
This research explores some modernistic soliton solutions to the (3+1)-dimensional periodic potential the Gross–Pitaevskii equation with a truncated M-fractional derivative plays a significant role in Bose–Einstein condensation, which describes the dynamics of the condensate wave function. The obtained results include trigonometric, hyperbolic trigonometric and exponential function solutions. Three techniques named: the expa function method, the Sardar sub-equation method, and the extended (G′/G)-expansion approach are employed to achieve a variety of new solutions for the governing model. More comprehensive information about the dynamical representation of some of the solutions is being presented by visualizing the 2D, 3D and contour plots. This work reveals a number of new types of traveling-wave solutions, such as the double periodic singular, the periodic singular, the dark singular, the dark kink singular, the periodic solitary singular, and the singular soliton solutions. These novel solutions are not the same as those that were previously studied for this governing equation. The presented techniques demonstrate clarity, efficacy, and simplicity, revealing their relevance to diverse sets of dynamic and static nonlinear equations pertaining to evolutionary events in computational physics, in addition to other real-world applications and a wide range of study fields for addressing a variety of other nonlinear fractional models that hold significance in the fields of applied science and engineering.
Citation: Haitham Qawaqneh, Ali Altalbe, Ahmet Bekir, Kalim U. Tariq. Investigation of soliton solutions to the truncated M-fractional (3+1)-dimensional Gross-Pitaevskii equation with periodic potential[J]. AIMS Mathematics, 2024, 9(9): 23410-23433. doi: 10.3934/math.20241138
[1] | Haifa Bin Jebreen . On the numerical solution of Fisher's equation by an efficient algorithm based on multiwavelets. AIMS Mathematics, 2021, 6(3): 2369-2384. doi: 10.3934/math.2021144 |
[2] | Jafar Biazar, Fereshteh Goldoust . Multi-dimensional Legendre wavelets approach on the Black-Scholes and Heston Cox Ingersoll Ross equations. AIMS Mathematics, 2019, 4(4): 1046-1064. doi: 10.3934/math.2019.4.1046 |
[3] | Ranbir Kumar, Sunil Kumar, Jagdev Singh, Zeyad Al-Zhour . A comparative study for fractional chemical kinetics and carbon dioxide CO2 absorbed into phenyl glycidyl ether problems. AIMS Mathematics, 2020, 5(4): 3201-3222. doi: 10.3934/math.2020206 |
[4] | Amjid Ali, Teruya Minamoto, Rasool Shah, Kamsing Nonlaopon . A novel numerical method for solution of fractional partial differential equations involving the ψ-Caputo fractional derivative. AIMS Mathematics, 2023, 8(1): 2137-2153. doi: 10.3934/math.2023110 |
[5] | Xiaoyong Xu, Fengying Zhou . Orthonormal Euler wavelets method for time-fractional Cattaneo equation with Caputo-Fabrizio derivative. AIMS Mathematics, 2023, 8(2): 2736-2762. doi: 10.3934/math.2023144 |
[6] | Kanagaraj Muthuselvan, Baskar Sundaravadivoo, Kottakkaran Sooppy Nisar, Fahad Sameer Alshammari . New technique for solving the numerical computation of neutral fractional functional integro-differential equation based on the Legendre wavelet method. AIMS Mathematics, 2024, 9(6): 14288-14309. doi: 10.3934/math.2024694 |
[7] | Amnah E. Shammaky, Eslam M. Youssef . Analytical and numerical techniques for solving a fractional integro-differential equation in complex space. AIMS Mathematics, 2024, 9(11): 32138-32156. doi: 10.3934/math.20241543 |
[8] | Emre Kırlı . A novel B-spline collocation method for Hyperbolic Telegraph equation. AIMS Mathematics, 2023, 8(5): 11015-11036. doi: 10.3934/math.2023558 |
[9] | Siqin Tang, Hong Li . A Legendre-tau-Galerkin method in time for two-dimensional Sobolev equations. AIMS Mathematics, 2023, 8(7): 16073-16093. doi: 10.3934/math.2023820 |
[10] | Ahmed Ayad Khudhair, Saeed Sohrabi, Hamid Ranjbar . Numerical solution of nonlinear complex integral equations using quasi- wavelets. AIMS Mathematics, 2024, 9(12): 34387-34405. doi: 10.3934/math.20241638 |
This research explores some modernistic soliton solutions to the (3+1)-dimensional periodic potential the Gross–Pitaevskii equation with a truncated M-fractional derivative plays a significant role in Bose–Einstein condensation, which describes the dynamics of the condensate wave function. The obtained results include trigonometric, hyperbolic trigonometric and exponential function solutions. Three techniques named: the expa function method, the Sardar sub-equation method, and the extended (G′/G)-expansion approach are employed to achieve a variety of new solutions for the governing model. More comprehensive information about the dynamical representation of some of the solutions is being presented by visualizing the 2D, 3D and contour plots. This work reveals a number of new types of traveling-wave solutions, such as the double periodic singular, the periodic singular, the dark singular, the dark kink singular, the periodic solitary singular, and the singular soliton solutions. These novel solutions are not the same as those that were previously studied for this governing equation. The presented techniques demonstrate clarity, efficacy, and simplicity, revealing their relevance to diverse sets of dynamic and static nonlinear equations pertaining to evolutionary events in computational physics, in addition to other real-world applications and a wide range of study fields for addressing a variety of other nonlinear fractional models that hold significance in the fields of applied science and engineering.
In the context of this article, all discussed graphs are assumed to be finite, simple, and undirected. In this regard, we use the notations: V(G) for the vertex set, E(G) for the edge set, F(G) for the face set, Δ(G) (or simply Δ if no confusion occurs) for the maximum degree, δ(G) for the minimum degree, and g(G) for the girth of a graph G. For a vertex x, NG(x) represents the set of vertices adjacent to x in G, and d(x) denotes the degree of vertex x, i.e., the number of vertices adjacent to x.
An injective k-coloring of a graph G refers to a mapping c that assigns a color from the set {1,2,…,k} to each vertex in V(G), this coloring satisfies the condition such that for any two vertices u1 and u2 in V(G), c(u1)≠c(u2) if N(u1)∩N(u2)≠∅. The injective chromatic number χi(G) of G is defined as the smallest integer k for which G has an injective k-coloring.
A list assignment of a graph G is a mapping L that assigns a color list L(x) to each vertex x∈V(G). For a list assignment L of G, if there is an injective coloring φ of G such that φ(x)∈L(x) for each x∈V(G), then φ is called an injective L-coloring. If a graph G can be injectively L-colored for any list assignment L with |L(x)|≥k for each x∈V(G), then G is said to be injectively k-choosable. The injective choosability number χli(G) of a graph G is defined as the minimum positive integer k for which the graph G is injectively k-choosable. It is important to note that χi(G)≤χli(G) holds for any graph G. For planar graphs, Borodin et al. [1] demonstrated that χli(G) and χi(G) are equivalent to Δ under certain conditions. These conditions are as follows: (1) Δ≥16 and g=7; (2) Δ≥10 and 8≤g≤9; (3) Δ≥6 and 10≤g≤11; (4) Δ=5 and g≥12.
The concept of injective coloring was introduced by Hahn et al. [11] in 2002. They showed the injective chromatic number of some special graphs such as paths, cycles, complete graphs, and stars. They also proved that for a connected graph G that is not K2, it holds that χ(G)≤χi(G)≤Δ(Δ−1)+1.
In 2010, Luˇzar proposed a conjecture for planar graphs in [13].
Conjecture A. Suppose G is a planar graph with maximum degree Δ.
(ⅰ) If Δ=3, then χi(G)≤5;
(ⅱ) If 4≤Δ≤7, then χi(G)≤Δ+5;
(ⅲ) If Δ≥8, then χi(G)≤⌊3Δ2⌋+1.
Luˇzar et al. [14] proved that if this conjecture is true, the upper bounds mentioned above are tight.
Several studies have focused on investigating the injective chromatic number of graphs considering the constraints of maximum degree Δ and girth g in [1,2,3,4,5,6,7,8,9,10,12], which can be described as follows:
Theorem 1. Let G be a planar graph with girth g≥g′ and maximum degree Δ≥Δ′.
(1) If (g′,Δ′)∈{(6,17),(7,7),(9,4)}, χi(G)≤Δ+1.
(2) If (g′,Δ′)∈{(6,9)}, χi(G)≤Δ+2.
(3) If (g′,Δ′)∈{(5,20)}, χi(G)≤Δ+3.
(4) For any Δ, if g′=6, χi(G)≤Δ+3; if g′=5, χi(G)≤Δ+6.
(5) If (g′,Δ′)∈{(6,24),(8,5)}, χli(G)≤Δ+1.
(6) If (g′,Δ′)={(6,8)}, χli(G)≤Δ+2.
(7) If (g′,Δ′)={(5,8)}, χli(G)≤Δ+6; If (g′,Δ′)={(5,10)}, χli(G)≤Δ+5; If (g′,Δ′)={(5,11)}, χli(G)≤Δ+4.
(8) For any Δ, if g′=6, χli(G)≤Δ+3; if g′=5, χli(G)≤Δ+5.
For planar graphs G with girth g≥5, Bu et al. [7] proved that if the maximum degree Δ≥20, the inequality χi(G)≤Δ+3 holds. Additionally, if the maximum degree Δ≥10, they established that χli(G)≤Δ+5 in [5], and if the maximum degree Δ≥11, they showed that χli(G)≤Δ+4 [2]. However, the best-known result for planar graphs G with girth g≥5 and any maximum degree Δ is χli(G)≤Δ+5, as shown in [12]. In this paper, for any maximum degree Δ, we present a proof that the list injective chromatic number of planar graphs without 3- and 4-cycles, as well as without intersecting 5-cycles, is at most Δ+4.
Theorem 2. If G is a planar graph with girth g≥5 and without intersecting 5-cycles, then χli(G)≤Δ+4.
In this section, we explore various structural properties of k-critical graphs, which are graphs that do not admit any injective L-coloring with |L(x)|≥k for every vertex x∈V(G), while every proper subgraph of G does allow such a coloring.
Let us introduce some notations. A vertex x in a graph G is defined to be k-vertex, k+-vertex, or k−-vertex if its degree is equal to k, at least k, or at most k, respectively. Similarly, a k-face, k+-face, or k−-face can be defined. A vertex adjacent to vertex x in a graph G is referred to as a k-neighbor, k+-neighbor, or k−-neighbor of x if it is a k-vertex, k+-vertex, or k−-vertex, respectively. Let Nk(x) denote the set of k-neighbors of vertex x, and let nk(x) and nk+(x) represent the counts of k-neighbors and k+-neighbors of x, respectively. We can define SG(x) as ∑y∈N(x)(d(y)−1)=∑y∈N(x)d(y)−d(x). It is evident that the number of vertices in graph G that share a common neighbor with vertex x is at most SG(x). Therefore, if the remaining vertices are injectively colored, vertex x has at most SG(x) forbidden colors. Consider a graph G that is (Δ+4)-critical. If a 3-vertex x of G has a 2-neighbor, we call x a bad vertex. Otherwise, if x does not have a 2-neighbor, we call it a good vertex. A vertex is referred to as a k(s)-vertex if it is a k-vertex and has exactly s 2-neighbors.
We can represent the adjacent vertices of a vertex v in graph G as v1,v2,…,vd(v) where d(v1)≤d(v2)≤…≤d(vd(v)) in ascending order of their degrees. Then, we call v a (d(v1),d(v2),…,d(vd(v)))-vertex. Additionally, if d(vi)=2, we denote the other neighbor of vi as v′i.
We present the following properties of (Δ+4)-critical graphs. Their proofs can be found in Reference [4].
Lemma 3. δ(G)≥2.
Lemma 4. For any edge uv∈E(G), max{SG(u),SG(v)}≥Δ+4.
Lemma 5. G has no adjacent 2-vertices.
Lemma 6. For a vertex v with 3≤d(v)≤5, if v1 is a 2-neighbor of v, l=n3+(v), ui(i=1,…,l) is the 3+-neighbor of v, then
(1) l≥2,
(2) ∑li=1d(ui)≥Δ+4+2l−d(v).
