Jumping unbounded nonlinearities and ALP condition
PETR TOMICZEK
Department of Mathematics
University of West Bohemia
Univerzitní 2732/8, 301 00 Pilsen
CZECH REPUBLIC
https://www.zcu.cz/en/Employees/person.html?personId=17020
Abstract: We investigate the existence of solutions to the nonlinear problem
u′′(x) + λ+u+(x)−λ−u−(x) + g(x, u(x)) = f(x), x ∈(0,2π),
u(0) = u(2π), u′(0) = u′(2π),
where the point [λ+, λ−]is a point of the Fučík spectrum Σ =
∞
S
m=0
Σm. We denote φmany nontrivial solution to
our problem with g=f= 0 corresponding to λ+, λ−∈Σm.We assume that g(x, s) = γ(x, s)s+h(x, s)and
the nonlinearity gsatisfies ALP type condition
Key-Words: Second order ODE, periodic, resonance, jumping nonlinearities, Dancer-Fucik spectrum, ALP
condition, saddle point theorem.
Received: May 26, 2021. Revised: February 23, 2022. Accepted: March 24, 2022. Published: April 21, 2022.
1 Introduction
The aim of this article is to provide new existence re-
sult for the periodic problem with unbounded jumping
nonlinearities
u′′(x) + λ+u+(x)−λ−u−(x) + g(x, u(x)) = f(x),
x∈(0,2π), u(0) = u(2π), u′(0) = u′(2π),
(1)
where nonlinearity g: [0,2π]×R→Ris a Cara-
théodory’s function, f∈L1(0,2π),u+=max{u, 0},
u−=max{−u, 0}.
To prove the existence results for nonjumping
problems (λ+=λ−) authors formulated several con-
ditions. In 1969, a paper by Landesman and Leach [1]
for a periodic problem opened the way towards what
today is usually called the Landesman-Lazer condi-
tion, introduced one year later in [2] for a semilinear
problem.
We can also study the periodic problems with fric-
tion u′′(x) + r(x)u′(x) + g(x, u(x)) = f(x)in [3] or
for positive solutions see [4]. One of the latest results
in this regard is [5]. The singular periodic problem is
investigate in [6] by lower and upper solution. The
authors of [7] use phase-plane analysis to prove the
existence of a periodic solution to a nonlinear impact
oscillator. The reader is referred to [8], [9] for the
problem with impulsive differential equations.
A significant alternative to the Landesman-Lazer
condition was proposed by Ahmad, Lazer and Paul
[10] (ALP condition) in 1976, but for the bounded
nonlinearity g. The ALP condition generalizes (see
[11]) the classical Landesman-Lazer condition and
also the potential Landesman-Lazer condition (see
[12]). Therefore to relax the boundedness of gis
a problem which attracted several authors’ attention
(see [13]). In [14] with f≡0, the nonlinearity g
is allowed to be unbounded and satisfies |g(x, s)| ≤
q(x)|s|α+h(x), where 0≤α < 1,q, h ∈L2(0,2π)
with assumption lim|s|→∞ R2π
0G(x, s)dx/|s|2α=∞,
where G(x, s) = Rs
0g(x, t)dt .
The existence results for jumping problems (λ+=
λ−) with bounded nonlinearities gare investigated in
[15], [16], with sublinear nonlinearities in [17]. In
this article we obtain a solution to (1) for gwith linear
growth.
For g≡0and f≡0problem (1) becomes
u′′(x) + λ+u+(x)−λ−u−(x) = 0, x ∈(0,2π),(2)
u(0) = u(2π), u′(0) = u′(2π).
It is well known (see [18]) that problem (2) has non-
trivial solutions only when the pairs (λ+, λ−)lies in
the set of points made up of the curves
Σ0={[λ+, λ−]∈R2|λ+λ−= 0 },
Σm={[λ+, λ−]∈R2|m1
√λ+
+1
√λ−= 2 },
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where m∈N. The set Σ =
∞
S
m=0
Σmis called the
Fučík spectrum.
Using the Landesman-Lazer type conditions au-
thors usually suppose that gsatisfies the linear growth
restriction |g(x, s)| ≤ q(x)|s|+h(x)and there are
functions a, A ∈L1(0,2π), constants r, R ∈Rsuch
that g(x, s)≥A(x)for a.e. x∈[0,2π]and all s≥R
and g(x, s)≤a(x)for a.e. x∈[0,2π]and all s≤r
(see [19]). These conditions imply our assumptions
(see also [20]), that is the function gcan be decom-
posed as
g(x, s) = γ(x, s)s+h(x, s),(3)
where
0≤γ(x, s)≤q1(x),|h(x, s)| ≤ q2(x)(4)
for a.e. x∈(0,2π), for all s∈R, with some q1, q2∈
L1(0,2π). Moreover λ+≥λ−,[λ+, λ−]∈Σm,
m∈Nand there exists ε > 0such that
lim sup
s→+∞
g(x,s)
s≤(m+ 1)2−λ+−ε ,
lim sup
s→−∞
g(x,s)
s≤(m+ 1)2−λ−−ε . (5)
We denote φmany nontrivial solution to (2) corre-
sponding to [λ+, λ−]∈Σm.We shall suppose the
following ALP type conditions
lim
|s|→∞ Z2π
0
[G(x, s φm(x))−f(x)s φm(x)] dx = +∞
(6)
and
lim inf
|s|→∞ Z2π
0
[H(x, s φm(x)) −f(x)s φm(x)] dx ≥c1
(7)
with some constant c1, where H(x, s) =
s
R0
h(x, t)dt.
