The obstacle problem for the p-laplacian via optimal stopping of tug-of-war games

Abstract

We present a probabilistic approach to the obstacle problem for the p-Laplace operator. The solutions are approximated by running processes determined by tug-of-war games plus noise, and letting the step size go to zero, not unlike the case when Brownian motion is approximated by random walks. Rather than stopping the process when the boundary is reached, the value function is obtained by maximizing over all possible stopping times that are smaller than the exit time of the domain.

Appendices

Appendix 1: A proof of Lemma 3.3: games end almost-surely

1. Consider a new “game-board” $Y=\mathbb {R}^N$ with the same initial token position $x_0\in \Omega$. By the same symbols $\sigma _I$ and $\sigma _{II}$ we denote the extensions on $\{Y^{n}\}_{n=0}^\infty$ of the given strategies $\sigma _I$ and $\sigma _{II}$, defined as in the formula (5.12), where in order to simplify notation we suppress the overline in $\bar{\sigma }$. Define also the new transition probabilities:

$$\begin{aligned} \gamma _n[x_0,\ldots , x_n] = \frac{\alpha }{2}\delta _{\sigma _I^n(x_0,\ldots , x_n)} + \frac{\alpha }{2}\delta _{\sigma _{II}^n(x_0,\ldots , x_n)} + \frac{\beta }{|B_\epsilon |} \mathcal {L}_N\lfloor B_\epsilon (x_n), \end{aligned}$$

and let $\mathbb {P}^{n, x_0}_{\sigma _I, \sigma _{II}}$ and $\mathbb {P}^{x_0}_{\sigma _I, \sigma _{II}}$ be the resulting probability measures on $Y^{\infty , x_0}$ as in Sect. 3.4. By Lemma 5.2 and since $\tau \le \tau _0$, it follows that:

$$\begin{aligned} \mathbb {P}^{x_0}_{\tau , \sigma _{I}, \sigma _{II}} (\{\tau <\infty \})= & {} \sum _{n=0}^\infty \mathbb {P}^{n, x_0}_{\tau , \sigma _I, \sigma _{II}} \left( \{\omega \in X^{x_0,\infty }; ~ \tau (\omega )=n\right\} ) \nonumber \\= & {} \sum _{n=0}^\infty \bar{\mathbb {P}}^{n, x_0}_{\bar{\sigma }_I, \bar{\sigma }_{II}} \big (\{\omega \in X^{x_0,\infty }; ~ \tau (\omega )=n\}\big ) \nonumber \\= & {} \sum _{n=0}^\infty \bar{\mathbb {P}}^{n, x_0}_{\bar{\sigma }_I, \bar{\sigma }_{II}} \big (\{\omega \in Y^{x_0,\infty }; ~ \tau (\omega )=n\}\big ) \nonumber \\= & {} {\mathbb {P}}^{x_0}_{\bar{\sigma }_{I}, \bar{\sigma }_{II}} (\{\tau <\infty \}) \ge {\mathbb {P}}^{x_0}_{\bar{\sigma }_{I}, \bar{\sigma }_{II}} (\{\tau _0<\infty \}). \end{aligned}$$

(5.20)

Let now $A_0$ be the sector in $B_\epsilon =B_\epsilon (0)$:

$$\begin{aligned} A_0=\Big \{x\in \mathbb {R}^N; ~ |x|\in (\epsilon /2, \epsilon ) ~~ \text{ and } ~~ \angle (x, e_1) \in (-\pi /8, \pi /8)\Big \}. \end{aligned}$$

For $M\in \mathbb {N}$ sufficiently large to ensure that M consecutive shifts of the token by vectors chosen from $A_0$ will get the token, originally located at any point in $\Omega$, out of $\Omega$, define:

$$\begin{aligned} S_{x_0}= & {} \Big \{\omega =(x_0, x_1,\ldots )\in Y^{x_0, \infty }; ~ \exists i_0 \\ \forall i= & {} i_0, i_0+1,\ldots , i_0+M \quad x_{i+1} - x_i\in A_0\Big \}. \end{aligned}$$

t is clear that:

$$\begin{aligned} \mathbb {P}^{x_0}_{\sigma _I, \sigma _{II}} (\{\tau _0<\infty \}) \ge \mathbb {P}^{x_0}_{\sigma _I, \sigma _{II}} (S_{x_0}). \end{aligned}$$

