Positive Solutions For A Singular Second Order Boundary Value Problem ∗

Positive Solutions For A Singular Second Order Boundary Value Problem ^∗

Wen-Shu Zhou

Received 28 January 2008

Abstract

In this paper we obtain sufficient conditions of existence of positive solutions for a singular second order boundary value problem. Our argument is based on regularization technique, upper and lower solutions method and the Arzel´a-Ascoli theorem.

1 Introduction

In [1], Bertsch and Ughi investigated the following BVP which arises in study of a class of degenerate parabolic equations (also see [2, 3]):







u⁰⁰+N−1

t u⁰−γ|u⁰|²

u + 1 = 0, 0< t <1, u(1) =u⁰(0) = 0,

(1)

whereN is a positive integer andγ >0, and obtained one decreasing positive solution via theories of ordinary differential equation. In the very recent paper [4], the authors considered the following BVP:





 u⁰⁰+λ

tu⁰−γ|u⁰|²

u +f(t) = 0, 0< t <1, u(1) =u⁰(0) = 0,

(2)

and proved, by the classical method of elliptic regularization, that BVP (2) has one positive solution which is not decreasing in the case: λ >0, γ > ^1+λ₂ , f ∈C[0,1] and f >0 on [0,1].

This paper considers the more general problem:







u⁰⁰+λu⁰

t^m −γ|u⁰|²

u^p +f(t) = 0, 0< t <1, u(1) =u⁰(0) = 0,

(3)

∗Mathematics Subject Classifications: 34B18

†Department of Mathematics, Dalian Nationalities University, 116600, P.R. China

154

whereλ, m, γ, p >0,f(t)∈C[0,1] andf(t)>0 on [0,1]. By a solution to BVP (3) we mean a functionu∈C²(0,1)∩C¹[0,1] which is positive in (0, 1) and satisfies (3). By an argument based on the regularization technique, upper and lower solutions and the Arzel´a-Ascoli theorem, we obtain sufficient conditions of existence of solutions. Our main result reads

THEOREM 1. Let λ∈ (0,+∞), p∈ [1,2), m∈ (0, p/(2−p)], and let f ∈C[0,1]

and f(t)>0 on [0,1]. Ifγ >inf

t>1G(t), whereG(t) :R⁺→R⁺ is defined by G(t) =p+λ(2−p)

2 t^p−1+(2−p)²max[0,1]f

4 t^p−2,

then BVP (3) has at least one solution.

REMARK 1. Ifp= 1, then inf

t>1G(t) = ^1+λ₂ . Clearly, Theorem 1 is an extension of the existence results of [1, 4].

REMARK 2. Letp∈(1,2), and denote T0= (2−p)³max[0,1]f

2(p−1)[p+λ(2−p)], T∗=

(T0, T0>1, 1, T0<1.

Then inf

t>1G(t) =G(T∗).Indeed, since lim

t→0⁺G(t) = lim

t→+∞G(t) = +∞,G(t) must reach a minimum at some pointt∈(0,∞) such thatG⁰(t) = 0, and then, solving this equation yields t=T0 and hence, inf

t>0G(t) =G(T0). SinceG⁰(t)>0 for all t>T0, we see that

t>1infG(t) = inf

t>0G(t) =G(T0) ifT0>1, and inf

t>1G(t) =G(1) ifT0<1.

2 Proof of Theorem 1

Let∈(0,1), and defineH(t, v, ξ) : (0,1)×R×R→Rby H(t, v, ξ) =−λ ξ

(t+^1/α)^m+γ |ξ|²

[I(v)]^p −f(t), where α= _2−p² , andI(v) =v+² ifv>0,I(v) =² ifv <0. We have

|H(t, v, ξ)|6 λ

^m/α|ξ|+γ|ξ|^{2 2p} + max

[0,1]f

6 λ

^m/α(1 +|ξ|²) + γ

^2p|ξ|²+ max

[0,1] f 6

λ ^m/α + γ

^2p + max

[0,1] f H(|ξ|)

(4)

for all (t, v, ξ) ∈ (0,1)×R×R, where H(s) = 1 +s² for s > 0. Define operator L :C²(0,1)→C(0,1) by

(Lu)(t) =−u⁰⁰+H(t, u, u⁰), 0< t <1.

