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A mathematical model for simulating the spread of infectious disease using the Caputo-Fabrizio fractional-order operator

  • We examined intraspecific infectious rivalry in a dynamic contagious disease model. A non-linear dynamic model that considers multiple individual categories was used to study the transmission of infectious diseases. The combined effect of parameter sensitivities on the model was simulated using system sensitivities. To investigate the dynamic behavior and complexity of the model, the Caputo-Fabrizio (C-F) fractional derivative was utilized. The behavior of the proposed model around the parameters was examined using sensitivity analysis, and fractional solutions included more information than the classical model. Fixed point theory was used to analyze the existence and uniqueness of the solution. The Ulam-Hyers (U-H) criterion was used to examine the stability of the system. A numerical approach based on the C-F fractional operator was utilized to improve comprehension and treatment of the infectious disease model. A more precise and valuable technique for solving the infectious disease model was used in MATLAB numerical simulations to demonstrate. Time series and phase diagrams with different orders and parameters were generated. We aimed to expedite patient recovery while reducing the frequency of disease transmission in the community.

    Citation: Parveen Kumar, Sunil Kumar, Badr Saad T Alkahtani, Sara S Alzaid. A mathematical model for simulating the spread of infectious disease using the Caputo-Fabrizio fractional-order operator[J]. AIMS Mathematics, 2024, 9(11): 30864-30897. doi: 10.3934/math.20241490

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  • We examined intraspecific infectious rivalry in a dynamic contagious disease model. A non-linear dynamic model that considers multiple individual categories was used to study the transmission of infectious diseases. The combined effect of parameter sensitivities on the model was simulated using system sensitivities. To investigate the dynamic behavior and complexity of the model, the Caputo-Fabrizio (C-F) fractional derivative was utilized. The behavior of the proposed model around the parameters was examined using sensitivity analysis, and fractional solutions included more information than the classical model. Fixed point theory was used to analyze the existence and uniqueness of the solution. The Ulam-Hyers (U-H) criterion was used to examine the stability of the system. A numerical approach based on the C-F fractional operator was utilized to improve comprehension and treatment of the infectious disease model. A more precise and valuable technique for solving the infectious disease model was used in MATLAB numerical simulations to demonstrate. Time series and phase diagrams with different orders and parameters were generated. We aimed to expedite patient recovery while reducing the frequency of disease transmission in the community.



    We only consider finite and undirected graphs throughout this paper. Let G=(V(G),E(G)) be a graph with n=|V(G)| vertices and m=|E(G)| edges. For any vertex uV(G), we use dG(u) (or du when no confusion can arise) to denote the degree of u in G, which is the number of edges incident to u. Such a maximal number (resp. minimal number) is called the maximal degree Δ(G) (resp. minimal degree δ(G)). For any vertex u in G, we use NG(u) to denote the set of all vertices adjacent with u, and the elements of NG(u) are called neighbors of u. A sequence of positive integers π=(d1,d2,,dn) is called the degree sequence of G if di=dvi for any vertex viV(G), where i=1,2,,n.

    The join of two graphs G1 and G2, denoted by G1+G2, is the graph with the vertex set V(G1)V(G2) and edge set E(G1)E(G2){xy|xV(G1),yV(G2)}. The cyclomatic number of G is the minimum number of edges in it whose removal makes it acyclic, denoted by γ=γ(G). Let Gγn be the set of n-vertex graphs with cyclomatic number γ. We use Kn and Pn to denote the complete graph and path of n vertices, respectively. As usual, we use the symbol (Pn) to denote the length of the path Pn, which equals to the number of edges in Pn. The cyclomatic number, denoted by γ, of a graph G is the minimum number of edges of G whose removal makes G acyclic. Let Gγn be the class of all connected graphs with order n and cyclomatic number γ. We use [4] for terminology and notation not defined here.

    The topological index is a real number that can be used to characterize the properties of the molecule graph. Nowadays, hundreds of topological indices have been considered and used in quantitative structure-activity and quantitative structure-property relationships. One of the well-known topological indices is the general Randić index, which was defined by Bollobás and Erdös [5] and Amic [1] independently:

    Rα(G)=uvE(G)[dudv]α,

    where α is a nonzero real number. This topological index has been extensively investigated. We encourage interested readers to consult [3,6,7,10,11,13] for more mathematical properties and their applications.

