# Boundary singularity for fractional elliptic and parabolic problems

David Gómez-Castro

October 17, 2023

## Co-authors

Juan Luis Vázquez

Nicola Abatangelo

U. Bologna

Hardy Chan

U. Basel

# Fractional Laplacian in $$\mathbb R^d$$

## Fractional Laplacian

$(-\Delta)^s u (x) = {C(d,s)} \,\, \underbrace{\lim_{\varepsilon \to 0} \int_{\mathbb R^d \setminus B_\varepsilon (x) }}_{\text{ P.V. }\int _{\mathbb R^d}} \frac{u(x) - u(y)}{|x-y|^{d+2s}} d y.$

Equivalent definitions (up-to ten )

• The unique operator such that

$\mathcal F[ (-\Delta)^s u ] = |\xi|^{2s} \mathcal F[u]$

i.e. the spectral fractional power of $$-\Delta$$.

• The infinitesimal generator of a Lévy process $$X_h$$, characterised by “long jumps”

$(-\Delta)^s u (x) = \lim_{h \to 0} \frac{ \mathbb E[ f(x) - f(x - X_h) ] }{ h }$

## Laplace equation

$\begin{cases} (-\Delta)^s u (x) = f(x) & \textrm{for all } x \in \mathbb R^d \\ u(x) \to 0 & \textrm{as } |x| \to \infty \end{cases}$

Transforming the equation $$|\xi|^{2s} \widehat u(\xi) = \widehat f(\xi)$$.

Formally, we can recover $$u = \mathcal F^{-1}[|\xi|^{-2s}] * f$$.

For $$d > 2s$$, $$\mathcal F^{-1}[|\xi|^{-2s}]$$ is the Riesz kernel so we get

$u(x) = C(d,-2s) \int_{\mathbb R^d} \frac{f(y)}{|x-y|^{d-2s}} dy.$

For any $$d, 2s$$ we can also look for this solution by the minisation of the energy functional

$J(u) = \frac {C(d,s)} 2 \int_{\mathbb R^d} \int_{\mathbb R^d } \frac{|u(x) - u(y)|^2}{|x-y|^{d+2s}} dx \, dy - \int_\Omega f u$

in homogeneous fractional Sobolev spaces .

## Fractional Elliptic/Heat/Porous Medium in $$\mathbb R^d$$

• Analysis of the operator: extension

• Obstacle: ,

• Heat: We can compute estimates for the heat kernel $$K(t,\cdot) = \mathcal F^{-1} [e^{-t|\xi|^{2s}}]$$.

• Porous Medium: ,

• Connection to geometry:

• Semilinear elliptic: ,

There is a different Porous-Medium-type fractional equation $\partial_t u = \mathrm{div} (u^{m-1} \nabla (-\Delta)^{-s} u)$

We will not discuss it:

$$m = 2$$ , $$m \ne 2$$

## Numerics

Finite Difference schemes $(-\Delta)^s_h u(x) = \sum_{i \in \mathbb Z^d} \omega_{i} (u(x) - u(x+ih))$

$$d = 1$$: .

Discontinuous Galerkin:

# Fractional LaplacianS in bounded domains

## Restricted fractional Laplacian

Singular integral operator: $(-\Delta)^s_{\mathrm {RFL}} u (x)= C(d,s) \, \mathrm{P.V.}\int_{ \mathbb R^d } \frac{u(x) - u(y)}{|x-y|^{d+2s}} \; dy.$

If we work only in $$\Omega$$, we must prescribe $$u$$ in $$\Omega^c = \mathbb R^d \setminus \Omega$$.

## Spectral fractional Laplacian

Operational power. $-\Delta \varphi_m = \lambda_m \varphi_m \textrm{ in } \Omega, \qquad \varphi_m = 0 \textrm{ on } \partial \Omega.$

one defines $u(x) = \sum_{m=1}^{+\infty} {u_m} \varphi_m(x) \qquad \longmapsto \qquad (-\Delta)_{\mathrm{SFL}}^s u (x) = \sum_{m=1}^{+\infty} \lambda_m^s u_m \varphi_m (x).$

The “boundary condition” is $$u = 0$$ on $$\partial \Omega$$.

## Censored fractional Laplacian (CFL)

For $$s > \frac 1 2$$ $$$\tag{CFL} (-\Delta)^s_{\mathrm{CFL}} u (x) = C(d,s) \, \mathrm{P.V.} \int_{ \Omega } \frac{u(x) - u(y)}{|x-y|^{n+2s}} \; dy,$$$

We do not integrate over $$\Omega^c$$, so it makes sense to pick simply $$u=0$$ on $$\partial\Omega.$$

## Laplace equation

$\begin{dcases} \mathcal L u = f & \Omega \\ u = 0 & \partial \Omega \text{ or } \mathbb R^d \setminus \overline \Omega \\ \end{dcases}$

We observe

• $$\mathcal L = (-\Delta)^s_{\mathrm{RFL}}, (-\Delta)^s_{\mathrm{CFL}}$$ are sub-differentials of energies $J(u) = \int_A \int_A \frac{|u(x) - u(y)|^2 }{|x-y|^{d+2s}} dx \, dy.$

• $$\mathcal L = (-\Delta)^s_{\mathrm{SFL}}$$ is just a power.

The inverse is naturally $$(-\Delta_\Omega)^{-s}$$, and in works between “powers” of $$H_0^1(\Omega)$$.

