Directory UMM :Journals:Journal_of_mathematics:BJGA:
Symmetry analysis of the cylindrical
Laplace equation
Mehdi Nadjafikhah, Seyed-Reza Hejazi
Abstract. The symmetry analysis for Laplace equation on cylinder is
considered. Symmetry algebra, the structure of the Lie algebra of the symmetries and some related topics such as invariant solutions, one-parameter
subgroups, one dimensional optimal system and differential invariants are
given.
M.S.C. 2000: 49J30, 49K30, 55S45.
Key words: Laplace equation, invariant solution, optimal system.
1
Introduction
Pierre-Simon, marquis de Laplace (March 23, 1749 - March 5, 1827) was a French
mathematician and astronomer whose work was pivotal to the development of mathematical astronomy. He summarized and extended the work of his predecessors in his
five volume Mcanique Cleste (Celestial Mechanics) (1799-1825). This seminal work
translated the geometric study of classical mechanics, used by Isaac Newton, to one
based on calculus, opening up a broader range of problems. He formulated Laplace’s
equation, and invented the Laplace transform which appears in many branches of
mathematical physics, a field that he took a leading role in forming. The Laplacian
differential operator, widely used in applied mathematics, is also named after him.
In mathematics and physics, the Laplace operator or Laplacian, denoted by ∆ or
∇2 and named after Pierre-Simon de Laplace, is a differential operator, specifically an
important case of an elliptic operator, with many applications. In physics, it is used in
modeling of wave propagation (the wave equation is an important second-order linear
partial differential equation that describes the propagation of a variety of waves, such
as sound waves, light waves and water waves. It arises in fields such as acoustics, electromagnetics, and fluid dynamics. Historically, the problem of a vibrating string such
as that of a musical instrument was studied by Jean le Rond d’Alembert, Leonhard
Euler, Daniel Bernoulli, and Joseph-Louis Lagrange.), an important partial differential equation which describes the distribution of heat (or variation in temperature) in
a given region over time. For a function of three spatial variables (x, y, z) and one time
∗
Balkan
Journal of Geometry and Its Applications, Vol.14, No.2, 2009, pp. 53-64.
c Balkan Society of Geometers, Geometry Balkan Press 2009.
°
54
Mehdi Nadjafikhah, Seyed-Reza Hejazi
variable t; heat flow, forming the Helmholtz equation. It is central in electrostatics and
fluid mechanics, anchoring in Laplace’s equation and Poisson’s equation. In quantum
mechanics, it represents the kinetic energy term of the Schrödinger equation. The kinetic energy of an object is the extra energy which it possesses due to its motion. It is
defined as the work needed to accelerate a body of a given mass from rest to its current
velocity. Having gained this energy during its acceleration, the body maintains this
kinetic energy unless its speed changes. Negative work of the same magnitude would
be required to return the body to a state of rest from that velocity. In mathematics,
functions with vanishing Laplacian are called harmonic functions; the Laplacian is at
the core of Hodge theory and the results of de Rham cohomology [9].
The aim is to analysis the point symmetry structure of the Laplace equation on
cylinder, i.e., cylindrical Laplace equation, which is
∇2 u =
(1.1)
1 ∂ ³ ∂u ´
1 ∂2u ∂2u
r
+ 2 2 + 2 = 0,
r ∂r ∂r
r ∂θ
∂z
where u is a smooth function of (r, θ, z).
The symmetry algebra of (1.1) is determined in the next section and some results
yield from the structure of the Lie algebra of symmetries are given.
2
Lie symmetries of the equation
The method of determining the classical symmetries of a partial differential equation
is standard which is described in [1, 4, 5] and [8]. To obtain the symmetry algebra of
the (1.1), we take an infinitesimal generator of symmetry algebra of the form
X = ξ1 (r, θ, z, u)
∂
∂
∂
∂
+ ξ2 (r, θ, z, u)
+ ξ3 (r, θ, z, u)
+ η(r, θ, z, u) .
∂r
∂θ
∂z
∂u
Using the invariance condition, i.e., applying the second prolongation X (2) to (1.1),
the following system of 18 determining equations yields:
ξ2 u = 0,
ξ3 u = 0,
ηuu = 0,
ξ3 rzz = 0,
ξ3 θzz = 0,
ξ3 zzz = 0,
ξ3 rr + ξ3 zz = 0,
ξ3 θ + r2 ξ2 z = 0,
ξ3 θ − rξ3 rθ = 0,
ξ3 rz + 2ηru = 0,
ξ3 θz + 2ηθu = 0,
ξ3 zz + 2ηzu = 0,
ξ1 + r(ξ2 θ − ξ3 z ) = 0,
ξ2 θ + r(ξ2 rθ − ξ3 rz ) = 0,
ξ2 θθ − rξ2 r − ξ3 θz = 0,
ξ3 θθ + r(ξ3 r + rξ3 zz ) = 0,
ξ3 θz + r(2ξ2 r − rξ2 rr ) = 0,
ηθθ + r(ηr + rηrr + rηzz ) = 0.
The solution of the above system gives the following infinitesimals,
∂
∂
∂
,
X2 =
,
X3 = u ,
∂θ
∂z
∂u
∂
∂
∂
1
∂
1
∂
= r
+z ,
X5 = rz
+ (z 2 − r2 )
− zu ,
∂r
∂z
∂r 2
∂z
2 ∂u
∂
1
∂
∂
1
∂
= sin θ
+ cos θ ,
X7 = cos θ
− sin θ ,
∂r
r
∂θ
∂r
r
∂θ
X1 =
X4
X6
55
Symmetry analysis of the cylindrical Laplace equation
∂
1
∂
∂
+ z cos θ
− r sin θ ,
∂r
r
∂θ
∂z
1
∂
∂
∂
− z sin θ
− r cos θ ,
= z cos θ
∂r
r
∂θ
∂z
1
∂
1
∂
∂
1
∂
= (z 2 − r2 ) sin θ
+ (z 2 + r2 ) cos θ
− rz sin θ
+ ru sin θ ,
2
∂r 2r
∂θ
∂z
2
∂u
1
∂
1
∂
∂
1
∂
= (z 2 − r2 ) cos θ
− (z 2 + r2 ) sin θ
− rz cos θ
+ ru cos θ .
2
∂r 2r
∂θ
∂z
2
∂u
X8 = z sin θ
X9
X10
X11
The commutation relations of the 11−dimensional Lie algebra g spanned by the vector
fields X1 , · · · , X11 , are shown in following table.
X1
X1
X2
X3
X4
X5
X6
X7
X8
X9
X10
X11
2.1
X2
0
0
0
0
0
0
0
−X2
0 12 X3 − X4
−X7
0
X6
0
−X9
−X6
X8
−X7
−X11 −X8
X10
−X9
X3
0
0
0
0
0
0
0
0
0
0
0
X4
X5
X6
X7
X8
X9
X10
X11
0
0
X7
−X6
X9
−X8
X11
−X10
X2 − 12 X3 + X4
0
0
X6
X7
X8
X9
0
0
0
0
0
0
0
0
0
X5
−X7
−X7
0
0
X10
X11
−X5
0
−X8
−X9
−X10 −X11
0
0
X7
X8
0
0
−X2
0 12 X3 − X4 −X1
X7
X9
0
0
0
−X2
X1 21 X3 − X4
0
X10
X2
0
0
X1
−X5
0
0
X11
0
X2
−X1
0
0
−X5
−X10
0
− 12 X3 + X4 −X1
X5
0
0
0
−X11
0
X1 − 21 X3 + X4 0
X5
0
0
Differential invariants
Suppose that G is a transformation group. It is well known that a smooth real differential function I : J n → R, where J n is the corresponding n−th jet space, is a
differential invariant for G if and only if for all X ∈ g, its n−th prolongation annihilates I, i.e., X (n) (I) = 0. Here we are going to find the differential invariants of order
0, 1 and 2 and some of their results. For more details and some practical examples
see [3] and [8].
