# Liouville Theorems for a class of degenerate or singular Monge-Ampère
equations

Ling Wang and Bin Zhou  School of Mathematical Sciences, Peking University,
Beijing 100871, China<EMAIL_ADDRESS>School of Mathematical
Sciences, Peking University, Beijing 100871, China<EMAIL_ADDRESS>
###### Abstract.

In this note, we classify solutions to a class of Monge-Ampère equations whose
right hand side may be degenerate or singular in the half space. Solutions to
these equations are special solutions to a class of fourth order equations,
including the affine maximal hypersurface equation, in the half space. Both
the Dirichlet boundary value and Neumann boundary value cases are considered.

###### Key words and phrases:

degenerate Monge-Ampère equation, Liouville theorems, partial Legendre
transform, method of moving spheres

###### 2020 Mathematics Subject Classification:

35J96, 35J70, 35B53, 35A09

This research is partially supported by NSFC grants 12271008 and National Key
R$\&$D Program of China SQ2020YFA0712800.

##  1\. Introduction

The main purpose of this paper is to investigate Liouville theorems for the
following class of Monge-Ampère equations

(1.1) $\det D^{2}u=(a+bx_{n})^{\alpha},\ \ \alpha\in\mathbb{R}$

in the half space
$\mathbb{R}^{n}_{+}:=\left\\{(x^{\prime},x_{n})\in\mathbb{R}^{n}:x_{n}>0\right\\}$,
where $a\geq 0$ and $b>0$. A motivation to consider (1.1) comes from the study
of the following class of fourth order equations

(1.2) $U^{ij}w_{ij}=0,$

where $(U^{ij})$ denotes the cofactor matrix of $(u_{ij})$ and $w=(\det
D^{2}u)^{-\theta}$, $\theta\in\mathbb{R}$ ($\theta\neq 0$). In particular,
when $\theta=\frac{n+1}{n+2}$, it is the affine mean curvature equation in
affine geometry [Ch] and when $\theta=1$, it is Abreu’s equation [Ab]. A first
breakthrough for the study of this class of equations is the Chern conjecture,
also known as the affine Bernstein theorem now, solved by Trudinger-Wang
[TW1], which says an entire strictly, uniformly convex solution to (1.2) on
$\mathbb{R}^{2}$ when $\theta=\frac{3}{4}$ must be a quadratic polynomial.
Later, it is shown that the Bernstein theorem also holds when
$\frac{3}{4}<\theta\leq 1$ [JL, Z] and $\theta<0$ [TW2]. When we consider
(1.2) in $\mathbb{R}_{+}^{n}$, one can easily find solutions where are not
quadratic polynomials. In particular, solutions to (1.1) with
$\alpha=-\frac{1}{\theta}$ give a class of special solutions to (1.2). Note
that in the entire space case, (1.1) admits no convex solutions if $b\neq 0$.

When $a=0$, (1.1) may be degenerate ($\alpha>0$) or singular ($\alpha<0$) on
$\partial\mathbb{R}^{n}_{+}$. When $\alpha\geq 0$, Savin [S1, S2] proved that
if the Dirichlet boundary value $u(x^{\prime},0)=\frac{1}{2}|x^{\prime}|^{2}$
is assigned, any convex continuous solution to (1.1) with the growth condition
$u=O(|x|^{3+\alpha-\varepsilon})$ as $|x|\to+\infty$ must be the form of

$u(Ax)=Bx_{n}+\frac{1}{2}|x^{\prime}|^{2}+\frac{x_{n}^{2+\alpha}}{(2+\alpha)(1+\alpha)}$

for some sliding $A$ along $x_{n}=0$, and some constant $B$. In particular,
when $\alpha=0$, the solution is a quadratic polynomial. This result was later
extended to the singular case with $\alpha\in(-1,0)$ by Savin and Zhang [SZ].
There are examples show that the growth condition at infinity is necessary in
general dimensions. When $\alpha=-1$, the local asymptotic behavior of the
solution near the boundary in dimension two was studied in [Ru].

In this paper, we concentrate on the two dimensional case. Our first result
classifies all solutions to (1.1) with Dirichlet condition in dimension two
when $\alpha>-2$.

######  Theorem 1.1.

Let $u(x,y)\in C^{2}({\mathbb{R}_{+}^{2}})\cap
C(\overline{\mathbb{R}_{+}^{2}})$ be a convex solution to

(1.3) $\left\\{\begin{aligned} \det D^{2}u&=(a+by)^{\alpha}&&\text{in
}\mathbb{R}^{2}_{+},\\\\[5.0pt] u(x,0)&=\frac{1}{2}x^{2}&&\text{on
}\partial\mathbb{R}^{2}_{+},\end{aligned}\right.$

where $a\geq 0$, $b>0$, and $\alpha>-2$. Then there exist
$A,\,B,\,C\in\mathbb{R}$ with $A\geq 0$ such that

(1.4) $u(x,y)=\left\\{\begin{aligned}
&\frac{(b-aA)(a+by)^{2+\alpha}}{b^{3}(1+\alpha)(2+\alpha)}+\frac{A(a+by)^{3+\alpha}}{b^{3}(2+\alpha)(3+\alpha)}-By\\\
&\quad-\frac{(b-aA)a^{2+\alpha}}{b^{3}(1+\alpha)(2+\alpha)}-\frac{Aa^{3+\alpha}}{b^{3}(2+\alpha)(3+\alpha)}+\frac{(x-Cy)^{2}}{2(1+Ay)},&&\alpha\neq-1;\\\\[7.0pt]
&\frac{b-aA}{b^{3}}(a+by)\ln(a+by)+\frac{A}{2b}y^{2}-By\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,-\frac{(b-aA)a\ln
a}{b^{3}}+\frac{(x-Cy)^{2}}{2(1+Ay)},&&\alpha=-1.\end{aligned}\right.$

######  Remark 1.2.

  1. (1)

When $a=0$, we improve the exponent in the results of [S1, S2, SZ] to
$\alpha>-2$ in two dimensional case. This exponent is sharp since (1.3) admits
no solutions continuous up to the boundary for $\alpha\leq-2$ (see details in
Remark 3.1). When $\alpha=0$, Theorem 1.1 can be also found in [Fi, Page
145-148].

