Next Article in Journal
Concise Review: Asymmetric Cell Divisions in Stem Cell Biology
Next Article in Special Issue
Convergence Rate of a Stable, Monotone and Consistent Scheme for the Monge-Ampère Equation
Previous Article in Journal
Centrally Extended Conformal Galilei Algebras and Invariant Nonlinear PDEs

Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

A Monge–Ampere Equation with an Unusual Boundary Condition

by
Marc Sedjro
RWTH Aachen University, Lehrstuhl für Mathematik (Analysis) Templergraben 55, Aachen 52062, Germany
Symmetry 2015, 7(4), 2009-2024; https://doi.org/10.3390/sym7042009
Submission received: 2 September 2015 / Revised: 21 October 2015 / Accepted: 27 October 2015 / Published: 5 November 2015

Abstract

:
We consider a class of Monge–Ampere equations where the convex conjugate of the unknown function is prescribed on a boundary of its domain yet to be determined. We show the existence of a weak solution.

1. Introduction

Let $Ω ⊂ R 2$ be open, convex and bounded. We are interested in the following Monge–Ampere equations:
$det D 2 u ( x , y ) = f ( x , y ) a 2 ( ∂ u ∂ x ) ( x , y ) ∈ Ω D u ( Ω ) ⊂ Λ h u * ( h ( z ) , z ) = a ( h ( z ) ) on h > 0$
where $a$ and f are prescribed and $f > 0$; the unknowns are $u : Ω ⟶ R$ and $h : [ 0 , 1 ] ⟶ [ 0 , 1 )$. For such h, we associate the set:
$Λ h = ( s , z ) ∈ Λ : 0 ≤ s ≤ h ( z ) with Λ : = [ 0 , 1 ) × [ 0 , 1 ] .$
The function $u * : Λ ⟶ R$ denotes the convex conjugate of u. Typically, the function $a : [ 0 , 1 ) ⟼ ( 0 , ∞ )$ is smooth and satisfies the following property:
$lim s → 1 a ( s ) = ∞ and lim s → 1 ( 1 - s ) a ( s ) = 1 .$
The Monge–Ampere equations are known to play an important role in the formulation of some problems in meteorology and fluid mechanics; semigeostrophic equations and their variants provide examples of such problems (see [1,2,3]). Recently, Cullen and this author have discovered that the so-called forced axisymmetric flows that arise in meteorology can be formulated as Monge–Ampere equations coupled with continuity equations. However, it is important to note that these Monge–Ampere equations come with a boundary condition that is unusual, as this condition is derived from the unique structure of forced axisymmetric flows. A treatment of forced axisymmetric flows can be found in [4]. We initiate a generalization of the problem by considering Equation (1). We note that the first boundary condition in Equation (1) is standard in the theory of optimal mass transport [5]. The second boundary condition in Equation (1) is unusual. More precisely, it requires the convex conjugate of the unknown in the Monge–Ampere equation to be prescribed on a boundary of its a priori undetermined domain. Our aim is to investigate a class of prescribed functions for which Equation (1) admits a solution. In this paper, we impose that $a$ satisfies the following condition:
$a ′ ( s ) ≥ L$
for some $L > 0$ and all $s ∈ [ 0 , 1 )$. In addition to the above constraints, we assume that
$f ∈ L 1 ( Ω ) , Ω ⊂ [ 0 , L 0 ] × [ η 0 , L 0 ]$
with
$0 < η 0 < L 0 < L$
and we require h to satisfy the balance of mass equation:
$∫ Ω f ( x , y ) d x d y = ∫ Λ h a 2 ( s ) d s d z$
We propose a variational approach to Equation (1). Inspired by the Hamiltonian that comes along with the axisymmetric flows, we introduce the following functional:
$J ( u ) = ∫ Ω - u ( x , y ) f ( x , y ) d x d y + inf h ∈ H ∫ 0 1 d z ∫ 0 h ( z ) a ( s ) - u * ( s , z ) a 2 ( s ) d s$
We show that the maximizer of J over the set:
$U = u ∈ C ( Ω ¯ ) : u = u * *$
provides a solution for Equation (1). This paper is organized in the following way: In Section 2, we give some definitions and fix the notation. In Section 3, we provide some well-known results on the convex conjugate of functions. In Section 4, we consider the minimization problem involved in Equation (7) and establish some stability results. In Section 5, we prove our main result.

