Maximum Principles for the General Operator of ...

Maximum and Anti – Maximum Principles for the General Operator of Second Order with Variable Coefficients by Irina V. Barteneva 1 , Alberto Cabada2∗and Alexander O. Ignatyev3 1

Donetsk Railway Institute Gornaya Street, 6 Donetsk – 83018 Ukraine

2

Departamento de An´alise Matem´atica Facultade de Matem´aticas

Universidade de Santiago de Compostela 15706 – Santiago Spain 3

Institute for Applied Mathematics and Mechanics R.Luxemburg Street,74 Donetsk – 83114 Ukraine

∗

Research partially supported by DGESIC, project PB97 – 0552 – C02, and by Xunta

de Galicia, project XUGA20701B98

1

Keywords: Maximum principles, Comparison results, lower and upper solutions, nonlinear boundary value problems. AMS subject classification: 34B15, 34A45

Mailing address: Alberto Cabada, Depto. de An´ alise Matem´ atica, Facultade de Matem´ aticas, Univ. de Santiago de Compostela, 15706, Santiago de Compostela, Spain FAX number: 34 981 59 70 54 E–mail: [email protected]

2

1

Introduction

The solvability of nonlinear boundary value problems via the method of lower and upper solutions represents a fundamental tool in the theory of ordinary differential equations. For the second order case involving different boundary value conditions as Neumann or periodic (among others), this study has been done classically when the lower solution is under the upper solution, see [4] and references therein. The non classical case, i. e. the lower solution is over the upper solution, has been treated in more recent papers, for instance in [6] the periodic case is considered, in [3] the Neumann one and in [2] the separated boundary conditions are studied. In all the cases, to obtain existence and approximation of solutions via the monotone method for problem u00 (t) + a u0 (t) + f (t, u(t)) = 0 with different boundary conditions, is translated to the study of maximum and anti–maximum principles for the operator u00 + a u0 + b u, with a, b real numbers (a = 0 in [2]). Thus, our main objective is to establish the positivity or negativity properties of the general operator defined by the following expression L(p, q) u(t) ≡ u00 (t) + p(t) u0 (t) + q(t) u(t),

t ∈ J = [0, R],

(1.1)

where p, q ∈ L1 (J) are given and the Neumann boundary conditions are considered u0 (0) = A;

u0 (R) = B.

(1.2)

This study is done in section 3, for it we study the sign of the unique solutions of some Cauchy homogeneous problems. This study is given in section 2. 3

In section 4, we obtain comparison results for operator L(p, q) with the following boundary conditions u0 (0) = A;

u(R) = B, Mixed conditions,

(1.3)

u(0) = A;

u0 (R) = B, Mixed conditions,

(1.4)

u(R) = B, Dirichlet conditions,

(1.5)

u(0) = A; and u(0) − u(R) = A;

u0 (0) − u0 (R) = B, Periodic conditions.

(1.6)

This results are derived from the ones given in section 3 for the Neumann case. Finally, in section 5 we apply the given results to obtain existence and approximation of solutions of the second order boundary value problem u00 (t) + p(t) u0 (t) + f (t, u(t)) = 0, for a. e. t ∈ J,

(1.7)

with the above mentioned boundary value conditions in presence of a pair of ordered lower and an upper solution.

2

Positive solutions of second order initial homogeneous problems

In this section we study the sign of the solutions v0 and vR of the Cauchy problems v000 (t) + p(t) v00 (t) + q(t) v0 (t) = 0, v0 (0) = 1, 4

v00 (0) = 0,

(2.1)

and 00 0 vR (t) + p(t) vR (t) + q(t) vR (t) = 0, vR (R) = 1,

0 vR (R) = 0.

