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volume 6, issue 5, article 137, 2005.

Received 07 April, 2005;

accepted 08 September, 2005.

Communicated by:D. Hinton

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Journal of Inequalities in Pure and Applied Mathematics

ON VECTOR BOUNDARY VALUE PROBLEMS WITHOUT GROWTH RESTICTIONS

CHRISTOPHER C. TISDELL AND LIT HAU TAN

School of Mathematics

The University of New South Wales Sydney 2052, Australia.

EMail:cct@maths.unsw.edu.au EMail:lithau@maths.unsw.edu.au

c

2000Victoria University ISSN (electronic): 1443-5756 113-05

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On Vector Boundary Value Problems Without Growth

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Christopher C. Tisdell and Lit Hau Tan

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Abstract

Herein, we consider the existence of solutions to second-order systems of two- point boundary value problems (BVPs). The methods used involve the topological transversality approach of Granas et. al. combined with a Bernstein- Nagumo condition from Gaines and Mawhin. The new results allow the treatment of systems of BVPs without growth restrictions in the third variable. The new results also are applicable to systems of BVPs that may have singularities in the right-hand side at the end-points of the interval of existence. Some examples are presented to illustrate the theory.

2000 Mathematics Subject Classification:Primary 34B15.

Key words: Boundary value problem, No growth restriction, Existence of solutions, a priori bounds.

This research was supported by the Australian Research Council’s Discovery Projects DP0450752.

This paper is based on the talk given by the first author within the “International Conference of Mathematical Inequalities and their Applications, I”, December 06- 08, 2004, Victoria University, Melbourne, Australia [http://rgmia.vu.edu.au/

conference]

1. Introduction

Consider the existence of solutions to the second-order, ordinary differential equation

(1.1) x⁰⁰ =F(t, x, x⁰), t∈[0,1], subject to some suitable boundary conditions.

Topological methods, used in proving the existence of solutions to boundary value problems, such as: the continuation method of Gaines and Mawhin [5], [6]; or the topological transversality method of Granas, Guenther and Lee [9], [10]; generally rely on guaranteeing a priori bounds on solutions (and their derivatives) to the BVP under consideration in such a way that the same a priori bounds apply to a certain family of BVPs.

A classical issue associated with the preceding discussion is the following question. How can we ensure an a priori bound on solutions’ derivatives x⁰ to (1.1) with the bound on x⁰ being in terms of an a priori bound on possible solutions x? A sufficient condition that guarantees the desired a priori bound onx⁰is traditionally known as a “Bernstein-Nagumo condition”.

For scalar-valued BVPs, many authors have formulated Bernstein-Nagumo conditions for (1.1), for example: [3], [18], [15], [14], [22], [1], [10], [16] and also see references therein.

However, for vector-valued BVPs (i.e. F : [0,1] × R²ⁿ → Rⁿ), less is known about sufficient Bernstein-Nagumo conditions, perhaps to the Bernstein- Nagumo question becoming more difficult than in the scalar-valued situation (see [2, Remark 1.41] or [12] for more discussion and some examples.)

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Authors such as: Hartman [11]; Schmitt and Thompson [21]; Gaines and Mawhin [6]; Fabry [4]; George and Sutton [7]; and George and York [8] have all presented interesting Bernstein-Nagumo conditions for vector BVPs. Their conditions involved growth-type conditions onF inx⁰ or the existence of suitable Lyapunov functions.

Herein, we consider vector equations of the type (1.2) x⁰⁰ =f(t, x, x⁰), t ∈[0,1],

where f : [0,1]×R²ⁿ → Rⁿ and (1.2) is subject to the following boundary conditions:

(1.3) x⁰(0) =g₁(x(0)), x⁰(1) =g₂(x(1)), (where each g_i :Rⁿ →Rⁿ).

