Riemann–Hilbert problem 193
5. Riemann-Hilbert problem and solvability
D
EFINITION
3. A rational function pz : = azz
− m
is said to be Riemann-Hilbert factorizable with respect to
|z| = 1 if the following factorization pz
= p
−
z p
+
z, holds, where p
+
z, being holomorphic in |z| 1 and continuous up to the boundary,
does not vanish in |z| ≤ 1, and p
−
z, being holomorphic in |z| 1 and continuous
up to the boundary, does not vanish in |z| ≥ 1.
The factorizability is equivalent to saying that the R–H problem for the jump func- tion p and the circle has a solution.
E
XAMPLE
5. We consider pz : = azz
− m
a0 6= 0 m ≥ 1. Let az be a
polynomial of order m + n n ≥ 1. Then we have
pz = cz − λ
1
· · · z − λ
m
z − λ
m+1
· · · z − λ
m+n
z
− m
= c1 − λ
1
z · · · 1 −
λ
m
z z
− λ
m+1
· · · z − λ
m+n
, where λ
j
∈ C. We can easily see that p is Riemann-Hilbert factorizable with respect to the unit circle if and only if
R H |λ
1
| ≤ · · · ≤ |λ
m
| 1 |λ
m+1
| ≤ · · · ≤ |λ
m+n
|. T
HEOREM
2. Suppose that RH is satisfied. Then the kernel and the cokernel of the map
∗ vanishes. Proof. We consider the kernel of
∗. By definition, π pu = 0 is equivalent to pe
iθ
ue
iθ
= ge
iθ
, where g consists of negative powers of e
iθ
. If |λ
j
| 1 the series 1 − λ
j
e
− iθ
− 1
consists of only negative powers of e
iθ
. Hence, if 1 − λ
j
e
− iθ
U e
iθ
= Fe
iθ
for some F consisting of negative powers it follows that U e
iθ
= 1 − λ
j
e
− iθ
− 1
Fe
iθ
consists of negative powers. By repeating this argument we see that z
− λ
m+1
· · · z − λ
m+n
uz, z
= e
iθ
consists of only negative powers. On the other hand, since this is a polynomial of z we obtain u
= 0. Next we study the cokernel. Let f
∈ H
2
T be given. For the sake of simplicity
we want to solve 1
− λ
1
e
− iθ
e
iθ
− λ
2
ue
iθ
≡ f e
iθ
modulo negative powers,
194 M. Yoshino
where |λ
1
| 1 |λ
2
|. Hence we have e
iθ
− λ
2
ue
iθ
≡ 1
− λ
1
e
− iθ
− 1
f = f
+
+ f
−
≡ f
+
modulo negative powers. Here f
+
resp. f
−
consists of Fourier coefficients of non- negative resp. negative part. Hence, we have
e
iθ
− λ
2
ue
iθ
= f
+
. The solution is given by ue
iθ
= e
iθ
− λ
2 −
1
f
+
. Hence the cokernel vanishes. This ends the proof.
6. Index formula of an ordinary differential operator
Kategori