FILIPPOV LEMMA FOR BEAM DIFFERENTIAL INCLUSIONS

GRZEGORZ BARTUZEL AND ANDRZEJ FRYSZKOWSKI

Abstract. In the paper we deal with beam differential inclusion

Dy = y′′′′ + 4k^{4}y ∈ F (x, y)

defined on [−1, 1] with boundary conditions

y (−1) = y (1) = y′ (−1) = y′ (1) = 0. We show that if the right-hand side is Lipschitz continuous in y then a version of the Filippov Lemma holds. Our considerations are based on the integral representation of the inverse operator to D.

1. Introduction

Differential inclusions of the form

(1.1)  Dy ∈ F (x, y) ,

where D is a given differential operator, are natural generalizations of differential equations as well as ordinary differential inclusions. The theory of properties of such mathematical problems has been developing since the early seventies and has paid attention of many mathematicians. Most of the effort was concentrated for differential inclusions of the first order, while the inclusions of higher order are less examined. However lately we observe a remarkable amount of interest in this field and in the field of partial differential inclusions.

For ordinary differential inclusions with Lipschitz continuous right-hand side the main effect was discovered by Filippov and Ważewski and shortly speaking it says that the solution set of

y′ ∈ F (x, y)

is dense in the solution set of the so called relaxed differential inclusions

y′ ∈ clcoF (x, y) ,

where clcoA stands for the closed convex hull of a set A. The Filippov-Ważewski theorem has been generalized for differential inclusions of many types including higher order situations and partial differential operators. In particular the authors have examined differential inclusions of the Sturm-Liouville type, Schrödinger type and BiLaplace type on a circle.

Achieving a Filippov-Ważewski type results is usually based on a version of Filippov Lemma. These type of lemmas give a construction of a solution of (1.1) which is sufficiently close to a given function.

Date: November 01, 2005.

1991 Mathematics Subject Classification. Primary 26A24, 28A15; Secondary 46G05, 39A05, 28A05;

In most of the proofs of Filippov Lemma type results for (1.1) there is exploited an integral form of the right inverse to the opertor D. This means that there exists an integral kernel G (x, t) defined on [−1.1]×[−1, 1] such that for every f the unigue solution y of the equation

Dy = f,

can be performed in the form y (x) = Rf = ∫_{−1}^{1} G (x, t) f (t) dt. Having known that

the integral kernel is nonnegative we can construct a required solution by the iteration procedure and then passing to the limit.

Having known the operator R we can attack the problem (1.1) by the fixed points results. Namely, we have an equivalence that y = Rf is a solution of (1.1) if and only if

f ∈ F (x, Rf) ,

i.e. f is a fixed point of the the substitution operator F (x, Rf) . This method is very efficient for a differential operator D with the right inverse R in the integral

form, especially when (Rf) (x) = ∫_{−1}^{1} G (x, t) f (t) dt with the integral kernel (Green

function) G (x, t) .

In the present paper we want to examine differential inclusions with the beam differential operator

(1.2)  Dy = y′′′′ + 4k^{4}y ∈ F (x, y)

defined on [−1, 1] with boundary conditions

(1.3)  y (−1) = y′ (−1) = y (1) = y′ (1) = 0,

which describes the clamped beam.

By a solution of (1.2) with boundary conditions (1.3) we mean a function

y ∈ V = W^{4,1} [−1, 1] ∩ H_{0}^{2} [−1, 1]

satisfying (1.2) and (1.3). In the study of the beam equation we use slightly more general form

Dy = y′′′′ − (α^{2} + β^{2}) y′′ + α^{2}β^{2}y = f,

α, β ∈ ℭ, with boundary conditions (1.3). We shall show that for any given f ∈ L^{1} [−1, 1] the equation

Dy = y′′′′ + 4k^{4}y = f

has the unique solution y = Rf in V, R : L^{1} [−1, 1] → V is the integral operator with the nonnegative Green function.

