- Email: [email protected]
A two-parametric three-point problem
Nonlinear Analysis 72 (2010) 3707–3721
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
A two-parametric three-point problem G. Bognár a , J. Čepička b , P. Drábek b,∗ , P. Nečesal b , E. Rozgonyi a a
Institute of Mathematics, University of Miskolc, 3515 Miskolc-Egyetemváros, Hungary
b
University of West Bohemia, Univerzitní 22, 306 14 Plzeň, Czech Republic
article
abstract
info
We study the existence of a solution x = x(t ), t ∈ R, of the three-point problem
Article history: Received 12 March 2009 Accepted 12 January 2010
x00 + λx + g (t , x) = h,
x(0) = 0,
Bη x = 0.
(P)
We give sufficient and necessary conditions on g and h in order for (P) to have a solution. Our conditions depend on the values of λ and η. © 2010 Elsevier Ltd. All rights reserved.
MSC: 34B10 34B15 Keywords: Landesman–Lazer type condition Leray–Schauder degree Spectrum of the three-point problem
1. Introduction In this paper we study the existence of a solution x = x(t ), t ∈ R, to the three-point problem x00 + λx + g (t , x) = h(t ),
x(0) = 0,
Bη x = 0,
(1)
where (η, λ) ∈ R2 are parameters, g and h are given functions and x(η) − x(π ) x0 (π )
Bη x :=
for η 6= π , for η = π .
In particular, if η = 0 then x satisfies homogeneous Dirichlet boundary conditions at 0 and π ; if η = π then x satisfies homogeneous Dirichlet–Neumann conditions at 0 and π . Necessary and sufficient conditions for the existence of a solution x = x(t ), t ∈ (0, π ), of equation x00 +λx + g (t , x) = h(t ) subject to the Dirichlet–Neumann and Dirichlet boundary conditions at 0 and π , respectively, are given by Landesman–Lazer conditions (cf. [1] and [5]). More general, the three-point boundary value problem (1) with η ∈ (0, π ) was studied recently in [2]. However, the investigation of the ‘‘global’’ solution of (1) is more involved. These facts motivate the detailed study of the set
σ := {(η, λ) ∈ R2 : x00 + λx = 0, x(0) = 0, Bη x = 0 has a non-zero solution x = x(t ), t ∈ R} as well as of corresponding non-zero solutions.
∗
Corresponding author. E-mail addresses: [email protected] (G. Bognár), [email protected] (J. Čepička), [email protected] (P. Drábek), [email protected] (P. Nečesal), [email protected] (E. Rozgonyi). 0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.01.007
3708
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
One of the main goals of this paper is to provide sufficient and also necessary conditions in terms of the set σ and the functions g and h in order for (1) to have at least one solution. In formulating these conditions we have to distinguish among several cases, depending on the position of parameter η. By a (global) solution to (1) we understand a function x = x(t ) defined on the whole of R such that x0 is absolutely continuous and the equation in (1) is satisfied pointwise almost everywhere in R. Our basic assumptions on g and h are the following ones: h ∈ L1loc (R);
(h1)
g is a Carathéodory function, i.e., g (t , ·) is continuous for almost all t ∈ R and g (·, s) is measurable for all s ∈ R, and there exist functions m, l ∈ L1loc (R) such that for all s ∈ R and almost all t ∈ R we have
|g (t , s)| ≤ m(t ) + l(t ) |s| .
(g1)
Next, we formulate the main results of this paper. In what follows, we set
( (η, π ) Iη := (0, π ) (0, η)
for η < 0, for 0 ≤ η ≤ π , for η > π .
The nonresonance case, (η, λ) 6∈ σ , is rather simple. Let us assume that, in addition to (g1) and (h1), also the following condition is satisfied: there exists δ ∈ [0, 1) such that for all s ∈ R and almost all t ∈ Iη we have
|g (t , s)| ≤ m(t ) + l(t ) |s|δ .
(g2)
Then problem (1) has at least one solution (cf. Theorem 5, Section 4). On the other hand, the resonance case, (η, λ) ∈ σ , is more complicated. Let ϕ be a non-zero solution (‘‘eigenfunction’’) of x00 + λx = 0, x(0) = 0, Bη x = 0 (cf. Sections 2 and 5) and ψ be a function associated with (η, λ) ∈ σ given in Section 2 (an ‘‘eigenfunction’’ of the adjoint problem; cf. Remark 3). We assume that for all s ∈ R and almost all t ∈ Iη ,
|g (t , s)| ≤ m(t ),
(g3)
and, moreover, for almost all t ∈ Iη the following limits exist: g (t , ±∞) := lim g (t , s).
(g4)
s→±∞
We also assume that
Z
g (t , +∞)ψ(t ) dt +
Iη+
<
Z Iη+
Z
g (t , −∞)ψ(t ) dt <
Iη−
Z
h(t )ψ(t ) dt Iη
g (t , −∞)ψ(t ) dt +
Z Iη−
g (t , +∞)ψ(t ) dt ,
(gh)
where Iη+ := {t ∈ Iη : ϕ(t ) > 0} and Iη− := {t ∈ Iη : ϕ(t ) < 0}. Then problem (1) has at least one solution (cf. Theorem 10). If for almost all t ∈ Iη and all s ∈ R we have g (t , +∞) < g (t , s) < g (t , −∞),
(g5)
then
Z
g (t , +∞)ψ(t ) dt +
b Iη+
<
Z b Iη+
Z
g (t , −∞)ψ(t ) dt <
b Iη−
g (t , −∞)ψ(t ) dt +
Z
h(t )ψ(t ) dt Iη
Z b Iη−
g (t , +∞)ψ(t ) dt ,
(2)
is necessary for (1) to have a solution. Here
b Iη+ := t ∈ Iη : ψ(t ) > 0
and b Iη− := t ∈ Iη : ψ(t ) < 0 .
In this paper we also study the bifurcation problem x00 + λx + g (λ, t , x) = 0,
x(0) = 0,
Bη x = 0,
(3)
where g represents ‘‘higher order terms’’. We point out that classical abstract bifurcation results of Rabinowitz and Dancer can be applied to get non-zero solutions of problem (3). A special case of the above equation is also considered, x00 + λ sin x = 0, in order to prove the existence of multiple solutions and to illustrate the role of the set σ in connection with the number of solutions (see Section 6 for the details).
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3709
Fig. 1. The structure of σ .
2. Non-zero solutions of linear problems As we already pointed out in the Introduction, detailed investigation of the set σ is important, i.e., we are looking for all pairs (η, λ) ∈ R2 for which x00 + λx = 0,
x(0) = 0,
Bη x = 0
(4)
has a non-zero solution. It is a matter of elementary calculation that linear problem (4) has a non-zero solution if and only if (η, λ) ∈ σ , where
σ :=
[
C2k ∪
k∈Z\{0}
[
C2j+1 ,
j∈Z
n
C2j+1
√
o λ = 2kπ , n o √ := (η, λ) ∈ R × R+ : (π + η) λ = (2j + 1)π .
C2k := (η, λ) ∈ R × R+ : (π − η)
√
The corresponding non-zero solution is then a non-zero multiple of ϕ(t ) = sin λt, t ∈ R (see Fig. 1). Let us give more detailed expression for the set σ depending on the position of η. 1.
If η < −π ,
σ =
∞ [
C2k ∪
If η = −π ,
σ =
∞ [
= (−π , k2 ) : k ∈ N .
C2k
k=1
3.
If − π < η < π ,
σ =
∞ [
C2k ∪
k=1
4.
If η = π ,
σ =
∞ [
C2j+1 .
j=−∞
k=1
2.
−1 [
∞ [
C2j+1 .
j =0
=
C2j+1
π,
(2j + 1)2 4
j =0
5.
If η > π ,
σ =
−1 [ k=−∞
C2k ∪
∞ [
: j ∈ N ∪ { 0} .
C2j+1 .
j =0
Note that for η = −π the set σ consists of (−π , k2 ) where k2 are the eigenvalues of the periodic problem x00 + λx = 0,
x(−π ) = x(π ),
x0 (−π ) = x0 (π ).
On the other hand, if η = π then σ consists of π ,
(2j+1)2 4
where
(2j+1)2 4
are the eigenvalues of the Dirichlet–Neumann
problem x00 + λx = 0,
x(0) = 0,
x0 (π ) = 0.
In general, if (η, λ) ∈ C2k for some k ∈ Z \ {0} then η 6= π and λ is an eigenvalue of the periodic problem x00 + λx = 0,
x(η) = x(π ),
x0 (η) = x0 (π );
3710
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
if (η, λ) ∈ C2j+1 for some j ∈ Z then η 6= −π and λ is an eigenvalue of the Dirichlet–Neumann problem x + λx = 0,
x(0) = 0,
00
x
0
η+π 2
= 0.
Let us also define the set
b σ := {(η, λ) ∈ σ : ∃j ∈ Z, k ∈ Z \ {0} such that (η, λ) ∈ C2k ∩ C2j+1 }. Notice that the set b σ consists of all intersections of the curves C2k and C2j+1 . Notice also that the projection of b σ onto the η-axis, projηb σ , consists of points
ηj,k =
2j − 2k + 1 2j + 2k + 1
π,
j ∈ Z, k ∈ Z \ {0}
(5)
and for any i = 0, 1, 2, . . . , we have
ηj,k = η(2i+1)j+i,(2i+1)k . In particular, for any (η, λ) ∈ b σ we also have (η, λi ) ∈ b σ , where λi = (2i + 1)2 λ, i = 1, 2, . . .. Moreover, we prove the following assertion. Lemma 1. The set projη b σ is dense in R. Proof. Clearly, we have −π , 0, π 6∈ projη b σ . Let us write j = m + n, k = n − m, m, n ∈ Z, m 6= n, in (5). Then
ηj,k =
2(j − k) + 1 2(j + k) + 1
π=
4m + 1 4n + 1
π.
It is enough to show that the set
M=
4m + 1 4n + 1
: m, n ∈ Z, m 6= n
p
is dense in R. To this end, let q , p ∈ Z, q ∈ N, be any rational number. For p 6= q we choose a sequence that 4ip + 1
lim
i→∞
4iq + 1
=
p q
lim
4(i + 1)p + 1 4iq + 1
4ip+1 4iq+1
o∞
⊂ M such
i=1
.
For p = q we choose a sequence
i→∞
n
n
4(i+1)p+1 4iq+1
o∞
⊂ M satisfying
i =1
= 1.
The density of M in R then follows.
Next we study necessary and sufficient conditions in order for x00 + λx = h,
x(0) = 0,
Bη x = 0
to have a solution. Namely, it follows from the variation of constants formula that (6) has a solution for any h ∈ whenever (η, λ) 6∈ σ . On the other hand, if (η, λ) ∈ σ then (6) has a solution if and only if
Z
(6) L1loc
(R)
h(t )ψ(t ) dt = 0, Iη
where both the function ψ and the domain of integration Iη depend on the position of η. Let η < 0, η 6= −π and (η, λ) ∈ σ . We have two different possibilities: either (η, λ) ∈ C2k or (η, λ) ∈ C2j+1 . Here k = 1, 2, 3, . . . for η < 0, η 6= −π ; j = −1, −2, −3, . . . for η < −π and j = 0, 1, 2, . . . for −π < η < 0. Then the variation of constants formula together with three-point condition from (1) imply that ψ = ψ2k if (η, λ) ∈ C2k , and ψ = ψ2j+1 if (η, λ) ∈ C2j+1 , where (see Figs. 2 and 3) 2kπ ψ2k (t ) = sin (π − t ) in (η, π ), π −η (2j + 1)π − sin π + η (π + t ) in (η, 0), ψ2j+1 (t ) = (2j + 1)π − sin (π − t ) in [0, π ). π +η
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3711
Fig. 2. The graphs of ϕ(t ) (dotted lines) and ψ5 (t ) with (η, λ) ∈ C5 .
