- Email: [email protected]
On the existence of positive solutions for a class of semilinear elliptic equations
Nonlinear Analysis 52 (2003) 1239 – 1247
www.elsevier.com/locate/na
On the existence of positive solutions for a class of semilinear elliptic equations Imed Bachar ∗ , Noureddine Zeddini Departement de Mathematiques, Faculte des Sciences de Tunis, Campus Universitaire, 1060 Tunis, Tunisia Received 1 December 2001; accepted 11 March 2002
Abstract We prove some existence and nonexistence results for the semilinear elliptic equation 0u = q(x)f(u) on ⊆ Rn (n ¿ 2) where u is required to blow up on the boundary of and f is a nonnegative function which is assumed to be Lipschitz continuous and bounded away from zero on each interval [; ∞) and have at worst linear growth. In particular, we extend some results already obtained in the case where f(u) = u , 0 ¡ 6 1. ? 2002 Elsevier Science Ltd. All rights reserved. Keywords: Di8erential operator; Existence; Nonexistence; Entire solutions; Large solutions
1. Introduction In this paper, we are concerned with the existence and nonexistence of positive solutions for the following nonlinear di8erential equation: 0u = q(x)f(u);
x ∈ ;
(*)
where ⊆ Rn (n ¿ 2), q and f satisfy some required hypotheses. In [3], Lair and Wood have considered the equation (∗) in the case where f(u) = u , 0 ¡ 6 1. They studied the existence and nonexistence of positive large solutions and positive bounded ones under adequate hypotheses on q. In this work, we extend the results of [3] by means of simple arguments to more general functions f and q. In particular we give a simple proof of Theorem 1 in [3]. ∗
Corresponding author. E-mail addresses: [email protected] (I. Bachar), [email protected] (N. Zeddini).
0362-546X/02/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 1 6 3 - 3
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I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
The second section of this paper deals with the existence and uniqueness of a positive solution to the following problem: 1 (Au ) − p(t)f(u) = 0; in (0; ∞); A lim A(t)u (t) = 0;
t→0
u(0) = ¿ 0;
where A, p and f are nonnegative functions and satisfy some appropriate hypotheses detailed below. These results will be used in Section 3, where the equation (∗) is studied. 2. Solutions of the equation (1=A)(Au ) − p(t)f (u) = 0 We consider the following second order di8erential operator L deEned by Lu:= (1=A)(Au ) , where A : [0; ∞) → [0; ∞) is a continuous function, di8erentiable and 1 t positive on (0; ∞) satisfying the following condition 0 (1=A(t)) ( 0 A(s) ds) dt ¡ ∞. The aim of this section is to investigate the existence and the uniqueness of a positive solution of the following problem: Lu − p(t)f(u) = 0; in (0; ∞); (P1 ) lim A(t)u (t) = 0; u(0) = ¿ 0: t→0
The following hypotheses on p and f are adopted: (H1) p; f : [0; ∞) → [0; ∞) are continuous. (H2) ∀c ¿ 0, ∃ ¿ 0 such that, ∀x; y ∈ [c; ∞) we have |f(x) − f(y)| 6 |x − y|. Remark 1. Under the hypotheses (H1) and (H2); there exist a ¿ 0 and b ¿ 0 such that for each x ¿ 0; 0 6 f(x) 6 ax + b. Theorem 1. Under the hypotheses (H1) and (H2); the problem (P1 ) has a unique 2 positive solution u ∈ C([0; ∞)) ∩ tC ((0; ∞)). s Moreover; for each t ¿ 0 we have a + b 6 au(t) + b 6 (a + b) exp(a 0 (1=A(s))( 0 A(r)p(r) dr) ds). Proof. Let (um )m¿0 be the sequence of functions deEned on [0; ∞) by u0 (t) = ;
s 1 um+1 (t) = + A(r)p(r)f(um (r)) dr ds; ∀m ∈ N: 0 A(s) 0 Let K be the operator deEned on C([0; ∞)) by s t 1 A(r)p(r)g(r) dr ds; for t ∈ [0; ∞): Kg(t):= 0 A(s) 0 t s We put h(t) = K1(t) = 0 (1=A(s))( 0 A(r)p(r) dr) ds; for t ∈ [0; ∞) and denote by C + ([0; ∞)) the set of nonnegative continuous functions in [0; ∞) and K j = K j−1 ◦ K for any integer j ¿ 2.
t
I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
1241
Clearly um ∈ C + ([0; ∞)) and we have u0 (t) = ; um+1 (t) = + K(f ◦ um )(t): Note that for each t ¿ 0 and m ¿ 0, um (t) ¿ . Then by (H2), there exists ¿ 0 such that |f(um+1 (t)) − f(um (t))| 6 |um+1 (t) − um (t)|: Next we intend to prove by induction that for each t ¿ 0 and m ∈ N, we have (i) K m 1(t) 6 (h(t))m =m!. (ii) |um+1 (t) − um (t)| 6 f()m ((h(t))m+1 =(m + 1)!). Indeed, if m = 0 or 1, (i) is valid. Now for a given m ∈ N suppose (i), then we have K m+1 1(t) = K(K m 1)(t) 1 1 K([h]m )(t) = 6 m! m!
