---
abstract: 'We consider a viscous approximation for a nonlinear degenerate convection-diffusion equations in two space dimensions, and prove an $L^1$ error estimate. Precisely, we show that the $L^1_{\mathrm{loc}}$ difference between the approximate solution and the unique entropy solution converges at a rate $\mathcal{O}({\varepsilon}^{1/2})$, where ${\varepsilon}$ is the viscous parameter.'
title: 'An error estimate for viscous approximate solutions to degenerate anisotropic convection-diffusion equations'
---

<span style="font-variant:small-caps;">Christian Klingenberg</span>

<span style="font-variant:small-caps;">Ujjwal Koley</span>

(Communicated by the associate editor name)

Introduction {#sec:intro}
============

In this paper, we are interested in certain “viscous” approximations of entropy solutions of the following Cauchy problem $$\label{eq:main}
  \begin{cases}
    u_t + f(u)_y = u_{xx}, &(x,y,t)\in {\Pi_T},\\
    u(x,y,0)=u_0(x,y), & (x,y)\in {\mathbb{R}}^2,
  \end{cases}$$ where ${\Pi_T}={\mathbb{R}}\times {\mathbb{R}}\times (0,T)$ with $T>0$ fixed, $u:{\Pi_T}\to {\mathbb{R}}$ is the unknown function and $f:{\mathbb{R}}\rightarrow {\mathbb{R}}$ is the convective flux function. The main characteristics of this type of equations is that it has mixed parabolic-hyperbolic type, due to the directional separation of the diffusion and convection effects: while matter is convected along the $y$ axis, it is simultaneously diffused along orthogonal direction.

The existence of solutions of equation can be obtained by the classical method of adding a vanishing viscosity, in other terms a diffusion, in the missing direction (along the $y$ axis). Since the most characteristics of Equation is that it has mixed parabolic-hyperbolic type, or in other words, it has strong degeneracy due to the lack of diffusion in the $x$-direction, it is difficult to establish the uniqueness of solutions of . In [@ez1], authors have proved the existence as well as uniqueness of such problems.

In this paper we are interested in certain approximate solutions of coming from solving the uniformly parabolic problem $$\begin{aligned}
 u^{{\varepsilon}}_t + f(u^{{\varepsilon}})_y =  u^{{\varepsilon}}_{xx} + {\varepsilon}u^{{\varepsilon}}_{yy}.
\end{aligned}
\label{eq:regular}$$ We refer to $u^{{\varepsilon}}$ as a “viscous” approximate solution of . Since the convergence of $u^{{\varepsilon}}$ to the unique entropy solution $u$ of as ${\varepsilon}{\downarrow}0$ is well known, so our interest here is to give an explicit rate of convergence for $u^{{\varepsilon}}$ as ${\varepsilon}{\downarrow}0$, i.e., an $L^1$ error estimate for viscous approximate solutions. There are several ways to prove such an error estimate. One way is to view it as a consequence of a continuous dependence estimate. Combining the ideas of [@ez1] with a variant of Kruzkov’s “doubling of variables” device for , we prove that ${\left\|u^{{\varepsilon}} -u\right\|}_{L^1({\Pi_T})} = \mathcal{O}({\varepsilon}^{1/2})$. Although our proof is of independent interest, it may also shed some light on how to obtain an error estimates for numerical methods.

Preliminaries {#sec:prelim}
=============

Independently of the smoothness of the initial data, due to the lack of diffusion in the $x$-direction, jumps may form in the solution $u$. Therefore we consider solutions in the weak sense, i.e.,

Set ${\Pi_T}={\mathbb{R}}\times {\mathbb{R}}\times (0,T)$. A function $$u(t,x) \in C\left([0,T];L^1({\mathbb{R}}^2)\right) \cap L^\infty({\Pi_T})$$ is a weak solution of the initial value problem if it satisfies:

-   For all test functions ${\varphi}\in
        \mathcal{D}({\mathbb{R}}^2 \times [0,T))$ \[def:w2\] $$\label{eq:weaksol}
          \iiint_{{\Pi_T}} \left( u{\varphi}_t + f(u) {\varphi}_y + u {\varphi}_{xx} \right)\,dx \,dy \,dt 
          + \iint_{{\mathbb{R}}^2} u_0(x,y) {\varphi}(x,y,0) \,dx \,dy= 0.$$

-   $u = u(x,y,t)$ is continuous at $t=0$ as a function: $[0,T) \rightarrow L^1({\mathbb{R}}^2)$.

