On three solutions of the second order periodic boundary-value problem
We consider the periodic boundary-value problem x'' + a(t)x' + b(t)x = f(t, x, x'), x(') =x(2π), x'(0) = x' (2π), where a, b are Lebesgue integrable functions and f fulfils the Caratheodory conditions. We extend results about the Leray – Schauder topological deg...
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irk-123456789-1747632021-01-28T01:27:34Z On three solutions of the second order periodic boundary-value problem Draessler, J. Rachůnková, I. We consider the periodic boundary-value problem x'' + a(t)x' + b(t)x = f(t, x, x'), x(') =x(2π), x'(0) = x' (2π), where a, b are Lebesgue integrable functions and f fulfils the Caratheodory conditions. We extend results about the Leray – Schauder topological degree and ´ present conditions implying nonzero values of the degree on sets defined by lower and upper functions. Using such results we prove the existence of at least three different solutions to the above problem. 2001 Article On three solutions of the second order periodic boundary-value problem / J. Draessler, I. Rachůnková // Нелінійні коливання. — 2001. — Т. 4, № 3. — С. 471-486. — Бібліогр.: 6 назв. — англ. 1562-3076 AMS Subject Classification: 34B15, 34C25 http://dspace.nbuv.gov.ua/handle/123456789/174763 en Нелінійні коливання Інститут математики НАН України |
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We consider the periodic boundary-value problem x'' + a(t)x' + b(t)x = f(t, x, x'), x(') =x(2π), x'(0) = x' (2π), where a, b are Lebesgue integrable functions and f fulfils the
Caratheodory conditions. We extend results about the Leray – Schauder topological degree and ´ present conditions implying nonzero values of the degree on sets defined by lower and upper
functions. Using such results we prove the existence of at least three different solutions to the
above problem. |
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Article |
| author |
Draessler, J. Rachůnková, I. |
| spellingShingle |
Draessler, J. Rachůnková, I. On three solutions of the second order periodic boundary-value problem Нелінійні коливання |
| author_facet |
Draessler, J. Rachůnková, I. |
| author_sort |
Draessler, J. |
| title |
On three solutions of the second order periodic boundary-value problem |
| title_short |
On three solutions of the second order periodic boundary-value problem |
| title_full |
On three solutions of the second order periodic boundary-value problem |
| title_fullStr |
On three solutions of the second order periodic boundary-value problem |
| title_full_unstemmed |
On three solutions of the second order periodic boundary-value problem |
| title_sort |
on three solutions of the second order periodic boundary-value problem |
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Інститут математики НАН України |
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2001 |
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http://dspace.nbuv.gov.ua/handle/123456789/174763 |
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On three solutions of the second order periodic boundary-value problem / J. Draessler, I. Rachůnková // Нелінійні коливання. — 2001. — Т. 4, № 3. — С. 471-486. — Бібліогр.: 6 назв. — англ. |
| series |
Нелінійні коливання |
| work_keys_str_mv |
AT draesslerj onthreesolutionsofthesecondorderperiodicboundaryvalueproblem AT rachunkovai onthreesolutionsofthesecondorderperiodicboundaryvalueproblem |
| first_indexed |
2025-07-15T11:55:38Z |
| last_indexed |
2025-07-15T11:55:38Z |
| _version_ |
1837713891947184128 |
| fulltext |
Nonlinear Oscillations, Vol. 4, No. 4, 2001
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC
BOUNDARY-VALUE PROBLEM*
J. Draessler
University of Education in Hradec Králové, Czech Republic
e-mail: Jan.Draessler@uhk.cz
I. Rachůnková
Palacký University , Tomkova 40, 779 00 Olomouc, Czech Republic
e-mail: rachunko@risc.upol.cz
We consider the periodic boundary-value problem x′′ + a(t)x′ + b(t)x = f(t, x, x′), x(0) =
x(2π), x′(0) = x′(2π), where a, b are Lebesgue integrable functions and f fulfils the
Carathéodory conditions. We extend results about the Leray – Schauder topological degree and
present conditions implying nonzero values of the degree on sets defined by lower and upper
functions. Using such results we prove the existence of at least three different solutions to the
above problem.
AMS Subject Classification: 34B15, 34C25
1. Introduction
We will study the periodic boundary-value problem
x′′ + a(t)x′ + b(t)x = f(t, x, x′), (1.1)
x(0) = x(2π), x′(0) = x′(2π), (1.2)
where a, b are Lebesgue integrable functions on J = [0, 2π] and f fulfils the Carathéodory
conditions on J × R2.
Having values of the Leray – Schauder topological degree of an operator which corresponds
to problem (1.1), (1.2) and which is defined on proper sets, we can decide whether there are
solutions of (1.1), (1.2) lying in these sets. In [1] and [2], where the special case of equation
(1.1) (with a = b = 0 on J and with f having an one-sided Lebesgue integrable bound) was
considered, such sets were found by means of lower and upper functions of problem (1.1), (1.2).
Here we extend results about the degreee of [1, 2] to equation (1.1) with nonzero a, b.
Moreover we present theorems which guarantee the existence of at least three solutions to
(1.1), (1.2).