2. Notation and Definitions

In this section, we introduce some notation and recall some standard definitions.
• $H$ denotes the set of all continuous functions $h : [ 0 , 1 ] ⟶ [ 0 , 1 )$.
• Let $X ⊂ R 2$ be a convex set, $Y 1 , Y 2 ∈ X$ and $t ∈ [ 0 , 1 ]$. A function $v : X ⟼ R$ is convex if
$v ( t Y 1 + ( 1 - t ) Y 2 ) ≤ t v ( Y 2 ) + ( 1 - t ) v ( Y 2 ) .$
• Let $X ⊂ R 2$ be a convex set. If $v : X ⟼ R$ is a convex function and $Y 0 ∈ X$, the subdifferential of v at $Y 0$, denoted by $∂ . v ( Y 0 )$, is defined as:
$∂ . v ( Y 0 ) : = Z ∈ R 2 : v ( Y ) ≥ v ( Y 0 ) + 〈 Z , Y - Y 0 〉 ∀ Y ∈ X .$
• Given two Borel measures μ and ν of the same finite total mass on $R 2$, we say that a Borel map T pushes forward μ onto ν, and we write $T # μ = ν$ if
$μ T - 1 ( A ) = ν ( A )$
for all Borel sets $A ⊂ R 2$.
• Given two Borel measures μ and ν of the same finite total mass on $R 2$, $Γ ( μ , ν )$ denotes the set of all transport plans γ, such that:
$Π 1 # γ = μ and Π 2 # γ = ν .$
Here, $Π 1$ and $Π 2$ denote, respectively, the first and second projection maps.
Definition 2.1. Let $u : Ω ¯ ⟶ R$. We say that v is the convex conjugate of u if
$v ( Y ) = sup X ∈ Ω ¯ 〈 X , Y 〉 - u ( X ) for all Y ∈ Λ$
and we write $v = u *$. Similarly, let $v : Λ ¯ ⟶ R .$ We say that u is the convex conjugate of v if:
$u ( X ) = sup Y ∈ Λ ¯ 〈 X , Y 〉 - v ( Y ) for all X ∈ Ω$
and we write $u = v *$.
Remark 2.2. If u is convex and lower semicontinuous then:
$Y ∈ ∂ . u ( X ) if and only if u ( X ) + u * ( Y ) = 〈 X , Y 〉 .$
We consider the Brenier solutions of the Monge–Ampere equation (see [6,7]).
Definition 2.3. (Solution in the sense of Brenier) We say that $( u , h )$ is a weak solution for Equation (1) if u is Lipschitz continuous, h is continuous and,
$D u # χ Ω f = χ Λ h a 2 u * ( h ( z ) , z ) = a ( h ( z ) ) on h > 0 .$
Remark 2.4. Note that for a solution u of Equation (1), in the sense of Brenier, with only Lipschitz regularity, “$D u ( Ω ) ⊂ Λ h$” is to be understood as $D u ( x , y ) ∈ Λ h for a . e ( x , y ) ∈ Ω$.

3. Preliminaries

In this section, we collect some standard results on convex conjugate functions. We will give a sketchy proof and refer the reader to relevant references. Let us consider the Lipschitz continuous functions $v : Λ ⟶ R$, such that:
$0 ≤ ∂ s v ≤ L 0 and η 0 < ∂ z v ≤ L 0$
Lemma 3.1. Let $λ ∈ R$ and $u ∈ C ( Ω )$. Then,
(i)
$( u + λ ) * = u * - λ$.
(ii)
$∂ . u * ⊂ Ω ¯$. As a consequence, $u *$ is Lipschitz continuous and satisfies Equation (12).
(iii)
If $u = u * *$, then $∂ . u ⊂ Λ ¯$. In this case, if we assume in addition that $u * ( 0 , 0 ) = 0$, then there exists a constant $C L$ only dependent on L, such that:
$| | u | | L ∞ ( Ω ) + | | D u | | L ∞ ( Ω ) ≤ C L 0 a n d | | u * | | L ∞ ( Λ ) + | | D u * | | L ∞ ( Λ ) ≤ C L 0 .$
Proof. (i) is trivial. To obtain (ii), we observe that $u *$ is the supremum of affine (and so, convex) functions. Therefore, $u *$ is convex and lower semicontinuous. In light of Remark 2.2, Equation (9) implies that $∂ . u * ⊂ Ω ¯$. In view of the second equation of Equation (4), $u *$ is Lipschitz continuous and satisfies Equation (12). If $u = u * *$, then a similar argument as in (ii) yields $∂ . u ⊂ Λ ¯$. Set
$C L : = 2 + 8 L 0 .$
Since $∂ . u * ⊂ Ω ¯$, we have:
$| | D u * | | L ∞ ( Λ ) ≤ 2 L 0$
And so,
$| u * ( Y ) = | u * ( Y ) - u * ( 0 , 0 ) | ≤ 2 L 0 | Y | ≤ 4 L 0$
for all $X ∈ Ω$. Thus,
$| | u * | | L ∞ ( Ω ) + | | D u * | | L ∞ ( Ω ) ≤ C L$
Similarly, as $∂ . u ⊂ Λ ¯$,
$| | D u | | L ∞ ( Ω ) ≤ 2$
We exploit Equations (10) and (14) to obtain that
$| u ( X ) | ≤ 8 L 0$
so that
$| | u | | L ∞ ( Ω ) + | | D u | | L ∞ ( Ω ) ≤ C L .$
This proves (iii). ☐
The proof of the following Lemma can be seen in [8,9].
Lemma 3.2. Let $u ∈ C ( Ω )$, $φ ∈ C c ( Ω )$ and $ε > 0$.
Then,
$lim ε → 0 ( u + ε φ ) * - u * ε = - φ ∘ D u * a . e .$
and
$| | ( u + ε φ ) * - u * ε | | ∞ ≤ ∥ φ ∥ ∞ .$

