CC BY
0Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 619358,11 pages http://dx.doi.org/10.1155/2013/619358
Research Article
The Modified Parseval Equality of Sturm-Liouville Problems with Transmission Conditions
Mudan Bai, Jiong Sun, and Siqin Yao
School of Mathematical Sciences, Inner Mongolia University, Hohhot 010021, China Correspondence should be addressed to Jiong Sun; masun@imu.edu.cn Received 5 July 2013; Revised 5 September 2013; Accepted 26 September 2013 Academic Editor: Alberto Cabada
Copyright © 2013 Mudan Bai et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider the Sturm-Liouville (S-L) problems with very general transmission conditions on a finite interval. Firstly, we obtain the sufficient and necessary condition for X being an eigenvalue of the S-L problems by constructing the fundamental solutions of the problems and prove that the eigenvalues of the S-L problems are bounded below and are countably infinite. Furthermore, the asymptotic formulas of the eigenvalues and eigenfunctions of the S-L problems are obtained. Finally, we derive the eigenfunction expansion for Green's function of the S-L problems with transmission conditions and establish the modified Parseval equality in the associated Hilbert space.
1. Introduction
The Sturm-Liouville (S-L) theory, as an active area of research in pure and applied mathematics, plays an important role in solving many problems in mathematical physics and is concerned in many publications [1-7]. It is well known that for the classical S-L problems, the solutions or the derivatives of the solutions are continuous on the interval, but these conditions cannot be satisfied in many practical physical problems. So, a class of S-L operators with "discontinuity," that is, the S-L problems with transmission conditions at an interior point, are concerned by many mathematical and physical researchers [8-10]. Such conditions are known by various names including transmission conditions [11, 12], interface conditions [13-15], jump conditions [16], and discontinuous conditions [17,18].
In this paper, we consider the following Sturm-Liouville equation:
ly:=-y" + q(x)y = Xy, xe/=[-1,0)u(0,1], (1)
with boundary conditions:
ky ■= a.iy{-1) + <X2y (-1) = 0, l2y ■= ßiy(1) + ß2y' (1) = 0,
(2) (3)
and transmission conditions:
l3y ■=y(0+)-Yly(0-)-y2y' (0-) = 0, lAy := y' (0+) - y3y (0-) - y4y' (0-) = 0,
where A is a complex eigenparameter and q e L(I, R); notice that the potential function q(x) guarantee y(0±) and y'(0±) in (4) makes sense (see Theorem 1); all coefficients of the boundary and transmission conditions are real numbers. Throughout this paper, we assume that a^ =0, ^ +p2 =0, and
We derive the eigenfunction expansion for Green's function of the S-L problem (1)-(4) and establish the modified Parseval equality of the S-L problem with very general transmission conditions at one inner point 0 of the finite interval [-1,1].
The organization of this paper is as follows. After the Introduction, we construct the basic solutions of S-L equation (1) with transmission conditions (4) and obtain the sufficient and necessary condition for A being an eigenvalue of the S-L problem in Section 2. In Section 3, the asymptotic formulas for eigenvalues and eigenfunctions of the S-L problem are
obtained by using the asymptotic expressions of the solutions, and we prove that the eigenvalues of the S-L problem are bounded below and are countably infinite. Section 4 contains the eigenfunction expansion for Green's function of the S-L problem (1)-(4) and we establish the modified Parseval equality in the associated Hilbert space by using the eigenfunction expansion for Green's function.
2. The Basic Solutions and Eigenvalues
We construct the basic solutions of S-L equation (1) with transmission conditions (4) and obtain the sufficient and necessary condition for X being an eigenvalue of the S-L problem in this section. At first, we prove the existence of finite limits for all solution y of (1) and its derivative at both sides of 0 point in the following theorem.
Theorem 1. Assume the coefficient function q(x) in S-L equation (1) is real-valued and Lebesgue integrable on [-1, 0) and (0,1], that is, q e L(I, R); then for the solution y of (1), the limits
y (0±) = lim y (x), y' (0±) = lim y' (x) (6)
exist and are finite.
In order to prove Theorem 1, we need the following lemma.
Lemma 2 (see [19]). Let J = (a,b), ->x> < a < b < >x>. Let Mnm(L(a, b), C) denote the set of n x m matrices with entries from L((a, b), C)(n, m e N).
