The eigenvalue Characterization for the constant Sign Green's Functions of (k,n−k) problems
TThe eigenvalue Characterization for theconstant Sign Green’s Functions of ( k, n − k ) problems. Alberto Cabada ∗ and Lorena Saavedra † Departmento de An´alise Matem´atica,Facultade de Matem´aticas,Universidade de Santiago de Compostela,Santiago de Compostela, Galicia, [email protected], [email protected]
Abstract
This paper is devoted to the study of the sign of the Green’s func-tion related to a general linear n th -order operator, depending on a realparameter, T n [ M ], coupled with the ( k, n − k ) boundary value conditions.If operator T n [ ¯ M ] is disconjugate for a given ¯ M , we describe the in-terval of values on the real parameter M for which the Green’s functionhas constant sign.One of the extremes of the interval is given by the first eigenvalue ofoperator T n [ ¯ M ] satisfying ( k, n − k ) conditions.The other extreme is related to the minimum (maximum) of the firsteigenvalues of ( k − , n − k + 1) and ( k + 1 , n − k −
1) problems.Moreover if n − k is even (odd) the Green’s function cannot be non-positive (non-negative).To illustrate the applicability of the obtained results, we calculatethe parameter intervals of constant sign Green’s functions for particularoperators. Our method avoids the necessity of calculating the expressionof the Green’s function.We finalize the paper by presenting a particular equation in which it isshown that the disconjugation hypothesis on operator T n [ ¯ M ] for a given¯ M cannot be eliminated. Key words: n order boundary value problem; Green’s functions; discon-jugation; Maximum Principles; Spectral Theory. AMS Subject Classification: ∗ Partially supported by Ministerio de Educaci´on y Ciencia, Spain and FEDER, projectsMTM2010-15314 and MTM2013-43014-P. † Supported by Plan I2C scholarship, Conselleria de Educaci´on, Cultura e O.U., Xunta deGalicia, and FPU scholarship, Ministerio de Educaci´on, Cultura y Deporte, Spain. a r X i v : . [ m a t h . C A ] S e p Introduction
It is very well known that the validity of the method of lower and upper solu-tions, coupled with the monotone iterative techniques [13, 21], is equivalent tothe constant sign of the Green’s function related to the linear part of the stud-ied problem [1, 2]. Moreover, by means of the celebrated Krasnosel’ski˘ı contrac-tion/expansion fixed point theorem [19], nonexistence, existence and multiplicityresults are derived from the construction of suitable cones on Banach spaces.Such construction follows by using adequate properties of the Green’s function,one of them is its constant sign [3, 17, 18, 26]. Recently, the combination of thetwo previous methods has been proved as a useful tool to ensure the existenceof solution [4, 5, 12, 16, 24].Having in mind the power of this constant sign property, we will describethe interval of parameters for which the Green’s function related to the generallinear n th -order equation T n [ M ] u ( t ) ≡ u ( n ) ( t )+ a ( t ) u ( n − ( t )+ · · · + a n − ( t ) u (cid:48) ( t )+( a n ( t )+ M ) u ( t ) = 0(1) t ∈ I ≡ [ a, b ], coupled with the so-called ( k, n − k ) two point boundary valueconditions: u ( a ) = u (cid:48) ( a ) = · · · = u ( k − ( a ) = u ( b ) = u (cid:48) ( b ) = · · · = u ( n − k − ( b ) = 0 , (2)1 ≤ k ≤ n −
2, has constant sign on its square of definition I × I .The main hypothesis consists on assuming that there is a real parameter ¯ M for which operator T n [ ¯ M ] is disconjugate on I .An exhaustive study of the general theory and the fundamental propertiesof the disconjugacy are compiled in the classical book of Coppel [11]. Differentsufficient criteria to ensure the disconjugacy character of the linear operator T n [0] has been developed in the literature, we refer the classical references [27,28]. Sufficient conditions for particular cases have been obtained in [15, 20, 25]and, more recently, in [14]. We mention that operator u ( n ) ( t ) + a ( t ) u ( n − ( t )is always disconjugate in I , see [11] for details, in particular the results herepresented are valid for operator u ( n ) ( t ) + M u ( t ).As it has been shown in [11], the disconjugacy character implies the constantsign of the Green’s function g M related to problem (1)–(2). However, as we willsee along the paper, the reciprocal property is not true in general: there arereal parameters M for which the Green’s function has constant sign but theequation (1) is not disconjugate. In other words, the disconjugacy characteris only a sufficient condition in order to ensure the constant sign of a Green’sfunction related to problem (1)–(2).In fact, from the disconjugacy character of operator T n [ ¯ M ] in I , it is shownin [11] that the Green’s function g M satisfies a suitable condition, stronger thanits constant sign. Such condition fulfills the one introduced in [1, Section 1.8].So, following the results given in that reference we conclude that the set ofparameters M for which g M has constant sign is an interval H T . Moreover if n − k is even then the maximum of H T is the opposed to the biggest negativeeigenvalue of problem (1)–(2), when n − k is odd the minimum of H T is theopposed to the least positive eigenvalue of such problem.Thus, the difficulty remains in the characterization of the other extreme ofthe interval H T . In this case, as it is shown in [1, Section 1.8], such extreme is2ot an eigenvalue of the considered problem, so to attain its exact value is notimmediate. In practical situations it is necessary to obtain the expression of theGreen’s function, which is, in general, a difficult matter to deal with. We pointout that this problem is not restricted to the ( k, n − k ) boundary conditions,the difficulty in obtaining the non eigenvalue extreme remains true for anykind of linear conditions [7, 22]. In [6], provided operator T n [ M ] has constantcoefficients, it has been developed a computer algorithm that calculates theexact expression of a Green’s function coupled with two-point boundary valueconditions. However, such expression is often too complicated to manage, andto describe the interval H T is really very difficult in practical situations. In factthere is not a direct method of construction for non constant coefficients.We mention that the disconjugacy theory has been used in [23] to obtainthe values for which the third order operators u (cid:48)(cid:48)(cid:48) + M u ( i ) , i = 0 , ,
2, coupledwith conditions (1 ,
2) and (2 ,
1) have constant sign Green’s function. Similarprocedure has been done in [8] for the fourth order operator u (4) + M u , coupledwith conditions (2 ,
2) and, more recently, in [9] with conditions (1 ,
3) and (3 , n th - order operator T n [ M ].It is for this that we make in this work a general characterization of theregular extreme of the interval of constant sign H T by means of the spectraltheory. We will show that it is an eigenvalue of the same operator T n [ M ] butrelated to different two-point boundary value conditions. In fact, if n − k iseven, it will be the minimum of the two least positive eigenvalues related toconditions ( k − , n − k + 1) and ( k + 1 , n − k − n − k is odd. So, wemake a general characterization for the general operator T n [ M ] and we avoidthe necessity of calculate the Green’s function and to study its sign dependenceon the real parameter M .We note that if operator T n [ M ] has constant coefficients, to obtain the cor-responding eigenvalues we only must to calculate the determinant of the matrixof coefficients of a linear homogeneous algebraic system. Numerical methodsare also valid for the non-constant case.It is important to mention that, as consequence of the obtained results,denoting by g M the Green’s function related to problem (1)–(2), we concludethat ( − n − k g M ( t, s ) cannot be negative on I × I for all M ∈ R .The paper is scheduled as follows: in a preliminary section 2 we introducethe fundamental concepts that are needed in the development of the paper.Next section is devoted to the proof of the main result in which the regularextreme is obtained via spectral theory. In section 4 some particular cases areconsidered where it is shown the applicability of the obtained results. In lastsection is introduced an example that shows that the disconjugacy hypothesison the main result cannot be eliminated.3 Preliminaries
In this section, for the convenience of the reader, we introduce the fundamentaltools in the theory of disconjugacy and Green’s functions that will be used inthe development of further sections.
Definition 2.1.
Let a k ∈ C n − k ( I ) for k = 1 , . . . , n . The n th -order lineardifferential equation (1) is said to be disconjugate on an interval I if every nontrivial solution has less than n zeros on I , multiple zeros being counted accordingto their multiplicity. Definition 2.2.
The functions u , . . . , u n ∈ C n ( I ) are said to form a Markovsystem on the interval I if the n Wronskians W ( u , . . . , u k ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u · · · u k ... · · · ... u ( k − · · · u ( k − k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , k = 1 , . . . , n , (3) are positive throughout I . The following result about this concept is collected on [11, Chapter 3].
Theorem 2.3.
The linear differential equation (1) has a Markov fundamentalsystem of solutions on the compact interval I if, and only if, it is disconjugateon I . In order to introduce the concept of Green’s function related to the n th -order scalar problem (1)-(2), we consider the following equivalent first ordervectorial problem: x (cid:48) ( t ) = A ( t ) x ( t ) , t ∈ I , B x ( a ) + C x ( b ) = 0 , (4)with x ( t ) ∈ R n , A ( t ) , B, C ∈ M n × n , defined by x ( t ) = u ( t ) u (cid:48) ( t )... u ( n − ( t ) , A ( t ) = I n − − ( a n ( t ) + M ) − a n − ( t ) · · · − a ( t ) ,B = (cid:18) I k
00 0 (cid:19) , C = (cid:18) I n − k (cid:19) . (5)Here I j , j = 1 , . . . , n −
1, is the j × j identity matrix. Definition 2.4.
We say that G is a Green’s function for problem (4) if itsatisfies the following properties: (G1) G ≡ ( G i,j ) i,j ∈ , . . . , n : ( I × I ) \ { ( t, t ) , t ∈ I } → M n × n . (G2) G is a C function on the triangles (cid:8) ( t, s ) ∈ R , a ≤ s < t ≤ b (cid:9) and (cid:8) ( t, s ) ∈ R , a ≤ t < s ≤ b (cid:9) . (G3) For all i (cid:54) = j the scalar functions G i,j have a continuous extension to I × I . For all s ∈ ( a, b ) , the following equality holds: ∂∂t G ( t, s ) = A ( t ) G ( t, s ) , for all t ∈ I \ { s } . (G5) For all s ∈ ( a, b ) and i ∈ { , . . . , n } , the following equalities are fulfilled: lim s → t + G i,i ( s, t ) = lim s → t − G i,i ( t, s ) = 1 + lim s → t + G i,i ( t, s ) = 1 + lim s → t − G i,i ( s, t ) . (G6) For all s ∈ ( a, b ) , the function t → G ( t, s ) satisfies the boundary conditions B G ( a, s ) + C G ( b, s ) = 0 . It is very well known that Green’s function related to this problem followsthe following expression [1, Section 1.4] G ( t, s ) = g ( t, s ) g ( t, s ) · · · g n − ( t, s ) g M ( t, s ) ∂∂t g ( t, s ) ∂∂t g ( t, s ) · · · ∂∂t g n − ( t, s ) ∂∂t g M ( t, s )... ... · · · ... ... ∂ n − ∂t n − g ( t, s ) ∂ n − ∂t n − g ( t, s ) · · · ∂ n − ∂t n − g n − ( t, s ) ∂ n − ∂t n − g M ( t, s ) , (6)where g M ( t, s ) is the scalar Green’s function related to problem (1)-(2).Using Definition 2.4 we can deduce the properties fulfilled by g M ( t, s ). Inparticular, g M ∈ C n − ( I ) and it satisfies, as a function of t , the two-pointboundary value conditions (2).We also mention a result which appears on [11, Chapter 3, Section 6] andthat connects the disconjugacy and the sign of the Green’s function related tothe problem (1)-(2). Lemma 2.5.
