Tải bản đầy đủ (.pdf) (59 trang)

Khóa luận tốt nghiệp: Không gian Sobolev phụ thuộc thời gian potx

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (594.07 KB, 59 trang )


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)


L
1
(0, T ; X)
W
1
p
(0, T ; X)
17
L
p
(0, T ; X), C([0, T ]; X),
L
p
(0, T ; X), W
1
p
(0, T ; X)
E
E
R C ρ E E
ρ
ρ(x) ≥ 0 ∀x ∈ E ρ(x) = 0 x = 0
ρ(λx) = |λ|ρ(x) ∀λ ∈ K ∀x ∈ E
ρ(x + y) ≤ ρ(x) + ρ(y) ∀x, y ∈ E
E ρ
ρ E
d(x, y) := ρ(x − y), (∀x, y ∈ E)
E






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
 > 0, ∃N > 0 u
k
− u
l
 < , k, l ≥ N.
E E
{u
k
}

k=1
⊂ E u ∈ E {u
k

}

k=1
u.
E
E
E
E
E E
E E
E.
E {x
n
}
n∈N

E
x ∈ E {x
n
k
}
n∈N

x.
{u
k
}

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
ϕ(x, x) ≥ 0, ∀x ∈ E
ϕ
E.
|ϕ(x, y)|

2
≤ ϕ(x, x).ϕ(y, y), ∀x, y ∈ E
a = ϕ(x, x), b = ϕ(x, y), c = ϕ(y, y). a, c
|b|
2
≤ ac.
λ ∈ C
0 ≤ ϕ(x + λy, x + λy) = ϕ(x, x) + λϕ(x, y) + λϕ(y, x) + λλϕ(y, y)
a + λb + λb + λλc ≥ 0 λ ∈ C
a c c > 0 λ = −
b
c
(1.2)
0 ≤ a −
b
c
b −
b
c
b +
bb
c
2
c = a −
bb
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)
=


ϕ(x, x) + ϕ(y, y)

2


ϕ(x + y, x + y) ≤

ϕ(x, x) +

ϕ(y, y) x, y ∈ E
E E
ϕ(x, x) = 0 ⇒ x = 0.
ϕ E ϕ(x, y) < x, y >
< x, y > x y.
E ., .
x =

(x, x) ∀x ∈ E E
E
E
E. f(x)
E E
E.
f(x + y) = f(x) + f(y) x, y ∈ E.
f(αx) = αf(x) x ∈ E α.
f C > 0
(∀x ∈ E) |f(x)| ≤ Cx.
{u
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
 u u ≤ lim
k→∞
inf u
k
− u
{u
k

}

k=1
H
u ≤ lim
k→∞
u
k

H
H H.
C
k
(Ω)
x = (x
1
, x
2
, , x
n
) R
n
C(Ω) Ω.
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)

C

(Ω) = C

(Ω) ∩

C
(Ω)

L
p
(Ω)
Ω R
k
µ
σ F R
k
p ≥ 1, L
p
(Ω) p Ω
L
p
(Ω) = {f : Ω −→ R :


|f|
p
dµ < +∞}
L
p
(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
+
α.β ≤
α
p
p
+
β
q
q
α = 0 β = 0
α > 0 β > 0

f(t) =
t
p
p
+
t
−q
q
, (t > 0)
f

(t) = t
−q−1
(t
p+q
− 1) = 0 t = 0 f

(t) < 0 (0; 1)
f

(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
+
β
q
p

−1
q
≥ 1
αβ
p
q
+ 1 = p
q

p
+ 1 = q,
α.β ≤
α
p
p
+
β
q
q
¨o p, q > 1
1
p
+
1
q
= 1.
f ∈ L
p
(Ω), g ∈ L
q
(Ω),


|f.g|dµ ≤



|f|
p



1
p



|g|
q


1
q
fg
1
≤ f
p
g
q
f = 0 g = 0 f = 0 g = 0
f.g = 0 fg
1
= 0.
f
p
> 0, g
q
> 0. x ∈ Ω 1.5
α =
|f(x)|

f
p
β =
|g(x)|
g
q
|f(x)g(x)|
f
p
g
q

1
p
|f(x)|
p
f
p
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
q


|g(x)|
q

fg
1
f
p
g
q


1
p
f
p
p
f
p
p
+
1
q
g
q
q
g
q
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
n
: |y
i
− x
i
| < R, i = 1, n

f ∈ L
p
(Ω)  > 0 f(x) = 0 x = Ω
L
p

(R
n
) R

R
n
\Q(0,R)
|f(x)|
p
dx < ε
p

Q(0,R+1)
|f(x) + g(x)|
p
dx < ε
p
g
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
)X
j
(x)
x
j
|g
R
(x) − h(x)| = |g
R
(x) − g
R
(x

j
)| < εR
−n
p
x x
j

Q(0,R)
|g
R
− h|
p
dx < ε
p
g
R
(x) = 0, h(x) = 0 x ∈ R
n
\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

