# Certain Energies of Graphs for Dutch Windmill and Double-Wheel Graphs

Authors: Jing Wu; Muhammad Arfan Ali; Hafiz Mutee ur Rehman; Yan Dou
Journal: Journal of Mathematics
 (2022)

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## Abstract

Energy of a graph is defined as the sum of the absolute values of the eigenvalues of the adjacency matrix associated with the graph. In this research work, we find color energy, distance energy, Laplacian energy, and Seidel energy for the Dutch windmill graph of cycle lengths 4, 5, and 6. Also, we find the lower bounds of the double-wheel graph for energy, Seidel energy, color energy, distance energy, Laplacian energy, and Harary energy.


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## Body

## 1. Introduction

Energy of a graph is one of the important concepts in graph theory. The energy of a graph is mostly studied in the context of spectral graph theory. Gutman [1] introduced the concept of energy of a graph for the first time. Gutman defined energy of a graph as the sum of the absolute values of the eigenvalues of the adjacency matrix associated with the graph. Adjacency matrix is used to represent a finite graph, and it is a square matrix. The entries of the adjacency matrix provide information whether any two vertices are joined with each other or not. The total pi electron energy of conjugated hydrocarbons in chemistry computed by using Huckel molecular orbital theory coincides with the energy defined by Gutman, so the energies calculated in graph theory have a special significance.Calculating the bounds for graph spectra has been the main area of research among graph theorists. McClelland [2] worked on the estimation of pi electron energies. Nageswari and Sarasija [3] calculated edge energy bounds of finite, simple, and undirected graphs. Das and Gutman [4] calculated some upper and lower bounds for EG in terms of number of vertices and number of edges. Adiga and Rakshith [5] calculated upper bounds for the extended energy of graphs. Jahanbani [6] calculated lower bounds for the energy of graphs. In this research work, we calculate lower bounds for various energies of the double-wheel graph for the first time. Now, we introduce some terminologies associated with our work.In mathematics, in graph theory, we study mathematical structures which are called graphs, and we use these graphs to represent pairwise relations between different objects. So, graph theory is the study of graphs and their characteristics. One of the most important concepts in graph theory is energy of a graph. This concept was introduced by Gutman [1] which defined energy as the sum of the absolute values of the roots of the characteristics equation, and these roots are called eigenvalues of the graph under consideration.In [7], energy, Seidel energy, Harary energy, distance energy, color energy, and Laplacian energy of the friendship graph have been calculated. In this research paper, we calculate these energies of the Dutch windmill graph of cycle lengths 4, 5, and 6, and also, we find lower bounds of these energies of the double-wheel graph.


## 2. Basic Definitions and Notations

Definition 1.
For a graphG, the adjacency matrix is the square matrix of order n×n denoted by AG and is defined as AG=aij, where(1)aij=1ifvivj∈E0ifvivj∉E..
The adjacency matrix gives us information whether a vertex in a pair of vertices is joined with the other vertex or not.Definition 2.
The Laplacian matrix is also a square matrix defined as follows:(2)LaG=DEGG−AG,where DEGG is the diagonal matrix of vertex degrees and AG is the adjacency matrix of graph G.Definition 3.
The Harary matrix of a graphG is the square matrix of order n whose i,j entry is defined as 1/dij where dij is the distance between the vertices vi and vj.Definition 4.
The color matrix ofG is the square matrix of order n denoted by AcG=aij where(3)aij=1ifviandvjareadjacentwithdifferentcolors−1ifviandvjarenonadjacentwiththesamecolor0,otherwise..Definition 5.
The distance matrix ofG is a matrix of order n, and its entries represent shortest distances between its different vertices.Definition 6.
The Seidel matrix of a graph is denoted bySG=aij, where(4)aij=1ifvivj∉E−1ifvivj∈E0,otherwise..
The Seidel matrix is a square, real symmetric matrix of ordern. The eigenvalues of the Seidel matrix are the eigenvalues of the graph G.Definition 7.
The sum of the absolute values of the roots of the characteristic equation associated with the adjacency matrix of the graph determines the energy of that graph. Ifλ1,λ2,λ3,…,λn are the eigenvalues of the adjacency matrix, then energy of graph EG is defined as follows:(5)EG=∑i=1nλi.
