2.2. SECOND-ORDER PDES 31
where A; B; C and P are arbitrary functions of t and x. Substituting further into (2.162c) gives
3PQ C
2
3
P
3
u
3
2P
.
2P
x
AP
/
u
2
C
.
P
t
C 3P
xx
2PA
x
2AP
x
2BP
/
u (2.164)
A
t
A
xx
C 2AA
x
C 2B
x
2CP D 0:
From (2.164) we find that if P ¤ 0 then we have a preliminary from of Q. us, two cases
emerge:
.i/ P ¤ 0;
.i i/ P D 0:
Case 1 P ¤ 0 Dividing (2.164) by 3P and isolating Q gives
Q D
2
9
P
2
u
3
C
2
3
.
2P
x
AP
/
u
2
P
t
C 3P
xx
2PA
x
2AP
x
2BP
3P
u
A
t
A
xx
C 2AA
x
C 2B
x
2CP
3P
: (2.165)
As Q D Q.u/ only, from (2.165) we deduce that P D p, A D a, B D b and C D c, all constant.
us, Q has the form
Q D
2
9
p
2
u
3
2
3
apu
2
C
2
3
bu C
2
3
c: (2.166)
With this choice of Q we find the remaining equation in (2.162) is automatically satisfied. us,
we find
X D pu C a; U D
1
3
p
2
u
3
pau
2
C bu C c; (2.167)
and the following two are compatible:
u
t
D u
xx
2
9
p
2
u
3
2
3
apu
2
C
2
3
bu C
2
3
c; (2.168a)
u
t
C .pu C a/u
x
D
1
3
p
2
u
3
pau
2
C bu C c: (2.168b)
Case 2 P D 0 In this case, (2.163) reduces to
X D A; U D Bu C C; (2.169)
while (2.162c) and (2.162d) become
A
t
C 2AA
x
C 2B
x
A
xx
D 0; (2.170)
32 2. COMPATIBILITY
and
.Bu C C /Q
0
C .2A
x
B/Q D .B
t
C 2BA
x
B
xx
/u C C
t
C 2CA
x
C
xx
: (2.171)
Since (2.171) should only be an ODE for Q, four cases arise:
(i) B D 0, C D 0,
(ii) B D 0, C ¤ 0,
(iii) B ¤ 0, C D 0,
(iv) B ¤ 0, C ¤ 0.
Each will be considered separately.
Subcase (i) B D 0, C D 0
In this case, (2.170) and (2.171) become
A
t
C 2AA
x
A
xx
D 0; 2A
x
Q.u/ D 0; (2.172)
from which we deduce that A D a, a constant, since Q ¤ 0. us, for arbitrary Q.u/, the fol-
lowing are compatible:
u
t
D u
xx
C Q.u/; u
t
C au
x
D 0: (2.173)
Subcase (ii) B D 0, C ¤ 0
In this case, equation (2.171) becomes
CQ
0
C 2A
x
Q D C
t
C 2CA
x
C
xx
; (2.174)
and dividing through by C gives
Q
0
.u/ C
2A
x
C
Q.u/ D
C
t
C 2CA
x
C
xx
C
: (2.175)
Since (2.175) should be only an equation involving u, then
2A
x
C
D m; (2.176a)
C
t
C 2CA
x
C
xx
C
D k
1
; (2.176b)
where m and k
1
are arbitrary constants. With these assignments, (2.175) becomes
Q
0
.u/ mQ.u/ D k
1
: (2.177)
We note that m ¤ 0, as m D 0 gives Q
00
D 0 which is inadmissible. From (2.176a), we solve for
C giving
C D
2A
x
m
; (2.178)
2.2. SECOND-ORDER PDES 33
and with this choice (2.176b) becomes
A
tx
A
xxx
C 2A
2
x
k
1
A
x
D 0: (2.179)
Within this subcase (2.170) becomes
A
t
C 2AA
x
A
xx
D 0: (2.180)
It is left as an exercise to the reader to show that if (2.179) and (2.180) are consistent, then
k
1
A
x
D 0; and since A
x
D 0 gives C D 0, then we require that k
1
D 0. We can readily solve
(2.179) and (2.180) for A giving
A D
c
1
x C c
2
2c
1
t C c
0
; (2.181)
where c
0
c
2
are arbitrary constants while solving (2.177) gives
Q.u/ D k
2
e
mu
; (2.182)
where k
2
is an additional arbitrary constant. us, the following are compatible:
u
t
D u
xx
C k
2
e
mu
; u
t
C
c
1
x C c
2
2c
1
t C c
0
u
x
D
2c
1
m.2c
1
t C c
0
/
: (2.183)
Subcase (iii) B ¤ 0, C D 0
In this case, equation (2.171) becomes
BuQ
0
.u/ C
.
2A
x
B
/
Q.u/ D
.
