ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu
BEHAVIOR AT INFINITY OF ψ-EVANESCENT SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS
PHAM NGOC BOI
Abstract. In this article we present some necessary and sufficient conditions for the existence ofψ-evanescent solution of the nonhomogeneous linear differ- ential equationx0=A(t)x+f(t), which is related to the notion ofψ-ordinary dichotomy for the equationx0 =A(t)x. We associate that with the condi- tion ofψ-ordinary dichotomy for the homogeneous linear differential equation x0=A(t)x.
1. Introduction
The existence of ψ-bounded and ψ-stable solutions on R+ for systems of ordi- nary differential equations has been studied by many authors; such as Akinyele [1], Avramescu [2], Boi [4, 5], Constantin [6], Diamandescu [9, 10, 11]. Also, in [5, 9, 10, 11] the authors prove several sufficient conditions of theψ-evanescence at
∞,−∞for the solutions of linear differential equations.
The purpose of this paper is to provide a condition for the existence ofψ- evanes- cent solution of the equationsx0=A(t)x+f(t), which is concerned with the notion of ψ-ordinary dichotomy for the equation x0 =A(t)x. We shall deal with the ex- istence of ψ-evanescent solution of nonhomogeneous equations, which have been studied in recent works, such as [5, 9, 11].
Denote by Rd the d-dimensional Euclidean space. Elements in this space are denoted byx= (x1, x2, . . . , xd)T and their norm bykxk= max{|x1|,|x2|, . . . ,|xd|}.
For reald×dmatricesA, we define the norm|A|= supkxk61kAxk. LetR+= [0,∞), R− = (−∞,0],J =R−,J =R+ orJ =R. Letψi :J →(0,∞),i= 1,2, . . . , d be continuous functions and let
ψ= diag{ψ1, ψ2, . . . , ψd}.
Definition 1.1. A functionf :J→Rd is said to be
• ψ-bounded onJ ifψf is bounded onJ.
• ψ-integrable onJ iff is measurable andψf is Lebesgue integrable onJ.
2000Mathematics Subject Classification. 34A12, 34C11, 34D05.
Key words and phrases. ψ-bounded solutions;ψ-ordinary dichotomy;ψ-evanescent solutions.
c
2010 Texas State University - San Marcos.
Submitted April 29, 2010. Published July 28, 2010.
1
InRd, consider the following equations onJ.
x0=A(t)x+f(t), (1.1)
x0=A(t)x. (1.2)
whereA(t) is a continuousd×dmatrix function andf(t) is a continuous function fort∈J.
By a solution of (1.1), we mean a continuous function satisfying (1.1) for almost t in J. LetY(t) be the fundamental matrix of (1.2) with Y(0) =Id, the identity d×dmatrix. Ad×dmatrixP is said to be a projection matrix ifP2=P. IfP is a projection, then so isId−P. Two projectionsP, Id−P are called supplementary.
Definition 1.2. Equation (1.2) is said to have a ψ-ordinary dichotomy on J if there exist positive constantsK, Land two supplementary projectionsP1, P2such that
|ψ(t)Y(t)P1Y−1(s)ψ−1(s)|6K fors6t;s, t∈J, (1.3)
|ψ(t)Y(t)P2Y−1(s)ψ−1(s)|6L fort6s;s, t∈J. (1.4) Also we say that (1.2) has a ψ-ordinary dichotomy on J with two supplementary projectionsP1, P2.
Remark 1.3. It is easily verified that if (1.2) has aψ-ordinary dichotomy onR+ and onR− with two supplementary projectionsP1, P2 then (1.2) has aψ-ordinary dichotomy onRwith two supplementary projectionsP1, P2. Note that forψ=Id, we obtain the notion of ordinary dichotomy (see [7, 8])
Theorem 1.4 ([4, 9]). (a) Equation (1.1) has at least oneψ-bounded solution on R+ for every ψ-integrable function f on R+ if and only if (1.2)has aψ-ordinary dichotomy onR+.
(b) Suppose that (1.2)has aψ-ordinary dichotomy andlimt→∞|ψ(t)Y(t)P1|= 0.
Letf be aψ-integrable function onR+. Then everyψ-bounded solution x(t) of (1.1) onR+ is such that limt→∞kψ(t)x(t)k= 0.