  2. (2)

If we assume $u=O(|(x,y)|^{3+\alpha-\varepsilon})$ as $|(x,y)|\to+\infty$,
then we have that $A$ must be $0$ in (1.4). Hence we can recover some of the
results in [S2, SZ] in dimension two.

  3. (3)

In a subsequent work, we are going to study the Liouville type theorem the
following problem

(1.5) $\left\\{\begin{aligned} U^{ij}w_{ij}&=0&&\text{ in
}\mathbb{R}_{+}^{n},\\\ u&=\frac{1}{2}|x^{\prime}|^{2}&&\text{ on
}\partial\mathbb{R}_{+}^{n},\\\ w&=1&&\text{ on
}\partial\mathbb{R}_{+}^{n}.\end{aligned}\right.$

The main idea to prove Theorem 1.1 is as follows. Let $u(x,y)$ be a uniformly
convex solution to (1.1). Then its partial Legendre transform in the
$x$-variable is

(1.6) $u^{\star}(\xi,\eta)=xu_{x}(x,y)-u(x,y),$

where $(\xi,\eta)=(u_{x},y)$. It is easy to check that $u^{\star}$ is a
solution to

(1.7) $(a+b\eta)^{\alpha}u_{\xi\xi}^{\star}+u_{\eta\eta}^{\star}=0.$

When $a=0$, this Grushin type equation was studied in [CS]. By a change of
variables $v(x_{1},x_{2})=u^{\star}\left(x_{1},f(x_{2})\right)$, where

$\xi=x_{1},\
\eta=f(x_{2})=b^{\frac{-\alpha}{\alpha+2}}\left(\frac{\alpha+2}{2}x_{2}\right)^{\frac{2}{\alpha+2}}-\frac{a}{b},$

we know that $v$ solves the following divergence type equation

(1.8) $\operatorname{div}\left(x_{2}^{\frac{\alpha}{\alpha+2}}\nabla
v\right)=0,$

which may be degenerate or singular. A Liouville theorem for (1.8) on the
upper half space has been obtained recently by [WZ]. However, in our case, the
domain may shift after the transformations. Hence, we need a generalization of
the result in [WZ] to general upper half spaces.

The above approach also works for the case of Neumann problem. As for the
Neumann boundary value case, we only consider the degenerate case, stated as
follows.

######  Theorem 1.3.

Let $u(x,y)\in C^{2}({\mathbb{R}_{+}^{2}})\cap
C^{1}(\overline{\mathbb{R}_{+}^{2}})$ be a convex solution to

(1.9) $\left\\{\begin{aligned} \det D^{2}u&=y^{\alpha}&&\text{in
}\mathbb{R}^{2}_{+},\\\ u_{y}(x,0)&=0&&\text{on
}\partial\mathbb{R}^{2}_{+},\end{aligned}\right.$

where $\alpha\geq 0$. Then there exist a constant $A>0$, and a linear function
$l(x)$ such that

(1.10)
$u(x,y)=\frac{1}{2A}x^{2}+\frac{A}{(2+\alpha)(1+\alpha)}y^{2+\alpha}+l(x).$

######  Remark 1.4.

  1. (1)

When $\alpha=0$, Theorem 1.3 is included in [JT, Theorem 1.1]. In fact, it is
proved in [JT] that any convex solution to Neumann problem of Monge-Ampère
equations in the half plane must be a quadratic polynomial for two dimensional
case, and the conclusion still holds for dimension $n\geq 3$ if either the
boundary value is zero or the solution restricted on some $n-2$ dimensional
subspace is bounded from above by a quadratic function. Here we extend this to
the degenerate case.

  2. (2)

It is still unknown whether Theorem 1.3 is true for $\alpha<0$. Although our
method doesn’t work for this case, we believe the conclusion is true for
$\alpha>-1$.

Finally, we turn to the Liouville theorem on the whole space. The celebrated
result of Jörgens [Jö], Calabi [Ca] and Pogorelov [Po] states that any entire
classical convex solution to the Monge-Ampère equation

$\det D^{2}u=1\quad\text{in }\mathbb{R}^{n}$

must be a quadratic polynomial. Caffarelli [Caf] extended this result to
viscosity solutions (the proof can be also found in [CL, Theorem 1.1]). For
another direction of extension, Jin and Xiong [JX] studied the class of
equations

(1.11) $\det D^{2}u(x,y)=|y|^{\alpha}$

on the whole plane $\mathbb{R}^{2}$, and established a Liouville theorem.

######  Theorem 1.5 ([JX, Theorem 1.1]).

Let $u(x,y)$ be convex generalized (or Alexandrov) solution to (1.11) with
$\alpha>-1$. Then there exist some constants $A>0$, $B\in\mathbb{R}$ and a
linear function $l(x,y)$ such that

(1.12)
$u(x,y)=\frac{1}{2A}x^{2}+\frac{AB^{2}}{2}y^{2}+Bxy+\frac{A}{(2+\alpha)(1+\alpha)}|y|^{2+\alpha}+l(x,y).$

At the end of this paper, we use the approach above to give a new proof of
this result in the case of $\alpha\geq 0$. The main idea of Jin and Xiong in
[JX] is that using the partial Legendre transform to change (1.11) into a
class of linearized Monge-Ampère equations, then applying the Harnack
inequality for linearized Monge-Ampère equations derived by Caffarelli and
Gutiérrez [CG] and the scaling argument to classify all solutions of the
transformed equation. Our new proof is similar to Theorem 1.1 and Theorem 1.3.

The structure of this paper is as follows. In Section 2, we derive the
Liouville theorems for a class of linear elliptic equations in divergence form
including (1.8). Then we prove Theorem 1.1, Theorem 1.3 and Theorem 1.5 in
Section 3.

