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Multiple positive solutions for some four-point boundary value problems with p-Laplacian
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Applied Mathematics and Computation 202 (2008) 413–426 www.elsevier.com/locate/amc
Multiple positive solutions for some four-point boundary value problems with p-Laplacian q Xiangfeng Li Department of Mathematics, Longdong University, Qingyang, Gansu, 745000, PR China
Abstract This paper is devoted to the existence of multiple positive solutions for some nonlinear singular four-point boundary value problems with p-Laplacian. Under appropriate assumptions on the nonlinear terms, by applying several known fixed point theorems, sufficient conditions for the existence of at least three positive solutions are established. Ó 2008 Elsevier Inc. All rights reserved. Keywords: p-Laplacian operator; Multiple positive solutions; Four-point boundary value problem; Fixed point theorem
1. Introduction In this paper, we study the existence of multiple positive solutions for the following quasilinear singular four-point boundary value problems: 0
ðup ðu0 ÞÞ þ aðtÞf ðuðtÞÞ ¼ 0;
0 < t < 1;
ð1:1Þ
and 0
ðup ðu0 ÞÞ þ aðtÞgðuðtÞ; u0 ðtÞÞ ¼ 0;
0 < t < 1;
ð1:10 Þ
subject to the nonlinear boundary conditions aup ðuð0ÞÞ bup ðu0 ðnÞÞ ¼ 0;
cup ðuð1ÞÞ þ dup ðu0 ðgÞÞ ¼ 0;
ð1:2Þ
where up ðxÞ ¼ jxjp2 x; p > 1; uq ðxÞ ¼ jxjq2 x is the inverse function to up ; 1p þ 1q ¼ 1; a > 0; b P 0; c > 0; d P 0; n; g 2 ð0; 1Þ is prescribed and n < g. aðtÞ may be singular at t ¼ 0 and/or t ¼ 1. In recent years, the study of the existence of positive solutions for boundary value problems with p-Laplacian operator has been an interesting topic; there are much current attentions focused on the study of nonlinear multi-point (at least three point) boundary value problem, see [2,4,9,12,13] and references therein. The q
This work is sponsored by the Tutorial Scientific Research Foundation of Education Department of Gansu Province (0710-04). E-mail address: [email protected]
0096-3003/$ - see front matter Ó 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2008.02.006
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X. Li / Applied Mathematics and Computation 202 (2008) 413–426
methods and techniques employed in these papers involve the use of the fixed point theorem in cones [2,13], the Avery–Peterson fixed point theorem [4,12], and the monotone iterative technique [9]. Recently, when the nonlinear term f does not depend on the first-order derivative, Eq. (1.1) together with some multi-point boundary conditions has been studied by researchers, for example, (see [3,5,7,8,10,11] and references therein). In [8], Liu studied the existence of positive solution of Eq. (1.1) subject to the three-point boundary value conditions u0 ð0Þ ¼ 0;
uð1Þ ¼ buðgÞ;
