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BY 4.0 license Open Access Published by De Gruyter May 8, 2019

Non-local Degenerate Diffusion Coefficients Break Down the Components of Positive Solutions

  • M. Delgado , C. Morales-Rodrigo , J. R. Santos Júnior and A. Suárez EMAIL logo

Abstract

This paper deals with nonlinear elliptic problems where the diffusion coefficient is a degenerate non-local term. We show that this degeneration implies the growth of the complexity of the structure of the set of positive solutions of the equation. Specifically, when the reaction term is of logistic type, the continuum of positive solutions breaks into two disjoint pieces. Our approach uses mainly fixed point arguments.

MSC 2010: 35J60; 35J70; 35J15

1 Introduction

We are interested in the structure of the set of positive solutions of the equation

(1.1) { - a ( Ω u p ( x ) 𝑑 x ) Δ u = f ( λ , x , u ) in  Ω , u = 0 on  Ω ,

where ΩN, N1, is a bounded and smooth domain, p>0, f:×Ω× is a nonlinear function that will be detailed along the paper, and the non-local diffusion coefficient a:+[0,) is a continuous function, with a(0)0, that verifies the following hypothesis (which will be assumed throughout the work):

  1. There exists s1>0 such that a(s1)=0 and a(s)>0 for all s>0 and ss1.

Equation (1.1) has received a lot of attention in the last years, especially when the function a is far away from zero, see, for example, [5, 6, 8, 14] and references therein. In some papers, for instance, [2] and [9], the hypotheses on a were weakened and it was allowed that

lim s + a ( s ) = 0 .

Recently, in [1] and [12], it was assumed that a can vanish in several points. In [12], by considering the same sort of equations as in (1.1), with p>1, and a reaction term similar (but not equal) to a logistic equation, Gasiński and Santos Júnior prove the existence of at least 2K positive solutions if the function a vanishes in K positive points and satisfies some appropriate additional conditions. The approach in [1] is based on fixed point techniques, sub-super solution arguments and variational methods.

In this paper, we study (1.1) for each of the following reaction terms:

  1. Linear case: f(λ,x,u)=f(x), where f is a bounded, non-negative and non-trivial function.

  2. Eigenvalue problem: f(λ,x,u)=λu, where λ.

  3. Sublinear case: f(λ,x,u)=λuq, where λ and q(0,1).

  4. Logistic case: f(λ,x,u)=λu-u2, where λ.

We start with the first three cases. In the local case, that is, a1 in (1.1), the problems are well understood. In the linear case, there exists a unique positive solution. In the eigenvalue case, there exists a increasing sequence {λn}n of eigenvalues with λn+; the first one, λ1, called principal eigenvalue, is the only one having an associated positive eigenfunction. Finally, it is also well known that the sublinear problem possesses a positive solution if and only if λ>0. In the case λ>0, there exists at most one positive solution.

In the non-local case, in the first three cases, basically we can compute directly the solutions. In these cases, the main result can be summarized as follows: the study of equation (1.1) is equivalent to the study of the following nonlinear real equation in R>0:

  1. a ( R ) R 1 / p = A in the linear case,

  2. a ( R ) = λ λ 1 in the eigenvalue problem,

  3. a ( R ) R 1 - q p = λ B in the sublinear case,

where A and B are positive constants related to f and q, respectively (see Section 2 for their concrete values) and λ1 is the principal eigenvalue associated to -Δ under homogeneous Dirichlet boundary conditions. Moreover, the solutions have the following expressions, respectively:

u = R 0 1 / p A e , u = R 0 1 / p φ 1 , u = R 0 1 / p B 1 / ( 1 - q ) ω ,

where e and ω are the unique positive solutions of the following equations, respectively:

- Δ e = 1 in  Ω ,    - Δ ω = ω q in  Ω ,    e = ω = 0 on  Ω ,

φ 1 is the positive eigenfunction associated to λ1 such that Ωφ1p=1 and R0 is any positive solutions of the above real equations.

Therefore, in view of the above results, in the linear case, there exists at least one positive solution if A=a(R)R1/p for some R>0.

Now, we focus our attention on the eigenvalue and sublinear cases. Our aim is to show the results with respect to the real parameter λ. In these cases, the results depend on the behavior of the functions a(R) and

b ( R ) := a ( R ) R 1 - q p ,

respectively. In Figures 14 we present a number of admissible bifurcation diagrams for the eigenvalue problem and the sublinear case. We have represented in the vertical axis the quantity Ωup(x)𝑑x of the positive solutions of (1.1).

