127
C H A P T E R 6
Functional Separability
In a standard course in PDEs, one encounters the heat equation
u
t
D u
xx
; (6.1)
and solutions in the form u D T .t/X.x/, often referred to as separable solutions. A natural
question is: do nonlinear PDEs admit separable solutions? In general, the answer is no but
sometimes they admit some sort of a variation of separable solution. For example,
u
t
D
.
uu
x
/
x
(6.2)
does admit solutions of the form
u
D
TX
, leading to
T
0
T
2
D
.
XX
0
/
X
: (6.3)
It also admits solutions of the form
u D a.t/x
2
C b.t /x C c.t/: (6.4)
Substitution of (6.4) into (6.2) and isolating coefficients of x gives
a
0
D 6a
2
;
b
0
D 6ab; (6.5)
c
0
D 2ac C b
2
: (6.6)
However,
u
t
D
u
xx
1 C u
2
x
(6.7)
doesn’t admit separable solutions of the form u D T .t/X.x/, since substitution of this form
into (6.7) leads to
T
0
X D
TX
00
1 C T
2
X
02
(6.8)
which doesn’t separate. However, Eq. (6.7) does admit separable solutions of the form u D
T .t/ C X.x/.
Sometimes it is not at all obvious that a particular PDE admits some form of a separable
solution. e following two examples illustrate this.
128 6. FUNCTIONAL SEPARABILITY
Example 6.1 Consider
u
tt
D u
xx
C e
u
(6.9)
and separable solutions of the form
u D ln.T .t/ C X.x//: (6.10)
Substitution of (6.10) into (6.9) gives
.T C X/T
00
T
02
.T C X/
2
D
.T C X/X
00
X
02
.T C X/
2
C .T C X/: (6.11)
Simplifying gives
.T C X/T
00
T
02
D .T C X/X
00
X
02
C .T C X/
3
; (6.12)
and differentiating (6.12) with respect to t and x gives
X
0
T
000
D T
0
X
000
C 6a.T CX/T
0
X
0
; (6.13)
or
T
000
T
0
D
X
000
X
0
C 6.T C X/: (6.14)
is leads to the two equations
T
000
6T T
0
D kT
0
; X
000
C 6XX
0
D kX
0
; (6.15)
where k is an arbitrary constant. Integrating each twice gives
1
2
T
02
T
3
D
1
2
kT
2
C c
1
T C c
2
;
1
2
X
02
C X
3
D
1
2
kX
2
C c
3
X C c
4
; (6.16)
where c
1
c
4
are constants of integration. Finally, substituting (6.16) back into (6.12) shows
that it is identically satisfied if c
1
C c
3
D 0 and c
2
c
4
D 0. us, any solution of
T
02
D 2T
3
C kT
2
2c
1
T C 2c
2
; X
02
D 2X
3
C kX
2
C 2c
1
X C 2c
2
; (6.17)
when substituted into (6.10), will give rise to an exact solution of the PDE (6.9).
Example 6.2 Consider
u
t t
u
xx
D sin u; (6.18)
and solutions of the form
u D 4 tan
1
X.x/
T .t/
: (6.19)
6. FUNCTIONAL SEPARABILITY 129
Substituting (6.19) into (6.18) and re-arranging terms gives
2T
02
T T
00
X
2
T
00
T
C 2X
02
T
2
X
00
X
XX
00
T
2
X
2
D 0: (6.20)
Differentiating (6.20) twice with respect to t and x gives
1
T T
0
T
00
T
0
C
1
XX
0
X
00
X
0
D 0; (6.21)
leading to the equations
1
T T
0
T
00
T
0
D k;
1
XX
0
X
00
X
0
D k; (6.22)
where k is again a constant. Each can be integrated twice, leading to
X
02
D c
1
X
4
C c
2
X
2
C c
3
; T
02
D c
1
T
4
C c
4
T
2
C c
5
: (6.23)
Substituting (6.23) into (6.20) leads to the conditions
c
4
c
2
D 1, c
3
C c
5
D 0. Any solution of
the Jacobi elliptic equations (6.23) with these constant restrictions will lead to an exact solution
of (6.18) in the form of (6.19).
A natural question is: how did one know that separable solutions of
u
t t
D u
xx
C e
u
; (6.24)
were of the form
u D ln.T .t/ C X.x//; (6.25)
or separable solutions of
u
tt
u
xx
D sin u; (6.26)
were of the form
u D 4 tan
1
.
