Electronic Journal of Differential Equations, Vol. 2004(2004), No. 137, pp.... ISSN: 1072-6691. URL: or

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Electronic Journal of Differential Equations, Vol. 2004(2004), No. 137, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF MULTIPLE SOLUTIONS FOR A CLASS OF
SECOND-ORDER ORDINARY DIFFERENTIAL EQUATIONS
XIAO-BAO SHU, YUAN-TONG XU
Abstract. By means of variational structure and Z2 group index theory, we
obtain multiple solutions for the second-order differential equation
d
du
(p(t) ) + q(t)u + f (t, u) = 0, 0 < t < 1,
dt
dt
subject to one of the following two sets of boundary conditions:
u0 (0) = u(1) + u0 (1) = 0
or
u(0) = u(1) = 0 .
1. Introduction
Erbe and Mathsen [6] study the boundary-value problem
−(ru0 )0 + qu = λf (t, u),
0 < t < 1,
0
αu(0) − βu (0) = 0 = γu(1) + δu0 (1),
where λ > 0 is a parameter, α, β, γ, δ ≥ 0 and αδ + αγ + βγ > 0, f ∈ C((0, 1) ×
R, R), r ∈ C([0, 1], (0, ∞)) and q ∈ C([0, 1], [0, ∞)).
In this paper we are interested in the study of second-order ordinary differential
equation
d
du
(p(t) ) + q(t)u + f (t, u) = 0, 0 < t < 1,
(1.1)
dt
dt
subject to one of the following two sets of boundary conditions
u0 (0) = 0 = γu(1) + u0 (1)
(1.2)
u(0) = u(1) = 0
(1.3)
or
By means of variational structure and Z2 group index theory, we obtain multiple
solutions of boundary-value problems for (1.1) and lower bound estimate of number
for the solutions.
2000 Mathematics Subject Classification. 34B15, 34B05, 65K10, 34B24.
Key words and phrases. Variational structure; Z2 group index theory; critical points;
boundary value problems.
c 2004 Texas State University - San Marcos.
Submitted October 5, 2004. Published November 25, 2004.
Supported by grant 10471155 from NNSF of China, grant 031608 from the Foundation
of the Guangdong province Natural Science Committee, and grant 20020558092 from
Foundation for PhD Specialities of Educational Department of China.
1
2
X. B. SHU & Y. T. XU
EJDE-2004/137
A critical point of f is a point x0 where f 0 (x0 ) = θ and a critical value is a number
c such that f (x0 ) = c for some critical point x0 . Next, we recall the definition of
the Palais-Smale condition.
Definition Let E be real Banach space and f ∈ C 1 (E, R). We say that f satisfies the Palais-Smale condition if every sequence {xn } ⊂ E such that {f (xn )} is
bounded and f 0 (xn ) → θ as n → ∞ has a converging subsequence.
Let K = {x ∈ E : f 0 (x) = θ}, Kc = {x ∈ E : f 0 (x) = θ, f (x) = c} and
fc = {x ∈ E : f (x) ≤ c}. The class of subsets ofP
X \ {θ} ⊂ E closed and symmetric
with respect to the origin will be denoted by . Next, we recall the concept of
genus.
Definition Let E be a real Banach space, and Σ = {A : A ⊂ E \ {θ} is a closed,
symmetric set}. Define γ : Σ → Z + ∪ {+∞} as follows

n

min{n ∈ Z : there exists an odd continuous map ϕ : A → R \ {θ}};
γ(A) = 0 If A = ∅;


+∞ If there is no odd continuous map ϕ : A → Rn \ {θ} for n ∈ Z.
P
Then we say γ is the genus of
. Denote i1 (f ) = lima→−0 γ(fa ) and i2 (f ) =
lima→−∞ γ(fa ).
P
We know that if A ∈
and if there exists an odd homeomorphism of n−sphere
onto A then γ(A) = n +P1; If X is a Hilbert space, and E is an n−dimensional
subspace of X, and A ∈
is such that A ∩ E ⊥ = ∅ then γ(A) ≤ n.
The following Lemmas play an important role in proving our main results.
Lemma 1.1 ([5]). Let f ∈ C 1 (E, R) be an even functional which satisfies the
Palais-Smale condition and f (θ) = 0. Then
(P1) If there exists an m-dimensional subspace X of E and ρ > 0 such that
sup f (x) < 0,
x∈X∩Sρ
then we have i1 (f ) ≥ m
e of E such that
(P2) If there exists a j-dimensional subspace X
inf f (x) > −∞,
e⊥
x∈X
we have i2 (f ) ≤ j
If m ≥ j, (P1) and (P2) hold, then f has at least 2(m − j) distinct critical points.
Lemma 1.2 ([9]). Let f ∈ C 1 (X, R) be an even functional which satisfies the
Palais-Smale condition and f (θ) = 0. If
(F1) There exists ρ > 0, α > 0 and a finite dimensional subspace E of X, such
that f |E ⊥ ∩Sρ ≥ α
e of X, there is a r = r(E)
e > 0, such
(F2) For all finite dimensional subspace E
e r
that f (x) ≤ 0 for x ∈ E\B
Then f possesses an unbounded sequence of critical values.
