The couple grade of Self-adjoint Domains of Singular【精品论文大全】 .doc

精品论文推荐the couple grade of self-adjoint domains of singulardifferential operators1sun jiong1, wang wanyi1,2, zheng zhiming31department of mathematics, inner mongolia university, hohhot, p.r. china（010021）2 department of mathematics, inner mongolia normal university, hohhot, p.r. china（010021）3department of mathematics, beihang university, beijing, p.r. china（100083）.email: , abstractthe self-adjoint domains associated with a scalar differential expression of an arbitrary even orderare investigated in this paper. there is a known one-to-one correspondence between these domains and the complete lagrangian subspaces of a naturally defined symplectic space. using this correspondence, a characterization of the self-adjoint domains is obtained for the situations where the differential expression is regular at one of the end points of its interval and in either the limit-point category or the limit-circle category at the other end point. moreover we get a complete describing on the couple-grade of boundary conditions, and point out there are how many classifications on the couple-grade of boundary conditions in limit-point and limit-circle cases.keywords: differential operator，couple-grade，boundary condition，self-adjoint，symplecticgeometry.2000 mathematics subject classification： primary 34b20，34b24；secondary 47b2.introduction2one of the major successes of operator theory is that it offers a framework for discussing and analyzing operators in quantum mechanics. for example, the schrdinger theory is based on properties of the differential operators in the wave equation. most operators appearing in quantum mechanics are self-adjoint operators. characterizing self-adjoint differential operators is a fundamental problem in operator theory, and a huge amount of effort has been devoted to it 3,13,18.letn ( k ) ( k ) -15-l ( y ) = pn k y , t 0, )(0.1)k = 0be a 2n -th order symmetric differential expression with real coefficients defined on i = 0, ) ,kwherepn k c( i ), p0 (t ) 0 for t i . how can one generate self-adjoint operators in lassociated with the symmetric differential expression? in 1976, w.n. everitt generalized titchmarsh-weyls theory to the high-order cases, and gave some self-adjoint domains with the maximum and the minimum deficiency indices 5,6,7, called everitts domains. we know that everitts domains are a complete solution to the problem of describing all the self-adjoint extensions of any given symmetric differential expression with the minimum deficiency index (i.e., the limit-point case), but in the limit-circle case (i.e., the maximum deficiency index) the everitts domains are only special self-adjoint domains. using the general theory of linear operators, cao zhi-jiang obtained a complete and direct characterization of all the self-adjoint extensions of symmetric differential expressions with the maximum deficiency index in 1985 1, and proved thatthe everitts domains are only special cases of his results. in 1986, sun jiong 17 extended everitt1 supported by the national nature science foundation of china (10561005) )(10661008), and thespecialized research fund for the doctoral program of higher education (20040126008).and caos results to the cases where the symmetric differential expressions have middle deficiencyindices. the key point of 17 is to prove a decomposition ofdm , the domain of maximal operatorlm , such that the conditions that elements of dmsatisfy at pointt = 0and at infinity areseparated. using this decomposition he obtained a complete characterization of all self-adjoint extensions of high-order symmetric differential operators with any