Lie-B\"acklund symmetry and non-invariant solutions of nonlinear evolution equations

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by  Ivan M. Tsyfra

Lie-B¨acklund symmetry and non-invariant solutions of
nonlinear evolution equations
I.M.Tsyfra1, W. Rzeszut2 and V.A. Vladimirov1
1 AGH University of Science and Technology, Faculty of Applied Mathematics, 30
Mickiewicza Avenue, 30-059, Krakow, Poland
2 Faculty of Mathematics and Computer Science, Jagiellonian University, Krakow, Poland
7 Abstract
1
0 We study the symmetry reduction of nonlinear partial differential equations which are used
2 for describing diffusion processes in nonhomogeneous medium. We find ansatzes reducing partial
n differentialequationstosystemsofordinarydifferentialequations. Theansatzesareconstructedby
a usingoperatorsofLie-B¨acklundsymmetryofthirdorderordinarydifferentialequation. Themethod
J gives the possibility to find solutions which cannot be obtained by virtue of classical Lie method.

3 Such solutions have been constructed for nonlinear diffusion equations which are invariant with
1
respecttoone-parameter,two-parameterandthree-parameterLiegroupsofpointtransformations.

]
P
1 Introduction
A
.
h
It is of common knowledge, that the most effective method for constructing solutions of nonlinear
t
a ODEs of mathematical physics is the symmetry reduction method which brings a PDE with several
m
independent variables down to another PDE with fewer independent variables, or even to ODE. The
[
method can be both classical [1] and non-classical [2, 3, 4, 5]. In these cases the construction of a

1 proper ansatz (by which we mean a general form of a invariant solution) boils down to solving a
v
quasilinear first order DE, therefore ansatz includes one arbitrary function and the initial equation
2
2 reduces to a single differential equation with fewer independent variables (especially an ODE). In [6]
7 and [7] the concept of conditional Lie-Ba¨cklund symmetry of evolution equations is proposed. By
3
using this method one can reduce nonlinear evolution equations with two independent variables to
0
. system of ODEs. The approach is used to construct exact solutions of nonlinear diffusion equations
1
0 in [8]. The relationship of generalized conditional symmetry of evolution equations to compatibility of
7 system of differential equations is studied in [11]. Svirshchevskii [9] put forward the reduction method
1
for evolution equations of the form
:
v
u = K[u]
i t
X
whereu= u(t,x). Themethodis applicableif K[u]∂ is aLie-Ba¨cklund symmetryoperator of alinear
r u
a homogeneous ODE. In this paper we use the method proposed in [10], which is a generalization of
the Svirshchevskii’s method, meaning we can analyze symmetries of a nonlinear (or nonhomogeneous
linear) equation together with ODEs which include, besides dependent and independent variables,
parametricvariables andderivatives withrespecttothem. Asitwas shownin[10], such generalization
is important. For example the equation
u a(t,x)u = 0
xx
−
is invariant w.r.t. the symmetry operator
u u +3uxxu ∂
t− xxx ux u
(cid:0) (cid:1)
provided that a(t,x) is a solution to the KdV equation
a = a +6aa .
t xxx x
−
1
It is clear then, that the method is related to the inverse scattering transformation method.
The idea is to use Lie-Ba¨cklund symmetries of ordinary differential equations (linear or nonlinear) for
constructing solutions of evolutionary equations. In this paper we consider a nonlinear evolutionary
equationthatdescribestransportphenomenaininhomogeneousmediumandapplyareductionmethod
based on the symmetries of third order nonlinear ODEs. Note that the proposed method could be
applied to not only evolution-type equations, but to differential equations of any type as well.
