Self referred equations with an integral boundary condition

Abstract

In this note we will study three differential problems with a dynamic which will be represented by a self referred equation and a boundary condition which will be expressed as an integral constraint. We will prove that, under certain assumptions, there exists at least one solution of for all of these problems by using a fixed point theorem.

Full text

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY
CONDITION
GIACOMO CHIRIATTI*, MATTEO FASIELLO**, RAFFAELE GRANDE***, AND EDUARDO PASCALI****
Abstract. In this note we will study three differential problems with a dynamic which will
be represented by a self referred equation and a boundary condition which will be expressed as
an integral constraint. We will prove that, under certain assumptions, there exists at least one
solution of for all of these problems by using Schauder’s fixed point theorem. In the end we will
propose briefly some open problems.
1. Introduction
The hereditary phenomena were widely studied in the past (see e.g. [2], [3], [11]) due to their
impact in applied sciences, for example in engineering and biology. Hence the so called self-referred
and hereditary equations were proposed in order to write a model to describe this type of events
(see e.g. [1], [5], [6], [7] and [10]). Formally we may represent this class of equations as follows: let
us consider Xa space of functions, A:X→R,B:X→Rtwo functional operators. Then a self
referred equation may be written as
Au(x, t) = u(Bu(x, t), t).
In our problems the boundary condition will be expressed in the form
Zβ
0
˙y2(t)dt =δβ > 0
or
Zβ
0
y2(x)dx =δβ > 0
where 0 < β ≤1 and 0 < δ < 1. This constraint may represent, roughly speaking, a quantity
which is preserved in an isolated system (e.g. the energy). This condition was implemented in
an ODE problem in [4] with a dynamic in the form ¨y(t) = f(t, y(t),˙y(t)),with fwas a positive,
bounded and globally Lipschitz function.
The goal of this paper is to study a differential problem which mix self referred equations and
integral boundary constraints. We will study the differential problem





¨y(t) = βy(y(t)), t ∈[0,1],
y(0) = α,
Rβ
0˙y2(t)dt =δβ > 0
(1.1)
2010 Mathematics Subject Classification. Primary 34A34, 45G10.
Key words and phrases. Ordinary differential equations, evolution equations, hereditary phenomena.
1

2 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
where 0 < δ < 1, 0 ≤α≤1 and 0 < β ≤1. We may also study a variation of (1.1) where we do
not know the initial state of our system. Hence the second problem which will be studied is
(˙y(t) = βyRt
0y(s)ds, t ∈[0,1],
Rβ
0y2(t)dt =δβ (1.2)
where 0 < δ < 1 and 0 < β ≤1. In the last problem we introduce also a space variable, i.e.
(∂
∂t y(x, t) = βyRt
0y(x, s)ds, t,(x, t)∈[0,1] ×[0,1],
Rβ
0y2(x, t)dt =βg(x)(1.3)
where 0 < β ≤1 and g: [0,1] →[0,1] is smooth function.
The technique to find the existence of a solution follows these steps: we will study the problems
heuristically, obtaining the operator which we want to study and the space where we will work on.
Then we will check that, under certain conditions, the space of functions which we have found is
bounded, closed and convex. Then we will prove that operator is continuous and compact and we
conclude applying the Schauder’s fixed point theorem.
The paper is organized as follows: in the Section 2 we will recall briefly some preliminaries
of functional analysis, such as Ascoli-Arzel`a theorem and Schauder’s fixed point theorem; in the
Section 3 we will prove that the problem (1.1) has at least one solution; in the Section 4 we will
prove a similar result for the problem (1.2); in the Section 5 we will study the problem (1.3) and
in the Section 6 we will state briefly some open problems.
2. Preliminaries
Let us recall some classical definitions and results of functional analysis. For further details we
refer to [8].
Definition 1. Let us consider a sequence {fn}n∈Nof continuous functions on an interval I⊂R;
•The sequence {fn}n∈Nis equibounded if there exists a real number M > 0 such that it
holds true
|fn(x)| ≤ M
for all n∈Nand for every x∈I.
•The sequence {fn}n∈Nis equicontinuous if, for every ε > 0, there exists a δ > 0 such that
|fn(x)−fn(y)|< ε
whenever |x−y|< δ and for all n∈N.
Definition 2. Let us consider X,Ytwo normed spaces and T:X→Yan operator. Then the
operator Tis compact if, for every bounded subsequence {xn}n∈N⊂Xit is possible to extract a
convergent subsequence of the sequence {T xn}n∈N⊂Y.
These definitions are necessary in order to state the following classical results of functional
analysis. These theorems are crucial to prove the existence of a fixed point for a functional and
then the existence of a solution for our differential problem.
Theorem 2.1 (Ascoli-Arzel`a theorem).Let us consider a sequence of real-valued continuous func-
tions {fn}n∈Ndefined on a closed and bounded interval [a, b]of the real line. If {fn}n∈Nis equi-
bounded and equicontinuous, then there exists a subsequence {fnk}nk∈Nthat converges uniformly.