Let η = −π and (−π , λ) ∈ σ . In this case λ = k2 , ψ = ψ2k , where (see Fig. 3)
ψ2k (t ) = sin k(π − t )
in (−π , π ).
Let 0 ≤ η < π and (η, λ) ∈ σ . We have two different possibilities: either (η, λ) ∈ C2k or (η, λ) ∈ C2j+1 . Here j = 0, 1, 2, . . . , k = 1, 2, 3, . . . , ψ = ψ2k if (η, λ) ∈ C2k , and ψ = ψ2j+1 if (η, λ) ∈ C2j+1 , where (see Figs. 2 and 3)
ψ2k (t ) =
0
in (0, η),
2kπ
(π − t ) in [η, π ), sin π −η (2j + 1)π (2j + 1)π 2 cos π + η π sin π + η t ψ2j+1 (t ) = (2j + 1)π (π − t ) − sin π +η Let η = π and (π , λ) ∈ σ . In this case λ =
ψ2j+1 (t ) = sin
2j + 1 2
t
(2j+1)2 4
in (0, η), in [η, π ). , ψ = ψ2j+1 , where (see Fig. 3)
in (0, π ).
Let η > π and (η, λ) ∈ σ . We have again two different possibilities: either (η, λ) ∈ C2k or (η, λ) ∈ C2j+1 . Here j = 0, 1, 2, . . . , k = −1, −2, −3, . . . , ψ = ψ2k if (η, λ) ∈ C2k , and ψ = ψ2j+1 if (η, λ) ∈ C2j+1 , where (see Figs. 2 and 3)
ψ2k (t ) =
0
in (0, π),
2kπ
(π − t ) in [π , η), sin π −η (2j + 1)π (2j + 1)π 2 cos π + η π sin π + η t ψ2j+1 (t ) = (2j + 1)π sin (π + t ) π +η
in (0, π ), in [π , η).
Lemma 2. Let (η, λ) ∈ σ , ϕ be a non-zero solution of (4) and ψ be as above. Then Proof. Let (η, λ) ∈ C2k , k ∈ Z \ {0}. Recall that η 6= π . For η < π , we have
Z
ϕ(t )ψ(t ) dt = Iη
π
Z η
ϕ(t )ψ(t ) dt =
η−π 2
cos
2kπ π =: J1 (η). π −η
R Iη
ϕ(t )ψ(t ) dt = 0 if and only if (η, λ) ∈ b σ.
3712
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
Fig. 3. The contour graphs of ψ4 (left) and ψ7 (right) as functions of (t , η).
On the other hand, for η > π , we have
Z
ϕ(t )ψ(t ) dt =
η
Z π
Iη
ϕ(t )ψ(t ) dt = −J1 (η).
π Finally, J1 (η) = 0 if and only if π2k−η = 12 + k + j, j ∈ Z, i.e., η = ηj,k . This is equivalent to (η, λ) ∈ b σ. Let (η, λ) ∈ C2j+1 , j ∈ Z. Recall that η 6= −π in this case. For η < 0, we have
Z
ϕ(t )ψ(t ) dt =
π
Z η
Iη
0
Z
ϕ(t )ψ2j+1 (t ) dt = −
sin η
(2j + 1)π (2j + 1)π t sin (π + t ) dt π +η π +η
π
(2j + 1)π (2j + 1)π t sin (π − t ) dt π +η π +η 0 π +η (2j + 1)π = cos η =: J2 (η). 2 π +η Z
−
sin
For 0 ≤ η < π , we have
Z
ϕ(t )ψ(t ) dt =
π
Z
ϕ(t )ψ2j+1 (t ) dt =
2 cos
0
Iη
η
Z 0
π
Z −
sin η
(2j + 1)π (2j + 1)π π sin2 t dt π +η π +η
(2j + 1)π (2j + 1)π t sin (π − t ) dt = J2 (η). π +η π +η
For η = π , we have
Z
ϕ(t )ψ(t ) dt =
π
Z
sin2
0
Iη
π (2j + 1) t dt = > 0. π +η 2
For η > π , we have
Z
ϕ(t )ψ(t ) dt = Iη
η
Z
ϕ(t )ψ2j+1 (t ) dt =
π
Z
2 cos
0
0
η
Z + π
(2j + 1)π (2j + 1)π sin t sin (η + t ) dt = J2 (η). π +η π +η (2j+1)π
Finally J2 (η) = 0, η 6= π , if and only if π +η
=
1 2
+ j − k, k ∈ Z \ {0}, i.e., η = ηj,k , which is equivalent to (η, λ) ∈ b σ.
Let h ∈ X := L1 (Iη ). Then the boundary value problem x00 = h,
x(0) = 0,
(2j + 1)π (2j + 1) π sin2 t dt π +η π +η
Bη x = 0
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3713
has a solution if and only if x = Tη h, where Tη : X → X is defined as follows:
(Tη h)(t ) :=
t
Z
(t − τ )h(τ ) dτ + 0
for η 6= π and
(Tη h)(t ) :=
π
Z
(τ − t )h(τ ) dτ −
π −η π
Z
0
Z
t
η
(τ − η)h(τ ) dτ −
π
Z
(π − τ )h(τ ) dτ
0
τ h(τ ) dτ
0
t
for η = π . The operator Tη : X → W 2,1 (Iη ) is linear, continuous and Im Tη = x ∈ W 2,1 (Iη ) : x(0) = 0, Bη x = 0 .
Moreover, thanks to the compact imbedding W 2,1 (Iη )⊂ ⊂ X the operator Tη is compact from X into itself. Problem (4) can be restated as an operator equation in X : x + λTη x = 0. In particular, (η, λ) ∈ σ if and only if − λ1 is a real eigenvalue of Tη . It is a matter of elementary calculation to see that (η, λ) ∈ σ \ b σ implies that the (algebraic) multiplicity of − λ1 is 1, i.e.,
dim Ker
1
λ
I + Tη
= dim Ker
1
λ
2 I + Tη
= 1,
and (η, λ) ∈ b σ implies that this multiplicity is 2, i.e.,
dim Ker
1
λ
I + Tη
= 1,
dim Ker
1
λ
2 I + Tη
= dim Ker
1
λ
3 I + Tη
= 2.
Let Br (0) be a ball in X centered at the origin and with radius r > 0. We denote as deg(I + λTη ; Br (0), 0)
(7)
the Leray–Schauder degree of the mapping I + λTη with respect to the ball Br (0) and the origin 0. It follows from the Leray–Schauder degree index formula that for (η, λ) ∈ R2 \ σ the degree (7) is well-defined, it is constant with respect to (η, λ) in the same component (i.e., the open and connected part) of R2 \ σ and it is equal to either +1 or −1. Remark 3. Let Tη∗ : X ∗ → X ∗ be the adjoint operator associated with Tη . Here X ∗ = L∞ (Iη ). Let (η, λ) ∈ σ be fixed. Then
the function ψ associated with (η, λ) is nothing but the eigenfunction of Tη∗ associated with the eigenvalue − λ1 . In particular, Tη∗ has an integral representation. For example, for η < 0, we have y = Tη∗ f if and only if
Z t Z π t −η sf (s) ds, (t − s)f (s) ds + π − η Zη y(t ) = Zη t π t −π sf (s) ds, (t − s)f (s) ds + π −η η π
t ∈ (η, 0), t ∈ (0, π ).
For a given f ∈ L∞ (Iη ) the function y ∈ L∞ (Iη ) can be interpreted as a solution of the equation y00 = f considered separately on the intervals (η, 0) and (0, π ) which satisfy the following ‘‘boundary and transition conditions’’: y(η) = y(π ) = 0,
lim y(t ) = lim y(t ),
t →0−
t →0+
lim y0 (t ) = lim y0 (t ).
t →π−
t →η+
1
In particular, the function y need not be a C -function at the point 0. For 0 ≤ η < π , η = π and η > π we have a similar integral representation of Tη∗ as well as a similar interpretation of y. 3. An initial value problem In this section we study the existence of a (global) solution to the initial value problem x00 + λx + g (t , x) = h(t ),
x(ξ ) = x0 ,
x0 (ξ ) = x1 ,
(8)
with ξ ∈ R being arbitrary but fixed. Namely, we show that under assumptions (g1) and (h1) stated in Section 1, problem (8) has a solution x = x(t ) defined on the whole of R. Indeed, notice that (8) can be written as a system x0 = y, x(ξ ) = x0 , y0 = −λx − g (t , x) + h(t ),
y(ξ ) = x1 ,
i.e., x0 = f(t , x),
x(ξ ) = (x0 , x1 ),
(9)
3714
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
where x = (x, y), f(t , x) = (y, −λx − g (t , x) + h(t )). Our assumptions on g and h imply that Carathéodory’s function f satisfies
|f(t , x)| ≤ M (t ) + L(t ) |x| for almost all t ∈ R and all x ∈ R 2 , where M, L ∈ L1loc (R) are fixed functions depending on λ and h, m, l ∈ L1loc (R). The existence of a global solution x = x(t ), t ∈ R, to (9) now follows from Schauder fixed point theorem combined with the usual extensibility procedure based on Gronwall’s lemma (cf. [3]). Hence (8) has a solution x = x(t ) which is defined on the whole of R. 4. A nonresonance problem In this section we assume that (η, λ) 6∈ σ . Under some additional assumptions on the growth of g we prove that (1) has at least one solution. We treat (1) separately on a bounded interval Iη and on its complement R \ Iη . Let us assume η < 0 and consider a three-point boundary value problem on the interval Iη = (η, π ): x00 + λx + g (t , x) = h(t ),
x(0) = 0,
x(η) = x(π ).
(10)
We prove later in this section that (10) has at least one solution xbv = xbv (t ), t ∈ [η, π]. According to Section 3 the equation x00 + λx + g (t , x) = h(t ) subject to the initial conditions x(η) = xbv (η),
x0 (η) = x0bv (η)
and x(π ) = xbv (π ),
x0 (π ) = x0bv (π )
has solutions x−∞ = x−∞ (t ), t ∈ (−∞, η] and x+∞ = x+∞ (t ), t ∈ [π , +∞), respectively. Putting x−∞ (t ), xbv (t ), x+∞ (t ),
( x(t ) =
t ∈ (−∞, η], t ∈ (η, π ), t ∈ [π , +∞)
we obtain a (global) solution of (1). Hence it remains to prove the existence of a solution of the boundary value problem (10) on the interval Iη . Proposition 4. Let h satisfy (h1) and g satisfy (g2). Then (10) has at least one solution x ∈ W 2,1 (Iη ). Proof. Let G : X → X be the Nemytskii operator associated with g, i.e., G(x)(t ) := g (t , x(t )), t ∈ Iη . Then (g2) implies that G is continuous. Problem (10) can then be written as an operator equation x = Tη h − λTη x − Tη (G(x)).
