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Article

Radially Symmetric Positive Solutions of the Dirichlet Problem for the p-Laplace Equation

Department of Mathematics, Kennesaw State University, Kennesaw, GA 30144, USA
Mathematics 2024, 12(15), 2351; https://doi.org/10.3390/math12152351
Submission received: 1 July 2024 / Revised: 19 July 2024 / Accepted: 25 July 2024 / Published: 27 July 2024

Abstract

:
We consider the p-Laplace boundary value problem with the Dirichlet boundary condition. A new lower estimate for positive solutions of the problem is obtained. As an application of this new lower estimate, some sufficient conditions for the existence and nonexistence of positive solutions for the p-Laplace problem are obtained.

1. Introduction

Differential equations involving p-Laplace operators have wide applications in physics, and they have received quite some attention recently. For example, in 2007, by using the theory of lower and upper solutions, Jin, Yin, and Wang [1] studied the existence of positive radial solutions for the p-Laplacian boundary value problem
div ( | u | p 2 u ) = f ( | x | , u ) , x Ω ,
u ( x ) = 0 , x Ω ,
where p > 1 and Ω R n is the unit open ball centered at the origin. We refer the reader to [2] for a historical account of the origin of the p-Laplace operator. For a very short list of some recent papers on p-Laplace boundary value problems, we refer the reader to the papers [3,4,5,6,7,8,9,10,11].
In this paper, we consider the boundary value problem
Δ p w ( x ) + g ( | x | ) f ( w ( x ) ) = 0 , x Ω ,
w = 0 , x Ω .
Here, Ω R n is the unit open ball centered at the origin, | x | denotes the Euclidean norm of x R n , and
Δ p w = div | w | p 2 w .
Throughout the paper, we assume that
(H)
n 1 is a positive integer, p ( n , + ) is a positive real number, g : [ 0 , 1 ] [ 0 , + ) and f : [ 0 , + ) [ 0 , + ) are continuous functions, and g ( t ) 0 on the interval [ 0 , 1 ] .
For convenience, we define the function Φ p : R R by
Φ p ( x ) = | x | p 2 x , x 0 , 0 , x = 0 .
It is clear that Φ p ( x ) is an increasing and continuous function, and x Φ p ( x ) 0 for all real x. The inverse function of Φ p ( x ) is denoted by Φ p 1 ( x ) in this paper.
It is well known (see [1], for example) that if we consider only radially symmetric positive solutions of problem (1),(2), then problem (1),(2) reduces to the following boundary value problem for a second-order ordinary differential equation:
t 1 n ( t n 1 Φ p ( u ( t ) ) ) + g ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 ,
u ( 0 ) = 0 , u ( 1 ) = 0 .
That is, if u ( t ) is a solution to problem (3),(4), then w ( x ) = u ( | x | ) solves the boundary value problem (1),(2), and vice versa.
Our main focus in this paper is on positive solutions to problem (3),(4). By a positive solution to problem (3),(4), we mean a solution u ( t ) such that u ( t ) > 0 on ( 0 , 1 ) . As has been repeatedly pointed out in the literature (see [6,12], for example), in the study of positive solutions to boundary value problems, a priori upper and lower estimates for positive solutions play a crucial role. In particular, once we obtain some a priori upper and lower estimates, we can use these estimates to approximate the first eigenvalue of the corresponding eigenvalue problem (see [13], for example). Also, by using these upper and lower estimates, we can establish some nice existence results for multiple positive solutions (see [12,14], for example). The main purpose of this paper is to present a new lower estimate for positive solutions to problem (3),(4).
Throughout this paper, we let X = C [ 0 , 1 ] be equipped with the supremum norm
v = max t [ 0 , 1 ] | v ( t ) | for all v X .
Clearly, X is a Banach space. We define
Y = v X | v ( t ) 0 for 0 t 1 .
It is clear that Y is a positive cone in X. It is also clear that the boundary value problem (3),(4) is equivalent to the integral equation
u ( t ) = t 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) f ( u ( s ) ) d s d r , 0 t 1 .
Define the operator T : Y X by
( T u ) ( t ) = t 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) f ( u ( s ) ) d s d r , 0 t 1 , u Y .
It is clear that if (H) holds, then T ( Y ) Y . By some standard arguments, we can show that T is a completely continuous operator. Also, it is now clear that u Y is a fixed point of T if and only if u is a non-negative solution to problem (3),(4).
This paper is organized as follows. In Section 2, we prove a new type of lower estimate for positive solutions of problem (3),(4). In Section 3 and Section 4, we prove some existence and nonexistence results for positive solutions for problem (3),(4). An example is included at the end of the paper to illustrate our existence and non-existence results.