B
t
C 2BA
x
B
xx
/
u; (2.184)
and dividing through by B gives
uQ
0
.u/ C
2A
x
B
B
Q.u/ D
B
t
C 2BA
x
B
xx
B
u: (2.185)
Since (2.185) should be only an equation involving u, then
2A
x
B
B
D m; (2.186a)
B
t
C 2BA
x
B
xx
B
D k
1
; (2.186b)
where m and k
1
are arbitrary constants and (2.185) becomes
uQ
0
.u/ mQ.u/ D k
1
u: (2.187)
We can solve (2.186a) for B explicitly provided that m ¤ 1. As this is a special case, we consider
it first.
34 2. COMPATIBILITY
Special Case: m D 1
If m D 1, then from (2.187)
uQ
0
.u/ Q.u/ D k
1
u; (2.188)
whose solution is
Q.u/ D
.
k
1
ln u C k
2
/
u; (2.189)
where k
2
is a constant of integration. For this case, (2.186a) becomes
A
x
D 0; (2.190)
which must be solved in conjunction with (2.170) and (2.186b). ese are easily solved giving
A D c
1
e
k
1
t
C c
2
; B D c
3
e
k
1
t
1
2
c
1
k
1
e
k
1
t
x; (2.191)
where c
1
; c
2
and c
3
are arbitrary constants. us, the following are compatible:
u
t
D u
xx
C
.
k
1
ln u C k
2
/
u; u
t
C
c
1
e
k
1
t
C c
2
u
x
D
c
3
e
k
1
t
1
2
c
1
k
1
e
k
1
t
x
u: (2.192)
In the case where m ¤ 1, we solve (2.186a) for B giving
B D
2A
x
1 m
: (2.193)
us, (2.170) and (2.186b) become
A
t
C 2AA
x
C
3 C m
1 m
A
xx
D 0; A
tx
C 2A
2
x
A
xxx
D k
1
A
x
: (2.194)
ese are compatible provided that either k
1
D 0 or m D 3. Of course there are other choices
but we require that B ¤ 0 which eliminate those like A D 0 and A
x
D 0 . e solution for A in
each case is
A D
c
1
x C c
3
2c
1
t C c
0
.k
1
D 0/; A D 3
S
0
S
.m D 3/; (2.195)
where S D S.x/ satisfies S
00
C
k
1
4
S D 0. For each case B is given through (2.193) and Q is
obtained through (2.187). Our final results:
u
t
D u
xx
C k
2
u
m
; u
t
C
c
1
x C c
3
2c
1
t C c
0
u
x
D
2c
1
u
.1 m/.2c
1
t C c
0
/
; (2.196)
and
u
t
D u
xx
C k
2
u
3
1
2
k
1
u; u
t
3
S
0
S
u
x
D 3
S
0
S
0
u; (2.197)
2.2. SECOND-ORDER PDES 35
are compatible.
Case (iv) B ¤ 0, C ¤ 0
In this case, dividing (2.187) by B gives
u C
C
B
Q
0
C
.2A
x
B/
B
Q D
.B
t
C 2BA
x
B
xx
/
B
u C
C
t
C 2CA
x
C
xx
B
: (2.198)
Since (2.198) should be only an equation involving u, then
C
B
D a; (2.199a)
.2A
x
B/
B
D m; (2.199b)
.B
t
C 2BA
x
B
xx
/
B
D k
1
; (2.199c)
C
t
C 2CA
x
C
xx
B
D k
2
; (2.199d)
where a, m, k
1
and k
2
are arbitrary constants. From (2.199) we deduce k
2
D ak
1
and from
(2.198) that Q satisfies
.u C a/Q
0
.u/ mQ.u/ D k
1
.u C a/; (2.200)
which is (2.187) with a translation in the argument u. Since we can translate u in the original
PDE without loss of generality, we can set a D 0 without loss of generality, and thus we are
led back to case (iii).
Exact Solutions
One of the main goals of deriving compatible equations is to use them to construct exact so-
lutions of a given PDE. As cubic source terms appear to be special we will focus on these and
consider PDEs of the form
u
t
D u
xx
C q
1
u
3
C q
2
u (2.201)
where q
1
and q
2
are constant. is PDE is know as the Newel-Whitehead-Segel equation and
was introduced by Newel and Whitehead [40] and Segel [41] to model various phenomena in
fluid mechanics.
As cubic source terms arose in two places in our study, we consider each separately. We
first consider (2.168). Here we will set a D 0; b D 3k and c D 0, where k is constant, and for
convenience we choose p D 3. In this case, (2.168) becomes
u
t
C 3uu
x
D 3u
3
3ku; (2.202a)
u
t
D u
xx
2u
3
2ku: (2.202b)
Eliminating u
t
in (2.202) gives
u
xx
C 3uu
x
C u
3
C ku D 0 (2.203)
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