Delta-funksiya

Delta - funksiya yoki Dirak delta funksiyasi

Delta-funksiyaning sxematik grafigi

Fanga kiritilish tarixi va matematik isbotiTahrirlash

Birinchi marta 1926-yilda Dirak tomonidan kiritilgan $\delta$ -funksiya odatda quyidagicha ta'riflanadi:

$\delta (x)=0$ , agar $x\neq 0;(1)$
$\delta (x)=\infty$ , agar $x=0;(2)$
$\int \limits _{-\infty }^{+\infty }\delta (x)dx=1$ , agar $-\infty <x<+\infty ;(3)$

Integrallash chegaralari $-\infty ,+\infty$ bo`lishi shart emas. Delta-funksiya cheksiz bo`lgan nuqtani o`z ichiga olgan har qanday soha integrallash sohasi bo`lishi mumkin. Delta-funksiyaning ma'nosi ham shundaki, integral uning argumenti bo`yicha olinadi.

Har qanday uzluksiz $f(x)$ funksiya uchun yozish mumkin:

$\int \limits _{-\infty }^{+\infty }f(x)\delta (x)dx=f(0)$ , agar $-\infty <x<+\infty ;(4)$

Haqiqatdan, $\delta (x)$ xususiyatiga muvofiq, yuqoridagi integralda faqat $x=0$ nuqta atrofigina ahamiyatlidir. Cheksiz kichik sohada uzluksiz funksiyani o`zgarmas hisoblash mumkin. U vaqtda $f(x)$ funksiyaning $f(0)$ qiymatini integral belgisining oldiga chiqariladi. Qolgan integral esa 3-formulaga asosan birga tengdir.

Delta-funksiya uchun muhim bo`lgan yana bir formulani qarab ko`raylik:

$\delta (kx)={\frac {1}{|k|}}\delta (x);(5)$ , agar $k=const\neq 0$

Haqiqatdan ham, agar $kx$ ni $\xi$ orqali belgilasak,

$\xi =kx;(6)$

U vaqtda (1) va (2) formulalarga asosan,

$\delta (\xi )=0,$ agar $\xi \neq 0;(7)$ $\delta (\xi )=\infty ,$ agar $\xi =0;(8)$

(5) da ifodalangan tenglik belgisining ma'nosi shundan iboratki, uning o`ng va chap tomonlari integral ostidagi ko`paytuvchilar sifatida olinib, bir xil natija beradi.

Tenglamaning chap tomonini ko`rib chiqaylik. Agar $k>0$ bo`lsa,

$\int \limits _{-\infty }^{+\infty }\delta (kx)dx={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }|k|\delta (kx)dx={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }k\delta (kx)dx={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }\delta (kx)d(kx)$ bo`ladi. Ammo (6) ga muvofiq,

$\int \limits _{-\infty }^{+\infty }\delta (kx)d(kx)={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }\delta (\xi )d\xi$

Endi (6)-(8) ifodalarni nazarda tutib, delta-funksiya ta'rifiga asosan yozishimiz mumkin

$\int \limits _{-\infty }^{+\infty }\delta (\xi )d\xi =1;(9)$

demak,

$\int \limits _{-\infty }^{+\infty }\delta (kx)dx={\frac {1}{|k|}}$

Agar $k<0$ bo`lsa,

$\int \limits _{-\infty }^{+\infty }\delta (kx)dx={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }|k|\delta (kx)dx=-{\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }k\delta (kx)dx=-{\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }\delta (kx)d(kx)=-{\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }\delta (\xi )d\xi ={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }\delta (\xi )d\xi ={\frac {1}{|k|}}$ bo`ladi.

(5) ning o`ng tomonidan olingan integral esa

$\int \limits _{-\infty }^{+\infty }{\frac {1}{|k|}}\delta (x)dx={\frac {1}{|k|}}\int \limits _{-\infty }^{+\infty }\delta (x)dx={\frac {1}{|k|}}$ bo`ladi.

Shunday qilib, (5) ifoda isbotlandi. Xususiy hol $k=-1$ uchun

$\delta (-x)=\delta (x);(10)$

XossalariTahrirlash

Delta-funksiya - juft funksiya hisoblanadi.
$x\delta ^{\prime }(x)=-\delta (x)$
$\delta (f(x))=\sum _{k}{\frac {\delta (x-x_{k})}{|f'(x_{k})|}}$ , bu yerda $x_{k}$ - $f(x)$ funksiyaning nollari
Bir o`lchamli Delta-funksiyadan olingan integral Heaviside funksiyasi beradi:

${\displaystyle \theta (x)=\left\{{\begin{array}{*{35}{l}}0,&x<0,\\1,&x>0.\\\end{array}}\right.}$

Delta-funksiyaning filtrlash xossasi:

$\int \limits _{-\infty }^{+\infty }f(x)\delta (x-x_{0})\,dx=f(x_{0}).$

Integral tasavvurlarTahrirlash

Amaliyotda, ko`pincha, Delta-funksiyaning integral ko`rinishidan foydalanish qulay:

$:\delta (t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{+\infty }e^{i\omega t}\,d\omega$

Isbot

Quyidagi integralni ko`rib chiqaylik:

I(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{\infty }e^{i\omega t}\,d\omega ,

(1)

Ushbu integralni quyidagi limit ko`rinishida ham yozishimiz mumkin:

I(t)=\lim _{N\to \infty }I_{N}(t),

bu yerda

I_{N}(t)={\frac {1}{2\pi }}\int \limits _{-N}^{N}e^{i\omega t}\,d\omega ={\frac {1}{\pi }}N{\frac {\sin {tN}}{tN}}.

(2)

Ma'lumki,

\int \limits _{-\infty }^{\infty }{\frac {\sin t}{t}}\,dt=\pi .

(3)

Agar ixtiyoriy $N$ uchun (3) dan foydalansak, quyidagi tenglik o`rinli:

\int \limits _{-\infty }^{\infty }I_{N}(t)\,dt={\frac {1}{\pi }}\int \limits _{-\infty }^{\infty }{\frac {\sin {tN}}{tN}}\,d(tN)=1.

(4)

Demak, $N$ ning cheksiz o`suvchi qiymatlarida (2) funksiya uchun Delta-funksiyaning barcha xossalari o`rinli va u $\delta (t)$ ga intiladi.

Yana qarangTahrirlash

AdabiyotlarTahrirlash

Mallin R.X., Klassik elektrodinamika, O`qituvchi, - T. 1974
Дирак П. А. М. Основы квантовой механики / Пер. с англ. — М., 1932 (есть много переизданий).