# Reversed S-Shaped Bifurcation Curve for a Neumann Problem.

1. Introduction

The existence and multiplicity of solutions of Neumann problems have been investigated by many authors; see, for example, [1-5]. 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 this paper, we study the bifurcation and the exact multiplicity of solutions for the Neumann problem

u" + a(x)u - b (x) [u.sup.3] = [lambda]f (x), x [member of] (0,1), u' (0) = u' (1) = 0, (1)

depending on a real parameter [lambda], where a(x), b(x), and f(x) are given continuous functions and b(x) > 0 and a(x) may change sign.

Llibre and Roberto  studied the existence and the stability of periodic solutions of Duffing equation

x" + cx' + a(t)x + b (t) [x.sup.3] = [lambda]h (t). (2)

Chen and Li  also studied the Duffing equation (2) in a very particular case, i.e., a(t) [equivalent to] a > 0,b(t) = 1, and a, c constants. They obtained that (2) has exactly three T-periodic solutions. Lomtatidze et al.  also studied the existence of periodic solutions of Duffing type equations.

Tzeng et al.  also studied the global bifurcation and exact multiplicity of positive solutions of Dirichlet problem with cubic nonlinearity

u" + [lambda] (-[epsilon][u.sup.3] + [sigma][u.sup.2] - [kappa]u + [rho]) = 0, - 1 < x < 1, u(-1) = u(1) = 0, (3)

using the time-map method, where [epsilon], [sigma], [kappa], [rho] are constants. Equation (3) is an autonomous system, and the time-map method has been successfully employed to solve the problem (3), but it is not applicable to study the nonautonomous system (1). There are many results on exact multiplicity of solutions for the Dirichlet problems; see [9-12].

In , 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 , 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.

2. Preliminaries

Definition 1 (see ). We call a solution u of the equation

u" + g (x, u) = f (x), x [member of] (0,1), u' (0) = u' (1) = 0 (4)

a stable solution if the principal eigenvalue [[mu].sub.1] ([g.sub.u] (x,u)) of the equation

w" + [g.sub.u] (x, u)w = -[mu]w, x [member of] (0,1), w' (0) = w' (1) = 0 (5)

is strictly positive. The solution u is unstable if the principal eigenvalue [[mu].sub.1]([g.sub.u](x, u)) is negative.

Definition 2 (see ). We call a solution u of the equation

u" + g (x,u) = f (x), x [member of] (0,1), u' (0) = u' (1) = 0 (6)

a nondegenerate solution if the linearized equation

w" + [g.sub.u] (x, u)w = 0, x [member of] (0,1), w' (0) = w' (1) = 0 (7)

does not admit any nontrivial solutions. The solution u is degenerate (singular) if the linearized equation (7) has nontrivial solutions.

Definition 3 (see ). A mapping G : X [right arrow] Y (X and Y are topological spaces) is said to be proper if for every compact set K [subset] Y, the set [G.sup.-1] (K) is compact in X.

Lemma 4 (see ). Let X be a real Banach space and L a linear, compact map in X. Suppose that [lambda] [not equal to] 0 and [[lambda].sup.-1] is not an eigenvalue of L. Let [OMEGA] [subset] X be an open bounded and 0 [member of] [OMEGA]. Then

deg (I - [lambda]L, [OMEGA], 0) = [(-1).sup.m([lambda])] , (8)

where m([lambda]) is the sum of the algebraic multiplicities of all the eigenvalues [mu] satisfying [mu][lambda] > 1, and m([lambda]) = 0 if L has no eigenvalues [mu] of this kind.

Lemma 5. Consider the eigenvalue problem

-u" (x) + u(x) = [mu]u, 0 < x < 1, u' (0) = u' (1) = 0. (9)

Then (9) has the Green function

[mathematical expression not reproducible] (10)

and

[mathematical expression not reproducible] (11)

where for any R > 0, [B.sub.R] = {u [member of] C[0,1] : [parallel]u[parallel] [less than or equal to] R} denotes a ball of radius R, and [mathematical expression not reproducible].

Proof. It is easy to obtain (10) and the proof is omitted. The eigenvalue problem (9) is equivalent to the equation [mu]-[mu]Lu = 0. The equation u - [mu]Lu = 0 has nontrivial solutions if and only if [[mu].sup.-1] is the eigenvalue of the operator L. For [mu] > 1, the general solution of equation (9) is

u(x) = c cos ([square root of [mu] - 1x]) + d sin ([square root of [mu] - 1x]). (12)

By Neumann boundary value condition, we have

-c[square root of [mu]-1] sin ([square root of [mu]-1]) = 0. (13)

When c [not equal to] 0, we have [mu] = [n.sup.2][[pi].sup.2] + 1, where n [greater than or equal to] 0, n [member of] N. Therefore, [[lambda].sub.n] = [([n.sup.2][[pi].sup.2] + 1).sup.-1] is the eigenvalue of L, where N [greater than or equal to] 0, N(L - [[lambda].sub.n]) is the one-dimensional subspace of Banach space X spanned by [u.sub.n] = cos(n[pi]x).

