Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 226, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF SOLUTIONS FOR (k, n − k − 2) CONJUGATE
BOUNDARY-VALUE PROBLEMS AT RESONANCE WITH
dim ker L = 2
WEIHUA JIANG
Abstract. By constructing suitable project operators and using the coincidence degree theory due to Mawhin, the existence of solutions for (k, n − k − 2)
conjugate boundary-value problems at resonance with dimkerL = 2 is obtained.
1. Introduction
The existence of solutions for (k, n − k) conjugate boundary-value problems at
nonresonance has been studied in many papers (see [1, 2, 3, 6, 7, 9, 10, 11, 16, 14,
21, 25, 27, 29, 30, 31, 32]). The solvability of boundary-value problems at resonance
has been investigated by many authors (see [4, 5, 8, 12, 13, 15, 17, 18, 19, 20, 26, 22,
24, 28, 33]). In [12], the existence of solutions for (k, n − k) conjugate boundaryvalue problems at resonance with dim ker L = 1 has been studied. To the best
of our knowledge, no paper discusses the existence of solutions for (k, n − k − 2)
conjugate boundary-value problems at resonance with dim ker L = 2. We will fill
this gap in the literature.
In this article, we investigate the existence of solutions for the (k, n − k − 2)
conjugate boundary-value problem at resonance
(−1)n−k y (n) (t) = f t, y(t), y 0 (t), . . . , y (n−1) (t) + ε(t), a.e. t ∈ [0, 1],
(1.1)
y (i) (0) = y (j) (1) = 0,
y (n−2) (1) =
l
X
0 ≤ i ≤ k − 1, 0 ≤ j ≤ n − k − 3,
βj y (n−2) (ηj ),
y (n−1) (1) =
j=1
m
X
αi y (n−1) (ξi ),
i=1
where 1 ≤ k ≤ n − 3, 0 < η1 < η2 < · · · < ηl < 1, 0 < ξ1 < ξ2 < · · · < ξm < 1.
In this article, we assume that the following conditions hold.
Pl
Pl
Pm
(H1)
i=1 αi = 1,
j=1 βj = 1,
j=1 βj ηj = 1.
2000 Mathematics Subject Classification. 35B34, 34B10.
Key words and phrases. Resonance; Fredholm operator; boundary value problem.
c
2013
Texas State University - San Marcos.
Submitted July 24, 2012. Published October 11, 2013.
1
(1.2)
2
W. JIANG
e
(H2) e = 1
e3
EJDE-2013/226
e2 6= 0, where
e4 e1 = 1 −
m
X
a i ξi ,
e2 =
i=1
e3 =
1
2
1−
m
X
ai ξi2 ,
i=1
l
X
1
1−
βj ηj2 ,
2
j=1
e4 =
l
X
1
1−
βj ηj3 .
6
j=1
∞
(H3) ε(t) ∈ L [0, 1], f : [0, 1] × Rn → R satisfies Carathéodory conditions;
i.e., f (·, x) is measurable for each fixed x ∈ Rn , f (t, ·) is continuous for
a.e. t ∈ [0, 1], and for each r > 0, there exists Φr ∈ L∞ [0, 1] such that
|f (t, x1 , x2 , . . . , xn )| ≤ Φr (t) for all |xi | ≤ r, i = 1, 2, . . . , n, a.e. t ∈ [0, 1].
2. Preliminaries
For convenience, we introduce some notation and a theorem. For more details
see [23]. Let X and Y be real Banach spaces and L : dom L ⊂ X → Y be a
Fredholm operator with index zero, P : X → X, Q : Y → Y be projectors such
that
Im P = ker L,
ker Q = Im L,
X = ker L ⊕ ker P,
Y = Im L ⊕ Im Q.
It follows that
Ldom L∩ker P : dom L ∩ ker P → Im L
is invertible. We denote its inverse by KP .
Let Ω be an open bounded subset of X, dom L ∩ Ω 6= ∅, the map N : X → Y
will be called L-compact on Ω if QN (Ω) is bounded and KP (I − Q)N : Ω → X is
compact.
