Existence theorem of nonlinear singular boundary value problem

Nonlinear Analysis 46 (2001) 465 – 473

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Existence theorem of nonlinear singular boundary value problem Habib Maagli, Syrine Masmoudi ∗ Departement de Mathematiques, Faculte des Sciences de Tunis, Universite de Tunis II, Campus Universitaire, 1060 Tunis, Tunisia Received 22 February 1999; accepted 20 September 1999

Keywords: Di0erential operator; Nonlinear singular problem; Positive solution; Schauder’s 3xedpoint theorem

1. Introduction In [6], Zhao considered the following problem in R+ = (0; ∞): 

 u + f(: ; u) = 0;  u¿0 on R+ ; (∗)   limt → 0+ u(t) = 0; where f is a measurable function on R+ ×R+ dominated by a convex positive function. Then he established the existence of in3nitely many solutions of (∗). On the other hand, Agarwal and O’Regan [1] studied the di0erential equation (1=A)(Au
)
+ f(: ; u; Au
) = 0 on (0,1), where A and f are two regular functions, with mixed boundary data limt → 0+ Au
(t) = u(1) = 0 or Dirichlet data u(1) = u(0) = 0. The goal is to generalize the result of Zhao to the more general boundary value problem in R+  1


   A (Au ) + f(: ; u; Au ) = 0; (P) u¿0 on R+ ;    limt → 0+ u(t) = 0; ∗

Corresponding author. Fax: +216-1-885-350.

0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 4 5 5 - 1

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H. Maagli, S. Masmoudi / Nonlinear Analysis 46 (2001) 465 – 473

where A is a positive and di0erentiable function on R+ and f is a measurable function on R+ ×R+ ×R+ dominated by a regular function. In particular, we prove that the multiple solutions result of Zhao is valid for a wider class of nonlinear singular problem. More precisely, if 1=A is integrable in a neigbourhood of 0 and if f satis3es some appropriate conditions, we use a 3xed point argument to prove the existence of a constant b ¿ 0 such that for each c ∈ (0; b], the 1 problem (P) has  t a solution u ∈ C([0; ∞))∩C ((0; ∞)) satisfying limt → ∞ u(t)= (t) = c, where (t) = 0 dr=A(r), for t ≥ 0. Moreover, we show that the result is also valid if we change the operator u → (1=A)(Au
)
to u → (1=A)(Au
)
− qu, where q is a measurable nonnegative function on R+ such that Aq is locally integrable on [0,∞). Note that various existence results for this type of equations have appeared in the literature in the special case f(x; y; z) ≡ f(x; y) (see [2– 6]).

2. Notations and hypotheses In this paper, we consider the di0erential operator of second order Lu = LA u =

1 (Au
)
; A

where A is a positive and di0erentiable function on R+ . Note that A(0) can be equal to 0. In addition, we assume that 1=A is integrable in a neigbourhood of 0 and we de3ne the function on [0; ∞) by  (t) =

t

0

dr : A(r)

We denote by (∞) := limt → ∞ (t). Furthermore, we assume f : R+ ×R+ ×R+ → R is a measurable function, continuous with respect to the second and the third variable and satisfying the following inequality: |f(x; y; z)| ≤ yh1 (x; y; z) + zh2 (x; y; z)

for all x; y; z ∈ R+ ;

where h1 and h2 are two nonnegative and measurable functions on R+ ×R+ ×R+ , nondecreasing with respect to the second and the third variable and satisfying lim

(y; z) → (0;0)

hi (x; y; z) = 0 for i = 1; 2:

Finally, we denote by t ∨ s = max(t; s) and t ∧ s = min(t; s).

H. Maagli, S. Masmoudi / Nonlinear Analysis 46 (2001) 465 – 473

467

3. The main result In this section, we prove an existence result for the following nonlinear problem in R+ :  Lu + f(: ; u; Au
) = 0;    u¿0 on R+ ; (P)    limt → 0+ u(t) = 0: We require the additional assumptions  ∞ A(t) (t)h1 (t; (t); 1) dt ¡ ∞ and 0

 0

∞

A(t)h2 (t; (t); 1) dt ¡ ∞:

Theorem 1. There exists a constant b ¿ 0 such that for each c ∈ (0; b]; the problem (P) has a solution u ∈ C([0; ∞)) ∩ C 1 ((0;  ∞∞)) which satis4es (i) for any t ≥ 0; u(t) = c (t) + 0 A(s) (t ∧ s) (1 − (t ∨ s)= (∞)) f(s; u(s); (Au
)(s)) ds. (ii) limt → ∞ u(t)= (t) = c. Proof. Let

