positive and maximal positive solutions of singular mixed boundary

Cent. Eur. J. Math. • 7(4) • 2009 • 694-716DOI: 10.2478/s11533-009-0049-9

Central European Journal of Mathematics

Positive and maximal positive solutions of singularmixed boundary value problem

Research Article

Ravi P. Agarwal12∗, Donal O’Regan3† , Svatoslav Stanek4‡

1 Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, Florida, USA

2 KFUPM Chair Professor, Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals, Dhahran31261, Saudi Arabia

3 Department of Mathematics, National University of Ireland, Galway, Ireland

4 Department of Mathematical Analysis, Faculty of Science, Palacký University, Olomouc, Czech Republic

Received 29 January 2009; accepted 2 September 2009

Abstract: The paper is concerned with existence results for positive solutions and maximal positive solutions of singularmixed boundary value problems. Nonlinearities h(t, x, y) in differential equations admit a time singularity att = 0 and/or at t = T and a strong singularity at x = 0.

MSC: 34B16, 34B18

Keywords: Singular mixed problem • Positive solution • Maximal positive solution • Time singularity • Space singularity •Lower and upper functions

© Versita Warsaw and Springer-Verlag Berlin Heidelberg.

1. Introduction

Let T > 0 and a ∈ [0, T ). Throughout the paper AC 1[a, T ] denotes the set of functions having absolutely continuousderivative on [a, T ], L1[0, T ] stands for the set of Lebesgue integrable functions on [0, T ] and Lloc [a, T ) denotes the set ofLebesgue integrable functions on any compact interval I ⊂ [a, T ). Finally, q ∈ Car((0, T ) × D ), D = (0, ∞) × R, meansthat for each compact interval J ⊂ (0, T ) and each compact set B ⊂ D there exists a function φJ,B ∈ L1(J) such that

(a) q(·, x, y) : J → R is measurable on J for all (x, y) ∈ B ,

(b) q(t, ·, ·) : B → R is continuous on B for a.e. t ∈ J,

∗ E-mail: [email protected]† E-mail: [email protected]‡ E-mail: [email protected]

694

R.P. Agarwal, D. O’Regan, S. Stanek

(c) |q(t, x, y)| ≤ φJ,B (t) for a.e. t ∈ J and all (x, y) ∈ B .

We say that q(t, x, y) ∈ Car((0, T ) × D ) has a time singularity at t = 0 (t = T ) if there exists (x, y) ∈ D such that

∫ ε

0|q(t, x, y)| dt = ∞

(∫ T

T−ε|q(t, x, y)| dt = ∞

)for all sufficiently small ε > 0.If limx→0+ |g(t, x, y)| = ∞ for a.e. t ∈ [0, T ] and some y ∈ R, then we say that x = 0 is a singular point of q or wealso say that q has a space singularity at x = 0. We distinguish two types of space singularities, namely a weak spacesingularity and a strong space singularity. We say that x = 0 is a weak (strong) space singularity of q if there existsy ∈ R such that for a.e. t ∈ [0, T ] and all sufficiently small ρ > 0 the relation

∫ ρ

0|q(t, x, y)| dx < ∞

(∫ ρ

0|q(t, x, y)| dx = ∞

)holds. For time and space singularities see [22–24]. A simple example of a function q having a time singularity att = 0, T and a strong space singularity at x = 0 is

q(t, x, y) = ytα (T − t)β + 1

xγ , α, β, γ ∈ [1, ∞).

If γ ∈ (0, 1), then q has a weak space singularity at x = 0.Singular mixed problems arose for example when searching for positive, radially symmetric solutions of the nonlinearelliptic partial differential equations

∆u + g(r, u) = 0 on Ω, u∣∣∣∂Ω

= 0,

where Ω is the open disc in Rn centered at the origin, and r is the radial distance from the origin. Radially symmetricsolutions of the above problem are solutions of the singular mixed problem (see, e.g., [6, 13, 16])

u′′ − n − 11 − t u′ + g(1 − t, u) = 0, u(0) = 0, u′(1) = 0.

In [8] the singular mixed problem

u′′ = n − 11 − t u′ + γ |u′|2

u − 1,

u(0) = 0, u′(1) = 0,

is considered, where the differential equation has a time singularity at t = 1 and a strong space singularity at x = 0.The authors proved the existence of an increasing solution. This result was used in [7] to study some properties ofsolutions for a class of degenerate parabolic equations.For the solvability of the singular mixed boundary value problem

u′′ = h(t, u, u′), u(0) = 0, u′(T ) = 0,

in the set C 0[0, T ] ∩ AC 1loc(0, T ] (AC 1

loc(0, T ] denotes the set of functions having absolutely continuous derivative on anycompact interval in (0, T ]) we refer to [2, 17, 19, 21, 27, 28]. The existence of solutions u ∈ AC 1[0, T ] to the above problemcan be found, e.g., in [1, 3, 4, 17, 22]. Note that papers [3, 4, 17] deal with problem allowing just space singularities butnot time ones, whereas papers [1, 22] consider both time and space singularities.We discuss the singular mixed boundary value problem

u′′ = p(u′)[f(t, u, u′) − r(t)], (1)

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Positive and maximal positive solutions of singular mixed boundary value problem

u(0) = 0, u′(T ) = 0, (2)

where p ∈ C 0(−a, a), r ∈ L1[0, T ], f ∈ Car((0, T ) × D ), D = (0, ∞) × R, 0 < a ≤ ∞, are positive and f(t, x, y) may havesingularities at t = 0, t = T and x = 0.We say that a function u ∈ AC 1[0, T ] is a positive solution of problem (1),(2) if u > 0 on (0, T ], u fulfils (2) andu′′(t) = p(u′(t))[f(t, u(t), u′(t)) − r(t)] for a.e. t ∈ [0, T ].The first aim of this paper is to give conditions on the functions p, f and r in equation (1) which guarantee the existenceof a positive solution u of problem (1), (2).We work with the following conditions on the functions p, f and r in (1):

(H1) p ∈ C 0(−a, a) is positive, 0 < a ≤ ∞,

(H2) f ∈ Car((0, T ) × D ), D = (0, ∞) × R, and

0 ≤ f(t, x, y) ≤ A(

xη0 + |y|γ0

tµ0+ |y|γ1

(T − t)µ1+ |y|γ

xη

)+ h(x, y) (3)

for a.e. t ∈ [0, T ] and all (x, y) ∈ D , where h ∈ C 0([0, ∞) × R) is nonnegative, h(x, 0) = 0 for x ∈ [0, ∞), andA, η0, µi, γi, γ and η are positive constants, µ0 < 2η0, µi ≤ γi, 2η ≤ γ (i = 0, 1),

(H3) r ∈ L1[0, T ] and there exists r∗ ∈ (0, ∞) such that

r(t) ≥ r∗ for a.e. t ∈ [0, T ]

and

min∫ 0

−a

dsp(s) ,

∫ a

0

dsp(s)

>∫ T

0r(t) dt,

(H4) if µ1 ≥ 1 in (H2), then there exist ν ∈ (0, T ) and ω ∈ C 0[0, a), ω(0) = 0, ω > 0 on (0, a) such that for a.e. t ∈ [ν, T ]and all x ∈ (0, 1 + aT ), y ∈ (−a, a), the estimate

f(t, x, y) ≥ ω(|y|)(T − t)µ1

(4)

is fulfilled.

Remark 1.1.If µ1 ∈ (0, 1) in (H2), then f ∈ Car((0, T ] × D ), which follows from inequality (3).

Remark 1.2.Under assumption (H1) the differential equation (1) can be written in the form

(φ(u′))′ = f(t, u, u′) − r(t),

where φ(y) =∫ y

0ds

p(s) for y ∈ (−a, a).

It follows from assumption (H2) that f(t, x, y) admits a time singularity at t = 0 and/or at t = T and a strong spacesingularity at x = 0. Hence problem (1), (2) admits both time and strong space singularities. We note that many papersdealing with mixed problems use explicitly the existence of lower and upper functions in assumptions. In contrast tothose papers we construct lower and upper functions using conditions (H1) − (H3). We show that conditions (H1) − (H4)guarantee the existence of a positive solution of problem (1), (2). The existence result (Theorem 3.1) is proved bya combination of regularization and sequential techniques with the method of lower and upper functions (see, e.g.,[19, 25, 26]). In limit processes the Fatou lemma is used ([5, 20]).

