Anti-Ramsey Problems in Complete Bipartite Graphs for t Edge-Disjoint Rainbow Spanning Trees

Abstract

Let $r(K_{p,q},t)$ be the maximum number of colors in an edge-coloring of the complete bipartite graph $K_{p,q}$ not having t edge-disjoint rainbow spanning trees. We prove that $r(K_{p,p},1)=p^2-2p+2$ for $p\ge 4$ and $r(K_{p,q},1)=pq-2q+1$ for $p>q\ge 4$. Let $t\ge 2$. We also show that $r(K_{p,p},t)=p^2-2p+t+1$ for $p \ge 2t+\sqrt{3t-3}+4$ and $r(K_{p,q},t)=pq-2q+t$ for $p > q \ge 2t+\sqrt{3t-2}+4$.

Appendix

Label the vertices in $V(G)-V(H)$ as $x_1,\ldots , x_{\ell _1}\in X{\setminus } V(H)$ and $y_1,\ldots , y_{\ell _2}\in Y{\setminus } V(H)$ such that (1) $|N_G(x_1)\cap V(H)|\ge \cdots \ge |N_G(x_{\ell _1-\ell _2+1})\cap V(H)| \ge \max \{ |N_G(x_{i})\cap V(H)|: \ell _1-\ell _2+2 \le i \le \ell _1\}$ (For vertex $x_i$, if i is greater than $\ell _1$, we ignore the corresponding item); (2) $y_i=\text{ arg }\max \{| N_G(y)\cap (V(H)\cup \{x_1,\ldots ,x_{\ell _1-\ell _2+i}\})|:y\in Y{\setminus } (V(H) \cup \{y_1,\ldots ,y_{i-1}\})\}$ for $1\le i\le \ell _2$ (We ignore (2), if $\ell _2=0$); (3) $x_{\ell _1-\ell _2+i}=\text{ arg }\max \{|N_G(x)\cap (V(H)\cup \{y_1,\ldots ,y_{i-1}\})|:x\in X{\setminus } (V(H) \cup \{x_1,\ldots ,x_{\ell _1-\ell _2+i-1}\})\}$ for $2\le i\le \ell _2$ (We ignore (3), if $\ell _2 \le 1$). That is we firstly choose $x_i \ (1 \le i \le \ell _1-\ell _2+1)$ in $X{\setminus } V(H)$ based on (1); then choose $y_i \ (1 \le i \le \ell _2)$ and $x_{\ell _1-\ell _2+i} \ (2 \le i \le \ell _2)$ alternately such that (2) and (3) hold.

Let $d_i^*=|N_G(x_i)\cap V(H)|$ for $1\le i\le \ell _1-\ell _2+1$, $d_{\ell _1-\ell _2+i}^*=|N_G(x_{\ell _1-\ell _2+i})\cap (V(H)\cup \{y_1,\ldots ,y_{i-1}\})|$ for $2\le i\le \ell _2$ and $D_j^*=|N_G(y_j)\cap (V(H)\cup \{x_1,\ldots ,x_{\ell _1-\ell _2+j}\})|$ for $1 \le j \le \ell _2$. Then we have $d_1^*\ge \cdots \ge d_{\ell _1-\ell _2+1}^*$ and $D_j^*\le (2t-1)+\ell _1-\ell _2+j$ for $1 \le j \le \ell _2$. We first have the following claims.

Claim $1'$. If $\ell _1\ge 3$, then $d_i^* \ge t+1$ for $1 \le i \le \ell _1-2$.

Proof of Claim $1'$. We first consider the case when $\ell _2 \le 1$. To the contrary, suppose that $d_{\ell _1-2}^* \le t$. Then

$$\begin{aligned} |E(G)| \le (p-\ell _1)q + (\ell _1-3)(2t-1)+3t+\ell _1\ell _2 \le (-q+2t)\ell _1 + pq-3t+3. \end{aligned}$$

Since $\ell _1 \ge 3$ and $q \ge 2t+\sqrt{3t-2}+4$, we have $|E(G)| \le pq-3q+3t+3$, a contradiction with the fact $|E(G)| = pq-2q+t+1$.

