function of $b_N$ in full generality. However the approach used in [@LevPer16] does not allow to find the exact asymptotic for the mixing time nor to prove cutoff, and does not answer our question concerning the pattern of relaxation to equilibrium.

Results
-------

We identify two main regimes for the pattern of relaxation to equilibrium. The *large bias* regime where $$\label{largebias}
\lim_{N\to \infty} \frac{N b_N}{(\log k_N)\vee 1}=\infty.$$ and the *small bias* regime where $$\label{smallbias}
\begin{cases}
\lim\limits_{N\to \infty} \frac{Nb_N}{\log k_N}=0,\\
\lim\limits_{N\to \infty} k_N=\infty. 
\end{cases}$$ We identify the asymptotic expression for the mixing time in both regimes. In the large bias regime we show that the mixing time coincides with the time needed by the particle density to reach equilibrium like in the constant bias case.

\[Th:largebias\]

When holds, and $\lim_{N\to \infty} k_N/N=\alpha\in [0,1]$, we have for every ${\varepsilon}\in (0,1)$ $$\lim_{N\to \infty} \frac{b_N {T_{\rm mix}}^{N,k_N}({\varepsilon})}{N}= \left(\sqrt{\alpha}+\sqrt{1-\alpha}\right)^2.$$

To state our result in the small bias regime, let us introduce the quantity $$\label{thegap}
\operatorname{\mathrm{gap}}_{N}:= (\sqrt{p_N}-\sqrt{q_N})^2+ 4 \sqrt{p_Nq_N} \sin\left( \frac{\pi}{2N}\right)^2,$$ which corresponds to the spectral gap associated with the generator . Notice that it does not depend on the number $k$ of particles in the system. The pattern of relaxation is similar to the one 