Electronic Journal of Differential Equations, Vol. 2000(2000), No. 40, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
ftp ejde.math.swt.edu ejde.math.unt.edu (login: ftp)
THREE SYMMETRIC POSITIVE SOLUTIONS FOR LIDSTONE
PROBLEMS BY A GENERALIZATION OF THE
LEGGETT-WILLIAMS THEOREM
RICHARD I. AVERY, JOHN M. DAVIS, & JOHNNY HENDERSON
Abstract. We study the existence of solutions to the fourth order Lidstone
boundary value problem
y (4) (t) = f (y(t), −y 00 (t)),
y(0) = y 00 (0) = y 00 (1) = y(1) = 0 .
By imposing growth conditions on f and using a generalization of the multiple
fixed point theorem by Leggett and Williams, we show the existence of at
least three symmetric positive solutions. We also prove analogous results for
difference equations.
1. Introduction
First we are concerned with the existence of multiple solutions for the fourth
order Lidstone boundary value problem (BVP)
y (4) (t) = f (y(t), −y 00 (t)),
00
0 ≤ t ≤ 1,
00
y(0) = y (0) = y (1) = y(1) = 0 ,
(1)
(2)
where f : R × R → [0, ∞) is continuous. We will impose growth conditions on f
which ensure the existence of at least three symmetric positive solutions of (1), (2).
There is much current attention focused on questions of positive solutions of
boundary value problems for ordinary differential equations, as well as for for finite
difference equations; see [2, 5, 6, 8, 10, 11, 12, 13, 17, 18, 19, 20] to name a few.
Much of this interest is due to the applicability of certain fixed point theorems of
Krasnosel’skii [15] or Leggett and Williams [16] to obtain positive solutions or multiple positive solutions which lie in a cone. The recent book by Agarwal, O’Regan,
and Wong [1] gives a good overview for much of the work which has been done and
the methods used.
Mathematics Subject Classification. 34B18, 34B27, 39A12, 39A99.
Key words. Lidstone boundary value problem, Green’s function,
multiple solutions, fixed points, difference equation.
c
2000
Southwest Texas State University and University of North Texas.
Submitted December 6, 1999. Published May 23, 2000.
1
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R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
EJDE–2000/40
In [3], Avery imposed conditions on g which yield at least three positive solutions
to the second order conjugate BVP
y 00 (t) + g(y(t)) = 0 ,
0 ≤ t ≤ 1,
y(0) = y(1) = 0 ,
(3)
(4)
using the Leggett-Williams Fixed Point Theorem. Henderson and Thompson [14]
improved these results by using the symmetry of the associated Green’s function
and then Avery and Henderson [5] established similar results by applying the Five
Functionals Fixed Point Theorem [4] (which is a generalization of the LeggettWilliams Fixed Point Theorem) to obtain the existence of three positive solutions
of certain BVPs. Davis, Eloe, and Henderson [8] imposed conditions on f to yield
at least three positive solutions to the 2mth order Lidstone BVPs by applying the
Leggett- Williams Fixed Point Theorem. Note that [8] is the only work which has
allowed f to depend on higher order derivatives of y. This paper is in the same
spirit as [5] and [8] since we apply the Five Functionals Fixed Point Theorem [4]
and also allow f to depend on y 00 . This derivative dependence generalizes [9] as
well.
In Section 2, we provide some background results and state the Five Functionals
Fixed Point Theorem. In Section 3, we impose growth conditions on f which allow
us to apply this theorem in obtaining three symmetric positive solutions of (1), (2).