Well-known special cases of the rather general boundary conditions (1.3) in- clude: the Sturm-Liouville boundary conditions

αx(0)−βx⁰(0) = C, γx(1) +δx⁰(1) =D, (1.4)

α, β, γ, δare constants inR;C, D are constants inRⁿ; and the homogenous Neumann boundary conditions

(1.5) x⁰(0) = 0, x⁰(1) = 0;

plus variations of the above (see Remarks1and3), including nonlinear boundary conditions.

In Section2we combine the topological transversality method of [10, The-

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Proposition 5.1]. The combination leads to novel and quite general existence theorems for solutions to the above systems of BVPs. In particular, the new results extend the workings of [10] and [6] in the sense that the new results herein allow the treatment of certain classes of BVPs whereas the theorems of [10] and [6] may not directly apply.

In Section3 we briefly consider systems of BVPs with singularities in the right-hand side.

Examples are presented throughout the paper to demonstrate the applicabil- ity of the new theorems. It appears that no existing theory in the literature is applicable to the examples given.

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2. Existence Results

To generate our new topological transversality-based existence theorems, we consider the following family of BVPs:

a(t)x⁰⁰+b(t)x⁰+c(t)x=λg(t, x, x⁰), t∈[0,1], (2.1)

a₀x(0) +a₁x⁰(0) +a₂x(1) +a₃x⁰(1) = λψ₁(x(0), x⁰(0), x(1), x⁰(1)), (2.2)

b0x(0) +b1x⁰(0) +b2x(1) +b3x⁰(1) = λψ2(x(0), x⁰(0), x(1), x⁰(1)), (2.3)

where: λ ∈ [0,1]; a, b, c are continuous functions witha(t) 6= 0 for any t ∈ [0,1];eacha_i andb_iare constants;g : [0,1]×R²ⁿ →Rⁿand eachψ_i :R⁴ⁿ→ Rⁿ.

Below we denote k · k as the usual Euclidean norm and h·,·i as the usual inner product onRⁿ.

To streamline the proofs of our results, we will use the following existence theorem, a vector-variant of [10, Theorem 6.1, Chap.II].

Theorem 2.1. Let g and each ψ_i be continuous and letR > 0be a constant independent ofλ.If:

the family (2.1)–(2.3) has only the zero solution forλ= 0; and (2.4)

forλ∈(0,1]all possible solutionsx∈C²([0,1];Rⁿ)to (2.1)–(2.3) (2.5)

satisfy max{kx(t)k, kx⁰(t)k, kx⁰⁰(t)k}< R, t ∈[0,1], then forλ= 1the BVP (2.1)–(2.3) has at least one solution.

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Theorem 2.2. Letf be continuous and letM, N be positive constants with

(2.6) N >max

kCk+|α|M

|β| , kDk+|γ|M

|δ|

.

If

(2.7) α/β >0, γ/δ <0, α(γ+δ) +βγ6= 0;

and

(2.8) hx, f(t, x, x⁰)i+kx⁰k² >0, for t∈[0,1], kxk ≥M, hx, x⁰i= 0;

and

(2.9) hx⁰, f(t, x, x⁰)i>0, for t∈[0,1], kxk ≤M, kx⁰k=N, then (1.2), (1.4) has at least one solution.

Proof. Consider the family of BVPs:

x⁰⁰=λf(t, x, x⁰), t∈[0,1], (2.10)

αx(0)−βx⁰(0) =λC, (2.11)

γx(1) +δx⁰(0) =λD, (2.12)

forλ∈[0,1]and see that this is in the form (2.1)–(2.2), withg =f.

Letxbe a solution to (2.10)–(2.12). Sinceα(γ+δ) +βγ 6= 0, note that, for λ = 0, the above family of BVPs only has the zero solution by direct calculation.

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We show that (2.8) and (2.7) imply (2.13) kx(t)k ≤M₁ := max

kCk

|α| , M, kDk

|γ|

, for t∈[0,1].

