Abstract
This paper investigates a P-function associated with solutions to boundary value problems of some generalized Monge-Ampère equations in bounded convex domains. It will be shown that P attains its maximum value either on the boundary or at a critical point of any convex solution. Furthermore, it turns out that such P-function is actually a constant when the underlying domain is a ball. Therefore, our results provide a best possible maximum principle in the sense of L. Payne. As an application, we will use these results to study an overdetermined boundary value problem. More specifically, we will show solvability of this overdetermined boundary value problem forces their P-function to be a constant.
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1 Introduction
We recall the notion of “best possible maximum principle” introduced by L. Payne a few decades ago [7, 8]. A function P that depends on solutions and their derivatives of a boundary value problem on bounded domains is said to satisfy a best possible maximum principle if P satisfies the weak maximum principle for every bounded domain \(\Omega \), and if there is a special domain on which it is a constant. As an example, let u satisfy
The function
satisfies
where \(u_i=\frac{\partial u}{\partial x_i},\) \(u_{ij}=\frac{\partial ^2 u}{\partial x_i\partial x_j}.\) It follows that P attains its maximum value on \(\partial \Omega \). If \(\Omega \) is a ball of radius R centered at the origin, we have
The corresponding function \(P=|Du|^2-\frac{2}{n}u\) is a constant in \(\Omega \).
In [2], the following problem is discussed. Let \(\Omega \subset {\mathbb {R}}^n\) be a bounded convex domain, and let u be a convex solution to the boundary value problem
Corresponding to this solution, consider the function
In [2], the authors prove that P satisfies a best possible maximum principle. Similar problems are discussed in [6, 9].
Let \(\Omega \subset {\mathbb {R}}^n\) be a bounded convex domain. For a smooth function u we have
where \(T_{(n-1)}(D^2u)=[T_{(n-1)}^{ij}(D^2u)]\) is the adjoint of the Hessian matrix \(D^2u\). Here and in what follows sub-indices denote partial differentiation, and the summation convention from 1 to n over repeated indices is in effect.
In the present paper, we find a best possible maximum principle relative to the following generalized Monge-Ampère equation. With \(p>1\), we consider the boundary value problem
Note that this generalization of the Monge-Ampère operator is similar to the p-Laplacian as the generalization of the Laplacian [13]. For a discussion of problems related to Monge-Ampère operators we refer to [1, 4, 5].
Problem (1) in case of \(p=2\) has been discussed in [2]. From now on we concentrate on the case \(p>1\), \(p\not =2\). We suppose \(u\in \Phi \), where
Let \(u_m=\min _\Omega u(x)=u(x_0)\). Clearly, \(|Du|=0\) at \(x_0\) only. We say that u is smooth in \({\mathcal {O}}\) if it is at least \(C^4\) in \({\mathcal {O}}\).
Note that the matrix \(D^2u\) is positive definite where \(|Du|>0\). We say that \(u\in \Phi \) is a solution to (1) if
We claim that
Indeed, let
If we multiply (1) by \((t-u)^+\) and we integrate over \(\Omega \) we find
It follows that
As \(t\rightarrow u_m\), the claim follows.
Define the P-function
and \(P(x_0)=-u_m\).
Our first result is the following
Theorem 1.1
Let \(\Omega \) be a bounded convex domain. If \(u\in \Phi \) is solution to Problem (1), the function P defined as in (2) attains its maximum value either on the boundary \(\partial \Omega \) or at the point where \(Du=0\). Moreover, if \(\Omega \) is a ball then P is a constant.
By using Theorem 1.1, we shall discuss the following overdetermined problem. Let \(u\in \Phi \) be a solution to Problem (1). Furthermore, if \(u_m=\min _\Omega u(x)\), suppose there is some constant c such that
where \({{\mathcal {H}}}_{(n-1)}\) is the Gauss curvature of \(\partial \Omega \).
Theorem 1.2
If there is a solution \(u\in \Phi \) to problem (1) which satisfies the additional condition (3), then the function P defined as in (2) is a constant in \(\Omega \).
Overdetermined problems for second order linear and quasilinear equations were discussed more than fifty years ago in the seminal papers [12, 15].
In case of \(n=2\), we shall prove the analogous of Theorem 1.1 for the minimum. Similar results are proved in [3]. As an application, we will prove Theorem 1.2 without the restriction \(\frac{p-1}{p}c\ge -u_m\).
2 A Best Possible Maximum Principle
Recall that, where \(|Du|>0\), the operator \(T_{(n-1)}^{ij}(D^2u)\) is divergence free (see, for example, [10, 11]), that is
Moreover, since \(T_{(n-1)}(D^2u)\) is the adjoint of the Hessian matrix \(D^2u\), we have
where I is the \(n\times n\) identity matrix. On using these facts, after some computation one finds
Therefore, Equation (1) for \(x\in \Omega \) such that \(|Du|>0\), can be written as
Proof of Theorem 1.1
Let \(x_0\in \Omega \) be the point where \(Du=0\), and let \(u_m=u(x_0)\). Arguing by contradiction, let \({\tilde{x}}\in {\Omega {\setminus } \{x_0\}}\) 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 \({\bar{x}}\in {\Omega {\setminus } \{x_0\}}\). We show that this cannot happen.