We present the results obtained for the model medium with exponential and polynomial hetero-
geneity. Within the method applied, nonlinear transport equation is reduced to a system of three
ODEs. After integrating (solving) the system of ODEs, we obtain exact solution of the initial equa-
tion. Since the method applied differs from the classical Lie method, it does not enable to construct
algorithms for generation of new solutions, or production of conservation laws. Its only advantage is
the preservation of the reduction property. In addition, it doesn’t ensure that none of the solutions
obtained could obtained within the classical method. Therefore there is a very important question of
distinguishing truly new solutions obtained within the method proposed.
Based on the fact that a set of point and Lie-Ba¨cklund symmetry operators (of the ODE) form a
Lie algebra, we distinguish a class of diffusion equations whose solutions, obtained with the help of
the aforementioned approach, cannot be obtained through the classical Lie method.
Furthermore, it can be used to construct a large class of nonlinear evolution equation all of which
are reduced to systems of ordinary differential equations by the same ansatz and possess solutions
which are not invariant in the classical Lie sense.
rd
2 Application of the method by using 3 order ODEs
We consider third order differential equations of the following normal form
u U(x,u,u ,u ) = 0,
xxx x xx
−
assuming that function U is a power series of each of its arguments except x:
U = a (x)uj0uj1uj2.
j0,j1,j2 x xx
j0,jX1,j2 Z
∈
Here a (x) are some smooth functions to be determined. We require that this ODE admits the
j0,j1,j2
symmetry operator X = F(x,u,u ,u )∂ , where F is the right hand side of the transport equation,
x xx u
but the functions are independent of t, i.e., u(x,t) u(x), u (x,t) u (x), etc. The corresponding
x x
→ →
symmetry condition reads as follows:
X( ) u U(x,u,u ,u ) = 0, (1)
∞ xxx x xx
(cid:0) − (cid:1)(cid:12)(cid:12)uxxx=U
(cid:12)
where X( ) is the prolongation of the Lie B¨acklund symmetry generator X. The algorithm of solving
∞
this equation is quite cumbersome and we omit it, presenting merely the summarizing results in
Appendix.
2.1 How to obtain solutions of equation u = (H(x)) +F(x,u,u )?
t u xx x
We use the fact that the PDE maintains the ansatz after we modify the PDE by adding the terms
corresponding to symmetries of the ODE that generates that ansatz. Unfortunately, we are unable
to present solutions to the majority of ODEs we deal within the method applied. Therefore we shall
restrict ourselves to the ansatzes solvable in explicit forms, and connected with H(x) = 1 .
c2x2+c1x+c0
For the same reason, the choices of F(x,u,u ) are limited to those, for which the system of reduction
x
equations is easily solvable in the real domain.
For H(x) = κ, (κ = const) equation
x
u = 9uxxux 12u3x + 6u 18u2x 18u 12u (2)
xxx u − u2 x xx− x u − x2 x− x3
2
admits the operator Q = ( κ ) ∂ . It has the solution
1 xu xx u
u(x) = 1
±x√ϕ2x2+ϕ1x+ϕ0
and a 10-parameter Lie group of point and contact symmetries:
X = u∂ , X = xu ∂ , X = x2u3∂ , X = x3u3∂ , X = x4u3∂ , (3)
1 u 2 x u 3 u 4 u 5 u
X = (u +u )∂ , X = (2xu+x2u )∂ , (4)
6 x x u 7 x u
X = x2u2x+4xuux+4u2∂ , X = x2u2x+3xuux+2u2∂ , X = x2u2x+2xuux+u2∂ . (5)
8 x2u3 u 9 x3u3 u 10 x4u3 u
so the reduction to a system of ODEs is possible for any equation fo the form
10
κ
u = +a X u, a = const, i= 1,...,10.
t i i i
(cid:16)xu(cid:17)xx Xi=1
Let’s consider equation
κ
u = +a u+a xu +a (xu)3, κ,a R.
