Nonlinear Analysis64(2006)1528–1542/locate/na Positive solutions for singular non-linear third-order periodic boundary value problemsJifeng Chu a,∗,Zhongcheng Zhou ba Department of Applied Mathematics,College of Science,Hohai University,Nanjing210098,PR Chinab School of Mathematics and Finance,Southwest Normal University,Chongqing400715,PR ChinaReceived26May2005;accepted5July2005AbstractIn this paper,we are concerned with the problem of the existence of positive solutions for non-linear third-order periodic boundary value problemu + 3u=f(t,u),0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1,2.Here, ∈(0,1√3)is a positive constant and our non-linearity f(t,u)may be singular at u=0.The proof relies on a non-linear alternative of Leray–Schauder type and on Krasnoselskiifixed point theorem on compression and expansion of cones.᭧2005Elsevier Ltd.All rights reserved.MSC:34B15Keywords:Periodic boundary value problem;Positive solutions;Singular;Leray–Schauder alternative;Fixed point theorem in cones∗Corresponding author.E-mail address:jifengchu@(J.Chu).0362-546X/$-see front matter᭧2005Elsevier Ltd.All rights reserved.doi:10.1016/j.na.2005.07.005J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421529 1.IntroductionIn this paper,we establish the existence of positive solutions for the third-order periodic boundary value problemu + 3u=f(t,u),0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1,2,(1.1)where ∈(0,1√3)is a positive constant and the type of the non-linear term f(t,u)we aremainly interested in the case that f(t,u)has a singularity near u=0.In recent years,the non-linear periodic boundary value problems have been widely studied by many authors.For details,see[3–10,12–15,17–20]and the references therein.For the boundary value problem(1.1),we recall the following two results.In[10],by using Schauder fixed point theorem,together with perturbation technique,it is established the existence of at least one positive solution under the following conditions:(A1)f(t,u)is a non-negative function defined on[0,2 ]×(0,+∞)and f(t,u)is inte-grable on[0,2 ]for eachfixed u∈(0,+∞);(A2)f(t,u)is non-increasing in u>0for almost all t∈[0,2 ]andlim u→0+f(t,u)=+∞,limu→+∞f(t,u)=0hold uniformly for t∈[0,2 ].(A3)For eachfixed constant >0,the inequality 2f(s, )d s<+∞holds.In[17],the condition of monotonicity on f(t,u)in(A2)is removed and the existence of multiple positive solutions under suitable conditions on f(t,u)is obtained by using thefixed point index theory.Instead of Schauderfixed point theorem used in[10]andfixed point index theory in[17],our main tool in this paper is a non-linear alternative of Leray–Schauder type and afixedpoint theorem in cones due to Krasnoselskii.We remark that it is sufficient to prove that T:K∩(¯ 2\ 1)→K is continuous and completely continuous in Theorem2.2(Section 2).This point is essential and advantageous for singular problems.The method used inthis paper was applied successfully by Agarwal and O’Regan in[1,2]for the existence ofsingular Dirichlet,Conjugate,Focal and(n,p)problems and in[7,8]for the multiplicityof positive periodic solutions to superlinear