(1) Suppose that P, F satisfy
PeMn (L(a,c), C); F e Mn,m (L(a,c), C)
for some c e (a, b), where Mn(L(a, c), C) = Mn n (L(a,c), C). For some u e J and C e Mnm(C), let Y be the solution of
Y' = PY + F, Y (U) = C. (8)
Then Y(a) = limt^a+ Y(t) exists and is finite.
(2) Suppose that
PeMn (L(c,b), C); F e Mn,m (L(c,b), C)
for some c e (a,b). For some u e J and C e Mn m(C), let Y be the solution of (8) on J. Then,Y(b) = limt^b_ Y(t) exists and is finite.
Proof of Theorem 1. Let a = 0, b = 1, and let Y = ( yy< P = ( q-xo and F = (0). Then (1) is equivalent to the equation Y' = PY
+ F on (0,1]. From Lemma 2, we knowthat Y(0+) = lim Y(t) exists and is finite. This implies that
y(0+)= lim y(x), y' (0+) = lim y' (x) (10) exist and are finite.
Let a = -l,b = 0. Then, (1) is equivalent to the equation Y' = PY + F on [-1, 0). Since Y(0-) = limt^0- Y(t) exists and is finite by Lemma 2, the limits
y (0-) = lim y (x),
exist and are finite.
y (0-) = lim y (x) (11)
x —» 0—
Let us define a new inner product in L2(I) as follows, which is associated with transmission conditions (4) and useful to investigate the S-L problem (1)-(4):
if' g) = p\ fiBidx + \ f2~g2dx> for f,g zL2 (I),
J-l J0
where f = f(x)\[-1fl)'f2 = f(x)|(01]. It is easyto verifythat (L2(I)' {•'•)) is a Hilbert space. For simplicity, it is denoted by H. The norm induced by the inner product is denoted by || • ||H. We consider the S-L problem (1)-(4) in the associated Hilbert space H.
Theorem 3 (see [14]). The S-L problem (1)-(4) is self-adjoint.
We construct two basic solutions <p(x, X) and %(x, X) of S-L equation (1) by the following procedure. At first, we consider the initial-value problem
-y + q(x) y = Xy, xe|-1,0), y(-\) = a2, y (-1) = -ai.
By virtue of Theorem 1.5 in [20], the problem (13) has a unique solution (x' X) for each X z C, which is an entire function ofX for each fixed x z [-1,0).
Similarly, the initial-value problem
-y" + q(x) y = Xy, x e (0,1], y(0+)=y1fa (0-X) + y2Û (0-,X), y (0+) = y3$1 (0-X) + y4Ú (0-,X)
has a solution 02(x, X) for each X z C. Moreover, 02(x, X) is an entire function of X for each fixed x z (0,1]. We define a function 0(x, X) onx z I = [-1,0) U (0' 1] by
0 (x, X) = -
(x, X), 02 (x, X),
x e [-1,0), x e (0,1].
Obviously, 0(x,X) satisfies S-L equation (1), the boundary condition (2), and both transmission conditions (4); that is, 0(x, X) is a solution of S-L equation (1).
As same as above, the initial-value problem
-y" + q(x) y = Xy, x e (0,1]. y(1) = ß2, y' (1) = -ß1
has a solution x2(x,X) which is an entire function of X for each fixed x e (0,1]. Similarly, the initial-value problem
-y" + l(x) y = ^y, x e [-1,0), y(0-)=l(Y4X2 (0+,V-Y2X2 (0+,V),
y (0-) = --{Y3X2 (0+,V-YIX2 (0+,*))
also has a unique solution x1 (x, X), which is an entire function of X for each fixed x e [-1, 0). Let
X(x,X) = ^
Xi (x,X), xe[-1,0), X2 (x,X), xe(0,1].
By (16) and (17), x(x,h) satisfies S-L equation (1) and conditions (3) and (4); that is, x(x, X) is another solution of S-L equation (1).
It is well known that, from the ordinary linear differential equation theory, the Wronskians wXX) := W(^j(x, X), Xj(x, X)) (j = 1,2) are independent of the variable x.
Lemma 4. The equality <m2(X) = pw1(X) holds for each X e C.
