Galaktionov V.A., Svirshchevskii S.R. Exact Solutions and Invariant Subspaces of Nonlinear Partial Differential Equations in Mechanics and Physics
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2 Invariant Subspaces: Mathematics in 1D 63
by Theorem 2.8, a nonlinear operator of the zero order cannot preserve a subspace
of dimension exceeding one. This completes the proof.
Likewise, for the case of a first-order operator, we shall carry out further analysis
by studying the quadratic operators
F[y] = b
2,2
q
2
+ b
1,2
pq +b
0,2
yq + b
1,1
p
2
+ b
0,1
yp + b
0,0
y
2
, (2.71)
restricting ourselves to the case of constant coefficients,
b
i, j
= constant for i, j = 0, 1, 2. (2.72)
In this case, the following result holds:
Lemma 2.21
If equation
(2.59)
is invariant under the operator
(2.71)
,
(2.72)
,then
the coefficients of the equation do not depend on
x
.
Proof. In (2.71), let b
22
=
C
2
and C
2
+ b
2
12
+ b
2
02
= 0. We deduce from (2.65) that
b
12
=−
7
10
a
1
C, (2.73)
and, hence, it suffices to consider the following two cases:
Case (i): C = 2, a
1
=−
5
7
b
12
= constant, and
Case (ii): C = 0, b
12
= 0.
Let us again concentrate on the invariance condition (2.60). In case of (i), equating
the coefficients of y
3
y
4
, y
2
y
4
, y
1
y
4
,andyy
4
to zero, we have
−5a
2
1
+ 24a
2
+ 15b
02
+ 20b
11
= 0,
20a
2
− 7a
3
1
− 14a
1
a
2
+ 20a
1
b
02
+ 10a
1
b
11
+ 100a
3
+ 75b
01
= 0,
2a
2
b
12
+ a
2
1
b
02
+ a
1
b
01
+ 2a
2
b
02
+ 20a
4
+ 10b
00
= 0,
a
2
b
02
+ 10a
5
= 0.
From these equalities, we derive successively that the coefficients a
2
, ..., a
5
are con-
stant, and, moreover, it turns out that a
5
= 0.
In the second case of (ii), equating the coefficients of y
3
y
4
, y
2
y
4
, y
1
y
4
,andy
2
3
to
zero yields
3b
02
+ 4b
11
= 0, 4a
1
b
02
+ 2a
1
b
11
+ 15b
01
= 0,
a
1
(5b
02
+ 2b
11
) + a
2
1
b
02
+ a
1
b
01
+ 2a
2
b
02
+ 10b
00
= 0,
−2a
1
b
02
− 3a
1
b
11
+ 5b
01
= 0.
(2.74)
The first equation gives b
11
=−
3
4
b
02
, and the second and fourth equation then take,
respectively, the form
a
1
b
02
+ 6b
01
= 0anda
1
b
02
+ 20b
01
= 0,
whence comes a
1
b
02
= b
01
= 0. Since, in this case, b
02
does not vanish (we con-
sider second-order operators), it follows that a
1
= 0, andthenthethirdequationof
the system (2.74) yields a
2
= const. By these conditions, studying the coefficients
of y
1
y
3
, y
1
y
2
,andyy
2
implies that a
3
= 0, a
4
= constant, and a
5
= 0, which
completes the proof.
© 2007 by Taylor & Francis Group, LLC
64 Exact Solutions and Invariant Subspaces
The next theorem gives a full description of operators (2.71), (2.72) preserving
5D subspaces. All operators are shown up to scaling in x and multiplication by an
arbitrary non-zero number. In cases of (1)–(3) below, where the same subspace is
invariant under arbitrary linear combinations
F[y] = C
1
F
1
[y] + C
2
F
2
[y], where C
1
, C
2
= constant, (2.75)
of linearly independent operators F
1
and F
2
, only these basis operators are given.
In all the cases, the corresponding invariant subspaces and (in parentheses) their
defining equations (2.59) are indicated.
