Volume 2010, Article ID 429813,18pages doi:10.1155/2010/429813
Research Article
Superlinear Singular Problems on the Half Line
Irena Rach ˚unkov ´a and Jan Tome ˇcek
Department of Mathematical Analysis and Applications of Mathematics, Faculty of Science, Palack´y University, tˇr . 17. listopadu 12, 771 46 Olomouc, Czech Republic
Correspondence should be addressed to Irena Rach ˚unkov´a,[email protected]
Received 19 October 2010; Accepted 7 December 2010
Academic Editor: Zhitao Zhang
Copyrightq2010 I. Rach ˚unkov´a and J. Tomeˇcek. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The paper studies the singular differential equationptuptfu, which has a singularity at
t0. Here the existence of strictly increasing solutions satisfying sup{|ut|:t∈0,∞} ≥L >0 is proved under the assumption thatfhas two zeros 0 andLand a superlinear behaviour near−∞. The problem generalizes some models arising in hydrodynamics or in the nonlinear field theory.
1. Introduction
Let us consider the problem
ptuptfu, 1.1
u0 0, u∞ L, 1.2
whereLis a positive real parameter.
Definition 1.1. Letc >0. A functionu∈C10, c∩C20, csatisfying1.1on0, cis called
a solution of 1.1on0, c.
Definition 1.2. Letube a solution of1.1on0, cfor eachc > 0. Thenuis calleda solution of 1.1on0,∞. Ifumoreover fulfils conditions1.2, it is calleda solution of problem1.1,
1.2.
In this paper we are interested in the existence of strictly increasing solutions and, in particular, of homoclinic solutions. In what follows we assume
f ∈LiplocR, f0 fL 0, 1.3
fx<0 forx∈0, L, 1.4
there existsB <0 such thatfx>0 for x∈B,0, 1.5
FBFL, whereFx −
x
0
fzdz, 1.6
p∈C0,∞∩ C10,∞, p0 0, 1.7
pt>0, t∈0,∞, lim t→ ∞
pt
pt 0. 1.8
Under assumptions1.3–1.8problem1.1,1.2generalizes some models arising in hydrodynamics or in the nonlinear field theorysee1–5. If a homoclinic solution exists, many important properties of corresponding models can be obtained. Note that if we extend the functionptin1.1from the half-line ontoRas an even function, then any solution of 1.1, 1.2 has the same limit L as t → −∞ and t → ∞. This is a motivation for
Definition 1.3. Equation1.1is singular att 0 becausep0 0. In6,7we have proved
that assumptions1.3–1.8are sufficient for the existence of strictly increasing solutions and homoclinic solutions provided
1
0
ds
ps <∞or there existsL0< B such thatfL0 0. 1.9
Here we assume that1.9is not valid. Then
fx>0 forx <0, 1.10
and the papers6,8provide existence theorems for problem1.1,1.2iffhas a sublinear or linear behaviour near−∞. The case thatfhas a superlinear behaviour near−∞is studied in this paper. To this aim we consider the initial conditions
u0 B, u0 0, 1.11
whereB <0, and introduce the following definition.
Definition 1.4. Letc > 0 and letube a solution of 1.1on0, csatisfying1.11. Thenuis calleda solution of problem1.1,1.11on0, c. Ifumoreover fulfils
ut>0 for t∈0, c, uc L, 1.12
We have proved in6,8that for sublinear or linearf the existence of a homoclinic solution follows from the existence of an escape solution of problem1.1,1.11. Therefore our first task here is to prove that at least one escape solution of1.1,1.11exists, provided
1.3–1.8,1.10, and
fx 0 forx > L 1.13
hold, andfhas a superlinear behaviour near−∞. This is done inSection 2. Using the results
of Section 2 “Theorem 2.10”, and of 6, Theroms 13, 14 and 20 we get the existence of a
homoclinic solution inSection 3.
Note that by Definitions1.3and1.4just the values of a solution which are less than
Lare important for a decision whether the solution is homoclinic or escape one. Therefore condition1.13can be assumed without any loss of generality.
Close problems about the existence of positive solutions have been studied in9–11.
2. Escape Solutions
In this section we assume that1.3–1.8,1.10, and1.13hold. We will need some lemmas.
Lemma 2.1see6, Lemma 3. For eachB < 0, problem1.1,1.11has a unique solutionuon
0,∞such that
ut≥B fort∈0,∞. 2.1
In what follows by a solution of1.1,1.11we mean a solution on0,∞.
