Generalized lower and upper functions for the φ-Laplacian

ISSN 0012-2661, Differential Equations, 2014, Vol. 50, No. 5, pp. 598–607. c© Pleiades Publishing, Ltd., 2014.Original Russian Text c© A.Ya. Lepin, L.A. Lepin, 2014, published in Differentsial’nye Uravneniya, 2014, Vol. 50, No. 5, pp. 601–610.

ORDINARY DIFFERENTIAL EQUATIONS

Generalized Lower and Upper Functionsfor the ϕ-Laplacian

A. Ya. Lepin and L. A. LepinInstitute for Mathematics and Computer Science, Latvia University, Riga, Latvia

e-mail: [email protected] December 16, 2013

Abstract—We give definitions of generalized lower and upper functions and generalized solu-tion, which differs from a solution in the conventional sense in that the derivative of a generalizedsolution can be equal to +∞ and −∞. We show how to use generalized solutions for obtainingclassical results.

DOI: 10.1134/S0012266114050036

The method of lower and upper functions is often used when proving the existence of a solutionof the Dirichlet problem

(ϕ(t, x, x′))′ = f(t, x, x′), x(a) = A, x(b) = B.

Let α and β be a lower and an upper function. Then, under additional conditions of Bernstein–Nagumo type, one can prove the existence of a solution x of the Dirichlet problem satisfying theinequalities α ≤ x ≤ β. For the case in which ϕ(t, x, x′) = x′, it was shown (see [1; 2, p. 46]) thatthere exists a generalized solution of the Dirichlet problem provided that there only exist lower andupper functions. We prove the corresponding result in the general case.

Comparing generalized solutions for the ϕ-Laplacian with viscosity solutions [3] for elliptic par-tial differential equations, one can note the following. Both generalized solutions for the ϕ-Laplacianand viscosity solutions for elliptic equations can be defined via lower and upper functions. If a lowerfunction coincides with an upper one, then in both the cases, we obtain a generalized (viscosity)solution. A generalized solution has good properties. For example, the set of generalized solutionsof the ϕ-Laplacian between a lower and an upper function is compact and has the minimum andmaximum generalized solutions, and the Dirichlet problem is solvable without additional conditionsof the Nagumo or Schrader type. For elliptic equations, there exists a unique viscosity solution,and it is stable. Both generalized and viscosity solutions may have singularities, which distin-guishes them from ordinary solutions. For example, the derivative of a generalized solution for theϕ-Laplacian can be equal to +∞ and −∞, and a viscosity solution is not necessarily differentiableeverywhere. A historical survey of and basic information on the theory of the ϕ-Laplacian can befound in [4–17].

MAIN DEFINITIONS

Consider the differential equation

(ϕ(t, x, x′))′ = f(t, x, x′), t ∈ I = [a, b], (1)

where the function ϕ belongs to C(I ×R2, R) and is strictly increasing with respect to x′ for fixed tand x and f : I×R2 → R satisfies the Caratheodory condition: the function f(·, x, y) is measurableon I for fixed x, y ∈ R, f(t, · , ·) is continuous on R2 for t ∈ I, and for any compact set P ⊂ R2,there exists a function g ∈ L1(I,R) such that |f(t, x, y)| ≤ g(t) for all (t, x, y) ∈ I × P .

598

GENERALIZED LOWER AND UPPER FUNCTIONS FOR THE ϕ-LAPLACIAN 599

Definition 1. A function x ∈ C1(I,R) is a solution of Eq. (1) if the function ϕ(t, x(t), x′(t)) :I → R is absolutely continuous and satisfies Eq. (1) almost everywhere on I.

Definition 2. A function α ∈ Lip(I,R) is a lower function of Eq. (1) if the inequality

ϕ(t2, α(t2), α′(t2)) − ϕ(t1, α(t1), α′(t1)) ≥t2∫

t1

f(s, α(s), α′(s)) ds (2)

holds for arbitrary t1 ∈ (a, b) and t2 ∈ (t1, b) provided that the derivatives α′(t1) and α′(t2) exist.A function β ∈ Lip(I,R) is an upper function of Eq. (1) is the inequality

ϕ(t2, β(t2), β′(t2)) − ϕ(t1, β(t1), β′(t1)) ≤t2∫

t1

f(s, β(s), β′(s)) ds (3)

holds for arbitrary t1 ∈ (a, b) and t2 ∈ (t1, b) provided that the derivatives β′(t1) and β′(t2) exist.

We denote the set of lower functions by A(I,R) and the set of upper functions by B(I,R).