The symmetry group of the cylindrical Laplace equation has not any nonconstant
ordinary differential invariants, i.e., it does not have any differential invariants of order
zero, thus the orbit of generic points (the points which lies in the orbit of maximal
dimension) has dimension four and its isotropy subgroup is seven dimensional.
The first prolonged action of the symmetry group has one differential invariant,
which is a function of (u, ur , uθ , uz ), so the action has a generic orbit of dimension
six, and its isotropy subgroup is five dimensional.
Finally the second prolonged action has another differential invariant which is a
function of (u, ur , uθ , uz , urr , urθ , urz , uθθ , uθz , uzz ), consequently the dimension of the
generic orbit is equal to the dimension of the group; thus, the second prolonged action
is transitive and has a discrete isotropy subgroup.
According to the relations between above subjects and [4], we can obtain the
number of functionally independent differential invariants up to order n.
56
Mehdi Nadjafikhah, Seyed-Reza Hejazi
Theorem 2.1. The number of functionally independent differential invariants
¢ up
¡
−8
to order n ≥ 3 for the symmetry group of cylindrical Laplace equation is 3+n
n
and
the
number
of
strictly
differential
invariants
which
occur
in
each
prolongation
is
¡3+n¢ ¡2+n¢
−
.
n
n−1
Proof. According to the above discussion for n ≥ 3 the action is transitive, so the
number of independent differential invariants in , is
¶
¶
µ
µ
3+n
3+n
− 8.
− 11 =
in = 3 +
n
n
If jn is the number of strictly differential invariants then,
¶
¶ µ
¶ µ
¶ µ
µ
2+n
3+n
3+n−1
3+n
.
−
=
−
jn =
n−1
n
n−1
n
2.2
✷
Reduction of the equation
The Eq. (1.1) maybe regarded as a submanifold M of the jet space J 2 (R3 , R). Doing
as section 3.1 of [5] and find (in a sense) the most general group-invariant solutions
to the Eq. (1.1).
Theorem 2.2. The one-parameter groups gi (t) : M → M generated by the Xi , i =
1, · · · , 11, are given in the following table:
g1 (t) : (r, θ, z, u) 7−→ (r, θ + t, z, u),
g2 (t) : (r, θ, z, u) 7−→ (r, θ, z + t, u),
g3 (t) : (r, θ, z, u) 7−→ (r, θ, z, uet ),
g4 (t) :
g5 (t) :
g6 (t) :
g7 (t) :
g8 (t) :
g9 (t) :
g10 (t) :
g11 (t) :
(r, θ, z, u) 7−→ (ret , θ, zet , u),
³
´
t
1
(r, θ, z, u) 7−→ rzt + r, θ, (z 2 − r2 ), − zut + u ,
2
2
´
³
t
(r, θ, z, u) 7−→ t sin θ + r, cos θ + θ, z, u ,
r
³
´
t
(r, θ, z, u) 7−→ t cos θ + r, sin θ + θ, z, u ,
r
´
³
t
(r, θ, z, u) 7−→ tz sin θ + r, z cos θ + θ, −rt sin θ + z, u ,
r
³
´
t
(r, θ, z, u) 7−→ tz cos θ + r, − z sin θ + θ, −rt cos θ + z, u ,
r
Ã
´
t 2
t ³ z2
(r, θ, z, u) 7−→
(z − r2 ) sin θ,
+ r cos θ + θ,
2
2 r
!
1
−rzt sin θ + z, rut sin θ + u ,
2
Ã
´
t 2
t ³ z2
(r, θ, z, u) 7−→
(z − r2 ) cos θ, −
+ r sin θ + θ,
2
2 r
!
1
−rzt cos θ + z, rut cos θ + u ,
2
Symmetry analysis of the cylindrical Laplace equation
57
where entries give the transformed point exp(tXi )(r, θ, z, u) = (r̃, θ̃, z̃, ũ).
The transformations g1 and g2 demonstrate the radius- and space-invariance and
radius- and angle-invariance solutions of the equation. The one-parameter g3 reflects
the linearity of the cylindrical laplace equation, and the well-known scaling symmetry
turns up in g4 . The most general one-parameter group of symmetries is obtained
by considering a general linear combination of the given vector fields; the explicit
formula for the group transformations is very complicated, but alternatively we can
use theorem 2.52 in [4], and represent an arbitrary group transformation g as the
composition of transformations in theorem (2.2).
Theorem 2.3. Let i = 1, · · · , 11 and t ∈ R. If u = U (r, θ, z) is a solution of Eq.
(1.1) so are the functions ui (r, θ, z) = U (r, θ, z), where
u1 = U (r, θ + t, z),
u2 = U (r, θ, z + t),
u3 = e−t U (r, θ, z),
u4 = U (et r, θ, et z),
³
´
t
1
u5 = 1 + zt U (rzt + r, θ, (z 2 − r2 )),
2
2
³
´
t
t
u6 = U (t sin θ + r, cos θ + θ, z),
u7 = U t cos θ + r, cos θ + θ, z ,
r
r
³
´
t
8
u = U tz sin θ + r, z cos θ + θ, −rt sin θ + z ,
r
´
³
t
9
u = U tz cos θ + r, − z sin θ + θ, −rt cos θ + z ,
Ãr
!
³
´
rt
t 2
t 2
10
2
2
u = 1 − sin θ U
(z − r ) sin θ, (z + r ) cos θ + θ, z − rzt sin θ ,
2
2
2r
Ã
!
³
´
rt
t 2
t 2
11
2
2
u = 1 − cos θ U
(z − r ) cos θ, (z + r ) sin θ + θ, z − rzt cos θ .
2
2
2r
For example, if u(r, θ, z) = 1 be a constant solution of Eq. (1.1), we conclude
−t
that the trivial functions gi (t) · 1 =³1 for
´ i = 1, 2, 4, 6, 7, 8, 9, g3 (t) · 1 = e and
g5 (t) · 1 = 2 + zt, and by applying g5 2a
· g3 (ln b) on u = 1, we conclude the linear
b
solution u = az + b. The two nontrivial solutions for Eq. (1.1) are:
g10 (t) · 1 = 2 − rt sin θ, g11 (t) · 1 = 2 − rt cos θ.
´
³
· g3 ln d · 1, g10 (−a) · 1 and g11 (−b) · 1, we conclude
Now by summing g1 , g5 −8+2c
d
the following result:
Corollary 2.4. The function u(r, θ, z) = ar sin θ + br cos θ + dz + c, is a solution of
Eq. (1.1), where a, b, c and d are arbitrary constants.
2.3
Some invariant solutions
The first advantage of symmetry group method is to construct new solutions from
known solutions. Neither the first advantage nor the second will be investigated here,
58
Mehdi Nadjafikhah, Seyed-Reza Hejazi
but symmetry group method will be applied to the Eq. (1.1) to be connected directly
to some order differential equations. To do this, a particular linear combinations of
infinitesimals are considered and their corresponding invariants are determined.
The cylindrical Laplace equation expressed in the coordinates (r, θ, z), so to reduce
this equation is to search for its form in specific coordinates. Those coordinates will
be constructed by searching for independent invariants (y, v) corresponding to an
infinitesimal generator. so using the chain rule, the expression of the equation in the
new coordinate allows us to the reduced equation. Here we will obtain some invariant
solutions with respect to symmetries.
1. First consider X6 . To determine the independent invariants I, we need to solve
the first order partial differential equation X6 {I(r, θ, z, u)} = 0, that is
sin θ
∂I
cos θ ∂I
∂I
∂I
+
+0
+0
= 0,
∂r
r ∂θ
∂z
∂u
which is a homogeneous first order PDE. Thus we solve the associated characteristic
ordinary differential equations
dr
cos θdθ
dz
du
=
=
=
.
sin θ
r
0
0
z
and v = u.
Hence we obtain two functionally independent invariants y =
r sin θ
If we treat v as a function of y, we can compute formulae for the derivatives
of u with respect to x and t in terms of y, v and the derivatives of v with respect
to y, along with a single parametric variable, which we designate to be t, so that
x will be the corresponding principal variable. Using the chain rule we find that if
u = v = v(y) = v(z/r sin θ), then
(2.1)
(2.2)
(2.3)
(2.4)
ur
urr
uθθ
uzz
=
=
=
=
vy yr ,
yrr vy + vyy yr ,
yθθ vy + vyy yθ ,
yzz vy + vyy yz .