∗ Supported by grant No. 201/01/1451 of the Grant Agency of Czech Republic.
c© D. Draessler, I. Rachůnková, 2001 471
472 J. DRAESSLER, I. RACHUNKOVA
Throughout the paper we keep the following notations. L(J) is the Banach space of Lebesgue
integrable functions on J equipped with the norm ||x||1 =
2π∫
0
|x(t)|dt and L∞(J) denotes the
Banach space of essentially bounded on J functions with the norm ||x||∞ = ess sup {|x(t)| :
t ∈ J}. For k ∈ N ∪ {0}, Ck(J) and ACk(J) are the Banach spaces of functions having con-
tinuous k-th derivatives on J and of functions having absolutely continuous k-th derivatives on
J , respectively. As usual, the corresponding norms are defined by ||x||Ck =
k∑
i=0
max{|x(i)(t)| :
t ∈ J} and ||x||ACk = ||x||Ck+||x(k+1)||1. The symbols C(J) or AC(J) are used instead of C0(J)
or AC0(J). Car(J×R2) is the set of functions f : J×R2 → R satisfying the Carathéodory con-
ditions on J ×R2, i.e., (i) for each (x, y) ∈ R2 the function f(·, x, y) : J → R is measurable, (ii)
for a.e. t ∈ J the function f(t, ·, ·) : R2 → R is continuous, (iii) sup(x,y)∈K |f(t, x, y)| ∈ L(J)
for each compact set K ⊂ R2. For a Banach space X and a set M ⊂ X, cl (M) stands for the
closure ofM and ∂M denotes the boundary ofM . If Ω is an open bounded subset in C1(J) and
the operator T : cl (Ω) → C1(J) is compact, then deg (I − T,Ω) denotes the Leray – Shauder
topological degree of I−T with respect to Ω, where I stands for the identity operator on C1(J).
For a definition and properties of the degree see e.g. [3 – 6].
By a solution of problem (1.1), (1.2) we understand a function u ∈ AC1(J) satisfying (1.1)
for a.e. t ∈ J and fulfilling conditions (1.2).
A function σ1 ∈ AC1(J) is said to be a lower function of (1.1), (1.2), if
σ′′1 + a(t)σ′1 + b(t)σ1 ≥ f(t, σ1, σ
′
1) a.e. on J,
σ1(0) = σ1(2π), σ′1(0) ≥ σ′1(2π).
A function σ2 ∈ AC1(J) is called an upper function of (1.1), (1.2), if
σ′′2 + a(t)σ′2 + b(t)σ2 ≤ f(t, σ2, σ
′
2) a.e. on J,
σ2(0) = σ2(2π), σ′2(0) ≤ σ′2(2π).
A lower function σ1 of (1.1), (1.2) is called strict, if σ1 does not satisfy (1.1) a.e. on J and if there
exists ε ∈ (0,∞) such that
σ′′1 + a(t)y + b(t)x ≥ f(t, x, y)
holds a.e. on J and for all (x, y) ∈ [σ1(t), σ1(t) + ε]× [σ′1(t)− ε, σ′1(t) + ε].
An upper function σ2 of (1.1), (1.2) is called strict, if σ2 does not satisfy (1.1) a.e. on J and
if there exists ε ∈ (0,∞) such that
σ′′2 + a(t)y + b(t)x ≤ f(t, x, y) (1.3)
holds a.e. on J and for all (x, y) ∈ [σ2(t)− ε, σ2(t)]× [σ′2(t)− ε, σ′2(t) + ε].
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 473
Now, let us define operators which will make possible to write problem (1.1), (1.2) in an
operator form. Denote
domL = {x ∈ AC1(J) : x satisfies (1.2)}. (1.4)
We can see that
L : domL → L(J), x 7→ x′′ + a(·)x′ + b(·)x (1.5)
is a linear bounded operator and
F : C1(J) → L(J), x 7→ f(·, x(·), x′(·)) (1.6)
is a continuous (nonlinear in general) operator, and problem (1.1), (1.2) is equivalent to the
operator equation
Lx = Fx. (1.7)
To determine an operator the degree of which will be studied we need to distinguish two cases:
KerL = {0} and KerL 6= {0}.
We will say that problem (1.7) is resonance if KerL 6= {0}. If KerL = {0} the problem is
called nonresonance.
Both cases are investigated in Section 2.
2. Nonresonance and Resonance Problems
I. First, let us consider the nonresonance case KerL = {0}. It means that the homogeneous
linear boundary-value problem corresponding to (1.1), (1.2),
x′′ + a(t)x′ + b(t)x = 0, x(0) = x(2π), x′(0) = x′(2π), (2.1)
has the trivial solution, only. One class of nonresonance problems (1.1), (1.2) is characterized
in the next lemma.
Lemma 2.1. Let us suppose that a, b ∈ L(J) and that b satisfies
b(t) ≤ 0 a.e. on J (2.2)
and
2π∫
0
b(t)dt 6= 0. (2.3)
Then problem (2.1) has only the trivial solution, i.e., KerL = {0}.
