The Extremal Function for Two Disjoint Cycles

Abstract

A theta graph is the union of three internally disjoint paths that have the same two distinct end vertices. We show that every graph of order $n\ge 9$ and size at least $\lfloor \frac{7n-13}{2}\rfloor$ contains two disjoint theta graphs. We also show that every 2-edge-connected graph of order $n\ge 6$ and size at least $3n-5$ contains two disjoint cycles, such that any specified vertex with degree at least three belongs to one of them. The lower bound on size in both is sharp in general.

1 Introduction

All graphs considered are finite, simple, and undirected and we use Bondy and Murty [2] for terminology and notation not defined here. For a graph $G$, we denote its vertex set, edge set, minimum degree by $V(G)$, $E(G)$, and $\delta (G)$, respectively. The order and size of a graph $G$ is defined by $|V(G)|$ and $|E(G)|$, respectively. A set of subgraphs is said to be vertex-disjoint or independent if no two of them have any common vertex in $G$, and we use disjoint to stand for vertex-disjoint throughout this paper. If $u$ is a vertex of $G$ and $H$ is either a subgraph of $G$ or a subset of $V(G)$, we define $N_{H}(u)$ to be the set of neighbors of $u$ contained in $H$, and $d_{H}(u)=|N_{H}(u)|$. If $H^{\prime }$ is also a subgraph of $G$ with $V(H)\cap V(H^{\prime })=\emptyset$, we define $N(H^{\prime },H)=\cup _{x\in V(H^{\prime })}N_H(x)$. For a subset $U$ of $V(G)$, $G[U]$ denotes the subgraph of $G$ induced by $U$. In particularly, if the context is clear, we may also use $[U]$ for $G[U]$. If $S$ is a set of subgraphs of $G$, we write $G\supseteq S$, it means that $S$ is isomorphic to a subgraph of $G$, in particular, we use $mS$ to represent a set of $m$ vertex-disjoint copies of $S$. For a subgraph or subset $H$ of $G$, $G-H=[V(G)-V(H)]$. For two disjoint subsets or subgraphs $S$ and $T$ of $G$, we let $E(S,T)$ denote the set of edges of $G$ joining a vertex in $S$ and a vertex in $T$. When $S=\{x_{1},x_{2},\ldots ,x_{t}\}$, we may also use $[x_{1},x_{2},\ldots ,x_{t}]$ to denote $[\{x_{1},x_{2},\ldots ,x_{t}\}]$. When one of $S$ and $T$ contains a single vertex, say $S=\{x\}$, we write $E(x,T)$ for $E(S,T)$. Let $n$ be a positive integer, let $K_n$ denote the complete graph of order $n$ and $K_{4}^{-}$ denote the graph obtained by removing exactly one edge from $K_{4}$. Throughout this paper, we consider that any cycle has a fixed orientation. Let $C$ be a cycle of $G$. For $x,y\in V(C)$, we denote by $\overrightarrow{C}[x,y]$ a path from $x$ to $y$ on $\overrightarrow{C}$. The reverse sequence of $C[x,y]$ is denoted by $\overleftarrow{C}[y,x]$. We write $C[x,y]-\{x,y\}$, $C[x,y]-\{x\}$, $C[x,y]-\{y\}$ by $C(x,y)$, $C(x,y]$, and $C[x,y)$, respectively.

The research for the existence of subgraphs of a graph has been considered in many context. Perhaps the most investigated structures are cycle, forest, and chorded cycle, for example see [13]. Given a cycle $C$ of graph $G$, a chord of $C$ is an edge of $G-E(C)$ which joins two vertices of $C$. A chorded cycle is a cycle which contains at least one chord, and we use $\tau (C)$ to denote the number of chords in $C$. Pósa [11] posed the question for chorded cycles and he proved that any graph $G$ with $\delta (G)\ge 3$ contains a chorded cycle. In view of this, Bialostocki et al. [1] proposed the following natural common generalization problem, and proved by Chiba et al. [4].

Theorem 1.1

[4] Let $r$, $s$ be two nonnegative integers and let $G$ be a graph with $|V(G)|\ge 3r+4s$. Suppose for any pair of nonadjacent $u$ and $v$ in $G$, $d_{G}(u)+d_{G}(v)\ge 4r+6s-1$. Then $G$ contains $r+s$ disjoint cycles, such that $s$ of them are chorded cycles.

A theta graph is a union of three internally disjoint paths that have the same two distinct end vertices. A chorded cycle is a simple example of a theta graph, but in general a theta graph needs not be a chorded cycle. It is obvious that $K_{4}^{-}$ is the theta graph with minimum order. In particular, every theta graph contains an even cycle, and the idea of theta graphs has been studied in a wide variety of situations (see [3, 5, 8, 10]).

Our research is motivated by a classic extremal result, which is obtained by Pósa and mentioned by Erdős in [6]. Note that it is also a basic fact that every graph $G$ with order $n\ge 3$ and size at least $n$ contains a cycle.

Theorem 1.2

[6] Every graph of order $n\ge 6$ and size at least $3n-5$ contains two disjoint cycles.

Similarly, we are interested in the existence of disjoint theta graphs, since if a graph $G$ contains specified number of disjoint theta graphs, then $G$ also contains the same number of disjoint even cycles. For a graph $F$ of order $k$ and an integer $n\ge k$, the extremal number $ex(n;F)$ of $F$ is the maximum number of edges in a graph of order $n$ that does not contain $F$ as a subgraph. Given a cycle of even length, say $C_{2k}$ (here $k$ is a positive integer), Erdős [7] conjectured that $ex(n;C_{2k})=\Theta (n^{1+\frac{1}{k}})$ and this problem is considered to be one of the key problems in extremal graph theory. Although the efforts of many leading researchers had been made, the general proof of this conjecture is still open and we refer the reader [12] for further progression. However, for two disjoint cycles of even length but without specified length, we obtain the following result.

