IOSR Journal of Mathematics (IOSRJM) www.iosrjournals.org
IOSR Journal of Mathematics (IOSRJM)
ISSN: 2278-5728 Volume 3, Issue 4 (Sep.-Oct. 2012), PP 01-03
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Precision Angular Measurements Using Scale of Chords
Dr A. M. Chandra
(Department of Civil Engineering, Sharda, University, India)
Abstract: Presently angular measurements are made using protractors having a normal accuracy of 1 or at
the most ½. The scale of chords, as linear scale, can be constructed for measurement and construction of
angles having accuracy equal to that of a normal protractor. This paper presents a new concept of construction
of a diagonal scale of chords that can have an accuracy of 10 or less, and thus, using diagonal scale of chords,
angles can be constructed or measured to a higher accuracy which is not possible using normal size protractors.
Keywords: Angle measurements, Protractor, Scale of chords, Diagonal scale of chords
I.
Introduction
For the measurement and construction of angles, protractor is the most convenient commonly used tool.
The normal size protractors have their least count as 1, but the bigger size protractors can have ½ or 30. To
increase the least count further, the size of the protractor has to be increased making its use inconvenient.
Another method of measuring or constructing angles is using a scale of chords which is a linear scale. The plain
scale of chords can have a least count which can be equal to that of a protractor. The least count of the plain
scale of chords can be increased by making it a diagonal scale of chords without increasing its length. It is a new
concept in measuring and constructing angles manually to higher accuracy which was not possible earlier.
II.
Plain Scale Of Chords
To construct a plain scale of chords, a 90 arc AB’ (the arc subtending an angle of 90 at its centre) as
shown in Fig. 1, of convenient radius is divided into nine equal parts A-1’, 1’-2’, 2’-3’, etc., each part
subtending 10 at the centre O [Chandra and Chandra, 2003]. The division of an arc into odd number of equal
divisions is done by the trial and error method using a bow divider. A new method of dividing an arc into odd
number of equal divisions has been developed by the author and is going to be published soon.
The line AO is extended to a convenient length AC. The points 1’, 2’, 3’, etc., on the arc AB’ are
transferred to their corresponding points 1, 2, 3, etc., on the line AC by drawing arcs of radii A-1’, A-2’, A-3’,
etc., having centre at A. Each division A-1, 1-2, 2-3, etc., corresponds to an angle of 10. By dividing each
division into five equal parts, the scale of chords, so constructed, has a least count of 2, and if it is possible to
divide these divisions into 10 equal parts, the least count will be 1. Similarly, if a bigger arc AB’ is taken it
may be possible to get a least count of ½ or 30, but practically it may not be possible to get a least count less
than 30 due to inconvenience in use of the scale as its size becomes too large, though theoretically it is possible.
6’
7’
8’
B’
5’
4’
3’
2’
1’
A
0
8
3
1
4
5
O 7
2
C
33
B
b
10 20 30 40 50 60 70 80 90
a
Fig. 1 Plain scale of chords
Now to construct an angle of say 33 using the scale of chords, draw a line PQ equal to AO, i.e., a, and then
draw the arc QM of radius a with centre at P (Fig. 2). On this arc, mark the point K at a distance equal to b from
Q, the distance corresponding to 33as shown in Fig. 1. Join P and K, and the angle KPQ is equal to 33.
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Precision Angular Measurements Using Scale of Chords
M
K
b
33
P
Q
a
Fig. 2 Constructing an angle using scale of chords
If it is required to measure an angle MPN () shown in Fig. 3, extend the line PN to Q such that PQ is equal to a.
Now draw an arc of radius a with centre at P, intersecting the extension of PM at K. Take a bow divider and
measure the distance QK equal to c. Lay down this measured distance c on the scale of chords which will give
the value of the angle as 41 in this case.
K
M
c
P
N
a
Q
Fig. 3 Measuring an angle using scale of chords
III.
Diagonal Scale Of Chords
Fig. 4 shows a diagonal scale of chords APQB. The base AB is a part of the plain scale of chords, and
its construction is explained above. Now to convert this plain scale of chords into a diagonal scale of chords,
draw a perpendicular AP to AB of a convenient length at A, and complete the rectangle APQB. Divide AP into
six equal divisions, and draw lines for each division parallel to AB. Also extend the lines of each division of 1of
the plain scale of chords up to the line PQ, and draw the diagonals for each 1 division as shown in the figure.
This construction now provides the least count of 10. The horizontal distances b and c more precisely indicate
the angles as 3340 and 4120, respectively.
6’
B’
8’
7’
5’
4’
3’
2’
1’
1 A
40
20
P
0
3
1
2
3340
b
4120
c
4
5
O
7
8
B
C
Q
10 20 30 40 50 60 70 80 90
a
Fig. 4 Diagonal scale of chords
Construction and measurement of angles to an accuracy of 10 can be done using the diagonal scale of chords
in a similar manner as in the case of plain scale of chords explained above.
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Precision Angular Measurements Using Scale of Chords
Practical Considerations
IV.
There are two important points which require attention. The first point to be noted is that the divisions
A-1, 1-2, 2-3, etc., are not equal in length. Though the chord lengths A-1’, 1’-2’, 2’-3’, etc., are equal, the
distance A-2 is not equal to [(A-1’) + (1’-2’)], and so on. With the same logic, the subdivisions of the divisions
A-1, 1-2, 2-3, etc., are also not equal. Since the difference between the lengths of two subdivisions is so small
that it cannot be measured, they may be treated to be equal from practical consideration of plotting accuracy.
Another point to be noted that the line AP shown in Fig. 4, can be divided into 12 equal parts giving the least
count of the scale as 5, but the difference between the linear distance for 5 and 10 will be so small that it
cannot be measured, and therefore, angles having difference of 5 cannot be measured. Thus dividing the line AP
into 12 equal parts is not going to improve the accuracy of the scale until AP and AB are taken sufficiently large,
which will make the scale inconvenient to use.
V.
Conclusions
A new concept of diagonal scale of chords presented here has its own practical use for construction and
measurement of angles manually to an accuracy which is not achievable using a normal protractor. The diagonal
scale of chords can be manufactured like other scales and can be made available in market to achieve higher
accuracy in angular measurements.
Acknowledgements
The author is grateful to the Arba Minch University, Ethiopia, where he was working teaching
assignment, for all help and conducive environment provided by the university to write this technical paper
during 2002-04.
References
[1]
A. M. Chandra and Satish Chandra, Engineering Graphics (Narosa Publishing House, New Delhi, 2003)
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IOSR Journal of Mathematics (IOSR-JM)
ISSN: 2278-5728. Volume 3, Issue 4 (Sep-Oct. 2012), PP 04-14
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Dominating Sets and Domination Polynomials of Square Of
Cycles
A. Vijayan1, K. Lal Gipson2
1
Assistant Professor, Department of Mathematics, Nesamony Memorial Christian College, Marthandam,
Kanayakumari District, Tamil Nadu, South India.
2
Assistant Professor, Department of Mathematics, Mar Ephraem College of Engineering and Technology,
Mathandam, Kanayakumari District, Tamil Nadu, South India.
Abstract: Let G = (V, E) be a simple graph. A set S V is a dominating set of G, if every vertex in V-S is
2
2
adjacent to atleast one vertex in S. Let Cn be the square of the Cycle Cn and let D(Cn , i ) denote the family of
2
2
2
all dominating sets of Cn with cardinality i. Let d (Cn , i ) = | D(Cn , i ) |. In this paper, we obtain a recursive
2
formula for d (Cn , i ) . Using this recursive formula, we construct the polynomial,
D(Cn2 , x) = i n d (Cn2 , i) xi , which we call domination polynomial of Cn2 and obtain some properties of
n
5
this polynomial.
Keywords: domination set, domination number, domination polynomials.
I.
Introduction
Let G = (V,E) be a simple graph of order |V| = n. For any vertex vV, the open neighbourhood of is
the set N (v) = {uV/uvE} and the closed neighbourhood of is the set N[v] = N (v) {v}. For a set S V,
the open neighbourhood of S is N(S) = US N (v) and the closed neighbourhood of S is N[S] = N(S) S. A set
S V is a dominating set of G, if N[S] = V, or equivalently, every vertex in V-S is adjacent to atleast one vertex
in S. The domination number of a graph G is defined as the minimum size of a dominating set of vertices in G
and it is denoted as G .A cycle can be defined as a closed path, and is denoted by Cn .
1.1. Definition
The square of a graph: The 2nd power of a graph with the same set of vertices as G and an edge
between two vertices if and only if there is a path of length atmost 2 between them.
2
2
Let Cn be the square of the cycle Cn (2nd power) with n vertices. Let D(Cn , i ) be the family of dominating sets
2
2
2
of the graph Cn with cardinality i and let d (Cn , i ) = | D(Cn , i ) |. We call the polynomial
D(Cn2 , x) =
n
n
i
5
d (Cn2 , i) x i the domination polynomial of the graph Cn2 .
In the next section, we construct the families of the dominating sets of the square of cycles by recursive method.
As usual we use x for the largest integer less than or equal to x and x for the smallest integer
greater than or equal to x. Also, we denote the set {1, 2… n} by [n], throughout this paper.
II.
Dominating Sets Of Square Of Cycles
2
n
2
Let D(C , i ) be the family of dominating sets of Cn with cardinality i. We investigate the
2
dominating sets of Cn . We need the following lemma to prove our main results in this section.
Lemma.2.1
n
2
( Cn ) =
5
By Lemma2.1 and the definition of domination number, one has the following lemma:
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Dominating Sets And Domination Polynomials Of Square Of Cycles
Lemma.2.2
n
D(Cn2 , i) = if and only if i n or i
5
A simple path is a path in which all its internal vertices have degree two, and the end vertices have degree one.
The following lemma follows from observation.
Lemma 2.3 [2]
If a graph G contains a simple path of length 5k -1, then every dominating set of G must contain at least
k vertices of the path.
2
In order to find a dominating set of Cn , with cardinality i, we only need to consider
D(Cn21 , i 1), D(Cn22 , i 1), D(Cn23 , i 1), D(Cn24 , i 1) and D(Cn25 , i 1) .The families of these
dominating sets can be empty or otherwise. Thus, we have eight combinations, whether these five families are
empty or not. Two of these combinations are not possible (Lemma 2.5 (i), & (ii)).
Also, the combination that
D(Cn21 , i 1) D(Cn22 , i 1) D(Cn23 , i 1) D(Cn24 , i 1) D(Cn25 , i 1)
does not need to be considered because it implies that D( Pn , i ) = (See lemma 2.5 (iii)). Thus we only need
to consider five combinations or cases. We consider those cases in theorem (2.7).
2
Lemma 2.4
If Y D(Cn6 , i 1) , and there exists x [n] such that Y {x} D(Cn , i) ,
2
2
then Y D(Cn5 , i 1) .
2
Proof:
Suppose that Y D(Cn5 , i 1) .
2
Since Y D(Cn6 , i 1) , Y contains at least one vertex labeled n–6, n–7 or n–8.
2
If n – 6 Y, then Y D(Cn5 , i 1) , a contradiction. Hence n –7 or n – 8 Y, but then in this case,
2
Y {x} D(Cn , i) , for any x [n], also a contradiction.
2
Lemma 2.5
i)
If D(Cn1 , i 1) = D(Cn3 , i 1) = then D(Cn2 , i 1) =
2
2
2
ii) If D(Cn1 , i 1) , D(Cn3 , i 1) then D(Cn2 , i 1) ,
2
2
2
iii) If D(Cn1 , i 1) = , D(Cn2 , i 1) = , D(Cn3 , i 1) = , D(Cn 4 , i 1) = ,
2
2
2
2
D(Cn25 , i 1) = , then D(Cn2 , i) =
Proof
i)
n 3
Since D(Cn1 , i 1) = D(Cn3 , i 1) = , by lemma 2.2, i -1 n-1 (or) i -1
.
5
2
2
ii) In either case, we have D(Cn2 , i 1) = .
2
n 2
iii) Suppose that D(Cn2 , i 1) = , so by Lemma 2.2, we have i – 1 n – 2 or i – 1
.
5
2
If i -1 n – 2, then i – 1 n – 3. Therefore D(Cn3 , i 1) = , a contradiction.
2
iv) Suppose that D(Cn , i) , Let Y D(Cn , i) ). Thus, atleast one vertex labeled n or n – 1 or n – 2 is
2
2
in Y. If n Y, then by Lemma 2.3, at least one vertex labeled n – 1, n – 2, n – 3, n – 4 or n – 5 is in Y.
If n – 1 Y or n – 2 Y, then Y – {n} D(Cn1 , i 1) , a contradiction. If n – 3 Y,
2
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Dominating Sets And Domination Polynomials Of Square Of Cycles
then Y – {n} D(C
2
n2
, i 1) , a contradiction. Now, suppose that n – 1 Y. Then, by Lemma 2.3, at least
one vertex labeled n – 2, n – 3, n – 4, n – 5 or n – 6 is in Y. If n – 2 Y or n–3 Y,
Y – {n – 1} D(C
2
n2
then
, i 1) a contradiction. If n – 4 Y, then Y – {n – 1} D(Cn23 , i 1) ,
contradiction. If n – 5 Y, then Y – {n – 1} D(C
2
n4
, i 1) , a contradiction. If n – 6 Y,
a
then
Y – {n – 1} D(Cn5 , i 1) , a contradiction. Therefore, D(Cn , i) = .
2
2
Lemma 2. 6
If D(Cn , i) ,then
2
D(Cn21 , i 1) = D(Cn22 , i 1) = D(Cn23 , i 1) = D(Cn24 , i 1) = and D(Cn25 , i 1)
i)
if and only if n=5k and i=k for some k N;
2
2
2
2
2
ii) D(Cn2 , i 1) = D(Cn3 , i 1) = D(Cn 4 , i 1) = D(Cn5 , i 1) = then D(Cn1 , i 1) if
and only if i = n;
iii) D(Cn1 , i 1) = , D(Cn2 , i 1) , D(Cn3 , i 1) , D(Cn 4 , i 1) ,
2
2
n 5
D(C
2
2
2
5k 2
, i 1) , if and only if n = 5k+2 and i = 5 for some kN;
iv) D(Cn1 , i 1) , D(Cn2 , i 1) , D(Cn3 , i 1) , D(Cn 4 , i 1) , and
2
2
2
2
D(Cn25 , i 1) = if and only if i = n – 3.
v)
D(Cn21 , i 1) , D(Cn22 , i 1) , D(Cn23 , i 1) , D(Cn24 , i 1) , and
n 1
+1 i n – 4.
D(Cn25 , i 1) if and only if
5
Proof
i)
()
Since D(Cn1 , i 1) ) = D(Cn2 , i 1) = D(Cn3 , i 1) = D(Cn 4 , i 1) = , by lemma 2.2,
2
2
2
2
n 2
i – 1 n – 1 or i – 1
5
If i – 1 n – 1, then i n and by lemma 2.2, D(Cn , i) = , a contradiction.
2
n 2
n
n 2
So i
+ 1, and since D(Cn , i) , we have i
+ 1, which implies that n = 5k
5
5
5
and i = k for some kN.
() If n = 5k and i = k for some kN, then by lemma 2.2,
2
D(Cn21 , i 1) = D(Cn22 , i 1) = D(Cn23 , i 1) = D(Cn24 , i 1) = , and D(Cn25 , i 1) .
ii)
() Since D(Cn2 , i 1) = D(Cn3 , i 1) = D(Cn 4 , i 1) = D(Cn5 , i 1) = , by lemma 2.2,
2
2
2
2
n 3
n 3
n 1
. If i – 1
, then i – 1
and hence
5
5
5
D(Cn22 , i 1) = , a contradiction. So i -1 n – 2. Also, D(Cn21 , i 1) .
Therefore i n-1.Therefore i n.But i n. Hence i = n.
n 1
2
iii)
() Since D(Cn1 , i 1) = , by lemma 2.2, i – 1 n – 1 or i – 1
5
If i – 1 n – 1, then i – 1 n – 2 ,by Lemma 2.2,
i – 1 n – 2 or i – 1 <
D(Cn22 , i 1) = D(Cn23 , i 1) = D(Cn24 , i 1) = D(Cn25 , i 1) = , a contradiction.
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Dominating Sets And Domination Polynomials Of Square Of Cycles
n 1
n 2
Therefore i
+1. But i –1
,because D(Cn2 , i 1) .
5
5
2
n 2
Therefore i
+1.
5
n 2
n 1
Hence,
+1 i
+1.
5
5
5k 2
This holds only if n = 5k +2 and i = k+1 = 5 ,for some kN.
5k 2
5 for some kN, then by lemma 2.2,
() If n = 5k+2 and i =
D(Cn21 , i 1) = , D(Cn22 , i 1) , D(Cn23 , i 1) , D(Cn24 , i 1) and
D(Cn25 , i 1) .
iv)
n 5
( ) Since D(Cn5 , i 1) = ,by lemma 2.2, i – 1 n – 5 or i – 1
.
5
2
n 2
n 5
Since D(Cn2 , i 1) , by lemma 2.2,
+ 1 i n – 1. Therefore i – 1
is not
5
5
possible. Therefore i – 1 n – 5. Therefore i n-4.Therefore i n-3.
2
Since D(Cn3 , i 1) = , i-1 n-4. Therefore i n-3. Hence i = n – 3.
2
() if i = n – 3 then by lemma 2.2, D(Cn1 , i 1) , D(Cn2 , i 1) , D(Cn3 , i 1) = ,
2
2
2
D(Cn24 , i 1) and D(Cn25 , i 1) = .
v) () Since D(Cn1 , i 1) , D(Cn2 , i 1) , D(Cn3 , i 1) , D(Cn 4 , i 1)
2
2
2
2
n 1
i – 1 n – 1,
and D(Cn5 , i 1) , then by applying lemma 2.2,
5
2
n 2
n 3
n 3
5 i – 1 n – 2, 5 i – 1 (n – 3), 5 i – 1 (n – 4) and
n 3
n 1
n 1
5 i – 1 n – 5. So 5 i – 1 (n – 5) and hence 5 + 1 i n – 4.
n 1
() If
5 + 1 i n – 4, then the result follows from lemma 2.2.
