On the evolutionary bifurcation curves for the one-dimensional prescribed mean curvature equation with logistic type

Jiajia Zhang; Yuanhua Qiao; Lijuan Duan; Jun Miao

doi:10.1515/math-2021-0070

Article Open Access

On the evolutionary bifurcation curves for the one-dimensional prescribed mean curvature equation with logistic type

Jiajia Zhang , Yuanhua Qiao , Lijuan Duan and Jun Miao

Published/Copyright: August 28, 2021

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$Open Mathematics$

From the journal Open Mathematics Volume 19 Issue 1

Abstract

We study the bifurcation diagrams and exact multiplicity of positive solutions for the one-dimensional prescribed mean curvature equation

− u ′ 1 + u ′ 2 ′ = λ u 1 + u p , − L < x < L , u ( − L ) = u ( L ) = 0 ,

where λ is a bifurcation parameter, and L , p > 0 are two evolution parameters. We prove that on the ( λ , ‖ u ‖ ∞ ) -plane, for 0 < p ≤ 2 4 , the bifurcation curve is ⊃ -shaped bifurcation starting from ( 0 , 0 ) . And for p = 1 , f ( u ) = u 1 + u is a logistic function, then the bifurcation curve is also ⊃ -shaped bifurcation starting from π 2 4 L 2 , 0 . While for p > 1 , the bifurcation curve is reversed ε -like shaped bifurcation if L > L ∗ , and is exactly decreasing for λ > λ ∗ if 0 < L < L ∗ .

Keywords: mean curvature; bifurcation curve; logistic function

MSC 2010: 34B18; 34C23; 34B15

1 Introduction

In this paper, we study the bifurcation curve and multiplicity of positive solutions for the one-dimensional prescribed mean curvature problem

(1.1) − u ′ 1 + u ′ 2 ′ = λ f ( u ) , − L < x < L , u ( − L ) = u ( L ) = 0 ,

where λ > 0 is a bifurcation parameter, and L > 0 is an evolution parameter. Many multiplicity results of the one-dimensional mean curvature-type equations like (1.1) have been investigated by a number of authors by using the time-map, bifurcation theory and other techniques [1,2, 3,4,5, 6,7,8, 9,10,11, 12,13,14, 15,16,17, 18,19].

Relative to the quasilinear equation (1.1), there is another semilinear equation

(1.2) − u ″ = λ f ( u ) , − L < x < L , u ( − L ) = u ( L ) = 0 ,

which has also received much attention in recent years [20,21,22, 23,24,25]. For equation (1.2), one parameter is actually enough because a scale of variable x incorporates the parameter λ in the length L of the domain [8]. Hence, we consider the bifurcation cure of equation (1.1) in this paper, which will lead to different bifurcation behavior in λ for different values of L .

General f ( u ) has many different types of nonlinearities, like e u , ( 1 + u ) p , ( 1 − u ) p , a u ( 0 < a < 1 , a = 1 , a > 1 ) , e a u a + u , 1 ( 1 − u ) 2 , which satisfied f ( 0 ) > 0 , hence the bifurcation curve starts from ( 0 , 0 ) , and they have been recently investigated [2,4, 5,6,7, 8,17,26]. Other functions satisfying f ( 0 ) = 0 , and f ′ ( 0 ) = 0 or f ′ ( 0 ) > 0 like e u − 1 , u p , u p + u q , u − u 3 , e a u a + u − 1 , have also been studied [2,3,9,12,16,19]. The bifurcation curve may start from any point on λ -axis.

In this paper, we consider the nonlinearity function

(1.3) f ( u ) = u 1 + u p ,

where the constant p > 0 . Function (1.3) satisfies f ( 0 ) = 0 ; however, f ′ ( 0 ) = + ∞ , which is different from previous research. Furthermore, when p = 1 , function (1.3) becomes f ( u ) = u 1 + u . It is a standard logistic function, which is widely used as a logarithmic linear model for classification and regression scenarios. The logistic function, also denoted as “ S ” shape function or sigmoid function appears in models of population growth, spread of epidemic diseases and wind turbine power. It is the most commonly used activation function in neural networks as it approaches zero while u approaches zero, and as u approaches infinity, it approaches one. We investigate the curvature problem of logistic type with the least number of parameters [27].

For (1.3), we have

f ′ ( u ) = p u ( u + 1 ) f ( u ) , f ″ ( u ) = p f ( u ) u 2 ( 1 + u ) 2 ( p − 1 − 2 u ) .

The nonlinearity f ( u ) satisfies f ( 0 ) = 0 , f ′ ( u ) > 0 on [ 0 , + ∞ ) , lim u → + ∞ f ( u ) = 1 > 0 . In addition,

If p ≤ 1 , then f ( u ) is concave on ( 0 , ∞ ) ;
If p > 1 , then f ( u ) is convex-concave on ( 0 , ∞ ) . More precisely, f ( u ) is convex on 0 , p − 1 2 and concave on p − 1 2 , + ∞ , then p − 1 2 is the unique inflection point of f ( u ) . So the concavity and convexity of f ( u ) is related to the sign of p − 1 , then we investigate problem (1.1)(1.3) at p > 1 , p = 1 and p < 1 , respectively.