deficiency indices. in subsequent years, a series of papers were accomplished by the group of inner mongolian university, directed by cao zhi-jiang, liu jing-lin and sun jiong, extending the results of cao and sun to various cases: two singular endpoints case by z. j. shang and r. y. zhu 16 and w. m. li 11, self-adjoint extensions in a direct sum space by w. m. li 12 and s. z. fu 10, the j -self-adjointcase by z. j. shang 14, and more. in 1990, evans extended the results of 17 to a more generalcases 4. all the above results come from the method of differential operators theory and functional analysis, i.e., from analysis methods.in 1999, everitt and markus studied the complete description of all self-adjoint extensions of symmetric differential operators by symplectic geometry firstly 9 and 8. they discovered a natural one-to-one correspondence between the set of all self-adjoint extensions of the minimal operator and the set of all complete lagrangian subspaces of an associated complex symplectic space. the characterization of the self-adjoint domains amounts to the characterization of the complete lagrangian subspaces of the complex symplectic space. even though interesting information about self-adjoint domains is obtained from this correspondence, 9 and 8 do not have any example of a singular differential expression for which the self-adjoint domains can be explicitly characterized using this correspondence.in this paper, using the above correspondence, we explicitly work out a characterization of theself-adjoint domains for the situations where l ( y)is regular at 0 and in either the limit-pointcategory or the limit-circle category at + . these two cases will be called the regular-limit-pointcase and the regular-limit-circle case, respectively. the situations where l ( y)is regular at +and in either the limit-point category or the limit-circle category at 0 can be handled similarly andhence are omitted. the characterization is parallel to the one given by cao in 1 when l ( y)is inthe regular-limit-circle case, and to the one obtained by sun in 17 whenl ( y)is in theregular-limit-point case. moreover we get a complete describing on the couple-grade of boundary conditions, and point out there are how many classifications on the couple-grade of boundary conditions in limit-point and limit-circle cases.the organization of this paper is as follows. in section 1, we introduce our notation and recall some basic results. the regular-limit-point case is treated in section 2, while the regular-limit-circle case in section 3.1. preliminariesdefinition 1.1 9 a complex symplectic space s is a complex linear space equipped with aprescribed symplectic form : . here, a form : : s s c(1) : is sesquilinear, i.e.,is symplectic ifc1u + c2v : w = c1 u : w + c2 v : w,(2) : is skew-hermitian, i.e.,u : v = v : u , so u : c1v + c2 w = c 1u : v + c 2 u : w,(3) : is non-degenerate, i.e.,where u, v, w s,c1, c2 cu : s = 0 implies u = 0 ,and u : s = u : v | v s.definition 1 2.9 a linear subspace l in the complex symplectic space s is called lagrangian incase l : l = 0 , that is, u : v = 0 , for any u, v l. a lagrangian subspace l sis complete ifu sand u : l = 0 imply u l .definition1.39 let s be a complex symplectic space. then linear subspace s and s+ are symplectic ortho-complements in s if(1) s = spans , s+ ,(2) s : s+ = 0 .in this case, we write s = s s+ .we define the maximal operator and minimal operator associated with l ( y) as follows:tmax ( y) = l( y),d(tmax) = y l2 (i ) | y k ac ( i )for k = 0,1, 2, 2n 1, y 2 n l2 (i );tmin ( y) = l ( y),d(t ) = y d(t) | y k (0) = 0 fork = 0, 2n 1, y, z() = 0minmaxfor anyz d(tmax ),where y, zis the langrange bilinear