If the nonlinearity gis L1-bounded (as in [10])
then clearly (6) implies (7). We obtain for example
the existence result to the equation (1) with the non-
linearity g(x, s) = s/(1 + s2) + f(x)or g(x, s) =
[(m+ 1)2−λ+−ε]|sin s|s+f(x)if λ+≥λ−.
2 Preliminaries
We shall use the Lebesgue space Lp(0,2π)with the
norm ∥u∥p. We denote by Hthe Sobolev space 2π-
periodic absolutely continuous functions u:R→R
such that u′∈L2(0,2π)endowed with the norm
∥u∥=R2π
0u2dx +R2π
0(u′)2dx1/2 .
By a solution to (1) we mean a function uin
W2,1(0,2π)such that the equation (1) is satisfied a.e.
on (0,2π)and u(0) = u(2π),u′(0) = u′(2π).
We study (1) by using of variational method. More
precisely, we look for critical points of the functional
I:H→R, which is defined by
I(u) = 1
2Z2π
0
[(u′)2−λ+(u+)2−λ−(u−)2]dx
−Z2π
0
[G(x, u)−fu]dx .
(8)
Every critical point u∈Hof the functional Isatisfies
Z2π
0
[u′v′−(λ+u+−λ−u−)v]dx
−Z2π
0
[g(x, u)v−fv]dx = 0 for all v∈H .
Then uis also a weak solution to (1) and vice versa.
The usual regularity argument for ODE yields im-
mediately (see Fučík [18]) that any weak solution to
(1) is also the solution in the sense mentioned above.
We say that Isatisfies Palais-Smale condition (PS)
if every sequence (un)for which Iis bounded in H
and I′(un)→0(as n→ ∞)contains a convergent
subsequence.
To obtain a critical point of the functional Iwe
will use the following variant of Saddle Point Theo-
rem (see [21]), which is proved in Struwe [21, Theo-
rem 8.4].
Theorem 1 Let V, H+be closed subsets in H,H=
V⊕H+and Qa bounded subset in Vwith boundary
∂Q. Set Γ = {h:h∈C(H, H), h(u)=uon ∂Q }.
Suppose I∈C1(H, R)and
(i)H+∩∂Q =∅,
(ii)H+∩h(Q)=∅, for every h∈Γ,
(iii)there are constants µ, ν such that
µ=infu∈H+I(u)>supu∈∂Q I(u) = ν,
(iv)Isatisfies Palais-Smale condition.
Then the number
γ=inf
h∈Γsup
u∈Q
I(h(u))
defines a critical value γ > ν of I.
We say that H+and ∂Q link if they satisfy condi-
tions i), ii) of the theorem above.
We use result from [16, section 2] to assert that any
nontrivial solution to the boundary-value problem (2)
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corresponding to [λ+, λ−]∈Σm, m ∈Nmust be a
translate, or phase shift, of a positive multiple of the
function φm:R→Rgiven by
φm(x) =
pλ−sin(pλ+x),
x∈[0,π
√λ+
),
−pλ+sin(pλ−(x−π
√λ+
)),
x∈[π
√λ+
,π
√λ+
+π
√λ−
),
pλ−sin(pλ+(x−π
√λ+−π
√λ−
)),
x∈[π
√λ+
+π
√λ−
,2π
√λ+
+π
√λ−
),
.
.
.
−pλ+sin(pλ−(x−(2π−π
√λ−
)),
x∈[2π−π
√λ−
,2π]
after it has been extended to be 2π-periodic over all
of R.
We denote θ1=π/(2pλ+)and
φθ(x) = φm(x+θ1−θ), x ∈[0,2π],(9)
where θ∈[0,2π], then φθ(x)is a nontrivial solution
to the boundary-value problem (2) corresponding to
[λ+, λ−]∈Σm, m ∈N.
Let H−be the subspace of Hspanned by
1,sin x, cos x, sin 2x, . . . , sin(m−1)x, cos(m−1)x.
For K > 0,L > 0, we define sets
V={u∈H:u=aφθ+w, θ ∈[0,2π], a ∈R+
0,
w∈H−},
Q={u∈V: 0 ≤a≤K, ∥w∥ ≤ L}.
(10)
Let H+be the subspace of Hspanned by sin(m+
1)x, cos(m+ 1)x, sin(m+ 2)x, cos(m+ 2)x, . . . .
Next, we verify the assumptions (i) of Theorem 1
and assumption H=V⊕H+.
Lemma 1 It holds
H+∩∂Q =∅.(11)
Proof We suppose for contradiction that there is
u∈∂Q ∩H+. We denote ⟨·,·⟩ the inner product
in L2(0,2π). Then
0u∈H+
=⟨u, sin mx⟩u∈∂Q
=⟨Kφθ+w, sin mx⟩w∈H−
=
K⟨φθ,sin mx⟩K>0
=⟨φθ,sin mx⟩.
Similarly ⟨φθ,cos mx⟩= 0 .It is easy to see
that ⟨φθ,sin mx⟩= 0 (see figure 1) only for θ=
kπ/m , k ∈Z. But ⟨φkπ/m,cos mx⟩ = 0 a contra-
diction.