 E = v ∈ C([0; ∞)) ∩ C 1 ((0; ∞)): lim+ (A( v)
)(t) and lim (A( v)
)(t) exist

t→0



t→∞

endowed with the norm v = A( v)
∞ = supt ≥ 0 |(A( v)
)(t)|. Then the map (E;  · ) → (C([0; ∞]);  · ∞ ); v → A( v)
is an isometry. It follows that (E;  · ) is a Banach space. In the sequel, we denote by h(x; y; z) = (x)h1 (x; y; z) + h2 (x; y; z), for x; y; z ∈ R+ . From the hypotheses on h1 and h2 and Lebesgue’s theorem, it follows that  ∞ lim+ A(s)h(s;  (s); ) ds = 0: →0

0

∞ Then there exists  ¿ 0 such that 0 A(s)h(s;  (s); ) ds ≤ 13 . Let b = 23  and c ∈ (0; b]. In order to apply a 3xed point argument, we set   c 3c
G = v ∈ E: ≤ (A( v) )(t) ≤ ; for t ∈ [0; ∞] : 2 2 We de3ne the operator T on G by

 ∞ 1 (t ∨ s) Tv(t) = c + A(s) (t ∧ s) 1 − (t) 0 (∞) ×f(s; ( v)(s); (A( v)
)(s)) ds:

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H. Maagli, S. Masmoudi / Nonlinear Analysis 46 (2001) 465 – 473

Now, we divide the proof into three steps. Step 1: We show that T G ⊂ G. First, we shall prove that T G ⊂ E. Let v ∈ G. So, for any t ¿ 0, we have c=2A(t) ≤ ( v)
(t) ≤ 3c=2A(t). Integrating from 0 to t, we obtain (c=2) (t) ≤ ( v)(t) ≤ (3c=2) (t) i.e. c=2 ≤ v(t) ≤ 3c=2 for each t ≥ 0. On the other hand, we have Tv(∞) := lim Tv(t) = c t→∞

and Tv(0) := lim+ Tv(t) t→0

 = c+

0

∞

(s) f(s; ( v)(s); (A( v)
)(s)) ds: A(s) 1 − (∞)

Then for all t1 ; t2 ∈ [0; ∞], we get |Tv(t1 ) − Tv(t2 )|   ∞  ≤ A(s) (s)  0

1 1 − (t1 ∨ s) (t2 ∨ s)



  f(s; ( v)(s); (A( v) )(s)) ds:


Since the maps y → h(x; y; z) and z → h(x; y; z) are nondecreasing, we obtain  

  1 1
 f(s; ( v)(s); (A( v) )(s)) A(s) (s)  − (t1 ∨ s) (t2 ∨ s) ≤ 3cA(s)h(s;  (s); ): This together with the fact that the map t → 1= (t ∨ s) is continuous on [0; ∞], for s ¿ 0, and the dominated convergence theorem imply that Tv ∈ C([0; ∞]). Furthermore, it is easy to verify that Tv ∈ C 1 ((0; ∞)) and for any t ¿ 0  t 1 A(s) (s)f(s; ( v)(s); (A( v)
)(s)) ds (A( Tv)
)(t) = c − (∞) 0

 ∞ (s) f(s; ( v)(s); (A( v)
)(s)) ds: + A(s) 1 − (∞) t So, from the dominated convergence theorem, we get (A( Tv)
)(0) := lim+ (A( Tv)
)(t) t→0

 = c+

0

∞

(s) f(s; ( v)(s); (A( v)
)(s)) ds A(s) 1 − (∞)

H. Maagli, S. Masmoudi / Nonlinear Analysis 46 (2001) 465 – 473

469

and (A( Tv)
)(∞) := lim (A( Tv)
)(t) t→∞

= c−

1 (∞)



∞

0

A(s) (s)f(s; ( v)(s); (A( v)
)(s)) ds:

Consequently Tv ∈ E. In addition, for any t ¿ 0, we have  ∞
|(A( Tv) )(t) − c| ≤ A(s)|f(s; ( v)(s); (A( v)
)(s))| ds 0

≤

3c 2



∞

0

c A(s)h(s;  (s); ) ds ≤ : 2

Then we conclude the invariance of G under T . Step 2. We show that T G is relatively compact in (E;  · ). For any t ¿ 0, we have [A( Tv)
]
(t) = − A(t)f(t; ( v)(t); (A( v)
)(t));

a:e on R+ :