Self-adjoint compact operators. Furthermore $$\lambda_1 > 0$$.

Nice theory of energy solutions.

Higher regularity: RFL , CFL ,

## Green kernels

The solution of the Laplace equation $\begin{dcases} \mathcal L u = f & \Omega \\ u = 0 & \partial \Omega \text{ or } \mathbb R^d \setminus \overline \Omega \\ \end{dcases}$

Allows us to define $$\mathrm G : L^2 (\Omega) \to L^2(\Omega)$$ we can represent it by a kernel $u(x) = \mathrm G[f] (x) = \int_\Omega \mathbb G(x,y) f(y) dy.$

The probabilistic approach provides in each of our cases a similar shape $\mathbb G(x,y) \asymp \frac{1}{|x-y|^{d-2s}} \left(1 \wedge \frac{\delta(x)}{|x-y|} \right)^\gamma\left(1 \wedge \frac{\delta(y)}{|x-y|} \right)^\gamma$

Since we deal only with self-adjoint $$\mathcal L$$, then $$\mathbb G(x,y) = \mathbb G(y,x)$$.

The probabilistic approach provides in each of our cases a similar shape $\mathbb G(x,y) \asymp \frac{1}{|x-y|^{d-2s}} \left(1 \wedge \frac{\delta(x)}{|x-y|} \right)^\gamma\left(1 \wedge \frac{\delta(y)}{|x-y|} \right)^\gamma$

And $$\gamma \in (0,1]$$ depends on the setting

• RFL: $$\gamma = s$$.

• SFL: $$\gamma = 1$$

• CFL: $$\gamma = 2s-1$$, only $$s \in (\frac 1 2, 1)$$.

## Numerics

Finite Differences

• For the RFL and CFL we can re-use the weights of the whole space.

• For the SFL: Make $$A$$ numerical matrix for $$-\Delta$$ problem and take $$A^s$$: $\textrm{e.g. } A = \begin{pmatrix} 2 & -1 & \\ -1 & 2 & - 1 \\ & \ddots & \\ &&-1&2 \end{pmatrix}$

Finite Elements

• For the RFL and CFL integrate by parts: ,

• For the SFL use semigroup formula:

# Functional set-up

## Optimal set of data

Given that $$\mathbb G(x,y) \asymp \frac{1}{|x-y|^{d-2s}} \left(1 \wedge \frac{\delta(x)}{|x-y|} \right)^\gamma\left(1 \wedge \frac{\delta(y)}{|x-y|} \right)^\gamma$$

For $$K \Subset \Omega$$ $\int_K |G(f)| \le C_K \int_\Omega |f| \delta^\gamma.$

So $$G: L^1 (\Omega, \delta^\gamma) \to L^1_{loc} (\Omega)$$

If $$f \ge 0$$ then, using the kernel estimates $\tag{Hopf} G[f] (x) \ge c \delta^\gamma(x) \int_\Omega f(y) \delta(y)^\gamma dy.$

If $$0\le f \notin L^1 (\Omega, \delta^\gamma)$$, then $$G[f] = +\infty$$.

## Laplace equation Weak dual formulation

Multiplying by $$\varphi$$ and integrating $\int_\Omega ( \mathcal L u ) \varphi = \int_\Omega f \varphi$

If $$\varphi$$ is in the suitable class of homogeneous boundary conditions $\int_\Omega u ( \mathcal L \varphi )= \int_\Omega f \varphi$

Choosing the set of test functions $$\varphi$$ depends on the problem.

In particular, if we take $$\varphi = \mathrm{G} (\psi)$$ we a weak dual solution if $$u \in L^1_{loc} (\Omega)$$ and

$$$\tag{WDF} \int_\Omega u \psi = \int_\Omega f \mathrm{G} (\psi) \qquad \forall \psi \in L^\infty_c(\Omega)$$$

# Large solutions

## Definition

By large solution we mean solutions that satisfy $\mathcal L u = F(x,u) \text{ in } \Omega$

By large solution we mean $$\| u \|_{L^\infty} = \infty$$.

We are mostly here in boundary blow-up: $u(x) \to \infty \text{ as } \mathrm{dist}(x, \partial \Omega) \to 0.$

From now on

$\delta(x) = \mathrm{dist}(x, \partial \Omega) .$

## For the usual Laplacian

For the problems $-\Delta u + F(u) = 0$

Keller-Osserman condition:

Canonical example: $$-\Delta u + u^p = 0$$ with certain $$p$$.

This kind of singular behaviour exists also in the fractional case: , , ,

## Large solutions with boundary blow-up for $$\mathcal L$$

denoting $$\delta(x) = \mathrm{dist}(x, \partial \Omega)$$

We can compute $$\mathrm G[\delta^\beta] \asymp \delta^\alpha$$ with

$\mathrm G[\delta^\beta] \asymp \begin{dcases} \delta^\gamma & \beta +2s > \gamma \\ \delta^{\gamma} \log|\delta| & \beta +2s = \gamma \\ \delta^{\beta + 2s} & \beta +2s < \gamma \\ &\text{ and } \beta > -1-\gamma \\ \infty & \beta \le -1-\gamma \end{dcases}$

$$\infty \not \equiv G[\delta^\beta] \notin L^\infty \iff -1-\gamma < \beta < -2s$$.

For $$-\Delta$$ we have $$\gamma = 1 = s$$. Either $$G[\delta^\beta] \in L^\infty$$ or $$G[\delta^\beta] \equiv + \infty$$