1(x+1)
sin(x)
232
2x
-6
-4
-2
1
2
y
Figure 1: Solution φθ(x) = φ1(x+θ1−θ)to (2) for
θ= 0
Lemma 2 It holds
H=V⊕H+.(12)
Proof To prove this lemma, we first need to show that
an arbitrary element uof Hcan be expressed in the
form
u=v+h , (13)
where v∈Vand h∈H+. To establish (13), we
observe that every u∈Hcan be written in the form
u(x) = u(x)+amcos mx+bmsin mx+eu(x),(14)
for all x∈[0,2π], and some constants am, bm, where
u∈H−and eu∈H+. We want to show that we can
also write uin the form
u(x) = u1(x) + ϱφθ(x) + eu1(x),(15)
for some constants ϱ > 0and θ∈[0,2π], where
u1∈H−and eu1∈H+. Taking inner products with
cos mx and sin mx in (14) and (15) gives rise to the
system
ϱ⟨φθ,cos mx⟩=πam
ϱ⟨φθ,sin mx⟩=πbm.(16)
We denote p(θ) = ⟨φθ,sin mx⟩then p(0) = 0 (see
figure 1) and
p(θ) =Z2π
0
φm(x+θ1−θ)sin mx dx
=ny=x+θ1−θo
=Z2π+θ1−θ
θ1−θ
φm(y)sin(m(y−θ1+θ)) dy
=Z2π
0
φm(y)sin(m(y−θ1+θ)) dy ,
(17)
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since the integrated functions are 2π-periodic. Hence
function psatisfies p′′(θ) = −m2p(θ), thus p(θ) =
csin mθ , c > 0.
Therefore we can rewrite (16) to the system
ϱ c cos mθ =πam
ϱ c sin mθ =πbm.(18)
Hence, the system in (16) is solvable for any amand
bmin Rand there exist ϱm≥0and θm∈[0,(2π)/m]
such that
ϱmφθm(x)=h1(x)+amcos mx+bmsin mx+h2(x),
for all x∈[0,2π],
(19)
where h1∈H−and h2∈H+.
Next, solve for amcos mx+bmsin mx in (19) and
substitute into the expansion for uin (14) to obtain the
representation in (15), where u1=u−hand eu1=
eu−e
h. We have therefore proved that H=V+H+.
To complete the proof of (12), we need to show that
V∩H+={0}.We can repeat the steps from the
proof of lemma 1. For u∈V∩H+we obtain:
0u∈H+
=⟨u, sin mx⟩u∈V
=⟨aφθ+w, sin mx⟩w∈H−
=
a⟨φθ,sin mx⟩
and similarly a⟨φθ,cos mx⟩= 0.Hence a= 0, u = 0
and V∩H+={0},the proof is complete. We have
proved that His spanned by Vand H+.
We denote the first integral in the functional Iby
J(u) = R2π
0[(u′)2−λ+(u+)2−λ−(u−)2]dx . and
formulate the following lemma, which is proved in
[12, Lemma 2.2].
Lemma 3 Let φbe a solution to (2) with [λ+, λ−]∈
Σm, m ∈N,λ+≥λ−. We put u=aφ +w,a≥0,
w∈H. Then it holds
Z2π
0
[(w′)2−λ+w2]dx≤J(u)≤Z2π
0
[(w′)2−λ−w2]dx.
(20)
We will also use the following nonexistence of par-
ticular nontrivial solution to a BVP like (1) (see [22,
Theorem 8, remarks 2]).
Lemma 4 Let γ±be two maps in L∞(0,2π). There
exists m∈N, two points [λ+,m, λ−,m]∈Σm,
[λ+,m+1, λ−,m+1]∈Σm+1 such that on [0,2π]
λ±,m γ±(x)λ±,m+1 (21)
(λ±,m =γ±(x)and also γ±(x)=λ±,m+1 on a set
of positive measure), then the problem
u′′(x) + γ+(x)u+(x)−γ−(x)u−(x) = 0 ,
u(0) = u(2π), u′(0) = u′(2π)(22)
has only the trivial solution u(x)≡0.
3 Main result
Theorem 2 Let [λ+, λ−]∈Σm,m∈N,λ+≥λ−.
Under the assumptions (3), (4),(5), (6) and (7) Prob-
lem (1) has at least one solution in H.
We shall prove that the functional Idefined by (8)
satisfies the assumptions in Theorem 1 (Saddle Point
Theorem).
i) We infer from Lemmas 1, 2 that H=V⊕H+and
∂Q ∩H+=∅.
ii) The proof of the assumption H+∩h(Q)=∅
∀h∈Γis similar to the proof in [13, example 8.2].
Let π:H→Vbe the continuous projection of H
onto V. We have to show that 0∈π(h(Q)). For t∈
[0,1],u∈Qwe define ht(u) = tπ(h(u))+(1−t)u .
Function htdefines a homotopy of h0=id with
h1=π◦h. Moreover, ht|∂Q =id for all t∈[0,1] .