For n [greater than or equal to] 0, Lemma 4 implies that

[mathematical expression not reproducible], (14)

This completes the proof.

Consider the eigenvalue problem

u" + a(x)u + [mu]u = 0, x [member of] (0,1), u' (0) = u' (1) = 0. (15)

We recall some propositions listed as follows.

Proposition 6. It is well known that the problem (15) has the eigenvalues [[mu].sub.1](a(x)) < [[mu].sub.2](a(x)) [less than or equal to] [[mu].sub.3](a(x)) [less than or equal to] .... The first eigenvalue [[mu].sub.1] (a(x)) is real and simple, and the corresponding eigenfunction does not change sign. When a(x) [equivalent to] 0, we denote [[mu].sub.i](0) by [[mu].sub.i]. It is obvious that [[mu].sub.1] = 0 is the first eigenvalue and [[mu].sub.2] = [[pi].sup.2] is the second eigenvalue of the eigenvalue equation

u" + [mu]u = 0, x [member of] (0,1), u' (0) = u' (1) = 0. (16)

Proposition 7. The comparison theorem of eigenvalues can be stated as follows: [[mu].sub.i](a(x)) is strictly decreasing in the sense that [a.sub.1](x) [much less than] [a.sub.2](x) implies that [[mu].sub.i]([a.sub.1](x)) > [[mu].sub.i]([a.sub.2](x)), where [a.sub.1](x) [much less than] [a.sub.2](x), namely, [a.sub.1](x) [less than or equal to] [a.sub.2](x), with the strict inequality on a set of positive measure.

Proposition 8. Suppose that [[mu].sub.k] [much less than] a(x) [much less than] [[mu].sub.k+1], k = 1, 2, .... Then the equation

u" + a(x)u = 0, x [member of] (0,1), u' (0) = u' (1) = 0 (17)

does not admit any nontrivial solutions.

Proof. Argue by contradiction, (17) admits nontrivial solutions. Let u(x) be a nontrivial solution of the equation (17). It follows from [[mu].sub.k] [much less than] a(x) [much less than] [[mu].sub.k+1] that [[mu].sub.k](a(x)) < [[mu].sub.k]([[mu].sub.k]) = 0 and [[mu].sub.k+1] (a(x)) > [[mu].sub.k+1]([[mu].sub.k+1]) = 0, for k = 1, 2, 3 .... It follows from (17) that 0 is the eigenvalue, contradiction. Therefore, (17) admits only trivial solution 0. Equation (17) does not admit any nontrivial solutions. This completes the proof.

In order to prove our main theorem using the topological degree theory, the following lemma is essential. In the following lemma, the weight function a(x) may change sign.

Lemma 9. Suppose that a(x) [much less than] [[pi].sup.2] satisfies the equation

[L.sub.a] u = u" + a(x)u = 0, x [member of] (0,1), u' (0) = u' (1) = 0. (18)

Then

[mathematical expression not reproducible] (19)

where [mathematical expression not reproducible] is defined in (10), for any R > 0, [B.sub.R] = {u [member of] C[0,1]: [parallel]u[parallel] [less than or equal to] R} denotes a ball of radius R.

Proof. Equation (18) is equivalent to the equation

[mathematical expression not reproducible]. (20)

In (11), let n = 0 and [mu] - 1 = [beta]; we have

[mathematical expression not reproducible], (21)

Next we calculate deg(I-A, [B.sub.R], 0). Let H(t, u) = u-(1-t)([beta]+ 1)Lu - tAu. Now we prove that H(t, u) [not equal to] 0 for [for all ]t [member of] [0,1] and x [member of] [partial derivative][B.sub.R]. Argue by contradiction and assume that H(t,u) = 0; that is, for [for all]t [member of] [0,1], the equation

u" + (1-t) [beta]u + ta (x)u = 0, x [member of] (0,1), u' (0) = u' (1) = 0 (22)

has solution u satisfying [parallel]u[parallel] = R on [partial derivative][B.sub.R].