(2.2)

Firstly, we define the function φ for a, b > 0, as follows

   π a 2   √ √ , − arctan    4b − a2 2 4b − a2   2/a, φ(a, b) = !  √  2 − 4b  1 a + a    √ log ,  √ 2 a − 4b a − a2 − 4b

if

4b > a2 ;

if

4b = a2 ;

if

4b < a2 ;

(2.3)

√ by continuity, we define φ(0, b) = π/(2 b) for all b > 0. We obtain the following results. Lemma 2.1 Suppose that q(t) < 0 for a. e. t ∈ J. Then v0 > 0 in J and v00 > 0 in (0; R]. Proof. Using the fact that (e

Rt 0

p(s) ds 0 v0 (t))0

= −q(t) e

Rt 0

p(s) ds

v0 (t) for a. e. t ∈ J,

(2.4)

we know that, for all t1 ∈ (0, R], such that v0 > 0 in [0, t1 ) (obviously, such a point exists), integrating between 0 and t1 we conclude that R t1

e

0

p(s) ds 0 v0 (t1 )

> v00 (0) = 0.

Thus, v00 > 0 in J and, clearly, v0 > 0 in J.

t u

Lemma 2.2 Let a ≥ 0 and b > 0 be such that p(t) ≥ −a and q(t) ≤ b for a. e. t ∈ J. 5

Consider function φ defined in (2.3) and Z0 (p, q) be the first positive zero of v0 . Then Z0 (p, q) ≥ φ(a, b). Furthermore, if q(t) > 0 for a. e. t ∈ J, we have that v00 < 0 in J. Proof. We know that v0 (t) > 0 on [0, Z0 (p, q)). Clearly v00 (Z0 (p, q)) < 0 because if it is equals to zero, we contradict the uniqueness of solution of problem L(p, q) u = 0 passing through Z0 (p, q). Denote γ(t) ≡

v00 (t) , v0 (t)

t ∈ J,

(2.5)

γ is a continuous function in [0, Z0 (p, q)) such that γ(0) = 0

and

lim

t→Z0 (p,q)−

γ(t) = −∞.

(2.6)

Clearly, function γ is an absolutely continuous function in [0, R1 ] for all R1 ∈ (0, Z0 (p, q)). Thus, from (2.5), we have that for all R1 ∈ (0, Z0 (p, q)) (see [1]) v00 = γv0 ,

v000 = γ 0 v0 + γv00 = (γ 0 + γ 2 )v0 for a. e. t ∈ [0, R1 ].

(2.7)

In consequence the previous equality holds in [0, Z0 (p, q)). After substitution of (2.7) into (2.1), we obtain γ 0 + γ 2 + pγ + q = 0 for a. e. t ∈ [0, Z0 (p, q)); that is: γ 0 (t) = −γ 2 (t) − p(t)γ(t) − q(t) for a.e. t ∈ [0, Z0 (p, q)). By equation (2.6), we know that there exists τ ∈ [0, Z0 (p, q)) such that γ(τ ) = 0 and γ(t) < 0 for all t ∈ (τ, Z0 (p, q)). On this interval dγ ≥ −(γ 2 − aγ + b). dt 6

Thus, γ2

dγ ≥ −dt, − aγ + b

and then Z

−∞

dγ ≥ −(Z0 (p, q) − τ ) ≥ −Z0 (p, q). − aγ + b 0 Z ∞ Z 0 dγ dγ Since φ(a, b) ≡ = , we verify the first 2 2 γ + aγ + b 0 −∞ γ − aγ + b assertion of the Lemma. γ2

The fact that v00 < 0 in J is deduced from expression (2.4) and integrating t u

between 0 and t. Analogously one can prove the following results.

