Fredholm's third theorem for second order singular Dirichlet problem

Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 http://www.boundaryvalueproblems.com/content/2014/1/59 R E S E A R C H Open Access Fredholm’s third theorem for second-order singular Dirichlet problem Alexander Lomtatidze1,2 and Zdenˇ ek Opluštil2* *Correspondence: [email protected].cz 2Institute of Mathematics, Faculty of Mechanical Engineering, Brno University of Technology, Technická 2, Brno, 616 69, Czech Republic Full list of author information is available at the end of the article Abstract Consider the singular Dirichlet problem u =p(t)u+q(t); u(a)=0, u(b)=0, where p,q:]a,b[→Rare locally Lebesgue integrable functions. It is proved that if b a (s–a)(b–s)p(s)–ds <+∞and b a (s–a)(b–s)q(s)ds <+∞, then Fredholm’s third theorem remains true. MSC: 34B05 Keywords: singular Dirichlet problem; Fredholm’s third theorem 1 Introduction Consider the boundary value problem u =p(t)u+q(t), () u(a)=, u(b)=, () where p,q∈Lloc(]a,b[). We are mainly interested in the case, when the functions pand qare not integrable on [a,b]. In this case, the problem (), ()issaidtobesingular.Itis proved in []thatif b a (s–a)(b–s)p(s)–ds <+∞() and b a (s–a)(b–s)q(s)ds <+∞,() then, for the singular problem (), (), the Fredholm alternative holds. More precisely, the following theorem is true. ©2014 Lomtatidze and Opluštil; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 2 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Theorem . ([, Theorem .]) Let () hold.Then the problem (), () is uniquely solvable for any q satisfying () iff the corresponding homogeneous equation u =p(t)u(a) has no nontrivial solution satisfying (). The aim of this paper is to show that, under the assumption (), the Fredholm’s third theorem remains true. Before formulation of the main results, we introduce the following notation. Ris the set of real numbers. For x∈R,weput[x]–= (|x|–x). C(I), where I⊂R, is a set of continuous functions u:I→R. For u∈C([α,β]), we put u[α,β]=max{|u(t)|:t∈[α,β]}. AC loc(]α,β[) is the set of functions u:]α,β[→R, which are absolutely continuous together with their first derivative on every closed subinterval of ]α,β[. Lloc(]α,β[) is the set of functions p:]α,β[→R,whichareLebesgueintegrableonevery closed subinterval of ]α,β[. By f(a)(resp.,f(b)) we denote the right (resp., left) limit of the function f:]a,b[→Rat the point a(resp., b). By a solution of equation () we understand a function u∈AC loc(]a,b[), which satisfies it almost everywhere in ]a,b[.Asolutionofequation()satisfying() is said to be a solution of the problem (), (). Wewillsaythatacertainpropertyholdsin]α,β[ if it takes place on every closed subinterval of ]α,β[. Recall that we consider the problem (), (), where p,q∈Lloc(]a,b[). Theorem . Let () hold.Then the homogeneous problem (a), () hasnomorethanone, up to a constant multiple,nontrivial solution. Remark . Belowwewillshow(seeProposition.)thatif() holds and uis a nontrivial solution of (a), (), then there exists r>suchthat u(t)≤r(t–a)(b–t)fort∈[a,b]. Theorem . Let () hold and the homogeneous problem (a), () have a nontrivial solution u.Then the problem (), (), where the function q satisfies (), is solvable iff the condition b a q(s)u(s)ds = () is fulfilled. Remark . In view of Remark . and condition (), the function quis integrable on [a,b] and, therefore, condition () is meaningful. 2 Auxiliary statements First of all, for convenience of references, we recall two lemmas from []. Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 3 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Lemma . ([, Lemma .]) Let () and () hold.Then,for any α∈[a,b[and β∈]α,b], every solution u of equation () satisfying u(α)=, u(β)= admits the estimate (t–a)(b–t)u(t)≤u[α,β]b–a+b a (s–a)(b–s)p(s)–ds +b a (s–a)(b–s)q(s)ds for t ∈]α,β[. Lemma . ([, Lemma .]) Let () hold.Then there exist a∈]a,b[, b∈]a,b[, and >such that,for any α∈[a,a[, β∈]b,b], and q satisfying (), every solution u of equation () satisfying u(α)= admits the estimate u(t)≤(t–a)u[α,a]+t a (s–a)q(s)ds +(t–a)a tq(s)dsfor t ∈]α,a], while every solution u of equation () satisfying u(β)= admits the estimate u(t)≤(b–t)u[b,β]+b t (b–s)q(s)ds +(b–t)t bq(s)dsfor t ∈[b,β[. Next proposition immediately follows from Lemma .. Proposition . Let () hold and ube a nontrivial solution of the homogeneous problem (a), (). Then there exists r>such that u(t)≤r(t–a)(b–t)for t ∈[a,b]. Proposition . Let () hold and ube a nontrivial solution of (a) satisfying u(a)= (respectively,u(b)=).Then there exists a∈]a,b[(respectively,b∈]a,b[) such that u(t)= for t ∈]a,a]respectively,u(t)= for t ∈[b,b[.() Proof In view of ()thereexistsa∈]a,b[(respectively,b∈]a,b[) such that a a (s–a)p(s)–ds < respectively, b b (b–s)p(s)–ds < . Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 4 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Hence, the inequality a a (s–a)(a–s)p(s)–ds <a–a respectively, b b (s–b)(b–s)p(s)–ds <b–b holds, as well. The latter inequality, by virtue of [,Lemma.],impliesthatforanya< t<t<a(respectively, b<t<t<b), the problem u =p(t)u;u(t)=, u(t)= has no nontrivial solution. Now suppose that uis a nontrivial solution of (a)satisfyingu(a)=(respectively, u(b) = ). Then it follows from the above that either u(t)= fort∈]a,a]respectively, u(t)= fort∈[b,b[,() or there is a t∈]a,a](respectively,t∈[b,b[) such that u(t)= fort∈]a,t[, u(t)= respectively, u(t)= fort∈]t,b[, u(t)= . () It is now clear that ()holdswitha=a(respectively, b=b)if()holds,andwith a=a+t (respectively, b=t+b )if()issatisfied.  Lemma . Let () and () hold.Let,moreover,u be a solution of the problem (), () and ube a solution of the problem (a), (). Then lim t→a+u(t)u(t)–u(t)u (t)=, lim t→b–u(t)u(t)–u(t)u (t)=. () Proof It is clear that u(t)u(t)–u(t)u (t)=q(t)u(t)fort∈]a,b[. Hence, u(t)u(t)–u(t)u (t)=δ–c t q(s)u(s)ds for t∈]a,b[, () where c=a+b and δ=u(c)u(c)–u(c)u (c). By virtue of Proposition . and condition (), the function quis integrable on [a,b]. Thus, it follows from () that there exists a finite limit lim t→a+u(t)u(t)–u(t)u (t)=ε.() Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 5 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Nowwewillshowthatε= . Suppose the contrary, let ε>. () Then there is α∈]a,b[suchthat u(t)u(t)–u(t)u (t)>ε for t∈]a,α]. () On account of Proposition ., we can assume without loss of generality that u(t)= fort∈]a,α]. () Then it follows from ()that u(t) u(t) >ε u (t)for t∈]a,α]. Hence μu(t)–u(t)>ε u(t)α t ds u (s)for t∈]a,α], () where μ=u(α) u(α). Taking now into account Proposition .,wegetfrom()that μu(t)–u(t)>εu(t) t–a– α–afor t∈]a,α], where ε=ε r (b–a). The latter inequality, in view of the conditions u(a)=andu(a)=, implies that lim t→a+ |u(t)| t–a=. () On the other hand, by virtue of Lemma .,thereisM>suchthat (t–a)u(t)≤Mfor t∈]a,α]. () In view of ()and(), we get lim t→a+u(t)u(t)=lim t→a+(t–a)u(t) |u(t)| t–a=, and therefore, on account of (), we obtain lim t→a+u(t)u (t)=ε. Now, let α∈]a,α[besuchthat u(t)u (t)>ε for t∈]a,α]. Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 6 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Then it is clear that u[a,b]u (t)>ε for t∈]a,α] and consequently u[a,b]u(t)>ε (t–a)fort∈]a,α]. However, the latter inequality and ()yieldthatε≤, which contradicts (). The contradiction obtained proves the first equality in (). By the same arguments one can prove the second equality in ().  