Directory UMM :Journals:Journal_of_mathematics:EJQTDE:

Electronic Journal of Qualitative Theory of Differential Equations
2011, No. 6, 1-18; http://www.math.u-szeged.hu/ejqtde/

Existence of Minimal and Maximal Solutions for a
Quasilinear Elliptic Equation With Integral Boundary
Conditions
Mohammed Derhab
Department of Mathematics
Faculty of Sciences
University Abou-Bekr Belkaid Tlemcen
B.P.119, Tlemcen
13000, Algeria
e-mail: derhab@yahoo.fr

Abstract: This work is concerned with the construction of the minimal
and maximal solutions for a quasilinear elliptic equation with integral boundary
conditions, where the nonlinearity is a continuous function depending on the
first derivative of the unknown function. We also give an example to illustrate
our results.

Keywords: Integral boundary conditions; upper and lower solutions; monotone iterative technique; p−Laplacian; Nagumo–Wintner condition

AMS Classification: 34B10, 34B15

EJQTDE, 2011 No. 6, p. 1

1

Introduction

This work is concerned with the construction of the minimal and the maximal
solutions of the following nonlinear boundary value problem

′
− (ϕp (u′ )) = f (x, u, u′ ) , x ∈ (0, 1) ,




Z1




′

 u (0) − a0 u (0) = g1 (x) u (x) dx,
(1)
0

1

Z





u (1) + a1 u′ (1) = g2 (x) u (x) dx,


0

p−2

where ϕp (y) = |y|
y, p > 1, f : [0, 1] × R2 → R and gi : [0, 1] → R+ are a
continuous functions (i = 1, 2 ) and a0 and a1 are two positive real numbers.
Problems with integral boundary conditions arise naturally in thermal conduction problems [13], semiconductor problems [21], hydrodynamics problems
[15], underground water flow [18] and medical sciences [see [14] and [23]].
It is well know that the method of upper and lower solutions coupled with
monotone iterative technique has been used to prove existence of solutions of
nonlinear boundary value problems by various authors ( see [3], [6], [9], [16] and
[17]).
The purpose of this work is to show that it can be applied successfully to
problems with integral boundary conditions of type (1). Our results improve
and generalize those obtained in [6], [7] and [9].
The plan of this paper is as follows: In section 2, we give some preliminary
results that will be used throughout the paper. In section 3, we state and prove
our main result. Finally in section 4, we give an example to illustrate our results.

2

Preliminary results

In this section, we give some preliminary results that will be used in the remainder of this paper.
We consider the following problem

 − (ϕp (u′ ))′ = F (x, u′ ) − b
h (x, u) , x ∈ (0, 1) ,
(2)
u (0) − a2 u′ (0) = a,

u (1) + a3 u′ (1) = b,

where F : [0, 1] × R → R is a continuous function, b
h : [0, 1] × R → R is a
continuous function and strictly increasing in its second variable, a2 and a3 are
a positive real numbers, a ∈ R and b ∈ R.
Lemma 1 (Weak comparison principle).
Let u1 , u2 are such that ui ∈ C 1 ([0, 1]), ϕp (u′i ) ∈ C 1 (0, 1), i = 1, 2, and

EJQTDE, 2011 No. 6, p. 2


′
 − (ϕp (u′1 ))′ − F (x, u′1 ) + b
h (x, u1 ) ≤ − (ϕp (u′2 )) − F (x, u′2 ) + b
h (x, u2 ) , x ∈ (0, 1) ,
u1 (0) − a2 u′1 (0) ≤ u2 (0) − a2 u′2 (0) ,

u1 (1) + a3 u′1 (1) ≤ u2 (1) + a3 u′2 (1) ,
then u1 (x) ≤ u2 (x), for all x ∈ [0, 1] .

Proof. Assume that there exists x0 ∈ [0, 1] such that
u2 (x0 ) − u1 (x0 ) = min (u2 (x) − u1 (x)) < 0.
0≤x0 ≤1

Then since (u2 − u1 ) ∈ C 1 ([0, 1]), we have (u2 − u1 )′ (x0 ) = 0.
If x0 = 0, we obtain the contradiction
 ′

′

0 > u2 (0) − u1 (0) ≥ a2 u2 (0) − u1 (0) = 0.
A similar argument holds if x0 = 1.
If x0 ∈ (0, 1), we have
ϕp (u′2 (x0 )) = ϕp (u′1 (x0 )) ,
then since ϕp is strictly increasing, we obtain that
′

−ϕp (u′2 ) (x) + ϕp (u′1 ) (x)
≤ 0.
x→x0
x − x0

′

− (ϕp (u′2 )) (x0 ) + (ϕp (u′1 )) (x0 ) = lim
But on this point, we have

− (ϕp (u′2 )) (x0 ) + b
h (x0 , u2 (x0 )) + (ϕp (u′1 )) (x0 ) − b
h (x0 , u1 (x0 ))

′
≥ F (x0 , u2 (x0 )) − F (x0 , u′1 (x0 )) = 0.
′

′

Which means that,
′

′

− (ϕp (u′2 )) (x0 ) + (ϕp (u′1 )) (x0 ) ≥ b
h (x0 , u1 (x0 )) − b
h (x0 , u2 (x0 )) .

