4.1-bo'lim: Laplas almashtirishi ta'rifi (the definition)
Bilasizmi, butun bir bo'limni faqat biror tushunchaning ta'rifiga bag'ishlash har doim biroz jiddiy va qo'rqinchli tuyuladi. Laplas almashtirishlari (yoki shunchaki integral almashtirishlar) birinchi marta o'rganilayotganda ancha murakkab ko'rinishi mumkin. Biroq, ushbu dars davomida ko'rib chiqishimizdek, ular aslida birinchi ko'ringandek qo'rqinchli emas.
Laplas almashtirishining bevosita ta'rifiga o'tishdan oldin, biz yana bir muhim tayanch tushunchani aniqlab olishimiz lozim.
Agar oraliqni chekli sondagi qism oraliqlarga ajratish mumkin bo'lsa va funksiya har bir ochiq qism oraliqda (ya'ni uning chekka nuqtalarisiz) uzluksiz bo'lib, har bir qism oraliqning chekkalarida chekli bir tomonlama limitlarga ega bo'lsa, bunday funksiya berilgan oraliqda bo'lakli uzluksiz funksiya (piecewise continuous) deyiladi. Quyida bo'lakli uzluksiz funksiya grafigi keltirilgan:
Boshqacha aytganda, bo'lakli uzluksiz funksiya — bu chekli sondagi sakrash uzilishlariga ega bo'lgan va hech bir nuqtada cheksizlikka qarab ketmaydigan (ya'ni chegaralangan) funksiyadir.
Endi Laplas almashtirishining qat'iy matematik ta'rifini keltiramiz.
Laplas almashtirishi ta'rifi
Faraz qilaylik, \(f(t)\) — \(t \ge 0\) sohada bo'lakli uzluksiz funksiya bo'lsin. Ushbu \(f(t)\) funksiyasining Laplas almashtirishi \(\mathcal{L}\left\{ {f\left( t \right)} \right\}\) kabi belgilanadi va quyidagi xosmas integral orqali ta'riflanadi:
\[\begin{equation}\mathcal{L}\left\{ {f\left( t \right)} \right\} = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - s\,t}}f\left( t \right)\,dt}}\label{eq:eq1}\end{equation}\]Laplas almashtirishlari uchun juda qulay muqobil belgilash ham mavjud. Odatda biz Laplas almashtirishini qisqacha quyidagicha belgilaymiz:
\[\mathcal{L}\left\{ {f\left( t \right)} \right\} = F\left( s \right)\]Ushbu muqobil yozuvda e'tibor bering: almashtirish natijasi aslida yangi o'zgaruvchi \(s\) ning funksiyasidir va aniq integrallash jarayonida barcha \(t\) o'zgaruvchilari yo'qolib ketadi.
Ta'rifdagi integral xosmas integral (improper integral) deb ataladi. Shuning uchun almashtirishlarni hisoblashga kirishishdan oldin bu kabi integrallar qanday hisoblanishini eslab olish foydalidir.
Yechim
Eslatib o'tamiz, xosmas integral yuqori chegarasi bo'yicha limitga aylantiriladi:
\[\int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{c\,t}}\,dt}} = \mathop {\lim }\limits_{n \to \infty } \int_{{\,0}}^{{\,n}}{{{{\bf{e}}^{c\,t}}\,dt}}\]Dastlab integralni hisoblaymiz, so'ngra limit qiymatini topamiz:
\[\begin{align*}\int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{c\,t}}\,dt}} & = \mathop {\lim }\limits_{n \to \infty } \int_{{\,0}}^{{\,n}}{{{{\bf{e}}^{c\,t}}\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left. {\left( {\frac{1}{c}{{\bf{e}}^{c\,t}}} \right)} \right|_0^n\\ & = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{c}{{\bf{e}}^{c\,n}} - \frac{1}{c}} \right)\end{align*}\]Endi bu yerda nihoyatda ehtiyot bo'lishimiz kerak: \(c\) ning ishorasi javobga bevosita ta'sir qiladi. Biz \(c \ne 0\) deb oldik. Agar \(c > 0\) bo'lsa, ko'rsatkichli ifoda \(e^{cn} \to \infty\) ga ketadi va integral uzoqlashadi. Agar \(c < 0\) bo'lsa, u holda \(e^{cn} \to 0\) ga intiladi va limit chekli bo'ladi.
