A game for the parabolic two membranes problem

Proof. First, we observe that the inequalites (36) and (37) holds for \((x,t)\in\mathbb{R}^{N}\backslash\Omega\times[0,T)\) and \((x,t)\in\Omega\times\{0\}\). Let us recall the definition of \(J_1\) and \(J_2\) \[\label{J11} \begin{array}{ll} \displaystyle J_1(w)(x,t)=\alpha_1\left[\frac{1}{2} \sup_{y \in B_{\varepsilon}(x)}w(y,t) + \frac{1}{2} \inf_{y \in B_{\varepsilon}(x)}w(y,t)\right] +(1-\alpha_1) {{\Huge{f}}_{B_{\varepsilon}(x)}}w(y,t)dy+\varepsilon^2h_1(x,t), \end{array} \tag{38}\] and \[\label{J22} \begin{array}{ll} \displaystyle J_2(w)(x,t)=\alpha_2\left[\frac{1}{2} \sup_{y \in B_{\varepsilon}(x)}w(y,t) + \frac{1}{2} \inf_{y \in B_{\varepsilon}(x)}w(y,t)\right]+(1-\alpha_2) {{\Huge{f}}_{B_{\varepsilon}(x)}}w(y,t)dy+\varepsilon^2h_2(x,t). \end{array} \tag{39}\]

Then, let us prove the following claim:
Claim. For \((x,t)\in\Omega\times(0,T)\) it holds that \[\label{dppz0} z_0(x,t)\leq \min\Big\{J_1(z_0)(x,t-\varepsilon^2),J_2(z_0)(x,t-\varepsilon^2)\Big\}-\varepsilon^2. \tag{40}\] Proof of the Claim. We aim to show that \[\label{rec1} \varepsilon^2 \leq \min\Big\{J_1(z_0)(x,t-\varepsilon^2)-z_0(x,t),J_2(z_0)(x,t-\varepsilon^2)-z_0(x,t)\Big\}. \tag{41}\] Using Taylor’s expansions we obtain \[\begin{align} \displaystyle J_1(z_0)(x,t\! -\! \varepsilon^2)\! -\! z_0(x,t)=&\alpha_1\left[\frac{1}{2}\sup_{y \in B_{\varepsilon}(x)}(z_0(y,t\! -\! \varepsilon^2)\! -\! z_0(x,t\! -\! \varepsilon^2))+\frac{1}{2}\inf_{y \in B_{\varepsilon}(x)}(z_0(y,t-\varepsilon^2)\! -\! z_0(x,t\! -\! \varepsilon^2))\right]\notag\\ &\displaystyle \ +(1\! -\! \alpha_1){\Huge{f}}_{B_{\varepsilon}(x)}(z_0(y,t\! -\! \varepsilon^2)\! -\! z_0(x,t\! -\! \varepsilon^2))dy+z_0(x,t\! -\! \varepsilon^2)\! -\! z_0(x,t)+\varepsilon^2 h_1(x,t\! -\! \varepsilon^2)\notag\\ \displaystyle =&\Big(\! -\! \frac{\partial z_0}{\partial t}(x,t)+\frac{\alpha_1}{2}\Delta^1_{\infty}z_0(x,t\! -\! \varepsilon^2)+\frac{(1\! -\! \alpha_1)}{2(N+2)}\Delta z_0(x,t\! -\! \varepsilon^2)\Big)\varepsilon^2+\varepsilon^2 h_1(x,t\! -\! \varepsilon^2) +o(\varepsilon^2). \end{align} \tag{42}\]

Analogously, \[\begin{aligned} \displaystyle J_2(z_0)(x,t-\varepsilon^2)-z_0(x,t) =&\Big(-\frac{\partial z_0}{\partial t}(x,t)+\frac{\alpha_2}{2}\Delta^1_{\infty}z_0(x,t-\varepsilon^2) +\frac{(1-\alpha_2)}{2(N+2)}\Delta z_0(x,t-\varepsilon^2)\Big)\varepsilon^2\notag\\ &+\varepsilon^2 h_2(x,t-\varepsilon^2) +o(\varepsilon^2). \end{aligned} \tag{43}\]

If we come back to (41) and we divide by \(\varepsilon^2\), we obtain that it is necesary to prove that it holds \[\label{maxdosop1} \begin{array}{ll} \displaystyle 1\leq\min\Big\{-\frac{\partial z_0}{\partial t}(x,t)+\Delta^1_p z_0(x,t-\varepsilon^2)+h_1(x,t-\varepsilon^2),-\frac{\partial z_0}{\partial t}(x,t)+\Delta^1_q z_0(x,t-\varepsilon^2)+h_2(x,t-\varepsilon^2)\Big\}+\frac{o(\varepsilon^2)}{\varepsilon^2}, \end{array} \tag{44}\] for \(\varepsilon>0\) small enough. Using the properties of \(z_0\) we have \[-\frac{\partial z_0}{\partial t}(x,t)+\Delta^1_p z_0(x,t-\varepsilon^2)+h_1(x,t-\varepsilon^2)\ge K+h_1(x,t-\varepsilon^2)\ge 2>1, \tag{45}\] and \[-\frac{\partial z_0}{\partial t}(x,t)+\Delta^1_q z_0(x,t-\varepsilon^2)+h_2(x,t-\varepsilon^2)\ge K+h_2(x,t-\varepsilon^2)\ge 2>1. \tag{46}\]

Thus, the inequality (44) holds for \(\varepsilon\) small enough.

This claim implies that \((z_0,z_0)\) is subsolution to the DPP (1). This ends the proof. ◻

Proof. (a) By induction: \[\displaystyle u_{1}^{\varepsilon}(x,t)= \frac{1}{2} J_1(u_0^{\varepsilon})(x,t-\varepsilon^2) \displaystyle +\frac{1}{2}\max\Big\{ J_1(u_0^{\varepsilon})(x,t-\varepsilon^2), J_2(v_0^{\varepsilon})(x,t-\varepsilon^2)\Big\}\ge u^\varepsilon_0(x,t), \tag{50}\] for \((x,t)\in\Omega\times(0,T)\). Here we used that \(u^\varepsilon_0=v^\varepsilon_0=z_0\) and that \((z_0,z_0)\) is a subsolution to the DPP.

Analogously, \[\displaystyle v_{1}^{\varepsilon}(x,t)= \frac{1}{2} J_2(v_0^{\varepsilon})(x,t-\varepsilon^2) \displaystyle +\frac{1}{2}\min\Big\{ J_1(u_0^{\varepsilon})(x,t-\varepsilon^2), J_2(v_0^{\varepsilon})(x,t-\varepsilon^2)\Big\}\ge v^\varepsilon_0(x,t). \tag{51}\]

Outside the domain it is clear that \[\begin{aligned} \begin{cases} u_{1}^{\varepsilon}(x,t) =& f(x,t)\ge u^\varepsilon_0(x,t), \qquad\qquad (x,t) \in (\mathbb{R}^{N} \backslash \Omega)\times[0,T),\\ v_{1}^{\varepsilon}(x,t)=& g(x,t)\ge v^\varepsilon_0(x,t) , \qquad\qquad (x,t) \in (\mathbb{R}^{N} \backslash \Omega)\times[0,T),\\ u_{1}^{\varepsilon}(x,0)=& u_0(x)\ge u^\varepsilon_0(x,0), \qquad\qquad~~ x \in \Omega, \\ v_{1}^{\varepsilon}(x,0)=& v_0(x)\ge v^\varepsilon_0(x,0), \qquad\qquad~~ x \in \Omega.\end{cases} \end{aligned} \tag{52}\]

Now, let us deal with the inductive step, \(u^\varepsilon_n\le u^\varepsilon_{n-1}\) and \(v^\varepsilon_n\le v^\varepsilon_{n-1}\). The definition of \(J_1\) and \(J_2\) implies that \[J_1(u^\varepsilon_n)(x,t-\varepsilon^2)\ge J_1(u^\varepsilon_{n-1})(x,t-\varepsilon^2) \quad \mbox{and} \quad J_2(v^\varepsilon_n)(x,t-\varepsilon^2)\ge J_2(v^\varepsilon_{n-1})(x,t-\varepsilon^2).\]

Hence, \[\begin{aligned} \displaystyle u_{n+1}^{\varepsilon}(x,t)=& \frac{1}{2} J_1(u_n^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u_n^{\varepsilon})(x,t-\varepsilon^2), J_2(v_n^{\varepsilon})(x,t-\varepsilon^2)\Big\}\notag\\ \displaystyle \ge& \frac{1}{2} J_1(u_{n-1}^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u_{n-1}^{\varepsilon})(x,t-\varepsilon^2), J_2(v_{n-1}^{\varepsilon})(x,t-\varepsilon^2)\Big\}=u^\varepsilon_n(x,t). \end{aligned} \tag{53}\]

Analogously, we get \(v^\varepsilon_{n+1}\ge v^\varepsilon_n\).

(b) \((z_0,z_0)\) is subsolution thanks to Lemma 1. Using (a) and the monotonicity of \(J_1\) and \(J_2\) we get \[\begin{aligned} \displaystyle \frac{1}{2} J_1&(u_{n+1}^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u_{n+1}^{\varepsilon})(x,t-\varepsilon^2), J_2(v_{n+1}^{\varepsilon})(x,t-\varepsilon^2)\Big\}\notag\\ \displaystyle &\ge \frac{1}{2} J_1(u_{n}^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u_{n}^{\varepsilon})(x,t-\varepsilon^2), J_2(v_{n}^{\varepsilon})(x,t-\varepsilon^2)\Big\}=u^\varepsilon_{n+1}(x,t). \end{aligned} \tag{54}\]

Analogously for \(v^\varepsilon_n\). This ends the proof. ◻

Proof. It is clear the the inequality holds outside the domain \(\Omega\times(0,T)\). Inside the domain we argue by contradiction. Suposse that there exists \(n_0\in\mathbb{N}\) such that \[\max\Big\{\sup_{\Omega\times(0,T)}(u^\varepsilon_{n_0}-w_0),\sup_{\Omega\times(0,T)}(v^\varepsilon_{n_0}-w_0)\Big\}=\sup_{\Omega\times(0,T)}(u^\varepsilon_{n_0}-w_0)=\theta>0.\]

Let \((x_k,t_k)\in\Omega\times(0,T)\) such that \[\theta-\frac{1}{k}<(u^\varepsilon_{n_0}-w_0)(x_k,t_k).\]

Using the inequalities we get \[\begin{aligned} \label{dossup} \displaystyle \theta-\frac{1}{k}<&(u^\varepsilon_{n_0}-w_0)(x_k,t_k)\notag\\ \le&\max \Big\{J_1(u^\varepsilon_{n_0})(x_k,t_k-\varepsilon^2),J_2(v^\varepsilon_0)(x_k,t_k-\varepsilon^2)\Big\} \displaystyle-\max \Big\{J_1(w_0)(x_k,t_k-\varepsilon^2),J_2(w_0)(x_k,t_k-\varepsilon^2)\Big\}-\varepsilon^2 \notag\\ \displaystyle \le& \max \Big\{(J_1(u^\varepsilon_{n_0})-J_1(w_0))(x_k,t_K-\varepsilon^2),(J_2(v^\varepsilon_{n_0})-J_2(w_0))(x_k,t_k-\varepsilon^2)\Big\}-\varepsilon^2. \end{aligned} \tag{57}\]