2. Preliminaries
Lemma 2.1. Equation (1.2) has a ψ-ordinary dichotomy on J with two supple- mentary projections P1, P2 if and only if two following conditions are satisfied for allξ∈Rd:
kψ(t)Y(t)P1ξk6Kkψ(s)Y(s)ξk fors6t;s, t∈J (2.1) kψ(t)Y(t)P2ξk6Lkψ(s)Y(s)ξk fort6s;s, t∈J (2.2) Proof. If (1.2) has aψ-ordinary dichotomy onJ then
kψ(t)Y(t)P1Y−1(s)ψ−1(s)yk6Kkyk fors6t;s, t∈J (2.3) kψ(t)Y(t)P2Y−1(s)ψ−1(s)yk6Lkyk fort6s;s, t∈J (2.4) for any vectory ∈Rd. Choose y=ψ(s)Y(s)ξ, we obtain (2.1), (2.2). Conversely, if (2.1) (2.2) are true, for any vector y ∈Rd, putting ξ =Y−1(s)ψ−1(s)y we get (2.3), (2.4). This implies that (1.2) has aψ-ordinary dichotomy onJ. The proof is
complete.
Remark 2.2. If (1.2) has aψ-ordinary dichotomy onR+ with two supplementary projectionsP1, P2 then there exist positive constantsKP, LP such that
kψ(t)Y(t)P1ξk6KPkξk, kψ(t)Y(t)ξk>LPkP2ξk for allξ∈Rd, allt>0.
Indeed, let s = 0, we deduce from (2.1) that kψ(t)Y(t)P1ξk 6 Kkψ(0)ξk 6 KPkξk for all t >0, whereKP =K|ψ(0)|. Lett = 0, we deduce from (2.2) that kψ(0)P2ξk 6Lkψ(s)Y(s)ξk, for all s>0. Then kψ(s)Y(s)ξk >LPkP2ξk, for all s>0, whereLP = [L|ψ−1(0)|]−1.
Now, letX1 ={u∈Rd|u=x(0), x(t) is aψ-bounded solution of (1.2) onR+} and let X0 = {u ∈ Rd|u = x(0), x(t) is a solution of (1.2) on R+ such that ψ(t)x(t)→0,ast→ ∞ }.
Lemma 2.3. If (1.2) has a ψ-ordinary dichotomy on R+ and Q1, Q2 are two supplementary projections, then (1.2)has aψ-ordinary dichotomy onR+ with two supplementary projections Q1, Q2 if and only if
X0⊂Q1Rd⊂X1 (2.5)
Proof. The “only if” part. Suppose that (1.2) has aψ-dichotomy with two supple- mentary projectionsQ1, Q2, we show that (2.5) holds. First, we proveQ1Rd⊂X1. For any u∈Q1Rd, there existsv ∈Rd such thatu=Q1v. Let y(t) be a solution of (1.2) such thaty(0) =u. It follows from Remark 2.2 that
kψ(t)y(t)k=kψ(t)Y(t)uk=kψ(t)Y(t)Q1vk6KQkvk fort>0,
whereKQ is a positive constant. This implies thatu∈X1. HenceQ1Rd⊂X1. We proveX0⊂Q1Rd. Foru∈X0, letx(t) be a solutions of (1.2) such thatx(0) =u.
It implies that
kψ(t)x(t)k →0, ast→ ∞ (2.6) From Remark 2.2, we have
kψ(t)x(t)k=kψ(t)Y(t)uk>LQkQ2uk, fort>0 (2.7) whereLQ is a positive constant. The relations (2.6) and (2.7) implyQ2u= 0, then u∈Q1Rd. Thus (2.5) holds.