Substituting
³ ur ,´urr , uθθ and uzz in (1.1) we obtain an ODE with solution u =
z
C1 arctan r sin
θ + C2 , which is called an invariant solution for Eq. (1.1).
z
and v = u, and the solution of
r cos θ
reduced equation
solution) by substituting (2.1), (2.2), (2.3) and (2.4) is
³ (invariant
´
z
u = C1 arctan r cos θ + C2 .
2.
Similarly, the invariants of X7 are y =
r2 + z 2
and v = u. The invariant solution constructed
r sin θ
³√
´
2
2
with (2.1), (2.2), (2.3) and (2.4) is u = C1 arctanh rrsin+zθ + C2 obtained from
ordinary differential equations
3.
The invariants of X8 are
59
Symmetry analysis of the cylindrical Laplace equation
dr
rdθ
dz
du
=
=
=
.
z sin θ
a cos θ
−r sin θ
0
The above mentioned method in sections 1 and 2 can be followed for X9 with invari³√
´
r2 + z 2
2 +z 2
and v = u, and invariant solution u = C1 arctanh rrcos
+ C2 . We
ants
θ
r sin θ
can proceed for other Xi s as above.
2.4
The Lie algebra of symmetries
In this section, we determine the structure of full symmetry Lie algebra g, of Eq.
(1.1).
Theorem 2.5. The full symmetry Lie algebra g, of Eq. (1.1) has the following semidirect decomposition:
(2.5)
g = R ⋉ g1 ,
where g1 is a simple Lie algebra.
Proof: The center z of g is SpanR {X3 }. Therefore the quotient algebra g1 = g/z is
SpanR {Y1 , · · · , Y10 }, where Yi := Xi + z for i = 1, · · · , 10. The commutator table of
this quotient algebra is given in following table:
Y1
Y2
Y3
Y4
Y5
Y6
Y7
Y8
Y9
Y10
Y1
Y2
Y3
Y4
Y5
Y6
Y7
Y8
Y9
Y10
0
0
0
0
−Y7
Y5
−Y8
Y7
−Y10
Y9
0
0
−Y2
−Y3
0
0
−Y5
−Y6
−Y7
−Y8
0
Y2
0
−Y4
Y5
Y6
0
0
−Y9
−Y10
0
Y3
Y4
0
Y7
Y8
Y9
Y10
0
0
Y7
0
−Y5
−Y7
0
0
Y2
0
Y3
Y1
−Y5
0
−Y6
−Y8
0
0
0
Y2
−Y1
Y3
Y8
Y5
0
−Y9
−Y2
0
0
−Y1
Y4
0
−Y7
Y6
0
−Y10
0
−Y2
Y1
0
0
−Y4
Y10
Y7
Y9
0
−Y3
Y1
−Y4
0
0
0
−Y9
Y8
Y10
0
−Y1
−Y3
0
−Y4
0
0
The Lie algebra g is non-solvable, because
½
¾
1
g(1) = [g, g] = SpanR X1 , X2 , X3 − X4 , X5 , · · · , X11 ,
2
g(2)
= [g(1) , g(1) ] = g(1) .
Similarly, g1 is semi-simple and non-solvable, because
(1)
g1 = [g1 , g1 ] = SpanR {Y1 , · · · , Y10 } = g1 .
The Lie algebra g admits a Levi decomposition as the following semi-direct product
g = r ⋉ s, where r = SpanR {X3 } is the radical of g (the largest solvable ideal contained
in g), and
60
Mehdi Nadjafikhah, Seyed-Reza Hejazi
s=
½
¾
1
X1 , X2 , X3 − X4 , X5 , · · · X11 .
2
r is a one-dimensional subalgebra of g, therefore it is isomorphic to R; Thus the
identity g = r ⋉ s reduces to g = R ⋉ g1 .
✷
2.5
Optimal system of sub-algebras
As is well known, the theoretical Lie group method plays an important role in finding
exact solutions and performing symmetry reductions of differential equations. Since
any linear combination of infinitesimal generators is also an infinitesimal generator,
there are always infinitely many different symmetry subgroups for the differential
equation. So, a mean of determining which subgroups would give essentially different
types of solutions is necessary and significant for a complete understanding of the
invariant solutions. As any transformation in the full symmetry group maps a solution
to another solution, it is sufficient to find invariant solutions which are not related by
transformations in the full symmetry group, this has led to the concept of an optimal
system [6]. The problem of finding an optimal system of subgroups is equivalent to
that of finding an optimal system of subalgebras. For one-dimensional subalgebras,
this classification problem is essentially the same as the problem of classifying the
orbits of the adjoint representation. This problem is attacked by the naive approach
of taking a general element in the Lie algebra and subjecting it to various adjoint
transformations so as to simplify it as much as possible. The idea of using the adjoint
representation for classifying group-invariant solutions is due to [5] and [6].
The adjoint action is given by the Lie series
(2.6)
Ad(exp(tYi )Yj ) = Yj − t[Yi , Yj ] +
t2
[Yi , [Yi , Yj ]] − · · · ,
2
where [Yi , Yj ] is the commutator for the Lie algebra, t is a parameter, and i, j =
1, · · · , 10. We can write the adjoint action for the Lie algebra g1 , and show that
Theorem 2.6. A one dimensional optimal system of the g1 is given by
1)
2)
3)
4)
5)
6)
7)
Y1 ,
Y3 ,
Y4 ,
Y7 ,
Y8 ,
Y9 ,
aY1 + bY3 ,
8)
9)
10)
11)
12)
13)
14)
aY1 + bY4 ,
aY1 + bY7 ,
aY1 + bY8
aY1 + bY9 ,
aY3 + bY7 ,
aY3 + bY8 ,
aY4 + bY9 ,
15)
16)
17)
18)
19)
20)
21)
aY7 + bY8 ,
aY1 + bY4 + cY9 ,
aY1 + bY3 + cY8 ,
aY1 + bY7 + cY8 ,
aY3 + bY7 + cY8 ,
aY1 + bY3 + cY7 ,
aY1 + bY3 + cY7 + dY8 ,
where a, b, c, d ∈ R are arbitrary numbers.