##  2\. Liouville theorems for linear elliptic equations in divergence form

In this section, we establish a Liouville theorem for a class of linear
elliptic equations in divergence form, which may be degenerate or singular
cases, in the half space. This theorem can be viewed as an extension of [WZ,
Theorem 1.1]. The proof is very similar to [WZ, Theorem 1.1], where the method
of moving sphere will be used. Denote
$\mathbb{R}_{l}^{n}=\\{x=(x^{\prime},x_{n}):x^{\prime}\in\mathbb{R}^{n-1},\,x_{n}>l\\}$
for $l\geq 0$.

######  Theorem 2.1.

For $n\geq 2$ and $a\in\mathbb{R}$, let $u\in C^{2}(\mathbb{R}_{l}^{n})\cap
C^{0}(\overline{\mathbb{R}_{l}^{n}})$ be a solution to

$\begin{cases}\operatorname{div}\left(x_{n}^{a}\nabla
u\right)=0,&u>-C_{0}\text{ in }\mathbb{R}_{l}^{n},\\\\[4.0pt]
u(x^{\prime},l)=0,&\text{ on }\mathbb{R}^{n-1}\times\\{x_{n}=l\\},\end{cases}$

where $l\geq 0$ and $C_{0}>0$. Then $u=C_{*}\left(x_{n}^{1-a}-l^{1-a}\right)$
for some nonnegative constant $C_{*}$. In particular, when $a\geq 1$,
$C_{*}=0$.

######  Remark 2.2.

When $l=0$, Theorem 2.1 is just the Theorem 1.1 of [WZ].

###### Proof of Theorem 2.1.

We extend $u$ to $\overline{\mathbb{R}^{n}_{+}}$ by letting
$u\left(x^{\prime},x_{n}\right)=0$ in $\mathbb{R}^{n-1}\times[0,l)$, and
denote it by $\widetilde{u}$. Hence, we know that $\widetilde{u}(x)\in
C\left(\overline{\mathbb{R}^{n}_{+}}\right)$.

Firstly, we show that $\widetilde{u}$ is weakly differentiable in
$\mathbb{R}_{+}^{n}$ and

$\nabla\widetilde{u}=\left\\{\begin{aligned} &\nabla
u,&&\mathbb{R}_{l}^{n},\\\
&0,&&\mathbb{R}^{n-1}\times(0,l).\end{aligned}\right.$

Indeed, $\forall\,\varphi\in C_{0}^{\infty}(\mathbb{R}_{+}^{n})$, by
integration by parts, we have

$\int_{\mathbb{R}_{+}^{n}}\widetilde{u}\,\partial_{x_{i}}\varphi\mathrm{~{}d}x=\int_{\mathbb{R}^{n-1}\times(l,+\infty)}{u}\,\partial_{x_{i}}\varphi\mathrm{~{}d}x=-\int_{\mathbb{R}^{n-1}\times(l,+\infty)}\partial_{x_{i}}{u}\,\varphi\mathrm{~{}d}x$

for $i\leq n-1$ and

$\displaystyle\int_{\mathbb{R}_{+}^{n}}\widetilde{u}\,\partial_{x_{n}}\varphi\mathrm{~{}d}x$
$\displaystyle=\int_{\mathbb{R}^{n-1}\times(l,+\infty)}{u}\,\partial_{x_{n}}\varphi\mathrm{~{}d}x$
$\displaystyle=-\int_{\mathbb{R}^{n-1}\times\\{x_{n}=l\\}}{u}\,\varphi\mathrm{~{}d}{x^{\prime}}-\int_{\mathbb{R}^{n-1}\times(l,+\infty)}\partial_{x_{n}}{u}\,\varphi\mathrm{~{}d}x$
$\displaystyle=-\int_{\mathbb{R}^{n-1}\times(l,+\infty)}\partial_{x_{n}}{u}\,\varphi\mathrm{~{}d}x,$

where we used ${u}(x^{\prime},l)=0$ in the last equality.

Next, we show that $\widetilde{u}\in
W^{1,2}_{loc}\left(\mathbb{R}_{+}^{n}\right)\cap
C\left(\overline{\mathbb{R}^{n}_{+}}\right)$ is a weak solution to

(2.1)
$\begin{cases}\operatorname{div}\left(x_{n}^{a}\nabla\widetilde{u}\right)=0,&\widetilde{u}>-C_{0}\text{
in }\mathbb{R}^{n}_{+},\\\\[4.0pt] \widetilde{u}=0,&\text{ on
}\partial\mathbb{R}^{n}_{+}.\end{cases}$

Indeed, for any $\varphi\in C_{0}^{\infty}(\mathbb{R}_{+}^{n})$, there is

$\displaystyle\int_{\mathbb{R}_{+}^{n}}\operatorname{div}\left(x_{n}^{a}\nabla\widetilde{u}\right)\varphi\mathrm{~{}d}x$
$\displaystyle=\int_{\mathbb{R}^{n-1}\times(l,+\infty)}\operatorname{div}\left(x_{n}^{a}\nabla\widetilde{u}\right)\varphi\mathrm{~{}d}x+\int_{\mathbb{R}^{n-1}\times(0,l)}\operatorname{div}\left(x_{n}^{a}\nabla\widetilde{u}\right)\varphi\mathrm{~{}d}x$
$\displaystyle=-\int_{\mathbb{R}^{n-1}\times(l,+\infty)}x_{n}^{a}\nabla{u}\cdot\nabla\varphi\mathrm{~{}d}x-\int_{\mathbb{R}^{n-1}\times\\{x_{n}=l\\}}\partial_{x_{n}}{u}\,\varphi\mathrm{~{}d}{x^{\prime}}$
$\displaystyle=\int_{\mathbb{R}^{n-1}\times(l,+\infty)}\operatorname{div}\left(x_{n}^{a}\nabla{u}\right)\varphi\mathrm{~{}d}x=0.$

It’s clear that $\widetilde{u}>-C_{0}$ in $\mathbb{R}^{n}_{+}$ and
$\widetilde{u}=0$ on $\partial\mathbb{R}^{n}_{+}$. Hence, $\widetilde{u}\in
W^{1,2}_{loc}\left(\mathbb{R}_{+}^{n}\right)\cap
C\left(\overline{\mathbb{R}^{n}_{+}}\right)$ is a weak solution to (2.1).

