Ground States for Logarithmic Schrödinger Equations on Locally Finite Graphs

Abstract

In this paper, we study the following logarithmic Schrödinger equation:

$$\begin{aligned} -\Delta u+a(x)u=u\log u^2\ \ \ \ \text{ in } V, \end{aligned}$$

where $\Delta$ is the graph Laplacian, $G=(V,E)$ is a connected locally finite graph, the potential $a: V\rightarrow {\mathbb {R}}$ is bounded from below and may change sign. We first establish two Sobolev compact embedding theorems in the case when different assumptions are imposed on a(x). This leads to two kinds of associated energy functionals, one of which is not well defined under the logarithmic nonlinearity, while the other is $C^1$. The existence of ground state solutions are then obtained by using the Nehari manifold method and the mountain pass theorem respectively.

Appendix

In this appendix, we present two examples to show that there exists $u\in H^1(V)$ but $\int _{V}u^2\log u^2dx=-\infty$.

Example 1

We consider a connected locally finite graph $G=(V,E)$ such that $V:={\mathbb {N}} \cup \{0\}$, where ${\mathbb {N}}$ denotes the set of natural numbers. Fixed $x_0=0\in V$. Let

$$\begin{aligned} u(x)= {\left\{ \begin{array}{ll} \left( \vert x\vert \log \vert x\vert \right) ^{-1}, ~ &{}\vert x\vert \ge 3,\\ 0, ~ &{}\vert x\vert \le 2, \end{array}\right. } \end{aligned}$$

where $\vert x\vert :=d(x,x_0)$, the measure

$$\begin{aligned} \mu (x)= {\left\{ \begin{array}{ll} x, ~ &{}x\ge 1,\\ 1, ~ &{}x=0. \end{array}\right. } \end{aligned}$$

For simplicity, we assume that $\omega _{xy}=1$.

Let’s recall the following fact

$$\begin{aligned} \sum _{n=2}^{\infty }\frac{1}{n(\log n)^p} {\left\{ \begin{array}{ll} <\infty ,~~~&{} \text{ if } p>1,\\ =\infty ,~~~&{} \text{ if } 0\le p\le 1. \end{array}\right. }\end{aligned}$$

First of all, we prove that $u\in H^1(V)$. By the definition of u and $\mu$, we have

$$\begin{aligned} \int _V u^2d\mu =\sum _{x\in {\mathbb {N}}}\mu (x)u^2(x)=\sum _{x\ge 3}\frac{x}{x^2(\log x)^2}=\sum _{x\ge 3}\frac{1}{x(\log x)^2}<\infty \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _V\vert \nabla u\vert ^2d\mu =\frac{1}{2}\sum _{x\in {\mathbb {N}}}\sum _{y\sim x}\left( u(y)-u(x)\right) ^2\\&\quad =\frac{1}{2}\left( u(3)-u(2)\right) ^2+\frac{1}{2}\left( u(2)-u(3)\right) ^2+\frac{1}{2}\left( u(4)-u(3)\right) ^2\\&\qquad +\frac{1}{2}\sum _{x\ge 4}\sum _{y\sim x}\left( u(y)-u(x)\right) ^2\\&\quad =\frac{1}{3^2(\log 3)^2}+\sum _{x\ge 3}\left( \frac{1}{(x+1)\log (x+1)}-\frac{1}{x\log x}\right) ^2\\&\quad \le \frac{1}{3^2(\log 3)^2}+\sum _{x\ge 3}\left( \frac{1}{(x+1)^2\left( \log (x+1)\right) ^2}+\frac{1}{x^2(\log x)^2}\right) \\&\quad \le \frac{1}{3^2(\log 3)^2}+\sum _{x\ge 3}\frac{2}{x^2(\log x)^2}\\&\quad <\infty . \end{aligned}\end{aligned}$$

Next, we prove that $\int _Vu^2\log u^2d\mu =-\infty$. Note that

$$\begin{aligned} u^2(x)\log u^2(x)=-\left( \frac{2}{x^2\log x}+\frac{2\log (\log x)}{x^2(\log x)^2}\right) , \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&\int _Vu^2\log u^2d\mu =\sum _{x\in {\mathbb {N}}}\mu (x)u^2(x)\log u^2(x)\\&\quad =\sum _{x\ge 3}x\cdot u^2(x)\log u^2(x)\\&\quad =-\left( \sum _{x\ge 3}\frac{2}{x\log x}+\sum _{x\ge 3}\frac{2\log (\log x)}{x(\log x)^2}\right) \\&\quad =-(I+II). \end{aligned}\end{aligned}$$

Since $II>0$, it suffices to prove the following fact

$$\begin{aligned} \sum _{x\ge 3}\frac{2}{x\log x}=\infty . \end{aligned}$$

But this is obvious, and thus we completes the proof.

