Galaktionov V.A., Svirshchevskii S.R. Exact Solutions and Invariant Subspaces of Nonlinear Partial Differential Equations in Mechanics and Physics
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1 Linear Invariant Subspaces: Examples 23
Consider the first case of β =
3
2
,where
u = C
1
+ C
2
x + C
3
x
2
+ C
4
x
3
∈ W
4
⇒ (1.65)
F
3
2
[u] = 3
C
3
C
4
− C
4
C
3
x
3
+
3C
2
C
4
− 6C
4
C
2
+ C
3
C
3
x
2
+
2C
2
C
3
− 9C
4
C
1
− C
3
C
2
x − 3C
3
C
1
+ C
2
C
2
.
For the second case β =
4
3
, choosing the PDE
u
tt
= u
x
u
tx
−
4
3
u
t
u
xx
+ 1,
we find solutions from the second invariant subspace,
u(x , t) = C
1
(t) + C
2
(t)x
2
+ C
3
(t)x
4
∈
ˆ
W
3
, (1.66)
C
1
=−
8
3
C
2
C
1
+ 1,
C
2
=
4
3
C
2
C
2
− 16C
3
C
1
,
C
3
=−8C
3
C
2
+
16
3
C
2
C
3
.
It is curious that, unlike in most of the previous cases, for β =
4
3
, the full subspace
of the fourth-order polynomials
ˆ
W
5
= L{1, x , x
2
, x
3
, x
4
}
is not invariant under F
4/3
[u]. The computations are easy:
u = C
1
+ C
2
x + C
3
x
2
+ C
4
x
3
+ C
5
x
4
∈
ˆ
W
5
⇒
F
4
3
[u] = 4
C
4
C
5
− C
5
C
4
x
5
+
1
3
16C
3
C
5
− 24C
5
C
3
+ 3C
4
C
4
x
4
+
2
3
6C
2
C
5
− 18C
5
C
2
+ 5C
3
C
4
− 3C
4
C
3
x
3
+
1
3
−48C
5
C
1
+ 9C
2
C
4
− 15C
4
C
2
+ 4C
3
C
3
x
2
+
2
3
−12C
4
C
1
+ 3C
2
C
3
− C
3
C
2
x +
1
3
−8C
3
C
1
+ 3C
2
C
2
.
The projection onto x
5
is 4(C
4
C
5
− C
5
C
4
) = 0 in general, so no invariance of
ˆ
W
5
exists. A unified approach to
∂
∂t
-dependent operators is developed Section 2.7.
Example 1.21 (Chaplygin gas equation) The following system in IR × IR :
ρ
t
+ ρ
x
u
x
+ ρu
xx
= 0,
u
t
+
1
2
(u
x
)
2
=
λ
ρ
2
,
(1.67)
consists of the equation of continuity for the density ρ(x, t),andEuler’s force equa-
tion for an ideal fluid of zero vorticity with the velocity potential u(x, t),inwhich
the pressure P is related to the density by
P =−
2λ
ρ
, where λ = constant = 0.
Expressing ρ from the first equation in (1.67) and substituting into the second yields
the Chaplygin gas equation (1904) [103]
1
u
t
+
1
2
(u
x
)
2
t
+
u
x
u
t
+
1
2
(u
x
)
2
x
= 0. (1.68)
© 2007 by Taylor & Francis Group, LLC
24 Exact Solutions and Invariant Subspaces
Its solutions are expressed via those of the linear wave equations X
tt
= X
ss
;see
[80]. On differentiation, we obtain from (1.68) a PDE with quadratic operators,
F[u] ≡ u
tt
+ 2u
x
u
xt
− 2u
xx
u
t
= 0. (1.69)
The basic polynomial subspace for F (a module) is still W
3
= L{1, x, x
2
},sothat,
for the solution (1.62) of (1.69), there occurs the DS
C
1
=−2C
2
C
2
+ 4C
3
C
1
,
C
2
=−4C
2
C
3
,
C
3
=−4C
3
C
3
.
For C
3
(t) =
1
4t
S
i
milarly to the previous example, for the more general equation
F
αβ
[u] ≡ u
tt
+ αu
x
u
xt
+ βu
xx
u
t
= 0, (1.70)
operator F
αβ
admits
W
4
= L{1, x , x
2
, x
3
}, if 3α + 2β = 0;
ˆ
W
3
= L{1, x
2
, x
4
}, if 4α + 3β = 0.