Proof. Since the Wronskians Wj(X) := W(4j(x,X),Xj(x,X)) (j = 1, 2) are independent of the variable x, then by (14) and (17),
W2 (X)
$2 (0+,X) X2 (0+,V $2 (0+,X) x'2 (0+,X)
yiti (0-,X) + Y2$[ (0-,X) yiXi (0-,X) + Y2X1 (0-,X)
Y301 (0-, X) + Y40l (0-, X) Y3Xi (0-, X) + Y4x1 (0-, X)
= p^i W ■
(19) □
Let w(X) = w1(X). Then w(X) = (1/p)w2(X) and is an entire function of X.
Theorem 5. The eigenvalues of the S-L problem (1)-(4) coincide with the zeros of the function w(X).
Proof. Let X0 be an eigenvalue of the S-L problem (1)-(4) and let y(x,X0) be any corresponding eigenfunction. Let us assume that <m(X0) = 0. Then, W($j(x, X0), Xj(x, X0)) = 0 (j = 1,2). Consequently, each pair of functions ^1(x, X0), x1(x, X0) and <p2(x,\0),x2(x,X0) is linearly independent. Therefore, the solution y(x, X0) of (1) maybe represented in the form
y(x,X0) =
CiQi (x,X0)+C2Xi ^ X0), difa {x,Xo) + d2X2 (x,^o)
x e [-1,0), X e (0, 1] ,
where at least one of the constants c1 ,c2,d1,d2 is not zero. Since y(x,X0) satisfies boundary conditions (2) and (3), we obtain c2 = 0, d1 = 0. And substituting (20) in condition (4), we get c1 and d2 satisfying the following equations:
d2X2 K) - Ci {Yi<Pi (0-, K) + 72(0-, Xo)) = 0 à2X2 X0) - Ci {h4i(0-, ^0) + Y4(0-, xo)) =
From (14), the determinant of the matrix of coefficients in the last equations is equal to <m2(X0) which is not zero by the assumption. Hence, c1 = 0 and d2 = 0. This is a contradiction. Thus, each eigenvalue of the S-L problem (1)-(4) is a zero of the function w(X).
Conversely, let <m(X0) = 0. Then, W(^1(x, X0), X1(x,X0)) = 0 for all x e [-1,0). Consequently, the functions (x,X0) and Xi_(x,h0) are linearly dependent. That is,
X1 (x,\0) = k1fa (x,X0), xe[-1,0), (22) for some k1 = 0. Then
«1X(-1,X0) + a2X H^
= «1^1 (-1,X0)+a2X1 H^ (23)
= k1 (a1$1 (-1, xo) + <h$1(-1, Ao)) = 0
So, x(x, X0) satisfies condition (2). And the solution x(x, X0) satisfies condition (3) and both transmission condition (4) from its construction. Thus x(x, X0) would be an eigenfunc-
tion of the S-L problem (1)-(4) for the eigenvalue X0.
Lemma 6. The eigenvalues of the S-L problem (1)-(4) are simple.
Proof. Let X = u + iv. We differentiate the equation lx(x, X) = Xx(x,X) with respect to X and have IX\(x,X) = Xxx(x,X) + X(x, X). Then
ik^ <t>) - (Xk H) = i^Xx + X,(t>)- <XX' x4) = {x,^)+2iv{xx ,4°).
By integration by parts and some calculations
dXx,4)-(Xx,m
= Pj (-Xx (x, x) + q (x) Xx (x, h)) 4(x, h)dx
+ j (-Xx (x, + q (x) XX (x, x)) 4(x, X)dx
- P j Xx (x, X) (-0" (x, X) + q (x) 0 (x, X))dx
- \ Xx (x, X) (-0" (x, X) + q (x) 0 (x, X))dx °
= P [(Xix (x>X) 0[ (x>X) - Xix (x>X) 0i (x>X))
+ (Xix (x, X) 02 (x, X) - x2x (x, X) 02 (x, X))I = p{a2Xx (-1,X)+aiXx (-1,X))
from (13), (14), (16), and (17) where Xj\(x, X), x'j\(x, X) are the derivatives of Xj(x,X), Xj(x,X) with respect to X(j = 1,2). Since w(X) is independent of the variable x and the value of 01(-1,X) and 0' (-1, X) given in (13),
dj1(X)
= dX(0i (-1'X)X1 (-l'X)
-01 (-1,X)X1 (-1,X))
= a2Xn (-1,X) + aiXix (-1,X)-
So, by (24), (25), and (26),
Let X0 be an eigenvalue of the S-L problem (1)-(4). Then X0 is real. Since <m(X0) = 0 for the eigenvalue of the S-L problem (1)-(4) from Theorem 5, Xj(x,X0) = Cj0j(x,X0) for some Cj = 0 (j = 1,2). And from (4),
X2 (0+' Xo) = ci (Yi0i(0-' Xo) + Ï20i(0-' Xo))
= Ci02 (0+'X0) '
X2 (0+' Xo) = ci (Ys0i(0-' X0) + Ï40[ (0-' Xo))
= C102 (0+'X0)'
we get c1 = c2 = 0. Thus, (27) becomes
«>' (Xo)=~(X'0) = cL {0'0) = 0. P P
Hence X0 is simple.
