# On the Solvability of a Resonant Third-Order Integralm-Point Boundary Value Problem on the Half-Line

Authors: O. F. Imaga; J. G. Oghonyon; P. O. Ogunniyi
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

In this work, the existence of at least one solution for the following third-order integral andm-point boundary value problem on the half-line at resonance ρtu′t″=wt,ut,u′t,u″t,t∈0,∞,u0=∑j=1mαj∫0ηjutdt,u′0=0,limt⟶∞ρtu′t′=0, will be investigated. The Mawhin’s coincidence degree theory will be used to obtain existence results while an example will be used to validate the result obatined.


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## Body

## 1. Introduction

This work studies the existence of solution for a resonant third-order boundary value problem with integral andm-point boundary conditions on the half-line
(1)ρtu′t″=wt,ut,u′t,u″t,t∈0,∞,(2)u0=∑j=1mαj∫0ηjutdt,u′0=0,limt⟶∞ρtu′t′=0,where w:0,∞×ℝ3⟶ℝ is an S-Carathéodory function, αi∈ℝ, i=1,2,⋯,m, 0<η1<⋯<ηm<1, ρ∈C0,∞∩C20,∞, ρt>0 for t∈0,∞, 1/ρ∈L10,∞ and the resonance condition is ∑j=1mαjηj=1.Boundary value problems on the half-line arise in the modeling of various physical processes like the flow of fluid over semi-infinite porous media and wet surfaces [1].Boundary value problem (1)-(2) is said to be at resonance since the corresponding homogeneous problem
(3)ρtu′t″=0,t∈0,∞,u0=∑j=1mαj∫0ηjutdt,u′0=0,limt⟶∞ρtu′t′=0,has a nontrivial solution ut=C, where C is a constant. Boundary value problems at resonance can be expressed in the abstract form as Lu=Nu, where L is a linear differential operator that is not invertible. Mawhin's coincidence degree theory [2] is an excellent tool for studying resonant problems of this type.The problem of existence of solutions for resonant boundary value problems has received the attention of many authors recently, both in the bounded domain and on the half-line. For instance, the authors in [3] studied the third-order problem
(4)u′′′t=ft,ut,u′t,u″t,t∈0,1,u0=∑i=1m−2αiuξi,u′0=0,u1=βuηunder the resonance conditions: β=1,∑i=1m−2αi=1,∑i=1m−2αiξi2=1 and β=1/η2,∑i=1m−2αi=1,∑i=1m−2αiξi2=1. They applied the coincidence degree arguments to obtain existence results in a bounded domain. For other works on a bounded domain, see [4–7].In [8], the authors considered the multipoint boundary value problem on the half-line
(5)u″t=ft,ut,u′t,t∈0,∞,u0=0,u′∞=∑i=1m−1αiu′ξi,where αi>0 and 0=ξ1<⋯<ξm−1<∞. They applied a perturbation technique in obtaining existence results under the resonant condition ∑i=1m−1αi=1.Iyase [9] used coincidence degree arguments to study existence of solutions for the multipoint boundary value problem at resonance on the half-line
(6)qtu″t′=gt,ut,u′t,u″t,0,∞,u′0=∑i=1m−1αi∫0ξiutdt,u0=0,limt⟶∞qtu″t=0under the resonant condition ∑i=1m−1αi=2 which is different from ours. For other literature on resonant multipoint boundary value problems on the half-line, see [10–14]. Motivated by the results mentioned above, we study the existence of solutions for the resonant third-order boundary value problem with integral and m-point boundary conditions on the half-line. In Section 2 of this work, necessary lemmas, theorems, and definitions will be given; Section 3 will be dedicated to stating and proving the condition for existence of solutions. An example will be given in Section 4 to corroborate the result obtained.


