Applied Mathematics E-Notes, 11(2011), 197–207 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Positive Solutions For Singular Boundary Value
Problems Involving p-Laplacian Operators
Changxiu Songy, Xuejun Gaoz
Received 6 September 2010
Abstract
The existence of positive solutions for singular boundary value problems involving p-Laplacian operators are investigated. By applying the …xed point theorem of cone expansion and compression of norm type, su¢ cient conditions are
established for the existence of positive solutions.
1
Introduction
In this paper, we study the following singular boundary value problem (BVP) involving
p-Laplacian operators
8
( (x0 ))0 + a(t)f (xt ; y t ) = 0; 0 < t < 1;
>
>
< p 0 0
( p (y )) + b(t)g(xt ; y t ) = 0; 0 < t < 1;
(1)
x(t) = '(t); 1 t 1 + ; x0 (0) = 0;
>
>
:
0
y(t) = (t); 1 t 1 + ; y (0) = 0;
where xt = x(t + ); 2 [0; ]; 0
< 1; p ( ) is the p-Laplacian operator; '; :
[1; 1 + ] ! [0; +1) are continuous, and '(1) = (1) = 0:
For p-Laplacian equations, many results have been obtained, for example see papers
[1-4]. But most of them are concerned with ordinary di¤erential equations. Recently,
the study of BVP of functional di¤erential equations [5-6] is of signi…cance since they
arise and have applications in variational problems of control theory and in other areas
of applied mathematics. In this paper, by constructing an integral equation which is
equivalent to BVP (1), we study the existence of positive solutions of nonlinear singular
BVP of the form (1).
Let C = C([0; ]; R) be a Banach space with a norm jj!jjC = sup0
j!( )j and
C + = f! 2 C : !( )
0; 2 [0; ]g:
We assume the following:
Mathematics Subject Classi…cations: 34B15
of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, P. R.
China
z School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, P. R.
China
y School
197
198
Positive Solutions for Singular Boundary Problems
(H1 ) f; g : C + C + ! (0; +1) are continuous;
(H2 ) a; b : (0; 1) ! [0; +1) are continuous, and
0<
Z
Z
a(t)
E
0<
Z
0
1
1
a(t)dt < +1; 0 <
0
Z
q(
Z
b(t)
E
s
a(r)dr)ds < +1; 0 <
0
Z
1
0
In this paper, we may choose a
Z
Z
1
b(t)dt < +1;
0
Z
q(
s
b(r)dr)ds < +1:
0
2 (0; minf 14 ; 1 4 g) by (H2 ) such that
1
a(t)dt > 0 and
Z
1
b(t)dt > 0:
De…ne C = f! 2 C + : 0 < jj!jjC
!( ); 2 [0; ]g and E = ft 2 [0; 1] : 0
t+
1; 0
g = [0; 1
]: Note that for t 2 [ ; 1
]
E; we have
xt0 = y0t = 0:
DEFINITION 1. A function (x; y) 2 C 1 [0; 1] C 1 [0; 1] is called a positive solution
of BVP (1) if it satis…es the following:
1. (x; y) satis…es BVP(1);
2. x(t) > 0; y(t) > 0; t 2 (0; 1); and
3. ( p (x0 ); p (y 0 )) is absolutely continuous on [0; 1]:
