76 4. POINT AND CONTACT TRANSFORMATIONS
and the second-order derivatives transform as
u
xx
D
U
XX
U
3
X
;
u
xy
D
U
Y
U
XX
U
X
U
XY
U
3
X
; (4.19)
u
yy
D
U
2
Y
U
XX
2U
X
U
Y
U
XY
C U
2
X
U
Y Y
U
3
X
:
4.1.2 LEGENDRE TRANSFORMATION
ese transformations are of the form
x D U
X
; y D U
Y
; u D X U
X
C Y U
Y
U: (4.20)
Using Jacobians as we did for point transformations, we find the first derivatives transform as
u
x
D X; u
y
D Y; (4.21)
and the second derivatives transform as
u
xx
D
U
Y Y
U
XX
U
Y Y
U
2
XY
; u
xy
D
U
XY
U
XX
U
Y Y
U
2
XY
; u
yy
D
U
XX
U
XX
U
Y Y
U
2
XY
: (4.22)
4.1.3 AMPERE TRANSFORMATION
ese transformations are of the form
x D U
X
; y D Y; u D XU
X
U: (4.23)
Using Jacobians, we find the first derivatives transform as
u
x
D X; u
y
D U
Y
; (4.24)
and the second derivatives transform as
u
xx
D
1
U
XX
; u
xy
D
U
XY
U
XX
; u
yy
D
U
XX
U
Y Y
U
2
XY
U
XX
: (4.25)
Note that we could easily have chosen
x D X; y D U
Y
; u D Y U
Y
U; (4.26)
and the derivatives would have transformed similarly.
We now consider some examples.
4.1. CONTACT TRANSFORMATIONS 77
Example 4.1 Consider
u
2
y
u
xx
2u
x
u
y
u
xy
C u
x
u
uu
C u
3
y
D 0: (4.27)
Under the Hodograph transformation (4.16), Equation (4.27) becomes
U
XX
1 D 0; (4.28)
which integrates giving
U D
1
2
X
2
C F .Y /X C G.Y /; (4.29)
where F and G are arbitrary functions; via (4.16) we obtain the exact solution
y D
1
2
x
2
C F .u/x C G.u/: (4.30)
Example 4.2 Consider the nonlinear diffusion equation
u
t
D
u
xx
u
2
x
: (4.31)
Under the Hodograph transformation
t D T; x D U; u D X; (4.32)
Equation (4.31) becomes
U
T
D U
XX
; (4.33)
the heat equation!
Example 4.3 Consider
u
xx
x
u
xx
u
yy
u
2
xy
D 0: (4.34)
Under the Legendre transformation (4.20), Eq. (4.34) becomes
U
Y Y
U
XX
U
Y Y
U
2
XY
U
X
U
Y Y
U
XX
U
Y Y
U
2
XY
U
XX
U
XX
U
Y Y
U
2
XY
U
2
XY
.U
XX
U
Y Y
U
2
XY
/
2
D 0;
(4.35)
which, after simplification, becomes
U
Y Y
U
X
D 0; (4.36)
the heat equation.
Example 4.4 Consider
u
xx
u
yy
u
x
u
y
u
xx
u
yy
u
2
xy
D 0: (4.37)
78 4. POINT AND CONTACT TRANSFORMATIONS
Under the Legendre transformation (4.20), Eq. (4.37) becomes
U
Y Y
U
XX
U
XX
U
Y Y
U
2
XY
.X Y /
U
Y Y
U
XX
U
2
XY
.U
XX
U
Y Y
U
2
XY
/
2
D 0; (4.38)
which, after simplification becomes
U
XX
U
Y Y
D X Y: (4.39)
A particular solution of (4.39) is U D .X
3
C Y
3
/=6; with this, we can transform (4.39) to the
standard wave equation
V
XX
V
Y Y
D 0; (4.40)
via
U D V C
1
6
X
3
C
1
6
Y
3
: (4.41)
e solution of (4.40) is
V D F .X C Y / C G.X Y /: (4.42)
Composing (4.42), (4.41) and (4.20) gives
x D F
0
.X C Y / C G
0
.X Y / C
1
2
X
2
;
y D F
0
.X C Y / G
0
.X Y / C
1
2
Y
2
; (4.43)
u D .X C Y /F
0
.X C Y / C .X Y /G
0
.X Y /
F .X C Y / G.X C Y / C
1
3
X
3
C
1
3
Y
3
;
the exact solution of (4.37)
Example 4.5 Consider
u
xx
u
yy
u
2
xy
D 1: (4.44)
Under the Ampere transformation (4.23), (4.44) becomes
1
U
XX
U
XX
U
Y Y
U
2
XY
U
XX
U
2
XY
U
2
XX
D 1; (4.45)
which, after simplification, becomes
U
XX
C U
Y Y
D 0; (4.46)
Laplace’s equation!
4.1. CONTACT TRANSFORMATIONS 79
It is possible to combine several transformations as the following example demonstrates.
Example 4.6 We consider the PDE
u
xx
u
yy
u
2
xy
D
u
2
x
C u
2
y
2
: (4.47)
is PDE appears in elasticity [64]. Under the Ampere transformation,
x D X; y D U
Y
; u D Y U
Y
U; u
x
D U
X
; u
y
D Y; (4.48)
and (4.47) becomes
U
XX
C
U
2
X
C Y
2
2
U
Y Y
D 0: (4.49)
If we let U D Y V , (4.49) becomes
V
XX
C Y
3
V
2
X
C 1
2
.
Y V
Y Y
C 2V
Y
/
D 0: (4.50)
Introducing the new variable Y D 1=S, the Y derivatives transform as
V
Y
D S
2
V
S
; V
Y Y
D S
4
V
SS
C 2S
3
V
S
;
so that (4.50) becomes
V
XX
C
V
2
X
C 1
2
V
SS
D 0: (4.51)
e transformation (4.48) under the change of variables so far is
x D X; y D V SV
S
; u D V
S
; u
x
D
V
S
S
; u
y
D
1
S
: (4.52)
Next, we perform a Legendre transformation on (4.51)
X D W
; S D W
; V D W
C W
W; V
X
D ; V
S
D ; (4.53)
giving
W
C
2
C 1
2
W
D 0: (4.54)
If we let
D tan ; W D sec Q; (4.55)
then (4.54) becomes
Q
C Q
C Q D 0: (4.56)
Composing all the transformations into one (with new variables) gives
x D cos XU
X
C sin XU; y D sin X U
X
cos XU; u D Y; (4.57)
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