3. Asymptotic Formulas for
Eigenvalues and Eigenfunctions
In this section, the asymptotic formulas for eigenvalues and eigenfunctions of the S-L problem (1)-(4) are obtained by using the asymptotic expressions of the solutions. At first, we calculate the asymptotic expressions of the solutions.
(29) □
Lemma 7. Let X = s2, s = a + it. If a2 = 0, then one has the following asymptotic expressions:
d-k0i (x, X) = «2d-k cos s (x + 1) + O^e^) ,
k = 0,1,
as |X| ^ <xi. Ifa2 = 0, then
jk d: s d xk
k k j-k0i (x, X) = -^d- sins (x + 1) + o(lslk-2e^) ,
k = 0,1, (31)
as |X| ^ œ>.
Proof. The asymptotic formulas for 01 (x, X) follow from the similar formulas of Lemma 1.7 in [20]. □
The asymptotic formulas for 02(x, X) are as follows.
Lemma 8. Let X = s2, s = a + it; then 02(x,X) has the following asymptotic formulas as |X| ^ >x>:
(1) ifa2 = 0,y2 = 0, then
02 (x, X) = -a2y2s sin s ddL- cos sx + O , (32)
(2) ifa2 = 0,y2 = 0, then
—k02 (x,X)
dk d = a2 (y1 cos s—k cos sx - y4 sms^-r sin sx
+ O(|sГ1eWx), (3) ifa2 = 0,y2 = 0, then
02 (x, X) = -a1y2 cos s-^-j cos sx + O (|s| 1e^x) ,
(4) ifa2 = 0,y2 = 0, then j-j2 (x,X)
dk dk y, sin s-T cos sx + y4 cos s-T sin sx
1 d xk 4 d xk
O^e1^ )
fork = 0,1.
Proof. Let X = s2, s = a + it. <p2(x,X) satisfies (14). By the method of variation of constants, we have $2(x, X) satisfying the following integral equation:
42 (x,X) = (y1^1 (0-,x) + Y2<p[ (0-,^)) cossx
+ 1 (y3$1 (0-, X) + (0-, X)) sin sx (36) 1 (x
+ - I sin [s (x - z)] q(z) 42 (z,X)dz. s J0
Let a2 = 0. Substituting (30), for k = 0,in (36), we have 42 (x, X) = - a2y2s sin s cos sx
+ a2 (Y1 cos s cos sx - y4 sin s sin sx)
+ -a2y3 cos s sin sx
- \ sin [s (x - z)] q (z) 42 (z,X)dz s Jo
+ o(Vl(*+1)
We consider the case y2 = 0. Let ip2(x,X) = isle1 tlxF(x,X). Multiplying (37) by isi-1e-^x, we have the following:
F (x, X) = - a2y2e sin s cos sx
+ —a2e~ltlx (y1 cos s cos sx - y4 sin s sin sx)
1 -ltlx
+---a2y3e " cos s sin sx
M2 1 cx
+ — \ sin [s (x - z)] e~w(x~z)q (z) F (z, X) dz jsj Jo
Let ^(X) = max_1SxSljF(x, A)j. Then
/4W < \«2Y2\ + ^
for some M0 > 0. Consequently, X) = 0(1) as |A| ^ ot.
So, ip2(x,X) = isle1 t,xF(x,X) = 0(|s|eltlx) as W ^ ot. Substituting the asymptotic equality in (37) gives (32) for k = 0. And the case for k = 1 is obtained by differentiating (37) and by some similar calculations.
Similarly, we can prove (33), (34), and (35). □
When X is big enough, the asymptotic formulas for <p2(x,X) are obtained in Lemma 8. Substituting the asymptotic formulas of <p2(x,X) and the value of x2(x,X) at x = 1, which is given in (16), into to(X) = (1/p)to2(X), we can establish the following lemma.
