Consider the following 2nd order nonlinear elliptic equation on $\mathbb{R}^n$: $$-\Delta \varphi_\varepsilon + \sum_i a_i(x, \varepsilon)\partial_i \varphi_\varepsilon + \varphi_\varepsilon = N(\varphi_\varepsilon),$$ where $N$ is a smooth function which gives the nonlinearity, $a$ is smooth, and $\varepsilon$ is a parameter (one can also think of the above pde as a one-parameter family of equations). Now, let us consider solutions $\varphi_\varepsilon \in H^1(\mathbb{R}^n)$, and suppose we know that if $\varphi_\varepsilon$ varies continuously with $\varepsilon$ in the $L^p(\mathbb{R}^n)$-norm, where $p \in [2, \frac{2n}{n - 2})$ (the range of Sobolev embedding). Let us also assume, if need be, nice decay properties on $\varphi_\varepsilon$, like vanishing at infinity. Can we somehow conclude from this that $\varphi_\varepsilon$ varies continuously with $\varepsilon$ in the $L^r(\mathbb{R}^n)$-norm, for some $r \geq \frac{2n}{n - 2}$? Any ideas would be appreciated.
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Is $N$ bounded ? Or $||N'||_\infty<1$ ? $\endgroup$– Jean DuchonCommented Aug 6, 2015 at 15:49
-
$\begingroup$ @JeanDuchon $N$ is not bounded, a model case would be power type nonlinearity, let's say, $N(x) = x^2$. But if there is something that can be said in the cases you mention, I would be really interested in learning about them as well. Thanks! $\endgroup$– daveCommented Aug 6, 2015 at 16:01
-
$\begingroup$ a few comments. Lets assume that $a$ doesn't overly effect the decay of $ \phi_\epsilon$ which we assume is positive and decays to zero at $ \infty$. I think the zero order terms probably implies that $ \phi$ decays to zero exponentially. I assume we are thinking of $N(u)=u^p$ and lets assume that $p$ is subcritical. Can one use a blow up argument and the classic Liouville theorem related to $ -\Delta v = v$ in $ R^N$ to say something? $\endgroup$– Math604Commented Aug 7, 2015 at 4:05
-
$\begingroup$ typo in last line...should be $ -\Delta v = v^p$ in $ R^N$. $\endgroup$– Math604Commented Aug 7, 2015 at 20:40
Add a comment
|