Open Access

Oscillating global continua of positive solutions of second order Neumann problem with a set-valued term

Boundary Value Problems20122012:47

DOI: 10.1186/1687-2770-2012-47

Received: 14 October 2011

Accepted: 23 April 2012

Published: 23 April 2012

Abstract

In this note, we study the oscillating global continua of the differential inclusion of the form

- u + q u λ F ( , u ) , u ( 0 ) = 0 , u ( 1 ) = 0 ,

where F is a "set-valued representation" of a function with jump discontinuities along the line segment [0, 1] × {0}, and λ [0, ∞) is a parameter. The proof of our main result relies on an approximation procedure.

Mathematics Subject Classification 2000: 34B16; 34B18.

Keywords

climate model differential inclusion eigenvalue positive solutions

1 Introduction

In recent years, nonsmooth analysis has come to play an important role in functional analysis [1], dynamical systems [2], control theory [3], optimization [4], mechanical systems [5], differential equation [6, 7] etc. Since many mathematical and physical problems may be reduced to ODES or PDES with discontinuous nonlinearities, the existence of multiple solutions for differential inclusion problems has been widely investigated [819].

In this article, we are concerned with the following differential inclusion problem which raises from a Budyko-North type energy balance climate models:
- u ( x ) + q ( x ) u ( x ) λ F ( x , u ( x ) ) , a .e . x ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0
(1.1)

(see [2025] and the references therein). In particular, the set-valued right hand side arise from a jump discontinuity of the albedo at the ice-edge in these models. By filling such a gap, one arrives at the set-valued problem (1.1). As in [25], we are here interested in a considerably simplify version as compared to the situation from climate modeling, e.g. a one-dimensional regular Sturm-Liouville differential operator substitutes for a two-dimensional Laplace-Beltrami operator or a singular Legendre-type operator, and the jump discontinuity is transformed to u = 0 in a way, which resembles only locally the climatological problem.

We are concerned with the set-valued problem (1.1) under the following assumptions

(H1) qC([0, 1],(0,+∞));

(H2) f+ C ([0, 1] × [0,+∞), (0,+∞)), inf x [ 0 , 1 ] f + ( x , 0 ) > 0 , lim s + f + ( x , s ) s = b ( x ) C ( [ 0 , 1 ] , ( 0 , ) ) .

Let the set-valued function F in (1.1) is given by
F ( x , y ) = { f + ( x , y ) } , x [ 0 , 1 ] , y > 0 , [ 0 , f + ( x , 0 ) ] , x [ 0 , 1 ] .
Notice that if f + (x, 0) ≡ 0, x [0, 1], then the differential inclusion problem (1.1) reduces to the BVP of differential equation
- u ( x ) + q ( x ) u ( x ) = λ f + ( x , u ( x ) ) , x ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0 .
(1.2)

In the last 20 years, the positive solutions of (1.2) have been studied by several authors, see Jiang and Liu [26], Chu et al. [27] and Sun et al. [28].

The purpose of this article is to investigate the oscillating global continua of positive solutions of the differential inclusion problem (1.1). The proof of our main result relies on an approximation procedure. The rest of the article is organized as follows. In Section 2, we state some notations and prove some preliminary results. In Section 3, we state and prove our main result. In Section 4, an example is given to illustrate the application of our main result.

2 Notations and preliminaries

Recall Kuratowski's notion of lower and upper limits of sequence of sets.

Definition 2.1. [29] Let X be a metric space and {Z l }l be a sequence of subsets of X. The set
lim sup l Z l : = x X : lim inf l dist ( x , Z l ) = 0
is called the upper limit of the sequence {Z l }, whereas
lim inf l Z l : = x X : lim l dist ( x , Z l ) = 0

is called the lower limit of the sequence {Z l }.

Definition 2.2. [29] A component of a set M is meant a maximal connected subset of M.

Lemma 2.1. [29] Suppose that Y is a compact metric space, A and B are non-intersecting closed subsets of Y, and no component of Y intersects both A and B. Then there exist two disjoint compact subsets Y A and Y B , such that Y = Y A Y B , A Y A , B Y B .

Using the above Whyburn Lemma, Ma and An [30] proved the following

Lemma 2.2. [30, Lemma 2.1] Let Z be a Banach space and let {A n } be a family of closed connected subsets of Z. Assume that

(i) there exist z n A n , n = 1, 2, ..., and z* Z, such that z n z*;

(ii) r n = sup {x | x A n } = ∞;

(iii) for every R > 0, n = 1 A n B R is a relatively compact set of Z, where B R = {x Z | xR}. Then there exists an unbounded component https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-47/MediaObjects/13661_2011_Article_157_IEq3_HTML.gif in lim sup l A l and z * C .