Case 1. When a(x) does not change sign:

(i) For a(x) [much less than] 0, choose [beta] < 0; that is, for [for all]t [member of] [0,1], ((1 - t)[beta] + ta(x)) < 0, by Proposition 7, we have [[mu].sub.1] ((1 t)[beta] + ta(x)) > [[mu].sub.1](0) = 0. It follows that (22) has only trivial solution u [equivalent to] 0. Contradiction, therefore, we have H(t, u) [not equal to] 0 for t [member of] [0,1], x [member of] [partial derivative][B.sub.R]. Using the homotopy invariance we get

deg (I-A, [B.sub.R], 0) = deg (H (1,x), [B.sub.R], 0) = deg (H(0,x),[B.sub.R],0) = deg (l-([beta]+1)L, [B.sub.R], 0)=1. (23)

Therefore, for a(x) [much less than] 0, that is, [[mu].sub.1](a(x)) > 0,

deg (I-A, [B.sub.R], 0) = 1. (24)

(ii) For 0 [much less than] a(x) [much less than] [[pi].sup.2], choose [beta] [member of] (0,[[pi].sup.2]), that is, for [for all]t [member of] [0,1], 0 [much less than] (1-t)[beta] + ta(x) [much less than] [[pi].sup.2], we have [[mu].sub.1]((1 - t)[beta] + ta(x)) < [[mu].sub.1](0) = 0 and [[mu].sub.2]((1 - t)[beta] + ta(x)) > [[mu].sub.2] ([[pi].sup.)] = 0. By Proposition 8, it follows that (22) has only trivial solution u = 0, contradiction. Therefore, we have H(t, u) [not equal to] 0 for t [member of] [0, 1], x [member of] [partial derivative][B.sub.R]. Using the homotopy invariance, we get

deg (I-A, [B.sub.R], 0) = deg (H(1,x), [B.sub.R], 0) = deg (H(0,x), [B.sub.R], 0) = deg (l-([beta]+1)L, [B.sub.R], 0) = -1. (25)

Therefore, for [[mu].sub.1] (a(x)) < 0, a(x) [much less than] [[pi].sup.2], we have

deg (I-A, [B.sub.R], 0) = -1. (26)

Case 2. When a(x) changes sign, let a(x) = [a.sup.+](x) - [a.sup.-](x), where [a.sup.+](x) = max(a(x), 0}, [a.sup.-](x) = max{-a(x), 0}.

(i) For [[mu].sub.1](a(x)) > 0, consider the following equation:

u" + (t[a.sup.+] (x) - [a.sup.-] (x)) u = 0, x [member of] (0,1), u' (0) = u' (1) = 0. (27)

where t [member of] [0,1]. Obviously,

t[a.sup.+] (x) - [a.sup.-] (x) [less than or equal to] [a.sup.+] (x) - [a.sup.-] (x) = a(x), [for all]t [member of] [0,1]. (28)

By Proposition 7 and (28), we have

[[mu].sub.1] (t[a.sup.+] (x) - [a.sup.-] (x)) [greater than or equal to] [[mu].sub.1] (a (x)) > 0, [for all]t [member of] [0,1]. (29)

When t = 0, the solution of (27) is

[mathematical expression not reproducible]. (30)

Let [mathematical expression not reproducible], we have

deg (I-[L.sub.1], [B.sub.R], 0)= 1. (31)

The solution of (27) is

[mathematical expression not reproducible]. (32)

Let [h.sub.1](t,u) = u - tAu - (1 - t[).sub.L1]u. It is easy to prove that [h.sub.1](t,u) [not equal to] 0 for [for all]t [member of] [0,1], u [member of] [partial derivative][B.sub.R]. Therefore, using the homotopy invariance, we get

deg(I-A, [B.sub.R], 0) = deg([h.sub.1] (1r), [B.sub.R], 0) = deg ([h.sub.1] (0r), [B.sub.R], 0) = deg (I-[L.sub.1,] [B.sub.R], 0)=1. (33)

(ii) For [[mu].sub.1](a(x)) < 0, a(x) [much less than] [[pi].sup.2], by Proposition 7, [[mu].sub.2](a(x)) > 0; similarly, let

[mathematical expression not reproducible]. (34)

Since [a.sup.+](s) [much less than] 0, we have

deg (I-[L.sub.2], [B.sub.R], 0) = -1. (35)

Let [h.sub.2](t,u) = u - tAu - (1 - t)[L.sub.2]u. It is easy to prove that [h.sub.2](t, u) [not equal to] 0 for [for all]t [member of] [0, 1], u [member of] [partial derivative][B.sub.R]. Therefore, using the homotopy invariance we get

deg(I-A, [B.sub.R], 0) = deg([h.sub.2] (1, *), [B.sub.R], 0) = deg ([h.sub.2] (0, *), [B.sub.R], 0) = deg (I-[L.sub.2], [B.sub.R], 0) = -1. (36)

This completes the proof.