## 2. Preliminaries

In this section, we will give definitions, theorems, and lemmas that are required for this work.Definition 1 (see [6]).
LetU and Z be Banach spaces. A linear operator L:domL⊂U⟶Z is called a Fredholm mapping of index zero if kerL and Z/ImL are finite dimensional.TakeU, Z to be normed spaces, L:domL⊂U⟶Z a Fredholm mapping of zero index and P:U⟶U,Q:Z⟶Z projectors that are continuous such that ImP=kerL, kerQ=ImL, ImQ=kerL and U=kerL⊕kerP, Z=ImL⊕ImQ, then LdomL∩kerP:domL∩kerP⟶ImL is invertible. The inverse of the mapping L will be denoted by Kp:ImL⟶domL∩kerP while the generalized inverse, KP,Q:Z⟶domL∩kerP is defined as KP,Q=KpI−Q.Definition 2 (see [13]).
A mapw:0,+∞×ℝ3⟶ℝ is S-Carathéodory, if the following conditions are satisfied:
(i)
For eachd,e,f∈ℝ3, the mapping t⟶qt,d,e,f is Lebesgue measurable(ii)
For a.e.t∈0,∞, the mapping d,e,f⟶wt,d,e,f is continuous on ℝ3(iii)
For eachk>0, there exists φkt, with tφkt∈L10,+∞ such that, for a.e. t∈0,∞ and every d,e,f∈−k,k, we have(7)∣wt,d,e,f∣≤φkt.Definition 3.
LetL:domL⊂U⟶Z be a Fredholm mapping, E a metric space, and N:E⟶Z a nonlinear mapping. N is said to be L-compact on E if QN:E⟶Z and KP,QN:E⟶U are compact on E. Also, N is L-completely continuous if it is L-compact on every bounded E⊂U.Theorem 4 (see [9]).
LetU be the space of all bounded continuous vector-valued functions on 0,∞ and M⊂U. Then, M is relatively compact on U if the following conditions hold:
(i)
M is a bounded subset of U(ii)
The functions inM are equicontinuous on any compact interval of 0,∞(iii)
The functions fromM are equiconvergent; that is, if given ε>0, there exists T=Tε>0 such that ∣ft−f∞∣<ε, for all t>T and f∈MTheorem 5 (see [2]).
LetL:domL⊂U⟶Z be a Fredholm map of index zero and let N:U⟶Z be L-compact on Ω¯. Assume that the following conditions are satisfied:
(i)
Lu≠λNu for every u,λ∈domL\kerL∩∂Ω×0,1(ii)
Nu∉ImL for every u∈kerL∩∂Ω(iii)
degQNkerL,Ω∩kerL,0≠0, where Q:Z⟶Z is a projection such that ImL=kerQThen, the abstract equationLu=Nu has at least one solution in domL∩Ω¯.Let(8)U=u∈C20,+∞:u,u′,ρu′′∈AC0,+∞,limt⟶∞e−tuitexist,i=0,1,2,ρu′′′∈L10,∞,where AC0,+∞ is the set of absolutely continuous functions. The norm defined on U is u=maxu∞,u′∞,u′′∞ where u∞=supt∈0,+∞e−t∣ut∣.LetZ=z∈L10,+∞∩C20,+∞:∫0+∞s∣z∣ds<+∞ and define the norm z=maxz∞,z1,zL1 where zL1=∫0∞∣zs∣ds, z1=∫0∞s∣zs∣ds and z∞ is the supremum norm on 0,∞. The linear operator L:domL⊂U⟶Z will be defined by