Haemers [7] introduced Seidel energy of a graph. To calculate the Seidel energy of a graph, we make its Seidel matrix as defined in equation (4) and find its characteristic equation; now, the sum of the absolute values of the roots of this characteristic equation determines the Seidel energy of the graph under consideration.Definition 8.
Ifλ1,λ2,λ3,…,λn are the eigenvalues of the Seidel matrix, then Seidel energy is defined as follows:(6)EG=λ1+λ2+λ3+⋯+λn.
Zhou [8] introduced Laplacian energy of a graph. To calculate Laplacian energy of a graph, first, we form its Laplacian matrix as defined in 2.3.7; then, we find its characteristic equation. Now, the sum of the absolute values of the roots of this characteristic equation determines the Laplacian energy of the graph under consideration.Definition 9.
Ifλ1,λ2,λ3,…,λn are the eigenvalues of the Laplacian matrix, then Laplacian energy is defined as follows:(7)L.EG=λ1+λ2+λ3+⋯+λn.
Indulal et al. [9] defined distance energy of a graph. To calculate distance energy of a graph, first, we form its distance matrix as defined in 2.3.3; then, we find its characteristic equation. Now, the sum of the absolute values of the roots of this characteristic equation determines the distance energy of the graph under consideration.Definition 10.
Ifλ1,λ2,λ3,…,λn are the eigenvalues of the distance matrix, then distance energy D.EG is defined as follows:(8)D.EG=λ1+λ2+λ3+⋯+λn.
Color energy was defined by Adiga and Smitha [10]. To calculate color energy of a graph, first, we form its color matrix, and then we find its characteristic equation. Now, the sum of the absolute values of the roots of this characteristic equation determines the color energy of the graph under consideration.Definition 11.
Ifλ1,λ2,λ3,…,λn are the eigenvalues of the color matrix, then color energy is defined as follows:(9)C.EG=λ1+λ2+λ3+⋯+λn.
Harary energy of a graph was defined in [11–13]. To calculate Harary energy of a graph, first, we form its Harary matrix, and then we find its characteristic equation. Now, the sum of the absolute values of the roots of this characteristic equation determines the Harary energy of the graph under consideration. For other useful results, we refer to the readers [14–17].Definition 12.
Ifλ1,λ2,λ3,…,λn are the eigenvalues of the Harary matrix, then Harary energy is defined as follows:(10)H.EG=λ1+λ2+λ3+⋯+λn.


## 3. Main Results

Theorem 1.
The energy of a Dutch windmill graphD4m is 22m−2+2m+2.Proof.
The adjacency matrix forD4m is(11)AD4m=0101101..011010000..000101000..001010000..001000010..000000101..001000010..00......................0000000..011000000..10mn−m+1×mn−m+1.
The characteristic equation isλ−0m+1λ2−2m−1λ2−2m+2=0.
The eigenvalues areλ1=0m+1times, λ2=2m−1times, λ3=−2m−1times, λ4=2m+2, and λ5=−2m+2.
Now, energy ofD4m can be calculated by using Definition 2.4.1 as follows:(12)ED4m=m+1λ1+m−1λ2+m−1λ3+λ4+λ5=m+10+m−12+m−1−2+2m+2+−2m+2=2m−2+2m−2+2m+2+2m+2=22m−2+2m+2.Theorem 2.
The energy ofD5m is 25m−5+χ.Proof.
The adjacency matrix forD5m is(13)AD5m=010011001..1001101000000..0000010100000..0000001010000..0000100100000..0000100000100..0000000001010..0000000000101..0000100000010..0000..............................100000000..0100000000000..1010000000000..0101100000000..0010mn−m+1×mn−m+1.
The characteristic equation of the above adjacency matrix is(14)λ2−λ−1m−1λ2+λ−1mλ3−λ2−2m+1λ+2m=0.
The eigenvalues areλ1=−1+5/2mtimes, λ2=−1−5/2mtimes, λ3=1−5/2m−1times, λ4=1+5/2m−1times, and λ5,λ6,λ7, where λ5,λ6,λ7 are the roots of the cubic equation given in the following:(15)λ3−λ2−2m+1λ+2m=0.
