we have $$\frac{d}{dt} \int_{-L}^L \Psi(n_{\epsilon})\,dx + \int_{-L}^L n_{\epsilon} \partial_x n_{\epsilon}\cdot \partial_x p_{\epsilon}\psi'(n_{\epsilon})\,dx = \int_{-L}^L ({n_1}_{\epsilon}  G_1(p_{\epsilon} ) +{n_2}_{\epsilon}  G_2(p_{\epsilon} ))\psi(n_{\epsilon} )\,dx,$$ where $\Psi$ is an antiderivative of $\psi$. We choose $\psi(n)= \epsilon (\ln(n)-\ln(1-n)+\frac{1}{1-n})$ so that $n_{\epsilon} \psi'(n_{\epsilon})=P'(n_{\epsilon})$. Inserting the expression of $\psi$, we get $$\begin{aligned}
&\frac{d}{dt} \int_{-L}^L \epsilon n_{\epsilon} \ln\Big(\frac{n_{\epsilon}}{1-n_{\epsilon}}\Big)\,dx 
+ \int_{-L}^L |\partial_x p_{\epsilon}|^2 dx  \leq G_{m} \int_{-L}^L \epsilon n_{\epsilon}\left|\ln(n_{\epsilon})-\ln(1-n_{\epsilon})+\frac{1}{1-n_{\epsilon}}\right|\,dx.\end{aligned}$$ After integrating in time and using the expression of the pressure , we have $$\begin{aligned}
\int_{-L}^L \epsilon n_{\epsilon} \ln\Big(\frac{p_{\epsilon}}{\epsilon}\Big)\,dx 
& -\int_{-L}^L \epsilon n_{\epsilon}^{\mbox{\scriptsize ini}} \ln\left(\frac{n_{\epsilon}^{\mbox{\scriptsize ini}}}{1-n_{\epsilon}^{\mbox{\scriptsize ini}}}\right)\,dx + \int_0^T \int_{-L}^L |\partial_x p_{\epsilon}|^2\,dxdt   \\
&\leq G_{m} \int_0^T \int_{-L}^L \left(\epsilon n_{\epsilon} \Big|\ln\Big(\frac{p_{\epsilon}}{\epsilon}\Big)\Big|+p_{\epsilon}\right) \,dx.\end{aligned}$$ Then, to prove that $\partial_x p_\epsilon \in L^2(Q_T)$, we are left to find a uniform bound on $ \int_{-L}^L \epsilon n_{\epsilon} |\ln(\frac{p_{\epsilon}}{\epsilon})|dx$. Using the expression of $p_\epsilon$ in , we have $$\begin{aligned}
\int_{-L}^L \epsilon n_{\epsilon} |\ln\big(\frac{p_{\epsilon}}{\epsilon}\big)|\,dx & \leq \int_{-L}^L \epsilon n_{\epsilon} |\ln p_{\epsilon}| \,dx  + \epsilon \ln(\epsilon) \int_{-L}^Ln_{\epsilon}\,dx \\
 & \leq  \int_{-L}^L (1-n_{\epsilon})p_{\epsilon} |\ln p_{\epsilon}|\,dx + \epsilon \ln(\epsilon) \int_{-L}^L n_{\epsilon}\,dx\end{aligned}$$ Since $n_\epsilon$ is bounded in $L^1$, the second term of the right hand side is uniformly bounded with respect to $\epsilon$. Moreover given that $0\leq p_\epsilon \leq P_M$ and $x\mapsto x|\ln x|$ is uniformly bounded on $[0,P_M]$, we get 