We thus assume that $\ell _2 \ge 2$ in the rest of the proof. First, if $\ell _1>\ell _2$, we show that $d_{\ell _1-\ell _2}^*\ge t+1$. Suppose that $d_{\ell _1-\ell _2}^*\le t$. Then

$$\begin{aligned} |E(G)|&\le (p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2-1)(2t-1)+t\\&+\sum _{j=1}^{\ell _2}(t+j-1)+\sum _{j=1}^{\ell _2}(2t-1+\ell _1-\ell _2+j)\\&= (p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1)+\ell _2(3t+\ell _1-1)-t+1. \end{aligned}$$

Recall that $|E(G)|=pq-2q+t+1$. So we should have

$$\begin{aligned}&f(\ell _1,\ell _2):=(p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1)+\ell _2(3t+\ell _1-1)\\&-t+1-(pq-2q+t+1)\ge 0. \end{aligned}$$

Consider the partial derivative of $f(\ell _1,\ell _2)$ with respect to $\ell _1$ and $\ell _2$. We have

$$\begin{aligned} \frac{\partial f(\ell _1,\ell _2)}{\partial \ell _1}= & {} 2\ell _2+2t-q-1,\\ ~~ \frac{\partial f(\ell _1,\ell _2)}{\partial \ell _2}= & {} 2\ell _1+t-p. \end{aligned}$$

Since $2 \le \ell _2 < \ell _1 \le p-2t$ and $\ell _2 \le q-2t < p-2t$, we have

$$\begin{aligned} f(\ell _1,\ell _2) \le \max \{f(p-2t,2),f(p-2t,q-2t),f(\ell _2+1,\ell _2) \} . \end{aligned}$$

Simple computation shows that

$$\begin{aligned} f(p-2t,2)&=(-q+2t+1)p+2tq-4t^2+2q-6t\\&\le (-q+2t+1)(q+1)+2tq-4t^2+2q-6t = -(q-2t-1)^2+2< 0,\\ f(p-2t,q-2t)&= -p-qt+2q+2t^2 \le -(q+1)-qt+2q+2t^2=(1-t)q+2t^2-1\\&\le (1-t)\Big (2t+\sqrt{3t-2}+3 \Big )+2t^2-1=(1-t)\sqrt{3t-2}+3-2t<0 \end{aligned}$$

and

$$\begin{aligned} f(\ell _2+1,\ell _2)= 2\ell _2^2+(-p-q+3t+1)\ell _2+q-1. \end{aligned}$$

Since $2 \le \ell _2 \le q-2t$, we have $f(\ell _2+1,\ell _2) \le \max \{ f(3,2), f(q-2t+1,q-2t)\}$. However $f(3,2) =-2p-q+6t+9 <0$ and

$$\begin{aligned} f(q-2t+1,q-2t)&= (-q+2t)p+q^2-3qt+2t^2+2q-2t-1 \\&\le (-q+2t)(q+1)+q^2-3qt+2t^2+2q-2t-1=(-t+1)q+2t^2-1 \\&\le (-t+1)(2t+4)+2t^2-1=-2t+3 <0. \end{aligned}$$

Thus $f(\ell _1,\ell _2) < 0$, a contradiction.

Now we will show that $d_i^* \ge t+1$ for $\ell _1-\ell _2+1 \le i \le \ell _1-2$. In this case, $\ell _2\ge 3$. To the contrary, we assume that $d_{\ell _1-\ell _2+i_0}^* \le t$ for some $1 \le i_0 \le \ell _2-2$. The choice of $x_{i_0}$ yields $d_{\ell _1-\ell _2+i_0+j}^* \le t+j$ for $0 \le j \le \ell _2-i_0$. Then we have

$$\begin{aligned} |E(G)|&\le (p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1)\\&+\sum _{j=0}^{i_0-2}(2t-1+j)+\sum _{j=0}^{\ell _2-i_0}(t+j)+\sum _{j=1}^{\ell _2}(2t-1+\ell _1-\ell _2+j)\\&= (p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1)+\ell _2(3t-i_0+\ell _1)+i_0^2+(i_0-1)t-3i_0+2. \end{aligned}$$