t 1 2 x 4 i
(cid:16)xu(cid:17)xx ∈
Symmetry classification of this equation can be written down as
X = ∂ ,
1 t
X = 2x∂ 3u∂ ,
2 x u
−
2a = 3a , a = 0 = X = e(2a1 3a2)t a x∂ +∂ +a u∂ ,
1 2 4 3 − 2 x t 1 u
6 ⇒ −
2a = 3a , a = 0 = X = a x∂ (cid:0)+t∂ +(a t+ 1)u∂ . (cid:1)
1 2 4 ⇒ 3 − 2 x t 1 2 u
Using the ansatz
u(x,t) = 1 ,
±x√ϕ2(t)x2+ϕ1(t)x+ϕ0(t)
we get the following reduction equations
ϕ +2(a 2a )ϕ = 0,
′2 1 − 2 2
ϕ +2(a a )ϕ = 0,
′0 1− 2 0
ϕ 1κϕ2 +2κϕ ϕ +2(a 3a )ϕ +2a = 0.
′1− 2 1 0 2 1 − 2 2 1 4
For 2a = 3a , a = 0
1 2 4
6
1
u(x,t) = .
±x c e2(2a2 a1)t 2√c c e(3a2 2a1)ttanh κ√c0c2 e(3a2 2a1)t x+c e2(a2 a1)t
q 2 − − 0 2 − (cid:0)3a2−2a1 − (cid:1) 0 −
For a = 3a = 3a, a = 0
1 2 2 2 4
1
u(x,t) = .
±
x c eatx2 2√c c tanh κ√c c (t+c ) x+c e at
2 0 2 0 2 1 0 −
q −
(cid:0) (cid:1)
For a , a = 0 and a arbitrary
1 2 4
1
u(x,t) = .
±x c x2 2√κ2c0c2+κa4 tanh √κ2c c +κa (t+c ) x+c
q 2 − κ 0 2 4 1 0
(cid:0) (cid:1)
For a = 0, a = 0, a = 0 we are unable to determine the solution. Let us note, that none of the
1 2 4
6 6 6
above solutions is invariant under any nonzero linear combination of X ,X ,X , as one can show
1 2 3
{ }
that
3
α X (u u(x,t)) = 0 = α = 0.
Xi=1 i i − (cid:12)u=u(x,t) ⇒∀j∈{1,2,3} j
(cid:12)
3
letusconsiderthecaseH(x) = κ , (κ = const).Itcanbeverified bydirectchecking, thatequation
x2
u = 9uxxux 12u3x + 12u 36u2x 60u 60u (6)
xxx u − u2 x xx− x u − x2 x− x3
admits the operator Q = ( κ ) ∂ . It has a solution
1 x2u xx u
u(x) = 1
±x2√ϕ2x2+ϕ1x+ϕ0
and a 10-parameter Lie group of point and contact symmetries:
X = u∂ , X = xu ∂ , X = x4u3∂ , X = x5u3∂ , X = x6u3∂ , (7)
1 u 2 x u 3 u 4 u 5 u
X = (2u +u )∂ , X = (3xu+x2u )∂ , (8)
6 x x u 7 x u
X = x2u2x+6xuux+9u2∂ , X = x2u2x+5xuux+6u2∂ , X = x2u2x+4xuux+4u2∂ , (9)
8 x4u3 u 9 x5u3 u 10 x6u3 u
so the reduction to a system of ODEs is possible for every equation from the class
10
κ
u = +a X u, a = const, i = 1,...,10.
t (cid:16)x2u(cid:17)xx iXi=1 i i
Let’s consider equation
κ
u = +a x5u3+a x6u3+a (3xu+x2u ), κ,a R.
t (cid:16)x2u(cid:17)xx 4 5 7 x i ∈
Symmetry classification of this equation can be written down as
Y = ∂ ,
1 t
a = 0 a = 0 = Y = ( 2x2a5 2x)∂ +t∂ + 3(4xa5 +3)u∂ ,
4 6 ∧ 7 ⇒ 2 − a4 − x t 2 a4 u
a = 0 = Y = x∂ +t∂ + 5u∂ , Y = x2∂ 3xu∂ ,
4 ⇒ 2 − x t 2 u 3 x− u
a = a = 0= Y = a tx2∂ +t∂ +(3a tx+ 1)u∂ .
