Boundary singularity for
fractional elliptic and parabolic problems
Young Researcher’s Workshop on Nonlocal PDEs and Applications.
Universidad de Granada
October 27, 2022.
Co-authors
Juan Luis Vázquez
U. Autónoma de Madrid, Real Academia de Ciencias
Nicola Abatangelo
U. Bologna
Hardy Chan
U. Basel
Fractional Laplacian in \(\mathbb R^d\)
Fractional Laplacian
\[ (-\Delta)^s u (x) = \frac{C(d,s)}2 \int_{\mathbb R^d } \frac{2u(x) - u(x+y) - u(x-y)}{|y|^{d+2s}} d y. \]
The popular choice is \[ {C(d,s)} = \left( \int_{\mathbb R^d} \frac{1 - \cos \omega_1}{|\omega|^{d+2s}} d \omega\right)^{-1} \]
Equivalent definitions (up-to ten (Kwasnicki, 2017))
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”
\[ (-\Delta)^s u (x) = \lim_{h \to 0} \frac{ \mathbb E[ f(x) - f(x - X_h) ] }{ h } \]
The Riesz potential
\[ \mathcal I_{\alpha} [u] (x) = C(d, \alpha) \int_{\mathbb R^d} \frac{u(y)}{|x-y|^{d-\alpha}} dy. \]
It is defined for \(0 < \alpha < d\).
It is known that with the right constant \[ \mathcal F[ \mathcal I_{\alpha} u ] = |\xi|^{-\alpha} \mathcal F[u] \]
Thus, \[ \mathcal I_{2s} = \Big( (-\Delta)^s \Big)^{-1} \]
All of the above works on tempered functions, in particular \[ u(x) \to 0, \qquad \text{ as } |x| \to \infty. \]
Laplace equation
\[ \begin{cases} (-\Delta)^s u (x) = f(x) & \textrm{for all } x \in \mathbb R^d \\ u(x) \to 0 & \textrm{as } |x| \to \infty \end{cases} \]
Then when \(d > 2s\) we can recover through the Riesz potential \[ u(x) = C(d,s) \int_{\mathbb R^d} \frac{f(y)}{|x-y|^{d-2s}} dy. \]
For any \(d, 2s\) we can also find this solution by the minisation of the energy functional
\[ J(u) = \frac 1 2 \int_{\mathbb R^d} \int_{\mathbb R^d } \frac{|u(x) - u(y)|^2}{|x-y|^{d+2s}} dx \, dy - \int_\Omega f u \]
Regularity: (Caffarelli & Silvestre, 2007) extension
The principal value
Notice that \[\begin{align*} \int_{ \mathbb R^d \setminus B_{\varepsilon} (0) } &\frac{ 2u(x) - u(x+y) - u(x-y) }{|y|^{d+2s}} dy \\ &= 2\int_{ \mathbb R^d \setminus B_{\varepsilon} (0) } \frac{ u(x) - u(y)}{|x-y|^{d+2s}} dy \\ \end{align*}\]
Thus, it makes sense to define \[\begin{equation*} \mathrm{P.V.}\int_{ \mathbb R^d } = \lim_{\varepsilon \to 0} \int_{ \mathbb R^d \setminus B_{\varepsilon} (0) } \end{equation*}\]
Numerics
There are both Finite Difference schemes \[ (-\Delta)^s_h u(x) = \sum_{i \in \mathbb Z^d} \omega_{i-j} (u(x) - u(x+ih)) \]
Some choice of weights for \(d = 1\) is given in (Huang & Oberman, 2014).
See also (Del Teso, Endal & Jakobsen, 2018)
Finite Element schemes are also possible. The un-bounded domain is tricky.
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 \(s > \frac 1 2\) \[\begin{equation} \tag{CFL} (-\Delta)^s_{\mathrm{CFL}} u (x) = C(d,s) \, \mathrm{P.V.} \int_{ \Omega } \frac{u(x) - u(y)}{|x-y|^{n+2s}} \; dy, \end{equation}\]
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 (Ros-Oton & Serra, 2014), CFL (Fall & Ros-Oton, 2021)
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: (Borthagaray, Nochetto & Salgado, 2019)
For the SFL use semigroup formula: (Cusimano, del Teso & Gerardo-Giorda, 2020)
Large solutions
Definition
By large solution we mean solutions that satisfy \[ \mathcal L u = F(x,u) \text{ in } \Omega \]
But that blow up on the boundary \[ u(x) \to \infty \text{ as } \mathrm{dist}(x, \partial \Omega) \to 0. \]
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:
Let \(f\) be continuous, \(F(0) = 0\) and \(F > 0\) otherwise,
then positive large solutions exist iff \[ \int_a^\infty \left( \int_0^t F(s) ds \right)^{-\frac 1 2} d t = \infty. \]
In radial coordinates this easy by shooting arguments
Canonical example: \(\Delta u = u^p\) with \(p > 1\).
