Петухова Н.Ю. Методические указания по подготовке к экзамену по курсу уравнения математической физики
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r
ϕ (x, y) ∈ R
2
r =
=
p
x
2
+ y
2
x = r cos ϕ y = r sin ϕ
½
∆u = 0, r < R,
u|
r=R
= cos
4
ϕ;
½
∆u = 0, r > 2,
(u + 2u
r
)|
r=2
= ϕ
2
, |u||
r=∞
< ∞;
½
∆u = 0, 1 < r < 2,
u|
r=1
= sin
2
ϕ + cos
3
ϕ, u|
r=2
= 0;
½
∆u = 0, 2 < r < 4,
u|
r=2
= 0, u
r
|
r=4
= ϕ.
(x, y) ∈ R
2
r =
=
p
x
2
+ y
2
x = r cos ϕ y = r sin ϕ
½
∆u = 0, r < 2,
u|
r=2
= 3 + sin ϕ;
(
∆u = 0, r > 1,
|u(r, ϕ)||
r=∞
< ∞, u
r
|
r=1
= sin
³
ϕ +
π
4
´
;
½
∆u = 0, r < R,
(u + u
r
)|
r=R
= 2 cos ϕ;
½
∆u = 0, r > 2,
u|
r=2
= cos
2
ϕ, |u(r, ϕ)||
r=∞
< ∞;
(
∆u = 0, r < 10,
(u + 3u
r
)|
r=10
= cos
³
ϕ −
π
6
´
;
½
∆u = 0, r > 3,
u
r
|
r=3
= 2 − sin
2
ϕ, |u(r, ϕ)||
r=∞
< ∞.
t r ϕ
r
ϕ t
J
0
(x) t > 0 (x, y) ∈ R
2
r =
=
p
x
2
+ y
2
x = r cos ϕ y = r sin ϕ
½
u
tt
= a
2
∆u, t > 0, r < 1,
u|
r=1
= 0, u|
t=0
= r
2
sin ϕ, u
t
|
t=0
= 0;
(
4u
tt
= ∆u, t > 0, r < 1,
u|
r=1
= 0, u|
t=0
= J
0
³
µ
(0)
1
r
´
, u
t
|
t=0
= 0,
µ
(0)
1
J
0
(x)
(
u
t
= 4∆u, t > 0, r < 2,
u|
r=2
= 0, u|
t=0
= J
1
³
µ
(1)
2
r
2
´
sin ϕ,
µ
(1)
2
J
1
(x)
(
u
tt
= ∆u, t > 0, r < 2,
u|
r=2
= 0, u|
t=0
= J
0
³
µ
(0)
1
r
2
´
, u
t
|
t=0
J
2
³
µ
(2)
1
r
2
´
cos 2ϕ,
µ
(0)
1
J
0
(x) µ
(2)
1
J
2
(x)
u
K(x, y)
(x, y) ∈ R
2
K
∗
(x, y)
K(x, y)
K(x, y)
K(x, y) =
½
1) x − y
3
x; 2) xy − x
2
y
2
; 3) y cos x − sin y;
4) e
x+y
5) sin(x − y); 6) e
xy
; 7) cos(xy);
8)
1
x + y
; 9)
1
(x − y)
3
¾
.
K(x, y)
ϕ(x)
ϕ(x) = λ
R
1
−1
(xe
−x
sh y + sin xy
4
) cos yϕ(y) dy + f(x)
ϕ(x) = λ
R
2
0
(xy − e
x
sh y)ϕ(y) dy + x
3
ϕ(x) = λ
R
1
0
sin(x + 3y)ϕ(y) dy + cos x
ϕ(x) = λ
R
2
1
³
e
x−y
+ sin x
x
´
ϕ(y) dy + x
2
ϕ(x) = λ
R
1
−1
³
xy +
x + 3y
2
´
ϕ(y) dy + 2x
K(x, y) = 1 (x, y) ∈ [0; 1]
K
∗
(x, y)
f(x)
ϕ(x) = λ
R
2
0
(xy + e
x
sh y)ϕ(y) dy + f(x)
ϕ(x) = λ
R
1
0
cos(x + 3y)ϕ(y) dy + f(x)
ϕ(x) = λ
R
1
−1
e
x−y
ϕ(y) dy + f (x)
ϕ(x) = λ
R
−1
−2
³
cos(x − y) + sin x
x
´
ϕ(y) dy + f (x)
y(x) Ly =
= −y
00
h
0;
π
6
i
y(0) = 0
y
³
π
6
´
= 0
y(x) Ly =
= −y
00
[0; 1] y(0) = 0
y
0
(1) = 0
y(x) Ly =
= −y
00
[0; π] y
0
(0) = 0
y
0
(π) + y(π) = 0
y(x) Ly =
= −y
00
h
1
4
;
1
2
i
y
³
1
4
´
=
= y
0
³
1
2
´
= 0
y(x)
Ly = −y
00
h
π
2
; π
i
y
0
³
π
2
´
= y(π) = 0
y(x) Ly =
= −y
00
h
0;
π
4
i
y(0) = 0
y
³
π
4
´
− y
0
³
π
4
´
= 0
L
½
Ly = −y
00
, 0 < x < 1,
y(0) = y(1) = 0;
½
Ly = −y
00
, 0 < x < 2,
y(0) = y
0
(2) = 0;
½
Ly = −y
00
+ y, 0 < x < 1,
y(0) + y
0
(0) = 0, y(1) = 0;
½
Ly = −y
00
+ 4y, 0 < x < 1,
2y(0) − y
0
(0) = 0, y
0
(1) = 0.