Theorem 1.3

Every graph of order $n\ge 8$ and size at least $f(n)$ contains two disjoint theta graphs, if

$$\begin{aligned} f(n)=\left\{ \begin{array}{l@{\quad }l} 23&{} \hbox {if }\quad n=8\\ \lfloor \frac{7n-13}{2}\rfloor &{}\hbox {if }\quad n\ge 9. \end{array}\right. \end{aligned}$$

When $n=8$, to see that the bound 23 presented in Theorem 1.3 is sharp, we construct the graph from $K_7$ and adding exactly one pendant edge, which has order $8$ and size $22$, but contains at most one theta graph. When $n\ge 9$, we construct the following examples: Let $n_{1}$ and $n_{2}$ be two integers with $n_{1}\ge 9$ and $n_{2}\ge 9$ such that $n_{1}$ is odd and $n_{2}$ is even, and let $l_{1}=\frac{n_{1}-3}{2}$ and $l_{2}=\frac{n_{2}-4}{2}$. Let $F=K_3$, $H_{1}=l_{1}K_{2}$, and $H_{2}=l_{2}K_{2}\cup K_{1}$, and let $G_i=F+H_i$ for each $i\in \{1,2\}$. It is obvious that the graph $G_i$ has order $n_i$ for each $i\in \{1,2\}$, $|E(G_{1})|=3+7l_{1}=\frac{7n_{1}-15}{2}=\lfloor \frac{7n_{1}-13}{2}\rfloor -1$ and $|E(G_{2})|=6+7l_{2}=\frac{7n_{1}-16}{2}=\lfloor \frac{7n_{2}-13}{2}\rfloor -1$. Furthermore, it follows from the construction of $G_{1}$ and $G_{2}$ that every theta graph in $G_i$ contains at least two vertices in $F$. Since $|V(F)|=3$, $G_i$ does not contain two disjoint theta graphs for each $i\in \{1,2\}$.

Corollary 1.4

Every graph of order $n\ge 8$ and size at least $f(n)$ contains two disjoint cycles of even length, if

$$\begin{aligned} f(n)=\left\{ \begin{array}{l@{\quad }l} 23&{}\hbox {if }\quad n=8\\ \lfloor \frac{7n-13}{2}\rfloor &{}\hbox {if }\quad n\ge 9. \end{array}\right. \end{aligned}$$

Our another motivation is Theorem 1.5 obtained by Bialostocki et al. [1], which determine the extremal number for the existence of two disjoint chorded cycles: Let $g(n)$ be the smallest number of edges in a graph of $n$ vertices that ensures the existence of two disjoint chorded cycles, Bialostocki et al. [1] obtained the following two theorems.

Theorem 1.5

[1] Every graph of order $n\ge 8$ and size at least $g(n)$ contains two disjoint chorded cycles, if

$$\begin{aligned} g(n)=\left\{ \begin{array}{ll} 23&{}\text{ if } \quad n=8\\ 25&{} \text{ if } \quad n=9\\ 28&{}\text{ if } \quad n=10\\ 32&{} \text{ if } \quad n=11\\ 5n-24&{}\text{ if } \quad n\ge 12. \end{array}\right. \end{aligned}$$

Theorem 1.6

[1] Let $G$ be a graph of order at least 8 and $\delta (G)\ge 6$, then $G$ contains two disjoint chorded cycles.

As a chorded cycle is a simple example of a theta graph, we can deduce the lower bound of edge condition in Theorem 1.5 to ensure the existence of disjoint theta graphs. In particular, Kawarabayashi [9] considered the minimum degree to ensure the existence of disjoint copies of $K_{4}^{-}$ in a general graph $G$, which can be seen the specified version of disjoint chorded cycles.

Theorem 1.7

[9] Let $k$ be a positive integer and $G$ be a graph with order $n\ge 4k$. If $\delta (G)\ge \frac{n+k}{2}$, then $G$ contains $k$ disjoint copies of $K_{4}^{-}$.

Finally, we are also interested in the following problem: Given a graph $G$ and let $u\in V(G)$ be any vertex in $V(G)$, determining the extremal number for the existence of two disjoint cycles in $G$, such that $u$ belongs to one of these two cycles.

Theorem 1.8

Let $G$ be a 2-edge-connected graph of order $n\ge 6$ and size at least $3n-5$. Then for each $u\in V(G)$ with degree at least three in $G$, $G$ contains two disjoint cycles, such that $u$ belongs to one of them.

The size bound of Theorem 1.8 is tight, which can been seen by the graph $K_{1,1,1,n-3}$, this graph does not contain two disjoint cycles and its size is $3n-6$. We show that 2-edge-connected condition is also necessary by following example: Let $n=3l+1$ with $l\ge 7$. Let $G_i\cong K_{l}$ for each $i\in \{1,2,3\}$. Then $G^{*}$ is obtained by attaching a vertex $u$ to $G_{1},G_{2}$, and $G_3$, such that $|E(u,V(G_i))|=1$ for each $i\in \{1,2,3\}$. It is obvious that $|E(G^{*})|=\frac{3l(l-1)+6}{2}>9l-2$, but $G^{*}$ does not contain two disjoint cycles such that the vertex of degree three belongs to one of these two cycles.

2 Proof of Theorem 1.3

If $n=8,9$, then Theorem 1.5 gives us the required conclusion. Hence, it is sufficient to prove that every graph of order $n\ge 10$ and size at least $\lfloor \frac{7n-13}{2}\rfloor$ contains two disjoint theta graphs. We employ induction on $n$.

Assume that for all integers $k$ with $9\le k<n$, every graph of order $k$ and size at least $\lfloor \frac{7k-13}{2}\rfloor$ contains two disjoint theta graphs. In the following proof, we always let $G$ be any graph of order $n$ and size at least $\lfloor \frac{7n-13}{2}\rfloor$. By the way of contradiction, we suppose that

$$\begin{aligned} G\,\,\text{ does } \text{ not } \text{ contain } \text{ two } \text{ disjoint } \text{ theta } \text{ graphs }. \end{aligned}$$

(1)

Claim 2.1

$4\le \delta (G)\le 5.$

Proof

By Theorem 1.6 and (1), $\delta (G)\le 5$. Suppose that $\delta (G)\le 3$ and let $u\in V(G)$ such that $d_{G}(u)=\delta (G)$. The graph $G-u$ is of order $n-1$ and size $\lfloor \frac{7n-13}{2}\rfloor -d_{G}(u)\ge \frac{7n-14}{2}-3=\frac{7(n-1)-13}{2}$, by induction hypothesis, $G-u$ contains two disjoint theta graphs, and so does $G$. This contradicts (1). Therefore, $\delta (G)\ge 4$. $\square$

Let $v_0$ be a vertex in $G$ such that $d_{G}(v_0)=\delta (G)$. In what following, we always assume that $N_G(v_0)=\{v_{1},\ldots ,v_l\}$ and $H=[v_{1},v_{2},\ldots ,v_l]$, where $l=d_{G}(v_0)$. By Claim 2.1, $4\le l\le 5$. If $l=4$, then let $\varepsilon _l=1$; if $l=5$, then let $\varepsilon _l=2$. Note that $l=3+\varepsilon _l$.