Since, A(t)|f(t; ( v)(t); (A( v)
)(t))| ≤

3c A(t)h(t;  (t); ) 2

and t → A(t)h(t;  (t); ) is integrable on [0; ∞], then the family {A( Tv)
; v ∈ G} is equicontinuous on [0; ∞]. In addition, for any t ∈ [0; ∞]; {(A( Tv)
)(t); v ∈ G} is uniformly bounded in R. From Ascoli’s theorem, it follows that {A( Tv)
; v ∈ G} is relatively compact in (C([0; ∞]);  · ∞ ). This implies that T G is relatively compact in (E;  · ). Step 3. We prove the continuity of T in G. Let {vn } be a sequence in G such that vn − v = A( vn )
− A( v)
∞ → 0 Then for any t ∈ [0; ∞], we have |(A( Tvn )
)(t) − (A( Tv)
)(t)| ≤

Since (s)vn (s) − (s)v(s) =

s

 0

∞

as n → ∞:

A(s)|f(s; ( vn )(s); (A( vn )
)(s))

− f(s; ( v)(s); (A( v)
)(s)) ds:

1 [(A( vn )
)(r) 0 A(r)

− (A( v)
)(r)] dr, it follows that

| (s)vn (s) − (s)v(s)| ≤ A( vn )
− A( v)
∞ (s) → 0

as n → ∞:

On the other hand, we have A(s)|f(s; ( vn )(s); (A( vn )
)(s)) − f(s; ( v)(s); (A( v)
)(s))| ≤ 3cA(s)h(s;  (s); ):

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H. Maagli, S. Masmoudi / Nonlinear Analysis 46 (2001) 465 – 473

So, by the continuity of f with respect to the second and the third variable and the dominated convergence theorem, we deduce that Tvn − Tv = A( Tvn )
− A( Tv)
∞ → 0

as n → ∞:

Now, from the Schauder’s 3xed-point theorem, there exists v0 ∈ G such that Tv0 = v0 . We put u(t) = (t)v0 (t). Then for any t ≥ 0, we have

 ∞ (t ∨ s) f(s; u(s); (Au
)(s)) ds: u(t) = c (t) + A(s) (t ∧ s) 1 − (∞) 0 This shows that u ∈ C([0; ∞)) ∩ C 1 ((0; ∞)) is a solution of the equation Lu + f(: ; u; Au
) = 0 on R+ , satisfying u(0) = 0 and limt → ∞ u(t)= (t) = c. Example 2. Let ;  ≥ 0 such that max(; ) ¿ 1 and  ∈ [0; 1). Let k be a measurable function on R+ satisfying  ∞ t +(1−−) |k(t)| dt ¡ ∞: 0

Then there exists a constant b ¿ 0 such that for each c ∈ (0; b], the following problem:  



 +    u (t) + t u (t) + k(t)(u(t)) (u (t)) = 0; t ∈ R ; u¿0 on R+ ;     limt → 0+ u(t) = 0; has a solution u ∈ C([0; ∞)) ∩ C 1 ((0; ∞)) which satis3es limt → ∞ t −1 u(t) = c. Example 3. Let ;  ≥ 0 such that max(; ) ¿ 1 and k be a measurable function on R+ satisfying  ∞ √ ( t(1 + t))(1−) (arctan t 2 ) inf (1; arctan t 2 )|k(t)| dt ¡ ∞: 0

Then there exists a constant b ¿ 0 such that for each c ∈ (0; b], the following problem:

 √ 1 1

  u u
(t) + k(t)(u(t)) (u
(t)) inf (u(t); t(1 + t)u
(t)) = 0; (t) + +   2t 1+t    t ∈ R+ ;    u ¿ 0 on R+ ;     limt → 0+ u(t) = 0; has a solution u ∈ C([0; ∞)) ∩ C 1 ((0; ∞)) which satis3es limt → ∞ u(t) = c.

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471

4. Generalization Let q be a measurable nonnegative function on R+ such that Aq is locally integrable on [0; ∞). We will give a more general result for the following problem in R+ :  Lu − qu + f(: ; u; Au
) = 0;    u¿0 on R+ ; (Q)    limt → 0+ u(t) = 0: So, we need the following result [4]. Proposition 4. The di5erential equation Lu − qu = 0 has a unique solution ’ ∈ C([0; ∞)) ∩ C 1 ((0; ∞)); nondecreasing on [0; ∞); satisfying ’(0) = 1 and limt → 0+ A’
(t) = 0. In the sequel, we denote by  t dr !(t) = ’(t) 2 (r) A(r)’ 0

for t ≥ 0;

the unique solution in C([0; ∞)) ∩ C 1 ((0; ∞)) of the following problem: Lu − qu = 0 on R+ ; limt → 0+ Au
(t) = 1;

u(0) = 0

and we assume that  ∞ A(t)’(t)!(t)h1 (t; !(t); A!
(t)) dt ¡ ∞ (H) 0

and

 0

∞

A2 (t)’(t)!
(t)h2 (t; !(t); A!
(t)) dt ¡ ∞:

