Excited states of Bose–Einstein condensates with degenerate attractive interactions

Abstract

We study the Bose–Einstein condensates (BEC) in two or three dimensions with attractive interactions, described by $L^{2}$ constraint Gross-Pitaevskii energy functional. First, we give the precise description of the chemical potential of the condensate $\mu$ and the attractive interaction a. Next, for a class of degenerate trapping potential with non-isolated critical points, we obtain the existence and the local uniqueness of the excited states by accurately analyzing the location of the concentrated points and the Lagrange multiplier. Our results on degenerate trapping potential with non-isolated critical points are new for ground states of BEC and singularly perturbed nonlinear Schrödinger equations.

Appendices

Appendix

A. The kernel of a linear operator

Lemma A.1

Let $\xi _0:=(\xi _1,\cdots ,\xi _k)$ be a bounded solution of following system:

$$\begin{aligned} -\Delta \xi _{i}(x)+\big (1-3Q^2(x)\big )\xi _{i}(x)=-\frac{2}{ka_*}Q(x) \Big (\sum ^k_{l=1}\int _{{\mathbb {R}}^N}Q^3(x)\xi _l(x)\Big ),~\text{ for }~i=1,\cdots ,k. \end{aligned}$$

(A.1)

Then for $N=2,3$, it holds

$$\begin{aligned} \xi _i(x)= \sum ^N_{j=0}\gamma _{i,j}\psi _j, \end{aligned}$$

(A.2)

where $\gamma _{i,j}$ are some constants,

$$\begin{aligned} \psi _0=Q+x\cdot \nabla Q,~\psi _j=\frac{\partial Q}{\partial x_j},~\text{ for }~j=1,\cdots ,N. \end{aligned}$$

(A.3)

Moreover, $\gamma _{i,0}=\gamma _{l,0}$ for all $i,l=1,\cdots ,k$.

Proof

Noting that Q(x) is a radial function, using the technique of the separation of variables, we can prove

$$\begin{aligned} \xi _i(x)= \sum ^N_{j=1}\gamma _{i,j}\psi _j+ \xi _{i, 0}, \end{aligned}$$

where $\gamma _{i, j}$ is some constant, and $\xi _{i, 0}$ is a radial function, satisfying

$$\begin{aligned} -\Delta \xi _{i,0}(x)+\big (1-3Q^2(x)\big )\xi _{i, 0}(x)=-\frac{2}{ka_*}Q(x) \Big (\sum ^k_{l=1}\int _{{\mathbb {R}}^N}Q^3(x)\xi _l(x)\Big ). \end{aligned}$$

We set $\bar{L}(u):=-\Delta u(x)+\big (1-3Q^2(x)\big )u(x)$. Then

$$\begin{aligned} \bar{L}\psi _0= -2 Q. \end{aligned}$$

Since $\bar{L}$ has no non-trivial bounded radially symmetric kernel, it holds

$$\begin{aligned} \xi _{i, 0}= \gamma _{i, 0} \psi _0. \end{aligned}$$

Using (3.12), we find that $\gamma _{i, 0}$ satisfies

$$\begin{aligned} -2Q(x)\gamma _{i,0}=-\frac{2}{ka_*}Q(x) \frac{4-N}{4} \int _{{\mathbb {R}}^N} Q^4 =-\frac{2}{k}Q(x) \sum ^k_{l=1}\gamma _{l,0}, \end{aligned}$$

which gives $\gamma _{i,0}=\gamma _{l,0}$ for all $i,l=1,\cdots ,k$. $\square$

B. Calculations involving curvatures

Now let $\Gamma \in C^2$ be a closed hypersurface in ${\mathbb {R}}^N$. For $y\in \Gamma$, let $\nu (y)$ and T(y) denote respectively the outward unit normal to $\Gamma$ at y and the tangent hyperplane to $\Gamma$ at y. The curvatures of $\Gamma$ at a fixed point $y_0\in \Gamma$ are determined as follows. By a rotation of coordinates, we can assume that $y_0=0$ and $\nu (0)$ is the $x_N$-direction, and $x_j$-direction is the j-th principal direction.

In some neighborhood ${\mathcal {N}}={\mathcal {N}}(0)$ of 0, we have

$$\begin{aligned} \Gamma =\bigl \{ x: x_N=\varphi (x')\bigr \}, \end{aligned}$$

where $x'=(x_1,\cdots ,x_{N-1})$,

$$\begin{aligned} \varphi (x') =\frac{1}{2} \sum _{j=1}^{N-1} \kappa _j x_j^2 + O(|x'|^3), \end{aligned}$$

where $\kappa _j$, is the j-th principal curvature of $\Gamma$ at 0. The Hessian matrix $[D^2 \varphi (0)]$ is given by

$$\begin{aligned} {[}D^2 \varphi (0)]=diag [\kappa _1,\cdots ,\kappa _{N-1}]. \end{aligned}$$

Suppose that W is a smooth function, such that $W(x)=a$ for all $x\in \Gamma$.