2. Main Results
In this paper, we use Lemma 1.1 and 1.2 to study the boundary-value problems
(1.1)-(1.2) and (1.1)-(1.3)
Theorem 2.1. Let f , p(t) and q(t) satisfy the following conditions:
EJDE-2004/137
(i)
(ii)
(iii)
(iv)
(v)
(vi)
MULTIPLE SOLUTIONS FOR ODE’S
3
p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1]
f ∈ C([0, 1] × R, R)
limu→0 f (t,u)
= ξ(t) > 0 uniformly for t ∈ [0, 1], λ = min0≤t≤1 ξ(t)
u
There exists α > 0 such that f (t, α) ≤ 0
f (t, u) is odd in u
− λ2 < q(t) + p(t) ≤ 0, for all 0 ≤ t ≤ 1.
Then (1.1)-(1.2), has at least 2n nontrivial solutions in C 2 [0, 1] whenever
2n2 (M + p(1)|γ|)(1 + π 2 ) < λ ≤ 2(n + 1)2 (M + p(1)|γ|)(1 + π 2 )
m
and γ > − 2p(1)
Proof. Set h : [0, 1] × R → R,


if u > α,
f (t, α)
h(t, u) = f (t, u)
if |u| ≤ α,


f (t, −α) if u < −α
Let us consider the functional defined on H01 (0, 1) by
Z 1
1
1
p(1) 2
[ p(t)|u0 (t)|2 − q(t)|u(t)|2 − G(t, u(t))]dt +
γu (1),
(2.1)
I(u) =
2
2
2
0
Ru
Where G(t, u) = 0 h(t, v)dv. The norm k · k and inner product (·, ·) can be defined
respectively by
Z 1
Z 1
kuk = ( (|u0 (t)|2 + |u(t)|2 )dt)1/2 ; (u, v) =
(u0 (t)v 0 (t) + u(t)v(t))dt .
0
0
Thus H01 (0, 1) = W01,2 (0, 1) will be a Hilbert space.
Let E = H01 (0, 1), since h(t, u) is an odd continuous map in u, we know that
I ∈ C 1 (E, R) is even in u and I(θ) = 0.
First, we will show that the critical points of the I(u) are the solutions of (1.1)(1.2) in C 2 [0, 1]. Since
Z 1
I(u + sv) = I(u) + s{ [p(t)u0 (t)v 0 (t) − q(t)u(t)v(t) − h(t, u + θ(t)sv)v(t)]dt
0
Z 1
s2
{ (p(t)|v 0 (t)|2
2 0
− q(t)|v(t)|2 )dt + p(1)γv 2 (1)} ∀u, v ∈ E, 0 < θ(t) < 1
+ p(1)γu(1)v(1)} +
(2.2)
We have, for all u, v ∈ E,
Z 1
(I 0 (u), v) =
[p(t)u0 (t)v 0 (t) − q(t)u(t)v(t) − h(t, u(t))v]dt + p(1)γu(1)v(1) . (2.3)
0
0
By I (u) = 0, one gets
Z 1
[p(t)u0 (t)v 0 (t) − q(t)u(t)v(t) − h(t, u(t))v]dt + p(1)γu(1)v(1) = 0
0
(2.4)
4
X. B. SHU & Y. T. XU
EJDE-2004/137
for all v ∈ E. On the other hand,
Z 1
Z 1
du
d
0
0
p(t)u (t)v (t)dt +
(p(t) )vdt
dt
0
0 dt
Z 1
Z 1
Z
=
p(t)u0 (t)v 0 (t)dt +
p(t)u00 (t)v(t)dt +
0
0
1
p0 (t)u0 (t)v(t)dt
(2.5)
0
= p(1)v(1)u0 (1) − p(0)u0 (0)v(0) = 0
So, it is easy to see that
Z 1
du
d
v[ (p(t) ) + q(t)u(t) + h(t, u(t))]dt
dt
dt
0
0
= p(1)v(1)(u (1) + γu(1)) − p(0)u0 (0)v(0) = 0
Hence we obtain
d
du
(p(t) ) + q(t)u(t) + h(t, u(t)) = 0
dt
dt
Thus the critical points of I(u) are the solutions of (1.1)-(1.2) in C 2 [0, 1].
For convenience, we transform (2.1) into
Z 1
1
I(u) =
[ p(t)(|u0 (t)|2 + |u(t)|2 )
0 2
1
p(1) 2
− (q(t) + p(t))|u(t)|2 − G(t, u(t))]dt +
γu (1) .
2
2
By condition (iv) of Theorem 2.1, we have uh(u) ≤ 0 when |u(t)| ≥ α. So
Z 1
Z 1Z u
Z 1Z α
G(t, u) =
h(t, v)dvdt ≤
|h(t, v)|dv dt .
0
0
0
0
−α
R1Rα
Denote by c the value of 0 −α |h(t, v)|dvdt. On the other hand, q(t) + p(t) ≤ 0,
R1
then − 0 (q(t) + p(t))|u(t)|2 dt ≥ 0. So, we have
p(1) 2
m
kuk2 − c +
γu (1) ∀u ∈ E.
