Lobachevskii Journal of Mathematics

http://ljm.ksu.ru
ISSN 1818-9962

Vol. 24, 2006, 135-142 \begin{matrix}c \\ ◯ \end{matrix} W.Zhou, S. Cai

W. S. Zhou, S. F. Cai

POSITIVE SOLUTIONS FOR A SINGULAR SECOND ORDER ORDINARY DIFFERENTIAL EQUATION

(submitted by A. V. Lapin)

Abstract. This paper is concerned with the positive solutions for a singular second order ordinary differential equation. Under appropriate conditions, by the classical method of elliptic regularization, we prove the existence of position solutions.

1. Introduction

This paper is concerned with the existence of positive solutions for a singular second order ordinary differential equation

\frac{φ′′}{\sqrt{1 + |φ′|^{2}}} − λ\frac{|φ′|^{2}}{φ(1 + |φ′|^{2})^{m}} + f(t) = 0,  0 < t < 1,  (1)

with one of the following boundary conditions

φ(1) = φ(0) = 0,  (2)
φ(1) = φ(0) = φ′(1) = φ′(0) = 0,  (3)

where λ > 0, m ≥ \frac{1}{2}, f(t) ∈ C^{1}[0, 1] and f(t) > 0 on [0, 1].

existence.

2000 Mathematical Subject Classification. 34B15.

Key words and phrases. singular differential equation, positive solution,

Supported by 985 Program of Jilin University.

It is well known that boundary value problems (BVPs) for singular second order ordinary differential equations arise in the field of gas dynamics, flow mechanics, theory of boundary layer, and so on. In recent years, singular ordinary differential equation with dependence on the first order derivative have been studied extensively, see for example [1–8] and references therein where some general existence results were obtained. We point out that the case considered here is not in their considerations since it does not satisfy some sufficient conditions of those papers. Our considerations were motivated by [9] in which the authors studied the following singular differential equation

φ′′ − λ\frac{|φ′|^{2}}{φ} + 1 = 0, 0 < t < 1,

with the boundary conditions: φ(1) = φ′(0) = 0. By ordinary differential equation theories, they obtained a decreasing positive solution. In the present paper, we consider (1) and will use the classical method of elliptic regularization to obtain positive solutions to BVP (1), (2) and BVP (1), (3). However, it is easy to see from the boundary conditions (2) (or (3)) that any positive solution to BVP (1), (2) (or BVP (1), (3)) must not be decreasing. Thus the existence results obtained here are not a simple extension of [9]

We say φ ∈ C^{2}(0, 1)∩C[0, 1] is a solution to BVP (1),(2) if it is positive in (0,1) and satisfies (1) and (2). Similarly, we say φ ∈ C^{2}(0, 1) ∩ C^{1}[0, 1] is a solution to BVP (1),(3) if it is positive in (0,1) and satisfies (1) and (3).

The main purpose of this paper is to prove the following theorems.

Theorem 1 Let m ≥ \frac{1}{2}, λ > 0, f(t) ∈ C^{1}[0, 1] and 0 < min_{[0,1]} f ≤ max_{[0,1]} f < 1. Then BVP (1),(2) admits at least a solution.

Theorem 2 Let m ≥ \frac{1}{2}, f(t) ∈ C^{1}[0, 1], 0 < min_{[0,1]} f ≤ max_{[0,1]} f < 1 and λ > \frac{1}{2}[1 − (max_{[0,1]} f)^{2}]^{1/2−m}. Then BVP (1),(3) admits at least a solution.

2. Proofs of Theorems

We will use the classical method of elliptic regularization to prove Theorem 1. For this, we consider the following regularized problem:

\frac{φ′′}{\sqrt{1 + |φ′|^{2}}} − λ\frac{|φ′|^{2}}{(1 + |φ′|^{2})^{m}} \frac{sgn_{ε}(φ)}{I_{ε}(φ)} + f(t) = 0, 0 < t < 1,

φ(1) = φ(0) = ε,

where ε ∈ (0, 1), I_{ε}(s) and sgn_{ε}(s) are defined as follows:

I_{ε}(s) = \{ \begin{matrix}s, & s ≥ ε, \\ \frac{s^{2} + ε^{2}}{2ε} , & |s| < ε, \\ − s, & s ≤ −ε, \end{matrix} sgn_{ε}(s) = \{ \begin{matrix}1, & s ≥ ε, \\ \frac{2s}{ε} − \frac{s^{2}}{ε^{2}} , & 0 ≤ s < ε, \\ \frac{2s}{ε} + \frac{s^{2}}{ε^{2}} , & − ε ≤ s < 0, \\ − 1, & s < −ε. \end{matrix}

Clearly, I_{ε}(s), sgn_{ε}(s) ∈ C^{1}(ℝ), and I_{ε}(s) ≥ ε/2, 1 ≥ |sgn_{ε}(s)|, sgn_{ε}(s) · sgn(s) ≥ 0 in ℝ.