(11)
It follows from the compactness of Tη that the operator x → Tη h − λTη x − Tη (G(x)) is compact from X into itself. Consider the homotopy H (τ , x) := x + λTη x + τ Tη (G(x)) − Tη h
and prove that there is an r > 0 such that H (τ , x) 6= 0
(12)
for all τ ∈ [0, 1] and x ∈ ∂ Br (0). Assume the opposite, i.e., there exist τn ∈ [0, 1], kxn k → ∞ such that H (τn , xn ) = 0. Multiply (13) by
1 kxn k
(13) xn kxn k .
and define yn :=
yn + λTη yn + τn
Tη (G(xn ))
kxn k
−
Tη h
kx n k
Then
= 0.
(14)
We have Tη h
kx n k
→ 0,
Tη (G(xn ))
kx n k
→0
in X as n → ∞
by (g2) and passing to a subsequence if necessary we can assume (due to the compactness of Tη and [0, 1]) that there exist τ ∈ [0, 1], τn → τ , w ∈ X , Tη yn → w. It follows from (14) that there is a y ∈ X such that yn → y in X . Continuity of Tη
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3715
implies w = Tη y and using (14) once again we conclude that y + λTη y = 0.
(15)
Since kyk = 1, Eq. (15) contradicts (η, λ) 6∈ σ . Hence (12) is established. It now follows from the homotopy invariance property of the Leray–Schauder degree and Section 2 that deg(I + λTη + Tη (G(.)) − Tη h; Br (0), 0) = deg(I + λTη ; Br (0), 0) = ±1. Hence the existence of a solution of (10) follows.
Similar arguments apply for Iη with 0 ≤ η ≤ π and η > π . We conclude this section with the following statement. Theorem 5. Let (η, λ) 6∈ σ , h and g satisfy (h1), (g1) and (g2). Then problem (1) has at least one solution. Conjecture 6. Let (η, λ) 6∈ σ and d = dist {(η, λ), σ } > 0. Assume that a : Iη → R is a function satisfying
|a(t )| < d for almost all t ∈ Iη . Then x00 + (λ + a(t ))x = 0,
x(0) = 0,
Bη x = 0
has only the zero solution. Remark 7. Assume that Conjecture 6 is true. Then Proposition 4 and Theorem 5 hold true also with (g2) replaced by
|g (t , s)| ≤ m(t ) + d |s| for almost all t ∈ Iη and all s ∈ R. 5. The resonance case In this section we assume that (η, λ) ∈ σ is fixed. We discuss here in detail the case Iη = (η, π ) (i.e., η < 0) leaving the remaining two cases Iη = (0, π ) and Iη = (0, η) to the reader. Like in Section 4 we prove the existence of a solution to (1) by pasting together the solutions of initial value problems, x−∞ and x+∞ , defined on intervals (−∞, η] and [π , +∞), respectively, together with the solution of the boundary value problem (10). Let ϕ and ψ be functions associated with (η, λ) ∈ σ (cf. Section 2). There exists a constant c 6= 0 such that φ = c ϕ satisfies kφk = 1 and
Z
φ(t )ψ(t ) dt ≥ 0.
(16)
Iη
Proposition 8. Let (η, λ) ∈ σ \ b σ , h satisfy (h1), g satisfy (g1), (g3), (g4) and (gh). Then (10) has at least one solution. Proof. As in the proof of Proposition 4 from Section 4 we apply the Leray–Schauder degree argument. Notice that for (η, λ) 6∈ b σ the strong inequality holds in (16), i.e., we have
Z
φ(t )ψ(t ) dt > 0.
(17)
Iη
Notice that there exists an ε > 0 (small enough) such that (η, λ) 6∈ b σ for all µ ∈ [λ−ε, λ). Hence for τ ∈ (0, 1] the equation x + (λ − τ ε)Tη x = 0 has only a trivial solution. In particular, the Leray–Schauder degree deg(I + µTη ; Br (0), 0) is well-defined for all r > 0, µ ∈ [λ − ε, λ) and it is equal to either +1 or −1 (cf. Section 2). We consider the homotopy H (τ , x) = x + λTη x − τ ε Tη x + (1 − τ )(Tη (G(x)) − Tη h) for τ ∈ [0, 1] and x ∈ Br (0). Our goal is to prove that there is an r > 0 such that for any τ ∈ [0, 1] and any x ∈ X such that kxk = r, we have H (τ , x) 6= 0.
(18)
The reader should notice that if (18) is established then the homotopy invariance property of the Leray–Schauder degree would imply deg(I + λTη + Tη (G(.)) − Tη h; Br (0), 0) = deg(I + (λ − ε)Tη ; Br (0), 0) = ±1.
3716
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
The existence of a solution x ∈ Br (0) of (10) would immediately follow. Hence it remains to prove (18). We argue via contradiction. Let τn ∈ [0, 1], xn ∈ X , kxn k → ∞ be such that H (τn , xn ) = 0, i.e., xn + λTη xn − τn ε Tη xn + (1 − τn )(Tη (G(xn )) − Tη h) = 0. xn kxn k
Let yn :=
(19)
and divide (19) by kxn k. We obtain
yn + λTη yn − τn ε Tη yn + (1 − τn )
Tη (G(xn ))
kx n k
−
Tη h
= 0.
kxn k
(20)
If we pass to a suitable subsequence, we may assume that τn → τ ∈ [0, 1]. Since h ∈ X is fixed and g is bounded on Iη × R by a fixed function m ∈ X , we have Tη h
kx n k
→ 0,
Tη (G(xn ))
→0
kx n k
in X as n → ∞.
The compactness of Tη : X → X implies that we can pass again to a subsequence and assume that there exists w ∈ X such that Tη yn → w in X . Passing to the limit for n → ∞ in (20) we obtain yn → −λw + τ εw
in X .
We set y = −λw + τ εw . Then kyk = 1 and continuity of Tη implies Tη yn → Tη y
in X .
Hence from (20) we conclude that y + (λ − τ ε)Tη y = 0, i.e., y00 + (λ − τ ε)y = 0,
y(0) = 0,
y(η) = y(π ).
For τ 6= 0 we obtain a contradiction. However, there is no contradiction for τ = 0 but in this case τn → 0 and y, kyk = 1, is a solution of y00 + λy = 0,
y(0) = 0,
y(η) = y(π ).
Hence y = ±φ . Notice that (19) is equivalent to x00n + λxn − τn ε xn + (1 − τn )(g (t , xn ) − h) = 0.
(21)
According to Section 2 we have
Z
(x00n + λxn )ψ dt = 0. Iη
Multiplying (21) by ψ and integrating over Iη , we then get
Z
− τn ε
xn (t )ψ(t ) dt + (1 − τn )
Z
Iη
g (t , xn (t ))ψ(t ) dt = (1 − τn ) Iη
Z
h(t )ψ(t ) dt .
(22)
Iη
Let us assume that yn → φ . Then xn → +∞ on Iη+ and xn → −∞ on Iη− . By the Lebesgue theorem
Z
yn (t )ψ(t ) dt → Iη
Z
φ(t )ψ(t ) dt > 0 Iη
(by (17)). Hence, taking n large enough in (22), we obtain
Z
g (t , xn (t ))ψ(t ) dt ≥ Iη
Z
h(t )ψ(t ) dt .
(23)
Iη
Using the Lebesgue theorem again, we get
Z lim
n→∞
Iη
g (t , xn (t ))ψ(t ) dt =
Z Iη+
g (t , +∞)ψ(t ) dt +
Z Iη−
g (t , −∞)ψ(t ) dt .
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3717
Combining with (23) we have
Z Iη+
g (t , +∞)ψ(t ) dt +
Z Iη−
g (t , −∞)ψ(t ) dt ≥
Z
h(t )ψ(t ) dt , Iη
a contradiction with the left inequality in (gh). A similar argument yields a contradiction with the right inequality in (gh) if yn → −φ . The proof is finished.
Before we prove that the assertion of Proposition 8 remains valid also for (η, λ) ∈ b σ , we need some auxiliary properties of the operator Tη , namely, we need a kind of continuous dependence of Tη on η. More precisely, let ηn & η monotonically. For each n ∈ N let us extend Tηn (keeping the same notation) to an operator from L1 (Iη ) into L1 (Iη ) as follows: for h ∈ L1 (Iη ) we set
(Tηn h)(t ) :=
for t ∈ (η, ηn ), for t ∈ Iηn .
0 (Tηn h)(t )
For a fixed h ∈ L1 (Iη ) it follows from the definition of Tη , Tηn that
T η h − T η h → 0 n
as n → ∞.
(24)
In fact, we have even stronger property: given arbitrary but fixed c > 0 and arbitrary sequence hn ∈ L (Iη ), khn k ≤ c, we have 1
T η hn − T η hn → 0 n
as n → ∞.
(25)
Proposition 9. Let (η, λ) ∈ b σ and h, g satisfy the assumptions from Proposition 8. Then (10) has at least one solution. Proof. We choose (ηn , λn ) ∈ σ \ b σ such that (ηn , λn ) → (η, λ) as n → ∞ and ηn & η monotonically. Set X = L1 (Iη ) and extend each Tηn to an operator from X into X as explained above. Let φn := φηn ,λn , ψn := ψηn ,λn be functions associated with (ηn , λn ). Keeping the same notation we extend φn and ψn onto Iη setting φn = ψn = 0 on (η, ηn ). Then φn → φ and ψn → ψ in L∞ (Iη ) as n → ∞. Hence there exists n0 ∈ N such that (gh) is satisfied by φ and ψ replaced by φn and ψn , respectively, for any n ≥ n0 . According to Proposition 8 for any n ≥ n0 , x + λn Tηn x + Tηn (G(x)) = Tηn h has a solution xn ∈ L1 (Iηn ). Keeping the same notation we extend xn by zero on (η, ηn ). Then xn ∈ X . We claim that there exists a constant c > 0 such that
kx n k ≤ c .
(26)
Assume the contrary, i.e., there is a subsequence, denoted again by
{ xn } ∞ n =1 ,
xn + λn Tηn xn + Tηn (G(xn )) = Tηn h. Let us set yn := yn =
xn kxn k
Tηn h
kx n k
such that kxn k → ∞ as n → ∞, and (27)
and divide (27) by kxn k. We get
− λn Tηn yn −
Tηn (G(xn ))
kx n k
.
(28)
Passing to the limit for n → ∞ we conclude that Tηn h
kxn k
→ 0,
Tηn (G(xn ))
kx n k
→0
in X as n → ∞.
Moreover, we have
λn Tη yn − λTη yn → 0. n Hence, passing to a subsequence if necessary, and using compactness of Tη , we may assume that there exists w ∈ X such that Tη yn → w in X . Finally, we conclude from (28) that there exists y ∈ X such that yn → y in X as n → ∞. Hence w = Tη y and y + λTη y = 0, i.e., y = ±φ . Let us assume y = φ . Eq. (27) is equivalent to x00n + λn xn + g (t , xn ) = h,
xn (0) = 0,
xn (ηn ) = xn (π ).
