Planar Turán Numbers of Cubic Graphs and Disjoint Union of Cycles

Abstract

The planar Turán number of a graph H, denoted by $ex_{_\mathcal {P}}(n,H)$, is the maximum number of edges in a planar graph on n vertices without containing H as a subgraph. This notion was introduced by Dowden in 2016 and has attracted quite some attention since then; those work mainly focus on finding $ex_{_\mathcal {P}}(n,H)$ when H is a cycle or Theta graph or H has maximum degree at least four. In this paper, we first completely determine the exact values of $ex_{_\mathcal {P}}(n,H)$ when H is a cubic graph. We then prove that $ex_{_\mathcal {P}}(n,2C_3)=\lceil 5n/2\rceil -5$ for all $n\ge 6$, and obtain the lower bounds of $ex_{_\mathcal {P}}(n,2C_k)$ for all $n\ge 2k\ge 8$. Finally, we also completely determine the exact values of $ex_{_\mathcal {P}}(n,K_{2,t})$ for all $t\ge 3$ and $n\ge t+2$.

1 Introduction

All graphs considered in this paper are finite and simple. We use $K_n$, $C_n$ and $P_n$ to denote the complete graph, cycle and path on n vertices, respectively. Given a graph G, we use |G| to denote the number of vertices, e(G) the number of edges, $\delta (G)$ the minimum degree, $\Delta (G)$ the maximum degree. For a vertex $v\in V(G)$, we will use $N_G(v)$ to denote the set of vertices in G which are adjacent to v. We define $N_G[v]:=N_G(v)\cup \{v\}$. For any $S,S'\subseteq V(G)$, we use $e_G(S,S')$ to denote the size of edge set $\{xy\in E(G)\mid x\in S \text { and } y\in S'\}$. For any set $S \subset V(G)$, the subgraph of G induced on S, denoted G[S], is the graph with vertex set S and edge set $\{xy \in E(G) \mid x, y \in S\}$. We denote by $G {{\setminus }}S$ the subgraph of G induced on $V(G) {{\setminus }}S$. If $S=\{v\}$, then we simply write $G{{\setminus }}v$. The join $G+H$ (resp. union $G\cup H$) of two vertex-disjoint graphs G and H is the graph having vertex set $V(G)\cup V(H)$ and edge set $E(G) \cup E(H)\cup \{xy\, |\, x\in V(G), y\in V(H)\}$ (resp. $E(G)\cup E(H)$). For a positive integer t and a graph H, we use tH to denote the disjoint union of t copies of H. Let $T_n$ denote a plane triangulation on $n\ge 3$ vertices. We use $T_n^-$ and $K_n^-$ to denote a graph obtained from $T_n$ and $K_n$ with one edge removed, respectively. Given two isomorphic graphs G and H, we may (with a slight but common abuse of notation) write $G = H$. For any positive integer k, we define $[k]:=\{1,2, \ldots , k\}$.

Given a graph H, a graph is H-free if it does not contain H as a subgraph. One of the best known results in extremal graph theory is Turán’s Theorem [12], which gives the maximum number of edges that a $K_t$-free graph on n vertices can have. The celebrated Erdős–Stone Theorem [3] then extends this to the case when $K_t$ is replaced by an arbitrary graph H with at least one edge, showing that the maximum number of edges possible is $(1+o(1)){n\atopwithdelims ()2}\left( \frac{\chi (H)-2}{\chi (H)-1}\right)$, where $\chi (H)$ denotes the chromatic number of H.

In this paper, we continue to study the topic of “extremal" planar graphs, that is, how many edges can an H-free planar graph on n vertices have? We define $ex_{_\mathcal {P}}(n,H)$ to be the maximum number of edges in an H-free planar graph on n vertices. Dowden [2] initiated the study of $ex_{_\mathcal {P}}(n,H)$ and proved the following result.

Theorem 1.1

(Dowden [2]) Let n be a positive integer.

• (a) $ex_{_\mathcal {P}}(n, C_3)=2n-4$ for all $n\ge 3$.

• (b) $ex_{_\mathcal {P}}(n, K_4)=3n-6$ for all $n\ge 4$.

• (c) $ex_{_\mathcal {P}}(n, C_4)\le {15}(n-2)/7$ for all $n\ge 4$, with equality when $n\equiv 30 (\textrm{mod}\, 70)$.

• (d) $ex_{_\mathcal {P}}(n, C_5)\le 12(n-2)/{5}$ for all $n\ge 5$.

• (e) $ex_{_\mathcal {P}}(n, C_5)\le (12n-33)/{5}$ for all $n\ge 11$. Equality holds for infinity many n.

This topic has attracted quite some attention since then. We refer the reader to a recent survey [11] of the present authors for more information. Let $\Theta _k$ denote the family of Theta graphs on $k\ge 4$ vertices, that is, graphs obtained from $C_k$ by adding an additional edge joining two non-consecutive vertices. The present authors [10] obtained tight upper bounds for $ex_{_\mathcal {P}}(n, \Theta _k)$ for $k\in \{4,5\}$ and an upper bound for $ex_{_\mathcal {P}}(n, \Theta _6)$.

Theorem 1.2

(Lan, Shi and Song [10]) Let n be a positive integer.

• (a) $ex_{_\mathcal {P}}(n, \Theta _4)\le {12(n-2)}/5$ for all $n\ge 4$, with equality when $n\equiv 12 (\textrm{mod}\, 20)$.

• (b) $ex_{_\mathcal {P}}(n, \Theta _5)\le {5(n-2)}/2$ for all $n\ge 5$, with equality when $n\equiv 50 (\textrm{mod}\, 120)$.

• (c) $ex_{_\mathcal {P}}(n, C_6)\le ex_{_\mathcal {P}}(n, \Theta _6) \le {18(n-2)}/7$ for all $n\ge 6$.

Theorem 1.2(c) has been strengthened by the authors in [5, 6] with tight upper bounds.

Theorem 1.3

(Ghosh et al. [5, 6]) Let n be a positive integer.

• (a) $ex_{_\mathcal {P}}(n, C_6)\le (5n-14)/2$ for all $n\ge 18$, with equality when $n\equiv 10 (\textrm{mod}\, 18)$.

• (b) $ex_{_\mathcal {P}}(n, \Theta _6)\le (18n-48)/7$ for all $n\ge 14$. Equality holds for infinitely many n.

