Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation
An eigenvalue problem on the half line for a singular nonlinear ordinary differential operator of the second order is considered. We find sufficient conditions under which this problem has a solution which has a prescribed number of zeroes and vanishes at infinity.
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| Цитувати: | Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation / I.O. Parasyuk, S.V. Pozur // Нелінійні коливання. — 2002. — Т. 5, № 3. — С. 346-368. — Бібліогр.: 18 назв. — англ. |
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irk-123456789-1758382021-02-03T01:29:40Z Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation Parasyuk, I.O. Pozur, S.V. An eigenvalue problem on the half line for a singular nonlinear ordinary differential operator of the second order is considered. We find sufficient conditions under which this problem has a solution which has a prescribed number of zeroes and vanishes at infinity. Розглядається задача на власнi значення на пiвосi для нелiнiйного сингулярного звичайного диференцiального рiвняння другого порядку. Знайдено достатнi умови, при яких ця задача має розв’язок iз заданою кiлькiстю нулiв, що прямує до нуля на нескiнченностi. 2002 Article Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation / I.O. Parasyuk, S.V. Pozur // Нелінійні коливання. — 2002. — Т. 5, № 3. — С. 346-368. — Бібліогр.: 18 назв. — англ. 1562-3076 http://dspace.nbuv.gov.ua/handle/123456789/175838 517.9 en Нелінійні коливання Інститут математики НАН України |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine |
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English |
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An eigenvalue problem on the half line for a singular nonlinear ordinary differential operator of the second
order is considered. We find sufficient conditions under which this problem has a solution which has a
prescribed number of zeroes and vanishes at infinity. |
| format |
Article |
| author |
Parasyuk, I.O. Pozur, S.V. |
| spellingShingle |
Parasyuk, I.O. Pozur, S.V. Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation Нелінійні коливання |
| author_facet |
Parasyuk, I.O. Pozur, S.V. |
| author_sort |
Parasyuk, I.O. |
| title |
Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation |
| title_short |
Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation |
| title_full |
Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation |
| title_fullStr |
Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation |
| title_full_unstemmed |
Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation |
| title_sort |
singular nonlinear eigenvalue problem for second order differential equation with energy dissipation |
| publisher |
Інститут математики НАН України |
| publishDate |
2002 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/175838 |
| citation_txt |
Singular nonlinear eigenvalue problem for second order differential equation with energy dissipation / I.O. Parasyuk, S.V. Pozur // Нелінійні коливання. — 2002. — Т. 5, № 3. — С. 346-368. — Бібліогр.: 18 назв. — англ. |
| series |
Нелінійні коливання |
| work_keys_str_mv |
AT parasyukio singularnonlineareigenvalueproblemforsecondorderdifferentialequationwithenergydissipation AT pozursv singularnonlineareigenvalueproblemforsecondorderdifferentialequationwithenergydissipation |
| first_indexed |
2025-07-15T13:17:07Z |
| last_indexed |
2025-07-15T13:17:07Z |
| _version_ |
1837719017969680384 |
| fulltext |
UDC 517.9
SINGULAR NONLINEAR EIGENVALUE PROBLEM
FOR SECOND ORDER DIFFERENTIAL EQUATION
WITH ENERGY DISSIPATION*
СИНГУЛЯРНА НЕЛIНIЙНА ЗАДАЧА НА ВЛАСНI ЗНАЧЕННЯ
ДЛЯ ДИФЕРЕНЦIАЛЬНОГО РIВНЯННЯ ДРУГОГО ПОРЯДКУ
З ДИСИПАЦIЄЮ ЕНЕРГIЇ
I. O. Parasyuk, S. V. Pozur
Kyiv Taras Shevchenko Nat. Univ.
Volodymyrs’ka St., 64, Kyiv, 01033, Ukraine
e-mail: pio@mechmat.univ.kiev.ua
An eigenvalue problem on the half line for a singular nonlinear ordinary differential operator of the second
order is considered. We find sufficient conditions under which this problem has a solution which has a
prescribed number of zeroes and vanishes at infinity.
Розглядається задача на власнi значення на пiвосi для нелiнiйного сингулярного звичайного ди-
ференцiального рiвняння другого порядку. Знайдено достатнi умови, при яких ця задача має
розв’язок iз заданою кiлькiстю нулiв, що прямує до нуля на нескiнченностi.
1. Introduction. The goal of this paper is to establish sufficient conditions for the solvability of
the following singular boundary-value problem:
y′′ + p(x, y, y′)y′ + q(x, y) = λy, (1)
lim
x→+0
y(k)(x) < ∞, k = 0, 1, 2, (2)
lim
x→+∞
y(x) = 0. (3)
Here p(x, y, z) ∈ C1((0,∞)× R× R 7→ R), q(x, y) ∈ C1((0,∞)× R 7→ R) are functions such
that
|p(x, y, z)|+ |q(x, y)| = O(x−1), x → +0,
and λ ≥ 0 is a positive ”spectral” parameter. We ask whether there exists λ > 0 for which
the system (1) – (3) has a nontrivial solution. Thus, the above system is treated as a nonlinear
eigenvalue problem. Moreover, we will seek for such a λ that there exists a solution to (1) – (3)
with a prescribed number of zeroes. The problems of such a kind naturally arise when studyi-
ng global spherically-symmetric solutions of multidimensional nonlinear evolution equations
of the classical field theory. E.g., let us search for a function u(t,x;λ) = eiλty(‖x‖), λ ∈ R,
x = (x1, . . . , xn), satisfying the nonlinear Schrödinger equation (NSE)
iut + ∆u+ Φ(|u|)u = 0,
∗ This work was partially supported by the State Fund of Fundamental Research (grant No. 01.07./00047).
c© I. O. Parasyuk, S. V. Pozur, 2002
346 ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 347
where ∆ is the n-dimensional Laplace operator and Φ(·) ∈ C(R+ 7→ R). We require that
u(t,x;λ) represents a global solution which vanishes as ‖x‖ → ∞. This requirement will be
fulfilled once the function y(x) satisfies the equation
y′′ +
n− 1
x
y′ + Φ(y)y = λy (4)
as well as the boundary conditions (2), (3). There are a lot of papers in which the nonlinear
singular boundary-value problems on the half line are studied (see [1 – 7]). In particular, in [1]
the authors considered the equation (4) in the case where Φ(y)y = |y|αsign y, λ = 1, n − 1 =
= γ. They have shown that the corresponding problem (1) – (3) has a nontrivial solution iff the
parameters α and γ satisfy one of the following conditions: (a) γ > 0, 1 < α <
< (γ + 3)/(γ − 1); (b) γ ∈ (0, 1], α > 1; (c) γ > 0, α < 1; (d) γ ≤ 0, α > 1. In the
cases (a), (b) any solution either is sign-preserving or has a finite number of zeroes; moreover,
there exists a solution with the prescribed number of zeroes. In the case (c) any solution is
oscillating, while in the case (d) it is always sign-preserving.
As it was shown in [8], the NSE with Φ(|u|) = |u|k, k = 4/n, by means of the ansätz
u =
(
2√
λ(1 + t2)
)n/4
exp
(
− itx2
2(t2 + 1)
)
y
(√
λx2
1 + t2
)
,
is reduced to the ODE
d2y
dx2
+
a
x
dy
dx
+
|y|ky
x
= λy, (5)
where a = n/2. In [7], it has been proved that for any l ∈ Z+ the singular problem (5), (2), (3)
has a nontrivial solution with exactly l zeroes provided that the following inequality holds:
max(0, 1− 1/k) < a < 1 +
2
k
. (6)
We will generalize this result for the problem (1) – (3).