Here we used that \(\max\{a,b\}-\max\{c,d\}\le\max\{a-c,b-d\}\). Let us consider the inequalities \[\inf_{y \in B_{\varepsilon}(x_k)}u^\varepsilon_{n_0}(y,t_k-\varepsilon^2)-\inf_{y \in B_{\varepsilon}(x_k)}w_0(y,t_k-\varepsilon^2)\le \sup_{y \in B_{\varepsilon}(x_k)}(u^\varepsilon_{n_0}-w_0)(y,t_k-\varepsilon^2)\le \theta, \tag{58}\] and \[\sup_{y \in B_{\varepsilon}(x_k)}u^\varepsilon_{n_0}(y,t_k-\varepsilon^2)-\sup_{y \in B_{\varepsilon}(x_k)}w_0(y,t_k-\varepsilon^2)\le \sup_{y \in B_{\varepsilon}(x_k)}(u^\varepsilon_{n_0}-w_0)(y,t_k-\varepsilon^2)\le\theta, \tag{59}\] and finally \[ {{\Huge{f}}_{B_{\varepsilon}(x_k)}}(u^\varepsilon_{n_0}-w_0)(y,t_k-\varepsilon^2)dy\le \theta. \tag{60}\]

Hence, using again the definition of \(J_1\) (38) and \(J_2\) (39) we get that \[J_1(u^\varepsilon_{n_0})(x,t-\varepsilon^2)-J_1(w_0)(x,t-\varepsilon^2)\le\theta\quad \mbox{and} \quad J_2(v^\varepsilon_{n_0})(x,t-\varepsilon^2)-J_2(w_0)(x,t-\varepsilon^2)\le\theta. \tag{61}\]

If we come back to (57) we get \[\theta-\frac{1}{k}+\varepsilon^2<(u^\varepsilon_{n_0}-w_0)(x_k,t_k)\le\theta, \tag{62}\] which is a contradiction if \(k\in\mathbb{N}\) is large enough. This ends the proof. ◻

Proof. Let us start proving that \(u^\varepsilon_0\) is a subsolution. That is \[,\] for \(x\in\Omega\). Here we used that \[-M\le\sup_{y \in B_{\varepsilon}(x)}u_0^\varepsilon(y) \quad \mbox{and} \quad -M\le\inf_{y \in B_{\varepsilon}(x)}u_0^\varepsilon(y),\] and finally \[-M\le {{\Huge{f}}_{B_{\varepsilon}(x)}}u_0^\varepsilon(y)dy.\]

On the other hand \(u^\varepsilon_0(x)=f(x)\) for \(x\in\mathbb{R}^N\backslash\Omega\). The sequence defined by (66) is nondecreasing and composed by subsolutions to the DPP (65) (see Lemma 2).

Let us prove that the sequence is uniformly bounded. In fact, we have that \[u^\varepsilon_n\le M, \tag{68}\] for all \(n\ge 0\). We use an inductive argument. It is clear that \(u^\varepsilon_0\le M\). Suppose that \(u_n\le M\). Using that \(\sup_{y \in B_{\varepsilon}(x)}u_{n}^\varepsilon(y)\le M\), \(\inf_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)\le M\) and \( {{\Huge{f}}_{B_{\varepsilon}(x)}}u_0^\varepsilon(y)dy\le M\) we get \(u_{n+1}\le M\).

Now, let us show that \(u^\varepsilon(x):=\lim_{n\rightarrow \infty}u_n^\varepsilon(x)\) is a solution to (65). It is clear that if \(x\in\mathbb{R}^N\backslash\Omega\) \[u^\varepsilon(x)=\lim_{n\rightarrow \infty}u_n^\varepsilon(x)=f(x).\]

For \(x\in\Omega\), let us consider \[\begin{aligned} \label{dosu} \displaystyle (u_{n+1}^{\varepsilon}-u_n^\varepsilon)(x)=& \alpha\left[\frac{1}{2}\sup_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)+\frac{1}{2}\inf_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)-\frac{1}{2}\sup_{y \in B_{\varepsilon}(x)}u_{n-1}^\varepsilon(y)-\frac{1}{2}\inf_{y \in B_{\varepsilon}(x)}u_{n-1}^\varepsilon(y)\right]\notag\\ & +(1-\alpha) {{\Huge{f}}_{B_{\varepsilon}(x)}}(u_n^\varepsilon-u_{n-1}^\varepsilon)(y) dy. \end{aligned} \tag{69}\]

If we define \[C_n:=\lVert u_n^\varepsilon-u_{n-1}^\varepsilon\mathbb{R}Vert_{L^\infty(\Omega)}.\] Using (69) and the inequalities \[\sup_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)-\sup_{y \in B_{\varepsilon}(x)}u_{n-1}^\varepsilon(y)\le\sup_{y \in B_{\varepsilon}(x)}(u^\varepsilon_n-u^\varepsilon_{n-1})(y), \tag{70}\] and \[\inf_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)-\inf_{y \in B_{\varepsilon}(x)}u_{n-1}^\varepsilon(y)\le\sup_{y \in B_{\varepsilon}(x)}(u^\varepsilon_n-u^\varepsilon_{n-1})(y), \tag{71}\] we get \[(u_{n+1}^{\varepsilon}-u_n^\varepsilon)(x)\le \alpha C_n+(1-\alpha)C_n.\]

Thus, \(C_{n+1}\le C_n\).

Now, let us consider the following set \[\Gamma_1=\Big\{x\in\Omega : d(x,\partial\Omega)<\frac{\varepsilon}{2}\Big\}. \tag{72}\]

Using the assumed regularirty on \(\partial\Omega\) (uniform exterior sphere condition) we have \[\eta_1=\sup_{x\in\Gamma_1}\frac{|B_\varepsilon(x)\cap\Omega|}{|B_\varepsilon(x)|}<1.\]

Given \(x\in\Gamma_1\), we get \[(u^\varepsilon_{n+1}-u^\varepsilon_n)(x)\le \alpha C_n+(1-\alpha) {{\Huge{f}}_{B_{\varepsilon}(x)\cap\Omega}}(u^\varepsilon_n-u^\varepsilon_{n-1})(y)dy\le \alpha C_n+(1-\alpha)\eta_1 C_n. \tag{73}\]

Here we used that \[ {{\Huge{f}}_{B_{\varepsilon}(x)}}(u^\varepsilon_n-u^\varepsilon_{n-1})(y)dy=\frac{1}{|B_{\varepsilon}|}\int_{B_{\varepsilon}(x)\cap\Omega}(u^\varepsilon_n-u^\varepsilon_{n-1})(y)dy,\] since \((u^\varepsilon_n-u^\varepsilon_{n-1})(x)=0\) when \(x\in\mathbb{R}^N\backslash\Omega\). Thus, \[\label{T1} (u^\varepsilon_{n+1}-u^\varepsilon_n)(x)\le (\alpha+(1-\alpha)\eta_1) C_n\le\theta_1 C_n, \tag{74}\] with \(\theta_1=\alpha+(1-\alpha)\eta_1<1\) for all \(x\in\Gamma_1\). Let us continue with \[\Gamma_2=\Big\{x\in\Omega : d(x,\Gamma_1)<\frac{\varepsilon}{2}\Big\}. \tag{75}\]

Notice that \(\Gamma_1\subset\Gamma_2\). Let us define \[\eta_2=\sup_{x\in\Gamma_2}\frac{|B_\varepsilon(x)\cap(\Omega\backslash\Gamma_1)|}{|B_\varepsilon(x)|}<1.\]

Here we used again the uniform exterior sphere condition. Given \(x\in\Gamma_2\) we obtain \[\begin{aligned} \displaystyle (u^\varepsilon_{n+2}-u^\varepsilon_{n+1})(x)\le&\alpha C_{n+1}+(1-\alpha)\frac{1}{|B_\varepsilon(x)|}\left[\int_{B_{\varepsilon}(x)\cap(\Omega\backslash\Gamma_1)}(u^\varepsilon_{n+1}-u^\varepsilon_n)(y)dy+\int_{B_{\varepsilon}(x)\cap\Gamma_1}(u^\varepsilon_{n+1}-u^\varepsilon_n)(y)dy\right]\notag\\ \displaystyle \le&\alpha C_n+(1-\alpha)\left[\eta_2 C_n+(1-\eta_2)\theta_1C_n\right]=\left[\alpha+(1-\alpha)[\eta_2+(1-\eta_2)\theta_1]\right]C_n=\theta_2 C_n. \end{aligned} \tag{76}\] where \(\theta_2=\alpha+(1-\alpha)[\eta_2+(1-\eta_2)\theta_1]<1\). Here we used (74). Notice that \(\theta_1<\theta_2<1\). Iterating this procedure we obtain \[\Gamma_k=\Big\{x\in\Omega : d(x,\Gamma_{k-1})<\frac{\varepsilon}{2}\Big\}, \quad \mbox{and}\quad \eta_k=\sup_{x\in\Gamma_k}\frac{|B_\varepsilon(x)\cap(\Omega\backslash\Gamma_{k-1})|}{|B_\varepsilon(x)|}<1 . \tag{77}\]

Then, for \(x\in\Gamma_k\) \[(u^\varepsilon_{n+k}-u^\varepsilon_{n+k-1})(x)\le\theta_k C_n, \tag{78}\] where \(\theta_k=\alpha+(1-\alpha)[\eta_k+(1-\eta_k)\theta_{k-1}]<1\). Notice that, if \(k_0=\lceil\frac{diam(\Omega)}{\varepsilon/2}\mathbb{R}ceil\) we obtain \(\Omega\subset \Gamma_{k_0}\). Thus \[\label{k0} C_{n+k_0}\le \theta_{k_0} C_n. \tag{79}\]

Notice that \(C_{k_0}\le \theta_{k_0} C_0\) and \(C_{k_0+j}\le C_{k_0}\le \theta_{k_0} C_0\) for \(0\le j\le k_0-1\). Moreover \[C_{lk_0+j}\le \theta_{k_0}^l C_0 \quad \rightarrow \quad \sum_{j=0}^{\infty}C_{lk_0+j}\le \sum_{i=0}^{\infty}k_0\theta_{k_0}^{l+i} C_0.\]

Finally, \[\begin{array}{ll} \displaystyle \lVert u^\varepsilon_{n+m}-u^\varepsilon_n\mathbb{R}Vert_{L^\infty(\Omega)}\le \sum_{j=1}^{m}\lVert u^\varepsilon_{n+j}-u^\varepsilon_{n+j-1}\mathbb{R}Vert_{L^\infty(\Omega)}\le \sum_{j=1}^{\infty}C_{n+j} \displaystyle \le\sum_{i=1}^{\infty}k_0\theta_{k_0}^{\lfloor\frac{n}{k_0}\mathbb{R}floor+i}C_0, \end{array} \tag{80}\] and this is small if \(n\in\mathbb{N}\) is large enough. Then, \((u^\varepsilon_n)_{n\ge 0}\) is a Cauchy sequence in \(L^\infty\), and this implies that \(u_n\rightrightarrows u\) uniformly in \(\Omega\) as \(n \to \infty\). Thus, we get \[\label{supinf} \sup_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)\rightarrow \sup_{y \in B_{\varepsilon}(x)}u^\varepsilon(y) \quad \mbox{and} \quad \inf_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)\rightarrow \inf_{y \in B_{\varepsilon}(x)}u^\varepsilon(y). \tag{81}\] Finally, we also have \[\label{convint} {{\Huge{f}}_{ B_{\varepsilon}(x)}}u_n^\varepsilon(y)dy\rightarrow {{\Huge{f}}_{B_{\varepsilon}(x)}}u^\varepsilon(y)dy. \tag{82}\]

