Georgian Mathematical Journal Volume 8 (2001), Number 4, 669–681

ON THE EXISTENCE OF SINGULAR SOLUTIONS

M. BARTUŠEK AND J. OSIČKA

Abstract. Sufficient conditions are given, under which the equation y^{(n)} = f(t, y, y′, …, y^{(l)})g(y^{(n−1)}) has a singular solution y [T, τ) → R, τ < ∞ satisfying lim_{t→τ_{−}} y^{(i)}(t) = c_{i} ∈ R, i = 0, 1, …, l and lim_{t→τ_{−}} |y^{(j)}(t)| = ∞ for j = l + 1, …, n − 1 where l ∈ {0, 1, …, n − 2}.

2000 Mathematics Subject Classification: 34C11.

Key words and phrases: Singular solution, fourth order differential equation, black hole solution.

1. Introduction

Consider the n-th order differential equation

y^{(n)} = f(t, y, y′, …, y^{(l)})g(y^{(n−1)}),  (1)

where n ≥ 2, l ∈ {0, 1, …, n−2}, f ∈ C^{0}(R_{+}×R^{l+1}), g ∈ C^{0}(R), R_{+} = [0, ∞), R = (−∞, ∞) and there exists α ∈ {−1, 1} such that

αf(t, x_{1}, …, x_{l+1})x_{1} > 0 for x_{1} ≠ 0.  (2)

A solution y defined on the interval [T, τ) ⊂ R_{+} is called singular if τ < ∞ and y cannot be defined for t = τ.

The problem of the existence of singular solutions satisfying the Cauchy initial-value problem and their asymptotic behaviour is thoroughly studied in [4] for the second order Emden–Fowler equation

y′′ = r(t)|y|^{λ} sgn y, r(t) ≥ 0.  (3)

In the common case for (1), the profound investigations are carried out in [5]. All these results concern the case α = 1. For α = −1, sufficient conditions are given in [2], under which singular solutions of

y^{(n)} = r(t)|y|^{λ} sgn y, n ≥ 2, r ≤ 0,

exist.

Another problem concerning singular solutions is solved in [3] (n = 2, f(t, x_{1}, …, x_{l}) ≡ r(t)|x_{1}|^{σ} sgn x_{1}, g(x) = |x|^{λ}) and in [1] in the case l = n − 2.

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

Let τ ∈ (0, ∞). Sufficient and/or necessary conditions are given there, under which a singular solution y exists with given asymptotic behaviour at the left-hand side point τ of the definition interval c_{i} ∈ R,

lim_{t→τ_{−}} y^{(i)}(t) = c_{i} for i = 0, 1, …, n − 2, lim_{t→τ_{−}} |y^{(n−1)}(t)| = ∞. (4)

In the present paper this result is generalized to the case in which we seek a singular solution y satisfying the condition

τ ∈ (0, ∞), c_{i} ∈ R; lim_{t→τ_{−}} y^{(i)}(t) = c_{i}, i = 0, 1, …, l,

lim_{t→τ_{−}} |y^{(j)}(t)| = ∞ for j = l + 1, …, n − 1.

(5)

Note that in [3] such solutions are called blackhole solutions (for n = 2 and l = 0).

Denote by [[a]] the entire part of the number a.

2. Main Results

Let y be a solution of (1) satisfying (5). Since, according to (2) f(t, c_{0}, c_{1}, …, c_{l}) ≠ 0 if and only if c_{0} ≠ 0, we ought to divide our investigation into two cases c_{0} ≠ 0 and c_{0} = 0 for which the results are different.

Let c_{0} ≠ 0. The following theorem gives a necessary condition for the existence of a solution of (1), (5).

Theorem 1. Let c_{0} ≠ 0, M ∈ (0, ∞), K ∈ (0, 1], λ ≤ 2 for l = n − 2,

\begin{matrix}λ¯∈(1 + \frac{1}{n − l − 1}, 1 + \frac{1}{n − l − 2}] for < n − 2, \\ K|x|^{λ} ≤ g(x) ≤ |x|^{λ} for |x| ≥ M. \end{matrix}  (6)

Then equation (1) has no singular solution y satisfying (5).

The next theorem shows that in the opposite case in (6) problem (1), (5) is solvable.

Theorem 2. Let τ ∈ (0, ∞), c_{0} ≠ 0, M ∈ (0, ∞), β = α sgn c_{0}, λ > 2 for l = n − 2,

1 + \frac{1}{n − l − 1} < λ ≤ 1 + \frac{1}{n − l − 2} for l < n − 2  (7)

and

g(x) ≥ |x|^{λ} for βx ≥ M.  (8)

Then there exists a singular solution y of (1) satisfying (5) which is defined in a left neighborhood of τ.