Large solutions
The solution of the Laplace equation \[ \begin{dcases} \mathcal L u = f & \Omega \\ u = 0 & \partial \Omega \text{ or } \mathbb R^d \setminus \overline \Omega \\ \end{dcases} \qquad u(x) = \mathrm G[f] (x) = \int_\Omega \mathbb G(x,y) f(y) dy. \]
where \(\mathbb G(x,y) \asymp \frac{1}{|x-y|^{d-2s}} \left(1 \wedge \frac{\delta(x)}{|x-y|} \right)^\gamma\left(1 \wedge \frac{\delta(y)}{|x-y|} \right)^\gamma\)
(Abatangelo, GC & Vázquez, 2022) : \[ \mathrm G[\delta^\beta] \asymp \begin{dcases} \delta^\gamma & \beta +2s > \gamma \\ \delta^{\gamma} \log|\delta| & \beta +2s = \gamma \\ \delta^{\beta + 2s} & \beta +2s < \gamma \text{ and } \beta > -1-\gamma \\ \infty & \beta \le -1-\gamma \end{dcases} \]
Summary
\[ \mathrm G[\delta^\beta] \asymp \begin{dcases} \delta^\gamma & \beta +2s > \gamma \\ \delta^{\gamma} \log|\delta| & \beta +2s = \gamma \\ \delta^{\beta + 2s} & \beta +2s < \gamma \text{ and } \beta > -1-\gamma \\ \infty & \beta \le -1-\gamma \end{dcases} \]
Therefore, the relation \(\mathrm G[\delta^\beta] \asymp \delta^\alpha\) is
Large solution of the Dirichlet problem
Large solution of the Dirichlet problem
No large solutions in the SFL
The Bogdan solution
Take \(s \in (0,1)\)
\[ u(x) = \begin{cases} (1-|x|^2)^{s-1} & \text{if } |x|<1 \\ 0 & \text{if } |x| \ge 1 \\ \end{cases} \]
satisfies
\[ (-\Delta)^s u(x) = 0 \qquad \text{if } |x| < 1. \]
Reviewing the usual Laplacian
From the interior to the boundary.
Usual Laplacian
Laplace equation
\[ \begin{dcases} -\Delta u = f & \Omega \\ u = 0 & \partial \Omega \\ \end{dcases} \]
Poisson
\[ \begin{dcases} -\Delta v = 0 & \Omega \\ v = h & \partial \Omega \\ \end{dcases} \]
For \(\varphi \in W_0^{1,\infty} (\Omega)\) we have that
\[ \int_\Omega u (-\Delta\varphi) = \int_\Omega f \varphi \]
\[ \int_\Omega v (-\Delta\varphi) = - \int_{\partial \Omega} h \frac{\partial \varphi}{\partial n} \]
We aim to prove \(u_m = \mathrm G[f_m] \to v\). Can we make \[ \int_\Omega f_m \varphi \to - \int_{\partial \Omega} h \frac{\partial \varphi}{\partial n} ? \]
Tubular neighbourhood of \(\partial \Omega\)
The map \[ \begin{aligned} (-R,R) \times \partial \Omega &\to U_R \\ (r,z) &\mapsto z - r \vec n(z) \end{aligned} \] is smooth and invertible for \(R\) small enough. This defines a tubular neighbourhood of \(\partial \Omega\).
