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
$$ \begin{align*}\mathrm{det}(D^2u)=\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)u_i\bigg)_j,\end{align*} $$
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:
$$ \begin{align*}T_{(n-1)}^{ij}(D^2u)=\frac{\partial \mathrm{det}(D^2u)}{\partial u_{ij}},\ \ \ i,j=1,\ldots,n.\end{align*} $$
Moreover, the tensor
$\bigg [T_{(n-1)}^{ij}(D^2u)\bigg ]$
is symmetric and divergence-free, that is,
$$ \begin{align}\bigg(T_{(n-1)}^{ij}(D^2u)\bigg)_j=0,\ \ \ i=1,\ldots,n.\end{align} $$
If I denotes the
$n\times n$
identity matrix, we have
$$ \begin{align}T_{(n-1)}(D^2u)D^2u=I\, \mathrm{det}(D^2u).\end{align} $$
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
$$ \begin{align}G(s^2):=g(s^2)+2s^2g'(s^2)>0.\end{align} $$
We also suppose that
$g(0)=G(0)>0$
(i.e., positive and finite). Note that
$$ \begin{align*}G(s^2)=\frac{d}{ds}\bigg(g(s^2)s\bigg).\end{align*} $$
Therefore, the function
$g(s^2)s$
is positive and strictly increasing for
$s>0$
. A typical example is
$$ \begin{align*}g(s^2)=(1+s^2)^{-\frac{1}{2}},\ \ \ G(s^2)=(1+s^2)^{-\frac{3}{2}}.\end{align*} $$
We define the g-Monge–Ampère operator as
$$ \begin{align*} \frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)g^n(|Du|^2)u_i\bigg)_j , \end{align*} $$
where
$Du$
denotes the gradient vector of the function u, whereas
$|\cdot |$
represents the euclidian norm, so that we have
$|Du|^2=u_iu_i$
.
$$ \begin{align}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)g^n(|Du|^2)u_i\bigg)_j=g^{n-1}(|Du|^2)G(|Du|^2)\mathrm{det}(D^2u).\end{align} $$
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
$$ \begin{align*}{\cal A}^{ij}=(g(|Du|^2)u_i)_j=\bigg(g(|Du|^2)\delta^{i\ell}+2g'(|Du|^2)u_{i} u_\ell \bigg)u_{\ell j}.\end{align*} $$
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
$$ \begin{align*}{\cal B}=\bigg[g(|Du|^2)\delta^{i\ell}+2g'(|Du|^2)u_{i} u_\ell\bigg]\end{align*} $$
are the following:
$$ \begin{align*}\Lambda^1=\cdots=\Lambda^{n-1}=g(|Du|^2), \ \Lambda^n=G(|Du|^2).\end{align*} $$
Since det
${\cal A}$
=det
${\cal B}\cdot $
det
$(D^2u)$
, we find
$$ \begin{align*}\text{det}{\cal A}=g^{n-1}(|Du|^2)G(|Du|^2)\mathrm{det}(D^2u).\end{align*} $$
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
$$ \begin{align*} \frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)g^n(|Du|^2)u_i\bigg)_j =f(-u)\ \ \mathrm{in}\ \Omega,\ \ u=0\ \ \mathrm{on}\ \ \partial\Omega, \end{align*} $$
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
$$ \begin{align*}\Psi(\Omega):=\{u\in C^2(\Omega)\cap C^{0,1}(\overline\Omega):\ u\ \text{is strictly convex in}\ \Omega\ \text{and} \ u=0 \ \text{on}\ \partial\Omega\}.\end{align*} $$
Recall from [Reference Chou and Wang5] that a minimizer
$u\in \Psi (\Omega )$
of the functional
$$ \begin{align*}\inf_{v\in \Psi(\Omega)}\int_\Omega\bigg[\frac{1}{n(n+1)}T_{(n-1)}^{ij}(D^2v)v_iv_j+\int_0^v f(-\tau)d\tau\bigg]dx\end{align*} $$
satisfies the equation
$$ \begin{align*}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)u_i\bigg)_j =f(-u).\end{align*} $$
We extend the above result to our g-Monge–Ampère equation.