In the first case, 3α +2β = 0, for solutions (1.65) of (1.70), the DS is
C
1
= α(3C
3
C
1
− C
2
C
2
),
C
2
= α(9C
4
C
1
+ C
3
C
2
− 2C
2
C
3
),
C
3
= α(6C
4
C
2
− C
3
C
3
− 3C
2
C
4
),
C
4
= 3α(C
4
C
3
− C
3
C
4
).
In the second case, 4α +3β = 0, the DS for solutions (1.66) takes the form
C
1
=
8
3
αC
2
C
1
,
C
2
=
4
3
α(12C
3
C
1
− C
2
C
2
),
C
3
=
8
3
α(3C
3
C
2
− 2C
2
C
3
).
Concerning other subspaces, we take the trigonometric subspace
u(x , t) = C
1
(t) + C
2
(t) cos x + C
3
(t) sin x ,
and find that equation (1.70) with β =−α (this is true for (1.69)) admits such
solutions with the DS
C
1
=−α
C
2
C
2
+ C
3
C
3
,
C
2
=−αC
2
C
1
,
C
3
=−αC
3
C
1
.
Example 1.22 (Born–Infeld equations) Consider now the Born–Infeld (BI) equa-
tions in IR × IR
ρ
t
+
ρ
2
c
2
+a
2
c
2
+(u
x
)
2
x
= 0,
u
t
+ ρc
2
c
2
+(u
x
)
2
ρ
2
c
2
+a
2
= 0,
(1.71)
which were introduced in 1934 as a nonlinear correction to the linear Maxwell equa-
tions for electromagnetism, [73]. At the limit c →∞(c is the speed of light), this
relativistic Born–Infeld model reduces to the above non-relativistic Chaplygin gas
© 2007 by Taylor & Francis Group, LLC
, explicit solutions can be found (cf. Example 1.6).
1 Linear Invariant Subspaces: Examples 25
equation with λ =
1
2
a
2
. The BI equations belong to the family of 6×6 systems of
hyperbolic conservation laws, together with two solenoidal constraints on the mag-
netic field and electric displacements; see [80] and [266] for details on physics and
mathematics. Excluding ρ from the system (1.71) yields a single PDE
u
t
√
c
4
+c
2
(u
x
)
2
−(u
t
)
2
t
+ c
2
u
x
√
c
4
+c
2
(u
x
)
2
−(u
t
)
2
x
= 0,
which gives a cubic polynomial equation for u(x , t),
c
2
c
2
+ (u
x
)
2
u
tt
+ c
2
c
4
+ (c
2
− 1)(u
x
)
2
− (u
t
)
2
u
xx
− (c
2
− 1)u
t
u
x
u
xt
= 0.
In particular, exact solutions exist on L{1, x },i.e.,u(x , t) = C
1
(t) + C
2
(t)x .
Example 1.23 (Galilean invariant PDEs) Another similar quadratic operator oc-
curs in general Galilean invariant PDEs
F[u] ≡ u
xx
u
t
− u
x
u
tx
= G[u] ≡ G(u, u
x
, u
xx
, ...), (1.72)
where G[u] is an arbitrary (possibly elliptic) operator. This equation is invariant
under the Galilean transformation of the reference frame,
x → x +vt for any v = constant. (1.73)
So, if u(x , t) is a solution of (1.72), u(x +vt, t) is also a solution. Galilean transfor-
mations and invarianceunder generalized Galilean algebrasplay an important role in
the analysis of various PDEs, and especially in many physical and mechanical mod-
els; see Remarks for references. Furthermore, besides (1.73), equation (1.72) admits
a stronger symmetry generating an exceptionallywide class of solutions: if u(x, t) is
a solution of (1.72), then
u(x + f (t), t) is a solution for any C
1
-function f (t). (1.74)
Without extra hypotheses (say, symmetry), the Cauchy problem for (1.72) makes no
sense, since by (1.74) for given initial data, it admits an infinite-dimensional set of
solutions. Equations such as (1.72) are fully nonlinear PDEs with unknown concepts
ofpropersolutionsand local regularityproperties.Exactsolutionsmay help to clarify
some evolution characteristics and possible singularities of such PDEs. Operator F
is related to the remarkable operator F
rem
[u] = uu
xx
− (u
x
)
2
to be studied later on;
2D invariant modules of F,suchasL{1, cosγ x} and L{1, coshγ x } for any γ = 0.