We prove the “if” part. Suppose that (1.2) has a ψ-ordinary dichotomy on R+ with two supplementary projections P1, P2. LetQ1, Q2 be two supplementary projections such that (2.5) holds. We will prove that (1.2) has a ψ-ordinary di- chotomy on R+ with two supplementary projections Q1, Q2. Let Qe1,Qe2 be two supplementary projections such thatQe1Rd=X0. Applying (2.5) toP1, P2 we get Qe1Rd=X0⊂P1Rd⊂X1. The setX00 = (P1−Qe1)Rd is a subset ofP1Rd, supple- mentary toX0. We will show that there exists a positive constant numberN such that
kψ(t)Y(t)uk>Nkuk, for allu∈X00, t>0 (2.8) In fact, otherwise there exists a sequence of unit vectors{vn} ⊂X00, n= 1,2, . . . and a sequence of numberstn>0 such thatkψ(tn)Y(tn)vnk →0. By the compactness of the unit sphere inX00, we may assume thatvn→v∈X00 as n→ ∞, wherev is a unit vector. By Remark 2.2 and (v−vn)∈X00 ⊂P1Rd, we obtain
kψ(tn)Y(tn)(v−vn)k=kψ(tn)Y(tn)P1(v−vn)k6KPkv−vnk
Letting n → ∞, we obtain kψ(tn)Y(tn)(v−vn)k → 0. Then kψ(tn)Y(tn)vnk+ kψ(tn)Y(tn)(v−vn)k → 0 as tn → ∞. Then kψ(tn)Y(tn)vk → 0 as tn → ∞.
Hencev∈X0. On the other hand,v∈X00, we havev= 0, which is a contradiction to the unit ofv. Thus (2.8) holds.
From (2.8) and (2.1) we obtain
Nk(P1−Qe1)uk6kψ(t)Y(t)(P1−Qe1)uk
6kψ(t)Y(t)P1uk+kψ(t)Y(t)Qe1uk 6Kkψ(s)Y(s)uk+kψ(t)Y(t)Qe1uk
(2.9)
for u∈ Rd, 0 6s 6t. Let t → ∞, we get kψ(t)Y(t)Qe1uk → 0. From (2.9), we have
Nk(P1−Qe1)uk6kψ(s)Y(s)uk fors>0 (2.10) From Remark 2.2 and (2.10), we have
kψ(t)Y(t)(P1−Qe1)uk6KPk(P1−Qe1)uk6KPN−1kψ(s)Y(s)uk fort, s>0 (2.11) Consequently,
kψ(t)Y(t)Qe1uk6kψ(t)Y(t)P1uk+kψ(t)Y(t)(P1−Qe1)uk
6(K+KPN−1)kψ(s)Y(s)uk for 06s6t (2.12) FromQe2=P2+P1−Qe1and (2.11), we obtain
kψ(t)Y(t)Qe2uk6kψ(t)Y(t)P2uk+kψ(t)Y(t)(P1−Qe1)uk
6(L+KPN−1)kψ(s)Y(s)uk for 06t6s (2.13) FromQe1Rd=X0⊂Q1Rd ⊂X1, we obtainQ2Qe1Rd⊂Q2Q1Rd= 0 thenQ1Qe1= (Id−Q2)Qe1=Qe1. Thus
Q1Qe2=Q1(Id−Qe1) =Q1−Qe1 (2.14) By the definition ofX1, there existsN0>0 such that
kψ(t)Y(t)uk6N0kuk, fort>0 (2.15) It follows from Lemma 2.1, (2.12), (2.13) that (2.2) has aψ-ordinary dichotomy on R+ with two supplementary projectionsQe1,Qe2. By Remark 2.2 we have
kψ(s)Y(s)uk>LeQkQe2uk fors>0. Combining this inequality, (2.14) and (2.15) we obtain
kψ(t)Y(t)(Q1−Qe1)uk6N0k(Q1−Qe1)uk
6N0kQ1Qe2uk6N0|Q1|kQe2uk 6K2kψ(s)Y(s)uk, fort, s>0
(2.16)
whereK2 is a positive constant. From (2.12), (2.16), we have kψ(t)Y(t)Q1uk6kψ(t)Y(t)Qe1uk+kψ(t)Y(t)(Q1−Qe1)uk
6(K+KPN−1+K2)kψ(s)Y(s)uk, for 06s6t (2.17)
FromQ2=Qe2+Qe1−Q1, (2.13) and (2.16), we obtain
kψ(t)Y(t)Q2uk6kψ(t)Y(t)Qe2uk+kψ(t)Y(t)(Qe1−Q1)uk
6(L+KPN−1+K2)kψ(s)Y(s)uk, for 06t6s (2.18) Lemma 2.1 and (2.17), (2.18) follow that (1.2) has aψ-ordinary dichotomy onR+
with two supplementary projectionsQ1, Q2. The proof is complete.
