Peierls ( 7 ) has gone into details , but his treatment , he admits , is non-rigorous .sx As Dolph ( 8 ) points out , the promised justification of this has never appeared .sx Schwartz ( 9 ) , in a very important and powerful paper , treats the Sturm-Liouville case ( and also certain singular cases ) , but only as a special case of a long and complicated function-theoretic argument .sx Keldysh ( 10 ) has also given a linear-operator approach to the problem .sx Altogether , there does seem a case for a direct justification of Peierls's work that does not depend on function-theoretic arguments , and this is particularly so when it appears that , without any great complication , it is possible at the same time to make a contribution to the singular case in which the range of x remains finite but q(x ) becomes discontinuous at one or other or both of the end-points .sx This contribution does not seem to be covered by the existing function-theoretic arguments .sx The problem we shall consider is the following .sx We take the equation where q(r ) may be complex but is continuous except at r = 0 , and where [FORMULA] exists .sx We suppose that l is a positive integer or zero .sx The reader will readily verify that the analysis is not restricted to those values of l , but this is the case of practical importance .sx ( The equation is the well-known equation that arises when a three-dimensional equation with spherical symmetry is solved by the method of separation of variables .sx ) The boundary conditions we impose are ( 1.3 ) , for some b 0 , together with the requirement that y(x ) be ) .sx This , as the analysis shows , is sufficient to define an eigenvalue problem , except in the case l = 0 , when we have to impose a further condition of the type ( 1.2 ) at a = 0 .sx Despite this , the case l = 0 is similar enough to the case l 0 , so that we can safely restrict ourselves to l 0 .sx The case l = 0 , with q(r ) continuous , is just the Sturm-Liouville case , which therefore comes out as a particular case of the argument .sx We shall examine the eigenfunctions associated with this eigenvalue problem .sx As usual , an eigenfunction is a non-trivial solution of the equation ( 1.4 ) which satisfies the boundary conditions .sx In the self-adjoint case , the set of eigenfunctions would be complete , i.e. any reasonable function could be expanded in a series of them .sx In the non-self-adjoint case , we shall see that in general this no longer holds , but that the set of eigenfunctions can be made complete by adding to it certain other functions which , though not eigenfunctions , are related to them .sx ( Their precise form will be found in @135 .sx ) I shall refer to these additional functions as adjoint functions .sx The problem can be extended to the case in which r = b is also a discontinuity of q(r ) , of the same type as at r = 0 .sx It will not be necessary to discuss in detail this extension , but it will be clear that the same general conclusions hold on the completeness of the set of eigenfunctions and adjoint functions .sx I have limited myself to proving completeness , but , at least in certain cases , much more can be proved .sx For example , in the Sturm-Liouville case , a very straightforward adaptation of ( 1 ) [Ch .sx =1] shows that not only is the set of eigenfunctions and adjoint functions complete , but also that , if f(r ) is any function of L(0,b ) , then the eigenfunction expansion of f(r ) ( an expansion which , of course , includes adjoint functions ) converges under Fourier conditions to f(r) .sx This analysis does not seem to extend to the singular cases considered in this paper .sx 2 .sx If [FORMULA] , then ( 1.4 ) has solutions [FORMULA] , of which [FORMULA] is ) .sx If we then write ( 1.4 ) in the form we see that it is formally equivalent to the integral equation [FORMULA] .sx Our first objective is to prove that , for [FORMULA] , and all [FORMULA] sufficiently large , the solution of ( 1.4 ) that is L :sx 2:(0,b ) is , apart from a multiplicative constant , [FORMULA] , where o(1 ) denotes a term small where [FORMULA] is large , uniformly for r in [0,b] , and where [FORMULA] .sx We do this by investigating ( ) .sx Let [FORMULA] .sx Then for all r , 5l , where A denotes various positive