Proof. Fit : g1 → g1 defined by Y →
7
Ad(exp(tYi )Y ) is a linear map, for i =
1, · · · , 10. The matrices Mit of Fit , i = 1, · · · , 10, with respect to basis {Y1 , · · · , Y10 }
are
61
Symmetry analysis of the cylindrical Laplace equation
M1t
M2t
M3t
M4t
M5t
=
=
=
=
=
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
−t
t2 /2
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
e −1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
t
t2 /2
1
0
−t2 /2
0
0
0
0
0
t3 /6
0
0
1
0
0
0
0
0
0
0
t
0
t
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
t
0
0
0
0
0
0
1
−t
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
sin t
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
−t
0
t2 /2
0
0
0
0
0
0
1
0
−t
0
t2 /2
0
0
0
e−t − 1
0
0
0
0
0
0
0
t2 /2
t
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
t
0
0
0
0
0
sin t
cos t − 1
cos t
cos t
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
et − 1
0
0
0
0
0
0
0
0
0
t
0
1
0
0
0
0
0
−t
0
1
0
0
0
−t2 /2
0
0
0
0
0
0
t
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1 − cos t
t − sin t
− sin t
0
0
0
0
0
0
0
0
0
1
0
−t
0
0
0
0
0
0
0
0
0
cos t
sin t
0
0
0
0
0
0
0
1
0
−t
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
et − 1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
t2 /2
0
t
0
1
0
0
0
0
t2 /2
0
t
0
1
0
t
0
0
−t
0
0
1
0
0
t2 /2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
,
0
0
0
0
0
0
0
0
0
1
,
0
0
0
0
0
0
0
,
− sin t
cos t
0
0
0
0
0
0
0
0
e−t − 1
0
,
0
0
0
0
0
0
0
0
0
e−t − 1
,
62
Mehdi Nadjafikhah, Seyed-Reza Hejazi
M6t
M7t
M8t
M9t
t
M10
=
=
=
=
=
1
0
0
0
0
0
0
0
−t
0
cos t
0
0
0
0
0
0
− sin t
0
0
cos t
0
0
0
0
0
sin t
0
0
0
0
cos t
0
0
0
− sin t
0
0
0
0
1
0
0
0
0
t
0
0
0
0
0
0
1
0
−t
0
0
0
0
0
1
0
0
0
−t
0
0
0
0
0
0
1
0
−t2 /2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
t
0
0
0
1
0
0
0
0
0
0
−t
0
0
0
1
0
0
0
t2 /2
0
0
0
−t
0
0
1
0
0
0
t2 /2
0
0
0
0
0
0
1
0
0
0
0
0
0
−t
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
cos t
0
0
− sin t
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
cos t
0
0
0
0
sin t
0
0
sin t
0
0
cos t
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
t
0
0
0
0
0
1
0
0
0
sin t
0
0
0
0
0
0
cos t
0
0
0
0
1
0
0
−t
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
cos t
0
0
0
0
0
sin t
0
−t2 /2
0
1
0
0
−t
0
0
0
0
−t2 /2
0
1
0
0
0
−t
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
sin t
0
0
0
cos t
0
0
0
0
0
t
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
t
0
0
0
0
0
1
0
0
− sin t
0
0
0
0
0
cos t
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
,
0
0
0
0
0
0
0
0
0
1
0
0
0
− sin t
0
0
0
0
0
cos t
0
0
0
− sin t
0
0
0
0
cos t
0
0
0
0
0
0
0
0
0
1
0
0
0
t
0
−t2 /2
0
0
0
1
0
t
0
0
0
0
t2 /2
0
0
0
1
−t
0
0
0
t2 /2
0
0
0
1
0
0
0
t
0
0
−t2 /2
0
0
0
1
,
.
,
,
Symmetry analysis of the cylindrical Laplace equation
Let Y =
10
X
63
ai Yi , then
i=1
t10
F10
◦
F9t9
◦ ··· ◦
³1
F1t1
: Y 7→
Ã
³
cos t7 cos t8 + t5 sin t8 +
´
³1
´ ´
t25 cos t7 + t6 sin t7 t9 sin t8 −
t25 sin t7 − t6 cos t7 t10 a1 + · · · +
2
2
³³ ³ 1
´
´
´ 1³1
−
t25 sin t7 − t6 cos t7 t9 t10 −
t25 cos t7 + t6 sin t7 t210 sin t8 +
2
2 2
!
³
´
´
1 1 2
t cos t7 + t6 sin t7 t29 sin t8 a10 Y1
(cos t7 cos t8 + t5 sin t8 )t9 +
2 2 5
..
.
("
³³³ 1
´
³1
t22 t24 (est − 1) cos t1 − t2 t4 cos t1 t5 −
t2 t2 (et3 − 1) sin t1 −
4
4 2 4
´ ´
t2 t4 sin t1 + (e−t3 − 1) sin t1 + (e−t3 − 1) cos t1 t6 cos t7 −
³1
´
´
t22 t4 (et3 − 1) cos t1 − t2 cos t1 sin t7 cos t8 + · · · −
Ã2
´
1 ³ 1 2 2 t3
t2 t4 (e − 1) cos t1 − t2 t4 cos t1 t35 sin t7 +
6 4
´
³ ³1
1
−
t22 t24 (et3 − 1) cos t1 − t2 t4 cos t1 t5 t6 + t22 (et3 − 1) sin t1 −
4
2
´
1 ³ 1 2 2 t3
t2 t4 (e − 1) sin t1 − t2 t4 sin t1 + (e−t3 − 1) sin t1 + (e−t3 − 1) cos t1 t25 +
2 4
1 ³ 1 2 2 t3
t t (e − 1) sin t1 − t2 t4 sin t1 + (et3 − 1) sin t1 +
2 4 2 4
"Ã
! #
³1
´ ´
2
−t3
t2 t2 (et3 − 1) sin t1 − t2 t4 sin t1 +
(e
− 1) cos t1 t6 cos t7 t10 a1 + · · · +
4 2 4
!
´
(e−t3 − 1) sin t1 + (e−t3 − 1) cos t1 t5 + · · · + (e−t3 − 1) sin t1 + · · · t10 +
+
··· +
(e−t3
³1
t22 t24 (et3 − 1) sin t1 − t2 t4 sin t1 + (e−t3 − 1) sin t1 +
# )
´
1 2 2 t3
− 1) cos t1 + t2 t4 (e − 1) cos t1 − t2 t4 cos t1 cos t8 a10 Y10 .
4
4
Then, if ai = 0, for i = 2, 3, 4, 5, 6, 9, 10, and a1 , a7 , a8 , are non-zero then we can
make the coefficients of Y8 , vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can
assume that a1 = a7 = a8 = 1. And Y is reduced to Case 1, 4, 5 and 18.
If ai = 0, for i = 2, 4, 5, 6, 9, 10, and a1 , a3 , a7 , a8 , are non-zero then we can make
the coefficients of Y8 , vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can assume
that a3 = a9 = 1. And Y is reduced to Case 2, 7, 9, 12, 13 and 15.
64
Mehdi Nadjafikhah, Seyed-Reza Hejazi
If ai = 0, for i = 2, 3, 5, 6, 7, 10, and a1 , a4 , a8 , a9 , are non-zero then we can make
the coefficients of Y4 and Y8 vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can
assume that a4 = 1. And Y is reduced to Case 3, 6, 8, 11, 14 and 16.
If ai = 0, for i = 2, 4, 5, 6, 9, 10, and a1 , a3 , a7 , a8 , are non-zero then we can make
the coefficients of Y8 , vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can assume
that a1 = 1. And Y is reduced to Case 10, 17, 19, 20, and 21.
✷
According to our optimal system of one-dimensional subalgebras of the full symmetry algebra g1 , we need only find group-invariant solutions for 21 one-parameter
subgroups generated by Y as Theorem (2.6).
3
Conclusions
In this paper by using the criterion of invariance of the equation under the infinitesimal
prolonged infinitesimal generators, we find the most Lie point symmetry group of
the cylindrical Laplace equation. Looking the adjoint representation of the obtained
symmetry group on its Lie algebra, we have found the preliminary classification of
group-invariants solutions.
References
[1] N. Bila, Symmetry Groups and Conservation Laws of Certain Partial Differential
Equations, Balkan Journal of Geometry and its Applications 5, 1(2000), 39-56.
[2] M. Nadjafikhah, Lie Symmetries of Burger’s Equation, Adv. Appl. Clifford Alg.,
DOI 10.1007/s00006-008-0127-2.
[3] M. Nadjafikhah and S.R. Hejazi, Differential Invariants of SL(2) and SL(3)Action on R2 , Mathematical Science, Vol. 1, No. 3 (2007) 75-84.
[4] P.J. Olver, Equivalence, Invariant and Symmetry, Cambridge University Press,
Cambridge, 1995.
[5] P.J. Olver, Applications of Lie Groups to Differential equations, Second Edition,
GTM, Vol. 107, Springer Verlage, New York, 1993.
[6] L.V. Ovsiannikov, Group Analysis of Differential Equations, Academic Press,
New York, 1982.
[7] A.D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers
and scientists, Chapman and Hall/CRC, 2002.
[8] C. Udriste and N. Bila, Symmetry Lie Group of Monge-Ampere Equation, Balkan
Journal of Geometry and its Applications 3, 2(1998), 121-133.
[9] http://www.wikipedia.com/
Authors’ addresses:
Mehdi Nadjafikhah and Seyed-Reza Hejazi
School of Mathematics,
Iran University of Science and Technology,
Narmak-16, Tehran, I.R.IRAN.
