1. MARCO TEÓRICO DE ESTUDIO
1.4. MARCO LEGAL
1.4.3. Ley Orgánica de Régimen Tributario Interno
Let us consider equations
dx
dt =A(t)x, (3.89)
d y
dt =B(t)y, (3.90)
whereAandB∈C(R; [En]).
Definition 3.38. One says that (3.89) satisfies the condition of Favard in the positive (resp., negative) direction, denoting it byΦ+(resp.,Φ−), if for everyB ∈ω
A(resp.,αA) (3.90)
Linear Differential Equations 67 Definition 3.39. If (3.89) satisfies the condition Φ+ and Φ−, then one say that (3.89)
satisfies the two-sided (or weak condition of Favard) and denote itΦ.
Definition 3.40. If for eachB ∈H(A) (3.90) has no nonzero bounded onRsolutions, then one says that (3.89) satisfies the condition of Favard (or reinforced condition of Favard) and denote itF.
Between the introduced notions there exists a close relation. We study this question below. Here we only note that from the weak condition of Favard, generally speaking, the condition of Favard does not follow (the inverse is obvious). The last is confirmed by the example
dx
dt =(arctant)x. (3.91)
It is easy to check that (3.91) satisfies the conditionΦbut does not satisfy the condition F, since (3.91) has nonzero bounded onRsolutions.
Definition 3.41. They say that a function f ∈ C(R;En) is stable by Lagrange in the
positive (resp., negative) direction, denoting itL+ (resp.,L−), ifH+(f) (resp.,H−(f))
is compact inC(R;En).
Definition 3.42. If the function f ∈C(R;En) is st.L+and st.L−, then they say that f is
stable by Lagrange and denote it st.L.
3.3.3.2. Some Properties of Equations Satisfying the Condition of Favard in the Positive Direction
Lemma 3.43. LetA∈C(R; [En]) be st.L+(resp.,L−) and (3.89) satisfy the conditionΦ+ (resp.,Φ−). Then, ifϕ(t,A,x) is bounded onR+(resp.,R−), then
lim
t→+∞ϕ(t,A,x)=0 resp., limt→−∞ϕ(t,A,x)=0
. (3.92)
Proof . Suppose the contrary, that is, there exist ε0 > 0 and tk → +∞ such that
|ϕ(tk,A,x)| ≥ ε0. Without loss of generality we can consider that the sequences {ϕ(tk,A,x)}and{A(tk)}are convergent inEnandC(R; [En]), respectively. Assumex0 = limk→+∞ϕ(tx,A,x) andB=limk→+∞A(tk). Then according toLemma 3.37,ϕ(t,B,x0)= limk→+∞(t+tk,A(tk),x) is a nontrivial bounded on Rsolution of (3.90) and B ∈ ωA.
The last contradicts to the condition ofLemma 3.43. The second case is considered in the
similar way.
Lemma 3.44. LetA ∈ C(R; [En]) be st. L+ and let (3.89) satisfy the conditionΦ+. If ϕ(t,A,x) is unbounded onR+, then limt→+∞|ϕ(t,A,x)| =+∞.
Proof . Let us suppose the contrary, that is, for someL > 0 there exist sequences{sk},
{tk}, and{lk}satisfying the next conditions:
(1) sk< tk< lk< sk+1; (2) {sk} →+∞ask→+∞; (3) |ϕ(τ,A,x)|> Lfor allτ∈(sk,lk); (4) |ϕ(sk,A,x)| = |ϕ(lk,A,x)| =L; (5) |ϕ(tk,A,x)| =max{|ϕ(t,A,x)|:t∈[sk,lk]}. Assume xk:= ϕ(tk,A,x) ϕ tk,A,x= ϕ tk,A,x− 1 ·ϕtk,A,x . (3.93)
Then there are held the relations: (a) |xk| =1;
(b) |ϕ(t,A(tk),xk)| = |ϕ(tk,A,x)|−1· |ϕ(t+tk,A,x)| ≤1
for allt∈[sk−tk,lk−tk]. Let us show that{sk−tk} → −∞({lk−tk} →+∞). In fact,
under the conditions ofLemma 3.44the sequences{A(tk)}and{xk}can be considered convergent. Putx0 =limk→+∞xkandB:=limk→+∞A(tk). If we suppose that{s
k−tk} →
−∞, then it can be considered convergent. Assumeτ0:=limk→+∞{sk−tk}. Passing to the
limit in the equality
ϕ sk−tk,A(tk),xk=ϕ tk,A,x− 1 ·ϕsk,A,x=Lϕ tk,A,x− 1 , (3.94) we obtain thatϕ(τ0,B,x0)=0. From the last equality it follows thatx0 =0 that contra- dicts to the condition (a). In the same way it is proved that{lk−tk} → +∞. From the
said above and the condition (b) it follows that|ϕ(t,B,x0)| ≤1 holds for allt∈R, and |x0| = 1. So, ϕ(t,B,x0) is a nontrivial bounded onRsolution of (3.90). The obtained
contradiction proves the lemma.