(29)
3718
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
We multiply (29) by ψn and integrate over Iη . Using the fact that
Z
(x00n + λn xn )ψn dt = 0 Iη
we obtain
Z
g (t , xn (t ))ψn (t ) dt =
Z
Iη
h(t )ψn (t ) dt .
Iη
Passing to the limit for n → ∞, it follows from the Lebesgue theorem and φn → φ , ψn → ψ in L∞ (Iη ) that
Z Iη+
g (t , +∞)ψ(t ) dt +
Z Iη−
g (t , −∞)ψ(t ) dt =
Z
h(t )ψ(t ) dt Iη
which contradicts (gh). Similarly, we arrive at a contradiction in the case y = −φ . The claim (26) is thus proved. With (26) in hand we proceed as follows. We have (24),
λn Tη xn − λTη xn → 0, n
Tη (G(xn )) − Tη (G(xn )) → 0 n
(30)
due to λn → λ, (25) and (g3). Compactness of Tη : X → X and (26) yield the existence of w1 , w2 ∈ X such that Tη xn → w1 ,
Tη (G(xn )) → w2
in X as n → ∞
(31)
(we pass to a subsequence if necessary). Now, we conclude from (27), (30) and (31) that there exists x ∈ X such that xn → x in X as n → ∞, and x + λTη x + Tη (G(x)) = Tη h, i.e., x solves (23). This completes the proof.
Now we formulate the main result of this section. Its proof follows immediately from Propositions 8 and 9. Theorem 10. Let (η, λ) ∈ σ , h and g satisfy (h1), (g1), (g3), (g4) and (gh). Then problem (1) has at least one solution. 6. A bifurcation problem; multiple solutions In this section we consider the bifurcation problem x00 + λx + g (λ, t , x) = 0,
x(0) = 0,
Bη x = 0,
(32)
where g = g (λ, t , s) : R3 → R is a function which is continuous with respect to λ and s for almost all t and which is measurable with respect to t for all λ and s. Moreover, we assume that for any bounded interval Λ ⊂ R there exist m, l ∈ L1loc (R) such that for all λ ∈ Λ, for all s ∈ R and for almost all t ∈ R, we have
|g (λ, t , s)| ≤ m(t ) + l(t )|s|.
(g5)
Moreover, we assume that given any bounded intervals Λ, T ⊂ R, g (λ, t , s) = o(|s|),
s→0
(g6)
uniformly with respect to λ ⊂ R and t ∈ T . In particular, g (λ, t , s) = 0,
t ∈ R, λ ∈ R
and clearly, (32) has always the zero solution x(t ) ≡ 0 in R. It follows from (g5) that for any ξ ∈ R and λ ∈ R the initial value problem x00 + λx + g (λ, t , x) = 0,
x(ξ ) = x0 ,
x0 (ξ ) = x1
(33)
has a solution defined on the whole of R (cf. Section 3). Hence the existence of a solution x = x(t ) to problem (32) can be pasted together from the solution of (32) on the bounded interval Iη and corresponding solutions of the initial value problems on the complement of Iη in R. Let us choose η < 0 and concentrate on the boundary value problem x00 + λx + g (λ, t , x) = 0,
x(0) = 0,
x(η) = x(π ).
(34)
Let G : R × X → X be the Nemytskii operator associated with g, i.e., G(λ, x)(t ) = g (λ, t , x(t )). It follows from (g5) that G is continuous. In particular, Tη G is compact and, due to (g6), given a bounded interval Λ ⊂ R, G(λ, x) = o(kxk),
kx k → 0
uniformly with respect to λ ∈ Λ. Since (34) is equivalent to the operator equation x + λTη x + Tη (G(λ, x)) = 0,
(35)
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3719
Fig. 4. The bifurcation diagram for (37) with (η, λ) ∈ [0, π] × [0, 50].
the global bifurcation theorems of Rabinowitz and Dancer (see e.g. [4], Th. 5.2.34 on p. 295 and Th. 5.2.38 on p. 300) can be applied. We thus get non-zero solutions of (35). Indeed, let us define f (λ, x) := x + λTη x + Tη (G(λ, x)) and
S = {(λ, x) ∈ R × X : x 6= 0, f (λ, x) = 0}. Let us fix (η, λ0 ) ∈ σ \ b σ . It follows from Section 2 that − λ10 is an eigenvalue of Tη of (algebraic) multiplicity equal to 1. It follows from above mentioned abstract results that there is a component C of S which contains nontrivial solutions of (35), C consists of two connected sets C + and C − , x0 (0) > 0 (or x0 (0) < 0) for (λ, x) ∈ C + (or (λ, x) ∈ C − )
C = C+ ∪ C−,
{(λ0 , 0)} ∈ C + ∩ C −
and such that C ± are either unbounded, or C + ∩ C − 6= {(λ0 , 0)} (see [4], p. 300, for more details). However, the latter case cannot occur if g = g (λ, t , s) is locally Lipschitz continuous with respect to the third variable s. The argument used to prove this fact relies on the uniqueness theorem of the initial value problem (33) with ξ = x0 = x1 = 0; see [4], Ex. 5.3.39 on p. 301. The bifurcation results of Rabinowitz and Dancer cannot be applied for (η, λ0 ) ∈ b σ , since the (algebraic) multiplicity of the eigenvalue − λ1 of Tη is equal to 2. However, even in this case the existence of local branches of nontrivial solutions of 0 (34) bifurcating from (λ0 , 0) can be proved under additional smoothness assumptions on g. The argument used to prove this fact is based on the combination of the Lyapunov–Schmidt reduction and the implicit function theorem. It is carried through in detail in [4], Ex. 4.3.24 on pp. 177, 178, for the special case of the equation x00 + λ sin x = 0
(36)
subject to the periodic boundary conditions. In this paper, we focus on (36) subject to the three-point conditions x00 + λ sin x = 0,
x(0) = 0,
Bη x = 0
(37)
and introduce numerical algorithms which demonstrate the existence of global branches of non-zero solutions which bifurcate from (λ, 0) ∈ R × X , where (η, λ) ∈ σ . The reader should notice that (37) reduces to (32) if we put g (λ, t , s) = λ(sin s − s). In fact, the existence of these branches and their unbounded character with respect to λ provide the multiplicity results for (37). Our numerical algorithm allows the classification of the number of solutions of (37) with prescribed number of nodes in Iη if (η, λ) belongs to a given component of R2 \ σ . We also present new algorithms in order to visualize 3D bifurcation diagrams and graphs of corresponding non-zero solutions. Let us describe the procedure for constructing the bifurcation diagram in Fig. 4, where the grey level represents the number of non-zero solutions of (37): Algorithm 11. Let us fix ηmin < ηmax , λmin < λmax and x1min < x1max . 1. Let us construct the mesh {(ηi , λj , x1k ) : i = 0, . . . , nη , j = 0, . . . , nλ , k = 0, . . . , nx1 } of [ηmin , ηmax ] × [λmin , λmax ] × [x1min , x1max ] with equidistant step sizes and with nη · nλ · nx1 number of elements.
3720
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
Fig. 5. Solution diagrams for (37) with fixed η = η127 , η160 , η172 , η190 , η220 .
Fig. 6. Solution diagrams for (37) with fixed λ = λ476 , λ578 , λ605 , λ640 , λ1000 .
2. For i = 0, . . . , nη , j = 0, . . . , nλ and k = 0, . . . , nx1 , let us define xi,j,k := sign(Bηi x), where x solves the following initial value problem: x00 + λj sin x = 0,
x(0) = 0,
x0 (0) = x1k .
3. Let us construct the solution diagrams in the (λ, x1 )-plane for fixed ηi such that we determine the zero contour lines of xi,, (see Fig. 5). Moreover, the zero contour lines of x,j, represent the solution diagrams in the (η, x1 )-plane for fixed λj (see Fig. 6). Let us note that the light (dark) grey color in the solution diagrams in Figs. 5 and 6 means that xi,j,k = 1 (xi,j,k = −1). 4. Finally, the bifurcation diagram in the (η, λ)-plane is given by the number of sign changes of xi,j, (see Fig. 4). More precisely, for i = 0, . . . , nη and j = 0, . . . , nλ , the number of sign changes of xi,j, determines the number of solutions of (37) with η = ηi and λ = λj . Let us note that the bifurcation diagram in Fig. 4 was constructed using 1.256 × 109 mesh elements.
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3721
Conjecture 12. Let (η, λ) ∈ [0, π ) × R be fixed. There exist exactly 2p non-zero solutions of the problem (37), where p is given by
n
o
˜ ∈ σ : λ˜ < λ . p = card (η, λ)
(38)
Remark 13. Let us point out several properties of the cardinality p from (38) in order to determine easily the number of non-zero solutions of the problem (37) for fixed (η, λ) ∈ [0, π ) × R: 1. The number p is bounded as follows: 0 ≤ p ≤ k¯ + ¯j + 1, where
(
2kπ
2
)
<λ , π −η ) ( 2 ( 2j + 1)π ¯j := max j ∈ N ∪ {0} : <λ . π +η Moreover, if η ∈ [0, π ) \ Θ , where Θ := ηj,k : j, k ∈ N, 2k < 2j + 1 , then p = k¯ + ¯j + 1 for an arbitrary value of λ. On the other hand, if η ∈ Θ then p < k¯ + ¯j + 1 for λ sufficiently large. If we proceed as in the proof of Lemma 1, it is possible to show that Θ is a dense subset of [0, π]. 2. The cardinality p from (38) is equal to 2(m − n) with m and n given by ( ) ( ) +∞ +∞ [ [ ˜ ∈ ˜ < λ + card (η, λ) ˜ ∈ ˜ <λ , m = card (η, λ) C2k : λ C2j+1 : λ k¯ := max k ∈ N :
k=1
j =0
n
o ˜ ∈b n = card (η, λ) σ : λ˜ < λ .
Acknowledgements The authors are grateful to the anonymous referee for valuable comments. The authors were supported by the Ministry of Education, Youth and Sports of the Czech Republic, grant no. ME09109, Program KONTAKT, and partially by MSM 4977751301 (J. Čepička, P. Drábek, P. Nečesal) and by the Hungarian–Czech Intergovernmental S&T Cooperation Programme CZ-14/2005 and the Hungarian Research Grant OTKA K61620 (G. Bognár, E. Rozgonyi). References [1] E.M. Landesman, A.C. Lazer, Linear eigenvalues and a nonlinear boundary value problem, Pacific J. Math. 33 (1970) 311–328. [2] G. Bognár, J. Čepička, P. Drábek, P. Nečesal, E. Rozgonyi, Necessary and sufficient conditions for the existence of solution to three-point BVP, Nonlinear Anal. 69 (2008) 2984–2995. [3] G. Teschl, Ordinary Differential Equations and Dynamical Systems, http://www.mat.univie.ac.at/~gerald/ftp/book-ode/ode.pdf. [4] P. Drábek, J. Milota, Methods of nonlinear analysis, in: Applications to Differential Equations, Birkhäuser, Basel, Boston, Berlin, 2007. [5] A.C. Lazer, D.E. Leach, On a nonlinear two-point boundary value problem, J. Math. Anal. Appl. 26 (1969) 20–27.
