On blow-up criterion for the nonlinear Schrödinger equation systems

Abstract

In this paper, we study the blow-up problem of nonlinear Schrödinger equations

　　　　　　　　 \{\begin{array}{ll}\text{i}{\mathrm{\partial }}_{t}v+\mathrm{\Delta }u+(|u{|}^2+|v{|}^2)u=0,\hfill & (t,x)\in {\mathbb{R}}^{1+n},\hfill \\ \mathrm{i}{\mathrm{\partial }}_{t}v+\mathrm{\Delta }v+(|u{|}^2+|v{|}^2)v=0,\hfill & (t,x)\in {\mathbb{R}}^{1+n},\hfill \\ u(0,x)=u_0\left(x\right),\hfill & v(0,x)=v_0\left(x\right),\hfill \end{array}

and prove that the solution of negative energy blows up in finite or infinite time.

Keywords

Nonlinear Schrödinger equations / blow up / negative energy

1 Introduction

Schrödinger's equation is the fundamental equation in hadron mechanics. Its general form is

\text{i}{\mathrm{\partial }}_{t}u+\mathrm{\Delta }u+Vu=0,

where V denotes the potential, while \frac{|u{|}^2}{\Vert u{\Vert }_{L^2({\mathbb{R}}^n)}^2} denotes the probability density of the particle appearing at the point (x,t), and the solution u is called the wave function. In this paper, we study the well-known focusing nonlinear Schrödinger equation system

(1.1)\{\begin{array}{l}\text{i}{\mathrm{\partial }}_{t}u+\mathrm{\Delta }u+(|u{|}^p+|v{|}^p)|u{|}^{p-2}u=0,(t,x)\in {\mathbb{R}}^{1+n},\\ \mathrm{i}{\mathrm{\partial }}_{t}v+\mathrm{\Delta }v+(|u{|}^p+|v{|}^p)|v{|}^{p-2}v=0,(t,x)\in {\mathbb{R}}^{1+n},\\ u(0,x)=u_0(x),v(0,x)=v_0(x).\end{array}

The equations have received a lot of attention from mathematicians especially because of their applications in nonlinear optics, see [1, 6]. For (1.1), the potentials of these two equations are V=(|u{|}^p+|v{|}^p)|u{|}^{p-2} and (|u{|}^p+|v{|}^p)|v{|}^{p-2}, respectively. Noting that the potential V depends on the wave functions u and v, which gives the nonlinear terms. When the probability density \frac{|u{|}^2}{\Vert u{\Vert }_{L^2({\mathbb{R}}^n)}^2} is very large, the potential V becomes a large negative value with a high absolute magnitude. (1.1) has the important stretch-invariant property, i.e.,

u_{\lambda }={\lambda }^{\frac{1}{p-1}}u({\lambda }^{2}t,\lambda x),v_{\lambda }={\lambda }^{\frac{1}{p-1}}v({\lambda }^{2}t,\lambda x).

In this sense, the equations and {\dot{H}}^{s}c norm are invariant under this telescoping transformation, where

s_c=\frac{n}{2}-\frac{1}{p-1}.

When p=2, the solution of (1.1) follows the conservation laws of mass, momentum, and energy, i.e.,

\begin{array}{rl}& M(u(t),v(t)):={\int }_{{\mathbb{R}}^n}(|u(t,x){|}^2+|v(t,x){|}^2)\mathrm{d}x=M(u_0,v_0),\\ & P(u(t),v(t)):=\text{Im}{\int }_{{\mathbb{R}}^n}(\overline{u(t,x)}\mathrm{\nabla }u(t,x)+\overline{v(t,x)}\mathrm{\nabla }v(t,x))\text{d}x=P(u_0,v_0),\\ & E(u(t),v(t)):=\frac{1}{2}{\int }_{{\mathbb{R}}^n}(|\mathrm{\nabla }u(t,x){|}^2+|\mathrm{\nabla }v(t,x){|}^2)\text{d}x-\frac{1}{4}{\int }_{{\mathbb{R}}^n}(|u(t,x){|}^2+|v(t,x){|}^2{)}^2\text{d}x\\ & =E(u_0,v_0).\end{array}

Regarding this system, the following results are well-known: , then the corresponding solution bursts at a finite moment, see Glassey [3]. Also, when u_0 is radially symmetric, , n=2 or n=3, the corresponding solution also bursts at a finite moment. Du et al. [2], Holmer and Roudenko [4] proved that when u_0 only belongs to H^1, , a single solution of the equation blasts at a finite or infinite time. These two papers give two different methods of proof, respectively, of which, the Profile decomposition method is used in [4], and only the local Virial's identity and the almost finite propagation speed are used in [2]. In this paper, the method of [2] is applied to study the system of equations case, and the corresponding results in the articles [2, 4] are obtained as follows.

