# Global Bifurcation for Second-Order Neumann Problem with a Set-Valued Term

Authors: Ruyun Ma; Jiemei Li
Journal: International Journal of Differential Equations
 (2009)

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## Abstract

We study the global bifurcation of the differential inclusion of the form−(ku′)′+g(⋅,u)∈μF(⋅,u),u′(0)=0=u′(1), where F is a “set-valued representation” of a function with jump discontinuities along the line segment [0,1]×{0}. The proof relies on a Sturm-Liouville version of Rabinowitz's bifurcation theorem and an approximation procedure.


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## Body

## 1. Introduction

We are concerned with the following differential inclusion which arises from a Budyko-North type energy balance climate models:(1.1)-(ku′)′(x)+g(x,u(x))∈μF(x,u(x)),x∈(0,1)a.e.u′(0)=0,u′(1)=0;
see [1–6] and the references therein. In particular, the set-valued right-hand side arises from a jump discontinuity of the albedo at the ice-edge in these models. By filling in such a gap, one arrives at the set-valued problem (1.1). As in [6], we are here interested in a considerably simplified version as compared to the situation from climate modeling; for example, 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.Assume that(H1)
k∈C1([0,1]), infk>0;(H2)
g∈C([0,1]×ℝ), g(x,·) strictly increasing for x∈[0,1],
(1.2)g1(x)≔lim|y|→0g(x,y)y
exists uniformly for x∈[0,1], and g1(x)>0 on [0,1],(H2′)
g satisfies that
(1.3)g2(x):=lim|y|→∞g(x,y)y
exists uniformly for x∈[0,1];(H3)
f+∈C([0,1]×ℝ+,(0,∞)), inff+>0, f-∈C([0,1]×ℝ-,(-∞,0)), supf-<0.LetF in (1.1) be given by
(1.4)F(x,y):={{f+(x,y)},x∈[0,1],y>0,[f-(x,0),f+(x,0)],x∈[0,1],{f-(x,y)}x∈[0,1],y<0,
and set(1.5)𝒮:={(μ,w)∈ℝ×C1([0,1])∣(μ,w)solves(1.1)}.
Throughout 𝒮 will be considered as subset of the Banach space Y:=ℝ×C1[0,1] under the norm(1.6)∥(μ,w)∥Y≔max{|μ|,∥w∥∞,∥w′∥∞}.
Let(1.7)ℤ+:={0,1,2,…}.Using a Sturm-Liouville version of Rabinowitz's bifurcation theorem and an approximation procedure, Hetzer [6] proved the following.Theorem A (see [6, Theorem]).
Let (H1)–(H3) be fulfilled. Then there exist sequences{Cn±}n∈ℤ+ of unbounded, closed, connected subsets of 𝒮 with (0,0)∈Cn± and the property that u has exactly n zeroes, which are all simple, if (μ,u)∈Cn±∖{(0,0)}. Moreover, u is positive (negative) on an interval (0,x̃) for some x̃∈(0,1], if (μ,u)∈Cn+ ((μ,u)∈Cn-) and u≢0.It is easy to see from Theorem A that the effect of the discontinuity at zero is a solution branch which consists of infinitely many subbranches all meeting in(0,0). Two subbranches are distinguished by the number of zeroes of the respective solutions. However, Theorem A provides no any information about the asymptotic behavior of Cn± at infinity.It is the purpose of this paper to study the asymptotic behavior ofCn± at infinity, and accordingly, to determine values of μ, for which there exist infinitely many nodal solutions of (1.1) (here and after, a function u∈AC1[0,1] is a nodal solution of (1.1) if all of zeroes of u are simple). To wit, we have the following.Theorem 1.1.
Let (H1)–(H3) and (H2′) be fulfilled. Assume that (H4)
(1.8)(f+)∞(x)=(f-)∞(x)=:b(x)∈C([0,1],(0,∞)),
where
(1.9)(f+)∞(x):=lims→+∞f+(x,s)s,(f-)∞(x):=lims→-∞f-(x,s)s.
Then for each n∈ℤ+, Cn+ joins (0,0) with (ηn,∞), Cn- joins (0,0) with (ηn,∞), where ηn, (n∈ℤ+), is the n-th eigenvalue of the linear problem:(1.10)-(ku′)′(x)+g2(x)u(x)=ηb(x)u(x),x∈[0,1],u′(0)=0,u′(1)=0.Corollary 1.2.
Let (H1)–(H4) and (H2′) be fulfilled. Let k∈ℕ be fixed. Then
(1) for eachμ∈[ηk-1,ηk), (1.1) has infinitely many solutions:
(1.11)ujν,ν∈{+,-},j∈{k,k+1…},
which satisfies that uj+ has exactly j simple zeroes and uj+ is positive on an interval (0,x̃) for some x̃∈(0,1], uj- has exactly j simple zeroes and uj- is negative on an interval (0,x̃) for some x̃∈(0,1);
(2) for eachμ∈(0,η0), (1.1) has infinitely many solutions:
(1.12)ujν,ν∈{+,-},j∈{0,1,2…}
which satisfies that uj+ has exactly j simple zeroes, and uj+ is positive on an interval (0,x̃) for some x̃∈(0,1], uj- has exactly j simple zeroes, and uj- is negative on an interval (0,x̃) for some x̃∈(0,1).