In Section 4, we prove discrete analogs of the results in Section 3.
2. Some Background Definitions and Results
In this section, we provide some background material from the theory of cones
in Banach spaces in order that this paper be self-contained. We also state the Five
Functionals Fixed Point Theorem for cone preserving operators.
Definition 1. Let E be a Banach space over R. A nonempty, closed set P ⊂ E is
a cone provided
(a) αu + βv ∈ P for all u, v ∈ P and all α, β ≥ 0, and
(b) u, −u ∈ P implies u = 0.
Definition 2. A Banach space E is a partially ordered Banach space if there exists
a partial ordering on E satisfying
(a) u v, for u, v ∈ E implies tu tv, for all t ≥ 0, and
(b) u1 v1 and u2 v2 , for u1 , u2 , v1 , v2 ∈ E imply u1 + u2 v1 + v2 .
Let P ⊂ E be a cone and define u v if and only if v − u ∈ P. Then is a
partial ordering on E and we will say that is the partial ordering induced by P.
Moreover, E is a partially ordered Banach space with respect to .
We also state the following definitions for future reference.
Definition 3. The map α is a nonnegative continuous concave functional on P
provided α : P → [0, ∞) is continuous and
α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y)
for all x, y ∈ P and 0 ≤ t ≤ 1. Similarly we say the map β is a nonnegative
continuous convex functional on P provided β : P → [0, ∞) is continuous and
β(tx + (1 − t)y) ≤ tβ(x) + (1 − t)β(y)
for all x, y ∈ P and 0 ≤ t ≤ 1.
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THREE SYMMETRIC POSITIVE SOLUTIONS
3
Definition 4. An operator, A, is completely continuous if A is continuous and
compact, i.e. A maps bounded sets into precompact sets.
Let γ, β, θ be nonnegative continuous convex functionals on P and α, ψ be
nonnegative continuous concave functionals on P. Then for nonnegative numbers
h, a, b, d, and c, we define the following convex sets:
P (γ, c) = {x ∈ P | γ(x) < c},
P (γ, α, a, c) = {x ∈ P | a ≤ α(x), γ(x) ≤ c},
Q(γ, β, d, c) = {x ∈ P | β(x) ≤ d, γ(x) ≤ c},
P (γ, θ, α, a, b, c) = {x ∈ P | a ≤ α(x), θ(x) ≤ b, γ(x) ≤ c},
Q(γ, β, ψ, h, d, c) = {x ∈ P | h ≤ ψ(x), β(x) ≤ d, γ(x) ≤ c}.
In obtaining multiple symmetric positive solutions of (1), (2) the following socalled Five Functionals Fixed Point Theorem will be fundamental.
Theorem 1. [4] Let P be a cone in a real Banach space E. Suppose α and ψ be
nonnegative continuous concave functionals on P and γ, β, and θ be nonnegative
continuous convex functionals on P such that, for some positive numbers c and m,
α(x) ≤ β(x) and kxk ≤ mγ(x) for all x ∈ P (γ, c).
Suppose further that A : P (γ, c) → P (γ, c) is completely continuous and there exist
constants h, d, a, b ≥ 0 with 0 < d < a such that each of following is satisfied:
(C1) {x ∈ P (γ, θ, α, a, b, c) | α(x) > a} 6= ∅
and α(Ax) > a for x ∈ P (γ, θ, α, a, b, c),
(C2) {x ∈ Q(γ, β, ψ, h, d, c) | β(x) < d} 6= ∅
and β(Ax) < d for x ∈ Q(γ, β, ψ, h, d, c),
(C3) α(Ax) > a provided x ∈ P (γ, α, a, c) with θ(Ax) > b,
(C4) β(Ax) < d provided x ∈ Q(γ, β, d, c) with ψ(Ax) < h.
Then A has at least three fixed points x1 , x2 , x3 ∈ P (γ, c) such that
β(x1 ) < d, a < α(x2 ), and d < β(x3 ) with α(x3 ) < a.
3. Three Symmetric Positive Solutions
In this section, we will impose growth conditions on f which allow us to apply
Theorem 1 in regard to obtaining three symmetric positive solutions of (1), (2). We
will apply Theorem 1 in conjunction with a completely continuous operator whose
kernel, G(t, s), is the Green’s function for −v 00 = 0, satisfying (4). In particular,