Claim 2.2

For each $1\le i\le l$, $d_{H}(v_i)\ge l-\varepsilon _l$.

Proof

Suppose that there exists $1\le i \le l$ such that $d_{H}(v_i)\le l-\varepsilon _l-1=(l-1)-\varepsilon _l$. Without loss of generality, we may assume that $i=l$, and we may also assume that $v_jv_l\notin E(G)$ for each $1\le j\le \varepsilon _l$ (Otherwise, we can relabel the index of $V(H)$). Define the edge set $X=\{v_jv_l:1\le j\le \varepsilon _l\}$ and construct the graph $G^{\prime }=(G-v_0)+X$, which is a graph with order $n-1$ and $|E(G^{\prime })|=\lfloor \frac{7n-13}{2}\rfloor -l+\varepsilon _l\ge \frac{7n-14}{2}-l+\varepsilon _l=\frac{7(n-1)-13}{2}$, because of $l=3+\varepsilon _l$. By induction hypothesis, $G^{\prime }$ contains two disjoint theta graphs, say $T_{1}$ and $T_{2}$. Clearly, at least one of $T_{1}$ and $T_{2}$, say $T_{1}$, does not contain vertex $v_l$, and of course, $E(T_{1})\cap X=\emptyset$. Then by (1), $E(T_{2})\cap X\not =\emptyset$.

Suppose that $|E(T_{2})\cap X|=1$. We may assume that $E(T_{2})\cap X=\{v_lv_{1}\}$. Then $T_{2}^{\prime }=(T_{2}-\{v_{1}v_l\})+\{v_{1}v_0,v_lv_0\}$ is a theta graph in $G$, and $T_{1}$ and $T_{2}^{\prime }$ are disjoint in $G$, which contradicts (1). Therefore, it remains the case $E(T_{2})\cap X=\{v_{1}v_l,v_{2}v_l\}$ as $\varepsilon _l\le 2$. Let

$$\begin{aligned} T_{2}^{\prime }=\left\{ \begin{array}{l@{\quad }l} (T_{2}-v_l)+\{v_0v_{1},v_0v_{2}\}, &{} {\mathrm{if }}\quad \,d_{T_{2}}(v_l)=2\\ (T_{2}-\{v_{1}v_l,v_{2}v_l\})+\{v_0v_{1},v_0v_l,v_0v_{2}\}, &{} {\mathrm{if }}\quad \,d_{T_{2}}(v_l)=3. \end{array}\right. \end{aligned}$$

Then it is obvious that $T_{1}$ and $T_{2}^{\prime }$ are two disjoint theta graphs in $G$, which contradicts (1). $\square$

By Claim 2.1 and the definition of $\varepsilon _l$, we have

$$\begin{aligned} \text{ for } \text{ each } \text{ subset }\,\,S\,\,\text{ of }\,\, V(H)\,\,\text{ with }\,\, |S|\ge 3,\, [\{v_0\}\cup S]\supseteq K_4^{-}, \end{aligned}$$

(2)

in particular,

$$\begin{aligned} \text{ If }\,\,l=4,\,\,\text{ then }\,\,[\{v_0\}\cup V(H)]\cong K_5. \end{aligned}$$

(3)

Now let $G^{*}=G-(V(H)\cup \{v_0\})$, and let ${\mathcal {F}}$ be the set of components of $G^{*}$. Since $[V(H)\cup \{v_0\}]\supseteq K_{4}^{-}$ by (2), it follows from (1) that every graph in ${\mathcal {F}}$ contains no theta graph.

Claim 2.3

$|V(F)|\le 2$ for each $F\in {\mathcal {F}}$.

Proof

Otherwise, suppose that $F\in {\mathcal {F}}$ and $|V(F)|\ge 3$. Since $F$ is a component of $G^{*}$ and $F$ contains no theta graph, each block of $F$ is either a $K_{2}$ or a cycle. Let $C$ denote the set of cut vertices of $F$.

We show that each block of $F$ is $K_{2}$. Otherwise, suppose that there exists a block $B$ of $F$, such that $B$ is a cycle. Assume for the moment that $B$ is an end block of $F$. Let $u_{1}$ and $u_{2}$ be two distinct vertices in $V(B)-C$. If $F=B$, then let $u_3\in V(F)-\{u_{1},u_{2}\}$; otherwise, $F$ contains at least two end blocks, let $u_3\in V(F)$ such that $u_3\notin C$ and $u_3$ belongs to some end block which is different from $B$. As $d_{F}(u_i)\le 2$ for each $i$ with $1\le i\le 3$, $|E(u_i,V(H))|\ge \delta (G)-2=l-2$ for each $i$ with $1\le i\le 3$. Since $4\le l\le 5$ by Claim 2.1, there exist $i,j$ with $1\le i,j\le 3$ and $i\not =j$ and a vertex $v\in V(H)$, such that $u_iv,u_jv\in E(G)$. Since $B$ is a cycle, it is easy to see that $[V(F)\cup \{v\}]$ contains a theta graph, and by applying (2), $[\{v_0\}\cup V(H)-\{v\}]$ contains a theta graph, that is, $G$ contains two disjoint theta graphs, which contradicts (1). Thus, $B$ is not a end block, and in particular, we see that every end block of $F$ is isomorphic to $K_{2}$. Therefore, we can take two distinct vertices $u_{1}$ and $u_{2}$, such that $u_{1},u_{2}\notin C$ and $u_{1}$ and $u_{2}$ belong to different end blocks of $F$, and there exists a path from $u_{1}$ to $u_{2}$ passing through at least two vertices in $V(B)\cap C$. Since $d_{F}(u_i)=1$ for each $1\le i\le 2$, it follows that $|E(u_i,V(H))|\ge l-1$ for each $i$ with $1\le i\le 2$. Hence, there exists a vertex $v\in V(H)$ such that $u_{1}v,u_{2}v\in E(G)$. Since $B$ is a cycle, it is easy to see that $[V(F)\cup \{v\}]$ contains a theta graph, as $[\{v_0\}\cup V(H)-\{v\}]$ also contains a theta graph by (2), $G$ contains two disjoint theta graphs, which contradicts (1). Thus, $F$ is a tree.