## 2. Notations and Preliminary Results

Recall Kuratowski's notion of lower and upper limits of sequences of sets.Definition 2.1 (see [7]).
LetX be a metric space and let {Zl}l∈ℕ be a sequence of subsets of X. The set
(2.1)lim supl→∞Zl:={x∈X:lim infl→∞dist(x,Zl)=0}
is called the upper limit of the sequence {Zl}, whereas
(2.2)lim infl→∞Zl:={x∈X:liml→∞dist(x,Zl)=0}
is called the lower limit of the sequence {Zl}.Definition 2.2 (see [7]).
Acomponent of a set M is meant a maximal connected subset of M.Lemma 2.3 (see [7]).
Suppose thatY 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 YA and YB, such that Y=YA∪YB, A⊂YA, B⊂YB.Using the above Whyburn Lemma, Ma and An [8] proved the following.Lemma 2.4 (see [8, Lemma 2.1]).
LetZ be a Banach space and let {An} be a family of closed connected subsets of Z. Assume that(i)
there existzn∈An, n=1,2,…, and z*∈Z, such that zn→z*;(ii)
rn=sup{∥x∥∣x∈An}=∞;(iii)
for everyR>0, (⋃n=1∞An)∩BR is a relatively compact set of Z, where
(2.3)BR={x∈Z∣∥x∥≤R}.
Then there exists an unbounded component𝒞 in limsupl→∞Al and z*∈𝒞.Remark 2.5.
The limiting processes for sets go back at least to the work of Kuratowski [9]. Lemma 2.4 will play an important role in the proof of Theorem 1.1. It is a slight generalization of the following well-known results due to Whyburn [7].Proposition 2.6 (Whyburn [7, page 12]).
LetZ be a Banach space and let {An} be a family of closed connected subsets of Z. Let liminfl→∞Al≠∅ and ⋃l∈ℕAl is relatively compact. Then lim supl→∞Al is nonempty, compact, and connected.Lemma 2.7.
Letq∈C([0,1],(0,∞)). Let pm∈C([0,1],(0,∞)) be such that
(2.4)pm(t)≥ρ,t∈[0,1]
for some ρ>0. Suppose that the sequence {(μm,ym)} satisfies
(2.5)-(kym′)′+q(t)ym=μmpm(t)ym,ym′(0)=ym′(1)=0
with either
(2.6)(ym|I)(t)>0∀msufficientlylarge
or
(2.7)(ym|I)(t)<0∀msufficientlylarge,
where I:=[α,β] with α<β being a given closed subinterval of (0,1). Then
(2.8)|μm|≤M0
for some positive constant M0.Proof.
We only deal with the case that(ym|I)(t)>0 for all m sufficiently large. The other case can be treated by the similar way. We may assume that (ym|I)(t)>0 for all m∈ℕ.
We divide the proof into three cases.
Case 1.
Let(αm,βm) be a subinterval of [0,1] satisfying(i)
I⊂(αm,βm);(ii)
ym(αm)=ym(βm)=0;(iii)
ym(t)>0 for all t∈(αm,βm).
Letψm(t) and φm(t) be the unique solution of the problems:
(2.9)-(ky′)′+q(t)y=0,t∈(αm,βm),y(αm)=0,y′(αm)=1,-(ky′)′+q(t)y=0,t∈(αm,βm),y(βm)=0,y′(βm)=-1,
respectively. Then it is easy to check ψm(·) is nondecreasing on (αm,βm), φm(·) is nonincreasing on (αm,βm), and that Green's function Gm(t,s) of
(2.10)-(ky′)′+q(t)y=0,t∈(αm,βm),y(αm)=y(βm)=0
is explicitly given by
(2.11)Gm(t,s)=1φm(αm){ψm(t)φm(s),αm≤t≤s≤βm,φm(t)ψm(s),αm≤s≤t≤βm.
