Scattering threshold for the focusing coupled Schrödinger system revisited

Abstract

This note investigates the coupled Schrödinger system

$$\begin{aligned} i\dot{u}_j +\Delta u_j= -\left( \sum \nolimits _{k=1}^{m}a_{jk}|u_k|^p\right) |u_j|^{p-2}u_j. \end{aligned}$$

Indeed, beyond the mass-energy threshold given in Saanouni (Appl Anal, 2020. https://doi.org/10.1080/00036811.2020.1808201), a scattering versus finite time blow-up dichotomy is obtained in the mass super-critical and energy sub-critical regime. Moreover, one extends the previous work [18] to the non-radial case.

Appendix: Proof of (4.5)

Let $u\in C({\mathbb {R}}, H)$, a solution to (1.1). Denote the quantity

$$\begin{aligned} V(t):=R^2\sum _{j=1}^m\int _{{\mathbb {R}}^N}\psi \left( \frac{x}{R}\right) |u_j(t,x)|^2\,dx. \end{aligned}$$

Multiplying the equation (1.1) by $2u_j$ and examining the imaginary parts,

$$\begin{aligned} \partial _t (|u_j|^2) =-2\mathfrak {I}({\bar{u}}_j \Delta u_j). \end{aligned}$$

Thus, one gets

$$\begin{aligned} V'= & {} -2R^2\sum _{j=1}^m\int _{{\mathbb {R}}^N}\psi \left( \frac{x}{R}\right) \mathfrak {I}({\bar{u}}_j \Delta u_j)\,dx\\= & {} 2R\sum _{j=1}^m\mathfrak {I}\int _{{\mathbb {R}}^N}\left( \nabla \psi \left( \frac{x}{R}\right) .\nabla u_j\right) {\bar{u}}_j \,dx\\= & {} 2R\sum _{j=1}^m\mathfrak {I}\int _{{\mathbb {R}}^N}\left( \partial _k\psi \left( \frac{x}{R}\right) \partial _ku_j\right) {\bar{u}}_j \,dx. \end{aligned}$$

Denote ${\mathcal {N}}:=\Big (\sum _{k=1}^{m}a_{jk}|u_k|^p\Big )|u_j|^{p-2}u_j$ and Compute using the Eq. (1.1),

$$\begin{aligned} \partial _t\mathfrak {I}(\partial _k u_j{\bar{u}}_j)= & {} \mathfrak {I}(\partial _k\dot{u}_j{\bar{u}}_j)+\mathfrak {I}(\partial _k u_j\bar{\dot{u}}_j)\\= & {} \mathfrak {R}(i\dot{u}_j\partial _k{\bar{u}}_j)-\mathfrak {R}(i\partial _k \dot{u}_j\bar{u}_j)\\= & {} \mathfrak {R}(\partial _k{\bar{u}}_j(-\Delta u_j-{\mathcal {N}}))-\mathfrak {R}({\bar{u}}_j\partial _k(-\Delta u_j-{\mathcal {N}}))\\= & {} \mathfrak {R}({\bar{u}}_j\partial _k\Delta u_j-\partial _k{\bar{u}}_j\Delta u_j)+\mathfrak {R}({\bar{u}}_j\partial _k{\mathcal {N}}-\partial _k{\bar{u}}_j{\mathcal {N}}). \end{aligned}$$

Recall the identity

$$\begin{aligned} \frac{1}{2}\partial _k\Delta (|u_j|^2)-2\partial _l\mathfrak {R}(\partial _{k}u_j\partial _l{\bar{u}}_j)=\mathfrak {R}({\bar{u}}_j\partial _k\Delta u_j-\partial _k{\bar{u}}_j\Delta u_j). \end{aligned}$$

Then,

$$\begin{aligned} (I)_j:= & {} \int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}({\bar{u}}_j\partial _k\Delta u_j-\partial _k{\bar{u}}_j\Delta u_j)\,dx\\= & {} \int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \left( \frac{1}{2}\partial _k\Delta (|u_j|^2)-2\partial _l\mathfrak {R}(\partial _ku_j\partial _l{\bar{u}}_j)\right) \,dx\\= & {} -\,\frac{1}{2R^3}\int _{{\mathbb {R}}^N}\Delta ^2\psi \left( \frac{x}{R}\right) |u_j|^2\,dx+\frac{2}{R}\int _{{\mathbb {R}}^N}\partial _l\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}(\partial _ku_j\partial _l{\bar{u}}_j)\,dx. \end{aligned}$$