Definition 10 (see ). Let [OMEGA] be open set in Banach space E. Suppose A : [bar.[OMEGA]] [right arrow] E is completely continuous operator, f = I-A. Let [x.sub.0] [member of] [OMEGA] be an isolated fixed point of f in [OMEGA]; that is, there exists r > 0, such that [B.sub.r] = {x : [parallel]x - [x.sub.0][parallel] < r} [subset] [OMEGA], and [x.sub.0] is only fixed point of f in [[bar.B].sub.r]. Define

ind (I - A, [x.sub.0]) = deg (I - A, [B.sub.r], 0). (37)

Lemma 11 (see ). Let [OMEGA] be open set in Banach space E. Suppose A : [bar.[OMEGA]] [right arrow] E is completely continuous operator; A has no fixed point on [partial derivative][OMEGA]. Suppose that there are finite isolated fixed points [x.sub.1], [x.sub.2], ... , [x.sub.m]. Then

[mathematical expression not reproducible]. (38)

Lemma 12 (see ). Let [OMEGA] be open set in Banach space E. Suppose A : [bar.[OMEGA]] [right arrow] E is completely continuous operator, [x.sub.0] [member of] [OMEGA], A[x.sub.0] = [x.sub.0]. Suppose that A is Frechet differentiable at [x.sub.0] and 1 is not the eigenvalue of derived operator A'([x.sub.0]). Then [x.sub.0] is an isolated fixed point of A and

ind (I - A, [x.sub.0]) = ind (I - A' ([x.sub.0]) ,0) = [(-1).sup.[beta]], (39)

where [beta] is the sum of the algebraic multiplicities of all the eigenvalues A'([x.sub.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 .

Lemma 13. Suppose that a (x), [a.sub.1](x), and [a.sub.2](x) [member of] C [0,1] such that

a(x),[a.sub.1] (x), [a.sub.2] (x) [much less than] [[pi].sup.2]. (40)

Then

(1) the possible solution u of (18) is either u(x) [equivalent to] 0 or u(x) [not equal to] 0 for each x [member of] [0, 1];

(2) [L.sub.a], u = 0 (i = 1,2) cannot both admit nontrivial solutions if [a.sub.1](x) [much less than] [a.sub.2](x).

Lemma 14. Let f(x) > 0 on [0,1] and a(x) satisfy

a(x) [much less than] [[pi].sup.2]/4. (41)

If u(x) is a solution of the nonhomogeneous differential equation

u" + a(x)u = f (x), x [member of] (0,1), u' (0) = u' (1) = 0, (42)

then the following statements hold:

(1) either u(x) > 0 or u(x) < 0 for all x [member of] [0, 1];

(2) maximum principle: u(x) < 0, if [[mu].sub.1] (a(x)) > 0;

(3) uniform antimaximum principle: u(x) > 0, if [[mu].sub.1] (a(x)) < 0.

For the reader's convenience, we denote the continuous function g(x, u) = a(x)u - b(x)[u.sup.3] in (1). The assumptions on function g(x, u) are listed as follows:

(f1) g(x, u) is locally differentiable with respect to the second variable with [g.sub.u] (x, u) [much less than] [[pi].sup.2];

(f2) [mathematical expression not reproducible], uniformly in x [member of] [0, 1].

Lemma 15. Let X be an order Banach space and [w, v] = {u [member of] X | w [less than or equal to] u [less than or equal to] v} an order set. Assume that the given continuous function g(x, u) satisfies (f1). Then

(1) the solutions of(1) are totally ordered;

(2) (1) cannot admit three distinct solutions in [w, v] if [g.sub.u](x, u) is strictly increasing or strictly decreasing in [w, v].

For the convenience, we recast (1) in the operator form

F ([lambda], u) [??] u" + g (x, u) - [lambda]f (x), R x [C.sup.2] [0, 1] [right arrow] C [0,1]. (43)

Lemma 16 (see ). Let X and Z be Banach spaces. Assume that F : R x U [right arrow] Z is continuously differentiable on V x U [subset] R x X satisfying the following three conditions:

(1) F([[lambda].sub.*], [u.sub.*]) = 0 for some ([[lambda].sub.*],[u.sub.*]) [member of] V x U, dim N([D.sub.u]F([[lambda].sub.*], [u.sub.*])) = 1 and the null space N([D.sub.u]F([[lambda].sub.*],[u.sub.*])) = span{w};

(2) the Fredholm index of [D.sub.u]F([[lambda].sub.*],[u.sub.*]) is zero, codim R([D.sub.u]F([[lambda].sub.*],[u.sub.*])) = 1;

(3) [D.sub.[lambda]]F([[lambda].sub.*],[u.sub.*])) [??] R([D.sub.u]F([[lambda].sub.*],[u.sub.*])).

