6. FUNCTIONAL SEPARABILITY 137
Requiring that (6.84) and (6.50) be compatible gives rise to the following set of determining
equations:
k
00
C F k
0
D 0; (6.85a)
F
000
C 9FF
00
C 6F
02
C 30F
2
F
0
D 0: (6.85b)
Since F D f
00
=f
0
, then (6.85a) integrates to give
k D c
1
f C c
2
; (6.86)
where c
1
and c
2
are arbitrary constants. Furthermore, (6.85b) becomes
f
.5/
f
0
C
13f
00
f
.4/
f
02
C
9f
0002
f
02
81f
002
f
000
f
03
C
72f
004
f
04
D 0; (6.87)
which possesses the integrating factor f
03
, which allows us to integrate, giving
f
.4/
f
04
9f
000
f
00
f
05
C
12f
003
f
06
C c
3
D 0: (6.88)
Equation (6.88) possesses the integrating factor f
0
leading to
f
000
f
03
3f
002
f
04
C c
3
f C c
4
D 0; (6.89)
where c
3
and c
4
are additional arbitrary constants. If we perform a Hodograph transformation
on (6.89) (interchanging the roles of f and u) we get the linear ODE
u
fff
.
c
3
f C c
4
/
u
f
D 0: (6.90)
If if c
3
¤ 0 this can be solved in terms of integrated Airy functions. If c
3
D 0, then (6.89) can
be integrated twice, giving
f
0
D
1
p
c
4
u
2
C c
5
u C c
6
: (6.91)
It can further be integrated, depending on the sign of c
4
. Here we consider the cases where
c
4
D 1, c
4
D 0; and c
4
D 1. We will also set c
1
D 1, c
2
D 0.
Case 1. c
4
D 1 In this case, (6.91) integrates to give
f D tan
1
u
1
2
c
5
p
u
2
C c
5
u C c
6
!
C c
7
: (6.92)
We will set c
2
D c
5
D c
7
D 0 and c
1
D c
6
D 1. From (6.83), (6.86), and (6.48), PDEs of the
form
u
t
C tan
1
u
p
1 u
2
u
x
C u
xxx
D 0 (6.93)
138 6. FUNCTIONAL SEPARABILITY
admit separable solutions of the form
u D sin
.
A.t/x C B.t/
/
; (6.94)
where A and B satisfy
A
0
C A
2
D 0; B
0
C AB A
3
D 0: (6.95)
ese are easily solved, giving
A D
1
t C a
0
; B D
b
0
t C a
0
1
.t C a
0
/
2
; (6.96)
a
0
; b
0
are constant, and via (6.94) (with a
0
D b
0
D 0) gives
u D sin
xt 1
t
2
: (6.97)
Case 2. c
4
D 0 In this case (6.91) integrates to give
f D
2
p
c
5
u C c
6
c
5
C c
7
: (6.98)
We will set c
1
D 1; c
2
D 0; c
5
D 4; c
6
D 0; c
7
D 0. From (6.83), (6.86), and (6.48), PDEs of the
form
u
t
C
p
uu
x
C u
xxx
D 0 (6.99)
admit separable solutions of the form
u D
.
A.t/x C B.t/
/
2
; (6.100)
where A and B satisfy
A
0
C A
2
D 0; B
0
C AB D 0: (6.101)
ese are easily solved, giving
A D
1
t C a
0
; B D
b
0
t C a
0
; (6.102)
a
0
; b
0
are constant, and via (6.100) (with a
0
D b
0
D 0) gives
u D
x
t
2
: (6.103)
6. FUNCTIONAL SEPARABILITY 139
Case 3. c
4
D 1 In this case, (6.91) integrates to give
f D ln
ˇ
ˇ
ˇ
ˇ
1
2
c
5
C u C
p
u
2
C c
5
u C c
6
ˇ
ˇ
ˇ
ˇ
C c
7
: (6.104)
We will set c
2
D c
5
D c
7
D 0 and c
1
D c
6
D 1. From (6.83), (6.86), and (6.48), PDEs of the
form
u
t
C ln ju C
p
u
2
C 1ju
x
C u
xxx
D 0 (6.105)
admit separable solutions of the form
u D sinh
.