2. A New Lower Estimate

In this section, we present a new lower estimate for positive solutions to problem (3),(4). This lower estimate (see (7) below) is called by some authors the norm-type, for the simple reason that its expression is the norm u times a function of t. To the best of our knowledge, there is no lower estimate of this type for problem (3),(4) in the literature.
For this purpose, we define the function a : [ 0 , 1 ] [ 0 , 1 ] by
a ( t ) = 1 t ( p n ) / ( p 1 ) , 0 t 1 .
The function a ( t ) is used to give the lower estimate for positive solutions of problem (3),(4). Since p > n , a ( t ) is continuous on [ 0 , 1 ] , it is clear that a ( 0 ) = 1 and a ( 1 ) = 0 . We leave it to the reader to verify that a ( t ) is decreasing on [ 0 , 1 ] . We begin with some technical lemmas.
Lemma 1.
If u C 2 ( 0 , 1 ] C 1 [ 0 , 1 ] satisfies the boundary conditions (4) and u is such that
( t n 1 Φ p ( u ( t ) ) ) 0 , 0 < t < 1 ,
then,
u ( t ) 0 and u ( t ) 0
on the interval [ 0 , 1 ] , and u ( 0 ) = u .
The proof of the lemma is quite straightforward and is, therefore, left to the reader. The next lemma gives a lower estimate for positive solutions of problem (3),(4).
Lemma 2.
Suppose that (H) holds. If u C 2 ( 0 , 1 ] C 1 [ 0 , 1 ] satisfies the boundary conditions (4) and the inequality (6) holds, then
u ( t ) u a ( t ) , 0 t 1 .
Proof. 
By Lemma 1, we have u ( t ) 0 on [ 0 , 1 ] and u ( 0 ) = u . We define an auxiliary function h ( t ) as follows:
h ( t ) = u ( t ) u a ( t ) , 0 t 1 .
It is easy to see that
h ( 0 ) = h ( 1 ) = 0 .
To prove the lemma, it suffices to show that h ( t ) 0 for 0 t 1 . We use the method of contradiction to prove the lemma. For this purpose, we assume, to the contrary, that h ( t 0 ) < 0 for some t 0 ( 0 , 1 ) .
Since h ( 0 ) = 0 > h ( t 0 ) , by the mean value theorem, there exists t 1 ( 0 , t 0 ) such that h ( t 1 ) < 0 . Since h ( t 0 ) < 0 = h ( 1 ) , there exists s 1 ( t 0 , 1 ) such that h ( s 1 ) > 0 .
Note that h ( t 1 ) < 0 and h ( s 1 ) > 0 imply that
u ( t 1 ) u a ( t 1 ) < 0 , u ( s 1 ) u a ( s 1 ) > 0 .
Since Φ p is strictly increasing, we have
Φ p ( u ( t 1 ) ) Φ p ( u a ( t 1 ) ) < 0 , Φ p ( u ( s 1 ) ) Φ p ( u a ( s 1 ) ) > 0 .
We now define another auxiliary function v ( t ) as follows:
v ( t ) = t n 1 Φ p ( u ( t ) ) Φ p ( u a ( t ) ) , 0 t 1 .
It is clear that v ( t 1 ) < 0 and v ( s 1 ) > 0 . Since v ( t 1 ) < 0 < v ( s 1 ) , there exists t 2 ( t 1 , s 1 ) ( 0 , 1 ) such that
v ( t 2 ) > 0 .
On the other hand, by Equations (5), (6), and (8), we have
v ( t ) = ( t n 1 Φ p ( u ( t ) ) ) 0 , 0 < t < 1 ,
which contradicts (9). The proof of the lemma is now complete. □
We now summarize our findings in the following theorems.
Theorem 1.
Suppose that (H) holds. If u C 2 ( 0 , 1 ] C 1 [ 0 , 1 ] satisfies the boundary conditions (4), and the inequality (6) holds, then u ( t ) 0 on [ 0 , 1 ] , and
a ( t ) u ( 0 ) u ( t ) u ( 0 ) , 0 t 1 .
In particular, if u C 2 ( 0 , 1 ] C 1 [ 0 , 1 ] is a nonnegative solution to the boundary value problem (3),(4), then u ( t ) satisfies the estimates (10).
The next theorem follows immediately.
Theorem 2.
Suppose that (H) holds. If w ( x ) is a radially symmetric positive solution to the p-Laplace boundary value problem (1),(2), then
w ( 0 ) w ( x ) w ( 0 ) a ( | x | ) , | x | < 1 .
Here, 0 = ( 0 , 0 , , 0 ) is the origin of the R n space.
Now, we define a subset P of Y as follows:
P = { v Y : a ( t ) v ( 0 ) v ( t ) v ( 0 ) on [ 0 , 1 ] } .
Clearly, P is a positive cone of the Banach space X. From now on, we restrict the operator T on the cone P. Again, T : P Y is a completely continuous operator. And, by the same arguments as those used to prove Theorem 1, we can show that T ( P ) P provided (H) holds. We also note that if v P , then
v = v ( 0 ) .
Now, it is clear that, in order to solve problem (3),(4) for a positive solution, we only need to find a fixed point u of T in P such that u > 0 .

