Let us consider a smooth bounded domain $\Omega \subset \mathbb R^n$ and the problem $$ (1) \quad \begin{cases} (\Delta)^s u +\lambda u= 0 & x \in \Omega \\ u = 1 & x \in \mathbb R^n \setminus \Omega \end{cases} $$ where $\lambda >0$ and the fractional Laplacian ia $$ (\Delta )^{s}u(x)=c_{n,s}\int \limits _{\mathbb {R} ^{n}}{{\frac {u(x)u(y)}{xy^{n+2s}}}\,dy}$$ with $$ {\displaystyle c_{n,s}={\frac {4^{s}\Gamma (n/2+s)}{\pi ^{n/2}\Gamma (s)}}}$$ I know several references with data $u(x) \equiv 0$ for $x \in \mathbb R^n \setminus \Omega$, but where can I find a proof for existence and uniqueness of viscosity solutions to the problem above? Also, is it true that the problem above is equivalent to $$ (2)\quad \begin{cases} (\Delta)^s v + \lambda v = \underbrace{ \lambda \mathbf{1}_{\Omega^c}}_{=0} + c_{n,s} \int_{\mathbb R^n \setminus \Omega} x  z^{n2s} dz & x \in \Omega \\ v = 0 & x \in \mathbb R^n \setminus \Omega \end{cases} $$ i.e. that the change of variables $v = u\mathbf{1}_{\Omega^c} $ can be performed to reduce the original problem to one with homogeneous data and a source term?

$\begingroup$ What is your motivation for this? Are you interested in viscosity solutions to equations involving the fractional Laplacian in general? I ask because $u \equiv 1$ is the (unique) strong solution $\endgroup$– JackTNov 3, 2021 at 4:13

$\begingroup$ @JackT Thank you! I've added a lower order term (which was the original model I had in mind) $\endgroup$– ZacNov 3, 2021 at 9:48

$\begingroup$ Why not simply consider $v = 1u$, which solves $(\Delta)^s v + \lambda v = \lambda$ in $\Omega$, with homogeneous Dirichlet condition $v = 0$ in $\Omega^c$? The unique solution of the latter is given by $$v(x) = \int_0^\infty \int_\Omega \lambda e^{\lambda t} p_t^\Omega(x, y) dy dt,$$ where $p_t^\Omega(x,y)$ is the corresponding heat kernel. $\endgroup$– Mateusz KwaśnickiNov 3, 2021 at 9:55

$\begingroup$ @MateuszKwaśnicki Thanks. Then, for $u$ we have $u(x) = 1 \int_0^\infty \int_\Omega \lambda e^{\lambda t} p_t^{\Omega}(x,y) dy dt$. Can we write this in a more compact way? Also, why the heat kernel and not the Green function of the fractional Laplacian? $\endgroup$– ZacNov 3, 2021 at 17:42

$\begingroup$ @Zac: The integral $\int_0^\infty e^{\lambda t} p_t^\Omega(x,y)dt$ is precisely the Green function for $(\Delta)^s + \lambda$, or the $\lambda$Green function for $(\Delta)^s$. A "more compact way" to write this is, for example, $u(x) = \mathbb E^x e^{\lambda \tau}$, where $\tau$ is the hitting time of $\Omega^c$ for the isotropic $2s$stable Lévy process. :) $\endgroup$– Mateusz KwaśnickiNov 3, 2021 at 18:53
1 Answer
I you are not really interested about viscosity solutions, but just the "philosophical reason" why the solution of the problem with inhomogeneous exterior condition can be written in terms of the heat kernel (as suggested by the comments), here is an answer.
Choose $z \notin \overline\Omega$ and define $$ u_z(x) = \int_0^\infty \int_\Omega e^{\lambda t} p^\Omega_t(x, y) c_{n,s} y  z^{n2s} dy dt , $$ where $p^\Omega$ is the heat kernel for $(\Delta)^s$ in $\Omega$, with zero condition in $\Omega^c$. Then, formally, $$ \begin{aligned} (\Delta)^s u_z(x) & = \int_0^\infty \int_\Omega e^{\lambda t} (\Delta_x)^s p^\Omega_t(x, y) c_{n,s} y  z^{n2s} dy dt \\ & = \int_0^\infty \int_\Omega e^{\lambda t} (\tfrac\partial{\partial t}) p^\Omega_t(x, y) c_{n,s} y  z^{n2s} dy dt . \end{aligned} $$ Add $\lambda u_z$: $$ \begin{aligned} ((\Delta)^s + \lambda) u_z(x) & = \int_0^\infty \int_\Omega e^{\lambda t} (\tfrac\partial{\partial t} + \lambda) p^\Omega_t(x, y) c_{n,s} y  z^{n2s} dy dt \\ & = \int_0^\infty \int_\Omega (\tfrac\partial{\partial t}) (e^{\lambda t} p^\Omega_t(x, y)) c_{n,s} y  z^{n2s} dy dt \\ & = \int_\Omega p^\Omega_0(x, y) c_{n,s} y  z^{n2s} dy = c_{n,s} x  z^{n2s} . \end{aligned} $$ The function $u_z$ is known as the $\lambda$Poisson kernel for $(\Delta)^s$ in $\Omega$; if $\lambda = 0$, this is just the Poisson kernel.
Now define $$v(x) = \int_{(\overline\Omega)^c} \lambda u_z(x) dz$$ for $x \in \Omega$. By the above calculation, $$((\Delta)^s + \lambda) v(x) = \int_{(\overline\Omega)^c} c_{n,s} \lambda x  z^{n2s} dz .$$ In other words, if $u(x) = v(x)$ for $x \in \Omega$ and $u(x) = \lambda$ otherwise, then $$((\Delta)^s + \lambda) u(x) = 0 ,$$ as desired.

$\begingroup$ Thank you so much! This was very helpful. I still have some followup questions, which I've included in the text of the bounty opened above $\endgroup$– ZacNov 4, 2021 at 10:51

$\begingroup$ there are some typos in the text unfortunately, I meant $\mathbf 1_{\Omega^c}$ and $v=u\mathbf{1}_{\Omega^c}$ above $\endgroup$– ZacNov 4, 2021 at 10:52