If the linear differential equation (1) is disconjugate and g M ( t, s ) is the Green’s function related to the problem (1)-(2), hence g M ( t, s ) p ( t ) ≥ , ( t, s ) ∈ I × I ,g M ( t, s ) p ( t ) > , ( t, s ) ∈ [ a, b ] × ( a, b ) . where p ( t ) = ( t − a ) k ( t − b ) n − k . The adjoint of the operator T n [ M ], is given by the following expression, seefor details [1, Section 1.4] or [11, Chapter 3, Section 5], T ∗ n [ M ] v ( t ) ≡ ( − n v ( n ) ( t ) + n − (cid:88) j =1 ( − j ( a n − j v ) ( j ) ( t ) + ( a n ( t ) + M ) v ( t ) , (7)and its domain of definition is D ( T ∗ n [ M ]) = v ∈ C n ( I ) | n (cid:88) j =1 j − (cid:88) i =0 ( − j − − i ( a n − j v ) ( j − − i ) ( b ) u ( i ) ( b ) (8)= n (cid:88) j =1 j − (cid:88) i =0 ( − j − − i ( a n − j v ) ( j − − i ) ( a ) u ( i ) ( a ) (with a = 1) , ∀ u ∈ D ( T n [ M ]) .
5n our case, because of boundary conditions (2), we can express the domainof the operator T n [ M ], D ( T n [ M ]), as X k = (cid:110) u ∈ C n ( I ) | u ( a ) = · · · = u ( k − ( a ) = u ( b ) = · · · = u ( n − k − ( b ) = 0 (cid:111) , so we can replace expression (8) with D ( T ∗ n [ M ]) = v ∈ C n ( I ) | n (cid:88) j = n − k +1 j − (cid:88) i = n − k ( − j − − i ( a n − j v ) ( j − − i ) ( b ) u ( i ) ( b )= n (cid:88) j = k +1 j − (cid:88) i = k ( − j − − i ( a n − j v ) ( j − − i ) ( a ) u ( i ) ( a ) (with a = 1) , ∀ u ∈ C n ( I ) . In order to simplify the previous expression, we choose a function u ∈ C n ( I )satisfying u ( σ ) ( a ) = 0 , σ = 1 , . . . , n − ,u ( µ ) ( b ) = 0 , µ = 1 , . . . , n − ,u ( n − ( b ) = 1 . Realizing that a = 1, we conclude that every function v ∈ D ( T ∗ n [ M ]) mustsatisfy v ( b ) = 0.Moreover, if we now choose a function in C n ( I ) that satisfies u ( σ ) ( a ) = 0 , σ = 1 , . . . , n − ,u ( µ ) ( b ) = 0 , µ = 1 , . . . , n − , µ (cid:54) = n − u ( n − ( b ) = 1 , we conclude that any function v ∈ D ( T ∗ n [ M ]) has to satisfy − v (cid:48) ( b ) + a ( b ) v ( b ) = 0 . Since a ∈ C n − ( I ) and v ( b ) = 0, we conclude that v (cid:48) ( b ) = 0.Repeating this process we achieve that the domain of the adjoint operatoris given by D ( T ∗ n [ M ]) = X n − k . (9)The next result appears in [11, Chapter 3, Theorem 9] Theorem 2.6.
The equation (1) is disconjugate on an interval I if, and onlyif, the adjoint equation, T ∗ n [ M ] y ( t ) = 0 is disconjugate on I . We denote g ∗ M ( t, s ) as the Green function of the adjoint operator, T ∗ n [ M ].In [1, Section 1.4] it is proved the following relationship g ∗ M ( t, s ) = g M ( s, t ) . (10)Defining now the following operator (cid:98) T n [( − n M ] := ( − n T ∗ n [ M ] , (11)6e deduce, from the previous expression, that (cid:98) g ( − n M ( t, s ) = ( − n g ∗ M ( t, s ) = ( − n g M ( s, t ) . (12)Obviously, Theorem 2.6 remains true for operator (cid:98) T n [( − n M ]. Definition 2.7.
Operator T n [ M ] is said to be inverse positive (inverse negative)on X k if every function u ∈ X k such that T n [ M ] u ≥ in I , must verify u ≥ ( u ≤ ) on I . Next results are proved in [1, Section 1.6, Section 1.8].
Theorem 2.8.
Operator T n [ M ] is inverse positive (inverse negative) on X k if,and only if, Green’s function related to problem (1)-(2) is non-negative (non-positive) on its square of definition. Theorem 2.9.
Let M , M ∈ R and suppose that operators T n [ M j ] , j = 1 , ,are invertible in X k . Let g j , j = 1 , , be Green’s functions related to operators T n [ M j ] and suppose that both functions have the same constant sign on I × I .Then, if M < M , it is satisfied that g ≤ g on I × I . In the sequel, we introduce two conditions on g M ( t, s ) that will be used alongthe paper.( P g ) Suppose that there is a continuous function φ ( t ) > t ∈ ( a, b ) and k , k ∈ L ( I ), such that 0 < k ( s ) < k ( s ) for a.e. s ∈ I , satisfying φ ( t ) k ( s ) ≤ g M ( t, s ) ≤ φ ( t ) k ( s ) , for a. e. ( t, s ) ∈ I × I . ( N g ) Suppose that there is a continuous function φ ( t ) > t ∈ ( a, b ) and k , k ∈ L ( I ), such that k ( s ) < k ( s ) < s ∈ I , satisfying φ ( t ) k ( s ) ≤ g M ( t, s ) ≤ φ ( t ) k ( s ) , for a. e. ( t, s ) ∈ I × I .
Finally, we introduce the following sets, which are going to particularize H T , P T = { M ∈ R , | g M ( t, s ) ≥ ∀ ( t, s ) ∈ I × I } ,N T = { M ∈ R , | g M ( t, s ) ≤ ∀ ( t, s ) ∈ I × I } . Next results describe the structure of the two previous parameter’s set.
Theorem 2.10. [1, Lemma 1.8.33] Let ¯ M ∈ R be fixed. Suppose that operator T n [ ¯ M ] is invertible on X k , its related Green’s function is non-negative on I × I ,it satisfies condition ( P g ), and the set P T is bounded from above. Then P T =( ¯ M − λ , ¯ M − ¯ µ ] , with λ > the least positive eigenvalue of operator T n [ ¯ M ] in X k and ¯ µ ≤ such that T n [ ¯ M − ¯ µ ] is invertible in X k and the related non-negative Green’s function g ¯ M − ¯ µ vanishes at some points on the square I × I . Theorem 2.11. [1, Lemma 1.8.25] Let ¯ M ∈ R be fixed. Suppose that operator T n [ ¯ M ] is invertible in X k , its related Green’s function is non-positive on I × I ,it satisfies condition ( N g ), and the set N T is bounded from below. Then N T =[ ¯ M − ¯ µ, ¯ M − λ ) , with λ < the biggest negative eigenvalue of operator T n [ ¯ M ] in X k and ¯ µ ≥ such that T n [ ¯ M − ¯ µ ] is invertible in X k and the related non-positive Green’s function g ¯ M − ¯ µ vanishes at some points on the square I × I . Main Result
This section is devoted to prove the eigenvalue characterization of the sets P T and N T . Such result is enunciated on the following Theorem Theorem 3.1.
Let ¯ M ∈ R be such that equation T n [ ¯ M ] u ( t ) = 0 is disconjugateon I . Then the two following properties are fulfilled:If n − k is even then the operator T n [ M ] is inverse positive on X k if, andonly if, M ∈ ( ¯ M − λ , ¯ M − λ ] , where: • λ > is the least positive eigenvalue of operator T n [ ¯ M ] in X k . • λ < is the maximum of: – λ (cid:48) < , the biggest negative eigenvalue of operator T n [ ¯ M ] in X k − . – λ (cid:48)(cid:48) < , the biggest negative eigenvalue of operator T n [ ¯ M ] in X k +1 .If n − k is odd then the operator T n [ M ] is inverse negative on X k if, andonly if, M ∈ [ ¯ M − λ , ¯ M − λ ) , where: • λ < is the biggest negative eigenvalue of operator T n [ ¯ M ] in X k . • λ > is the minimum of: – λ (cid:48) > , the least positive eigenvalue of operator T n [ ¯ M ] in X k − . – λ (cid:48)(cid:48) > , the least positive eigenvalue of operator T n [ ¯ M ] in X k +1 . In order to prove this result, we separate the proof in several subsections. T n [ ¯ M ] We are interested into put operator T n [ ¯ M ] as a composition of suitable operatorsof order h ≤ n . Such expression allow us to control the values of such operatorsat the extremes of the interval a and b .We recall the following result proved in [11, Chapter 3] Theorem 3.2.