Since u1 (x0 ) > u2 (x0 ) and the function h is strictly increasing in its second
variable, we obtain that
′

′

− (ϕp (u′2 )) (x0 ) + (ϕp (u′1 )) (x0 ) > 0.
Which is a contradiction.
Definition 2.1: We say that α is a lower solution of (2) if
i) α ∈ C 1 ([0, 1]) and ϕp (α′ ) ∈ C 1 (0, 1) .

′
− (ϕp (α′ )) ≤ F (x, α′ ) − b
h (x, α) , x ∈ (0, 1) ,
ii)
α (0) − a2 α′ (0) ≤ a, α (1) + a3 α′ (1) ≤ b.
EJQTDE, 2011 No. 6, p. 3

Definition 2.2: We say that β is an upper solution of (2) if
i) β ∈ C 1 ([0, 1]) and ϕp (β ′ ) ∈ C 1 (0, 1) .

′
− (ϕp (β ′ )) ≥ F (x, β ′ ) − b
h (x, β) , x ∈ (0, 1) ,
ii)

β (0) − a2 β ′ (0) ≥ a, β (1) + a3 β ′ (1) ≥ b.
Now, if moreover F is a bounded function, then we have the following result.
Theorem 2 Suppose that α and β are lower and upper solutions of problem
(2) such that α (x) ≤ β (x) for all 0 ≤ x ≤ 1. Then the problem (2) admits a
unique solution u ∈ C 1 ([0, 1]) with ϕp (u′ ) ∈ C 1 (0, 1) such that
α (x) ≤ u (x) ≤ β (x) , for all 0 ≤ x ≤ 1.
Proof. Using a proof similar to that of Theorem 1 in [25], we can prove that
the problem (2) admits at least one solution and by Lemma 1, it follows that
this problem admits a unique solution.
Now, we consider the following problem

′ ′
u′ ) , x ∈ (0, 1) ,

 − (ϕp (u )) = f (x, u,

1

Z





 u (0) − a0 u′ (0) = g1 (x) u (x) dx,
(3)
0

1

Z





u (1) + a1 u′ (1) = g2 (x) u (x) dx,


0

2

where f : [0, 1] × R → R is a continuous function, gi : [0, 1] → R+ are a
continuous functions (i = 0, 1 ) and a0 and a1 are two positive real numbers.
Definition 2.3: We say that u is a solution of (3) if
i) u ∈ C 1 ([0, 1]) and ϕp (u′ ) ∈ C 1 (0, 1) .
ii) u satisfies (3).

Definition 2.4: We say that u is a lower solution of (3) if
i) u ∈ C 1 ([0, 1]) and ϕp (u′ ) ∈ C 1 (0, 1) .

′
− (ϕp (u′ )) ≤ f (x, u, u′ ) , x ∈ (0, 1) ,




Z1



′

 u (0) − a0 u (0) ≤ g1 (x) u (x) dx,
ii)
0


Z1





u (1) + a1 u′ (1) ≤ g2 (x) u (x) dx.


0

Definition 2.5: We say that u is an upper solution of (3) if
EJQTDE, 2011 No. 6, p. 4

i) u ∈ C 1 ([0, 1]) and ϕp (u′ ) ∈ C 1 (0, 1) .

′
− (ϕp (u′ )) ≥ f (x, u, u′ ) , x ∈ (0, 1) ,




Z1



′

 u (0) − a0 u (0) ≥ g1 (x) u (x) dx,
ii)
0


Z1




 u (1) + a1 u′ (1) ≥ g2 (x) u (x) dx.


0

Now, we define the Nagumo–Wintner condition.
Definition 2.6: We say that f : [0, 1] × R2 → R satisfies a Nagumo–
Wintner condition relative to the pair u and u, if there exist C ≥ 0 and a
functions Q ∈ Lp ([0, 1]) and Ψ : [0, +∞) → (0, +∞) continuous, such that


1
|f (x, u, v)| ≤ Ψ (|v|) Q (x) + C |v| p−1 ,
(4)
for all (x, u, v) ∈ D, where

and



D = (x, u, v) ∈ [0, 1] × R2 : u (x) ≤ u (x) ≤ u (x)
+∞
Z
0

1

sp

 ds = +∞.
1
Ψ |s| p−1

(5)

We have the following result
Lemma 3 Let f : [0, 1] × R2 → R satisfying Nagumo–Wintner conditions (4)
and (5) in D. Then there exists a constant K > 0, such that every solution of
problem (3) verifying u (x) ≤ u (x) ≤ u (x), for all x ∈ [0, 1] , satisfies ku′ k0 ≤
K.
Proof.
Since u (x) ≤ u (x) ≤ u (x), for all x ∈ [0, 1] , we have
u (1) − u (0) ≤ u (1) − u (0) ≤ u (1) − u (0) .
Let
η := max {|u (1) − u (0)| , |u (1) − u (0)|} .
By the mean value theorem, there exists x0 ∈ (0, 1) such that
u (1) − u (0) = u′ (x0 ) ,
and then,
|u′ (x0 )| ≤ η.

