ABSTRACT
On the other hand, if 1 E O'(K), but 1 ¢ O'(L)UO'(M), then 1 = Oi+{Jj, where i runs over some finite set P C lN and j runs over some finite set Q C IN. It follows from (15.37) that Zij = (1-0i - {Jj)-l/ij if i ¢ P and j ¢ Q, and lij = 0 if i E P or j E Q. We conclude that the condition
(15.39) lij =0 (i E P, j E Q) is necessary and sufficient for the solvability of equation (15.32). This condition simply means that I is orthogonal to the nullspace (15.40) N(I - K) = span{4>i @,pj: i E P,j E Q}
(15.41)
of the operator 1-K. If (15.39) is satisfied, the equation (15.32) has a finite number of linearly independent solutions of the form
Z(t,s) = E Cij4>i(t),pj(s) (iJ)epxQ
where Cij (i E P,j E Q) are arbitrary constants. Now suppose that 1 E O'(L) U O'(M), but 1 is not an eigenvalue of K. Then it follows from (15.37) and (15.38) that equation (15.32) is solvable if and only if the series
(15.42) 00 ( It. )2.~ 1-o~'-/J'1,,=1 1 J converges: If the condition (15.42) holds, then the equation (15.32) has unique solution which may be found by formula (15.38). On the other hand, if 1 E O'(L) U O'(M) and 1 is an eigenvalue of K, then it follows from (15.37) that equation (15.31) is solvable if and only if
