Electronic Journal of Qualitative Theory of Differential Equations 2006, No. 8, 1-13; http://www.math.u-szeged.hu/ejqtde/

On Singular Solutions of a Second Order Differential Equation

Miroslav Bartušek

Abstract

In the paper, sufficient conditions are given under which all nontrivial solutions of (g(a(t)y′))′+r(t)f(y) = 0 are proper where a > 0, r > 0, f(x)x > 0, g(x)x > 0 for x ≠ 0 and g is increasing on R. A sufficient condition for the existence of a singular solution of the second kind is given.

2000 MSC: 34C11

Key words and phrases: Nonlinear second order differential equation, singular solutions, noncontinuable solutions.

1 Introduction

Consider the differential equation

(g(a(t)y′))′ + r(t)f(y) = 0,  (1)

where a ∈ C^{0}(R_{+}), r ∈ C^{0}(R_{+}), g ∈ C^{0}(R), f ∈ C^{0}(R), R_{+} = [0, ∞), R = (−∞, ∞), g is increasing on R and

a > 0, r > 0 on R_{+}, f(x)x > 0 and g(x)x > 0 for x ≠ 0. (2)

Sometimes the following condition will be assumed.

lim_{z→∞} g(z) = − lim_{z→−∞} g(z) = ∞.  (3)

Definition. A function y defined on J ⊂ R_{+} is called a solution of (1) if y ∈ C^{1}(J), g(a(t)y′) ∈ C^{1}(J) and (1) holds on J.

It is clear that (1) is equivalent to the system y_{1} = y, y_{2} = g(a(t)y′),

y′_{1} = \frac{g^{−1}(y_{2})}{a(t)}, y′_{2} = −rf(y_{1}),  (4)

where g^{−1} is the inverse function to g. Hence, as the right-hand sides of (4) are continuous, the Cauchy problem for (1) has a solution.

Definition. Let y be a continuous function defined on [0, τ) ⊂ R_{+}. Then y is called oscillatory if there exists a sequence {t_{k}}_{k=1}^{∞}, t_{k} ∈ [0, τ), k = 1, 2, … of zeros of y such that lim_{k→∞} t_{k} = τ and y is nontrivial in any left neighbourhood of τ.

Definition. A solution y of (1) is called proper if it is defined on R_{+} and sup_{τ≤t<∞} |y(t)| > 0 for every τ ∈ (0, ∞). It is called singular of the first kind if it is defined on R_{+}, there exists τ ∈ (0, ∞) such that y ≡ 0 on [τ, ∞) and sup_{T ≤t<τ} |y(t)| > 0 for every T ∈ [0, τ). It is called singular of the second kind if it is defined on [0, τ), τ < ∞, and cannot be defined at t = τ. A singular solution y is called oscillatory if it is an oscillatory function on [0, τ).

In the sequel we will investigate only solutions that are defined either on R_{+} or on [0, τ), τ < ∞ and cannot be defined at = τ.

Remark 1. (i) According to (2) every nontrivial solution of (1) is either proper, singular of the first kind, or singular of the second kind. (ii) A solution is singular of the second kind if and only if

lim_{t→τ−} sup |y′(t)| = ∞.

(iii) If y is a singular solution of the first kind then y(τ) = y′(τ) = 0. Consider the equation with p-Laplacian

(A(t)|y′|^{p−1}y′)′ + r(t)f(y) = 0,  (5)

where p > 0, A ∈ C^{0}(R_{+}) and A > 0 on R_{+}. This is a special case of (1) with g(z) = |z|^{p−1}z and a = A^{\frac{1}{p}}. It is widely studied now; see e.g. [3], [4],

[8] and the references therein.

Recall the following sufficient conditions for the nonexistence of singular solutions of (5). Theorem A. (i) If M > 0, M_{1} > 0 and |f(x)| ≤ M_{1}|x|^{p} for |x| ≤ M, then there exists no singular solution of the first kind of (5). (ii) If M > 0, M_{1} > 0 and |f(x)| ≤ M_{1}|x|^{p} for |x| ≥ M, then there exists no singular solution of the second kind of (5). (iii) Let the function A^{\frac{1}{p}} r be locally absolutely continuous on R_{+}. Then every solution of (5) is proper.