Theorem 2.1 ([23]). Let L : dom L ⊂ X → Y be a Fredholm operator of index
zero and N : X → Y L-compact on Ω. Assume that the following conditions are
satisfied:
(1) Lx 6= λN x for every (x, λ) ∈ [(dom L \ ker L) ∩ ∂Ω] × (0, 1);
(2) N x ∈
/ Im L for every x ∈ ker L ∩ ∂Ω;
(3) deg(QN |ker L , Ω ∩ ker L, 0) 6= 0, where Q : Y → Y is a projection such that
Im L = ker Q.
Then the equation Lx = N x has at least one solution in dom L ∩ Ω.
Take X = C n−1 [0, 1] with norm kuk = max{kuk∞ , ku0 k∞ , . . . , ku(n−1) k∞ },
R1
where kuk∞ = maxt∈[0,1] |u(t)|, Y = L1 [0, 1] with norm kxk1 = 0 |x(t)|dt. Define operator Ly(t) = (−1)n−k y (n) (t) with
n
dom L = y ∈ X : y (n) ∈ Y, y (i) (0) = y (j) (1) = 0, 0 ≤ i ≤ k − 1,
0 ≤ j ≤ n − k − 3, y (n−2) (1) =
l
X
j=1
y (n−1) (1) =
m
X
i=1
o
αi y (n−1) (ξi ) .
βj y (n−2) (ηj ),
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EXISTENCE OF SOLUTIONS
Let N : X → Y be defined as
N y(t) = f t, y(t), y 0 (t), . . . , y (n−1) (t) + ε(t),
3
t ∈ [0, 1].
Then problem (1.1), (1.2) becomes Ly = N y.
We use convention that 1/k! = 0, for k = −1, −2, . . . . By simple calculation, we
can get the following results.
1
1
1
...
k!
(k+1)!
(n−3)!
1
1
1
.
.
.
k!
(n−4)!
(k−1)!
...
...
1
1
1
. . . [n−3−(n−k−3)]! [k−(n−k−3)]!
[k+1−(n−k−3)]!
=
(n − k − 3)! (n − k − 4)!
1
·
...
6= 0.
k!
(k + 1)!
(n − 3)!
So, the following lemmas hold.
Lemma 2.2. The system of linear equations
xk
xk+1
xn−3
1
+
+ ··· +
+
= 0,
k!
(k + 1)!
(n − 3)! (n − 2)!
xk+1
xn−3
1
xk
+
+ ··· +
+
= 0,
(k − 1)!
k!
(n − 4)! (n − 3)!
...
xk
xk+1
+
+ ...
[k − (n − k − 3)]! [k + 1 − (n − k − 3)]!
1
xn−3
+
=0
+
[n − 3 − (n − k − 3)]! [n − 2 − (n − k − 3)]!
has only one solution, its denoted by (ak , ak+1 , . . . , an−3 ).
Lemma 2.3. The system of linear equations
xk
xk+1
xn−3
1
+
+ ··· +
+
= 0,
k!
(k + 1)!
(n − 3)! (n − 1)!
xk
xk+1
xn−3
1
+
+ ··· +
+
= 0,
(k − 1)!
k!
(n − 4)! (n − 2)!
...
xk
xk+1
+
+ ...
[k − (n − k − 3)]! [k + 1 − (n − k − 3)]!
xn−3
1
+
+
=0
[n − 3 − (n − k − 3)]! [n − 1 − (n − k − 3)]!
has only one solution, it is denoted by (bk , bk+1 , . . . , bn−3 ).
Lemma 2.4. For given u ∈ Y , the system of linear equations
Z
xk
xk+1
xn−3
(−1)n−k 1
+
+ ··· +
+
(1 − s)n−1 u(s)ds = 0,
k!
(k + 1)!
(n − 3)!