Suppose (x(t); y(t)) is a solution of BVP (1). Then
8
R1
Rs
r r
t 1;
>
q [ 0 a(r)f (x ; y )dr]ds; 0
t
>
< x(t) =
'(t);
1 t 1+ ;
R1
Rs
(2)
r r
>
t 1;
>
q [ 0 b(r)g(x ; y )dr]ds; 0
t
: y(t) =
(t);
1 t 1+ :
Suppose that (x0 (t); y0 (t)) is the solution of
8
0;
0
>
>
< x0 (t) =
'(t); 1
0;
0
>
>
: y0 (t) =
(t); 1
BVP (1) with f
t
t
t
t
0; g
0: Then
1;
1+ ;
1;
1+ :
(3)
If (x(t); y(t)) is the solution of BVP (1) and (u(t); v(t)) = (x(t) x0 (t); y(t) y0 (t)),
noting that (u(t); v(t)) = (x(t); y(t)) for 0 t 1; we have from (2) that
8
R1
Rs
r
r r
r
t 1;
>
q [ 0 a(r)f (u + x0 ; v + y0 )dr]ds; 0
t
>
u(t)
=
<
0;
1 t 1+ ;
R1
Rs
(4)
r
r
r r
>
;
v
+
y
)dr]ds;
0
t
1;
[
b(r)g(u
+
x
>
q
0
0
t
0
: v(t) =
0;
1 t 1+ :
Let K be a cone in Banach space X = C[0; 1 + ]
K = f(u; v) 2 X : u(t)
0; v(t)
0; u(t) + v(t)
C[0; 1 + ] de…ned by
g(t)jj(u; v)jj; t 2 [0; 1]g;
C. X. Song and X. J. Gao
199
where jj(u; v)jj = jjujj + jjvjj; jjujj = sup ju(t)j; jjvjj = sup jv(t)j; and
t2[0;b]
t2[0;b]
1 t; 0
0;
1
g(t) =
t
t
1;
1+ :
De…ne
A(u; v)(t) =
R1
Rs
q[ 0
a(r)f (ur + xr0 ; v r + y0r )dr]ds; 0
1
t
t
1;
1+ ;
R1
Rs
q[ 0
b(r)g(ur + xr0 ; v r + y0r )dr]ds; 0
1
t
t
1;
1+ ;
t
0;
B(u; v)(t) =
t
0;
and
(u; v)(t) = (A(u; v)(t); B(u; v)(t)); 0
t
1+ :
(5)
Under assumptions (H1 ) and (H2 ), BVP (1) has a solution if and only if has a
…xed point (u; v), that is, (u; v) = (u; v):
The following lemma will play an important role in the proof of our results and can
be found in the book [7].
LEMMA 1. Assume that X is a Banach space and K X is T
a cone in X; 1 , 2
are open subsets of X, and 0 2 1
:
Furthermore,
let
:
K
( 2 n 1 ) ! K be
2
a completely continuous operator satisfying one of the following conditions:
(i) jj (x)jj
(ii) jj (x)jj
jjxjj; 8 x 2 K
T
@ 1 ; jj (x)jj
T
jjxjj; 8 x 2 K @ 2 ; jj (x)jj
Then there is a …xed point of
LEMMA 2. The map
in K
T
(
2
n
jjxjj; 8 x 2 K
T
@ 2;
T
jjxjj; 8 x 2 K @ 1 :
1 ).
: X ! X in (5) is completely continuous and
The proof of Lemma 2 can be found in [5-6].
2
Main Results
In the sequel, we let
f0 :=
f0 :=
! 1 ;! 2 2C
f1 :=
f1 :=
lim
(jj! 1 jjC +jj! 2 jjC )!0
f (! 1 ; ! 2 )
(jj! 1 jjC + jj! 2 jjC )p
;
f (! 1 ; ! 2 )
;(jj! 1 jjC +jj! 2 jjC )!0 (jj! 1 jjC + jj! 2 jjC )p
lim
f (! 1 ; ! 2 )
(jj! 1 jjC +jj! 2 jjC )!1 (jj! 1 jjC + jj! 2 jjC )p
! 1 ;! 2 2C
1
lim
1
;
;
f (! 1 ; ! 2 )
;(jj! 1 jjC +jj! 2 jjC )!1 (jj! 1 jjC + jj! 2 jjC )p
lim
1
1
;
(K)
K:
200
Positive Solutions for Singular Boundary Problems
g0 :=
g0 :=
lim
(jj! 1 jjC +jj! 2 jjC )!0
1
g(! 1 ; ! 2 )
(jj! 1 jjC +jj! 2 jjC )!1 (jj! 1 jjC + jj! 2 jjC )p
lim
and
! 1 ;! 2 2C
;
f (! 1 ; ! 2 )
(jj! 1 jjC + jj! 2 jjC )p
lim
! 1 ;! 2 2C ;(jj! 1 jjC +jj! 2 jjC )!0
g1 :=
g1 :=
f (! 1 ; ! 2 )