If, moreover, ε > 0, g(x) > 0 for βx ∈ (0, ε],

l + \frac{1 − α}{2} is odd, (−1)^{i}c_{i}c_{0} ≥ 0 for i = 1, 2, …, l,  (9)

and

|∫_{0}^{βε} \frac{ds}{g(s)}| = ∞,  (10)

then y is defined on the interval [0, τ).

Corollary 1. Let c_{0} ≠ 0, M ∈ (0, ∞) and

g(x) = |x|^{λ} for |x| ≥ M.

Then (1) has a singular solution y satisfying (5) if and only if (7) is valid.

Corollary 2. Let λ > 1 + \frac{1}{n−1} and M ∈ R_{+} be such that

g(x) ≥ x^{λ} for x ≥ M.

Then (1) has a singular solution.

Remark. For α = 1 the conclusion of Corollary 2 is known, see, e.g., [6, Theorem 11.3]. For α = −1 it generalizes Corollary 1 in [1].

The following two theorems solve the same problem in the case

β ∈ {−1, 1}, c_{0} = 0, (−1)^{i}βc_{i} ≥ 0 for i = 1, 2, …, l.  (11)

Theorem 3. Let τ ∈ (0, ∞), σ > 0, ε > 0,M ∈ (0, ∞), M̄ ∈ (0, ∞),

l − \frac{1 − α}{2} be odd,  (12)
2 + (n − 2)σ < λ for l = n − 2,
1 + \frac{lσ + 1}{n − l − 1} < λ ≤ 1 + \frac{(l + 1)σ + 1}{n − l − 2} for l < n − 2,  (13)

(8) and (11) hold. Further, let

|f(t, x_{1}, …, x_{l+1})| ≥ M̄|x_{1}|^{σ}  (14)

for t ∈ [0, τ], βx_{1} ∈ [0, ε], (−1)^{j}βx_{j+1} ∈ [(−1)^{j}βc_{j}, (−1)^{j}βc_{j} + ε], j = 1, …, l.

Then there exists a singular solution y of (1) satisfying (5), which is defined in a left neighborhood of τ.

If, moreover, g(x) > 0 for βx ∈ (0, ε] and (10) holds, then y is defined on the interval [0, τ).

Theorem 4. Let σ > 0, c_{i} = 0 for i = 0, 1, …, l, M ∈ (0, ∞), ε > 0, α ∈ {−1, 1}, r ∈ C^{0}(R_{+}), αr(t) > 0 on R_{+}, (12) hold and

g(x) = |x|^{λ} for |x| ≥ M.

Then the equation

y^{(n)} = r(t)|y|^{σ}g (y^{(n−1)}) sgn y  (15)

has a solution y satisfying (1), (5) if and only if (13) is valid.

The following proposition shows that assumption (12) in Theorems 3 and 4 is important.

Proposition. Let c_{i} = 0, i = 0, 1, …, l and l − \frac{α−1}{2} be even. Let g(x) ≥ 0 on R. Then equation (1) has no solution satisfying (1), (5).

In this paper the main assumptions are imposed on the function g depending on y^{(n−1)}. But solutions of (1), (5) may exist for the equation

y^{(n)} = f(t, y, …, y^{(j)}), j ∈ {l + 1, …, n − 1},  (16)

too. From this we formulate an open problem.

Open problem. To study the existence of a solution satisfying (1), (5) of equation (16).

3. Lemmas and Proofs

We need the next two lemmas.

Lemma 1. Let [a, b] ⊂ R_{+}, ϕ ∈ C^{0}[a, b] and f̃ ∈ C^{0}([a, b]×R^{n}) be such that

f̃(t, x_{1}, …, x_{n})| ≤ ϕ(t), t ∈ [a, b], x_{i} ∈ R, i = 1, …, n.

Then for arbitrary γ_{i} ∈ R, i = 0, 1, …, n − 1, the equation

u^{(n)} = f̃(t, u, u′, …, u^{(n−1)})

has at least one solution satisfying the boundary value conditions

u^{(i)}(b) = γ_{i} for i = 0, 1, …, l + 1;
u^{(j)}(a) = γ_{j+1} for j = l + 1, …, n − 2.

Proof. It follows, e.g., from [6, Lemma 10.1] since the homogeneous problem

u^{(n)} = 0, u^{(i)}(b) = u^{(j)}(a) = 0 for i = 0, 1, …, l + 1; j = l + 1, …, n − 2,

has a trivial solution only.