Let Ψ(t) and Φ(t) be the unique solution of the problems:
(2.12)-(ky′)′+q(t)y=0,t∈(0,1),y(0)=0,y′(0)=1,-(ky′)′+q(t)y=0,t∈(0,1),y(1)=0,y′(1)=-1,
respectively. Then it is easy to check that Ψ(·) is nondecreasing on (0,1) and Φ(·) is nonincreasing on (0,1), and
(2.13)Φ(0)≥φm(αm),Ψ(1)≥ψm(βm).
Let ψI(t) and φI(t) be the unique solution of the problems
(2.14)-(ky′)′+q(t)y=0,t∈(α,β),y(α)=0,y′(α)=1,-(ky′)′+q(t)y=0,t∈(α,β),y(β)=0,y′(β)=-1,
respectively. Then, for (t,s)∈[α+(β-α)/4,β-(β-α)/4]×[α+(β-α)/4,β-(β-α)/4],(2.15)Gm(t,s)≥1Φ(0)ψI(α+β-α4)φI(β-β-α4).
Since(2.16)Gm(t,s)Gm(s,s)≥{ψm(t)ψm(s),αm≤t≤s≤βm,φm(t)φm(s),αm≤s≤t≤βm,≥{ψm(t)Ψ(1),αm≤t≤s≤βm,φm(t)Φ(0),αm≤s≤t≤βm,≥min{ψm(t)Ψ(1),φm(t)Φ(0)}=:δm(t),
it follows that for t∈[α+(β-α)/4,β-(β-α)/4],
(2.17)ym(t)=μm∫αmβmGm(t,s)pm(s)ym(s)ds≥δm(t)μm∫αmβmGm(s,s)pm(s)ym(s)ds≥δm(t)∥(ym|[αm,βm])∥∞≥δm(t)∥(ym|[α+(β-α)/4,β-(β-α)/4])∥∞≥δI(t)∥(ym|[α+(β-α)/4,β-(β-α)/4])∥∞,
where
(2.18)δI(t)≔min{ψI(t)Ψ(1),φI(t)Φ(0)}.
Set(2.19)δ0:=min{δI(t)∣t∈[α+β-α4,β-β-α4]}.
Then
(2.20)mint∈[α+(β-α)/4,β-(β-α)/4]ym(t)≥δ0∥(ym|[α+(β-α)/4,β-(β-α)/4])∥∞.
By (2.5), we have that
(2.21)ym(t)=μm∫αmβmGm(t,s)pm(s)ym(s)ds,
which together with (2.15) and (2.20) imply that for t∈[α+(β-α)/4,β-(β-α)/4],
(2.22)ym(t)≥μm∫IGm(t,s)ρym(s)ds≥μm∫α+(β-α)/4β-(β-α)/4Gm(t,s)ρym(s)ds≥δ0μm∫α+(β-α)/4β-(β-α)/4Gm(t,s)ρds·∥(ym|[α+(β-α)/4,β-(β-α)/4])∥∞≥δ0μmΦ(0)ψI(α+β-α4)φI(β-β-α4)ρ∫α+(β-α)/4β-(β-α)/4ds·∥(ym|[α+(β-α)/4,β-(β-α)/4])∥∞.
Therefore
(2.23)|μm|≤(δ0ρΦ(0)ψI(α+β-α4)φI(β-β-α4)·β-α2)-1.Case 2.
Let(0,βm) be a subinterval of [0,1] satisfying(i)
I⊂(0,βm);(ii)
ym′(0)=0,ym(βm)=0;(iii)
ym(t)>0 for all t∈(0,βm).
Letψ̅m(t) and φ̅m(t) be the unique solution of the problems:
(2.24)-(ky′)′+q(t)y=0,t∈(0,βm),y′(0)=0,y(βm)=1,-(ky′)′+q(t)y=0,t∈(0,βm),y(βm)=0,y′(βm)=-1,
respectively. Then it is easy to check that ψ̅m(·) is nondecreasing on (0,βm), φ̅m(·) is nonincreasing on (0,βm), and Green's function G*(t,s) of
(2.25)-(ky′)′+q(t)y=0,t∈(0,βm),y′(0)=y(βm)=0
is explicitly given by
(2.26)G*(t,s)=1φ̅m(0){ψ̅m(t)φ̅m(s),0≤t≤s≤βm,φ̅m(t)ψ̅m(s),0≤s≤t≤βm.
By the similar method to prove Case 1, we may get the desired results.Case 3.
Let(αm,1) be a subinterval of [0,1] satisfying(i)
I⊂(αm,1);(ii)
ym(αm)=0,ym′(1)=0;(iii)
ym(t)>0 for all t∈(αm,1).
Using the same method to prove Case2, with obvious changes, we may show that (2.8) is true.Case 4.
Let(αm,βm)=(0,1). We may assume that ym(t)>0 for all (0,1).
Letψ(t) and φ(t) be the unique solution of the problems
(2.27)-(ky′)′+q(t)y=0,t∈(0,1),y(0)=0,y′(0)=1,-(ky′)′+q(t)y=0,t∈(0,1),y(1)=0,y′(1)=-1,
respectively. Then, it is easy to verify that ψ is strictly increasing on [0,1] and φ is strictly decreasing on [0,1]. Using the same method to deal with Case 1, we may get the desired results.