Consider the problem:

((Lu)(t) = 0, 0< t <1,

u(1) =u(0) = 0. (5)

We calluan upper solution (lower solution) of problem (5) ifLu>(6)0 in (0,1), and u(t)>(6)0 fort= 0,1.

We will apply the upper and lower solutions method (see [5, pp.153, Theorem 2.5.4]

or [6, Theorem 1 and Remark 2.4]) to obtain one positive solution of problem (5). Note thatR+∞

0 s

H(s)ds= +∞.Then it suffices to find a lower solution and an upper solution to obtain a solution.

LEMMA 1. LetU =C1W^α with α= _2−p² , where W(t) =t(1−t) andC1∈(0,1) such that 2C1α+C1αλ2^α−1−m+γC₁^2−pα²6min

[0,1]f(t). ThenU is a lower solution of problem (5).

PROOF. Note thatW⁰⁰=−2, W 6tand|W⁰|61 on [0,1]. Since U >0 in (0,1), some calculations give by noticingα>1 +m

LU=−U⁰⁰−λ U⁰

(t+^1/α)^m +γ |U⁰|²

(U+²)^p −f(t) 6−U⁰⁰−λ U⁰

(t+^1/α)^m +γ|U⁰|² U^p −f(t)

= 2C1αW^α−1−C1α(α−1)W^α−2|W⁰|² +C1αλ W^α−1W⁰

(t+^1/α)^m +γC₁^2−pα²|W⁰|²−f(t) 62C1αW^α−1+C1αλ W^α−1W⁰

(t+^1/α)^m+γC₁^2−pα²|W⁰|²−f(t) 62C1α+C1αλ(t+^1/α)^α−1−m+γC₁^2−pα²−f(t) 62C1α+C1αλ2^α−1−m+γC₁^2−pα²−min

[0,1]f(t) 60, 0< t <1.

Thus, U is a lower solution of problem (5). The lemma follows.

Let inf

s>1H(s)≡ δ. Then it follows from the definition of infimum and γ > δ that forδ0=^γ−δ₂ >0, there existsC∗>1, such thatH(C∗)< δ+δ0< γ.

LEMMA 2. There exists a positive constant0∈(0,1), such that for any∈(0, 0), V=C∗(t+^1α)^αis an upper solution of problem (5).

PROOF. Noticingα>2 and 1 +m6α, we have LV=−V⁰⁰−λ V⁰

(t+^1/α)^m+γ |V⁰|²

(V+²)^p −f(t)

=−C∗α(α−1)(t+^1/α)^α−2−λαC∗(t+^1/α)^α−1−m +γC_∗^2−pα²[1 +²C_∗⁻¹(t+^1/α)^−α]^−p−f(t)

>−C∗α(α−1)[1 +^1/α]^α−2−λαC∗[1 +^1/α]^α−1−m +γC_∗^2−pα²[1 +C_∗⁻¹]^−p−max

[0,1] f(s)

=γC_∗^2−pα²−C∗α(α−1)−λαC∗−max

[0,1] f(s) +e

=C_∗^2−pα²(γ− G(C∗)) +e, 0< t <1,

wheree=C∗α(α−1)[1−(1+^1/α)^α−2]+λαC∗[1−(1+^1/α)^α−1−m]+[1+C_∗⁻¹]^−p−1.

Clearly,e→0, (→0).Sinceγ >G(C∗), there exists 0∈(0,1) such that C_∗^2−pα²(γ− G(C∗)) +e>0.

This shows that for any∈(0, 0), LV>0,0< t <1.The lemma follows.

According to [5, pp.153, Theorem 2.5.4] or [6, Theorem 1 and Remark 2.4], for any fixed∈(0, 0), problem (5) has a solutionu∈C¹[0,1] satisfyingu⁰∈C¹(0,1) and

V>u>U >0, t∈(0,1). (6) Hence u satisfies

u⁰⁰+λ u⁰

(t+^1/α)^m−γ |u⁰|²

(u+²)^p+f(t) = 0, 0< t <1. (7) LEMMA 3. There exists a positive constantC2 independent of, such that for all ∈(0, 0)

|u⁰(t)|6C2, t∈[0,1]. (8) PROOF. It follows fromu(1) =u(0) = 0 andu>0 for allt∈[0,1] that

u⁰(0)>0>u⁰(1). (9) Integrating (7) over (0, 1) and integrating by parts give

u⁰(t)

1

0+ λu(t) (t+^1/α)^m

1

0

+mλ Z 1

0

u

(t+^1/α)^1+mdt

−γ Z 1

0

|u⁰|² (u+²)^pdt+

Z 1 0

f(t)dt= 0, and then, we obtain by (9)

γ Z 1

0

|u⁰|²

(u+²)^pdt6 λu(t) (t+^1/α)^m

1

0

+mλ Z 1

0

u

(t+^1/α)^1+mdt+ Z 1

0

f(t)dt.