4. A Minimization Problem and Some Stability Results

For any $v ∈ C ( Λ )$, we define
$S v ( s , z ) : = ∫ 0 s a ( t ) - v ( t , z ) a 2 ( t ) d t , ( s , z ) ∈ Λ .$
Lemma 4.1. Let v and $v n n = 0 ∞ ⊂ C ( Λ ¯ )$ satisfy Equation (12).
(i)
The sub-levels of $S v ( · , z )$ are bounded, uniformly for all $z ∈ [ 0 , 1 ]$: for $m ∈ R$, there exists a constant $c v , m$, such that $0 < c v , m < 1$ and
$inf 0 ≤ z ≤ 1 S v ( · , z ) ≤ m ⊂ [ 0 , c v , m ]$
Moreover, if $v n n = 1 ∞$ is uniformly bounded, then, for $m ∈ R$, there exists a constant $C m$, such that:
$0 < C m < 1$
and
$⋃ n = 0 ∞ inf 0 ≤ z ≤ 1 S v n ( · , z ) ≤ m ⊂ [ 0 , C m ]$
(ii)
Fix $z ∈ [ 0 , 1 ]$. There exists $λ ∈ [ 0 , 1 )$ such that
$S v ( λ , z ) ≤ S v ( s , z ) f o r a l l s ∈ [ 0 , 1 )$
Furthermore, either $λ = 0$ with $a ( 0 ) ≥ v ( 0 , z )$ or $λ ∈ ( 0 , 1 )$ and satisfies
$a ( λ ) = v ( λ , z )$
(iii)
Let $z n n = 0 ∞ ⊂ [ 0 , 1 ]$, such that $z n n = 1 ∞$ converges to $z 0$. Let $λ n$ satisfy Equation (19) with z replaced by $z n$ and v replaced by $v n$for $n ≥ 0$. Assume that $λ 0$ satisfies Equation (19) uniquely with z replaced by $z 0$, v replaced by $v 0$ and that $v n n = 1 ∞$ converges uniformly to $v 0$. Then, $λ n n = 1 ∞$ converges to $λ 0$.
Proof. 1. Since $lim t → Â ´ 1 ( 1 - t ) a ( t ) = 1$, we can choose $s 0 ∈ [ 0 , 1 )$, such that
$1 ≤ 2 a ( t ) ( 1 - t )$
for all $s 0 ≤ t < 1$. Setting $d = | | v | | ∞$ and invoking the fact that $lim s → 1 a ( s ) = ∞$, we can further choose $s 0$, such that
$a ( t ) ≥ 2 d$
for all $s 0 ≤ t < 1$. We exploit Equations (21) and (22) to obtain
$1 16 ( 1 - t ) 3 ≤ a 3 ( t ) 2 ≤ a 3 ( t ) - d a 2 ( t )$
for all $s 0 ≤ t < 1$. Note that:
$a 3 ( t ) - d a 2 ( t ) ≤ a ( t ) - v ( t , z ) a 2 ( t )$
for all $s 0 ≤ t < 1$. We combine Equations (23) and (24) to get
$∫ s 0 s 1 2 ( 1 - t ) 3 d t ≤ ∫ s 0 s a ( t ) - v ( t , z ) a 2 ( t ) d t$
for all $s 0 ≤ s ≤ 1$ and $0 ≤ z ≤ 1$. Therefore,
$∫ 0 s 0 a ( t ) - v ( s , z ) a 2 ( t ) d t + ∫ s 0 s 1 2 ( 1 - t ) 3 d t ≤ S v ( s , z )$
Note that the first term of Equation (26) is finite and that
$lim s → 1 - ∫ s 0 s 1 2 ( 1 - t ) 3 d t = ∞$
Let $m ∈ R$. In view of Equation (27), the Equation (26) implies that if
$inf 0 ≤ z ≤ 1 S v ( s , z ) ≤ m ,$
then there exists a constant $c v , m$, such that
$0 ≤ s ≤ c v , m < 1 .$
In other words, the sub-levels of $S v ( · , z )$ are bounded, uniformly for all $z ∈ [ 0 , 1 ]$.
2. Consider $v n n = 1 ∞ ⊂ C ( Λ )$. Following the reasoning above, we obtain
$∫ 0 s 0 a ( t ) - v n ( s , z ) a 2 ( t ) d t + ∫ s 0 s 1 2 ( 1 - t ) 3 d t ≤ S v n ( s , z )$
Assume that $v n n = 1 ∞$ is uniformly bounded. Then, the first term in Equation (28) is bounded. In view of Equation (27), Equation (28) implies that if
$inf z ∈ ( 0 , 1 ) S v n ( s , z ) ≤ m$
then there exists a constant $C m$, such that
$0 ≤ s ≤ C m < 1 .$
3. Fix $z ∈ [ 0 , 1 ]$. The continuity of $S v ( · , z )$ ensures that:
$s : S v ( · , z ) ≤ m$
is closed and then compact in view of Equation (16). We use again the continuity of $S v ( · , z )$ to obtain that $S v ( · , z )$ has a minimizer in $[ 0 , c v , m ]$. This ensures the existence of λ in Equation (19). If $λ > 0$, we use the differentiability of $S v ( · , z )$ on $( 0 , 1 )$ to obtain that $∂ s S v ( λ , z ) = 0$, that is $a ( λ ) = v ( λ , z ) .$ If $λ = 0$, then $a ( 0 ) - v ( 0 , z ) = ∂ s S v ( 0 , z ) ≥ 0$. This proves (ii).
