Instability of ground states for the NLS equation with potential on the star graph

Abstract

We study the nonlinear Schrödinger equation with an arbitrary real potential $V(x)\in (L^1+L^\infty )(\Gamma )$ on a star graph $\Gamma$. At the vertex an interaction occurs described by the generalized Kirchhoff condition with strength $-\gamma <0$. We show the existence of ground states $\varphi _{\omega }(x)$ as minimizers of the action functional on the Nehari manifold under additional negativity and decay conditions on V(x). Moreover, for $V(x)=-\dfrac{\beta }{x^{\alpha }}$, in the supercritical case, we prove that the standing waves $e^{i\omega t}\varphi _{\omega }(x)$ are orbitally unstable in $H^{1}(\Gamma )$ when $\omega$ is large enough. Analogous result holds for an arbitrary $\gamma \in {\mathbb {R}}$ when the standing waves have symmetric profile.

Appendix

Below we show some properties of the operator $H_{\gamma , V}$ introduced by (1.2).

Lemma 4.10

Let $\gamma \in {\mathbb {R}}$ and $V(x)=\overline{V(x)}\in L^{1}(\Gamma )+L^{\infty }(\Gamma )$. The quadratic form ${F}_{\gamma , V}$ given by (1.3) is semibounded and closed, and the operator $H_{\gamma , V}$ defined by

$$\begin{aligned} \begin{aligned}&( H_{\gamma , V}v)_{e}=-v''_{e}+V_ev_{e},\\&{\hbox {dom}}( H_{\gamma , V})=\left\{ v\in H^{1}(\Gamma ):\,\,\,-v''_{e}+V_ev_{e}\in L^{2}({\mathbb {R}}^{+}),\,\,\,\sum ^{N}_{e=1}v^{\prime }_{e}(0)=-\gamma v_{1}(0)\right\} . \end{aligned} \end{aligned}$$

is the self-adjoint operator associated with ${F}_{\gamma , V}$ in $L^{2}(\Gamma )$.

Proof

We can write $V(x)=V_{1}(x)+V_{2}(x)$, with $V_{1}\in L^{1}(\Gamma )$ and $V_{2}\in L^{\infty }(\Gamma )$. Thus, using the Gagliardo–Nirenberg inequality (see formula (2.1) in [10]) and the Young inequality, we have

$$\begin{aligned} \left| \int _{\Gamma }V(x)\left| v(x)\right| ^{2}\mathrm{d}x\right|&\le \left\| V_{1}\right\| _1\left\| v\right\| ^{2}_\infty +\left\| V_{2}\right\| _\infty \left\| v\right\| ^{2}_2\nonumber \\&\le C\left\| V_{1}\right\| _1\left\| v'\right\| _2\left\| v\right\| _2+\left\| V_{2}\right\| _\infty \left\| v\right\| ^{2}_2\nonumber \\&\le \varepsilon \left\| v'\right\| ^{2}_2+C_{\varepsilon }\left\| v\right\| ^{2}_2,\,\,\,\varepsilon >0. \end{aligned}$$

(4.18)

Similarly, by the Sobolev embedding, we obtain

$$\begin{aligned} \left| \gamma \left| v_{1}(0)\right| ^{2}\right| \le |\gamma |\Vert v\Vert _\infty ^{2}\le C\Vert v'\Vert _2\left\| v\right\| _2\le \varepsilon \left\| v'\right\| ^{2}_2+C_{\varepsilon }\left\| v\right\| ^{2}_2. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| \gamma \left| v_{1}(0)\right| ^{2}+\int _{\Gamma }V(x)|v(x)|^{2}\mathrm{d}x\right| \le 2\varepsilon \left\| v'\right\| ^{2}_2+C_\varepsilon \left\| v\right\| ^{2}_2,\,\,\text{ for } \text{ every }\,\,\varepsilon >0.\nonumber \\ \end{aligned}$$

(4.19)