For any fixed $x\in\partial\mathbb{R}_{+}^{n}$ and $\lambda>0$, by Kelvin
transformation

$y^{x,\lambda}=x+\frac{\lambda^{2}(y-x)}{|y-x|^{2}},\quad\forall
y\in\overline{\mathbb{R}_{+}^{n}},$

we define

$\widetilde{u}_{x,\lambda}(y)=\frac{\lambda^{n-2+a}}{|y-x|^{n-2+a}}\widetilde{u}\left(y^{x,\lambda}\right),\quad\forall
y\in\overline{\mathbb{R}_{+}^{n}}.$

By [YD, Theorem 2.1], we know that $\widetilde{u}_{x,\lambda}(y)\in
W^{1,2}_{loc}\left(\mathbb{R}_{+}^{n}\right)$ satisfies
$\operatorname{div}\left(y_{n}^{a}\nabla\widetilde{u}_{x,\lambda}\right)=0$ in
the weak sense, i.e. $\widetilde{u}_{x,\lambda}$ satisfies the same equation.

For $a>2-n$, we consider $\bar{u}=\widetilde{u}+C_{0}$ instead of
$\widetilde{u}$. Then $\displaystyle\lim_{|y|\rightarrow 0}\bar{u}(x+y)=C_{0}$
for $x\in\partial\mathbb{R}_{+}^{n}$. Let

$w_{x,\lambda}(y)=\bar{u}(y)-\bar{u}_{x,\lambda}(y),\quad\forall\,y\in\mathbb{R}_{+}^{n}.$

We have

$\varliminf_{|y|\rightarrow+\infty}w_{x,\lambda}(y)\geq
0-\lim_{|y|\rightarrow+\infty}\frac{\lambda^{n-2+a}}{|y-x|^{n-2+a}}\bar{u}\left(x+\frac{\lambda^{2}(y-x)}{|y-x|^{2}}\right)=0.$

By the maximum principle, we have
$\widetilde{u}_{x,\lambda}(y)\leq\widetilde{u}(y),\,\,\forall\,y\in\mathbb{R}^{n}_{+}\backslash
B_{\lambda}(x)$. Hence by Lemma 2.3 below, we know that
$\widetilde{u}(y^{\prime},y_{n})=\widetilde{u}(y_{n})$. Then solving the
corresponding ODE gives us the desired result.

For $a<2-n$, we consider $\bar{u}=\widetilde{u}-1$ instead of $\widetilde{u}$.
Then $\displaystyle\lim_{|y|\rightarrow 0}\bar{u}(x+y)=-1$ for
$x\in\partial\mathbb{R}_{+}^{n}$. Let

$w_{x,\lambda}(y)=\bar{u}(y)-\bar{u}_{x,\lambda}(y),\quad\forall\,y\in\mathbb{R}_{+}^{n}.$

We have

$\displaystyle\varliminf_{|y|\rightarrow+\infty}w_{x,\lambda}(y)$
$\displaystyle=\varliminf_{|y|\rightarrow+\infty}\bar{u}(y)-\lim_{|y|\rightarrow+\infty}\frac{|y-x|^{2}}{\lambda^{2}}\bar{u}\left(x+\frac{\lambda^{2}(y-x)}{|y-x|^{2}}\right)$
$\displaystyle\geq-1-C_{0}+\lim_{|y|\rightarrow+\infty}\frac{|y-x|^{2}}{\lambda^{2}}$
$\displaystyle=+\infty.$

Again by the maximum principle, we have
$\widetilde{u}_{x,\lambda}(y)\leq\widetilde{u}(y),\,\,\forall\,y\in\mathbb{R}^{n}_{+}\backslash
B_{\lambda}(x)$. Similarly, by Lemma 2.3, we also have
$\widetilde{u}(y^{\prime},y_{n})=\widetilde{u}(y_{n})$, then we can obtain the
conclusion.

As for $a=2-n$, we need to modify $\widetilde{u}_{x,\lambda}(y)$ to be

$\widetilde{u}_{x,\lambda}(y)=\widetilde{u}\left(y^{x,\lambda}\right)+\ln\frac{\lambda}{|y-x|}.$

Then by similar arguments, we also have
$\widetilde{u}_{x,\lambda}(y)\leq\widetilde{u}(y),\,\,\forall\,y\in\mathbb{R}^{n}_{+}\backslash
B_{\lambda}(x)$. The result follows by applying Lemma 2.4. ∎

In the proof of Theorem 2.1, we used two crucial lemmas of moving spheres
[WZ]. For readers’ convenience, we include a proof here, which is very similar
to the proof of [Li, Lemma 5.7].

######  Lemma 2.3 ([WZ, Lemma 3.3]).

Assume $f(y)\in C^{0}\left(\overline{\mathbb{R}_{+}^{n}}\right)$, $n\geq 2$,
and $\tau\in\mathbb{R}$. Suppose

(2.2)
$\left(\frac{\lambda}{|y-x|}\right)^{\tau}f\left(x+\frac{\lambda^{2}(y-x)}{|y-x|^{2}}\right)\leq
f(y)$

for $\lambda>0$, $x\in\partial\mathbb{R}_{+}^{n}$, $y\in\mathbb{R}_{+}^{n}$
satisfying $|y-x|\geq\lambda$. Then

$f(y)=f\left(y^{\prime},y_{n}\right)=f\left(0^{\prime},y_{n}\right),\quad\forall
y=\left(y^{\prime},y_{n}\right)\in\mathbb{R}_{+}^{n}.$

###### Proof.

For any fixed $y^{\prime},z^{\prime}\in\mathbb{R}^{n-1}$ with $y^{\prime}\neq
z^{\prime}$ and $y_{n}>0$, we denote $y=(y^{\prime},y_{n})$ and
$z=(z^{\prime},z_{n})$, where $z_{n}=\frac{b-1}{b}y_{n}$ for $b>1$. Then we
have