(9)L:u↦Lut=ρtu′t′′,where
(10)domL=u∈U∩C30,+∞:u0=∑j=1mαj∫0ηjutdt,u′0=0,limt⟶∞ρtu′t′=0.Also, the nonlinear operatorN:U⟶Z will be defined by Nut=wt,ut,u′t,u′′t, t∈0,+∞; thus, problem (1)-(2) may be written in the form Lu=Nu.Lemma 6.
The following conditions hold:(i)
kerL=u∈domL:ut=h,h∈ℝ,t∈0,∞(11)ImL=z∈Z:∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞zvdvdτdsdt=0(ii)
L:domL⊂U⟶U is a Fredholm operator of index zero, while the continuous linear projector Q:Z⟶Z may be defined as
(12)Qy=θt∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞zvdvdτdsdt,where
(13)θt=e−t∑j=1mαj∫0ηj∫0t1−e−s/ρsdsdt(iii)
The generalized inverseKp:ImL⟶domL∩kerP of L may be written as
(14)KPy=−∫0t1ρs∫0s∫τ∞zvdvdτds.
Furthermore,KPy≤A1zL1 where
(15)A1=max1ρ1,1ρ∞,1.Proof.
Let(16)ρtu′t′′=0,t∈0,∞,then
(17)ut=h+ρ0u′0∫0t1ρsds+ρ∞u′∞′∫0tsρsds.
☐From (2), we have u′0=0,limt⟶∞ρtu′t′=0; then, (17) becomes
(18)ut=h.Therefore,kerL=u∈domL:ut=h,h∈ℝ,t∈0,∞; hence, (i) holds. Next, we show that (ii), (iii), and (iv) also hold.Forz∈ImZ, consider the problem
(19)ρtu′t′′=zt,which has a solution ut defined as
(20)ut=u0+ρ0u′0∫0t1ρsds+ρ∞u″∞′∫0tsρsds−∫0t1ρs∫0s∫τ∞zvdvdτds.Applying the boundary conditions and using∑j=1mαjηj=1, one obtains
(21)∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞zvdvdτdsdt=0,ut=h−∫0t1ρs∫0s∫τ∞zvdvdτds,where h is an arbitrary constant and ut is a solution of (19) satisfying (2). Therefore, condition (ii) holds.For anyz∈Z, let the projector
(22)Qz=θt∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞zvdvdτdsdt,t∈0,∞.Letz1=z−Qz, then z1=zI−Q, where I is the identity operator. Since
(23)∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞θvdvdτdsdt=1,then z1∈ImL and Z=ImL+ImQ. From ImL∩ImQ=0, we have Z=ImL⊕ImQ. Therefore, dimkerL=dimImQ=1. Thus, L is a Fredholm operator of index zero and (iii) holds.GivenP:U⟶U defined as
(24)Pu=u0,t∈0,∞,the generalized inverse KP:ImL⟶domL∩kerP of L can then be written as
(25)−∫0t1ρs∫0s∫τ∞zvdvdτds.In fact, for anyz∈ImL,
(26)LKPzt=ρtKPz′t′′=zt,and for u∈domL∩kerP, it follows that
(27)KPLut=−∫0t1ρs∫0s∫τ∞ρvu′v′′dvdτds=−∫0t1ρs∫0sρ∞u′∞′−ρτu′τ′dτds=∫0t1ρsρsu′s−ρ0u′0ds=ut−u0.SincePu=u0=0, then
(28)KPLut=ut,t∈0,∞;thus, KP=LdomL∩kerP−1. In addition,
