110 5. FIRST INTEGRALS
5.4 FIRST INTEGRALS AND LINEARIZATION
As we’ve seen in the preceding section, some PDEs admit very general classes of first integrals.
A natural question is: can they be used to simplify the form of a PDE?
5.4.1 HYPERBOLIC MA EQUATIONS
Lie [72] and [73] considered PDEs of the form
Ar C Bs C C t C D
rt s
2
D E; (5.154)
that admitted two general first integrals, say
F
1
.
˛
1
; ˇ
1
/
D 0; F
2
.
˛
2
; ˇ
2
/
D 0: (5.155)
In (5.155), it is assumed that F
1
and F
2
are arbitrary functions of ˛
i
and ˇ
i
that are functions
of x; y; u; p; and q. Lie was able to show that when these first integrals exist, it is possible to
transform (5.154) to U
XY
D 0:
We define
X D ˛
1
; Y D ˛
2
; P D ˇ
1
; Q D ˇ
2
(5.156)
and we’ll assume that U exists, such that the contact conditions are satisfied. We consider
U
XY
D 0:
is can be rewritten as
@P
@Y
D
@.X; P /
@.X; Y /
D
@.X; P /
@.x; y/
=
@.X; Y /
@.x; y/
D 0;
giving
@.X; P /
@.x; y/
D 0 (5.157)
or
X
x
P
y
X
y
P
x
D 0: (5.158)
We now bring in (5.156) and suppress subscripts. us, (5.158) becomes
˛
x
C p˛
u
C r˛
p
C s˛
q
ˇ
y
C qˇ
u
C sˇ
p
C tˇ
q
˛
y
C q˛
u
C s˛
p
C t˛
q
ˇ
x
C pˇ
u
C rˇ
p
C sˇ
q
D 0: (5.159)
Expanding (5.159) gives
˛
p
ˇ
y
C qˇ
u
ˇ
p
˛
y
C q˛
u

r
C
ˇ
q
.
˛
x
C p˛
u
/
C ˛
q
ˇ
y
C qˇ
u
ˇ
q
˛
y
C q˛
u
˛
p
.
ˇ
x
C pˇ
u
/
s
C
ˇ
q
.
˛
x
C p˛
u
/
˛
q
.
ˇ
x
C pˇ
u
/
t (5.160)
C
˛
p
ˇ
q
˛
q
ˇ
p
rt s
2
D
˛
y
C q˛
u
.
ˇ
x
C pˇ
u
/
.
˛
x
C p˛
u
/
ˇ
y
C qˇ
u
:
5.4. FIRST INTEGRALS AND LINEARIZATION 111
As ˛ and ˇ are first integrals, they satisfy the following:
D
.
F
x
C pF
u
/
CF
p
C
1
DF
q
D 0; (5.161a)
D
F
y
C qF
u
C
2
DF
p
AF
q
D 0; (5.161b)
where
D
2
2
BD C AC CDE D 0: (5.162)
As we are assuming that D ¤ 0, then using (5.161), we eliminate the terms
˛
x
C p˛
u
; ˛
y
C q˛
u
; ˇ
x
C pˇ
u
; ˇ
y
C qˇ
u
(5.163)
from (5.160). Simplifying gives
A
D
˛
p
ˇ
q
ˇ
p
˛
q
r C
.
1
C
2
/
˛
p
ˇ
q
ˇ
p
˛
q
s
C
C
D
˛
p
ˇ
q
ˇ
p
˛
q
t C
˛
p
ˇ
q
˛
q
ˇ
p
rt s
2
D
1
2
AC
D
2
˛
p
ˇ
q
ˇ
p
˛
q
:
As we have assumed that ˛
p
ˇ
q
ˇ
p
˛
q
¤ 0, this term cancels, giving
A
D
r C
.
1
C
2
/
s C
C
D
t C
rt s
2
D
1
2
AC
D
2
:
From (5.162), we deduce that
1
C
2
D B=D and
1
2
D .AC C DE/=D
2
, giving
A
D
r C
B
D
s C
C
D
t C
rt s
2
D
AC C DE
U
2
AC
D
2
: (5.164)
Simplifying gives rise to (5.154).
Recall the contact conditions from the previous chapter. ey are:
U
p
PX
p
QY
p
D 0;
U
q
PX
q
QY
q
D 0;
U
u
PX
u
QY
u
D ; (5.165)
U
x
PX
x
QY
x
D p;
U
y
PX
y
QY
y
D q:
ese will be needed for the following examples.