EJQTDE, 2011 No. 6, p. 5

We put by definition
L := |u′ (x0 )| and δe := 2 max (kuk0 , kuk0 ) .

Take K > (η, ku′ k0 , ku′ k0 ) such that
ϕpZ(K)
ϕp (η)

1

p−1
p−1
sp
 ds > kQkp δe p + C δe p .

1
Ψ |s| p−1

(6)

Now, we are going to prove that |u′ (x)| ≤ K, for all x ∈ [0, 1] .
Suppose, on the contrary that there exists x1 ∈ [0, 1] such that |u′ (x1 )| > K.
Then by the continuity of u′ , we can choose x2 ∈ [0, 1] verifying one of the
following situations:
i) u′ (x0 ) = L, u′ (x2 ) = K and L ≤ u′ (x) ≤ K, for all x ∈ (x0 , x2 ) .
ii) u′ (x2 ) = K, u′ (x0 ) = L and L ≤ u′ (x) ≤ K, for all x ∈ (x2 , x0 ) .
iii) u′ (x0 ) = −L, u′ (x2 ) = −K and −K ≤ u′ (x) ≤ −L, for all x ∈ (x0 , x2 ) .
iv) u′ (x2 ) = −K, u′ (x0 ) = −L and −K ≤ u′ (x) ≤ −L, for all x ∈ (x2 , x0 ) .
Assume that the case i) holds. The others can be handled in a similar way.
Since u is a solution of the problem (3) and by the Nagumo–Wintner condition (4), we have


1
′
(ϕp (u′ )) (x) ≤ Ψ (u′ (x)) Q (x) + C. (u′ (x)) p−1 , for all x ∈ (x0 , x2 ) . (7)
Since L ≤ η and ϕp is increasing, we have
ϕpZ(K)
ϕp (η)

1

sp
 1  ds ≤
Ψ s p−1

ϕpZ(K)
ϕp (L)

1

sp
 1  ds.
Ψ s p−1

(8)

Now if we put s = ϕp (u′ (x)), we obtain that
ϕpZ(K)
ϕp (L)

1

sp
 1  ds =
Ψ s p−1

Zx2

1

(ϕp (u′ (x))) p
′
(ϕp (u′ (x))) dx.
Ψ (u′ (x))

x0

EJQTDE, 2011 No. 6, p. 6

Then by (7) and (8), it follows that
ϕpZ(K)
ϕp (η)

≤

Zx2

(ϕp (u′ (x))) p
′
(ϕp (u′ (x))) dx
Ψ (u′ (x))

≤

Zx2

h
i
1
1
(ϕp (u′ (x))) p Q (x) + C. (u′ (x)) p−1 dx

=

Zx2

(u′ (x))

=

Zx2

1

sp
 1  ds
Ψ s p−1

1

x0

x0

p−1
p

x0
′

(u (x))

p−1
p

h

i
1
Q (x) + C. (u′ (x)) p−1 dx

Q (x) dx + C

x0

≤

≤

1

(u′ (x)) p dx

x0

 p−1
x
p
Z2 
p
 p−1
p−1
′
dx
+
kQkp 
(u (x)) p

x0
 p1   x
 p−1
p
Zx2 
Z2

p
1
p


(u′ (x)) p dx  .  1 p−1 dx
+C. 
x0

=

Zx2

kQkp (u (x2 ) − u (x0 ))
p−1
p−1
kQk δe p + C δe p .

x0

p−1
p

1

+ C. (u (x2 ) − u (x0 )) p . (x2 − x0 )

p

Which a contradiction with (6).

