Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 10s (2025) 

  

561 
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∪ 

A Study on the Differential Value of Total Graph 

 

D. Muralidharan,1 M.S. Paulraj2 and D. Yokesh3 

1Department of Mathematics, Sri Sairam Institute of Technology, Chennai, Tamil Nadu, India. 

2Department of Mathematics, A.M. Jain College, Chennai, Tamil Nadu, India. 

3Department of Mathematics, Anand Institute of Technology, Chennai, Tamil Nadu, India. 

E-Mail: murali.maths@sairamit.edu.in 

 

Article History: 

Received: 12-01-2025 

Revised: 15-02-2025 

Accepted: 01-03-2025 

Abstract:   

Let 𝐺 = (𝑉, 𝐸) be a graph and X be a subset of V. Let 𝐵(𝑋) be the set of vertices 

in V − X that has a neighbour in a set X. The differential of a set X, is defined as 

∂(X) which is |B(X)| − |X| and the differential of a graph is ∂(G) = max {∂(X)/X ⊂ 

V}. The total graph T (G) of a  g raph  G is the graph whose vertex set is V (G) 

∪ E(G) with two vertices of T (G) being adjacent if and only if the corresponding 

elements of G are either adjacent or incident. In this paper, we study the differential value 

of total graph for some standard graphs and its bounds. 

Keywords: total graph, domination number 

Mathematics Subject Classification 05C38, 05C69 

 

1. Introduction 

Throughout this paper, 𝐺 = (𝑉, 𝐸) is a simple finite graph of 𝑛 vertices. For theoretical 

terminology about graph which is not given here, we refer to Harary [7]. For a vertex v ∈ V, 

the open neighbourhood of 𝑣 is 𝑁 (𝑣)  = {𝑢 ∈  𝑉/𝑢𝑣 ∈  𝐸} and the closed neighbourhood 

of the set 𝑁 [𝑣]  =  𝑁 (𝑣)  ∪  {𝑣}. For a set X ⊂ V, its open neighbourhood 𝑁(𝑋) =

⋃ 𝑁(𝑣)𝑣𝜖𝑉   and 𝑁[𝑋] = 𝑁(𝑋) ∪ 𝑋 is the closed neighbourhood.  A set 𝐷 ⊆  𝑉 is a dominating 

set [6,10] of 𝐺 if every vertex in 𝑉 −  𝐷 is adjacent to some vertex in 𝐷. The boundary 

𝐵(𝑋) of a set 𝑋 is defined to be the set of vertices in 𝑉 − 𝑋 dominated by vertices in 𝑋, that 

is 𝐵(𝑋) = (𝑉 − 𝑋) ∩ 𝑁(𝑋). The differential 𝜕(𝑋) of 𝑋 is defined as |𝐵(𝑋)| − |𝑋|. The 

differential of a graph G is 𝜕(𝐺) = 𝑚𝑎𝑥{𝜕(𝑋)/𝑋 ⊂ 𝑉}. If 𝑆 ⊂  𝑉 and 𝜕(𝐺) = 𝜕(𝑆), then 

S is a ∂-set. The differential of a set was first defined by Hedetniemi and later studied by 

Mashburn et al. and Goddard and Henning [3,4,9,13]. A graph 𝐺 is complete graph if every 

distinct pair of vertices are adjacent. A complete bipartite graph is a special type of bipartite graph 

where every vertex of one set is connected to every other vertex of other set. A complete binary 

tree is a special type of binary tree where all the levels of the tree are filled completely except 

the lowest level nodes which are filled from as left as possible. A graph 𝐺 is said to be dominant 

differential if it contains a 𝜕 − set which is also a dominating set. Some examples are complete 

graph and wheel graph.  

mailto:murali.maths@sairamit.edu.in


Communications on Applied Nonlinear Analysis 

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2 

The total graph 𝑇 (𝐺) [1,2,5,8,11,12,14] of 𝐺 is the graph whose vertex set is 𝑉 (𝐺)  ∪

 𝐸(𝐺) with two vertices of 𝑇 (𝐺) being adjacent if and only if the corresponding elements of 

𝐺 are either adjacent or incident and |𝑉 (𝑇 (𝐺))| = 𝑙. 

