48
BRUCE K. DRIVER
†
4.
Cauchy — Kovalevskaya Theorem
As a warm up we will start with the corresponding result for ordinary di
ff
erential
equations.
Theorem 4.1
(ODE Version of Cauchy — Kovalevskaya, I.)
.
Suppose
a >
0
and
f
: (
−
a, a
)
→
R
is real analytic near
0
and
u
(
t
)
is the unique solution to the ODE
(4.1)
˙
u
(
t
) =
f
(
u
(
t
))
with
u
(0) = 0
.
Then
u
is also real analytic near
0
.
We will give four proofs. However it is the last proof that the reader should focus
on for understanding the PDE version of Theorem 4.1.
Proof.
(First Proof.)If
f
(0) = 0
,
then
u
(
t
) = 0
for all
t
is the unique solution
to Eq. (4.1) which is clearly analytic. So we may now assume that
f
(0)
6
= 0
.
Let
G
(
z
) :=
R
z
0
1
f
(
u
)
du,
another real analytic function near
0
.
Then as usual we have
d
dt
G
(
u
(
t
)) =
1
f
(
u
(
t
))
˙
u
(
t
) = 1
and hence
G
(
u
(
t
)) =
t.
We then have
u
(
t
) =
G
−
1
(
t
)
which is real analytic near
t
= 0
since
G
0
(0) =
1
f
(0)
6
= 0
.
Proof.
(Second Proof.) For
z
∈
C
let
u
z
(
t
)
denote the solution to the ODE
(4.2)
˙
u
z
(
t
) =
zf
(
u
z
(
t
))
with
u
z
(0) = 0
.
Notice that if
u
(
t
)
is analytic, then
t
→
u
(
tz
)
satis
fi
es the same equation as
u
z
.
Since
G
(
z, u
) =
zf
(
u
)
is holomorphic in
z
and
u,
it follows that
u
z
in Eq.
(4.2)
depends holomorphically on
z
as can be seen by showing
¯
∂
z
u
z
= 0
,
i.e. showing
z
→
u
z
satis
fi
es the Cauchy Riemann equations. Therefore if
± >
0
is chosen small
enough such that Eq. (4.2) has a solution for

t

< ±
and

z

<
2
,
then
(4.3)
u
(
t
) =
u
1
(
t
) =
∞
X
n
=0
1
n
n
!
∂
n
z
u
z
(
t
)

z
=0
.
Now when
z
∈
R
, u
z
(
t
) =
u
(
tz
)
and therefore
∂
n
z
u
z
(
t
)

z
=0
=
∂
n
z
u
(
tz
)

z
=0
=
u
(
n
)
(0)
t
n
.
Putting this back in Eq. (4.3) shows
u
(
t
) =
∞
X
n
=0
1
n
!
u
(
n
)
(0)
t
n
which shows
u
(
t
)
is analytic for
t
near
0
.
Proof.
(Third Proof.) Go back to the original proof of existence of solutions,
but now replace
t
by
z
∈
C
and
R
t
0
f
(
u
(
τ
))
dτ
by
R
z
0
f
(
u
(
ξ
))
dξ
=
R
1
0
f
(
u
(
tz
))
zdt.
Then the usual Picard iterates proof work in the class of holomorphic functions to
give a holomorphic function
u
(
z
)
solving Eq. (4.1).
Proof.
(Fourth Proof: Method of Majorants) Suppose for the moment we have
an analytic solution to Eq. (4.1). Then by repeatedly di
ff
erentiating Eq. (4.1) we
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PDE LECTURE NOTES, MATH 237AB
49
learn
¨
u
(
t
) =
f
0
(
u
(
t
))
˙
u
(
t
) =
f
0
(
u
(
t
))
f
(
u
(
t
))
u
(3)
(
t
) =
f
00
(
u
(
t
))
f
2
(
u
(
t
)) + [
f
0
(
u
(
t
))]
2
f
(
u
(
t
))
.
.
.
u
(
n
)
(
t
) =
p
n
³
f
(
u
(
t
))
, . . . , f
(
n
−
1)
(
u
(
t
))
´
where
p
n
is a polynomial in
n
variables with all nonnegative integer coe
ﬃ
cients.
The
fi
rst few polynomials are
p
1
(
x
) =
x, p
2
(
x, y
) =
xy, p
3
(
x, y, z
) =
x
2
z
+
xy
2
.
Notice that these polynomials are universal, i.e. are independent of the function
f
and
¯
¯
¯
u
(
n
)
(0)
¯
¯
¯
=
¯
¯
¯
p
n
³
f
(0)
, . . . , f
(
n
−
1)
(0)
´¯
¯
¯
≤
p
n
³

f
(0)

, . . . ,
¯
¯
¯
f
(
n
−
1)
(0)
¯
¯
¯
´
≤
p
n
³
g
(0)
, . . . , g
(
n
−
1)
(0)
´
where
g
is any analytic function such that
¯
¯
f
(
k
)
(0)
¯
¯
≤
g
(
k
)
(0)
for all
k
∈
Z
+
.
(We
will abbreviate this last condition as
f
¿
g.
)
Now suppose that
v
(
t
)
is a solution
to
(4.4)
˙
v
(
t
) =
g
(
v
(
t
))
with
v
(0) = 0
,
then we know from above that
v
(
n
)
(0) =
p
n
³
g
(0)
, . . . , g
(
n
−
1)
(0)
´
≥
¯
¯
¯
u
(
n
)
(0)
¯
¯
¯
for all
n.
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 Three '10
 Smith
 Equations, Analytic function, Holomorphic function, Eq., Smooth function, BRUCE K. DRIVER

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