Despite of the fact that the beam equations are widely used in physical and engineering problems and described in many publications we did not have seen the Green function in such form. So we we have decided to present our considerations, however the methods used are quite standard.

2. A beam equation

In this section we consider, for every given f ∈ L^{1} [−1, 1] and α, β ∈ ℂ, the differential equation

(2.1)  y′′′′ − (α^{2} + β^{2}) y′′ + α^{2}β^{2}y = f,

α^{2} ≠ β^{2}, with boundary conditions on the interval [−1, 1]

(2.2)  y (−1) = y′ (−1) = 0,  y (1) = y′ (1) = 0.

By a solution we mean a function y ∈ H^{4,1} [−1, 1] satisfying (2.1) and (2.2).

Observe first that solution of (2.1) can be build up with functions z, w ∈ H^{2,1} [−1, 1] satisfying

z′′ − α^{2}z = f and w′′ − β^{2}w = f.

Indeed. Taking y = z − w we have

y′′ = z′′ − w′′ = α^{2}z − β^{2}w ∈ H^{2,1} [−1, 1]

and this implies that y ∈ H^{4,1} [−1, 1] . Hence

y′′′′ = α^{2}z′′ − β^{2}w′′ = α^{2} (α^{2}z + f) − β^{2} (β^{2}w + f )
= α^{4}z − β^{4}w + (α^{2} − β^{2}) f

and

y′′′′ − (α^{2} + β^{2}) y′′ + α^{2}β^{2}y =
= α^{4}z − β^{4}w + (α^{2} − β^{2}) f − (α^{2} + β^{2}) (α^{2}z − β^{2}w) + α^{2}β^{2} (z − w) =
= (α^{2} − β^{2}) f

Therefore

y = \frac{z − w}{α^{2} − β^{2}}

is a solution of (2.1). Moreover the boundary conditions (2.2) can be translated as

z (1) = w (1) , z (−1) = w (−1) ,  ((1d))
(2.3)  z′ (1) = w′ (1) = A and z′ (−1) = w′ (−1) = B.

We start with the equation

(2.4)  z′′ − α^{2}z = f

with boundary conditions

(2.5)  z′ (−1) = z′ (1) = 0.

Let z (x, α) = z_{1} (x, α) = \frac{i cosh(α−αx)}{\sqrt{α sinh(2α)}} be a solution of z′′ −α^{2}z = 0 with conditions

z (1, α) = \frac{i}{\sqrt{α sinh (2α)}}, z′ (1, α) = 0.

Then z_{2} (x, α) = z (−x, α) = \frac{i cosh(α+αx)}{\sqrt{α sinh(2α)}} solves z′′ − α^{2}z = 0 with conditions

z (−1, α) = \frac{i}{\sqrt{α sinh (2α)}}, z′ (−1, α) = 0

and by the general Sturm-Liouville theory the solution of (2.4) with boundary conditions (2.5) is an integral operator

z (x) = ∫_{−1}^{1} H (x, t, α) f (t) dt,

4 GRZEGORZ BARTUZEL AND ANDRZEJ FRYSZKOWSKI

with the Green function

H (x, t, α) = \{ \begin{matrix}z (x, α) z (−t, α)  for  − 1 ≤ t ≤ x ≤ 1 \\ z (t, α) z (−x, α)  for  − 1 ≤ x ≤ t ≤ 1 \end{matrix}
= −\frac{1}{α sinh (2α)} \{ \begin{matrix}cosh (α + tα) cosh (α − xα)  for  − 1 ≤ t ≤ x ≤ 1 \\ cosh (α + αx) cosh (α − αt)  for  − 1 ≤ x ≤ t ≤ 1 \end{matrix} .

