On a Nonlocal Boundary Value Problem for Second Order Nonlinear Singular Differential Equations

Abstract Criteria for the existence and uniqueness of a solution of the boundary value problem are established, where ƒ :]a, b[×R 2 → R satisfies the local Carathéodory conditions, and μ : [a, b] → R is the function of bounded variation. These criteria apply to the case where the function ƒ has nonintegrable singularities in the first argument at the points a and b.

where µ : [a, b] → R is the function of bounded variation.The criteria for the unique solvability of the problem in the linear case are contained in [7,8].In the nonlinear case a problem of type (1.1), (1.2) has been considered in [4][5][6].However, in these works µ is assumed to be a piecewise constant function (µ(t) = 0 for a ≤ t ≤ t 0 and µ(t) = 1 for t 0 < t ≤ b).
The theorems of the existence and uniqueness of a solution of problem (1.1), (1.2) given in the present paper cover the case where µ is, generally speaking, not piecewise constant, and f is not integrable in the first argument on the segment [a, b], having singularities at the points t = a and t = b.
Before we pass to formulating the main results let us introduce the following definitions.
Remark 1.4.In the case where µ = 0 (see [2]) or f does not depend on the third argument and is integrable in the first argument in the neighborhood of the point t = b, then we can neglect condition (1.11) ((1.12)) in Theorems 1.1 and 1.2.

Some Auxiliary Propositions
In this paragraph we shall prove some properties of solutions of the equations and where Note that in this case the well-known Green's theorem is valid (see [8], Theorem 1.1).More precisely, the following lemma is true.
Lemma 2.1.Let condition (2.2) be fulfilled.Then for problem (2.1), (1.2) to be uniquely solvable, it is necessary and sufficient that the corresponding homogeneous problem (2.1 0 ), (1.2) have only the zero solution.If the latter condition is fulfilled, then there exists the unique Green's function where Remark 2.1.Taking into account Lemma 2.1 of [5] (see also Lemmas 1.1 and 1.1 of the monograph [3]), we can easily see that the Green's function G admits the estimates where c > 0 is a constant.