Hence the topological degree deg(ht, Q, 0) is well-
defined and by homotopy invariance we have deg(π◦
h, Q, 0) = deg(id, Q, 0) = 1 .Hence 0∈π(h(Q)),
as was to be shown.
iii) Firstly, we note that by (4), (5), we get
0≤lim inf
|s|→∞
g(x, s)
s,
0≤lim inf
|s|→∞
G(x, s)
s2
≤lim sup
s→±∞
G(x, s)
s2≤(m+ 1)2−λ±−ε
2
(23)
for a.e. x∈[0,2π]. Now we estimate the functional
Ion the space H+, we prove that
lim
∥u∥→∞ I(u) = ∞for all u∈H+.(24)
Since u∈H+, we have
Z2π
0
(u′)2dx ≥(m+ 1)2Z2π
0
u2dx . (25)
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The definition of I, (23), and (25) yield
lim inf
∥u∥→∞
I(u)
∥u∥2=lim inf
∥u∥→∞
1
∥u∥21
2Z2π
0
[(u′)2−λ+(u+)2
−λ−(u−)2]dx −Z2π
0
[G(x, u)−fu]dx
≥lim inf
∥u∥→∞
1
∥u∥21
2Z2π
0
[(m+ 1)2u2−λ+(u+)2
−λ−(u−)2]dx−Z2π
0
G(x, u)
u2u2dx
≥lim inf
∥u∥→∞
ε
2∥u∥2
2
∥u∥2.
(26)
If lim inf∥u∥→∞ ∥u∥2
2/∥u∥2= 0 then it follows from
the definition of Iand (23) that
lim inf
∥u∥→∞
I(u)
∥u∥2=1
2.(27)
Then (26) and (27) imply lim inf∥u∥→∞ I(u) = ∞. It
follows from (24) and the fact that H+is compactly
embedded in C[0,2π]that there exists a real number,
µ, such that I(u)≥µfor all u∈H+; in fact, we may
take µto be defined by
µ=inf
u∈H+I(u).(28)
We will next show that we can pick K > 0and L > 0
such that supu∈∂Q I(u)< µ, where Q={u∈H:
u=aφθ+w, 0≤a≤K, w ∈H−,∥w∥ ≤ L,
θ∈[0,2π]}, where φθis given in (9). We argue by
contradiction. Suppose that sup∥u∥→∞ I(u) = −∞
for u∈∂Q is not true. Then there is a sequence
(un)⊂∂Q such that ∥un∥ → ∞ and a constant c−
satisfying
lim inf
n→∞ I(un)≥c−.(29)
Due to (23)
lim infn→∞ R2π
0(G(x, un)−fun)/∥un∥2dx ≥0.
Hence from the definition of Iand (29) we have
lim inf
n→∞
1
2Z2π
0
(u′
n)2−λ+(u+
n)2−λ−(u−
n)2
∥un∥2dx ≥0.
(30)
We denote vn=un/∥un∥and we proceed as in [16,
pg.24]. Then,
vn∈∂B ∩V, for all n∈N,(31)
where Bdenotes the closed unit ball in H, and Vis
as defined in (10) (V={u∈H:u=aφθ+w, 0≤
a, w ∈H−}); so that ∂B ∩Vlives in a finite dimen-
sional subspace of H(see [16, Remark 3.4]). We also
have, that
vn=anφθn+zn,(32)
where
zn∈B∩H−, an∈[0,1/r],(33)
where r=∥φθ∥. Using the compactness of B∩H−
and the closed intervals [0,1/r]and [0,2π], we may
assume, as a consequence of (32), (33), that
vn→v0in H, (34)
where
v0=a0φθ0
+z0, a0∈[0,1/r], θ0∈[0,2π], z0∈B∩H−
.
Therefore, letting n→ ∞, using (30) and (34) we
obtain
Z2π
0
[(v′
0)2−λ+(v+
0)2−λ−(v−
0)2]dx ≥0.(35)
By lemma 3 we have for v0∈V, v0=a0φθ0+z0
Z2π
0
[(v′
0)2−λ+(v+
0)2−λ−(v−
0)2]dx
≤Z2π
0
[(z′
0)2−λ−z2
0]dx , z0∈H−.
(36)
By (35), (36) we get
0≤Z2π
0
[(z′
0)2−λ−z2
0]dx . (37)
We note that 0≤lim inf|s|→∞ g(x, s)/s≤
lim sup|s|→∞ g(x, s)/s, thus (5) implies λ+≤(m+
1)2−εwith some ε > 0. Since 1/pλ++1/pλ−=
2/mwe obtain
1
pλ−
<2
m−1
m+ 1 =m+ 2
m(m+ 1)
⇒pλ−>m(m+ 1)
m+ 2 > m −1.
(38)
We denote δ=λ−−(m−1)2>0. Therefore by
(37) we get
0≤Z2π
0
[(z′
0)2−((m−1)2+δ)z2
0]dx . (39)
We note that for z0∈H−it holds
Z2π
0
[(z′
0)2−(m−1)2z2
0]dx ≤0.(40)
Combining (39) with (40) we deduce that z0≡0and
v0=a0φθ0,where a0= 1/∥φθ0∥and φθ0is a non-
trivial solution to the homogeneous boundary-value
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problem (2) corresponding to [λ+, λ−]∈Σm, we de-
note φm0=a0φθ0.
Because of the compact imbedding H⊂C(0,2π)
and (34), we have vn→φm0(x)in C(0,2π)and
lim
n→∞ un(x) = (+∞where φm0(x)>0,
−∞ where φm0(x)<0.
