Добавил:

Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Ивано-Франковский национальный технический университет нефти и газа

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

Гургула, Мойсишин "Розрахи з матана",

.pdf

Скачиваний:

1384

Добавлен:

17.02.2016

Размер:

2.45 Mб

Скачать

☆

<<< < Предыдущая 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 3738 / 4638 39 40 41 42 43 44 45 46 > Следующая >>>

371

8.9.	∂2u		= 9	u ,	0 ≤ r <17 , 0 < t < ∞ ,
	∂t	2

r 2

u(r , 0)=

1 −

= 0 ,

∂u(r , 0)

∂t

8.10.

∂2u

=10

u ,

0 ≤ r <16 ,

∂t

r 2

u(r , 0)=

−

= 0 ,

∂u(r , 0)

∂t

8.11.

∂2u

=11

u ,

0 ≤ r <15 ,

∂t

r 2

u(r , 0)=

−

= 0 ,

∂u(r , 0)

∂t

8.12.

∂2u

=12

u ,

0 ≤ r <14 ,

∂t

r 2

u(r , 0)=

−

= 0 ,

∂u(r , 0)

∂t

8.13.

∂2u

=13

u ,

0 ≤ r <13 ,

∂t

u(17, t)= 0 .

0 < t < ∞ ,

u(16, t)= 0 .

0 < t < ∞ ,

u(15, t)= 0 .

0 < t < ∞ ,

u(14, t)= 0 .

0 < t < ∞ ,

372

u(r , 0)=

−

∂u(r , 0)

= 0 ,

u(13, t)= 0 .

∂t

8.14.

∂2u

=14

u ,

0 ≤ r <12 ,

0 < t < ∞ ,

∂t2

u(r , 0)=

−

∂u(r , 0)

= 0 ,

u(12, t)= 0 .

∂t

8.15.

∂2u

=15

u ,

0 ≤ r <11 ,

0 < t < ∞ ,

∂t2

r 2

u(r , 0)=

−

∂u(r , 0)

= 0 ,

u(11, t)= 0 .

∂t

8.16.

∂2u

=16

u ,

0 ≤ r <10 ,

0 < t < ∞ ,

∂t2

r 2

u(r , 0)=

−

∂u(r , 0)

= 0 ,

u(10, t)= 0 .

∂t

8.17.

∂2u

=17

u ,

0 ≤ r < 9 ,

0 < t < ∞ ,

∂t2

r 2

u(r , 0)=

−

∂u(r , 0)

= 0 ,

u(9, t)= 0 .

∂t

373

8.18.	∂2u		=18	u ,		0 ≤ r < 8 ,
8.18.	∂t	2	=18	u ,		0 ≤ r < 8 ,
	∂t	2
				1		r 2
	u(r , 0)=				1	−		,
	u(r , 0)=			8	1	−		,
				8		8

				=	0 ,
	∂u(r , 0)			=	0 ,
			∂t
8.19.	∂2u		=19	u ,		0 ≤ r < 7 ,
8.19.	∂t	2	=19	u ,		0 ≤ r < 7 ,
	∂t	2
				1		r 2
	u(r , 0)=				1	−		,
	u(r , 0)=			8	1	−		,
				8		7

				= 0 ,
	∂u(r , 0)			= 0 ,
			∂t
8.20.	∂2u		= 20	u ,			0 ≤ r < 6 ,
8.20.	∂t	2	= 20	u ,			0 ≤ r < 6 ,
	∂t	2
				1		r 2
	u(r , 0)=				1	−		,
	u(r , 0)=			8	1	−		,
				8		6

				=	0 ,
	∂u(r , 0)			=	0 ,
			∂t
8.21.	∂2u		= 21	u ,		0 ≤ r < 5 ,
8.21.	∂t	2	= 21	u ,		0 ≤ r < 5 ,
	∂t	2
				1		r 2
	u(r , 0)=				1	−		,
	u(r , 0)=			8	1	−		,
				8		5

				= 0 ,
	∂u(r , 0)			= 0 ,
			∂t
8.22.	∂2u		= 22	u ,			0 ≤ r < 4 ,
8.22.	∂t	2	= 22	u ,			0 ≤ r < 4 ,
	∂t	2

0 < t < ∞ ,

u(8, t)= 0 .