Remark 2.1. The limiting processes for sets go back at least to the work of Kuratowski [31]. Lemma 2.2 is a slight generalization of the following well-know result due to Whyburn [29]:

Proposition 2.1. (Whyburn [29, p. 12]) Let Z be a Banach space and {A n } be a family of closed connected subsets of Z. Let lim inf l A l and l A l is relatively compact. Then lim sup l A l is nonempty, compact and connected.

Next, we introduce the result of global solution behavior of the bifurcation branches of the equation
x = μ ( L x + N x ) , μ R , x X ,
(2.1)

to wit the following lemma.

Lemma 2.3. [32] (Dancer (1974)) Assume that

(C1) The operators L, N: XX are compact on the Banach space X over R. Furthermore, L is linear and Nx/x → 0 as x → 0;

(C2) The real number μ 0 is a characteristic number of L of odd algebraic multiplicity;

(C1+) The real Banach space X has an order cone K with X = K-K, i.e., every x X can be represented as x = x1 - x2, where x1, x2 K. Furthermore, L + N is positive, i.e., L + N maps K into K;

(C2+) The spectral radius r(L) of L is positive. We set μ0 = (r(L))-1.

Then (μ0, 0) is a bifurcation point of equation (2.1) and
S + : = { ( μ , x ) R × X : ( μ , x ) i s a s o l u t i o n o f ( 2 . 1 ) w i t h μ > 0 , x > 0 } ¯

contains an unbounded solution component C 1 + which passes through (μ0, 0).

If additionally

(C3+) The linear operator L is strongly positive, then ( μ , x ) C 1 + and μμ0 always implies x > 0 and μ > 0.

Remark 2.2. This result is often called the nonlinear Krein-Rutman theorem. It will play an important role in the proof of our main result.

Let φ and ψ be the unique solution of the problems
- u ( x ) + q ( x ) u ( x ) = 0 , x ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 1
and
- u ( x ) + q ( x ) u ( x ) = 0 , x ( 0 , 1 ) , u ( 1 ) = 0 , u ( 1 ) = 1
respectively. Then it is easy to check φ(·) is nondecreasing on (0,1), ψ(·) is nonincreasing on (0,1), and the Green's function G(x, s) of
- u ( x ) + q ( x ) u ( x ) = 0 , x ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0
is explicitly given by
G ( x , s ) = - 1 ψ ( 0 ) ψ ( x ) φ ( s ) , 0 s x 1 , φ ( x ) ψ ( s ) , 0 x s 1 .
(2.2)
Moreover, we have that
0 < G ( x , s ) G ( s , s ) , ( x , s ) [ 0 , 1 ] × [ 0 , 1 ] ; σ G ( s , s ) G ( x , s ) , ( x , s ) [ 0 , 1 ] × [ 0 , 1 ]
(2.3)

with σ : = min 1 ψ ( 0 ) , 1 φ ( 1 ) . .

3 The main result

Let Σ be the closure of the set of positive solutions of (1.1) in [0, ∞) × C1[0, 1], and * := {1, 2,..., N}. The main result of this article is the following theorem.

Theorem 3.1. Assume that (H1)-(H2) hold. If

(H3) there is an increasing sequence of positive numbers ξ j 1 N and a small enough constant δ such that ξ1 < σ(ξ2 - δ) and
Φ ( ξ 2 j - 1 ) < 1 2 0 1 G ( s , s ) d s - 1 ( ξ 2 j - 1 - δ ) , j * ; Ψ ( ξ 2 j ) > 2 0 1 G 1 2 , s d s - 1 ( ξ 2 j + δ ) , j * ,
where
Φ ( l ) : = max { f + ( t , c ) : 0 t 1 , 0 c l + δ } , Ψ ( l ) : = min { f + ( t , c ) : 0 t 1 , σ ( l - δ ) c l + δ } ,

then there exits an unbounded component C 1 + in Σ with ( 0 , 0 ) C 1 + . Moreover,

(i) ( λ , u ) C 1 + with u = ξ2j- 1 for some j * implies that λ ≥ 2;

(ii) ( λ , u ) C 1 + with u = ξ2j for some j * implies that λ 1 2 .

