A singular perturbation problem for a nonlinear Schrödinger system with three wave interaction

Abstract

In this paper, we consider the locations of spikes of ground states for the following nonlinear Schrödinger system with three wave interaction

as $\varepsilon \rightarrow +0$. In addition, we study the asymptotic behavior of a quantity $\inf _{x \in {\mathbb {R}}^N} {\tilde{c}}({{\textbf{V}}}(x);\gamma )$ as $\gamma \rightarrow \infty$ which determines locations of spikes. In particular, we give the sharp asymptotic behavior of a ground states of (${{\mathcal {P}}}_\varepsilon$) for $\gamma$ sufficiently large and small, respectively. Furthermore, we consider when all the ground states of (${{\mathcal {P}}}_\varepsilon$) are scalar or vector.

Appendix

In this Appendix, we consider the radial symmetry and monotonicity of classical solutions of elliptic system of the following type:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_i'' + f_i(x,u_1,\dots ,u_k) = 0 &{} \text { in }\ {\mathbb {R}},\quad i =1,\dots ,k,\\ u_i > 0 &{} \text { in }\ {\mathbb {R}},\\ u_i(x) \rightarrow 0, &{} \text { as }\ \left|x\right| \rightarrow \infty , \end{array}\right. } \end{aligned}$$

(5.1)

where $k \ge 1$.

Busca-Sirakov [5] studied the radial symmetry and monotonicity of classical solutions of elliptic systems for $N \ge 2$. Moreover, Ikoma [16] considered the symmetry and monotonicity of the solutions in the case $N=1$ and $k=2$.

Here, we show the symmetry result for $N=1$ after slight modification.

Let us note ${{\textbf{u}}} = (u_1,\dots ,u_k) \in (0,\infty )^k$ and

$$\begin{aligned} A(x,{{\textbf{u}}}) = \left( \frac{\partial f_i}{\partial u_j}(x,{{\textbf{u}}}) \right) _{1 \le i,j \le k}. \end{aligned}$$

We suppose that $f_i \in C^1({\mathbb {R}} \times (0,\infty )^k,{\mathbb {R}})$ for all $i=1,\dots ,k$ and (A0)–(A4):

• (A0) $f_i(-x,{{\textbf{u}}}) = f_i(x,{{\textbf{u}}})$ for all $x \in {\mathbb {R}}$, ${{\textbf{u}}} \in (0,\infty )^k$ and $i=1,\dots ,k$.

• (A1) $(\partial f_i/\partial x)(x,{{\textbf{u}}}) \le 0$ for all $x \ge 0$, ${{\textbf{u}}} \in (0,\infty )^k$ and $i=1,\dots ,k$.

• (A2) $(\partial f_i / \partial u_j)(x,{{\textbf{u}}}) \ge 0$ for all $x \in {\mathbb {R}}$, ${{\textbf{u}}} \in (0,\infty )^k$ and $i,j \in \{1,\dots ,k\},\ i \ne j$.

• (A3) there exist constants $\varepsilon > 0$ and $R_1 > 0$ such that for any $I,J \subset \{1,\dots ,k\}$, $I \cap J = \emptyset$, $I \cup J = \{1,\dots ,k\}$, there exist $i_0 \in I$ and $j_0 \in J$ such that $(\partial f_{i_0} / \partial u_{j_0})(x,{{\textbf{u}}}) > 0$ for all $(x,{{\textbf{u}}}) \in {{\mathcal {O}}}$, where

  $$\begin{aligned} {{\mathcal {O}}} = \{(x,{{\textbf{u}}}) \in {\mathbb {R}} \times (0,\infty )^k \mid \left|x\right| > R_1,\ \left|{{\textbf{u}}}\right| < \varepsilon \}. \end{aligned}$$

• (A4) all k-principal minors of $-A(x,u_1,\dots ,u_k)$ have positive determinants, for all $(x,{{\textbf{u}}}) \in {{\mathcal {O}}}$, $1 \le i \le k$. We recall that the k-principal minors of a matrix $(m_{ij})_{1 \le i,j \le k}$ are the submatrices $(m_{ij})_{1 \le i,j \le k'}$ with $1 \le k' \le k$.

