Integrable and continuous solutions of a nonlinear quadratic

This addendum concerns the paper of the above title found in EJQTDE No. 25 (2008). There are some misprints in that paper: (i) Page 3, line 5 should be k : [0, 1] × [0, 1] → R + satisfies Carathèodory condition (i.e. measurable in t for all s ∈ [0, 1] and continuous in s for all t ∈ [0, 1]) such that 1 0 k(t, s)m 2 (s)ds is bounded ∀t ∈ [0, 1]. (ii) Page 6, line 6 should be k : [0, 1]×[0, 1] → R + satisfies Carathèodory condition (i.e. measurable in s for all t ∈ [0, 1] and continuous in t for all s ∈ [0, 1]) such that k(t, s)m 2 (s) ∈ L 1 ∀t ∈ [0, 1].


Introduction
Quadratic integral equations are often applicable in the theory of radiative transfer, kinetic theory of gases, in the theory of neutron transport and in the traffic theory.Especially, the so-called quadratic integral equation of Chandraskher type can be very often encountered in many applications (see [1]- [4]).Let I = [0, T ], L 1 = L 1 [0, T ] be the space of Lebesgue integrable functions on I and C[0, T ] be the space of continuous functions defined on I. Consider now the following integral equation It is proved that ( [5], Caratheodory Theorem ) if f is measurable in t ∈ [t 0 , T ] for each fixed x ∈ B ⊂ R and continuous in x ∈ B for each fixed t ∈ [t 0 , T ] and there exists a function m ∈ L 1 such that |f (t, x)| ≤ m(t) then, there exists a solution x ∈ C[t 0 , T ] of the integral equation ( 1) and the equivalent initial value problem This result has been generalized (by the authors [7]) for the fractional-order integral equation EJQTDE, 2008 No. 25, p. 1 In this work, we generalize this result for the nonlinear quadratic integral equation Firstly we prove the existence of at least one positive solution x ∈ L 1 of the quadratic integral equation (3) where the functions f and g satisfy Carathèodory condition.
Secondly we prove the existence of at least one positive continuous solution for equation (3) where g satisfies Peano condition and f satisfies Carathèodory condition.Also the existence of the maximal and minimal ( continuous ) solutions will be proved
Tychonov's Theorem.Suppose B is a complete, locally convex linear space and S is a closed convex subset of B. Let the mapping T : B → B be continuous and T (S) ⊂ S. If the closure of T (S) is compact, then T has a fixed-point in S.
Schauder fixed-point Theorem.Let S be a convex subset of a Banach space B, let the mapping T : S → S be compact, continuous.Then T has at least one fixed-point in S.
Arzelà-Ascoli Theorem.Let E be a compact metric space and C(E) the Banach space of real or complex valued continuous functions normed by If A = {f n } is a sequence in C(E) such that f n is uniformly bounded and equicontinuous.Then the closure of A is compact.

L 1 −positive solution
Let I = [0, 1], and consider the assumptions: (i) a : For the existence of at least one L 1 −positive solution of the quadratic integral equation ( 3) we have the following theorem.
Theorem 3.1 Let the assumptions (i)-(iii) be satisfied.Then the nonlinear quadratic integral equation ( 3) has at least one where The set Q can be shown to be nonempty, bounded, closed and convex in L 1 .
Let H be the operator defined by the formula We shall prove that H : so Hx ∈ Q and hence HQ ⊂ Q.
To apply Schauder fixed-point Theorem, we shall prove that HQ is relatively compact in L 1 .
By using Lusin Theorem and Scorza Dragoni Theorem , we can find a closed subset This means that the sequence {Hx h } is sequence of equi-continuous functions on A n and we can prove that this sequence is uniformly bounded.Now Hence by Arzelà-Ascoli Theorem Hx h is relatively compact subset of C(A n ) and this can be done for each n ∈ N.This implies the existence of convergent subsequence {x h j } EJQTDE, 2008 No. 25, p. 4 of {x h } in each C(A n ).Given > 0 and choose n 1 ∈ N so that meas(A n 1 ) < , then Since C(A n ) is complete metric space, hence this subsequence is a Cauchy sequence in each C(A n ), n = 1, 2, 3, ... That is for given > 0 and j, l are arbitrary large we have But we want to prove that the set HQ is relatively compact in L 1 , that is HQ is compact in L 1 .
To do this, we will prove that the sequence { Hx h } is convergent in L 1 , since L 1 is complete metric space, then it is sufficient to prove that the subsequence { Hx h j } is a Cauchy sequence in L 1 .
i.e. ∀ η > 0, ∃ N (η) and An x 0 (t) dt < η/4 such that Now from ( 4) and ( 5) we have Choose N such that l, j > N, then (6) implies that || Hx h j − Hx h l || C(An) ≤ η/2.This means that the subsequence { Hx h j } is a Cauchy sequence in L 1 which implies that HQ is relatively compact in L 1 .Then H has at least one fixed point.Hence there exists at least one solution x ∈ L 1 of (1).Since all conditions of Shauder's fixed-point Theorem hold, then H has a fixed point in Q.

