Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 61, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLUTIONS TO THIRD-ORDER MULTI-POINT
BOUNDARY-VALUE PROBLEMS AT RESONANCE WITH
THREE DIMENSIONAL KERNELS
SHUANG LI, JIAN YIN, ZENGJI DU
Abstract. In this article, we consider the boundary-value problem
x000 (t) = f (t, x(t), x0 (t), x00 (t)),
x00 (0) =
m
X
αi x00 (ξi ),
x0 (0) =
i=1
l
X
γk x0 (σk ),
t ∈ (0, 1),
x(1) =
n
X
βj x(ηj ),
j=1
k=1
where f : [0, 1] × R3 → R is a Carathéodory function, and the kernel to the
linear operator has dimension three. Under some resonance conditions, by
using the coincidence degree theorem, we show the existence of solutions. An
example is given to illustrate our results.
1. Introduction
This concerns the third-order nonlinear differential equation
x000 (t) = f (t, x(t), x0 (t), x00 (t)),
t ∈ (0, 1),
(1.1)
with the boundary conditions
x00 (0) =
m
X
i=1
αi x00 (ξi ),
x0 (0) =
l
X
γk x0 (σk ),
x(1) =
n
X
βj x(ηj ),
(1.2)
j=1
k=1
where f : [0, 1] × R3 → R is a Carathéodory function, 0 < ξ1 < · · · < ξm < 1,
0 < σ1 < · · · < σl < 1, 0 < η1 < · · · < ηn < 1, αi , γk , βj ∈ R (i = 1, . . . , m;
k = 1, . . . , l; j = 1, . . . , n) and σ1 > {ξ1 , . . . , ξm }.
The existence of solutions for multi-point boundary-value problems at resonance
case has been extensively studied by many authors [1, 2, 3, 4, 5, 6, 7, 8, 10]. When
the linear equation Lx = x000 = 0 with the boundary conditions (1.2) has a nontrivial solution, i.e. dim ker L ≥ 1. we say that boundary value problem (BVP for
short) (1.1) and (1.2) is a resonance problem.
The case of dim ker L = 1 has been discussed by many authors [1, 2, 4, 6]. For the
case of dim ker L = 2, there are some results in [3, 7, 8, 10]., Lin, Du and Meng [7]
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. Multi-point boundary-value problem; coincidence degree theory;
resonance.
c
2014
Texas State University - San Marcos.
Submitted October 13, 2013. Published March 5, 2014.
1
2
S. LI, J. YIN, Z. DU
EJDE-2014/61
discussed the third-order multi-point boundary value-problem with dim ker L = 2,
x000 (t) = f (t, x(t), x0 (t), x00 (t)),
x00 (0) =
m
X
αi x00 (ξi ),
x0 (0) = 0,
t ∈ (0, 1),
x(1) =
i=1
n
X
βj x(ηj ).
(1.3)
(1.4)
j=1
Zhao, Liang and Ren [11] studied the nonlinear third-order boundary-value problems with dim ker L = 3,
x000 (t) = f (t, x(t), x0 (t), x00 (t)) + e(t),
x0 (0) =
n
X
αj x0 (ξj ),
x00 (0) = x00 (η),
t ∈ (0, 1),
x00 (1) = x00 (ζ).
j=1
In [11], the results are obtained under
m11
M = m21
m31
the assumption that
m12 m13 m22 m23 6= 0.
m32 m33 (1.5)
This condition is difficult to verify and looks superfluous.
Inspired by the above results, in this paper, we study (1.1)-(1.2) at resonance
and show that the assumption
Λ(p, q, r)
(p + 1)(p + 2) Pm α ξ p
i i
Pi=1
= (q + 1)(q + 2) m
α
ξiq
i
i=1
Pm
r
(r + 1)(r + 2)
i=1 αi ξi
Pl
(p + 2) k=1 γk σkp+1
Pl
(q + 2) k=1 γk σkq+1
Pl
(r + 2) k=1 γk σkr+1
Pn
2(1 − j=1 βj ηjp+2 )
Pn
2(1 − j=1 βj ηjq+2 ) 6= 0
Pn
2(1 − j=1 βj ηjr+2 )
can replace (1.5) from [5, 11], by using Lemma 3.1 below. If there exists σk (k =
Pl
1, 2, . . . , l) satisfying x0 (0) = k=1 γk x0 (σk ) = 0, then BVP (1.3)-(1.4) is a special
case of BVP (1.1)-(1.2).
The remaining part of this article is organized as follows: In section 2, we will
state some definitions and lemmas which would be useful in the proving of main
results of this paper. In section 3, by applying Mawhin coincidence degree theory,
we obtain some sufficient conditions which guarantee the existence of solution for
BVP (1.1)-(1.2) at resonance case. In the last section, an example is given to
illustrate our results.