Now, energy can be calculated by using Definition 2.4.1 as follows:(16)ED5m=mλ1+mλ2+m−1λ3+m−1λ4+λ5+λ6+λ7,ED5m=m−1+52+m−1−52+m−11−52+m−11+52+λ5+λ6+λ7,ED5m=m−1+52+m−1−52+m−11−52+m−11+52+χ,ED5m=25m−5+χ,χ=λ5+λ6+λ7.Theorem 3.
The energy of a Dutch windmill graphD6m is(17)21+3m−23+22m+3+4m2−4m+92+22m+3−4m2−4m+92.Proof.
The adjacency matrix forD6m is(18)AD6m=01000110001..1000110100000000..0000001010000000..0000000101000000..0000000010100000..0000010001000000..0000010000001000..0000000000010100..0000000000001010..0000000000000101..0000010000000010..00000....................................10000000000..0100000000000000..1010000000000000..0101000000000000..0010110000000000..00010mn−m+1×mn−m+1.
The characteristic equation isλ+1mλ−1mλ−0m−1λ2−3m−1λ4−2m+3λ2+4m=0.
The eigenvalues of the above characteristic equation areλ1=−1mtimes, λ2=1mtimes, λ3=0m−1timesλ4=3m−1times, λ5=−3m−1times, λ6=2m+3+4m2−4m+9/2, λ7=2m+3−4m2−4m+9/2, λ8=−2m+3+4m2−4m+9/2, and λ9=−2m+3−4m2−4m+9/2.
Now, energy can be calculated by using Definition 2.4.1 as follows:(19)ED6m=mλ1+mλ2+m−1λ3+m−1λ4+m−1λ5+λ6+λ7+λ8+λ9,=m−1+m1+m−10+m−13+m−1−3+2m+3+4m2−4m+92+−2m+3+4m2−4m+92+2m+3−4m2−4m+92+−2m+3−4m2−4m+92,=21+3m−23+22m+3+4m2−4m+92+22m+3−4m2−4m+92.Theorem 4.
LetG be a double-wheel graph; then, EG>2+21+2n.Proof.
The adjacency matrix of order2n+1 for a double-wheel graph Wn, n≥3, is(20)AWn=010...100...01101...000...01110...000...01..........................................100...000...01000...001...11000...010...01...........................................000...010...01111111111111102n+1×2n+1.
The characteristic equation of the above adjacency matrix is(21)λ−2λ2−2λ−2nλ2n−2+λ2n−1+⋯λ=0.
The three roots of the above equation areλ1=2,λ2=1+1+2n,andλ3=1−1+2n.
Now, the energy of the double-wheel graph can be calculated by using Definition 2.4.1 as follows:(22)EWn=λ1+λ2+λ3+⋯+λ2n+1>λ1+λ2+λ3=2+1+1+2n−1+1+2n=2+21+2n=21+1+2n.Theorem 5.
The Seidel energy ofD4m is 42m−42+m+χ.Proof.
The Seidel matrix forD4m is(23)SD4m=0−11−1−11−1−11−1−11−1−10−111111111111−10−1111111111−11−10111111..111−11110−11111..1111111−10−1111..111−11111−10111..111−11111110−11..1111111111−10−1..111−11111111−10..111..............................−1111111111..0−111111111111..−10−1−1111111111..1−10mn−m+1×mn−m+1.
The characteristic equation of the above Seidel matrix is(24)λ+1mλ2+2λ−7m−1λ3−3m−2λ2+2m−7λ+8m2−11m=0.
Eigenvalues areλ1=−1mtimes, λ2=−1+22m−1times, λ3=−1−22m−1times, and λ4,λ5,λ6, where λ4,λ5,λ6 are the roots of the cubic equation given in the following:(25)λ3−3m−2λ2+2m−7λ+8m2−11m=0.
Now, Seidel energy can be calculated by using Definition 2.4.2 as follows:(26)S.E.G=mλ1+m−1λ2+m−1λ3+λ4+λ5+λ6=m−1+m−1−1+22+m−1−1−22+λ4+λ5+λ6=m−1+m−1−1+22+m−1−1−22+χ=m−m+1+22m−22+m−1+22m−22+χχ=λ4+λ5+λ6=42m−42+m+χ,χ=λ4+λ5+λ6.Theorem 6.
The Seidel energy ofD5m is given by 25+14m−14+χ.Proof.