Taking limits in (66), and using (81) and (82) we conclude that \[\begin{aligned} \displaystyle \underbrace{u_{n+1}^{\varepsilon}(x)}_{\downarrow}=& \alpha\left[\frac{1}{2}\underbrace{\sup_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)}_{\downarrow}+\frac{1}{2}\underbrace{\inf_{y \in B_{\varepsilon}(x)}u_n^\varepsilon(y)}_{\downarrow}\right]+(1-\alpha)\underbrace{ {{\Huge{f}}_{B_{\varepsilon}(x)}}u_n^\varepsilon(y)dy}_{\downarrow}\notag\\ \displaystyle u^{\varepsilon}(x)=& \alpha\left[\frac{1}{2}\sup_{y \in B_{\varepsilon}(x)}u^\varepsilon(y)+\frac{1}{2}\inf_{y \in B_{\varepsilon}(x)}u^\varepsilon(y)\right]+(1-\alpha) {{\Huge{f}}_{B_{\varepsilon}(x)}}u^\varepsilon(y)dy. \end{aligned} \tag{83}\]

This ends the proof of the lemma. ◻

Proof of Theorem 2. We know that \((u_n^{\varepsilon},v_n^{\varepsilon})\) is a subsolution to the DPP (1) for all \(n\geq 0\), then, we have \[u_{n}^{\varepsilon}(x,t)\le \frac{1}{2} J_1(u_n^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u_n^{\varepsilon})(x,t-\varepsilon^2),J_2(v_n^{\varepsilon})(x,t-\varepsilon^2)\Big\}.\]

Taking limit as \(n\rightarrow\infty\) on the right side we get \[u_{n}^{\varepsilon}(x,t)\le \frac{1}{2} J_1(u^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u^{\varepsilon})(x,t-\varepsilon^2),J_2(v^{\varepsilon})(x,t-\varepsilon^2)\Big\}.\]

Taking limit on the left side we arrive to \[u^{\varepsilon}(x,t)\le \frac{1}{2} J_1(u^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u^{\varepsilon})(x,t-\varepsilon^2),J_2(v^{\varepsilon})(x,t-\varepsilon^2)\Big\}.\]

Analogously, we obtain \[v^{\varepsilon}(x,t)\le \frac{1}{2} J_2(v^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\min\Big\{ J_1(u^{\varepsilon})(x,t-\varepsilon^2),J_2(v^{\varepsilon})(x,t-\varepsilon^2)\Big\}.\]

Thus, \((u^{\varepsilon},v^{\varepsilon})\) is a subsolution to the DPP (1).

Now, we will use ideas form the computation used in the proof of Lemma 4. Let us define \[C_n=\max\{\lVert u^\varepsilon_n-u^\varepsilon_{n-1}\lVert_{L^\infty(\Omega\times(0,T))}, \lVert v^\varepsilon_n-v^\varepsilon_{n-1}\lVert_{L^\infty(\Omega\times(0,T))}\}. \tag{84}\]

Let us start with \(u_n\): \[\begin{aligned} \displaystyle (u^\varepsilon_{n+1}-u^\varepsilon_{n})(x,t)=&\frac{1}{2} J_1(u^\varepsilon_n)(x,t-\varepsilon^2)-\frac{1}{2} J_1(u^\varepsilon_{n-1})(x,t-\varepsilon^2) +\frac{1}{2}\max\{J_1(u^\varepsilon_n)(x,t\! – \! \varepsilon^2), J_2(v^\varepsilon_n)(x,t\! -\! \varepsilon^2)\} \notag\\ & – \! \frac{1}{2}\max\{J_1(u^\varepsilon_{n-1})(x,t \! -\! \varepsilon^2), J_2(v^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2)\}\notag\\ \displaystyle \le&\frac{1}{2} C_n+\frac{1}{2}\max\{ J_1(u^\varepsilon_n)(x,t\! -\! \varepsilon^2)\! – \! J_1(u^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2), J_2(v^\varepsilon_n)(x,t\! -\! \varepsilon^2)\! – \! J_2(v^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2)\}\notag\\ \displaystyle\le& C_n. \end{aligned} \tag{85}\]

Here we used again that \(\max\{a,b\}-\max\{c,d\}\le\max\{a-c, b-d\}\). Now, for \(v^\varepsilon_n\) we have \[\begin{aligned} \displaystyle (v^\varepsilon_{n+1}-v^\varepsilon_{n})(x,t)=&\frac{1}{2} J_2(v^\varepsilon_n)(x,t-\varepsilon^2)-\frac{1}{2} J_2(v^\varepsilon_{n-1})(x,t-\varepsilon^2) +\frac{1}{2}\min\{J_1(u^\varepsilon_n)(x,t\! -\! \varepsilon^2), J_2(v^\varepsilon_n)(x,t\! -\! \varepsilon^2)\}\notag\\ &- \frac{1}{2}\min\{J_1(u^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2), J_2(v^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2)\}\notag\\ \displaystyle \le&\frac{1}{2} C_n+\frac{1}{2}\max\{ J_1(u^\varepsilon_n)(x,t\! -\! \varepsilon^2)\! -\! J_1(u^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2), J_2(v^\varepsilon_n)(x,t\! -\! \varepsilon^2)\! -\! J_2(v^\varepsilon_{n-1})(x,t\! -\! \varepsilon^2)\}\notag\\ \displaystyle\le& C_n. \end{aligned} \tag{86}\]

Here we used that \(\min\{a,b\}-\min\{c,d\}\le\max\{a-c, b-d\}\). Thus, we get \[C_{n+1}\le C_n. \tag{87}\]

Let us consider again the set \[\Gamma_1=\Big\{x\in\Omega : d(x,\partial\Omega)<\frac{\varepsilon}{2}\Big\}. \tag{88}\]

As before, using the assumed regularity on the boundary of \(\Omega\) we have that \[\eta_1=\sup_{x\in\Gamma_1}\frac{|B_\varepsilon(x)\cap\Omega|}{|B_\varepsilon(x)|}<1.\]

Given \(x\in\Gamma_1\) we get \[J_1(u^\varepsilon_n)(x,t-\varepsilon^2)- J_1(u^\varepsilon_{n-1})(x,t-\varepsilon^2)\le (\alpha_1+(1-\alpha_1)\eta_1)C_n,\] and \[J_2(v^\varepsilon_n)(x,t-\varepsilon^2)- J_2(v^\varepsilon_{n-1})(x,t-\varepsilon^2)\le (\alpha_2+(1-\alpha_2)\eta_1)C_n.\]

Thus \[(u^\varepsilon_{n+1}-u^\varepsilon_n)(x,t-\varepsilon^2)\le \theta_1 C_n \quad \mbox{and} \quad (v^\varepsilon_{n+1}-v^\varepsilon_n)(x,t-\varepsilon^2)\le \theta_1 C_n,\] whit \(\theta_1=\max\{\alpha_1+(1-\alpha_1)\eta_1, \alpha_2+(1-\alpha_2)\eta_1\}<1\). Proceeding as before, we obtain \(k_0=k_0(\Omega)\in\mathbb{N}\) and \(\theta_0<1\) such that \[C_{n+k_0}\le \theta_0 C_n.\]

Arguing as before we get the uniform convergence \(u^\varepsilon_n\rightrightarrows u^\varepsilon\) and \(v^\varepsilon_n\rightrightarrows v^\varepsilon\). Then, we get \[J_1(u^\varepsilon_n)(x,t)\rightarrow J_1(u^\varepsilon)(x,t) \quad , \quad J_2(v^\varepsilon_n)(x,t)\rightarrow J_2(v^\varepsilon)(x,t).\]

Finally, using the definition (47) and taking limit we obtain \[\begin{aligned} \begin{cases} \displaystyle \underbrace{u^\varepsilon_{n+1}(x,t)}_{\downarrow}=&\frac{1}{2} \underbrace{J_1(u^\varepsilon_n)(x,t-\varepsilon^2)}_{\downarrow}+\frac{1}{2}\max\{\underbrace{J_1(u^\varepsilon_n) (x,t-\varepsilon^2)}_{\downarrow}, \underbrace{J_2(v^\varepsilon_n)(x,t-\varepsilon^2)}_{\downarrow}\}\\ \displaystyle u^\varepsilon(x,t)=&\frac{1}{2} J_1(u^\varepsilon)(x,t-\varepsilon^2)+\frac{1}{2}\max\{J_1(u^\varepsilon)(x,t-\varepsilon^2), J_2(v^\varepsilon)(x,t-\varepsilon^2)\},\end{cases} \end{aligned} \tag{89}\] and \[\begin{aligned} \begin{cases} \displaystyle \underbrace{v^\varepsilon_{n+1}(x,t)}_{\downarrow}=&\frac{1}{2} \underbrace{J_2(v^\varepsilon_n)(x,t-\varepsilon^2)}_{\downarrow}+\frac{1}{2}\min\{\underbrace{J_1(u^\varepsilon_n) (x,t-\varepsilon^2)}_{\downarrow}, \underbrace{J_2(v^\varepsilon_n)(x,t-\varepsilon^2)}_{\downarrow}\}\\ \displaystyle v^\varepsilon(x,t)=&\frac{1}{2} J_2(v^\varepsilon)(x,t-\varepsilon^2)+\frac{1}{2}\min\{J_1(u^\varepsilon)(x,t-\varepsilon^2), J_2(v^\varepsilon)(x,t-\varepsilon^2)\}.\end{cases} \end{aligned} \tag{90}\]

This ends the proof of the theorem. ◻

Alternative proof of existence of a solution to the DPP (1). We look for a pair \((u,v)\) that solves (1). It is clear that we need to impose the boundary conditions \[\label{BCInd.88} \left\lbrace \begin{array}{ll} u^{\varepsilon}(x,t) = f(x,t), & (x,t) \in (\mathbb{R}^{N} \backslash \Omega)\times[0,T), \\[8pt] v^{\varepsilon}(x,t)= g(x,t), & (x,t) \in (\mathbb{R}^{N} \backslash \Omega)\times[0,T), \end{array} \right. \tag{91}\] and the initial conditions \[\label{ICInd.99} \left\lbrace \begin{array}{ll} u^{\varepsilon}(x,0)= u_0(x), & x \in \Omega, \\[8pt] v^{\varepsilon}(x,0)= v_0(x), & x \in \Omega. \end{array} \right. \tag{92}\] Hence, we are left with determining \((u,v)\) in \(\Omega \times (0,T)\) in such a way that the equations in (1) are satisfied.