(
G(t, s) =
t(1 − s), 0 ≤ t ≤ s ≤ 1,
s(1 − t), 0 ≤ s ≤ t ≤ 1.
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R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
EJDE–2000/40
We will make use of various properties of G(t, s), namely
Z 1
t(1 − t)
, 0 ≤ t ≤ 1,
G(t, s) ds =
2
0
Z r1
Z 1
1
G(1/2, s) ds =
G(1/2, s) ds = 2 , 2 < r,
1
4r
0
1− r
Z 12
Z 1− 1r
r2 − 4
G(1/2, s), ds =
G(1/2, s) ds =
, 2 < r,
1
1
16r2
r
2
Z 1−t1
Z t2
G(t1 , s) ds +
G(t1 , s) ds = t1 (t2 − t1 ), 0 < t1 < t2 < 1/2,
t1
(5)
(6)
(7)
1−t2
1
G(1/2, r)
= , 0 < t ≤ 1/2,
G(t, r)
2t
t1
G(t1 , r)
min
= , 0 < t1 < t2 ≤ 1/2.
0≤r≤1 G(t2 , r)
t2
max
0≤r≤1
(8)
(9)
Let E = C[0, 1] be endowed with the maximum norm,
kvk = max |v(t)| ,
0≤t≤1
and for 0 < t3 < 1/2 define the cone P ⊂ E by
P = v ∈ E | v(t) = v(1 − t), v(t) ≥ 0, v(t) is concave for all t ∈ [0, 1],
and
min |v(t)| ≥ 2t3 kvk .
t∈[t3 ,1−t3 ]
Finally, we define the nonnegative continuous concave functionals α, ψ and the
nonnegative continuous convex functionals β, θ, γ on P by
γ(v) =
ψ(v) =
β(v) =
α(v) =
θ(v) =
max
t∈[0,t3 ]∪[1−t3 ,1]
v(t) = v(t3 ),
min
v(t) = v(1/r),
max
v(t) = v(1/2),
t∈[ r1 ,1− r1 ]
t∈[ r1 ,1− r1 ]
min
v(t) = v(t1 ),
max
v(t) = v(t2 ),
t∈[t1 ,t2 ]∪[1−t2 ,1−t1 ]
t∈[t1 ,t2 ]∪[1−t2 ,1−t1 ]
where t1 , t2 , and r are nonnegative numbers such that
0 < t1 < t2 <
1
1
and ≤ t2 .
2
r
We observe here that for each v ∈ P,
α(v) = v(t1 ) ≤ v(1/2) = β(v),
kvk = v(1/2) ≤
v(t3 )
1
=
γ(v),
2t3
2t3
(10)
(11)
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THREE SYMMETRIC POSITIVE SOLUTIONS
5
and also that y ∈ P is a solution of (1), (2) if and only if there exists a v ∈ P such
that
Z 1
y(t) =
G(t, s)v(s) ds, 0 ≤ t ≤ 1,
0
where v is of the form
Z
Z 1
G(t, s)f
v(t) =
0
1
G(s, τ )v(τ )dτ, v(s) ds,
0
0 ≤ t ≤ 1.
In light of these preliminaries, we are now ready to present the main result of
this section.
Theorem 2. Suppose there exist 0 < a < b < tt21 b ≤ c such that f satisfies each of
the following growth conditions:
h
i 2 c
a
2a
a − r2 t3 (1−t
for all (z, w) ∈ a(r−2)
(G1) f (z, w) < r8r
2 −4
r3 , 8 × r , a ,
3)
i
h
c(t2 +t (1−2t ))
bt (t2 −t2 )
(G2) f (z, w) ≥ t1 (t2b−t1 ) for (z, w) ∈ bt1 (t2 − t1 ), 1 24t3 2 + 2 2t21 1 ×
h
i
b, tt21 b ,
i h
i
h
2c
c
c
×
0,
.
(G3) f (z, w) ≤ t3 (1−t
for
(z,
w)
∈
0,
)
16t
2t
3
3
3
Then the Lidstone BVP (1), (2) has at least three symmetric positive solutions y1 ,
y2 , y3 , such that
00
max
t∈[0,t3 ]∪[1−t3 ,1]
−yi (t) ≤ c,
for i = 1, 2, 3,
00
min
t∈[t1 ,t2 ]∪[1−t2 ,1−t1 ]
−y1 (t) > b,
00
max −y2 (t) < a,
t∈[ 1r ,1− 1r ]
and
min
00
t∈[t1 ,t2 ]∪[1−t2 ,1−t1 ]
−y3 (t) < b
with
for some v1 , v2 , v3 ∈ P satisfying
Z 1
G(t, s)vi (s) ds,
yi (t) =
00
max −y3 (t) > a,
t∈[ r1 ,1− r1 ]
i = 1, 2, 3.
0
Proof. Define the completely continuous operator A by
Z 1
G(t, s)f (Bv(s), v(s)) ds
Av(t) =
0
where
Z
1
Bv(s) =
G(s, τ )v(τ ) dτ.
0
We seek three fixed points v1 , v2 , v3 ∈ P of A which satisfy the conclusion of the
theorem. We note first that if v ∈ P, then from the properties of G(t, s), Av(t) ≥ 0,
Bv(t) ≥ 0, (Av)00 (t) = −f (B(v(t), v(t)) ≤ 0, 0 ≤ t ≤ 1, Av(t3 ) ≥ 2t3 Av(1/2), and
Av(t) = Av(1 − t), 0 ≤ t ≤ 1/2. Consequently, A : P → P.
Also, for all v ∈ P, by (10) we have α(v) ≤ β(v) and by (11), kvk ≤ 2t13 γ(v). If
v ∈ P (γ, c), then kvk ≤ 2t13 γ(v) ≤ 2tc3 , which implies that, for s ∈ [0, 1],
c
c
and Bv(s) ∈ 0,
.
v(s) ∈ 0,
2t3
16t3
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R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
EJDE–2000/40