If there exist three distinct leaves in $V(F)$, say $u_{1},u_{2}$, and $u_3$, then likewise the proof as above, we obtain that $|E(u_i,V(H))|\ge l-1$ for each $1\le i\le 3$, this implies that there exists a vertex $v\in V(H)$, such that $u_iv\in E(G)$ for each $i$ with $1\le i\le 3$. Combining with (2), $G$ contains two disjoint theta graphs, which contradicts (1). Therefore, $F$ is exactly a path of order at least 3.

Let $u_{1}$ and $u_{2}$ be two end vertices in $F$ and let $u_3\in V(F)-\{u_{1},u_{2}\}$. Suppose that there exists $v\in V(H)$ such that $u_iv\in E(G)$ for each $i$ with $1\le i\le 3$, then by the similar arguments as above, $G$ contains two disjoint theta graphs, a contradiction. Therefore, since $|E(u_i,V(H))|\ge l-1$ for each $i$ with $1\le i\le 2$ and $|E(u_3,V(H))|\ge l-2$, we have $l=4$. Without loss of generality, we may assume that $N_{H}(u_{1})=\{v_{1},v_{2},v_3\}$, $N_{H}(u_{2})=\{v_{2},v_3,v_4\}$, and $N_{H}(u_3)=\{v_{1},v_4\}$. Then by (3), $[v_0,v_{1},v_4,u_3]\supseteq K_{4}^{-}$ and $[v_{2},v_3,u_{1},u_{2}]\supseteq K_{4}^{-}$, that is, $G\supseteq 2K_{4}^{-}$, which contradicts (1), this completes the proof of Claim 2.3. $\square$

Since $n\ge 10$ and $4\le |V(H)|\le 5$, it follows from Claim 2.3 that $|{\mathcal {F}}|\ge 2$.

Claim 2.4

For each graph $F\in {\mathcal {F}}$ such that $|V(F)|=2$, there exists $S\subset V(H)$ with $|S|=2$ and $[V(F)\cup S]\supseteq K_{4}^{-}$.

Proof

Let $F\in {\mathcal {F}}$ such that $|V(F)|=2$, label $V(F)=\{u_{1},u_{2}\}$. Since $|E(u_i,V(H))|\ge l-1$ for each $i$ with $1\le i\le 2$, it follows from the pigeonhole principle that there exists a subset $S\subset V(H)$ with $|S|=2$ and $S\subseteq N_{H}(u_{1})\cap N_{H}(u_{2})$. As $u_{1}u_{2}\in E(G)$, $[V(F)\cup S]\supseteq K_{4}^{-}$. $\square$

Claim 2.5

$|E(u,\{v_0\}\cup V(H))|=|E(u,V(H))|\le l-1$ for all $u\in V(G^{*})$.

Proof

Suppose that there exists $u\in V(G^{*})$ such that $|E(u,V(H))|\ge l$. Assume for the moment that there exists a graph $F\in {\mathcal {F}}$ such that $u\notin V(F)$ and $|V(F)|=2$. By Claim 2.4, there exists $S\subset V(H)$ such that $|S|=2$ and $[V(F)\cup S]\supseteq K_{4}^{-}$. Since $V(H)\subseteq N_{G}(u)$, it follows from (2) and (3) that $[\{v_0,u\}\cup (V(H)-S)]\supseteq K_{4}^{-}$, which contradicts (1). Hence, it follows from Claim 2.3 that $|V(F)|=1$ for each graph $F\in {\mathcal {F}}$ such that $u\notin V(F)$, in particular, by Claim 2.3, there exists at least two components $F_{1}$ and $F_{2}$ of $G^{*}$, such that $u\notin V(F_i)$ for each $i\in \{1,2\}$. Write $V(F_i)=\{u_i\}$ for each $i$ with $i\in \{1,2\}$. Then both $u_{1}$ and $u_{2}$ are adjacent to all vertices in $V(H)$. Hence by (2) and (3), we see that $[\{v_0,u,u_{1},u_{2}\}\cup V(H)]\supseteq 2K_{4}^{-}$, which contradicts (1). $\square$

By Claims 2.3 and 2.5, $|V(F)|=2$ for all $F\in {\mathcal {F}}$ and $\sum _{F\in {\mathcal {F}}}|E(F)|=\frac{n-1-l}{2}$. If $l=4$, then it follows from Claim 2.5 that $|E(u,V(H))|\le 3$ for all $u\in V(G^{*})$. Then we have

$$\begin{aligned} |E(G)|&= |E([\{v_0\}\cup V(H)])|+|E(V(G^{*}),\{v_0\}\cup V(H))|+\sum _{F\in {\mathcal {F}}}|E(F)|\\&\le 10+3|V(G^{*})|+\sum _{F\in {\mathcal {F}}}|E(F)|\\&= 10+3(n-5)+\frac{n-5}{2}\\&= \frac{7n-15}{2}, \end{aligned}$$

this is a obvious contradiction. Hence, $l=5$. Then for each $u\in V(G^{*})$, $|E(u,V(H))|=4$ by Claim 2.5 and the fact that $d_{G}(u)\ge l$, by combining (2) and Claim 2.4, we obtain that $G$ contains two disjoint theta graphs, which contradicts (1). This completes the proof of Theorem 1.3.

3 Proof of Theorem 1.8

If $n=6$, it is obvious that Theorem 1.8 is true. Hence, $n\ge 7$. We employ induction on $n$. Assume that for all integers $k$ with $6\le k<n$, every 2-edge-connected graph of order $k$ and size at least $3k-5$ contains two disjoint cycles, such that $u$ belongs to one of them, where $u$ is any specified vertex with degree at least three.

Let $G$ be a 2-edge-connected graph and with size at least $3n-5$, let $u\in V(G)$ denote any vertex with degree at least three in $G$ and fix it in this section. By the way of contradiction, we may assume that $G$ does not contain two disjoint cycles, such that $u$ belongs to one of them.

Claim 3.1

$d_{G}(x)\ge 3$ for each $x\in V(G-u)$.

Proof

By contradiction. Suppose that there exists $x\in V(G-u)$ such that $d_{G}(x)\le 2$. As $G$ is bridgeless and connected, thus, $G$ is 2-edge-connected and then $d_{G}(x)=2$. Let $u_{1},u_{2}\in N_{G}(x)$. Define $G^{*}=(G-x)+\{u_{1}u_{2}\}$ if $u_{1}u_{2}\notin E(G)$; Otherwise, define $G^{*}=G-x$.