Lemma 9. Let X = s2, s = a + it; then to(X) has the following asymptotic formulas for large enough X.
(1) If a2 = 0,y2 = 0, = 0, then
(X) = —^2Y2*2sin2s + O (jsj e1 ll). (40)
(2) If a2 = 0,y2 = 0, ß2 = 0, then
to (X) = —a2ß1y2ssins coss + O . (41)
(3) If a2 = 0,y2 = 0, ß2 = 0, then
to (X) = —a2ß2 (y1 + y4)ssins coss + O . (42)
(4) If a2 = 0,y2 = 0, ß2 = 0, then
to(X) = —a2ß1 (y4sin2s - y^os^) + O (jsj (43)
(5) If a2 = 0,y2 = 0, ß2 = 0, then
to (X) = -—a1ß2y2s sins cos s + O (e^) . (44) p
(6) If a2 = 0, y2 = 0, ß2 = 0, then
to (X) = —a1ß1y2cos2s + O (jsj-1eltl). (45) p
(7) If a2 = 0,y2 = 0, ß2 = 0, then
to(X) = —a1ß2 (y^in2s - y4cos2s) + O (jsj ). (46)
(8) If a2 = 0,y2 = 0, = 0, then
to (X) = —a161 (v1 + y4) sins cos s + O (|s|-2e't') . (47) ps y '
Theorem 10. The eigenvalues of the S-L problem (1)-(4) are bounded below and countably infinite; the unique cluster point is infinity.
Proof. From Theorem 5, the eigenvalues of the S-L problem (1)-(4) are zeros of the function to(X) which is an entire function of X. And the asymptotic formulas of the function to(X) are obtained in Lemma 9. Let
to (X) = to* (X) +S(X), 22
where to*(X) = -(1/p)a2ß2y2s2sin2s,8(X) = 0(jsjeltl) for the case a2 = 0,y2 = 0, ß2 = 0 from Lemma 9. Let X = s2, s = a + it and let
r'n = {a = s2 = (a + it)2 j jaj = (n + n,0 < t < n} , r" = {a = s2 = (a + it)2 jjaj<(n+—)n,t = n},
and let rn = ^ U r". Then rn is a closed curve on the plane of X. Next, we prove that |«*(X)| > ^(X^ on Tn. When W is big enough, on ^
1 I is _->s| 1 I i(n+(1/2))n -t -i(n+(1/2))n t\
|sin s| = — \e - e = - e 1 1 2 1 1 2
1|. -t .ti 1 tN -2t\^em = - ie + ie = —e 1 + e > —. 21 ' 2 ' ' 4
Similarly, when |X| is big enough, on r 1
. . ion -n -ion n
sins| = —\e e - e e 1 2 1 1
2ion -2n\ ^ e
- n 1 21U/1 2n ^ "
= —e 1 - e e > —
We obtain
(X)lr„ =
--a2ß2y2s sin s
> 4pe \a2ß2y2s sins|
|J (X)\T > |<5(X)|r_.
By Rouche's Theorem in [21], w(X) and w* (X) have the same zeros interior of L, = T' U T'' as follows:
02,n2,(2n)2,
, (nn)
Letting s = it, we can prove that |w*(X)|r > |<5(X)|r . Hence, w(X) = w(-t2) = 0 as |i| ^ rn. Thus, w(X) only has finite negative zeros. The proof of other cases in Lemma 9 is similar to the proof of the case a2 = 0,y2 = 0, = 0. □
Theorem 11. Let X1, X2, X3,..., be the collection of all eigenvalues of the S-L problem (1)-(4) and let (p1 (x),(2(x),... be the corresponding normalized eigenfunctions. Then
-o < X1 < X 2 <■■■ Xn ■ And {(pn; ne N} is complete in H and
W^ <Pm) = {
1, n = m, 0, n=m.
Theorem 12. The following asymptotic formulas hold for eigenvalues and eigenfunctions of the S-L problem (1)-(4) for large enough n z N.
(1) Ifa2 = 0, y2 = 0, p2 = 0, then
■fn = (n- 1)n + ,
a2 cos (n-1)n(x+1)+O (n) , xe [-1,0),
x e (0,1].
(2) Ifa2 = 0, y2 = 0, ß2 = 0, then
0(x,Xn) =
a2 cos ■
-n + O
n(x+ 1)
+ O(rJ-
(n-1) (n-1) a2y2—2—n cos —2—nx
+ O(1) (k is even), O(1)(k is odd),
x e [-1,0),
x e (0,1].