Since weak solutions are not uniquely determined by their initial data, one must impose an additional entropy condition to single out the physically relevant solution.

A weak solution $u$ of the initial value problem is called an entropy solution, if the following entropy inequality holds for all test functions $0 \le {\varphi}\in \mathcal{D}({\mathbb{R}}^2 \times (0,T))$: $$\label{eq:entropysol}
 \begin{aligned}
 &\iiint_{{\Pi_T}} {\left|u -\psi(x)\right|}{\varphi}_t +
   {\mathrm{sign}}(u -\psi(x)) (f(u) -f(\psi(x))) {\varphi}_y \,dxdydt \\
  &\qquad \ge \iiint_{{\Pi_T}} - {\left|u- \psi(x)\right|} {\varphi}_{xx}  - {\mathrm{sign}}(u- \psi(x)) \psi_{xx} {\varphi}\,dxdydt. 
 \end{aligned}$$

Note that the above entropy condition is inspired by Kružkov’s entropy condition for scalar conservation laws. However, due to the presence of diffusion term $u_{xx}$ in , Kružkov’s entropy condition has to be modified in this case. This is done by introducing as entropy test functions all functions of the form ${\left|u -\psi(x)\right|}$ and $\psi$ smooth, while Kružkov’s definition asks for $\psi$ to be a constant.

Regarding the regularized problem , it is well known [@ez1] that for all ${\varepsilon}> 0$ this problem has a unique solution $$\begin{aligned}
 u^{{\varepsilon}} \in C([0,T]; L^1({\mathbb{R}}^2)) \cap L^{\infty}({\Pi_T}).\end{aligned}$$ In fact, $$\begin{aligned}
 u^{{\varepsilon}} \in C([0,T]; W^{2,p}({\mathbb{R}}^2)) \cap C^{1}([0,T]; L^{p}({\mathbb{R}}^2)),\end{aligned}$$ for all $1 < p < \infty$. In addition, $u^{{\varepsilon}}$ satisfies the following properties

-   $\iint_{{\mathbb{R}}^2} u^{{\varepsilon}}(x,y,t) \,dxdy = \iint_{{\mathbb{R}}^2} u_0(x,y) \,dxdy$,

-   $ \text{TV}\, (u^{{\varepsilon}}(\cdot,\cdot, t)) \le \text{TV}\, (u_0) $,

-   $ {\left\|\partial_t u^{{\varepsilon}}(t)\right\|}_1 \le C [\text{TV}\, (u_0) + \text{TV}\, ((u_0)_x)]$.

Now we are in a position to state our main result, which is the following

Let $u$ be the unique entropy solution to and $u^{{\varepsilon}}$ be as defined by . Assume that $u_0 \in BV$, $f$ is Lipschitz continuous. Choose a constant $$M>\max_{{\left|u\right|}<{\left\|u_0\right\|}_{L^\infty({\mathbb{R}}^2)}} {\left|f'(u)\right|},$$ and another constant $L>MT$, where $T>0$. Then there exists a constant $C$, independent of ${\varepsilon}$, but depending on $f$, $L$, $T$ and $u_0$, such that $$\int_{{\mathbb{R}}}\int_{-L+Mt}^{L-Mt}{\left|u(t,x,y) - u^{{\varepsilon}}(t,x,y)\right|}\,dxdy  \le C {\varepsilon}^{1/2}
    \quad\text{for $t\le T$.}$$

Proof of the main theorem {#sec:proof}
=========================

The theorem will be proved by a “doubling of the variables” argument, which was introduced by Kruzhkov [@kr1; @kr2] in the context of hyperbolic conservation laws.

