In the original article we proved existence and uniqueness of entropy solutions for the nonhomogeneous Dirichlet problem associated to the relativistic heat equation. In the proof of uniqueness of both elliptic and parabolic entropy solutions (Theorems 2 and 3, respectively) we used that \(\mathrm{sign}^+_0(u - \overline{u}) u \in BV(\Omega )\) for two entropy solutions \(u,\overline{u}\). However, we did not prove this but we wrongly stated that this was true for any given functions \(u,\overline{u}\in { TBV}(\Omega )\), which is not the case, in general.
The expression \(\mathrm{sign}^+_0(u - \overline{u}) u\) appears in the process of applying the doubling variables method, known as Kruzhkov’s method, if one lets the parameter \({\varepsilon }\rightarrow 0^+\), which serves to approximate the sign function, before joining the space variables. In this note, we correct the aforementioned mistake by joining first the space variables and at the very end, letting \({\varepsilon }\rightarrow 0^+\).
We assume here all notations in the original article. Moreover, due to the fact that \(u(t,x)\) is a solution to \(u_t=\nu \mathrm{div}\Big (\frac{u\nabla u}{\sqrt{u^2+\frac{\nu ^2}{c^2}|\nabla u|^2}}\Big )\) if and only if \(v(t,x)=u\left( \frac{\nu t}{c^2},\frac{\nu }{c}x\right) \) is a solution to \(v_t=\mathrm{div}(\mathbf{a}(v,Dv))\) with \(\mathbf{a}(z, \xi ): = \frac{z \xi }{\sqrt{z^2 + \vert \xi \vert ^2}}\), for the sake of simplicity, we assume that \(\nu =c=1\).
In order to be able to join first the space variables in Kruzkov’s method, we need first some auxiliary results.
Lemma 1
Let \(U\subset \mathbb {R}^k\) be an open set. Let \(\mu \in \mathcal M(U)\) be such that \(|\mu |(U)<+\infty \). Then, given an uncountable collection of \(\mu \)-measurable disjoint subsets \(\{B_{\varepsilon }\}\) of \(U\) it holds
$$\begin{aligned} \#\{{\varepsilon }: |\mu |(B_{\varepsilon })>0\}\le \aleph _0. \end{aligned}$$
Proof
By contradiction, suppose that \(A:=\{{\varepsilon }: |\mu |(B_{\varepsilon })>0\}\) is uncountable. Then, there is \(n\in {\mathbb N}\) such that the set \(A_n:=\left\{ {\varepsilon }: |\mu |(B_{\varepsilon })>\frac{1}{n}\right\} \subseteq A\) is uncountable (otherwise one could write \(A=\cup _{n=1}^\infty A_n\) and \(A\) would be countable). This implies that there exists a sequence \(\{{\varepsilon }_k\}_{k=0}^\infty \subset A_n\). By \(\sigma \)-additivity of \(\mu \) we would have a contradiction since
$$\begin{aligned} +\infty =\sum _{k=0}^\infty |\mu |(B_{{\varepsilon }_k})=|\mu |\left( \bigcup _{k=0}^\infty B_{{\varepsilon }_k}\right) \le |\mu |(U)<+\infty . \end{aligned}$$
\(\square \)
Let \(w,\overline{w}\in L^1(\Omega )\), \(\psi \in \mathcal D(\Omega )\) and \(F \in W^{1,\infty }({\mathbb R})\). Given \(x\in \Omega {\setminus } S_{\overline{w}}\), we define \(f_x:\Omega \rightarrow {\mathbb R}\) as
$$\begin{aligned} f_x(y):=\psi \left( \frac{y+x}{2} \right) {\chi }_{\{0<w(y)-\overline{w}(x)<{\varepsilon }\} }(F(w)(y)-F(\overline{w})(x)). \end{aligned}$$
Lemma 2
Let \(\rho _n\) be a sequence of classical mollifiers in \(\Omega \) and let \(w,\overline{w}\in BV(\Omega )\). Then, if \(|D^c \overline{w}|(\{\tilde{w}-\tilde{\overline{w}}={\varepsilon }\})=0\) it holds
$$\begin{aligned} \int _{\Omega }(f_x*\rho _n)(x)d|D^c \overline{w}|\mathop {\rightarrow }\limits ^{n\rightarrow \infty }\int _\Omega f_x(x)d|D^c \overline{w}|. \end{aligned}$$
Proof
Since \(J_w\) is \(\mathcal H^{N-1}\) rectifiable, then \(J_w\) is \(\sigma \)-finite with respect to the Haussdorff measure \(\mathcal H^{N-1}\). Then, by [1], Proposition 3.92], it follows that \(|D^c\overline{w}|(S_w)=|D^c\overline{w}|(J_w)=0\). Hence,
$$\begin{aligned} \int _{\Omega }(f_x*\rho _n)(x)d|D^c \overline{w}| = \int _{\Omega {\setminus } ( S_{\overline{w}} \cup S_w) }(f_x*\rho _n)(x)d|D^c (\overline{w})|. \end{aligned}$$
Since \(f_x\in L^1(\Omega )\), we have that \((f_x*\rho _n)(y)\rightarrow f_x(y)\) for all \(y \in \Omega {\setminus } S_{f_x}\). On the other hand, if \(x\in \Omega {\setminus }(S_w\cup S_{\overline{w}})\), then
$$\begin{aligned} x\in \Omega {\setminus } S_{f_x} \ \iff x\in \Omega {\setminus } S_{{\chi }_{\{0<w(\cdot )-\overline{w}(x)<{\varepsilon }\} }}=\Omega {\setminus } \partial ^*{\{0<w(\cdot )-\overline{w}(x)<{\varepsilon }\} }. \end{aligned}$$
Now, \(\partial ^*({\{0<w(\cdot )-\overline{w}(x)<{\varepsilon }\} })\subset \{w(\cdot )=\overline{w}(x)\}\cup \{w(\cdot )-\overline{w}(x)={\varepsilon }\} )\). Observe also that \(x\in \{w(\cdot )-\overline{w}(x)={\varepsilon }\}\) if and only if \(x\in \{w-\overline{w}={\varepsilon }\}\).