4 5 ⇒ 4 − 7 x t 7 2 u
Since the ansatz is
u(x,t) = 1 ,
±x2√ϕ2(t)x2+ϕ1(t)x+ϕ0(t)
the reduction equations are
ϕ +2κϕ ϕ 1κϕ2 +2a +a ϕ = 0,
′2 0 2− 2 1 5 7 1
ϕ +2a +2a ϕ = 0,
′1 4 7 0
ϕ = 0,
′0
with a solution
ϕ = (c0a7+a4)2t2 (c a +a )(c1 + a4 )t+ c21 + c1a4 2c0a5 + a4(c0a7+a4) +c e 2c0κt,
2 c0 − 0 7 4 c0 κc20 4c0 2−κc20 2κ2c30 2 −
ϕ = 2(c a +a )t+c ,
1 0 7 4 1
−
ϕ =c ,
0 0
c = 0,
0
6
or
ϕ = 2κa2t3+a (a κc )t2+(1c2κ c a 2a )t+c ,
2 3 4 4 7− 1 2 1 − 1 7− 5 2
ϕ = 2a t+c ,
1 4 1
−
ϕ = 0.
0
4
Theabove solutions are invariant underthe nontrivial linear combination of Y ,Y ,Y ,Y only when
1 2 3 4
{ }
a = a = 0 or a = ϕ = 0.
4 5 4 0
Let’s consider equation
κ
u = +a u+a x5u3+a x6u3, κ,a R, a = 0.
t (cid:16)x2u(cid:17)xx 1 4 5 i ∈ 1 6
Symmetry classification of this equation can be written down as
Z = ∂ ,
1 t
a = 0 = Z = x2∂ 3xu∂ ,
4 2 x u
⇒ −
a = a = 0 = Z = x∂ 2u∂ , Z = e2a1t(∂ +a u∂ ).
4 5 3 x u 4 t 1 u
⇒ −
Since the ansatz is
u(x,t) = 1 ,
±x2√ϕ2(t)x2+ϕ1(t)x+ϕ0(t)
the reduction equations are as follows:
ϕ +2κϕ ϕ 1κϕ2 +2a ϕ +2a = 0,
′2 0 2− 2 1 1 2 5
ϕ +2a ϕ +2a = 0,
′1 1 1 4
ϕ +2a ϕ = 0,
′0 1 0
The bove system is fulfilled if
ϕ2 = e−2a1t(cid:18)eca01κe−2a1t(cid:16)c2− 4κa21 2c1a4−c0(4a5 − aa2421κ) Γ 0, ca01κe−2a1t (cid:17)+ 4cc210(cid:19)− aa51 + κ4aa3124,
(cid:0) (cid:1) (cid:0) (cid:1)
ϕ = a4 +c e 2a1t,
1 −a1 1 −
ϕ = c e 2a1t,
0 0 −
c = 0,
0
6
or
ϕ = (c c1a4κt)e 2a1t κc21e 4a1t a5 + κa24,
2 2 − a1 − − 4a1 − − a1 4a31
ϕ = a4 +c e 2a1t,
1 −a1 1 −
ϕ = 0.
0
Thesolutionu(x,t)withfunctionsϕ asaboveisnotinvariantinclassical sensewhena = 0ora = 0.
i 4 5
6 6
Let’s consider equation
κ
u = +a x4u3+a x5u3+a x6u3, κ,a R, a = 0.
t (cid:16)x2u(cid:17)xx 3 4 5 i ∈ 3 6
Symmetry classification of this equation can be written down as
W = ∂ ,
1 t
a = a = 0 = W = x∂ 2u∂ .
