Concave power solutions of the dominative p-Laplace equation

In this paper, we study properties of solutions of the Dominative p-Laplace equation with homogeneous Dirichlet boundary conditions in a bounded convex domain Ω\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega $$\end{document}. For the equation -Dpu=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-{\mathcal {D}}_p u= 1$$\end{document}, we show that u\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sqrt{u}$$\end{document} is concave, and for the eigenvalue problem Dpu+λu=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {D}}_p u + \lambda u=0$$\end{document}, we show that logu\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\log {u}$$\end{document} is concave.


Introduction
The Dominative p-Laplace operator was defined as D p u := Δu + (p − 2)λ max (D 2 u), p ≥ 2, by Brustad in [3] and later studied in [4]. See also [6] for a stochastic interpretation and a game-theoretic approach of the equation. Here, λ max denotes the largest eigenvalue of the Hessian matrix We shall study the two equations −D p u = 1 and D p u + λu = 0 in a bounded convex domain Ω ⊂ R n . The positive solutions with zero boundary values have the property for −D p u = 1 that √ u is concave, see Theorem 1.1 below. In Theorem 1.2 we show that for D p u + λu = 0, log u is concave. Problems related to concave solutions have been studied for p-Laplace type equations, and we give a quick review of the results. The operator is closely related to the normalized p-Laplace operator, Δ N p u = |∇u| 2−p div |∇u| p−2 ∇u , F. Høeg NoDEA which describes a Tug-of-war game with noise, see [18]. Due to this, the operator has been studied extensively over the last 15 years, and we refer to [2,10,11] for an introduction and some regularity results. The solutions are weak and appear in the form of viscosity solutions and we refer to [8] for an introduction of viscosity solutions. If u is a solution of the problem in a bounded convex domain Ω ⊂ R n , one can show that √ u is concave. This problem, including more complex right-hand sides, was studied in the 1970's and 1980's by [13][14][15]17]. For n = 1 and n = 2, a brute force calculation shows that √ u is concave. For n ≥ 3 the proofs are more complicated. For the ordinary p-Laplacian, [19] showed that u p−1 p is concave. One should note that simply setting p = 2 does not simplify the proof. Thus, the papers [14,15] are still of great value. For the infinity Laplacian, Δ ∞ u = D 2 u∇u, ∇u , [7] showed that u 3 4 is concave. Our result for the Dominative p-Laplace equation can be formulated in the following theorem. We say that Ω satisfies the interior sphere condition if for all y ∈ ∂Ω there is an x ∈ Ω and an open ball B r (x) such that B r (x) ⊂ Ω and y ∈ ∂B r (x).
in a bounded convex domain Ω ⊂ R n which satisfies the interior sphere condition. Then √ u is concave.
Further, we study the eigenvalue problem and give the following result.
This equation has the solution u = 0, which is obviously already concave. This is better than the square root being concave, so for p = ∞ a stronger result is obtained. (For a less trivial result, another normalization with p is needed.) For the Helmholtz equation Δu + λu = 0, the problem related to concave logarithmic solutions has been studied in [5,9,15]. The nonlinear eigenvalue problem associated with the p-Laplace equation has been studied for example in [16,19]. In [19], Sakaguchi showed that log u is a concave function.

Preliminaries and notation
The gradient of a function f : and its Hessian matrix is We will use the operator and if applied to a matrix X ∈ S n , we use Also, the normalized p-Laplace operator is referred to, Viscosity solutions The Dominative p-Laplace operator is uniformly elliptic. Therefore, it is convenient to use viscosity solutions as a notion of weak solutions. Throughout the text, we always keep p ≥ 2. In the definition below, g is assumed to be continuous in all variables.
if it is a viscosity sub-and supersolution of −D p u = g(x, u, ∇u) and u = 0 on ∂Ω.
When defining viscosity solutions to Δ N p u = g(x, u, ∇u), one has to be careful at points where the gradient vanishes.
if it is a viscosity sub-and supersolution of −Δ N p u = 1 and u = 0 on ∂Ω. We also need an equivalent definition of viscosity solutions using the suband superjets. For functions u : Ω → R n they are given by if it is a viscosity sub-and supersolution of −D p u = g(x, u, ∇u) and u = 0 on ∂Ω.
We mention some results in [12] obtained for the normalized p-Laplace equation, which we will use together with the relationship between the nor-    = g(x, u, ∇u).
Hence, u is a viscosity supersolution of −Δ N p u = g(x, u, ∇u). If u is a viscosity subsolution of −Δ N p u = g(x, u, ∇u) and u − φ obtains a maximum at x ∈ Ω, On the other hand, if ∇φ(x) = 0, −D p φ ≤ g(x, u, 0) by definition. Hence, u is a viscosity supersolution of −D p u = g(x, u, ∇u).
The following Lemma will be applied in the proof of the concavity, and it relies on the fact that the mapping (q, A) → q, A −1 q is convex in S + for each q ∈ R n . Here, S + consists of the symmetric positive definite matrices.
Proof. In the appendix of [1] it was shown that (q, A) → q, A −1 q is convex, With these choices, Using inequality (2.1) we find .
By induction, the inequality in Lemma 2.7 holds.

Convex envelope
The convex envelope of a function u : Ω → R n is defined as We are interested in the convex envelope of the square root, v = − √ u, and we have the following result on what happens near the boundary of Ω. .
Proof. Since u is, in particular a viscosity supersolution to −Δ N p u = 1, Lemma 3.2 in [12] gives the result.
A second application of Lemma 2.6 shows that u is a viscosity subsolution of We now focus our attention on the convex envelope, v * * . It turns out that v * * is a viscosity supersolution to the same equation as v. with v * * = 0 on ∂Ω.
Proof. According to ([1], Lemma 4) we have v * * = v = 0 on ∂Ω so we only have to show that v * * is a viscosity supersolution. To this end, let (q, A) ∈ J 2,− v * * (x). By Lemma 2.8 we can decompose x in a convex combination of interior points, with x 1 , . . . , x k ∈ Ω. By Proposition 1 in [1] there are A 1 , . . . , A k ∈ S + such that (q, A i ) ∈J 2,− v(x i ) and for all > 0 small enough. Since v is a viscosity supersolution, Multiplying both sides with μ i v(x i ) and a summation i = 1, . . . , k yields .