Properties of the set U
(2.3) Proof.Let us assume the contrary that the lemma is invalid.Then there exists satisfies the inequalities Introduce the function where u 1 is the solution of equation (2.1 0 ) satisfying the initial conditions Clearly, v is the solution of equation (2.1 0 ) and According to the condition of the lemma and to inequality (2.4) we easily find that Consequently, for some point t 0 ∈]a 1 , b[ we have v(t 0 ) < 0. Taking now into account (2.6), we obtain that the function v has at least two zeros on ]a 1 , b], which contradicts the condition of the lemma.
The following lemma is proved analogously.
Lemma 2.3.Let condition (2.3) be fulfilled.Then there exists 3) be fulfilled, and let ε 0 be the number occurring in Lemma 2.3.For any ε ∈]0, ε 0 [ denote by u ε the solution of equation (2.1 0 ) satisfying the conditions Assume that δ i (ε) → 0 for ε → 0, i = 1, 2. Then we can easily see that and Then for some function has a solution u such that u(t) > 0 for a < t < b, u(a+) = 0, u(b−) = 0; (2.11) moreover, we may assume that g 1 ≡ p 1 in some neighborhood of the point b.
Proof.Because of (2.8) there exists a 1 ∈ [a, b[ such that equation (2.1 0 ) has the solution u satisfying the conditions If a 1 = a, then we have nothing to prove.Therefore we shall assume that a 1 > a.
Let v be the solution of the boundary value problem Suppose Bearing this in mind, we can easily verify that the function satisfies conditions (2.11) and is the solution of equation (2.10), where Lemma 2.5.Let µ be nondecreasing, conditions (2.7) be fulfilled, and (2.12) Then for some function In the case where (2.8) is fulfilled, by Lemma 2.4 we can assume without loss of generality that equation (2.1 0 ) has the solution u satisfying conditions (2.11).Thus we may assume that if (2.12) is fulfilled, then equation (2.1 0 ) has the solution u satisfying conditions (2.14).
Let us choose (2.15) Then (cf., for example, [3], Lemma 1.5 1 ) Hence the boundary value problem has the unique solution u 0 .Moreover, u 0 (t) ≥ 1 for a 0 ≤ t ≤ b, since the mapping t −→ u 0 (t) σ(p 2 )(t) does not increase.Suppose It is easily seen that for some α 0 > 0 the inequalities Let u 1 be the solution of the singular Cauchy problem where Let us show that from (2.20) we find that which, by (2.17), implies Because of (2.15) and (2.24) we find from (2.23) that Consequently, (2.20) is fulfilled.
Owing to (2.16) and (2.21), we easily get Thus we have shown that the solution u 1 of problem (2.20) satisfies the conditions According to the comparison theorem (see [7], Theorem 1) and Remark 1, the equation v = p 0 (t)v + p 2 (t)v , where has the solution v satisfying the conditions for a ≤ b, where Clearly, u is the solution of equation (2.10), where where u 1 is the solution of problem (2.1 0 ), (2.5).
Proof.Let us assume the contrary that the lemma is invalid.Then for any natural k there exists a measurable function where u 1 (•, k) is the solution of the singular Cauchy problem Introduce the functions On the other hand, from Lemma 2.1 of [5] (see also [2], the proof of Lemma 2.2) it follows that the sequences In the sequel we shall need the following notations: Because of Lemmas 2.1, 2.2 and 2.6, and taking into consideration Lemma 2.1 of [5], we easily see that the lemma below is valid.
Lemma 2.8.Let (2.26) be fulfilled.Then there exists a positive number r such that no matter what the functions g, g i ∈ L loc (]a, b[), i = 1, 2, satisfying the conditions has the unique solution u, and where Because of conditions (2.26), (2.29) and (2.30) from Theorem 1 of [7] it follows that problem (2.31), (2.32) has the unique solution u.
Let v be the solution of the problem According to Lemma 2.
moreover, either Hence w satisfies the conditions where c 1 is a positive number not depending on u.
Estimate now |u (t)|.Let t 1 ∈]a, b[ be an arbitrary point, where u (t 1 ) = 0.For definiteness, we shall assume that u(t 1 )u (t 1 ) ≥ 0. Then either or there exists First we assume that condition (2.39) is fulfilled.Multiplying both parts of equality (2.31) by σ b−ε (g 2 )(t), where ε ∈]0, b − t 1 [, and then integrating from t 1 to b − ε, we obtain whence, owing to (2.30), we find As ε tends to zero, from the last inequality we get (2.41) On the other hand, according to (2.38), Therefore from (2.41) we have where Consequently, Similarly, we can show that estimate (2.42) also holds in the case where (2.40) is fulfilled.
Because t 1 is taken to be arbitrary, we have from (2.42) that (2.33) holds, where r = max{ c 1 , c 2 }.Lemma 2.9.Let (2.26) be fulfilled and let t 1 ∈]a, b[.Then there exists a positive number r 2 such that no matter what the functions g, g i ∈ L loc (]a, b[) satisfying conditions (2.29), (2.30) and g 1 ≡ p 1 might be, the solution u of the boundary value problem (2.31), (2.32) admits the estimate where Let r be the number occurring in Lemma 2.8 and let u be the solution of problem (2.31), (2.32), Let v be the solution of the boundary value problem By Lemma 2.7 (in the case where µ ≡ 0) we have that v(t) ≥ 0 for t 1 ≤ t ≤ b.Similarly to that done in proving Lemma 2.8, we can show that (2.44) Obviously, where G 0 is the Green's function of the problem and u 2 is the solution of the singular Cauchy problem ds, i = 1, 2, maps continuously the space C [a, b]; R 2 into its own compact subset.Therefore, owing to Schauder's principle, there exists a vector-function ( From this we find that x 2 (t) = x 1 (t)σ ab (p 2 )(t), and the function u(t) = x 1 (t) for a ≤ t ≤ b is the solution of problem (1.1), (1.2).