LetXe1 ={u∈Rd|u=x(0), x(t) is a ψ-bounded solution of (1.2) onR− }, and letXe0={u∈Rd|u=x(0), x(t) is a solution of (1.2) onR−such thatψ(t)x(t)→0, as t→ −∞ }. From Theorem 1.4 and Lemma 2.3, we obtain the following results on half-lineR−.
Lemma 2.4. (a) Equation (1.1) has at least one ψ-bounded solution on R− for everyψ-integrable functionf onR− if and only if (1.2)has aψ-ordinary dichotomy onR−.
(b) If (1.2)has aψ-ordinary dichotomy onR−andQe1,Qe2are two supplementary projections, then (1.2)has aψ-ordinary dichotomy onR− with two supplementary projectionsQe1,Qe2 if and only if
Xe0⊂Qe2Rd⊂Xe1 (2.19) Proof. The proof of this Lemma is similar to that of Theorem 1.4 and Lemma 2.3 with the corresponding replacement (t> s>0 with 0 >s> t, P1 with −P2, P2
with−P1,∞with−∞,−∞with∞. . . ).
Definition 2.5. A functionx(t) is said to be
• ψ-evanescent at∞if limt→∞kψ(t)x(t)k= 0.
• ψ-evanescent at−∞if limt→−∞kψ(t)x(t)k= 0.
• ψ-evanescent at±∞if limt→±∞kψ(t)x(t)k= 0.
Note that forψ=Id, we obtain the notion of evanescent solution of (1.1) at±∞
(see [3])
Lemma 2.6. If (1.1)has at least one solution onR,ψ-evanescent at∞for every ψ-integrable functionf onRthen every solution of (1.2)is the sum of two solution of (1.2), one of which is ψ-bounded on R−, and the other is defined on R+, ψ- evanescent at∞.
Proof. Set
h(t) =
0 for|t|>1 1 fort= 0
linear fort∈[−1,0], t∈[0,1]
Fix a solution x(t) of (1.2). Then h(t)x(t) is a ψ-integrable function on R. Set y(t) =x(t)Rt
0h(s)ds, we have y0(t) =A(t)x(t)
Z t 0
h(s)ds+h(t)x(t) =A(t)y(t) +h(t)x(t).
By hypothesis, the equation
y0(t) =A(t)y(t) +h(t)x(t)
has a solutiony(t) one R, ψ-evanescent at∞. Set x1(t) = ey(t)−y(t) +12x(t) and x2(t) = −y(t) +e y(t) + 12x(t). It follows from R0
−1h(t)dt = R1
0 h(t)dt = 12 that
x1(t) =y(t) fore t>1;x2(t) =−y(t) fore t6−1. Then x2 is the solution of (1.2), ψ-bounded on R−, x1 is the solution of (1.2) on R+, ψ-evanescent at ∞. The solutionx(t) is the sum of two solutions x1(t) and x2(t) of (1.2), these solutions satisfy the conditions of Lemma. The proof is complete.
3. the main results
Theorem 3.1. Suppose thatf is aψ-integrable function onR+. Then(1.1)has at least one solution onR+,ψ-evanescent at∞if and only if (1.2)has aψ-ordinary dichotomy onR+.
Proof. First, we prove the “if” part. By Lemma 2.3, we can consider (1.2) has a ψ-ordinary dichotomy onR+ with two supplementary projectionsP1, P2such that P1Rd=Xo. Let
g(t) = Z t
0
Y(t)P1Y−1(s)f(s)ds− Z ∞
t
Y(t)P2Y−1(s)f(s)ds.
It is easy to see thatg(x) is a solution of (1.1) on R+. We shall prove thatg(x) is ψ-evanescent at ∞on R+. Sincef is ψ-integrable on R+, it follows that for a givenε >0, there existsT >0 such that
(K+L) Z ∞
T
kψ(s)f(s)kds < ε/2.
ByP1Rd=Xo, there existst1> T such that, fort>t1,
|ψ(t)Y(t)P1| Z T
0
kY−1(s)f(s)kds < ε/2.
Then fort>t1, we have kψ(t)g(t)k6
Z T 0
|ψ(t)Y(t)P1|.kY−1(s)f(s)kds +
Z t T
|ψ(t)Y(t)P1Y−1(s)ψ−1(s)|.kψ(s)f(s)kds +
Z ∞ t
|ψ(t)Y(t)P2Y−1(s)ψ−1(s)|.kψ(s)f(s)kds 6|ψ(t)Y(t)P1|
Z T 0
kY−1(s)f(s)kds+ (K+L) Z ∞
T
kψ(s)f(s)kds
< ε/2 +ε/2 =ε
This shows thatg(x) isψ-evanescent at∞. The “only if” part evidently holds, by
Theorem 1.4(a).