    Even though the mathematical and chemical theory of the general Randić index has been well considered, some extremal graph-theoretical problems concerning this graph invariant are still open. In this paper, we focus on exploring the extremal graphs in Gγn with respect to the general Randić index.

    It is interesting to explore the extremal graphs for some topological indices in the class of graphs with a given cyclomatic number. In this section, we focus on determining the extremal graphs in Gγn with the minimum general Randić index. Before proceeding, we shall prove or list several facts as preliminaries.

    Lemma 2.1. The function P(x,α)=2αxα+1(x1)α[2α(x2)+3α]+xα(2α3α)6α>0 for x4 and 1α3925.

    Proof. It is routine to check that

    P(x,α)=2αxα+1(x1)α[2α(x2)+3α]+xα(2α3α)6α=2αxα+1(x1)α[2α(x1)2α+3α]+xα(2α3α)6α=2α[xα+1(x1)α+1]+(2α3α)(x1)α+(2α3α)xα6α=2α[xα+1(x1)α+1]+(2α3α)[xα+(x1)α]6α=2α[xα+(x1)α]+2αx(x1)[xα1(x1)α1]+(2α3α)[xα+(x1)α]6α=(22α3α)[xα(x1)α]+2αx(x1)[xα1(x1)α1]6α.

    Note that ρ(t)=tα(t1)α is an increasing function for t[4,+), and 22α>3α if, and only if, α<ln2ln3ln21.709, then we have

    P(x,α)=(22α3α)[xα(x1)α]+2αx(x1)[xα1(x1)α1]6α(22α3α)(4α+3α)+2α12(4α13α1)6α=58α36α9α12α.

    For simplicity, let H(α)=58α36α9α12α. To continue our proof, we first verify the following fact.

    Claim 1. The function ϱ(t)=k1atk2btk3ct has a unique zero point in the interval [0,+) for any positive real numbers k1,k2,k3,a,b,c such that k1k2k3>0 and 1<a<b<c.

    Proof of Claim 1. It is routine to check that ϱ(t)=k1lnaatk2lnbbtk3lncct. Note that ϱ(0)=k1k2k3>0 and ϱ(M)=at[k1k2(ba)tk3(ca)t]t=M, and it follows that ϱ(t) has zero points in the interval [0,+). Without loss of generality, we assume that t1,t2=t1+h[0,+) are the two distinct zero points of ϱ(t) for h>0, which is equivalent to k1at1k2bt1k3ct1=0 and k1at2k2bt2k3ct2=0. Besides, we know that ϱ(t)=k1lnaatk2lnbbtk3lncct, which implies that

    ϱ(t1)=k1lnaat1k2lnbbt1k3lncct1<lna(k1at1k2bt1k3ct1)=0.

    In addition, we have

    ϱ(t2)=ϱ(t1+h)=k1at1ahk2bt1bhk3ct1ch=(k2bt1+k3ct1)ahk2bt1bhk3ct1ch<(k2bt1+k3ct1)ah(k2bt1+k3ct1)bh=(k2bt1+k3ct1)(ahbh)<0,

    which contradicts to the fact that ϱ(t2)=0. Hence, there must exist a unique number t0[0,+) such that ϱ(t0)=0. As desired, we have completed the proof of Claim 1.

    Claim 2. The function H(α)=58α36α9α12α has a unique zero point in the interval (1,2).

    Proof of Claim 2. It is routine to check that H(α)=5ln88α3ln66αln99αln1212α. Note that H(1)=1>0 and H(2)=13<0, and it follows that H(α) has zero points in the interval (1,2). Without loss of generality, we assume that α0,α1,,αl are the zero points of H(α) such that 1<α0<α1<<αl. Hence, H(α0)=58α036α09α012α0=0. Furthermore,

    H(α0)=5ln88α03ln66α0ln99α0ln1212α0=ln8(36α0+9α0+12α0)3ln66α0ln99α0ln1212α0=3(ln8ln6)k16α0(ln9ln8)k29α0(ln12ln8)k312α0.

    It follows from Claim 1 that ϱ(t)|a=6,b=9,c=12 has a unique zero point in the interval t0[0,+). Consequently, we know that the unique zero point of ϱ(t)|a=6,b=9,c=12 must lie in the interval (0,1) since ϱ(0)|a=6,b=9,c=12>0 and ϱ(1)|a=6,b=9,c=12=0.7473<0. Hence, ϱ(t)|a=6,b=9,c=12<0 always holds for any real number t1. This implies that H(αi)=ϱ(αi)|a=6,b=9,c=12<0 for αi>1 and i=0,1,,l, which contradicts to the continuity of the function H(α). As desired, we have completed the proof of Claim 2.