(n − 1)! 0
Z
xk
xk+1
xn−3
(−1)n−k 1
+
+ ··· +
+
(1 − s)n−2 u(s)ds = 0,
(k − 1)!
k!
(n − 4)!
(n − 2)! 0
...
4
W. JIANG
EJDE-2013/226
xk
xk+1
xn−3
+
+ ··· +
[k − (n − k − 3)]! [k + 1 − (n − k − 3)]!
[n − 3 − (n − k − 3)]!
Z 1
(−1)n−k
(1 − s)n−1−(n−k−3) u(s)ds = 0
+
[n − 1 − (n − k − 3)]! 0
has only one solution, its denoted by (Bk (u), Bk+1 (u), . . . , Bn−3 (u)).
Define the operators T1 , T2 , Q1 , Q2 : Y → R as follows:
Z 1
m
X
T1 u(t) =
αi
u(s)ds,
i=1
T2 u(t) =
l
X
j=1
βj
hZ
ξi
1
Z
(1 − s)u(s)ds + (1 − ηj )
ηj
1
(e4 T1 u − e3 T2 u), Q2 u =
e
Obviously, e1 = T1 (1), e2 = T2 (1), e3 = T1 (t), e4
Q1 u =
ηj
u(s)ds ,
0
1
(−e2 T1 u + e1 T2 u).
e
= T2 (t).
Lemma 2.5. Assume that (H1) holds, then L : dom L ⊂ X → Y is a Fredholm
operator of index zero and the linear continuous projector Q : Y → Y can be defined
as
Qu = Q1 u + t · Q2 u,
and the linear operator KP : Im L → dom L ∩ ker P can be written as
Z
n−3
X Bi (u)
(−1)n−k t
ti +
KP u =
(t − s)n−1 u(s)ds.
i!
(n − 1)! 0
i=k
Pn−1
Proof. Take y ∈ ker L. We obtain y = i=k xi!i ti satisfying
xk
xk+1
xn−2
xn−1
+
+ ··· +
+
= 0,
k!
(k + 1)!
(n − 2)! (n − 1)!
xk+1
xn−2
xn−1
xk
+
+ ··· +
+
= 0,
(k − 1)!
k!
(n − 3)! (n − 2)!
...
xk
xk+1
+
+ ...
[k − (n − k − 3)]! [k + 1 − (n − k − 3)]!
xn−2
xn−1
+
+
= 0.
[n − 2 − (n − k − 3)]! [n − 1 − (n − k − 3)]!
Setting xn−2 = 1, xn−1 = 0, and xn−2 = 0, xn−1 = 1, respectively, by Lemmas
2.2, 2.3, we have
y=
n−3
X
i=k
cai + dbi i
c
d
t +
tn−2 +
tn−1 , c, d ∈ R.
i!
(n − 2)!
(n − 1)!
Therefore,
n−3
X cai + dbi
c
d
ti +
tn−2 +
tn−1 , c, d ∈ R .
ker L = y : y =
i!
(n − 2)!
(n − 1)!
i=k
Define the linear operator P : X → X as follows
P y(t) =
n−3
X
i=k
y (n−2) (0)ai + y (n−1) (0)bi i y (n−2) (0) n−2 y (n−1) (0) n−1
t +
t
+
t
.
i!
(n − 2)!
(n − 1)!
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EXISTENCE OF SOLUTIONS
5
Obviously, Im P = ker L and P 2 y = P y. For any y ∈ X, it follows from y =
(y − P y) + P y that X = ker P + ker L. By simple calculation, we can get that
ker L ∩ ker P = {0}. So, we have
X = ker L ⊕ ker P.
We will show that
Z
m
X
Im L = u ∈ Y :
αi
βj
hZ
1
u(s)ds = 0,
ξi
i=1
l
X
(2.1)
1
Z
ηj
j=1
ηj
(1 − s)u(s)ds + (1 − ηj )
i
u(s)ds = 0 .