We always allow that solutions u ∈ C 2 ( − L , L ) ∩ C [ − L , L ] for problem (1.1)(1.3) satisfy u ′ ∈ C ( [ − L , L ] , [ − ∞ , + ∞ ] ) [3,10,15]. Then we define the bifurcation diagram S p of (1.1)(1.3) by

S p ≡ { ( λ , ‖ u ‖ ∞ ) : λ > 0 and u λ ∈ C 2 ( − L , L ) ∩ C [ − L , L ] is a positive solution of (1.1)(1.3) } .

We focus on the quasilinear problem (1.1)(1.3) as it is different from the following semilinear problems on the bifurcation curves:

− u ″ = λ u 1 + u p , − L < x < L u ( − L ) = u ( L ) = 0 .

While p ≤ 1 , f ″ ( u ) < 0 , p is not a bifurcation parameter for the semilinear problems by applying the study by Laetsch [28]. Before going further to discussions on quasilinear problem (1.1)(1.3), we first introduce the following terminologies [21,29], which will be used throughout the paper.

⊃ -shaped: We say that, on the ( λ , ‖ u ‖ ∞ ) -plane, the bifurcation curve S p is ⊃ -shaped if S p has at least one turning point, say ( λ ∗ , ‖ u λ ∗ ‖ ∞ ) , satisfying

at ( λ ∗ , ‖ u λ ∗ ‖ ∞ ) the bifurcation curve S p turns to the left,
S p initially continues to the right and eventually continues to the left.

Reversed S-shaped: We say that, on the ( λ , ‖ u ‖ ∞ ) -plane, the bifurcation curve S is reversed S -shaped if S p has at least two turning points, say ( λ ∗ , ‖ u λ ∗ ‖ ∞ ) and ( λ ∗ , ‖ u λ ∗ ‖ ∞ ) , satisfying λ ∗ < λ ∗ and ‖ u λ ∗ ‖ ∞ > ‖ u λ ∗ ‖ ∞ , and

at ( λ ∗ , ‖ u λ ∗ ‖ ∞ ) the bifurcation curve S p turns to the left,
at ( λ ∗ , ‖ u λ ∗ ‖ ∞ ) the bifurcation curve S p turns to the right,
S p initially continues to the left and eventually continues to the left.

Reversed S-like shaped: On the ( λ , ‖ u ‖ ∞ ) -plane, the bifurcation curve S p is reversed S-like shaped if S p is ⊃ -shaped, and it has a turning point which results in turning to the right. Thus, S p has at least three turning points and it initially goes to the right, and eventually it turns to the left.

The paper is organized as follows. Section 2 contains statements of main results. Section 3 contains several lemmas needed to prove the main results. Finally, Section 4 contains the proofs of the main results.

2 Main results

The main results in this paper are Theorems 1–3 of (1.1)(1.3) with a > 0 .

First, let F ( u ) = ∫ 0 u f ( t ) d t , then F ′ ( u ) = f ( u ) , F ( 0 ) = 0 , and f ( u ) is from equation (2.1). Due to the symmetry, we just consider the differential equation

(2.1) − u ′ 1 + u ′ 2 ′ = λ f ( u ) , x ∈ ( 0 , L ) ,

with the initial-boundary conditions u ( 0 ) = r , u ′ ( 0 ) = 0 , u ( L ) = 0 , or u ′ ( L ) = − ∞ .

Then we have the energy conservation relation

1 − 1 1 + ( u ′ ) 2 + λ F ( u ) = λ F ( r )

and

− u ′ = 1 − [ 1 + λ F ( u ) − λ F ( r ) ] 2 1 + λ F ( u ) − λ F ( r ) .

Here, r ≥ F − 1 1 λ if and only if ∣ u ′ ( L ) ∣ = − ∞ . Hence, we refer to the curve r = F − 1 1 λ as “the gradient blow-up curve” for the bifurcation problem (1.1). F − 1 1 λ becomes a threshold of the initial value. When the initial value r reaches or exceeds this threshold level, solutions of (1.1) have to be nonclassical [30].

In terms of f ( u ) = u 1 + u p , the following theorems are obtained, according to the different values of the parameter p .

Theorem 1

(Figure 1) Consider positive solutions u ∈ C 2 ( − L , L ) ⋂ C [ − L , L ] of (1.1)(1.3) with 0 0 , there exist two constants λ ∗ and λ ∗ with 0 < λ ∗ < λ ∗ such that the bifurcation curve is ⊃ -shaped on the ( λ , ‖ u ‖ ∞ ) -plane. More precisely,

for any λ ∈ ( 0 , λ ∗ ) ∪ { λ ∗ } , problem (1.1)(1.3) has exactly one positive solution,
for any λ ∈ [ λ ∗ , λ ∗ ) , problem (1.1)(1.3) has exactly two positive solutions,
for any λ ∈ ( λ ∗ , + ∞ ) , problem (1.1)(1.3) has no positive solution.