Theorem 2.1 Let
$$ \begin{align} h^n(t^2)=t^{-n-1}\int_0^t\tau^n g^n(\tau^2)\, d\tau.\end{align} $$
A minimizer
$u\in \Psi (\Omega )$
of the functional
$$ \begin{align}\inf_{v\in \Psi(\Omega)}\int_\Omega\bigg[\frac{1}{n}T_{(n-1)}^{ij}(D^2v)h^n(|Dv|^2)v_iv_j+\int_0^v f(-\tau)d\tau\bigg]dx\end{align} $$
satisfies
$$ \begin{align*}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)g^n(|Du|^2)u_i\bigg)_j =f(-u).\end{align*} $$
Proof By integration by parts, we can write the integral in (2.2) as
$$ \begin{align}\int_{\Omega}\bigg[\frac{-v}{n}\bigg(T_{(n-1)}^{ij}(D^2v)h^n(|Dv|^2)v_i\bigg)_j+\int_{0}^vf(-\tau)d\tau\bigg]dx.\end{align} $$
Arguing as in the proof of (1.4), we find
$$ \begin{align}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2v)h^n(|Dv|^2)v_i\bigg)_j=h^{n-1}(|Dv|^2)H(|Dv|^2)\mathrm{det}(D^2v),\end{align} $$
where
$$ \begin{align*}H(s^2)=h(s^2)+2s^2h'(s^2).\end{align*} $$
In view of (2.4), the expression in (2.3) reads as
$$ \begin{align}\int_{\Omega}\bigg[(-v)h^{n-1}(|Dv|^2)H(|Dv|^2)\mathrm{det}(D^2v)+\int_{0}^vf(-\tau)d\tau\bigg]dx.\end{align} $$
If u is a minimizer of (2.5), we have
$$ \begin{align*} &\frac{d}{dt}\int_{\Omega}\bigg[(-u-tv)h^{n-1}(|Du+tDv|^2)H(|Du+tDv|^2)\mathrm{det}(D^2u+tD^2v)\\ &+ \int_{0}^{u+tv}f(-\tau)d\tau\bigg]dx\bigg|_{t=0}=0. \end{align*} $$
By computation, we find
$$ \begin{align} &\int_{\Omega}(-v)h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{det}(D^2u)\, dx \nonumber\\ &+\int_{\Omega}(-u)\bigg(h^{n-1}(|Du|^2)H(|Du|^2)\bigg)'2 Du\cdot Dv\, \mathrm{det}(D^2u)\, dx\nonumber\\ &+\int_{\Omega}(-u)h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{trace}\bigg(T_{(n-1)}(D^2u)D^2v\bigg)\, dx\\ &+ \int_\Omega f(-u)v\, dx=0.\nonumber \end{align} $$
Let us compute
$$ \begin{align*} &\int_{\Omega}(-u)h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{trace}\bigg(T_{(n-1)}(D^2u)D^2v\bigg)\, dx\\ &=\int_{\Omega}(-u)h^{n-1}(|Du|^2)H(|Du|^2)T_{(n-1)}^{ij}(D^2u)v_{ij}\, dx\\ &=\int_{\Omega}\bigg(u\, h^{n-1}(|Du|^2)H(|Du|^2)T_{(n-1)}^{ij}(D^2u)\bigg)_jv_{i}\, dx\\ &=\int_{\Omega} h^{n-1}(|Du|^2)H(|Du|^2)T_{(n-1)}^{ij}(D^2u)v_{i}u_j\, dx\\ &\quad+\int_{\Omega}u\, \bigg(h^{n-1}(|Du|^2)H(|Du|^2)\bigg)'2u_{jh}u_hT_{(n-1)}^{ij}(D^2u)v_{i}\, dx. \end{align*} $$
Integrating by parts and recalling (1.2), from the latter equation, we find
$$ \begin{align} &\int_{\Omega}(-u)h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{trace}\bigg(T_{(n-1)}(D^2u)D^2v\bigg)\, dx \nonumber\\ &=\int_{\Omega}(-v) \bigg(h^{n-1}(|Du|^2)H(|Du|^2)T_{(n-1)}^{ij}(D^2u)u_j\bigg)_i\, dx\\ &\quad+\int_{\Omega}u\, \bigg(h^{n-1}(|Du|^2)H(|Du|^2)\bigg)'2Du\cdot Dv\, \mathrm{det}(D^2u)\, dx. \nonumber \end{align} $$
Insertion of (2.7) into (2.6) yields
$$ \begin{align*} &\int_{\Omega}(-v)h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{det}(D^2u)\, dx\\ &+\int_{\Omega}(-v) \bigg(h^{n-1}(|Du|^2)H(|Du|^2)T_{(n-1)}^{ij}(D^2u)u_j\bigg)_i\, dx=\int_\Omega (-v)f(-u)\, dx. \end{align*} $$
Since v is arbitrary, we find