First, observe that from we have for any test function ${\varphi}$ with compact support in ${\mathbb{R}}\times {\mathbb{R}}\times (0,T)$ and $u = u(x,z,s)$, $$\begin{aligned}
 &\iiint_{{\Pi_T}} {\left|u -\psi(x)\right|}{\varphi}_s +
   {\mathrm{sign}}(u -\psi(x)) (f(u) -f(\psi(x))) {\varphi}_z \,dxdzds \\
  &\qquad \ge \iiint_{{\Pi_T}} - {\left|u- \psi(x)\right|} {\varphi}_{xx}  - {\mathrm{sign}}(u- \psi(x)) \psi_{xx} {\varphi}\,dxdzds. 
 \end{aligned}
\label{eq:main_1}$$ For the regularized equation , we start not with the entropy condition, but in the argument leading up to this condition. To do that, first define the regularized counterpart of the signum function as $${\mathrm{sign}}_\eta(\sigma) =
\begin{cases}
  {\mathrm{sign}}(\sigma) & {\left|\sigma\right|}>\eta,\\
  \sin\left(\frac{\pi\sigma}{2\eta}\right) &\text{otherwise,}
\end{cases}$$ where $\eta>0$ and the signum function is defined as $${\mathrm{sign}}(\sigma) =
\begin{cases}
  -1 & \sigma<0,\\
  0 & \sigma = 0,\\
  1 & \sigma>0,
\end{cases}$$