(jj! 1 jjC + jj! 2 jjC )p
1
1
;
;
g(! 1 ; ! 2 )
;(jj! 1 jjC +jj! 2 jjC )!1 (jj! 1 jjC + jj! 2 jjC )p
lim
1
:
THEOREM 1. Assume (H1 ) and (H2 ) hold. Then BVP (1) has at least one positive
solution if one of the following conditions is satis…ed:
(H3 ) f0 = 0; f1 = +1; g0 = 0; '(t) = (t) 0; t 2 [1; 1 + ]; or
(H4 ) f0 = +1; f1 = 0; g1 = 0:
PROOF. Suppose that (H3 ) is satis…ed. By (t)
0; '(t)
0; t 2 [1; 1 + ]; we
know xt0 = y0t = 0; t 2 [0; 1 + ]: Since f0 = 0; for " > 0 (we choose " satisfying
R1
Rs
" 0 q [ 0 a(r)dr]ds 12 ), there is a 1 > 0 such that
f (! 1 ; ! 2 )
("(jj! 1 jjC + jj! 2 jjC ))p
1
;0
jj! 1 jjC + jj! 2 jjC
1:
De…ne
1
For (u; v) 2 @
1
= f(u; v) 2 X : jj(u; v)jj <
\ K; we deduce that jjur jjC + jjv r jjC
jjA(u; v)jj =
Z
1
0
Z
Z
[
q
1
for r 2 [0; 1] and thus
s
a(r)f (ur ; v r )dr]ds
0
1
0
Z
[
q
s
0
a(r)("(jjur jj + jjv r jj))p
"(jjujj + jjvjj)
Z
1
(jjujj + jjvjj):
2
1
2 (jjujj
Similarly, we have B(u; v)
1 g:
0
1
Z
[
q
1
dr]ds
s
a(r)dr]ds
0
+ jjvjj): This implies
jj (u; v)jj = jjA(u; v)jj + jjB(u; v)jj
jj(u; v)jj; (u; v) 2 @
1
\ K:
On the other hand, since f1 = +1; for M > 0 we can choose M satisfying
R1
Rs
a(r)dr]ds 1, there exists a 2 > 1 such that
M
q[
f (! 1 ; ! 2 )
(M (jj! 1 jjC + jj! 2 jjC ))p
1
; ! 1 ; ! 2 2 C ; jj! 1 jjC + jj! 2 jjC
De…ne
2
= f(u; v) 2 X : jj(u; v)jj <
2 g:
2:
C. X. Song and X. J. Gao
For (u; v) 2 @
2
201
\ K; we deduce
jjur jjC
jjujj
g(t)jjujj
u(t); t 2 [ ; 1
]; r 2 [0; 1];
jjv r jjC
jjvjj
g(t)jjvjj
v(t); t 2 [ ; 1
]; r 2 [0; 1];
which implies that ur ; v r 2 C for r 2 [ ; 1
jjur jjC
jjujj =
Thus, for (u; v) 2 @
2
jjA(u; v)jj =
] and
jjv r jjC
2;
jjvjj =
2;
r 2 [ ;1
]:
\ K; we have
Z
Z
q[
1
0
Z
s
a(r)f (ur ; v r )dr]ds
0
1
Z
[
q
s
a(r)(M (jjur jjC + jjv r jjC ))p
M (jjujj + jjvjj)
jjujj + jjvjj
= jj(u; v)jj:
Z
Z
[
q
1
1
dr]ds
s
a(r)dr]ds
That is,
jj (u; v)jj
jj(u; v)jj; (u; v) 2 @
2
\ K:
According
to the …rst part of Lemma 1, it follows that
has a …xed point(u; v) 2
T
K ( 2 n 1 ):
Now, suppose that (H4 ) is satis…ed. Since f0 = +1; for M > 0 (choose M satisfying
R1
Rs
M
a(r)dr]ds 1), there exists a 1 > 0 such that
q[
(M (jj! 1 jjC + jj! 2 jjC ))p
f (! 1 ; ! 2 )
1
; ! 1 ; ! 2 2 C jj! 1 jjC + jj! 2 jjC
1:
De…ne
1
For (u; v) 2 @
1
= f(u; v) 2 X : jj(u; v)jj <
1 g:
\ K; we deduce
jjur jjC
jjujj
g(t)jjujj
u(t); t 2 [ ; 1
]; r 2 [0; 1];
jjv r jjC
jjvjj
g(t)jjvjj
v(t); t 2 [ ; 1
]; r 2 [0; 1];
which implies that ur ; v r 2 C for r 2 [ ; 1
jjur jjC
jjujj =
1;
jjv r jjC
] and
jjvjj =
1;
r 2 [ ;1
]:
(6)
202
Positive Solutions for Singular Boundary Problems
]; we have xr0 = y0r = 0: Thus, for (u; v) 2 @
For r 2 [ :1
jjA(u; v)jj =
Z
Z
q[
1
0
Z
a(r)f (ur ; v r )dr]ds
Z
[
q
s