The linear differential equation (1) has a Markov system ofsolutions if, and only if, the operator T n [ M ] has a representation T n [ M ] y ≡ v v . . . v n ddt (cid:18) v n ddt (cid:18) · · · ddt (cid:18) v ddt (cid:18) v y (cid:19)(cid:19)(cid:19)(cid:19) , (13) where v k > on I and v k ∈ C n − k +1 ( I ) for k = 1 , . . . , n . It is obvious that for any real parameter M , denoting λ = M − ¯ M , we canrewrite operator T n [ M ] as follows: T n [ M ] u ( t ) ≡ T n [ ¯ M ] u ( t ) + λ u ( t ) . If we assume that equation T n [ ¯ M ] u ( t ) = 0 is disconjugate on I , because ofTheorems 2.3 and 3.2, we can express T n [ ¯ M ] as T n [ ¯ M ] u ( t ) ≡ v ( t ) . . . v n ( t ) T n u ( t ) , T k are built as T u ( t ) = u ( t ) , T k u ( t ) = ddt (cid:18) v k ( t ) T k − u ( t ) (cid:19) , k = 1 , . . . , n − , t ∈ I, (14)with v k > I , v k ∈ C n − k +1 ( I ), for k = 1 , . . . , n .Let us see now that T h u ( t ) is given as a linear combination of u ( t ) , u (cid:48) ( t ) , . . . , u ( h ) ( t )with the form T h u ( t ) = 1 v ( t ) . . . v h ( t ) u ( h ) ( t ) + p h ( t ) u ( h − ( t ) + · · · p h h ( t ) u ( t ) , (15)where p h i ∈ C n − h ( I ).Indeed, we are going to prove this equality by induction.For h = 1 T u ( t ) = ddt (cid:18) v ( t ) u ( t ) (cid:19) = 1 v ( t ) u (cid:48) ( t ) − v (cid:48) ( t ) v ( t ) u ( t ) . Assume, by induction hypothesis, that equation (15) is satisfied for some h ∈ { , . . . , n − } , therefore T h +1 u ( t ) = ddt (cid:18) v h +1 ( t ) (cid:18) v ( t ) . . . v h ( t ) u ( h ) ( t ) + p h ( t ) u ( h − ( t ) + · · · p h h ( t ) u ( t ) (cid:19)(cid:19) = ddt (cid:18) v ( t ) . . . v h +1 ( t ) u ( h ) ( t ) (cid:19) + ddt (cid:18) v h +1 ( t ) (cid:16) p h ( t ) u ( h − ( t ) + · · · p h h ( t ) u ( t ) (cid:17)(cid:19) , which, clearly has the form of equation (15).Finally, taking into account boundary conditions (2) and the regularity offunctions p h i , we conclude that T u ( a ) = 0 , . . . , T k − u ( a ) = 0 , T u ( b ) = 0 , . . . , T n − k − ( b ) = 0 . Moreover T k u ( a ) = 1 v ( a ) . . . v k ( a ) u ( k ) ( a ) , (16) T n − k u ( b ) = 1 v ( b ) . . . v n − k ( b ) u ( n − k ) ( b ) . (17)So, from the positiveness of v h on I , h ∈ { , . . . , n } , we have that T k u ( a )and u ( k ) ( a ) have the same sign. The same property holds for T n − k u ( b ) and u ( n − k ) ( b ). This subsection is devoted to express, as functions of g M ( t, s ), the functions g ( t, s ) , . . . , g n − ( t, s ), defined on (6), as the first row componentes of the Green’sfunction of the vectorial system (4).By studying the adjoint operator as in [1, Section 1.3], we know that therelated Green’s function of the adjoint operator G ∗ satisfies that G ∗ ( t, s ) = G T ( s, t ). Moreover, the following equality holds:9 ∂t ( − G ∗ ( t, s )) = − A T ( t ) ( − G ∗ ( t, s )) , t ∈ I \ { s } . So, we can transform previous equality in (cid:18) − ∂∂t G ( s, t ) (cid:19) T = − ∂∂t G T ( s, t ) = − A T ( t ) (cid:16) − G T ( s, t ) (cid:17) = A T ( t ) G T ( s, t ) = ( G ( s, t ) A ( t )) T . Hence ∂∂t G ( s, t ) = − G ( s, t ) A ( t ) , or, which is the same, ∂∂s G ( t, s ) = − G ( t, s ) A ( s ) . (18)Using this equality, we are going to prove by induction the following ones g n − j ( t, s ) = ( − j ∂ j ∂s j g M ( t, s )+ j − (cid:88) k =0 α jk ( s ) ∂ k ∂s k g M ( t, s ) , j = 1 , . . . , n − . (19)Here α ji ( s ) are functions of a ( s ) , . . . , a j ( s ) and of its derivatives until order( j −
1) and follow the recurrence formula α j +1 k ( s ) = 0 , k ≥ j + 1 , (20) α j +10 ( s ) = a j +1 ( s ) − (cid:16) α j (cid:17) (cid:48) ( s ) , j ≥ , (21) α j +1 k ( s ) = − (cid:18) α jk − ( s ) + (cid:16) α jk (cid:17) (cid:48) ( s ) (cid:19) , ≤ k ≤ j. (22)Using equality (18), we deduce that the Green’s matrix’ terms which are onposition (1 , i ), i = 1 , . . . , n , satisfy the following equality g i − ( t, s ) = − ∂∂s g i ( t, s ) + a n − i +1 ( s ) g M ( t, s ) , i = 2 , . . . , n , (23)where g M ( t, s ) ≡ g n ( t, s ).If we take i = n in equation (23) we deduce g n − ( t, s ) = − ∂∂s g M ( t, s ) + a ( s ) g M ( t, s ) , which give us equation (19) for j = 1.Assume now that equalities (19) – (22) are fulfilled for some j ∈ { , . . . , n − } given. Let us see that they hold again for j + 1.10 n − j − ( t, s ) = − ∂∂s (cid:32) ( − j ∂ j ∂s j g M ( t, s ) + j − (cid:88) k =0 α jk ( s ) ∂ k ∂s k g M ( t, s ) (cid:33) + a j +1 ( s ) g M ( t, s )= a j +1 ( s ) g M ( t, s ) + ( − j +1 ∂ j +1 ∂s j +1 g M ( t, s ) − j − (cid:88) k =0 (cid:16) α jk (cid:17) (cid:48) ( s ) ∂ k ∂s k g M ( t, s ) − j − (cid:88) k =0 α jk ( s ) ∂ k +1 ∂s k +1 g M ( t, s )= ( − j +1 ∂ j +1 ∂s j +1 g M ( t, s ) + a j +1 ( s ) g M ( t, s ) − j − (cid:88) k =0 (cid:16) α jk (cid:17) (cid:48) ( s ) ∂ k ∂s k g M ( t, s ) − j (cid:88) k =1 α jk − ( s ) ∂ k ∂s k g M ( t, s )= ( − j +1 ∂ j +1 ∂s j +1 g M ( t, s ) + j (cid:88) k =0 α j +1 k ( s ) ∂ k ∂s k g M ( t, s ) . Now, we can express Green’s matrix related to problem (4), G ( t, s ), as ( − n − ∂ n − ∂s n − g M ( t, s ) + n − (cid:88) k =0 α n − k ( s ) ∂ k ∂s k g M ( t, s ) · · · g M ( t, s )( − n − ∂ n ∂t ∂s n − g M ( t, s ) + n − (cid:88) k =0 α n − k ( s ) ∂ k +1 ∂t ∂s k g M ( t, s ) · · · ∂∂t g M ( t, s ) · · · ... ... · · · ( − n ∂ n − ∂t n − ∂s n − g M ( t, s ) + n − (cid:88) k =0 α n − k ( s ) ∂ n − k ∂t n − ∂s k g M ( t, s ) · · · ∂ n − ∂t n − g M ( t, s ) (24) If coefficients a ( s ) , . . . , a n − ( s ) , a n ( s ) are constants, a , . . . , a n − , a n , wecan solve explicitly the recurrence form (20) – (22) and deduce that α jk ( s ) = ( − k a j − k . So, we have that g n − j ( t, s ) = j (cid:88) k =0 ( − k a j − k ∂ k ∂s k g M ( t, s ) , with a = 1 , and we can rewrite G ( t, s ) as n − (cid:88) k =0( − k an − − k ∂k∂sk gM ( t, s ) · · · (cid:88) k =0( − k a − k ∂k∂sk gM ( t, s ) gM ( t, s ) n − (cid:88) k =0( − k an − − k ∂k +1 ∂t∂sk gM ( t, s ) · · · (cid:88) k =0( − k a − k ∂k +1 ∂t∂sk gM ( t, s ) ∂∂t gM ( t, s )... ... n − (cid:88) i =0 ( − k an − − k ∂n − k∂tn − ∂sk gM ( t, s ) · · · (cid:88) k =0( − k a − k ∂n − k∂tn − ∂sk gM ( t, s ) ∂n − ∂tn − gM ( t, s ) .
11n particular, if T n [ M ] u ( t ) ≡ u ( n ) ( t ) + M u ( t ) we conclude that g n − j ( t, s ) = ( − j ∂ j ∂s j g M ( t, s ) , so Green’s matrix, G ( t, s ), is given by expression ( − n − ∂ n − ∂s n − g M ( t, s ) · · · − ∂∂s g M ( t, s ) g M ( t, s )( − n − ∂ n ∂t∂s n − g M ( t, s ) · · · − ∂ ∂t∂s g M ( t, s ) ∂∂t g M ( t, s )... ...( − n − ∂ n − ∂t n − ∂s n − g M ( t, s ) · · · − ∂ n ∂t n − ∂s g M ( t, s ) ∂ n − ∂t n − g M ( t, s ) . Remark 3.3.
We note that in the general case it is possible to obtain some ofthe components of system (20) – (22) . α j ( s ) = j − (cid:88) i =0 ( − i a ( i ) j − i ( s ) ,α j ( s ) = j − (cid:88) i =1 ( − i i a ( i − j − i ( s ) ,α j +1 j ( s ) = ( − j +1 a ( s ) . Now we will proceed with the proof of the Main Theorem 3.1. To this end, wewill divide the proof in several steps.First, we are going to show a previous lemma.
Lemma 3.4.