Let $g(i_0)=(p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1) +\ell _2(3t-i_0+\ell _1)+i_0^2+(i_0-1)t-3i_0+2$. Then $g'(i_0)=-\ell _2+2i_0+t-3$ and $g''(i_0)=2$. Thus $g(i_0)\le \max \{g(1),g(\ell _2-2)\}$. Denote

$$\begin{aligned} f_1(\ell _1,\ell _2)=g(1)=(p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1)+\ell _2(3t-1+\ell _1) \end{aligned}$$

and

$$\begin{aligned} f_2(\ell _1,\ell _2)=g(\ell _2-2)= & {} (p-\ell _1)(q-\ell _2)\\&+(\ell _1-\ell _2)(2t-1)+\ell _2(4t-5+\ell _1)-3t+12. \end{aligned}$$

Recall that $|E(G)|=pq-2q+t+1$. Therefore we should have $F_i(\ell _1,\ell _2)=f_i(\ell _1,\ell _2)-(pq-2q+t+1)\ge 0$ for $i=1,2$. Consider the partial derivative of $F_i(\ell _1,\ell _2)$ with respect to $\ell _1$ and $\ell _2$ for $i=1,2$. We have

$$\begin{aligned} \frac{\partial F_i(\ell _1,\ell _2)}{\partial \ell _1} = 2\ell _2+2t-q-1, \frac{\partial F_i(\ell _1,\ell _2)}{\partial \ell _2} = 2\ell _1+t-p+s, \end{aligned}$$

where $s=0$ if $i=1$ and $s=t-4$ if $i=2$. Since $3 \le \ell _2 \le \ell _1 \le p-2t$ and $\ell _2 \le q-2t < p-2t$, for $i=1,2$, we have

$$\begin{aligned} F_i(\ell _1,\ell _2) \le \max \{F_i(p-2t,3),F_i(p-2t,q-2t),F_i(\ell _2,\ell _2)\} . \end{aligned}$$

Simple computation shows that

$$\begin{aligned} F_1(p-2t,3)&=F_2(p-2t,3) \\&= (-q+2t+2)p+2qt-4t^2+2q-8t-1 \\&\le (-q+2t+2)(q+1)+2qt-4t^2+2q-8t-1\\&= -\Big (q-2t-\frac{3}{2}\Big )^2+\frac{13}{4}<0,\\ F_1(p-2t,q-2t)&= -p-qt+2q+2t^2+t-1 \le -(q+1)-qt+2q+2t^2+t-1\\&=(1-t)q+2t^2+t-2 \le (1-t)(2t+\sqrt{3t-2}+3)+2t^2+t-2\\&=(1-t)\sqrt{3t-2}+1<0,\\ F_2(p-2t,q-2t)&= -p-2q+6t+11 <0,\\ F_1(\ell _2,\ell _2)&= 2\ell _2^2+(-p-q+3t-1)\ell _2+2q-t-1 \end{aligned}$$

and

$$\begin{aligned} F_2(\ell _2,\ell _2) = 2\ell _2^2+(-p-q+4t-5)\ell _2+2q-4t+11. \end{aligned}$$

Since $3 \le \ell _2 \le q-2t$, we have $F_i(\ell _2,\ell _2) \le \max \{F_i(3,3),F_i(q-2t,q-2t)\}$. However, $F_i(3,3)= -3p-q+8t+14 <0$ for $i=1,2$,

$$\begin{aligned}&F_1(q-2t,q-2t) \\&= (-q+2t)p+q^2-3qt+2t^2+q+t-1 \\&\le (-q+2t)(q+1)+q^2-3qt+2t^2+q+t-1 \\&=-qt+2t^2+3t-1 \le -t(2t+3)+2t^2+3t-1=-1<0,\\ F_2(q-2t,q-2t)&= (-q+2t)p+q^2-2qt-3q+6t+11 \\&\le (-q+2t)(q+1)+q^2-2qt-3q+6t+11=-4q+8t+11<0 \end{aligned}$$

Thus $F_i(\ell _1,\ell _2) < 0$ for $i=1,2$, a contradiction. $\square$

Claim $2'$. If $\ell _2\ge 3$, then $D_i^* \ge t+1$ for $1 \le i \le \ell _2-2$.