The structure of the paper is as follows. In Section 2 we construct a solution y(x, λ) to
the problem (1), (2), and show that y(x, λ) ∈ C([0,∞) × [0,∞) 7→ R) if the equation (1)
has a property of energy dissipation and the function q(x, y) satisfies certain nonlinearity and
convexity conditions.
The main results, — Theorems 1 and 2, — are proved in the Section 3. Here we present a
“supremum principle” of finding eigenvalues. This principle is applicable under the assumpti-
on that the function y(x, 0) has at least one zero. Although some additional requirements
concerning functions p(x, y, z) and q(x, y) are quite numerous, they are satisfied by a wide
class of equations (1). Besides, they can be considerably simplified if we restrict ourselves to
more special and concrete, but important types of functions p(x, y, z), q(x, y), e.g., p(x, y, z) =
= P1(x)P2(y)P3(z), q(x, y) = q1(x)q2(y). As an example, in Section 4 we consider the
eigenvalue problem of the form (1) – (3) for a generalized Emden – Fowler equation.
Section 5 contains auxiliary results which are referred to in the proofs of the main theorems.
2. Existence of solution on the half line. Assume that the following singularity condition at
x = 0 is fulfilled:
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
348 I. O. PARASYUK, S. V. POZUR
H1: For any compact set K ⊂ R2 there exist the limits
p0(y, z) := lim
x→+0
xp(x, y, z), q0(y) := lim
x→+0
xq(x, y),
uniform in (y, z) ∈ K. It is not hard to show that if for a given λ ≥ 0 there exists a solution y(x)
of the system (1), (2) then its initial values
y0 := lim
x→+0
y(x), z0 := lim
x→+0
y′(x)
must satisfy the equation
p0(y, z)z + q0(y) = 0. (7)
Basing on this fact, let us adopt the following hypothesis:
H2: There exists a solution y = y0, z = z0 of the equation (7) such that y0 6= 0, and in some
neighborhood B(y0, z0) of the point (y0, z0) the functions p0(y, z), q0(y) are twice continuously
differentiable and satisfy the nondegeneracy condition
∂p0(y0, z0)
∂z
z0 + p0(y0, z0) > 0; (8)
besides, the functions p(x, y, z), q(x, y) admit the representation
p(x, y, z) =
p0(y, z)
x
+ p1(x, y, z), q(x, y, z) =
q0(y)
x
+ q1(x, y),
where p1(x, y, z), q1(x, y) are continuous on the set [0,∞)×B(y0, z0), together with their partial
derivatives in y, z up to the second order.
Proposition 1. Let the hypotheses H1, H2 be true. Then for any λ0 > 0 there exists ρ > 0
such that for each λ ∈ [0, λ0] the equation (1) has a solution y(x, λ), x ∈ [0, ρ], satisfying the
initial conditions y(0, λ) = y0, y
′
x(0, λ) = z0. As a function of the variables (x, λ), this solution
is continuous on [0, ρ]× [0, λ0].
Proof. After the change of dependent variable y = y0 + z0x+ v in (1), we get the equation
v′′ +
a
x
v′ +
b
x
v = f(x, λ) + g(x, λ)v + h(x)v′ + F (x, v, v′) (9)
where
a = p0(y0, z0) + p′0z(y0, z0)z0, b = p′0y(y0, z0)z0 + q′0(y0),
f(x, λ) = λ(y0 + z0x)− p(x, y0 + z0x, z0)z0 − q(x, y0 + z0x),
g(x, λ) = λ+
b
x
− ∂p(x, y0 + z0x, z0)
∂y
z0 −
∂q(x, y0 + z0x)
∂y
,
h(x) =
a
x
− ∂p(x, y0 + z0x, z0)
∂z
z0 − p(x, y0 + z0x, z0),
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 349
and
F (x, v, v′) = p(x, y0 + z0x, z0)z0 + q(x, y0 + z0x)+
+
(
∂p(x, y0 + z0x, z0)
∂y
z0 +
∂q(x, y0 + z0x)
∂y
)
v+
+
(
∂p(x, y0 + z0x, z0)
∂z
z0 + p(x, y0 + z0x, z0)
)
v′−
− p(x, y0 + z0x+ v, z0 + v′)(z0 + v′)− q(x, y0 + z0x+ v).
In view of H1, H2, for any λ ∈ [0, λ0] the functions f, h, g can be extended by continuity on the
half line [0,∞). Now it is easy to see that the assertion to be proved follows from Lemma 1 and
Remark 1 below in Section 5. The proposition is proved.
We say that the equation (1) satisfies the energy dissipation condition if
H3: q′x(x, y) ≤ 0, p(x, y, z) ≥ 0 ∀(x, y, z) ∈ (0,∞)× R2.
Note that the equation (1), with p ≡ 0, is a Lagrangian one with the energy function
V [y] :=
y′2
2
+Q(x, y)− λy2
2
, where Q(x, y) :=
y∫
0
q(x, s) ds.
When the hypothesis H3 holds, the derivative V ′[y] of this function, in virtue of the equation
(1), is nonpositive,
V ′[y] := −p(x, y, y′)y′2 +Q′x(x, y) ≤ 0.
In the sequel we shall assume that the function q(x, y) satisfies the following nonlinearity and
convexity conditions:
H4: For any x ∈ (0,∞) there exist the limits
lim
y→0
q(x, y)
y
= 0, lim
y→±∞
q(x, y)
y
= ∞;
besides, there exist r > 0 and R > r such that for all x ∈ (0, r) ∪ (R,∞), (y, z) ∈ R2, and
y 6= 0 there exists the derivative q′′yy(x, y) which satisfies the inequality
∂2q(x, y)
∂y2
y > 0 ∀y 6= 0.
Proposition 2. If the hypotheses H1 –H4 are true, then for each λ ≥ 0 the solution y(x, λ)
from Proposition 1 is continuable on the half line [0,∞). Moreover, as a function of the variables
(x, λ), this solution is continuous on [0,∞)× [0,∞).
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
350 I. O. PARASYUK, S. V. POZUR
Proof. Let, for some λ∗ ∈ [0,∞), the function y∗(x) := y(x, λ∗) be continuable only on a
finite interval [0, x∗). Then |y∗(x)| → ∞, x → x∗ − 0. Thus,
V [y∗(x)] ≥ Q(x, y∗(x))− λy2∗(x)
2
≥ Q(x∗, y∗(x))− λy2∗(x)
2
→ ∞, x → x∗ − 0.
But on the other hand, from H3 we get V [y∗(x)] ≤ V [y∗(x∗/2)] < ∞ if x ∈ (x∗/2, x∗). The
contradiction obtained implies boundedness of y∗(x) on any bounded set of the half line [0,∞).
But then, in view of H3, y
′
∗(x) has the same property. Thus the natural domain of definition
for y(x, λ) is D := [0,∞) × [0,∞). The continuity of y(x, λ) at any point (0, λ) ∈ ∂D, where
λ ∈ [0,∞), has already been proved in Proposition 1. The continuity of this function at any
point (x, λ), where x > 0, λ ≥ 0, follows from standard theorems of the general theory of
ODEs.