On using the equations (5) and (6) we find
where \((D^2u)^{-1}\) is the inverse matrix of \(D^2u\). Therefore, if \([u^{kl}]= (D^2u)^{-1}\) we have
We compute
Since
we find
Further differentiation yields
We note that our equation (1) is invariant under a rigid rotation. Let us make a suitable rotation around the point \({\bar{x}}\) such that \(D^2u\) has a diagonal form at this point. With some abuse of notation, \({\tilde{P}}_i\) denote derivatives of \({\tilde{P}}\) with respect to the new variables. Then,
Clearly, also \((D^2u)^{-1}\) will have a diagonal form at \({\bar{x}}\). Furthermore, for i fixed we have \( u^{ii}u_{ii}=1.\) Multiplying by \( u^{ii}\) the equation in above and adding from \(i=1\) up to \(i=n\) we get
Hence, since the matrix \(T_{(n-1)}(D^2u)\) is divergence free, we have
from which we find (recall that we are adding over repeated indices)
Therefore,
Insertion of this equation into (8) yields
Now we evaluate the quantity \( u^{ii}u^{kl}_{ii}u_ku_l\). Unfortunately, our computations are quite long. Since \([u^{kl}]\) is the inverse matrix of \([u_{kl}]\), we have
Differentiating with respect to \(x^i\) we find
Since \(D^2u\) has a diagonal form at \({\bar{x}}\), from the latter equation we find (here we do not add with respect to i, k or l)
Multiplying by \(u^{ii}\) and adding from \(i=1\) up to \(i=n\) we get
To evaluate the last quantity in (12), let us differentiate Equation (6) with respect to \(x^k\). We find
By (9) we have
By the last two equations we find
Simplifying we get
Further differentiation with respect to \(x^l\) yields
Note that we are using the condition that \(D^2u\) has a diagonal form at the point \({\bar{x}}\). Since \(u^{ij}_l=-u^{ii}u^{jj}u_{ijl}\) (with i and j fixed), by the previous equation we find
Insertion of (13) into (12) leads to
Simplifying we find
Multiplying by \(u_ku_l\) (and adding with respect to k and l) we get,
Inserting this equation into (11) we find
Let us evaluate the first quantity of the right hand side in (14). By using the inequality
with \(a_i=b_i(\sum _kc_{ik})\) we find
Hence,
Since
on using (10) we find
Therefore,
Insertion of (15) into (14) yields
To finish, we must evaluate the last quantity of the right hand side in (16). This is easy. By (7), at \({\bar{x}}\) (the point of maximum of \({\tilde{P}}\)) we have
whence,
Multiplying by \(u_i\) and adding from \(i=1\) up to \(i=n\) we find
Insertion of this equation into (16) yields
After simplification we get
contradicting the assumption that \({\bar{x}}\) is a point of maximum for \({\tilde{P}}\). It follows that P must attain its maximum value either on the boundary \(\partial \Omega \) or at the point where \(Du=0\).
Now consider the case \(\Omega \) is a ball of radius R, centered at the origin. If \(u=u(r)\), \(r=|x|\), we may assume \(u_1=u'\), \(u_i=0,\ \ 2\le i\le n.\) Then,
Therefore, Equation (6) reads as
or
Integrating we find
Integrating again and using the condition \(u(R)=0\) we find
Recall that our P-function reads as
where \([u^{ij}]\) is the inverse matrix of \([u_{ij}].\) Then,
Hence P is a constant, and the theorem is proved.
3 An Overdetermined Problem
Proof of Theorem 1.2
We note that (see [14])
where \({{\mathcal {H}}}_{(n-1)}\) is the Gauss curvature of \(\partial \Omega \). Therefore, we can write condition (3) as
By Theorem 1.1, the maximum of the function P is either \(\frac{p-1}{p}c\) (attained on \(\partial \Omega \)) or \(-u_m\) (attained where \(Du=0\)). Hence, since \(\frac{p-1}{p}c\ge -u_m\), we have
Recall that \(x_0\) is the point of minimum for u, and that \(u_m\) is the minimum value of u. For \(u_m\le t<0\) we define
Clearly, \(\Omega _{u_m}=\Omega .\) Moreover we have
Let \(e=(e^1,\cdots ,e^n)\) be the exterior unit normal to \(\partial \Omega _t\). On \(\partial \Omega _t\) we have \(u_k=|Du|e^k\). Therefore, using Equation (1) we find
On \(\Sigma _t\) we have (see [14])
where \({{\mathcal {H}}}_{(n-1)}\) is the Gauss curvature of \(\Sigma _t\). Hence,
On noting that
and that
we have
Hence, by (27) we find
and
Now we use a sort of Pohozaev identity. We find
Since
we have
Hence,
Finally, on using Equation (1) once more we find
Therefore,
Hence, letting \(t\rightarrow u_m\) in (22) and using (21), (23) and (24) we find
On the other hand, using condition (17) we find
From this equation and (24) it follows that
This together with (18) shows that \(P(x)=\frac{p-1}{p}c\) in \(\Omega \).