Definition 3. A function α belongs to the class BB+(I,R) if and only if for each t ∈ (a, b]there exists a left derivative α′

l(t) and a limit limτ→t− α′l(τ) satisfying α′

l(t) ≥ limτ→t− α′l(τ), for

each t ∈ [a, b), there exists a right derivative α′r(t) and a limit limτ→t+ α′

r(τ) satisfying α′r(t) ≤

limτ→t+ α′r(τ), and α′

l(t) ≤ α′r(t) for each t ∈ (a, b).

A function β belongs to the class BB−(I,R) if and only if for each t ∈ (a, b] there existsa left derivative β′

l(t) and a limit limτ→t− β′l(τ) satisfying β′

l(t) ≤ limτ→t− β′l(τ), for each t ∈ [a, b),

there exists a right derivative β′r(t) and a limit limτ→t+ β′

r(τ) satisfying β′r(t) ≥ limτ→t+ β′

r(τ), andβ′

l(t) ≥ β′r(t) for each t ∈ (a, b).

Definition 4. A bounded function α ∈ BB+(I,R) is called a generalized lower function ofEq. (1) and we write α ∈ AG(I,R) if α is a lower function of Eq. (1) on any compact interval J ⊂ Ion which α satisfies the Lipschitz condition.

A bounded function β ∈ BB−(I,R) is called a generalized upper function of Eq. (1) and wewrite β ∈ BG(I,R) if β is an upper function of Eq. (1) on any compact interval J ⊂ I on which itsatisfies the Lipschitz condition.

Definition 5. A function x : I → R is called a generalized solution of Eq. (1) if it is simul-taneously a generalized lower function and a generalized upper function of Eq. (1). The set ofgeneralized solutions is denoted by SG(I,R).

MAIN LEMMAS

The following assertion (see [2, p. 13]) describes the properties of functions of bounded bendingintroduced in Definition 3.

Lemma 1. A function α ∈ BB+(I,R) has the following properties.1. limt→τ− α′

r(t) exists for τ ∈ (a, b] and is equal to limt→τ− α′l(t).

2. limt→τ+ α′l(t) exists for τ ∈ [a, b) and is equal to limt→τ+ α′

r(t).3. limt→τ− α(t) exists for τ ∈ (a, b].4. limt→τ+ α(t) exists for τ ∈ [a, b).5. α(τ) ≤ max{limt→τ− α(t), limt→τ+ α(t)}, τ ∈ (a, b).6. For τ ∈ (a, b], the inequality α(τ) ≤ limt→τ− α(t) implies that α′

l(τ) = limt→τ− α′l(t).

7. For τ ∈ [a, b), the inequality α(τ) ≤ limt→τ+ α(t) implies that α′r(τ) = limt→τ+ α′

r(t).

DIFFERENTIAL EQUATIONS Vol. 50 No. 5 2014

600 A.Ya. LEPIN, L.A. LEPIN

8. For τ ∈ (a, b), the inequality limt→τ− α(t) < limt→τ+ α(t) implies that α′r(τ) = +∞.

9. For τ ∈ (a, b), the inequality limt→τ− α(t) > limt→τ+ α(t) implies that α′l(τ) = −∞.

10. The set of points of discontinuity of the function α is at most countable.11. The set of points at which limt→τ− α′

l(t) = −∞ and limt→τ+ α′r(t) = +∞ is finite.

12. The set of points at which max{|α′l(τ)|, |α′

r(τ)|} = ∞ is closed and has the zero measure.

A function β ∈ BB−(I,R) has similar properties.

Lemma 2. For arbitrary M > 0, c1 ∈ R, and c2 ∈ (−∞, c1), there exists a number δ > 0 suchthat , for arbitrary α ∈ AG(I,R) and t1, t2 ∈ I, the estimates |α(t)| < M, t ∈ I, t1 < t2, α′

r(t1) ≥ c1,and α′

l(t2) ≤ c2 imply the inequality t2 − t1 ≥ δ.

Proof. Let

t3 = sup{t ∈ [t1, t2) : α′r(t) ≥ c1}, t4 = inf{t ∈ (t3, t2] : α′

l(t) ≤ c2}.

It follows from Definition 3 and Lemma 1 that

limt→t2−

α′r(t) = lim

t→t2−α′

l(t) ≤ α′l(t2) ≤ c2 < c1.

Consequently, t3 < t2. In a similar way, we obtain t4 > t3. Let us show that α′r(t3) = c1.