Actually, such continua C 1 + can be obtained as upper limits in the sense of Kura-towski of sequence of solution continua from associated continuous problems. To this end one sets
d g : = inf { f + ( x , 0 ) : x [ 0 , 1 ] } ,
(3.1)

fixes l0 such that d g l 0 < ξ 1 , and selects an approximation sequence {f l } C ([0, 1] × , ) (l > l0) of F satisfying:

(A1) f l (x, y) = ly for x [0, 1] and y 0 , d g 2 l ;

(A2) d g 2 f l ( x , y ) f + ( x , y ) for x [0, 1] and y d g 2 l , d g l ;

(A3) f l (x,y) = f + (x, y) for x [0, 1] and y d g l ;

(A4) {f l (x, y)}l is nondecreasing in l for (x, y) [0, 1] × (0,∞).

Next, we show that the continuous problem
- v ( x ) + q ( x ) v ( x ) = λ f l ( x , v ( x ) ) , x ( 0 , 1 ) , v ( 0 ) = 0 , v ( 1 ) = 0
(3.2l)
has an unbounded closed subsets C 1 , l + of the positive solutions set of (3.2 l ) with
  1. (a)

    λ 1 l , 0 is the bifurcation point contained in C 1 , l + ;

     
  2. (b)

    If ( μ , ϑ ) C 1 , l + and ϑ 0, then ϑ is positive on (0,1).

     
It is easy to see that (3.2 l ) equivalent to
v ( x ) = λ 0 1 G ( x , s ) f l ( s , v ( s ) ) d s .
(3.3)
Let
( L v ) ( x ) : = l 0 1 G ( x , s ) v ( s ) d s , ( N v ) ( x ) : = 0 1 G ( x , s ) ( f l ( s , v ( s ) ) - l v ( s ) ) d s , v C [ 0 , 1 ] .
Then according to (3.3), (3.2 l ) can be written as the following operator equation
v = λ ( L v + N v ) .

Clearly, the operators L, N : C[0, 1] → C[0, 1] are compact on the Banach space

C[0, 1]. Furthermore, L is linear and thanks to (2.3)(A1) that
N v v = 0 1 G ( x , s ) f l ( s , v ( s ) ) - l v ( s ) v d s 0 1 G ( s , s ) f l ( s , v ( s ) ) - l v ( s ) v d s 0 , as v 0 ,

which implies that the condition (C1) of Lemma 2.3 is satisfied.

Denote the principal eigenvalue of
- ω ( x ) + q ( x ) ω ( x ) = λ ω ( x ) , x ( 0 , 1 ) , ω ( 0 ) = 0 , ω ( 1 ) = 0 ,
(3.4)
by λ1, then we know that λ1 > 0 (see [33]). Since (3.4) is equivalent to operator equation
ω = λ l L ω ,

we have that ( r ( L ) ) - 1 = λ 1 l . Therefore, the conditions (C2)(C2+) of Lemma 2.3 are satisfied.

Let the cone K in C[0, 1] is given by
K = u C [ 0 , 1 ] u ( x ) 0 , 0 x 1 .

It is easy to see thanks to (A1)-(A4) and (2.3) that the (C1+)(C3+) conditions of Lemma 2.3 are satisfied.

According to Lemma 2.3, we obtain that λ 1 l , 0 is a bifurcation point of the positive solutions set of (3.2 l ) for every l {l0 + 1, l0 + 2, ...} =: 0, and for each l 0 there exits an unbounded closed subsets C 1 , l + of the positive solutions set of (3.2 l ) with (a) and (b).

Combining the above with the fact
lim l λ 1 l , 0 = ( 0 , 0 )
and utilizing Lemma 2.2, it concludes that there exits an unbounded component C 1 + with
( 0 , 0 ) C 1 +
(3.5)
and
C 1 + lim sup l C 1 , l + .
(3.6)
Denote the cone P in C[0, 1] by
P = u C [ 0 , 1 ] min 0 x 1 u ( x ) σ u .
Define an operator Tλ : PC[0, 1] by
T λ u ( x ) = λ 0 1 G ( x , s ) f l ( s , u ( s ) ) d s , x [ 0 , 1 ] .

It is easy to get the following lemma.

Lemma 3.1. Assume that (H1), (H2) and (A1)-(A4) hold. Then T λ : PP is completely continuous.