Theorem 1.1　 Assume that s>s_c, n⩾2. Let the initial values u_0\in H^s({\mathbb{R}}^n), v_0\in H^s({\mathbb{R}}^n) and , let (u,v) be the solution corresponding to equations (1.1) on [O,T_{\mathrm{m}\mathrm{a}\mathrm{x}}) when p=2. Then one of the following two conclusions holds:

(1) the solution blasts in finite time when , and

\underset{t\uparrow T_{max}}{lim}(\Vert u(t){\Vert }_{H^s}+\Vert v(t){\Vert }_{H^s})=\mathrm{\infty };

(2) when T_{\text{max}}=\mathrm{\infty }, there exists a time series t_n such that when t_n\to \mathrm{\infty }, for any q>4,

\underset{t_n\uparrow \mathrm{\infty }}{lim}(\Vert u(t_n){\Vert }_{L^2}+\Vert v(t_n){\Vert }_{L^2})=\mathrm{\infty }.

The same conclusion holds when time is negative.

Remark 1.1　Roughly speaking, Theorem 1.1(1) refers to finite-time blasting and Theorem 1.1(2) refers to infinite-time blasting (in which case, the L^q norm can be replaced by the H^s norm with the Sobolev embedding method). It is not clear whether Theorem 1.1(2) can be ruled out at this stage, or whether it can actually occur (even for the single equation case, this phenomenon is unclear).

2 Proof of Theorem 1.1

To prove Theorem 1.1, we first prove one of its lemma, i.e., Lemma 1.1.

Let

K(u,v):=8\int (|\mathrm{\nabla }u{|}^2+|\mathrm{\nabla }v{|}^2)\text{d}x-2n\int (|u{|}^2+|v{|}^2{)}^2\text{d}x.

We know that when p=2, the solution of (1.1) satisfies the Viry equation, i.e.,

\frac{{\text{d}}^2}{\text{d}{t}^2}\int |x{|}^2(|u(t,x){|}^2+|v(t,x){|}^2)\text{d}x=K(u,v).

This leads to the Glassey’s conclusion, see [3], which means if x{u}_0\in L^2({\mathbb{R}}^n),x{v}_0\in L^2({\mathbb{R}}^n), and there exsits such at

(2.1)

then the solution (u,v) explodes in a finite time.

Lemma 1.1　 Assume that n, s are the same as the n, s in Theorem 1.1. If there exists such that Eq. (2.1) holds, there is no global solution u\in C({\mathbb{R}}^{+};H^s), v\in C({\mathbb{R}}^{+};H^s), making when q>4,

Before proving Lemma 2.1, we first introduce a local result (see [2]).

Proposition 2.1 (local existence)　 Let s⩾s_c, and n, s, s_c are the same as the n, s, s_c in Theorem 1.1. Then for any (u_0,v_0)\in H^s({\mathbb{R}}^n), when p=2, (1.1) has a unique local solution (u,v)\in C([0,T);H^s({\mathbb{R}}^n)). Additionally, if s>s_c, then the time T only depends on \Vert u_0{\Vert }_{H^s}+\Vert v_0{\Vert }_{H^s}.

Proof of Proposition 2.1　This paper mainly uses the method given in [2] to consider the local Virial identity ream. Let

I(t)=\int \mathrm{\Phi }(x)|u(t,x){|}^2\text{d}x,J(t)=\int \mathrm{\Phi }(x)|v(t,x){|}^2\text{d}x.

The following proposition can be obtained by direct calculation.