Lemma B.1

We have

$$\begin{aligned}&\frac{\partial W(x)}{\partial x_l}\Bigr |_{x=0}=0, ~ l=1,\cdots ,N-1, \end{aligned}$$

(B.1)

$$\begin{aligned}&\frac{\partial ^2 W(x)}{\partial x_m\partial x_l }\Bigr |_{x=0} =-\frac{\partial W\big (x\big )}{\partial x_N}\Bigr |_{x=0} \kappa _i \delta _{ml}, ~\text{ for }~m, l=1,\cdots , N-1, \end{aligned}$$

(B.2)

where $\kappa _1,\cdots ,\kappa _{N-1}$, are the principal curvatures of $\Gamma$ at 0.

Proof

First, we have $W\big (x',\varphi (x')\big )=0$. And then we find

$$\begin{aligned} \frac{\partial W\big (x',\varphi (x')\big )}{\partial x_m} + \frac{\partial W\big (x',\varphi (x')\big )}{\partial x_N} \frac{\partial \varphi (x')}{\partial x_m} =0, ~\text{ for }~m=1,\cdots ,N-1. \end{aligned}$$

(B.3)

Letting $x'=0$ in (B.3), we obtain (B.1).

Differentiating (B.3) with respect to $x_l$ for $l=1,\cdots ,N-1$, we get

$$\begin{aligned}&\frac{\partial ^2 W\big (x',\varphi (x')\big )}{\partial x_m\partial x_l} +\frac{\partial ^2 W\big (x',\varphi (x')\big )}{\partial x_m\partial x_N} \frac{\partial \varphi (x')}{\partial x_l} + \frac{\partial W\big (x',\varphi (x')\big )}{\partial x_N} \frac{\partial ^2 \varphi (x')}{\partial x_m x_l}\nonumber \\&\quad + \Big (\frac{\partial ^2 W \big (x',\varphi (x')\big )}{\partial x_N \partial x_l} +\frac{\partial ^2 W\big (x',\varphi (x')\big )}{\partial x_N\partial x_N} \frac{\partial \varphi (x')}{\partial x_l}\Big ) \frac{\partial \varphi (x')}{\partial x_m}=0. \end{aligned}$$

(B.4)

Let $x'=0$ in (B.4), then we get (B.2). $\square$

C. An example

In this section, we use the above results to the following potential V(x). Let

$$\begin{aligned} F_1(x)= \sum ^N_{j=1}\frac{x_j^2}{a_j^2}-1,~F_2(x)= \sum ^N_{j=1}\big (\frac{x_j}{a_j}-3\big )^2-1, \end{aligned}$$

where $a_j>0$, $a_j\ne a_l$ for $j\ne l$. Let $\Gamma _i$ is defined by $F_i(x)=0$ with $i=1,2$. Take

$$\begin{aligned} V(x)= {\left\{ \begin{array}{ll} F_1^2+1,~\text{ in }~W_1,\\ F_2^2+1,~\text{ in }~W_2,\\ \text{ else },~\text{ in }~{\mathbb {R}}^N\backslash \bigcup ^2_{i=1}W_i. \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} W_i=\Big \{x\in {\mathbb {R}}^N; |F_i|\le \delta _0\Big \}, ~\text{ with } \text{ some } \text{ small } \text{ fixed }~\delta _0>0~\text{ and }~i=1,2. \end{aligned}$$

Lemma C.1

All critical points $\Delta V$ on $\Gamma _1$ are $(\pm a_1,0,\cdots , 0), \cdots , (0,\cdots , 0, \pm a_N)$.

Proof

First, we find

$$\begin{aligned} \nabla V(x) = 2 F \nabla F,~~\Delta V= 2 F \Delta F +2 |\nabla F|^2 =\Big (\sum ^N_{l=1}\frac{x_l^2}{a_l^2}-1\Big ) \sum _{l=1}^N\frac{4}{a_i^2} +\sum _{l=1}^N \frac{8x_l^2}{a_l^4}. \end{aligned}$$

To find a critical point $\Delta V$ on $\Gamma$, we need to study the following equation

$$\begin{aligned} \nabla (\Delta V) =\lambda \nabla F, \end{aligned}$$

for some unknown constant $\lambda$. That is,

$$\begin{aligned} \frac{8x_l}{a_l^2} \sum _{k=1}^N \frac{1}{a_k^2} + \frac{16 x_i}{a_l^4} = \frac{2\lambda x_i}{a_l^2}, \quad l=1,\cdots , N. \end{aligned}$$