Then there is a continuously differentiable curve through ([[lambda].sub.*],[u.sub.*]); that is, there exists {([lambda](s),u(s)) = ([[lambda].sub.*] + [tau]T(s), [u.sub.*] + sw + z(s)) | s [member of] (-[delta], [delta]), ([lambda](0),u(0)) = ([[lambda].sub.*],[u.sub.*])} such that

F ([lambda](s), u(s)) = 0, for s [member of] (-[delta],[delta]), (44)

and all solutions of F([lambda],u) = 0 in a neighborhood of ([[lambda].sub.*],[u.sub.*]) belong to the curve, [tau](0) = [tau]'(0) = z(0) = z' (0) = 0.

3. Main Results

Lemma 17. Assume (f2) holds. Then for every fixed [lambda], (1) has at least a solution and

deg (I-A, [B.sub.R], 0) = 1, (45)

for R large enough, where [mathematical expression not reproducible] and G is defined in (10).

Proof. It follows from (f2) that there exists R >0 large enough such that g(x, R) - [lambda]f(x) < 0 and g(x, -R) - [lambda]f(x) > 0. Therefore, -R and R are subsolution and supersolution of (1), respectively. Therefore, there exists at least one solution of (1) between -R and R. Next, we show that all the solutions of (1) are between -R and R for R > 0 large enough. Argue by contradiction, suppose that there exists [x.sub.0] such that u(x) attains its maximum value and u([x.sub.0]) > R. We have

u" ([x.sub.0]) + a ([x.sub.0]) u ([x.sub.0]) - b ([x.sub.0]) [u.sup.3] ([x.sub.0]) - [lambda]([x.sub.0]) = 0. (46)

It follows from u' ([x.sub.0]) = 0 and u"([x.sub.0]) < 0 that

a ([x.sub.0]) u ([x.sub.0]) - b ([x.sub.0]) [u.sup.3] ([x.sub.0]) - [lambda]f ([x.sub.0]) [greater than or equal to] 0. (47)

From the above there exists R > 0 large enough such that g(x, R) - [lambda]f(x) < 0, contradiction. Similar, suppose that there exists [x.sub.1] such that u(x) attains its minimum value and u([x.sub.1]) < -R; we will obtain contradiction. Therefore, there exists R > 0 large enough such that [parallel]u[parallel] < R for all the solutions u(x). (1) is equivalent to the following equation:

u (x) = (Au) (x). (48)

Next we calculate deg(I - A, [B.sub.R], 0). Let h(t, u) = u- tAu, [for all]t [member of][0, 1]. The equation h(t, u) = 0 is equivalent to the following equation:

u" + tg (x, u)-(1-t)u- t[lambda]f (x) = 0. (49)

It is evident that -R and R are subsolution and supersolution of (49), respectively. Therefore, all the solutions of (49) must be between -R and R; that is, the equation h(t, u) = 0 has no solution on [partial derivative][B.sub.R]. Therefore, h(t, u) [not equal to] 0 for t [member of] [0, 1], u [member of] [partial derivative][B.sub.R]. By the homotopy invariance properties of the topological degree we have

deg (I-A, [B.sub.R], 0) = deg (h(*, 1), [B.sub.R], 0) = deg (h((*, 0), [B.sub.R], 0) = deg (I, [B.sub.R], 0) = 1. (50)

This completes the proof.

Lemma 18. Suppose that [[mu].sub.1](a(x)) < 0, a(x) [much less than] [[pi].sup.2], and f(x) > 0 for all x [member of] [0,1]. Then (1) has exactly three solutions [u.sub.0](x), 0, [v.sub.0](x) for [lambda] = 0, where [u.sub.0](x) is the unique Positive stable solution and [v.sub.0] (x) is the unique negative stable solution.

Proof. For the mapping F defined in (43), let [lambda] = 0, and we have

F(u) = u" + a(x)u-b(x)[u.sup.3]. (51)

Next we prove that any solution u(x) of F(u) = 0 is nondegenerate.

First, we have the Frechet derivative of (51)

[F.sub.u] (u) [w] = w" + (a (x) - 3b (x) [u.sup.2]) w. (52)

It is obvious that u = 0 is the solution of F(u) = 0, and therefore, when u = 0, a(x)-3b(x)[u.sup.2] = a(x) in (52). It follows from [[mu].sub.1](a(x)) < 0 that 0 is nondegenerate.