3

Main result

In this section, we state and prove our main result.
On the nonlinearity f , we shall impose the following condition:
H) There exists a continuous function h : R → R strictly increasing such
that s 7→ f (x, s, z) + h (s) is increasing for all x ∈ [0, 1] and all z ∈ R.
The main result of this work is:
Theorem 4 Let u and u be a lower and upper solution solution respectively
for problem (3) and such that u ≤ u in [0, 1] . Assume that H) is satisfied and
the Nagumo–Wintner conditions relative to u and u holds. Then the problem
(3) has a maximal solution u∗ and a minimal solution u∗ such that for every
solution u of (3) with u ≤ u ≤ u in [0, 1], we have u ≤ u∗ ≤ u ≤ u∗ ≤ u in
[0, 1] .
For the proof of this theorem, we need a preliminary lemma.
Let w, w ∈ C 1 ([0, 1]) be fixed such that
i) ϕp (w′ ), ϕp (w ′ ) ∈ C 1 (0, 1) .
EJQTDE, 2011 No. 6, p. 7

p−1
p

ii) u ≤ w ≤ w ≤ u in [0, 1] .
Let δ (v) := max (−K, min (v, K)), for all v ∈ R, where K is the constant
defined in the proof of lemma 3. Then the function δ is continuous and bounded.
In fact, we have δ (v) = v for all v such that |v| ≤ K; and |δ (v)| ≤ K for all
v ∈ R.
We consider the following problems:

′
− (ϕp (u′ )) + h (u) = f (x, w, δ (u′ )) + h (w) , x ∈ (0, 1) ,




Z1



′

 u (0) − a0 u (0) = g0 (s) w (s) ds,
(9)
0

1

Z





u (1) + a1 u′ (1) = g1 (s) w (s) ds,


0

and


′
− (ϕp (u′ )) + h (u) = g (x, w, δ (u′ )) + h (w) , x ∈ (0, 1) ,




Z1



′

 u (0) − a0 u (0) = g0 (s) w (s) ds,
0


Z1





u (1) + a1 u′ (1) = g1 (s) w (s) ds.



(10)

0

Lemma 5 Let w and w be a lower and upper solution respectively for problem
(3). Assume that H) is satisfied and the Nagumo–Wintner conditions relative
e and u
b of (9) and (10)
to u and u holds. Then there exists a unique solution u
such that u ≤ w ≤ u
e≤u
b ≤ w ≤ u.
Proof.
The proof will be given in several steps.
Step 1: w is a lower solution of (9).
Proof: Let x ∈ (0, 1) , we have
′

− (ϕp (w ′ )) + h (w) ≤
≤

f (x, w, w′ ) + h (w)
f (x, w, w′ ) + h (w) .

This means that,
′

∀ x ∈ (0, 1) , − (ϕp (w ′ )) + h (w) ≤ f (x, w, w ′ ) + h (w) .
Now since w is a lower solution of (3) and u ≤ w ≤ u in [0, 1], then by using
a proof similar to that of lemma 3, we prove that kw′ k0 ≤ K. Hence δ (w ′ ) = w ′
and we obtain that
′

∀x ∈ (0, 1) , − (ϕp (w ′ )) + h (w) ≤ f (x, w, δ (w ′ )) + h (w) .

(11)

EJQTDE, 2011 No. 6, p. 8

On the other hand, we have
Z1

′

w (0) − a0 w (0) ≤

0
Z1

≤

g0 (s) w (s) ds

g0 (s) w (s) ds.

0

That is,
′

w (0) − a0 w (0) ≤

Z1

g0 (s) w (s) ds.

(12)

Z1

g1 (s) w (s) ds.

(13)

0

Similarly, we have
′

w (1) + a1 w (1) ≤

0

Then by (11), (12) and (13), it follows that w is a lower solution of (9).
Step 2: w is an upper solution of (9).
Proof: Let x ∈ (0, 1), we have
′

− (ϕp (w ′ )) + h (w) ≥ f (x, w, w′ ) + h (w) .
Now by using a proof similar to that of lemma 3, we prove that kw′ k0 ≤ K.
Hence δ (w′ ) = w′ and we obtain that
′

∀ x ∈ (0, 1) , − (ϕp (w ′ )) + h (w) ≥ f (x, w, δ (w ′ )) + h (w) .

(14)

Also, we have
′

w (0) − a0 w (0) ≥

Z1

g0 (s) w (s) ds,

(15)

Z1

g1 (s) w (s) ds.

(16)

0

and
′

w (1) + a1 w (1) ≥

0

Then by (14), (15) and (16), it follows that w is an upper solution of (9).
By Steps 1 and 2 and since the functions (x, u′ ) 7→ f (x, w, δ (u′ )) + h (w)
is a bounded continuous function and u 7→ h (u) is continuous and strictly
increasing, then by theorem 2, it follows the existence of a unique solution u
e of
e ≤ w.
(9) such that w ≤ u
Similarly, we can prove the existence and uniqueness of a solution to (10),
which we call u
b such that w ≤ u
b ≤ w.
EJQTDE, 2011 No. 6, p. 9

Finally, by using a proof similar to that of lemma 1, we prove that u
b≤u
e in
[0, 1] .
Proof. of Theorem 4
The proof will be given in several steps.
We take u0 = u, u0 = u and define the sequences (un )n≥1 , (un )n≥1 by



′



− ϕp u′n+1 + h (un+1 ) = f x, un , δ u′n+1 + h (un ) , x ∈ (0, 1) ,




Z1



′

 un+1 (0) − a0 un+1 (0) = g0 (s) un (s) ds,

(Pn+1 )