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
A two-parametric three-point problem G. Bognár a , J. Čepička b , P. Drábek b,∗ , P. Nečesal b , E. Rozgonyi a a
Institute of Mathematics, University of Miskolc, 3515 Miskolc-Egyetemváros, Hungary
b
University of West Bohemia, Univerzitní 22, 306 14 Plzeň, Czech Republic
article
abstract
info
We study the existence of a solution x = x(t ), t ∈ R, of the three-point problem
Article history: Received 12 March 2009 Accepted 12 January 2010
x00 + λx + g (t , x) = h,
x(0) = 0,
Bη x = 0.
(P)
We give sufficient and necessary conditions on g and h in order for (P) to have a solution. Our conditions depend on the values of λ and η. © 2010 Elsevier Ltd. All rights reserved.
MSC: 34B10 34B15 Keywords: Landesman–Lazer type condition Leray–Schauder degree Spectrum of the three-point problem
1. Introduction In this paper we study the existence of a solution x = x(t ), t ∈ R, to the three-point problem x00 + λx + g (t , x) = h(t ),
x(0) = 0,
Bη x = 0,
(1)
where (η, λ) ∈ R2 are parameters, g and h are given functions and x(η) − x(π ) x0 (π )
Bη x :=
for η 6= π , for η = π .
In particular, if η = 0 then x satisfies homogeneous Dirichlet boundary conditions at 0 and π ; if η = π then x satisfies homogeneous Dirichlet–Neumann conditions at 0 and π . Necessary and sufficient conditions for the existence of a solution x = x(t ), t ∈ (0, π ), of equation x00 +λx + g (t , x) = h(t ) subject to the Dirichlet–Neumann and Dirichlet boundary conditions at 0 and π , respectively, are given by Landesman–Lazer conditions (cf. [1] and [5]). More general, the three-point boundary value problem (1) with η ∈ (0, π ) was studied recently in [2]. However, the investigation of the ‘‘global’’ solution of (1) is more involved. These facts motivate the detailed study of the set
σ := {(η, λ) ∈ R2 : x00 + λx = 0, x(0) = 0, Bη x = 0 has a non-zero solution x = x(t ), t ∈ R} as well as of corresponding non-zero solutions.
∗
Corresponding author. E-mail addresses: [email protected] (G. Bognár), [email protected] (J. Čepička), [email protected] (P. Drábek), [email protected] (P. Nečesal), [email protected] (E. Rozgonyi). 0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.01.007
3708
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
One of the main goals of this paper is to provide sufficient and also necessary conditions in terms of the set σ and the functions g and h in order for (1) to have at least one solution. In formulating these conditions we have to distinguish among several cases, depending on the position of parameter η. By a (global) solution to (1) we understand a function x = x(t ) defined on the whole of R such that x0 is absolutely continuous and the equation in (1) is satisfied pointwise almost everywhere in R. Our basic assumptions on g and h are the following ones: h ∈ L1loc (R);
(h1)
g is a Carathéodory function, i.e., g (t , ·) is continuous for almost all t ∈ R and g (·, s) is measurable for all s ∈ R, and there exist functions m, l ∈ L1loc (R) such that for all s ∈ R and almost all t ∈ R we have
|g (t , s)| ≤ m(t ) + l(t ) |s| .
(g1)
Next, we formulate the main results of this paper. In what follows, we set
( (η, π ) Iη := (0, π ) (0, η)
for η < 0, for 0 ≤ η ≤ π , for η > π .
The nonresonance case, (η, λ) 6∈ σ , is rather simple. Let us assume that, in addition to (g1) and (h1), also the following condition is satisfied: there exists δ ∈ [0, 1) such that for all s ∈ R and almost all t ∈ Iη we have
|g (t , s)| ≤ m(t ) + l(t ) |s|δ .
(g2)
Then problem (1) has at least one solution (cf. Theorem 5, Section 4). On the other hand, the resonance case, (η, λ) ∈ σ , is more complicated. Let ϕ be a non-zero solution (‘‘eigenfunction’’) of x00 + λx = 0, x(0) = 0, Bη x = 0 (cf. Sections 2 and 5) and ψ be a function associated with (η, λ) ∈ σ given in Section 2 (an ‘‘eigenfunction’’ of the adjoint problem; cf. Remark 3). We assume that for all s ∈ R and almost all t ∈ Iη ,
|g (t , s)| ≤ m(t ),
(g3)
and, moreover, for almost all t ∈ Iη the following limits exist: g (t , ±∞) := lim g (t , s).
(g4)
s→±∞
We also assume that
Z
g (t , +∞)ψ(t ) dt +
Iη+
<
Z Iη+
Z
g (t , −∞)ψ(t ) dt <
Iη−
Z
h(t )ψ(t ) dt Iη
g (t , −∞)ψ(t ) dt +
Z Iη−
g (t , +∞)ψ(t ) dt ,
(gh)
where Iη+ := {t ∈ Iη : ϕ(t ) > 0} and Iη− := {t ∈ Iη : ϕ(t ) < 0}. Then problem (1) has at least one solution (cf. Theorem 10). If for almost all t ∈ Iη and all s ∈ R we have g (t , +∞) < g (t , s) < g (t , −∞),
(g5)
then
Z
g (t , +∞)ψ(t ) dt +
b Iη+
<
Z b Iη+
Z
g (t , −∞)ψ(t ) dt <
b Iη−
g (t , −∞)ψ(t ) dt +
Z
h(t )ψ(t ) dt Iη
Z b Iη−
g (t , +∞)ψ(t ) dt ,
(2)
is necessary for (1) to have a solution. Here
b Iη+ := t ∈ Iη : ψ(t ) > 0
and b Iη− := t ∈ Iη : ψ(t ) < 0 .
In this paper we also study the bifurcation problem x00 + λx + g (λ, t , x) = 0,
x(0) = 0,
Bη x = 0,
(3)
where g represents ‘‘higher order terms’’. We point out that classical abstract bifurcation results of Rabinowitz and Dancer can be applied to get non-zero solutions of problem (3). A special case of the above equation is also considered, x00 + λ sin x = 0, in order to prove the existence of multiple solutions and to illustrate the role of the set σ in connection with the number of solutions (see Section 6 for the details).
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3709
Fig. 1. The structure of σ .
2. Non-zero solutions of linear problems As we already pointed out in the Introduction, detailed investigation of the set σ is important, i.e., we are looking for all pairs (η, λ) ∈ R2 for which x00 + λx = 0,
x(0) = 0,
Bη x = 0
(4)
has a non-zero solution. It is a matter of elementary calculation that linear problem (4) has a non-zero solution if and only if (η, λ) ∈ σ , where
σ :=
[
C2k ∪
k∈Z\{0}
[
C2j+1 ,
j∈Z
n
C2j+1
√
o λ = 2kπ , n o √ := (η, λ) ∈ R × R+ : (π + η) λ = (2j + 1)π .
C2k := (η, λ) ∈ R × R+ : (π − η)
√
The corresponding non-zero solution is then a non-zero multiple of ϕ(t ) = sin λt, t ∈ R (see Fig. 1). Let us give more detailed expression for the set σ depending on the position of η. 1.
If η < −π ,
σ =
∞ [
C2k ∪
If η = −π ,
σ =
∞ [
= (−π , k2 ) : k ∈ N .
C2k
k=1
3.
If − π < η < π ,
σ =
∞ [
C2k ∪
k=1
4.
If η = π ,
σ =
∞ [
C2j+1 .
j=−∞
k=1
2.
−1 [
∞ [
C2j+1 .
j =0
=
C2j+1
π,
(2j + 1)2 4
j =0
5.
If η > π ,
σ =
−1 [ k=−∞
C2k ∪
∞ [
: j ∈ N ∪ { 0} .
C2j+1 .
j =0
Note that for η = −π the set σ consists of (−π , k2 ) where k2 are the eigenvalues of the periodic problem x00 + λx = 0,
x(−π ) = x(π ),
x0 (−π ) = x0 (π ).
On the other hand, if η = π then σ consists of π ,
(2j+1)2 4
where
(2j+1)2 4
are the eigenvalues of the Dirichlet–Neumann
problem x00 + λx = 0,
x(0) = 0,
x0 (π ) = 0.
In general, if (η, λ) ∈ C2k for some k ∈ Z \ {0} then η 6= π and λ is an eigenvalue of the periodic problem x00 + λx = 0,
x(η) = x(π ),
x0 (η) = x0 (π );
3710
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
if (η, λ) ∈ C2j+1 for some j ∈ Z then η 6= −π and λ is an eigenvalue of the Dirichlet–Neumann problem x + λx = 0,
x(0) = 0,
00
x
0
η+π 2
= 0.
Let us also define the set
b σ := {(η, λ) ∈ σ : ∃j ∈ Z, k ∈ Z \ {0} such that (η, λ) ∈ C2k ∩ C2j+1 }. Notice that the set b σ consists of all intersections of the curves C2k and C2j+1 . Notice also that the projection of b σ onto the η-axis, projηb σ , consists of points
ηj,k =
2j − 2k + 1 2j + 2k + 1
π,
j ∈ Z, k ∈ Z \ {0}
(5)
and for any i = 0, 1, 2, . . . , we have
ηj,k = η(2i+1)j+i,(2i+1)k . In particular, for any (η, λ) ∈ b σ we also have (η, λi ) ∈ b σ , where λi = (2i + 1)2 λ, i = 1, 2, . . .. Moreover, we prove the following assertion. Lemma 1. The set projη b σ is dense in R. Proof. Clearly, we have −π , 0, π 6∈ projη b σ . Let us write j = m + n, k = n − m, m, n ∈ Z, m 6= n, in (5). Then
ηj,k =
2(j − k) + 1 2(j + k) + 1
π=
4m + 1 4n + 1
π.
It is enough to show that the set
M=
4m + 1 4n + 1
: m, n ∈ Z, m 6= n
p
is dense in R. To this end, let q , p ∈ Z, q ∈ N, be any rational number. For p 6= q we choose a sequence that 4ip + 1
lim
i→∞
4iq + 1
=
p q
lim
4(i + 1)p + 1 4iq + 1
4ip+1 4iq+1
o∞
⊂ M such
i=1
.
For p = q we choose a sequence
i→∞
n
n
4(i+1)p+1 4iq+1
o∞
⊂ M satisfying
i =1
= 1.
The density of M in R then follows.
Next we study necessary and sufficient conditions in order for x00 + λx = h,
x(0) = 0,
Bη x = 0
to have a solution. Namely, it follows from the variation of constants formula that (6) has a solution for any h ∈ whenever (η, λ) 6∈ σ . On the other hand, if (η, λ) ∈ σ then (6) has a solution if and only if
Z
(6) L1loc
(R)
h(t )ψ(t ) dt = 0, Iη
where both the function ψ and the domain of integration Iη depend on the position of η. Let η < 0, η 6= −π and (η, λ) ∈ σ . We have two different possibilities: either (η, λ) ∈ C2k or (η, λ) ∈ C2j+1 . Here k = 1, 2, 3, . . . for η < 0, η 6= −π ; j = −1, −2, −3, . . . for η < −π and j = 0, 1, 2, . . . for −π < η < 0. Then the variation of constants formula together with three-point condition from (1) imply that ψ = ψ2k if (η, λ) ∈ C2k , and ψ = ψ2j+1 if (η, λ) ∈ C2j+1 , where (see Figs. 2 and 3) 2kπ ψ2k (t ) = sin (π − t ) in (η, π ), π −η (2j + 1)π − sin π + η (π + t ) in (η, 0), ψ2j+1 (t ) = (2j + 1)π − sin (π − t ) in [0, π ). π +η
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3711
Fig. 2. The graphs of ϕ(t ) (dotted lines) and ψ5 (t ) with (η, λ) ∈ C5 .
