độ trơn, tính giải tích, tính chính quy gevrey của nghiệm của phương trình nửa tuyến tính elliptic suy biến - Pdf 19

class="bi x0 y0 w1 h1"
G
a,b
k,c
f + ψ

x, y, f,
∂f
∂x
, x
k
∂f
∂y

= 0,
a, b, c ∈ C, k
G
a,b
k,c
=


∂x
− iax
k

∂y


∂x
− ibx

k
∂f
∂y

= 0.
a, b, c ∈ C, k
G
a,b
k,c
=


∂x
− iax
k

∂y


∂x
− ibx
k

∂y

+ icx
k−1

∂y
.

(Ω) : k Ω,
D(Ω) : Ω,
D

(Ω) : D(Ω),
C

(Ω) : Ω,
G
s
(Ω) : s Ω,
L
p
loc
: p,
R
n
: n .
¨o
P (D)
P (x, D)
¨o
G
k,λ
=

2
∂x
2
+ x

a = −1, b = 1.
X
2
=

∂x
− iax
k

∂y
, X
1
=

∂x
− ibx
k

∂y
.
G
a,b
k,c
a, b ab < 0
G
a,b
k,c
k
k
G


ψ

ψ
G
a,b
k,c
f + ψ

x, y, f,
∂f
∂x
, x
k
∂f
∂y

= 0.
a = −1, b = 1, c = λ + k G
a,b
k,c
= G
k,λ
.
a = −1, b = 1
k
k k k k
a, b
a = −1, b = 1
a, b, c k

k,c
G
m
k,loc
(Ω) S
m
loc
(Ω)
k
k
k
G
a,b
k,c
G
a,b
k,c
k
k
L
1
loc
(R
2
) x, y
G
a,b
k,c
m Ω ⊂ R
n

n
, a
α
(x) ∈ C

(Ω), D
α
=

|α|
i
|α|
∂x
α
1
1
∂x
α
2
2
∂x
α
n
n
.
F (x) ∈ L
1
loc
(Ω)
x

M
(Ω

) P(x, D)u ∈
C

(Ω

) u ∈ C

(Ω

).
P (x, D) Ω


 Ω, u ∈ D

(Ω

) P (x, D)u ∈ C

(Ω

) u ∈ C

(Ω

).
P (x, D) a

(Ω

).
G
s
(Ω) 1 ≤ s < ∞
f(x) ∈ C

(Ω) K  Ω C
α x ∈ K
|∂
α
f(x)| ≤ C
|α|+1
(α!)
s
.
(x, τ
α
)
|α|≤m
∈ (Ω

×
˜
Ω) Φ(x, ∂
α
)
|α|≤m
m

α
f)
|α|≤m
∈ C

(Ω

) f ∈ C

(Ω

)
Φ(x, τ
α
)
|α|≤m
∈ A(Ω ×
˜
Ω)(G
s
(Ω ×
˜
Ω))
Φ(x, ∂
α
)
|α|≤m




(Ω ×
˜
Ω)),
Φ(x, ∂
α
)
|α|≤m
Ω Ω

 Ω, f ∈ C

(Ω

)
Φ(x, ∂
α
f)
|α|≤m
∈ A(Ω

)(G
s
(Ω

)) f ∈ A(Ω

)(G
s
(Ω


k

∂y
, X
2
=

∂x
− iax
k

∂y
.
f(x, y) Ω

α
f(x, y)
∂x
α
,

β
f(x, y)
∂y
β
,

α+β
f(x, y)
∂x

+
= −ax
k+1
+ bu
k+1
+ i(k + 1)(y − v)
A

= bx
k+1
− au
k+1
− i(k + 1)(y − v)

R = (x
k+1
+ u
k+1
)
2
+ (k + 1)
2
(y − v)
2
R
1
= (x
k+1
− u
k+1

k+1
)
p =

(a − b)
2
x
k+1
u
k+1
R
−1
xu = 0,
0 xu = 0,
M = A

c
(k+1)(b−a)
+
A

k(b−a)−c
(k+1)(b−a)

.
k Re(a) < 0 Re(b) > 0
p /∈ (1, +∞)
p = 1 ⇔ y = v, x = ± u u = 0
A, B, C, D ∈ R, C
2

u
k+1

− ab(x
2k+2
+ u
2k+2
) + (a
2
+ b
2
)x
k+1
u
k+1
+ (k + 1)
2
(y − v)
2
+ i(b + a)(x
k+1
− u
k+1
)(k + 1)(y − v)