2. Preliminaries
Now, some notation and an abstract existence result [9] are introduced. Let Y, Z
be real Banach spaces and let L : dom L ⊂ Y → Z be a linear operator which is a
Fredholm map of index zero and P : Y → Y , Q : Z → Z be continuous projectors
such that Im P = ker L, ker Q = Im L and Y = ker L ⊕ ker P , Z = Im L ⊕ Im Q.
It follows that L|dom L∩ker P : dom L ∩ ker P → Im L is invertible, we denote the
inverse of that map by KP . Let Ω be an open bounded subset of Y such that
dom L ∩ Ω 6= ∅, the map N : Y → Z is said to be L-compact on Ω̄ if the map
QN (Ω̄) is bounded and KP (I − Q)N : Ω̄ → Y is compact.
Lemma 2.1 ([9, Theorem IV]). Let L be a Fredholm map of index zero and let N
be L-compact on Ω̄. Assume that the following conditions are satisfied:
(i) Lx 6= λN x for every (x, λ) ∈ ((dom L\ ker L) ∩ ∂Ω) × [0, 1];
EJDE-2014/61
SOLUTIONS TO THIRD-ORDER MULTI-POINT BVPS
3
(ii) N x ∈
/ Im L for every x ∈ ker L ∩ ∂Ω;
(iii) deg(QN |ker L , Ω ∩ ker L, 0) 6= 0, where Q : Z → Z is a continuous projector
as above with Im L = ker Q.
Then the abstract equation Lx = N x has at least one solution in dom L ∩ Ω̄.
We use the classical spaces C[0, 1], C 1 [0, 1], C 2 [0, 1] and L1 [0, 1]. For x in C 2 [0, 1],
we use the norms kxk∞ = maxt∈(0,1) {|x(t)|} and
kxk = max{kxk∞ , kx0 k∞ , kx00 k∞ }.
For L1 [0, 1] we denote the norm by k · k1 . We define the Sobolev space
W 3,1 (0, 1) = x : [0, 1] → R : x, x0 , x00 are absolutely continuous on [0, 1]
with x000 ∈ L1 [0, 1] .
Let Y = C 2 [0, 1], Z = L1 [0, 1], and define the linear operator L : dom L ⊂ Y → Z
as Lx = x000 , x ∈ dom L, where
dom L = {x ∈ W 3,1 (0, 1) : x satisfies boundary conditions (1.2)}.
We define N : Y → Z as
N x = f (t, x(t), x0 (t), x00 (t)),
t ∈ (0, 1),
then (1.1)-(1.2) can be written as Lx = N x.
The resonance conditions of (1.1)-(1.2) are as follows
m
X
αi = 1,
i=1
n
X
βj ηj = 1,
j=1
l
X
γk = 1,
j=1
βj ηj2
βj = 1,
j=1
k=1
n
X
n
X
= 1,
l
X
(2.1)
γk σk = 0.
k=1
Define the following symbols:
Λ(p, q, r)
(p + 1)(p + 2) Pm α ξ p (p + 2) Pl γ σ p+1 2(1 − Pn β η p+2 )
i=1 i i
k=1 k k
j=1 j j
P
Pl
Pn
q
q+1
q+2 = (q + 1)(q + 2) m
α
ξ
(q
+
2)
γ
σ
2(1
−
β
η
)
,
i
k
j
i
j
k
Pi=1
Pk=1
Pj=1
(r + 1)(r + 2) m αi ξir (r + 2) l γk σ r+1 2(1 − n βj ηjr+2 )
i=1
k=1
j=1
k
(q + 2) Pl γ σ q+1 2(1 − Pn β η q+2 )
k=1 k k
j=1 j j
P
P
M11 = ,
(r + 2) lk=1 γk σkr+1 2(1 − nj=1 βj ηjr+2 )
(q + 1)(q + 2) Pm α ξ q 2(1 − Pn β η q+2 )
j
i
i
j
Pi=1
Pnj=1
M12 = − r+2 ,
r
(r + 1)(r + 2) m
α
ξ
2(1
−
β
η
)
i
j
i
j
i=1
j=1
Pm
Pl
q
q+1 (q + 1)(q + 2) i=1 αi ξi (q + 2) k=1 γk σk Pm
P
M13 = ,
(r + 1)(r + 2) i=1 αi ξir (r + 2) lk=1 γk σkr+1 (p + 2) Pl γ σ p+1 2(1 − Pn β η p+2 )
k=1 k k
j=1 j j
P
P
M21 = − ,
(r + 2) lk=1 γk σkr+1 2(1 − nj=1 βj ηjr+2 )
(p + 1)(p + 2) Pm α ξ p 2(1 − Pn β η p+2 )
j j
Pi=1 i ir
Pj=1
M22 = n
r+2 ,
(r + 1)(r + 2) m
α
ξ
2(1
−
β
η
)
i=1 i i
j=1 j j
4
S. LI, J. YIN, Z. DU
EJDE-2014/61
Pl
Pm
p+1 p
(p + 1)(p + 2) i=1 αi ξi (p + 2) k=1 γk σk P
Pl
M23 = − ,
r
(r + 1)(r + 2) m
(r + 2) k=1 γk σkr+1 i=1 αi ξi
(p + 2) Pl γ σ p+1 2(1 − Pn β η p+2 )
k=1 k k
j=1 j j
P
P
M31 = ,
(q + 2) lk=1 γk σkq+1 2(1 − nj=1 βj ηjq+2 )
(p + 1)(p + 2) Pm α ξ p 2(1 − Pn β η p+2 )
j
i
j
i
Pi=1
Pj=1
M32 = − n
q
q+2 ,
(q + 1)(q + 2) m
2(1 − j=1 βj ηj )
i=1 αi ξi
Pm
Pl
p
p+1 (p + 1)(p + 2) i=1 αi ξi (p + 2) k=1 γk σk Pl
P
M33 = q
(q + 1)(q + 2) m
(q + 2) k=1 γk σkq+1 i=1 αi ξi
3. Main results
Lemma 3.1. Assume condition (2.1) holds, then there exist p ∈ {1, 2, . . . , n},
q ∈ Z + , q ≥ p + 1 and r ∈ Z + large enough number, such that Λ(p, q, r) 6= 0.
