The asymptotic p-Poisson equation as $p \rightarrow \infty$ in Carnot-Carathéodory spaces

Abstract

In this paper we study the asymptotic behavior of solutions to the subelliptic p-Poisson equation as $p\rightarrow +\infty$ in Carnot-Carathéodory spaces. In particular, introducing a suitable notion of differentiability, extend the celebrated result of Bhattacharya et al. (Rend Sem Mat Univ Politec Torino Fascicolo Speciale 47:15–68, 1989) and we prove that limits of such solutions solve in the sense of viscosity a hybrid first and second order PDE involving the $\infty$-Laplacian and the Eikonal equation.

Appendix

As already pointed out, Proposition 2.14 can still be proved assuming

(D1):

$(\Omega ,d_\Omega )$ is a Carnot-Carathéodory space,

(D2):

$d_\Omega$ is continuous with respect to the Euclidean topology,

(LIC):

The vectors $X_1(x),\ldots ,X_m(x)$ are linearly independent for any $x\in \Omega$

instead of (LIC) and (2.4). The previous set of conditions embraces many relevant families of vector fields, such as for instance Carnot Groups. However, when considering the two sets of hypotheses given by the Hörmander condition and (D1), (D2), (LIC), one can show that neither of the two implies the other. Indeed, from one hand it is well known that the Grushin plane, i.e. ${\mathbb {R}}^2$ equipped with the Carnot-Carathéodory distance generated by the vector fields

$$\begin{aligned} X=\frac{\partial }{\partial x}\qquad Y=x\frac{\partial }{\partial y}, \end{aligned}$$

satisfies the Hörmander condition, while X and Y are clearly linearly dependent in $\{(0,y)\ |\ y\in {\mathbb {R}}\}$. On the other hand, there are examples of (even smooth) families of vector fields satisfying (D1), (D2), (LIC) which does not satisfies the Hörmander condition. Let us consider the two linearly independent vector fields X, Y defined on ${\mathbb {R}}^3$ by

$$\begin{aligned} X=\frac{\partial }{\partial x}\qquad Y=\frac{\partial }{\partial y}+\varphi (x)\frac{\partial }{\partial z}, \end{aligned}$$

where $\varphi (x):=\psi (x)+\psi (-x)$ and $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is defined by

$$\begin{aligned} \psi (x)= {\left\{ \begin{array}{ll}e^{-\frac{1}{x}}&{}\text { if }x>0\\ 0 &{}\text { otherwise}. \end{array}\right. } \end{aligned}$$

Since $\varphi ^{(k)}(0)=0$ for any $k\in {\mathbb {N}}$, it is easy to see that

$$\begin{aligned}{}[X,[\ldots ,[X,Y]\ldots ](0,y,z)=[Y,[\ldots ,[X,Y]\ldots ](0,y,z)=0 \end{aligned}$$

for any $y,z\in {\mathbb {R}}$ so X, Y do not satisfy the Hörmander condition in $\{(0,y,z)\ |\ y,z\in {\mathbb {R}}\}$.

It is not difficult to show that they induce a Carnot-Carathéodory distance d on ${\mathbb {R}}^3$, and that the identity map

$$\begin{aligned} Id :({\mathbb {R}}^3,d_e)\longrightarrow ({\mathbb {R}}^3,d) \end{aligned}$$

is continuous. Indeed, let $A=(x,y,z)$ and $B=(x_1,y_1,z_1)$ in ${\mathbb {R}}^3$. We construct a horizontal curve joining them whose horizontal length tends to zero as A tends to B in the Euclidean topology. First, notice that moving along the X direction the induced Carnot-Carathéodory distance is comparable with the Euclidean one. Hence, without loss of generality, we can assume that $x=x_1=0$. Moreover, since $Y=\frac{\partial }{\partial y}$ on $\{x=0\}$, then moving along the Y direction inside $\{x=0\}$ the induced Carnot-Carathéodory distance is comparable with the Euclidean one. Hence we assume that $y_1=y$. The last step is to join (0, y, z) and $(0,y, z_1)$. We assume, without loss of generality, that $z_1>z$. Let us set

$$\begin{aligned} \delta :=-\frac{1}{\log (\sqrt{z_1-z})} \end{aligned}$$

then $\delta \rightarrow 0^+$ as $z_1\rightarrow z$. Let us define the curves $\gamma _1,\ldots ,\gamma _4:[0,1]\rightarrow {\mathbb {R}}^3$ by

$$\begin{aligned} \gamma _1(t)= & {} (0,y,z)+t(\delta ,0,0), \\ \gamma _2(t)= & {} (\delta ,y,z)+t\left( 0,\frac{z_1-z}{\varphi (\delta )},z_1-z\right) , \\ \gamma _3(t)= & {} \left( \delta ,y+\frac{z-z_1}{\varphi (\delta )},z_1\right) +t(-\delta ,0,0) \end{aligned}$$

and

$$\begin{aligned} \gamma _4(t)=\left( 0,y+\frac{z_1-z}{\varphi (\delta )},z_1\right) +t\left( 0,\frac{z-z_1}{\varphi (\delta )},0\right) \end{aligned}$$

it is easy to see that they are horizontal and that they connect (0, y, z) and $(0,y,z_1)$. Moreover, a quick computation shows that

$$\begin{aligned} d((0,y,z),(0,y,z_1))\le 2\delta +\frac{z_1-z}{\varphi (\delta )}=-\frac{2}{\log (\sqrt{z_1-z})}+\sqrt{z_1-z}. \end{aligned}$$

As the right hand side tends to zero as $z_1\rightarrow z$, the conclusion follows.