    Now, we continue to our proof. Note that H(3925)=583925363925939251239250.01857>0 and H(1.57)=581.57361.5791.57121.570.07428<0. Hence, P(x,α)H(α)>0 for α[1,3925]. As desired, we have completed the proof of Lemma 2.1.

    Figure 1.  The graph of the function H(α)=58α36α9α12α for α[0,3925), where α and H(α) denote the horizontal and vertical axes, respectively.

    Lemma 2.2. The function Q(x,α)=3α(xα+1(x1)α+1)9α+2xα(2α3α)>0 for x4 and α1.

    Proof. For simplicity, we distinguish the following two cases.

    Case 1. α[1,3).

    Note that h(t)=tα is an increasing function in the interval [11x,1] for α1, and it follows from Lagrange's mean value formula that h(1)h(11x)=[1(11x)]h(ξ)=1xh(ξ)=1xαξα1>0, where ξ(11x,1). Hence, x[1(11x)α]=x[h(1)h(11x)]=αξα1. Thus, we have

    Qx(x,α)=3α(α+1)[xα(x1)α]+2αxα1(2α3α)=xα1{3α(α+1)x[1(11x)α]+2α(2α3α)}=xα1[3αα(α+1)ξα1+2α(2α3α)]=αxα1[3α(α+1)ξα1+2(2α3α)].

    By our initial hypothesis, it is routine to check that ξα1>(11x)α1, then we have

    Qx(x,α)>αxα1[3α(α+1)(11x)α1+2(2α3α)]>αxα1[3α(α+1)(34)α1+2(2α3α)](because11x>34)=2α3αxα1[12(α+1)(34)α1+(23)α1]>2α3αxα1[932(α+1)+(23)α1](because(34)α1>(34)2).

    Let p(α)=932(α+1)+(23)α1, then we have p(α)=932+(23)αln23 and p(α)=(23)α(ln23)2>0. Hence, p(α)p(1)=932+23ln(23)1100>0, which implies that p(α) is increasing in the interval [1,+). Hence, p(α)p(1)=1148>0. It immediately yields that Qx(x,α)>2α3αxα1p(α)>0. Therefore, we have

    Q(x,α)f(4,α)=3α(4α+13α+1)9α+24α(2α3α)=29α[(129)α+(89)α2]>0,

    as desired, and we have completed the proof.

    Case 2. α[3,+).

    Note that

    Qx(x,α)=3α(α+1)[xα(x1)α]+2αxα1(2α3α)=xα1{3α(α+1)x[1(11x)α]+2α(2α3α)}.

    Let g(α)=12x(11x)α be a function defined in the interval [3,+), then we have g(α)=(11x)αln(1+1x1)>0. Hence, g(α)g(3)=12x(11x)3=x23x+1x3>0, implying that 1(11x)α>2x. Thus, we have

    Qx(x,α)=xα1{3α(α+1)x[1(11x)α]+2α(2α3α)}>2αxα1[3αα+1α(3α2α)](becauseα+1α>1)>0.

    Let l(α)=(129)α+(89)α2. It is routine to check that l(α)=(89)α[(128)αln(129)+ln(89)]>(89)α[ln(129)+ln(89)]=(89)αln15>0. Hence, l(α)l(3)=782729>0. It then follows that

    Q(x,α)f(4,α)=3α(4α+13α+1)9α+24α(2α3α)=29α[(129)α+(89)α2]>0,

    as desired, and we have completed the proof.

    Proposition 2.3. Let GGγn be a graph with γ3 and n2(γ1), then there is no pendent vertex in G if it has a minimum general Randić index for α1.

    Proof. Suppose to the contrary that there exists a pendent vertex in G. Let u be a vertex of degree at least three and NG(u)={u1,u2,,uk}. In what follows, we use P=uu1^u2^ur to denote a pendent path in G. Assume that u2u1 is another neighbor of u with du22. We consider the graph ^G1=Guu2+u2^ur (depicted in Figure 2), which is an element of Gγn. Let l2 be the number of vertices in {u3,u4,,uk}, whose degree is greater than or equal to two. Clearly, l2 and ki=3dαui2α(l2). For simplicity, we distinguish the following two cases:

    Figure 2.  The transformation G^G1.