We show that $G^{*}$ is a 2-edge-connected graph. Otherwise, $u_{1}u_{2}\in E(G)$ and $xu_{1}u_{2}x$ forms a block of $G$. If $xu_{1}u_{2}x$ is not an end block of $G$, then it is obvious that $G$ contains two disjoint cycles, such that $u$ belongs to one of them, a contradiction. Thus, we may assume that $xu_{1}u_{2}x$ is an end block of $G$. Without loss of generality, we may assume that $u_{1}$ is a cut vertex of $V(G)$. Suppose that $u\not =u_{2}$, then $n\ge 8$, otherwise, $n=7$ and $16\le |E(G)|\le 1+d_{G}(u_{1})+\frac{4\times 3}{2}\le 13$, a contradiction. If $u=u_{1},$ then remove the triangle $xu_{1}u_{2}x$, the order of the graph $G-\{x,u_{1},u_{2}\}$ is $n-3$ and its size is at least $3n-5-(n-1+1)=2n-5\ge n-3$, thus, $G-\{x,u_{1},u_{2}\}$ contains a cycle, that is, $G$ contains two desired disjoint cycles, a contradiction. Hence, $u\not =u_{1}$. Now consider $G-x-u_{2}$, which is a bridgeless graph of order $n-2$ and size $3n-8$, by induction hypothesis, $G-x-u_{2}$ contains two disjoint cycles, such that $u$ belongs to one of them, a contradiction. Hence, this forces $u=u_{2}$ and $d_{G}(u)=2$, a contradiction. This shows that $G^{*}$ is a 2-edge-connected graph indeed.

Now consider the above defined graph $G^{*}$, which is a 2-edge-connected graph with order $n-1$ and size is at least $3n-7=3(n-1)-4$, by induction hypothesis, $G^{*}$ contains two disjoint cycles, say $Q_{1}$ and $Q_{2}$, such that $u\in V(Q_{1})$. It is obvious that the edge $u_{1}u_{2}$ belongs to one of the $Q_{1}$ and $Q_{2}$, since if not, then $G-x$ contains two desired cycles and so does $G$, a contradiction. Without loss of generality, say $u_{1}u_{2}\in E(Q_{1})$, by replacing $u_{1}u_{2}$ in $Q_{1}$ by $u_{1}xu_{2}$, we obtain a new cycle $Q_{1}^{\prime }$, which disjoints $Q_{2}$, a contradiction. This proves Claim 3.1. $\square$

Let $\mathcal {B}$ denote the set of blocks of $G$.

Claim 3.2

$|\mathcal {B}|=1$.

Proof

Otherwise, suppose that $|\mathcal {B}|\ge 2$ and let $B_{1},B_{2}$ are two blocks in $G$. Since $G$ is a bridgeless graph, each block of $G$ is 2-connected. Without loss of generality, say $u\in V(B_{1})$ and $V(B_{1})\cap V(B_{2})\not =\emptyset$. Let $C_{1}$ be the cycle in $B_{1}$ such that $u\in V(C_{1})$. We may assume that $B_{2}$ is an end block of $G$, otherwise, let $B_3$ denote another end block which is different from $B_{1}$ (Note that $G$ contains at least two end blocks). It is obvious that $B_3$ contains a cycle, which disjoints from $C_{1}$, a contradiction. Therefore, for any $z\in V(B_{2})-V(B_{1})\cap V(B_{2})$, $d_{B_{2}-V(B_{1})\cap V(B_{2})}(z)\ge 2$, and $|V(B_{2})-V(B_{1})\cap V(B_{2})|\ge 3$ by applying Claim 3.1, therefore, $[V(B_{2})-V(B_{1})\cap V(B_{2})]$ contains a cycle, say $C_{2}$, which disjoints from $C_{1}$, then, $C_{1}$ and $C_{2}$ are two disjoint cycles in $G$, a contradiction. This completes the proof of Claim 3.2. $\square$

By Claim 3.2, $G$ is a 2-connected graph. Therefore, there exists a cycle in $G$, say $Q$, such that $u\in V(Q)$. Subject to this requirement, we choose $Q$ such that

$$\begin{aligned} \tau (Q)\,\,\text{ is } \text{ maximalized. } \end{aligned}$$

(4)

Claim 3.3

$\tau (Q)\ge 1$.

Proof

By the way of contradiction, we may assume that $\tau (Q)=0$. If $Q$ is a hamiltonian cycle in $G$, it follows from Claim 3.1 that $\tau (Q)\ge 1$, a contradiction. Hence, we may assume that $V(G-Q)\not =\emptyset$ and let $S^{\prime }$ denote any one component of $G-V(Q)$. If $|V(S^{\prime })|=1$, say $V(S^{\prime })=\{w\}$, then $|E(w,V(Q))|\ge 3$ by applying Claim 3.1, this implies that $[V(Q)\cup \{w\}]$ contains a chorded cycle containing $u$, by applying our choice (4), $\tau (Q)\ge 1$, a contradiction. Hence, $|V(S^{\prime })|\ge 2$. Since $G$ does not contain two disjoint cycles, such that $u$ belongs to one of them, $S^{\prime }$ is a tree and contains at least two leaves, say $u_{1}$ and $u_{2}$. Clearly, there exists a path in $S^{\prime }$ connecting $u_{1}$ and $u_{2}$, say $P$. By Claim 3.1, $|E(u_{1},V(Q))|\ge 2$ and $|E(u_{2},V(Q))|\ge 2$. Then $[V(Q\cup P)]$ contains a chorded cycle containing $u$, by applying our choice (4), $\tau (Q)\ge 1$, a contradiction once again. $\square$

Claim 3.4

There exists no triangle containing $u$, such that $d_{G}(u)=3$.

Proof

Otherwise, we assume that $T=up_{1}p_{2}u$ is a triangle in $G$, such that $d_{G}(u)=3$. Now, remove this triangle from $G$, we consider the graph $G-T$, which is a graph with order $n-3$ and size at least $3n-5-(2n-2)=n-3$, this implies that $G-T$ contains a cycle, and so $G$ contains two desired cycles, a contradiction.

Claim 3.5

$Q$ is not a hamiltonian cycle in $G$.