Similarly, we have the following Theorem.
Theorem 3.2. Suppose thatf is aψ-integrable function onR−. Then(1.1)has at least one solution onR−,ψ-evanescent at−∞if and only if (1.2)has aψ-ordinary dichotomy onR−.
Theorem 3.3. Suppose that (1.2) has aψ-ordinary dichotomy on R+ andf is a ψ-integrable function on R+. Then following statements are equivalent
(a) every ψ-bounded solution of (1.2)on R+ isψ- evanescent at∞.
(b) every ψ-bounded solution of (1.1)on R+ isψ-evanescent at ∞.
Proof. By Lemma 2.3, we consider (1.2) has a ψ-ordinary dichotomy on R+ with two supplementary projections P1, P2 such that P1Rd = Xo. Let S1 be the set of allψ-bounded solutions of (1.1) on R+ and let S2 be the set of all ψ-bounded solutions of (1.2) onR+. We establish a mappinghfrom S2 toS1:
(hx)(t) =x(t) +g(t), whereg(t) as in the proof of Theorem 3.1. We obtain
t→∞lim kψ(t)(hx)(t)−ψ(t)x(t)k= lim
t→∞kψ(t)g(t)k= 0
Thus h(x) is ψ-bounded on R+. Hence h(x) belongs to S1. It is easily to verify thathis one-to-one mapping betweenS2 andS1.
Suppose that statement (a) is satisfied. Letzbe arbitraryψ-bounded solution of (1.1) onR+. The foregoing follow that there exists ψ-bounded solutionxof (1.2) onR+such that h(x) =z and
t→∞lim kψ(t)z(t)−ψ(t)x(t)k= 0
By hypothesis,xisψ-evanescent at∞. Thuszisψ-evanescent at∞. Suppose that statement (b) is satisfied, the proof is similarly. The proof is complete.
Note that the above Theorem is a supplement to Theorem 1.4(b). Similarly, we have the following Theorem.
Theorem 3.4. Suppose that (1.2)has a ψ-ordinary dichotomy on R− andf is a ψ-integrable function on R−. Then following statements are equivalent
(a) every ψ-bounded solution of (1.2)on R− isψ- evanescent at−∞.
(b) every ψ-bounded solution of (1.1)on R− isψ-evanescent at −∞.
Corollary 3.5. Suppose that (1.2)has a ψ-ordinary dichotomy on R and f is a ψ-integrable function on R. Then following statements are equivalent
(a) every ψ-bounded solution of (1.2)on R+ isψ- evanescent at∞ and every ψ-bounded solution of (1.2)onR− isψ- evanescent at−∞.
(b) every ψ-bounded solution of(1.1)on Risψ-evanescent at ±∞.
Note that the above corollary is a supplement to [11, Theorem 3.3].
Theorem 3.6. Suppose that (1.2)has no non-trivial solution onR,ψ-evanescent at ∞. Then (1.1) has a unique solution on R, ψ-evanescent at ∞ for every ψ- integrable function f onRif and only if (1.2)has a ψ-ordinary dichotomy onR. Proof. First, we prove the “if” part. By Lemma 2.3, we can consider (1.2) has a ψ-ordinary dichotomy onR+ with two supplementary projectionsP1, P2such that P1Rd=Xo. Let
x(t) = Z t
−∞
Y(t)P1Y−1(s)f(s)ds− Z ∞
t
Y(t)P2Y−1(s)f(s)
Then the functionx(t) is aψ-bounded solution of (1.1) onR. We shall prove that x(t) isψ-evanescent at∞. We have, fort >0,
ψ(t)x(t) =ψ(t)Y(t)P1 Z 0
−∞
P1Y−1(s)f(s)ds+ψ(t)g(t), whereg(t) as in the proof of Theorem 3.1. Since
kP1Y−1(s)f(s)k6|Y−1(0)|.|ψ−1(0)|.|ψ(0)Y(0)P1Y−1(s)ψ−1(s)|.kψ(s)f(s)k
and f is ψ-integrable on R, we have that P1Y−1(s)f(s) is integrable onR−. Let a=R0
−∞P1Y−1(s)f(s)ds. It follows fromP1Rd=X0 that
t→∞lim kψ(t)Y(t)P1ak= 0.
