# Orbital stability for NLS

Jordan Bell
April 3, 2014

Let $n=3$, and take $p<\frac{4}{3}$. Some of the material we will present for general $n$ when it doesn’t simplify our work to use $n=3$.

The (defocusing) nonlinear Schrödinger equation is

 $i\phi_{t}+\Delta\phi+|\phi|^{p-1}\phi=0.$

$\phi(x,0)=\phi_{0}\in H^{1}$.

For a function $\psi$ on $\mathbb{R}^{n}$, the orbit of the function under the symmetries of NLS is

 $\mathscr{G}_{\psi}=\{\psi(\cdot+x_{0})e^{i\gamma}:(x_{0},\gamma)\in\mathbb{R}^% {n}\times\mathbb{T}\}.$

We say that $\psi$ is orbitally stable if initial data being near it implies that the solution of NLS is near it always.

We define

 $\rho(\phi(t),\mathscr{G}_{\psi})=\inf_{(x_{0},\gamma)\in\mathbb{R}^{n}\times% \mathbb{T}}\|\phi(\cdot+x_{0},t)e^{i\gamma}-\psi\|_{H^{1}}.$

The ground state equation is

 $\Delta u-u+|u|^{p-1}u=0.$

The ground state equation comes from the solution $\phi(x,t)=e^{it}u(x)$ of NLS. It is a fact that there is a positive bounded solution $R$ of the ground state equation, which we call a ground state.

###### Theorem 1.

The ground state $R$ is orbitally stable: for any $\epsilon>0$ there is a $\delta(\epsilon)>0$ such that if

 $\rho(\phi_{0},\mathscr{G}_{R})<\delta(\epsilon)$

then for all $t>0$

 $\rho(\phi(t),\mathscr{G}_{R})<\epsilon.$

We define the energy functional $\mathscr{E}$ by

 $\mathscr{E}[\phi]=\int|\nabla\phi|^{2}+|\phi|^{2}-\frac{2}{p+1}|\phi|^{p+1}dx,$

so $\mathscr{E}[\phi]$ is a function of time but not of space.

It is a fact that for each $t$ there are $x_{0}=x_{0}(t)$ and $\gamma=\gamma(t)$ such that

 $\|\phi(\cdot+x_{0},t)e^{i\gamma}-R\|_{H^{1}}=\rho(\phi(t),\mathscr{G}_{R}).$

Let $w=\phi(\cdot+x_{0},t)e^{i\gamma}-R$; so $\|w(t)\|_{H^{1}}=\rho(\phi(t),\mathscr{G}_{R})$.

Let $\Delta\mathscr{E}=\mathscr{E}[\phi_{0}]-\mathscr{E}[R]$. We have

 $\displaystyle\Delta\mathscr{E}$ $\displaystyle=$ $\displaystyle\mathscr{E}[\phi(\cdot,t]-\mathscr{E}[R]$ $\displaystyle=$ $\displaystyle\mathscr{E}[\phi(\cdot+x_{0},t)e^{i\gamma}]-\mathscr{E}[R]$ $\displaystyle=$ $\displaystyle\mathscr{E}[R+w]-\mathscr{E}[R].$

We shall express $\mathscr{E}[R+w]$ as a Taylor expansion about $R$. We compute the first variation as follows:

 $\displaystyle d\mathscr{E}[R]w$ $\displaystyle=$ $\displaystyle\int\nabla w\nabla\overline{R}+\nabla R\nabla\overline{w}+w% \overline{R}+R\overline{w}-|R|^{p-1}(w\overline{R}+R\overline{w})$ $\displaystyle=$ $\displaystyle 2\Re\int\nabla w\nabla R+wR-w|R|^{p-1}R$ $\displaystyle=$ $\displaystyle 2\Re\int w(-\Delta R+R-|R|^{p-1}R)$ $\displaystyle=$ $\displaystyle 0,$

where we used the fact that $R$ is real valued, integration by parts, and the fact that $R$ is a solution of the ground state equation. So the first variation of $\mathscr{E}$ at $R$ is $0$.

We now compute the second variation of $\mathscr{E}$.

 $\displaystyle d^{2}\mathscr{E}[R][w]$ $\displaystyle=$ $\displaystyle 2\Re\int-w\Delta\overline{w}+|w|^{2}-\frac{p-1}{2}R^{p-1}w^{2}-% \frac{p-1}{2}R^{p-1}|w|^{2}$ $\displaystyle-R^{p-1}|w|^{2}$ $\displaystyle=$ $\displaystyle 2\Re\int-w\Delta\overline{w}+|w|^{2}-\frac{p-1}{2}R^{p-1}w^{2}-% \frac{p+1}{2}R^{p-1}|w|^{2}$

Write $w=u+iv$. Then we have

 $\displaystyle d^{2}\mathscr{E}[R][w]$ $\displaystyle=$ $\displaystyle 2\int-u\Delta u-v\Delta v+u^{2}+v^{2}-R^{p-1}(pu^{2}+v^{2})$

Define

 $L_{+}=-\Delta+1-pR^{p-1}\qquad L_{-}=-\Delta+1-R^{p-1},$

which gives

 $(L_{+}u,u)_{L^{2}}=\int-u\Delta u+u^{2}-pu^{2}R^{p-1}$

and

 $(L_{-}v,v)_{L^{2}}=\int-v\Delta v+v^{2}-v^{2}R^{p-1}.$

Thus

 $d^{2}\mathscr{E}[R][w]=2(L_{+}u,u)_{L^{2}}+2(L_{-}v,v)_{L^{2}}.$

And we assert that the remainder term of the Taylor series is $O(\int|w|^{3})$, because $R$ is bounded. Therefore

 $\Delta\mathscr{E}=(L_{+}u,u)_{L^{2}}+(L_{-}v,v)_{L^{2}}+O\Big{(}\int|w|^{3}% \Big{)}.$

We can bound $\int|w|^{3}$ using the Gagliardo-Nirenberg inequality, which gives us (for $n=3$)

 $\|w\|_{L^{3}}^{3}\leq C_{0}\|\nabla w\|_{L^{2}}^{3/2}\|w\|_{L^{2}}^{3/2}\leq C% _{0}\|w\|_{H^{1}}^{3},$

for some $C_{0}$ that doesn’t depend on $w$. Therefore

 $\Delta\mathscr{E}=(L_{+}u,u)_{L^{2}}+(L_{-}v,v)_{L^{2}}+O(\|w\|_{H^{1}}^{3}),$

so there is some $C$ such that

 $\Delta\mathscr{E}\geq(L_{+}u,u)_{L^{2}}+(L_{-}v,v)_{L^{2}}-C\|w\|_{H^{1}}^{3}.$