In case \(w(x)=\overline{w}(x)\),
$$\begin{aligned} |(f_x*\rho _n)(x)|\le & {} \int _\Omega \rho _n(x-y)|f_x(y)| dy\\\le & {} \Vert \nabla F \Vert _\infty \int _\Omega \rho _n(x-y)\psi \left( \frac{x+y}{2}\right) |w(y)-\overline{w}(x)|\ dy\rightarrow 0=f_x(x) \end{aligned}$$
Then, if \(x\in \Omega {\setminus }(S_w\cup S_{\overline{w}}\cup \{\tilde{w}-\tilde{\overline{w}}={\varepsilon }\})\), we obtain that \(f_x*\rho _n(x)\rightarrow f_x(x)\) pointwise. Moreover, since \(\Vert f_x*\rho _n\Vert _\infty \le \Vert f_x\Vert _\infty \), we can apply the dominated convergence Theorem to obtain that
$$\begin{aligned} \int _{\Omega {\setminus } (S_w\cup S_{\overline{w}}\cup \{\tilde{w}-\tilde{\overline{w}}={\varepsilon }\})}(f_x*\rho _n)(x)d|D^c \overline{w}| \mathop {\rightarrow }\limits ^{n\rightarrow \infty }\displaystyle \int _{\Omega {\setminus } (S_w\cup S_{\overline{w}}\cup \{\tilde{w}-\tilde{\overline{w}}={\varepsilon }\})}f_x(x)d|D^c \overline{w}| \end{aligned}$$
Therefore,
$$\begin{aligned} \int _{\Omega }(f_x*\rho _n)(x)d|D^c \overline{w}|\mathop {\rightarrow }\limits ^{n\rightarrow \infty }\int _\Omega f_x(x)d|D^c \overline{w}|. \end{aligned}$$
\(\square \)
Let \(w,\overline{w}\in BV(\Omega )\) and \(\psi \in \mathcal D(\Omega )\). Given \(x\in \Omega {\setminus } (S_{\overline{w}}\cup S_{\nabla \overline{w}})\), we define \(g_x:\Omega \rightarrow {\mathbb R}\) as
$$\begin{aligned} g_x(y):=\psi \left( \frac{y+x}{2}\right) {\chi }_{\{0<w(y)-\overline{w}(x)<{\varepsilon }\} }(w(y)-\overline{w}(x))|\nabla w(y)-\nabla _x\overline{w}(x)|. \end{aligned}$$
Lemma 3
Let \(\rho _n\) be a sequence of classical mollifiers in \(\Omega \). If \(\mathcal L^N(\{w=\overline{w}+{\varepsilon }\})=0\), then
$$\begin{aligned} \int _{\Omega }(g_x*\rho _n)(x)dx\mathop {\rightarrow }\limits ^{n\rightarrow \infty }\int _\Omega g_x(x)dx. \end{aligned}$$
Proof
Let us denote by \(\tilde{\Omega }:=\Omega {\setminus }(S_w\cup S_{\nabla w})\). Then,
$$\begin{aligned} \int _{\Omega }(g_x*\rho _n)(x)dx= & {} \int _{\tilde{\Omega }}(g_x*\rho _n)(x)dx\\= & {} \int _{\tilde{\Omega }\cap \{w=\overline{w}\}}(g_x*\rho _n)(x)dx+\int _{\tilde{\Omega }{\setminus }\{w=\overline{w}\}}(g_x*\rho _n)(x)dx. \end{aligned}$$
For the first term,
$$\begin{aligned} \left| \int _{\tilde{\Omega }\cap \{w=\overline{w}\}}(g_x*\rho _n)(x)dx\right|\le & {} C\int _{\tilde{\Omega }\cap \{w=\overline{w}\}}n^N\int _{\Omega \cap B_{\frac{1}{n}}(x)}|\nabla w(y)-\nabla _x\overline{w}(x)|dy dx\\\rightarrow & {} \frac{C}{\omega _N}\int _{\tilde{\Omega }\cap \{w=\overline{w}\}}|\nabla w(x)-\nabla \overline{w} (x)| \, dx \mathop = \limits ^{{[1,Remark\,\,\,3.93]}}\,0 \end{aligned}$$
For the second term, we split \(\tilde{\Omega }{\setminus }\{w=\overline{w}\}\) into two Borel sets: \(\tilde{\Omega }_1 :=\tilde{\Omega }{\setminus }(\{w=\overline{w}\}\cup \{w=\overline{w}+{\varepsilon }\})\), \(\tilde{\Omega }_2:=\tilde{\Omega }\cap \{w=\overline{w}+{\varepsilon }\}\). Observe that in \(\tilde{\Omega }_2\) both the integral and its limit are equal to \(0\) and we do not have to prove anything. In \(\tilde{\Omega }_1\) instead, all points are Lebesgue points of \(g_x\). Therefore, applying dominated convergence Theorem (note that \(\Vert g_x*\rho _n\Vert _1\le \Vert g_x\Vert _1<+\infty \)) we obtain the result. \(\square \)
With an easy adaptation of the proofs of the preceding results, one can obtain their time dependent counterparts.
Lemma 4
Let \(\rho _n\), \(\rho _m\) be sequences of classical mollifiers in \(\Omega \) and \((0,T)\), respectively, \(w,\overline{w}\in L^1_{loc,w}(0,T;BV(\Omega ))\), \(\psi \in \mathcal D(\Omega )\) and
$$\begin{aligned} f_x(s,t)(y):=\rho _m(s-t)\psi \left( \frac{y+x}{2} \right) {\chi }_{\{0<w(s,y)-\overline{w}(t,x)<{\varepsilon }\} }(F(w)(s,y)\!-\!F(\overline{w})(t,x)). \end{aligned}$$
Then, if \(\int _{(0,T)^2}|D^c \overline{w}(s)|(\{\widetilde{w(s)}-\widetilde{\overline{w}(t)}={\varepsilon }\}))ds dt=0\) it holds
$$\begin{aligned}&\int _{(0,T)^2}\int _{\Omega }(f_x(s,t)*\rho _n)(x)d|D^c \overline{w}(s)| ds dt \\&\quad \mathop {\rightarrow }\limits ^{n\rightarrow \infty }\displaystyle \int _{(0,T)^2}\int _\Omega f_x(s,t)(x)d|D^c \overline{w}(s)|ds dt. \end{aligned}$$
Lemma 5
Let \(\rho _n\)
\(\rho _m\) be sequences of classical mollifiers in \(\Omega \) and \((0,T)\), respectively, \(w,\overline{w}\in L^1_{loc,w}(0,T;BV(\Omega ))\), \(\psi \in \mathcal D(\Omega )\) and
$$\begin{aligned} g_x(s,t)(y):= & {} \rho _m(s-t)\psi \left( \frac{y+x}{2}\right) {\chi }_{\{0<w(s,y)-\overline{w}(t,x)<{\varepsilon }\} } \\&\times \, (w(s,y)-\overline{w}(t,x))|\nabla w(s,y)-\nabla _x\overline{w}(t,x)|. \end{aligned}$$
If \(\mathcal L^{N+2}(\{w(s)=\overline{w}(t)+{\varepsilon }\})=0\), then
$$\begin{aligned} \int _{(0,T)^2}\int _{\Omega }(g_x(s,t)*\rho _n)(x)dx ds dt \mathop {\rightarrow }\limits ^{n\rightarrow \infty }\int _{(0,T)^2}\int _\Omega g_x(s,t)(x)dx ds dt. \end{aligned}$$
We have now all the ingredients for the proofs of uniqueness. Note that in the proof of Theorem 2 of the orginial article the mistake takes place in passing from (41) up to (42) while in the proof of Theorem 3 of the original article from (124) up to (130). Since both of the proofs are similar, we will give the whole proof for the elliptic case (up to (42) in the original article) and we will only sketch the one for the parabolic case.
In the statement of Theorem 2 of the original article, there was also a mistake. The correct statement is the following one.