4. Now, let us prove (iii). Note that $λ n n = 1 ∞ ⊂ [ 0 , C 0 ]$, and so, there exists a subsequence of $λ n n = 1 ∞$ still denoted by $λ n n = 1 ∞$ that converges to some $λ * .$ For $s ∈ [ 0 , 1 )$, we have:
$S v n ( λ n , z n ) ≤ S v 0 ( s , z n )$
Let M be a constant, such that $m a x ( s , C m ) ≤ M < 1$. As $v n n = 1 ∞$ converges uniformly to $v 0$, we have that $S v n n = 1 ∞$ converges uniformly to $S v 0$ on $[ 0 , M ] × [ 0 , 1 ]$. This, along with the continuity of $S v n$ and Equation (29), yields:
$S v 0 ( λ * , z 0 ) = lim n → ∞ S v n ( λ n , z n ) ≤ lim n → ∞ S v n ( s , z n ) = S v 0 ( s , z 0 )$
As s is arbitrary and $λ 0$ is the unique solution of Equation (19) with z replaced by $z 0$, we see Equation (30) to conclude that $λ 0 = λ *$, and so, the whole sequence $λ n n = 1 ∞$ converges to $λ 0$. ☐
Lemma 4.2. We assume that v satisfies Equation (12).
(1)
Let $z ∈ [ 0 , 1 ]$ and $λ i$ satisfy Equation (19), $i = 1 , 2$. Then,
$λ 1 = λ 2 .$
(2)
Let $z 1 , z 2 ∈ [ 0 , 1 ]$ and $λ 1 , λ 2$ satisfy Equation (19), respectively, for z replaced, respectively, by $z 1$ and $z 2$. Then:
$z 1 < z 2 ⟹ λ 1 ≤ λ 2$
Proof. 1. Fix $z ∈ [ 0 , 1 ]$, and note that $S v ( 0 , z ) = 0$. If λ is as in Equation (19) and $c v , 0$ as in Equation (16), then
$0 ≤ λ ≤ c v , 0 .$
By Lemma 4.1 (ii), either $λ = 0$ with $a ( 0 ) - v ( 0 , z ) ≥ 0$ or $λ ∈ ( 0 , 1 )$ with $a ( λ ) = v ( λ , z )$. Assume $λ = 0$. In view of Equations (3), (5) and (12), we have that $∂ s ( a - v ( · , s ) ) > 0$, so that:
$0 ≤ a ( 0 ) - v ( 0 , z ) < a ( s ) - v ( s , z ) for 0 < s < 1 .$
And so,
$S v ( 0 , z ) = 0 < ∫ 0 s a ( t ) - v ( t , z ) a 2 ( t ) d s = S v ( s , z ) for 0 < s < 1 .$
It follows that if $λ = 0$, then Equation (19) holds uniquely for $λ = 0$.
Let $λ 1 , λ 2 ∈ ( 0 , 1 )$, such that
$a ( λ i ) = v ( λ i , z ) , i = 1 , 2 .$
Since $∂ s v ( s , z ) ≤ L 0$, we have
$a ( λ 2 ) - a ( λ 1 ) = v ( λ 2 , z ) - v ( λ 1 , z ) ≤ L 0 ( λ 2 - λ 1 )$
On the other hand, we use Equation (3) to obtain:
$a ( λ 2 ) - a ( λ 1 ) ≥ L ( λ 2 - λ 1 )$
We combine Equations (32) and (33) to get that
$λ 1 = λ 2 .$
2. Let $z 1 , z 2 ∈ [ 0 , 1 ]$, such that $z 1 < z 2$, and let $λ 1 , λ 2$ satisfy Equation (19) for z replaced respectively by $z 1$ and $z 2$. As $∂ z v ( s , · ) > 0$, we have
$v ( s , z 1 ) ≤ v ( s , z 2 ) for 0 ≤ s < 1 .$
and so,
$S v ( s , z 2 ) ≤ S v ( s , z 1 ) for 0 ≤ s < 1$
If $λ 1 = 0$, then Equation (31) trivially holds. Assume $λ 2 = 0$. Then, we use the fact that $λ 2$ satisfies Equation (19) for z replaced by $z 2$ and Equation (34) to get
$0 = S v ( λ 2 , z 2 ) ≤ S v ( λ 1 , z 2 ) ≤ S v ( λ 1 , z 1 )$
Again, as $λ 1$ satisfies Equation (19) for z replaced by $z 1$, we have
$S v ( λ 1 , z 1 ) ≤ S v ( 0 , z 1 ) = 0$
We combine Equations (35) and (36) to obtain that
$S v ( λ 1 , z 1 ) = 0$
By the uniqueness result in Part (1),
$λ 1 = 0$
Thus, Equation (31) holds. Assume next that $λ 1 , λ 2 ∈ ( 0 , 1 )$. Then,
$v ( λ i , z i ) = a ( λ i ) , i = 1 , 2$
We use again the fact that $∂ z v ( s , · ) > 0$ to obtain:
$v ( λ 1 , z 1 ) ≤ v ( λ 1 , z 2 )$
In light of Equation (39), the equation in Equation (40) becomes
$a ( λ 1 ) ≤ v ( λ 1 , z 2 )$
and so,