## 3. Proof of the Results

Recall the proof of Theorem A.By [6, Remark 1], the hypotheses (H1)–(H3) imply that(3.1)𝒮∩((-∞,0]×C1([0,1]))=(-∞,0]×{0}.Actually, such continua can be obtained as upper limits in the sense of Kuratowski of sequences of solution continua from associated continuous problems. To this end one sets(3.2)df≔min{inff+,inf|f-|}
and selects an approximation sequence {fl}∈C([0,1]×ℝ,ℝ)ℕ of F satisfying(A1)
fl(x,y)=ly for x∈[0,1] and y∈[-df/2l,df/2l];(A2)
fl(x,y)×sgn(y)≥df/2 for x∈[0,1] and |y|≥df/2l; fl≤f+ on [0,1]×[df/2l,df/l]; fl≥f- on [0,1]×[-df/l,-df/2l];(A3)
fl(x,y)=f+(x,y) for x∈[0,1] and y≥df/l; fl(x,y)=f-(x,y) for x∈[0,1] and y≤-df/l;(A4)
{fl(x,y)}l∈ℕ is nondecreasing in l for (x,y)∈[0,1]×(0,∞); {fl(x,y)}l∈ℕ is nonincreasing in l for (x,y)∈[0,1]×(-∞,0).Remark 3.1.
Let(3.3)ξ(x,u):=g(x,u)-g1(x)u.
We may show that there exists a positive constant γ̅, independent of l, such that for each l∈ℕ,
(3.4)fl(x,u)u-ξ(x,u)γu≥ρ0,∀γ≥γ̅
for some constant ρ0>0.
In fact, it is easy to see from the definition offl that
(3.5)fl(x,u)u≥ρ1,u≠0
for some positive constant ρ1, independent of l.
Applying (H2) and (H2′), it concludes that
(3.6)0≤|ξ(x,u)u|≤ρ2
for some positive constant ρ2. Therefore, if we take
(3.7)γ̅:=2ρ2ρ1,ρ0=ρ12,
then (3.4) holds.It is easy to see thanks to (H2) and (A1) that(3.8l)-(kv′)′(x)+g(x,v(x))=μfl(x,v(x)),x∈[0,1],v′(0)=0,v′(1)=0
falls into the scope of the Sturm-Liouville version of the celebrated Rabinowitz bifurcation theorem (cf. [10] for a more general, but somewhat different setting).Indeed, denote the strictly increasing sequence of simple eigenvalues of(3.9)-(kψ′)′(x)+g1(x)ψ(x)=λψ(x),x∈[0,1],ψ′(0)=0,ψ′(1)=0,
by {λn}n∈ℤ+ and set(3.10)μn,l≔λnl.
Then (μn,l,0) is a bifurcation point of the solution set of (3.8l) for every n∈ℤ+, and for each (n,l)∈ℤ+×ℕ, there exist two unbounded closed connected subsets Cn,l± of the solution set of (3.8l) with the following.(a)
Cn,l+∩Cn,l-={(μn,l,0)}. Moreover, (μn,l,0) is the only bifurcation point contained in Cn,l±.(b)
If(μ,ϑ)∈Cn,l+ and ϑ≢0, then ϑ possesses exactly n simple zeroes (and no multiple zeroes) in (0,1) and is positive on (0,δ) for some δ>0.(c)
If(μ,ϑ)∈Cn,l- and ϑ≢0, then ϑ possesses exactly n simple zeroes (and no multiple zeroes) in (0,1) and is negative on (0,δ) for some δ>0.Combining the above with the fact(3.11)liml→∞(μn,l,0)=(0,0)
and utilizing Lemma 2.4, it concludes that there exists an unbounded component Cnν with(3.12)(0,0)∈Cnν,Cnν⊆limsupl→∞Cn,lν,ν∈{+,-}.As an immediate consequence of [6, Lemma 4-6], we have the followingLemma 3.2.