Let ¯ M ∈ R , such that T n [ ¯ M ] u ( t ) = 0 is disconjugate on I . Thenthe following properties are fulfilled: • If n − k is even, then T n [ ¯ M ] is a inverse positive operator on X k and itsrelated Green’s function, g ¯ M ( t, s ) , satisfies ( P g ). • If n − k is odd, then T n [ ¯ M ] is a inverse negative operator on X k and itsrelated Green’s function satisfies ( N g ).Proof. By Lemma 2.5 we have that for all s ∈ ( a, b ) ∃ lim t → a + g ¯ M ( t, s ) p ( t ) = (cid:96) ( s ) > , ∃ lim t → b − g ¯ M ( t, s ) p ( t ) = (cid:96) ( s ) > , s ∈ ( a, b ), we have that function g ¯ M ( t, s ) p ( t ) is a strictly positive andcontinuous function in I , thus0 < k ( s ) = min t ∈ I g ¯ M ( t, s ) p ( t ) < max t ∈ I g ¯ M ( t, s ) p ( t ) = k ( s ) , s ∈ ( a, b ) . (25)Since g M is a continuous function, we have that k and k are continuousfunctions too.If n − k is even, we take φ ( t ) = p ( t ) and condition ( P g ) is trivially fulfilled.If n − k is odd, we take φ ( t ) = − p ( t ) and multiplying equation (25) by − N g ) holds immediately.First, notice that, as a direct corollary of the previous Lemma the assertionfor λ in Theorem 3.1 follows directly from Theorems 2.10 and 2.11.Now, we are going to prove the assertion in Theorem 3.1 concerning λ .The proof will be done in several steps. In a first moment we will show that,if n − k is even, the Green’s function changes sign for all M > ¯ M − λ and forall M < ¯ M − λ when n − k is odd.After that we will prove that such estimation is optimal in both situations.In order to make the paper more readable, along the proofs of this subsectionit will be assumed that n − k is even. The arguments with n − k odd will bepointed out at the end of the subsection.Step 1. Behavior of Green’s function on a neighborhood of s = a and s = b . First, we construct two functions that will characterize the values of M ∈ R for which Green’s function oscillates, or not, on a neighborhood of s = a and s = b .In order to do that, we denote Green’s function related to problem (1)-(2)as follows g M ( t, s ) = g M ( t, s ) , a ≤ t < s ≤ b,g M ( t, s ) , a ≤ s ≤ t ≤ b. Since g M ( t, s ) is a Green’s function, it is satisfied that T n [ M ] g M ( t, s ) = 0 , t ∈ [ a, b ] , t (cid:54) = s , where g M ( t, s ) is acting as a function of t .Therefore, differentiating the previous expression, we deduce that T n [ M ] (cid:18) ∂ h g M ( t, s ) ∂s h (cid:19) = ∂ h ∂s h ( T n [ M ] g M ( t, s )) = 0 , h = 0 , . . . , n − , t (cid:54) = s . (26)In particular, we can define the functions u ( t ) = ∂ k ∂s k g M ( t, s ) | s = b ≡ g M s k ( t, b ) , t ∈ I , (27) v ( t ) = ∂ n − k ∂s n − k g M ( t, s ) | s = a ≡ g M s n − k ( t, a ) , t ∈ I . (28)13ecause of the relation between g M ( t, s ) and g ∗ M ( t, s ), shown in (10), andtaking into account the boundary conditions of the adjoint operator, it is notdifficult to deduce that g M s h ( t, a ) = g ∗ M t h ( a, s ) = 0 , ≤ h ≤ n − k − ,g M s (cid:96) ( t, b ) = g ∗ M t (cid:96) ( b, s ) = 0 , ≤ (cid:96) ≤ k − . So, we are interested in to know the values of M for which functions u ( t ) and v ( t ) oscillate on I . Such property guarantees that Green’s function oscillates ona neighborhood of s = a or s = b for such values. Moreover it provides a higherbound for the set of parameters where Green’s function does not oscillate.Step 1.1. Boundary conditions of v ( t ) . Because of equality (26) we know that T n [ M ] v ( t ) = 0 on I . In this step we aregoing to see which boundary conditions satisfies function v .We have that G ( t, s ) as it appears on (24) is Green’s matrix related tovectorial problem (4). Using the expressions of matrices B and C given by (5),if we consider first row of resultant matrix, we obtain for s ∈ ( a, b ) the followingexpression g M ( a, s ) = 0 , − g M s ( a, s ) + α ( s ) g M ( a, s ) = 0 , ...( − n − k g M s n − k ( a, s ) + n − k − (cid:88) i =0 α n − ki ( s ) g M s i ( a, s ) = 0 . Thus, while k >
1, none of the previous elements belongs to the diagonalof the matrix Green’s function. Since it has discontinuities only at its diagonalentries, see Definition 2.4, by considering the limit of s to a , we deduce that theprevious equalities hold for g M ( a, a ), i.e. : g M ( a, a ) = 0 , − g M s ( a, a ) + α ( a ) g M ( a, a ) = 0 , ...( − n − k g M s n − k ( a, a ) + n − k − (cid:88) i =0 α n − ki ( a ) g M s i ( a, a ) = 0 , so, we conclude that g M ( a, a ) = g M s ( a, a ) = · · · = g M s n − k ( a, a ) = 0 , hence v ( a ) = 0.Analogously, since we do not reach any diagonal element, we deduce that v (cid:48) ( a ) = · · · = v ( k − ( a ) = 0. 14et us see what happens for v ( k − ( a ) with k >
1. We arrive at the followingsystem written as a function of g M ( t, s ): g M t k − ( a, s ) = 0 , − g M t k − s ( a, s ) + α ( s ) g M t k − ( a, s ) = 0 , ...( − n − k g M t k − s n − k ( a, s ) + n − k − (cid:88) i =0 α n − ki ( s ) g M t k − s i ( a, s ) = 0 . This system remains true for s = a , and because of the continuity of Green’smatrix at t = s on the non-diagonal elements and the break which is producedon its diagonal, we arrive at the following system for g M ( a, a ): g M t k − ( a, a ) = 0 , − g M t k − s ( a, a ) + α ( a ) g M t k − ( a, a ) = 0 , ...( − n − k g M t k − s n − k ( a, a ) + n − k − (cid:88) i =0 α n − ki ( a ) g M t k − s i ( a, a ) = 1 , hence g M t k − ( a, a ) = · · · = g M t k − s n − k − ( a, a ) = 0 , and v ( k − ( a ) = g M t k − s n − k ( a, a ) = ( − n − k . Obviously, taking k = 1, the same argument tell us that v ( a ) = ( − n − .To see the boundary conditions at t = b , we have the following system for s ∈ ( a, b ), written as a function of g M ( t, s ) g M ( b, s ) = 0 , − g M s ( b, s ) + α ( s ) g M ( b, s ) = 0 , ...( − n − k g M s n − k ( b, s ) + n − k − (cid:88) i =0 α n − ki ( s ) g M s i ( b, s ) = 0 , hence g M ( b, s ) = · · · = g M s n − k ( b, s ) = 0 .
15y continuity, this is satisfied at s = a , so v ( b ) = g M s n − k ( b, a ) = 0 . Using (24) and (5), since there is no jump in this case, it is immediate toverify that v (cid:48) ( b ) = · · · = v ( n − k − ( b ) = 0.As consequence v is the unique solution of the following problem, which wedenote as ( P v ): T n [ M ] v ( t ) = 0 , t ∈ I ,v ( a ) = · · · = v ( k − ( a ) = 0 ,v ( b ) = · · · = v ( n − k − ( b ) = 0 ,v ( k − ( a ) = ( − n − k . Remark 3.5.
We note that, to attain the previous expression, we have not usedany disconjugacy hypotheses on operator T n [ M ] . Moreover the proof is valid for n − k even or odd. In other works, function v solves problem ( P v ) for any linearoperator defined in (1) and any k ∈ { , . . . , n − } . We know, because of g ¯ M ( t, s ) is of constant sign on I × I (see Lemma 3.4),that if M = ¯ M function v must be of constant sign in I .Step 1.2. If v is of constant sign in I then it can not have any zero in ( a, b ) . We are now going to see that while v ( t ) is of constant sign in I it can not haveany zero in ( a, b ). So the sign change comes on at t = a or t = b .In order to do that, we are going to consider the decomposition of operator T n [ M ] made in Subsection 3.1.Since n − k is even, using Lemma 3.4, we know that operator T n [ ¯ M + λ ] is,for λ = 0, inverse positive on X k . So, the characterization of λ < λ > v ∈ C n ( I ) is a solution of a linear differential equation, hence itis only allowed to have a finite number of zeros on I . Therefore, if v ( t ) ≥ v ( t ) > t ∈ I \ { t , . . . , t (cid:96) } . In particular v ( t ) > t ∈ I . Thus T n [ ¯ M ] v ( t ) = − λ v ( t ) < t ∈ I . (29)As we have shown in Subsection 3.1, we know that T n [ ¯ M ] v ( t ) = v ( t ) . . . v n ( t ) ddt (cid:18) v n ( t ) T n − v ( t ) (cid:19) . Since for every k = 1 , . . . , n , v k ∈ C n − k +1 ( I ) and v k ( t ) > I , we concludethat 1 v n ( t ) T n − v ( t ) must be decreasing on I .Therefore, since v n ( t ) > I we have that T n − v ( t ) can vanish at mostonce in I .Arguing by recurrence, we have that T v ( t ) = v ( t ) can have at most n zeros on I (multiple zeros being counted according to their multiplicity) while v ( t ) ≥ v vanishes n − a and b , hence it can not have a double zero on ( a, b ).This implies that sign change can not come from ( a, b ).16tep 1.3 Change sign of v at t = a and t = b . We are now going to see that the sign change cannot come from a neighbor-hood of t = a .Since n − k is even, as we have proved before, v ( k − ( a ) > M ∈ R ,which implies, since v ( a ) = · · · = v ( k − ( a ) = 0, that v ( t ) = g M s n − k ( t, a ) isalways positive on a neighborhood of t = a . This allows us to affirm thatGreen’s function, g M ( t, s ), is positive on a neighborhood of ( a, a ).Using Step 1.2, we have that v will keep constant sign on I while v ( n − k ) ( b ) =0 is not satisfied, i.e., while an eigenvalue of T n [ ¯ M ] on X k − is not attained.Or equivalently, if M ∈ [ ¯ M , ¯ M − λ (cid:48) ] then g M ( t, s ) remains positive on a rightneighborhood of s = a . Moreover, by Theorem 2.10, we deduce that g M ( t, s )oscillates in I × I for all M > ¯ M − λ (cid:48) .Step 1.4. Study of function u . In order to analyse the behaviour of the Green’s function on a left neighbor-hood of s = b , we work now with the function u defined in (27).Using the same arguments than of v , we conclude that u is the uniquesolution of the following problem, which we denote as ( P u ): T n [ M ] u ( t ) = 0 , t ∈ I ,u ( a ) = · · · = u ( k − ( a ) = 0 ,u ( b ) = · · · = u ( n − k − ( b ) = 0 ,u ( n − k − ( b ) = ( − k − . As in Remark 3.5, we have that this property does not depend either on thedisconjugacy of operator T n [ M ] nor if n − k is even or odd.Using analogous arguments to the ones done with v , we can prove that signchange cannot come on the open interval ( a, b )Moreover, from condition u ( n − k − ( b ) = ( − k − , sign change of u cannotappear on b .So u is of constant sign in I until u ( k ) ( a ) = 0 is verified, i.e., while aneigenvalue of T n [ ¯ M ] on X k +1 does not exist. Or, equivalently, while M ∈ [ ¯ M , ¯ M − λ (cid:48)(cid:48) ].Thus we have that if M is on that interval, Green’s function g M ( t, s ) hasconstant sign on a left neighborhood of s = b , but once M > ¯ M − λ (cid:48)(cid:48) Green’sfunction oscillates in I × I .As a consequence of Step 1, we deduce that interval ( ¯ M − λ , ¯ M − λ ]cannot be enlarged. Moreover we have also proved that the Green’s functionhas constant sign on a neighborhood of s = a and of s = b for all M in suchinterval.Step 2. Behavior of Green’s function on a neighborhood of t = a and t = b . Now, let us see what happens on a neighborhood of t = a and t = b . Inorder to do that, we are going to use the operator (cid:98) T n [( − n ¯ M ] defined in (11)and the relation between g M ( t, s ) and (cid:98) g ( − n M ( t, s ) given in (12).Arguing as in Step 1, we will obtain the values of the real parameter M forwhich (cid:98) g ( − n M ( t, s ) is of constant sign on a neighborhood of s = a and s = b .17nce we have done it, we will be able to apply such property to the behaviourof g M ( t, s ) on a neighborhood of t = a or t = b .The analogous problem for operator (cid:98) T n [( − n M ] related to problem (1)-(2)is given by (cid:98) T n [( − n M ] v ( t ) = 0 , t ∈ I ,v ( a ) = · · · = v ( n − k − ( a ) = 0 ,v ( b ) = · · · = v ( k − ( b ) = 0 . Theorem 2.6 implies that equation T ∗ n [ ¯ M ] u ( t ) = 0 is disconjugate on I . So,the same holds with (cid:98) T n [( − n ¯ M ] u ( t ) = 0. Reasoning as in Step 1, we areable to prove that (cid:98) g ( − n M ( t, s ) has constant sign on a neighborhood of s = a ,while an eigenvalue of (cid:98) T n [( − n ¯ M ] on X n − k − , let it be denoted as (cid:98) λ (cid:48)(cid:48) , is notattained.This fact is equivalent to the existence of an eigenvalue of T ∗ n [ ¯ M ] on X n − k − ,that will be ( − n (cid:98) λ (cid:48)(cid:48) . Now, using the fact that the real eigenvalues of anoperator coincide with those of the adjoint operator, we conclude that λ (cid:48)(cid:48) =( − n (cid:98) λ (cid:48)(cid:48) is the biggest negative eigenvalue of T n [ ¯ M ] on X n − ( n − k − = X k +1 and (cid:98) g ( − n M ( t, s ) is of constant sign on a right neighborhood of s = a while M ∈ [ ¯ M , ¯ M − λ (cid:48)(cid:48) ]. So, Green’s function of problem (1)-(2), g M ( t, s ), does notoscillate on a right neighborhood of t = a .Analogously, arguing as before, we know that (cid:98) g ( − n M ( t, s ) does not oscil-late on a left neighborhood of s = b while an eigenvalue of (cid:98) T n [ M ] on X n − k +1 is not attained, which is equivalent to the existence of an eigenvalue of T n [ M ]on X k − . If M ∈ [ ¯ M , ¯ M − λ (cid:48) ] we can affirm that Green’s function of operator (cid:98) T n [( − n M ], (cid:98) g ( − n M ( t, s ), will not oscillate on a left neighborhood of s = b ,as consequence Green’s function of problem (1)-(2), g M ( t, s ), will not oscillateon a left neighborhood of t = b .As a consequence of the two previous Steps, we have already proved that if M ∈ [ ¯ M , ¯ M − λ ] then Green’s function remains of constant sign on a neighbor-hood of the boundary of I × I . And if M > ¯ M − λ Green’s function oscillateson I × I .Step 3. The Green’s function does not become to change sign on ( a, b ) × ( a, b ) . In this Step we will prove that the oscillation of Green’s function related toproblem (1)-(2) must begin on the boundary of I × I . Using Theorem 2.9 wehave that, provided it has non-negative sign on I × I , g M decreases in M .As consequence, once we prove that g M cannot have a double zero on( a, b ) × ( a, b ), the change of sign must start on the boundary of I × I .Let us see that if g M ( t, s ) ≥ I × I then g M ( t, s ) > a, b ) × ( a, b ).Denote, for a fixed s ∈ ( a, b ), w s ( t ) = g M ( t, s ). By definition, denoting, asin Step 1, λ = M − ¯ M , we have that T n [ ¯ M ] w s ( t ) + λ w s ( t ) = 0 , t ∈ I , t (cid:54) = s . Since g ¯ M ≥ I × I , the behaviour for M < ¯ M has been characterized inLemma 3.4 and Theorem 2.10. 18o we must pay our attention on the situation M > ¯ M , i.e. λ >
0. In sucha case, since, as in Step 1.2, we have that w s ( t ) ≥ I , we know that T n [ ¯ M ] w s ( t ) = − λ w s ( t ) < t ∈ I .
Using (13) and (14), we have that T n [ ¯ M ] w s ( t ) = v ( t ) . . . v n ( t ) T n w s ( t ) , with v k > I for k = 1 , . . . , n . In particular, it is satisfied that T n w s ( t ) < I .Notice that, for all s ∈ ( a, b ), it is satisfied that w s ∈ C n − ( I ) and w ( n − s ( s + ) − w ( n − s ( s − ) = 1. Therefore, due to the definition of T n [ ¯ M ] and expression (15),we have that 1 v n ( t ) T n − w s ( t ) is a continuous function on [ a, s ) ∪ ( s, b ].Since T n w s ( t ) = ddt (cid:18) v n ( t ) T n − w s ( t ) (cid:19) < t (cid:54) = s , we can affirm that1 v n ( t ) T n − w s ( t ) is a decreasing function on I with a positive jump at t = s . So,it can have, at most, two zeros in I , (see Figure 1).Figure 1: v n ( t ) T n − w s ( t ), maximal oscillation with I = [0 , Even we can not guarantee that T n − w s ( t ) is decreasing, since v n > I , we conclude that it has the same sign as 1 v n ( t ) T n − w s ( t ), i.e, it can have atmost two zeros on I .By the other hand, using equation (15) again, we conclude that 1 v n − ( t ) T n − w s ( t )is a continuous function on I . Now, (14) tell us that 1 v n − ( t ) T n − w s ( t ) can reachat most 4 zeros on I (see Figure 2).As before, we do not know intervals where T n − w s ( t ) is increasing or de-creasing, but since v n − ( t ) > v n − ( t ) T n − w s ( t ), so it can reach at most 4 zeros.Following this argument, since v k > I for k = 1 , . . . , n , we know that T n − − h w s ( t ) can have not more than 4 + h zeros on I (multiple zeros beingcounted according to their multiplicity). In particular, w s ( t ) = T w s ( t ) canhave n + 2 zeros at most, having n in the boundary.19igure 2: v n − ( t ) T n − w s ( t ), maximal oscillation with I = [0 , This fact allows w s to have a double zero on ( a, b ). So, to show that suchdouble root cannot exist, we need to prove that maximal oscillation is not pos-sible. To this end, we point out that if for any h it is verified that the sign of T n − − h w s ( a ) is equal to the sign of T n − − h +1 w s ( a ) we lose a possible oscillation.Therefore, for maximal oscillation it must be satisfied T n − h w s ( a ) > , if h odd ,T n − h w s ( a ) < , if h even.However, since w s ( t ) ≥ I and w s ( a ) = w (cid:48) s ( a ) = · · · = w ( k − s ( a ) = 0, wededuce that w ( k ) s ( a ) ≥ w ( k ) s ( a ) > w ( k ) s ( a ) = 0 wewould have n + 2 zeros at most, having n + 1 in the boundary. So, only a simplezero is allowed in the interior, which is not possible without oscillation.Therefore w ( k ) s ( a ) = w ( n − ( n − k )) s ( a ) >
0. Since n − k is even, using now (16),we also know that T k w s ( a ) >
0, which inhibits maximal oscillation.So we conclude that if g M ( t, s ) ≥ I × I then g M ( t, s ) > a, b ) × ( a, b ),as we wanted to prove.As a consequence of the three previous Steps, we have described the set ofthe real parameters M for which the Green’s function is non-negative on I × I when n − k is even.If n − k is odd we can do similar arguments to achieve the proof. In thesequel, we enumerate the main ideas to be developedStep 1. 1Step 1.1. It has no modifications.Step 1.2. In equality (29) we have λ < v ( t ) < I , so it remainstrue and we can proceed analogously.Step 1.3. In this case, we have that v ( k − ( a ) <
0. Our attainment in thisStep is that g M ( t, s ) remains negative while M ∈ [ ¯ M − λ (cid:48) , ¯ M ] in aneighborhood of s = a and oscillates for all M < ¯ M − λ (cid:48) .Step 1.4. The arguments are not modified, but the final achievement is that g M ( t, s ) is negative in a neighborhood of s = b for M ∈ [ ¯ M − λ (cid:48)(cid:48) , ¯ M ]an oscillates for all M < ¯ M − λ (cid:48)(cid:48) .20tep 2. Using the same arguments we conclude that the interval where g M ( t, s ) isnon-positive on the boundary of I × I is [ ¯ M − λ , ¯ M ].Step 3. In this case we have that w ( k ) s ( a ) = w ( n − ( n − k )) s ( a ) <
0, with n − k oddcontradicting maximal oscillation too.Thus, our result is proved.As a direct consequence of the arguments used in Step 1.3, without assumingthe existence of ¯ M ∈ R for which equation T n [ ¯ M ] u ( t ) = 0 is disconjugate on I ,we arrive at the following result. Corollary 3.6.
Let T n [ M ] be defined as in (1) . Then the two following prop-erties hold:If n − k is even, then it does not exist M ∈ R such that operator T n [ M ] isinverse negative in X k .If n − k is odd, then it does not exist M ∈ R such that operator T n [ M ] isinverse positive in X k .Proof. It is enough to take into account that v , defined in (28), is the uniquesolution of problem ( P v ). Since v ( k − ( a ) = ( − n − k we conclude that, if n − k is even, Green’s function has positive values in any neighborhood of ( a, a ) andnegative when n − k is odd.So, the result holds from Theorem 2.8. In order to obtain the eigenvalues of particular problems we calculate a funda-mental system of solutions y [ M ]( t ) , . . . , y n [ M ]( t ) of equation (1) where every y k [ M ]( t ) verifies the initial conditions y ( n − k ) k [ M ]( a ) = 1 , y ( n − j ) k [ M ]( a ) = 0 , j = 1 , . . . , n , j (cid:54) = k . Then we denote the n − W nk [ M ]( t ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y [ M ]( t ) . . . y k [ M ]( t ) y (cid:48) [ M ]( t ) . . . y (cid:48) k [ M ]( t )... y ( k − [ M ]( t ) . . . y ( k − k [ M ]( t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , k = 1 , . . . , n − . As a consequence of the characterization done in [11, Chapter 3, Lemma 12],we deduce that the eigenvalues of problem (1) in X k are given as the λ ∈ R forwhich W n − k [ − λ ]( b ) = 0. So, in the sequel, we will use this method to find theeigenvalues of the different considered problems. T n [ M ] u ( t ) ≡ u ( n ) ( t ) + M u ( t ) First of all, we are going to consider problems where T n [ M ] u ( t ) ≡ u ( n ) ( t ) + M u ( t ), with [ a, b ] = [0 , M = 0, u ( n ) ( t ) = 0 is always disconjugate, see[11, Chapter 3]. So, hypotheses of Theorem 3.1 are satisfied.21 emark 4.1. Note that adjoint equation to problem T n [ M ] u = 0 , u ∈ X k isgiven by T ∗ n [ M ] u ( t ) = ( − n u ( n ) ( t ) + M u ( t ) = 0 , u ∈ X n − k . So, if we have that λ i is an eigenvalue of u ( n ) in X k , it is also an eigenvalueof ( − n u ( n ) in X n − k . Thus, ( − n λ i is an eigenvalue of u ( n ) in X n − k .As consequence, we only need to obtain first (cid:106) n (cid:107) Wronskians, where (cid:98)·(cid:99) means the floor function. - Order 2The eigenvalues of operator u (cid:48)(cid:48) ( t ) in X must satisfy W [ λ ](1) = 0, whichcan be replaced by the following equationsin( √− λ ) = 0 , (30)so it closest to zero negative eigenvalue is (cid:0) λ (cid:1) = − π .And so, we can affirm that Green’s function related to operator u (cid:48)(cid:48) ( t )+ M u ( t )is negative if, and only if, M ∈ (cid:0) −∞ , π (cid:1) .This result has been already obtained in different references (See [1] andreferences therein), but here it is not necessary to have the expression of theGreen’s function.- Order 3 λ (cid:117) . W [ λ ] = 0, which is equivalentto the equation cos (cid:18) √ λ (cid:19) − √ (cid:18) √ λ (cid:19) = e − λ . Then, the least positive eigenvalue of operator u (3) ( t ) in X is (cid:0) λ (cid:1) and thebiggest negative eigenvalue of operator u (3) ( t ) in X is − (cid:0) λ (cid:1) .So, we can affirm that Green’s function of operator u (3) ( t ) + M u ( t ) • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) , • in X is positive if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) .This result has been obtained by means of the explicit form of Green’s func-tion in [23].- Order 4 λ (cid:117) .
