Symmetry Analysis and Conservation Laws of some third-order Difference Equations

0.11 MB·

by  S. Mamba

Symmetry Analysis and Conservation Laws of some third-order
Difference Equations
∗
S Mamba, MK Folly-Gbetoula and AH Kara
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School of Mathematics, University of the Witwatersrand, Johannesburg,

n South Africa.
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J
Abstract
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We derive a method for finding Lie Symmetries for third-order differ-

] ence equations. We use these symmetries to reduce the order of the
I
S differenceequationsandhenceobtainthesolutionsofsomethird-order
.
n differenceequations. Wealsointroduceatechniqueforobtainingtheir
li first integrals.
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[
Key words: Difference equation; symmetry; reduction; group invariant so-

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lutions
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0
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5 1 Introduction
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0
. The concept of Lie Symmetry analysis applied onto difference equations
1
0 (∆E’s) was studied by P. Hydon [1] and others see [2–5]. In [1], the au-
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thor introduced a method for obtaining symmetries and first integrals of
1
: second-order difference equations. In this paper, we extend these findings
v
i and introduce a technique for obtaining symmetries and first integrals for
X
third-order ∆E’s.
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a
1.1 Mathematical tools to be used
The definitions andnotationsused inthis paper followfromthe onesadopted
by Hybon in [1]. Assume we have an Nth order difference equation of the
−
form
un+N = ω(n,un,un+1,...,un+N−1), (1)
∗[email protected]
1
for some function ω. We will assume that ∂ω = 0 and that the point
∂un 6
transformations are given by
Γ : x x˜(x;ǫ), (2)
ǫ
7→
where x = x , i = 1,...,p are continuous variables. We then refer to Γ
i
as one-parameter Lie group of transformations if it satisfies the following
transformations (see [6])
Γ is the identity map if x˜ = x when ǫ = 0
0
•
Γ Γ = Γ for every a and b sufficiently close to 0
a b a+b
•
Each x˜ can be represented as a Taylor series (in a neighbourhood of
i
•
ǫ = 0 that is determined by x), and therefore
x˜(x : ǫ) = x +ǫξ (x)+O(ǫ2),i = 1,...,p. (3)
i i i
We shall assume that the Lie point symmetries in this article take the fol-
lowing form
n˜ = n; u˜ u +ǫQ(n,u ) (4)
n n n
≃
and that the analogous infinitesimal symmetry generator takes the form
X = Q(n,un)∂un+SQ∂un+1+S2Q(n,un)∂un+2+...+SN−1Q(n,un)∂un+N−1,
(5)
where S is defined as the shift operator, i.e.
S : n n+1 (6)
7→
andQ(n,u )isreferredtoasthecharacteristic. Weshalldefinethesymmetry
n
condition as
u˜n+N = ω(n,u˜n,u˜n+1,...,u˜n+N−1) (7)
each time (1) is true. Substituting the Lie point symmetries (4) into the
symmetry condition (7) yields linearized symmetry condition
S(N)Q Xω = 0, (8)
−
whenever (1) is true. We then define the conservation laws of (1) as follows
(S id)φ(n,un,un+1,...,un+N−1) = 0, (9)
−
where id is the identity function each time (1) holds.
