Multiplicity of normalized solutions for dipolar Gross-Pitaevskii equation with a mass subcritical perturbation

In this paper, we study the multiplicity of normalized solutions to the following Gross-Pitaevskii equation arising in trapped dipolar quantum gases: \(\begin{aligned} -\frac{1}{2}\Delta u+\lambda _{1}|u|^2u+\lambda _2(K*|u|^2)u-h(\varepsilon x)|u|^{p-2}u+\mu u=0\,\,\,\,{\textrm{in}}\,\,\mathbb {R}^3 \end{aligned}\) $$\begin{array}{c}\hfill -\frac{1}{2}\mathrm{\Delta }u+{\lambda }_1{|u|}^{2}u+{\lambda }_2{(K\ast |u|}^2{)u-h(\epsilon x)|u|}^{p-2}u+\mu u=0\text{in}{\mathbb{R}}^3\end{array}$$ under the mass constraint \(\begin{aligned} \int _{\mathbb {R}^3}|u|^2dx=c^2, \end{aligned}\) $$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^3}{|u|}^{2}dx=c^2,\end{array}$$ where \((\lambda _1,\lambda _2)\in \mathbb {R}^2, c, \varepsilon >0\) $({\lambda }_1,{\lambda }_2)\in {\mathbb{R}}^2,c,\epsilon >0$ , \(2<p<\frac{10}{3}, h:\mathbb {R}^3\rightarrow [0,\infty )\) $2<p<\frac{10}{3},h:{\mathbb{R}}^3\to [0,\infty )$ is a continuous function and K is the classical dipole-dipole interaction kernel. In the unstable regime, that is, \((\lambda _1,\lambda _2)\in \mathcal {D}=\big \{(\lambda _1, \lambda _2)\in \mathbb {R}^2: \lambda _1<\frac{4\pi }{3}\lambda _2\le 0\ {\textrm{or}}\ \lambda _1<-\frac{8\pi }{3}\lambda _2\le 0\big \},\) $({\lambda }_1,{\lambda }_2)\in \mathcal{D}=\{({\lambda }_1,{\lambda }_2)\in {\mathbb{R}}^2:{\lambda }_1<\frac{4\pi }{3}{\lambda }_2\le 0\text{or}{\lambda }_1<-\frac{8\pi }{3}{\lambda }_2\le 0\},$ we establish the existence of a normalized solution that corresponds to an interior local minimizer of the constraint functional provided that \(\inf \limits _{x\in \mathbb {R}^3}h(x)=\lim \limits _{|x|\rightarrow \infty }h(x)=h_{\infty }>0.\) $\underset{x\in {\mathbb{R}}^3}{inf}h(x)=\underset{|x|\to \infty }{lim}h(x)=h_{\infty }>0.$ Moreover, if the origin is a maximum point of function h, we obtain that the number of normalized solutions is not less than the number of global maximum points of h when \(\varepsilon >0\) $\epsilon >0$ is sufficiently small.