PERIODIC SOLUTIONS OF VOLTERRA INTEGRAL EQUATIONS

Consider the system of equations x ( t ) = f ( t ) + ∫ − ∞ t k ( t , s ) x ( s ) d s , ( 1 ) and x ( t ) = f ( t ) + ∫ − ∞ t k ( t , s ) g ( s , x ( s ) ) d s . ( 2 ) Existence of continuous periodic solutions of (1) is shown using the resolvent function of the kernel k . Some important properties of the resolvent function including its uniqueness are obtained in the process. In obtaining periodic solutions of (1) it is necessary that the resolvent of k is integrable in some sense. For a scalar convolution kernel k some explicit conditions are derived to determine whether or not the resolvent of k is integrable. Finally, the existence and uniqueness of continuous periodic solutions of (1) and (2) are btained using the contraction mapping principle as the basic tool.


I. INTRODUCTION.
In thls paper we study the existence and uniqueness of perlodlc solutions of the Integral equations t x(t) f(t) + f k(t,s)x(s)ds, -(R)<t<" (1.1) and t x(t) f(t) + ] k(t,s)g(s,x(s))ds, -<t<', (1.2) M.N.ISLAM where x, g and f are vectors in R n k is an n by n matrix function with elements in R, and R n is the vector space of n-dimenslonal column vectors.We llst our basic assumptions in Section 2. The results and their proofs are presented in Sections 3, 4 and 5.
In Section 3 we present two basic results, Theorems and 2, which are used in Theorem 3 of Section 4 to obtain the existence of a continuous periodic solution of (I.I).
Theorem deals with the resolvent kernel associated with the Volterra equation t R + y(t) f(t) + k(t s)y(s)ds, In Theorem 2 we obtain (I.I) as a limit equation of (1.3).The resolvent equation corresponding to (1.3) is t r(t,s) ffi-k(t,s) + f k(t,u)r(u,s)du, O6s't, s (1.4) an its solution r(t,s) is called the resolvent kernel.The importance of the resolvent derives from the fact that the solution y(t) of (1.3) is given by t y(t) f(t) f r(t,s)f(s)ds, t)0.0 The existence of continuous r(t,s) as a solution of (1.4) is a known result (see [I, Chapter IV, Theorem 3.1]).In Theorem we prove the uniqueness of r(t,s) which is used to establish an important property, (4.4), of r(t,s).We use (4.4) together with (1.6) and other properties derived in Lemmas 2 and 3 in obtaining periodic solutions of (I.I).Notice that all the properties of the resolvent function derived in this paper including the integrabillty properties obtained in Theorem 4 are significant results by themselves.
We assume that r(t,s) is Integrable in the sense that t sup I Ir(t,s) ds ' < for all complex number z satisfying Re z 0.
Some results regarding the property (1.6) are available in [2,3].In case k(t,s)=a(ts) is of convolution type for which the resolvent r(t,s)fb(t-s) is also of convolution type, it can be verified that if both a(t) and b(t) are of class LI(R+) then all the The integrability of b(t) (i.e.b(t) is in the class LI(R+)) is also studied in [5][6][7][8][9].
Analyzing the transcendental relation (1.7) we derive in Theorem 4 a few explicit conditions regarding the Integrabfllty of b(t).
In Section 5 we use the familiar contraction mapping principle to show the existence and uniqueness of periodic solutions of (I.I) and (1.2).We obtain these results in Theorems 5 and 6.
Some results related but different from the results of the present paper on periodic solutions are available in [I0-17].
a vector x in R n For let x denote a norm of x equivalent to the Euclidean which corresponds to the vector norm Ixl.
Throughout this paper we make the following assumptions of f, k, and g: (AI) f(t) is continuous and T-perlodlc on R for some T > O; (A2) k(t,s) is continuous in (t,s) for -<sgt<', k(t,s)=0 for s>t; (A3) k(t+T,s+T)--k(t,s) for -<st<; (A4) there exists a constant 8 > 0 such that t 0 tO for each >0 there exists a d>O such that whenever lhl(d then t t+h 0 t for all tO; (Note that the second integral becomes zero for h<0 since k(t,s)-0 for s>t).
(A6) g(t,x) is defined on RxRn, for each x in R n the function g(t,x) is T- periodic in t, and g(t,O)--O for all -<t<; (AT) for each a>O, there exists an n>O such that It may seem that (A2)-(A4) possibly imply (AS).To see that (AS) is independent of (A2)-(A4), consider the following example suggested by C.E. Langenhop:
It can be shown that (i) k(t,s) is continuous for -<st<with t k(T,s)=2(s), (ll) there exists a constant 6>0 such that k(t,s)ds6 for all 0 t R + t'0, and (Iii)  k(t,s)ds is not uniformly continuous on The definition of 0 k(t,s) along with (1) and (li) show that k(t,s) satisfies (A2)-(A4).However, k(t,s) t does not satisfy (AS) since (AS) would imply f k(t,s)ds is unlformly continuous on + 0 R 3. TWO BASIC RESULTS.
Although the existence of a continuous solution r(t,s) of (1.4) is a known result the uniqueness of such r(t,s) does not seem to be explicitly shown anywhere.
In Theorem we establish the uniqueness of r(t,s).
If k(t,s) is continuous on Ost< , then there exists a unique continuous solution r(t,s) of (1.4) on Ost< .
PROOF.We only prove the uniqueness of r(t,s).By way of contradiction, suppose there are two solutions r( with r(t,s)#w(t,s) for all Then for any continuous q we have t y(t) q(t) f r(t,s)q(s)ds, t,0, 0 and t y(t) q(t) f w(t,s)q(s)ds, t)0.0 as the unique solution of the Volterra integral equation t y(t) The uniqueness of the solution y(t) is a well known result (see e.g., [18, Theorem
Since Um(t,s) is continuous in (t,s) and Um(t |,sl)-s>0 there exists an (3.2) with e<s such that Um(tl,S)>O for O<Sl-SSl+e.Let us choose a continuous function q such that qm(sl)=l, qm(s))0, qm(S)--0 for 0sCsl-e s)sl+, and qj(s)=0 for J=l,2,...,n, JCm.Then it follows from the choice of q and from the property of that the left side of (3.2) is nonzero, which is a contradiction.um LEMMA I. Suppose k(t,s) satisfies (A2) and (A3).Then (A4) holds if and only if holds.
