80 4. POINT AND CONTACT TRANSFORMATIONS
and (4.47) is transformed to the Helmholtz equation
U
XX
C U
Y Y
C U D 0: (4.58)
ese results are presented in Arrigo and Hill [64].
It is important to realize that not all transformations of the form (4.14) are contact trans-
formations. Consider, for example,
x D X; y D U; u D U
X
: (4.59)
First derivatives transform as
u
x
D
U
Y
U
XX
U
X
U
XY
U
Y
; u
y
D
U
XY
U
Y
; (4.60)
showing that they possess second order terms. us, we wish to establish conditions to guarantee
that a transformation is, in fact, a contact transformation.
4.2 CONTACT CONDITION
We first consider the problem with ODEs before moving to PDEs. Suppose we have a trans-
formation
X D X.x; y; p/; Y D Y.x; y; p/; P D P .x; y; p/; (4.61)
where p D
dy
dx
and P D
d Y
dX
. We wish to calculate
d Y
dX
. From (4.61) we have
d Y
dX
D
Y
x
C Y
y
dy
dx
C Y
p
d
2
y
dx
2
X
x
C X
y
dy
dx
C X
p
d
2
y
dx
2
D P; (4.62)
or
Y
x
C Y
y
dy
dx
C Y
p
d
2
y
dx
2
D P
X
x
C X
y
dy
dx
C X
p
d
2
y
dx
2
: (4.63)
As X and Y are independent of
d
2
y
dx
2
, then from (4.63) we have
Y
x
C pY
y
D P
X
x
C pX
y
; (4.64a)
Y
p
D PX
p
: (4.64b)
e set of equations (4.64) is know as the contact conditions. In 1872, Lie [63] gave the following
definition of a contact transformation: if X; Y; P are independent functions of x; y; p such that
d Y P dX D .dy pdx/ (4.65)
4.2. CONTACT CONDITION 81
for some D .x; y; p/, then X D X.x; y; p/; Y D Y.x; y; p/; P D P .x; y; p/ is a contact
transformation. Inserting the appropriate differentials into (4.65) gives
Y
x
dx C Y
y
dy C Y
p
dp P .X
x
dx C X
y
dy C X
p
dp/ D .dy pdx/: (4.66)
Isolating coefficients with respect to dx, dy and dp gives
Y
x
PX
x
D p; (4.67a)
Y
y
PX
y
D ; (4.67b)
Y
p
PX
p
D 0; (4.67c)
and it is a simple matter to show that eliminating in (4.67) gives (4.64). With Lie’s definition
in hand, we can extend this to PDEs.
If
d U P dX Qd Y D
.
du pdx qdy
/
; (4.68)
then under the transformations
X D X.x; y; u; p; q/; Y D Y.x; y; u; p; q/; U D U.x; y; u; p; q/; (4.69)
(4.68) becomes
U
x
dx C U
y
dy C U
u
du C U
p
dp C U
q
dq (4.70)
P
X
x
dx C X
y
dy C X
u
du C X
p
dp C X
q
dq
(4.71)
Q
Y
x
dx C Y
y
dy C Y
u
du C Y
p
dp C Y
q
dq
(4.72)
D
.
d U U
X
dX U
Y
d Y
/
: (4.73)
Expanding and equating differentials gives
U
p
PX
p
QY
p
D 0;
U
q
PX
q
QY
q
D 0;
U
u
PX
u
QY
u
D ; (4.74)
U
x
PX
x
QY
x
D p;
U
y
PX
y
QY
y
D q:
e relations (4.74) are the contact conditions.
Example 4.7 Consider the Hodograph transformation
X D x; Y D u U D y: (4.75)
Calculating the first derivatives, we find
P D p=q; Q D 1=q: (4.76)
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