KHÓA LUẬN TỐT NGHIỆP TOÁN: KHÔNG GIAN SOBOLEV PHỤ THUỘC V€ÀO THỜI GIAN - Pdf 14

class="bi x0 y0 w1 h1"

class="bi x1 y1 w2 h2"
C
k
(Ω)
L
p
(Ω)
W
k
p
(Ω)
W
k
p
(Ω), (1 ≤ p < ∞), (k ∈ Z
+
.)

W
k
p
(Ω), (1 ≤ p < ∞)
H
−1
(Ω)
L
p
(0, T ; X)
C([0, T ]; X)





d(x + y, y + z) = d(x, y)
d(λx, λy) = |λ|d(x, y)
∀x, y, z ∈ E, ∀λ ∈ K.
d (1.1)
ρ.
E
d(x, y) := x − y, x, y ∈ E
{u
k
}

k=1
⊂ E
E
ǫ > 0, ∃N > 0 u
k
− u
l
 < ǫ, k, l ≥ N.
E E
{u
k
}

k=1
⊂ E


k=1
⊂ E
u ∈ E
lim
k→∞
u
k
− u = 0
u
k
→ u k → ∞.
E E
ϕ : E × E −→ C
ϕ(x
1
+ x
2
, y) = ϕ(x
1
, y) + ϕ(x
2
, y), ∀x
1
, x
2
, y ∈ E
ϕ(λx, y) = λϕ(x, y), ∀x, y ∈ E; ∀λ ∈ C
ϕ(x, y) = ϕ(x, y), ∀x, y ∈ E
ϕ E ϕ ≥ 0

c
|b|
2
≤ a.c
a = c = 0 λ = −b, a = c = 0 (1.2) −2|b|
2
≥ 0
b = 0 |b|
2
≤ a.c.
|b|
2
≤ a.c.
ϕ
E

ϕ(x + y, x + y) ≤

ϕ(x, x) +

ϕ(y, y) x, y ∈ E
ϕ(x + y, x + y) = ϕ(x, x) + ϕ(x, y) + ϕ(x, y) + ϕ(y, y)
= ϕ(x, x) + 2
ϕ(x, y) + ϕ(y, y)
ϕ(x, y) ≤ |ϕ(x, y)| ≤

ϕ(x, x)ϕ(y, y).
ϕ(x + y, x + y) ≤ ϕ(x, x) + 2

ϕ(x, x)ϕ(y, y) + ϕ(y, y)

k
}

k=1
⊂ E
u ∈ E < u

, u
k
>−→< u

, u >
u

∈ E

u
k
⇀ u.
E

E,
E.
u
k
→ u u
k
⇀ u.
u
k

C
k
(Ω) Ω
C

(Ω)
C
c
(Ω) Ω.
Ω R
n
u ∈ C

(Ω) {x ∈ Ω |u(x) = 0}
u supp u supp u u(x)

C
(Ω)
C(Ω)


C
k
(Ω) = C
k
(Ω) ∩

C
(Ω)


(Ω) 1 ≤ p < +∞
f
p
=



|f|
p


1
p
Ω R
n
C
c
(Ω)
L
p
(Ω) p > 1
p, q > 1
1
p
+
1
q
= 1
α, β ∈ R
+

(t) > 0
(1; +∞) f f(1) =
1
p
+
1
q
= 1.
t
p
p
+
t
−q
q
≥ 1
t > 0
t = α
1
q

−1
p
α
p
q

−1
p
+

p
(Ω), g ∈ L
q
(Ω),


|f.g|dµ ≤



|f|
p


1
p



|g|
q


1
q
fg
1
≤ f
p
g

p
+
1
q
|g(x)|
q
g
q
q
2 µ
1
f
p
g
q


|f(x)g(x)|dµ ≤
1
p f
p
p


|f(x)|
p
dµ+
1
q g
q

=
1
p
+
1
q
= 1
fg
1
≤ f
p
g
q
L
p
(Ω) p > 1
p ≥ 1 Ω R
n
L
p
(Ω)
L
p
(Ω)
R x ∈ R
n
Q(x, R)
Q(x, R) =

y ∈ R

R
Q(0, R + 1) Q(0, R)
∃δ > 0
|g
R
(x) − g
R
(y)| < εR
−n
p
; x, y ∈ Q(0, R), |x − y| < δ
δ = R

n2

N N δ
Q(0, R) R2

N
S X
j
(x)
N
h(x) =

j
g
R
(x
j

\Q(0, R)


