Pub
.
Ma
.
UAB
Vol
.
26
N° 1
Ma a
1982
ABOUT
SOME
QUESTIONS
OF
DIFFERENTIAL
ALGEBRA
CONCERN
I
NG TO
ELEMFJNTARY
FUIJCTIONS
San iago
Za zuelaA mengou
.
Dp o
.
de
Algeb a
y
Fundamen os,
Uni e sidad
de
Ba celona
Rebu
el
2 de
juny
del

l081
The
z udy
o6
elemen any
6unc
.íon4,
ha
.íz,
o6
hoee
6unc
.íon4
buíl
up by
u4
.íng
xa
.íonal
6unc
.íonz
o en
Q,
Exponen-
íalz,
Logan
.í hmz
and
Algebna
.íc
ope a
.íon4,
begán
zomewha
z
.íz
ema
.cally
w
.í h
he
d
.í enze
wohk4
ha
Jozeph
L
.íou .ílle
d
.íd
in
he
1830'4
.
Al hough
h
.íz
capi al
a,ím
waz
o
ob a
.ín
ome
ezul
.
abou
he
.ín egna
.íon
by
mean4
o6
elemen any
6une
.íona,
along
-
h
.íz
way
he
had
o
4 udy azpec 4
mo se
c
.í cunzc ibed
o
he
z nuc-
une
o6
hese
6une
.íonz
.
Abou
-
he
6
.ínz
po .ín
we
muz
zay ha
.í,
ce a
.ínly,
L
.íou .ílle
ob a
.íned
a
e4ul
ha ,
e en
.ímp o ed
z
.ínce
ha
ime,
haz
non
changed
e44en íally
.
I
4u6
.íce4,
6o~
ezemple,
o
compa e
he
wonk
o6
Líou ílle
g
.í en
in
121
and
Rozenl
.ích 'e
-
in
[6)
.
lde
can
obaen e
hen
ha
he
ín oduc íon
o6
new
langua-
ge
and new
hecníque
haz
only
línealízed
he
p oblem,
makíng
clean
whích
pnopen íez
o6
elemen a y
6une íon4
cha ac eníze
hem
.
U4íng
hí4
new
language
and new
hecníque
.í
íz
po4zí-
ble
deal mo e
cleanly
wí h 4ome
quez íon4 abou
elemen any
6unc-
íonz
.
Fon
exemple,
heí
í nedundancy,
ha
waz
alneady
ez abl
.í
4hed
by
Líou ílle
hímzel6
in
[21
and
Handy
in
[11
.
Now,
he
ínne-
dundancy
íz a
conzequence
o6
an
S cuc une
heonem
(1
.2)
ha
we
can
quíkly gí e
by
u4íng
Ro4enlích 'z hecníquez
.
Alzo,
he
non
zol abílí y
by'meanz
o6
elemen any
6unc íonz
o6
cen aín
clazeí-
cal
nanzcenden al
equa íon4
can
be
eazíly
ez ablízhed
.
(oe
z u-
dy
híz
in
Sec íon
2
.
Ano hen
ea men
o6
he
S nuc une
heo e ,
and
ínnedundance
quez íonz
(bu
non
u4íng
Rozenlích 'z
hecníque4)
can
be
6ound
in
he
Rízch'4 papen
[5)
.
1
.-
AN
STRUCTURE
THEOREM
.
Fi s
some
de ini ions
.
All
ields
will
be
conmu a i e
and
o
cha ac e is ic
ze o
.
1
.A
.

Remembe
ha i
E is a
ield,
a
map
D :E
-»
Eis a
de i a ion
o
E
i
:

(1)
V
x,

y
E
E
D(x+y)

=
D(x)
+
D(y),

(2)
M
x,y
EE
D(xy)

=
xD(y)
+
yD(x)
.
I
ollows
om
(2)

ha D(1)
=
0,
hence
by

(1)
D(z)
-0

y
z
E
z
.