In this part, we will prove Theorem 2 by contradiction, and the graphs discussed below are planar graphs. Suppose that Theorem 2 is not true. Let G be a (Δ+4)-critical graph and L be the corresponding list assignment of G with |L(x)|≥Δ+4 for each x∈V(G). It is, G does not admit any injective L-coloring, while every proper subgraph of G does allow such a coloring. Then G is connected and δ(G)≥2.
First, we assign an initial charge ω(v) to each vertex v such that ω(v)=32d(v)−5 and a charge ω(f)=d(f)−5 to each face f. By Euler's formula, |V(G)|−|E(G)|+|F(G)|=2, we have
∑v∈V(G)(32d(v)−5)+∑f∈F(G)(d(f)−5)=−10.
We then proceed by transferring charges from one element to another, resulting in a new charge ω′(t) for each t∈V(G)∪F(G) such that ∑t∈V(G)∪F(G)ω′(t)=−10 still holds. Our goal is to show that if all transfers result in ω′(t)≥0 for each t∈V(G)∪F(G), then we will reach a contradiction:
0≤∑t∈V(G)∪F(G)ω′(t)=∑t∈V(G)∪F(G)ω(t)=−10<0.
Hence, the theorem is proved.
Now, we prove the following lemma.
Lemma 7. Δ≥4.
Proof. According to Lemma 3 and Lemma 5, we have Δ≥3. If Δ=3, let d(v)=3, then SG(v)=d(v1)+d(v2)+d(v3)−3<Δ+4. However, we can observe that SG(v1)≤Δ+Δ+3−3=2Δ<Δ+4, which contradicts Lemma 4. Therefore, Δ≥4.
We will then prove the theorem by distinguishing several cases based on the maximum degree Δ. In the rest, we always give a proper subgraph of G, denoted by G′. Since G is critical, G′ has an injective L-coloring c. After deleting the colors of some vertices, let L′c(x) denote the set of available colors of x.
Claim 8. A 4(2)-vertex is not adjacent to a 4-vertex with at least one 2-neighbor.
Proof. Let us consider a 4(2)-vertex denoted as u, with its neighbors u1 and u2 having degrees of 2, and another neighbor u3 as a 4-vertex. Let v be the adjacent 2-vertex of u3. For convenience, assume that d(u4)=Δ. Let G′=G−uu1. Now we delete the colors on u,u1 and v. Obviously, |L′c(u)|≥Δ+4−(2×2+4+Δ−d(u)−1)≥1, |L′c(u1)|≥Δ+4−(Δ+4−d(u1))≥2, |L′c(v)|≥Δ+4−(Δ+4−d(v)−1)≥3. So we can recolor u,u1,v, in turn and an injective L-coloring of G is obtained, a contradiction.
We use the following discharging rules.
R1-1. Each 2-vertex receives 1 from each adjacent 3+-vertex.
R1-2. A 4-vertex sends 14 to each adjacent 3-vertex.
R1-3. A (3,4,4)-vertex sends 118 to each adjacent (3,3,3)-vertex.
R1-4. Each 6+-face equally distributes its positive charge to each incident 3+-vertex.
R1-5. A 4(0)-vertex sends 14 to each of its 4(2)-neighbors.
First, we check ω′(v)≥0 for each vertex v∈V(G).
Case 1. If d(v)=2, then ω(v)=32×2−5=−2. By R1-1, ω′(v)=−2+1×2=0.
Case 2. If d(v)=3, then ω(v)=32×3−5=−12. According to Lemma 4, n2(v)=0. If n3(v)≤1, then ω′(v)≥−12−118+2×14+2×16=518 by R1-2, R1-3, and R1-4. If n3(v)=2, then ω′(v)≥−12+14+2×16=112 by R1-2 and R1-4. If n3(v)=3, then according to Lemma 4, each 3-neighbor of v must be (3,4,4)-vertex. Therefore, using R1-3 and R1-4, we have ω′(v)≥−12+3×118+2×16=0.
Case 3. If d(v)=4, then ω(v)=32×4−5=1. According to Lemma 4, n2(v)≤2. If n2(v)=0, then ω′(v)≥1−4×14+3×16=12 by R1-2, R1-4, and R1-5. If n2(v)=1, then according to Lemma 4, n3(v)≤2. So ω′(v)≥1−1−2×14+3×16=0 by R1-1, R1-2, and R1-4. If n2(v)=2, then n4(v)=2 by Lemma 4, and each 4-neighbor of v must be a 4(0)-vertex by Claim 8. It follows that ω′(v)≥1−2+2×14+3×16=0 by R1-1, R1-5, and R1-4.
We now check ω′(f)≥0 for each f∈F(G).
If f is a 5-face, then ω′(f)=d(f)−5=0 since no charge is discharged to or from f. If f is a 6+-face, according to R1-4, f gives away its positive charge, so ω′(f)=0.
We have checked ω′(x)≥0 for all x∈V(G)∪F(G). Therefore, the proof is completed when Δ=4.
Claim 9. The following configurations are forbidden.
(1) A (2,2,5,5)-vertex adjacent to a 5-vertex with at least two 2-neighbors.
(2) A (2,2,4,5)-vertex adjacent to a 5-vertex with at least one 2-neighbor.
(3) A (2,3,3,3,3)-vertex adjacent to a bad 3-vertex.
(4) A 5(3)-vertex adjacent to a bad 3-vertex.
Proof. (1) Let v be a (2, 2, 5, 5)-vertex with d(v1)=d(v2)=2 and d(v3)=d(v4)=5. Suppose v4 is adjacent to at least two 2-vertices u and w. Let G′=G−vv1. Now we delete the colors on v,v1,u, and w. It is clear that |L′c(v)|≥Δ+4−(2×2+5×2−d(v)−2)≥1. If v1 and u (or w) have no common neighbor, then we have |L′c(v1)|≥Δ+4−(Δ+4−d(v1))≥2, |L′c(u)|≥Δ+4−(Δ+5−d(u)−2)≥3, and |L′c(w)|≥Δ+4−(Δ+5−d(w)−2)≥3. Thus, we can recolor v,u,w, and v1 in turn to obtain an injective L-coloring of G. If v1 and u have a common neighbor, then we have |L′c(v1)|≥Δ+4−(Δ+4−d(v1)−1)≥3, |L′c(u)|≥Δ+4−(Δ+5−d(u)−3)≥4, and |L′c(w)|≥Δ+4−(Δ+5−d(w)−2)≥3. Thus, we can recolor v,v1,w, and u in turn to obtain an injective L-coloring of G.
(2) Let v be a (2, 2, 4, 5)-vertex with d(v1)=d(v2)=2, d(v3)=4, d(v4)=5, and u be a 2-neighbor of v4. Let G′=G−vv1. Now we delete the colors on v,v1, and u. It is clear that |L′c(v)|≥Δ+4−(2×2+4+5−d(v)−1)≥1. If N(v1)∩N(u)=∅, then we have |L′c(v1)|≥Δ+4−(Δ+4−d(v1))≥2 and |L′c(u)|≥Δ+4−(Δ+5−d(u)−1)≥2. Thus, we can recolor v,u, and v1 in turn to obtain an injective L-coloring of G. If N(v1)∩N(u)≠∅, then we have |L′c(v1)|≥Δ+4−(Δ+4−d(v1)−1)≥3 and |L′c(u)|≥Δ+4−(Δ+5−d(u)−2)≥3. Thus, we can recolor v,v1, and u in turn to obtain an injective L-coloring of G.
(3) Let v be a (2, 3, 3, 3, 3)-vertex with d(v1)=2, d(v2)=d(v3)=d(v4)=d(v5)=3, and u be a 2-neighbor of v2. Let G′=G−vv1. Now we delete the colors on v,v1, and u. It is clear that |L′c(v)|≥Δ+4−(2+3×4−d(v)−1)≥1, |L′c(v1)|≥Δ+4−(Δ+5−d(v1))≥1, and |L′c(u)|≥Δ+4−(Δ+3−d(u)−1)≥4. So we can recolor v,v1, and u in turn, and an injective L-coloring of G is obtained.
(4) Let v be a 5(3)-vertex with d(v1)=d(v2)=d(v3)=2, d(v4)=3, and u be a 2-neighbor of v4. For convenience, assume d(v5)=Δ. Let G′=G−vv1. Now we delete the colors on v,v1, and u. It is clear that |L′c(v)|≥Δ+4−(2×3+3+Δ−d(v)−1)≥1, |L′c(v1)|≥Δ+4−(Δ+5−d(v1))≥1, and |L′c(u)|≥Δ+4−(Δ+3−d(u)−1)≥4. So we can recolor v,v1, and u in turn, and an injective L-coloring of G is obtained.
We use the following discharging rules.
R2-1. Each 2-vertex receives 1 from each adjacent 3+-vertex.
R2-2. A 4-vertex sends 118 to each adjacent 3-vertex.
R2-3. A 5-vertex sends 14 to each adjacent (2,2,5,5)-vertex, 12 to each adjacent (2,2,4,5)-vertex.
R2-4. A 5-vertex sends 712 to each adjacent bad 3-vertex, 16 to each adjacent good 3-vertex.
R2-5. A (3,4+,5)-vertex sends 118 to each adjacent (3,3,3+)-vertex or adjacent (3,4,4)-vertex.
R2-6. Each 6+-face equally distributes its positive charge to each incident 3+-vertex.
First, we check ω′(v)≥0 for each vertex v∈V(G).
Case 1. d(v)=2, ω(v)=32×2−5=−2. By R2-1, ω′(v)=−2+1×2=0.
Case 2. d(v)=3, ω(v)=32×3−5=−12. Let N(v)={v1,v2,v3}. By Lemma 6(1), n2(v)≤1.
If n2(v)=1, then according to Lemma 6(2), n5(v)=2. By R2-1, R2-4, and R2-6, ω′(v)≥−12−1+2×712+2×16=0.
If n2(v)=0 and n3(v)≥2, let d(v1)=d(v2)=3; then v1 and v2 are both (3,4+,5)-vertices by Lemma 4. So ω′(v)≥−12+2×118+2×16+min{118,118,16}=0.
If n2(v)=0 and n3(v)=1, then v must be either a (3,4,4)-vertex or a (3,4+,5)-vertex. When v is a (3,4,4)-vertex, the 3-vertex adjacent to v must be (3,4+,5)-vertex by Lemma 4, so ω′(v)≥−12+118+2×118+2×16=0; When v is a (3,4+,5)-vertex, ω′(v)≥−12−118+118+16+2×16=0.
If n2(v)=0 and n3(v)=0, then ω′(v)≥−12+3×118+2×16=0.
Case 3. d(v)=4, ω(v)=32×4−5=1. Let N(v)={v1,v2,v3,v4}. By Lemma 6(1), n2(v)≤2.
If n2(v)=2, let d(v1)=d(v2)=2, then by Lemma 6(2), d(v3)+d(v4)≥Δ+4. So v must be a (2,2,4+,5)-vertex. Hence, ω′(v)≥1−2×1+min{12,2×14}+3×16=0 by R2-1, R2-3, and R2-6.
If n2(v)≤1, then ω′(v)≥1−1−3×118+3×16=13 by R2-1, R2-2, and R2-6.
Case 4. d(v)=5, ω(v)=32×5−5=52. Let N(v)={v1,v2,v3,v4,v5}. By Lemma 6, n2(v)≤3.
If n2(v)=3, then n3(v)≤1 by Lemma 4. When n2(v)=3 and n3(v)=1, then v is a (2,2,2,3,5)-vertex. By Claim 9(4), v is not adjacent to a bad 3-vertex. Therefore, by R2-1, R2-4 and R2-6, we have ω′(v)≥52−3×1−16+4×16=0. When n2(v)=3 and n3(v)=0, by Claim 9(1)(2), v is not adjacent to a (2,2,4,5)-vertex or a (2,2,5,5)-vertex. Thus, ω′(v)≥52−3×1+4×16=16.