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2 Finding symmetries and first integrals
2.1 Symmetries
Consider the following third-order difference equation
u = ω(n,u ,u ,u ). (10)
n+3 n n+1 n+2
The linearized symmetry condition then amounts to
S3Q Xω = 0, (11)
−
where
X = Q(n,u )∂u +SQ(n,u )∂u +S2Q(n,u )∂u , (12)
n n n n+1 n n+2
istheanalogoussymmetrygenerator. Inordertoobtainthecharacteristic
Q(n,u ), we apply (11) to (10) to obtain
n
∂ω ∂ω ∂ω
Q(n+3,ω) Q(n,u ) SQ(n,u ) S2Q(n,u ), (13)
n n n
− ∂u − ∂u − ∂u
n n+1 n+2
where ω represents the left hand side of (10). We then differentiate (13) with
respect to u , keeping ω and taking u as a function of u ,u ,u and
n n+1 n n+1 n+2
ω. We end up with the following determining equation
ω
,un ω Q(n,u )+ω SQ′(n,u )+ω SQ(n,u )
ω ,unun+1 n ,un+1 n ,un+1un+1 n
,un+1
+ω (cid:2) S2Q(n,u ) ω Q′(n,u ) ω Q(n,u ) (14)
,un+1un+2 n − ,un n − ,unun n
ω SQ(n,u ) ω S2Q(n,u ) = 0,
− ,unun+1 n −(cid:3) ,unun+2 n
where g denotes the derivative of g with respect to x. Note that (10)
,x
maysometimes beindependent ofu oru ,andsotheabovedetermining
n+1 n+2
equationchanges. Foranexampleif (10)isindependent ofu ,theresulting
n+1
determining equation is
ω
,un ω Q(n,u )+ω S2Q′(n,u )+ω S2Q(n,u )
ω ,unun+2 n ,un+2 n ,un+2un+2 n
,un+2 (15)
ω (cid:2)Q′(n,u ) ω Q(n,u ) ω S2Q(n,u ) = 0. (cid:3)
− ,un n − ,unun n − ,unun+2 n
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2.2 Conservation Laws
We now introduce the general case for finding first integrals of third-order
∆E’s. Assume that
φ = φ(n,u ,u ,u ) (16)
n n+1 n+2
is the first integral of (10). According to the equation
Sφ = φ, (17)
we have that
φ(n+1,u ,u ,ω) = φ(n,u ,u ,u ). (18)
n+1 n+2 n n+1 n+2
We then define
∂φ ∂φ ∂φ
P = , P = , P = . (19)
0 1 2
∂u ∂u ∂u
n n+1 n+2
It then follows that
∂ω
P = S(P )
0 2
∂u ·
n
(cid:18) (cid:19)
∂ω
P = S(P )+ S(P ) (20)
1 0 2
∂u ·
(cid:18) n+1(cid:19)
∂ω
P = S(P )+ S(P ).
2 1 2
∂u ·
(cid:18) n+2(cid:19)
We then join (20) to form a single equation involving only P , which will
2
be the determining equation for computing first integrals. The equation is
as follows
S2(ω )S3(P )+S(ω )S2(P )+ω S(P ) P = 0. (21)
,un 2 ,un+1 2 ,un+2 2 − 2
We can then solve for P in (21), hence solve (20) and integrate to obtain
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the first integrals.
3 Applications
We are going to use the determining equation (14) to obtain the symmetries
of two third-order ∆E’s.
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3.1 Symmetries and exact solutions
3.1.1 Example 1
Consider the third-order difference equation
au u
n+1 n+2
u = , (22)
n+3
u +bn
n+2 n
where a and b are arbitrary constants. The determining equation (14) be-
comes
u bu u bu
− n+1 n+2 Q+ n+2 SQ′ + n S2Q
bu +u −(bu +u )2 bu +u (bu +u )2
n n+2 (cid:20) n n+2 n n+2 n n+2 (cid:21)
u u 2bu u u
n+1 n+2 ′ n+1 n+2 n+2
+ Q Q+ SQ
(bu +u )2 − (u +bu )3 (bu +u )2
n n+2 n+2 n n n+2
u (bu u )
n+1 n n+2
+ − = 0.
(bu +u )3
n n+2
(23)
Multiplying by (bu + u )3 throughout and differentiating twice with
n n+2
respect to u we get the following differential equation
n
bQ(2)(n,u )+(bu +u )Q(3) = 0. (24)
n n n+2
Separating (24) with respect to u we get that
n+2
Q(2) = 0, (25)
whose most general solution is
Q(n,u ) = c (n)+u c (n). (26)
n 1 n 2
To obtain the nature of the constants c , and c , which are functions of n,
1 2
we substitute (26) back into (13) and (23). We find that
c1(n) = 0 c2(n) = k0 +k1bn2 +( √b)nk2, (27)
−
where k and k are arbitrary constants. It then follows after setting b = 1
1 2
without loss of generosity that
Q(n,u ) = u k +( 1)nu k . (28)
n n 1 n 2
−
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We therefore have that the symmetries of (22) are
X = u ∂u +u ∂u +u ∂u
1 n n n+1 n+1 n+2 n+2
(29)
X = u ( 1)n∂u +u ( 1)n+1∂u +u ( 1)n+2∂u
2 n n n+1 n+1 n+2 n+2
− − −
We note that X is a symmetry for any value of b. From X in (29), the
1 1
resulting invariants are
du du du dv
n n+1 n+2 n
= = = . (30)
u u u 0
n n+1 n+2
Workingwiththefirst dun andsecond dun+1 invariants, first dun
• un un+1 un
and third dun+2 inva(cid:16)riant(cid:17)s we get that(cid:16) (cid:17) (cid:16) (cid:17)
un+2
(cid:16) (cid:17)
u u
n n
K = K = v = f(K ,K ), (31)
1 2 n 1 2
u u
n+1 n+2
where K and K are constants.