repulsive singular equations.In this paper,wediscuss both positone case and semi-positone case.The remaining part of the paper is organized as follows.In Section2,some preliminaryresults will be given,including a famousfixed point theorem in cones due to Krasnoselskii.In Section3,we are devoted to the positone case,i.e.,f(t,u)is positive.In this case,weestablish the existence of at least two positive solutions.In Section4,the semi-positonecase,i.e.,f(t,u) −L for some L>0,is studied.In this case,we obtain the existence ofat least one positive solution.Some illustrating examples will be given.1530J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15422.PreliminariesIn this section,we present some preliminary results which will be needed in Sections3 and4.First,as in[17],we transform our problem into an integral equation.For any function u∈C[0,2 ],we define the operator(J u)(t)=2g(t,x)u(x)d x,(2.1)whereg(t,x)=⎧⎪⎨⎪⎩e− (t−x)1−e−2,0 x t 2 ,e− (2 +t−x)1−e−2,0 t x 2 .(2.2)By a direct calculation,we can easily obtain20g(t,x)d x=1.Now,we consider the problemu − u + 2u=f(t,J u),u(i)(0)=u(i)(2 ),i=0,1.(2.3) If u is a positive solution of problem(2.3),i.e.,u(t)>0for t∈[0,2 ],it is easy to verify that y(t)=(J u)(t)is a positive solution of problem(1.1).Therefore,we will concentrate our study on problem(2.3).Lemma2.1(Sun and Liu[17]).The boundary value problem(2.3)is equivalent to integral equationu(t)=2G(t,s)f(s,(J u)(s))d s,(2.4)whereG(t,s)=⎧⎪⎨⎪⎩2e( /2)(t−s)[sin√32(2 −t+s)+e− sin√32(t−s)]√3 (e +e− −2cos√3 ),0 s t 2 , 2e( /2)(2 +t−s)[sin√32(s−t)+e− sin√32(2 −s+t)]√3 (e +e− −2cos√3 ),0 t s 2 .(2.5)Moreover,for G(t,s),we have the estimates0<m=2sin(√3 )√3 (e +1)2G(t,s) 2√3 sin(√3 )=M∀s,t∈[0,2 ].(2.6)The technique we present is based on Krasnoselskii’sfixed point theorem in a cone, which we state here for the convenience of the reader.J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421531 Theorem2.2(Krasnosel’skii[11]).Let X be a Banach space and K(⊂X)be a cone. Assume that 1, 2are open subsets of X with0∈ 1,¯ 1⊂ 2,and letT:K∩(¯ 2\ 1)→Kbe a continuous and compact operator such that either(i) T u u ,u∈K∩j 1and T u u ,u∈K∩j 2;or(ii) T u u ,u∈K∩j 1and T u u ,u∈K∩j 2.Then T has afixed point in K∩(¯ 2\ 1).In applications below,we take X=C[0,2 ]with the supremum norm · and defineK=u∈X:u(t) 0for all t∈[0,2 ]and min0 t 2u(t) u,(2.7)where =m/M.One may readily verify that K is a cone in X.Now,suppose that f:[0,2 ]×R→[0,∞) is a continuous function.Define an operator T:X→X by(T u)(t)=2G(t,s)f(s,(J u)(s))d s(2.8) for u∈X and t∈[0,2 ].Lemma2.3.T is well defined and maps X into K.Moreover,T is continuous and completely continuous.Proof.It is easy to see that T is continuous and completely continuous since f:[0,2 ]×R→[0,∞)is a continuous function.Thus,we only need to show that T(X)⊂K.Let u∈X,then we havemin 0 t 2 (T u)(t)=min0 t 22G(t,s)F(s,(J u)(s))d s m2F(s,(J u)(s))d smax0 t 22G(t,s)F(s,(J u)(s))d s= T u ,i.e.,T u∈K.This implies that T(X)⊂K and the proof is completed.3.Positone caseIn this section,we establish the existence and multiplicity of positive solutions to(1.1) in the positone case.Recall that(1.1)is positone if f(t,u)>0for all t∈[0,2 ]and all u>0.In this paper,the notation 0means that (t) 0for all t∈[0,2 ]and (t)>0 for t in a subset of positive measure.1532J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–1542Theorem3.1.Suppose f(t,u)satisfies the following conditions:(F1)For each constant H>0,there exists a function H 0such that f(t,u) H(t) for all(t,u)∈[0,2 ]×(0,H].