Pm
Proof. Clearly there exists p ∈ {1, 2, . . . , n} such that i=1 αi ξip+2 6= 0. Otherwise,
Pm
we have i=1 αi ξip+2 = 0 and p ∈ {1, 2, . . . , n}, then
m
X
αi ξij (1 − ξi ) = 0,
j = 1, 2, . . . , n − 1;
i=1
i.e.,

ξ1 (1 − ξ1 )
...

..
..

.
.
ξ1n−1 (1 − ξ1 ) . . .
Because
ξ1 (1 − ξ1 )
...
..
..
.
n−1 .
ξ
(1
−
ξ
)
.
.
.
1
1
   
ξm (1 − ξm )
α1
0
  ..   .. 
..
  .  = . .
.
n−1
αm
0
ξm (1 − ξm )
m
Y
Y
ξi (1 − ξi )
(ξj − ξi ) 6= 0.
=
n−1
1≤i≤j≤m−1
(1 − ξm ) i=1
ξm
Pm
So we have αi = 0, i ∈ {1, 2, . . . , m}. Which is a contradiction to i=1 αi = 1 of
condition (2.1).
Pm
Similarly, there exists q ∈ {1, 2, . . . , n + 1}, such that i=1 αi ξiq 6= 0. And for
each s ∈ Z, s ≥ 0, there exists ks ∈ {sn + 1, . . . , (s + 1)n} such that
ξm (1 − ξm )
..
.
l
X
k=1
γk σkks +1 6= 0,
m
X
αi ξiks 6= 0.
i=1
Set
Pm
Pl
l
n
X
(q + 1) i=1 αi ξiq k=1 γk σkks +1 o
S = ks ∈ Z :
γk σkq+1 =
,
Pm
(ks + 1) i=1 αi ξiks
k=1
then S is a finite set. If else, there exists a monotone sequence {kst }, t = 1, 2, . . . ,
kst ≤ kst+1 , such that
Pm
Pl
k +1
l
X
(q + 1) i=1 αi ξiq k=1 γk σk st
q+1
γk σk =
.
Pm
k
(kst + 1) i=1 αi ξi st
k=1
EJDE-2014/61
SOLUTIONS TO THIRD-ORDER MULTI-POINT BVPS
5
For σ1 > {ξ1 , . . . , ξm },
l
X
γk σkq+1
= lim
kst →∞
k=1
Pl
k +1
αi ξiq k=1 γk σk st
= ∞,
Pm
k
(kst + 1) i=1 αi ξi st
(q + 1)
Pm
which is a contradiction. Then
Pm
p
(p + 1)(p + 2) i=1 αi ξi
Pm
q
(q + 1)(q + 2) i=1 αi ξi
i=1
Pl
(p + 2) k=1 γk σkp+1 Pl
q+1 6= 0
(q + 2) k=1 γk σk Thus, we have
lim Λ(p, q, r)
(p + 1)(p + 2) Pm α ξ p (p + 2) Pl γ σ p+1 2(1 − Pn β η p+2 )
i
k
j
i
j
k
Pi=1
Plk=1
Pj=1
n
q
q+1
q+2 = (q + 1)(q + 2) m
(q + 2) k=1 γk σk
2(1 − j=1 βj ηj )
i=1 αi ξi
0
0
2
Pm
Pl
p
p+1 (p + 1)(p + 2) i=1 αi ξi (p + 2) k=1 γk σk Pm
P
= 2
6= 0.
(q + 1)(q + 2) i=1 αi ξiq (q + 2) lk=1 γk σkq+1 r→∞
So there exist p ∈ {1, 2, . . . , n}, q ∈ Z + , q ≥ p + 1 and r ∈ Z + large enough
number, such that Λ(p, q, r) 6= 0.