474 J. DRAESSLER, I. RACHUNKOVA
Proof. Suppose on the contrary that KerL 6= {0}. Then there exists a nontrivial solution
u of (2.1) and, having in mind condition (1.2) and the fact that −u ∈ KerL, we can assume
without loss of generality that
max
t∈J
u(t) = u(t0) > 0, u′(t0) = 0, t0 ∈ [0, 2π). (2.4)
Further, if we extend the functions a, b and u to function that are 2π-periodic on R, we get for
all t ∈ R
u′(t) = −e−A(t)
t∫
t0
b(s)u(s)eA(s)ds, (2.5)
where A(t) =
∫ t
t0
a(s)ds. Conditions (2.4) and (2.5) yield
u(t) > 0 and u′(t) ≥ 0 for all t ∈ [t0,∞). (2.6)
On the other hand, in view of conditon (2.3) we see that u cannot be a constant function. This
together with the periodicity of u imply that u′ has to change its sign on each interval of the
length 2π, which contradicts (2.6). Thus problem (2.1) has only the trivial solution.
Remark 2.1. Condition (2.3) in Lemma 2.1 cannot be omitted because problem (2.1) with
b(t) = 0 a.e. on J has constant nontrivial solutions.
If KerL = {0}, then the Green function G of (2.1) exists and we can find the inverse (to L)
operator
L−1 : L(J) → dom L, y 7→
2π∫
0
G(t, s)y(s)ds. (2.7)
If we denote
L+ = iL−1 : L(J) → C1(J), (2.8)
where i : AC1(J) → C1(J) is the embedding operator, then the operator L+F is absolutely
continuous and problem (1.1), (1.2) is equivalent to the operator equation (I−L+F )x = 0, x ∈
domL. The degree theory implies that provided for some open bounded set Ω ⊂ C1(J) the
relation
deg (I − L+F,Ω) 6= 0 (2.9)
is true, the operator L+F has a fixed point in Ω. This means, in view of (2.7), (2.8), that this fixed
point belongs to domL and so problem (1.1), (1.2) has a solution in Ω. We will see in Section
4 that such a set Ω can be found by means of strict lower and upper functions of problem
(1.1), (1.2).
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 475
II. Now, we will consider resonance problems having KerL 6= {0}. Using Lemma 2.1 we
can transform such problems to nonresonance ones by means of auxiliary operators Lµ andHµ.
So, let KerL 6= {0} and let dom L be given by (1.4). Then for a µ ∈ (−∞, 0) we define a
linear operator
Lµ : dom L → L(J), x 7→ x′′ + a(·)x′ + µx (2.10)
and an operator
Hµ : C1(J) → L(J), x 7→ hµ(·, x(·), x′(·)), (2.11)
where
hµ(t, x, y) = f(t, x, y) + (µ− b(t))x.
We see that Lµ andHµ are continuous and problem (1.1), (1.2) is equivalent to the operator
equation
Lµx = Hµx. (2.12)
According to Lemma 2.1 problem (2.12) is nonresonance, i.e., KerLµ = {0}. Therefore we can
argue as in Part I and get the inverse (to Lµ) operator
L−1µ : L(J) → dom L, y 7→
2π∫
0
Gµ(·, s)y(s)ds,
where Gµ is the Green function of
x′′ + a(t)x′ + µx = 0, x(0) = x(2π), x′(0) = x′(2π). (2.13)
As before, denoting
L+
µ = iL−1µ : L(J) → C1(J), (2.14)
we arrive to the operator equation
(I − L+
µHµ)x = 0, x ∈ dom L, (2.15)
which is equivalent to (1.1), (1.2). Since L+
µHµ is absolutely continuous, we can use the degree
theory again and deduce that if
deg (I − L+
µHµ,Ω) 6= 0 (2.16)
for some open bounded set Ω ⊂ C1(J), then equation (2.15) has a solution in Ω∩dom L, which
implies that problem (1.1), (1.2) has a solution in Ω.
To summarize, for the existence of a solution to (1.1), (1.2) in Ω we need to prove:
(I) deg (I − L+F,Ω) 6= 0 if KerL = {0}.
(II) deg (I − L+
µHµ,Ω) 6= 0 for some negative µ if KerL 6= {0}.
476 J. DRAESSLER, I. RACHUNKOVA
3. Values of the Leray – Shauder Degree
In this section we prove several theorems with statements of the type (2.9) or (2.16). For
definitions of operators see (1.5), (1.6), (2.8), (2.10), (2.11) and (2.14).
Proposition 3.1. Let Ker L = {0}. Further suppose that there exist numbers c, r1 ∈ (0,∞)
such that for any λ ∈ [0, 1] each solution u of the equation
(I − λL+F )x = 0, x ∈ dom L (3.1)
satisfies
|u(tu)| < c for some tu ∈ J, ||u′||C < r1. (3.2)
Denote r0 = c+ 2πr1 and
Ω = {x ∈ C1(J) : ||x||C < r0, ||x′||C < r1}. (3.3)
Then
deg (I − L+F,Ω) = 1.
Proof. Let us choose λ ∈ [0, 1] and let u be a corresponding solution of (3.1) with this λ.
Then u fulfils (3.2) and so |u(t)| ≤ |u(tu)| +
∣∣∣ t∫
tu
u′(s)ds
∣∣∣ < c+
2π∫
0
|u′(s)|ds < r0 for each t ∈ J .