A.t/x C B.t/
/
; (6.106)
where A and B satisfy
A
0
C A
2
D 0; B
0
C AB C A
3
D 0: (6.107)
ese are easily solved, giving
A D
1
t C a
0
; B D
b
0
t C a
0
C
1
.t C a
0
/
2
; (6.108)
a
0
; b
0
are constant, and via (6.106) (with a
0
D b
0
D 0) gives
u D sinh
xt C 1
t
2
: (6.109)
If we were to chose c
6
D 1, we would obtain the exact solution
u D cosh
xt C 1
t
2
(6.110)
for the PDE
u
t
C ln ju C
p
u
2
1ju
x
C u
xxx
D 0: (6.111)
HIGHER-DIMENSIONAL FUNCTIONAL SEPARABLE
SOLUTIONS
In this section we extend our results to PDEs in three dimensions, and, in particular, we consider
the nonlinear elliptic equation
u
xx
C u
yy
C u
zz
D Q.u/: (6.112)
We will seek separable solutions of the form
f .u/ D A.x/ C B.y/ C C.z/: (6.113)
140 6. FUNCTIONAL SEPARABILITY
e reader can verify that (6.113) arises on solving
u
xy
D F .u/u
x
u
y
; u
xz
D F .u/u
x
u
z
; u
yz
D F .u/u
y
u
z
; (6.114)
where F D f
00
=f
0
. We wish to make (6.112) and (6.114) compatible. By differentiating
(6.112) and (6.114) with respect to x; y; and z, we calculate all third-order derivatives. Re-
quiring that these be compatible with each other gives rise to
2
F
0
C F
2
u
xx
F u
2
x
F
00
C 2FF
0
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q D 0;
(6.115a)
2
F
0
C F
2
u
yy
F u
2
y
F
00
C 2FF
0
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q D 0;
(6.115b)
2
F
0
C F
2
u
zz
F u
2
z
F
00
C 2FF
0
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q D 0;
(6.115c)
and the first-order condition
3F
00
C 8FF
0
C 2F
3
u
2
x
C u
2
y
C u
2
z
3Q
00
C 3FQ
0
C
7F
0
C 4F
2
Q D 0: (6.116)
We can solve (6.115) for u
xx
; u
yy
; and u
zz
, provided that F
0
C F
2
¤ 0. As F
0
C F
2
D 0 is a
special case, we consider this first. is gives F D u
1
, with the constant of integration omitted.
Equation (6.115) reduce to
u
2
Q
00
uQ
0
C Q D 0: (6.117)
is integrates to
Q D q
1
u C q
2
u ln juj; (6.118)
where q
1
and q
2
are arbitrary constants, with q
2
¤ 0 as we are interested in NLPDEs. However,
as we can scale u in our original PDE (6.112) with this particular Q, we can set q
1
D 0 without
loss of generality. Since F D f
00
=f
0
, then from (6.113) we obtain
u D e
ACBCC
: (6.119)
Substitution of (6.119) into (6.112) with (6.118) gives
A
00
C A
02
C B
00
C B
02
C C
00
C C
02
e
ACBCC
D q
2
e
ACBCC
.
A C B C C
/
; (6.120)
which we see separates into
A
00
C A
02
D q
2
A C a; B
00
C B
02
D q
2
B C b; C
00
C C
02
D q
2
C C c; (6.121)
where a; b and c are constants such that a C b C c D 0. We can further transfer the constants
a; b; and c to the solution (6.119), allowing us to set a D b D c D 0 and the ODEs in (6.121)
6. FUNCTIONAL SEPARABILITY 141
become the single ODE w
00
C w
02
D q
2
w; w D w./ which can be solved to quadature. If the
first constant of integration is omitted, we obtain the explicit solution
w D
1
2
C
q
2
4
. w
0
/
2
; (6.122)
where w
0
is a constant of integration. us, the forms of A; B, and C are
A D
1
2
C
q
2
4
.x x
0
/
2
; B D
1
2
C
q
2
4
.y y
0
/
2
; C D
1
2
C
q
2
4
.z z
0
/
2
: (6.123)
As we have the flexibility of translational and scaling symmetry, we arrive at our final solution
u D e
q
.
x
2
Cy
2
Cz
2
/
C3=2
(6.124)
as an exact solution to the PDE
u
xx
C u
yy
C u
zz
D 4qu ln juj: (6.125)
Note that we set q
2
D 4q.
We now assume that F
0
C F
2
¤ 0. We now can solve for all second-order derivatives.
From (6.115) we obtain
u
xx
D F u
2
x
C
.F
00
C 2FF
0
/
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q
2
.
F
0
C F
2
/
; (6.126a)
u
yy
D F u
2
y
C
.F
00
C 2FF
0
/
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q
2
.
F
0
C F
2
/
; (6.126b)
u
zz
D F u
2
z
C
.F
00
C 2FF
0
/
u
2
x
C u
2
y
C u
2
z
C Q
00
FQ
0
3F
0
C 2F
2
Q
2
.
F
0
C F
2
/
: (6.126c)
to which we add those given in (6.114). Requiring (6.114) and (6.126) be compatible gives
F
000
C 3FF
00
C 2F
02
C 2F
2
F
0
u
2
x
C u
2
y
C u
2
z
Q
000
C 2FQ
00
C
4F
0
C F
2
Q
0
C
3F
00
C FF
0
2F
3
Q D 0: (6.127)
Eliminating u
2
x
C u
2
y
C u
2
z
between (6.116) and (6.127) gives rise to our first constraint
on F and Q.
3F
00
C 8FF
0
C 2F
3
Q
000
C 2FQ
00
C .3F
0
C F
2
/Q
0
C .3F
00
C FF
0
2F
3
/Q
F
000
C 3FF
00
C 2F
02
C 2F
2
F
0
3Q
00
C 3FQ
0
C .7F
0
C 4F
2
/Q
D 0: (6.128)
If we differentiate (6.116) with respect to either x; y, or z and use (6.114), (6.116), and (6.126)
we obtain a second constraint
3
3F
00
C 8FF
0
C 2F
3
Q
000
9F
000
C 45FF
00
C 24F
02
C 74F
2
F
0
Q
00
C
9FF
000
30F
0
F
00
C 24F
2
F
00
56FF
02
2F
3
F
0
Q
0
C
21F
0
C 12F
2
F
000
27F
002
C
26F
3
28FF
0
F
00
C 56F
03
C 42F
2
F
02
C 48F
4
F
0
C 8F
6
D 0: (6.129)
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