Lemma 2.2　 For any \mathrm{\Phi }\in C^4({\mathbb{R}}^n), we have

(2.2)I^{\prime }(t)=2\text{Im}\int \mathrm{\nabla }\mathrm{\Phi }\cdot \mathrm{\nabla }u\overline{u}\text{d}x,J^{\prime }(t)=2\text{Im}\int \mathrm{\nabla }\mathrm{\Phi }\cdot \mathrm{\nabla }v\overline{v}\text{d}x,

(2.3)I^{\mathrm{\prime }\mathrm{\prime }}(t)=4\text{Re}\sum _{i,k}^n\int {\mathrm{\partial }}_j{\mathrm{\partial }}_k\mathrm{\Phi }\cdot {\mathrm{\partial }}_{j}u{\mathrm{\partial }}_k\overline{u}\text{d}x+2\int \mathrm{\nabla }\mathrm{\Phi }\cdot \mathrm{\nabla }(|u{|}^2+|v{|}^2)|u{|}^2\text{d}x-\int {\mathrm{\Delta }}^2\mathrm{\Phi }|u{|}^2\text{d}x,

(2.4)J^{\mathrm{\prime }\mathrm{\prime }}(t)=4\text{Re}\sum _{j,k}^n\int {\mathrm{\partial }}_j{\mathrm{\partial }}_k\mathrm{\Phi }\cdot {\mathrm{\partial }}_{j}v{\mathrm{\partial }}_k\overline{v}\text{d}x+2\int \mathrm{\nabla }\mathrm{\Phi }\cdot \mathrm{\nabla }(|u{|}^2+|v{|}^2)|e{|}^2\text{d}x-\int {\mathrm{\Delta }}^2\mathrm{\Phi }|v{|}^2\text{d}x.

While from Eqs. (2.2)–(2.4), there is

(2.5)\begin{array}{rl}(I+J{)}^{\mathrm{\prime }\mathrm{\prime }}(t)=& 4\text{Re}\sum _{j,k}^n\int {\mathrm{\partial }}_j{\mathrm{\partial }}_k\mathrm{\Phi }\cdot ({\mathrm{\partial }}_{j}u{\mathrm{\partial }}_k\overline{u}+{\mathrm{\partial }}_{j}v{\mathrm{\partial }}_k\overline{v})\text{d}x-\int \mathrm{\Delta }\mathrm{\Phi }(|u{|}^2+|v{|}^2{)}^2\text{d}x\\ & -\int {\mathrm{\Delta }}^2\mathrm{\Phi }(|u{|}^2+|v{|}^2)\text{d}x.\end{array}

Lemma 2.3　 Let {\eta }_0>0. For any t⩽\frac{{\eta }_{0}R}{4m_0\overline{C_0}}, we have

(2.6){\int }_{|x|⩾R}(|u(t,x){|}^2+|v(t,x){|}^2)\text{d}x⩽{\eta }_0+o_R(1).

Proof　When \mathrm{\Phi } is radial symmetry, from (2.2), we can get

(2.7)(I+J{)}^{\prime }(t)=2\text{Im}\int \left({\mathrm{\Phi }}^{\prime }\frac{x\cdot \mathrm{\nabla }u}{r}\overline{u}+{\mathrm{\Phi }}^{\prime }\frac{x\cdot \mathrm{\nabla }v}{r}\overline{v}\right)\text{d}x,

where r=|x|. Let \Vert u_0{\Vert }_{L^2}=\Vert v_0{\Vert }_{L^2}=m_0. From (2.7), there is

\begin{array}{rl}(I+J)(t)& =(I+J)(0)+{\int }_0^t(I+J{)}^{\mathrm{\prime }}(t^{\mathrm{\prime }})\text{d}{t}^{\mathrm{\prime }}\\ & ⩽(I+J)(0)+t\Vert {\mathrm{\Phi }}^{\prime }{\Vert }_{L^{\mathrm{\infty }}}\underset{t^{\prime }\in [0,t]}{sup}(\Vert u{\Vert }_{L^2}\Vert \mathrm{\nabla }u{\Vert }_{L^2}+\Vert v{\Vert }_{L^2}\Vert \mathrm{\nabla }v{\Vert }_{L^2}).\end{array}

Fixed the large constant R>0. Let \mathrm{\Phi } satisfied

\mathrm{\Phi }=\{\begin{array}{ll}0,& 0⩽r⩽{\displaystyle \frac{R}{2}},\\ 1,& r⩾R,\end{array}

and

0⩽\mathrm{\Phi }⩽1,0⩽{\mathrm{\Phi }}^{\mathrm{\prime }}⩽\frac{4}{R}.

Then

(I+J)(t)⩽{\int }_{|x|⩾\frac{R}{2}}(|u_0{|}^2+|v_0{|}^2)\text{d}x+\frac{4m_0\overline{C_0}t}{R}.