Demak, ushbu integral faqatgina \(c < 0\) bo'lgandagina yaqinlashuvchi (ya'ni chekli limitga ega) bo'ladi. Bu holda biz quyidagini olamiz:
\[\begin{equation}\int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{c\,t}}\,dt}} = - \frac{1}{c}\hspace{0.25in}\hspace{0.25in}{\text{faqat }}\,c < 0\, {\text{bo'lganda}}\label{eq:eq2}\end{equation}\]Buni eslab olganimizdan so'ng, endi aniq funksiyalarning Laplas almashtirishlarini hisoblaymiz. Eng sodda va muhim funksiyadan boshlaymiz.
Yechim
Buning uchun \(f(t) = 1\) funksiyasini to'g'ridan-to'g'ri ta'rifga qo'yamiz:
\[\mathcal{L}\left\{ 1 \right\} = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - st}}\,dt}}\]Ushbu integral 1-Misoldagi integral bilan aynan bir xil bo'lib, bu yerda \(c = -s\). Demak, yuqoridagi formuladan foydalanish kifoya:
\[\mathcal{L}\left\{ 1 \right\} = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - \,s\,t}}\,dt}} = - \frac{1}{{ - s}}\hspace{0.25in}\hspace{0.25in}{\text{agar }} - s < 0\; {\text{bo'lsa}}\]Soddalashtirish natijasida quyidagi fundamental natijaga ega bo'lamiz:
\[\mathcal{L}\left\{ 1 \right\} = \frac{1}{s}\hspace{0.25in}{\text{faqat }}s > 0\; {\text{bo'lganda}}\]E'tibor bergan bo'lsangiz, almashtirish mavjud bo'lishi uchun biz \(s\) o'zgaruvchisiga cheklov qo'yishimizga to'g'ri keldi: \(s > 0\). Barcha Laplas almashtirishlari \(s\) uchun ma'lum yaqinlashish cheklovlariga ega bo'ladi. Dastlabki bosqichda bu cheklovlarga unchalik chuqur to'xtalmasak-da, ularning mavjudligini aslo unutmaslik kerak.
Navbatdagi misolni ko'rib chiqamiz.
Yechim
Funksiyani Laplas almashtirishi ta'rifiga qo'yamiz va daraja xossalaridan foydalanib soddalashtiramiz:
\[\mathcal{L}\left\{ {{{\bf{e}}^{a\,t}}} \right\} = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - s\,t}}{{\bf{e}}^{a\,t}}\,dt}} = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{\left( {a - s} \right)t}}\,dt}}\]Yana bir bor, bu integral 1-Misoldagi integralga to'liq mos keladi, bunda \(c = a - s\). Demak:
\[\begin{align*}\mathcal{L}\left\{ {{{\bf{e}}^{a\,t}}} \right\} & = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{\left( {a - s} \right)t}}\,dt}}\\ & = - \frac{1}{{a - s}}\hspace{0.25in}\hspace{0.25in}{\text{agar }}a - s < 0\\ & = \frac{1}{{s - a}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{\text{faqat }}s > a\; {\text{bo'lganda}}\end{align*}\]Endi bevosita sodda eksponensial integraldan farq qiluvchi trigonometrik funksiyaning almashtirishini hisoblaymiz.