(F2)There exist continuous,non-negative functions g(u)and h(u)on(0,∞)such that f(t,u) g(u)+h(u)for all(t,u)∈[0,2 ]×(0,∞),and g(u)>0is non-increasing and h(u)/g(u)is non-decreasing in u∈(0,∞). (F3)There exists a positive number r such thatrg( r/ )(1+h(r/ )/g(r/ ))>1 2,where =m/M,m and M are given as in Section2.Then boundary value problem(1.1)has at least one positive solution with0< u <r/ . Proof.We only need to show problem(2.3)has at least one positive solution with0< u <r. If this is true,then(J u)(t)is a positive solution of(1.1)with0< J u <r/ since0< J u =max0 t 2 2g(t,x)u(x)d x u2g(t,x)d x=1u <r.To do so,we will use the Leray–Schauder alternative principle,together with a truncation technique.Let N0={n0,n0+1,...},where n0∈{1,2,...}is chosen such that1 2gr1+h(r/ )g(r/ )+1n0<r.(3.1)See(F3).Fix n∈N0,the idea isfirst to show that the“modified”problemu − u + 2u=f n(t,J u)+ 2n,0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1(3.2) has a solution u n∈C2[0,2 ]with0< u n r,wheref n(t,x)=f(t,1/n ),x 1/n ,f(t,x),x 1/n .(3.3)For ∈(0,1),consider the family of problemsu − u + 2u= f n(t,J u)+ 2n,0 t 2 ,u(i)(0)=u(i)(2 ),i=0,1.(3.4)J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421533 Let u be a solution of problem(3.4),then it follows from Lemma2.1thatu(t)=20G(t,s)f n(s,(J u)(s))+2nd s= 2G(t,s)f n(s,(J u)(s))d s+1n,(3.5)where G(t,s)is the Green function given by(2.5).First,it is claimed that any solution u of problem(3.5)verifiesu =max0 t 2|u(t)|=r(3.6)independently of .Otherwise,assume that u is a solution of(3.5)for some ∈(0,1)such that u =r. Note that f n(t,(J u)(s)) 0.By Lemma2.3,for all t∈[0,2 ],u(t) 1/n andu(t)−1n=2G(t,s)f n(s,(J u)(s))d s m2f n(s,(J u)(s))d smMmax0 t 22G(t,s)f n(s,(J u)(s))d s=u−1n.Thus,u(t)u−1n+1nu −1n+1n> r.Therefore,we have from condition(F2),for all t∈[0,2 ],u(t)=20G(t,s)f n(s,(J u)(s))d s+1n= 2G(t,s)f(s,(J u)(s))d s+1n2G(t,s)f(s,(J u)(s))d s+12G(t,s)g((J u)(s))1+h((J u)(s))g((J u)(s))d s+1ngr1+hrgr2G(t,s)d s+1n0=1gr1+hrgr+1(3.7)since r/ (J u)(t) r/ . Therefore,r= u 12gr1+hrgr+1n0.This is a contradiction to the choice of n0and the claim is proved.1534J.Chu,Z.Zhou /Nonlinear Analysis 64(2006)1528–1542Define the operator T :X →X by(T u)(t)= 2 0G(t,s)f n (s,(J u)(s))d s +1n ,(3.8)then (3.5)is equivalent to the fixed point problem u =(1− )1n+ T u .