a(r)(M (jjur jjC + jjv r jjC ))p
M (jjujj + jjvjj)
jjujj + jjvjj
= jj(u; v)jj;
which implies jj (u; v)jj
Z
Z
[
q
1
jj(u; v)jj; 8(u; v) 2 @
1
("(jj! 1 jjC + jj! 2 jjC ))p
Choose a positive constant
2
> 1+maxff
q 1
2
(! 1 ; ! 2 ) : 0
1
1
dr]ds
s
a(r)dr]ds
\ K:
On the other hand, since f1 = 0, for 8 " > 0; 9N >
f (! 1 ; ! 2 )
\ K; we have
s
0
1
2
1
such that
; jj! 1 jjC + jj! 2 jjC > N:
such that
jj! 1 jjC +jj! 2 jjC
Z 1
N +jju0 jj+jjv0 jjg q [ (a(r)+b(r))dr]:
0
De…ne
2
= f(u; v) 2 X : jj(u; v)jj <
For (u; v) 2 @ 2 \ K; we have from the facts: x0 (t)
0; t 2 [0; 1 + ]; that for r 2 [0; 1];
jjur + xr0 jjC + jjv r + y0r jjC
2 g:
0; u(t)
0; y0 (t)
jjur jjC + jjv r jjC > N; for jjur jjC + jjv r jjC > N;
and
jjur + xr0 jjC + jjv r + y0r jjC
for jjur jjC + jjv r jjC
jjur jjC + jjxr0 jjC + jjv r jjC + jjy0r jjC
N + jjx0 jj + jjy0 jj;
N: Let
:= maxff q
1
0; v(t)
(! 1 ; ! 2 ) : 0
jj! 1 jjC + jj! 2 jjC
N + jjx0 jj + jjy0 jjg:
C. X. Song and X. J. Gao
203
R1
q[ 0
Thus, for " satisfying 0 < "(1 + jjx0 jj + jjy0 jj)
jjA(u; v)jj =
Z
1
s
a(r)f (ur + xr0 ; v r + y0r )dr]ds
0
0
Z
Z
[
q
a(r)dr] < 21 ; we have
1
0
Z
[
q
1
a(r)f (ur + xr0 ; v r + y0r )dr]ds
0
Z
=
a(r)f (ur + xr0 ; v r + y0r )dr
q[
jjur jjC +jjv r jjC >N
Z
+
a(r)f (ur + xr0 ; v r + y0r )dr]
0 jjur jjC +jjv r jjC
N
maxf"(jjur + xr0 jjC + jjv r + y0r jjC ); g
Z
maxf"(jju + x0 jj + jjv + y0 jj); g q [
1
1
maxf (jjujj + jjvjj) + ;
2
2
1
<
(jjujj + jjvjj)
2
1
=
:
2 2
Similarly, we have B(u; v)
1
2 (jjujj
Z
q[
Z
[
q
1
a(r)dr]
0
1
a(r)dr]
0
1
a(r)dr]g
0
+ jjvjj): This implies
jj (u; v)jj = jjA(u; v)jj + jjB(u; v)jj
jj(u; v)jj; (u; v) 2 @
2
\ K:
According
to the second part of Lemma 1, it follows that has a …xed point(u; v) 2
T
K ( 2 n 1 ):
T
Suppose that (u(t); v(t)) is the …xed point of in K ( 2 n 1 ); then (x(t); y(t)) =
(u(t) + x0 (t); v(t) + y0 (t)) is a positive solution of BVP (1). This completes the proof.
Similarly, we have the next theorem.
THEOREM 2. Assume (H1 ) and (H2 ) hold. Then BVP (1) has at least a positive
solution if one of the following conditions is satis…ed:
(H03 ) f0 = 0; g1 = +1; g0 = 0; '(t) = (t) 0; t 2 [1; 1 + ]; or
(H04 ) f1 = 0; g1 = 0; g0 = +1:
In what follows, we shall consider the existence of multiple positive solutions for
BVP (1).
THEOREM 3. Assume (H1 ); (H2 ) hold and the following conditions are satis…ed:
(H5 ) f0 = +1; f1 = +1;
(H6 ) 9 a p1 > 0 such that for 8 0 jj! 1 jjC + jj! 2 jjC p1 + p0 ; one has
and g(! 1 ; ! 2 ) (l1 p1 )p
R1
Rs
max f (t); '(t)g; l1 = f2 0 q [ 0 a(r)dr]dsg 1 :
f (! 1 ; ! 2 )
where p0 =
1 t 1+
(l1 p1 )p
1
1
;
204
Positive Solutions for Singular Boundary Problems
Then BVP (1) has at least two positive solutions.