3. Existence of Positive Solutions

As an application of the lower estimate obtained in the last section, we now establish some existence and nonexistence results for positive solutions to problem (3),(4). We use the following fixed point theorem, which is due to Krasnosel’skii [15], to prove our existence results.
Theorem 3.
Let X be a Banach space over the reals, and let P X be a cone in X. Let ≤ be the partial order on X determined by P. Assume that Ω 1 and Ω 2 are bounded open subsets of X with 0 Ω 1 and Ω ¯ 1 Ω 2 . Let
L : P ( Ω ¯ 2 Ω 1 ) P
be a completely continuous operator such that, either
(K1)
L u u if u P Ω 1 , and L u u if u P Ω 2 ; or
(K2)
L u u if u P Ω 1 , and L u u if u P Ω 2 .
Then, L has a fixed point in P ( Ω ¯ 2 Ω 1 ) .
Remark 1.
In Theorem 3, ≤ is the partial order on X determined by P. That is, if f , g X , then
f g ( g f ) P .
Hence, the inequality L u u means that ( L u u ) P , and the inequality L u u means that ( u L u ) P .
We begin by defining constants A and B by
A = 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( a ( s ) ) p 1 d s d r
and
B = 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) d s d r .
Also, we define the following constants that are related to the function f:
F 0 = lim sup x 0 + f ( x ) x p 1 , f 0 = lim inf x 0 + f ( x ) x p 1 , F = lim sup x + f ( x ) x p 1 , f = lim inf x + f ( x ) x p 1 .
These constants are used in the statements of our existence and nonexistence theorems. Our first existence result is given below. Though Krasnosel’skii’s fixed point theorem has become quite a standard tool for finding positive solutions; the proof of the next theorem is included here for completeness and reference purposes.
Theorem 4.
If
B F 0 1 / ( p 1 ) < 1 < A f 1 / ( p 1 ) ,
then problem (3),(4) has at least one positive solution.
Proof. 
Choose ϵ > 0 such that B ( F 0 + ϵ ) 1 / ( p 1 ) < 1 . Then, there exists H 1 > 0 such that
f ( x ) ( F 0 + ϵ ) x p 1 for 0 < x H 1 .
For each u P with u = H 1 , we have
