Lyapunov type inequality for extremal Pucci's equations
J. Tyagi, R.B. Verma

TL;DR
This paper derives a Lyapunov inequality for extremal Pucci's equations, extending classical results to fully nonlinear elliptic equations, which could impact stability analysis and eigenvalue estimates.
Contribution
It introduces a Lyapunov type inequality for extremal Pucci's equations, generalizing classical inequalities to a broader class of nonlinear elliptic PDEs.
Findings
Established Lyapunov inequality for extremal Pucci's equations
Extended classical Lyapunov inequalities to fully nonlinear elliptic equations
Provides tools for stability and eigenvalue analysis in nonlinear PDEs
Abstract
In this article, we establish Lyapunov type inequality for the following extremal Pucci's equation \begin{equation*} \left\{ \begin{aligned}{} \mathcal{M}^{+}_{\lambda,\Lambda}(D^{2}u)+a(x)u&=0~\text{in}~\Omega,\\ u&=0~\text{on}~\partial\Omega, \end{aligned} \right. \end{equation*} where is a smooth bounded domain in . This works generalize the well-known works on Lyapunov inequalities to fully nonlinear elliptic equations.
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Taxonomy
TopicsAdvanced Mathematical Modeling in Engineering · Numerical methods in inverse problems · Nonlinear Partial Differential Equations
Lyapunov type inequality for extremal Pucci’s equations
J.Tyagi, R.B. Verma
J. Tyagi Indian Institute of Technology Gandhinagar
Palaj, Gandhinagar Gujarat, India-382355.
[email protected], [email protected]
R.B.Verma Indian Institute of Technology Gandhinagar
Palaj, Gnadhinagar Gujarat India-382355.
(Date: 14–06–2017)
Abstract.
In this article, we establish Lyapunov type inequality for the following extremal Pucci’s equation
[TABLE]
where is a smooth bounded domain in . This works generalize the well-known works on Lyapunov inequalities to fully nonlinear elliptic equations.
Key words and phrases:
Pucci’s extremal operator, Nontrivial solutions, Viscosity solutions, Lyapunov inequality
2010 Mathematics Subject Classification:
Primary 35J60, 35B09; Secondary 35J75, 49L25.
Submitted 14–06–2017. Published—–.
1. Introduction
The aim of this article is to establish Lyapunov type inequality for the following Pucci’s extremal equation:
[TABLE]
where is a bounded smooth domain in . Here, is called Pucci extremal operator. For a given , Pucci extremal operator is defined as follows:
[TABLE]
where is an real symmetric matrix. Let us recall the earlier developments on this subject. The classical Lyapunov type inequality says that the necessary condition to have a nontrivial solution to the following boundary value problem
[TABLE]
is
[TABLE]
see [23, 4]. Later, A. Winter [35] improved this inequality by replacing by and also proved that is the optimal constant. Further, this result was generalised to differential equations containing the term as well as qusilinear equations, see [17] and [24], respectively. In [24], Pinasco considered the following one dimensional -Laplace boundary value problem
[TABLE]
where and is a bounded positive function and proved the Lyapunov inequality as well as the lower bound for the eigenvalue as an application of Lyapunov inequality. For the recent work in this direction, see [2]. In [16], Elbert considered (1.5) with and proved that the necessary condition for (1.5) with to have a nontrivial solution is
[TABLE]
Further, Lee et al. [11] considered more general operator than a -Laplace operator. In fact, they replace the -Laplace operator in (1.5) with the following operator
[TABLE]
where is positive, integrable and is integrable function on and established the Lyapunov type inequality. Finally, in this direction, we would like to mention the works of de Nápoli and Pinasco [15], where the authors considered the -Laplace operator. In fact, they considered the following operator
[TABLE]
where is an odd nondecreasing function, such that and is a convex function, see [30] for similar kind of works. Lyapunov inequality has also been generalised in the context of fractional differential equations, see[20, 21, 22, 27, 28, 29].
We remark that there are interesting works on the Lyapunov inequality for partial differential equation (in short, PDE). In fact, the establishment of Lyapunov type inequality for PDE was actively started by generalising the corresponding result for the ODE in [9]. Cañada et al. [9] considered the Neumann problem corresponding to (1.3), that is,
[TABLE]
and defined the following set
[TABLE]
They also defined the following quantity
[TABLE]
and studied the qualitative properties of and obtained the explicit expression for as a function of and There are a good number of applications of Lyapunov inequality, see for instance [1, 3, 4, 8, 10, 17, 25]. The result in [9] has been generalised in the context of PDE in [7]. More precisely, the authors considered the following problem
[TABLE]
where is a bounded and smooth domain and the function belongs to the set defined as follows:
[TABLE]
[TABLE]
They also defined the quantity similar to as in (1.10), that is,
[TABLE]
and studied the properties of More precisely, they proved the following theorem.