0
In fact, if u ∈ Im L, there exists y ∈ dom L such that u = Ly ∈ Y . This, together
with y i (0) = 0, 0 ≤ i ≤ k − 1, implies that
Z
n−1
X ci
(−1)n−k t
ti +
(t − s)n−1 u(s)ds.
y(t) =
i!
(n − 1)! 0
i=k
Pm
Pm
(n−1)
Since i=1 αi = 1 and y
(1) = i=1 αi y (n−1) (ξi ), we obtain
Z 1
m
X
αi
u(s)ds = 0.
(2.2)
i=1
Since
Pl
j=1
ξi
Pl
βj ηj = 1 and y (n−2) (1) =
Z
l
hZ 1
X
βj
(1 − s)u(s)ds + (1 − ηj )
βj = 1,
j=1
j=1
ηj
Pl
j=1
ηj
βj y (n−2) (ηj ), we obtain
i
u(s)ds = 0.
(2.3)
0
On the other hand, if u ∈ Y satisfies (2.2) and (2.3), take
Z
n−3
X Bi (u)
(−1)n−k t
ti +
(t − s)n−1 u(s)ds.
y=
i!
(n − 1)! 0
i=k
It follows from (2.2), (2.3) and Lemma 2.4 that y ∈ dom L. Obviously, Ly = u. So,
we get u ∈ Im L.
Now we will prove that Q : Y → Y is a projector such that ker Q = Im L,
Y = Im L ⊕ Im Q. For u ∈ Y , since
1
1
Q1 (1) = [e4 T1 (1) − e3 T2 (1)] = 1, Q1 (t) = [e4 T1 (t) − e3 T2 (t)] = 0,
e
e
1
1
Q2 (1) = [−e2 T1 (1) + e1 T2 (1)] = 0, Q2 (t) = [−e2 T1 (t) + e1 T2 (t)] = 1,
e
e
we have
Q1 (Qu) = Q1 (Q1 u + t · Q2 u) = Q1 u · Q1 (1) + Q2 u · Q1 (t) = Q1 u,
Q2 (Qu) = Q2 (Q1 u + t · Q2 u) = Q1 u · Q2 (1) + Q2 u · Q2 (t) = Q2 u.
Thus,
Q2 u = Q1 (Qu) + t · Q2 (Qu) = Q1 u + t · Q2 u = Qu.
Since u ∈ ker Q, we have
e4 T1 u − e3 T2 u = 0,
−e2 T1 u + e1 T2 u = 0.
6
W. JIANG
EJDE-2013/226
It follows from (H2) that T1 u = T2 u = 0. So, u ∈ Im L; i.e., ker Q ⊂ Im L. Clearly,
Im L ⊂ ker Q. So, Im L = ker Q. This, together with Q2 y = Qy, means that
Im L ∩ Im Q = {0}. Thus, we have Y = Im L ⊕ Im Q. Considering (2.1), we know
that L is a Frdholm operator of index zero.
Define the operator KP : Y → X as follows
Z
n−3
X Bi (u)
(−1)n−k t
(t − s)n−1 u(s)ds.
KP u =
ti +
i!
(n − 1)! 0
i=k
For u ∈ Im L, by Lemma 2.4, we have KP u ∈ dom L. Clearly, KP u ∈ ker P . So,
we get that KP (Im L) ⊂ dom L ∩ ker P . Now we will prove that KP is the inverse
of L|dom L∩ker P .
Obviously, LKP u = u, for u ∈ Im L. On the other hand, for y ∈ dom L ∩ ker P ,
we have
Z
n−3
X Bi (Ly)
(−1)n−k t
ti +
(t − s)n−1 (−1)n−k y (n) (s)ds
KP Ly(t) =
i!
(n − 1)! 0
i=k
=
n−3
X
i=k
Bi (Ly) − y (i) (0) i
t + y(t).
i!