Therefore u 6∈ ∂Ω and so the operator I − λL+F is the homotopy on cl (Ω) × [0, 1], which
implies that deg (I − L+F,Ω)=deg (I,Ω) = 1.
Proposition 3.2. Let Ker L 6= {0} and let µ ∈ (−∞, 0). Moreover, let us suppose that there
are positive numbers c, r1 such that for any λ ∈ [0, 1] each solution u of the equation
(I − λL+
µHµ)x = 0, x ∈ dom L
satisfies (3.2). Then
deg (I − L+
µHµ,Ω) = 1,
where Ω is given by (3.3) and r0 = c+ 2πr1.
Proof. We can argue as in the proof of Proposition 3.1.
Using the homotopy argument as before we get the following modification of Proposi-
tion 3.1.
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 477
Proposition 3.3. Let Ker L = {0} and let there exist ρ∗ ∈ (0,∞) such that for any λ ∈ [0, 1]
each solution u of (3.1) satisfies ||u||C1 ≤ ρ∗. Then for each ρ > ρ∗,
deg (I − L+F,K(ρ)) = 1, (3.4)
where
K(ρ) = {x ∈ C1(J) : ||x||C1 < ρ}. (3.5)
We see that a priori estimates of solutions of problems under consideration are essential for
the determination of Ω and for the degree computation. In contrast to Propositions 3.1 – 3.3,
where we assumed such estimates directly, now, we will show conditions which can be imposed
on f to ensure the needed estimates.
Theorem 3.1. Let Ker L= {0} and let there exist e ∈ L(J) such that
|f(t, x, y)| ≤ e(t) for a.e. t ∈ J and each x, y ∈ R. (3.6)
Then there exists ρ∗ ∈ (0,∞) such that (3.4), (3.5) are true for each ρ > ρ∗.
Proof. Let u be a solution of (3.1) for some λ ∈ [0, 1]. Then
u(t) = λ
2π∫
0
G(t, s)f(s, u(s), u′(s))ds,
where G is the Green function of (2.1). Denote
γ = max{|G(t, s)| : t, s ∈ J}, δ = max
{∣∣∣∣∂G(t, s)
∂t
∣∣∣∣ : t, s ∈ J
}
.
Then ||u||C1 ≤ (γ + δ)||e||1 = ρ∗ and we can use Proposition 3.3.
Remark 3.1. In the case KerL 6= {0}, condition (3.6) need not be sufficient for the exis-
tence of solutions of (1.1), (1.2), which is obvious if we choose (1.1) in the form x′′ = 1. (Clearly,
the problem x′′ = 0, x(0) = x(2π), x′(0) = x′(2π) has nontrivial solutions and the problem
x′′ = 1, x(0) = x(2π), x′(0) = x′(2π) is not solvable.) Moreover, having KerL 6= {0}, the
Green function G of (2.1) does not exist and we cannot argue as in the proof of Theorem 3.1
and hence, without additional assumptions, we are not able to get an assertion about the degree
as before. In this case, the method of lower and upper functions, which is used in Section 4, can
be a profitable instrument.
4. The Leray – Shauder Degree and Lower and Upper Functions
Let us consider problem (1.1), (1.2) and functions σ1, σ2 ∈ AC1(J). Further, for any
µ ∈ (−∞, 0) let Gµ be the Green function of (2.13) and let the operators Lµ, L+
µ , Hµ be given
by (2.10), (2.14) and (2.11). We denote
ri = max{||σ(i)1 ||C, ||σ
(i)
2 ||C}, i = 0, 1, γµ = max
J×J
∣∣∣∣∂Gµ(t, s)
∂t
∣∣∣∣ . (4.1)
478 J. DRAESSLER, I. RACHUNKOVA
Proposition 4.1. Let σ1, σ2 be strict lower and upper functions of (1.1), (1.2) such that
σ1 < σ2 on J, (4.2)
and let there exist e ∈ L(J) satisfying
|f(t, x, y)| < e(t) for a.e. t ∈ J and each (x, y) ∈ [σ1(t), σ2(t)]× R. (4.3)
Then for any µ ∈ (−∞, 0)
deg (I − L+
µHµ,Ωµ) = 1, (4.4)
where
Ωµ = {x ∈ C1(J) : σ1 < x < σ2 on J, ||x′||C < Mµ}, (4.5)
and Mµ ≥ γµ(3||e||1 + (||b||1 − 2πµ)r0 + ||a||1).
Proof. Let us choose µ ∈ (−∞, 0) and put, for a.e. t ∈ J and for each (x, y) ∈ R2,
qµ(t, x, y) = f(t, σ(x), y) + (µ− b(t))σ(x),
where
σ(x) =
σ2(t), if σ2(t) < x;
x, if σ1(t) ≤ x ≤ σ2(t);
σ1(t), if x < σ1(t).
Further, define
pµ(t, x, y) =
qµ(t, x, y) + ω
(
t,
x− σ2(t)
x− σ2(t) + 1
)
, if σ2(t) < x;
qµ(t, x, y), if σ1(t) ≤ x ≤ σ2(t);
qµ(t, x, y)− ω
(
t,
σ1(t)− x
σ1(t)− x+ 1
)
, if x < σ1(t),
(4.6)
and for ε ∈ [0, 1],
ω(t, ε) = sup
(x,y,z)∈Dt,ε
{|f(t, x, y)− f(t, x, z)|+ |a(t)(y − z)|},
where Dt,ε = {(x, y, z) ∈ R3 : σ1(t) ≤ x ≤ σ2(t), |y| ≤ 1 + |σ′1(t)| + |σ′2(t)|, |y − z| ≤ ε}.