If α′r(t3) < c1, then limt→t3− α′

r(t) = limt→t3− α′l(t) ≤ α′

l(t3) ≤ α′r(t3) < c1, which contradicts the

definition of t3. In a similar way, we obtain α′l(t4) = c2. Obviously, c2 < α′

l(t) ≤ α′r(t) < c1

for t ∈ (t3, t4). By using the Caratheodory condition, for the set P = [−M,M ] × [c2, c1], we findthe corresponding point g. Let c = max{|c1|, |c2|} and

min{ϕ(t, x, c1) − ϕ(t, x, c2) : t ∈ I, x ∈ [−M,M ]} = 2Δ,

and let the number δ be chosen so as to ensure that |∫ t6

t5g(t) dt| < Δ for |t6 − t5| < δ and

ϕ(t5, x1, c1) − ϕ(t6, x2, c2) > Δ, |t6 − t5| < δ, |x2 − x1| < cδ.

Then from the inequality

ϕ(t4, α(t4), α′l(t4)) − ϕ(t3, α(t3), α′

r(t3)) ≥t4∫

t3

f(s, α(s), α′(s)) ds,

we obtain a contradiction for t4 − t3 < δ. The proof of the lemma is complete.

Lemma 3. Let M > 0 and αn ∈ AG(I,R), n ∈ {1, 2, . . .}, satisfy the condition

sup{|αn(t)| : t ∈ I, n ∈ {1, 2, . . .}} < M,

and let the sequence αn converge pointwise to a function α : I → R. Then α ∈ AG(I,R), and

lim supn→∞

α′nr(t) ≤ α′

r(t), t ∈ [a, b), lim infn→∞

α′nl(t) ≥ α′

l(t), t ∈ (a, b].

If α satisfies the Lipschitz condition with constant L > 0 on the interval [a1, b1] ⊂ I, then, forarbitrary c ∈ (a1, b1), d ∈ (c, b1), and L1 ∈ (L,∞), there exists a k ∈ {1, 2, . . .} such that the αn,n > k, satisfy the Lipschitz condition with constant L1 on the interval [c, d].



Proof. Let us show that α ∈ BB+(I,R). In accordance with Definition 3, we prove the existenceof the right derivative α′

r(t) for t ∈ [a, b). Suppose the contrary. Then there exist c1, c2 ∈ R suchthat D+α(t) < c2 < c1 < D+α(t). Consequently, for each δ > 0, there exist t1, t2, t3, t4 ∈ (t, b)such that

t1 < t2 < t3 < t4 < t + δ, α(t2) − α(t1) > c1(t2 − t1), α(t4) − α(t3) < c2(t4 − t3).

We choose a number n ∈ {1, 2, . . .} such that αn(t2) − αn(t1) > c1(t2 − t1) and αn(t4) − αn(t3) <c2(t4 − t3). It follows from Definition 3 that there exist points t5 ∈ [t1, t2] and t6 ∈ [t3, t4] such thatα′

nr(t5) ≥ c1 and α′nl(t6) ≤ c2, but this contradicts Lemma 2. The existence of the left derivative,

the inequalities α′l(t) ≤ α′

r(t), the existence of the limits limτ→t− α′l(τ) and limτ→t+ α′

r(τ), and theinequalities α′

l(t) ≥ limτ→t− α′l(τ) and α′

r(t) ≤ limτ→t+ α′r(τ) can be proved in a similar way.

Let us show that lim supn→∞ α′nr(t) ≤ α′

r(t) for t ∈ [a, b). Suppose the contrary. Then thereexist c1, c2 ∈ R such that

α′r(t) < c2 < c1 < lim sup

n→∞α′

nr(t).

By using Lemma 2, we find δ. If α(t) ≤ limτ→t+ α(τ), then, by Lemma 1, α′r(t) = limτ→t+ α′

r(τ), andthere exist t1, t2 ∈ (t, b) such that t1 < t2 < t+ δ and α(t2)−α(t1) < c2(t2− t1). Take n ∈ {1, 2, . . .}such that α′

nr(t) > c1 and αn(t2) − αn(t1) < c2(t2 − t1). Then there exists a t3 ∈ [t1, t2] such thatα′

nl(t3) ≤ c2. By using Lemma 2, we arrive at a contradiction.If α(t) > limτ→t+ α(τ), then there exists a t1 ∈ (t, b) such that t1 < t + δ and α(t1) − α(t) <

c2(t1 − t). Consequently, there exists an n ∈ {1, 2, . . .} and a t2 ∈ (t, t1) such that α′nr(t) > c1 and

αn(t1) − αn(t2) < c2(t1 − t2). Then there exists a t3 ∈ [t2, t1] such that α′nl(t3) ≤ c2. By using

Lemma 2, we arrive at a contradiction. The inequality lim infn→∞ α′nl(t) ≥ α′

l(t), t ∈ (a, b], can beproved in a similar way.