Remark 2.2see6, Remark 4. Choosea≥0 andA≤L, and consider the initial conditions
ua A, ua 0. 2.2
Problem1.1,2.2has a unique solutionuona,∞. In particular, forA0 andAL, we getu≡0 andu≡L, respectively. Clearly, fora > 0,u≡0 andu≡Lare solutions of1.1on the whole interval0,∞.
Lemma 2.3. LetB <0and letube a solution of problem1.1,1.11which is not an escape solution. Let us denote
θsup{t >0 :u <0 in0, t}, bsupt >0 :u>0in 0, t . 2.3
Then0< θ≤b≤ ∞holds andpuis increasing on0, θ. Ifθ <∞, thenθ < band
maxptut:t∈0, b pθuθ. 2.4
Letθ <∞. Thenθis the first zero ofuanduθ>0.Remark 2.2yields thatuθ 0 is not possible. This implies thatθ < b. Asuis strictly increasing onθ, banduis not an escape solution, we have 0< u < Lonθ, b. Thuspupfu<0 onθ, band hencepu
is decreasing onθ, b. This gives2.4.
Lemma 2.4. LetB <0and letube a solution of problem1.1,1.11which is not an escape solution. Assume thatbis given byLemma 2.3. Then
ub∈0, L, ub 0. 2.5
Proof. From1.1, we have
ut p
t
ptu
t fut, t >0, 2.6
and, by multiplication and integration over0, t,
u2t
2
t
0
ps psu
2sdsFu0−Fut, t >0. 2.7
1Assume thatb ∞. The definition ofbyieldsu > 0 on0,∞. Sinceuis not an escape solution, it is bounded above and there exists
ub lim
t→ ∞ut∈B, L. 2.8
Therefore the following integral is bounded and, since it is increasing, it has a limit
0≤ lim t→ ∞
t
0
ps psu
2sds≤FB−Fub<∞. 2.9
So, by2.7, limt→ ∞u2texists. By virtue of2.8, we get
ub lim t→ ∞u
t lim t→ ∞u
2t 0.
2.10
Ifub ∈ {/ 0, L}, then by1.4,1.10 and2.6we get limt→ ∞ut ub fub/0, which contradicts2.10. Hence,ub ∈ {0, L}. In particular, ifθis defined as inLemma 2.3, then
ub 0 forθ∞, ub L forθ <∞. 2.11
2Assume thatb <∞. Then the continuity ofugivesub 0 andθofLemma 2.3
fulfils 0< θ < b. We deduce that 0< u < Lonθ, bas in the proof ofLemma 2.3.Remark 2.2
yields that if ub 0, then neither ub 0 nor ub L can occur. Therefore ub ∈
Denote
Pt
t
0
psds, t∈0,∞. 2.12
Lemma 2.5. LetB <0and letube a solution of problem1.1,1.11. Further assume maximalb >0
such thatut>0andut∈B, Lfort∈0, b. Then
t
0
2FuspspsdsFutp2t 1
2p
2tu2t, t∈0, b. 2.13
Fora∈0,1, let us denote
Qx:2Fx−aL−xfx, x∈−∞, L. 2.14
Then
t
0
Quspsds < Pt2Fut u2t
t
0 2
psPs
p2s −a 1
psu2sds, t∈0, b.
2.15
Proof. For equality2.13 see Lemma 4.6 in 8. Let us prove2.15. Using the per partes integration, we get fort∈0, b
t
0
Quspsds
t
0
2Fus−aL−usfuspsds
2FutPt I1 I2,
2.16
where
I12 t
0
fususPsds, I2−a t
0
L−usfuspsds. 2.17
By multiplication and integration of1.1we obtain
2
t
0
ususPsds 2
t
0
u2sp
s
psPsdsI1, 2.18
and by the per partes integration,
I1 Ptu2t t
0 2
ps
ps Ps−ps
To computeI2, we use1.1and get
I2−a t
0
L−uspsusds. 2.20
By the per partes integration we derive
I2−aptutL−ut−a t
0
psu2sds <−a t
0
psu2sds. 2.21
We have proved that2.15is valid.
Lemma 2.6. LetC < B,{Bn}∞n1 ⊂ −∞, Cand let{un}∞n1 be solutions of problem1.1,1.11
withBBn,n∈N. Let us denote
bnsup
t >0 :un∈B, L, un>0 in0, t , n∈N. 2.22
Then for eachn∈Nthere exists a uniqueγn∈0, bnsatisfying
un
γn
C. 2.23
If the sequence{γn}∞n1is unbounded, then there exists an escape solution in{un}∞n1.