½
−y
00
= λy, 0 < x < 2,
y(0) = y(2) = 0;
½
−y
00
+ xy = λy + x
3
, 1 < x < 2,
y(1) = 0, y
0
(2) = 0;
½
−x
2
y
00
− 2xy
0
= λy, 1 < x < 2,
y(1) + y
0
(1) = 0, y(2) = 0;
½
−e
x
y
00
− e
x
y
0
+ x
3
y = λy + sin x, 0 < x < 1
y(0) + 2y
0
(0) = 0, y
0
(1) = 0.
R
3
R
3
R
3
∆u
(r, ϕ, θ) (x, y, z) ∈ R
3
x = r cos ϕ sin θ y =
= r sin ϕ sin θ z = r cos θ r =
p
x
2
+ y
2
+ z
2
ϕ ∈ [0; 2π] θ ∈
∈ [0, π]
∆u = 0
u(r, ϕ, θ) =
= Z(r)Y (ϕ, θ)
Z(r)
Y (ϕ, θ)
Y (ϕ, θ) = Φ(ϕ)X(θ)
Φ(ϕ)
X(θ)
P
n
(ξ) n = 0
Z
1
−1
P
n
(ξ)P
m
(ξ) dξ, n 6= m.
P
n
(ξ) n = 0
P
(m)
n
(ξ) m = 0
n
∆u = 0
u(r, ϕ, θ) =
=
∞
P
n=0
Z
n
(r)Y
n
(ϕ, θ) Z
n
(r) Y
n
(ϕ, θ)
u(r, ϕ, θ)
½
∆u = 0, r < R,
u|
r=R
= f(ϕ, θ);
½
∆u = 0, r > R,
|u| |
r>∞
< ∞, u|
r=R
= f(ϕ, θ);
½
∆u = 0, R
1
< r < R
2
,
u|
r=R
1
= f(ϕ, θ), u|
r=R
2
= f(ϕ, θ).
(x, y, z) ∈ R
3
x =
= r cos ϕ sin θ y = r sin ϕ sin θ z = r cos θ r =
p
x
2
+ y
2
+ z
2
½
∆u = 0, r < R,
u|
r=R
= cos θ;
½
∆u = 0, r > R,
|u| →
r→∞
0, u|
r=R
= cos θ;
½
∆u = 0, r < R,
u|
r=R
= cos
2
θ;
½
∆u = 0, r > R,
|u| →
r→∞
0, u|
r=R
= cos
2
θ;
(
∆u = 0, r < R,
u|
r=R
= sin
³
ϕ +
π
4
´
sin θ;
(
∆u = 0, r > R,
|u| →
r→∞
0, u|
r=R
= sin
³
ϕ +
π
4
´
sin θ;
½
∆u = 0, r < 1,
(u + u
r
)|
r=1
= sin θ cos ϕ + sin
2
θ;
½
∆u = 0, r > 1,
|u| →
r→∞
0, (u − 2u
r
)|
r=R
= 1 + sin ϕ sin θ.
L
Lu = −
d
dx
µ
p(x)
du
dx
¶
+ ε(x)u, a < x < b,
p(x) ∈ C
2
(a; b); ε(x) ∈ C(a, b), p(x) > 0.