Consider the Galilean invariant equation with the porous medium operator on the
right-hand side (for convenience,the original equation was divided by u
xx
)
u
t
−
1
u
xx
u
x
u
tx
= (uu
x
)
x
in IR × IR
+
. (1.75)
The invariant subspace for both left and right-hand sides is W
3
= L{1, x , x
2
}. Bear-
ing in mind the translational nonuniqueness,we take the symmetric expression
u(x , t) = C
1
(t) + C
2
(t)x
2
,
© 2007 by Taylor & Francis Group, LLC
see Examples 1.36 and 2.45. Equation (1.72) may admit exact solutions on various
26 Exact Solutions and Invariant Subspaces
which, similar to the ZKB solution, may be expected to exhibit the properties of the
source-type pattern for (1.75), and obtain the DS
C
1
= 2C
1
C
2
,
C
2
=−6C
2
2
.
Integrating yields the following compactly supported solution:
u(x , t) =
A(T − t)
1
3
−
1
6(T −t )
x
2
+
≡ (T − t)
1
3
A −
1
6
y
2
+
, (1.76)
where y = x/(T − t)
2/3
is the spatial rescaled variable and A > 0 is a constant. The
positive part in (1.76) formally mimics a typical finite propagation property for the
PME and needs a special setting of a free boundary problem (FBP). We will deal with
several problems like that for more mathematically reliable PDEs. Hence, (1.76) is
a self-similar solution of (1.75), exhibiting another nonlinear phenomenon, such as
the finite-time extinction: u(x , t) vanishes identically as t → T
−
.
Stability analysis of theseunusualsingularpatternsuses the correspondingrescaled
solution given by
u(x , t) = (T − t)
1
3
v(y,τ) with τ =−ln(T − t),
satisfies the following rescaled PDE:
v
τ
−
v
y
v
yy
v
τ y
+
2
3
v
yy
y +
1
3
v
y
= (vv
y
)
y
−
2
3
v
y
y +
1
3
v. (1.77)
The profile g(y) = (A −
1
6
y
2
)
+
that is obtained in (1.76) from the exact solution
may be treated as a “weak” stationary solution of (1.77), possibly, in an FBP frame-
work, which is not known. For general rescaled solutions, the passage to the limit
τ →∞in (1.77) is an
OPEN PROBLEM (recall that the general well-posedness of
such problems is also obscure). Local regularity and any potential or gradient prop-
erties of such flows are unknown. Inserting other operators on the right-hand side
of (1.75) may yield a different type of solutions on invariant subspaces, including
trigonometric or exponential ones, leading to non-similarity solutions (see further
examples below for better justifies models).
1.3.2 General polynomial operators
Let ν ={ν
1
, ..., ν
K
}∈IR
K
for2 ≤ K ≤ n−1 be a row(a multiindex)of nonnegative
integers of fixed length M =|ν|=ν
1
+ ... + ν
K
≥ 2. The corresponding primitive
monomial operator is given by
F
ν
[u] =
K
j =1
D
j
x
u
ν
j
. (1.78)
We next define the polynomial operator with arbitrary constants a
ν
∈ IR ,
F[u] =
(|ν|=M)
a
ν
F
ν
[u]. (1.79)
Proposition 1.24 (i)
Each operator
(1.78)
and, hence
(1.79)
, admits the subspace
W
n
in
(1.55)
if
K
j =1
(n − 1 − j )ν
j
≤ n − 1
for all
ν
such that
{|ν|=M}. (1.80)
© 2007 by Taylor & Francis Group, LLC
1 Linear Invariant Subspaces: Examples 27
(ii)
If
(1.80)
is not valid for
ν
from some subset
N ⊂ IR
K
and
(ν∈N )
(n − 1 − j )ν
j
= n,
then
W
n
is invariant under the extra condition
(ν∈N )
a
ν
( j )
(n − 1)(n − 2)...(n − j )
ν
j
= 0. (1.81)
Proof. Fixanarbitrary
u = C
1
+ C
2
x + ... + C
n
x
n−1
∈ W
n
.
