Delta - funksiya yoki Dirak delta funksiyasi
Delta-funksiyaning sxematik grafigi Fanga kiritilish tarixi va matematik isboti tahrir Birinchi marta 1926-yilda Dirak tomonidan kiritilgan δ {\displaystyle \delta } -funksiya odatda quyidagicha ta'riflanadi:
δ ( x ) = 0 {\displaystyle \delta (x)=0} , agar x ≠ 0 ; ( 1 ) {\displaystyle x\neq 0;(1)} δ ( x ) = ∞ {\displaystyle \delta (x)=\infty } , agar x = 0 ; ( 2 ) {\displaystyle x=0;(2)} ∫ − ∞ + ∞ δ ( x ) d x = 1 {\displaystyle \int \limits _{-\infty }^{+\infty }\delta (x)dx=1} , agar − ∞ < x < + ∞ ; ( 3 ) {\displaystyle -\infty <x<+\infty ;(3)} Integrallash chegaralari − ∞ , + ∞ {\displaystyle -\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 ) {\displaystyle f(x)} funksiya uchun yozish mumkin:
∫ − ∞ + ∞ f ( x ) δ ( x ) d x = f ( 0 ) {\displaystyle \int \limits _{-\infty }^{+\infty }f(x)\delta (x)dx=f(0)} , agar − ∞ < x < + ∞ ; ( 4 ) {\displaystyle -\infty <x<+\infty ;(4)}
Haqiqatdan, δ ( x ) {\displaystyle \delta (x)} xususiyatiga muvofiq, yuqoridagi integralda faqat x = 0 {\displaystyle x=0} nuqta atrofigina ahamiyatlidir. Cheksiz kichik sohada uzluksiz funksiyani o`zgarmas hisoblash mumkin. U vaqtda f ( x ) {\displaystyle f(x)} funksiyaning f ( 0 ) {\displaystyle 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:
δ ( k x ) = 1 | k | δ ( x ) ; ( 5 ) {\displaystyle \delta (kx)={\frac {1}{|k|}}\delta (x);(5)} ,agar k = c o n s t ≠ 0 {\displaystyle k=const\neq 0}
Haqiqatdan ham, agar k x {\displaystyle kx} ni ξ {\displaystyle \xi } orqali belgilasak,
ξ = k x ; ( 6 ) {\displaystyle \xi =kx;(6)}
U vaqtda (1) va (2) formulalarga asosan,
δ ( ξ ) = 0 , {\displaystyle \delta (\xi )=0,} agar ξ ≠ 0 ; ( 7 ) {\displaystyle \xi \neq 0;(7)} δ ( ξ ) = ∞ , {\displaystyle \delta (\xi )=\infty ,} agar ξ = 0 ; ( 8 ) {\displaystyle \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 {\displaystyle k>0} bo`lsa,
∫ − ∞ + ∞ δ ( k x ) d x = 1 | k | ∫ − ∞ + ∞ | k | δ ( k x ) d x = 1 | k | ∫ − ∞ + ∞ k δ ( k x ) d x = 1 | k | ∫ − ∞ + ∞ δ ( k x ) d ( k x ) {\displaystyle \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,
∫ − ∞ + ∞ δ ( k x ) d ( k x ) = 1 | k | ∫ − ∞ + ∞ δ ( ξ ) d ξ {\displaystyle \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
∫ − ∞ + ∞ δ ( ξ ) d ξ = 1 ; ( 9 ) {\displaystyle \int \limits _{-\infty }^{+\infty }\delta (\xi )d\xi =1;(9)}
demak,
∫ − ∞ + ∞ δ ( k x ) d x = 1 | k | {\displaystyle \int \limits _{-\infty }^{+\infty }\delta (kx)dx={\frac {1}{|k|}}}
Agar k < 0 {\displaystyle k<0} bo`lsa,
∫ − ∞ + ∞ δ ( k x ) d x = 1 | k | ∫ − ∞ + ∞ | k | δ ( k x ) d x = − 1 | k | ∫ − ∞ + ∞ k δ ( k x ) d x = − 1 | k | ∫ − ∞ + ∞ δ ( k x ) d ( k x ) = − 1 | k | ∫ − ∞ + ∞ δ ( ξ ) d ξ = 1 | k | ∫ − ∞ + ∞ δ ( ξ ) d ξ = 1 | k | {\displaystyle \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
∫ − ∞ + ∞ 1 | k | δ ( x ) d x = 1 | k | ∫ − ∞ + ∞ δ ( x ) d x = 1 | k | {\displaystyle \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 {\displaystyle k=-1} uchun
δ ( − x ) = δ ( x ) ; ( 10 ) {\displaystyle \delta (-x)=\delta (x);(10)}
Delta-funksiya - juft funksiya hisoblanadi. x δ ′ ( x ) = − δ ( x ) {\displaystyle x\delta ^{\prime }(x)=-\delta (x)} δ ( f ( x ) ) = ∑ k δ ( x − x k ) | f ′ ( x k ) | {\displaystyle \delta (f(x))=\sum _{k}{\frac {\delta (x-x_{k})}{|f'(x_{k})|}}} , bu yerda x k {\displaystyle x_{k}} - f ( x ) {\displaystyle f(x)} funksiyaning nollariBir o`lchamli Delta-funksiyadan olingan integral Heaviside funksiyasi beradi: θ ( x ) = { 0 , x < 0 , 1 , x > 0. {\displaystyle {\displaystyle \theta (x)=\left\{{\begin{array}{*{35}{l}}0,&x<0,\\1,&x>0.\\\end{array}}\right.}}
Delta-funksiyaning filtrlash xossasi: ∫ − ∞ + ∞ f ( x ) δ ( x − x 0 ) d x = f ( x 0 ) . {\displaystyle \int \limits _{-\infty }^{+\infty }f(x)\delta (x-x_{0})\,dx=f(x_{0}).}
Amaliyotda, ko`pincha, Delta-funksiyaning integral ko`rinishidan foydalanish qulay:
: δ ( t ) = 1 2 π ∫ − ∞ + ∞ e i ω t d ω {\displaystyle :\delta (t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{+\infty }e^{i\omega t}\,d\omega }
Isbot
Quyidagi integralni ko`rib chiqaylik:
I ( t ) = 1 2 π ∫ − ∞ ∞ e i ω t d ω , {\displaystyle 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 → ∞ I N ( t ) , {\displaystyle I(t)=\lim _{N\to \infty }I_{N}(t),} bu yerda
I N ( t ) = 1 2 π ∫ − N N e i ω t d ω = 1 π N sin t N t N . {\displaystyle 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,
∫ − ∞ ∞ sin t t d t = π . {\displaystyle \int \limits _{-\infty }^{\infty }{\frac {\sin t}{t}}\,dt=\pi .} (3)Agar ixtiyoriy N {\displaystyle N} uchun (3) dan foydalansak, quyidagi tenglik o`rinli:
∫ − ∞ ∞ I N ( t ) d t = 1 π ∫ − ∞ ∞ sin t N t N d ( t N ) = 1. {\displaystyle \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 δ ( t ) {\displaystyle \delta (t)} ga intiladi.