Set $$\psi_\eta(u,\psi)=\int_{\psi}^{u} {\mathrm{sign}}_\eta (z-\psi)\,dz.$$ This is a convex entropy for all $\psi$. Set $u^{{\varepsilon}}=u^{{\varepsilon}}(x,y,t)$ and rewrite as $$u^{{\varepsilon}}_t + \left(f(u^{{\varepsilon}})-f(\psi)\right)_y = u^{{\varepsilon}}_{xx} + {\varepsilon}(u^{{\varepsilon}} -\psi(x))_{yy},$$ and multiply this with $(\psi_\eta)_u(u,\psi) \, {\varphi}$ where ${\varphi}$ is a test function with compact support in ${\mathbb{R}}\times {\mathbb{R}}\times (0,T)$. Observe that the solution $u^{{\varepsilon}}$ of is smooth. Hence after a partial integration, we arrive at $$\begin{aligned}
  &\iiint_{{\Pi_T}} \psi_\eta(u^{{\varepsilon}},\psi) {\varphi}_t +
   Q_{\eta} (u^{{\varepsilon}},\psi) {\varphi}_y \,dxdydt \\
  &\qquad = \iiint_{{\Pi_T}} - {\mathrm{sign}}_\eta (u^{{\varepsilon}}- \psi) u^{{\varepsilon}}_{xx}{\varphi}- {\varepsilon}{\mathrm{sign}}_\eta (u^{{\varepsilon}}- \psi) (u^{{\varepsilon}} -\psi(x))_{yy} {\varphi}\,dxdydt,\end{aligned}$$ where we have used $Q'_{\eta}(u,\psi) = \psi'_\eta(u,\psi) f'(u)$. Next, taking limit as $\eta \rightarrow 0$, we end up with the parabolic equality $$\begin{aligned}
  &\iiint_{{\Pi_T}} {\left|u^{{\varepsilon}} -\psi(x)\right|}{\varphi}_t +
   {\mathrm{sign}}(u^{{\varepsilon}} -\psi(x)) (f(u^{{\varepsilon}}) -f(\psi(x))) {\varphi}_y \,dxdydt \\
  &\qquad = \iiint_{{\Pi_T}} - {\mathrm{sign}}(u^{{\varepsilon}}- \psi(x)) u^{{\varepsilon}}_{xx} {\varphi}- {\varepsilon}\, {\mathrm{sign}}(u^{{\varepsilon}}- \psi(x)) (u^{{\varepsilon}} -\psi(x))_{yy} {\varphi}\,dxdydt, \\
& \qquad = \iiint_{{\Pi_T}} - {\mathrm{sign}}(u^{{\varepsilon}}- \psi(x)) u^{{\varepsilon}}_{xx} {\varphi}- {\varepsilon}\, {\left|u^{{\varepsilon}} -\psi(x)\right|}_{yy} {\varphi}\,dxdydt.
\end{aligned}
\label{eq:main_2}$$ At this point we are ready to use “doubling of the variables” device. First, using the entropy inequality for the solution $u=u(x,z,s)$ with $\psi(x)= u^{{\varepsilon}}(x,y,t)$, we get for $(y,t) \in {\mathbb{R}}\times (0,T)$ $$\begin{aligned}
 &\iiint_{{\Pi_T}} {\left|u -u^{{\varepsilon}}\right|}{\varphi}_s +
   {\mathrm{sign}}(u -u^{{\varepsilon}}) (f(u) -f(u^{{\varepsilon}})) {\varphi}_z \,dxdzds \\
  &\qquad \ge \iiint_{{\Pi_T}} - {\left|u- u^{{\varepsilon}}\right|} {\varphi}_{xx}  - {\mathrm{sign}}(u- u^{{\varepsilon}}) u^{{\varepsilon}}_{xx} {\varphi}\,dxdzds. 
 \end{aligned}
\label{eq:main_3}$$ Similarly, from the parabolic equality for the solution $u^{{\varepsilon}} = u^{{\varepsilon}}(x,y,t)$ with $\psi(x)= u(x,z,s)$, we get for $(z,s) \in {\mathbb{R}}\times (0,T)$ $$\begin{aligned}
  &\iiint_{{\Pi_T}} {\left|u^{{\varepsilon}} -u\right|}{\varphi}_t +
   {\mathrm{sign}}(u^{{\varepsilon}} -u) (f(u^{{\varepsilon}}) -f(u)) {\varphi}_y \,dxdydt \\
  & \qquad = \iiint_{{\Pi_T}} - {\mathrm{sign}}(u^{{\varepsilon}}- u) u^{{\varepsilon}}_{xx} {\varphi}- {\varepsilon}\, {\left|u^{{\varepsilon}} -u\right|}_{yy} {\varphi}\,dxdydt.
\end{aligned}
\label{eq:main_4}$$ We now integrate over $(y,t) \in {\mathbb{R}}\times (0,T)$ and over $(z,s) \in {\mathbb{R}}\times (0,T)$. Addition of those two results yields $$\begin{aligned}
 & \int\iiiint_{Q_T} {\left| u^{{\varepsilon}} -u \right|} ({\varphi}_t +{\varphi}_s) +
   {\mathrm{sign}}( u^{{\varepsilon}} -u ) ( f(u^{{\varepsilon}}) -f(u)) ( {\varphi}_y + {\varphi}_z) \,dX \\
  &\qquad \ge \int\iiiint_{Q_T} - {\left|u^{{\varepsilon}} - u\right|} {\varphi}_{xx}  - {\varepsilon}\, {\left|u^{{\varepsilon}} -u\right|}_{yy} {\varphi}\,dX,\\
& \qquad := \int \iiiint_{Q_T} \mathcal{Q}_1 + \mathcal{Q}_2 \,dX,
 \end{aligned}
\label{eq:main_5}$$ where $dX=dx\,dy\,dz\,dt\,ds$ and $Q_T = {\mathbb{R}}\times {\mathbb{R}}\times {\mathbb{R}}\times (0,T) \times (0,T)$.