(i) Taking into account the higher-degreeterms only, due to (1.80)
‡
,
F
ν
[x
n−1
] ∼ x
ρ
∈ W
n
, with ρ =
(n − 1 − j )ν
j
.
(ii) For any ν ∈ N , we need to control the coefficient of x
n
in each monomial,
F
ν
[... + C
n
x
n−1
] = ... + C
M
n
( j )
(n − 1)(n − 2)...(n − j )
ν
j
x
n
.
Summing up over all ν ∈ N yields that the condition (1.81) cancels the term with
x
n
in F[u].
Remark 1.25 (Extensions) It is not difficult to extend the result in (ii) to the case
(ν∈N
1
)
(n − 1 − j )ν
j
= n + 1forsomeN
1
⊂ IR
K
.
Then F[u] will contain terms with x
n+1
having the common multipliers C
M
n
,and
the next lower-order term on L{x
n
} with the common multiplies C
n−1
C
M−1
n
.There-
fore, we will need two hypotheses that guarantee that W
n
is still invariant. The first
hypothesis is similar to (1.81), and the second one is slightly more complicated but
still easy to derive in general. This analysis can be continued to include three or more
hypotheses on the operator.
We finish with the following application to more general equations.
Remark 1.26 (Polynomial PDEs)Nowletν =#ν
i, j
# be a K×K matrix of non-
negative integers, with 2 ≤ K ≤ n−1, and let the corresponding monomial
F
ν
[u] =
(i, j )
(D
i
t
D
j
x
u)
ν
i, j
be defined in the space of sufficiently smooth functions u = u(x, t) of two indepen-
dent variables x and t. The polynomial operator is then
F[u] =
(ν)
a
ν
(t)F
ν
[u].
Assume that, similar to (1.80),
(i, j )
(n − 1 − j )ν
i, j
≤ n − 1forallν.
Then the PDE
F[u] = L[u] + f (x, t) on W
n
,
‡
Here, “∼” means equality up to a non-zero constant multiplier.
© 2007 by Taylor & Francis Group, LLC
28 Exact Solutions and Invariant Subspaces
where f (·, t) ∈ W
n
for all t and L is a linear operator
L =
b
i, j
(x , t) D
i
t
D
j
x
,
with b
i, j
(·, t) ∈ W
j +1
for 1 ≤ i, j ≤ K and t ∈ IR , reduces to an ODE system.
1.3.3 One-dimensional polynomial subspaces
The 1D invariant subspaces are easy to deal with. For instance, consider the second-
order operator of nonlinear diffusion
F
2
[u] = (k(u)u
x
)
x
, with k(u) = αu
p
+ βu
2( p+1)
, (1.82)
where p = −1 is a constant. In this case, (1.82) admits the 1D subspace
W
1
= L{x
µ
}, with µ =
1
p+1
(x > 0) (1.83)
(and L{e
x
} if p =−1), since the first term (αu
p
u
x
)
x
, corresponding to αu
p
,van-
ishes. We will plug this operator into some PDEs.
Example 1.27 (Parabolic, KdV, and Cahn–Hilliard PDEs) First, we add to F
2
a
general first-order operator preserving subspace (1.83),
u
t
= F
2
[u] + γ u +
( j ≥0)
a
j
u
j
(u
x
)
−
µ( j −1)
µ−1
(µ = 1).
Then there exists the solution
u(x , t) = C(t)x
µ
, (1.84)
where C(t) satisfies the ODE
C
= βµ(µ + 1)C
2p+3
+ γ C +
a
j
µ
−
µ( j −1)
µ−1
C
µ−j
µ−1
.
For p =−
1
2
, i.e., µ = 2, we introduce a third-order KdV-type equation (odd-order
nonlinear PDEs are the subject of Chapter 4)
u
t
= u
xxx
+
α
√
u
+ βu
u
x
x
+ γ u + δuu
xx
,
where the ODE takes the form C
= 2(3β +δ)C
2
+γ C. For the fourth-orderCahn–
Hilliard equation (see Section 3.1 for details), we take µ = 3forp =−
2
3
,so
u
t
=−u
xxxx
+
αu
−
2
3
+ βu
2
3
u
x
x
+γ u + δu
x
u
xx
+ εuu
xxx
+ ...