$Figure 1 ⊃ \supset -shaped bifurcation starting from ( 0 , 0 0,0 ) with 0 < p < 2 4 0\lt p\lt \frac{\sqrt{2}}{4} .$

Figure 1

⊃ -shaped bifurcation starting from ( 0 , 0 ) with 0 < p < 2 4 .

Theorem 2

(Figure 2) Consider positive solutions u ∈ C 2 ( − L , L ) ⋂ C [ − L , L ] of (1.1)(1.3) with p = 1 . Then the bifurcation curve is ⊃ -shaped on the ( λ , ‖ u ‖ ∞ ) -plane. More precisely,

when λ > 1 , for L < π 2 , there exist two constants 1 < λ ∗ < π 2 L 2 < λ ∗ such that
1. for any λ ∈ λ ∗ , π 2 L 2 ∪ { λ ∗ } , problem (1.1)(1.3) has exactly one positive solution,
2. for any λ ∈ π 2 L 2 , λ ∗ , problem (1.1)(1.3) has exactly two positive solutions,
3. for any λ ∈ ( 0 , λ ∗ ) ∪ ( λ ∗ , + ∞ ) , problem (1.1)(1.3) has no positive solution,
when 0 < λ ≤ 4 π 2 , for L > π 2 4 , there exist two constants π 2 L 2 < λ ∗ < λ ∗ < 4 π 2 such that
1. for any λ ∈ π 2 L 2 , λ ∗ ∪ { λ ∗ } , problem (1.1)(1.3) has exactly one positive solution,
2. for any λ ∈ [ λ ∗ , λ ∗ ) , problem (1.1)(1.3) has exactly two positive solutions,
3. for any λ ∈ 0 , π 2 L 2 ∪ ( λ ∗ , + ∞ ) , problem (1.1)(1.3) has no positive solution.

$Figure 2 ⊃ \supset -shaped bifurcation starting from π 2 L 2 , 0 \left({\left(\frac{\pi }{2L}\right)}^{2},0\right) with p = 1 p=1 . (a) λ > 1 \lambda \gt 1 , 0 < L < π 2 0\lt L\lt \frac{\pi }{2} , (b) 0 < λ ≤ 4 π 2 0\lt \lambda \le \frac{4}{{\pi }^{2}} , L > π 2 4 L\gt \frac{{\pi }^{2}}{4} .$

Figure 2

⊃ -shaped bifurcation starting from π 2 L 2 , 0 with p = 1 . (a) λ > 1 , 0 < L < π 2 , (b) 0 < λ ≤ 4 π 2 , L > π 2 4 .

Theorem 3

(Figure 3) Consider positive solutions u ∈ C 2 ( − L , L ) ⋂ C [ − L , L ] of (1.1)(1.3) with p > 1 . There exist two positive numbers L ∗ < L ∗ such that the following statements hold:

If 0 < L ≤ L ∗ , the bifurcation curve is exactly decreasing shaped. There exists λ ∗ > 0 such that problem (1.1)(1.3) has exactly one positive solution on 0 , F − 1 1 λ for λ ≥ λ ∗ , and has no root on 0 , F − 1 1 λ for λ < λ ∗ .
If L > L ∗ , the bifurcation curve is reversed S-like shaped on the ( λ , ‖ u ‖ ∞ ) -plane. There exist two constants λ ∗ and λ ∗ with 0 < λ ∗ < λ ∗ such that problem (1.1)(1.3) has exactly one positive solution on 0 , F − 1 1 λ for λ > λ ∗ , has no positive solution on 0 , F − 1 1 λ for λ < λ ∗ , and has at least one positive solution on 0 , F − 1 1 λ for λ ∗ < λ < λ ∗ .

$Figure 3 Bifurcation curves with p = 1 p=1 . (a) Exactly decreasing shaped, for 0 < L ≤ L ∗ 0\lt L\le {L}_{\ast } (b) Reversed S-like shaped, for L > L ∗ L\gt {L}^{\ast } .$

Figure 3

Bifurcation curves with p = 1 . (a) Exactly decreasing shaped, for 0 < L ≤ L ∗ (b) Reversed S-like shaped, for L > L ∗ .

3 Lemmas

To prove Theorems 1–3, we make a detailed analysis of the time map for one-dimensional prescribed mean curvature problem (1.1)(1.3). The time-map method was applied by many authors of [2,9]. In this section, we define the time map formula for (1.1) by

(3.1) T λ ( r ) = ∫ 0 r 1 + λ F ( u ) − λ F ( r ) 1 − [ 1 + λ F ( u ) − λ F ( r ) ] 2 d u , r ∈ 0 , F − 1 1 λ .