1 2
- We choose f(K ,K ) = K K , therefore
1 2 1 2
·
u
n
v = , (32)
n
u
n+2
hence
1+bv
n
v = . (33)
n+1
a
Solving (33) results in
v = c b n−1 1−(ab)n if a = b (34a)
n 1 a − b−a 6
v = c + n if a = b. (34b)
n 1(cid:0) (cid:1)b
We have reduced a third-order difference equation (10) into a
second-order difference equation (34a).
Case a=b: Combining (32) and (34b), we have
1 a
u = u = u . (35)
n+2 n n
v ac +n
n 1
Therefore, the solution to (22), when a = b, is given by
21−n2bn2−1(c1 +c2( 1)n)
u = − , c Z, (36)
n d(b2c1 +1)n2−1 2 ∈
6
where v in given in (34b) and (a) is the Pochhammer symbol
k1 n
(rising factorial.)
Case a = b: Combining (32) and (34a), we have
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1 1
u = u = u . (37)
n+2 vn n c b n−1 1−(ab)n n
1 a − b−a
(cid:0) (cid:1)
Therefore, the solution to (22) is given by
n−1 k3−1
u =c exp ( 1)k3+1 ( 1)−k1+1lnv +
n 2 − − k1
!
kY3=1 kX1=0(cid:0) (cid:1)
n−1 k4−1 exp ( 1)k2 k2 ( 1)−k1+1lnv
k2=1− − k1=0 − k1
c
3kX4=0 Qkk43=1exp ((cid:16)−1)k3+1Pkk31−=10(−1)−k1+1lnvk1(cid:17)
n−1 Q (cid:16)k3−1 P (cid:17)
exp ( 1)k3+1 ( 1)−k1+1lnv ,
− − k1
!
kY3=1 kX1=0(cid:0) (cid:1)
(38)
where v in given in (34a).
k1
- Suppose that we choose f(K ,K ) = K , following the same
1 2 1
method above, we end up with v = un , i.e.,
n un+1
1 bv 1+bv v
n n n+2
v = + = . (39)
n+2
av a av
n+1 n+1
One can solve for v from (39) and hence solve for u .
n n
Working with the first dun and third invariant dun+2 , second in-
• un un+2
variant dun+1 and thi(cid:16)rd in(cid:17)variant dun+2 , we g(cid:16)et that(cid:17)
un+1 un+2
(cid:16) (cid:17) (cid:16) (cid:17)
u u
n n+1
K = K = v = f(K ,K ). (40)
1 2 n 1 2
u u
n+2 n+2
- Choosing f(K ,K ) = K /K we get that
1 2 2 1
u
n+1
v = (41)
n
u
n
7
and therefore
av (a/b)v
n n
v = = . (42)
n+2
b+v v 1+(1/b)v v
n n+1 n n+1
The solution of ((42)) is given in [7]( put A = a/b and B=1/b) by
v =exp ( 1)n−1ln v +
n 1
− | |
"
(43)
n−1 v v a/bk1
( 1)n−1 ( 1)−k1ln 0 1
− kX1=1 − (cid:12)(cid:12)1+(1/b)v0v1( ki=1−01a/bi)(cid:12)(cid:12)#
(cid:12) (cid:12)
with 1/((1/b)v v ) / 1,1+(cid:12)(cid:12)(a/b),..., n−P2(a/b)i . Co(cid:12)(cid:12)mbining
− 0 1 ∈ { i=0 }
(41) and (43) we get that
P
n−1
u = c v , (44)
n 4 k1
kY1=1
where v is given in (43), is also a solution to (22).