2
2
Next, we show that I(u) has lower bound. We divide our proof into two parts
(I) When γ ≥ 0, it is easy to see
m
I(u) ≥ kuk2 − c ∀u ∈ E
(2.6)
2
m
(II) When − 2p(1)
< γ < 0, we divide again our proof into two parts in order to
show I(u) has lower bound: (a) If there exist t0 ∈ [0, 1] such that u(t0 ) = 0, then
Z 1
Z 1
√
|u(1)| = |
u0 (s)ds| ≤
|u0 (s)|ds ≤ 2kuk .
I(u) ≥
t0
0
So, we get
1
(m + 2γp(1))kuk2 − c ∀u ∈ E,
(2.7)
2
(b) If does not exist t1 ∈ [0, 1] such that u(t1 ) = 0, then u(t) > 0 or u(t) < 0, for
all t ∈ [0, 1]. We might as well let u(t) > 0 for all t ∈ [0, 1].
When max0≤t≤1 u(t) ≤ 1,we have u(1) ≤ 1 and
I(u) ≥
I(u) ≥
m
1
kuk2 − c + γp(1),
2
2
EJDE-2004/137
MULTIPLE SOLUTIONS FOR ODE’S
5
i.e., I(u) has lower bound.
When max0≤t≤1 |u(t)| > 1, since u ∈ C 2 [0, 1], there exist t2 ∈ [0, 1] such that
u(t2 ) = min0≤t≤1 u(t). So
Z 1
Z 1
0
u(1) − u(t2 ) = |
u (s)ds| ≤
|u0 (s)|ds
t2
0
i.e.
Z
u(1) ≤ u(t2 ) +
1
Z
|u (s)|ds ≤ (
0
Z
u (t)dt) + (
2
1
2
1
u2 (t)dt +
0
Z
1
1
|u0 (t)|2 ) 2
0
0
0
√ Z
≤ 2(
1
1
|u0 (t)|2 )1/2 =
√
2kuk
0
As in the proof of (I), we have
1
I(u) ≥ (m + 2γp(1))kuk2 − c. ∀u ∈ E.
(2.8)
2
m
< γ < 0. From
By (a) and (b), it is easy to see I(u) has lower bound when − 2p(1)
1
(I) and (II), we get that I(u) has lower bound for all u ∈ H0 (0, 1), i.e., i2 (I) = 0.
Next, we verify that I(u) satisfies the Palais-Smale condition. Suppose that
{un } ⊂ E with and
c1 ≤ I(un ) ≤ c2 ,
0
I (un ) → 0 as
n → ∞.
(2.9)
(2.10)
Then
Z
sup{
1
[p(t)u0n v 0 − q(t)un v − h(t, un (t))v]dt + γp(1)un (1)v(1)} → 0,
0
as n → ∞,
(2.11)
∀u, v ∈ E, kvk = 1
with kzn k = kI 0 (xn )k. Let us denote εn = kzn k, then εn → 0 as n → ∞. Replace
v by un in above equality. By (2.3) and (2.11), we have
Z 1
Z 1
[p(t)|u0n (t)|2 − q(t)|un (t)|2 ]dt =
h(t, un )un (t)dt + hzn , un i.
0
0
The above equality is equivalent to
Z 1
p(t)[|u0n (t)|2 + |un (t)|2 ]dt
0
Z
=
1
[(q(t) + p(t))|un (t)|2 + h(t, un )un (t)]dt + hzn , un i
0
So, there exist ξ ∈ [0, 1] such that
Z 1
2
p(ξ)kun k =
[(q(t) + p(t))|un (t)|2 + h(t, un )un (t)]dt + hzn , un i .
(2.12)
0
Next, we show that {un } satisfying condition (2.9) and (2.10) is bounded. We
divide again our proof into two parts.
(c) When γ ≥ 0, by (2.6), one gets
2
2
kun k2 ≤ (I(un ) + c) ≤ (c2 + c),
m
m
q
2
(c2 + c).
i.e., kun k ≤ m
6
X. B. SHU & Y. T. XU
EJDE-2004/137
m
(d) When − 2p(1)
< γ < 0, by the above proof and (2.7) and (2.8), we have
kun k2 ≤
2
2
(I(un ) + c) ≤
(c2 + c)
m + 2γp(1)
1 + 2γp(1)
or
kuk2 ≤
2
1
[c2 + c − γp(1)]
m
2
i.e.,
s
kun k ≤
2
(c2 + c)
m + 2γp(1)
r
or kuk ≤
2
1
(c2 + c − γp(1)).
m
2
By (c) and (d), it is easy to see {un } is bounded in the space H01 (0, 1). Reflexivity of
H01 (0, 1) implies that there exists a subsequence of {un } which is weak convergent
in H01 (0, 1). We still denote it by {un } and suppose that un * u0 in H01 (0, 1) as
n → ∞. On the one hand, by boundedness of {un } and (2.12), we have
Z 1
p(ξ)kun k2 −
[(q(t) + p(t))|un (t)|2 + h(t, un )un (t)]dt → 0 as n → ∞
0
Note that the weak convergent of {un } in H01 (0, 1) implies the uniform convergence
of {un } in C([0, 1], R) [8, Proposition 1.2 ]. Hence
Z 1
p(ξ)kun k2 →
[(q(t) + p(t))|u0 (t)|2 + h(t, u0 )u0 (t)]dt as n → ∞
0
This means that {un } converges in H01 (0, 1). So the P.S. condition holds.
