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

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?

• 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 Nov 3, 2021 at 4:13
• @JackT Thank you! I've added a lower order term (which was the original model I had in mind)
– Zac
Nov 3, 2021 at 9:48
• Why not simply consider $v = 1-u$, 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. Nov 3, 2021 at 9:55
• @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?
– Zac
Nov 3, 2021 at 17:42
• @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. :-) Nov 3, 2021 at 18:53

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.
• there are some typos in the text unfortunately, I meant $\mathbf 1_{\Omega^c}$ and $v=u-\mathbf{1}_{\Omega^c}$ above