If n2(v)=2, then n3(v)≤2 by Lemma 4. By Claim 9(1)(2), v is not adjacent to a (2,2,4,5)-vertex or a (2,2,5,5)-vertex. Therefore, ω′(v)≥52−2×1−2×712+4×16=0.
If n2(v)=1 and n3(v)=4, then by Claim 9(3), v is not adjacent to bad 3-vertices. Thus, ω′(v)≥52−1−4×16+4×16=32.
If n2(v)=1 and n3(v)≤3, then by Claim 9(2), v is not adjacent to a (2,2,4,5)-vertex. Therefore, ω′(v)≥52−1−3×712−14+4×16=16.
If n2(v)=0, then ω′(v)≥52−5×712+4×16=14 by R2-3, R2-4, and R2-6.
We now check ω′(f)≥0 for each f∈F(G).
If f is a 5-face, then ω′(f)=d(f)−5=0 since no charge is discharged to or from f. If f is a 6+-face, according to R2-6, f gives away its positive charge, so ω′(f)=0.
We have verified that ω′(x)≥0 for all x∈V(G)∪F(G). This completes the proof when Δ=5.
Claim 10. The following configurations are forbidden.
(1) A 5(3)-vertex adjacent to a 4(2)-vertex.
(2) A 5(3)-vertex adjacent to a bad 3-vertex.
Proof. (1) Let v be a 5(3)-vertex with d(v1)=d(v2)=d(v3)=2, d(v4)=4, and x,y as 2-neighbors of v4. For convenience, assume d(v5)=Δ. Let G′=G−vv1. Now we delete the colors on v,v1,x, and y. Obviously, |L′c(v)|≥Δ+4−(2×3+4+Δ−d(v)−2)≥1, |L′c(v1)|≥Δ+4−(Δ+5−d(v1))≥1, |L′c(x)|≥Δ+4−(Δ+4−d(x)−2)≥4, and |L′c(y)|≥Δ+4−(Δ+4−d(y)−2)≥4. So we can recolor v,v1,x, and y in turn, and an injective L-coloring of G is obtained.
(2) The proof can be seen from Claim 9(4).
Claim 11. Let f1=v′1v1vv2v′2x, f2=v′3v3vv2v′2y be two 6-faces. Suppose d(v)=6 and SG(v)<Δ+4. If d(v1)=d(v2)=d(v3)=2, then d(x)≥3 and d(y)≥3. Additionally, if d(x)=3 and d(y)=3, then at most one of x or y is a bad 3-vertex. (The configuration composed of f1 and f2 is called Configuration A of v. See H1 in Figure 1)
Proof. By Lemma 4, we conclude that d(v′1)=d(v′2)=d(v′3)=Δ. The subsequent steps of the proof follow a similar approach as presented in Claim 3 of the reference paper [8] by Chen et al.
Claim 12. Let f1=v′1v1vv2v′2x be a 6-face, f2=v′3v3vv2v′2yz be a 7-face. Suppose d(v)=6 and SG(v)<Δ+4. If d(v1)=d(v2)=d(v3)=2 and d(z)≤5, then min{d(x),d(y)}≥3. (The configuration composed of f1 and f2 is called Configuration B of v. See H2 in Figure 1)
Proof. To prove this claim, we will use a proof by contradiction. Let G′=G−vv1. According to Lemma 4, we can deduce that d(v′1)=d(v′2)=d(v′3)=Δ. We will now consider three cases.
Case A. d(x)=2 and d(y)=2. In this case, erase the colors on vertices v,v1,v2,v3,x, and y. Now, we have |L′c(v)|≥(Δ+4)−SG(v)≥1, |L′c(v1)|≥(Δ+4)−(Δ+6−2−3)=3, |L′c(v2)|≥(Δ+4)−(Δ+6−2−4)=4, |L′c(v3)|≥(Δ+4)−(Δ+6−2−2)=2, |L′c(x)|≥(Δ+4)−(Δ+Δ−2−3)=9−Δ=3, and |L′c(y)|≥(Δ+4)−(Δ+5−2−2)=3. Therefore, we can recolor vertices v,v3,v1,v2,x, and y sequentially, obtaining an injective L-coloring of G, which leads to a contradiction.
Case B. d(x)=2 and d(y)≥3. In this case, erase the colors on v,v1,v2,v3, and x. We have |L′c(v)|≥(Δ+4)−SG(v)≥1, |L′c(v1)|≥(Δ+4)−(Δ+6−2−3)=3, |L′c(v2)|≥(Δ+4)−(Δ+6−2−3)=3, |L′c(v3)|≥(Δ+4)−(Δ+6−2−2)=2, and |L′c(x)|≥(Δ+4)−(Δ+Δ−2−2)=8−Δ=2. If L′c(x)∩L′c(v3)≠∅, let α∈L′c(x)∩L′c(v3), then recolor x and v3 with α, and recolor vertices v,v1, and v2 sequentially. Suppose L′c(x)∩L′c(v3)=∅. If L′c(x)∩L′c(v1)=∅, we can simply recolor vertices v,v3,v1,v2, and x sequentially. If L′c(x)∩L′c(v1)≠∅, let β∈L′c(x)∩L′c(v1). Note that β∉L′c(v3). We can recolor vertex v1 with color β, and then recolor vertices v,x, and v2 sequentially. Since |L′c(v3)|≥2 and β∉L′c(v3), there exists at least one color γ∈L′c(v3) that is different from the color of v2 and γ≠β. Therefore, we can recolor vertex v3 with color γ.
Case C. d(y)=2 and d(x)≥3. In this case, erase the colors on v,v1,v2,v3, and y. We have|L′c(v)|≥(Δ+4)−SG(v)≥1, |L′c(v1)|≥(Δ+4)−(Δ+6−2−2)=2, |L′c(v2)|≥(Δ+4)−(Δ+6−2−3)=3, |L′c(v3)|≥(Δ+4)−(Δ+6−2−2)=2, and |L′c(y)|≥(Δ+4)−(Δ+5−2−1)=2. Therefore, we can recolor vertices v,v1,v3,v2, and y sequentially.
In all three cases, we obtain an injective L-coloring of G, which leads to a contradiction. Therefore, the claim is proved.
Claim 13. Let f1=xv1vv2y be a 5-face. Suppose d(v)=6 and SG(v)<Δ+4. If d(v1)=2 and d(v2)=3, then d(y)≥3. (See H3 in Figure 1)
Proof. The proof is carried out by contradiction. Let G′=G−vv1. Suppose d(y)=2. Erase the colors on vertices v,v1, and y. Then we have |L′c(v)|≥(Δ+4)−SG(v)≥1, |L′c(v1)|≥(Δ+4)−(Δ+6−2−1)=1, and |L′c(y)|≥(Δ+4)−(Δ+3−2−2)=5. We can then recolor v,v1, and y in turn.
We use the following discharging rules.
R3-1. Each 2-vertex receives 1 from each adjacent 3+-vertex.
R3-2. A bad 3-vertex receives 12 from each adjacent 5-vertex, 1930 from each adjacent 6-vertex.
R3-3. A good 3-vertex receives 16 from each adjacent 5-vertex, 13 from each adjacent 6-vertex.
R3-4. Suppose d(v)=3. If SG(v)<Δ+4, then v receives 118 from each adjacent 3-vertex or 4-vertex.
R3-5. A 4(2)-vertex receives 1360 from each adjacent 5-vertex, 1330 from each adjacent 6-vertex.
R3-6. Each 6+-face equally distributes its positive charge to each incident 3+-vertex.
R3-7. In Configuration A or B, v receives 112 along edge v′2x from v′2, and 112 along edge v′2y from v′2, for a total of 16 received from v′2. (See Fig. 2)
R3-8. For a 5-face f=vv1xyv2, if d(v)=6, d(v1)=d(v2)=2, d(x)=d(y)=Δ, we call it Configuration C1 of v. In Configuration C1, v receives 16 along edge xv1 from x, and 16 along edge yv2 from y, for a total of 13 from x and y. On the other hand, if d(v)=6, d(v1)=2, d(v2)=3, d(x)=Δ, and d(y)≥5, we call it Configuration C2 of v. In Configuration C2, v receives 16 along edge xv1 from x.(See Figure 2)
R3-9. For a 7-face f=yv′2v2vv3v′3z, if d(v)=6, d(v2)=d(v3)=2 and d(v′2)=d(v′3)=d(z)=Δ, we call it Configuration D of v. In Configuration D, v receives 112 along edge zy from z, and 112 along edge v′3v3 from v′3, for a total of 16 from z and v′3. (See Figure 2)
Remark 1. Let d(v)=6. In Configuration A, v receives 16 from v′2, and receives 14+14 from f1 and f2, so v receives 23 in total; in Configuration B, v receives 16 from v′2, receives 14 from f1 and receives 25 from f2, so v receives 4960 in total; in Configuration D, v receives 16 from v′3 and z, receives 25 from f, so v receives 1730 in total.
First, we check ω′(v)≥0 for each vertex v∈V(G).
Case 1. d(v)=2. We have ω(v)=32×2−5=−2. By applying R3-1, we obtain ω′(v)=−2+1×2=0.
Case 2. d(v)=3. We have ω(v)=32×3−5=−12. Let N(v)={v1,v2,v3}. By Lemma 6(1), n2(v)≤1.
If n2(v)=1, let d(v1)=2, then d(v2)+d(v3)≥Δ+5 by Lemma 6(2). Using R3-1, R3-2, and R3-6, we have ω′(v)≥−12−1+min{12+1930,2×1930}+16+15=0.
If n2(v)=0 and n3(v)≥2, then SG(v)<Δ+4. By applying R3-3, R3-4, and R3-6, we have ω′(v)≥−12+min{3×118,2×118+16,2×118+13}+2×16=0.
Suppose n2(v)=0 and n3(v)=1; let d(v1)=3. If SG(v)<Δ+4, then d(v2)+d(v3)<Δ+4=10. By applying R3-4, R3-3 and R3-6, we have ω′(v)≥−12+min{3×118,2×118+16}+2×16=0. If SG(v)≥Δ+4, then d(v2)+d(v3)≥Δ+4=10. By applying R3-4, R3-3, and R3-6, we have ω′(v)≥−12−118+min{13,2×16}+2×16=19.
Suppose n2(v)=0 and n3(v)=0. If SG(v)<Δ+4, then d(v1)=d(v2)=d(v3)=4. By applying R3-4 and R3-6, we have ω′(v)≥−12+3×118+2×16=0. If SG(v)≥Δ+4, v is adjacent to at least one 5+-vertex. By applying R3-3 and R3-6, we have ω′(v)≥−12+16+2×16=0.
Case 3. d(v)=4. We have ω(v)=32×4−5=1. Let N(v)={v1,v2,v3,v4}. By Lemma 6(1), n2(v)≤2.
If n2(v)=2, let d(v1)=d(v2)=2, then d(v3)+d(v4)≥Δ+4 by Lemma 6(2). Using R3-1, R3-5, and R3-6, we have ω′(v)≥1−2+min{1330,2×1360}+16+2×15=0.
If n2(v)≤1, then ω′(v)≥1−1−3×118+3×16=13 by R3-1, R3-4, and R3-6.
Case 4. d(v)=5. We have ω(v)=32×5−5=52. Let N(v)={v1,v2,v3,v4,v5}. By Lemma 6(1), we have n2(v)≤3.
If n2(v)=3, let d(v1)=d(v2)=d(v3)=2. Then, using Claim 10 and Lemma 6(2), we see that v is not adjacent to a bad 3-vertex or a 4(2)-vertex, and d(v4)+d(v5)≥Δ+3. This means that v is adjacent to at most one good 3-vertex. Using R3-1, R3-3, and R3-6, we have ω′(v)≥52−3×1−16+4×16=0.
If n2(v)=2, by Lemma 4, we have n3(v)≤2.
Suppose n2(v)=2 and n3(v)=2. Then, another neighbor of v must be 5+-vertex. Using R3-1, R3-2, and R3-6, we have ω′(v)≥52−2×1−2×12+4×16=16.
If n2(v)=2 and n3(v)≤1, using R3-1, R3-2, R3-5, and R3-6, we have ω′(v)≥52−2×1−12−2×1360+4×16=730.