Remark 1. Introducing for x ∈ [−1, 1] a noncommutative convolution

(f ∗ g) (x) = ∫_{−1}^{x} f (t) g (x − t) dt

we have

(2.6) (f ∗ g)′ (x) = ∫_{−1}^{x} f (t) g′ (x − t) dt + f (x) g (0) = (f ∗ g′) (x) + f (x) g (0)

Observe that using (2.6) we get that for z (x) = \frac{1}{α} (f ∗ sinh (αx)) is

z′ (x) = f ∗ cosh (αx)
and z′′ (x) = α (f ∗ sinh (αx)) + f (x) .

So the function z (x) = \frac{1}{α} (f ∗ sinh (αx)) is a PS of z′′ − α^{2}z = f. The GS is of the form

z (x) = \frac{1}{α} (f ∗ sinh (αx)) + C cosh (γ + αx)

with

z′ (x) = (f ∗ cosh (αx)) + Cα sinh (γ + αx) .

Using the boundary conditions we get

Cα sinh (γ − α) = 0,
Cα sinh (γ + α) = − (f ∗ cosh (αx)) (1) .

Hence

γ = α,
C = −(f ∗ cosh (αx)) (1) ; α sinh (2α).

FILIPPOV LEMMA FOR BEAM 5

and this means that

z (x) = \frac{(f ∗ sinh (αx)) sinh (2α) − (f ∗ cosh (αx)) (1) cosh (α + αx)}{α sinh (2α)}

=

∫_{−1}^{x} f (t) sinh (αx − αt) sinh (2α) dt − ∫_{−1}^{1} f (t) cosh (α + αx) cosh (α − αt) dt

α sinh (2α)

= \frac{∫_{−1}^{x} f (t) [sinh (αx − αt) sinh (2α) − cosh (α + αx) cosh (α − αt)] dt − ∫_{x}^{1} z (t) cosh (α + αx) cosh (α − αt) dt}{α sinh (2α)}

= \frac{− ∫_{−1}^{x} f (t) cosh (α + tα) cosh (α − xα) dt − ∫_{x}^{1} f (t) cosh (α + αx) cosh (α − αt) dt}{α sinh (2α)} = ∫_{−1}^{1} H (x, t, α) f (t) dt,

where

H (x, t, α) = −\frac{1}{α sinh (2α)} \{ \begin{matrix}cosh (α + tα) cosh (α − xα) for − 1 ≤ t ≤ x ≤ 1 \\ cosh (α + αx) cosh (α − αt) for − 1 ≤ x ≤ t ≤ 1 \end{matrix}

For the equation

z′′ − α^{2}z = f

with boundary conditions

z′ (1) = A, z′ (−1) = B

we obtain the solution

z (x) = ∫_{−1}^{1} H (x, t, α) f (t) dt + \frac{iz (x, α) B}{\sqrt{α sinh (2α)}} − \frac{iz (−x, α) A}{\sqrt{α sinh (2α)}}
= ∫_{−1}^{1} H (x, t, α) f (t) dt − AH (x, 1, α) + BH (x, −1, α) .

Analogously the problem

w′′ − β^{2}w = f

with boundary conditions

w′ (1) = A, w′ (−1) = B

possess the solution

w (x) = ∫_{−1}^{1} H (x, t, β) f (t) dt − AH (x, 1, β) + BH (−1, x, β)

By the introductory remarks we obtain a family of solutions of (2.1) with the boundary conditions

y′ (−1) = y′ (1) = 0

given by

y (x) = \frac{z − w}{α^{2} − β^{2}} = ∫_{−1}^{1} \frac{W (x, t, α, β)}{α^{2} − β^{2}}f (t) dt − \frac{AW (x, 1, α, β) − BW (x, −1, α, β)}{α^{2} − β^{2}}
= ∫_{−1}^{1} \frac{W (x, t, α, β)}{α^{2} − β^{2}}f (t) dt − \frac{1}{α^{2} − β^{2}} [ \begin{matrix}W (x, 1, α, β) \\ −W (x, −1, α, β) \end{matrix} ]^{T} [ \begin{matrix}A \\ B \end{matrix} ] ,

where

W (x, t, α, β) = H (x, t, α) − H (x, t, β) .