PROOF. By (H5 ), there exists a 1 : 0 < 1 < p1 such that
(M (jj! 1 jjC + jj! 2 jjC ))p 1 ; jj! 1 jjC + jj! 2 jjC
R1
R1
where M satis…es M 1
a(r)dr]ds 1: De…ne
q[
f (! 1 ; ! 2 )
= f(u; v) 2 X : jj(u; v)jj <
1
For (u; v) 2 @
1
!1 ; !2 2 C ;
1 g:
\ K; similar to (6) one has ur ; v r 2 C and
jjur jjC + jjv r jjC
1
1;
(jjujj + jjvjj) =
1;
r 2 [ ;1
]:
Hence, we obtain an analogous inequality:
jj (u; v)jj = jjA(u; v)jj + jjB(u; v)jj
Similarly, there exists a
3
1
\ K:
> p1 such that
(M (jj! 1 jjC + jj! 2 jjC ))p
f (! 1 ; ! 2 )
jj(u; v)jj; (u; v) 2 @
1
; jj! 1 jjC + jj! 2 jjC
3;
!1 ; !2 2 C ;
M is chosen as above. De…ne
3
For (u; v) 2 @
3
= f(u; v) 2 X : jj(u; v)jj <
3 g:
\ K; one has ur ; v r 2 C and
jjur jjC + jjv r jjC
(jjujj + jjvjj) =
3;
r 2 [ ;1
]:
Furthermore, we have
jj (u; v)jj = jjA(u; v)jj + jjB(u; v)jj
By (H6 ), let
one has
2
= p1 , de…ne
2
jjA(u; v)jj =
= f(u; v) 2 X : jj(u; v)jj <
Z
1
0
l 1 p1
Z
q[
Z
0
=
=
=
Similarly, we have B(u; v)
jj(u; v)jj; (u; v) 2 @
0
1
2 g:
3
\ K:
For (u; v) 2 @
s
a(r)f (ur + xr0 ; v r + y0r )dr]ds
Z
[
q
s
a(r)dr]ds
0
1
p1
2
1
2 2
1
jj(u; v)jj:
2
1
2 (jjujj
+ jjvjj): This implies
jj (u; v)jj = jjA(u; v)jj + jjB(u; v)jj
jj(u; v)jj; (u; v) 2 @
2
\ K:
2
\ K;
C. X. Song and X. J. Gao
205
According to Lemma 1, it follows that has two …xed points, that is to say, BVP (1)
has at least two positive solutions. This completes the proof.
From above, the following theorems are obvious.
THEOREM 4. Assume (H1 ) and (H2 ) hold and the following conditions are satis…ed:
(H7 ) f0 = 0; f1 = 0; g0 = 0; g1 = 0; (t) = '(t) 0:
(H8 ) 9 a p2 > 0 such that for 8 p2 jj! 1 jjC + jj! 2 jjC p2 ; one has
f (! 1 ; ! 2 ) (l2 p2 )p 1 ;
nR
o 1
Rs
1
where l2 = 0
:
q [ 0 a(r)dr]dt
Then BVP (1) has at least two positive solutions.
THEOREM 5. Assume (H1 ) and (H2 ) hold and the following conditions are satis…ed:
(H05 ) g0 = +1; g1 = +1;
(H6 ) 9 a p1 > 0 such that for 8 0 jj! 1 jjC + jj! 2 jjC p1 + p0 ; one has
and g(! 1 ; ! 2 ) (l1 p1 )p
n R
o
Rs
1
max f (t); '(t)g and l1 = 2 0 q [ 0 a(r)dr]ds
f (! 1 ; ! 2 )
where p0 =
(l1 p1 )p
1
1
;
1
1 t 1+
:
Then BVP (1) has at least two positive solutions.
THEOREM 6. Assume (H1 ) and (H2 ) hold and the following conditions are satis…ed:
(H7 ) f0 = 0; f1 = 0; g0 = 0; g1 = 0; (t) = '(t) 0:
(H09 ) 9 a p2 > 0 such that for 8 p2 jj! 1 jjC + jj! 2 jjC p2 ; one has
g(! 1 ; ! 2 ) (l2 p2 )p
nR
o 1
Rs
1
where l2 = 0
:
q [ 0 b(r)dr]dt
Then BVP (1) has at least two positive solutions.