Lemma 3.45. LetA∈C(R; [En]) be st.L+and (3.89) satisfy the conditionΦ+. If a solution ϕ(t,A,x) is unbounded onR+, then there existsc >0 such that
max 0≤t≤τ
ϕ(t,A,x)≤cϕ(τ,A,x) (3.95)
for allτ∈R+.
Proof . Suppose the contrary. Then for everyk∈Nthere isLk≥ksuch that
max 0≤t≤Lk
ϕ(t,A,x)≥kϕ
Lk,A,x. (3.96)
Chooseτk∈[0,Lk] so that the equality
ϕ
τk,A,x= max
0≤t≤Lk
ϕ(t,A,x) (3.97)
holds. Then inequality (3.96) takes the form
ϕ
Linear Differential Equations 69 Put xk:= ϕ(τk,A,x) ϕ τk,A,x= ϕ τk,A,x−1·ϕ τk,A,x . (3.99) Then xk=1, ϕ t,A(τk),x k=ϕτk,A,x|−1·ϕt+τk,A,x (3.100)
for allt∈[τk,Lk−τk]. Reasoning in the same way that inLemma 3.44and taking into
consideration the relations
(a) |ϕ(Lk−τk,A(τk),xk)| = |ϕ(τk,A,x)|−1· |ϕ(Lk,A,x)| ≤k−1,
(b) |ϕ(−τk,A(τk),x)| = |ϕ(τk,A,x)|−1· |x|,
we obtain that{τk} → +∞and {Lk −τk} → +∞. Without loss of generality we can
consider that the sequences{A(τk)}and{x
k}are convergent. Letx0 := limk→+∞xk and
B :=limk→+∞A(τk). It is clear thatB ∈ω
A,|x0| = 1 and|ϕ(t,B,x0)| ≤1 for allt∈ R. The last contradicts to the condition of the lemma. 3.3.3.3. Scalar Equations
Below we suppose that the space En is one-dimensional (En = R), that is, A = a ∈
C(R;R).
Lemma 3.46. Let a ∈ C(R;R) be st. L+. If (3.89) satisfies the condition Φ+ and the solutionsϕ(t,a,x) of (3.90) asx = 0 are unbounded fort ∈ R+. Then solutions of every (3.90) withB=b∈ωaare bounded onR−.
Proof . Let b ∈ ωa. Then there existstk → +∞such thatb = limk→+∞a(tk). define a sequence
xk:=
ϕtk,a,x
αk , (3.101)
whereαk:=max{ϕ(t,a,x) :t∈[0,tk]}. For it there are fulfilled the following conditions:
(1) |xk| =1;
(2) |ϕ(t,a(tk),x
k)| =α−k1|ϕ(t+tk,a,x)| ≤1
for allt ∈ [−tk, 0]. In virtue of (1) the sequence {xk} can be considered convergent.
Assumex0:=limk→+∞xk. Passing to the limit in inequality (2) we get|ϕ(t,b,x0)| ≤1 for allt∈R−. In this case, byLemma 3.45,|xk| ≥c−1 >0 and, consequently,|x0| ≥c−1 > 0. To finish the proof of the lemma it is sufficient to use the fact that (3.89) is a linear
homogeneous scalar equation.
Lemma 3.47. Letabe st.L+and (3.89) satisfy the conditionΦ+. If all solutions of (3.89) are bounded onR+, then there exist numbersN,ν >0 such that
ϕ(t,b,x)≤N e−νt|x| (3.102)
Proof . The proof of the formulated statement may be done by repeating exactly the
reasoning from the work [122].
Lemma 3.48. Letabe st.L+and (3.89) satisfy the conditionΦ+. Then (3.89) is hyperbolic