(41)
We return to (29) and firstly estimate by lemma 3 us-
ing (40) (with z0=wn∈H−) the first integral in
I(un)
Z2π
0
(u′
n)2−λ+(u+
n)2−λ−(u−
n)2dx
≤Z2π
0
[(w′
n)2−λ−w2
n]dx
=Z2π
0
[(w′
n)2+w2
n−(λ−+ 1)w2
n]dx
=∥wn∥2−((m−1)2+δ+ 1)∥wn∥2
2
≤ ∥wn∥2−(m−1)2+δ+ 1
(m−1)2+ 1 ∥wn∥2
=−δ
(m−1)2+ 1∥wn∥2
(42)
since ∥wn∥2≤((m−1)2+ 1)∥wn∥2
2. By (29) and
(42) we obtain
lim inf
n→∞ −δ
2((m−1)2+ 1) ∥wn∥2
−Z2π
0
[G(x, un)−fun]dx≥c−.
We denote cm=δ
2((m−1)2+1) >0, then equivalently
lim sup
n→∞ cm∥wn∥2+Z2π
0
[G(x, un)−fun]dx≤−c−.
(43)
We use the decomposition (3) of g(x, s) = γ(x, s)s+
h(x, s)and denote Γ(x, s) = Rs
0γ(x, t)t dt , we
rewrite (43) into
lim sup
n→∞ cm∥wn∥2+Z2π
0
[Γ(x, un)
+H(x, un)−fun]dx≤ −c−.
(44)
By the mean value theorem, (3),(4) and the compact
embedding Hinto C([0,2π]) (∥·∥C([0,2π]) ≤c2∥·∥)
we obtain
Z2π
0
[H(x, un)−H(x, anφm0)] dx
=Z2π
0
[h(x, ξn(x)) wn)] dx ≤ ∥q2∥1c2∥wn∥,
(45)
where ξn(x)∈(anφm0(x), un(x)).
Similarly R2π
0fwn≤ ∥f∥1c2∥wn∥. Therefore by
(44), (45) we get lim supn→∞cm∥wn∥2−(∥f∥1+
∥q2∥1)c2∥wn∥+R2π
0[Γ(x, un) + H(x, anφm0)−
fanφm0]dx
≤−c−and consequently there exists a
constant c3such that
lim sup
n→∞ Z2π
0
[Γ(x, un)
+H(x, anφm0)−fanφm0]dx ≤c3.
(46)
For a.e. x∈(0,2π)function Γ(x, s)is nonincreasing
for s < 0;Γ(x, 0) = 0 and Γ(x, s)is nondecreasing
for s > 0. Hence we get
lim
n→∞ Z2π
0
Γ(x, un)dx =lim
n→∞ Z2π
0
Γ(x, anφm0)dx
(47)
since lim
n→∞ un(x) = lim
n→∞ anφm0(x)=+∞for
x∈(0,2π)such that φm0(x)>0, and lim
n→∞ un(x) =
lim
n→∞ anφm0=−∞ for x∈(0,2π)such that
φm0(x)<0. We rewrite condition (6) in the follo-
wing form
lim
n→∞ Z2π
0
[Γ(x, anφm0(x))
+H(x, anφm0(x)) −fanφm0(x)] dx =∞.
(48)
If the limit in (47) is finite we obtain a contradiction
to (46), (48). If the limit in (47) is infinite we obtain
a contradiction to (46) and assumption (7). Hence
sup∥u∥→∞ I(u) = −∞ for u∈∂Q and we have
showed that we can pick K > 0and L > 0such that
µ=inf
u∈H+I(u)>sup
u∈∂Q
I(u) = ν .
iv) For Assumption (iv) of theorem 1, we show that
functional Isatisfies the Palais-Smale condition.
For contradiction we suppose that the sequence
(un)is unbounded and there exists a constant c4such
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that
1
2Z2π
0
(u′
n)2−λ+(u+
n)2−λ−(u−
n)2dx
−Z2π
0
[G(x, un)−fun]dx≤c4
(49)
and lim
n→∞ ∥I′(un)∥= 0 .(50)
Let (wk)be an arbitrary sequence bounded in H. It
follows from (50) and the Schwarz inequality
lim
n→∞
k→∞ Z2π
0
[u′
nw′
k−(λ+u+
n−λ−u−
n)wk]dx
−Z2π
0
[g(x, un)wk−fwk]dx
=|lim
n→∞
k→∞ ⟨I′(un), wk⟩| ≤ lim
n→∞
k→∞ ∥I′(un)∥·∥wk∥= 0 .
(51)
Since R2π
0[(f/∥un∥)wk]dx →0we obtain by (51)
lim
n→∞
m→∞
k→∞ Z2π
0 u′
n
∥un∥−u′
m
∥um∥w′
k
−λ+u+
n
∥un∥−u+
m
∥um∥−λ−u−
n
∥un∥−u−
m
∥um∥wkdx
−Z2π
0g(x, un)
∥un∥−g(x, um)
∥um∥wkdx= 0 .
(52)
We put vn=un/∥un∥and wk=vn−vmin (52), we
conclude
lim
n→∞
m→∞ Z2π
0
(v′
n−v′
m)2dx
−Z2π
0h(λ+(v+
n−v+
m)−λ−(v−
n−v−
m))(vn−vm)idx
−Z2π
0hg(x, un)
∥un∥−g(x, um)
∥um∥(vn−vm)idx
=0.