Theorem 2.7
n
For every n 6 and i ,
5
i)
If D(Cn1 , i 1) = D(Cn2 , i 1) = D(Cn3 , i 1) = D(Cn 4 , i 1) = and
2
2
2
2
D(Cn25 , i 1) then D(Cn2 , i) = {1, 6,…………….. n – 4 },{2, 7,………….. n – 3 },
{3, 8,………..… n – 2},{4, 9,…………….. n – 1 },{5, 10,…………….. n }
2
n2
ii) If D(C
2
, i 1) = D(C
then D(Cn , i) =
2
n 3
[n]
,
,
, i 1) = D(C
, i 1) = and D(Cn21 , i 1)
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, i 1) = D(C
2
n4
2
n 5
Dominating Sets And Domination Polynomials Of Square Of Cycles
2
n 1
iii) If D(C
, i 1) = , D(C
2
n2
, i 1) , D(Cn23 , i 1) , D(Cn24 , i 1) and
D(Cn25 , i 1) then
D(Cn2 , i) = {1, 6,……….. n – 4 },{2, 7,……….. n – 3 },{3, 8,……… … n – 2},
{4, 9……… n – 1 },{5, 10,…………….. n }
, X3 n .X4 n 1 ,
n 2 if 2 X 5
2
2
2
, X3 D(Cn3 , i 1) , X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1) }…… … (2.1)
X5
n 3 if 2 X 5
2
2
2
iv) If D(Cn3 , i 1) = , D(Cn2 , i 1) , D(Cn1 , i 1) , then
D(Cn2 , i) = [n] - {x}/x[n]
.
v) If D(Cn1 , i 1) , D(Cn22 , i 1) , D(Cn3 , i 1) , D(Cn 4 , i 1) ,
2
2
2
D(Cn25 , i 1) then
D(Cn2 , i) = X1 {n}, X2 {n - 1}, X3 {n-2}, X4 {n - 3}, X5 {n-4}/ X1 D(Cn21 , i 1) ,
X2 D(Cn22 , i 1) , X3 D(Cn3 , i 1) , X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1)
2
2
2
……….…. ...(2.2)
Proof
i)
Since, D(Cn1 , i 1) = D(Cn22 , i 1) = D(Cn3 , i 1) = D(Cn 4 , i 1) = and
2
2
2
D(Cn25 , i 1) by lemma (2.6) (i) n = 5k, and i=k for some kN.
The set
{1, 6,…………….. n – 4 },{2, 7,………….. n – 3 },{3, 8,………..… n – 2},
{4, 9,…………….. n – 1 },{5, 10,…………….. n }
has
n
n
elements and it covers all vertices. The other sets with cardinality
5
5
{1, 6,…………….. n },{2, 7,………….. n },{3, 8,………..… n },
{4, 9,…………….. n },{5, 10,…………….. n }
,
Clearly, all these are not dominating sets. Therefore
{2, 7,………….. n – 3 },{3, 8,………..… n – 2},
{4, 9,…………….. n – 1 },{5, 10,…………….. n }
is the only dominating set of cardinality
ii)
,
We
have D(Cn22 , i 1) =,
{1, 6,…………….. n – 4 },
,
n
= k = i.
5
D(Cn23 , i 1)
=
D(Cn24 , i 1)
and D(Cn1 , i 1) . By lemma (2.6) (ii). We have i = n. So, D(Cn , i) =
2
iii)
2
=
[n]
D(Cn25 , i 1)
=
.
We have D(Cn1 , i 1) = , D(Cn22 , i 1) ≠ , D(Cn3 , i 1) , D(Cn 4 , i 1) and
2
2
2
D(Cn25 , i 1) .
5k 2
5 = k+1 for some kN.
By lemma (2.6). (iii), n = 5k+2 and i =
We denote the families Y1 =
{1, 6,……,5k-4 ,5k+1 },{2, 7,……, 5k-3 ,5k+2 },
{3, 8,……… 5k-2 ,5k+3},{4, 9……5k-1, 5k+4 },{5, 10,…………5k,5k+5 }
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and
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Dominating Sets And Domination Polynomials Of Square Of Cycles
if 2 X 5
5k
2
, X3 D(Cn3 , i 1) ,
Y2 = X3 5k 2 .X4 5k 1 , X 5
5k 1 if 2 X 5
2
2
X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1)
We shall prove that D(C5k 2 , k 1) = Y1 Y2 .Since D(C5k , k ) = {1, 6,……,5k-4 ,5k+1 },
2
2
{2, 7,……, 5k-3 ,5k+2 },{3, 8,…… 5k-2 ,5k+3},{4, 9……5k-1, 5k+4 },{5, 10,…………5k,5k+5 }
,then
Y1 D(C52k 2 , k 1) .Also it is obvious that Y2 D(C52k 2 , k 1) .
Therefore Y1 Y2 D(C5k 2 , k 1) …………………………………………………………………….(2.3)
2
) Now let Y D(C5k 2 , k 1) . Then 5k + 2, 5k+1 or 5k is in Y. If 5k+2 Y, then by lemma (3), atleast one
vertex labeled 1,2 or 3 and 5k+1, 5k or 5k-1 is in Y. If 5k+1 or 5k is in Y, then
2
Y – {5k+2} D(C5k 1 , k ) , a contradiction; because D(C5k 1 , k ) = . Hence 5k – 1 Y, 5k Y and
2
2
5k + 1 Y.
Therefore, Y = X {5k+2} for some X D(C5k , k ) .Hence Y Y1 . Now suppose that 5k + 1 Y and 5k+2
2
Y. BY lemma (2.3) at least one vertex labeled 5k,
5k–1or 5k–2 is in Y. If 5kY then Y–{5k+1} D(C5k , k ) = {1, 6,……,5k-4 ,5k+1 },{2, 7,……, 5k-3 ,5k+2
2
},{3, 8,……… 5k-2 ,5k+3},{4, 9……5k-1, 5k+4 },{5, 10,…………5k,5k+5 }
a contradiction because 5 k
X for all X D(C5k , k ) .
2
Therefore 5k – 1, or 5k – 2 is in Y, but 5kY.
Thus Y= X {5k+1} for some of X D(C5k 1 , k ) . So
2
D(C52k 2 , k 1) Y1 Y2 ………………………………………………………………………………... (2.4)
From (2.3) and (2.4),
D(Cn2 , i) = {1, 6,……….. n – 4 },{2, 7,……….. n – 3 },{3, 8,……… … n – 2},
, X3 n .X4 n 1 ,
{4, 9……… n – 1 },{5, 10,…………….. n }
n 2 if 2 X 5
2
2
2
, X3 D(Cn3 , i 1) , X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1) }
X5
n 3 if 2 X 5
2
2
iv)
If D(Cn3 , i 1) = , D(Cn22 , i 1) , D(Cn1 , i 1) , by lemma (2.6) (iv) i = n –3.
2
Therefore D( Pn , i ) = {[n] - {x} / x[n]}.
2
For example D( P6 ,5) = {[6] - {x} / x[6]}.
{1,2,3,4,5,6}-{1,2,3,4,5,6} /6[6]} .
= {1,2,3,4,5 }-{{1,2,3,4 ,6}-{{1,2 ,4,5,6}-{{1, 3,4,5,6}-{{2,3,4,5,6}
v) D(Cn1 , i 1) , D(Cn22 , i 1) , D(Cn3 , i 1) , D(Cn 4 , i 1) ,
2
2
2
D(Cn25 , i 1) . Let X1 D(Cn21 , i 1) ,so atleast one vertex labeled n-1, n – 2 or n – 3 is in X1.
2
If n-1, n-2 or n-3 X1, then X1 {n} D(Cn , i) .
Let X2 D(Cn22 , i 1) , then n – 2 or n – 3 or n – 4 is in X2.
If n-2, n-3 or n-4 X2, then X2 {n – 1} = D(Cn , i) . Now let X3 D(Cn3 , i 1) then n – 3,
n – 4 or n – 5 is in X3.
2
2
If n – 3 or n – 4 or n – 5 X3 ,then X3 {n – 2} = D(Cn , i) . Now let X4 D(Cn 4 , i 1) ,
then n – 4, or n – 5 or n – 6 is in X4.
2
2
2
2
If n-4 X4 ,then X4{n-3} D(Cn , i) , for X {n, n-1}. If n-5 X5 ,then X5 {n-4} D(Cn , i) .
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Dominating Sets And Domination Polynomials Of Square Of Cycles
Thus we have X1 {n}, X2 {n - 1}, X3 {n-2}, X4 {n - 3}, X5 {n-4}/ X1 D(Cn1 , i 1) ,
2
X2 D(Cn2 , i 1) , X3 D(Cn3 , i 1) , X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1)
2
2
2
2
D(Cn , i) ....(2.5)
2
Now suppose that n-1 Y, n Y then by lemma (3), atleast one vertex labeled n-2, n-3 or in Y. If n-2 Y then
Y=(X3) {n-2} for some, X3 D(Cn3 , i 1) .
2
Now suppose then, n-5 Y & n-4 Y, then by lemma (3) one vertex labeled n-6, n-7, in Y.
If n-4Y, then Y = (X2) {n-1} for X2 , D(Cn2 , i 1) ,
2
So D(Cn , i)
2
X1 {n}, X2 {n - 1}, X3 {n-2}, X4 {n - 3}, X5 {n-4}/ X1 D(Cn1 , i 1) ,
2
X2 D(Cn2 , i 1) , X3 D(Cn3 , i 1) , X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1)
2
2
2
2
….………..(2.6)
From (2.5) & (2.6)
D(Cn2 , i) = X1 {n}, X2 {n - 1}, X3 {n-2}, X4 {n - 3}, X5 {n-4}/ X1 D(Cn21 , i 1) ,
X2 D(Cn2 , i 1) , X3 D(Cn3 , i 1) , X4 D(Cn 4 , i 1) , X5 D(Cn5 , i 1)
2
2
2
2
.
Domination Polynomial Of A Square Cycle ( Cn2 )
III.
Let D(Cn , x) =
2
n
n
i
5
d (Cn2 , i) xi be the domination polynomial of a cycle Cn2 . In this section, we study
this polynomial.
Theorem 3.1
i)
2
2
If D(Cn , i) is the family of dominating sets with cardinality i of Cn , then
d (Cn2 , i) d (Cn21 , i 1) d (Cn22 , i 1) d (Cn23 , i 1) d (Cn24 , i 1) d (Cn25 , i 1)
ii) For every n 6,
D(Cn2 , x) = x D(Cn21 , x) D(Cn22 , x) D(Cn23 , x) D(Cn24 , x) D(Cn25 , x) with the initial values
D(C12 , x) x, D(C22 , x) x 2 2 x , D(C32 , x) x3 3x 2 3x ,
D(C42 , x) x 4 4 x3 6 x 2 4 x, D(C52 , x) x5 5 x 4 10 x3 10 x 2 5 x
Proof
i) It follows from theorem 2.7
ii) It follows from part (i) and the definition of the domination polynomial.
2
Using Theorem 3.1, we obtain d (Cn , i ) for 1≤n≤15 as shown in table 1.
2
2
Table 1: d (Cn , i ) , the number of dominating set of Cn with cardinality i.
i
n
1
2
3
4
5
6
7
8
9
10
11
1
2
3
4
5
1
2
3
4
5
0
0
0
0
0
0
1
3
6
10
15
14
12
9
5
0
1
4
1
10 5
1
20 15
6
35 35 21
45 70 56
57 115 114
34 126 126
27 150 328
6
7
8
9
10
11
1
7
28
82
196
402
1
8
36
118
314
1
9
45
163
1
10
55
1
11
1
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13
14 15
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Dominating Sets And Domination Polynomials Of Square Of Cycles
12
13
14
15
0
0
0
0
0
0
0
0
18 159 462 723 715 477 218
66
12
1
10 150 588 1164 1431 1191 695 284
78
13
1
4 126 678 1698 2567 2614 1885 979 362 91 14 1
1 93 711 2262 4183 5145 4490 2863 1341 453 105 15
2
In the following Theorem, we obtain some properties of d (Cn , i ) .
Theorem 3.2
2
The following properties hold for the coefficients of D(Cn , x) ;
i)
d (C52n , n) 5
ii)
d (Cn2 , n) 1
iii)
d (C , n 1) n
iv)
d (C , n 2) nC2 for every n 3
v)
d (Cn2 , n 3) nC3
vi)
d (Cn2 , n 4) nC4 for every n 5
1
for every n N.
for every n N
for every n 2
2
n
2
n
for every n 4
vii) d ( P5n i , n) n for i = 1, …………….., n = 1,…………….
2
Proof
D(C52n , n) = {1, 6,……….. n – 4 },{2, 7,………….. n – 3 },{3, 8,………..… n – 2},
i)
, we have d (C52n , n) 5 .
2
Since D(Cn , i) = [n] ,we have the result.
{4, 9… n – 1}, {5, 10… n}
ii)
iii)
Since D( Pn , i ) = {[n] - {x} /x[n]}, we have d (Cn , n 1) n .
iv)
By induction on n
The result is true for n = 3
2
2
2
L.H.S. = d (C3 ,1) = 3 (from table)
3.2
=3
2
R.H.S =
Therefore, the result is time for n = 3
Now suppose that the result is true for all numbers less than „n‟ and we prove it for n.
By theorem (3.1),
d (Cn2 , n 2) d (Cn21 , n 3) d (Cn22 , n 3) d (Cn23 , n 3) d (Cn24 , n 3) d (Cn25 , n 3)
(n 1)(n 2)
(n 2) 1
2
n2 3n 2 2n 4 2
=
2
2
n n
=
2
n(n 1)
=
.
2
=
v)
By induction on n
The result is true for n = 4
2
L.H.S = d (C4 ,1) = 4 (from table)
2
R.H.S = d (C4 ,1) =
4.3.2
6
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Dominating Sets And Domination Polynomials Of Square Of Cycles
Therefore the result is true for n = 1
Now suppose the result is for all natural numbers less than n.
By theorem 3.1,
d (Cn2 , n 3) d (Cn21 , n 4) d (Cn22 , n 4) d (Cn23 , n 4) d (Cn24 , n 4) d (Cn25 , n 4)
(n 1)(n 2)(n 3) (n 2)(n 3)
=
(n 3) 1
6
2
1
= (n 1)(n 2)(n 3) 3(n 2)(n 3) 6(n 3) 6
6
1
=
(n 1)(n 2)(n 3) 3(n 2)(n 3) 6(n 3 1)
6
1
= (n 1)(n 2)(n 3) 3(n 2)(n 3) 6(n 2)
6
1
= (n 1)(n 2)(n 3) 3(n 2)(n 3 2) .
6
1
= (n 1)(n 2)(n 3) 3(n 2)(n 1)
6
1
= (n 1)(n 2)(n 3 3)
6
1
= n(n 1)(n 2)
6
vi)
Let n = 5
By induction on n
2
L.H.S = d (C5 ,1) = 5 (from table)
n(n 1)(n 2)(n 3)
1.2.3.4
5.4.3.2
=
1.2.3.4
R.H.S =
=5
Therefore the result is true for n =1
Now suppose that the result is true for all natural numbers less than or equal to n.
d (Cn2 , n 4) d (Cn21 , n 5) d (Cn22 , n 5) d (Cn23 , n 5) d (Cn24 , n 5) d (Cn25 , n 5)
n(n 1)(n 2)(n 3)(n 4) (n 2)(n 3)(n 4) (n 3)(n 4)
=
(n 4) 1
24
6
2
1
(n 1)(n 2)(n 3)(n 4) 4(n 2)(n 3)(n 4) 12(n 3)(n 4) 24(n 4) 24
=
24
1
(n 1)( n 2)( n 3)( n 4) 4( n 2)( n 3)( n 4) 12( n 3)( n 4) 24( n 4 1)
=
24
1
(n 1)(n 2)(n 3)(n 4) 4(n 2)(n 3)(n 4) 12(n 3)(n 4) 24(n 3)
=
24
1
(n 1)(n 2)(n 3)(n 4) 4(n 2)(n 3)(n 4) 12(n 3)(n 4 2)
=
24
1
(n 1)(n 2)(n 3)(n 4) 4(n 2)(n 3)(n 4) 12(n 3)(n 2)
=
24
1
(n 1)(n 2)(n 3)(n 4) 4(n 2)(n 3)(n 4 3)
=
24
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12 | Page
Dominating Sets And Domination Polynomials Of Square Of Cycles
1
(n 1)(n 2)(n 3)(n 4) 4(n 2)(n 3)(n 1)
24
1
=
(n 1)(n 2)(n 3)(n 4 4)
24
1
=
n(n 1)(n 2)(n 3)
24
=
Theorem 3.3
d Ci2 , n = 5i 2 d Ci2 , n 1
i)
ii)
For every j ≥ n , d Cn1 , j 1
iii)
If Sn
5n
i n
5 n 5
5
n
n
i
5
2
– d Cn2 , j 1 d Cn2 , j – d Cn25 , j
d Cn2 , j , then for every n ≥ 6, Sn= Sn-1+ Sn-2+ Sn-3+ Sn-4+ Sn-5 with initial values
S1=1, S2=3, S3=7,S4=15 and S5 = 31
Proof
i)
Proof by induction on n
First suppose that n =2 then,
10
i 2
d (Ci2 , 2) 75 5 i 2 d Ci2 ,1
5
d (Ci2 , k ) i k d Ci21 , k 1 i k d Ci22 , k 1 i k d Ci23 , k 1 i k d Ci24 , k 1
i k
5k
5k
5k
5k
5k
i k d Ci25 , k 1
5k
5i k 1 d Ci21 , k 2 5 i k 1 d Ci22 , k 2 5i k 1 d Ci23 , k 2 5i k 1 d Ci24 , k 2
5( k 1)
5( k 1)
5( k 1)
5( k 1)
5i k 1 d Ci25 , k 2
5( k 1)
=5
5 k 5
i k 1
d Ci2 , k 1
We have the result.
ii)
By Theorem 3.1, We have
d (Cn21 , j 1) d (Cn2 , j 1)
=
d (Cn2 , j ) d (Cn21 , j ) d (Cn22 , j ) d (Cn23 , j ) d (Cn24 , j ) d (Cn21 , j )
d (Cn2 , j ) d (Cn21 , j ) d (Cn22 , j ) d (Cn23 , j ) d (Cn24 , j ) d (Cn21 , j )
d (Cn21 , j ) d (Cn22 , j ) d (Cn23 , j ) d (Cn24 , j ) d (Cn25 , j )
d (Cn21 , j 1) d (Cn2 , j 1) d (Cn2 , j ) d (Cn25 , j )
Therefore we have the result
iii)
By theorem 3.1, we have
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Dominating Sets And Domination Polynomials Of Square Of Cycles
Sn i n d Cn2 , j
n
5
Sn i n d Cn21 , j 1 d (Cn22 , j 1) d (Cn23 , j 1) d (Cn24 , j 1) d (Cn25 , j 1)
n
+
5
n 1
n
i 1
5
n 1
n
i 1
5
d Cn21 , j 1 i n 1 d Cn22 , j 1 i n 1 d C n23 , j 1
n 1
n 1
5
5
d Cn24 , j 1 i n 1 d Cn25 , j 1
n 1
5
Sn = Sn-1 + Sn-2 + Sn-3 + Sn-4 + Sn-5
References
[1].
