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HARDY-SOBOLEV-MAZ'YA INEQUALITIES: SYMMETRY AND BREAKING SYMMETRY OF EXTREMAL FUNCTIONS

机译:HARDY-SOBOLEV-MAZ'YA不等式:极值函数的对称性和破裂对称性

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Denote points in R-k x RN-k as pairs xi = (x, y), and assume 2 <= k < N. In this paper, we study the problem {-Delta upsilon = lambda vertical bar x vertical bar(-2) upsilon + vertical bar x vertical bar(-b) upsilon(p-1) in R-N, x not equal 0 upsilon > 0, where p > 2, b = N - pN-2/2 and lambda <= (k-2/2)(2), the Hardy constant. Our results are the following: (i) Let p < 2(N-k+1)*. Then there exists at least an entire cylindrically symmetric solution. (ii) Let p <= 2(N)* and lambda >= 0. Then any solution upsilon is an element of L-p(RN; vertical bar x vertical bar(-b) d xi) is cylindrically symmetric. (iii) Let p < 2(N)* and lambda <= (k-2/2)(2) - k-1/p-2. Then ground state solutions are not cylindrically symmetric, and therefore there exist at least two distinct entire solutions. We prove also similar results for the degenerate problem {-div(vertical bar x vertical bar(-2 alpha)del u = vertical bar x vertical bar(-beta p)u(p-1), x not equal 0 u > 0, namely, for the Euler-Lagrange equations of the Maz'ya inequality with cylindrical weights.
机译:将Rk x RN-k中的点表示为xi =(x,y)对,并假定2 <= k 0,其中p> 2,b = N-pN-2 / 2和lambda <=(k-2 / 2)(2),Hardy常数。我们的结果如下:(i)令p <2(N-k + 1)*。然后至少存在整个圆柱对称的解决方案。 (ii)令p <= 2(N)*且lambda> =0。那么任何解决方案upsilon都是L-p(RN;竖线x竖线(-b)d xi)的元素是圆柱对称的。 (iii)令p <2(N)*,λ<=(k-2 / 2)(2)-k-1 / p-2。然后,基态解不是圆柱对称的,因此存在至少两个不同的整体解。我们还证明了退化问题的相似结果{-div(垂直线x垂直线(-2 alpha)del u =垂直线x垂直线(-beta p)u(p-1),x不等于0 u> 0 ,即具有圆柱权重的Maz'ya不等式的Euler-Lagrange方程。

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