Theorem 1
Given \(0\le g\in L^\infty (\Omega )\), \(v, \overline{v} \in L^{1}(\Omega )\), \(v\ge 0\), \(\overline{v} \ge 0\), let \(u, \overline{u}\) be two bounded entropy solutions of the problems
$$\begin{aligned} \left\{ \begin{array}{ll} u - \mathrm{div}\ \mathbf{a}(u, Du) = v &{}\quad \hbox {in} \quad \Omega \\ u=g &{}\quad \hbox {on}\quad \partial \Omega \end{array}\right. \end{aligned}$$
(1)
and
$$\begin{aligned} \left\{ \begin{array}{ll} \overline{u} - \mathrm{div}\ \mathbf{a}(\overline{u}, D\overline{u}) = \overline{v} &{}\quad \hbox {in}\quad \Omega \\ \overline{u}=g &{}\quad \hbox {on}\quad \partial \Omega \end{array}\right. \end{aligned}$$
(2)
respectively. Then,
$$\begin{aligned} \int _\Omega (u - \overline{u})^+ \le \int _\Omega (v - \overline{v})^+. \end{aligned}$$
Proof
For \(a > 0\), we denote by \(T^\infty _a\) the truncature function
$$\begin{aligned} T^\infty _a(s):= \left\{ \begin{array}{ll} a &{}\quad \hbox {if} \quad s \le a \\ s &{}\quad \hbox {if} \quad s \ge a. \end{array} \right. \end{aligned}$$
Let \(b > a > 0\). We note that by Lemma 1 , we can choose \({\varepsilon }\rightarrow 0\) such that \({\varepsilon }<\frac{a}{2}\),
$$\begin{aligned} \mathcal L^N\left\{ T_a^\infty \tilde{u}-{T_\frac{a}{2}^\infty \tilde{\overline{u}}}={\varepsilon }\right\} =0 \end{aligned}$$
(3)
and \(\left( |D^c T_{a}^\infty u|+|D^c T_{\frac{a}{2}}^\infty \overline{u}|\right) \left\{ T_a^\infty \tilde{u}-{T_\frac{a}{2}^\infty \tilde{\overline{u}}}={\varepsilon }\right\} =0\). Observe now that
$$\begin{aligned} \left\{ T_a^\infty \tilde{u}-{T_\frac{a}{2}^\infty \tilde{\overline{u}}}={\varepsilon }\right\} \cap \{\tilde{u}>a\}=\left\{ T_a^\infty \tilde{u}-{T_{a-{\varepsilon }}^\infty \tilde{\overline{u}}}={\varepsilon }\right\} \cap \{\tilde{u}>a\}, \end{aligned}$$
which implies that \(|D^c T_{a}^\infty u|\{T_a^\infty \tilde{u}-{T_{a-{\varepsilon }}^\infty \tilde{\overline{u}}}={\varepsilon }\}=0.\) On the other hand,
$$\begin{aligned} |D^c T_{a-{\varepsilon }}^\infty \overline{u}|\{T_a^\infty \tilde{u}-{T_{ a - {\varepsilon }}^\infty \tilde{\overline{u}}}={\varepsilon }\}= & {} |D^c T_{a-{\varepsilon }}^\infty \overline{u}|\left\{ T_a^\infty \tilde{u}-{T_\frac{a}{2}^\infty \tilde{\overline{u}}}={\varepsilon }\right\} \\\le & {} |D^c T_\frac{a}{2}^\infty \overline{u}|\left\{ T_a^\infty \tilde{u}-{T_\frac{a}{2}^\infty \tilde{\overline{u}}}={\varepsilon }\right\} =0. \end{aligned}$$
Therefore,
$$\begin{aligned} (|D^c T_{a}^\infty u|+|D^c T_{a-{\varepsilon }}^\infty \overline{u}|)\{T_a^\infty \tilde{u}-{T_{a-{\varepsilon }}^\infty \tilde{\overline{u}}}={\varepsilon }\}=0. \end{aligned}$$
(4)
We consider \(T(r):=T_{a,b}(r)-a\), \(S_{{\varepsilon }, l}(r):= T_{{\varepsilon }}(r - l)^+ = T_{l, l + {\varepsilon }}(r) - l \in \mathcal{P}^+\) and \(S_{\varepsilon }^l(r):= T_{{\varepsilon }}(r - l)^- + {\varepsilon }= T_{l- {\varepsilon }, l}(r) + {\varepsilon }- l \in \mathcal{P}^+,\) where \(l \ge 0\). Let us denote
$$\begin{aligned} J^+_{T,{\varepsilon },l}(r) = \int _l^r T(s) T_{{\varepsilon }}(s-l)^+ \, ds, \ J^-_{T,{\varepsilon },l}(r) = \int _l^r T(s) T_{{\varepsilon }}(s-l)^- \, ds. \end{aligned}$$
Given \(0\le g\in L^\infty (\Omega )\), \(v, \overline{v} \in L^{1}(\Omega )\), \(v\ge 0\), \(\overline{v} \ge 0\), let \(u, \overline{u}\) be bounded entropy solutions of the problems (1) and (2), respectively. Let \(\rho _n\) be a sequence of classical mollifiers in \(\Omega \), \(0 \le \psi \in \mathcal {D}(\Omega )\) and \(b > a> 2 {\varepsilon }> 0\). We write \(\displaystyle \xi _{n}(x,y) = \rho _n(x-y)\psi \left( \frac{x+y}{2}\right) . \)
If we denote \(\mathbf{z}(y) = \mathbf{a}(u(y),\nabla u(y))\) and \(\overline{\mathbf{z}}(x) = \mathbf{a}(\overline{u}(x), \nabla \overline{u}(x))\), we have
$$\begin{aligned} u - \mathrm{div}(\mathbf{z}) = v \, \quad \mathrm{and}\quad \overline{u} - \mathrm{div}(\overline{\mathbf{z}}) = \overline{v}\quad \mathrm{in}\quad \mathcal{D}^{\prime }(\Omega ). \end{aligned}$$
Multiplying Eq. (1) by \(T(u)S_{{\varepsilon }, \overline{u}}(u) \xi _n\) and (2) by \(T(\overline{u}) S_{\varepsilon }^{u}(\overline{u}) \xi _n\) and integrating by parts, integrating again in \(x\) and \(y\) respectively and adding both equations, we obtain
$$\begin{aligned}&\displaystyle \int _{\Omega } \int _{\Omega } (uT(u)-\overline{u}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\xi _n \, dx \, dy + {\varepsilon }\displaystyle \int _{\Omega } \int _{\Omega } (\overline{u}- \overline{v})T(\overline{u}) \xi _n \, dx dy\nonumber \\&\qquad + \displaystyle \int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z}, D_y(T(u)S_{{\varepsilon },\overline{u}}(u)) \right) \, \, dx + \displaystyle \int _{\Omega \times \Omega } T(u) S_{{\varepsilon },\overline{u}}(u) \mathbf{z}\cdot \nabla _y \xi _n \, dy \, dx\nonumber \\&\qquad + \displaystyle \int _{\Omega } \left( \int _{\Omega } \xi _{n} (\overline{\mathbf{z}}, D_x(T(\overline{u}) S_{\varepsilon }^{u}(\overline{u}))) \right) \, dy + \displaystyle \int _{\Omega \times \Omega }T(\overline{u}) S_{\varepsilon }^{u}(\overline{u}) \overline{\mathbf{z}} \cdot \nabla _x \xi _n \, dx \, dy \nonumber \\&\quad = \displaystyle \int _{\Omega } \int _{\Omega } (vT(u)-\overline{v}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\xi _n \, dx \, dy. \end{aligned}$$
(5)
Let \(I_1\), \(I_2\) be, respectively, the first term and the rest of the terms at the left hand side of the above identity, and let \(I_3\) be the right hand side term.