X.x/=T .t/
/
‹ (6.27)
We also ask: are there others? If an equation admits functional separable solutions, then it would
be of the form
f .u/ D T .t/X.x/; or f .u/ D T .t/ C X.x/:
However, these are equivalent, as we can take natural logarithms of both sides of the first and
reset f; T and X. us, we only consider
f .u/ D T .t/ C X.x/: (6.28)
Differentiating (6.28) with respect to t and x gives
f
0
u
tx
C f
00
u
t
u
x
D 0;
130 6. FUNCTIONAL SEPARABILITY
or
u
tx
D F .u/u
t
u
x
; (6.29)
where we set F D f
00
=f
0
. If a PDE admits solutions of the form (6.28) then it will be compat-
ible with (6.29). As an example, we will reconsider (6.9). Taking t and x derivatives of both (6.9)
and (6.29) gives
u
tt t
u
txx
D e
u
u
t
; (6.30a)
u
ttx
u
xxx
D e
u
u
x
; (6.30b)
u
ttx
D F
0
u
2
t
u
x
C F u
tt
u
x
C F u
t
u
tx
; (6.30c)
u
txx
D F
0
u
t
u
2
x
C F u
tx
u
x
C F u
t
u
xx
: (6.30d)
Solving (6.30) for the third-order derivatives u
tt t
; u
ttx
; u
txx
; and u
xxx
gives
u
t t t
D F u
x
u
tx
C F u
t
u
xx
C F
0
u
t
u
2
x
C e
u
u
t
(6.31a)
u
t tx
D F u
x
u
t t
C F u
t
u
tx
C F
0
u
2
t
u
x
(6.31b)
u
txx
D F u
x
u
tx
C F u
t
u
xx
C F
0
u
t
u
2
x
(6.31c)
u
xxx
D F u
x
u
t t
C F u
t
u
tx
C F
0
u
2
t
u
x
e
u
u
x
: (6.31d)
Requiring compatibility
.
u
t tx
/
t
D
.
u
tt t
/
x
;
.
u
txx
/
t
D
.
u
ttx
/
x
;
.
u
xxx
/
t
D
.
u
txx
/
x
; (6.32)
and using both (6.9) and (6.29) gives
F
00
C 2FF
0
u
3
t
u
x
C u
3
x
C
3F
0
C 2F
2
C F 1
e
u
u
t
u
x
D 0; (6.33)
from which we obtain two equations for F
F
00
C 2FF
0
D 0; (6.34a)
3F
0
C 2F
2
C F 1 D 0: (6.34b)
Requiring that (6.34) be compatible gives
F D 1; F D
1
2
: (6.35)
Since F D f
00
=f
0
, upon integrating we get
f D c
1
e
u
C c
2
; f D c
1
e
u=2
C c
2
; (6.36)
where c
1
and c
2
are arbitrary constants. us,
u
tt
D u
xx
C e
u
(6.37)
6. FUNCTIONAL SEPARABILITY 131
admits separable solutions of the form
c
1
e
u
C c
2
D T .t / C X.x/; (6.38a)
c
1
e
u=2
C c
2
D T .t / C X.x/: (6.38b)
Note that we can set c
1
D 1 and c
2
D 0 without loss of generality. Solving for u, we obtain
u D ln jT .t / C X.x/j; (6.39a)
u D 2 ln jT .t/ C X.x/j; (6.39b)
and the first solution in (6.39) we saw earlier.
Example 6.3 Seek functional separable solutions to the nonlinear diffusion equation
u
t
D
u
x
u
2
x
: (6.40)
We now wish that (6.40) and
u
tx
D F .u/u
t
u
x
.noting that F D f
00
=f
0
/ (6.41)
are compatible. From our original PDE (6.40), we have
u
xx
D u
2
u
t
C
2u
2
x
u
: (6.42)
From cross differentiation
.
u
xx
/
t
D
.
u
tx
/
t
and (6.42) and (6.41), we obtain
u
2
u
tt
C 2uu
2
t
C
4u
x
u
tx
u
2u
2
x
u
t
u
2
D F
0
u
t
u
2
x
C F u
x
u
tx
C F u
t
u
xx
: (6.43)
Using (6.42) and (6.41) to eliminate derivatives u
tx
and u
xx
, we obtain
u
t t
D
F
0
u
t
u
2
x
u
2
2u
2
t
u
C
2u
t
u
2
x
u
4
C
F
2
u
t
u
2
x
u
2
C F u
2
t
2F u
t
u
2
x
u
3
: (6.44)
We now further require that (6.41) and (6.44) be compatible. is leads to
2
F
0
C F
2
3F
u
C
3
u
2
u
2
t
u
x
C
F
00
u
2
C
2FF
0
u
2
C
2F
2
u
3
2F
u
4
u
t
u
3
x
D 0; (6.45)
from which we obtain two equations for F
F
0
C F
2
3F
u
C
3
u
2
D 0; (6.46a)
F
00
u
2
C
2FF
0
u
2
C
2F
2
u
3
2F
u
4
D 0: (6.46b)
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