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY CONDITION 3
Theorem 2.2 (Fixed point Schauder’s theorem).Let us consider Xa bounded, closed and convex
Banach spaces and a continuous and compact operator T:X→X. Thus the operator will have a
fixed point.
3. First problem
As stated in the Introduction, we will divide our proof in smaller steps. We will start finding
by heuristic methods the operator which we want to study (namely T) and the space where we
will work on (namely X). We will prove then that, under certain conditions, Xis bounded, closed
and convex. Then we will prove that Tis continuous and compact and in the end we conclude
applying the Schauder’s theorem.
3.1. The definition of the functional T.In this section we will study heuristically the ODE
problem stated in (1.1).
Let us start observing that the dynamic of the system is easily integrable by using standard
technique of ODEs theory. By multiplying both sides with ˙y(t) we get immediately
¨y(τ) ˙y(τ) = βy(y(τ)) ˙y(τ)⇒1
2
d
dτ ˙y2(τ) = βd
dτ Zy(τ)
0
y(s)ds.
Then, by integrating on the interval [0, t] both sides of the equation, it is straightforward to get
1
2Zt
0
d
dτ ˙y2(τ)dτ =βZt
0
d
dτ Zy(τ)
0
y(s)dsdτ
⇒1
2˙y2(t) = 1
2˙y2(0) + βZy(t)
0
y(s)ds −Zα
0
y(s)ds
which implies
˙y2(t) = ˙y2(0) + 2βZy(t)
α
y(s)ds. (3.1)
Thus, by applying the condition (3.1) on our integrodifferential condition, we obtain
δβ =Zβ
0
˙y2(t)dt = ˙y2(0)β+ 2βZβ
0Zy(t)
α
y(s)dsdt
from which we deduce
˙y(0) = ±sδ−2Zβ
0Zy(t)
α
y(s)dsdt. (3.2)
The existence condition on (3.2) leads us naturally to the definition of the operator Tand of the
space of function X.
Definition 3. Let us consider the space of continuous functions C([0,1],[0,1]). Then we define
the space of function X⊂C([0,1],[0,1]) as
X:= y
y∈C([0,1],[0,1]) and δ−2Zβ
0Zy(t)
α
y(s)dsdt ≥0,(3.3)
and the operator Tas
T y(t) = α+tsδ−2Zβ
0Zy(t)
α
y(s)dsdt +βZt
0Zτ
0
y(y(s))dsdτ, t ∈[0,1].(3.4)
Let us remark some immediate properties related to the boundedness of our operator.

4 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
Remark 1. Since y∈Xthen
T y(t)≥α,
for all y∈Xand for all t∈[0,1].
Remark 2. Let us recall that, for all y∈Xand for all t∈[0,1], it holds true that |y(t)| ≤ 1.
Hence βRt
0Rτ
0y(y(s))dsdτ ≤β
2. Therefore we obtain
T y(t)≤α+sδ−2Zβ
0Zy(t)
α
y(s)dsdt +β
2≤α+sδ+ 2 Z{t∈[0,β ]|y(t)<α}Zα
y(t)
y(s)dsdt +β
2
≤α+sδ+ 2 Z{t∈[0,β ]|y(t)<α}Zα
0
y(s)dsdt +β
2.
Thus we can write
T y(t)≤α+pδ+ 2βα +β
2.
Assuming that α+√δ+ 2βα +β
2≤1, we get that α≤T y(t)≤1 for all y∈Xand for all t∈[0,1].
Remark 3. We estimate
δ−2Zβ
0ZT y(t)
α
T y(s)dsdt =δ−2Z{t∈[0,β]|α<T y (t)}ZT y(t)
α
T y(s)dsdt
+ 2 Z{t∈[0,β ]|α≥T y(t)}Zα
T y(t)
T y(s)dsdt
≥δ−2Z{t∈[0,β]|α<T y(t)}ZT y (t)
α
T y(s)dsdt
≥δ−2Z{t∈[0,β]|α<1}Z1
α
T y(s)dsdt
≥δ−2β(1 −α).
Hence, if we consider δ−2β(1 −α)≥0, we have also T y ∈Xfor all y∈X.
Remark 4. The Remarks 1, 2, 3 suggest us the following assumptions
(α+√δ+ 2βα +β
2≤1,
δ−2β(1 −α)≥0.(3.5)
Let us show that the system (3.5) has at least one solution. Let us consider α= 0. Then
(√δ≤2−β
2,
2β≤δ⇒2β≤δ≤2−β
22
.
This system of inequalities is well defined since it is equivalent to 8β≤4−4β+β2which implies
that 0 < β ≤6−4√2. Hence it is possible to find a triple (α, β, δ) which solves our system of
inequalities.
Proposition 3.1. Under the assumption (3.5), the operator T:X→Xis well defined.
Proof. The proof is an immediate consequence of the Remarks 1, 2, 3 and 4. □