We can see that ω ∈ Car(J × [0, 1]) is nonnegative and nondecreasing in the second variable,
ω(t, 0) = 0 a.e. on J . Moreover, for a.e. t ∈ J and any y ∈ R satisfying |y − σ′i(t)| ≤ 1 the
inequality
|f(t, σi, σ
′
i)− f(t, σi, y)|+ |a(t)(y − σ′i)| ≤ ω(t, |y − σ′i|), i = 1, 2, (4.7)
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 479
is true. In view of (4.6), for a.e. t ∈ J and for all (x, y) ∈ R2, we have
|pµ(t, x, y)| < 3e(t) + (|b(t)| − µ)r0 + |a(t)|. (4.8)
Recall that Lµ is defined by (2.10) and define an operator
Pµ : C1(J) → L(J), x 7→ pµ(·, x(·), x′(·)).
With respect to Lemma 2.1, we have KerLµ = {0}. Therefore, according to (4.8), Theorem 3.1
ensures the existence of ρ∗ ∈ (r0 +Mµ,∞) such that for each ρ > ρ∗,
deg (I − L+
µPµ,K(ρ)) = 1, (4.9)
where K(ρ) = {x ∈ C1(J) : ||x||C1 < ρ}. Let us consider an arbitrary solution u ∈ dom L of
the equation (I−L+
µPµ)x = 0 and let us prove that u ∈ Ωµ. Since u(t) =
2π∫
0
Gµ(t, s)pµ(s, u(s),
u′(s))ds for all t ∈ J , we have that
u′′ + a(t)u′ + µu = pµ(t, u, u′)
for a.e. t ∈ J . By (4.1) and (4.8), we get
||u′||C ≤ max
t∈J
2π∫
0
∣∣∣∣∂Gµ(t, s)
∂t
∣∣∣∣ |pµ(s, u(s), u′(s))|ds < Mµ. (4.10)
Let us show that
σ1 < u < σ2 on J. (4.11)
Put v = u− σ2 on J and assume on the contrary that
max
t∈J
{v(t)} = v(t0) ≥ 0.
Then, having in mind conditions (1.2), we can assume without loss of generality that v′(t0) = 0
and t0 ∈ [0, 2π).
First, let v(t0) > 0. Then there is δ > 0 such that for a.e. t ∈ (t0, t0 + δ)
v(t) > 0, |v′(t)| < v(t)
v(t) + 1
< 1. (4.12)
Therefore we have, for a.e. t ∈ (t0, t0 + δ),
v′′(t) =u′′(t)− σ′′2(t) ≥ f(t, σ2, u
′) + (µ− b(t))σ2 + ω
(
t,
x− σ2
x− σ2 + 1
)
− a(t)u′ − µu− f(t, σ2, σ
′
2) + a(t)σ′2 + b(t)σ2,
480 J. DRAESSLER, I. RACHUNKOVA
and using (4.7), (4.12), we get v′′(t) > 0 for a.e. t ∈ (t0, t0 + δ). Hence,
0 <
t∫
t0
v′′(s)ds ≤ v′(t) for all t ∈ (t0, t0 + δ),
which contradicts the fact that v(t0) is the maximal value of v on J . Thus, u ≤ σ2 on J . The
inequality σ1 ≤ u on J can be proved analogously putting v = σ1 − u on J . So, we have
σ1 ≤ u ≤ σ2 on J. (4.13)
It remains to prove that the inequalities in (4.13) must be strict. Suppose that v(t0) = 0.
Since σ2 is a strict upper function of (1.1), (1.2), there is ε > 0 such that (1.3) is valid a.e. on J
and for all x ∈ [σ2(t)−ε, σ2(t)], y ∈ [σ′2(t)−ε, σ′2(t)+ε]. Moreover, since σ2 is not a solution of
(1.1), there is δ > 0 such that for each t ∈ [t0, t0+δ) the inequalities−ε ≤ v(t) ≤ 0, |v′(t)| ≤ ε
are satisfied and we can assume without loss of generality that there exists ξ ∈ (t0, t0 + δ) such
that v′(ξ) < 0. On the other hand, according to (1.3), we have
v′′(t) = u′′(t)− σ′′2(t) = f(t, u, u′)− a(t)u′(t)− b(t)u(t)− σ′′2(t) ≥ 0
for a.e. t ∈ (t0, t0 + δ), thus
0 ≤
ξ∫
t0
v′′(s)ds = v′(ξ) < 0,
a contradiction. Therefore u < σ2 on J . The inequality σ1 < u on J can be proved similarly
for v = σ1 − u on J . Thus, we have proved (4.10) and (4.11), which means that u belongs to
Ωµ. But then, by (4.9) and the excission property of the degree, we get
deg (I − L+
µPµ,Ωµ) = 1,
and since Pµ = Hµ on cl (Ωµ), assertion (4.4) is valid.
