# Results for Twin Singular Nonlinear Problems with Damping Term

Authors: Youwei Zhang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

By using the functional type cone expansion and compression fixed-point theorem in cones, some new and general results on the existence of positive solution for twin singular boundary value problems with damping term are obtained. An example is given to illustrate our results.


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## Body

## 1. Introduction

The study of multipoint boundary value problem for linear second-order ordinary differential equation was initiated by I1'in and Moiseev [1] motivated by the work of Bitsadze and Samarskiĭ [2–4] on nonlocal linear elliptic boundary value problem, which is a new area of still fairly theoretical exploration in mathematics. Several authors have expounded on various aspects of this theory; see the survey paper by Gupta et al. [5–7] and the references cited therein. Thereamong, the study of singular boundary value problem for ordinary differential equation has led to several important applications in applied mathematics and physical science, such as the Thomas-Fermi problem
(1.1)x′′(t)-t-1/2x3/2=0,0<t<1,x(0)=0=x(1),
which appears in determining the electrical potential in an atom. For other application results, we refer to [8–10]. With regards to this, increasing attention is paid to question of singular boundary value problem and has obtained many excellent results of the existence of the positive solution for two multiple points nonlinear singular boundary value problem [11–15]. The main techniques are the upper and lower solutions method [16], the Leray-Schauder continuation theory [17], and the fixed-point theory in cones [18] et al. We would like to mention some results of Ma and O'Regan [13] and Yao [15], which motivated us to consider the singular boundary value problems. In [13], the authors have studied a multipoint boundary value problem
(1.2)x′′(t)=f(t,x(t),x′(t))+e(t),0<t<1,x′(0)=0,x(1)=∑i=1m-2aix(ξi),
where ξi∈]0,1[, ai∈ℝ, i=1,2,…,m-2, 0<ξ1<ξ2<⋯<ξm-2<1, f:[0,1]×ℝ2→ℝ is a function satisfying Carathéodory's conditions and (1-t)e(t)∈L1]0,1[; the Leray-Schauder continuation theorem leads to the existence of single C1[0,1[ solution. The literature [15] has discussed a second-order boundary value problem
(1.3)x′′(t)+h(t)f(x(t))=0,0<t<1,ax(0)-bx′(0)=0,ax(1)+bx′(1)=0,
where h(t) is symmetric on ]0,1[ and may be singular at both end points t=0 and t=1. The author has proved the existence of n symmetric positive solutions and established a corresponding iterative scheme, the main tool being the monotone iterative technique.A powerful tool for proving existence of solution to boundary value problem is the fixed-point theory. In many cases, it is possible to find single, double, or multiple solutions for boundary value problem, and for the same problem, using the various methods, one can obtain different results under some appropriate conditions. To my best knowledge, very little work has been done on the existence of positive solution for boundary value problem by using the functional type cone expansion and compression fixed-point theorem. The aim of this paper is to establish some new and general results on the existence of positive solution to singular boundary value problems with damping term(1.4)u′′(t)-λu′(t)+h(t)f(t,u(t))=0,a<t<b,(1.5)u′(a)-λu(a)=0,γu(b)+δu′(b)=∑i=1naiu(ti),(1.6)γu(a)+δu′(a)=∑i=1naiu(ti),u′(b)-λu(b)=0,
where λ, γ, δ, ai, ti, (i=1,2,…,n), h, and f satisfy(H1)
-∞<a<t1<⋯<tn<b<∞, λ,γ,δ∈]0,∞[, ai∈]0,∞[, (i=1,2,…,n),(H2)
d1=(γ+λδ)exp(λb)-∑i=1naiexp(λti)>0, d2=∑i=1naiexp(λti)-(γ+λδ)exp(λa)>0,(H3)
h:]a,b[→[0,∞[ is continuous function, h(t) may be singular at t=a and/or t=b, and 0<∫abh(s)ds<∞,(H4)
f:[a,b]×[0,∞[→[0,∞[ satisfies the Carathéodory condition, that is, for each x∈[0,∞[, the mapping t→f(t,x) is Lebesgue measurable on [a,b] and for a.e. t∈[a,b], the mapping x→f(t,x) is continuous on [0,∞[, and for each r>0, there exists ϕr∈L1[a,b] such that f(t,x)≤ϕr(t) for all u∈[0,r] and for a.e. t∈[a,b].We will impose some advisable conditions on the nonlinearityf to ensure the existence of at least one positive solution for the above problems. In order to obtain our results, we construct special operator which is the base for further discussion and provide two crucial functionals on cones. Applying the functional type cone expansion and compression fixed-point theorem to the operator and functionals, we obtain some new and general results on the existence of at least one positive solution for the twin singular problems (1.4), (1.5) and (1.4), (1.6). Our results improve and generalize those in [15, 19].Letα and β be nonnegative continuous functionals on a cone 𝒫 in real Banach space ℬ. For positive numbers r and L, we define the sets
(1.7)P(α,r)={x∈P:r<α(x)},P(β,L)={x∈P:β(x)<L},P(β,α,r,L)={x∈P:r<α(x),β(x)<L}.