Proof of Claim $2'$. To the contrary, we assume that $D_{i_0}^* \le t$ for some $1 \le i_0 \le \ell _2-2$. Note that $D_{i_0+j}^* \le D_{i_0}^*+j$ for $0 \le j \le \ell _2-i_0$ and $d_{\ell _1-\ell _2+j}^* \le (2t-1)+j-1$ for $1 \le j \le \ell _2$. Hence we have

$$\begin{aligned} |E(G)|&\le (p-\ell _1)(q-\ell _2)+(\ell _1-\ell _2)(2t-1)+\sum _{j=0}^{\ell _2-i_0}(t+j)\\&\quad +\sum _{j=1}^{i_0-1}(2t-1+\ell _1-\ell _2+j)+\sum _{j=1}^{\ell _2}(2t-1+j-1)\\&=(p-\ell _1)(q-\ell _2)+(2\ell _1-1)(t-1)+\ell _2(t+1)+i_0(t-2+\ell _1-2\ell _2)+\ell _2^2+i_0^2. \end{aligned}$$

Let $g(i_0)=(p-\ell _1)(p-\ell _2)+(2\ell _1-1)(t-1)+\ell _2(t+1)+i_0(t-2+\ell _1-2\ell _2) +\ell _2^2+i_0^2$. By the same argument as above, we have $g(i_0)\le \max \{g(1),g(\ell _2-2)\}$. Denote

$$\begin{aligned} f_1(\ell _1,\ell _2)=g(1)=(p-\ell _1)(q-\ell _2)+\ell _1(2t-1)+\ell _2(t-1+\ell _2) \end{aligned}$$

and

$$\begin{aligned} f_2(\ell _1,\ell _2)=g(\ell _2-2)=(p-\ell _1)(q-\ell _2)+2\ell _1(t-2)+\ell _2(2t-1+\ell _1)-3t+9. \end{aligned}$$

Recall that $|E(G)|=pq-2q+t+1$. Therefore we should have $F_i(\ell _1,\ell _2)=f_i(\ell _1,\ell _2)-(pq-2q+t+1)\ge 0$ for $i=1,2$.

For $F_1(\ell _1,\ell _2)$, it is easy to obtain that

$$\begin{aligned} \frac{\partial ^2 F_1(\ell _1,\ell _2)}{\partial \ell _2^2}=2. \end{aligned}$$

Moreover, $3 \le \ell _2 \le \ell _1 \le p-2t$ and $\ell _2\le q-2t$, it follows that $F_1(\ell _1, \ell _2) \le \max \{ F_1(\ell _1, 3), F_1(\ell _1, \min \{\ell _1, q-2t \})\}$. Simple computation shows that

$$\begin{aligned} F_1(\ell _1, 3)&= (-q+2t+2)\ell _1-3p+2q+2t+5\\&\le 3(-q+2t+2)-3p+2q+2t+5\\&=-3p-q+8t+11 <0; \end{aligned}$$

if $\ell _1 \ge q-2t$, then

$$\begin{aligned} F_1(\ell _1, q-2t)&= -\ell _1-pq+2pt+q^2-3qt+2t^2+q+t-1\\&\le -(q-2t)-pq+2pt+q^2-3qt+2t^2+q+t-1 =(-q+2t)p+q^2-3qt+2t^2+3t-1\\&\le (-q+2t)(q+1)+q^2-3qt+2t^2+3t-1 = -q(t+1)+2t^2+5t-1 \\&\le -(2t+\sqrt{3t-2}+3)(t+1)+2t^2+5t-1=-(t+1)\sqrt{3t-2}-4<0; \end{aligned}$$

if $\ell _1 < q-2t$, then

$$\begin{aligned} F_1(\ell _1, \ell _1)&= 2\ell _1^2+(-p-q+3t-2)\ell _1+2q-t-1, \end{aligned}$$

since $3 \le \ell _1 < q-2t$, we have $F_1(\ell _1, \ell _1) \le \max \{F_1(3, 3), F_1(q-2t, q-2t)\}$. However $F_1(3, 3) =-3p-q+8t+11 < 0$ and

$$\begin{aligned} F_1(q-2t, q-2t)&=(-q+2t)p+q^2-3qt+2t^2+3t-1 \\&\le (-q+2t)(q+1)+q^2-3qt+2t^2+3t-1 = (-t-1)q+2t^2+5t-1 \\&\le (-t-1)(2t+3)+2t^2+5t-1 = -4 < 0. \end{aligned}$$

Thus $F_1(\ell _1, \ell _2) <0$, a contradiction.