The
se
CD
=
{x
E
E/ D(x)

= 0} is
he
se
o
cons an s
o
D
.
Gi en
ha

~
x
E
E
D(x
n
)
=
nD(x)x
n-1
,
D(x
-1
)
_
-D(x)x
_
1
we
ha e
ha
CD is a
sub ield
o
E
.
A
ield
E
wi h
a
amily o
de i a ions
A is a
di e en ial
ield
;
hen,
C =
l
C is
he
ield
o
cons an s
o
he
dí e en ial
ield
E
.
DEA
D
Le
E
C
F
wo di e en ial
ields
.
The ex ension
E
C
F is
di e en ial
i
Y
D
E
AF
,
DI
E
E
AE
.
Al hough
wo
di e en
de i a ions
o
AF
can
coincide
o e
E,
we
won'
dis inguish
be wen
AE
and
AF
.
Le
CE
,
CF
be
he
espec-
i e
cons an
ields
.
We
ha e
CE
CC
F
.
When
he
equali yholds
we
say
ha
he
ex ension
is
wi h
he
same
ield
o
cons an s
.
.
EXEMPLES
:
¢(X
1 ,

. .
.,
X
n
)
wi h
A=
(ó/d
x
.
)i=1
..
.n
is
a
di e en ial
ield
.
1
C(X)
C
O(X,
eX
)
is
a
di e en ialex ension
wi h
he
same
ield
o
cons an s
.
1
.B
.

The
elemen a y
na u e
ís
hen
o mula ed
in
he
nex
way
:
le
E
be
a
di e en ial
ield
;
x,
y
E
E
.
Then
- y =
Log(x)

-

Dy
=
Dx/
x
D
E
A

(y
is
Loga i hm
o
x)
-
y
=
Exp(x)

b

Dy/
y =
Dx

D
E
A

(y
is
Exponen ial
o x)
I E
C
Fisa
di e en ial
ex ension,
y
E
F is
Elemen a y
o e
E i
and
on-
ly
i
-
ei he
y is
algeb aic
o e
E
-
o
y =
Log(x)
being
x
E
E
-
o
y
=
Exp(x)
being
x
E
E
.
The di e en ial
ex ension
E
C
F is
Elemen a y
i F -
E(O
1
,

. .
.,
O
n
)
wi h
0
1
elemen a y
o e
E,
and
Oi
elemen a y
o e
E(O
1
,

. .
.,
O
i-1
)

i
>2
.
Then,
Ca dá
E =
Ca dAF
.
1
.C
.

The
ool
wich
allow
us
o
liea ize
he
a gumen s
is
he
Module
o
Di e en ials
.
A
as
cons uc ion
o
i
(su icien
o
us) is
he
ollowing
:
le
E
C
F
be íelds
and
conside
he
F- ec o
space
gene a ed
by
he
symbols
{dx}x
E
F
.
Le
us
impose
hem
he
ollowing
ela ions
:
(1)
d
x,

y
E
F

d(x+y)
=
dx
+
dy
(2)

x,

y
E
F

d(xy)

=
xdy
+
ydx
(3)
~xEE
dx
=
0
.
Then
we
ge
a
F- ec o space
called
he
Module
o
he
Di e en ials
o
E
CF
.
I s
symbol
is
S2F
E'
Remembe
oo
ha
i
{xi}i=1
..
.
a e
elemen s
o
F,
hen
hey
a e
algeb aical
ly
independen
o e
Ei
and
only
i he
amily
{dxi}i=l
. .
.
is
F-linea y
independen
on 2F/E
"
So T
.deg
.
E
F
=
dimF(nF/E)
"
(see [6],
P op
.3
)
his
wo k
:
The
nex
esul ,
due
o
Rosenlich ,
is
a
undamen al
one o
1
.1 .-
THEOREM
.

Le
E
C
F
be
a
di e en ialex ension
wi h
he
same
ield
o
cons an s
.
Le
C
be
his
ield
and
ake
y
l
,
.
.
.,
Yn
E
F,

z
i
s

. .