$$ \begin{align} &h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{det}(D^2u) \nonumber\\ &+\bigg(h^{n-1}(|Du|^2)H(|Du|^2)T_{(n-1)}^{ij}(D^2u)u_j\bigg)_i=f(-u). \end{align} $$
Arguing as in the proof of (1.4), one proves that
$$ \begin{align*}h^{n-1}(|Du|^2)H(|Du|^2)\mathrm{det}(D^2u)=\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)h^n(|Du|^2)u_i\bigg)_j.\end{align*} $$
On using the latter equation and the symmetry of
$T_{(n-1)}^{ij}(D^2u)$
, from (2.8), we find
$$ \begin{align*}\frac{1}{n}\bigg(\bigg[h^{n}(|Du|^2)+nh^{n-1}(|Du|^2)H(|Du|^2)\bigg]T_{(n-1)}^{ij}(D^2u)u_i\bigg)_j=f(-u).\end{align*} $$
Finally, recalling that
$H(s^2)=h(s^2)+2s^2h'(s^2)$
, by (2.1), we find
$$ \begin{align*}h^{n}(|Du|^2)+nh^{n-1}(|Du|^2)H(|Du|^2)=g^n(|Du|^2).\end{align*} $$
Hence,
$$ \begin{align*}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)g^n(|Du|^2)u_i\bigg)_j =f(-u).\end{align*} $$
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
$$ \begin{align*}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)u_i\bigg)_j=1\ \ \text{in}\ \ \Omega,\ \ u=0\ \ \text{on}\ \ \partial\Omega,\end{align*} $$
and let
$$ \begin{align*}P(x)=\frac{1}{2}|Du|^2-u.\end{align*} $$
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
$$ \begin{align}\frac{1}{n}\bigg(T_{(n-1)}^{ij}(D^2u)g^n(|Du|^2)u_i\bigg)_j=1\ \ \text{in}\ \ \Omega,\ \ u=0\ \ \text{on}\ \ \partial\Omega,\end{align} $$
and define the P-function
$$ \begin{align} P(x)=\int_0^{|Du|}G(t^2)t\, dt-u,\end{align} $$
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
$$ \begin{align*}P(r)=\int_0^{v'}G(\tau^2)\tau\, d\tau-v.\end{align*} $$
Differentiation yields
$$ \begin{align}P'(r)=G((v')^2)v'v"-v'.\end{align} $$
On using (1.4), we can write the equation in (3.1) (in the radial case) as
$$ \begin{align*}g^{n-1}((v')^2)G((v')^2)\mathrm{det}(D^2v)=1.\end{align*} $$
Since
$$ \begin{align*}\mathrm{det}(D^2v)=v"\bigg(\frac{v'}{r}\bigg)^{n-1},\end{align*} $$
we find
$$ \begin{align*}g^{n-1}((v')^2)G((v')^2)(v')^{n-1}v"=r^{n-1}.\end{align*} $$
Since
$$ \begin{align*}G(s^2)=\frac{d}{ds}\bigg(g(s^2)s\bigg),\end{align*} $$
we can write the previous equation as
$$ \begin{align*} \bigg(g((v')^2)v'\bigg)^{n-1}\frac{d}{dr}\bigg(g((v')^2)v'\bigg)=r^{n-1}, \end{align*} $$
or, equivalently,
$$ \begin{align*} \frac{1}{n}\frac{d}{dr}\bigg(g((v')^2)v'\bigg)^{n}=r^{n-1}. \end{align*} $$
Recalling that g is continuous on
$[0,r)$
and that
$v'(0)=0$
, we integrate the above identity over
$(0, r)$
, to find
$$ \begin{align*} \bigg(g((v')^2)v'\bigg)^{n}=n\int_0^rt^{n-1}=r^n, \end{align*} $$
or, equivalently,
$$ \begin{align*} g((v')^2)v'=r. \end{align*} $$
Differentiation yields
$$ \begin{align*}G((v')^2)v"=1.\end{align*} $$
By (3.3) and the latter equation, we find
$$ \begin{align*}P'(r)=v'\bigg[G((v')^2)v"-1\bigg]=0.\end{align*} $$
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
$$ \begin{align*}P(\tilde x)=\int_0^{|Du(\tilde x)|}G(t^2)t\, dt-u(\tilde x)>\max_{x\in\partial\Omega}\int_0^{|Du(x)|}G(t^2)t\, dt.\end{align*} $$
Choose
$0<\tau <1$
close enough to
$1$
