# Comment on “Existence Theorem for Integral and Functional Integral Equations with Discontinuous Kernels”

Authors: Krzysztof A. Topolski
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We present a counterexample to the main result of the abovementioned paper showing that this result
is false and cannot be improved in a simple way.


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## Body

## 1. Introduction

In [1] author considers the nonlinear Volterra integral equation (VIE)(1)xt=ut+∫0tft,τ,xτdτand the nonlinear functional Volterra integral equation (FVIE)(2)xt=ut+∫0tft,τ,xτ,xdτ.Theorem 2.1 of [1] states the following.Theorem 1.
Letu:[0,1]→R and let f:[0,1]×[0,1]×R→R be given. Suppose that (C1)–(C4) are fulfilled.(C1)
u is continuous.(C2)
For each(t,x)∈[0,1]×R, the function τ→f(t,τ,x) is Lebesgue measurable. For all (t,x)∈[0,1]×R and for almost all τ∈[0,1],(3)ft,τ,x<Mτ,whereM:[0,1]→[0,∞] is a Lebesgue integrable function.(C3)
For each(t,τ,x)∈[0,1]×[0,1]×R(4)limsupy↑x⁡ft,τ,y≤ft,τ,x=liminfy↓x⁡ft,τ,y.(C4)
LetF={y∈R:y≤u+∫01‍Mτdτ}, where u=max{ut;t∈[0,1]}. For every y∈F and all n∈N the functions(5)t⟶∫0tsupx-y≤1/3n⁡ft,τ,xdτare equicontinuous and tend to zero ast↓0.
Under the above assumptions VIE expressed by (1) has extremal solutions in the interval [0,1].In the following we present a counterexample showing that this result is false.


## 2. Comment on the Assumption (C4)

Define(6)hy,nt=∫0tsupx-y≤1/3n⁡ft,τ,xdτ,y∈F,n∈N,t∈0,1.(C4) states that hy,n(t) is equicontinuous in [0,1] and tends to zero as t↓0. In fact the last seems to be superfluous since it follows from equicontinuity (or from (C2)).Assume thatf does not depend on τ; that is, we set f(t,τ,x)=f(t,x) (with a small violation of notation). Now (C2) gives ft,x≤M in [0,1]×R for some M≥0. We also have(7)hy,nt=tsupx-y≤1/3n⁡ft,x=supx-y≤1/3n⁡tft,x,y∈F,n∈N,t∈0,1.Proposition 2.
Iff is Lipschitz continuous in t, that is, if there exists L≥0 such that(8)ft,x-ft¯,x≤Lt-t¯for all (t,x),(t¯,x)∈[0,1]×R then (C4) is satisfied.Proof.
For allt,t¯∈[0,1] and y∈F, n∈N, we have(9)hy,nt-hy,nt¯=supx-y≤1/3n⁡tft,x-supx-y≤1/3n⁡t¯ft¯,x≤supx-y≤1/3n⁡tft,x-t¯ft¯,x≤supx-y≤1/3n⁡tft,x-t¯ft,x+supx-y≤1/3n⁡t¯ft,x-t¯ft¯,x≤Mt-t¯+Lt-t¯=M+Lt-t¯.This gives equicontinuity of hy,n(t).


## 3. The Counterexample

Our example is a modification of this given in [2].Consider VIE(10)zs=-s∫-1szτ1/2dτs∈-1,1,where z1/2=z1/2sgn⁡z for any z∈R (see [2]).Consider VIE(11)xt=2-4t∫0txτ1/2dτt∈0,1.Proposition 3.
Sets=2t-1, t∈[0,1], and z(s)=x(t), s∈[-1,1]. A function z(s) is a solution of (10) if and only if x(t) is a solution of (11).Proof.
Suppose thatz(s) is a solution of (10). Set s=2t-1, t∈[0,1], in (10). We have(12)xt=z2t-1=1-2t∫-12t-1zτ1/2dτ.By making a substitution τ=2r-1 in the integral we get(13)xt=2-4t∫0tz2r-11/2dr=2-4t∫0txr1/2dr.Hence, x(t) satisfies (11). Similarly setting t=(s+1)/2, s∈[-1,1], in (11) we obtain that z(s) satisfies (10) if x(t) satisfies (11).Corollary 4.
VIE (10) has a maximal (minimal) solution if and only if (11) has a maximal (minimal) solution.Consider VIE(14)xt=2-4t∫0tIxτ1/2dτt∈0,1,where I(x)=(sgn⁡x)min{x,4} for x∈R.Proposition 5.
VIE’s (11) and (14) have the same (nonempty) sets of solutions.Proof.
The statement follows from the fact that every solution of (11) and (14) takes its values in the interval [-4,4] where I(x)=x. Indeed, if x satisfies (11) and x=max{xt;t∈[0,1]} then we have(15)xt≤2-4t∫0txτ1/2dτ≤2∫01x1/2dτ≤2x1/2,t∈0,1which implies x≤2x1/2 and x≤4. Since Ix≤x a similar estimation holds for (14). Of course a zero function is a solution of both equations.Setf(t,τ,x)=f(t,x)=(2-4t)[I(x)]1/2. Of course, f is Lipschitz continuous in t. It is not difficult to verify (see Proposition 2) the following.Remark 6.
VIE (14) satisfies all the assumptions of Theorem 1.Proposition 7.
VIE (14) has no extremal solution in [-1,1].Proof.
In view of Corollary4 and Proposition 5 we only need to show that (10) has no extremal solutions. This was in fact done in [2] where the proof is rather long and complicated. For the reader’s convenience, we present an original and short explanation.
Suppose thatv is not a trivial solution of the problem(16)v′s=-s1/2vs1/2,v-1=0in-1,1.Such solution exists since this problem, in view of the classical theory, has many solutions. Suppose that zM(s) is a maximal solution of (10). Since 0, sv(s), -sv(s) are all solutions of (10) we have zM(s)≥max{0,sv(s),-sv(s)}; hence zM(s)≥0 and it is not identically zero. This gives zM(s)=-s∫-1s‍zM(τ)1/2dτ<0 for some s∈(0,1]. This leads to a contradiction. We finish the proof by observing that the negative of a minimal solution of (10) must be its maximal solution.Remark 8.
Of course, we can improve Theorem1 by assuming that f is nondecreasing in x. In this case however, (C3) is not necessary and (C4) can be reduced to a simpler one and the result is well-known.Remark 9.
Theorem 3.1 [1] (FVIE (2)) and Theorem 4.1 [1] (system of Volterra integral equation) are false since they generalize Theorem 1 (Theorem 2.1 [1]).



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*Source: 698121-2015-07-08.xml*