AI 中文总结
本文利用伯恩斯坦技术替代原有迭代方法，重新得到四阶Hénon方程的Fazly-Wei-Xu逐点估计，放宽了解的有界性条件，证明系数最优并给出共形度量存在性结果。
AI 中文摘要
在《Analysis and PDE》2015年第8卷第1541-1563页的一项重要工作中，M. Fazly、J. Wei和X. Xu针对四阶Hénon方程Δ²u = |x|^σ u^p（在Rⁿ中，n≥5，σ≥0且隐含接近0，p>(n+4+2σ)/(n-4)）的任意有界正C⁴解u，建立了如下逐点估计：在Rⁿ中，-Δu ≥ [2/(n-4)]·(|∇u|²/u) + √[2/(p+1 - 8/(n(n-4)))]·|x|^(σ/2)·u^((p+1)/2)。他们的论证依赖于标准Moser证明形式的复杂迭代方法。本文提供了一种替代方法，该方法依赖于最大值原理的伯恩斯坦技术。论证的核心是合适辅助函数的选择，这不仅使我们得到相同的逐点估计（该估计甚至对p=(n+4+2σ)/(n-4)也成立），还放宽了解的有界性条件；此外，该逐点不等式实际对任意σ≥0均成立。该估计引出了Rⁿ上具有正标量曲率和Q曲率的共形度量存在性的有趣结果，我们还证明系数2/(n-4)是最优的，从而对上述文献中的问题给出了首个答案，我们的证明似乎更简单且具有构造性。
英文摘要
In a remarkable work published in Analysis and PDE 8 (2015) 1541-1563, M. Fazly, J. Wei, and X. Xu establish the following pointwise estimate \[-Δu \geq \frac 2{n-4} \frac{|\nabla u|^2}u + \sqrt{\frac 2{p+1- \frac 8{n(n-4)}}} |x|^{\frac σ2} u^\frac{p+1}2 \quad \text{in } \mathbf R^n\] for any bounded, positive, $C^4$-solution $u$ to the the fourth-order Hénon equation \[Δ^2 u = |x|^σu^p \quad \text{in } \mathbf R^n\] with $n \geq 5$, $σ\geq 0 $ but implicitly near $0$, and $p>(n+4+2σ)/(n-4)$. Their argument relies on a sophisticated iteration argument in the fashion of the standard Moser proof. In this paper, we provide an alternative approach which relies on the Bernstein technique via the maximum principle. The heart of our argument is a suitable choice of an auxiliary function allowing us not only achieving the same pointwise estimate, which still holds even for $p = (n+4+2σ)/(n-4)$, but also relaxing the boundedness of solutions. In addition, the pointwise inequality actually holds for any $σ\geq 0$. This estimate leads to an interesting consequence of the existence of conformal metric of $\mathbf R^n$ having positive scalar and $Q$ curvatures. We also show that the coefficient $2/(n-4)$ is sharp, hence providing the first answer to the question in the op. cit. paper. Our proof appears to be simpler and constructive.
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