Theorem 2.22
The set of operators
(2.71)
with constant coefficients preserving 5D
subspaces is exhausted by:
(1) F
1
[y] = y
y −
3
4
(y
)
2
, F
2
[y] = (y
)
2
,
with the subspace
W
5
= L{1, x , x
2
, x
3
, x
4
}
y
(5)
= 0
;
(2) F
1
[y] = y
y −
3
4
(y
)
2
+ y
2
, F
2
[y] =
y
+ 4y
2
,
with
W
5
= L{1, cos x , sin x, cos2x , sin2x}
y
(5)
=−5y
− 4y
;
(3) F
1
[y] = y
y −
3
4
(y
)
2
− y
2
, F
2
[y] =
y
− 4y
2
,
with
W
5
= L{1, cosh x, sinh x , cosh2x, cosh2x}
y
(5)
= 5y
− 4y
;
(4) F[y] =
y
−
7
2
y
+ 3y
2
,
with
W
5
= L
1, e
1
2
x
, e
x
, e
3
2
x
, e
2x
4y
(5)
= 20y
(4)
− 35y
+ 25y
− 6y
;
(5) F[y] =
5y
−
27
2
y
+ 7y
y
−
3
2
y
− y
,
with
W
5
= L
1, e
−
1
2
x
, e
1
2
x
, e
x
, e
2x
4y
(5)
= 12y
(4)
− 7y
− 3y
+ 2y
;
(6) F[y] = (y
− 5y
+ 6y)(y
− 2y
− 3y),
with
W
5
= L
1, e
−x
, e
x
, e
2x
, e
3x
y
(5)
= 5y
(4)
− 5y
− 5y
+ 6y
.
Note that operators F
2
k = 2a
n
d˜x → −
1
2
˜x. The equivalence transformation for operators F
1
in (1) and
F
1
in (3) is y = x
2
˜y, ˜x = ln x,i.e.,α = x
2
and β = ln x in Remark 2.11.
Proof. We consider in greater detail cases (i) and (ii) that appeared in the proof of
Lemma 2.21.
Case (i): C = 2, a
1
=−
5
7
b
12
. Then, as has been shown above, a
5
= 0. Consider
two possibilities: b
12
= 0andb
12
= 0.
(i.1) In the subcase where b
12
= 0, using the scaling in x implies b
12
=−7. The first
coefficient on the right-hand side of (2.59) is then a
1
= 5.
Equating the coefficients of y
3
y
4
, y
2
3
, y
2
y
4
, y
1
y
4
, y
2
y
3
, y
1
y
3
, y
2
2
, y
1
y
2
,andy
2
1
to
zero in the identity (2.60) (the rest of the coefficients vanish in view of the assumed
© 2007 by Taylor & Francis Group, LLC
in (1) and F in (4) are equivalent; see Example 2.13, where
2 Invariant Subspaces: Mathematics in 1D 65
hypotheses), we obtain the system
15b
02
+ 20b
11
+ 24a
2
− 125 = 0,
−10b
02
− 15b
11
+ 5b
01
− 37a
2
+ 4a
3
= 0,
20b
02
+ 10b
11
+ 15b
01
− 14a
2
+ 20a
3
− 175 = 0,
(2a
2
+ 25)b
02
+ 5b
01
+ 10b
00
+ 20a
4
= 0,
7a
2
b
02
+ 4a
2
b
11
− 50b
01
+ 20b
00
− 35a
2
− 116a
3
+ 10a
4
= 0,
(5a
2
+ 3a
3
)b
02
+ 2a
2
b
01
− 40b
00
− 95a
4
= 0,
10a
3
b
02
+ 8a
3
b
11
− 3a
2
b
01
− 30b
00
− 35a
3
− 35a
4
= 0,
5(a
3
+ 3a
4
)b
02
+ 10a
4
b
11
+ 3a
3
b
01
− 6a
2
b
00
− 35a
4
= 0,
5a
4
b
02
+ 5a
4
b
01
− 2a
3
b
00
= 0.
Using the first four equations yields the expressions for the coefficients of operator
(2.71), with b
22
=
C
2
, in terms of the coefficients of (2.59),
b
02
=
1
35
(−124a
2
− 32a
3
− 675),
b
11
=
1
35
(51a
2
+ 24a
3
+ 725), b
01
=
1
35
(164a
2
− 20a
3
+ 825),
b
00
=
1
175
(124a
2
2
+ 32a
2
a
3
+ 1815a
2
+ 450a
3
− 350a
4
+ 6375).