(53)
Due to compact imbedding H⊂L2(0,2π), C([0,2π])
there is v0∈Hsuch that (up to subsequence) vn⇀
v0weakly in H,vn→v0strongly in L2(0,2π),
C([0,2π]). Due to assumption (3), (4) the sequence
(g(x, un)/∥un∥)is L1-bounded, thus (53) implies
vn→v0strongly in H.
It follows from assumptions (3), (4), (5) (up to sub-
sequence) that
g(x, un)
∥un∥=γ(x, un)un
∥un∥+h(x, un)
∥un∥
⇀ γ+
0(x)v+
0−γ−
0(x)v−
0in L1(0,2π),
(54)
where 0≤γ+
0(x)≤(m+ 1)2−λ+−ε, 0≤
γ−
0(x)≤(m+ 1)2−λ−−εfor a.e. x∈(0,2π),
since the sequence γn(x) := γ(x, un(x)) is both
bounded and equi-integrable in L1(0,2π)(see Dun-
ford, Schwarz [24]). We get from (51) and (54)
Z2π
0
[v′
0w′−((λ++γ+
0)v+
0
−(λ−+γ−
0)v−
0)w]dx = 0 for all w∈H.
(55)
It follows from (54), (55) and from the usual regular-
ity argument for ordinary differential equations (see
Fučík [18]) that v0is a solution with norm ∥v0∥= 1
to the periodic BVP
v′′
0−(λ++γ+
0)v+
0+ (λ−+γ−
0)v−
0= 0
x∈(0,2π), v0(0) = v0(2π), v′
0(0) = v′
0(2π),
(56)
where by (38)
m2≤λ+≤λ++γ+
0(x)≤(m+ 1)2−ε,
(m−1)2<(m−1)2+δ=λ−
≤λ−+γ−
0(x)≤(m+ 1)2−ε
(57)
for a.e. x∈(0,2π). Therefore using lemma 4 with
[λ+, λ−]∈Σm,[(m+ 1)2,(m+ 1)2]∈Σm+1 equa-
tion (56) and inequalities (57) we obtain
γ(x, un(x)) →γ0(x) = 0 for a.e x∈(0,2π)
and vn(x)→v0(x) = φm(x)
∥φm∥,
(58)
where φmis a solution to (2) with [λ+, λ−]∈Σm.
Now we estimate the first integral in (51). We set
un=anφm+u⊥
n, where an≥0and u⊥
n∈H−⊕
H+. We remark that u=u+−u−and using (21) in
the first integral in (51) we denote
Iw≡Z2π
0
[(anφm+u⊥
n)′w′
k
−(λ+u+
n−λ−u−
n)wk]dx
and we obtain
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Iw=Z2π
0
[(anφm+u⊥
n)′w′
k
−(λ+u+
n−λ−u−
n)wk]dx
=Z2π
0
[anφ′
mw′
k+ (u⊥
n)′w′
k
−((λ+−λ−)u+
n+λ−un)wk]dx
=Z2π
0
[an(λ+φ+
m−λ−φ−
m)wk+ (u⊥
n)′w′
k
−((λ+−λ−)u+
n+λ−un)wk]dx
=Z2π
0{an[(λ+−λ−)φ+
m+λ−φm]wk
+(u⊥
n)′w′
k−[(λ+−λ−)(anφm+u⊥
n)+
+λ−(anφm+u⊥
n)] wk}dx
=Z2π
0
[(λ+−λ−)(anφ+
m−(anφm+u⊥
n)+)wk
+(u⊥
n)′w′
k−λ−u⊥
nwk]dx .
(59)
Similarly
Iw=Z2π
0
[(λ+−λ−)(anφ−
m−(anφm+u⊥
n)−)wk
+(u⊥
n)′w′
k−λ+u⊥
nwk]dx .
(60)
We add (59) and (60), thus
2Iw=Z2π
0
[(λ+−λ−)(|anφm|−|anφm+u⊥
n|)wk
+ 2(u⊥
n)′w′
k−(λ++λ−)u⊥
nwk]dx .
(61)
We set u⊥
n=un+eunwhere un∈H−,eun∈H+
and we put wk=un−eun+anφm, an≥0,(k=n)
in (61) , we get
2In≡Z2π
0
[(λ+−λ−)(|anφm|−|anφm+un+eun|)
·(un−eun) + 2(u′
n)2−2(eu′
n)2
−(λ++λ−)(u2
n−eu2
n)] dx
+
Z2π
0
[(λ+−λ−)(|anφm|−|anφm+u⊥
n|)anφm
+2(u⊥
n)′anφ′
m−(λ++λ−)u⊥
nanφm]dx.
(62)
Hence using |x|−|y| ≤ |x−y|and (21) we obtain
2In≤Z2π
0
[ (λ+−λ−)|un+eun||un−eun|
+ 2(u′
n)2−2(eu′
n)2
−(λ++λ−)((un)2−(eun)2) ] dx
+Z2π
0
[(λ+−λ−)(|anφm|−|anφm+u⊥
n|)anφm
+2an(λ+φ+
mu⊥
n−λ−φ−
mu⊥
n)
−(λ++λ−)u⊥
nanφm]dx
=Z2π
0
[ (λ+−λ−)|u2
n−eu2
n|+ 2(u′
n)2
−(λ++λ−)(un)2−2(eu′
n)2
+(λ++λ−)(eun)2]dx
+Z2π
0
[(λ+−λ−)(|anφm|−|anφm+u⊥
n|)anφm
+u⊥
n(λ+−λ−)|anφm|]dx .