0
t
1 A(s)
s
0
m
A(r)p(r)(h(r)) dr
ds:
Since the function h is nondecreasing, it follows that t s 1 1 m+1 m K 1(t) 6 (h(s)) A(r)p(r) dr ds m! 0 A(s) 0 t 1 1 = (h(s))m h (s) ds = (h(t))m+1 : m! 0 (m + 1)! Clearly (ii) is satisEed for m = 0. Assume that (ii) holds for some m ∈ N, then we have |um+2 (t) − um+1 (t)| = |K(f ◦ um+1 )(t) − K(f ◦ um )(t)| 6 K(|f ◦ um+1 − f ◦ um |)(t) 6 K(|um+1 − um |)(t) 6 f()
m+1 K([h]m+1 )(t) (m + 1)!
6 f()
m+1 (h(t))m+2 : (m + 1)! (m + 2)
This completes our aJrmation. Therefore, the sequence (um )m¿0 converges locally uniformly to a function u satisfying for each t ¿ 0, s t 1 u(t) = + A(r)p(r)f(u(r)) dr ds: 0 A(s) 0 Hence u ∈ C + ([0; ∞)) ∩ C 2 ((0; ∞)) and u is a positive solution of (P1 ).
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Now, we intend to show the uniqueness. Indeed, let u and v be two positive solutions of the problem (P1 ). Then for each R ∈ (0; ∞) and t ∈ [0; R] we have s t 1 A(r)p(r)|f(u(r)) − f(v(r))| dr ds |u(t) − v(t)| 6 0 0 A(s) s t 1 6 A(r)p(r)|u(r) − v(r)| dr ds 0 0 A(s) 6 K(|u − v|)(t): Since K is a nondecreasing operator we deduce by induction that for each m ¿ 0, |u(t) − v(t)| 6 m K m (|u − v|)(t) 6 m sup |u(r) − v(r)|K m 1(R) r∈[0;R]
6 sup |u(r) − v(r)| r∈[0;R]
(h(R))m : m!
Letting m tends to inEnity, we get |u(t) − v(t)| = 0, ∀t ∈ [0; R]. So u = v on [0; ∞). Finally, for each t ∈ (0; ∞) we have t 1 u (t) = A(r)p(r)f(u(r)) dr ¿ 0: A(t) 0 Using Remark 1, we get au(t) + b t A(r)p(r) dr: u (t) 6 A(t) 0 Consequently t u (s) ds 6 h(t); 0 au(s) + b This yields that for each t ¿ 0, a + b 6 au(t) + b 6 (a + b) exp(ah(t)): ∞ s Corollary 1. (1) If (H1) and (H2) are satis6ed and 0 (1=A(s))( 0 A(r)p(r) dr) ds ¡ ∞; then the positive solution u of the problem (P1 ) is bounded. (2) Assume moreover that (H3): inf t∈[; ∞) f(t) = " ¿ 0. t s Then, we have for each t ¿ 0, u(t) ¿ + " 0 (1=A(s))( 0 A(r)p(r) dr) ds. In particular ∞ s u is bounded on [0; ∞) if and only if 0 (1=A(s))( 0 A(r)p(r) dr) ds ¡ ∞.
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1243
Remark 2. For A(t) = t n−1 (n ¿ 3); we deduce a simple proof of Theorem 1 in [3]. Example 1. Let A(t) = t 2"+1 for t ∈ [0; ∞) and " ¿ 0. Assume that p(t) =
1 1 or p(t) = 2 1 + t #+1 (1 + t ) [Log(2 + t)]#
with # ¿ − 1:
Then; for ∈ (0; 1) and $ ¿ 0; the following problem: u (t) +
2" + 1 u (t) − p(t)u (t) Log(1 + u$ (t)) = 0 t
u (0) = 0;
in (0; ∞)
u(0) = ¿ 0;
has a unique positive solution u ∈ C([0; ∞)) ∩ C 2 ((0; ∞)). Moreover; this solution is bounded if and only if # ¿ 1.