4 5 2 x u
⇒ −
Since the ansatz is
u(x,t) = 1 ,
±x2√ϕ2(t)x2+ϕ1(t)x+ϕ0(t)
the the functions ϕ , ν = 0,1,2 satisfy the system
ν
ϕ +2κϕ ϕ 1κϕ2+2a = 0,
′2 0 2− 2 1 5
ϕ +2a = 0,
′1 4
ϕ +2a = 0,
′0 3
5
having the solution
ϕ2 = e2κ(a3t2−c0t)(cid:18)−√π(cid:0)κ(c1−4c√0aa324a)32κ+(aa324−4a5)(cid:1)e2κac320erf κ(−√22aa33t+κc0) +c2(cid:19)+ 4aa43(−2a4t+2c1− c0aa34),
(cid:0) (cid:1)
ϕ = 2a t+c ,
1 4 1
−
ϕ = 2a t+c .
0 3 0
−
The solution u(x,t) with functions ϕ as above is not invariant under the translations generated by
i
W = ∂ when a = or a = 0. It is not not invariant under nonzero linear combination of W ,W
1 t 4 5 1 2
6 6 { }
for a = a = 0.
4 5
6
3 Appendix
Third order ODEs admitting Lie-Ba¨cklund symmetries with infinitesimal generator X = H(x) ∂
u xx∂u
(k , k , k , k , k , γ, n are all real arbitrary constants, H is differentiable) are as follows:(cid:0) (cid:1)
1 2 3 4 5
u = 7uxxux 8u3x 3Hxu +8Hxu2x 4Hxxu + Hxxxu
xxx u − u2 − H xx H u − H x H
for arbitrary H,
u = u2xx +4uxxux 6u3x u uuxx +6u2x 3u +u,
xxx ux u − u2 − xx− ux u − x
for H = ex
uxxx = uu2xxx +4uxxuux −6uu3x2 + x1−+nγuxx+ n(x(+1−γn)2)uuuxxx + 6xn+−γ2uu2x − n((x3+nγ−)52)ux+ n(n(−x+1)γ(n)3−3)u
for H = (x+γ)n. Also if H has the form H = 1 , then
c2x2+c1x+c0
u = 9uxxux 12u3x + 6(2c2x+c1) u 18(2c2x+c1) u2x
xxx u − u2 c2x2+c1x+c0 xx− c2x2+c1x+c0 u
−6(10c22x(2c+2x102c+1cc12xx+−c20c)02c2+3c21)ux− 6(2c2x+c1)((5cc222xx22++c15xc1+cc20x)−33c0c2+2c21)u,
1
if H = (x+γ)−2, then
u = 10uxxux 15u3x + 5k2(x+γ)2+3k1 u 35k2(x+γ)2+23k1 u2x
xxx u − u2 (x+γ) k2(x+γ)2+k1 xx− 2(x+γ) k2(x+γ)2+k1 u
(cid:0) (cid:1) (cid:0) (cid:1)
25k2(x+γ)2+15k1 u 45k2(x+γ)2+25k1 u+ k3(x+γ)3/2 u4, k = 0 k =0
−2(x+γ)2 k2(x+γ)2+k1 x − 8(x+γ)3 k2(x+γ)2+k1 k2(x+γ)2+k1 1 6 ∨ 2 6
(cid:0) (cid:1) (cid:0) (cid:1)
3
if H = (x+γ)−2, then
u = 10uxxux 15u3x + 12k2(x+γ)2+10k1 u 87k2(x+γ)2+75k1 u2x
xxx u − u2 (x+γ) k2(x+γ)2+k1 xx− 2(x+γ) k2(x+γ)2+k1 u
(cid:0) (cid:1) (cid:0) (cid:1)
135k2(x+γ)2+105k1 u 455k2(x+γ)2+315k1 u+ k3(x+γ)7/2 u4, k = 0 k =0
−2(x+γ)2 k2(x+γ)2+k1 x − 8(x+γ)3 k2(x+γ)2+k1 k2(x+γ)2+k1 1 6 ∨ 2 6
(cid:0) (cid:1) (cid:0) (cid:1)
if H = (x+γ) 2, then
−
u =9uxxux 12u3x + 12 u 36 u2x 60 u 60 u
xxx u − u2 x+γ xx− x+γ u − (x+γ)2 x − (x+γ)3
+k ((x+γ)5u3u +3(x+γ)4u4)+k ((x+γ)4u3u +2(x+γ)3u4)
1 x 2 x
+k ((x+γ)uu 2(x+γ)u2 4uu 6u2 )
3 xx− x − x− x+γ
u =10uxxux 15u3x + 15 u 55 u2x 105 u 105 u
