Giáo trình giải tich 3 part 4 ppt

R

b
a
f(x)dx =

β
α
f(ϕ(t))ϕ

(t)dt
ϕ (α, β) (a, b)
ω = f(x)dx
ϕ
∗
ω = f(ϕ(t))ϕ

(t)dt

b
a
ω =

β
α
ϕ
∗
ω
k
V R k
V

ω : V ×···×V
  
k
→ R
v
1
, ··· ,v
k
∈ V α ∈ R 1 ≤ i<j≤ k
ω(v
1
, ··· ,v
i
+ v

i
, ··· ,v
k
)=ω(v
1
, ··· ,v
i
, ··· ,v
k
)+ω(v
1
, ··· ,v

i
, ··· ,v

k
).
ω(v
1
, ··· ,αv
i
, ··· ,v
k
)=αω(v
1
, ··· ,v
i
, ··· ,v
k
).
ω(v
1
, ··· ,v
i
, ··· ,v
j
, ··· ,v
k
)= − ω(v
1
, ··· ,v
j
, ··· ,v
i
, ··· ,v

k
).
ω
ω(v
1
, ··· ,v
i
··· ,v
j
, ··· ,v
k
)=0 v
i
= v
j
i = j
ω(v
σ(1)
, ··· ,v
σ(k)
)=(σ)ω(v
1
, ··· ,v
k
)
σ {1, ··· ,k} (σ)

i<j
(σ(j) −σ(j))
⇒ v

i
= v
j
2ω(v
1
, ··· ,v
i
··· ,v
i
, ··· ,v
k
)=0
⇒ v
i
= v
j
= v + w
ω(v
1
, ··· ,v,··· ,w,··· ,v
k
)+ω(v
1
, ··· ,w,··· ,v,··· ,v
k
)=0.
⇒
−1
k
⇒ σ i j

F R
3
W
F
(v)=<F,v>, v∈ R
3
1 R
3
F v
ω
F
(v
1
,v
2
)=<F,v
1
×v
2
>, v
1
,v
2
∈ R
3
2
R
3
F v
1

,v
2
n R
n
det(v
1
, ··· ,v
n
)
v
1
, ··· ,v
n
∈ R
n
Λ
k
(V ) Λ
k
(V ) k
V
(ω + γ)(v
1
, ··· ,v
k
)=ω(v
1
, ··· ,v
k
)+γ(v

1
, ··· ,v
k
)
(αω)(v
1
, ··· ,v
k
)=αω(v
1
, ··· ,v
k
) ω, γ ∈ Λ
k
(V ),α∈ R
(Λ
k
(V ), +, ·) R
Λ
1
(V ) V Λ
1
(V )=V
∗
= L(V, R)
ϕ
1
,ϕ
2
∈ V

∗
2 ϕ
1
∧ ϕ
2
: V × V → R,
(ϕ
1
∧ ϕ
2
)(v
1
,v
2
)=ϕ
1
(v
1
)ϕ
2
(v
2
) − ϕ
2
(v
1
)ϕ
1
(v
2

) = det

ϕ
1
(v
1
) ϕ
1
(v
2
)
ϕ
2
(v
1
) ϕ
2
(v
2
)

R
2
ϕ(v
1
),ϕ(v
2
) ϕ =(ϕ
1
,ϕ

2
):V → R
2
ϕ
1
, ··· ,ϕ
k
∈ V
∗
k
ϕ
1
∧···∧ϕ
k
∈ Λ
k
(V )
ϕ
1
∧···∧ϕ
k
(v
1
, ··· ,v
k
)=

σ
(σ)ϕ
σ(1)

(v
1
) ···ϕ
σ(k)
(v
k
)=det(ϕ
i
(v
j
)),v
1
, ··· ,v
k
∈ V,
ϕ
1
∧···∧ϕ
k
=

σ
(σ)ϕ
σ(1)
⊗···⊗ϕ
σ(k)
ϕ
1
, ··· ,ϕ
k

,ϕ

i
∈ Λ
1
(V ),α,β∈ R i =1, ··· ,k
ϕ
1
∧···∧(αϕ
i
+βϕ

i
)∧···∧ϕ
k
= αϕ
1
∧···∧ϕ
i
∧···∧ϕ
k
+βϕ
1
∧···∧ϕ

i
∧···∧ϕ
k
.
ϕ

σ(1)
∧···∧ϕ
σ(k)
= (σ)ϕ
1
∧···∧ϕ
k
, σ
k V
R ϕ
1
, ··· ,ϕ
n
V
∗
Λ
k
(V )
{ϕ
i
1
∧···∧ϕ
i
k
, 1 ≤ i
1
< ···<i
k
≤ n}
ω ∈ Λ