By condition (G3) we have
Z
γ(Av) =
1
max
t∈[0,t3 ]∪[1−t3 ,1]
Z
=
1
G(t, s)f (Bv(s), v(s)) ds
0
G (t3 , s) f (Bv(s), v(s)) ds
Z 1
2c
G (t3 , s) ds
t3 (1 − t3 )
0
0
≤
= c.
Therefore, A : P (γ, c) → P (γ, c).
It is immediate that
t2
v ∈ P γ, θ, α, b, b, c | α(v) > b 6= ∅,
t1
2a
v ∈ Q γ, β, ψ, , a, c | β(v) < a 6= ∅,
r
and thus the first parts of (C1) and (C2) are satisfied.
In order to show the second part of (C1) holds, let v ∈ P (γ, θ, α, b, tt21 b, c). For
each s ∈ [t1 , t2 ] ∪ [1 − t2 , 1 − t1 ], we have b ≤ v(s) ≤ tt21 b. Hence
bt1 (t2 − t1 ) ≤ Bv(s) ≤
c(t21 + t2 (1 − 2t2 )) bt2 (t22 − t21 )
+
,
4t3
2t1
and by condition (G2) and (7) we see
Z
α(Av) =
1
min
G(t, s)f (Bv(s), v(s)) ds
t∈[t1 ,t2 ]∪[1−t2 ,1−t1 ]
Z
1
G (t1 , s) f (Bv(s), v(s)) ds
=
0
Z
t1
=
Z
t2
>
≥
0
G (t1 , s) f (Bv(s), v(s)) ds +
b
t1 (t2 − t1 )
b
t1 (t2 − t1 )
Z
t1
1−t2
t2
G (t1 , s) ds +
1−t1
G (t1 , s) f (Bv(s), v(s)) ds
b
t1 (t2 − t1 )
t1 [(1 − t1 )2 − (1 − t2 )2 ]
+
2
Z
1−t1
G (t1 , s) ds
1−t2
t1 (t22 − t21 )
2
= b.
To verify the second
part of (C2), let v ∈ Q(γ, β, ψ, 2a
r , a, c). This implies that
1
1
for each s ∈ r , 1 − r , we have
2a
,a
v(s) ∈
r
a(r − 2) a
and Bv(s) ∈
.
,
r3
8
EJDE–2000/40
THREE SYMMETRIC POSITIVE SOLUTIONS
7
Thus, by condition (G1) and the calculations in (5) and (6),
Z 1
β(Av) = max
G(t, s)f (Bv(s), v(s)) ds
t∈[ 1r ,1− r1 ] 0
Z 1
G (1/2, s) f (Bv(s), v(s)) ds
=
0
Z
=2
1
r
Z
G (1/2, s) f (Bv(s), v(s)) ds + 2
0
c
+
r2 t3 (1 − t3 )
= a.
<
8r2
2
r −4
a−
1
2
1
r
G (1/2, s) f (Bv(s), v(s)) ds
c
r2 t3 (1 − t3 )
r2 − 4
8r2
To show (C3) holds, suppose v ∈ P (γ, α, b, c) with θ(Av) > tt21 b. Using (9), we
get
Z 1
G(t, s)f (Bv(s), v(s)) ds
α(Av) =
min
t∈[t1 ,t2 ]∪[1−t2 ,1−t1 ]
Z
G (t1 , s) f (Bv(s), v(s)) ds
=
0
Z
0
1
1
G (t1 , s)
G (t2 , s) f (Bv(s), v(s)) ds
0 G (t2 , s)
Z
t1 1
G (t2 , s) f (Bv(s), v(s)) ds
≥
t2 0
t1
= θ(Ay)
t2
> b.
=
Finally, to show (C4), we take v ∈ Q(γ, β, a, c) with ψ(Av) <
have
Z 1
G(t, s)f (Bv(s), v(s)) ds
β(Av) = max
t∈[ r1 ,1− r1 ] 0
Z 1
G (1/2, s) f (Bv(s), v(s)) ds
=
Z
0
1
=
0
2a
r .
Using (8), we
G (1/2, s)
G (1/r, s) f (Bv(s), v(s)) ds
G (1/r, s)
Z
r 1
≤
G (1/r, s) f (Bv(s), v(s)) ds
2 0
r
= ψ(Ay)
2
< a.
Therefore the hypotheses of Theorem 1 are satisfied and there exist three positive
solutions y1 , y2 , and y3 for the Lidstone BVP (1), (2). Moreover, these solutions
are of the form
Z
1
yi (t) =
for some v1 , v2 , v3 ∈ P.
0
G(t, s)vi (s) ds,
i = 1, 2, 3,
8
R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
EJDE–2000/40
Remark. We have chosen to perform the analysis when f is autonomous. However,
if f = f (t, w, z) and in addition, for each (w, z), f (t, w, z) is symmetric about t = 12 ,
then an analogous theorem would be valid with respect to the same cone P.
4. Discrete Analogs
Motivated by the early multiple solutions results for difference equations [1, 2, 6]
and specifically papers involving difference equations satisfying Lidstone boundary
conditions [7], we want to extend Theorem 2 to discrete problems. To this end,
we will again impose growth conditions on f which allow us to apply Theorem 1
and obtain three symmetric positive solutions of the discrete fourth order Lidstone
BVP
∆4 y(t − 2) = f (y(t), −∆2 y(t − 1)),
2
a+2≤t≤b+2
2
y(a) = ∆ y(a) = 0 = ∆ y(b + 2) = y(b + 4),
(12)
(13)
where f : R × R → [0, ∞) is continuous. We will apply Theorem 1 in conjunction with a completely continuous operator whose kernel, G1 (t, s), is the Green’s
function for
−∆2 w(t − 1) = 0,
w(a + 1) = 0 = w(b + 3).
In particular,