xxx u − u2 x+γ xx− x+γ u − (x+γ)2 x− (x+γ)3
+k (x+γ)uu 3(x+γ)u2 10uu 15 u2
1 xx − x− x − x+γ
+k (cid:0)(x+γ)2u2u +3(x+γ)u3 +k (x+γ)3u4(cid:1)+k (x+γ)5u5+k (x+γ)6u5,
2 x 3 4 5
(cid:0) (cid:1)
u =10uxxux 15u3x + 14k1+15k2(x+γ) u 52k1+55k2(x+γ) u2x 95k1+105k2(x+γ) u
xxx u − u2 (x+γ)(k1+k2(x+γ)) xx− (x+γ)(k1+k2(x+γ)) u − (x+γ)2(k1+k2(x+γ)) x
90k1+105k2(x+γ) u+ k3(x+γ)4 u4+k (x+γ)2u2u + 2k1+3k2(x+γ)(x+γ)u3
− (x+γ)3(k1+k2(x+γ)) k1+k2(x+γ) 4 x k1+k2(x+γ)
(cid:16) (cid:17)
+k (x+γ)uu 3(x+γ)u2 9k1+10k2(x+γ)uu 12k1+15k2(x+γ) u2 , k = 0 k = 0,
5 xx− x − k1+k2(x+γ) x− (x+γ)(k1+k2(x+γ)) 1 6 ∨ 2 6
(cid:16) (cid:17)
and finally, if H = const, then
u = 10uxxux 15u3x + uxx 3 u2x +k u2u + 1 u3 + k3 u4+k uu 3u2 uux ),
xxx u − u2 x+k1 − x+k1 u 2 x x+k1 x+k1 4 xx− x − x+k1
(cid:0) (cid:1) (cid:0)
u = 10uxxux 15u3x +k u2u +k u5+k xu5+k u4+k uu 3u2 ,
xxx u − u2 1 x 2 3 4 5 xx− x
u =9uxxux 12u3x +k (xu3u +u4)+k u3u(cid:0), (cid:1)
xxx u − u2 1 x 2 x
u = u2xx +4uxxux 6u3x.
xxx ux u − u2
7
4 Conclusions
We have constructed solutions of nonlinear evolution equations describing the diffusion processes in
nonhomogeneous medium by using the generalization of Svirshchevskii method given in [10]. We
show that the method gives us the possibility to obtain solutions which are not invariant ones in
the classical Lie sense. We use the Lie-Ba¨cklund symmetry operators of the third order ordinary
differential equations. Thecorrespondingansatzes reduce nonlinear diffusion equations to the systems
of three ordinary differential equations. This way one is able to obtain the solutions which cannot be
constructed by classical Lie method in the cases when the dimension of invariance Lie algebra is equal
to 1,2,3. If the Lie algebra of Lie invariance group of the transport equation under consideration is
four-dimensional, then the solutions obtained by using our method could also beobtained via classical
Lie symmetry method as is shown in Section 2. These results agree with the ones given in [12] if
the solutions are found with the help of point conditional symmetry operators. The approach can
be also applied to construct other classes of diffusion-type equations (and exact solutions) by using
Lie-Ba¨cklund symmetry of other ordinary differential equations given in Appendix.
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