Considerr₁(t) = kx(t)k² fort ∈ [0,1]and lett₀ ∈ [0,1]be such thatr₁(t₀) = maxt∈[0,1]r(t). If r1(t0) = 0 then r1(t) = 0 for all t ∈ [0,1] and obviously kx(t)k = 0< M for all t ∈ [0,1]and allM > 0, so assumer₁(t₀) > 0from now on.

Ift₀ = 0then

0≥r₁⁰(t0) = 2hx(0), x⁰(0)i

= 2hx(0),αx(0)−λC

β i from (2.11)

= 2α

βkx(0)k²

1− hx(0), λCi αkx(0)k²

.

Thus, by (2.7),

1≤ hx(0), λCi

αkx(0)k² ≤ kx(0)kkCk

|α| kx(0)k²,

giveskx(0)k ≤ kCk/|α|and we must havekx(t)k ≤ kCk/|α|for allt ∈[0,1].

If t0 = 1then 0 ≤ r⁰₁(1) and a similar argument to the case t0 = 0gives kx(t)k ≤ kDk/|γ|for allt ∈[0,1].

Ift₀ ∈ (0,1)andr₁(t₀)≥ M² then0 = r⁰₁(t₀) = 2hx(t₀), x⁰(t₀)i. We also

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have

0≥r⁰⁰₁(t₀) = 2

hx(t₀), x⁰⁰(t₀)i+kx⁰(t₀)k²

= 2

hx(t₀), λf(t₀, x(t₀), x⁰(t₀))i+kx⁰(t₀)k²

≥2λ

hx(t₀), f(t₀, x(t₀), x⁰(t₀))i+kx⁰(t₀)k²

>0,

by (2.8), a contradiction. Hence we havekx(t₀)k< M for allt₀ ∈(0,1).

Combining all of the above bounds we obtain (2.13).

Letx∈C²([0,1];Rⁿ)be a solution to (1.2) withkx(t)k ≤M₁fort∈[0,1].

We now show that (2.6) and (2.9) implykx⁰(t)k< N for allt∈[0,1].

Argue by contradiction by assumingr(t₀) = kx⁰(t₀)k² −N² ≥ 0for some t0 ∈ [0,1] such that maxt∈[0,1]r(t) = r(t0). If t0 = 0 then rearranging the boundary conditions we obtain

kx⁰(0)k=

λC−αx(0) β

≤ kCk+|α|M

|β| < N,

and thusr(0)<0.Similarly,

kx⁰(1)k ≤ kDk+|γ|M

|δ| < N,

and thusr(1)<0.So we see thatt0 ∈(0,1).

Now sincer(1) < 0 andr(t₀) ≥ 0 we must have a pointt₁ ∈ [t₀,1) such

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thatr(t₁) = 0and

0≥r⁰(t₁) = 2hx⁰(t₁), x⁰⁰(t₁)i

= 2hx⁰(t1), λf(t1, x(t1, x⁰(t1))i

>0,

for allλ∈(0,1]by (2.9), a contradiction.

It is clear to see that once bounds on x and x⁰ are found, a bound on x⁰⁰ follows naturally, as

kx⁰⁰(t)k=kλf(t, x, x⁰)k ≤ kf(t, x, x⁰)k

≤P for t∈[0,1], kxk ≤M, kx⁰k ≤N, for someP ≥0.

So we see that there exists anR >0with R = max

max

kCk

|α| , M, kDk

|γ|

, N, P

+ 1 such that (2.5) holds.

Thus, by Theorem2.1, the family (2.10)–(2.12) has a solution forλ = 1.For λ = 1,(2.10)–(2.12) is equivalent to (1.2)–(1.4) and hence the result follows.

Example 2.1. Let x = (x₁, x₂) and p = (p₁, p₂). Consider (1.2), (1.4) for

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n = 2, where

f(t, x, p) =f(t, x₁, x₂, p₁, p₂)

=

x₁e^x²^p² + x²₁p³₁ + p₁ x₂e^x²^p² + x²₂p³₂ + p₂

!