(3) Ifa2 = 0,y2 = 0, ß2 = 0, then
Xn ==l^n + O(n
a9 cos -
n(x+1)+O(1 n
x e [-1,0),
x e (0, 1] .
(4) Ifa2 = 0, y2 = 0, ß2 = 0, then
■Xn = (n - 1)n± arctan A — +o(
0(x,Xn
a9 cos
(n - 1)n ± arctan
h y4 -I
x(x+1)+O(n), x e [-1,0),
a21 y1 cos
(n - 1)n ± arctan
h y4 J
(n - 1)n ± arctan J—
(n - 1)n± arctan J —
(n - 1)n ± arctan J —
x e (0,1].
(5) If a2 = 0,y2 = 0, ß2 = 0, then
2a1 . (n-1)
sin-n (x + 1)
(n- 1)n + O' 1
t(x,\n) = a1?2 cos ^^nx
+ o(-) (k is odd) O ( — ) (k is even),
(6) If a2 = 0,y2 = 0, ß2 = 0, then
= (n-\)n + °Cn
x e (0,1].
4(x,Xn) =
-, , n— sin ( n--
(n - (1/2)) n \ 2
x h (x + 1) + O (-1) , %e[-1,0), x e (0,1].
(7) If a2 = 0, y2 = 0, ß2 = 0, then
■J\l =(n-1)n± arctan A — + o(i v V y1 \n
4(^ K)
(n- 1)n ± arctan ^y4/y1
(n- 1)n ± arctan J-
x(x+1) + o(-12), xe[-1,0),
(n- 1)n ± arctan ^y4/y1 x ( sin
(n- 1)n± arctan J — U1-I
(n- 1) n ± arctan J —
+ y4 cos
x sin 1
(n- 1)n± arctan
(n- 1)n ± arctan
x e (0,1].
(8) If a2 = 0,y2 = 0, ß2 = 0, then
xe[-h0), HX>K) =
2<X1 . (n-1)
■ sin-n(x + 1)
(n- 1)n
X e [-1,0), X e (0,1].
Proof. By Theorem 5, we know that the eigenvalues of the S-L problem (1)-(4) coincide with the zeros of the function to(X). From Lemma 9 and Theorem 10, the entire function to(X) has a sequence of zeros:
■fn = (n-1)" + o(n),
for the case a2 = 0,y2 = 0, p2 = 0 when n is big enough. By using the asymptotic formulas of eigenvalues Xn, the corresponding eigenfunctions <p(x,\n) have the following asymptotic formulas:
a2 cos (n-1)n (x+1)+0 , xe[-1,0),
0(1), %e(0,1]
for the case a2 = 0,y2 = 0, p2 = 0. Similarly, we can obtain the other cases. □
By the asymptotic formulas for the eigenvalues and eigenfunctions of the S-L problem (1)-(4) in the above theorem, we know that the series X^=1(9n(x)9n(^)/^n) converges absolutely and uniformly where <pn(x) is the normalized eigenfunction.
Green's function of the classical S-L problem is studied in [22]. Here, we obtain Green's function of the S-L problem (1)-(4) as follows:
G(x, $,X)
1$1 (SA) X1 (X,x) P to1
(x,X)X1 (S,\) P to1 1& (x,X)^(^,X)
P to $1 (£ A) Xi (x,x) to1
02 ($, X2 (x, X) to2
02 (X, %2 (Z,
-1<$<x<0,
-1 < x < 0, < 0,
-1 < x < 0, 0<^<1,
-1 <0, 0<x<1,
0<$<x<1,
0<x<$<1.
Let us consider the nonhomogeneous differential equa-
In order to prove Theorem 14, we need the following lemma.
ly - Xy = f (x)
Lemma 15. The S-L problem (1)-(4) is equivalent to
y (x, X) =
together with boundary and transmission conditions (2)-(4) where f e H. Green's functions provide a powerful method for solving the linear nonhomogeneous equations.
Theorem 13. Let f e H. Then thefunction
y(x,X) = p\° G(x,S,X)f(S)dS + f G(x,S,X)f(S)dS
J-1 J0
satisfies (69) and both boundary and transmission conditions (2)-(4), where G(x, S, X) is the Green function defined in (68).