Since m 6 _2−p^p , 1 +m 6 α = _2−p² . From (6), it is easy to see that _(t+^λu1/α^(t))^m

1 0+ mλR1

0 u

(t+^1/α)^1+mdt is uniformly bounded and hence, there exists a positive constant C3 independent of, such that

Z 1 0

|u⁰|²

(u+²)^pdt6C3. (10)

By the inequality: a6a²+ 1 (a∈R), we obtain

|u⁰|

(t+^1/α)^m 6 |u⁰|²

(t+^1/α)^2m+ 1, t∈[0,1]. (11) By (6), we haveu+² 62C∗(t+^1/α)^α, t∈[0,1].Noticingαp>2m, we see that there exists a positive constant C4independent of, such that

(u+²)^p6C4(t+^1/α)^2m, t∈[0,1].

Combining this and (11) we obtain

|u⁰|

(t+^1/α)^m 6C4

|u⁰|²

(u+²)^p + 1, t∈[0,1], which and (10) imply that

Z 1 0

|u⁰|

(t+^1/α)^mdt6C3C4+ 1≡C5. (12) On the other hand, integrating (7) over (t1, t2), we have

u⁰(t)

t2

t1

=−λ Z t2

t1

u⁰

(t+^1/α)^mdt+γ Z t2

t1

|u⁰|² (u+²)^pdt−

Z t2

t1

f(t)dt.

Combining this with (10) and (12) we obtain for all∈(0, 0)

|u⁰(t2)−u⁰(t1)|6C6, ∀t1, t2∈[0,1], (13) where C6 =λC5+γC3+R1

0 f(t)dt. Noticing u(1) =u(0) = 0 and using the mean value theorem, there exists t ∈(0,1), such that u⁰(t) = 0. Then takingt1 =t in (13), we obtain the desired result.

By (6) and (8), we derive from (7) that there exists for anyδ∈(0,1/2) a positive constant Cδ independent of, such that for all∈(0, 0)

|u⁰⁰(t)|6Cδ, δ6t61−δ.

From this and (8) and using Arzel´a-Ascoli theorem, there exist a subsequence of{u}, still denoted by{u}, and a functionu∈C¹(0,1)∩C[0,1] such that, as→0,

u→u, uniformly in C[0,1], u→u, uniformly in C¹[δ,1−δ],

and hence, by u(1) =u(0) = 0 and (6), usatisfies u(1) =u(0) = 0,C∗t^α >u(t)>

C1[t(1−t)]^αfor allt∈[0,1], thereforeu(t)>0 for allt∈(0,1), andu⁰(0) = lim

t→0 u(t)

t = 0. Then usatisfies the boundary conditions in (3).

Below, we show thatusatisfies the equation in (3). Integrating (7) over [t0, t] yields u⁰(t) =γ

Z t t0

|u⁰|²

(u+²)^pds−λ Z t

t0

u⁰

(s+^1/α)^mds− Z t

t0

f(s)ds+u⁰(t0), and letting→0 and using Lebesgue dominated convergence theorem, we have

u⁰(t) =γ Z t

t0

|u⁰|² u^p ds−λ

Z t t0

u⁰ s^mds−

Z t t0

f(s)ds+u⁰(t0), 0< t <1. (14) From this, we see that u∈C²(0,1) and satisfies the equation in (3).

It remains to show thatu⁰ is continuous att= 0 andt= 1. Letting→0 in (10) and (12) and using Fatou’s Lemma, we have

Z 1 0

|u⁰|²

u^p dt6C3,

Z 1 0

|u⁰|

t^mdt6C5,

which show that ^|u_u⁰p^|2,^|u_tm^0| ∈ L¹[0,1]. By the absolute continuity of integral, we see from (14) that u⁰ ∈C[0,1]. Theorem 1 is proved.

Acknowledgment.The author wants to thank the referee for his important com- ments which improve this paper. This research is supported by Dalian Nationalities University (no.20076209).

References

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