Goal

To analyze whether it is stable against environment changes.

Results

Basing on the equations, since applying their Taylor series makes the equations formed in a linear one keeping the topology of the solution to the original equations, the characteristic equations can be presented: $$E(\mu) = \left(\mu+\lambda\frac{\gamma}{(Ce+a)^2}\right)(\mu+d_{a/r})\left(\mu+k_{fa}+\lambda\frac{\gamma}{(Ce+a_{uf})^2}\right)-k_{fa}T_acopy_a(k_3E_1+k_4E_2+k_5E_3) = 0$$ $$\begin{cases} E_1 = \dfrac{\frac{2a}{a_0}e^{-2\mu \tau}\left(1+\frac{r^4}{r_0}e^{-4\mu \tau}\right)}{\left(1+\frac{r^4}{r_0}e^{-4 \mu \tau}+\frac{a^2}{a_0}e^{-2 \mu\tau}\right)^2}\\ E_2 = \dfrac{2ae^{-2 \mu\tau}\frac{r^2}{r_0}e^{-2 \mu\tau}}{\left(1+\frac{r^4}{r_0}e^{-4 \mu\tau}+\frac{a^2}{a_0}e^{-2\mu\tau}\right)^2}\\ E_3 = \dfrac{\frac{2a}{a_0}e^{-2\mu \tau}}{\left(1+\frac{r^4}{r_0}e^{-4 \mu\tau}+\frac{a^2}{a_0}e^{-2\mu\tau}\right)^2} \end{cases}$$ If there were at least one periodical solution, the equation should have at least one imaginary root.
Denote$ \begin{cases} c_1 = \lambda\frac{\gamma}{(Ce+a)^2}\\ c_2 = d_{a/r}\\ c_3 = k_{fa}+\lambda\frac{\gamma}{(Ce+a)^2}\\ c_4 = k_{fa}T_a(k_3E_1+k_4E_2+k_5E_3)copy_a \end{cases} $
Namely, $E(\mu) = (\mu+c_1)(\mu+c_2)(\mu+c_3)-c_4 = 0$ has imaginary root(s).
Set $\mu = iy$, $y\in R$ Thus $iy(c_1c_2+c_2c_3+c_1c_3-y^2)-(c_1+c_2+c_3)y^2+c_1c_2c_3 = c_4$ is required for imaginary root(s).
Namely, $ \begin{cases} y(c_1c_2+c_2c_3+c_1c_3-y^2) = Im(c_4) \\ -(c_1+c_2+c_3)y^2+c_1c_2c_3 = Re(c_4) \end{cases} $
Namely $\|y(c_1c_2+c_2c_3+c_1c_3-y^2\|^2+\|-(c_1+c_2+c_3)y^2+c_1c_2c_3\|^2=\|c_4\|^2$
Set $x=y^2$
Thus the existence of periodical solution is equal to: $$x^3+(c_1^2+c_2^2+c_3^2)x^2+(c_1^2c_2^2+c_2^2c_3^2+c_1^2c_3^2)x+c_1^2c_2^2c_3^2 = (k_{fa}T_acopy_a)^2\left(\dfrac{\left(\frac{2a}{a_0}\right)^2\left(k_3^2+k_5^2\right)}{\left(1+\frac{r^4}{r_0}+\frac{a^2}{a_0}\right)^4}+\dfrac{\left(\frac{2ar^2}{r_0}\right)^2\left(r^4k_3^2+k_4^2\right)}{\left(1+\frac{r^4}{r_0}+\frac{a^2}{a_0}\right)^4}\right)$$ has at least one positive real root. Denote p, q, $\delta$ as coefficients related to parameters in DDEs. Due to the limited space, these coefficients can be found in supplementary data.
According to the Cardano's Formula, If $q<0$, $\delta>0$,then there exists a center of oscillation. In this case, derivative of period with respect to both Arabinose and IPTG are not 0.
If $q = 0$, $p<0$, then also exists a center of oscillation. In this case, derivative of period with respect to both Arabinose and IPTG are 0.
In fact, in our case - when Arabinose $\in [0,10] $ and IPTG $\in [0,10]$ - belongs to the first aspect, which means both Arabinose and IPTG contribute to the period.