(29)KPz∞=supt∈0,+∞e−t∣−∫0t1ρs∫0s∫τ∞zvdvdτds∣≤zL1∫0∞s∣ρs∣ds≤1ρ1zL1,(30)KPz′∞=supt∈0,∞e−t−1ρt∫0t∫τ∞zvdvdτ≤1e1ρ∞zL1≤1ρ∞zL1,(31)KPz′′∞=supt∈0,+∞e−t∣−∫τ∞zvdv∣≤∫0∞∣zv∣dv≤zL1.Hence, (29), (30), and (31) give
(32)KPz≤max1ρ1,1ρ∞,1zL1.Lemma 7.
The nonlinear operatorN is L-compact if w is an S-Carathéodory function.Proof.
To prove this lemma, we have to show that Theorem4 satisfied the operator KP,QNu. This we will do in three steps. Given that D⊂U is bounded and k=sup∥u∥:u∈D, consider KP,QND. Since w:0,∞×ℝ3 satisfies the S-Carathéodory conditions with respect to L10,∞, there exists a Lebesgue integrable function φk, with tφk∈L10,∞ such that
(33)Nut=wt,ut,u′t,u′′t≤φkt,a.e.t∈0,∞.
☐Hence, for allu∈D,
(34)QNu∞=supt∈0,∞e−t∣QNuv∣=supt∈0,∞e−tθt∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,,uv,u′v,u′′vdvdτdsdt≤θ∞φkL1∑j=1m∣αj∣∫0ηj∫0∞s∣ρs∣dsdt≤∥θ∥∞∥φk∥L1∥1ρ∥1∑j=1m∣αj∣ηj<∞,∥QNu∥L1=∫0∞∣QNuv∣dv=∫0∞θt∑j=1mαj∫0ηj∫0ζ1ρs∫0s∫τ∞wv,uv,u′v,u′′vdvdτdζdsdt≤∥θ∥L1∥φk∥L1∑j=1m∣αj∣∫0ηj∫0∞s∣ρs∣dsdζ≤∥θ∥L1∥φk∥L1∥1ρ∥1∑j=1m∣αj∣ηj<∞,∥QNu∥1=∫0∞v∣QNuv∣dv=∫0∞tθt∑j=1mαj∫0ηj∫0ζ1ρs∫0s∫τ∞wv,uv,u′v,u′′vdvdτdsdζdt≤∥θ∥1∥φk∥L1∑j=1m∣αj∣∫0ηj∫0∞s∣ρs∣dsdζ≤∥θ∥1∥φk∥L1∥1ρ∥1∑j=1m∣αj∣ηj<∞.Hence,∥QNu∥=max∥QNu∥∞,∥QNu∥L1,∥QNu∥1 is bounded. Next, we prove that KPI−QND¯ is compact. For u∈D, we have
(35)∥KP,QNu∥∞=supt∈0,∞e−t∣KP,QNu∣≤∥φk∥L1∥1ρ∥11+∥θ∥L1∥1ρ∥1∑j=1m∣αj∣ηj<∞,∥KP,QNu′∥∞=supt∈0,∞e−t∣KP,QNu′∣≤1e∥φk∥L1∥1q∥∞1+∥θ∥L1∥1ρ∥1∑j=1m∣αj∣ηj<∞,∥KP,QNu′′∥∞=supt∈0,∞e−t∣KP,QNu′′∣≤∥φk∥L11+∥θ∥L1∥1ρ∥1∑j=1m∣αj∣ηj<∞.Hence,KP,QND is uniformly bounded in U; thus, (i) of Also, we will prove that (ii) of Theorem 4 holds. Also, we will prove that (ii) of Theorem 4 holds, that is, KP,QND is equicontinuous on 0,T. Let t1,t2∈0,T, where T∈0,+∞. Then,
(36)KP,QNut2−KP,QNut1≤∥φk∥L11+∥θ∥L1∥1ρ∥1∑j=1mαjηj∫t1t2sρsds⟶0,ast1⟶t2,KP,QNu′t2−KP,QNu′t1≤∥φk∥L11+∥θ∥L1∥1ρ∥1∑j=1mαjηj1ρt1−1ρt2∫0t1dτ+∥φk∥L11+∥θ∥L1∥1ρ∥1∑j=1mαjηj1ρt2∫t1t1dτ⟶0,ast1⟶t2,KP,QNu′′t2−KP,QNu′′t1≤∫t1t2φkv+∥θ∥L1∥1ρ∥1∑j=1mαjηj⟶0,ast1⟶t2.Thus,KP,QND is equicontinuous every compact subset of 0,∞. Finally, we show that (iii) of Theorem 4 holds; that is, KP,QND is equiconvergent at ∞:
(37)KP,QNut−KP,QNu∞≤∥φk∥L11+∥θ∥L1∥1ρ∥1∑j=1mαjηj∫t∞sρsds⟶0,uniformlyast⟶∞,∣KP,QNu′t−KP,QNu′∞∣+∥φk∥L11+∥θ∥L1∥1ρ∥1∑j=1m∣αj∣ηj1ρ∞∫t∞dτ⟶0,uniformlyast⟶∞,KP,QNu′′t−KP,QNu′′∞≤∫t∞φkv+θv∥φk∥L1∥1ρ∥1∑j=1mαjηj⟶0,uniformlyast⟶∞.Hence,KP,QND is equiconvergent at ∞. Thus, from Definition 3 and Theorem 4, we see that the nonlinear operator N is L-compact.