For $F_2(\ell _1,\ell _2)$, it is easy to obtain that

$$\begin{aligned} \frac{\partial F_2(\ell _1,\ell _2)}{\partial \ell _1}= & {} 2\ell _2-q+2t-4,\\ \frac{\partial F_2(\ell _1,\ell _2)}{\partial \ell _2}= & {} 2\ell _1-p+2t-1 . \end{aligned}$$

Moreover, $3 \le \ell _2 \le \ell _1 \le p-2t$ and $\ell _2\le q-2t$, it follows that $F_2(\ell _1, \ell _2) \le \max \{ F_2(p-2t, 3), F_2(p-2t, q-2t), F_2(\ell _2, \ell _2)\}$. Simple computation shows that

$$\begin{aligned} F_2(p-2t, 3)&= (-q+2t-1)p+2qt-4t^2+2q-2t+5\\&\le (-q+2t-1)(q+1)+2qt-4t^2+2q-2t+5\\&=-(q-2t)^2+4<0.\\ F_2(p-2t, q-2t)&= -4p+q+6t+8 <0 \end{aligned}$$

and

$$\begin{aligned} F_2(\ell _2, \ell _2)&= 2\ell _2^2+(-p-q+4t-5)\ell _2+2q-4t+8. \end{aligned}$$

Since $3 \le \ell _2 \le q-2t$, we have $F_2(\ell _2, \ell _2) \le \max \{F_2(3, 3), F_2(q-2t, q-2t) \}$. However, $F_2(3, 3)=-3p-q+8t+11 < 0$ and

$$\begin{aligned} F_2(q-2t, q-2t)&= (-q+2t)p+q^2-2qt-3q+6t+8 \\&\le (-q+2t)(q+1)+q^2-2qt-3q+6t+8=-4q+8t+8 <0. \end{aligned}$$

Thus $F_2(\ell _1, \ell _2) <0$, a contradiction. $\Box$

Denote $A=\{x_{\ell _1-1}, x_{\ell _1}\}$ and $B=\{ y_{\ell _2-1}, y_{\ell _2}\}$. In the next, if the subscript of the vertex x or y is not a positive integer, we mean that such vertex does not exist. For example, $B=\emptyset$ if $\ell _2=0$. Let $G'=G-(A\cup B)$. Since H is 2t-edge-connected, Lemma 5 implies that H contains t edge-disjoint rainbow spanning trees. By Claims 3 and 4, we can iteratively add $x_1,\ldots ,x_{\ell _1-\ell _2}$,$x_{\ell _1-\ell _2+1},y_1,x_{\ell _1-\ell _2+2}, y_2,\ldots ,x_{\ell _1-2}, y_{\ell _2-2}$ to H to obtain t edge-disjoint rainbow spanning trees of $G'$; call them $T_1, \ldots , T_{t}$. If $d^*(x) \ge t$ for any $x\in A$ and $D^*(y) \ge t$ for any $y\in B$, we can further augment $T_1, \ldots , T_{t}$, by adding t edges incident to $x_{\ell _1-1}$, then incident to $y_{\ell _2-1}$, $x_{\ell _1}$ and $y_{\ell _2}$ iteratively to obtain t edge-disjoint rainbow spanning trees of $K_{p,q}$. So we assume that at least one of $d_{\ell _1-1}^*, d_{\ell _1}^*,D_{\ell _2-1}^*,D_{\ell _2}^*$ is less than t in the following proof. Let $m=\#\{s\le t|s\in \{d_{\ell _1-1}^*, d_{\ell _1}^*,D_{\ell _2-1}^*,D_{\ell _2}^*\}\}$. Then $m\ge 1$.

Claim $3'$. $m\le 2$.