Next, we prove that any nontrivial solution u(x) of F(u) = 0 is nondegenerate. F(u) = 0 is equivalent to the following equation:

u (x) = (Au) (x), (53)

where [mathematical expression not reproducible] and G is defined in (10). The equation

w - A' (u)w = 0 (54)

is equivalent to [F.sub.u](u)[w] = 0, where A'(u) denotes the derivative operator of A. When [[mu].sub.1](a(x)) < 0, a(x) [much less than] [[pi].sup.2], by Lemma 9, Definition 10, and Lemma 12, there exists R >0, and we have

ind (I - A, 0) = ind (i - A' (0), 0) = deg (I-A, [B.sub.R], 0) = -1. (55)

If u(x) is a nontrivial solution of F(u) = 0, it follows from Lemma 13 that u(x) [not equal to] 0 for all x [member of] [0,1]. Therefore,

[mathematical expression not reproducible] (56)

Let [q.sub.1](x) = g(x,u)/u and [q.sub.2](x) = [q.sub.u](x, u). It follows from the above hypothesis that u(x) is a nontrivial solution of [mathematical expression not reproducible]. By the second conclusion of Lemma 13, we have that [mathematical expression not reproducible] does not have nontrivial solution, which implies that u(x) is a nondegenerate solution.

Finally, we prove the positive solution is unique and stable. We denote by [u.sub.0] the nontrivial solution of (51). It is obvious that -[u.sub.0](x) is also the nontrivial solution of (51). It follows from (56) and the comparison of eigenvalues that [[mu].sub.1]([q.sub.2](x)) > [[mu].sub.1] ([q.sub.1] (x)) = 0. By Lemma 9, Definition 10, and Lemma 12, we have

ind (I - A,u) = ind (i - A' (u) , 0) = deg (I-A , [B.sub.R], 0) = 1. (57)

Let m be the number of nontrivial solutions of F(u) = 0. Hence, F is proper. Since 0 is a regular value of F, m must be finite. According to Lemma 17, we have deg(I - A, [B.sub.R], 0) = 1 for sufficiently large R. By Lemmas 9 and 11 and the index formula, we have that

1 = deg (I-A, [B.sub.R], 0) = ind (I - A, 0) + [summation over (u [not equal to] 0)] ind (I - A, [u.sub.i]) = -1 + m. (58)

It is obvious that m = 2. Let [v.sub.0](x) = -[u.sub.0](x) for x [member of] [0,1]. Therefore, F(u) = 0 has exactly three solutions [u.sub.0],0, and [v.sub.0]. Since [[mu].sub.1]([q.sub.2]) > [[mu].sub.1]([q.sub.1]) = 0, we have that the positive solution [u.sub.0](x) is unique and stable. The negative solution [v.sub.0](x) is also stable. This completes the proof.

Lemma 19. Suppose that a(x) [much less than] [[pi].sup.2]/4. For some [lambda] < 0, (1) has a unique positive solution [u.sub.1](x) with [u.sub.1](x) > [u.sub.0](x), where [u.sub.0](x) is the unique positive solution of (1) for [lambda] = 0.

Proof. By Lemma 18, for [lambda] = 0 (1) has the unique positive solution [u.sub.0], which provided a subsolution of (1) for [lambda] < 0. There exists R > 0 large enough such that g(x, R) - [lambda]f(x) < 0. Thus, R is a supersolution of (1). Therefore, we prove that there exists a positive solution [u.sub.1](x) of (1) for [lambda] < 0 such that [u.sub.0](x) < [u.sub.1](x) < R.

Next we will prove that the positive solution [u.sub.1] is unique. Assume, by contradiction that (1) has another positive solution u(x) for [lambda] < 0. Let v(x) = u(x) - [u.sub.1](x) [not equal to] 0; then w satisfies the following equation:

v" + a(x) v -b (x) [[u.sup.2] (x) + u (x) [u.sub.1] (x) + [u.sup.2.sub.1] (x)] v = 0. (59)

Clearly, a(x) - b(x)[[u.sup.2](x) + u(x)[u.sub.1](x) + [u.sup.2.sub.1](x)] < a(x) [much less than] [[pi].sup.2]/4. By Lemma 13, v(x) [equivalent to] 0 is the solution of (59), which is a contradiction. Therefore, the solution [u.sub.1] obtained above is only positive solution. For [lambda] < 0, [u.sub.1] satisfies (1) and

[u".sub.1] + q (x) [u.sub.1] = [lambda]f (x), (60)

where q(x) = a(x) - b(x)[u.sup.2.sub.1][(x). Again since f(x) > 0 and [lambda] < 0, it follows from Lemma 14 that [[mu].sub.1](q(x)) > 0. Consider the linearization associated with (1)

w" + (a(x)-3b(x)[u.sup.2] (x))w = 0, (61)

where [g.sub.u](x, u) = a(x) - 3b(x)[u.sup.2](x) < q(x). Thus, [[mu].sub.1]([g.sub.u](x, u)) > [[mu].sub.1](q(x)) > 0. Therefore, by Lemma 13 we have [u.sub.1](x) is nondegenerate. This completes the proof.