As observed in [2], for all $n\ge 6$, the planar triangulation $2K_1+C_{n-2}$ is $K_4$-free. Hence, $ex_{_\mathcal {P}}(n,H)=3n-6$ for all graphs H which contains $K_4$ as a subgraph and $n\ge 6$. The present authors [9] also investigated a variety of sufficient conditions on $K_4$-free planar graphs H such that $ex_{_\mathcal {P}}(n,H)=3n-6$ for all $n\ge |H|$.

Theorem 1.4

(Lan, Shi and Song [9]) Let H be a $K_4$-free planar graph and let $n\ge |H|$ be an integer. Then $ex_{_\mathcal {P}}(n,H)=3n-6$ if one of the following holds, where $n_{_k}(H)$ denotes the number of vertices of degree k in H for a positive integer k.

• (a) $\chi (H)=4$ and $n\ge |H|+2$.

• (b) $\Delta (H)\ge 7$.

• (c) $\Delta (H)=6$ and either $n_{_6}(H)+n_{_5}(H)\ge 2$ or $n_{_6}(H)+n_{_5}(H)=1$ and $n_{_4}(H)\ge 5$.

• (d) $\Delta (H)=5$ and either H has at least three 5-vertices or H has exactly two adjacent 5-vertices.

• (e) $\Delta (H)=4$ and $n_{_4}(H)\ge 7$.

• (f) H is 3-regular with $|H|\ge 9$ or H has at least three vertex-disjoint cycles or H has exactly one vertex u of degree $\Delta (H) \in \{4,5,6\}$ such that $\Delta (H[N_H(u)])\ge 3$.

• (g) $\delta (H)\ge 4$ or H has exactly one vertex of degree at most 3.

In the same paper, the present authors [9] also determined the values of $ex_{_\mathcal {P}}(n, H)$ when H is a star, or wheel or (t, r)-fan. Ghosh et al. [7] recently determined the values of $ex_{_\mathcal {P}}(n,H)$ when H is a double star. Theorem 1.4 implies that $ex_{_\mathcal {P}}(n,H)$ remains wide open when H is subcubic. In particular, it seems quite non-trivial to determine $ex_{_\mathcal {P}}(n, C_k)$ for all $k\ge 7$. Very recently, Cranston et al. [1] proved that for each $k\ge 11$ and n sufficiently large (as a function of k),

$$\begin{aligned} ex_{_\mathcal {P}}(n, C_k) >\left( 3-\frac{3}{k}\right) n-6-\frac{6}{k}. \end{aligned}$$

They further proposed the following conjecture.

Conjecture 1.5

(Cranston et al. [1]) There exists a constant D such that for all k and for all sufficiently large n, we have

$$\begin{aligned} ex_{_{\mathcal {P}}}(n,C_k)\le \left( 3-\frac{3}{Dk^{lg_2^3}}\right) n. \end{aligned}$$

In this paper, we continue to study the planar Turán numbers of k-regular graphs, disjoint union of cycles and complete bipartite graphs. We prove the following main results.

Theorem 1.6

Let H be a k-regular planar graph with $k\ge 3$ and let $n\ge |H|$ be an integer. Then

$$\begin{aligned}ex_{_\mathcal {P}}(n,H)= {\left\{ \begin{array}{ll} \, 3n-6 &{} \text { if } \, |H|\ge 8, \text { or } |H|=6 \text { and } n\ge 10;\\ \, 3n-7 &{} \text { if } \, |H|=6 \text { and } n\le 9. \end{array}\right. } \end{aligned}$$

Theorem 1.7

For integers n and t with $n\ge 3t\ge 3$, we have

Theorem 1.8

Let n and k be positive integers.

• (a) Suppose $n\ge 2k\ge 8$, and r is the remainder of $n-3$ when divided by $k-2$. Then

  $$\begin{aligned}{} & {} ex_{_\mathcal {P}}(n,2C_k)\ge \left( 3-\frac{1}{k-2}\right) n+\frac{3+r}{k-2} -5\\{} & {} \quad +\max \{1-r, 0\}. \end{aligned}$$

• (b) Suppose $n\ge 2k\ge 14$, and $\varepsilon _1$ and $\varepsilon _2$ are the remainder of $n-(2k-1)$ when divided by $k-4+\frac{k-1}{2}$ (k is odd) and $k-6+\frac{k}{2}$ (k is even), respectively.

• (b1) If k is odd, then $ex_{_\mathcal {P}}(n,2C_k)=3n-6$ for all $n\le 3k-4$, and

  $$\begin{aligned}{} & {} ex_{_\mathcal {P}}(n,2C_k)\ge \left( 3-\frac{1}{k-4+\lfloor k/2\rfloor }\right) n\\{} & {} \quad +\frac{5+\varepsilon _1}{k-4+\lfloor k/2\rfloor }-\frac{17}{3}+\max \{1-\varepsilon _1, 0\}\, \text { for all } n\ge 3k-3.\end{aligned}$$

• (b2) If k is even, then $ex_{_\mathcal {P}}(n,2C_k)=3n-6$ for all $n\le 3k-7$, and

  $$\begin{aligned}{} & {} ex_{_\mathcal {P}}(n,2C_k)\ge \left( 3-\frac{1}{k-6+ k/2}\right) n\\{} & {} \quad +\frac{7+\varepsilon _2}{k-6+ k/2}-\frac{17}{3}+\max \{1-\varepsilon _2, 0\} \, \text { for all } n\ge 3k-6.\end{aligned}$$

Theorem 1.9

For integers $t\ge 3$ and $n\ge t+2$, we have

$$\begin{aligned}ex_{_{\mathcal {P}}}(n, K_{2,t})= {\left\{ \begin{array}{ll} 3n-6 &{}\, \text { if }\; t\ge 5\; \text { and }\;n\ge t+2, \;\text { or } \;t=4 \;\text { and } \;n\ge 9, \;\text { or } \;t=3 \;\text { and } \;n\ge 12;\\ 3n-7 &{}\, \text { if }\; t=4 \; \text { and }\; n\le 8;\\ 3n-8 &{}\, \text { if } \; t=3 \; \text { and } \; n\le 11.\\ \end{array}\right. } \end{aligned}$$

The remainder of the paper is organized as follows: we prove Theorem 1.6 in Sect. 2, Theorems 1.7 and  1.8 in Sect. 3, and Theorem 1.9 in Sect. 4.

2 Planar Turán Number of Regular Graphs

We begin this section with two lemmas that will be essential in determining the planar Turán numbers of regular planar graphs. Proofs of Lemmas 2.1 and 2.2 are given in the Appendix.