Let us start with \(t\in (0,\varepsilon^2]\). Since \(t-\varepsilon^2 \le 0\), for those times we have that \[\label{Ind.22} \left\lbrace \begin{array}{ll} \displaystyle u^{\varepsilon}(x,t)= \frac{1}{2} J_1(u_0)(x) +\frac{1}{2}\max\Big\{ J_1(u_0)(x), J_2(v_0)(x)\Big\} , & (x,t) \in \Omega\times(0,\varepsilon^2], \\[8pt] \displaystyle v^{\varepsilon}(x,t)= \frac{1}{2} J_2(v_0)(x) +\frac{1}{2}\min\Big\{ J_1(u_0)(x), J_2(v_0)(x)\Big\}, & (x,t) \in \Omega\times(0,\varepsilon^2], \end{array} \right. \tag{93}\] solves the equations in the DPP. Once we have defined \((u,v)\) in \(\Omega \times (0,\varepsilon^2]\) we look for \(t\in (\varepsilon^2, 2\varepsilon^2]\) and we get that the pair of functions given by \[\label{Ind.66} \left\lbrace \begin{array}{ll} \displaystyle u^{\varepsilon}(x,t)= \frac{1}{2} J_1(u^{\varepsilon})(x,t-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u^{\varepsilon})(x,t-\varepsilon^2), J_2(v^{\varepsilon})(x,t-\varepsilon^2)\Big\} , & (x,t) \in \Omega\times (\varepsilon^2,2\varepsilon^2], \\ \displaystyle v^{\varepsilon}(x,t)= \frac{1}{2} J_2(v^{\varepsilon})(x,t-\varepsilon^2) +\frac{1}{2}\min\Big\{ J_1(u^{\varepsilon})(x,t-\varepsilon^2), J_2(v^{\varepsilon})(x,t-\varepsilon^2)\Big\}, & (x,t) \in \Omega\times (\varepsilon^2,2\varepsilon^2], \end{array} \right. \tag{94}\] solves the DPP in \(\Omega \times (\varepsilon^2,2\varepsilon^2]\).

Iterating this procedure \([T/\varepsilon^2]\) times we obtain a pair of functions \((u,v)\) that is a solution to (1) in the whole \(\Omega \times (0,T)\). ◻

Proof. We only include a sketch of the proof. We refer to [23] (Theorem 18) where the authors proved a similar result for the elliptic case.

Fix \(\delta>0\), and take \((u^{\varepsilon},v^{\varepsilon})\) solution to the DPP (1). Assume that we start at a point in the first board, \((x_0,t_0,1)\). Then, we choose a strategy \(S_{I}^{\ast}\) for Player I using the solution to the (DPP) (1) as follows: Whenever \(j_k =1\) Player I decides to stay in the first board if \[\max \Big\{J_1(u^{\varepsilon})(x_k,t_k-\varepsilon^2), J_2(v^{\varepsilon})(x_k, t_k-\varepsilon^2)\Big\} =J_1(u^{\varepsilon})(x_k,t_k-\varepsilon^2) ,\] and in this case Player I chooses a point \[x_{k+1}^I=S_{I}^{\ast}\left(( x_k,t_k,j_k)\right) \quad \mbox{such that} \sup_{y \in B_{\varepsilon}(x_k,t_k-\varepsilon^2)}u^{\varepsilon}(y,t_k-\varepsilon^2)-\frac{\delta}{2^{k+1}}\leq u^{\varepsilon}(x_{k+1}^I,t_k-\varepsilon^2), \tag{96}\] to play the Tug-of-War game.

On the other hand, Player I decides to jump to the second board if \[\max \Big\{J_1(u^{\varepsilon})(x_k,t_k-\varepsilon^2), J_2(v^{\varepsilon})(x_k,t_k-\varepsilon^2)\Big\} =J_2(v^{\varepsilon})(x_k,t_k-\varepsilon^2) ,\] and in this case Player I chooses a point \[x_{k+1}^I=S_{I}^{\ast}\left(( x_k,t_k,j_k)\right) \quad \mbox{such that} \sup_{y \in B_{\varepsilon}(x_k)}v^{\varepsilon}(y,t_k-\varepsilon^2)-\frac{\delta}{2^{k+1}}\leq v^{\varepsilon}(x_{k+1}^I,t_k-\varepsilon^2), \tag{97}\] to play the Tug-of-War game in the second board.

Given \(S^\ast_I\) strategy for Player I, and \(S_{II}\) any strategy for Player II, we consider the sequence of random variables \[M_k=w^{\varepsilon}(x_k,t_k,j_k)-\varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l-\varepsilon^2)\chi_{\{j=1\}}(j_{l+1})-h_2(x_l,t_l-\varepsilon^2)\chi_{\{j=2\}}(j_{l+1})\right)-\frac{\delta}{2^k}. \tag{98}\] where \(w^{\varepsilon}(x_k,t_k,1)=u^{\varepsilon}(x_k,t_k)\), \(w^{\varepsilon}(x_k,t_k,2)=v^{\varepsilon}(x_k,t_k)\) and \[\chi_{\{j=i\}}(j)=\left\lbrace \begin{array}{ll} \displaystyle 1 & j=i ,\\[8pt] \displaystyle 0 & j\neq i. \end{array} \right. \tag{99}\]

It holds that \((M_k)_{k\geq 0}\) is a submartingale. That is \[\displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0,1)}[M_{k+1}|M_0,\dots,M_k]\ge M_k. \tag{100}\]

To prove this fact we need to consider several cases.

Suppose that \(j_k=1\) and \(j_{k+1}=1\) (that is, the token remains on the first board at the \(k\) and the \(k+1\) plays). Then

\[\begin{aligned} \displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0,1)}&[M_{k+1}|M_0,\dots,M_k]\notag\\ \displaystyle =&\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,1)}\left[u^{\varepsilon}(x_{k+1},t_{k+1})\! – \! \varepsilon^2\sum_{l=0}^{k}\left(h_1(x_l,t_l)\chi_{\{j=1\}}(j_{l+1})\! -\! h_2(x_l,t_l)\chi_{\{j=2\}}(j_{l+1})\right)-\frac{\delta}{2^{k+1}}|M_0,\dots ,M_k\right]\notag\\ \displaystyle =&\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,1)}\! \left[u^{\varepsilon}(x_{k+1},t_{k+1})\! -\! \varepsilon^2h_1(x_k,t_k)\! -\! \varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l)\chi_{\{j=1\}}(j_{l+1})\! -\! h_2(x_l,t_l)\chi_{\{j=2\}}(j_{l+1})\right)\right.\notag\\ &\left.-\frac{\delta}{2^{k+1}}|M_0,\dots ,M_k\right]\notag\\ \displaystyle =&\alpha_1\Big(\frac{1}{2} u^\varepsilon(x_{k+1}^I,t_{k+1})+\frac{1}{2} u^\varepsilon(x_{k+1}^{II},t_{k+1})\Big)+(1-\alpha_1) { {\Huge{f}}_{B_{\varepsilon}(x_k)}}u^\varepsilon(y)dy-\varepsilon^2h_1(x_k,t_k)\notag\\ &\displaystyle -\varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l)\chi_{\{j=1\}}(j_{l+1})-h_2(x_l,t_l)\chi_{\{j=2\}}(j_{l+1})\right)-\frac{\delta}{2^{k+1}}\notag\\ \displaystyle \ge&\alpha_1\Big(\frac{1}{2} \sup_{y \in B_{\varepsilon}(x_k)}u^\varepsilon(y,t_{k+1})-\frac{\delta}{2^{k+1}}+\frac{1}{2} \inf_{y \in B_{\varepsilon}(x_k)}u^\varepsilon(y,t_{k+1})\Big)+(1-\alpha_1) {{\Huge{f}}_{B_{\varepsilon}(x_k)}}u^\varepsilon(y)dy\notag\\ \displaystyle & -\varepsilon^2h_1(x_k,t_k)-\varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l)\chi_{\{j=1\}}(j_{l+1})-h_2(x_l,t_l)\chi_{\{j=2\}}(j_{l+1})\right)-\frac{\delta}{2^{k+1}}\notag\\ \displaystyle\ge& \frac{1}{2} J_1(u^\varepsilon)(x_{k},t_k-\varepsilon^2)+\frac{1}{2}\max\Big\{J_1(u^\varepsilon)(x_{k},t_k-\varepsilon^2),J_2(v^\varepsilon)(x_{k},t_k-\varepsilon^2)\Big\}\notag\\ \displaystyle & -\varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l)\chi_{\{j=1\}}(j_{l+1})-h_2(x_l,t_l)\chi_{\{j=2\}}(j_{l+1})\right)-\frac{\delta}{2^{k}}\notag\\ \displaystyle & u^\varepsilon(x_k,t_k)-\varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l)\chi_{\{j=1\}}(j_{l+1})-h_2(x_l,t_l)\chi_{\{j=2\}}(j_{l+1})\right)-\frac{\delta}{2^{k}}=M_k. \end{aligned} \tag{101}\]

Here we used that \(j_{k+1}=1\), \(t_{k+1}=t_k-\varepsilon^2\) and \(\max\{J_1(u^\varepsilon),J_2(v^\varepsilon)\}=J_1(u^\varepsilon)\).

We omit proof of the cases, \(j_k=1\) and \(j_{k+1}=2\), \(j_k=2\) and \(j_{k+1}=1\), \(j_k=2\) and \(j_{k+1}=2\), because the computations are similar. Thus, we get that \(M_k\) is a submartingale. Using the OSTh we obtain \[\displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0,1)}[M_{\tau\wedge k}]\geq M_0 \quad \mbox{for any} \ k\in\mathbb{N}. \tag{102}\]

Taking limit as \(k\rightarrow\infty\) we get \[\displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0,1)}[M_{\tau}]\geq M_0. \tag{103}\]

If we take \(\inf_{S_{II}}\) and then \(\sup_{S_{I}}\) we arrive to \[\displaystyle \sup_{S_{I}}\inf_{S_{II}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,1)}[M_{\tau}]\geq M_0. \tag{104}\]

This inequality says that \[\displaystyle \sup_{S_{I}}\inf_{S_{II}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,1)}[\mbox{total payoff}]\geq u(x_0,t_0)-\delta. \tag{105}\]

To prove an inequality in the opposite direction we fix a strategy for Player II as follows: Whenever \(j_k=2\) Player II decides to stay in the second board if \[\min\Big\{J_1(u^{\varepsilon})(x_k,t_k-\varepsilon^2),J_2(v^{\varepsilon})(x_k,t_k-\varepsilon^2)\Big\}=J_2(v^{\varepsilon})(x_k,t_k-\varepsilon^2), \tag{106}\] and Player II decides to jump to the first board when \[\min\Big\{J_1(u^{\varepsilon})(x_k,t_k-\varepsilon^2),J_2(v^{\varepsilon})(x_k,t_k-\varepsilon^2)\Big\}=J_1(v^{\varepsilon})(x_k,t_k-\varepsilon^2). \tag{107}\]

If we play Tug-of-War (in both boards) Player II chosses \(x_{k+1}^{II}=S_{II}^{\ast}\left(( x_k,t_k,j_k)\right)\) such that \[\inf_{y \in B_{\varepsilon}(x_k,t_k-\varepsilon^2)}w^{\varepsilon}(y,t_k-\varepsilon^2,j_{k+1})+\frac{\delta}{2^{k+1}}\geq w^{\varepsilon}(x_{k+1}^{II},t_k-\varepsilon^2,j_{k+1}). \tag{108}\]

Given this strategy for Player II and any strategy for Player I, using similar computations like the ones we did before, we can prove that the sequence of random variables \[N_k=w^{\varepsilon}(x_k,t_k,j_k)-\varepsilon^2\sum_{l=0}^{k-1}\left(h_1(x_l,t_l-\varepsilon^2)\chi_{\{j=1\}}(j_{l+1})-h_2(x_l,t_l-\varepsilon^2)\chi_{\{j=2\}}(j_{l+1})\right)+\frac{\delta}{2^k}, \tag{109}\] is a supermartingale. Finally, using the OSTh we arrive to \[\inf_{S_{II}}\sup_{S_{I}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,1)}[\mbox{total payoff}]\leq u^{\varepsilon}(x_0,t_0)+\delta. \tag{110}\]

Then, we have obtained \[u^{\varepsilon}(x_0,t_0)-\delta\leq \sup_{S_{I}}\inf_{S_{II}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,1)}[\mbox{total payoff}]\leq \inf_{S_{II}}\sup_{S_{I}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,1)}[\mbox{total payoff}]\leq u^{\varepsilon}(x_0,t_0)+\delta, \tag{111}\] for any \(\delta>0\).