The following Kolmogorov–Horny type inequality is a very useful tool (see, e.g., the proof of Lemma 5.2 in [6]).

Lemma 2. Let [a, b] ⊂ R_{+}, a < b, m ≥ 2 be an integer, u ∈ C^{m}[a, b], and let u^{(j)} have zero in the interval [a, b] for j = 1, …, m − 1. Then

ρ_{i} ≤ 2^{i(m−i)}ρ0^{\frac{m−i}{m}} ρm^{\frac{i}{m}} , i = 1, 2, …, m − 1,

where

ρ_{i} = max {|u^{(i)}(t)| a ≤ t ≤ b}, i = 0, 1, …, m.

Proof of Theorem 1. Let for simplicity c_{0} > 0 and α = 1. Put λ_{1} = \frac{1}{λ−1} and let y [τ_{1}, τ) → R be a solution of (1), (5). Then, according to (1) and (2) lim_{t→τ_{−}} y^{(j)}(t) = ∞ for j = l + 1, …, n. Let T ∈ [τ_{1}, τ) be such that

\begin{matrix}y(t) ≥ \frac{c_{0}}{2} on [T, τ), y^{(j)}(T) ≥ 0, j = l + 1, l + 2, …, n − 2, \\ y^{(n−1)}(T) ≥ M. \end{matrix} (17)

By this and the boundedness of y^{(i)}(t), i = 0, 1, …, l, we obtain from (1)

y^{(n)}(t) ≤ M_{1} [y^{(n−1)}(t)]^{λ} , t ∈ [T, τ),

where M_{1} is a suitable constant. Let λ ≤ 1 + \frac{1}{n−l−1}. Hence the integration on the interval [t, τ) yields

y^{(n−1)}(t) ≥ [(λ − 1)M_{1}(τ − t)]^{−λ_{1}} , t ∈ [T, τ).  (18)

Hence n − l − 1 ≤ λ_{1}, and the Taylor Series Theorem, (17) and (18) yield

c_{l} = y^{(l)}(τ) = ∑_{i=0}^{n−l−2} \frac{y^{(l+i)}(T)}{i!}(τ − T)^{i} + ∫_{T}^{τ} \frac{(τ − s)^{n−l−2}}{(n − l − 2)!} y^{(n−1)}(s) ds

≥ y^{(l)}(T) + M_{2} ∫_{T}^{τ}(τ − s)^{n−l−2−λ_{1}} ds = ∞,

where

M_{2} = \frac{[(λ − 1)M_{1}]^{−λ_{1}}}{(n − l − 2)!} .

Hence a solution y satisfying (1), (5) does not exist in this case.

Let l < n − 2 and λ > 1 + \frac{1}{n−l−2}. Hence n − l − 2 − λ_{1} > 0. Then, similarly to (18), we can prove that

y^{(n−1)}(t) ≤ [(λ − 1)M_{3}(τ − t)]^{−λ_{1}}, t ∈ [T, τ),  (19)

where M_{3} = min {Kf (t, y(t), …, y^{(l)}(t)) T ≤ t ≤ τ} > 0.

From this the Taylor Series Theorem yields

∞ = y^{(l+1)}(τ) = ∑_{i=0}^{n−l−3} \frac{y^{(l+i+1)}}{i!} (τ − T)^{i} + ∫_{T}^{τ} \frac{(τ − s)^{n−l−3}}{(n − l − 3)!} y^{(n−1)}(s) ds

≤ M_{4} + M_{5} ∫_{T}^{τ}(τ − s)^{n−l−3−λ_{1}} ds < ∞

as n − l − 3 − λ_{1} > −1; M_{4} and M_{5} are positive constants. The contradiction obtained proves that a singular solution does not exist. Proof of Theorem 2. For l = n − 2 we proved the statement in [1]. Thus let l < n − 2 and, first, we prove the result for (7) with λ ≠ 1 + \frac{1}{n−l−2}.

We prove the statement for α = 1 and c_{0} > 0; thus β = 1. For the other cases the proof is similar.

Let

N > 2 max (c_{0}, |c_{1}|, …, |c_{l}|), k_{0} > [[2M]] ,  (20)

D = \{[x_{1}, …, x_{l+1}] \frac{c_{0}}{2} ≤ x_{1} ≤ c_{0}, |x_{j}| ≤ N for j = 2, …, l + 1\}, M_{1} = min{f(t, x_{1}, …, x_{l+1}) t ∈ [0, τ], [x_{1}, …, x_{l+1}] ∈ D} > 0, M_{2} = max{f(t, x_{1}, …, x_{l+1}) t ∈ [0, τ], [x_{1}, …, x_{l+1}] ∈ D}, M_{3} = 2[(λ − 1)M_{1}]^{−λ_{1}}, λ_{1} = \frac{1}{λ−1}, λ̄ = n − l − 1 − λ_{1} > 0,

N_{1} = 2^{n−l−2}M3^{\frac{1}{n−l−1}} [1 − \frac{λ_{1}}{n−l−1}]^{−1}.