3. The main results. In what follows we shall suppose that
H5: The function y(x, 0) has at least l + 1 zeroes on (0,∞) where l is a natural number.
Here we do not discuss additional conditions which the functions p(x, y, z) and q(x, y) must
satisfy to guarantee the fulfillment of the hypothesis H5. There are a lot of papers where such
conditions has been established for nonlinear second order differential equations with damping
(see, e.g., [9 – 16]).
Let Lm be the set of all numbers Λ > 0 such that the solution y(x, λ) has at least m + 1
zeroes for any λ ∈ (0,Λ). Obviously,
Lm+1 ⊆ Lm.
Let the hypotheses H1 –H5 be true. Then each set Lm, m = 0, 1, . . . , l+ 1, is nonempty and
bounded (Proposition 4 in Section 5). Introduce the numbers
λm = supLm and ym(x) := y(x, λm), m = 0, . . . , l.
Obviously, λl ≤ · · · ≤ λ0.
We are going to show that each λm is an eigenvalue and ym(x) is the corresponding ei-
genfunction of the problem (1) – (3). We will prove this fact under some additional technical
assumptions described below.
The following assumption concerns the behavior of p(x, y, z) for large values of x:
H6: The function p(x, 0, 0) has a finite limit when x → +∞, and
lim sup
x→+∞
sup
(y,z)∈K
(
|p′y(x, y, z)|+ |p′z(x, y, z)|
)
< ∞
for any compact set K ⊂ R2.
We also impose a technical condition on the rate of energy dissipation. Namely,
H7: For any ε > 0, there exist functions α±(x, ε) which are continuous with respect to the
variable x ∈ [R,∞), satisfy the inequalities
0 ≤ α±(x, ε) ≤ p(x, y, z)z2 −Q′x(x, y) ∀(y, z) ∈ M±(x, ε), ∀x ≥ R,
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 351
where
M±(x, ε) :=
{
(y, z) ∈ R2 : ± y ≥ εx, ± z ≥ ε,
z2
2
+Q(x, y) ≥ εx2
}
,
and have the following property:
∞∫
R
α±(x, ε) dx = ∞.
Observe that if
∂p(x, y, z)
∂y
y ≥ 0,
∂p(x, y, z)
∂z
z ≥ 0 ∀x ≥ R, (10)
then each of the conditions
∞∫
R
p(x, 0, 0) dx = ∞ or
∞∫
R
p(x,± εx,± ε) dx = ∞ ∀ε > 0
obviously guarantee the fulfillment of H7.
From H4 it follows that for any fixed x ∈ (0, r) ∪ (R,∞) and any λ > 0 the equation
q(x, y) = λy
has exactly one positive solution y := η+(x, λ) and one negative solution y := η−(x, λ). The
minimal values of the function Q(x, y)− λy2/2 on the semi-axes y > 0 and y < 0 are attained
at the points y = η+(x, λ) and y = η−(x, λ) respectively. This minimal values we designate by
µ±(x, λ) :=
[
Q(x, y)− λy2
2
]
y=η±(x,λ)
.
Denote by ξ+(x, λ) (ξ−(x, λ)) the unique positive (negative) root of the equation Q(x, y) =
= λy2/2.
Next we introduce the sets
N±(x, λ) =
{
(y, z) ∈ R2 : ± y > ± η±(x, λ), ± z > 0,
z2
2
+Q(x, y) >
λy2
2
}
and adopt the following assumption:
H8: For any λ > 0, there exist functions β±(x, λ) which are continuous with respect to
variable x ∈ (R,∞), satisfy the inequalities
0 ≤ β±(x, λ) ≤ p(x, y, z) ∀(y, z) ∈ N±(x, λ), ∀x > R,
and for these functions either
∞∫
R
exp
( ∫
β±(s, λ) ds
)
|µ±(s, λ)|
|ξ±(s, λ)− η±(s, λ)|
ds = ∞ ∀λ > 0, (11)
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
352 I. O. PARASYUK, S. V. POZUR
or
∞∫
R
∣∣Q′x(x, y)
∣∣
y=η±(x,λ)
dx = ∞ ∀λ > 0. (12)
In view of the fact that both |µ±(x, λ)| and |η±(x, λ)| are nondecreasing, this condition
is satisfied by a wide class of functions q(x, y) which obey H3, H4. It should also be noted that
if (10) holds then it is naturally to check the fulfillment of (11) setting β±(x, λ) :=
:= p(x, η±(x, λ), 0).
Theorem 1. If the hypotheses H1 –H8 are true, then for any m = 0, . . . , l the pair
(λm, ym(x)), where ym(x) := y(x, λm), is a solution of the problem (1) – (3). Moreover,
ym(x)y′m(x) < 0 for all sufficiently large x, and
|ym(x)|+ |y′m(x)| = O(e−γmx), x → ∞, (13)
where γm is a positive number.
Proof. Let us show that V [ym(x)] > 0 for all x ≥ 0. Indeed, if we suppose that there are
x0 > 0 and ε > 0 such that V [ym(x0)] < −ε, then, according to Proposition 6 (Section 5), for
all λ ∈ Lm sufficiently close to λm, the (m+ 1)-st zero, xm+1(λ), of ym(x) satisfies the inequali-
ties xm+1(λ) > x0 and V [y(xm+1(λ), λ)] ≤ V [y(x0, λ)] ≤ −ε/2. This, however, contradicts
Proposition 3 (Section 5).
According to Proposition 5 (Section 5) the function ym(x) does not vanish for all sufficiently
large x. Then from Proposition 7, it follows that ym(x)y′m(x) < 0 and |ym(x)| +
+ |y′m(x)| → 0, x → ∞.
Observe now that ym(x) can be regarded as a solution of the linear equation
y′′ + a(x)y′ + b(x)y = 0 (14)
in which
a(x) :=
{
p(x, 0, 0) +
[
p(x, ym(x), y′m(x))− p(x, 0, 0)
]}
, b(x) := λm −
qm(x, ym(x))
ym(x)
.
From H4, H6 it follows that this equation satisfy conditions of the Lemma 2 (Section 5).
The theorem is proved.
Theorem 1 asserts that each λm is an eigenvalue of the problem (1) – (3), but it does not
guarantees that λi 6= λj for i 6= j. Now we will analyze this question.
Introduce the sets
P±(x, λ) :=
{
(y, z) ∈ R2 :
± η±(x, λ)
2
≤ ± y ≤ ± η(x, λ), ± z >
√
λy2 − 2Q(x, y)
}
and the following assumptions:
H9: For any λ > 0, there exist functions π±(x, λ) which are continuously differentiable with
respect to the variable x ∈ (R,∞), satisfy the inequalities
0 ≤ π±(x, λ) ≤ p(x, y, z) ∀(y, z) ∈ P±(x, λ), ∀x > R,
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 353
and for which
1
π2±(x, λ)
∂π±(x, λ)
∂x
≥ −ω±(λ) ∀x ≥ R
with some functions ω± : (0,∞) 7→ (0,∞) bounded on any compact subset of the half line
(0,∞).
Note that if the condition 10 holds then it is natural to set
π±(x, λ) := p
(
x,
η±(x, λ)
2
,±
√
|µ±(x, λ)|
2
)
.