, t∈[0,1], x₁(0)

x₂(0)

!

−2

x⁰₁(0) x⁰₂(0)

!

=





√1 2



,

x₁(1) x2(1)

! + 2

x⁰₁(1) x⁰₂(1)

!

=





√1 2



.

There is no growth condition applicable to f and thus the theorems of [11], [21], [4] do not apply. We will apply Theorem2.2.

Firstly, forkxk ≥M, withM to be chosen below, andhx, pi= 0,consider hx, f(t, x, p)i=x²₁e^x²^p² + (x₁p₁)³+x₁p₁+x²₂e^x²^p² + (x₂p₂)³+x₂p₂

=e^x²^p²[x²₁+x²₂] (sincex₁p₁ =−x₂p₂)

>0 for any positive choice ofM.

For convenience, choose M = 1,thus (2.8) holds. Now, forkxk ≤ 1, kpk = N = 2we have

hp, f(t, x, p)i=p₁x₁e^x²^p² +x²₁p⁴₁+p²₁+p₂x₂e^x²^p² +x²₂p⁴₂+p²₂

≥4 +e^x²^p²[p₁x₁+p₂x₂]

≥2>0 for kxk ≤1, kpk= 2,

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thus (2.9) holds.

It is easy to see that (2.6) holds for our choice ofM = 1andN = 2and for the given boundary conditions. Thus Theorem 2.2 is applicable and the BVP has a solution.

Theorem 2.3. Letf be continuous and letM, N be positive constants with (2.14) 2N² ≥ −hx, x⁰i, for kxk ≤M, kx⁰k=N.

If (2.8) and (2.9) hold then (1.2), (1.5) has at least one solution.

x⁰⁰−2x⁰−x=λ[f(t, x, x⁰)−2x⁰ −x], t ∈[0,1], (2.15)

x⁰(0) = 0, (2.16)

x⁰(1) = 0, (2.17)

for λ ∈ [0,1] and see that this is in the form (2.1)–(2.3) with g(t, x, x⁰) = f(t, x, x⁰)−2x⁰−x.

Letxbe a solution to (2.15)–(2.17). By direct calculation, the only solution to (2.15)–(2.17) forλ = 0isx= 0,so (2.4) holds.

Now rearranging (2.15) we obtain

x⁰⁰ =λf(t, x, x⁰) + 2(1−λ)x⁰+ (1−λ)x, t∈[0,1], (2.18)

:=q_λ(t, x, x⁰).

We show thatkx(t)k < M for allt ∈[0,1]and allλ ∈(0,1]. Considerr(t) = kx(t)k² fort ∈ [0,1]and let t₀ ∈ [0,1]be such that r(t₀) = maxt∈[0,1]r(t) ≥

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Ift₀ = 0then the boundary conditions givehx(0), x⁰(0)i= 0. Therefore, by (2.8) we have

0<2

hx(0), f(0, x(0), x⁰(0))i+kx⁰(0)k²

, and so 0<2λ

hx(0), f(0, x(0), x⁰(0))i+kx⁰(0)k²

, forλ ∈(0,1]

≤2 [hx(0), λf(0, x(0), x⁰(0))i+ 2(1−λ)hx(0), x⁰(0)i +(1−λ)kx(0)k²+kx⁰(0)k²

= 2

hx(0), q_λ(0, x(0), x⁰(0))i+kx⁰(0)k²

= 2

hx(0), x⁰⁰(0)i+kx⁰(0)k²

=r⁰⁰(0),

sor⁰(t)is strictly increasing fortnear 0. Therefore0 = r⁰(0) < r⁰(t)fortnear 0. This means thatr(t)is increasing fortnear 0, that is,r(0)< r(t)and hence r(0)6= max_t∈[0,1]r(t).

Ift₀ = 1then a similar argument to the case fort₀ = 0giveskx(1)k< M. Ift₀ ∈ (0,1)then an identical argument to the proof of Theorem 2.2gives kx(t0)k< M. Hence we havekx(t)k< M for allt∈[0,1].