3. Large solutions of the equation (u = q(x)f (u) In this section, we are concerned with the existence and nonexistence of a large solution for the following nonlinear di8erential equation: (P2 )
0u = q(x)f(u);
x ∈ ;
where ⊆ Rn (n ¿ 2), q : → [0; ∞) is a Borel measurable function and f : [0; ∞) → [0; ∞) is a continuous function. We recall that for bounded domain , a solution u of (P2 ) is called large solution if u(x) → ∞ as x → @. Whenever is unbounded, u is called a large solution if u satisEes further u(x) → ∞ as |x| → ∞ within . Theorem 2. Let = Rn ; n ¿ 3. Assume that f satis6es (H2) and (H3) and q(x) = p(|x|) is a radial continuous function. Then the equation (P2 ) has a large positive solution in Rn if and only if ∞ rp(r) dr = ∞: 0
Proof. SuJciency follows from Corollary 1 with A(t) = t n−1 . To prove the necessity, we assume that ∞ rp(r) dr ¡ ∞: 0
Thus, we suppose We intend to prove that equation (P2 ) has no large positive solution. that equation (P2 ) has a positive solution u. DeEne v(t):= S u(t!) d((!), where S is the unit sphere in Rn and d( denotes the normalized measure on S. Then by
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[2, Section 1, Proposition 6] and Remark 1, we have for each t ¿ 0 1 0v(t) = n−1 (t n−1 v ) = 0u(t!) d((!) t S = p(t) f(u(t!)) d((!) 6 p(t) (au(t!) + b) d((!) S
and v (t) =
1 t n−1
B(0; t)
0u(x) d x =
S
1 t n−1
B(0; t)
p(|x|)f(u(x)) d x ¿ 0:
Therefore v satisEes the following di8erential inequality: 1 n−1 (t v ) 6 p(t)(av(t) + b); t n−1 v (0) = 0: Since v is nondecreasing, we deduce that t v (t) 1 6 n−1 sn−1 p(s) ds: (av(t) + b) t 0 Consequently
(av(t) + b) 6 (av(0) + b) exp a
0
t
1
sn−1 ∞
a 6 (av(0) + b) exp n−2
0
0
s
r n−1 p(r) dr
rp(r) dr :
Thus, v is bounded and hence u cannot be a large solution. ∞ Example 2. Let ∈ (0; 1); $ ¿ 0 and p ∈ C + ([0; ∞)) satisfying 0 rp(r) dr=∞. Then the equation 0u = p(|x|)u Log(1 + u$ ); x ∈ Rn (n ¿ 3); has a large positive solution in Rn . Theorem 3. Suppose that is a bounded domain in Rn (n ¿ 2); q is continuous in L and there exist a; b ¿ 0 such that for each x ¿ 0 f(x) 6 ax + b: Then the equation (P2 ) has no positive large solution in . Proof. It is similar to that of Theorem 2 in [3]. In the sequel let = Rn , n ¿ 3. For each nonnegative measurable function g, we deEne the potential kernel Vg on Rn by g(y) +(n=2 − 1) Vg(x):=cn dy; where cn = : n−2 |x − y| 2,n=2 n R Note that Vg is a lower semicontinuous function.
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1245
Theorem 4. Assume that Vq is a continuous bounded function and f satis6es the following hypothesis: (H4): ∀ ¿ 0; ∃k ¿ 0 such that; ∀x ¿ y ¿ we have f(x) − f(y) 6 k(x − y). Then the equation (P2 ) has at least one positive continuous bounded solution in Rn . Proof. Let ¿ 0. Using (H4); we may choose # ¿ 0 such that (1) The function x → #x − f(x) is nondecreasing in [; ∞). (2) ∀x ¿ ; f(x) 6 #x. ˜ F; Ft ; Xt ; /t ; P x ) be the n-dimensional process of the Brownian In the sequel; let (; x motion and E the expectation on P x . Furthermore; let Vq be the kernel deEned for each nonnegative measurable function g by (cf. [1;4]) ∞ t Vq (g)(x):= E x (e− 0 q(Xs ) ds g(Xt )) dt; ∀x ∈ Rn : 0
Note that the following resolvent equation is satisEed: (3) V = V#q + #V#q (qV ) = V#q + #V (qV#q ). So we have the fundamental identity (4) [I + #V (q:)][I − #V#q (q:)] = I ; on Bb (Rn ); where Bb (Rn ) denotes the set of bounded Borel measurable functions in Rn . Let c ¿ e#Vq∞ and ’ be the function deEned in Rn by ’(x):=c(1 − #V#q (q))(x) = cE x (e−#
∞ 0
q(Xs ) ds
):
We consider the following set: 4 = {u ∈ Bb (Rn ): ’ 6 u 6 c}: For each u ∈ 4 and x ∈ Rn , we deEne Tu(x):=’(x) + V#q [q(#u − f(u))](x): Using the Jensen-inequality, we have for each u ∈ 4 and x ∈ Rn u(x) ¿ ’(x) = cE x (e−# x ¿ ce−#E (