Proof. Choosen ∈ N. The monotonicity and continuity of un in0, bngive a uniqueγn ∈ 0, bn. If{γn}∞n1is unbounded we argue as in the proof of Lemma 4.8 in8.
LetC < Band let{Bn}∞n1,{un}∞n1,{bn}∞n1and{γn}∞n1be sequences fromLemma 2.6.
Assume that for anyn∈N,unis not an escape solution of problem1.1,1.11.Lemma 2.6 implies that
Γ:supγn:n∈N <∞. 2.24
We can assume that that either there existsb0>0 such that
bn≤b0, n∈N, 2.25
or
bn>Γ 1, n∈N. 2.26
Otherwise we take a subsequence. Some additional properties of{un}∞n1are given in the next
Lemma 2.7. Denote
θn:sup{t >0 :un<0 in0, t}, n∈N, 2.27
and assume that the sequence{θn}∞n1is bounded above. Then there existsK >0such that
ptunt≤K fort∈0, bn, n∈N. 2.28
Proof. ByLemma 2.4we have
unbn∈0, L, unbn 0, n∈N. 2.29
Step 1sequence{pγnunγn}∞n1is bounded. Assume on the contrary that{pγnunγn}∞n1
is unbounded. We may write
lim n→ ∞p
γn
un
γn
∞ 2.30
otherwise we take a subsequence. Equality2.13yields forn∈Nandt∈γn, bn,
0<
t
γn
2Funspspsds
Funtp2t 1 2p
2tu2 nt−F
un
γn
p2γn
−1 2p
2γ n
un2
γn
.
2.31
Using1.4,1.6,1.10,C < Band the fact thatunt∈C, Lfort∈γn, bn, we get
Funt< FC fort∈
γn, bn
. 2.32
Consequently, inequality in2.31leads to
FCp2γn 1
2p
2γ n
un2
γn
< FCp2t 1
2p
2tu2
nt 2.33
fort∈γn, bn. Therefore
p2γn
un2
γn
−2FCp2t< p2tun2t, t∈
γn, bn
, n∈N. 2.34
We will consider two cases.
Case 1. If2.25holds, then2.34gives forn∈N
p2γn
un2
γn
−2FCp2b0< p2b0un2t, t∈
γn, bn
By2.30, for each sufficiently largen∈N, we get
p2γn
un2
γn
>2FC 1p2b0. 2.36
Putting it to2.35, we have 1≤unbn, contrary to2.29.
Case 2. If2.26holds, then2.34gives forn∈N
p2γn
un2
γn
−2FCp2Γ 1< p2Γ 1un2t, t∈
γn,Γ 1
. 2.37
Due to2.30, we have
p2γn
un2
γn
>2FC L−C2p2Γ 1 2.38
for each sufficiently largen∈N. Putting it to2.37, we getL−C < untfort ∈γn,Γ 1. Integrating it overγn,Γ 1, we obtainL < unΓ 1. Equation1.1and condition 1.13 yieldunt>0 fort≥Γ 1, and soL < unbn, contrary to2.29.
We have proved that there existsK0>0 such that
pγn
un
γn
≤K0, n∈N. 2.39
Step 2estimate forpun. Choosen∈N. By2.32we get
t
γn
2Funspspsds <2FC t
γn
pspsds < FCp2t, t∈γn, bn
. 2.40
This together with2.31and2.39imply
1 2p
2tu2
nt<2FCp2t 1 2K
2 0, t∈
γn, bn
. 2.41
According to2.27andLemma 2.3we see thatθn ∈γn, bnis the first zero ofun. Since the sequence{θn}∞n1 is bounded above, there existsΓ0 <∞such thatθn ≤Γ0,n∈N. Then1.8 and2.41give
1 2p
2θ
nun2θn<2FCp2Γ0 1 2K
2
Put
1 2K
22FCp2Γ0 1
2K
2
0. 2.43
Then, by virtue of2.4, inequality2.28is valid.
Lemma 2.8. ConsiderC < BandΓsatisfying2.23and2.24. Letθn,n∈Nbe given by2.27.