In this paper, we study the differential value of total graph of some standard graphs and its bounds. 

2. Results 

Theorem 1. For any graph G of n vertices, 1 ≤ ∂(T(G)) ≤
n(n−1)

2
 

Theorem 2. If T (G) is the total graph of a graph G, then ∂(T(G)) ≤
n(n−1)

2
            for 

n ≥ 2. 

Proof. We have to prove by induction method. We have to prove that the result is true for n 

= 2. When n = 2, then 𝜕(𝑇 (𝐺))  ≤  
2(2−1)

2
=  1. It is always true. We assume that the result 

is true for n = k. Then, 𝜕(𝑇 (𝐺))  ≤  
𝑘(𝑘−1)

2
. We have to prove that the result is true for 𝑛 =

 𝑘 +  1. Consider 𝐺 with 𝑛 =  𝑘 +  1 vertices. Removing a vertex v ∈ T (G) and |V (T (G)) 

− {v}| = k, hence by induction hypothesis, 𝜕(𝑇(𝐺) − {𝑣}) + k ≤
k(k−1)

2
+ k = 

k2−k+2k

2
=

k2+k

2
=

k(k+1)

2
. So, 𝜕(𝑇 (𝐺))  ≤  

𝑘(𝑘+1)

2
. Therefore, the result is true for 𝑛 =  𝑘 +  1 and the 

result is true for all 𝑛. So, ∂(T(G)) ≤
n(n−1)

2
.  

Theorem 3. If G is a complete graph, then ∂(T(G)) =
𝑛(𝑛−1)

2
, for n ≥ 2. 

Proof. Given that 𝐺 is a complete graph. Consider 𝑆 is a 𝜕− set of 𝑇(𝐺). When 𝑛 is even, 

choose any arbitrary vertex 𝑢1in 𝑆 and choose next vertex 𝑢2 which is not adjacent with 𝑢1. Choose 

next vertex 𝑢3 which is not adjacent with 𝑢1 and 𝑢2. Continuing this process until |S| =
n

2
. 

Clearly, 𝑆 is a dominant differential of 𝑇 (𝐺). Therefore, ∂(T(G)) = |B(S)| − |S| =

[
n(n+1)

2
−
n

2
] −

n

2
=

n(n−1)

2
. If 𝑛 is odd, we choose the vertices in S as we discussed in the 

above case and S dominates all the vertices except one in the case. Therefore, 

∂(T(G)) = [
n(n+1)

2
−
n−1

2
− 1] −

n−1

2
=

n(n−1)

2
. In both the cases, ∂(T(G)) =

𝑛(𝑛−1)

2
. 

Hence the proof.  

Theorem 4. Given a positive integer k, there exist a graph on n vertices whose total graph 

on 𝑙 vertices with 𝜕(𝑇(𝐺)) = 𝑘 

Proof. If 𝑛 is odd, we consider the graph with 𝑛 =
𝑘+5

2
 vertices whose total graph has 

𝑘+3

2
 

copies of 𝑘3 with exactly one vertex as common, say 𝑣. Clearly, 𝜕− set of 𝑇(𝐺) contains 𝑣 

only. Hence 𝜕(𝑇(𝐺)) = 𝑘. If 𝑛 is even, when 𝑛 = 4, we consider the circulant graph C4,2. 

For other cases, we consider the circulant graph G = C4,2 in which 
𝑘−6

2
  copies of P2 which is 

attached with a vertex, say u of C4,2. Let S be the 𝜕− set of above graphs. Consider S = {u, v} 

where 𝑣 ∉ 𝑁(𝑢). So, 𝜕(𝑇(𝐺)) = 𝑘. 