Now, since \(\overline{u}-\overline{v} = \mathrm{div}\, \overline{\mathbf{z}}\) and \(\nabla _y \xi _n(x,y) +\nabla _x \xi _n(x,y) = \rho _n\nabla \psi \), we have
$$\begin{aligned} I_2= & {} {\varepsilon }\displaystyle \int _{\Omega } \int _{\Omega } T(\overline{u})(\mathrm{div}\, (\overline{\mathbf{z}}) \, \xi _n +\overline{\mathbf{z}}\cdot \nabla _x\xi _n)\, dx dy\\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z}, D_y(T(u)S_{{\varepsilon },\overline{u}(x)}(u))) \right) \, \, dx + \int _{\Omega \times \Omega }T(\overline{u}) T_{\varepsilon }(u-\overline{u})^+ \overline{\mathbf{z}} \cdot \nabla _y \xi _n \, dx \, dy\\&+ \int _{\Omega } \left( \int _{\Omega } \xi _{n} (\overline{\mathbf{z}}, D_x(T(\overline{u})S_{\varepsilon }^{u(y)}(\overline{u}))) \right) \, dy - \int _{\Omega \times \Omega } T(u)T_{{\varepsilon }}(u-\overline{u})^+ \mathbf{z}\cdot \nabla _x \xi _n \, dy \, dx\\&+\int _{\Omega \times \Omega }\rho _n T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi ={\varepsilon }\displaystyle \int _{\Omega } \int _{\Omega }{ T(\overline{u}) \mathrm{div}\, (\overline{\mathbf{z}} \, \xi _n)} \, dx dy \\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z}, D_y(T(u)S_{{\varepsilon },\overline{u}(x)}(u))) \right) dx - \int _{\Omega \times \Omega } \xi _n \overline{\mathbf{z}} \cdot D_y(T(\overline{u}) T_{\varepsilon }{ (u-\overline{u})^+}) \, dx dy\\&+ \int _{\Omega } \left( \int _{\Omega } \xi _{n} (\overline{\mathbf{z}}, D_x(T(\overline{u})S_{\varepsilon }^{u(y)}(\overline{u}))) \right) \, dy + \int _{\Omega \times \Omega } \xi _n \mathbf{z}\cdot D_x (T(u)T_{\varepsilon }(u - \overline{u})^+)) \, dy \, dx\\&+\int _{\Omega \times \Omega }\rho _n T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi ={ -}{\varepsilon }\displaystyle \int _{\Omega } \int _{\Omega } \xi _n (\overline{\mathbf{z}},DT(\overline{u})) dy \\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z},D_y J_{T'S_{{\varepsilon },\overline{u}(x)})}(u)) \right) \, dx + \int _{\Omega } \left( \int _{\Omega } \xi _{n}(\overline{\mathbf{z}},D_x J_{T'S_{\varepsilon }^{u(y)}}(\overline{u})) \right) \, dy\\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z},D_y J_{TS^{\prime }_{{\varepsilon },\overline{u}(x)}}(u)) \right) \, dx -\int _{\Omega } T(\overline{u}) \left( \int _{\Omega } \xi _{n} \overline{\mathbf{z}} \cdot D_y \, T_{\varepsilon }(u-\overline{u})^+ \right) \, dx\\&+ \int _{\Omega }\left( \int _{\Omega } \xi _{n}(\overline{\mathbf{z}},D_x J_{TS_{\varepsilon }^{u(y)\prime }} (\overline{u})) \right) \, dy + \int _{\Omega } T(u(y)) \left( \int _{\Omega } \xi _{n} \mathbf{z}\cdot D_x T_{\varepsilon }(u-\overline{u})^+ \right) \, dy\\&+\int _{\Omega \times \Omega }\rho _n T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi = I^1_2 + I^2_2, \end{aligned}$$
where \(I_2^1\) denotes the sum of the first three terms and the last one while \(I_2^2\) denotes the sum from the fourth to the seventh terms.
Let us consider the second and third terms in \(I_2^1\). Since by Definition 1 in the original article,
$$\begin{aligned} h_{S_{{\varepsilon },\overline{u}(x)}}(u, DT(u))&\le (\mathbf{z},D_y J_{T'S_{{\varepsilon },\overline{u}(x)}}(u)) \quad \mathrm{and \ }\\ h_{S_{\varepsilon }^{u(y)}}(\overline{u}, DT(\overline{u}))&\le (\overline{\mathbf{z}},D_x J_{T'S_{\varepsilon }^{u(y)}}(\overline{u})) \end{aligned}$$
as measures in \(\Omega \) (see Eq. (30) in the original article for the Definitions of the measures \(h_S\)), we have
$$\begin{aligned}&\int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z},D_y J_{T'S_{{\varepsilon },\overline{u}(x)})}(u)) \right) \, dx \ge 0\quad \mathrm{and \ }\\&\int _{\Omega } \left( \int _{\Omega } \xi _{n}(\overline{\mathbf{z}},D_x J_{T'S_{\varepsilon }^{u(y)}}(\overline{u})) \right) \, dy \ge 0. \end{aligned}$$
Hence,
$$\begin{aligned} I_2^1 \ge {\varepsilon }\displaystyle \int _{\Omega } \int _{\Omega } \xi _n (\overline{\mathbf{z}},DT(\overline{u})) dy+\int _{\Omega ^2}\rho _n T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi . \end{aligned}$$
(6)
We split \(I^2_2\) into \(I^2_2= I^2_2(ac) + I^2_2(s),\) where \(I^2_2(ac)\) contains the absolutely continuous parts of the integrands in \(I^2_2\) and \(I^2_2(s)\) contains their singular parts. Now,
$$\begin{aligned} I^2_2(ac)= & {} \int _{\Omega } \int _{\Omega } \xi _{n} T(u)\, \mathbf{z}\cdot \nabla _y T_{\varepsilon }(u-\overline{u})^+ \, dy\, dx \\&- \int _{\Omega } \int _{\Omega } \xi _{n} T(\overline{u})\, \overline{\mathbf{z}} \cdot \nabla _y \, T_{\varepsilon }(u-\overline{u})^+ \, dy \, dx\\&- \int _{\Omega } \int _{\Omega } \xi _{n} T(\overline{u}) \, \overline{\mathbf{z}} \cdot \nabla _x T_{\varepsilon }(u-\overline{u})^+ \, dx \, dy \\&+ \int _{\Omega } \int _{\Omega } \xi _{n} T(u)\, \mathbf{z}\cdot \nabla _x \, T_{\varepsilon }(u-\overline{u})^+ \, dx \, dy\\= & {} \int _{\Omega } \int _{\Omega } \xi _n \left( \mathbf{z}T(u)-\overline{\mathbf{z}} T(\overline{u}) \right) \left( \nabla _y T_{\varepsilon }(u-\overline{u})^+ + \nabla _x T_{\varepsilon }(u-\overline{u})^+ \right) \, dx \, dy\\= & {} \int _{\Omega } \int _{\Omega } \xi _n (\mathbf{z}-\overline{\mathbf{z}})T(u) (\nabla _y T_{\varepsilon }(u-\overline{u})^+ + \nabla _x T_{\varepsilon }(u-\overline{u})^+)\, dx \, dy\\&+\int _{\Omega } \int _{\Omega } \xi _n \overline{\mathbf{z}}( T(u)-T(\overline{u})) (\nabla _y T_{\varepsilon }(u-\overline{u})^+ + \nabla _x T_{\varepsilon }(u-\overline{u})^+)\, dx \, dy\\=: & {} A^1 + A^2. \end{aligned}$$
Let us estimate \(A^1\). First, observe that