Then choose a positive integer n O such that t+nT>0 for all n)n 0. It follows from (A3) that r t+nT for all n)n O.This implies (3.3).
By virtue of Lemma the integrals involved in Theorem 2 and in subsequent results of this paper are defined and finite.
THEOREM 2. Suppose (AI)-(A5) hold.If y(t) is the continuous bounded solution of R + (1.3) on then there exists a sequence of integers nj/ as J+ such that y(t+njT)/x(t), a continuous solution of (I.I) on R, as j+', and the convergence is uniform on compact subsets of R. + PROOF.Since y(t) is a continuous and bounded function on R it follows from (A4) and (AS) that f0t k(t,s)y(s)ds is bounded and uniformly continuous on R+.Thus, from (1.3) and (AI) we see that the function y(t) is bounded and uniformly continuous on R+.
Hence, for any a>0 the sequence {y(t+nT), nT>a, nEN} of translated functions is equicontlnuous and uniformly bounded on -t< , where N denotes the set of positive integers.
Therefore, by Ascoli's theorem there exists a sequence of integers nj and a continuous function x(t) such that max ly(t+njT)-x(t) < This proves that y(t+njT)/x(t) as J+, and the convergence is uniform in t on each compact subset of R.
Let L be a bound for Ix<.,I then a few calculations yield t t M.N.ISLAM t This last expression tends to zero as J+.
Therefore, taking the limit in the sequence of translated equations (obtained from (1.3)) y(t+njT) f(t) + ft k(t s)y(s+njT)ds -nit as j+-, we get (l.l) as required to show that x(t) satisfies (I.I) on R.
4. PERIODIC SOLUTIONS USING THE RESOLVENT KERNEL.
The proof of Lemma 3 involves the use of (1.4) and the application of Fubini's theorem.We omit its proof because a parallel result is available in [2, Theorem 2].
In Lemma 2 we proved that r satisfies the relation r(t+T,s+T)-r(t,s) for 0st<'.
Let us extend this r using the relation r(t,s)ffir(t+nT,s+nT) for -'<st<0 where n is a positive integer and large enough so that t+nT,s+nT>0.This extended r is now defined and continuous for -'<st<'.Also, r(t,s) satisfies the relation r(t+T, s+T) r(t,s) for-(R) < s t < -. holds.Thus, the integrals involved in Theorem 3 make sense.
Suppose also r(t,s) satisfies (1.6).Then (I.I) has a continuous periodic solution x(t) on R. (We use the term "Periodic solution" to refer to T-periodlc solution).
PROOF.It follows from (1.5), (1.6) and (A1) that the solution y(t) of (1.3) is t r(t s)f(s)ds is uniformly bounded on R+.
Again, (AI) and Lemma 3 imply that 0 + continuous on R So, by Theorem 2 there exists a sequence of integers nj such that y(t+njT)/x(t), a continuous solution of (I.1) on R, as J+.
Let H be a bound for If(s) when -<s<'.For--<t< , if t+njT>O with J>Itl then ,which tends to zero as j.Taking the limit in the sequence of translated equations (obtained from (1.5)) y(t+njT) f(t) ft -niT r(t,s)f(s)ds as j/-, we obtain x(t) f(t) fL r(t,s)f(s)ds. (4.5) if follows from (A1) and (4.4) that x(t) in (4.5) is T-perlodic.REMARK I.In the proof of Theorem 3 one may notice that it is only the continuity instead of uniform continuity of r(t,s)f(s)ds that is needed.which is relatively weaker version of condition (4.3).Note that for condition (4.6) to hold, assumption (A5) could be replaced by the following property: h'O for each t 0. (4.7) However, to prove Theorem 2 which is used in Theorem 3 we need (AS) so that t f k(t,s)y(s)ds can be uniformly continuous.The uniform continuity of 0 788 M.N.ISLAM t I k(t,s)y(s)ds is needed for the equlcontinulty of {y(t+nT), nT>a, neN} on 0 L REMARK 2.
If k(t,s) a(t-s) with a(t) in the class (R+) then a(t) satisfies (A3) and (AS).Similarly, if the resolvent b(t) of a(t) is of class LI(R+) then b(t) satisfies (4.3) and (4.4).Therefore, the results of Theorems 2 and 3 include convolution equations as special cases.
The following are a few condltlons derived from Theorem 0 to determlne whether or L not b(t) is of class (R+).
THEOREM 4. Suppose a(t) is a real valued continuous function on R + with a(t) in the class LI(R+).Let b(t) be the resolvent of a(t).
If f0 [a(t)Idt < I, then b(t) is in the class LI(R+).
(ill) Suppose a(t) does not change its sign on R + If -I 0 a(t)dt < I, then b(t) is in the class LI(R+).
PROOF.Since a(t) is a scalar function, the condition (1.7) becomes l-a*(z) 0 for Re z ) 0 fOe t)dt , let q(z) l-a (z).PROOF of (i).It is sufficient to prove that there exists at least one root of q(z) in the closed right half plane.If 10 a(t)dt-I then q(0) l-a (0) -0.So, z-0 is a root of q(z).If f0 a(t)dt > then q(0) < 0. Considering y-0, x > 0 where , * =a* x+lyfz, we obtain a (z) (x) which tends to zero as x ".Thus, q(x)l-a (x) R + as x .Since q(x) is a real valued continuous function on q(0) < 0, R + and q(x) as x , it follows that there is a real positive root of q(x) on PROOF of (ll).
From the hypothesis we get la*(z) < for Re z )0.So, l,<z>#-l-.*<z>l> ,-l.*<z>l > o ..> O. erefore, q(z) has no root z, Re z) 0. PROOF of (ill).We assume that a(t) O. Otherwise b(t) 50.Since a(t) does not change its sign, the condition -I < f0 a(t)dt < is the same as the one in (ll).So, we consider only the case f;a(t)dt--I.
This shows that the condition (4.9) holds.
For yffiO, x 0 the function q(z) q(x) I. From (1) we know that fe-Xt(t)dt-tends to zero as x / (R).This shows that q(x) / as x / .So, q(z) has no root for y=0, x > 0.
5. PERIODIC SOLUTIONS USING THE CONTRACTION MAPPING PRINCIPLE.
Let X {x(t): R/Rn, x(t) is continuous and bounded on R} For x in X let llxllsup {[x(t)l:-< t < (R)}.Then (X, II.II)is a Banach space.For simplicity we write X instead of <X, ll.ll )" Let ,P'f{x in X: x<t+T)fx<t) for all < t < "}.THEOREM 6. Suppose (AI) (A7) hold.Consider 8 of (A4).For any e > 0 with 8 < choose of (AT).Then for each f satisfying If(t) (I-8)   as well as assumption (AI) there exists a unique continuous periodic solution x(t) of (1.2) with Ix(t)l on R.Moreover, the function x(t) is the only continuous solution of (1.2) with Ix(t) on R.