553 is the least positive solution of W [ λ ](1) = 0, simplifying thatexpression we have tan (cid:18) λ √ (cid:19) = tanh (cid:18) λ √ (cid:19) .λ (cid:117) . W [ − λ ](1) = 0, which can beexpressed as cos( λ ) cosh( λ ) = 1 . u (4) ( t ) in X and X is given by − (cid:0) λ (cid:1) .The least positive eigenvalue of operator u (4) ( t ) in X is (cid:0) λ (cid:1) .Therefore, we can affirm without calculating it explicitly, that Green’s func-tion related to the operator u (4) ( t ) + M u ( t ) • in X and X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) .These results have been obtained using the explicit form of Green’s functionin [8] and [9].- Order 5We can obtain λ (cid:117) . λ (cid:117) . W [ λ ](1) = 0 and W [ − λ ](1) = 0, respectively. But the equations obtainedare too complicate to show here and they have not so much interest.The least positive eigenvalue of operator u (5) ( t ) in X is (cid:0) λ (cid:1) .The biggest negative eigenvalue of operator u (5) ( t ) in X is − (cid:0) λ (cid:1) .The least positive eigenvalue of operator u (5) ( t ) in X is (cid:0) λ (cid:1) .The biggest negative eigenvalue of operator u (5) ( t ) in X is − (cid:0) λ (cid:1) .Therefore, we conclude without calculating it explicitly, that Green’s func-tion related to the operator u (5) ( t ) + M u ( t ) • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) .- Order 6 λ (cid:117) . W [ λ ](1) = 0, which is equiva-lent to sin( λ ) − √ (cid:18) λ (cid:19) sinh (cid:32) √ λ (cid:33) + sin (cid:18) λ (cid:19) cosh (cid:32) √ λ (cid:33) = 0 .λ (cid:117) . W [ − λ ](1) = 0, which we canexpress as − e λ/ (cid:16) e λ + 1 (cid:17) + √ (cid:16) e λ − (cid:17) sin (cid:32) √ λ (cid:33) + (cid:16) e λ + 1 (cid:17) cos (cid:32) √ λ (cid:33) − e λ/ cos (cid:16) √ λ (cid:17) = 0 . (cid:117) . is the least positive solution of W [ λ ](1) = 0, which can berepresented as the first positive root of the following equationsin( λ ) (cid:16) − cos( λ ) + cosh (cid:16) √ λ (cid:17) + 4 (cid:17) − (cid:18) λ (cid:19) cosh (cid:32) √ λ (cid:33) = 0 . The biggest negative eigenvalue of operator u (6) ( t ) in X and X is given by − (cid:0) λ (cid:1) .The least positive eigenvalue of operator u (6) ( t ) in X and X is (cid:0) λ (cid:1) .The biggest negative eigenvalue of operator u (6) ( t ) in X is − (cid:0) λ (cid:1) .Hence, we can affirm without calculating it explicitly, that Green’s functionrelated to operator u (6) ( t ) + M u ( t ) • in X or in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X or in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) .- Order 7We are not able to obtain analytically the eigenvalues of operator u (7) ( t ),but we can obtain them numerically.The least positive eigenvalue of this operator in X is (cid:0) λ (cid:1) , where λ (cid:117) . X is − (cid:0) λ (cid:1) , where λ (cid:117) . X is (cid:0) λ (cid:1) , where λ (cid:117) . X is − (cid:0) λ (cid:1) .The least positive eigenvalue in X is (cid:0) λ (cid:1) .The biggest negative eigenvalue in X is − (cid:0) λ (cid:1) .So, we conclude without calculating it explicitly, that Green’s function re-lated to the operator u (7) ( t ) + M u ( t ) • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) .24 Order 8 λ (cid:117) . λ (cid:117) . λ (cid:117) . λ (cid:117) . W [ λ ](1) = 0, W [ − λ ](1) = 0, W [ λ ](1) = 0 and W [ − λ ](1) = 0 respectively, but their expressions are toobig to show it here and they do not bring any important information.The biggest negative eigenvalue of operator u (8) ( t ) in X and X is given by − ( λ ) .The least positive eigenvalue of operator u (8) ( t ) in X and X is given by( λ ) .The biggest negative eigenvalue of operator u (8) ( t ) in X and X is given by − ( λ ) .The least positive eigenvalue of operator u (8) ( t ) in X is ( λ ) .So, we can affirm without calculating it explicitly, that Green’s functionrelated to the operator u (8) ( t ) + M u ( t ) • in X or in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X or in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) . • in X or in X is negative if, and only if, M ∈ (cid:104) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:17) . • in X is positive if, and only if, M ∈ (cid:16) − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) (cid:105) .As we have said before, third-order problems were explicitly calculated on[23]. And fourth-order problems were calculated on [8] in X and on [9] in X and X , respectively. But, in all of these cases were necessary to obtain theexpression of Green’s function and analyse it.Moreover, in all the problems treated on [8, 9, 23] it is also satisfied that theopen optimal interval where Green’s function is of constant sign coincide withthe optimal interval where the equation (1) is disconjugate.However, in [10, Theorem 2.1] it is proved the following characterization ofthe interval of disconjugacy: Theorem 4.2.
Let ¯ M ∈ R and n ≥ be such that T n [ ¯ M ] u ( t ) = 0 is a discon-jugate equation on I . Then, T n [ M ] u ( t ) = 0 is a disconjugate equation on I if,and only if, M ∈ ( ¯ M − λ , ¯ M − λ ) , where • λ = + ∞ if n = 2 and, for n > , λ > is the minimum of the leastpositive eigenvalues on T n [ ¯ M ] in X k , with n − k even. • λ < is the maximum of the biggest negative eigenvalues on T n [ ¯ M ] in X k , with n − k odd. As consequence we have that the interval of constant sign of the Green’sfunction and the one of the disconjugacy for the linear operator are not thesame in general. We have already proved (see Lemma 3.4) that while equation(1) is disconjugate its related Green’s function must be of constant sign. So,if both intervals do not coincide, the optimal interval where the equation (1)25s disconjugate must be contained in the open optimal interval where Green’sfunction is of constant sign .If, using the characterization given in Theorem 4.2, we calculate the optimalinterval on M of disconjugacy for equation u (5) ( t ) + M u ( t ) = 0 , t ∈ [0 , . We have that it is given by ( − (cid:0) λ (cid:1) , (cid:0) λ (cid:1) ).But, as we have shown before, Green’s function related to the problem onthe space X remains positive on the interval ( − (cid:0) λ (cid:1) , − (cid:0) λ (cid:1) ]. So, its biggestopen interval is strictly bigger than the optimal interval of disconjugacy. Remark 4.3.
In this kind of problems, if λ is an eigenvalue on [0 , , then λ ( b − a ) n is an eigenvalue on [ a, b ] .So, we can obtain our conclusions about Green’s function’ sign on any arbi-trary interval [ a, b ] . This characterization of the interval where Green’s function is of constant signis also useful for those problems which have more non-nulls coefficients .For example we can consider the operator of fourth order T n [ M ] u ( t ) ≡ u (4) ( t ) + 10 u (3) ( t ) + 10 u (cid:48)(cid:48) ( t ) + 10 u (cid:48) ( t ) + M u ( t ) , t ∈ [0 , . (31)We can show, using the characterization given in Theorem 2.3, that T n [0] u ( t ) =0 is a disconjugate equation on [0 ,
1] and, so, Theorem 3.1 holds.First, we calculate numerically the closest to zero eigenvalues in each X k , k = 1 , , • The biggest negative eigenvalue in X is − (7 . . • The least positive eigenvalue in X is (5 . . • The biggest negative eigenvalue in X is − (5 . .Realize that in this case we need to obtain the three correspondents Wron-skians because it is not possible to connect eigenvalues in X with those in X by means of its corresponding adjoint equation.So, we conclude that Green’s function related to operator T n [ M ] u ( t ) definedin (31) satisfies that • in X is negative if, and only if, M ∈ [ − (5 . , (7 . ), • in X is positive if, and only if, M ∈ ( − (5 . , (5 . ], • in X is negative if, and only if, M ∈ [ − (5 . , (5 , ).Notice that in this case the interval of disconjugation is ( − (5 . , (5 . ).So, we have obtained an example of a fourth order equation in which its intervalof disconjugation does not coincide with the biggest open interval where Green’sfunction is of constant sign in X . 26n the sequel, we show an example where operator T n [ M ] does not verifydisconjugation hypothesis for ¯ M = 0.If we choose the operator T n [ M ] u ( t ) ≡ u (4) ( t ) + 10 u (3) ( t ) + 550 u (cid:48) ( t ) + M u ( t ) , t ∈ [0 , . (32)We obtain that equation T n [0] u ( t ) = 0 is not disconjugate on [0 , T n [ − u ( t ) = 0 we can affirm, by means of Theorem2.3, that it is disconjugate on [0 , T n [ − u ( t ).If we calculate the closest to zero eigenvalues we have • The biggest negative eigenvalue of T n [ − u ( t ) in X is − . • The least positive eigenvalue in X is 11 . • The biggest negative eigenvalue in X is − . T n [ M ] u ( t ), definedin (32), satisfies that • in X is inverse negative if, and only if, M ∈ [ − − . , −
600 +9565 .