H10: On any compact set of positive values of the parameter λ, the function
sup
x≥R
|η±(x, λ)|√
|µ±(x, λ)|
is bounded and the limit
lim
x→+∞
ln |η±(x, λ)|
x
= 0
is uniform.
By using a technique which is quite similar to that of the paper [1] we are able to prove the
following theorem.
Theorem 2. Let the hypotheses H1 –H10 be true. Then
0 < λl < · · · < λ1 < λ0,
and for any m = 0, 1, . . . , l the eigenfunction ym(x) := y(x, λm) has exactly m zeroes (the
function |y0(x)| is positive).
Proof. Let m be an arbitrary nonnegative integer not exceeding l. According to Propositi-
on 6 (Section 5) the (m+ 1)-st zero xm+1(λ) of the function y(x, λ) tends to infinity when λ →
→ λm − 0. For this reason there exists an integer k ∈ [1,m + 1] such that
lim supλ→λm−0 xk(λ) = ∞ but, if k > 1, lim supλ→λm−0 xk−1(λ) < ∞. Our goal is to show
that one can choose such a positive λ < λm near λm that the function y(x, λ) has only k zeroes.
Obviously, this is possible only if k = m+ 1, and hence, λm+1 < λm for any m < l.
Let {λ(i)}i=1,2,... be a sequence such that λ(i) → λm − 0 and xk(λ
(i)) → ∞ as i → ∞.
Suppose, on the contrary, that for each natural i the function y(x, λ(i)) has a zero xk+1(λ
(i)) >
> xk(λ
(i)). Without loss of generality we may assume that
|y(x, λ(i))| > 0 ∀x ∈
(
xk(λ
(i)), xk+1(λ
(i))
)
.
Under this assumption we are going to show that for sufficiently large i the function
V [y(x, λ(i))] vanishes on (xk(λ
(i)), xk+1(λ
(i))). In view of Proposition 3 this will give us the
desired contradiction.
Let us choose a sufficiently small δ > 0 in such a way that
δ < |η±(R, λm)|, Q(R, y) <
λmy
2
4
∀y ∈ (−δ, δ).
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354 I. O. PARASYUK, S. V. POZUR
Then for all sufficiently large i we have the inequalities
δ < |η±(x, λ(i))| ∀x ≥ R,
λ(i)y2
2
−Q(x, y) >
λmy
2
4
∀y ∈ (−δ, δ).
Next we choose xδ > R so large that xδ > supi≥1 xk−1(λ
(i)) if k > 1 and
|ym(x)| < δ, ym(x)y′m(x) < 0, V [ym(x)] < δ ∀x ≥ xδ.
Then for all sufficiently large i the following inequalities holds:
|y(xδ, λ
(i))| < δ, y(xδ, λ
(i))y′(xδ, λ
(i)) < 0, V [y(xδ, λ
(i))] < δ, xk(λ
(i)) > xδ.
Now it is not hard to show that |y(x, λ(i))| is monotone decreasing to zero on [xδ, xk(λ
(i))] and
simultaneously
y′2(x, λ(i))
2
>
λmy
2(x, λ(i))
4
∀x ∈ [xδ, xk(λ
(i))].
From this it follows that
|y(x, λ(i))| < e−λm(x−xδ)/
√
2δ ∀x ∈ [xδ, xk(λ
(i))].
Besides, since |y(x, λ(i))| < |η±(x, λ(i))| for x ∈ [xδ, xk(λ
(i))], we have y′′(x, λ(i))y′(x, λ(i)) <
< 0 and
|y′(x, λ(i))| < |y′(xδ, λ(i))| <
√
2(λmδ2 + δ) ∀x ∈ [xδ, xk(λ
(i))].
Now for sufficiently small δ and all sufficiently large i one can extend the functions
a(x) = ai(x) := p(x, y(x, λ(i)), y′(x, λ(i))), b(x) = bi(x) := −λ(i) +
q(x, y(x, λ(i)))
y(x, λ(i))
on the semi-axis [xδ,∞) in such a way that they satisfy the conditions of Lemma 2 (Section 5)
with x0 = xδ and with the numbers α, β, ε independent of i. But then taking into account that
y(x, λ(i)) is a solution of the equation
y′′ + ai(x)y′ + bi(x)y = 0,
we get the estimate
|y′(xk(λ(i)), λ(i))| ≤ Kδe
−γmxk(λ(i))
with some positive constants Kδ and γm. Thus,
V [y(xk(λ
(i)))] ≤
K2
δ
2
e−2γmxk(λ
(i)). (15)
Without loss of generality we assume that y′(xk(λ(i))) > 0. Taking into account that
V [y(x, λ(i))] > 0 for x ∈
[
xk(λ
(i)), xk+1(λ
(i))
]
, it is easy to establish the existence of the nearest
to xk(λ(i)) value x = x∗(λ
(i)) ∈ (xk(λ
(i)), xk+1(λ
(i))) for which[
y(x, λ(i)) = η+(x, λ(i))
]
x=x∗(λ(i))
.
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SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 355
Then there exists x∗(λ(i)) such that[
y(x, λ(i)) =
η+(x, λ(i))
2
]
x=x∗(λ(i))
,
and
η+(x, λ(i))
2
< y(x, λ(i)) < η+(x, λ(i)), y′(x, λ(i)) >
√
λ(i)y2 − 2Q(x, y)
∣∣∣
y=y(x,λ(i))
for all x ∈ (x∗(λ
(i)), x∗(λ(i))).
Now for such values of x, using the convexity property of the function f(y) :=
:=
√
λy2 − 2Q(x, y) (Lemma 3, Section 5) and inequality (15), we have
V ′[y(x, λ)] ≤ −π+(x, λ)y′
2
(x, λ) ≤
≤ −π+(x, λ)
|µ+(x∗(λ), λ)|V [y(x, λ)]
4V [y(x∗(λ), λ)]
−
− π+(x, λ)
√
|µ+(x∗(λ), λ)|
8
y′(x, λ) ≤
≤ −ω+(λ)π+(x, λ)V [y(x, λ)]− π+(x, λ)
√
|µ+(R, λ)|
3
y′(x, λ),
provided that λ = λ(i) and i is sufficiently large. Integrating the above differential inequality
we get
V [y(x∗(λ), λ)] ≤ C(λ)
[
V [y(x∗(λ), λ)]−
−
√
|µ+(R, λm)|
3
x∗(λ)∫
x∗(λ)
e
ω+(λ)
∫ x
x∗(λ) π+(s,λ) ds
π+(x, λ)y′(x, λ) dx
]
,
where
C(λ) = e
−ω+(λ)
∫ x∗(λ)
x∗(λ)
π+(s,λ) ds
.
Since H9 is true, the function eω+(λ)
∫ x
x∗(λ) π+(s,λ) ds
π+(x, λ) is nondecreasing. Thus,
V [y(x∗(λ), λ)] ≤ C(λ)
[
K2
δ
2
e−2γmxk(λ)−
−
√
|µ+(R, λm)|π+(x∗(λ), λ)
3
(
y(x∗(λ), λ)− y(x∗(λ), λ)
)]
≤
≤ C(λ)
[
K2
δ
2
e−2γmxk(λ) −
√
|µ+(R, λ)|η+(R, λ)
6
π+(x∗(λ), λ)
]
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356 I. O. PARASYUK, S. V. POZUR
where λ = λ(i).
