ON UNIQUE SOLVABILITY OF THE INITIAL VALUE PROBLEM FOR ...

52 N. Dilna
6.1. Existence of a solution of (18), (19). To show the existence of a
solution of (18), (19), it is sufficient to impose some conditions on µ 1 and
ω 1 only. More precisely, the following statement is true.
Proposition 1. The problem (18), (19) is solvable for any c ∈ R if at
least one of the following conditions is satisfied:
∫ b
a
[µ 1 (s)] − ds < 1,
ω 1 (t) ≤ t for a.e. t ∈ [a, b];
∫ b
√
[µ 1 (s)] + ds < 1 + 2√1 −
a
∫ b
a
(21a)
[µ 1 (s)] − ds . (21b)
Here, by definition, [u(t)] + :=max {u(t), 0} and [u(t)] − :=max {−u(t), 0}
for t ∈ [a, b] and u : [a, b] → R. To prove Proposition 1, the following
lemmata can be used.
Let f : D([a, b], R) → L 1 ([a, b], R) be an operator and h : D([a, b], R) → R
be a continuous linear functional.
Lemma 1 ([6, Lemma 3.1]). Let there exist a linear operator l :C([a, b], R)
→ L 1 ([a, b], R) such that the problem
u ′ (t) + (lu)(t) = 0, u(a) = 0 (22)
has only the trivial solution and for all v ∈ C([a, b], R)
|(lv)(t)| ≤ η(t) max |v(s)|, t ∈ [a, b], (23)
s∈[a,b]
where η ∈ L 1 ([a, b], R) does not depend on v. Moreover, assume that there
exists a positive number ρ such that for every δ ∈ (0, 1) and for an arbitrary
solution u ∈ D([a, b], R) of the problem
the estimate
u ′ (t) + (lu)(t) = δ[(fu)(t) + (lu)(t)], u(a) = δh(u) (24)
holds. Then the problem
has at least one solution.
max |u(t)| ≤ ρ (25)
t∈[a,b]
u ′ (t) = (fu)(t), t ∈ [a, b],
u(a) = h(u)
Lemma 1 is, in fact, [6, Theorem 1] in the formulation of [5].
Lemma 2 (Lemma 3.4 from [5]). Let there exist a linear operator l such
that the condition (23) is satisfied and the homogeneous problem (22) has
only the trivial solution. Then there exists a positive number r such that for
any q ∈ L([a, b], R) and real c every solution u of the problem
u ′ (t) = (lu)(t) + q(t),
u(a) = c
(26)