(t − a − 1)(b + 3 − s)


, a +1 ≤ t ≤ s ≤ b+3,
b+2−a
G1 (t, s) =

 (s − a − 1)(b + 3 − t) , a + 1 ≤ s ≤ t ≤ b + 3 .
b+2−a
We will write our solutions of (12), (13) in the form
y(t) =
b+1
X
G0 (t, s)v(s)
s=a+1
where G0 (t, s) is the Green’s function for
−∆2 w(t − 1) = 0,
w(a) = 0 = w(b + 4),
and v is a fixed point of a completely continuous operator with kernel G1 (t, s). In
particular,

(t − a)(b + 4 − s)


, a ≤ t ≤ s ≤ b + 4,
b+4−a
G0 (t, s) =

 (s − a)(b + 4 − t) , a ≤ s ≤ t ≤ b + 4.
b+4−a
We will make use of various properties of G0 (t, s) and G1 (t, s), namely
b+3
X
G0 (t, s) =
(t − a)(b + 4 − t)
,
2
G1 (t, s) =
(t − a − 1)(b + 3 − t)
,
2
s=a+1
b+2
X
s=a+2
a ≤ t ≤ b + 4,
a + 1 ≤ t ≤ b + 3,
EJDE–2000/40
THREE SYMMETRIC POSITIVE SOLUTIONS
b+4−k
X3
G(tm , s) = C1 ,
9
(14)
s=a+k3
a+k
3 −1
X
G1 (tm , s) +
s=a+2
a+k
X2
G1 (tm , s) = C2 ,
(15)
s=b+5−k3
b+4−k
X1
G1 (t1 , s) +
s=a+k1
b+4−k
X3
b+3
X
G1 (t1 , s) = C3 ,
(16)
s=b+4−k2
G0 (t3 , s) = C4 ,
s=a+k3
a+k
X2
s=a+k1
a+k
1 −1
X
b+4−k
X1
G0 (t1 , s) +
G0 (tm , s) +
s=a+1
a+k
X2
G0 (t1 , s) = C5 ,
s=b+4−k2
b+3−k
X2
s=a+k2 +1
G0 (tm , s) +
s=a+k1
b+3
X
G0 (tm , s) +
b+4−k
X1
G0 (tm , s) = C6 ,
s=b+5−k1
G0 (tm , s) = C7 ,
s=b+4−k2
tm − a − 1
G1 (tm , r)
=
, a + 2 ≤ t ≤ tm ,
a+2≤r≤b+2 G1 (t, r)
t−a−1
t0 − a − 1
G1 (t0 , r)
=
, a + 2 ≤ t0 ≤ t00 ≤ tm ,
min
a+2≤r≤b+2 G1 (t00 , r)
t00 − a − 1
max
(17)
(18)
For completeness, we have calculated and included the values of the above constants.
(tm − a − 1)(b + 3 − tm ) − (t3 − a − 1)(2)
,
2
(t3 − a − 1)(2)
,
C2 =
2
(t1 −a)[(t2 −a)(2) +(b+4−t1)(2) −(t1 −a−1)(2) −(b+3−t2)(2) ]
,
C3 =
2(b + 2 − a)
C1 =
C4 =
(t3 − a)[(b + 5 − t3 )(2) − (t3 − a)(2) ]
,
2(b + 4 − a)
C5 =
(t1 −a)[(t2 +1−a)(2) +(b+5−t1)(2) −(t1 −a)(2) −(b+4−t2)(2) ]
,
2(b + 4 − a)
(b + 4 − tm )(tm − a) + (t1 − a)(2) − (t2 − a + 1)(2)
,
2
(t2 − a + 1)(2) − (t1 − a)(2)
,
C7 =
2
C6 =
Define
tm
b+4+a
=
2
10
R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
and
EJDE–2000/40
b+4−a
.
M=
2
Let E1 = {y | y : [a, b + 4] → R} be endowed with the maximum norm,
kyk1 =
max |y(t)| ,
a≤t≤b+4
and for 2 ≤ k0 ≤ M , let t0 = a + k0 and define the cone P1 ⊂ E1 by