Since KP (Ly) ∈ dom L and y ∈ dom L, we obtain (KP Ly)(j) (1) = y (j) (1) = 0,
0 ≤ j ≤ n − k − 3. Thus (Bk (Ly) − y (k) (0), Bk+1 (Ly) − y (k+1) (0), . . . , Bn−3 (Ly) −
y (n−3) (0)) is the only zero solution of the system of linear equations
xk
xk+1
xn−3
+
+ ··· +
= 0,
k!
(k + 1)!
(n − 3)!
xk
xk+1
xn−3
+
+ ··· +
= 0,
(k − 1)!
k!
(n − 4)!
...
xk
xk+1
+
+ ...
[k − (n − k − 2)]! [k + 1 − (n − k − 2)]!
xn−3
= 0.
+
[n − 3 − (n − k − 3)]!
So, we have KP Ly = y, for y ∈ dom L ∩ ker P . Thus, KP = (L|dom L∩ker P )−1 . The
proof is complete.
3. Main results
Lemma 3.1. Assume Ω ⊂ X is an open bounded subset and dom L ∩ Ω 6= ∅, then
N is L-compact on Ω.
Proof. By (H3), we have that QN (Ω) is bounded. Now we will show that KP (I −
Q)N : Ω → X is compact.
It follows from (H3) that there exists constant M0 > 0 such that |(I − Q)N y| ≤
M0 , a.e. t ∈ [0, 1], y ∈ Ω. Thus, KP (I −Q)N (Ω) is bounded. By (H3) and Lebesgue
Dominated Convergence theorem, we get that KP (I − Q)N : Ω → X is continuous.
Rt
Since { 0 (t − s)j (I − Q)N y(s)ds, y ∈ Ω}, j = 0, 1 . . . , n − 1 are equi-continuous, and
tj , j = 0, 1 . . . , n−1 are uniformly continuous on [0,1], using Ascoli-Arzela theorem,
we obtain that KP (I − Q)N : Ω → X is compact. The proof is complete.
To obtain our main results, we need the following assumptions.
EJDE-2013/226
EXISTENCE OF SOLUTIONS
7
(H4) There exist constants M1 > 0, M2 > 0 such that if |y (n−1) (t)| > M1 ,
t ∈ [ξm , 1] then
Z 1
m
X
αi
N y(s)ds 6= 0,
i=1
and if |y
ξi
(n−2)
(t)| > M2 , t ∈ [0, η1 ] then
Z
l
hZ 1
X
βj
(1 − s)N y(s)ds + (1 − ηj )
ηj
j=1
ηj
i
N y(s)ds 6= 0.
0
(H5) There exist functions g, h, ψi ∈ L1 [0, 1], i = 1, 2, . . . , n, with kψn k1 := r1 <
Pn−1
1
1/2, i=1 kψi k1 := r2 < 1−2r
4 , θ ∈ [0, 1), and some 1 ≤ j ≤ n − 1 such
that
n
X
ψi (t)|xi | + h(t)|xj |θ .
|f (t, x1 , x2 , . . . , xn )| ≤ g(t) +
i=1
(H6) There exist constants c0 > 0, d0 > 0 such that, for
y=
n−3
X
i=k
cai + dbi i
c
d
t +
tn−2 +
tn−1 ∈ ker L,
i!
(n − 2)!
(n − 1)!
one of the following two conditions holds
(1) c · T1 N y < 0, if |c| > c0 , d · T2 N y < 0, if |d| > d0 ,
(2) c · T1 N y > 0, if |c| > c0 , d · T2 N y > 0, if |d| > d0 ,
Lemma 3.2. Suppose (H1)–(H5) hold, then the set
Ω1 = {y ∈ dom L \ ker L : Ly = λN y, λ ∈ (0, 1)}
is bounded.
Proof. Take y ∈ Ω1 . By N y ∈ Im L, we have
Z 1
m
X
αi
N y(s)ds = 0,
i=1
l
X
βj
hZ
1
Z
(1 − s)N y(s)ds + (1 − ηj )
ηj
j=1
(3.1)
ξi
ηj
i
N y(s)ds = 0.