Proof. Cases (i) and (ii) are simple applications of results in [8, Theorems 1.1 and 1.2] (also see [1]). Case (iii) is proved in [3, Theorem 2] .

Theorem A (iii) shows that if A and r are smooth enough, singular solutions do not exist. But the following theorem shows that singular solutions may exist. Theorem B ([3] Theorem 4). Let 0 < λ < p (0 < p < λ). Then there exists a positive continuous function r defined on R_{+} such that the equation

(|y′|^{p−1}y′)′ + r(t)|y|^{λ} sgn y = 0  (6)

has a singular solution of the first (of the second) kind.

Note that the proof of Theorem B uses ideas from [5] and [6] for the case p = 1.

The goal of this paper is to generalize results of Theorems A and B to Eq. (1).

2 Main results

We begin our investigations with simple properties of singular solutions.

Lemma 1. Let y be a singular solution of (1) and τ be the number in its definition. Then y is oscillatory if and only if y′ is an oscillatory function on [0, τ).

Proof. It follows directly from system (4) since, due to (2), y′ is an oscillatory function on [0, τ) if and only if y_{2} = g(a(t)y′) is oscillatory on the same interval.

Theorem 1. (i) Every singular solution of the first kind of (1) is oscillatory. (ii) If (3) holds, then every singular solution of the second kind of (1) is oscillatory.

Proof. (i) Let y be a singular solution of the first kind of (1) and τ < ∞ be the number from its definition. Suppose, contrarily, that y > 0 in a left neighborhood of τ (the case y < 0 can be studied similarly). Then (1) and (2) yield g(ay′) is decreasing and hence, ay′ is decreasing on I. From this and from Remark 1 (iii), we have y′(τ) = 0 and hence y′ > 0 on I; this contradicts the fact that y > 0 on I and y(τ) = 0. (ii) Let y be a singular solution of the second kind of (1) defined on [0, τ), τ < ∞. Suppose, contrarily, that y > 0 in a left neighbourhood I = [τ_{1}, τ) of τ (the case y < 0 can be studied similarly). Then (1) and (2) yield ay′ is decreasing on I and according to Remark 1 (ii) and Lemma 1 lim_{t→τ_{−}} y′(t) = −∞. Hence y is positive and decreasing in a left neighbourhood of τ and rf(y) is bounded on I. From this, we have

−∞ = g(a(τ)y′(τ)) − g(a(τ_{1})y′(τ_{1})) = − ∫_{τ_{1}}^{τ} r(t)f(y(t))dt > −∞.

This contradiction proves the statement.

The following example shows that singular solutions of the second kind may be nonoscillatory if (3) does not hold.

Example 1. The differential equation

((1 − \frac{1}{(|y′| + 1)^{2}} ) sgn y′)′ + r(t)y = 0

with r(t) = \frac{8}{(2\sqrt{1−t}+1)^{4}} for t ∈ [0, 1] and r(t) = 8 for t > 1 has a nonoscillatory singular solution of the second kind of the form y = \frac{1}{2} + \sqrt{1 − t}.

The first result for the nonexistence of singular solutions follows from more common results of Mirzov [8] that are specified for (1).

Theorem 2. Let d_{1}(z) = max(|g^{−1}(z)|, |g^{−1}(−z)|) and

d_{2}(z) = max ( max_{0≤s≤|z|} f(s), − min_{0≤s≤|z|} f(−s)) for z ∈ R.

(i) If for every t^{∗} ∈ R_{+} the problem

z′ = \frac{1}{a(t)}d_{1}(d_{2}(z) ∫_{t∗}^{t} r(s)ds), y(t^{∗}) = 0  (7)

has the trivial solution on [t^{∗}, ∞) only, then (1) has no singular solution of the first kind. (ii) If for every c_{1} ≥ 0 and c_{2} ≥ 0 the Cauchy problem

z′ = \frac{1}{a(t)}d_{1} (c_{1} + d_{2}(z) ∫_{0}^{t} r(s)ds) , z(0) = c_{2}  (8)

has the upper solution defined on R_{+}, then (1) has no singular solution of the second kind.