Suppose that the function α satisfies the Lipschitz condition with constant L > 0 on the interval[a1, b1], c ∈ (a1, b1), d ∈ (c, b1), and L1 ∈ (L,∞). Let us prove the existence of a k ∈ {1, 2, . . .} suchthat the function αn, n > k, satisfies the Lipschitz condition with constant L1 on the interval [c, d].Suppose the contrary. Without loss of generality, we assume that all functions αn do not satisfythe Lipschitz condition with constant L1 on the interval [c, d]. Then there exists a tn ∈ [c, d] suchthat either α′

nr(tn) > L1 or α′nl(tn) < −L1. Indeed, if we suppose the contrary, then it follows

from Lemma 1 that the function αn is continuous on [c, d] and satisfies the Lipschitz condition withconstant L1.

Consider the case in which α′nr(tn) > L1, n ∈ {1, 2, . . .}. Let t0 ∈ [c, d] be a limit point of

the sequence tn, and let L2 ∈ (L,L1). For each δ > 0, there exist τ1, τ2 ∈ (a1, b1) such thatt0 < τ1 < τ2 < t0 + δ. Obviously, α(τ2) − α(τ1) < L2(τ2 − τ1). Take an n ∈ {1, 2, . . .} suchthat tn < τ1 and αn(τ2) − αn(τ1) < L2(τ2 − τ1). By following the above argument and by usingLemma 2, we arrive at a contradiction. The remaining cases can be considered in a similar way.

For t ∈ (a, b), we have

α′l(t) ≤ lim inf

n→∞α′

nl(t) ≤ lim supn→∞

α′nl(t) ≤ lim sup

n→∞α′

nr(t) ≤ α′r(t),

α′l(t) ≤ lim inf

n→∞α′

nl(t) ≤ lim infn→∞

α′nr(t) ≤ lim sup

n→∞α′

nr(t) ≤ α′r(t).

If the derivative α′(t) exists at the point t ∈ (a, b), then these inequalities imply that

limn→∞

α′nl(t) = lim

n→∞α′

nr(t) = α′(t).

Suppose that the function α has derivative at points t1, t2 ∈ (a1, b1), t1 < t2. Then for the interval[t1, t2], there exists a k ∈ {1, 2, . . .} such that the inequality

ϕ(t2, αn(t2), α′nl(t2)) − ϕ(t1, αn(t1), α′

nr(t1)) ≥t2∫

t1

f(s, αn(s), α′n(s)) ds



holds for n > k. Now, by passing to the limit, we obtain

ϕ(t2, α(t2), α′(t2)) − ϕ(t1, α(t1), α′(t1)) ≥t2∫

t1

f(s, α(s), α′(s)) ds,

which completes the proof of the lemma.

Lemma 4. If α1, α2 ∈ AG(I,R), then α = max{α1, α2} ∈ AG(I,R).

Proof. The condition α ∈ BB+(I,R) was proved in [2, p. 34] and readily follows from Lemma 1.Let the function α satisfy the Lipschitz condition on the interval [a1, b1] ⊂ I. Let us show thatthe set J = {t ∈ (a1, b1) : α1(t) < α2(t)} is open. Let the inequality α1(t) < α2(t) be satis-fied for t ∈ (a1, b1); then α(t) = α2(t). It follows from the Lipschitz condition for the function αand Lemma 1 that limτ→t+ α1(τ) < α(t). By virtue of the continuity of the function α, we havelimτ→t+ α2(τ) = α(t). Consequently, α2 > α1 in some right neighborhood of the point t. Similarconsiderations hold for the left half-neighborhood. We have thereby shown that J is open. Ob-viously, J consists of finitely or countably many intervals. If either J = ∅ or J = (a1, b1), thencondition (2) is obvious.

Let J = (a2, b2). Consider the case in which a1 < a2, t1 < a2, and t2 ∈ (a2, b2). Then we have

ϕ(a2, α1(a2), α′1l(a2)) − ϕ(t1, α1(t1), α′

1(t1)) + ϕ(t2, α2(t2), α′2(t2)) − ϕ(a2, α2(a2), α′

2r(a2))

≥a2∫

t1

f(s, α1(s), α′1(s)) ds +

t2∫

a2

f(s, α2(s), α′2(s)) ds =

t2∫

t1

f(s, α(s), α′(s)) ds.

