Improved Hardy inequalities in the Grushin plane
The Grushin operator is the operator defined on \(\mathbb{R}^2=\mathbb{R}_{x}\times\mathbb{R}_{y}\) by NEWLINE\[NEWLINE \Delta_{L}=\frac{\partial^2}{\partial x^2}+4x^2\,\frac{\partial^2}{\partial y^2}. NEWLINE\]NEWLINE Denote by \(\nabla_{L}=(\partial_{x},\,2x\partial_{y})\). Then \(\Delta_{L}=\langle \nabla_{L},\,\nabla_{L}\rangle\). Let \(\rho:=\rho(x,\,y)=(x^4+y^2)^{\frac{1}{4}}\). Let \(\Omega\) be a bounded domain in the Grushin plane \(\mathbb{R}^2\) with \(0\in \Omega\). The authors prove the following results:NEWLINENEWLINE1) There exists a constant \(C_{1}>0\) such that for all \(f\in C_{0}^{\infty}(\Omega)\), NEWLINE\[NEWLINE \int_{\Omega}|\nabla_{L}f|^2\,dxdy-\frac{1}{4}\int_{\Omega}\frac{f^2}{\rho^2}\,|\nabla_{L}\rho|^2\,dxdy\geq C_{1}\,\int_{\Omega}|f|^2\,dxdy. NEWLINE\]NEWLINE 2) Let \(1\leq q <2\). There exists a constant \(C_{2}>0\) such that for all \(f\in C_{0}^{\infty}(\Omega)\), NEWLINE\[NEWLINE \int_{\Omega}|\nabla_{L}f|^2\,dxdy-\frac{1}{4}\int_{\Omega}\frac{f^2}{\rho^2}\,|\nabla_{L}\rho|^2\,dxdy\geq C_{2}\,\Big(\int_{\Omega}|\nabla_{L}f|^q\,dxdy\Big)^{\frac{2}{q}}. NEWLINE\]NEWLINE 3) There exists a constant \(C_{3}>0\) such that for all \(f\in C_{0}^{\infty}(\Omega)\), NEWLINE\[NEWLINE \int_{\Omega}|\nabla_{L}f|^2\,dxdy-\frac{1}{4}\int_{\Omega}\frac{f^2}{\rho^2}\,|\nabla_{L}\rho|^2\,dxdy\geq C_{3}\,\Big(\int_{\Omega}|f|^6\,X^4\Big(\frac{\rho(x,\,y)}{D}\Big)dxdy\Big)^\frac{1}{3}, NEWLINE\]NEWLINE where \(D>\sup_{(x,y)\in \Omega}\rho(x,\,y)\) and \(X(s):=(-\ln s)^{-1},\;\;0<s\leq 1\).
scientific article; zbMATH DE number 6052874
scientific article