The Seidel matrix forD5m is(27)S.ED5m=0−111−1−111−1..−111−1−10−1111111...11111−10−111111..111111−10−11111..1111−111−101111..1111−111110−111..111111111−10−11..1111111111−10−1..1111−1111111−10..1111..............................−111111111..0−111111111111..−10−11111111111..1−10−1−111111111..11−10mn−m+1×mn−m+1.
The characteristic equation of the above Seidel matrix isλ2−5mλ2+4λ−10m−1λ3−4m−4λ2−4m+1λ+16m2−16=0.
The eigenvalues of the above characteristic equation areλ1=5mtimes, λ2=−5mtimes, λ3=−2+14m−1times, λ4=−2−14m−1times, and λ5,λ6,λ7, where λ5,λ6,λ7 are the roots of the cubic equation given in the following:(28)λ3−4m−4λ2−4m+1λ+16m2−16=0.
Now, Seidel energy can be calculated by using Definition 2.4.2 as follows:(29)S.E.D5m=mλ1+mλ2+m−1λ3+m−1λ4+λ5+λ6+λ7=m5+m−5+m−1−2+14+m−1−2−14+λ5+λ6+λ7=m5+m−5+m−1−2+14+m−1−2−14+χ=2m5+14−14+χ,χ=λ5+λ6+λ7.Theorem 7.
The Seidel energy ofD6m is 5+43m−43−1+ℑ.Proof.
The Seidel matrix forD6m is(30)SD6m=0−1111−1−1111−1..−1111−1−10−111111111..111111−10−11111111..1111111−10−1111111..11111111−10−111111..11111−1111−1011111..11111−1111110−1111..11111111111−10−111..111111111111−10−11..1111111111111−10−1..11111−111111111−10..11111....................................−11111111111..0−111111111111111..−10−11111111111111..1−10−1111111111111..11−10−1−11111111111..111−10mn−m+1×mn−m+1.
The characteristic equation for the above Seidel matrix isλ+1m−1λ−1mλ+3mλ2+2λ−11m−1λ4−5m−3λ3−9−mλ2+24m2+5m−11λ−m40m−31=0.
The eigenvalues of the above Seidel matrix areλ1=−1m−1times, λ2=1mtimes, λ3=−1+23m−1times, λ4=−1−23m−1times, and λ5,λ6,λ7,λ8 where λ5,λ6,λ7,λ8 are roots of the biquadratic equation given in the following:(31)λ4−5m−3λ3−9−mλ2+24m2+5m−11λ−m40m−31=0.
Now, Seidel energy can be calculated by using Definition 2.4.2 as follows:(32)=m−11+m1+m3+m−1−1+23+m−11+23+ℑ=m−1+m+3m−m+1+23m−23+23m−23−1+m+ℑ=5m+43m−43−1+ℑ=5+43m−43−1+ℑ,ℑ=λ5+λ6+λ7+λ8.Theorem 8.
LetG be a double-wheel graph; then, S.EWn>5+4n2−12n+25.Proof.
A Seidel matrix of order2n+1 for a double-wheel graph Wn is(33)SWn=0−111...−111...1−1−10−11...111...1−11−10−1...111...1−111−10...111...1−1............................................................−1111...011...1−11111...10−1...−1−11111...1−10...1−1.............................................−1−1−1−1...−1−1−1...−102n+1×2n+1.
The characteristic equation of the Seidel matrix is(34)λ+5λ2−2n−5λ−2nλ2n−2+λ2n−1+⋯λ=0.
The three roots of this characteristic equation areλ1=−5, λ2=2n−5−4n2−12n+25/2, and λ3=2n−5−4n2−12n+25/2. Now, the Seidel energy can be calculated by using Definition 4 as follows:(35)S.EWn=λ1+λ2+λ3+⋯λ2n+1>λ1+λ2+λ3=−5+2n−5+4n2−12n+252+2n−5−4n2−12n+252=5+2n−5+4n2−12n+252+−2n+5+4n2−12n+252=5+4n2−12n+25.Theorem 9.
The Laplacian energy of Dutch windmill graphD4m is 8 m.Proof.