form associated withl ( y), andyk is the k thquasi-derivative of y defined bykyk = d y , 0 k n 1,dxkyn = pd n y,n k 0 dxnyn + k p dyd yn+ k 1 1kn=k dxn k , .dx by the theory of ordinary differential operators, tminand tmax are closed operators, tmax = tmin ,tmin = tmax and tminis a symmetric operator. sets = d (tmax ) /d (tmin ) ,and define a symplectic form by f : g = f + d (tmin ) : g + d (tmin ) = f , g ,where f =f + d (tmin ) , 0g = g + d (tmin ) s.we haved(tmin ) = f d(tmax ) | f : d(tmax ) = 0.lemma 1 19 s = d (tmax ) /d (tmin ) equipped with : is a complex symplectic space.lemma 1 2.9 s = s s+ , where s+ = f s | f k (0) = 0, k = 0,1, 2n 1,s = f s | for anyg d(tmax ), f , g () = 0 .lemma 1 39(balanced intersection principle) let l be a complete lagrangian subspace, then0 1 dim s2 dim l s= 1 dim s2+ dim l s+ 1 mindim s2 , dim s+ .definition 1 49 let l be a complete lagrangian subspace, ifk = 1 dim s2 dim l s= 1 dim s2+ dim l s+ ,then l is called of the k -th grade, d (tl ) is also called of the k -th grade.remark 1 1. a similar number of “grade” was introduced in 1996 by z. j. shang 15 who tried toclassify the j -self-adjoint domains of j -symmetric differential expressions in the limit-circle case.lemma 1 4 9 (gkn theorem)2(1) there exists a self-adjoint extension of tmin , if and only if there exists acomplete lagrangian subspace l s .(2) the lagrangian subspace l sis complete if and only if dim l = 1 dim s.(3.) there exists a natural one-to-one correspondence between the set t of all self-adjiontextentions of t 0 and the set lt of all complete lagrangian subspaces lt slt = d (t ) /d (tmin ) .such that(4) let l sbe a complete lagrangian subspace. then, the unique self-adjoint extension tlcorresponding to l is given byd (tl ) = c1 f1 + + cm fm + d (tmin ) ,where f1 , f2 , fm is a basis for l withnumbers.f1 , fm d (tmax ) , and c1, c2 , cm are complexremark 1 2. from lemma 4, we know that characterizing the self-adjoint extensions of tmin isequivalent to the problem of discussing the complete lagrangian subspaces of s = d( max ) /d(min ) , i.e.,the description and classification of complete lagrangian subspaces of s is equivalent to thedescription and classification of the self-adjoint domains of l ( y ) .2. regular-limit-point casethroughout this section let l ( y) be in the regular-limit-point case. recall that in this case, l ( y) isregular at 0 and belongs to the limit-point category at + .lemma 2 1. let l ( y) be in the regular-limit-point case, thend(t ) = f d(t) | f k (0) = 0, k = 0,1, 2n 1.minmaxproofsincel ( y )is in limit-point case and the deficiency indices are (n, n) , we have f , g ( ) = 0 for any f , g d (tmax ) .we havelemma 2 2. s = s,s+ = 0 .lemma 2 3. dim s = dim d (tmax ) /d (tmin ) = 2n .theorem 2 18 in the regular-limit-point case, all complete lagrangian subspaces are of the 0-thgrade.because dim s = 2n , s and c 2 n are isomorphic as complex vectorspaces, let e1 = (1, 0, 0) ,e2 = ( 0,1, 0, 0) , ,e2 n = ( 0, 0,1)betheusualbasisforc 2 n,thens = spane1, e2 , e2 n . for f s , we choose the coordinates of f as:f = ( f (0), f 1 (0), f 2 n 1 (0) =f (0)e1 + f 1 (0)e2 + + f 2 n 1 (0)e2 n . (2.1)theorem 2 2. for any f , g s , f : g = fhg where 001 h = 0h1 ,h = 010 . h0 1 1100 nn proof since l ( y ) is the regular-limit-point case, by the definition of the symplectic form of s ,we have0f : g = f , g = f , g (0) = fhg .theorem 2 3. l is a 0 -th grade complete lagrangian subspace of s if and only if there existaij c,andi = 1, 2, n,j = 1, 2, 2n, such thatl = span1e, 2 e, n e(2.2)(1) rank a = n;ij(2) h = 0,1 i, j n ,122 n twherea = (aij )n2 n ,i = (ai1 , ai 2 , ai,2 n ),i = 1, 2, n , and e = (e , e , e ) .proof ()for any f , g l, there exist si , ti c,f = s11e + s2 2 e + + sn n ei = 1, 2, n, such thatnnn= (i =1si ai1)e1 + (i =1si ai 2)e2 + + (i =1si ai ,2 n)e2 n ,g = t11e + t2 2 e + + tn n ennn1= (i =1ti ai1 )e + (i =1ti ai 2)e2 + + (i =1ti ai, 2 n)e2 n .from theorem 2 and (2), we getnnnnnnf : g = ( si ai1, si ai 2 , si ai ,2 n )h ( ti ai1 , ti ai 2 , ti ai ,2 n )= 0.i =1i =1i =1i =1i =1i =1so l : l = 0 , that is, l is a lagrangian subspace of s . from (1), dim l = n , so l is a 0 -thgrade complete lagrangian subspace of s .