If n2(v)≤1, using R3-1, R3-2 and R3-6, we have ω′(v)≥52−1−4×12+4×16=16.
Case 5. d(v)=6. We have ω(v)=32×6−5=4. Let N(v)={v1,v2,v3,v4,v5,v6}.
Case 5.1. n2(v)=0. If v is incident with a Configuration D, v receives at least −112+25=1960 from the configuration. However, if v is incident with a 6-face f, v receives at least 16 from f. Therefore, in the worst case scenario, we assume that v is not incident with any Configurations D. Using R3-2 and R3-6, we have ω′(v)≥4−6×1930+5×16=3130.
Case 5.2. n2(v)=1. Similarly to Case 5.1, we assume that v is not incident with any Configurations D. Then there exists at most one Configuration C1, or one Configuration C2, or one Configuration A, or one Configuration B. Using R3-1, R3-2, R3-3, R3-4, R3-7, R3-8, and R3-6, we have ω′(v)≥4−1−5×1930−max{16,2×112}+5×16=12.
Case 5.3. n2(v)=2. Similarly to Case 5.1, we assume that v is not incident with any Configurations D. Then there exists at most one Configuration C1, or one Configuration C2, or two Configurations A, or two Configurations B. Using R3-1, R3-2, R3-3, R3-4, R3-7, R3-8, and R3-6, we have ω′(v)≥4−2−4×1930−max{16,16+2×112,4×112}+3×16+2×15=130.
Case 5.4. n2(v)=3. Similarly to Case 5.1, we assume that v is not incident with any Configurations D. If v is incident with at most one bad 3-vertex, then there exist at most one Configuration C1 (or Configuration C2) and two Configurations A (or Configurations B), or at most three Configurations A (or Configurations B). When v is incident with a Configuration C1 (or Configuration C2) that requires charges to be sent to it, v must be adjacent to at least one 5+-vertex. Therefore, in the worst-case scenario, v is incident with three Configurations A, ω′(v)≥4−3×1−1930−2×1330−3×16+6×15=15.
If v is incident with two bad 3-vertices, there exist at most two Configurations A incident with v by Claim 11. So ω′(v)≥4−3×1−2×1930−1330−2×16+4×14+min{−16+14+25,15}=16.
If v is incident with three bad 3-vertices, there exist at most three Configurations B incident with v by Claim 11 and Claim 12. So ω′(v)≥4−3×1−3×1930+min{−3×16+3×14+3×25, −2×16+2×14+2×25+15,−1×16+14+25+2×15+16,4×15+16, 14+2×15+2×16,5×15}=115.
Case 5.5. n2(v)=4. Let d(vi)=2, i=1,…,4. We consider three subcases.
Subcase 5.5.1. v is adjacent to two bad 3-vertices. Let d(v5)=d(v6)=3. Note that SG(v)=2×4+3×2−6=8<Δ+4. By Lemma 4, we have d(v′i)=Δ for i=1,…,4 and another neighbor of vj must be a 5+-vertex for j=5,6.
If v is incident with one Configuration C1, say f1, there are three cases shown in Figure 3(1)–(3). In (1) and (2), we have ω′(v)≥4−4×1−2×1930+2×16+min{2×14+2×15+16,14+4×15}=760. In (3), there is at most one Configuration B. So ω′(v)≥4−4×1−2×1930+2×16+min{−2×112+2×14+25+2×15,3×14+2×15}=15.
If v is incident with one Configuration C2, say f6, there are three cases shown in Figure 3(4)–(6). We have ω′(v)≥4−4×1−2×1930+16+4×14+15=110.
If v is incident with a 5-face, but not Configuration C1 or Configuration C2, there is only one case where d(f5)=5 according to Claim 13 (see Figure 3(7)). If v is incident with at least one Configuration A or B, then ω′(v)≥4−4×1−2×1930+2×112+4×14+15=110. If v is not incident with any Configurations A or B, then max{d(f1),d(f2)}≥7 and max{d(f2),d(f3)}≥7. Therefore, either d(f2)≥7 or min{d(f1),d(f3)}≥7. Suppose d(f2)=7 and min{d(f1),d(f3)}=6. Since v is not incident with any Configuration B, f2 must be Configuration D. Thus, v receives (2×112+25) from f2, and ω′(v)≥4−4×1−2×1930+(2×112+25)+3×14+15=14. If d(f2)=7 and min{d(f1),d(f3)}≥7, we have ω′(v)≥4−4×1−2×1930+3×25+14+15=2360. If d(f2)≥8, then ω′(v)≥4−4×1−2×1930+36+3×14+15=1160. Now suppose min{d(f1),d(f3)}≥7. In this case, ω′(v)≥4−4×1−2×1930+2×25+2×14+15=730.
If v is not incident with any 5-faces, then v is incident with at most one Configuration B to which v needs to send charges. So ω′(v)≥4−4×1−2×1930+min{−2×112+25+5×14,2×14+4×15}=130.
Subcase 5.5.2. v is adjacent to one bad 3-vertex. Let v5 be a bad 3-vertex. If d(v6)≥5, then v is incident with at most one Configuration C1 (or C2), or at most one Configuration A (or B) that requires charges to be sent to it. Therefore, ω′(v)≥4−4×1−1930−16+2×14+3×15=1130. If d(v6)≤4, then according to Lemma 4, d(v′i)=Δ, i=1,…,4. We consider three cases (see Figure 4(1)–(3)). In (1) and (2), we only need to consider the worst case where v6 is a 4(2)-vertex and v is incident with a 5-face but not Configuration C1 or C2. In this case, ω′(v)≥4−4×1−1930−1330+min{3×14+15+16,2×14+3×15}=130. In (3), f4 and f5 may form a Configuration A or B. We consider the worst case where v6 is a 4(2)-vertex and v is incident with a 5-face but not Configuration C1 or C2. If f1 and f2 cannot form configuration A, then max{d(f1),d(f2)}≥7. In this case, ω′(v)≥4−4×1−1930−1330−2×112+14+25+3×15=160. If f1 and f2 form configuration A, then ω′(v)≥4−4×1−1930−1330−2×112+2×112+2×14+3×15=130.
Subcase 5.5.3. v is not adjacent to any bad 3-vertices. Let us consider the worst case where v is adjacent to two 4(2)-vertices and is incident with a 5-face, but not Configuration C1 or C2. In this case, ω′(v)≥4−4×1−2×1330+min{3×14+15+16,2×14+3×15,−2×112+4×14+15}=16.
Case 5.6. n2(v)=5. Let d(vi)=2, i=1,…,5. We consider four subcases.
Subcase 5.6.1. v6 is a bad 3-vertex. Let N(v6)={v,x,y} and d(x)=2. According to Lemma 4, we have d(v′i)=Δ for i=1,…,5, and d(x)+d(y)≥Δ+1=7. Thus, d(y)≥5. By Claim 13, v is not incident with a 5-face that is not C1 or C2.
Suppose v is incident with a Configuration C1. If v is incident with at least one 7+-face, then ω′(v)≥4−5×1−1930+2×16+4×14+26=130. If v is not incident with any 7+-faces, then v is incident with at least one Configuration A. So ω′(v)≥4−5×1−1930+2×16+2×112+4×14+15=115.
Suppose v is incident with a Configuration C2. In this case, C2 must be f5 (see Fig. 5(1)). If v is incident with at least two 7+-faces, then ω′(v)≥4−5×1−1930+16+3×14+2×25=112. If v is incident with one 7+-face, then v is incident with at least one Configuration A. So ω′(v)≥4−5×1−1930+16+2×112+4×14+25=110. If v is not incident with any 7+-faces, then v is incident with at least three Configurations A. So ω′(v)≥4−5×1−1930+16+5×14+3×16=1760.
Suppose v is not incident with any 5-faces. If v is incident with at least two 7+-faces, then ω′(v)≥4−5×1−1930+min{2×25+3×14+15,26+25+4×14}=110. If v is incident with one 7+-face, then v is incident with at least one Configuration A. So ω′(v)≥4−5×1−1930+25+4×14+15+2×112=215. If v is not incident with any 7+-faces, then v is incident with at least three Configurations A. So ω′(v)≥4−5×1−1930+15+5×14+3×16=1960.
Subcase 5.6.2. v6 is a good 3-vertex. Let N(v6)={v,x,y}. According to Lemma 4, we have d(v′i)=Δ for i=1,…,5, and d(x)+d(y)≥Δ+1=7.
If v is not incident with a 5-face, then ω′(v)≥4−5×1−13+4×14+2×15=115.
If v is incident with a Configuration C1 or C2, then ω′(v)≥4−5×1−13+min{2×15+3×14+2×16,15+4×14+16}=130.
If v is incident with a 5-face but not Configuration C1 or C2, then the 5-face must be f5 or f6 (see Figure 5(2)). If v is incident with at least one 7+-face, then ω′(v)≥4−5×1−13+min{15+3×14+25,4×14+26}=0. If v is not incident with 7+-faces, then f1 and f2, f2 and f3, and f3 and f4 form three Configurations A. Therefore, ω′(v)≥4−5×1−13+4×14+15+3×16=1130.
Subcase 5.6.3. v6 is a 4(2)-vertex. Let N(v6)={v,x,y,z} and d(x)=d(y)=2. According to Lemma 4, we know that d(v′i)=Δ for i=1,…,5, and d(z)≥4. It is clear that v is not incident with a Configuration C2.
If v is not incident with a 5-face, then ω′(v)≥4−5×1−1330+5×14+15=160.
If v is incident with a Configuration C1, then ω′(v)≥4−5×1−1330+4×14+15+2×16=110.
Suppose v is incident with a 5-face but not Configuration C1. Then the 5-face must be f5 or f6 (see Figure 5(3)). If v is incident with at least two 7+-faces, ω′(v)≥4−5×1−1330+min{2×14+15+2×25,3×14+25+26}=120. If v is incident with one 7+-face, then v is incident with at least one Configuration A. So ω′(v)≥4−5×1−1330+min{4×14+26+2×112,3×14+15+25+2×112}=115. If v is not incident with 7+-faces, then f1 and f2, f2 and f3, and f3 and f4 form three Configurations A. So ω′(v)≥4−5×1−1330+4×14+15+3×16=415.
Subcase 5.6.4. v6 is not a 3-vertex or a 4(2)-vertex. If v is incident with a Configuration D, v receives at least −112+25=1960 from the configuration. However, if v is incident with a 6-face f, v receives at least 16 from f. Therefore, in the worst-case scenario, we assume that v is not incident with any Configurations D. If v is incident with a Configuration C1 or C2, then ω′(v)≥4−5×1−16+4×14+15=130. If v is not incident with a Configuration C1 or C2, then ω′(v)≥4−5×1+4×14+15=15.
Case 5.7. n2(v)=6. Let d(vi)=2, i=1,…,6. By Lemma 4, we have d(v′i)=Δ for i=1,…,6.
If v is incident with at least four 7+-faces, then ω′(v)≥4−6×1+min{4×25+2×14,4×25+14+2×16}=110 by R3-1, R3-6, and R3-8.
Suppose v is incident with three 7+-faces. If at least one of the 7+-faces is an 8+-face, then ω′(v)≥4−6×1+min{36+2×25+3×14,36+2×25+2×14+2×16}=120. Now, let us consider the case where v is exactly incident with three 7-faces. If at least one of the three 7-faces is Configuration D, then ω′(v)≥4−6×1+3×25+2×112+min{3×14,2×14+2×16}=760. Assume that none of the three 7-faces is Configuration D. If v is incident with a 5-face, then the 5-face must be Configuration C1. Hence, ω′(v)≥4−6×1+3×25+2×16+2×14=130. If v is not incident with a 5-face, then v is incident with either six Configurations B, or one Configuration A and four Configurations B, or two Configurations A and two Configurations B. In this case, ω′(v)≥4−6×1+3×25+3×14+4×16=3760.