Lemma 2.1

Let G be a cubic planar graph on 8 vertices. If G is $K_4$-free, then $G\in \{G_1,G_2,G_3\}$, where graphs $G_1, G_2, G_3$ are depicted in Fig. 1.

Fig. 1

Graphs $G_1$, $G_2$ and $G_3$

Lemma 2.2

Let G be a cubic planar graph on 6 vertices. Then G is the graph given in Figure 2.

Fig. 2

The unique cubic planar graph H on 6 vertices

Proof of Theorem 1.6:

Let H and n be given as in the statement. By Theorem 1.1(b) and Theorem 1.4(f, g), we see that $ex_{_\mathcal {P}}(n,H)=3n-6$ for all $n\ge |H|$ if H contains a copy of $K_4$, or $k=3$ and $|H|\ge 9$, or $k\ge 4$. For the remainder of the proof, let H be a $K_4$-free cubic planar graph with $|H|\le 8$. Assume first that $|H|=8$. By Lemma 2.1, we see that $H\in \{G_1,G_2,G_3\}$. Then the planar triangulation $2K_1+C_{n-2}$ is H-free when $n>|H|$, and the planar triangulation $K_2+P_{n-2}$ is H-free when $n=|H|$. Hence, $ex_{_\mathcal {P}}(n,H)=3n-6$ for all $n\ge |H|$ and $|H|=8$.

It remains to consider the case when $|H|=6$. Then H is the graph given in Fig. 2. We next show that $ex_{_\mathcal {P}}(n,H)=3n-6$ for all $n\ge 10$. Let $n:=4k+2+\ell$ for some $\ell \in \{0,1,2,3\}$ and integer $k\ge 2$. Let $Q_k$ be a plane triangulation on $n=4k+2$ vertices constructed as follows: for each $i\in [k]$, let $C^i$ be a cycle with vertices $u_{i,1},u_{i,2},u_{i,3},u_{i,4}$ in order, let $Q_k$ be the plane triangulation obtained from disjoint union of $C^1,\ldots ,C^k$ by adding edges $u_{i,j}u_{i+1,j}$ and $u_{i,j}u_{i+1,j+1}$ for all $i\in [k-1]$ and $j\in [4]$, where all arithmetic on the index $j+1$ here is done modulo 4, and finally adding two new nonadjacent vertices u and v such that u is adjacent to all vertices of $C^1$ and v is adjacent to all vertices of $C^k$. The graph $Q_k$ when $k=3$ is depicted in Fig. 3. Let $Q_k^\ell =Q_k$ if $\ell =0$. For $\ell \in \{1,2,3\}$, let $F_j$ be the face of $Q_k$ with vertices $u_{k-1,j},u_{k,j},u_{k,j+1}$ for each $j\in [\ell ]$, and let $Q_k^\ell$ be the plane triangulation on n vertices obtained from $Q_k$ by adding one new vertex, say $x_j$, adjacent to the three vertices on the boundary of $F_j$ for each $j\in [\ell ]$. It can be checked that $Q_k^\ell$ is H-free. Therefore, $ex_{_\mathcal {P}}(n,H)=3n-6$ for all $n\ge 10$. $\square$

Fig. 3

The plane triangulation $Q_k$ when $k=3$, where $C^1,C^2,C^3$ are in red, blue and green, respectively

We next show that $ex_{_\mathcal {P}}(n,H)=3n-7$ for each $n\in \{6,7,8,9\}$. To obtain the desired upper bound, it suffices to show that every plane triangulation T on $n\in \{6,7,8,9\}$ vertices contains a copy of H. Suppose not. Let T be an H-free plane triangulation such that $|T|\in \{6,7,8,9\}$ is minimum. Note that every plane triangulation on 6 vertices contains a copy of H. Thus $|T|\in \{7,8,9\}$. Let $x\in V(T)$ with $d_T(x)=\delta (T)$. Then $d_T(x)\in \{3,4\}$. If $d_T(x)=3$, then $T-x$ is a plane triangulation and contains a copy of H by the minimality of |T|, a contradiction. Thus $d_T(x)=4$. Clearly, T[N[x]] is a wheel on 5 vertices. Let $Y=V(T){{\setminus }}N[x]$. Note that $|Y|\le 4$. Then $d_{N(x)}(y)\le 2$ for every $y\in Y$, else T contains a copy of H. It follows that $|Y|=4$ and so $|T|=9$, else $e(T)=e(T[N[x]])+e(N(x), Y)+e(T[Y])\le 8+2|Y|+2|Y|-3<3(5+|Y|)-6$. Since $\delta (T)=4$, we see that $e(T[Y])\le 5$. Then $e(T)=e(T[N[x]])+e(N(x), Y)+e(T[Y])\le 8+2|Y|+5=21$, which yields $T\cong Q_2-u+u_{1,1}u_{1,3}$ and so T contains a copy of H, a contradiction. Thus, every plane triangulation T on $n\in \{6,7,8,9\}$ vertices contains a copy of H. On the other hand, for each $n\in \{6,7, 8\}$, the planar graph $K_2+(P_3\cup P_{n-5})$ is H-free with $3n-7$ edges because every induced subgraph of $K_2+(P_3\cup P_{n-5})$ on 6 vertices contains a vertex of degree two or is isomorphic to $K_2+2P_2$; for $n=9$, the planar graph $Q_2{{\setminus }}u$ is H-free with $3n-7$ edges because every induced subgraph of $Q_2{{\setminus }}u$ on 6 vertices either contains a vertex of degree two or is isomorphic to a wheel on 6 vertices. Hence, $ex_{_\mathcal {P}}(n,H)= 3n-7$ for each $n\in \{6,7,8,9\}$, as desired. $\square$

3 Planar Turán Number of Disjoint Union of Cycles

Given a plane graph G and an integer $i\ge 3$, an i-face in G is a face of order i. Let $f_i$ and f(G) denote the number of i-faces and all faces in G, respectively. In this section we study the planar Turán number of disjoint union of cycles. We first consider $tC_k$, the t vertex-disjoint copies of $C_k$, and give a tight bound for $ex_{_\mathcal {P}}(n,tC_3)$ for all $n\ge 3t\ge 3$. It is worth noting that $ex_{_\mathcal {P}}(n,H)=3n-6$ if H has three vertex-disjoint cycles, due to Theorem 1.4(f).