,

z
C,F-{0}
and

{ciJ
"
}
i=l
. .
.nC
C
such ha
Y
i =
i,

. .
.,
.n,

Y
D
E
p
j=l
. .
.
cijDzj/z
.
+
Dyi
E
J
E
.
Then
ei he
T
.deg
.
E
E(y
l
, . .
.,
Y
n
,
z
l
,
. .
.,
z
)
>n
o
he
n
elemen s
o
A
F/E
.
JLl
c
ij 1/
z
dz
j
+dyi
,
i -
1,
.
.
.,
n a e
C-linea y

dependen
.
P oo
:
see
Theo em
1
.
o
[6j
.
1
.D
.

Le
F
be
a
di e en ial
ield
.
We
say
ha
he
equalí y
Y -
Log
X
has
a
solu ion
in
F
i
he e
a e
elemen s
x, y
É
F
e í ying
¡
.
I is
na-
u al,
hen,
o
ask haw
many
solu ions
o
his
equalí y he e
a e
in an
elemen a y
ex ension
E
C
F
.
The ollowing
heo em, om
wich
Risch
gi es
ano he
e sion
in [S],
answe s
his
ques ion
.
P e iously
some
no a ion
:
le
E
C
F
be
an
elemen a y
di e en ial
ield
ex ension
wi h
he
same
ield
o
cons an s
:

E
C
F =
E(O
1
,

. .
.,
e
n
) .

Le
y
l
=
Log
x
l
, . .
.,
y -
Log
x
he
no algeb aic
casesamong
he
O
i
's
;
ha
is,
= T
.deg
.
EF
and
o
each
O
i
no algeb aic
(
o e
he
p eceeding
subex ension
)
he e
exis s
x
j
o y
j
such ha
Oi-
x
j
o y
j
dependeng
on
whe he
0
i
is
Exponen ial
o
Loga i hm
.
Suposse
hey
a e
a anged
acco ding
o
he o de
o
appe-
ance
and
ha
E is
an
algeb aic
closu e
o
E
.
1
.2
.-
Theo em
.

On
he
abo emen ioned
hypo hesis
i
he
equali y
Y -
=
Log
X
holds
in
F,
o any
solu ion
x, y
he eexis
c
1
,

.
.
.,
c
E
C

g
E
E
nF,

and
n
1
,

....
n
,

n
E
Z
such ha
P oo
:
i
he
equali y
holds
in F
we
can
conside
he
sys em
n n
y
+
c
l
y
l
+
...
+
c
y
=
,
x
nx
l
l . . .
x
=
Dy
i-
Dxi
./
x

=
0EE
.
1
Dy
-1/x
Dx
= 0
E
E
By
Theo em
1
.1
we
ge
DEA
-
ei he
T
.deg
.
E
E(y
1
, . .
.,
Y
,
y, x
l
,
.
.
.,
-
o
he
elemen s
o
n
F/E
:
(dy
.i
-1/x
.
dx
i
), i -
1,
. .
.,
,
i
(dy
- 1/xdx)
a e
C-linea ydependen
.
So
he eexis
c
l
,

.
.
.,
c
,
c
E
C
no
all
ze o such ha
We
can
also ake
c ~ 0
since
o he wise
Bu
i y = 0
J
.
o
some
j,
because
o
he elemen a i y
o
E
C
F,
each dy
i
,
dx
.
excep dy
is
a
linea
combina ion
o
he
p eceeding
-1
dO
s
wi h
coe-
J
icien s
in F
.
Bu
hey
a e F-linea y
independen
beacuse
o
1
.C
.
So
c
=0
.
The
same
happens
i
x
=0i
.
o
some
i
.
Appliyng
epea dly
his
a gumen
we
ge
c
1
-
.
.
.
-c
=
0,
no
possible
.
He e
i is
clea
ha
only
he secondcondi ion
is
possible
.
(1)
c(dy
-
1/
x
dx)
+

ci(dyi
.-
1/x
.
dx
i
)
=0
.
ei (dyi
-
1/
xidx
i
)
= 0
.
i=1
Hence,
di iding
by
c,
we
can assume
g
,
(2)
dy
+
c
1
dy
1
+
...
+
c
dy
-
1/
x
dx
+
c
1
1/
xldx
1
+
. . .
+
c 1/x dx
Conside
now
a
maximal
Q-linea y
independen
sys em
among
he
{1,

cl
,

. .
.,

c
}

:

{e
l
,

. .
.,

e
k
}

such
ha
e
l
-

1
.

Then
k
d
i

:

c
i
-

L

qijej

,
qij
E
Q
Ji,
j
.

The e o e
j=1
k
1/
x
dx
+ c
1
1/
x
dx
1
+
.
. .
+
c l/x
dx
- e
1
1/
x
dx
+

1
g
lj
e
j
dx
1
+
.
. .
+
1


j=1
k


+ 1
g j e
j
l/
x
dx
-
e
l
(1/ x dx
+

gill/xidxi)
+
. .
.
+
e k
(
z
gikl/x
dxi)
j=1

i=1

i=1
i
-
e
l
l/
d
l
+
...
+
e
k
l/
d
k
,
1

k
Then
(2')
d(y
+
c1dy1
+
. .
.
+
c
dy
)
- e
1
1/
d
1
+
...
+
e
k
l/
d
k
.
1

k
By
P op
4
.
o
16]
we ha e
- y
+
cly1
+
...
+
c
y - g
EE
1F
-
i
CE 1F
~i
.
So
xx
g11
.. .
.,
xg l
EE¡1F
.
Bu
i
~i
qij
-
mil/1
'
mil'
mEZ
we
ge
m
l-l

m

_
xmxl
..
.
x
l =
EE
1F,
q .e .d
.
Some ímes
i
is
possible
o
gi e
a
comple e
desc ip ion
o he
solu ion
o
Y
=
Log
X
.
This
happens
when
E is
a
classical
di e en ial
ield
1
.3
.-
Theo em
.

.
On
he
hypo hesis
o
Theo em
1
.2,
supose
mo eo e
ha
E =
C(z),
C
he
ield
o
cons an s
o
E
and
z
iÉC
such ha
~
D
=,5

Dz
E
C
.
Then,
any
solu ion
o
he
equali y
can
be
w i cn
in
he
o m
y=clyl+
. .
.
+c y +c
being
1
q
11
q l
xx
1
.
.,
x
"
qlk q k
k - x
1
. . .
x
x = x
l
l
. .
.

x
c'

,

being
c
l
,

. .
.,

c
E
Q,

c,

c'
EC
.
P oo
:
applying
he
same
a gumen
used
in
1
.2
and
aking he
sys em

we
ge
he e exis
q
l
,

. .
.,
q
E
Q
such
ha
xx
l
1 . . .

x
E

l
F
.
Bu
any
de i á ion
has
only
one
ex ension
o
an
algeb aic ex ension
o
E
([8]
Cap
.
2,
17,
Co
.
2)
.
so
C 1F
is a
ield
o
cons an s
and
gi en
ha
E
C
F
is an
ex ension
wi h
he
same
ield
o
cons an s
we
ha e
C
=
C 1F
.
The e o e
(1)
x
=
x
1 1
.
.
.
x
c'

,

c'
E
C
.
De i ing
(1)
yields
q
l q
D
E
p

,

Dy
=
Dx/
x °
D(x
1

. .
.,
x

)1(x11
'..

xq )
1
g
1
Dx
1
/
x
+
.
.
.
+
g
Dx
/x
.
So
1

y - g
l
y
l
+
"
'

+
g y
+
c

,

c
E
C,

q
.e
.d
.
Rema k
:
i
can
happen
ha
x
1
1 . . .
x
5E
F
.
Howe e ,
i
is an
algeb aic
poin
ha
doesn'
dis u b
he
elemen a i y
o
he
p ocesa
.
2
.-
SOME
CowsEQUFNCEs
.
2 .A
.

