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Fix $ \Omega$ a bounded smooth domain in $ R^N$ and suppose $0<w(x)$ is a smooth solution of $ -\Delta w(x)=w(x)^2$ in $ \Omega$ with $ w=0$ on $ \partial \Omega$ (were are assuming $2< \frac{N+2}{N-2}$. I am attempting to show the kernel of some linear operator is empty. After playing around I have arrived at the linear equation

$$- \Delta \hat{\phi}(x,\tau) + C_\tau(x) \hat{\phi}(x,\tau)=0 \quad x \in \Omega$$ with $ \hat{\phi}(x,\tau)=0$ on $x \in \partial \Omega$ where $ C_\tau(x):=\tau^2 - 2 w(x)$; and where $ \tau \in R$ is a parameter. (The actually linear equation I am solving involves $ \phi(x,t)$ where $ x \in \Omega$ and $ t \in R$ and where $ | \phi(x,t)| \le e^{-\sigma |t|}$ where $ \sigma>0$ is fixed. Here $ \hat{\phi}$ is the Fourier transform of $ \phi(x,t)$ in the $t$ variable.

So here is my question. I would like to show that $ \phi(x,t)=0$ and hence I would like to show that $ \hat{\phi}(x,\tau)=0$ for all $ x \in \Omega$ and $ \tau \in R$. Note that $ C_\tau(x) \ge 0$ for $ \tau^2 \ge 2 \max_\Omega w$ and hence for this range of $ \tau$ we can apply the maximum principle to see that $ \hat{\phi}(x,\tau)=0$ for all $ x \in \Omega$ and $ \tau^2 \ge 2 \max_\Omega w$. \

QUESTION. Consider $ \tau \mapsto \hat{\phi}(x,\tau)= \hat{\phi}(\tau) (x) $ to be a mapping from $R$ to some Banach space of functions. If we happened to know that this mapping was real analytic in $ \tau$ then trivially we would have that $ \hat{\phi}(x,\tau)=0$ for all $ x \in \Omega$ and for all $ \tau \in R$. (of course I have no clue what ``real analytic'' means for some abstract mapping into a Banach space, but I have seen some papers from Norman Dancer who proved various related results). I guess I am curious if an approach like that could possibly work and hence it would be worth attempting to learn more about real analytic functions or maybe this approach can be easily seen to fail the way i have posed it.

thanks

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  • $\begingroup$ Here is a reference for the above mentioned paper: E.N. Dancer, Real analyticity and non-degeneracy, Mathematische Annalen E.N. Dancer $\endgroup$ – Math604 Jun 30 '15 at 7:55
  • $\begingroup$ Why is it linear PDE? $\endgroup$ – Fan Zheng Oct 19 '15 at 3:41

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