Thirdly, we show that Theorem 2.1 holds by Lemma 1.1. Denote βk (t) =
√
2
kπ cos kπt, k = 1, 2, 3, . . . , n, . . . , then
Z 1
Z 1
1
2
|βk (t)| dt = 2 2 ,
|βk0 (t)|2 dt = 1
k π
0
0
Definite n-dimensional linear space
En = span{β1 (t), β2 (t), . . . βn (t)}.
It is obvious that En is a symmetric set. Suppose ρ > 0, then
E n ∩ Sρ =
n
X
bk βk :
k=0
n
X
k=0
b2k (1 +
1
) = ρ2
k2 π2
1
Let g(t, u) = ξ(t)
h(t, u) − u, by condition (iii) of Theorem 2.1, limu→0
uniformly for t ∈ [0, 1]. We choose ε such that
g(u)
u
= 0,
1
2(M + p(1)|γ|)(1 + π 2 )
−
.
n2
λ
By condition (iii) of Theorem 2.1, there exist δ > 0 such that |g(t, u)| ≤ ε|u|
whenever |u| ≤ δ. We can choose ρ such that 0 < ρ < min{α, δ}, and have
n
n √
X
X
2
max u(t) ≤
|bk | ≤ kuk = k
bk βk k
0≤t≤1
kπ
0<ε<
k=0
n
X
=(
k=0
k=0
b2k (1 +
1
))1/2 = ρ < min{α, δ}
k2 π2
EJDE-2004/137
MULTIPLE SOLUTIONS FOR ODE’S
7
when u ∈ En ∩ Sρ . So
Z
u
G(t, u) = ξ(t)
[v + g(t, v))]dv
Z u
1
= ξ(t)|u(t)|2 + ξ(t)
g(t, v)dv
2
0
Z u
1
≥ ξ(t)|u(t)|2 − ξ(t)
εvdv
2
0
1
= ξ(t)(1 − ε)|u(t)|2
2
≥ λ(1 − ε)|u(t)|2
0
From q(t) + p(t) ≥ − λ2 , we get that
n
1
Z
(q(t) + p(t))|u(t)|2 dt ≤
−
0
λ X b2k
2
k2 π2
k=0
So
1
Z
1
1
[ p(t)(|u0 (t)|2 + |u(t)|2 ) − (q(t) + p(t))|u(t)|2 ]dt
2
2
I(u) =
0
1
Z
−
G(t, u)dt +
0
Z
p(1) 2
γu (1)
2
1
1
1
[ (|u0 (t)|2 + |u(t)|2 ) − λ(1 − ε)|u(t)|2 ]dt
2
0 2
n
2
X
λ
bk
p(1)
+
+
|γ|kuk2C
4
k2 π2
2
≤M
k=0
n
n
n
X
X
M X 2
1
1
b2k
p(1)
)
+
|γ|(
bk (1 + 2 2 ) − λ(1 − 2ε)(
≤
2
k π
4
k2 π2
2
k=0
k=0
k=0
n
√
2|bk | 2
)
kπ
n
X b2
1
1
M + p(1)|γ| X 2
k
bk (1 + 2 2 ) − λ(1 − 2ε)
≤
2
k π
4
k2 π2
<
M + p(1)|γ|
2
k=0
n
X
k=0
k=0
n
X
1
1
b2k
2
bk (1 + 2 ) − λ(1 − ε)
π
4
n2 π 2
k=0
n
X
λ M + p(1)|γ| π 2 + 1
1
≤ (
b2k
− 2 2 + ε)
2
2
λ
π
2n π
k=0
n
2
≤
X
λ (M + p(1)|γ|)(1 + π )
1
(
− 2 + ε)
b2k
2
2π
λ
2n
2
≤
λ 2(M + p(1)|γ|)(1 + π )
1
(
− 2 + ε)
4π 2
λ
n
k=0
n
X
b2k < 0
k=0
By Lemma 1.1 and the above result, we have i1 (I) ≥ n and I has 2n distinct critical
points, i.e., boundary-value problem (1.1)-(1.2) has at least 2n nontrivial solutions
in C 2 [0, 1].
8
X. B. SHU & Y. T. XU
EJDE-2004/137
Next we consider the boundary-value problem (1.1)-(1.3). Similar to Theorem
2.1, we have the following result.
Theorem 2.2. Let f , p(t) and q(t) satisfy the following conditions:
(i) p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1]
(ii) f ∈ C([0, 1] × R, R)
(iii) limu→0 f (t,u)
= ξ(t) > 0 uniformly for t ∈ [0, 1], λ = min0≤t≤1 ξ(t)
u
(iv) There exists α > 0, such that f (t, α) ≤ 0
(v) f (t, u) is odd in u
(vi) − λ2 < q(t) + p(t) ≤ 0 for all 0 ≤ t ≤ 1.