Lemma 3.2. If condition (2.1) holds, then L : dom L ⊂ Y → Z is a Fredholm
operator of index zero. Furthermore, the continuous projector operator Q : Z → Z
can be defined by
Q(y) = (T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1 ,
where p, q, r are given by Lemma 3.1 and
p(p + 1)(p + 2)
[M11 Q1 y + M12 Q2 y + M13 Q3 y],
Λ(p, q, r)
q(q + 1)(q + 2)
[M21 Q1 y + M22 Q2 y + M23 Q3 y],
T2 y =
Λ(p, q, r)
r(r + 1)(r + 2)
T3 y =
[M31 Q1 y + M32 Q2 y + M33 Q3 y],
Λ(p, q, r)
Z σk
m Z ξi
l
X
X
Q1 y =
y(s)ds, Q2 y =
γk
(σk − s)y(s)ds,
T1 y =
0
i=1
Z
Q3 y =
1
(1 − s)2 y(s)ds −
0
k=1
n
X
0
Z
βj
j=1
ηj
(ηj − s)2 y(s)ds.
0
The linear operator KP : Im L → dom L ∩ ker P can be written as
Z
1 t
KP y(t) =
(t − s)2 y(s)ds, y ∈ Im L.
2 0
Furthermore kKP yk ≤ kyk1 , y ∈ Im L.
Proof. It is clear that
ker L = {x ∈ dom L : x = a + bt + ct2 , a, b, c ∈ R}.
6
S. LI, J. YIN, Z. DU
EJDE-2014/61
Now we show that
Im L = {y ∈ Z : Q1 y = Q2 y = Q3 y = 0}.
(3.1)
x000 = y
(3.2)
Since the problem
has a solution x(t), such that boundary conditions (1.2) hold; i.e.,
x00 (0) =
m
X
αi x00 (ξi ),
x0 (0) =
i=1
l
X
γk x0 (σk ),
x(1) =
n
X
βj x(ηj ),
j=1
k=1
if and only if
Q1 y = Q2 y = Q3 y = 0.
(3.3)
In fact, if (3.2) has a solution x(t) that satisfies the boundary conditions (1.2), then
we have
Z
1 t
1
(t − s)2 y(s)ds.
x(t) = x(0) + x0 (0)t + x00 (0)t2 +
2
2 0
According to condition (2.1), we have Q1 y = Q2 y = Q3 y = 0.
On the other hand, we let
Z
1 t
x(t) = a + bt + ct2 +
(t − s)2 y(s)ds,
2 0
where a, b, c are arbitrary constants. If (3.3) holds, then x(t) is a solution of (3.2)
and (1.2). Hence (3.1) holds. By Lemma 3.1, there exists p ∈ {1, 2, . . . , n}, q ∈
Z + , q ≥ p + 1 and r ∈ Z + is a large enough number, such that Λ(p, q, r) 6= 0. Set
p(p + 1)(p + 2)
[M11 Q1 y + M12 Q2 y + M13 Q3 y],
Λ(p, q, r)
q(q + 1)(q + 2)
[M21 Q1 y + M22 Q2 y + M23 Q3 y],
T2 y =
Λ(p, q, r)
r(r + 1)(r + 2)
T3 y =
[M31 Q1 y + M32 Q2 y + M33 Q3 y].
Λ(p, q, r)
T1 y =
And we define
Q(y) = (T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1 ,
then dim Im Q = 3. So we have
T1 ((T1 y)tp−1 )
p(p + 1)(p + 2)
[M11 Q1 ((T1 y)tp−1 ) + M12 Q2 ((T1 y)tp−1 ) + M13 Q3 ((T1 y)tp−1 )]
Λ(p, q, r)
p(p + 1)(p + 2)
=
[M11 Q1 (tp−1 ) + M12 Q2 (tp−1 ) + M13 Q3 (tp−1 )](T1 y)
Λ(p, q, r)
=
=
m
l
X
X
1
[M11 (p + 1)(p + 2)
αi ξip + M12 (p + 2)
γk σkp+1
Λ(p, q, r)
i=1
k=1
+ 2M13 (1 −
n
X
j=1
= T1 y,
βj ηjp+2 )](T1 y)
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7
T1 ((T2 y)tq−1 )
q(q + 1)(q + 2)
[M11 Q1 ((T2 y)tq−1 ) + M12 Q2 ((T2 y)tq−1 ) + M13 Q3 ((T2 y)tq−1 )]
Λ(p, q, r)
q(q + 1)(q + 2)
=
[M11 Q1 (tq−1 ) + M12 Q2 (tq−1 ) + M13 Q3 (tq−1 )](T2 y)
Λ(p, q, r)
=
=
m
l
X
X
1
[M11 (q + 1)(q + 2)
αi ξiq + M12 (q + 2)
γk σkq+1
Λ(p, q, r)
i=1
k=1
+ 2M13 (1 −
n
X
βj ηjq+2 )](T2 y) = 0,
j=1
T1 ((T3 y)tr−1 )
r(r + 1)(r + 2)
[M11 Q1 ((T3 y)tr−1 ) + M12 Q2 ((T3 y)tr−1 ) + M13 Q3 ((T3 y)tr−1 )]
Λ(p, q, r)
r(r + 1)(r + 2)
[M11 Q1 (tr−1 ) + M12 Q2 (tr−1 ) + M13 Q3 (tr−1 )](T3 y)
=
Λ(p, q, r)
=
=
l
m
X
X
1
γk σkr+1
αi ξir + M12 (r + 2)
[M11 (r + 1)(r + 2)
Λ(p, q, r)
i=1
k=1
+ 2M13 (1 −
n
X
βj ηjr+2 )](T3 y) = 0.
