On the maximum atom-bond sum-connectivity index of graphs

Lemma 1

[12] Let G be a graph. If u and v are non-adjacent vertices in G, then A B S ( G + u v ) > A B S ( G ) .

Lemma 2

Let

where min { x , y } ≥ 1 . For every positive real number s, define the function g s ( x , y ) = f ( x + s , y ) − f ( x , y ) . Then f is strictly increasing in x and in y. The function g s is strictly decreasing and convex in x and in y.

Proof

The first and second partial derivatives of f with respect to x and y are calculated as

Clearly for x > 1 , ∂ f ∂ x ( x , y ) > 0 . Thus, f is strictly increasing in x and in y . Since ∂ 2 f ∂ x 2 ( x , y ) < 0 , whenever x > 1 , we obtain ∂ f ∂ x ( x , y ) is strictly decreasing in x when x ≥ 1 . So

and thus g s ( x , y ) = f ( x + s , y ) − f ( x , y ) is strictly decreasing in x when x ≥ 1 . Additionally ∂ 2 f ∂ x 2 ( x , y ) is strictly increasing when x ≥ 1 . So

and hence g s ( x , y ) , is convex in x when x ≥ 1 .□

Lemma 3

Let M and N be real numbers satisfying 1 ≤ M ≤ N . Then for every positive real number s, the function h s ( x ) = g s ( x , N ) − g s ( x , M ) is increasing in x when x ≥ 1 .

Proof

When x ≥ 1 , we have ∂ 2 f ∂ x ∂ y is strictly increasing in x . So

Thus, ∂ g s ∂ x is increasing in y , and hence

Therefore, h s ( x ) is increasing in x when x ≥ 1 .□

Lemma 4

[16] Let n and r be integers such that 2 ≤ r < n .

1. If nr is even, then there is a connected r-regular graph of order n.

2. If nr is odd, then there is a connected nearly r-regular graph of order n.

Theorem 1

Let n ≥ 2 and δ ≥ 1 . If G ∈ ϒ n , δ , then

with equality if and only if G ≅ D n , δ .

Proof

Let G ∗ ∈ ϒ n , δ be having the maximum A B S . Let u ∈ V ( G ) such that d u = δ . Assume there are two nonadjacent vertices v , w ∈ V ( G ) \ { u } . Then G ′ = G ∗ + { v w } ∈ ϒ n , δ and A B S ( G ′ ) > A B S ( G ∗ ) , which is a contradiction. Therefore, G ∗ ≅ D n , δ .□

Theorem 2

Let n ≥ 2 and Δ ≥ 1 be integers.

1. If Δ n is even, then for each G ∈ Ψ n , Δ ,

   (2) A B S ( G ) ≤ n 2 Δ ( Δ − 1 )

   with equality if and only if G is Δ -regular.

2. If Δ n is odd, then for each G ∈ Ψ n , Δ ,

   (3) A B S ( G ) ≤ ( Δ − 1 ) 2 Δ − 3 2 Δ − 1 + Δ n − 2 Δ + 1 2 Δ − 1 Δ

   with equality if and only if G is nearly Δ -regular.

Proof

(1) Since Δ n is even, by Lemma 4, there is a Δ -regular graph of order n . Let G ∈ Ψ n , Δ . Then for each u v ∈ E ( G ) , we have f ( d u , d v ) ≤ f ( Δ , Δ ) . Additionally, we have

Thus,

Moreover, the equality holds if and only if ∣ E ( G ) ∣ = Δ n 2 and for all u v ∈ E ( G ) , f ( d u , d v ) = f ( Δ , Δ ) . Consequently, the equality holds if and only if G is Δ -regular.

(2) Since Δ n is odd, by Lemma 4, there is nearly Δ -regular graph of order n . Let G ∈ Ψ n , Δ . Clearly G is not Δ -regular because Δ n is odd. Let v ∈ V ( G ) such that d v < Δ and let M be the set of all edges incident with v . Then we have for each u ∈ N ( v ) , f ( d u , d v ) ≤ f ( Δ , Δ − 1 ) and for each u w ∈ E ( G ) \ M , f ( d u , d w ) ≤ f ( Δ , Δ ) . Additionally,

Thus,

Since 2 Δ − 3 2 Δ − 1 − Δ − 1 4 Δ > 0 and d v ≤ Δ − 1 , we obtain

Moreover, the equality holds if and only if d v = Δ − 1 , f ( d u , d v ) = f ( Δ − 1 , Δ ) for all u ∈ N ( v ) , and f ( u , w ) = f ( Δ , Δ ) for all u w ∈ E ( G ) \ M . Thus, the equality holds if and only if G is nearly Δ -regular.□

Theorem 3

Let n ≥ 5 and χ ≥ 3 and let q = ⌊ n ∕ χ ⌋ and r = n − q χ . If G ∈ Π n , χ , then

with equality if and only if G ≅ T n , χ .