URL: webpages.iust.ac.ir\m nadjafikhah,
E-mail: m nadjafikhah@iust.ac.ir, reza hejazi@iust.ac.ir
Laplace equation
Mehdi Nadjafikhah, Seyed-Reza Hejazi
Abstract. The symmetry analysis for Laplace equation on cylinder is
considered. Symmetry algebra, the structure of the Lie algebra of the symmetries and some related topics such as invariant solutions, one-parameter
subgroups, one dimensional optimal system and differential invariants are
given.
M.S.C. 2000: 49J30, 49K30, 55S45.
Key words: Laplace equation, invariant solution, optimal system.
1
Introduction
Pierre-Simon, marquis de Laplace (March 23, 1749 - March 5, 1827) was a French
mathematician and astronomer whose work was pivotal to the development of mathematical astronomy. He summarized and extended the work of his predecessors in his
five volume Mcanique Cleste (Celestial Mechanics) (1799-1825). This seminal work
translated the geometric study of classical mechanics, used by Isaac Newton, to one
based on calculus, opening up a broader range of problems. He formulated Laplace’s
equation, and invented the Laplace transform which appears in many branches of
mathematical physics, a field that he took a leading role in forming. The Laplacian
differential operator, widely used in applied mathematics, is also named after him.
In mathematics and physics, the Laplace operator or Laplacian, denoted by ∆ or
∇2 and named after Pierre-Simon de Laplace, is a differential operator, specifically an
important case of an elliptic operator, with many applications. In physics, it is used in
modeling of wave propagation (the wave equation is an important second-order linear
partial differential equation that describes the propagation of a variety of waves, such
as sound waves, light waves and water waves. It arises in fields such as acoustics, electromagnetics, and fluid dynamics. Historically, the problem of a vibrating string such
as that of a musical instrument was studied by Jean le Rond d’Alembert, Leonhard
Euler, Daniel Bernoulli, and Joseph-Louis Lagrange.), an important partial differential equation which describes the distribution of heat (or variation in temperature) in
a given region over time. For a function of three spatial variables (x, y, z) and one time
∗
Balkan
Journal of Geometry and Its Applications, Vol.14, No.2, 2009, pp. 53-64.
c Balkan Society of Geometers, Geometry Balkan Press 2009.
°
54
Mehdi Nadjafikhah, Seyed-Reza Hejazi
variable t; heat flow, forming the Helmholtz equation. It is central in electrostatics and
fluid mechanics, anchoring in Laplace’s equation and Poisson’s equation. In quantum
mechanics, it represents the kinetic energy term of the Schrödinger equation. The kinetic energy of an object is the extra energy which it possesses due to its motion. It is
defined as the work needed to accelerate a body of a given mass from rest to its current
velocity. Having gained this energy during its acceleration, the body maintains this
kinetic energy unless its speed changes. Negative work of the same magnitude would
be required to return the body to a state of rest from that velocity. In mathematics,
functions with vanishing Laplacian are called harmonic functions; the Laplacian is at
the core of Hodge theory and the results of de Rham cohomology [9].
The aim is to analysis the point symmetry structure of the Laplace equation on
cylinder, i.e., cylindrical Laplace equation, which is
∇2 u =
(1.1)
1 ∂ ³ ∂u ´
1 ∂2u ∂2u
r
+ 2 2 + 2 = 0,
r ∂r ∂r
r ∂θ
∂z
where u is a smooth function of (r, θ, z).
The symmetry algebra of (1.1) is determined in the next section and some results
yield from the structure of the Lie algebra of symmetries are given.
2
Lie symmetries of the equation
The method of determining the classical symmetries of a partial differential equation
is standard which is described in [1, 4, 5] and [8]. To obtain the symmetry algebra of
the (1.1), we take an infinitesimal generator of symmetry algebra of the form
X = ξ1 (r, θ, z, u)
∂
∂
∂
∂
+ ξ2 (r, θ, z, u)
+ ξ3 (r, θ, z, u)
+ η(r, θ, z, u) .
∂r
∂θ
∂z
∂u
Using the invariance condition, i.e., applying the second prolongation X (2) to (1.1),
the following system of 18 determining equations yields:
ξ2 u = 0,
ξ3 u = 0,
ηuu = 0,
ξ3 rzz = 0,
ξ3 θzz = 0,
ξ3 zzz = 0,
ξ3 rr + ξ3 zz = 0,
ξ3 θ + r2 ξ2 z = 0,
ξ3 θ − rξ3 rθ = 0,
ξ3 rz + 2ηru = 0,
ξ3 θz + 2ηθu = 0,
ξ3 zz + 2ηzu = 0,
ξ1 + r(ξ2 θ − ξ3 z ) = 0,
ξ2 θ + r(ξ2 rθ − ξ3 rz ) = 0,
ξ2 θθ − rξ2 r − ξ3 θz = 0,
ξ3 θθ + r(ξ3 r + rξ3 zz ) = 0,
ξ3 θz + r(2ξ2 r − rξ2 rr ) = 0,
ηθθ + r(ηr + rηrr + rηzz ) = 0.
The solution of the above system gives the following infinitesimals,
∂
∂
∂
,
X2 =
,
X3 = u ,
∂θ
∂z
∂u
∂
∂
∂
1
∂
1
∂
= r
+z ,
X5 = rz
+ (z 2 − r2 )
− zu ,
∂r
∂z
∂r 2
∂z
2 ∂u
∂
1
∂
∂
1
∂
= sin θ
+ cos θ ,
X7 = cos θ
− sin θ ,
∂r
r
∂θ
∂r
r
∂θ
X1 =
X4
X6
55
Symmetry analysis of the cylindrical Laplace equation
∂
1
∂
∂
+ z cos θ
− r sin θ ,
∂r
r
∂θ
∂z
1
∂
∂
∂
− z sin θ
− r cos θ ,
= z cos θ
∂r
r
∂θ
∂z
1
∂
1
∂
∂
1
∂
= (z 2 − r2 ) sin θ
+ (z 2 + r2 ) cos θ
− rz sin θ
+ ru sin θ ,
2
∂r 2r
∂θ
∂z
2
∂u
1
∂
1
∂
∂
1
∂
= (z 2 − r2 ) cos θ
− (z 2 + r2 ) sin θ
− rz cos θ
+ ru cos θ .
2
∂r 2r
∂θ
∂z
2
∂u
X8 = z sin θ
X9
X10
X11
The commutation relations of the 11−dimensional Lie algebra g spanned by the vector
fields X1 , · · · , X11 , are shown in following table.
X1
X1
X2
X3
X4
X5
X6
X7
X8
X9
X10
X11
2.1
X2
0
0
0
0
0
0
0
−X2
0 12 X3 − X4
−X7
0
X6
0
−X9
−X6
X8
−X7
−X11 −X8
X10
−X9
X3
0
0
0
0
0
0
0
0
0
0
0
X4
X5
X6
X7
X8
X9
X10
X11
0
0
X7
−X6
X9
−X8
X11
−X10
X2 − 12 X3 + X4
0
0
X6
X7
X8
X9
0
0
0
0
0
0
0
0
0
X5
−X7
−X7
0
0
X10
X11
−X5
0
−X8
−X9
−X10 −X11
0
0
X7
X8
0
0
−X2
0 12 X3 − X4 −X1
X7
X9
0
0
0
−X2
X1 21 X3 − X4
0
X10
X2
0
0
X1
−X5
0
0
X11
0
X2
−X1
0
0
−X5
−X10
0
− 12 X3 + X4 −X1
X5
0
0
0
−X11
0
X1 − 21 X3 + X4 0
X5
0
0
Differential invariants
Suppose that G is a transformation group. It is well known that a smooth real differential function I : J n → R, where J n is the corresponding n−th jet space, is a
differential invariant for G if and only if for all X ∈ g, its n−th prolongation annihilates I, i.e., X (n) (I) = 0. Here we are going to find the differential invariants of order
0, 1 and 2 and some of their results. For more details and some practical examples
see [3] and [8].
The symmetry group of the cylindrical Laplace equation has not any nonconstant
ordinary differential invariants, i.e., it does not have any differential invariants of order
zero, thus the orbit of generic points (the points which lies in the orbit of maximal
dimension) has dimension four and its isotropy subgroup is seven dimensional.