For any m ≥ \frac{1}{2}, it follows from Theorem 4.1 of Chapter 7 in [10] that for any fixed ε ∈ (0, 1), the above regularized problem admits a classical solution φ_{ε} ∈ C^{2}(0, 1) ∩ C^{1}[0, 1]. By the maximal principle, it is easy to see that φ_{ε}(t) ≥ ε on [0, 1]. Thus φ_{ε} satisfies

\frac{φ′′_{ε}}{\sqrt{1 + |φ′_{ε}|^{2}}} − λ\frac{|φ′_{ε}|^{2}}{φ_{ε}(1 + |φ′_{ε}|^{2})^{m}} + f(t) = 0, 0 < t < 1,  (4)
φ_{ε}(0) = φ_{ε}(1) = ε.

Note that (4) is equivalent to

( ∫_{0}^{φ′_{ε}(t)} \frac{1}{\sqrt{1 + s^{2}}} ds)′ − λ\frac{|φ′_{ε}|^{2}}{φ_{ε}(1 + |φ′_{ε}|^{2})^{m}} + f(t) = 0, 0 < t < 1. (5)

Lemma 1 Under the assumptions of Theorem 1, for all ε ∈ (0, 1) there holds

|φ′_{ε}(t)| ≤ \frac{Λ}{\sqrt{1 − Λ^{2}}} , t ∈ [0, 1],

where Λ =^{Δ} max_{[0,1]} f < 1. Proof. Noticing φ_{ε}(1) = φ_{ε}(0) = ε and φ_{ε}(t) ≥ ε for all t ∈ [0, 1], we have

φ′_{ε}(0) = lim_{t→0^{+}} \frac{φ_{ε}(t) − ε}{t} ≥ 0,

φ′_{ε}(1) = lim_{t→1^{−}} \frac{φ_{ε}(t) − ε}{t − 1} ≤ 0.

On the other hand, it follows from (5) that

( ∫_{0}^{φ′_{ε}} \frac{1}{\sqrt{1 + s^{2}}} ds)′ + Λ ≥ 0, 0 < t < 1,

i.e.

( ∫_{0}^{φ′_{ε}(t)} \frac{1}{\sqrt{1 + s^{2}}} ds + Λt)′ ≥ 0, 0 < t < 1.

Thus the function ∫_{0}^{φ′_{ε}(t)} \frac{1}{\sqrt{1+s^{2}}} ds+Λt is non-decreasing on [0, 1], therefore

\begin{matrix}Λ ≥ & ∫_{0}^{φ′_{ε}(1)} \frac{1}{\sqrt{1 + s^{2}}} ds + Λ \\ ≥ & ∫_{0}^{φ′_{ε}(t)} \frac{1}{\sqrt{1 + s^{2}}} ds + Λt \\ ≥ & ∫_{0}^{φ′_{ε}(0)} \frac{1}{\sqrt{1 + s^{2}}} ds ≥ 0, t ∈ [0, 1], \end{matrix}

and hence

| ∫_{0}^{φ′_{ε}(t)} \frac{1}{\sqrt{1 + s^{2}}} ds| ≤ Λ, t ∈ [0, 1].

From this and using the inequality

|\frac{z}{\sqrt{1 + z^{2}}} | ≤ | ∫_{0}^{z} \frac{1}{\sqrt{1 + s^{2}}} ds|, z ∈ ℝ,

we obtain

|\frac{φ′_{ε}(t)}{\sqrt{1 + |φ′_{ε}(t)|^{2}}} | ≤ Λ, t ∈ [0, 1],

Noticing Λ < 1, we obtain

|φ′_{ε}(t)| ≤ \frac{Λ}{\sqrt{1 − Λ^{2}}} , t ∈ [0, 1].

This completes the proof of Lemma 1.

Denote Λ_{m} by Λ_{m} = (1 − Λ^{2})^{1/2−m}, where Λ is the same as that of Lemma 1. From (4) and Lemma 1 we obtain obtain

−φ′′_{ε} + λ\frac{|φ′_{ε}|^{2}}{φ_{ε}} − min_{[0,1]} f ≥ 0, t ∈ (0, 1).  (6)
−φ′′_{ε} + \frac{λ}{Λ_{m}} \frac{|φ′_{ε}|^{2}}{φ_{ε}} − \frac{Λ}{(1 − Λ^{2})^{1/2}} ≤ 0, t ∈ (0, 1).  (7)

To obtain the uniform bounds of φ_{ε}, the following comparison theorem will be proved to be very useful.