## 3. Existence Result

Theorem 8.
Letw:0,+∞×R3⟶R be an S-Carathéodory function. If the following hold:
(E1)There exists constant σ∈0,1 and functions b,c,d,e,g∈L10,∞, satisfying ∫0+∞sbsds<+∞,∫0+∞scsds<+∞,∫0+∞sdsds<+∞,∫0+∞sesds<+∞,∫0+∞s∣gs∣ds<+∞, such that for all x,y,z∈ℝ3 and t∈0,∞,
(38)∣wt,x,y,z∣≤e−tbt∣x∣+ct∣y∣+dt∣z∣+etzσ+gt.
(E2)There exists a constant B>0, such that for u∈domL, if ∣ut∣>B, for all t∈0,∞, then
(39)θt∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,uv,u′v,u′′vdvdτdsdt≠0.
(E3)There exists a constant B∗>0, such that for any ut=h∈ℝ, if ∣h∣>B∗, then
(40)h·∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,h,0,0dvdsdτdt<0(41)orh·∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,h,0,0dvdsdτdt>0.
Then, problem (1) subject to (2) has at least one solution if
(42)bL1+cL1+dL1<1A,where A=max1/ρ1+1/ρ∞,21/ρ∞,1/ρ∞+1.In order to prove Theorem8, the following lemmas are required. These lemmas will help us prove that condition (iii) of Theorem 5 holds.Lemma 9.
The setΩ1=u∈domL\kerL:Lu=λNuforsomeλ∈0,1 is bounded.Proof.
Assume thatu∈Ω1 and Lu=λNu, then λ≠0, QNu=0 and by condition (E2) of Theorem 8, there exists t0∈0,∞, such that ∣ut0∣≤B. By the absolute continuity of u, it follows from u0=ut0−∫0t0u′tdt that
(43)∣u0∣≤∣ut0∣+∫0t0∣u′v∣dv≤B+u′∞.
Fromu′t=1/ρt∫0tu′′vdv, one gets
(44)u′∞≤supt∈0,∞e−t1∣ρt∣∫0t∣u″∣vdvds≤1ρ∞u″∞,and from u′′=∫t∞u′′′vdv, we obtain
(45)∥u′′∥∞≤supt∈0,∞e−t∫t∞∣u′′′v∣dv≤∫0∞∣Luv∣dv≤∫0∞∣Nuv∣dv≤NuL1.
ence, from (43) and (45)
(46)Pu=∣u0∣≤B+u′∞≤B+u′L1≤B+1ρ∞NuL1.
Also, foru∈Ω1, u∈domL\kerL, then I−Pu∈domL∩kerP and LPu=0; then, from Lemma 6, one gets
(47)I−Pu=KPLI−Pu≤max1ρ1,1ρ∞,1LI−PuL1≤max1ρ1,1ρ∞,1NuL1.
Hence, from (46) and (47),
(48)u≤Pu+I−Pu≤B+max1ρ1+1ρ∞,21ρ∞,1ρ∞+1NuL1.
Settingmax1/ρ1+1/ρ∞,21/ρ∞,1/ρ∞+1=A gives
(49)u≤B+ANuL1.
From (38), we see that
(50)∥Nu∥L1≤∫0∞∣wt,uv,u′v,u′′z∣dv≤ubL1+cL1+dL1+eL1u′′∞σ.
From (49), we obtain
(51)∥u∥≤B+A∥e∥L1u′′∞σ1−A∥b∥L1+∥c∥L1+∥d∥L1.Hence, Ω1 is bounded.☐Lemma 10.
Given thatΩ2=u∈kerL:J−1u−1−λ/λQNu=0,λ∈0,1. Suppose that (40) holds, for all ∣h∣>B∗ where J:ImQ⟶kerL is the linear isomorphism defined by Jh=h, then Ω2 is bounded.Proof.
For allut=h∈Ω2, one obtains
(52)h=1−λλ∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,h,0,0dvdsdτdt.
Whenλ=1, then h=0. However, when ∣h∣>B∗, then in view of (40), one gets
(53)λh2=h1−λ∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,h,0,0dvdsdτdt<0,contradicting λh2>0. Then, u=∣x∣≤B∗, implying that u≤B∗. Hence, Ω2 is bounded.☐Proof of Theorem 8.
We will now show that all the conditions of Theorem5 hold. Take ∪k=12Ω¯k as an open bounded subset of U. Lemma 6 shows that L is a Fredholm operator of index zero, and from Lemma 7, N is L-compact on Ω¯. Thus, conditions (i) and (ii) of Theorem 5 are satisfied based on the definition of Ω. Finally, we show that condition (iii) of Theorem 5 holds. Set Eu,λ=−λJu+1−λQNu. Since Ω¯2⊂Ω, then for every u∈kerL∩∂Ω, Eu,λ≠0. Therefore, the homotopy property of the Brouwer degree gives
(54)degQNkerL,Ω∩kerL,0=degE·,0,Ω∩kerL,0=degE·,1,Ω∩kerL,0=deg±J,Ω∩kerL,0≠0.
☐Hence, condition (iii) of Theorem5 holds, and problem (1)-(2) has at least one solution in U.