The first prolonged action of the symmetry group has one differential invariant,
which is a function of (u, ur , uθ , uz ), so the action has a generic orbit of dimension
six, and its isotropy subgroup is five dimensional.
Finally the second prolonged action has another differential invariant which is a
function of (u, ur , uθ , uz , urr , urθ , urz , uθθ , uθz , uzz ), consequently the dimension of the
generic orbit is equal to the dimension of the group; thus, the second prolonged action
is transitive and has a discrete isotropy subgroup.
According to the relations between above subjects and [4], we can obtain the
number of functionally independent differential invariants up to order n.
56
Mehdi Nadjafikhah, Seyed-Reza Hejazi
Theorem 2.1. The number of functionally independent differential invariants
¢ up
¡
−8
to order n ≥ 3 for the symmetry group of cylindrical Laplace equation is 3+n
n
and
the
number
of
strictly
differential
invariants
which
occur
in
each
prolongation
is
¡3+n¢ ¡2+n¢
−
.
n
n−1
Proof. According to the above discussion for n ≥ 3 the action is transitive, so the
number of independent differential invariants in , is
¶
¶
µ
µ
3+n
3+n
− 8.
− 11 =
in = 3 +
n
n
If jn is the number of strictly differential invariants then,
¶
¶ µ
¶ µ
¶ µ
µ
2+n
3+n
3+n−1
3+n
.
−
=
−
jn =
n−1
n
n−1
n
2.2
✷
Reduction of the equation
The Eq. (1.1) maybe regarded as a submanifold M of the jet space J 2 (R3 , R). Doing
as section 3.1 of [5] and find (in a sense) the most general group-invariant solutions
to the Eq. (1.1).
Theorem 2.2. The one-parameter groups gi (t) : M → M generated by the Xi , i =
1, · · · , 11, are given in the following table:
g1 (t) : (r, θ, z, u) 7−→ (r, θ + t, z, u),
g2 (t) : (r, θ, z, u) 7−→ (r, θ, z + t, u),
g3 (t) : (r, θ, z, u) 7−→ (r, θ, z, uet ),
g4 (t) :
g5 (t) :
g6 (t) :
g7 (t) :
g8 (t) :
g9 (t) :
g10 (t) :
g11 (t) :
(r, θ, z, u) 7−→ (ret , θ, zet , u),
³
´
t
1
(r, θ, z, u) 7−→ rzt + r, θ, (z 2 − r2 ), − zut + u ,
2
2
´
³
t
(r, θ, z, u) 7−→ t sin θ + r, cos θ + θ, z, u ,
r
³
´
t
(r, θ, z, u) 7−→ t cos θ + r, sin θ + θ, z, u ,
r
´
³
t
(r, θ, z, u) 7−→ tz sin θ + r, z cos θ + θ, −rt sin θ + z, u ,
r
³
´
t
(r, θ, z, u) 7−→ tz cos θ + r, − z sin θ + θ, −rt cos θ + z, u ,
r
Ã
´
t 2
t ³ z2
(r, θ, z, u) 7−→
(z − r2 ) sin θ,
+ r cos θ + θ,
2
2 r
!
1
−rzt sin θ + z, rut sin θ + u ,
2
Ã
´
t 2
t ³ z2
(r, θ, z, u) 7−→
(z − r2 ) cos θ, −
+ r sin θ + θ,
2
2 r
!
1
−rzt cos θ + z, rut cos θ + u ,
2
Symmetry analysis of the cylindrical Laplace equation
57
where entries give the transformed point exp(tXi )(r, θ, z, u) = (r̃, θ̃, z̃, ũ).
The transformations g1 and g2 demonstrate the radius- and space-invariance and
radius- and angle-invariance solutions of the equation. The one-parameter g3 reflects
the linearity of the cylindrical laplace equation, and the well-known scaling symmetry
turns up in g4 . The most general one-parameter group of symmetries is obtained
by considering a general linear combination of the given vector fields; the explicit
formula for the group transformations is very complicated, but alternatively we can
use theorem 2.52 in [4], and represent an arbitrary group transformation g as the
composition of transformations in theorem (2.2).
Theorem 2.3. Let i = 1, · · · , 11 and t ∈ R. If u = U (r, θ, z) is a solution of Eq.
(1.1) so are the functions ui (r, θ, z) = U (r, θ, z), where
u1 = U (r, θ + t, z),
u2 = U (r, θ, z + t),
u3 = e−t U (r, θ, z),
u4 = U (et r, θ, et z),
³
´
t
1
u5 = 1 + zt U (rzt + r, θ, (z 2 − r2 )),
2
2
³
´
t
t
u6 = U (t sin θ + r, cos θ + θ, z),
u7 = U t cos θ + r, cos θ + θ, z ,
r
r
³
´
t
8
u = U tz sin θ + r, z cos θ + θ, −rt sin θ + z ,
r
´
³
t
9
u = U tz cos θ + r, − z sin θ + θ, −rt cos θ + z ,
Ãr
!
³
´
rt
t 2
t 2
10
2
2
u = 1 − sin θ U
(z − r ) sin θ, (z + r ) cos θ + θ, z − rzt sin θ ,
2
2
2r
Ã
!
³
´
rt
t 2
t 2
11
2
2
u = 1 − cos θ U
(z − r ) cos θ, (z + r ) sin θ + θ, z − rzt cos θ .
2
2
2r
For example, if u(r, θ, z) = 1 be a constant solution of Eq. (1.1), we conclude
−t
that the trivial functions gi (t) · 1 =³1 for
´ i = 1, 2, 4, 6, 7, 8, 9, g3 (t) · 1 = e and
g5 (t) · 1 = 2 + zt, and by applying g5 2a
· g3 (ln b) on u = 1, we conclude the linear
b
solution u = az + b. The two nontrivial solutions for Eq. (1.1) are:
g10 (t) · 1 = 2 − rt sin θ, g11 (t) · 1 = 2 − rt cos θ.
´
³
· g3 ln d · 1, g10 (−a) · 1 and g11 (−b) · 1, we conclude
Now by summing g1 , g5 −8+2c
d
the following result:
Corollary 2.4. The function u(r, θ, z) = ar sin θ + br cos θ + dz + c, is a solution of
Eq. (1.1), where a, b, c and d are arbitrary constants.
2.3
Some invariant solutions
The first advantage of symmetry group method is to construct new solutions from
known solutions. Neither the first advantage nor the second will be investigated here,
58
Mehdi Nadjafikhah, Seyed-Reza Hejazi
but symmetry group method will be applied to the Eq. (1.1) to be connected directly
to some order differential equations. To do this, a particular linear combinations of
infinitesimals are considered and their corresponding invariants are determined.
The cylindrical Laplace equation expressed in the coordinates (r, θ, z), so to reduce
this equation is to search for its form in specific coordinates. Those coordinates will
be constructed by searching for independent invariants (y, v) corresponding to an
infinitesimal generator. so using the chain rule, the expression of the equation in the
new coordinate allows us to the reduced equation. Here we will obtain some invariant
solutions with respect to symmetries.
1. First consider X6 . To determine the independent invariants I, we need to solve
the first order partial differential equation X6 {I(r, θ, z, u)} = 0, that is
sin θ
∂I
cos θ ∂I
∂I
∂I
+
+0
+0
= 0,
∂r
r ∂θ
∂z
∂u
which is a homogeneous first order PDE. Thus we solve the associated characteristic
ordinary differential equations
dr
cos θdθ
dz
du
=
=
=
.
sin θ
r
0
0
z
and v = u.
Hence we obtain two functionally independent invariants y =
r sin θ
If we treat v as a function of y, we can compute formulae for the derivatives
of u with respect to x and t in terms of y, v and the derivatives of v with respect
to y, along with a single parametric variable, which we designate to be t, so that
x will be the corresponding principal variable. Using the chain rule we find that if
u = v = v(y) = v(z/r sin θ), then
(2.1)
(2.2)
(2.3)
(2.4)
ur
urr
uθθ
uzz
=
=
=
=
vy yr ,
yrr vy + vyy yr ,
yθθ vy + vyy yθ ,
yzz vy + vyy yz .