Proof of Theorem 1.7:

By Theorem 1.1(a) and Theorem 1.4(f), $ex_{_\mathcal {P}}(n,tC_3)=2n-4$ if $t=1$, and $ex_{_\mathcal {P}}(n,tC_3)=3n-6$ if $t\ge 3$. We may assume that $t=2$. Then $n\ge 6$. We first show that $ex_{_\mathcal {P}}(n,2C_3)\ge \lceil 5n/2\rceil -5$. Let P be a path on $n-2$ vertices and S be a maximum independent set of P containing the two ends of P. Let G be the planar graph on n vertices obtained from P by adding two new adjacent vertices u and v such that u is joined to every vertex in V(P) and v is joined to every vertex in S. Then G is $2C_3$-free with $|G|=n$ and $e(G)=(n-3)+(n-1)+\lceil (n-2)/2\rceil =\lceil 5n/2\rceil -5$. Hence, $ex_{_\mathcal {P}}(n,2C_3)\ge e(G)=\lceil 5n/2\rceil -5$.

We next show that $ex_{_\mathcal {P}}(n,2C_3)\le \lceil 5n/2\rceil -5$. It can be easily checked that every plane graph $T_n^-$ contains a copy of $2C_3$, where $n\in \{6,7\}$. Hence, $ex_{_\mathcal {P}}(n,2C_3)\le e(T_n^-)-1=3n-8=\lceil 5n/2\rceil -5$ when $n\in \{6,7\}$. We may assume that $n\ge 8$. Let G be a $2C_3$-free plane graph on $n \ge 8$ vertices. We first prove that

$(*)$ $f_3\le n-1$.

To prove $(*)$, suppose $f_3\ge n\ge 8$. Let $\mathcal {F}$ be the set of all 3-faces of G. Then $|\mathcal {F}|=f_3$. For each $v\in V(G)$, let $\mathcal {F}(v):=\{F\in \mathcal {F}\mid v\in V(F)\}$. Then $|\mathcal {F}(v)|\le n-1$ and so $\mathcal {F}{\setminus }\mathcal {F}(v)\ne \emptyset$. Since G is $2C_3$-free, we see that $V(F)\cap V(F')\ne \emptyset$ for every pair $F, F'\in \mathcal {F}$. Since $f_3\ge n\ge 8$, there exist $F', F''\in \mathcal {F}$ such that $|V(F')\cap V(F'')|=1$. We may assume that $V(F')=\{x,y,z\}$ and $V(F'')=\{x,u,w\}$, where x, y, z, u, w are pairwise distinct. It follows that for every $F\in \mathcal {F}{\setminus } \mathcal {F}(x)$, we have $|V(F)\cap \{y,z\}|\ge 1$ and $|V(F)\cap \{u,w\}|\ge 1$.

Suppose $|V(F)\cap \{y,z\}|=1$ and $|V(F)\cap \{u,w\}|=1$ for every $F\in \mathcal {F}{\setminus } \mathcal {F}(x)$. Then $|\mathcal {F}{\setminus } \mathcal {F}(x)|\le 4$. In addition, if $|\mathcal {F}{\setminus } \mathcal {F}(x)|=1$, then $|\mathcal {F}(x)|\le 6$ (see Figure 4(a) when $|\mathcal {F}(x)|=6$); if $|\mathcal {F}{\setminus } \mathcal {F}(x)|=2$, then $|\mathcal {F}(x)|\le 4$ (see Fig. 4(b) when $|\mathcal {F}(x)|=4$, where vertices a, b, c are not necessary distinct); if $3\le |\mathcal {F}{\setminus } \mathcal {F}(x)|\le 4$, then $|\mathcal {F}(x)|\le 3$ (see Figure 4(c, d) when $|\mathcal {F}(x)|=3$, where vertices a and z in Figure (c) are not necessary distinct). Thus $f_3=|\mathcal {F}{{\setminus }} \mathcal {F}(x)| +|\mathcal {F}(x)|\le 7$, contrary to the assumption that $f_3\ge n\ge 8$. $\square$

Fig. 4

a $\mathcal {F}(x)=\{F',F'',F_1,F_2,F_3,F_4\}$; b $\mathcal {F}(x)=\{F',F'',F_1,F_2\}$; c, d $\mathcal {F}(x)=\{F',F'',F_1\}$

Suppose for some $F^*\in \mathcal {F}{\setminus } \mathcal {F}(x)$, we have $y, z\in V(F^*)$ or $u, w\in V(F^*)$, say the former. Recall that for every $F\in \mathcal {F}{\setminus } \mathcal {F}(x)$, we have $|V(F)\cap \{y,z\}|\ge 1$ and $|V(F)\cap \{u,w\}|\ge 1$. We may further assume that $u\in V(F^*)$. Then $|\mathcal {F}{{\setminus }} \mathcal {F}(x)|\le 3$. In addition, if $|\mathcal {F}{{\setminus }} \mathcal {F}(x)|=1$, then $|\mathcal {F}(x)|\le 5$ (see Fig. 5(a) when $|\mathcal {F}(x)|=5$, where vertices a and w, or b and c are not necessary distinct); if $2\le |\mathcal {F}{\setminus } \mathcal {F}(x)|\le 3$, then $|\mathcal {F}(x)|\le 4$ (see Fig. 5(b, c), where vertices a, b, c in Figure (b) and a, z in Figure (c) are not necessary distinct). It follows that $f_3=|\mathcal {F}{{\setminus }} \mathcal {F}(x)| +|\mathcal {F}(x)|\le 7$, contrary to the assumption that $f_3\ge n\ge 8$. This completes the proof of $(*)$.

Fig. 5

a $\mathcal {F}(x)=\{F',F'',F_1,F_2,F_3\}$; b $\mathcal {F}(x)=\{F',F'',F_1,F_2\}$; c $\mathcal {F}(x)=\{F',F'',F_1\}$

By $(*)$, we see that

$$\begin{aligned} 2e(G)=3f_3+\sum _{i\ge 4}if_i\ge 3f_3+4(f(G)-f_3)=4f(G)-f_3\ge 4f(G)-(n-1), \end{aligned}$$

which implies that $f(G)\le (2e(G)+n-1)/4$. By Euler’s formula,

$$\begin{aligned}n-2=e(G)-f(G)\ge e(G)/2-(n-1)/4.\end{aligned}$$

Hence, $e(G)\le \lceil 5n/2\rceil -5$, as desired. This completes the proof of Theorem 1.7. $\square$

For general $k\ge 4$, we give lower bound constructions for $ex_{_\mathcal {P}}(n,2C_k)$. It would be interesting to know whether the bounds in Theorem 1.8(b) are desired tight upper bounds for $ex_{_\mathcal {P}}(n,2C_k)$ for all $k\ge 7$ and n sufficiently large.