$a ( λ 1 ) - v ( λ 1 , z 2 ) ≤ 0 = a ( λ 2 ) - v ( λ 1 , z 2 )$
As $∂ s ( a - v ( · , z 2 ) ) > 0$, we have that $a - v ( · , z 2 )$ is monotone increasing. Thus, Equation (42) yields:
$λ 1 ≤ λ 2$
so that Equation (31) holds. ☐
Proposition 4.3. Let $v n n = 0 ∞ ⊂ C ( Λ ¯ )$ satisfying Equation (12).
(i)
The functional
$F v 0 : h ⟼ ∫ 0 1 S v 0 ( h ( z ) , z ) d z$
has a unique minimizer $h 0$ over the set of all continuous functions $h : [ 0 , 1 ] ⟶ [ 0 , 1 )$. Moreover, $h 0$ is monotone, and $h 0 ( z )$ satisfies Equation (19) for v replaced by $v 0$.
(ii)
Assume that $v n n = 1 ∞$ is uniformly convergent to $v 0$ and $h n$ is the minimizer of $F v n$. Then,
$F v n ( h n ) ⟶ F v 0 ( h 0 )$
Proof. Define $h n$, $n ≥ 0$, in the following way:
$h n ( z ) is the minimizer of S v n ( · , z ) over [ 0 , 1 ) .$
Lemma 4.2 (2) shows that $h n$ is monotone increasing; Lemma 4.1 (iii) ensures that each $h n$ is continuous. In order to prove (i), we claim that $h 0$ is the unique solution for the following minimization problem:
$inf h ∫ 0 1 S v ( h ( z ) , z ) d z$
The fact that $h 0$ is a solution for Equation (45) is straightforward as a result of Equation (19). Assume that $h ˜$ is another minimizer of $F v 0$ as above. Then,
$∫ 0 1 S v ( h 0 ( z ) , z ) d z = ∫ 0 1 S v ( h ˜ ( z ) , z ) d z$
and
$S v ( h 0 ( z ) , z ) ≤ S v ( h ˜ ( z ) , z )$
We use Equations (46) and (47) to obtain that
$S v ( h 0 ( z ) , z ) = S v ( h ˜ ( z ) , z ) a . e .$
By the uniqueness of the minimizer in Lemma 4.2 (1), we use Equation (48) to conclude that $h 0 = h ˜ a . e .$, and the continuity of $h 0$ and $h ˜$ yields $h 0 = h ˜$.
3. Since $S v 0 ( 0 , z ) = 0$ for all $z ∈ [ 0 , 1 ]$, we have that $0 ≤ h n , h 0 ≤ C 0$, where $C 0$ is provided by Equation (17). As $h n$ is monotone, the Helly theorem implies that there exists a subsequence $h n k k = 1 ∞$ of $h n n = 1 ∞$, such that $h n k k = 1 ∞$ converges pointwise to some function g. In view of Lemma 4.1 (iii), we have that $g = h 0$.
Observe that
$| S v n k ( h n k ( z ) , z ) - S v ( h 0 ( z ) , z ) | ≤ | ∫ 0 h 0 ( z ) | v ( s , z ) - v n k ( s , z ) | a 2 ( s ) d s | + | ∫ h 0 ( z ) h n k ( z ) a ( s ) - v n k ( s , z ) a 2 ( s ) d s |$
Note that
$lim sup k → ∞ | ∫ 0 h 0 ( z ) v ( s , z ) - v n k ( s , z ) a 2 ( s ) d s | ≤ lim sup n → ∞ ∫ 0 C 0 | v ( s , z ) - v n k ( s , z ) | a 2 ( s ) d s ≤ lim sup k → ∞ ∫ 0 C 0 a 2 ( s ) d s | | v - v n k | | ∞ = 0$
As ${ v n } n = 1 ∞$ converges uniformly, it is bounded in the uniform norm by a constant, say $C 0$. Note that
$M 0 : = | | a | | L ∞ ( [ 0 , C 0 ] ) + C 0 | | a | | L ∞ ( [ 0 , C 0 ] ) 2 < ∞ .$
We use the fact that ${ v n } n = 1 ∞$ is bounded along with the pointwise convergence of $h n k k = 1 ∞$ to obtain
$lim sup n → ∞ | ∫ h 0 ( z ) h n k ( z ) a ( s ) - v n k ( s , z ) a 2 ( s ) d s | ≤ M 0 lim sup n → ∞ | ∫ h 0 ( z ) h n k ( z ) d s | ≤ M 0 lim sup n → ∞ | h 0 ( z ) - h n k ( z ) | = 0$
We combine Equations (49)–(51) to obtain that
$lim k → ∞ S v n k ( h n k ( z ) , z ) = S v ( h 0 ( z ) , z )$
Note that
$| S v n k ( h n ( z ) , z ) | ≤ ∫ 0 C 0 | | a | | L ∞ ( [ 0 , C 0 ] ) + | | v n | | ∞ | | a | | L ∞ ( [ 0 , C 0 ] ) 2 d s ≤ C 0 M 0 | | a | | L ∞ ( [ 0 , C 0 ] ) 2$
Invoking the Lebesgue-dominated convergence theorem, we use Equations (53) and (52) to prove Equation (44).