(5.21)

2. We now show that the probability in the right hand side of (5.21) equals 1. Recall that for a bounded $\mathcal {F}^{x_0}$-measurable function $f:Y^{x_0,\infty }\rightarrow \mathbb {R}$, its conditional expectation $\mathbb {E}^{x_0}_{\sigma _I, \sigma _{II}}\{f|\mathcal {F}_1^{x_0}\}$ is the function: $(x_0, x_1)\mapsto \mathbb {E}^{x_1}_{\sigma _I', \sigma _{II}'}[f']$, where $\sigma _I'$, $\sigma _{II}'$ are strategies on $Y^{x_0, \infty }$ given by:

$$\begin{aligned} (\sigma _I')^n(x_1,\ldots , x_{n+1})= & {} \sigma _I^{n+1}(x_0, x_1,\ldots , x_{n+1}), \qquad (\sigma _{II}')^n(x_1,\ldots , x_{n+1}) \\= & {} \sigma _{II}^{n+1}(x_0, x_1,\ldots , x_{n+1}), \end{aligned}$$

while the Borel random variable $f':Y^{x_1, \infty }\rightarrow \mathbb {R}$ is similarly set to be: $f'(x_1,x_2,\ldots ) = f( x_0, x_1,x_2,\ldots )$. Consequently:

(5.22)

where each set $S_{x_1}^{x_0} = \{(x_1,x_2,\ldots )\in Y^{x_1,\infty };~ (x_0, x_1,x_2,\ldots )\in S_{x_0}\}$ clearly contains $S_{x_1}$. Let now:

$$\begin{aligned} q(x) = \inf _{\tilde{\sigma }_I,\tilde{\sigma }_{II}} \mathbb {P}^x_{\tilde{\sigma }_I,\tilde{\sigma }_{II}} (S_x). \end{aligned}$$

(5.23)

By an easy translation invariance argument, $q(x) = q$ is actually independent of $x\in \mathbb {R}^N$. Hence, in view of (5.22) we obtain:

$$\begin{aligned} \mathbb {P}^{x_0}_{\sigma _{I}, \sigma _{II}} (S_{x_0})\ge & {} \frac{\alpha }{2}\mathbb {P}^{\sigma _I^0(x_0)}_{\sigma _I', \sigma _{II}'}(S_{\sigma _I^0(x_0)}) + \frac{\alpha }{2}\mathbb {P}^{\sigma _{II}^0(x_0)}_{\sigma _I', \sigma _{II}'}(S_{\sigma _{II}^0(x_0)}) \nonumber \\&+ \frac{\beta }{|B_\epsilon |} \int _{B_\epsilon (x_0){\setminus } (x_0+A_0)} \mathbb {P}^{x_1}_{\sigma _I', \sigma _{II}'}(S_{x_1})~\text{ d }x_1 \nonumber \\&+ \frac{\beta }{|B_\epsilon |} \int _{x_0+A_0} \mathbb {P}^{x_1}_{\sigma _I', \sigma _{II}'}(S^{x_0}_{x_1})~\text{ d }x_1 \nonumber \\\ge & {} \left( \frac{\alpha }{2} + \frac{\alpha }{2} + \frac{\beta }{|B_\epsilon |} |B_\epsilon {\setminus } A_0|\right) q + \frac{\beta }{|B_\epsilon |} \int _{x_0+A_0} \mathbb {P}^{x_1}_{\sigma _I', \sigma _{II}'}(S^{x_0}_{x_1})~\text{ d }x_1 \nonumber \\= & {} \theta q + \frac{\beta }{|B_\epsilon |} \int _{x_0+A_0} \mathbb {P}^{x_1}_{\sigma _I', \sigma _{II}'}(S^{x_0}_{x_1})~\text{ d }x_1, \end{aligned}$$

(5.24)

where we defined:

$$\begin{aligned} \theta = \alpha + \beta \left( 1-\frac{|A_0|}{|B_\epsilon |}\right) . \end{aligned}$$