$x=y+b(z-y)\in\partial\mathbb{R}_{+}^{n}$

and

$z=x+\frac{\lambda^{2}(y-x)}{|y-x|^{2}},$

where $\lambda=\sqrt{|z-x|\cdot|y-x|}$. By (2.2), we have

(2.3) $\left(\frac{\lambda}{|y-x|}\right)^{\tau}f(z)\leq f(y).$

Since

$\displaystyle\lim_{b\to+\infty}\frac{\lambda}{|y-x|}=\lim_{|x|\to\infty}\sqrt{\frac{|z-x|}{|y-x|}}=1,\quad\lim_{b\to+\infty}z_{n}=\lim_{b\to+\infty}\frac{b-1}{b}y_{n}=y_{n}.$

and $f$ is continuous, we have $f(z^{\prime},y_{n})\leq f(y^{\prime},y_{n})$.
By the arbitrariness of $y^{\prime}\neq z^{\prime}$, the proof is completed. ∎

######  Lemma 2.4.

Suppose that $f\in C^{0}\left(\overline{\mathbb{R}_{+}^{n}}\right)$ satisfies
that for all $x\in\partial\mathbb{R}_{+}^{n}$ and $\lambda>0$,

$f(y)\geq
f\left(x+\frac{\lambda^{2}(y-x)}{|y-x|^{2}}\right)+\ln\frac{\lambda}{|y-x|},\quad\forall
y\in\mathbb{R}_{+}^{n}\backslash B_{\lambda}(x).$

Then

$f(y)=f\left(y^{\prime},y_{n}\right)=f\left(0^{\prime},y_{n}\right),\quad\forall
y=\left(y^{\prime},y_{n}\right)\in\mathbb{R}_{+}^{n}.$

###### Proof.

The proof is the same as Lemma 2.3. It suffices to replace (2.3) by

$\ln\frac{\lambda}{|y-x|}+f(z)\leq f(y).$

∎

A Liouville theorem for the Neumman boundary value is also derived in [WZ].

######  Theorem 2.5 ([WZ, Theorem 1.2]).

Assume $n\geq 2$ and $\max\\{-1,2-n\\}<a<1$. Suppose $u(x)\in$
$C^{2}\left(\mathbb{R}_{+}^{n}\right)\cap
C^{1}\left(\overline{\mathbb{R}_{+}^{n}}\right)$ satisfies

(2.4) $\begin{cases}\operatorname{div}\left(x_{n}^{a}\nabla
u\right)=0,&u>0,\quad\text{ in }\mathbb{R}_{+}^{n},\\\\[3.0pt]
x_{n}^{a}\frac{\partial u}{\partial x_{n}}=0&\text{ on
}\partial\mathbb{R}_{+}^{n}.\end{cases}$

Then $u=C$ for some positive constant $C$. The boundary condition in (2.4)
holds in the following sense:

$\displaystyle\lim_{x_{n}\rightarrow 0^{+}}x_{n}^{a}\frac{\partial u}{\partial
x_{n}}=0.$

##  3\. Proof of main theorems

In this section, we first derive the new equation under the partial Legendre
transform. Let $\Omega\subset\mathbb{R}^{2}$ and $u(x,y)$ be a uniformly
convex function on $\Omega$. The partial Legendre transform in the
$x$-variable is

(3.1) $u^{\star}(\xi,\eta)=xu_{x}(x,y)-u(x,y),$

where

(3.2)
$(\xi,\eta)=\mathcal{P}(x,y):=\left(u_{x},y\right)\in\mathcal{P}(\Omega):=\Omega^{\star}.$

We have

$\frac{\partial(\xi,\eta)}{\partial(x,y)}=\left(\begin{array}[]{ccc}u_{xx}&&u_{xy}\\\
0&&1\end{array}\right),\quad\text{ and
}\quad\frac{\partial(x,y)}{\partial(\xi,\eta)}=\left(\begin{array}[]{ccc}\frac{1}{u_{xx}}&&-\frac{u_{xy}}{u_{xx}}\\\
0&&1\end{array}\right).$

Hence,

(3.3) $\displaystyle u_{\xi}^{\star}=x,\ \ u_{\eta}^{\star}=-u_{y},$ (3.4)
$\displaystyle u_{\xi\xi}^{\star}=\frac{1}{u_{xx}},\ \
u_{\eta\eta}^{\star}=-\frac{\operatorname{det}D^{2}u}{u_{xx}},\ \
u_{\xi\eta}^{\star}=-\frac{u_{xy}}{u_{xx}}.$

Then if $u$ is a solution to

$\det D^{2}u=(a+bx)^{\alpha},$

$u^{\star}$ is a solution to

$(a+b\eta)^{\alpha}u_{\xi\xi}^{\star}+u_{\eta\eta}^{\star}=0.$

We will apply the results in Section 2 related to this equation to prove the
main theorems.

###  3.1. The case of Dirichlet boundary value

We use Theorem 2.1 to prove Theorem 1.1.

###### Proof of Theorem 1.1.

We consider the the partial Legendre transform $u^{\star}$ of $u$ on
$\mathbb{R}^{2}_{+}$. Note that $\xi=u_{x}$, $\eta=y$ and $u_{x}=x$ on
$\\{y=0\\}$ by (1.3), which gives us that
$\mathcal{P}(\\{y=0\\})=\\{\eta=0\\}$. Hence, we have
$\mathcal{P}\left(\mathbb{R}_{+}^{2}\right)=\mathbb{R}_{+}^{2}$. Then if $u$
is a solution to (1.3), $u^{\star}$ is a solution to

(3.5) $\left\\{\begin{aligned}
(a+b\eta)^{\alpha}u_{\xi\xi}^{\star}+u_{\eta\eta}^{\star}&=0&&\text{ in
}\mathbb{R}\times(0,+\infty),\\\ u^{\star}(\xi,0)&=\frac{\xi^{2}}{2}&&\text{
on }\mathbb{R}\times\\{0\\},\end{aligned}\right.$

where we used the fact that the Legendre transform of
$x\mapsto\frac{1}{2}x^{2}$ is $\xi\mapsto\frac{1}{2}\xi^{2}$.