constants .sx Let [FORMULA] .sx Then , if [FORMULA] , ( 2.1 ) gives since [FORMULA] exists , the o(1 ) term denoting a quantity which tends to zero as [FORMULA] .sx Also , if [FORMULA] , where [FORMULA] .sx But , for [FORMULA] , we have [FORMULA] .sx For , for all z , [FORMULA] , so that [FORMULA] .sx The required estimate for G(r,t,5l ) follows from this by using the asymptotic expressions for [FORMULA] .sx Substituting this estimate in ( 2.2 ) , we obtain The first of the two integrals in the last line is [FORMULA] since [FORMULA] in the range of integration and [FORMULA] .sx The second integral is [FORMULA] , by a similar type of argument .sx ( The second integral will not , of course , appear if [FORMULA] .sx ) It thus follows from ( 2.1 b ) and ( 2.4 ) that , for [FORMULA] , if [FORMULA] is large enough , i.e. that [FORMULA] .sx If we substitute this result back in the integral in ( 2.1 ) and re-estimate this integral on the same lines as has just been done , we emerge with ( 2.1 a) .sx Thus any solution of ( 2.1 ) satisfies ( 2.1 a) .sx That there is one ( and just one ) solution of ( 2.1 ) can be proved by the usual iteration process , of which the work above is effectively the first step .sx Then ( 2.1 ) can be differentiated back to show that the solution is a solution of ( ) .sx We have thus found a solution of ( 1.4 ) that is ) .sx If we denote this solution by 15f(r,5l ) , then any other solution apart from a constant multiple of 15f(r,15l ) is given by a constant multiple of [FORMULA] , and knowing now the behaviour of 15f(r,15l ) near r = 0 , we can readily verify that 15ps(r,15l ) is not ) .sx The L :sx 2:(0,b ) solution is therefore ( apart from a multiplicative constant ) unique .sx We remark finally that , since [FORMULA] is an integral function of 5l , the process of solving ( 2.1 ) by iteration shows that [FORMULA] is also an integral function of 5l .sx 3 .sx We now consider the solution 15ch(r,15l ) which satisfies ( 1.4 ) and the boundary conditions [FORMULA] .sx As in ( 1 ) [Ch .sx =1] 15ch(r,15l ) is an integral function of 5l .sx The Wronskian of 15f , ch is independent of r and so may be written as 15o(l ) , and [FORMULA] will be an integral function of 5l .sx Further , the vanishing of 15o(l ) is a necessary and sufficient condition for 15f , ch to be multiples the one of the other , i.e. for 5l to be an eigenvalue .sx For large values of [FORMULA] , ( The asymptotic behaviour of 15f@7(r,15l ) is obtained by differentiating ( 2.1 ) with respect to r and proceeding as before .sx ) Hence , for large values of [FORMULA] , the zeros of 15o(l ) must be near the zeros of [FORMULA] , which are , of course , independent of q(r) .sx Further , for large [FORMULA] , the zeros of 15o(l ) are simple .sx This is best seen by writing where C is a circle with centre 5l , and by using the asymptotic expression ( 3.1 ) for 15o(l ) to give an asymptotic expression for 15o@7(l) .sx It is then clear that values of 5l near the zeros of [FORMULA] do not satisfy 15o@7(l ) = 0 .sx We now construct the function 15F(r,15l ) , where [FORMULA] , and f(t ) is any function which is ) .sx This is a meromorphic function of 5l , having poles at the zeros of 15o(l) .sx It will be our object in the next section to show that , if f(t ) is such that all the residues of 15F(r,15l ) at its poles vanish , then f(t ) = 0 almost everywhere .sx 4 .sx If all the residues vanish , 15F(r,15l ) becomes an integral function of 5l .sx Let us suppose that we can prove ( as we shall do ) that we can find a sequence of circles [FORMULA] , with [FORMULA] , such that 15F(r,15l ) is bounded on the circles , with the bound possibly dependent on r , but independent of n. Then , by Liouville's theorem , 15F(r,15l ) is a constant , independent of 5l .sx Suppose then that 15F(r,15l ) = g(r) .sx It follows by differentiation that [FORMULA] , with the result holding at least almost everywhere .sx By varying 5l , we have g(r ) = 0 , and hence f(r ) = 0 almost everywhere .sx It remains to prove the boundedness of 15F(r,15l ) , with r fixed , but [FORMULA] , on the circles [FORMULA] .sx Since we are concerned only with results 'almost everywhere' , we may exclude r = 0 .sx The differential equation is thus non-singular in the interval [r,b] , and we can appeal to ( 1 ) [equation ( 1.7.8)] to get an asymptotic form of 15ch(r,15l ) for sufficiently large [FORMULA] .sx In fact , we have [FORMULA] , where A denotes various positive constants independent of 5l .sx From @132 we have , again for fixed r and sufficiently large [FORMULA] , Finally , if we choose the sequence [FORMULA] to be such that [FORMULA] , we see that on each of the circles [FORMULA] , and so , on those circles , for n sufficiently large , we have from ( 3.1 ) that If we now substitute ( 4.1 ) , ( 4.2 ) , ( 4.3 ) in the definition of 15F(r,15l ) , and use Schwarz's inequality to estimate the integrals , we see readily that , on the circles [FORMULA] , 15F(r,15l ) is bounded with bound independent of n. 5 .sx From this , we can deduce the completeness of the eigenfunctions and adjoint functions .sx Before we do this , however , we must examine the nature of these eigenfunctions and adjoint functions .sx In the real self-adjoint case , it is well known that the zeros of 15o(l ) are real and simple , and , if 15l;n ; is such a zero , 15ch(r,15l;n; ) is a multiple of 15f(r,15l;n; ) , so that we may write [FORMULA] .sx Then , near 5l = 15l;n ; , the singular part of 15F(r,15l ) is [FORMULA] .sx Hence the residue at 5l = 15l;n ; is [FORMULA] , and this vanishes for all r if and only if the Fourier coefficient of f(t ) with respect to the eigenfunction 15f(t,15l;n; ) vanishes .sx The argument remains valid even in the non-self-adjoint case provided that 15l;n ; is a simple zero of 15o(l) .sx However , there is no longer any guarantee that the eigenvalues of 15o(l ) will be simple , and counterexamples are easily provided .sx Suppose now that 15l;n ; is a zero of order p of 15o(l) .sx Then , at 5l = 15l;n ; , 15F(r,15l ) has a residue of the form where the A;s;(15l;n; ) are constants depending on the derivatives of 15o(l ) at 5l = 15l;n ; and whose precise value will not concern us .sx Now 15o(l ) can be written in the form [FORMULA] , and we know that 15o(l;n; ) = 15o@7(l;n; ) = 0 .sx Hence [FORMULA] , and interchange of the order of differentiation gives that is independent of r. If we repeat this process with higher differentiations with respect to 5l , we obtain finally that is independent of r for s = 0 , 1, .sx . , p-1 .sx This implies that , for these values of s , so that ( 5.1 ) can be expressed as a linear combination of the p functions [FORMULA] , the coefficients being homogeneous linear combinations of the p expressions [FORMULA] , or , what is the same thing , homogeneous linear combinations of the p expressions [FORMULA] .sx For the residue to vanish it is therefore sufficient that all the Fourier coefficients of f(t ) with respect to the p functions [FORMULA] should vanish .sx Hence , if all the Fourier coefficients of f(t ) vanish at all zeros of 15o(l ) , then all the residues of 15F(r,15l ) vanish , and so , as already proved , f(t ) = 0 almost everywhere .sx This shows , by application of a standard theorem , that the system of eigenfunctions and adjoint functions , where the adjoint functions are [FORMULA] , is complete .sx The question does arise whether the adjoint functions are indeed necessary for completeness , or whether on the contrary they themselves can be expressed as linear combinations of the eigenfunctions , and so be eliminated from the expansion of an arbitrary function .sx It is a standard theorem in the theory of orthogonal functions that all the eigenfunctions and adjoint functions are necessary if they form an orthonormal set , and we shall prove that they are substantially orthonormal in @13 6 .sx What we shall actually prove ( and it is clear that this will be sufficient ) is ( =1 ) that all the eigenfunctions and adjoint functions associated with an eigenvalue 15l;n ; are orthogonal to all the eigenfunctions and adjoint functions associated with an eigenvalue 15l;m ; , where [FORMULA] ; ( =2 ) that the eigenfunctions and adjoint functions associated with an eigenvalue 15l;n ; of multiplicity p can be expressed by a non-singular transformation as linear combinations of p orthonormal functions .sx It should be remarked that the number of multiple eigenvalues is at most finite , and so the number of adjoint functions is at most finite .sx