Reversed S-Shaped Bifurcation Curve for a Neumann Problem

We study the bifurcation and the exact multiplicity of solutions for a class of Neumann boundary value problem with indefinite weight. We prove that all the solutions obtained form a smooth reversed S-shaped curve by topological degree theory, CrandallRabinowitz bifurcation theorem, and the uniform antimaximum principle in terms of eigenvalues. Moreover, we obtain that the equation has exactly either one, two, or three solutions depending on the real parameter. The stability is obtained by the eigenvalue comparison principle.


Introduction
The existence and multiplicity of solutions of Neumann problems have been investigated by many authors; see, for example, [1][2][3][4][5].It is well known that determining the exact number of solutions of semilinear equations is usually a very difficult and challenging task.The results on exact multiplicity of solutions for Neumann problems are very few in the previous literature.
In [13], under Neumann boundary value conditions, the bifurcation of solutions to a logistic equation with harvesting has been investigated using the uniform antimaximum principle and Crandall-Rabinowitz bifurcation theorem.The uniform antimaximum principle plays an important role in proving the main results.More theories and applications of antimaximum principle can be seen, for example, [14,15].As continuation of [13], in this paper, the bifurcation of solutions for a Neumann problem with cubic nonlinearity is investigated using the uniform antimaximum principle, Crandall-Rabinowitz bifurcation theorem, the topological degree theory, and the continuation method.More detailed results on bifurcation theory can be seen in [16,17].The topological degree theory and the uniform antimaximum principle play an important role in proving the main results in this paper.

Preliminaries
Definition 1 (see [18]).We call a solution  of the equation a stable solution if the principal eigenvalue  1 (  (, )) of the equation is strictly positive.The solution  is unstable if the principal eigenvalue  1 (  (, )) is negative.
Definition 2 (see [18]).We call a solution  of the equation a nondegenerate solution if the linearized equation does not admit any nontrivial solutions.The solution  is degenerate (singular) if the linearized equation ( 7) has nontrivial solutions.Definition 3 (see [19]).A mapping  :  →  ( and  are topological spaces) is said to be proper if for every compact set  ⊂ , the set  −1 () is compact in .
Lemma 4 (see [19]).Let  be a real Banach space and  a linear, compact map in .Suppose that  ̸ = 0 and  −1 is not an eigenvalue of .Let Ω ⊂  be an open bounded and 0 ∈ Ω.Then where () is the sum of the algebraic multiplicities of all the eigenvalues  satisfying  > 1, and () = 0 if  has no eigenvalues  of this kind.
Proof.It is easy to obtain (10) and the proof is omitted.The eigenvalue problem ( 9) is equivalent to the equation − = 0.The equation  −  = 0 has nontrivial solutions if and only if  −1 is the eigenvalue of the operator .For  > 1, the general solution of equation ( 9) is By Neumann boundary value condition, we have When  ̸ = 0, we have  =  2  2 + 1, where  ≥ 0,  ∈ N. Therefore,   = ( 2  2 + 1) −1 is the eigenvalue of , where  ≥ 0, ( −   ) is the one-dimensional subspace of Banach space  spanned by   = cos().
For  ≥ 0, Lemma 4 implies that This completes the proof.
does not admit any nontrivial solutions.
Let Ω be open set in Banach space .Suppose  : Ω →  is completely continuous operator,  0 ∈ Ω,  0 =  0 .Suppose that  is Fréchet differentiable at  0 and 1 is not the eigenvalue of derived operator   ( 0 ).Then  0 is an isolated fixed point of  and where  is the sum of the algebraic multiplicities of all the eigenvalues   ( 0 ) in (0, 1).Now we state three known results, which plays a key role in proving the main results in this paper.We do not provide their proof which can be seen in [21].
Finally, we prove the positive solution is unique and stable.We denote by  0 the nontrivial solution of (51).It is obvious that − 0 () is also the nontrivial solution of (51).It follows from (56) and the comparison of eigenvalues that  1 ( 2 ()) >  (57) Let  be the number of nontrivial solutions of () = 0. Hence,  is proper.Since 0 is a regular value of ,  must be finite.According to Lemma 17, we have deg( − ,   , 0) = 1 for sufficiently large .By Lemmas 9 and 11 and the index formula, we have that It is obvious that  = 2. Let V 0 () ≡ − 0 () for  ∈ [0, 1].Therefore, () = 0 has exactly three solutions  0 , 0, and V 0 .Since  1 ( 2 ) >  1 ( 1 ) = 0, we have that the positive solution  0 () is unique and stable.The negative solution V 0 () is also stable.This completes the proof.
Proof.The proof is similar to the proof of Lemma 19.
Theorem 21.Assume that the first eigenvalue  1 (()) < 0 and () > 0 for all  ∈ [0,1].Suppose that () ≪  2 /4.Then all the solutions of ( 1) are of one sign and lie on a unique reversed S-shaped solution curve, which is symmetric with respect to the origin.More precisely, there exists  * > 0, such that (i) For  >  * , (1) has no positive solution and has a unique negative solution which is stable.
(ii) For  = ± * , (1) has exactly two solutions.Moreover, when  =  * , the negative solution is stable and the positive solution is degenerate.When  = − * , the positive solution is stable and the negative solution is degenerate.
(iii) For − * <  <  * , (1) has exactly three ordered solutions at the same  and the middle solution is unstable and the remaining two are stable.Moreover, when  < 0, the maximal solution is positive and the other two are negative.When  > 0, the minimal solution is negative and the other two are positive.
(iv) For  < − * , (1) has no negative solution and has a unique positive solution which is stable.
Proof.It follows from Lemma 19 that (1) has a unique nondegenerate positive solution  1 () for some  < 0. The solution curve can be continued a little bit such that  1 () remains positive for increasing  when  < 0. By Lemma 18, (1) has a unique nondegenerate positive solution  0 () for  = 0.The positive solution curve can pass through  1 () and  0 () and can be continued further for increasing  until the linearized equation (61) admits the nontrivial solutions.We claim that the curve of positive solutions cannot be continued for  >  * .Next we will prove the existence of  * .Since  1 (()) < 0, let V() > 0 be the first eigenfunction of the problem V  +  () V + V = 0,  ∈ (0, 1) , V  (0) = V  (1) = 0. (62)
by   .It is obvious that  1 = 0 is the first eigenvalue and  2 =  2 is the second eigenvalue of the eigenvalue equation