It follows from the inequality α′1l(a2) ≤ α′

2r(a2) that ϕ(a2, α(a2), α′1l(a2))−ϕ(a2, α(a2), α′

2r(a2)) ≤ 0.Therefore, condition (2) is satisfied. The remaining cases can be considered in a similar way.

Let J = (a2, b2) ∪ · · · ∪ (ak, bk). The case k = 2 has already been considered. In a similar way,by induction, one can prove inequality (2) for k > 2. Consider the case of countably many intervals(ai, bi). Fix t1, t2 ∈ (a1, b1), t1 < t2, and set

T = {t ∈ [t1, t2] : |α′1l(t)| = ∞∨ |α′

1r(t)| = ∞}.

Obviously, α1(t) < α2(t) for t ∈ T . By Lemma 1, T is a closed set. Consequently, it can becovered by finitely many intervals (ai, bi). Let T ⊂ (a2, b2)∪ · · · ∪ (ak, bk) = Tk and αk+1(t) = α1(t)for t ∈ [t1, t2]\Tk, and let αk+1(t) = α2(t) for t ∈ [t1, t2] ∩ Tk. It follows from the precedingconsiderations that inequality (2) holds for αk+1. Then

ϕ(t2, αi(t2), α′i(t2)) − ϕ(t1, αi(t1), α′

i(t1)) ≥t2∫

t1

f(s, αi(s), α′i(s)) ds

for sufficiently large i. By passing to the limit as i → ∞ in this inequality, we obtain condition (2).The proof of the lemma is complete.

GENERALIZED SOLVABILITY OF THE DIRICHLET PROBLEM

Theorem 1. If A ⊂ AG(I,R), A �= ∅, α = supA, and sup{α(t) : t ∈ I} < ∞, thenα ∈ AG(I,R).

Proof. If A is a finite set, then the assertion of the theorem follows from Lemma 4. Let A bean infinite set, and let G be the set of rational points of the interval I. From elements of the setA, we form a sequence {α1

n}, n = 1, 2, . . . , with the property

sup{α1n(t) : n ∈ {1, 2, . . .}} = α(t), t ∈ G.



We construct a sequence α2n, n = 1, 2, . . . , as follows:

α21 = α1

1, α2n = max{α2

n−1, α1n}, n = 2, 3, . . .

It follows from Lemma 4 that α2n ∈ AG(I,R), n = 1, 2, . . . The sequence α2

n is bounded andmonotone. Consequently, it pointwise converges to a function α∗ : I → R. Obviously, α∗ ≤ α. Itfollows from Lemma 3 that α∗ ∈ AG(I,R). Let us prove the inequality

α(t) ≤ max{ limτ→t−

α∗(τ), limτ→t+

α∗(τ)}, t ∈ (a, b). (4)

If this inequality fails, then there exists an α1 ∈ A such that

α1(t) > max{ limτ→t−


α∗(τ)}.

It follows from Lemma 1 that

α1(t) ≤ max{ limτ→t−

α1(τ), limτ→t+

α1(τ)} ≤ max{ limτ→t−


α∗(τ)} < α1(t).

The contradiction thus obtained implies inequality (4). If the function α∗ is continuous at a pointt ∈ (a, b), then it follows from inequality (4) that α∗(t) ≤ α(t) ≤ α∗(t). Therefore, the functionsα and α∗ coincide at the points of continuity of the function α∗. By Lemma 1, there are at mostcountably many points of discontinuity of the function α∗. Consequently, the set

G1 = {t ∈ I : α(t) �= α∗(t)}

is at most countable.Let the sequence {α3

n}, n = 1, 2, . . . , satisfy the condition

sup{α3n(t) : n ∈ {1, 2, . . .}} = α(t), t ∈ G1.

We define a sequence {α4n}, n = 1, 2, . . . , by setting

α41 = max{α2

1, α31}, α4

n = max{α4n−1, α

2n, α3

n}, n = 2, 3, . . .

It follows from Lemma 4 that α4n ∈ AG(I,R), n = 1, 2, . . . Then the sequence {α4

n} convergespointwise to the function α. By Lemma 3, α ∈ AG(I,R). The proof of the theorem is complete.