(2.76)
The fifth equation then implies the following:
a
4
=−
1
25
4a
2
2
+ 95a
2
+ 30a
3
+ 375
. (2.77)
Using these conditions, we reduce the remaining four equations to
428a
2
2
+ 276a
2
a
3
+ 5415a
2
+ 32a
2
3
+ 1875a
3
+ 175a
4
+ 17000 = 0,
1236a
2
2
+ 964a
2
a
3
+ 13365a
2
+ 128a
2
3
+ 4875a
3
− 875a
4
+ 38250 = 0,
372a
3
2
+ 96a
2
2
a
3
+ 5445a
2
2
+ 1670a
2
a
3
+ 2325a
2
a
4
+19125a
2
+ 550a
2
3
+ 600a
3
a
4
+ 2250a
3
+ 10250a
4
= 0,
124a
2
2
a
3
+ 32a
2
a
2
3
+ 1815a
2
a
3
− 500a
2
a
4
+450a
2
3
+ 300a
3
a
4
+ 6375a
3
− 1875a
4
= 0.
Resolving the first three equations with respect to the quadratic terms yields:
a
2
2
=
5
36
(−93a
2
+ 2a
3
− 275),
a
2
a
3
=
5
36
(30a
2
− 41a
3
+ 125),
a
2
3
=
5
36
(−165a
2
− 46a
3
− 875).
(2.78)
It is not difficult to check that, in view of these relations, the last equation becomes
the identity. From the first equation of system (2.78), it follows that
a
3
=
1
2
36
5
a
2
2
+ 93a
2
+ 275
, (2.79)
and the second and third equations then take the form
144a
3
2
+ 2680a
2
2
+ 15925a
2
+ 30625 = 0 and (2.80)
1296a
4
2
+ 33480a
3
2
+ 317525a
2
2
+ 1310750a
2
+ 1990625 = 0. (2.81)
The roots of the cubic equation (2.80) are
−
35
4
, −
175
36
, and − 5,
© 2007 by Taylor & Francis Group, LLC
66 Exact Solutions and Invariant Subspaces
simultaneously satisfying equation (2.81).
Foreach of the obtained values of a
2
, usingthe formulae(2.79), (2.77),and (2.76),
we determine the rest of the coefficients,
a
2
=−
35
4
, a
3
=
25
4
, a
4
=−
3
2
,
b
02
= 6, b
11
=
49
4
, b
01
=−21, b
00
= 9;
(2.82)
a
2
=−
175
36
, a
3
=
125
36
, a
4
=−
725
162
,
b
02
=−
110
21
, b
11
=
1345
84
, b
01
=−
25
21
, b
00
=
9950
567
;
(2.83)
a
2
=−5, a
3
=−5, a
4
= 6,
b
02
= 3, b
11
= 10, b
01
= 3, b
00
=−18.
(2.84)
Collections of the coefficients(2.82) and (2.84) correspond,respectively,to items (4)
and (6) in the statement of the theorem, while (2.83) corresponds to item (5) up to
scaling in x.
(i.2) In the subcase where b
12
= 0, from (2.73), it follows that a
1
= 0. Equating the
coefficients of y
3
y
4
, y
2
y
4
, y
1
y
4
, y
2
3
, y
2
y
3
, y
1
y
3
, y
2
2
, y
2
1
,andy
1
y
2
in (2.60) to zero
(the rest of the coefficients are identically zero), we find
24a
2
+ 15b
02
+ 20b
11
= 0, 4a
3
+ 3b
01
= 0,
a
2
b
02
+ 10a
4
+ 5b
00
= 0, 4a
3
+ 5b
01
= 0,
7a
2
b
02
+ 4a
2
b
11
+ 10a
4
+ 20b
00
= 0, 2a
2
b
01
+ 3a
3
b
02
= 0,
−3a
2
b
01
+ 10a
3
b
02
+ 8a
3
b
11
= 0, −2a
3
b
00
+ 5a
4
b
01
= 0,
−6a
2
b
00
+ 3a
3
b
01
+ 15a
4
b
02
+ 10a
4
b
11
= 0.
The second and the fourth equations yield a
3
= b
01
= 0, so the system is simplified
and has the form
24a
2
+ 15b
02
+ 20b
11
= 0,
a
2
b
02
+ 10a
4
+ 5b
00
= 0,
7a
2
b
02
+ 4a
2
b
11
+ 10a
4
+ 20b
00
= 0,
−6a
2
b
00
+ 15a
4
b
02
+ 10a
4
b
11
= 0.