Assume that
θn>Γ 2, n∈N. 2.44
Then there existsK∈0,∞such that
ptunt≤K fort∈0,Γ 1, n∈N. 2.45
Proof. Assume on the contrary that
supptunt:t∈0,Γ 1, n∈N ∞. 2.46
ByLemma 2.3,punis increasing on0, θn,n∈N. Therefore
suppΓ 1unΓ 1:n∈N ∞, 2.47
and therefore there existsn0∈Nsuch that
pΓ 1un0Γ 1>|C|pΓ 2. 2.48
Moreover2.23,2.24,2.27,2.44, and the monotonicity ofun0andpun0yield
un0t∈C,0, ptun0t>|C|pΓ 2 fort∈Γ 1,Γ 2. 2.49
Integrating the last inequality overΓ 1,Γ 2, we obtainun0Γ 2−un0Γ 1> |C|, so
|un0Γ 1|>|C|, a contradiction.
Lemma 2.9. Let real sequences{Bn}∞n1,{κn}∞n1,{σn}∞n1be given and assume that
lim
n→ ∞Bn−∞, {κn} ∞ n1⊂
1 2,1
, {σn}∞n1⊂
1 2,1
. 2.50
Letk≥2and
1< r < k 2
(fork2we assumer∈1,∞) be such that
0< lim x→ −∞
|x|r
fx <∞. 2.52
Assume thatQis given by2.14witha∈0,2/r 1. Then
lim
n→ ∞QκnBn |
Bn| fσnBn
k/2
∞. 2.53
Proof. By2.50, limn→ ∞κnBn limn→ ∞σnBn −∞. Condition2.52yields that there exists
λ∈0,∞such that
lim x→ −∞
|x|r
fx λ, xlim→ −∞
Fx
|x|r 1
1
r 1λ. 2.54
Therefore
lim x→ −∞
2Fx
L−xfx 2 limx→ −∞ |x| L−x
Fx
|x|r 1
|x|r fx
2
r 1. 2.55
Hence
lim
x→ −∞Qx xlim→ −∞L−xfx
2Fx
L−xfx−a
a0xlim→ −∞L−xfx,
2.56
wherea0:2/r 1−a >0. Consequently,
lim
n→ ∞QκnBn |
Bn| fσnBn
k/2
a0lim
n→ ∞L−κnBnfκnBn |
Bn| fσnBn
k/2
a0nlim→ ∞L−κnBn
fκnBn |κnBn|r
|
σnBn|r
fσnBn k/2
κrn 1
σnrk/2
|Bn|r−k/2r−1
≥a0
1 2
r 1
λk/2−1lim n→ ∞|Bn|
r0,
2.57
wherer0 r 1−kr−1/2>0, becauser is less than the critical valuek 2/k−2. We
Now we are ready to prove the following main result of this paper.
Theorem 2.10. Assume that
lim t→0
pt
tk−2 ∈0,∞ 2.58
for somek ≥2. Further, letrandfbe such that2.51and2.52are valid. Then there existsB < B such that the corresponding solution of problem1.1,1.11is an escape solution.
Proof. Assumption2.51impliesk−2/k < 2/r 1 < 1, and hence we can choosea ∈
k−2/k,2/r 1and defineQby2.14. According to1.4,1.10, and2.56, there exists
C < Bsuch thatQx>0 forx∈−∞, C∪0, L. Consequently, we can findQ ∈0,∞such that
Qx≥ −Q forx∈−∞, L. 2.59
Let{Bn}∞n1,{un}∞n1,{bn}∞n1,{γn}∞n1be sequences defined inLemma 2.6. Moreover, let
lim
n→ ∞Bn−∞. 2.60
Assume that for anyn∈N,unis not an escape solution of problem1.1,1.11. ByLemma 2.4 we have
unbn∈0, L, unbn 0, n∈N. 2.61
Condition2.60givesn0∈Nsuch that
Bn<2C forn∈N, n≥n0. 2.62
Choose an arbitraryn≥n0. We will construct a contradiction.
Step 1inequality forun. Sinceunis increasing on0, bn,2.62gives a uniqueγn ∈ 0, γn satisfying
un
γn 1
2Bn. 2.63
By2.59we have
t
0
Qunspsds > γn
0
Qunspsds−Q t
γn
psds, t∈γn, bn
becauseQunt>0 fort∈γn, γn. Further, there existsκn∈1/2,1satisfying
QκnBn min
Qx:x∈
Bn,B2n
. 2.65
Therefore, according to2.12,
t
0
Qunspsds > QκnBnP
γn−QP t, t∈γn, bn
. 2.66
Let us put
Int t
0 2
psPs
p2s −a 1
psun2sds, t∈0, bn. 2.67
Then inequalities2.15and2.66imply
QκnBnP
γn−QP t< Pt2Funt un2t
Int, t∈
γn, bn
. 2.68
Step 2estimate ofPγnfrom below. Sinceunis a solution of1.1on0,∞, we have
unt 1
pt
t
0
psfunsds, t∈
0, γn. 2.69
Therefore
fσnBnPt
pt ≥u
nt≥f
ρnBn Pt
pt, t∈
0, γn, 2.70
whereρn, σn∈1/2,1, are such that
fρnBn
min
fx:x∈
Bn,B2n
,
fσnBn max
fx:x∈
Bn,Bn 2
.