Further, let T ∈ [0, τ) be such that

τ − T < (\frac{M_{3}}{M} )^{λ−1} , (τ − T)^{n−l−2} < \frac{k_{0}}{M} ,  (21)

τ − T < \frac{1}{M_{2}} ∫_{M}^{2M} \frac{ds}{g(s)},  (22)
(τ − T)λ̄ ≤ (2N_{1})^{−n+l+1}N,  (23)
∑_{r=i+1}^{l−1} |c_{r}|\frac{(τ − T)^{r−i}}{(r − i)!} + N \frac{(τ − T)^{l−i}}{(l − i)!} ≤ \frac{N}{2} , i = 0, 1, …, l − 1,  (24)

∑_{r=1}^{l−1} |c_{r}|\frac{(τ − T)^{r}}{r!} + N \frac{(τ − T)^{l}}{l!} ≤ \frac{c_{0}}{2} .  (25)

Denote J = [T, τ) and note that due to λ̄ > 0, T exists.

Consider the auxilliary two-point boundary-value problem k ∈ {k_{0}, k_{0} + 1, … },

\begin{matrix}y^{(n)} = f (t, Φ_{0}(y), Φ_{1}(y′), …, Φ_{1}(y^{(l)})) g (Φ_{2}(t, y^{(n−1)})) , \\ y^{(i)}(τ) = c_{i}, i = 0, 1, …, l; y^{(l+1)}(τ) = k; \\ y^{(j)}(T) = 0, j = l + 1, …, n − 2, t ∈ J, \end{matrix}  (26)

where

Φ_{0}(s) = \begin{cases}s & for \frac{c_{0}}{2} ≤ s ≤ N, \\ N & for s > N, \\ \frac{c_{0}}{2} & for s < \frac{c_{0}}{2} , \end{cases}  (27)

Φ_{1}(s) = \begin{cases}s & for |s| ≤ N, \\ N sgn s & for |s| > N \end{cases}  (28)

and

Φ_{2}(t, s) = \begin{cases}s & for M ≤ s ≤ M_{3}(τ − t)^{−λ_{1}}, \\ M_{3}(τ − t)^{− \frac{1}{λ−1}} & for s > M_{3}(τ − t)^{−λ_{1}}, \\ M & for s < M. \end{cases}  (29)

Note that due to (21) Φ_{2} is well defined.

To prove the existence of a solution of (26), let us consider the sequence of boundary value problems

\begin{matrix}m̄_{0} > \frac{1}{τ − t}, m ∈ \{m̄_{0}, m̄_{0} + 1, … \}, τ_{m} = τ − \frac{1}{m}, \\ z^{(n)} = F (t, z, z′, …, z^{(l)}, z^{(n−1)}), \\ z^{(i)}(τ_{m}) = c_{i}, i = 0, 1, …, l, z^{(l+1)}(τ_{m}) = k, \\ z^{(j)}(T) = 0, j = l + 1, …, n − 2, \end{matrix}  (30)

where

F (t, x_{1}, …, x_{l+2}) = f (t, Φ_{0}(x_{1}), Φ_{1}(x_{2}), …, Φ_{1}(x_{l+1})) g (Φ_{2}(t, x_{l+2})) .

Since

|F (t, x_{1}, …, x_{l+2})| ≤ M_{2} max_{T ≤t̄≤t} max_{M≤s≤M_{3}(τ−t̄)^{−λ_{1}}} g(t̄, s), t ∈ [T, τ_{m}],

(30) has a solution z_{m} according to Lemma 1.

Further, we estimate z_{m}^{(n−1)}. Let J_{m} = [T, τ_{m}]. First we prove that

z_{m}^{(n−1)}(t) < M_{3}(τ − t)^{−λ_{1}}, t ∈ [T, τ_{m}),  (31)

for large m, say m ≥ m̄_{0}. If (31) is not valid, then either (i) there exists t_{1} ∈ [T, τ_{m}) such that

z_{m}^{(n−1)}(t_{1}) = M_{3}(τ − t_{1})^{−λ_{1}} and z_{m}^{(n−1)}(τ_{m}) ≤ M_{3}(τ − τ_{m})^{−λ_{1}} (32)

or (ii)

z_{m}^{(n−1)}(t) > M_{3}(τ − t)^{−λ_{1}}  (33)

in a left neighborhood of t = τ_{m}.

