4
$\begingroup$

I am interested in the critical equation $- \Delta w(x) = w(x)^p $ in $ R^N$ where $p=\frac{N+2}{N-2}$. After translation the solutions of this equation are all radial with maximum at the origin (and there is an explicit formula). Lets assume $ w(x)>0$ is the radial solution of the above problem with the added condition that $w(0)=1$.

Define $ L(\phi)= - \Delta \phi - p w(x)^{p-1} \phi$. My interest is in the kernel of $L$ and lets not worry about the exact domain of $L$.

So we expect that $ \phi_i:=w_{x_i}$ is in the kernel of $L$ for each $ 1 \le i \le N$. But also note the original equation is scale invariant and lets assume that $ w_\lambda(x)$ satisfies the equation (without the constraint $w_\lambda(0)=1$)) for all $ \lambda >0$ with $ w_1(0)=1$. Then $ \phi_{N+1}:=\partial_\lambda w_\lambda(x)|_{\lambda=1}$ is also in the kernel of $L$.

So my question is: is the kernel of $L$ given by $ \{\phi_i : 1 \le i \le N\}$ or $ \{ \phi: 1 \le i \le N+1\}$ or neither.

This answer is well known (just not by me and I haven't been able to locate the answer). I attempted to see if $\phi_{N+1}$ is really a linear combination of the other terms (recall there is an explicit formula for $w(x)$ but I am getting bog down in computations).

thanks

NEW QUESTION

Here is the full question I am really interested in. Again I believe this is known material but I can't seem to find the explicit references and my linear analysis is too weak to figure this out. Let $W(x)=w(x)$ be as above and we let $ W_{\lambda,z}(x)$ denote $W(x)$ dialated by $ \lambda>0$ and translated so its maximum is at $z \in R^N$. Let $ V_{\lambda,z}$ dennote the projection of $W_{\lambda,z}$ onto $H_0^1(\Omega)$. So $ V_{\lambda,z}(x)$ solves $-\Delta V_{\lambda,z}(x)= W_{\lambda,z}(x)^p$ in $ \Omega$ with $V_{\lambda,z} =0$ on \partial \Omega$. \

I am interested in the following linear operator: $$ L_{\lambda,z}(\phi)= -\Delta \phi - p V_{\lambda,z}^{p-1} \phi,$$ on certain weighted spaces which i now define.
$$ \| \phi\|_{X_\lambda}:=\sup_{x \in \Omega} (1 + \lambda^2 |x-z|^2 )^\frac{N-2}{4} \lambda^\frac{-(N-2)}{2} | \phi(x)| $$ $$\| f\|_{X_\lambda}:=\sup_{x \in \Omega} (1 + \lambda^2 |x-z|^2 )^\frac{N+2}{4} \lambda^\frac{-(N+2)}{2} | f(x)| $$ and we let $ X_\lambda$ and $Y_\lambda$ denote the completion of some suitable spaces under the appropriate norms; the functions in $ X_\lambda$ should satisfy a zero Dirichlet boundary condition.

This problem is coming up since I am attempting to solve some nonlinear PDE and I need to understand the linear theory on these spaces. In particular I would like a theory which is uniform in $ \lambda$ for large $\lambda$. Lets assume $ B_1 \subset \subset \Omega$.

So I would like a theory along the lines of: there is some $C>0$ and $ \lambda_0>0$ such that for all $ |z|<1$, $ \lambda > \lambda_0$ and $ f \in L^\infty(\Omega)$ there is some $ \phi_{\lambda,z} \in X_\lambda$ such that $L_{\lambda,z}(\phi_{\lambda,z})=f$ in $ \Omega$ and $$ \| \phi_{\lambda,z}\|_{X_\lambda} \le C \|f \|_{Y_\lambda}.$$ The important property I would like is the uniform bounds in $ \lambda$ and $z$.

Now I would assume this is infact not true. Define $$ \phi_i := \partial_{z_i} V_{\lambda,z} \quad 1 \le i \le N, \qquad \phi_{N+1}:=\partial_{\lambda} V_{\lambda,z}$$.

It appears that the functions $ \phi_i \; 1 \le i \le N+1$ should almost be in the kernel of $L_{\lambda,z}$ for large $ \lambda$. Lets decompose $X_\lambda$ by $ X_\lambda= \tilde{X}_\lambda + span\{ \phi_i: 1 \le i \le N+1\}$ and let $\tilde{Y}_\lambda$ denote the range of $L_{\lambda,z}$ restricted to $\tilde{X}_\lambda $ and we should be able to write $ Y_\lambda= \tilde{Y}_\lambda + span\{ \psi_i: 1 \le i \le N+1\}$.

QUESTION.

Is it plausable that one would have the following linear theory: there is some $ C>0$ and $ \lambda_0 >0$ such that for all $ f \in L^\infty(\Omega)$, $\lambda > \lambda_0$ and $ |z|<1$ there is some $ \phi_{\lambda,z} \in \tilde{X}_\lambda$ and $ c_i \in R$ such that $$L_{\lambda,z}( \phi_{\lambda,z})= f - \sum_{i=1}^{N+1} c_i \psi_i$$ and one has a uniform estimate of the form $$ \| \phi_{\lambda,z}\|_{X_\lambda} \le C \|f\|_{Y_\lambda}$$. ($c_i$ and $\psi_i$ will also depend on $z$ and $ \lambda$).

I realize there are a lot of details. The main issue I am having trouble with is exactly how to mod out the terms which are almost in the kernel for large $ \lambda$ and also how to mod out on the range... Anyhow comments would be greatly appreciated. thanks

$\endgroup$
4
  • $\begingroup$ It is easy to see that $\phi_{N+1}$ cannot be a linear combination of the others. Note that $\phi_{N+1}$ is radially symmetric, and all the others have zero average over the sphere. $\endgroup$ Jul 18, 2013 at 19:33
  • $\begingroup$ Honestly, I'm having trouble following your definition of $\phi_{N+1}$. However, it sounds like you mean the result of the dilation operator acting on the solution $w$. If that's the case, then all of these $\phi_i$ should be in the kernel of $L$. This should follow from the general fact that symmetry generators acting on solutions create linearized solutions, which is fairly straight forward to work out for yourself. $\endgroup$ Jul 18, 2013 at 20:22
  • $\begingroup$ Michael. Thanks, that is a nice way to see it. $\endgroup$
    – Craig
    Jul 18, 2013 at 21:31
  • $\begingroup$ Igor, I am not sure of the correct way of saying this in words... $w_\lambda(x)$ is a dilated version of $w(x)$ and one can check that $-\Delta w_\lambda(x)=w_\lambda(x)^p$. Also $ w_1(x)=w(x)$ and hence taking derivative in $ \lambda$ and setting $ \lambda=1$ shows that $ \partial_\lambda w_\lambda(x)|_{\lambda=1}$ is in the kernel of $L$. I was confused as to whether it was independent of the other terms. $\endgroup$
    – Craig
    Jul 18, 2013 at 21:36

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.