A function in \(h \in L^1(\partial \Omega)\) can be extend to \(L^1(U_R)\) by \[\widetilde h(x) = h(z).\]
For integration, there is a Jacobian such that \(J(z,0) = 1\) and \[ \int_{a < \delta < b} f(x) = \int_{a}^b \int_{\partial \Omega} f(z - r n(z)) \, J(z,r) d z \, d r \]
Localising to the boundary. Usual Laplacian
Can we make \(\int_\Omega f_m \varphi \to -\int_{\partial \Omega} h \frac{\partial \varphi}{\partial n} ?\)
Extend \(h\) towards the interior by the tubular neighbourhood mapping and \[ f_m (x) = \widetilde h (x) \frac{|\partial \Omega|\chi_{A_m}}{ |A_m| } \frac{1}{\delta (x)} \qquad \text{ where } A_m = \left\{ x: \delta(x) < \frac 1 m \right\} \]
We get
\(\displaystyle\int_\Omega f_m \varphi = \frac{|\partial \Omega|}{|\{ \delta < \frac 1 m \}|} \int_{ \{ \delta < \frac 1 m \} } \widetilde h \frac{\varphi}{\delta}\)
\(\displaystyle \phantom{\int_\Omega f_m \varphi} = \frac 1 m \int_0^{\frac 1 m} \int_{\partial \Omega} h(z) \frac { \varphi( z - r n(z) ) }{ r } J(z,r) \, dz \, dr\)
\(\displaystyle \phantom{\int_\Omega f_m \varphi} \longrightarrow -\int_{\partial \Omega} h \frac{\partial \varphi}{\partial n}\)
as \(m \to \infty\).
It works!
From the kernel side
\[ \begin{aligned} \mathrm G[f_m] (x) &= \frac{|\partial \Omega|}{|A_m|}\int_{A_m} \frac{\mathbb G(x,y)}{\delta(y)} \widetilde h(y) dy \end{aligned} \]
As \(m \to \infty\)
\[ \mathrm G[f_m] (x) \longrightarrow v(x) = -\int_{\partial \Omega} \frac{\partial \mathbb G}{\partial n_y} (x,\zeta) h(\zeta) d\zeta \]
This is the Poisson kernel.
Rigurous proof. Usual Laplacian
The first eigenfunction is \(\varphi_1 \asymp \delta\).
If \(f, g \ge 0\), then \(\mathrm G[f] / \mathrm G[g] \in L^\infty(\Omega)\).
If \(f \in L^\infty (\Omega)\), then \(\frac{\mathrm G[f]}{\delta} \in L^\infty (\Omega)\). Thus \(\mathrm G: L^\infty(\Omega) \to \delta L^\infty (\Omega)\) is compact.
Let \(\| f \|_{L^1(\Omega, \delta)} = \int_\Omega |f| \delta\)
We deduce \(\mathrm G: L^1(\Omega, \delta) \to L^1 (\Omega)\) is compact.
We have precisely we have a bounded sequence in this norm \[ f_m (x) = \widetilde h (x) \frac{|\partial \Omega|\chi_{A_m}}{ |A_m| } \frac{1}{\delta (x)} \]
So \(u_m = \mathrm G[f_m] \to \xi\) in \(L^1 (\Omega)\). Passing to the limit in the weak formulation \(\xi = v\). \(\square\)
Martin problem for fractional operators
Martin problem
The will show that we can pass \(\mathrm{supp}(f_m) \to \partial \Omega\) and \(\mathrm G[f_m] \to v \ne 0\).
We construct \(u^\star \in L^1_{loc} (\Omega)\) such that
for any \(h \in L^1 (\partial \Omega)\) the problem
\[ \begin{dcases} \mathcal L u = 0 & \Omega \\ u = 0 & \mathbb R^d \setminus \overline \Omega \\ \lim_{x \to z} \frac{u(x)}{u^\star (x)} = h(z) & \text{for all } z \in \partial \Omega \end{dcases} \]
admits a solution.
We show that \[ u^\star \asymp \delta^{2s-\gamma -1 }. \]
Martin problem. Literature
(Abatangelo, 2015) for the RFL
(Abatangelo & Dupaigne, 2017) for the SFL
(Chen, 2018) for the CFL
(Abatangelo, GC & Vázquez, 2022) unified theory for the RFL, CFL, and SFL.
Laplace equation
Weak dual formulation
Multiplying by \(\varphi\) and integrating \[ \int_\Omega \varphi \mathcal L u = \int_\Omega f \varphi \]
If \(\varphi\) is in the suitable class of homogeneous boundary conditions \[ \int_\Omega u \mathcal L \varphi = \int_\Omega f \varphi \]
In particular, if we take \(\varphi = \mathrm{G} (\psi)\) then we have the weak dual formulation
\[\begin{equation} \tag{WDF} \int_\Omega u \psi = \int_\Omega f \mathrm{G} (\psi) \end{equation}\]
This presentation was introduced by (Bonforte, Figalli & Vázquez, 2018)
Functional set-up
We recall that \(\mathbb G(x,y) \asymp \frac{1}{|x-y|^{d-2s}} \left(1 \wedge \frac{\delta(x)}{|x-y|} \right)^\gamma\left(1 \wedge \frac{\delta(y)}{|x-y|} \right)^\gamma\)
Since \(\mathrm G(\delta^\gamma) \asymp \delta^\gamma\), we know that \[ \mathrm G : L^\infty_{c} (\Omega) \to \delta^\gamma L^\infty( \Omega ). \]
By duality
\[ \mathrm G : L^1 (\Omega, \delta^\gamma ) \to L^1_{loc} ( \Omega ). \]
Weak compactness in \(L^1 (K)\): equi-integrability for \(A \subset K \Subset \Omega\).