$$\begin{aligned} \nabla _y T_{\varepsilon }(u-\overline{u}(x))^+(y)= & {} {\chi }_{(\overline{u}(x),\overline{u}(x)+{\varepsilon })}(u(y))\nabla _y u(y),\\ \nabla _x T_{\varepsilon }(u(y)-\overline{u})^+(x)= & {} - {\chi }_{(u(y)-{\varepsilon },u(y))}(\overline{u}(x)) \nabla _x \overline{u}(x)\\= & {} - {\chi }_{(\overline{u}(x),\overline{u}(x)+{\varepsilon })}(u(y)) \nabla _x \overline{u}(x). \end{aligned}$$
Since
$$\begin{aligned} (\mathbf{a}(z, \xi ) - \mathbf{a}( \hat{z}, \hat{\xi })) \cdot (\xi - \hat{\xi }) \ge - C \vert z - \hat{z} \vert \, \Vert \xi - \hat{\xi } \Vert \end{aligned}$$
(7)
for any \((z,\xi ), (\hat{z},\xi )\in {\mathbb R}\times {\mathbb R}^N\), \(\vert z\vert ,\vert \hat{z}\vert \le R\), we have
$$\begin{aligned} A^1= & {} \int _{\Omega } \int _{\Omega } \xi _n (\mathbf{z}-\overline{\mathbf{z}})T(u) (\nabla _y u - \nabla _x \overline{u}) {\chi }_{(\overline{u}(x),\overline{u}(x)+{\varepsilon })}(u)\, dx \, dy\\\ge & {} - C \Vert T(u)\Vert _\infty \int _{\Omega } \int _{\Omega } {\chi }_{[u> a]}\xi _n {\chi }_{(\overline{u}(x),\overline{u}(x)+{\varepsilon })}(u) \, \vert u- \overline{u}\vert \, \Vert \nabla _y u - \nabla _x \overline{u} \Vert \, dx\, dy\\\ge & {} - \tilde{C} \int _{\Omega ^2} \xi _n {\chi }_{(T_\frac{a}{2}^\infty \overline{u}(x),T_\frac{a}{2}^\infty \overline{u}(x)+{\varepsilon })}(T_a^\infty u) \, \vert T_a^\infty u- T_\frac{a}{2}^\infty \overline{u}\vert \, \Vert \nabla _y T_a^\infty u\\&- \nabla _x T_\frac{a}{2}^\infty \overline{u} \Vert \, dx dy. \end{aligned}$$
Having in mind (3) and applying Lemma 3 we obtain that
$$\begin{aligned} \lim _{n\rightarrow \infty } A^1\ge & {} -\tilde{C}\int _\Omega \psi (y){\chi }_{\{0<T_a^\infty u-T_\frac{a}{2}^\infty \overline{u}<{\varepsilon }\}}(T_a^\infty u-T_\frac{a}{2}^\infty \overline{u})|\nabla T_a^\infty u-\nabla T_\frac{a}{2}^\infty \overline{u}|dy\\\ge & {} -\tilde{C}{\varepsilon }\int _\Omega \psi (y){\chi }_{\{0<T_a^\infty u-T_\frac{a}{2}^\infty \overline{u}<{\varepsilon }\}}|\nabla T_a^\infty u-\nabla T_\frac{a}{2}^\infty \overline{u}|dy\ge -{\varepsilon }o({\varepsilon }), \end{aligned}$$
where \(o({\varepsilon }) \rightarrow 0\) when \({\varepsilon }\rightarrow 0^+\), and we have used coarea formula in the last inequality. Similarly,
$$\begin{aligned} \vert A^2 \vert= & {} \left| \int _{\Omega } \int _{\Omega } \xi _n \overline{\mathbf{z}}( T(u)-T(\overline{u})) (\nabla _y u - \nabla _x \overline{u}) {\chi }_{[\overline{u}(x),\overline{u}(x)+{\varepsilon }]}(u) \, dx \, dy \right| \\\le & {} \int _{\Omega }\int _{\Omega } {\chi }_{[u\ge a-{\varepsilon }]} {\chi }_{[\overline{u} \ge a-{\varepsilon }]} {\chi }_{[0\le u - \overline{u} \le {\varepsilon }]} \xi _n \vert u - \overline{u}\vert \, \Vert \nabla _y u - \nabla _x \overline{u} \Vert \, dx\, dy \le {\varepsilon }o({\varepsilon }). \end{aligned}$$
Hence,
$$\begin{aligned} \lim _{n\rightarrow +\infty } I_2^2(ac) \ge - {\varepsilon }o({\varepsilon }). \end{aligned}$$
Finally, let us compute \(I_2^2(s)\).
$$\begin{aligned} I_2^2(s)= & {} \int _{\Omega } \left( \int _{\Omega } \xi _{n} (\mathbf{z}, D_y J_{TS'{_{\varepsilon }, \overline{u}(x)}}(u))^s \right) \, dx\\&- \int _{\Omega } \left( \int _{\Omega } \xi _{n} T(\overline{u})\, \overline{\mathbf{z}} \cdot D_y^s \, T_{\varepsilon }(u-\overline{u})^+ \right) \, dx\\&+ \int _{\Omega } \left( \int _{\Omega } \xi _{n} (\overline{\mathbf{z}}, D_x J_{TS_{\varepsilon }^{u(y)\prime }}(\overline{u}))^s\right) \, dy\\&+ \int _{\Omega } \left( \int _{\Omega } \xi _{n} T(u)\, \mathbf{z}\cdot D_x^s T_{\varepsilon }(u -\overline{u})^+\right) \, dy\\\ge & {} \int _{\Omega } \left( \int _{\Omega } \xi _{n} ((h_T(u,D_y T_{\overline{u}}^{\overline{u}+{\varepsilon }}(u)))^s-T(\overline{u})\, \overline{\mathbf{z}} \cdot D_y^s \, T_{\varepsilon }(u-\overline{u})^+)\right) \, dx\\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} ((h_T(\overline{u},D_x T_{u-{\varepsilon }}^{u}(\overline{u})))^s+T({u})\, {\mathbf{z}} \cdot D_x^s \, T_{\varepsilon }(u-\overline{u})^+)\right) \, dy\\= & {} \int _{\Omega } \left( \int _{\Omega } \xi _{n} (|D_y^s J_{T\varphi }(T_{\overline{u}}^{\overline{u}+{\varepsilon }}(u))|-T(\overline{u})\, \overline{\mathbf{z}} \cdot D_y^s \, T_{\varepsilon }(u-\overline{u})^+)\right) \, dx\\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} (|D_x^s J_{T\varphi }(T_{u-{\varepsilon }}^{u}(\overline{u}))|+T({u})\, {\mathbf{z}} \cdot D_x^s \, T_{\varepsilon }(u-\overline{u})^+)\right) \, dy \\= & {} I_2^2(s,c)+I_2^2(s,j), \end{aligned}$$
where \(\varphi (s)=s\), \(I_2^2(s,c)\) collects all the Cantor parts of the measures and \(I_2^2(s,j)\) their jump parts. For the Cantor part,
$$\begin{aligned} I_2^2(s,c)= & {} \int _\Omega \left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}}(uT(u)|D^c u|-\overline{u}T(\overline{u})\overline{z}_b\cdot D^c u)\right) dx\\&+\int _\Omega \left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}}(\overline{u}T(\overline{u})|D^c \overline{u}|-uT(u)z_b\cdot D^c \overline{u})\right) dy\\\ge & {} \int _\Omega \left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}}(uT(u)-\overline{u}T(\overline{u}))|D^c u|\right) dx\\&+\int _\Omega \left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}}(\overline{u}T(\overline{u})-uT(u)) |D^c \overline{u}|\right) dy\\= & {} \int _\Omega \left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}}(uT(u)-\overline{u}T(\overline{u}))dx\right) d|D^c u|\\&+\int _\Omega \left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}}(\overline{u}T(\overline{u})-uT(u)) dy\right) d|D^c \overline{u}|\\= & {} \int _{\Omega \cap \{u>a\}}\left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}\cap \{\overline{u}>a-{\varepsilon }\}}(uT(u)-\overline{u}T(\overline{u}))dx\right) d|D^c u|\\&+\int _{\Omega \cap \{\overline{u}>a-{\varepsilon }\}}\left( \int _\Omega \xi _n{\chi }_{\{0<u-\overline{u}<{\varepsilon }\}\cap \{u>a\}}(\overline{u}T(\overline{u})-uT(u)) dy\right) d|D^c \overline{u}|. \end{aligned}$$