696


The second aim of this paper is to discuss the existence of the maximal positive solution of problem (1), (2) in theset AC 1[0, T ]. We say that a solution u ∈ AC 1[0, T ] of problem (1), (2) is its maximal positive solution if u ≥ u on[0, T ] for any positive solution u of problem (1), (2). In the literature extremal solutions (that is, maximal and minimalsolutions) are studied for regular and discontinuous boundary value problems, and extremal solutions usually ”lie” ina sector bounded by some functions α and β (as a rule α and β are lower and upper functions of the problems underdiscussion). The existence of extremal solutions is proved by monotone iterative techniques (see, e.g., [10]-[12], [14]), anapproximation method ([9, 11]), and a technique which works with ordered normed spaces and assumes that the set of allsolutions is upward and downward directed ([15]). In this paper we prove the existence of the maximal positive solutionof problem (1), (2) by a method which is a combination of the method of generalized lower and upper functions ([19]) withregularization and sequential techniques. Unfortunately, we are not able to prove the existence of a minimal positivesolution of problem (1), (2).The rest of the paper is organized as follows. Section 2 deals with a sequence of auxiliary regular mixed problems toproblem (1), (2). In Section 3, the main existence result for problem (1), (2) is proved by the results in Section 2. Section 4is devoted to the study of the maximal positive solution of problem (1), (2). The results are demonstrated with examples.

2. Auxiliary regular problems

Suppose (H1) − (H3). Then there exists S ∈ (0, a) such that

min∫ 0

−S

dsp(s) ,

∫ S

0

dsp(s)

>∫ T

0r(t) dt. (5)

By means of S define p, χ ∈ C 0(R) by the formulas

p(y) =

p(S) for y > S,p(y) for |y| ≤ S,p(−S) for y < −S,

χ(y) =

1 for |y| ≤ S,2 − |y|

S for S < |y| ≤ 2S,0 for |y| > 2S.

Let ε∗ be a positive constant such that

ε∗ < min

1, T4 , 2η0−µ0

√r∗

4A

, (6)

where constants A, r∗, η0 and µ0 are taken from conditions (H2) and (H3). For each ε ∈ (0, ε∗) define σε : R → [ε2, ∞)by the formula

σε(x) =

1 + ST for x > 1 + ST ,x for ε2 < x ≤ 1 + ST ,ε2 for x ≤ ε2.

Let tn ⊂ (T − ε∗, T ) be an increasing sequence, limn→∞ tn = T , and let inequality (3) hold for t = tn, (x, y) ∈ D andn ∈ N. Set

εn = T − tn for n ∈ N. (7)

Then εn ⊂ (0, ε∗) ⊂ (0, T4 ) and limn→∞ εn = 0. Put

fn(t, x, y) =

f(t, x, y) for t ∈ (0, tn), (x, y) ∈ D ,

f(tn, x, y) for t ∈ [tn, T ], (x, y) ∈ D .

Then fn ∈ Car((0, T ] × D ) and (H2) gives

0 ≤ fn(t, x, y)

≤ A[

xη0 + |y|γ0

(mint, tn)µ0+ |y|γ1

(maxT − t, εn)µ1+ |y|γ

xη

]+ h(x, y) (8)

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for a.e. t ∈ [0, T ] and all (x, y) ∈ D . Consider the regular mixed problem

u′′ = χ(u′)p(u′)[fn(t, σεn (u), u′) − r(t)], (9)

u(εn) = ε2n, u′(T ) = 0, (10)

on the interval [εn, T ].In order to prove the existence of a solution of problem (9), (10), we use the method of lower and upper functions. Wesay that α ∈ AC 1[εn, T ] is a lower function of problem (9), (10) if α ′′(t) ≥ χ(α ′(t))p(α ′(t))[fn(t, σεn (α(t)), α ′(t)) − r(t)] fora.e. t ∈ [εn, T ] and α(εn) ≤ ε2

n, α ′(T ) ≤ 0. Similarly, α ∈ AC 1[εn, T ] is an upper function of problem (9), (10) if thereverse inequalities are satisfied.The following result follows from [19, Lemma 3.7].

Lemma 2.1.Let conditions (H1) − (H3) hold. Let α and β be lower and upper function of problem (9), (10), α(t) ≤ β(t) for t ∈ [εn, T ].Let there exist q ∈ L1[εn, T ] such that

|χ(y)p(y)[fn(t, σεn (x), y) − r(t)]| ≤ q(t)

for a.e. t ∈ [εn, T ] and all α(t) ≤ x ≤ β(t), y ∈ R. Then problem (9), (10) has a solution un ∈ AC 1[εn, T ] such that

α(t) ≤ un(t) ≤ β(t) for t ∈ [εn, T ].

The following result gives a priori estimates for solutions of problem (9), (10).

Lemma 2.2.Let conditions (H1) − (H3) hold. Let un ∈ AC 1[εn, T ] be a solution of problem (9), (10). Then

ε2n ≤ un(t) < 1 + ST for t ∈ [εn, T ], (11)

|u′n(t)| < S for t ∈ [εn, T ], (12)

where S is a positive constant satisfying inequality (5).

Proof. Suppose that minun(t) : εn ≤ t ≤ T = un(ξ) < ε2n. Then ξ ∈ (εn, T ] and u′

n(ξ) = 0, which is clear forξ < T and it follows from (10) if ξ = T . Due to un(εn) = ε2

n, there exists τ ∈ (εn, ξ) such that un < ε2n on [τ, ξ ]. Now,

let us choose ρ ∈ (0, S) such that the inequalities

ργ0

εµ0n

+ ργ1

εµ1n

+ ργ

ε2ηn

< r∗

4A, (13)

h(ε2n, y) < r∗

4 for y ∈ [−ρ, ρ] (14)

are satisfied. In addition, since u′n(ξ) = 0, there exists τ1 ∈ [τ, ξ ] such that

|u′n(t)| ≤ ρ for t ∈ [τ1, ξ ]. (15)

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Then, by condition (H3) and inequalities (6), (8) and (13)–(15), we have χ(u′n(t)) = 1, σεn (un(t)) = ε2

n and

u′′n(t) = χ(u′

n(t))p(u′n(t))[fn(t, σεn (un(t)), u′

n(t)) − r(t)]

≤ p(u′n(t))

[A(

ε2η0n + |u′

n(t)|γ0

(mint, tn)µ0+ |u′

n(t)|γ1

(maxT − t, εn)µ1+ |u′

n(t)|γ

ε2ηn

)+ h(ε2

n, u′n(t)) − r(t)

]

< p(u′n(t))

[A(

ε2η0n + ργ0

εµ0n

+ ργ1

εµ1n

+ ργ

ε2ηn

)+ r∗

4 − r∗

]< p(u′

n(t))(

3r∗

4 − r∗

)= − p(u′

n(t))r∗

4

for a.e. t ∈ [τ1, ξ ]. In particular, u′′n(t) < − p(u′

n(t))r∗4 almost everywhere on [τ1, ξ ]. Hence

u′n(t) = −

∫ ξ

tu′′

n(s)ds > r∗

4

∫ ξ

tp(u′

n(s)) ds > 0, t ∈ [τ1, ξ).

Consequently, u′n(t) > 0 for t ∈ [τ1, ξ), contrary to minun(t) : t ∈ [εn, T ] = un(ξ). Therefore un ≥ ε2

n on [εn, T ] andu′

n(εn) ≥ 0 because un(εn) = ε2n.

We next prove inequality (12). Since f is nonnegative by (H2), we have

u′′n(t)

p(u′n(t)) ≥ −χ(u′

n(t))r(t) ≥ −r(t),

and therefore, (∫ u′n(t)

0

dsp(s) +

∫ t

0r(s) ds

)′

≥ 0 for a.e. t ∈ [εn, T ].

Hence the function∫ u′

n(t)0

dsp(s) +

∫ t0 r(s) ds is nondecreasing on [εn, T ]. From this and from u′

n(T ) = 0, u′n(εn) ≥ 0 we obtain

0 <∫ u′

n(εn)

0

dsp(s) +

∫ εn

0r(s) ds ≤

∫ u′n(t)

0

dsp(s) +

∫ t

0r(s) ds ≤

∫ T

0r(t) dt

for t ∈ [εn, T ]. Therefore,

−∫ T

0r(t) dt <

∫ u′n(t)

0

dsp(s) <

∫ T

0r(t) dt, t ∈ [εn, T ].

If u′n(ν) ≥ S for some ν ∈ [εn, T ], then

∫ T

0r(t) dt >

∫ u′n(ν)

0

dsp(s) ≥

∫ S

0

dsp(s) =

∫ S

0

dsp(s) ,

contrary to (5). Similarly, u′n(ν1) ≤ −S for some ν1 ∈ [εn, T ] leads to a contradiction. To summarize, we have verified

that |u′n(t)| < S for t ∈ [εn, T ].

Finally, it follows from (12) and from un(εn) = ε2n that

un(t) = ε2n +

∫ t

εn

u′n(s) ds < 1 + S(t − εn) < 1 + ST

for t ∈ [εn, T ]. In particular, un < 1 + ST on [εn, T ].