Now we need to find A and B such that

y (−1) = y (1) = 0.

This leads to the relations

AW (1, 1, α, β) − BW (−1, 1, α, β) = ∫_{−1}^{1} W (1, t, α, β) f (t) dt
AW (−1, 1, α, β) − BW (1, 1, α, β) = ∫_{−1}^{1} W (−1, t, α, β) f (t) dt

In matrix form

[ \begin{matrix}W (1, 1, α, β) & −W (1, −1, α, β) \\ W (1, −1, α, β) & −W (1, 1, α, β) \end{matrix} ] [ \begin{matrix}A \\ B \end{matrix} ]
= ∫_{−1}^{1} [ \begin{matrix}W (1, t, α, β) \\ W (−1, t, α, β) \end{matrix} ] f (t) dt.

This system possess the solution

[ \begin{matrix}A \\ B \end{matrix} ] = −\frac{∫_{−1}^{1} [ \begin{matrix}W (1, 1, α, β) & −W (1, −1, α, β) \\ W (1, −1, α, β) & −W (1, 1, α, β) \end{matrix} ] [ \begin{matrix}W (1, t, α, β) \\ W (−1, t, α, β) \end{matrix} ] f (t) dt}{det [ \begin{matrix}W (1, 1, α, β) & −W (1, −1, α, β) \\ W (1, −1, α, β) & −W (1, 1, α, β) \end{matrix} ]}.

Finally

y (x) = Rf = ∫_{−1}^{1} \frac{W (x, t, α, β)}{α^{2} − β^{2}}f (t) dt − [ \begin{matrix}W (x, 1, α, β) \\ −W (x, −1, α, β) \end{matrix} ]^{T} [ \begin{matrix}A \\ B \end{matrix} ]

= ∫_{−1}^{1} \frac{W (x, t, α, β)}{α^{2} − β^{2}}f (t) dt

+ ∫_{−1}^{1} \begin{matrix}[ \begin{matrix}W (x, 1, α, β) \\ −W (x, −1, α, β) \end{matrix} ] & T & [ \begin{matrix}W (1, 1, α, β) & −W (−1, 1, α, β) \\ W (−1, 1, α, β) & −W (1, 1, α, β) \end{matrix} ] & [ \begin{matrix}W (1, t, α, β) \\ W (−1, t, α, β) \end{matrix} ] \\ (α^{2} − β^{2}) & det & [ \begin{matrix}W (1, 1, α, β) & −W (−1, 1, α, β) \\ W (−1, 1, α, β) & −W (1, 1, α, β) \end{matrix} ] \end{matrix}f (t) dt

= ∫_{−1}^{1} G (x, t, α, β) f (t) dt,

where the Green function is given by

G (x, t, α, β)

= \begin{matrix}( & \begin{matrix}W (x, t, α, β) det [ \begin{matrix}W (1, 1, α, β) & −W (−1, 1, α, β) \\ W (−1, 1, α, β) & −W (1, 1, α, β) \end{matrix} ] \\ +\frac{[ \begin{matrix}W (x, 1, α, β) \\ −W (x, −1, α, β) \end{matrix} ]^{T} [ \begin{matrix}W (1, 1, α, β) & −W (−1, 1, α, β) \\ W (−1, 1, α, β) & −W (1, 1, α, β) \end{matrix} ][ \begin{matrix}W (1, t, α, β) \\ W (−1, t, α, β) \end{matrix} ]}{1} \end{matrix} & ) \\ (α^{2} − β^{2}) & det & [ \begin{matrix}W (1, 1, α, β) & −W (−1, 1, α, β) \\ W (−1, 1, α, β) & −W (1, 1, α, β) \end{matrix} ] \end{matrix},

One can check that for

β = \overline{α}

is

\overline{G (x, t, α, β)} = G (x, t, \overline{β}, \overline{α}) = G (x, t, α, β) ,

so in this case G (x, t, α, β) is real.