Following Kruzkov and Kuznetsov [@kr1; @kr2] we now specify a nonnegative test function $ {\varphi}=
{\varphi}(x,y,t,z,s)$ defined in $Q_T$. To this end, let $\omega \in
C_{0}^{\infty} ({\mathbb{R}})$ be a function satisfying $$\mathrm{supp}(\omega) \subset [-1,1], \qquad
  \omega(\sigma) \ge 0 , \qquad \int_{{\mathbb{R}}} \omega(\sigma)\, d\sigma = 1,$$ and define $\omega_r(x)=\omega(x/r)/r$. Next, let us choose $\phi \in C_c^{\infty} ({\mathbb{R}})$ such that $$\phi =
\begin{cases}
 1, & {\left|x\right|}<1,\\ 0 & {\left|x\right|}\ge 2,
\end{cases}$$ and $ 0  \le \phi \le 1$ when $1 \le {\left|x\right|} \le 2$. Then we define $K_{\beta}(x) = \phi(x/\beta)$. We will let $\beta \rightarrow \infty$ later. Furthermore, let $h(z)$ be defined as $$h(z)=
  \begin{cases}
    0, &z<-1,\\ z+1 & z\in [-1,0],\\ 1 &z>0.
  \end{cases}$$ and set $h_\alpha(z)=h(\alpha z)$. Let $\nu<\tau$ be two numbers in $(0,T)$, for any $\alpha> 0$ define $$\begin{gathered}
    H_{\alpha}(t)= \int_{-\infty}^{t}
    \omega_{\alpha}(\xi)\, d\xi, \\
    \begin{aligned}
      \Psi(y,t) &= \left(H_{\alpha_0} (t - \nu) - H_{\alpha_0} (t -
        \tau)\right) \left(h_\alpha(y-L_l(t)) -
        h_\alpha(y-L_r(t)-\frac{1}{\alpha})\right)\\
      &=:{\chi^{\alpha_0}_{(\nu,\tau)}(t)}\,{\chi^\alpha_{(L_l,L_r)}(y,t)}
    \end{aligned}
  \end{gathered}$$ where the lines $L_{l,r}$ are given by $$L_l(t)=-L+Mt,\ L_r(t)=L-Mt$$ where $M$ and $L$ are positive numbers, $M$ will be specified below. With $0<r< \min{\left\{\nu, T - \tau\right\}}$ and $\alpha_0\in(0,\min{\left\{\nu-r,T-\tau-r\right\}})$ we set $$\label{eq:testfn}
  {\varphi}(x,y,t,z,s)= K_{\beta}(x)\, \Psi(y,t)\,\omega_r(y-z)\,\omega_{r_0}(t-s).$$ We note that ${\varphi}$ has compact support and also that we have, $$\begin{aligned}
  {\varphi}_t + {\varphi}_s &= K_{\beta}(x) \, \Psi_t(y,t)\,
  \omega_r(y-z)\,\omega_{r_0}(t-s),\\
  {\varphi}_y+{\varphi}_z&= K_{\beta}(x)\, \Psi_y(y,t)\, \omega_r(y-z)\,\omega_{r_0}(t-s).\end{aligned}$$ For the record, we note that $$\label{eq:Psider}
  \begin{aligned}
    \Psi_t(y,t)&=-\chi^{\alpha_0}_{(\nu,\tau)}(t) M
    \left(h_\alpha'(y-L_l(t)) + h_\alpha'(y-L_r(t)-\frac{1}{\alpha})\right) \\
    & \qquad + 
    \left(\omega_{\alpha_0}(t-\nu) - \omega_{\alpha_0}(t-\tau)\right)
    \chi^\alpha_{(L_l,L_r)}(y,t),\\
    \Psi_y(y,t)  &= \chi^{\alpha_0}_{(\nu,\tau)}(t) 
    \left(h_\alpha'(y-L_l(t)) - h_\alpha'(y-L_r(t)-\frac{1}{\alpha})\right).
  \end{aligned}$$ We shall let all the “small parameters” $ \alpha$, $\alpha_0$, $r$, $r_0$, ${\varepsilon}$ and ${{\Delta x}}$ be sufficiently small and the “large parameter” $\beta$ be sufficiently big, but fixed.