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Remark 2.1.Inequalities (11) and (12) show that any solution un of problem (9), (10) fulfils the equalities σεn (un(t)) = un(t) andχ(u′

n(t)) = 1 on [εn, T ]. Consequently, any solution un of the above problem is a solution of the differential equation

u′′ = p(u′)[fn(t, u, u′) − r(t)] (16)

on the interval [εn, T ].

The existence of a lower function of problem (9), (10) is given in the next lemma. We note that εn = T − tn by (7) andεn ∈ (0, ε∗) ⊂ (0, T

4 ) for n ∈ N, where ε∗ > 0 satisfies inequality (6).

Lemma 2.3.Let (H1) − (H3) hold. Then there exist n∗ ∈ N and a positive constant C∗ ∈ (0, 1) such that the function

αn(t) =

C∗[(t − εn)(2T − t − 3εn) + εn]2 for t ∈ [εn, tn],

C∗[(2tn − T )2 + T − tn]2 for t ∈ (tn, T ](17)

is a lower function of problem (9), (10) with n ≥ n∗.

Proof. Let S ∈ (0, a) satisfy inequality (5) and let

p∗ = minp(y) : −S ≤ y ≤ S, p∗ = maxp(y) : −S ≤ y ≤ S. (18)

Define Q ∈ C 0[0, ∞) by the formula

Q(C ) = Cη0− µ02 (2T + 1)µ0 (2T 2 + 1)2η0−µ0 + [4CT (2T + 1)]µ0 Sγ0−µ0

+ [4C (2T 2 + 1)]µ1 Sγ1−µ1 + (16CT 2)ηSγ−2η, C ∈ [0, ∞).

Since Q is nonnegative and Q(0) = 0, there exists C1 ∈ (0, 1) such that

0 ≤ AQ(C ) + 4C (2T 2 + 1)p∗ ≤ p∗r∗

2p∗ for C ∈ [0, C1]. (19)

By (H2), h ∈ C 0([0, ∞) × R) is nonnegative and h(x, 0) = 0 for x ∈ [0, ∞). Therefore, there exists ρ ∈ (0, min1, S]such that

0 ≤ h(x, y) < p∗r∗

2p∗ for (x, y) ∈ [0, ρ] × [−ρ, ρ]. (20)

Put

C2 = ρ min

14T (T 2 + 1) , 1

(2T 2 + 1)2

,

C∗ = min

C1, C2,1

T 2(T + 1)2

(3T4

) µ0η0(

r∗p∗

Ap∗

) 1η0

.

Since limn→∞ εn = 0 there exists n0 ∈ N such that

εn <√

C∗(2T 2 + 1) for n ≥ n0. (21)

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Let us choose an arbitrary n ≥ n0 and let αn be given as in (17). Then αn ∈ AC 1[εn, T ], αn(εn) = C∗ε2n < ε2

n andα ′

n(T ) = 0. We now show that the inequality

α ′′n (t) − χ(α ′

n(t))p(α ′n(t))[fn(t, σεn (αn(t)), α ′

n(t)) − r(t)] ≥ 0 (22)

holds for a.e. t ∈ [εn, tn]. It follows from (17) that

0 < αn(t) < C∗(2T 2 + 1)2 ≤ ρ for t ∈ [εn, tn] (23)

α ′n(t) = 4C∗[(t − εn)(2T − t − 3εn) + εn](T − t − εn) for t ∈ [εn, tn] (24)

andα ′′

n (t) = 8C∗(T − t − εn)2 − 4C∗[(t − εn)(2T − t − 3εn) + εn].

In particular,α ′

n(t) < 4C∗T (2T 2 + 1) ≤ ρ for t ∈ [εn, tn], (25)

α ′′n (t) > −4C∗(2T 2 + 1) for t ∈ [εn, tn]. (26)

Since ε2n < C2(2T 2 + 1)2 ≤ ρ by (21), due to (23) we have σεn (αn(t)) < ρ, which together with (20) and (25) gives

h(σεn (αn(t)), α ′n(t)) < p∗r∗

2p∗ for t ∈ [εn, tn]. (27)

In view of the inequalities αn(t) < [(t − εn)(2T − t − 3εn) + εn]2 and [(t − εn)(2T − t − 3εn) + εn]2 ≥ ε2n, we have

σεn (αn(t)) < [(t − εn)(2T − t − 3εn) + εn]2 for t ∈ [εn, tn]. (28)

Next, we deduce from αn(t) < C∗(2T 2 + 1)2 and ε2n < C∗(2T 2 + 1)2 that

σεn (αn(t)) < C∗(2T 2 + 1)2 for t ∈ [εn, tn]. (29)

Using relations (24), (25), (28) and (29) we have (for t ∈ [εn, tn])

(σεn (αn(t)))η0

tµ0<(

(t − εn)(2T − t − 3εn) + εn

t

)µ0

[σεn (αn(t))]η0− µ02

< Cη0− µ02

∗

(2Tt + εn

t

)µ0

(2T 2 + 1)2η0−µ0

≤ Cη0− µ02

∗ (2T + 1)µ0 (2T 2 + 1)2η0−µ0 ,

(α ′n(t))γ0

tµ0=(

α ′n(t)t

)µ0

(α ′n(t))γ0−µ0

< (4C∗T )µ0

((t − εn)(2T − t − 3εn) + εn

t

)µ0

Sγ0−µ0

< [4C∗T (2T + 1)]µ0 Sγ0−µ0 ,

(α ′n(t))γ1

(T − t)µ1< [4C∗(2T 2 + 1)]µ1

(T − t − εn

T − t

)µ1

(α ′n(t))γ1−µ1

< [4C∗(2T 2 + 1)]µ1 Sγ1−µ1

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and, since σεn (αn(t)) ≥ αn(t),

(α ′n(t))γ

(σεn (αn(t)))η ≤ (α ′n(t))γ

(αn(t))η

< (4C∗T )2η[(t − εn)(2T − t − 3εn) + εn]2η

Cη∗ [(t − εn)(2T − t − 3εn) + εn]2η (α ′

n(t))γ−2η

≤ (16C∗T 2)ηSγ−2η.

In particular, for t ∈ [εn, tn],

(σεn (αn(t)))η0

tµ0< Cη0− µ0

2∗ (2T + 1)µ0 (2T 2 + 1)2η0−µ0 ,

(α ′n(t))γ0

tµ0< [4C∗T (2T + 1)]µ0 Sγ0−µ0 ,

(α ′n(t))γ1

(T − t)µ1< [4C∗(2T 2 + 1)]µ1 Sγ1−µ1

(α ′n(t))γ

(σεn (αn(t)))η < (16C∗T 2)ηSγ−2η.

The above inequalities together with inequalities (8), (19), (20), (26) and with the equality χ(α ′n(t)) = 1 give

α ′′n (t) − χ(α ′


n(t)) − r(t)]

> −4C∗(2T 2 + 1) − p∗A(

(σεn (αn(t)))η0 + (α ′n(t))γ0

tµ0+ (α ′

n(t))γ1

(T − t)µ1+ (α ′

n(t))γ

(σεn (αn(t)))η

)− p∗h(σεn (αn(t)), α ′

n(t)) + p∗r∗

≥ −p∗(

4C∗(2T 2 + 1)p∗ + AQ(C∗)

)− p∗r∗

2 + p∗r∗

≥ 0

for a.e. t ∈ [εn, tn]. Hence inequality (22) holds for a.e. t ∈ [εn, tn]. We have proved that for each n ≥ n0 the functionαn fulfils inequality (22) a.e. on [εn, tn].

Put n = min

n ∈ N : εn <( 3T

4) µ0

2η0(

r∗p∗Ap∗

) 12η0

and n∗ = maxn, n0. Then for each n ≥ n∗ the function αn satisfies

inequality (22) a.e. on [εn, tn]. In order to finish the proof it remains to verify that αn fulfills inequality (22) a.e. on [tn, T ]

for all n ≥ n∗. Let us choose an arbitrary n ≥ n∗. Since εn <( 3T

4) µ0

2η0(

r∗p∗Ap∗

) 12η0 and

αn(t) = C∗[(2tn − T )2 + T − tn]2 < C∗T 2(T + 1)2 ≤(

3T4

) µ0η0(

r∗p∗

Ap∗

) 1η0

, t ∈ [tn, T ],

we have

σεn (αn(t)) = maxε2n, αn(t) <

(3T4

) µ0η0(

r∗p∗

Ap∗

) 1η0

, t ∈ [tn, T ].

Consequently (we note that tn > 3T4 ),

[σεn (αn(t))]η0

tµ0n

< p∗r∗

Ap∗ , t ∈ [tn, T ].