If(μ,u)∈Cn±, then (μ,u) is a solution of (1.1) and u∈W2,∞([0,1]). Moreover, if (μ,u)∈Cn+ with u≢0, u has exactly n simple zeroes in [0,1], and u is positive on an interval (0,x̃) for some x̃∈(0,1]; if (μ,u)∈Cn- with u≢0, u has exactly n simple zeroes in [0,1], and u is negative on an interval (0,x̃) for some x̃∈(0,1].Lemma 3.3.
Let (H1)–(H4), (H2′) and (A1)–(A4) be fulfilled. Then for each (n,l)∈ℤ+×ℕ, the connected component Cn,l± joins (μn,l,0) with (ηn,∞).Proof.
Assume that{(rk,yk)}⊂Cn,l+ for some fixed (n,l)∈ℤ+×ℕ with
(3.13)|rk|+∥yk∥C1→∞.
The case {(rk,yk)}⊂Cn,l- can be treated by the same way.
We divide the proof into two steps.
Step 1.
We show that if there exists a constant numberM>0 such that
(3.14)rk∈(0,M],
then Cn,l+ joins (μn,l,0) with (ηn,∞). In this case it follows that
(3.15)∥yk∥C1→∞.
Define
(3.16)ζl(r,x,u):=r[fl(x,u)-b(x)u]-[g(x,u)-g2(x)u].
Then {(rk,yk)} satisfies the problem:
(3.17l)-(kyk′)′(x)+g2(x)yk(x)=rkb(x)yk(x)+ζl(rk,x,yk(x)),x∈[0,1],yk′(0)=0,yk′(1)=0.
Set
(3.18)ζ̃l(u)=max0≤|s|≤u,x∈[0,1],r∈[0,M]|ζl(r,x,s)|
then ζ̃ is nondecreasing, and (H4) and (H2′) yields
(3.19)limu→∞ζ̃(u)u=0.
Now, we divide (3.17l) by ||yk||C1 and set y̅k=(yk/||yk||C1). Since y̅k is bounded in C2[0,1], after taking a subsequence if necessary, we have that y̅k→y̅ for some y̅∈C1[0,1] with ||y̅||C1=1. Moreover, from the definition of fl and (3.19) and the fact that ζ̃ is nondecreasing, we have that
(3.20)limk→∞|ζl(rk,x,yk(x))|∥yk∥C1=0
since
(3.21)|ζl(rk,x,yk(x))|∥yk∥C1≤ζ̃(|yk(x)|)∥yk∥C1≤ζ̃(∥yk∥∞)∥yk∥C1≤ζ̃(∥yk∥C1)∥yk∥C1.
By standard limit procedure, we get
(3.22)-(ky̅′)′(x)+g2(x)y̅(x)=r̅b(x)y̅(x),x∈[0,1],y̅′(0)=0,y̅′(1)=0,
where r̅:=limk→∞rk, again choosing a subsequence and relabeling if necessary. Moreover, the fact that yk, k∈ℤ+, has exactly n simple zeroes in [0,1] implies that y̅ has exactly n simple zeroes in [0,1], too. Therefore r̅=ηn.Step 2.
We show that there exists a constantM such that rk∈(0,M], for all n. Suppose there is no such M, choosing a subsequence and relabeling if necessary, it follows that
(3.23)limk→∞rk=∞.
Let(3.24)τ(1,k)<⋯<τ(n,k)
denote the zeroes of yk, and set
(3.25)0=τ(0,k),τ(n+1,k)=1.
Then, after taking a subsequence if necessary,
(3.26)limk→∞τ(l,k):=τ(l,∞),l∈{0,1,…,n+1}.
We claim that for all l∈{0,1,…,n}(3.27)τ(l+1,∞)-τ(l,∞)=0.
Suppose on the contrary that there existsl0∈{0,1,…,n} such that
(3.28)τ(l0,∞)<τ(l0+1,∞).
Define a function p:[0,1]→ℝ by
(3.29)pl(x):={fl(x,yk(x))yk(x)-ξ(x,yk(x))rkyk(x),x∈[0,1],yk(x)≠0,l,yk(x)=0.
Then by Remark 3.1, there exists ρ0, such that
(3.30)pl(x)≥ρ0,x∈[0,1].