Taking b
11
and b
00
from the first two equations and plugging into the rest of them, a
single additional condition is obtaineda
4
=−
4
25
a
2
2
. The final solutionof theoriginal
system is then written as
b
11
=−
3
5
a
2
−
3
4
b
02
, b
00
=−
1
5
a
2
b
02
+
4
25
a
2
2
, a
4
=−
4
25
a
2
2
,
with arbitrary parameters a
2
and b
02
. Further analysis of these equations leads to
those operators (2.75) from (1)–(3) of the theorem that are singled out by the condi-
tion C
2
= 0.
Case (ii): If C = 0, b
12
= 0, b
02
= 0, then a
1
= a
3
= a
5
= 0. See the proof of
Lemma 2.21. Studying the coefficient of y
2
y
4
in (2.60) yields b
01
= 0, while van-
ishing the coefficients of y
3
y
4
, y
1
y
4
,andy
2
y
3
, y
1
y
2
(other coefficients are cancelled
© 2007 by Taylor & Francis Group, LLC
2 Invariant Subspaces: Mathematics in 1D 67
by the above conditions) leads to the system
3b
02
+ 4b
11
= 0, a
2
b
02
+ 5b
00
= 0,
7a
2
b
02
+ 4a
2
b
11
+ 20b
00
= 0, −6a
2
b
00
+ 15a
4
b
02
+ 10a
4
b
11
= 0.
The solutions of this system are
b
11
=−
3
4
b
02
, b
00
=−
1
5
a
2
b
02
, a
4
=−
4
25
a
2
2
that lead to the last case of operators (2.75) with C
2
= 0. The theorem is proved.
Notice that the description of all translation-invariant second-order operators that
preserve 5D subspaces is made by adding arbitrary linear operators of the second
order with constant coefficients.
By Theorem 2.22, it is easy to reconstruct a large number of quasilinear and fully
nonlinear heat and wave equations composed of linear combinations of operators,
u
t
= F[u]andu
tt
= F[u], (2.85)
exhibiting interesting finite-dimensional evolution on the corresponding invariant
subspaces. Various singularity formation phenomena, such as quenching, extinction,
blow-up, and propagation of finite interfaces, can be traced out by using such ex-
act solutions. We postpone more detailed singularity analysis until the next chapter,
where we begin the study of invariant subspaces and solutions of fourth-order thin
film equations with similar quadratic DSs.
2.5 First and second-order quadratic operators with subspaces of lower
dimensions
In this section, we consider quadratic operators
F[y] = b
11
(y
)
2
+ 2b
10
y
y + b
00
y
2
and (2.86)
F[y] = b
22
(y
)
2
+ 2b
21
y
y
+ 2b
20
y
y + b
11
(y
)
2
+ 2b
10
y
y + b
00
y
2
(2.87)
that admit invariant subspaces of the dimension which is less than maximal. The
coefficients of the operators, as well as the equations of invariant subspaces
y
(n)
= r
n−1
y
(n−1)
+ ... + r
1
y
+r
0
y
(notice the change in notation), are assumed to be constant. Such operators are
plugged into a number of nonlinear evolution PDEs of different types, including
reaction-diffusion, combustion, flame propagation, Boussinesq equations of water
wave interaction, and others, which typically take the evolution form (2.85). Notice
that many partial differential operators in IR
N
reduce to ordinary differential opera-
tors in radial geometry, so that they also fall into the scope of the present analysis.
2.5.1 First-order operators
We begin with the operators (2.86). Omitting most of technical calculus details ob-
tained by Reduce, let us present the final results. For completeness, the 3D subspaces
are also included.
© 2007 by Taylor & Francis Group, LLC
68 Exact Solutions and Invariant Subspaces
2D subspaces. Subspaces W
2
are defined by the ODE
y
= r
1
y
+r
0
y.
The invariance condition for (2.86) gives the following two cases:
Case I: r
0
= 0, i.e.,
y
−r
1
y
= 0, so that
W
2
= L{1, x } if r
1
= 0, or W
2
= L{1, e
r
1
x
} if r
1
= 0. (2.88)
This equation (and the corresponding subspace) is invariant under a 2D linear space
of operators spanned by
F
1
[y] = y(y
−r
1
y) and F
2
[y] = y
(y
−r
1
y). (2.89)
For r
1
= 0, one can choose another basis, such as
F
2
[y] +r
1
F
1
[y] = (y
)
2
−r
2
1
y
2
and F
2
[y] − r
1
F
1
[y] = (y
−r
1
y)
2
.