Analogously, we can prove that \[v^{\varepsilon}(x_0,t_0)-\delta\leq \sup_{S_{I}}\inf_{S_{II}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,2)}[\mbox{total payoff}]\leq \inf_{S_{II}}\sup_{S_{I}}\mathbb{E}_{S_{I},S_{II}}^{(x_0,t_0,2)}[\mbox{total payoff}]\leq v^{\varepsilon}(x_0,t_0)+\delta, \tag{112}\] for any \(\delta>0\). This ends the proof. ◻

Proof. Let us start with the following definition: Let \(w:[(\mathbb{R}^N\backslash\Omega\times [0,T))\cup(\Omega\times\{0\})]\rightarrow\mathbb{R}\) be given by, \[\label{funcionw} w(x,t) =\left\lbrace \begin{array}{ll} F(x,t) & \ \ \mbox{if} \quad x\notin\Omega, t\geq 0, \\[8pt] \displaystyle \mu_0(x) & \ \ \mbox{if} \quad x\in\Omega, t= 0. \end{array} \right. \tag{118}\]

From our conditions on the data, the function \(w\) is well defined and is Lipschitz in both variables, that is \[\label{Lw1} |w(x,t)-w(y,s)|\leq L(|x-y|+|t-s|). \tag{119}\]

Let us proceed with the proof of the lemma.

Case 1. If \((x,t), (y,s)\in\left(\mathbb{R}^N\backslash\Omega\times [0,T)\right)\cup\left(\Omega\times\{0\}\right)\) we have \[|\mu^{\varepsilon}(x,t)-\mu^{\varepsilon}(y,s)|=|w(x,t)-w(y,s)|\leq L(|x-y|+|t-s|)<\eta, \tag{120}\] if \(r_0<\frac{\eta}{2L}\).

Case 2. Suppose now that \((x,t)\in\Omega\times(0,T)\) and \((y,s)\in\partial\Omega\times[0,T)\). Without loss of generality we can suppose that \(\Omega\subset B_R(0)\backslash B_{\delta}(0)\) and \(y\in\partial B_{\delta}(0)\). Let us call \((x_0,t_0)=(x,t)\) the first position in the game. Assume that Player I uses the strategy of pulling towards \(0\), denoted by \(S_{I}^{\ast}\). That is, for \(x_k\neq 0\) \[x_{k+1}^I=S^\ast_I(x_0,\dots,x_k)=x_k-\varepsilon\frac{x_k}{|x_k|}.\]

Let us consider the sequence of random variables using \(S_I^\ast\) for Player I and any \(S_{II}\) for Player II, \[M_k=|x_k|-C\varepsilon^2 k. \tag{121}\]

If \(C>0\) is large enough \(M_k\) is a supermartingale. Indeed \[\begin{array}{ll} \displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[|x_{k+1}| |x_0,\dots x_k]\leq \beta\left[\frac{1}{2}(|x_k|+\varepsilon)+\frac{1}{2}(|x_k|-\varepsilon)\right]+(1-\beta) {{\Huge{f}}_{B_{\varepsilon}(x_k)}}|z|dz \leq |x_k|+C\varepsilon^2. \end{array} \tag{122}\]

The first inequality follows form the choice of the strategy, and the second from the estimate \[ {{\Huge{f}}_{B_{\varepsilon}(x)}}|z|dz\leq |x|+C\varepsilon^2. \tag{123}\]

Using the OSTh we obtain \[\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[|x_{\tau}|]\leq |x_0|+C\varepsilon^2\mathbb{E}_{S_{I}^{\ast},S_{II}}[\tau]. \tag{124}\]

Now, we use the following estimate (115) to get \[\begin{array}{ll} \displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[|x_{\tau}|]\leq |x_0|+C(\frac{R}{\delta})dist(\partial B_\delta (0),x_0)+o(1) \displaystyle \le \delta+C|x_0-y|+o(1). \end{array} \tag{125}\]

Using the continuity property for \(w\) (119) we have \[|w(x_{\tau},t_\tau)-w(0,s)|\leq L\left(|x_{\tau}|+|t_\tau -s|\right). \tag{126}\]

But, \(t_\tau=t_0-\varepsilon^2\tau\). Hence, we get \[\begin{aligned} \displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_{\tau},t_\tau)]\geq& w(0,s)-L\left[\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[|x_{\tau}|]+|t_0-s|+\varepsilon^2\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[\tau]\right]\notag\\ \displaystyle \geq & w(y,s)-L\delta-L\left[2(\delta+C|x_0-y|)+|t_0-s|+ o(1)\right]\notag\\ \displaystyle \geq & w(y,s)-L\left[3\delta+2Cr_0+o(1)\right]. \end{aligned} \tag{127}\]

Then \[\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_{\tau},t_\tau)+\varepsilon^2\sum_{j=0}^{\tau -1}h(x_j,t_j)]\geq w(y,s)-L\left[3\delta+2Cr_0\right]- \lVert h\mathbb{R}Vert_{\infty} C r_0 – o(1). \tag{128}\]

Thus, taking \(\inf_{S_{II}}\), and then \(\sup_{S_{I}}\) we obtain \[\mu^{\varepsilon}(x_0,t_0)\ge w(y,s)-L\left[3\delta+2Cr_0\right]- \lVert h\mathbb{R}Vert_{\infty} C r_0 – o(1)>w(y,s)-\eta. \tag{129}\]

We take \(\delta>0\) such that \(3L\delta<\frac{\eta}{3}\), then take \(r_0>0\) such that \(\left(2LC+\lVert h\mathbb{R}Vert_{\infty}C\right)r_0<\frac{\eta}{3}\) and then \(\varepsilon\) small such that \(o(1)<\frac{\eta}{3}\). Analogously, we can obtain the estimate \[\mu^{\varepsilon}(x_0,t_0)<w(y,s)+\eta, \tag{130}\] if player II use the strategy that pull towards \(0\). This ends the proof in this case.

Case 3. Suppose now that \((x,t)\in\Omega\times(0,T)\), \(y\in\overline{\Omega}\) and \(s=0\). Now we consider the \(S_I^\ast\) strategy for Player I pulling towards \(y\). That is \[x_{k+1}=S_I^\ast(x_0,\dots,x_k)=x_k-\varepsilon\frac{y-x_k}{|y-x_k|}, \tag{131}\] if \(|y-x_k|\ge \varepsilon\) and \(x_{k+1}=y\) in other case. Suppose that \(0<t=t_0<r_0\) for \(r_0\) small (to be chosen latter). Then, the stopping time is bounded. In fact, \(\tau\le \lceil\frac{r_0}{\varepsilon^2}\mathbb{R}ceil\) with probability one. Let us call \(M=\lceil\frac{r_0}{\varepsilon^2}\mathbb{R}ceil\). Now we will prove the following claim:
Claim. Given \(\theta>0\) and \(a>0\), there exists \(r_0>0\) and \(\varepsilon_0>0\) such that if Player I use \(S_I^\ast\) the strategy defined before, and Player II use any strategy \(S_{II}\), we get \[¶(\tau \geq \frac{a}{\varepsilon^2})<\theta \quad \mbox{ and } \quad ¶(|x_{\tau}-y|\ge a)<\theta. \tag{132}\] Proof of the claim. The first inequality holds if \(r_0<a\). To obtain the other inequality let us define the following sequence of random variables. \[X_{k}= \left\{ \begin{array}{ll} 1 \quad & \quad \mbox{if Player II wins,} \\[8pt] -1 \quad & \quad \mbox{if Player I wins,} \end{array} \right.\] for \(k\geq 1\), and \[Z_k=\sum_{j=1}^{k}X_k.\]

Observe that \(X_k\) are independent with \(\mathbb{E}[X_k]=0\) and \(\mathbb{V}[X_{k}]=1\). Then, \(\mathbb{E}[Z_k]=0\) and \(\mathbb{V}[Z_k]=k\). If we use Chebyshev’s Theorem we obtain \[¶(|Z_M|\geq \frac{a}{2\varepsilon})\leq \frac{\mathbb{V}[Z_M]}{(\frac{a}{2\varepsilon})^2}=\frac{4M\varepsilon^2}{a^2}\leq \frac{(\frac{4r_0}{\varepsilon^2}+1)\varepsilon^2}{a^2}\leq\frac{4r_0}{a^2}+\frac{\varepsilon^2}{a^2}<\theta,\] if \(\frac{4r_0}{a^2}<\frac{\theta}{2}\) and \(\frac{\varepsilon^2}{a^2}<\frac{\theta}{2}\). This says that the probability that Player II wins \(\frac{a}{2\varepsilon}\) more times than Player I is small. Then, if we take \(r_0<\frac{a}{2}\), we deduce that \[¶(|x_{\tau}-x_0|\geq \frac{a}{2})<\theta.\]

Here we use that the maximum distance that the position of the token can get away from \(x_0\) is \(\varepsilon\) at each step. Now, let us consider \[|x_{\tau}-y|\leq |x_{\tau}-x_0|+|x_0-y|<|x_{\tau}-x_0|+\frac{a}{2}.\]

Hence, we have \[\Big\{|x_{\tau}-y|\geq a \Big\}\subseteq \Big\{|x_{\tau}-x_0|\geq \frac{a}{2}\Big\},\] and then we conclude that \[¶({|x_{\tau}-y|\geq a})<\theta.\]

This ends the proof of the claim.

Using the definion (118) and the fact that \(s= 0\), we get \[|w(x_\tau,t_\tau)-\mu_0(y)|=|w(x_\tau,t_\tau)-w(y,s)|\le L\left(|x_\tau-y|+|t_\tau|\right)\le L\left(|x_\tau-y|+r_0\right).\]

Let us define \(A=\big\{|x_{\tau}-y|\geq a \big\}\), Using the claim, we obtain \[\begin{aligned} \displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_\tau,t_\tau)]=& \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_\tau,t_\tau)| A^c]¶(A^c)+ \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_\tau,t_\tau) | A]¶(A)\notag\\ \displaystyle \ge&\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_\tau,t_\tau) | A^c](1-\theta)-\lVert w\mathbb{R}Vert_{\infty}\theta\ge \mu_0(y)(1-\theta)-L\left(a+r_0\right)(1-\theta)-\lVert w\mathbb{R}Vert_{\infty}\theta. \end{aligned} \tag{133}\]

Adding the runing payoff we get \[\begin{aligned} \displaystyle \mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[w(x_\tau,t_\tau)+\varepsilon^2\sum_{j=0}^{\tau-1}h(x_j,t_j)] \displaystyle \ge&\mu_0(y)(1-\theta)-L\left(a+r_0\right)-\lVert w\mathbb{R}Vert_{\infty}\theta-\lVert h\mathbb{R}Vert_{\infty}\varepsilon^2\mathbb{E}_{S_{I}^{\ast},S_{II}}^{(x_0,t_0)}[\tau]\notag\\ \displaystyle\ge&\mu_0(y)(1-\theta)-L\left(a+r_0\right)-\lVert w\mathbb{R}Vert_{\infty}\theta-\lVert h\mathbb{R}Vert_{\infty}Cr_0+o(1). \end{aligned} \tag{134}\]

Thus, taking infimum over all possible strategies \(S_{II}\), and then supremum over \(S_I\) we get \[\mu^\varepsilon(x_0,t_0)\ge \mu_0(y)(1-\theta)-L(a+r_0)-\lVert w\mathbb{R}Vert_{\infty}\theta-\lVert h\mathbb{R}Vert_{\infty}Cr_0+o(1)>\mu_0(y)-\eta, \tag{135}\] if \(a>0\), \(\theta>0\), \(r_0>0\) and \(\varepsilon>0\) are small enough.