k
(V )
ω =

1≤i
1
<···<i
k
≤n
a
i
1
···i
k
ϕ
i
1
∧···∧ϕ
i
k
dim Λ
k
(V )=C
k
n
=
n!
(n − k)!k!
{ϕ
1

, ··· ,ϕ
n
} {e
1
, ··· ,e
n
} ϕ
i
(e
j
)=δ
ij
ω ∈ Λ
k
(V ) v
1
, ··· ,v
k
∈ V
v
1
=

i
1
ϕ
i
1
(v
1

)e
i
1
, ··· ,v
k
=

i
k
ϕ
i
k
(v
k
)e
i
k
,
ω(v
1
, ··· ,v
k
)=ω(

i
1
ϕ
i
1
(v

1
)e
i
1
, ··· ,

i
k
ϕ
i
k
(v
k
)e
i
k
)
=

i
1
,··· ,i
k
ϕ
i
1
(v
1
) ···ϕ
i

k
(v
k
)ω(e
i
1
, ··· ,e
i
k
)
=

i
1
<···<i
k

σ
ϕ
i
σ(1)
(v
1
) ···ϕ
i
σ(k)
(v
k
)(σ)ω(e
i

1
, ··· ,e
i
k
)
=

i
1
<···<i
k
ω(e
i
1
, ··· ,e
i
k
)ϕ
i
1
∧···∧ϕ
i
k
(v
1
, ··· ,v
k
)
{ϕ
i

1
∧···∧ϕ
i
k
, 1 ≤ i
1
< ···<i
k
≤ n}
ϕ
i
1
∧···∧ϕ
i
k
(e
j
1
, ··· ,e
j
k
)=

1 (i
1
, ··· ,i
k
)=(j
1
, ··· ,j

k
)
0 (i
1
, ··· ,i
k
) =(j
1
, ··· ,j
k
)
ω =

i
1
<···<i
k
a
i
1
·i
k
ϕ
i
1
∧···∧ϕ
i
k
=0,
ω(e

i
1
, ··· ,e
i
k
)=a
i
1
···i
k
=0
Λ
k
(V )=0 k>n Λ
n
(V ) C
n
n
=1 ω ∈ Λ
n
(V )
ω = aϕ
1
∧···∧ϕ
n
a ∈ R
U R
n
k k
U

ω : U → Λ
k
(R
n
).
ω C
p
C
p
Ω
k
p
(U) k C
p
U Ω
k
(U)=Ω
k
∞
(U)
Ω
k
p
(U)
U ⊂ R
3
F : U → R
3
F
W

F
: U → Λ
1
(R
3
),W
F
(x, y, z)(v)=<F(x, y, z),v >
ω
F
: U → Λ
2
(R
3
),ω(x, y, z)(v
1
,v
2
)=<F(x, y, z),v
1
× v
2
>
f : U → R C
p+1
x ∈ U f

(x):R
n
→ R

f 1
df : U → Λ
1
(R
n
),x→ df (x)=f

(x).
ix
i
: R
n
→ R, (x
1
, ··· ,x
n
) → x
i
dx
i
(x)(v)=x

i
(x)v = v
i
,v=(v
1
, ··· ,v
n
) ∈ R

n
.
df (x)(v)=f

(x)v =
∂f
∂x
1
(x)v
1
+ ···+
∂f
∂x
n
(x)v
n
=
∂f
∂x
1
(x)dx
1
(x)(v)+···+
∂f
∂x
n
(x)dx
n
(x)(v).
df =

n

i=1
∂f
∂x
i
dx
i
1 ϕ
1
, ··· ,ϕ
k
∈ Ω
1
(U)
(ϕ
1
∧···∧ϕ
k
)(x)=ϕ
1
(x) ∧···∧ϕ
k
(x),x∈ U,
k U 1 dx
1
, ··· ,dx
n
Ω
1