Proof. This follows from [8, Theorems 1.1 and 1.2 and Remark 1.1] setting φ_{1}(t, z) = \frac{1}{a(t)}d_{1}(z) and φ_{2}(t, z) = r(t)d_{2}(z).

Corollary 1. Let g(z) = −g(−z), f(z) = −f(−z), and let f be nondecreasing on R_{+}.

(i) If there exists a continuous function R(t) and a right neighbourhood I of z = 0 such that

f(z) ∫_{0}^{t} r(s) ds ≤ g(R(t)z)

for t ∈ R_{+} and for z ∈ I, then (1) has no singular solution of the first kind. (ii) For any c > 0 let there exist a continuous function R_{1}(c, t) and a neighbourhood I_{1}(c) of ∞ such that c + f(z) ∫_{0}^{t} r(s) ds ≤ g(R_{1}(c, t)z), t ∈ R_{+}, z ∈ I_{1}(c). Then there exists no singular solution of the second kind of (1).

Proof. In our case, d_{1}(z) = g^{−1}(z) and d_{2}(z) = f(z), z ∈ R_{+}. Moreover,

d_{1}(z) = d_{1}(−z) and d_{2}(z) = d_{2}(−z).  (9)

(i) It is clear that (7) can be studied only for |z| ∈ I. Then

0 ≤ d_{1}(d_{2}(z) ∫_{t^{∗}}^{t} r(s)ds) = g^{−1}(f(z) ∫_{t^{∗}}^{t} r(s)ds) ≤ g^{−1}(f(z) ∫_{0}^{t} r(s)ds)

≤ R(t)z,

t ∈ R_{+} and z ∈ I. From this and from (9), Eq. (7) is sublinear in I, the trivial solution z ≡ 0 is unique, and the statement follows from Th. 2 (i). (ii) We have 0 ≤ d_{1}(c_{1} + d_{2}(z) ∫_{0}^{t} r(s)ds) = g^{−1}(c_{1} + f(z) ∫_{0}^{t} r(s)ds) ≤ R_{1}(c_{1}, t)z, t ∈ R_{+}, z ∈ I_{1}(c_{1}). From this and from (9), Eq. (8) is sublinear for large values of z, (8) has the upper solution defined on R_{+}, and the statement follows from Theorem 2 (ii).

Corollary 2. Let p > 0, M > 0 and M_{1} > 0.

(i) Let

|g(z)| ≥ M|z|^{p} and |f(z)| ≤ M_{1}|z|^{p}  (10)

hold in a neighbourhood I of z = 0. Then (1) has no singular solution of the first kind. (ii) Let z_{0} ∈ R_{+} be such that (10) holds for |z| ≥ z_{0}. Then (1) has no singular solution of the second kind.

Proof. Let d_{1} and d_{2} be defined as in Theorem 2.

(i) Since (10) yields d_{1}(z) ≤ ^{\frac{|z|^{\frac{1}{p}}}{M}} and d_{2}(z) ≤ M_{1}|z|^{p} for z ∈ I, we have

0 ≤ d_{1}(d_{2}(z) ∫_{t^{∗}}^{t} r(s)ds) ≤ \frac{1}{M} (M_{1}|z|^{p} ∫_{t^{∗}}^{t} r(s)ds)^{\frac{1}{p}}

= M^{−1}(M_{1} ∫_{t^{∗}}^{t} r(s)ds)^{\frac{1}{p}} |z|, z ∈ I, t^{∗} ∈ R_{+}.

The remainder of the proof is similar to that of Cor. 1 (i).