In view of α ′n = 0 on [tn, T ], we have χ(α ′

n) = 1, h(σεn (αn), α ′n) = 0 on [tn, T ] and (see (8))

α ′′n (t) − χ(α ′


n(t)) − r(t)]

≥ −p(0)(

A [σεn (αn(t))]η0

tµ0n

+ h(σεn (αn(t)), α ′n(t)) − r(t)

)> −p∗A [σεn (αn(t))]η0

tµ0n

+ p∗r∗ > −p∗r∗ + p∗r∗ = 0

for a.e. t ∈ [tn, T ]. Hence inequality (22) holds for a.e. t ∈ [εn, T ] and all n ≥ n∗. Since αn(εn) < ε2n and α ′

n(T ) = 0 forn ≥ n∗, it follows that αn is a lower function of problem (9), (10) with n ≥ n∗.

702


The existence result for problem (16), (10) is given in the following lemma.

Lemma 2.4.Let conditions (H1) − (H3) hold and let n∗ ∈ N and C∗ ∈ (0, 1) be given as in Lemma 2.3. Then for each n ≥ n∗ thereexists a solution un ∈ AC 1[εn, T ] of problem (16), (10) satisfying inequalities (11), (12) and

un(t) ≥

C∗[(t − εn)(2T − t − 3εn) + εn]2 for t ∈ [εn, tn],C∗[(2tn − T )2 + T − tn]2 for t ∈ (tn, T ].

(30)

Proof. Let us choose an arbitrary n ≥ n∗. Then the function αn given as in (17) is a lower function of problem (9), (10)by Lemma 2.3. Put

β(t) = (2T 2 + 1)2 + 2St for t ∈ [εn, T ],

where S ∈ (0, a) satisfies inequality (5). Then β(εn) > ε2n and β′(t) = 2S for t ∈ [εn, T ]. Consequently, χ(β′) = 0 on

[εn, T ] andβ′′(t) − χ(β′(t))p(β′(t))[fn(t, σεn (β(t)), β′(t)) − r(t)] = 0

for a.e. t ∈ [εn, T ]. Hence β is an upper function of problem (9), (10) and αn(t) < C∗(2T 2 + 1)2 < (2T 2 + 1)2 < β(t) fort ∈ [εn, T ]. Put

L = (2T 2 + 1)2 + 2ST , W = maxh(x, y) : (x, y) ∈ [0, L] × [−2S, 2S].

Since χ(y) = 0 for any |y| ≥ 2S, inequality (8) gives

|χ(y)p(y)[fn(t, σεn (x), y) − r(t)]| < p∗[A(

Lη0 + (2S)γ1

εµ0n

+ (2S)γ1

εµ1n

+ (2S)γ

ε2ηn

)+ W + r(t)

]=: wn(t)

for a.e. t ∈ [εn, T ] and each αn(t) ≤ x ≤ β(t), y ∈ R, where p∗ is given in (18). As wn ∈ L1[εn, T ], there exists asolution un ∈ AC 1[εn, T ] of problem (9), (10) and αn(t) ≤ un(t) ≤ β(t) for t ∈ [εn, T ] by Lemma 2.1. In addition, un fulfilsinequalities (11) and (12) by Lemma 2.2. Hence χ(u′

n) = 1, p(u′n) = p(u′

n) and σεn (un) = un. Consequently, un is asolution of equation (16) on the interval [εn, T ].

3. Existence results for positive solutions

In this section the existence results for positive solutions of the singular mixed problem (1), (2) are presented.

Theorem 3.1.Assume that conditions (H1) − (H4) hold. Then there exists a positive solution u ∈ AC 1[0, T ] of problem (1), (2).

Proof. By Lemma 2.4, there exist n∗ ∈ N and C∗ ∈ (0, 1) such that for each n ≥ n∗ problem (16), (10) has a solutionun ∈ AC 1[εn, T ] satisfying inequalities (11), (12) and (30). Consider the sequence unn≥n∗ . Let us choose an arbitraryρ ∈ (0, ε∗

2 ) ⊂ (0, T8 ) and put nρ = minn ∈ N : n ≥ n∗, εn ≤ ρ. We note that ε∗ > 0 satisfies inequality (6) and

εn = T − tn by (7). Then for t ∈ [2ρ, tn] and n ≥ nρ, we have

C∗[(t − εn)(2T − t − 3εn) + εn]2 ≥ C∗[(2ρ − εn)(2T − 2ρ − 3εn) + εn]2

> C∗[ρ(2T − 5ρ)]2 > C∗ρ2T 2

and

C∗[(2tn − T )2 + T − tn]2 = C∗[(T − 2εn)2 + εn]2 > C∗(T − 2εn)4

> C∗T 4

16 > C∗ρ2T 2.

703


From these inequalities and from inequality (30) we obtain

un(t) > C∗ρ2T 2 =: Mρ for t ∈ [2ρ, T ] and n ≥ nρ.

We break the next part of the proof into two cases if µ1 ≥ 1, or µ1 ∈ (0, 1), where µ1 is taken from (H2).Case 1. Let µ1 ≥ 1. By (H2), f ∈ Car((0, T ) × D ), and consequently, there exists qρ ∈ L1[2ρ, T − 2ρ] such that

0 ≤ f(t, x, y) ≤ qρ(t)

for a.e. t ∈ [2ρ, T − 2ρ] and all (x, y) ∈ [Mρ, 1 + ST ] × [−S, S].

(31)

Then|u′′

n(t)| < p(u′n(t))[f(t, un(t), u′

n(t)) + r(t)] ≤ p∗[qρ(t) + r(t)] (32)

for a.e. t ∈ [2ρ, T − 2ρ] and all n ≥ nρ, where p∗ is given in (18). Hence the sequence u′nn≥nρ is equicontinuous

on [2ρ, T − 2ρ]. In addition, inequalities (11) and (12) show that the sequence unn≥nρ is bounded in C 1[2ρ, T ].Consequently, unn≥nρ is equicontinuous on [2ρ, T ]. Using the fact that ρ ∈ (0, ε∗

2 ) is arbitrary and using the Arzelà-Ascoli theorem and the diagonalization principle, we conclude that there exist u ∈ C 0(0, T ]∩C 1(0, T ) and a subsequenceukn of un such that

limn→∞

ukn (t) = u(t) locally uniformly on (0, T ],

limn→∞

u′kn (t) = u′(t) locally uniformly on (0, T ).

Due to inequalities (11), (12) and (30) we have

0 ≤ u(t) ≤ 1 + ST for t ∈ (0, T ], |u′(t)| ≤ S for t ∈ (0, T ), (33)

and

u(t) ≥ limn→∞

C∗[(t − εkn )(2T − t − 3εkn ) + εkn ]2 = C∗t2(2T − t)2, t ∈ (0, T ),

u(T ) ≥ limn→∞

C∗[(2tkn − T )2 + T − tkn ]2 = C∗T 4.

In particular,u(t) ≥ C∗t2(2T − t)2 for t ∈ (0, T ]. (34)

We claim that there exists a continuation u of u to the interval [0, T ] such that u ∈ AC 1[0, T ] and u(0) = 0, u′(T ) = 0.Letting n → ∞ in

ukn (t) = ε2kn +

∫ t

εkn

u′εkn

(s) ds ≤ ε2kn + S(t − εkn )

yields u(t) ≤ St for t ∈ (0, T ]. The last inequality and (33) imply limt→0+ u(t) = 0. It follows from the equality

limn→∞

p(u′kn (t))fkn (t, ukn (t), u′

kn (t)) = p(u′(t))f(t, u(t), u′(t))

for a.e. t ∈ [0, T ], from the inequality

∫ T

tp(u′

kn (s))fkn (s, ukn (s), u′kn (s)) ds ≤ u′

kn (T ) − u′kn (t) + p∗

∫ T

tr(s) ds

< 2S + p∗∫ T

0r(s) ds =: K

704


for t ∈ (0, T ] and n ∈ N, and from the Fatou lemma that

∫ T

tp(u′(s))f(s, u(s), u′(s)) ds ≤ K for t ∈ (0, T ],

which means that p(u′(t))f(t, u(t), u′(t)) ∈ L1[0, T ]. Passing to the limit as n → ∞ in

u′kn (T

2 ) − u′kn (t) =

∫ T/2

tp(u′

kn (s))[fkn (s, ukn (s), u′kn (s)) − r(s)] ds

we have (see (31))

u′(T2 ) − u′(t) =

∫ T/2

tp(u′(s))[f(s, u(s), u′(s)) − r(s)] ds (35)

for t ∈ [2ρ, T − 2ρ] by the Lebesgue dominated convergence theorem. Since ρ ∈ (0, ε∗2 ) is arbitrary, equality (35) holds

for all t ∈ (0, T ). Keeping in mind that p(u′(t))[f(t, u(t), u′(t)) − r(t)] belongs to the set L1[0, T ], the function in theright–hand side of (35) is absolutely continuous on [0, T ]. Set

z(t) = u′kn (T

2 ) −∫ T/2

tp(u′(s))[f(s, u(s), u′(s)) − r(s)] ds for t ∈ [0, T ].