\[ \begin{aligned} \int_A |\mathrm G(f)| &= \int_\Omega f \mathrm G( \mathrm{sign}(u) \, \chi_A) \le \| f \|_{L^1 (\Omega, \delta^\gamma)} \left\| \frac{\mathrm G( \chi_A )} {\delta^\gamma} \right\|_{L^\infty} \end{aligned} \]
With some sharp computations \[ \int_A |\mathrm G(f)| \le C \| f \|_{L^1 (\Omega, \delta^\gamma)} |A|^\alpha \mathrm{dist} (K, \partial \Omega)^{-\beta} \]
Hopf
Not difficult to check that \(\mathbb G (x,y) \ge c \delta(x)^\gamma \delta(y)^\gamma\).
Then, for \(f \ge 0\)
\[ u(x) \ge c \delta(x)^\gamma \int_\Omega f(x) \delta(y)^\gamma dy. \]
If \(0 \le f \notin L^1(\Omega, \delta^\gamma)\), then \(u \equiv +\infty\).
Localisation to the boundary
General setting
Therefore, if \(f_m\) are bounded in \(L^1(\Omega, \delta^\gamma)\), then \(\mathrm G[f_m]\) weak compact in \(L^1(K)\).
Through a diagonal argument define \[ u^\star = \lim_{m \to \infty} \mathrm G \left[ \delta^{-\gamma} \frac{|\partial \Omega|}{|A_m|} \chi_{A_m} \right ], \qquad A_m = \{ \frac 1 m < \delta < \frac 2 m \} \]
\(u_j \to u^\star\)
\(u_j \to u^\star\)
\(u_j \to u^\star\)
Localisation to the boundary
The kernel point of view
As we did before \[ \mathrm G[f_m] = \frac{|\partial \Omega|}{|A_j|} \int_{A_j} \frac{\mathbb G(x,y)}{\delta(y)^\gamma} \widetilde h (y) d y \]
as \(m \to \infty\) we get
\[ \mathrm G[f_m] \to \int_{\partial \Omega} \mathbb M(x,\zeta) h(\zeta) d\zeta = \mathrm M[h] . \]
where we have to assume that the following limit happens suitably (it does in the examples)
\[\begin{equation} \tag{H} \mathbb M(x,\zeta) = \lim_{y \to \zeta} \frac{\mathbb G(x,y)}{\delta(y)^\gamma} = D_\gamma \mathbb G(x,\zeta). \end{equation}\]
using the Green kernel estimates, we deduce that
\[ \mathbb M(x,\zeta) \asymp \frac{\delta(x)^\gamma}{|x-\zeta|^{d+\gamma - (2s-\gamma)}}. \]
The critical solution \(u^\star\)
When we take \(h = 1\) we get
\[ u^\star (x) = \int_{\partial \Omega} \mathbb M(x,\zeta) d \zeta. \]
Using the Martin kernel estimates
\[ u^\star (x) \asymp \delta^{2s-\gamma - 1 }. \]
This is
RFL: \(s-1\)
SFL: \(2(s-1)\)
CFL: \(0\), i.e. bounded.
Satisfying the interior equation
In this setting, we have not defined \(\mathcal L\). However, formally we have that \[ \mathcal L u_m (x) = 0 \qquad \text{if } \delta(x) > \frac 2 m. \]
in any kind of distributional setting where we can pass to the limit \(u = \mathcal M[h]\) satisfies
\[ \mathcal L u (x) = 0. \]
Satisfying the boundary equation.