Since Legendre transform does not change the convexity, we have that
$u^{\star}_{\xi\xi}\geq 0$. Denote $v:=u^{\star}_{\xi\xi}-1$. Differentiating
(3.5) twice respect to $\xi$, we have that $v\geq-1$ solves

(3.6) $\left\\{\begin{aligned}
(a+b\eta)^{\alpha}v_{\xi\xi}+v_{\eta\eta}&=0&&\text{ in
}\mathbb{R}\times(0,+\infty),\\\\[3.0pt] v(\xi,0)&=0&&\text{ on
}\mathbb{R}\times\\{\eta=0\\}.\end{aligned}\right.$

Let $\xi=x_{1}$,
$\eta=f(x_{2})=b^{\frac{-\alpha}{\alpha+2}}\left(\frac{\alpha+2}{2}x_{2}\right)^{\frac{2}{\alpha+2}}-\frac{a}{b}$
and

$\widetilde{v}(x_{1},x_{2})=v\left(x_{1},f(x_{2})\right).$

A direct calculation yields

$\displaystyle\widetilde{v}_{11}=v_{\xi\xi},$
$\displaystyle\widetilde{v}_{2}=b^{\frac{-\alpha}{\alpha+2}}\left(\frac{\alpha+2}{2}x_{2}\right)^{\frac{-\alpha}{\alpha+2}}v_{\eta},$
$\displaystyle\widetilde{v}_{22}=-\frac{\alpha}{\alpha+2}x_{2}^{-1}\widetilde{v}_{2}+(a+b\eta)^{-\alpha}v_{\eta\eta}.$

$\eta=0$ gives us that $x_{2}=\frac{2}{b(\alpha+2)}a^{\frac{\alpha+2}{2}}$.
Denote $l=\frac{2}{b(\alpha+2)}a^{\frac{\alpha+2}{2}}$. Hence by (3.6), we
know that $\tilde{v}\geq-1$ solves

$\left\\{\begin{aligned}
\widetilde{v}_{11}+\widetilde{v}_{22}+\frac{\alpha}{\alpha+2}x_{2}^{-1}\widetilde{v}_{2}&=0&&\text{
in }\mathbb{R}\times(l,+\infty),\\\\[3.0pt] \widetilde{v}(x_{1},0)&=0&&\text{
on }\mathbb{R}\times\\{x_{2}=l\\},\end{aligned}\right.$

i.e.,

$\left\\{\begin{aligned}
\operatorname{div}\left(x_{2}^{\frac{\alpha}{\alpha+2}}\nabla\widetilde{v}\right)&=0&&\text{
in }\mathbb{R}\times(l,+\infty),\\\\[3.0pt] \widetilde{v}(x_{1},0)&=0&&\text{
on }\mathbb{R}\times\\{x_{2}=l\\}.\end{aligned}\right.$

Applying Theorem 2.1 with $n=2$ and $a=\frac{\alpha}{\alpha+2}<1$, we know
that
$\widetilde{v}(x_{1},x_{2})=C_{*}\left({x_{2}}^{\frac{2}{\alpha+2}}-l^{\frac{2}{\alpha+2}}\right)$
for some nonnegative constant $C_{*}$. Transforming back to $(\xi,\eta)$, we
have $v(\xi,\eta)=A\eta$ for some $A\geq 0$, i.e.,
$u_{\xi\xi}^{\star}(\xi,\eta)=1+A\eta$. Then

$u^{\star}(\xi,\eta)=h_{1}(\eta)+\xi h_{2}(\eta)+\frac{\xi^{2}}{2}(1+A\eta)$

for some functions $h_{1},h_{2}:[0,+\infty)\rightarrow\mathbb{R}$. Recalling
(3.5), we have $h_{1}(0)=h_{2}(0)=0$ and

$h_{1}^{\prime\prime}(\eta)+\xi
h_{2}^{\prime\prime}(\eta)+(1+A\eta)(a+b\eta)^{\alpha}=0$

on $\mathbb{R}\times(0,+\infty)$. This implies that
$h_{1}^{\prime\prime}(\eta)+(1+A\eta)(a+b\eta)^{\alpha}=0$ and
$h_{2}^{\prime\prime}(\eta)=0$. By solving the ODEs, we obtain

$u^{\star}(\xi,\eta)=\left\\{\begin{aligned}
&B\eta-\frac{(b-aA)(a+b\eta)^{2+\alpha}}{b^{3}(1+\alpha)(2+\alpha)}-\frac{A(a+b\eta)^{3+\alpha}}{b^{3}(2+\alpha)(3+\alpha)}+C\xi\eta\\\
&\quad+\frac{(b-aA)a^{2+\alpha}}{b^{3}(1+\alpha)(2+\alpha)}+\frac{Aa^{3+\alpha}}{b^{3}(2+\alpha)(3+\alpha)}+\frac{\xi^{2}}{2}(1+A\eta),&&\alpha\neq-1;\\\\[8.0pt]
&B\eta-\frac{b-aA}{b^{3}}(a+b\eta)\ln(a+b\eta)-\frac{A}{2}\eta^{2}+C\xi\eta\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,+\frac{(b-aA)a\ln
a}{b^{3}}+\frac{\xi^{2}}{2}(1+A\eta),&&\alpha=-1,\end{aligned}\right.$

for some constants $B,C\in\mathbb{R}$. Recalling that the Legendre transform
is an involution on convex functions, we recover $u$ by taking the partial
Legendre transform of $u^{\star}$ :

$u(x,y)=\left\\{\begin{aligned}
&\frac{(b-aA)(a+by)^{2+\alpha}}{b^{3}(1+\alpha)(2+\alpha)}+\frac{A(a+by)^{3+\alpha}}{b^{3}(2+\alpha)(3+\alpha)}-By\\\
&\quad-\frac{(b-aA)a^{2+\alpha}}{b^{3}(1+\alpha)(2+\alpha)}-\frac{Aa^{3+\alpha}}{b^{3}(2+\alpha)(3+\alpha)}+\frac{(x-Cy)^{2}}{2(1+Ay)},&&\alpha\neq-1;\\\\[8.0pt]
&\frac{b-aA}{b^{3}}(a+by)\ln(a+by)+\frac{A}{2b}y^{2}-By\\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,-\frac{(b-aA)a\ln
a}{b^{3}}+\frac{(x-Cy)^{2}}{2(1+Ay)},&&\alpha=-1.\end{aligned}\right.$