Consider solutionsx ∈ C²([0,1];Rⁿ)with kx(t)k ≤ M for t ∈ [0,1].We now show that (2.9) implykx⁰(t)k< N for allt∈[0,1].

Argue by contradiction by assumingr₁(t₀) =kx⁰(t₀)k²−N² ≥ 0for some t₀ ∈ [0,1] such that max_t∈[0,1]r₁(t) = r₁(t₀). The boundary conditions give r₁(0) <0andr₁(1) < 0.So we see thatt₀ ∈ (0,1). Now sincer₁(1) <0and r₁(t₀)≥0we must have a pointt₁ ∈[t₀,1)such thatr₁(t₁) = 0and

0≥r⁰₁(t₁) = 2hx⁰(t₁), x⁰⁰(t₁)i

= 2hx⁰(t₁), q_λ(x(t₁), x(t₁), x⁰(t₁))i

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=λhx⁰(t₁), f(t₁, x(t₁), x⁰(t₁))i+ 2(1−λ)kx⁰(t₁)k²+ (1−λ)hx(t₁), x⁰(t₁)i

=λhx⁰(t₁), f(t₁, x(t₁), x⁰(t₁))i+ (1−λ)[2N²− hx(t₁), x⁰(t₁)i]

>0,

for allλ∈(0,1], a contradiction.

Hence we havekx⁰(t)k< N fort∈[0,1].

Since a priori bounds are now obtained onxandx⁰, the a priori bound on x⁰⁰naturally follows as in the proof of Theorem2.2.

Hence, by Theorem2.1, the family (2.15)–(2.17) has a solution for λ = 1, which is identical to the BVP (1.2), (1.5) and hence the result follows.

Example 2.2. Consider the scalar BVP (1.2), (1.5) where f is given by the right-hand side of

(2.19) x⁰⁰ = (x+ 1 +x⁰)e^x⁰, t∈[0,1].

It is not difficult to show that (2.19) satisfies (2.8), (2.9) and (2.14) forM = 3/2 andN = 2. Thus, by Theorem2.3we conclude that the scalar BVP (2.19), (1.5) has at least one solution.

Theorem 2.4. Letf, g₁andg₂be continuous and letM, N be positive constants such that

(2.20) N >max

kxk≤Mmax kg1(x)k, max

kxk≤Mkg2(x)k

.

If

(2.21) hz, g₁(z)i>0, hz, g₂(z)i<0, for all kzk ≥M,

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x⁰⁰−2x⁰−x=λ[f(t, x, x⁰)−2x⁰ −x], t ∈[0,1], (2.22)

x⁰(0) =λg1(x(0)), (2.23)

x⁰(1) =λg₂(x(0)), (2.24)

forλ∈[0,1].

Letxbe a solution to (2.22)–(2.24). See that, forλ= 0, the above family of BVPs only has the zero solution.

We show that kx(t)k ≤ M, for t ∈ [0,1]. Consider r₁(t) = kx(t)k² for t ∈[0,1]and lett₀ ∈[0,1]be such thatr₁(t₀) = maxt∈[0,1]r₁(t)≥M².

Ift₀ = 0then

0≥r⁰₁(t₀) = 2hx(0), x⁰(0)i

= 2hx(0), λg₁(x(0))i from (2.11)

>0 a contradiction.

Ift₀ = 1then 0 ≤ r⁰₁(1) and a similar arguement to the caset₀ = 0 gives another contradiction.

Ift₀ ∈ (0,1)such thatr₁(t₀) ≥ M² then0 = r⁰₁(t₀) = 2hx(t₀), x⁰(t₀)iand 0≥r₁⁰⁰(t₀)with a contradiction arising by (2.8) as in the proof of Theorem2.3.

Hence we havekx(t₀)k< M for allt₀ ∈(0,1).