Then (1.1)-(1.3) has at least 2n nontrivial solutions in C 2 [0, 1] whenever
2n2 M (1 + π 2 ) < λ ≤ 2(n + 1)2 M (1 + π 2 ).
Next, we consider the boundary-value problem (1.1)-(1.2).
Theorem 2.3. Let f , p(t) and q(t) satisfy the following conditions:
(i) p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1]
(ii) f ∈ C([0, 1] × R, R)
(iii) There exists T such that lim supu→0 f (t,u)
≤T
u
≥
2
and
α
>
0
such
that
(iv) There exists θ > 2M
m
Z u
1
0 < G(t, u) =
f (t, v)dv ≤ uf (t, u), ∀|u| ≥ α
θ
0
(v) f (t, u) is odd in u
(vi) q(t) ∈ C[0, 1], q(t) + p(t) ≤ 0 for all 0 ≤ t ≤ 1.
mθ−2M
, then (1.1)-(1.2) has infinite nontrivial solutions in C 2 [0, 1].
If γ > − 2(θ−2)p(1)
Proof. It is easy to see that for u ∈ H01 (0, 1), the functional
Z 1
1
1
p(1) 2
I(u) =
[ p(t)|u0 (t)|2 − q(t)|u(t)|2 G(t, u(t))]dt +
γu (1)
2
2
2
0
(2.13)
is well defined. The solutions of boundary-value problems (1.1)-(1.2) are the critical
points of the functional I(u). Note that I(u) is equivalent to
Z 1
1
1
p(1) 2
I(u) =
[ p(t)(|u0 (t)|2 + |u(t)|2 ) − (q(t) + p(t))|u(t)|2 − λG(t, u)]dt +
γu (1)
2
2
2
0
We show that Theorem 2.3 holds by using Lemma 1.2. Since f (t, u) is an odd
continuous map in u, we know that I ∈ C 1 (E, R) is even in u and I(θ) = 0.
Moreover, As in the proof of Theorem 2.1, one gets
Z 1
0
(I (u), v) =
[p(t)u0 (t)v 0 (t) − q(t)u(t)v(t)) − f (t, u)v(t)]dt + γp(1)u(1)v(1),
0
for all u, v ∈ E. The above equality is equivalent to
Z 1
Z 1
f (t, u)v(t)dt =
p(t)(u0 (t)v 0 (t) + u(t)v(t))dt
0
0
Z
−
0
1
(p(t) + q(t))u(t)v(t)dt − (I 0 (u), v) + p(1)γu(1)v(1).
EJDE-2004/137
MULTIPLE SOLUTIONS FOR ODE’S
9
Next, we verify that I(u) satisfies the Palais-Smale condition. Suppose that un ⊂ E
with
c1 ≤ I(un ) ≤ c2 ,
0
I (un ) → 0
as n → ∞
(2.14)
(2.15)
Now, we show that {un } of satisfying (2.14) and (2.15) is bounded. Denote E1 =
{t ∈ [0, 1]||un (t)| ≥ α}, E2 = [0, 1] \ E1 . We divide our proof into two parts .
mθ−2M
(A) When − 2(θ−2)p(1)
< γ < 0, by (iv), we have
1
p(t)
(|un (t)|2 + |u0n (t)|2 )dt −
2
Z
1
Z
I(un ) =
0
Z
0
1
1
(q(t) + p(t))|un (t)|2 dt
2
p(1) 2
−
G(t, un (t))dt +
γun (1)
2
0
Z
Z
m
≥ kun k2 −
G(t, un (t))dt −
G(t, un (t))dt
2
E1
E2
Z 1
1
p(1) 2
γun (1) −
(q(t) + p(t))|un (t)|2 dt
+
2
0 2
Z
m
1
≥ kun k2 −
un (t)f (t, un (t))dt − c3
2
E1 θ
Z 1
1
p(1) 2
γun (1) −
(q(t) + p(t))|un (t)|2 dt
+
2
0 2
Z 1
m
1
2
≥ kun k −
un (t)f (t, un (t))dt − c4
2
0 θ
Z 1
p(1) 2
1
+
γun (1) −
(q(t) + p(t))|un (t)|2 dt
2
0 2
Z 1
m
1
= kun k2 − (
p(t)(|u0n (t)|2 + |un (t)|2 )dt − (I 0 (un ), un )
2
θ 0
Z 1
p(1) 2
1 1
+ p(1)γu2n (1)) − c4 +
γun (1) − ( − )
(q(t) + p(t))|un (t)|2 dt
2
2 θ 0
m M
1
1 1
≥( −
)kun k2 + (I 0 (un ), un ) − c4 + ( − )p(1)γu2n (1)
2
θ
θ
2 θ
m M
1 0
1 1
2
≥( −
)kun k + kI (un )kkun k − c4 + ( − )p(1)γu2n (1)
2
θ
θ
2 θ
Remarks: (1) for the rest of this article, ci > 0. (2) The above equality makes use
R1
of −( 12 − θ1 ) 0 (q(t) + p(t))|un (t)|2 dt ≥ 0
To show that {un } is bounded, we divide again our proof into two parts.