(3.2)
0
Since Ly = λN y and y ∈ dom L, we obtain
y(t) =
n−1
X
i=k
ci i (−1)n−k
t +
λ
i!
(n − 1)!
t
Z
(t − s)n−1 N y(s)ds,
0
where ck , ck+1 , . . . , cn−1 satisfy
n−1
X
i=k
n−1
X
i=k
ci
(−1)n−k
=−
λ
i!
(n − 1)!
Z
1
(1 − s)n−1 N y(s)ds,
0
n−k
ci
(−1)
=−
λ
(i − 1)!
(n − 2)!
...
Z
0
1
(1 − s)n−2 N y(s)ds,
(3.3)
8
W. JIANG
n−1
X
i=k
ci
(−1)n−k
=−
λ
[i − (n − k − 3)]!
[i − (n − k − 3)]!
EJDE-2013/226
Z
1
(1 − s)i−(n−k−3) N y(s)ds.
0
It follows from y (i) (0) = y (j) (1) = 0, 0 ≤ i ≤ k − 1, 0 ≤ j ≤ n − k − 3 that there
exist points δi ∈ [0, 1] such that y (i) (δi ) = 0, i = 0, 1, . . . , n − 3. So, we have
Z t
y (i+1) (s)ds, i = 0, 1, . . . , n − 3.
y (i) (t) =
δi
Therefore,
ky (i) k∞ ≤ ky (i+1) k1 ≤ ky (i+1) k∞ , i = 0, 1, . . . , n − 3.
(3.4)
(n−1)
By (3.1) and (H4), there exists t0 ∈ [ξm , 1] such that |y
(t0 )| ≤ M1 . This,
together with (3.3), implies that
Z 1
f (s, y(s), y 0 (s), . . . , y (n−1) (s))ds + kεk1 .
|cn−1 | ≤ M1 +
0
By (3.2) and (H4), we get that there exists t1 ∈ [0, η1 ] such that |y (n−2) (t1 )| ≤ M2 .
It follows from (3.3) that
Z 1
f (s, y(s), y 0 (s), . . . , y (n−1) (s))ds + kεk1
|cn−2 | ≤ M2 + |cn−1 | +
0
1
Z
f (s, y(s), y 0 (s), . . . , y (n−1) (s))ds + 2kεk1 .
≤ M1 + M2 + 2
0
Thus,
ky (n−1) k∞ ≤ M1 + 2
Z
1
f (s, y(s), y 0 (s), . . . , y (n−1) (s))ds + 2kεk1 ,
0
ky (n−2) k∞ ≤ 2M1 + M2 + 4
Z
1
f (s, y(s), y 0 (s), . . . , y (n−1) (s))ds + 4kεk1 .
0
By (H5) and (3.4) we have
ky (n−1) k∞ ≤ r3 + 2r2 ky (n−2) k∞ + 2r1 ky (n−1) k∞ + 2khk1 ky (n−2) kθ∞
and
ky (n−2) k∞ ≤ 2r3 + M2 + 4r2 ky (n−2) k∞ + 4r1 ky (n−1) k∞ + 4khk1 ky (n−2) kθ∞ , (3.5)
where r3 = M1 + 2kgk1 + 2kεk1 . So, we obtain
ky (n−1) k∞ ≤
1
[r3 + 2r2 ky (n−2) k∞ + 2khk1 ky (n−2) kθ∞ ].
1 − 2r1
(3.6)
By (3.5) and (3.6), we have
ky (n−2) k∞ ≤
2r3
4r2
4khk1 (n−2) θ
+ M2 +
ky (n−2) k∞ +
ky
k∞ .
1 − 2r1
1 − 2r1
1 − 2r1
Therefore,
ky (n−2) k∞ ≤
1
[2r3 + (1 − 2r1 )M2 + 4khk1 ky (n−2) kθ∞ ].
1 − 2r1 − 4r2
It follows from θ ∈ [0, 1) that {ky (n−2) k∞ : y ∈ Ω1 } is bounded. By (3.4) and (3.6),
we get that Ω1 is bounded.