Then z ∈ AC [0, T ], z(t) = u′(t) for t ∈ (0, T ) and

z′(t) = p(u′(t))[f(t, u(t), u′(t)) − r(t)] for a.e. t ∈ [0, T ]. (36)

Let u(t) =∫ t

0 z(s) ds for t ∈ [0, T ]. Then u ∈ AC 1[0, T ], u(0) = 0 and u′(t) = z(t) for t ∈ [0, T ]. Consequently,u′(t) = u′(t) for t ∈ (0, T ) and since limt→0+ u(t) = 0, we have u(t) = u(t) for t ∈ (0, T ]. Hence u is the continuation ofu to the interval [0, T ]. Further, according to inequalities (33), (34) and equality (36), we have

C∗t2(2T − t)2 ≤ u(t) ≤ 1 + ST , |u′(t)| ≤ S for t ∈ [0, T ] (37)

andu′′(t) = p(u′(t))[f(t, u(t), u′(t)) − r(t)] for a.e. t ∈ [0, T ].

We have proved that u is a solution of (1), u(0) = 0 and, by inequality (37), u > 0 on (0, T ]. It remains to show thatu′(T ) = 0. To prove this fact, suppose the contrary, that is suppose that u′(T ) 6= 0. Then |u′(t)| ≥ ρ > 0 on an interval[ν1, T ], and therefore, by condition (H4),

f(t, u(t), u′(t)) ≥ b(T − t)µ1

for a.e. t ∈ [ν∗, T ],

where b = minω(s) : ρ ≤ s ≤ S and ν∗ = maxν, ν1. Then

u′(T ) − u′(ν∗) =∫ T

ν∗

p(u′(t))[f(t, u(t), u′(t)) − r(t)] dt

> bp∗

∫ T

ν∗

dt(T − t)µ1

− p∗∫ T

0r(t) dt = ∞,

which is impossible. Here p∗ and p∗ are given in (18). We have u′(T ) = 0.

705


Case 2. Let µ1 ∈ (0, 1). In view of f ∈ Car((0, T ] × D ) (see Remark 1.1), there exists qρ ∈ L1[2ρ, T ] such that

0 ≤ f(t, x, y) ≤ qρ(t)

for a.e. t ∈ [2ρ, T ] and all (x, y) ∈ [Mρ, 1 + ST ] × [−S, S].

(38)

Then inequality (32) holds for a.e. t ∈ [2ρ, T ] and all n ≥ nρ, and consequently, the sequence u′nn≥nρ is equicon-

tinuous on [2ρ, T ]. We also know that unn≥nρ is bounded in C 1[2ρ, T ]. Hence, by the Arzelà-Ascoli theorem and thediagonalization principle, there exist a function u ∈ C 1(0, T ] and a subsequence ukn of un such that

limn→∞

u(j)kn (t) = u(j)(t) locally uniformly on (0, T ] for j = 0, 1.

Then C∗t2(2T − t)2 ≤ u(t) ≤ 1 + ST , |u′(t)| ≤ S for t ∈ (0, T ]. Essentially, the same reasoning as in Case 1 shows thatp(u′(t))f(t, u(t), u′(t)) ∈ L1[0, T ] and limt→0+ u(t) = 0. Letting n → ∞ in

−u′kn (t) =

∫ T

tp(u′

kn (s))[fkn (s, ukn (s), u′kn (s)) − r(s)] ds

gives

− u′(t) =∫ T

tp(u′(s))[f(s, u(s), u′(s)) − r(s)] ds, t ∈ [2ρ, T ], (39)

by the Lebesgue dominated convergence theorem. Since ρ ∈ (0, ε∗2 ) is arbitrary, equality (39) holds for all t ∈ (0, T ].

Put

u(t) = −∫ t

0

∫ T

sp(u′(v))[f(v, u(v), u′(v)) − r(v)] dv ds for t ∈ [0, T ].

Then u ∈ AC 1[0, T ], u(0) = 0 = limt→0+ u(t), u′(t) = u′(t) for t ∈ (0, T ], u′(T ) = 0 and u fulfils (37). Hence u = u on(0, T ], and consequently,

u′′(t) = p(u′(t))[f(t, u(t), u′(t)) − r(t)] for a.e. t ∈ [0, T ].

We have proved that u is a positive solution of problem (1), (2).

Corollary 3.1.Let the assumptions of Theorem 3.1 be satisfied. Then for each λ > 0 the differential equation

u′′ = p(u′)[λf(t, u, u′) − r(t)] (40)

has a positive solution u ∈ AC 1[0, T ] fulfilling the boundary condition (2).

Proof. Let us choose an arbitrary λ ∈ (0, ∞). The differential equation (40) can be written in the equivalent form

u′′ = λp(u′)[f(t, u, u′) − r(t)

λ

].

Set p1 = λp and r1 = rλ . Then p1 ∈ C 0(−a, a) is positive, r1 ∈ L1[0, T ], r∗

λ ≤ r1(t) for a.e. on [0, T ] and

min∫ 0

−a

dsp1(s) ,

∫ a

0

dsp1(s)

>∫ T

0r1(t) dt.

Hence the functions p1 and r1 fulfil conditions (H1) and (H3) (with p and r replaced by p1 and r1). Consequently, byTheorem 3.1, there exists a positive solution u ∈ AC 1[0, T ] of problem (40), (2).

706


Example 3.1.Let ε = ±1. Consider the differential equation

u′′ = 11 + ε(u′)2

(uα

tµ0+ |u′|β

(T − t)µ1+ |u′|γ

uη + |u′|νeu − r(t))

(41)

where α, µ0, µ1, β, γ, η and ν are positive numbers, µ0 < 2α , µ1 ≤ β, 2η ≤ γ, and r ∈ L1[0, T ], r(t) ≥ r∗ > 0 for a.e.t ∈ [0, T ]. In addition, if ε = −1 let

∫ T0 r(t) dt < 2

3 . Put p(y) = 11+εy2 for y ∈ R if ε = 1 so a = ∞ here and for

y ∈ (−1, 1) if ε = −1 so a = 1 here, and f(t, x, y) = xα

tµ0 + |y|β(T−t)µ1 + |y|γ

xη + h(x, y) for (t, x, y) ∈ (0, T ) × (0, ∞) × R,where h(x, y) = |y|νex . Then f fulfils condition (H4) with w(s) = sβ in (4). Equation (41) satisfies the assumptions ofTheorem 3.1. Therefore, there exists a positive solution u ∈ AC 1[0, T ] of problem (41), (2) for ε = ±1.

4. The maximal positive solution

Let conditions (H1) − (H4) be satisfied. Then Theorem 3.1 guarantees the existence of a positive solution u of problem(1), (2). Set

A = u ∈ AC 1[0, T ] : u is a positive solution of problem (1), (2).

Then A is a nonempty set. We say that u ∈ A is the maximal positive solution of problem (1), (2) if u ≥ u on [0, T ] foreach u ∈ A.It is obvious that if problem (1), (2) has a maximal positive solution, then is unique. In order to prove the existence of themaximal positive solution of problem (1), (2), we use generalized lower and upper functions of the differential equation

u′′ = χ(u′)p(u′)[f(t, u, u′) − r(t)] (42)

on intervals of the type [t1, t2] ⊂ [0, T ], where the functions χ and p are given in Section 2.Let [t1, t2] ⊂ [0, T ]. In accordance with [19, Definition 3.2] we say that a positive function γ ∈ AC [t1, t2] is a generalizedlower (upper) function of the differential equation (42) on the interval [t1, t2] if

(i) γ′ can be written in the form γ′(t) = ξ(t) + ξ0(t), where ξ ∈ AC [t1, t2] and ξ0 : [t1, t2] → R is nondecreasing(nonincreasing) and its derivative vanishes a.e. on [t1, t2],

(ii) the inequality

γ′′(t) ≥ χ(γ′(t))p(γ′(t))[f(t, γ(t), γ′(t)) − r(t)]

(γ′′(t) ≤ χ(γ′(t))p(γ′(t))[f(t, γ(t), γ′(t)) − r(t)])

holds for a.e. t ∈ [t1, t2].

We will need the following result due to Kiguradze and Shekhter [19, Lemma 3.7].

Lemma 4.1.Let α and β be generalized lower and upper functions of equation (42) on an interval [t1, t2] ⊂ [0, T ], α(t) ≤ β(t) fort ∈ [t1, t2] and let limt→t−

2α ′(t) ≤ limt→t−

2β′(t). Let there exist q ∈ L1[t1, t2] such that

|χ(y)p(y)[f(t, x, y) − r(t)]| ≤ q(t)

for a.e. t ∈ [t1, t2] and all α(t) ≤ x ≤ β(t), y ∈ R.