5. A Maximization Problem and Main Result

We recall that
$J ( u ) = ∫ Ω - u ( x , y ) f ( x , y ) d x d y + inf h ∈ H ∫ 0 1 d z ∫ 0 h ( z ) a ( s ) - u * ( s , z ) a 2 ( s ) d s$
and:
$U = u ∈ C ( Ω ¯ ) : u = u * * .$
Lemma 5.1. The functional J is bounded above on $C ( Ω )$.
Proof. Let $h 0 ∈ H$ be a constant function, such that Equation (6) holds, and let $γ 0 ∈ Γ χ Ω f , χ Λ h 0 a 2$. Observe that:
$| 〈 X , Y 〉 | ≤ | X | | Y | ≤ 4 L 0 for all X ∈ Ω , Y ∈ Λ ,$
and
$γ 0 ( Ω × Λ h 0 ) = | | f | | L 1 ( Ω ) .$
Note that
$M 0 : = 4 L 0 | | f | | L 1 ( Ω ) + ∫ 0 h 0 a 3 ( s ) d s < ∞$
Let $u ∈ C ( Ω )$. Then, using Equation (9), we have
$- u ( X ) - u * ( Y ) ≤ - 〈 X , Y 〉 with X : = ( x , y ) ∈ Ω , Y : = ( s , z ) ∈ Λ .$
and so,
$∫ Ω - u ( x , y ) f ( x , y ) d x d y + ∫ 0 1 d z ∫ 0 h 0 - u * ( s , z ) a 2 d s = ∫ Ω × Λ h 0 - u ( X ) - u * ( Y ) = ∫ Ω × Λ h 0 - 〈 X , Y 〉 d γ 0 .$
Therefore,
$∫ Ω - u ( x , y ) f ( x , y ) d x d y + ∫ 0 1 d z ∫ 0 h 0 a ( s ) - u * ( s , z ) a 2 ( s ) d s ≤ ∫ Ω × Λ h 0 - 〈 X , Y 〉 d γ 0 + ∫ 0 h 0 a 3 ( s ) d s .$
In light of Equation (54), we have
$∫ Ω - u ( x , y ) f ( x , y ) d x d y + ∫ 0 1 d z ∫ 0 h 0 a ( s ) - u * ( s , z ) a 2 ( s ) d s ≤ M 0 .$
Thus, in view of the infimum term in $J ( u )$, we have
$J ( u ) ≤ M 0 .$
which proves the Lemma. ☐
Proposition 5.2. The functional $u ⟼ J ( u )$ admits a maximizer on $U$.
Proof. Let $u ∈ U$. In light of Lemma 5.1, set
$j = sup u ∈ U J ( u )$
Let $u n n ⊂ C ( Ω ¯ )$ be a maximizing sequence for the maximization problem in Equation (55). In what follows, we show in Step 1 that $u n n ∈ C ( Λ ¯ )$ converges up to a subsequence and in Step 2; we show that its limit is a maximizer in Equation (55).
Step 1.
Let $m ∈ [ 0 , 1 )$, and as $S u n * ( 0 , z ) = 0$, note that
$inf h ∈ H ∫ 0 1 d z ∫ 0 h ( z ) a ( s ) - u n * ( s , z ) a 2 ( s ) d s ≤ ∫ 0 1 d z ∫ 0 m a ( s ) - u n * ( s , z ) a 2 ( s ) d s$
Therefore,
$J ( u n ) ≤ ∫ Ω - u n ( x , y ) f ( x , y ) d x d y + ∫ 0 1 d z ∫ 0 m a ( s ) - u n * ( s , z ) a 2 ( s ) d s$
Next, we set
$u ¯ n : = u n + u n * ( 0 , 0 ) .$
By Lemma 3.1 (i),
$u ¯ n * : = u n * - u n * ( 0 , 0 ) .$
These, combined with Equation (57), yield that
$J ( u n ) ≤ ∫ Ω - u ¯ n ( x , y ) f ( x , y ) d x d y + u n * ( 0 , 0 ) | | f | | L 1 ( Ω ) + ∫ 0 1 d z ∫ 0 m a ( s ) - u ¯ n * ( s , z ) - u n * ( 0 , 0 ) a 2 ( s ) d s$
As $u ¯ n * ( 0 , 0 ) = 0$ and $u ¯ n * * = u ¯ n$, we use Lemma 3.1 (iii) to obtain
$| | u ¯ n | | L ∞ ( Ω ) + | | D u ¯ n | | L ∞ ( Ω ) ≤ C L 0 and | | u ¯ n * | | L ∞ ( Ω ) + | | D u ¯ n * | | L ∞ ( Ω ) ≤ C L 0$
Therefore, Equations (58) and (59) imply that
$J ( u n ) ≤ C L | | f | | L 1 ( Ω ) + u n * ( 0 , 0 ) | | f | | L 1 ( Ω ) + ∫ 0 m a ( s ) + C L - u n * ( 0 , 0 ) a 2 ( s ) d s$
That is,
$J ( u n ) - C L | | f | | L 1 ( Ω ) - ∫ 0 m a ( s ) + C L a 2 ( s ) d s ≤ u n * ( 0 , 0 ) | | f | | L 1 ( Ω ) - ∫ 0 m a 2 ( s ) d s$
In view of Equation (2), we use Equation (21) to get
$1 4 ( 1 - m ) + 1 4 ( 1 - s 0 ) = ∫ s 0 m 1 4 ( 1 - s ) 2 d s ≤ ∫ s 0 m a 2 ( s ) d s$
for some $s 0 ∈ [ 0 , 1 )$. We use Equation (62) to obtain:
$lim m → 1 ∫ 0 m a 2 ( s ) d s = + ∞ .$
Therefore, we can choose $m 0 ∈ [ 0 , 1 )$, such that:
$| | f | | L 1 ( Ω ) < ∫ 0 m 0 a 2 ( s ) d s .$
Setting first $m = 0$ and then $m = m 0$ in Equation (61), we obtain:
$J ( u n ) - C L 0 | | f | | L 1 ( Ω ) | | f | | L 1 ( Ω ) ≤ u n * ( 0 , 0 ) ≤ J ( u n ) - C L | | f | | L 1 ( Ω ) - ∫ 0 m 0 a ( s ) + C L 0 a 2 ( s ) d s | | f | | L 1 ( Ω ) - ∫ 0 m 0 a 2 ( s ) d s$
We next assume, without loss of generality, that
$j - 1 ≤ J ( u n ) for all n ≥ 1 .$
Then,
$j - 1 - C L | | f | | L 1 ( Ω ) | | f | | L 1 ( Ω ) ≤ u n * ( 0 , 0 ) ≤ j - C L | | f | | L 1 ( Ω ) - ∫ 0 m 0 a ( s ) + C L a 2 ( s ) d s | | f | | L 1 ( Ω ) - ∫ 0 m 0 a 2 ( s ) d s$
In view of Equation (64), $u n * ( 0 , 0 ) n = 1 ∞$ is bounded, and so, there exists a subsequence still denoted $u n * ( 0 , 0 ) n = 1 ∞$ that converges to some $C * ∈ R$. In light of Equation (59), we use Ascoli–Azerla to conclude that there exists a subsequence of $u n k k = 1 ∞$ still denoted by $u n k k = 1 ∞$, such that
$u ¯ n k k = 1 ∞ converges uniformly to u ¯ 0$
and
$u ¯ n k * k = 1 ∞ converges uniformly to u ¯ 0 * .$
We use the last two displayed convergence results and the convergence of $u n k ( 0 , 0 ) k = 1 ∞$ to obtain that:
$u n k k = 1 ∞ converges uniformly to u 0 = u ¯ 0 + C *$
and