Then, by the KLMN theorem [26, Theorem X.17], we infer that the quadratic form ${F}_{\gamma , V}$ is associated with a semibounded self-adjoint operator $T_{\gamma , V}$ defined by (observe that $A=H_{0,0}$ in [26, Theorem X.17], i.e., $V\equiv 0, \gamma =0$)

$$\begin{aligned} \begin{aligned}&{\hbox {dom}}( T_{\gamma , V})=\left\{ u\in H^1(\Gamma ):\,\exists \, y\in L^{2}(\Gamma )\, s.t.\, \forall v\in H^1(\Gamma ),\,\, F_{\gamma , V}(u, v)=(y, v)_2\right\} ,\\&T_{\gamma , V}u=y. \end{aligned} \end{aligned}$$

It is easily seen that ${\hbox {dom}}( H_{\gamma , V})\subseteq {\hbox {dom}}( T_{\gamma , V})$ and $T_{\gamma , V}u= H_{\gamma , V}u, \,\,u\in {\hbox {dom}}( H_{\gamma , V})$. Hence, it is sufficient to prove that ${\hbox {dom}}( T_{\gamma , V})\subseteq {\hbox {dom}}( H_{\gamma , V})$.

Let $\tilde{u}\in {\hbox {dom}}( T_{\gamma , V})$ and $\tilde{v}\in H^1(\Gamma )$, then there exists $\tilde{y}\in L^{2}(\Gamma )$ such that

$$\begin{aligned} F_{\gamma , V}(\tilde{u}, \tilde{v})=\int _\Gamma (\tilde{u}' \overline{\tilde{v}'}+V\tilde{u}\overline{\tilde{v}})\mathrm{d}x-\gamma \tilde{u}_1(0)\overline{\tilde{v}_1(0)}= (\tilde{y}, \tilde{v})_2. \end{aligned}$$

(4.20)

Observe that $\tilde{y}-V\tilde{u}\in L^1_{loc}(\Gamma )$ and set

$$\begin{aligned}z=(z_e)_{e=1}^N,\quad z_e(x)=\int _0^x\left( \tilde{y}_e(t)-V_e(t)\tilde{u}_e(t)\right) \mathrm{d}t.\end{aligned}$$

Suppose now additionally that $\tilde{v}$ has a compact support, then

$$\begin{aligned} \int _\Gamma (\tilde{y}-V\tilde{u})\overline{\tilde{v}}\mathrm{d}x=\int _\Gamma z'\overline{\tilde{v}}\mathrm{d}x=-\overline{\tilde{v}_1(0)}\sum \limits _{e=1}^Nz_e(0)-\int _\Gamma z \overline{\tilde{v}'}\mathrm{d}x. \end{aligned}$$

(4.21)

From (4.20), we deduce

$$\begin{aligned} \int _\Gamma (\tilde{y}-V\tilde{u})\overline{\tilde{v}}\mathrm{d}x=\int _\Gamma \tilde{u}' \overline{\tilde{v}'}\mathrm{d}x-\gamma \tilde{u}_1(0)\overline{\tilde{v}_1(0)}. \end{aligned}$$

(4.22)

Combining (4.21) and (4.22), we get

$$\begin{aligned} \int _\Gamma (\tilde{u}'+z)\overline{\tilde{v}'}\mathrm{d}x+\overline{\tilde{v}_1(0)}\left( -\gamma \tilde{u}_1(0)+\sum \limits _{e=1}^Nz_e(0)\right) =0. \end{aligned}$$

(4.23)

Choose $\tilde{v}=(\tilde{v}_e)_{e=1}^N$ such that $\tilde{v}_1(x)\in C_0^\infty ({\mathbb {R}}^+)$ and $\tilde{v}_2(x)\equiv \ldots \equiv \tilde{v}_N(x)\equiv 0.$ Then we obtain

$$\begin{aligned} \int _0^\infty (\tilde{u}'_1+z_1)\overline{\tilde{v}'_1}\mathrm{d}x=0, \end{aligned}$$