The Laplacian matrix forD4m is given as follows:(36)LD4m=2m−10−1−10−1−10−1..−10−1−10−1−12−10000000..0000000−12−1000000..000000−10−12000000..000000−10002−10000..0000000000−12−1000..000000−10000−12000..000000−10000002−10..0000000000000−12−1..000000−10000000−12..000000....................................−1000000000..2−100000000000000..−12−1000−1000000000..0−12000−1000000000..0002−100000000000..000−12−1−1000000000..0000−12mn−m+1×mn−m+1.
The characteristic equation of the above Laplacian matrix is given as follows:(37)λ−0λ−2mλ2−4λ+2m−1λ2−2m+4λ+6m+2=0.
Eigenvalues areλ1=0, λ2=2mtimes, λ3=2+2m−1times, λ4=2−2m−1times, λ5=m+2+m2−2m+2, and λ6=m+2−m2−2m+2.(38)L.EG=m+2+m2−2m+2+m+2−m2−2m+2+m−12+2+m−12−2+m2.
So, Laplacian energy of Dutch windmill graphL.ED4m=8m.Theorem 10.
The Laplacian energy of Dutch windmill graphD5m is 10 m.Proof.
The Laplacian matrix forD5m is given as follows:(39)LD5m=2m−100−1−100−1..−100−1−12−1000000..00000−12−100000..000000−12−10000..0000−100−120000..0000−100002−100..000000000−12−10..0000000000−12−1..0000−1000000−12..0000..............................−100000000..2−100000000000..−12−10000000000..0−12−1−100000000..00−12mn−m+1×mn−m+1.
The characteristic equation of the above Laplacian matrix is given as follows:(40)λ−0λ2−3λ+1m−1λ2−5λ+5mλ2−2m+3λ+4m+1=0.
The eigenvalues of the above Laplacian matrix are given as follows.λ1=0, λ2=3+5/2m−1times, λ3=3−5/2m−1times, λ4=5+5/2mtimes, λ5=5−5/2mtimes, λ6=2m+3+4m2−4m+5/2, and λ7=2m+3−4m2−4m+5/2.
So, the Laplacian energy is given by(41)L.ED5m=0+m5+52+m5−52+m−13+52+m−13−52+2m+3+4m2−4m+52+2m+3−4m2−4m+52,L.ED5m=10m.Theorem 11.
The Laplacian energy of the Dutch windmill graph isL.ED6m=10m−6+χ.Proof.
The Laplacian matrix forD6m is given as follows:(42)LD6m=2m−1000−1−1000−1..−1000−1−12−100000000..000000−12−10000000..0000000−12−1000000..00000000−12−100000..00000−1000−1200000..00000−1000002−1000..00000000000−12−100..000000000000−12−10..0000000000000−12−1..00000−100000000−12..00000....................................−10000000000..2−100000000000000..−12−10000000000000..0−12−1000000000000..00−12−1−10000000000..000−12mn−m+1×mn−m+1.
The characteristic equation of the above Laplacian matrix is given as follows:(43)λ−0λ−1mλ−2m−1λ−3mλ2−4λ+1m−1λ3−2m+6λ2+10m+9λ−10m+2=0.
The eigenvalues areλ1=0, λ2=1mtimes, λ3=2m−1times, λ4=3mtimes, λ5=2+3m−1times, λ6=2−3m−1times, and λ7,λ8,λ9, where λ7,λ8,λ9 are the roots of the cubic equation given in the following:(44)λ3−2m+6λ2+10m+9λ−10m+2=0.
Now, Laplacian energy=λ5+λ6+λ7+λ8+λ5+λ6+λ7+λ8+λ9.
So, the Laplacian energy ofD6m is given by(45)L.ED6m=0+m1+m−12+m3+m−12+3+m−12−3+χ,where χ=λ7+λ8+λ9.(46)L.ED6m=10m−6+χ.Theorem 12.
LetG be a double-wheel graph Wn; then, L.EG>2n.Proof.
A Laplacian matrix of order2n+1 for a double-wheel graph Wn, n≥3, is given as follows:(47)LWn=3−100....−1000−1−13−10...0000−10−13−1...000..0−100−13....000..0−1............................................−1000...300..0−10000...03−1...−1−10000...0−13..0−1..............................0000...0−10...3−1−1−1−1−1...−1−1−1..−12n.2n+1×2n+1.
The characteristic equation for the Laplacian matrix is given as follows:(48)detA−λI=0,λ−0λ−2n−1λ−1λ2n−2+λ2n−1+⋯λ=0.