From H9 it follows that
π+(x, λ) ≥ π+(R, λ)
π+(R, λ)ω+(λ)(x−R) + 1
and Proposition 8 (Section 5) allows to obtain the estimate x∗(λ(i)) = O
(
xk(λ
(i))
)
as i → ∞
provided that H10 is true. Now it is easy to show that V
[
y(x∗(λ(i)), λ(i))
]
< 0 for sufficiently
large i. This is just the required contradiction. The theorem is proved.
4. Singular eigenvalue problem for generalized Emden – Fowler equation. In this section we
consider the following boundary-value problem:
y′′ +
a
x
y′ + f(x)|y|ky = λy, (16)
lim
x→+0
y(x) = y0, lim
x→+0
y(k)(x) < ∞, k = 1, 2, (17)
lim
x→+∞
y(x) = 0 (18)
where a and k are positive constants and f(x) ∈ C1((0,∞) 7→ (0,∞)).
By applying Theorem 2, we arrive at the following generalization of results obtained in [7].
Theorem 3. Let the function f(x) satisfy the following conditions:
1) there exists the limit lim
x→+0
xf(x) = f0 ≥ 0;
2) f ′(x) ≤ 0;
3)
∞∫
1
xf(x) dx = ∞ if a ∈ (0, 1);
∞∫
1
xf(x) lnx dx = ∞ if a = 1;
∞∫
1
x1−(a−1)kf(x) dx =
= ∞ if a > 1;
4) lim
x→∞
ln f(x)
x
= 0.
Then for any y0 6= 0 and arbitrary natural l there exists a solution of the problem (16) – (18)
with exactly l zeroes.
Proof. From the assumptions 1 and 2 it follows that the hypotheses H1 –H4, H6 and H9 are
true. In fact, the equation (7) takes the form az + f0|y|ky = 0, and thus z0 = −f0|y0|ky0/a for
any y0. To verify H7 and H9 it is sufficient to put α±(x, ε) = π±(x, λ) = a/x.
As it was shown in [17], any nontrivial solution of the equation (r(x)y′)′+s(x)|y|γsgnx = 0,
where r(x), s(x) ∈ C([1,∞) 7→ (0,∞)), γ > 1, is oscillatory on (0,∞) provided that
∞∫
1
1
r(x)
dx = ∞ and
∞∫
1
s(x)
x∫
1
1
r(ξ)
dξ dx = ∞
or
∞∫
1
1
r(x)
dx < ∞ and
∞∫
1
s(x)
∞∫
x
1
r(ξ)
dξ
γ
dx = ∞.
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SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 357
Now letting r(x) = xa, s(x) = xaf(x) one can easily verify that the assumption 3 guarantees
the oscillatory property for any solution of the equation (16) with λ = 0. Thus the hypothesis
H5 is true.
Next, one can easily calculate that
η±(x, λ) = ±
(
λ
f(x)
)1/k
, ξ±(x, λ) = ±
(
(k + 2)λ
kf(x)
)1/k
,
|µ±(x, λ)| =
λk
2(k + 2)
(
λ
f(x)
)2/k
.
Thus, in view of condition 2, we get infx≥1 |ξ±(x, λ)− η±(x, λ)| > 0, and H8 is true.
Lastly, H10 follows from the assumption 4. The theorem is proved.
It should be noted that to satisfy the condition 3 it is necessary to require a ≤ 1 + 2/k. In
the special case f(x) = 1/x the assumption 3 takes the form 0 < a ≤ 1 + 1/k. This shows that
the condition a > 1 − 1/k in (6) is superfluous. On the other hand, Theorem 3 does not allow
us to obtain the best upper bound for a. The reason is that our requirement that all solutions of
(5) are oscillatory is stronger then the hypothesis H5.
5. Auxiliary results.
Lemma 1. Consider the equation
v′′ +
a
x
v′ +
b
x
v = f(x) + g(x)v + h(x)v′ + F (x, v, v′) (19)
where a > 0, b ∈ R, f, g, h ∈ C([0, r] 7→ R), F ∈ C((0, r]×B(0, 0) 7→ R), F (x, 0, 0) ≡ 0, and
B(0, 0) is a neighborhood of the origin in R2. Suppose that there exists a constant L > 0 such
that
|xF (x, v1, w1)− xF (x, v2, w2)| ≤ Lmax{|v1|, |v2|, |w1|, |w2|}
(
|v1 − v2|+ |w1 − w2|
)
for all (x, v, w), (x, v1, w1), (x, v2, w2) ∈ (0, r]×B(0, 0). Then there exist numbers ρ ≤ r, H > 0
and a solution v(x) ∈ C2([0, ρ] 7→ R) of the equation (19) satisfying the inequalities
|v(x)| ≤ Hx2, |v′(x)| ≤ Hx ∀x ∈ [0, ρ]. (20)
Proof. The equation
v′′ +
a
x
v′ +
b
x
v = 0 (21)
has a regular singular point at x = 0. The roots of the corresponding indicial equation are 0
and 1− a.
As is well known, if a /∈ N, the equation (21) has a pair of linearly independent solutions
v1(x), v2(x) with asymptotic representation
v1(x) = 1 +O(x), v2(x) = x1−a (1 +O(x)) , x → 0.
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358 I. O. PARASYUK, S. V. POZUR
In the resonant case where a ∈ N there exists a pair of linearly independent solutions with
asymptotic representation
v1(x) = 1 +O(x), v2(x) = αv1(x) lnx+ 1 +O(x), x → 0,
where α is a real number. In both cases the Wronskian W (x) of these solutions equals x−a/ω
where ω is a real number.
It is not hard to show that if for some ρ ∈ (0, r], we have that H > 0, a function v(x) ∈
∈ C1[0, ρ] satisfies the inequalities (20) and the integral equation
v(x) = A[v](x) :=
x∫
0
G(x, s)
[
f(s) + g(s)v(s) + h(s)v′(s) + F (s, v(s), v′(s))
]
ds, (22)
where
G(x, s) :=
v2(x)v1(s)− v1(x)v2(s)
W (s)
≡ ωsa
(
v2(x)v1(s)− v1(x)v2(s)
)
,
then this function is a solution on (0, ρ] of differential equation (19).
Next we show that there exists such H1 > 0 that
x∫
0
|G(x, s)| ds ≤ H1x
2,
x∫
0
|G′x(x, s)| ds ≤ H1x ∀x ∈ [0, r].
Now it turns out that we can choose sufficiently large H > 0 and sufficiently small ρ > 0 in
such a way that the mapping A : MH,ρ 7→ MH,ρ be a contraction in the space
MH,ρ = {v(x) ∈ C1[0, ρ] : |v(x)| ≤ Hx2, |v′(x)| ≤ Hx, x ∈ [0, ρ]}
supplied with the standard metric
%(v(x), w(x)) = max
x∈[0,ρ]
[
|v(x)− w(x)|+ |v′(x)− w′(x)|
]
.
(Note that for any v(x) ∈ MH,ρ we set, by definition, A[v](0) = 0.)