therefore $\tilde{u}'_1+z_1\equiv const\equiv c_1$. We have used that $\tilde{u}'_1+z_1\in \mathrm {Ran}(A)^{\perp },$ where $Av=v'$ with ${\hbox {dom}}(A)=C_0^\infty ({\mathbb {R}}^+)$ in $L^{2}({\mathbb {R}}^+)$. Analogously $\tilde{u}'_e+z_e\equiv const\equiv c_e,\,\, e=2,\ldots , N.$ Finally, from (4.23) we deduce

$$\begin{aligned} \overline{\tilde{v}_1(0)}\left( -\gamma \tilde{u}_1(0)-\sum \limits _{e=1}^N(\tilde{u}'_e(0)+z_e(0))+\sum \limits _{e=1}^Nz_e(0)\right) =0. \end{aligned}$$

Assuming that $\tilde{v}_1(0)\ne 0,$ we arrive at $\sum \limits _{e=1}^N\tilde{u}'_e(0)=-\gamma \tilde{u}_1(0).$ Moreover, $-\tilde{u}''+V\tilde{u}=z'+V\tilde{u}=\tilde{y}-V\tilde{u}+V\tilde{u}=\tilde{y}\in L^{2}(\Gamma ).$ Hence, $\tilde{u}\in {\hbox {dom}}( H_{\gamma , V})$ and ${\hbox {dom}}( T_{\gamma , V})\subseteq {\hbox {dom}}(H_{\gamma , V}).$ $\square$

Lemma 4.11

Suppose that $V(x)=\overline{V(x)}\in L^{2}_{\varepsilon }(\Gamma )+L^\infty (\Gamma )$, i.e., for any $\varepsilon >0$ and $V\in L^{2}_{\varepsilon }(\Gamma )+L^\infty (\Gamma )$ there exists a representation $V=V_1+V_2, \, V_1\in L^{2}(\Gamma ), V_2\in L^{\infty }(\Gamma )$, with $\Vert V_1\Vert _2^{2}\le \varepsilon$. Then, we have

$$\begin{aligned} {\hbox {dom}}( H_{\gamma , V})=\left\{ v\in H^{1}(\Gamma ):\,\,\,v_{e}\in H^{2}({\mathbb {R}}^{+}),\,\,\,\sum ^{N}_{e=1}v^{\prime }_{e}(0)=-\gamma v_{1}(0)\right\} :=D_{H^{2}}.\nonumber \\ \end{aligned}$$

(4.24)

Moreover, for m sufficiently large, $H_{\gamma , V}$-norm $\Vert (H_{\gamma , V}+m)\cdot \Vert _2$ is equivalent to $H^{2}$-norm on $\Gamma .$

Proof

Observe that, by $V(x)\in L^{2}_{\varepsilon }(\Gamma )+L^\infty (\Gamma )$, the Sobolev and the Young inequalities we get

$$\begin{aligned} \Vert Vv\Vert _2^{2}\le \Vert V_1\Vert _2^{2}\Vert v\Vert _\infty ^{2}+\Vert V_2\Vert _\infty ^{2}\Vert v\Vert _2^{2}\le \varepsilon \Vert v\Vert _{H^{2}(\Gamma )}^{2}+C\Vert v\Vert _2^{2} \end{aligned}$$

(4.25)

and

$$\begin{aligned} |(v'', Vv)_2|\le & {} \Vert v''\Vert _2 \Vert Vv\Vert _2\le \Vert v''\Vert _2 \Vert V_1\Vert _2\Vert v\Vert _\infty + \Vert v''\Vert _2 \Vert V_2\Vert _\infty \Vert v\Vert _2 \nonumber \\\le & {} C_1\Vert v''\Vert _2\Vert V_1\Vert _2\Vert v\Vert _{H^{2}(\Gamma )}+C_2\Vert v''\Vert _2 \Vert v\Vert _2 \le \varepsilon \Vert v\Vert ^{2}_{H^{2}(\Gamma )}+\varepsilon \Vert v''\Vert ^{2}_2\nonumber \\+ & {} C_\varepsilon \Vert v\Vert _2^{2} \le 2\varepsilon \Vert v\Vert ^{2}_{H^{2}(\Gamma )}+C_\varepsilon \Vert v\Vert _2^{2}. \end{aligned}$$

