1§ Boshlang’ich funksya va uning tasviri

Misollar. 1)

y ^  4 y

 s in 3 x

tenglamaning

y ( 0 )  0 ,

y ^( 0 )  0

boshlang‟ich

shartni qanoatlantruvchi yechimi topilsin.

Yechish. Yordamchi tenglamani tuzamiz:

y ( p )( p ²
 4 ) 

3 _;

p ²  9
y ( p ) 
( p ²

3

 9 )( p ²

 4 )

yoki

3 3



y ( p ) 

5 <sub>

5  </sub> 1 3 _ 3 2

p ²  9

p ²  4 ⁵ p ²  9 ^{1 0} p ²  4

bundan tenglamaning yechimini topamiz:

y ( t ) 

3 1

s in 2 t 

1 0 5

s in 3t

2) y ^ 
y  0

tenglamaning
y ( 0 )  1,

y ^( 0 )  3 ,

y ^( 0 )  6

boshlang‟ich shartlarni

qanoatlantruvchi yechimi topilsin.

Yechish. Yordamchi tenglama:


p ²  3 p  8



p ²  3 p  8

y ( p )( p ³  1)  p ²  1  p  3  8 ; yoki

y ( p )  

p ³  1 ( p  1)( p ² 

p  1)

2  p  6 1

1

2

2
^{p } ₂ 1 1 ₂

y ( p )    2

  .

p  1

p ² 

p  1

p  1

²   <sup>3

2</sup>  

 ¹   ¹ 

_ p  _  <sub> 

</sub> p  _  _{ }

 ²  _{ }

 ² 

 ² 

7. Ko’paytirish teoremasi

Teorema. Agar

f₁ ( t )

va f ₂ ( t ) funksiyalarni tasvirlari

F₁ ( p )

^{va F} ₂ ^{( p )} bo‟lsa,

ya‟ni
F₁ ( p ) 
f₁ ( t )

va ^F ₂ ^{( p ) }

f ₂ ( t )

u holda

t

_ f₁ ( ) f ₂ ( t   ) d 

0

funksiyani

tasviri ko‟paytmadan iborat bo‟ladi; ya‟ni

F₁ ( p ) F₂ ( p ) 

t

_ f₁ ( ) f₁ ( t   ) d 

0

(1.2.3)

Isbot.

t

_ f₁ ( ) f₁ ( t   ) d 

0

funksiyani tasvirini topamiz, tasvirni ta‟rifiga asosan:

 ^t 

^  pt ^ t ^

^{L } 

 0

f₁ ( ) f₁ ( t   ) d  _  _ e

 0

_{ } f₁ ( ) f ₂ ( t   ) d  _d t

 0 

O‟ng tomondagi ikki karrali integral

  0,   t

chiziqlar bilan chegaralangan soha bo‟yicha olinadi (2 rasm)

Bu integralda integrallash tartibini o‟zgartiramiz:

 ^t  ^ 



• 
  
  
  pt
  

^{L } 

f₁ ( ) f ₂ ( t   ) d  _ 

_{ } f₁ ( ) _ e f ₂ ( t   ) d t _d 

(1.2.4)

 0  0   

O‟zgaruvchini almashtramiz:

t  
 z ,
d t 

d z ichki integralni hisoblaymiz:

  

_{ e}  p t _f

( t   ) d t 

_{ e}  p ( z   ) _f

( z )d z

_{ e}  p

_{ e}  p z _f

( z ) d z

_{ e}  p _F

( p )

2 2 2 2

0 0 0
Ichki integralni qiymatini (1.2.4) ga qo‟yamiz:

_ t _ 
 p



 p

^{L } 

f₁ ( ) f ₂ ( t   ) d  _  _

f₁ ( ) e F ₂ ( p )d 

 F ₂ ( p ) _ e f₁ ( ) d 

 F ₂ ( p ) F₁ ( p )

 0  0 0

Demak

t

_ f₁ ( ) f ₂ ( t   ) d 

0

ifoda berilgan ikkita

f₁ ( t ) va
f ₂ ( t )

funksiyani

o‟ramasi deyiladi.
t t

Bunda

_ f₁ ( ) f ₂ ( t   ) d 

 _ f₁ ( t   ) f ₂ ( ) d 

tenglik kuchga egadir.