Then for each c1 ∈ [α(t1), β(t1)], c2 ∈ [α(t2), β(t2)] and limt→t−2

α ′(t) ≤ c3 ≤ limt→t−2

β′(t), equation (42) has solutionsu1, u2 ∈ AC 1[t1, t2] fulfilling the boundary conditions

u1(t1) = c1, u1(t2) = c2,

u2(t1) = c1, u′2(t2) = c3,

707


and the inequalityα(t) ≤ uj (t) ≤ β(t) for t ∈ [t1, t2], j = 1, 2.

Let us choose u∗ ∈ A and putA∗ = u ∈ A : u(t) ≥ u∗(t) for t ∈ [0, T ].

We now give the important properties of the functions belonging to the set A∗.

Lemma 4.2.Let (H1) − (H4) hold and let S ∈ (0, a) satisfy (5). Then each u ∈ A satisfies the inequality

|u′(t)| < S for t ∈ [0, T ]. (43)

In addition, if un ⊂ A∗, then there exist its subsequence ukn and u ∈ A∗ such that limn→∞ ukn (t) = u(t) uniformlyon [0, T ].

Proof. Let u ∈ A. It follows from the inequality

(∫ u′(t)

0

dsp(s) +

∫ t

0r(s) ds

)′

≥ 0 for a.e. t ∈ [0, T ]

that the function∫ u′(t)

0ds

p(s) +∫ t

0 r(s) ds is nondecreasing on [0, T ]. Since u′(T ) = 0 by (2), and u′(0) ≥ 0 which followsfrom u(0) = 0 and u > 0 on (0, T ], we have

0 ≤∫ u′(t)

0

dsp(s) +

∫ t

0r(s) ds ≤

∫ T

0r(s) ds, t ∈ [0, T ].

Hence ∣∣∣ ∫ u′(t)

0

dsp(s)

∣∣∣ ≤∫ T

0r(s) ds, t ∈ [0, T ],

which means that u′ fulfils inequality (43).Let un ⊂ A∗. Then

|u′n(t)| < S for t ∈ [0, T ], n ∈ N. (44)

Consequently, p(u′n) = p(u′

n), χ(u′n) = 1 on [0, T ] for n ∈ N, and

un(t) =∫ t

0u′

n(s) ds ≤ ST for t ∈ [0, T ], n ∈ N. (45)

We break the next part of the proof into two cases if µ1 ∈ (0, 1), or µ1 ≥ 1. Note that µ1 is a positive constant incondition (H2).Case (i). Let µ1 ∈ (0, 1). Let us choose an arbitrary ρ ∈ (0, T

2 ) and put mρ = minu∗(t) : t ∈ [ρ, T ]. Then mρ > 0,un(t) ≥ mρ for t ∈ [ρ, T ] and n ∈ N. Since f ∈ Car((0, T ] × D ) by Remark 1.1, there is an hρ ∈ L1[ρ, T ] such that

0 ≤ f(t, x, y) ≤ hρ(t)

for a.e. t ∈ [ρ, T ] and all (x, y) ∈ [mρ, ST ] × [−S, S].

(46)

Due to (44) and (45) we have

0 ≤ f(t, un(t), u′n(t)) ≤ hρ(t) for a.e. t ∈ [ρ, T ] and all n ∈ N.

708


Therefore,|u′′

n(t)| ≤ p∗[hρ(t) + r(t)] for a.e. t ∈ [ρ, T ] and all n ∈ N,

where p∗ is given in (18). As a result the sequence u′n is equicontinuous on [ρ, T ]. By (44) and (45), un is bounded

in C 1[0, T ]. Since ρ ∈ (0, T2 ) is arbitrary, by the Arzelà-Ascoli theorem and the diagonalization principle, there exist

u ∈ C 0[0, T ] ∩ C 1(0, T ] and a subsequence ukn of un such that

limn→∞

ukn (t) = u(t) uniformly on [0, T ],

limn→∞

u′kn (t) = u′(t) locally uniformly on (0, T ].

Clearly, u ≥ u∗ on [0, T ], u(0) = 0, and

limn→∞

p(u′kn (t))f(t, ukn (t), u′

kn (t)) = p(u′(t))f(t, u(t), u′(t)) a.e. on [0, T ]. (47)

Also, it follows from the relation

∫ T

0p(u′

kn (t))f(t, ukn (t), u′kn (t)) dt =

∫ T

0u′′

kn (t)) dt +∫ T

0p(u′

kn (t))r(t) dt

≤ −u′kn (0) + p∗

∫ T

0r(t) dt

< S + p∗∫ T

0r(t) dt =: K

that ∫ T

0p(u′

kn (t))f(t, ukn (t), u′kn (t)) dt < K for n ∈ N. (48)

From (47), (48), and from the fact that p(u′kn (t))f(t, ukn (t), u′

kn (t)) ∈ L1[0, T ] is nonnegative, we havep(u′(t))f(t, u(t), u′(t)) ∈ L1[0, T ] by the Fatou lemma. In view of (46), (47) and since ρ ∈ (0, T

2 ) is arbitrary, lettingn → ∞ in

u′kn (t) =

∫ T

tp(u′

kn (s))[f(s, ukn (s), u′kn (s)) − r(s)] ds

yields

u′(t) =∫ T

tp(u′(s))[f(s, u(s), u′(s)) − r(s)] ds, t ∈ (0, T ],

by the Lebesgue dominated convergence theorem. Since p(u′(t))f(t, u(t), u′(t)) ∈ L1[0, T ], it follows that u ∈ AC 1[0, T ],u′(T ) = 0 and u is a solution of (1). Hence u ∈ A∗.Case (ii). Let µ1 ≥ 1. Choose ρ ∈ (0, T

2 ) and put wρ = minu∗(t) : t ∈ [ρ, T − ρ]. Then wρ > 0, un(t) ≥ wρ fort ∈ [ρ, T − ρ] and n ∈ N. Since f ∈ Car((0, T ) × D ) by (H2), there exists qρ ∈ L1[ρ, T − ρ] such that

0 ≤ f(t, x, y) ≤ qρ(t)

for a.e. t ∈ [ρ, T − ρ] and all (x, y) ∈ [wρ, ST ] × [−S, S].

It follows from (44) and (45) that

0 ≤ f(t, un(t), u′n(t)) ≤ qρ(t) for a.e. t ∈ [ρ, T − ρ] and all n ∈ N,

and therefore,|u′′

n(t)| ≤ p∗[qρ(t) + r(t)] for a.e. t ∈ [ρ, T − ρ] and all n ∈ N.

709


Hence u′n is equicontinuous on [ρ, T −ρ]. Since un is bounded in C 1[0, T ] and ρ ∈ (0, T

2 ) is arbitrary, we deduce fromthe Arzelà-Ascoli theorem and the diagonalization principle that there exist u ∈ C 0[0, T ] ∩ C 1(0, T ) and a subsequenceukn of un such that

limn→∞

ukn (t) = u(t) uniformly on [0, T ],

limn→∞

u′kn (t) = u′(t) locally uniformly on (0, T ).

Then u∗ ≤ u ≤ S on [0, T ], u(0) = 0 and (47) holds. Arguing as in Case (i) we prove that p(u(t))f(t, u(t), u′(t)) ∈ L1[0, T ].Letting n → ∞ in

u′kn (t) = u′

kn (T2 ) +

∫ t

T /2p(u′

kn (s))[f(s, ukn (s), u′kn (s)) − r(s)] ds

gives

u′(t) = u′(T2 ) +

∫ t

T /2p(u′(s))[f(s, u(s), u′(s)) − r(s)] ds (49)

for t ∈ (0, T ), by the Lebesque dominated convergence theorem. Define u′ at t = 0 and t = T by (49). Then u ∈ AC 1[0, T ]and u is a solution of (1). In order to prove that u ∈ A∗ it remains to show that u′(T ) = 0. Suppose that u′(T ) 6= 0.Let ν ∈ (0, T ) be taken from (H4). Then there exists ν1 ∈ (ν, T ) such that m := min|u′(t)| : t ∈ [ν1, T ] > 0 and, bycondition (H4),

f(t, u(t), u′(t)) ≥ c(T − t)µ1

for a.e. t ∈ [ν1, T ],

where c = minw(s) : s ∈ [m, S] > 0. Consequently,

u′(T ) − u′(ν1) =∫ T

ν1

p(u′(t))[f(t, u(t), u′(t)) − r(s)] dt

≥∫ T

ν1

p(u′(t))[

c(T − t)µ1

− r(t)]

dt

≥∫ T

ν1

[p∗c

(T − t)µ1− p(u′(t))r(t)

]dt = ∞,

which is impossible. Hence u′(T ) = 0, and therefore, u ∈ A∗.

Theorem 4.1.Assume that conditions (H1) − (H4) hold. Then there exists the maximal positive solution of problem (1), (2).