Lemma 3.2. Assume that (H1), (H2) and (A1)-(A4) hold. If 0 ≤ u(x) ≤ r, r > 0, for x [0, 1], then
T λ u λ M r 0 1 G ( s , s ) d s ,

where M r = max 0 x 1 , 0 s r { f l ( x , s ) } .

Proof. Since f l (x, u(x)) ≤ M r for x [0, 1], it follows from (2.3) that
T λ u = λ 0 1 G ( x , s ) f l ( s , u ( s ) ) d s λ 0 1 G ( s , s ) f l ( s , u ( s ) ) d s λ M r 0 1 G ( s , s ) d s .
Lemma 3.3. Assume that (H1), (H2) and (A1)-(A4) hold. If σ(r - δ) ≤ u(x) ≤ r + δ, r > δ, for x [0, 1], then
T λ u λ m r 0 1 G 1 2 , s d s ,

where m r = min 0 x 1 , σ ( r - δ ) s r + δ { f l ( x , s ) } .

Proof. Since f l (x, u(x)) ≥ m r for x [0, 1], it follows that
T λ u λ 0 1 G 1 2 , s f l ( s , u ( s ) ) d s λ m r 0 1 G 1 2 , s d s .

Lemma 3.4. Assume that (H1), (H2), (H3) and (A1)-(A4) hold. then

(i) ( λ , u ) C 1 , l + with u 2j-1 - δ2j-1 + δ) for some j * implies that λ > 2;

(ii) ( λ , u ) C 1 , l + , with u ( ξ2j - δ,ξ2j + δ) for some j * implies that λ < 1 2 .

Proof. (i) Assume that ( λ , u ) C 1 , l + with u 2j-1 - δ, ξ2j-1 + δ) for some j *, then u = Tλ u and
0 u ( x ) ξ 2 j - 1 + δ for x [ 0 , 1 ] .
By Lemma 3.2 and (H3), it follows that
u = T λ u λ M ξ 2 j - 1 + δ 0 1 G ( s , s ) d s = λ max 0 x 1 , 0 s ξ 2 j - 1 + δ { f l ( x , s ) } 0 1 G ( s , s ) d s λ Φ ( ξ 2 j - 1 ) 0 1 G ( s , s ) d s < λ 1 2 0 1 G ( s , s ) d s - 1 ( ξ 2 j - 1 - δ ) 0 1 G ( s , s ) d s . = 1 2 λ ( ξ 2 j - 1 - δ ) .
Thus λ > 2.
  1. (ii)
    Assume that ( λ , u ) C 1 , l + with u 2j - δ, ξ2j + δ) for some j *, then u = Tλu and
    σ ( ξ 2 j - δ ) u ( x ) ξ 2 j + δ for x [ 0 , 1 ] .
     
By Lemma 3.3 and the assumption (H3), it follows that
u = T λ u λ m ξ 2 j 0 1 G 1 2 , s d s = λ min 0 x 1 , σ ( ξ 2 j - δ ) s ξ 2 j + δ { f l ( x , s ) } 0 1 G 1 2 , s d s = λ Ψ ( ξ 2 j ) 0 1 G 1 2 , s d s > λ 2 0 1 G 1 2 , s d s - 1 ( ξ 2 j + δ ) 0 1 G 1 2 , s d s . = 2 λ ( ξ 2 j + δ ) .

Thus λ < 1 2 .

Lemma 3.5. If ( λ , u ) C 1 + , then (λ, u) is a solution of (1.1) and u W2,∞(0, 1).

Proof. Let ( λ , u ) C 1 + . By the definition of C 1 + there exists a sequence {l k } 0 strictly increasing, and ( λ l k , v l k ) [ 0 , ) × C 1 [ 0 , 1 ] with ( λ l k , v l k ) C 1 , l k + for k and
( λ l k , v l k ) ( λ , u ) .
Since { f l k ( , v l k ( ) ) } is uniformly bounded, i.e.
f l k L 2 M ,
(3.7)

we can assume after passing to a subsequence, if necessary, that it converges weekly in L2(0, 1) to some ϕ. We claim that ϕ(x) F(x, u(x)) a.e. on (0, 1).

Let x0 (0, 1) with u(x0) > 0. Then there exist ρ > 0 and τ (0, min{x0, 1-x0}) with u(x) > ρ for all x (x0 - τ, x0 + τ), hence there is a k0 with v l k ( x ) > ρ 2 for all k > k0 and x (x0 - τ, x0 + τ). Choose k1 > k0 with d g l k 1 < ρ 2 . Then f l k ( x , v l k ( x ) ) = f + ( x , v l k ( x ) ) for all kk1 and x (x0 - τ, x0 + τ), which yields ϕ(x) = f + (x, u(x)) for x (x0 - τ, x0 + τ) a.e.