(I’) When max0≤t≤1 |u(t)| ≤ 1, as in the proof of Theorem 2.1, we have u2 (1) ≤ 1
and
m M
1
1 1
−
)kun k2 + kI 0 (un )kkun k − c4 + ( − )γ
2
θ
θ
2 θ
m M
1 0
1 1
2
( −
)kun k ≤ I(un ) − kI (un )kkun k + c4 − ( − )γ
2
θ
θ
2 θ
I(un ) ≥ (
Using θ >
2M
m ,
(2.14) and (2.15), it is not difficulty to see that {kun k} is bounded.
10
X. B. SHU & Y. T. XU
EJDE-2004/137
(II’) When max0≤t≤1 |u(t)| > 1, as in the proof of Theorem 2.1, we have u2 (1) ≤
2kuk2 and
m M
1 1
1
−
) + ( − )2p(1)γ]kun k2 ≤ I(un ) − kI 0 (un )kkun k + c4 ≤ c5 kun k + c5 .
2
θ
2 θ
θ
2M
mθ−2M
Since θ > m ≥ 2 and γ > − 2(θ−2)p(1) , it follows that {kun k} is bounded. From
mθ−2M
(I’) and (II’), we get that {kun k} is bounded when − 2(θ−2)p(1)
< γ < 0.
(B) when γ > 0, as in the proof of (A), it is not difficult to see that
Z 1
Z 1
p(t)
I(un ) =
(|un (t)|2 + |u0n (t)|2 dt −
G(t, un (t))dt
2
0
0
Z 1
p(1) 2
1
+
γun (1) −
(q(t) + p(t))|un (t)|2 dt
2
2
0
Z 1
m
1
2
≥ kun k −
un (t)f (t, un (t))dt
2
0 θ
Z 1
p(1) 2
1
+
γun (1) −
(q(t) + p(t))|un (t)|2 dt
2
0 2
m M
1
1 1
≥( −
)kun k2 + kI 0 (un )kkun k − c4 + ( − )p(1)γu2n (1)
2
θ
θ
2 θ
m M
1 0
2
)kun k + kI (un )kkun k − c4
≥( −
2
θ
θ
(We remark that the above equality use ( 12 − θ1 )p(1)γu2 (1) ≥ 0 ) and
[(
m M
1
−
)kun k2 ≤ I(un ) − kI 0 (un )kkun k + c4 ≤ c5 kun k + c5 .
2
θ
θ
So, we get that {kun k} is bounded when γ > 0.
From (A) and (B), we obtain that {un } satisfying (2.14) and (2.15) is bounded.
So the P.S. condition holds.
Thirdly, we show that Theorem 2.3 holds by using Lemma 1.2. First, we verify
condition (F1) of Lemma 1.2. Let βj (t) = cos jt, j = 1, 2, . . . . Consider the ndimensional subspace
(
En = span{β1 (t), β2 (t), . . . , βn (t)}
⊥
and let X = V . By (ii), we have δ > 0 such that |f (t, u(t))| ≤ T |u|, whenever
|u| ≤ δ.
Let ρ with ρ = δ. For any u ∈ Sρ ∩ X, we have kukC ≤ kuk = ρ = δ. From
Z 1
Z 1
1
|u(t)|2 dt ≤ 2
|u0 (t)|2 dt
n
0
0
it is easy to see
Z
0
1
|u(t)|2 dt ≤
ρ2
n2 + 1
So, when γ > 0, we have
Z 1
Z 1
p(t)
2
0
2
I(un ) =
(|un (t)| + |un (t)| )dt −
G(t, un (t))dt
2
0
0
Z 1
p(1) 2
1
+
γun (1) −
(q(t) + p(t))|un (t)|2 dt
2
0 2
EJDE-2004/137
MULTIPLE SOLUTIONS FOR ODE’S
11
Z 1
m
kuk2 −
G(t, u(t))dt
2
0
Z 1 Z |u(t)|
m 2
(
T vdv)dt
≥ ρ −
2
0
0
T
1
T
m
ρ2 = (m − 2
)ρ2 > 0 .
≥ ρ2 −
2
2(n2 + 1)
2
n +1
≥
T
T
Note that in the above equality, we use n2 > max{ m
, m+p(1)γ
} and
Z 1
1
−
(q(t) + p(t))|un (t)|2 dt > 0.
2
0
mθ−2M
When − 2(θ−2)p(1)
< γ < 0, we get
Z 1
Z 1
p(t)
(|un (t)|2 + |u0n (t)|2 )dt −
G(t, un (t))dt
I(un ) =
2
0
0
Z 1
p(1) 2
1
+
γun (1) −
(q(t) + p(t))|un (t)|2 dt
2
2
0
Z 1
m
p(1) 2
≥ kuk2 −
γun (1)
G(t, u(t))dt +
2
2
0
Z 1 Z |u(t)|
m + p(1)γ 2
ρ −
≥
(
T vdv)dt
2
0
0
m + p(1)γ 2
T
ρ2
≥
ρ −
2
2(n2 + 1)
1
T
= (m + p(1)γ − 2
)ρ2 > 0 .
