Remarks on the strong maximum principle involving $p$-Laplacian
Remarks on the strong maximum principle involving $p$-Laplacian
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关于涉及$p$-拉普拉斯算子的强极大值原理的评论
DOI：
10.32917/hmj/1487991624
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发表时间：
2016
影响因子：
0.2
通讯作者：
T. Horiuchi
中科院分区：
文献类型：
--
作者：
Xiaojing Liu;T. Horiuchi
现在，为了简单起见，让我们回顾一下关于强最大值原理的一些相关结果，假设Q(T)=|t|p−2t。经典的拉普拉斯强极大值原理断言：如果u是光滑的，且u≥0和−∆u≥0在区域(连通开集)Ω⊂R中，则u≡0或u>；0在Ω中。当−∆u被−∆+a(X)替换为∈L(Ω)，S>；N/2时，同样的结论也成立。后来，这些结果被推广到拟线性算子−∆Pu+a(X)up−1，N/p，这是一个弱Harnack不等式的结果。见[......]和[...]对于p=2和[...]对于p>；1。同一事实的另一个公式是，如果对某一点x∈Ω，u(X)=0，则u≡0在Ω中。然而，当a/∈L，对于任何S；N/p时，类似的结论都不成立。
Now let us recall some relating known results on the strong maximum principle assuming that Q(t) = |t|p−2t for simplicity. The classical strong maximum principle for a Laplacian asserts that if u is smooth, u ≥ 0 and −∆u ≥ 0 in a domain (a connected open set ) Ω ⊂ R , then either u ≡ 0 or u > 0 in Ω. The same conclusion holds when −∆u is replaced by −∆ + a(x) with a ∈ L(Ω), s > N/2. Later these results were extended to the quasilinear operators −∆pu + a(x)up−1 with 1 N/p. These are consequences of a weak Harnack’s inequality. See [ ........] and [... ] for p = 2 and [ ... ] for p > 1. Another formulation of the same fact says that if u(x) = 0 for some point x ∈ Ω, then u ≡ 0 in Ω. However a similar conclusion does not hold when a / ∈ L, for any s > N/p.