(ii) Similarly, d_{1}(z) ≤ ^{\frac{|z|^{\frac{1}{p}}}{M}} and d_{2}(z) ≤ M_{1}|z|^{p} for |z| ≥ z_{0}, and so

0 ≤ d_{1}(c_{1} + d_{2}(z) ∫_{0}^{t} r(s)ds) ≤ \frac{1}{M} (c_{1} + M_{1}|z|^{p} ∫_{0}^{t} r(s)ds)^{\frac{1}{p}} ,

t ∈ R_{+}, |z| ≥ z_{0}, c_{1} ≥ 0.

From this, equation (8) is sublinear for large |z|, the problem (8) has the upper solution defined on R_{+}, and the statement follows from Theorem 2 (ii).

Remark 2. Theorem A (i), (ii) is special case of Corollary 2 with g(z) = |z|^{p−1}z, a = A^{\frac{1}{p}}, and M = 1.

The following theorem generalizes Theorem A (iii); sufficient conditions for the nonexistence of singular solutions are posed on the functions a and r only.

Theorem 3. Let the function ar be locally absolute continuous on R_{+}, y be a nontrivial solution of (1) defined on [0, b), b ≤ ∞, ar = r_{0} − r_{1} on R_{+}, and

ρ(t) = ∫_{0}^{g(a(t)y′(t))} g^{−1}(σ)dσ + a(t)r(t) ∫_{0}^{y(t)} f(σ)dσ ≥ 0,  (11)

where r_{0} and r_{1} are nonnegative, nondecreasing and continuous functions. Then, for 0 ≤ s < t < b,

ρ(s) exp \{− ∫_{s}^{t} \frac{r′_{1}(σ)dσ}{a(σ)r(σ)}\} ≤ ρ(t) ≤ ρ(s) exp \{∫_{s}^{t} \frac{r′_{0}(σ)dσ}{a(σ)r(σ)}\} .  (12)

Moreover, y is not singular of the first kind, and if (3) holds, then y is proper.

Proof. Since ar is of locally bounded variation, the continuous nondecreasing functions r_{0} and r_{1} exist such that ar = r_{0} − r_{1}, and they can be chosen to be nonnegative on R_{+}. Moreover, r′_{0} ∈ L_{loc}(R_{+}) and r′_{1} ∈ L_{loc}(R_{+}). Then ρ is absolute continuous on [s, t] and

ρ′(τ) = (a(τ)r(τ))′ ∫_{0}^{y(τ)} f(σ)dσ, τ ∈ [s, t] a.e.

Let ε > 0 be arbitrary. Then (2) implies ρ(τ) ≥ 0 on [s, t], both terms in (11) are nonnegative, and

\frac{ρ′(τ)}{ρ(τ) + ε} = \frac{a(τ)r(τ)}{ρ(τ) + ε} ∫_{0}^{y(τ)} f(σ)dσ \frac{r′_{0}(τ) − r′_{1}(τ)}{a(τ)r(τ)};

hence,

−\frac{r′_{1}(τ)}{a(τ)r(τ)} ≤ \frac{ρ′(τ)}{ρ(τ) + ε} ≤ \frac{r′_{0}(τ)}{a(τ)r(τ)} a.e. on [s, t].

An integration and (11) yield

exp \{− ∫_{s}^{t} \frac{r′_{1}(σ)dσ}{a(σ)r(σ)}\} ≤ \frac{ρ(t) + ε}{ρ(s) + ε} ≤ exp \{∫_{s}^{t} \frac{r′_{0}(σ)ds}{a(σ)r(σ)}\} .

Since ε > 0 is arbitrary, (12) holds.

Let y be singular of the first kind. Then according to its definition and Remark 1 (iii), there exists τ ∈ (0, ∞) such that y(τ) = 0, y′(τ) = 0, and

sup_{T ≤t<τ} |y(t)| > 0 for every T ∈ [0, τ).  (13)

Hence, (11) and (12) yield ρ(τ) = 0 and ρ(t) = 0 on [0, τ]. From this and from (2), we have y = 0 on [0, τ]. This contradiction to (13) proves that y is not singular of the first kind.