Let η = −π and (−π , λ) ∈ σ . In this case λ = k2 , ψ = ψ2k , where (see Fig. 3)
ψ2k (t ) = sin k(π − t )
in (−π , π ).
Let 0 ≤ η < π and (η, λ) ∈ σ . We have two different possibilities: either (η, λ) ∈ C2k or (η, λ) ∈ C2j+1 . Here j = 0, 1, 2, . . . , k = 1, 2, 3, . . . , ψ = ψ2k if (η, λ) ∈ C2k , and ψ = ψ2j+1 if (η, λ) ∈ C2j+1 , where (see Figs. 2 and 3)
ψ2k (t ) =
0
in (0, η),
2kπ
(π − t ) in [η, π ), sin π −η (2j + 1)π (2j + 1)π 2 cos π + η π sin π + η t ψ2j+1 (t ) = (2j + 1)π (π − t ) − sin π +η Let η = π and (π , λ) ∈ σ . In this case λ =
ψ2j+1 (t ) = sin
2j + 1 2
t
(2j+1)2 4
in (0, η), in [η, π ). , ψ = ψ2j+1 , where (see Fig. 3)
in (0, π ).
Let η > π and (η, λ) ∈ σ . We have again two different possibilities: either (η, λ) ∈ C2k or (η, λ) ∈ C2j+1 . Here j = 0, 1, 2, . . . , k = −1, −2, −3, . . . , ψ = ψ2k if (η, λ) ∈ C2k , and ψ = ψ2j+1 if (η, λ) ∈ C2j+1 , where (see Figs. 2 and 3)
ψ2k (t ) =
0
in (0, π),
2kπ
(π − t ) in [π , η), sin π −η (2j + 1)π (2j + 1)π 2 cos π + η π sin π + η t ψ2j+1 (t ) = (2j + 1)π sin (π + t ) π +η
in (0, π ), in [π , η).
Lemma 2. Let (η, λ) ∈ σ , ϕ be a non-zero solution of (4) and ψ be as above. Then Proof. Let (η, λ) ∈ C2k , k ∈ Z \ {0}. Recall that η 6= π . For η < π , we have
Z
ϕ(t )ψ(t ) dt = Iη
π
Z η
ϕ(t )ψ(t ) dt =
η−π 2
cos
2kπ π =: J1 (η). π −η
R Iη
ϕ(t )ψ(t ) dt = 0 if and only if (η, λ) ∈ b σ.
3712
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
Fig. 3. The contour graphs of ψ4 (left) and ψ7 (right) as functions of (t , η).
On the other hand, for η > π , we have
Z
ϕ(t )ψ(t ) dt =
η
Z π
Iη
ϕ(t )ψ(t ) dt = −J1 (η).
π Finally, J1 (η) = 0 if and only if π2k−η = 12 + k + j, j ∈ Z, i.e., η = ηj,k . This is equivalent to (η, λ) ∈ b σ. Let (η, λ) ∈ C2j+1 , j ∈ Z. Recall that η 6= −π in this case. For η < 0, we have
Z
ϕ(t )ψ(t ) dt =
π
Z η
Iη
0
Z
ϕ(t )ψ2j+1 (t ) dt = −
sin η
(2j + 1)π (2j + 1)π t sin (π + t ) dt π +η π +η
π
(2j + 1)π (2j + 1)π t sin (π − t ) dt π +η π +η 0 π +η (2j + 1)π = cos η =: J2 (η). 2 π +η Z
−
sin
For 0 ≤ η < π , we have
Z
ϕ(t )ψ(t ) dt =
π
Z
ϕ(t )ψ2j+1 (t ) dt =
2 cos
0
Iη
η
Z 0
π
Z −
sin η
(2j + 1)π (2j + 1)π π sin2 t dt π +η π +η
(2j + 1)π (2j + 1)π t sin (π − t ) dt = J2 (η). π +η π +η
For η = π , we have
Z
ϕ(t )ψ(t ) dt =
π
Z
sin2
0
Iη
π (2j + 1) t dt = > 0. π +η 2
For η > π , we have
Z
ϕ(t )ψ(t ) dt = Iη
η
Z
ϕ(t )ψ2j+1 (t ) dt =
π
Z
2 cos
0
0
η
Z + π
(2j + 1)π (2j + 1)π sin t sin (η + t ) dt = J2 (η). π +η π +η (2j+1)π
Finally J2 (η) = 0, η 6= π , if and only if π +η
=
1 2
+ j − k, k ∈ Z \ {0}, i.e., η = ηj,k , which is equivalent to (η, λ) ∈ b σ.
Let h ∈ X := L1 (Iη ). Then the boundary value problem x00 = h,
x(0) = 0,
(2j + 1)π (2j + 1) π sin2 t dt π +η π +η
Bη x = 0
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3713
has a solution if and only if x = Tη h, where Tη : X → X is defined as follows:
(Tη h)(t ) :=
t
Z
(t − τ )h(τ ) dτ + 0
for η 6= π and
(Tη h)(t ) :=
π
Z
(τ − t )h(τ ) dτ −
π −η π
Z
0
Z
t
η
(τ − η)h(τ ) dτ −
π
Z
(π − τ )h(τ ) dτ
0
τ h(τ ) dτ
0
t
for η = π . The operator Tη : X → W 2,1 (Iη ) is linear, continuous and Im Tη = x ∈ W 2,1 (Iη ) : x(0) = 0, Bη x = 0 .
Moreover, thanks to the compact imbedding W 2,1 (Iη )⊂ ⊂ X the operator Tη is compact from X into itself. Problem (4) can be restated as an operator equation in X : x + λTη x = 0. In particular, (η, λ) ∈ σ if and only if − λ1 is a real eigenvalue of Tη . It is a matter of elementary calculation to see that (η, λ) ∈ σ \ b σ implies that the (algebraic) multiplicity of − λ1 is 1, i.e.,
dim Ker
1
λ
I + Tη
= dim Ker
1
λ
2 I + Tη
= 1,
and (η, λ) ∈ b σ implies that this multiplicity is 2, i.e.,
dim Ker
1
λ
I + Tη
= 1,
dim Ker
1
λ
2 I + Tη
= dim Ker
1
λ
3 I + Tη
= 2.
Let Br (0) be a ball in X centered at the origin and with radius r > 0. We denote as deg(I + λTη ; Br (0), 0)
(7)
the Leray–Schauder degree of the mapping I + λTη with respect to the ball Br (0) and the origin 0. It follows from the Leray–Schauder degree index formula that for (η, λ) ∈ R2 \ σ the degree (7) is well-defined, it is constant with respect to (η, λ) in the same component (i.e., the open and connected part) of R2 \ σ and it is equal to either +1 or −1. Remark 3. Let Tη∗ : X ∗ → X ∗ be the adjoint operator associated with Tη . Here X ∗ = L∞ (Iη ). Let (η, λ) ∈ σ be fixed. Then
the function ψ associated with (η, λ) is nothing but the eigenfunction of Tη∗ associated with the eigenvalue − λ1 . In particular, Tη∗ has an integral representation. For example, for η < 0, we have y = Tη∗ f if and only if
Z t Z π t −η sf (s) ds, (t − s)f (s) ds + π − η Zη y(t ) = Zη t π t −π sf (s) ds, (t − s)f (s) ds + π −η η π
t ∈ (η, 0), t ∈ (0, π ).
For a given f ∈ L∞ (Iη ) the function y ∈ L∞ (Iη ) can be interpreted as a solution of the equation y00 = f considered separately on the intervals (η, 0) and (0, π ) which satisfy the following ‘‘boundary and transition conditions’’: y(η) = y(π ) = 0,
lim y(t ) = lim y(t ),
t →0−
t →0+
lim y0 (t ) = lim y0 (t ).
t →π−
t →η+
1
In particular, the function y need not be a C -function at the point 0. For 0 ≤ η < π , η = π and η > π we have a similar integral representation of Tη∗ as well as a similar interpretation of y. 3. An initial value problem In this section we study the existence of a (global) solution to the initial value problem x00 + λx + g (t , x) = h(t ),
x(ξ ) = x0 ,
x0 (ξ ) = x1 ,
(8)
with ξ ∈ R being arbitrary but fixed. Namely, we show that under assumptions (g1) and (h1) stated in Section 1, problem (8) has a solution x = x(t ) defined on the whole of R. Indeed, notice that (8) can be written as a system x0 = y, x(ξ ) = x0 , y0 = −λx − g (t , x) + h(t ),
y(ξ ) = x1 ,
i.e., x0 = f(t , x),
x(ξ ) = (x0 , x1 ),
(9)
3714
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
where x = (x, y), f(t , x) = (y, −λx − g (t , x) + h(t )). Our assumptions on g and h imply that Carathéodory’s function f satisfies
|f(t , x)| ≤ M (t ) + L(t ) |x| for almost all t ∈ R and all x ∈ R 2 , where M, L ∈ L1loc (R) are fixed functions depending on λ and h, m, l ∈ L1loc (R). The existence of a global solution x = x(t ), t ∈ R, to (9) now follows from Schauder fixed point theorem combined with the usual extensibility procedure based on Gronwall’s lemma (cf. [3]). Hence (8) has a solution x = x(t ) which is defined on the whole of R. 4. A nonresonance problem In this section we assume that (η, λ) 6∈ σ . Under some additional assumptions on the growth of g we prove that (1) has at least one solution. We treat (1) separately on a bounded interval Iη and on its complement R \ Iη . Let us assume η < 0 and consider a three-point boundary value problem on the interval Iη = (η, π ): x00 + λx + g (t , x) = h(t ),
x(0) = 0,
x(η) = x(π ).
(10)
We prove later in this section that (10) has at least one solution xbv = xbv (t ), t ∈ [η, π]. According to Section 3 the equation x00 + λx + g (t , x) = h(t ) subject to the initial conditions x(η) = xbv (η),
x0 (η) = x0bv (η)
and x(π ) = xbv (π ),
x0 (π ) = x0bv (π )
has solutions x−∞ = x−∞ (t ), t ∈ (−∞, η] and x+∞ = x+∞ (t ), t ∈ [π , +∞), respectively. Putting x−∞ (t ), xbv (t ), x+∞ (t ),
( x(t ) =
t ∈ (−∞, η], t ∈ (η, π ), t ∈ [π , +∞)
we obtain a (global) solution of (1). Hence it remains to prove the existence of a solution of the boundary value problem (10) on the interval Iη . Proposition 4. Let h satisfy (h1) and g satisfy (g2). Then (10) has at least one solution x ∈ W 2,1 (Iη ). Proof. Let G : X → X be the Nemytskii operator associated with g, i.e., G(x)(t ) := g (t , x(t )), t ∈ Iη . Then (g2) implies that G is continuous. Problem (10) can then be written as an operator equation x = Tη h − λTη x − Tη (G(x)).