∞ 0
∞ 0
q(Xs ) ds
)
q(Xs ) ds)
¿ ce−#Vq∞ ¿ : So by (2) we have for each u ∈ 4, #u − f(u) ¿ 0 and hence T is nondecreasing on 4. Furthermore we have for each u ∈ 4, ’(x) 6 Tu(x) = ’(x) + V#q [q(#u − f(u))](x)
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I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
6 ’(x) + V#q (#qu)(x) 6 ’(x) + cV#q (#q)(x) = c: It follows that T4 ⊆ 4. Now we consider the sequence (um )m¿0 deEned on Rn by u0 = ’;
um+1 = Tum :
It is clear that (um )m¿0 ⊆ 4 and since T is nondecreasing on 4, we have u0 = ’ 6 u1 6 u2 6 · · · 6 um 6 um+1 6 c: Thus, from the monotone convergence theorem, the sequence (um )m¿0 converges to a positive bounded function u which satisEes u = ’ + V#q [q(#u − f(u))]: Applying [I + #V (q:)] on both side of the above equality and using (3) and (4), we deduce that (5) u = c − V (qf(u)). Moreover u satisEes ce−#Vq∞ 6 ’ 6 u 6 c: Since Vq is continuous bounded function and f(u) 6 #c, then by writing 1 1 Vq = V (qf(u)) + V [(#c − f(u))q]; #c #c we deduce that u is a continuous function. Now, from (5), u is a solution (in the weak sense) of (P2 ). Remark 3. (1) Assume that q is locally HOolder continuous in Rn and satisEes ∞ t6(t) dt ¡ ∞; where 6(t) = max q(x): 0
∞
|x|=t
n−1
Then; since V6(x) = (1=(n − 2)) 0 (r =(|x| ∨ r)n−2 )6(r) dr; is a continuous bounded function and V6 =Vq +V (6 −q); we deduce that Vq is a continuous bounded function. Thus Theorem 4 improves Theorem 3 in [3]. (2) Assume that the functions q; f are locally HOolder continuous with exponent ∈ (0; 1) and Vq is a continuous bounded function. 2+ Then the equation (P2 ) has at least one positive bounded solution u ∈ Cloc (Rn ). Example 3. Let q ∈ Lp (Rn ) ∩ L1 (Rn ) for p ¿ (n=2) and n ¿ 3. Then Vq ∈ Cb (Rn ). Hence, for 0 6 6 1 and 0 6 $ 6 1 with +$ 6 1, the following equation 0u = q(x)u (1 + u)$ , has a positive continuous bounded solution in Rn . Theorem 5. Assume that q is continuous in Rn ; f satis6es the following hypothesis:
∞ 0
r(min|x|=r q(x)) dr = ∞ and that
(H5) ∀a ¿ 0; ∃c ¿ 0 such that; for each x ∈ [0; a] we have f(x) ¿ cx.
I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
1247
Then the equation (P2 ) has no nonnegative nontrivial entire bounded solution in Rn . Proof (See [3]): Assume that u is a nonnegative nontrivial entire bounded solution of equation (P2 ) with 0 6 u 6 M . By (H5), there exists c ¿ 0 such that for each x ∈ [0; M ], we have f(x) ¿ cx. Let v be as in the proof of Theorem 2. Then for each t ¿ 0, we have t 1 n−1 v (t) = n−1 s q(s!)f(u(s!)) d((!) ds ¿ 0: (6) t S 0 Therefore there exist R ¿ 0 and ¿ 0 such that v(t) ¿ , ∀t ¿ R. Using (6), we get t r 1−n n−1 M ¿ v(t) = v(0) + r s q(s!)f(u(s!)) d((!) ds dr 0
¿ v(0) +
0
t
0
r
¿ v(0) + c
t
0
1−n
r 1−n t
R
= v(0) + c:
r
0
¿ v(0) + c:
R
t
s
r
1−n
s
n−1
n−1
r
S
min q(x)
|x|=s
S
f(u(s!)) d((!)
ds dr
sn−1 min q(x) v(s) ds dr |x|=s
0
r
R
s
n−1
min q(x)
|x|=s
ds dr
t 1−n min q(x) r dr ds:
|x|=s
s
Or by hypothesis, the right-hand side of this inequality tends to inEnity as t → ∞. This yields to a contradiction. Acknowledgements We are indebted to Professor H. Mˆaagli for his guidance, the useful discussions and all his encouragement. We also thank the referee for a careful reading of the paper. References [1] K.L. Chung, Z. Zhao, From Brownian Motion to SchrOodinger’s Equation, Springer, Berlin, 1995. [2] R. Dautray, J-L. Lions, et al., Analyse mathSematique et calcul numSerique pour les sciences et les techniques, Coll.C.E.A Vol. 2, L’opSerateur de Laplace, Masson, 1987. [3] A.V. Lair, A.W. Wood (Shaker), Large solutions of sublinear elliptic equations, Nonlinear Anal. 39 (2000) 745–753. [4] H. Maˆagli, Perturbation semi-linSeaire des rSesolvantes et des semi-groupes, Potential Anal. 3 (1994) 61–87.