Corollary 4.1. Let the assumptions of Proposition 4.1 be fulfilled and moreover let Ker L=
{0}. Further, suppose that G is the Green function of (2.1) and the operators L+, F are given by
(1.5), (1.6). Then
deg (I − L+F,Ω) = 1, (4.14)
where
Ω = {x ∈ C1(J) : σ1 < x < σ2 on J, ||x′||C < M}
and M = maxJ×J
{∣∣∣∣∂G(t, s)
∂t
∣∣∣∣} ||e||1.
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 481
Proof. We can argue similarly as in the proof of Proposition 4.1, working with G,L, F and
q(t, x, y) = f(t, σ(x), y) instead of Gµ, Lµ, Hµ and qµ.
Remark 4.1. Comparing Theorem 3.1 and Corollary 4.1 we see that we have used different
sets in their assertions (3.5) and (4.14) about degree values. In (3.5) we work with a ball K(ρ)
the radius of which is not specified, it is sufficiently large, only, while the set Ω in (4.14) is
described by means of lower and upper functions σ1 and σ2. Such specification of the set Ω will
be useful for the multiplicity result in Section 5.
Using a proper lemma on a priori estimates, we can weaken condition (4.3) in Proposi-
tion 4.1. Let us show one of such lemmas.
Lemma 4.1. Suppose that r ∈ (0,∞), q ∈ L∞(J), a, b, p ∈ L(J), q, p positive a.e. on J.
Further, let a constant r∗ satisfy r∗ ≥ (eM − A)A, where A = exp(||a||1) and M = r(2||q||∞ +
||b||1) +||a||1 + ||p||1. Then for each x ∈ AC1(J) fulfilling conditions (1.2),
||x||C < r (4.15)
and
x′′ + a(t)x′ + b(t)x ≤ (1 + |x′|)(q(t)|x′|+ p(t)) for a.e. t ∈ J, (4.16)
the estimate
||x′||C < r∗ (4.17)
is valid.
Proof. Suppose that x ∈ AC1(J) satisfies conditions (1.2), (4.15) and (4.16) and extend
x, q, a, b, p on R as 2π-periodic functions. Let us assume that max{x′(t) : t ∈ J} = x′(t0) > 0.
Then we can find τ0 < t0 such that t0 − τ0 < 2π, x′(τ0) = 0 and x′(t) > 0 on (τ0, t0]. With
respect to (4.16) we have, for a.e. t ∈ [τ0, t0],
x′′ + a(t)x′ ≤ (1 + x′)(q(t)x′ + p(t) + |b(t)|r).
Multiply this inequality by exp
t∫
τ0
a(s)ds
and put z(t) = x′(t) exp
t∫
τ0
a(s)ds
. Then, inte-
grating from τ0 to t0, we get
t0∫
τ0
z′(t)dt
A+ z(t)
< 2r||q||∞ + ||p||1 + ||b||1r.
Therefore z(t0) < eM −A and so x′(t0) < r∗.
482 J. DRAESSLER, I. RACHUNKOVA
Similarly, if we assume that min{x′(t) : t ∈ J} = x′(t1) < 0, we can find τ1 > t1 with
τ1 − t1 < 2π, x′(τ1) = 0, x′(t) < 0 on [t1, τ1). Then (4.16) yields a.e. on [t1, τ1]
x′′ + a(t)x′ ≤ (1− x′)(−q(t)x′ + p(t) + |b(t)|r).
Multiply this inequality by exp
t∫
τ1
a(s)ds
and put z(t) = −x′(t) exp
t∫
τ1
a(s)ds
. Then,
integrating from t1 to τ1, we get
−
τ1∫
t1
z′(t)dt
A+ z(t)
< 2r||q||∞ + ||p||1 + ||b||1r.
Therefore z(t1) < eM −A, and so x′(t1) > −r∗. The lemma is proved.
Consider the constant r∗ from Lemma 4.1 and put
e∗(t) = sup{|f(t, x, y)| : x ∈ [σ1(t), σ2(t)], y ∈ [−2r∗, 2r∗]}. (4.18)
Clearly e∗ ∈ L(J) and using Proposition 4.1 and Lemma 4.1 we can prove the following theo-
rem.
Theorem 4.1. Let σ1 and σ2 be strict lower and upper functions of (1.1), (1.2) satisfying (4.2).
Further, suppose that there exist functions q ∈ L∞(J), d ∈ L(J) which are positive a.e. on J and
such that for a.e. t ∈ J and for all x ∈ [σ1(t), σ2(t)], y ∈ R
f(t, x, y) ≤ (1 + |y|)(q(t)|y|+ d(t)). (4.19)
Then for any µ ∈ (−∞, 0)
deg (I − L+
µHµ,Ω
∗) = 1, (4.20)
where
Ω∗ = {x ∈ C1(J) : σ1 < x < σ2 on J, ||x′||C < r∗}, (4.21)
with r∗ from Lemma 4.1. (For L+
µ and Hµ see (2.14) and (2.11).)