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY CONDITION 5
3.2. The properties of the space X.Let us focus now on the topological properties on space X.
In order to apply our fixed point theorem we have to check that the space Xis closed and convex.
Proposition 3.2. Under the assumptions (3.5), the space Xas defined in (3.3) is closed.
Proof. Let us consider {yn}n∈Na Cauchy sequence in X⊂C([0,1],[0,1]). Let us recall that
C([0,1],[0,1]) is complete. Hence there will exists y∞∈C([0,1],[0,1]) such that yn→y∞uni-
formly in [0,1]. We obtain, of course, that y∞∈C([0,1],[0,1]). We have to prove that y∞verifies
also the conditions (3.5). To remark it let us observe that
Zyn(t)
α
yn(s)ds −Zy∞(t)
α
y∞(s)ds≤Zyn(t)
α
yn(s)ds −Zy∞(t)
α
yn(s)ds
+Zy∞(t)
α
yn(s)ds −Zyn(t)
α
y∞(s)ds
≤Zyn(t)
y∞(t)
yn(s)ds
+
+Zy∞(t)
α|yn(s)−y∞(s)|ds
≤2∥yn−y∞∥∞.
where ∥ ∥∞is the standard L∞norm. Hence it is straightforward to get
lim
n→∞ Zyn(t)
α
yn(t) = Zy∞(t)
α
y∞(s)ds (3.6)
and, by using the standard limit theorem for integrals, we obtain that, since for all n∈Nit holds
yn∈X,then y∞∈X.□
Proposition 3.3. Under the assumptions given in (3.5), the space Xis convex.
Proof. Let us consider y1,y2∈Xand µ,λ∈[0,1] such that λ+µ= 1. It is immediate to observe
that λy1+µy2∈C([0,1],[0,1]). Now our goal is to prove that λy1+µy2∈X, which is immediate
remarking
δ−2Zβ
0Zλy1+µy2
α
(λy1+µy2)dsdt =δ+
−2Z{t∈[0,β]|α≤λy1(t)+µy2(t)}Zλy1(t)+µy2(t)
α
(λy1(s) + µy2(s))dsdt
+ 2 Z{t∈[0,β ]|α>λy1(t)+µy2(t)}Zα
λy1(t)+µy2(t)
(λy1(s) + µy2(s))dsdt
≥δ−2β(1 −α)≥0.
Hence the space Xis convex. □
Hence we may conclude stating the following theorem which sum up all our results.
Theorem 3.4. Let us consider the operator T:X→Xas defined in Definition 3. Under the
assumptions (3.5), the operator T:X→Xis well defined and the space Xis closed, bounded and
convex.
Proof. The statement is a direct consequence of Propositions 3.1, 3.2 and 3.3. □