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Observation 5. For any graph G with n vertices, ∂(T(G)) ≠ 2. 

Theorem 6. For any graph G with n vertices, 1 ≤ ∂(T(G)) − ∂(G) ≤
n2−3n+4

2
 

Proof. Given that 𝐺 is a graph of 𝑛 vertices. The maximum value of the differential of any 

total graph is less than or equal to 
𝑛(𝑛−1)

2
 and clearly 𝜕(𝐺) ≤ 𝑛 − 2. Therefore, ∂(T(G)) −

∂(G) ≤
𝑛(𝑛−1)

2
− (𝑛 − 2) =

n2−3n+4

2
.  

Observation 7. For any graph G with n vertices, ∂(T(G)) − ∂(G) =
n2−3n+4

2
 if and only if G 

is a complete graph. 

 

Theorem 8. If G = 𝐾1,𝑛−1 is a star graph, then ∂ (T(𝐾1,𝑛−1)) = 2𝑛 − 3 

Proof. Let 𝑉(𝐾1,𝑛−1) = {𝑣, 𝑣1 , 𝑣2…𝑣𝑛−1} and 𝐸(𝐾1,𝑛−1) = {𝑒1, 𝑒2…𝑒𝑛−1} where 𝑒𝑖 = 𝑣𝑣𝑖, 

𝑖 = 1,2, …𝑛 − 1 and 𝑣 is a head vertex of the star graph. By the definition of total graph, 

𝑉(𝐾1,𝑛−1) = {𝑣1, 𝑣2…𝑣𝑛−1, 𝑒1, 𝑒2…𝑒𝑛−1} and |𝑉(𝐾1,𝑛−1)| = 2𝑛 + 1. Since 𝑑𝑒𝑔(𝑣) = 2𝑛 

and 𝑆 = {𝑣} is the differential set, then  ∂ (T(𝐾1,𝑛−1)) = 2𝑛 − 3 

 

Theorem 9. For any graph G = 𝐶𝑛, then 𝜕(𝑇(𝑃𝑛)) =

{
 
 

 
 3 ⌊

2𝑛

5
⌋ , 𝑛 ≡ 0,1,3 (𝑚𝑜𝑑 5)

3 ⌊
2𝑛

5
⌋ + 1, 𝑛 ≡ 4 (𝑚𝑜𝑑 5)

3 ⌊
2𝑛

5
⌋ + 2, 𝑛 ≡ 2 (𝑚𝑜𝑑 5)

 

Proof. let 𝑉(𝐶𝑛) = {𝑣1, 𝑣2, … . . , 𝑣𝑛} be the vertices of cycle of length 𝑛 (𝑛 ≥ 3)   and 𝐸(𝐶𝑛) =

{𝑒1, 𝑒2, … . . , 𝑒𝑛} be the edges of the corresponding vertices {𝑣1, 𝑣2, … . . , 𝑣𝑛}. Then, 𝑉(𝑇(𝐶𝑛)) =

{𝑣1, 𝑣2, … . . , 𝑣𝑛, 𝑒1, 𝑒2, … . . , 𝑒𝑛} and 𝐸(𝑇(𝐺)) = {𝑣𝑖𝑣𝑖+1/1 ≤ 𝑖 ≤ 𝑛 − 1} ∪ {𝑒𝑖𝑒𝑖+1/1 ≤ 𝑖 ≤ 𝑛 − 1} ∪

{𝑣𝑖𝑒𝑖/1 ≤ 𝑖 ≤ 𝑛} ∪ {𝑒𝑖𝑣𝑖+1/1 ≤ 𝑖 ≤ 𝑛 − 1} ∪ {𝑒𝑛𝑣1, 𝑒𝑛𝑒1, 𝑣𝑛𝑣1} and hence  |𝑉(𝑇(𝐶𝑛))| = 2𝑛. Let 𝑆 

be the differential set of 𝑇(𝐶𝑛). When 𝑛 ≡ 0 (𝑚𝑜𝑑 5). Consider 𝑆 = {𝑣1, 𝑒3, 𝑣6, … . . , 𝑣𝑛−4, 𝑒𝑛−2} be a 