On the other hand, as in the proof of Theorem 3.1, we have
t→∞lim kψ(t)g(t)k= 0.
Consequentlyx(t) is defined onR,ψ-evanescent at∞. The uniqueness of solution x(t) result from (1.2) has no non-trivial onR,ψ-evanescent solution at∞. Indeed, suppose that y is a solution onRof (1.1),ψ-evanescent at∞thenx−yis a solution solution on Rof (1.2),ψ-evanescent at∞. We conclude x=y sincex−y is the trivial solution of (1.2).
Now, we prove the “only if” part. Suppose that (1.1) has a unique ψ-bounded solution on R for every ψ- integrable function f on R. For eachu ∈ Rd, denote byx=x(t) the solution of (1.2), x(0) =u. By Lemma 2.6, we getx=x1+x2, where x2 is a ψ-bounded solution of (1.2) on R−, x1 is a solutions of (1.2) on R+ and ψ-evanescent at∞. Thus x1(0) ∈ X0 and x2(0) ∈ Xe1. It follows from u=x1(0) +x2(0) that
Rd=X0+Xe1. (3.1)
For anyv∈X0∩Xe1, denote byx(t) the solution of (1.2) such thatx(0) =v. Thus x(t) is a solution on R of (1.2), ψ-evanescent at ∞. By hypothesis, (1.2) has no non-trivial solution onR,ψ-evanescent at∞, thenx(t) is the trivial solution. This impliesv= 0. Consequently
X0∩Xe1= 0 (3.2)
The relations (3.1) and (3.2) imply thatRd is the direct sum ofX0andXe1. Every ψ-integrable functionf onR+, or onR−is the restriction of aψ-integrable function f onR, it follows that (1.2) satisfies Theorem 1.4(a) and Lemma 2.4(a). Hence (1.2) has a ψ-ordinary dichotomy on R+ and has a ψ-ordinary dichotomy on R−. Let P1, P2 be two projections such that ImP1 = X0, ImP2 =Xe1. Lemmas 2.3 and 2.4(b) follow that (1.2) has a ψ-ordinary dichotomy on R+ and has aψ-ordinary dichotomy onR− with two supplementary projections P1, P2. Remark 1.3 follows that (1.2) has a ψ-ordinary dichotomy onR with two supplementary projections
P1, P2. The proof is complete.
Similarly, we have the following Theorem.
Theorem 3.7. Suppose that (1.2)has no non-trivial solution onR,ψ-evanescent at −∞. Then (1.1) has a unique solution on R, ψ-evanescent at −∞ for every ψ-integrable functionf onRif and only if (1.2)has aψ-ordinary dichotomy onR.
Now, consider the equations
x0(t) = [A(t) +B(t)]x(t), (3.3) x0(t) = [A(t) +B(t)]x(t) +f(t) (3.4) where B(t) is a d×d continuous matrix function on R+ and f is aψ-integrable function onR+. We have the following result.
Theorem 3.8. Suppose that (1.2) has a ψ-ordinary dichotomy on R+. If δ = supt>0|ψ(t)B(t)ψ−1(t)| is sufficiently small, then following statements are equiva- lent
(a) every ψ-bounded solution of (3.3)on R+ isψ- evanescent at+∞.
(b) every ψ-bounded solution of (3.4)on R+ isψ-evanescent at +∞.
Proof. By [4, Theorem 3.7], equation (3.3) has aψ-ordinary dichotomy onR+. By
Theorem 3.3, we have the conclusion.
With similar proof, we can conclude thatJ =R−.
Theorem 3.9. Suppose that (1.2) has a ψ-ordinary dichotomy on R− and δ = supt60|ψ(t)B(t)ψ−1(t)| is sufficiently small. Then the following statements are equivalent
(a) every ψ-bounded solution of (3.3)on R− isψ- evanescent at−∞.
(b) every ψ-bounded solution of (3.4)on R− isψ-evanescent at −∞.
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Pham Ngoc Boi
Department of Mathematics, Vinh University, Vinh City, Vietnam E-mail address:pnboi [email protected]