(3.9)It is easy to see that T :X →X is continuous and completely continuous.Define the set U ={u ∈C [0,2 ]: u <r }.Since (3.6)holds,the non-linear alternative of Leray–Schauder [16]guarantees that T has a fixed point,denoted by u n ,in U ,i.e.,problem (3.2)has a solution u n with u n <r .Since u n satisfies u n =T (u n ),u n (t) 1/n for all t ∈[0,2 ]and u n is actually a positive solution of problem (3.2).Next,we claim that these solutions u n have a uniform positive lower bound,i.e.,there exists a constant >0,independent of n ∈N 0,such thatmin t ∈[0,2 ]u n (t) (3.10)for all n ∈N 0.To see this,we know from (F 1)that there exists a function r 0such that f (t,J u) r (t)for (t,u)∈[0,2 ]×(0,r ].Now,let u r (t)be the unique solution to the problem u − u + 2u = r (t),0 t 2 ,u (i)(0)=u (i)(2 ),i =0,1,(3.11)thenu r (t)= 20G(t,s) r (s)d s m r 1>0,where · 1denotes the usual L 1-norm over [0,2 ].So,we haveu n (t)= 2 0G(t,s)f n (s,(J u n )(s))d s +1/n= 2 0G(t,s)f (s,(J u n )(s))d s +1/n 2 0G(t,s) r (s)d s +1/n =u r (t)+1/n m r 1=: .In order to pass the solutions u n of the truncation problems (3.2)to that of the original problem (2.3),we need the following fact:u n K(3.12)for some constant K >0(independent of n ∈N 0)and for all n ∈N 0.J.Chu,Z.Zhou/Nonlinear Analysis64(2006)1528–15421535 In fact,there exists t0∈[0,2 ]such that u n(t0)=0since u n(0)=u n(2 ).Integrate(3.2) in[0,2 ],it is obtained2 2u n(t)d t=2f n(t,(J u n)(t))+2nd t,(3.13)since u n(0)=u n(2 ). Thenu n =max0 t 2 |u n(t)|=max0 t 2tt0u n(s)d s=max0 t 2tt0[f n(s,(J u n)(s))+ 2/n− 2u n(s)+ u n(s)]d s2[f n(s,(J u n)(s))+ 2/n]d s+ 22u n(s)d s+ |u n(t)−u n(t0)|=2 2 2u n(s)d s+ |u n(t)−u n(t0)|<4 2r+2 r=:K.As u n <r,by(3.12),{u n}n∈N0is a bounded and equi-continuous family on[0,2 ].Bythe Arzela–Ascoli Theorem,{u n}n∈N0has a subsequence,{u nk}k∈N,converging uniformlyon[0,2 ]to a function u∈X.From the fact u n <r and(3.10),u satisfies u(t) rfor all t∈[0,2 ].Moreover,u nksatisfies the integral equationu nk (t)=2G(t,s)f(s,(J u nk)(s))d s+1/n k.Letting k→∞,we arrive atu(t)=2G(t,s)f(s,(J u)(s))d s,where the uniform continuity of f(t,J u)on[0,2 ]×[ ,r]is used.Therefore,u is a positive solution to problem(2.3).Finally,it is not difficult to show that u <r,by noting that if u =r,the argument similar to the proof of thefirst claim(3.6)will yield a contradiction.Corollary3.2.Let the non-linearity in(1.1)bef(t,u)=b(t)u− + c(t)u +e(t),0 t 2 ,(3.14)where >0, 0,b(t),c(t),e(t)∈X are non-negative functions and b(t)>0for all t∈[0,2 ],and >0is a positive parameter.Then(i)if <1,problem(1.1)has at least one positive solution for each >0;and(ii)if 1,problem(1.1)has at least one positive solution for each0< < ∗,where ∗is some positive constant.1536J.Chu,Z.Zhou /Nonlinear Analysis 64(2006)1528–1542Proof.Remark that assumption (F 1)is fulfilled by H (t)=H − min t b(t).To verify (F 2),one may simply takeg(u)=b 0u − ,h(u)= c 0u +e 0,whereb 0=max t b(t)>0,c 0=max t c(t) 0,e 0=max te(t) 0.Now,the existence condition (F 3)becomes< r +1 +2−b 0 + −e 0r c 0r +for some r >0.So,problem (1.1)has at least one positive solution for0< < ∗:=sup r>0r +1 +2−b 0 + −e 0r c 0r + .Note that ∗=∞if <1and ∗<∞if 1.We have the desired results.Next,we will find another positive solution to problem (1.1)by using Theorem 2.2for certain non-linearities.Theorem 3.3.Suppose that (F 1)–(F 3)are satisfied.Furthermore ,assume that(F 4)There exist continuous ,non-negative functions g 1(u)and h 1(u)on (0,∞)such thatf (t,u)g 1(u)+h 1(u)for all (t,x)∈[0,2 ]×(0,∞)and g 1(u)>0is non-increasing and h 1(u)/g 1(u)is non-decreasing in u ∈(0,∞);and(F 5)There exists a positive number R >r such that Rg 1(R/ ) 1+h 1( R/ )g 1( R/ ) 1,where is as in Section 2.Then problem (1.1)has another positive solution ˜u with r/ < ˜u R/ .Proof.Let X =C [0,2 ]and K be the cone in X defined by (2.7).Let1={u ∈X : u <r }, 2={u ∈X : u <R }and define the operator T :K ∩(¯2\ 1)→K by (T u)(t)= 2 0G(t,s)f (s,(J u)(s))d s,0 t 2 .