𝜕 − set and hence. |𝑆| =
2𝑛

5
 and |𝐵(𝑆)| =

8𝑛

5
. Then, 𝜕(𝑇(𝐺)) = |𝐵(𝑆)| − |𝑆| = 3 ⌊

2𝑛

5
⌋. When 𝑛 ≡

1 (𝑚𝑜𝑑 5). Consider the differential set 𝑆 = {𝑣1, 𝑒3, 𝑣6, … . . , 𝑒𝑛−3, 𝑣𝑛−5} and 𝐶(𝑆) = {𝑒𝑛−1, 𝑣𝑛−1}. 

Clearly, |𝑆| = ⌊
2𝑛

5
⌋ and |𝐵(𝑆)| = ⌊

8𝑛

5
⌋. So, 𝜕(𝑇(𝐺)) = 3 ⌊

2𝑛

5
⌋. When 𝑛 ≡ 3 (𝑚𝑜𝑑 5). Consider the 

differential set 𝑆 = {𝑣1, 𝑒3, 𝑣6, … . . , 𝑒𝑛−5, 𝑣𝑛−2} and 𝐶(𝑆) = {𝑒𝑛−1}. Clearly, |𝑆| = ⌊
2𝑛

5
⌋ and |𝐵(𝑆)| =

⌊
8𝑛

5
⌋. So, 𝜕(𝑇(𝐺)) = |𝐵(𝑆)| − |𝑆| = 3 ⌊

2𝑛

5
⌋. When 𝑛 ≡ 4 (𝑚𝑜𝑑 5) Consider the set 𝑆 = 𝑆1 ∪ 𝑆2 where 

𝑆1 = {𝑣1, 𝑒3, 𝑣6, … . . , 𝑒𝑛−6, 𝑣𝑛−3} and 𝑆2 = {𝑣𝑛−1}.  Then 𝜕(𝑆2) = 1. So, 𝜕(𝑇(𝐺)) = 3 ⌊
2𝑛

5
⌋ + 1. 

When 𝑛 ≡ 2 (𝑚𝑜𝑑 5). Consider the set 𝑆 = 𝑆1 ∪ 𝑆2 where 𝑆1 = {𝑣1, 𝑒3, 𝑣6, … . . , 𝑒𝑛−4, 𝑣𝑛−6} and 𝑆2 =

{𝑣𝑛−1} . Then 𝜕(𝑆2) = 2. So, 𝜕(𝑇(𝐺)) = 3 ⌊
2𝑛

5
⌋ + 2. 

 

Theorem 10. For any path 𝐺 = 𝑃𝑛(𝑛 ≥ 5), then                                                  



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𝜕(𝑇(𝑃𝑛)) =

{
 
 

 
 3 ⌊

2𝑛 − 1

5
⌋ , 𝑛 ≡ 1,3,4 (𝑚𝑜𝑑 5)

3 ⌊
2𝑛 − 1

5
⌋ + 1, 𝑛 ≡ 2(𝑚𝑜𝑑 5)

3 ⌊
2𝑛 − 1

5
⌋ + 2, 𝑛 ≡ 0 (𝑚𝑜𝑑 5)

 

Proof. let 𝑉(𝑃𝑛) = {𝑣1, 𝑣2, … . 𝑣𝑛} be the vertices of path of length 𝑛 (𝑛 ≥ 3)   and       𝐸(𝑃𝑛) =

{𝑒1, 𝑒2, … . . , 𝑒𝑛−1}. Then, 𝑉(𝑇(𝑃𝑛)) = {𝑣1, 𝑣2, … . . , 𝑣𝑛, 𝑒1, 𝑒2, … . . , 𝑒𝑛−1} and |𝑉(𝑇(𝑃𝑛))| = 2𝑛 − 1. 