0


Z1





u
(1) + a1 u′n+1 (1) = g1 (s) un (s) ds,

 n+1
0

and



′





− ϕp u′n+1 + h un+1 = g x, un , δ u′n+1 + h (un ) , x ∈ (0, 1) ,




Z1




 un+1 (0) − a0 u′n+1 (0) = g0 (s) un (s) ds,

(Qn+1 )

0


Z1





u
(1) + a1 u′n+1 (1) = g1 (s) un (s) ds.

 n+1
0

Step 1: For all n ∈ N, we have

u ≤ un ≤ un+1 ≤ un+1 ≤ un ≤ u in [0, 1] .

Proof:
i) For n = 0, we have

′
− (ϕp (u′1 )) + h (u1 ) = f (x, u, δ (u′1 )) + h (u) , x ∈ (0, 1) ,




Z1



′

 u1 (0) − a0 u1 (0) = g0 (s) u (s) ds,
0


Z1





u (1) + a1 u′1 (1) = g1 (s) u (s) ds,

 1

(P1 )

0


′
− (ϕp (u′1 )) + h (u1 ) = g (x, u, δ (u′1 )) + h (u) , x ∈ (0, 1) ,




Z1



′

 u1 (0) − a0 u1 (0) = g0 (s) u (s) ds,
0


Z1





u′ (1) + a1 u1 (1) = g1 (s) u (s) ds.

 1

(Q1 )

0

EJQTDE, 2011 No. 6, p. 10

Since u and u are lower and upper solutions of problem (3), then by lemma
5, it follows that
u = u0 ≤ u1 ≤ u1 ≤ u0 = u in [0, 1] .
ii) Assume for fixed n > 1, we have
u ≤ un−1 ≤ un ≤ un ≤ un−1 ≤ u in [0, 1] ,
and we show that
u ≤ un ≤ un+1 ≤ un+1 ≤ un ≤ u in [0, 1] .
Let x ∈ (0, 1), we have
′

− (ϕp (u′n )) + h (un ) = f (x, un−1 , δ (u′n )) + h (un−1 ) .
Since un−1 ≥ un and using the hypothesis
f

(x, un−1 , δ (u′n ))

(17)

H), we obtain

+ h (un−1 ) ≥ f (x, un , δ (u′n )) + h (un ) .

(18)

Then by (17) and (18), it follows that
′

∀ x ∈ (0, 1) , − (ϕp (u′n )) ≥ f (x, un , δ (u′n )) .
Now by using a proof similar to that of lemma 3, we can prove that
ku′n k0 ≤ K. Hence δ (u′n ) = u′n and we obtain that
′

∀ x ∈ (0, 1) , − (ϕp (u′n )) ≥ f (x, un , u′n ) .

(19)

On the other hand, we have
un (0) −

a0 u′n

(0) =

Z1

g0 (s) un−1 (s) ds

≥

Z1

g0 (s) un (s) ds.

0

0

That is,
un (0) −

a0 u′n

(0) ≥

Z1

g0 (s) un (s) ds,

(20)

0

and
un (1) +

a1 u′n

(1) =

Z1

g1 (s) un−1 (s) ds

≥

Z1

g1 (s) un (s) ds.

0

0

EJQTDE, 2011 No. 6, p. 11

That is,
un (1) +

a1 u′n

(1) ≥

Z1

g1 (s) un (s) ds

(21)

0

Then by (19), (20) and (21), it follows that un is an upper solution of (3).
Similarly, we can prove that un is a lower solution of (3). Then by lemma
5, there exists a unique solution un+1 and un+1 of (Pn+1 ) and (Qn+1 )
such that
u ≤ un+1 ≤ un ≤ un ≤ un+1 ≤ u in [0, 1] .
Hence, we have
∀n ∈ N, u ≤ un+1 ≤ un ≤ un ≤ un+1 ≤ u in [0, 1] .

Step 2: The sequence (un )n∈N converge to a maximal solution of (3).
Proof: By Step 1 and since ku′n k0 ≤ K, for all n ∈ N, it follows that the

sequence (un )n∈N is uniformly bounded in C 1 ([0, 1]) .
Now let ε1 > 0 and t, s ∈ [0, 1] such that t < s, then for each n ∈ N, we have

s

Z












′
ϕp u′n+1 (s) − ϕp u′n+1 (t)  = 

f
τ,
u
(τ
)
,
δ
u
(τ
)
+
h
(u
(τ
))
−
h
(u
(τ
))
dτ
n
n
n+1
n+1




t
Zs






 f τ, un (τ ) , δ u′ (τ ) + h (un (τ )) − h (un+1 (τ ))  dτ
≤
n+1
t

≤ (M1 (f ) + 2M2 (h)) |s − t| ,

where
M1 (f ) := max {|f (x, s, z)| : x ∈ [0, 1] , u ≤ u ≤ u and |z| ≤ K} ,
and
M2 (h) := max {|h (u)| : u ≤ u ≤ u} .
If we put K1 := M1 (f ) + 2M2 (h) , one has





ϕp u′n+1 (s) − ϕp u′n+1 (t)  ≤ K1 |s − t| .
ε1
, we obtain
K1 + 1




u′n+1 (s) − ϕp u′n+1 (t)  < ε1 .