(63)
Inequality |a2−b2| ≤ a2+b2and (63) yield
2In≤2Z2π
0
[ (u′
n)2−λ−(un)2]dx
+Z2π
0
[−(eu′
n)2+λ+(eun)2]dx
+(λ+−λ−)Z2π
0
[(|anφm|−|anφm+u⊥
n|)anφm
+u⊥
n|anφm|]dx
≤2Z2π
0
[ (u′
n)2−λ−(un)2]dx
+Z2π
0
[−(eu′
n)2+λ+(eun)2]dx
+ 2 (λ+−λ−)ZMn
(u⊥
n)2dx ,
(64)
where Mn={x∈[0,2π] : φm(φm+u⊥
n/an)<0}.
The last inequality in (64) follows from the following
estimates
(|anφm|−|anφm+u⊥
n|)anφm+u⊥
n|anφm|
=0 (if anφm(anφm+u⊥
n)>0) x∈ Mn
sign (φm) 2 (anφm+u⊥
n)anφmx∈Mn
≤2(u⊥
n)2
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since anφm<0and anφm+u⊥
n>0imply u⊥
n>
anφm+u⊥
n,u⊥
n>−anφm>0and therefore
−(anφm+u⊥
n)anφm≤(u⊥
n)2.
We use |x|−|y| ≥ −|x−y|in (62) obtain similarly
In≥Z2π
0
[ (u′
n)2−λ+(un)2−(eu′
n)2+λ−(eun)2]dx
−(λ+−λ−)ZMn
(u⊥
n)2dx .
(65)
Using ∥ · ∥C([0,2π]) ≤c2∥·∥we get
ZMn
(u⊥
n)2dx ≤µ(Mn)c2∥u⊥
n∥2and µ(Mn)→0.
(66)
Since by (58) we have
un
∥un∥=(φm+u⊥
n/an)
∥φm+u⊥
n/an∥→φm
∥φm∥and u⊥
n
an
⇒0.
We write un=un+anφm+eun,un∈H−,eun∈H+.
We put wk= (un+anφm−eun)/(an∥u⊥
n∥1
2)in (51)
then using (64) we obtain
lim inf
n→∞
1
an∥u⊥
n∥1
2Z2π
0
[ (u′
n)2−λ−(un)2]dx
+Z2π
0
[−(eu′
n)2+λ+(eun)2]dx
+ (λ+−λ−)ZMn
(u⊥
n)2dx+Z2π
0
[γ(x, un)(eun)2]dx
−Z2π
0
[γ(x, un)( un+anφm)2
+(h(x, un)−f) ( un+anφm−e
un)] dx≥0.
(67)
We note that it holds ∥un∥2≤((m−1)2+ 1)∥un∥2
2,
∥eun∥2≥((m+ 1)2+ 1)∥eun∥2
2and using (66) we get
Z2π
0
[(u′
n)2−λ−(un)2]dx+Z2π
0
[−(eu′
n)2+λ+(eun)2]dx
+ (λ+−λ−)ZMn
(u⊥
n)2dx +Z2π
0
[γ(x, un)(eun)2]dx
=∥un∥2−(λ−+ 1)∥un∥2
2−∥eun∥2+(λ++ 1)∥eun∥2
2
+ (λ+−λ−)ZMn
(u⊥
n)2dx +Z2π
0
[γ(x, un)(eun)2]dx
≤(m−1)2−λ−
(m−1)2+ 1 ∥un∥2+λ+−(m+ 1)2
(m+ 1)2+ 1 ∥eun∥2
+ (λ+−λ−)µ(Mn)c2∥u⊥
n∥2
+Z2π
0
γ(x, un)dx c2∥eun∥2.
Hence and from (57), (58) and (66) it follows
Z2π
0
[ (u′
n)2−λ−(un)2]dx +Z2π
0
[−(eu′
n)2
+λ+(eun)2]dx + (λ+−λ−)ZMn
(u⊥
n)2dx
+Z2π
0
[γ(x, un)(eun)2]dx
≤−δ/2
(m−1)2+ 1∥un∥2+−ε/2
(m+ 1)2+ 1∥eun∥2
≤ −ϱ∥u⊥
n∥2
(68)
with some ϱ > 0. Therefore (67) and (68) imply
lim inf
n→∞
1
an∥u⊥
n∥1
2n−Z2π
0
[γ(x, un)( un+anφm)2
+ ( h(x, un)−f) ( un+anφm−eun)] dxo≥0.
(69)
Consequently
lim inf
n→∞ Z2π
0hh(x, un)−f)
∥u⊥
n∥1
2
((un−eun)/an+φm)idx
≥lim inf
n→∞ Z2π
0hγ(x, un)
∥u⊥
n∥1
2
an(un/an+φm)2dxi≥0.
(70)
Now we put wk= ( un−eun)/(∥u⊥
n∥2)in (51) to
obtain
lim inf
n→∞
1
∥u⊥
n∥2nZ2π
0
[ (u′
n)2−λ−(un)2]dx
+Z2π
0
[−(eu′
n)2+λ+(eun)2]dx
+(λ+−λ−)ZMn
(u⊥
n)2dx
+Z2π
0
[γ(x, un)((eun)2−(un)2)] dx
−Z2π
0
[(γ(x, un)anφm+h(x, un)−f)
·(un−eun)] dxo≥0.