so that
$$ \begin{align*}\int_0^{|Du(\tilde x)|}G(t^2)t\, dt-\tau u(\tilde x)>\max_{x\in\partial\Omega}\int_0^{|Du(x)|}G(t^2)t\, dt.\end{align*} $$
Then, also the function
$$ \begin{align*}\bar P(x)=\int_0^{|Du(x)|}G(t^2)t\, dt-\tau u(x)\end{align*} $$
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,
$$ \begin{align*}\bar P_i=G(|Du|^2)u_{ih}u_h-\tau u_i.\end{align*} $$
Further differentiation and computation at
$Du=0$
yields
$$ \begin{align*}\bar P_{ii}=G(0)u_{ih}u_{ih}- \tau u_{ii},\ \ \ i=1,\ldots,n.\end{align*} $$
Let us make a rigid rotation around the point
$\bar x$
so that
$$ \begin{align}D^2u=\text{diag}\{u_{11},\ldots,u_{nn}\}.\end{align} $$
Then,
$$ \begin{align}\bar P_{ii}= G(0)u_{ii}u_{ii}- \tau u_{ii},\ \ \ i=i,\ldots,n.\end{align} $$
Clearly, if
$(D^2u)^{-1}$
is the inverse of
$D^2u$
, also
$(D^2u)^{-1}$
will be diagonal, and
$$ \begin{align*}(D^2u)^{-1}=\text{diag}\{u^{11},\ldots,u^{nn}\},\end{align*} $$
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
$$ \begin{align}u^{ii}\bar P_{ii}= G(0)\Delta u- \tau n.\end{align} $$
On the other hand, from equations (3.1) and (1.4), we find (recall that
$g(0)=G(0)$
)
$$ \begin{align*}\text{det}(D^2u)=\frac{1}{G^n(0)}.\end{align*} $$
By using this equation, from (3.6), we find
$$ \begin{align}u^{ii}\bar P_{ii}= \frac{\Delta u}{\bigg(\text{det}(D^2u)\bigg)^{\frac{1}{n}}}- \tau n.\end{align} $$
Finally, since the matrix
$D^2u$
is diagonal and positive definite, we have (we also use the arithmetic–geometric mean inequality)
$$ \begin{align*}\bigg(\text{det}(D^2u)\bigg)^{\frac{1}{n}}= (u_{11}\cdots u_{nn})^{\frac{1}{n}}\le \frac{\Delta u}{n}.\end{align*} $$
By the latter inequality and (3.7), we find
$$ \begin{align*}u^{ii}\bar P_{ii}\ge n(1-\tau)>0.\end{align*} $$
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
$$ \begin{align}\bar P_{i}=G(|Du|^2)u_{ih}u_h-\tau u_i,\end{align} $$
and
$$ \begin{align*}\bar P_{ii}=2G'(u_{ih}u_h)^2+Gu_{iih}u_h+Gu_{ih}u_{ih}-\tau u_{ii},\ \ \ i=1,\ldots,n. \end{align*} $$
Let us make a rigid rotation around the point
$\bar x$
so that (3.4) holds. Then (for i fixed), we have
$$ \begin{align*}\bar P_{ii}=2G'u_{ii}^2u_i^2+Gu_{iih}u_h+Gu_{ii}^2-\tau u_{ii}. \end{align*} $$
Multiplying by
$u^{ii}$
and adding from
$i=1$
to
$i=n$
, we find
$$ \begin{align}u^{ii}\bar P_{ii}=2G'u_{ii}u_i^2+Gu^{ii}u_{iih}u_h+G\Delta u-n\tau.\end{align} $$
By using (1.4), let us write the equation in (3.1) as
$$ \begin{align}\mathrm{det}(D^2u)=\frac{1}{g^{n-1}(|Du|^2)G(|Du|^2)}.\end{align} $$
Differentiation with respect to
$x^h$
yields
$$ \begin{align}T^{ij}_{(n-1)}(D^2u)u_{ijh}=-\frac{1}{\bigg(g^{n-1}G\bigg)^2}\bigg[(n-1)g^{n-2}g'G+g^{n-1}G'\bigg]2u_{hk}u_k.\end{align} $$
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
$$ \begin{align*}T^{ij}_{(n-1)}(D^2u)=\frac{u^{ij}}{g^{n-1}G},\ \ \ \ i,j=1,\ldots,n.\end{align*} $$
Therefore, recalling that
$D^2u$
has a diagonal form, from (3.11), we find
$$ \begin{align*}u^{ii}u_{iih}=-\bigg[(n-1)\frac{g'}{g}+\frac{G'}{G}\bigg]2u_{hh}u_h.\end{align*} $$
