Non-variational radial solutions to a singular elliptic Dirichlet problem on the disk with Leray potential

Abstract

We prove the existence of non-$H^1_0(B_1(0))$ solutions of a class of singular elliptic problem in two dimensions.

Appendices

Appendix

A remark for Shen’s inequality

Let $\Omega$ be a bounded domain in $\mathbb {R}^N(N\ge 3)$ with smooth boundary and $0\in \Omega$. The classical Hardy inequality asserts that

$$\begin{aligned} \frac{(N-2)^2}{4}\int _\Omega \frac{u^2}{|x|^2}dx\le \int _\Omega |\nabla u|^2dx,\ \ \ \forall u\in H^1_0(\Omega ). \end{aligned}$$

This inequality and its various improvements are used in many contexts, see [2, 5, 6, 25] for more details. Brezis and Vázquez [5] obtained the following improved Hardy inequality

$$\begin{aligned} \lambda (\Omega )\int _\Omega u^2dx+\frac{(N-2)^2}{4}\int _\Omega \frac{u^2}{|x|^2}dx\le \int _\Omega |\nabla u|^2dx,\ \ \ \forall u\in H^1_0(\Omega ), \end{aligned}$$

where $\lambda (\Omega )$ is given by

$$\begin{aligned} \lambda (\Omega )=z^2_0\left( \frac{\omega _N}{|\Omega |}\right) ^{\frac{2}{N}}, \end{aligned}$$

$z_0\approx 2.4048$ is the first zero of the Bessel function $J_0$, $\omega _N$ and $|\Omega |$ denote the volume of the unit ball and $\Omega$, respectively. Now that we have the inequality (1.3), one may ask what is the analogy if $N=2$. Shen et al. [22] suggested the following inequality: there exists $\lambda _1\ge 0$ such that

$$\begin{aligned} \lambda _1\int _{B_1(0)} u^2dx+\frac{1}{4}\int _{B_1(0)} V(x)u^2dx\le \int _{B_1(0)}|\nabla u|^2dx,\ \ \ \forall u\in H^1_0(B_1(0)). \end{aligned}$$

We prove that

Theorem A.1

$$\begin{aligned} \lambda _1=\inf _{u\in \mathcal {U}}\int _{B_1(0)}\left( |\nabla u|^2-\frac{1}{4}V(x)u^2\right) dx=0, \end{aligned}$$

where

$$\begin{aligned} \mathcal {U}=\left\{ u\in H^1_0(B_1(0)\Big |\int _{B_1(0)} u^2dx=1\right\} . \end{aligned}$$

Proof

Define $\hat{u}_\varepsilon : [0,1]\rightarrow \mathbb {R}$ as follows

$$\begin{aligned} \hat{u}_\varepsilon (r)=\left\{ \begin{array}{ll} \displaystyle \left( \ln \frac{1}{\varepsilon }\right) ^{\beta _1} \left( \ln \frac{1}{r}\right) ^{\frac{1}{2}-\beta _1}, &{}0<r<\varepsilon ,\\ \displaystyle \sqrt{\ln \frac{1}{r}},&{}\varepsilon<r<1-\varepsilon ,\\ \displaystyle \left( \ln \frac{1}{1-\varepsilon }\right) ^{-\beta _2} \left( \ln \frac{1}{r}\right) ^{\frac{1}{2}+\beta _2}, &{}1-\varepsilon <r\le 1, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} u_\varepsilon (x):=\hat{u}_\varepsilon (|x|),\ \ \ \forall x\in B_1(0). \end{aligned}$$

where $\beta _1,\beta _2\in (0,1)$.

Direct computations yield that

$$\begin{aligned} (\hat{u}_\varepsilon )_r=\left\{ \begin{array}{ll} \displaystyle \left( \ln \frac{1}{\varepsilon }\right) ^{\beta _1}\left( \frac{1}{2}-\beta _1\right) \left( \ln \frac{1}{r}\right) ^{-\frac{1}{2}-\beta _1}\left( -\frac{1}{r}\right) , &{}0<r<\varepsilon ,\\ \displaystyle \frac{1}{2}\left( \sqrt{\ln \frac{1}{r}}\right) ^{-\frac{1}{2}}\left( -\frac{1}{r}\right) , &{}\varepsilon<r<1-\varepsilon ,\\ \displaystyle \left( \ln \frac{1}{1-\varepsilon }\right) ^{-\beta _2}\left( \frac{1}{2}+\beta _2\right) \left( \ln \frac{1}{r}\right) ^{-\frac{1}{2}+\beta _2}\left( -\frac{1}{r}\right) , &{}1-\varepsilon <r\le 1. \end{array} \right. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} \int ^1_0\frac{\hat{u}_\varepsilon ^2}{r\ln ^2\frac{1}{r}}dr\nonumber \\{} & {} \quad =\int ^\varepsilon _0\left( \ln \frac{1}{\varepsilon }\right) ^{2\beta _1} \left( \ln \frac{1}{r}\right) ^{-1-2\beta _1}\frac{1}{r}dr +\int ^{1-\varepsilon }_\varepsilon \frac{1}{r}\left( \ln \frac{1}{r}\right) ^{-1}dr \nonumber \\{} & {} \qquad +\int ^1_{1-\varepsilon }\left( \ln \frac{1}{1-\varepsilon }\right) ^{-2\beta _2} \frac{1}{r}\left( \ln \frac{1}{r}\right) ^{-1+2\beta _2}dr\nonumber \\= & {} \frac{1}{2\beta _1}+\ln \frac{\ln \frac{1}{\varepsilon }}{\ln \frac{1}{1-\varepsilon }} +\frac{1}{2\beta _2}, \end{aligned}$$