(71)
We suppose for contradiction that the sequence (u⊥
n)
is unbounded then due to (68) and (71) there exists
ϱ > 0such that
−ϱ+lim inf
n→∞ n−Z2π
0hγ(x, un)
∥u⊥
n∥1
2
anφm
un−eun
∥u⊥
n∥3
2idxo≥0
(72)
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or equivalently
−ϱ≥lim sup
n→∞ Z2π
0hγ(x, un)
∥u⊥
n∥1
2
anφm
un−eun
∥u⊥
n∥3
2idx .
(73)
We note that u⊥
n/an⇒0and we get by (70) (for
∥u⊥
n∥ → ∞)
lim
n→∞ Z2π
0
γ(x, un)
∥u⊥
n∥1
2
anφ2
mdx = 0 .(74)
We denote Sn=nx∈[0,2π]| |φm(x)| ≤ (un(x)−
eun(x))/(∥u⊥
n∥3/2)othen lim
n→∞ µ(Sn) = 0 and
Z[0,2π]\Snhγ(x, un)
∥u⊥
n∥1
2
anφm
un−eun
∥u⊥
n∥3
2idx
≤Z[0,2π]\Sn
γ(x, un)
∥u⊥
n∥1
2
anφ2
m.
(75)
By (51) (with wk= ( un−eun)/∥u⊥
n∥2), (65) we ob-
tain
lim sup
n→∞
1
∥u⊥
n∥2nZ2π
0
[ (u′
n)2−λ+(un)2]dx
+Z2π
0
[−(eu′
n)2+λ−(eun)2]dx
−(λ+−λ−)ZMn
(u⊥
n)2dx
+Z2π
0
[γ(x, un)((eun)2−(un)2)] dx
−Z2π
0
[γ(x, un)anφm+ (h(x, un)−f)
·(un−eun)] dxo≤0.
Hence there exists a constant c5such that
lim inf
n→∞
1
∥u⊥
n∥2R2π
0[γ(x, un)anφm(un−eun)] dx ≥c5.
Thus lim supn→∞ RSnh(γ(x, un)/∥u⊥
n∥1/2)anφm
(un−eun)/(∥u⊥
n∥3/2)idx ≥0since µ(Sn)→0.
Hence and by (74), (75) we get
lim sup
n→∞ Z2π
0hγ(x, un)
∥u⊥
n∥1
2
anφm
un−e
un
∥u⊥
n∥3
2idx ≥0
(76)
a contradiction to (73). This implies that the sequence
(u⊥
n)is bounded. We use (20) from Lemma 3 with
w=u⊥
nand we obtain
Z2π
0
[((u⊥
n)′)2−λ+(u⊥
n)2]dx
≤J(un)≤Z2π
0
[((u⊥
n)′)2−λ−(u⊥
n)2]dx
(77)
where J(un) = R2π
0(u′
n)2−λ+u2
n−λ−u2
ndx.
Hence boundedness of (u⊥
n)implies with (49) that
there exists a constant c6such that
Z2π
0
[G(x, un)−fun]dx≤c6for all n∈N.
(78)
We again use the decomposition G(x, s) = Γ(x, s) +
H(x, s)to rewrite (78) into
Z2π
0
[Γ(x, un)+H(x, un)−f(u⊥
n+anφm)dx≤c6
for all n∈N.
(79)
We use (45) boundedness of (u⊥
n)and (79) to obtain
a constant c7such that
Z2π
0
[Γ(x, un) + H(x, anφm)−fanφmdx≤c7
for all n∈N.
(80)
Using (47) and (80) we obtain a contradiction to as-
sumptions (6) (see (48)), (7), hence sequence (un)
is bounded. Then there exists u0∈Hsuch that
un⇀ u0in H,un→u0in L2(0,2π),C(0,2π)(tak-
ing a subsequence if it is necessary). It follows from
equality (39) that
lim
n→∞
m→∞
k→∞ nZ2π
0
[(un−um)′w′
k
−(λ+(u+
n−u+
m)−λ−(u−
n−u−
m))wk]dx
−Z2π
0
[g(x, un)−g(x, um)]wkdxo= 0 .
(81)
The nonlinearity gis the Carathéodory’s function,
thus strong convergence un→u0in C(0,2π)imply
lim
n→∞
m→∞ Z2π
0
[g(x, un)−g(x, um)](un−um)dx = 0 .
(82)
If we set wk=un,wk=umin (81) and subtract
these equalities, then by (82) we obtain
lim
n→∞
m→∞ Z2π
0
[(u′
n−u′
m)2−(λ+(u+
n−u+
m)
−λ−(u−
n−u−
m))(un−um)] dx = 0 .
(83)
Hence the strong convergence un→u0in L2(0,2π)
implies the strong convergence un→u0in H. This
shows that Jsatisfies Palais-Smale condition and the
proof of Theorem 2 is complete.
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ACKNOWLEDGMENTS
This work was supported by the project LO1506 of
the Czech Ministry of Education, Youth and Sports.
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2022.21.26
Petr Tomiczek
E-ISSN: 2224-2880
206
Volume 21, 2022
