3.3. GENERALIZED KDV EQUATION 55
found that solutions of the KdV equation (3.40) can be found using solutions of the modified
Korteweg–deVries (MKdV) equation
v
t
6v
2
v
x
C v
xxx
D 0; (3.41)
via the transformation
u
D
v
x
v
2
; (3.42)
which today is known as the Miura transformation. We ask whether it’s possible to connect two
general KdV-type equations.
3.3 GENERALIZED KDV EQUATION
In this section we connect the following general NLPDEs:
u
t
C A.u/u
x
C u
xxx
D 0; (3.43a)
v
t
C B.v/v
x
C v
xxx
D 0; (3.43b)
via the substitution
u D F .v; v
x
/; F
v
x
¤ 0: (3.44)
Substituting (3.44) into (3.43a) and imposing (3.43b) leads to the following determining equa-
tions:
F
pp
D 0; (3.45a)
F
vp
D 0; (3.45b)
A.F /F
p
B.v/F
p
C 3pF
vv
D 0; (3.45c)
A.F /F
v
B.v/F
p
pB
0
F
p
C p
2
F
vvv
D 0; (3.45d)
where p D v
x
. Differentiating (3.45c) with respect to p twice (using (3.45a) and (3.45b)) gives
F
3
p
A
00
.F / D 0: (3.46)
From (3.46), (3.45a) and (3.45b) we see
A.F / D c
1
F Cc
2
; F D c
3
p CF
1
.v/; (3.47)
where c
0
c
3
are arbitrary constants and F
1
an arbitrary function. With these assignments,
returning to (3.45c) and isolating coefficients with respect to p gives
3F
00
1
C c
1
c
2
3
D 0; (3.48a)
B c
1
F
1
c
2
D 0; (3.48b)
which we solve as
F
1
D
c
1
c
2
3
6
v
2
C c
4
v Cc
5
; B D c
1
F
1
C c
2
; (3.49)
56 3. DIFFERENTIAL SUBSTITUTIONS
(c
4
and c
5
additional constants) and shows that (3.45d) is automatically satisfied. As we have
the flexibility of translation and scaling in the variables, without loss of generality we can set
c
0
D 0; c
1
D 6; c
2
D 1; c
3
D a; c
4
D b. us, solutions of
u
t
C 6uu
x
C u
xxx
D 0 (3.50)
can be obtained via
u D v
x
v
2
C av C b; (3.51)
where v satisfies
v
t
C 6
v
2
C av C b
v
x
C v
xxx
D 0: (3.52)
Setting a D b D 0 gives the Muira transformation (3.42); the usual KdV-MKdV connection,
setting b D 0, gives a connection between the KdV and what is known as the Gardner
equation [59].
Example 3.3 We consider the NLPDE
u
xx
C 2uu
xy
C u
2
u
yy
D 0: (3.53)
is particular PDE arises in the study of highly frictional granular materials [60]. We consider
a differential substitution
u D F .v
x
; v
y
/; (3.54)
and the target PDE
v
xx
C A.v; v
x
; v
y
/v
xy
C B.v; v
x
; v
y
/v
yy
C C.v; v
x
; v
y
/ D 0: (3.55)
Substitution of (3.54) into (3.53) and imposing (3.55), on isolating the coefficients of
v
xy
; v
yy
; v
xyy
; v
yyy
and various products gives rise to the following determining equations:
C
2
F
pp
C .C C
p
pC
v
C qAC
v
2qF C
v
/F
p
qC
v
F
q
D 0;
(3.56a)
2C.A F /F
pp
2CF
pq
.C
q
C qA
v
/F
q
C
.qAA
v
pA
v
2F C
p
2qFA
v
C
q
C CA
p
C 2AC
p
/F
p
D 0; (3.56b)
2BCF
pp
2F CF
pq
.C
q
C qB
v
/F
q
C
.AC
q
2F C
q
pB
v
C CB
p
2qFB
v
C BC
p
C qAB
v
/F
p
D 0; (3.56c)
.F A/
2
F
pp
C 2.F A/F
pq
C F
qq
C .2AA
p
2FA
p
A
q
/F
p
A
p
F
q
D 0; (3.56d)
2B.A F /F
pp
C 2.F
2
AF B/F
pq
C 2FF
qq
.A
q
C B
p
/F
q
C
.BA
p
B
q
2FB
p
2FA
q
C AA
q
C 2AB
p
/F
p
D 0; (3.56e)
B
2
F
pp
2FBF
pq
C F
2
F
qq
C .AB
q
2FB
q
C BB
p
/F
p
B
q
F
q
D 0; (3.56f )
.F
2
C A
2
B 2AF /F
p
C .2F A/F
q
D 0; (3.56g)
B.A 2F /F
p
C .F
2
B/F
q
D 0: (3.56h)
3.3. GENERALIZED KDV EQUATION 57
Eliminating F
p
from (3.56g) and (3.56h) gives
B D AF F
2
; (3.57)
and returning to either (3.56g) and (3.56h) gives
.
A 2F
/
.F A/F
p
C F
q
D 0; (3.58)
leading to two cases: (i) A D 2F and (ii) A D
FF
p
C F
q
F
p
, but each case ultimately leads to the
same result and thus, we will only pursue the first case. On setting A D 2F in equations (3.56d),
(3.56e) and (3.56f), we obtain
F
2
F
pp
2FF
pq
C F
qq
C 4FF
2
p
4F
p
F
q
D 0; (3.59a)
F
3
F
pp
2F
2
F
pq
C FF
qq
C 3F
2
F
2
p
2FF
p
F
q
F
2
q
D 0; (3.59b)
F
4
F
pp
2F
3
F
pa
C F
2
F
qq
C 2F
3
F
2
p
2FF
2
q
D 0; (3.59c)
and on the elimination of the second-order derivatives we obtain
F
q
FF
p
D 0; (3.60)
and with this shows that (3.59) is identically satisfied. As F satisfies (3.60), both (3.56b) and
(3.56c) reduce
C
q
F C
p
D 0: (3.61)
With F and C satisfying (3.59) and (3.61), and the last equation in (3.56), (3.56a) becomes
C C
p
pC
v
qF C
v
F
p
C C
2
F
pp
D 0: (3.62)
If we perform the operation @
q
F @
p
on (3.62) and use (3.60) and (3.61), we obtain
2CF
p
C
p
.p C qF /F
p
C
v
C 3C
2
F
pp
D 0: (3.63)
From (3.62) and (3.63), (upon subtraction), we have
C
F
p
C
p
C 2CF
pp
D 0: (3.64)
If C D 0 then we have
u D F .p; q/; v
xx
C 2F v
xy
C F
2
v
yy
D 0; F
q
FF
p
D 0: (3.65)
If C ¤ 0 then the compatibility of F
p
C
p
C 2CF
pp
D 0 with (3.60) and (3.61) leads to
F
pp
D 0; (3.66)
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