When 𝑛 ≡ 1 (𝑚𝑜𝑑 5). Consider the differential set of a total graph 𝑆 = {𝑣2, 𝑒4, … . . , 𝑒𝑛−2, 𝑣𝑛−4} and 

𝐶(𝑆) = {𝑣𝑛}. Then, |𝐵(𝑆)| = ⌊
8𝑛−4

5
⌋ and |𝑆| = ⌊

2𝑛−1

5
⌋. So,  𝜕(𝑇(𝐺)) = 3 ⌊

2𝑛−1

5
⌋. When 𝑛 ≡

3 (𝑚𝑜𝑑 5), Consider the differential set of a total graph  𝑆 = {𝑣2, 𝑒4, … . . , 𝑒𝑛−4, 𝑣𝑛−1}. Then 

𝜕(𝑇(𝐺)) = 3 ⌊
2𝑛−1

5
⌋. When 𝑛 ≡ 4 (𝑚𝑜𝑑 5). Consider the differential set 𝑆 = {𝑣2, 𝑒4, … . . , 𝑒𝑛−5, 𝑣𝑛−2} 

and 𝐶(𝑆) = {𝑣𝑛, 𝑒𝑛−1}. Then 𝜕(𝑇(𝐺)) = 3 ⌊
2𝑛−1

5
⌋. When 𝑛 ≡ 2 (𝑚𝑜𝑑 5), there are three possible 

differential sets 𝑆 = 𝑆1 ∪ {𝑣𝑛−1} or 𝑆 = 𝑆1 ∪ {𝑣𝑛}  or 𝑆 = 𝑆1 ∪ {𝑒𝑛−1} where 𝑆1 =

{𝑣2, 𝑒4, … . , 𝑒𝑛−3, 𝑣𝑛−5}. Then 𝜕(𝑇(𝐺)) = 3 ⌊
2𝑛−1

5
⌋ + 1. When 𝑛 ≡ 0 (𝑚𝑜𝑑 5), the possible 

differential set 𝑆 = 𝑆1 ∪ 𝑆2 where 𝑆1 = {𝑣2, 𝑒4, … . , 𝑒𝑛−6, 𝑣𝑛−3} and 𝑆2 = {𝑒𝑛−2}. Then, 𝜕(𝑇(𝐺)) =

3 ⌊
2𝑛−1

5
⌋ + 2. 

 

Theorem 11. If 𝐺 = 𝐾𝑚1×𝑛1 is a complete bipartite graph, then 𝜕 (𝑇(𝐾𝑚1×𝑛1  )) = (𝑚1 + 1)𝑛 −

𝑚1. 

Proof. Since 𝐺 = 𝐾𝑚1×𝑛1 is a complete bipartite graph, the vertex set can be partitioned into two 

disjoint non empty sets 𝑉1(𝐺) = {𝑢1, 𝑢2…𝑢𝑚1
} and 𝑉2(𝐺) = {𝑣1, 𝑣2, … 𝑣𝑛1}.  Here, 𝑉 (𝑇(𝐾𝑚1×𝑛1

)) =

{𝑢𝑖/1 ≤ 𝑖 ≤ 𝑚1} ∪ {𝑣𝑗/1 ≤ 𝑗 ≤ 𝑛1} ∪ {𝑒𝑖𝑗/1 ≤ 𝑖 ≤ 𝑚1, 1 ≤ 𝑗 ≤ 𝑛1}  and |𝑉 (𝑇(𝐾𝑚1×𝑛1
))| = 𝑚1 +

𝑛1 +𝑚1𝑛1 . Clearly, the differential set  𝑆 = {𝑢1, 𝑢2…𝑢𝑚1
}. Since 𝑢1 is adjacent with 2𝑛1 vertices 

and 𝑢2 is adjacent with 𝑛1 vertices and so on. Therefore 𝜕 (𝑇(𝐾𝑚1×𝑛1
)) = (2𝑛1 − 1) + (𝑛1 − 1) +

⋯+ (𝑛1 − 1) = (𝑚1 + 1)𝑛1 − 𝑚1. 

