Nonlinear Instability of Inhomogeneous Steady States Solutions to the HMF Model

Abstract

In this work we prove the nonlinear instability of inhomogeneous steady states solutions to the Hamiltonian mean field (HMF) model. We first study the linear instability of this model under a simple criterion by adapting the techniques developed in Lin (Math Res Lett 8:521–534, 2001). In a second part, we extend to the inhomogeneous case some techniques developed in Grenier (Commun Pure Appl Math 53(9):1067–1091, 2000), Han-Kwan and Hauray (Commun Math Phys 334(2):1101–1152), Han-Kwan and Nguyen (J Stat Phys 162(6):1639–1653) and prove a nonlinear instability result under the same criterion.

Appendix: Existence of Unstable Steady States

In this section, we prove that the set of steady states satisfying the assumptions of Theorems 1.1 and 1.2 is not empty. More precisely, we prove the following

Lemma A.1

Let $m>0$. There exist $m>0$, $e_*<m$ and there exists a nonincreasing function F, ${\mathcal {C}} ^{\infty }$ on ${\mathbb {R}}$, such that $F(e)>0$ for $e<e_*$, $F(e)=0$ for $e\ge e_*$ and $|F'(e)|\le C|e_*-e|^{-\alpha }F(e)$ in the neighborhood of $e_*$, for some $\alpha \ge 1$, and such that the function $f(\theta ,v)=F(\frac{v^2}{2}-m\cos \theta )$ is a steady state solution to the HMF model (1.1) and such that $\kappa (m,F)>1$, where $\kappa (m,F)$ is given by

$$\begin{aligned} \kappa (m,F)=\int ^{2 \pi }_{0}\!\!\int ^{+\infty }_{-\infty } \left| F'\!\left( e(\theta ,v)\right) \right| \left( \frac{\displaystyle \int _{{\mathcal {D}}_{e(\theta ,v)}}(\cos \theta -\cos \theta ')(e(\theta ,v)+m\cos \theta ')^{-1/2}d\theta '}{\displaystyle \int _{\mathcal D_{e(\theta ,v)}}(e(\theta ,v)+m\cos \theta ')^{-1/2}d\theta '} \right) ^2d\theta dv, \end{aligned}$$

with

$$\begin{aligned} e(\theta ,v)=\frac{v^2}{2}-m\cos \theta ,\qquad {\mathcal D_e}=\left\{ \theta '\in {\mathbb {T}}\,:\,\,m\cos \theta '>-e\right\} . \end{aligned}$$

Proof

Let $m>0$ and F a nonincreasing ${\mathcal {C}} ^{\infty }$ function on ${\mathbb {R}}$ supported in $(-\infty ,m)$, which is not identically zero on $(-m,m)$. We first observe that $f(\theta ,v)=F(\frac{v^2}{2}-m\cos \theta )$ is a steady state solution to the HMF model (1.1) if and only if m and F satisfy $\gamma (m,F)=m$ with

$$\begin{aligned} \gamma (m,F):=\int _0^{2\pi }\int _{\mathbb {R}}F\left( \frac{v^2}{2}-m\cos \theta \right) \cos \theta d\theta dv>0. \end{aligned}$$

By using the linearity of $\gamma$ in F we deduce that $\frac{m}{\gamma (m,F)}F(\frac{v^2}{2}-m\cos \theta )$ is a steady state.

We proceed by a contradiction argument. Assume that

$$\begin{aligned} \kappa \left( m,\frac{m}{\gamma (m,F)}F\right) \le 1 \end{aligned}$$

for all $m>0$ and all nonincreasing ${\mathcal {C}} ^{\infty }$ function F supported in $(-\infty ,m)$ such that, denoting by $(-\infty ,e_*]$ the support of F, we have $|F'(e)|\le C|e_*-e|^{-\alpha }F(e)$ in the neighborhood of $e_*$, for some $\alpha \ge 1$. This is equivalent to

$$\begin{aligned} \kappa \left( m,F\right) \le \frac{\gamma (m,F)}{m}, \end{aligned}$$

or, after straightforward calculation and an integration by parts,

$$\begin{aligned} -\iint F'\left( e(\theta ,v)\right) g_m(e(\theta ,v))d\theta dv\le 0 \end{aligned}$$