Note that it can be proved that T λ ( r ) ∈ C 2 0 , F − 1 1 λ for (1.1), see [1, Lemma 3.1]. Observe that positive solutions u λ for (1.1)(1.3) correspond to

‖ u ‖ ∞ = λ and T λ ( r ) = L .

First, we determine the limits of T λ ( r ) and T λ ′ ( r ) as r → 0 + in the following lemmas. As f ( 0 ) = 0 , and lim r → 0 + f ( r ) r p = lim r → 0 + 1 ( 1 + r ) p = 1 , in addition, when p = 1 , f ″ ( u ) = − 2 ( 1 + u ) 3 , so f ″ ( 0 ) = − 2 < 0 . Hence, following from [2, Propositions 2.6 and 2.7], we have

Lemma 3.1

For (1.1)(1.3), for any fixed λ > 0 ,

lim r → 0 + T λ ( r ) = 0 , if 0 1

and

lim r → 0 + T λ ′ ( r ) = + ∞ , if 0 0 , if p = 1 , − ∞ , if p > 1 .

Next, we determine T λ ′ ( r ) at the endpoint F − 1 1 λ for λ > 0 .

Lemma 3.2

For (1.1)(1.3), T λ ′ F − 1 1 λ < 0 for λ > 0 .

Proof

Since

lim u → 0 + F ( u ) f ( u ) = lim u → 0 + ∫ 0 u s 1 + s p d s u 1 + u p = lim u → 0 + u ( u + 1 ) p = 0

and

f ′ ( u ) = p u ( u + 1 ) f ( u ) > 0 on ( 0 , + ∞ ) ,

using [31, Lemma 2.5], we get T λ ′ F − 1 1 λ < 0 .□

Denote

(3.2) g ( λ ) = T λ F − 1 1 λ .

Lemma 3.3

For (1.1)(1.3), g ( λ ) is a strictly decreasing function for λ > 0 . Moreover, lim λ → 0 + g ( λ ) = ∞ and lim λ → ∞ g ( λ ) = 0 .

Proof

First, according to [2, Proposition 2.8], g ( λ ) is strictly decreasing with respect to λ > 0 if

(3.3) f 2 ( u ) − f ′ ( u ) F ( u ) ≥ ( ≢ 0 ) .

Let Φ ( u ) ≡ f ( u ) + p 1 u + 1 − 1 u F ( u ) for u > 0 .

Then

Φ ′ ( u ) = f ′ ( u ) + p − 1 ( u + 1 ) 2 + 1 u 2 F ( u ) + p 1 u + 1 − 1 u f ( u ) = p 2 u + 1 u 2 ( u + 1 ) 2 F ( u ) > 0 ,

and

lim u → 0 + Φ ( u ) = lim u → 0 + u 1 + u p − p u ( u + 1 ) ∫ 0 u s 1 + s p d t = 0 − lim u → 0 + p u 1 + u p 2 u + 1 = 0 .

Therefore, Φ ( u ) > 0 . Then f 2 ( u ) − f ′ ( u ) F ( u ) > 0 , and inequality (3.3) holds. Then g ( λ ) is strictly decreasing for λ > 0 .

Second, since lim t → 0 + F ( t ) f ( t ) = 0 and lim t → ∞ F ( t ) f ( t ) = + ∞ , we obtain that lim λ → 0 + g ( λ ) = ∞ and lim λ → ∞ g ( λ ) = 0 by applying [2, Proposition 2.9].□

Define h ( λ ) = sup T λ ( r ) : r ∈ 0 , F − 1 1 λ . Following [2, Proposition 2.13] and [3, Lemma 3.4], we have

Lemma 3.4

For (1.1)(1.3), we have

for fixed r > 0 , T λ ( r ) is a continuous, strictly decreasing function of λ ∈ 0 , F − 1 1 λ ,
h ( λ ) is a continuous, strictly decreasing function of λ > 0 . Moreover, lim λ → 0 + h ( λ ) = ∞ and lim λ → ∞ h ( λ ) = 0 .

Next, we will prove the general diagram of T λ ( r ) .

Lemma 3.5

For (1.1)(1.3), T λ ( r ) has exactly one critical point, a local maximum, on 0 , F − 1 1 λ if 0 < p ≤ 2 4 .

Proof

If 0 0 by Lemmas 3.1 and 3.2. Thus, T λ ( r ) has at least one critical point, a local maximum, on 0 , F − 1 1 λ . Next, we need to prove T λ ( r ) has at most one critical point, a local maximum, on 0 , F − 1 1 λ . By [4, Lemma 3.6], we have

T λ ″ ( r ) − f ′ ( r ) f ( r ) T λ ′ ( r ) = f 2 ( r ) λ F 2 ( r ) ∫ 0 λ F ( r ) − I ( y , u ) ( 2 − y ) 2 1 − ( 1 − y ) 2 f 5 ( t ) d y ,

where y ∈ [ 0 , 1 ] , t = F − 1 λ F ( r ) − y λ .