Theorem 5.1

Suppose $f_1,\dots ,f_k$ satisfy (A0)–(A4), and ${{\textbf{u}}} = (u_1,\dots ,u_k)$ is a classical solution of (5.1). Then there exists a point $y_0 \in {\mathbb {R}}$ such that the functions $u_i$ are radially symmetric with respect to the origin $y_0$, that is $u_i(x) = u_i(\left|x - y_0\right|)$, $i=1,\dots ,k$. Moreover,

$$\begin{aligned} \frac{d u_i}{d r} < 0\quad \text { for all }\ r = \left|x - y_0\right| > 0. \end{aligned}$$

For $\lambda \in {\mathbb {R}}$, set $\Sigma _\lambda := (\lambda ,\infty )$ and for $x \in \Sigma _\lambda$, we define

$$\begin{aligned} x^\lambda&:= 2 \lambda - x,\\ U_i^\lambda (x)&:= u_i(x^\lambda ) - u_i(x) = u_i(2 \lambda - x) - u_i(x). \end{aligned}$$

Outline of the proof of Theorem 5.1

We define

$$\begin{aligned} \Lambda := \inf \{\lambda > 0 \mid U_i^\mu \ge 0\ \text { in }\ \Sigma _\mu \ \text { for }\ i=1,\dots ,k,\ \text { and all }\ \mu \ge \lambda \}. \end{aligned}$$

Since $u_i(x) \rightarrow 0$ as $\left|x\right| \rightarrow \infty$, there exists $R_0 \ge R_1$ such that $u_i(x) < \varepsilon /\sqrt{k}$ if $\left|x\right| \ge R_0$ for all $i=1,\dots ,k$. We take $\lambda ^* > R_0$, for which

$$\begin{aligned} \max _{\begin{array}{c} 1 \le i \le k\\ x \in [2 \lambda - R_0, 2 \lambda + R_0] \end{array}} u_i(x) < \min _{\begin{array}{c} 1 \le i \le k\\ x \in [- R_0, R_0] \end{array}} u_i(x), \end{aligned}$$

for all $\lambda > \lambda ^*$. Hence $U_i^\lambda > 0$ in $[2 \lambda - R_0, 2 \lambda + R_0] \subset \Sigma _\lambda$ for all $\lambda > \lambda ^*$. We notice that the functions $U_i^\lambda$ satisfy the following system

$$\begin{aligned} (U_i^\lambda )'' + \frac{\partial f_i}{\partial x}(\mathbf {\eta }) (\left|x^\lambda \right| - x) + \sum _{j=1}^k \frac{\partial f_i}{\partial u_j}(x,\xi _{i1},\dots ,\xi _{ik}) U_j^\lambda = 0,\quad i = 1,\dots ,k, \end{aligned}$$

where $\mathbf {\eta } = \mathbf {\eta }(x,\lambda ) \in (0,\infty )^{k+1}$ and

$$\begin{aligned} \xi _{ij} = \xi _{ij}(x,\lambda ) \in (\min \{u_j(x),u_j(x^\lambda )\},\max \{u_j(x),u_j(x^\lambda )\}). \end{aligned}$$

Since $\left|x^\lambda \right| < x$ for $x \in \Sigma _\lambda$, we obtain from (A1) the following systems of inequality for $U_i^\lambda$

$$\begin{aligned} (U_i^\lambda )'' + \sum _{j=1}^k \frac{\partial f_i}{\partial u_j}(x,\xi _{i1},\dots ,\xi _{ik}) U_j^\lambda \le 0,\quad i = 1,\dots ,k. \end{aligned}$$

(5.2)

We want to show that $U_i^\lambda \ge 0$ in $\Sigma _\lambda$, for all $\lambda > \lambda ^*$. We argue by contradiction. Suppose there exist $\lambda > \lambda ^*$ and $i_0 \in \{1,\dots ,k\}$ such that $\inf _{\Sigma _\lambda } U_{i_0}^\lambda < 0$. We set $J = \{j \mid U_j^\lambda \ge 0\ \text { in }\ \Sigma _\lambda \}$, and $I = \{1,\dots ,k\} {\setminus } J$ (note that $i_0 \in I$).