Proof. Putting (68) in (70), we have 1 (x
-Xi (x,X)\ 01 (S,X)f(S)dS
1 -1 10
-01 (x,X)\ X1 (S,X)f(S)dS
-01 (x,X)\ X2 (S,X)f(S)dS, xe[-1,0), 20 10
-X2 (x,X)\ 01 (S,X)f(S)dS
— X2 (x,X)\ 02 (Z,X)f(Z)dt w2 )0
-02 (x,X)\ X2 (S,X)f(S)dS, xe(0,1].
By differentiating twice y(x, X) and by the definitions of solutions 0j(x,X) and Xj(x,X) (j = 1,2), we obtain y' (x,X) = (q(x) - X)y(x, X) - f(x). So, the function y(x, X) defined in (70) is the solution of (69). And by calculations, we may prove that (71) satisfies both boundary and transmission conditions (2)-(4). □
4. Eigenfunction Expansion for Green's Function and the Modified Parseval Equality
In this section, we derive the eigenfunction expansion for Green's function of the S-L problem (1)-(4) and establish the modified Parseval equality in the associated Hilbert space H. Without loss of generality, we assume that X = 0 is not an eigenvalue. Let G(x, S) = G(x, S, 0).
Theorem 14. Let Xn be the eigenvalue of the S-L problem (1)-(4) and let (n(x) be the corresponding normalized eigenfunction. Then,
GM = -l*n Mfn®
n=1 Xn
y (x, X)-X (p G (x, S) y (S) dt+ j G (x, S) y (S) dt) = 0.
-1 ° (73)
Proof. By Theorem 13, we know that
y(x,X) = p\° G(x,S)f(S)dS+ f G(x,S)f(S)dS
satisfies -y"(x) + q(x)y(x) = f(x) and both boundary and transmission conditions (2)-(4). Equation (69)canbewritten in the form -y"(x) + q(x)y(x) = f(x) where f(x) = f(x) + Xy. Similar to (74), the new form of (69) has a solution
y (x, X) = p\° G (x, VfWdS+Ç G (x, S) f(S) (75)
which satisfies both boundary and transmission conditions (2)-(4). If f(x) = 0, then the corresponding homogeneous case is the S-L problem (1)-(4). Consequently, the S-L problem (1)-(4) is equivalent to
y (x, X)-X (p G (x, S) y (S) dS+ £ G (x, S) y (S) ds) = 0.
(76) □
Proof of Theorem 14. Let Xn be the eigenvalue of the S-L problem (1)-(4) and let (n(x) be the corresponding normalized eigenfunction as in Theorem 11. Let P(x, S) =
G(x,S) + iZMxfrnW/Xn). Then P(x,0 is continuous and symmetric. We assume that P(x, S) = 0. Then, by the Fredholm integral equation, there are a number X and a function y(x) = 0 in H such that
y(x) = \(p^P (x, S) y (S) dS + P (x, S) y (S) di) .
By Lemma 15,
<Pn (x)-Xn (pÇ^G(x,S)<Pn (S)dS
+ \°G(x,S)cpn (S)dt) = 0.
Putting G(x,S) = P(x,S) - !Zi(fn(x)fn(S)/Xn) in (78), we obtain
p\° P(x,S) <Pn (S)dS+\'p (x, S) <Pn (S) dS = 0. (79)
In the following, we prove that (y,fn) = 0 and y is an eigenfunction of the S-L problem (1)-(4). By (77) and (79)
ky, fn) = PI y(x) 9n(x)dx + | y(x) cpn(x)dx
+ Ii (PI° p(x,t)y(Vdi;
+ |0 P(x,Z)y(Z)dt)^jdx = ^p\ 1 (p I 1 p(x,^)?n (x)dx
+ |0 P(x,0<pn (x)dx)y(Z)d$ + 1 P(x,£)cpn(x)dx
+ I0 P(x,Z)^Wdx)y(Z)dZ = 0.
