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I have been trying to understand some linearization argument of a recent paper on a wave-like equation. In order to give a bit of context to my question, let us consider the following equation $$ u_{tt}+2\alpha u_t-\Delta u+u-f(u)=0, \qquad \alpha>0, \quad (t,x)\in\mathbb{R}\times\mathbb{R}^n, \qquad \qquad (1)$$ where $f(u)=\vert u\vert^{p-1}u$ with $p>2$ satisfying some conditions (I don't think that it is important or this purposes). This equation has very interesting time-independent solutions also called ground-states, which are solutions of the form $\vec{Q}=(Q,0)$, where $Q$ is the unique $H^1$ positive radial solution of $$-\Delta Q+Q-f(Q)=0$$

Now let me try to explain my question. According to the authors, if you linearize equation $(1)$ around $\vec Q$ you will get the operator $\mathcal{L}:=-\Delta+1-pQ^{p-1}. $ So I tried myself to linearize the equation to see if I got the same result (it's the first time I work with a wave-type equation). So I considered perturbations of the form $(u,u_t)=(Q(x)+\varepsilon\eta(t,x),0+\varepsilon \widetilde{\eta}(t,x))$. Then, I replace this function on equation $(1)$ and I derived with respect to $\varepsilon$ and evaluate at $\varepsilon=0$. My problem is that doing that I got an extra term $-2\alpha$ on the linearize operator $\mathcal{L}$ what makes me think that I am not understanding how to linearize. Does anyone know how to properly linearize this equation in order to get this operator? My guess is that since you are considering time-independent functions then the perturbations have to be of the form $(\varepsilon\eta,0)$, or something like that but I am not sure if that is correct. Has anyone an explanation for this?

Note: $H^1(\mathbb{R}^n)$ denotes the Sobolev space (also denoted by $W^{1,2}$).

Note2: The parameter $n$ denotes the dimension.

Sharik
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    I am pretty sure you did it right. The linear part of the equation should be preserved and so if $u = Q(x) + \epsilon\eta(t,x)$ then to leading order $\eta$ should solve the same equation as $u$ but with $f(u)$ replaced by $p Q^{p-1} \eta$. The only thing I can think of is that maybe you are using a first order formalism, in which case the $\mathcal{L}$ you wrote is only the top-left block of the operator, with the $2\alpha$ appearing as an off-diagonal term? – Willie Wong Feb 21 '20 at 19:05
  • @WillieWong I replaced $(u,u_t)=(Q(x)+\varepsilon\eta(t,x),\varepsilon\widetilde{\eta}(t,x))$ on the system, from where I got (after deriving with respecto to epsilon): First equation of the system $\eta_t=\widetilde{\eta}$. Second equation: $\widetilde{\eta}_t=\Delta\eta-\eta-2\alpha\eta+f'(Q)\eta$. So I don't understand why the authors got a different result. – Sharik Feb 21 '20 at 22:09
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    @WillieWong Oh, I just realized that my $2\alpha\eta$ term is actually $2\alpha\widetilde{\eta}$. Now everything has sense for me! Thank you for your answer – Sharik Feb 21 '20 at 22:26

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For time independent solutions, you simply set all time derivatives equal to $0$, so you obtain $$-\Delta u-u+f(u)=0.$$ Steady state implies $u_t=0$, so the point $(0,Q)$ is indeed a critical point of this system and we can linearize the steady state equation to obtain $$-\Delta Q+Q-pQ^{p-1}=0,$$ as desired. This can also be obtained by linearizing the system, as you have done, which results in a factor of $2\alpha u_t$ in the equation,then look for steady state solutions of the linearized equations. If you do this you will get the same result since the equation is linear in the time derivatives.

whpowell96
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  • Thank you very much for your answer, but I still don't see where I am missing this $2\alpha$ factor. In the comments I wrote my solution, do you understand why the authors don't have this factor? – Sharik Feb 21 '20 at 22:12