Starting the first term on the left of , we write $$\begin{aligned}
  \int_{Q_T} & {\left|u^{{\varepsilon}}-u\right|}  \left({\varphi}_s  +{\varphi}_t\right)\,dX \le
  \underbrace{\int_{{\Pi_T}} {\left|u^{{\varepsilon}}(x,y,t)-u(x,y,t)\right|} K_{\beta}(x)\Psi_t \,dxdydt}_\delta \\
  &\quad + \underbrace{\int_{{\Pi_T}}\int_{{\mathbb{R}}} {\left|u(x,y,t)-u(x,y,s)\right|} K_{\beta}(x) {\left|\Psi_t(x,t)\right|}
  \omega_{r_0}(t-s) \,dxdydsdt}_{\beta} \\ 
  &\quad + \underbrace{\int_{Q_T}  {\left|u(x,y,s)-u(x,z,s)\right|} K_{\beta}(x) {\left|\Psi_t(x,t)\right|}
  \omega_{r_0}(t-s)\,\omega_r(x-y)\, dX}_\gamma . \end{aligned}$$ Following [@uk1], it is easy to find that $$\beta+\gamma \le C\left(r_0+r\right).\label{eq:betagamma}$$ To continue the estimate with the first term on the left of , we split $\delta$ as follows $$\begin{aligned}
  \delta &= -\iiint_{{\Pi_T}} \chi_{(\nu,\tau)}^{\alpha_0}(t) M
    \left(h'_\alpha(y-L_l(t)) + h'_\alpha(y-L_r(t)-\frac{1}{\alpha})\right) \\
& \qquad \qquad \qquad \qquad K_{\beta}(x) {\left|u^{{\varepsilon}}(x,y,t)-u(x,y,t)\right|} \, dxdydt \\
 &\quad \quad + \iiint_{{\Pi_T}} \chi^\alpha_{(L_l,L_r)}(y,t)
  {\left|u^{{\varepsilon}}(x,y,t)-u(x,y,t)\right|} \\
& \qquad \qquad \qquad \qquad K_{\beta}(x)
  \left(\omega_{\alpha_0}(t-\nu)-\omega_{\alpha_0}(t-\tau)\right) \,dxdydt, \\
& := \delta_1 + \delta_2.\end{aligned}$$ The term $\delta_1$ will be balanced against the first order derivative term on the left hand side of . To estimate $\delta_2$ we set $e(x,y,t)={\left|u^{{\varepsilon}}(x,y,t)-u(x,y,t)\right|}$ and following [@uk1], we find $$\label{eq:delta2est}
  \begin{aligned}
    \delta_2 &\le \iint \chi^\alpha_{(L_l,L_r)}(y,\nu) K_{\beta}(x) {\left|u^{{\varepsilon}}(x,y,\nu)-u(x,y,\nu)\right|}
    \,dxdy \\ & \qquad - \iint \chi^\alpha_{(L_l,L_r)}(y,\tau) K_{\beta}(x)
    {\left|u^{{\varepsilon}}(x,y,\tau)-u(x,y,\tau)\right|} \,dxdy +
    C\alpha_0.
  \end{aligned}$$ Now we rewrite the “first derivative term” on the left hand side of . Doing this, we get $$\begin{aligned}
  \int_{Q_T} & K_{\beta}(x) {\mathrm{sign}}(u^{{\varepsilon}} -u)
    \left(f(u^{{\varepsilon}})-f(u)\right)
    \left({\varphi}_y+{\varphi}_z\right) \,dX \\ &= 
    \int_{Q_T} K_{\beta}(x) \operatorname*{sg}(x,y,z,t,s)\left(f(u^{{\varepsilon}}(x,y,t))-f(u(x,y,t))\right) \\
   & \qquad \qquad \qquad \qquad  \Psi_y(y,t)\,\omega_r(y-z)\, \omega_{r_0}(t-s)\,dX\\
    &\quad \quad + 
    \int_{Q_T} K_{\beta}(x) \operatorname*{sg}(x,y,z,t,s)\left(f(u(x,y,t))-f(u(x,z,s))\right) \\
 & \qquad \qquad \qquad \qquad \Psi_y(y,t)\,
    \omega_r(y-z)\,\omega_{r_0}(t-s)\,dX \\
    &=:\delta_3 + \delta_4,\end{aligned}$$ where we have set $\operatorname*{sg}(x,y,z,t,s)={\mathrm{sign}}(u^{{\varepsilon}}(x,y,t)- u(x,z,s))$. We proceed as follows $$\begin{aligned}
  {\left|\delta_4\right|} \le \int_{Q_T} &{\left|f(u(x,y,t))-f(u(x,z,s))\right|}
  \chi^{\alpha_0}_{(\nu,\tau)}(t) K_{\beta}(x) \\
  &\qquad \omega_{r_0}(t-s)\,\omega_r(y-z)
  \left(
    h'_\alpha(y-L_l(t)) + h'_\alpha(y-L_r(t)-\frac{1}{\alpha})\right) \,dX.\end{aligned}$$ We follow [@uk1] to estimate each of these two terms to conclude $$\label{eq:delta4bnd}
  {\left|\delta_4\right|} \le C\left(r_0+r\right).$$ Again, choosing $M$ larger than the Lipschitz norm of $f$ implies that $$\label{eq:dl1dl2bnd}
  \delta_1+\delta_3\le 0.$$