0 < t < ∞ ,

u(7, t)= 0 .

0 < t < ∞ ,

u(6, t)= 0 .

0 < t < ∞ ,

u(5, t)= 0 .

0 < t < ∞ ,

374

r 2

u(r , 0)=

−

0 ,

u(4, t)= 0 .

∂u(r , 0)

∂t

8.23.

∂2u

= 23

u ,

0 ≤ r < 3 ,

0 < t < ∞ ,

∂t

r 2

u(r , 0)=

−

= 0 ,

u(3, t)= 0 .

∂u(r , 0)

∂t

8.24.

∂2u

= 24

u ,

0 ≤ r < 2 ,

0 < t < ∞ ,

∂t

r 2

u(r , 0)=

−

0 ,

u(2, t)= 0 .

∂u(r , 0)

∂t

8.25.

∂2u

= 25

u ,

0 ≤ r <1 ,

0 < t < ∞ ,

∂t

[1 − r2 ],

u(r , 0)=

∂u(r , 0)

u(1, t)= 0 .

∂t

= 0 ,

8.26.

∂2u

= u ,

0 ≤ r < 6 ,

0 < t < ∞ ,

∂t

r 2

u(r , 0)=

1 −

= 0 ,

u(6, t)= 0 .

∂u(r , 0)

∂t

375

8.27.	∂2u			= 4 u ,				0 ≤ r < 5 ,
8.27.	∂t		2	= 4 u ,				0 ≤ r < 5 ,
	∂t		2
					1			r			2
	u(r , 0)=						1	−				,
	u(r , 0)=				8		1	−				,
					8			5
		∂u(r , 0)
					=		0 ,
					=		0 ,
				∂t
8.28.	∂2u			= 9 u ,				0 ≤ r < 4 ,
8.28.	∂t		2	= 9 u ,				0 ≤ r < 4 ,
	∂t		2
					1			r			2
	u(r , 0)=						1	−				,
	u(r , 0)=				8		1	−				,
					8			4
		∂u(r , 0)
					= 0 ,
					= 0 ,
				∂t
8.29.	∂2u			=16	u ,				0 ≤ r < 3 ,
8.29.	∂t		2	=16	u ,				0 ≤ r < 3 ,
	∂t		2
					1			r			2
	u(r , 0)=						1	−				,
	u(r , 0)=				8		1	−				,
					8			3
		∂u(r , 0)
					=		0 ,
					=		0 ,
				∂t
8.30.	∂2u			= 25		u ,				0 ≤ r < 2 ,
8.30.	∂t		2	= 25		u ,				0 ≤ r < 2 ,
	∂t		2
					1			r			2
	u(r , 0)=						1	−				,
	u(r , 0)=				8		1	−				,
					8			2
		∂u(r , 0)
					= 0 ,
					= 0 ,
				∂t

0 < t < ∞ ,

u(5, t)= 0 .

0 < t < ∞ ,

u(4, t)= 0 .

0 < t < ∞ ,

u(3, t)= 0 .

0 < t < ∞ ,

u(2, t)= 0 .

Задача 9. Сформулювати і розв’язати задачу математичної фізики.

9.1.	∂u	=16	∂2u	, 0	< x < 3 ,	t > 0 ,
	∂t		∂x2

376

0 ≤ x ≤

u(0, t)=

0 ,

u(x, 0)=

0 .

u(3, t)=

3 − x ,

< x

≤ 3 ,

9.2.

∂u

∂2u

0 < x < 2 ,

t > 0 ,

∂t

∂x2

u(0, t)= 0 ,

u(x, 0)=

0 ≤ x ≤ 1 ,

2 − x , 1 < x ≤ 2 ,

u(2, t)= 0 .

9.3.

∂u

∂2u

0 < x < 5 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)=

u(5, t)= 0 .

5 − x ,

< x ≤ 5 ,

9.4.

∂u

∂2u

0 < x < 4 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 2 ,

u(0, t)= 0 ,

u(x, 0)=

u(4, t)= 0 .

4 − x , 2 < x ≤ 4 ,

9.5.