(U)
k U
ω =

1≤i
1
<···<i
k
≤n
a
i
1
···i
k
dx
i
1
∧···∧dx
i
k
,
a
i
1
···i
k
U C
p
ω C
p

U ⊂ R
3
(x, y, z)
0 f : U → R
1
Pdx+ Qdy + Rdz
2 Adx ∧ dy + Bdy ∧ dz + Cdz ∧dx
3 fdx∧ dy ∧dz
U ⊂ R
3
F : U → R
3
F =(P, Q, R)
W
F
= Pdx+ Qdy + Rdz
ω
F
= Pdy ∧ dz + Qdz ∧ dx + Rdx ∧dy
U, V R
m
, R
n
ϕ : U → V, u =(u
1
, ··· ,u
m
) → x =(ϕ
1
(u), ··· ,ϕ

n
(u))
ϕ
∗
:Ω
k
(V ) → Ω
k
(U),ω→ ϕ
∗
ω
ω =

1≤i
1
<···<i
k
≤n
a
i
1
···i
k
(x)dx
i
1
∧···∧dx
i
k
,

ϕ
∗
ω(u)=

1≤i
1
<···<i
k
≤n
a
i
1
···i
k
(ϕ(u))dϕ
i
1
∧···∧dϕ
i
k
.
ϕ : R → R
2
,ϕ(t)=(x =cost, y =sint) ω(x, y)=xdy −ydx
ϕ
∗
ω(t)=costd(sin t) − sin td(cos t)=dt
ϕ : R
2
→ R

2
,ϕ(r, θ)=(x = r cos θ, y = r sin θ) ω(x, y)=dx ∧ dy
ϕ
∗
ω(r, θ)=d(r cos θ) ∧ d(r sin θ)
=(cosθdr − r sin θdθ) ∧(sin θdr + r cos θdθ)
= rdr ∧ dθ (do dr ∧dr = dθ ∧dθ =0,dθ∧dr = −dr ∧ dθ).
ϕ
∗
(ω
1
+ ω
2
)=ϕ
∗
(ω
1
)+ϕ
∗
(ω
2
),ω
1
,ω
2
∈ Ω
k
(V )
ϕ
∗

(γ
1
∧···∧γ
k
)=ϕ
∗
(γ
1
) ∧···∧ϕ
∗
(γ
k
),γ
1
, ··· ,γ
k
∈ Ω
1
(V )
ϕ
∗
(dx
i
)=dϕ
i
=
m

j=1
∂ϕ

i
∂u
j
du
j
ϕ : R
n
→ R
n
ϕ
∗
(f(x)dx
1
∧···∧dx
n
)=f(ϕ(u)) det ϕ

(u)du
1
∧···∧du
n
.
ϕ
∗
ω(u)(v
1
, ··· ,v
k
)=ω(ϕ(u))(ϕ


(u)v
1
, ··· ,ϕ

(u)v
k
).
k ∈ N
d :Ω
k
(U) → Ω
k+1
(U),
d(

1≤i
1
<···<i
k
≤n
a
i
1
···i
k
dx
i
1
∧···∧dx
i

k
)=

1≤i
1
<···<i
k
≤n
da
i
1
···i
k
∧ dx
i
1
∧···∧dx
i
k
.
n =2 (x, y)
d (Pdx+ Qdy)=dP ∧ dx + dQ ∧dy
=

∂P
∂x
dx +
∂P
∂y
dy


∧ dx +

∂Q
∂x
dx +
∂Q
∂y
dy

∧ dy
=

∂Q
∂x
−
∂P
∂y

dx ∧ dy
dx ∧dx = dy ∧dy =0,dy∧ dx = −dx ∧ dy
R
3
ω(x, y, z)=sinxydx + e
x
2
+y
dy +arctgxdz
dω =(d sin xy) ∧ dx + d(e
x

2
+y
) ∧ dy + d(arctgx) ∧dz
=(y cos xydx + x cos xydy) ∧ dx +(2xe
x
2
+y
dx + e
x
2
+y
dy) ∧dy +
1
1+x
2
dx ∧ dz
=(2xe
x
2
+y
− x cos xy)dx ∧dy −
1
1+x
2
dz ∧ dx.
d (P (x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz)
d (P (x, y, z)dx ∧ dz + Q(x, y, z)dz ∧ dx + Q(x, y, z)dx ∧dy) .
ω ∈ Ω
k
(R

n
) k ≥ n dω =0
d(ω
1
+ ω
2
)=dω
1
+ dω
2
, ∀ω
1
,ω
2
∈ Ω
k
(U)
d(γ
1
∧ γ
2
)=dγ
1
∧ γ
2
− γ
1
∧ dγ
2
, ∀γ