(4.26)

It is immediate from (4.25), (4.26) that

$$\begin{aligned} \Vert H_{\gamma , V}v\Vert _2^{2}=\Vert v''\Vert _2^{2}+2{\hbox {Re}}(v'', Vv)_2+ \Vert Vv\Vert _2^{2}\le C_1\Vert v\Vert ^{2}_{H^{2}(\Gamma )}.\end{aligned}$$

(4.27)

And for m sufficiently large, inequalities (4.25) and (4.26) imply

$$\begin{aligned} \Vert H_{\gamma , V}v\Vert _2^{2}+m^{2}\Vert v\Vert _2^{2}= & {} \Vert v''\Vert _2^{2}+2{\hbox {Re}}(v'', Vv)_2+ \Vert Vv\Vert _2^{2}+m^{2}\Vert v\Vert _2^{2}\nonumber \\\ge & {} C_2\Vert v\Vert ^{2}_{H^{2}(\Gamma )}. \end{aligned}$$

(4.28)

Thus, we get (4.24).

The second assertion follows from (4.27),(4.28), and

$$\begin{aligned}&\Vert ( H_{\gamma , V} + m)v\Vert _2^{2}=\Vert H_{\gamma , V}v\Vert _2^{2}+m^{2}\Vert v\Vert _2^{2}+2m(H_{\gamma , V}v,v)_2,\quad \\&|(H_{\gamma , V}v,v)_2|\le \Vert H_{\gamma , V}v\Vert _2\Vert v\Vert _2\le \varepsilon \Vert H_{\gamma , V}v\Vert _2^{2}+C_\varepsilon \Vert v\Vert _2^{2}. \end{aligned}$$

$\square$

Remark 4.12

Observe that there exists potential V(x) satisfying Assumptions 1–4 such that ${\hbox {dom}}( H_{\gamma , V})\ne D_{H^{2}}.$ For example, consider $V(x)=-1/x^\alpha ,\,1/2\le \alpha <1,$ and $N=\gamma =2$, then $v=(e^{-x}, e^{-x})\in D_{H^{2}}$, but

$$\begin{aligned} \Vert H_{\gamma , V}v\Vert _2^{2}=2\Vert -v''_1-\frac{v_1}{x^\alpha }\Vert _2^{2}>2e^{-2\varepsilon }\int _0^\varepsilon \frac{\mathrm{d}x}{x^{2\alpha }}=\infty . \end{aligned}$$

Lemma 4.13

Let $\gamma >0$ and $V(x)=\overline{V(x)}$ satisfy Assumptions 1–3. Then, the following assertions hold.

(i) The number $-\omega _0$ defined by (1.5) is negative.

(ii)  Let also $m>\omega _0$, then $\sqrt{ {F}_{\gamma , V}(v)+m\Vert v\Vert _2^{2}}$ defines a norm equivalent to the $H^1$-norm.

(iii)  The number $-\omega _0$ is the first eigenvalue of $H_{\gamma , V}$. Moreover, it is simple, and there exists the corresponding positive eigenfunction $\psi _0\in {\hbox {dom}}(H_{\gamma , V})$, i.e., $H_{\gamma , V}\psi _0=-\omega _0\psi _0.$

Proof

(i) To show $-\omega _0<0$, observe that

$$\begin{aligned} -\omega _0=\inf \sigma (H_{\gamma ,V})=\inf \left\{ {F}_{\gamma , V}(v)\,:\,v\in H^{1}(\Gamma ),\,\,\left\| v\right\| ^{2}_2=1\right\} . \end{aligned}$$

(4.29)

Consider $v^{\lambda }(x)=\lambda ^{\frac{1}{2}}v(\lambda x)$ with $\lambda >0$. Hence,

$$\begin{aligned} {F}_{\gamma , V}(v^{\lambda })&=\lambda ^{2}\left\| v'\right\| ^{2}_2-\lambda \gamma \left| v_{1}(0)\right| ^{2}+(Vv^{\lambda }, v^{\lambda })_2. \end{aligned}$$

For $\lambda$ small enough, we have ${F}_{\gamma , V}(v^{\lambda })<0$. Finally, $-\omega _0$ is finite since ${F}_{\gamma , V}(v)$ is lower semibounded.