Theorem 5. There exists a constant b ¿ 0 such that for each c ∈ (0; b]; the problem (Q) has a solution u ∈ C([0; ∞)) ∩ C 1 ((0; ∞)) which satis4es limt → ∞ u(t)=!(t) = c. Proof. We de3ne the di0erential operator LA’2 by 1 (A’2 u
)
: LA’2 u = A’2 So, we have the following equality: (L − q)(’u) = ’LA’2 u: Then, it’s obvious to see that u = ’v is a solution of (Q) if and only if v satis3es  L 2 v + g(: ; v; A’2 v
) = 0 on R+ ;    A’ (∗) v¿0 on R+ ;    limt → 0+ v(t) = 0;

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H. Maagli, S. Masmoudi / Nonlinear Analysis 46 (2001) 465 – 473

where g(x; y; z) =



z 1 f x; ’(x)y; A(x)’
(x)y + ’(x) ’(x)

for x; y; z ∈ R+ :

Then, to establish the desired result, we will apply Theorem 1 for the problem (∗). We note that 1=A’2 is integrable in a neigbourhood of 0. Furthermore, in view of the above hypotheses on f; g is a measurable function on R+ ×R+ ×R+ , continuous with respect to the second and the third variable and satisfying |g(x; y; z)| ≤ yk1 (x; y; z) + zk2 (x; y; z) where

for any x; y; z ∈ R+ ;

k1 (x; y; z) = h1 x; ’(x)y; (A’
)(x)y +

z ’(x)



’
(x) z
+ A(x) h2 x; ’(x)y; (A’ )(x)y + ’(x) ’(x) and k2 (x; y; z) =

1 z
: x; ’(x)y; (A’ )(x)y + h 2 ’2 (x) ’(x)

This implies that k1 and k2 are two nonnegative measurable functions on R+ ×R+ ×R+ , nondecreasing with respect to the second and the third variable and satisfying lim(y; z) → (0; 0) ki (x; y; x) = 0, for i = 1; 2. t In addition, if we denote by #(t) = 0 dr=A(r)’2 (r), it’s easy to see, by hypothesis (H) that  ∞ (A’2 )(t)#(t)k1 (t; #(t); 1) dt ¡ ∞ 0

and

 0

∞

(A’2 )(t)k2 (t; #(t); 1) dt ¡ ∞:

Consequently, by virtue of Theorem 1, we deduce the result. Example 6. Let ;  ≥ 0 such that max(; ) ¿ 1. Let k be a measurable function on R+ satisfying  ∞ (1 + t)3+2+3 |k(t)| dt ¡ ∞: 0

Then, there exists a constant b ¿ 0 such that for each c ∈ (0; b], the following problem  6   u

(t) − u(t) + k(t)(u(t)) (u
(t)) = 0; t ∈ R+ ;   (1 + t)2 u¿0 on R+ ;     lim + u(t) = 0; t→0

has a solution u ∈ C([0; ∞)) ∩ C 1 ((0; ∞)) which satis3es limt → ∞ u(t)=t 3 = c.

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References [1] R.P. Agarwal, D. O’Regan, Nonlinear superlinear and nonsingular second order boundary value problems, J. Di0erential Equations 143 (1998) 60–95. [2] R. Dalmasso, Solutions d’Jequations elliptiques semi-linJeaires singuliJeres, Ann. Math. Pura. Appl. SJer. 4 (153) (1988) 191–202. [3] R. Dalmasso, On a singular nonlinear elliptic problems of second and fourth orders, Bull. Sci. Math. SJer. 2 (116) (1992) 95–110. [4] H. Maˆagli, S. Masmoudi, Sur les solutions d’un opJerateur di0Jerentiel singulier semi-linJeaire, Potential Anal. 10 (1999) 289–304. [5] H. Usami, On a singular elliptic boundary value problem in a ball, Nonlinear Anal. 13 (1989) 1163–1170. [6] Z. Zhao, Positive solutions of nonlinear second order ordinary di0erential equations, Proc. Amer. Math. Soc. 121 (1994) 465–469.