Next, if u ≡ 0, let K: = {x (0, 1) : ϕ(x) > f+(x, 0)}. We claim that meas(K) = 0. Suppose that meas(K) > 0. Then ε := ∫ K [ϕ(x) - f+(x, 0)] dx > 0, and one finds η (0, ∞) with meas ( K ) f + ( x , y ) - f + ( x , 0 ) ε 2 for x [0, 1] and y [0, η]. Choosing k2 with v l k - u < η for kk2. One obtains for kk2:
K [ ϕ ( x ) - f l k ( x , v l k ( x ) ) ] d x = K [ ϕ ( x ) - f + ( x , 0 ) ] d x + K [ f + ( x , 0 ) - f l k ( x , v l k ( x ) ) ] d x = ε + K [ f + ( x , 0 ) - f + ( x , v l k ( x ) ) ] d x ε 2 ,

which contradicts f l k ( , v l k ( ) ) ϕ . Thus, meas(K) = 0.

Now, let A be the closed linear operator in L2(0, 1) defined by
dom ( A ) : = { φ W 2 , 2 [ 0 , 1 ] : φ ( 0 ) = 0 = φ ( 1 ) }
and := -φ" + . Clearly,
f l k ( x , v l k ( x ) ) ϕ ( x ) ,
(3.8)
hence v l k u and the fact that A is weakly closed yields
A u = λ ϕ ,
(3.9)
i.e.
A u λ F ( , u ( ) ) a .e .
Finally, we show that u W2,∞(0,1). In fact, from (3.9) we have
u ( x ) = q ( x ) u ( x ) - λ ϕ ( x ) .
(3.10)
According to (H1) and the boundedness of u we have
q u L ( 0 , 1 ) .
(3.11)
We claim that ϕ L(0,1). Suppose on the contrary that there exists a set E [0, 1], meas(E) > 0 such that |ϕ| is unbounded on E. Without loss of generality, we assume that
ϕ ( s ) > M w L 2 + 1 - [ 0 , 1 ] \ E ϕ ( x ) w ( x ) d x E w ( x ) d x , s E ,
(3.12)
where M is given by (3.7) and w L2(0,1). On the one hand, for k larger enough from (3.7), (3.8) and (H2) we have
0 1 ϕ ( x ) w ( x ) d x 0 1 f l k ( x , v l k ( x ) ) w ( x ) d x + 1 M w L 2 + 1 .
(3.13)
On the other hand, from (3.12) we have
0 1 ϕ ( x ) w ( x ) d x = E ϕ ( x ) w ( x ) d x + [ 0 , 1 ] \ E ϕ ( x ) w ( x ) d x > M v L 2 + 1 - [ 0 , 1 ] \ E ϕ ( x ) w ( x ) d x E w ( x ) d x E w ( x ) d x + [ 0 , 1 ] \ E ϕ ( x ) w ( x ) d x = M w L 2 + 1 ,
which contradicts (3.13). Thus,
ϕ L ( 0 , 1 ) .
(3.14)

Therefore, from (3.10), (3.11) and (3.14) we obtain u W2,∞(0,1).

Now we are in the position to prove Theorem 3.1.

Proof of Theorem 3.1.

Assume that ( λ , u ) C 1 + . We divide the proof into two cases.

Case l. If u = ξ2j- 1 for some j *, then λ ≥ 2.

Since ( λ , u ) C 1 + , there exists a sequence ( λ k i , z k i ) C 1 , k i + , such that
lim i λ k i = λ , lim i z k i = u .
Hence, for δ > 0 there exists i0 , such that
z k i - u < δ , i > i 0 ,
i.e.
ξ 2 j - 1 - δ < z k i < ξ 2 j - 1 + δ , i > i 0 .
By using Lemma 3.4, we obtain that
λ k i > 2 , i > i 0 .
Hence, we get
λ = lim i λ k i 2 .

Case 2. If u = ξ2j for some j *, then λ 1 2 .