Suppose v is incident with two 7+-faces. If both of the two 7+-faces are 8+-faces, then ω′(v)≥4−6×1+min{2×36+4×14,2×36+3×14+2×16}=0. If one of the two 7+-faces is an 8+-face and the other is Configuration D, then ω′(v)≥4−6×1+36+(25+2×112)+min{4×14,3×14+2×16}=115. If one of the two 7+-faces is an 8+-face and the other is not Configuration D, then the sum of Configurations A and Configurations B incident with v is at least two. So ω′(v)≥4−6×1+36+min{4×14+25+2×16,3×14+2×16+25+2×16}=730. If the two 7+-faces are exactly two 7-faces and at least one of them is Configuration D, then v is incident with one Configuration C1 or at least two Configurations A. So ω′(v)≥4−6×1+2×25+2×112+min{3×14+2×16,4×14+2×16}=120. If both of the two 7+-faces are exactly two 7-faces and neither of them is Configuration D, then the sum of Configurations A and Configurations B incident with v is at least two. So ω′(v)≥4−6×1+2×25+min{4×14+2×16,3×14+2×16+2×16}=215.
Suppose v is incident with one 7+-face. Then v is incident with one Configuration C1 and at least two Configurations A, or at least four Configurations A. So ω′(v)≥4−6×1+25+min{4×14+2×16+2×16,5×14+4×16}=115.
Suppose v is not incident with 7+-faces. Then v is incident with one Configuration C1 and at least four Configurations A, or at least six Configurations A. So ω′(v)≥4−6×1+min{5×14+2×16+4×16,6×14+6×16}=14.
We now check ω′(f)≥0 for each f∈F(G).
If f is a 5-face, then ω′(f)=d(f)−5=0 since no charge is discharged to or from f. If f is a 6+-face, according to R3-6, f gives away its positive charge, so ω′(f)=0.
We have verified that ω′(x)≥0 for all x∈V(G)∪F(G). This completes the proof when Δ=6.
Claim 14. Let f1=v′1v1vv2v′2x, f2=v′3v3vv2v′2y be two 6-faces. Suppose d(v)=6 and SG(v)<Δ+4. If d(v1)=d(v2)=d(v3)=2, then max{d(x),d(y)}≥3 (The configuration composed of f1 and f2 is called Configuration E of v. See H4 in Figure 6.)
Proof. By Lemma 4, we conclude that d(v′1)=d(v′2)=d(v′3)=Δ. The subsequent steps of the proof follow a similar approach as presented in Claim 4 of the reference paper [8] by Chen et al.
Claim 15. Let f1=v′1v1vv2v′2x be a 6-face, f2=v′3v3vv2v′2yz be a 7-face. Suppose d(v)=6 and SG(v)<Δ+4. If d(v1)=d(v2)=d(v3)=2 and d(z)≤5, then max{d(x),d(y)}≥3. (The configuration composed of f1 and f2 is called Configuration F of v. See H5 in Figure 6.)
Proof. The proof proceeds by contradiction. Let G′=G−vv1. According to Lemma 4, we can deduce that d(v′1)=d(v′2)=d(v′3)=Δ.
Suppose d(x)=2 and d(y)=2. Erase the colors on v,v1,v2,v3,x,y. Then we have |L′c(v)|≥(Δ+4)−SG(v)≥1, |L′c(v1)|≥(Δ+4)−(Δ+6−2−3)=3, |L′c(v2)|≥(Δ+4)−(Δ+6−2−4)=4, |L′c(v3)|≥(Δ+4)−(Δ+6−2−2)=2, |L′c(x)|≥(Δ+4)−(Δ+Δ−2−3)=9−Δ=2, |L′c(y)|≥(Δ+4)−(Δ+5−2−2)=3. Thus, we can sequentially recolor v,x,v3,v1,v2,y and obtain an injective L-coloring of G, which leads to a contradiction.
Claim 10 and Claim 13 remain valid in this portion.
We use the following discharging rules.
R4-1. Each 2-vertex receives 1 from each adjacent 3+-vertex.
R4-2. A bad 3-vertex receives 23 from each adjacent 7-vertex, 1730 from each adjacent 6-vertex, 715 from each adjacent 5-vertex.
R4-3. A good 3-vertex receives 13 from each adjacent 7-vertex, 16 from each adjacent 5-, 6-vertex.
R4-4. Suppose d(v)=3. If SG(v)<Δ+4, then v receives 118 from each adjacent 3-, 4-vertex.
R4-5. A 4(2)-vertex receives 1330 from each adjacent 7-vertex, 1360 from each adjacent 5-, 6-vertex.
R4-6. Each 6+-face equally distributes its positive charge to each incident 3+-vertex.
R4-7. In configuration E or F, v receives 112 along edge v′2x from v′2, and 112 along edge v′2y from v′2, for a total of 16 received from v′2. (See Figure 7)
R4-8. For a 5-face f=vv1xyv2, if d(v)=6, d(v1)=d(v2)=2, d(x)≥Δ−1 and d(y)≥Δ−1, we call it configuration G of v. In configuration G, v receives 16 along edge xv1 from x, and 16 along edge yv2 from y, for a total of 13 from x and y. (See Figure 7)
R4-9. For a 7-face f=yv′2v2vv3v′3z, if d(v)=6, d(v2)=d(v3)=2, d(v′2)=d(v′3)=Δ and d(z)≥6, we call it Configuration H of v. In Configuration H, v receives 112 along edge zy from z, and 112 along edge v′3v3 from v′3, for a total of 16 from z and v′3. (See Figure 7)
First, we check ω′(v)≥0 for each vertex v∈V(G).
Case 1. d(v)=2. We have ω(v)=32×2−5=−2. By R4-1, ω′(v)=−2+1×2=0.
Case 2. d(v)=3. We have ω(v)=32×3−5=−12. Let N(v)={v1,v2,v3}. By Lemma 6(1), n2(v)≤1. If n2(v)=1, let d(v1)=2, then by Lemma 6(2), we have d(v2)+d(v3)≥Δ+5. So ω′(v)≥−12−1+min{715+23,2×1730}+16+15=0 by R4-1, R4-2, and R4-6. If n2(v)=0 and n5+(v)=0, then SG(v)<Δ+4. By R4-4 and R4-6, ω′(v)≥−12+3×118+2×16=0. Suppose n2(v)=0 and n5+(v)≥1. If v is not adjacent to a 3-vertex u with SG(u)<Δ+4, then ω′(v)≥−12+16+2×16=0. If v is adjacent to a 3-vertex u with SG(u)<Δ+4, then by Lemma 4, the sum of degrees of the other two neighbors of v is at least Δ+4. So ω′(v)≥−12−118+min{13,2×16}+2×16=19.
Case 3. d(v)=4. We have ω(v)=32×4−5=1. Let N(v)={v1,v2,v3,v4}. By Lemma 6(1), n2(v)≤2. If n2(v)=2, let d(v1)=d(v2)=2, then by Lemma 6(2), we have d(v3)+d(v4)≥Δ+4. So ω′(v)≥1−2×1+min{1330,2×1360}+16+2×15=0 by R4-1, R4-5, and R4-6. If n2(v)≤1, then ω′(v)≥1−1−3×118+3×16=13.
Case 4. d(v)=5. We have ω(v)=32×5−5=52. Let N(v)={v1,v2,v3,v4,v5}. By Lemma 6, n2(v)≤3. If n2(v)=3, then by Claim 10, v is not adjacent to a bad 3-vertex or a 4(2)-vertex. Additionally, by Lemma 4, v is adjacent to at most one 3-vertex. Thus, ω′(v)≥52−3×1−16+4×16=0. If n2(v)=2, then by Lemma 4, we know that n3(v)≤2. Therefore, ω′(v)≥52−2×1−2×715−1360+4×16=160 by our rules. If n2(v)≤1, then ω′(v)≥52−1−4×715+4×16=310.
Case 5. d(v)=6. We have ω(v)=32×6−5=4. Let N(v)={v1,v2,v3,v4,v5,v6}. We consider four cases.
Case 5.1. n2(v)≤3. In this case, ω′(v)≥4−3×1−3×1730+5×16=215.
Case 5.2. n2(v)=4. Let d(v1)=d(v2)=d(v3)=d(v4)=2.
If at most one of v5 and v6 is a bad 3-vertex, then ω′(v)≥4−4×1−1730−1360+5×16=120.
Suppose both v5 and v6 are bad 3-vertices. By Lemma 4, we know that d(v′i)=Δ for i=1,…,4. If v is not incident with a 5-face, then ω′(v)≥4−4×1−2×1730+min{5×14+16,4×14+2×15}=415. If v is incident with a Configuration G, then ω′(v)≥4−4×1−2×1730+2×16+min{4×14+16,3×14+2×15}=2360. If v is incident with a 5-face but not Configuration G, then ω′(v)≥4−4×1−2×1730+min{5×14,4×14+16,3×14+2×15,4×14+15}=160.
Case 5.3. n2(v)=5, let d(v1)=d(v2)=d(v3)=d(v4)=d(v5)=2.
Subcase 5.3.1. v6 is not a 3-vertex or a 4(2)-vertex. Then v may incident with a Configuration G that requires charges from v. In this case, we have ω′(v)≥4−5×1−16+4×14+15=130.
Subcase 5.3.2. v6 is a bad 3-vertex. Let N(v6)={v,x,y} and d(x)=2. By Lemma 4, we have d(v′i)=Δ for i=1,…,5, and d(x)+d(y)≥Δ+1=8, so d(y)≥6.
Suppose v is incident with a Configuration G. If v is incident with at least one 7+-face, then ω′(v)≥4−5×1−1730+2×16+min{4×14+26,3×14+15+25}=110. If v is not incident with any 7+-faces, then v is incident with at least one Configuration E. So ω′(v)≥4−5×1−1730+2×16+2×112+4×14+15=215.
Suppose v is incident with a 5-face but not Configuration G. Then the 5-face must be f5 by Claim 13 (see Figure 8). If v is incident with at least two 7+-face and at least one of them is an 8+-face, then ω′(v)≥4−5×1−1730+3×14+25+36=112. So, assume v is incident with at least two 7-faces. If at least one of them is Configuration H, then ω′(v)≥4−5×1−1730+3×14+2×25+2×112=320. If neither of the two 7-faces is Configuration H, then v is incident with at least two Configurations E or F. So ω′(v)≥4−5×1−1730+3×14+2×25+2×16=1960. If v is incident with one 7+-face, then v is incident with at least one Configuration E. So ω′(v)≥4−5×1−1730+4×14+25+2×112=0. If v is not incident with any 7+-faces, then v is incident with at least three Configurations E. So ω′(v)≥4−5×1−1730+5×14+3×16=1160.
Suppose v is not incident with any 5-faces. If v is incident with at least one 7+-face, then ω′(v)≥4−5×1−1730+min{26+5×14,25+4×14+15}=160. If v is not incident with any 7+-faces, then v is incident with at least three Configurations E. So ω′(v)≥4−5×1−1730+5×14+15+3×16=2360.
Subcase 5.3.3. v6 is a good 3-vertex. By Lemma 4, d(v′i)=Δ for i=1,…,5.
If v is not incident with a 5-face, then ω′(v)≥4−5×1−16+4×14+2×15=730.
If v is incident with a Configuration G, then ω′(v)≥4−5×1−16+2×15+3×14+2×16=1960.
If v is incident with a 5-face but not Configuration G, then ω′(v)≥4−5×1−16+4×14+15=130.
Subcase 5.3.4. v6 is a 4(2)-vertex. Let N(v6)={v,x,y,z} and d(x)=d(y)=2. By Lemma 4, d(v′i)=Δ for i=1,…,5, and d(z)≥5.
If v is not incident with a 5-face, then ω′(v)≥4−5×1−1360+5×14+15=730.
If v is incident with a Configuration G, then ω′(v)≥4−5×1−1360+4×14+15+2×16=1960.
If v is incident with a 5-face but not Configuration G, then if v is incident with at least one 7+-face, ω′(v)≥4−5×1−1360+min{3×14+15+25,4×14+26,4×14+25}=760; if v is not incident with any 7+-faces, then v is incident with at least three Configurations E. So ω′(v)≥4−5×1−1360+4×14+15+3×16=2960.
