The graphs of the functions $v_1$ and $v_2$ are given in Figure 3(a) and Figure 3(b) respectively.

**Case A.III.** In addition to the constraints (\[caseA\]), the last set of the conditions obtained from the equation (\[relationnonlocalib\]) is $\displaystyle \{ k_1=-k_3, e^{\delta_3}=-b_1e^{\delta_1}\frac{(k_3+k_4)(k_2+k_3)}{(k_3-k_4)(k_2-k_3)}\}$. Therefore one-soliton solution of the system (\[nonlocalreducediAN=4\]) becomes $$\begin{aligned}
\displaystyle
&v_1=\frac{e^{-k_3x+\frac{k_3^3}{\mu}t+\delta_1}-\sigma_1\sigma_2\frac{(k_3+k_4)(k_2+k_3)}{(k_3-k_4)(k_2-k_3)}e^{(k_2-k_3+k_4)x-\frac{(k_2^3-k_3^3+k_4^3)}{\mu}t+\delta_1}   }{1+\sigma_1\sigma_2e^{(k_2+k_4)x-\frac{(k_2^3+k_4^3)}{\mu}t}},\\
&v_2=\frac{\sigma_1(k_2+k_4)e^{k_2x-\frac{k_2^3}{\mu}t}   }{\sqrt{a_1c_{24}}
[1+\sigma_1\sigma_2e^{(k_2+k_4)x-\frac{(k_2^3+k_4^3)}{\mu}t}]},\end{aligned}$$ for $\sigma_i=\pm 1, i=1, 2$.

**Example 4.** If we choose the parameters as $k_2=2, k_3=-1, k_4=3, e^{\delta_1}=\frac{1}{10}, a_1=1, \mu=c_{24}=\sigma_1=\sigma_2=-1$ one-soliton solution of (\[nonlocalreducediAN=4\]) becomes $$\displaystyle
v_1=\frac{6e^{x+t}+e^{6x+36t}}{60(1+e^{5x+35t})},\quad |v_2|^2=\frac{25e^{4x+16t}}{(1+e^{5x+35t})^2}.$$ Both of the solutions $v_1$ and $v_2$ are finite. The solution $v_2$ is bounded but $v_1$ not. The graphs of the functions $v_1$ and $|v_2|^2$ are given in Figure 4(a) and Figure 4(b) respectively.

Note that if we start with the conditions $c_{12}=c_{23}=c_{24}=0$ then the system (\[nonlocalreducediN=4\]) turns to be $$\begin{aligned}
&\mu v_{1,t}(t,x)+6c_{13}b_1v_1(t,x)v_1(-t,-x)v_{1,x}(t,x)+v_{1,xxx}(t,x)=0,\nonumber\\
 &\mu v_{2,t}(t,x)+6c_{13}b_1v_1(t,x)v_1(-t,-x)v_{2,x}(t,x)+v_{2,xxx}(t,x)=0,\label{nonlocalreducediAsimilar}\end{aligned}$$ which is very similar to the Case A.

**Case B.** Another set of the constraints solving both of the equations (\[relationnonlocalia\]) and (\[relationnonlocalib\]) is $$\begin{aligned}
\label{caseB}\displaystyle
&1)\, c_{12}=c_{23}=k_2=0,\quad 2)\,k_1=\frac{k_4[a_1k_4^2(k_3+k_4)+e^{2\delta_4}c_{24}(k_3-k_4)  ]}{-k_4^2a_1(k_3+k_4)+e^{2\delta_4}c_{24}(k_3-k_4)},\nonumber\\
& 3)\, e^{\delta_2}=\sigma_1\frac{k_4}{\sqrt{a_1c_{24}}},\quad 4)\, e^{\delta_3}=\sigma_2\frac{\sqrt{b_1(k_3^2-k_4^2)}}{\sqrt{c_{13}}}\frac{(e^{2\delta_4}c_{24}-a_1k_4^2)(k_3+k_4)}{a_1k_4^2(k_3+k_4)
-e^{2\delta_4}c_{24}(k_3-k_4)},\nonumber\\
&5)\,  e^{\delta_1}=\frac{\sigma_1e^{\delta_4}c_{24}(k_3-k_4)^2(k_3+k_4)[a_1k_4^2-e^{2\delta_4}c_{24}]}
{\sigma_2[a_1k_4^2(k_3+k_4)-e^{2\delta_4}c_{24}(k_3-k_4)]k_4\sqrt{a_1b_1c_{13}c_{24}(k_3^2-k_4^2)}},\end{aligned}$$ where $\sigma_i=\pm 1, i=1, 2$. In this case the system (\[nonlocalreducediN=4\]) reduces to $$\begin{aligned}
&\mu v_{1,t}(t,x)+6(c_{13}b_1v_1(t,x)v_1(-t,-x)+c_{24}a_1v_2(t,x)v_2(-t,-x))v_{1,x}(t,x)+v_{1,xxx}(t,x)=0,\nonumber\\
 &\mu v_{2,t}(t,x)+6(c_{13}b_1v_1(t,x)v_1(-t,-x)+c_{24}a_1v_2(t,x)v_2(-t,-x))v_{2,x}(t,x)+v_{2,xxx}(t,x)=0.\label{nonlocalreducediBN=4}\end{aligned}$$ One-soliton solution of the reduced 