The roots of this characteristic equation areλ1,λ2,λ3,…,λ2n+1. The three roots of the above characteristic equation are λ1=0,λ2=1,andλ3=2n−1. The Laplacian energy is defined as(49)L.E.G=λ1+λ2+λ3+⋯+λ2n+1.
Clearly,λ1+λ2+λ3+⋯⋯+λ2n+1>λ1+λ2+λ3(50)=2n.
So,L.EG>2n.Theorem 13.
The color energy forD4m is given by 1+22m−22+χ.Proof.
The color matrix forD4m is given as follows:(51)CD4m=01−111−11..1−11101−1−10−1..−10−1−11010−10..0−101−110−10−1..−10−11−10−101−1..−10−1−10−10101..0−101−10−1−110..−10−1........................1−10−1−10−1..01−1−10−100−10..1011−10−1−10−1..−110mn−m+1×mn−m+1.
The characteristic equation of the above color matrix is given as follows:(52)λ−1mλ2−2λ−1m−1λ3+3m−2λ2+2m2−6m−1λ−4m2−7m=0.
λ1=1mtimes, λ2=1+2m−1times, λ3=1−2m−1times, and λ4,λ5, λ6 are the eigenvalues of the above characteristic equation, where λ4,λ5,λ6 are the roots of the cubic equation given in the following:(53)λ3+3m−2λ2+2m2−6m−1λ−4m2−7m=0.
So, the color energy of(54)D4m=m1+m−11+2+m−11−2+λ4+λ5+λ6,=m1+m−11+2+m−11−2+χ,where χ=λ4+λ5+λ6.
So,C.ED4m=1+22m−22+χ.Theorem 14.
The color energy ofD5m is given by m−13+5+ℑ.Proof.
The color matrix forD5m of order mn−m+1 is given as follows:(55)CD5m=01−1011−101...1−101100−10−10−10...−10−10−111100−100...0−1000−1101−10−10...−10−1010010000−1...000−11−10−1001−10...−10−10−10−1001010...0−1000−10−10−1101...−10−101000−10010...000−1................................................1−10−10−10−10...01−10−10−1000−100...10100−10−10−10−10...−11011000−1000−1...0010mn−m+1×mn−m+1.
The characteristic equation is given as follows:(56)λ2−λ−1m−1λ2−3λ+1m−1λ5+4m−1λ4+5m2−15m+3λ3+−2m3+19m2−19m−2λ2+−9m3+26m2−6m−1λ+m−m3+9m2−10m−6=0.
The eigenvalues areλ1=1+5/2m−1times, λ2=1−5/2m−1times, λ3=3+5/2m−1times, λ4=3−5/2m−1times, and λ5,λ6,λ7,λ8,λ9. Here, λ5,λ6,λ7,λ8,λ9 are the roots of the equation given in the following:(57)λ5+4m−1λ4+5m2−15m+3λ3+−2m3+19m2−19m−2λ2+−9m3+26m2−6m−1λ+m−m3+9m2−10m−6=0.
Now, color energy=λ1+λ2+λ3+λ4+λ5+λ6+λ7+λ8+λ9.(58)C.ED5m=m−11+52+m−11−52+m−13−52+m−13+52+λ5+λ6+λ7+λ8+λ9,(59)C.ED5m=m−11+52+m−11−52+m−13−52+m−13+52+ℑ,where ℑ=λ5+λ6+λ7+λ8+λ9.
So,C.ED5m=m−13+5+ℑ.Theorem 15.
The color energy forD6m is given by m3+23−23−1+ℑ.Proof.
The color matrix forD6m is given as follows:(60)CD6m=01−10−111−10−11..1−10−11101−10−1−10−10−1..−10−10−1−1101−100−10−10..0−10−100−1101−1−10−10−1..−10−10−1−10−11010−10−10..0−10−101−10−110−10−10−1..−10−10−11−10−10−101−10−1..−10−10−1−10−10−10101−10..0−10−100−10−10−1−1101−1..−10−10−1−10−10−100−1101..0−10−101−10−10−1−10−110..−10−10−1..................................1−10−10−1−10−10−1..01−10−1−10−10−100−10−10..101−100−10−10−1−10−10−1..−1101−1−10−10−100−10−10..0−11011−10−10−1−10−10−1..−10−110.mn−m+1×mn−m+1.