Theorem 2. If α ∈ AG(I,R), β ∈ BG(I,R), α ≤ β, A ∈ [α(a), β(a)], and B ∈ [α(b), β(b)],then there exists a generalized solution of the Dirichlet problem

(ϕ(t, x, x′))′ = f(t, x, x′), x(a) = A, x(b) = B, α ≤ x ≤ β. (5)

Proof. Let us show that the function

x = sup{α∗ ∈ AG(I,R) : α ≤ α∗ ≤ β, α∗(a) ≤ A, α∗(b) ≤ B}

is a generalized solution of the Dirichlet problem (5). Obviously, α ≤ x ≤ β. Let α0(t) = α(t),t ∈ (a, b), α0(a) = A, and α0(b) = B. It follows from Definitions 3 and 4 that α0 ∈ AG(I,R).Consequently, x(a) = A and x(b) = B. Let us show that, on each interval [c, d] ⊂ I on which xsatisfies the Lipschitz condition, x is a solution of Eq. (1). If limτ→c+ β(τ) > x(c), then the one-sidedlocal solvability proved in [18] implies that there exists a t1 ∈ (c, d] and a solution y : [c, t1] → Rof Eq. (1) such that x(τ) ≤ y(τ) ≤ β(τ), τ ∈ [c, t1], y(c) = x(c), and y(t1) = x(t1).

We define a function z : [c, d] → R as follows: z(τ) = y(τ), τ ∈ [c, t1], and z(τ) = x(τ),τ ∈ [t1, d]. It follows from the proof of Lemma 4 that z ∈ AG([c, d], R). Therefore, x(τ) = y(τ),τ ∈ [c, t1]. The case in which limτ→c+ β(τ) = x(c) is similar to the preceding one. The coincidenceof x with a solution of Eq. (1) in a neighborhood of the point d can be proved in a similar way.



Let t ∈ (c, d). If limτ→t− β(τ) > x(t) and limτ→t+ β(τ) > x(t), then β(t) > x(t), and, byfollowing the lines of the preceding argument and by using the one-sided local solvability, we findthat the function x coincides with a solution of Eq. (1) in some neighborhood of the point t.

Consider the case in which limτ→t+ β(τ) = x(t). If limτ→t− β(τ) > x(t), then it follows fromLemma 1 that β′

r(t) = −∞ ≥ limτ→t+ β′r(τ). Therefore, limτ→t+ β′

r(τ) = −∞, which contradictsthe condition β ≥ x. From the relation limτ→t− β(τ) = limτ→t+ β(τ), we obtain β(t) = x(t). Thenwe have the inequalities

β′l(t) ≥ β′

r(t) ≥ x′r(t) ≥ x′

l(t) ≥ β′l(t),

which imply the existence of the derivatives β′(t), x′(t), and β′(t) = x′(t). By Lemma 1, thefunction β satisfies the Lipschitz condition in some neighborhood of the point t. The one-sided localsolvability implies that there exist points t1 ∈ (c, t) and t2 ∈ (t, d) and a solution y : [t1, t2] → Rof Eq. (1) such that x(τ) ≤ y(τ) ≤ β(τ), τ ∈ [t1, t2], y(t1) = x(t1), and y(t2) = x(t2). Just asabove, one can show that x(τ) = y(τ), τ ∈ [t1, t2]. It follows that x is a solution of Eq. (1) on theinterval [c, d].

To show that x is a generalized solution of Eq. (1), it suffices to prove the inclusion x ∈BB−(I,R). Since x ∈ BB+(I,R), it suffices to prove the conditions

limτ→t−

x′l(τ) = lim

τ→t+x′

r(τ), t ∈ (a, b), (6)

x′r(a) = lim

τ→a+x′

r(τ), (7)

x′l(b) = lim

τ→b−x′

l(τ). (8)

If the derivatives x′r(t) and x′

l(t) are finite at some point t ∈ (a, b), then, by Lemma 1, they arebounded in some neighborhood of the point t. Then the function x satisfies the Lipschitz conditionin that neighborhood and is a solution of Eq. (1) there. The validity of condition (6) in this caseis obvious. It remains to consider the case in which either x′

r(t) or x′l(t) is either +∞ or −∞.

Let x′r(t) = +∞. If we show that limτ→t− x′

l(τ) = +∞, then the inequality

limτ→t−

x′l(τ) ≤ x′

l(t) ≤ x′r(t) ≤ lim

τ→t+x′

r(τ)

will imply condition (6). Suppose the contrary. Then there exists a c1 ∈ R such that x′l(τ) < c1.