j=1
Similarly, we obtain
T2 ((T1 y)tp−1 ) = 0,
T2 ((T2 y)tq−1 ) = T2 y,
T3 ((T1 y)tp−1 ) = 0,
T3 ((T2 y)tq−1 ) = 0,
T2 ((T3 y)tr−1 ) = 0,
T3 ((T3 y)tr−1 ) = T3 y.
So we have
Q2 y = Q(T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1 )
= T1 ((T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1 )tp−1
+ T2 ((T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1 )tq−1
+ T3 ((T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1 )tr−1
= (T1 y(t))tp−1 + (T2 y(t))tq−1 + (T3 y(t))tr−1
= Qy.
Thus, Q is a well defined operator.
Now we need to show that ker Q = Im L. If y ∈ ker Q, then Qy = 0. By the
definition of Qy, we have
M11 Q1 y + M12 Q2 y + M13 Q3 y = 0,
M21 Q1 y + M22 Q2 y + M23 Q3 y = 0,
M31 Q1 y + M32 Q2 y + M33 Q3 y = 0.
Because of
M11
M21
M31
M12
M22
M32
M13 M23 = Λ2 (p, q, r) 6= 0,
M33 8
S. LI, J. YIN, Z. DU
EJDE-2014/61
we have Q1 y = Q2 y = Q3 y = 0, i.e. y ∈ ImL.
If y ∈ Im L, we have Q1 y = Q2 y = Q3 y = 0. From the definition of Qy, it is
obvious that Qy = 0, thus y ∈ KerQ. Hence, ker Q = Im L.
For y ∈ Z, let y = (y − Qy) + Qy, since Q(y − Qy) = Qy − Q2 y = 0, we know
y − Qy ∈ ker Q = Im L, and we have Qy ∈ Im Q. Thus Z = Im L + Im Q. And for
any y ∈ Im L ∩ Im Q, from y ∈ Im Q, there exists constants a, b, c ∈ R, such that
y(t) = atp−1 + btq−1 + ctr−1 . From y ∈ Im L, we obtain
m
m
m
bX
cX
aX
αi ξip +
αi ξiq +
αi ξir = 0,
p i=1
q i=1
r i=1
l
l
l
k=1
k=1
k=1
X
X
X
b
c
a
γk σkp+1 +
γk σkq+1 +
γk σkr+1 = 0,
p(p + 1)
q(q + 1)
r(r + 1)
n
n
X
X
1
2
1
2
1
1
βj ηjp+2 ) + b( −
βj ηjq+2 )
+
)(1 −
+
)(1 −
a( −
p p+1 p+2
q
q
+
1
q
+
2
j=1
j=1
n
X
1
2
1
+ c( −
+
)(1 −
βj ηjr+2 ) = 0.
r
r+1 r+2
j=1
(3.4)
In view of
Pm
p
1
i=1 αi ξi
p
1 Pl
p(p+1) k=1 γk σkp+1
A31
Pm
Pm
q
1
r
i=1 αi ξi
i=1 αi ξi ,
r
P
P
l
l
q+1
r+1 1
1
γ
σ
γ
σ
k
k
k=1
k=1
k
k
q(q+1)
r(r+1)
A32
A33
1
=
Λ(p, q, r) 6= 0,
p(p + 1)(p + 2)q(q + 1)(q + 2)r(r + 1)(r + 2)
1
q
where the entries of the third row are
n
X
2
1
1
βj ηjp+2 )
+
)(1 −
A31 = ( −
p p+1 p+2
j=1
n
X
1
2
1
A32 = ( −
+
)(1 −
βj ηjq+2 )
q
q+1 q+2
j=1
A33
I split the above matrix, please check it
n
X
1
2
1
=( −
+
)(1 −
βj ηjr+2 ) .
r
r+1 r+2
j=1
Therefore (3.4) has an unique solution a = b = c = 0, which implies Im Q ∩
Im L = {0} and Z = Im L ⊕ Im Q. Since dim ker L = dimImQ = codimImL = 3,
thus L is a Fredholm map of index zero.
Let P : Y → Y be defined by
1
P x(t) = x(0) + x0 (0)t + x00 (0)t2 ,
2
t ∈ (0, 1).
then, the generalized inverse operator KP : Im L → dom L ∩ ker P be defined by
Z
1 t
(t − s)2 y(s)ds, y ∈ Im L.
