Problem Background

Birinchi tartibli Differensal tenglama

Theorem 3.1 Boshlang’ich shartli (3.3) masala oraliqda yagona x(t) yechimga ega bo’lib, u quyidagi ko’rinishda aniqlanadi

(3.6)

Bu erda ushbu rekurent formula orqali aniqlanadi:

	(3.7)
,	(3.8)

Isbot. Let then for . Therefore equation (3.1) has a form

,

(3.9)

where

.

(3.9) tenglamadagi o`zgaruvchilarni ajratish boshlang`ich qiymat muammosiga olib keladi.

,
,
.

Ikkala tomonni integrallash orqali

,

we have

,
,
, .

Shuning uchun (3.9) tenglamaning yechimi

for .

(3.10)

Chunki x(t) uzliksiz funksiya

,

where

,
,
,
.

Shuningdek,

,
,	(3.11)

or

.

(3.12)

Using continuity condition x(t) at t = 2, by similar way, we have

,
,
.	(3.13)

or

,
,
.	(3.14)

If we let for n = k – 1, then

,
,
,	(3.15)

or

.

(3.16)

Then from

,
we get
,
,	(3.17)
Va
,
,
,
,	(3.18)
.	(3.19)
Bu esa isbotni tuldiradi.

6. Tenglama davriy yichimi haqida

Teorema 3.2. (3.1) tenglamaning yechimi funksiya n-davrli bo’lishi uchun tenglamaning yechimi ushbu

tenglik bajarilishi zarur va yetarli.


Isbot. Agar , u holda
3.1 teoremaga ko`ra

,

Bu yerda c_k quyidagicha aniqlanadi.

.

Keyin x(k) sifatida ifodalanishi mumkin.

.

(3.20)

Let x(t) be n-periodic function, i.e.

for .

This gives

as t = 0.

(3.21)

According to the equation (3.20) we have

.

Soddalashtiring

.

Har ikkala tomonning natural logarifmlarini olamiz.

.

Shu yerdan olamiz.

.

(3.22)

Aksincha (3.20) tenglama bajarilsin, ya`ni

.

x(t) tenglamaning yechimi bo`lsin.

keyin

u holda .

(3.22) tenglamaga muvofiq oxirgi tenglama teng bo`ladi.

u holda .

Bu beradi

,

keyin

,
for ,

Ya`ni

.

Bu degani

for , r = t – n,

Shunga o`xshash tarzda biz isbotlashimiz mumkin.

for .

bunda
,
keyin
,
va
,
, for .
Bizda mavjud bo`lgan oralig`ida
va ,
Buning yechimi
,
, for .
Bizda mavjud bo`lgan oralig`ida
va ,
Buning yechimi
,
, uchun .
Bu dalilni tuldiradi.