From

{\int }_{|x|⩾\frac{R}{2}}(|u_0(x){|}^2+|v_0(x){|}^2)\text{d}x=o_R(1)

and

{\int }_{|x|⩾R}(|u(t,x){|}^2+|v(t,x){|}^2)\text{d}x⩽(I+J)(t),

Lemma 2.3 is obtained.□

Lemma 2.4　 There exists a constant \tilde{C}(s,n,m_0,C_0)>0, {\theta }_q>0, such that

(I+J{)}^{\mathrm{\prime }\mathrm{\prime }}(t)⩽8K(u,v)+\tilde{C}(\Vert u{\Vert }_{L^2(|x|>R)}+\Vert v{\Vert }_{L^2(|x|>R)}{)}^{{\theta }_q}.

Proof　If \mathrm{\Phi } is radial symmetry, from (2.5), we can get

(2.8)\begin{array}{rl}(I+J{)}^{\mathrm{\prime }\mathrm{\prime }}(t)=& 4\int \frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r}(|\mathrm{\nabla }u{|}^2+|\mathrm{\nabla }v{|}^2)\text{d}x+4\int \left(\frac{{\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}}{r^2}-\frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r^3}\right)(|x\cdot \mathrm{\nabla }u{|}^2+|x\cdot \mathrm{\nabla }v{|}^2)\text{d}x\\ & -\int \left({\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}+(n-1)\frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r}\right)(|u{|}^2+|v{|}^2{)}^2\text{d}x-\int {\mathrm{\Delta }}^2\mathrm{\Phi }(|u{|}^2+|v{|}^2)\text{d}x.\end{array}

Eq. (2.8) can be rewritten as

(I+J{)}^{\mathrm{\prime }\mathrm{\prime }}(t)=8K(u,v)+R_1+R_2+R_3,

where

\{\begin{array}{l}{R}_1=4{\displaystyle \int \left({\displaystyle \frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r}}-2\right)(|\mathrm{\nabla }u{|}^2+|\mathrm{\nabla }v{|}^2)\text{d}x+4{\displaystyle \int \left({\displaystyle \frac{{\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}}{r^2}}-{\displaystyle \frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r^3}}\right)(|x\cdot \mathrm{\nabla }u{|}^2+|x\cdot \mathrm{\nabla }v{|}^2)\text{d}x,}}\\ R_2=-{\displaystyle \int \left({\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}+(n-1){\displaystyle \frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r}}-2n\right)(|u{|}^2+|v{|}^2{)}^2\text{d}x,}\\ R_3=-{\displaystyle \int {\mathrm{\Delta }}^2\mathrm{\Phi }(|u{|}^2+|u{|}^2)\text{d}x.}\end{array}

Choose \mathrm{\Phi } and let

(2.9)0⩽\mathrm{\Phi }⩽r^2,{\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}⩽2,{\mathrm{\Phi }}^{(4)}⩽\frac{4}{R^2},

where r=|x|,

\mathrm{\Phi }=\{\begin{array}{ll}{r}^2,& 0⩽r⩽R,\\ 0,& r⩾2R.\end{array}

First prove

(2.10)R_1⩽0.

The \mathbb{R} can be divided into two parts: \{\frac{{\mathrm{\Phi }}^{″}}{r^2}-\frac{{\mathrm{\Phi }}^{\prime }}{r^3}⩽0\},\{\frac{{\mathrm{\Phi }}^{″}}{r^2}-\frac{{\mathrm{\Phi }}^{\prime }}{r^3}>0\}. When \frac{{\mathrm{\Phi }}^{″}}{r^2}-\frac{{\mathrm{\Phi }}^{\prime }}{r^3}⩽0, from {\mathrm{\Phi }}^{\prime }⩽2r, there is R_1⩽0. When \frac{{\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}}{r^2}-\frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r^3}>0, from {\mathrm{\Phi }}^{\prime }⩽2, there is R_1⩽4\int ({\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}-2)(|\mathrm{\nabla }u{|}^2+|\mathrm{\nabla }v{|}^2)\text{d}x⩽0.