Collecting all the terms we see that $$\label{eq:estimate-middle}
  \begin{aligned}
    \iint K_{\beta}(x) \chi^\alpha_{(L_l,L_r)}&(y,\tau) {\left|u^{{\varepsilon}}(x,y,\tau)-u(x,y,\tau)\right|}
    \,dxdy \\
    &\le \iint K_{\beta}(x) \chi^\alpha_{(L_l,L_r)}(y,\nu)
    {\left|u^{{\varepsilon}}(x,y,\nu)-u(x,y,\nu)\right|} \,dxdy\\
    &\qquad + C\left(r_0+ r +
      \alpha_0 + \alpha\right) + \biggl|{\int_{Q_T} \mathcal{Q}_1 + \mathcal{Q}_2 \, dX }\biggr|.
  \end{aligned}$$ In order to estimate $\mathcal{Q}_1$ and $\mathcal{Q}_2$, we proceed as follows $$\begin{aligned}
 {\left|\int_{Q_T}\mathcal{Q}_1\right|} \,dX & = \int_{Q_T} {\left|u^{{\varepsilon}} - u\right|}{\varphi}_{xx} \,dX \\
& = \int_{Q_T} {\left|u^{{\varepsilon}} - u\right|} K_{\beta}''(x) \Psi(y,t)\,\omega_r(y-z)\,\omega_{r_0}(t-s) \,dX \\
& \le {\left\|u^{{\varepsilon}} +u\right\|}_{L^{\infty}} \int_{Q_T} \frac{1}{\beta^2} \phi''(\frac{x}{\beta}) \Psi(y,t)\,\omega_r(y-z)\,\omega_{r_0}(t-s) \,dX \\
& \le \frac{C}{\beta^2} \iiint \phi''(\frac{x}{\beta}) {\chi^{\alpha_0}_{(\nu,\tau)}(t)}\,{\chi^\alpha_{(L_l,L_r)}(y,t)} \,dxdydt \\
& \le \frac{CK}{\beta \alpha},\end{aligned}$$ where $K := \int_{{\mathbb{R}}} {\left|\phi''(y)\right|} \,dy$.

Next, $$\begin{aligned}
& {\left|\int_{Q_T}\mathcal{Q}_2\right|} \,dX  = {\varepsilon}\int_{Q_T} {\left|u^{{\varepsilon}} - u\right|}_y {\varphi}_{y} \,dX \\
& = {\varepsilon}\int_{Q_T} {\mathrm{sign}}({u^{{\varepsilon}} - u}) (u^{{\varepsilon}})_y  K_{\beta}(x)\, \Psi_y(y,t)\,\omega_r(y-z)\,\omega_{r_0}(t-s) \\
& + {\varepsilon}\int_{Q_T} {\mathrm{sign}}({u^{{\varepsilon}} - u}) (u^{{\varepsilon}})_y K_{\beta}(x)\, \Psi(y,t)\,\omega_r'(y-z)\,\omega_{r_0}(t-s) \,dX \\
& := \mathcal{Q}_{2,1} + \mathcal{Q}_{2,2}.\end{aligned}$$ Each of the above terms can be approximated as follows $$\begin{aligned}
& {\left|\int_{Q_T}\mathcal{Q}_{2,1}\right|} \,dX   = {\varepsilon}\int_{Q_T} {\mathrm{sign}}({u^{{\varepsilon}} - u}) (u^{{\varepsilon}})_y K_{\beta}(x)\,\Psi_y(y,t)\,\omega_r(y-z)\,\omega_{r_0}(t-s) \,dX \\
& \le {\varepsilon}K \int_{{\Pi_T}}(u^{{\varepsilon}})_y (x,y,t) \chi^{\alpha_0}_{(\nu,\tau)}(t) 
    \left(h_\alpha'(y-L_l(t)) - h_\alpha'(y-L_r(t)-\frac{1}{\alpha})\right) \,dxdydt \\
& \le K {\varepsilon}\alpha \iint {\left|u^{{\varepsilon}}_x\right|} \,dxdy \\
& \le C K \alpha {\varepsilon},\end{aligned}$$ where $K = {\left\|K_{\beta}\right\|}_{\infty}$ and $C = \text{TV}\, (u^{{\varepsilon}})$. Similarly for the other term $$\begin{aligned}
{\left|\int_{Q_T}\mathcal{Q}_{2,2}\right|} \,dX & = {\varepsilon}\int_{Q_T} {\mathrm{sign}}({u^{{\varepsilon}} - u}) (u^{{\varepsilon}})_y  K_{\beta}(x)\, \Psi(y,t)\,\omega_r'(y-z)\,\omega_{r_0}(t-s) \,dX \\
& \qquad \le \frac{K {\varepsilon}}{r} \int_{{\Pi_T}}(u^{{\varepsilon}})_y (x,y,t) {\chi^{\alpha_0}_{(\nu,\tau)}(t)}\,{\chi^\alpha_{(L_l,L_r)}(y,t)}\,dxdydt \\ 
& \qquad  \le C K \frac{{\varepsilon}}{r},\end{aligned}$$ where again $K = {\left\|K_{\beta}\right\|}_{\infty}$ and $C = \text{TV}\, (u^{{\varepsilon}})$.