(A.1)

with

$$\begin{aligned} g_m(e)=(\Pi _m \cos ^2\theta )(e)-\left( (\Pi _m\cos \theta )(e)\right) ^2-(\Pi _m \sin ^2\theta )(e) \end{aligned}$$

and for all function $h(\theta )$,

$$\begin{aligned} (\Pi _m h)(e)=\frac{\displaystyle \int _{{\mathcal {D}}_e}(e+m\cos \theta )^{-1/2}h(\theta )d\theta }{\displaystyle \int _{{\mathcal {D}}_e}(e+m\cos \theta )^{-1/2}d\theta }. \end{aligned}$$

Now, we choose the functions F as follows. We first pick a nonincreasing ${\mathcal {C}} ^{\infty }$ function $\Psi$ on ${\mathbb {R}}$ with support $(\infty ,e_\sharp ]\subset (-\infty ,m)$, then we set $e_*=\frac{e_\sharp +m}{2}$ and define

$$\begin{aligned} F_\varepsilon (e)=\Psi (e)+\varepsilon \exp \left( -(e_*-e)^{-1}\right) , \quad \text{ for } e<e_*, \end{aligned}$$

the parameter $\varepsilon >0$ being arbitrary. Since $F_\varepsilon$ satisfies the assumptions, it satisfies (A.1). Then, letting $\varepsilon \rightarrow 0$, we get

$$\begin{aligned} -\iint \Psi '\left( e(\theta ,v)\right) g_m(e(\theta ,v))d\theta dv\le 0. \end{aligned}$$

The function $\Psi$ being arbitrary, this is equivalent to

$$\begin{aligned} g_m(e)\le 0,\qquad \forall m>0,\quad \forall e\in (-m,m), \end{aligned}$$

or,

$$\begin{aligned} g_1(e)\le 0,\qquad \forall e\in (-1,1). \end{aligned}$$

(A.2)

Let us now prove that the function $g_1(e)$ is in fact positive in the neighborhood of $e=1$, which contradicts (A.2).

Indeed, we introduce

$$\begin{aligned} \alpha (e)=\int _{{\mathcal {D}}_e}(e+\cos \theta )^{-1/2}d\theta ,\qquad \beta (e)=\int _{{\mathcal {D}}_e}(e+\cos \theta )^{-1/2}\sin ^2\theta d\theta . \end{aligned}$$

We have

$$\begin{aligned}\alpha (e) g_1(e)&=\alpha (e)-2\beta (e)-\frac{1}{\alpha (e)}\left( \int _{{\mathcal {D}}_e}(e+\cos \theta )^{1/2}d\theta -e\alpha (e)\right) ^2\\&=(1-e^2)\alpha (e)-2\beta (e)+2e\int _{\mathcal {D}_e}(e+\cos \theta )^{1/2}d\theta \\&\quad -\frac{1}{\alpha (e)}\left( \int _{\mathcal {D}_e}(e+\cos \theta )^{1/2}d\theta \right) ^2. \end{aligned}$$

From [23], we have

$$\begin{aligned} \alpha (e)\sim -\sqrt{2}\log (1-e)\quad \text{ as } e\rightarrow 1^-, \end{aligned}$$

and direct calculations yield

$$\begin{aligned} \int _0^{2\pi }(1+\cos \theta )^{1/2}d\theta =4\sqrt{2},\qquad \beta (1)=\frac{8\sqrt{2}}{3}. \end{aligned}$$

This means that

$$\begin{aligned}&\alpha (e)g_1(e)\rightarrow \frac{8\sqrt{2}}{3}>0 \quad \text{ as } e\rightarrow 1^-, \\&g_1(e)\sim \frac{8\sqrt{2}}{\alpha (e)}\text{ as } e\rightarrow 1^-. \end{aligned}$$

This proves the claim. $\square$