Proof. Let

M = sup∫ T

0u(t) dt : u ∈ A∗

.

Since u < ST on [0, T ] for each u ∈ A by (43), we have M < ST 2. Let zn ⊂ A∗ be such that limn→∞∫ T

0 zn(t) dt = M.In view of Lemma 4.2 there exist u ∈ A∗ and a subsequence zkn of zn such that limn→∞ zkn (t) = u(t) uniformly on[0, T ]. Hence M = limn→∞

∫ T0 zkn (t) dt =

∫ T0 u(t) dt. In particular,

M =∫ T

0u(t) dt.

We now prove that u is the maximal positive solution of problem (1), (2). To prove this, suppose the contrary. Then thereexists z ∈ A∗ such that either (i) z > u on (t1, t2), 0 < t1 < t2 < T , and z(tj ) = u(tj ) for j = 1, 2, or (ii) z > u on (0, t3),0 < t3 < T , and z(t3) = u(t3), or (iii) z > u on (t4, T ), 0 < t4 < T , and z(t4) = u(t4). We consider separately cases(i)-(iii).

710


Case (i). Let z(t) > u(t) for t ∈ (t1, t2), 0 < t1 < t2 < T , and z(tj ) = u(tj ) for j = 1, 2. Put

α(t) =

z(t) for t ∈ [t1, t2],

u(t) for t ∈ [0, t1) ∪ (t2, T ].

Then α ∈ AC [0, T ], α ≥ u on [0, T ] andα ′(t) = ξ(t) + ξ0(t),

where

ξ0(t) =

0 for t ∈ [0, t1),

z′(t1) − u′(t1) for t ∈ [t1, t2],

z′(t1) − u′(t1) + u′(t2) − z′(t2) for t ∈ (t2, T ],

ξ(t) =

u′(t) for t ∈ [0, t1),

z′(t) − ξ0(t) for t ∈ [t1, t2],

u′(t) − ξ0(t) for t ∈ (t2, T ].

It is easy to check that ξ ∈ AC [0, T ] and ξ0 is a nondecreasing finite step function. As |z′| < S and |u′| < S on [0, T ]by Lemma 4.2, we have |α ′| < S a.e. on [0, T ]. Therefore, χ(α ′) = 1 and

α ′′(t) = χ(α ′(t))p(α ′(t))[f(t, α(t), α ′(t)) − r(t)] a.e. on [0, T ].

Hence α is a generalized lower function of equation (42) on any interval [c1, c2] ⊂ [0, T ]. Let β(t) = S(T + 2t) fort ∈ [0, T ]. Then χ(β′) = 0 on [0, T ] and thus

β′′(t) − χ(β′(t))p(β′(t))[f(t, β(t), β′(t)) − r(t)] = 0 a.e. on [0, T ],

which means that β is a generalized upper function of equation (42) on any [c1, c2] ⊂ [0, T ]. Put ∆ = mint1, T − t2and let εn ⊂ (0, ∆) be a decreasing sequence, limn→∞ εn = 0. The next part of the proof is divided into two subcasesif µ1 ∈ (0, 1), or µ1 ≥ 1, where µ1 is taken from (H2).Case (ia). Let µ1 ∈ (0, 1). Put mn = minu(t) : t ∈ [εn, T ]. Then mn > 0 and since f ∈ Car((0, T ] × D ) by Remark 1.1,there exists qn ∈ L1[εn, T ] such that

0 ≤ χ(y)f(t, x, y) ≤ qn(t)

for a.e. t ∈ [εn, T ] and all (x, y) ∈ [mn, 3ST ] × R.

Hence|χ(y)p(y)[f(t, x, y) − r(t)]| ≤ p∗[qn(t) + r(t)]

for a.e. t ∈ [εn, T ] and all (x, y) ∈ [mn, 3ST ] × R,

(50)

where p∗ is given in (18). Since limt→T − α ′(t) = u′(T ) = 0 and limt→T − β′(t) = 2S, Lemma 4.1 guarantees that for eachn ∈ N there exists a solution un ∈ AC 1[εn, T ] of equation (42) satisfying the inequality

α(t) ≤ un(t) ≤ β(t) for t ∈ [εn, T ], n ∈ N, (51)

and the boundary conditionsun(εn) = u(εn), u′

n(T ) = 0, n ∈ N.

711


We now verify that|u′

n(t)| ≤ 2S for t ∈ [εn, T ], n ∈ N. (52)

If not, there exists n0 ∈ N such that max|u′n0

(t)| : t ∈ [εn0 , T ] > 2S. Since u′n0

(T ) = 0 we have |u′n0

(t)| > 2S on aninterval (τ1, τ2) ⊂ (εn0 , T ) and |u′

n0(τ2)| = 2S. Then χ(u′

n0) = 0 on [τ1, τ2], and therefore u′′

n0= 0 a.e. on [τ1, τ2]. Hence

u′n0

is a constant function on [τ1, τ2], which is impossible.By (50),

|u′′m(t)| ≤ p∗[qn(t) + r(t)] for a.e. t ∈ [εn, T ] and all m ≥ n. (53)

Put

vn(t) =

u(εn) + u′n(ε+

n )(t − εn) for t ∈ [0, εn),

un(t) for t ∈ [εn, T ],

where u′n(ε+

n ) denotes the right-hand derivative of un at t = εn. We conclude from (51)-(53) that the sequence vn isbounded in C 1[0, T ] and v ′

n is equicontinuous on any compact subinterval of (0, T ]. Hence there exist v ∈ C 0[0, T ] ∩C 1(0, T ] and a subsequence vkn of vn such that limn→∞ vkn = v in C 0[0, T ] and limn→∞ v ′

kn (t) = v ′(t) locally uniformlyon (0, T ] by the Arzelà-Ascoli theorem and the diagonalization principle. Then v(0) = limn→∞(u(εkn ) − u′

kn (ε+kn )εkn ) =

u(0) = 0 since |u′kn (ε+

kn )| ≤ 2S for n ∈ N, v ′(T ) = 0, v(t) ≥ α(t) for t ∈ [0, T ], and

limn→∞

χ(v ′kn (t))p(v ′

kn (t))f(t, vkn (t), v ′kn (t)) = χ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) (54)

for a.e. t ∈ [0, T ]. Choose ρ ∈ (0, ∆). Using equality (54) and the fact that χ(v ′kn (t))p(v ′

kn (t))f(t, vkn (t), v ′kn (t)) ∈ L1[εkn , T ]

is nonnegative on [εkn , T ] and that for each n ∈ N fulfilling εkn < ρ the relation

∫ T

ρχ(v ′

kn (t))p(v ′kn (t))f(t, vkn (t), v ′

kn (t)) dt = −v ′kn (ρ) +

∫ T

ρχ(v ′

kn (t))p(v ′kn (t))r(t) dt < 2S + p∗

∫ T

0r(t) dt

holds, we have ∫ T

ρχ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) dt ≤ 2S + p∗

∫ T

0r(t) dt

by the Fatou lemma. Since ρ > 0 is arbitrary, the last inequality implies χ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) ∈ L1[0, T ]. Keepingin mind (50) and (54), letting n → ∞ in

v ′kn (t) = −

∫ T

tχ(v ′

kn (s))p(v ′kn (s))[f(s, vkn (s), v ′

kn (s)) − r(s)] ds,

gives

v ′(t) = −∫ T

tχ(v ′(s))p(v ′(s))[f(s, v(s), v ′(s)) − r(s)] ds for t ∈ (0, T ], (55)

by the Lebesgue dominated convergence theorem. Define

v ′(0) = −∫ T

0χ(v ′(s))p(v ′(s))[f(s, v(s), v ′(s)) − r(s)] ds.

It follows from χ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) ∈ L1[0, T ] and from (55) that v ∈ AC 1[0, T ] and v is a solution of (42). Weclaim that

|v ′(t)| < S for t ∈ [0, T ]. (56)

Indeed, v ′′(t) ≤ −p(v ′(t))r(t) for a.e. t ∈ [0, T ], which implies

(∫ v ′(t)

0

dsp(s) +

∫ t

0r(s) ds

)′

≥ 0 for a.e. t ∈ [0, T ].

712


We can now proceed analogously to the first part of the proof of Lemma 4.2 to show that (56) is true. In view of (56)we have χ(v ′) = 1 on [0, T ], and consequently, v ∈ A∗. Since v ≥ α on [0, T ], it follows that

∫ T0 v(t) dt > M, which is

impossible.Case (ib). Let µ1 ≥ 1. Put dn = minu(t) : t ∈ [εn, T − εn]. Then dn > 0. Since f ∈ Car((0, T ) × D ) by (H3), thereexists hn ∈ L1[εn, T − εn] such that

0 ≤ χ(y)f(t, x, y) ≤ hn(t)

for a.e. t ∈ [εn, T − εn] and all (x, y) ∈ [dn, 3ST ] × R.