y ∈ E1 such that y(a + k) = y(b + 4 − k) for all k ∈ [0, M ], 







y(t) ≥ 0, ∆2 y(t − 1) ≤ 0 for all a + 1 ≤ t ≤ b + 3,
.
P1 =


t0 − a




 and

kyk1
min
|y(t)| ≥
t∈[a+k0 ,b+4−k0 ]
tm − a
Similarly let E0 = {y | y : [a + 1, b + 3] → R} be endowed with the maximum norm,
kvk0 =
max
a+1≤t≤b+3
|v(t)| ,
and define the cone P0 ⊂ E0 by


v ∈ E0 such that v(a + 1 + k) = v(b + 3 − k) for all k ∈ [0, M − 1], 







v(t) ≥ 0, ∆2 v(t − 1) ≤ 0 for all a + 2 ≤ t ≤ b + 2,
.
P0 =


t0 − a − 1




 and

kvk0
min
|v(t)| ≥
t∈[a+k0 ,b+4−k0 ]
tm − a − 1
Finally, we define the nonnegative continuous concave functionals α, ψ and the
nonnegative continuous convex functionals β, θ, γ on P0 by
γ(v) =
ψ(v) =
β(v) =
α(v) =
θ(v) =
max
t∈[a+1,a+k0 ]∪[b+4−k0 ,b+3]
v(t) = v(t0 ),
min
v(t) = v(t3 ),
max
v(t) = v(tm ),
t∈[a+k3 ,b+4−k3 ]
t∈[a+k3 ,b+4−k3 ]
min
v(t) = v(t1 ),
max
v(t) = v(t2 ),
t∈[a+k1 ,a+k2 ]∪[b+4−k2 ,b+4−k1 ]
t∈[a+k1 ,a+k2 ]∪[b+4−k2 ,b+4−k1 ]
where t1 = a + k1 , t2 + a + k2 ,and t3 = a + k3 are nonnegative numbers such that
k1 , k2 , k3 ∈ [2, M ] and k1 ≤ k2 .
We observe here that for each v ∈ P0 ,
α(v) = v(t1 ) ≤ v(tm ) = β(v),
tm − a − 1
tm − a − 1
v(t0 ) =
γ(v),
kvk0 = v(tm ) ≤
t0 − a − 1
t0 − a − 1
(19)
(20)
and also that y ∈ P1 is a solution of (12), (13) if and only if there exists a v ∈ P0
such that
b+3
X
y(t) =
G0 (t, s)v(s), a ≤ t ≤ b + 4,
s=a+1
EJDE–2000/40
THREE SYMMETRIC POSITIVE SOLUTIONS
where v is of the form
v(t) =
b+2
X
!
b+3
X
G1 (t, s)f
s=a+2
11
G0 (s, τ )v(τ ), v(s) ,
a + 1 ≤ t ≤ b + 3.
τ =a+1
In light of these preliminaries, we are now ready to present the main result of
this section.
0
0
Theorem 3. Suppose there exist 0 < a0 < b0 < tt21 −a−1
−a−1 b ≤ c such that f
satisfies each of the following growth conditions:
0
0
2c C2
(Γ1) f (z, w) < Ca1 − (t0 −a−1)(b+3−t
for all (z, w) in
0 )C1
0
0
a (t3 − a − 1) 0
a (t3 − a − 1)C4 a0 (tm − a)(b + 4 − tm )
,
×
,a
(tm − a − 1)
2
tm − a − 1
0
(Γ2) f (z, w) < Cb 3 for all (z, w) in
2c0 C6
t2 − a − 1
b0 (t2 − a − 1)C7
b0 C5 ,
+
× b,
b
2(t1 − a − 1)
(t0 − a − 1)(b + 3 − t0 )
t1 − a − 1
0
2c
(Γ3) f (z, w) ≤ (t0 −a−1)(b+3−t
for all (z, w) in
0)
0
0
c (tm − a − 1)
c (tm − a)(2) (b + 4 − tm )
× 0,
.
0,
2(t0 − a − 1)
t0 − a − 1
Then the Lidstone BVP (12), (13) has at least three symmetric positive solutions
y1 , y2 , y3 ∈ P1 , such that
max
t∈[a+1,a+k0 ]∪[b+4−k0 ,b+3]
min
t∈[a+k1 ,a+k2 ]∪[b+4−k2 ,b+4−k1 ]
max
t∈[a+k3 ,b+4−k3 ]
−∆2 yi (t − 1) ≤ c0 ,
for i = 1, 2, 3,