Example 5.9 Consider
3r C s C t C rt s
2
C 9 D 0: (5.166)
112 5. FIRST INTEGRALS
is PDE admits the first integrals
F
1
.x C 2y C p; 3x 3y q/ D 0; F
2
.x 3y C p; 2x C3y C q/ D 0: (5.167)
Here we try
X D x C2y Cp; Y D x 3y C p; (5.168a)
P D 3x 3y q; Q D 2x C3y C q: (5.168b)
At this point, we need to find U such that the contact conditions (5.165) are satisfied, namely
U
p
D 5x;
U
q
D 0;
U
u
D ; (5.169)
U
x
D 5x p;
U
y
D 15y 5q q:
Cross differentiation of (5.169) shows they are consistent; thus, U (and ) exist. Integrating
(5.169) gives
U D 5xp C
5
2
x
2
15
2
y
2
5u; (5.170)
and calculating U
XY
using (5.168) and (5.170) gives
U
XY
D
3u
xx
C u
xy
C u
yy
C u
xx
u
yy
u
2
xy
C 9
5
.
u
xx
C 1
/
D 0: (5.171)
Example 5.10 Consider
u
2
y
u
xx
3u
x
u
y
u
xy
C 2u
2
x
u
yy
D u
x
u
2
y
: (5.172)
is PDE admits the first integrals
F
1
y C 2
p
q
; e
x
p
q
D 0; F
2
u; e
x
p
q
2
D 0: (5.173)
Here we try
X D y C 2
p
q
; Y D u; (5.174a)
P D e
x
p
q
; Q D e
x
p
q
2
: (5.174b)
5.4. FIRST INTEGRALS AND LINEARIZATION 113
At this point, we need to find U such that the contact conditions (5.165) are satisfied, namely
U
p
D 2e
x
p
q
2
; (5.175a)
U
q
D 2e
x
p
2
q
3
; (5.175b)
U
u
D e
x
p
q
2
C ; (5.175c)
U
x
D p; (5.175d)
U
y
D e
x
p
q
q: (5.175e)
Cross differentiation of (5.175) shows they are not consistent, so we modify our choice. We now
pick
X D y C 2
p
q
; Y D u; (5.176a)
P D a e
x
p
q
; Q D b e
x
p
q
2
; (5.176b)
with hopefully suitably chosen constants a and b. e contact conditions become
U
p
D 2ae
x
p
q
2
; (5.177a)
U
q
D 2ae
x
p
2
q
3
; (5.177b)
U
u
D be
x
p
q
2
C ; (5.177c)
U
x
D p; (5.177d)
U
y
D ae
x
p
q
q: (5.177e)
Compatibility between these gives
p
C be
x
1
q
2
D 0; (5.178a)
p
p
C 2ae
x
p
q
2
D
0; (5.178b)
q
p
ae
x
1
q
D 0; (5.178c)
q
2be
x
p
q
3
D 0; (5.178d)
(5.178e)
114 5. FIRST INTEGRALS
p
q
C 2ae
x
p
2
q
3
D 0; (5.178f)
q
q
C C ae
x
p
q
2
D 0; (5.178g)
x
C p
u
be
x
p
q
2
D 0; (5.178h)
y
C q
u
D 0; (5.178i)
q
x
p
y
C ae
x
p
q
D
0: (5.178j)
From (5.178a) and (5.178c) we see that b D a, and further from (5.178a) and (5.178b) we get
D ae
x
p
q
2
:
is assignment shows that the entire system (5.178) is satisfied; thus, U exists satisfying the
contact conditions. Integrating (5.177) (with a D 1) gives
U D e
x
p
2
q
2
; (5.179)
and under the contact transformation
X D y C 2
p
q
; Y D u; U D e
x
p
2
q
2
; (5.180)
gives U
XY
as
U
XY
D
e
x
u
2
y
u
xx
3u
x
u
y
u
xy
C 2u
2
x
u
yy
u
x
u
2
y
u
y
2z
2
y
z
xx
4z
x
z
y
z
xy
C 2z
2
x
z
yy
z
x
u
2
z
D 0: (5.181)
Example 5.11 Consider
u
y
u
y
C 1
u
xx
2u
x
u
y
C u
x
C u
y
C 1
u
xy
C u
x
.
u
x
C 1
/
u
yy
D 0: (5.182)
is PDE admits the first integrals
F
1
x C u;
q C 1
p
D 0; F
2
y C u;
p C1
q
D 0: (5.183)
Here we try
X D x Cu; Y D y C q; (5.184a)
P D
q C 1
p
; Q D
p C1
q
: (5.184b)
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