1 Introduction
Throughout this paper, we assume $\Omega \subset \mathbb R^n$ to be a bounded strictly convex domain with a smooth boundary $\partial \Omega $ . For $x\in \Omega $ , we write $x=(x^1,\ldots ,x^n)$ . We use subscripts to denote partial differentiation. For example, we write $u_i=\frac {\partial u}{\partial x^i}$ , $u_{ij}=\frac {\partial ^2 u}{\partial x^i\partial x^j}$ , etc. We consider smooth strictly convex functions u defined in $\Omega $ . The Monge–Ampère operator can be written as
where $D^2u$ denotes the Hessian matrix of the function u, det $(D^2u)$ is the determinant of $D^2u$ , and $T_{(n-1)}=T_{(n-1)}(D^2u)$ is the adjoint matrix of $D^2u$ (i.e., the cofactor matrix of $D^2u$ ). Here and in what follows, the summation convention from $1$ to n over repeated indices is in effect.
A useful equation is the following:
Moreover, the tensor $\bigg [T_{(n-1)}^{ij}(D^2u)\bigg ]$ is symmetric and divergence-free, that is,
If I denotes the $n\times n$ identity matrix, we have
The proof of these results can be found in [Reference Reilly14, Reference Reilly15].
Let $g: [0,\infty )\rightarrow (0,\infty )$ be a smooth real function satisfying
We also suppose that $g(0)=G(0)>0$ (i.e., positive and finite). Note that
Therefore, the function $g(s^2)s$ is positive and strictly increasing for $s>0$ . A typical example is
We define the g-Monge–Ampère operator as
where $Du$ denotes the gradient vector of the function u, whereas $|\cdot |$ represents the euclidian norm, so that we have $|Du|^2=u_iu_i$ .
Since the operator $\mathrm {det}(D^2u)$ (in the framework of strictly convex functions) is elliptic, then our g-Monge–Ampère operator is also elliptic.
A motivation for the definition of the g-Monge–Ampère operator (1.4) is the following. Using the Kronecker delta $\delta ^{i\ell }$ , define the $n\times n$ matrix ${\cal A} =[{\cal A}^{ij}]$ , where
The trace of the matrix ${\cal A}$ is the familiar operator $(g(|Du|^2)u_i)_i$ . We claim that the determinant of the matrix ${\cal A}$ coincides with our operator (1.4). Indeed, the eigenvalues $\Lambda ^1,\ldots ,\Lambda ^n$ of the $n\times n$ matrix
are the following:
Since det ${\cal A}$ =det ${\cal B}\cdot $ det $(D^2u)$ , we find
The claim follows from the latter equation and (1.4).
Note that ${\cal A}$ is not symmetric, in general. However, since ${\cal A}$ is the product of two symmetric matrices, it is similar to a diagonal matrix (see [Reference Horn and Johnson9, p. 487, Theorem 7.6.4]).
This paper is organized as follows. In Section 2, we show that the solution u of the g-Monge–Ampère problem
is the minimum of a suitable functional (depending on g). For $g=1$ , this result is well known (see, for example, [Reference Chou and Wang5]).
In Section 3, we consider a boundary value problem involving our g-Monge–Ampère operator in a bounded convex domain and introduce a P-function depending on the solution and its derivatives. We will show that this P-function attains its maximum value on the boundary of the underline domain. Furthermore, we will also show that such a P-function is identically constant when the underlying domain is a ball. Therefore, our P-function satisfies a best possible maximum principle in the sense of L. E. Payne [Reference Chen, Ma and Shi4, Reference Enache6, Reference Payne11, Reference Philippin12].
In Section 4, we consider the case when $n=2$ . In this case, we prove a best possible minimum principle. As a corollary, we solve a Serrin’s type overdetermined boundary value problem (see [Reference Brandolini, Nitsch, Salani and Trombetti2, Reference Serrin16, Reference Weinberger18]) for the corresponding g-Monge–Ampère equation. Similar problems are discussed in [Reference Barbu and Enache1, Reference Enache, Marras and Porru8, Reference Mohammed and Porru10, Reference Porru, Safoui and Vernier-Piro13] and the references therein.
Results of existence, uniqueness, and regularity for Monge–Ampère equations can be found in [Reference Caffarelli, Nirenberg and Spruck3, Reference Tso17].
2 Minimizing a functional
Define
Recall from [Reference Chou and Wang5] that a minimizer $u\in \Psi (\Omega )$ of the functional
satisfies the equation
We extend the above result to our g-Monge–Ampère equation.