And

(3.4) I ( y , u ) = ( 1 + y ) f 4 ( u ) + 2 ( 2 − y ) ( 1 − 3 y + y 2 ) F ( u ) f 2 ( u ) f ′ ( u ) + ( 1 − y ) ( 2 − y ) 2 F 2 ( u ) [ f ( u ) f ″ ( u ) − 3 ( f ′ ( u ) ) 2 ] ,

where y ∈ [ 0 , 1 ] , u ∈ ( 0 , + ∞ ) .

Since f ( u ) = u 1 + u p , f ′ ( u ) = p u ( 1 + u ) f ( u ) , and f ″ ( u ) = p ( p − 1 − 2 u ) u 2 ( 1 + u ) 2 f ( u ) , the equation I ( y , u ) in (3.4) can be simplified to

I ( y , u ) = ( 1 + y ) f 4 ( u ) + 2 ( 2 − y ) ( 1 − 3 y + y 2 ) F ( u ) f 3 ( u ) p u ( 1 + u ) − ( 1 − y ) ( 2 − y ) 2 F 2 ( u ) f 2 ( u ) p 2 u + 1 u 2 ( 1 + u ) 2 − 2 ( 1 − y ) ( 2 − y ) 2 F 2 ( u ) f 2 ( u ) p 2 u 2 ( 1 + u ) 2 .

Let

(3.5) ψ ( y , u ) = u 2 ( u + 1 ) 2 f 2 ( u ) I ( y , u ) = ( 1 + y ) u 2 ( u + 1 ) 2 f 2 ( u ) + 2 ( 2 − y ) ( 1 − 3 y + y 2 ) F ( u ) f ( u ) p u ( 1 + u ) − ( 1 − y ) ( 2 − y ) 2 F 2 ( u ) p ( 2 u + 1 ) − 2 ( 1 − y ) ( 2 − y ) 2 F 2 ( u ) p 2 .

In terms of ψ ( y , u ) , we have

∂ ψ ∂ y = u 2 ( u + 1 ) 2 f 2 ( u ) − 2 ( y − 1 ) ( 3 y − 7 ) F ( u ) f ( u ) p u ( 1 + u ) + ( y − 2 ) ( 3 y − 4 ) F 2 ( u ) p ( 2 u + 1 ) + 2 ( y − 2 ) ( 3 y − 4 ) F 2 ( u ) p 2 ,

then

( y − 2 ) ( 3 y − 4 ) ψ ( y , u ) + ( 1 − y ) ( 2 − y ) 2 ∂ ψ ∂ y = ( 2 y 3 − 2 y 2 − 10 y + 12 ) f 2 ( u ) + ( 4 y 3 − 22 y 2 + 40 y − 24 ) F ( u ) f ( u ) p u ( 1 + u ) = 2 ( y − 2 ) f ( u ) [ ( y 2 + y − 3 ) f ( u ) + ( y − 2 ) ( 2 y − 3 ) F ( u ) p u ( 1 + u ) ] .

Note that

φ ( y , u ) = ( y 2 + y − 3 ) f ( u ) + ( y − 2 ) ( 2 y − 3 ) F ( u ) p u ( 1 + u ) ,

then

∂ φ ∂ y = ( 2 y + 1 ) f ( u ) + ( 4 y − 7 ) F ( u ) p u ( 1 + u ) .

Therefore,

(3.6) ( y − 2 ) ( 2 y − 3 ) ∂ φ ∂ y + ( 7 − 4 y ) φ ( y , u ) = − 3 ( y − 1 ) ( 3 y − 5 ) f ( u ) < 0 .

As 0 < p < 1 , we have

φ ( 0 , u ) = − 3 f ( u ) + 6 F ( u ) p ( u + 1 ) u = 6 [ F ( u ) − p 2 2 f ′ ( u ) ] > 6 [ F ( u ) − f ′ ( u ) ] , φ ( 1 , u ) = − f ( u ) + F ( u ) p ( u + 1 ) u = F ( u ) − p 2 f ′ ( u ) > F ( u ) − f ′ ( u ) .

Next, we need to determine the sign of γ ( u ) ≡ F ( u ) − f ′ ( u ) .

γ ′ ( u ) = f ( u ) − f ″ ( u ) = u 2 ( u + 1 ) 2 + p ( 2 u + 1 − p ) u 2 ( u + 1 ) 2 f ( u ) > 0 ,

and γ ( 0 ) = 0 , so γ ( u ) > 0 . Then φ ( 0 , u ) > 0 , φ ( 1 , u ) > 0 .