To prove that the fixed point v(x) of the mapping A belongs to C2[0, ρ], it remains only
to observe the following property of the kernel G(x, s): for any function ϕ(x) ∈ C[0, ρ] the
function
x∫
0
G′′xx(x, s)ϕ(s) ds =
(
−av
′
2(x)
x
− bv2(x)
x
) x∫
0
ωsav1(s)ϕ(s) ds+
+
(
−av
′
1(x)
x
− bv1(x)
x
) x∫
0
ωsav2(s)ϕ(s) ds
can be extended by continuity onto the whole segment [0, ρ]. The lemma is proved.
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SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 359
Remark 1. If in equation (19) the functions f = f(x, λ), g = g(x, λ) ∈ C([0, r] × [0, λ0])
additionally depend on a parameter λ then the solution v(x, λ) satisfying (20) is a continuous
function of the variables (x, λ) on the set [0, ρ]× [0, λ0].
Proposition 3. Let the hypotheses H3, H4 be true. If for a nontrivial solution y(x), x ≥ 0, of
the equation (1) there exists x0 ≥ 0 such that V [y(x0)] ≤ 0 then y(x) does not vanish on [x0,∞).
Proof. In fact, if there exists x1 ≥ x0 such that y(x1) = 0, then V [y(x1)] =
(
y′(x1)
)2
/2 >
> 0. But V [y(x)] is nonincreasing, and thus we arrive at the contradiction, V [y(x0)] > 0.
Proposition 4. If the hypotheses H1 –H5 are true, then for m = 0, 1, . . . , l + 1 each set Lm is
nonempty and bounded.
Proof. From q(x, 0) ≡ 0 it follows that y ≡ 0 is a solution of (1). As a nontrivial solution,
the function y(x, 0) has only simple zeroes: if y(xi, 0) = 0 then y′(xi, 0) 6= 0, i = 1, . . . , l +
+ 1. Then by implicit function theorem there exists λl > 0 such that for any λ ∈ (0, λl) the
function y(x, λ) has at least l + 1 simple zeroes. Thus Ll 6= ∅. Now we are going to show that
one can choose λ∗ ≥ λl in such a way that the function y∗(x) := y(x, λ∗) does not vanish on
[0,∞). If z0 = 0, then q0(y0) = 0 and q0(y) = 0 for y ∈ [0, y0]. So limx→+0Q(x, y∗(x)) =
=
∫ y0
0
q1(0, y) dy, and
lim
x→+0
V [y∗(x)] =
y0∫
0
q1(0, y) dy − λ∗y0
2
< 0
if λ∗ is sufficiently large. From this it follows that V [y∗(x)] < 0 when x > 0 and, in virtue of
Proposition 3, y∗(x) does not vanish on [0,∞). Now let z0 6= 0. Without loss of generality we
assume that z0 < 0. Then y0 > 0. Note that in view of (8) there is a neighborhood of (y0, z0)
in which those points of the (y, z)-plane whose coordinates satisfy the equation (7) form the
graph of a function Z(y) ∈ C1([y0 − δ, y0 + δ] 7→ R), where δ is a positive number. We can
choose δ ≤ y0/2 in such a way that Z(y) < 0 for all y ∈ [y0− δ, y0]. For a fixed λ∗, the function
y∗(x) := y(x, λ∗) is decreasing on some interval [0, x1). Choose x1 as such a maximal number
that y∗′(x) < 0 and y0 − δ < y∗(x) ≤ y0 for all x ∈ [0, x1). Analyzing the nontrivial case
where x1 < ∞, let us show that if λ∗ is sufficiently large, then
y∗
′(x) > Z(y∗(x)) (23)
for all x ∈ (0, x1), i.e., the curve given by the equations
y = y∗(x), z = y∗
′(x), x ∈ (0, x1),
is contained inside the figure F bounded by the lines y = y0 − δ, y = y0, z = 0, z = Z(y).
It is not hard to calculate that
y′′xx(+0, λ) ∼ λy0
∂p0(y0,z0)
∂z z0 + p0(y0, z0)
, λ → ∞.
This implies
d
dx
∣∣∣
x=+0
(
y′∗(x)
)
>
d
dx
∣∣∣
x=+0
(
Z(y∗(x))
)
= Z ′(y0)z0,
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360 I. O. PARASYUK, S. V. POZUR
provided that λ∗ is sufficiently large. Taking into account that y∗′(+0) = z0 = Z(y∗(+0)), one
can assert that there exists a maximal number x2 ∈ (0, x1] such that the inequality (23) holds
for all x ∈ (0, x2). It turns out that x2 = x1. In fact, if we suppose that x2 < x1, we would get
y∗
′(x2) = Z(y∗(x2)) and, in view of p0(y, Z(y))Z(y) + q0(y) = 0,
y∗
′′(x2) = λ∗y∗(x2)− p1
(
x2, y∗(x2), y∗
′(x2)
)
y∗
′(x2)− q1
(
x2, y∗(x2)
)
≥
≥ λ∗y0/2− max
[0,x1]×F
|p1(x, y, z)z + q1(x, y)| >
> max
y0−δ≤y≤y0
Z ′(y)Z(y) ≥ dZ(y∗(x2))
dx
provided that λ∗ was chosen sufficiently large. On the other hand, by the definition of x2,
y∗
′′(x2) ≤
dZ(y∗(x2))
dx
. This contradiction proves that x2 = x1.
Now it is not hard to see that if y∗(x1) ∈ (y0 − δ, y0) then y∗′(x1) = 0; otherwise, y∗(x1) =
= y0 − δ. In the first case we have y′′∗(x1) ≥ 0 and thus q(x1, y∗(x1))− λ∗y∗(x1) ≤ 0. Together
with H4, this implies Q(x1, y∗(x1)) − λ∗y
2
∗(x1)/2 < 0 and V [y∗(x1)] < 0. Now the required
property of y∗(x) follows from Proposition 3 . In the second case, by means of the inequality
y′∗(x)
Z(y∗(x))
≤ 1, x ∈ [0, x1],
we get for x1 the lower bound which is independent of λ∗,
x1 ≥
y0−δ∫
y0
dy
Z(y)
≥ δ
max
y0−δ≤y≤y0
|Z(y)|
.
Now
V [y∗(x1)] ≤
(
max
y0−δ≤y≤y0
|Z(y)|
)2
/2 + max
y0−δ≤y≤y0
Q(x1, y)− λ∗y
2
0
8
< 0
provided that λ∗ was chosen sufficiently large. Again we can apply Proposition 3.
Proposition 5. For any m = 0, . . . , l the number of zeroes of each function ym(x) does not
exceed m.
Proof. In fact, otherwise ym(x) would have at least m+ 1 simple zeroes. Then y(x, λ) would
have the same property for λ belonging to some neighborhood of λm. This contradicts the
definition of λm as an exact upper bound.
Proposition 6. Let λ ∈ Lm and x1(λ) < · · · < xm+1(λ) be successive zeroes of y(x, λ). Then
xm+1(λ) → ∞, λ → λm − 0.
Proof. Should this assertion be false, there would exist a sequence λ(k) → λm − 0, k → ∞,
such that each sequence {xj(λ(k))}k=1,2,..., j = 1, . . . ,m + 1, be convergent to a finite limit x∗j
which is a zero of ym(x). It is not hard to see that in such a case
x∗1 < · · · < x∗m+1,
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SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 361
otherwise ym(x) would have at least one multiple zero. Hence, ym(x) hasm+1 different simple
zeroes. This, however, contradicts Proposition 5.
