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 }
δ
(
x
)
=
0
{\displaystyle \delta (x)=0}
x
≠
0
;
(
1
)
{\displaystyle x\neq 0;(1)}
δ
(
x
)
=
∞
{\displaystyle \delta (x)=\infty }
x
=
0
;
(
2
)
{\displaystyle x=0;(2)}
∫
−
∞
+
∞
δ
(
x
)
d
x
=
1
{\displaystyle \int \limits _{-\infty }^{+\infty }\delta (x)dx=1}
−
∞
<
x
<
+
∞
;
(
3
)
{\displaystyle -\infty <x<+\infty ;(3)}
Integrallash chegaralari
−
∞
,
+
∞
{\displaystyle -\infty ,+\infty }
Har qanday uzluksiz
f
(
x
)
{\displaystyle f(x)}
∫
−
∞
+
∞
f
(
x
)
δ
(
x
)
d
x
=
f
(
0
)
{\displaystyle \int \limits _{-\infty }^{+\infty }f(x)\delta (x)dx=f(0)}
−
∞
<
x
<
+
∞
;
(
4
)
{\displaystyle -\infty <x<+\infty ;(4)}
Haqiqatdan,
δ
(
x
)
{\displaystyle \delta (x)}
x
=
0
{\displaystyle x=0}
uzluksiz funksiyani o`zgarmas hisoblash mumkin. U vaqtda
f
(
x
)
{\displaystyle f(x)}
f
(
0
)
{\displaystyle f(0)}
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)}
k
=
c
o
n
s
t
≠
0
{\displaystyle k=const\neq 0}
Haqiqatdan ham, agar
k
x
{\displaystyle kx}
ξ
{\displaystyle \xi }
ξ
=
k
x
;
(
6
)
{\displaystyle \xi =kx;(6)}
U vaqtda (1) va (2) formulalarga asosan,
δ
(
ξ
)
=
0
,
{\displaystyle \delta (\xi )=0,}
ξ
≠
0
;
(
7
)
{\displaystyle \xi \neq 0;(7)}
δ
(
ξ
)
=
∞
,
{\displaystyle \delta (\xi )=\infty ,}
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}
∫
−
∞
+
∞
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)}
∫
−
∞
+
∞
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}
∫
−
∞
+
∞
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|}}}
(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|}}}
Shunday qilib, (5) ifoda isbotlandi. Xususiy hol
k
=
−
1
{\displaystyle k=-1}
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})|}}}
x
k
{\displaystyle x_{k}}
f
(
x
)
{\displaystyle f(x)}
Bir 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 ,}
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}}.}
Ma'lumki,
∫
−
∞
∞
sin
t
t
d
t
=
π
.
{\displaystyle \int \limits _{-\infty }^{\infty }{\frac {\sin t}{t}}\,dt=\pi .}
Agar ixtiyoriy
N
{\displaystyle N}
∫
−
∞
∞
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.}
Demak, N
δ
(
t
)
{\displaystyle \delta (t)}