From (3.6), we have φ ( y , u ) > 0 , so ( y − 2 ) ( 3 y − 4 ) ψ ( y , u ) + ( 1 − y ) ( 2 − y ) 2 ∂ ψ ∂ y < 0 . Since F ( u ) = ∫ 0 u f ( t ) d t ≤ u f ( u ) . From (3.5), we have

ψ ( 1 , u ) = 2 u 2 ( u + 1 ) 2 f 2 ( u ) − 2 F ( u ) f ( u ) p u ( u + 1 ) ≥ 2 u 2 ( u + 1 ) 2 f 2 ( u ) − 2 u 2 ( u + 1 ) f 2 ( u ) p = 2 u 2 ( u + 1 ) f 2 ( u ) ( u + 1 − p ) > 0

and

ψ ( 0 , u ) = u 2 ( u + 1 ) 2 f 2 ( u ) + 4 F ( u ) f ( u ) p u ( u + 1 ) − 4 F 2 ( u ) p ( 2 u + 1 ) − 8 F 2 ( u ) p 2 ≥ ( u + 1 ) 2 F 2 ( u ) + 4 p ( u + 1 ) F 2 ( u ) − 4 p ( 2 u + 1 ) F 2 ( u ) − 8 p 2 F 2 ( u ) = [ u 2 + 2 ( 1 − 2 p ) u + 1 − 8 p 2 ] F 2 ( u ) .

Then the discriminant 4 ( 1 − 2 p ) 2 − 4 ( 1 − 8 p 2 ) ≤ 0 , we have 0 < p ≤ 2 4 .

By the above analysis, when 0 < p ≤ 2 4 , ψ ( y , u ) ≥ 0 , then I ( y , u ) ≥ 0 . So T λ ( r ) has exactly one critical point, a local maximum, on 0 , F − 1 1 λ if 0 < p ≤ 2 4 .□

If p = 1 , f ( u ) = u 1 + u is a standard logistic function, we have

Lemma 3.6

For (1.1)(1.3) with p = 1 , T λ ( r ) has exactly one critical point, a local maximum, on 0 , F − 1 1 λ .

Proof

When p = 1 , similar to the analysis of Lemma 3.5, we have

( y − 2 ) ( 3 y − 4 ) ψ ( y , u ) + ( 1 − y ) ( 2 − y ) 2 ∂ ψ ∂ y < 0 ,

and ψ ( 1 , u ) = 2 u 2 ( u + 1 ) f 2 ( u ) u > 0 .

Note that ψ ( 0 , u ) = u 4 + 4 u 2 F ( u ) − 4 F 2 ( u ) ( 2 u + 3 ) ≜ g ( u ) , then

g ′ ( u ) = 4 u 3 ( u + 2 ) u + 1 − 8 u ( u + 2 ) u + 1 F ( u ) − 8 F 2 ( u ) , g ″ ( u ) = 12 u 4 + 24 u 3 + 8 u 2 ( u + 1 ) 2 − 24 u 2 + 32 u + 16 ( u + 1 ) 2 F ( u ) , g ‴ ( u ) = 24 u 4 + 48 u 3 + 24 u 2 ( u + 1 ) 3 + 16 u 2 l n ( u + 1 ) ( u + 1 ) 3 > 0 ,

and g ( 0 ) = g ′ ( 0 ) = g ″ ( 0 ) g ‴ ( 0 ) = 0 . Therefore, g ( u ) ≡ ψ ( 0 , u ) > 0 , so we have ψ ( y , u ) > 0 .

The proof of Lemma 3.6 is completed.□

For (3.1), we have

T λ ′ ( r ) = 1 r ∫ 0 r I 1 ( r , u , λ ) + I 2 ( r , u , λ ) { 1 − [ 1 + λ F ( u ) − λ F ( r ) ] 2 } 3 2 d u ,

where

I 1 ( r , u , λ ) = − λ 2 [ F ( r ) − F ( u ) ] 2 [ 3 + λ F ( u ) − λ F ( r ) ] < 0 , I 2 ( r , u , λ ) = λ [ 2 F ( r ) − 2 F ( u ) − r f ( r ) + u f ( u ) ] .

Note that θ ( u ) = 2 F ( u ) − u f ( u ) , then

(3.7) θ ′ ( u ) = f ( u ) − u f ′ ( u ) = u + 1 − p u + 1 f ( u ) < 0 , on ( 0 , p − 1 ) , = 0 , when u = p − 1 , > 0 , on ( p − 1 , + ∞ )

and

θ ″ ( u ) = − u f ″ ( u ) = p f ( u ) u ( u + 1 ) 2 ( 2 u + 1 − p ) < 0 , on 0 , p − 1 2 , = 0 , when u = p − 1 2 , > 0 , on p − 1 2 , + ∞ .

The graph of θ ( u ) is given in Figure 4.

$Figure 4 The graph of θ ( u ) \theta \left(u) .$

Figure 4

The graph of θ ( u ) .

Let λ p ≡ λ > 0 : F − 1 1 λ = p − 1 .

Lemma 3.7

For (1.1)(1.3), when p > 1 and λ ≥ λ p , T λ ( r ) is strictly decreasing on 0 , F − 1 1 λ .

Proof

Since λ ≥ λ p , we have 0 , F − 1 1 λ ⊂ ( 0 , p − 1 ] . From (3.7), θ ( u ) is decreasing on ( 0 , p − 1 ) , so θ ( u ) > θ ( r ) , when u < r . As I 2 ( r , u , λ ) = λ ( θ ( r ) − θ ( u ) ) < 0 , then T λ ′ ( r ) < 0 .