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EXISTENCE OF SOLUTIONS
9
Lemma 3.3. Suppose (H1)–(H3), (H6) hold. Then the set
Ω2 = {y ∈ ker L : N y ∈ Im L}
is bounded.
Proof. Take y ∈ Ω2 , then
y(t) =
n−3
X
i=k
c
d
cai + dbi i
t +
tn−2 +
tn−1 .
i!
(n − 2)!
(n − 1)!
By N y ∈ Im L, we have T1 N y = 0, T2 N y = 0. By (H6), we get that |c| ≤ c0 , |d| ≤
d0 . This means that Ω2 is bounded.
Lemma 3.4. Suppose (H1)–(H3), (H6) hold. Then the set
Ω3 = {y ∈ ker L : λJy + (1 − λ)ωQN y = 0, λ ∈ [0, 1]}
is bounded, where J : ker L → Im Q is a linear isomorphism given by
J
n−3
1
X cai + dbi
c
d
1
ti +
tn−2 +
tn−1 = (e4 c − e3 d) + (−e2 c + e1 d)t,
i!
(n − 2)!
(n − 1)!
e
e
i=k
where c, d ∈ R and
(
−1,
ω=
1,
if (H6)(1) holds,
if (H6)(2) holds.
Proof. Take y ∈ Ω3 . y ∈ ker L implies that
y=
n−3
X
i=k
c
d
cai + dbi i
t +
tn−2 +
tn−1 , c, d ∈ R.
i!
(n − 2)!
(n − 1)!
Since λJy + (1 − λ)ωQN y = 0, we obtain
λc = −(1 − λ)ωT1 N y,
λd = −(1 − λ)ωT2 N y.
If λ = 0, by (H6), we get |c| ≤ c0 , |d| ≤ d0 . If λ = 1, then c = d = 0. For λ ∈ (0, 1),
if |c| > c0 or |d| > d0 , then
λc2 = −(1 − λ)ωc · T1 N y < 0
or
λd2 = −(1 − λ)ωd · T2 N y < 0.
A contradiction. So, Ω3 is bounded.
Theorem 3.5. Suppose (H1)–(H6) hold. Then (1.1)–(1.2) has at least one solution
in X.
Proof. Let Ω ⊃ ∪3i=1 Ωi ∪ {0} be a bounded open subset of X. It follows from
Lemma 3.1 that N is L-compact on Ω. By Lemmas 3.2 and 3.3, we obtain
(1) Ly 6= λN y for every (y, λ) ∈ [(dom L \ ker L) ∩ ∂Ω] × (0, 1);
(2) N y ∈
/ Im L for every y ∈ ker L ∩ ∂Ω.
10
W. JIANG
EJDE-2013/226
We need to prove only that:
deg(QN |ker L , Ω ∩ ker L, 0) 6= 0.
Take
H(y, λ) = λJy + ω(1 − λ)QN y.
According to Lemma 3.4, we know that H(y, λ) 6= 0 for y ∈ ∂Ω ∩ ker L, λ ∈ [0, 1].
By the homotopy of degree, we obtain
deg(QN |ker L , Ω ∩ ker L, 0) = deg(ωH(·, 0), Ω ∩ ker L, 0)
= deg(ωH(·, 1), Ω ∩ ker L, 0)
= deg(ωJ, Ω ∩ ker L, 0) 6= 0.
By Theorem 2.1, we can obtain that Ly = N y has at least one solution in dom L∩Ω;
i.e., (1.1)–(1.2) has at least one solution in X. The prove is complete.
Acknowledgments. The author is grateful to anonymous referees for their constructive comments and suggestions which led to improvement of the original manuscript.
This work is supported by the Natural Science Foundation of China (11171088),
the Natural Science Foundation of Hebei Province (A2013208108) and the Doctoral
Program Foundation of Hebei University of Science and Technology (QD201020).
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Weihua Jiang
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018,
Hebei, China
E-mail address: weihuajiang@hebust.edu.cn