Substituting
³ ur ,´urr , uθθ and uzz in (1.1) we obtain an ODE with solution u =
z
C1 arctan r sin
θ + C2 , which is called an invariant solution for Eq. (1.1).
z
and v = u, and the solution of
r cos θ
reduced equation
solution) by substituting (2.1), (2.2), (2.3) and (2.4) is
³ (invariant
´
z
u = C1 arctan r cos θ + C2 .
2.
Similarly, the invariants of X7 are y =
r2 + z 2
and v = u. The invariant solution constructed
r sin θ
³√
´
2
2
with (2.1), (2.2), (2.3) and (2.4) is u = C1 arctanh rrsin+zθ + C2 obtained from
ordinary differential equations
3.
The invariants of X8 are
59
Symmetry analysis of the cylindrical Laplace equation
dr
rdθ
dz
du
=
=
=
.
z sin θ
a cos θ
−r sin θ
0
The above mentioned method in sections 1 and 2 can be followed for X9 with invari³√
´
r2 + z 2
2 +z 2
and v = u, and invariant solution u = C1 arctanh rrcos
+ C2 . We
ants
θ
r sin θ
can proceed for other Xi s as above.
2.4
The Lie algebra of symmetries
In this section, we determine the structure of full symmetry Lie algebra g, of Eq.
(1.1).
Theorem 2.5. The full symmetry Lie algebra g, of Eq. (1.1) has the following semidirect decomposition:
(2.5)
g = R ⋉ g1 ,
where g1 is a simple Lie algebra.
Proof: The center z of g is SpanR {X3 }. Therefore the quotient algebra g1 = g/z is
SpanR {Y1 , · · · , Y10 }, where Yi := Xi + z for i = 1, · · · , 10. The commutator table of
this quotient algebra is given in following table:
Y1
Y2
Y3
Y4
Y5
Y6
Y7
Y8
Y9
Y10
Y1
Y2
Y3
Y4
Y5
Y6
Y7
Y8
Y9
Y10
0
0
0
0
−Y7
Y5
−Y8
Y7
−Y10
Y9
0
0
−Y2
−Y3
0
0
−Y5
−Y6
−Y7
−Y8
0
Y2
0
−Y4
Y5
Y6
0
0
−Y9
−Y10
0
Y3
Y4
0
Y7
Y8
Y9
Y10
0
0
Y7
0
−Y5
−Y7
0
0
Y2
0
Y3
Y1
−Y5
0
−Y6
−Y8
0
0
0
Y2
−Y1
Y3
Y8
Y5
0
−Y9
−Y2
0
0
−Y1
Y4
0
−Y7
Y6
0
−Y10
0
−Y2
Y1
0
0
−Y4
Y10
Y7
Y9
0
−Y3
Y1
−Y4
0
0
0
−Y9
Y8
Y10
0
−Y1
−Y3
0
−Y4
0
0
The Lie algebra g is non-solvable, because
½
¾
1
g(1) = [g, g] = SpanR X1 , X2 , X3 − X4 , X5 , · · · , X11 ,
2
g(2)
= [g(1) , g(1) ] = g(1) .
Similarly, g1 is semi-simple and non-solvable, because
(1)
g1 = [g1 , g1 ] = SpanR {Y1 , · · · , Y10 } = g1 .
The Lie algebra g admits a Levi decomposition as the following semi-direct product
g = r ⋉ s, where r = SpanR {X3 } is the radical of g (the largest solvable ideal contained
in g), and
60
Mehdi Nadjafikhah, Seyed-Reza Hejazi
s=
½
¾
1
X1 , X2 , X3 − X4 , X5 , · · · X11 .
2
r is a one-dimensional subalgebra of g, therefore it is isomorphic to R; Thus the
identity g = r ⋉ s reduces to g = R ⋉ g1 .
✷
2.5
Optimal system of sub-algebras
As is well known, the theoretical Lie group method plays an important role in finding
exact solutions and performing symmetry reductions of differential equations. Since
any linear combination of infinitesimal generators is also an infinitesimal generator,
there are always infinitely many different symmetry subgroups for the differential
equation. So, a mean of determining which subgroups would give essentially different
types of solutions is necessary and significant for a complete understanding of the
invariant solutions. As any transformation in the full symmetry group maps a solution
to another solution, it is sufficient to find invariant solutions which are not related by
transformations in the full symmetry group, this has led to the concept of an optimal
system [6]. The problem of finding an optimal system of subgroups is equivalent to
that of finding an optimal system of subalgebras. For one-dimensional subalgebras,
this classification problem is essentially the same as the problem of classifying the
orbits of the adjoint representation. This problem is attacked by the naive approach
of taking a general element in the Lie algebra and subjecting it to various adjoint
transformations so as to simplify it as much as possible. The idea of using the adjoint
representation for classifying group-invariant solutions is due to [5] and [6].
The adjoint action is given by the Lie series
(2.6)
Ad(exp(tYi )Yj ) = Yj − t[Yi , Yj ] +
t2
[Yi , [Yi , Yj ]] − · · · ,
2
where [Yi , Yj ] is the commutator for the Lie algebra, t is a parameter, and i, j =
1, · · · , 10. We can write the adjoint action for the Lie algebra g1 , and show that
Theorem 2.6. A one dimensional optimal system of the g1 is given by
1)
2)
3)
4)
5)
6)
7)
Y1 ,
Y3 ,
Y4 ,
Y7 ,
Y8 ,
Y9 ,
aY1 + bY3 ,
8)
9)
10)
11)
12)
13)
14)
aY1 + bY4 ,
aY1 + bY7 ,
aY1 + bY8
aY1 + bY9 ,
aY3 + bY7 ,
aY3 + bY8 ,
aY4 + bY9 ,
15)
16)
17)
18)
19)
20)
21)
aY7 + bY8 ,
aY1 + bY4 + cY9 ,
aY1 + bY3 + cY8 ,
aY1 + bY7 + cY8 ,
aY3 + bY7 + cY8 ,
aY1 + bY3 + cY7 ,
aY1 + bY3 + cY7 + dY8 ,
where a, b, c, d ∈ R are arbitrary numbers.