Now we pass to the equation

(2.7)  y′′′′ + 4k^{4}y = y′′′′ − α^{4}y = f,

i.e in 2.4 the case α = k − ki and β = \overline{α} = αi . By the previous remarks, we know that in this case the Green function

G (x, t, k) = G (x, t, k − ki, k + ki)

for 2.7 with boundary conditions 2.2 is real. Moreover

Lemma 1. The Green function G (x, t, k) for (2.7) with boundary conditions (2.2) is nonnegative for sufficiently small k > 0.

Proof. By the standard calculation we can show that the uniform limit

G_{0} (x, t) = lim_{α→0}G (x, t, α)

= \{ \begin{matrix}\frac{(1+t)^{2}(1−x)^{2}((x−t)+(1−t)(1+x))}{24} for − 1 ≤ t ≤ x ≤ 1 \\ \frac{(1−t)^{2}(1+x)^{2}((t−x)+(1+t)(1−x))}{24} for − 1 ≤ x ≤ t ≤ 1 \end{matrix}

= \{ \begin{matrix}\frac{(1+t)^{2}(1−x)^{2}(2x−2t−tx+1)}{24} for − 1 ≤ t ≤ x ≤ 1 \\ \frac{(1−t)^{2}(1+x)^{2}(2t−2x−tx+1)}{24} for − 1 ≤ x ≤ t ≤ 1 \end{matrix} ≥ 0,

(picture for −1 ≤ t ≤ x ≤ 1), what is the Green function for

y′′′′ = f

with boundary conditions

y (−1) = y′ (−1) = y (1) = y′ (1) = 0.

Since G_{0} (x, t) is positive on each interval [−1 + ε, 1 − ε] therefore for sufficiently small k > 0 the Green G (x, t, α) function is nonnegative as well. □

Corollary 1. Let y = Rf is the solution (2.7) with boundary conditions (2.2). Then, for sufficiently small k > 0,

|Rf| ≤ R |f|

3. A Filippov Lemma

The nonnegativity of the Green function for sufficiently small k > 0 is the main tool in considerations concerning an existence of solution for the problem

(3.1)  y′′′′ + 4k^{4}y ∈ F (x, y)

with boundary conditions

(3.2)  y (−1) = y′ (−1) = y (1) = y′ (1) = 0.

We shall pose the following assumptions on F : [−1, 1] × R → cl (R) , where cl (R) stands for the family of all nonempty compact subsets of R :

Condition 1. For every y ∈ R the multifunction F (·, y) is Lebesque measurable in x;

Condition 2. There exists an m ∈ L^{1} [−1, 1] such for a.e. x ∈ R the multifunction F (x, ·) is Lipschitz continuous in y with a constant m (x) , i.e. for every y_{1}, y_{2} ∈ R the inequality

d_{H} (F (x, y_{1}) , F (x, y_{2})) ≤ m (x) |y_{1} − y_{2}|

holds, where d_{H} (K, L) stands for the Hausdorff distance between sets K, L ∈ c (R) ;

Condition 3. The multivalued mapping x → F (x, 0) is integrably bounded by an a ∈ L^{1} [−1, 1], ie.

sup {|y| : y ∈ F (x, 0)} ≤ a (x) a.e. in [−1, 1] ;

Condition 4. Differential operator Dy = y′′′′ + 4k^{4}y has nonnegative Green function.