Lemma 20. For [lambda] > 0, (1) has a unique negative solution [v.sub.1](x) with [v.sub.1](x) < [v.sub.0](x), where [v.sub.0](x) is the unique negative solution of (1) for [lambda] = 0.

Proof. The proof is similar to the proof of Lemma 19.

Theorem 21. Assume that the first eigenvalue [[mu].sub.1](a(x)) < 0 and f(x) > 0 for all x [member of] [0,1]. Suppose that a(x) [much less than] [[pi].sup.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 [[lambda].sub.*] > 0, such that

(i) For [lambda] > [[lambda].sub.*] (1) has no positive solution and has a unique negative solution which is stable.

(ii) For [lambda] = [+ or -][[lambda].sub.*], (1) has exactly two solutions. Moreover, when [lambda] = [[lambda].sub.*], the negative solution is stable and the positive solution is degenerate. When [lambda] = -[[lambda].sub.*] the positive solution is stable and the negative solution is degenerate.

(iii) For -[[lambda].sub.*] < [lambda] < [[lambda].sub.*], (1) has exactly three ordered solutions at the same [lambda] and the middle solution is unstable and the remaining two are stable. Moreover, when [lambda] < 0, the maximal solution is positive and the other two are negative. When [lambda] > 0, the minimal solution is negative and the other two are positive.

(iv) For [lambda] < - [[lambda].sub.*], (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 [u.sub.1] (x) for some [lambda] < 0. The solution curve can be continued a little bit such that [u.sub.1](x) remains positive for increasing [lambda] when [lambda] <0. By Lemma 18, (1) has a unique nondegenerate positive solution [u.sub.0](x) for [lambda] = 0. The positive solution curve can pass through [u.sub.1] (x) and [u.sub.0](x) and can be continued further for increasing [lambda] until the linearized equation (61) admits the nontrivial solutions. We claim that the curve of positive solutions cannot be continued for [lambda] > [[lambda].sub.*]. Next we will prove the existence of [[lambda].sub.*]. Since [[mu].sub.1](a(x)) < 0, let v(x) > 0 be the first eigenfunction of the problem

v" + a (x) v + [mu]v = 0, x [member of] (0,1), v' (0) = v' (1) = 0. (62)

Multiplying (1) by v(x) and subtracting from (62) multiplied by u, after that integrating over [0, 1], we obtain

[mathematical expression not reproducible] (63)

Applying the mean-value theorem for (63), there exists [xi] [member of] [0, 1] such that

[mathematical expression not reproducible] (64)

For all b([xi]) [less than or equal to] [b.sub.0] = [max.sub.x[member of][0,1]]b(x), the only root of h(u([xi]), b([xi])) = m is negative for m < h([square root of - ([[mu].sub.1]/3[b.sub.0])], [b.sub.0]) < 0. Let M = h([square root of (-[[mu].sub.1]/3[b.sub.0])], [b.sub.0]) and

[mathematical expression not reproducible] (65)

and it follows from (64) that u([xi]) < 0 for [lambda] > [[lambda].sub.*]. Therefore, there exists a [[lambda].sub.*] > 0 such that (1) has no positive solution for [lambda] > [[lambda].sub.*].

By applying Lemma 14, we obtain that [u.sub.*](x) is still a positive solution of (1) for [lambda] = [[lambda].sub.*]. We denote the degenerate solution ([[lambda].sub.*], [u.sub.*]). At ([[lambda].sub.*], [u.sub.*]), we verify that Lemma 16 can be applied here. It follows from (43) that [D.sub.u]F([[lambda].sub.*], [u.sub.*])[v] = v" + [a(x) - 3b(x)[u.sup.2]*]v = 0. In fact, 0 is simple and the principal eigenvalue of (5), and the first eigenfunction v(x) > 0; therefore, dimN([D.sub.u]F([[lambda].sub.*], [u.sub.*])) = codimR([D.sub.u]F([[lambda].sub.*],[u.sub.*])) = 1 and the null space N([D.sub.u]F([[lambda].sub.*],[u.sub.*])) = span{v}. Therefore, the Fredholm index of [D.sub.u]F([[lambda].sub.*], [u.sub.*]) is zero. Conditions (1) and (2) of Lemma 16 are satisfied. Next we verify Condition (3) of Lemma 16. Suppose on the contrary that [D.sub.[lambda]]F([[lambda].sub.*], [u.sub.*]) = -f(x) [member of] R([D.sub.u]F([[lambda].sub.*], [u.sub.*])); namely, there is a continuous function w satisfying

w" + [a (x) - 3b (x) [u.sup.2.sub.*]] w = -f (x). (66)

Consider the linearized equation of (1) at ([[lambda].sub.*], [u.sub.*])

v" + [a (x) - 3b (x) [u.sup.2.sub.*]] v = 0. (67)

