Finite spectrum of 2n th order boundary value problems

Accepted Manuscript Finite spectrum of 2nth order boundary value problems Ji-jun Ao, Jiong Sun, Anton Zettl PII: DOI: Reference:

S0893-9659(14)00329-2 http://dx.doi.org/10.1016/j.aml.2014.10.003 AML 4643

To appear in:

Applied Mathematics Letters

Received date: 31 August 2014 Revised date: 11 October 2014 Accepted date: 11 October 2014 Please cite this article as: J.-j. Ao, J. Sun, A. Zettl, Finite spectrum of 2nth order boundary value problems, Appl. Math. Lett. (2014), http://dx.doi.org/10.1016/j.aml.2014.10.003 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

*Manuscript Click here to view linked References

Finite Spectrum of 2nth Order Boundary Value Problems Ji-jun Ao, Jiong Sun, and Anton Zettl Abstract. For any even positive integer 2n and any positive integer m we construct a class of regular self-adjoint and non-self-adjoint boundary value problems whose spectrum consists of at most (2n 1)m + 1 eigenvalues. Our main result reduces to previously known results for the cases n = 1 and n = 2: In the self-adjoint case with separated boundary conditions this upper bound can be improved to n(m + 1):

1. Introduction We study boundary value problems consisting of the equation (1.1)

(py (n) )(n) + qy = wy; on J = (a; b);

1 < a < b < +1

with boundary conditions (1.2)

AY (a) + BY (b) = 0; A; B 2 M2n (C):

Here and below Mk;m (F ) denotes the k m matrices over F = C; the complex numbers, or F = R the reals, Mk;m (F ) = Mk (F ) when m = k; Y = [y [0] ; y [1] ; ; y [2n 1] ]T ; where T denotes transpose, is the spectral parameter, n 2 N =f1; 2; 3; g; y [j] denotes the quasi-derivatives: y [j] = y (j) ; j = 0; : : : ; n 1; y [n+j] = (py (n) )(j) ; j = 0; : : : ; n 1; where y (j) denotes the ordinary derivative. The coe¢ cients are assumed to satisfy the minimal condition (1.3)

r = 1=p; q; w 2 L(J; C);

where L(J; C) is the set of complex valued functions which are Lebesgue integrable on J: It is well known that condition (1.3) implies that the quasi-derivatives y [j] have …nite limits at the endpoints a; b and thus (1.1), (1.2) is a well de…ned boundary value problem. Condition (1.3) is minimal in the sense that it is necessary and su¢ cient for all initial value problems of Eq.(1.1) to have a unique solution on [a; b]; see [8]. The next lemma is well known. It de…nes the characteristic function ( ) which characterizes the eigenvalues of the boundary value problem (1.1), (1.2) as zeros (roots) of ( ): 1991 Mathematics Subject Classi…cation. Primary 34B24, 34L15;Secondary 34L05. Key words and phrases. 2nth order boundary value problems, eigenvalues, …nite spectrum. This paper is in …nal form and no version of it will be submitted for publication elsewhere. 1

2

JI-JUN AO, JIONG SUN, AND ANTON ZETTL

Lemma 1. Let (1.3) hold and let (x; ) = [ ij (x; )]; x 2 J denote the fundamental matrix of the system representation of equation (1.1), (see (3.1) below) determined by the initial condition (a; ) = I, 2 C: Then a complex number is an eigenvalue of the BVP (1.1), (1.2) if and only if (1.4)

( ) = det[A + B (b; )] = 0:

The function ( ) = det[A + B (b; )]; 2 C is an entire function and is called the characteristic function of equation (1.1) and of its system representation (3.1). Proof. This is well known. It follows from Lemma 1 that - unless ( ) is identically zero - the eigenvalues, if any, of problem (1.1), (1.2) are isolated with no …nite accumulation point in C. Thus there are exactly four possibilities: (1) There are no eigenvalues. (2) Every complex number is an eigenvalue. (3) There are a countably in…nite number of eigenvalues. (4) There are n eigenvalues for some n 2 N: Clearly case (2) holds when A = B = 0 and case (1) holds when A = 0; B = I; the identity matrix, since the fundamental matrix (b; ) is nonsingular for every 2 C. Thus we call cases (1) and (2) degenerate. Definition 1. We call problem (1.1), (1.2) degenerate if 2 C or ( ) 6= 0 for every 2 C:

( )

0 for all

In the case when the coe¢ cients satisfy: (1.5)

r = 1=p; q; w 2 L(J; R); p > 0; w > 0 a:e: on J;

and the boundary conditions (1.2) are self-adjoint: (1.6)

AE2n A = BE2n B ; rank(A : B) = 2n;

where E2n is the simplectic matrix with k = 2n given by (1.7)

Ek = (( 1)r

k r;k+1 s )r;s=1 ;

k = 2; 3; 4; : : : ;

the boundary value problem (1.1), (1.2) is self-adjoint (has a representation as a self-adjoint operator S in the Hilbert space H = L2 (J; w)) and there are a countably in…nite number of eigenvalues, i. e., case (3) holds. For n = 1 Atkinson [5] proved that the positivity conditions on p and w can be considerably weakened to r = 1=p 0; w 0 and r and w are both positive on some common subinterval of J: We conjecture that this result of Atkinson holds also for n > 1 but this is an open problem. In this paper we …nd a class of problems for case (4). This we do by constructing a characteristic function ( ) which is a polynomial in : This construction is complicated and involves a partition of the interval J and a construction of nonnegative coe¢ cients r and w which are not positive on any common subinterval of J: Our main results reduce to known results when n = 1 [11] and n = 2 [1–4, 6]. Our construction has its roots in these papers and also uses an inductive scheme: The organization of this paper is as follows: Following this Introduction, Section 2 contains the statement of the main results and Section 3 their proofs. These are based on several lemmas, some of which may be of independent interest.

M ATRIX REPRESENTATIONS

3

2. Higher Order Boundary Value Problems with Finite Spectrum In this section we state our main results. For l = 2m + 1, m 2 N consider a partition of the interval J = (a; b) : (2.1)

a = a0 < a1 < a2 <

< al

1

< al = b;

such that Z

(2.2)

a2k+1

r

=

1 = 0 on [a2k ; a2k+1 ]; p

w(x)xs dx

6=

0; s = 0; 1; : : : ; n

a2k

1; k = 0; 1; : : : ; m;

and Z

(2.3)

q a2k+2

r(x)xs dx

a2k+1

Let (s)

rk

(s)

qk (2.4)

(s)

wk

= = =

Z

a2k+2

a2k+1 Z a2k+1 a Z 2k a2k+1

=

w = 0 on [a2k+1 ; a2k+2 ];

6= 0; s = 0; 1; : : : ; n

1; k = 0; 1; : : : ; m

r(x)xs dx; s = 0; 1; : : : ; n

1; k = 0; 1; : : : ; m

q(x)xs dx; s = 0; 1; : : : ; n

1; k = 0; 1; : : : ; m;

w(x)xs dx; s = 0; 1; : : : ; n

1; k = 0; 1; : : : ; m:

1:

1;

a2k

Now we can state our main results. The …rst theorem rules out an in…nite number of eigenvalues. Theorem 1. Let n; m 2 N and let l = 2m + 1: Suppose (1.3) and (2.1) to (2.4) hold. Then the boundary value problem (1.1), (1.2) is degenerate or has at most (2n 1)m + 1 eigenvalues. Proof. The proof consists of constructing a characteristic function which is a polynomial of order at most (2n 1)m + 1: The details will be given in the next section. Next we specialize to separated self-adjoint boundary conditions. Recently, Hao, Sun, Wang, and Zettl [10, 12, 13] have shown that these have a canonical representation of the following form: A1 U (a) + A2 V (a) = 0 B1 U (b) + B2 V (b) = 0;

(2.5) where U = (y [0] ; y [1] ;

; y [n

1] T

) ; V = (y [n] ; y [n+1] ;

T denotes transpose, and A1 ; A2 ; B1 ; B2 are n (2.6)

; y [2n

1] T

) ;

n matrices satisfying

n

A1 En A2 + ( 1) A2 En A1 = 0 = B1 En B2 + ( 1)n B2 En B1 ;

where En is given by (1.7) and the n have rank n:

2n matrices A = (A1 : A2 ); B = (B1 : B2 )

4

JI-JUN AO, JIONG SUN, AND ANTON ZETTL

Theorem 2. Let n; m 2 N and let l = 2m+1: Suppose (1.3), (2.1) to (2.4) hold. Then the boundary value problem (1.1), (1.2), (2.5), (2.6) has at most n(m + 1) eigenvalues. Proof. This will be given in the next section. Corollary 1. Let n; m 2 N and let l = 2m + 1: Suppose (2.7)

r = 1=p; q; w 2 L(J; R);

(2.1) to (2.4) hold and w satis…es: w 0 on J and w is positive on some subintervals of J: Then the boundary value problem (1.1), (1.2), (2.5), (2.6) has at most n(m + 1) eigenvalues and these are all real. Proof. The existence of at most n(m + 1) eigenvalues follows directly from Theorem 2. The proof that all eigenvalues are real is essentially the same as the proof for the classical case when p and w are both positive on J; it follows from the Lagrange identity, the hypothesis on w and the self-adjointness of the boundary conditions (2.5), (2.6). 3. Construction of Polynomial Characteristic Functions and Proofs Let ui = y [i 1] ; i = following system [7, 9]: (3.1) 0 0 1 0 B 0 0 1 B .. .. .. B B . . . B B 0 0 B 0 0 U0 = B B B 0 0 B B .. .. .. B . . . B @ 0 0 w

u01

q

1; 2; : : : ; 2n on J; then Eq.(1.1) can be written as the

.. . 1 0 0 .. .

.. . 0

.. .

1 p

0 1 .. .

0 .. .

0 0 .. .

.. .

0 0 .. .

0 0 0 0 0 0 .. .. . . 0 1 0 0

.. .

0

u02

1

0

u1 u2 .. .

B C B C B C B C B C B un 1 C B C C U; U = B un B C B un+1 C B C B C .. B C . B C @ u2n 1 A u2n

1

C C C C C C C C on J; C C C C C C A

; u02n = where = u2 ; = u3 ; ; u0n 1 = un ; u0n = run+1 ; u0n+1 = un+2 ; ( w q)u1 : With the next lemma we start our analysis of the characteristic function. Lemma 2. Let (1.3) hold and let (x; ) = [ ij (x; )]; x 2 J denote the fundamental matrix of the system representation (3.1) of equation (1.1), determined by the initial condition (a; ) = I, 2 C: Let ( ) be given by (1.4) for 2 C: Then ( ) = det(A) + det(B) +

2n X 2n X

cij

ij

i=1 j=1

(3.2)

+

+

X

1 i1 ;j1 ; ;i2n 1 ;j2n 1 2n j1 6=j2 6= 6=j2n 1

X

+

ci1 j1 i2 j2

i1 j1 i2 j2

1 i1 ;j1 ;i2 ;j2 2n j1 6=j2

ci1 j1

i2n

1 j2n

1

i1 j1

i2n

1 j2n

1

;

M ATRIX REPRESENTATIONS

5

where cij ; 1 i; j 2n; ci1 j1 i2 j2 ; 1 i1 ; j1 ; i2 ; j2 2n; j1 6= j2 ; ; ci1 j1 i2n 1 j2n 1 ; 1 i1 ; j1 ; ; i2n 1 ; j2n 1 2n; j1 6= j2 6= 6= j2n 1 are constants which depend only on the matrices A and B. Proof. We outline the proof here. The details are similar to those given in [14] for the second order case and in [1, 6] for the fourth order case and hence omitted. Note that for any A = (aij ); B = (bij ) 2 M2n (C) det(A + B) = det(A) + det(B) + P (A; B);

where P (A; B) denotes the sum of the possible products of the elements of A and B belonging to di¤erent rows and di¤erent columns: Hence we have ( ) = det[A + B (b; )] = det(A) + det(B (b; )) + P (A; B (b; )): Since

(a; ) = I, det( (b; )) = 1 and P (A; B (b; )) can be written in the form P (A; B (b; )) =