The main result of the paper is the following

Theorem 1. Assume that F : [−1, 1] × R → cl (R) satisfies Conditions 1,2,3 and 4.. Moreover assume that the Lipschitz constant is small enough, i.e. there exists function b ∈ L^{∞} [−1, 1] such the inequality involving the integral kernels

∑_{n=0}^{∞} ∫_{−1}^{1} p (t) G^{(n)} (x, t) dt ≤ b (x) a.e. in [−1, 1]

holds. Let u_{0} ∈ L^{1} [−1, 1] be an arbitrary function such that we have

d_{H} (u_{0} (x) , F (x, (Ru_{0}) (x))) ≤ p (x) a.e. in [−1, 1]

with p ∈ L^{1} [−1, 1] . Then for sufficiently small k > 0 there exists a solution y = Ru ∈ V of (3.1) and (3.2) such that

|u (x) − u_{0} (x)| ≤ m (x) b (x) a.e. in [−1, 1]

and

|(Ru − Ru_{0}) (x)| ≤ b (x) a.e. in [−1, 1]

Proof. We begin with an observation that for every φ ∈ C [−1, 1] the multivalued mapping x → F (x, φ (x)) is measurable with compact values and integrably bounded by a + m |φ| , i.e.

(3.3) sup {|y| : y ∈ F (x, φ (x))} ≤ a (x) + m (x) |φ (x)| a.e. in [−1, 1] .

10 GRZEGORZ BARTUZEL AND ANDRZEJ FRYSZKOWSKI

The measurability of x → F (x, φ (x)) is obvious. Therefore the Castaing Theorem yields the existence of measurable functions {f_{n}}_{n∈N} such that

cl {f_{n} (x) : n ∈ N} = F (x, φ (x)) a.e. in [−1, 1] .

Observe that for every n ∈ N we have |f_{n} (x)| ≤ d_{H} (f (x) , F (x, 0))+d_{H} (0, F (x, 0)) ≤ m (x) |φ (x)|+a (x) a.e. in [−1, 1] and hence

sup {|y| : y ∈ F (x, φ (x))} ≤ m (x) |φ (x)| + a (x) a.e. in [−1, 1] , what shows (3.3).

Take any u ∈ L^{1} [−1, 1] and denote by

K (u) = {v ∈ L^{1} [−1, 1] : v (x) ∈ F (x, (Ru) (x)) a.e. in [−1, 1]} . Since Ru ∈ C [−1, 1] then by (3.3) each K (u) is nonempty. Moreover for every u, v ∈ L^{1} [−1, 1] and any f ∈ K (u) there is g ∈ K (v) such that

|f (x) − g (x)| ≤ m (x) |(Ru) (x) − (Rv) (x)| a.e. in [−1, 1]

what in virtue of nonnegativity of R means that

(3.4)  |f (x) − g (x)| ≤ m (x) (R |u − v|) (x) a.e. in [−1, 1] .

In what follows we shall adopt the Filippov technique with necessary changes. Starting with u_{0} we choose u_{1} ∈ K (u_{0}) such that |u_{1} (x) − u_{0} (x)| = d_{H} (u_{0} (x) , F (x, (Ru_{0}) (x))) ≤ p (x) a.e. in [−1, 1] .

The Corollary 1 yields the inequality

(R |u_{1} − u_{0}|) (x) ≤ (Rp) (x) a.e. in [−1, 1] .

Define

y_{0} (x) = ∫_{−1}^{1} G (x, t) p (t) dt = (Rp) (x)

and observe that

R |u_{1} − u_{0}| (x) ≤ (Rp) (x) = ∫_{−1}^{1} p (t) G (x, t) dt = y_{0} (x) a.e. in [−1, 1]

By (3.4) there exists u_{2} ∈ K (u_{1}) such that |u_{2} (x) − u_{1} (x)| ≤ m (x) (R |u_{1} − u_{0}|) (x) ≤ m (x) (Rp) (x) = m (x) y_{0} (x) a.e. in [−1, 1] . Aplying Corollary 1 again we have the inequality

(R |u_{2} − u_{1}|) (x) ≤ (R [my_{0}]) (x) a.e. in [−1, 1] .

Take y_{1} (x) = ∫_{−1}^{1} m (t) G_{0} (x, t) y_{0} (t) dt = (R [my_{0}]) (x) and notice that

(R |u_{2} − u_{1}|) (x) ≤ R ([m (Rp)]) (x) = y_{1} (x) .