By Lemma 2, taking \(F(s) = s T(s)\), \(w = T^\infty _a(u)\) and \(\overline{w} = T^\infty _{a - {\varepsilon }}(\overline{u} )\) and having in mind (4), we have that
$$\begin{aligned}&\lim _{n\rightarrow +\infty } I_2^2(s,c)\\&\quad \ge \int _\Omega \psi {\chi }_{\{0<u-\overline{u}<{\varepsilon }\}\cap \{u>a\}\cap \{\overline{u}>a-{\varepsilon }\}}(u T(u)-\overline{u} T(\overline{u})) d(|D^c u|-|D^c\overline{u}|)\\&\quad \ge -C{\varepsilon }\int _\Omega {\chi }_{\{0<T_a^\infty u-T_{a-{\varepsilon }}^\infty \overline{u}<{\varepsilon }\}}d|D^c(T_a^\infty (u)-T_{a-{\varepsilon }}^\infty (\overline{u}))|\ge -{\varepsilon }o({\varepsilon }). \end{aligned}$$
For the jump part,
$$\begin{aligned} I_2^2(s,j)&\ge \int _{\Omega } \left( \int _{\Omega } \xi _{n} (|D_y^j J_{T\varphi }(T_{\overline{u}}^{\overline{u}+{\varepsilon }}(u))|-\overline{u} T(\overline{u})\, |D_y^j T_{\overline{u}}^{\overline{u}+{\varepsilon }}(u)|)\right) \, dx\\&+\int _{\Omega } \left( \int _{\Omega } \xi _{n} (|D_x^j J_{T\varphi }(T_{u-{\varepsilon }}^{u}(\overline{u}))|-u T({u})|D_x^j T_{u-{\varepsilon }}^u(\overline{u})|)\right) \, dy. \end{aligned}$$
Observe now that
with \(\xi _{u^\pm ,\overline{u},{\varepsilon }} \in [T_{\overline{u}}^{\overline{u}+{\varepsilon }}(u^-),T_{\overline{u}}^{\overline{u}+{\varepsilon }}(u^+)]\). Similarly, as measures,
Therefore,
$$\begin{aligned} I_2^2(s,j)\ge -C{\varepsilon }^2\int _\Omega \int _{J_{T_{a-{\varepsilon }}^\infty (\overline{u})}} \xi _n d\mathcal H^{N-1} dy\ge -C{\varepsilon }^2\int _{J_{T_{a-{\varepsilon }}^\infty (\overline{u})}}d\mathcal H^{N-1}\ge -C{\varepsilon }^2. \end{aligned}$$
Collecting all these facts, we obtain
$$\begin{aligned} \lim _{n\rightarrow \infty }I_2^2 \ge {\varepsilon }o({\varepsilon }). \end{aligned}$$
Letting \(n\rightarrow \infty \) in (5), we have
$$\begin{aligned}&\int _{\Omega } \psi (uT(u)-\overline{u}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\, dx \\&\qquad + \int _{\Omega }T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi \left( x\right) \,dx \\&\quad \le \int _{\Omega }\psi (vT(u)-\overline{v}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\, dx - {\varepsilon }\int _{\Omega } \psi (\overline{\mathbf{z}}, \, DT(\overline{u}))+{\varepsilon }o({\varepsilon }). \end{aligned}$$
We take now a sequence \(\psi _m\uparrow {\chi }_\Omega \), \(\psi _m\in \mathcal D(\Omega )\) in the above formula. Then,
$$\begin{aligned}&\int _{\Omega } (uT(u)-\overline{u}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\, dx \\&\qquad + \limsup _{m\rightarrow \infty }\int _{\Omega }T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi _m\,dx\\&\quad \le \int _{\Omega } (vT(u)-\overline{v}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\, dx - {\varepsilon }\int _{\Omega } (\overline{\mathbf{z}}, \, DT(\overline{u}))+{\varepsilon }o({\varepsilon }). \end{aligned}$$
Let us see that the second term in the above expression is nonnegative.
$$\begin{aligned}&\int _{\Omega }T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}- T(\overline{u})\overline{\mathbf{z}}) \cdot \nabla \psi _m\left( x\right) \,dx \\&\quad = -\int _\Omega \psi _m T_{\varepsilon }(u-\overline{u})^+T(u)\mathrm{div}(\mathbf{z})\, dx - \int _\Omega \psi _m (\mathbf{z},D(T_{\varepsilon }(u-\overline{u})^+T(u)))\\&\quad \quad + \int _\Omega \psi _m T_{\varepsilon }(u-\overline{u})^+T(\overline{u})\mathrm{div}(\overline{\mathbf{z}})\, dx +\int _\Omega \psi _m (\overline{\mathbf{z}},D(T_{\varepsilon }(u-\overline{u})^+T(\overline{u}))). \end{aligned}$$
Now, since \(T_{\varepsilon }(u-\overline{u})T(u) \in BV(\Omega )\), \(u \ge g\), \(\overline{u} \ge g\) in \(\partial \Omega \),
$$\begin{aligned} \{ x \in \partial \Omega \ : \ u(x) > \overline{u} (x) \} \subset \{ x \in \partial \Omega \ : \ u(x) > g (x) \}, \end{aligned}$$
\([\mathbf{z},\nu ] = u \kappa \) with \(\kappa \in \mathrm {sign} (g-u)\) in \(\partial \Omega \,\cap \, \{u > 0\}\), and \([\overline{\mathbf{z}},\nu ] = \overline{u} \, \overline{\kappa }\) with \( \overline{\kappa }\in \mathrm {sign}(g-\overline{u})\) in \(\partial \Omega \cap \{\overline{u} > 0\}\), we have
$$\begin{aligned}&\lim _{m\rightarrow \infty } \int _{\Omega }T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}- T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi _m\left( x\right) \,dx\\&\quad =-\int _\Omega T_{\varepsilon }(u-\overline{u})^+T(u)\mathrm{div}(\mathbf{z})\, dx -\int _\Omega (\mathbf{z},D(T_{\varepsilon }(u-\overline{u})^+T(u)))\\&\qquad +\int _\Omega T_{\varepsilon }(u-\overline{u})^+T(\overline{u})\mathrm{div}(\overline{\mathbf{z}})\, dx+\int _\Omega (\overline{\mathbf{z}},D(T_{\varepsilon }(u-\overline{u})^+T(\overline{u})))\\&\quad =-\int _{\partial \Omega } \left( [\mathbf{z},\nu ]T(u)-[\overline{\mathbf{z}}, \nu ]T(\overline{u}) \right) T_{\varepsilon }(u-\overline{u})^+\,d\mathcal H^{N-1}\\&\quad = \int _{[u > \overline{u}] \cap \partial \Omega } \left( [\overline{\mathbf{z}}, \nu ]{\chi }_{\{\overline{u} > a\}}T(\overline{u})-[\mathbf{z},\nu ] {\chi }_{\{u > a\}}T(u) \right) T_{\varepsilon }(u-\overline{u})^+\,d\mathcal H^{N-1}\\&\quad = \int _{[u > \overline{u}] \cap \partial \Omega } \left( \overline{u}\, \overline{\kappa } T(\overline{u})- u \kappa T(u) \right) T_{\varepsilon }(u-\overline{u})^+ \,d\mathcal H^{N-1}\\&\quad \ge \int _{[u > \overline{u}] \cap \partial \Omega } (u T(u) - \overline{u}T(\overline{u}))T_{\varepsilon }(u-\overline{u})^+ \,d\mathcal H^{N-1} \ge 0. \end{aligned}$$
Therefore,