∂u

∂2u

0 < x < 5 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)=

u(5, t)= 0 .

5 − x ,

< x

≤

5 ,

9.6.

∂u

∂2u

0 < x < 3 ,

t > 0 ,

∂t

∂x2

377

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)

u(3, t)= 0 .

− x

< x ≤ 3 ,

9.7.

∂u

= 25

∂2u

0 < x < 8 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 4 ,

u(0, t)

= 0 ,

u(x, 0)

= 0 .

u(8, t)

− x

, 4 < x ≤ 8 ,

9.8.

∂u

= 9

∂2u

0 < x < 2 ,

t > 0 ,

∂t

∂x2

u(0, t)

u(x, 0)

x2 ,

0 ≤ x ≤ 1 ,

= 0 ,

− x

, 1 < x ≤ 2 ,

= 0 .

u(2, t)

9.9.

∂u

=16

∂2u

0 < x <1 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)

1 − x

< x ≤ 1 ,

u(1, t)= 0 .

9.10.

∂u

= 4

∂2u

0 < x < 4 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤

2 ,

u(0, t)= 0 ,

u(x, 0)=

u(4, t)= 0 .

4 ,

4 − x , 2 < x ≤

9.11.

∂u

= 9

∂2u

0 < x <10 , t > 0 ,

∂t

∂x2

0 ≤ x ≤

5 ,

u(0, t)= 0 ,

u(x, 0)=

u(10, t)= 0 .

10 ,

10 − x , 5 < x ≤

378
9.12.	∂u	=	25		∂2u		,							0 < x < 9 ,											t > 0 ,
9.12.	∂t	=	25		∂x2		,							0 < x < 9 ,											t > 0 ,
	∂t				∂x2
							2x		2				,			0 ≤ x ≤						9			,	u(0, t)= 0 ,
							2x															9

	u(x, 0)=						9												2
	u(x, 0)=						9									9			2
						9 − x ,										9	< x ≤ 9 ,									u(9, t)= 0 .
						9 − x ,											< x ≤ 9 ,
														2
														2
9.13.	∂u	=	9	∂2u			,			0				< x < 3 ,								t > 0 ,
9.13.	∂t	=	9	∂x2			,			0				< x < 3 ,								t > 0 ,
	∂t			∂x2
							2x		2			,			0 ≤ x ≤						3			,		u(0, t)= 0 ,
							2x														3

	u(x, 0)=						3												2
	u(x, 0)=						3							3					2
														3												u(3, t)= 0 .
						3 − x ,											< x ≤ 3 ,
						3 − x ,											< x ≤ 3 ,
														2
														2
9.14.	∂u	=	∂2u			,				0 < x < 5 ,								t > 0 ,
9.14.	∂t	=	∂x2			,				0 < x < 5 ,								t > 0 ,
	∂t		∂x2
							2x		2		,				0 ≤ x ≤					5			,			u(0, t)= 0 ,
							2x													5

	u(x, 0)=						5												2
	u(x, 0)=						5								5				2
						5 − x ,									5		< x ≤ 5 ,									u(5, t)= 0 .
						5 − x ,											< x ≤ 5 ,
														2
														2
9.15.	∂u	=	4	∂2u			,						0 < x < 7 ,									t > 0 ,
9.15.	∂t	=	4	∂x2			,						0 < x < 7 ,									t > 0 ,
	∂t			∂x2
							2x		2				,			0 ≤ x ≤						7			,	u(0, t)= 0 ,
							2x															7

	u(x, 0)=						7												2
	u(x, 0)=						7									7			2
						7 − x ,										7	< x ≤ 7 ,									u(7 , t)= 0 .
						7 − x ,											< x ≤ 7 ,
														2
														2
9.16.	∂u	=	25		∂2u		,							0 < x <1 ,										t > 0 ,
9.16.	∂t	=	25		∂x2		,							0 < x <1 ,										t > 0 ,
	∂t				∂x2														1
								2											1
						2x				,				0 ≤ x ≤										,		u(0, t)= 0 ,
	u(x, 0)=					2x				,				0 ≤ x ≤					2					,		u(0, t)= 0 ,
	u(x, 0)=													1					2
														1												u(1, t)= 0 .
						1 − x ,											< x	≤1 ,
						1 − x ,											< x	≤1 ,
														2
														2