Having already constructed y_{n−1} and u_{n} ∈ K (u_{n−1}) such that |u_{n} (x) − u_{n−1} (x)| ≤ m (x) (R |u_{n−1} − u_{n−2}|) (x) ≤ m (x) y_{n−1} (x) a.e. in [−1, 1]

FILIPPOV LEMMA FOR BEAM 11

we define y_{n} (x) = ∫_{−1}^{1} m (t) G_{0} (x, t) y_{n−1} (t) dt = (R [my_{n−1}]) (x) and notice that

(R |u_{n} − u_{n−1}|) (x) ≤ R [m (Ry_{n−1})] (x) = y_{n} (x) a.e. in [−1, 1] Then, by the Lipschitzness, we have

d_{H} (u_{n} (x) , F (x, (Ru_{n}) (x))) ≤ m (x) (R |u_{n} − u_{n−1}|) (x) = m (x) y_{n} (x) . So one can find u_{n+1} ∈ K (u_{n}) such that |u_{n+1} (x) − u_{n} (x)| ≤ m (x) (R |u_{n} − u_{n−1}|) (x) = m (x) y_{n} (x) a.e. in [−1, 1] . Now we can define y_{n+1} (x) = ∫_{−1}^{1} m (t) G (x, t) y_{n} (t) dt = (R [my_{n}]) (x), what ends the induction step. So we have constructed elements u_{n+1} ∈ K (u_{n}) such that

|u_{n+1} (x) − u_{n} (x)| ≤ m (x) y_{n} (x) = m (x) ∫_{−1}^{1} p (t) G^{(n)} (x, t) dt a.e. in [−1, 1]

and hence

|(Ru_{n+1}) (x) − (Ru_{n}) (x)| (x) ≤ y_{n+1} (x) = ∫_{−1}^{1} p (t) G^{(n+1)} (x, t) a.e. in [−1, 1] .

Therefore we have

|u_{n+k} (x) − u_{n} (x)| ≤ m (x) ∫_{−1}^{1} (p (t) { ∑_{i=n}^{n+k−1} G^{(i)} (x, t)} dt) ≤ m (x) b (x) a.e. in [−1, 1]

and

|(Ru_{n+k}) (x) − (Ru_{n}) (x)| ≤ ∫_{−1}^{1} (p (t) { ∑_{i=n+1}^{n+k} G^{(i)} (x, t)} dt) ≤ b (x) a.e. in [−1, 1] .

So both sequences {u_{n} (·)} and (Ru_{n}) (·) are convergent a.e. and in L^{1} [−1, 1] . Denote

lim u_{n} = u

and therefore

lim Ru_{n} = Ru.

Taking in the previous inequality for n = 0 and passing to the limit with k we obtain relations

|u (x) − u_{0} (x)| ≤ m (x) b (x) a.e. in [−1, 1]

and

|(Ru) (x) − (Ru_{0}) (x)| ≤ b (x) a.e. in [−1, 1] .

Moreover, since u_{n+1} (x) ∈ F (x, [Ru_{n}] (x)) and F is Lipschitz continuous in y, we have

u (x) ∈ F (x, [Ru] (x)) a.e. in [−1, 1] .

Checking that just costructed function u satisfies the required conditions is straightforward. □

Acknowledgement 1. Some parts of the paper were prepared during the visit in the Technische Hochschule in Darmstadt and partially supported by that institution. The authors are grateful to Professor Hans-Dieter Alber for his hospitality and discusions on the subject of the paper.

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Faculty of Mathematics and Information Science, Warsaw University of Technology, Plac Politechniki 1, 00-661 Warsaw, Poland

E-mail address: grgb@alpha.mini.pw.edu.pl
URL: http://www.tku.edu.tw

Faculty of Mathematics and Information Science, Warsaw University of Technology, Plac Politechniki 1, 00-661 Warsaw, Poland

E-mail address: fryszko@alpha.mini.pw.edu.pl
URL: http://www.alpha.mini.pw.edu.pl