(1.85)
admits solutions (1.84) with µ = 3.
Example 1.28 (Thin film models) In a similar fashion, the fourth-order thin film
operator (see Section 3.1)
F
4
[u] =−[(αu
p
+ βu
q
)u
xxx
]
x
, where q =
4( p+1)
3
,
preserves subspace (1.83) with µ =
3
p+1
. Hence, the thin film equation (TFE)
u
t
= F
4
[u] + γ u + δu
5
(u
x
)
−
4µ
µ−1
( p = 2)
© 2007 by Taylor & Francis Group, LLC
1 Linear Invariant Subspaces: Examples 29
has the solution (1.84) with the ODE
C
=−µ(µ
2
− 1)(µ − 2)βC
q+1
+ γ C + δµ
−
4µ
µ−1
C
µ−5
µ−1
.
1.3.4 Operators of the same total order: criterion of invariance
Quadratic case. Fix a natural k ≥ 1, and consider a general quadratic operator of
the following form:
F[u] =
(i+j =k)
a
i, j
D
i
x
uD
j
x
u, (1.86)
where the coefficients satisfy a
i, j
= a
j,i
for all {i, j ≥ 0:i + j = k}. The total
differential order of each monomial is equal to k. Note that, for polynomial invariant
subspaces, it is actually suffices to consider operators of the form (1.86); see Section
2.6 for more details and references. The particular case k = 4 is associated with thin
film operators
F[u] =−uu
xxxx
+ βu
x
u
xxx
+ γ(u
xx
)
2
(α, β ∈ IR ),
which will be systematically studied in Chapter 3. The symmetric form of this oper-
ator is as follows:
F[u] =−
1
2
uu
xxxx
+
1
2
βu
x
u
xxx
+ γ(u
xx
)
2
+
1
2
βu
xxx
u
x
−
1
2
u
xxxx
u.
Taking first u from the standard polynomial subspace,
u =
n−1
µ=0
C
µ
x
µ
∈ W
n
= L{1, x, ..., x
n−1
}, (1.87)
substituting into (1.86), and introducing the notation σ
i
µ
= µ(µ − 1)...(µ − i + 1),
we obtain
F[u] =
(i, j )
a
i, j
(µ)
C
µ
x
µ
(i)
(ν)
C
ν
x
ν
( j )
=
(i, j )
a
i, j
(µ)
C
µ
σ
i
µ
x
µ−i
(ν)
C
ν
σ
j
ν
x
ν−j
=
(µ,ν)
C
µ
C
ν
(i, j )
a
i, j
σ
i
µ
σ
j
ν
x
µ+ν−k
.
Since constants {C
µ
} are independent, we conclude that F : W
n
→ W
n
iff
(i+j =k)
a
i, j
σ
i
µ
σ
j
ν
= 0forallµ ≤ ν such that µ + ν − k > n − 1. (1.88)
Similarly, choosing any polynomial subspace with “gaps”,
u =
(µ∈N )
C
µ
x
µ
∈ W
n
, (1.89)
where the summation is performed over some subset of nonnegative integers N ,the
invariance condition in (1.88) must be valid for all µ + ν − k ∈ N . Dealing with
invariant subspaces for quadratic operators (1.86)pose a number of interesting math-
ematical problems. Later on, we perform a detailed study of operators like (1.86) on
general operators), and some other subspaces.
Cubic case. Consider the cubic operators from thin film theory (Chapter 3)
F[u] =
(i+j +l=k)
a
i, j,l
D
i
x
uD
j
x
uD
l
x
u, (1.90)
© 2007 by Taylor & Francis Group, LLC
polynomial(Section 2.6.4), exponential(see Example1.49, and Section 7.1 for more
30 Exact Solutions and Invariant Subspaces
where {a
i, j,l
} are elements of an absolutely symmetric tensor of rank 3 (symmetric
relative to any pair of indices). Substituting (1.87) yields
F[u] =
(µ,ν,ω)
C
µ
C
ν
C
ω
(i, j,l)
a
i, j,l
σ
i
µ
σ
j
ν
σ
l
ω
x
µ+ν+ω−k
.
Therefore, the invariance criterion of W
n
is
(i, j,l)
a
i, j,l
σ
i
µ
σ
j
ν
σ
l
ω
= 0forallµ ≤ ν ≤ ω, µ + ν + ω − k > n − 1. (1.91)
Similar conditions occur for quartic, quintic and higher-degree operators.