Now let us prove the inequality

min{ limτ→t−

x(τ), limτ→t+

x(τ)} < min{ limτ→t−

β(τ), limτ→t+

β(τ)}. (9)

If inequality (9) fails, then

limτ→t−

x(τ) = limτ→t−

β(τ) ≤ limτ→t+

x(τ) ≤ limτ→t+

β(τ) (10)

orlim

τ→t+x(τ) = lim

τ→t+β(τ) ≤ lim

τ→t−x(τ) ≤ lim

τ→t−β(τ). (11)

Let condition (10) be satisfied. It follows from Lemma 1 that

β′r(t) ≤ β′

l(t) ≤ limτ→t−

β′l(τ) = lim

τ→t−β′

r(τ) ≤ limτ→t−

x′r(τ) < c1. (12)

If limτ→t+ x(τ) < β(τ), then it follows from condition (10) that β′l(t) = +∞, which contradicts

condition (12). For β(t) < limτ→t+ x(τ), from condition (10), we obtain β′r(t) = +∞, which

contradicts condition (12). Consequently, β(t) = limτ→t+ x(τ).Let us show that x(t) = β(t). Indeed, it follows from the inequality x(t) < β(t) that the

function x0, which differs from x only at the point t and is equal to β(t) at this point, is a generalizedlower function, which contradicts the definition of the function x. From the conditions x(t) = β(t),(10), and (12), we have x′

r(t) ≤ β′r(t) < c1, which contradicts the relation x′

r(t) = +∞. Let



condition (11) be satisfied. It follows from Definitions 3 and 4 that limτ→t+ x′r(τ) = +∞. Therefore,

limτ→t+ β′r(τ) = +∞. By using Definitions 3 and 4 again, we obtain the inequalities

+∞ = limτ→t+

β′r(τ) ≤ β′

r(t) ≤ β′l(t) ≤ lim

τ→t−β′

l(τ).

Consequently, β is a monotone function, and limτ→t+ β(τ) = limτ→t− β(τ); i.e., condition (10) issatisfied. This completes the proof of inequality (9).

Consider the case in which limτ→t− x(τ) ≤ limτ→t+ x(τ). Take points t1 ∈ (a, t) and t2 ∈ (t, b)close enough to t to ensure that the solution u of the Cauchy problem

(ϕ(t, u, u′))′ = f(t, u, u′), u(t1) = x(t1), u′(t1) = c1,

exists on the interval [t1, t2] and satisfies the conditions x′r(t1) < u′(t1), u(t2) < x(t2), and u(τ) ≤

β(τ), τ ∈ [t1, t2]. It follows from the results obtained above that the function x0(τ) = x(τ),τ ∈ I\[t1, t2], x0(τ) = max{x(τ), u(τ)}, τ ∈ [t1, t2], is a generalized lower function, and thiscontradicts the definition of the function x. The case of limτ→t− x(τ) > limτ→t+ x(τ) can beconsidered in a similar way. This completes the study of the case in which x′

r(t) = +∞. The caseof x′

l(t) = −∞ can be considered in a similar way. If x′r(t) = −∞, then x′

l(t) ≤ x′r(t) = −∞. Now

it follows from the relation x′l(t) = +∞ that x′

r(t) = +∞. This completes the proof of relation (6).Now let us prove relation (7). If x(a) ≤ limτ→a+ x(τ), then condition (7) follows from Lemma 1.

Let x(a) > limτ→a+ x(τ). Then x′r(a) = −∞. If limτ→a+ x(τ) < limτ→a+ β(τ), then from the

relations−∞ = β′

r(a) ≥ limτ→a+

β′r(τ) ≥ lim

τ→a+x′

r(τ) > −∞,

we arrive at a contradiction. The proof of condition (7) is complete. Condition (8) can be provedin a similar way. Consequently, x ∈ BB−(I,R), which completes the proof of the theorem.

Remark 1. Under the assumptions of Theorem 1, there exist a maximum and minimumgeneralized solutions of the Dirichlet problem (5). Indeed, let X be the set of generalized solutionsof the Dirichlet problem, and let

α1(t) = sup{x(t) : x ∈ X}, β1(t) = inf{x(t) : x ∈ X}, t ∈ I.

Then α1 ∈ AG(I,R) and β1 ∈ BG(I,R). By Theorem 2, there exist generalized solutions y and zof the Dirichlet problems

(ϕ(t, y, y′))′ = f(t, y, y′), y(a) = A, y(b) = B, α ≤ y ≤ β1,

(ϕ(t, z, z′))′ = f(t, z, z′), z(a) = A, z(b) = B, α1 ≤ z ≤ β.