Then if we choose |s − t| ≤

ϕp

Therefore the sequence (ϕp (u′n ))n∈N is equicontinuous on [0, 1] .
Now since the mapping ϕ−1
p is an increasing homeomorphism from R onto
R, we deduce from


′
−1
′

|u′n (s) − u′n (t)| = ϕ−1
p (ϕp (un (s))) − ϕp (ϕp (un (t)))
EJQTDE, 2011 No. 6, p. 12

that the sequence (u′n )n∈N is equicontinuous on [0, 1] .


Hence by the Arz´ela-Ascoli theorem, there exists a subsequence unj of
(un )n∈N which converges in C 1 ([0, 1]) .
Let
u := lim unj .
nj →+∞

Then
u′ =

lim u′ .
nj →+∞ nj

But by Step 1 the sequence (un )n∈N is decreasing and bounded from below,
then the pointwise limit of this sequence exists and it is denoted by u∗ . Hence,
we have u = u∗ and moreover, the whole sequence converges in C 1 ([0, 1]) to u∗ .
Let x ∈ (0, 1) , we have
−ϕp u′n+1




(x) =

ϕp u′n+1



(0) +

Zx

f τ, un (τ ) , δ u′n+1 (τ )

0







+ h (un (τ )) − h (un+1 (τ )) dτ.

Now, as n tends to +∞, we obtain that



f τ, un (τ ) , δ u′n+1 (τ ) + h (un (τ )) − h (un+1 (τ )) → f (τ, u∗ (τ ) , δ (u∗′ (τ ))) .
Also, we have






∃ K4 > 0, ∀ n ∈ N, ∀ τ ∈ [0, 1] , f τ, un (τ ) , δ u′n+1 (τ ) + h (un (τ )) − h (un+1 (τ )) ≤ K4 .
Hence, the dominated convergence theorem of Lebesgue implies that
∗′

∗′

−ϕp (u (x)) = ϕp (u (0))+

Zx

(f (τ, u∗ (τ ) , δ (u∗′ (τ ))) + h (u∗ (τ )) − h (u∗ (τ ))) dτ.

0

Thus, we obtain
′

∀ x ∈ (0, 1) , − (ϕp (u∗′ )) = f (x, u∗ , δ (u∗′ )) .

(22)

Also, by the dominated convergence theorem of Lebesgue, we have
∗

∗′

u (0) − a0 u (0) =

Z1

g0 (s) u∗ (s) ds,

(23)

Z1

g1 (s) u∗ (s) ds.

(24)

0

and
∗

∗′

u (1) + a1 u (1) =

0

EJQTDE, 2011 No. 6, p. 13

By (22), (23) and (24), it follows that u∗ is a solution of the following problem

′
− (ϕp (u∗′ )) = f (x, u∗ , δ (u∗′ )) , x ∈ (0, 1) ,




Z1



∗
∗′

u (0) − a0 u (0) = g0 (s) u∗ (s) ds,

(25)
0

1

Z





u∗ (1) + a1 u∗′ (1) = g1 (s) u∗ (s) ds.


0

Now using a proof similar to that of lemma 3, we prove that ku∗ k ≤ K.
Hence δ (u∗′ ) = u∗′ and consequently u∗ is a solution of (3).
Now, we prove that if u is another solution of (3) such that u ≤ u ≤ u in
[0, 1], then u ≤ u∗ in [0, 1] .
Since u is a lower solution of (3), then by Step 1, we have
∀ n ∈ N, u ≤ un .
Letting n → +∞, we obtain that
u ≤ lim un = u∗ .
n→+∞

Which means that u∗ is a maximal solution of problem (3).
Step 3: The sequence (un )n∈N converges to a minimal solution of (3).
Proof: The proof is similar to that of Step 2, so it is omitted.
The proof of our result is complete.