PROOF.
Fix s > 0 with e8 < I. Then from (A7) it follows that there exists an Choose f satisfying (AI) and the condition If(t) (l-eM) for all < t < ".
Consider the set.
For # in S, define a map A on S by A(t) f(t) + L k(t,s)g(s,#(s))ds. (5.3) It follows from (AT) and Lemma 5 that fSk(t,s)g(s,(s))ds is continuous on R.
Therefore, the function A(t) in (5.3) is continuous on R. Again, lAw(t) g (1-aB)n+aBn-n.So, A maps from S into itself.Finally, for , in S, IA-Agll 811-911" Since = < 1, mapping A is a strict contraction.This proves that there exists a unique continuous solution x(t) of (1.2) with Ix(t)t n on R.
It follows from Lemma 4 that the function x(t+T) is also a continuous solution of (1.2) with Ix(t+T)t n on R.This shows that x(t)-x(t+T) for all t in R as required.

REMARK.
Theorems 5 and 6 hold even if we replace assumption (AS) by (4.7).
Condition (4.7) will provide the required continuity of A(t) in (5.2) and (5.3).

ACKN OWLEDGMENT S
This work is part of the author's Ph.D. dissertation written under the direction of Professor T.A. Burton.
The author gratefully acknowledges Professor Button's guldance throughout this work.
4 hold (Remark 2).A necessary and sufficient condition for b(t) L to be of class (R+) was obtained by Paley and Wiener [4] in the following result: L THEOREM 0. Suppose a(t) is in the class (R+).Then the resolvent b(t) is in L the class (R+) if and only if the determinant det(I f e-Zta(t)dt) # 0 0 (1.7)