Theorem 1.1**.**
*The following statements hold:
- (1)
If then If then . 2. (2)
If then is attained. In this case, any function on which is attained is of the form: 3. (i)
, , where is the first eigenvalue. 4. (ii)
* if where is a solution of the problem*
[TABLE] 5. (3)
The mapping is continuous and the mapping defined by
[TABLE]
is strictly increasing. 6. (4)
There exists always the limits and and takes the following values:
- (i)
* if * 2. (ii)
*, if *
* if *
In [7], attainability question in the case (i.e, critical case) was left open. This question was settled in [32], by showing that it does not attain. While in contrast to Dirichlet boundary case, in Neumann boundary case for it is attained, see [18]. Further, the results of [7] have been extended to the -Laplace operator with Robin boundary condition in [19]. More precisely, they considered the following boundary value problem
[TABLE]
and proved the similar results as in [7]. For the Lyapunov type inequality to -Laplace operator with the Dirichlet boundary condition, we refer to [14].
Motivated by the above mentioned research and recent works on fully nonlinear elliptic equations, see [5, 12, 13, 26, 31, 33, 34], there is a natural question to ask.
Question: Can we establish Lyapunov type inequality for fully nonlinear elliptic equations?
The aim of this article is to answer this question. More precisely, we establish Lyapunov type inequality for (1.1). We remark that the techniques used in earlier research works are not applicable due to the non-divergence nature of the problem under consideration. Here, we use another notion of the weak solution, so-called -viscosity solution. We employ Aleksandrov-Bakelman-Pucci estimate for viscosity solutions to get the desired results. For the definition of -viscosity solution, see Definition 2.2. In order to formulate our results, let us introduce some notations. Let us define as follows:
[TABLE]
and set
[TABLE]
The main result of this paper is the following theorem which we will prove in the next section.
Theorem 1.2**.**
The following statements hold:
- (i)
* for * 2. (ii)
* for , and for * 3. (iii)
* is not positive, in general, for *
We prove Theorem 1.2(iii) through an example. This example also suggests that if we remove a specific class of functions, then we get for all . For these specific class of functions, see Remark 3.2. Let us consider the following sets
[TABLE]
[TABLE]
and set
[TABLE]
It is clear that so if then either or . Now, for , set
[TABLE]
and we also prove the following:
Theorem 1.3**.**
For we have .
The organisation of this paper is as follows: In Section 2, we present important auxiliary results which are used in this article. Section 3 is devoted to the proof of main Thoerem 1.2 while Section 4 contains the proof of Theorem 1.3.
2. Auxiliary Results and Statements
We begin this section by recalling the definition of Pucci’s extremal operator. For given , Pucci extremal operator is defined as follows:
[TABLE]
where is a symmetric matrix of size . In general, it is very difficult to find the eigenvalues of the Hessian of a function. But if the given function is radial, that is, there is some such that , then the eigenvalues of the Hessian are given by the following lemma.
Lemma 2.1** (Lemma 3.1[13]).**
Let be function such that . Then for any the eigenvalues of the Hessian are with multiplicity and with multiplicity 1.
Definition 2.2**.**
A function is called -viscosity subsolution (resp. supersolution) to (1.1) in if for any and any point at which has local maximum (resp. minimum), we have
[TABLE]
In the proof of our results, Aleksandrov-Bakelman-Pucci (in short, ABP) estimate for viscosity solutions plays an important role. In the context of the viscosity solution this result first of all was proved by Luis A. Caffarelli in the context of continuous viscosity solution, see [5] and in the context of -viscosity solution it appears in [12]. Further, this result has been generalised in many ways. Here, we adopt ABP estimate from [[31], see Theorem 3]. In order to state the theorem, let us set
[TABLE]
Theorem 2.3**.**
Suppose is an -viscosity solution of
[TABLE]
in (resp. ) where Then
[TABLE]
(resp. ), where is a positive constant which depends on .
3. Proof of main Theorem
Proof of Theorem 1.2(i): Let us take an arbitrary and be a corresponding nontrivial solution to (1.1), that is, satisfies
[TABLE]
This implies that satisfies
[TABLE]
so by Theorem 2.3, we get
[TABLE]
Since is a nontrivial solution to (1.1), so
[TABLE]
[TABLE]
Now, if then by taking the infimum for all we get the required result, that is,
[TABLE]
Now on the other hand, if then
[TABLE]
So again by (3.1), we get
[TABLE]
[TABLE]
Since (3.2) is true for any so by taking infimum, we again get required result.
Remark 3.1*.*
In the above proof, we have used the fact that is nonempty. However, if is negative then . In this case, we define a function , then satisfies following equation
[TABLE]
for the details, see Remark 2.14 [12]. Here is positive so the set is nonempty, in fact, in this case . Note also that satisfies
[TABLE]
Now, using for any symmetric matrix so we find that satisfies the following inequality:
[TABLE]
in -viscosity sense. Now, repeating the same arguments as in (i), we obtain the required result.
