Bu erda biz hal qilishimiz kerak bo'lgan asosiy farqlardan biri shundaki, bizda ikkita dastlabki shart mavjud. Bu biz hozirgacha ko'rgan narsa emas, lekin vaqt o'tganda buni hal qilish unchalik qiyin bo'lmaydi.

Biz bu muammo uchun o'zgaruvchilarni ajratishni allaqachon amalga oshirganmiz, ammo keling, buni davom ettiramiz va bu erda takrorlaymiz, shuning uchun bizda deyarli to'liq hal qilingan boshqa muammo borligini aytishimiz mumkin.

Shunday qilib, keling, mahsulot yechimidan boshlaylik.

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

Buni ikkita chegara shartlariga ulash,

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

Mahsulot yechimini differensial tenglamaga ulash, ajratish va ajratish konstantasini kiritish,

\[\begin{align*}\frac{{{\partial ^2}}}{{\partial {t^2}}}\left( {\varphi \left( x \right)h\left( t \right)} \right) & = {c^2}\frac{{{\partial ^2}}}{{\partial {x^2}}}\left( {\varphi \left( x \right)h\left( t \right)} \right)\\ \varphi \left( x \right)\frac{{{d^2}h}}{{d{t^2}}} & = {c^2}h\left( t \right)\frac{{{d^2}\varphi }}{{d{x^2}}}\\ \frac{1}{{{c^2}h}}\frac{{{d^2}h}}{{d{t^2}}} & = \frac{1}{\varphi }\frac{{{d^2}\varphi }}{{d{x^2}}} = - \lambda \end{align*}\]

Biz qulaylik uchun \({c^2}\) ni chap tomonga o'tkazdik va ajratish doimiysi uchun \( - \lambda \) ni tanladik, shuning uchun \(\varphi \) uchun differentsial tenglama ma'lum (va echilgan) holatga mos keladi.

O'zgaruvchilarni ajratish natijasida olingan ikkita oddiy differentsial tenglamalar,

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

Yuqoridagi chegara masalasini hal qildik 1-misol Ushbu bobning Issiqlik tenglamasini echish bo'limi va shuning uchun bu muammoning o'z qiymatlari va o'z funktsiyalari:

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

Birinchi oddiy differensial tenglama endi,

\[\frac{{{d^2}h}}{{d{t^2}}} + {\left( {\frac{{n\pi c}}{L}} \right)^2}h = 0\]

va \(h\) koeffitsienti aniq ijobiy bo'lgani uchun, buning yechimi,

\[h\left( t \right) = {c_1}\cos \left( {\frac{{n\pi c\,t}}{L}} \right) + {c_2}\sin \left( {\frac{{n\pi c\,t}}{L}} \right)\]

Yuqoridagi koeffitsientlarning har biri nolga teng deb o'ylash uchun hech qanday sabab yo'qligi sababli biz ikkita mahsulot echimini olamiz,

\[\begin{array}{*{20}{c}}{{u_n}\left( {x,t} \right) = {A_n}\cos \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)}\\{{u_n}\left( {x,t} \right) = {B_n}\sin \left( {\frac{{n\pi c\,t}}{L}} \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)}\end{array}\hspace{0.25in}n = 1,2,3, \ldots \]

Yechim shundan iboratki,

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

Endi, ikkinchi boshlang'ich shartni qo'llash uchun biz buni \(t\) ga nisbatan farqlashimiz kerak, shuning uchun,

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

Agar biz olgan dastlabki shartlarni qo'llasak,

\[\begin{align*}& u\left( {x,0} \right) = f\left( x \right) = \sum\limits_{n = 1}^\infty {\left[ {{A_n}\cos \left( 0 \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right) + {B_n}\sin \left( 0 \right)\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \right]} = \sum\limits_{n = 1}^\infty {{A_n}\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \\ & \frac{{\partial u}}{{\partial t}}\left( {x,0} \right) = g\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{n\pi c}}{L}{B_n}\sin \left( {\frac{{n\,\pi \,x}}{L}} \right)} \end{align*}\]

Bularning ikkalasi ham Furye sinus qatoridir. Birinchisi \(0 \le x \le L\) da \(f\left( x \right)\) uchun, ikkinchisi esa biroz tartibsiz koeffitsientli \(0 \le x \le L\) da \(g\left( x \right)\) uchun. So'nggi bir necha bo'limda bo'lgani kabi, biz \({A_n}\) va \({B_n}\) uchun formulalarni olish uchun sinuslarning ortogonalligidan foydalanish yoki oldingi ishdagi formuladan qayta foydalanishni tanlash bilan duch keldik.

Formulalarni qayta ishlatish osonroq bo'ladi, shuning uchun formulalar yordamida shakl hosil qiladi Furye sinus qatori biz olgan bo'lim,

\[\begin{align*}{A_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \\ \frac{{n\pi c}}{L}{B_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{g\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \end{align*}\]

Ikkinchisini yechib, biz olamiz,

\[\begin{align*}{A_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \\ {B_{\,n}} & = \frac{2}{{n\pi c}}\int_{{\,0}}^{{\,L}}{{g\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \end{align*}\]