3
$\begingroup$

Dear all, I would like to prove the exponential decay of the derivatives of a solution to the following equation in $\mathbb{R}^N$: $$ \sqrt{-\Delta+m^2} u +u= f(u), $$ where I can assume that $m \neq 0$, that $f$ is smooth and that the solution $u \in H^{1/2}(\mathbb{R}^N)$ is at least Hölder-continuous. I also know that $u(x) = O(e^{-C|x|})$ at infinity. If the equation were local (like $-\Delta u + u = f(u)$), the usual approach would consist in using interior Schauder estimates for $\nabla u$ (or some Harnack-like estimate); I did not find any precise reference for Schauder theory of non-local equations, except in the case $m=0$ (the fractional laplace equation).

Any help is welcome.

$\endgroup$
7
  • $\begingroup$ I think it would be helpful to define what you mean by $\sqrt{−\Del+m^2}$ (via some integral, Fourier transform, etc.) for preciseness and to best facilitate a response. The theory of regularity for operators like the fractional Laplacian have been treated relatively recently (last 5-10 years, from what I know) by Caffarelli, Silvestre, and perhaps their techniques apply? $\endgroup$ Dec 8, 2012 at 16:07
  • $\begingroup$ The usual way to do something like this is to try to adapt the proof that $u$ is exponentially decaying to $\nabla u$. What's convenient here is that $\nabla$ commutes with your operator, so the vector-valued function $V = (u, \nabla u)$ satisfies roughly the same equation as $u$ itself, except that it is a system instead of a single equation. You would also use elliptic regularity to establish a priori regularity of $\nabla u$. $\endgroup$
    – Deane Yang
    Dec 8, 2012 at 16:14
  • $\begingroup$ @DanielSpector It seems to me that the pure fractional laplacian is rather different, for example solutions decay like powers. Moreover, many results by Silvestre are based on the representation via a convolution-like integral, which is not the case for "my" operator, which is often defined by functional calculus. $\endgroup$
    – Siminore
    Dec 8, 2012 at 17:02
  • $\begingroup$ Hmm. If you are in $\mathbb{R}^N$, this should be equivalent (functional calculus or integral), since the fractional Laplacian's symbol is multiplication by the Fourier variable? I guess my question is then this: What do you mean when you write the above equation? $\endgroup$ Dec 8, 2012 at 17:52
  • $\begingroup$ I mean: Silvestre defines $$(-\Delta)^u(x) = \operatorname{PV}\int \frac{u(y)}{|x-y|^{n+2s}}dy$$ but I do not think such a definition can be extended to $(-\Delta +1)^{1/2}$. And I am unable to use Fourier analysis to produce Silvestre's regularity results. $\endgroup$
    – Siminore
    Dec 8, 2012 at 18:20

1 Answer 1

2
$\begingroup$

There has recently been work by M. Cappiello, T. Gramchev, and L. Rodino on related problems, see e.g. Entire extensions and exponential decay for semilinear elliptic equations, J. Anal. Math. 111 (2010), 339-367 http://link.springer.com/article/10.1007%2Fs11854-010-0021-4.

An exemplary result reads: Suppose that the nonlinearity $f$ is a polynomial in $u$ and that $u\in \langle x \rangle^{-\delta} H^s({\mathbb R}^N)$, where $s>N/2$, $\delta>0$, is a solution of the semilinear elliptic equation shown above. Then $u$ belongs to the Gelfand-Shilov space $S_1^1({\mathbb R}^N)$. Amongst others, this implies that there is an $\varepsilon>0$ so that, for all $\alpha\in{\mathbb N}_0^N$, $$ (\partial^\alpha u)(x)= O(e^{-\varepsilon\,|x|}) \enspace \text{as $|x|\to\infty$.} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.