Theorem 2.1 Let
A minimizer $u\in \Psi (\Omega )$ of the functional
satisfies
Proof By integration by parts, we can write the integral in (2.2) as
Arguing as in the proof of (1.4), we find
where
In view of (2.4), the expression in (2.3) reads as
If u is a minimizer of (2.5), we have
By computation, we find
Let us compute
Integrating by parts and recalling (1.2), from the latter equation, we find
Insertion of (2.7) into (2.6) yields
Since v is arbitrary, we find
Arguing as in the proof of (1.4), one proves that
On using the latter equation and the symmetry of $T_{(n-1)}^{ij}(D^2u)$ , from (2.8), we find
Finally, recalling that $H(s^2)=h(s^2)+2s^2h'(s^2)$ , by (2.1), we find
Hence,
The theorem follows.
3 A best possible maximum principle
Let u be a solution to some boundary value problem in a domain $\Omega $ . Following Payne [Reference Payne11], we say that a function $P(x)$ , depending on u and its derivatives, satisfies a best possible maximum principle if it satisfies a maximum principle for every convex domain $\Omega $ and, in addition, it is a constant for some special domain $\Omega $ (a ball in our case).
For a discussion on the best possible maximum principles related to second-order linear (or quasi-linear) elliptic equations, we refer to [Reference Philippin12]. Concerning Monge–Ampère equations, we recall a special case of Theorem 2.3 of [Reference Enache, Marras and Porru8]. Let u be a strictly convex smooth solution to the problem
and let
By Theorem 2.3 of [Reference Enache, Marras and Porru8], $P(x)$ attains its maximum value on $\partial \Omega $ ; furthermore, in case $\Omega $ is a ball, $P(x)$ is a constant.
We are going to extend this result to our g-Monge–Ampère equation. Consider the problem
and define the P-function
where G is defined as in (1.3). We note that our result is already proved in [Reference Porru, Safoui and Vernier-Piro13] by using a quite complicate argument. We give here a different and more clean proof. Moreover, our method allows us to prove that if $P(x)$ is identically constant, then $\Omega $ must be a ball.
Theorem 3.1 Let u be a strictly convex smooth solution to Problem (3.1). If $P(x)$ is defined as in (3.2), we have the following.
-
(i) If $\Omega $ is a ball, then $P(x)$ is identically constant.
-
(ii) For any convex $\Omega $ , $P(x)$ attains its maximum value on $\partial \Omega $ .
-
(iii) If $P(x)$ is identically constant in $\Omega $ , then $\Omega $ must be a ball.
Proof (i) If $\Omega $ is a ball, $u(x)$ is radial. If $v(r)=u(x)$ for $|x|=r$ , we have
Differentiation yields
On using (1.4), we can write the equation in (3.1) (in the radial case) as
Since
we find
Since
we can write the previous equation as
or, equivalently,
Recalling that g is continuous on $[0,r)$ and that $v'(0)=0$ , we integrate the above identity over $(0, r)$ , to find
or, equivalently,
Differentiation yields
By (3.3) and the latter equation, we find
It follows that $P(r)$ is identically constant.
(ii) Let $\Omega $ be a bounded convex domain. Arguing by contradiction, let $\tilde x\in \Omega $ be a point such that
Choose $0<\tau <1$ close enough to $1$ so that
Then, also the function
attains its maximum value at some point $\bar x\in \Omega $ . At the point $\bar x$ , we have either $Du=0$ or $|Du|>0$ . Consider first the case $Du=0$ . Then,
Further differentiation and computation at $Du=0$ yields
Let us make a rigid rotation around the point $\bar x$ so that
Then,
Clearly, if $(D^2u)^{-1}$ is the inverse of $D^2u$ , also $(D^2u)^{-1}$ will be diagonal, and
where $u^{ij}$ is the $(i,j)$ -entry of the matrix $(D^2u)^{-1}.$ Multiplying (3.5) by $u^{ii}$ and adding from $i=1$ to $i=n$ , we find
On the other hand, from equations (3.1) and (1.4), we find (recall that $g(0)=G(0)$ )
By using this equation, from (3.6), we find
Finally, since the matrix $D^2u$ is diagonal and positive definite, we have (we also use the arithmetic–geometric mean inequality)
By the latter inequality and (3.7), we find
Hence, $\bar P$ cannot have a maximum point at $\bar x$ with $Du(\bar x)=0$ .