3. Similarly as in the previous step, for every $x_1\in x_0+A_0$ there holds:

$$\begin{aligned} \mathbb {P}^{x_1}_{\sigma _{I}', \sigma _{II}'} \left( S^{x_0}_{x_1}\right)= & {} \frac{\alpha }{2}\mathbb {P}^{\sigma _I^1(x_0, x_1)}_{\sigma _I'', \sigma _{II}''}\left( S^{x_0, x_1}_{\sigma _I^1(x_0, x_1)}\right) + \frac{\alpha }{2}\mathbb {P}^{\sigma _{II}^1(x_0, x_1)}_{\sigma _I'', \sigma _{II}''}\left( S^{x_0, x_1}_{\sigma _{II}^1(x_0, x_1)}\right) \\&+ \frac{\beta }{|B_\epsilon |} \int _{B_\epsilon (x_0)} \mathbb {P}^{x_2}_{\sigma _I'', \sigma _{II}''}\left( S^{x_0,x_1}_{x_2}\right) ~\text{ d }x_2, \end{aligned}$$

where the set $S^{x_0,x_1}_{x_2}=\{(x_2, x_3,\ldots )\in Y^{x_2,\infty }; ~ (x_2, x_2, x_2, x_3\ldots )\in S_{x_0}\}$ contains the set $S_{x_2}$. By (5.23) we see that:

$$\begin{aligned} \inf _{\tilde{\sigma }_I,\tilde{\sigma }_{II}} \mathbb {P}^{x_2}_{\tilde{\sigma }_I,\tilde{\sigma }_{II}} \left( S^{x_0, x_1}_{x_2}\right) \ge q, \end{aligned}$$

and hence:

$$\begin{aligned} \mathbb {P}^{x_1}_{\sigma _{I}', \sigma _{II}'} \left( S^{x_0}_{x_1}\right) \ge \theta q + \frac{\beta }{|B_\epsilon |} \int _{x_1+ A_0} \mathbb {P}^{x_2}_{\sigma _I'', \sigma _{II}''}\left( S^{x_0,x_1}_{x_2}\right) ~\text{ d }x_2. \end{aligned}$$

Since $1-\theta = \beta |A_0|/|B_\epsilon |$, the estimate in (5.24) becomes:

$$\begin{aligned} \mathbb {P}^{x_0}_{\sigma _{I}, \sigma _{II}} (S_{x_0}) = \theta q + (1-\theta )\theta q + \left( \frac{\beta }{|B_\epsilon |} \right) ^2\int _{x_0+A_0} \int _{x_1+A_0} \mathbb {P}^{x_2}_{\sigma _I'', \sigma _{II}''}\left( S^{x_0,x_1}_{x_2}\right) ~\text{ d }x_2 ~\text{ d }x_1. \end{aligned}$$

Iterating the same argument as above M times, we arrive at:

(5.25)

But each probability under the iterated integrals equals to 1, because: $S^{x_0,\ldots ,x_{M-1}}_{x_M} = Y^{x_M,\infty }$ for $x_1\in x_0+A_0$, $x_2\in x_1+A_0$, $\ldots , x_M\in x_{M-1}+A_0$. Consequently, by (5.25) we get:

$$\begin{aligned} \mathbb {P}^{x_0}_{\sigma _{I}, \sigma _{II}} (S_{x_0}) \ge \sum _{n=0}^{M-1}(1-\theta )^n\theta q + (1-\theta )^M = \big (1-(1-\theta )^M\big ) q + (1-\theta )^M. \end{aligned}$$

Infimizing over all strategies $\sigma _I, \sigma _{II}$, it follows that $q\ge 1$, since $\theta <1$ because of $\beta >0$. Further:

$$\begin{aligned} \mathbb {P}^{x_0}_{\sigma _{I}, \sigma _{II}} (S_{x_0}) \ge q=1. \end{aligned}$$

This achieves (3.5) in view of (5.20) and (5.21).$\square$

Appendix 2: A proof of Lemma 4.2: uniqueness of viscosity solutions

1. Firstly, note that the continuous function u is a viscosity p-supersolution to (1.7). Thus, by the classical result in [7], u is p-superharmonic in $\Omega$, and consequently (see [10]) we have $u\in W^{1,p}_{loc}(\Omega )$. In the same manner, it follows from Definition 4.1 that u is a viscosity p-subsolution on the open set $\mathcal {V}=\{x\in \Omega ; ~ u(x)>\Psi (x)\}$, hence u is p-subharmonic in $\mathcal {V}$.