Proof of the Main Results
Proof of Theorem 1.1.Let r, r 2 , H 0 , H 1 and H be respectively the numbers and the functions appearing in Lemmas 2.8 and 2.9.Assume that Therefore, by Lemma 2.10, for any natural k the boundary value problem has at least one solution u k .On account of (3.2), u k is the solution of the equation where and  Let g 1 be the function chosen by Lemma 2.5, and let u 0 = 0 be the solution of problem (2.10), (1.2).The function f (t, x, y) = g 1 (t)x + p 2 (t)y + u 0 (t) satisfies conditions (1.8), (1.11).On the other hand, we can easily see that (cf., for example, [7], Theorem 1) the equation u = g 1 (t)u + p 2 (t)u + u 0 (t) has no solution satisfying the boundary conditions (1.2).Let g 1 be the function chosen by Lemma 2.5.Clearly, the function g 1 (t)x + p 2 (t)y satisfies conditions (1.20), (1.21) and (1.12), where p 2 (t, x, y) = p 2 (t, x, y) = p 2 (t) and

1 . 2 p
ON A NONLOCAL BOUNDARY VALUE PROBLEM FOR SECOND ORDER NONLINEAR SINGULAR DIFFERENTIAL EQUATIONS A. LOMTATIDZE AND L. MALAGUTI Abstract.Criteria for the existence and uniqueness of a solution of the boundary value problem u = f (t, u, u ); u(a+) = 0, u(b−) = b a u(s) dµ(s) are established, where f :]a, b[×R 2 → R satisfies the local Carathéodory conditions, and µ : [a, b] → R is the function of bounded variation.These criteria apply to the case where the function f has nonintegrable singularities in the first argument at the points a and b.Statement of the Main Results Below we shall use the following notations: R is a set of real numbers.L([a, b]) is the set of functions p :]a, b[→ R which are Lebesgue integrable on [a, b].L loc (]a, b[) is the set of functions p :]a, b[→ R which are Lebesgue integrable on [a + ε, b − ε] for arbitrarily small ε > 0. K 0 (]a, b[×R 2 ) is the set of functions g :]a, b[×R 2 → R for which the mapping t −→ g(t, x 1 (t), x 2 (t)) is measurable no matter what the continuous functions x i :]a, b[→ R, i = 1, 2, might be.σ : L loc (]a, b[) → L loc (]a, b[) is the operator defined by the equality σ(p)(t) = exp t a+b (s) ds .If σ(p) ∈ L([a, b]), α ∈ [a, b] and β ∈]α, b], then σ α (p)(t) = 1 σ(p)(t) )(s) ds.u(s+) and u(s−) are the limits of the function u at the point s from the right and from the left.If µ : [a, b] → R is a function of bounded variation, then for each t ∈ [a, b] by µ * (t) we denote a total variation of the function µ on the segment [a, t].Under the solution of the equation u = f (t, u, u ), (1.1) where f :]a, b[×R 2 → R satisfies the Carathéodory conditions on every compact contained in ]a, b[×R 2 , we understand the function u :]a, b[→ R which is absolutely continuous together with its first derivative on every segment from ]a, b[, satisfying (1.1) a.e.In the present paper we concern ourselves with the problem of the existence and uniqueness of a solution of equation (1.1) satisfying the boundary conditions u(a+) = 0, u(b−) = b a u(s) dµ(s), ( 9), equation (2.10) has a solution u such that u(t) > 0 for a < t < b, u(a+) = 0, u(b−) = b a u(s)dµ(s); (2.13) moreover, we may assume that g 1 ≡ p 1 in a neighborhood of the point b.Proof.Owing to (2.12), either (2.8) is fulfilled or (p 1 , p 2 ) ∈ U 0 (]a, b[), and equation (2.1 0 ) has the solution u satisfying the conditions u(t) > 0 for a < t ≤ b, u(a+) = 0, u(b−) < b a u(s)dµ(s).(2.14)
Then there exists a function f satisfying the conditions of Theorem 1.1 for which problem (1.1), (1.2) has no solution.
[8]and p 3 (t) ≡ 0 in a neighborhood of the point b.Taking into account Theorem 1.2 in[8], we obtain from Theorem 1.1 that in this case problem (1.1), (1.2) has at least one solution.As it is seen from the example, the function f may have nonintegrable singularities for t = a and t = b.
b[ and a function p :]a, b[→]0, +∞[ such that conditions (1.15)-(1.17)and (1.22) are fulfilled, where p 1 and p 2 are the functions defined by equalities (1.18) and (1.19).Moreover, let b a Because the sequence (g k ) +∞k=1 is uniformly bounded and equicontinuous in ]a, b[ 1 , we may assume without loss of generality that it converges uniformly in ]a, b[.So we can easily see that the function 2, . . . .