Let (i) be valid. As (26)–(30) yield z_{m}^{(n)}(t) > 0 and z_{m}^{(n−1)} is increasing on J_{m}, it follows from (32) and (21) that

M ≤ z_{m}^{(n−1)}(t), t ∈ [t_{1}, τ_{m}].  (34)

Hence

z_{m}^{(n)}(t) ≥ M_{1} (z_{m}^{(n−1)}(t))^{λ} , t ∈ J_{m},

and the integration and (32) yield

\frac{τ_{m} − t_{1}}{M_{3}^{λ−1}} > \frac{1}{[z_{m}^{(n−1)}(t_{1})]^{λ−1}} − \frac{1}{[z_{m}^{(n−1)}(τ_{m})]^{λ−1}} ≥ M_{1}(λ − 1)(τ_{m} − t_{1}),

which contradicts the definition of M_{3}.

Let (33) be valid and let t_{1}, T ≤ t_{1} < τ_{m} be such that z_{m}^{(n−1)}(t) > M_{3}(τ −t)^{−λ_{1}} on the interval [t_{1}, τ_{m}). Then the Taylor Series Theorem yields

k = z_{m}^{(l+1)}(τ_{m}) ≥ ∫_{t_{1}}^{τ_{m}} \frac{(τ_{m} − s)^{n−l−3}}{(n − l − 3)!} z_{m}^{(n−1)}(s) ds

≥ \frac{M_{3}}{(n − l − 3)!} ∫_{t_{1}}^{τ_{m}}(τ_{m} − s)^{n−l−3}(τ − s)^{−λ_{1}} ds

≥ \frac{−M_{3}}{(n − l − 2)!} ∫_{t_{1}}^{τ_{m}}(τ − s)^{n−l−1−λ_{1}} \frac{d}{ds} ((1 − \frac{1}{m(τ − s)})^{n−l−2}) ds

≥ \frac{M_{3}}{(n − l − 2)!} (\frac{1}{m})^{n−l−2−λ_{1}} (1 − \frac{1}{m(τ − t_{1})})^{n−l−2} → ∞ for m → ∞.

Hence (31) holds.

Further, we prove indirectly the following estimation from bellow

M < z_{m}^{(n−1)}(t), t ∈ J_{m}.  (35)

Note that z_{m}^{(n−1)} is increasing, and first we prove that (35) is valid for t = τ_{m}. Let, conversely, z_{m}^{(n−1)}(τ_{m}) ≤ M. Then

k_{0} ≤ k = z_{m}^{(l+1)}(τ_{m}) = ∫_{T}^{τ_{m}} \frac{(τ_{m} − s)^{n−l−3}}{(n − l − 3)!} z_{m}^{(n−1)}(s) ds ≤ \frac{M}{(n − l − 2)!}(τ_{m} − T)^{n−l−2},

which contradicts (21). Thus (35) holds. Let T_{1} ∈ [T, τ_{m}) exist such that z_{m}^{(n−1)}(T_{1}) = M. Then M < z_{m}^{(n−1)}(t) on J_{m} and

z_{m}^{(n)}(t) ≤ M_{2}g (z_{m}^{(n−1)}(t)) , t ∈ J_{m}.

From this, by the integration, we have

∫_{M}^{2M} \frac{ds}{g(s)} ≤ ∫_{M}^{k} \frac{ds}{g(s)} ≤ M_{2}(τ_{m} − T) < M_{2}(τ − T).

The contradiction with (22) proves that (35) is valid and according to (30)

z_{m}^{(j)}(t) ≥ 0 on J_{m}, j = l + 1, l + 2, …, n.  (36)

Denote ρ = max_{t∈J_{m}} |z_{m}^{(l)}(t)|. Then, by virtue of (31), (36) and Lemma 2 with [a, b] = [τ, t], u = z_{m}^{(l)} and m = n − l − 1, we have

0 ≤ z_{m}^{(i+1)}(t) ≤ 2^{n−l−2}ρ^{\frac{n−l−2}{n−l−1}} (z_{m}^{n−1}(t))^{\frac{1}{n−l−1}} ≤ 2^{n−l−2}M3^{\frac{1}{n−l−1}} ρ^{\frac{n−l−2}{n−l−1}} (τ − t)^{− \frac{λ_{1}}{n−l−1}} ,

and hence, as \frac{λ_{1}}{n−l−1} ∈ (0, 1), the integration on J_{m} yields

0 ≤ c_{l} − z_{m}^{(l)}(T) ≤ N_{1}(τ − T)^{1− \frac{λ_{1}}{n−l−1}} ρ^{\frac{n−l−2}{n−l−1}} .  (37)

Since z_{m}^{(l)} is increasing on J_{m}, either z_{m}^{(l)}(T) ≥ −|c_{l}| and ρ = |c_{l}| or z_{m}^{(l)}(T) < −|c_{l}| and (23) and (37) yield

c_{l} + ρ ≤ N_{1}(τ − T)^{1− \frac{λ_{1}}{n−l−1}} ρ^{\frac{n−l−2}{n−l−1}} ≤ \frac{1}{2}N^{\frac{1}{n−l−1}} ρ^{\frac{n−l−2}{n−l−1}} .