Analogously, we obtain \[\mu^\varepsilon(x,t)\le \mu_0(y)+\eta. \tag{136}\]

In this case, we use the strategy \(S_{II}^\ast\) pulling towards \(0\).

Case 4. Now, given two points \((x,t), (y,s)\in\Omega\times(0,T)\) with \(|x-y|<r_0\), and \(|t-s|<r_0\), we couple the game starting at \(x_0=x\) and \(t_0=t\), with the game starting at \(y_0=y\) and \(s_0=s\) making the same movements. This means that \(x_{k+1}-y_{k+1}=x_{k}-y_k\) for \(k\ge 0\) (it is clear that \(t_{k+1}-s_{k+1}=t_k-s_k\)). We can think the two games position mimic each other. This coupling generates two sequences of positions \(x_i\) and \(y_i\) such that \(|x_i – y_i|<r_0\) and \(j_i=k_i\). It is clear that \(t_i=t-\varepsilon^2i\) and \(s_i=s-\varepsilon^2i\), then \(|t_i-s_i|<r_0\). This continues until one of the game ends. Here we have two possibilities:

– If the game ends leaving the domain \(\Omega\) (say, for example \(y_\tau\notin\Omega\)) . At this point for the game starting at \((x_0,t_0)\) we arrived to the position \((x_\tau,t_\tau)\), with \(x_\tau\) close to the exterior point \(y_\tau \not\in \Omega\) (since we have \(|x_\tau – y_\tau|<r_0\)) and hence we can use our previous estimates for points close to the boundary to conclude that \[|\mu^{\varepsilon}(x_0,t_0)- \mu^\varepsilon (y_0,s_0)|< \eta.\]

– If the game ends leaving the doman from the bottom (say \(s_\tau= 0\) ), we have that \(t_\tau\le r_0\), then we can use the estimate obtained in case 3 to conclude that \[|\mu^{\varepsilon}(x_0,t_0)- \mu^\varepsilon (y_0,s_0)|< \eta.\]

This ends the proof. ◻

Proof. We will proceed using ideas similar to the ones used in Lemma 7. We start again with the following definition: Let us consider \(w_1:(\mathbb{R}^N\backslash\Omega\times [0,T))\cup(\Omega\times\{0\})\rightarrow\mathbb{R}\), \[w_1(x,t) =\left\lbrace \begin{array}{ll} f(x,t) & \ \ \mbox{if} \quad x\notin\Omega, t\geq 0, \\[8pt] \displaystyle u_0(x) & \ \ \mbox{if} \quad x\in\Omega, t= 0, \end{array} \right. \tag{141}\] and \(w_2:(\mathbb{R}^N\backslash\Omega\times [0,T))\cup(\Omega\times\{0\})\rightarrow\mathbb{R}\), \[w_2(x,t) =\left\lbrace \begin{array}{ll} g(x,t) & \ \ \mbox{if} \quad x\notin\Omega, t\geq 0, \\[8pt] \displaystyle v_0(x) & \ \ \mbox{if} \quad x\in\Omega, t= 0. \end{array} \right. \tag{142}\]

It is clear that \(w_1(x,t)\ge w_2(x,t)\). Also from the conditions on the data we have that \[\label{Lw.44} |w_i(x,t)-w_i(y,s)|\leq L(|x-y|+|t-s|), \tag{143}\] for \(i=1,2\).

Let us proceed with the proof of the lemma. We consider two cases.

Case 1. Suppose that \((x,t), (y,s)\in[(\mathbb{R}^N\backslash\Omega\times [0,T))\cup(\Omega\times\{0\})]\), then we have \[|u^{\varepsilon}(x,t)-u^{\varepsilon}(y,s)|=|w_1(x,t)-w_1(y,s)|\leq L\left(|x-y|+|t-s|\right)<\eta,\] and \[|v^{\varepsilon}(x,t)-v^{\varepsilon}(y,s)|=|w_2(x,t)-w_2(y,s)|\leq L\left(|x-y|+|t-s|\right)<\eta,\] if \(2Lr_0<\eta\).

Case 2. Let us begin with the estimate of \(u^{\varepsilon}\). Suppose now that \((x,t)\in\Omega\times(0,T)\) and \((y,s)\in\partial\Omega\times(0,T)\) in the first board (we denote \((x,t,1)\) and \((y,s,1)\)). Without loss of generality we suppose again that \(\Omega\subset B_R(0)\backslash B_{\delta}(0)\) and \(y\in\partial B_{\delta}(0)\). Let us call \(x_0=x\) the first position in the game. Player I uses the following strategy called \(S_{I}^{\ast}\): the token always stay in the first board (Player I decides not to change boards), and pulls towards \(0\) when Tug-of-War is played. In this case we have that \(u^\varepsilon\) is a supersolution to the DPP that appears in Lemma 7 (with \(\beta=\alpha_1\)). Notice that the game is always played in the first board. As Player I wants to maximize the expected value we get that the first component for our system, \(u^\varepsilon\), satisfies \[u^\varepsilon(x,t)\ge \mu^\varepsilon(x,t), \tag{144}\] (the value function when the player that wants to maximize is allowed to choose to change boards is bigger than or equal to the value function of a game where the player does not have the possibility of making this choice). From this bound and Lemma 7, a lower bound for \(u^\varepsilon\) close to the boundary follows. That is, from the estimate obtained in that lemma, we get \[u^\varepsilon(x,t)\ge w_1(y,s)-\eta, \tag{145}\] if \(|x-y|<r_0\), \(|t-s|<r_0\) and \(\varepsilon<\varepsilon_0\) for some \(r_0\) and \(\varepsilon_0\).

Now, the next estimate requires a particular strategy for Player II, called \(S_{II}^\ast\): when play the Tug-of-War game, Player II pulls towards 0 (in both boards) and if in some step Player I decides to jump to the second board, then Player II decides to stay always in this board and then the position never comes back to the first board. Using that \(w_1\ge w_2\) we will repeat the ideas used in Lemma 7: Suppose that \(j_{\tau}=1\). This means that \(j_k=1\) for all \(0\leq k \leq \tau\). Then we obtain \[\mathbb{E}_{S_{I},S_{II}^{\ast}}^{(x,t,1)}[\mbox{final payoff}]\leq w_1(y,s)+\eta, \tag{146}\] for \(r_0\) and \(\varepsilon_0\) small enough. On the other hand, if \(j_{\tau}=2\), we have \[\mathbb{E}_{S_{I},S_{II}^{\ast}}^{(x,t,1)}[w_2(x_{\tau},t_\tau)]\leq w_2(y,s)+\eta\leq w_1(y,s)+\eta. \tag{147}\]

In both cases, taking \(\sup_{S_{I}}\) and then \(\inf_{S_{II}}\) we arrive to \[u^\varepsilon(x,t)\le w_1(y,s)+\eta, \tag{148}\] taking \(\delta>0\), \(r_0>0\) and \(\varepsilon>0\) small enough.

Case 3. Now, given two points \((x,t,j), (y,s,l)\in\Omega\times(0,T)\times\{1,2\}\) with \(j=l\) (that is, both position are in the same board). Also we assume \(|x-y|<r_0\), and \(|t-s|<r_0\). Then, we couple the game starting at \((x_0,t_0,j_0)=(x,t,j)\), with the game starting at \((y_0,s_0,l_0)=(y,s,l)\) making the same movements, and changing boards at the same time. This means that, \(j_k=l_k\), and \(x_{k+1}-y_{k+1}=x_{k}-y_k\) for \(k\ge 0\) (it is clear that \(t_{k+1}-s_{k+1}=t_k-s_k\)). We can think the two games position mimic each other. This coupling generates two sequences of positions \(x_i\) and \(y_i\) such that \(|x_i – y_i|<r_0\) and \(j_i=k_i\). It is clear that \(t_i=t-\varepsilon^2i\) and \(s_i=s-\varepsilon^2i\), then \(|t_i-s_i|<r_0\). Using the same computation as in Lemma 7 we get \[|u^{\varepsilon}(x,t)-u^{\varepsilon}(y,s)|<\eta, \tag{149}\] if \(r_0>0\) and \(\varepsilon>0\) are small enough.

Analogously we can obtain the estimates for \(v^{\varepsilon}\) and complete the proof. ◻

Proof. We divide the proof in several cases.

1) \(u\) and \(v\) are ordered: From the fact that \[u^{\varepsilon_j} \geq v^{\varepsilon_j},\] in \(\mathbb{R}^N\times[0,T)\) we get \[u \geq v,\] in \(\overline{\Omega}\times[0,T)\).

2) The lateral boundary conditions: As we have that \[u^{\varepsilon_j} = f, \qquad v^{\varepsilon_j} = g,\] in \(\mathbb{R}^N \setminus \Omega\times [0,T)\) we get \[u|_{\partial \Omega\times [0,T)} = f, \qquad v|_{\partial \Omega\times [0,T)} = g.\]

(3) The initial conditions: As we have that \[u^{\varepsilon_j}(x,0) = u_0(x), \qquad v^{\varepsilon_j}(x,0) = v_0(x),\] we obtain \[u(x,0) = u_0(x), \qquad v(x,0) = v_0(x).\]

(4) The equation for \(u\): First, let us show that \(u\) is a viscosity supersolution to \[\label{Supu} \frac{\partial u}{\partial t}(x,t)-\Delta_p^1 u(x,t)= h_1(x,t), \tag{150}\] for \((x,t)\in \Omega\times (0,T)\). To this end, consider a point \((x_0,t_0)\in \Omega\times (0,T)\) and a smooth function \(\varphi\in C^{2,1}(\Omega\times(0,T))\) such that \((u-\varphi)(x_0,t_0)=0\) is a strict minimum of \((u-\varphi)\). Then, from the uniform convergence there exists a sequence of points, that we will denote by \(\{(x_{\varepsilon},t_{\varepsilon})\}_{\varepsilon>0}\), such that \(x_{\varepsilon}\rightarrow x_0\) and \(t_{\varepsilon}\rightarrow t_0\), and it holds \[(u^{\varepsilon}-\varphi)(x_{\varepsilon},t_{\varepsilon})\leq(u^{\varepsilon}-\varphi)(y,s)+o(\varepsilon^2), \tag{151}\] for all \((y,s)\in\Omega\times[0,T)\), that is, \[\label{ump} u^{\varepsilon}(y,s)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\geq\varphi(y,s)-\varphi(x_{\varepsilon},t_{\varepsilon})-o(\varepsilon^2). \tag{152}\]

From the DPP (1) we have \[\begin{aligned} \label{desDPP} \displaystyle 0=&\frac{1}{2} J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})+\frac{1}{2}\max\Big\{ J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-u(x_{\varepsilon},t_{\varepsilon}), J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon}) \Big\}\notag\\ \geq & J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon}). \end{aligned} \tag{153}\]

Using (152) we get \[0\ge J_1(\varphi)(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon})-o(\varepsilon^2), \tag{154}\] if we add and subtract \(\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)\) and we obtain \[\label{J1phi} 0\ge \varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon})+J_1(\varphi)(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-o(\varepsilon^2). \tag{155}\]