−1
.
u = 0 p = 0
u = 0 :
y = v


(−mc+nd)(X
2
+1)
+(m
2
+ c
2
− n
2
− d
2
)X

+ i

(−md − nc)(X
2
+ 1) + 2(mn + cd)X


−1
.
p
p =
[(m − c)
2
− (n − d)
2
]X

= 2(m−c)(n−d)X

(−mc+nd)(X
2
+1)+(m
2
+c
2
−n
2
−d
2
)X

.
X = 0 p = 0 ∈ (1, +∞).
X = 0
(md − nc)(c
2
+ d
2
− m
2
− n
2
)(X − 1)
2
= 0,
md −nc = 0 c
2

X

a
b

X
2
+ 1

+

a
b

2
+ 1

X
.
a, b md − nc = 0
a
b
=
m
c
p =

m
c
− 1


2
X

m
c
2X +

m
c

2
+ 1

X
= 1.
X = 1 x
k+1
= u
k+1
x = ±u
c
2
+ d
2
−m
2
−n
2
= 0 c

2
,
p =
r
2
(e

1
− e

2
)
2
X
−r
2
e

1
+iϕ
2
(X
2
+ 1) + r
2
(e
2iϕ
1
+ e
2iϕ

2
− 2 sin(ϕ
1
+ ϕ
2
)]X
−sin(ϕ
1
+ ϕ
2
)(X
2
+ 1) + (sin 2ϕ
1
+ sin 2ϕ
2
)X
.
p =
2[1 − cos(ϕ
1
− ϕ
2
)]X
(X
2
+ 1) − 2 cos(ϕ
1
− ϕ
2

2
)XU + 1 + 2i(a + b)(X − U)
.
X =
x
k+1
(k + 1)(y − v)
, U =
u
k+1
(k + 1)(y − v)
, XU ≥ 0.
a = m + in, b = c + id, p
p =
[(m − c)
2
− (n − d)
2
]XU
(−md+nc)(X
2
+U
2
)+(m
2
+c
2
−n
2
−d

)
− 2(n − d)(nc − md)

(X − U) − 2(m −c)(n − d) = 0.
(m
2
+ n
2
− c
2
− d
2
)(nc − md) = 0
X − U =
c − m
nc − md
X − U =
2(n − d)
m
2
+ n
2
− c
2
− d
2
.
nc − md = 0, m
2
+ n

(nc − md) = m
2
+ n
2
− c
2
− d
2
= 0,
−2(m − c)(n − d) = 0.
−2(m − c)(n − d) = 0 m = c n = d = 0
m = −c a = −b ∈ R.
p =
4b
2
XU
b
2
(X + U)
2
+ 1
< 1.
X − U =
2(n − d)
m
2
+ n
2
− c
2

c − m
nc − md
,
p =
XU
XU +
−mc
(nc − md)
2
.
−mc > 0, XU ≥ 0, p ≤ 1 m = 0
c = 0
p p ≤ 1; p = 1
y = v, x = ±u, u = 0. 
E
a,b
k,c
(x, y, u, v) G
a,b
k,c
.
M = M(x, y, u, v), F (p) = F
a,b
k,c
(p(x, y, u, v)),
E
a,b
k,c
= E
a,b

(a − b)
2
u
k+1
x
k+1
R
−2

− ab(u
k+1
− x
k+1
)
2
+
+ (k + 1)
2
(y − v)
2
+ i(k + 1)(y − v)(x
k+1
− u
k+1
)(a + b)

F

(p)
+

2k + 1
k + 1
p

F

(p) +
c(c − k(b − a))
(k + 1)
2
(b − a)
2
F (p) = 0,
F (p)
p(1 − p)F

(p) +

γ −(1 + α + β)p

F

(p) − αβF (p) = 0,
α =
c
(k + 1)(b − a)
, β =
k(b − a) − c
(k + 1)(b − a)
, γ =

k + 1
, p

:= C
1
F
a,b
k,c;1
(p) + C
2
F
a,b
k,c;2
(p).
F (α, β, γ, p) C
1
, C
2
k F
a,b
k,c
(p)
p /∈ (1, +∞) u = 0 p = 0
G
a,b
k,c
E(x, y, 0, 0) = G
a,b
k,c
C