(ii) Let $\varepsilon >0$. Firstly, notice that from (4.19) one easily gets

$$\begin{aligned} {F}_{\gamma , V}(v)+m\Vert v\Vert _2^{2}\le (1+2\varepsilon )\Vert v'\Vert _2^{2}+(C+m)\Vert v\Vert _2^{2}\le C_1\Vert v\Vert ^{2}_{H^1(\Gamma )}.\end{aligned}$$

Secondly, for $\varepsilon$ and $\delta$ sufficiently small,

$$\begin{aligned}&{F}_{\gamma , V}(v)+m\Vert v\Vert _2^{2}= \delta \Vert v'\Vert _2^{2}+(1-\delta )\left( \Vert v'\Vert _2^{2}+\frac{1}{1-\delta }(Vv,v)_2-\frac{\gamma }{1-\delta }|v_1(0)|^{2}\right) \\&\quad +m\Vert v\Vert _2^{2}\ge \delta \Vert v'\Vert _2^{2}-(1+\varepsilon )(1-\delta )\omega _0\Vert v\Vert _2^{2}+m\Vert v\Vert _2^{2}\ge C_2\Vert v\Vert ^{2}_{H^1(\Gamma )}. \end{aligned}$$

Indeed, the family of sesquilinear forms

$$\begin{aligned} {\mathrm {t}}(\kappa )[u,v]=(u',v')_2+\frac{1}{1-\kappa }(Vu,v)_2-\frac{\gamma }{1-\kappa }(u_1(0)\overline{v_1}(0)) \end{aligned}$$

is holomorphic of type (a) in the sense of Kato in the complex neighborhood of zero (see [21, Chapter VII, §4] for the definition and [21, Chapter VI, §1, Example 1.7] for the proof of sectoriality). Using inequality (4.7) in [21, Chapter VII] with $\kappa =\kappa _2=0, \kappa _1=\delta$, we obtain $|{\mathrm {t}}(\delta )[v]-{\mathrm {t}}(0)[v]|\le \varepsilon |{\mathrm {t}}(0)[v]|.$ Hence,

$$\begin{aligned}{\mathrm {t}}(\delta )[v]\ge {\mathrm {t}}(0)[v]-\varepsilon |{\mathrm {t}}(0)[v]|= {F}_{\gamma , V}(v)-\varepsilon | {F}_{\gamma , V}(v)|\ge -(1+\varepsilon )\omega _0\Vert v\Vert _2^{2}. \end{aligned}$$

(iii)   Step 1. Let $\{v_{n}\}$ be a minimizing sequence, that is, ${F}_{\gamma , V}(v_{n})\underset{n\rightarrow \infty }{\longrightarrow } -\omega _{0}$, $\left\| v_{n}\right\| ^{2}_2=1$ for all $n\in \mathbb {N}$. From (ii), we deduce that $\{v_{n}\}$ is bounded in $H^{1}(\Gamma )$. Then, there exist a subsequence $\{v_{n_k}\}$ of $\{v_n\}$ and $v_0\in H^{1}(\Gamma )$ such that $\{v_{n_k}\}$ converges weakly to $v_0$ in $H^1(\Gamma )$. Observe that, by the weak lower semicontinuity of $L^{2}$-norm and ${F}_{\gamma , V}(\cdot )$, we get $\left\| v_0\right\| _2\le 1$ and

$$\begin{aligned} F_{\gamma , V}(v_0)\le \lim _{k\rightarrow \infty }F_{\gamma , V}(v_{n_k})=-\omega _{0}<0. \end{aligned}$$

We have $\left\| v_0\right\| _{2}=1$, since, otherwise, there would exist $\lambda >1$ such that $\left\| \lambda v_0\right\| _{2}=1$ and $F_{\gamma , V}(\lambda v_0)=\lambda ^{2}F_{\gamma , V}(v_0)<-\omega _0$, which is a contradiction. Consequently, $v_0$ is a minimizer for (4.29).