Continuous data
When \(h \in \mathcal C(\partial \Omega)\) it is a direct computation that
\(\displaystyle \frac{\mathcal M[h](x) }{u^\star (x)} - h (\zeta_0)\) \(\displaystyle = \frac{\int_{\partial \Omega} \mathbb M(x,\zeta) h(\zeta) d \zeta }{ \int_{\partial \Omega} \mathbb M (x, \zeta) d \zeta } - \frac{\int_{\partial \Omega} \mathbb M(x,\zeta) d \zeta }{ \int_{\partial \Omega} \mathbb M (x, \zeta) d \zeta } h(\zeta_0)\)
Thus
\[ \left| \frac{\mathcal M[h] (x) }{u^\star (x)} - h (\zeta_0) \right| \le \int_{\partial \Omega} \frac{ \mathbb M(x,\zeta) }{\int_{\partial \Omega} \mathbb M (x, z) d z} | h(\zeta) - h(\zeta_0)| d \zeta \]
For any \(\varepsilon > 0\), due the kernel estimates
\[ \limsup_{x \to \zeta_0} \left| \frac{\mathcal M[h] (x) }{u^\star (x)} - h (\zeta_0) \right| \le \int_{|\zeta - \zeta_0| \le \varepsilon} \frac{ \mathbb M(x,\zeta) }{\int_{\partial \Omega} \mathbb M (x, z) d z} | h(\zeta) - h(\zeta_0)| d \zeta \]
As \(\varepsilon \to 0\) the RHS vanishes due to the continuity of \(h\).
Satisfying the boundary equation.
\(L^1\) data
A more involved argument works for \(h \in L^1 (\partial \Omega)\), and gives integral convergence.
For \(\psi \in \mathcal C (\overline \Omega)\) we have
\[ \frac{1}{\eta} \int_{\delta < \eta} \frac{\mathrm M[h]}{u^\star} \phi \longrightarrow \int_{\partial \Omega} h \phi, \qquad \text{ as } \eta \searrow 0. \]
Very weak formulation of singular boundary value problem
Let \(u_m = \mathrm G[f_m]\). Then, for \(\psi \in L^\infty_c (\Omega)\)
\[ \int_\Omega u_m \psi = \frac{|\partial \Omega|}{ |A_m| } \int_{A_m} \widetilde h \frac{\mathrm G[\psi]}{\delta^\gamma}. \]
Notice that as \(x \to \zeta \in \partial \Omega\)
\[ \lim_{x \to \zeta} \frac{\mathrm G[\psi] (x)}{\delta(x)^\gamma} = \lim_{x \to \zeta} \int_\Omega \frac{\mathbb G(y,x)}{\delta(x)^\gamma} \psi(y) dy \]
by the previous hypothesis
\[ D_\gamma \mathrm G[\psi] (\zeta) = \int_\Omega \mathbb M (y, \zeta) \psi(y) dy \]
Given that
\[ \int_\Omega u_m \psi = \frac{|\partial \Omega|}{ |A_m| } \int_{A_m} \widetilde h \frac{\mathrm G[\psi]}{\delta^\gamma}. \]
by compactness we know \(u_m \to u\) in \(L^1 (\mathrm{supp} \, \psi)\). Hence
\[\begin{equation} \tag{WDF$_s$} \int_\Omega u \psi = \int_{\partial \Omega} h D_\gamma {\mathrm G[\psi]}. \end{equation}\]
By Fubini
\[ \int_\Omega u \psi = \int_{\partial \Omega} h(\zeta) \int_\Omega \mathbb M(y, \zeta) \psi (y) dy \, d \zeta . \]
\[ \int_\Omega \left( u(y) - \int_{\partial \Omega} h(\zeta) \mathbb M(y, \zeta) dy \right ) \psi(y) dy = 0 , \qquad \forall \psi \in L^\infty_c (\Omega) . \]
Extensions:
Schrödinger and parabolic setting
Schrödinger-type problems
(Chan, GC & Vázquez, 2021) : we study
\[ \begin{cases} \mathcal L u = \lambda u + f & \text{in }\Omega, \\ \frac{u}{u^\star} = h & \text{on } \partial \Omega. \end{cases} \]
We use the eigenfunction expansion to understand the blow-up as
\[\lambda \to \mathrm{spectrum} (\mathcal L).\]
Parabolic setting
\[ \begin{dcases} \frac{\partial u}{\partial t} + \mathcal L u = f & \text{in } (0,\infty) \times \Omega, \\[2ex] \frac{u}{u^\star} = h &\text{on } (0,\infty) \times \partial \Omega \\[2ex] u = u_0 & \text{at } t = 0. \end{dcases} \]
We construct the heat kernel, and localise \(f\) to the boundary.
Fractional time derivative
Ongoing
\[ \begin{dcases} \frac{\partial^\alpha u}{\partial t^\alpha} + \mathcal L u = f & \text{in } (0,\infty) \times \Omega, \\[2ex] \frac{u}{u^\star} = h &\text{on } (0,\infty) \times \partial \Omega \\[2ex] \end{dcases} \]
either in the Caputo or Riemann-Liouville setting.
Thank you!