Proof

Let G ∗ ∈ Π n , χ be having the maximum A B S . The vertex set V ( G ) of G can be partitioned into χ subsets, say Y 1 , Y 2 , … , Y χ such that ∣ Y i ∣ = k i for i = 1 , 2 , … , χ , provided that k 1 ≤ k 2 ≤ ⋯ ≤ k χ . Consequently, G ∗ is isomorphic to a χ -partite graph. Thus, by Lemma 1, it must be isomorphic to the complete χ -partite graph K k 1 , k 2 , … , k χ . It is remaining to show that k χ − k 1 ≤ 1 . Seeking a contradiction, assume that k χ − k 1 ≥ 2 . Let H ≅ K d 1 , d 2 , … , d χ , where d 1 = k 1 + 1 , d χ = k χ − 1 , and d i = k i for every i ∈ { 2 , … , χ − 1 }

Since n − k 1 − 1 ≥ n − k χ + 1 , from Lemma 2 we obtain that for each i = 2 , … , χ − 1 , f ( n − k 1 − 1 , n − k i ) − f ( n − k χ + 1 , n − k i ) ≥ 0 and k χ g 1 ( n − k χ , n − k i ) − k 1 g 1 ( n − k 1 − 1 , n − k i ) > k 1 [ g 1 ( n − k χ , n − k i ) − g 1 ( n − k 1 − 1 , n − k i ) ] ≥ 0 . So A B S ( H ) − A B S ( G ∗ ) > 0 , which is a contradiction. Thus, k χ − k 1 ≤ 1 .□

Theorem 4

Let n and α be positive integers. If G ∈ Σ n , α , then

with equality if and only if G ≅ N α + K n − α .

Proof

Let G ∗ be having the maximum A B S value in Σ n , α . Let W be an independent set in G ∗ with ∣ W ∣ = α . Assume that there are two non-adjacent vertices u and v such that u ∈ W and v ∈ V ( G ∗ ) − W . Then G ∗ + u v ∈ Σ n , α and A B S ( G ∗ + u v ) > A B S ( G ∗ ) , which is a contradiction. So, each vertex in W is adjacent to every vertex in V ( G ∗ ) − W . Furthermore, every pair of vertices in V ( G ∗ ) − W are adjacent, yielding G [ V ( G ∗ ) − W ] ≅ K n − α . Thus, G ∗ ≅ N α + K n − α . Therefore,

Theorem 5

Let G ∈ Γ n , p .

1. If p = n − 1 , then G ≅ A B S ( S n − 1 ) , and thus A B S ( G ) = ( n − 1 ) n − 2 n .

2. If p = n − 2 , then A B S ( G ) ≤ 1 3 + n − 2 n + ( n − 3 ) n − 3 n − 1 , with equality if and only if G ≅ S 2 , n − 2 .

3. If p ≤ n − 3 , then

   A B S ( G ) ≤ p n − 2 n + ( n − p − 1 ) 2 n − 2 p − 3 2 n − 2 p − 1 + n − p − 1 ( n − p − 2 ) 3 2 2 ,

   with equality if and only if G ≅ K n − p p .

Proof

(1) Straightforward.

(2) Let u and v be the internal vertices of G . We may assume that there are t pendent vertices adjacent to u and p − t pendent vertices adjacent to v . Thus,

Consider the function h ( t ) = t f ( 1 , t + 1 ) + ( p − t ) f ( 1 , p − t + 1 ) .

where M = ( ( p − 1 ) t − t 2 + 3 p + 6 ) p t − t 2 + 2 t and N = ( p − t + 3 ) ( t + 2 ) ( p − t ) ( t + 2 ) . Clearly, both M > 0 and N > 0 when 1 ≤ t ≤ p − 1 . Thus the sign of h ′ ( t ) is determined by the sign of ( M − N ) ( M + N ) = M 2 − N 2 . Now

Hence, h ′ ( t ) < 0 when 1 ≤ t < p ∕ 2 and h ′ ( t ) > 0 when p ∕ 2 < t ≤ p − 1 . So h ( t ) has maximum value at t = 1 or t = p − 1 . Thus,

with equality if and only if G ≅ S 2 , p = S 2 , n − 2 .

(3) Let P be the set of pendent vertices in G . If there are two nonadjacent vertices u , v ∈ V ( G ) \ P , then G + { u v } ∈ Γ n , p and by Lemma 1, A B S ( G + { u v } ) > A B S ( G ) , which is a contradiction. Thus, the induced subgraph G [ V ( G ) \ P ] ≅ K n − p . Label the vertices of G [ V ( G ) \ P ] by u 1 , … , u n − p and for each j = 1 , … , n − p , let d j = ∣ N ( u j ) ∩ P ∣ so that d 1 ≥ d 2 ≥ … ≥ d n − p . To obtain the desired result we need to show that d 1 = p and d 2 = … = d n − p = 0 . So seeking a contradiction assume that d j ≥ 1 for some j ≥ 2 . Then d 1 ≥ d 2 ≥ 1 . Let x ∈ P ∩ N ( u 2 ) and take G ′ = G − { x u 2 } + { x u 1 } . Note that for each j = 1 , … , n − p , d e g G ( u i ) = d i + n − p − 1 . Then

Since d j ≥ 0 for j = 3 , … , n − p , by Lemma 3, we have

Thus,

where

Our next aim is to show that w ( t ) is increasing in t . Note that

Since ∂ g 1 ∂ x ( x , y ) is increasing in y when y ≥ 1 , we obtain

So

where

Since

we obtain w ′ ( t ) ≥ L − K > 0 , and thus w ( t ) is increasing in t as desired. This implies that A B S ( G ′ ) − A B S ( G ) > w ( d 1 ) − w ( d 2 − 1 ) > 0 , which is a contradiction. So d 1 = p and d j = 0 for all j = 2 , … , n − p − 1 , and hence G ≅ K n − p p .□