Proof of Theorem 1.8:

(a) Let n, k, r be given as in the statement. Let $t\ge 0$ be an integer satisfying

$$\begin{aligned}(2k-1)+t(k-2)+r=n.\end{aligned}$$

Let $P_1, P_2,\ldots ,P_{t+1}$ be vertex-disjoint paths with $|P_{t+1}|=2k-3$ and $|P_i|=k-2$ for each $i\in [t]$. Let $H=P_1\cup \cdots \cup P_{t+1}$. Then $|H|=(2k-3)+t(k-2)=n-2-r$ and

$$\begin{aligned}e(H)=2k-4+t(k-3)=2k-4+(n-2k+1-r)\left( 1-\frac{1}{k-2}\right) .\end{aligned}$$

Let Q be null graph when $r=0$ and a path on r vertices such that $V(Q)\cap V(H)=\emptyset$ when $r\ge 1$. Let G be the planar graph obtained from $H\cup Q$ by adding two new adjacent vertices such that each new vertex is joined to each vertex of $H\cup Q$. Clearly, G is $2C_k$-free with $|G|=|H|+| Q|+2=(n-2-r)+r+2=n$ and

$$\begin{aligned} e(G)&=2n-3+e(H)+e(Q)\\&=2n-3+2k-4+(n-2k+1-r)\left( 1-\frac{1}{k-2}\right) +\max \{r-1,0\}\\&=\left( 3-\frac{1}{k-2}\right) n+\frac{3+r}{k-2}-4-r+\max \{r-1,0\}\\&=\left( 3-\frac{1}{k-2}\right) n+\frac{3+r}{k-2}-5+\max \{1-r,0\}. \end{aligned}$$

Hence, $ex_{_\mathcal {P}}(n,2C_k)\ge e(G')=\left( 3-\frac{1}{k-2}\right) n+\frac{3+r}{k-2}-5+\max \{1-r,0\}$, as desired.

This completes the proof of Theorem 1.8(a). $\square$

(b) Let $n,k,\varepsilon _1, \varepsilon _2$ be given as in the statement. Throughout the proof, let $\mathcal {T}_m:=K_2+P_{m-2}$ be the plane triangulation on $m\ge 2$ vertices, with x and y on the outer face of $\mathcal {T}_m$, where x and y are the two adjacent vertices of degree $m-1$ in $\mathcal {T}_m$. Note that we allow m to be 2 here for a simpler proof later on; $\mathcal {T}_m$ has exactly $2m-4$ 3-faces. For each integer s satisfying $m\le s\le 2m-4$, let $\mathcal {T}_s^m$ denote a plane triangulation on s vertices obtained from $\mathcal {T}_m$ by adding $s-m$ new vertices (no two new vertices are added to the same face): each to a 3-face F of $\mathcal {T}_m$ and then joining it to all vertices on the boundary of F.

To prove (b1), let $k:=2p+1$, where $p\ge 3$ is an integer. We first show that $ex_{_\mathcal {P}}(n,2C_k)=3n-6$ for all $n\le 3k-4$. Note that $\mathcal {T}_{2p+1}$ has exactly $2(2p+1)-4=4p-2$ 3-faces. For each $n\le 3k-4=6p-2=(2p+1)+(4p-2)$, $\mathcal {T}^{2p+1}_n$ is $2C_k$-free because each $C_k$ in $\mathcal {T}^{2p+1}_n$ must contain at least $p+1$ vertices of $\mathcal {T}_{2p+1}$. Hence, $ex_{_\mathcal {P}}(n,2C_k)=3n-6$ for all $n\le 3k-4$. We next consider the case $n\ge 3k-3$. Let $t\ge 0$ be an integer satisfying

$$\begin{aligned} t(3p-3)+\varepsilon _1=n-(2k-1). \end{aligned}$$

Let $H_i:=\mathcal {T}^{p+1}_{3p-1}$ for each $i\in [t]$ when $t\ge 1$; $H_{t+1}:=\mathcal {T}_{\varepsilon _1+2}$ when $\varepsilon _1+2\le k-1$ and the plane triangulation $\mathcal {T}^{p+1}_{\varepsilon _1+2}$ when $\varepsilon _1+2\ge k$; and $H_{t+2}:=\mathcal {T}_{2k-1}$. For each $j\in [t+1]$, it is easy to check that each cycle of $H_j$ has length at most k; each cycle of length exactly k must contain the vertices x and y; $H_j$ is $2C_k$-free. Note that $H_{t+2}$ is $2C_k$-free, and each cycle of length exactly k in $H_{t+2}$ must contain either x or y (or possibly both). Finally, let G be the planar graph obtained from disjoint copies of $H_1, H_2, \ldots , H_{t+2}$ by pasting along the subgraph $K_2$ induced by x and y. It follows that G is $2C_k$-free and $|G|=t(3p-3)+(2k-1)+\varepsilon _1=n$. Note that $|H_{t+1}|\ge 2$ with equality when $\varepsilon _1=0$. Therefore,

$$\begin{aligned}&ex_{_\mathcal {P}}(n,2C_k)\ge e(G)\\ {}&\quad =\sum _{i=1}^t(e(H_i)-1)+(e(H_{t+1})-1)+(e(H_{t+2})-1)+1\\&=t[3(3p-1)-7]+ \max \{3(\varepsilon _1+2)-7, 0\}+(3(2k-1)-7)+ 1\\&=3n-t-7+\max \{1-\varepsilon _1, 0\} \\&=3n-\frac{n- (2k-1)-\varepsilon _1}{3p-3}-7+\max \{1-\varepsilon _1, 0\}\\&=\left( 3-\frac{1}{3p-3}\right) n+\frac{2k-1+\varepsilon _1}{3p-3}-7+\max \{1-\varepsilon _1, 0\}\\&=\left( 3-\frac{1}{k-4+\frac{k-1}{2}}\right) n+\frac{2k-1+\varepsilon _1}{k-4+\frac{k-1}{2}}-7+\max \{1-\varepsilon _1, 0\}\\&=\left( 3-\frac{1}{k-4+\lfloor k/2\rfloor }\right) n+\frac{5+\varepsilon _1}{k-4+\lfloor k/2\rfloor }-\frac{17}{3}+\max \{1-\varepsilon _1, 0\}.\\ \end{aligned}$$