4

Application

In this section, we apply the previous result to the following problem

1
′

− (ϕp (u′ )) = λ1 uk1 − uk2 + λ2 uk1 |u′ | p−1 in (0, 1) ,


Z1
′
′

u (0) − a0 u (0) = 0, u (1) + a1 u (1) = g1 (s) u (s) ds,



(26)

0

where 0 < k1 < p − 1, k2 > k1 , λ1 , and are a positive real parameters and
Z1
g1 (s) ds ≤ 1.
0

To study this problem, we need first consider the following problem:

′
− (ϕp (u′ )) = uk3 in (0, 1) ,
u (0) = 0, u (1) = 0,

(27)

where k3 > 0 and k3 6= p − 1.
Theorem 6 The problem (27) admits a unique positive solution Φk3 ,p .
EJQTDE, 2011 No. 6, p. 14

Proof.
Multiplying the differential equation in (27) by u′ and integrating the resulting equation over [0, x], we obtain
p

p

|u′ (x)| = |u′ (0)| −

p
uk3 +1 (x) , x ∈ [0, 1] .
(p − 1) (k3 + 1)

(28)



1
1
′
We note that u is symmetric about x = and u (x) > 0, for all x ∈ 0,
.
2
2
 
1
If we put by definition ρ = max u (x), then u
= ρ and ρ > 0.
x∈[0,1]
2
1
Now substituting x = in (28), we obtain
2
1




1
p
p
k3 +1
k3 +1
′
,
ρ
−u
(x)
, for all x ∈ 0,
u (x) =
(p − 1) (k3 + 1)
2


and thus,
u′ (x)


p
(ρk3 +1 − uk3 +1 (x))
(p − 1) (k3 + 1)



1
.
= 1, for all x ∈ 0,
2
1
p




Integrating the last equality on (0, x) , where x ∈ 0, 12 , we obtain
u(x)
Z
0

dv


p
(p − 1) (k3 + 1)

1
p
k
+1
k
+1
(ρ 3 − v 3 )

= x.

(29)

1
in (29), we obtain
2
Zρ
dv
1
G (ρ) :=
= .
1
2


0
p
p
k
+1
k
+1
3
3
(ρ
−v
)
(p − 1) (k3 + 1)

Letting x →

Thus, if the problem (27) admits a positive solution u, with max u (x) =
x∈[0,1]
 
1
1
u
= ρ, then we have G (ρ) = .
2
2
1
Conversely, if G (ρ) = . Defining u via equation (29), we can prove that
2
 
1
= ρ.
problem (27) admits a unique positive solution u, with max u (x) = u
2
x∈[0,1]
EJQTDE, 2011 No. 6, p. 15

Hence the problem (27) admits a unique positive solution u, with max u (x) =
x∈[0,1]
 
1
1
= ρ if and only if G (ρ) = .
u
2
2
Some easy computations shows that


1
p−1
p − 1 − k3
1 B

,
(p − 1) (k3 + 1) p
p
k3 + 1
p
G (ρ) =
,
ρ
p
k3 + 1
where B (k, l) is the Euler beta function defined by
B (k, l) =

Z1

k−1 l−1

(1 − t)

t

dt, k > 0 and l > 0.

0

1
admits a unique solution.
2
Hence the problem (27) admits a unique positive solution.

It is not difficult to see that the equation G (ρ) =

Theorem 7 Assume that λ1 > 1, then the problem (26) admits a maximal
solution u∗ and a minimal solution u∗ .
Proof. We put (u, u) = (εΦk1 ,p , L) where ε and L are a positive constants.
Z1
′
′
First, since Φk1 ,p (0) > 0, Φk1 ,p (1) < 0 and g1 (s) ds ≤ 1 and, it is easy to
0

check
 that
′
εΦk1 ,p (0) − a0 εΦk1 ,p (0) ≤ 0,




Z1


′


 εΦk1 ,p (1) + a1 εΦk1 ,p (1) ≤ g1 (s) εΦk1 ,p (s) ds,


Z1





L ≥ g1 (s) Lds.



0

0

Now
 u and u-are lower and upper solutions of (26), if we have 
1
 p−1
 p−1 k1


ε Φk1 ,p (x) ≤ λ1 εk1 Φkk11 ,p (x) − εk2 Φkk21 ,p (x) + λ2 εk1 Φkk11 ,p (x) εΦ′k1 ,p (x)
in (0, 1) ,
 0 ≥ λ Lk1 − Lk2 .
1
That
is

1

 p−1
 p−1−k1
 ′

1
≤ λ1 − ε1k2 −k1 Φkk21 −k
ε
in (0, 1) ,
,p (x) + λ2 εΦk1 ,p (x)
 Lk2 ≥ λ Lk1 .
1
Since k2 > k1 , if we choose λ1 > 1, ε sufficiently small and L sufficiently
large, we obtain that u and u is a lower and upper solutions of (26).
This implies that the problem (26) admits a maximal solution u∗ and a
minimal solution u∗ .