Proof. Fit : g1 → g1 defined by Y →
7
Ad(exp(tYi )Y ) is a linear map, for i =
1, · · · , 10. The matrices Mit of Fit , i = 1, · · · , 10, with respect to basis {Y1 , · · · , Y10 }
are
61
Symmetry analysis of the cylindrical Laplace equation
M1t
M2t
M3t
M4t
M5t
=
=
=
=
=
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
−t
t2 /2
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
e −1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
t
t2 /2
1
0
−t2 /2
0
0
0
0
0
t3 /6
0
0
1
0
0
0
0
0
0
0
t
0
t
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
t
0
0
0
0
0
0
1
−t
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
sin t
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
−t
0
t2 /2
0
0
0
0
0
0
1
0
−t
0
t2 /2
0
0
0
e−t − 1
0
0
0
0
0
0
0
t2 /2
t
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
t
0
0
0
0
0
sin t
cos t − 1
cos t
cos t
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
et − 1
0
0
0
0
0
0
0
0
0
t
0
1
0
0
0
0
0
−t
0
1
0
0
0
−t2 /2
0
0
0
0
0
0
t
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1 − cos t
t − sin t
− sin t
0
0
0
0
0
0
0
0
0
1
0
−t
0
0
0
0
0
0
0
0
0
cos t
sin t
0
0
0
0
0
0
0
1
0
−t
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
et − 1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
t2 /2
0
t
0
1
0
0
0
0
t2 /2
0
t
0
1
0
t
0
0
−t
0
0
1
0
0
t2 /2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
,
0
0
0
0
0
0
0
0
0
1
,
0
0
0
0
0
0
0
,
− sin t
cos t
0
0
0
0
0
0
0
0
e−t − 1
0
,
0
0
0
0
0
0
0
0
0
e−t − 1
,
62
Mehdi Nadjafikhah, Seyed-Reza Hejazi
M6t
M7t
M8t
M9t
t
M10
=
=
=
=
=
1
0
0
0
0
0
0
0
−t
0
cos t
0
0
0
0
0
0
− sin t
0
0
cos t
0
0
0
0
0
sin t
0
0
0
0
cos t
0
0
0
− sin t
0
0
0
0
1
0
0
0
0
t
0
0
0
0
0
0
1
0
−t
0
0
0
0
0
1
0
0
0
−t
0
0
0
0
0
0
1
0
−t2 /2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
t
0
0
0
1
0
0
0
0
0
0
−t
0
0
0
1
0
0
0
t2 /2
0
0
0
−t
0
0
1
0
0
0
t2 /2
0
0
0
0
0
0
1
0
0
0
0
0
0
−t
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
cos t
0
0
− sin t
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
cos t
0
0
0
0
sin t
0
0
sin t
0
0
cos t
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
t
0
0
0
0
0
1
0
0
0
sin t
0
0
0
0
0
0
cos t
0
0
0
0
1
0
0
−t
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
cos t
0
0
0
0
0
sin t
0
−t2 /2
0
1
0
0
−t
0
0
0
0
−t2 /2
0
1
0
0
0
−t
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
sin t
0
0
0
cos t
0
0
0
0
0
t
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
t
0
0
0
0
0
1
0
0
− sin t
0
0
0
0
0
cos t
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
,
0
0
0
0
0
0
0
0
0
1
0
0
0
− sin t
0
0
0
0
0
cos t
0
0
0
− sin t
0
0
0
0
cos t
0
0
0
0
0
0
0
0
0
1
0
0
0
t
0
−t2 /2
0
0
0
1
0
t
0
0
0
0
t2 /2
0
0
0
1
−t
0
0
0
t2 /2
0
0
0
1
0
0
0
t
0
0
−t2 /2
0
0
0
1
,
.
,
,
Symmetry analysis of the cylindrical Laplace equation
Let Y =
10
X
63
ai Yi , then
i=1
t10
F10
◦
F9t9
◦ ··· ◦
³1
F1t1
: Y 7→
Ã
³
cos t7 cos t8 + t5 sin t8 +
´
³1
´ ´
t25 cos t7 + t6 sin t7 t9 sin t8 −
t25 sin t7 − t6 cos t7 t10 a1 + · · · +
2
2
³³ ³ 1
´
´
´ 1³1
−
t25 sin t7 − t6 cos t7 t9 t10 −
t25 cos t7 + t6 sin t7 t210 sin t8 +
2
2 2
!
³
´
´
1 1 2
t cos t7 + t6 sin t7 t29 sin t8 a10 Y1
(cos t7 cos t8 + t5 sin t8 )t9 +
2 2 5
..
.
("
³³³ 1
´
³1
t22 t24 (est − 1) cos t1 − t2 t4 cos t1 t5 −
t2 t2 (et3 − 1) sin t1 −
4
4 2 4
´ ´
t2 t4 sin t1 + (e−t3 − 1) sin t1 + (e−t3 − 1) cos t1 t6 cos t7 −
³1
´
´
t22 t4 (et3 − 1) cos t1 − t2 cos t1 sin t7 cos t8 + · · · −
Ã2
´
1 ³ 1 2 2 t3
t2 t4 (e − 1) cos t1 − t2 t4 cos t1 t35 sin t7 +
6 4
´
³ ³1
1
−
t22 t24 (et3 − 1) cos t1 − t2 t4 cos t1 t5 t6 + t22 (et3 − 1) sin t1 −
4
2
´
1 ³ 1 2 2 t3
t2 t4 (e − 1) sin t1 − t2 t4 sin t1 + (e−t3 − 1) sin t1 + (e−t3 − 1) cos t1 t25 +
2 4
1 ³ 1 2 2 t3
t t (e − 1) sin t1 − t2 t4 sin t1 + (et3 − 1) sin t1 +
2 4 2 4
"Ã
! #
³1
´ ´
2
−t3
t2 t2 (et3 − 1) sin t1 − t2 t4 sin t1 +
(e
− 1) cos t1 t6 cos t7 t10 a1 + · · · +
4 2 4
!
´
(e−t3 − 1) sin t1 + (e−t3 − 1) cos t1 t5 + · · · + (e−t3 − 1) sin t1 + · · · t10 +
+
··· +
(e−t3
³1
t22 t24 (et3 − 1) sin t1 − t2 t4 sin t1 + (e−t3 − 1) sin t1 +
# )
´
1 2 2 t3
− 1) cos t1 + t2 t4 (e − 1) cos t1 − t2 t4 cos t1 cos t8 a10 Y10 .
4
4
Then, if ai = 0, for i = 2, 3, 4, 5, 6, 9, 10, and a1 , a7 , a8 , are non-zero then we can
make the coefficients of Y8 , vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can
assume that a1 = a7 = a8 = 1. And Y is reduced to Case 1, 4, 5 and 18.
If ai = 0, for i = 2, 4, 5, 6, 9, 10, and a1 , a3 , a7 , a8 , are non-zero then we can make
the coefficients of Y8 , vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can assume
that a3 = a9 = 1. And Y is reduced to Case 2, 7, 9, 12, 13 and 15.
64
Mehdi Nadjafikhah, Seyed-Reza Hejazi
If ai = 0, for i = 2, 3, 5, 6, 7, 10, and a1 , a4 , a8 , a9 , are non-zero then we can make
the coefficients of Y4 and Y8 vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can
assume that a4 = 1. And Y is reduced to Case 3, 6, 8, 11, 14 and 16.
If ai = 0, for i = 2, 4, 5, 6, 9, 10, and a1 , a3 , a7 , a8 , are non-zero then we can make
the coefficients of Y8 , vanish, by F2t1 and F3t2 . Scaling Y if necessary, we can assume
that a1 = 1. And Y is reduced to Case 10, 17, 19, 20, and 21.
✷
According to our optimal system of one-dimensional subalgebras of the full symmetry algebra g1 , we need only find group-invariant solutions for 21 one-parameter
subgroups generated by Y as Theorem (2.6).
3
Conclusions
In this paper by using the criterion of invariance of the equation under the infinitesimal
prolonged infinitesimal generators, we find the most Lie point symmetry group of
the cylindrical Laplace equation. Looking the adjoint representation of the obtained
symmetry group on its Lie algebra, we have found the preliminary classification of
group-invariants solutions.
References
[1] N. Bila, Symmetry Groups and Conservation Laws of Certain Partial Differential
Equations, Balkan Journal of Geometry and its Applications 5, 1(2000), 39-56.
[2] M. Nadjafikhah, Lie Symmetries of Burger’s Equation, Adv. Appl. Clifford Alg.,
DOI 10.1007/s00006-008-0127-2.
[3] M. Nadjafikhah and S.R. Hejazi, Differential Invariants of SL(2) and SL(3)Action on R2 , Mathematical Science, Vol. 1, No. 3 (2007) 75-84.
[4] P.J. Olver, Equivalence, Invariant and Symmetry, Cambridge University Press,
Cambridge, 1995.
[5] P.J. Olver, Applications of Lie Groups to Differential equations, Second Edition,
GTM, Vol. 107, Springer Verlage, New York, 1993.
[6] L.V. Ovsiannikov, Group Analysis of Differential Equations, Academic Press,
New York, 1982.
[7] A.D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers
and scientists, Chapman and Hall/CRC, 2002.
[8] C. Udriste and N. Bila, Symmetry Lie Group of Monge-Ampere Equation, Balkan
Journal of Geometry and its Applications 3, 2(1998), 121-133.
[9] http://www.wikipedia.com/
Authors’ addresses:
Mehdi Nadjafikhah and Seyed-Reza Hejazi
School of Mathematics,
Iran University of Science and Technology,
Narmak-16, Tehran, I.R.IRAN.
URL: webpages.iust.ac.ir\m nadjafikhah,
E-mail: m nadjafikhah@iust.ac.ir, reza hejazi@iust.ac.ir