Therefore, using the variational definitions of p-super- and p-subharmonic functions, we have that for any open, Lipschitz domain $\mathcal {U}\subset \subset \Omega$ there holds:

$$\begin{aligned} \int _{\mathcal {U}}|\nabla u|^p\le & {} \int _{\mathcal {U}}|\nabla (u+\phi )|^p \qquad \forall \phi \in \mathcal {C}_0^\infty (\mathcal {U},\mathbb {R}_+), \end{aligned}$$

(5.26)

$$\begin{aligned} \int _{\mathcal {U}\cap \mathcal {V}}|\nabla u|^p\le & {} \int _{\mathcal {U}\cap \mathcal {V}}|\nabla (u+\phi )|^p \qquad \forall \phi \in \mathcal {C}_0^\infty (\mathcal {U}\cap \mathcal {V},\mathbb {R}_-). \end{aligned}$$

(5.27)

Let now $\phi \in \mathcal {C}_0^\infty (\mathcal {U},\mathbb {R})$ be such that $\Psi \le u+\phi$. We write: $\phi = \phi ^+ +\phi ^-$ as the sum of the positive and negative parts of $\phi$. Denote:

$$\begin{aligned} D^+ = \{x\in \mathcal {U}; ~ \phi (x)>0\}\quad \text{ and } \quad D^- = \{x\in \mathcal {U}; ~ \phi (x)<0\}\subset \mathcal {V}. \end{aligned}$$

Then we have, in view of (5.26) and (5.27):

(5.28)

The above means precisely that u is a variational solution of the obstacle problem on $\mathcal {U}$, with the lower obstacle $\Psi$ and boundary data $f=u_{|\partial \mathcal {U}}$; we denote this problem by $\mathcal {K}_{\Psi , f}(\mathcal {U})$. Existence and uniqueness of such variational solution is an easy direct consequence of the strict convexity of the functional $\int _{\mathcal {U}} |\nabla u|^p$. It is also quite classical that such solutions obey a comparison principle [10].

2. Let now u and $\bar{u}$ be as in the statement of the Lemma. Fix $\epsilon >0$. By the uniform continuity of u, $\bar{u}$ on $\Omega$ and the fact that they coincide on $\partial \Omega$, there exists $\delta >0$ such that:

$$\begin{aligned} |u(x)-\bar{u}(x)|\le \epsilon \qquad \forall x\in \mathcal {O}_\delta (\partial \Omega ) := \big (\partial \Omega +B(0,\delta )\big )\cap \bar{\Omega }. \end{aligned}$$

(5.29)

Consider an open, Lipschitz set $\mathcal {U}$ satisfying: $\Omega {\setminus } \mathcal {O}_\delta (\partial \Omega )\subset \subset \mathcal {U} \subset \subset \Omega$. By the argument in Step 1, u is the variational solution of the problem set $\mathcal {K}_{\Psi , u_{\mid \partial \mathcal {U}}}(\mathcal {U})$, and it is also easy to observe that $\bar{u}+\epsilon$ is the variational solution of the problem $\mathcal {K}_{\Psi , \bar{u}_{\mid \partial \mathcal {U}}+\epsilon }(\mathcal {U})$. Since $u<\bar{u}+\epsilon$ on $\partial \mathcal {U}$ in view of (5.29), the comparison principle implies that $u\le \bar{u}+\epsilon$ in $\bar{\mathcal {U}}$.

Reversing the same argument and taking into account (5.29), we arrive at:

$$\begin{aligned} |u(x) - \bar{u}(x)|\le \epsilon \qquad \forall x\in \bar{\Omega }. \end{aligned}$$

We conclude that $u=\bar{u}$ in $\bar{\Omega }$ by passing to the limit $\epsilon \rightarrow 0$ in the above bound. $\square$