Lemma 2.3 Suppose that q(t) < 0 for a. e. t ∈ J. Then vR > 0 in J and 0 < 0 in [0, R). vR

Lemma 2.4 Let a ≥ 0 and b > 0 be such that p(t) ≤ a and q(t) ≤ b for a. e. t ∈ J. Consider the function φ defined in (2.3) and let ZR (p, q) be the biggest zero of vR such that ZR (p, q) < R. Then R − ZR (p, q) ≥ φ(a, b). 0 > 0 in J. If, in addition, q(t) > 0 for a. e. t ∈ J, we have that vR

Remark 2.1 The results exposed in Lemmas 2.2 and 2.4 are optimal in the sense that, as it is proved in [3], in the particular case of p and q are two constant functions, (q ≡ b, and p ≡ a or p ≡ −a) Z0 (±a, b) = φ(a, b) = R − ZR (±a, b). Note that in Lemmas 2.1 and 2.3 we do not impose any condition in the sign of function p and even in its boundedness. 7

Remark 2.2 Note that if q takes negative values in a positive measure set 0 (0) > 0. of J, we cannot assure that v00 (R) < 0 or vR

To see this, consider q as a negative function in [0, R1 ) for some R1 < R and q > 0 in (R1 , R]. We know, by Lemma 2.1, that v00 > 0 in [0, R1 ]. Thus, for R near enough to R1 we have that v00 (R) > 0. 0 (0). A similar argument is valid to study the sign of vR

The following result is proved in [3]: Proposition 2.1 For all a ≥ 0 function φ(a, ·) is continuous and strictly decreasing in (0, +∞) with φ(a, 0+ ) = +∞ and φ(a, +∞) = 0. Furthermore the equation φ(a, b) = R defines a function b ≡ b(R, a) strictly decreasing with respect to R > 0.

3

Neumann Problem

In this section we study the existence of constant sign solutions of the problem L(p, q) u(t) = h(t), for a. e. t ∈ J, u0 (0) = A, u0 (R) = B.

(3.1)

Here p, q, h ∈ L1 (J) and operator L(p, q) is defined in (1.1). Of course, (3.1) has a unique solution u ∈ W 2,1 (J) provided that the associated homogeneous problem (h ≡ 0 and A = B = 0) has only the trivial solution. We are interested in the set of functions q for which (3.1) is uniquely solvable and whenever h ≥ 0 and A ≥ 0 ≥ B we have u ≥ 0 or u ≤ 0 in J. To be concise, we look for functions p and q such that every u ∈ Wl (J) verifying L(p, q) u ≥ 0 for a. e. t ∈ J and u0 (0) ≥ 0 ≥ u0 (R) implies u ≥ 0 8

or u ≤ 0 in J, where l ≥ 0 and by Wl (J) we denote Wl (J) = {u ∈ C(J), u0 |[Tk ,Tk+1 ] ∈ AC([Tk , Tk+1 ]), u0 (Tk+ ) ≥ u0 (Tk− ); k = 0, . . . , l} and 0 = T0 < T1 < . . . < Tl < Tl+1 = R. This study continues the one done in [3] with p and q constant functions and u ∈ C 2 (J). There, optimal results were obtained. If (3.1) holds, write the equation in self – adjoint form

(e

Rt 0

Rt

p(s) ds 0

u (t))0 + q(t) e

0

p(s) ds

Rt

u(t) = e

p(s) ds

0

h(t), for a.e. t ∈ J,

(3.2)

and proceed in the same way with (2.1) and (2.2). Multiplying (3.2) by v0 and (2.1) by u, integrating by parts and subtracting we are left with

RR

−e

0

p(s) ds 0 v0 (R) u(R)

R Rt

Z =

e

0

p(s) ds

h(t) v0 (t) dt +

0

(3.3) +A − e

RR 0

p(s) ds

v0 (R) B +

l X

R Tk

e

Tk−1

p(s) ds

v0 (Tk ) (u0 (Tk+ ) − u0 (Tk− )).

k=1

In the same way, using (2.2), we obtain

0 vR (0) u(0)

R Rt

Z =

e

0

p(s) ds

h(t) vR (t) dt +

0

(3.4) +vR (0) A − e

RR 0

p(s) ds

B+

l X

R Tk

e

Tk−1

p(s) ds

k=1

9

v0 (Tk ) (u0 (Tk+ ) − u0 (Tk− )).