Case II: r
0
=−
2
9
r
2
1
= 0.Then,uptoscalinginx, the ODE and the subspace are
y
− 3y
+ 2y = 0andW
2
= L{e
x
, e
2x
}. (2.90)
The 1D space of operators is spanned by
F[y] = (y
− 2y)
2
. (2.91)
3D subspaces. These are given by
y
= r
2
y
+r
1
y
+r
0
y.
There are two cases:
Case I: r
0
= r
2
= 0, i.e.,
y
−r
1
y
= 0.
Up to scaling, we can consider r
1
= 0, ±1 and obtain the subspaces
W
3
= L{1, x , x
2
} (r
1
= 0),
W
3
= L{1, cos x , sin x} (r
1
=−1),
W
3
= L{1, cosh x, sinh x } (r
1
= 1).
The basis of operators is
F[y] = (y
)
2
−r
1
y
2
. (2.92)
Case II: r
0
= 0andr
1
=−
2
9
r
2
2
= 0. As above, we set r
2
= 3, r
1
=−2 to obtain
y
− 3y
+ 2y
= 0andW
3
= L{1, e
x
, e
2x
}.
The basis is
F[y] = (y
− 2y)
2
. (2.93)
It is easy to reduce operator (2.93) to (2.92) with r
1
= 0 (see Lemma 2.15).
© 2007 by Taylor & Francis Group, LLC
2 Invariant Subspaces: Mathematics in 1D 69
2.5.2 Second-order operators
Consider now the operators (2.87).
2D subspaces. The ODE for W
2
is
y
= r
1
y
+r
0
y.
For arbitrary r
1
and r
0
, there exists the basis
F
1
[y] = y(y
−r
1
y
−r
0
y), F
2
[y] = y
(y
−r
1
y
−r
0
y),
and F
3
[y] = y
(y
−r
1
y
−r
0
y),
(2.94)
composedof trivial operators annihilating W
2
. The subspaceof operatorsis extended
in two cases:
Case I: r
0
= 0. The ODE is
y
−r
1
y
= 0,
and the subspaces are as shown in (2.88). In addition to (2.94), two operators are
included:
F
4
[y] = y
(y
−r
1
y) and F
5
[y] = y(y
−r
1
y).
As above, for r
1
= 0, we can choose the basis
F
4
[y] +r
1
F
5
[y] = (y
)
2
−r
2
1
y
2
and F
4
[y] −r
1
F
5
[y] = (y
−r
1
y)
2
.
Case II: r
0
=−
2
9
r
2
1
= 0(r
1
= 3, r
0
=−2), i.e., (2.90) holds. The additional
operator is (2.91). Denoting f [y] = y
−2y, we can write down the above ODE and
the operator as follows:
f
− f = 0andF
4
[y] = ( f [y])
2
.
3D subspaces. Such subspaces W
3
are given by
y
= r
2
y
+r
1
y
+r
0
y.
There are two cases:
Case I: r
0
= 0, i.e.,
y
−r
2
y
−r
1
y
= 0.
For arbitrary r
2
and r
1
, there exist three linear independent operators
F
1
[y] = y(y
−r
2
y
−r
1
y), F
2
[y] = y
(y
−r
2
y
−r
1
y),
and F
3
[y] = y
(y
−r
2
y
−r
1
y).
The extensions are:
(i) If r
2
= 0, then F
4
[y] = (y
)
2
−r
1
y
2
;
(ii) If r
2
= 3andr
1
=−2, the extension is given by F
4
[y] = (y
− 2y)
2
.
These results are similar to those given above. The next case is new.
Case II: r
0
= 0. There exists a family of invariant subspaces given by the ODE with
the parameter α ∈ IR :
f
− f ≡ y
− αy
+ (3α −7)y
− 2(α −3)y = 0, where
© 2007 by Taylor & Francis Group, LLC
70 Exact Solutions and Invariant Subspaces
f [y] = y
+ (1 − α)y
+ 2(α −3)y.
The correspondingsubspaces are:
α = 4, 5: W
3
= L{e
x
, e
2x
, e
(α−3)x
},
α = 4: W
3
= L{e
x
, xe
x
, e
2x
},
α = 5: W
3
= L{e
x
, e
2x
, xe
2x
}.
For any α ∈ IR , the following operator is admitted:
F
1
[y] = ( f [y])
2
= [y
+ (1 − α)y
+ 2(α −3)y]
2
.