Thus ρ ≤ 2|c_{l}| or \frac{ρ}{2} ≤ c_{l} + ρ ≤ \frac{1}{2}N^{\frac{1}{n−l−1}} ρ^{\frac{n−l−2}{n−l−1}} and according to (20) in all cases we have

|z_{m}^{(l)}(t)| ≤ N, t ∈ J_{m}.  (38)

From this, (31), (36) and Lemma 2 with [a, b] = [T, t], u = z_{m}^{(l)} and m = n−l−1 we have

\begin{matrix}|z_{m}^{(j)}(t)| ≤ 2^{(j−l)(n−j−1)}N^{\frac{n−j−1}{n−l−1}} M3^{\frac{j−l}{n−l−1}} (τ − t)^{\frac{j−l}{n−l−1}} , \\ t ∈ J_{m}, j = l + 1, …, n − 2. \end{matrix}  (39)

Further, (20), (24), (25), (38) and the Taylor Series Theorem yield

c_{i} − z_{m}^{(i)}(t) = ∑_{r=i+1}^{l−1} \frac{c_{r}(t − τ_{m})^{r−i}}{(r − i)!} + ∫_{τ_{m}}^{t} \frac{(t − s)^{l−i−1}}{(l − i − 1)!} z_{m}^{(l)}(s) ds,

\begin{matrix}|z_{m}^{(i)}(t)| ≤ ∑_{r=i+1}^{l−1} \frac{c_{r}|}{(r − i)!}(τ − T)^{r−i} + \frac{N}{(l − i)!}(τ − T)^{l−i} + |c_{i}| ≤ N, \\ i = 0, 1, …, l − 1, t ∈ J_{m}, \end{matrix} (40)

|z_{m}(t)| ≥ c_{0} − ∑_{r=1}^{l−1} |c_{r}|\frac{(τ − T)^{r}}{r!} − \frac{N}{l!} (τ − T)^{l} ≥ \frac{c_{0}}{2} , t ∈ J_{m}.  (41)

Estimations (38), (39) and (40) show that {z_{m}^{(j)}}, j = 0, 1, …, n − 1, m = m_{0}, m_{0} + 1, …, are uniformly bounded with respect to j and m and hence according to the Arzelá–Ascoli Theorem (see [6], Lemma 10.2) there exists a subsequence that converges uniformly to the solution y_{k} of (26). At the same time, it is clear that (see (41), too)

\frac{c_{0}}{2} ≤ y_{k}(t) ≤ N, |y_{k}^{(i)}(t)| ≤ N, i = 1, 2, …, l,  (42)

\begin{matrix}|y_{k}^{(j)}(t)| ≤ 2^{(j−l)(n−j−1)}N^{\frac{n−j−1}{n−l−1}} M3^{\frac{j−l}{n−l−1}} (τ − t)^{\frac{j−l}{n−l−1}} , \\ t ∈ J, j = l + 1, …, n − 1. \end{matrix}  (43)

Moreover, (31), (35), (42) yield

Φ_{0} (y_{k}(t)) = y_{k}(t), Φ_{1} (y_{k}^{(i)}(t)) = y_{k}^{(i)}(t) for i = 1, 2, …, l,
Φ_{2} (t, y_{k}^{(n−1)}(t)) = y_{k}^{(n−1)}(t), t ∈ J,

and hence y_{k}(t) is a solution of (1) satisfying

y_{k}^{(i)}(τ) = c_{i}, i = 0, 1, …, l; y_{k}^{(l+1)}(τ) = k.