6 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
3.3. The continuity and compactness of T.Let us study now the properties of the operator
T. In the following proposition we show that Tis continuous.
Proposition 3.5. Under the assumptions given in (3.5), the operator T:X→Xas defined in
Definition 3 is continuous.
Proof. Let us consider the sequence of functions {yn}n∈N⊂Xand such that yn→y∞uniformly
in the set X. Let us consider the sequence {T yn}n∈Nand T y∞. Let us remark we may deduce
from (3.6) that
lim
n→∞ Zβ
0Zyn(t)
α
yn(s)dsdt =Zβ
0Zy∞(t)
α
y∞(s)dsdt.
Then it is obvious to obtain by continuity that
lim
n→∞ sδ−2Zβ
0Zyn(t)
α
yn(s)dsdt =sδ−2Zβ
0Zy∞(t)
α
y∞(s)dsdt.
Let us recall that y∞∈C([0,1],[0,1]) is uniformly continuous in the set [0,1]. Then we have that
∀ε > 0∃δ > 0 such that ξ, η ∈[0,1] |ξ−η|< δ ⇒ |y∞(ξ)−y∞(η)|< ε.
We estimate
|yn(yn(t)) −y∞(y∞(t))|≤|yn(yn(t)) −y∞(yn(t))|+|y∞(yn(t)) −y∞(y∞(t))|
≤ ∥yn−y∞∥+|y∞(yn(t)) −y∞(y∞(t))|
≤ ∥yn−y∞∥+ε(3.7)
for a sufficiently large n. The inequality (3.7) implies that, for all ε > 0 and for all t∈[0,1]
lim
n→∞ |yn(yn(t)) −y∞(y∞(t))| ≤ ε
and, as consequence
lim
n→∞ |yn(yn(t)) −y∞(y∞(t))|= 0.
We deduce immediately that
lim
n→∞ Zt
0Zτ
0
yn(yn(s))dsdτ =Zt
0Zτ
0
y∞(y∞(s))dsdτ
and then we conclude that Tis continuous. □
In the following proposition we prove that the operator Tis compact.
Proposition 3.6. Under the assumption (3.5), the operator T:X→Xas defined in Definition 3
is compact.
Proof. Let we remark that, for all y∈X, it holds true
0≤T y(t)≤α+β
2+pδ+ 2βα ≤1.
Thus if we consider an equibounded sequence {yn}n∈Nthen also we get that {T yn}n∈Nis equi-
bounded. Moreover, from the definition of T, we remark that, for every y∈X, there exists the
derivative d
dt T y(t). In particular we get

d
dt T y(t)≤α+pδ+ 2β+β.
Then the sequence {T yn}n∈Nis equicontinuous. Thus, by Ascoli-Arzel`a theorem, it is possible to
extract a convergent subsequence, which implies that Tis a compact operator. □

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY CONDITION 7
We conclude this section proving the theorem which allows us to state that there is at least one
solution of the differential problem (1.1).
Theorem 3.7. Under the assumptions (3.5), the differential problem





¨y(t) = βy(y(t)), t ∈[0,1],
y(0) = α,
Rβ
0˙y2(t)dt =δβ > 0
has at least one solution.
Proof. Let us recall that, by Theorem 3.4, the space Xis bounded, closed and convex. Furthermore
the operator T:X→Xis continuous and compact. Then, by Schauder’s fixed point theorem, we
obtain immediately that there exists a y∈Xsuch that T(y) = y.□
Example 1. By Theorem 3.7 we get that the integrodifferential problem





¨y(t) = βy(y(t)), t ∈[0,1],
y(0) = 0,
R1
10
0˙y2(t)dt =1
50
(3.8)
has at least one solution.
4. Second problem
We prove now that the problem (1.2) has at least one solution. As in the previous case, we will
find an operator T1, a space of function X1⊂C([0,1],[0,1]) and some assumptions which will allow
us to apply the Schauder’s fixed point theorem and recover the existence of a solution. Also in this
case we have to show that X1is bounded, closed and convex and our operator T1is continuous
and compact.
4.1. The definition of the operator T1.We start studying heuristically the differential problem
(1.2).
Let us remark that, multiplying both sides our ODE by y(t) and applying standard techniques, we
deduce
˙y(τ)y(τ) = βyZτ
0
y(s)dsy(τ)⇒d
dτ Zt
0
y2(τ)
2dτ =βZt
0
d
dτ ZRs
0y(τ)dτ
0
y(s)ds.
This implies
y2(t) = y2(0) + 2βZRt
0y(τ)dτ
0
y(s)ds.
Hence, by using the boundary condition of the problem (1.2), we obtain
δβ =Zβ
0
y2(t)dt =βy2(0) + 2βZβ
0ZRt
0y(τ)dτ
0
y(s)dsdt.
Hence we may rewrite
y(0) = ±sδ−2Zβ
0ZRt
0y(τ)dτ
0
y(s)dsdt.
We define the space of functions X1and the operator T1which will be studied in order to find the
existence of a solution of (1.2).

