3.5. DARBOUX TRANSFORMATIONS 61
Substituting (3.94) into (3.93) and imposing (3.95) gives
.
A
t
A
xx
fA
/
v
x
.
2A
x
C f
/
v D 0: (3.96)
Since (3.96) must be true for all v, we obtain
2A
x
C f D 0; (3.97a)
A
t
A
xx
fA D 0: (3.97b)
From (3.97a), we find f D 2A
x
, and eliminating f in (3.97b) gives
A
t
C 2AA
x
A
xx
D 0: (3.98)
As we saw previously, this is Burgers’ equation which was linearized via the Hopf–Cole trans-
formation
A D
x
=: (3.99)
is, in turn, gives f D 2
.
ln
/
xx
. us, we have the following: solutions of
u
t
D u
xx
C 2
.
ln
/
xx
u (3.100)
are given by
u D v
x
x
v; (3.101)
where both and v satisfy the heat equation. We now consider a few examples, choosing
various simple solutions of the heat equation.
Example 3.4 If we choose D x; then solutions of
u
t
D u
xx
2u
x
2
(3.102)
are obtained by
u D v
x
v
x
; (3.103)
where here, and in the examples that follow, v is any solution of the heat equation.
Example 3.5 If we choose D x
2
C 2t; then solutions of
u
t
D u
xx
.x
2
2t/
.x
2
C 2t/
2
u (3.104)
are obtained by
u D v
x
2x
x
2
C 2t
v: (3.105)