KP (y(t)) =
2 0
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9
If y ∈ Im L, we have (LKP )(y(t)) = (KP y)000 = y(t). If x(t) ∈ dom L ∩ ker P , we
have
(KP L)x(t) = (KP )(x000 (t))
Z
1 t
(t − s)2 x000 (s)ds
=
2 0
1
= x(t) − [x(0) + x0 (0)t + x00 (0)t2 ]
2
= x(t) − P x(t).
Because x ∈ dom L ∩ ker P , we know that P x(t) = 0. Thus (KP L)(x(t)) = x(t). It
is obviously that kKP yk ≤ kyk1 .
For the next theorem we use the assumptions:
(H1) There exists α(t), β(t), γ(t), θ(t) ∈ L1 [0, 1], such that for all (x1 , x2 , x3 ) in
R3 , t ∈ (0, 1),
|f (t, x1 , x2 , x3 )| ≤ α(t)|x1 | + β(t)|x2 | + γ(t)|x3 | + θ(t).
(H2) There exists a constant A > 0 such that for x(t) ∈ dom L, if |x(t)| > A
or |x0 (t)| > A or |x00 (t)| > A for all t ∈ (0, 1), then Q1 N (x(t)) 6= 0 or
Q2 N (x(t)) 6= 0 or Q3 N (x(t)) 6= 0.
(H3) There exists a constant B > 0 such that for a, b, c ∈ R, if |a| > B, |b| > B,
|c| > B, then either
Q1 N (a + bt + ct2 ) + Q2 N (a + bt + ct2 ) + Q3 N (a + bt + ct2 ) > 0,
(3.5)
or
Q1 N (a + bt + ct2 ) + Q2 N (a + bt + ct2 ) + Q3 N (a + bt + ct2 ) < 0.
(3.6)
Theorem 3.3. Let the conditions (2.1), (H1), (H2), (H3) hold. Then BVP (1.1)(1.2) have at least one solution in C 2 [0, 1], provided that kαk1 + kβk1 + kγk1 < 1.
Proof. We divide the proof into four steps.
Step 1: Let
Ω1 = {x ∈ dom L\ ker L : Lx = λN x, λ ∈ [0, 1]}.
Then Ω1 is bounded. Suppose that x ∈ Ω1 , we have Lx = λN x. Thus λ 6= 0, N x ∈
Im L = ker Q, hence
Q1 (N x(t)) = Q2 (N x(t)) = Q3 (N x(t)) = 0.
From (H2), there exists t1 , t2 , t3 ∈ (0, 1), such that |x(t1 )| ≤ A, |x0 (t2 )| ≤ A,
|x00 (t3 )| ≤ A. Since x, x0 , x00 are absolutely continuous for all t ∈ (0, 1), and
Z t
Z t
0
0
00
00
00
x (t) = x (t2 ) +
x (s)ds, x (t) = x (t3 ) +
x000 (s)ds.
t2
t3
Thus
kx00 k∞ ≤ A + kx000 k1 , kx0 k∞ ≤ 2A + kx000 k1 .
And
kxk∞ ≤ k(I − P )xk∞ + kP xk∞ = kP xk∞ + kKP L(I − P )xk∞
= kP xk∞ + kKP Lxk∞ ≤ kP xk∞ + kKP kLxk1
≤ kP xk∞ + kLxk1 = kP xk∞ + kx000 k1 .
10
S. LI, J. YIN, Z. DU
EJDE-2014/61
From (H1 ), we obtain
kx000 k1 = kLxk1 ≤ kN xk1
≤ kαk1 kxk∞ + kβk1 kx0 k∞ + kγk1 kx00 k∞ + kθk1
≤ (kαk1 + kβk1 + kγk1 )kx000 k1 + (2kβk1 + kγk1 )A + kαk1 kP xk∞ + kθk1 .
Then
kx000 k1 ≤
1
[(2kβk1 + kγk1 )A + kαk1 kP xk∞ + kθk1 ].
1 − (kαk1 + kβk1 + kγk1 )
So there exists a constant M1 > 0 such that kxk ≤ M1 . Hence we show that Ω1 is
bounded.
Step 2: Let Ω2 = {x ∈ ker L : N x ∈ Im L}. Then Ω2 is bounded. Since x ∈ Ω2 ,
x ∈ ker L = {x ∈ dom L : x = a + bt + ct2 , a, b, c ∈ R}, and QN x = 0, thus,
Q1 (N (a + bt + ct2 )) = Q2 (N (a + bt + ct2 )) = Q3 (N (a + bt + ct2 )) = 0.
From (H3),
kxk ≤ |a| + |b| + |c| ≤ 3B.
So Ω2 is bounded.
Step 3: Let
Ω3 = {x ∈ ker L : λJx + (1 − λ)QN x = 0, λ ∈ [0, 1]}.