    Case 1. (P)=1.

    Direct calculations show that

    Rα(G)Rα(^G1)=dαudαu2+dαu2α(du1)α2αdαu2+[dαu(du1)α]ki=3dαuidαudαu2+dαu2α(du1)α2αdαu2+2α(l2)[dαu(du1)α]=[(dαu22α+1)(dαu2α)]A1+2α{(l1)[dαu(du1)α]+(12α)}A2.

    It is not difficult to find the first term of the previous equality A1=[(dαu22α+1)(dαu2α)]>0 for α1, du3 and du22. To continue the proof, it remains to verify that A2>0. For simplicity, we let H(x)=(l1)[xα(x1)α](2α1) for α1 and x3. It is routine to check that H(x)>[(xα(x1)α)(3α2α)]+[(3α2α)(2α1)] since l3. Note that f1(x)=xα(x1)α is increasing in the interval [3,], then we have xα(x1)α3α2α. In addition, we know that 3α2α2α1 always holds for α1. Hence, H(x)>0 and, consequently, we have A2>0. It then immediately deduces that Rα(G)Rα(^G1)>0, a contradiction. This implies that there is no pendent vertex in G.

    Case 2. (P)2.

    Direct calculations show that

    Rα(G)Rα(^G1)=dαudαu2(du1)α2α+2αdαu2α(du1)α+2α(12α)2αdαu2+[dαu(du1)α]ki=3dαuidαudαu2(du1)α2α+2αdαu2α(du1)α+2α(l2)[dαu(du1)α]+2α(12α)2αdαu2=dαu2(dαu2α)A3+2α(l1)[dαu(du1)α]A4+2α[dαu(du1)α+12α]A5.

    Note that A3>0 and A4>0, and it is also not difficult to find that A5=1l1A2 is positive under the initial assumptions. Hence, Rα(G)Rα(^G1)>0. Again a contradiction. This implies that there is no pendent vertex in G.

    As desired, we complete the proof of Proposition 2.3.

    Proposition 2.4. Let GGγn be a graph with γ3 and n2(γ1), then the maximum vertex degree is three in G if it has minimum general Randić index for 1α3925.

    Proof. It follows from Proposition 2.3 that G contains at least one cycle as its induced subgraph, and the n-vertex cycle is the only connected graph for the which minimum and maximum vertex degree is two. Hence, in conjunction with the assumption γ3, we have Δ=Δ(G)3. To complete the proof, it suffices to show that Δ=3. If Δ>3, then it is routine to check that

    n=2iΔni2(γ1)=2(mn)=2(2iΔini22iΔni),

    which is equivalent to

    n24iΔ(i3)ni>4iΔ(43)ni>0.

    Hence, there at least exists a vertex of degree two. For simplicity, we suppose that u is the vertex in G with maximum degree and NG(u)={u1,u2,,uΔ}. We distinguish the following two cases.

    Case 1. i{1,2,,Δ} such that dui=2.

    For convenience, we suppose that u1 is the neighbor of u with degree two and du2du32.

    Subcase 1.1. du2=2 and u1 is not adjacent to u2.

    Let ^G2=Guu2+u1u2Gγn. t is the neighbor of u1, different from u, depicted in Figure 3. Hence, we have

    Rα(G)Rα(^G2)=dαu2α+1(du1)α3α6α+dαt(2α3α)+[dαu(du1)α]Δi=3dαui=dαu2α+1(du1)α3α6α+dαt(2α3α)+2α(du2)[dαu(du1)α]=2αdα+1u(du1)α[2α(du2)+3α]+dαt(2α3α)6α2αdα+1u(du1)α[2α(du2)+3α]+dαu(2α3α)6α.

    For simplicity, we let f2(x)=2αxα+1(x1)α[2α(x2)+3α]+xα(2α3α)6α. It follows from Lemma 2.1 that f2(x)=P(x,α)>0 for x4 and 1α3925. Hence, Rα(G)Rα(^G2)>0, which contradicts to the choice of G. Hence, the maximum vertex degree of G is three.

    Figure 3.  The transformation G^G2.

    Subcase 1.2. du2=2 and u1 is adjacent to u2.