We state the functional type cone expansion and compression fixed-point theorem [20].Lemma 1.1.
Let𝒫 be a cone in a real Banach space ℬ, and let α and β be nonnegative continuous functionals on 𝒫. Let 𝒫(β,α,r,L) be a nonempty bounded subset of 𝒫,
(1.8)T:P(β,α,r,L)¯⟶P
is a completely continuous operator with
(1.9)inf⁡x∈∂P(β,α,r,L)‖Tx‖>0,P(α,r)¯⊆P(β,L),
for 𝒫(β,L). If α(𝒯x)≥r for all x∈∂𝒫(α,r), β(𝒯x)≤L for each x∈∂𝒫(β,L), and for each y∈∂𝒫(α,r), z∈∂𝒫(β,L), θ∈]0,1], and μ∈[1,∞[, the functionals satisfy the properties
(1.10)α(θy)≤θα(y),β(μz)≥μβ(z),β(0)=0,
then 𝒯 has at least one positive fixed-point x such that
(1.11)r≤α(x),β(x)≤L.


## 2. Main Results

Letℬ be the Banach space C1[a,b] with the norm ∥u∥=max{∥u∥0,∥u′∥1}, where ∥u∥0=supt∈[a,b]|u(t)| and ∥u′∥1=supt∈]a,b[|u′(t)|, and let a cone in ℬ(2.1)P1={v∈B:visnondecreasingon⁡[a,b],v(a)=0}.
For v∈𝒫1, define the operator 𝒯1 by
(2.2)T1v(t)=∫ath(s)f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds,t∈[a,b],
where c1(v)=(1/d1)(δv(b)+∑i=1naiexp(λti)∫tibv(ω)exp(-λω)dω).Lemma 2.1.
Ifv∈𝒫1 is a fixed-point 𝒯1, then
(2.3)u(t):=exp⁡(λt)(c1(v)+∫tbv(ω)exp⁡(-λω)dω)
is one solution of the problem (1.4), (1.5).Proof.
Suppose thatv∈𝒫1 is a fixed-point 𝒯1 and u(t)=exp(λt)(c1(v)+∫tbv(ω)exp(-λω)dω), thus we have
(2.4)u′(t)-λu(t)=-v(t)=-T1v(t)=-∫ath(s)f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds.
Further,
(2.5)u′′(t)-λu′(t)=-h(t)f(t,exp⁡(λt)(c1(v)+∫ttnv(ω)exp⁡(-λω)dω))=-h(t)f(t,u(t)).
The boundary condition (1.5) is satisfied due to the relation between u, v, and c1(v).Forv∈𝒫1, we define the nonnegative continuous functionals α and β on 𝒫1 by
(2.6)α(v)=∫t1bv(ω)exp⁡(-λω)dω,β(v)=v(t1).Lemma 2.2.
Letr1>0. If v∈∂𝒫1(α,r1), then
(2.7)v(t1)≤exp⁡(λb)b-t1r1,∫abv(ω)exp⁡(-λω)dω≥r1.Proof.