1.4 Examples: trigonometric subspaces
1.4.1 Some classification results and examples
We next move to quadratic and cubic operators preserving
W
3
= L{1, cos x , sin x}. (1.92)
This 3D subspace and its 5D extensionsinduce finite Fourierexpansions of solutions
of several nonlinear PDEs. For such subspaces, typical computations are more in-
volved than those for the polynomial expansions. We restrict our attention to the two
most important cases n = 3andn = 5. The general results in Section 2.2 will be
devoted to subspaces of arbitrary dimensions.
Proposition 1.29
Subspace
(1.92)
is invariant under the quadratic operator
(1.56)
iff the coefficients
{b
j
}
satisfy the linear system
b
4
= b
6
+ b
1
− b
3
,
b
5
= b
2
.
The set of such operators is a 4D linear space spanned by
F
1
[u] = u
xx
(u
xx
+ u), F
2
[u] = u
x
(u
xx
+ u),
F
3
[u] = u(u
xx
+ u), F
4
[u] = (u
x
)
2
+ u
2
.
Proposition 1.30
Subspace
(1.92)
is invariant under the cubic operator
(1.57)
iff
b
1
=
1
2
(b
5
+ b
7
− b
10
), b
2
=
1
2
b
6
,
b
3
=
1
2
(−b
5
+ 3b
7
+ 2b
8
− 3b
10
), b
4
= b
9
−
1
2
b
6
.
The set of such operators is a 6D linear space spanned by
F
1
[u] = u
xx
[(u
xx
)
2
+ (u
x
)
2
], F
2
[u] = u
x
[(u
xx
)
2
+ (u
x
)
2
],
F
3
[u] = u[(u
xx
)
2
+ (u
x
)
2
], F
4
[u] = u
xx
[2uu
xx
− (u
x
)
2
+ u
2
],
F
5
[u] = u
x
[2uu
xx
− (u
x
)
2
+ u
2
], F
6
[u] = u[2uu
xx
− (u
x
)
2
+ u
2
].
Next, consider the 5D trigonometric subspace
W
5
= L{1, cos x , sin x, cos2x, sin2x }. (1.93)
Proposition 1.31
Subspace
(1.93)
is invariant under the quadratic operator
(1.56)
iff the coefficients
{b
j
}
satisfy
b
2
− 2b
5
= 0, b
2
− 4b
5
= 0,
b
1
− 4b
3
− 4b
4
+ 16b
6
= 0,
2b
1
− 4b
3
− 5b
4
+ 8b
6
= 0.
© 2007 by Taylor & Francis Group, LLC
1 Linear Invariant Subspaces: Examples 31
The set of such operators is a 2D linear space spanned by
F
1
[u] = uu
xx
−
3
4
(u
x
)
2
+ u
2
and
F
2
[u] = (u
xx
+ 4u)
2
.
Proposition 1.32
No nontrivial cubic operators
(1.57)
admit subspace
(1.93)
.
Proof. Subspace (1.93) is invariant under (1.57) iff
b
2
= b
4
= b
6
= b
9
= 0, b
3
= 4b
7
, b
5
= b
8
− 9b
5
+ 8b
7
= 0,
b
10
− b
1
+ 3b
7
= 0, −3b
1
+ b
5
+ 8b
7
= 0,
15b
1
− 5b
5
− 44b
7
= 0, 21b
1
− 4b
5
− 64b
7
= 0.
This system admits the trivial solution only.
Some examples of PDEs with standard basic trigonometric subspaces have been
studied in Section 1.1. Let us consider another example motivated by the structure of
the quadratic operator of the GT equation from Example 1.20.
Example 1.33 (Periodic solutions of the Gibbons–Tsarev equation) Similar to
Example 1.20, the quadratic operator of the GT equation (1.64) is convenient to
consider on the module (1.92). Hence, there exist exact solutions
u(x , t) = C
1
(t) + C
2
(t) cos x + C
3
(t) sin x ,
C
1
= C
2
C
2
+ C
3
C
3
+ 1,
C
2
= C
1
C
2
,
C
3
= C
1
C
3
.