(3.15)J.Chu,Z.Zhou /Nonlinear Analysis 64(2006)1528–15421537Since r u R for u ∈K ∩(¯ 2\ 1),thus 0< r/ (J u)(t) R/ .Since f :[0,2 ]×[ r/ ,R/ ]→[0,∞)is continuous,it follows from Lemma 2.3that the operator T :K ∩(¯ 2\ 1)→K is well defined and is continuous and completely continuous.First,we showT u < u for u ∈K ∩j 1.(3.16)In fact,if u ∈K ∩j 1,then u =r and r u(t) r for 0 t 2 .So we have(T u)(t)= 2 0G(t,s)f (s,(J u)(s))d s2 0G(t,s)g((J u)(s)) 1+h((J u)(s))g((J u)(s)) d s g r 1+h r g r 2 0G(t,s)d s =1 2g r 1+h r g r<r = u .This implies T u < u ,i.e.,(3.16)holds.Next,we show thatT u u for u ∈K ∩j 2.(3.17)To see this,let u ∈K ∩j 2.Then u =R and R u(t) R .As a result,it follows from (F 4)and (F 5)that,for 0 t 2 ,(T u)(t)= 2 0G(t,s)f (s,(J u)(s))d s2 0G(t,s)g 1((J u)(s)) 1+h 1((J u)(s))g 1((J u)(s)) d s 2 0G(t,s)g 1 R 1+h 1( R/ )g 1( R/ ) d s =1 2g 1 R 1+h 1( R/ )g 1( R/ )R = u .Now,(3.16),(3.17)and Theorem 2.2guarantee that T has a fixed point u ∈K ∩(¯ 2\ 1)with r u R .Clearly,˜u =J (u)is a positive solution of (1.1)and actually satisfiesr/ < ˜u R/ .Let us consider again the example (3.14)in Corollary 3.2for the superlinear case,i.e., >1.We assume also that c(t)>0for all t ∈[0,2 ].To verify (F 4),one may simply takeg 1(u)=b 1u − ,h 1(u)= c 1u +e 1,whereb 1=min t b(t)>0,c 1=min t c(t)>0,e 1=min t e(t) 0.Now,the existence condition (F 5)becomesR +1 +2−b 1 + −e 1 R c 1 + R + .(3.18)Since >1,the right-hand side goes to 0as R →+∞.Thus,for any given 0< < ∗,where ∗is as in Corollary 3.2,it is always possible to find such R ?r that (3.18)is satisfied.Thus,problem (1.1)has an additional positive solution ˜u .Corollary 3.4.Assume in (3.14)that >1and b(t)>0,c(t)>0for all t ∈[0,2 ].Then ,for each with 0< < ∗,with ∗given as in Corollary 3.2,the corresponding problem value problem (1.1)has at least two different positive solutions .4.Semi-positone caseIn this section,we establish the existence of at least one positive solutions of (1.1)in the semi-positone case.By the semi-positone case of (1.1),we mean that f (t,u)may change sign and satisfies(G 1)There exists a constant L >0such that F (t,u):=f (t,u)+L 0for all (t,u)∈[0,2 ]×(0,∞).Theorem 4.1.Suppose that f (t,u)satisfies (G 1).In addition ,we assume that the following conditions are satisfied :(G 2)There exist continuous non-negative functions g(u)and h(u)on (0,∞)such thatF (t,u) g(u)+h(u)for all (t,u)∈[0,2 ]×(0,∞)and g(u)>0is non-increasing and h(u)/g(u)is non-decreasing in u ∈(0,∞);(G 3)There exist continuous ,non-negative functions g 1(u)and h 1(u)on (0,∞)such thatF (t,u) g 1(u)+h 1(u)for all (t,u)∈[0,2 ]×(0,∞)and g 1(u)>0is non-increasing and h 1(u)/g 1(u)is non-decreasing in u ∈(0,∞);(G 4)There exists r > / such that r gr − 1+h(r/ − )g(r/ − ) >1 2,where =L/ 3, =m/M is the same as in Section 2.