99) = [ − . , . • in X is inverse positive if, and only if, M ∈ ( − − . , −
600 +28 . − . , − . • in X is inverse negative if, and only if, M ∈ [ − − . , −
600 +28 . − . , − . We have already seen that applying Theorem 3.1 is much easier to calculateoptimal intervals for M where Green’s function related to operator T n [ M ] u ( t )than obtain Green’s function expression explicitly. But, if we are referring to anoperator with non-constant coefficients this characterization is even more usefulbecause in the majority of the situations we are not able to obtain the explicitexpression for the Green’s function.Consider now the third order operator T n [ M ] u ( t ) ≡ u (3) ( t ) + t u (cid:48) ( t ) + M u ( t ) , t ∈ [0 , , (33)for which, by means of Theorem 2.3, we can verify that equation T n [0] u ( t ) = 0is disconjugate on [0 , • (4 . is the least positive eigenvalue of operator T n [0] u ( t ) in X . • − (4 . is the biggest negative eigenvalue of operator T n [0] u ( t ) in X .So, we can affirm • Green’s function related to operator T n [ M ] u ( t ) in X is positive if, andonly if, M ∈ ( − (4 . , (4 . ],27 Green’s function related to operator T n [ M ] u ( t ) in X is negative if, andonly if, M ∈ [ − (4 . , (4 . ).We can also apply it to a fourth order operator whose eigenvalues were alsoobtained numerically. T n [ M ] ≡ u (4) ( t ) + e t u (cid:48) ( t ) + M u ( t ) , t ∈ [0 , . (34)We can verify, by means of Theorem 2.3 again, that T n [0] u ( t ) = 0 is discon-jugate on [0 , • The biggest negative eigenvalue in X is − (5 . . • The least positive eigenvalue in X is (4 . . • The biggest negative eigenvalue in X is − (5 . .So, applying Theorem 3.1, we conclude that • Green’s function related to operator T n [ M ] u ( t ) in X is negative if, andonly if, M ∈ [ − (4 . , (5 . ), • Green’s function related to operator T n [ M ] u ( t ) in X is positive if, andonly if, M ∈ ( − (4 . , (5 . ], • Green’s function related to operator T n [ M ] u ( t ) in X is negative if, andonly if, M ∈ [ − (4 . , (5 . ). In this last section we show that the disconjugacy hypothesis on Theorem 3.1for some M = ¯ M cannot be avoided in general.To this end, we consider the operator T [ M ] u ( t ) = u (4) ( t ) − u (cid:48) ( t ) + M u ( t ) , t ∈ [0 , , (35)coupled with two-point boundary value conditions u (0) = u (cid:48) (0) = u (cid:48)(cid:48) (0) = u (1) = 0 . (36)The equation (35) is not disconjugate for M = 0, indeed: u ( t ) = − e t − − e − t cos (cid:0) √ t − (cid:1) + 33000 , is a solution of T [0] u ( t ) = 0 with 5 zeros on [0 , N g ) for ¯ M = 0. So, by means of Theorem 2.11, weknow that N T = [ − µ, − λ ) for some µ ≥ µ (cid:54) = λ , with λ the first eigenvaluerelated to operator T [0] on the space X .28s a consequence, we deduce that the validity of Theorem 3.1 is not ensuredwhen the disconjugacy assumption fails.We point out that, since the existence of at least one ¯ M for which operator T [ ¯ M ] is disconjugate on [0 ,
1] implies the validity of Theorem 3.1, operator T [ M ] cannot be disconjugate on [0 ,
1] for any real parameter M and not onlyfor ¯ M = 0.First, we obtain the Green’s function expression related to the operator T [0] u ( t ) in X , g ( t, s ). By means of the Mathematica package developed in[6], we have that if follows the expression e t − s ) − e − s + t ) (cid:16) − e s +5+2 e s cos (cid:16) √ s − (cid:17) + e (cid:17)(cid:16) − e t + e t +2 cos (cid:16) √ t (cid:17)(cid:17) − e e (cid:16) √ (cid:17) +2 e s − t cos (cid:16) √ t − s ) (cid:17) − , ≤ s ≤ t ≤ , − e − s + t ) (cid:16) − e s +5+2 e s cos (cid:16) √ s − (cid:17) + e (cid:17)(cid:16) − e t + e t +2 cos (cid:16) √ t (cid:17)(cid:17) (cid:16) − e e (cid:16) √ (cid:17)(cid:17) , < t < s ≤ . Let us see now that g ( t, s ) ≤ , × [0 ,
1] and that it satisfies condition( N g ), i.e., the following inequality is satisfied g ( t, s ) t ( t − > , for all ( t, s ) ∈ [0 , × (0 , . To study the behaviour on a neighborhood of t = 0 and t = 1, we define thefollowing functions k s ) = lim t → g t, s ) t t −
1) = e − s (cid:16) − e s +5 + 2 e s cos (cid:16) √ s − (cid:17) + e (cid:17) (cid:16) − e e
15 + 2 cos (cid:16) √ (cid:17)(cid:17) ,k s ) = lim t → − g t, s ) t t −
1) = 1300 e − s − e s (cid:16) √ (cid:16) √ s − (cid:17) − cos (cid:16) √ s − (cid:17)(cid:17) + e e e (cid:16) − e s +5 + 2 e s cos (cid:16) √ s − (cid:17) + e (cid:17) (cid:16) − e
15 + √ (cid:16) √ (cid:17) + cos (cid:16) √ (cid:17)(cid:17) − e e
15 + 2 cos (cid:16) √ (cid:17) . In the sequel we will prove that both functions are strictly positive on (0 , k (1) = k (cid:48) (1) = k (cid:48)(cid:48) (1) = 0 and that k (3)1 (1) = − e − e + e + 2 cos (cid:0) √ (cid:1) < . If we prove that k (3)1 ( s ) is strictly negative on [0 , k (cid:48)(cid:48) ( s ) would be positive and k (cid:48) ( s ) negative, we will deduce that k ( s ) > s ∈ (0 , k (3)1 ( s ) = − e − s (cid:0) e s cos (cid:0) √ s − (cid:1) + e (cid:1) (cid:0) − e + e + 2 cos (cid:0) √ (cid:1)(cid:1) , we only must check that k ( s ) := 2 e s cos (cid:16) √ s − (cid:17) + e > , s ∈ [0 , . s ∈ [0 , k ( s ) = e (cid:16) − e − π √ (cid:17) > , s ∈ [0 , . Consider now function k . We have that k (0) = 0 and k (cid:48) (0) = 1 + e (cid:0) e − √ (cid:0) e − (cid:1) sin (cid:0) √ (cid:1) − (cid:0) e (cid:1) cos (cid:0) √ (cid:1)(cid:1) e (cid:0) − e + e + 2 cos (cid:0) √ (cid:1)(cid:1) > . So, we study the sign of its first derivative k (cid:48) ( s ) = e − s − (cid:0) − e + e + 2 cos (cid:0) √ (cid:1)(cid:1) k ( s ) , with k ( s ) = e s (cid:16) √ (cid:16) e (cid:0) e − (cid:1) sin (cid:16) √ s − (cid:17) + sin (cid:16) √ s (cid:17)(cid:17) − e cos (cid:16) √ s − (cid:17) +3 cos (cid:16) √ s (cid:17)(cid:17) + e (cid:16) e − √ (cid:16) √ (cid:17) − (cid:16) √ (cid:17)(cid:17) . It is clear that such function satisfies k ( s ) > (cid:16) − − e + √ (cid:0) − e − e (cid:1)(cid:17) e s + e (cid:16) e − √ (cid:16) √ (cid:17) − (cid:16) √ (cid:17)(cid:17) , which is positive for s < (cid:16) log (cid:16) e − √ e sin (cid:16) √ (cid:17) − e cos (cid:16) √ (cid:17)(cid:17) − log (cid:16) √ e − √ e + 2 √ e (cid:17)(cid:17) (cid:117) . .Moreover, for s ∈ [1 − π √ , − π √ ] (cid:117) [0 . , . k ( s ) > (cid:0) − − e (cid:1) e s + e (cid:16) e − √ (cid:16) √ (cid:17) − (cid:16) √ (cid:17)(cid:17) , and right part of previous equality is positive for s < (cid:16) log (cid:16) e − √ e sin (cid:16) √ (cid:17) − e cos (cid:16) √ (cid:17)(cid:17) − log (cid:0) e (cid:1)(cid:17) (cid:117) . . Then, we have that k (cid:48) ( s ) > s ∈ [0 , − π √ ], and, as consequence, thesame holds for k ( s ).On the other hand, we have that k (1) = k (cid:48) (1) = 0 and k (cid:48)(cid:48) (1) = 1, moreover k (cid:48)(cid:48) ( s ) = e − s − (cid:0) − e + e + 2 cos (cid:0) √ (cid:1)(cid:1) k ( s ) , where k s ) = e s (cid:16) √ (cid:16) e
15 sin (cid:16) √ s − (cid:17) − sin (cid:16) √ s (cid:17)(cid:17) + 3 e (cid:16) e − (cid:17) cos (cid:16) √ s − (cid:17) + 3 cos (cid:16) √ s (cid:17)(cid:17) + e (cid:16) − e √ (cid:16) √ (cid:17) + 3 cos (cid:16) √ (cid:17)(cid:17) . Now, we must verify that k ( s ) > s > . e s (cid:32) − − √ (cid:16) e (cid:17) + 3 e (cid:16) e − (cid:17) cos (cid:32) √ (cid:33)(cid:33) + e (cid:16) − e √ (cid:16) √ (cid:17) + 3 cos (cid:16) √ (cid:17)(cid:17) . It is clear that it is positive for s ∈ ( s , s − e
20 + √ e
15 sin (cid:16) √ (cid:17) + 3 e
15 cos (cid:16) √ (cid:17) √ √ e
15 + 6 e (cid:18) √ (cid:19) − e
15 cos (cid:18) √ (cid:19) (cid:117) . , which ensures that k ( s ) > . , s ∈ [0 , e s +5 k ( s ) is boundedfrom below by k = 1100 (cid:0) − e s + 303 e s +5 − e (cid:1) , which is positive for s ∈ ( s , s ), where s = 1+ 15 log (cid:32) √ (cid:32)
13 tan − (cid:32) √ (cid:33)(cid:33) − (cid:32)
13 tan − (cid:32) √ (cid:33)(cid:33)(cid:33) (cid:117) . , and s = 1 + 15 log (cid:32) (cid:32)
13 tan − (cid:32) √ (cid:33)(cid:33)(cid:33) (cid:117) . . So, we conclude that k ( s ) > s ∈ (0 , N g ), we only have to verify that g ( t, s ) < t, s ) ∈ (0 , × (0 , t < s we can express g ( t, s ) = − e − s + t ) (cid:96) ( s ) (cid:96) ( t )3000 (cid:0) − e + e + 2 cos (cid:0) √ (cid:1)(cid:1) , where (cid:96) ( s ) = (cid:16) − e s +5 + 2 e s cos (cid:16) √ s − (cid:17) + e (cid:17) ,(cid:96) ( t ) = (cid:16) − e t + e t + 2 cos (cid:16) √ t (cid:17)(cid:17) . So, we must prove that both functions are positive on (0 , (cid:96) ( s ) is a positive multiple of k ( s ), so, as we have proved before, it is positivefor s ∈ (0 , (cid:96) , since it satisfies that (cid:96) (0) = (cid:96) (cid:48) (0) = (cid:96) (cid:48)(cid:48) (0) = 0, fromthe following expressions, valid for all t ∈ [0 , (cid:96) (3)2 ( t ) = 375 (cid:16) − e t + 9 e t + 2 √ (cid:16) √ t (cid:17)(cid:17) ≥ (cid:16) − e t + 9 e t − √ (cid:17) > ,