 

Theorem 12. If 𝐺 is a complete binary tree, then          

                               𝜕(𝑇(𝐺)) = {

2𝑘+3−7

3
, 𝑘 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

2𝑘+3−5

3
 , 𝑘 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

  

Proof. Let Sk be the set of all vertices in level 𝑘 and |𝑆𝑘| = 2
𝑘  where 0 ≤ 𝑘 ≤ 𝑚 and 𝑚 is a 

positive integer.  

case(i) 𝑘 is an odd integer. Clearly, 𝑆𝑘−1 ∪ 𝑆𝑘−3 ∪…∪ 𝑆2 ∪ 𝑆0 is a 𝜕 −set. 

𝜕(𝑇(𝐺)) =  2𝑘+1 + 2𝑘−1 + 2𝑘−3 +⋯+ 24 + 22 − 1 

    = 2𝑘+1 [1 +
2𝑘−1

2𝑘+1
+
2𝑘−3

2𝑘+1
+⋯+

24

2𝑘+1
+

22

2𝑘+1
] − 1 



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    = 2𝑘+1[1 + 2−2 + 2−4 +⋯+ 2−𝑘+3 + 2−𝑘+1] − 1 

    = 2𝑘+1 [1 + (
1

22
)
1

+ (
1

22
)
2

+ …+ (
1

22
)

𝑘−3

2
+ (

1

22
)

𝑘−1

2
] − 1 

    = 2𝑘+1 [ 
1−(

1

22
)

𝑘+1
2

1−(
1

22
)
] − 1 

    = 2𝑘+1 [ 
1−(

1

22
)

𝑘+1
2

(
3

22
)

] − 1 

    = 2𝑘+1 ×
22

3
× [  1 −  

1

2𝑘+1
   ] − 1 

    = [  
2𝑘+3−4

3
 ] − 1 = [  

2𝑘+3−7

3
 ] 

case(ii) 𝑘 is an even integer. Clearly, 𝑆𝑘−1 ∪ 𝑆𝑘−3 ∪…∪ 𝑆3 ∪ 𝑆1 is a 𝜕 −set. 

𝜕(𝑇(𝐺)) =  2𝑘+1 + 2𝑘−1 + 2𝑘−3 +⋯+ 25 + 23 + 1 

    = 2𝑘+1 [1 +
2𝑘−1

2𝑘+1
+
2𝑘−3

2𝑘+1
+⋯+

25

2𝑘+1
+

23

2𝑘+1
] + 1 

    = 2𝑘+1[1 + 2−2 + 2−4 +⋯+ 2−𝑘+4 + 2−𝑘+2] + 1 

    = 2𝑘+1 [1 + (
1

22
)
1

+ (
1

22
)
2

+ …+ (
1

22
)

𝑘−4

2
+ (

1

22
)

𝑘−2

2
] + 1 

    = 2𝑘+1 [ 
1−(

1

22
)

𝑘
2

1−(
1

22
)
] + 1 

    = 2𝑘+1 ×
22

3
× [  1 −  

1

2𝑘
   ] + 1 

    = [  
2𝑘+3−8

3
 ] + 1 = [  

2𝑘+3−5

3
 ] 

 

Theorem 13. If 𝐺 = 𝑃2 × 𝑃𝑛1 is a grid graph with then 𝜕(𝑇(𝑃2 × 𝑃𝑛)) = 𝑙 − 2𝑛1when 𝑛1 ≥ 2. 