$u n k * k = 1 ∞ converges uniformly to u 0 * = u ¯ 0 * + C * .$
Step 2.
To show the existence of a maximizer, it will be enough to study the continuity in the second term in the expression of J. Let $h ˜ n k$ and $h ˜$ denote respectively the minimizers in the second term of $J ( u n k )$ and $J ( u 0 )$. As $u n k * n ⊂ C ( Λ ¯ )$ satisfies Equation (12) and converges uniformly to $u 0 *$, we use Equation (44) to get:
$F u n k * ( h ˜ n k ) ⟶ F u * ( h ˜ )$
It follows that
$J ( u n k ) ⟶ J ( u 0 )$
As $u n n$ is a maximizing sequence, we have
$J ( u 0 ) = sup u ∈ U J ( u ) ,$
which concludes the proof. ☐
Theorem 5.3. Let $u 0 ∈ U$, such that:
$J ( u 0 ) = sup u ∈ U J ( u ) ,$
and $h 0$ the minimizer in Equation (43) for $v 0$ replaced by $u 0 *$. Then, $( u 0 , h 0 )$ provides a weak solution for Equation (1).
Proof. Let $ε > 0$ and $φ ∈ C c ( Ω )$. We define the functions $h 0$ and $h ε$ from $[ 0 , 1 ]$ to $[ 0 , 1 )$, such that for each $z ∈ [ 0 , 1 ]$ fixed, $h 0 ( z )$ and $h ε ( z )$ satisfy respectively Equation (19) for v replaced by $u 0 *$ and Equation (19) for v replaced by $( u 0 + ε φ ) *$. Then, in light of Proposition 4.3 (i),
$J ( u 0 ) = ∫ Ω - u 0 f ( x , y ) d x d y + ∫ 0 1 d z ∫ 0 h 0 ( z ) a ( s ) - u 0 * ( s , z ) a 2 ( s ) d s$
and
$J ( u 0 + ε φ ) = ∫ Ω - ( u 0 + ε φ ) f d x d y + ∫ 0 1 d z ∫ 0 h ε ( z ) a - ( u 0 + ε φ ) * a 2 ( s ) d s$
We use the definition of $h 0$ and $h ε$ to establish that
$S u 0 ( h 0 ( z ) , z ) - S u 0 + ε φ ( h ϵ ( z ) , z ) ≤ S u 0 ( h ϵ ( z ) , z ) - S u 0 + ε φ ( h ϵ ( z ) , z ) = ∫ 0 h ε ( z ) u 0 * - ( u 0 + ε φ ) * a 2 ( s ) d s .$
Likewise,
$S u 0 + ε φ ( h ϵ ( z ) , z ) - S u ( h 0 ( z ) , z ) ≤ S u + ε φ ( h 0 ( z ) , z ) - S u ( h 0 ( z ) , z ) = ∫ 0 h 0 ( z ) ( u 0 + ε φ ) * - u * a 2 ( s ) d s .$
Combining the last two displayed equations, we obtain:
$∫ 0 h ϵ ( z ) ( u 0 + ε φ ) * - u 0 * a 2 ( s ) d s ≤ S u 0 + ε φ ( h ϵ ( z ) , z ) - S u 0 ( h 0 ( z ) , z ) ≤ ∫ 0 h 0 ( z ) ( u 0 + ε φ ) * - u 0 * a 2 ( s ) d s$
Note that
$lim sup ε → 0 1 ε | ∫ 0 h ϵ ( z ) ( u 0 + ε φ ) * - u 0 * a 2 ( s ) d s - ∫ 0 h 0 ( z ) ( u 0 + ε φ ) * - u 0 * a 2 ( s ) d s | ≤ | ∫ h 0 ( z ) h ϵ ( z ) ( u 0 + ε φ ) * - u 0 * ε a 2 ( s ) d s | ≤ | | φ | | ∞ ∫ 0 1 a 2 ( s ) d s lim sup ε → 0 | h ϵ ( z ) - h 0 ( z ) |$
In view of Lemma 4.1 (iii), we use the fact that $( u 0 + ε φ ) *$ uniformly converges to $u 0 *$ as obtained Lemma 3.2 and standard arguments on sequences to show that
$lim ϵ → 0 h ϵ ( z ) = h 0 ( z )$
for all $z ∈ [ 0 . 1 ]$. Combining Equations (70) and (71), it follows that
$lim sup ε → 0 1 ε | ∫ 0 h ϵ ( z ) ( u + ε φ ) * - u 0 * a 2 ( s ) d s - ∫ 0 h 0 ( z ) ( u + ε φ ) * - u * a 2 ( s ) d s | = 0$
We use Equations (69) and (72) and Lemma 3.2 to obtain:
$lim ε → 0 S u 0 + ε φ ( h ϵ ( z ) , z ) - S u 0 ( h 0 ( z ) , z ) ε = lim ε → 0 ∫ 0 h 0 ( z ) ( u 0 + ε φ ) * - u 0 * ε a 2 ( s ) d s = - ∫ 0 h 0 ( z ) φ ∘ D u 0 * a 2 ( s ) d s$
Using Lemma 3.2, we note that
$| ∫ 0 h 0 ( z ) ( u 0 + ε φ ) * - u 0 * ε a 2 ( s ) d s | ≤ | | φ | | ∞ ∫ 0 C 0 a 2 ( s ) d s$
where $C 0$ is provided as in Lemma 4.1. Thus, in light of Equations (73) and (74), we use the Lebesgue-dominated convergence theorem,
$lim ε → 0 ∫ 0 1 S u + ε φ ( h ϵ ( z ) , z ) - S u 0 ( h 0 ( z ) , z ) ε d z = - ∫ 0 1 ∫ 0 h 0 ( z ) φ ∘ D u 0 * a 2 ( s ) d s d z .$
Note that
$J ( u 0 + ε φ ) - J ( u 0 ) ε = ∫ Ω φ f d x d y + ∫ 0 1 S u 0 + ε φ ( h ϵ ( z ) , z ) - S u 0 ( h ( z ) , z ) ε d z$
As $u 0$ is a maximizer for J, we have
$0 = lim ε → 0 J ( u 0 + ε φ ) - J ( u 0 ) ε = ∫ Ω φ f ( x , y ) d x d y - ∫ 0 1 ∫ 0 h 0 ( z ) φ ∘ D u 0 * a 2 ( s ) d s d z$
Since φ is arbitrary, we obtain from Equation (76) that
$D u 0 * # χ Λ h 0 a 2 = χ Ω f$
As $u 0 = u 0 * *$ and $D u 0$, $D u 0 *$ exist almost everywhere with respect to the Lebesgue measure, we have
$u 0 ( X ) + u 0 * ( D u 0 ( X ) ) = 〈 X , D u 0 ( X ) 〉 χ Ω f a . e . ,$
and
$u 0 ( D u 0 * ( Y ) ) + u 0 * ( Y ) = 〈 D u 0 * ( Y ) , Y 〉 χ Λ a 2 a . e .$
It follows that
$D u 0 * ∘ D u 0 ( X ) = X χ Ω f a . e . and D u 0 ∘ D u 0 * ( Y ) = Y χ Λ a 2 a . e .$
As a consequence, Equation (77) implies that:
$D u # χ Ω f = χ Λ h 0 a 2$
Note that as $h 0 ( z )$ satisfies Equation (19), the Equation (20) holds, as well, whenever $h 0 > 0$, that is,
$u 0 * ( h 0 ( z ) , z ) = a ( h 0 ( z ) ) h 0 > 0$
We combine Equations (78) and (79) to obtain a weak solution of Equation (1). ☐