Figure 1 shows both bifurcation diagrams in the case where a(R) as R and a(0)>0, as well as the graph of an admissible graph of a(s).

Figure 2 represents the case a(R) as R and a(0)=0. In this case, we have drawn only the bifurcation diagram of the eigenvalue problem due to the fact that the sublinear case is similar to the one already sketched in Figure 1.

Figure 3 represents the eigenvalue problem in the case a(R)a>0 as R. Observe that in this case, there does not exist any positive solution for λ large. The sublinear case is again similar to the one shown in Figure 1.

In Figure 4, we have drawn the case a(R)0 and b(R)0 as R. Observe that in this case a bifurcation to infinity scenario occurs as λ0.

Figure 1 
          Bifurcation diagrams in the case that a⁢(R)→∞{a(R)\to\infty} as R→∞{R\to\infty} and a⁢(0)>0{a(0)>0}. On the left-hand side, the graph of a;in the center and on the right hand side the bifurcation diagrams of the eigenvalue and sublinear cases, respectively.
Figure 1

Bifurcation diagrams in the case that a(R) as R and a(0)>0. On the left-hand side, the graph of a;in the center and on the right hand side the bifurcation diagrams of the eigenvalue and sublinear cases, respectively.

Figure 2 
          Bifurcation diagram of the eigenvalue problem in the case a⁢(R)→∞{a(R)\to\infty} as R→∞{R\to\infty} and a⁢(0)=0{a(0)=0}.
Figure 2

Bifurcation diagram of the eigenvalue problem in the case a(R) as R and a(0)=0.

Figure 3 
          Bifurcation diagram of the eigenvalue problem in the case a⁢(R)→a∞>0{a(R)\to a_{\infty}>0} as R→∞{R\to\infty} and a⁢(0)>0{a(0)>0}.
Figure 3

Bifurcation diagram of the eigenvalue problem in the case a(R)a>0 as R and a(0)>0.

Figure 4 
          Bifurcations diagrams of the eigenvalue problem and sublinear case, respectively, when a⁢(R)→0{a(R)\to 0} and b⁢(R)→0{b(R)\to 0}as R→∞{R\to\infty}.
Figure 4

Bifurcations diagrams of the eigenvalue problem and sublinear case, respectively, when a(R)0 and b(R)0as R.

The logistic case is more difficult and interesting. We recall that in the local case, that is, a1 in (1.1), there exists a positive solution if and only if λ>λ1. In the case λ>λ1, there exists at most one positive solution. Moreover, from the trivial solution emanates at λ=λ1 an unbounded continuum in ×C(Ω¯) of positive solutions. In fact, from another point of view, we have the following result (see [3, 15] and references therein): given α>0, there exists a unique positive solution (λ,u)=(λ(α),uα)×C2(Ω¯) of (1.1) with uα2=α. Moreover, αλ(α) is an increasing map of class C1 and λ(α)λ1 as α0 and λ(α) as α. Actually, this result is also true with p for any p[1,+].

In the non-local case, we cannot calculate directly the solution or applying any classical methods due to the degeneracy of the diffusion. Hence, we apply a fixed point argument to prove the following results:

  1. Fix α>0 and αs1. Then there exists a unique positive solution of (1.1), (λ,u)=(λα,uα)×C2(Ω¯), such that

    Ω u α p ( x ) 𝑑 x = α .

  2. The map α(0,s1)(s1,)λα is continuous and it verifies

    lim α 0 + λ α = a ( 0 ) λ 1 , lim α s 1 λ α = Λ 1 , lim α λ α = ,

    where

    Λ 1 := ( s 1 | Ω | ) 1 / p .

  3. Equation (1.1) cannot admit a positive solution, u, such that

    Ω u p ( x ) 𝑑 x = s 1 .

Roughly speaking, it can be said that the continuum of positive solutions obtained in the local case is broken in the non-local one. Indeed, if we consider the bifurcation diagram (λ,Ωup), then the continuum of positive solutions “breaks” at Ωup=s1 (see Figure 5).

Figure 5 
          Bifurcations diagrams in the logistic case.
Figure 5

Bifurcations diagrams in the logistic case.

As a consequence of the above results, we provide a more accurate description of the structure of the set of λ in such that (1.1) possesses at least one positive solution. We obtain the following results:

  1. Equation (1.1) possesses at least one positive solution for every

    λ ( min { a ( 0 ) λ 1 , Λ 1 } , ) { Λ 1 } .