(G 5)There exists a positive number R >r such that R g 1 R − 1+h 1( R/ − )1 1 2.Then problem (1.1)has a solution u ∈C 2[0,2 ]with u(t)>0for t ∈[0,2 ]and r/ < u + <R/ .Proof.We only need to show that problem (2.3)has a positive solution y(t)with r < y + R .To do so,we will show that problemu − u + 2u =F (t,(J u)(t)− ),0 t 2 ,u(0)=u(2 ),u (0)=u (2 )(4.1)has a positive solution u ∈C 2[0,2 ]with (J u)(t)> for t ∈[0,2 ]and r < u R .Let X =C [0,2 ]and K be a cone in X defined by (2.7).Let1={u ∈X : u <r }, 2={u ∈X : u <R }and define the operator T :K ∩(¯2\ 1)→K by (T u)(t)=2 0G(t,s)F (s,(J u)(s)− )d s,0 t 2 ,(4.2)where G(t,s)is the Green function given by (2.5).Since r u R for u ∈K ∩(¯ 2\ 1),thus 0< r/ − (J u)(s)− R/ − .Since F :[0,2 ]×[ r/ − ,R/ − ]→[0,∞)is continuous,it follows from Lemma 2.3that the operator T :K ∩(¯ 2\ 1)→K is well defined and is continuous and completely continuous.First,we showT u < u for u ∈K ∩j 1.(4.3)In fact,if u ∈K ∩j 1,then u =r and (J u)(t) r/ > for 0 t 2 .So we have from conditions (G 2)and (G 4)that,for 0 t 2 ,(T u)(t)= 2 0G(t,s)F (s,(J u)(s)− )d s2 0G(t,s)g((J u)(s)− ) 1+h((J u)(s)− )g((J u)(s)− ) d s 2 0G(t,s)g r − 1+h(r/ − )g(r/ − ) d s =1 2g r − 1+h(r/ − )g(r/ − )<r = usince r/ − (J u)(s)− r/ − .This implies T u < u ,i.e.,(4.3)holds.Next,we showT u u for u ∈K ∩j 2.(4.4)To see this,let u ∈K ∩j 2,then u =R and (J u)(t) R/ > for 0 t 2 .As a result,it follows from (G 3)and (G 5)that,for 0 t 2 ,(T u)(t)= 20G(t,s)F (s,(J u)(s)− )d s2 0G(t,s)g 1((J u)(s)− ) 1+h 1((J u)(s)− )g 1((J u)(s)− ) d s 2 0G(t,s)g 1 R − 1+h 1( R/ − )g 1( R/ − ) d s =1g 1 R − 1+h 1( R/ − )1 R = u .This implies T u u ,i.e.,(4.4)holds.Now,(4.3),(4.4)and Theorem 2.2guarantee that T has a fixed point u ∈K ∩(¯ 2\ 1)with r u R .Note u =r by (4.3).Clearly,this u is a positive solution of (4.1).Let y(t)=u(t)− ,then y(t)is a positive solution of (2.3)and r < y + R sincey (t)− y (t)+ 2y(t)=u (t)− u (t)+ 2(u(t)− )=F (t,(J u)(t)− )− 3=F (t,(J u)(t)− )−L=f (t,(J u)(t)− )=f (t,(Jy)(t))for all t ∈[0,2 ].Corollary 4.2.Let the non-linearity in (1.1)be (3.14),where 1, >1,b(t),c(t):[0,2 ]→(0,∞)are continuous functions ,e(t)∈C [0,2 ]and >0is a positive pa-rameter.Then (1.1)has at least one positive solution for each 0< < ∗,where ∗is some positive constant .Proof.We will apply Theorem 4.1with L =e 0=max 0 t 2 |e(t)|andg(u)=b 0u − ,h(u)= c 0u +2e 0,g 1(u)=b 1u − ,h 1(u)= c 1u .Here,b 1=min t b(t)>0,c 1=min t c(t)>0,b 0=max t b(t)>0,c 0=max tc(t)>0.Then conditions (G 1)–(G 3)are satisfied.The existence condition (G 4)becomes< 2r( r/ − ) −b 0−2e 0(r/ − )c 0(r/ − ) +for some r> / .Let∗=supr> / 2r( r/ − ) −b0−2e0(r/ − )c0(r/ − ) +.Note that ∗<∞since >1.Another existence condition(G5)becomes2R(R/ − ) −b1c1( R/ − ) +.(4.5)Since >1,the right-hand side of(4.5)goes to0as R→+∞.Thus,for any given 0< < ∗,it is always possible tofind such R?r that(4.5)is satisfied.Thus,problem (1.1)has at least one positive solution.5.AcknowledgmentThe authors express their thanks to the referee for his valuable suggestions. 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