Linearized operator around a solution by conserved quantities

Question

Let $$u_{tt}-u_{xx}= u-u^3 ,\: (t,x)\in \mathbb{R}\times \mathbb{R}.$$

I know that this equation admits wave traveling solution of the form $u(x,t)=\varphi(x-ct)$, where $ c\in \mathbb{R}$ is a constant and $\varphi$ is $L>0$ periodic. Moreover, using this question I found the operator linearized around $\varphi$ given by $$\mathcal{L}=(1-c^2)\frac{d^2}{dx^2}+(1-\varphi^2).$$

However, I want to find $ \mathcal{L} $ using conserved quantities $E$ and $F$, that is, write $$G''(\varphi):=E''(\varphi) -cF''(\varphi)=\mathcal{L}.$$ But I was unable to find such conserved quantities. A candidate for conservede quantitie $F$ I think is $$F(u)=\frac{1}{2}\int_{0}^{L} u^2\;dx.$$

Please explicitly write what are $E$ and $F$. Otherwise, your question is completely ambiguous. — Sharik, Apr 22 '20 at 14:43
I want to find $E$ and $F$. And I think $F$ has the form given above. — Guilherme, Apr 22 '20 at 14:52
I also tried to use $E$ as $E=\frac{1}{2}\int u_t^2+u_x^2+\frac{1}{2}(1-u^2)^2;dx$. — Guilherme, Apr 22 '20 at 16:06

Ivo Terek · Answer 1 · 2020-04-22T16:35:05.060

2

This is more of a comment than an actual answer, but: the idea is to use that derivatives by themselves are linear operators and to take the total derivative of the differential equation with the variable $u$. So the linearization of your ODE centered at $u$ is $$v_{tt} - v_{xx} = v- 3u^2v,$$seen as an equation for the "tangent vector" $v$ at $u$. I do not know how you got a single-variable differential operator for the linearization (nor how one would find a linearization via "conserved quantities", this seems like overcomplicating it). Using your notation, the linearized equation written in the form $\mathcal{L}v=0$ should be $$\mathcal{L} = (\partial_t)^2 - (\partial_x)^2 + (3u^2-1).$$

edited Apr 22 '20 at 16:35

answered Apr 22 '20 at 14:18

Ivo Terek

80,301

The operator I found is given by applying the process given here around a solution $\tilde{u}=u=\varphi$. – Guilherme Apr 22 '20 at 15:57
Incidentally, I think the operator you gave is the same as mine for $v=\varphi$ – Guilherme Apr 22 '20 at 16:02
He did the same thing as I did, just computing the directional derivative explicitly with an $\epsilon$ and taking the limit $\epsilon \to 0$. You must have a computation mistake. – Ivo Terek Apr 22 '20 at 16:03
Is that I did around the solution $u(x,t)=\varphi(x-ct)$. In this case, $u_{tt}=c^2\varphi''$ and $u_{xx}=\varphi''$. – Guilherme Apr 22 '20 at 16:28
Yes, I just fixed it. It seems off in your linearization that you don't have this factor $3$ anywhere. – Ivo Terek Apr 22 '20 at 16:36
Understand. You are right, I will redo my calculations. – Guilherme Apr 22 '20 at 16:40

Linearized operator around a solution by conserved quantities

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