    In fact, (1.1) possesses at least one positive solution uλ for each

    λ ( min { a ( 0 ) λ 1 , Λ 1 } , max { a ( 0 ) λ 1 , Λ 1 } ) , with  Ω u λ p ( x ) 𝑑 x < s 1 .

    In addition, (1.1) possesses at least one positive solution wλ for all

    λ ( Λ 1 , ) , with  Ω w λ p ( x ) 𝑑 x > s 1  for all  λ > Λ 1 .

  2. As a consequence, if Λ1<a(0)λ1, then (1.1) possesses at least two positive solutions for every

    λ ( Λ 1 , a ( 0 ) λ 1 ) .

In Figure 5 we have represented the bifurcation diagrams in both cases: a(0)λ1<Λ1 and Λ1<a(0)λ1.

It is worth pointing out that all the above results are true in the case a(0)=0. In such a case, the bifurcation point λ=a(0)λ1 transforms into λ=0.

Let us mention that all previous results can be extended to the case where a vanishes in a finite number of points. In Figure 6 we show an example where a possesses two zeros s1<s2. We have represented various diagrams depending on the relative position of a(0)λ1 with respect to Λ1<Λ2.

Figure 6 
          Bifurcations diagrams when a vanishes in two points.
Figure 6

Bifurcations diagrams when a vanishes in two points.

An outline of the work is as follows. In Section 2 we analyze the linear, eigenvalue and sublinear cases. The logistic equation is studied in Section 3.

2 Computable Cases

In this section we analyze different cases of reaction functions where we can characterize the set of positive solutions of (1.1). Basically, we study the linear and sublinear cases and the eigenvalue problem.

The main result is the following.

Proposition 2.1.

1.  The study of the positive solutions of (1.1) with f(λ,x,u)=f(x), fL(Ω), f0, f0, is equivalent to the study of the real equation

(2.1) a ( R ) R 1 / p = A , R > 0 ,

where

A := ( Ω e p ( x ) 𝑑 x ) 1 / p ,

and e is the unique positive solution of

- Δ e = f ( x ) in  Ω ,    e = 0 on  Ω .

In this case, for each R0>0, solving (2.1), a positive solution u is given by

u = R 0 1 / p A e .

2.  The study of the positive solutions of (1.1) with f(λ,x,u)=λu, λR, is equivalent to the study of the real equation

(2.2) a ( R ) = λ λ 1 , R > 0 .

Moreover, for each R0>0, solving (2.2), a positive solution u is given by

u = R 0 1 / p φ 1 ,

where φ1 is the positive eigenfunction associated to λ1 such that Ωφ1p=1.

3.  The study of the positive solutions of (1.1) with f(λ,x,u)=λuq, λR, q(0,1), is equivalent to the study of the real equation

(2.3) a ( R ) R 1 - q p = λ B , R > 0 ,

where

B := ( Ω ω ( x ) p 𝑑 x ) 1 - q p ,

and ω is the unique positive solution of

- Δ ω = ω q in  Ω ,    ω = 0 on  Ω .

Moreover, for each R0>0, solving (2.3), a positive solution u is given by

u = R 0 1 / p B 1 / ( 1 - q ) ω .

In all cases, if u>0 is a positive solution of (1.1), then

R = Ω u p ( x ) 𝑑 x

is solution of (2.1), (2.2) or (2.3).

Proof.

1.  It is clear that u=Ke is solution of (1.1) if

a ( K p Ω e p ( x ) 𝑑 x ) K = 1 .

Under the change of variable

R = K p Ω e p ( x ) 𝑑 x ,

we reach (2.1).

Conversely, if u is a positive solution of (1.1), we have that

- Δ ( a ( Ω u p ( x ) 𝑑 x ) u ) = f in  Ω ,    u = 0 on  Ω ,

and then

e = a ( Ω u p ( x ) 𝑑 x ) u .

As a consequence, taking the p-th power and integrating, we get

Ω e p ( x ) 𝑑 x = a ( Ω u p ( x ) 𝑑 x ) p Ω u p ( x ) 𝑑 x .

Hence, R=Ωup(x)𝑑x is solution of (2.1). This proves the first paragraph.

2. Denote by φ1 the positive eigenfunction associated to λ1 such that

Ω φ 1 p ( x ) 𝑑 x = 1 .

With a similar reasoning to the first paragraph, u=Kφ1 is solution of (1.1), provided that

a ( K p ) = λ λ 1 .