0 0

Misol.

y ^  y 
f ( t )

tenglamaning
y ( 0 ) 

y ^( 0 )  0

boshlang‟ich shartni

qanoatlantruvchi yechimi topilsin.

Yechish. Yordamchi tenglama tuzamiz:

bundan

y ( p )( p ²

 1) 

F ( p )

y ( p ) 

F ( p )

p ²  1

1



p ²  1
s in t

va F ( p ) 

f ( t )

bo‟lgani uchun
F ( p ) 
F₁ ( p )

orqali belgilab

ko‟paytrish teoremasini tadbiq etsak,

kelib chiqadi.

t

y ( t )  _

0
f ( ) s in ( t   ) d 

Izoh. 2) Agar

f₁ ( t ) 

^{f ( t )} va

f ₂ ( t )  1

desak, u holda

F₁ ( p ) 

^{F ( p )} va

1

F ₂ ( p ) 

p

bo‟ladi. Ko‟paytirish teoremasiga asosan:
1 ^{t t} F ( p )

F ( p )  _

p ₀

f ( ) d 

yoki

_ f ( ) d  

₀ p

va f ₂ ( t )  1

originalni integrallash haqidagi teorema kelib chiqadi.

8. Kechikish teoremasi

Agar

f _ t _

funksiya t<0 bo‟lganda aynan nolga teng bo‟lsa, u holda

f ( t  t ₀ )

funksiya t  t ₀

bo‟lganda aynan nolga teng bo‟ladi (rasm 3 a) va b))

3 rasm Quydagi kechikish teoremasini isbot qilamiz.

Teorema. Agar

f _ t _

funksiyani tasviri F(p) bo‟lsa, u holda

f ( t  t ₀ )

funksiya tasviri

e ^{ pt0} F ( p )

bo‟ladi, ya‟ni:

f ( t ) 

F ( p )

bo'lsa

f (t  t₀ ) 

e ^{ pt0} F ( p )

Isbot. Tasvirni ta‟rifiga asosan.

^L  <sup>f ( t  t

)</sup> ^



0 0 0 0
_ e ^{ p t} f ( t  t
) d t 

t <sub>0

</sub> e ^{ p t} f ( t  t
) d t 



_ e ^{ p t} f ( t  t

) d t

0 0 t ₀

Tenglikni o‟ng tomonidagi birinchi integral nolga teng, chunki

t  t ₀

bo‟lganda

f ( t  t ₀ ) 

U holda

0. 
   . Keyingi integralda o‟zgaruvchini almashtramiz: ^{t  t 0 }
   

^z , d t  dz ;

 

^L  ^{f ( t  t} ₀

⁾ ^

_{ e}  p ( z  t₀ ) _{f ( z ) d z}

 e ^{ p t 0} _ e ^{ pz} f ( z ) d z

 e ^{ p t 0} F ( p ) .

demak
f (t  t₀ ) 

0 0
e ^{ pt0} F ( p ) .

Misollar. I. § 1.1 da Xevisaydning birlik funksiyasi

uchun

1

 ₀ ( t ) 

p

topilgan edi. Kechikish teoremasiga

asosan  ₀
( t  h )

funksiya (rasm 4) uchun  ₀

( t  h ) 

0. 
   _e  ph
   

p

boladi.

II. Quyidagi funksiyani qaraymiz:

1

_ 0 , t  0 b o ' lg a n d a

 ₁

 ( t , h )  _ ( t )  


( t  h ) _ 



, 0  t 

h b o ' lg a n d a

1 0 0

h



_ h

_ ^{0 ,}
h  t b o ' lg a n d a

Agar bu funksiya 0 dan h gacha bo‟lgan vaqt oralig‟ida ta‟sir etuvchi kuch deb qaralsa (boshqa vaqtlarda nolga teng), u holda bu kuch impulsi birga teng bo‟ladi.