)Γ(
k(b−a)−c
(k+1)(b−a)
)
4(b − a)
1
k+1
πΓ(
1
k+1
)
:= C
a,b
k,c
.
u = 0, E
a,b
k,c
(x, y, u, v) F
a,b
k,c
(p)
F
a,b
k,c
(p) p /∈ (1, +∞) p → 1 p → 1
F
a,b
k,c;1
(p) = −

k,c
(x, y, u, v) x = u, y = v.
p
1
k+1
= ((b −a)
2
R
−1
)
1
k+1
xu → −1 (x, y) → (−u, v)
C
2
= −
Γ(
c+b−a
(k+1)(b−a)
)Γ(
(k+1)(b−a)−c
(k+1)(b−a)
)
4(b − a)
1
k+1
πΓ(
k+2
k+1
)

k,c;1
(p) + D
a,b
k,c
F
a,b
k,c;2
(p))
= −
Γ

c
(k+1)(b−a)
)Γ(
k(b−a)−c
(k+1)(b−a)
)F (
c
(k+1)(b−a)
,
k(b−a)−c
(k+1)(b−a)
,
k
k+1
, p

4(b − a)
1
k+1

1
k+1
πΓ(
k+2
k+1
)A
c+b−a
(k+1)(b−a)
+
A
(k+1)(b−a)−c
(k+1)(b−a)

G
a,b
k,c
k a, b, c, k
G
a,b
k,c
E
a,b
k,c
(x, y, u, v) = δ(x − u, y − v).
(u, v) ∈ R
2
.
u = 0,
x = u + r cos ϕ; y = v + r sin ϕ.
B

(x, y)).
R
2
ε
(u, v), E
a,b
k,c
(x, y, u, v)
E
a,b
k,c
(x, y, u, v) ∈ L
1
(R
2
ε
(u, v)).

B
ε
(u,v)
E
a,b
k,c
(x, y, u, v)dxdy.
(x, y) (u, v) u = 0
x
l
= u
l

+ o(r
2
),
A

= (b − a)u
k+1
+ (k + 1)(bu
k
cos ϕ − i sin ϕ)r
+
b(k + 1)ku
k−1
cos
2
ϕ
2
r
2
+ o(r
2
),
M = (b −a)

k
k+1
u
−k
+ o(1),
X

2

− abu
2k
cos
2
ϕ + sin
2
ϕ
+ i(b − a)u
k
cos ϕ sin ϕ


r
2
+ o(r
2
),
1 − p =
(k + 1)
2
(−au
k
cos ϕ + i sin ϕ)(bu
k
cos ϕ − i sin ϕ)r
2
(b − a)
2

C
2(b − a)
1
k+1
π
log r + O(1)

rdrdϕ < ∞.

B
ε
(u,v)
E
a,b
k,c
(x, y, u, v)dxdy < ∞.
u = 0 p = 0
F (p) = 1, E
a,b
k,c
(x, y, u, v) = M(x, y, u, v).
(x, y) → (0, v)
x
k+1
= r
k+1
cos
k+1
ϕ,
A

2
ϕ


k
k+1
+ o(1)

.

B
ε
(u,v)
E
a,b
k,c
(x, y, u, v)dxdy = C


0
ε

0
r

k
k+1

sin


B
ε
(u,v)
E
a,b
k,c
(x, y, u, v)dxdy < ∞.
E
a,b
k,c
(x, y, u, v) ∈ L
1
(R
2
(x, y))
E
a,b
k,c
G
a,b
k,c
.
G
a,b
k,c
E
a,b
k,c
(x, y, u, v) = δ(x − u, y − v),
(G

ω(x, y)) = lim
ε→0

R
2
ε
(u,v)
E
a,b
k,c
(x, y, u, v)G
b,a
k,−c
ω(x, y)dxdy.

R
2
ε
(u,v)
E
a,b
k,c
(x, y, u, v)G
b,a
k,−c
ω(x, y)dxdy.

R
2
ε


R
2
ε
(u,v)
ω(x, y)G
a,b
k,c
E
a,b
k,c
(x, y, u, v)dxdy
+

r=ε
E
a,b
k,c
(x, y, u, v)

X
2
ω(x, y)(ν
1
− ibx
k
ν
2
) − icx
k−1

) − icx
k−1
ν
2
ω(x, y),
B
2
(E
a,b
k,c
(x, y, u, v), a, b, c, k) = (ν
1
− iax
k
ν
2
)X
1
E
a,b
k,c
(x, y, u, v).

R
2
ε
(u,v)
E
a,b
k,c

(E
a,b
k,c
(x, y, u, v), a, b, c, k)ds
:= I
1
+ I
2
+ I
3
.
G
a,b
k,c
E
a,b
k,c
(x, y, u, v) = 0 R
2
ε
(u, v) I
1
= 0


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