This gives us a complete classification of all solutions to (1.3). ∎

######  Remark 3.1.

$\alpha>-2$ in Theorem 1.1 is sharp since (1.3) has no convex solutions
continuous up to boundary in $\mathbb{R}^{2}_{+}$ when $\alpha\leq-2$. Indeed,
if there exists a convex function $u\in C^{2}({\mathbb{R}_{+}^{2}})\cap
C(\overline{\mathbb{R}_{+}^{2}})$ solves (1.3), by [S2, Theorem 5.1], we will
have a Pogorelov type estimate

$(1-u)u_{xx}\leq C(\max|u_{x}|)$

in $S_{1}$, where $S_{h}=\\{x\in\mathbb{R}^{2}_{+}:u(x)<u(0)+\nabla u(0)\cdot
x+h\\}$ for $h>0$. Since $u(x,0)=\frac{1}{2}x^{2}\text{on
}\partial\mathbb{R}^{2}_{+}$, we know that $|u_{x}|$ is bounded in $S_{1}$
(depends on $\|u\|_{L^{\infty}(S_{2})}$). Then there exists a small $c_{0}>0$
such that $u_{xx}\leq C(\|u\|_{L^{\infty}(S_{2})})$ in $B_{c_{0}}^{+}$. Hence,
we have

$\displaystyle Cu_{yy}\geq u_{xx}u_{yy}\geq
u_{xx}u_{yy}-u_{xy}^{2}=y^{\alpha}\text{ in }B_{c_{0}}^{+},$

i.e. in $B_{c_{0}}^{+}$. Then it holds

$u(x,y)\geq\left\\{\begin{aligned}
&\frac{1}{C(1+\alpha)(2+\alpha)}y^{2+\alpha}+D(x)y+E(x),&&\alpha<-2,\\\\[4.0pt]
&-\frac{1}{C}\ln y+D(x)y+E(x),&&\alpha=-2,\end{aligned}\right.$

which means that $\displaystyle\lim_{y\to 0+}u(x,y)=+\infty$. This contradicts
with $u\in C(\overline{\mathbb{R}_{+}^{2}})$.

###  3.2. The case of Neumann boundary value

We prove Theorem 1.3 in this section.

###### Proof of Theorem 1.3.

For $\alpha=0$, it has been proved by Jian and Tu [JT, Theorem 1.1]. In the
following, we mainly prove the case for $\alpha>0$ .

Similarly as in the last section, we know that if $u$ is a solution to (1.9),
$u^{\star}$ is a solution to

(3.7) $\left\\{\begin{aligned}
\eta^{\alpha}u_{\xi\xi}^{\star}+u_{\eta\eta}^{\star}&=0&&\text{ in
}\mathbb{R}\times(0,+\infty),\\\\[4.0pt] u^{\star}_{\eta}(\xi,0)&=0&&\text{ on
}\mathbb{R}\times\\{0\\}.\end{aligned}\right.$

Denote $v:=u^{\star}_{\xi\xi}$. Differentiating (3.7) twice respect to $\xi$,
we have that $v>0$ solves

(3.8) $\left\\{\begin{aligned} \eta^{\alpha}v_{\xi\xi}+v_{\eta\eta}&=0&&\text{
in }\mathbb{R}\times(0,+\infty),\\\ v_{\eta}(\xi,0)&=0&&\text{ on
}\mathbb{R}\times\\{\eta=0\\}.\end{aligned}\right.$

Let $\xi=x_{1}$,
$\eta=\left(\frac{\alpha+2}{2}\right)^{\frac{2}{\alpha+2}}x_{2}^{\frac{2}{\alpha+2}}$
and

$\widetilde{v}(x_{1},x_{2})=v\left(x_{1},\left(\frac{\alpha+2}{2}\right)^{\frac{2}{\alpha+2}}x_{2}^{\frac{2}{\alpha+2}}\right).$

Then (3.8) gives us that $\tilde{v}>0$ solves

$\left\\{\begin{aligned}
\widetilde{v}_{11}+\widetilde{v}_{22}+\frac{\alpha}{\alpha+2}x_{2}^{-1}\widetilde{v}_{2}&=0&&\text{
in }\mathbb{R}\times(0,+\infty),\\\
x_{2}^{\frac{\alpha}{\alpha+2}}\widetilde{v}_{2}(x_{1},0)&=0&&\text{ on
}\mathbb{R}\times\\{x_{2}=0\\},\end{aligned}\right.$

i.e.,

$\left\\{\begin{aligned}
\operatorname{div}\left(x_{2}^{\frac{\alpha}{\alpha+2}}\nabla\widetilde{v}\right)&=0&&\text{
in }\mathbb{R}\times(0,+\infty),\\\
x_{2}^{\frac{\alpha}{\alpha+2}}\widetilde{v}_{2}(x_{1},0)&=0&&\text{ on
}\mathbb{R}\times\\{x_{2}=0\\}.\end{aligned}\right.$

Applying Theorem 2.5 with $n=2$ and $a=\frac{\alpha}{\alpha+2}\in(0,1)$, we
know that $\widetilde{v}=C$ for some positive constant $C$. Transforming back
to $(\xi,\eta)$, we have $v(\xi,\eta)=A$ for some $A>0$, i.e.,
$u_{\xi\xi}^{\star}(\xi,\eta)=A$ for some $A>0$. Then

$u^{\star}(\xi,\eta)=h_{1}(\eta)+\xi h_{2}(\eta)+\frac{A}{2}\xi^{2}$

for some functions $h_{1},h_{2}:[0,+\infty)\rightarrow\mathbb{R}$. Recalling
(3.7), we have

$h^{\prime}_{1}(0)=h^{\prime}_{2}(0)=0\quad\text{and}\quad
h_{1}^{\prime\prime}(\eta)+\xi
h_{2}^{\prime\prime}(\eta)+A\eta^{\alpha}=0\quad\text{on
}\mathbb{R}\times(0,+\infty).$

This implies that $h_{1}^{\prime\prime}(\eta)+A\eta^{\alpha}=0$ and
$h_{2}^{\prime\prime}(\eta)=0$. By solving these ODEs, we obtain

$u^{\star}(\xi,\eta)=\frac{A}{2}\xi^{2}+B\xi-\frac{A\eta^{2+\alpha}}{(1+\alpha)(2+\alpha)}+C$

for some constants $A,C\in\mathbb{R}$. Recalling that $u=(u^{\star})^{\star}$,
we have

$u(x,y)=\frac{1}{2A}(x-B)^{2}+\frac{Ay^{2+\alpha}}{(1+\alpha)(2+\alpha)}-C,$

which yields (1.10). ∎

###  3.3. The entire space case

Before proving Theorem 1.5, we first recall two theorems for (1.11) in [JX].