This completes the proof of Lemma 3.7.□

Then we study the profile of T λ ( r ) on 0 , F − 1 1 λ for 0 < λ < λ p in the following lemma. Obviously, there exists κ > p − 1 , satisfying θ ( r ) > 0 , then

I 2 ( κ , u , λ ) = λ ( θ ( κ ) − θ ( u ) ) > 0 , for 0 < u < κ .

From [4, Lemma 3.10], we have I 1 ( r , u , λ ) = O ( I 2 ( r , u , λ ) ) , so T λ ′ ( κ ) > 0 holds for sufficiently small λ > 0 .

From Lemmas 3.1 and 3.2, lim r → 0 + T λ ( r ) = + ∞ , lim r → 0 + T λ ′ ( r ) = − ∞ , and T λ ′ F − 1 1 λ < 0 , then T ( r ) has at least two critical points, a local minimum and a local maximum on 0 , F − 1 1 λ .

Lemma 3.8

For (1.1)(1.3), if p > 1 there exists κ > p − 1 , T ′ ( κ ) > 0 for sufficiently small λ > 0 . Moreover, T λ ( r ) has at least two critical points, a local minimum and a local maximum on 0 , F − 1 1 λ .

4 Proof of main results

Proof of Theorem 1

If 0 < p ≤ 2 4 , we obtain the following properties by Lemmas 3.1–3.5:

lim r → 0 + T λ ( r ) = 0 for all λ > 0 .
lim r → 0 + T λ ′ ( r ) = + ∞ for all λ > 0 .
T λ ′ F − 1 1 λ < 0 for all λ > 0 .
For any λ > 0 , T λ ( r ) has exactly one critical point, a local maximum, on 0 , F − 1 1 λ .
For any fixed r > 0 , T λ ( r ) is a continuous, strictly decreasing function of λ on 0 , F − 1 1 λ .
h ( λ ) is a continuous, strictly decreasing function on ( 0 , + ∞ ) .
lim λ → 0 + F − 1 1 λ = ∞ and lim λ → ∞ F − 1 1 λ = 0 .
lim λ → 0 + g ( λ ) = ∞ and lim λ → ∞ g ( λ ) = 0 .

So we have that there exists a unique λ ∗ > 0 such that g ( λ ∗ ) = L for any fixed L > 0 , then T λ ( r ) = L has only one root on 0 , F − 1 1 λ for 0 < λ < λ ∗ .

Moreover, there exists a unique λ ∗ ∈ ( λ ∗ , + ∞ ) such that

h ( λ ∗ ) = g ( λ ∗ ) = L .

Then T λ ( r ) = L has no root on 0 , F − 1 1 λ for λ > λ ∗ , and has two roots on 0 , F − 1 1 λ for λ ∗ < λ < λ ∗ .

The proof of Theorem 1 is completed.□

Proof of Theorem 2

If p = 1 , f ( u ) = u u + 1 , we obtain the following properties by Lemmas 3.1–3.4, 3.6:

lim r → 0 + T λ ( r ) = π 2 λ for all λ > 0 .
lim r → 0 + T λ ′ ( r ) > 0 for all λ > 0 .
T λ ′ F − 1 1 λ < 0 for all λ > 0 .
For any λ > 0 , T λ ( r ) has exactly one critical point, a local maximum, on 0 , F − 1 1 λ .
For any fixed r > 0 , T λ ( r ) is a continuous, strictly decreasing function of λ on 0 , F − 1 1 λ .
h ( λ ) is a continuous, strictly decreasing function on ( 0 , + ∞ ) .
lim λ → 0 + F − 1 1 λ = ∞ and lim λ → ∞ F − 1 1 λ = 0 .
lim λ → 0 + g ( λ ) = ∞ and lim λ → ∞ g ( λ ) = 0 .

From (3.1) and (3.2), we have

g ( λ ) = ∫ 0 F − 1 1 λ λ F ( u ) 1 − [ λ F ( u ) ] 2 d u .

Let y = λ F ( u ) , then

(4.1) g ( λ ) = ∫ 0 1 1 λ y 1 − y 2 f F − 1 y λ d y .

Let t = F − 1 y λ , and from F ( u ) ≤ u f ( u ) , we compute

g ( λ ) = ∫ 0 1 1 1 − y 2 F ( t ) f ( t ) d y ≤ ∫ 0 1 1 1 − y 2 t d y ≤ ∫ 0 1 1 1 − y 2 F − 1 1 λ d y = π 2 F − 1 1 λ .