Proof. Let 𝑇(𝑃2 × 𝑃𝑛1) be the total graph of the grid graph and the vertex set of the graph is 

𝑉 (𝑇(𝑃2 × 𝑃𝑛1)) = {𝑣11, … 𝑣1𝑛1, 𝑣21, … 𝑣2𝑛1 , 𝑢11, . . , 𝑢1𝑛1 , 𝑒11, … 𝑒1(𝑛1−1), 𝑒21, … . 𝑒2(𝑛1−1)}  

where 𝑢𝑖𝑗 = 𝑣𝑖𝑗𝑣(𝑖+1)𝑗, 𝑖, 𝑗 = 1,2,… , 𝑛1 and |𝑉 (𝑇(𝑃2 × 𝑃𝑛1))| = 5𝑛1 − 2 = 𝑙. Consider the 

differential set of the given grid graph 𝐺 is 𝑆 = {𝑣11, 𝑣14, 𝑢12, 𝑢15, 𝑣23, … . . 𝑣1(𝑛1−1), 𝑢1𝑛1 , 𝑣2(𝑛1−2)} 

which is also an 𝛾 −set. Therefore, 𝜕 (𝑇(𝑃2 × 𝑃𝑛1)) = 𝑙 − 2𝛾(𝐺) = 1 − 2𝑛1. 

3. References 

[1]  Athul,T.B., and Suresh Singh, G. Total graph of Regular graphs. Advances in 

Mathematics:Scientific Journal. 2020, 9, 4213-4220. ISSN 1857-8365. 

[2]  Behzad,M. The connectivity of Total graphs. Bulletin Australian Mathematical Society, 1969, 

1, 175-181 

[3]  Bermudo, S., and Fernau, H. Lower bound on the differential of a graph. Discret. Math. 2012, 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 10s (2025) 

  

566 
https://internationalpubls.com 

312, 32363250, doi:10.1016/j.disc.2012.07.021. 

[4]  Bermudo, S., Rodrguez, J.M., and Sigarreta, J.M. On the differential in graphs. Util. Math. 

2015, 97, 257270. 

[5]  Ganghua Xie., Yinkui Li. Cycle Multiplicity of Total graph of complete bipartite graph. Open 

Journal of Discrete Mathematics, 2013, 13, 95-99. 

[6]  Goddard,W and Henning,M.A Generalized domination and independence in graphs, Congr. 

(1997) 161-171. 

[7] Harary,F Graph Theory, Addison Wesley, Reading Mass., 1972. 

[8]  Li,Y.K. Cycle Multiplicity of Some Total graphs. Applied Mathematics and 

Computation, 2017, 107-113. 

[9] Mashburn.J.L., Haynes,T.W., Hedetniemi, S.M., Hedeteniemi, S.T., and Slater, P.J., 

Differential in graphs, Utilitas Mathematica 69, 43-54, 2006. 

[10] Mohammed Alatif, Puttaswamy and Nayaka, S.R, Boundary domination in total graphs, 

Gen.Math.Notes, Vol. 32, No. 1, January 2016, pp.12-20, ISSN 2219-7184 

[11] Murugesan.N, and Nair,D,S, (1,2) - domination in total graph of K1,n, Cn and Pn, 

Advances in Mathematics, 4(3) (2013), 516-526 

[12] Muthuramakrishnan,D., and Jayaraman,G.,Total Chromatic number of Middle and Total 

graph of Path and Sunlet graph, International Journal of Scientific and Innovative 

Mathematical Research,m2018, 4, 1-9. 

[13] Pushpam,P.R.L., and Yokesh.D, Differential in certain classes of graphs, Tamkang J. 

Math., 41 (2) (2010) 129–138. 

[14] Rajeswari,V., and Thiagarajan,K., SIBEDE Approach for Total graph of Path and 

Cycle graphs, Middle-East Journal of Scientific Research, 2017, 25, 1553-1558.