by applying the fixed point index theory. In [3], using a three-functional fixed point theorem due to Avery and Henderson, considered the existence of double positive solutions of Eq. (1.1) subject to one of the following two pairs of nonlinear three-point boundary value conditions: uð0Þ B0 ðu0 ðgÞÞ ¼ 0;
uð1Þ B1 ðu0 ð1ÞÞ ¼ 0;
uð0Þ B0 ðu0 ð0ÞÞ ¼ 0;
uð1Þ B1 ðu0 ðgÞÞ ¼ 0:
In [11], Su et al. studied the existence of multiple positive solutions of Eq. (1.1) subject to the nonlinear twopoint boundary value conditions aup ðuð0ÞÞ bup ðu0 ð0ÞÞ ¼ 0;
cup ðuð1ÞÞ þ dup ðu0 ð1ÞÞ ¼ 0:
The main tool is the fixed point theorem of cone expansion and compression of norm type. Very recently in [5,7,10], Ji et al., Liang et al. and Su et al. discussed the existence of a positive solution and double positive solutions of Eqs. (1.1) and (1.2), respectively. The main tools is the fixed point index theory and the Avery and Henderson fixed point theorem. More recently, when the nonlinear term f depends on the first-order derivative, Eq. (1.1)0 subject to the boundary value conditions u0 ð0Þ ¼ uð1Þ ¼ 0
or
uð0Þ ¼ ð1Þ ¼ 0;
has been studied by Wang et al. and Zhao et al., by means of the Leggett–Williams fixed point theorem and the fixed point theorem in cones, respectively, see [14,15]. However, for Eqs. (1.1), (1.2) and (1.1)0 , (1.2), there are currently few papers dealing with the existence of multiple (at least three) positive solutions. Motivated by the papers mentioned above, in this paper we shall consider the existence of multiple positive solutions for quasilinear singular boundary value problem (1.1), (1.2) and (1.1)0 , (1.2) by applying the Leggett–Williams fixed point theorem, the Schauder fixed point theorem and the Avery–Peterson fixed point theorem, respectively. In the rest of the paper, we make the following assumptions: (H1) f : Cð½0; 1Þ ! ½0; 1ÞÞ; g 2 Cð½0; R 1þ1Þ ð1; þ1Þ; ½0; þ1ÞÞ; (H2) aðtÞ 2 Cðð0; 1Þ; ½0; 1ÞÞ and 0 < 0 aðtÞdt < 1. Furthermore, aðtÞ is not an identical zero on any compact subinterval of ð0; 1Þ. 2. Preliminaries and main results In this section, we provide some background materials from the theory of cones in Banach spaces and Lemmas, then give out the main results of this paper. Definition 2.1. Let ðE; k kÞ be a real Banach space. A nonempty, closed, convex set P E is called a cone if it satisfies the following two conditions: (i) u 2 P ; k P 0, implies ku 2 P ; (ii) u 2 P ; u 2 P , implies u ¼ 0.
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Definition 2.2. A map a is said to be a nonnegative continuous concave functional on cone P of real Banach space E if a : P ! ½0; 1Þ; is continuous and aðtu þ ð1 tÞvÞ P taðuÞ þ ð1 tÞaðvÞ; for all u; v 2 P and t 2 ½0; 1. Similarly, we say that the map b is a nonnegative continuous convex functional on a cone P of real Banach space E if b : P ! ½0; 1Þ; is continuous and bðtu þ ð1 tÞvÞ 6 tbðuÞ þ ð1 tÞbðvÞ; for all u; v 2 P and t 2 ½0; 1. Let 0 < a < b; r > 0 be real numbers, aðuÞ be a nonnegative continuous concave functional on P, then we define the following convex sets: P r ¼ fu 2 P : kuk < rg; P r ¼ fu 2 P : kuk 6 rg; oP r ¼ fu 2 P : kuk ¼ rg; P ða; a; bÞ ¼ fu 2 P : a 6 aðuÞ; kuk 6 bg: Let c and h be nonnegative continuous convex functional on P ; a be a nonnegative continuous concave functional on P, and w be a nonnegative continuous functional on P. Then, for positive real numbers a; b; c; and d, we define the following convex sets: P ðc; dÞ ¼ fu 2 P : cðuÞ < dg; P ðc; dÞ ¼ fu 2 P : cðuÞ 6 dg; P ðc; a; b; dÞ ¼ fu 2 P : b 6 aðuÞ; cðuÞ 6 dg; P ðc; h; a; b; c; dÞ ¼ fu 2 P : b 6 aðuÞ; hðuÞ 6 c; cðuÞ 6 dg; and a convex closed set Rðc; w; a; dÞ ¼ fu 2 P : a 6 wðuÞ; cðuÞ 6 dg: Lemma 2.1. Suppose condition ðH2 Þ holds, then there exists a constant x 2 ð0; 12Þ such that Z 1x 0< aðtÞdt < 1: x
Furthermore, the function Z t Z AðtÞ ¼ uq aðtÞdt þ uq x
1x
aðtÞdt ;
x 6 t 6 1 x;
t
is a positive continuous function on ½x; 1 x, and therefore AðtÞ has a minimum on ½x; 1 x; we suppose that there exists ‘ > 0 such that ‘ ¼ mint2½x;1x AðtÞ. Proof. By condition ðH2 Þ, it is easily seen that Z 1x 0< aðtÞdt < 1; x
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is satisfied, and it is easy to prove that AðtÞ is continuous on ½x; 1 x. Next, let Z t Z 1x aðtÞdt ; A2 ðtÞ ¼ uq aðtÞdt : A1 ðtÞ ¼ uq x
t
Then, it follows from condition ðH2 Þ that the function A1 ðtÞ is strictly monotone increasing on ½x; 1 x and A1 ðxÞ ¼ 0; A2 ðtÞ is strictly monotone decreasing on ½x; 1 x and A2 ð1 xÞ ¼ 0; this shows ‘ ¼ mint2½x;1x AðtÞ > 0. The proof is complete. h In the sequel, let E ¼ C½0; 1 be endowed with the maximum norm kuk ¼ max juðtÞj: 06t61
1
E1 ¼ C ½0; 1 be endowed with the maximum norm kuk1 ¼ max max juðtÞj; max ju0 ðtÞj : 06t61
06t61
Clearly E; E1 are two Banach spaces. From the fact 0
ðup ðu0 ðtÞÞÞ ¼ aðtÞf ðuðtÞÞ 6 0;
0
ðup ðu0 ðtÞÞÞ ¼ aðtÞgðuðtÞ; u0 ðtÞÞ 6 0;
we know that u is concave on [0, 1]. So, we define the cone P E by P ¼ fu 2 E : uðtÞ P 0;
uðtÞ is the concave function on ½0; 1g:
we define the cone P 1 E1 by P 1 ¼ fu 2 E : uðtÞ P 0; aup ðuð0ÞÞ bup ðu0 ðnÞÞ ¼ 0; cup ðuð1ÞÞ þ dup ðu0 ðgÞÞ ¼ 0; uðtÞ is the concave function on ½0; 1g; Obviously, E1 E; P 1 P . Let the nonnegative continuous concave functional a1 , the nonnegative continuous convex functional h1 ; c1 , and the nonnegative continuous functional w1 be defined on the cone P 1 by c1 ðuÞ ¼ max ju0 ðtÞj;
w1 ðuÞ ¼ h1 ðuÞ ¼ max juðtÞj; 06t61 1 a1 ðuÞ ¼ min juðtÞj; x 2 0; : x6t6ð1xÞ 2 06t61
Lemma 2.2 [10]. Let u 2 P and x be as in Lemma 2.1, then uðtÞ P xkuk;
t 2 ½x; 1 x:
Apparently, if u 2 P 1 , then Lemma 2.2 also holds. Lemma 2.3. Let u 2 P 1 , then there exists a positive constant L, such that max juðtÞj 6 L max ju0 ðtÞj:
06t61
06t61
Proof. By uðtÞ uð0Þ ¼
Rt 0
u0 ðsÞds, we have
max juðtÞj 6 juð0Þj þ max ju0 ðtÞj:
06t61
06t61
On the other hand, by the first equation of (1.2), we have b b 0 ju ðnÞj 6 uq max ju0 ðtÞj: juð0Þj ¼ uq a a 06t61 Thus, we have
b max juðtÞj 6 1 þ uq max ju0 ðtÞj: 06t61 06t61 a
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417