Conflicts of Interest

The author declares no conflict of interest.

References

1. Benamou, J.-D.; Brenier, Y. Weak existence for the semigeostrophic equations formulated as a coupled Monge Ampere/Transport problem. SIAM J. Appl. Math. 1998, 58, 1450–1461. [Google Scholar] [CrossRef]
2. Cullen, M.J.P.; Gangbo, W. A variational approach for the 2-D semi-geostrophic shallow water equations. Arch. Ration. Mech. Anal. 2001, 156, 241–273. [Google Scholar]
3. Ambrosio, L.; Gangbo, W. Hamiltonian ODEs in the Wasserstein space of probability measures. Commun. Pure Appl. Math. 2008, 61, 18–53. [Google Scholar] [CrossRef]
4. Cullen, M.; Sedjro, M. On a model of forced axisymmetric flows. SIAM J. Appl. Math. 2014, 6, 3983–4013. [Google Scholar] [CrossRef]
5. Villani, C. Topics in Optimal Transportation, Graduate Studies in Mathematics 58; American Mathematical Society: Providence, RI, USA, 2003. [Google Scholar]
6. Brenier, Y. Polar factorization and monotone rearrangement of vector-valued functions. Commun. Pure Appl. Math. 1991, 44, 375–417. [Google Scholar] [CrossRef]
7. Figal, A. Regularity properties of optimal maps between nonconvex domains in the plane. Commun. Partial Differ. Equ. 2010, 35, 465–479. [Google Scholar] [CrossRef]
8. Gangbo, W. An elementary proof of the polar decomposition of vector-valued functions. Arch. Ration. Mech. Anal. 1995, 128, 380–399. [Google Scholar]
9. Gangbo, W. Quelques problemes d’analyse convexe. Rapport d’habilitation à Diriger des Recherches. Available online: http://people.math.gatech.edu/∼gangbo/publications/habilitation.pdf (accessed on 2 September 2015).

Share and Cite

MDPI and ACS Style

Sedjro, M. A Monge–Ampere Equation with an Unusual Boundary Condition. Symmetry 2015, 7, 2009-2024. https://doi.org/10.3390/sym7042009

AMA Style

Sedjro M. A Monge–Ampere Equation with an Unusual Boundary Condition. Symmetry. 2015; 7(4):2009-2024. https://doi.org/10.3390/sym7042009

Chicago/Turabian Style

Sedjro, Marc. 2015. "A Monge–Ampere Equation with an Unusual Boundary Condition" Symmetry 7, no. 4: 2009-2024. https://doi.org/10.3390/sym7042009