Since F is singular at ([[lambda].sub.*], [u.sub.*]), that is, (67) has a nontrivial solution v(x) such that v(x) > 0. Multiplying (66) by v, subtracting from (67) multiplied by w and integrating by parts on [0, 1], we have

[mathematical expression not reproducible], (68)

a contradiction, since both v(x) and f(x) are positive. Therefore, Condition (3) of Lemma 16 is satisfied. Near ([[lambda].sub.*], [u.sub.*]), the solutions of

F ([lambda] (s), u(s)) = 0, for all s [member of] [-[delta], [delta]] (69)

form a curve

[mathematical expression not reproducible] (70)

Differentiating (69) twice in s, setting s = 0 and [u.sub.s][|.sub.s=0] = v(x), and evaluating at ([[lambda].sub.*], [u.sub.*]), we have

[mathematical expression not reproducible] (71)

Multiplying the linearized equation (67) by [u.sub.ss], subtracting from (71) multiplied by v, and integrating by parts over [0, 1], we obtain

[mathematical expression not reproducible]. (72)

Therefore, ([[lambda].sub.*], [u.sub.*]) is a fold point of F([lambda],u) to the left. It follows from the above that the curve of positive solutions cannot be continued to the right indefinitely for all [lambda] > 0. Hence, the positive solution curve will make a left turn at ([[lambda].sub.*], [u.sub.*]). Near the critical point ([[lambda].sub.*], [u.sub.*]), by the Crandall- Rabinowitz bifurcation theorem, there are two branches of positive solutions denoted by the upper branch [u.sub.+](x,[lambda]) and the lower branch [u.sub_](x, [lambda]) with [u.sub._](x, [lambda]) < [u.sub.+](x, [lambda]). It follows from Lemmas 14, 15, and 18 that the fold point ([[lambda].sub.*], [u.sub.*]) is unique, and the upper branch [u.sub.+](x, [lambda]) is monotone decreasing for all [lambda] < [[lambda].sub.*] and lower branch [u.sub._](x, [lambda]) is monotone increasing for all 0 < [lambda] < [[lambda].sub.*]. Therefore, the lower branch curve is monotone increasing and continues to the left without any turnings. Rewriting (1) in the following form:

u" + (a (x) - b (x) [u.sup.2]) u = [lambda]f (x), (73)

since a(x) - b(x)[u.sup.2] < a(x) [much less than] [[pi].sup.2]/4, it follows from Lemma 14 that u(x) > 0 or u(x) < 0. This shows that the solution changes its sign only at ([lambda], u) = (0,0). Since the nonlinearity g(x, u) = a(x)u - b(x)[u.sup.3] is an odd function in u, it follows from the symmetry that if ([lambda], u(x)) is a solution of (1), so is (-[lambda], -u(x)). Thus, the component of solutions we constructed above forms a smooth reversed S-shaped curve with exactly two turning points ([[lambda].sub.*], [u.sub.*]) and (- [[lambda].sub.*], -[u.sub.*]).

It follows from Lemma 18 that the positive solution [u.sub.0] is stable. For [lambda] < [[lambda].sub.*], the upper branch u+(x, [lambda]) remains stable until it reaches the degenerate solution ([[lambda].sub.*], [u.sub.*]). Next we prove the lower branch [u.sub._](x, [lambda]) is unstable. Let [u.sub.m] be any one solution of lower branch [u.sub._](x, [lambda]) such that 0 < [u.sub.m] < [u.sub.*]. Since a(x) - 3b(x)[u.sup.2.sub.m] > a(x) - 3b(x)[u.sup.2.sub.*, we Have [[mu].sub.1](a(x) - 3b(x)[u.sup.2.sub.m]) < [[mu].sub.1](a(x) - 3b(x)[u.sup.2.sub.*) = 0. Therefore, the lower branch [u.sub._] (x, [lambda]) is unstable. Since the solution set is symmetric with respect to the origin, the stability of negative solutions u(x) is obtained by using the property of symmetry. The upper branch of the negative solutions is unstable and the lower branch of negative solutions is stable. Therefore, all solutions of (1) lie on a unique reversed S-shaped solution curve. This completes the proof.

https://doi.org/10.1155/2018/5376075

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work was completed with the support of Tian Yuan Special Funds of the National Science Foundation of China (no. 11626182).

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Hui Xing (iD), (1) Hongbin Chen, (2) and Ruofei Yao (2)

(1) Department of Mathematics, Xi'an Polytechnic University, Xi'an 710048, China

(2) School of Mathematics and Statistics, Xi'an Jiaotong University, Xi'an 710049, China

Correspondence should be addressed to Hui Xing; xinghui210@163.com

Received 24 April 2018; Revised 21 June 2018; Accepted 4 July 2018; Published 1 August 2018