8 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
Definition 4. Let us consider the space of continuous functions C([0,1],[0,1]). We define the
space of functions X1⊂C([0,1],[0,1]) as
X1=y∈C([0,1],[0,1]) 
δ−2Zβ
0ZRt
0y(τ)dτ
0
y(s)dsdt ≥0(4.1)
and the operator T1as
T1y(t) = sδ−2Zβ
0ZRt
0y(τ)dτ
0
y(s)dsdt +βZt
0yZτ
0
y(s)dsdτ.
Remark 5. Let us remark that the function F(t) = Rt
0y(τ)dτ is increasing in tsince y∈
C([0,1],[0,1]) and that, since |y(t)| ≤ 1 for all t∈[0,1], it holds true that 0 ≤F(t)≤t.
Remark 6. If δ−2β≥0 then it is immediate to get
T1y(t)≥pδ−2β≥0
for all y∈X1and t∈[0,1].
Remark 7. Since y∈C([0,1],[0,1]) we observe that, by Remark 5, we obtain
Zβ
0ZRt
0y(τ)dτ
0
y(s)dsdt ≤βt and βZt
0yZτ
0
y(s)dsdτ ≤β
and, if √δ+β≤1,then
T1y(t) = sδ−2Zβ
0ZRt
0y(τ)dτ
0
y(s)dsdt +βZt
0yZτ
0
y(s)dsdτ
≤√δ+βt ≤√δ+β≤1.
Remark 8. The conditions for which our operator is well defined are
(√δ+β≤1,
δ−2β≥0.(4.2)
The system (4.2) has at least one solution. In fact it may be solved, for instance, by the couple
(δ, β) = ( 1
4,1
8).
Hence we may conclude stating the following proposition.
Proposition 4.1. Under the conditions (4.2), the operator T1:X1→X1as in Definition 4 is
well defined.
Proof. It follows immediately from the Remarks 6, 7, 8. □
4.2. The properties of the space X1.In this section we will prove that X1is closed and convex,
as did previously with the space X.
Proposition 4.2. Under the assumptions (4.2), the space X1is closed.
Proof. Let us consider {yn}n∈Na Cauchy sequence in X1⊂C([0,1],[0,1]). Then, by the complete-
ness of this space, there will exist a y∞∈C([0,1],[0,1]) such that yn→y∞uniformly in [0,1]. We

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY CONDITION 9
get immediately that y∞∈C([0,1],[0,1]). Furthermore we remark that
ZRt
0yn(τ)
0
yn(s)ds −ZRt
0y∞(τ)dτ
0
y∞(s)ds
≤ZRt
0yn(s)ds
0
yn(τ)dτ −ZRt
0y∞(s)ds
0
yn(τ)dτ
+ZRt
0y∞(s)ds
0
yn(τ)dτ −ZRt
0y∞(s)ds
0
y∞(τ)dτ
≤Zt
0
yn(τ)dτ −Zt
0
y∞(τ)dτ
+ZRt
0y∞(s)ds
0
(yn(τ)−y∞(τ))dτ
(4.3)
which will converge to zero as n→ ∞. Hence
lim
n→∞ ZRt
0yn(τ)
0
yn(s)ds =ZRt
0y∞(τ)dτ
0
y∞(s)ds (4.4)
which concludes our proof. □
Proposition 4.3. Under the assumptions (4.2), the space X1is convex.
Proof. Let us consider y1, y2∈X1and λ, µ ∈[0,1] and λ+µ= 1. It is immediate to deduce that
λy1+µy2∈C([0,1],[0,1]). To prove that λy1+µy2∈X1is sufficient to remark that
δ−2Zβ
0ZRt
0λy1(τ)+µy2(τ)dτ
0
λy1(s) + µy2(s)dsdt
≥δ−2Zβ
0ZRt
0(λ+µ)dτ
0
(λ+µ)ds≥δ−2β≥0.
□
Hence we deduce the following theorem which sum up all our results about the space X1.
Theorem 4.4. Under the assumptions (4.2), the operator T1:X1→X1is well defined and the
space X1is bounded, closed and convex.
Proof. It is an immediate consequence of the Propositions 4.2 and 4.3. □
4.3. The continuity and compactness of T1.Our goal now is to check that our operator T1is
continuous and compact. This is a necessary step in order to apply the Schauder’s theorem.
Proposition 4.5. Under the assumptions (4.2), the operator T1:X1→X1as defined in Definition
4 is continuous.
Proof. We have obtained from (4.3) that, if {yn}n∈Nconverge uniformly to y∞, then
lim
n→∞ Zβ
0ZRt
0yn(τ)
0
yn(s)ds =Zβ
0ZRt
0y∞(τ)dτ
0
y∞(s)ds (4.5)
as n→ ∞. Hence, by standard theorems of convergence and continuity, it is straightforward to
remark that
lim
n→∞ sδ−2Zβ
0ZRt
0yn(τ)dτ
0
yn(s)dsdt =sδ−2Zβ
0ZRt
0y∞(τ)dτ
0
y∞(s)dsdt.
Let us recall that y∞∈C([0,1],[0,1]) is uniformly continuous in the set [0,1]. Then we have that
∀ε > 0∃δ > 0 such that ∀ξ, η ∈[0,1] |ξ−η|< δ ⇒ |y∞(ξ)−y∞(η)|< ε.