Proposition 2 Let φ_{i} ∈ C^{2}(0, 1) ∩ C[0, 1] and φ_{i} > 0 on [0, 1](i = 1, 2). If φ_{2} ≥ φ_{1} for t = 0, 1, and

−φ′′_{2} + ϱ\frac{|φ′_{2}|^{2}}{φ_{2}} − θ ≥ 0, t ∈ (0, 1),  (8)
−φ′′_{1} + ϱ\frac{|φ′_{1}|^{2}}{φ_{1}} − θ ≤ 0, t ∈ (0, 1),  (9)

where ϱ and θ are positive constants, then φ_{2}(t) ≥ φ_{1}(t), t ∈ [0, 1].

Proof. From (8) and (9), we have

(\frac{φ_{2}^{1−ϱ}}{1 − ϱ})′′ ≤ −\frac{θ}{φ_{2}^{ϱ}} , (ϱ ≠ 1)

( ln(φ_{2}))′′ ≤ −\frac{θ}{φ_{2}} . (ϱ = 1)

and

(\frac{φ_{1}^{1−ϱ}}{1 − ϱ})′′ ≥ −\frac{θ}{φ_{1}^{ϱ}} , (ϱ ≠ 1)

( ln(φ_{1}))′′ ≥ −\frac{θ}{φ_{1}} . (ϱ = 1)

Combining the above inequalities, we obtain

w′′ ≤ θ(\frac{1}{φ_{1}^{ϱ}} − \frac{1}{φ_{2}^{ϱ}} ), 0 < t < 1,  (10)

where w : [0, 1] → ℝ is defined by

w = \{ \begin{matrix}\frac{φ_{2}^{1−ϱ}}{1 − ϱ} − \frac{φ_{1}^{1−ϱ}}{1 − ϱ}, & (ϱ ≠ 1) \\ ln(φ_{2}) − ln(φ_{1}). & (ϱ = 1) \end{matrix}

Clearly, w ∈ C^{2}(0, 1) ∩ C[0, 1].

To prove the proposition, we argue by contradiction and assume that there exists a point t_{0} of (0, 1) such that φ_{2}(t_{0}) − φ_{1}(t_{0}) < 0. From the assumption, it is easy to see that w reaches a minimum at some point t_{∗} of (0, 1) such that

w(t_{∗}) = min_{t∈[0,1]} w(t) < 0,  (11)
w′′(t_{∗}) ≥ 0.  (12)

Combining (12) with (10), we have

θ(\frac{1}{φ_{1}^{ϱ}(t_{∗})} − \frac{1}{φ_{2}^{ϱ}(t_{∗})}) ≥ 0.

This implies φ_{2}(t_{∗}) ≥ φ_{1}(t_{∗}). However, from (11) we find that φ_{2}(t_{∗}) < φ_{1}(t_{∗}), a contradiction. Thus the proof of Proposition 2 is completed. Lemma 3 Under the assumptions of Theorem 1, for all ε ∈ (0, 1) there exists a positive constant C independent of ε such that

φ_{ε}(t) ≥ C[t(1 − t) + ε^{1/2}]^{2}, t ∈ [0, 1].

Proof. Let w_{ε} = C[t(1 − t) + ε^{1/2}]^{2}, where C ∈ (0, 1] will be determined later. By Proposition 2 and noticing (6), it suffices to show that

−w′′_{ε} + λ\frac{|w′_{ε}|^{2}}{w_{ε}} − min_{[0,1]} f ≤ 0, t ∈ (0, 1),  (13)

for some sufficiently small positive constant C independent of ε. Simple calculation shows that

w′_{ε} = 2C[t(1 − t) + ε^{1/2}](1 − 2t),
w′′_{ε} = 2C(1 − 2t)^{2} − 4C[t(1 − t) + ε^{1/2}],

and

−w′′_{ε} + λ\frac{|w′_{ε}|^{2}}{w_{ε}} − min_{[0,1]} f = − 2C(1 − 2t)^{2} + 4C[t(1 − t) + ε^{1/2}]

+ 4Cλ(1 − 2t)^{2} − min_{[0,1]} f

≤4C(2 + λ) − min_{[0,1]} f, 0 < t < 1.

Choosing a positive constant C such that

C ≤ min \{1, \frac{min_{[0,1]} f}{4(2 + λ)} \},

we find that (13) holds. Thus the proof of Lemma 3 is completed.

From (6), (7), Lemma 1 and Lemma 3, we derive that for any δ ∈ (0, 1/2) there exists a positive constant C_{δ} independent of ε such that

|φ′′_{ε}(t)| ≤ C_{δ}, δ ≤ t ≤ 1 − δ.  (14)

From (4), we have

φ′′_{ε} =λ\frac{(1 + |φ′_{ε}|^{2})^{1/2−m}|φ′_{ε}|^{2}}{φ_{ε}} − f(t)(1 + |φ′_{ε}|^{2})^{1/2}, 0 < t < 1.