(A.1)

and

$$\begin{aligned} \int ^\varepsilon _0r[(\hat{u}_\varepsilon )_r]^2dr= & {} \left( \ln \frac{1}{\varepsilon }\right) ^{2\beta _1} \left( \frac{1}{2}-\beta _1\right) ^2\frac{1}{2\beta _1} \left( \ln \frac{1}{r}\right) ^{-2\beta _1}\Big |^\varepsilon _0\nonumber \\ {}{} & {} =\frac{1}{2\beta _1} \left( \frac{1}{2}-\beta _1\right) ^2\end{aligned}$$

(A.2)

$$\begin{aligned} \int ^{1-\varepsilon }_\varepsilon r[(\hat{u}_\varepsilon )_r]^2dr= & {} -\frac{1}{4}\ln \ln \frac{1}{r}\Big |^{1-\varepsilon }_\varepsilon =\frac{1}{4}\ln \frac{\ln \frac{1}{\varepsilon }}{\ln \frac{1}{1-\varepsilon }},\end{aligned}$$

(A.3)

$$\begin{aligned} \int ^1_{1-\varepsilon }r[(\hat{u}_\varepsilon )_r]^2dr= & {} \left( \ln \frac{1}{1-\varepsilon }\right) ^{-2\beta _2}\left( \frac{1}{2}+\beta _2\right) ^2 \frac{-1}{2\beta _1} \left( \ln \frac{1}{r}\right) ^{2\beta _2}\Big |^1_{1-\varepsilon }\nonumber \\= & {} \frac{1}{2\beta _1} \left( \frac{1}{2}+\beta _2\right) ^2. \end{aligned}$$

(A.4)

By (A.1)-(A.4), we have

$$\begin{aligned} \int ^1_0\left[ r[(\hat{u}_\varepsilon )_r]^2-\frac{\hat{u}^2_\varepsilon }{4r\ln ^2\frac{1}{r}}\right] dr= \frac{\left( \frac{1}{2}-\beta _1\right) ^2}{2\beta _1} +\frac{\left( \frac{1}{2}+\beta _2\right) ^2}{2\beta _2} -\frac{1}{8\beta _1}-\frac{1}{8\beta _2} =\frac{\beta _1+\beta _2}{2} \end{aligned}$$

(A.5)

and

$$\begin{aligned} \int ^1_0r\hat{u}^2_\varepsilon dr= & {} \int ^\varepsilon _0\left( \ln \frac{1}{\varepsilon }\right) ^{2\beta _1}r \left( \ln \frac{1}{r}\right) ^{1-2\beta _1}dr+\int ^{1-\varepsilon }_\varepsilon r\ln \frac{1}{r}dr\nonumber \\{} & {} \quad +\int ^1_{1-\varepsilon }r\left( \ln \frac{1}{1-\varepsilon }\right) ^{-2\beta _2} \left( \ln \frac{1}{r}\right) ^{1+2\beta _2}dr\nonumber \\{} & {} =\frac{1}{4}+o(\varepsilon ). \end{aligned}$$

(A.6)

By (A.5) and (A.6), we get

$$\begin{aligned} \frac{\displaystyle \int _{B_1(0)}|\nabla u_\varepsilon |^2dx-\frac{1}{4}\int _{B_1(0)}V(x)u^2_\varepsilon dx}{\displaystyle \int _{B_1(0)}u^2_\varepsilon dx}= & {} \frac{\displaystyle \int ^1_0r(\hat{u}_\varepsilon )^2_rdr -\frac{1}{4}\int ^1_0\frac{\hat{u}^2_\varepsilon }{r\ln ^2\frac{1}{r}}dr}{\displaystyle \int ^1_0r\hat{u}^2_\varepsilon dr}\\= & {} \frac{\displaystyle \frac{1}{2}(\beta _1+\beta _2)}{\displaystyle \frac{1}{4}+o(\varepsilon )} \rightarrow 0\ \ \ \text {as}\ \beta _1,\beta _2\rightarrow 0. \end{aligned}$$

$\square$