$$\begin{aligned}&\int _{\Omega } (uT(u)-\overline{u}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\, dx \\&\quad \le \int _{\Omega } (vT(u)-\overline{v}T(\overline{u})) T_{\varepsilon }(u-\overline{u})^+\, dx - {\varepsilon }\int _{\Omega } (\overline{\mathbf{z}}, \, DT(\overline{u}))+{\varepsilon }o({\varepsilon }). \end{aligned}$$
Dividing the last expression by \({\varepsilon }\) and letting \({\varepsilon }\rightarrow 0\), letting \(a\rightarrow 0^+\), and finally dividing by \(b\) and letting \(b\rightarrow 0^+\) we obtain (42) in the original article. As explained before, the rest of the proof is already written in the 150–151 pages of the original article. \(\square \)
Sketch of proof of uniqueness of Theorem 3 of the original article: First of all, we note that given \(a>0\), by Lemma 1 there exist \({\varepsilon }\rightarrow 0\) such that
$$\begin{aligned} \mathcal L^{N+2}(\{(x,s,t): T_a^\infty (u(s,x)-T_{\frac{a}{2}}^\infty \overline{u}(t,x))={\varepsilon }\})=0 \end{aligned}$$
and
$$\begin{aligned} \displaystyle \int _{(0,T)^2} (|D^cT_a^\infty u(t)|+|D^cT_\frac{a}{2}^\infty \overline{u}(s)|)(\{{\widetilde{T_a^\infty u(s)}-\widetilde{T_{\frac{a}{2}}^\infty \overline{u}(t)}={\varepsilon }}\})\,ds\,dt=0. \end{aligned}$$
As stated before, we only sketch here how to pass from (124) to (130) in the original article. (124) reads as
$$\begin{aligned}&- \displaystyle \int _{(Q_T)^2} \left( J^+_{T,{\varepsilon },\overline{u}}(u) (\eta _{m,n})_t + J^-_{T,{\varepsilon },u}( \overline{u}) (\eta _{m,n})_s \right) - \displaystyle {\varepsilon }\int _{(Q_T)^2} J_{T}(\overline{u}) (\eta _{m,n})_s \ dy ds \nonumber \\&\quad + \displaystyle \int _{(Q_T)^2} \eta _{m,n} h_T(u, D_x S_{{\varepsilon }, \overline{u}}(u))+ \displaystyle \int _{(Q_T)^2} \eta _{m,n} h_T(\overline{u}, D_y S^{u}_{{\varepsilon }}(\overline{u}))\nonumber \\&\quad - \displaystyle \int _{(Q_T)^2} \overline{\mathbf{z}} \cdot \nabla _x \eta _{m,n} T(\overline{u}) S^{u}_{{\varepsilon }}(\overline{u}) - \displaystyle \int _{(Q_T)^2} \mathbf{z}\cdot \nabla _y \eta _{m,n} T(u) S_{{\varepsilon }, \overline{u}}(u ) \nonumber \\&\quad +\displaystyle \int _{(Q_T)^2}T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot (\nabla _x \eta _{m,n}+\nabla _y\eta _{m,n})\nonumber \\&\quad + \displaystyle {\varepsilon }\int _{(Q_T)^2} T(\overline{u})\overline{\mathbf{z}}\cdot (\nabla _x \eta _{m,n}+\nabla _y\eta _{m,n}) \le 0, \end{aligned}$$
(8)
with \(\displaystyle \eta _{m,n}(t, x, s, y):= \rho _m(x - y) \tilde{\rho }_n(t - s) \phi \bigg (\frac{t+s}{2} \bigg )\psi \left( \frac{x+y}{2}\right) \), being \(\rho _m\) a sequence of mollifiers in \(\Omega \) and \(\tilde{\rho }_n\) a sequence of mollifiers in \({\mathbb R}\). Let \(I_1,I_2\) be, respectively, the sum of the first two terms and the sum of the third up to the sixth terms of the above inequality. Working as in the proof of Theorem 1, (using Lemmas 4 and 5) we get
$$\begin{aligned} \lim _{m\rightarrow +\infty } I_2\ge {\varepsilon }o({\varepsilon })-{{\varepsilon }\int _{(0,T)^2\times \Omega } T(\overline{u})\overline{\mathbf{z}}\cdot \nabla \chi _{n}}. \end{aligned}$$
Hence, by (8), it follows that
$$\begin{aligned}&- \displaystyle \int _{(0,T)^2 \times \Omega } \left( J^+_{T,{\varepsilon },\overline{u}}(u) (\chi _{n})_t + J^-_{T,{\varepsilon },u}( \overline{u}) (\chi _{n})_s \right) \\&\qquad +\int _{(0,T)^2\times \Omega }T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \chi _{n} -{\varepsilon }\int _{(0,T)^2\times \Omega } T(\overline{u})\overline{\mathbf{z}}\cdot \nabla \chi _{n}\\&\quad \le {\varepsilon }o({\varepsilon })+{\varepsilon }\int _{(0,T)^2\times \Omega } J_{T}(\overline{u}) (\chi _{n})_s, \end{aligned}$$
where \(\chi _n(t,s,x):=\tilde{\rho }_n(t-s)\phi (\frac{t+s}{2})\psi (x)\). Letting now \(\psi =\psi _m\uparrow \chi _\Omega \) we get,
$$\begin{aligned}&-\displaystyle \int _{(0,T)^2\times \Omega }\left( J^+_{T,{\varepsilon },\overline{u}}(u)(\kappa _n)_t+J^-_{T,{\varepsilon },u}( \overline{u})(\kappa _n)_s\right) \nonumber \\&\quad +\displaystyle \liminf _{m\rightarrow +\infty }\int _{(0,T)^2\times \Omega }\kappa _n T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi _m \nonumber \\&\quad -\displaystyle {\varepsilon }\liminf _{m\rightarrow +\infty }\int _{(0,T)^2\times \Omega }\kappa _nT(\overline{u}(s,x))\overline{\mathbf{z}}(s,x)\cdot \nabla \psi _{m} \le {\varepsilon }\int _{(0,T)^2\times \Omega }J_{T}(\overline{u}(s,x))(\kappa _n)_s,\nonumber \\ \end{aligned}$$
(9)
where \(\kappa _n(t,s):=\tilde{\rho }_n(t-s)\phi (\frac{t+s}{2})\). Using now Steklov’s type averages, it is possible to prove the following
Claim
$$\begin{aligned} \liminf _{m\rightarrow +\infty }\int _{(0,T)^2\times \Omega }\kappa _nT(\overline{u})\overline{\mathbf{z}}\cdot \nabla \psi _{m}= - \int _0^T \int _0^T \int _{\partial \Omega } [\overline{\mathbf{z}}, \nu ] \kappa _nT(\overline{u}) \, d {\mathcal H}^{N-1} dt ds,\nonumber \\ \end{aligned}$$
(10)
and
$$\begin{aligned}&\liminf _{m\rightarrow +\infty }\int _{(0,T)^2\times \Omega }\kappa _n T_{\varepsilon }(u-\overline{u})^+(T(u)\mathbf{z}-T(\overline{u})\overline{\mathbf{z}})\cdot \nabla \psi _m\nonumber \\&\quad \displaystyle =-\int _{(0,T)^2}\int _{\partial \Omega }\kappa _n T_{\varepsilon }(u-\overline{u})^+([\mathbf{z},\nu ]T(u)-[\overline{\mathbf{z}},\nu ]T(\overline{u}))d\mathcal H^{N-1}. \end{aligned}$$
(11)
Once the claim is proved, by (10) and (11), dividing (9) by \({\varepsilon }\) and letting \({\varepsilon }\rightarrow 0^+\) we obtain (130) in the original article and the proof finishes. To finish this note we give the proof of (10) for the sake of completeness (the proof of (11) is analogous).