As estimations (42) a (43) do not depend on k, i and j, the Arzelá–Ascoli Theorem implies the existence of a subsequence of {y_{k}(τ)}_{k_{0}}^{∞} that converges uniformly to the solution of (1) satisfying

y^{(j)}(T) = 0, j = l + 1, …, n − 2,  (44)
lim_{t→τ_{−}} y^{(i)}(t) = c_{i}, i = 0, 1, …, l, lim_{t→τ_{−}} y^{(l)}(t) = ∞.  (45)

Let λ = 1 + \frac{1}{n−l−2}. Then there exists a sequence of \{λ_{s}\}_{1}^{∞} such that λ_{s} satisfies (7) and lim_{s→∞} λ_{s} = 1 + \frac{1}{n−l−2}. Denote by y_{s} a solution of (1), (5) with λ = λ_{s}. It follows from (21)–(25) that there exists T ∈ [0, τ) such that y_{s}, s ∈ {1, 2, … } is defined on the interval [T, τ). At the same time, since (38)–(41) do not depend on λ, there exists Φ such that

|y_{s}^{(i)}(t) ≤ Φ(t), t ∈ [T, τ), i = 0, 1, …, n − 1, s = 1, 2, ….

Hence, according to the Arzelá–Ascoli Theorem, there exists a subsequence of {y_{s}}_{1}^{∞} that converges uniformly to a solution of (1), satisfying (5).

Let (9) and (10) be valid. Let y be defined on the interval (τ̄, τ) ⊂ [0, τ) and not be extendable to t = τ̄. Then

lim sup_{t→τ̄^{+}} |y^{(n−1)}(t)| = ∞.  (46)

First we prove that

y^{(n−1)}(t) > 0 on (τ̄, τ).  (47)

Suppose that there exists τ_{1} ∈ (τ̄, τ) such that y^{(n−1)}(τ_{1}) = 0 and y^{(n−1)}(t) > 0 on the interval (τ_{1}, τ). As τ_{1} < T, it follows from this and (45) that y^{(j)}, j = 0, 1, …, l, are bounded on the interval (τ_{1}, τ). Let τ_{2} ∈ (τ_{1}, τ) be such that y^{(n−1)}(τ_{2}) = ε. Then by the integration of (1) and (10)

∞ = ∫_{0}^{ε} \frac{ds}{g(s)} = ∫_{τ_{1}}^{τ_{2}}f (t, y(t), …, y^{(l)}(t)) dt < ∞.

Hence (47) holds. As τ_{1} < T, it follows from (9), (44) and (45) that y(t) > 0 on the interval (τ̄, τ) (y^{(i)}, i = 0, 1, …, l change their signs). Thus (1) yields y^{(n)}(t) > 0 on the interval (τ̄, τ), which, together with (47), contradicts (46). Hence y is defined at t = τ̄ and τ̄ = 0. Proof of Theorem 3. Let α = 1 and β = 1. The proof is similar to the that of Theorem 2. Since (11) and (12) are valid, we can restrict our investigation to the case

D = {[x_{1}, …, x_{l+1}] 0 ≤ x_{1} ≤ ε, (−1)^{j}x_{j+1} ∈ [(−1)^{j}c_{j}, (−1)^{j}c_{j} + ε]}.

The only problem is that due to c_{0} = 0, we have M_{1} = 0 and M_{3} = ∞, where M_{1} and M_{3} are given as in the proof of Theorem 2. Thus (31) gives us no information and it must be proved in a different way. Hence we prove that (31) is valid with the new values of λ_{1} and M_{3} given by

λ_{1} = \frac{(n − 1)σ + 1}{λ + σ − 1}, M_{3} = [\frac{2(n − 1)σ + 2}{(λ + σ − 1)M_{1}} ]^{\frac{1}{λ+σ−1}}

where M_{1} = \frac{M̄}{[l!(n−l−2)!(n−1)]^{σ}}. Note that, similarly to the proof of Theorem 2, z_{m}^{(n−1)} is positively increasing on the interval J_{m} = [T, τ_{m}], τ_{m} = τ − \frac{1}{m}. Note that (13) yields

n − l − 2 ≤ λ_{1} < n − l − 1.

If (31) is not valid, then either (32) or (33) holds.

Let (32) be valid. It follows similarly to (34) that

M ≤ z_{m}^{(n−1)}(t), t ∈ [t_{1}, τ_{m}].  (48)

Now we will estimate z_{m}. According to (48) and the Taylor Series Theorem we have

z_{m}^{(l+1)}(s) = ∑_{r=0}^{n−l−3} \frac{z_{m}^{(l+1+r)}(t)}{r!}(s − t)^{r} + ∫_{t}^{s} \frac{(s − σ)^{n−l−3}}{(n − l − 3)!} z_{m}^{(n−1)}(σ) dσ

≥ z_{m}^{(n−1)}(t)\frac{(s − t)^{n−l−2}}{(n − l − 2)!} , t ≤ s ≤ τ_{m}.  (49)

Similarly, the Taylor Series Theorem, (11), (12) and (49) yield

z_{m}(t) ≥ ∫_{τ_{m}}^{t} \frac{(t − s)^{l}}{l!} z_{m}^{(l+1)}(s) ds ≥ z_{m}^{(n−1)}(t) ∫_{τ_{m}}^{t} \frac{(s − t)^{n−2}(−1)^{l}}{l!(n − l − 2)!} ds

= \frac{(τ_{m} − t)^{n−1}}{l!(n − l − 2)!(n − 1)}z_{m}^{(n−1)}(t), t ∈ [t_{1}, τ_{m}].