    Let ^G3=Guu3+u1u3Gγn, depicted in Figure 4. Hence, we have

    Rα(G)Rα(^G3)=3×dαu2α+4α(du1)α3α2×6α2α(du1)α+[dαu(du1)α]Δi=4dαui=3×dαu2α+4α(du1)α3α2×6α2α(du1)α+2α(du3)[dαu(du1)α]=2αdα+1u+4α(du1)α[2α(du2)+3α]2×6α.
    Figure 4.  The transformation G^G3.

    Let f3(x)=2αxα+1+4α(x1)α[2α(x2)+3α]2×6α, and then we have f3(x)=f2(x)+(3α2α)(xα2α)>0. Hence, Rα(G)Rα(^G3)>0 for x4 and 1α3925, which contradicts to the choice of G. Hence, the maximum vertex degree of G is three.

    Subcase 1.3. du2>2 and u1 is adjacent to u2 and du3>2.

    Let ^G4=Guu3+u1u3Gγn, depicted in Figure 5. Hence, we have

    Rα(G)Rα(^G4)=dαu2α+2αdαu2+dαudαu2+dαudαu3(du1)α3α3αdαu2dαu2(du1)α3αdαu3+[dαu(du1)α]Δi=4dαuidαu2α+dαu2(2α3α)+dαu3(dαu3α)(du1)α3α+dαu2(dαu33α)(du1)α3α+2α(du3)[dαu(du1)α]>2αdα+1u(du1)α[2α(du2)+3α]+dαu(2α3α)6α>0,

    and the last inequality holds because f2(x)>0 for x4 and 1α3925. Hence, Rα(G)Rα(^G4)>0. Again, a contradiction. Hence, the maximum vertex degree of G is three.

    Figure 5.  The transformation G^G4.

    Subcase 1.4. du2>2, u1 is adjacent to u2 and du3=2.

    Let ^G5=Guu3+u1u3Gγn, depicted in Figure 6. Hence, we have

    Rα(G)Rα(^G5)=dαu2α+1+2αdαu2+dαudαu26α(du1)α3αdαu23α(du1)αdαu2+[dαu(du1)α]Δi=4dαuidαu2α+16α(du1)α3α+dαu2[(2α3α+dαu(du1)α)]+2α(du3)[dαu(du1)α]>2αdα+1u(du1)α[2α(du2)+3α]+4α2×6α>0,

    and the last inequality holds because f3(x)>0 for x4 and 1α3925. Hence, Rα(G)Rα(^G5)>0. Again, a contradiction. Thus, we have completed that the maximum vertex degree of G is three.

    Figure 6.  The transformation G^G5.

    Subcase 1.5. du2>2, u1 is not adjacent to u2.

    Let ^G6=Guu2+u1u2Gγn, depicted in Figure 7. Hence, we have

    Rα(G)Rα(^G6)=dαu2α+2αdαt+dαudαu2(du1)α3α3αdαt3αdαu2+[dαu(du1)α]Δi=3dαuidαu2α+dαu(2α3α)+2α(dαu3α)(du1)α3α+2α(du2)[dαu(du1)α]=2αdα+1u(du1)α[2α(du2)+3α]+dαu(2α3α)6α>0,

    and the last inequality holds because f2(x)>0 for x4 and 1α3925. Hence, Rα(G)Rα(^G6)>0, a contradiction to the choice of G. Hence, the maximum vertex degree of G is three.

    Figure 7.  The transformation G^G6.

    Case 2. i{1,2,,Δ} such that dui>2.

    Note that there is a vertex u0V(G)NG(u) of degree two, which is not adjacent to at least one neighbor, say u1, of u. Let ^G7=Guu1+u0u1Gγn (depicted in Figure 8). Hence,

    Rα(G)Rα(^G7)=dαu1(dαu3α)+(2α3α)zNG(u0)dαz+[dαu(du1)α]Δi=2dαui3α(dαu3α)+2×dαu(2α3α)+3α(du1)[dαu(du1)α]=3α[dα+1u(du1)α+1]9α+2×dαu(2α3α).
    Figure 8.  The transformation G^G7.

    For simplicity, we let f4(x)=3α[xα+1(x1)α+1]9α+2xα(2α3α)=Q(x,α), which is positive for x4 and α1 by Lemma 2.2. Hence, we have Rα(G)Rα(^G7)>0. Thus, there would be a contradiction to the choice of G, and the maximum vertex degree of G is three.

    This completes the proof of Proposition 2.4.

    Let φij be the number of edges in G joining the vertices of degree i and j, and we use ni and nj to denote the number of vertices of degree i and j, respectively.