It is obvious that the condition 0 v0 (t) > 0 in [0, R), v00 (R) > 0, vR (t) > 0 in (0, R], vR (0) < 0, and

(3.5) u0 (Tk+ ) ≥ u0 (Tk− ); k = 1, . . . , l implies that u(0) ≤ 0 and u(R) ≤ 0 whenever h(t) ≥ 0 in [0, R] and B ≤ 0 ≤ A. In the same way, we have that 0 v0 (t) > 0 in [0, R), v00 (R) < 0, vR (t) > 0 in (0, R], vR (0) > 0, and

(3.6) u0 (Tk+ ) ≥ u0 (Tk− ); k = 1, . . . , l implies that u(0) ≥ 0 and u(R) ≥ 0 whenever h(t) ≥ 0 in [0, R] and B ≤ 0 ≤ A. In both cases the inequalities are strict if h 6≡ 0 in J. Thus, we are in a position to prove the following comparison results. Theorem 3.1 Suppose that q(t) < 0 for a. e. t ∈ J. Then if u ∈ Wl (J) for some l ≥ 0, L(p, q) u ≥ 0 for a. e. t ∈ J and u0 (0) ≥ 0 ≥ u0 (R), we conclude that u ≤ 0 in J. Proof. We do the proof for the case l = 1, as we can see the general case holds similarly. We denote T1 as T . Lemmas 2.1 and 2.3 imply the inequalities (3.5), in consequence u(0) ≤ 0 and u(R) ≤ 0. Now, let t1 < t2 be such that u(t1 ) = u(t2 ) = 0 and u > 0 in

10

0 − (t1 , t2 ). Clearly, u0 (t+ 1 ) ≥ 0 ≥ u (t2 ), in consequence,



e

Rt 0 p(s) ds q(t)u(t) > 0 for a. e t ∈ (t1 , t2 ). u (t) ≥ −e t1

Rt

p(s) ds 0

t1

Then, if T ∈ (t1 , t2 ), we have that

R t2

0≥e

t1

p(s) ds 0

RT

u (t2 ) + e

t1

p(s) ds

(u0 (T − ) − u0 (T + )) > u0 (t1 ) ≥ 0,

which is a contradiction. If T ∈ (0, t1 ] ∪ [t2 , R), the proof follows similarly.

t u

Theorem 3.2 Let a ≥ 0 and b > 0 be such that φ(a, b) ≥ R, | p(t) |≤ a and 0 < q(t) ≤ b for a. e. t ∈ J. Then, if there exits l ≥ 0 such that u ∈ Wl (J), L(p, q) u ≥ 0 for a. e. t ∈ J and u0 (0) ≥ 0 ≥ u0 (R), we have that u ≥ 0 in J. Proof. By Lemmas 2.2 and 2.4 we know that inequalities (3.6) hold and then u(0) ≥ 0 and u(R) ≥ 0. Firstly, we suppose h 6≡ 0 in J, by (3.3) and (3.4) we know that u(0) > 0 and u(R) > 0. In an analogous way to the proof of Lemma 2.2, using that | p(t) |≤ a for a. e. t ∈ J, we prove that the first positive zero zλ of the unique solution of the Cauchy problem vλ00 (t) + p(t) vλ0 (t) + (q(t) + λ) vλ (t) = 0, vλ (0) = 1,

vλ0 (0) = 0,

satisfies zλ ≤ φ(a, λ) for all λ ≥ 0. Thus, by Lemma 2.2, we know that z0 ≥ φ(a, b) ≥ R, Proposition 2.1 shows that there exists λ ≥ 0 such that zλ = R, that is, problem 11

