3.5. DARBOUX TRANSFORMATIONS 59
Substituting (3.74) into (3.73) (using (3.71) and (3.72) appropriately) gives
ˆ
tx
ˆ
1
C ˆ
x
ˆ
1
ˆ
t
ˆ
1
2
ˆ
xx
ˆ
1
C ˆ
x
ˆ
1
ˆ
x
ˆ
1
ˆ
x
ˆ
1
D
ˆ
xxx
ˆ
1
C 2ˆ
xx
ˆ
1
ˆ
x
ˆ
1
C ˆ
x
ˆ
1
ˆ
xx
ˆ
1
2ˆ
x
ˆ
1
ˆ
x
ˆ
1
ˆ
x
ˆ
1
: (3.75)
Expanding and canceling appropriately gives
ˆ
tx
ˆ
1
C ˆ
x
ˆ
1
ˆ
t
ˆ
1
D ˆ
xxx
ˆ
1
C ˆ
x
ˆ
1
ˆ
xx
ˆ
1
; (3.76)
which becomes on multiplying on the left by ˆ
1
and on the right by ˆ
ˆ
1
ˆ
t
x
D
ˆ
1
ˆ
xx
x
: (3.77)
Integrating with respect to x gives the linear heat matrix equation
ˆ
t
D ˆ
xx
: (3.78)
Note that, like the scalar case, the matrix function of integration can be set to zero without loss
of generality.
3.5 DARBOUX TRANSFORMATIONS
In 1882, Darboux considered the following problem: is it possible to find functions f .x/ such
that solutions of
y
00
C f .x/y D 0; z
00
C kz D 0 (3.79)
can be connected through
y D z
0
C A.x/z‹ (3.80)
A direct substitution of (3.80) into the first of (3.79) gives
z
000
C Az
00
C 2A
0
z
0
C A
00
z C f .x/
z
0
C Az
D 0; (3.81)
where prime denoted differentiation with respect to their argument. Imposing the second of
(3.79) gives
f C 2A
0
k
z
0
C
A
00
C fA kA
z D 0; (3.82)
and since this must be true for all z leads to
f C 2A
0
k D 0; (3.83a)
A
00
C fA kA D 0: (3.83b)
If f D k 2A
0
then (3.83b) becomes
A
00
2AA
0
D 0; (3.84)
60 3. DIFFERENTIAL SUBSTITUTIONS
which integrates once to become
A
0
A
2
D c (3.85)
where c is an arbitrary constant. If we let A D
0
= where D .x/, (3.85) becomes
00
C c D 0: (3.86)
Composing the result gives the solution of
y
00
C
2
.
ln
/
00
C k
y D 0; (3.87)
obtained via the Darboux transformation
y D z
0
0
z; (3.88)
where and z satisfy
00
C c D 0; z
00
C kz D 0; (3.89)
respectively.
For example, if we choose solutions D cosh x and z D ax C b (a; b constant) as solu-
tions of (3.89), then via (3.87) and (3.88) solutions of
y
00
C
2y
cosh
2
x
D 0 (3.90)
are given by
y D a .ax Cb/ tanh x: (3.91)
A natural question is: does this method apply to PDEs? e answer: Yes! A very important class
of PDEs is the .1 C 1/ dimensional Schrödinger equation
i
t
D
„
2m
xx
C V .x/ (3.92)
from Quantum Mechanics. Here, we will consider
u
t
D u
xx
C f .t; x/u; (3.93)
and seek Darboux transformations of the form
u D v
x
C A.t; x/v; (3.94)
where v satisfies the standard heat equation
v
t
D v
xx
: (3.95)
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)
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