Consider solutions x ∈ C²([0,1];Rⁿ) to (2.22) with kx(t)k ≤ M for t ∈ [0,1].We now show that (2.20) and (2.9) implykx⁰(t)k< N for allt ∈[0,1].

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Argue by contradiction by assumingr(t₀) = kx⁰(t₀)k² −N² ≥ 0for some t0 ∈ [0,1]such that maxt∈[0,1]r(t) = r(t0). By (2.20) we have r(0) < 0 and r(1) < 0in a similar fashion to the argument in the proof of Theorem2.2. So we see that t₀ ∈ (0,1). Now sincer(1) < 0 and r(t₀) ≥ 0 we must have a pointt1 ∈ [t0,1)such thatr(t1) = 0and0≥ r⁰(t1)with a contradiction being reached as in the proof of Theorem2.3.

Hence we havekx⁰(t)k< N fort∈[0,1].

Since a priori bounds are now obtained onxandx⁰, the a priori bound on x⁰⁰naturally follows as in the proof of Theorem2.2.

Hence, by Theorem2.1, the family (2.15)–(2.17) has a solution for λ = 1, which is just the BVP (1.2), (1.4) and hence the result follows.

Remark 1. Theorem2.4may be generalised to treat (1.2) subject to x⁰(0) =g₃(x(0), x⁰(0), x(1), x⁰(1))

(2.25)

x⁰(1) =g₄(x(0), x⁰(0), x(1), x⁰(1)) (2.26)

in the following way.

Letf,g3 andg4be continuous and letM,N be positive constants. Suppose eachg_i(h, i, j, k)is bounded on the setD, where

D:={(h, i, j, k)∈R⁴ⁿ :khk ≤M, kjk ≤M, (i, k)∈R²ⁿ} and

(2.27) N >max (

sup kg (h, i, j, k)k, sup kg (h, i, j, k)k )

.

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If (2.8), (2.9) hold and

hh, g₃(h, i, j, k)i>0, for h6= 0, (i, j, k)∈R³ⁿ hj, g4(h, i, j, k)i<0, for j 6= 0, (h, i, k)∈R³ⁿ. then the BVP (1.2), (2.25), (2.26) has at least one solution.

Remark 2. It is also clear that by combining the relevant bounding inequalities used in each of the Theorems in this section, the treatment of (1.2) subject to any of the following boundary conditions is possible:

x⁰(0) = 0, γx(1) +δx⁰(1) =D, γ/δ <0, αx(0)−βx⁰(0) =C, x⁰(0) = 0, α/β >0,

x⁰(0) = 0, x⁰(1) =g₂(x(0)), x⁰(0) =g₁(x(0)), x⁰(1) = 0, and so on.

Remark 3. In Theorems2.2–2.4the inequality (2.9) could be reversed and the existence theorems would still hold. However, for brevity we omit the statement of these new results.

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3. On BVPs with Singularities

In this final section we consider systems of BVPs that may have singularities in the right-hand side. Consider

(3.1) x⁰⁰=η(t)f(t, x, x⁰), t∈[0,1],

subject to any of the boundary conditions (1.3)–(1.5). Here 1/η : [0,1] → [0,∞)is continuous with η > 0on (0,1), η is integrable on[0,1]and η may be singular at t = 0 or at t = 1. Probably the most famous type of BVP involving singularities in the right-hand side is the Thomas-Fermi equation, (n(t) = 1/√

t, f(t, x, x⁰) = x^3/2) which appears in the study of electron distri- bution in an atom [13].