Proof of Claim $3'$. Suppose $m\ge 3$. The degree sum of those three corresponding vertices in G is less than and equal to $t+2+2(t+1)=3t+4$. Furthermore, there are at most two edges contributing 2 to the degree sum of these three vertices in $K_{p,q}$. Therefore, we have $|E(K_{p,q})|-|E(G)| \ge 2q+p-(3t+4)-2=2q+p-3t-6$. However, $|E(K_{p,q})|-|E(G)| = 2q-t-1$, a contradiction. $\square$

Claim $4'$. If $m=2$ and $\ell _2\ge 2$, then $\max \{d_{\ell _1-1}^* ,D_{\ell _2-1}^*\} \ge t+1$, $\max \{d_{\ell _1-1}^* ,D_{\ell _2}^*\} \ge t+1$ and $\max \{d_{\ell _1}^* ,D_{\ell _2-1}^*\} \ge t+1$. If $m=2$ and $\ell _2=1$, then $\max \{d_{\ell _1-1}^* ,D_{1}^*\} \ge t+1$.

Proof of Claim $4'$. We only consider the case when $m=2$ and $\ell _2\ge 2$. The other case when $m=2$ and $\ell _2=1$ is similar.

Clearly, $d_{\ell _1}^* \le d_{\ell _1-1}^*+1$, $D_{\ell _2}^* \le D_{\ell _2-1}^*+1$. Since $m=2$ and at least one of $d_{\ell _1-1}^*, d_{\ell _1}^*,D_{\ell _2-1}^*,D_{\ell _2}^*$ is less than t, we thus obtain $\max \{d_{\ell _1-1}^* ,D_{\ell _2-1}^*\} \ge t+1$.

Suppose $\max \{d_{\ell _1-1}^* ,D_{\ell _2}^*\} \le t$. Then $d_{\ell _1}^* \ge t+1$ and $D_{\ell _2-1}^*\ge t+1$ as $m=2$. Note that at least one of $d_{\ell _1-1}^*$ and $D_{\ell _2}^*$ is less than t. We have $d_{\ell _1-1}^* =t$, $d_{\ell _1}^* =t+1$ and $D_{\ell _2}^* \le t-1$. It follows that $d_{G}(x_{\ell _1-1}) \le d_{\ell _1-1}^*+2=t+2$, $d_{G}(x_{\ell _1}) \le d_{\ell _1}^*+1=t+2$ and $d_{G}(y_{\ell _2}) = D_{\ell _2}^* \le t-1$. Therefore $d_{K_{p,q}}(x_{i}) - d_G(x_{i}) \ge q-t-2$ for $i =\ell _1-1, \ell _1$ and $d_{K_{p,q}}(y_{\ell _2}) - d_G(y_{\ell _2}) \ge p-t+1$, which implies that $|E(K_{p,q})|-|E(G)| \ge 2(q-t-2)+(p-t+1)-2=2q+p-3t-5$. However $|E(K_{p,q})|-|E(G)|=pq- (pq-2q+t+1) =2q-t-1 < 2q+p-3t-5$, a contradiction.

Suppose $\max \{d_{\ell _1}^* ,D_{\ell _2-1}^*\} \le t$. Then $D_{\ell _2}^* \ge t+1$ and $d_{\ell _1-1}^* \ge t+1$. Note that at least one of $d_{\ell _1}^*$ and $D_{\ell _2-1}^*$ is less than t, we have $D_{\ell _2-1}^* =t$, $D_{\ell _2}^* =t+1$ and $d_{\ell _1}^* \le t-1$. It follows that $d_{G}(y_{\ell _2-1}) \le D_{\ell _2-1}^*+1=t+1$, $d_{G}(y_{\ell _1}) = D_{\ell _1}^*=t+1$ and $d_{G}(x_{\ell _1}) \le d_{\ell _1}^*+1 \le t$. Therefore $d_{K_{p,q}}(y_{i}) - d_G(y_{i}) \ge p-t-1$ for $i =\ell _2-1, \ell _2$ and $d_{K_{p,q}}(x_{\ell _1}) - d_G(x_{\ell _1}) \ge q-t$, which implies that $|E(K_{p,q})|-|E(G)| \ge 2(p-t-1)+(q-t)-2=2p+q-3t-4$. However $|E(K_{p,q})|-|E(G)|=2q-t-1 < 2p+q-3t-4$, a contradiction. $\square$

Claim $5'$. There are t edge-disjoint rainbow pairs $E_k=\{x_{\ell _1-1}y_{k1},x_{\ell _1}y_{k2}\}$ ($E_k'=\{x_{k1}y_{\ell _2-1},x_{k2}y_{\ell _2}\})$ for $1 \le k \le t$ in $K_{p,q}$ if $\ell _1\ge 2$ ($\ell _2\ge 2)$.