(11)
It follows from the compactness of Tη that the operator x → Tη h − λTη x − Tη (G(x)) is compact from X into itself. Consider the homotopy H (τ , x) := x + λTη x + τ Tη (G(x)) − Tη h
and prove that there is an r > 0 such that H (τ , x) 6= 0
(12)
for all τ ∈ [0, 1] and x ∈ ∂ Br (0). Assume the opposite, i.e., there exist τn ∈ [0, 1], kxn k → ∞ such that H (τn , xn ) = 0. Multiply (13) by
1 kxn k
(13) xn kxn k .
and define yn :=
yn + λTη yn + τn
Tη (G(xn ))
kxn k
−
Tη h
kx n k
Then
= 0.
(14)
We have Tη h
kx n k
→ 0,
Tη (G(xn ))
kx n k
→0
in X as n → ∞
by (g2) and passing to a subsequence if necessary we can assume (due to the compactness of Tη and [0, 1]) that there exist τ ∈ [0, 1], τn → τ , w ∈ X , Tη yn → w. It follows from (14) that there is a y ∈ X such that yn → y in X . Continuity of Tη
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3715
implies w = Tη y and using (14) once again we conclude that y + λTη y = 0.
(15)
Since kyk = 1, Eq. (15) contradicts (η, λ) 6∈ σ . Hence (12) is established. It now follows from the homotopy invariance property of the Leray–Schauder degree and Section 2 that deg(I + λTη + Tη (G(.)) − Tη h; Br (0), 0) = deg(I + λTη ; Br (0), 0) = ±1. Hence the existence of a solution of (10) follows.
Similar arguments apply for Iη with 0 ≤ η ≤ π and η > π . We conclude this section with the following statement. Theorem 5. Let (η, λ) 6∈ σ , h and g satisfy (h1), (g1) and (g2). Then problem (1) has at least one solution. Conjecture 6. Let (η, λ) 6∈ σ and d = dist {(η, λ), σ } > 0. Assume that a : Iη → R is a function satisfying
|a(t )| < d for almost all t ∈ Iη . Then x00 + (λ + a(t ))x = 0,
x(0) = 0,
Bη x = 0
has only the zero solution. Remark 7. Assume that Conjecture 6 is true. Then Proposition 4 and Theorem 5 hold true also with (g2) replaced by
|g (t , s)| ≤ m(t ) + d |s| for almost all t ∈ Iη and all s ∈ R. 5. The resonance case In this section we assume that (η, λ) ∈ σ is fixed. We discuss here in detail the case Iη = (η, π ) (i.e., η < 0) leaving the remaining two cases Iη = (0, π ) and Iη = (0, η) to the reader. Like in Section 4 we prove the existence of a solution to (1) by pasting together the solutions of initial value problems, x−∞ and x+∞ , defined on intervals (−∞, η] and [π , +∞), respectively, together with the solution of the boundary value problem (10). Let ϕ and ψ be functions associated with (η, λ) ∈ σ (cf. Section 2). There exists a constant c 6= 0 such that φ = c ϕ satisfies kφk = 1 and
Z
φ(t )ψ(t ) dt ≥ 0.
(16)
Iη
Proposition 8. Let (η, λ) ∈ σ \ b σ , h satisfy (h1), g satisfy (g1), (g3), (g4) and (gh). Then (10) has at least one solution. Proof. As in the proof of Proposition 4 from Section 4 we apply the Leray–Schauder degree argument. Notice that for (η, λ) 6∈ b σ the strong inequality holds in (16), i.e., we have
Z
φ(t )ψ(t ) dt > 0.
(17)
Iη
Notice that there exists an ε > 0 (small enough) such that (η, λ) 6∈ b σ for all µ ∈ [λ−ε, λ). Hence for τ ∈ (0, 1] the equation x + (λ − τ ε)Tη x = 0 has only a trivial solution. In particular, the Leray–Schauder degree deg(I + µTη ; Br (0), 0) is well-defined for all r > 0, µ ∈ [λ − ε, λ) and it is equal to either +1 or −1 (cf. Section 2). We consider the homotopy H (τ , x) = x + λTη x − τ ε Tη x + (1 − τ )(Tη (G(x)) − Tη h) for τ ∈ [0, 1] and x ∈ Br (0). Our goal is to prove that there is an r > 0 such that for any τ ∈ [0, 1] and any x ∈ X such that kxk = r, we have H (τ , x) 6= 0.
(18)
The reader should notice that if (18) is established then the homotopy invariance property of the Leray–Schauder degree would imply deg(I + λTη + Tη (G(.)) − Tη h; Br (0), 0) = deg(I + (λ − ε)Tη ; Br (0), 0) = ±1.
3716
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
The existence of a solution x ∈ Br (0) of (10) would immediately follow. Hence it remains to prove (18). We argue via contradiction. Let τn ∈ [0, 1], xn ∈ X , kxn k → ∞ be such that H (τn , xn ) = 0, i.e., xn + λTη xn − τn ε Tη xn + (1 − τn )(Tη (G(xn )) − Tη h) = 0. xn kxn k
Let yn :=
(19)
and divide (19) by kxn k. We obtain
yn + λTη yn − τn ε Tη yn + (1 − τn )
Tη (G(xn ))
kx n k
−
Tη h
= 0.
kxn k
(20)
If we pass to a suitable subsequence, we may assume that τn → τ ∈ [0, 1]. Since h ∈ X is fixed and g is bounded on Iη × R by a fixed function m ∈ X , we have Tη h
kx n k
→ 0,
Tη (G(xn ))
→0
kx n k
in X as n → ∞.
The compactness of Tη : X → X implies that we can pass again to a subsequence and assume that there exists w ∈ X such that Tη yn → w in X . Passing to the limit for n → ∞ in (20) we obtain yn → −λw + τ εw
in X .
We set y = −λw + τ εw . Then kyk = 1 and continuity of Tη implies Tη yn → Tη y
in X .
Hence from (20) we conclude that y + (λ − τ ε)Tη y = 0, i.e., y00 + (λ − τ ε)y = 0,
y(0) = 0,
y(η) = y(π ).
For τ 6= 0 we obtain a contradiction. However, there is no contradiction for τ = 0 but in this case τn → 0 and y, kyk = 1, is a solution of y00 + λy = 0,
y(0) = 0,
y(η) = y(π ).
Hence y = ±φ . Notice that (19) is equivalent to x00n + λxn − τn ε xn + (1 − τn )(g (t , xn ) − h) = 0.
(21)
According to Section 2 we have
Z
(x00n + λxn )ψ dt = 0. Iη
Multiplying (21) by ψ and integrating over Iη , we then get
Z
− τn ε
xn (t )ψ(t ) dt + (1 − τn )
Z
Iη
g (t , xn (t ))ψ(t ) dt = (1 − τn ) Iη
Z
h(t )ψ(t ) dt .
(22)
Iη
Let us assume that yn → φ . Then xn → +∞ on Iη+ and xn → −∞ on Iη− . By the Lebesgue theorem
Z
yn (t )ψ(t ) dt → Iη
Z
φ(t )ψ(t ) dt > 0 Iη
(by (17)). Hence, taking n large enough in (22), we obtain
Z
g (t , xn (t ))ψ(t ) dt ≥ Iη
Z
h(t )ψ(t ) dt .
(23)
Iη
Using the Lebesgue theorem again, we get
Z lim
n→∞
Iη
g (t , xn (t ))ψ(t ) dt =
Z Iη+
g (t , +∞)ψ(t ) dt +
Z Iη−
g (t , −∞)ψ(t ) dt .
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3717
Combining with (23) we have
Z Iη+
g (t , +∞)ψ(t ) dt +
Z Iη−
g (t , −∞)ψ(t ) dt ≥
Z
h(t )ψ(t ) dt , Iη
a contradiction with the left inequality in (gh). A similar argument yields a contradiction with the right inequality in (gh) if yn → −φ . The proof is finished.
Before we prove that the assertion of Proposition 8 remains valid also for (η, λ) ∈ b σ , we need some auxiliary properties of the operator Tη , namely, we need a kind of continuous dependence of Tη on η. More precisely, let ηn & η monotonically. For each n ∈ N let us extend Tηn (keeping the same notation) to an operator from L1 (Iη ) into L1 (Iη ) as follows: for h ∈ L1 (Iη ) we set
(Tηn h)(t ) :=
for t ∈ (η, ηn ), for t ∈ Iηn .
0 (Tηn h)(t )
For a fixed h ∈ L1 (Iη ) it follows from the definition of Tη , Tηn that
T η h − T η h → 0 n
as n → ∞.
(24)
In fact, we have even stronger property: given arbitrary but fixed c > 0 and arbitrary sequence hn ∈ L (Iη ), khn k ≤ c, we have 1
T η hn − T η hn → 0 n
as n → ∞.
(25)
Proposition 9. Let (η, λ) ∈ b σ and h, g satisfy the assumptions from Proposition 8. Then (10) has at least one solution. Proof. We choose (ηn , λn ) ∈ σ \ b σ such that (ηn , λn ) → (η, λ) as n → ∞ and ηn & η monotonically. Set X = L1 (Iη ) and extend each Tηn to an operator from X into X as explained above. Let φn := φηn ,λn , ψn := ψηn ,λn be functions associated with (ηn , λn ). Keeping the same notation we extend φn and ψn onto Iη setting φn = ψn = 0 on (η, ηn ). Then φn → φ and ψn → ψ in L∞ (Iη ) as n → ∞. Hence there exists n0 ∈ N such that (gh) is satisfied by φ and ψ replaced by φn and ψn , respectively, for any n ≥ n0 . According to Proposition 8 for any n ≥ n0 , x + λn Tηn x + Tηn (G(x)) = Tηn h has a solution xn ∈ L1 (Iηn ). Keeping the same notation we extend xn by zero on (η, ηn ). Then xn ∈ X . We claim that there exists a constant c > 0 such that
kx n k ≤ c .
(26)
Assume the contrary, i.e., there is a subsequence, denoted again by
{ xn } ∞ n =1 ,
xn + λn Tηn xn + Tηn (G(xn )) = Tηn h. Let us set yn := yn =
xn kxn k
Tηn h
kx n k
such that kxn k → ∞ as n → ∞, and (27)
and divide (27) by kxn k. We get
− λn Tηn yn −
Tηn (G(xn ))
kx n k
.
(28)
Passing to the limit for n → ∞ we conclude that Tηn h
kxn k
→ 0,
Tηn (G(xn ))
kx n k
→0
in X as n → ∞.
Moreover, we have
λn Tη yn − λTη yn → 0. n Hence, passing to a subsequence if necessary, and using compactness of Tη , we may assume that there exists w ∈ X such that Tη yn → w in X . Finally, we conclude from (28) that there exists y ∈ X such that yn → y in X as n → ∞. Hence w = Tη y and y + λTη y = 0, i.e., y = ±φ . Let us assume y = φ . Eq. (27) is equivalent to x00n + λn xn + g (t , xn ) = h,
xn (0) = 0,
xn (ηn ) = xn (π ).
