lambda } \right) + \sqrt \lambda \,{c_2}\cos \left( {L\sqrt \lambda } \right)\\ \sqrt \lambda \,{c_1}\sin \left( {L\sqrt \lambda } \right) & = - \sqrt \lambda \,{c_1}\sin \left( {L\sqrt \lambda } \right)\\ 2\sqrt \lambda \,{c_1}\sin \left( {L\sqrt \lambda } \right) & = 0\end{align*}\]

Biz shunga o'xshash narsani olamiz. Biroq, e'tibor bering, agar \(\sin \left( {L\sqrt \lambda } \right) \ne 0\), biz \({c_1} = {c_2} = 0\) ga ega bo'lishga majbur bo'lamiz va bu bizga istamagan arzimas yechimni beradi.

Bu shuni anglatadiki, bizda \(\sin \left( {L\sqrt \lambda } \right) = 0\) bo'lishi kerak, bu esa o'z navbatida (oldingi misollarimizdagi ishdan) bu muammoning ijobiy xos qiymatlari ekanligini anglatadi:

\[{\lambda _{\,n}} = {\left( {\frac{{n\pi }}{L}} \right)^2}\hspace{0.25in}n = 1,2,3, \ldots \]

Endi \({c_1} = 0\) yoki \({c_2} = 0\) deb ishonish uchun hech qanday sabab yo'q. Biz bilamizki, ularning ikkalasi ham nolga teng bo‘la olmaydi, demak, bu muammo uchun ijobiy xos qiymatlarga mos keladigan ikkita o‘z funksiyalari to‘plami mavjud. Ular,

\[{\varphi _n}\left( x \right) = \cos \left( {\frac{{n\pi x}}{L}} \right)\hspace{0.25in}{\varphi _n}\left( x \right) = \sin \left( {\frac{{n\pi x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \]

\(\underline {\lambda = 0} \)
Bu holatda umumiy yechim:

\[\varphi \left( x \right) = {c_1} + {c_2}x\]

Birinchi chegara shartini qo'llash,

\[\begin{align*}{c_1} + {c_2}\left( { - L} \right) & = {c_1} + {c_2}\left( L \right)\\ 2L{c_2} & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_2} = 0\end{align*}\]

Umumiy yechim shunday bo'ladi:

\[\varphi \left( x \right) = {c_1}\]

va bu ikkinchi chegara shartini ahamiyatsiz darajada qondiradi. Shuning uchun \(\lambda = 0\) bu BVP uchun xos qiymat va bu xos qiymatga mos keladigan xos funksiyalar:

\[\varphi \left( x \right) = 1\]

\(\underline {\lambda < 0} \)
Ushbu yakuniy ish uchun umumiy yechim bu erda:

\[\varphi \left( x \right) = {c_1}\cosh \left( {\sqrt { - \lambda } \,x} \right) + {c_2}\sinh \left( {\sqrt { - \lambda } \,x} \right)\]

Birinchi chegara shartini qo‘llash va giperbolik kosinusning juft va giperbolik sinusning toq ekanligidan foydalanish,

\[\begin{align*}{c_1}\cosh \left( { - L\sqrt { - \lambda } } \right) + {c_2}\sinh \left( { - L\sqrt { - \lambda } } \right) & = {c_1}\cosh \left( {L\sqrt { - \lambda } } \right) + {c_2}\sinh \left( {L\sqrt { - \lambda } } \right)\\ - {c_2}\sinh \left( { - L\sqrt { - \lambda } } \right) & = {c_2}\sinh \left( {L\sqrt { - \lambda } } \right)\\ 2{c_2}\sinh \left( {L\sqrt { - \lambda } } \right) & = 0\end{align*}\]

Endi, bu holda biz \(\lambda < 0\) va shunga o'xshash \(L\sqrt { - \lambda } \ne 0\) deb taxmin qilamiz. Bu burilish bizga \(\sinh \left( {L\sqrt { - \lambda } } \right) \ne 0\) ekanligini aytadi. Shuning uchun bizda \({c_2} = 0\) bo'lishi kerak.