Since ( λ , u ) C 1 + , there exists a sequence ( λ k i , z k i ) C 1 , k i + , such that
lim i λ k i = λ , lim i z k i = u .
Hence, for δ > 0 there exists i0 , such that
z k i - u < δ , i > i 0 ,
i.e.
ξ 2 j - δ < z k i < ξ 2 j + δ , i > i 0 .
By using lemma 3.4, we obtain that
λ k i < 1 2 , i > i 0 .
Hence, we get
λ = lim i λ k i 1 2 .

Corollary 3.1. Assume that (H1)-(H3) hold. Then

(i) for each λ ( 0 , 1 2 ) , (1.1) has at least one positive solution: u 0 C 1 + ;

(ii) for each λ [ 1 2 , 2 ] , (1.1) has N positive solutions:
u k , k = 1 , 2 , , N ,

which satisfy that u k C 1 + , k = 1 , 2 , , N .

Proof. According to Theorem 3.1, the boundary value problem (1.1) has an unbounded component C 1 + in Σ with ( 0 , 0 ) C 1 + . Moreover,

( ĩ ) ( λ , u ) C 1 + with u = ξ2j-1 for some j * implies that λ ≥ 2;

( i i ̃ ) ( λ , u ) C 1 + with u = ξ2j for some j * implies that λ 1 2 .

From the facts ( 0 , 0 ) C 1 + , and ( λ , u ) C 1 + with u = ξ1 implies that λ ≥ 2 and the connectivity of C 1 + , we obtain
C 1 + λ × C 1 [ 0 , 1 ] , λ 0 , 1 2 ,

which implies for each λ ( 0 , 1 2 ) , (1.1) has at least one positive solution: u 0 C 1 + .

Let
C 1 + , k : = { ( λ , u ) C 1 + | ξ k - 1 u < ξ k } , k = 1 , 2 , , N ,
where ξ0 = 0, ξ k (k = 1, 2,..., N) is given by (H3). Then according to ( i ̃ ) ( i i ̃ ) and the connectivity of C 1 + , we obtain
C 1 + , k λ × C 1 [ 0 , 1 ] , λ 1 2 , 2 , k = 1 , 2 , , N ,
which implies for each λ ( 1 2 , 2 ) , (1.1) has N positive solutions:
u k , k = 1 , 2 , , N ,

and u k C 1 + , k C 1 + , k = 1 , 2 , , N .

4 Example

In this section, an example is given to illustrate the application of our main result (Theorem 3.1). Consider second order Neumann differential inclusion problem
- u ( x ) + u ( x ) λ F ( u ( x ) ) , a .e . x ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0 ,
(4.1)
where the set-valued function F in (4.1) is given by
F ( y ) = 49 2 y - 195 2 , y 4 , 1 16 y + 1 4 , 0 < y < 4 , [ 0 , 1 4 ] , y = 0 .
Obviously, (H 1), (H 2) conditions of Theorem 3.1 are satisfied. Moreover, Green's function of the associated linear problem
- u + u = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 ,
can be explicitly expressed by
G ( x , s ) = 1 2 ( e - e - 1 ) ( e x - 1 + e 1 - x ) ( e s + e - s ) , 0 s x 1 , ( e x + e - x ) ( e s - 1 + e 1 - s ) , 0 x s 1 .

By calculation we can get 0 1 G ( s , s ) d s = e e - e - 1 , 0 1 G 1 2 , s d s = 1 and σ = 2 e + e - 1 .

Let ξ1 = 3, ξ2 = 11, δ = 1, then we can check that ξ1 = 3 < 5 < σ(ξ2 - δ), and
Φ ( ξ 1 ) = 1 2 < 3 4 < 1 - e - 2 = 1 2 0 1 G ( s , s ) d s - 1 ( ξ 1 - δ ) , Ψ ( ξ 2 ) 25 > 24 = 2 0 1 G 1 2 , s d s - 1 ( ξ 2 + δ ) .
So that (H 3) condition of Theorem 3.1 is satisfied. Therefore, according to Theorem 3.1 the differential inclusion problem (4.1) has an unbounded component C 1 + in Σ with ( 0 , 0 ) C 1 + . Moreover,
  1. (i)

    ( λ , u ) C 1 + with u = 3 implies that λ ≥ 2;

     
  2. (ii)

    ( λ , u ) C 1 + with u = 11 implies that λ 1 2 .

     

Declarations

Acknowledgements

The authors express their gratitude to Professors Ma Tian and Ma Ruyun for their guidance and encouragement, also to an anonymous referee for a number of valuable comments and suggestions.

Authors’ Affiliations

(1)
Department of Mathematics, Sichuan University

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