The characteristic equation for the above color matrix is given as follows:(61)λ−0mλ−1m−1λ−2mλ2−2λ−2m−1λ4+5m−3λ3+6m2−14mλ2−18m2−16m−2λ−2m−22m=0.
The eigenvalues areλ1=0mtimes, λ2=1m−1times, λ3=2mtimes, λ4=1+3m−1times, λ5=1−3m−1times, and λ6,λ7,λ8,λ9, where λ6,λ7,λ8,λ9 are the roots of the biquadratic equation given in the following:(62)λ4+5m−3λ3+6m2−14mλ2−18m2−16m−2λ−2m−22m=0.
Now, color energy is(63)C.ED6m=λ1+λ2+λ3+λ4+λ5+λ6+λ7+λ8+λ9,C.ED6m=m0+m−11+m2+m−11+3+m−11−3+λ6+λ7+λ8+λ9,C.ED6m=m0+m−11+m2+m−11+3+m−11−3+ℑ,where ℑ=λ6+λ7+λ8+λ9.
So,C.ED6m=m3+23−23−1+ℑ.Theorem 16.
LetG be a double-wheel graph; then, C.EWn>3+1+8n.Proof.
The color matrix for a double-wheel graph is given as follows:(64)CWn=01−1...1−10...01101...00−1...01−110...000...01..........................................100...000...−11−100...001...11........................................................000...−110...01111111111111102n+1×2n+1.
The characteristic equation of the color matrix is given as follows:(65)detA−λI=0,λ−3λ2−λ−2nλ2n−2+λ2n−1+⋯λ=0.
The roots of this characteristic equation areλ1,λ2,λ3,…,λn. The three roots of this characteristic equation are λ1=3, λ2=1+1+8n/2, and λ3=1−1+8n/2.
Clearly,λ1+λ2+λ3+⋯+λ2n+1>λ1+λ2+λ3(66)=3+1+1+8n2+−1+1+8n2,=3+1+8n.
C.EWn=λ1+λ2+λ3+⋯λn.
So,C.EWn>3+1+8n.Theorem 17.
The distance energy of Dutch windmill graphD4m is 24m−3+32m2−2m+1.Proof.
The distance matrix for Dutch windmill graphD4m is(67)DD4m=0121121121101223223221013433431210232232123201223223431013431232210232123223201223433431011232232210mn−m+1×mn−m+1.
The characteristic equation isλ+6m−1λ+2mλ−0mλ2−8m−6λ−2m2+6m=0.
The eigenvalues areλ1=−6m−1times, λ2=−2mtimes, λ3=0mtimes, λ4=4m−3+32m2−2m+1, and λ5=4m−3−32m2−2m+1.
The distance energy can be calculated by using Definition 2.4.3 as follows:(68)D.E.D4m=m−1λ1+mλ2+mλ3+λ4+λ5=m−1−6+m−2+m0+4m−3+32m2−2m+1+4m−3−32m2−2m+1=6m−6+2m+4m−3+32m2−2m+1−4m+3+32m2−2m+1=8m+62m2−2m+1−6=24m−3+32m2−2m+1.Theorem 18.
The distance energy ofD5m is 12m−9+ℑ, where ℑ is the sum of the absolute values of the roots of the equation λ3−12m−9λ2−4m+32−6m−41λ−6m=0.Proof.
The distance matrix forD5m is(69)DD5m=012211221..1221101222332..2332210123443..3443221013443..3443122102332..2332123320122..2332234431012..3443234432101..3443123322210..2332..............................123322332..0122234433443..1012234433443..2101123322332..2210mn−m+1×mn−m+1.
The characteristic equation of the above distance matrix is(70)λ2+9λ+5m−1λ2+3λ+1mλ3−12m−9λ2−4m+32−6m−41λ−6m=0.
The eigenvalues of the above characteristic equation areλ1,λ2,λ3, λ4=−9+61/2m−1times, λ5=−9−61/2m−1times, λ6=−3+5/2mtimes, and λ7=−3−5/2mtimes, where λ1,λ2,λ3 are the roots of the cubic equation given in the following:(71)λ3−12m−9λ2−4m+32−6m−41λ−6m=0.