Taking R=Kp, we get (2.2).

On the other hand, if u is a positive solution of (1.1), we obtain

- Δ u = λ a ( Ω u p ( x ) 𝑑 x ) u in  Ω , u = 0 on  Ω ,

whence, since u>0 in Ω, it follows that

λ 1 = λ a ( Ω u p ( x ) 𝑑 x ) and u = K φ 1 in  Ω ,

for some K>0. Hence, R=Ωup(x)𝑑x is a solution of (2.2). 3. Again u=Kω is a solution of (1.1) if

a ( K p Ω ω p ( x ) 𝑑 x ) K 1 - q = λ .

This leads us to (2.3).

On the other hand, if u is a positive solution of (1.1), we have that

- Δ u = ( λ a ( Ω u p ( x ) 𝑑 x ) ) u q ,

and then

u = ( λ a ( Ω u p ( x ) 𝑑 x ) ) 1 / ( 1 - q ) ω .

Now, the result follows promptly by taking the p-th power of the previous equality and integrating in Ω. ∎

3 The Logistic Equation

In this section we study the more interesting case f(λ,x,u)=λu-u2, λ. Before proving the main result for this case, we recall some well-known results about the local logistic equation

(3.1) { - d Δ u = λ u - u 2 in  Ω , u = 0 on  Ω ,

where d>0 and λ. Basically, the following results can be found in [4, 10, 11, 13].

Proposition 3.1.

Equation (3.1) has a positive solution if and only if λ>dλ1. In such case, there exists at most one positive solution, denoted by u[d,λ]. Moreover, the map

( d λ 1 , + ) C 2 ( Ω ¯ ) , λ u [ d , λ ] ,

is differentiable, strictly increasing and we have

(3.2) lim λ d λ 1 u [ d , λ ] = 0 .

Furthermore, the map

( 0 , λ λ 1 ) C 2 ( Ω ¯ ) , d u [ d , λ ] ,

is differentiable, strictly decreasing and

(3.3) u [ d , λ ] λ uniformly in compact sets of  Ω as  d 0 .

Finally, the following estimates hold:

(3.4) ( λ - d λ 1 ) φ 1 φ 1 u [ d , λ ] λ in  Ω ,

where φ1 is a positive eigenfunction associated to λ1.

In the following result, we complete the singular perturbation problem (3.3) (see [7] for a wide generalization of (3.3)).

Proposition 3.2.

Consider two sequences dn>0 and λn such that dn0 and λnλ*>0. Then

u [ d n , λ n ] λ * uniformly in compact sets of  Ω as  d n 0 .

Proof.

Basically, we follow the proof of [11, Theorem 2.1]. Firstly, observe that since λnλ*>0 and dn0, it follows that for n large enough, it holds that λn>dnλ1. Hence, there exists the positive solution u[dn,λn] of (3.1).

Let KΩ be a compact subset of Ω and fix ε>0. We have to show that there exists n0 such that for nn0,

(3.5) λ * - ε u [ d n , λ n ] ( x ) λ * + ε , x K .

Since λnλ*, by using the upper bound of (3.4), we have that for nn0,

u [ d n , λ n ] λ n λ * + ε .

We prove the left-hand side inequality of (3.5). Consider x0K and take r0>0 such that B=B(x0,r0)KΩ. Take d0>0, which will be chosen later. Then for dnd0, using Proposition 3.1, we get that

u [ d n , λ n ] u [ d 0 , λ * - ε / 4 ] u [ d 0 , λ * - ε / 4 , B ] in  B ,

where u[d0,λ*-ε,B] denotes the solution of (3.1) in B. Now, by (3.4), we obtain

u [ d n , λ n ] u [ d 0 , λ * - ε / 4 , B ] ( λ * - ε / 4 - d 0 λ 1 B ) φ 1 B in  B ,

where λ1B and φ1B denote the principal eigenvalue and the positive eigenfunction associated to -Δ in B under homogeneous Dirichlet boundary conditions and such that φ1B=1. Taking d0 small,

u [ d n , λ n ] u [ d 0 , λ * - ε , B ] ( λ * - ε 2 ) φ 1 B in  B .

Since φ1B is radially symmetric, 1=φ1B(x0)=φ1B, it follows that for R1 small,

φ 1 B λ * - ε λ * - ε 2 in  B ( x 0 , R 1 ) ,

and then

u [ d n , λ n ] λ * - ε , x B ( x 0 , R 1 ) .