[2].
[3].
[4].
[5].
[6].
S.Alikhani and Y.H.Peng, Introduction to domination polynomial of a graph. arXiv:0905.2251v1[math.CO] 14 May 2009.
S.Alikhani and Y.H.Peng, 2009, Domination sets and Domination Polynomials of paths, International journal of Mathematics and
Mathematical Sciences. Article ID 542040.
G.Chartand and P.Zhang, Introduction to Graph Theory, McGraw-Hill, Boston, Mass, USA, 2005.
T.W.Haynes ,S.T.hedetniemi,and P.J.Slater, Fundamental of Domination in graphs,vol.208 of Monographs and Textbooks in Pure
and Applied Mathematics, Marcel Dekker, New York,NY,USA,1998.
S.Alikhani and Y.H.Peng, Domination sets and Domination Polynomials of cycles, arXiv: 0905.3268v [math.CO] 20 May 2009.
A.Vijayan and K.Lal Gipson, Domination sets and Domination Polynomials of Square paths, accepted in “Open Access journal of
Discrete Mathematics.” – USA.
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14 | Page
IOSR Journal of Mathematics (IOSR-JM)
ISSN: 2278-5728. Volume 3, Issue 4 (Sep-Oct. 2012), PP 15-18
www.iosrjournals.org
One Dimensional Coupled Thermoelastic Problem Due To
Periodic Heating In A Semi-Infinite Rod.
1
Dr. Ashoke Das, 2Dr. Bimalendu Das,
1
Asst. Prof., Raiganj College(University College), Raiganj, U/Dinajpur.W.B.India.
2
Retired Professor, North Bengal University, Siliguri, 734430.W.B. India
Abstract: This paper is concerned with the determination of the distribution of temperature and displacement
in a thin semi-infinite elastic rod when its free end is subjected to periodic heating. It has been pointed out by
P.Chadwick(1960) that the rigorous approach,i.c. the approach by way of the coupled equations, to the thermal
boundary value problem. In this paper the one dimensional problem of the periodic heating of the free surface
of a semi-infinite rod has been solved by a perturbation procedure, approximations upto the first order being
retained.
I.
Introduction
Approach of solving the coupled thermoelastic boundary value problems meets with severe analytical
difficulties till 1960, only one partial solution(8) had been published. It has been possible to study the behavior
of transient solutions, only for very small or very large values of the tome. In view of this, Lessen .M.(1956)(9)
Suggested a procedure to study the approximate solutions of the coupled thermoelastic equations. This one
dimensional coupled problem of the periodic heating of the free surface of a semi- infinite rod has been solved
by perturbation procedure. Approximations upto the first order being retained.
II.
Method Of Solution.
Coupled thermoelastic equations in one dimensional case(10) is taken in the form
𝜕2 𝜃
𝜕𝜃
𝜕2 𝑢
=
𝑓
+
𝑔
𝜕𝑥 2
𝜕𝑡
𝜕𝑥𝜕𝑡
𝜕2 𝑢
𝜕𝜃
−𝑏
2
𝜕𝑥
𝛿𝑥
When
𝑎 =
1
𝑢𝜏
2
,
𝑓 =
= 𝑎
𝜌𝑐 𝑙 2
𝑘𝜏
1
𝜕2 𝑢
𝜕𝑡 2
, 𝑔 =
𝛼𝐸 𝑙 2
𝑘𝜏
,
𝑏 = 𝛼𝑇
𝐿 𝑏𝑒𝑖𝑛𝑔 𝑡𝑒 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑙𝑒𝑛𝑔𝑡, 𝜏 𝑡𝑒 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑡𝑖𝑚𝑒 , 𝑇 𝑡𝑒 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒,
𝑐 𝑏𝑒𝑖𝑛𝑔 𝑡𝑒 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 , 𝐸 , 𝑡𝑒 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑛𝑑 𝑉 =
𝐸
𝜌
, 𝑢, 𝑡, 𝑥, 𝜃 𝑎𝑣𝑒 𝑢𝑠𝑢𝑎𝑙 𝑚𝑒𝑎𝑛𝑖𝑛𝑔.
𝐿𝑒𝑡 𝜃 𝑥, 𝑡
= 𝜃0 𝑥, 𝑡 + 𝑏𝜃1 𝑥, 𝑡 + 𝑏 2 𝜃2 𝑥, 𝑡 + … … … … … … … … … … …
𝑢 𝑥, 𝑡 = 𝑢0 𝑥, 𝑡 + 𝑏 𝑢1 𝑥, 𝑡 + 𝑏 2 𝑢2 𝑥, 𝑡 + … … … … … … … … … … … … ..
Making use of (2) into (1) and equating to zero, the different powers of b,
We get
𝜕 2 𝜃0
𝜕𝜃0
𝜕 2 𝑢0
=
𝑓
+
𝑔
𝜕𝑥 2
𝜕𝑡
𝜕𝑥𝜕𝑡
𝜕 2 𝑢0
𝜕 2 𝑢0
= 𝑎
𝜕𝑥2 2
𝜕𝑡 2
𝜕 𝜗1
𝜕𝜗1
𝜕 2𝑢1
= 𝑓
+ 𝑔
2
𝜕𝑥
𝜕𝑡
𝜕𝑥𝜕𝑡
𝜕 2 𝑢𝑖
𝜕𝜃𝑖−1
𝜕 2 𝑢𝑖
−
𝑏
=
𝛼
,
𝑖 = 1,2,3, … … … . .
𝜕𝑥 2
𝜕𝑥
𝜕𝑡 2
.
(2)
(3)
(4)
The boundary and initial conditions are
𝜃 0, 𝑡 = 𝜑0 sin 𝑛𝑡 𝐻 𝑡
𝑈,𝑥 0, 𝑡 = 0
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15 | Page
One Dimensional Coupled Thermoelastic Problem Due To Periodic Heating In A Semi-Infinite Rod.
𝑢 𝑥, 0 = 𝜃 𝑥. 0 = 𝑢,𝑡 𝑥, 0 = 0
U(x ,t) → 0 𝑎𝑠 𝑥 → ∞
𝜃 𝑥, 𝑡 → 0 𝑎𝑠 𝑥 →
These are satisfied if 𝜃𝑖 𝑥, 𝑡 , 𝑢𝑖 𝑥, 𝑡 , 𝑖 =
𝜃0 0, 𝑡 = 𝜑0 sin 𝑛𝑡 𝐻
𝜃𝑖 0, 𝑡 = 0
𝑢𝑖,𝑥 0, 𝑡 = 𝑢𝑖,𝑡 𝑥, 0
𝑢𝑖 𝑥, 𝑡 → 0 , 𝜃𝑖
(5)
∞
0,1,2, … … … … . . 𝑠𝑎𝑡𝑖𝑠𝑓𝑦 𝑡𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑡
= 𝑢𝑖, 𝑥, 0 = 𝜃𝑖 𝑥, 0 = 0
𝑥, 𝑡 → 0 𝑎𝑠 𝑥 → ∞ 𝑖 = 1,2,3
The terms
𝜃0 𝑥, 𝑡 𝑎𝑛𝑑 𝑢0 𝑥, 𝑡 𝑖𝑛 𝑡𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 2 𝑓𝑜𝑟𝑚 𝑡𝑒 𝑠𝑜 − 𝑐𝑎𝑙𝑙𝑒𝑑 𝑢𝑛𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚
𝑤𝑖𝑙𝑒 𝜃𝑖 𝑥, 𝑡 , 𝑢𝑖 𝑥, 𝑡 𝑔𝑖𝑣𝑒 𝑡𝑒 𝑖𝑡 𝑜𝑟𝑑𝑒𝑟 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑓 𝑏 𝑖𝑠 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑡𝑒 𝑝𝑒𝑟𝑡𝑢𝑟𝑏𝑎𝑡𝑖𝑜𝑛
Parameters. From the table(11) values of b for Al, Pb, Fe, and Cu are seen to be less than 0.01 in each case.
This shows that the solution containing terms upto the first power of b will give a fairly accurate approximate
results to the exact solution.
III.
Unperturbed Solution.
The unperturbed solutions 𝑢0 𝑥, 𝑡 , 𝜃0 (𝑥, 𝑡) are the solution of
𝜕 2 𝑢0
𝜕 2 𝑢0
= 𝑎
2
𝜕𝑥
𝜕𝑡 2
And
𝜕 2 𝜃0
𝜕𝜃0
𝜕 2 𝑢0
= 𝑓
+ 𝑔
2
𝜕𝑥
𝜕𝑡
𝜕𝑥𝜕𝑡
Respectively, Subject to the initial and boundary conditions
𝑢0,𝑥 0, 𝑡 = 𝑢0 𝑥, 0 = 𝑢0,𝑡 𝑥, 0 = 𝜃0 𝑥, 0 = 0
𝜃0 0, 𝑡 = 𝜑 sin 𝑛𝑡 𝐻 𝑡
𝑢0 𝑥, 𝑡 → 0,
𝜃0 𝑥, 𝑡 → 0,
𝑎𝑠 𝑥 → ∞
(12)
Taking Laplace transforms
on these equations and on the boundary conditions give with the help of the
other conditions
𝑑 2𝑢0
𝑑𝑥2
= 𝑎𝑝2 𝑢0 ;
𝑢0,𝑥 0, 𝑝 = 0 … … … … … … … … … … ….
𝑑2 𝜃0
𝑑𝑢0
= 𝑓𝑝𝜃0 + 𝑔𝑝
;
2
𝑑𝑥
𝑑𝑥
Where 𝑢0 , 𝜃0 denote Laplace transform of 𝑢0 , 𝜃0 .
𝜃0 𝑜, 𝑝 =
𝜑0 𝑛
………………
+ 𝑝2
𝑛2
6
(7)
The solution of the differential equation (6), subject to the condition that 𝑢0 𝑥, 𝑝 → 0 𝑎𝑠 𝑥 → ∞, 𝑖𝑠
𝑢0 = 𝑐1 𝑒 − 𝑎𝑝𝑥 … … … … … … … … … … … … … … … … … … … … … … … … ..
8
The boundary condition 𝑢0,𝑥 0, 𝑝 = 0 gives
𝑐1 = 0
(9)
Hence 𝑢0 𝑥, 𝑝 = 0
Taking inverse , 𝑢0 𝑥, 𝑡 = 0
(10)
Substituting from (9) in the differential equation of (7), the solution of the equation with the boundary condition
of (7) is
𝑛𝜑
𝜃0 0, 𝑝 = 𝑛 2 + 0𝑝 2 𝑒 − 𝑓𝑝𝑥
(11)
Taking inverse Laplace transform of (11) (3)
𝜃0 0, 𝑥 =
2𝜑 0
𝜋
∞
𝑥 𝑓
2 𝑡
𝑥 2𝑓
sin 𝑛(𝑡 − 4𝜇 2 ) 𝑒 −𝜇 𝑑𝜇
IV.
(12)
First Order Perturbation.
The first order perturbation is the solution of
𝜕 2𝑢1
𝜕𝑥 2
𝜕 2 𝜃1
–𝑎
𝜕 2𝑢1
𝜕𝑡 2
𝜕𝜃1
= 𝑓
𝜕𝑡
Subject to the boundary conditions
𝜕𝑥 2
=
𝜕𝜃0
𝜕𝑥
+ 𝑔
(13)
𝜕2𝑢1
𝜕𝑥𝜕𝑡
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(14)
16 | Page
One Dimensional Coupled Thermoelastic Problem Due To Periodic Heating In A Semi-Infinite Rod.
𝑢1,𝑥 0, 𝑡 = 𝑢1,𝑡 𝑥, 0 = 𝑢1 𝑥, 0 = 𝜃1 𝑥, 0 = 0
𝑢1 𝑥, 𝑡
→
0 𝑎𝑠 𝑥 → ∞
(15)
𝜃1 0, 𝑡
→ 0 , 𝜃1 𝑥, 𝑡
→ 0 , 𝑎𝑠 𝑥 → ∞
(16)
Laplace transform of the equation (13) and of the boundary conditions give with the help of the other conditions
𝑑 2𝑢1
𝑑𝜃
− 𝑎𝑝2 𝑢1 = 𝑑𝑥0 ;
𝑢1,𝑥 0, 𝑝 = 0
(17)
Substituting from (11) in the differential equation of (17) and solving , we obtain , on using the condition that
𝑢1 𝑥, 𝑝 → 0 𝑎𝑠 𝑥 → ∞,
𝑑𝑥2
𝑢1 = 𝑐 𝑒 −
Where 𝛽0 = 𝑛 𝜑0 𝑓
The boundary condition
𝑎𝑝𝑥
𝑒−
𝑓
𝑝 𝑝−
𝑎
𝛽
𝑎
+
𝑢1,𝑥 0, 𝑝 = 0
𝑓𝑝𝑥
𝑝2 + 𝑛 2
gives
𝛽0
𝑓
𝑎𝑓 𝑝− (𝑝 2 + 𝑛 2 )
𝑎
𝑐 =
(19)
Using the results (5,9) , the inverse Laplace transform of (18) is obtained as
𝑢1 𝑥, 𝑡 =
𝑓
𝛽 0 𝐻 𝑡− 𝑎𝑥
𝑒𝑎
𝑓2
𝑎𝑓 𝑛 2 + 2
𝑎
𝛽0
1
𝑥 2𝑓
𝑒 −𝜇
4𝜇 2
𝜇2
𝑓𝑡
𝑎
𝑒𝑎
2
𝑓2
𝑎 𝑛 2+ 2
𝑎
𝑡− 𝑎𝑥
𝑒
𝑓
3
2
𝑑𝜇 +
−
𝑓𝑥
𝑎
−
𝑓
𝑎𝑛
𝑒𝑟𝑓𝑐
𝑓2
∞
𝑎𝑛 𝜋
𝑥 𝑓
2 𝑡
sin 𝑛 𝑡 −
𝑥
𝑓
2
𝑡
𝑎𝑥 − cos 𝑛 𝑡 −
𝑓𝑡
−
− 𝑒
𝑎
𝑥 2𝑓
sin 𝑛 𝑡 − 4𝜇 2
𝑒 −𝜇
𝑓𝑥
𝑎
𝑒𝑟𝑓𝑐
𝑎𝑥
𝑥
𝑓
2
𝑡
−
+
𝑓𝑡
− 𝑥
𝑎
𝑓
∞
𝜋
𝑥 𝑓
2 𝑡
cos 𝑛 𝑡 −
2
𝜇2
𝑑𝜇
(20)
On taking Laplace transforms of the equation (14) and of the boundary condition (16), we obtain with the help
of (18) and (19)
𝑑2 𝜃1
𝑔𝛽0
𝑝2 𝑒 − 𝑎𝑝𝑥
𝑔𝛽0 𝑓
−
𝑓𝑝𝜃
=
−
1
𝑓
𝑑𝑥 2
𝑓 𝑝−
𝑎
2
2
𝑎 (𝑝 + 𝑛 )
𝑝𝑒 − 𝑓𝑝𝑥
𝑓
𝑝 − 𝑎 (𝑝2 + 𝑛2 )
And
𝜃1 0, 𝑝 = 0 , 𝜃1 𝑥, 𝑝 → 0 𝑎𝑠 𝑥 → ∞
The solution of the above equation subject to these boundary conditions is
𝜃1 𝑥, 𝑝 =
𝑔𝛽 0
𝑎 𝑓
𝑝𝑒 − 𝑓𝑝𝑥
𝑓 2 2
𝑝−
(𝑝 + 𝑛 2 )
𝑎
−
𝑔𝛽 0 𝑥
𝑝𝑒 − 𝑓𝑝𝑥
2 𝑎
𝑓
𝑝− (𝑝 2 +𝑛 2 )
𝑎
−
𝑝𝑒 − 𝑎 𝑝𝑥
𝑔𝛽 0
𝑎 𝑓
𝑝−
(21)
𝑓 2 2
(𝑝 +𝑛 2 )
𝑎
Taking inverse Laplace transform, we obtain from (21),
𝜃1 𝑥, 𝑡 =
𝑡
𝛽0 𝑔
2𝑎 𝑓
𝑓𝑢
cos 𝑛 𝑡 − 𝑢 𝑒 𝑎
0
+
−
−
Where H(t)
𝑢−
𝑎𝑥 𝑓𝑥
𝑥 𝑓
)𝑒 𝑎 𝑒𝑟𝑓𝑐
+
2
2 𝑢
𝑥𝑔𝛽0
4 𝑎
𝑓𝑢
𝑎
𝑡
𝑓𝑢
cos 𝑛 𝑡 − 𝑢 𝑒 𝑎
0
𝑓𝑥
𝑒 𝑎 𝑒𝑟𝑓𝑐
𝑥 𝑓
+
2 𝑢
𝑎𝑥 −
𝑒
2
𝑓𝑢
𝑎
𝑓𝑥
𝑎
𝑒𝑟𝑓𝑐
𝑥 𝑓
−
2 𝑢
𝑓𝑢
+ (𝑢
𝑎
𝑒𝑟𝑓𝑐
𝑥 𝑓
−
2 𝑢
𝑓𝑢
𝑎
𝑑𝑢
𝑎 −
𝑒
𝑓
𝑑𝑢 −
𝑓𝑥
𝑎
𝛽0 𝑔𝐻(𝑡 − 𝑎𝑥)
𝑎 𝑓
= 0, t < 0
= 1, t > 0
𝑡− 𝑎𝑥
𝑓𝑢
cos 𝑛 𝑡 − 𝑢 − 𝑎𝑥 𝑢𝑒 𝑎 𝑑𝑢
0
(22)
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17 | Page
One Dimensional Coupled Thermoelastic Problem Due To Periodic Heating In A Semi-Infinite Rod.
