UNIQUE SOLUTION TO PERIODIC BOUNDARY VALUE PROBLEMS

Existence of unique solution to periodic boundary value problems of differential equations with continuous or discontinuous righthand side is considered by utilizing the method of lower and upper solutions and the monotone properties of the operator. This is subject to discussion in the present paper.

We now present the main result of this section.
Setting F(t,z) = f(t,z) Mx for (t,z) e J x R. Then it is easy to verify that is a solution of PBVP (I).From Lemma 2.1 it follows that PBVP (I) possesses unique solution.Therefore u(t) defined above is the unique solution of PBVP (I).
And hence we define n operator A on the sector [/, c] by Av = u, where u is the unique solution of PBVP (I).We now show the following two conclusions.
So Lemma 2.1 implies p <_ 0. Thus fl _< Aft.Similarly, we can show that We now show that A is increasing on the sector [3, or].In fact, if Av = u and Av = u, where v, v [, cz] and v <_ v. Setting p = u u we see that p' = u' u' = Mu + f(t, v) Mv.Mu: f(t, v:) + Mv:z > Mp and p(0) = p(2r).This leads to p _< 0 from Lemma 2.1.So ux <_ u2, and hence A is increasing on the sector [fl, o].
On the other hand, we define an operator B on the sector [fl, cz] as follows" Then B is decreasing on the sector [, cz since F(t, x) is decreasing in x from the conditions imposed on f.But obviously Bv(t) is a solutio of PBVP (I).So operator B is identical with operator A on the sector [fl, o] from the uniqueness of solutioa to PBVP (I).And hence operator A is both increasing and decreasing on the sector [/, cz] and satisfies i).Thus A transforms the sector [/, cz] onto a point u* This implies that u" is the unique fed point of A on the sector [, c].However solving PBVP (2.1)-(2.2) is equivalent to finding fixed poings of operator A. I-Ienee u" is the unique solugion of PBVP (2.1)-(2.2) on ghe sector [/,   Pemark It is obvious that the above theorem can not be proved by applying either comparision theorem or operator theory.So it should be noted that it is effective to combine operator theory with comparision results.

DISCONTINUOUS RIGHT-HAND SIDE
Let us consider the the following periodic boundary value problem (3.1) u' = u), a.e.J; (3.2) = where J = [0, 27r].A function a e AC([0, 27r], R) is said to be a lower solution if c' < f(t, c) -% for almost all t 6 J, where 0 7 = M[a(0)-o( Similarly a function/3 AC([0, 27r],R) is said to be an upper solution if/3' > f(t,/3) 7t for almost all t in J, where In order to present the main result of the section we first show a result that is similar to Lemma 2.1.where M > 0. Then re(t) <_ 0 on [0, 27r].
Setting F(t,x) = f(t,x) Mx for (t,x) e d x R. Then it is easy to verify that On the other hand, if we define n operator B on the sector [fl, a] as follows: By(t) = Bv(O)e Mt + f F(s, v(s))eM(t-')ds and By(O) Bv(2?r)---1 Then B is decreasing on the sector [fl, c] since F(t, x) is decreasing in x from the conditions imposed on f.But obviously Bv() is a solution of PBVP (II).So operator B is identical with operator A oa the sector [/, c].And hence operator A is both increasing and decreasing on the sector [fl, a] and satisfies (i).From this it follows that A transforms the sector [fl, c] onto a point u* [fl, c].This implies that u* is the unique fixed point of A on the sector [fl, ].However solving PBVP is equivaleng go finding fixed poings of operagor A. Hence u" is ghe unique solution of PBVP (3.1)-(3.2)on the sector [fl, ll:teceived: December 1990, Revised: April 1991 YONG SUN 2. CONTINUOUS RIGHT-HAND SIDE Consider ghe firs order periodie boundary value problem (2.1) ([O,27r],R) and m' >_ Mm + 7 for almost all t in 0 "Y' = M[m(2r)-m(0)] if m(O) >_ m(27r), if re(O) < m(2r)