The theorem is proved.
4 The Case \(n=2\)
Theorem 4.1
Let \(\Omega \subset {\mathbb {R}}^2\) be a bounded convex domain. If \(u\in \Phi \) is a solution to Problem (1) in \(\Omega \), the function P defined as in (2) for \(n=2\) attains its minimum value either on the boundary \(\partial \Omega \) or at the point where \(Du=0\).
Proof
Let \(x_0\in \Omega \) be the point where \(Du=0\), and let \(u_m=u(x_0)\). Arguing by contradiction, let \({\tilde{x}}\in {\Omega {\setminus } \{x_0\}}\) be a point such that
Choose \(1<\tau <2-\frac{1}{p}\) with \(\tau \) close enough to 1 so that
Then, also the function
attains its minimum value at some point \({\bar{x}}\in {\Omega {\setminus } \{x_0\}}\). Choose \(\tau \) such that \(\tau <2-\frac{1}{p}\). We show that this cannot happen.
Let us write \({\tilde{P}}\) as
We perform a rigid rotation around \({\bar{x}}\) so that \(D^2u\) has a diagonal form at this point. By the same computation as in the proof of Theorem 1.1 we find Equation (14) with \(n=2\), that is,
To evaluate the last quantity in (28) we differentiate \(\tilde{P}\) with respect to \(x^s\). Since \({\bar{x}}\) is a point of minimum, by (27) we find
Recalling that \(D^2u\) has a diagonal form, from the latter equation we find
If we multiply by \(u_s\) we get
Inserting this into (27) we find
To finish, we must evaluate the first quantity in (30). From Equation (1) with \(n=2\) we find (recall that \(D^2u\) is assumed to be diagonal at \({\bar{x}}\))
Putting \(2(2-p)|Du|^{-2}:=\alpha \) we have the two equations
Moreover, from (29) we have the two more equations
The system of four equations (31)-(32) is linear with respect to \(u_{111}, u_{112}, u_{122}\) and \(u_{222}\), and the determinant of the coefficients is equal to \((u^{11}u^{22})^2S^2\), where
By elementary computation we find
We start computing \(u^{kk}u_{11k}u_k\). On using (33) and (34) we find
Since
we get
Since
and
by (43) we get
We find
Recalling that \(D^2u\) is diagonal, we have \(u_{11}u^{11}=u_{22}u^{22}=1\). Therefore, from (39) we get
Insertion of the latter result into (38) yields
Now we compute \(u^{kk}u_{22k}u_k\). By using (35) and (36) (or changing the index 1 and 2 in (40)) we find
Finally, let us compute \(u^{kk}u_{12k}u_k\). By using (34) and (35), we find
After some simplification we find
Further simplification and use of the equations \(u_{11}u^{11}=u_{22}u^{22}=1\) yields
By using (40), (41) and (42) we find
If we expand the powers in above and use the equations
we find
Since \(\alpha |Du|^{2}=2(2-p)\), we have
Therefore,
Inserting the latter result into (30), after simplification we find
Since \(1<\tau <2-\frac{1}{p}\) we have \(p(2-\tau )-1>0\) and \(u^{ii} P_{ii}<0\), contradicting the assumption that \({\bar{x}}\) was a point of minimum. The theorem is proved.
If \(n=2\), we prove Theorem 1.2 by using the following condition
where c is some positive constant and \({{\mathcal {H}}}_{(1)}\) is the curvature of the boundary \(\partial \Omega \).
Theorem 4.2
Let \(\Omega \subset {\mathbb {R}}^2\) be a bounded convex domain. If there is a solution \(u\in \Phi \) to problem (1) in \(\Omega \) which satisfies the additional condition (43), then the function P defined as in (2) with \(n=2\) is a constant in \(\Omega \).
Proof
If \(\frac{p-1}{p}c\ge -u_m\), then the conclusion follows by Theorem 1.2. So, in what follows, we suppose \(\frac{p-1}{p}c< -u_m\). By Theorem 4.1, the function P has its minimum value on \(\partial \Omega \). Therefore
By the same computation as in the proof of Theorem 1.2 we find (21) with \(n=2\), that is,
On the other hand, by the Pohozaev identity (26) for \(n=2\) we have
This result together with (44) shows that \(P(x)=\frac{p-1}{p}c\) in \(\Omega \).
The theorem is proved.
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Mohammed, A., Porru, G. A Best Possible Maximum Principle and an Overdetermined Problem for a Generalized Monge-Ampère Equation. J Geom Anal 34, 49 (2024). https://doi.org/10.1007/s12220-023-01500-w
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DOI: https://doi.org/10.1007/s12220-023-01500-w
Keywords
- Monge-Ampère type equations
- P-function
- Best possible maximum principle
- Overdetermined boundary-value problem