Differentiating the above equation with respect to t we get

φ′′′_{ε} =\frac{2λφ′_{ε}φ′′_{ε}}{φ_{ε}} [(1 + |φ′_{ε}|^{2})^{1/2−m} + (\frac{1}{2} − m)(1 + |φ′_{ε}|^{2})^{−m−1/2}|φ′_{ε}|^{2}]

− \frac{λ(1 + |φ′_{ε}|^{2})^{1/2−m}(φ′_{ε})^{3}}{φ_{ε}^{2}} − f(t)(1 + |φ′_{ε}|^{2})^{−1/2}φ′_{ε}φ′′_{ε}

− f′(t)(1 + |φ′_{ε}|^{2})^{1/2}, 0 < t < 1.

By (14), Lemma 1 and Lemma 3, we derive that for any δ ∈ (0, 1/2), there exists a positive constant C_{δ} independent of ε such that

|φ′′′_{ε}(t)| ≤ C_{δ}, δ ≤ t ≤ 1 − δ.

From this and Lemma 1 and using Alzelá-Ascoli theorem and diagonal sequential process, we see that there exists a subsequence {φ_{ε_{n}}} of {φ_{ε}} and a function φ ∈ C^{2}(0, 1) ∩ C[0, 1] such that, as ε_{n} → 0,

φ_{ε_{n}} → φ, uniformly in C[0, 1],

φ_{ε_{n}} → φ, uniformly in C^{2}[δ, 1 − δ].

Combining these with (4) (or (5)) and the boundary conditions satisfied by φ_{ε_{n}}, we find that φ satisfies (1) and (2). By Lemma 3, we have

φ(t) ≥ C[t(1 − t)]^{2}, t ∈ [0, 1],  (15)

therefore φ > 0 in (0,1), and thus φ is a solution to BVP (1), (2). This completes the proof of Theorem 1.

Proof of Theorem 2. From Theorem 1, we see that for any λ > 0, BVP (1),(2) admits a solution φ which can be approximated by φ_{ε_{n}} satisfying (4) (or (5)) with ε = ε_{n}. Hence it suffices to show φ satisfies φ′(1) = φ′(0) = 0 for λ > \frac{1}{2}[1 − (max_{[0,1]} f)^{2}]^{1/2−m} = \frac{1}{2}(1 − Λ^{2})^{1/2−m} = \frac{1}{2}Λ_{m}. We claim that if λ > \frac{1}{2}Λ_{m}, then there exist positive constants C independent of ε_{n} such that

φ_{ε_{n}}(t) ≤ C(1 + ε_{n}^{1/2} − t)^{2} on [0, 1],  (16)
φ_{ε_{n}}(t) ≤ C(t + ε_{n}^{1/2})^{2} on [0, 1].  (17)

We first show (16). Let v_{ε_{n}} = C(1 + ε_{n}^{1/2} − t)^{2}, where C ≥ 1 will be determined later. A calculation shows that

\begin{matrix}− v′′_{ε_{n}} + \frac{λ}{Λ_{m}} \frac{|v′_{ε_{n}}|^{2}}{v_{ε_{n}}} − \frac{Λ}{(1 − Λ^{2})^{1/2}} \\ = 2C(\frac{2λ}{Λ_{m}} − 1) − \frac{Λ}{(1 − Λ^{2})^{1/2}} , 0 < t < 1. \end{matrix}

Choosing a positive constant C such that

C ≥ max \{1, \frac{Λ_{m}Λ}{2(2λ − Λ_{m})(1 − Λ^{2})^{1/2}} \}

and noticing λ > \frac{1}{2}Λ_{m}, we find that

−v′′_{ε_{n}} + \frac{λ}{Λ_{m}} \frac{|v′_{ε_{n}}|^{2}}{v_{ε_{n}}} − \frac{Λ}{(1 − Λ^{2})^{1/2}} ≥ 0, 0 < t < 1,

and then, by Proposition 2 and noticing (7), we obtain (16). Similarly we can prove the claim (17). Letting ε_{n} → 0 in (16) and (17) to yield

φ(t) ≤ C min{t^{2}, (1 − t)^{2}}, t ∈ [0, 1].

Combining this with (15) we immediately obtain φ′(1) = φ′(0) = 0. Thus the proof of Theorem 2 is completed.

142 W.ZHOU, S. CAI

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Department of Mathematics, Jilin University, Changchun 130012, P.R.China E-mail address: pdezhou@126.com

School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, P. R. China

Received August 26, 2006