www.elsevier.com/locate/na
On the existence of positive solutions for a class of semilinear elliptic equations Imed Bachar ∗ , Noureddine Zeddini Departement de Mathematiques, Faculte des Sciences de Tunis, Campus Universitaire, 1060 Tunis, Tunisia Received 1 December 2001; accepted 11 March 2002
Abstract We prove some existence and nonexistence results for the semilinear elliptic equation 0u = q(x)f(u) on ⊆ Rn (n ¿ 2) where u is required to blow up on the boundary of and f is a nonnegative function which is assumed to be Lipschitz continuous and bounded away from zero on each interval [; ∞) and have at worst linear growth. In particular, we extend some results already obtained in the case where f(u) = u , 0 ¡ 6 1. ? 2002 Elsevier Science Ltd. All rights reserved. Keywords: Di8erential operator; Existence; Nonexistence; Entire solutions; Large solutions
1. Introduction In this paper, we are concerned with the existence and nonexistence of positive solutions for the following nonlinear di8erential equation: 0u = q(x)f(u);
x ∈ ;
(*)
where ⊆ Rn (n ¿ 2), q and f satisfy some required hypotheses. In [3], Lair and Wood have considered the equation (∗) in the case where f(u) = u , 0 ¡ 6 1. They studied the existence and nonexistence of positive large solutions and positive bounded ones under adequate hypotheses on q. In this work, we extend the results of [3] by means of simple arguments to more general functions f and q. In particular we give a simple proof of Theorem 1 in [3]. ∗
Corresponding author. E-mail addresses: [email protected] (I. Bachar), [email protected] (N. Zeddini).
0362-546X/02/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 0 2 ) 0 0 1 6 3 - 3
1240
I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
The second section of this paper deals with the existence and uniqueness of a positive solution to the following problem: 1 (Au ) − p(t)f(u) = 0; in (0; ∞); A lim A(t)u (t) = 0;
t→0
u(0) = ¿ 0;
where A, p and f are nonnegative functions and satisfy some appropriate hypotheses detailed below. These results will be used in Section 3, where the equation (∗) is studied. 2. Solutions of the equation (1=A)(Au ) − p(t)f (u) = 0 We consider the following second order di8erential operator L deEned by Lu:= (1=A)(Au ) , where A : [0; ∞) → [0; ∞) is a continuous function, di8erentiable and 1 t positive on (0; ∞) satisfying the following condition 0 (1=A(t)) ( 0 A(s) ds) dt ¡ ∞. The aim of this section is to investigate the existence and the uniqueness of a positive solution of the following problem: Lu − p(t)f(u) = 0; in (0; ∞); (P1 ) lim A(t)u (t) = 0; u(0) = ¿ 0: t→0
The following hypotheses on p and f are adopted: (H1) p; f : [0; ∞) → [0; ∞) are continuous. (H2) ∀c ¿ 0, ∃ ¿ 0 such that, ∀x; y ∈ [c; ∞) we have |f(x) − f(y)| 6 |x − y|. Remark 1. Under the hypotheses (H1) and (H2); there exist a ¿ 0 and b ¿ 0 such that for each x ¿ 0; 0 6 f(x) 6 ax + b. Theorem 1. Under the hypotheses (H1) and (H2); the problem (P1 ) has a unique 2 positive solution u ∈ C([0; ∞)) ∩ tC ((0; ∞)). s Moreover; for each t ¿ 0 we have a + b 6 au(t) + b 6 (a + b) exp(a 0 (1=A(s))( 0 A(r)p(r) dr) ds). Proof. Let (um )m¿0 be the sequence of functions deEned on [0; ∞) by u0 (t) = ;
s 1 um+1 (t) = + A(r)p(r)f(um (r)) dr ds; ∀m ∈ N: 0 A(s) 0 Let K be the operator deEned on C([0; ∞)) by s t 1 A(r)p(r)g(r) dr ds; for t ∈ [0; ∞): Kg(t):= 0 A(s) 0 t s We put h(t) = K1(t) = 0 (1=A(s))( 0 A(r)p(r) dr) ds; for t ∈ [0; ∞) and denote by C + ([0; ∞)) the set of nonnegative continuous functions in [0; ∞) and K j = K j−1 ◦ K for any integer j ¿ 2.
t
I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
1241
Clearly um ∈ C + ([0; ∞)) and we have u0 (t) = ; um+1 (t) = + K(f ◦ um )(t): Note that for each t ¿ 0 and m ¿ 0, um (t) ¿ . Then by (H2), there exists ¿ 0 such that |f(um+1 (t)) − f(um (t))| 6 |um+1 (t) − um (t)|: Next we intend to prove by induction that for each t ¿ 0 and m ∈ N, we have (i) K m 1(t) 6 (h(t))m =m!. (ii) |um+1 (t) − um (t)| 6 f()m ((h(t))m+1 =(m + 1)!). Indeed, if m = 0 or 1, (i) is valid. Now for a given m ∈ N suppose (i), then we have K m+1 1(t) = K(K m 1)(t) 1 1 K([h]m )(t) = 6 m! m!