10 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
Thus we estimate

ynZτ
0
yn(s)ds−y∞Zτ
0
y∞(s)ds≤
ynZτ
0
yn(s)ds−y∞Zτ
0
yn(s)ds
+
y∞Zτ
0
yn(s)ds−y∞Zτ
0
y∞(s)ds
≤ ∥yn−y∞∥∞+ε
where ∥ ∥∞is the standard L∞norm. Thus it is immediate to get the continuity. □
Proposition 4.6. Under the assumptions (4.2), the operator T1:X1→X1as defined in Definition
4 is compact.
Proof. Let us consider the sequence {yn}n∈N⊂X1. Then it is straightforward to see that also the
sequence {T1yn}n∈Nwill be equibounded. Let us remark that
d
dt T1yn(t) = βynZt
0
yn(τ)dτyn(τ)
and then d
dt T1yn(t)≤β. Hence the sequence {T yn}n∈Nis equicontinuous. Hence, by the Ascoli-
Arzel`a theorem it is possible to extract a subsequence. Thus Tis compact. □
We conclude this section showing the existence of a solution of the problem (1.2).
Theorem 4.7. Under the assumptions (4.2), the integrodifferential problem
(˙y(t) = βyRt
0y(s)ds, t ∈[0,1],
Rβ
0y2(t)dt =δβ
with 0< β ≤1has at least one solution.
Proof. Let us recall that, by Theorem 4.4, the space X1is bounded, closed and convex. Furthermore
the operator T1:X1→X1is continuous and compact. Then, by Schauder’s theorem, we obtain
immediately that there exists a y∈X1such that T1(y) = y.□
Example 2. By Theorem 4.7 we get that the integrodifferential problem
(˙y(t) = 1
8yRt
0y(s)ds, t ∈[0,1],
R1
8
0y2(t)dt =1
32
has at least one solution.
5. Third problem
5.1. The definition of the operator TL.Finally let us study the problem (1.3).
Let us start remarking that, by multiplying both sides of the integrodifferential equation of (1.3)
by y(x, t), we get
∂
∂t y2(x, t)=2βy(x, t)yZt
0
y(x, s)ds, t
which implies, by integrating both sides w.r.t. the time variable between 0 and t
y2(x, t) = y2(x, 0) + 2βZt
0
y(x, s)yZs
0
y(x, τ )dτ, sds.

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY CONDITION 11
Integrating one more time w.r.t the time variable both the sides of the integrodifferential equation
(1.3) and by using the boundary condition of (1.3) we get
βg(x) = βy2(x, 0) + 2βZβ
0Zt
0
y(x, s)yZτ
0
y(x, s)ds, τ dτdt.
Then it is straightforward to deduce
y2(x, 0) = g(x)−2Zβ
0Zt
0
y(x, τ )yZτ
0
y(x, s)ds, τ dτdt.
We define now the space of functions and the operator that we need to prove the existence of a
solution of (1.3).
Definition 5. Let us consider the space of continuous functions C([0,1] ×[0,1],[0,1]). We define
the space XL⊂C([0,1] ×[0,1],[0,1]) as
XL=y∈C[0,1]×[0,1],[0,1]
g(x)−2Zβ
0Zt
0
y(x, τ )yZτ
0
y(x, s)ds, τ dτdt ≥ε0
with |y(x2, t)−y(x2, t)| ≤ L|x2−x1|for a L > 0 and for all x1, x2∈[0,1] and t∈[0,1]
with ε0>0. We define the functional TLas
TLy(x, t) = sg(x)−2Zβ
0Zt
0
y(x, τ )yZτ
0
y(x, s)ds, τ dτdt +βZt
0
yZτ
0
y(x, s)ds, τ dτ.
Remark 9. Let us remark that, for all y∈XL, we get
TLy(x, t)≥√ε0>0 and TLy(x, t)≤pg(x) + β.
If we assume as condition pg(x) + β≤1 for all x∈[0,1] then it is straightfoward to deduce that
0<√ε0≤TLy(x, t)≤1.
Remark 10. Let us remark that, for all y∈XL, there exists the derivative w.r.t. the time varible
of the operator TLy. In particular
0≤∂
∂t TLy(x, t)≤βyZt
0
y(x, s)ds, t≤β.
Remark 11. Since g: [0,1] →[0,1] is smooth, then we remark that there exists a constant Lg>0
such that |g(x2)−g(x1)| ≤ Lg|x2−x1|. Thus we may estimate the difference
TLy(x2, t)−TLy(x1, t) = sg(x2)−2Zβ
0Zt
0
y(x2, τ )yZτ
0
y(x2, s)ds, τ dτdt
−sg(x1)−2Zβ
0Zt
0
y(x1, τ )yZτ
0
y(x1, s)ds, τ dτdt
+βZt
0yZτ
0
y(x2, s)ds, τ −yZτ
0
y(x1, s)ds, τ dτ.
We remark, using the Lipschitz property of the function y, that
Zt
0yZτ
0
y(x2, s)ds, τ −yZτ
0
y(x1, s)ds, τ dτ≤LZt
0Zτ
0|y(x2, s)−y(x1, s)|dsdτ
≤L2
2|x2−x1|