( T u ) ( 0 ) = 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) f ( u ( s ) ) d s d r 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( F 0 + ϵ ) ( u ( s ) ) p 1 d s d r = ( F 0 + ϵ ) 1 / ( p 1 ) 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( s ) ) p 1 d s d r ( F 0 + ϵ ) 1 / ( p 1 ) 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( 0 ) ) p 1 d s d r = ( F 0 + ϵ ) 1 / ( p 1 ) u ( 0 ) 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) d s d r = ( F 0 + ϵ ) 1 / ( p 1 ) u ( 0 ) B < u ( 0 ) = u ,
that is, ( T u u ) ( 0 ) < 0 , which implies that ( T u u ) P . So, if we let
Ω 1 = u X | u < H 1 ,
then,
T u u , for any u P Ω 1 .
To construct Ω 2 , we first choose a positive real number f ^ such that f ^ < f and
1 < A f ^ 1 / ( p 1 ) .
Then, we choose c ( 3 / 4 , 1 ) and δ > 0 such that
( f ^ δ ) 1 / ( p 1 ) 0 c Φ p 1 r 1 n 0 r s n 1 g ( s ) ( a ( s ) ) p 1 d s d r > 1 .
Now, there exists H 3 > 0 such that f ( x ) ( f ^ δ ) x p 1 for x H 3 . Let H 2 = H 1 + H 3 / a ( c ) . If u P with u = H 2 , then, for 0 t c , we have
u ( t ) a ( t ) u a ( c ) H 2 > H 3 .
So, if u P with u = H 2 , then
( T u ) ( 0 ) = 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) f ( u ( s ) ) d s d r 0 c Φ p 1 r 1 n 0 r s n 1 g ( s ) f ( u ( s ) ) d s d r 0 c Φ p 1 r 1 n 0 r s n 1 g ( s ) ( f ^ δ ) ( u ( s ) ) p 1 d s d r = ( f ^ δ ) 1 / ( p 1 ) 0 c Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( s ) ) p 1 d s d r ( f ^ δ ) 1 / ( p 1 ) 0 c Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( 0 ) a ( s ) ) p 1 d s d r = ( f ^ δ ) 1 / ( p 1 ) u ( 0 ) 0 c Φ p 1 r 1 n 0 r s n 1 g ( s ) ( a ( s ) ) p 1 d s d r > 1 · u ( 0 ) = u ( 0 ) = u ,
which means T u u . So, if we let Ω 2 = u X : u < H 2 , then Ω 1 ¯ Ω 2 , and
T u u , for any u P Ω 2 .
Therefore, condition (K1) of Theorem 3 is satisfied, and so there exists a fixed point of T in P. This completes the proof of the theorem. □
Our next theorem is a companion result to the one above.
Theorem 5.
If
B F 1 / ( p 1 ) < 1 < A f 0 1 / ( p 1 ) ,
then the boundary value problem (3),(4) has at least one positive solution.
The proof of Theorem 5 is similar to that of Theorem 4 and is, therefore, left to the reader.