Proof of Claim $5'$. We only consider the case when $\ell _1 \ge 2$, the other case when $\ell _2 \ge 2$ is similar. We construct an auxiliary bipartite graph $Q=(Z,W;E(Q))$, where $Z=\{e_1,\ldots ,e_{q}\}$ and $W=\{f_1,\ldots ,f_{q}\}$ are the sets of edges in $K_{p,q}$ incident to $x_{\ell _1-1}$ and $x_{\ell _1}$, respectively, and $e_if_j \in E(Q)$ if and only if $e_i$ and $f_j$ have different colors. We say $e_i\in Z$ (resp. $f_j\in W$) has color a in Q if the edge $e_i$ (resp. $f_j$) incident to $x_{\ell _1-1}$ (resp. $x_{\ell _1}$) has color a in $K_{p,q}$. Now we need to show that $\alpha '(Q) \ge t$.

If $\delta (Q) \ge t/2$, Lemma 6 implies that $\alpha '(Q) \ge t$ as $q > t$. Suppose that $\delta (Q) \le (t-1)/2$. Without loss of generality, we can let $e \in Z$ be a vertex with minimum degree in Q, and let a be its color. By the definition of Q, W contains $q-d_Q(e) \ge q-(t-1)/2 > t$ vertices with color a. Let k be the number of vertices in Z with color a. If $q-k \ge t$, then $\alpha '(Q) \ge t$ by matching t vertices with color a in W into Z. Hence we suppose $k> q-t > t$. Now each part contains at least t vertices with color a. Since $|E(G)|=q(p-2)+t+1$, $d_G(x_{\ell _1-1})+d_G(x_{\ell _1})\ge t+1$ which implies that there are at least t edges incident to $x_{\ell _1-1}$ or $x_{\ell _1}$ in $K_{p,q}$ with distinct colors other than color a. No matter how those vertices are distributed to Z and W, we can obtain $\alpha '(Q) \ge t$. $\square$

Claim $6'$. There are t edge-disjoint rainbow pairs $E_k=\{x_{\ell _1}y_{k_1},y_{\ell _2}x_{k_2}\}$ for $1 \le k \le t$ in $K_{p,q}$.

Proof of Claim $6'$. We construct an auxiliary bipartite graph $Q=(Z,W;E(Q))$, where $Z=\{z_1,\ldots ,z_{q}\}$ is the set of edges in $K_{p,q}$ incident to $x_{\ell _1}$, and $W=\{w_1,\ldots ,w_{p-1}\}$ is the set of edges in $K_{p,q}$ incident to $y_{\ell _2}$ except the edge $x_{\ell _1}y_{\ell _2}$, and $z_iw_j \in E(Q)$ if and only if $z_i$ and $w_j$ have different colors. We say $z_i\in Z$ (resp. $w_j\in W$) has color a in Q if the edge $z_i$ (resp. $w_j$) incident to $x_{\ell _1}$ (resp. $y_{\ell _2}$) has color a in $K_{p,q}$.

If $\delta (Q) \ge t/2$, Lemma 6 implies that $\alpha '(Q) \ge t$ as $q > t$. Suppose that $\delta (Q) \le (t-1)/2$. Let $z \in Z$ be a vertex with minimum degree in Q, and let a be its color. The other case when the vertex with minimum degree in Q belongs to W is similar. By the definition of Q, W contains $p-1-d_Q(z) \ge p-1-(t-1)/2 > t$ vertices with color a. Let k be the number of vertices in Z with color a. If $q-k \ge t$, then $\alpha '(Q) \ge t$ by matching t vertices with color a in W into Z. Hence $k> q-t > t$. Now each part contains at least t vertices with color a. Since $|E(G)|=pq-2q+t+1\ge (p-1)(q-1)+t+1$, there are at least t edges with distinct colors other than a incident to $x_{\ell _1}$ or $y_{\ell _2}$ in $K_{p,q}$. Now t such vertices in Q can be matched to t vertices with color a to obtain $\alpha '(Q) \ge t$, no matter how those vertices are distributed to Z and W. $\square$