The space of operators is extended for:
(i) If α = 2, then the additional operator is
F
2
[y] = (y
− y
− 2y)(y
− 5y) + 2[(y
)
2
− yy
− 2y
2
];
(ii) If α = 6, then F
2
[y] = (y
− 5y
+ 6y)(y
− 3y); and
(iii) If α = 7, then F
2
[y] = (y
− 5y
+ 4y)
2
.
The case α = 3 coincides with case I(ii).
4D subspaces. Invariant subspaces W
4
are given by the ODE
y
(4)
−r
3
y
−r
2
y
−r
1
y
−r
0
y = 0.
There are several cases:
Case I: r
0
= 0.
I.1. r
1
= r
3
= 0. For any r
2
∈ IR , there exists the operator F
1
[y] = (y
)
2
−r
2
(y
)
2
.
For r
2
= 0, there exist additional operators
F
2
[y] = y
y
and F
3
[y] = yy
−
2
3
(y
)
2
.
I.2. A one-parameter family of invariant subspaces is given by
f
− f
≡ y
(4)
− αy
+ (3α −7)y
− 2(α − 3)y
= 0, (2.95)
where f [y] = y
+(1−α)y
+2(α −3)y. For any α ∈ IR , there exists the operator
F
1
[y] = ( f [y])
2
= [y
+ (1 − α)y
+ 2(α −3)y]
2
.
The invariant subspaces are:
α = 3, 4, 5: W
4
= L{1, e
x
, e
2x
, e
(α−3)x
},
α = 3: W
4
= L{1, x , e
x
, e
2x
},
α = 4: W
4
= L{1, e
x
, xe
x
, e
2x
},
α = 5: W
4
= L{1, e
x
, e
2x
, xe
2x
}.
Extra extensions are available in the following cases:
(i) If α = 2, there exist
F
2
[y] = (y
− y
− 2y)(y
− 3y
+ 2y), F
3
[y] = (y
− y
− 2y)(y
− 2y);
(ii) If α = 6, then F
2
[y] = (y
− 5y
+ 6y)(y
− 3y); and
© 2007 by Taylor & Francis Group, LLC
2 Invariant Subspaces: Mathematics in 1D 71
(iii) If α = 7, then F
2
[y] = (y
− 5y
+ 4y)
2
.
The above cases partially coincide with similar ones above. The next case, as well as
others below, are new.
(iv) If α = 1, then F
2
[y] = (y
− y
)
2
− 4(y
− y)
2
.
I.3. Another one-parameter family of subspaces given by
y
(4)
− y
− βy
+ βy
= 0
β =
1
4
, 4
is invariant under the operator F[y] = (y
− y
)
2
− β(y
− y)
2
. The corresponding
subspaces are given by
β = 0, 1: W
4
= L
1, e
x
, e
√
βx
, e
−
√
βx
,
β = 0: W
4
= L{1, x , x
2
, e
x
},
β = 1: W
4
= L{1, e
x
, xe
x
, e
−x
}.
For β =
1
4
and β = 4, this yields equation (2.95) with α = 2 (up to scaling x → 2x)
and α = 1, respectively. We also obtain one of the operators admitted by (2.95).
I.4. The one-parameter family of subspaces given by
y
(4)
− 2y
+ (γ +1)y
− γ y
= 0
γ = −2,
2
9
admits the operator F[y] = (y
− 2y
+ y)(y
− y
) + γ(y
− y)
2
. For γ =
2
9
and
γ =−2, these yield equation (2.95) with α = 6 (up to scaling x → 3x)andα = 2
respectively, and one of the admitted operators. Subspaces are
γ = 0,
1
4
: W
4
= L
1, e
x
, e
1
2
(1−
√
1−4γ)x
, e
1
2
(1+
√
1−4γ)x
,
γ = 0: W
4
= L{1, x , e
x
, xe
x
},
γ =
1
4
: W
4
= L{1, e
x
, e
x
2
, xe
x
2
}.
Case II: r
0
= 0. The only subspace W
4
and the operator are given by
f
− 3 f
+ 2 f ≡ y
(4)
− 10y
+ 35y
− 50y
+ 24y = 0,
F[y] = (y
− 7y
+ 12y)
2
≡ ( f [y])
2
.
The ODE is easily integrated and yields W
4
= L{e
x
, e
2x
, e
3x
, e
4x
}.