Moreover,

$$\begin{aligned} (II)_j:= & {} \int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}({\bar{u}}_j\partial _k{\mathcal {N}}-\partial _k{\bar{u}}_j{\mathcal {N}})\,dx\\= & {} \int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}(\partial _k[{\bar{u}}_j{\mathcal {N}}]-2\partial _k{\bar{u}}_j{\mathcal {N}})\,dx\\= & {} -\,\int _{{\mathbb {R}}^N}\left( \frac{1}{R}\Delta \psi \left( \frac{x}{R}\right) {\bar{u}}_j{\mathcal {N}}+2\mathfrak {R}\left( \partial _k\psi \left( \frac{x}{R}\right) \partial _k{\bar{u}}_j{\mathcal {N}}\right) \right) \,dx\\= & {} -\,\frac{1}{R}\sum _{k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\Delta \psi \left( \frac{x}{R}\right) |u_ku_j|^p\,dx-2\int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}\left( \partial _k{\bar{u}}_j{\mathcal {N}}\right) \,dx. \end{aligned}$$

Now, write

$$\begin{aligned} \mathfrak {R}(\partial _k{\bar{u}}_j{\mathcal {N}})= & {} \displaystyle \sum _{l=1}^{m}a_{jl}\mathfrak {R}(\partial _k{\bar{u}}_j|u_l|^p|u_j|^{p-2}u_j)\\= & {} \frac{1}{p}\displaystyle \sum _{l=1}^{m}a_{jl}\partial _k(|u_j|^p)|u_l|^p. \end{aligned}$$

Then,

$$\begin{aligned} \displaystyle \sum _{j=1}^{m}\mathfrak {R}(\partial _k{\bar{u}}_j{\mathcal {N}})= & {} \frac{1}{p}\displaystyle \sum _{j,l=1}^{m}a_{jl}\partial _k(|u_j|^p)|u_l|^p\\= & {} \frac{1}{2p}\displaystyle \sum _{j,l=1}^{m}a_{jl}\partial _k(|u_ju_l|^p). \end{aligned}$$

So,

$$\begin{aligned} \sum _{j=1}^m(II)_j= & {} -\,\frac{1}{R}\sum _{j,k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\Delta \psi \left( \frac{x}{R}\right) |u_ku_j|^p\,dx-2\sum _{j=1}^m\int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}(\partial _k{\bar{u}}_j{\mathcal {N}})\,dx\\= & {} -\,\frac{1}{R}\sum _{j,k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\Delta \psi \left( \frac{x}{R}\right) |u_ku_j|^p\,dx-\frac{1}{p}\sum _{j,k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \partial _k(|u_ju_k|^p)\,dx\\ \end{aligned}$$

Finally

$$\begin{aligned} V''(t)= & {} 2R\left( \sum _{j=1}^m(I)_j+\sum _{j=1}^m(II)_j\right) \\= & {} 2R\left( \sum _{j=1}^m-\frac{1}{2R^3}\int _{{\mathbb {R}}^N}\Delta ^2\psi \left( \frac{x}{R}\right) |u_j|^2\,dx+\frac{2}{R}\int _{{\mathbb {R}}^N}\partial _l\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}(\partial _ku_j\partial _l{\bar{u}}_j)\,dx\right) \\&+\,2R\left( -\frac{1}{R}\sum _{j,k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\Delta \psi (\frac{x}{R})|u_ku_j|^p\,dx-\frac{1}{p}\sum _{j,k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\partial _k\psi \left( \frac{x}{R}\right) \partial _k(|u_ju_k|^p)\,dx\right) \\= & {} -\frac{1}{R^2}\int _{{\mathbb {R}}^N}\Delta ^2\psi \left( \frac{x}{R}\right) |u|^2\,dx+4\int _{{\mathbb {R}}^N}\partial _l\partial _k\psi \left( \frac{x}{R}\right) \mathfrak {R}(\partial _ku\partial _l{\bar{u}})\,dx\\&-\,2\left( 1-\frac{1}{p}\right) \sum _{j,k=1}^{m}a_{jk}\int _{{\mathbb {R}}^N}\Delta \psi \left( \frac{x}{R}\right) |u_ku_j|^p\,dx. \end{aligned}$$

This finishes the proof.