We will need the next lemma in the proof of the sufficiency part of Theorem . and thus, we will suppose that Theorem . and the necessity part of Theorem . are true. Lemma . Let () hold and the homogeneous problem (a), () have a nontrivial solution u.Then there exist n∈Nand r >such that,for any q satisfying () and () and every n >n,the solution u of the problem u =p(t)+  np(t)–u+q(t); u(a)=, u(b)= admits the estimate u(t)≤rb a (s–a)(b–s)q(s)ds for t ∈[a,b]. Proof Suppose the contrary, let the assertion of the lemma be violated. Then, for any n∈ N,thereexistkn≥n,qn∈Lloc(]a,b[), and un∈AC loc(]a,b[) such that b a (s–a)(b–s)qn(s)ds <+∞,b a qn(s)u(s)ds =, u n(t)=p(t)+  knp(t)–un(t)+qn(t)fort∈]a,b[, un(a)=, un(b)= and un[a,b]>nb a (s–a)(b–s)qn(s)ds. Introduce the notation ˜ un(t)=  un[a,b] un(t), ˜ qn(t)=  un[a,b] qn(t)fort∈]a,b[. Then it is clear that ˜ u n(t)=p(t)+  knp(t)–˜ un(t)+˜ qn(t)fort∈]a,b[, () Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 7 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 ˜ un(a)=, ˜ un(b)=, ˜ un[a,b]=, () b a (s–a)(b–s)˜ qn(s)ds < n() and b a ˜ qn(s)u(s)ds = . () By virtue of Lemma . (with q(t)=  kn[p(t)]–˜ un(t)+˜ qn(t)) and (), we have (t–a)(b–t)˜ u n(t)≤b–a+n+ nb a (s–a)(b–s)p(s)–ds +b a (s–a)(b–s)˜ qn(s)ds for t∈]a,b[, () while, by virtue of Lemma . (with q(t)=  kn[p(t)]–˜ un(t)+˜ qn(t)), there exist a∈]a,b[, b∈]a,b[, and >suchthat ˜ un(t)≤t–a+a a (s–a)  knp(s)–˜ un(s)+˜ qn(s) dsfor t∈]a,a], ˜ un(t)≤b–t+b t (b–s)  knp(s)–˜ un(s)+˜ qn(s) dsfor t∈[b,b[. () On account of ()and(), the sequence {un}+∞ n= is uniformly bounded and equicontinuous in ]a,b[. Thus, by virtue of the Arzelà-Ascoli lemma, we can assume without loss of generality that lim n→+∞˜ un(t)=v(t)uniformlyin]a,b[, () where v∈C(]a,b[) and, moreover, lim n→+∞˜ u na+b =c.() In view of ()itisclearthat ˜ un(t)=˜ una+b +t–a+b ˜ u na+b  +t a+b s a+b p(ξ)+  knp(ξ)–˜ un(ξ)+˜ qn(ξ)dξds for t∈]a,b[. Hence, on account of (), (), (), and (), we get v(t)=va+b +ct–a+b +t a+b s a+b  p(ξ)v(ξ)dξds for t∈]a,b[. Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 8 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Therefore, v∈AC loc(]a,b[) and vis a solution of equation (a). On the other hand, it follows from (), in view of (), (), and (), that v(t)≤(t–a)fort∈]a,a]andv(t)≤(b–t)fort∈[b,b[, and thus vis a solution of the problem (a), (). By virtue of ()and(), it is clear that there are n∈N,a∈]a,a], and b∈[b,b[ such that ˜ un(t)< fort∈[a,a]∪[b,b],n>n. Therefore, ˜ un[a,b]=forn>n. Taking now into account (), we get that v[a,b]= and, therefore, vis a nontrivial solution of the problem (a), (). By virtue of Theorem .,thereisλ=suchthat v(t)=λu(t)fort∈[a,b]. () Moreover, in view of the necessity part of Theorem . (with q(t)=  kn[p(t)]–˜ un(t)+˜ qn(t)), (), (), (), and (), we get b ap(s)–˜ un(s)v(s)ds =. () Let now α∈]a,b[andβ∈]α,b[ be arbitrary. Then, in view of (), we have lim n→+∞β αp(s)–˜ un(s)v(s)ds =β αp(s)–v (s)ds.() On account of (), (), and Proposition ., the function [p]–vis integrable on [a,b]. Taking into account (), we get α ap(s)–˜ un(s)v(s)ds ≤α ap(s)–v(s)ds and b βp(s)–˜ un(s)v(s)ds ≤b βp(s)–v(s)ds. Hence, () implies the inequality β αp(s)–˜ un(s)v(s)ds ≤α ap(s)–v(s)ds +b βp(s)–v(s)ds, which, together with (), results in β αp(s)–v (s)ds ≤α ap(s)–v(s)ds +b βp(s)–v(s)ds. Lomtatidze and Opluštil Boundary Value Problems 2014, 2014:59 Page 9 of 12 http://www.boundaryvalueproblems.com/content/2014/1/59 Since αand βwere arbitrary, we get from the latter inequality that b ap(s)–v (s)ds =. Taking now into account that v≡ , we get [p]–≡, i.e.,p(t)≥fort∈]a,b[. However, in this case the problem (a), () has no nontrivial solution, which contradicts the assumption of the lemma.  3Proofs Proof of Theorem . Let uand vbe any nontrivial solutions of (a). By virtue of Lemma . (with u≡vand q≡), we get lim t→a+u (t)v(t)–u(t)v (t)=. On the other hand, clearly u (t)v(t)–u(t)v (t)= fort∈]a,b[, and, therefore, u (t)v(t)–u(t)v (t)= fort∈[a,b]. () Choose t∈]a,b[suchthat u (t)=. It is clear that u(t)=sinceotherwiseu≡. Then it follows from ()that v (t)= and as above v(t)=.Putλ=u(t) v(t)and w(t)=u(t)–λv(t)fort∈[a,b]. Evidently, wis a solution of equation (a)andw(t) = . However, it follows from ()that w(t)=.Consequently,w≡andthusu≡λv. Proof of Theorem . Let ube a nontrivial solution of (a), ()whileube a solution of (), (). Put f(t)=u(t)u(t)–u(t)u (t)fort∈]a,b[. It is clear that f(t)=q(t)u(t)fort∈]a,b[.