Since all k-principal minors of $- A(x,{{\textbf{u}}})$ have positive determinants, for all $(x,{{\textbf{u}}}) \in {{\mathcal {O}}}$, we do not need to introduce a function g as in [5]. Note the following lemma stated as Lemma 2.2 in [13].

Lemma 5.2

Let $M = (m_{ij})_{1 \le i,j \le k}$ be a matrix such that $m_{ij} \le 0$ for $i \ne j$. Assume all k-principal minors of M have positive determinants. Then

• (i) all minors of M obtained by dropping lines and columns of the same order have positive determinants.

• (ii) if $M_{ij}$ is the minor of M obtained by dropping the i th line and j th column we have $(-1)^{i+j} \det M_{ij} \ge 0$.

Hence we may assume $I = \{1,\dots ,p\}$. Since $U_j^\lambda \ge 0$ in $\Sigma _\lambda$ for $j \in J$, from (A2) and (5.2), we have

$$\begin{aligned} (U_i^\lambda )'' + \sum _{j=1}^p \frac{\partial f_i}{\partial u_j}(x,\xi _{i1},\dots ,\xi _{ik}) U_j^\lambda \le 0,\quad i = 1,\dots ,p. \end{aligned}$$

Since $\inf _{\Sigma _\lambda } U_i^\lambda < 0$ for all $i=1,\dots ,p$, $U_i^\lambda > 0$ in $[2 \lambda - R_0, 2 \lambda + R_0]$, $U_i^\lambda (\lambda ) = 0$ and $U_i^\lambda (x) \rightarrow 0$ as $\left|x\right| \rightarrow \infty$, there exist $x_1,\dots ,x_p \in \Sigma _\lambda {\setminus } [2 \lambda - R_0, 2 \lambda + R_0]$ such that $U_i^\lambda (x_i) = \min _{\Sigma _\lambda } U_i^\lambda < 0$. Then $(U_i^\lambda )''(x_i) \ge 0$ and $(U_i^\lambda )'(x_i) = 0$. Substituting $x=x_i$ at i-th equation and using the fact that $U_j^\lambda (x_j) \le U_j^\lambda (x_i)$, we have

$$\begin{aligned} \sum _{j=1}^p \frac{\partial f_i}{\partial u_j}(x_i,\xi _{i1},\dots ,\xi _{ik}) U_j^\lambda (x_j) \le 0,\quad i = 1,\dots ,p. \end{aligned}$$

(5.3)

(5.3) can be written as

$$\begin{aligned} M {{\textbf{U}}} = Y, \end{aligned}$$

where

$$\begin{aligned}&{{\textbf{U}}} := (U_1^\lambda (x_1),\dots ,U_p^\lambda (x_p)),\quad M = (m_{ij})_{1 \le i,j \le p},\quad Y = (y_1,\dots ,y_p),\\&y_i \ge 0,\quad m_{i,j} := - \frac{\partial f_i}{\partial u_j}(x_i,\xi _{i1},\dots ,\xi _{ik}) \end{aligned}$$

Since $x_i \not \in [2 \lambda - R_0, 2 \lambda + R_0]$, then $x_i^\lambda \not \in [-R_0, R_0]$, that is, $\left|x_i^\lambda \right| > R_0$. Noting that $x_i > R_0$, we have $u_j(x_i),u_j(x_i^\lambda ) < \varepsilon /\sqrt{k}$. Thus we have $\xi _{i1}(x_i,\lambda )^2 + \dots + \xi _{ik}(x_i,\lambda )^2 < \varepsilon ^2$. From (A2) and (A4), we have $m_{ij} \le 0$ for $i \ne j$, and all p-principal minors of M have positive determinants. Since $\det M > 0$, it follows that ${{\textbf{U}}} = M^{-1} Y$. From Lemma 5.2, $U_i^\lambda (x_i) \ge 0$ for all $i=1,\dots ,p$. This contradicts the fact that $U_i(x_i) < 0$. Hence $\Lambda < \infty$.

The rest of the proof of Theorem 5.1 can be showed by the same argument as in [5]. $\square$