2
$\begingroup$

I initially posted this question on Stackexchange Mathematics (see here) but as I got no answer, maybe someone here will be able to help me.

I am watching a serie of lectures on "Blow up solution for the energy critical heat equation" from Monica Musso on YouTube and at some point she states a result I do not understand. Let met recall the setting.

We are studying the following system: $$u_t = \Delta u + u^{N+2/N-2} \quad \text{in }\Omega \times (0,T), \quad \text{and} \quad u = 0 \quad \text{on } \partial \Omega \times (0,T).$$ We also require $u(\cdot, 0) = u_0$ in $\Omega$. Steady states are solutions of the problem that do not depend on time, i.e. they satisfy $$0 = \Delta u + u^{N+2/N-2}.$$ We may show that, steady states are given by $$U_{\lambda, \xi}(x) = \frac{1}{\lambda^{N - 2/2}}U\left(\frac{x - \xi}{\lambda}\right), \quad\text{where } U(y) = \frac{\alpha_N}{(1 + |y|^2)^{N - 2/2}}.$$ We linearized this equation around $U$ and we define $$L(\varphi) = \Delta \varphi + \frac{N + 2}{N - 2} U^{\frac{N+2}{N-2} - 1}\varphi.$$ At this point, she claimed that, by Noether's theorem, all the invariances of the solutions of the initial problem will generate an element of the kernel of the operator $L$, namely $$\text{span}\left(\partial_1 U, \ldots, \partial_N U, \frac{N - 2}{2} U + \nabla U \cdot y\right) \subset \text{ker }L,$$ where the first $N$ elements come from the invariance by translation and the last one comes from the scaling invariance. Moreover, we may show that this kernel is in fact non-degenerate, in the sense that we have the equality instead of the inclusion.

My question is about this use of Noether's theorem. I tried to read its statement in Lee's Introduction to Smooth Manifold, but it was really differential geometry oriented and I admit that I do not totally get the link with the above problem. Could one of you try to explain it to me ? Or at least giving me a reference with a more "Analysis-version" of this theorem ? I have some background in differential geometry but I am not totally comfortable with it.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ I would guess that the lecturer misspoke; this claim has nothing to do with Noether's theorem, rather it is just a statement of what it means to linearize a continuous symmetry of a PDE. Given a continuous symmetry $U \mapsto U_\lambda = U + \xi_U \lambda + O(\lambda^2)$ of a nonlinear PDE, the lineariziation (or the corresponding infinitesimal symmetry) $\xi_U$ will quite obviously satisfy the linearized PDE at $U$. The only relation to Noether's theorem I can see is that it has infinitesymal symmetries in its hypotheses. $\endgroup$
    – Igor Khavkine
    Nov 9, 2022 at 13:34
  • $\begingroup$ Could you explain why the linearized symmetry will satisfy the linearized PDE? I'm not sure to see why $\endgroup$
    – Falcon
    Nov 9, 2022 at 15:26
  • $\begingroup$ Take the equation $PDE[U_\lambda] = 0$ and expand to first order in $\lambda$. $\endgroup$
    – Igor Khavkine
    Nov 9, 2022 at 16:08
  • $\begingroup$ Ah okay, and each order must be zero? $\endgroup$
    – Falcon
    Nov 9, 2022 at 16:10
  • 1
    $\begingroup$ Yes, you are expanding both sides, so the right-hand side is $0+0\lambda + 0\lambda^2 + \cdots$, while the equality of power series implies the equality of the coefficients. $\endgroup$
    – Igor Khavkine
    Nov 9, 2022 at 16:44

0

Reset to default

Your Answer

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