1
,γ
2
∈ Ω
1
(U).
d(dω)=0 d ◦d =0
d(ϕ
∗
ω)=ϕ
∗
(dω) dϕ
∗
= ϕ
∗
d
γ
1
= adx
i
,γ
2
= bdx
j
d(γ
1
∧ γ
2
)=d(adx
i

∧ bdx
j
)=d(abdx
i
∧ dx
j
)
= d(ab) ∧ dx
i
∧ dx
j
=(bda + adb) ∧dx
i
∧ dx
j
= bda ∧dx
i
∧ dx
j
+ adb ∧ dx
i
∧ dx
j
=(da ∧dx
i
) ∧ bdx
j
− adx
i
∧ db ∧dx

j
= dγ
1
∧ γ
2
− γ
1
∧ γ
2
.
dx
I
= dx
i
1
∧···∧dx
i
k
I =(i
1
, ··· ,i
k
) k {1, ···n}
ω = a
I
dx
I
d(dω)=d(da
I
∧ dx

I
)=d


i
∂a
I
∂x
i
dx
i
∧ dx
I

=

i
d

∂a
I
∂x
i

∧ dx
i
∧ dx
I
=


i



j
∂
2
a
I
∂x
j
∂x
i
dx
j


∧ dx
i
∧ dx
I
= −

i

j
∂
2
a
I

∂x
i
∂x
j
dx
i
∧ dx
j
∧ dx
I
( dx
i
∧ dx
j
= −dx
j
∧ dx
i
)
= −d(dω)( i, j)
2d(dω)=0
ω = a
I
dx
I
∈ Ω
k
(V )
d(ϕ
∗

ω)=d(a
I
◦ ϕdϕ
I
)=d(a
I
◦ ϕ) ∧ dϕ
I
.
ϕ
∗
(dω)=ϕ
∗
(da
I
∧ dx
I
)=ϕ
∗
(da
I
) ∧ ϕ
∗
(dy
I
)=ϕ
∗
(da
I
) ∧ dϕ

I
.
d(a
I
◦ ϕ)=ϕ
∗
(da
I
)
ϕ
∗
(da
I
)=ϕ
∗



j
∂a
I
∂x
j
dx
j


=

j

∂a
I
◦ ϕ
∂x
j
dϕ
j
=

j
∂a
I
◦ ϕ
∂x
j
(

i
∂ϕ
j
∂u
i
du
i
)=d(a
I
◦ϕ).
d
ω ∈ Ω
k

(U)
ω
U dω =0 U
ω
U η ∈ Ω
k−1
(U) ω = dη
ω ω d(dη)=0
ω(x, y)=
ydx − xdy
x
2
+ y
2
∈ Ω
1
(R
2
\ 0)
ω dω =
x
2
− y
2
(x
2
+ y
2
)
2

dy ∧dx −
y
2
− x
2
(x
2
+ y
2
)
2
dx ∧ dy =0
ω f ∈ Ω
0
(R
2
\ 0) ω = df
ϕ(t) = (sin t, cos t)
ϕ
∗
ω = ϕ
∗
(df )=d(ϕ
∗
f)=d(f ◦ ϕ)=(f ◦ϕ)

dt.
ϕ
∗
ω =

cos td(sin t) −sin td(cos t)
sin
2
t +cos
2
t
= dt (f ◦ ϕ)

(t) ≡ 1
f ◦ ϕ(t)=t+ f ◦ ϕ 2π
ω = a
1
dx
1
+ ···+ a
n
dx
n
∈ Ω
1
(U) f ∈ Ω
0
(U)
df = ω f
ω
f
∂f
∂x
1
= a