Here J : ker L → Im Q is the linear isomorphism given by
1
(a1 tp−1 + b1 tq−1 + c1 tr−1 ),
J(a + bt + ct2 ) =
Λ(p, q, r)
where
a, b, c ∈ R
a1 = p(p + 1)(p + 2)(M11 |a| + M12 |b| + M13 |c|),
b1 = q(q + 1)(q + 2)(M21 |a| + M22 |b| + M23 |c|),
c1 = r(r + 1)(r + 2)(M31 |a| + M32 |b| + M33 |c|).
Then Ω3 is bounded.
Set
p(p + 1)(p + 2)
B1 =
,
Λ(p, q, r)
B2 =
q(q + 1)(q + 2)
,
Λ(p, q, r)
X1 = λ|a| + (1 − λ)Q1 N (a + bt + ct2 ),
B3 =
r(r + 1)(r + 2)
,
Λ(p, q, r)
X2 = λ|b| + (1 − λ)Q2 N (a + bt + ct2 ),
X3 = λ|c| + (1 − λ)Q3 N (a + bt + ct2 ).
Since x(t) = a + bt + ct2 ∈ Ω3 , then we have λJx + (1 − λ)QN x = 0; i.e.,
B1 M11 X1 + B1 M12 X2 + B1 M13 X3 = 0,
B2 M21 X1 + B2 M22 X2 + B2 M23 X3 = 0,
B3 M31 X1 + B3 M32 X2 + B3 M33 X3 = 0.
Because
B1 M11
B2 M21
B3 M31
B1 M12
B2 M22
B3 M32
M11
B1 M13 B2 M23 = (B1 B2 B3 ) M21
M31
B3 M33 =
M12
M22
M32
M13 M23 M33 p(p + 1)(p + 2)q(q + 1)(q + 2)r(r + 1)(r + 2)
6= 0.
Λ(p, q, r)
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SOLUTIONS TO THIRD-ORDER MULTI-POINT BVPS
11
Then X1 = X2 = X3 = 0. If λ = 1, then |a| = |b| = |c| = 0. If λ 6= 1 and |a| > B
or |b| > B or |c| > B and (3.5) hold, then
λ(|a| + |b| + |c|) = −(1 − λ) Q1 (N (a + bt + ct2 )) + Q2 (N (a + bt + ct2 ))
+ Q3 (N (a + bt + ct2 )) < 0,
which contradicts λ(|a| + |b| + |c|) > 0. Thus
kxk ≤ |a| + |b| + |c| ≤ 3B.
So Ω3 is bounded.
If (3.6) holds, we let
Ω3 = {x ∈ ker L : −λJx + (1 − λ)QN x = 0, λ ∈ [0, 1]}.
Similarly, we can proof that Ω3 is bounded.
Step 4: In the following, we shall prove that all the conditions of Lemma 2.1 are
satisfied. Let Ω ⊂ Y be a bounded open set such that ∪3i=1 Ωi ⊂ Ω. By the AscoliArzela theorem, we can show that KP (I − Q)N : Ω → Y is compact, thus N is
L−compact on Ω. Then by the above argument, we obtain
(i) Lx 6= λN x for each (x, λ) ∈ ((dom L\ ker L) ∩ ∂Ω) × [0, 1];
(ii) N x ∈
/ Im L for each x ∈ ker L ∩ ∂Ω.
At last we prove that (iii) of the Lemma 2.1 is satisfied. Let H(x, λ) = ±λJx +
(1 − λ)QN x. According to the above argument, we have H(x, λ) 6= 0, for x ∈
ker L ∩ ∂Ω. Thus, by the homotopy property of degree, we get
deg(QN |ker L , Ω ∩ ker L, 0) = deg(H(·, 0), Ω ∩ ker L, 0)
= deg(H(·, 1), Ω ∩ ker L, 0)
= deg(±J, Ω ∩ ker L, 0) 6= 0.
Then by Lemma 2.1, Lx = N x has at least one solution in dom L ∩ Ω, so that BVP
(1.1)-(1.2) has at least one solution in C 2 [0, 1].
4. Applications
We consider the boundary value problem
x000 (t) = x00 (t) + sin x(t)(1 − cos x0 (t)), t ∈ (0, 1),
1
1
1
1
x00 (0) = 2x00 ( ) − x00 ( ), x0 (0) = 3x0 ( ) − 2x0 ( ),
5
4
3
2
9 1
1
16 3
x(1) = x( ) − 4x( ) + x( ).
5 3
2
5 4
(4.1)
(4.2)
(4.3)
So f (t, x(t), x0 (t), x00 (t)) = x00 (t) + sin x(t)(1 − cos x0 (t)), α1 = 2, α2 = −1, ξ1 = 51 ,
1
ξ2 = 14 , γ1 = 3, γ2 = −2, σ1 = 13 , σ2 = 12 , β1 = 95 , β2 = −4, β3 = 16
5 , η1 = 3 ,
1
3
η2 = 2 , η3 = 4 . Then we have α1 − α2 = 1, γ1 + γ2 = 1, β1 + β2 + β3 = 1,
β1 η1 + β2 η2 + β3 η3 = 1, β1 η12 + β2 η22 + β3 η32 = 1, γ1 σ1 + γ2 σ2 = 0. So the condition
(2.1) holds.