Further extensions of trigonometric subspaces are possible for a three-parameter
family of such quadratic operators,
F[u] = u
x
u
tx
− βu
t
u
xx
+ γ uu
txx
+ δuu
t
(β,γ,δ ∈ IR ). (1.94)
Proposition 1.34
The only operator
(1.94)
associated with the 5D module
(1.93)
is
as follows:
F[u] = u
x
u
tx
−
2
3
u
t
u
xx
−
2
3
uu
txx
−
4
3
uu
t
. (1.95)
Proof. We begin with the necessity. Using the subspace of cos-functions,
u = C
1
+ C
2
cosx + C
3
cos2x ∈ W
3
,
we give the results of final calculations, keeping only the suspicious terms that may
not belong to W
3
,
F[u] =
4sin
2
2x + (4β − 4γ + δ) cos
2
2x
C
3
C
3
+
(4β − γ + δ) cos x cos2x + 2sinx sin 2x
C
3
C
2
+
(β − 4γ + δ) cos x cos2x + 2sinx sin 2x
C
2
C
3
+ ... .
These three terms belong to W
5
for arbitrary values C
2,3
and C
2,3
iff
4β − 4γ + δ = 4,
4β − γ + δ = 2,
β − 4γ + δ = 2,
from which come β =−γ =
2
3
and δ =−
4
3
yielding operator (1.95).
© 2007 by Taylor & Francis Group, LLC
32 Exact Solutions and Invariant Subspaces
To see the sufficiency, we write down the operator (1.95) in the form
F[u] =−
2
3
uu
xx
−
3
4
(u
x
)
2
+ u
2
t
.
The operator in square brackets, up to scaling x →
√
3
2
x, coincides with the operator
in Example 1.15 and admits the subspace (1.93), so the same subspace is available
for F,whereW
5
becomes a module.
Example 1.35 (Fast diffusion equation) Consider the following quasilinear fast
diffusion equation with source and absorption:
v
t
=
v
−
4
3
v
x
x
−
3
4
v
−
1
3
−
3
2
av −
3
2
bv
5
3
, (1.96)
where a and b are constants. Setting u = v
−2/3
(notice that u is not the standard
pressure) yields a PDE with a cubic operator,
u
t
= u
2
u
xx
−
1
2
u(u
x
)
2
+
1
2
u
3
+ au + b.
It follows from Proposition 1.30 (see operator F
6
therein) that (1.96) has solutions
v(x , t) =
C
1
(t) + C
2
(t) cos x + C
3
(t) sin x
−
3
2
, (1.97)
C
1
=
1
2
C
1
(C
2
1
− C
2
2
− C
2
3
) + aC
1
+ b,
C
2
=
1
2
C
2
(C
2
1
− C
2
2
− C
2
3
) + aC
2
,
C
3
=
1
2
C
3
(C
2
1
− C
2
2
− C
2
3
) + aC
3
.
These solutions can be derived by using a quadratic representation of (1.96). Setting
u = v
−4/3
(the correct pressure transformation) yields a family of solutions on the
Example 1.36 (Riabouchinsky–Proudman–Johnson (RPJ) equation) Consider
the RPJ equation [488, 473]
u
xxt
= εu
xxxx
− uu
xxx
+ u
x
u
xx
, (1.98)
with the third-order quadratic operator F[u] =−uu
xxx
+ u
x
u
xx
. This PDE arises
in the so-called Berman problem and is derived from the Navier–Stokes equations
in IR
2
describing the unsteady plane flow of a viscous incompressible fluid confined
between parallel walls. In this case, ε =
1
Re
, where Re is the Reynolds number.The
stationary solutions of (1.98) were studied by A.S. Berman in 1953, [43].
Equation (1.98) can be integrated once to reveal the special quadratic operator
u
xt
= εu
xxx
−
uu
xx
− (u
x
)
2
+ g(t),
where g(t) is an arbitrary smooth function of integration related to the fluid pressure.
We will refer to
F
rem
[u] = uu
xx
− (u
x
)
2
(1.99)
as the remarkable operator, in view of its outstanding invariant properties that will
be used a few times later on. Namely, F
rem
admits the 3D subspace
W
3
= L{1, cosγ x, sin γ x} for any γ = 0. (1.100)
© 2007 by Taylor & Francis Group, LLC
5D subspace (1.93) that includes those given by (1.97); cf. Example 1.15.