Forv∈∂𝒫1(α,r1), that is, α(v)=r1. Since v(t) nondecreasing on [a,t1], we have
(2.8)r1=α(v)=∫t1bv(ω)exp⁡(-λω)dω≤∫t1bv(b)exp⁡(-λω)dω≤v(b)exp⁡(-λt1)(b-t1),
so we get (2.7). Furthermore,
(2.9)∫abv(ω)exp⁡(-λω)dω=∫at1v(ω)exp⁡(-λω)dω+∫t1bv(ω)exp⁡(-λω)dω≥r1.Lemma 2.3.
LetL1>0. If v∈∂𝒫1(β,L1), then
(2.10)∫t1bv(ω)exp⁡(-λω)dω≥L1exp⁡(-λb)(b-t1).Proof.
Forv∈∂𝒫1(β,L1), that is to say, β(v)=L1. In view of v(t) is nondecreasing on [a,b], for ω∈[t1,b], we have
(2.11)v(ω)≥v(t1)=β(v)=L1.
Hence,
(2.12)∫t1bv(ω)exp⁡(-λω)dω≥L1∫t1bexp⁡(-λω)dω≥L1exp⁡(-λb)(b-t1).Lemma 2.4.
Let (H1)–(H4) hold, then 𝒯1:𝒫1→𝒫1 is completely continuous.Proof.
By (H1)–(H4), we observe that 𝒯1v∈C1[a,b], (𝒯1v)(t) is nondecreasing on [a,b], and (𝒯1v)(a)=0, so 𝒯1:𝒫1→𝒫1. Since h(t) may be singular at t=a and/or t=b, we take the arguments to show that 𝒯1 is completely continuous.
Assume thatvn,v0∈𝒫1. In view of f satisfying the Carathéodory condition, it is easy to see that
(2.13)‖vn-v0‖0⟶0impliesthatsup⁡s∈Ω|f(s,exp⁡(λs)(c1(vn)+∫sbvn(ω)exp⁡(-λω)dω))-f(s,exp⁡(λs)(c1(v0)+∫sbv0(ω)exp⁡(-λω)dω))(s,exp⁡(λs)(c1(vn)+∫sbvn(ω)exp⁡(-λω)dω))|ds⟶0
as n→∞, where Ω=[a,b] or ]a,b[. Thus, we have
(2.14)‖T1vn-T1v0‖0=sup⁡t∈[a,b]|(T1vn)(t)-(T1v0)(t)|≤∫abh(s)|f(s,exp⁡(λs)(c1(vn)+∫sbvn(ω)exp⁡(-λω)dω))-f(s,exp⁡(λs)(c1(v0)+∫sbv0(ω)exp⁡(-λω)dω))|ds,‖(T1vn)′-(T1v0)′‖1=sup⁡t∈]a,b[|(T1vn)′(t)-(T1v0)′(t)|≤sup⁡t∈]a,b[(h(t)|f(t,exp⁡(λt)(c1(vn)+∫sbvn(ω)exp⁡(-λω)dω))-f(t,exp⁡(λt)(c1(v0)+∫sbv0(ω)exp⁡(-λω)dω))|).
Therefore,
(2.15)‖T1vn-T1v0‖⟶0(n⟶∞).
This means that the operator 𝒯1:𝒫1→𝒫1 is continuous.
Choose two sequences{φn}n=1∞, {ψn}n=1∞⊂]a,b[ satisfying φn≤ψn for any n≥1, such that φn→a and ψn→b as n→∞, respectively. Define
(2.16)hn(t)={inf⁡a≤t≤φnh(t),h(t),φn<t<ψninf⁡ψn≤t≤bh(t),
and an operator sequence {𝒯1n}n=1∞ by
(2.17)T1nv=∫athn(s)f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds.
Clearly, hn:[a,b]→[0,∞) is a piecewise continuous function, and the operator 𝒯1n:𝒫1→𝒫1 is well defined. Further, we can see that 𝒯1n:𝒫1→𝒫1 is completely continuous.