It remains to prove (b2). Let $k: =2p$, where $p\ge 4$ is an integer. Similar to the proof of (I), we see that $\mathcal {T}^{2p-1}_n$ is $2C_k$-free for each $n\le 3k-7$, because each $C_k$ in $\mathcal {T}^{2p-1}_n$ must contain at least p vertices of $\mathcal {T}_{2p-1}$. Hence, $ex_{_\mathcal {P}}(n,2C_k)=3n-6$ for all $n\le 3k-7$. We next consider the case $n\ge 3k-6$. Let $t\ge 0$ be an integer satisfying

$$\begin{aligned} t(3p-6)+\varepsilon _2=n-(2k-1). \end{aligned}$$

Let $H_i:=\mathcal {T}^{p}_{3p-4}$ for each $i\in [t]$ when $t\ge 1$; $H_{t+1}:=\mathcal {T}_{\varepsilon _2+2}$ when $\varepsilon _2+2\le k-1$ and the plane triangulation $\mathcal {T}^{p}_{\varepsilon _2+2}$ when $\varepsilon _2+2\ge k$; and $H_{t+2}:=\mathcal {T}_{2k-1}$. For each $j\in [t+1]$, it is easy to check that each cycle of $H_j$ has length at most k; each cycle of length exactly k must contain the vertices x and y; $H_j$ is $2C_k$-free. Note that $H_{t+2}$ is $2C_k$-free, and each cycle of length exactly k in $H_{t+2}$ must contain either x or y (or possibly both). Finally, let G be the planar graph obtained from disjoint copies of $H_1, H_2, \ldots , H_{t+2}$ by pasting along the subgraph $K_2$ induced by x and y. It follows that G is $2C_k$-free and $|G|=t(3p-6)+(2k-1)+\varepsilon _2=n$. Note that $|H_{t+1}|\ge 2$ with equality when $\varepsilon _2=0$. Therefore,

$$\begin{aligned}&ex_{_\mathcal {P}}(n,2C_k)\ge e(G)\\ {}&\quad =\sum _{i=1}^t(e(H_i)-1)+(e(H_{t+1})-1)+(e(H_{t+2})-1)+1\\&=t[3(3p-4)-7]+\max \{3(\varepsilon _2+2)-7, 0\}+(3(2k-1)-7)+1\\&=3n-t-7+\max \{1-\varepsilon _2, 0\} \\&=3n-\frac{n- (2k-1)-\varepsilon _2}{3p-6}-7+\max \{1-\varepsilon _2, 0\}\\&=\left( 3-\frac{1}{3p-6}\right) n+\frac{2k-1+\varepsilon _2}{3p-6}-7+\max \{1-\varepsilon _2, 0\}\\&=\left( 3-\frac{1}{k-6+ k/2}\right) n+\frac{2k-1+\varepsilon _2}{k-6+ k/2}-7+\max \{1-\varepsilon _2, 0\}\\&=\left( 3-\frac{1}{k-6+ k/2}\right) n+\frac{7+\varepsilon _2}{k-6+ k/2}-\frac{17}{3}+\max \{1-\varepsilon _2, 0\}, \end{aligned}$$

as desired. This completes the proof of Theorem 1.8(b). $\square$

We end this section with a result showing that $ex_{_\mathcal {P}}(n, C_3^+)=ex_{_\mathcal {P}}(n,C_3)$ for all $n\ge 4$, where $C_k^+$ denotes the graph on $k+1$ vertices obtained from $C_k$ by adding a pendant edge.

Proposition 3.1

For all $n\ge 4$, $ex_{_{\mathcal {P}}}(n, C_3^+)=ex_{_{\mathcal {P}}}(n,C_3)$.

Since $C_3$ is a subgraph of $C_3^+$, we have $ex_{_{\mathcal {P}}}(n,C_3^+)\ge ex_{_{\mathcal {P}}}(n,C_3)$. Let G be a $C_3^+$-free planar graph with at least four vertices. If G is connected, then G is also $C_3$-free; if G is disconnected, then we may prove $e(G)\le ex_{_{\mathcal {P}}}(n,C_3)$ by induction on n, which is left to readers.

4 Planar Turán Number of Complete Bipartite Graphs

Finally, we study the planar Turán number of $K_{m,t}$, where $t\ge m\ge 1$. Note that $ex_{_{\mathcal {P}}}(n, K_{m,t})=3n-6$ when $m\ge 3$; Theorem 6 in [9] completely determines the values of $ex_{_{\mathcal {P}}}(n, K_{m,t})$ when $m=1$; Theorem 1.1(c) settles $ex_{_{\mathcal {P}}}(n, K_{m,t})$ for the case $m=t=2$. We prove the remaining cases for $ex_{_{\mathcal {P}}}(n, K_{2,t})$. Let $O_n$ denote the unique outerplane graph with $2n-3$ edges, maximum degree 4, and the outer face of order n. The graph $O_n$ when $n=12$ is depicted in Fig. 6. It is easy to see that $O_n$ is $K_{1,5}$-free because the maximum degree of $O_n$ is four.

Fig. 6

The graph $O_{12}$

Proof of Theorem 1.9:

Let n and t be given as in the statement. Note that $O_{n-1}$ is $K_{2,3}$-free because $O_{n-1}$ is an outerplane graph. It follows that $K_1+O_{n-1}$ is $K_{2,5}$-free. Hence, $ex_{_{\mathcal {P}}}(n, K_{2,t})=3n-6$ for all $t\ge 5$ and $n\ge t+2$. We may assume that $t\in \{3,4\}$. We first consider the case $n\ge 12$. Let $H_{\ell }$ be the plane graph on $6\ell$ vertices constructed as follows: for each $i\in [\ell ]$, let $C^i$ be a cycle with vertices $u_{i,1},u_{i,2},\ldots ,u_{i,6}$ in order, let $H_{\ell }$ be the plane graph obtained from disjoint union of $C^1,\ldots ,C^{\ell }$ by adding edges $u_{i,j}u_{i+1,j}$ and $u_{i,j}u_{i+1,j+1}$ for all $i\in [\ell -1]$ and $j\in [6]$, where all arithmetic on the index $j+1$ here is done modulo 6. For $i\in \{1,\ell \}$, let $W_5^i$ be a wheel on 6 vertices $v_{i,0},v_{i,1},v_{i,2},v_{i,3},v_{i,4},v_{i,5}$, where $v_{i,0}$ is the center vertex of $W_5^i$; $K_p^i$ be a complete graph with vertices $w_{i,1},w_{i,2},\ldots ,w_{i,p}$. $\square$