Proposition 7. Let the hypotheses H3, H4, H6 –H8 be true. If for some fixed λ > 0 the equati-
on (1) has a solution y(x) which possesses only a finite number of zeroes and satisfies the condi-
tion V [y(x)] > 0 ∀x > 0, then y(x)y′(x) < 0 for all sufficiently large positive x and
|y(x)|+ |y′(x)| → 0, x → +∞.
Proof. Without loss of generality we analyze the case where there exists such a positive
x0 that y(x) > 0 ∀x > x0. First we suppose that y′(x0) > 0 and show that there is x1 >
> x0 such that y′(x1) = 0. Let, on the contrary, y′(x) do not vanish on [x0,∞). Put ν :=
:= infx≥x0 y
′(x) and consider each of the following possible cases: (a) ν > 0; (b) ν = 0.
In the case (a) we would have y(x) ≥ y(x0) + νx > νx and, under an appropriate choice of
ε = ε(ν, x0),
V [y(x)] = V [y(x0)] +
x∫
x0
V ′[y(s)] ds ≤ V [y(x0)]−
x∫
x0
α±(s, ε) ds → −∞, x → ∞.
Thus the case (a) is impossible.
In the case (b), for any given ε > 0, there exists x̃ > max(x0, R) such that 0 < y′(x̃) < ε.
From H4 it follows that for any sufficiently small ε > 0 the equation
Q(x, y)− λy2
2
= −ε
2
2
has a pair of positive roots ξ1(x, λ, ε), ξ2(x, λ, ε) satisfying the inequalities
0 < ξ1(x, λ, ε) < η+(x, λ) < ξ2(x, λ, ε) < ξ+(x, λ).
It is not hard to see that the function ξ1(x, λ, ε) is nonincreasing while the functions ξ2(x, λ, ε),
η+(x, λ), ξ+(x, λ) are nondecreasing with respect to x. Moreover, ξ1(x, λ, ε) → 0, ε → 0, and
ξ2(x, λ, ε) → ξ+(x, λ), ε → 0. We may think that ξ1(x, λ, ε) < y(x0) for all x ≥ x̃. Now it is
clear that until x ≥ x̃ and 0 < y′(x) < ε the following inequalities hold:
Q(x, y(x))− λy2(x)
2
> −ε
2
2
, (24)
y(x) > ξ2(x, λ, ε), q(x, y(x))− λy(x) > 0. (25)
Hence,
y′′(x) ≤ λy(x)− q(x, y(x)) < 0
and y′(x) is monotone decreasing. Obviously, in case (b) we have y′(x) → 0, x → ∞. Thus,
(25) must be realized for all x ≥ x̃.
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362 I. O. PARASYUK, S. V. POZUR
To obtain a contradiction from this fact, let us estimate y′′(x) more accurately. Rewriting the
equality
Q(x, ξ2)−
λξ22
2
= −ε
2
2
in the form
µ+(x, λ) +
ξ2(x,λ,ε)∫
η+(x,λ)
(q(x, y)− λy) dy = −ε
2
2
and assuming that ε is so small that ε < |µ+(R, λ)|/2 we get
ξ2(x,λ,ε)∫
η+(x,λ)
(q(x, y)− λy) dy ≥ |µ+(x, λ)|/2.
Observe that [q(x, y)−λy]y=η+(x,λ) = 0 and for any fixed x the derivative q′y(x, y)−λ is increasi-
ng in y. Taking into account that q(x, y)− λy > 0 for y > η+(x, λ) we see that q′y(x, y)− λ > 0
once y > η+(x, λ). Hence, q(x, y) − λy is monotone increasing in y on (η+(x, λ),∞). Now by
the mean value theorem we get
(q(x, y(x))− λy(x))(ξ2 − η+) ≥ (q(x, y)− y)
∣∣
y=ξ2
(ξ2 − η+) ≥ |µ+|
2
and thus
q(x, y(x))− λy(x) ≥ |µ+(x, λ)|
2(ξ2(x, λ, ε)− η+(x, λ))
≥ |µ+(x, λ)|
2(ξ+(x, λ)− η+(x, λ))
. (26)
Solving for x > x̃ the differential inequality
y′′(x) ≤ −β+(x, λ)y′(x)− |µ+(x, λ)|
2(ξ+(x, λ)− η+(x, λ))
we obtain the estimate
y′(x) ≤ e−
∫ x
x̃ β+(s,λ) ds
y′(x̃)−
x∫
x̃
e
∫ τ
x̃ β+(s,λ) ds |µ+(τ, λ)|
2(ξ+(τ, λ)− η+(τ, λ))
dτ
.
Now if (11) holds, then y′(x) → −∞ as x → ∞ and we arrive at a contradiction. Otherwise,
(12) holds, and hence, V [y(x)] → −∞, x → ∞. This is again impossible.
Thus we have proved that there exists x1 > x0 such that y′(x1) = 0, y(x1) > ξ+(x1, λ) and
y′′(x1) ≤ λy(x1)− q(x1, y(x1)) < 0.
From this it follows that y′(x) < 0 for all x > x1. In fact, should the first moment x2 > x1
existed for which y′(x2) = 0 we would have y′′(x2) ≥ 0 and hence q(x2, y(x2)) − λy(x2) ≤ 0.
This would imply the contradiction: V [y(x2)] ≤ 0.
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SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 363
Let y∗ := limx→∞ y(x). In view of H3 the function Q(x, y(x)) is nonincreasing,
Q(x′′, y(x′′)) ≤ Q(x′′, y(x′)) ≤ Q(x′, y(x′)) ∀x′′ > x′.
As Q(x, y(x)), y(x) and V [y(x)] have finite limits when x → ∞, y′(x) has the same property,
and obviously limx→∞ y
′(x) = 0.
Let us now show that y∗ = 0. Suppose on the contrary that y∗ > 0. We may think that
ε was chosen so small that ξ1(R, λ, ε) < y∗ and |y′(x)| < ε for all sufficiently large x. Then
y(x) ≥ y∗ ≥ ξ1(x, λ, ε), V [y(x)] > 0, and hence, the inequality (24) holds. This yields (25) and
boundedness of the difference ξ2(x, λ, ε)− η+(x, λ). For this reason in view of (26),
lim inf
x→∞
(q(x, y(x))− λy(x)) > 0
and hence, taking into account H6,
lim sup
x→∞
y′′(x) < 0.
This inequality contradicts existence of zero limit for y′(x). The proposition is proved.
Lemma 2. Let the coefficients a(x), b(x) ∈ C[x0,∞) of the equation
y′′ + a(x)y′ + b(x)y = 0 (27)
satisfy the inequalities
|a(x)− α|2 ≤ 4β − ε, −B ≤ b(x) ≤ −β ∀x ≥ x0
with some constants α 6= 0, B > β > 0 and ε ∈ (0, 4β).