Then prove R_2. As \text{supp}({\mathrm{\Phi }}^{\mathrm{\prime }\mathrm{\prime }}+(n-1)\frac{{\mathrm{\Phi }}^{\mathrm{\prime }}}{r}-2n)\subset [R,\mathrm{\infty }), by interpolation, there exist such that

\begin{array}{rl}{R}_2& ⩽C(\Vert u{\Vert }_{L^q(|x|>R)}+\Vert v{\Vert }_{L^q(|x|>R)}{)}^{1-{\theta }_q}(\Vert u{\Vert }_{L^2(|x|>R)}+\Vert v{\Vert }_{L^2(|x|>R)}{)}^{{\theta }_q},\\ & ⩽C{C}_0^{1-{\theta }_q}(\Vert u{\Vert }_{L^2(|x|>R)}+\Vert v{\Vert }_{L^2(|x|>R)}{)}^{{\theta }_q},\end{array}

where C=C(s,n)>0.

Next prove R_3.

\begin{array}{rl}{R}_3& ⩽\int {\mathrm{\Delta }}^2\mathrm{\Phi }|u{|}^2\text{d}x+\int {\mathrm{\Delta }}^2\mathrm{\Phi }|v{|}^2\text{d}x\\ & ⩽C{R}^{-2}\Vert u{\Vert }_{L^2(|x|>R)}^2+C{R}^{-2}\Vert v{\Vert }_{L^2(|x|>R)}^2.\end{array}

Combine Eqs. (2.8)–(2.10), for R>1, we can get

(I+J{)}^{\mathrm{\prime }\mathrm{\prime }}(t)⩽8K(u,v)+\tilde{C}(\Vert u{\Vert }_{L^2(|x|>R)}+\Vert v{\Vert }_{L^2(|x|>R)}{)}^{{\theta }_Q},

where \tilde{C}=\tilde{C}(s,n,m_0,C_0)>0. Lemma 2.4 is proved.□

By (1.6) and Lemma 2.4, we notice that for any t⩽T:=\frac{{\eta }_{0}R}{4m_0\overline{C}}, there is

(I+J{)}^{\mathrm{\prime }\mathrm{\prime }}(t)⩽8K(u,v)+\tilde{C}({\eta }_0^{{\theta }_q}+o_R(1)).

Integrating from 0 to T and applying (2.1), we have

\begin{array}{c}(I+J)(t)⩽(I+J)(0)+(I+J{)}^{\mathrm{\prime }}(0)T+{\int }_0^T{\int }_0^t(8K(u,v)+\tilde{C}{\eta }_0^{{\theta }_q}+o_R(1))\text{d}{t}^{\mathrm{\prime }}\text{d}t\\ ⩽(I+J)(0)+(I+J{)}^{\mathrm{\prime }}(0)T+(8{\beta }_0+\tilde{C}{\eta }_0^{{\theta }_q}+o_R(1))\cdot \frac{1}{2}{T}^2.\end{array}

Let \tilde{C}{\eta }_0^{{\theta }_q}=-{\beta }_0, and R is large enough. Then for T=\frac{{\eta }_{0}R}{4m_0\overline{C}}, we have

(2.11)(I+J)(t)⩽(I+J)(0)+(I+J{)}^{\mathrm{\prime }}(0)\frac{{\eta }_{0}R}{4m_0\overline{C}}+{\alpha }_0{R}^2,

where the constant , and {\alpha }_0 does not depend on R. Now, we need to prove the following two equations:

(2.12)(I+J)(0)=o_R(1)R^2,(I+J{)}^{\prime }(0)=o_R(1)R.

In fact,

Using the same method, we can get the second estimate, and then prove (2.12). Using (2.11)–(2.12), choose the sufficiently large R, we can get when T=\frac{{\eta }_{0}R}{4m_0\overline{C}}, there is

(I+J)(T)⩽o_R(1)R^2+{\alpha }_0{R}^2⩽\frac{1}{2}{\alpha }_0{R}^2.

Since , we end up with , which contradicts the definition. Thus, Lemma 2.1 is proved.□

Now, assuming the opposite condition, i.e., the equation has global solutions and is uniformly bounded, then

Then, we prove that there exists , such that

\overline{C_0}=\underset{t\in {\mathbb{R}}^{+}}{sup}\Vert \mathrm{\nabla }u(t){\Vert }_{L^2}+\underset{t\in {\mathbb{R}}^{+}}{sup}\Vert \mathrm{\nabla }v(t){\Vert }_{L^2}.

In fact, there is the L^4 bounded norm formed by interpolation between L^2 and L^q, and then the bounded property of H^1 norm follows the energy conservation law. Thus, Theorem 1.1 is proved.□

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