Linearization of a PDE

Question

I have been struggling with some linearization argument of the following paper: "M. Weinstein: Modulational stability of ground states of NLS". In order to give a bit of context to my question, let us consider the NLS equation $$ 2i\phi_t+\Delta\phi+\vert\phi\vert^{2\sigma}\phi=0, \quad 0<\sigma<\tfrac{2}{n-2}.$$ This equation has very interesting "localized" solutions of the form: $$\phi(t,x)=u(x)e^{it/2},$$ where $u(t,x)$ solves $\Delta u-u+\vert u\vert^{2\sigma}u=0$. Besides, the latter equation has an even more interesting real, positive and radial $H^1(\mathbb{R}^n)$ solution called "Ground state" and denoted by $R(x)$.

Now let me try to explain my question. Consider the perturbed Initial Valued Problem (IVP): $$2i\phi_t^\varepsilon+\Delta \phi^\varepsilon+\vert \phi^\varepsilon\vert^{2\sigma}\phi^\varepsilon=\varepsilon F(\vert \phi^\varepsilon\vert)\phi^\varepsilon, \quad \phi^\varepsilon(t=0,x)=R(x)+\varepsilon S(x)$$ We will seek solutions of the previous equation of the form $$\phi^\varepsilon(t,x)=(R(x)+\varepsilon w_1+\varepsilon^2 w_2+...)e^{it/2}.$$ According to Weinstein if you reeplace this function into the perturbed equation and linearize you will get the following IVP for the linearized perturbation $w$: $$ 2iw_t+\Delta w-w+(\sigma+1)R^{2\sigma}w+\sigma R^{2\sigma}\overline{w}=F(R)R, \quad w(0,x)=0.$$ Now my problem is: I do not really understand how to obtain this linearization, can someone explain a little bit how to do it? Or recommend some references to learn about it. I tried replacing $\phi^\varepsilon$ (truncated after $\varepsilon w$) and then I took the derivative with respect to $\varepsilon$ and evaluate at $\varepsilon=0$, but I cannot recover the equation claimed by Weinstein, so I think that I'm not undersitanding how to linearize.

Note2: The parameter $n$ denotes the dimension.

Answer 1

If we have a nonlinear operator acting on some space of (weakly) differentiable functions, linearizing it at a given "point" means simply verify that its Gateaux derivative exists at that "point". In the case under analysis, we have that the nonlinear operator has the following form: $$ \mathfrak{F}(t,x,\phi^\varepsilon)=2i\phi_t^\varepsilon+\Delta \phi^\varepsilon+\vert \phi^\varepsilon\vert^{2\sigma}\phi^\varepsilon-\varepsilon F(\vert \phi^\varepsilon\vert)\phi^\varepsilon $$ Define $w_1(t,x)\triangleq w(t,x)$: noting that $$ \begin{split} \left.\phi^\varepsilon(t,x)\right|_{\varepsilon=0}&=R(x)e^{it/2}\iff \left. \overline{\phi^\varepsilon}(t,x)\right|_{\varepsilon=0}=R(x)e^{-it/2},\\ \\ \left.\frac{\partial}{\partial \varepsilon} \phi^\varepsilon(t,x)\right|_{\varepsilon=0} &= w(t,x)e^{it/2}\iff \left.\frac{\partial}{\partial \varepsilon} \overline{\phi^\varepsilon}(t,x)\right|_{\varepsilon=0} = \overline{w}(t,x)e^{-it/2},\\ \\ \left.\frac{\partial}{\partial \varepsilon} |\phi^\varepsilon(t,x)|^{2\sigma}\right|_{\varepsilon=0} & = \left.\frac{\partial}{\partial \varepsilon} \big(\phi^\varepsilon(t,x)\overline{\phi^\varepsilon}(t,x)\big)^{\sigma}\right|_{\varepsilon=0}\\ &=\left.\sigma |\phi^\varepsilon(t,x)|^{2\sigma-2}\left(\overline{\phi^\varepsilon}(t,x)\frac{\partial}{\partial \varepsilon}\phi^\varepsilon(t,x)+\phi^\varepsilon(t,x)\frac{\partial}{\partial \varepsilon}\overline{\phi^\varepsilon}(t,x)\right)\right|_{\varepsilon=0}\\ &=\left.\sigma |\phi^\varepsilon(t,x)|^{2\sigma-2}\left(R(x) w(t,x) + R(x)\overline{w}(t,x)\right)\right|_{\varepsilon=0}\\ &=\sigma R(x)^{2\sigma-1}\left(w(t,x) + \overline{w}(t,x)\right), \end{split} $$ and assuming that $F\in C^1$, the calculation of the Gateaux derivative of $\mathfrak{F}$ goes as follows $$ \begin{split} \left.\frac{\partial}{\partial \varepsilon}\right|_{\varepsilon=0}\mathfrak{F}(t,x,\phi^\varepsilon) & =2iw_te^{it/2}-we^{it/2}+\Delta w e^{it/2} \\ & \quad+\left.\left(\phi^\varepsilon \frac{\partial}{\partial \varepsilon} \vert \phi^\varepsilon\vert^{2\sigma} + \vert \phi^\varepsilon\vert^{2\sigma}\frac{\partial}{\partial \varepsilon} \phi^\varepsilon\right)\right|_{\varepsilon=0} -F(R)Re^{it/2}\\ &= 2iw_te^{it/2}-we^{it/2}+\Delta w e^{it/2} \\ & \quad + \sigma R^{2\sigma}(w + \overline{w})e^{it/2} + R^{2\sigma}we^{it/2} -F(R)Re^{it/2}\\ &= \big[2iw_t-w+\Delta w + (\sigma+1) R^{2\sigma}w + \sigma R^{2\sigma}\overline{w} -F(R)R\big]e^{it/2}\\ \end{split} $$ As we can see, the operator $\partial_\varepsilon\mathfrak{F}$ is linear (or perhaps is better to say affine) in the arguments $w$ and $\overline{w}$, thus it is the sought for linearization, and the PDE in the paper of Weinstein is simply obtained by equating it to $0$.