Let (3) be valid and y be a singular solution of the second kind. Then according to Remark 1 (ii), there exists a sequence {t_{k}}_{k=1}^{∞} such that t_{k} ∈ [0, b), lim_{k→∞} t_{k} = b, and lim_{k→∞} |y′(t_{k})| = ∞. Hence, (3) yields lim_{k→∞} g(a(t_{k})y′(t_{k})) = ∞. From this and from (12) we have for s = 0 and t = t_{k}, k = 1, 2, …, that

∞ = lim_{k→∞} ρ(t_{k}) ≤ ρ(0) exp \{∫_{0}^{τ} \frac{r′_{0}(σ)dσ}{a(σ)r(σ)}\} .

The contradiction proves that y is not singular of the second kind and, according to Remark 1 (i), it is proper.

Theorem 4. Let the assumptions of Theorem 3 be valid and let

ρ_{1}(t) = \frac{1}{a(t)r(t)} ∫_{0}^{g(a(t)y′(t))} g^{−1}(σ)dσ + ∫_{0}^{y(t)} f(σ)dσ.  (14)

Then for 0 ≤ s < t < b we have

ρ_{1}(s) exp \{− ∫_{s}^{t} \frac{r′_{0}(σ)dσ}{a(σ)r(σ)}\} ≤ ρ_{1}(t) ≤ ρ_{1}(s) exp \{∫_{s}^{t} \frac{r′_{1}(σ)dσ}{a(σ)r(σ)}\} . (15)

Proof. The proof is similar to that of Theorem 3 since

ρ′_{1}(τ) = −\frac{(a(τ)r(τ))′}{(a(τ)r(τ))^{2}} ∫_{0}^{g(a(τ)y′(τ))} g^{−1}(σ)dσ = \frac{r′_{1}(τ) − r′_{0}(τ)}{a(τ)r(τ)} \frac{∫_{0}^{g(a(τ)y′(τ))} g^{−1}(σ)dσ}{a(τ)r(τ)}

a.e. on [s, t].

Remark 3. Inequalities (12) and (15) are proved in [7] for Equation (5) with p = 1 and a ≡ 1, in [3] for g(z) = |z|^{p−1}z with p > 0, and in [8] for Equation (6).

Corollary 3. Let ar be locally absolute continuous on R_{+}. Let ρ and ρ_{1} be given by (11) and (14), respectively.

(i) If ar is nondecreasing on R_{+}, then for an arbitrary solution y of (1), ρ is nondecreasing and ρ_{1} is nonicreasing on R_{+}. (ii) If ar is nonincreasing on R_{+}, then for an arbitrary solution y of (1), ρ is nonincreasing and ρ_{1} is nondecreasing on R_{+}.

Proof. It follows from (12) and (15) as r_{0} ≡ r and r_{1} ≡ 0 in case (i), and r_{0} ≡ r(0), r_{1} = r(0) − r in case (ii).

In [1] there is an example of Eq. (1) with a ≡ 1, g(z) ≡ z, f(z) = |z|^{λ} sgn z and 0 < λ < 1 for which there exists a proper solution y with infinitely many accumulation points of zeros. The following corollary gives a sufficient condition under which every solution of (1) has no accumulation point of zeros in its interval of definition.

Corollary 4. If ar is locally absolute continuous on R_{+}, then every nontrivial solution y of (1) has no accumulation point of its zeros and has no double zero in its interval of definition.

Proof. Let τ be an accumulation point of zeros or a double zero of a solution y of (1) lying in its definition interval. Hence, y(τ) = 0 and y′(τ) = 0. Then, ȳ(t) = y(t) on [0, τ] and ȳ(t) = 0 for t > τ is a singular solution of the first kind of (1) that contradits Theorem 3.

Corollary 5. Let ar be locally absolute continuous and nondecreasing (nonincreasing) on R_{+}. Let y be a solution of (1) defined on [0, b), b ≤ ∞, and {t_{k}}_{k=1}^{N}, N ≤ ∞, be a (finite or infinite) increasing sequence of zeros of y′ lying in [0, b). Then the sequence of local extrema {|y(t_{k})|}_{k=1}^{N} is nonincreasing (nondecreasing).