Fig. 7

The plane triangulations $R_k^0$ and $R_k^1$ when $k=4$, where $C^1,C^2,C^3$ are in blue

Fig. 8

The plane triangulations $R_k^2$ and $R_k^3$ when $k=4$, where $C^1,C^2,C^3,C^4$ are in blue

Fig. 9

The plane triangulations $R_k^4$ and $R_k^5$ when $k=4$, where $C^1,C^2,C^3,C^4$ are in blue

Let $n:=6k+r$ for some $r\in \{0, 1, \dots , 5\}$ and integer $k\ge 2$. When $r=0$, let $R_k^0$ be the plane triangulation on n vertices obtained from $H_{k-2}\cup W_5^1\cup W_5^{k-2}$ by adding edges $u_{i,j}v_{i,j}$, $u_{i,j}v_{i,j+1}$ and $u_{i,6}v_{i,1}$ for all $i\in \{1,k-2\}$ and $j\in [5]$, where all arithmetic on the index $j+1$ here and henceforth is done modulo 5. When $r=1$, let $R_k^1$ be the plane triangulation on n vertices obtained from $H_{k-1}\cup W_5^{k-1}\cup K_1^1$ by adding $w_{1,1}u_{1,m}$ for any $m\in [6]$ and adding edges $v_{k-1,j}u_{k-1,j}$, $v_{k-1,j}u_{k-1,j+1}$ and $u_{k-1,6}v_{k-1,1}$ for all $j\in [5]$, where all arithmetic on the index $j+1$ here is done modulo 5. When $r=2$, let $R_k^2$ be the plane triangulation on n vertices obtained from $H_{k}\cup K_1^1\cup K_1^k$ by adding edges $u_{i,j}w_{i,1}$ for all $i\in \{1,k\}$ and $j\in [6]$. When $r=3$, let $R_k^3$ be the plane triangulation on n vertices obtained from $H_{k}\cup K_1^1\cup K_2^k$ by adding edges $u_{1,i}w_{1,1}$ and $u_{k,j}w_{k,1}$ and $u_{k,s}w_{k,2}$ for all $i\in [6]$, $j\in [4]$ and $s\in \{1,4,5,6\}$. When $r=4$, let $R_k^4$ be the plane triangulation on n vertices obtained from $H_{k}\cup K_2^1\cup K_2^k$ by adding edges $u_{i,j}w_{i,1}$ and $u_{i,s}w_{i,2}$ for all $i\in \{1,k\}$, $j\in [4]$ and $s\in \{1,4,5,6\}$. When $r=5$, let $R_k^5$ be the plane triangulation on n vertices obtained from $H_{k}\cup K_2^1\cup K_3^k$ by first adding edges $u_{1,j}w_{1,1}$ and $u_{1,s}w_{1,2}$ for all $j\in [4]$ and $s\in \{1,4,5,6\}$, and then joining $w_{k,1}$ to $u_{k, 1}, u_{k, 2},u_{k, 3}$; $w_{k,2}$ to $u_{k, 3}, u_{k, 4},u_{k, 5}$; $w_{k,3}$ to $u_{k, 1}, u_{k, 5},u_{k, 6}$. For all $r\in \{0,1,\ldots ,5\}$, the graphs $R_k^r$ when $k=4$ are depicted in Figs. 7-9. One can check that $R_k^{r}$ is $K_{2,t}$-free and so $ex_{_{\mathcal {P}}}(n, K_{2,t})=3n-6$ for all $t\in \{3,4\}$ and $n\ge 12$.

We then consider the case $t=4$ and $6\le n\le 11$. Let $Q_2$ be the plane graph defined in the proof of Theorem 1.6. Let $Q_2'$ be the plane triangulation on 9 vertices obtained from $Q_2{{\setminus }}v$ by adding edge $u_{2,1}u_{2,3}$; $Q_2''$ be the plane triangulation on 11 vertices obtained from $Q_2$ by adding one vertex w joining to $v,u_{2,1},u_{2,2}$. Then $Q_2',Q_2,Q_2''$ are $K_{2,4}$-free. Hence, $ex_{_{\mathcal {P}}}(n, K_{2,4})=3n-6$ for all $9\le n\le 11$. Let $G_1$ and $G_2$ be the unique plane graphs with degree sequence 5, 4, 4, 3, 3, 3 and 6, 4, 4, 4, 4, 3, 3 respectively. Then the plane graphs $G_1,G_2,Q_2'{{\setminus }}u$ are $K_{2,4}$-free. Hence, $ex_{_{\mathcal {P}}}(n, K_{2,4})\ge 3n-7$ for all $n\in \{6,7,8\}$. Clearly, for $n\in \{6,7\}$, every plane triangulation on n vertices is not $K_{2,4}$-free. Hence, $ex_{_{\mathcal {P}}}(n, K_{2,4})=3n-7$ for all $n\in \{6,7\}$. We next prove that $ex_{_{\mathcal {P}}}(n, K_{2,4})\le 3n-7$ when $n=8$. Suppose not. Let G be a $K_{2,4}$-free plane triangulation on n vertices. Let $v\in V(G)$ with $d_G(v)=\delta (G)$. Then $\delta (G)=4$, else $e(G)\ge {5n}/{2}>3n-6$ because $n=8$ when $\delta (G)\ge 5$, or $e(G)=e(G{{\setminus }}v)+d_G(v)\le 3(n-1)-7+3=3n-7$ when $\delta (G)\le 3$. Then there exists $u\in V(G){{\setminus }}N_G[v]$ such that $N_G(u)\cap N_G(v)\ge 3$. But then G contains a $K_{2,4}$ as a subgraph, a contradiction, as desired. Hence, $ex_{_{\mathcal {P}}}(n, K_{2,4})=3n-7$ when $n=8$.