2.71
Integrating2.70over0, γn, we get
fσnBn γ
n
0
Ps psds≥
1
2|Bn| ≥f
ρnBn γn
0
Ps
Hence
2r−1 ρnBn r
fρnBn 1
|Bn|r−1 ≥
γ n
0
Ps
psds. 2.73
By2.52,2.60andr >1, we deduce that
lim n→ ∞
γ n
0
Ps
psds0. 2.74
SincePt/pt>0 fort >0, we get
lim
n→ ∞γn0. 2.75
Due to 2.58, there exists μ ∈ 0,∞ such that limt→0 pt/k − 1tk−2 μ. Then
limt→0 pt/tk−1 μand limt→0 kPt/tk μ. Hence for each ∈0, μthere existsδ > 0
such that, fort∈0, δ,
μ−k−1tk−2< pt<μ k−1tk−2,
μ−tk−1< pt<μ tk−1,
μ−t
k
k < Pt<
μ t
k
k.
2.76
Consequently
Pt pt <
tμ kμ−,
ptPt p2t <
k−1
k
μ μ−
2
, t∈0, δ. 2.77
Having in mind2.75, we can choosen0in2.62such that for alln≥n0the inequalityγn≤δ holds. Hence2.72and the first inequality in2.77yield
|Bn|
2fσnBn < γ
n
0
sμ kμ−ds
γ2nμ
2kμ−. 2.78
Putμ2
0kμ−/μ . Thenμ20|Bn|/fσnBn≤γ 2 n, and
γn≥μ0 |
Bn| fσnBn
1/2
. 2.79
On the other hand, by2.76,
Pγn
γ n
0
ptdt >
γ n
0
μ−tk−1dt μ−
k γ
k
By2.79, this yields
Pγn> μ− k μ
k 0
|
Bn| fσnBn
k/2 :μ0
|
Bn| fσnBn
k/2
. 2.81
Step 3estimate of{In}∞nn0. The inequalitya > k−2/k givesa 1 > 2k−1/k. Hence there exists∈0, μsuch that
a 1 2k−1
μ 2
kμ−2 . 2.82
Having in mind2.76, we chooseδto thisand then, by the second inequality in2.77, we obtain
2ptPt
p2t < a 1, fort∈0, δ. 2.83
Therefore
Int<0 for t∈0, δ, n∈N, n≥n0. 2.84
By Lemmas2.7and2.8there existsK >0 such that
ptunt≤K, fort∈Jn, n∈N. 2.85
HereJn 0, bn,n∈N, if2.25holds andJn 0,Γ 1,n∈N, if2.26holds. In addition there existsb >Γ 1 such thatJn ⊂ 0,b,n∈ N.Note that if{θn}∞n1 inLemma 2.8is not
bounded but does not fulfil2.44, we work with a proper subsequence fulfilling2.44.By virtue of2.84and2.85we get
Int≤K2 t
δ 2
psPs
p2s −a 1
1
psds, fort∈Jn, t≥δ, n∈N, n≥n0. 2.86
Inequalities2.84and2.86yield
Int< K2b 1
pδ
⎛ ⎜ ⎝2p
bPb
p2δ a 1 ⎞ ⎟
Step 4final contradictions. Putting2.81 and2.87to 2.68and using1.6,1.10and
C < B, we obtain
QκnBn |
Bn| fσnBn
k/2
μ0−Pt
Q 2FC−K
< Ptun2t, fort∈
γn, bn
, n∈N, n≥n0.
2.88
First, let us assume that2.26holds andJn 0,Γ 1,n∈N. So, conditions2.85, and2.88 yield
QκnBn |
Bn| fσnBn
k/2
μ0−PΓ 1
Q 2FC−K
< PΓ 1un2Γ 1< K2
PΓ 1
p2Γ 1 <∞, forn∈N, n≥n0.
2.89
Lettingn → ∞we get a contradiction to2.53.