−∆2 y1 (t − 1) > b0 ,
−∆2 y2 (t − 1) < a0 ,
and
min
t∈[a+k1 ,a+k2 ]∪[b+4−k2 ,b+4−k1 ]
−∆2 y3 (t − 1) < b0
with
max
t∈[a+k3 ,b+4−k3 ]
−∆2 y3 (t − 1) > a0 ,
for some v1 , v2 , v3 ∈ P0 satisfying
yi (t) =
b+2
X
G0 (t, s)vi (s),
i = 1, 2, 3.
s=a+2
Proof. Define the completely continuous operator A by
Av(t) =
b+2
X
G1 (t, s)f (Bv(s), v(s)),
a + 1 ≤ t ≤ b + 3,
s=a+2
where
Bv(s) =
b+3
X
τ =a+1
G0 (s, τ )v(τ ).
12
R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
EJDE–2000/40
We seek three fixed points v1 , v2 , v3 ∈ P0 of A which satisfy the conclusion of the
theorem. We note first that if v ∈ P0 , then from the properties of G1 (t, s) and
G0 (t, s), it follows that Av(t) ≥ 0, Bv(t) ≥ 0, and
∆2 (Av)(t − 1) = −f (Bv(t), v(t)) ≤ 0, a + 2 ≤ t ≤ b + 2,
∆2 (Bv)(t − 1) = −v(t) ≤ 0,
a + 2 ≤ t ≤ b + 2,
Moreover, Av(t0 ) ≥ ttm0 −a−1
−a−1 Av(tm ), and Av(a+k) = Av(b+4−k) for 1 ≤ k ≤ M .
Consequently, A : P0 → P0 .
Also, for all v ∈ P0 , by (19) we have α(v) ≤ β(v) and by (20),
tm − a − 1
γ(v).
kvk0 ≤
t0 − a − 1
If v ∈ P (γ, c0 ), then
kvk0 ≤
tm − a − 1
t0 − a − 1
γ(v) ≤
c0 (tm − a − 1)
,
(t0 − a − 1
which implies that, for s ∈ [a + 1, b + 3],
0
0
c (tm − a)(2) (b + 4 − tm )
c (tm − a − 1)
and Bv(s) ∈ 0,
.
v(s) ∈ 0,
t0 − a − 1
2(t0 − a − 1)
By condition (Γ3) we have
γ(Av) =
=
b+2
X
max
t∈[a+1,a+k0 ]∪[b+4−k0 ,b+3]
b+2
X
G1 (t0 , s)f (Bv(s), v(s))
s=a+2
≤
G1 (t, s)f (Bv(s), v(s))
s=a+2
2c0
t0 − a − 1)(b + 3 − t0 )
X
b+2
G1 (t0 , s)
s=a+2
= c0 .
Therefore, A : P (γ, c0 ) → P (γ, c0 ).
It is immediate that
0
0 b (t2 − a − 1) 0
0
, c ) : α(v) > b 6= ∅,
v ∈ P (γ, θ, α, b ,
t1 − a − 1
a0 (t3 − a − 1) 0 0
0
v ∈ Q(γ, β, ψ,
, a , c ) : β(v) < a 6= ∅,
tm − a − 1
and thus the first parts of (C1) and (C2) are satisfied.
0
2 −a−1)
0
In order to show the second part of (C1) holds, let v ∈ P (γ, θ, α, b0 , b (t
t1 −a−1 , c ).
For each s ∈ [a + k1 , a + k2 ] ∪ [b + 4 − k2 , b + 4 − k1 ], we have
b0 ≤ v(s) ≤
Hence
b0 C5 ≤ Bv(s) ≤
b0 (t2 − a − 1)
.
t1 − a − 1
2c0 C6
b0 (t2 − a − 1)C7
+
,
2(t1 − a − 1)
(t0 − a − 1)(b + 3 − t0 )
EJDE–2000/40
THREE SYMMETRIC POSITIVE SOLUTIONS
13
and by condition (Γ2) and (16) we see
α(Av) =
b+2
X
min
t∈[a+k1 ,a+k2 ]∪[b+4−k2 ,b+4−k1 ]
b+2
X
=
G1 (t, s)f (Bv(s), v(s))
s=a+2
G1 (t1 , s)f (Bv(s), v(s))
s=a+2
a+k
X2
≥
G1 (t1 , s)f (Bv(s), v(s)) +
s=a+k1
>
b0
C3
b+4−k
X1
s=b+4−k2
a+k
X2
G1 (t1 , s) +
s=a+k1
b+4−k
X1
G1 (t1 , s)f (Bv(s), v(s))
!
G1 (t1 , s)
s=b+4−k2
= b0 .
0
−a−1)
, a0 , c0 ). This
