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Московский государственный индустриальный университет

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[НЕСОРТИРОВАННОЕ]

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Kurs_vysshei_matematiki_UP_Berkov_N.A._2007-2

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t\x

t = 0 t = 0, 20 t = 0, 40 t = 0, 60 t = 0, 80 t = 1, 00

2
0	t=0
	t=0,2
-2	t=0,4
	t=0,6
-4	t=0,8
	t=1,0

-6
-8
0	0,2	0,4	0,6	0,8	1

F unB(x, t) ∂U

∂x

F 9

←

m M hP r <>

<> <> <>

U U <0> U

(U<0>) m

(U<1>) m

(U<2>) m -2

(U<3>) m -4

(U<4>) m

-6

(U<5>) m

0,2

0,4 xm

0,6

0,8

L := 1 a := 1 T max := 1

h :=	L	τ := r ·		h 2
	M			a
τ P r := 0.2			:= 0.1
		hP r

r := 0.5 M := 20

N := T max

M P r :=	hP r	F unB(x, t) := 2 · et − x
	h

U :=

← x

f or

0..M

←

m h

←

ngr

←

0, M P r..M

f or ←m

Uk,0

k + 1

←

1..N

f or←n

←

−

F unB(x0, t)

ynp10

(y0

y1)

f or

1..M

−

←

F unB(xm, t)

←

ynp1m

ym +

(ym 1

ym)

ynp1

−

τ P r,

←

τ P r

ngr

t < 0.1

←

k + 1

f or

m 0, M P r..M

←

Uk,ngr

ngr

ngr + 1

M hP r =	M	m := 0..M hP r xm	:= m · hP r
	M P r

				u(x, y)
	∂2u				∂2u			∂	2u
A(x, y)			+ 2B(x, y)				+ C(x, y)			+
	∂2x				∂x∂y			∂	2y
+D(x, y)		∂u		+ E(x, y)	∂u	+ G(x, y) = 0.
		∂x
					∂y

x y ξ η

ξ = ϕ(x, y),

η= ψ(x, y).

x y

ϕ(x, y)

ψ(x, y)

∂u

∂2u

∂u ∂u ∂2u ∂2u ∂2u

∂x

∂y

∂x2

∂x∂y

∂y2

∂ξ ∂η

∂ξ2

∂ξ∂η

∂η2

∂u

∂ξ

∂u ∂η

;

∂u

∂ξ

∂u ∂η

;

∂x

∂ξ ∂x

∂y

∂η ∂y

∂η ∂x

∂ξ ∂y

∂2u

∂

∂u

∂

∂u

∂ξ

∂u ∂

∂ξ

∂x2

∂x

∂ξ

∂x

∂ξ

∂x

∂

∂u

∂η

∂u ∂

∂η

∂2u ∂ξ

∂x

∂η

∂x

∂η

∂x

∂ξ2

∂x

∂2u ∂η

∂ξ

∂u ∂2ξ

∂2u ∂η

∂ξ∂η

∂x

∂ξ

∂x2

∂η2

∂x

∂2u ∂ξ

∂η

∂u ∂2η

∂2u

∂ξ

∂ξ∂η

∂x

∂η

∂x2

∂2ξ

∂x

∂2u ∂ξ ∂η ∂2u

∂η

∂u ∂2ξ

∂u ∂2η

∂ξ∂η ∂x ∂x

∂2η

∂x

∂ξ

∂x2

∂η

∂x2

Uyy

Uxy

∂2u

∂ξ

∂2u ∂ξ ∂η ∂2u

∂η

∂u ∂2ξ

∂u ∂2η

Uyy =

+ 2

;

∂ξ2

∂y

∂ξ∂η ∂y ∂y

∂η2

∂y

∂ξ ∂y2

∂η ∂y2

Uxy

∂2u ∂ξ ∂ξ

∂2u

∂ξ ∂η

∂2u ∂η ∂η

∂ξ2 ∂x ∂y

∂ξ∂η

∂x

∂y

∂y ∂x

∂η2

∂x

∂y

∂u ∂2u

∂ξ ∂x∂y

∂η ∂x∂y

∂2u

∂ξ

∂ξ ∂ξ

∂ξ

+ 2B

+ C

∂ξ2

∂x

∂x ∂y

∂y

∂2u

∂ξ ∂η

+ B

+ C

∂ξ∂η

∂x ∂x

∂x

∂y

∂x

∂y

∂2u

∂η

∂η ∂η

∂η

+ 2B

+ C

∂η2

∂x

∂y

ξ, η, u,

∂u

= 0.

+F1

∂ξ

∂η

∂2u

ϕ(x, y)

ψ(x, y)

∂ξ2

∂η2

A(x, y)(zx)2 + 2B(x, y)zxzy + C(x, y)(zy)2 = 0.

z = ϕ(x, y)

G ϕ(x, y) = C

A(x, y)(dy)2 − 2B(x, y)dxdy + C(x, y)(dx)2 = 0.

ϕ(x, y) = C

= ϕ(x, y)

(dx)2

A(x, y)(

)2 − 2B(x, y)

+ C(x, y) = 0.