0
t
1 A(s)
s
0
m
A(r)p(r)(h(r)) dr
ds:
Since the function h is nondecreasing, it follows that t s 1 1 m+1 m K 1(t) 6 (h(s)) A(r)p(r) dr ds m! 0 A(s) 0 t 1 1 = (h(s))m h (s) ds = (h(t))m+1 : m! 0 (m + 1)! Clearly (ii) is satisEed for m = 0. Assume that (ii) holds for some m ∈ N, then we have |um+2 (t) − um+1 (t)| = |K(f ◦ um+1 )(t) − K(f ◦ um )(t)| 6 K(|f ◦ um+1 − f ◦ um |)(t) 6 K(|um+1 − um |)(t) 6 f()
m+1 K([h]m+1 )(t) (m + 1)!
6 f()
m+1 (h(t))m+2 : (m + 1)! (m + 2)
This completes our aJrmation. Therefore, the sequence (um )m¿0 converges locally uniformly to a function u satisfying for each t ¿ 0, s t 1 u(t) = + A(r)p(r)f(u(r)) dr ds: 0 A(s) 0 Hence u ∈ C + ([0; ∞)) ∩ C 2 ((0; ∞)) and u is a positive solution of (P1 ).
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Now, we intend to show the uniqueness. Indeed, let u and v be two positive solutions of the problem (P1 ). Then for each R ∈ (0; ∞) and t ∈ [0; R] we have s t 1 A(r)p(r)|f(u(r)) − f(v(r))| dr ds |u(t) − v(t)| 6 0 0 A(s) s t 1 6 A(r)p(r)|u(r) − v(r)| dr ds 0 0 A(s) 6 K(|u − v|)(t): Since K is a nondecreasing operator we deduce by induction that for each m ¿ 0, |u(t) − v(t)| 6 m K m (|u − v|)(t) 6 m sup |u(r) − v(r)|K m 1(R) r∈[0;R]
6 sup |u(r) − v(r)| r∈[0;R]
(h(R))m : m!
Letting m tends to inEnity, we get |u(t) − v(t)| = 0, ∀t ∈ [0; R]. So u = v on [0; ∞). Finally, for each t ∈ (0; ∞) we have t 1 u (t) = A(r)p(r)f(u(r)) dr ¿ 0: A(t) 0 Using Remark 1, we get au(t) + b t A(r)p(r) dr: u (t) 6 A(t) 0 Consequently t u (s) ds 6 h(t); 0 au(s) + b This yields that for each t ¿ 0, a + b 6 au(t) + b 6 (a + b) exp(ah(t)): ∞ s Corollary 1. (1) If (H1) and (H2) are satis6ed and 0 (1=A(s))( 0 A(r)p(r) dr) ds ¡ ∞; then the positive solution u of the problem (P1 ) is bounded. (2) Assume moreover that (H3): inf t∈[; ∞) f(t) = " ¿ 0. t s Then, we have for each t ¿ 0, u(t) ¿ + " 0 (1=A(s))( 0 A(r)p(r) dr) ds. In particular ∞ s u is bounded on [0; ∞) if and only if 0 (1=A(s))( 0 A(r)p(r) dr) ds ¡ ∞.
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Remark 2. For A(t) = t n−1 (n ¿ 3); we deduce a simple proof of Theorem 1 in [3]. Example 1. Let A(t) = t 2"+1 for t ∈ [0; ∞) and " ¿ 0. Assume that p(t) =
1 1 or p(t) = 2 1 + t #+1 (1 + t ) [Log(2 + t)]#
with # ¿ − 1:
Then; for ∈ (0; 1) and $ ¿ 0; the following problem: u (t) +
2" + 1 u (t) − p(t)u (t) Log(1 + u$ (t)) = 0 t
u (0) = 0;
in (0; ∞)
u(0) = ¿ 0;
has a unique positive solution u ∈ C([0; ∞)) ∩ C 2 ((0; ∞)). Moreover; this solution is bounded if and only if # ¿ 1.