2n X 2n X

cij

ij

i=1 j=1

+

X

X

+

ci1 j1 i2 j2

i1 j1 i2 j2

+

1 i1 ;j1 ;i2 ;j2 2n j1 6=j2

ci1 j 1

i2n

1 j2n

1

i1 j1

i2n

1 j2n

1

;

1 i1 ;j1 ; ;i2n 1 ;j2n 1 2n j1 6=j2 6= 6=j2n 1

where cij ; 1 i; j 2n; ci1 j1 i2 j2 ; 1 i1 ; j1 ; i2 ; j2 2n; j1 6= j2 ; ; ci1 j1 i2n 1 j2n 1 ; 1 i1 ; j1 ; ; i2n 1 ; j2n 1 2n; j1 6= j2 6= 6= j2n 1 are constants which depend only on the matrices A and B. Corollary 2. Let (1.3) hold and let (x; ) = [ ij (x; )]; x 2 J denote the fundamental matrix of the system representation of equation (1.1) determined by the initial condition (a; ) = I, 2 C: Let ( ) be the characteristic function given by (1.4) for 2 C: Consider the problem (1.1) with separated self-adjoint BCs (2.5), (2.6). Then X ci1 j1 in jn i1 j1 (3.3) ( )= in jn : 1 i1 ;j1 ; ;in ;jn 2n j1 6=j2 6= 6=jn

Proof. Note that the last n rows of A and the …rst n rows of B in the boundary conditions (1.2) are zero. Hence we have det(A) = 0; det(B) = 0 and the forms other then ci1 j1 in jn i1 j1 in jn vanish in (3.2), and the result follows. Next we study the structure of the principal fundamental matrix (x; ) for coe¢ cients satisfying conditions (2.1)-(2.4). This is basic to our results. Lemma 3. Let (1.3), (2.1)-(2.4) hold. Let (x; ) = [ ij (x; )] be the fundamental matrix solution of the system (3.1) determined by the initial condition (a; ) = I for each 2 C. Let 8 " # (1) > F O > k > ; k = 0; 2; : : : ; 2m; > (2) (1) > > Fk Fk < (3.4) Fk (x; ; ak ) = " # > > (1) (3) > F F > k k > ; k = 1; 3; : : : ; 2m 1; > (1) : O Fk

6

JI-JUN AO, JIONG SUN, AND ANTON ZETTL

where O denotes the n de…ned as follows:

(1)

(3.5)

(3.6)

(2)

Fk

Fk

2

6 6 6 =6 6 6 4

1

x

0 .. . 0 0

1 .. . 0 0

Rx

(3)

Fk

Then we have (3.8)

ak )dt

w

q)(x

t)n

2 (t

ak )dt

q)(x

Rx

w

ak (

1

r(x

t)n

2

r(x Rx

(ak ; ) = Fk

ak

1 (ak ;

a ( R xk ak (

Rx

(t

ak )dt

Rx

Raxk ak

r(x r(x

Rx

ak )dt

ak

ak )dt 1)

(ak

1;

w

q)(x

w

q)(x

ak (

ak )dt

ak )dt

; ak

Rx

ak )dt

(t

t)(t r(t

t)(t

q)(t

t)n

ak

ak

1 (t

r(x Rx ak

w

(a2k+1 ; ) = Tk Tk

1

n 1 1 (t ak ) dt (n 1)! (t ak )n 1 n 2 t) dt (n 1)!

t)n

(t ak )n 1 dt (n 1)! (t a )n 1 q) (n k 1)! dt

q)(x

Rx

ak (

w

t)

n 1 (t ak ) (n 1)! n t)n 2 (t (nak )1)!

t)n

n

t) (t (nak )1)! n

1

r (t (nak )1)! dt

1

dt

1 );

k= m; we

T0 ; k = 1; 2; : : : ; m:

Proof. Observe from (3.1) that un is constant on each subinterval [a2k ; a2k+1 ]; k = 0; 1; : : : ; m; where r is identically zero, and on each subinterval [a2k ; a2k+1 ]; k = 0; 1; : : : ; m we have that (3.10) un (x) = un (a2k ); un 1 (x) = un 1 (a2k ) + un (a2k )(x a2k ); a2k )2 un 2 (x) = un 2 (a2k ) + un 1 (a2k )(x a2k ) + un (a2k ) (x 2! ; 2

n

1

a2k ) ) u1 (x) = u1 (a2k ) + u2 (a2k )(x a2k ) + u3 (a2k ) (x 2! + + un (a2k ) (x (na2k1)! ; Rx Rx u2n (x) = u2n (a2k ) + u1 (a2k ) a2k ( w q)dt + u2 (a2k ) a2k ( w q)(t a2k )dt Rx )n 1 + + un (a2k ) a2k ( w q) (t (na2k1)! dt; n