Continuous solutions
Let I = [0, 1], and consider the assumptions: (i) a : (iii) g : where M 2 is a positive constant.It is clear that the set S is closed and convex.Let H be the operator defined by the formula Assumptions (ii) and (iii) imply that H : S → C is continuous operator in x .We shall prove that HS ⊂ S.
For every x ∈ S we have where | a(t) | ≤ M 1 .Then, Hx ∈ S and hence HS ⊂ S. Also for x ∈ S and t 1 and Then we have This means that the functions of HS are equi-continuous on [0, 1], then by Arzelá-Ascoli Theorem the closure of HS is compact .Hence, all conditions of Tychonov fixed-point Theorem hold, then H has a fixed point in S.
5 Maximal and minimal solutions Definition 5.1 [8] Let q(t) be a solution of the nonlinear quadratic integral equation (3).Then q(t) is said to be a maximal solution of (3) if every solution of (3) satisfies the inequality x(t) < q(t), ∀t ∈ I.A minimal solution s(t) can be defined by similar way by reversing the above inequality i.e. x(t) > s(t), ∀t ∈ I.
We shall use the following lemma to prove the existence of the maximal and minimal solutions.Proof.Firstly we shall prove the existence of the maximal solution of (3).Let > 0 be given.Now consider the quadratic integral equation where Clearly the functions f (t, x ) and g (t, x ) satisfy assumptions (ii) and (iii) and therefore equation ( 8) has at least a positive solution x (t) ∈ C(I).Let 1 and 2 be such that 0 < 2 < 1 < .Then x Applying Lemma 5.1 on ( 9) and (10), we have x 2 (t) < x 1 (t) f or t ∈ I.
As shown before the family of functions x (t) is equi-continuous and uniformly bounded.
Hence by Arzelá-Ascoli Theorem, there exists a decreasing sequence n such that → 0 EJQTDE, 2008 No. 25, p. 8 as n → ∞ and lim n→∞ x n (t) exists uniformly in I and denote this limit by q(t) .
From the continuity of the functions f (t, x ) and g (t, x ) in the second argument, we get q(t) = lim n→∞ x n (t) = a(t) + g(t, q(t)) t 0 k(t, s) f (s, q(s)) ds which implies that q(t) as a solution of (3).Finally, we shall show that q(t) is the maximal solution of (3).To do this, let x(t) be any solution of (3).Then from the uniqueness of the maximal solution (see [8]), it is clear that x (t) tends to q(t) uniformly in t ∈ I as → 0. By similar way as done above we can prove the existence of the minimal solution.
I × R + → R + satisfy Caratheodory condition (i.e.measurable in t for all x ∈ R + and continuous in x for all t ∈ [0, 1] ) and there exist two functions m 1 , m 2 ∈ L 1 such that [6]isfies Carathèodory condition (i.e.measurable in t for all s and continuous in s for all t.) Now for the existence of at least one positive continuous solution of the nonlinear quadratic integral equation (3) we have the following theorem.Theorem 4.1 Let the assumptions (i)-(iv) be satisfied.Then the equation (3) has at least one positive solution x ∈ C(I).Proof.We shall use Tychonov's fixed point Theorem to prove this theorem It can be verified that[6]C is complete locally convex linear space.Define a subset S of C by Let the assumptions of Theorem 4.1 be satisfied and f (t, x), g(t, x) are nondecreasing in x on I .Then there exist maximal and minimal solutions of equation (3).