Yechim
Funksiyani ta'rifga qo'yamiz va qulaylik uchun \(F(s)\) belgilashidan foydalanamiz:
\[\begin{align*}\mathcal{L}\left\{ {\sin \left( {at} \right)} \right\} & = F\left( s \right)\\ & = \int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - st}}\sin \left( {at} \right)\,dt}}\\ & = \mathop {\lim }\limits_{n \to \infty } \int_{{\,0}}^{{\,n}}{{{{\bf{e}}^{ - st}}\sin \left( {at} \right)\,dt}}\end{align*}\]Ushbu integralni bo'laklab integrallash usuli bilan hisoblaymiz:
\[F\left( s \right) = \mathop {\lim }\limits_{n \to \infty } \left( { - \left. {\left( {\frac{1}{a}{{\bf{e}}^{ - st}}\cos \left( {at} \right)} \right)} \right|_0^n - \frac{s}{a}\int_{{\,0}}^{{\,n}}{{{{\bf{e}}^{ - st}}\cos \left( {at} \right)\,dt}}} \right)\]Birinchi hadni hisoblaymiz va hosil bo'lgan integralga yana bir bor bo'laklab integrallashni qo'llaymiz:
\[F\left( s \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{a}\left( {1 - {{\bf{e}}^{ - sn}}\cos \left( {an} \right)} \right) - \frac{s}{a}\left( {\left. {\left( {\frac{1}{a}{{\bf{e}}^{ - st}}\sin \left( {at} \right)} \right)} \right|_0^n + \frac{s}{a}\int_{{\,0}}^{{\,n}}{{{{\bf{e}}^{ - st}}\sin \left( {at} \right)\,dt}}} \right)} \right)\]Endi ikkinchi hadni hisoblab, limitga o'tamiz va soddalashtiramiz:
\[\begin{align*}F\left( s \right) & = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{a}\left( {1 - {{\bf{e}}^{ - sn}}\cos \left( {an} \right)} \right) - \frac{s}{a}\left( {\frac{1}{a}{{\bf{e}}^{ - sn}}\sin \left( {an} \right) + \frac{s}{a}\int_{{\,0}}^{{\,n}}{{{{\bf{e}}^{ - st}}\sin \left( {at} \right)\,dt}}} \right)} \right)\\ & = \frac{1}{a} - \frac{s}{a}\left( {\frac{s}{a}\int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - st}}\sin \left( {at} \right)\,dt}}} \right)\\ & = \frac{1}{a} - \frac{{{s^2}}}{{{a^2}}}\int_{{\,0}}^{{\,\infty }}{{{{\bf{e}}^{ - st}}\sin \left( {at} \right)\,dt}}\end{align*}\]E'tibor bering, limitlarni hisoblashda biz quyidagi ikkita limit nolga teng bo'lishi uchun \(s > 0\) deb oldik:
\[\begin{align*}\mathop {\lim }\limits_{n \to \infty } {{\bf{e}}^{ - sn}}\cos \left( {an} \right) & = 0\\ \mathop {\lim }\limits_{n \to \infty } {{\bf{e}}^{ - sn}}\sin \left( {an} \right) & = 0\end{align*}\]O'ng tomonda hosil bo'lgan integral aynan biz hisoblashni boshlagan dastlabki \(F(s)\) integralining o'zidir. Demak:
\[F\left( s \right) = \frac{1}{a} - \frac{{{s^2}}}{{{a^2}}}F\left( s \right)\]Endi ushbu tenglamadan \(F(s)\) ni topamiz:
\[F(s) \left( 1 + \frac{s^2}{a^2} \right) = \frac{1}{a} \implies F(s) \left( \frac{a^2 + s^2}{a^2} \right) = \frac{1}{a}\] \[\mathcal{L}\left\{ {\sin \left( {at} \right)} \right\} = F\left( s \right) = \frac{a}{{{s^2} + {a^2}}}\hspace{0.25in}\hspace{0.25in}{\text{faqat }}s > 0\; {\text{bo'lganda}}\]Ushbu misol ko'rsatib turibdiki, hatto eng sodda funksiyalar uchun ham Laplas almashtirishini bevosita ta'rif orqali integrallab topish ancha mehnat talab qiladigan jarayondir. Shuning uchun amaliyotda standart funksiyalarning Laplas almashtirishlari jadvallaridan va almashtirishning asosiy xossalaridan foydalaniladi.
Keyingiga o'tishdan oldin bitta muhim izoh berib o'tamiz. Ayrim adabiyotlarda Laplas almashtirishi quyidagicha ikki tomonlama integral ko'rinishida beriladi:
\[\mathcal{L}\left\{ {f\left( t \right)} \right\} = \int_{{\, - \infty }}^{{\,\infty }}{{{{\bf{e}}^{ - st}}f\left( t \right)\,dt}}\]Bunda quyi chegara noldan emas, balki minus cheksizlikdan boshlanadi. Bunday hollarda deyarli doim \(f(t)\) funksiyasi \(t < 0\) da nolga teng deb qabul qilinadi:
\[f\left( t \right) = \left\{ {\begin{array}{*{20}{l}}{f\left( t \right)}&{{\text{agar }}t \ge 0}\\0&{{\text{agar }}t < 0}\end{array}} \right.\]Boshqacha aytganda, \(t < 0\) da funksiya nol bo'lgani uchun integralning birinchi yarmi nolga aylanib tushib qoladi va natijada biz yuqoridagi bir tomonlama standart Laplas almashtirishi formulasiga qaytamiz. Funksiyani \(t < 0\) da nolga aylantirish uchun odatda Xevisayd pog'onaviy funksiyasi (Heaviside step function) qo'llaniladi. Biz bu funksiyani keyingi bo'limlarda batafsil o'rganamiz.