Thus, either $x_l=0$, or $\lambda = 4 \displaystyle \sum _{k=1}^N \frac{1}{a_k^2} + \frac{8}{a_l^2}$. If $\lambda =4 \displaystyle \sum _{k=1}^N \frac{1}{a_k^2} + \frac{8}{a_l^2}$, then $x_j=0$ for all $j\ne l$. This shows that all critical points $\Delta V$ on $\Gamma$ are $(\pm a_1,0,\cdots , 0), \cdots , (0,\cdots , 0, \pm a_N)$. $\square$

Without loss of generality, we consider the point $b_1=(0,\cdots , 0, a_N)$. In this case, $\tau _j$ is the $x_j$ direction, $j=1, \cdots , N-1$, and $\nu$ is the $x_N$ direction.

Lemma C.2

If $a_l\ne a_j$ for $l\ne j$. Thus, $b_1$ is non-degenerate on $\Gamma _1$.

Proof

We have

$$\begin{aligned} \frac{\partial ^2 V(b_1)}{\partial x_N^2}=\frac{8}{a_N^2}>0, ~~ \frac{\partial \Delta V(b_1)}{\partial x_N}=\frac{8}{a_N} \sum _{l=1}^N\frac{1}{a_l^2}+ \frac{16}{a_N^3}>0. \end{aligned}$$

On $\Gamma _1$, it holds

$$\begin{aligned} \Delta V(x)=\sum _{l=1}^N \frac{8 x_l^2}{a_l^4}= 8 \Bigl (\sum _{l=1}^{N-1} \frac{ x_l^2}{a_l^4}+\frac{1}{a_N^2} \Bigl ( 1- \sum _{l=1}^{N-1} \frac{ x_l^2}{a_l^2}\Bigr ) \Bigr ). \end{aligned}$$

This gives

$$\begin{aligned} \Big (\frac{\partial ^2 \Delta V(b_1)}{\partial x_lx_j}\Big )_{1\le l,j\le N-1}= \;diag~\Bigl (\frac{16}{a_1^2} \Big (\frac{1}{a_1^2}-\frac{1}{a_N^2}\Big ),\cdots ,\frac{16}{a_{N-1}^2} \Big (\frac{1}{a_{N-1}^2}-\frac{1}{a_N^2}\Big )\Bigr ), \end{aligned}$$

which is non-singular since $a_l\ne a_j$ for $l\ne j$. Thus, $b_1$ is also non-degenerate on $\Gamma _1$. $\square$

Lemma C.3

The matrix

$$\begin{aligned} \Big (\frac{\partial ^2 \Delta V(b_1)}{\partial x_ix_j}\Big )_{1\le i,j\le N-1}+\frac{\partial \Delta V(b_1)}{\partial x_N} diag (\kappa _1, \cdots , \kappa _{N-1}) \end{aligned}$$

(C.1)

is non singular if one of the following conditions holds

• (1) $a_N<a_l$, $l=1,\cdots , N-1$.

• (2) $a_N>a_l$, $l=1,\cdots , N-1$ and all the $a_i$ are close to a constant.

Proof

Near $b_1=(0,\cdots , 0, a_N)$, $\Gamma$ is given by

$$\begin{aligned} x_N= a_N \sqrt{ 1-\sum _{l=1}^{N-1} \frac{x_l^2}{a_l^2}}= a_N -\frac{1}{2} \sum _{l=1}^{N-1} \frac{ a_N x_l^2}{a_l^2}+ O(|x'|^3). \end{aligned}$$

Thus, $\kappa _l = -\frac{ a_N }{a_l^2},\quad l=1,\cdots , N-1$. So we have

$$\begin{aligned} \frac{\partial \Delta V(b_1)}{\partial x_N} \kappa _j(b_1)=-\Bigl (\frac{8}{a_j^2} \sum _{l=1}^N\frac{1}{a_l^2} +\frac{16}{a_j^2a_N^2}\Bigr ). \end{aligned}$$

If $x_0$ is a maximum point of $\Delta V$ on $\Gamma$, that is $a_N<a_l$, $l=1,\cdots , N-1$, then $\Big (\frac{\partial ^2 \Delta V(x_0)}{\partial x_lx_j}\Big )_{1\le l,j\le N-1}$ is negative. Thus, (C.1) is also a negative matrix. On the other hand, if $b_1$ is a minimum point of $\Delta V$ on $\Gamma _1$ and all the $a_j$ are close to a constant, that is $a_N>a_l$, $l=1,\cdots , N-1$, then (C.1) is negative. $\square$

Finally, for $b_2=(0,\cdots ,0,4a_N)\in \Gamma _2$, we have the similar results.