Abstract
In this note, we study the oscillating global continua of the differential inclusion of the form
where F is a "setvalued 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 solutions1 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 BudykoNorth type energy balance climate models:
(see [2025] and the references therein). In particular, the setvalued right hand side arise from a jump discontinuity of the albedo at the iceedge in these models. By filling such a gap, one arrives at the setvalued 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 onedimensional regular SturmLiouville differential operator substitutes for a twodimensional LaplaceBeltrami operator or a singular Legendretype 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 setvalued problem (1.1) under the following assumptions
(H1) q∈C([0, 1],(0,+∞));
(H2) f^{+}∈ C ([0, 1] × [0,+∞), (0,+∞)),
Let the setvalued function F in (1.1) is given by
Notice that if f^{+}(x, 0) ≡ 0, x ∈ [0, 1], then the differential inclusion problem (1.1) reduces to the BVP of differential equation
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∈ℕ}be a sequence of subsets of X. The set
is called the upper limit of the sequence {Z_{l}}, whereas
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 nonintersecting 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,
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 wellknow 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
Next, we introduce the result of global solution behavior of the bifurcation branches of the equation
to wit the following lemma.
Lemma 2.3. [32] (Dancer (1974)) Assume that
(C1) The operators L, N: X → X 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 = KK, i.e., every x ∈ X can be represented as x = x_{1}  x_{2}, where x_{1}, x_{2} ∈ 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 μ_{0} = (r(L))^{1}.
Then (μ_{0}, 0) is a bifurcation point of equation (2.1) and
contains an unbounded solution component
If additionally
(C3+) The linear operator L is strongly positive, then
Remark 2.2. This result is often called the nonlinear KreinRutman 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
and
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
is explicitly given by
Moreover, we have that
with
3 The main result
Let Σ be the closure of the set of positive solutions of (1.1) in [0, ∞) × C^{1}[0, 1], and ℕ* := {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
where
then there exits an unbounded component
(i)
(ii)
Actually, such continua
fixes l_{0} ∈ ℕ such that
(A1) f_{l} (x, y) = ly for x ∈ [0, 1] and
(A2)
(A3) f_{l}(x,y) = f^{+}(x, y) for x ∈ [0, 1] and
(A4) {f_{l} (x, y)}_{l∈ℕ} is nondecreasing in l for (x, y) ∈ [0, 1] × (0,∞).
Next, we show that the continuous problem
has an unbounded closed subsets
(a)
(b) If
It is easy to see that (3.2_{l}) equivalent to
Let
Then according to (3.3), (3.2_{l}) can be written as the following operator equation
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
which implies that the condition (C1) of Lemma 2.3 is satisfied.
Denote the principal eigenvalue of
by λ_{1}, then we know that λ_{1}> 0 (see [33]). Since (3.4) is equivalent to operator equation
we have that
Let the cone K in C[0, 1] is given by
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
Combining the above with the fact
and utilizing Lemma 2.2, it concludes that there exits an unbounded component
and
Denote the cone P in C[0, 1] by
Define an operator T_{λ} : P → C[0, 1] by
It is easy to get the following lemma.
Lemma 3.1. Assume that (H1), (H2) and (A1)(A4) hold. Then T_{λ}: P → P 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
where
Proof. Since f_{l}(x, u(x)) ≤ M_{r} for x ∈ [0, 1], it follows from (2.3) that
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
where
Proof. Since f_{l}(x, u(x)) ≥ m_{r} for x ∈ [0, 1], it follows that
Lemma 3.4. Assume that (H1), (H2), (H3) and (A1)(A4) hold. then
(i)
(ii)
Proof. (i) Assume that
By Lemma 3.2 and (H3), it follows that
Thus λ > 2.
(ii) Assume that
By Lemma 3.3 and the assumption (H3), it follows that
Thus
Lemma 3.5. If
Proof. Let
Since
we can assume after passing to a subsequence, if necessary, that it converges weekly in L^{2}(0, 1) to some ϕ. We claim that ϕ(x) ∈ F(x, u(x)) a.e. on (0, 1).
Let x_{0} ∈ (0, 1) with u(x_{0}) > 0. Then there exist ρ > 0 and τ ∈ (0, min{x_{0}, 1x_{0}}) with u(x) > ρ for all x ∈ (x_{0}  τ, x_{0} + τ), hence there is a k_{0} ∈ ℕ with
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
which contradicts
Now, let A be the closed linear operator in L^{2}(0, 1) defined by
and Aφ := φ" + qφ. Clearly,
hence
i.e.
Finally, we show that u ∈ W^{2,∞}(0,1). In fact, from (3.9) we have
According to (H1) and the boundedness of u we have
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
where M is given by (3.7) and w ∈ L^{2}(0,1). On the one hand, for k larger enough from (3.7), (3.8) and (H2) we have
On the other hand, from (3.12) we have
which contradicts (3.13). Thus,
Therefore, from (3.10), (3.11) and (3.14) we obtain u ∈ W^{2,∞}(0,1).
Now we are in the position to prove Theorem 3.1.
Proof of Theorem 3.1.
Assume that
Case l. If ∥u∥_{∞} = ξ_{2j1} for some j ∈ ℕ*, then λ ≥ 2.
Since
Hence, for δ > 0 there exists i_{0} ∈ ℕ, such that
i.e.
By using Lemma 3.4, we obtain that
Hence, we get
Case 2. If ∥u∥_{∞} = ξ_{2j} for some j ∈ ℕ*, then
Since
Hence, for δ > 0 there exists i_{0} ∈ ℕ, such that
i.e.
By using lemma 3.4, we obtain that
Hence, we get
Corollary 3.1. Assume that (H1)(H3) hold. Then
(i) for each
(ii) for each
which satisfy that
Proof. According to Theorem 3.1, the boundary value problem (1.1) has an unbounded component
From the facts
which implies for each
Let
where ξ_{0} = 0, ξ_{k} (k = 1, 2,..., N) is given by (H3). Then according to
which implies for each
and
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
where the setvalued function F in (4.1) is given by
Obviously, (H1), (H2) conditions of Theorem 3.1 are satisfied. Moreover, Green's function of the associated linear problem
can be explicitly expressed by
By calculation we can get
Let ξ_{1} = 3, ξ_{2} = 11, δ = 1, then we can check that ξ_{1} = 3 < 5 < σ(ξ_{2} δ), and
So that (H3) condition of Theorem 3.1 is satisfied. Therefore, according to Theorem 3.1 the
differential inclusion problem (4.1) has an unbounded component
(i)
(ii)
Competing interests
The authors declare that they have no competing interests.
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.
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