EJQTDE, 2011 No. 6, p. 16

References
[1] B. Ahmed, A. Alsaedi & B. Alghamdi, Analytic approximation of
solutions of the forced Duffing equation with integral boundary conditions,
Nonlinear Anal. Real World Appl. 9 (2008), pp. 1727-1740.
[2] B. Ahmed & A. Alsaedi, Existence of approximate solutions of the
forced Duffing equation with discontinuous type integral boundary conditions, Nonlinear Anal. Real World Appl. 10 (2009), pp. 358-367.
[3] S. R. Bernfeld & J. Chandra, Minimal and maximal solutions of nonlinear boundary value problems, Pacific Journal of Mathematics, Vol. 71,
N◦ 1 (1977), pp. 13-20.
[4] A. V. Bitsadze, On the theory of nonlocal boundary value problems,
Soviet Math. Dock. 1964, 30, pp. 8-10.
[5] A. V. Bitsadze & A. A. Samarskii, Some elementary generalizations
of linear elliptic boundary value problems, Dokl. Akad. Nauk SSSR 1969,
185, pp. 739-740.
[6] A. Boucherif, Differential equations with nonlocal boundary conditions,
Nonlinear Analysis T. M. A. 47 (2001), pp. 2419-2430.
[7] A. Boucherif, Positive solutions of second order differential equations
with integral boundary conditions, Discrete and Continuous Dynamical
Systems Supplement 2007, pp. 155-159.
[8] A. Boucherif, Second order boundary value problems with integral
boundary conditions, Nonlinear Analysis T. M. A. 70 (2009), pp. 364371.
[9] A. Boucherif & S. M. Bouguima, Solvability of non local multipoint
boundary value problems, Nonlinear Studies Vol. 8 N◦ 4 (2001), pp. 395406.
[10] A. Boucherif & J. Henderson, Positive solutions of second order
boundary value problems with changing signs Carath´eodory nonlinearities,
Electronic Journal of Qualitative Theory of Differential equations N◦ 2
(2006), pp. 1-14.
′

[11] A. Cabada & R. L. Pouso, Existence results for the problem (Φ (u′ )) =
f (t, u, u′ ) with nonlinear boundary conditions, Nonlinear Analysis T. M.
A. 35 (1999), pp. 221-231.
˜ada, P. Dra
´bek & J. L. Ga
´mez, Existence of positive solutions
[12] A. Can
for some problems with nonlinear diffusion, Transactions of the American
Mathematical Society, Volume 349, Number 10, October 1997, pp. 42314249.

EJQTDE, 2011 No. 6, p. 17

[13] J. R. Cannon, The solution of the heat equation subject to the specification of energy, Quarterly Applied Math. Vol. 21, N0 2 (1963), pp.
155-160.
[14] V. Capasso & K. Kunisch, A reaction-diffusion system arising in modeling man-environment diseases, Q. Appl. Math. 46 (1988), pp. 431-449.
[15] R. Yu Chegis, Numerical solution of a heat conduction problem with
an integral boundary conditions, Litovsk. Mat. Sb. Vol. 24 (1984), pp.
209-215.
[16] M. Cherpion, C. De Coster & P. Habets, Monotone iterative methods for boundary value problems, Differential and Integral Equations 12
(1999), pp. 309-338.
[17] M. Derhab, A quasilinear elliptic system with integral boundary conditions, Proceedings of the Jangjeon Mathematical Society, Vol. 12,( 2009),
Number 2, pp. 165-187.
[18] R. E. Ewing & T. Lin, A class of parameter estimation techniques for
fluid flow in porous media, Adv. Water Resources 14 (1991), pp. 89-97.
[19] M. G. Huidobro & P. Ubilla, Multiplicity of solutions for a class of
nonlinear second order equations, Nonlinear Analysis T. M. A. Vol. 28, N◦
9 (1997), pp. 1509-1520.
[20] V. Il′ in & E. Moiseev, Nonlocal boundary value problems of the first
kind for a Sturm-Liouville operator in its differential and finite difference
aspects, Differential Equations 23 (1987), pp. 803-810.
[21] N. I. Ionkin, Solution of a boundary value problem in heat conduction
theory with nonlocal boundary conditions, Differential Equations Vol. 13
(1977), pp. 294-304.
´tson, Gradient method in Sobolev spaces for nonlocal boundary[22] J. Kara
value problems, Electronic J. Diff. Eqns. 51 (2000), pp. 1-17.
¨ nfeld, Integral boundary conditions for unsteady
[23] F. Nicoud & T. Scho
biomedical CFD applications, Int. J. Numer. Meth. Fluids 40 (2002), pp.
457-465.
´, S. Stane
ˇk & M. Tvrdy
´ . Singularities and Laplacians
[24] I. Rach˚
unkova
in Boundary Value Problems for Nonlinear Ordinary Differential Equations.
In: Handbook of Differential Equations (Ordinary Differential Equations)
Vol.3 pp. 605-721.
[25] J. Wang & D. Jiang, A unified approach to some two-point, three-point
and four-point boundary value problems with Carath´eodory functions, J.
Math. Anal. & Appl. 211 (1997), pp. 223-232.
(Received July 9, 2010)
EJQTDE, 2011 No. 6, p. 18