Asymptotic formula for fractional Laplacian

For the solution of $$\begin{cases} \lambda u^\epsilon - \frac{\epsilon^2}{2} \Delta u^\epsilon = 0 &\text{in } \Omega \\ u^\epsilon=1 & \text{on } \partial \Omega \end{cases}$$ Varadhan proved that $$\lim_{\epsilon \to 0} - \epsilon \log u^\epsilon = \sqrt{2\lambda} \mathrm{dist} (x,\partial \Omega).$$

Can we prove that a similar asymptotic formula holds for $$\begin{cases} u^\epsilon + \epsilon (-\Delta)^s u^\epsilon = 0 &\text{in } \Omega \\ u^\epsilon=1 & \text{on } \mathbb R^N \setminus \Omega \end{cases}$$ where $$(-\Delta)^s$$ is the fractional Laplacian operator?

• Just out of curiosity: what hypotheses are imposed on $\Omega$ - is some regularity required? Apr 9 at 20:26
• @LeoMoos In Varadhan's paper it says that $\Omega$ is open, but I don't know if this is truly enough or if you also need to assume more regularity (which I would be ok with)
– Jun
Apr 9 at 20:35

Probabilistic approach

I have to think about the right reference, but things are very much different in the non-local case.

The solution $$u = u^\epsilon$$ can be written in probabilistic terms as $$u(x) = \mathbb E^x e^{-k \tau_D} ,$$ where $$k = \lambda / \epsilon$$, $$\tau_D$$ is the first exit time from the open set $$D = \Omega$$ (when probability enters, I prefer to use $$D$$ rather than $$\Omega$$ for the domain) for the corresponding isotropic stable Lévy process $$X_t$$, and $$\mathbb E^x$$ is the expectation corresponding to the process started at $$x$$.

In the local case, the Brownian motion is very unlikely to exit $$D$$ quickly. Roughly speaking, for $$t$$ small, the probability that $$\tau_D < t$$ is comparable to the probability that $$X_t \notin D$$, which in turn is roughly $$\exp(-\operatorname{dist}(x,\partial\Omega)^2 / (4 t)).$$ Thus, $$u(x)$$ can be estimated from below by (again roughly) $$\exp(-k t - \operatorname{dist}(x,\partial\Omega)^2 / (4 t)) .$$ Optimizing with respect to $$t$$ leads to a lower bound of the form $$\exp(-\sqrt{k} \operatorname{dist}(x,\partial\Omega) k t) ,$$ which is off by a factor $$2$$ in the exponent compared to Varadhan's result.

Now let us consider the non-local case and the jump-type stable process. In this case the probability that $$\tau_D < t$$ is quite large: it is of the order $$c_{N,s} t \int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-2s} dz + O(t^2)$$ (this is comparable to $$t \operatorname{dist}(x,\partial\Omega)^{-2s}$$ given some regularity of the complement of $$\Omega$$). Therefore, we immediately get a lower bound for $$u(x)$$ of the form $$t e^{-k t} c_{N,s} \int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-2s} dz ,$$ and optimization leads to $$c_{N,s} k^{-1} \int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-2s} dz ,$$ which behaves in a completely different way.

Conjecture

So a natural conjecture is that $$\lim_{\epsilon \to 0^+} \frac{u^\epsilon(x)}{\epsilon} = \frac{c_{N,s}}{k} \int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-2s} dz .$$ It should not be very difficult to write down a rigorous argument. In particular, I expect that the non-local case is far simpler than the local one.

My guess is that the above result follows relatively easily from what is known and can be found in literature, but, again, I do not have any particular reference in mind at this moment.

Final remark: my guess is that Varadhan's result, as well as the above conjecture, require an exterior cone condition, or something similar.

PDE approach

Edit: Here is a more PDE-oriented approach.

Suppose that $$u$$ is the solution of $$u + \epsilon (-\Delta)^s u = 0 \qquad \text{in } \Omega$$ and $$u = 1$$ in the complement of $$\Omega$$. Let $$w = u$$ in $$\Omega$$ and $$w = 0$$ in the complement of $$\Omega$$. Then, at least formally, we have $$(\epsilon^{-1} I + (-\Delta)^s) w(x) = c_{N,s} \int_{\mathbb R^N \setminus \Omega} \frac{1}{|y - x|^{N + 2 s}} \, dy =: g(x)$$ in $$\Omega$$, and therefore $$w$$ is the resolvent for $$(-\Delta)^s$$ in $$\Omega$$ applied to $$g$$: $$u(x) = w(x) = \int_\Omega \biggl(\int_0^\infty e^{-t/\epsilon} p_t^\Omega(x, y) dt\biggr) g(y) dy ;$$ here $$p_t^\Omega(x,y)$$ is the heat kernel for $$(-\Delta)^s$$ in $$\Omega$$. (This is a very informal derivation of what is essentially the Ikeda–Watanabe formula in probability.)

Now let us rewrite the expression for $$u$$ a bit. We have $$\frac{u(x)}{\epsilon} = \int_\Omega \biggl(\int_0^\infty e^{-s} p_{\epsilon s}^\Omega(x, y) dt\biggr) g(y) dy .$$ Now let us write $$u = u^\epsilon$$ and consider $$\epsilon \to 0^+$$. Intuitively it is clear that the expression in brackets converges to the Dirac delta at $$x$$ as $$\epsilon \to 0^+$$. Making this rigorous requires some effort, but since the first part was already informal, let me ignore all details. We find that $$\lim_{\epsilon \to 0^+} \frac{u^\epsilon(x)}{\epsilon} = \int_\Omega g(y) \delta_x(dy) = g(x) ,$$ which is precisely the conjectured result.

• Time permitting, I'll think about references. As $s \to 1$, the constant $C(s)$ will converge to zero, so in a sense we do recover the local case in a very weak form. :-) This is one of the numerous results where there is a "phase transition" at $s = 1$. The simplest one (and one that is closely related to this question) is the estimate of the heat kernel: for a fixed $t$, it is $(1+|x|)^{-N-2s}$ when $s < 1$, but $\exp(-|x|^2)$ when $s = 1$. Apr 10 at 8:00
• Thanks! Isn't it also strange that the right-hand side in the conjecture does not seem to satisfy the boundary condition $u^\epsilon =1$ in $\mathbb R^N \setminus \Omega$?
– Jun
Apr 10 at 8:54
• I think this works as expected: $u^\epsilon / \epsilon$ is equal to $1/\epsilon$ in the complement of $\Omega$, and this goes to infinity as $\epsilon \to 0^+$. Apr 10 at 9:18
• Thanks! Formally, is it clear that $\lambda^{-1}\mathrm{dist}(x,\partial\Omega)^{-2s}$ solves the limit equation $\lambda u = 0$ in $\Omega$ and $u=\infty$ in $\mathbb R^N \setminus \Omega$?
– Jun
Apr 14 at 8:21
• I cannot parse your question: the only solution of $\lambda u = 0$ is of course $u = 0$, is it not? By the way, I realise the "$\dist(x,\partial \Omega)$" term is not quite correct, I'll update the answer momentarily. Apr 14 at 8:36