(1)
L
L
Lu = −u
00
0 < x < 1 u ∈
˚
C
1
[0; 1]
Lu = −
d
dx
(e
x
· u
0
(x)) −1 < x < 1 u ∈
˚
C
1
[−1; 1]
Lu = −
d
dx
(x
2
u
0
(x)) − u(x) 1 < x < 2 u ∈
˚
C
1
[1; 2]
Lu = −
d
dx
((1+x
2
)u
0
(x))+sin u(x) −1 < x < 1 u ∈
˚
C
2
[−1; 1]
u(x) ∈
˚
C
1
(a; b]
J(u) =
Z
b
a
(p(x)u
02
(x) + ε(x) · u) dx,
p(x) ∈ C
1
(a, b), ε(x) ∈ C(a, b), p(x) > 0.
u(x)
inf
u∈D
J(u)
J(u) =
R
1
0
u
02
(x) dx D = {u(x) ∈
˚
C
1
[0; 1] kuk = 1}
J(u) =
R
π
0
u
02
(x) dx D = {u(x) ∈
˚
C
1
[0; π] kuk = 1}
J(u) =
R
π
0
u
02
(x) dx D = {u(x) ∈ C
1
[0; π] u
0
(0) = u
0
(π) = 0}
J(u) =
R
1
0
u
02
(x) dx D = {u(x) ∈ C
1
[0; 1] u(0) = 0 u
0
(1) =
= 0}
u
0
(x) ∈ D
J(u) u
0
(x)
D = {u(x) ∈
˚
C
1
[0; 1]} J(u) =
R
1
0
(u
02
+ 2u) dx
D = {u(x) ∈
˚
C
1
[0, π]} J(u) =
R
π
0
(u
02
− 2u sin x) dx
D = {u(x) ∈ C
1
(0, π) u(0) = 0 = u
0
(π)} J (u) =
R
π
0
(u
02
−
− 2u cos x) dx
D = {u(x) ∈ C
1
(0, 1) u(0) = 0 = u
0
(1)} J(u) =
R
1
0
(u
02
+
+ 2xu) dx
u
0
(x) ∈ D
J(u) =
Z
Ω
(|∇u|
2
+ 2f(x)u) dx, x = (x
1
, x
2
, x
3
),
Ω R
3
D = {u(x) ∈ C
1
(Ω) u|
∂Ω
= ϕ}
f(x) ∈ C(Ω)
u
0
(x)
u
0
(x) ∈ D
J(u) x = (x
1
, x
2
) u
0
(x)
D = {u(x) ∈ C
1
(|x| 6 1) u|
|x|=1
= x
1
J(u) =
R
|x|<1
u
02
dx
D = {u(x) ∈ C
1
(|x| 6 1) u|
|x|=1
= x
2
1
} J(u) =
R
|x|<1
(u
02
+
+ 2u) dx
D = {u(x) ∈ C
1
(|x| 6 2) u|
|x|=2
= x
1
+x
2
J(u) =
R
|x|<2
u
02
dx
D = {u(x) ∈ C
1
(|x| 6 2) u|
|x|=2
= 0} J(u) =
R
|x|<2
(u
02
+
+ 2
p
x
2
1
+ x
2
2
u) dx
inf
u∈D
J(u)
J(u) =
R
π
0
u
02
dx D = {u(x) ∈
˚
C
1
(0, π] kuk
2
=
R
π
0
u
2
(x) dx =
= 1
R
π
0
u(x) sin(kx) dx = 0 k = 1 }
J(u) =
R
π
0
(u
02
+ u
2
)dx D = {u(x) ∈ C
1
(0, π) kuk
2
= 1
u
0
(0) = u
0
(π) = 1
R
π
0
u(x) cos kx dx = 0 k = 1 }
x ∈ R
n
D D
0
D
L(R
n
) L
0
(R
n
)
{f
k
(x)} f
k
(x) ∈ D
0
f(x) ∈ D
0
m m > 1 f
(m)
(x)
m
f(x) ∈ D
0
(R
1
) θ(x)
f(x) = θ(x) sin x m = 1 f(x) = θ(x) cos x m = 1
f(x) = x
2
θ(x) m = 1
f(x) = (1 − cos x)θ(x) m = 1
f(x) = sign x m = 1
f(x) = x sign x m = 1
f(x) = xθ(x − 1) m = 1
f(x) = θ(1 − |x|) m = 1
f(x) = xθ(x) m = 1
f(x) = (1 − cos x)
2
δ(x) m = 1
f(x) ∈ D
0
(R
1
)
f(x) = δ(x) f(x) = δ(x − x
0
)
f(x) = x
3
f(x) = θ(1 − |x|)
f(x) = 1 f(x) = cos x
Lu = f(x)
L Lu =
=
m
P
k=0
a
k
u
(k)
(x)
f(x, t) ∈ D
0
(R
2
)
ε(x, t)
ε(x, t)
f(x, t) ∈
∈ D
0
(R
n+1
) ε(x, t)