By calculations, we obtain
Z 1/5
Z 13
Z 1/2
1
1
Q1 y = 2
y(s)ds, Q2 y = 3
( − s)y(s)ds − 2
( − s)y(s)ds,
3
2
0
0
0
12
S. LI, J. YIN, Z. DU
EJDE-2014/61
Z 1
9 3 1
(1 − s) y(s)ds −
Q3 y =
( − s)2 y(s)ds
5 0 3
0
Z 1/2
Z
1
16 3/4 3
( − s)2 y(s)ds −
( − s)2 y(s)ds,
+4
2
5 0
4
0
1225231
Λ(1, 2, 3) =
,
1834400
5731015
45787
3429
M11 = −
, M12 =
, M13 = −
,
1508544
311040
38880
70758
3089
8649
M21 =
, M22 =
, M23 =
,
279935
9600
2260
1269
79
477
, M32 = −
, M33 = −
.
M31 = −
3888
3600
200
6
[M11 Q1 y + M12 Q2 y + M13 Q3 y],
T1 y =
Λ(p, q, r)
24
T2 y =
[M21 Q1 y + M22 Q2 y + M23 Q3 y],
Λ(p, q, r)
60
[M31 Q1 y + M32 Q2 y + M33 Q3 y].
T3 y =
Λ(p, q, r)
Z
1
2
Define Qy by Qy = T1 y + (T2 y)t + (T3 y)t2 , and we take KP y(t) as in Lemma 3.2,
then Lemma 3.2 holds.
On the other hand, we have
|f (t, x(t), x0 (t), x00 (t))| ≤ |x00 (t)| + 2,
t ∈ (0, 1).
And let α(t) = 0, β(t) = 0, γ(t) = 1, θ(t) = 2, then the condition (H1) of Theorem
3.3 is satisfied. If x00 (t) > 8 = A, t ∈ (0, 1), then
Z 1/5
Q1 y = 2
(x00 (t) + sin x(t)(1 − cos x0 (t)))dt
0
Z
1/4
(x00 (t) + sin x(t)(1 − cos x0 (t)))dt
−
Z
0
1/5
Z
1/4
7dt −
>2
0
10dt >
0
9
,
10
If x00 (t) < −8 = −A, t ∈ (0, 1), then
Z 1/5
Q1 y = 2
(x00 (t) + sin x(t)(1 − cos x0 (t)))dt
0
Z
1/4
−
Z
(x00 (t) + sin x(t)(1 − cos x0 (t)))dt
0
1/5
Z
(−6)dt −
<2
0
1/4
(−9)dt < −
0
3
,
20
So condition (H2) is satisfied.
If |a| > 16 = B, |b| > 16 = B, |c| > 16 = B, then
Q1 N (a + bt + ct2 ) + Q2 N (a + bt + ct2 ) + Q3 N (a + bt + ct2 )
Z 1/5
=2
(|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
0
EJDE-2014/61
SOLUTIONS TO THIRD-ORDER MULTI-POINT BVPS
13
1/4
Z
(|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
−
0
1
3
Z
+3
0
1
( − t)(|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
3
1/2
Z
−2
0
1
Z
1
( − t)(|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
2
(1 − t)2 (|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
+
0
−
9
5
1
3
Z
Z
0
1/2
+4
0
16
−
5
1
( − t)2 (|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
3
Z
0
1
( − t)2 (|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt
2
3/4
3
( − t)2 (|2c| + sin(a + bt + ct2 )[1 − cos(b + 2ct)])dt > 0.
4
So condition (H3) is satisfied. Thus all the conditions of Theorem 3.3 are satisfied.
Hence BVP (4.1)-(4.3) has at least one solution in C 2 [0, 1].
Acknowledgments. This research was sponsored by the Natural Science Foundation of China (11071205, 11101349) and by PAPD of Jiangsu Higher Education
Institutions, the Humanistic and Social Scientific Research Planning Program in
Ministry of Education of China (No. 12YJA630063), the Social Science Foundation of Jiangsu Province of China (No. 10GLB001).
The authors express their sincere thanks to the anonymous reviewer for his or
her valuable comments and suggestions for improving the quality of this article.
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14
S. LI, J. YIN, Z. DU
EJDE-2014/61
Shuang Li
School of Management, China University of Mining and Technology, Xuzhou, Jiangsu
221116, China
E-mail address: lishuangchina@163.com
Jian Yin
School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou, Jiangsu
221116, China
E-mail address: yinjian719@163.com
Zengji Du (corresponding author)
School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou, Jiangsu
221116, China
E-mail address: duzengji@163.com, Fax 0086-516-83403299