(29)
3718
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
We multiply (29) by ψn and integrate over Iη . Using the fact that
Z
(x00n + λn xn )ψn dt = 0 Iη
we obtain
Z
g (t , xn (t ))ψn (t ) dt =
Z
Iη
h(t )ψn (t ) dt .
Iη
Passing to the limit for n → ∞, it follows from the Lebesgue theorem and φn → φ , ψn → ψ in L∞ (Iη ) that
Z Iη+
g (t , +∞)ψ(t ) dt +
Z Iη−
g (t , −∞)ψ(t ) dt =
Z
h(t )ψ(t ) dt Iη
which contradicts (gh). Similarly, we arrive at a contradiction in the case y = −φ . The claim (26) is thus proved. With (26) in hand we proceed as follows. We have (24),
λn Tη xn − λTη xn → 0, n
Tη (G(xn )) − Tη (G(xn )) → 0 n
(30)
due to λn → λ, (25) and (g3). Compactness of Tη : X → X and (26) yield the existence of w1 , w2 ∈ X such that Tη xn → w1 ,
Tη (G(xn )) → w2
in X as n → ∞
(31)
(we pass to a subsequence if necessary). Now, we conclude from (27), (30) and (31) that there exists x ∈ X such that xn → x in X as n → ∞, and x + λTη x + Tη (G(x)) = Tη h, i.e., x solves (23). This completes the proof.
Now we formulate the main result of this section. Its proof follows immediately from Propositions 8 and 9. Theorem 10. Let (η, λ) ∈ σ , h and g satisfy (h1), (g1), (g3), (g4) and (gh). Then problem (1) has at least one solution. 6. A bifurcation problem; multiple solutions In this section we consider the bifurcation problem x00 + λx + g (λ, t , x) = 0,
x(0) = 0,
Bη x = 0,
(32)
where g = g (λ, t , s) : R3 → R is a function which is continuous with respect to λ and s for almost all t and which is measurable with respect to t for all λ and s. Moreover, we assume that for any bounded interval Λ ⊂ R there exist m, l ∈ L1loc (R) such that for all λ ∈ Λ, for all s ∈ R and for almost all t ∈ R, we have
|g (λ, t , s)| ≤ m(t ) + l(t )|s|.
(g5)
Moreover, we assume that given any bounded intervals Λ, T ⊂ R, g (λ, t , s) = o(|s|),
s→0
(g6)
uniformly with respect to λ ⊂ R and t ∈ T . In particular, g (λ, t , s) = 0,
t ∈ R, λ ∈ R
and clearly, (32) has always the zero solution x(t ) ≡ 0 in R. It follows from (g5) that for any ξ ∈ R and λ ∈ R the initial value problem x00 + λx + g (λ, t , x) = 0,
x(ξ ) = x0 ,
x0 (ξ ) = x1
(33)
has a solution defined on the whole of R (cf. Section 3). Hence the existence of a solution x = x(t ) to problem (32) can be pasted together from the solution of (32) on the bounded interval Iη and corresponding solutions of the initial value problems on the complement of Iη in R. Let us choose η < 0 and concentrate on the boundary value problem x00 + λx + g (λ, t , x) = 0,
x(0) = 0,
x(η) = x(π ).
(34)
Let G : R × X → X be the Nemytskii operator associated with g, i.e., G(λ, x)(t ) = g (λ, t , x(t )). It follows from (g5) that G is continuous. In particular, Tη G is compact and, due to (g6), given a bounded interval Λ ⊂ R, G(λ, x) = o(kxk),
kx k → 0
uniformly with respect to λ ∈ Λ. Since (34) is equivalent to the operator equation x + λTη x + Tη (G(λ, x)) = 0,
(35)
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3719
Fig. 4. The bifurcation diagram for (37) with (η, λ) ∈ [0, π] × [0, 50].
the global bifurcation theorems of Rabinowitz and Dancer (see e.g. [4], Th. 5.2.34 on p. 295 and Th. 5.2.38 on p. 300) can be applied. We thus get non-zero solutions of (35). Indeed, let us define f (λ, x) := x + λTη x + Tη (G(λ, x)) and
S = {(λ, x) ∈ R × X : x 6= 0, f (λ, x) = 0}. Let us fix (η, λ0 ) ∈ σ \ b σ . It follows from Section 2 that − λ10 is an eigenvalue of Tη of (algebraic) multiplicity equal to 1. It follows from above mentioned abstract results that there is a component C of S which contains nontrivial solutions of (35), C consists of two connected sets C + and C − , x0 (0) > 0 (or x0 (0) < 0) for (λ, x) ∈ C + (or (λ, x) ∈ C − )
C = C+ ∪ C−,
{(λ0 , 0)} ∈ C + ∩ C −
and such that C ± are either unbounded, or C + ∩ C − 6= {(λ0 , 0)} (see [4], p. 300, for more details). However, the latter case cannot occur if g = g (λ, t , s) is locally Lipschitz continuous with respect to the third variable s. The argument used to prove this fact relies on the uniqueness theorem of the initial value problem (33) with ξ = x0 = x1 = 0; see [4], Ex. 5.3.39 on p. 301. The bifurcation results of Rabinowitz and Dancer cannot be applied for (η, λ0 ) ∈ b σ , since the (algebraic) multiplicity of the eigenvalue − λ1 of Tη is equal to 2. However, even in this case the existence of local branches of nontrivial solutions of 0 (34) bifurcating from (λ0 , 0) can be proved under additional smoothness assumptions on g. The argument used to prove this fact is based on the combination of the Lyapunov–Schmidt reduction and the implicit function theorem. It is carried through in detail in [4], Ex. 4.3.24 on pp. 177, 178, for the special case of the equation x00 + λ sin x = 0
(36)
subject to the periodic boundary conditions. In this paper, we focus on (36) subject to the three-point conditions x00 + λ sin x = 0,
x(0) = 0,
Bη x = 0
(37)
and introduce numerical algorithms which demonstrate the existence of global branches of non-zero solutions which bifurcate from (λ, 0) ∈ R × X , where (η, λ) ∈ σ . The reader should notice that (37) reduces to (32) if we put g (λ, t , s) = λ(sin s − s). In fact, the existence of these branches and their unbounded character with respect to λ provide the multiplicity results for (37). Our numerical algorithm allows the classification of the number of solutions of (37) with prescribed number of nodes in Iη if (η, λ) belongs to a given component of R2 \ σ . We also present new algorithms in order to visualize 3D bifurcation diagrams and graphs of corresponding non-zero solutions. Let us describe the procedure for constructing the bifurcation diagram in Fig. 4, where the grey level represents the number of non-zero solutions of (37): Algorithm 11. Let us fix ηmin < ηmax , λmin < λmax and x1min < x1max . 1. Let us construct the mesh {(ηi , λj , x1k ) : i = 0, . . . , nη , j = 0, . . . , nλ , k = 0, . . . , nx1 } of [ηmin , ηmax ] × [λmin , λmax ] × [x1min , x1max ] with equidistant step sizes and with nη · nλ · nx1 number of elements.
3720
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
Fig. 5. Solution diagrams for (37) with fixed η = η127 , η160 , η172 , η190 , η220 .
Fig. 6. Solution diagrams for (37) with fixed λ = λ476 , λ578 , λ605 , λ640 , λ1000 .
2. For i = 0, . . . , nη , j = 0, . . . , nλ and k = 0, . . . , nx1 , let us define xi,j,k := sign(Bηi x), where x solves the following initial value problem: x00 + λj sin x = 0,
x(0) = 0,
x0 (0) = x1k .
3. Let us construct the solution diagrams in the (λ, x1 )-plane for fixed ηi such that we determine the zero contour lines of xi,, (see Fig. 5). Moreover, the zero contour lines of x,j, represent the solution diagrams in the (η, x1 )-plane for fixed λj (see Fig. 6). Let us note that the light (dark) grey color in the solution diagrams in Figs. 5 and 6 means that xi,j,k = 1 (xi,j,k = −1). 4. Finally, the bifurcation diagram in the (η, λ)-plane is given by the number of sign changes of xi,j, (see Fig. 4). More precisely, for i = 0, . . . , nη and j = 0, . . . , nλ , the number of sign changes of xi,j, determines the number of solutions of (37) with η = ηi and λ = λj . Let us note that the bifurcation diagram in Fig. 4 was constructed using 1.256 × 109 mesh elements.
G. Bognár et al. / Nonlinear Analysis 72 (2010) 3707–3721
3721
Conjecture 12. Let (η, λ) ∈ [0, π ) × R be fixed. There exist exactly 2p non-zero solutions of the problem (37), where p is given by
n
o
˜ ∈ σ : λ˜ < λ . p = card (η, λ)
(38)
Remark 13. Let us point out several properties of the cardinality p from (38) in order to determine easily the number of non-zero solutions of the problem (37) for fixed (η, λ) ∈ [0, π ) × R: 1. The number p is bounded as follows: 0 ≤ p ≤ k¯ + ¯j + 1, where
(
2kπ
2
)
<λ , π −η ) ( 2 ( 2j + 1)π ¯j := max j ∈ N ∪ {0} : <λ . π +η Moreover, if η ∈ [0, π ) \ Θ , where Θ := ηj,k : j, k ∈ N, 2k < 2j + 1 , then p = k¯ + ¯j + 1 for an arbitrary value of λ. On the other hand, if η ∈ Θ then p < k¯ + ¯j + 1 for λ sufficiently large. If we proceed as in the proof of Lemma 1, it is possible to show that Θ is a dense subset of [0, π]. 2. The cardinality p from (38) is equal to 2(m − n) with m and n given by ( ) ( ) +∞ +∞ [ [ ˜ ∈ ˜ < λ + card (η, λ) ˜ ∈ ˜ <λ , m = card (η, λ) C2k : λ C2j+1 : λ k¯ := max k ∈ N :
k=1
j =0
n
o ˜ ∈b n = card (η, λ) σ : λ˜ < λ .
Acknowledgements The authors are grateful to the anonymous referee for valuable comments. The authors were supported by the Ministry of Education, Youth and Sports of the Czech Republic, grant no. ME09109, Program KONTAKT, and partially by MSM 4977751301 (J. Čepička, P. Drábek, P. Nečesal) and by the Hungarian–Czech Intergovernmental S&T Cooperation Programme CZ-14/2005 and the Hungarian Research Grant OTKA K61620 (G. Bognár, E. Rozgonyi). References [1] E.M. Landesman, A.C. Lazer, Linear eigenvalues and a nonlinear boundary value problem, Pacific J. Math. 33 (1970) 311–328. [2] G. Bognár, J. Čepička, P. Drábek, P. Nečesal, E. Rozgonyi, Necessary and sufficient conditions for the existence of solution to three-point BVP, Nonlinear Anal. 69 (2008) 2984–2995. [3] G. Teschl, Ordinary Differential Equations and Dynamical Systems, http://www.mat.univie.ac.at/~gerald/ftp/book-ode/ode.pdf. [4] P. Drábek, J. Milota, Methods of nonlinear analysis, in: Applications to Differential Equations, Birkhäuser, Basel, Boston, Berlin, 2007. [5] A.C. Lazer, D.E. Leach, On a nonlinear two-point boundary value problem, J. Math. Anal. Appl. 26 (1969) 20–27.