1
, ··· ,
∂f
∂x
n
= a
n
.
ω dω =0 a
1
, ··· ,a
n
∂a
j
∂x
i
=
∂a
i
∂x
j
i, j =1, ··· ,n.
U U
U R
n
x
0
∈ U
C
1

h : U × [0, 1] → U, (x, t) → h(x, t)
h(x, 0) = x
0
h(x, 1) = x, ∀x ∈ U
U ∀x, y ∈ U [x, y]={x + t(y − x):t ∈ [0, 1]}⊂U
R
n
U ∃x
0
∈ U : ∀x ∈ U, [x
0
,x] ⊂ U
h(x, t)=x
0
+ t(x − x
0
)
U R
n
U
U
ω ∈ Ω
k
(U),dω =0 ⇔∃η ∈ Ω
k−1
(U),ω= dη.
J
t
: U → U × [0, 1],J
t

(x)=(x, t) k =1, 2, ···
K :Ω
k
(U × [0, 1]) → Ω
k−1
(U)
∗ Kd + dK = J
∗
1
− J
∗
0
Ω
k
(U × [0, 1])
a(x, t)dx
I
b(x, t)dt ∧dx
J
, I =(i
1
, ··· ,i
k
),J =(j
1
, ··· ,j
k−1
).
K
K(a(x, t)dx

I
)=0
K(b(x, t)dt ∧dx
J
)=


1
0
b(x, t)dt

dx
J
∗
(Kd + dK)(adx
I
)=K(da ∧ dx
I
)+d(0) = (

1
0
∂a
∂t
dt)dx
I
=(a(x, 1) − a(x, 0)dx
I
=(J
∗

1
− J
∗
0
)(adx
I
).
∗
(Kd + dK)(bdt ∧ dx
J
)=K(db ∧ dt ∧ dx
J
)+d((

1
0
bdt) ∧ dx
J
)
= K(

i
∂b
∂x
i
dx
i
∧ dt ∧ dx
J
)+d((


1
0
bdt) ∧ dx
J
)
= −

1
0
(

i
∂b
∂x
i
)dt ∧ dx
i
∧ dx
J
+ d((

1
0
bdt) ∧ dx
J
)
= −d((

1

0
bdt) ∧ dx
J
)+d((

1
0
bdt) ∧ dx
J
)=0.
(J
∗
1
− J
∗
0
)(bdt ∧ dx
J
)=b(x, 1)d(1) ∧dx
J
− b(x, 0)d(0) ∧dx
J
=0.
h : U ×[0, 1] → U x
0
ω ∈ Ω
k
(U)
dω =0 η = Kh
∗

ω (k −1) dη = ω
∗
(Kd + dK)h
∗
ω =(J
∗
1
− J
∗
0
)h
∗
ω.
⇔ Kdh
∗
ω + dKh
∗
ω =(h ◦ J
1
)
∗
ω −(h ◦J
0
)
∗
ω.
⇔ Kh
∗
dω + dKh
∗

ω =(id
U
)
∗
ω −(x
0
)
∗
ω.
⇔ 0+dKh
∗
ω = ω +0.
η = Kh
∗
ω
U ω
1
,ω
2
∈ Ω
k
(U) dω
1
= dω
2
η ∈ Ω
k−1
dη = ω
1
− ω

2
R
2
\0
η
η dη = ω η = Kh
∗
ω
ω =(x
2
− 2yz)dx +(y
2
− 2zx)dy +(z
2
− 2xy)dz ∈ Ω
1
(R
3
)
dω =0 f df = ω
R
3
0 h(x, y, z, t)=(tx, ty, tz)
h
∗
ω = t
2
(x
2
− 2yz)(xdt + tdx)+t

2
(y
2
− 2zx)(ydt + tdy)+t
2
(z
2
− 2xy)(zdt + tdz).
Kh
∗
ω =

1
0
t
2
(x
2
− 2yz)xdt +

1
0
t
2
(y
2
− 2zx)ydt +

1
0

t
2
(z
2
− 2xy)zdt.
f = Kh
∗
ω =
1
3
(x
3
+ y
3
+ z
3
−6xyz) ω df = ω
f df = ω
(1)
∂f
∂x
= x
2
− 2yz
(2)
∂f
∂y
= y
2
− 2zx

(3)
∂f
∂z
= z
2
− 2xy
f
(1) f =
x
3
3
− 2xyz + ϕ(y, z)
(2)
∂ϕ
∂y
= y
2
ϕ =
y
3
3
+ ψ(z)
(3)
∂ψ
∂z
= z
2
ψ =
z
3

3
+
f =
1
3
(x
3
+ y
3
+ z
3
) − 2xyz+