Keling, olish uchun ikkinchi chegara shartini qo'llaymiz,

\[\begin{align*}\sqrt { - \lambda } \,{c_1}\sinh \left( { - L\sqrt { - \lambda } } \right) & = \sqrt { - \lambda } \,{c_1}\sinh \left( {L\sqrt { - \lambda } } \right)\\ 2\sqrt { - \lambda } \,{c_1}\sinh \left( {L\sqrt { - \lambda } } \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = 0\end{align*}\]

\(\lambda \) haqidagi taxminimizga ko'ra, bizda \({c_1} = 0\) ga ega bo'lishdan boshqa ilojimiz yo'q va shuning uchun bu chegaraviy masala uchun salbiy xos qiymatlar yo'q.

Xulosa qilib aytadigan bo'lsak, bizda quyidagi o'z qiymatlari va o'ziga xos funktsiyalar to'plami bor va biz \(\lambda = 0\) korpusini kosinus korpusiga birlashtirganimizni e'tiborga olamiz, chunki u narsalarni biroz soddalashtirish uchun bo'lishi mumkin.

\[\begin{align*}{\lambda _{\,n}} & = {\left( {\frac{{n\pi }}{L}} \right)^2} & \hspace{0.25in}{\varphi _n}\left( x \right) & = \cos \left( {\frac{{n\,\pi x}}{L}} \right)\hspace{0.25in}n = 0,1,2,3, \ldots \\ {\lambda _{\,n}} & = {\left( {\frac{{n\pi }}{L}} \right)^2} & \hspace{0.25in}{\varphi _n}\left( x \right) & = \sin \left( {\frac{{n\,\pi x}}{L}} \right)\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]

Vaqt muammosi yana biz bu erda ishlagan ikkita muammo bilan bir xil va bizda shunday bo'ladi:

\[G\left( t \right) = c{{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\]

Endi bu misol biz ko'rib chiqqan oldingi ikkita issiqlik muammosidan biroz farq qiladi. Bunday holda, bizda qisman differentsial tenglama va chegara shartlarini qondiradigan ikkita turli xil mahsulot echimlari mavjud. Ular,

\[\begin{align*}{u_n}\left( {x,t} \right) & = {A_n}\cos \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\hspace{0.25in}n = 0,1,2,3, \ldots \\ {u_n}\left( {x,t} \right) & = {B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]

Superpozitsiya printsipi hali ham amal qiladi va shuning uchun ulardan har qandayining yig'indisi ham yechim bo'ladi va shuning uchun bu qisman differentsial tenglamaning yechimi:

\[u\left( {x,t} \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \]

Agar biz boshlang'ich shartni qo'llasak, natijaga erishamiz.

\[u\left( {x,0} \right) = f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right)} \]

va oldingi ikkita misolda ko'rganimizdek, biz Furye seriyasini olamiz. Bu safargi farq shundaki, biz to'liq olamiz Furye seriyasi \( - L \le x \le L\)-da parcha-parcha silliq boshlang'ich holat uchun. Oldingi ikkita misolda ta'kidlanganidek, biz sinuslar va kosinuslarning ortogonalligidan foydalangan holda koeffitsientlar uchun formulalarni qayta tiklashimiz mumkin yoki biz allaqachon bajargan ishni eslashimiz mumkin. Hozirgi vaqtda bajarilgan ishni qayta tiklash uchun hech qanday sabab yo'q, shuning uchun koeffitsientlar quyidagicha berilgan:

\[\begin{align*}{A_0} & = \frac{1}{{2L}}\int_{{\, - L}}^{L}{{f\left( x \right)\,dx}}\\ {A_n} & = \frac{1}{L}\int_{{\, - L}}^{L}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\hspace{0.25in}n = 1,2,3, \ldots \\ {B_n} & = \frac{1}{L}\int_{{\, - L}}^{L}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]

E'tibor bering, bu muammoning boshida qilganimiz kabi \(x\) ni o'rnatishning sababi. To'liq Furye seriyasi \( - L \le x \le L\) oralig'ini talab qiladi, biz birinchi ikkita muammoda ko'rgan Furye sinusi va kosinuslar seriyasiga \(0 \le x \le L\) kerak.