The i s conclusion
we d aw om
1
.
is
ha
we'll
name
The
I e-
dundance
o
Elemen a y
Func ions
.
This
means
ha
building
up
elemen a y
ex ensions
by
means
o
algeb aic
elemen s,
loga i hm
elemen s
o
exponen-
ial
elemen s
a e
comple ly
independen
p ocesses
:
no
one
o hem
can
be
ob ained
om
he
o he s
.
In
o de
o
se he
p oblem
we'll
use
an
adecua e language
;
we
say
ha
he
di e en ial ex ension
E
C
F is
Algeb aic
i
he
ield
ex en-
sion
E
C
F so
is
;

i
la
Loga i hmic
i F =
E(0
1
,

.
.
.,

0
n
)

such
ha
0
1
=
LogT
1
,
T 1
EE,
0
1
=
Log
i
,
T i
EE(0
11

. .
.,
0
i-1
)

'Vi
>2
.
Changing
Log
by
Exp
we
ha e
an
Exponen ial
ex ension
.
2
.1
.-
Lemma
.

Le
E
C
F =
E(0)
be
a
di e en iaiex ension
wi h
he
same
ield
o
cons an s
C
and
0
Y-
E
.
(1)
I
VD
=á

DO(
-
=E,
hen
0 is
anscenden al
o e
E
.
(2)
I
VD
E
p

DO/
0
C=E,
hen
0
is
.algeb aic
o e
E
i
and
only
i
he e
exis s
n
C
=N such
ha
O n
E
E,
and
he
i educible
polynomial
o
0
o e
E
is X n - On
,
n
being
he
leas
o
he e
na u als
.
P oo
:
assume
0
o
be
algeb aic
o e
E
and
le
P(X)
-
)In
+a
1
X
n-1
+
.
.
.
+
+ a
n-1
X +
an
be
he
i educible
polynomial
o
0
.
Then
.
(*) On
+
a
1
0
n-1
+
...
+
a
n-1
0
+
a
n = 0
.
(1)
De i ing
(*)
we
ge
1
D
E
p,

(Da
1
+
nD0)O
n-1
+
...
- 0
.
Gi en
ha P(X)
is
he
i educiblepolynomial
o
0
o e
E
we
ha e ha
V
D
E
p
Da
1
+
nDO
=0
.

So

V
D
C=¿
DO
=
D(-a
1
/n)
and
o+al/n is a
cons an
.
Due
o
E
C
F
ís
wi h
he
same
íeld
o
cons an s
we
ge O
E
E,
no
possible
.
(2)
Now
i
su ices
o
p o e
ha
0
n
C=E
.
De i ing
(*)
we
ge
Y
D
E
A
nDO/
OOn
+
(Da
1
+
(n-1)D0/0)O
n-1
+
...
+
Da
n
=
0
.'Bu
an¢ 0,

so
Dan =
nDO/
0
aJ
D
E
A
.
Hence
Da
n/an -
nDO/
0
.*
Dan
/a
n =
DO
n
/
On
-
D(a
n
/
On) - 0
YDEA
.
So a
n
/
O
nECCE,
and
On
EE,
q
.e.d
.
2
.2 .-
Theo em
.

Le
E
be
a
di e en ial
ield
wi h
ield
o
cons an s
C
.
Le
E
C
F
= (0
1
,

.
.
.,
0
)
be
an
elemen aldi e en ialex ension
wi h
he
same
ield
o
cons an s
.
Then
:
(a)
W
hen
F is
Loga í hmic,
E
C
F is a
pu ely
anscenden al
ex-
ension
.
I E
C
F is
Exponen ial,
E
C
F is
pu ely
anscenden al
unless
he-
e
exis
n
i
,

. .
.,
n
E
=-
Z
such ha

0n
1

. .