) un+1 (x) = un+1 (a2k ) + un+2 (a2k )(x a2k ) + + u2n (a2k ) (x (na2k1)! Rx n 1 +u1 (a2k ) a2k ( w q)(x t) dt + Rx n 1 n 1 (t a2k ) +un (a2k ) a2k ( w q)(x t) (n 1)! dt:

1

3

7 dt 7 7 7 7: 7 1 dt 7 5 1

):

And, if we let T0 = F0 (a1 ; ; a0 ); Tk = F2k (a2k+1 ; ; a2k )F2k 1 (a2k ; ; a2k 1; 2; : : : ; m; then (a1 ; ) = F0 (a1 ; ; a0 ) = T0 and, in general, for 1 k have (3.9)

are

7 7 7 7; 7 7 5

2

1

t)n

r(x

Rx

x

q)(x

w

(3)

3

1

.. .

w

ak (

Raxk ak

ak 1 0

a ( R xk ak (

Rx

(x ak )n (n 1)! (x ak )n (n 2)!

.. .

Rx

k

2 R x r(x t)n 1 dt 6 Raxk 6 n 2 dt 6 ak r(x t) 6 =6 6 Rx 6 r(x t)dt ak 4 Rx rdt ak

x

(2)

n matrices Fk ; Fk ; Fk

(x ak )2 2!

ak

2 R x ( w q)(x t)n 1 dt 6 Rak 6 x 6 a ( w q)(x t)n 2 dt k 6 =6 6 Rx 6 6 q)(x t)dt ak ( w 4 Rx ( w q)dt a

(3.7)

(1)

n zero matrix and the n

3

7 7 7 7 7; 7 7 7 5

M ATRIX REPRESENTATIONS

7

Similarly, because q and w are both identically zero, u2n is constant on each subinterval [a2k 1 ; a2k ]; k = 1; 2; : : : ; m; and so we have (3.11) u2n (x) = u2n (a2k 1 ); u2n 1 (x) = u2n 1 (a2k 1 ) + u2n (a2k 1 )(x a2k 1 ); 2 2k 1 ) u2n 2 (x) = u2n 2 (a2k 1 ) + u2n 1 (a2k 1 )(x a2k 1 ) + u2n (a2k 1 ) (x a2! ; n

1

1) ; un+1 (x) = un+1 (a2k 1 ) + un+2 (a2k 1 )(x a2k 1 ) + + u2n (a2k 1 ) (x a(n2k 1)! Rx Rx un (x) = un (a2k 1 ) + un+1 (a2k 1 ) a2k 1 rdt + un+2 (a2k 1 ) a2k 1 r(t a2k 1 )dt Rx n 1 1) + + u2n (a2k 1 ) a2k 1 r (t a(n2k 1)! dt;

+ u2 (a2k 1 )(x a2k 1 ) + + un (a2k 1 ) (x Rx +un+1 (a2k 1 ) a2k 1 r(x t)n 1 dt + Rx n 1 1) dt: +u2n (a2k 1 ) a2k 1 r(x t)n 1 (t a(n2k 1)!

u1 (x) = u1 (a2k

1)

a2k 1 )n (n 1)!