    Proposition 2.5.. ([2]) Let GGγn, γ3, be a graph such that it contains only vertices of degrees two and three, then the following holds:

    (i) at least two vertices of degree two are adjacent if n>5(γ1).

    (ii) φ22=0 implies φ33=0 (or φ33=0 implies φ22=0) if n=5(γ1).

    (iii) at least two vertices of degree three are adjacent if 2(γ1)n5(γ1).

    Proposition 2.6. Let GGγn be a graph with γ3 and n>5(γ1), then at least one of the vertices x and y for any edge e=xy has the degree two in G if it has a minimum general Randić index for 1α3925.

    Proof. By Proposition 2.3 and Proposition 2.4, we know 2du3 holds for any vertex u in G. Simultaneously, it follows from Proposition 2.5 that there at least exist two vertices, say u1 and u2, such that φ22>0. Suppose to the contrary that there exists two adjacent vertices v1 and v2 of degree three (i.e., φ33>0). Let u0u1 be the vertex adjacent with u2, which may coincide with v1 or v2. For convenience, we distinguish the following two cases.

    Case 1. NG(u1)NG(u2)=.

    Let ^G8=G{u1u2,u2u0,v1v2}+{u1u0,v1u2,v2u2} (depicted in Figure 9), which is an element in Gγn. By direct calculations, we have Rα(G)Rα(^G8)=4α+9α2×6α>0. This contradicts to the assumption of G. Hence, φ33=0. As desired, we have completed the proof.

    Figure 9.  The transformation G^G8.

    Case 2. NG(u1)NG(u2).

    Without loss of generality, we let u0NG(u1)NG(u2). In what follows, we consider the following three subcases.

    If u0{v1,v2} and u0 is not adjacent to v1 and v2, we let ^G9=G{u2u0,v1v2}+{u2v2,v1u0} (depicted in Figure 10), which is an element in Gγn. By direct calculations, we have Rα(G)Rα(^G9)=0. Note that N^G9(u1)N^G9(u2)=, by the analogous method as in Case 1, and there exists a new graph ~G1 such that Rα(G)Rα(~G1)=Rα(^G9)Rα(~G1)>0. This contradicts to the assumption of G. As desired, we have completed the proof.

    Figure 10.  The transformation G^G9.

    If u0{v1,v2} and u0 is adjacent to v1, we let ^G10=G{u1u2,v1v2}+{u1v1,u2v2}, which is an element in Gγn. By direct calculations, we have Rα(G)Rα(^G10)=4α+9α26α>0. This contradicts to the assumption of G. As desired, we have completed the proof.

    If u0=v2, then we consider a neighbor ˜v of v1 different from v2. Let ^G11=G{u2v2,˜vv1}+{v1u2,˜vv2}Gγn, and again we obtain that Rα(G)Rα(^G11)=0. Note that N^G11(u1)N^G11(u2)=, by the analogous method as in Case 1, and there exists a new graph ~G2 such that Rα(G)Rα(~G2)=Rα(^G11)Rα(~G2)>0. This contradicts to the assumption of G. We have completed the proof.

    Proposition 2.7. Let GGγn be a graph with γ3 and n=5(γ1), then one of the vertices x and y for any edge e=xy has the degree two and the other has the degree three in G if it has the minimum general Randić index for 1α3925.

    Proof. It follows from Propositions 2.3 and 2.4 that 2du3 holds for any vertex u in G. If φ22=0 and φ330, then one can find that φ23=0 by Proposition 2.5, a contradiction. If φ220 and φ330, then from the proof of Proposition 2.6 we conclude that there exists a graph ^G12Gγn such that Rα(G)Rα(^G12)>0. This contradicts to the initial assumption of G. Hence, φ22=φ33=0. As desired, we complete the proof of Proposition 2.7.

    In a similar way, we obtain the following fact.

    Proposition 2.8. Let GGγn be a graph with γ3 and 2(γ1)<n<5(γ1), then G does not contain any edge connecting the vertices of degree two if it has a minimum general Randić index for 1α3925.

    Denote by ¯Gij[φij0] that the graphs contain only vertices of degree i and j, such that for every edge in a one end-vertex has the degree i and the other end-vertex has the degree j, and we use ¯Gij[φii=0] (resp. ¯Gij[φjj=0]) to denote the graphs containing only vertices of degree i and j, such that no vertices of degree i (resp. j) are adjacent.