On
E
E
.
1
'

Le
x
E
E
.
(b)
The
equali y
Y
=
Log(x)
ne e
holds
in
F-E
i E
C
F is
Alge-
b aic
o
Exponen ial
.
(c)
The
equali y
Y
e
Exp(x)
ne e
holds
in
F-E
i E
C
F
is
Loga-
i hmic,
and
i
he e
is a
solu ion
when
E
C
F
is
Algeb aic
hen
he e
e-
xis s
n
E
N
such
ha
y
nE
E
.
Mo eo e ,
i E =
C(z),
z 9E
C,
Dz
E
C
E
A
being
C
he
ield
o
cons an s
o
E, C
algeb aica ly,closed,
hen
he e
a e
no
excep ions
o
he
case
(c)
.
P oo
:

(a)
The
s amen
is an
easy
consequence
o
Lemma
2
.1
o he
Lo-
ga i hmic
case
.
Assume
ha
E
C
F
is
Exponen ial
and no
pu ely
anscen-
den al
ex ension
.
By
Lemma
2
.1
he e
exis s
Os
,
p
E
N
such ha
OS
E
E(O1,
.
.
.,
0
s-1
) .

Le
Ok
he
i s
o
hem,
ha
is,

0
1
,

. .
.,
O
k-1 a e algeb aic
independen
o e
E
and
Ok
E
E(O
1 , .
.
.,
0k-l)
.
Then
by
Theo em
1
.2
we
ge
he
s amen
.
(b)
Le
y
be
a
solu ion
.
Then,

Y
D
E
p
Dy
-
Dx/x
.
Since
x
E
E
we
can
Cake
Che di e en ial
ex ension
E
C
E(y)
.
By
2
.1
y is
no algeb aic
o-
e
E
.
Suposse
now
E
C
F is
Exponen ial
.
Then,
by
1
.2
we
ge
he e
exis
. .
.,
c
:

C, n, n
, . .
.,
n
E
.

Z
such
ha
cl
1
l
k

i
E

1

lk

n
.

n
.
_

i 1
y
+
c
.
.
+
. . .
+
c
.
.
EEnF,
xn0
.
1 . . .
0
.
k
EEnF,
i
l
i1

l
k
lk

il

lk
being
0
.
,
. .
.,
0
.
a
maximal
algeb aically
independen
sys em
o e
E
among
il lk
0
1
9

. .
.,
.O
líke
in

1
.D
.
Bu
xEE
:

so n

,
.
. .
.,
n
.

a e
0,

and
looking
in
1
.2

o Che
i1

l
k
cons uc íon
o
he ena u als
we
ha e
c
.
_
...
= c
.
=
11

lk
0
.
Hence
y
E
E
nF,
no
possible
as
we
ha e p o ed
abo e
.
(c)
The
Lemma
2
.1
assu e
us
ha
i
y is
algeb aic
o e
E
hen
he e exis
n
E
N
such
ha
y
n
CE
.
Suposse
E
C
F
is
Loga i hmic
.
By
1
.2
we
ha e
he eexis
n
1
,

. .
.,

n
,

n
E
Z,

c
l
,

. .
.,

c
E
=-
C
such
ha
_

nn
_
x
+
c
1
0i
+
. . .
+
c 0
EEnF
,
y
n
1
1
.. .
EEnF
.
Now,
by
Lemma
2
.1,
0
1
,
.
.
.,
O is
an
algeb aically
independen
sys em
o e
E, so c
1
-
...
= c
- 0,
and
n
1
-
.
. .
= n - 0
(
look
o Che
cons uc ion
o
n
1
,

. .
.,
n
in
1
.2
) .
Hence
y
E
Én
F=
E,
no
possible
.
On
Che assum íon
ha
E=
C(z)
.....
ce
o
aplying
Theo em
1
.3 o yn
E=-
E =
CM
.
Rema k
:
an
example
ha
gi e
us an
excep ion
o
(a) is
:
E =
Q(z)(Exp(2z+2z
2 )), F -
E(Expz,
Expz
2
)
.
Then,
Exp(z+z
2
)
E
F-E
and
is
algeb aic
o e
E
.
2 .B
.