This finishes the proof. ∎

To prove the main result of this section, we are going to use a fixed point argument. The following result will be crucial in our argument.

Proposition 3.3.

Let

Λ 1 := ( s 1 | Ω | ) 1 / p .

Take R>0 and Rs1. There exists a unique λR>a(R)λ1 such that

(3.6) R = Ω u [ a ( R ) , λ R ] p ( x ) 𝑑 x .

Define the map H:(0,){s1}R, H(R)=λR. Then H is continuous. Furthermore,

(3.7) lim R 0 + ( R ) = a ( 0 ) λ 1 , lim R ( R ) = + , lim R s 1 ( R ) = Λ 1 .

Finally, if the function a is increasing on (R1,R2), then H(R) is also increasing on (R1,R2).

Proof.

Fix R>0 and Rs1. Then a(R)0 and hence u[a(R),λ] is well defined for each λ>a(R)λ1. Moreover, by Proposition 3.1, the map λu[a(R),λ] is continuous and strictly increasing. Thus,

( λ ) := Ω u [ a ( R ) , λ ] p ( x ) 𝑑 x

is continuous, strictly increasing and by (3.2) and (3.4), we get

lim λ a ( R ) λ 1 ( λ ) = 0 , lim λ ( λ ) = + .

Therefore, there exists a unique λR>a(R)λ1 such that

( λ R ) = R .

We show now the continuity of the map (R). Take a sequence RnR0(0,s1)(s1,). We have that

R n = Ω u [ a ( R n ) , λ R n ] p ( x ) 𝑑 x , R 0 = Ω u [ a ( R 0 ) , λ R 0 ] p ( x ) 𝑑 x .

Assume by contradiction that there exists ε1>0 such that for all n, we have

λ R n λ R 0 + ε 1 .

Observe that

(3.8) u [ a ( R n ) , λ R n ] ( x ) u [ a ( R n ) , λ R 0 + ε 1 ] ( x ) , x Ω .

Since a(Rn)a(R0), we get that

(3.9) u [ a ( R n ) , λ R 0 + ε 1 ] u [ a ( R 0 ) , λ R 0 + ε 1 ] in  C 2 ( Ω ¯ ) .

On the other hand, by the monotony of the map λu[a(R0),λ], we get that there exists ε2>0 such that

(3.10) Ω u [ a ( R 0 ) , λ R 0 + ε 1 ] p ( x ) 𝑑 x > ε 2 + Ω u [ a ( R 0 ) , λ R 0 ] p ( x ) 𝑑 x = ε 2 + R 0 .

Then, using (3.8), (3.9) and (3.10), we conclude Rnε2+R0, a contradiction because RnR0.

Now, we prove (3.7). By the definition of λR, see (3.6), and using the upper bound of (3.4), we get

(3.11) R λ R p | Ω | .

On the other hand, by the lower bound of (3.4), we obtain

(3.12) R ( λ R - a ( R ) λ 1 ) p Ω φ 1 p ( x ) 𝑑 x φ 1 p .

Consequently, using (3.11), we get that (R) as R. In addition, by (3.12), λR-a(R)λ10 as R0. Thus, (R)a(0)λ1 as R0+.

Now, assume that Rns1. Inequality (3.12) enables us to assure that λRn is bounded and hence, at least for a subsequence, λRnλ0 for some λ00. Moreover, thanks to (3.11), we get that λ0>0. Then, by Proposition 3.2, we have that

u [ a ( R n ) , λ R n ] λ 0 uniformly in compact sets.

Thus,

u [ a ( R n ) , λ R n ] λ 0 almost everywhere  x Ω ,

and for some ε>0 and n large,

u [ a ( R n ) , λ R n ] p ( λ 0 + ε ) p .

Hence, by the dominated convergence theorem,

Ω u [ a ( R n ) , λ R n ] p 𝑑 x λ 0 p | Ω | .

Then, regarding (3.6), it follows that

s 1 = λ 0 p | Ω | .

This proves that (Rn)Λ1 as Rns1.

Finally, assume that a is increasing in (R1,R2). Take R3,R4(R1,R2) such that R3<R4 and assume by contradiction that (R3)(R4), that is, λR3λR4. Since a is increasing, a(R3)<a(R4), and then

u [ a ( R 3 ) , λ ] > a [ a ( R 4 ) , λ ] ,

and hence, by Proposition 3.1,

u [ a ( R 3 ) , λ R 3 ] > u [ a ( R 4 ) , λ R 3 ] u [ a ( R 4 ) , λ R 4 ] .