    Theorem 2.9. Let GGγn be a graph with γ3 and n vertices, then the following holds for 1α3925:

    (i) Rα(G)9α(n+γ1) for n=2(γ1), with equality if, and only if, G is isomophic to cubic graphs.

    (ii) Rα(G)6α(2n4γ+4)9α(n5γ+5) for 2(γ1)<n<5(γ1), with equality if, and only if, G is isomophic to ¯G23[φ22=0].

    (iii) Rα(G)6α(n+γ1) for n=5(γ1), with equality if, and only if, G is isomophic to ¯G23[φ230].

    (iv) Rα(G)4α(n5γ+5)+6α(6γ6) for n>5(γ1), with equality if, and only if, G is isomophic to ¯G23[φ33=0].

    Proof. Let ^G13Gγn be a graph that achieves the minimum general Randić index. We only give the proof of (ii); the rest could be proved in a similar way. It follows from Propositions 2.3 and 2.4 that 2du3 holds for any vertex u in ^G13. Hence, we have n2+n3=n and 2n2+3n3=2(n+γ1) by the Handshaking Theorem. Besides, by Proposition 2.8, it is easily seen that φ22=0. Hence, φ23=2n2 and φ23+2φ33=3n3. Direct calculations show that φ23=2n4γ+4 and φ33=5γn5. Thus, Rα(G)Rα(^G13)=6α(2n4γ+4)9α(n5γ+5). The corresponding extremal graph is ¯G23[φ22=0].

    The second Zagreb index is another well-known vertex degree-based graph invariant in chemical graph theory, which was introduced in 1972 by Gutman and Trinajstić [8]. We encourage the interested reader to consult [9,12] for more information for this graph invariant. Undoubtedly, the second Zagreb index is the special case of the general Randić index when α=1. It is easily seen that Theorem 2.9 extends one of the main results proved by Ali et al. [2].

    We begin with the following auxiliary result, which plays an important part in our proofs.

    Proposition 3.1. Let GGγn be a graph with a maximum general Randić index for α1, then Δ(G)=n1.

    Proof. Suppose to the contrary that there exists a vertex u in G with Δ(G)=du<n1. Note that there exists vV(G) such that uv, dudv and NG(v)NG(u)={v1,v2,,vp}. We can construct a new graph ^G14 the following way

    ^G14=G{vv1,vv2,,vvp}+{uv1,uv2,,uvp}Gγn.

    It is routine to check that

    Rα(^G14)Rα(G)=xNG(u)NG(v)[(du+p)αdαu]dαx+pi=1[(du+p)αdαv]dαvi+yNG(u)NG(v)[(du+p)α+(dvp)αdαudαv]dαy.

    Note that H(t)=tα is an increasing function for α1, and the first and second terms of the previous equality are nonnegative. By the Lagrange mean value theorem, we have (du+p)αdαu=αpξα1 (resp. dαv(dvp)α=αpηα1) for ξ(du,du+p) (resp. η(dvp,dv)). Hence, A6=(du+p)α+(dvp)αdαudαv>0, and, consequently, we have Rα(^G14)>Rα(G), a contradiction. This completes the proof.

    Proposition 3.2. ([14]) Let x1,x2,,xn,p,t1 be integers, α be any real number such that α{0,1} and x1+x2++xn=p.

    (1) The function f(x1,x2,,xn)=ni=1xαi is the minimum for α<0 or α>1 (maximum for 0<α<1, respectively) if, and only if, x1,x2,,xn are almost equal, or |xixj|1 for every i,j=1,2,,n.

    (2) If x1x2t, the maximum of the function f(x1,x2,,xn) is reached for α<0 or α>1 (minimum for 0<α<1,respectively) only for x1=ptn+2,x2=t,x3=x4==xn=1. The second maximum (the second minimum, respectively) is attained only for x1=ptn+1,x2=t+1,x3=x4==xn=1.

    Theorem 3.3. Let GGγn be a graph with \gamma = {k-1\choose 2} and k\geq 4 , then for \alpha\geq 1 we have

    \begin{equation*} \begin{split} R_{\alpha}(G)\leq {k-1\choose 2}(k^{2}-2k+3)^{2\alpha}+(n-1)^{\alpha}(k^{2}-2k+3)^{\alpha}+(n-2)(n-1)^{\alpha}, \end{split} \end{equation*}

    with equality if, and only if, G\cong(K_{1}^{\gamma}\cup (n-2)K_{1})+K_{1}\cong K_{n}^{\gamma} , depicted in Figure 11.