LetR>0, BR:={v∈𝒫1:∥v∥0≤R}, and MR=sup{f(t,u):(t,u)∈[a,b]×[0,R¯]}, where R¯=(Rexp(λb)/d1)(δ+(1/λ)∑i=1nai(1-exp(-λ(b-ti))))+(R/λ)(expλ(b-a)-1)>0. We will prove that 𝒯1n approach 𝒯1 uniformly on BR. From the absolute continuity of integral, we obtain
(2.18)lim⁡n→∞∫l(n)h(s)ds=0,
where l(n)=[a,φn]∪[ψn,b]. For each v∈BR, t∈[a,φn], we have
(2.19)‖T1nv-T1v‖0=sup⁡t∈[a,φn]|∫at(hn(s)-h(s))f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds≤MR∫aφn|hn(s)-h(s)|ds⟶0(n⟶∞).
For each v∈BR, t∈[ψn,b], we have
(2.20)‖T1nv-T1v‖0=sup⁡t∈[ψn,b]|∫at(hn(s)-h(s))f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds≤MR∫ab|hn(s)-h(s)|ds⟶0(n⟶∞).
It is easy to see that, for each v∈BR and φn<t<ψn, there is ∥𝒯1nv-𝒯1v∥0→0 as n→∞. Similarly, for any v∈BR, and t∈[a,φn], ]φn,ψn[, [ψn,b], respectively, we can obtain that
(2.21)‖(T1nv)′-(T1v)′‖1=sup⁡t|(hn(t)-h(t))f(t,exp⁡(λt)(c1(v)+∫tbv(ω)exp⁡(-λω)dω))|≤MR|hn(t)-h(t)|⟶0(n⟶∞).
From the above argument, we obtain
(2.22)‖T1nv-T1v‖=max⁡{‖T1nv-T1v‖0,‖(T1nv)′-(T1v)′‖1}⟶0(n⟶∞).
That is to say, the sequence 𝒯1n is uniformly approximate 𝒯1 on any bounded subset of 𝒫1. Therefore, 𝒯1:𝒫1→𝒫1 is completely continuous.For convenience, we set(2.23)m1=∫t1bexp⁡(-λt)(∫ath(s)ds)dt,M1=∫abh(s)ds,u1=r1exp⁡(λa)(δexp⁡(λa)d1(b-a)+1),u2=L1exp⁡(λt1)δ+Σi=1nai(b-ti)exp⁡(λ(ti-b))d1.We are now ready to apply a functional type cone expansion and compression fixed-point theorem to the operator𝒯1 to give the sufficient conditions for the existence of at least one positive solution to the problem (1.4), (1.5).Theorem 2.5.
Suppose that (H1)–(H4) hold. Assume that there exist positive numbers k1, r1, and L1 with (exp(λb))/(b-t1)r1<L1 such that(A1)
f(t,w)≥k1, (t,w)∈[a,b]×[r1,∞[,(A2)
f(t,w)≥r1/m1, (t,w)∈[a,t1]×[u1,∞[,(A3)
f(t,w)≤L1/M1,(t,w)∈[t1,b]×[u2,∞[,then the operator 𝒯1 has at least one fixed-point v such that r1≤α(v) and β(v)≤L1, and the problem (1.4), (1.5) has at least one positive solution u such that
(2.24)u(t)=exp⁡(λt)(c1(v)+∫tbv(ω)exp⁡(-λω)dω).Proof.