## 4. Example

Consider the following problem(55)ρtu′t″=e−tu20+110sinu″1/16+120u″,(56)u0=18∫01/6utdt−2∫01/2utdt,u′0=0,limt⟶∞ρtu′t′=0,where
(57)ρt=et,t∈0,∞,α1=18,α2=−2,η1=16,η2=12,∑j=12αjηj=1816−212=3≠1,∑j=12αj∫0ηj∫0t1−e−sesdsdt=∑j=12αjηj2+e−ηj−e−2ηj4−34=−0.01684≠0.Given thatt,x,y,z∈0,∞×ℝ3, then wt,x,y,z>0 and
(58)θt∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,uv,u′v,u″vdvdτdsdt≠0;hence, condition (E2) is satisfied. For any h∈ℝ such that ∣h∣>B∗, where B∗>0,
(59)h·∑j=1mαj∫0ηj∫0t1ρs∫0s∫τ∞wv,h,0,0dvdτdsdt=h∑j=1mαj∫0ηj∫0t1es∫0s∫τ∞h20e−vdvdτdsdt=−0.01684h220≠0since∣h∣>0.Now,∣wt,x,y,z∣≤e−t∣x∣/20+1/10·11/16+∣z∣/20. Taking bt=1/20e−t, ct=0, dt=1/20e−t, et=0, t∈0,∞, then b,c,d,e,g∈L10,∞ satisfy condition (E1). Finally,
(60)bL1=∫0∞120∣e−t∣dt=120,dL1=∫0∞120e−tdt=120,A=max1ρ1+1ρ∞,21ρ∞,1ρ∞+1=max2,2,2=2,bL1+cL1+dL1=120+120=110<12.All the conditions are satisfied, and we conclude from Theorem8 that there exists at least one solution ut for problem (55) subject to (56).



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