Insertion of this equation into (3.9) leads to
$$ \begin{align*}u^{ii}\bar P_{ii}=2G'u_{ii}u_i^2-\bigg[(n-1)\frac{g'G}{g}+G'\bigg]2u_{hh}u_h^2+G\Delta u-n\tau.\end{align*} $$
Simplifying, we find
$$ \begin{align}u^{ii}\bar P_{ii}=-2(n-1)\frac{g'G}{g}u_{hh}u_h^2+G\Delta u-n\tau,\end{align} $$
Since
$\bar x$
is assumed to be a point of maximum, we have
$\bar P_i=0$
, and from (3.8), we find
$$ \begin{align}Gu_{hh}u_h^2=\tau |Du|^2.\end{align} $$
Insertion of (3.13) into (3.12) yields
$$ \begin{align}u^{ii}\bar P_{ii}=2(1-n)\frac{g'}{g}\tau |Du|^2+G\Delta u-n\tau.\end{align} $$
If
${\cal A}=[{\cal A}^{ij}]$
with
${\cal A}^{ij}=\bigg (g(|Du|^2)u_i\bigg )_j$
, we know that
$$ \begin{align*}\mathrm{det}{\cal A}=g^{n-1}(|Du|^2)G(|Du|^2)\mathrm{det}(D^2u).\end{align*} $$
Therefore, by (1.4), the equation in (3.1) can be written as
$$ \begin{align*}\mathrm{det}{\cal A}=1.\end{align*} $$
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
$$ \begin{align*}1=\bigg(\mathrm{det}{\cal A}\bigg)^{\frac{1}{n}}\le \bigg({\cal A}^{11}\cdots {\cal A}^{nn}\bigg)^{\frac{1}{n}}\le \frac{{\cal A}^{11}+\cdots + {\cal A}^{nn}}{n},\end{align*} $$
with equality sign if and only if
$$ \begin{align}{\cal A}^{11}=\cdots ={\cal A}^{nn},\ \ \text{and}\ \ {\cal A}^{ij}=0\ \ \forall i\not=j.\end{align} $$
Therefore,
$$ \begin{align*}{\cal A}^{11}+\cdots + {\cal A}^{nn}\ge n\end{align*} $$
and
$$ \begin{align*}\bigg(g(|Du|^2)u_i\bigg)_i\ge n.\end{align*} $$
Recalling that
$D^2u$
has a diagonal form, this inequality can be written as
$$ \begin{align*}g\Delta u+2g'u_{ii}u_i^2\ge n.\end{align*} $$
On using (3.13), the latter inequality reads as
$$ \begin{align*}g\Delta u+2\frac{g'}{G}\tau |Du|^2\ge n,\end{align*} $$
from which we find
$$ \begin{align*}G\Delta u+2\frac{g'}{g}\tau |Du|^2\ge n\frac{G}{g}=n+2n\frac{g'}{g}|Du|^2.\end{align*} $$
Hence,
$$ \begin{align*}G\Delta u+2(\tau-n)\frac{g'}{g}|Du|^2\ge n.\end{align*} $$
Inserting this estimate into (3.14), we find
$$ \begin{align*} u^{ii}\bar P_{ii}&\ge n(1-\tau)+\frac{g'}{g}|Du|^2 2n(1-\tau)\\ &=n(1-\tau)\bigg(1+\frac{g'}{g}2|Du|^2\bigg)\\ &=n(1-\tau)\frac{G}{g}>0. \end{align*} $$
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
$$ \begin{align*}u^{ii} P_{ii}=0\ \ \ \text{in}\ \ \Omega.\end{align*} $$
Therefore, by the argument used to prove (ii), all equations in (3.15) must hold. This means that
$$ \begin{align*}\bigg(g(|Du|^2)u_1\bigg)_1=\cdots=\bigg(g(|Du|^2)u_n\bigg)_n, \ \text{and}\ \ \bigg(g(|Du|^2)u_i\bigg)_j=0,\ \ \forall i\not=j.\end{align*} $$
Then, for some
$x_0\in \Omega $
, we have
$$ \begin{align*} &g(|Du|^2)u_i=x^i-x^i_0,\ \ i=1,\ldots,n,\\ &g^2(|Du|^2)u_i^2=(x^i-x^i_0)^2,\\ &g^2(|Du|^2)\sum_1^nu_i^2=\sum_1^n(x^i-x^i_0)^2=r^2,\\ &g^2(|Du|^2)|Du|^2=r^2,\\ &g(|Du|^2)|Du|=r. \end{align*} $$
Since
$g(s^2)s$
is strictly increasing,
$|Du|$
must be radially symmetric around the point
$x_0$
. Finally, since
$$ \begin{align*}\int_0^{|Du|}G(t^2)t\, dt-u=c,\end{align*} $$
also u will be radially symmetric. Statement (iii) follows.