5D subspaces. This case corresponds to the maximal dimension of invariant sub-
spaces admitted by nonlinear quadratic second-order differential operators studied
in the previous section. For completeness, we now briefly comment on these conclu-
sions using a slightly different representation of the results. As a new feature, some
lower-dimensional invariant subspaces from W
5
are also presented. This study of lin-
ear invariant manifolds on invariant subspaces is important for future applications to
nonlinear evolution PDEs. In general, such questions are not straightforward at all.
Subspaces W
5
are given by the ODE
y
(5)
−r
4
y
(4)
−r
3
y
−r
2
y
−r
1
y
−r
0
y = 0.
Invariant subspaces are known to occur only if r
0
= 0. There are six cases that have
already been indicated in Theorem 2.22. We present these up to scaling in x.
© 2007 by Taylor & Francis Group, LLC
72 Exact Solutions and Invariant Subspaces
Cases (1)–(3). The ODE is y
(5)
− 5ry
+ 4r
2
y
= 0, where, up to scaling, we set
r = 0, ±1 and obtain the subspaces
W
5
= L{1, x , x
2
, x
3
, x
4
} (r = 0),
W
5
= L{1, cos x , sin x, cos2x, sin2x } (r =−1),
W
5
= L{1, cosh x , sinh x , cosh2x, sinh2x} (r = 1).
The basis of operators is F
1
[y] = yy
−
3
4
(y
)
2
−ry
2
and F
2
[y] = (y
− 4ry)
2
.
Case (4). The ODE is
f
− 3 f
+ 2 f
≡ y
(5)
− 10y
(4)
+ 35y
− 50y
+ 24y
= 0,
where f [y] = y
− 7y
+ 12y and W
5
= L{1, e
x
, e
2x
, e
3x
, e
4x
}. The operator is
F[y] = (y
− 7y
+ 12y)
2
≡ ( f [y])
2
. Note that the subspace L{e
3x
, e
4x
} is the
kernel (the null-set) of F.
There exist other lower-dimensional subspaces from W
5
,suchas
W
4
= L{e
x
, e
2x
, e
3x
, e
4x
} and W
3
= L{e
2x
, e
3x
, e
4x
},
as well as the 2D ones W
2
= L{e
3x
, e
4x
} and W
2
= L{e
2x
, e
4x
}. Such subspaces are
easily found by the change
˜x = e
−x
, ˜y = e
−4x
y, implying that
˜
F[˜y] = e
−4x
F[y] = ( ˜y
˜x ˜x
)
2
.
Therefore, the subspaces W
5
, W
4
, W
3
, W
2
,andW
2
are transformed into
˜
W
5
=
L{1, ˜x, ˜x
2
, ˜x
3
, ˜x
4
},
˜
W
4
= L{1, ˜x , ˜x
2
, ˜x
3
},
˜
W
3
= L{1, ˜x , ˜x
2
},
˜
W
2
= L{1, ˜x},and
˜
W
2
= L{1, ˜x
2
}, respectively.
Case (5). The ODE and the subspace are
y
(5)
− 6y
(4)
+ 7y
+ 6y
− 8y
= 0andW
5
= L{1, e
−x
, e
x
, e
2x
, e
4x
},
with the operator F[y] = (5y
− 27y
+ 28y)(y
− 3y
− 4y). Here, the kernel is
ker F = L{1, e
−x
, e
4x
}. It is easy to see that W
4
= L{1, e
−x
, e
x
, e
4x
}⊂W
5
is also
invariant.
Case (6). The ODE and the subspace are
y
(5)
− 5y
(4)
+ 5y
+ 5y
− 6y
= 0andW
5
= L{1, e
−x
, e
x
, e
2x
, e
3x
}.
The operatoris F[y] = (y
−5y
+6y)(y
−2y
−3y). Then ker F = L{1, e
−x
, e
3x
}
and W
4
= L{1, e
x
, e
2x
, e
3x
}⊂W
5
is invariant.
This completes the classification of invariant subspaces for the nonlinear second-
order operators, since the subspaces of the dimension higher than five can be pre-
served by linear differential operators only (Section 2.2).
2.6 Operators preserving polynomial subspaces
In this section, we study ordinary differential operators preserving polynomial sub-
spaces. Our goal is a complete description of the whole set of such operators, as well
as of subsets of operators of a given order and translation-invariant operators.
© 2007 by Taylor & Francis Group, LLC