4. Nonexistence Results and Example

In this section, we give some sufficient conditions for the nonexistence of positive solutions.
Theorem 6.
Suppose that (H) holds. If f ( x ) < ( x / B ) p 1 for all x ( 0 , + ) , then problem (3),(4) has no positive solutions.
Proof. 
Assume, on contrary, that u ( t ) is a positive solution of problem (3),(4). Then, u P , u ( t ) > 0 for 0 < t < 1 , and
u ( 0 ) = 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) f ( u ( s ) ) d s d r < 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( s ) / B ) p 1 d s d r = B 1 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( s ) ) p 1 d s d r B 1 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) ( u ( 0 ) ) p 1 d s d r = u ( 0 ) B 1 0 1 Φ p 1 r 1 n 0 r s n 1 g ( s ) d s d r = u ( 0 ) B 1 B = u ( 0 ) ,
which is a contradiction. The proof of the theorem is now complete. □
In a similar fashion, we can prove the next theorem.
Theorem 7.
Suppose that (H) holds. If f ( x ) > ( x / A ) p 1 for all x ( 0 , + ) , then problem (3),(4) has no positive solutions.
We conclude this section with an example.
Example 1.
Consider the following p-Laplace boundary value problem:
Δ p w ( x ) + λ g ( | x | ) f ( w ( x ) ) = 0 , x Ω ,
w = 0 , x Ω ,
where λ > 0 is a parameter, p = 4 , Ω R 2 is the unit open ball centered at the origin, and
g ( t ) = 1 t 2 , 0 t 1 , f ( u ) = λ u 3 · 1 + 8 u 1 + u , u 0 .
It is clear that, if we seek a radially symmetric solution only, then problem (11),(12) reduces to the following problem:
t 1 t Φ 4 ( u ( t ) ) + g ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 ,
u ( 0 ) = 0 , u ( 1 ) = 0 .
Here, Φ 4 ( x ) = x 3 .
We easily see that problem (13),(14) is a special case of problem (3),(4) in which n = 2 and p = 4 . In this case, we have a ( t ) = 1 t 2 / 3 . Also, we have F 0 = f 0 = λ and F = f = 8 λ . It is clear that
λ x 3 < f ( x ) < 8 λ x 3 , x > 0 .
Calculations by using a standard Computer Algebra System (CAS) indicate that
A 0.317485 , B 0.550302 .
From Theorem 4, we see that if
3.906 1 8 A 3 < λ < 1 B 3 6.0006 ,
then problem (13),(14) has at least one positive solution. From Theorems 6 and 7, we see that if
e i t h e r λ < 1 8 B 3 0.7501 or λ > 1 A 3 31.2487
then problem (13),(14) has no positive solutions.
It follows that, if (15) holds, then problem (11),(12) has at least one radially symmetric positive solution. And if (16) holds, then problem (11),(12) has no radially symmetric positive solutions.

5. Conclusions

In summary, we present a new lower estimate for radially symmetric positive solutions to the Dirichlet boundary value problem for the p-Laplace equation. The proof of this new lower estimate is elementary, making it accessible to undergraduate students. As an application, some sufficient conditions for the existence and nonexistence of positive solutions are obtained. In proving the existence results, we apply Krasnosel’skii’s fixed point theorem on cones.
Some future developments we would like to see include
  • Using the lower estimate in conjunction with other fixed-point theorems to establish new existence results;
  • Using the lower estimate to solve the corresponding singular boundary value problem.

Funding

This research received no external funding.

Data Availability Statement

No data sets were generated during this research.

Acknowledgments

The author is grateful to the anonymous referee for their valuable comments and suggestions.

Conflicts of Interest

The author declares no conflicts of interest.

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Yang, B. Radially Symmetric Positive Solutions of the Dirichlet Problem for the p-Laplace Equation. Mathematics 2024, 12, 2351. https://doi.org/10.3390/math12152351

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Yang B. Radially Symmetric Positive Solutions of the Dirichlet Problem for the p-Laplace Equation. Mathematics. 2024; 12(15):2351. https://doi.org/10.3390/math12152351

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Yang, Bo. 2024. "Radially Symmetric Positive Solutions of the Dirichlet Problem for the p-Laplace Equation" Mathematics 12, no. 15: 2351. https://doi.org/10.3390/math12152351

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Yang, B. (2024). Radially Symmetric Positive Solutions of the Dirichlet Problem for the p-Laplace Equation. Mathematics, 12(15), 2351. https://doi.org/10.3390/math12152351

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