Example 2

Assume that $G=(V,E)$ is a connected locally finite graph, and there exist some constants $\mu _{\min },\ \mu _{\max }>0$ such that $0<\mu _{\min }\le \mu (x)\le \mu _{\max }$ for all $x\in V$.

Fixed $x_0\in V$. Let

$$\begin{aligned} u(x)= {\left\{ \begin{array}{ll} \left( \vert x\vert ^{\frac{1}{2}}\log \vert x\vert \right) ^{-1}, ~ &{}\vert x\vert \ge 3,\\ 0, ~ &{}\vert x\vert \le 2, \end{array}\right. } \end{aligned}$$

where $\vert x\vert :=d(x,x_0)$. Then $u(x)\in H^1(V)$ and $\int _Vu^2\log u^2d\mu =-\infty$. In fact, from the definition of u and $\mu$, we have

$$\begin{aligned} \int _V u^2d\mu =\sum _{x\in V}\mu (x)u^2(x)=\sum _{\vert x\vert \ge 3}\frac{\mu (x)}{\vert x\vert \left( \log \vert x\vert \right) ^2}\le \mu _{\max }\sum _{\vert x\vert \ge 3}\frac{1}{\vert x\vert \left( \log \vert x\vert \right) ^2}<\infty \end{aligned}$$

and

$$\begin{aligned}\begin{aligned}&\int _V\vert \nabla u\vert ^2d\mu \\&\quad =\frac{1}{2}\sum _{x\in V}\sum _{y\sim x}\omega _{xy}\left( u(y)-u(x)\right) ^2\\&\quad =\frac{1}{2}\sum _{\vert x\vert =2}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =3 \end{array}}\frac{\omega _{xy}}{3(\log 3)^2}+\frac{1}{2}\sum _{\vert x\vert =3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =2 \end{array}}\frac{\omega _{xy}}{3(\log 3)^2}\\&\qquad +\frac{1}{2}\sum _{\vert x\vert =3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =4 \end{array}}\omega _{xy}\left( \frac{1}{\sqrt{4}\log 4}-\frac{1}{\sqrt{3}\log 3}\right) ^2+\frac{1}{2}\sum _{\vert x\vert \ge 4}\sum _{y\sim x}\omega _{xy}\left( u(y)-u(x)\right) ^2\\&\quad =\sum _{\vert x\vert =3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =2 \end{array}}\frac{\omega _{xy}}{3(\log 3)^2}+\sum _{\vert x\vert \ge 3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =\vert x\vert +1 \end{array}}\omega _{xy}\left( \frac{1}{\left( \vert x\vert +1\right) ^{\frac{1}{2}}\log \left( \vert x\vert +1\right) }-\frac{1}{\vert x\vert ^{\frac{1}{2}}\log \vert x\vert }\right) ^2\\&\quad \le \sum _{\vert x\vert =3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =2 \end{array}}\frac{\omega _{xy}}{3(\log 3)^2}+\omega _{\max }\sum _{\vert x\vert \ge 3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =\vert x\vert +1 \end{array}}\left( \frac{1}{\left( \vert x\vert +1\right) \left( \log \left( \vert x\vert +1\right) \right) ^2}+\frac{1}{\vert x\vert \left( \log \vert x\vert \right) ^2}\right) \\&\quad \le \sum _{\vert x\vert =3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =2 \end{array}}\frac{\omega _{xy}}{3(\log 3)^2}+\omega _{\max }\sum _{\vert x\vert \ge 3}\sum _{\begin{array}{c} y\sim x \\ \vert y\vert =\vert x\vert +1 \end{array}}\frac{2}{\vert x\vert \left( \log \vert x\vert \right) ^2}\\&\quad <\infty . \end{aligned}\end{aligned}$$

In what follows, we prove that $\int _Vu^2\log u^2d\mu =-\infty$. By direct calculations, we have

$$\begin{aligned}\begin{aligned} \int _Vu^2\log u^2d\mu&=\sum _{x\in V}\mu (x)u^2(x)\log u^2(x)\\&=\sum _{\vert x\vert \ge 3}\mu (x)u^2(x)\log u^2(x)\\&=-\left( \sum _{\vert x\vert \ge 3}\frac{\mu (x)}{\vert x\vert \log \vert x\vert }+\sum _{\vert x\vert \ge 3}\frac{2\mu (x)\log \left( \log \vert x\vert \right) }{\vert x\vert \left( \log \vert x\vert \right) ^2}\right) \\&=-\infty . \end{aligned}\end{aligned}$$