Now we choose a closed interval I⊂(τ(l0,∞),τ(l0+1,∞)) with positive length, then we know from Lemma 2.7 that yk (after taking a subsequence if necessary) must change sign on I. However, this contradicts the fact that for all k sufficiently large, we have I⊂(τ(l0,k),τ(l0+1,k)) and
(3.31)(-1)l0νyk(x)>0,x∈(τ(l0,k),τ(l0+1,k)).
Therefore, (3.27) holds.
On the other hand, it follows(3.32)1=τ(n+1,k)-τ(0,k)=∑l=0n(τ(l+1,k)-τ(l,k))
that
(3.33)1=∑l=0n(τ(l+1,∞)-τ(l,∞))
which contradicts (3.27).
Therefore(3.34)|rk|≤M
for some constant number M>0, independent of k∈ℕ.Now we are in the position to prove Theorem1.1.Proof of Theorem1.1.
We only prove thatCn+ has the desired property, the case of Cn- can be treated by the same way.
Assume that{(μk,zk)}⊂Cn+ is a sequence with
(3.35)|μk|+∥zk∥C1→∞.
We claim that
(3.36)limk→∞(μk,zk)=(ηn,∞).
Assume on the contrary that (3.36) is not true. We divide the proof into two cases.
Case 1.
limk→∞μk≠ηn.In this case, we may take a subsequence of {μk}, denote it by {μk} again, with the property that there exists ε0>0, such that for each k∈ℕ,
(3.37)|μk-ηn|≥ϵ0.
Since{(μk,zk)}⊂Cn+, it follows that for each k∈ℤ+, there exists a sequence {(γkj,zkj)}⊂Cn,kj+, such that
(3.38)limj→∞γkj=μk,limj→∞zkj=zk.
Now let us consider the sequence{(γkk,zkk)}. Obviously, we have that
(3.39)(γkk,zkk)∈Cn,kk+,|γkk|+∥zkk∥C1→∞.
Equation (3.39) implies that
(3.40)-(kzkk′)′(x)+g2(x)zkk(x)=γkkb(x)zkk(x)+ζkk(γkk,x,zkk(x)),x∈[0,1],zkk′(0)=0,zkk′(1)=0,
Noticing that ρ0 in (3.30) is independent of l and using Remark 3.1 and the method to prove Lemma 3.3 and with obvious changes, we may show that {γkk} is bounded, and subsequently
(3.41)limk→∞γkk=ηn.
However, this contradicts (3.37).
Case 2.
limk→∞||zk||C1≠∞. In this case, after taking a subsequence of {zk} and relabeling if necessary, we may assume that
(3.42)∥zk∥C1≤M0
for some constant M0>0. Equation (3.35) together with (3.42) implies
(3.43)limk→∞μk=+∞.
Using the same notations as those in Case 1, we have from (3.43) that
(3.44)limk→∞γkk=+∞.
Combining this with (3.40) and using Remark 3.1 and the similar method to prove Step 2 of Lemma 3.3 and noticing that ρ0 in (3.30) is independent of l, it concludes that {γkk} is bounded. This is a contradiction.Remark 3.4.
It is easy to see from Theorem1.1 and its proof that the “jumping” of F at u=0: f+(x,0)-f-(x,0)(=:Δ(x)) does not affect the asymptotic behavior of Cn± at infinity. In other words, for any nonnegative function Δ(x), the asymptotic behavior of Cn± at infinity is the same.



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*Source: 373851-2009-11-24.xml*