Next, we need to compare the endpoint value g ( λ ) and π 2 λ . Note that α ( λ ) = F − 1 1 λ − 1 λ , then α ( ∞ ) = 0 and

α ′ ( λ ) = 1 f 1 λ − 1 λ 2 + 1 2 λ 3 2 = λ ( λ + 1 ) − 2 2 λ 2 ( λ + 1 ) = ( λ 2 + 3 λ + 4 ) ( λ − 1 ) 2 λ 2 ( λ + 1 ) [ λ ( λ + 1 ) + 2 ] ,

so when λ > 1 , α ′ ( λ ) > 0 , then α ( λ ) < 0 , hence g ( λ ) < π 2 λ . When λ > 1 and L < π 2 , there exist two constants 1 < λ ∗ < π 2 L 2 < λ ∗ such that

h ( λ ∗ ) = g ( λ ∗ ) = L ,

then T λ ( r ) = L has two roots on 0 , F − 1 1 λ for π 2 L 2 ≤ λ < λ ∗ , has one root on 0 , F − 1 1 λ for λ ∗ < λ < π 2 L 2 and λ = λ ∗ , and has no root on 0 , F − 1 1 λ for λ < λ ∗ and λ > λ ∗ .

The proof of statement (1) is complete.

As f ( u ) < 1 , from (4.1), we compute

g ( λ ) > ∫ 0 1 1 λ y 1 − y 2 d y = 1 λ

so when 0 < λ ≤ 4 π 2 , we have g ( λ ) > π 2 λ . When 0 < λ ≤ 4 π 2 , L > π 2 4 , there exist two constants π 2 L 2 < λ ∗ < λ ∗ such that

h ( λ ∗ ) = g ( λ ∗ ) = L ,

then T λ ( r ) = L has two roots on 0 , F − 1 1 λ for λ ∗ ≤ λ < λ ∗ , has one root on 0 , F − 1 1 λ for π 2 L 2 < λ < λ ∗ and λ = λ ∗ , and has no root on 0 , F − 1 1 λ for λ < π 2 L 2 and λ > λ ∗ .

The proof of Theorem 2 is completed.□

Proof of Theorem 3

If p > 1 , we obtain the following properties by Lemmas 3.1–3.4, 3.7:

lim r → 0 + T λ ( r ) = + ∞ .
lim r → 0 + T λ ′ ( r ) = − ∞ .
T λ ′ F − 1 1 λ < 0 for all λ > 0 .
For any λ > 0 , T λ ( r ) has exactly one critical point, a local maximum, on 0 , F − 1 1 λ .
For any fixed r > 0 , T λ ( r ) is a continuous, strictly decreasing function of λ on 0 , F − 1 1 λ .
h ( λ ) is a continuous, strictly decreasing function on ( 0 , + ∞ ) .
lim λ → 0 + F − 1 1 λ = ∞ and lim λ → ∞ F − 1 1 λ = 0 .
lim λ → 0 + g ( λ ) = ∞ and lim λ → ∞ g ( λ ) = 0 .
T λ ( r ) is strictly decreasing on 0 , F − 1 1 λ for λ ≥ λ p .

Note that L ∗ = T F − 1 1 λ p , when 0 < L ≤ L ∗ , T λ ( λ ) = L has exactly one positive solution on 0 , F − 1 1 λ for λ ≥ λ ∗ , and has no root on 0 , F − 1 1 λ for λ < λ ∗ . Statement (1) holds. From Lemma 3.8, we have
there exists L ∗ > L ∗ , then when L > L ∗ , there exists a unique constant λ ∗ > 0 such that sup T λ ∗ ( r ) : r ∈ 0 , F − 1 1 λ ∗ = L ∗ , so the equation T λ ( r ) = L has no root on 0 , F − 1 1 λ for 0 < λ < λ ∗ .
There exists a unique constant λ ∗ > λ ∗ > 0 , such that the maxima value of the local maximum value of T λ ∗ ( r ) is L ∗ , then the equation T λ ( r ) = L has exactly one root on 0 , F − 1 1 λ for λ > λ ∗ .

The proof of Theorem 3 is completed.□

Remark 1

It may be suggested from Theorems 1–3 that (a) ⊃ -shaped bifurcation diagrams do not change qualitatively for 0 1 when parameter L changes if g ( λ ) does not decrease on ( 0 , + ∞ ) . Further there may exist two positive numbers L ∗ < L ∗ such that the following statements hold:

If 0 < L ≤ L ∗ , the bifurcation curve is exactly decreasing shaped.
If L ∗ < L ≤ L ∗ , the bifurcation curve is ⊃ -shaped.
If L > L ∗ , the bifurcation curve is reversed S-shaped.

In future research, we will focus on such topics.

Acknowledgements

This research was supported by Beijing Municipal Natural Science Foundation (No. 4202025), partially sponsored by the National Natural Science Foundation of China (No. 61672070), Beijing Municipal Education Commission (No. KZ201910005008, KM201911232003), and the Research Fund from Beijing Innovation Center for Future Chips (No. KYJJ2018004).

Conflict of interest: Authors state no conflict of interest.

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Received: 2021-02-18

Revised: 2021-04-20

Accepted: 2021-06-28

Published Online: 2021-08-28

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https://doi.org/10.1515/math-2021-0070

Keywords for this article

mean curvature; bifurcation curve; logistic function

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