Let $\psi _{0}=\left| v_0\right|$, then $\psi _{0}\ge 0$ on $\Gamma$ and $\left\| \psi _{0}\right\| ^{2}_2=\left\| v_{0}\right\| ^{2}_2=1$. Notice that $\left\| \psi '_{0}\right\| ^{2}_2\le \left\| v'_0\right\| ^{2}_2,$ therefore ${F}_{\gamma , V}(\psi _{0})\le {F}_{\gamma , V}(v_0)$. Then, $\psi _{0}$ is a minimizer of (4.29). This implies the existence of the Lagrange multiplier $-\mu$ such that

$$\begin{aligned} F'_{\gamma , V}(\psi _0)=-\mu Q'(\psi _0),\quad Q(v)=\Vert v\Vert _2^{2}. \end{aligned}$$

Repeating the arguments from the proof of [2, Theorem 4], we get $\psi _0\in {\hbox {dom}}(H_{\gamma , V})$ and

$$\begin{aligned} H_{\gamma , V}\psi _0=-\mu \psi _0. \end{aligned}$$

Multiplying the above equation by $\overline{\psi _0}$ and integrating, we conclude $\mu =\omega _0.$ Recalling that $V(x)\le 0$ a.e. on $\Gamma$, and arguing as in the proof of Proposition 1.1, one can show that $\psi _0>0$ on $\Gamma .$ Notice that one needs to apply [28, Theorem 1] with $\beta (s)=\omega _0 s.$

Step 2. Suppose that $u_0$ is a nonnegative solution of

$$\begin{aligned} H_{\gamma , V}u_0=-\omega _0 u_0. \end{aligned}$$

(4.30)

Let us show that there exists $C>0$ such that $u_0(x)=C\psi _0(x)$. Assume that this is false. Then, there exists $C>0$ such that $\widetilde{u}_0(x)=u_0(x)-C\psi _0(x)$ takes both positive and negative values. We have $H_{\gamma , V}\widetilde{u}_0=-\omega _0 \widetilde{u}_0;$ consequently, $\widetilde{v}_0=\widetilde{u}_0/\Vert \widetilde{u}_0\Vert _2$ is the minimizer of (4.29). Arguing as in Step 1, one can show that $|\widetilde{v}_0|$ is also a minimizer and $|\widetilde{v}_0|>0$. Therefore, $\widetilde{u}_0(x)$ has a constant sign. This is a contradiction.

Suppose now that $u_0$ is an arbitrary solution to (4.30) such that $\Vert u_0\Vert _2^{2}=1$ (that is, $u_0$ is a minimizer of (4.29)). Define $w_0=|{\hbox {Re}}u_0|+i|{\hbox {Im}}u_0|,$ then $|w_0|=|u_0|$ and $|w'_0|=|u'_0|$; consequently, $F_{\gamma , V}(u_0)=F_{\gamma , V}(w_0)$ and $\Vert w_0\Vert _2^{2}=1.$ Therefore, $w_0$ is a minimizer of (4.29). This implies that $w_0$ satisfies (4.30), and, in particular, $|{\hbox {Re}}u_0|$ and $|{\hbox {Im}}u_0|$ satisfy (4.30). Thus, $|{\hbox {Re}}u_0|=C_1 \psi _0$ and $|{\hbox {Im}}u_0|=C_2\psi _0,\, C_1,C_2>0$; consequently, ${\hbox {Re}}u_0=\widetilde{C}_1 \psi _0$ and ${\hbox {Im}}u_0=\widetilde{C}_2\psi _0,\, \widetilde{C}_1,\widetilde{C}_2\in {\mathbb {R}},$ since ${\hbox {Re}}u_0$ and ${\hbox {Im}}u_0$ do not change the sign. Finally, $u_0=\widetilde{C}_1\psi _0+i\widetilde{C}_2\psi _0=\widetilde{C}\psi _0, \, \widetilde{C}\in {\mathbb {C}},$ and therefore, $-\omega _0$ is simple.

$\square$