and $\mu$.

=6.5 cm =4.5 cm =6.5 cm =4.5 cm -0.05cm 0.15 cm **( a )** 6.5 cm **( b )**\

Firstly we check whether the spontaneous symmetry breaking is described well. Therefore, we take $m_q=0$, i.e. in chiral limit. Then, we take different values of $v_4$ and $\mu$ and solve out $\sigma$ as a function of temperature in Fig.\[simplev4v6\](a)($v_4=2,\mu=0.1\rm{GeV}$(black solid line), $v_4=2,\mu=0.5\rm{GeV}$( black dashed line), $v_4=10, \mu=0.1\rm{GeV}$(blue solid line), $v_4=10, \mu=0.5\rm{GeV}$(blue dashed line)). From the figure, one could see that, for different values of $v_4$, the qualitative results for $\sigma$ are the same. At low temperature, chiral symmetry is broken by finite condensate $\sigma$, while above a certain temperature $T_c$, $\sigma$ turns to zero and chiral symmetry is restored. The phase transition is of second order type. It could also be seen that $T_c$ is not depend on $v_4$ for a fixed $\mu$[^1]. Also, we find that a larger $v_4$ leads to a smaller $\sigma$ at the same temperature when $v_4>0$, while there is no condensate when $v_4\leq0$.

![Chiral condensate $\sigma$ in chiral limit($m_q=0$) as a function of the temperature T for $V(\chi, F^2)=-\frac{3}{2}\chi^2+v_4\chi^4+v_6 \chi^6$ with $v_4=-2, v_6=100$ and $\mu=0.5\rm{GeV}$.[]{data-label="simplev4+v6"}](simplev4v6.eps){width="8.5cm" height="6.5cm"}

Up to finite temperature, it seems chiral phase 