31e deduce that (cid:96) ( t ) > t ∈ (0 , < s ≤ t < g ( t, s ) as follows g ( t, s ) = 13000 ( p ( t − s ) − p ( t, s )) , < s ≤ t < , where p ( t, s ) = e − s + t ) (cid:96) ( s ) (cid:96) ( t ) − e + e + 2 cos (cid:0) √ (cid:1) and p ( r ) = e r + 2 e − r cos (cid:16) √ r (cid:17) − . From the previously proved positiveness of (cid:96) and (cid:96) , we know that p ( t, s ) >
0. On the other hand, since p (0) = p (cid:48) (0) = p (cid:48)(cid:48) (0) = 0, if we verify that p (3)2 ( r ) > r ∈ [0 , p on(0 , p (3)2 ( r ) = 1000 e r + 2000 e − r cos (cid:16) √ r (cid:17) . This function is trivially positive whenever 0 ≤ r ≤ π √ (cid:117) . r ∈ [0 , p (3)2 ( r ) > e r − e − r , which is positive if, and only if, r > log(2)15 (cid:117) . p ( r ) > r ∈ (0 , p ( t − s ) < p ( t, s ) for 0 < s ≤ t <
1, we can concludethat g ( t, s ) < g (1 , s ) = 0, we know that p (1 , s ) = p (1 − s ), for everyfixed s ∈ (0 , k > , g ( t, s ) < t = 1 for every s ∈ (0 , p ( t, s ) > p ( t − s ) on aneighborhood of t = 1 for every s ∈ (0 , s ∈ (0 , p ( t, s ) and p ( t − s ) are convexfunctions of t By direct calculation, we have that ∂ ∂t p ( t, s ) = 100 e − s + t ) (cid:0) e t + √ (cid:0) √ t (cid:1) − cos (cid:0) √ t (cid:1)(cid:1) (cid:96) ( s ) − e + e + 2 cos (cid:0) √ (cid:1) ,
32o we only need to verify that p ( t ) = (cid:16) e t + √ (cid:16) √ t (cid:17) − cos (cid:16) √ t (cid:17)(cid:17) > , t ∈ (0 , . The following inequality is trivially fulfilled p ( t ) > e t + √ (cid:16) √ t (cid:17) − q ( t ) , t ∈ [0 , . We have that q (cid:48) ( t ) = 15 e t + 15 cos(5 √ t ) > e t − > , since q (0) = 0, we conclude that q > p ( t ) > ,
1] and also ∂ ∂t p ( t, s ) > p (3)2 ( r ) >
0, for r ∈ [0 , p (cid:48)(cid:48) (0) = 0, so forevery fixed s ∈ (0 , p (cid:48)(cid:48) ( t − s ) > t ∈ ( s, s ∈ (0 , p ( t, s ) and p ( t − s ) are convexfunctions of t .From the fact that p ( t, s ) > p ( t − s ) on a neighborhood of t = 1, p (1 , s ) = p (1 − s ) and, also, p ( s, s ) > p (0), we can affirm that p ( t, s ) > p ( t − s )for t ∈ [ s, g ( t, s ) < < s ≤ t <
1, and condition ( N g ) isfulfilled.Now, as a consequence of Theorem 2.11, we know that g M ( t, s ) ≤ M ∈ [0 , − λ ), where λ < T [0] u ( t ) in X .To verify that Theorem 3.1 does not hold in this case we will prove thatfor M < T [0] u ( t ) in X .As in the previous section, we can obtain numerically the first eigenvaluesof T [0], which can be given by the following approximation values: • The biggest negative eigenvalue in X is λ (cid:117) − (12 . . • The least positive eigenvalue in X is λ (cid:117) (10 . . • The biggest negative eigenvalue in X in λ (cid:117) − (9 . . Remark 5.1.
Realize that, since T [0] u ( t ) = 0 is not disconjugate on [0 , ,we have no a priori information about the sign of the eigenvalues λ and λ .However, since g satisfies ( N g ) , we can ensure, without calculate it, that λ < . Finally, let’s see that there exists M ∗ > − λ for which g M ∗ has not constantsign on I × I .We are going to study the following function v ( t ) = ∂∂s g M ∗ ( t, s ) | s =0 .
33s we have proved in the proof of Theorem 3.1, if this function has not con-stant sign on I then the Green’s function must necessarily change sing in aneighborhood of s = 0.For M ∗ = − (cid:117) − (9 . , we have that v ( t ) follows the next expres-sion e − (cid:16) √ (cid:17) t e t (cid:16) √ − (cid:17) e (cid:113) t − − √ + 446 e (cid:113) t (cid:18) √
831 sin (cid:18)(cid:113) t (cid:19) + 34625 cos (cid:18)(cid:113) t (cid:19)(cid:19) − (cid:18) √
831 sin (cid:18)(cid:113) (cid:19) + 34625 cos (cid:18)(cid:113) (cid:19)(cid:19) + 277 e (cid:18) √
669 sinh (cid:18)(cid:113) (cid:19) − (cid:18)(cid:113) (cid:19)(cid:19) − (cid:16) √ (cid:17) e (cid:16) √ − (cid:17) e (cid:113) e (cid:113) (cid:18) (cid:18)(cid:113) (cid:19) − √
831 sin (cid:18)(cid:113) (cid:19)(cid:19) e (cid:113) − (cid:16) √ (cid:17) t e t (cid:16) √ − (cid:17) e (cid:113) t − − √ + 446 e (cid:113) t (cid:115) t − √
831 sin (cid:115) t , which, see Figure 3, changes sign on I . (cid:45) (cid:45) (cid:45) Figure 3:
Graph of v As consequence the Green’s function has not constant sign for a value of M bigger than − λ .Even more, we can verify numerically which is the interval for M , where g M ( t, s ) is non-positive on I × I . We observe that change sign come first onthe interior of I × I . It comes in ( t, s ) (cid:117) (0 . , . ∈ (0 , × (0 ,
1) for M (cid:117) − (7 . . So we deduce that it is given by [ − (7 . , − λ ).As consequence we conclude the example that show us that if we suppressthe disconjugacy hypothesis, Theorem 3.1 is not true in general. References [1] A. Cabada,
Green’s Functions in the Theory of Ordinary Differential Equa-tions , Springer Briefs in Mathematics, 2014.[2] A. Cabada,
The method of lower and upper solutions for second, third,fourth, and higher order boundary value problems , J. Math. Anal. Appl.185 (1994) 302-320.[3] A. Cabada, J. A. Cid,
Existence and multiplicity of solutions for a periodicHill’s equation with parametric dependence and singularities , Abstr. Appl.Anal. 2011, Art. ID 545264, 19 pp.[4] A. Cabada, J. A. Cid,
Existence of a non-zero fixed point for non-decreasingoperators via Krasnoselskii’s fixed point theorem,
Nonlinear Anal., 71(2009), 2114–2118.[5] A. Cabada, J. A. Cid, G. Infante,
New criteria for the existence of non-trivial fixed points in cones , Fixed Point Theory and Appl., 2013:125,(2013), 12 pp. 346] A. Cabada, J.A. Cid, B. M´aquez-Villamar´ın,
Computation of Green’s func-tions for boundary value problems with Mathematica , Applied Mathematicsand Computation 219 (2012) 1919-1936.[7] A. Cabada, J.A. Cid, L. Sanchez,
Positivity and lower and upper solutionsfor fourth order boundary value problems , Nonlinear Anal. 67 (2007), 1599-1612.[8] A. Cabada, R. R. Engui¸ca
Positive solutions of fourth order problems withclamped beam boundary conditions , Nonlinear Anal. 74 (2011), 3112-3122.[9] A. Cabada, C. Fern´andez-G´omez
Constant Sign Solutions of Two-PointFourth Order Problems , Appl. Math. Comput. 263 (2015), 122-133.[10] A. Cabada, L. Saavedra
Disconjugacy characterization by means of spectral ( k, n − k ) problems , Appl. Math. Lett. 52 (2016), 21-29.[11] W. A. Coppel, Disconjugacy , Lecture Notes in Mathematics, Vol. 220.Springer-Verlag, Berlin-New York, 1971.[12] J. A. Cid, D. Franco, F. Minh´os,
Positive fixed points and fourth-orderequations , Bull. Lond. Math. Soc., 41 (2009), 72–78.[13] C. De Coster, P. Habets,
Two-Point Boundary Value Problems: Lower andUpper Solutions , Mathematics in Science and Engineering 205, Elsevier B.V., Amsterdam, 2006.[14] U. Elias,
Eventual disconjugacy of y ( n ) + µ p ( x ) y = 0 for every µ , Arch.Math. (Brno) 40 (2004), 2, 193–200.[15] L. Erbe, Hille-Wintner type comparison theorem for selfadjoint fourth orderlinear differential equations , Proc. Amer. Math. Soc. 80 (1980), 3, 417–422.[16] D. Franco, G. Infante, J. Per´an,
A new criterion for the existence of multiplesolutions in cones , Proc. Roy. Soc. Edinburgh Sect. A, 142 (2012), 1043–1050.[17] J. R. Graef, L. Kong, H. Wang,
A periodic boundary value problem withvanishing Green’s function , Applied Mathematics Letters 21 (2008), 176-180.[18] J. R. Graef, L. Kong, H. Wang,
Existence, multiplicity, and dependence ona parameter for a periodic boundary value problem , J. Differential Equations245 (2008), 1185-1197.[19] M. A. Krasnosel’ski˘ı,
Positive Solutions of Operator Equations , Noordhoff,Groningen, 1964.[20] M. K. Kwong, A. Zettl,
Asymptotically constant functions and second orderlinear oscillation , J. Math. Anal. Appl. 93 (1983), 2, 475–494.[21] G. S. Ladde, V. Lakshmikantham, A. S. Vatsala,
Monotone Iterative Tech-niques for Nonlinear Differential Equations , Pitman, Boston M. A. 1985.3522] H. Li, Y. Feng, C. Bu,
Non-conjugate boundary value problem of a thirdorder differential equation , Electron. J. Qual. Theory Differ. Equ. 2015, 21,1-19[23] R. Ma, Y. Lu
Disconjugacy and monotone iteration method for third-orderequations , Commun. Pure Appl. Anal., 13, 3, (2014), 1223-1236.[24] H. Persson,
A fixed point theorem for monotone functions , Appl. Math.Lett., 19 (2006) 1207–1209.[25] W. Simons,
Some disconjugacy criteria for selfadjoint linear differentialequations , J. Math. Anal. Appl. 34 (1971) 445–463.[26] P. J. Torres,
Existence of One-Signed Periodic Solutions of Some Second-Order Differential Equations via a Krasnoselskii Fixed Point Theorem , J.Differential Equations 190 (2003), 2, 643 – 662.[27] A. Zettl,
A constructive characterization of disconjugacy , Bull. Amer.Math. Soc. 81 (1975), 145–147.[28] A. Zettl,