3
1
;
Examples
We have several examples.
EXAMPLE 1. Consider BVP
8
1
1
( (x0 ))0 4 (t + 31 ) + y 3 (t + 31 ) = 0; 0 < t < 1;
>
>
< p 0 01
1
( p (y )) 3 (t + 31 ) + y 4 (t + 13 ) = 0; 0 < t < 1;
>
x(t) = 0; 1 t 1 + 31 ; x0 (0) = 0;
>
:
y(t) = 0; 1 t 1 + 31 ; y 0 (0) = 0:
(7)
Here, a(t) = b(t) = 1; xt = x(t + ) x(t + 31 ); y t = y(t + ) y(t + 31 ); = 13 ; p = 98 ;
and
1 1
1 1
1 1
1 1
f (! 1 ; ! 2 ) = ! 14 ( ) + ! 23 ( ); g(! 1 ; ! 2 ) = ! 13 ( ) + ! 24 ( ):
3
3
3
3
206
Positive Solutions for Singular Boundary Problems
As jj! 1 jjC + jj! 2 jjC ! 0, we have
1
f (! 1 ; ! 1 )
(jj! 1 jjC + jj! 2 jjC )p
1
1
! 14 ( 31 ) + ! 23 ( 13 )
=
1
(jj! 1 jjC + jj! 2 jjC ) 8
1
1
jj! 1 jjC4 + jj! 2 jjC3
1
(jj! 1 jjC + jj! 2 jjC ) 8
1
(jj! 1 jjC + jj! 2 jjC ) 8
! 0
and
1
g(! 1 ; ! 1 )
(jj! 1 jjC + jj! 2 jjC )p
1
1
! 13 ( 31 ) + ! 24 ( 13 )
=
1
(jj! 1 jjC + jj! 2 jjC ) 8
1
1
jj! 1 jjC3 + jj! 2 jjC4
1
(jj! 1 jjC + jj! 2 jjC ) 8
1
(jj! 1 jjC + jj! 2 jjC ) 8
! 0;
that is to say f0 = 0 and g0 = 0 hold.
On the other hand, suppose ! 1 ; ! 2 2 C : Then ! 1 ( )
As ! 1 ; ! 2 2 C ; jj! 1 jjC + jj! 2 jjC ! 1; we get
1
f (! 1 ; ! 1 )
(jj! 1 jjC + jj! 2 jjC )p
1
=
jj! 1 jjC ; ! 2 ( )
jj! 2 jjC :
1
! 14 ( 13 ) + ! 23 ( 13 )
1
(jj! 1 jjC + jj! 2 jjC ) 8
1
1
( jj! 1 jjC ) 4 + ( jj! 2 jjC ) 3
1
1
3
(jj! 1 jjC + jj! 2 jjC ) 8
1
(jj! 1 jjC + jj! 2 jjC ) 8
! +1;
which means that f0 = +1 holds. According to Theorem 1, it follows that BVP (7)
has at least one positive solution.
Acknowledgment. Supported by grants (no. 10871052) and (no. 109010600)
from NNSF of China, and by grant (no.10151009001000032) from NSF of Guangdong.
References
[1] D. Q. Jiang and W. J. Gao, Upper and lower solution method and a singular
boundary value problems for the one-dimensional singular p-Laplacian, J. Math.
Anal. Appl., 252(2)(2000), 631–648.
C. X. Song and X. J. Gao
207
[2] F. H. Wong, Existence of positive solutions for m-Laplacian BVPs, Appl. Math.
Letters, 12(1999), 11–17.
[3] R. Manasevich and F. Zanolin, Time-mapping and multiplicitiy of solutions for the
one-dimensional p-Laplacian, Nonlinear Analysis, 21(4)(1993), 269–291.
[4] X. M. He and W. G. Ge, Existence of positive solutions for the one-dimensional
p-Laplacian equations, Acta Mathematica Sinica 46(4)(2003), 805–810.
[5] P. X. Weng and D. Q. Jing, Existence of positive solutions for boundary value
problem of second-order FDE, Computers and Mathematics with Applications,
37(1999), 1–9.
[6] C. Song and P. Weng, Existence of positive solutions for p-Laplacian singular boundary value problems of functional di¤erential equation, Appl. Math. J. Chinese Univ.
Ser. B, 20(1)(2005), 51–58.
[7] M. A. Krasnosel’skii, Positive Solutions of Operator Equations, Noordhoof, Groningen, 1964.