Proof. Let us take r0 and r1 according to (4.1), put
r = r0, p = d a.e. on J, (4.22)
and assume that r∗ from Lemma 4.1 satisfies r∗ > r1. For y ∈ R define
χ(y, r∗) =
1, if |y| ≤ r∗;
2− |y|/r∗, if r∗ < |y| < 2r∗;
0, if |y| ≥ 2r∗,
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 483
and consider the equation
x′′ + a(t)x′ + b(t)x = f∗(t, x, x′), (4.23)
where f∗(t, x, y) = χ(y, r∗)f(t, x, y) for a.e. t ∈ J and all x, y ∈ R. We can see that σ1 and σ2
are strict lower and upper functions for (4.23), (1.2), and that
|f∗(t, x, y)| < e∗(t) for a.e. t ∈ J and for all x ∈ [σ1(t), σ2(t)], y ∈ R,
where e∗ is given by (4.18). So, for any µ ∈ (−∞, 0), we can define an operator
H∗µ : C1(J) → L(J), x 7→ f∗(·, x(·), x′(·)) + (µ− b(·))x
and a set Ωµ by (4.5) with Mµ = r∗ + γµ(3||e∗||1 + (||b||1 − 2πµ)r0 + ||a||1). Then, applying
Proposition 4.1 to problem (4.23), (1.2), we get
deg (I − L+
µH
∗
µ,Ωµ) = 1. (4.24)
Let u ∈ Ωµ be a solution of (4.23), (1.2). Then, by (4.22), (4.19), we have ||u||C < r and
u′′ + a(t)u′ + b(t)u = χ(u′, r∗)f(t, u, u′) ≤ (1 + |u′|)(q(t)|u′|+ p(t)) a.e. on J.
Therefore, by Lemma 4.1, ||u′||C < r∗ and so, in view of (4.21), u ∈ Ω∗. Using (4.24) and the
excission property of the degree we get deg (I − L+
µH
∗
µ,Ω
∗) = 1 which, together with the fact
that Hµ = H∗µ on cl (Ω∗), imply (4.20).
Corollary 4.2. Let the assertions of Theorem 4.1 be fulfilled and moreover let KerL = {0}.
Further, suppose that the operators L+, F are given by (1.5), (1.6). Then
deg (I − L+F,Ω∗) = 1,
with Ω∗ by Theorem 1.1.
Proof. We can argue similarly as in the proof of Theorem 4.1, working with L,F , F ∗ :
C1(J) → L(J), x 7→ f∗(·, x(·), x′(·)) and Corollary 4.1 instead of Lµ, Hµ, H∗µ and Proposi-
tion 4.1, respectively.
5. Main Results
Using properties of the Leray – Shauder degree we get the following existence result as the
direct consequence of Theorem 4.1 or Corollary 4.2.
Theorem 5.1. Let σ1 and σ2 be strict lower and upper functions of (1.1), (1.2) satisfying (4.2).
Further, suppose that there exist functions q ∈ L∞(J), d ∈ L(J) which are positive a.e. on J and
such that for a.e. t ∈ J and for all x ∈ [σ1(t), σ2(t)], y ∈ R condition (4.19) is satisfied. Then
problem (1.1), (1.2) has at least one solution x such that σ1 < x < σ2 on J .
484 J. DRAESSLER, I. RACHUNKOVA
Remark 5.1. The existence of a solution to (1.1), (1.2) can be proved under weaker as-
sumptions than those in Theorem 5.1. Particularly, σ1 and σ2 need not be strict and we can
assume that σ1 ≤ σ2 on J . Then (1.1), (1.2) has a solution x satisfying σ1 ≤ x ≤ σ2 on J . For
the proof of this generalization we can modify the corresponding proofs in [2].
Now, we will prove our main result about the existence of three solutions of problem
(1.1), (1.2). To this aim we will consider reverse ordered lower and upper functions σ1 and
σ2 of this problem , i.e., we will suppose
σ2 < σ1 on J. (5.1)
Theorem 5.2. Let σ1 and σ2 be strict lower and upper functions of (1.1), (1.2) satisfying (5.1).
Let the inequalities
lim inf
x→∞
(f(t, x, 0)− b(t)x) > 0, lim sup
x→−∞
(f(t, x, 0)− b(t)x) < 0 (5.2)
be fulfilled uniformly for a.e. t ∈ J . Finally, suppose that there exist functions q ∈ L∞(J),
d ∈ L(J) which are positive a.e. on J and such that condition (4.19) holds for a.e. t ∈ J and for
all x, y ∈ R. Then problem (1.1), (1.2) has at least three different solutions.
Proof. According to inequalities (5.2) we can find a number ρ > max{||σ1||C, ||σ2||C} such
that
f(t, ρ, 0)− b(t)ρ > 0 f(t,−ρ, 0) + b(t)ρ < 0, a.e. on J. (5.3)
For a.e. t ∈ J and for all x, y ∈ R define functions
g(t, x, y) = f(t, x, y)− a(t)y − b(t)x,
h(t, x, y) =
g(t,−ρ, y)− ω1
(
t,
−ρ− x
−ρ− x+ 1
)
, if x < −ρ;
g(t, x, y), if |x| ≤ ρ;
g(t, ρ, y) + ω2
(
t,
x− ρ
x− ρ+ 1
)
, if x > ρ,
and, for ε > 0, put
ωi(t, ε) = sup
z∈[−ε,ε]
{|g(t, (−1)iρ, 0)− g(t, (−1)iρ, z)|}, i = 1, 2.