The cone𝒫1 and operator 𝒯1 are defined by (2.1) and (2.2), respectively. By the properties of operator 𝒯1, it suffices to show that the conditions of Lemma 1.1 hold with respect to 𝒯1. In view of Lemma 2.1, it is not difficult to prove that a fixed point of 𝒯1 is coincident with the solution of the boundary value problem (1.4), (1.5), so we concentrate on the existence of the fixed point of the operator 𝒯1. Set 𝒫1(β,α,r1,L1) is a nonempty bounded subset of 𝒫1. From Lemma 2.4, it can be shown that
(2.25)T1:P1(β,α,r1,L1)¯⟶P1
is completely continuous by the Arzela-Ascoli lemma. For v∈∂𝒫1(β,α,r1,L1), the assumption (A1) implies that
(2.26)‖T1v‖0=∫abh(s)f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds≥k1∫abh(s)ds,‖(T1v)′‖1=sup⁡t∈]a,b[h(t)f(t,exp⁡(λt)(c1(v)+∫tbv(ω)exp⁡(-λω)dω))≥k1sup⁡t∈]a,b[h(t).
the hypotheses of (H3) lead to
(2.27)inf⁡v∈∂P1(β,α,r,L)‖T1v‖≥k1min⁡{sup⁡t∈]a,b[h(t),∫abh(s)ds}>0.
If v∈𝒫1(α,r1)¯, then β(v)=v(t1)≤(exp(λb))/(b-t1)r1<L1, and so v∈𝒫1(β,L1)¯, that is, 𝒫1(α,r1)¯⊆𝒫1(β,L1). It follows that the conditions of Lemma 1.1 hold with respect to 𝒯1. By the definition of functionals α and β, we can check that the functionals satisfy the properties α(θy)=∫t1bθy(ω)exp(-λω)dω=θ∫t1by(ω)exp(-λω)dω=θα(y) for y∈∂𝒫1(α,r1) and θ∈]0,1], β(μz)=μz(t1)=μβ(z) for z∈∂𝒫1(β,L1) and μ∈[1,∞[, β(0)=0.
We now prove thatα(𝒯1v)≥r1, in Lemma 1.1, holds. In fact, if v∈∂𝒫1(α,r1), by the properties of c1(v) and Lemma 2.2, for each t∈[a,t1],
(2.28)u(t)=exp⁡(λt)(c1(v)+∫tbv(ω)exp⁡(-λω)dω)≥exp⁡(λa)(c1(v)+r1)≥r1exp⁡(λa)(δexp⁡(λa)d1(b-a)+1).
Hence, by the assumption (A2) and (2.28), there is
(2.29)α(T1v)=∫t1b(T1v)(t)exp⁡(-λt)dt=∫t1bexp⁡(-λt)(∫ath(s)f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds)dt≥r1m1∫t1bexp⁡(-λt)(∫ath(s)ds)dt=r1.
Finally, we assert that β(𝒯1v)≤L1, in Lemma 1.1, also holds. If v∈∂𝒫1(β,L1), by Lemma 2.3, fort∈[t1,b],
(2.30)u(t)=exp⁡(λt)(c1(v)+∫tbv(ω)exp⁡(-λω)dω)≥v(t1)exp⁡(λt1)δ+Σi=1nai(b-ti)exp⁡(λ(ti-b))d1=L1exp⁡(λt1)δ+Σi=1nai(b-ti)exp⁡(λ(ti-b))d1.
The assumption (A3) and (2.30) imply that
(2.31)β(T1v)=(T1v)(b)=∫abh(s)f(s,exp⁡(λs)(c1(v)+∫sbv(ω)exp⁡(-λω)dω))ds≤L1M1∫abh(s)ds=L1.
To sum up, the hypotheses of Lemma1.1 are satisfied. Hence, the operator 𝒯1 has at least one fixed point, that is, the problem (1.4), (1.5) has at least one positive solution.Let the cone(2.32)P2={v∈B:visnonincreasingon⁡[a,b],v(b)=0}.
Evidently, 𝒫2⊂ℬ. For v∈𝒫2, define the operator 𝒯2 by
(2.33)T2v(t)=∫tbh(s)f(s,exp⁡(λs)(c2(v)+∫asv(ω)exp⁡(-λω)dω))ds,t∈[a,b],
where c2(v)=(1/d2)(δv(a)-∑i=1naiexp(λti)∫ativ(ω)exp(-λω)dω).We only give the preliminary lemmas and result of the problem (1.4), (1.6), the proofs are similar to the above argument.Lemma 2.6.