It remains to consider the case $t=3$ and $5\le n \le 11$. Since $T_5^-$ has a copy of $K_{2,3}$ and $K_1+P_4$ is $K_{2,3}$-free, we have $ex_{_{\mathcal {P}}}(n, K_{2,3})=3n-8$ when $n=5$. Let $O_7'$ be the near 4-regular plane graph obtained from $O_7$ by adding edges between vertices of degree i, where $i\in \{2,3\}$; $O_8'$ be the 4-regular plane graph obtained from $O_8$ by adding edges between vertices of degree at most 3. Let J be the plane graph given in Fig. 10. Let $J'$ be the plane graph obtained from $J{{\setminus }}x_2$ by adding edge $x_1x_3$, $J''$ be the plane graph obtained from $J{{\setminus }}\{x_1,x_3\}$ by joining $x_2$ to $x_4,x_5$. Then the plane graphs $K_1+C_5,O_7',O_8',J'',J',J$ are $K_{2,3}$-free. Hence, $ex_{_{\mathcal {P}}}(n, K_{2,3})\ge 3n-8$ for all $6\le n\le 11$. We shall show that $ex_{_{\mathcal {P}}}(n, K_{2,3})\le 3n-8$ for all $6\le n\le 11$. Suppose this is not true. Let G be a $K_{2,3}$-free plane graph on n vertices with $e(G)\ge 3n-7$, where $6\le n\le 11$. We choose such a G with n minimum. Let $v\in V(G)$ with $d_G(v)=\delta (G)$. Then $\delta (G)\le 4$, else $e(G)\ge {5n}/{2}>3n-6$ because $n\le 11$, a contradiction. Next, if $\delta (G)\le 3$, then $e(G{{\setminus }}v)\le 3(n-1)-8$ by the minimality of n and the fact that $ex_{_{\mathcal {P}}}(n, K_{2,3})\le 3n-8$ when $n=5$. Thus, $e(G)=e(G{{\setminus }}v)+d_G(v)\le 3(n-1)-8+3=3n-8$, a contradiction. This proves that $\delta (G)=4$. We see G is not plane triangulation because G is $K_{2,3}$-free. So G has at least three vertices of degree four, else $e(G)\ge \lceil {(5n-2)}/{2}\rceil \ge 3n-6$ because $n\le 11$, a contradiction. This implies that G contains $K_1+C_4$ as a subgraph and so G is not $K_{2,3}$-free, a contradiction. Therefore, $ex_{_{\mathcal {P}}}(n, K_{2,3})=3n-8$ for all $6\le n\le 11$.

This completes the proof of Theorem 1.9.

Fig. 10

The graph J

Appendix

Proof of Lemma 2.1

Let G be given as in the statement. To establish the desired result, we first assume that there exists a vertex, say $v_1\in V(G)$, such that $v_1$ belongs to no triangle in G. Then $e(G[N_G[v_1]])=3$. Let $N_G(v_1)=\{v_2,v_3,v_4\}$ and $A=V(G){{\setminus }}N_G[v_1]$. Then $e_G(N_G(v_1),A)=6$ because G is a cubic graph. Hence, $e(G[A])=e(G)-e(G[N_G[v_1]])-e_G(N_G(v_1),A)=3$, which implies that $G[A]\in \{K_{1,3},P_{4},K_3\cup K_1\}$. Since G is cubic, we see $G[A]\ne K_3\cup K_1$ by the planarity of G. That is either $G[A]= K_{1,3}$ or $G[A]= P_{4}$. If $G[A]= K_{1,3}$ and $x\in A$ is the center of $K_{1,3}$, then $G{{\setminus }}\{v_1,x\} = C_6$ and so $G=G_1$. So we next assume that $G[A]= P_{4}$. Let G[A] be a path with vertices $v_5,v_6,v_7,v_8$ in order. We claim that $v_6$ and $v_7$ have no common neighbour in G. Suppose $v_6$ and $v_7$ have a common neighbour, say $v_2$. Then $N_G(v_3)=N_G(v_4)=\{v_1,v_5,v_8\}$ because G is a cubic graph. But then G has a $K_{3,3}$-minor with one part $\{v_2,v_3,v_4\}$ and the other part $\{v_1,\{v_5,v_6\},\{v_7,v_8\}\}$, a contradiction. Thus, $v_6$ and $v_7$ have no common neighbour in G. Without loss of generality, we may assume that $v_6v_2,v_7v_3\in E(G)$. Notice that $N_G(v_4)=\{v_1,v_5,v_8\}$ because G is a cubic graph. Then we see that either $v_5v_2\in E(G)$ or $v_5v_3\in E(G)$. If $v_5v_3\in E(G)$, then $v_8v_2\in E(G)$ and so G contains a $K_{3,3}$-minor with one part $\{v_2,v_3,v_4\}$ and the other part $\{v_1,\{v_5,v_6\},\{v_7,v_8\}\}$, a contradiction. Hence, $v_5v_2\in E(G)$ and so $v_8v_3\in E(G)$, which implies that $G=G_2$.

Next we assume that every vertex in G belongs to at least one triangle. We claim that there must exist a vertex such that it belongs to two triangles. Suppose that every vertex in G belongs to exactly one triangle. Let $v_1\in V(G)$. Then $G[N_G[v_1]]= K_1+(K_2\cup K_1)$. Let $A=V(G){{\setminus }}N_G[v_1]$. Then $e_G(N_G(v_1),A)=4$ because G is a cubic graph. Hence, $e(G[A])=e(G)-e(G[N_G[v_1]])-e_G(N_G(v_1),A)=4$, which implies that either $G[A]= C_4$ or $G[A]= K_1+(K_2\cup K_1)$. If $G[A]= C_4$, there exist two vertices in A which does not belong to any triangle, a contradiction. If $G[A]= K_1+(K_2\cup K_1)$, then there exists one vertex such that it belongs to either two triangles or no triangle, a contradiction. Thus, there must exist a vertex belonging to two triangles. Assume that $v_1$ belongs to two triangles in G. Since G is $K_4$-free, we have $G[N_G[v_1]]= K_4^-$. Let $A=V(G){{\setminus }}N_G[v_1]$. Then $e_G(N_G(v_1),A)=2$ because G is a cubic graph. Hence, $e(G[A])=e(G)-e(G[N_G[v_1]])-e_G(N_G(v_1),A)=5$, which implies that $G[A]= K_4^-$. Thus, $G=G_3$. $\square$

Proof of Lemma 2.2

Let G be given as in the statement. Let v be an arbitrary vertex in G and $V(G){{\setminus }}N_G[v]=\{v_1,v_2\}.$ Then $v_1v_2\in E(G)$, else $N_G(v_i)=N_G(v)$ for any $i\in [2]$, which yields that $G\cong K_{3,3}$. Since G is a cubic graph, we see that $|N_{N_G(v)(v_i)}|=2$ for any $i\in [2]$, and $N_{N_G(v)}(v_1)\notin N_{N_G(v)}(v_2)$. Hence, $G[N_G(v)]\cong K_2\cup K_1$, which implies that G is the graph given in Figure 2. $\square$