Then there exist positive numbers K = K(x0, α, β,B, ε), γ = γ(α, β,B, ε) and linearly
independent solutions y1(x), y2(x) of (27) for which the following estimates hold
|y1(x)|+ |y′1(x)| ≤ Ke−γx, |y2(x)| ≥ eγx
K
, x ≥ x0. (28)
If y(x) is a nontrivial solution of (27) vanishing at a point x1 > x0, then
|y′(x1)| ≤ K|y(x0)y
′
2(x0)− y′(x0)y2(x0)|e
−γx1−
∫ x1
x0
a(s) ds
. (29)
Proof. The derivative of the quadratic form W (y, z) := −αy2− 2yz, in virtue of the system
y′ = z, z′ = −b(x)y − a(x)z (30)
being equivalent to the equation (27), equals W ′(y, z) = −2(z2 − b(x)y2 +
(
α − a(x)
)
yz). For
eigenvalues λ1(x), λ2(x) of the corresponding matrix(
2b(x) a(x)− α
a(x)− α −2
)
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
364 I. O. PARASYUK, S. V. POZUR
the following inequalities holds:
λ1(x) + λ2(x) ≤ −2(1 + β), λ1(x)λ2(x) ≥ ε.
Hence, these eigenvalues are negative and separated from zero. Since the matrix of W (y, z)
is nondegenerate and indefinite, the system (30) is dichotomic (see, e.g. [18]). Hence, it has a
fundamental system (y1(x), z1(x)), (y2(x), z2(x)) such that
W (y1(x), z1(x)) > 0, W (y2(x), z2(x)) < 0 ∀x ∈ [x0,∞), (31)
and for certain K1 > 0 and γ > 0 the following inequalities holds:
|y1(x)|+ |z1(x)| ≤ K1e
−γx, |y2(x)|+ |z2(x)| ≥ eγx/K1 ∀x ≥ x0.
Observe that we may choose as (y2(x), z2(x)) an arbitrary solution of the system (30) with
initial values satisfying the condition W (y2(x0), z2(x0)) < 0. Set, e.g., z2(x0) = (1 + |α|)y2(x0)
and y2(x0) > 0.Now to establish (28) it remains only to prove that for someK > 0 the function
y2(x) obeys |y2(x)| ≥ eγx/K for all x ≥ x0.
The second inequality in (31) implies y2(x) > 0 and y′2(x) = z2(x) > −α
2
y2(x) for all
x ≥ x0. Let us show that y′2(x) < Ay2(x), where A is a positive number such that
A > y′2(x0)/y2(x0) and A2 > A sup
x≥x0
|a(x)|+ sup
x≥x0
|b(x)|.
In fact, should there exist such a first “moment” x̄ > x0 that y′2(x̄) = Ay2(x̄), then y′′2(x̄) ≥
≥ Ay′2(x̄) and hence, in virtue of (27),
−a(x̄)Ay2(x̄)− b(x̄)y2(x̄) ≥ A2y2(x̄).
This, however, is impossible for the choice of A.
Thus, we have proved that |y′2(x)| = |z2(x)| ≤ max
(∣∣∣α
2
∣∣∣ , A) |y2(x)|.Consequently, we may
set K = K1(1 + max(|α|, A)).
Now we will deduce (29). Let w0 stand for the value of the Wronskian of y1(x), y2(x) at
x = x0. Then y(x) = c1y1(x) + c2y2(x) where
c1 =
y(x0)y
′
2(x0)− y′(x0)y2(x0)
w0
, c2 = −c1y1(x1)
y2(x1)
,
and hence,
y′(x1) = c1
y′1(x1)y2(x1)− y1(x1)y′2(x1)
y2(x1)
=
= [y(x0)y
′
2(x0)− y′(x0)y2(x0)]e
−
∫ x1
x0
a(s) ds
/y2(x1).
This, obviously, implies (29).
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SINGULAR NONLINEAR EIGENVALUE PROBLEM FOR SECOND ORDER DIFFERENTIAL EQUATION . . . 365
Lemma 3. Let f(y) ∈ C2[a, b] ∩ C3(a, b), f(0) = f ′(0) = 0, and f(y) > 0, f ′′′(y) < 0 for
all y ∈ (a, b). Then
d2
√
f(y)
dy2
< 0 ∀y ∈ (a, b).
Proof. We have
d2
√
f(y)
dy2
=
1
2
(f(y))−3/2
[
f(y)f ′′(y)− 1
2
f ′
2
(y)
]
.
Now let us show that the function F (y) = f(y)f ′′(y) − 1
2
f ′
2
(y) takes only negative values on
(a, b). In fact, F (0) = 0 and F ′(y) = f(y)f ′′′(y) < 0 for all y ∈ (a, b).
Proposition 8. Let the hypotheses H3, H4, H6, H10 be true and for some λ > 0 the equation
(1) have a solution y(x) for which there exist x0 > R and x∗ > x0 such that
y(x0) = 0, 0 < y′(x0) <
η+(R, λ)
2
, y(x∗) = η(x∗, λ),
V [y(x)] > 0, 0 < y(x) < η+(x, λ) ∀x ∈ (x0, x
∗).
Then
x∗ −M(λ)| ln η+(x∗, λ)| ≤ x0 + eP (λ)M(λ) +M(λ)| ln y′(x0)|
where
M(λ) := sup
x≥R
η+(x, λ)√
2|µ+(x, λ)|
, P (λ) := sup
x≥R
0≤y≤η+(R,λ)M(λ)
0≤z≤η+(R,λ)
p(x, y, z).
Proof. Let x1 be a point at which the function y′(x) attains its minimal value on the
segment [x0, x
∗]. In virtue of Lemma 3, for any fixed x and λ, the function y 7→
7→
√
λy2 − 2Q(x, y) is convex on the interval (0, ξ+(x, λ)). For this reason
y′(x) ≥
√
λy2(x)− 2Q(x, y(x)) ≥
√
2|µ+(x, λ)|
η+(x, λ)
y(x) ∀x ∈ [x0, x
∗].
Thus,
y′(x1) ≥
√
2|µ+(x, λ)|
η+(x, λ)
y′(x1)(x1 − x0)
and
x1 − x0 ≤ M(λ), y(x1) ≤
η+(R, λ)M(λ)
2
. (32)
Let x∗ ∈ (x0, x
∗) be a point such that
y(x∗) =
η+(x∗, λ)
2
,
η+(x, λ)
2
< y(x) < η+(x, λ) ∀x ∈ (x∗, x
∗).
ISSN 1562-3076. Нелiнiйнi коливання, 2002, т . 5, N◦ 3
366 I. O. PARASYUK, S. V. POZUR
Since y(x∗) ≥
η+(R, λ)
2
≥ y′(x0), there exists x2 ∈ (x0, x∗] such that y(x2) = y′(x0). But
y(x2) ≥ y′(x1)(x2 − x0). Hence, x2 − x0 ≤ y′(x0)/y
′(x1). To estimate y′(x1) observe that in
view of (32)
y′′(x) ≥ −p(x, y(x), y′(x))y′(x) ≥ −P (λ)y′(x) ∀x ∈ [x0, x1].
This implies y′(x1) ≥ y′(x0)e
−P (λ)M(λ) and thus
x2 − x0 ≤ eP (λ)M(λ).
Since the inequality y′(x) ≥ y(x)/M(λ) holds on [x0, x
∗] we get
η(x∗, λ) = y(x∗) ≥ e(x
∗−x2)/M(λ)y(x2) = e(x
∗−x2)/M(λ)y′(x0)
and finally
x∗ ≤ x2 +M(λ)
[
| ln η+(x∗, λ)|+ | ln y′(x0)|
]
≤
≤ x0 + eP (λ)M(λ) +M(λ)
[
| ln η+(x∗, λ)|+ | ln y′(x0)|
]
.
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Received 02.04.2002
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