NONLINEAR BOUNDAR Y VALU E PROBLEM S 7 We shall denot e b y KPQ. Z — dom L O ker P the generalized inverse of L define d b y KPtQ = K P (I - 0 . EXAMPLES. (1 ) I f X an d Z have the same finite dimensio n an d L = 0 , one ca n tak e P = / i n X, Q = / i n Z, an d the n Ay 7 i s the zero mapping fro m Z t o X. (2) I f L = / : ^ — • X, the n P is the zer o mapping in X, Q the zer o mapping in Z an d K0 0 th e identity i n X. (3) I f L\ do m L C X — • Z i s one-to-one an d onto , then P is the zer o mapping in Xf Q the zero mapping in Z an d K0 0 = L ~"!. REMARKS. (1 ) I n Example s (2) and (3) above, the choic e o f P and Q is unique, which is of cours e not th e cas e when ke r L an d coke r L ar e not reduce d t o {0} . (2) I f L i s a Fredholm mappin g of index zero then , fo r ever y isomorphism /: I m Q —* ker Lt th e mapping JQ 4 - Kp Q i s an (algebraic ) isomorphism fro m Z ont o do m L and , for ever y x G dom Z, (/Q + K PQ ylx = (Z , + /- 1 /*)*. In fact , i f z G Z, we easily obtai n th e followin g chai n of equivalences : {JQ + #i,e)z = x * /Q z = Ac, JC Pffi z = ( / - P) x » Qz = J~ l Px, LpKPQz = ! ( / - P ) x *Qz= J~ l Px, (I-Q)z = Lx*z = (J~ l P + L)x. 2. Boundar y value problems for ordinary differential equations . I n this sectio n we shall give some examples of Fredhol m mapping s related t o boundar y valu e problems for or - dinary differentia l equations . Le t I = [0 , 1] an d conside r th e linea r boundar y valu e prob- lem x'(f)=/(f), tGI, (I.l) Mx(0) + Nx(l) = c, where x\i) = (d/dt)x(t), f E Ll(I, R n )9 th e spac e of (equivalenc e classe s of) Lebesgu e in- tegrable mapping s from I int o Rn, c € R n an d M an d TV ar e (n x «)-reai matrices. B y solu- tion o f (I.l) w e shall mean a n absolutely continuou s mappin g x fro m I int o R n whic h satisfies th e first equatio n i n (I.l ) almos t everywher e o n I an d verifies th e secon d conditio n in (I.l) , i.e. the boundar y condition . I f w e take X = C(I, Rn), th e spac e o f continuou s mappings from I int o Rn9 an d defin e do m L = {x G X: x i s absolutely continuou s o n / } , L: x M (x', Afr(0) + Nx(l)l Z = L l (I, R n ) x Rn,g = (/ , c) , then th e boundar y valu e problem (I.l ) i s equivalent t o th e abstrac t equatio n Lx = g. Moreover , ker L = { x G I : ^ i s a constan t mappin g and (M + N)x(0) - 0} , so that dim ker L = di m ker(M + Af) . On th e othe r hand, the first equatio n i n (I.l ) i s equivalent t o x(0 = *(0) + fj(s)ds, tGI, so that th e secon d equation i n (I.l) ca n b e writte n

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