    Figure 11.  The graph G\cong(K_{1}^{\gamma}\cup (n-2)K_{1})+K_{1}\cong K_{n}^{\gamma} with degree sequence (n-1, 2\gamma+1, 1, 1, \ldots, 1) .

    Proof. It follows from Proposition 3.1 that there at least exists one vertex with a maximum degree n-1 . Hence, we have G = \widehat{G_{15}}+K_{1} , which contains |V(\widehat{G_{15}})| = n-1 vertices and m(\widehat{G_{15}}) = {k-1\choose 2} edges. For simplicity, let \pi = (d_{1}, d_{2}, \ldots, d_{n}) and \widehat{\pi} = (\widehat{d_{1}}, \widehat{d_{2}}, \ldots, \widehat{d_{n-1}}) be the nonincreasing degree sequence of G and \widehat{G_{15}} , respectively. Hence, d_{i} = \widehat{d_{i-1}}+1 for i = 2, 3, \ldots, n and d_{1} = n-1. Thus, we have

    \begin{equation*} \begin{split} R_{\alpha}(G) = &\sum\limits_{v_{1}v_{j}\in E(G)}\left[d_{1}d_{j}\right]^{\alpha}+\sum\limits_{v_{i}v_{j}\in E(G), 2\leq i < j\leq n}\left[d_{i}d_{j}\right]^{\alpha}\\ = &(n-1)^{\alpha}\sum\limits_{i = 2}^{n}d_{i}^{\alpha}+\sum\limits_{v_{i}v_{j}\in E(\widehat{G_{15}})}\left[\left(\widehat{d_{i}}+1\right)\left(\widehat{d_{j}}+1\right)\right]^{\alpha}. \end{split} \end{equation*}

    By Proposition 3.2 for \alpha\geq 1 , we have

    \begin{equation*} \begin{split} \mathscr{A}_{7} = &\sum\limits_{i = 1}^{n}d_{i}^{\alpha}-d_{1}^{\alpha}\\ \leq&1\cdot(n-1)^{\alpha}+1\cdot t^{\alpha}+(n-2)\cdot 1^{\alpha}-1\cdot(n-1)^{\alpha}\\ = &(n-1)^{\alpha}+(k^{2}-3k+3)^{\alpha}+(n-2)-(n-1)^{\alpha}\\ = &(k^{2}-3k+3)^{\alpha}+(n-2), \end{split} \end{equation*}

    where d_{1} = n-1 = 2m-t-n+2, d_{2} = t = 2\gamma+1 and d_{3} = d_{4} = \ldots = d_{n} = 1 . In addition, we find the maximum value of

    \begin{equation*} \begin{split} \mathscr{A}_{8} = &\sum\limits_{v_{i}v_{j}\in E(\widehat{G_{15}})}\left[\left(\widehat{d_{i}}+1\right)\left(\widehat{d_{j}}+1\right)\right]^{\alpha}\\ \leq& {k-1\choose 2}\left[\left(k^{2}-3k+3\right)\left(k^{2}-3k+3\right)\right]^{\alpha}, \end{split} \end{equation*}

    with equality if, and only if, \widehat{G_{15}}\cong K_{1}^{\gamma}\cup (n-2)K_{1} .

    It follows from the previous that

    \begin{equation*} \begin{split} R_{\alpha}(G) = &(n-1)^{\alpha}\mathscr{A}_{7}+\mathscr{A}_{8}\\ \leq &{k-1\choose 2}(k^{2}-2k+3)^{2\alpha}+(n-1)^{\alpha}(k^{2}-2k+3)^{\alpha}+(n-2)(n-1)^{\alpha}, \end{split} \end{equation*}

    Hence, G = (K_{1}^{\gamma}\cup (n-2)K_{1})+K_{1}\cong K_{n}^{\gamma} . This completes the proof of Theorem 3.3.

    The authors declare they have not used Artificial Intelligence(AI) tools in the creation of this article.

    We would like to show our great gratitude to the anonymous referees for carefully reading the manuscript and improving its presentation and accuracy. The first author is supported by the Natural Science Foundation of Beijing (No.1222012) and National Key Research and Development Project (No.2019YFB2006602).

    The authors declare that they have no conflicts of interest.



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