12 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
and, by considering standard algebraic computations, the Remark 9 and the Lipschitz condition
of the functions gand y, we get
sg(x2)−2Zβ
0Zt
0
y(x2, τ )yZτ
0
y(x2, s)ds, τ dτdt
−sg(x1)−2Zβ
0Zt
0
y(x1, τ )yZτ
0
y(x1, s)ds, τ dτdt
≤1
2√ε0g(x2)−g(x1)
+Zβ
0Zt
0
y(x2, τ )yZτ
0
y(x2, s)ds, τ −y(x1, τ)yZτ
0
y(x1, s)ds, τ 
dτdt
≤1
2√ε0Lg|x2−x1|+ 2 Zβ
0Zt
0|y(x2, τ )−y(x1, τ)|dτ dt
+ 2 Zβ
0Zt
0
yZτ
0
y(x2, s)ds, τ −yZτ
0
y(x1, s)ds, τ 
dτdt
≤1
2√ε0Lg+Lβ2+L2β3
3|x2−x1|.
If we consider L > 0 such that 1
2√ε0Lg+Lβ2+L2β3
3+βL2
2≤Lthen we have that |TLy(x2, t)−
TLy(x1, t)| ≤ L|x2−x1|for all y∈XL.
Remark 12. Let us observe that, if we assume that g(x)−2β2≥ε0for all x∈[0,1], we observe
g(x)−2Zβ
0Zt
0
TLy(x, τ )TLyZτ
0
TLy(x, s)ds, τ dτdt ≥g(x)−2β2≥ε0.
Remark 13. Remarks 9, 10, 11, 12 allows us to define as set of assumptions











pg(x) + β≤1,
|g(x2)−g(x1)| ≤ Lg|x2−x1|,
1
2√ε0Lg+Lβ2+L2β3
3+βL2
2≤L,
g(x)−2β2≥ε0.
(5.1)
By the previous remarks we get immediately the following result.
Proposition 5.1. Under the assumptions (5.1), the operator TL:XL→XLas in Definition 5 is
well defined.
Proof. It is an immediate consequence of the Remarks 9, 10, 11, 12. □
5.2. The properties of the space XL.As did for the two previous problems, we prove now that
XLis closed and convex.
Proposition 5.2. Under the assumptions (5.1), the space XLis closed.
Proof. Let us suppose that {yn}n∈N⊂XLand, due to the completeness of C([0,1] ×[0,1],[0,1]),
yn→y∞uniformly with y∞∈C([0,1] ×[0,1],[0,1]). It is immediate to remark that, since it holds
true that |yn(x2, t)−yn(x1, t)| ≤ L|x2−x1|, then we may estimate y∞as |y∞(x2, t)−y∞(x1, t)| ≤