3. Large solutions of the equation (u = q(x)f (u) In this section, we are concerned with the existence and nonexistence of a large solution for the following nonlinear di8erential equation: (P2 )
0u = q(x)f(u);
x ∈ ;
where ⊆ Rn (n ¿ 2), q : → [0; ∞) is a Borel measurable function and f : [0; ∞) → [0; ∞) is a continuous function. We recall that for bounded domain , a solution u of (P2 ) is called large solution if u(x) → ∞ as x → @. Whenever is unbounded, u is called a large solution if u satisEes further u(x) → ∞ as |x| → ∞ within . Theorem 2. Let = Rn ; n ¿ 3. Assume that f satis6es (H2) and (H3) and q(x) = p(|x|) is a radial continuous function. Then the equation (P2 ) has a large positive solution in Rn if and only if ∞ rp(r) dr = ∞: 0
Proof. SuJciency follows from Corollary 1 with A(t) = t n−1 . To prove the necessity, we assume that ∞ rp(r) dr ¡ ∞: 0
Thus, we suppose We intend to prove that equation (P2 ) has no large positive solution. that equation (P2 ) has a positive solution u. DeEne v(t):= S u(t!) d((!), where S is the unit sphere in Rn and d( denotes the normalized measure on S. Then by
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[2, Section 1, Proposition 6] and Remark 1, we have for each t ¿ 0 1 0v(t) = n−1 (t n−1 v ) = 0u(t!) d((!) t S = p(t) f(u(t!)) d((!) 6 p(t) (au(t!) + b) d((!) S
and v (t) =
1 t n−1
B(0; t)
0u(x) d x =
S
1 t n−1
B(0; t)
p(|x|)f(u(x)) d x ¿ 0:
Therefore v satisEes the following di8erential inequality: 1 n−1 (t v ) 6 p(t)(av(t) + b); t n−1 v (0) = 0: Since v is nondecreasing, we deduce that t v (t) 1 6 n−1 sn−1 p(s) ds: (av(t) + b) t 0 Consequently
(av(t) + b) 6 (av(0) + b) exp a
0
t
1
sn−1 ∞
a 6 (av(0) + b) exp n−2
0
0
s
r n−1 p(r) dr
rp(r) dr :
Thus, v is bounded and hence u cannot be a large solution. ∞ Example 2. Let ∈ (0; 1); $ ¿ 0 and p ∈ C + ([0; ∞)) satisfying 0 rp(r) dr=∞. Then the equation 0u = p(|x|)u Log(1 + u$ ); x ∈ Rn (n ¿ 3); has a large positive solution in Rn . Theorem 3. Suppose that is a bounded domain in Rn (n ¿ 2); q is continuous in L and there exist a; b ¿ 0 such that for each x ¿ 0 f(x) 6 ax + b: Then the equation (P2 ) has no positive large solution in . Proof. It is similar to that of Theorem 2 in [3]. In the sequel let = Rn , n ¿ 3. For each nonnegative measurable function g, we deEne the potential kernel Vg on Rn by g(y) +(n=2 − 1) Vg(x):=cn dy; where cn = : n−2 |x − y| 2,n=2 n R Note that Vg is a lower semicontinuous function.
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Theorem 4. Assume that Vq is a continuous bounded function and f satis6es the following hypothesis: (H4): ∀ ¿ 0; ∃k ¿ 0 such that; ∀x ¿ y ¿ we have f(x) − f(y) 6 k(x − y). Then the equation (P2 ) has at least one positive continuous bounded solution in Rn . Proof. Let ¿ 0. Using (H4); we may choose # ¿ 0 such that (1) The function x → #x − f(x) is nondecreasing in [; ∞). (2) ∀x ¿ ; f(x) 6 #x. ˜ F; Ft ; Xt ; /t ; P x ) be the n-dimensional process of the Brownian In the sequel; let (; x motion and E the expectation on P x . Furthermore; let Vq be the kernel deEned for each nonnegative measurable function g by (cf. [1;4]) ∞ t Vq (g)(x):= E x (e− 0 q(Xs ) ds g(Xt )) dt; ∀x ∈ Rn : 0
Note that the following resolvent equation is satisEed: (3) V = V#q + #V#q (qV ) = V#q + #V (qV#q ). So we have the fundamental identity (4) [I + #V (q:)][I − #V#q (q:)] = I ; on Bb (Rn ); where Bb (Rn ) denotes the set of bounded Borel measurable functions in Rn . Let c ¿ e#Vq∞ and ’ be the function deEned in Rn by ’(x):=c(1 − #V#q (q))(x) = cE x (e−#
∞ 0
q(Xs ) ds
):
We consider the following set: 4 = {u ∈ Bb (Rn ): ’ 6 u 6 c}: For each u ∈ 4 and x ∈ Rn , we deEne Tu(x):=’(x) + V#q [q(#u − f(u))](x): Using the Jensen-inequality, we have for each u ∈ 4 and x ∈ Rn u(x) ¿ ’(x) = cE x (e−# x ¿ ce−#E (