To verify the second part of (C2), let v ∈ Q(γ, β, ψ, a t(tm3−a−1
implies that for each s ∈ [a + k3 , b + 4 − k3 ], we have
a0 (t3 − a − 1)
≤ v(s) ≤ a0
tm − a − 1
and
a0 (tm − a)(b + 4 − tm )
a0 (t3 − a − 1)C4
≤ Bv(s) ≤
.
(tm − a − 1)
2
Thus, by condition (Γ1) and the calculations in (14) and (15),
β(Av) =
=
max
t∈[a+k3 ,b+4−k3 ]
b+2
X
b+2
X
G1 (t, s)f (Bv(s), v(s))
s=a+2
G1 (tm , s)f (Bv(s), v(s))
s=a+2
=
a+k
3 −1
X
G1 (tm , s)f (Bv(s), v(s)) +
s=a+2
b+2
X
+
G1 (tm , s)f (Bv(s), v(s))
s=a+k3
G1 (tm , s)f (Bv(s), v(s))
s=b+5−k3
b+4−k
X3
2c0
<
(t0 − a − 1)(b + 3 − t0 )
0
=a.
C2 +
2c0 C2
a0
−
C1
(t0 − a − 1)(b + 3 − t0 )C1
C1
14
R.I. AVERY, J.M. DAVIS, AND J. HENDERSON
To show (C3) holds, suppose v ∈ P (γ, α, b0 , c0 ) with θ(Av) >
(18), we get
α(Av) =
=
b+2
X
min
t∈[a+k1 ,a+k2 ]∪[b+4−k2 ,b+4−k1 ]
b+2
X
EJDE–2000/40
b0 (t2 −a−1)
t1 −a−1 .
Using
G1 (t, s)f (Bv(s), v(s))
s=a+2
G1 (t1 , s)f (Bv(s), v(s))
s=a+2
=
b+2 X
G1 (t1 , s)
G1 (t2 , s)f (Bv(s), v(s))
G1 (t2 , s)
s=a+2
≥
b+2
t1 − a − 1 X
G1 (t2 , s)f (Bv(s), v(s))
t2 − a − 1 s=a+2
t1 − a − 1
θ(Av)
t2 − a − 1
> b0 .
=
Finally, to show (C4), we take v ∈ Q(γ, β, a0 , c0 ) with ψ(Av) <
(17), we have
β(Av) =
=
max
b+2
X
t∈[a+k3 ,b+4−k3 ]
b+2
X
a0 (t3 −a−1)
tm −a−1 .
Using
G1 (t, s)f (Bv(s), v(s))
s=a+2
G1 (tm , s)f (Bv(s), v(s))
s=a+2
b+2 X
G1 (tm , s)
G1 (t3 , s)f (Bv(s), v(s))
=
G1 (t3 , s)
s=a+2
X
b+2
tm − a − 1
≤
G1 (t3 , s)f (Bv(s), v(s))
t3 − a − 1 s=a+2
tm − a − 1
ψ(Av)
=
t3 − a − 1
< a0 .
The hypotheses of Theorem 1 are satisfied and as a result there exist three positive
solutions y1 , y2 , and y3 ∈ P1 for the Lidstone BVP (12), (13). Moreover, these
solutions are of the form
yi (t) =
b+1
X
G0 (t, s)vi (s),
i = 1, 2, 3,
s=a+1
for some v1 , v2 , v3 ∈ P0 .
Remark. We have chosen to perform the analysis when f is autonomous. However,
if f = f (t, w, z) and in addition, for each (w, z), f (t, w, z) is symmetric about
t = tm , then an analogous theorem would be valid with respect to the same cone
P0 .
EJDE–2000/40
THREE SYMMETRIC POSITIVE SOLUTIONS
15
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Richard I. Avery
College of Natural Sciences, Dakota State University, Madison, SD 57042, USA
E-mail address: averyr@pluto.dsu.edu
John M. Davis
Department of Mathematics, Baylor University, Waco, TX 76798, USA
E-mail address: John M Davis@baylor.edu
Johnny Henderson
Department of Mathematics, Auburn University, Auburn, AL 36849, USA
E-mail address: hendej2@mail.auburn.edu