This is a contradiction because

R 3 = Ω u [ a ( R 3 ) , λ R 3 ] p ( x ) 𝑑 x < R 4 = Ω u [ a ( R 4 ) , λ R 4 ] p ( x ) 𝑑 x .

This completes the proof. ∎

Now, we are ready to establish the main result:

Theorem 3.4.

  1. Equation ( 1.1 ) possesses at least one positive solution for every

    λ ( min { a ( 0 ) λ 1 , Λ 1 } , ) { Λ 1 } .

  2. In fact, ( 1.1 ) possesses at least one positive solution u λ for each

    λ ( min { a ( 0 ) λ 1 , Λ 1 } , max { a ( 0 ) λ 1 , Λ 1 } ) , with  Ω u λ p ( x ) 𝑑 x < s 1 .

  3. In addition, ( 1.1 ) possesses at least one positive solution w λ for all

    λ ( Λ 1 , ) , with  Ω w λ p ( x ) 𝑑 x > s 1 for all  λ > Λ 1 .

  4. Equation ( 1.1 ) does not possess any positive solution u such that

    Ω u p ( x ) 𝑑 x = s 1 .

Proof.

First, we prove item 2. Since the map (R) is continuous and (0,s1) is connected, we conclude that (0,s1)=Λ, where Λ is an interval. Moreover, since (R)a(0)λ1 as R0 and (R)Λ1 as Rs1, we obtain that

I := ( min { a ( 0 ) λ 1 , Λ 1 } , max { a ( 0 ) λ 1 , Λ 1 } ) Λ = ( 0 , s 1 ) .

Hence, we claim that for each λI, there exists at least one positive solution uλ such that

Ω u λ p ( x ) 𝑑 x < s 1 .

Indeed, take λI; then λΛ. Hence, there exists R1(0,s1) such that (R1)=λ, that is, λR1=λ, or equivalently,

R 1 = Ω u [ a ( R 1 ) , λ R 1 ] p ( x ) 𝑑 x .

Hence, denoting uλ=u[a(R1),λR1], we have that

- a ( R 1 ) Δ u λ = λ R 1 u λ - u λ 2 ,

and then

- a ( Ω u λ p ( x ) 𝑑 x ) Δ u λ = λ u λ - u λ 2

and

Ω u λ p ( x ) 𝑑 x = R 1 < s 1 .

Now, we prove item 3. With similar reasoning to that used in the above case and taking into account that (R)+ as R, we conclude that (R) transforms the interval (s1,+) into a interval denoted by Γ, (s1,)=Γ, such that

J := ( Λ 1 , + ) Γ .

Moreover, for each λJ, there exists at least one positive solution wλ such that

Ω w λ p ( x ) 𝑑 x > s 1 .

It is clear that item 1 is consequence of items 2 and 3.

Finally, observe that (1.1) does not admit a positive solution u such that

Ω u p ( x ) 𝑑 x = s 1 .

Indeed, in such a case u must verify 0=u(λ-u) in Ω, which is impossible due to the Dirichlet homogeneous boundary condition. ∎

As a direct consequence of Theorem 3.4, we obtain the following corollary.

Corollary 3.5.

If Λ1<a(0)λ1, then (1.1) possesses at least two positive solutions for

λ ( Λ 1 , a ( 0 ) λ 1 ) .

Remark 3.6.

We could apply [9, Theorem 4] and conclude that for a(0)>0, the point λ=a(0)λ1 is a local bifurcation point from the trivial solution u0. Moreover, we could ascertain the bifurcation direction depending on p and a(0), see [9, Theorem 4].


Communicated by Julian Lopez Gomez


Funding source: MINECO/FEDER/UE

Award Identifier / Grant number: MTM2015-69875-P

Funding source: CNPq

Award Identifier / Grant number: 302698/2015-9

Funding source: CAPES

Award Identifier / Grant number: 88881.120045/2016-01

Funding statement: M. Delgado, C. Morales-Rodrigo and A. Suárez were partially supported by MTM2015-69875-P (MINECO/ FEDER, UE). J. R. Santos Júnior was partially supported by CNPq-Proc. 302698/2015-9 and CAPES-Proc. 88881.120045/2016-01, Brazil.

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Received: 2019-03-28
Accepted: 2019-04-04
Published Online: 2019-05-08
Published in Print: 2020-02-01

© 2019 Walter de Gruyter GmbH, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 International License.

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