Hence|χ(y)p(y)[f(t, x, y) − r(t)| ≤ p∗[hn(t) + r(t)]

for a.e. t ∈ [εn, T − εn] and all (x, y) ∈ [dn, 3ST ] × R,

(57)

and consequently, by Lemma 4.1, for each n ∈ N there exists a solution un ∈ AC 1[εn, T − εn] of equation (42) fulfillingthe inequality

α(t) ≤ un(t) ≤ β(t) for t ∈ [εn, T − εn], n ∈ N, (58)

and the boundary conditionsun(εn) = u(εn), un(T − εn) = u(T − εn). (59)

In order to prove that|u′

n(t)| ≤ 2S for t ∈ [εn, T − εn], n ∈ N,

suppose the contrary, that is suppose max|u′n0

(t)| : t ∈ [εn0 , T − εn0 ] > 2S for some n0 ∈ N. Since un0 (T − εn0 ) −un0 (εn0 ) = u′

n0(ν1)(T − 2εn0 ), u(T − εn0 ) − u(εn0 ) = u′(ν2)(T − 2εn0 ) for some ν1, ν2 ∈ (εn0 , T − εn0 ) and, by (59),

un0 (εn0 ) = u(εn0 ), un0 (T − εn0 ) = u(T − εn0 ), we have u′n0

(ν1) = u′(ν2). Consequently, |u′n0

(ν1)| < S because |u′| < S on[0, T ] by Lemma 4.2. Therefore, there exist εn0 ≤ τ1 < τ2 ≤ T − εn0 , such that |u′

n0| > 2S on (τ1, τ2) and |u′

n0(τj )| = 2S

for some j ∈ 1, 2, which means that χ(u′n0

) = 0 on [τ1, τ2]. Hence u′′n0

(t) = 0 for a.e. t ∈ [τ1, τ2], and so u′n0

is aconstant function on [τ1, τ2], which is impossible.For each n ∈ N define vn ∈ C 1[0, T ] by the formula

vn(t) =

u(εn) + u′

n(ε+n )(t − εn) for t ∈ [0, εn),

un(t) for t ∈ [εn, T − εn],

u(T − εn) + u′n((T − εn)−)(t − T + εn) for t ∈ (T − εn, T ],

where u′n(ε+

n ) and u′n((T − εn)−) denotes the right-hand side derivative of un at εn and the left-hand derivative of

un at T − εn, respectively. Then |vn(t)| ≤ 3ST , |v ′n(t)| ≤ 2S for t ∈ [0, T ], |v ′′

m(t)| = |u′′m(t)| ≤ p∗[hn(t) + r(t)] for a.e.

t ∈ [εn, T −εn] and all m ≥ n. Hence vn is bounded in C 1[0, T ] and v ′n is equicontinuous on any compact subinterval

of (0, T ). By the Arzelà-Ascoli theorem and the diagonalization principle, there exist v ∈ C 0[0, T ] ∩ C 1(0, T ) and asubsequence of vn which we also denote by vn such that limn→∞ vn = v in C 0[0, T ] and limn→∞ v ′

n(t) = v ′(t) locallyuniformly on (0, T ). Then v(0) = limn→∞(u(εn) − u′

n(ε+n )εn) = u(0), v(T ) = limn→∞(u(T − εn) + u′

n((T − εn)−)εn = u(T )and

v(t) ≥

z(t) for t ∈ [t1, t2],

u(t) for t ∈ [0, t1) ∪ (t2, T ].(60)

In addition,lim

n→∞χ(v ′

n(t))p(v ′n(t))f(t, vn(t), v ′

n(t)) = χ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) for a.e. t ∈ [0, T ], (61)

713


and

∫ T−εn

εn

χ(v ′n(t))p(v ′

n(t))f(t, vn(t), v ′n(t)) dt =

∫ T−εn

εn

χ(u′n(t))p(u′

n(t))f(t, un(t), u′n(t)) dt

= u′n(T − εn) − u′

n(εn) +∫ T−εn

εn

χ(u′n(t))p(u′

n(t))r(t) dt

< 4S + p∗∫ T

0r(t) dt

for n ∈ N. Choose ρ ∈ (0, T2 ). Using equality (61), the inequality

∫ T−εn

εn

χ(v ′n(t))p(v ′

n(t))f(t, vn(t), v ′n(t)) dt < 4S + p∗

∫ T

0r(t) dt

and the Fatou lemma, we obtain

∫ T−ρ

ρχ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) dt ≤ 4S + p∗

∫ T

0r(t) dt.

It follows from the last inequality and since ρ ∈ (0, T2 ) is arbitrary that the function χ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) belongs

to the set L1[0, T ]. Due to (57) and (61), letting n → ∞ in

v ′n(t) = v ′

n(T2 ) +

∫ t

T /2χ(v ′

n(s))p(v ′n(s))[f(s, vn(s), v ′

n(s)) − r(s)] ds, t ∈ [εn, T − εn],

gives

v ′(t) = v ′(T2 ) +

∫ t

T /2χ(v ′(s))p(v ′(s))[f(s, v(s), v ′(s)) − r(s)] ds, t ∈ (0, T ),

by the Lebesgue dominated convergence theorem. Define

v ′(t) = v ′(T2 ) +

∫ t

T /2χ(v ′(s))p(v ′(s))[f(s, v(s), v ′(s)) − r(s)] ds for t = 0, T .

Since χ(v ′(t))p(v ′(t))f(t, v(t), v ′(t)) ∈ L1[0, T ] we have v ∈ AC 1[0, T ] and v is a solution of equation (42). If v ′(T ) 6= 0,then using condition (H4) and the same reasoning as in the proof Theorem 3.1 (see Case 1) we obtain a contradiction.Hence v ′(T ) = 0 and v is a solution of problem (42), (2). We can now proceed analogously to Case (ia) to show thatinequality (56) is true. Hence χ(v ′) = 1 on [0, T ], and therefore, v ∈ A∗. It follows from (60) that

∫ T0 v(t) dt > M, which

is impossible.Case (ii). Let z(t) > u(t) for t ∈ (0, t3), t3 ∈ (0, T ), and z(t3) = u(t3). Put

α(t) =

z(t) for t ∈ [0, t3],

u(t) for t ∈ (t3, T ].

Then α ∈ AC [0, T ], α ≥ u on [0, T ] andα ′(t) = ξ(t) + ξ0(t),

where

ξ0(t) =

0 for t ∈ [0, t3],

u′(t3) − z′(t3) for t ∈ (t3, T ],ξ(t) =

z′(t) for t ∈ [0, t3],

u′(t) − ξ0(t) for t ∈ (t3, T ].

714


Since ξ ∈ AC [0, T ], ξ0 is nondecreasing on [0, T ], ξ ′0 = 0 a.e. on [0, T ] and

α ′′(t) = χ(α ′(t))p(α ′(t))[f(t, α(t), α ′(t)) − r(t)] a.e. on [0, T ],

we see that α is a generalized lower function of equation (42) on any interval [c1, c2] ⊂ [0, T ]. The function β(t) = S(T +2t)for t ∈ [0, T ] is a generalized upper function of (42) on any [c1, c2] ⊂ [0, T ]. Put ∆ = mint3, T − t3 and let εn ⊂ (0, ∆)be a decreasing sequence, limn→∞ εn = 0. Essentially, the same reasoning as in Case (i) shows that there exists v ∈ A∗

and v ≥ α . Then∫ T

0 v(t) dt > M, which is impossible.Case (iii). Let z(t) > u(t) for t ∈ (t4, T ), t4 ∈ (0, T ), and z(t4) = u(t4). We can proceed analogously to Case (ii) nowwith

α(t) =

u(t) for t ∈ [0, t4],

z(t) for t ∈ (t4, T ],

to show that there exists v ∈ A∗, v ≥ α on [0, T ]. Then∫ T

0 v(t) dt > M, which is impossible.

Example 4.1.Let α, µ0, µ1, β, γ, η, ν be positive numbers, µ0 < 2α , µ1 ≤ β, 2η ≤ γ and let ε = ±1. Let r ∈ L1[0, T ], r(t) ≥ r∗ > 0 fora.e. t ∈ [0, T ]. In addition, if ε = −1, let

∫ T0 r(t) dt < 2

3 . Example 3.1 shows that the differential equation

u′′ = 11 + ε(u′)2

(uα

tµ0+ |u′|β

(T − t)µ1+ |u′|γ

uη + |u′|νeu − r(t))

(62)

satisfies conditions (H1) − (H4). Hence problem (62), (2) has the maximal positive solution by Theorem 4.1.

Acknowledgements

Third author is supported by grant of the Council of Czech Government MSM 6198959214.

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