V.
Discussion.
It is observed that a factor of the form 𝑒
perturbation solution is valid only so long as 𝑏𝑒
𝑓
(𝑡− 𝑎𝑥 )
𝑎
𝑓
(𝑡− 𝑎𝑥)
𝑎
is present in
𝑢1 (𝑥, 𝑡) and in 𝜃1 (𝑥, 𝑡) . Hence the
< 1.
This appears to be a characteristic of the one- dimensional problem.
The equations (10) and (12) and (20) and (22), where
𝑢 𝑥, 𝑡 = 𝑢0 𝑥, 𝑡 + 𝑏 𝑢1 𝑥, 𝑡
𝜃 𝑥, 𝑡 = 𝜃0 𝑥, 𝑡 + 𝑏 𝜃1 𝑥, 𝑡
Will give the required approximate solution for the displacement and temperature. Also if g = 0, we obtain the
solution of the uncoupled equations of thermoelasticity.
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
ATKINSON,K.E.(1976); A survey of Numerical Methods for the Solution of Fredholm- Integral Equations of the Second – Kind.
Society of Industial and applied Mathematics, Philadelphia,Pa.
CARSLAW,H.S. AND JAEGER, J.C. (1959); Conduction of heat in Solids, 2 nd Edn. O.U.P.
SNEDDON, I.N.(1972); The Use of Integral Transform, McGraw Hill, New-York.
SOKOLNIKOFF,I.S.(1956); Mathematical theory of Elasticity, McGraw Hill Book Co.
NOWACKI, W.(1986); Thermoelasticity, 2 nd End. Pergamon Press.
WASTON,G.N. (1978) ATreatise on the Theory of Bessel Functions, 2 nd Edn. C.U.P.
LOVE,A.E.H.(1927) A Treatise on the Mathematical Theory of Elasticity, 4 th Edn. Dover Publication.
PARIA.G. (1968): Instantaneous heat sources in an infinite solid, India, J.Mech.Math.(spl. Issue), partI,41.
LESSEN. M.(1968) ; j.Mech.Phys.solids.5,p.
SNEDDON,I.N.(1958); Prog.Roy.Soc.Edin.1959,pp 121-142.
INTERNATIONAL Critical Table of Numerical data.(1927); Vol-II, N.R.C.,U.S.A., McGraw Hill
Book Co.
ERDELYI. A.(1954); Tables of Integral Transform,Vol-II, McGraw Hill Book Co. Inc.N.Y.
www.iosrjournals.org
18 | Page
IOSR Journal of Mathematics (IOSR-JM)
ISSN: 2278-5728. Volume 3, Issue 4 (Sep-Oct. 2012), PP 19-23
www.iosrjournals.org
Two Fixed Point Theorems in Topological vector space Valued
Cone Metric Spaces with Complete Topological Algebra Cones
Tadesse Bekeshie 1, G.A Naidu 2 and K.P.R Sastry 3
1,2
(Department of Mathematics Andhra University, Visakhapatnam-530 003, India,
3
(8-28-8/1, Tamil Street, Chinna Waltair, Visakhapatnam-530 017, India,
Abstract: In this paper we prove two fixed point theorems in topological vector space valued cone metric
spaces (briefly TVS-CMS). To that end we introduce the concept of complete topological algebra cone (briefly
CTA cone). Our theorems are generalizations of corresponding theorems in [8] and [13]. The paper also gives
answers to the open problems posed in [15]. Finally we give an application of our first theorem.
AMS Mathematics Subject Classification (2010): 47H10, 54H25, 46A40
Key terms: topological vector space, ordered topological vector space, algebra over a field, topological
algebra over a field, topological vector space valued cone metric space, scalarization function, CTA cone
I.
Introduction
The famous Banach Contraction Principle states that every contraction in a complete metric space has a
unique fixed point. It has two core hypotheses: completeness and contractivity. Both notions depend on the
definition of the underlying metric. Much recent work has focused on the extension of the notion of metric
spaces and the related notion of contractivity. One extension of metric spaces is the so called cone metric space.
In cone metric spaces, the metric is no longer a positive number but a vector, in general an element of a Banach
space equipped with a cone. An overview of a historical background of metric spaces and cone metric spaces is
given in the next few paragraphs.
In 1906, the French mathematician Maurice Frechet [4, 6] introduced the concept of metric spaces,
although the name “metric” is due to Hausdorff[4, 7].
In 1934, the Serbian mathematician Duro Kurepa[10], a PhD student of Frechet, introduced metric
spaces in which an ordered vector space is used as the codomain of a metric instead of the set of real numbers.
In the literature the metric spaces with vector valued metrics are known under various names such as
pseudometric spaces, k-metric spaces, generalized metric spaces, cone-valued metric spaces, cone metric spaces,
abstract metric spaces and vector valued metric spaces. Fixed point theory in K-metric spaces was developed by
A.I.Perov in 1964[11, 12]. For more details on fixed point theory in K-metric and K-normed spaces, we refer the
reader to [18].
In 2007, Huang and Zhang [8] reintroduced such metric spaces under the name cone metric spaces
without mentioning the previous works. But they went further, defining convergent and Cauchy sequences in
terms of interior points of the underlying cone. They also proved some fixed point theorems in such spaces in
the same work. Subsequently, several authors have studied fixed point theory of cone metric spaces.
In 2009 I.Beg et al [3] and in 2010 Du [5] generalized cone metric spaces to topological vector space
valued cone metric spaces (TVS-CMS). In this approach ordered topological vector spaces are used as the
codomain of the metric, instead of Banach spaces. While I.Beg et al used Hausdorff TVS; Du used locally
convex Hausdorff TVS. However, a result in [16] shows that if the underlying cone of an ordered TVS is solid
and normal it must be an ordered normed space. So, proper generalizations from Banach space valued cone
metric spaces to TVS-CMS can be obtained only in the case of nonnormal cones.
Several authors had showed that some fixed point theorems in usual metric spaces and their TVS-CMS
counterparts are equivalent. See [2], [3] and [5]. One of the tools used to prove such equivalences is the so
called nonlinear scalarization function. This method was used earlier in optimization theory to convert
optimization problems in vectorial forms to their scalar forms. This function was introduced into fixed point
theory by Du [5].
Our objectives are as follows. We introduce the notion of complete topological algebra (CTA) cone.
Then we prove Banach’s fixed point theorem and Kannan’s fixed point theorem in TVS-CMS with CTA cones.
The contraction constants used are taken to be elements of the CTA cone, instead of being scalars. The results
obtained generalize those in [8] and [13] hence the classical Banach’s and Kannan’s fixed point theorems. More
over our theorems answer the open problems raised in [15].
In line with the methods used in [2] and [5], we used the nonlinear scalarization method of Du to prove
our theorems. This means that we converted our theorem statements (which are given in TVS-CMS (X, d)) to
equivalent statements in the metric space (X, de) where the metric de is defined by de: = 𝜉𝑒 o d while
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19 | Page
Two Fixed Point Theorems in Topological vector space Valued Cone Metric Spaces with Complete
𝜉𝑒 ∶ E → ℝ 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝜉𝑒 (y): = inf {t ∈ ℝ| 𝑦 ≤ 𝑡𝑒 }, e being any element in int P. After converting the
statements from “vectorial” form (TVS-CMS) to equivalent statements in “scalar” form (metric spaces(X, d e)),
then we used known results available in the literature of metric fixed point theory to complete the proofs. We
also used a new property of the function 𝜉𝑒 (See Lemma 3.2) in our arguments.
The paper is organized as follows. In section 2, we consider the notations, notions and known results
which we use later in sections 3, 4 and 5. In section 3, we present our main results. The fourth section presents
an application and the fifth a conclusion.
II.
Preliminaries
Definition 2.1[1, 17]: A vector space E over a field K (ℝ or ℂ) is said to be a topological vector space (TVS)
over K if it is furnished with a topology 𝜏 such that the vector space operations are continuous, i.e.,
i) The addition operation (x, y)→ 𝑥 + 𝑦 as a function from E × E to E is continuous.
ii) The scalar multiplication operation (a, x)→ 𝑎 . 𝑥 as a function from K×E to E is continuous.
In this case one says that 𝜏 is a vector topology or a linear topology on the vector space E, or that 𝜏 is
compatible with the linear structure of E.
Here, ℝ and ℂ are equipped with their usual topologies where as E×E and K×E are endowed with their
respective product topologies.
Caution: Some authors (e.g., Rudin [14]) require the topology on a TVS E to be Hausdorff, an approach which
we do not follow in this manuscript.
Definition 2.2 [3, 5]: Let (E,𝜏) be a topological vector space. A subset P of E is called a cone if:
i) P is closed, nonempty and nontrivial (i.e., P ≠ {0});
ii) ax + by ∈ P for all x, y ∈ 𝑃 and nonnegative real numbers a and b and
iii) P ∩ (-P) = {0}.
If, in addition, the interior of P is nonempty, we say that P is a solid cone.
Definition 2.3 [3, 5]: Let (E,𝜏) be a topological vector space and P ⊆ E be a cone.
We define a partial ordering ≤ 𝑜𝑛 𝐸 with respect to P by x ≤y if and only if y-x ∈ P. We write x < y if x ≤ y and
x ≠ y. Likewise, we write 𝑥 ≪ y if y – x ∈ 𝑖𝑛𝑡 P, where 𝑖𝑛𝑡 P denotes the interior of P. If ambiguity is possible
we can use the notations≤𝑃 , <𝑃 and≪𝑃 . The pair (E, P) consisting of a TVS E and a solid cone P of E is called
a partially ordered TVS.
In [2], it is proved that the existence of a solid cone in a TVS ensures that the vector topology is
Hausdorff. In [1], it is mentioned that any TVS possessing a closed cone is Hausdorff. In view of [1] and from
the assumption that P is both solid and closed, we observe that the proof in [2] is an extra effort.
Throughout this paper (E, P) denotes a partially ordered real TVS E with a solid cone P.
Definition 2.4 [3, 5] Let X be a nonempty set. A mapping d: X×X →E satisfying
(d1) d(x, y) ≥ 0;
(d2) d(x, y) = 0 ⇔ x = y;
(d3) d(x, y) = d(y, x) and
(d4) d(x, y) ≤ d(x, z) + d (z, y) for all x, y, z ∈ X,
is called a TVS- valued cone metric on X. The pair (X, d) is called a TVS-valued cone metric space (written
briefly as TVS-CMS).
Definition 2.5 [3, 5] Let (X, d) be a TVS-cone metric space, x ∈X and {x n}, n=1,2,…be a sequence in X. Then
we say
i) {x n} TVS-converges to x if for every c ∈ int P there exists a natural number N such that d (x n, x) ≪ c for all
n ≥ N.
ii) {x n} is a TVS-Cauchy sequence if for every c ∈ int P there exists a natural number N such that
d(x n, x
m) ≪ c for all n, 𝑚 ≥ N.
iii) (X, d) is a TVS-complete cone metric space if every Cauchy sequence in (X, d) is convergent in (X, d).
Next we define the term algebra over a field and its concomitants.
Definition 2.6 [17]: An algebra A over a field K is a vector space over K such that for each ordered pair of
elements x, y ∈ A a unique product xy ∈A is defined with the properties
1) (xy)z=x(yz);
2) x(y+z)= xy + xz and (x+y)z=xz+yz and
3) α(xy)= (αx)y= x(αy) for all α ∈K and x, y, z ∈ 𝐴.
The algebra A is called a real algebra if K = ℝ and a complex algebra if K =ℂ.
Caution: In the literature, there are a few other variants of the definition of the term algebra over a field.
Definition 2.7 [17]: A topological algebra A over a field K is a topological vector space together with a
continuous multiplication that makes A an algebra over K.
We denote a topological algebra A with vector topology 𝜏 and multiplication “ ∙ ” by (A,𝜏, ∙) .
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Two Fixed Point Theorems in Topological vector space Valued Cone Metric Spaces with Complete
Definition 2.8 [17]: A complete topological algebra A is a topological algebra which is complete, considered as
a uniform space.
Definition 2.9 [5]: Let e ∈ int P. The formula 𝜉𝑒 (y): = inf {t ∈ ℝ| 𝑦 ≤ 𝑡𝑒}, where y ∈ E, defines a function
from E into R, and is called the nonlinear scalarization function on E (w.r.t P and e).
The following lemma states properties of the nonlinear scalarization function 𝜉𝑒 .
Lemma 2.10 [5, 2]: For any e ∈ int P, the function 𝜉𝑒 has the following properties:
1) 𝜉𝑒 (0) = 0;
2) y ∈ P implies 𝜉𝑒 (y) ≥ 0;
3) If y2 ≤ y1, then 𝜉𝑒 (y2) ≤ 𝜉𝑒 (y1) for any y1, y2 ∈ E;
4) 𝜉𝑒 is subadditive on E, i.e. 𝜉 e (x + y) ≤ 𝜉 e (x ) + 𝜉 e (y) for all x, y ∈ E;
5) 𝜉𝑒 is positively homogeneous on E , i.e., 𝜉 e ( βx ) = β 𝜉 e ( x ) for every x ∈ E and positive real number β, and
6) 𝜉𝑒 is continuous on E.
Lemma 2.11[5, 2]: Let (X, 𝑑) be a TVS-CMS over an ordered TVS (E, P) and e ∈ int P. Then
a) the function de :E×E→[0, ∞) defined by de = 𝜉𝑒 o d is a metric.
b) (X, d) is TVS-complete if and only if (X, de) is complete.
III.
Main Results
We start with introducing the notion of CTA cone.
Definition 3.1: Let (E,𝜏,∙ ) be a real complete topological algebra with unity e. A subset P of E is called a
complete topological algebra cone (briefly CTA cone) if P is a cone of E with the additional properties that:
(i) e ∈ P and (ii) x, y ∈ P implies xy ∈ P
(3.1.1)
Lemma 3.2: Let (X, d) be a complete TVS-CMS with unity e and P ⊆ E be a CTA cone. Then the nonlinear
scalarization function 𝜉𝑒 satisfies the following property:
𝜉𝑒 (xy) ≤ 𝜉𝑒 (𝑥) 𝜉𝑒 (y) for all x,y ∈ P.
(3.2.1)
Proof: By definition, x ≤ 𝜉𝑒 (x) e. This implies 𝜉𝑒 (x) e – x ∈ P. It follows from (3.1.1(ii)) and from the fact that
e is the unity of E that 𝜉𝑒 (x) y – xy ∈ P. This in turn implies xy ≤ 𝜉𝑒 (𝑥) y. On the other hand, 𝜉𝑒 (𝑥) y ≤ 𝜉𝑒 (𝑥)
𝜉𝑒 (y) e. Using the transitivity property of the partial ordering ≤ we get xy ≤ 𝜉𝑒 (𝑥) 𝜉𝑒 (y) e. Hence, the result
follows. ∎
The following theorem is an extension of the cone metric space version of Banach’s Fixed point Theorem (i.e.,
Theorem 1 of [8] and Theorem 2.3 of [13]).
Theorem 3.3: Let (X, d) be a complete TVS-CMS and P ⊆ E be a complete topological algebra cone. Let T be a
self mapping of X satisfying
d (Tx, Ty) ≤ k d(x, y) ,
(3.3.1)
for some k ∈ P with 0 ≤ k ≪ e and for all x, y ∈ X. Then T has a unique fixed point.
Proof: From (3.2.1), Lemma 2.10(3) and Lemma 3.2 we get
d (Tx, Ty) ≤ k d(x, y) ⇒ 𝜉𝑒 𝑑 𝑇𝑥, 𝑇𝑦 y ≤ 𝜉𝑒 𝑘𝑑 𝑥, 𝑦
⇒ de(Tx, Ty) ≤ 𝜉𝑒 (𝑘) de(x, y)
(3.3.2)
We show that 0≤ 𝜉𝑒 𝑘 < 1. From the assumption 0 ≤ k ≪ e and Lemma 2.10(3) it follows that 0 ≤ 𝜉𝑒 𝑘 ≤ 1.
Suppose 𝜉𝑒 𝑘 = 1. Since e- k 𝜖 int P, so there exists an open set U such that e- k 𝜖 U ⊆int P. Since k –e + U is
−1
a neighborhood of 0 and the sequence 𝑒 converges to 0, so –β e ∈ k – e + U for some β > 0. Hence,k ≤ (1 – β)
𝑛
e. This is a contradiction to the fact that 𝜉𝑒 𝑘 =1. By Lemma 2.11(b) the metric space (X, de) is complete and
hence using (3.2.2) we can conclude that X has a unique fixed point. ∎
Remark 3.4: If the cones used in [8], Theorem 1 and [13], Theorem 2.3 are CTA cones (so that E is a complete
topological algebra), then these theorems are easy consequences of our Theorem 3.3 above. This can be justified
as follows. Assume that the hypotheses of Theorem 2.3, [13] are true, i.e, let (X, d) be a complete cone metric
space with cone P ⊆ E. Let T be a self mapping of X satisfying
d(Tx, Ty) ≤ c d(x, y) ,
for some real number c such that 0 ≤ c < 1 and for all x, y ∈ X. Put k:= ce, where e is the unity of the T.V.S
algebra E. Clearly 0 ≤ k ≪ e and d(Tx, Ty) ≤ kd(x, y). Thus the hypotheses of Theorem 2.3, [13] imply the
hypotheses of Theorem 3.3 above. The rest is trivial.
Our next theorem is an extension of the cone metric space version of one of the Kannan type Fixed Point
theorems [Theorem 2 [8] and Theorem 2.6 [13]].
Theorem 3.5 Let (X, d) be a complete TVS-CMS and P⊆ E be a complete topological algebra cone. Let T be a
self mapping of X satisfying
d (Tx, Ty) ≤ k [ d (Tx, x) + d(Ty, y) ] ,
(3.5.1)
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Two Fixed Point Theorems in Topological vector space Valued Cone Metric Spaces with Complete
1
for some k ∈ P; where 0 ≤ k ≪ 2 𝑒 for all x, y ∈ X. Then T has a unique fixed point.