v¯00 (t) + p(t) v¯0 (t) + (q(t) + λ)¯ v (t) = 0, v¯0 (0) = 0, v¯(R) = 0, has a solution v¯ > 0 in [0, R). Suppose that there exits t1 ∈ (0, R] such that u(t1 ) = 0 and u > 0 on [0, t1 ). Obviously u0 (t− 1 ) ≤ 0 and if T ∈ (0, t1 ), Z

t1 R t

e

0

p(s)ds

Z h(t) v¯(t) dt = −λ

e

e

0

p(s)ds

0

p(s)ds

u(t) v¯(t) dt − u0 (0) v¯(0)+

0

0 RT

t1 R t

R t1

v¯(T )(u0 (T − )−u0 (T + ))+e

0

p(s)ds

R t1

u0 (t1 ) v¯(t1 )−e

0

p(s)ds

u(t1 ) v¯0 (t1 )

Now, if h 6≡ 0 in [0, t1 ] we have that the second part of the equality is less or equals to zero and the first one is strictly positive. If h ≡ 0 in [0, t1 ] and λ > 0, the second part of the equality is strictly negative. When h ≡ 0 in [0, t1 ] and λ = 0, we know that u0 (T − ) = u0 (T + ) which implies that u ∈ W 2,1 (J) and u0 (t1 ) = 0. From these two properties we deduce immediately that u ≡ 0 in this interval which is not true. To finish the proof we consider the case h ≡ 0 in J. Using expression (3.4), we have that if u0 (T + ) > u0 (T − ) or u0 (R) < 0 then u(0) > 0 and we attain a contradiction as previously. Thus, if u0 (T + ) = u0 (T − ) and u0 (R) = 0 we have that u(0) = 0, in such a case if u0 (0) > 0, we deduce from (3.3) that u(R) > 0 and the proof follows the same steps that in the previous case. The other case, that is, u ∈ W 2,1 (J) and u(0) = u0 (0) = 0 implies u ≡ 0 in J and the result is completed. When T ∈ [t1 , R] the proof holds in a similar way.

12

t u

Remark 3.1 This result is optimal in the sense that for all a ≥ 0 and b > 0 such that φ(a, b) < R formulas (3.3) and (3.4) show that there exist positive functions h such that u(0) < 0 or u(R) < 0.

4

Dirichlet, mixed and periodic conditions.

As a consequence of the results given in the previous section for the Neumann problem, we deduce analogous results for the operator L(p.q) with different boundary conditions which we have denoted by (1.3), (1.4), (1.5) and (1.6). The results are the following. Theorem 4.1 Suppose that q(t) < 0 or | p(t) |≤ a,

0 < q(t) ≤ b for a. e. t ∈ J and φ(a, b) ≥ R,

for a pair of a ≥ 0 and b > 0 given. Then, if u ∈ Wl (J) for some l ≥ 0, L(p, q) u ≥ 0 for a. e. t ∈ J, u0 (0) ≥ 0 and u(R) ≤ 0, we have that u ≤ 0 in J. Proof. Let t1 ≤ R be such that u(t1 ) = 0 and u < 0 in (t1 , R]. 0 + If q(t) < 0 for a. e. t ∈ J, we know that u0 (t− 1 ) ≤ u (t1 ) ≤ 0 and then

Theorem 3.1 implies u ≤ 0 in [0, t1 ] and the result is proved. In the other case, if h 6≡ 0 in [0, t1 ] we know, by (3.3), that u(t1 ) > 0 which is not possible. If h ≡ 0 in [0, t1 ], since φ(a, b) > R ≥ t1 , using Lemma 2.2 we have that u0 (t1 ) = 0 and then u ≡ 0 in [0, t1 ].

t u

Analogously, we have

13

Theorem 4.2 Suppose that q(t) < 0 for a. e. t ∈ J or | p(t) |≤ a,

0 < q(t) ≤ b for a. e. t ∈ J and φ(a, b) ≥ R,

for a pair of a ≥ 0 and b > 0 given. Then, if u ∈ Wl (J) for some l ≥ 0, L(p, q) u ≥ 0 for a. e. t ∈ J, u(0) ≤ 0 and u0 (R) ≤ 0, we have that u ≤ 0 in J. Remark 4.1 Similar comments to Remark 3.1 are valid for these two previous theorems. Theorem 4.3 Suppose that q(t) < 0 for a. e. t ∈ J or | p(t) |≤ a,