Proof

Otherwise, suppose that $Q$ is a hamiltonian cycle in $G$ and label $Q=uv_{1}v_{2}\ldots v_{n-1}u$ such that $u=v_0$ and $\overrightarrow{Q}$ is consistent with the increasing order of the indices of $v_i$ ($0\le i\le n-1$). As $d_{G}(u)\ge 3$, there exists $2\le t\le n-2$, such that $uv_{t}\in E(G)$. Then by our assumption, both $[\overrightarrow{Q}[v_{1},v_{t})]$ and $[\overrightarrow{Q}(v_{t},v_{n-1}]]$ are acyclic. Since $|E(G)|\ge 3n-5$, there exists $y\in V(Q)$, such that $d_{Q}(y)\ge 5$. Otherwise, the degree sum formula gives us

$$\begin{aligned} 6n-10\le 2|E(G)|=\sum _{x\in V(G)}d_{G}(x)\le 4n, \end{aligned}$$

which contradicts $n\ge 7$. Therefore, it is natural to consider the following three cases.

Case 1

$y=u=v_0$.

Suppose that there exist two integers $2\le a<t<b\le n-1$, such that $uv_{a},uv_b\in E(G)$. Then both of $[\overrightarrow{Q}[v_{1},v_b)]$ and $[\overrightarrow{Q}(v_{a},v_{n-1}]]$ are acyclic. By the same reason, $|E(\overrightarrow{Q}[v_{1},v_{a}],\overrightarrow{Q}(v_b,v_{n-1}])|\le 1$ and $|E(\overrightarrow{Q}[v_{1},v_{a}),\overrightarrow{Q}[v_b,v_{n-1}])|\le 1$. This implies that $|E(\overrightarrow{Q}[v_{1},v_{a}],\overrightarrow{Q}[v_b,v_{n-1}])|\le 3$. If the equality holds, then by Claim 3.1 and our assumption, $n=6$, a contradiction. Hence, $|E(\overrightarrow{Q}[v_{1},v_{a}],\overrightarrow{Q}[v_b,v_{n-1}])|\le 2$. Consequently,

$$\begin{aligned} 3n-5 \le |E(G)|&= n+d_{G}(u)+ |E(\overrightarrow{Q}[v_{1},v_{a}],\overrightarrow{Q}[v_b,v_{n-1}])|\nonumber \\&\le n+n-1+2 \nonumber \\&= 2n+1, \end{aligned}$$

(5)

which contradicts the fact that $n\ge 7$. Therefore, by symmetry, we may assume that there exist two integers $2\le a<b<t$, such that $uv_{a},uv_b\in E(G)$. Suppose that $v_{1}v_{t}\in E(G)$, then by our assumption and Claim 3.1, $v_{a}v_{n-1}\in E(G)$, this forces $n=6$, otherwise, by Claim 3.1, $G$ contains two desired cycles in each case, a contradiction. But this contradicts the fact that $n\ge 7$. Hence, $v_{1}v_{t}\notin E(G)$. By Claim 3.1 again, we may assume that there exists $t<c\le n-1$, such that $v_{1}v_c\in E(G)$. If $c\not =n-1$, then by our assumption and Claim 3.1, $E(v_{n-1},\overrightarrow{Q}[v_{1},v_{a}])\not =\emptyset$, say $v_{n-1}v_m\in E(G)$, where $1\le m\le a$. However, $v_{1}\overrightarrow{Q}[v_{1},v_m]v_{n-1}\overleftarrow{Q}[v_{n-1},v_c]v_{1}$ and $uv_b\overrightarrow{Q}[v_b,v_{t}]u$ are two disjoint cycles, a contradiction. Hence, $c=n-1$. Now, by the above arguments and Claim 3.1, $n=6$, otherwise, $G$ contains two desired cycles, a contradiction. But this contradicts the fact that $n\ge 7$. This completes the proof of Case 1.

Case 2

$y=v_{t}$.

In this case, our proof is similar with the proof of Case 1. Suppose that there exist two integers $1\le a<t<b\le n-1$, such that $v_{t}v_{a},v_{t}v_b\in E(G)$. Then $a\not =t-1$ and $b\not =t+1$. Now replace the segments $[\overrightarrow{Q}[v_{1},v_b)]$ and $[\overrightarrow{Q}(v_{a},v_{n-1}]]$ in Case 1 by $[\overrightarrow{Q}(v_{a},v_{t-1}]]$ and $[\overrightarrow{Q}[v_{t+1},v_b)]$, respectively, by the same arguments, we can obtain a contradiction. Therefore, by symmetry, we may assume that there exist two integers $1\le a<b<t$, such that $v_{t}v_{a},v_{t}v_b\in E(G)$. Suppose that $v_{t+1}v_b\in E(G)$, then $uv_{t-1}\in E(G)$ by our assumption and Claim 3.1, but this forces $n=6$, a contradiction. Hence, $v_{t+1}v_b\notin E(G)$ and we may assume that there exists $b<c\le t-1$, such that $v_{t+1}v_c\in E(G)$. If $c\not =t-1$, then by Claim 3.1, $E(v_{t-1},\overrightarrow{Q}[v_{t+1},u))\not =\emptyset$, say $v_{t-1}v_m\in E(G)$, where $t+1\le m\le n-1$. However, $v_c\overrightarrow{Q}[v_c,v_{t-1}]v_{m}\overleftarrow{Q}[v_{m},v_{t+1}]v_c$ and $v_{t}u\overrightarrow{Q}[u,v_{a}]v_{t}$ are two disjoint cycles, a contradiction. Hence, $c=t-1$. Now, by Claim 3.1, $n=6$, otherwise, $G$ contains two desired cycles, a contradiction. But this contradicts the fact that $n\ge 7$. This completes the proof of Case 2.

Case 3

$y\not =u$ and $y\not =v_{t}$.