()if l is a 0 -th grade complete lagrangian subspace of s , then dim l = n,dim l s+ = 0, dim l s = nand l : l = 0 . so there existaij c,i = 1, n,j = 1, 2, 2n , such that (2.2) holds and conditions (1) and (2) are satisfied.corollary 2 1. l is a 0 -th grade complete lagrangian subspace of s if and only if there existaij c,i = 1 , 2, n, j = 1, 2, ,n, such that1 2 n 1tttandl = f ssi c, i = 1, n, ( f (0), f(0), f(0)= a (s1, s2 , sn ) (1) rank a = n;ij(2) h = 0, 1 i, j n,wherea = aij ,i = ai1 , ai 2 , ai,2 n , i = 1, 2, n. n2 n3. regular-limit-circle casethroughout this section, let l( y) be in the regular-limit-circle case. recall that in this case, l( y)is regular at 0 and belongs to the limit-circle category at + . then, the deficiency indices of l( y) are(2n, 2n) , and there exists a linear independent system of solutions i (i = 1, 2n) of l( y) = 0satisfying (i , j (0) = j , wherej = iin0 . 0iin by 2,9,13, we have the following conclusions:lemma 3 1 for anyf , g d(tmax ) , we haven2 n f , g ( x) = i f ,r ( x) g, r ( x) i f ,r ( x) g, r ( x).r =1r = n+1lemma 3 2 if all solutions of l( y) = 0 belong to l2 (i ) , thenk +1d(tmin ) = f d(tmax ) | f k (0) = 0, f ,() = 0, k = 0,1, 2n 1.lemma 3 3 dim s = 4n, dim s = dim s+ = 2n.theorem 3 1. in the regular-limit-circle case, there exists a k -th grade ( k = 0,1, n ) completelagrangian subspace of s , and the grade k of any complete lagrangian subspace is among 0,1, n .since dim s = 4n, s and c 4 n are linear isomorphic. let2ne1 = (1, 0, 0, 0), e2 = (0,1, 0, 0), , e2 n = (0su,u0u,uuuu,ut1,0.0),2n+12n + 2f 1 = (0su,.u.u.u,u0u,u1ut,0, 0), f 2 = (0su,u.u.u,u0uu,t1, 0, 0), f 2 n = (0, 0,1).then s = spane1, e2 , e2 n , f 1 , f 2 , f 2 n .for f s , we choose the coordinates of f as:1 2 n 1f = ( f (0), f(0), f(0), f ,1 (), f ,2 (), f ,2 n ()= f (0)e1 + f 1 (0)e2 + + f 2 n1 (0)e2 n122 n+ f ,1 () f+ f ,2 () f+ + f ,2 n () f. (3.1)theorem 3 2. for any f , g s, f : g = fhg , where0 0 h100 001 h h1000 h010= , =, 00iin1 000ii n 100 n n 100 010 i = .n 001 nn proof by the definition of the symplectic form and lemma 1, we havef : g = f , g 0n= f , g () f , g (0)2 n= i f ,r () g, r () i f ,r () g, r ()r =1nr = n +1 f k 1 (0) g 2 n k (0) f 2 n k (0) g k 1(0)k =1= fhg .+theorem 3 3. s= spane1, e2 , e2 n ,s = span f 1, f 2 , f 2 n .proof first we prove that s= spane1, e2 , e2 n . for any f s, we havef d(tmax )and f ,k () = 0,k = 1, 2, 2n. we getf = f (0)e1 + f 1 (0)e2 + + f 2 n 1 (0)e2 n spane1, e2 , e2 n .thus, s spane1, e2 , e2 n . if f spane1, e2 , e2 n , thenf = f (0)e1 + f 1 (0)e2 + + f 2 n1 (0)e2 n ,i.e., f ,k () = 0,k = 1, 2, 2n. so, spane1, e2 , e2 n122 n s . therefore,s =spane , e , e .similarly, we get s+= span f 1, f 2 , f 2 n .theorem 3 4. l is a 0-th grade complete lagrangian subspace of s if and only if there exist aij ,bij c,andi = 1, 2, n,j = 1, 2, 2n, such thatl = span1e, 2 e, 2 n e(3.2)(1) rank an = rank bn = n,ij(2) h = 0, 1 i, j 2n,122 n122 n twherean = (aij )n2 n , bn = (bij )n2 n ,e = (e , e , e, f , f, f) and = (ai1 , ai, 2 n , 0, 0),1 i n,i (0, 0, bi n,1 , bi n,2 n ), n + 1 i 2n.proof ()for any f , g l, there exist s1i , s2i , t1i , t2i c,i = 1, 2, n, such thatnnnnnnf = s1ii