(rasm 5)

Bu funksiya tasviri
1 1 1 1 1  e ^{ ph}

• 
  ph
  

 ₁ ( t , h )  (  e )  ( )

h p p p h
Mehanikada bazi qisqa vaqt oralig‟ida ta‟sir etuvchi oniy ta‟sir etuvchi va chekli

ipulsga ega kuch deb qarash qulay bo‟ladi. Shunga ko‟ra h  0

funksiyaning limitiga teng bo‟lgan  ( t ) funksiya qaraladi:

bo‟lganda  ₁ ( t , h )

 ( t ) 

lim  ₁ ( t , h )

h  0

Bu funksiya impulsli funksiya yoki delta funksiya deyiladi. Ba‟zan fizikada uni

Dirak funksiyasi ham deyiladi. tasvirining

 ( t ) funksiyani tasvirini

 ₁ ( t , h ) funksiya

h  0 dagi limiti deb topamiz :




1 1 1

 ( t , h )  _  ( t , h ) e ^{ p t} d t 

_e  p t _{ }

(1  e ^{ p h} )

1 1

₀ p h

p h p h

0

t  0

bo‟lganda ^{ 1}

^{( t , h )  0} . Shuning uchun
 ( t ) 
lim

h  0

¹ (1  e ^{ ph} )

ph

Lopital qoidasiga asosan:
lim

h  0

_{1  e}  p t

p h

 lim

h  0

_{p e}  p h

p

 lim e ^{ ph}  1 .

h  0

Demak  ( t )  1 . Delta funksiya mehanikada, matematikani ko‟pgina
bo‟limlarida, hususan matematik – fizikaning ko‟p masalalarida uchraydi. Agar

 ( t )

funksiya massasi 1 ga teng bo‟lgan moddiy nuqta

t  0

momentda 1 ga teng

tezlik beradigan kuch deb qarash mumkin.

Shunga o‟xshash

 ( t  t ₀ ) funksiyani

t  t ₀

momentida birlik massaga 1 ga

teng tezlik beruvchi kuch deb qarash mumkin. Kechikish teoremasiga asosan:

 (t  t₀ ) 

t <sub>0

e</sub>  pt

d ² x

Yuqoridagi kabi

_  ( t  t ₀ )d t

t

 1 . Endi

dt ²

 f ( t

)  t

differensial

tenglamaning
t  0

bo‟lganda

x ₀  0 ,

x ₀^  0

boshlang‟ich shartlarni

qanoatlantruvchi yechimni ko‟ramiz.

Yordamchi tenglama
p ² x ( p ) 

F ( p )  1 . Bundan
x ( p ) 

F ( p ) 1


va


_p 2 _p 2

t

x ( t )  _

0

f ( )( t   ) d  .

Delta funksiyani quyidagi xossalarini ko‟rib o‟tamiz:

   t  0

integral

bo‟lganda 0 ga 0  t   bo‟lganda 1 ga teng bo‟lgan quyidagi

t

_  ( ) d 

_ 0 ,

 _

   t  0

_{ }1, 0  t  

Xevisaydning birlik funksiyasini ^{ 0 ( t )} ga tengdir:

t

 ₀ ( t )  _


 ( ) d 

Bu tenglikni ikki tomonini differensiallab, quyidagi shartli tenglikni olamiz:

 ₀^ ( t )   ( t )

Bu tenglikni ma‟nosini tushuntirish uchun quyidagi funksiyani olamiz:

(1.2.5)

_ 0 ( a g a r t  0 b o 'ls a )



 ₀ ( t , h )  _ h

( a g a r 0  t  h

b o 'ls a )


^1 ( a g a r

t  h

b o 'ls a )

Bu holda:

bunda

 ₀^ (t , h )   ₁ (t , h )
(1.2.6)

lim  ₀ ( t , h )   ₀ ( t ) va

h  0

lim  ₀^ ( t , h )   ₀^ ( t )

h  0

(1.2.7)

(1.2.4) va (5 ) tenglika asosan (1.2.5) shartli tenglik kelib chiqadi:

 ₀^ ( t )   ( t ) .