The theorem is proved.
Remark From Theorem 3.1, we get the following estimate:
$$ \begin{align*}-u_m\le \int_0^{|Du|_M}G(t^2)t\, dt,\end{align*} $$
where
$$ \begin{align*}u_m=\min_\Omega u(x),\ \ \ |Du|_M=\max_{\partial\Omega}|Du|.\end{align*} $$
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
$$ \begin{align*}P(\tilde x)=\int_0^{|Du(\tilde x)|}G(t^2)t\, dt-u(\tilde x)<\min_{x\in\partial\Omega}\int_0^{|Du(x)|}G(t^2)t\, dt.\end{align*} $$
Choose
$\tau>1$
close enough to
$1$
so that
$$ \begin{align*}\int_0^{|Du(\tilde x)|}G(t^2)t\, dt-\tau u(\tilde x)<\min_{x\in\partial\Omega}\int_0^{|Du(x)|}G(t^2)t\, dt.\end{align*} $$
Then, also the function
$$ \begin{align*}\bar P(x)=\int_0^{|Du(x)|}G(t^2)t\, dt-\tau u(x)\end{align*} $$
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,
$$ \begin{align}u^{ii}\bar P_{ii}=-2\frac{g'G}{g}u_{hh}u_h^2+G\Delta u-2\tau.\end{align} $$
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
$$ \begin{align}Gu_{hh}u_h^2=\tau |Du|^2.\end{align} $$
Insertion of (4.2) into (4.1) yields
$$ \begin{align}u^{ii}\bar P_{ii}=-2\frac{g'}{g}\tau |Du|^2+G\Delta u-2\tau.\end{align} $$
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
$$ \begin{align*}u_{11}=\frac{\tau}{G}.\end{align*} $$
Since
$n=2$
, equation (3.1) at
$\bar x$
reads as
$gGu_{11}u_{22}=1$
, and then, by our last equation, we find
$$ \begin{align*}u_{22}=\frac{1}{\tau g}.\end{align*} $$
Hence,
$$ \begin{align}\Delta u=\frac{\tau}{G}+\frac{1}{\tau g}.\end{align} $$
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
$$ \begin{align*}u^{ii}\bar P_{ii}=\frac{G}{g} \bigg(\frac{1}{\tau}-\tau\bigg)<0.\end{align*} $$
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
$$ \begin{align}\bar P_{kk}=Gu_{kk}^2-\tau u_{kk}\ge 0,\ \ \ k=1,2.\end{align} $$
Since
$\bar x$
is a point of minimum (also) for u, we have
$u_{11}\ge 0$
and
$u_{22}\ge 0$
. But since
$$ \begin{align*}u_{11}u_{22}=\frac{1}{g(0)G(0)}>0,\end{align*} $$
we must have
$u_{11}> 0$
and
$u_{22}> 0$
. Hence, (4.5) implies that
$$ \begin{align*}Gu_{11}-\tau \ge 0\ \ \mathrm{and}\ \ Gu_{22}-\tau \ge 0.\end{align*} $$
It follows that
$$ \begin{align}G^2u_{11}u_{22}\ge \tau^2.\end{align} $$
On the other hand, our equation at
$\bar x$
(where
$Du=0$
, so
$g=G$
) reads as
$$ \begin{align*}G^2u_{11}u_{22}=1,\end{align*} $$
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
$$ \begin{align*}|Du|=c\ \ \ \mathrm{on}\ \partial\Omega ,\end{align*} $$
then
$\Omega $
must be a ball.
