Harnack inequality and self-similar solutions for fully nonlinear fractional parabolic equations

Abstract

We find bounds for the principal eigenvalue and eigenfunction associated to the existence of self-similar solutions to a fully nonlinear parabolic problem. In contrast with the local case, the upper and lower bounds for the eigenfunction are of the same polynomial order $|x|^{-(N+2s)}$. In order to accomplish this we prove a Harnack inequality for general nonlocal elliptic equations with zero order terms.

Appendix

Here we collect some technical results that are used above.

For $R > 0$ denote $w_{s,R}(y) = (R + |y|)^{-(N + 2s)}$ and we omit the subscript R when $R = 1$. Also denote

$$\begin{aligned} \Vert u \Vert _{L^1(w_{s, R})} = \int _{\mathbb {R}^N} u(y) w_{s,R}(y)dy. \end{aligned}$$

Lemma 5.1

Let $b \in L^\infty _{loc}(B_{2R})$. Assume that u is a bounded, nonnegative viscosity solution of

$$\begin{aligned} \mathcal I(u) + b(x) \cdot Du \le c_1 u \quad \text{ in } \ B_{2R}. \end{aligned}$$

Then, there exists a constant $C > 0$ such that

$$\begin{aligned} \Vert u \Vert _{L^1(w_{s, R})} \le C\Big ( R^{1-2s} ||b||_{L^\infty (B_{2R})} + R^{-2s}||c||_{L^\infty (B_{2R})} \Big ) \inf _{B_R} u+C||u||_{L^\infty (B_{2R})} \end{aligned}$$

Proof

We assume $R =1$ and conclude the general result by considering the usual rescaling $u_R(x) = u(Rx)$. We follow the ideas of [2].

Let $\varphi \in C_0^\infty (B_{3/2})$ such that $0 \le \varphi \le 1$ and $\varphi \equiv 1$ in $B_1$. If u is nontrivial, the strong maximum principle implies that $u > 0$ in $B_2$. Then, there exists $0 < t \le \inf _{B_1} u$ such that $u \ge t\varphi$ in $\mathbb {R}^N$. Moreover, enlarging t if necessary, we can consider a point $z_0 \in B_{3/2}$ for which $u(z_0) = t\varphi (z_0)$. Then, the viscosity inequality for u allows us to write

$$\begin{aligned} t \mathcal {I}[B_{1/4}(z_0)](\varphi , z_0) + \mathcal {I}[B_{1/4}(z_0)^c](u,z_0) - C t \Vert b\Vert _\infty \le C t \varphi (z_0), \end{aligned}$$

where $C > 0$ just depends on $c_1$ and universal constants. From here, by the smoothness of $\varphi$ it is direct to see the existence of a universal constants $C > 0$ just depending on N, s such that

$$\begin{aligned} \gamma \int \limits _{B_{1/4}(z_0)^c} \frac{u(y)}{|z_0 - y|^{N + 2s}}dy \le C \Gamma \Vert u \Vert _{L^\infty (B_2)} + C t. \end{aligned}$$

The result follows by rearranging terms and using the definition of t. $\square$

Next lemma states a known fact that solutions of equation of the type (5.1) are Hölder continuous and localizes the $L^\infty$ dependence of the right hand side.

Lemma 5.2

Let $\Theta \subset \mathbb {R}^N$ be a domain and assume $B_{R}\subset \Theta$. Let u be a solution of

$$\begin{aligned} \mathcal I u+b \cdot Du = \lambda u \quad \text{ in } \ \Theta . \end{aligned}$$

Then, there exists $\alpha \in (0,1)$ and a constant $C=C_R$ not depending on the domain such that

$$\begin{aligned} {[}u]_{C^\alpha (B_{R/2})} \le C_R(1 + \Vert u\Vert _{L^\infty (B_R)} + \Vert u\Vert _{L^1(w_s)}). \end{aligned}$$

Proof

Without loss of generality we can assume that $R=1$ and then conclude by scaling. Note first that u satisfies (in the viscosity sense)

$$\begin{aligned} \mathcal {M}^+ u - C_1 |Du|&\le C_2,\\ \mathcal {M}^- u + C_1 |Du|&\ge -C_2, \end{aligned}$$

in $B_1$, for some constants $C_1, C_2 > 0$. It is direct to check that the function $\tilde{u} := u(x) \textbf{1}_{B_4}(x)$, $x \in \mathbb {R}^N$, satisfies the inequalities

$$\begin{aligned}&\mathcal {M}^+ \tilde{u} - C_1 |D \tilde{u}| \le C_2 + C \Vert u \Vert _{L^1(w_s)} \quad \text{ in } \ B_1 \\&\mathcal {M}^- \tilde{u} + C_1 |D \tilde{u}| \ge -C_2 \quad \text{ in } \ B_1, \end{aligned}$$

where the constant $C > 0$ in the right-hand side of the first inequality just depends on N, s and the ellipticity constants. Standard regularity theory asserts the existence of $\alpha \in (0,1)$ and a constant $C > 0$ such that

$$\begin{aligned} {[}u]_{C^\alpha (B_{1/2})} = [\tilde{u}]_{C^\alpha (B_{1/2})} \le C(1+ \Vert u\Vert _{L^\infty (B_1)} + \Vert u \Vert _{L^1(w_s)}) \end{aligned}$$

$\square$

Lemma 5.3

Let $\alpha < \frac{N + 2s}{2s}$. Let u be a bounded viscosity solution of

$$\begin{aligned} \mathcal {M}^+ u + \frac{1}{2s} x \cdot Du \ge -\alpha u\quad \text{ in }\quad B_{R}, \end{aligned}$$

with $u\le C_0|x|^{-(N + 2s)}$ for some $C_0 > 0$, and $u\le 0$ in $B_R$. Then there exists $R_0$ so that if $R>R_0$, then $u\le 0$ in $\mathbb {R}^N$.

Proof

The result for $\alpha \le 0$ is direct. Assume $0 < \alpha$ and let $\beta$ such that $2s \alpha< \beta < N + 2s$ and denote $\epsilon = \beta - 2s\alpha$. Consider $\Phi$ the function of Lemma 4.3 defined with such an exponent $\beta$.

Take $R_0 > r_c$ as in Lemma 4.3 and fix c small enough in order to have

$$\begin{aligned} \mathcal {M}^+ \Phi + \frac{1}{2s} x \cdot \Phi \le -\frac{\beta - \epsilon /2}{2s} \Phi \quad \text{ in } \ B_{r_c}^c. \end{aligned}$$

Suppose now by contradiction that for some $R \ge R_0$, there exists $x_R$ with $|x_R|>R$ such that $u(x_R)>0$. Then, since $u \le 0$ in $B_R$ and u decays faster than $\Phi$ at infinity, there exists $\eta > 0$ such that $u \le \eta \Phi$ in $\mathbb {R}^N \setminus \{ 0 \}$ and a point $x \in B_{R}^c$ such that $u(x) = \eta \Phi (x)$.

Then, using $\eta \Phi$ as a test function for u and the estimate above, we obtain

$$\begin{aligned} \alpha \ge \frac{\beta - \epsilon /2}{2s}, \end{aligned}$$

but this is a contradiction. $\square$