The
ques ion
o
whe he
some
anscenden al
equa
i
o
ns
can
be
sol ed
by
means
o
elemen a y
u
n
c
i
ons
some imes
can
be
answe edusing
Che S uc u e
heo em
1
.2
.
Le
us
see wo classical
examples
:
assume
E
C
F
is
a
di en ial
ex ension
wi h
Che
same
ield
o
cons an s
.
Le
C
be his
ield
and
E
-
C(z)
such ha
1
D
E
á
Dz
E
C,
z
§E
C
.
Suposse
C is
algeb aically
closed
and
E
C
F
Elemen a y
.
Conside
Che
equa ion

aY
=
Log(BY)

a,
B
C
-
E
.
Che
s amen
is
consequen-
Suposse
he e
ís
a
solu ion
in
F,
y
.
Using
he
same
no a ion
o
1
.3
e
ge
c, c
E
C, c
1
,

.
.
.,
c
n
E
Q
.
Passing
o
he
Module
o
di e en ials,
2F,/E,
we
ha e
(-e
í
)1/
x
.
dx
í
.
1
1
y =
cx
1
. . .
x
nn
,

y - c
+
ely1
+
...
+
cnyn
,
'((x
1 1
.
.
x
n
cn)e
1
1/x1
dx
1

+

. . .

+

(x
1
c1

. .

x
n
.
cn
a

)c
n
1/
xn
dx
n
)°
a
=
e1dy1
+
...
+
cndyn
,
a' -
ac/B
.
Bu
aking
he
Module
o
di e en ials
espec
on
he
penul ima-
e
subex ensionno algeb aic
and
aking
in o
accoun
1
.C
e ha e ha
a'
(x
1
1. . .
x
n
n)cn1/x
dx
n= c
n
dy
n
,
whé e
dx
n
n
se o
he
elemen a i y
o
E
C
F,
being
one
o hem
no
ze o
.
The e o e
cn=
=
0
;
epea ing
his
a gumen
we
ha e
cí
= 0

i
.
Consequen ly,
any
solu-
íon
is
i ial
.
As
a
pa icula
case
and
aking
E _
¢(z) e
ha e
ha
he
equa-
ion

Log(Y)
=
Y/
z

has no
solu ion
by
means
o
elemen a y
unc ions
.
Wi h
he
same
hypo esis
conside
now
he
equa ion
(*)

Y +
a =
BExp(yY),+
V
Exp(-YY),

a,

0,0',
Y
E
E
.
Le
y
be
a
solu ion,
y
E
F
.
We
can
suposse
also ha

Exp(yy)
EF
.
Then
by
1
.3
and
wi h
he
same
no a ion
we
ha e
(**)

Yy
=
c
+
cly1
+

. . .

+
cn
yn
,

c
E
C,

c
1
,

. .
.,

en
E
Q
.
Subs uíng
o
yy
in
(*)
we
ge
ha
y
+
a =
Bx
1
1. . .
x
n
n.+
.B'x
1
c1
...
x
n
cn
(whe e
e
ha e
ope a ed
adequa ly
B,
B')
.
Passing
now
o
he
Module
o
di e en ials
RF/E
e
ge
0
o dy
n
= 0
becau-
dy
=

B(x
c1
...
xnn)cí1/x
i
dx
i
+

B'(x
1
c1
...
xn
cn
)(-c
i
)1/
x
.
dx
i
.
i

i

i
Bu
aking
ín o
accoun
(**)
we
ha e
ha
1/
Y
(c1dy1
+
. . .
+
c
n
dy
n
)
_

B(x
e1
. .
.
x
en
)c
i
1/
xi
dx
í
+

B~(x
1
c1
. . .
xn
cn
)
1
3