We will study the auxiliary equation
x′′ = h(t, x, x′). (5.4)
Choose an arbitrary number η > 0 and put σ̃2(t) = ρ+ η, σ̃1(t) = ρ− η for all t ∈ J . Then, in
view of (5.3),
h(t, ρ+ η, 0) = g(t, ρ, 0) + ω2
(
t,
η
η + 1
)
> 0
ON THREE SOLUTIONS OF THE SECOND ORDER PERIODIC BOUNDARY-VALUE PROBLEM 485
is valid for a.e. t ∈ J . This means that σ̃2 is an upper function of (5.4), (1.2) and that it is not
a solution of (5.4). Further, put ε = (η/2)(η/2 + 1)−1 and choose arbitrary x ∈ [σ̃2 − ε, σ̃2],
y ∈ [σ̃′2 − ε, σ̃′2 + ε]. Then
x ∈
(
ρ+
η
2
, ρ+ η
)
, y ∈ [−ε, ε], |y| < x− ρ
x− ρ+ 1
, (5.5)
whence
ω2(|y|) ≤ ω2
(
t,
x− ρ
x− ρ+ 1
)
.
Thus, according to (5.5), we have
h(t, x, y) =g(t, ρ, y) + ω2
(
t,
x− ρ
x− ρ+ 1
)
≥g(t, ρ, 0)− |g(t, ρ, y)− g(t, ρ, 0)|+ ω2(t, |y|) > 0,
and we proved that σ̃2 is a strict upper function of (5.4), (1.2). Similarly we can get that σ̃1 is a
strict lower function of (5.4), (1.2).
Equation (5.4) can be written in the form
x′′ + a(t)x′ + b(t)x = f̃(t, x, x′),
where f̃(t, x, y) = h(t, x, y) + a(t)y + b(t)x. Put p(t) = d(t) + |b(t)|η + ω2(η/(η + 1)) a.e. on
J . Then, by (4.19), for a.e. t ∈ J and for all (x, y) ∈ [σ̃1, σ̃2] × R, the inequality f̃(t, x, y) ≤
(1 + |y|)(q(t)|y|+ p(t)) is satisfied.
Therefore any solution x of problem (5.4), (1.2) which fulfils ||x||C ≤ ρ+ η, satisfies condi-
tion (4.16). So, if we put r = ρ+ η, we can use Lemma 4.1 and get r∗ such that estimate (4.17)
is valid. According to this r∗ we define the sets
D = {x ∈ C1(J) : ||x||C < ρ+ η, ||x′||C < r∗},
D1 = {x ∈ D : σ1 < x on J}, D2 = {x ∈ D : x < σ2 on J},
and
D3 = {x ∈ D : σ2(tx) < x(tx) < σ1(tx) for all tx ∈ J}.
Choose µ ∈ (−∞, 0) and define an operator
H̃µ : C1(J) → L(J), x 7→ f̃(·, x(·), x′(·)) + (µ− b(·))x.
Then Theorem 4.1 guarantees that
deg (I − L+
µ H̃µ, D1) = 1, deg (I − L+
µ H̃µ, D2) = 1, (5.6)
and
deg (I − L+
µ H̃µ, D) = 1.
486 J. DRAESSLER, I. RACHUNKOVA
(For L+
µ see (2.14).) Now, we use the aditivity of the degree. Since D3 = D − cl (D1 ∪D2),
where D1, D2 ⊂ D are disjoint sets, we have
deg (I − L+
µ H̃µ, D) = deg (I − L+
µ H̃µ, D1) + deg (I − L+
µ H̃µ, D2) + deg (I − L+
µ H̃µ, D3).
Therefore
deg (I − L+
µ H̃µ, D3) = −1. (5.7)
Conditions (5.6) and (5.7) imply that problem (5.4), (1.2) has solutions xi ∈ Di, i = 1, 2, 3.
Since D1, D2 and D3 are disjoint, the solutions x1, x2 and x3 are different.
It remains to prove that any solution x of (5.4), (1.2) satisfies
||x||C ≤ ρ. (5.8)
Suppose that x is an arbitrary solution of (5.4), (1.2) and that maxt∈J x(t) = x(t0) > ρ.
Without loss of generality we can suppose that there is an interval [t0, τ ] ⊂ [0, 2π) such that
x′(t0) = 0, x(t) > ρ and |x′(t)| < x(t)− ρ
x(t)− ρ+ 1
for all t ∈ [t0, τ ].
Then for a.e. t ∈ [t0, τ ],
x′′ = h(t, x, x′) = g(t, ρ, x′) + ω2
(
t,
x(t)− ρ
x(t)− ρ+ 1
)
> g(t, ρ, 0)− |g(t, ρ, x′)− g(t, ρ, 0)|+ ω2(t, |x′|) > 0,
which implies that x′(t) > 0 for all t ∈ (t0, τ ]. But this contradicts the fact that x(t0) is the
maximum value on J . The estimate x ≥ −ρ on J can be proved analogously. Thus the solutions
x1, x2 and x3 satisfy estimate (5.8) and so they are solutions of problem (1.1), (1.2), as well. This
completes the proof.
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Received 19.10.2001
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