Ifv∈𝒫2 is a fixed-point 𝒯2, then
(2.34)u(t)=exp⁡(λt)(c2(v)+∫atv(ω)exp⁡(-λω)dω)
is one solution of the problem (1.4), (1.6).Forv∈𝒫2, the nonnegative continuous functionals α and β on 𝒫2 are defined by
(2.35)α(v)=∫atnv(ω)exp⁡(-λω)dω,β(v)=v(tn).Lemma 2.7.
Letr2>0. If v∈∂𝒫2(α,r2), then
(2.36)v(tn)≤exp⁡(λtn)tn-ar2,∫abv(ω)exp⁡(-λω)dω≥r2.Lemma 2.8.
LetL2>0. If v∈∂𝒫2(β,L2), then
(2.37)∫atnv(ω)exp⁡(-λω)dω≥L2exp⁡(-λtn)(tn-a).Lemma 2.9.
Let (H1)–(H4) hold, then 𝒯2:𝒫2→𝒫2 is completely continuous.For convenience, we set(2.38)m2=∫atnexp⁡(-λt)(∫tbh(s)ds)dt>0,M2=∫abh(s)ds>0,u3=r2exp⁡(λb),u4=L2exp⁡(λa)δ-Σi=1nai(ti-a)exp⁡(λ(ti-a))d2.Theorem 2.10.
Suppose that (H1)–(H4) hold. Assume that δ-Σi=1naiexp(λ(ti-a))(ti-a)>0, then there exist positive numbers k2, r2, and L2 with ((exp(λtn))/(tn-a))r2<L2 such that(B1)
f(t,w)≥k2,  (t,w)∈[a,b]×[r2,∞[,(B2)
f(t,w)≥(r2/m2), (t,w)∈[tn,b]×[u3,∞[,(B3)
f(t,w)≤(L2/M2), (t,w)∈[a,tn]×[u4,∞[.
Then the operator𝒯2 has at least one fixed-point v such that r2≤α(v) and β(v)≤L2, and the problem (1.4), (1.6) has at least one positive solution u such that
(2.39)u(t)=exp⁡(λt)(c2(v)+∫atv(ω)exp⁡(-λω)dω).


## 3. Examples

Consider the problems(3.1)u′′(t)-u′(t)+h(t)f(t,u(t))=0,0<t<1,(3.2)u′(0)-u(0)=0,u(1)+2u′(1)=2u(12),(3.3)u(0)+2u′(0)=2u(12),u′(1)-u(1)=0.Let(3.4)h(t)=1t(1-t),f(t,w)={10-3t+2w+2,w<2,10-3t+2-2,w≥2.
It is easy to check that hypotheses (H1)–(H4) hold. For the problem (3.1), (3.2), by some calculations, we have d1≈4.858, m1≈0.348, and M1≈3.142. Taking k1=0.001, r1=0.2, and L1=2, satisfying the following conditions: f(t,w)≥0.001, (t,w)∈[0,1]×[0.2,∞[, f(t,w)≥r1/m1≈0.575, (t,w)∈[0,1/2]×[0.141,∞[, f(t,w)≤L1/M1≈0.637, and (t,w)∈[1/2,1]×[2.242,∞[. Thus, the hypotheses of Lemma 1.1 are fulfilled, and so the operator 𝒯1 has at least one fixed point, that is to say, the problem (3.1), (3.2) has at least one positive solution. For the problem (3.1), (3.3), by some calculations, we obtain d2≈0.297, m2≈0.348, and M2≈3.142. Taking k2=0.001, r2=0.1, and L2=2, combining with the following conditions: f(t,w)≥0.001, (t,w)∈[0,1]×[0.1,∞[, f(t,w)≥r2/m2≈0.287, (t,w)∈[1/2,1]×[0.272,∞[, f(t,w)≤L2/M2≈0.637, and (t,w)∈[0,1/2]×[2.364,∞[. So the problem (3.1), (3.3) has at least one positive solution.



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*Source: 416173-2011-05-02.xml*