Consider \[\begin{aligned} \label{J1conphi} \displaystyle J_1(\varphi)&(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2) \notag\\[8pt] \displaystyle =& \alpha_1\underbrace{\left[\frac{1}{2} \sup_{y \in B_{\varepsilon}(x_{\varepsilon})}(\varphi(y,t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)) + \frac{1}{2} \inf_{y \in B_{\varepsilon}(x_{\varepsilon})}(\varphi(y,t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2))\right]}_{I} \notag\\[8pt] & +(1-\alpha_1)\underbrace{ {{\Huge{f}}_{B_{\varepsilon}(x_{\varepsilon})}}(\varphi(y,t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2))dy}_{II}+\varepsilon^2h_1(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2). \end{aligned} \tag{156}\]

Let us analyze \(I\) and \(II\). We begin with \(I\): Assume that \(\nabla\varphi(x_0,t_0)\neq 0\). Let define \(z_{\varepsilon}=\frac{\nabla\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)}{|\nabla\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)|}\neq 0\). If \(\varepsilon>0\) is small enough, it holds that \[\sup_{y \in B_{\varepsilon}(x_{\varepsilon})}\varphi(y,t_{\varepsilon}-\varepsilon^2)\sim\varphi(x_{\varepsilon}+\varepsilon z_{\varepsilon},t_{\varepsilon}-\varepsilon^2) \quad \mbox{and} \quad \inf_{y \in B_{\varepsilon}(x_{\varepsilon})}\varphi(y,t_{\varepsilon}-\varepsilon^2)\sim\varphi(x_{\varepsilon}-\varepsilon z_{\varepsilon},t_{\varepsilon}-\varepsilon^2).\]

Then, we have \[I\sim\frac{1}{2}(\varphi(x_{\varepsilon}+\varepsilon z_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2))+\frac{1}{2}(\varphi(x_{\varepsilon}-\varepsilon z_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)). \tag{157}\]

From a simple Taylor expansion we conclude that \[\begin{aligned} \displaystyle\frac{1}{2}&(\varphi(x_{\varepsilon}+\varepsilon z_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2))+\frac{1}{2}(\varphi(x_{\varepsilon}-\varepsilon z_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2))\notag\\ \displaystyle&=\frac{1}{2}\varepsilon^2\langle D^2\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2 )z_{\varepsilon},z_{\varepsilon}\mathbb{R}angle +o({\varepsilon^2}). \end{aligned} \tag{158}\]

Dividing by \(\varepsilon^2\) the first inequality and taking the limit as \(\varepsilon\rightarrow 0\) we obtain \[\label{limI} \frac{1}{2}\langle D^2\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2 )z_{\varepsilon},z_{\varepsilon}\mathbb{R}angle\rightarrow \frac{1}{2}\langle D^2\varphi(x_0,t_0 )z_0,z_0\mathbb{R}angle, \tag{159}\] where \(z_0=\frac{\nabla\varphi(x_0,t_0)}{|\nabla\varphi(x_0,t_0)|}\). Thus \[I\rightarrow\frac{1}{2}\Delta^1_{\infty}\varphi(x_0,t_0).\]

See [17]) for more details. When \(\nabla\varphi=0\) arguing again using Taylor’s expansions we get \[\limsup_{\varepsilon\to 0} I \geq \frac{1}{2} \lambda_1 (D^2 \varphi (x_0,t_0)).\]

See [9] for the details.

Now, we look at \(II\): Using again Taylor’s expansions we obtain \[{ {\Huge{f}}_{B_{\varepsilon}(x_{\varepsilon})}}(\varphi(y,t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2))dy=\frac{\varepsilon^2}{2(N+2)}\Delta \varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)+o(\varepsilon^2).\]

Dividing by \(\varepsilon^2\) and taking limits as \(\varepsilon \rightarrow 0\) we get \[\label{limII} \displaystyle II\rightarrow \frac{1}{2(N+2)}\Delta \varphi(x_{0},t_0). \tag{160}\]

Therefore, if we come back to (155), dividing by \(\varepsilon^2\) and taking limit \(\varepsilon\rightarrow 0\) we obtain \[0\geq -\frac{\partial\varphi}{\partial t}(x_0,t_0)+\frac{\alpha_1}{2}\Delta^1_{\infty}\varphi(x_0,t_0)+\frac{1-\alpha_1}{2(N+2)}\Delta\varphi(x_0,t_0)+h_1(x_0,t_0), \tag{161}\] when \(\nabla \varphi (x_0,t_0) \neq 0\) and \[0\geq -\frac{\partial\varphi}{\partial t}(x_0,t_0)+\frac{\alpha_1}{2}\lambda_1(D^2\varphi(x_0,t_0))+\frac{1-\alpha_1}{2(N+2)}\Delta\varphi(x_0,t_0)+h_1(x_0,t_0), \tag{162}\] when \(\nabla \varphi (x_0,t_0) = 0\).

Using the definition of the normalized \(p-\)Laplacian we have arrived to \[\frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_p^1\varphi(x_0,t_0)\geq h_1(x_0,t_0). \tag{163}\]

Thus we proved that \(u\) is a viscosity supersolution of (150).

Now we are going to prove that \(u\) is viscosity solution to \[\label{parmonica} \frac{\partial u}{\partial t}(x,t)-\Delta_p^1 u(x,t)= h_1(x,t), \tag{164}\] in the set \(\left(\Omega\times (0,T)\right)\cap\{ u>v \}\). Let us consider \((x_0,t_0)\in\left(\Omega\times (0,T)\right)\cap\{ u>v \}\). Let \(\eta>0\) be such that \[u(x_0,t_0)\geq v(x_0,t_0)+3\eta. \tag{165}\]

Then, using that \(u\) and \(v\) are continuous functions, there exists \(\delta>0\) such that \[u(y,t)\geq v(y,t)+2\eta \quad \mbox{for all} \quad (y,t)\in B_{\delta}(x_0)\times(t_0-\delta,t_0+\delta), \tag{166}\] and, using that \(u^{\varepsilon}\rightrightarrows u\) and \(v^{\varepsilon}\rightrightarrows v\) we have \[u^{\varepsilon}(y,t)\geq v^{\varepsilon}(y,t)+\eta \quad \mbox{for all} \quad (y,t)\in B_{\delta}(x_0)\times(t_0-\delta,t_0+\delta) , \tag{167}\] for \(0<\varepsilon<\varepsilon_0\) for some \(\varepsilon_0>0\). Given \((z,t)\in B_{\frac{\delta}{2}}(x_0)\times(t_0-\frac{\delta}{2},t_0+\frac{\delta}{2})\) and \(\varepsilon<\min\{\varepsilon_0,\frac{\delta}{2}\}\) we obtain \[B_{\varepsilon}(z)\times\{t-\varepsilon^2\}\subset B_{\delta}(x_0)\times(t_0-\frac{\delta}{2},t_0+\frac{\delta}{2}). \tag{168}\]

Using that \(u^{\varepsilon}\rightrightarrows u\) we have the following limits: \[\label{Claim1} \displaystyle \sup_{y \in B_{\varepsilon}(z)}u^{\varepsilon}(y,t-\varepsilon^2)\rightarrow u(z,t), \qquad \mbox{ as } \varepsilon\rightarrow 0. \tag{169}\]

In fact, from our previous estimates we have that \[\Big|\sup_{y \in B_{\varepsilon}(z)}u^{\varepsilon}(y,t-\varepsilon^2)-u(z,t) \Big|\leq \sup_{y \in B_{\varepsilon}(z)}|u^{\varepsilon}(y,t-\varepsilon^2)-u(y,t-\varepsilon^2)|+ \sup_{y \in B_{\varepsilon}(z)}|u(y,t-\varepsilon^2)-u(z,t)|. \tag{170}\]

Using that \(u^{\varepsilon}\rightrightarrows u\), there exists \(\varepsilon_1>0\) such that if \(\varepsilon<\varepsilon_1\) \[|(u^{\varepsilon}-u)(x,t)|<\frac{\theta}{2} \quad \mbox{for all} \quad (x,t)\in\Omega\times(0,T]. \tag{171}\]

Now, using that \(u\) is continuous, there exists \(\varepsilon_2>0\) such that \[|u(y,t)-u(z,s)|<\frac{\theta}{2} \quad \mbox{if} \quad |(y,t)-(z,s)|<\varepsilon_2, \tag{172}\] thus, if we take \(\varepsilon<\frac{1}{2}\min\{\varepsilon_0,\varepsilon_1,\varepsilon_2,\frac{\delta}{2} \}\) we obtain \[\Big|\sup_{y \in B_{\varepsilon}(z)}u^{\varepsilon}(y,t-\varepsilon^2)-u(z,t)\Big|<\theta. \tag{173}\]

This proves (169).

Also, with a similar argument, we get, \[\label{Claim2} \displaystyle \lim_{\varepsilon\to 0}\inf_{y \in B_{\varepsilon}(z)}u^{\varepsilon}(y,t-\varepsilon^2)= u(z,t). \tag{174}\]

Finally, we also have, \[\label{Claim3} \displaystyle \lim_{\varepsilon\to 0} {{\Huge{f}}_{B_{\varepsilon}(z)}}u^{\varepsilon}(y,t-\varepsilon^2)dy= u(z,t). \tag{175}\]

In fact, let us compute \[\begin{array}{ll} \displaystyle\left| {{\Huge{f}}_{B_{\varepsilon}(z)}}u^{\varepsilon}(y,t-\varepsilon^2)dy-u(z,t)\right| \leq {{\Huge{f}}_{B_{\varepsilon}(z)}} \left| u^{\varepsilon}(y,t-\varepsilon^2)-u(y,t-\varepsilon^2) \right| dy+ {{\Huge{f}}_{B_{\varepsilon}(z)}} \left| u(y,t-\varepsilon^2)-u(z,t)\right| dy. \end{array} \tag{176}\]

Now we use again that \(u^{\varepsilon}\rightrightarrows u\) and that \(u\) is a continuous function to obtain \[ {{\Huge{f}}_{B_{\varepsilon}(z)}}|u^{\varepsilon}(y,s)-u(y,s)|dy<\frac{\theta}{2} \qquad \mbox{ and }\qquad {{\Huge{f}}_{B_{\varepsilon}(z)}}|u(y,t-\varepsilon^2)-u(z,t)|dz<\frac{\theta}{2},\] for \(\varepsilon>0\) small enough. Thus we get \[\left| {{\Huge{f}}_{B_{\varepsilon}(z)}}u^{\varepsilon}(y,t-\varepsilon^2)dy-u(z,t) \right|<\theta. \tag{177}\]

Using the previous limits, (169), (174) and (175) we obtain \[J_1(u^{\varepsilon})(z,t-\varepsilon^2)\rightarrow u(z,t) \qquad \mbox{ as } \varepsilon\rightarrow 0. \tag{178}\]

Analogously, we can prove that \[J_2(v^{\varepsilon})(z,t-\varepsilon^2)\rightarrow v(z,t), \qquad \mbox{ as } \varepsilon\rightarrow 0. \tag{179}\]