2
n +1
T
T
Note that the above equality uses n2 > max{ m
, m+p(1)γ
}.
We sum up the conclusions above to obtain that I(u) > 0 for all u ∈ Sρ ∩ X,
i.e., condition (F1) of Lemma 1.2 holds.
Finally, we verify condition (F2) of Lemma 1.2. By (iv), one gets
G(t, u(t)) ≥ c7 |u|θ − c8 .
For all finite dimensional subspace E1 of E, there exist c9 such that
Z 1
1/θ
|u(t)|θ dt
≥ c9 kuk, ∀u ∈ E1 .
0
On the other hand, since p(t) ∈ C 1 [0, 1], q(t) ∈ C[0, 1] and p(t) + q(t) ≤ 0, there
exist a positive number Q such that −Q = min0≤t≤1 p(t) + q(t), so
Z 1
Z 1
2
−
(p(t) + q(t))|u(t)| dt ≤ Q
|u(t)|2 dt < Qkuk2 .
0
When u ∈ E1 and
0
mθ−2M
− 2(θ−2)p(1)
Z
I(u) =
1
< γ < 0, from the above result, it is easy to obtain
p(t) 0 2
(|u (t)| + |u(t)|2 ) − G(t, u(t))]dt
2
0
Z 1
p(1) 2
1
+
γu (1) −
(q(t) + p(t))|u(t)|2 dt
2
2
0
[
12
X. B. SHU & Y. T. XU
≤
M +Q
kuk2 −
2
Z
EJDE-2004/137
1
G(t, u(t))dt
0
Z 1
M +Q
2
≤
kuk − c7
|u(t)|θ dt + c8
2
0
M +Q
kuk2 − c7 cθ9 kukθ + c8
≤
2
M +Q
=(
− c7 cθ9 kukθ−2 )kuk2 + c8 .
2
When u ∈ E1 and γ ≥ 0, as in the proof of (I’) and (II’), we have the following two
results.
(1) when max0≤t≤1 |u(t)| ≤ 1, we have
Z 1
p(t) 0 2
I(u) =
[
(|u (t)| + |u(t)|2 ) − G(t, u(t))]dt
2
0
Z 1
p(1) 2
1
γu (1) −
(q(t) + p(t))|u(t)|2 dt
+
2
2
0
Z 1
M +Q
p(1)
2
≤
kuk − c7
|u(t)|θ dt + c8 +
γ
2
2
0
M +Q
p(1)
≤
kuk2 − c7 cθ9 kukθ + c8 +
γ
2
2
M +Q
p(1)
=(
− c7 cθ9 kukθ−2 )kuk2 + c8 +
γ
2
2
(2) when max0≤t≤1 |u(t)| > 1, we have u2 (1) ≤ 2kuk2 and
Z 1
p(t) 0 2
(|u (t)| + |u(t)|2 ) − G(t, u(t))]dt
I(u) =
[
2
0
Z 1
p(1) 2
1
+
γu (1) −
(q(t) + p(t))|u(t)|2 dt
2
0 2
Z 1
M + Q + 2γ
≤
kuk2 − c7
|u(t)|θ dt + c8
2
0
M + Q + 2γ
≤
kuk2 − c7 cθ9 kukθ + c8
2
M + Q + 2γ
=(
− c7 cθ9 kukθ−2 )kuk2 + c8
2
We sum up the conclusions above to obtain that I(u) ≤ 0, for all u ∈ E1 \BR
when R = R(E1 ) is adequately big, i.e., condition (F2) of Lemma 1.2 holds. So
I possesses infinite critical point, i.e. the boundary-value problem (1.1)-(1.2) has
infinitely many nontrivial solutions in C 2 [0, 1].
Using a technique similar to the one above, we can show that the following
theorem.
Theorem 2.4. Let f , p(t) and q(t) be the function satisfying the following conditions:
(i) p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1]
(ii) f ∈ C([0, 1] × R, R)
(iii) There exists T such that lim supu→0 f (t,u)
≤T
u
EJDE-2004/137
MULTIPLE SOLUTIONS FOR ODE’S
2M
m
≥ 2 and α > 0 such that
Z u
1
0 < G(t, u) =
f (t, v)dv ≤ uf (t, u),
θ
0
(iv) There exists θ >
13
∀|u| ≥ α
(v) f (t, u) is odd in u
(vi) q(t) ∈ C[0, 1], q(t) + p(t) ≤ 0 for 0 ≤ t ≤ 1.
Then boundary-value problem (1.1)-(1.3) has infinitely solutions in C 2 [0, 1].