Proof: Using (3.5.1), Lemma 2.10 (3 and 4) and Lemma (3.2) we obtain
d (Tx, Ty) ≤ k [ d (Tx, x) + d(Ty, y) ] ⇒ 𝜉𝑒 𝑑 𝑇𝑥, 𝑇𝑦 y ≤ 𝜉𝑒 𝑘[𝑑 𝑇𝑥, 𝑥 + 𝑑 𝑇𝑦, 𝑦 ]
⇒ de(Tx, Ty) ≤ 𝜉𝑒 (𝑘) [𝑑𝑒 𝑇𝑥, 𝑥 + 𝑑𝑒 𝑇𝑦, 𝑦 ]
1
1
1
1
By assumption we have 0 ≤ k ≪ 𝑒. This implies 0 ≤ 𝜉𝑒 𝑘 ≤ . Suppose 𝜉𝑒 𝑘 = . Since 𝑒 − 𝑘 ∈ int P, so
2
2
1
2
2
there exists an open set U such that 𝑒 − 𝑘 ∈ U ⊆ P. Since k – ½ e+U is a neighbourhood of 0 and
2
−1
1
1
2
2
𝑛
𝑒
convergs to 0, so there exists β > 0 such that -βe ∈W. This implies that ( – β)e – k ∈ U ⊆ P. Hence k ≤ ( – β
1
)e . This contradicts our supposition. Hence 0 ≤ 𝜉𝑒 𝑘 < . Since the TVS-CMS (X, d) is complete, so is the
2
metric space (X, de) (by Lemma 2.11(b)). The rest follows from Kannan’s Fixed Point Theorem. ∎
Remark 3.6 Theorem 3.5 is a generalization of the corresponding theorems in [8] and [13] provided that the
cones used there in are CTA cones. The justification is analogous to that of Remark 3.4.
Remark 3.7: Theorems 3.3 and 3.5 give positive answers to one of the open problems raised in [15]. We insist
1
1
that the conditions k < e and k < e cannot replace the corresponding conditions, 𝑘 < 1 and 𝑘 < , used in
2
2
[15] in view of the following counter example.
Example 3.8: Let X: = C [0, 1] (the real vector space of all continuous real valued functions on [0, 1] equipped
with the supremum norm). Then X is a real Banach space and P = {f ∈ 𝑋|0 ≤ 𝑓} is a cone of X. See [14].
Moreover, int P = {f ∈ 𝑋|0 < 𝑓 𝑥 for all 𝑥 ∈ [0,1]} is nonempty. Thus P is a solid cone.
Define d : X×X → X by d(f, g):= |f - g| ;i.e., d(f, g):[0, 1]→ ℝ given by d(f, g)(x): =|f(x)-g(x)|, x ∈ X. Note that
(X, d) is a complete CMS (hence a complete TVS-CMS) and the map Tf:= kf + e (where k is the identity map on
[0, 1] and ex=1,x ∈X ) satisfies the conditions fe=ef=e for all f in X, 0 ≤ k< e and d( Tf, Tg) ≤ kd(f, g) for all f,
g in X but T has no fixed point. Hence the condition k < e cannot replace the corresponding condition, 𝑘 <
1
1, used in Theorem 3.2[15]. By similar argument we can show that the condition k < 2e cannot replace the
1
corresponding condition, 𝑘 < 2, used in Theorem 3.4[15].
Remark 3.9: In [2] the authors suggested that when contraction constants are taken from P, then fixed point
statements in TVS-CMS may not be convertible to equivalent statements in usual metric spaces. But our results
show that even when contraction constants are in P the equivalences mentioned in [2] can be valid.
IV.
Application
Even though some fixed point statements stated in TVS-CMS can be converted to equivalent statements in
usual metric spaces and vice versa, we suggest that it is still important to investigate fixed point theory of TVSCMS in greater depth. This is because unexpected applications can arise. For instance, it may happen that there
can be cases where a fixed point problem posed in TVS-CMS is solvable with less effort than an equivalent
problem posed in metric spaces or vice versa. We leave this as an open problem. If this happens to be true, then
working in TVS-CMS can give us opportunity to solve more problems.
As an illustration we present an easy application to Theorem 3.3.
𝑠𝑖𝑛𝑡
Example 4.1: Let (X, d) be as described in Example 3.8. Define the operator T by (Tf)(t): = (1+𝑠𝑖𝑛𝑡 )𝑓 𝑡 , 0 ≤
𝑡 ≤ 1. Then T is a self mapping of X. Moreover it is a contraction with respect to d with “contraction constant”
the function k(t)=sint, 0 ≤ t ≤ 1.Therefore T has a unique fixed point.
V.
Conclusion
In this paper and in [15] extensions of Banach’s and Kannan’s fixed point theorems to TVS-CMS (X,
d) are discussed. In [15], (X, d) is a cone metric space. The contraction constants k used in both papers are no
longer scalars but vectors. The results in these two papers can be summarized as follows. When ||k|| < 1 or k ≪ e
( for Banach’s fixed point theorem) and ||k|| < ½ or k ≪ ½ e ( for Kannan’s fixed point theorem), then the
conclusions of the corresponding theorems are valid. But when ||k|| = 1 or 𝑘 < 1 (for Banach’s fixed point
theorem) and ||k|| = ½ or k < ½ e ( for Kannan’s fixed point theorem), then the conclusions of the
corresponding theorems are invalid.
References
[1]
[2]
[3]
[4]
[5]
[6]
C.D. Aliprantis and R. Tourky , Cones and Duality (American Mathematical Society, 2007).
I. D.Arandelovic and D. J. Keckic, TVS-Cone Metric Spaces as a Special case of Metric Spaces, arXiv: 1202.5930v1 [math.FA],
2012.
I. Beg, A. Azam and M. Arshad, Common fixed points for maps on topological vector space valued cone metric spaces, Internat. J.
Math. Math. Sciences, 2009 (2009).
M. M. Deza and E. Deza, Encyclopedia of Distances (Springer-Verlag, 2009), i – x.
W. S.Du, A note on cone metric fixed point theory and its equivalence, Nonlinear Analysis, 72 (5) (2010), 2259-2261.
M. Fréchet, Sur quelques points du calcul fonctionnel. Rendi. Circ. Mat. Palermo, 22(1906), 1- 74.
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Two Fixed Point Theorems in Topological vector space Valued Cone Metric Spaces with Complete
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
[18]
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Company, New York (1949).
L.G. Huang and X.Zhang, Cone Metric Spaces and Fixed Point Theorems of Contractive mappings, J. Math. Anal. Appl. , 332
(2007), 1467 - 1475.
M.C. Joshi and R.K. Bose, Some Topics in Nonlinear Functional Analysis (Wiley Eastern Ltd., New Delhi, 1985).
D.R. Kurepa, Tableaux ramifies d’ensembles. Espaces pseudo-distancies, C. R. Acad. Sci. Paris, 198 (1934), 1563–1565.
A.I. Perov, On Cauchy problem for a system of ordinary differential equations, Pviblizhen. Met. Reshen. Differ. Uvavn., 2(1964),
115-134.
P. D. Proinov, “A unified theory of cone metric spaces and its applications to the fixed point,” In press,
http://arxiv.org/abs/1111.4920.
Sh. Rezapour, R.Hamlbarani, Some notes on the paper: Cone metric spaces and fixed point theorems of contractive mappings, J.
Math. Anal. Appl. , 345 (2008), 719-724.
W. Rudin, Functional Analysis (McGraw-Hill Science, 1991).
K.P.R. Sastry, G.A. Naidu and Tadesse Bekeshie, Fixed Point Theorems in Cone Metric Spaces with Banach Algebra Cones,
IJMSEA, (6)(V)(2012), 129 – 136.
J. S. Vandergraft, “Newton’s method for convex operators in partially ordered spaces,” SIAM Journal on Numerical Analysis, 4
(1967), 406–432.
A. Wilansky, Modern Methods in Topological Vector Spaces (McGraw-Hill, 1978).
P.P.Zabrejko, “K-metric and K-normed linear spaces: survey,” Collectanea Mathematica, 48 (4-6) (1997), 825–859.
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23 | Page
IOSR Journal of Mathematics (IOSR-JM)
ISSN: 2278-5728. Volume 3, Issue 4 (Sep-Oct. 2012), PP 24-30
www.iosrjournals.org
On the generalized bilinear differential equations
1
M. Y. Adamu, 2E. Suleiman
1,2
Mathematical Sciences Programme, Abubakar Tafawa Balewa University, Bauchi, Nigeria.
Abstract: By using the generalized Hirrota bilinear operators a kind of bilinear differential equations is
established and examined when the linear super position principle can apply to the resulting generalized
bilinear differential equations. Examples of generalized bilinear differential equations together with an
algorithm using weights are computed using a 1+1 and 2+1 dimensional equations in order to shed more lights
on the presented general scheme for the construction of the bilinear differential equations which posses linear
subspaces of solutions.
I.
Introduction
Integrable systems and nonlinear evolution equations have attracted much attention of mathematicians
as well as physicist for the last 2 decades. The analysis of exact travelling wave solutions to nonlinear evolution
equations plays a vital role in the study of nonlinear physical phenomena. Single solitons are most beneficial
solutions among travelling wave solutions. The existence of multi-soliton solutions, especially two-soliton and
three-solton is useful in establishing optical communication systems. However, besides solitons , another
attractive set of multi-exponential wave solutions [1] is a linear combination of exponential waves. It was shown
that some of the nonlinear equations can posses such a linear superposition principle[2,3]. Also, special
solutions by combining exponential functions and trigonometric functions were presented and called
complexitons [4].
A variety of powerful methods have been used to study integrable systems and nonlinear evolution
equations, such as Hirota bilinear method [5], the pfaffian technique, the dressing method [6], the inverse
scattering method,the Backlund transformation method [8, 9], the Darboux transformation and the generalized
symmetry method [10]. These approaches possess powerful features that make it possible to create multiple
soliton solutions for a wide range of integrable systems and nonlinear evolution equations.
Many important equations of mathematical physics are rewritten in the Hirota bilinear form through dependent
variable transformations [5]. For example the KdV equation:
u xxx 6uu x ut 0
The Jimbo-Miwa equation
u xxxy 3u y u xx 3u x u xy 2u yt 3u xz 0
which can express as
D
4
x
Dt Dx F .F 0
under the transformation
u 2 ln f xx
( Dx3 Dy 2Dt Dy 3D x Dz ) f . f 0
under the transformation
u 2 ln f xx respectively, where Dx , Dy , and Dt are Hirota bilinear operators
and are generally defined by:
m
D D D f .g ' ' ' g x, y, t f x ' y ' , t ' x x ' , t t '. y y '
x x y y t t
n
n
x
m
t
k
k
y
for nonnegative integers m,n and k
In this paper we would like tofurther confirm the kind of generalized bilinear differential operator and
examine when is the linear superposition principle will be applied to the corresponding bilinear differential
equations that is established in [11]. The resulting theory paves a way to construct a new kind of bilinear
differential equations which possess linear subspaces of solutions. The considered solutions are linear
combinations of exponential travelling wave solutions, and the involved exponential wave solutions may or may
not satisfy the corresponding dispersion relations. All the obtained results will exhibit that there are bilinear
differential equations different from Hirota bilinear equations, which share some common features with the
linear differential equations.
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On the generalized bilinear differential equations
The paper is organized as follows. In section 2, we will visit the generalized Hirota bilinear operators
as introduced [11], and see how we can establish a kind of generalized bilinear differential equations. The
analyzed linear superposition principle for exponential travelling waves and the established criterion for
guaranteeing the existence of linear subspaces of exponential travelling wave solutions to the generalized
bilinear differential equations is employed as in [11].while in section four we will present two examples of
newly introduced bilinear differential equations, together with an algorithm using weights to compute. Finally,
concluding remarks will be drawn in section 5.
II.
Bilinear Differential Operators And Equations
Let M, P N be given,. We introduce a kind of bilinear differential equation :
M
Dpn1j x1 ...DpnMj xM f .g (
i 1
' )ni f ( x1...xm ) g ( x1' ...xM' ) | x' x , x' x
1
M
M
xi
xi
(2.1)
Where n1...nM are arbitrary nonnegative integers and for an integer m the mth power of
is defined by:
m ()r ( m) , if m r (m) mod p, with 0 r (m) p
(2.2)
For example if p=2k (k N), All the above bilinear differential operators are Hirota bilinear operators, since
D2 k , x Dx
(see for example [11]) for the detailed analysis of the algorithm)
Ie if p = 2, then we have
1, 2 1, 3 1, 4 1,...
which gives the pattern of symbols for the Hirota bilinear operators
, , , , , , , ,...( p 2)
And if p = 3, we have
1, 2 1, 3 1, 4 1, 5 1, 6 1,...
which gives the pattern of symbols for the Hirota bilinear operators
, , , , , , , ,...( p 3)
And if p = 5, we have
1, 2 1, 3 1, 4 1, 5 1, 6 1, 7 1, 8 1, 9 1, 10 1,...
which gives the pattern of symbols for the Hirota bilinear operators
, , , , , , , , ,... (p = 5)
And when p = 7, the pattern of symbol is
, , , , , , , , , , , ,...( p 7)
Following those patterns of symbols, some new bilinear differential operators can be computed as:
D3, x f .g f x g fg x
D3, x D3,t f xt g f x gt ft g x fg xt
D3,3 x f .g f xxx g 3 f xx g x 3 f x g xx fg xxx
D3,2 x D3,t f. g f xxt g f xx gt 2 f xt g x 2 f x g xt f t g xx fg xxt
D3,3 x D3,t f .g f xxxt g f xxx gt 3 f xxt g x 3 f xt g xx 3 f xx g xt 3 f x g xxt f t g xxx fg xxxt
D3,4 x f. g f xxxx g 4 f xxx g x 6 f xx g xx 4 f x g xxx fg xxxx
5
D3,
x f .g f xxxxx g 5 f xxxx g x 10 f xxx g xx 10 f xx g xxx 5 f x g xxxx fg xxxxx
D3,6 x f .g f xxxxxx g 6 f xxxxxx g x 15 f xxxx g xx 20 f xxx g xxx 15 f xx g xxxx 6 f x g xxxxxx fg xxxxxx
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On the generalized bilinear differential equations
D5, x f .g f x g fg x
D5, x D5,t f xt g f x gt ft g x fg xt
D5,3 x f. g f xxx g 3 f xx g x 3 f x g xx fg xxx
D5,2 x D5,t f .g f xxt g f xx gt 2 f xt g x 2 f x g xt f t g xx fg xxt
D5,3 x D3,t f. g f xxxt g f xxx gt 3 f xxt g x 3 f xt g xx 3 f xx g xt 3 f x g xxt f t g xxx fg xxxt
D5,5 x f .g f xxxxx g 5 f xxxx g x 10 f xx g xxx 5 f x g xxxx fg xxxxx
We
can
observe
that
from
those
formulas
we
see
that
except
3
for D5, x f . f
0, D3,3 x f . f ,
D3,5 x f . f and D5,5 x f . f do not equal zero, which is different from the Hirota case: Dx3 f . f Dx5 f . f 0
Now let p be a polynomial in M variables and introduce a generalized bilinear differential equation:
P( Dp, x1 ,..., Dp, xM ) f . f 0
(2.3)
Observe particularly that when p = 3, we have the generalized bilinear KdV equation:
( D3, x D3,t D3,4 x ) f . f 2 f xt f 2 f x ft 6 f xx2 0
(2.4)
The generalized Sawada-Kotera equation:
2
( D3, x D3,t D3,6 x ) f . f 2 f xt f 2 f x ft 2 ff xxxxxx 20 f xxx
0
We will like, following the approach of [10, 12] to discuss the linear subspaces of solutions to the generalized
bilinear differential equations defined by (2.3). to be precise, as in Hirota case [5], and also show when the
linear superposition principle will apply to the generalized bilinear differential equation (2.3).
III.
Linear Superposition Principle
Let us now fix N and in introduce N waves variables:
i k1,i x1 ... kM ,i xM ,
1 i N
(3.1)
and N exponential wave functions:
k x ... kM ,i xM
fi ei e 1,i 1
, 1 i N,
(3.2)
Where the k j ,i ' s are constants. Note also that we have a bilinear identity:
P( Dp. x1 ,..., Dp, xM )ei .e j P(k1,i k1, j ,..., kM ,i kM , j )ei e j ,
(3.3)
Where the power of α obey the rule (2.2). Then we consider a linear combination solution to the linear
differential equation (2.3):
f 1 f1 ... N f N 1e1 ... N eN ,
(3.4)
Where
i , 1 i N , are arbitrary constants. Using (3.3) we can compute that
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On the generalized bilinear differential equations
P ( D p , x1 ,..., D p , xM ) f . f
N
i , j 1
j
j
P ( D p , x1 ,..., D p , xM )ei .e
j
P (k1,i k1, j ,..., k M .i k M , j )e
i
N
i , j 1
i
i j
(3.5)
N
i2 [ P ( k1,i k1, j ,..., k M .i k M , j )e 2i
i 1
1 i j N
i j [ P (k1,i k1, j ,..., k M .i k M , j )
P (k1,i k1, j ,..., k M .i k M , j )e i j
It thus follows that a linear combination function f defined by (3.4)solves the generalized bilinear differential
equations (2.3) if and only if:
P(k1,i k1, j ,..., kM ,i kM , j ) P(k1, j k1,i ,..., kM , j kM ,i ) 0, 1 i j N ,
(3.6)
are satisfied, where the powers of α obey the rule (2.2). the condition in (3.6) present a system of nonlinear
algebraic equations on the wave related numbers k j ,i ' s and the coefficients of the polynomial P. generally, it is
not easy to solve (3.6). but in many cases, such systems have various sets of solutions .
However, the whole result was summarized in a theorem (criterion for linear superposition principle) in [13].
The theorem tells us that the linear superposition principle can apply to generalized bilinear differential
equations defined by (2.3). it also paves a way of constructing N-wave solutions to the generalized bilinear
differential equations.
IV.
Application
Let us now compute examples of the generalized bilinear differential equations, defined by (2.3), with
linear subspaces of solutions, by applying the theorem in [11] .The problem is how to construct a multivariate
polynomial P( x1 ,..., xM ) such that:
P(k1,1 k1,2 ,..., kM ,1 kM ,2 ) P(k1,2 k1,1,..., kM ,2 kM ,1 ) 0,
(4.1)
Holds for two sets of constants k1,i ,..., kM ,i , i 1, 2, where the powers of α obey the rule (2.2). our basic idea
is to introduce weights of independent variables and then use parameterization of wave numbers and
frequencies.