R
n
|f(x) −h(x)|
p
dx

1
p



Q(0,R)
|f(x) −h(x)|
p
dx

1
p
+


R
n
\Q(0,R )
|f(x)|
p
dx

R
n
\Q(0,R)
|f(x)|
p
dx

1
p



Q(0,R+1)
|f(x) − g
R
(x)|
p
dx

1
p
+


Q(0,R)
|g
R
(x) − h(x)|
p
dx

p
(Ω), p ≥ 1, f(x) = 0
Ω ǫ > 0 δ > 0


|f(x) −f(x + y)|
p
dx < ε
y |y| < δ
Ω R
n
x
0
, x ∈ Ω, x
0
x Ω.
L
p
(ω).
Ω f ∈
L
p
(Ω), p ≥ 1, f(x) = 0
Ω. ǫ > 0,


|f(x) −f(λx)|
p
dx < ǫ,
|λ − 1| < δ.

u
u ∈ L
p
(Ω), p ≥ 1 lim
h→0
u
h
− u
L
p
(Ω)
= 0
u(x) = 0 x ∈ R
n
\Ω
u
h
(x) = h
−n


θ(
x − y
h
)u(y)dy =

R
n
θ(z)u(x + hz)dz
u

p
dz
h → 0
f, g ∈ L
1
(Ω)


f
h
(x)g(x)dx =


f(x)g
h
(x)dx.


f
h
(x)g(x)dx = h
−n





θ

x − h


C

(Ω)
f = 0.
W
k
p
(Ω)
u, v ∈ L
1
loc
(U)
α
v α u

U
uD
α
φdx = (−1)
α

U
vφdx
φ ∈ C

c
(U).
D
α

α
n
α
u
v
1
, v
2
∈ L
1
loc
(U)
u
v
1
= v
2
v
1
, v
2
∈ L
1
loc
(U)
u

U
uD
α

2
= 0
n = 1, U = (0, 2) u(x), v(x)
u(x) =





x
0 < x ≤ 1
1 1 < x < 2
v(x) =





1 0 < x ≤ 1
0
1 < x < 2
u

= v v u
φ ∈ C

c
(U)
2


φdx + φ(1) − φ(1) = −
2

0
vφdx.
n = 1, U = (0, 2)
u(x) =





x
0 < x ≤ 1
2
1 < x < 2
u

v ∈ L
1
loc
(U)
2

0


dx = −
2


= −
1

0
φdx − φ(1)

m
}

m=1
0 ≤ φ
m
≤ 1, φ
m
(1) = 1, φ
m
(x) → 0, x = 1.
φ φ
m
(1.4) m → ∞
1 = lim
x→∞
φ
m
(1) = lim
x→∞

2

0

C

(Ω) ψ
v(x)
u
1
(x) u
2
(x) v(x)


(u
1
(x) − u
2
(x))ψ(x)dx = 0; ∀ψ(x) ∈

C

(Ω)
u
1
(x) − u
2
(x) ∈ L
1,loc
(Ω) u
1
(x) − u
2

α
α Ω
α Ω

⊂ Ω f ∈ L
1
(Ω)
v ∈

C

(Ω)



f

α
v

α
1
x
1

α
n
x
n
dx =

|α|


ωvdx = −1
|α|



ωvdx
ω ∈ L
1
(Ω) ω ∈ L
1
(Ω

)



f

α
v

α
1
x
1

α



D
α
v
α
f(x) = |x| (−1; 1)
∀x = 0 x = 0
f

(0
+
) = 1, f

(0
+
) = −1 f(x) = |x|
1

−1
|x|
dv
dx
dx = −
1

−1
ωvdx; ∀v ∈

C

dx +
1

0
dx = 2
1

−1
|x|
∂v
∂x
dx =
0

−1
|x|
∂v
∂x
dx +
1

0
|x|
∂v
∂x
dx
1

−1
|x|

1

0
1vdx)
= −
1

−1
ωvdx
f(x) = |x| (−1; 1)
(−1; 1)
U ∈ R
2
f(x) = f
1
(x) + f
2
(x) f
1
, f
2
ϕ ∈ C

0
(U)

U
f
1
(x)

dxdy =
b

a
f
2
(x)
∂ϕ
∂x




y=β(x)
y=α(x)
dx = 0

U
f(x)

2
ϕ
∂x∂y
dxdy =

U
f
1
(x)




x
x ≥ 0
0
x < 0
u(x) /∈ C
1
(R)
u

(0
+
) = lim
x→0
+
u(x) − u(0)
x
= lim
x→0
+
x − 0
x
= 1
u

(0

) = lim
x→0

+∞

0


(x)dx = xφ(x)


+∞
0

+∞

0
φ(x)dx
= −
+∞

0
φ(x)dx = −
0

−∞
0φ(x)dx −
+∞

0
1φ(x)dx
= −


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