Viscosity solutions of $(-\Delta)^s u = 0$ in $\Omega $ with non-homogeneous data $u = 1$ in $\mathbb R^n \setminus \Omega$

Question

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 $$ (-\Delta )^{s}u(x)=c_{n,s}\int \limits _{\mathbb {R} ^{n}}{{\frac {u(x)-u(y)}{|x-y|^{n+2s}}}\,dy}$$ 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 $$ (2)\quad \begin{cases} (-\Delta)^s v + \lambda v = \underbrace{- \lambda \mathbf{1}_{\Omega^c}}_{=0} + c_{n,s} \int_{\mathbb R^n \setminus \Omega} |x - z|^{-n-2s} dz & x \in \Omega \\ v = 0 & x \in \mathbb R^n \setminus \Omega \end{cases} $$ 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?

Answer 1

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 $$ u_z(x) = \int_0^\infty \int_\Omega e^{-\lambda t} p^\Omega_t(x, y) c_{n,s} |y - z|^{-n-2s} dy dt , $$ where $p^\Omega$ is the heat kernel for $(-\Delta)^s$ in $\Omega$, with zero condition in $\Omega^c$. Then, formally, $$ \begin{aligned} (-\Delta)^s u_z(x) & = \int_0^\infty \int_\Omega e^{-\lambda t} (-\Delta_x)^s p^\Omega_t(x, y) c_{n,s} |y - z|^{-n-2s} dy dt \\ & = \int_0^\infty \int_\Omega e^{-\lambda t} (-\tfrac\partial{\partial t}) p^\Omega_t(x, y) c_{n,s} |y - z|^{-n-2s} dy dt . \end{aligned} $$ Add $\lambda u_z$: $$ \begin{aligned} ((-\Delta)^s + \lambda) u_z(x) & = \int_0^\infty \int_\Omega e^{-\lambda t} (-\tfrac\partial{\partial t} + \lambda) p^\Omega_t(x, y) c_{n,s} |y - z|^{-n-2s} dy dt \\ & = \int_0^\infty \int_\Omega (-\tfrac\partial{\partial t}) (e^{-\lambda t} p^\Omega_t(x, y)) c_{n,s} |y - z|^{-n-2s} dy dt \\ & = \int_\Omega p^\Omega_0(x, y) c_{n,s} |y - z|^{-n-2s} dy = c_{n,s} |x - z|^{-n-2s} . \end{aligned} $$ 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, $$((-\Delta)^s + \lambda) v(x) = \int_{(\overline\Omega)^c} c_{n,s} \lambda |x - z|^{-n-2s} dz .$$ 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.