And we have
y(x)—x(pf° G(x,V у (V + G (x, !■) у в) dt
= y(x)-X( p
iu / TO
[PM—i
fn (x) fn №
л/ TO
I (PM-I Ju V
fn(x) fn (Q
y(Z)dt
y(x)—x((p | ° P (x, $ у (?) j P (x, $ у ß)
TO fn (X) /- \
- I -j- \У,9п)
n=1 лп
y(x)—x(p|° P(x,Vy(Z)dt
+ I1 P(x,Z)y(Z)dt
by (77). This implies that у is the eigenfunction of the S-L problem (1)-(4) by Lemma 15. So, by (y,<pn) = 0 and the completeness of the eigenfunctions in Theorem 11, we know that у = 0. Consequently, P(x, = 0. The proof is completed. □
Lemma 16. Let be the set of all functions defined by
y(x) =
f1 (x) x e [-1,0), y2 (x) xe(0,1],
where (x) e C™[-1,0), y2(x) e C™(0,1]. Then <0™ is dense in H.
Proof. Let f be any function in L2(I) with f(x) = f1(x), x e [-1,0), and f(x) = f2(x),x e (0,1]. Since c™°[-1,0) is dense in L2[-1,0) as in [23], for f1 e L2[-1,0), there exists a function g1 e Cj^-1,0) such that
fU 2 e
Pj \fi (x) — 0i (*)| dx< 2-
Similarly, for f2 e L2(0,1], there exists a function g2 e C™(0,1] such that
j \f2 (X) — 02 (X)\2dx<^. Ju 2
Then, for any f e H and e > 0, there exists g e with 0(x) = &Xx%^ such that
i \fi(x)— 0i (x)\2dx + j \f2 (X)—02 (x)\2dx
Thus, CTO is dense in H.
In the following theorem, we prove the modified Parseval equality in the associated Hilbert space H.
Theorem 17. Letf e H. Then, the modified Parseval equality holds; that is,
rn =It (J),
where \\f\\2H = (f, f) and
cn (f) = p\ f (x) fn(x)dx + I f(x) <pn(x)dx. (87)
Proof. Let c™ be as in Lemma 16. At first, we prove that (86) holds for f e C^. Denote g(x) = -f" (x) + q(x)f. Then by Theorems 13 and 14,
f(x)=p\° G(x,Vg(Z)dt+ I G(x,$)g($)dt,
TO1 = (x)
T=i Xn
pf 9„(S)9(S)d!; + j <pn(t)0(z)dz).
Multiplying by (m(x) and integrating the new equation, we have
P I Pm(x)f (x) dx + j0 Pm(x)f (x) dx
--t (p j° vmm&g & #+j vmm&g ® .
f(x) = tCn (f)Pn (x),
where cjf) = {f, <pj = p J- f(x)<pn(x)dx + J f(x)
<pn(x)dx. Thus, for f e C;,
= C (f )■
Next, we prove that (86) holds for f e H. For any f e H, there exists a sequence jfklkeN c C™ converging to f in H since C™ is dense in H. Firstly, we prove that c%(f) < <x>
andlimfc^m TZ1 c^(fk) = ZZ1 c2(f).
By the Cauchy-Schwartz inequality, ^(f/) - cn(f) = Kfk -f,(Pn)| * \\fk -f\\H. Henc^ Kmk^c^fk) = Cn(f).
Since TZ=1(cn(fk )-cn(fm))2 = ZZ1 c2(fk -fm) = \\fk - fm\\H
for fk -fm eC~,
Z (cn (fk) - Cn (fm)) * \\fk - fm
Let k ^ rn, then inequality (92) becomes '^N=1(cn(f) -Cn(fm))2 * Wf - fmfH. Letting N ^ ra,we have
Rcn (f)-Cn (fm))2 *Wf-f,
Then by the Minkowski inequality
^Z/Ci (f)=Z(Cn (f)-Cn (fm) + Cn (fm))2
n=1 n=1
Z(Cn (f)-Cn (fm))'
_1/2• 2
Zch (fm
and by Holder's inequality,
^ZC (f)-lC2n (fk)
Z(Cn (f)-Cn (fk))(Cn (f) + Cn (fk))
0, as k —> rn.
*(I(Cn (f)-Cn (fk))2) (I(Cn (f)+Cn (fk))2)
This means that limk^m TZ1 Cl(fk) = ZZ1 Cn(f).
Since fk ^ f inH ask ^ rn, lim^^ WfkWH = WfWH. We obtain
\\f\\H = klim \\fk\\H = klim C2 (fk) = Zt (f). (96)
Acknowledgments
The authors thank the reviewers for their careful reading of the paper and for the constructive comments and suggestions which have led to the improvement of the presentation of the paper. This work was supported by the National Nature Science Foundation of China (11161030) and the Specialized Research Fund for the Doctoral Program of Higher Education of China (20121501120003).
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