n
3.1.2 Example 2
We now shift our attention to focus on a second third-order difference equa-
tion
u = a(n)u +c(n)u , (45)
n+3 n n+2
where a and c are arbitrary functions of n. We note that (45) is independent
of u , so we use determining equation (15). One can easily establish that
n+1
the solution to the determining equation is
Q(n,u ) = k u +k (n), (46)
n 1 n 2
where k is a constant and k is a function of n. Substituting (45) into (11)
1 2
we get that
k (n+3) = a(n)k (n)+c(n)k (n+2), (47)
2 2 2
where we notice that k (n) satisfies the original equation (45). Assuming
2
that U , U and U are solutions of (47), the resulting symmetries are
1 2 3
X = u ∂u X = U ∂u X = U ∂u X = U ∂u . (48)
0 n n 1 1 n 2 2 n 3 3 n
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The characteristic equation, using X , is given by
0
du du du dV
n n+1 n+2 n
= = = . (49)
u u u 0
n n+1 n+2
The invariant V is then given by
n
u u
n+1 n+2
V = g , .
n
u u
(cid:18) n n+1(cid:19)
Assume g(x,y) = x, we have
u
n+1
V = (50)
n
u
n
and this implies that
a(n)
V = +c(n). (51)
n+2
V V
n n+1
On the other hand, by assuming that g(x,y) = 1/x, we have
u
n
V = (52)
n
u
n+1
and this implies that
1
V = . (53)
n+2
c(n)+a(n)V V
n n+1
With these change of variables we have obtained two different reduced forms
(in terms of order), given by ((50)) and ((51)), of (45).
3.2 Conservation Laws
3.2.1 Example 1
Consider the equation (45), assume that one required to obtain the first
integrals. By letting P = P (n,u ), one can find that the solution to the
2 2 n
determining equation (21) is
P = B(n). (54)
2
Substituting (54), we find that B(n) satisfies the equation
B(n) = a(n+2)B(n+3)+c(n)B(n+1). (55)
9
We can then shown that
P =a(n)B (n+1), (56)
0 i
P =a(n+1)B (n+2), (57)
1 i
P =B (n), (58)
2 i
′
where B s, i = 1,...3 are solutions to (55). We can therefore deduce that the
i
first integrals of such equations (45) are
Φ = P u +P u +P u +G (n) (59)
i 0 n 1 n+1 2 n+2 i
= a(n)B (n+1)u +a(n+1)B (n+2)u +B (n)u +G (n), (60)
i n i n+1 i n+2 i
for some functions G ,i = 1,2,3, to be determined. By substituting (60)
i
into (17), we were able to find that G (n + 1) = G (n) = K, where K is a
i i
constant. For simplicity, we shall assume that K is zero. It has to be noted
that now that we know the symmetries and first integrals of (45), we can
create new first integrals. Below are new first integrals of (45):
Φ =X Φ = U P +P SU +(S2U )P (61)
3i 1 i 1 0 1 1 1 2
Φ =X Φ = U P +P SU +(S2U )P (62)
4i 2 i 2 0 1 2 2 2
Φ =X Φ = U P +P SU +(S2U )P . (63)
5i 3 i 3 0 1 3 3 2
Also, from
(S id)Φ = (SP )[u c(n)u a(n)u ], (64)
i 2 n+3 n+2 n
− − −
we obtain new integrals
P
0
Φ =SP = B (n+1) = . (65)
6i 2 i
a(i)
3.2.2 Example 2
Now consider the equation (22), i.e.,
au u
n+1 n+2
u = .
n+3
u +bu
n+2 n
Assuming one required to compute the first integrals, one would hence use
(21) to come up with the equation
abu u au abu u
− n+3 n+4 S3P + n+3 S2P + n n+1 SP P = 0. (66)
(u +bu )2 2 u +bu 2 (u +bu )2 2− 2
n+4 n+2 n+3 n+1 n+2 n
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