0 < q(t) ≤ b for a. e. t ∈ J and φ(a, b) ≥ R

for a pair of a ≥ 0 and b > 0 given. Then, if u ∈ Wl (J) for some l ≥ 0, L(p, q) u ≥ 0 for a. e. t ∈ J, u(0) ≤ 0 and u(R) ≤ 0, we have that u ≤ 0 in J. Proof. If q(t) < 0 for a. e. t ∈ J, then the proof is the same as in Theorem 3.1. In the other case, let t1 < t2 be such that u(t1 ) = u(t2 ) = 0 and u > 0 0 − in (t1 , t2 ). Since u0 (t+ 1 ) ≥ 0 ≥ u (t2 ) and t2 − t1 ≤ φ(a, b), we know that if

h 6≡ 0 in [t1 , t2 ] then, proceeding in an analogous way to equations (3.3) and (3.4), u(t1 ) > 0 and u(t2 ) > 0 which is not possible. If h ≡ 0 in [t1 , t2 ] we have u0 (t1 ) = 0 and, if it is the case, u0 (T − ) = u0 (T + ). This implies u ≡ 0 in [t1 , t2 ] and the result holds.

t u

Finally, we prove the following result for the periodic problem, the idea is based in [5] 14

Theorem 4.4 Let a ≥ 0 and b > 0 be given for which | p(t) |≤ a,

0 < q(t) ≤ b for a. e. t ∈ J and 2 φ(a, b) ≥ R.

Then, if u ∈ Wl (J) for some l ≥ 0, L(p, q) u ≥ 0 for a. e. t ∈ J, u(0) = u(R) and u0 (0) ≥ u0 (R), we have that u ≤ 0 in J. Proof.

Let t1 , t2 ∈ [0, R] be such that u(t1 ) = maxt∈J {u(t)} and u(t2 ) =

mint∈J {u(t)}, without loss of generality, we can suppose t1 < t2 . By the 0 + boundary conditions we know that u0 (t1 ) = 0 and u0 (t− 2 ) ≤ 0 ≤ u (t2 ).

If t2 − t1 ≤ R/2 Theorem 3.2 implies u ≥ 0 in [t1 , t2 ] and also in [0, R]. When t2 − t1 > R/2 we define function w(t) = u(t) in [0, R] and w(t) = u(t − R) if t ∈ (R, 2 R]. Clearly function w satisfies equation L(P, Q) w(t) = H(t), for a. e. t ∈ [0, 2 R], with P (t) = p(t) for all t ∈ [0, R] and P (t) = p(t − R) when t ∈ (R, 2 R], and the obvious notation for Q and H. In this case R + t1 − t2 ≤ R/2 and we can apply Theorem 3.2 to function w in the interval [t2 , t1 + R] and the result is proved.

t u

A simpler proof admits the following result: Theorem 4.5 Suppose that q(t) < 0 for a. e. t ∈ J, then, if u ∈ Wl (J) for some l ≥ 0 and L(p, q) u ≥ 0 for a.e. t ∈ J, u(0) = u(R) and u0 (0) ≥ u0 (R), we have that u ≤ 0 in J. Remark 4.2 In Theorem 4.3 for the Dirichlet problem and in Theorem 4.4 for the periodic case, the estimates obtained are not the best possible. In [2] and [6] are obtained the optimal estimates for some particular cases of 15

operator L(p, q) with constant coefficients.