1

From this we can see the piecewise construction of ui (x); i = 1; 2; : : : ; 2n as functions on [a; b]. If we let U (x) = [u1 (x); u2 (x); ; u2n (x)]T on [a; b], and let Uj (x; ) = U (x; ; ej ), where ej ; j = 1; 2; : : : ; 2n are the 2n dimensional identity vectors, then we see that (x; ) = [U1 (x; ) U2 (x; ) U2n (x; )]. This establishes (3.8). Note that b = a2m+1 . Thus the structure of following:

given in Lemma 3 yields the

Corollary 3. For the fundamental matrix = ( ij ) we have that (3.12) m + ~ ij ( ); i; j = 1; 2; : : : ; n or i; j = n + 1; n + 2; : : : ; 2n; ij (b; ) = Rij m+1 + ~ ij ( ); i = n + 1; n + 2; : : : ; 2n; j = 1; 2; : : : ; n; ij (b; ) = Rij m 1 + ~ ij ( ); i = 1; 2; : : : ; n; j = n + 1; n + 2; : : : ; 2n; ij (b; ) = Rij (i)

where Rij ; 1 i; j 2n are constants depending on rk ; k = 0; 1; : : : m 1; i = (i) 0; 1; : : : ; n 1; wk ; k = 0; 1 : : : m; i = 0; 1; : : : ; n 1 and the endpoints a; b; ~ ij ( ) are functions of , in which the degrees of are less than m; m + 1; m 1; respectively. For example, 11 (b; ) = R11 m + ~ 11 ( ); and so the degree of in ~ 11 ( ) is < m: Corollary 4. If Rij 6= 0; 1 i; j 2n in (3.12), then (1) The degree of ij (b; ) is m for i; j = 1; 2; : : : ; n or i; j = n + 1; n + 2; : : : ; 2n: (2) The degree of ij (b; ) is m+1 for i = n+1; n+2; : : : ; 2n; j = 1; 2; : : : ; n: (3) The degree of ij (b; ) is m 1 for i = 1; 2; : : : ; n; j = n + 1; n + 2; : : : ; 2n: Now we prove our main results. P roof of T heorem 1: Since ( ) = det[A + B (b; )] where (b; ) = [ ij (b; )], from Lemma 2 and Corollary 3 we know that the characteristic function ( ) is a polynomial in and has the form of (3.2). We denote the maximum of degree of in ij (b; ) by dij ; 1 i; j 2n, by Corollary 3 the maximum of degree of in the matrix (b; ) can be written as the following matrix (3.13)

(dij ) =

(m)n n (m + 1)n

n

(m 1)n (m)n n

n

;

8

JI-JUN AO, JIONG SUN, AND ANTON ZETTL

where (m)n n denotes the n n matrix in which all the elements are m. From (3.2) and (3.13) we conclude that the maximum of the degree of in ( ) is (2n 1)m + 1. Thus from the Fundamental Theorem of Algebra, ( ) has at most (2n 1)m + 1 roots or ( ) = 0 for all 2 C: Remark 1. For n = 1 Theorem 1 reduces to a result obtained in [11]; and for n = 2 it reduces to a result in [1]. P roof of T heorem 2: From Corollary 2, Corollary 3 and (3.13) we know that the maximum of degree of the characteristic function ( ) is n(m + 1). Hence from the Fundamental Theorem of Algebra, ( ) has at most n(m + 1) roots and the result follows. Finally, we give an example to illustrate our main results. Example 1. Let 8 > > > > > > < (3.14) > > > > > > :

n = 2 and consider the fourth order boundary value problem (py 00 )00 + qy = wy; on J = ( 1; 4); y( 1) + (py 00 )( 1) = 0; y 0 ( 1) (py 00 )0 ( 1) = 0; 2y(4) (py 00 )0 (4) = 0; y 0 (4) (py 00 )(4) = 0:

Choose m = 2 and suppose p; q; w are piecewise polynomial functions de…ned as follows: (3.15) 8 8 8 1; ( 1; 0) 2; ( 1; 0) > > > 1; ( 1; 0) > > > > > > > > > 0; (0; 1) 1; (0; 1) < < 0; (0; 1) < 2; (1; 2) 3; (1; 2) 1; (1; 2) w(x) = q(x) = p(x) = > > > > > > 0; (2; 3) 0; (2; 3) 1=2; (2; 3) > > > > > > : : : 4; (3; 4): x 1; (3; 4); 1; (3; 4); From the conditions 0 1 B 0 B (3.16) A=B B 0 @ 0

given we know that 1 0 0 1 0 B 1 0 1 C C B B 0 0 0 C ; B = C B @ 0 0 0 A

0 0 2 0

0 0 0 1

0 0 0 1

Then we can deduce that the characteristic function ( )