Now, the distance energy ofD5m can be calculated by using Definition 2.4.3 as follows:(72)D.E.D5m=λ1+λ2+λ3+m−1λ4+m−1λ5+mλ6+mλ7=λ1+λ2+λ3+m−1−9+612+m−1−9−612+m−3+52+m−3−52=λ1+λ2+λ3+m−19−612+m−19+612+m3−52+m3+52=λ1+λ2+λ3+9m−9−61m+61+9m−9+61m−61+3m−5m+3m+5m2=λ1+λ2+λ3+12m−9=ℑ+12m−9,where ℑ=λ1+λ2+λ3.Theorem 19.
The distance energy ofD6m is 18m−14+ℑ, where ℑ is the sum of the absolute values of the roots of the equation λ3−18m−14λ2−14m2+43m−16λ−6m2m+4=0.Proof.
The distance matrix forD6m of order mn−m+1 is(73)DD6m=01232112321...1232110123223432...2343221012334543...3454332101245654...4565423210134543...3454312321023432...2343212343201232...2343223454310123...3454334565421012...4565423454332101....3454312343223210...23432.........................................................12343223432...0123223454334543...1012334565445654...2101223454334543...3210112343223432...23210mn−m+1×mn−m+1.
The characteristic equation of the above distance matrix is(74)λ−02mλ+4mλ2+14λ+16m−1λ3−18m−14λ2−14m2+43m−16λ−6m2m+4=0.
The eigenvalues of the above distance matrix areλ1=02mtimes, λ2=−4mtimes, λ3=−7+33m−1times, λ4=−7−33m−1times, and λ5,λ6,λ7, where λ5,λ6,λ7 are the roots of the cubic equation given in the following:(75)λ3−18m−14λ2−14m2+43m−16λ−6m2m+4=0.
Now, distance energy can be calculated by using Definition 2.4.3 as follows:(76)D.E.D6m=2mλ1+mλ2+m−1λ3+m−1λ4+λ5+λ6+λ7=2m0+m−4+m−1−7+33+m−1−7−33+λ5+λ6+λ7=2m0+m−4+m−1−7+33+m−1−7−33+λ5+λ6+λ7=2m0+m−4+m−1−7+33+m−1−7−33+λ5+λ6+λ7=4m+m−17−33+m−17+33+λ5+λ6+λ7=4m+7m−7−33m+33+7m−7+33m−33+λ5+λ6+λ7=18m−14+ℑ,where ℑ=λ5+λ6+λ7.Theorem 20.
LetG be a double-wheel graph; then, Harary energy is given as follows:(77)H.EG>0.5+24n2+33n+1.Proof.
A Harary matrix of order2n+1 for a double-wheel graph Wn is given as follows:(78)HWn=011/21/2...11/21/2..1/211011/2...1/21/21/2..1/211/2101...1/21/21/2..1/211/21/210...1/21/21/2..1/21..........................................11/21/21/2...01/21/2..1/211/21/21/21/2...1/201..111/21/21/21/2...1/210..1/21............................1/21/21/21/2...1/211/2..011111...111..102n+1×2n+1.
The characteristic equation of the Harary matrix is given as follows:(79)detA−λI=0,λ−0.5λ2−2n+12λ−2nλ2n−2+λ2n−1+⋯λ=0.
The characteristic equation of this matrix has degree2n+1. So, λ1,λ2,…λ2n+1 are the eigenvalues of the given Harary matrix. The Harary energy is defined as(80)H.EG=λ1+λ2+λ3+⋯+λ2n+1.
Out of these2n+1 eigenvalues, three eigenvalues are 0.5,2n+1+4n2+33n+1, and 2n+1−4n2+33n+1.
Clearly,λ1+λ2+λ3+⋯⋯+λ2n+1>λ1+λ2+λ3(81)=0.5+24n2+33n+1.
So,H.EG>0.5+24n2+33n+1.


## 4. Conclusions

In this paper, we find different energies, for example, color energy, distance energy, Laplacian energy, and Seidel energy, for the Dutch windmill graph of cycle lengths 4, 5, and 6. Also, we find the lower bounds of the double-wheel graph for energy, Seidel energy, color energy, distance energy, Laplacian energy, and Harary energy. It is interesting to find out the computed energies for other well-known families of graphs. It is also interesting to compute other energies for the same families of graphs.



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