From this, (8), (14), (48) and (49)

z_{m}^{(n)}(t) ≥ M̄z_{m}^{σ}(t) (z_{m}^{(n−1)}(t))^{λ} ≥ M_{1}(τ_{m} − t)^{(n−1)σ} (z_{m}^{(n−1)}(t))^{λ+σ} .

The integration on the interval [t_{1}, τ_{m}] yields

\frac{2(τ_{m} − t_{1})^{(n−1)σ+1}}{M_{3}^{λ+σ−1}} ≥ \frac{(τ − t_{1})^{(n−1)σ+1} − m^{−(n−1)σ+1}}{M_{3}^{λ+σ−1}} ≥

≥ \frac{1}{[z_{m}^{(n−1)}(t_{1})]^{λ+σ−1}} − \frac{1}{[z_{m}^{(n−1)}(τ_{m})]^{λ+σ−1}} ≥ \frac{M_{1}(λ + σ − 1)}{(n − 1)σ + 1}(τ_{m} − t_{1})^{(n−1)σ+1}

for large m. The contradiction obtained with the definition of M_{3}, shows that (32) does not hold. The fact that (33) is impossible can be proved similarly to the same case in the proof of Theorem 2.

Proof of Theorem 4. (i) Let y [T, τ) → R be a solution of (1), (5) with α = 1 and, for simplicity, y^{(n−1)}(t) ≥ M on the interval [T, τ). Put λ_{1} = \frac{(n−1)σ+1}{λ+σ−1} and M_{1} as in the proof of Theorem 3.

Let λ ≥ 1 + \frac{(l+1)σ+1}{n−l−2} for l < n − 2; hence n − l − 2 − λ_{1} ≥ 0. We can prove similarly to (44) − (46) that

y^{(n)}(t) ≥ M_{1}(τ − t)^{(n−1)σ}[y^{(n−1)}(t)]^{λ+σ}, t ∈ [T, τ).

From this and by the integration we obtain an estimation from above of y^{(n−1)} similar to (19) and the proof is similar to the second part of the proof of Theorem 1.

Let λ < 1 + \frac{lσ+1}{n−l−1}; hence n − l − 1 − λ_{1} < 0. Then

y(t) = ∫_{τ}^{t} \frac{(t − s)^{l}}{l!} y^{(l+1)}(s) ds ≤ \frac{|y^{(l)}(t)|}{l!}(τ − t)^{l}, t ∈ [T, τ).

From this

y^{(n)}(t) ≤ M_{2}y^{σ}(t) [y^{(n−1)}(t)]^{λ} ≤ M_{2}(τ − t)^{lσ} [y^{(n−1)}(t)]^{λ} ,

and the integration on the interval [t, τ) yields

y^{(n−1)}(t) ≥ M_{3}(τ − t)^{− \frac{lσ+1}{λ−1}} ,

where

M_{2} = max_{t∈[0,τ]} r(t), M_{3} = [\frac{M_{2}(λ − 1)}{lσ + 1}]^{− \frac{1}{λ−1}} .

The proof is similar to the first part of the proof of Theorem 1, only in (18) we take λ_{1} = \frac{lσ+1}{λ−1}.

(ii) The existence problem is solved by Theorem 3.

Proof of Proposition. Let y [T, τ) → R be a solution of (1) and (5) with α = 1; hence l is even. Let lim_{t→τ^{−}} y^{(n−1)}(t) = ∞. Then y^{(l)} < 0 in a left neighborhood I of τ. From this and from l being even we can conclude that y < 0 and y^{(n)} ≤ 0 on I. The contradiction to lim_{t→τ^{−}} y^{(n−1)} = ∞ proves the statement. Other possible cases can be proved similarly.

Acknowledgement

The research of the second author was supported by Grant 201/99/0295 of the Grant Agency of the Czech Republic.

References

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(Received 10.04.2001)

Authors’ address:

Department of Mathematics Masaryk University Janáčkovo nám. 2a, 662 95 Brno Czech Republic E-mail: Bartusek@math.muni.cz

Osicka@math.muni.cz