5

Monotone method

In this section we apply the results given in the previous sections to obtain existence and approximation of extremal solutions of the nonlinear problem (1.7) with different boundary conditions. To do this, we shall say that α is a lower solution of (1.7) if α ∈ W 2,1 (J) and α00 (t) + p(t) α0 (t) + f (t, α(t)) ≥ 0, for a. e. t ∈ J. Analogously, an upper solution of (1.7) is a function β ∈ W 2,1 (J) satisfying the reversed inequalities. Now, following standard arguments (see the proof of [3, Theorem 3.1]), it is not difficult to prove the following results for problem (1.7) - (1.2). Theorem 5.1 Let α and β be respectively a lower and an upper solution of (1.7) with α ≤ β in J such that α0 (0) ≥ A ≥ β 0 (0) and α0 (R) ≤ B ≤ β 0 (R). Assume that the following conditions are satisfied: (H1 ) f : J × R −→ R is a Carath´eodory function, i.e. f (·, u) is measurable on I for each u ∈ R, f (t, ·) is continuous on R for a.e. t ∈ I and for every R > 0 there exists hR ∈ L1 (I) such that |f (t, u)| ≤ hR (t)

for a.e. t ∈ I and all u with |u| ≤ R.

(H2 ) q(t) u−f (t, u) ≤ q(t) v−f (t, v) for a. e. t ∈ I and α(t) ≤ v ≤ u ≤ β(t), with q ∈ L1 (J) such that q(t) < 0 in a. e. t ∈ J. 16

Then there exist two monotone sequences {αn } and {βn } such that α0 = α and β0 = β, which converge uniformly to the minimal and the maximal solutions of (1.7) - (1.2) on [α, β]. Theorem 5.2 Let a ≥ 0 and b > 0 be such that | p(t) |≤ a for a. e. t ∈ J, α and β be, respectively a lower and an upper solution of (1.7) with α ≥ β in J such that α0 (0) ≥ A ≥ β 0 (0) and α0 (R) ≤ B ≤ β 0 (R). Assume that condition (H1 ) is satisfied and (H3 ) q(t) u−f (t, u) ≥ q(t) v−f (t, v)for a. e. t ∈ I and β(t) ≤ v ≤ u ≤ α(t), with q ∈ L1 (J) such that 0 < q(t) ≤ b for a. e. t ∈ J. Then, if φ(a, b) ≥ R there exist two monotone sequences {αn } and {βn } such that α0 = α and β0 = β, which converge uniformly to the maximal and the minimal solutions of (1.7) - (1.2) on [β, α]. Theorem 5.3 The assertion proved in Theorem 5.2 is optimal in the sense that for all a ≥ 0 and b > 0 such that φ(a, b) < R, we can find f a Carath´eodory function satisfying condition (H3 ), q ∈ L1 (J) such that 0 < q(t) ≤ b for a. e. t ∈ J, p ∈ L1 (J) such that | p(t) |≤ a for a. e. t ∈ J and α ≥ β lower and upper solutions of problem (1.7) - (1.2) (with A = 0 = B) for which there exist no solution lying between α and β. Analogous results to Theorem 5.1 hold for problem (1.7) when α ≤ β in J satisfy α0 (0) ≥ A ≥ β 0 (0) and α(R) ≤ B ≤ β(R) if condition (1.3) is considered.

17

α(0) ≤ A ≤ β(0) and α0 (R) ≥ B ≥ β 0 (R), when we study condition (1.4). α(0) ≤ A; α(R) ≤ B and β(0) ≤ A; β(R) ≤ B, for problem (1.7) – (1.5). When α(0) − α(R) = A = β(0) − β(R) and α0 (0) − α0 (R) ≥ B ≥ β 0 (0) − β 0 (R), Theorem 5.1 if α ≤ β, and Theorem 5.2 for α ≥ β (with 2 φ(a, b) ≥ R) hold. Since the estimation in the length of the interval is not the best possible, we cannot obtain for this case an analogous result to Theorem 5.3.

References [1] H. Brezis

Analyse fonctionnelle. Th´eorie et applications,

Masson,

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