In view of the proof of [17, Theorem 1.5] and [17, Theorem 0.1], in order to prove the existence of solutions to (3.1) subject to (2.2), (2.3), it is sufficient to show that:

(i) all solutions to

a(t)x⁰⁰+b(t)x⁰+c(t)x=λη(t)g(t, x, x⁰), t ∈[0,1], (3.2)

a₀x(0) +a₁x⁰(0) +a₂x(1) +a₃x⁰(1) = λψ₁(x(0), x⁰(0), x(1), x⁰(1)), (3.3)

b₀x(0) +b₁x⁰(0) +b₂x(1) +b₃x⁰(1) = λψ₂(x(0), x⁰(0), x(1), x⁰(1)), (3.4)

satisfy

max{kx(t)k,kx⁰(t)k}< R, t ∈[0,1], for someR >0, independent ofλ∈(0,1]; and

(ii) that forλ = 0, the family (3.2)–(3.4) has only the zero solution. Then, for

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(This solution will also be inC²((0,1);Rⁿ)because solutions to (3.2) are abso- lutely continuous on[0,1]and satisfy (3.2) almost everywhere.) Above,g and eachψ_i are as in Section2.

There are only minor modifications needed in the proofs of Section 2 to obtain the a priori bounds on solutions to (3.1) and therefore we only present the statement of the new theorems for brevity.

Theorem 3.1. Letηbe as above and letf be continuous. LetM, N be positive constants satisfying (2.6) If (2.7), (2.8) (2.9) hold then (3.1), (1.4) has at least one solution.

Theorem 3.2. Letηbe as above and letf be continuous. LetM, N be positive constants satisfying (2.14). If (2.8) and (2.9) hold then (3.1), (1.5) has at least one solution.

Theorem 3.3. Letηbe as above and letf, g₁ andg₂ be continuous. LetM, N be positive constants such (2.20) holds. If (2.21) holds and (2.8), (2.9) hold, then (3.1), (1.3) has at least one solution.

Example 3.1. Let f and the boundary conditions be defined as in Example 2.1. Consider η(t) = 1/√

t. Then by Theorem 3.1 the singular BVP under consideration has at least one solutionx∈C¹([0,1];Rⁿ)∩C²((0,1);Rⁿ).

Example 3.2. Let x = (x₁, x₂) and p = (p₁, p₂). Consider (3.1), (1.4) for n = 2, whereη(t) = 1/√

t

f(t, x, p) =f(t, x₁, x₂, p₁, p₂)

= (x₁+p₁)k(t, x₁, x₂, p₁, p₂) (x₂+p₂)k(t, x₁, x₂, p₁, p₂)

!

, t∈[0,1].

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There is no growth condition applicable to f and thus the theorems of [11], [21], [4], [19], [20] do not apply.

We claim that the singular BVP has a solution ifkis continuous and satisfies k(t, x₁, x₂, p₁, p₂)>0, for all(t, x₁, x₂, p₁, p₂)∈[0,1]×R²×R², for someM ≤N such that (2.6) and (2.7) hold.

We will apply Theorem3.1.

Firstly, forkxk ≥M andhx, pi= 0,consider hx, f(t, x, p)i=k(t, x₁, x₂, p₁, p₂)

x²₁+x²₂+x₁p₁ +x₂p₂

=k(t, x₁, x₂, p₁, p₂)[x²₁+x²₂] (sincex₁p₁ =−x₂p₂)

>0 for any positive choice ofM.

Thus (2.8) holds. Now, forkxk ≤M, kpk=N we have hp, f(t, x, p)i=k(t, x₁, x₂, p₁, p₂)

p₁x₁+p₂x₂+p²₁+p²₂

≥k(t, x₁, x₂, p₁, p₂)

N²+p₁x₁+p₂x₂

>0 for kxk ≤M, kpk=N, for any choice ofN such thatN ≥M, thus (2.9) holds.

Thus by Theorem3.1the singular BVP under consideration has at least one solutionx∈C¹([0,1];Rⁿ)∩C²((0,1);Rⁿ).

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References

[1] K. AKÔ, Subfunctions for ordinary differential equations. V., Funkcial.

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[9] A. GRANAS, R. GUENTHER AND J. LEE, Nonlinear boundary value problems for some classes of ordinary differential equations, Rocky Moun- tain J. Math., 10(1) (1980), 35–58.

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