Following the approach of [11], we introduce the weights for the independent variables:
(w( x1 ),..., w( xM )) (n1 ,..., nM )
(4.2)
Where each weight is an integer, and then form a polynomial P( x1 ,..., xM ) being homogenous
in some weight. Second, for i = 1,2, we parameterization the constants k1,i ,..., kM ,i , consisting
of wave numbers and frequencies, using a free parameter ki as follows:
k j ,i b j kini ,1 j M
Where the b j ' s
(4.3)
are constants to be determined. The parameterization balances the degree of the free
parameters in the system (4.1). detailed procedures could also be found in [11]
In what follows we present two illustrative examples in a 1+1 and a 2+1 dimensions, which apply to the above
parameterization achieved by using one free parameter.
Example1. Examples with positive weights:
Let us introduce the weights of independent variables:
(w(x), w(t)) = (1,2)
(4.4)
Then a general even polynomial being homogenous in weights 4 reads:
P c1 x4 c2 x2 c3t 2
(4.5)
Following the parameterization of wave numbers and frequency in (4.3), the wave variables read
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27 | Page
On the generalized bilinear differential equations
i ki x b k t ,1 i N ,
Where the ki , 1 i N are arbitrary constants, but b1 is a constant to be determined. In this example, the
3
1 i
corresponding generalized bilinear differential equation reads:
P( D3, x , D3,t ) f . f 6c1 f xx2 2c2 f xxt f 2c3 ftt f 2c3 ft 2 0
(4.6)
And the corresponding linear subspace of N-wave solutions is given by:
N
N
i 1
i 1
f i fi i eki x b1ki t ,
2
i , 1 i N
3c1 c3b12 0
c2b1 c3b12 0
Where
(4.7)
are arbitrary constants, but b1 need to satisfy:
(4.8)
Therefore, the coefficients of the polynomial P need to satisfy:
c22 3c1c3
(4.9)
Then the non-trivial solution of b1 is given:
b1
3c1
c2
(4.10)
Example 2
Let us introduce the weight of independent variables:
(w(x), w(y), w(t)) = (1, 3, 2)
(4.11)
Then an even polynomial being polynomial in weight 3 reads:
P c1 x3 c2 y c3 xt
(4.12)
Following the parameterization of wave numbers and frequency in (4.3), the wave variables
readi ki x b1ki y b2 ki t ,
3
1 i N,
2
Where the ki , 1 i N are arbitrary constants, but b1 and b2 are constants to be determined. In this
example, the corresponding generalized bilinear differential equation reads:
P( D3, x , D3, y , D3,t ) f . f 2c1 f xxx f 2c3 f xt f 2c3 f x ft 0
(4.13)
the corresponding linear subspace of N-wave solutions is given by:
N
N
i 1
i 1
f i fi i eki x b1ki y b2ki t
(4.14)
Where
i , 1 i N
3
2
are arbitrary constants, but b1 need to satisfy:
c1 c3b2 c2b1 0
c3b2 c2b1
0
if c1 0 , then the solution becomes:
b1 c3 , b2 c2
(4.16)
Examples with positive and negative weights
Example 1
Let us introduce the weight of independent variables:
(w(x), w(t)) = (1, -2)
(4.15)
(4.17)
Then an even polynomial being polynomial in weight 3 reads:
P c1 x c2 x3t
(4.18)
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On the generalized bilinear differential equations
Following the parameterization of wave numbers and frequency in (4.3), the wave variables
readi ki x b1ki t ,
2
1 i N,
Where the ki , 1 i N are arbitrary constants, but b1 is a constant to be determined. In this example, the
corresponding generalized bilinear differential equation reads:
P( D3, x , D3,t ) f . f 6c2 f xx f xt 0
(4.19)
The corresponding linear subspace of N-wave solutions is given by:
N
N
f i fi i eki x b2ki
i 1
Where
2
t
i 1
i , 1 i N
(4.20)
are arbitrary constants, but b1 need to satisfy
6c2b1 0
Which gives c2 0
While b1 is arbitrary and can be written as:
b1 c1
Example 2:
(4.21)
Introducing the weights of independent variables as:
(w(x), w(y), w(t)) = (1,-1,2)
(4.22)
Then a general even polynomial being homogenous in weights 3 reads:
P c1 x3 c2 x 4 y c3 xt c4 yt 2
(4.23)
Following the parameterization of wave numbers and frequency in (4.3), the wave variables read
i ki x b1ki1 y b2 ki2t , 1 i N ,
Where the ki , 1 i N are arbitrary constants, but b1 and b2 are constants to be determined. In this
example, the corresponding generalized bilinear differential equation reads:
P( D3, x , D3, y , D3,t ) f . f 2c1 f xxx f 2c2 f xxxxy f 8c2 f xxxy f x 12c2 f xxy f xx 8c2 f xxx f xy
2c2 f xxxx f y 2c3 f xt f 2c3 f x ft 2c4 f ttty f 0
(4.24)
And the corresponding linear subspace of N-wave solutions is given by:
N
N
i 1
i 1
f i fi i eki x b1ki
1
y b2 ki2t
,
(4.25)
Where i , 1 i N are arbitrary constants, but b1 and b2 need to satisfy
c4b1b22 c3b2 6c2b1 0
c1 3c2b1 0
c3b2 3c2b1 0
(4.26)
which gives a solution of
b1
c1
c
, b2 1
3c2
c3
(4.27)
c12c4 3c2c32
(4.28)
V.
Conclusion
We were able to ascertain newly introduced kind of bilinear differential operators established in [10]
and analyzed when the corresponding generalized bilinear equation posses superposition principle. In particular
we computed two examples using 1+1 and a 2+1 dimensional equation by an algorithm using weights and their
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On the generalized bilinear differential equations
linear subspaces of exponential travelling wave solutions. The balance requirements of the weights allow us to
present a class of parameterization of wave numbers and frequencies.
Our results further confirm the generalized Hirota bilinear operators and the established bridge between bilinear
differential equations and linear differential equations [11]. The existence of linear subspaces of solutions
amends the diversity of exact solution generated by various analytical methods (see for example, [5]. The
generalized bilinear operators (2.1), definitely bring more chance to generate non trivial trilinear differential
equations.
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
Ma, W. X., Huang,T. W. Zhang, Y., A multiple Exp-function Method for nonlinear differential equations and its application
Physica Scipta 82 2010 065003.
Ma, W. X. and Fan, E., Linear superposition principle applying to Hirota bilinear equations. Computers And Mathematics With
Appllications, 61,(2011 950-959
W. X., Ma, Y .Zhang,, Y .Tang ,. and J . Tu,. Hirota bilinear equations with linear subspaces of solutions. Appllied Mathematics
And Computations 218, 2012, 7174-7183
W. X, Ma, , and Y . You,Solving Korteweg de-Vries equation by its bilinear form: Wronskian solution. Trans. Amer. Math. Soc.,
357,2005:1753-1778
R. Hirota.. The Direct Method in Soliton Theory. (Cambridge University Press) (2004)
Asaad, M. G and Ma, W. X., , Pfaffian solution to a (3+1)-dimensional generalized B.typeKadomtsev-Petviashvilli equation and
its modified counterpart, applied Math. And Comp.218, 2012, 5524-5542
R. Hirota,., a new form of Baclund Transformation and its relation to the inverse scattering problem. Progr. Of Theoret. Phys., 52,
1974, 329-338
N. C, Freeman, and J.J.C Nimmo, , the use of Backlund transformation in obtaining N-soliton solutions in Wronskian form. Phys.
lett. A 95, 1983, 1-3
S . Zangh, and T. C, Xia,., A generalized new auxiliary equation and its application to nonlinear partial differential equations.
Phys. Lett. A, 363, 2007, 356-360
M. Jimbo. and T. Miwa, solitons and infinite dimensional Lie algebra, publications in research Institute for Mathematical
Sciences. 19, 1983, 943-1001
W. X. Ma Generalized bilinear differential equations Diversity of exact solution to a restricted Boiti-Leon-pempinelli dispersive
long wave system. Physics Letter. A 310 2012,325-333
A. Wazwaz, multiple soliton solutions for Calogero-Bogoyavlensskii-Schiff, Jimbo-Miwa and YTSF equations, Appllied
Mathematics and Computations 203, 2008, 592-597.
M. Y. Adamu, and E. Suleiman,. linear subspaces of solutions applied to Hirota bilinear equations, Aceh, Journ. Sci. and Techn.1
(2) 2012, 45-51
www.iosrjournals.org
30 | Page
IOSR Journal of Mathematics (IOSR-JM)
ISSN: 2278-5728. Volume 3, Issue 4 (Sep-Oct. 2012), PP 31-35
www.iosrjournals.org
Neutrosophic Set and Neutrosophic Topological Spaces
1
A.A.Salama, 2S.A.Alblowi
1
Egypt, Port Said University, Faculty of Sciences Department of Mathematics and Computer Science
2
Department of Mathematics, King Abdulaziz University, Saudi Arabia
Abstract: Neutrosophy has been introduced by Smarandache [7, 8] as a new branch of philosophy. The
purpose of this paper is to construct a new set theory called the neutrosophic set. After given the fundamental
definitions of neutrosophic set operations, we obtain several properties, and discussed the relationship between
neutrosophic sets and others. Finally, we extend the concepts of fuzzy topological space [4], and intuitionistic
fuzzy topological space [5, 6] to the case of neutrosophic sets. Possible application to superstrings and
space–time are touched upon.
Keywords: Fuzzy topology; fuzzy set; neutrosophic set; neutrosophic topology
I.
Introduction
The fuzzy set was introduced by Zadeh [9] in 1965, where each element had a degree of membership.
The intuitionstic fuzzy set (Ifs for short) on a universe X was introduced by K. Atanassov [1, 2, 3] in 1983 as a
generalization of fuzzy set, where besides the degree of membership and the degree of non- membership of each
element. After the introduction of the neutrosophic set concept [7, 8]. In recent years neutrosophic algebraic
structures have been investigated. Neutrosophy has laid the foundation for a whole family of new mathematical
theories generalizing both their classical and fuzzy counterparts, such as a neutrosophic set theory.
II.
Terminologies
We recollect some relevant basic preliminaries, and in particular, the work of Smarandache in [7, 8],
and Atanassov in [1, 2, 3]. Smarandache introduced the neutrosophic components T, I, F which represent the
membership, indeterminacy, and non-membership values respectively, where
interval.
2.1 Definition. [3,4]
Let T, I,F be real standard or nonstandard subsets of
Sup_T=t_sup, inf_T=t_inf
Sup_I=i_sup, inf_I=i_inf
Sup_F=f_sup, inf_F=f_inf
n-sup=t_sup+i_sup+f_sup
n-inf=t_inf+i_inf+f_inf,
T, I, F are called neutrosophic components
III.
0 ,1
is nonstandard unit
0 ,1
, with
Neutrosophic Sets and Its Operations
We shall now consider some possible definitions for basic concepts of the neutrosophic set and its
operations.
3.1 Definition
Let X be a non-empty fixed set. A neutrosophic set ( NS for short) A is an object having the form
A x , A x , A x , A x : x X
Where A x, A x and
A
x which represent the degree of member
ship function (namely A x ), the degree of indeterminacy (namely
A x ), and the degree of non-member
ship (namely A x ) respectively of each element x X to the set A .
3.1 Remark
A neutrosophic A x , A x , A x , A x : x X can be identified to an ordered triple
A, A, A
in 0,1 on. X .
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Neutrosophic Set And Neutrosophic Topological Spaces
3.2 Remark
For the sake of simplicity, we shall use the symbol A x , A , A , A for the
NS A x , A x , A x , A x : x X
3.1 Example
Every IFS A a non-empty set X is obviously on NS having the form
A x , A x ,1 A x A x , A x : x X
Since our main purpose is to construct the tools for developing neutrosophic set and neutrosophic topology, we
must introduce the NSS 0N and 1N in X as follows:
0N may be defined as:
01
02
03
04
x ,0,1,1 : x X
x ,0,1,0 : x X
x ,0,0,0 : x X
0N x ,0,0,1 : x X
0N
0N
0N
1N may be defined as:
11
12
13
14
x ,1,0,1 : x X
x ,1,1,0 : x X
x ,1,1,1 : x X
1N x ,1,0,0 : x X
1N
1N
1N
3.2 Definition
Let A A , A , A
a NS on X , then the complement of the set A
C A , for short
maybe defined as
three kinds of complements
x ,1 x ,1 x : x X ,
C C A x , , x , x : x X
C C A x , ,1 x , x : x X
C 1
C A
A
2
A
3
A
A
A
A
A
A
One can define several relations and operations between NSS follows:
3.3 Definition
B in the form
Let x be a non-empty set, and NSS A and
B x , B x , B x , B x
A B
A x , A x , A x , A x ,
, then we may consider two possible definitions for subsets A B
may be defined as
(1) A B A x B x , A x and A x B x x X
(2) A B A x B x , A x B x and A x B x
3.1 Proposition
For any neutrosophic set A the following are holds
(1) 0N A , 0N 0N
(2) A 1N , 1N 1N
3.4. Definition
Let X be a non-empty set, and A x , A x , A x , A x , B x , B x , B x , B x are NSS . Then
(1) A B maybe defined as:
I 1 A B x , A x .B x , A x . B x ,
A x . B x
I 2 A B x , A x B x , A x B x ,
A x B x
I 3 A B x , A x B x , A x B x , A x B x
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Neutrosophic Set And Neutrosophic Topological Spaces
(2) A B may be defined as:
U 1 A B x , A x B x , A x B x ,
A x B x
U 2 A B x , A x B x , A x B x ,
A x B x
(3) A x , A x , A x ,1 A x
(4) A x ,1 A x , A x , A x
We can easily generalize the operations of intersection and union in definition 3.4 to arbitrary family of NSS as
follow:
3.5 Definition
Let Aj : j J be a arbitrary family of NSS in X , then
(1) Aj maybe defined as:
(i) Aj x , Aj x , Aj x , Aj x
j J
j J
(ii) Aj x , Aj x , Aj x , Aj x
(2) Aj maybe defined as:
(i) Aj x , , ,
(ii) Aj x , , ,
3.6. Definition
Let A and B are neutrosophic sets then
A B may be defined as
A B x , A B , A x B x , A B x
3.2. Proposition
For all A , B two neutrosophic sets then the following are true
(1) C A B C A C B
(2) C A B C A C B
IV.
Neutrosophic Topological Spaces
Here we extend the concepts of fuzzy topological space [4], and intuitionistic fuzzy topological space [5,
7] to the case of neutrosophic sets.
4.1 Definition
A neutrosophic topology ( NT for short) an a non empty set X is a family of neutrosophic subsets in X
satisfying the following axioms
NT1 ON ,1N ,
NT 2 G1 G 2 for any G1 ,G 2 ,
NT 3 G G : i J
In this case the pair X , is called a neutrosophic topological space ( NTS for short) and any neutrosophic
set in is known as neuterosophic open set ( NOS for short) in X . The elements of are called open
i
i
neutrosophic sets, A neutrosophic set F is closed if and only if it C (F) is neutrosophic open.
4.1 Example
Any fuzzy topological space X , 0 in the sense of Chang is obviously a NTS in the form A : A 0
wherever we identify a fuzzy set in X whose members ship function is A with its counterpart.
4.1. Remark Neutrosophic topological spaces are very natural generalizations of fuzzy topological spaces
allow more general functions to be members of fuzzy topology.
4.3 Example
Let X x and
B x ,0.4,0.6,0.8 : x X
D x ,0.5,0.6,0.4 : x X
A x ,0.5,0.5,0.4 : x X
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Neutrosophic Set And Neutrosophic Topological Spaces
C x ,0.4,0.5,0.8 : x X
Then the family On ,1n , A , B ,C , D of N Ss in X is neutrosophic topology on X
4.4 Example
Let X , 0 be a fuzzy topological space in changes sense such that 0 is not indiscrete suppose now that
0 0 N ,1N V j : j J then we can construct two NTSS on X as follows
a) 0 0 N ,1N x,V j , ( x),0 : j J .
b) 0 0 N ,1N x,V j ,0, ( x),1 V j : j J .
4.1 Proposition
Let X , be a NTS on X , then we can also construct several NTSS on X in the following way:
a) o,1 [ ]G : G ,
b) o,2 G : G ,
Proof
a) NT1 and NT 2 are easy.
NT 3 Let [ ]G j :
j J , G j 0,1 .Since
G j x, G j , G j , G j or x, G j , G j , G j or x, G j , G j , G j , we have
[ ]G j x, G j , G j ,(1 G j ) or x, G j , G j , (1 G j ) 0,1
b) This similar to (a)
4.2 Definition
Let X ,1 , X , 2 be two neutrosophic topological spaces on X . Then 1 is said be contained in 2 (in
symbols 1 2 ) if
G 2
for each G 1 . In this case, we also say that
4.2 Proposition
Let j : j J be a family of NTSS on X . Then
j is the coarsest NT on X containing all.
j,s
j
1
is coarser than 2 .
is A neutrosophic topology on X .Furthermore,
Proof. Obvious
4.3 Definition
The complement of A (C (A) for short) of NOS. A is called a neutrosophic closed set ( NCS for short) in X .
Now, we define neutrosophic closure and interior operations in neutrosophic topological spaces:
4.4 Definition
Let X , be NTS and A x , A x , A x , A x be a NS in X .
Then the neutrosophic closer and neutrosophic interior of Aare defined by
NCl ( A) K : K is an NCS in X and A K
NInt ( A) G : G is an NOS in X and G A.It can be also shown that
It can be also shown that NCl (A) is NCS and NInt (A) is a NOS in X
a) A is in X if and only if NCl ( A) .
b) A is NCS in X if and only if NInt ( A) A .
4.2 Proposition
For any neutrosophic set
A
in x , we have
(a) NCl (C ( A) C ( NInt ( A),
(b) NInt (C ( A)) C ( NCl ( A)).
Proof.
a) Let A x, A , A , A : x X and suppose that the family of neutrosophic subsets contained in
A are indexed by the family if NSS contained in
A
are indexed by the
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Neutrosophic Set And Neutrosophic Topological Spaces
family A x, G , G ,G : i J . Then we see that NInt ( A)
x, Gi , G i ,Gi
i
i
i
and hence C ( NInt ( A)) x,G , G i ,G . Since C ( A) and
i
i
Gi A and Gi A
for
each i J , we obtaining C (A) . i.e NCl (C ( A)) x,G , G , G . Hence
i
i
i
NCl (C ( A) C ( NInt ( A), follows immediately
b) This is analogous to (a).