Case 5.4. n2(v)=6. Let d(vi)=2, i=1,…,6. According to Lemma 4, we have d(v′i)=Δ for i=1,…,6. Therefore, if v is incident with a 5-face, it must be Configuration G.
If v is incident with at least four 7+-faces, then ω′(v)≥4−6×1+min{4×25+2×14,4×25+14+2×16}=110.
Suppose v is incident with three 7+-faces. If at least one of the 7+-faces is an 8+-face, then ω′(v)≥4−6×1+min{36+2×25+3×14,36+2×25+2×14+2×16}=120. So we consider the case where v is exactly incident with three 7-faces. If at least one of the three 7-faces is Configuration H, then ω′(v)≥4−6×1+3×25+2×112+min{3×14,2×14+2×16}=760. Now let us assume that none of the three 7-faces is Configuration H. If v is incident with a 5-face, then ω′(v)≥4−6×1+3×25+2×16+2×14=130. If v is not incident with a 5-face, then v is incident with either six Configurations F, or one Configuration E and four Configurations F, or two Configurations E and two Configurations F. So ω′(v)≥4−6×1+3×25+3×14+4×16=3760.
Suppose v is incident with two 7+-faces. If both of the two 7+-faces are 8+-faces, then ω′(v)≥4−6×1+min{2×36+4×14,2×36+3×14+2×16}=0. If one of the two 7+-faces is an 8+-face, and the other is Configuration H, then ω′(v)≥4−6×1+36+(25+2×112)+min{4×14,3×14+2×16}=115. If one of the two 7+-faces is an 8+-face, and the other is not Configuration H, then v is incident with at least two Configurations F, or one Configuration F and one Configuration E, or two Configurations E. Therefore, ω′(v)≥4−6×1+36+min{4×14+25+2×16,3×14+2×16+25+2×16}=730. If the two 7+-faces are both 7-faces and at least one of them is Configuration H, then v must be incident with one Configuration G or at least two Configurations E. Thus, ω′(v)≥4−6×1+2×25+2×112+min{3×14+2×16,4×14+2×16}=120. If the two 7+-faces are both 7-faces and neither of them is Configuration H, then v must be incident with at least two Configurations E or two Configurations F. Hence, ω′(v)≥4−6×1+2×25+min{4×14+2×16,3×14+2×16+2×16}=215.
Suppose v is incident with one 7+-face. Then v is incident with one Configuration G and at least two Configurations E, or at least four Configurations E. Hence, ω′(v)≥4−6×1+25+min{4×14+2×16+2×16,5×14+4×16}=115.
Suppose v is not incident with any 7+-faces. Then v is incident with one Configuration G and at least four Configurations E, or at least six Configurations E. Therefore, ω′(v)≥4−6×1+min{5×14+2×16+4×16,6×14+6×16}=14.
Case 6. d(v)=7. We have ω(v)=32×7−5=112. Let N(v)={v1,v2,v3,v4,v5,v6,v7}.
Case 6.1. n2(v)≤3. There are at most n2(v) Configurations E or F that require charges from v. Therefore, we have ω′(v)≥112−n2(v)×1−(d(v)−n2(v))×23−n2(v)×16+6×16≥13.
Case 6.2. n2(v)=4. There are at most four Configurations E or F that require charges from v. Therefore, we have ω′(v)≥112−4×1−3×23−4×16+min{2×15+2×14+2×16,6×15,4×15+14+16}=130.
Case 6.3. n2(v)=5. Let d(vi)=2 for i=1,…,5. We consider three subcases.
Subcase 6.3.1. Both v6 and v7 are bad 3-vertices. Let N(v6)={v,x1,y1} and N(v7)={v,x2,y2} where d(x1)=d(x2)=2. By Lemma 4, we have d(v′i)≥Δ−1 for i=1,…,5, and d(y1)≥5, d(y2)≥5.
If v is incident with a Configuration G, without loss of generality, let us say it is f2. Then v is incident with at most three Configurations E (or F) to which v needs to send charges (see Figure 9. In (1), f7 and f1, f4 and f5 may form two Configurations E or F; In (2), f7 and f1, f3 and f4, and f5 and f6 may form three Configurations E or F; In (3), f7 and f1, f4 and f5, and f5 and f6 may form three Configurations E or F). So we have ω′(v)≥112−5×1−2×23−3×16+2×16+6×16=0.
If v is incident with a 5-face, but not Configuration G, without loss of generality, let us say it is f6. Then v is incident with at most three Configurations E (or F) to which v needs to send charges (see Figure 9. In (1), f7 and f1, f4 and f5 may form two Configurations E or F; in (2), f7 and f1, f3 and f4 may form two Configurations E or F; in (3), f7 and f1, f2 and f3, f4 and f5 may form three Configurations E or F). So, we have ω′(v)≥112−5×1−2×23−3×16+min{4×14+2×15,3×14+3×15}=160.
If v is not incident with any 5-faces, then v is incident with at most four Configurations E (or F) to which v needs to send charges (see Figure 9. In (1), f7 and f1, f4 and f5 may form two Configurations E or F; in (2), f7 and f1, f3 and f4, f5 and f6 may form three Configurations E or F; in (3), f7 and f1, f2 and f3, f4 and f5, f5 and f6 may form four Configurations E or F). So, we have ω′(v)≥112−5×1−2×23−4×16+4×14+3×16=110.
Subcase 6.3.2. One of v6 and v7 is a bad 3-vertex. If v is incident with a 5-face, as discussed in Subcase 6.3.1, v is incident with at most three Configurations E (or F) that require charges from v (see Figure 9). Therefore, ω′(v)≥112−5×1−23−max{1330,13}−3×16+min{4×14+2×15,3×14+3×15}=14. If v is not incident with any 5-faces, as discussed in Subcase 6.3.1, v is incident with at most four Configurations E (or F) that require charges from v (see Figure 9). Hence, ω′(v)≥112−5×1−23−max{1330,13}−4×16+min{4×14+2×15+16,3×14+4×15}=1760.
Subcase 6.3.3. Neither v6 nor v7 is a bad 3-vertex. If v is incident with a 5-face, then v is incident with at most three Configurations E (or F) that require charges from v (see Figure 9). So ω′(v)≥112−5×1−max{2×1330,1330+13,2×13}−3×16+6×16=215. If v is not incident with any 5-faces, then v is incident with at most four Configurations E (or F) that require charges from v (see Figure 9). So ω′(v)≥112−5×1−max{2×1330,1330+13,2×13}−4×16+3×14+4×16=2360.
Case 6.4. n2(v)=6. Let d(vi)=2, i=1,…,6.
If d(v7)≥5 and v is incident with a Configuration G that requires charges from v, then v is incident with at most one Configuration H, or one Configuration E (or F) that requires charges from v. Therefore, ω′(v)≥112−6×1−16−max{112,2×112}+5×14+15=3760.
If d(v7)≥5 and v is not incident with a Configuration G, then v is incident with at most two Configurations H, or two Configurations E (or F) that require charges from v. Therefore, ω′(v)≥112−6×1−max{2×112,2×16}+5×14+15=4960.
Suppose v7 is a bad 3-vertex or a 4(2)-vertex. If v is incident with a 5-face but not Configuration G, then v is incident with at most one Configuration E (or F) that requires charges from v. Therefore, we have ω′(v)≥112−6×1−max{23,1330}−16+5×14+15=760. If v is not incident with a 5-face, then v is incident with at most two Configurations E (or F) that require charges from v. Therefore, we have ω′(v)≥112−6×1−max{23,1330}−2×16+5×14+2×15=320.
Suppose v7 is a good 3-vertex or a 4-vertex (not a 4(2)-vertex). Then v is incident with at most two Configurations E (or F) that require charges from v. Therefore, we have ω′(v)≥112−6×1−13−2×16+4×14+2×15=730.
Case 6.5. n2(v)=7. Then ω′(v)≥112−7×1+6×14=0.
We now check ω′(f)≥0 for each f∈F(G).
If f is a 5-face, then ω′(f)=d(f)−5=0 since no charge is discharged to or from f. If f is a 6+-face, according to R4-6, f gives away its positive charge, so ω′(f)=0.
We have verified that ω′(x)≥0 for all x∈V(G)∪F(G). This completes the proof when Δ=7.
If p=uwv is a path in G and d(w)=2, we say that u is pseudo-adjacent to v if uv∉E(G).
We use the following discharging rules.
R5-1. Each 2-vertex receives 1 from each adjacent 3+-vertex.
R5-2. Let v be a bad 3-vertex, uv∈E(G). If d(u)≥8, v receives 1 from u; if d(u)=7, v receives 23 from u; if d(u)=6, v receives 715 from u; if d(u)=5, v receives 215 from u.
R5-3. Let v be a good 3-vertex, uv∈E(G). If d(u)=Δ, v receives 29 from u; if 5≤d(u)≤Δ−1, v receives 16 from u.
R5-4. Let d(v)=3, uv∈E(G). If SG(v)<Δ+4 and 3≤d(u)≤4, then v receives 118 from u.
R5-5. Let v be a 4(2)-vertex, uv∈E(G). If d(u)≥7, v receives 1330 from u; if d(u)=6, v receives 1360 from u.
R5-6. Each 6+-face equally distributes its positive charge to each of its incident 3+-vertices.
R5-7. Each 6-vertex receives 112 from each of its pseudo-adjacent Δ-vertices.
R5-8. Let f=vv1xyv2 be a 5-face. If d(v)=6, d(v1)=d(v2)=2, d(x)=d(y)=Δ, f is called configuration I of v. In configuration I, v receives 18 along edge xv1 from x, and 18 along edge yv2 from y, for a total of 14 from x and y.
First, we check ω′(v)≥0 for each vertex v∈V(G).
Case 1. d(v)=2, ω(v)=32×2−5=−2. By R5-1, ω′(v)=−2+1×2=0.
Case 2. d(v)=3, ω(v)=32×3−5=−12. Let N(v)={v1,v2,v3}. By Lemma 6(1), we know that n2(v)≤1. If n2(v)=1, let d(v1)=2. Then, by Lemma 6(2), we have d(v2)+d(v3)≥Δ+5. Therefore, ω′(v)≥−12−1+min{1+215,23+715}+16+15=0 by R5-1, R5-2, and R5-6. If n2(v)=0 and n5+(v)=0, then SG(v)<Δ+4. By applying R5-4 and R5-6, we can conclude that ω′(v)≥−12+3×118+2×16=0. Suppose n2(v)=0 and n5+(v)≥1. If v is not incident with a 3-vertex u with SG(u)<Δ+4, then we have ω′(v)≥−12+16+2×16=0. If v is incident with a 3-vertex u with SG(u)<Δ+4, then by Lemma 4, the sum of the degrees of the other two neighbors of v is at least Δ+4. Therefore, ω′(v)≥−12−118+min{29,2×16}+2×16=0.
Case 3. d(v)=4, ω(v)=32×4−5=1. Let N(v)={v1,v2,v3,v4}. By Lemma 6(1), we know that n2(v)≤2. If n2(v)=2, let d(v1)=d(v2)=2. Then, by Lemma 6(2), we have d(v3)+d(v4)≥Δ+4. Therefore, by using R5-1, R5-5, and R5-6, we can conclude that ω′(v)≥1−2×1+min{1330,2×1360}+16+2×15=0. If n2(v)≤1, then ω′(v)≥1−1−3×118+3×16=13.
Case 4. d(v)=5, ω(v)=32×5−5=52. Let N(v)={v1,v2,v3,v4,v5}. According to Lemma 6, we know that n2(v)≤3. If n2(v)=3, using Lemma 4, it follows that v is adjacent to at most one 3-vertex. Therefore, ω′(v)≥52−3×1−max{215,16}+4×16=0. If n2(v)=2, then by Lemma 4, we have n3(v)≤2. So ω′(v)≥52−2×1−max{2×16,2×215,16+215}+4×16=56. If n2(v)≤1, then ω′(v)≥52−1−4×16+4×16=32.
Case 5. d(v)=6, ω(v)=32×6−5=4. Let N(v)={v1,v2,v3,v4,v5,v6}. We consider four cases.
Case 5.1. n2(v)≤3. In this case, ω′(v)≥4−3×1−3×715+5×16=1330.