By Case 2, we have $d_{Q}(v_{t})\le 4$. By symmetry, it suffices to consider the case $y\in V(\overrightarrow{Q}[v_{1},v_{t}))$. Firstly, we prove $|E(y,\overrightarrow{Q}(v_{t},u))|\ge 2$. Otherwise, it forces that $yu,yv_{t}\in E(G)$, $|E(y,\overrightarrow{Q}(v_{t},u))|\ge 1$, $y\not =v_{1}$, and $y\not =v_{t}$, as $d_{Q}(y)\ge 5$. As $d_{Q}(v_{1})\ge 3$ by Claim 3.1, this forces $v_{1}v_{t}\in E(G)$, otherwise, $G$ contains two desired cycles. However, we see that $d_{Q}(v_{t})\ge 5$, a contradiction. Secondly, we prove that $|E(y,\overrightarrow{Q}(v_{t},u))|=2$. Otherwise, there exists $t<a<b<c\le n-1$ such that $yv_{a},yv_b,yv_c\in E(G)$, see Fig. 1a. If $y\not =v_{1}$, then by our assumption, $E(v_{1},\overrightarrow{Q}(v_{a},u))=\emptyset$ and $v_{1}v_{t}\notin E(G)$. This implies that there exists $v_m\in V(\overrightarrow{Q}(v_{t},v_{a}])$, such that $v_{1}v_m\in E(G)$. However, $uv_{t}\overrightarrow{Q}[v_{t},v_m]v_{1}\overleftarrow{Q}[v_{1},u]$ and $yv_b\overleftarrow{Q}[v_b,v_c]y$ are two disjoint cycles, a contradiction. Hence, $y=v_{1}$ and $y=v_{t-1}$ by symmetry. This implies that $uyv_{t}u$ forms a triangle. By Claim 3.4, $d_{Q}(u)\ge 4$. See Fig.  1b, we can easily find two desired cycles, a contradiction.

Fig. 1

The structure in Case 3

Now, label $v_{a},v_b\in V(\overrightarrow{Q}(v_{t},u))$ such that $yv_{a},yv_b\in E(G)$, where $t<a<b\le n-1$. Since $d_{Q}(y)\ge 5$, then either $yu\in E(G)$ or $yv_{t}\in E(G)$. Suppose that $yu\in E(G)$, this implies that $y\not =v_{1}$, see Fig.  1c. By our assumption and Claim 3.1, $v_{1}v_{a}\in E(G)$ there at most six chords in $Q$, this gives us $3n-5\le |E(G)|\le n+6$, a contradiction. Hence, $yu\notin E(G)$ and so $yv_{t}\in E(G)$. Then again, $y\not =v_{t-1}$, $v_bv_{t-1}\in E(G)$ and there at most five chords in $Q$, this gives us $3n-5\le |E(G)|\le n+5$, a contradiction. This proves Case 3 and Claim 3.5. $\square$

By Claim 3.5, $G-V(Q)\not =\emptyset$. Let $S_{1}$ denote arbitrary one component of $G-V(Q)$. For convenience, label $Q=uv_{1}v_{2}\ldots v_qu$ such that $u=v_0$ and $\overrightarrow{Q}$ is consistent with the increasing order of the indices of $v_i$ ($0\le i\le q$). Note that $q\ge 3$ by Claim 3.3. By our assumption, $S_{1}$ is a tree.

Claim 3.6

For each $w\in V(S_{1})$ with $d_{S_{1}}(w)\le 1$, we have $d_{G}(w)\ge 4$.

Proof

Otherwise, suppose that no such vertex exists in $V(S_{1})$. Let $u^{\prime }\in V(S_{1})$ such that $d_{S_{1}}(u^{\prime })\le 1$, then $d_{G}(u^{\prime })=3$ by Claim 3.1. Let $v_l,v_j$ denote the neighbors of $u^{\prime }$ on $Q$, where $l<j$.

Suppose that $u\not =v_l$ and $u\not =v_j$, then consider the graph $G^{\prime }=G-u^{\prime }$, we prove that $G^{\prime }$ is 2-edge-connected. This is obviously true if $V(S_{1})=\{u^{\prime }\}$. Thus, we consider the case $|V(S_{1})|\ge 2$. If $S_{1}$ contains at least three leaves, we have nothing to prove, hence, $S_{1}$ contains exactly two leaves, which implies that $S_{1}$ is exactly a path. Now, by Claim 3.1, $G^{\prime }$ is 2-edge-connected. This implies that $G^{\prime }$ is a 2-edge-connected graph with order $n-1$ and size at least $3(n-1)-5$, by induction hypothesis, $G^{\prime }$ contains two disjoint cycles, such that $u$ belongs to one of them, and so does $G$, a contradiction. Hence, without loss of generality, we may assume that $u=v_l$. Then consider the graph $G^{\prime }=G-u^{\prime }+v_lv_j$ if $v_lv_j\notin E(G)$; otherwise, consider $G^{\prime }=G-u^{\prime }$, which is a 2-edge-connected graph of order $n-1$ and size at least $3(n-1)-5$ and $d_{G^{\prime }}(u)\ge 3$, by induction hypothesis, $G^{\prime }$ contains two disjoint cycles, such that $u$ belongs to one of them, then we can extend these two cycles to $G$ by replacing $v_lv_j$ by $v_lu^{\prime }v_j$, a contradiction. $\square$

Throughout the rest of this paper, choose any $u_{1}\in V(S_{1})$ such that $d_{S_{1}}(u_{1})\le 1$. By Claim 3.6, $|E(u_{1},V(Q))|\ge 3$.

Claim 3.7

$u$ is not incident with any chord of $Q$.

Proof

Otherwise, we may assume that there exists $2\le t\le q-1$, such that $uv_{t}\in E(G)$. Note that $|E(V(S_{1}),\overrightarrow{Q}[v_{1},v_{t-1}])|\le 1$, and $|E(V(S_{1}),\overrightarrow{Q}[v_{t+1},v_{q}])|\le 1$. Since $|E(u_{1},V(Q))|\ge 3$, then only three cases occur by symmetry, see Fig.  2, where $1\le a<t<b\le q$.