Proof of the claim For \(\tau > 0\), we define the function \([\kappa _n(s)]^{\tau }\), as the Dunford integral
$$\begin{aligned}{}[\kappa _n(s)]^{\tau }(t)= \frac{1}{\tau }\int _t^{t+\tau }\kappa _n(r,s)T(\overline{u}(r,x)) \, dr. \end{aligned}$$
Then, since \(\overline{\xi }= \mathrm{div}(\overline{\mathbf{z}})\) in the sense of Definition \(4\) in the original article, we have
$$\begin{aligned} { I_m}&:= \int _{(0,T)^2\times \Omega }\kappa _nT(\overline{u}(s,x))\overline{\mathbf{z}}(s,x)\cdot \nabla \psi _{m}\\&= \lim _{\tau \rightarrow 0} \int _{(0,T)^2\times \Omega } [\kappa _n(s)]^{\tau }(t)\overline{\mathbf{z}}(t,x) \cdot \nabla (\psi _m(x)- 1)\\&= - \lim _{\tau \rightarrow 0} \left\{ \int _0^T \int _{Q_T} (\psi _m(x)- 1) (\overline{\mathbf{z}}, D_x([\kappa _n(s)]^{\tau })) \, ds\right. \\&\left. \quad + \int _0^T \int _0^T \langle \overline{\xi }(t), [\kappa _n(s)]^{\tau }(t)(\psi _m- 1) \rangle \, dt ds\right\} \\&\quad +\lim _{\tau \rightarrow 0} \int _0^T \int _0^T \int _{\partial \Omega } [\overline{\mathbf{z}}(t), \nu ](\psi _m- 1) [\kappa _n(s)]^{\tau }(t) \, d {\mathcal H}^{N-1} dt ds =:\! I^{1}_m+I^2_m \!+\! I^3_m. \end{aligned}$$
Observe that limit in the l.h.s. above as \(\tau \rightarrow 0^+\) exists. We prove that the limit of \(I^3_m\) exists, hence also the limit of \(I^{1}_m+I^2_m\). It is not difficult to see (see the proof of (109) in the original article) that
$$\begin{aligned} |D([\kappa _n(s)]^{\tau }(t))|(\Omega ) \mathop {\longrightarrow }\limits ^{\tau \rightarrow 0}\vert D(\kappa _nT(\overline{u}(t))) \vert (\Omega ). \end{aligned}$$
(12)
Hence,
$$\begin{aligned} I^3_m = - \int _0^T \int _0^T \int _{\partial \Omega } [\overline{\mathbf{z}}(t), \nu ] \kappa _n T(\overline{u}(t)) \, d {\mathcal H}^{N-1} dt ds. \end{aligned}$$
On the other hand, by (12),
$$\begin{aligned} \vert I^1_m \vert\le & {} \limsup _{\tau \rightarrow 0} {\Vert \overline{\mathbf{z}}\Vert _\infty } \int _0^T \int _{Q_T} (1 - \psi _m) \vert D([\kappa _n(s)]^{\tau }(t) ) \vert \\= & {} { \Vert \overline{\mathbf{z}}\Vert _\infty } \int _0^T \int _{Q_T}(1 - \psi _m) \vert D(\kappa _nT(\overline{u}(t))) \vert , \end{aligned}$$
which implies that \(\lim _{m \rightarrow \infty } I^1_m = 0.\) Then,
$$\begin{aligned} I^2_m= & {} - \lim _{\tau \rightarrow 0} \int _0^T \int _0^T \langle \overline{\xi }(t), [\kappa _n(s)]^{\tau }(t)(\psi _m- 1) \rangle \, dt ds \\= & {} \lim _{\tau \rightarrow 0} \int _0^T \int _0^T \int _{\Omega } \overline{u}(t,x) \frac{T(\overline{u}(t+\tau ))\kappa _n(t+\tau )- T(\overline{u}(t))\kappa _n(t)}{\tau } (\psi _m - 1). \end{aligned}$$
Let \(\displaystyle Q(r):= \int _0^r T(\tau ) \, d\tau .\) Therefore, \(Q(r) - Q(s) \le T(r)(r -s).\) Thus,
$$\begin{aligned} I^2_m= & {} \lim _{\tau \rightarrow 0} \int _0^T \int _0^T \int _{\Omega } (1 - \psi _m) \frac{\overline{u}(t) - \overline{u}(t - \tau )}{\tau } T(\overline{u}(t))\kappa _n(t) \\\ge & {} \lim _{\tau \rightarrow 0} \int _0^T \int _0^T \int _{\Omega } (1 - \psi _m) \kappa _n(t,s) \frac{Q(\overline{u}(t)) - Q(\overline{u}(t-\tau ))}{\tau }\\= & {} \lim _{\tau \rightarrow 0} \int _0^T \int _0^T \int _{\Omega } (1 - \psi _m) Q(\overline{u}(t)) \frac{\kappa _n(t,s) - \kappa _n(t+ \tau ,s)}{\tau }\\= & {} - \int _0^T \int _0^T \int _{\Omega } (1 - \psi _m) Q(\overline{u}(t)) \frac{d\kappa _n(t,s) }{dt}\mathop {\longrightarrow }\limits ^{m\rightarrow +\infty }0. \end{aligned}$$
Therefore,
$$\begin{aligned} \liminf _{m \rightarrow \infty } I^2_m \ge 0. \end{aligned}$$
Taking into account the above facts, we get
$$\begin{aligned} \liminf _{m \rightarrow \infty } I_m \ge - \int _0^T \int _0^T \int _{\partial \Omega } [\mathbf{z}(t), \nu ] \kappa _nT(\overline{u}(t)) \, d {\mathcal H}^{N-1} dt ds. \end{aligned}$$
(13)
In order to obtain the opposite inequality, we work similarly with
$$\begin{aligned}{}[\kappa _n(s)]_{\tau }(t):= \frac{1}{\tau }\int _{t-\tau }^{t}\kappa _n(r,s)T(\overline{u}(r,x)) \, dr, \end{aligned}$$
and the claim is proved.