4.3 Proposition
Let x , be a NTS and
(a)
(b)
(c)
(d)
(e)
(f)
(g)
A , B be two neutrosophic sets in
X . Then the following properties hold:
NInt ( A) A,
A NCl ( A),
A B NInt ( A) NInt ( B),
A B NCl ( A) NCl ( B),
NInt ( NInt ( A)) NInt ( A) NInt ( B),
NCl ( A B) NCl ( A) NCl ( B),
NInt (1N ) 1N ,
(h)
NCl (ON ) ON ,
Proof (a), (b) and (e) are obvious (c) follows from (a) and Definitions.
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
K. Atanassov, intuitionistic fuzzy sets, in V.Sgurev, ed.,Vii ITKRS Session, Sofia(June 1983 central Sci. and Techn. Library, Bulg.
Academy of Sciences( 1984)).
K. Atanassov, intuitionistic fuzzy sets, Fuzzy Sets and Systems 20(1986)87-96.
K. Atanassov, Review and new result on intuitionistic fuzzy sets , preprint IM-MFAIS-1-88, Sofia, 1988.
C.L. Chang, Fuzzy Topological Spaces, J. Math. Anal. Appl. 24 (1968)182-1 90.
Dogan Coker, An introduction to intuitionistic fuzzy topological spaces, Fuzzy Sets and Systems. 88(1997)81-89.
Reza Saadati, Jin HanPark, On the intuitionistic fuzzy topological space, Chaos, Solitons and Fractals 27(2006)331-344 .
Florentin Smarandache , Neutrosophy and Neutrosophic Logic , First International Conference on Neutrosophy , Neutrosophic Logic
, Set, Probability, and Statistics University of New Mexico, Gallup, NM 87301, USA(2002) , [email protected]
F. Smarandache. A Unifying Field in Logics: Neutrosophic Logic. Neutrosophy, Neutrosophic Set, Neutrosophic Probability.
American Research Press, Rehoboth, NM, 1999.
L.A. Zadeh, Fuzzy Sets, Inform and Control 8(1965)338-353
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35 | Page
IOSR Journal of Mathematics (IOSR-JM)
ISSN: 2278-5728. Volume 3, Issue 4 (Sep-Oct. 2012), PP 36-51
www.iosrjournals.org
Operator Regularization in Evaluating Feynman Diagrams in
QED in (3+1) Dimensional Space-Time
M. Forkan1, M. Abul Mansur Chowdhury2
1
2
Department of Mathematics, University of Chittagong, Chittagong-4331, Bangladesh.
Research Centre for Mathematical and Physical Sciences, University of Chittagong, Chittagong-4331,
Bangladesh.
Abstract: Operator regularization is one of the excellent prescriptions for studying gauge theories. Among the
many regularization prescriptions Dimensional regularization and Pre-regularization are the best methods for
evaluating loop diagrams perturbatively. On the other hand Operator regularization can also be said one of the
best methods for studying gauge theories because of its two-fold use. With this prescription one can adopt pathintegral method with the combination of background field quantization and Schwinger expansion to find the
result of the required problem without considering any Feynman diagrams. Also from this prescription one can
consider Feynman diagrams and evaluating these diagrams using the Operator regularization prescription. In
this paper we have shown how one can use both the options of Operator regularization method to evaluate
Feynman diagrams in QED in (3+1) dimensional space-time.
Keywords: Operator regularization, Dimensional regularization, Feynman diagrams in QED, Path-integral
method, Background field quantization and Generating functional.
I.
Introduction
Feynman diagrammatic technique is a standard way of studying of gauge theories in a perturbative
way. The problem is when one tries to evaluate the loop diagrams arising from the theory in consideration using
Feynman integrals. Most of the time divergencies arise in these integrals. So one has to use some regularization
method to overcome this problem.
Evaluating loop integrals using different regularization procedures give results which are very often
dependent on the regulating parameters which are not expected. However in some cases choosing the
appropriate value of the parameter one can get the exact result. In this paper three basic diagrams in QED in
(3+1) dimensional space-time is studied by the method of Operator regularization [1] in two different
approaches and Dimensional regularization [2]. We have shown here how the results in these three approaches
are same.
Operator regularization is a little bit different than that of other regularization methods. Because all the
methods mentioned here are perturbative method. That is one has to draw all possible Feynman graphs and
following any regularization method one can find the transition amplitude of the required problem. But Operator
regularization method can be used in a two-fold way. That is without considering Feynman graphs [1] which
depend only on path integral method and also considering Feynman graphs [4]. In the first case one has to use
back-ground field quantization in the Lagrangian then the operators and inverse operators have to regulated
following a given prescription [1, 3]. After some simplification Schwinger expansion [5] has to be taken. From
the expansion one can choose appropriate terms for the problem in consideration. It means that we do not have
to consider Feynman graphs. In this method we do not have to face any divergencies at any stage of calculation.
However, after quantizing with operator regularization, there is a way to consider Feynman graphs. Then
following any regularization method one can find transition amplitudes of the problem. In this case this is a
combination of Operator regularization and other regularization methods.
In this paper at first we will show how the second option of Operator regularization can be used in
evaluating loop diagrams. That means from the path integral form of the Operator regularization how one can
choose the part of the prescription which can be applied to find the amplitude of the basic Feynman graphs and
then we will show how these problems can be obtained from the first option that is from path integral form of
the method and the results will be compared later on.
II.
Operator Regularization Prescription
Operator regularization is an alternative way of computing quantum correction in quantum field theory
in context of background-field quantization, which was given by D.G.C. McKeon et.al. [1, 3]. In this method the
Feynman diagrams of the usual perturbation series can be avoided because this method depends on path
integrals. But at one stage there is an option to consider Feynman diagrams. That is from this prescription one
can choose either path integral method or Feynman diagrammatic approach. In this approach we regulate
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Spaceoperators and inverse operators rather than the initial Lagrangian. To one-loop order this scheme reduces to a
perturbative expansion of the well-known - function regularization [6-9] associated with the superdeterminant
of an operator.
This prescription is given in ref. [3] but for completeness let us briefly describe it here. The
background-field method in context of path-integral quantization [10-12] is the starting point of this procedure.
We consider the general case where a field i x which may be either fermionic or bosonic. Let us consider a
field i x is quantized into its background classical part f i x and quantum part qi x .That is
(2.1)
i x = f i x + qi x
The general form of the Lagrangian L f i qi that we will consider is
1
1
1
(2.2)
L f i , qi = qi M ij f j q j + aijk f j qi q j q k + bijkl qi q j q k ql
2
3!
4!
The generating functional for Green’s functions in the theory in the presence of a source function J i x is
given by
dqk exp dx [ L f i qi + J i qi ]
Z fi , J j =
(2.3)
For simplicity let us deal only with one-loop effects. Then from the generating function we have to restrict our
attention only to those terms in Eq. (2.2) that are bilinear in q i .We thus consider only
1
(2.4)
q i M ij f j q j
2
Upon substituting Eq. (2.4) into Eq. (2.3) we arrive at the one-loop generating functional
1
(2.5)
dq k exp dx qi M ij f i q j
Z f i ,0 =
2
Evaluation of the functional integral in Eq. (2.5) involves a determinant which we call as superdeterminant
of M ij , as q i may be either fermionic or bosonic [13-16].
L
(2)
=
That means from equation (2.5) the one-loop generating functional for Green’s functions is
Z1 f i ,0= s det
1
2
M ij f i
(2.6)
Equation (2.6) tells us that we have to regularize the superdeterminant of the operator M ij and it’s inverse.
The superdeterminant of an operator can be written as
det = exptr ln
(2.7)
Let us regularize ln in the following way:
d n n1
(2.8)
ln lim n
, (n = 1, 2, 3, ... )
0 d
n!
In facing no divergences we can always choose n to be greater than or equal to the number of “loop momentum
integrals” or in other words order in .
Hence,
dn
det exp tr lim
0 d n
and
n 1
n!
1m1 d m ln
m 1! d m
d n n1 m m
lim n
0 d
n! m
(2.9a)
m
(2.9b)
If we now rewrite as
=
1
dt t
1
exp t
(2.10)
0
in eq. (2.9) we arrive at the result
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Spacedet exp 0
where we have defined the
=
This is the usual
1
(2.11a)
- function
dtt
tr exp t
1
(2.11b)
0
- function regularization of the determinant of an operator.
Equations (2.8) and (2.9) are the main steps of the Operator regularization which is used in (2.6) to
evaluate the Green’s function of any problem. From this point we can divide the prescription in two fold way.
That means if we use Schwinger expansion for the operator like
1
d 1
t2
0t
0t
1
dtt tr e
te I +
due 1u 0t I e u0t I
det = exp lim
0 d
2
0
0
t3
3
1
duu dve 1u 0t I e u 1v 0t I e uv0t I .......
0
0
1
(2.12)
where, 0 I with 0 is independent of the background field f i and I is at least linear in f i .
Then following the steps described in ref.[3] we can find the result of the problems in consideration.
Also these equations can be used in evaluating Feynman loop-diagrams. For one-loop take n =1, for twoloops take n = 2 and so on.
Following (2.9b) we can write the general prescription of Operator regularization for the Feynman diagrams as
follows [4]:
dn
n m
(2.13)
m lim n 1 1 2 2 ... n n
0 d
n!
where the n s are arbitrary. For one-loop diagrams it is enough to use n = 1.When m = 2 and n = 1, then
eq. (2.13) taken the form
d
(2.14)
2 lim
1 2
0 d
Now applying this to the three divergent one loop Feynman diagrams in QED.
2.1 One Loop Correction to the Fermion Line in (3+1) Dimensions
Starting with the Feynman diagram for the one loop correction to the fermions line which is represented
by
p :
Fig.-1: One loop Feynman diagram for external fermion lines.
Using the Feynman rules one can write
p
= ie 2
p as,
p l m
d 4l
2 p l
4
2
m2 l 2
Using the Feynman identity, we can write
p l m
p = ie dx d l
2 p l x m x l 1 x
1
4
2
4
0
2
2
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2
2
38 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceShifting the variable of integration as l l px and simplifying we get
p = ie dx d l
2 l
1
4
2
4
0
p 1 x m l
2
2
m 2 x p 2 x1 x
The term linear in l integrates to zero because of symmetric integration, so
p = ie dx d l p 1 x m
2 l m x p x1 x
1
4
2
4
2
0
2
(2.1.1)
2
2
Which is taken as the common starting point for both Dimensional and Operator regularization.
Now proceeding with operator regularization, following the rule cited in Eqs. (2.11) and (2.12), the above
result becomes,
p = ie dx lim dd d l 1 p 1 x m
2 l m x p x1 x
1
4
2
0
4
0
Using the standard integral
d 2w l
2 l
2w
1
2
M
2
1
1
=
2 A
4 A
w
2
2
2
2
(2.1.2)
A w
(2.1.3)
M
2 A w
we get,
p = ie
1
4 2
d 1
.
0 d 2
m 2 x p 2 x1 x
dx p 1 x m lim
0
(2.1.4)
Here,
d 1
.
2
2
0 d 2
m
x
p
x
1
x
d 1
2
2
= lim
, Taking u m x p x1 x
0 d 1u
1 u 1 u 1 u ln u
= lim
2
0
1
u
= 1 ln u
lim
(2.1.5)
Therefore Eqn. (2.1.5) becomes,
p = ie
2
dx p 1 x m 1 ln m x p
1
1
2
4 2
0
The advantage of this method is that here we can use 4-dimensional
n as in dimensional regularization.
Doing the -algebra we arrive at
p = 2ie
2
dx p 1 x 2m 1 ln m
1
1
4 2
2
2
x1 x
-algebra. We do not have to go from 4 to
x p 2 x1 x
(2.1.6)
0
Let us now separate the finite part and divergent part from (2.1.6) as follows:
p = 2ie
2
4
2
= 2ie
= 2ie
2
1
dx p 1 x 2m 1 ln m
1
1
2
4 2
2
x p 2 x1 x ln 2 ln 2
0
1
4 2
m 2 x p 2 x1 x
dx p 1 x 2m ln 2 1 ln
2
0
1
m 2 x p 2 x1 x
dx p 1 x 2m 1 ln
2
0
1
where 2 taken from the arbitrary .
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Space-
p is,
Thus the finite part of
2i
m 2 x p 2 x1 x
dx p 1 x 2m ln
2
0
1
e2
4 2
(2.1.7)
and the divergent part is,
2i
dx p 1 x 2m 1
4 2
e
= i
1
e2
2
0
1 p 4m
4 2
(2.1.8)
Comparing this to the result from Dimensional regularization [17-18], the finite part is
e2
4 2
p 2 x1 x m 2 x
dx p 1 x 2m ln
4 2
0
1
(2.1.9)
and the divergent part is
e2
(
2
1 ) p 4m -
e2
(2.1.10)
2m
4 2
where 0 .5772 is the Euler- Mascheroni constant.
There is a constant difference between these two methods that stems from dimensionally continuing the gamma
matrices, but the resulting over all phase should not effect the physics.
4 2
2.2 One Loop Correction to the Boson (Photon) Line in (3+1) Dimensions
Let us consider the Feynman diagram for the one loop correction to the photon line which is represented by
p :
Fig.-2: One loop Feynman diagram for external boson lines.
The QED one loop correction to the photon line in 4-dimensions is
l p m l m
Tr
2 l p 2 m 2 l 2 m 2
Combining the denominator using the Feynman identity and simplifying, we get
p = e 2
p = e 2
d 4l
4
1
Tr l p m l m
d 4l
2 l p
dx
x m 2 x l 2 m 2 1 x
Now putting l l px in Eq. (2.2.1), then we get,
4
0
p = e 2
1
d 4l
Here
(2.2.1)
2
Tr l p x p m l p x m
2 l p1 x x m x l px
dx
4
2
0
=
2
2
2
m 2 1 x
2
N
D
D = l p1 x 2 x m 2 x l px 2 m 2 1 x 2
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40 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Space-
= l 2 x 2l p1 x x p 2 1 x 2 x m 2 x l 2 1 x 2l p1 x p 2 x 2 1 x m 2 m 2 x
= l 2 m 2 p 2 x1 x
and
2
2
N = Tr l p 1 x m l p x m
l px m 2
=4 l p1 x l px m 2
= Tr l p1 x
=4 2l l 2 x1 x p p p 2 l 2 m 2 p 2 x1 x
Hence equation (2.2.1) with l l becomes,
1
2l l
2 x1 x p p p 2
d 4l
p =4 e 2 dx
2
4
2
2
2
2
l m p x1 x
2 l 2 m 2 p 2 x1 x
l 2 m 2 p 2 x1 x
0
(2.2.2)
If we apply the following integrals in the first and third terms in the integrand of equation (2.2.2),
I.
II.
d l
d dl
d
l
l
l l
2
2lq m
2
1
2
2lq m 2
=
i
d
2
q
= 1 i
d
1
m
2
2
2
q 2 m 2
d
x q q d
d
2 2
d
2
1
g q 2 m2 1 d
2
2
2
We arrive at,
x1 x
2
2
2
2
p = -8 e p p p dx
(2.2.3)
4 l m p x1 x
2
0
Which is again taken as the common starting point for both Dimensional and Operator regularization for one
loop correction to the photon lines.
2
1
2
d 4l
Using the operator regularization rule which describe in section-2 in above eq., we obtain,
p =8 e
p p p dx lim dd d l
2 l
1
2
1 x1 x
4
2
4
0
0
2
m 2 p 2 x1 x
(2.2.4)
2
Due to the momentum integral (2.1.3), from eq. (2.2.4) we get, A 2 , w 2 , M 2 m 2 p 2 x1 x , then
Eq. (2.2.4) becomes,
1
1
d 1
2
p =-8 e 2
p
p
p
.
(2.2.5)
dx x1 x lim
2
2
2
0 d 2
4
m
p
x
1
x
0
From Eq. (2.1.5) we get,
d 1
.
0 d 2
m 2 p 2 x1 x
Thus equation (2.1.4) becomes,
lim
p =8 e 2
=8 e 2
4
2
4 2
= 1 ln u , where u m 2 p 2 x1 x .
p p p dx x1 x 1 ln m
1
1
1
2
0
1
2
p p p dx x1 x 1 ln m
2
0
2
p 2 x1 x
p 2 x1 x
4 2
where 2 taken from the arbitrary .
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Space-
Therefore the finite part of p is,
8e
and
2
1
4 2
p p p dx x1 x ln p
1
2
2
0
x1 x m 2
4 2
(2.2.6)
the divergent part is,
-8 e 2
1
2
4 2
2
=-
p p p dx x1 x 1
1
4 e
3 4 2
0
p p p 1
2
(2.2.7)
Comparing this against the result of Dimensional regularization [17-18], the finite part is
8ie 2 p p p 2
4
2
and the divergent part of p is,
8ie 2 p p p 2
4 2
p 2 x1 x m 2
dx x1 x ln
2 2
0
1
1
2
(2.2.8)
dx x1 x
0
4ie 2 p p p 2 2
=
2
34
We see that both results are the same in form.
(2.2.9)
2.3 One Loop Correction to the Vertex in (3+1) Dimensions
Let us now consider the Feynman diagram for the one loop correction to the vertex which is represented
by p, q .
Fig.-3: One loop Feynman diagram for vertex function.
The QED one loop correction to the vertex in 4-dimensions is
d 4l
i
i
ie
ie 2
ie
p, q =
4
p l m
q l m
l
2
= ie 3
p l m q l m
d 4l
2
l 2 p l 2 m 2 q l 2 m 2
4
(2.3.1)
Now we introduce the 3-parameter Feynman formula for combining the denominator,
1
1 x
1
1
in the eqn. (2.3.1), we obtain,
2 dx dy
abc
a1 x y bx cy 3
0
0
1
p, q = 2ie3 dx
1 x
0
0
p l m q l m
d 4l
2 l
dy
4
2
1 x y p l 2 m 2 x q l 2 m 2 y
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3
42 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceIf we change the variables l l px qy and simplify the denominator and numerator, we obtain,
1
p, q = 2ie3 dx
1 x
l p 1 x q y m l q 1 y p x m
d 4l
2 l
dy
0
4
2
0
m 2 x y p 2 x1 x q 2 y1 y 2 p.qxy
3
This integral contains convergent and divergent pieces. The part of the numerator quadratic in l is divergent,
the rest convergent, so separating the divergent piece 1 p, q and convergent piece 2 p, q , i.e.
p, q = 1 p, q + 2 p, q .