379

9.17.	∂u	=	9	∂2u		,			0 < x < 4 ,	t > 0 ,
9.17.	∂t	=	9	∂x2		,			0 < x < 4 ,	t > 0 ,
	∂t			∂x2
						x	2	,	0 ≤ x ≤	2 ,	u(0, t)= 0 ,
	u(x, 0)=					x
						2
						2
											u(4, t)= 0 .
									, 2 < x ≤	4 ,
					4 − x				, 2 < x ≤	4 ,
9.18.	∂u	=	∂2u		,			0	< x <10 ,	t > 0 ,
9.18.	∂t	=	∂x2		,			0	< x <10 ,	t > 0 ,
	∂t		∂x2

				2
	u(x, 0)=			x	,	0 ≤ x ≤ 5 ,
	u(x, 0)=			5	,	0 ≤ x ≤ 5 ,

			10 − x , 5 < x ≤ 10 ,
9.19.	∂u	= 4	∂2u	,		0 < x < 2 , t >
9.19.	∂t	= 4	∂x2	,		0 < x < 2 , t >
	∂t		∂x2
	u(x, 0)= x2 ,					0 ≤ x ≤1 ,
			2 − x , 1 < x ≤ 2 ,

u(0, t)= 0 ,u(10, t)= 0 .

0 ,

u(0, t)= 0 ,u(2, t)= 0 .

9.20.

∂u

=16

∂2u

0 < x < 8 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 4 ,

u(0, t)= 0 ,

u(x, 0)

4 ,

u(8, t)= 0 .

− x

, 4 < x ≤ 8 ,

9.21.

∂u

∂2u

< x <1 , t > 0 ,

∂t

∂x2

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)

u(1, t)= 0 .

1 − x ,

< x ≤1 ,

9.22.

∂u

= 25

∂2u

0 < x < 4 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 2 ,

u(0, t)= 0 ,

u(x, 0)

2 ,

u(4, t)= 0 .

− x

, 2 < x ≤ 4 ,

380

9.23.

∂u

∂2u

0 < x < 6 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 3 ,

u(0, t)

= 0 ,

u(x, 0)=

= 0 .

u(6, t)

, 3 < x ≤ 6 ,

6 − x

9.24.

∂u

∂2u

0 < x <1 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)

u(1, t)= 0 .

1 − x ,

< x ≤1 ,

9.25.

∂u

∂2u

0 < x < 5 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤

u(0, t)= 0 ,

u(x, 0)

5 − x

< x ≤ 5 ,

u(5, t)= 0 .

9.26.

∂u

∂2u

0 < x < 6 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 3 ,

u(0, t)

= 0 ,

u(x, 0)=

= 0 .

u(6, t)

, 3 < x ≤ 6 ,

6 − x

9.27.

∂u

∂2u

< x <12 ,

t > 0 ,

∂t

∂x2

0 ≤ x ≤ 6 ,

u(0, t)= 0 ,

u(x, 0)=

u(12, t)= 0 .

12 − x , 6 < x ≤12 ,

9.28.

∂u

∂2u

0 < x < 2 ,

t > 0 ,

∂t

∂x2

<<< < Предыдущая 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 3738 / 4638 39 40 41 42 43 44 45 46 > Следующая >>>

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]

#
16.07.2019179.2 Кб8ГПЗ_Технологічний процес.doc
#
17.02.20162.3 Mб38граві-магніто розв.посібник.doc
#
01.05.2025609.28 Кб0Графи.doc
#
01.04.202549.84 Кб0Гроші та кредит.docx
#
14.11.20181.07 Mб25грунтознавство (Дутчин М.М.) 310111.doc
#
17.02.20162.45 Mб1384Гургула, Мойсишин "Розрахи з матана",.pdf
#
01.05.202543.89 Кб0дієслово.docx
#
07.12.20181.13 Mб3дільниця (1).doc
#
13.11.2019445.95 Кб38движок_60.doc
#
12.08.2019166.4 Кб1Дейдей.doc
#
12.11.2019103.94 Кб1Декларація про валютні цінності, доходи та майн...doc