3. Examples
Example 3.1. For 0 < t < 1, consider the boundary-value problem
d
du
((6 + sin t) ) + (−100 + cos t)u + (1000(1 + t2 ) sin u − 10u3 ) = 0,
dt
dt
u0 (0) = 0, u(1) + u0 (1) = 0
(3.1)
Note that
f (t, u) = 1000(1 + t2 ) sin u − 10u3 ,
p(t) = 6 + sin t,
q(t) = −100 + cos t
So f (t, u) satisfy conditions (ii) and (v) of Theorem 2.1. In addition,
0 < 5 ≤ p(t) = 6 + sin t ≤ 7 ∀t ∈ [0, 1].
then (i) and (vi) of Theorem 2.1 hold. When |u(t)| = 4, we have
f (t, 4) = 1000(1 + t2 ) sin 4 − 10 × 43 < 0 ,
= 1 + t2 uniformly
i.e., (iv) of Theorem 2.1 holds. On the other that limu→0 f (t,u)
u
2
for t ∈ [0, 1], λ = min0≤t≤1 1 + t = 1000, i.e., (iii) holds, and
2 × 12 × (13 + sin 1)(1 + π 2 ) < λ < 2 × 22 × (13 + sin 1)(1 + π 2 )
By Theorem 2.1 we have (3.1) has at least 2 nontrivial solutions in C 2 [0, 1].
Example 3.2. For 0 < t < 1, consider boundary-value problem
du
d
((6 + sin t) ) + (−100 + cos t)u + (1000(1 + t2 ) sin u − 10u3 ) = 0,
dt
dt
u(0) = u(1) = 0,
(3.2)
As in Example 3.1, it is easy to verify all conditions of Theorem 2.2 hold and
2 × 22 × 7(1 + π 2 ) < λ < 2 × 32 × 7(1 + π 2 )
By Theorem 2.2, we have (3.2) has at least 4 nontrivial solutions in C 2 [0, 1].
Example 3.3. Consider the boundary-value problem
d
du
((6 + sin t) ) + (−9 + t2 )u(t) + t(u3 (t) + u(t)) = 0,
dt
dt
u0 (0) = 0, u(1) + u0 (1) = 0
0 < t < 1,
It is easy to see that
f (t, u) = t(u3 (t) + u(t)),
p(t) = 6 + sin t
q(t) = −9 + t2
So, f (t, u) satisfies conditions (ii) and (v) of Theorem 2.3. In addition
lim sup
u→0
f (t, u)
t(u3 + u)
= lim sup
=1
u
u
u→0
(3.3)
14
X. B. SHU & Y. T. XU
EJDE-2004/137
i.e., conditions (iii) of Theorem 2.3 hold. Moreover,
Z u
Z u
√
1
1
1
f (t, v)dv =
t(v 3 + v)dv = t( u4 + u2 ) ≤ ut(u3 + u) ∀|u(t)| > 2
4
2
3
0
0
2M
14
=
0 < 5 ≤ p(t) ≤ 7, θ = 3 >
m
5
So conditions (iv) holds. On the other hand, it is easy to see that
p(t) + q(t) = 6 + sin t + (−9 + t2 ) = −3 + t2 + sin t ≤ 0
∀t ∈ [0, 1]
So conditions (i) and (vi) of Theorem 2.3 hold. By Theorem 2.3, we obtain that
(3.3) has infinitely many nontrivial solutions in C 2 [0, 1].
Example 3.4. Consider boundary-value problem
d
du
((6 + sin t) ) + (−9 + t2 )u(t) + t(u3 (t) + u(t)) = 0,
dt
dt
u(0) = u(1) = 0,
0 < t < 1,
(3.4)
As in Example 3.3, it is easy to verify that all conditions of Theorem 2.4 hold.
By Theorem 2.4, we obtain that (3.4) has infinitely many nontrivial solutions in
C 2 [0, 1].
References
[1] I. Addou and Shin-Hwa Wang. Exact multiplicity results for a p-Laplacian positone problem
with concave-convex-concave nonlinearities, Electron J. diff. Eqns., Vol. 2004(2004), No. 72,
1-24.
[2] R. P. Agarwal and D. O’Regan; Existence theory for single and multiple solutions to singular
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[3] G. Anello; A multiplicity theorem for critical points of functionals on reflexive Banach spaces.
Arch. Math. 82 (2004), no. 2, 172-179.
[4] L. E. Bobisud, J. E. Calvert, and W. D. Royalty; Some existence results for singular boundary
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Jishu Chubanshe, Shanghai, 1986.
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problems. Nonlinear Anal. 46 (2001), no. 7, Ser. A: Theory Methods, 979-986.
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211-222.
[8] J. Mawhin and M. Willem; Critical point theory and Hamiltonian systems. Applied Mathematical Sciences, 74. Springer-Verlag, New York, 1989.
[9] P. H. Rabinowitz; Minimax methods in critical point theory with applications to differential
equations. CBMS Regional Conference Series in Mathematics, 65. by the American Mathematical Society, Providence, RI, 1986.
[10] Xu Yuan Tong and Guo Zhi-Ming; Applications of a Zp index theory to periodic solutions for
a class of functional differential equations. J. Math. Anal. Appl. 257 (2001), no. 1, 189-205.
[11] Xu Yuan Tong; Subharmonic solutions for convex nonautonomous Hamiltonian systems.
Nonlinear Anal. 28 (1997), 1359-1371.
Xiao-Bao Shu
Department of Mathematics, Sun Yat-Sen University, Guangzhou, 510275, China
E-mail address: [email protected]
Yuan-Tong Xu
Department of Mathematics, Sun Yat-Sen University, Guangzhou, 510275, China
E-mail address: [email protected]
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