Case 5.2. n2(v)=4. We then have ω′(v)≥4−4×1−2×715+min{14+4×15,2×14+2×15+16}=760.
Case 5.3. n2(v)=5. Let d(v1)=d(v2)=d(v3)=d(v4)=d(v5)=2. If d(v6)≥5, then ω′(v)≥4−5×1+3×14+2×15=320. If d(v6)≤4, then by Lemma 4, we have d(v′i)=Δ, i=1,…,5. Therefore, ω′(v)≥4−5×1−715+3×14+2×15+5×112=110 by R5-1, R5-2, R5-3, R5-5, R5-6, and R5-7.
Case 5.4. n2(v)=6. Let d(vi)=2, i=1,…,6. By Lemma 4, we have d(v′i)=Δ, i=1,…,6. If v is incident with a 5-face, it must be Configuration I. Therefore, ω′(v)≥4−6×1+5×14+14+6×112=0 by R5-1, R5-6, R5-7 and R5-8.
Case 6. d(v)=7, ω(v)=32×7−5=112.
If n2(v)≤5, then ω′(v)≥112−5×1−2×23+6×16=16.
If n2(v)=6, then ω′(v)≥112−6×1−23+4×14+2×15=730.
If n2(v)=7, then ω′(v)≥112−7×1+6×14=0.
Case 7. d(v)=8, ω(v)=32×8−5=7.
Note that the worst case is that v is adjacent to t 2-vertices and (8−t) bad 3-vertices.
If t=8, then ω′(v)≥7−8×1−8×112+7×14=112 by R5-1, R5-6 and R5-7.
If t=7, then ω′(v)≥7−8×1−7×112+5×14+2×15=115 by R5-1, R5-2, R5-3, R5-5, R5-6, and R5-7.
If t=6, then ω′(v)≥7−8×1−6×112+min{4×14+2×15+16,3×14+4×15}=115.
If t=5, then ω′(v)≥7−8×1−5×112+min{3×14+2×15+2×16,2×14+4×15+16,14+6×15}=130.
If t=4, then ω′(v)≥7−8×1−4×112+min{7×15,2×14+2×15+3×16,14+4×15+2×16,16+6×15}=710.
If t=3, then ω′(v)≥7−8×1−3×112+min{14+2×15+4×16,4×15+3×16,2×16+5×15}=120.
If t≤2, then ω′(v)≥7−8×1−2×112+7×16=0.
Case 8. d(v)≥9.
If n2(v)=d(v), then ω′(v)≥32d(v)−5−d(v)×1−112n2(v)+14(d(v)−1)=23d(v)−214>0.
If n2(v)=d(v)−1, then ω′(v)≥32d(v)−5−d(v)×1−112n2(v)+14(d(v)−3)+2×15=23d(v)−7915>0.
If n2(v)=d(v)−2, then ω′(v)≥32d(v)−5−d(v)×1−112n2(v)+min{14(d(v)−4)+2×15+16,14(d(v)−5)+4×15}=23d(v)−7915>0.
If n2(v)≤d(v)−3, then ω′(v)≥32d(v)−5−d(v)×1−112n2(v)+16(d(v)−1)=712d(v)−5912>0.
We now check ω′(f)≥0 for each f∈F(G).
If f is a 5-face, then ω′(f)=d(f)−5=0 since no charge is discharged to or from f. If f is a 6+-face, according to R5-6, f gives away its positive charge, so ω′(f)=0.
We have checked ω′(x)≥0 for all x∈V(G)∪F(G). This completes the proof when Δ≥8, and hence the proof of the whole Theorem 2.
In this paper, we consider the list injective chromatic index of planar graphs without intersecting 5-cycles and proved that such graphs have χli(G)≤Δ+4 if g(G)≥5. Based on the result of Theorem 2, the following question is meaningful, namely: for a planar graph G with g(G)≥5, explore the upper bound of χli(G) when G has no adjacent 5-cycles.
The authors declare that no Artificial Intelligence was used.
Hongyu Chen devised the project, the main ideas, proof outline, and wrote the manuscript. Li Zhang verified the results and polished the paper.
This work was supported by National Natural Science Foundations of China (Grant No.11401386)
The authors declare no conflicts of interest.
[1] |
A. Neirameh, M. Eslami, New solitary wave solutions for fractional Jaulent–Miodek hierarchy equation, Mod. Phys. Lett. B, 36 (2022), 2150612. https://doi.org/10.1142/S0217984921506120 doi: 10.1142/S0217984921506120
![]() |
[2] |
H. Rezazadeh, M. S. Osman, M. Eslami, M. Ekici, A. Sonmezoglu, M. Asma, et al., Mitigating internet bottleneck with fractional temporal evolution of optical solitons having quadratic–cubic nonlinearity, Optik, 164 (2018), 84–92. https://doi.org/10.1016/j.ijleo.2018.03.006 doi: 10.1016/j.ijleo.2018.03.006
![]() |
[3] |
M. S. Osman, One-soliton shaping and inelastic collision between double solitons in the fifth-order variable-coefficient Sawada–Kotera equation, Nonlinear Dyn., 96 (2019), 1491–1496. https://doi.org/10.1007/s11071-019-04866-1 doi: 10.1007/s11071-019-04866-1
![]() |
[4] |
F. Badshah, K. U. Tariq, M. Inc, F. Mehboob, On lump, travelling wave solutions and the stability analysis for the (3+1)-dimensional nonlinear fractional generalized shallow water wave model in fluids, Opt. Quant. Electron., 56 (2024), 244. https://doi.org/10.1007/s11082-023-05826-1 doi: 10.1007/s11082-023-05826-1
![]() |
[5] |
K. K. Ali, A. Wazwaz, M. S. Osman, Optical soliton solutions to the generalized nonautonomous nonlinear Schrödinger equations in optical fibers via the sine-Gordon expansion method, Optik, 208 (2020), 164132. https://doi.org/10.1016/j.ijleo.2019.164132 doi: 10.1016/j.ijleo.2019.164132
![]() |
[6] |
M. S. Aktar, M. A. Akbar, M. S. Osman, Spatio-temporal dynamic solitary wave solutions and diffusion effects to the nonlinear diffusive predator-prey system and the diffusion-reaction equations, Chaos Soliton. Fract., 160 (2022), 112212. https://doi.org/10.1016/j.chaos.2022.112212 doi: 10.1016/j.chaos.2022.112212
![]() |
[7] |
B. Inan, M. S. Osman, A. k. Turgut, D. Baleanu, Analytical and numerical solutions of mathematical biology models: The Newell-Whitehead-Segel and Allen-Cahn equations, Math. method. Appl. Sci., 43 (2020), 2588–2600. https://doi.org/10.1002/mma.6067 doi: 10.1002/mma.6067
![]() |
[8] |
M. Adel, D. Baleanu, U. Sadiya, M. A. Arefin, M. H. Uddin, M. A. Elamin, et al., Inelastic soliton wave solutions with different geometrical structures to fractional order nonlinear evolution equations, Results in Physics, 38 (2022), 105661. https://doi.org/10.1016/j.rinp.2022.105661 doi: 10.1016/j.rinp.2022.105661
![]() |
[9] |
S. Kumar, S. K. Dhiman, D. Baleanu, M. S. Osman, A. Wazwaz, Lie symmetries, closed-form solutions, and various dynamical profiles of solitons for the variable coefficient (2+1)-dimensional KP equations, Symmetry, 14 (2022), 597. https://doi.org/10.3390/sym14030597 doi: 10.3390/sym14030597
![]() |
[10] |
F. Badshah, K. U. Tariq, A. Bekir, R. N. Tufail, H. Ilyas, Lump, periodic, travelling, semi-analytical solutions and stability analysis for the Ito integro-differential equation arising in shallow water waves, Chaos Soliton. Fract., 182 (2024), 114783. https://doi.org/10.1016/j.chaos.2024.114783 doi: 10.1016/j.chaos.2024.114783
![]() |
[11] |
K. K. Ali, R. Yilmazer, M. S Osman, Dynamic behavior of the (3+1)-dimensional KdV–Calogero–Bogoyavlenskii–Schiff equation, Opt. Quant. Electron., 54 (2022), 160. https://doi.org/10.1007/s11082-022-03528-8 doi: 10.1007/s11082-022-03528-8
![]() |
[12] |
S. Tarla, K. K. Ali, R. Yilmazer, M. S. Osman, The dynamic behaviors of the Radhakrishnan–Kundu–Lakshmanan equation by Jacobi elliptic function expansion technique, Opt. Quant. Electron., 54 (2022), 292. https://doi.org/10.1007/s11082-022-03710-y doi: 10.1007/s11082-022-03710-y
![]() |
[13] |
S. Rashid, K. T. Kubra, S. Sultana, P. Agarwal, M. S. Osman, An approximate analytical view of physical and biological models in the setting of Caputo operator via Elzaki transform decomposition method, J. Comput. Appl. Math., 413 (2022), 114378. https://doi.org/10.1016/j.cam.2022.114378 doi: 10.1016/j.cam.2022.114378
![]() |
[14] | H. F. Ismael, I. Okumuş, T. Aktürk, H. Bulut, M. S. Osman, Analyzing study for the 3D potential Yu–Toda–Sasa–Fukuyama equation in the two-layer liquid medium, J. Ocean Eng. Sci., 2022. In press. https://doi.org/10.1016/j.joes.2022.03.017 |
[15] |
A. R. Seadawy, N. Cheemaa, A. Biswas, Optical dromions and domain walls in (2+1)-dimensional coupled system, Optik, 227 (2021), 165669. https://doi.org/10.1016/j.ijleo.2020.165669 doi: 10.1016/j.ijleo.2020.165669
![]() |
[16] |
S. El-Ganaini, M. O. Al-Amr, New abundant solitary wave structures for a variety of some nonlinear models of surface wave propagation with their geometric interpretations, Math. Method. Appl. Sci., 45 (2022), 7200–7226. https://doi.org/10.1002/mma.8232 doi: 10.1002/mma.8232
![]() |
[17] |
M. O. Al-Amr, H. Rezazadeh, K. K. Ali, A. Korkmaz, N1-soliton solution for Schrödinger equation with competing weakly nonlocal and parabolic law nonlinearities, Commun. Theor. Phys., 72 (2020), 065503. https://doi.org/10.1088/1572-9494/ab8a12 doi: 10.1088/1572-9494/ab8a12
![]() |
[18] | M. N. Rasheed, M. O. Al-Amr, E. A. Az-Zo'bi, M. A. Tashtoush, L. Akinyemi, Stable optical solitons for the Higher-order Non-Kerr NLSE via the modified simple equation method, Mathematics, 9 (2021) 1986. https://doi.org/10.3390/math9161986 |
[19] |
M. Eslami, H. Rezazadeh, The first integral method for Wu–Zhang system with conformable time-fractional derivative, Calcolo, 53 (2016), 475–485. https://doi.org/10.1007/s10092-015-0158-8 doi: 10.1007/s10092-015-0158-8
![]() |
[20] |
H. Rezazadeh, D. Kumar, A. Neirameh, M. Eslami, M. Mirzazadeh, Applications of three methods for obtaining optical soliton solutions for the Lakshmanan–Porsezian–Daniel model with Kerr law nonlinearity, Pramana, 94 (2020), 39. https://doi.org/10.1007/s12043-019-1881-5 doi: 10.1007/s12043-019-1881-5
![]() |
[21] |
A. Zafar, M. Raheel, M. Mirzazadeh, M. Eslami, Different soliton solutions to the modified equal-width wave equation with Beta-time fractional derivative via two different methods, Rev. Mex. Fis., 68 (2022), 010701. https://doi.org/10.31349/revmexfis.68.010701 doi: 10.31349/revmexfis.68.010701
![]() |
[22] |
S. Sahoo, S. S. Ray, M. A. Abdou, New exact solutions for time-fractional Kaup-Kupershmidt equation using improved (G′/G)−expansion and extended (G′/G)−expansion methods, Alex. Eng. J., 59 (2020), 3105–3110. https://doi.org/10.1016/j.aej.2020.06.043 doi: 10.1016/j.aej.2020.06.043
![]() |
[23] |
Z. Zou, R. Guo, The Riemann–Hilbert approach for the higher-order Gerdjikov–Ivanov equation, soliton interactions and position shift, Commun. Nonlinear Sci., 124 (2023), 107316. https://doi.org/10.1016/j.cnsns.2023.107316 doi: 10.1016/j.cnsns.2023.107316