Fig. 2

The structure in Claim 3.7

Now, suppose that $|E(u_{1},V(Q))|\ge 4$. That is, $u_{1}u\in E(G)$ in Fig. 2a. Note that $\overrightarrow{Q}(v_{a},v_{t})$ (possibly $\overrightarrow{Q}(v_{a},v_{t})=\emptyset$) does not make contribution to the chords of $Q$, thus, replace $Q$ by $u_{1}v_{t}\overrightarrow{Q}[v_{t},v_{a}]u_{1}$, we arrive at a contradiction to (4). Hence, $|E(u_{1},V(Q))|=3$ and $|V(S_{1})|\ge 2$ by Claim 3.6. Now, choose another leave vertex in $S_{1}$, say $u_{2}$, such that $u_{1}\not =u_{2}$ and let $P$ be path connecting $u_{1}$ and $u_{2}$ in $S_{1}$. However, by our assumption and the symmetry role between $u_{1}$ and $u_{2}$, we have $d_{G}(u_{2})\le 3$ in Fig.  2a and c, which contradicts Claim 3.6. Now, it remains the case as in Fig.  2b. However, Note that $\overrightarrow{Q}(v_{a},v_{t})$ (possibly $\overrightarrow{Q}(v_{a},v_{t})=\emptyset$) does not make contribution to the chords of $Q$, thus, replace $Q$ by $u_{2}v_{t}\overrightarrow{Q}[v_{t},v_{a}]u_{1}Pu_{2}$, we arrive at a contradiction to (4). $\square$

Fig. 3

The structure between $u_{1}$ and $Q$

Since $\tau (Q)\ge 1$ by Claim 3.3, we may assume that there exists $1\le a<b\le q$, such that $v_{a},v_b\not =u$, and $v_{a}v_b\in E(G)$. Without loss of generality, we may assume that $v_0=u\in V(\overrightarrow{Q}(v_b,v_{a}))$. Since $d_{G}(u)\ge 3$, without loss of generality, by Claim 3.7, we may assume that $E(u,V(S_{1}))\not =\emptyset$. Note that $|E(V(S_{1}),\overrightarrow{Q}(v_{a},v_b))|\le 1$ by our assumption. In the following proof, when $|S_{1}|\ge 2$, we always assume that $u_{2}$ is another leave vertex in $S_{1}$, such that $u_{1}\not =u_{2}$. Let $z\in N(u,V(S_{1}))$.

Claim 3.8

$E(u_{1},\overrightarrow{Q}(v_b,u))=\emptyset$ and $E(u_{1},\overrightarrow{Q}(u,v_{a}))=\emptyset$.

Proof

Otherwise, suppose that there exists $v_c\in \overrightarrow{Q}(v_b,u)$, such that $u_{1}v_c\in E(G)$. If $z=u_{1}$, then $u_{1}\overrightarrow{Q}[v_c,u]u_{1}$ and $\overrightarrow{Q}[v_{a},v_b]v_{a}$ are two desired cycles, a contradiction. Hence, $z\not =u_{1}$. Since $S_{1}$ is a tree, there exists a path $P$ in $S_{1}$ connecting $z$ and $u_{1}$, then $zPu_{1}\overrightarrow{Q}[v_c,u]z$ and $\overrightarrow{Q}[v_{a},v_b]v_{a}$ are two desired cycles, a contradiction. $\square$

Claim 3.9

$z\not =u_{1}$.

Proof

By the way of contradiction. Suppose that $z=u_{1}$. If $V(S_{1})=\{u_{1}\}$, then by Claim 3.8 and our assumption, the situation between $u_{1}$ and $Q$ is as in Fig.  3a. Note that $\overrightarrow{Q}(u,v_{a})$ (possibly $\overrightarrow{Q}(u,v_{a})=\emptyset$) does not make contribution to the chords of $Q$, thus, replace $Q$ by $u_{1}\overrightarrow{Q}[v_{a},u]u_{1}$, we arrive at a contradiction to (4). Hence, $|V(S_{1})|\ge 2$, then by Claim 3.8 and our assumption, the neighbors of $u_{2}$ in $V(Q)$ is as in Fig.  3b, Note that $\overrightarrow{Q}(u,v_{a})$ (possibly $\overrightarrow{Q}(u,v_{a})=\emptyset$) does not make contribution to the chords of $Q$, thus, replace $Q$ by $u_{2}\overrightarrow{Q}[v_{a},u]u_{1}Pu_{2}$, we arrive at a contradiction to (4). $\square$

By Claim 3.9 and Claim 3.6, the situation between $u_{1}$ and $Q$ is as in Fig.  3c, where $1\le a<c<b\le q$. Note that $\overrightarrow{Q}(u,v_{a})$ (possibly $\overrightarrow{Q}(u,v_{a})=\emptyset$) does not make contribution to the chords of $Q$, thus, replace $Q$ by $u_{1}Pz\overleftarrow{Q}[u,v_{a}]u_{1}$, we arrive at a contradiction to (4). This completes the whole proof of Theorem 1.8.

4 Conclusion

In this paper, we determine the extremal number for a graph to contain two disjoint theta graphs, and we also determine the extremal number for a bridgeless graph to contain two disjoint cycles, such that any specified vertex belongs to one of them. As a natural extension, for any positive integer $k\ge 2$, we consider the extremal number of $k$ disjoint theta graphs and we conjecture as follows:

Conjecture 4.1

Let $k\ge 2$ be an integer. Every graph of order $n$ and size at least $f(n,k)+1$ contains $k$ disjoint theta graphs, when

$$\begin{aligned} f(n,k)&= \max \left\{ \left( {\begin{array}{c}4k-1\\ 2\end{array}}\right) \right. \\&\left. +\,(n-4k+1), \left\lfloor \frac{2(k-1)(2k-1)+(4k-1)(n-2k+1)}{2} \right\rfloor \right\} \end{aligned}$$

If Conjecture 4.1 is true, then the bound on size is best possible, which can be seen as follows: $G_{1}$ is obtained by $K_{4k-1}$ and an isolated vertex set of order $n-4k+1$, such that there exists a matching of size $4k-1$ between $K_{4k-1}$ and the isolated vertex set. The order of $G_{1}$ is $n$ and size $\left( {\begin{array}{c}4k-1\\ 2\end{array}}\right) +(n-4k+1)$, but $G_{1}$ does not contain $k$ disjoint theta graphs. Also, let $n$ be an integer such that $n-(2k-1)$ is even. Let $l_{1}=\frac{n-(2k-1)}{2}$, $F=K_{2k-1}$, $H_{1}=l_{1}K_{2}$, and $G_{2}=F+H_{1}$. It is obvious that the graph $G_{2}$ has order $n$, $|E(G_{1})|=(k-1)(2k-1)+(4k-1)l_{1}=(k-1)(2k-1)+\frac{(4k-1)(n-2k+1)}{2}=\left\lfloor \frac{2(k-1)(2k-1)+(4k-1)(n-2k+1)}{2} \right\rfloor$. Clearly, $G_{2}$ does not contain $k$ disjoint theta graphs.

Note that Theorem 1.3 implies that Conjecture 4.1 is true for $k=2$.