Proof. Let ar be nondecreasing on R_{+}. As all assumptions of Corollary 3 are fulfilled, ρ_{1} is nonincreasing and the statement follows from ρ_{1}(t_{k}) = ∫_{0}^{y(t_{k})} f(σ)dσ and (2). If ar is nonincreasing, the proof is similar.

The following corollary generalizes Theorem B and it shows that singular solutions may exist if ar is not locally absolutely continuous on R_{+}.

Corollary 6. Let A ≡ 1, 0 < λ < p (0 < p < λ) and lim_{z→0} \frac{f(z)}{|z|^{λ} sgn z} = M ∈ (0, ∞). Then there exists a positive continuous function r such that Equation (5) has a singular solution of the first (second) kind.

Proof. Let 0 < λ < p. Then Theorem B yields the existence of a positive continuous function r̄ defined on R_{+} such that (6) (with r = r̄) has a singular solution y of the first kind. Put

r(t) = r̄(t)\frac{|y(t)|^{λ} sgn y(t)}{f(y(t))} if y(t) ≠ 0

and r(t) = \frac{r̄(t)}{M} if y(t) = 0. From this and from (2), the function r is positive and continuous on R_{+}, and

(|y′(t)|^{p−1}y′(t))′ = −r̄(t)|y(t)|^{λ} sgn y(t) = −r(t)f(y(t));

hence y is also a solution of (5). If 0 < p < λ, then the proof is similar.

Example 1 shows that the statement of Theorem 3 does not hold if (3) is not valid; singular solutions of the second kind may exist. The following theorem gives sufficient conditions for the existence of such solutions.

Theorem 5. Let M ∈ (0, ∞), a ≡ 1 on R_{+}, and g ∈ C^{1}(R).

(i) If β ∈ {−1, 1}, λ > 2, and

0 < g′(z) ≤ |z|^{−λ} for βz ≥ M,  (16)

then (1) possesses a singular solution of the second kind. (ii) If

g′(z) ≥ |z|^{−2} for |z| ≥ M,  (17)

then (1) has no nonoscillatory singular solution of the second kind.

Proof. (i) Let β = 1; if β = −1, the proof is similar. Consider the differential equation

y′′ = −r(t)f(y)G(y′),  (18)

where G ∈ C^{0}(R), G(z)z > 0 for z ≠ 0, and

G(z) = (g′(z))^{−1} for z ≥ M.  (19)

Put M_{1} = [(λ − 1) min_{0≤s≤1} r(s) min_{−3≤s≤− \frac{1}{2}} |f(s)|]^{− \frac{1}{λ−1}} . Let τ be such that

0 < τ ≤ 1, τ ≤ 2M_{1}^{− \frac{λ−1}{λ−2}} , τ ≤ [ max_{0≤s≤1} r(s) max_{−3≤s≤− \frac{1}{2}} |f(s)|]^{−1} ∫_{M}^{2M} \frac{ds}{G(s)}

and

τ ≤ g(M) [ max_{0≤s≤1} r(s) max_{−4≤s≤−3} |f(s)|]^{−1} .  (20)

Then (16) and (19) yield G(z) ≥ z^{λ} for z ≥ M and according to Theorem 1 in [2] (with n = 2, M = M, β = 1, c_{0} = −1, α = −1, T = \frac{τ}{2}, N = 3; see the proof of Theorem 1 and (13) – (17) as well), there exists a solution y of (18) defined in [\frac{τ}{2}, τ) such that

lim_{t→τ_{−}} y(t) = −1, lim_{t→τ_{−}} y′(t) = ∞,

and

−3 ≤ y(t) ≤ −\frac{1}{2}, M ≤ y′(t) ≤ M_{1}(τ − t)^{− \frac{1}{λ−1}} , t ∈ [\frac{τ}{2}, τ).  (21)