SELF REFERRED EQUATIONS WITH AN INTEGRAL BOUNDARY CONDITION 13
L|x2−x1|.
Furthermore let us remark
Zβ
0Zt
0
yn(x, t)ynZτ
0
yn(x, s)ds, τ dτdt −Zβ
0Zt
0
y∞(x, t)y∞Zτ
0
y∞(x, s)ds, τ dτdt
≤Zβ
0Zt
0|yn(x, τ )−y∞(x, τ)|dτ dt
+Zβ
0Zt
0
ynZτ
0
yn(x, s)ds, τ −y∞Zτ
0
y∞(x, s)ds, τ 
dτdt
≤β2
2∥yn−y∞∥+Zβ
0Zt
0
ynZτ
0
yn(x, s)ds, τ −ynZτ
0
y∞(x, s)ds, τ 
+ynZτ
0
y∞(x, s)ds, τ −y∞Zτ
0
y∞(x, s)ds, τ 
dτdt
≤
β2
2+Lβ3
3+β2
2∥yn−y∞∥∞.(5.2)
where ∥ ∥∞is the standard L∞norm. Finally, remarking that
ε0≤lim
n→∞ g(x)−2Zβ
0Zt
0
yn(x, τ )ynZτ
0
yn(x, s)ds, τ dτdt
then it is immediate to conclude that XLis closed. □
Proposition 5.3. Under the assumptions (5.1), the space XLis convex.
Proof. Let us consider y1, y2∈XLand λ, µ ∈[0,1] and λ+µ= 1. It is easy to observe that
λy1+µy2∈C([0,1] ×[0,1],[0,1]). To prove that λy1+µy2∈XLis sufficient to remark that
g(x)−2Zβ
0Zt
0
(λy1+µy2)(x, τ )(λy1+µy2)Zτ
0
(λy1+µy2)(x, s)ds, τ dτdt
≥g(x)−2β2≥ε0.
□
Thus we may state the following theorem.
Theorem 5.4. Under the assumptions (5.1), the operator TL:XL→XLas defined in Definition
5 is well defined and XLis a bounded, closed and convex space.
Proof. It is an immediate consequence of the Propositions 5.1, 5.2 and 5.3. □
5.3. The continuity and compactness of TL.We show now that the operator TLis continuous
and compact.
Proposition 5.5. Under the assumptions (5.1), the operator TL:XL→XLas defined in Defini-
tion 5 is continuous.
Proof. From Proposition 5.2 we get
lim
n→∞ Zt
0
ynZτ
0
yn(x, s)ds, τ dτ =Zt
0
y∞Zτ
0
y∞(x, s)ds, τ dτ.
From this estimate it is immediate to deduce the continuity of our operator TL.□
Proposition 5.6. Under the assumptions (5.1), the operator TL:XL→XLdefined as in the
Definition 5 is compact.

14 GIACOMO CHIRIATTI, MATTEO FASIELLO, RAFFAELE GRANDE, AND EDUARDO PASCALI
Proof. Let us consider the sequence {yn}n∈N⊂XL. It is straightforward to remark that {TLyn}n∈N
is equibounded by Remark 9. Let us observe that also that {TLyn}n∈Nis equicontinuous, since,
by Remarks 10 and 11, it holds true
|TLyn(x2, t)−TLyn(x1, t)| ≤ L|x2−x1|and 0 ≤∂
∂t TLyn(x, t)≤β(5.3)
for all n∈N. Hence, by Ascoli-Arzel`a theorem, it is possible to extract a convergent subsequence
of our seqeunce. Thus the operator TLis compact. □
We conclude stating and proving that our differential problem (1.3) has at least one solution.
Theorem 5.7. Under the assumptions (5.1), the integrodifferential problem
(∂
∂t y(x, t) = βyRt
0y(x, s)ds, t,(x, t)∈[0,1] ×[0,1],
Rβ
0y2(x, t) = βg(x)
with 0< β ≤1has at least one solution.
Proof. Let us recall that, by Theorem 5.4, the space XLis bounded, closed and convex. Further-
more the operator TL:XL→XLis continuous and compact. Then, by Schauder’s theorem, we
obtain immediately that there exists a y∈XLsuch that TL(y) = y.□
6. Open problems
We propose some open problems which are a further generalization of (1.3).
(1)
(∂
∂t y(x, t) = βyRx
0y(ξ, t)dξ, t, x ∈[0,1], t ∈[0,1]
Rβ
0y2(x, t)dt =βg(x)
where 0 < β ≤1 and g: [0,1] →[0,1] smooth function
(2)





∂2
∂t2y(x, t) = βyRx
0y(ξ, s)dξ, t, x ∈[0,1], t ∈[0,1],
y(x, 0) = f(x),
Rβ
0∂
∂s y(x, s)2ds =βg(x)
where f: [0,1] →[0,1] and g: [0,1] →[0,1] suitable smooth functions and 0 < β ≤1.
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* Universit`
a del Salento,
Ex Collegio Fiorini,
Via per Arnesano,
Lecce,
ITALY
Email address:giacomo.chiriatti@studenti.unisalento.it
** Scuola Superiore ISUFI,
Complesso Ecoteckne,
Lecce,
ITALY
Email address:matteo.fasiello1@studenti.unisalento.it
*** The Czech Academy of Sciences,
Institute of Information Theory and Automation,
Prague,
Czech Republic,
Email address:grande@utia.cas.cz
**** Universit`
a del Salento,
Ex Collegio Fiorini,
Via per Arnesano,
Lecce,
ITALY
Email address:eduardo.pascali@unisalento.it