Proof of Theorem 1.2 (ii). The proof is based on the construction of an example. Here, this example is a modification of an example given for linear case, see[ Lemma 3.1, [7]]. First of all, note that if we define (for arbitrary ), then . On the other hand, if we define (for arbitrary ), then . Hence
[TABLE]
Further, for let us define two numbers
[TABLE]
which are frequently used in the construction of examples, below. The proof is divided into two cases separately; and
Case . In view of (3.5), without loss of generality, we can suppose that Let us take two arbitrary real numbers satisfying and choose 0<\epsilon<\big{(}\frac{c}{d}\big{)}^{\frac{1}{d-c}}. Define the following radial function:
[TABLE]
In the above expression of , and are given as follows:
[TABLE]
Note that, for 0<\epsilon\leq\Big{(}\frac{c\big{(}1+\frac{d}{2}\big{)}}{d\big{(}1+\frac{c}{2}\big{)}}\Big{)}^{\frac{1}{d-c}}, By noting that it is easy to observe that the following holds:
[TABLE]
Now, since 0<\epsilon<\big{(}\frac{c}{d}\big{)}^{\frac{1}{d-c}} so in view of (3.6), the functions and are positive. It is easy to see that is a continuous function and in view of Lemma 2.1, (as in Lemma3.1 [7]), it satisfies (1.1), where is given by the following expression
[TABLE]
Here, obviously and so and
[TABLE]
Now, for , let us calculate:
[TABLE]
Thus, we get
[TABLE]
Now, by definition of ,
[TABLE]
Now, for fixed real numbers with , we can take limit in (3.8) and find that
[TABLE]
Finally, taking limit when tends to zero in (3.9), we conclude that and this complete the proof for case .
Case . Note that for for any and we have
[TABLE]
Thus, again without loss of generality, we can suppose that Also, for
[TABLE]
Now, take an arbitrary real number and satisfying and consider the following radial function:
[TABLE]
As in case , again it is easy to see that the function defined above satisfies (1.1) with given as follows:
[TABLE]
It is easy to see that and Hence, . Let us estimate the norm of
[TABLE]
The first integral can be evaluated to get the following:
[TABLE]
The second integral in (3.10) can be estimated as follows:
[TABLE]
On combining (3.10),(3.11) and (3.12), we find that
[TABLE]
Thus,
[TABLE]
But, for fixed real number we can take limit tending to zero in (3.13) to get
[TABLE]
Finally, taking limit as approaching to we conclude that
Proof of Theorem 1.2(iii). We prove this part by constructing a simple example. Let us consider the following problem
[TABLE]
Next onwards, we denote by , so . Let us also define a number , and consider the following function:
[TABLE]
It is easy to verify that for each satisfies (3.15), where is given by
[TABLE]
Thus, it is clear that for each
[TABLE]
Now, let us assume that , and compute:
[TABLE]
Thus, we find that as Consequently, for
[TABLE]
Remark 3.2*.*
In the above example, we have shown that for each the following problem:
[TABLE]
has a nontrivial solution and for , as .
Now, choose large enough such that for the following hold:
- (i)
2. (ii)
Note that, if then by (ii), we have
[TABLE]
In view of (3.17), for any we have
[TABLE]
Thus, for any definition of and (3.18), yields the following
[TABLE]
Now for , consider the following set
[TABLE]
Note that, in view of assumption (i), for , we have so
[TABLE]
Also notice that for any we have
[TABLE]
Thus, in view of (3.19) and (3.20), takes the following form
[TABLE]
Hence the Lebesgue measure of i.e,
[TABLE]
for Of course the Lebesgue measure of the following set
[TABLE]
is not zero.
In view of the above, it is natural ask that if we remove those functions from for which Lebesgue measure and , then, whether the modified quantity corresponding to is positive or not? In fact, the answer to this question is affirmative. Next, in the proof of Theorem 1.3, we answer to this question.
4. Proof of Theorem 1.3
If then from the definition of , so in this case, the result follows form Theorem 1.2. Now, we consider the case . Let us take an arbitrary and let be a corresponding nontrivial solution to (1.1). Now, if then for a.e, we have
[TABLE]
So an integration yields that
[TABLE]
Now, otherwise, if so we have a.e . Since there is a nontrivial solution to
[TABLE]
That is,
[TABLE]
In order to get the required result, we need to adjust the right hand side of (4.2) before applying the (ABP) estimate. So let us proceed. Since so
[TABLE]
In particular, since so . Now by (4.2), we find that
[TABLE]
Noting that , and we conclude that . Therefore by (2.3), we get
[TABLE]
Now
[TABLE]
Now (4.4) and (4.5) yield that
[TABLE]
i.e,
[TABLE]
Now on combining (4.6) and (4.1), we find that
[TABLE]
Since is arbitrary so by taking infimum over we find , and this completes the proof.
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