∞ 0
∞ 0
q(Xs ) ds
)
q(Xs ) ds)
¿ ce−#Vq∞ ¿ : So by (2) we have for each u ∈ 4, #u − f(u) ¿ 0 and hence T is nondecreasing on 4. Furthermore we have for each u ∈ 4, ’(x) 6 Tu(x) = ’(x) + V#q [q(#u − f(u))](x)
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I. Bachar, N. Zeddini / Nonlinear Analysis 52 (2003) 1239 – 1247
6 ’(x) + V#q (#qu)(x) 6 ’(x) + cV#q (#q)(x) = c: It follows that T4 ⊆ 4. Now we consider the sequence (um )m¿0 deEned on Rn by u0 = ’;
um+1 = Tum :
It is clear that (um )m¿0 ⊆ 4 and since T is nondecreasing on 4, we have u0 = ’ 6 u1 6 u2 6 · · · 6 um 6 um+1 6 c: Thus, from the monotone convergence theorem, the sequence (um )m¿0 converges to a positive bounded function u which satisEes u = ’ + V#q [q(#u − f(u))]: Applying [I + #V (q:)] on both side of the above equality and using (3) and (4), we deduce that (5) u = c − V (qf(u)). Moreover u satisEes ce−#Vq∞ 6 ’ 6 u 6 c: Since Vq is continuous bounded function and f(u) 6 #c, then by writing 1 1 Vq = V (qf(u)) + V [(#c − f(u))q]; #c #c we deduce that u is a continuous function. Now, from (5), u is a solution (in the weak sense) of (P2 ). Remark 3. (1) Assume that q is locally HOolder continuous in Rn and satisEes ∞ t6(t) dt ¡ ∞; where 6(t) = max q(x): 0
∞
|x|=t
n−1
Then; since V6(x) = (1=(n − 2)) 0 (r =(|x| ∨ r)n−2 )6(r) dr; is a continuous bounded function and V6 =Vq +V (6 −q); we deduce that Vq is a continuous bounded function. Thus Theorem 4 improves Theorem 3 in [3]. (2) Assume that the functions q; f are locally HOolder continuous with exponent ∈ (0; 1) and Vq is a continuous bounded function. 2+ Then the equation (P2 ) has at least one positive bounded solution u ∈ Cloc (Rn ). Example 3. Let q ∈ Lp (Rn ) ∩ L1 (Rn ) for p ¿ (n=2) and n ¿ 3. Then Vq ∈ Cb (Rn ). Hence, for 0 6 6 1 and 0 6 $ 6 1 with +$ 6 1, the following equation 0u = q(x)u (1 + u)$ , has a positive continuous bounded solution in Rn . Theorem 5. Assume that q is continuous in Rn ; f satis6es the following hypothesis:
∞ 0
r(min|x|=r q(x)) dr = ∞ and that
(H5) ∀a ¿ 0; ∃c ¿ 0 such that; for each x ∈ [0; a] we have f(x) ¿ cx.
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Then the equation (P2 ) has no nonnegative nontrivial entire bounded solution in Rn . Proof (See [3]): Assume that u is a nonnegative nontrivial entire bounded solution of equation (P2 ) with 0 6 u 6 M . By (H5), there exists c ¿ 0 such that for each x ∈ [0; M ], we have f(x) ¿ cx. Let v be as in the proof of Theorem 2. Then for each t ¿ 0, we have t 1 n−1 v (t) = n−1 s q(s!)f(u(s!)) d((!) ds ¿ 0: (6) t S 0 Therefore there exist R ¿ 0 and ¿ 0 such that v(t) ¿ , ∀t ¿ R. Using (6), we get t r 1−n n−1 M ¿ v(t) = v(0) + r s q(s!)f(u(s!)) d((!) ds dr 0
¿ v(0) +
0
t
0
r
¿ v(0) + c
t
0
1−n
r 1−n t
R
= v(0) + c:
r
0
¿ v(0) + c:
R
t
s
r
1−n
s
n−1
n−1
r
S
min q(x)
|x|=s
S
f(u(s!)) d((!)
ds dr
sn−1 min q(x) v(s) ds dr |x|=s
0
r
R
s
n−1
min q(x)
|x|=s
ds dr
t 1−n min q(x) r dr ds:
|x|=s
s
Or by hypothesis, the right-hand side of this inequality tends to inEnity as t → ∞. This yields to a contradiction. Acknowledgements We are indebted to Professor H. Mˆaagli for his guidance, the useful discussions and all his encouragement. We also thank the referee for a careful reading of the paper. References [1] K.L. Chung, Z. Zhao, From Brownian Motion to SchrOodinger’s Equation, Springer, Berlin, 1995. [2] R. Dautray, J-L. Lions, et al., Analyse mathSematique et calcul numSerique pour les sciences et les techniques, Coll.C.E.A Vol. 2, L’opSerateur de Laplace, Masson, 1987. [3] A.V. Lair, A.W. Wood (Shaker), Large solutions of sublinear elliptic equations, Nonlinear Anal. 39 (2000) 745–753. [4] H. Maˆagli, Perturbation semi-linSeaire des rSesolvantes et des semi-groupes, Potential Anal. 3 (1994) 61–87.