p_{\epsilon}|\,dx \\
  &\leq  C \int_{n_{\epsilon}\leq 1/2} |\partial_{x} n_{\epsilon}|\,dx + 2 \int_{n_{\epsilon}\geq 1/2} \frac{1}{2}|\partial_{x} p_{\epsilon}|\,dx \\
   &\leq  C \|\partial_{x} n_{\epsilon}(t)\|_{L^1(-L,L)} + 2 \int_{n_{\epsilon}\geq 1/2} n_{\epsilon}|\partial_{x} p_{\epsilon}|\,dx.
  \end{aligned}$$ Then, we integrate in time and we deduce using $$\| \partial_{x} p_{\epsilon} \|_{L^1(Q_T)} \leq C'e^{G_mT} (\| \partial_{x} n_{\epsilon} \|_{L^1(-L,L)}\textcolor{magenta}{+1)}.$$ Hence we have an uniform bound on $\partial_x p_\epsilon$ in $L^1(Q_T)$. To recover the estimate on $ \partial_{x} {n_1}_{\epsilon}$ and $ \partial_{x} {n_2}_{\epsilon}$ we deduce from , $$\partial_{x} {n_1}_{\epsilon}= \partial_{x} {n}_{\epsilon} \mathbf{1}_{ x \leq  \zeta_{\epsilon}(t)}+n_{\epsilon} \delta_{x =  \zeta_{\epsilon}(t)},$$ $$\partial_{x} {n_2}_{\epsilon}= \partial_{x} {n}_{\epsilon} \mathbf{1}_{x \leq  \zeta_{\epsilon}(t)}-n_{\epsilon} \delta_{ x =  \zeta_{\epsilon}(t)}.$$ So $$\| \partial_{x} {n_1}_{\epsilon} \|_{L^1(-L,L)} = \int_{x \leq  \zeta_{\epsilon}(t)} \partial_{x_i} {n_1}_{\epsilon} ~dx+ {n}_{\epsilon}(t, \zeta_{\epsilon}(t))    \leq \| \partial_{x} n_{\epsilon} \|_{L^1(-L,L)} + \| {n_1}_{\epsilon}\|_{\infty},$$ and $$\| \partial_{x} {n_2}_{\epsilon} \|_{L^1(-L,L)} = \int_{x \geq  \zeta_{\epsilon}(t)} \partial_{x_i} {n_2}_{\epsilon}  ~dx- {n}_{\epsilon}(t, \zeta_{\epsilon}(t))   \leq \| \partial_{x} n_{\epsilon} \|_{L^1(-L,L)} + \| {n_2}_{\epsilon}\|_{\infty}$$ This concludes the proof.

###  L2 estimate for dxp

\[lem:L2dp2\] Let us assume that and hold. Let $({n_1}_\epsilon,{n_2}_\epsilon,p_\epsilon)$ be a solution to –. Then, for all $T>0$ we have a uniform bound on $\partial_x p_\epsilon$ in $L^2(Q_T)$.

For a given function $\psi$ we have, multiplying by $\psi(n_\epsilon)$, $$\partial_t n_{\epsilon} \psi(n_\epsilon) -\partial_x(n_{\epsilon} \partial_x p_{\epsilon})\psi(n_{\epsilon})= ({n_1}_{\epsilon}  G_1(p_{\epsilon} ) +{n_2}_{\epsilon}  G_2(p_{\epsilon} ))\psi(n_{\epsilon}).$$ Integrating on $(-L,L)$, 