Hence, (16), (19) and (21) yield y is the solution of Eq. (1) on [\frac{τ}{2}, τ). We will prove that y can be defined on [0, τ) and, thus, y is singular of the second kind. Let, to the contrary, y be defined on (τ̄, τ) ⊂ [0, τ) so that it cannot be defined at τ̄. Then

lim sup_{t→τ̄_{+}} |y′(t)| = ∞.  (22)

First, we prove that

y′(t) > 0 on (τ̄, τ).  (23)

Suppose, that τ_{1} ∈ (τ̄, τ) exists such that y′(τ_{1}) = 0 and y′(t) > 0 on (τ_{1}, τ); according to (21), τ_{1} < \frac{τ}{2}. Hence, y is increasing on (τ_{1}, τ) and negative. From this, (1), and (2), the functions g(y′) and y′ are increasing on (τ_{1}, τ). Further, we estimate y on [τ_{1}, \frac{τ}{2}] using (21) and the definition of τ. We have

\begin{matrix}− 3 ≥ y(t) = y(\frac{τ}{2}) + ∫ t_{\frac{τ}{2}} y′(s)ds ≥ y (\frac{τ}{2}) − y′ (\frac{τ}{2}) (\frac{τ}{2} − t) \\ ≥ −3 − M_{1}(\frac{τ}{2})^{− \frac{1}{λ−1}} \frac{τ}{2} ≥ −3 − M_{1}(\frac{τ}{2})^{1− \frac{1}{λ−1}} ≥ −4, t ∈ [τ_{1}, \frac{τ}{2}]. \end{matrix}  (24)

An integration of (1) on [τ_{1}, \frac{τ}{2}], (2), (21), (24), and τ ≤ 1, yield

\begin{aligned}g(M) & ≤ g (y′(\frac{τ}{2})) − g(y′(τ_{1})) = − ∫τ_{1}^{\frac{τ}{2}} r(s)f (y(s)) ds \\ & ≤ max_{0≤s≤1} r(s) max_{−4≤s≤−3} |f(s)|\frac{τ}{2}. \end{aligned}

This contradiction to (20) proves that (23) is valid. From this and from (21), y < 0 on (τ̄, τ), and (1) yields g(y′) and y′ are incerasing on this interval. Thus, according to (23), y′ is bounded in a right neighbourhood of τ̄ which contradicts (22), and so y is defined on so [0, τ). (ii) Suppose, that y is a nonoscillatory singular solution of (1) of the second kind defined on [0, τ). Then Lemma 1 and Remark 1 (ii) yield lim_{t→τ_{−}} |y′(t)| = ∞ and lim_{t→τ_{−}} y(t) = C ∈ [−∞, ∞]. Suppose that

lim_{t→τ_{−}} y′(t) = ∞  (25)

(the opposite case can be studied similarly).

Let C ∈ (−∞, ∞). Due to (1) and (17), y is a solution of Eq. (18) and (19) on [T, τ) ∈ [0, τ) where T is such that y′(t) ≥ M on [T, τ). But this contradicts a result in [2, Theorem 2].

Let C = ∞. Then lim_{t→τ_{−}} y(t) = ∞. But according to (1) and (2), the functions g(y′) and y′ are decreasing in a left neighbourhood of τ, which contradicts (25). Clearly, the case C = −∞ is impossible due to (25).

Acknowledgement. This work was supported by the Grant No A1163401/04 of the Grant Agency of the Academy of Science of the Czech Republic and by the Ministry of Education of the Czech Republic under the project MSM 0021622409.

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[8] Mirzov D. D., Asymptotic Properties of Solutions of Systems of Nonlinear Nonautonomous Ordinary Differential Equations. Adygea, Maikop (1993) (in Russian); Folia F.S.N. Univ. Masar. Brunen. Math. 14 (2004).

Author’s address: Department of Mathematics, Masaryk University, Janáčkovo nám 2a, 602 00 Brno, Czech Republic e-mail: bartusek@math.muni.cz

(Received March 8, 2006)