Thus the divergent piece is,
1
1 p, q = 2ie3 dx
1 x
l l
d 4l
2 l
0
dy
4
2
0
M2
(2.3.2)
3
where, M 2 m 2 x y p 2 x1 x q 2 y1 y 2 p.qxy .
Which is taken as the common starting point for both Dimensional and Operator regularization for one-loop
correction to the vertex.
Again, using the operator regularization rule which describe in section-2 in above eq., we obtain,
1
1 x
l l
d 4l
d
1
1 p, q = 2ie3 dx dy lim
3
0 d
2 4
l2 M 2
0
0
Now performing the momentum integral, we get
1
4 2
d 1
.
0 d 3
M2
1 x
1
e2
p, q = ie
dx
0
dy lim
0
(2.3.3)
(2.3.4)
Here,
lim
0
d 1
. , where u = M2.
d 3 u
1
d
2
0 d
3 2 u
2 3 2 u 1 2 3u
= lim
0
[ 2 3 2 u ]2
= lim
=
3 2u ln u
[ 3 2u ]
2
2
2
2 3 2 ln u
4
Again performing the
1 p, q = ie
-algebra in 4-dimentions, eq. (2.3.4) reduces to,
1 x
1
e2
4
2
dy 4 3 2 2 ln u 4
1
dx
0
0
Hence the finite part of 1 p, q is,
1 x
1
e2
M2
dy ln 2
0
4 0
where, M 2 m 2 x y p 2 x1 x q 2 y1 y 2 p.qxy
2 ie
2
dx
(2.3.5)
and the divergent part is
- ie
e2
4 2
1
1 x
0
0
dx dy 3 2
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Space 3
2
e2
=- ie
4
2
(2.3.6)
Comparing this result with the Dimensional regularization [17-18] result, the finite part is
8e 3
1 x
1
M2
dy ln
2
4
0
dx
4 2
0
(2.3.7)
where, M 2 m 2 x y p 2 x1 x q 2 y1 y 2 p.qxy
and the divergent part is,
8e 3
1 x
1
dy ( 1 )
4
2
0
4e
2
dx
3
0
2
(2.3.8)
1)
Which agree in form, recalling that Operator regularization goes further than Dimensional regularization in so
much as that it actually removes the divergences.
=
4
2
(
III. Path Integral Form of Operator Regularization for One Loop
Generating Functional in QED
Let us consider the QED Lagrangian,
2
1
1
L = 2 i e m
(3.1)
4
2
Let us expand this Lagrangian taking background field quantization of the fields. Let the background field
expansion of gauge field and fermionic field are respectively,
V Q
q
where V and are the classical fields and Q and q are the quantum fields.
Therefore Eq. (3.1) becomes,
1
1
L = V Q V Q 2 q i e V Q m q
V Q
4
2
1
=
Q Q V V 2 + i m + i m q
4
+ q i m + e V V q + Q + Q q q V
q V q q Q q Q q
1
V Q
2
2
2
(3.2)
To find the one-loop 1PI generating functional we need to consider only the terms in the Lagrangian bilinear in
the quantum fields; then from Eq b .(3.2) we obtain
1
1
L (2) = q i eV q Q Q 2
Q 2 e Q q eq Q
2
4
1
1
= qD
q Q p 2 1 p p Q e Q
q eq Q
2
i eV p eV .
where, D
(3.3)
The formalism of Section-2 cannot be directly applied to the bilinear Lagrangian L (2) of Eq.(3.3) as q and q
are independent quantum fields in the associated path integral. However, it is possible to rewrite L (2) in the
form of Eq. (2.4) by the following device. We introduce the notation
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44 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Space q1
q 2
q
T
T
T , q , q and we identify the quantum field . of the original formalism of the background
q
.
q
n
Q
Q
field method with i.e. q .
T
q
Thus the Lagrangian (3.3) can be written as,
2
1
e e T T Q
p 1 p p
T
0
q
T T
(2) 1
T
D
L = Q q q e
T
2
D
0
q
e
Q
1 T
= hi M ij h j , where, h j q
(3.4)
2
T
q
Evaluation of the path integral (2.5) leads at once to the one-loop generating functional
1
(3.5)
Z 1 = dq k exp dx hi M ij f i h j
2
Evaluation of the functional integral in Eq. (3.5) involves the “superdeterminant’’ [13-16] of M ij as hi may be
either fermionic or bosonic.
Now we can proceed in two ways: either (a) complete the square in the fermionic variables f and f or (b)
complete the square in the bosonic variable b [1, 3].
Following ref. [1] let us complete the square in fermionic variables in the argument of the exponential on the
right-hand side of Eq. (3.5), then we get
1
I dfdfdb exp b T M bbb b T M bf M ff1 M fb b f bM bf M ff1 M ff f M ff1M fb b
2
The change of variables,
(3.6)
f f M ff1 M fb b and f f bM bf M ff1 , then we get,
1
f db
exp b T M bb 2M bf M ff1 M fb b f M
ff f
I df d
2
Now using the standard Gaussian integrals
1
1
(3.7)
db exp 2 b Ab = det A
and df df exp f Bf = det B
T
2
in Eq. (3.7), we obtain,
I=
det
1
2
M
det M ff
bb
2M bf M ff1 M fb
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceThus one-loop generating functional for Green’s functions Z 1 in Eq. (3.5) becomes,
Z1 =
det A
det D
=
(say)
det B
1
2
2
1
2
1
2
det p 1 p p e D e D
(3.8)
1
Here we see that Z 1 is the ratio of determinant of operators. Each of the determinants occurring in Eq.
(3.8) requires regularization and a corresponding -function. The numerator and denominator separately
contribute to Green’s functions with only external boson lines and with both external fermions lines and vertex
function in massless QED respectively.
3.1 One-Loop Generating Functional and Loop Corrections for External Boson Lines
To find the loop corrections or to write the generating functional for external boson lines one has to
make a close look at the numerator of eq. (3.8) and on the other hand for external fermion lines one has to take
care of the denominator of eq. (3.8). So for bosonic case we have to regulate the det A through the use of function in Eq. (2.11a) yielding
Lit
,
(3.1.1)
tr exp t with p eV .
(3.1.2)
A
det A = Z 1A = exp
where, =
1
dt t
0
1
0
As we mentioned in section-II, after regularization we have to consider Schwinger expansion, to this view let
us now identify the operator 0 and I with p and eV , respectively, then by Eq. (2.10), Eq. (3.1.1) can be
written as,
d
Z 1A =epx Lim
0 d
1
t2
dtt 1tr e p t te p t + eV
2
0
3 1
1
t
duu dve 1u p t
3
0
1
due
1u p t
eV e up t eV
0
eV eu 1 v p t eV e uvp t eV +... ...
0
(3.1.3)
To one-loop order this series plays the same role as Feynman rules in the usual perturbation theory. Here
we want to evaluate the one-loop correction to the two-point function for external photon in QED; we restrict
our attention to the term bilinear in V on the right-hand side of Eq. (3.1.3). This leaves us with
1
2
t 1
e
tr due 1u p t V e up t V
(3.1.4)
dt
2
0
0
Now let us complete the functional trace
1
T= tr due 1u p t V e up t V
(3.1.5)
0
Schwinger has pointed out that such traces are most easily evaluated in momentum space. We introduce a
complete orthonormal set of states p that are eigenstates of the operator p , where, in n dimensions,
Z1AVV
d
= exp Lim
0 d
x p = e
and
p f q =
ip . x
(3.1.6a)
2 n 2
f p q
2 n 2
( 3.1.6b)
On the right-hand side of Eq. (3.1.6b), f p q is the Fourier transform of f x :
f p q =
d nx
2
n
f x e ix. p q
(3.1.7)
2
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Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceEquation (3.1.5) takes the form,
T= d 4 p d 4 q d 4 r d 4 s p e 1u p t q
Upon inserting the complete set 1 d 4 p p p
(3.1.8) as,
V p q
T = d 4 p d 4 q d 4 r d 4 s
=
=
d 4 p d 4 q d 4 r
d 4 p d 4 q
2 2
4
2
2
2
r e up t s
q V r
s V r
(3.1.8)
at the appropriate places, and using (3.1.6), we rewrite Eq.
e ir .q
e 1u p t .
2
2
r q .
V r s
2
2
e urt .
2 2
s p
e 1u p tV p q V q s e uq t e is . p s p
e 1u p t uq tV p q V q p
2 4
e is . p
(3.1.9)
After shifting the variable of integration p p q , Eq. (3.1.9) becomes,
T=
d 4 pd 4 q
2
4
e q 1u p t V p V p
(3.1.10)
Upon substituting Eq. (3.1.10) into Eq. (3.1.4), we find that
A
Z1AVV = exp LimVV
0
where,
e2
2
A
=
VV
dt t 1
0
1
(3.1.11a)
d 4 pV p V p
du
0
d 4q
2 e
q 1u p t
4
(3.1.11b)
We use Eq. (2.10) to integrate over t, then (3.1.11b) becomes,
A
=
VV
=
e2
2
1
2
4
0
e 2 2
2
2
d 4q
d 4 pV p V p du
1
d 4 pV p V p du
0
q 1 u p 2
q 1 u p 2
2 4 q 2 1 u 2 p 2 2
d 4q
(3.1.12)
Now the last integral I 1 (say) of Eq. (3.1.12) can be calculated as follows:
q 1 u p 2
2 4 q 2 1 u 2 p 2 2
d 4 q q s 2 2q 1 1 u p ...... 1 u 2 p 2
=
2 4
q 2 1 u 2 p 2 2
d 4q
I1 =
Differentiating eq. (3.1.11b) with respect to and taking 0 , we see that the product terms in will
vanish. Hence in the numerator of I 1 only the first and last term will contribute.
2
d 4q
q 2 2
I1 =
2 4 q 2 1 u 2 p 2
2
1 u 2 p 2
q 2 1 u 2 p 2 2
To evaluate this integral let us consider the standard integral,
r
r n mr n
n
1
d nq
q2
2 2 r m
2
2
c
=
n
n
m
n
2
2
m
2 q c
16 2 4
2
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(3.1.13)
(3.1.14)
47 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceUsing eq. (3.1.14) in (3.1.13) we get,
I1 =
1 u 2 p 2
1
16
4
2
4
2
2
2
2
2
2
2 2
2
2
2
2 2
+ 1 u 2 p 2 1 u 2 p 2
=
2
1
11 2 1 u
2
1
16
2
20 2
.
2 2 2
2 2
1 + 1 1 u 2 p 2 p 2
2
p 2 2 3
Thus Eq. (3.1.12) becomes,
VV
=
1
e 2 2 1
2 2
2
16
2
1
1
d pV p V p du 1 2 1 u
0
4
p 2 2 3
2
1
+ 1 1 u 2 p 2 p 2
=
e
2
32
2
1
d
4
+ 1
=
e2
pV p V p
1
32 2
2
p 2 4 2
d 4 pV p V p 12 2
2
2
2
2
3
2
+ 1 1
=
e2
32
2
2
1 2
e2
4
1
p 2 p 2
3 s
1 1
1
p 2 2 1 2 2 2 + 1 1 1 p 2 p 2
3
3
2
2 2
1
2
=
d
1
32 2
1
1
p 2 2 3 1
pV p V p 1 2
3
2
1
p 2 p 2
3
4 6 3 8 2
p 2 + 11 31 p 2 p2
d 4 pV p V p 12 2
16 2 5 6
2
(3.1.15)
2
Now differentiating Eq. (3.1.15) w. r. to , we get
=
VV
e2
32
2
4
d 4 pV p V p +
1
12 2
16
where we have used .
4
2
6 3 8 2 2 5
1
2
2
p
1
+
2
2
16
2 5 6
2
3
2
5 6 4 18 16
.
2
2 5 6
4 6 3 8 2
.
2
5 6
p 2 ln p 1
+ 11 ln 1 1 31 p 2 p 2 + 12 32 p 2 p 2
+ 11 31 p 2 ln p p 2 + 11 31 p 2
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p 2 ln p 2
48 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional Space A 0 = LimVV
A
VV
0
=
=
e2
32
2
d
2
d
e2
96
4
4
1
1
2
1
pV p V p p 2 ln 1 p 2 p 2 ln p p 2 ln p
9
3
3
3
1
1
pV p V p p 2 ln 1 ln p 2
3
2
(3.1.16)
A
Substituting of Eq. (3.1.16) into Eq. (3.1.11a) yields our final expression for Z 1VV as,
e2
1
1
Z1AVV =exp
d 4 pV p V p p 2 ln 1 ln p 2
2
2
3
64
This contributes to the to the one–loop generating functional for external bosons (photon) lines.
(3.1.17)
To find one-loop correction for external boson lines from above generating functional, we have to take
logarithm on Eq. (3.1.17) and then functional differentiation of the expansion with respect to momentum p .
Thus the one-loop correction for the external boson lines is,
=
e2
64
2
e2
96
2
1
1
p 2 ln 1 ln p 2
3
2
1
1
p 2 ln p 2 ln 1
3
2
(3.1.18)
where,
the finite part is
e2
96 2
1
1
p 2 ln p 2
3
2
(3.1.19)
e2
(3.1.20)
p 2 ln 1
96 2
The result in (3.1.18) is of the same form as we obtained by the diagrammatic form of Operator
regularization and Dimensional regularization methods in section-II. In this section we have shown and
explained how one can choose the appropriate terms from the Schwinger expansion for the problem in hand.
and the divergent part is
3.2 One-Loop Generating Functional and Loop Corrections for External Fermion Lines and Vertex
Function
In this case we focus on the denominator in Eq. (3.8), so that let us regulate the det B through use of
the - function in Eq. (3.11a) yielding
1
det B = Z 1B = exp Lit B ,
0
2
where,
B
=
dt t
1
0
1
(3.2.1)
1
1
tr exp t p 2 1 p p e 2
p eV
e 2
1
p eV
(3.2.2)
In Eq. (3.2.2) it is understood that the exponential is trexp tB, where
1
1
1
B p 2 1 p p e 2
e 2
p eV
p eV
B0 B I
(3.2.3)
where, B0 is independent of the background field and , and B I is at least linear in and .
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49 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceNow as before to use Schwinger expansion in this case let us use the eq. (2.10) and then taking bilinear in and
on the on the right-hand side of Eq. (3.2.1), we end up with
d 1
1
1
(3.2.4)
dt t s trexp p 2 1 p p t 2e 2 D
1
Z 1B = exp Lim
s
0
2
ds
s
0
The exponential factor in the trace of Eq. (3.2.4) can be simplified using the complete set of orthonormal
projections operators:
p p
(3.2.5a)
T p
p 2
p p
(3.2.5b)
L p
p2
These allows us to write e tB0
as
1
exp p 2 1 p p
t =
n 0
1 2
1
p t T L
n!
= e tp T + e
2
tp 2
n
L
(3.2.6)
and let us expand D
1 in powers of the back-ground field in the -function Eq. (3.2.4):
D
1 =
1
1
1
1 eV
=
p
p
p eV
1
1 1
1 1
1
1
= eV eV eV .... ....
p
p
p
p
p p
It is interesting to note that at this stage this is straightforward to apply the perturbative expansion of Eq.
(2.10) to this -function and to select from the expansion those terms appropriate for any particular Greens
function. This means that from the expansion we can choose appropriate terms that are associated with the
related problems that we are interested in. Let us consider here the -function for the fermion two-point
function and the vertex function, we find
tp 2
tp 2
1 1
1
e2
(3.2.7)
e
V
e
T
e
L
B
dt t tr
p p
p
0
Following ref. [19] in the approach-A, we compute from Eq. (3.2.7) the -function in the limit of zero
momentum transfer to the photon:
B
e2
8
2
d
4
1 1
2
1
1
p 2
2
1
p 3
p
e
p p 2 V 0 2 2 p .V 0 p
4
p
Therefore by Eq. (3.2.1) the contributions to the one-loop generating functional is
2
e
e3
3
1 e
Z 1B exp 2 d 4 p ln ln p 2 p p 2 V 0 p
4
16 4
2
2 8
(3.2.8)
p.V 0
(3.2.9)
2
p
This contributes to the one-loop generating functional for external fermion (electron) lines and vertex function
in QED.
d 4 p p p p
To obtain the one-loop correction for external fermion lines and vertex function, we have to take logarithm
of Eq.(3.2.9) and then functional differentiation with respect to momentum p .
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50 | Page
Operator Regularization in Evaluating Feynman Diagrams in QED in (3+1) Dimensional SpaceHence from Eq. (3.2.9), we get
=
e2
16 2
e3 3
e3
3
2
2
ln
ln
p
p
V
0
p
ln ln p p p p
64 4 2
32 4
2
p p p
=
=
e2
16 2
e2
16
2
p V 0
p2
e3 3
3
2
2
ln ln p p p p
ln ln p p V 0 p
64 4 2
2
3
p p
e
p V 0 p
4
p2
32
e3
3
2
ln ln p p p p
64 4
2
e3
32 4
3
2
ln ln p p V 0 p
2
p
p
p V 0 p
p2
e3 3
3
2
2
ln
ln
p
p
p
p
ln ln p 2 p V 0 p
4
2
64 2
16 2
(3.2.10)
From the expansion (3.2.10) we can find the one-loop correction for the external fermion lines and one-loop
vertex function by choosing the appropriate terms. This expression is of the same form as obtained by DR and
OR methods with Feynman diagrams in section- II.
Thus the one-loop correction for the external fermion lines is,
=
e2
3
p ln ln p 2
16
2
and the one-loop correction to the vertex function is,
e2
2
(3.2.11)
e3 3
2
(3.2.12)
ln ln p 2
4 2
64
The result in (3.2.11) and (3.2.12) is of the same form as we obtained by the diagrammatic form of Operator
regularization and Dimensional regularization methods in section-II.
IV. Conclusion
In this paper we have evaluated basic QED loop diagrams in (3+1) dimensions with and without
considering Feynman diagrams by the Operator regularization. We have compared the results of both procedure
of Operator regularization and that of Dimensional regularization. We have seen that in both cases the result is
of the same form with Dimensional regularization except a finite constant term difference. This will not affect
the renormalization procedure.
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51 | Page
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