Therefore $$\label{eq:qsum}
  {\left|\int_{{\Pi_T}^2} \mathcal{Q}_1 + \mathcal{Q}_2 \,dX \right|} \le 
  C {\varepsilon}+ \frac{C}{\beta \alpha } + \frac{C {\varepsilon}}{r} .$$ where $C$ depends on (among other things) $L$ and $T$, but not on the parameters $\alpha_0$, $\alpha$, $r_0$, $r$, $\beta$ or ${\varepsilon}$.

Now we have proved the follwoing Lemma:

\[lem:estimates\] Assume that $u$ and $u^{{\varepsilon}}$ take values in the interval $[-K,K]$ for some positive $K$. Let $M>\max_{v\in
    [-K,K]} {\left|f'(v)\right|}$. Then if $T\ge\tau>\nu\ge 0$ and $L-M \tau >0$, we have $$\label{eq:estimates}
    \begin{aligned}
    \int_{{\mathbb{R}}}\int_{-L+M\tau}^{L-M\tau} &{\left|u^{{\varepsilon}}(x,y,\tau)-u(x,y,\tau)\right|}\,dxdy \\&\le 
    \int_{{\mathbb{R}}^2} {\left|u^{{\varepsilon}}(x,y,\nu)-u(x,y,\nu)\right|} \,dxdy\\
    &\quad + C\biggl[
    r_0 + r  +\alpha + {\varepsilon}+ \frac{1}{\alpha \beta} + \frac{{\varepsilon}}{r}\biggr].
    \end{aligned}$$

This follows from and , observing that we can send $\alpha_0$ to zero. Now we let $u(x,y,t)$ be the unique entropy solution of . Also note that since we let $\beta$ tends to $\infty$ so $ \gamma = \frac{1}{\beta}$ is small. We set $\alpha=r=r_0$ and ${\varepsilon}= \gamma$, and assume that $\alpha$ is sufficiently small, then $$\label{eq:vudx-estimate}
  \int_{{\mathbb{R}}}\int_{-L+Mt}^{L-Mt} {\left|{u_{{{\Delta x}}}}(x,t)-v(x,t)\right|}\,dx \,dy\le
  C\left(\alpha+\frac{{\varepsilon}}{\alpha}\right),$$ for some constant $C$ which is independent of the small parameters $\alpha$, ${\varepsilon}$. This follows from . Then setting ${\varepsilon}=\alpha^2$ proves the main theorem.

Acknowledgments {#acknowledgments .unnumbered}
===============

The second author acknowledges support from the Alexander von Humboldt Foundation, through a Humboldt Research Fellowship for postdoctoral researchers.

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Received xxxx 20xx; revised xxxx 20xx.