######  Theorem 3.2 ([JX, Theorem 4.1]).

Let $\Omega$ be an open convex set in $\mathbb{R}^{2}$, and $u$ be the
generalized solution of

$\operatorname{det}D^{2}u(x)=\left|y\right|^{\alpha}\quad\text{in
}\Omega\text{, }$

with $u=0$ on $\partial\Omega$. Then $u$ is strictly convex in $\Omega$, and
$u\in C_{\text{loc }}^{1,\delta}(\Omega)$ for some $\delta>0$ depending only
on $\alpha$. Furthermore, the partial Legendre transform $u^{\star}$ of $u$ is
a strong solution of

$|\eta|^{\alpha}u^{\star}_{\xi\xi}+u^{\star}_{\eta\eta}=0\quad\text{in
}\mathcal{P}(\Omega),$

where the map $\mathcal{P}$ is given in (3.2).

######  Theorem 3.3 ([JX, Theorem 4.2]).

Let $u$ be a generalized solution of

$\operatorname{det}D^{2}u(x)=\left|y\right|^{\alpha}\quad\text{in
}\mathbb{R}^{2}.$

Then $u$ is strictly convex.

Hence Theorem 3.2 and Theorem 3.3 give us that $u$ is strictly convex, and
then $u$ is smooth away from $\left\\{y=0\right\\}$. Furthermore, we know that
$u\in C_{\text{loc }}^{1,\delta}\left(\mathbb{R}^{2}\right)$ and the partial
Legendre transform $u^{\star}$ of $u$ is a strong solution of (3.10).

Next, we need a Liouville theorem for degenerate elliptic equations in
divergence form. This theorem is a partial extension of [WZ, Corollary 1.4],
where they assumed stronger conditions.

######  Theorem 3.4.

Assume that $n=2$ and $a\geq 0$. Then any positive $C^{1}(\mathbb{R}^{2})$
solution to

(3.9) $\operatorname{div}\left(|x_{2}|^{a}\nabla u\right)=0\quad\text{in
}\mathbb{R}^{2}$

is a constant function.

###### Proof.

Note that $u$ only belongs to $C^{1}(\mathbb{R}^{2})$. We need to prove this
theorem in the weak sense. Hence we can repeat the same process as in the
proof of Theorem 2.1. Due to its similarity, we omit the details here. ∎

Now, we are ready to give the proof of Theorem 1.5.

###### Proof of Theorem 1.5.

Our proof only works for the case $\alpha\geq 0$. We consider the partial
Legendre transform $u^{\star}$ of $u$. $u^{\star}$ is a solution to

(3.10) $|\eta|^{\alpha}u^{\star}_{\xi\xi}+u^{\star}_{\eta\eta}=0\quad\text{in
}\mathbb{R}^{2}.$

Let $v:=u^{\star}_{\xi\xi}\geq 0$. Differentiating (3.10) twice respect to
$\xi$, we have that $v\geq 0$ solves

(3.11) $|\eta|^{\alpha}v_{\xi\xi}+v_{\eta\eta}=0\quad\text{in
}\mathbb{R}^{2}.$

By a change of variables, we let

$\widetilde{v}(x_{1},x_{2})=\left\\{\begin{aligned}
&v\left(x_{1},\left(\frac{\alpha+2}{2}\right)^{\frac{2}{\alpha+2}}x_{2}^{\frac{2}{\alpha+2}}\right),&&\eta\geq
0,\\\\[5.0pt]
&v\left(x_{1},-\left(\frac{\alpha+2}{2}\right)^{\frac{2}{\alpha+2}}(-x_{2})^{\frac{2}{\alpha+2}}\right),&&\eta<0.\end{aligned}\right.$

A direct calculation yields

$\displaystyle\widetilde{v}_{11}$ $\displaystyle=v_{\xi\xi},$
$\displaystyle\widetilde{v}_{22}$
$\displaystyle=-\frac{\alpha}{\alpha+2}x_{2}^{-1}\widetilde{v}_{2}+|\eta|^{-\alpha}v_{\eta\eta}.$

By (3.11), we know that $\widetilde{v}\geq 0$ solves

$\widetilde{v}_{11}+\widetilde{v}_{22}+\frac{\alpha}{\alpha+2}x_{2}^{-1}\widetilde{v}_{2}=0\quad\text{in
}\mathbb{R}^{2},$

i.e.,

$\operatorname{div}\left(|x_{2}|^{\frac{\alpha}{\alpha+2}}\nabla\widetilde{v}\right)=0\quad\text{in
}\mathbb{R}^{2}.$

Hence, by Theorem 3.4 with $a=\frac{\alpha}{\alpha+2}\geq 0$ in (3.9), we
obtain $\widetilde{v}\equiv$ constant. Thus $u^{\star}_{\xi\xi}\equiv A$,
where $A$ is a constant. Similar to the proofs of Theorem 1.1 and Theorem 1.3,
by solving these ODEs, we have

$u^{\star}(\xi,\eta)=\frac{A}{2}\xi^{2}-\frac{A}{(1+\alpha)(2+\alpha)}|\eta|^{2+\alpha}+B\xi\eta+l(\xi,\eta).$

Again by $u=(u^{\star})^{\star}$, we have (1.12). ∎

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