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Московский государственный индустриальный университет

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Kurs_vysshei_matematiki_UP_Berkov_N.A._2007-2

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x = 0 x = l

u(x, t)|x=0 = ψ1(t),

u(x, t)|x=l = ψ2(t).

u(x, t)|x=0 = 0,

u(x, t)|x=l = 0.

		dQ\|x=0 = dQ\|x=l = 0.
	∂u		∂u			∂u

dQ = −k	∂n	dS · dt,	∂n	x=0	=	∂n	x=l	= 0.

x
∂u				∂u
∂u				∂u

∂u		= −	∂u			= 0;
∂n	x=0	= −		∂x	x=0	= 0;


n					n
n					n
∂n	x=l	=	∂x x=l	= 0
0				l	x

∂u			∂u

∂x	x=0	= 0,	∂x	x=l	= 0.

	∂u				∂u
k	∂x		= q1(t),	qi > 0			= q2(t).
k	∂x	x=0	= q1(t),	−k	∂x	x=l	= q2(t).

qi < 0

α(u − u)|x=0 = q1(t), α(u − u)|x=l = q2(t),

ucp

∂u

= 0, u|x=l = 0. ∂x x=0

		u = u(x, y, t)				σ
L
L
			S			V
u						t
	∂u/∂t = 0
	∂2u		∂2u		∂2u
u =		+		+		= 0.
u =	∂x2	+	∂y2	+	∂z2	= 0.

u(x, y, z) S = ψ(x, y, z),

ψ(x, y, z)

∂c∂t = D c, c = c(x, y, z)

a20 + ∞ (an cos nx + bn sin nx)

n=1

a0, a1, b1, · · · , an, bn, · · ·

2π

[−π, π] (−∞, ∞)

f (x) = a20 + ∞ (an cos nx + bn sin nx)

n=1

2π

sin x sin y = 12 (cos(x − y) − cos(x + y)) ; cos x cos y = 12 (cos(x − y) + cos(x + y)) ; sin x cos y = 12 (sin(x − y) + sin(x + y)) ;

sin2 x = 12 (1 − cos 2x); cos2 x = 12 (1 + cos 2x); sin x cos x = 12 sin 2x.

n k n = k

π				π
π		π
cos nx cos kx dx = 0,		sin nx sin kx dx = 0,
−π		−π
−πsin nx cos kx dx = 0,		π		−πcos nx dx = 0,
−πsin nx cos kx dx = 0,		−πsin nx dx = 0,		−πcos nx dx = 0,
n = k
π	π			π
−πcos2 kx dx = π,	−πsin2 kx dx = π,		−πsin kx cos kx dx = 0.

[−π, π]

π	π
f (x) dx =		a0
f (x) dx =		2
−π	−π

		an	−π	π	πsin nx dx .
dx +	∞	an	πcos nx dx + bn		πsin nx dx .
			−π	−π
	n=1

a0 = 1 πf (x) dx. π −π

f (x) cos

−π ∞

+ an

π a0 dx = a0π

−π 2

			cos kx	k N
	a0	π
kx dx =	a0	cos kx dx+
kx dx =		cos kx dx+
π	2	−π	π
cos nx cos kx dx + bn			sin nx cos kx dx .

n=1	−π	−π

n = k

ak cos2 kx dx = akπ.

−π

f (x) cos kx dx = akπ.

		−π
	1 π
ak =		−πf (x) cos kx dx.
	π

		−π π	sin kx
		−π π	bk
			bk
		π
bk =	1	f (x) sin kx dx.

	π
		−π
		f (x)	2π
			[−π, π]

		π			π			π
a0 =	1	f (x) dx,	an =	1	f (x) cos nx dx,	bn =	1	f (x) sin nx dx.

	π			π			π
		−π			−π			−π

f (x) [a, b]

f (x)

[a, b]

f (x)

2π

f (x)

S(x)

x0 x → x0

S(x0) =	1	lim	lim


	2 x→x0− f (x) + x→x0+ f (x) .

x (−π, π] f (x) = x

(−π, π]	x
f1(x) = f x − Int		2π	Int(x)
	2π

f1(x)

(−π, π]

1 x2

x dx =

−

= 0;

π 2

−π

x cos nx dx =

an =

xd sin nx =

x sin nx

−π−

−π

πn

πn −π

nπ

− − 1 π

π−π πn

− − π

πsin nx dx

cos nx

cos nπ

cos(

nπ)

= 0;

x sin nx dx =

xd cos nx =

x cos nx

−nπ

−π

−

− −π

π −π

−πn −π

−πn

− − n

−π

cos nx dx

π cos nπ + π cos(

nπ)

sin nx

−

−n

cos nπ =

−

a = a =

· · ·

= a =

· · ·

= 0; b

2(−1)n−1

−

− n −

f1(x) = 2

+ · · · +

sin nx + . . . .

sin x

sin 2x

sin 3x

sin 4x

(

1)n

f1(x) = 2

∞

(−1)n−1 sin nx .

n=1

ϕ(x) ϕ(−x) = ϕ(x)

−π		−ππ		0 π			0
π			0		π			π		π
	ϕ(x) dx = ϕ(x) dx + ϕ(x) dx = ϕ(−x) dx + ϕ(x) dx =
π					0					0
= 0	ϕ(x) dx + 0		ϕ(x) dx = 2		0	ϕ(x) dx.
−πϕ(xπ		−π π		0		ϕ(−x) = −ϕ(x)				0
−πϕ(xπ		−π π		0			0		−	0
π		0		π				π		π
	) dx = ϕ(x) dx + ϕ(x) dx = ϕ( x) dx + ϕ(x) dx =
= − 0 ϕ(x) dx + 0 ϕ(x) dx = 0.										f (x)
										f (x)
										a0
ak								π	bk
ak									bk

	a0 = 0,		an = 0,		bn =		2	0 f (x) sin nx dx.

							π