Let $\bar x\in \Omega $ be a point of maximum for $\bar P$ , and let $|Du|>0$ at $\bar x$ . We have
and
Let us make a rigid rotation around the point $\bar x$ so that (3.4) holds. Then (for i fixed), we have
Multiplying by $u^{ii}$ and adding from $i=1$ to $i=n$ , we find
By using (1.4), let us write the equation in (3.1) as
Differentiation with respect to $x^h$ yields
Since $T_{(n-1)}(D^2u)D^2u=\mathrm {det}(D^2u)I$ , on using (3.10) and recalling that $u^{ij}$ is the $(i,j)$ -entry of the matrix $(D^2u)^{-1},$ we get
Therefore, recalling that $D^2u$ has a diagonal form, from (3.11), we find
Insertion of this equation into (3.9) leads to
Simplifying, we find
Since $\bar x$ is assumed to be a point of maximum, we have $\bar P_i=0$ , and from (3.8), we find
Insertion of (3.13) into (3.12) yields
If ${\cal A}=[{\cal A}^{ij}]$ with ${\cal A}^{ij}=\bigg (g(|Du|^2)u_i\bigg )_j$ , we know that
Therefore, by (1.4), the equation in (3.1) can be written as
On the other hand, since ${\cal A}$ is positive definite, by the Hadamard inequality (see Theorem 7.8.1 of [Reference Horn and Johnson9]) and the arithmetic–geometric mean inequality, we have
with equality sign if and only if
Therefore,
and
Recalling that $D^2u$ has a diagonal form, this inequality can be written as
On using (3.13), the latter inequality reads as
from which we find
Hence,
Inserting this estimate into (3.14), we find
It follows that $\bar P$ cannot have a maximum point at $\bar x$ with $|Du(\bar x)|>0$ . We conclude that P must attain its maximum value on the boundary $\partial \Omega $ .
(iii) If $P(x)$ is a constant, we have
Therefore, by the argument used to prove (ii), all equations in (3.15) must hold. This means that
Then, for some $x_0\in \Omega $ , we have
Since $g(s^2)s$ is strictly increasing, $|Du|$ must be radially symmetric around the point $x_0$ . Finally, since
also u will be radially symmetric. Statement (iii) follows.
The theorem is proved.
Remark From Theorem 3.1, we get the following estimate:
where
Note that this estimate is sharp, in the sense that the equality sign holds when $\Omega $ is a ball.
4 The case $n=2$
Here, we prove a minimum principle for our P-function, which extend the result obtained in the particular case $g\equiv 1$ in [Reference Enache7].
Theorem 4.1 Let u be a strictly convex smooth solution to Problem (3.1) in case $n=2$ , and let P(x) be defined as in (3.2). Then P attains its minimum value on the boundary $\partial \Omega $ .
Proof Arguing by contradiction, let $\tilde x\in \Omega $ be a point such that
Choose $\tau>1$ close enough to $1$ so that
Then, also the function
attains its minimum value at some point $\bar x\in \Omega $ . We may have either $|Du(\bar x)|>0$ or $Du(\bar x)=0$ . Consider first the case $|Du(\bar x)|>0$ . By the same computations as in the proof of Theorem 3.1, we find (3.12) with $n=2$ , that is,
As in the proof of Theorem 3.1, we assume that (3.4) holds at $\bar x$ . Since $\bar x$ is a point of minimum, we have $\bar P_i=0$ , and from (3.8), we find
Insertion of (4.2) into (4.1) yields
Since $|Du|>0$ , we have either $u_1\not =0$ or $u_2\not =0$ . If $u_1\not =0$ , by (3.8), we have
Since $n=2$ , equation (3.1) at $\bar x$ reads as $gGu_{11}u_{22}=1$ , and then, by our last equation, we find
Hence,
Note that (4.4) continues to holds if $u_1=0$ and $u_2\not =0$ . Insertion of (4.4) into (4.3) leads to
It follows that $\bar P$ cannot have a minimum point at any $x\in \Omega $ with $|Du|>0$ .
Consider now the case $Du(\bar x)=0$ . At $\bar x$ , we have
Since $\bar x$ is a point of minimum (also) for u, we have $u_{11}\ge 0$ and $u_{22}\ge 0$ . But since
we must have $u_{11}> 0$ and $u_{22}> 0$ . Hence, (4.5) implies that
It follows that
On the other hand, our equation at $\bar x$ (where $Du=0$ , so $g=G$ ) reads as
in contradiction with (4.6) because $\tau>1$ .
We have proved that $\bar P$ cannot have a minimum point at $\bar x$ with $|Du(\bar x)|=0$ . We conclude that P must attain its minimum value on the boundary $\partial \Omega $ . The theorem is proved.
Corollary 4.2 Let u be a strictly convex smooth solution to Problem (3.1) in case $n=2$ . If u satisfies the additional condition
then $\Omega $ must be a ball.