Obviously, y is a minimum generalized solution of the Dirichlet problem (5), and z is a maxi-mum one.

Remark 2. If α ∈ AG(I,R), β ∈ BG(I,R), and α ≤ β, then generalized global solvabilitytakes place between the functions α and β: for arbitrary c ∈ [a, b), d ∈ (c, b], C ∈ [α(c), β(c)], andD ∈ [α(d), β(d)], there exists a generalized solution x : [c, d] → R of Eq. (1) such that x(c) = C,x(d) = D, and α(t) ≤ x(t) ≤ β(t), t ∈ [c, d].

Theorem 3. If α, β : I → R are bounded functions, α ≤ β, and generalized global solvabilitytakes place between the functions α and β, then α ∈ AG(I,R) and β ∈ BG(I,R).

Proof. Let us show that α ∈ BB+(I,R). Let us prove the existence of the derivative α′r(t)

at the point t ∈ [a, b). Suppose the contrary. Then there exist c1, c2 ∈ R such that D+α(t) <c2 < c1 < D+α(t). Consequently, for each δ > 0, there exist points t1, t2, t3, t4 ∈ I such thatt < t1 < t2 < t3 < t4 < t + δ, α(t2) − α(t1) > c1(t2 − t1), and α(t4) − α(t3) < c2(t4 − t3). Letx : [t1, t4] → R be a generalized solution lying between α and β and satisfying the conditionsx(t1) = α(t1) and x(t4) = α(t4). Consequently, x(t2)−x(t1) > c1(t2− t1), x(t4)−x(t3) < c2(t4− t3),and there exist points t5, t6 ∈ [t1, t4] such that t5 < t6, x′(t5) = c1, x′(t6) = c2, and c2 < x′(τ) < c1,



τ ∈ (t5, t6). By using Lemma 2, we arrive at a contradiction. In a similar way, one can show thatthere exists a derivative α′

l(t), t ∈ (a, b], limits limτ→t− α′l(τ), t ∈ (a, b], and limτ→t+ α′

r(τ), t ∈ [a, b),and

α′l(t) ≤ α′

r(t), t ∈ (a, b), α′l(t) ≥ lim

τ→t−α′

l(τ), t ∈ (a, b], α′r(t) ≤ lim

τ→t+α′

r(τ), t ∈ [a, b).

Consequently, α ∈ BB+(I,R).Let the function α satisfy the Lipschitz condition with constant L > 0 on the interval [c, d] ⊂ I.

Let us construct a function sequence

αn ∈ AG([c, d], R), n = 1, 2, . . .

Let x1 ∈ [c, d] → R be a generalized solution of Eq. (1) lying between α and β and satisfying theconditions x1(c) = α(c) and x1(d) = α(d). Then α1 = x1. We set t1 = (d − c)/2 and define x21

and x22 by analogy with x1 for the intervals [c, t1] and [t1, d]. Then α2(t) = x21(t), t ∈ [c, t1], andα2(t) = x22(t), t ∈ [t1, d]. Next, by dividing the intervals [c, t1] and [t1, d] into halves, we obtain thefunction α3. The remaining functions αn are constructed in a similar way. Obviously,

αn ∈ AG([c, d], R), n = 1, 2, . . .

By virtue of above argument and Lemma 2, there exists a number k ∈ {1, 2, . . .} such that, forall n > k, the functions αn satisfy the Lipschitz condition with constant L + 1. Consequently,the sequence αn uniformly converges to α, and the function α satisfies condition (2) on the intervalby Lemma 3. The proof for the function β can be carried out in a similar way. The proof of thetheorem is complete.

Definition 6. If α ∈ AG(I,R) and β ∈ BG(I,R), α ≤ β, and for arbitrary A ∈ [α(a), β(a)]and B ∈ [α(b), β(b)], each generalized solution of the Dirichlet problem

(ϕ(t, x, x′))′ = f(t, x, x′), x(a) = A, x(b) = B, α ≤ x ≤ β, (13)

is its solution (in the ordinary sense), then we say that the compactness condition is satisfied.

Obviously, this condition is less restrictive than the Schrader condition in [19].

Theorem 4. If α ∈ AG(I,R), β ∈ BG(I,R), α ≤ β, and the compactness condition is satisfied ,then the Dirichlet problem (13) is solvable.

Proof. The Dirichlet problem (13) has a generalized solution by Theorem 2, and it follows fromthe compactness condition that it is its solution.

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Generalized lower and upper functions for the φ-Laplacian

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