Finally, let us assume that2.25holds andJn 0, bn,n∈N. Then2.61,2.88, and
Jn⊂0,byield
QκnBn |
Bn| fσnBn
k/2
μ0−P
bQ 2FC−K
< Pbun2bn 0, forn∈N, n≥n0,
2.90
contrary to2.53.
Remark 2.11. We assume that k ≥ 2 inTheorem 2.10. In particular fork 2 andpt ct,
c >0, the functionfcan behave in neighbourhood of−∞as a function|x|r for arbitraryr >1. Now, let2.58hold fork <2. Then limt→0 tk−1/pt∈0,∞and therefore
1
0
ds ps
1
0
sk−1
ps
1
sk−1ds <∞, 2.91
which is the first condition in1.9. We have proved in6,7that, in this case, assumptions
1.3–1.8are sufficient for the existence of an escape solution.
Example 2.12. Letα∈ −1,1andpt t3 αsint3,t ∈0,∞. Thenpt 3t21 αcost3
and
lim t→0
pt
t2 tlim→0 3
1 αcost331 α. 2.92
Hence, fork4 condition2.58is satisfied. The critical valuek 2/k−2is equal to 3. By
Example 2.13. Letpt tlnt 1,t∈0,∞. Thenpt lnt 1 t/t 1and
lim t→0
pt t t→lim0
lnt 1
t
1
t 1
2. 2.93
Hence, fork3 condition2.58is satisfied. The critical valuek 2/k−2is equal to 5. By
Theorem 2.10, ifffulfils2.52withr ∈1,5, problem1.1,1.11has an escape solution.
3. Homoclinic Solutions
Having an escape solution we can deduce the existence of a homoclinic solution by the same arguments as in6. For completeness we bring here the main ideas. Remember that our basic assumptions1.3–1.8,1.10and1.13are fulfilled in this section.
By Lemma 11 in6, a solutionuof problem1.1,1.11is homoclinic if and only if
sup{ut:t∈0,∞}L. 3.1
By Theorem 16 in6, a solutionuof problem1.1,1.11is an escape solution if and only if
sup{ut:t∈0,∞}> L. 3.2
The third type of solutions of problem1.1,1.11is characterized in the next definition.
Definition 3.1. A solutionuof problem1.1,1.11is called damped, if
sup{ut:t∈0,∞}< L. 3.3
The following properties of damped and escape solutions are important for the existence of homoclinic solutions.
Theorem 3.2see6, Theorem 13 on damped solutions. LetBbe of 1.5and1.6. Assume
thatuis a solution of problem1.1,1.11withB∈B,0. Thenuis damped.
Theorem 3.3see6, Theorem 14. LetMdbe the set of allB <0such that corresponding solutions of problem1.1,1.11are damped. ThenMdis open in−∞,0.
Theorem 3.4see6, Theorem 20. LetMebe the set of allB <0such that corresponding solutions of problem1.1,1.11are escape ones. ThenMeis open in−∞,0.
Having these theorems we get the main result of this section.
Theorem 3.5 On a homoclinic solution. Assume that the assumptions of Theorem 2.10 are
satisfied. Then problem1.1,1.2has a homoclinic solution.
of problem1.1,1.11is neither damped nor an escape solution. Therefore sup{ut : t ∈
0,∞}L, and by Lemma 11 in6, such solutionuis homoclinic.
The proof ofTheorem 3.5implies that if problem1.1,1.11has an escape solution, then it has also a homoclinic solution. Hence the following corollary is true.
Corollary 3.6. Assume that the assumptions ofTheorem 2.10are satisfied. Let problem1.1,1.11
have no homoclinic solution. Then it has no escape solution.
If we assume2.51and2.52, then the growth offat−∞is less than the critical value
k 2/k−2. This is necessary for the existence of homoclinic solutions of some types of
1.1. See the next example.
Example 3.7. Letk, r ∈N,k >2,r >1. Consider1.1, wherept tk−1andfx 1−xr − 1−xforx ≤1 andfx 0 forx >1. Thenpandf satisfy conditions1.3–1.8,1.10,
1.13,2.52and2.58withL1. ByTheorem 3.5, if
r < k 2
k−2, 3.4
then problem1.1,1.11has a homoclinic solution. But if
r≥ k 2
k−2, 3.5
then we have proved in 12 that problem 1.1, 1.11 has no homoclinic solution and consequently no escape solution.
Acknowledgment
This paper was supported by the Council of Czech Government MSM 6198959214.
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