1337=3888

5

+510814321=373248

2

= 1=972

6

+ 2046073=93312

4

33600737=15552

0 0 1 0

1

C C C: C A

29748595=93312

3

+ 1443137=1296:

Hence the BVP (3.14), (3.15) has exactly n(m + 1) = 6 eigenvalues 0

= 1:1006;

1

= 1:6442;

2

= 3:0943;

3

= 12:5296;

4

= 60:3816;

5

= 255:4998:

Acknowledgments. We thank the referee for her/his comments and suggestions. These resulted in a correction of the statement of Corollary 4, a more illustrative Example 1, and other improvements. This work was supported by National Natural Science Foundation of China (Grant Nos.11301259 and 11161030), Natural Science Foundation of Inner Mongolia (Grant Nos.2013MS0105 and 2014BS0105) and a Grant-in-Aid for Scienti…c Research from Inner Mongolia University of Technology (Grant No.ZD201310). The

M ATRIX REPRESENTATIONS

9

third author was supported by the Ky and Yu-fen Fan US-China Exchange fund through the American Mathematical Society. This made his visit to Inner Mongolia University possible. The third author also thanks the School of Mathematical Sciences of Inner Mongolia University for its hospitality. References [1] J. J. Ao, F. Z. Bo and J. Sun, Fourth order boundary value problems with …nite

spectrum, App. Math. Comput., 244 (2014), 952-958. [2] J. J. Ao, J. Sun, Matrix representations of fourth order boundary value problems

[3] [4] [5] [6]

[7] [8]

[9] [10] [11] [12] [13] [14]

with coupled or mixed boundary conditions, Linear and Mltilinear Algebra, Published online( doi:10.1080/03081087.2014.959515.) J. J. Ao, J. Sun and A. Zettl, Matrix representations of fourth order boundary value problems with …nite spectrum, Linear Algebra and Its Appl., 436 (2012), 2359-2365. J. J. Ao, J. Sun and A. Zettl, Equivalence of fourth order boundary value problems and matrix eigenvalue problems, Result. Math., 63 (2013), 581-595. F. V. Atkinson, Discrete and Continuous Boundary Problems, Academic Press, New York/London, 1964. F. Z. Bo and J. J. Ao, The …nite spectrum of fourth order boundary value problems with transmission conditions, Abstract and Applied Analysis, 2014(2014), Article ID 175489, 7 pages. B. Chanane, Accurate solutions of fourth order Sturm-Liouville problems, J. Comp. and Appl. Math., 234, (2010), 3064-3071 W. N. Everitt and D. Race, On necessary and su¢ cient conditions for the existence of Caratheodory solutions of ordinary di¤erential equations, Quaest. Math., 3 (1976), 507-512. L. Greenberg and M. Marletta, Numerical methods for higher order Sturm-Liouville problems, J. Comp. and Appl. Math., 125 (2000), 367-383. X. Hao, J. Sun, A. Wang and A. Zettl, Characterization of domains of self-adjoint ordinary di¤erential operaters II, Result. Math., 61 (2012), 255-281. Q. Kong, H. Wu and A. Zettl, Sturm-Liouville problems with …nite spectrum, J. Math. Anal. and Appl., 263 (2001), 748-762. A. Wang, J. Sun and A. Zettl, Characterization of domains of self-adjoint ordinary di¤erential operaters, J. Di¤erential Equations, 246 (2009), 1600-1622. A. Wang, J. Sun and A. Zettl, The classi…cation of self-adjoint boundary conditions: Separated, coupled, and mixed, J. Funct. Analysis, 255 (2008), 1554-1573. A. Zettl, Sturm-Liouville Theory, Amer. Math. Soc., Mathematical Surveys and Monographs 121, 2005.

Ji-jun Ao: College of Sciences, Inner Mongolia University of Technology, Hohhot, 010051, China. E-mail address : [email protected] Jiong Sun: School of Mathematical Sciences, Inner Mongolia University, Hohhot, 010021, China. E-mail address : [email protected] Anton Zettl: Mathematics Department, Northern Illinois University, DeKalb, IL. 60115, USA E-mail address : [email protected]