−

∂ϕ

−

∂x

∂y

∂ϕ

+ 2B

∂ϕ ∂ϕ

∂ϕ

∂x

+ C =

= 0,

∂ϕ

∂y

z = ϕ(x, y)

ϕ(x, y) = C

dy/dx

= B(x, y) + B2(x, y) − A(x, y)C(x, y) /A(x, y);

= B(x, y) −

B2(x, y) − A(x, y)C(x, y) /A(x, y).

ϕ(x, y) = C1

ψ(x, y) = C2

B2(x, y) −A(x, y)C(x, y) > 0 D

B2(x, y) −A(x, y)C(x, y) < 0 D

B2(x, y)−A(x, y)C(x, y) = 0 D

∂2u

2 ∂2u

∂u

= a

2 ∂2u

∂2u

= a

= 0.

∂x

∂t

∂x

∂y

∂t

∂2u

2 ∂2u

−

∂2u

= a

= 0

∂t2

∂x2

∂t2

A = a2 B = 0 C = −1 B2 − AC = a2 > 0

∂u

2 ∂

= a

2 ∂2u

∂u

∂t

∂x2

−

= 0

−AC = 0

A = a B = 0 C = 0 B

∂x2

∂t

∂2u

= 0

∂x2

∂y2

A = 1 B = 0 C = 1 B2 − AC = −1 < 0

A B C

ϕ(x, y) = C1

ψ(x, y) = C2

ξ = ϕ(x, y) η = ψ(x, y)

ϕ(x, y)

ψ(x, y)

∂2u ∂2u

∂ξ

∂ξ ∂ξ

+ 2B

∂ξ2

∂η2

∂x

∂x ∂y

∂ξ

∂η

∂η ∂η

∂η

= 0 A

+ 2B

+ C

= 0

∂y

∂x

∂x ∂y

∂y

∂2u

ξ, η, u,

∂u

= 0 A1

A1 + F1

∂ξ∂η

∂ξ

∂η

∂2u

∂ξ∂η

∂2u

ξ, η, u,

∂u

∂ξ∂η

∂ξ

∂η

u(ξ, η)

u(ϕ(x, y), ψ(x, y))

∂2u

2 ∂2u

= a

∂x

∂t

2 ∂2u

∂2u

−

= 0

A = a2 B = 0 C = −1

∂x2

∂t2

− 1 = 0

− 1 = 0 −a

− 1 = 0

− 1 = 0 a

= 1

at + C1 = x x − at = C1.

−a

− 1 = 0 −a

= 1

−at + C2 = x x + at = C2.

ξ = x − at

η = x + at

ξ η

ξ = x − at

∂ξ

= 1;

∂ξ

= −a;

∂x

∂t

η = x + at

∂η

= 1;

∂η

= a;

∂x

∂t

∂u

∂u ∂ξ

∂u ∂η

−a

∂u

+ a

;

∂t

∂ξ ∂t

∂η ∂t

∂ξ

∂η

∂u

∂u ∂ξ

∂u ∂η

∂u

;

∂x

∂ξ ∂x

∂ξ

∂η

∂η ∂x

∂2u

∂

∂u

∂

∂u

∂2u ∂ξ

∂2u ∂η

= −a

+ a

= −a

∂t2

∂t

∂ξ

∂t

∂η

∂ξ2

∂t

∂ξ∂η

∂t

∂2u ∂t ∂2u ∂η

∂2u

= −a −

a +

∂ξ∂η

∂η

∂η2

∂t

∂ξ2

∂ξ∂η

∂2u

+a −

a +

a = a2

−

;

∂ξ∂η

∂η2

∂ξ2

∂ξ∂η

∂η2

∂2u

∂ ∂u ∂u

∂2u ∂ξ

∂2u ∂η

∂2u ∂ξ

∂x2

∂x

∂ξ

∂η

∂ξ2 ∂x

∂ξ∂η

∂x

∂ξ∂η ∂x

∂2u ∂η

∂2u

+ 2

∂η2 ∂x

∂ξ2

∂ξ∂η

∂η2

∂2u

+ 2

− a2

− 2

= 0,

∂ξ2

∂ξ∂η

∂η2

∂ξ2

∂ξ∂η

∂η2

∂2u

= 0.

∂ξ∂η

ξ η

u(ξ, η) = F (ξ) + (η)

ξ = x − at, η = x + at

u(t, x) = F (x − at) + (x + at).

ϕ(x, y) = C

ξ = ϕ(x, y)

ϕx

ϕy

= 0

η = y

∂2u

ϕ(x, y)

∂ξ2

∂ξ

∂ξ ∂ξ

∂ξ

+ 2B

+ C

= 0.

∂x

∂x ∂y

∂y

√

B2 − AC = 0

B = AC

∂ξ

∂ξ ∂ξ

∂ξ

∂ξ ∂ξ

0 = A

+ 2B

+ C

= A

+ 2√A√C

∂x

∂x ∂y

∂y

∂x

∂x ∂y

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