On local properties 285
3. The case Re a
+
0, Re a
−
0, Re b
+
0, Re b
−
T
HEOREM
2. Assume that Re a
+
0, Re a
−
0, Re b
+
0, Re b
−
0. Then G
c a,b
is not hypoelliptic nor local solvable at the origin. Proof. We separate the proof into some cases
I The non-resonance case a
+
6= b
+
, a
−
6= b
−
. We retain all notations used previously. For η 0 we define the following solution of 2 V x , η
= u
+
x , η is defined as in 5. Note that when x
≤ 0 the solutions ˆu
− 1
x , η, ˆu
− 2
x , η exponentially decrease when η
→ +∞. A If
c
+
b
+
−a
+
∈ Z
−
∪ 0 then we set f = f
λ
, v = v
λ
as in section 2, with U x , η replaced by V x , η. B If
c
+
b
+
−a
+
∈ Z
−
∪ 0 then we set f = ∂
x
f
λ
, v = v
λ
as in section 2, with U x , η replaced by V x , η.
Then in a similar way as in section 2 we can contradict 9 by using f, v with large enough λ.
II The resonance case a
+
= b
+
, a
−
= b
−
. A When c
+
6= 0, c
−
6= 0 by taking the limit when a
+
→ b
+
, a
−
→ b
−
in 3 it is easy to see that the following pair is solutions of 2 see [10], p. 266
when x ≥ 0 :
˜u
+ 1
x , η : = xe
−a
+
η e
− 1
|x|
−2c
+
η e
− 1
|x| 1
2
9
1 2
, 1, 4 c
+
η e
−
1 |x|
1 2
, ˜u
+ 2
x , η : = xe
−a
+
η e
− 1
|x|
+2c
+
η e
− 1
|x| 1
2
9
1 2
, 1,
−4 c
+
η e
−
1 |x|
1 2
, when x
≤ 0 : ˜u
− 1
x , η : = xe
−a
−
η e
− 1
|x|
−2c
−
η e
− 1
|x| 1
2
9
1 2
, 1, 4 c
−
η e
−
1 |x|
1 2
, ˜u
− 2
x , η : = xe
−a
−
η e
− 1
|x|
+2c
−
η e
− 1
|x| 1
2
9
1 2
, 1,
−4 c
−
η e
−
1 |x|
1 2
. Next for η 0 let us define the following solution of 2
˜ V x , η
= ˜u
+ 1
x , η if x ≥ 0,
c
˜ V
1, −
η ˜u
− 1
x , η + c
˜ V
2, −
η ˜u
− 2
x , η if x ≤ 0,
where c
˜ V
1, −
η, c
˜ V
2, −
η are chosen as in section 2. Now we can repeat the proof in
section 2. B When c
+
6= 0, c
−
= 0 then we have solutions ˜u
+ 1
x , η, ˜u
+ 2
x , η when x ≥ 0, and
e
−a
−
η e
1 |x|
, x e
−a
−
η e
1 |x|
when x ≤ 0.
C When c
+
= 0, c
−
6= 0 then we have solutions e
−a
+
η e
1 |x|
, x e
−a
+
η e
1 |x|
when x ≥ 0,
and ˜u
− 1
x , η, ˜u
− 2
x , η when x ≤ 0.
D When c
+
= 0, c
−
= 0 then we have solutions e
−a
+
η e
1 |x|
, x e
−a
+
η e
1 |x|
when x ≥ 0,
and e
−a
−
η e
1 |x|
, x e
−a
−
η e
1 |x|
when x ≤ 0.
III The half-resonance case a
+
6= b
+
, a
−
= b
−
. This case can be treated by using the solutions in part I when x
≥ 0, and the solutions in part II when x ≤ 0.
286 N.M. Tri
IV The half-resonance case a
+
= b
+
, a
−
6= b
−
. This case can be treated by using the solutions in part II when x
≥ 0, and the solutions in part I when x ≤ 0. R
EMARK
2. The following case Re a
Kategori