1
p



Q(0,R)
|f(x) −g
R
(x)|
p
dx


1
p
+


Q(0,R)
|g
R
(x) − h(x)|
p
dx

1
p
+


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

1
p
+


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


1
p
≤ 3ε
X
j
L
p
(Ω)
L
p
(Ω)
L
p
(Ω), p ≥ 0
Ω R
n
, f ∈ L
p
(Ω), p ≥ 1, f(x) = 0
Ω  > 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.  > 0,


|f(x) − f(λx)|
p
dx < ,
|λ − 1| < δ.
f(λx) = f(x + (λ − 1)x). Ω
(λ − 1)x ∈ Ω. 1.4
θ(x)

C

(R
n
)
θ(x) = θ(−x) θ(x) ≥ 0, θ(x) = 0 |x| > 1

R
n

θ(x) = 1
θ(x)
u ∈ L
p
(Ω), p ≥ 1,
u
h
(x) = h
−n


θ(
x − y
h
)u(y)dy
R
n
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
h
(x) − u(x) =

R
n
θ(z) [u(x + hz) − u(x)] dz
|u
h
(x) − u(x)|
p
≤ C

|z|<1

[u(x + hz) − u(x)]
p
dz
x


|u
h
(x) − u(x)|
p
dx ≤ C

|z|<1
dz


[u(x + hz) − u(x)]
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
y

f(y)dy

g(x)dx
= h
−n


f(y)dy


θ


x − h
y

g(x)dx
=


f(x)g
h
(x)dx
f ∈ L
1
(Ω)


f(x)ϕ(x)dx = 0 ϕ ∈

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
u = v.
α = (α
1
, α
2
, , α
n
) |α| = α
1
+ α
2
+ + α
n
= k

D
α
φ =

α
1
∂x
1
α
1


α
n
∂x
n
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
φdx = (−1)
α

U
v
1
φdx = (−1)
α

U
v
2
φdx
φ ∈ C

c

(U)

U
(v
1
− v
2
)φdx = 0
φ ∈ C

c
(U) v
1
− v
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

0


dx = −
2

0
vφdx.
2

0


dx =
1

0



dx +
2

1
φ

dx
= −
1

0
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

0
vφdx
φ ∈ C

c
(U)
v φ (1.3)

2

0
vφdx =
2

0


dx =
1


0


dx + 2
2

1
φ

dx
= −
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

m
dx −
1

0

m
dx

= 0,
Ω R
n
u(x) ∈ L
p
(Ω)
α v(x) ∈ L
p
(Ω)



u(x)ψ(x)dx = (−1)
|α|


v(x)D
α
ψ(x)dx
ψ ∈

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
(x) = 0 Ω
u
1
(x) = u
2
(x) Ω
ψ ∈

C

(Ω)


u(x)ψ(x)dx = (−1)
|α|


v(x)D
α
ψ(x)dx

ψ ∈

C

(Ω) v(x)
u(x) D
α
v(x)
v(x) Ω
α
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 =



f

α
v
∂x
1
α
1
∂x
n
α
n
dx
v ∈

C


(Ω

) v ∈

C

(Ω) Ω

⊂ Ω
−1
|α|


ωvdx = −1
|α|



ωvdx
ω ∈ L
1
(Ω) ω ∈ L
1
(Ω

)



f


α
v

α
1
x
1

α
n
x
n
dx = −1
|α|



ωvdx
∀v ∈

C

(Ω

)

α
f


α
1
x
1

α
n
x
n
= ω


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

(R)
ω =





1 0 ≤ x < 1
−1 − 1 < x < 0
ω ∈ L
1
(−1; 1)
1

−1
ωdx =
0

−1

ωdx +
1

0
ωdx = −
0

−1
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|
∂v
∂x
dx =
0

−1
x
∂v
∂x
dx +
1

0
x
∂v
∂x
dx
=
0

−1
vdx −

1

0
vdx = −(
0

−1
(−1)vdx +
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)

2
ϕ
∂x∂y
dxdy =
b

a
f
1
(x)
∂ϕ
∂x




y=β(x)
y=α(x)
dx = 0
supp ϕ ⊂ U


U
f
2
(x)

2
ϕ
∂x∂y
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)

2
ϕ
∂x∂y
dxdy +

U
f
2
(x)

2
ϕ
∂x∂y
dxdy

2
ϕ
∂x∂y
≡ 0 U

2
ϕ

∂y∂x
≡ 0
Ω = R,
u(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

u(x) − u(0)
x
= lim
x→0

0 − 0
x
= 0
u

(0
+
) = u

(0

)
u(x) x = 0
u(x)
φ ∈

C

(R)


R
u(x)φ

(x)dx =
+∞

0


(x)dx = xφ(x)


+∞
0

+∞

0
φ(x)dx
= −
+∞

0
φ(x)dx = −
0

−∞
0φ(x)dx −
+∞


0
1φ(x)dx
= −

R
φ(x)dx

×