Now, if we recall that \(u(z,t)\geq v(z,t)+2\eta\), we obtain \[\label{river-plate} J_1(u^{\varepsilon})(z,t-\varepsilon^2)\geq J_2(v^{\varepsilon})(z,t-\varepsilon^2)+\eta, \tag{180}\] if \(\varepsilon>0\) is small enough. Then, using de DPP and (180) we have \[\label{DPPu} \begin{array}{ll} \displaystyle u^{\varepsilon}(z,t)=\frac{1}{2} J_1(u^{\varepsilon})(z,t-\varepsilon^2)+\frac{1}{2}\max\{J_1(u^{\varepsilon})(z,t-\varepsilon^2),J_2(v^{\varepsilon})(z,t-\varepsilon^2)\} =J_1(u^{\varepsilon})(z,t-\varepsilon^2), \end{array} \tag{181}\] for all \((z,t)\in B_{\frac{\delta}{2}}(x_0)\times(t_0-\frac{\delta}{2},t_0+\frac{\delta}{2})\) and for every \(\varepsilon>0\) small enough. Let us prove that \(u\) is viscosity subsolution to the equation (164). Given now \(\varphi\in\mathscr{C}^{2,1}(\Omega\times(0,T))\) such that \((u-\varphi)(x_0,t_0)=0\) is maximum of \(u-\varphi\). Then, from the uniform convergence there exists sequence of points \((x_{\varepsilon},t_{\varepsilon})_{\varepsilon>0}\subset B_{\frac{\delta}{2}}(x_0)\times(t_0-\frac{\delta}{2},t_0+\frac{\delta}{2})\), such that \(x_{\varepsilon}\rightarrow x_0\), \(t_{\varepsilon}\rightarrow t_0\) and \[(u^{\varepsilon}-\varphi)(x_{\varepsilon},t_{\varepsilon})\geq(u^{\varepsilon}-\varphi)(y,t)-o(\varepsilon^2), \tag{182}\] for all \((y,t)\in\Omega\times(0,T)\), that is, \[\label{ump2} u^{\varepsilon}(y,t)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\leq\varphi(y,t)-\varphi(x_{\varepsilon},t_{\varepsilon})+o(\varepsilon^2). \tag{183}\]

From the DPP (1) and using (183) we have \[\begin{aligned} \label{igualDPP} \displaystyle 0=&\frac{1}{2} J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)+\frac{1}{2}\max\Big\{ J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2), J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2) \Big\}-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon}) \notag\\[8pt] \displaystyle \qquad =& J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\le J_1(\varphi)(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon})+o(\varepsilon^2). \end{aligned} \tag{184}\]

If we add and subtract \(\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)\) we get \[\label{subphi} 0\le \varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon})+J_1(\varphi)(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)+o(\varepsilon^2). \tag{185}\]

Passing to the limit as before we obtain \[0\leq -\frac{\partial\varphi}{\partial t}(x_0,t_0)+\frac{\alpha_1}{2}\Delta^1_{\infty}\varphi(x_0,t_0)+\frac{(1-\alpha_1)}{2(N+2)}\Delta\varphi(x_0,t_0)+h_1(x_0,t_0), \tag{186}\] when \(\nabla \varphi (x_0,t_0) \neq 0\) and \[0\leq -\frac{\partial\varphi}{\partial t}(x_0,t_0)+\frac{\alpha_1}{2}\lambda_N (D^2 \varphi(x_0,t_0))+\frac{(1-\alpha_1)}{2(N+2)}\Delta\varphi(x_0,t_0)+h_1(x_0,t_0), \tag{187}\] if \(\nabla \varphi (x_0,t_0) = 0\). Hence we arrived to \[\frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_p^1\varphi(x_0,t_0)\leq h_1(x_0,t_0). \tag{188}\]

This proves that \(u\) is viscosity subsolution of the equation (164) inside the open set \(\{u>v\}\).

As we have that \(u\) is a viscosity supersolution in the whole \(\Omega\times(0,T)\), we conclude that \(u\) is viscosity solution to \[\frac{\partial u}{\partial t}(x_0,t_0)-\Delta_p^1 u(x_0,t_0)=h_1(x_0,t_0) , \tag{189}\] in the set \(\{u>v\}\).

(5) The equation for \(v\): The case that \(v\) is a viscosity subsolution to \[\frac{\partial v}{\partial t}(x,t)-\Delta_q^1 v(x,t)= h_2(x,t),\] is analogous. Here we use that \[\begin{aligned} \label{desDPP-v} \displaystyle 0=&\min\Big\{ J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-v(x_{\varepsilon},t_{\varepsilon}), J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon}) \Big\}\notag\\[8pt] \leq& J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon}). \end{aligned} \tag{190}\]

To show that \(v\) is a viscosity solution to \[\frac{\partial v}{\partial t}(x_0,t_0)-\Delta_q^1 v(x_0,t_0)= h_2(x_0,t_0), \tag{191}\] if \((x_0,t_0)\in\left(\Omega\times(0,T)\right)\cap\{u>v\}\) we proceed as before.

(6) Extra condition: Now let us prove the extra condition \[\displaystyle \left(\frac{\partial u}{\partial t}(x,t)-\Delta_{p}^{1}u(x,t)\right)+\left(\frac{\partial v}{\partial t}(x,t)-\Delta_q^1 v(x,t)\right)= h_1(x,t)+h_2(x,t) ,\] for \((x,t)\in \Omega\times (0,T)\). Notice that it is necesary to prove only the case \(u=v\), because in the set \(\{u>v\}\) we have \[\frac{\partial u}{\partial t}(x,t)-\Delta_{p}^{1}u(x,t)=h_1(x,t) \quad \mbox{and} \quad \frac{\partial v}{\partial t}(x,t)-\Delta_q^1 v(x,t)=h_2(x,t). \tag{192}\]

Let us start proving the subsolution case. Given \((x_0,t_0)\in\{u=v\}\) and \(\varphi\in\mathscr{C}^{2,1}\) such that \((u-\varphi)(x_0,t_0)=0\) is maximum of \(u-\varphi\). Notice that since \(v(x_0,t_0)=u(x_0,t_0)\) and \(v\leq u\) in \(\Omega\times(0,T)\) we also have that \((v-\varphi)(x_0,t_0)=0\) is maximum of \(v-\varphi\). Then, from the uniform convergence there exists a sequence of points \(\{(x_{\varepsilon},t_{\varepsilon})\}_{\varepsilon>0}\subset B_{\frac{\delta}{2}}(x_0)\times(0,T)\), such that \(x_{\varepsilon}\rightarrow x_0\), \(t_{\varepsilon}\rightarrow t_0\), and \[\label{umenosphi} (u^{\varepsilon}-\varphi)(x_{\varepsilon},t_{\varepsilon})\geq(u^{\varepsilon}-\varphi)(y,t)+o(\varepsilon^2), \tag{193}\] for all \((y,t)\in\Omega\times(0,T)\). Let us consider two cases:

Case 1. Suppose that \(u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})>v^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\) for a subsequence such that \(\varepsilon_j\rightarrow 0\). Let us notice that, if \[J_1(u^{\varepsilon})(z,t)<J_2(v^{\varepsilon})(z,t),\] we have that \[u^{\varepsilon}(z,t)=\frac{1}{2} J_1(u^{\varepsilon})(z,t)+\frac{1}{2} J_2(v^{\varepsilon})(z,t) \quad \mbox{and} \quad v^{\varepsilon}(z,t)=\frac{1}{2} J_1(u^{\varepsilon})(z,t)+\frac{1}{2} J_2(v^{\varepsilon})(z,t), \tag{194}\] and then we get \[u^{\varepsilon}(z,t)=v^{\varepsilon}(z,t),\] in this case.

This remark implies that when \(u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})>v^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\) we have \[J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j})\geq J_2(v^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}). \tag{195}\]

If we use the DPP (1) we get \[\begin{aligned} \displaystyle 0=&\frac{1}{2}\left(J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\right)\notag\\ &+\frac{1}{2} \max\{J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j}) , J_2(v^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\} \notag\\[8pt] \displaystyle =&\frac{1}{2}\left(J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\right)+\frac{1}{2} \left(J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\right)\notag\\[8pt] \displaystyle =&J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j}), \end{aligned} \tag{196}\] and using (193) we obtain \[0=J_1(u^{\varepsilon})(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-u^{\varepsilon}(x_{\varepsilon_j},t_{\varepsilon_j})\leq J_1(\varphi)(x_{\varepsilon_j},t_{\varepsilon_j}-\varepsilon_j^2)-\varphi(x_{\varepsilon_j},t_{\varepsilon_j}). \tag{197}\]

Taking limit as \(\varepsilon_j\rightarrow 0\) as before we get \[\label{ine-1} \frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_{p}\varphi(x_0,t_0)\leq h_1(x_0,t_0). \tag{198}\]

We have proved before that \(v\) is a subsolution to \[\frac{\partial v}{\partial t}(x,t)-\Delta_{q}v(x,t) = h_2(x,t),\] in the whole \(\Omega\times(0,T)\). Therefore, as \((v-\varphi)(x_0,t_0)=0\) is a maximum of \(v-\varphi\) we get \[\label{ine-2} \frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_{q}\varphi(x_0,t_0)\leq h_2(x_0,t_0). \tag{199}\]

Thus, from (198) and (199) we conclude that \[\left(\frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_{p}\varphi(x_0,t_0)\right)+ \left(\frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_{q}\varphi(x_0,t_0)\right)\leq h_1(x_0,t_0)+h_2(x_0,t_0). \tag{200}\]

Case 2. If \(u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})=v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\) for \(\varepsilon<\varepsilon_0\). Using the DPP (1) we have \[\begin{aligned} \displaystyle u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})=&\frac{1}{2} J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)+\frac{1}{2} J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2), \notag\\[8pt] \displaystyle v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})=&\frac{1}{2} J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2) +\frac{1}{2} J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2), \end{aligned} \tag{201}\] then we get \[\begin{aligned} \displaystyle \max\{J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2), J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)\}=&J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2), \notag\\[8pt] \displaystyle \min\{J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2), J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)\}=&J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2). \end{aligned} \tag{202}\]

If we use again (193) we obtain \[\begin{array}{ll} \varphi(y,t)-\varphi(x_{\varepsilon},t_{\varepsilon})\geq u^{\varepsilon}(y,t)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})+o(\varepsilon^2) \geq v^{\varepsilon}(y,t)-v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})+o(\varepsilon^2), \end{array} \tag{203}\] here we used that \(u^{\varepsilon}\geq v^{\varepsilon}\) and \(u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})= v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\). Thus \[\begin{aligned} \displaystyle 0=&\frac{1}{2} \left(J_1(u^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-u^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\right)+\frac{1}{2} \left(J_2(v^{\varepsilon})(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-v^{\varepsilon}(x_{\varepsilon},t_{\varepsilon})\right)\notag\\[8pt] \displaystyle \quad \leq& \frac{1}{2} \left(J_1(\varphi)(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon})\right)+\frac{1}{2} \left(J_2(\varphi)(x_{\varepsilon},t_{\varepsilon}-\varepsilon^2)-\varphi(x_{\varepsilon},t_{\varepsilon})\right). \end{aligned} \tag{204}\]

Taking limit as \(\varepsilon\rightarrow 0\) we conclude that \[\left(\frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_{p}\varphi(x_0,t_0)\right)+ \left(\frac{\partial\varphi}{\partial t}(x_0,t_0)-\Delta_{q}\varphi(x_0,t_0)\right)\leq h_1(x_0,t_0)+h_2(x_0,t_0). \tag{205}\]

Then, we get the subsolution case in the viscosity sense (taking care of the semicontinuous envelopes when the gradient of \(\varphi\) vanishes). We have just proved that the extra condition is verified with an inequality when we touch \(u\) and \(v\) from above at some point \((x_0,t_0)\) with a smooth test function.

The proof that the other inequality holds when we touch \(u\) and \(v\) from below is analogous and hence we omit the details. ◻