![]() |
[24] |
N. Nasreen, U. Younas, T. A. Sulaiman, Z. Zhang, D. Lu, A variety of M-truncated optical solitons to a nonlinear extended classical dynamical model, Results Phys., 51 (2023), 106722. https://doi.org/10.1016/j.rinp.2023.106722 doi: 10.1016/j.rinp.2023.106722
![]() |
[25] |
F. Badshah, K. U. Tariq, A. Bekir, S. M. R. Kazmi, Stability, modulation instability and wave solutions of time-fractional perturbed nonlinear Schrödinger model, Opt. Quant. Electron., 56 (2024), 425. https://doi.org/10.1007/s11082-023-06058-z doi: 10.1007/s11082-023-06058-z
![]() |
[26] |
O. M. Braun, Y. S. Kivshar, Nonlinear dynamics of the Frenkel–Kontorova model, Phys. Rep., 306 (1998), 1–108. https://doi.org/10.1016/S0370-1573(98)00029-5 doi: 10.1016/S0370-1573(98)00029-5
![]() |
[27] |
S. Shen, Z. J. Yang, Z. G. Pang, Y. R. Ge, The complex-valued astigmatic cosine-Gaussian soliton solution of the nonlocal nonlinear Schrödinger equation and its transmission characteristics, Appl. Math. Lett., 125 (2022), 107755. https://doi.org/10.1016/j.aml.2021.107755 doi: 10.1016/j.aml.2021.107755
![]() |
[28] |
L. M. Song, Z. J. Yang, X. L. Li, S. M. Zhang, Coherent superposition propagation of Laguerre–Gaussian and Hermite–Gaussian solitons, Appl. Math. Lett., 102 (2020), 106114. https://doi.org/10.1016/j.aml.2019.106114 doi: 10.1016/j.aml.2019.106114
![]() |
[29] |
S. Shen, Z. J. Yang, X. L. Li, S. Zhang, Periodic propagation of complex-valued hyperbolic-cosine-Gaussian solitons and breathers with complicated light field structure in strongly nonlocal nonlinear media, Commun. Nonlinear Sci., 103 (2021), 106005. https://doi.org/10.1016/j.cnsns.2021.106005 doi: 10.1016/j.cnsns.2021.106005
![]() |
[30] |
Z. Y. Sun, D. Deng, Z. G. Pang, Z. J. Yang, Nonlinear transmission dynamics of mutual transformation between array modes and hollow modes in elliptical sine-Gaussian cross-phase beams, Chaos Soliton. Fract., 178 (2024), 114398. https://doi.org/10.1016/j.chaos.2023.114398 doi: 10.1016/j.chaos.2023.114398
![]() |
[31] |
Z. Y. Sun, J. Li, R. Bian, D. Deng, Z. J. Yang, Transmission mode transformation of rotating controllable beams induced by the cross phase, Opt. Express, 32 (2024), 9201–9212. https://doi.org/10.1364/OE.520342 doi: 10.1364/OE.520342
![]() |
[32] |
I. M. Batiha, S. A. Njadat, R. M. Batyha, A. Zraiqat, A. Dababneh, Sh. Momani, Design fractional-order PID controllers for single-joint robot arm model, Int. J. Advance Soft Compu. Appl., 14 (2022), 96–114. 10.15849/IJASCA.220720.07 doi: 10.15849/IJASCA.220720.07
![]() |
[33] | H. Qawaqneh, A. Zafar, M. Raheel, A. A. Zaagan, E. H. M. Zahran, A. Cevikel, et al., New soliton solutions of M-fractional Westervelt model in ultrasound imaging via two analytical techniques, Opt. Quant. Electron. 56 (2024), 737. https://doi.org/10.1007/s11082-024-06371-1 |
[34] |
A. Zafar, K. K. Ali, M. Raheel, K. S. Nisar, A. Bekir, Abundant M-fractional optical solitons to the pertubed Gerdjikov–Ivanov equation treating the mathematical nonlinear optics, Opt. Quant. Electron., 54 (2022), 25. https://doi.org/10.1007/s11082-021-03394-w doi: 10.1007/s11082-021-03394-w
![]() |
[35] |
A. Zafar, A. Bekir, M. Raheel, H. Rezazadeh, Investigation for optical soliton solutions of two nonlinear Schrödinger equations via two concrete finite series methods, Int. J. Appl. Comput. Math., 6 (2020), 65. https://doi.org/10.1007/s40819-020-00818-1 doi: 10.1007/s40819-020-00818-1
![]() |
[36] |
N. Ullah, M. I. Asjad, J. Awrejcewicz, T. Muhammad, D. Baleanu, On soliton solutions of fractional-order nonlinear model appears in physical sciences, AIMS Mathematics, 7 (2022), 7421–7440. https://doi.org/10.3934/math.2022415 doi: 10.3934/math.2022415
![]() |
[37] | N. Taghizadeh, S. R. M. Noori, S. B. M. Noori, Application of the extended (G′/G)-expansion method to the improved Eckhaus equation, Appl. Appl. Math., 9 (2014), 24. |
[38] |
M. Ekici, Soliton and other solutions of nonlinear time fractional parabolic equations using extended (G′/G)-expansion method, Optik, 130 (2017), 1312–1319. https://doi.org/10.1016/j.ijleo.2016.11.104 doi: 10.1016/j.ijleo.2016.11.104
![]() |
[39] |
W. Wu, X. Ma, B. Zeng, H. Zhang, P. Zhang, A novel multivariate grey system model with conformable fractional derivative and its applications, Comput. Ind. Eng., 164 (2022), 107888. https://doi.org/10.1016/j.cie.2021.107888 doi: 10.1016/j.cie.2021.107888
![]() |
[40] |
R. Khalil, M. A. Horani, A. Yousef, M. Sababheh, A new defination of fractional derivative, J. Comput. Appl. Math., 264 (2014), 65–70. https://doi.org/10.1016/j.cam.2014.01.002 doi: 10.1016/j.cam.2014.01.002
![]() |
[41] |
K. S. Nisar, A. Ciancio, K. K. Ali, M. S. Osman, C. Cattani, D. Baleanu, et al., On beta-time fractional biological population model with abundant solitary wave structures, Alex. Eng. J., 61 (2022), 1996–2008. https://doi.org/10.1016/j.aej.2021.06.106 doi: 10.1016/j.aej.2021.06.106
![]() |
[42] | N. Al-Salti, E. Karimov, K. Sadarangani, On a differential equation with Caputo-Fabrizio fractional derivative of order 1≤ beta ≤2 and application to mass-spring-damper system, 2016, arXiv: 1605.07381. https://doi.org/10.48550/arXiv.1605.07381 |
[43] |
A. S. T. Tagne, J. M. E. Ema'a, G. H. Ben-Bolie, D. Buske, A new truncated M-fractional derivative for air pollutant dispersion, Indian J. Phys., 94 (2020), 1777–1784. https://doi.org/10.1007/s12648-019-01619-z doi: 10.1007/s12648-019-01619-z
![]() |
[44] |
A. Zafar, A. Bekir, M. Raheel, W. Razzaq, Optical soliton solutions to Biswas-Arshed model with truncated M-fractional derivative, Optik, 222 (2020), 165355. https://doi.org/10.1016/j.ijleo.2020.165355 doi: 10.1016/j.ijleo.2020.165355
![]() |
[45] |
A. Neirameh, Exact analytical solutions for 3D- Gross-Pitaevskii equation with periodic potential by using the Kudryashov method, J. Egypt. Math. Soc., 24 (2016), 49–53. https://doi.org/10.1016/j.joems.2014.11.004 doi: 10.1016/j.joems.2014.11.004
![]() |
[46] |
M. Ma, Z. Huang, Bright soliton solution of a Gross–Pitaevskii equation, Appl. Math. Lett., 26 (2013), 718–724. https://doi.org/10.1016/j.aml.2013.02.002 doi: 10.1016/j.aml.2013.02.002
![]() |
[47] |
A. A. Bastami, M. R. Belić, D. Milović, N. Z. Petrović, Analytical chirped solutions to the (3+1)-dimensional Gross-Pitaevskii equation for various diffraction and potential functions, Phys. Rev. E, 84 (2011), 016606. https://doi.org/10.1103/PhysRevE.84.016606 doi: 10.1103/PhysRevE.84.016606
![]() |
[48] |
T. A. Sulaiman, G. Yel, H. Bulut, M-fractional solitons and periodic wave solutions to the Hirota-Maccari system, Mod. Phys. Lett. B, 33 (2019), 1950052. https://doi.org/10.1142/S0217984919500520 doi: 10.1142/S0217984919500520
![]() |
[49] | J. V. da C. Sousa, E. C. de Oliveira, A new truncated M-fractional derivative type unifying some fractional derivative types with classical properties, 2017, arXiv: 1704.08187v4. https://doi.org/10.48550/arXiv.1704.08187 |
[50] |
A. T. Ali, E. R. Hassan, General Expa-function method for nonlinear evolution equations, Appl. Math. Comput., 217 (2010), 451–459. https://doi.org/10.1016/j.amc.2010.06.025 doi: 10.1016/j.amc.2010.06.025
![]() |
[51] | E. M. E. Zayed, A. G. Al-Nowehy, Generalized Kudryashov method and general expa function method for solving a high order nonlinear schrödinger equation, J. Space Explor., 6 (2017), 120. |
[52] |
K. Hosseini, Z. Ayati, R. Ansari, New exact solutions of the Tzitzéica-type equations in non-linear optics using the expa function method, J. Mod. Optic., 65 (2018), 847–851. https://doi.org/10.1080/09500340.2017.1407002 doi: 10.1080/09500340.2017.1407002
![]() |
[53] |
A. Zafar, The expa function method and the conformable time-fractional KdV equations, Nonlinear Eng., 8 (2019), 728–732. https://doi.org/10.1515/nleng-2018-0094 doi: 10.1515/nleng-2018-0094
![]() |
1. | H. Bin Jebreen, Hijaz Ahmad, A Novel and Efficient Numerical Algorithm for Solving 2D Fredholm Integral Equations, 2020, 2020, 2314-4785, 1, 10.1155/2020/6662721 | |
2. | Haifa Bin Jebreen, Fairouz Tchier, On the Numerical Simulation of HPDEs Using θ-Weighted Scheme and the Galerkin Method, 2020, 9, 2227-7390, 78, 10.3390/math9010078 | |
3. | Neslişah İmamoğlu Karabaş, Sıla Övgü Korkut, Gurhan Gurarslan, Gamze Tanoğlu, A reliable and fast mesh-free solver for the telegraph equation, 2022, 41, 2238-3603, 10.1007/s40314-022-01927-x | |
4. | M. Abdelhakem, Mona Fawzy, M. El-Kady, Hanaa Moussa, An efficient technique for approximated BVPs via the second derivative Legendre polynomials pseudo-Galerkin method: Certain types of applications, 2022, 43, 22113797, 106067, 10.1016/j.rinp.2022.106067 | |
5. | Ishtiaq Ali, Maliha Tehseen Saleem, Azhar ul Din, Special Functions and Its Application in Solving Two Dimensional Hyperbolic Partial Differential Equation of Telegraph Type, 2023, 15, 2073-8994, 847, 10.3390/sym15040847 | |
6. | Şuayip Yüzbaşı, Gamze Yıldırım, A numerical approach to solve hyperbolic telegraph equations via Pell–Lucas polynomials, 2023, 17, 1658-3655, 10.1080/16583655.2023.2255404 | |
7. | Haifa Bin Jebreen, Beatriz Hernández-Jiménez, An approach based on the pseudospectral method for fractional telegraph equations, 2023, 8, 2473-6988, 29221, 10.3934/math.20231496 |