Similarly, by the second equation of (1.2), we have d max juðtÞj 6 1 þ uq max ju0 ðtÞj: 06t61 06t61 c Therefore, setting b d L ¼ min 1 þ uq ; 1 þ uq ; a c h
the proof of Lemma 2.3 is complete. Next, we define
1 ð2:1Þ /ðuÞ ¼ ðuðxÞ þ uð1 xÞÞ for all u 2 P : 2 Obviously, /ðuÞ is the concave functional, and it is easy to prove that /ðuÞ 6 kuk, for arbitrary u 2 P . For convenience, we introduce the following constants: Z g Z g Z g b k1 ¼ u q aðrÞdr þ uq aðrÞdr ds; a n 0 s Z g Z 1 Z s d aðrÞdr þ uq aðrÞdr ds: k2 ¼ u q c n n n Obviously, k1 ; k2 > 0 Z N ¼ max uq
g
Z aðrÞdr ; uq
0
1
aðrÞdr
;
n
Z g Z g Z g b aðrÞdr þ uq aðrÞdr ds; M ¼ max uq a n 0 s Z g Z 1 Z s d uq aðrÞdr þ uq aðrÞdr ds ; c n n n Z n Z n Z 1 Z s uq aðrÞdr ds; uq aðrÞdr ds : m ¼ min 0
g
s
g
Here are the main results of this paper. Theorem 2.1. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist positive constants a; b; d, such that 0 < a < xb < b < b=x 6 d. Also assume that f satisfies (H3) f ðuÞ < up kai ; i ¼ 1; 2 for 0 6 u 6 a; (H4) one of the following conditions holds: e (i) there exists constant e > d, such that f ðuÞ 6 u p ki ; i ¼ 1; 2 for 0 6 u 6 e or, f ðuÞ (ii) lim supu!1 up ðuÞ < up k1i ; i ¼ 1; 2. (H5) f ðuÞ > up
2b x‘
for xb 6 u 6 d.
Then, the boundary value problem (1.1) and (1.2) has at least three positive solutions u1 ; u2 , and u3 satisfying 0 < ku1 k < a;
/ðu2 Þ > b;
ku3 k > a;
and
/ðu3 Þ < b:
Corollary 2.1. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist some positive constants ai ; bi ; d i ; i 2 N ; N is the positive integer number set, such that 0 < a1 < xb1 < b1 < b1 =x 6 d 1 < a2 < xb2 < b2 < b2 =x 6 d 2 < < an , and assume that f satisfies
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(H6) f ðuÞ < up kaji ; j ¼ 1; 2; for 0 6 u 6 ai ; 1 6 i 6 n; i (H7) f ðuÞ > up 2b for xbi 6 u 6 d i ; 1 6 i 6 n 1. x‘ Then, the boundary value problem (1.1) and (1.2) has at least 2n 1 positive solutions. Corollary 2.2. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist some positive constants ai ; bi ; d i :i 2 N ; N is the positive integer number set, such that 0 < a1 < xb1 < b1 < d 1 < a2 < xb2 < b2 < d 2 < < an < , and assume that f satisfies (H8) f ðuÞ < up ðkaji Þ; j ¼ 1; 2 for 0 6 u 6 ai ; i 2 N ; i (H9) f ðuÞ > up ð2b Þ for xbi 6 u 6 d i ; i 2 N . x‘
Then, the boundary value problem (1.1) and (1.2) has infinitely many positive solutions. Theorem 2.2. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist positive constants a; b; d, such that 0 < a < b < xLd. Also assume that g satisfies the following conditions: (H10) ðgðu; u0 Þ 6 up Nd for ðu; u0 Þ 2 ½0; Ld ½d; d;
b for ðu; u0 Þ 2 b; xb ½d; d; (H11) gðu; u0 Þ > up xm (H12) gðu; u0 Þ < up Ma for ðu; u0 Þ 2 ½0; a ½d; d. Then, the boundary value problem (1.1)0 and (1.2) has at least three positive solutions u1 ; u2 , and u3 satisfying max ju0i ðtÞj 6 d
06t61
for i ¼ 1; 2; 3;
b < min ju1 ðtÞj; x6t61x
b 1 < max ju2 ðtÞj < ; 06t61 x
with max ju1 ðtÞj 6 Ld; 06t61
with d min ju2 ðtÞj < b; x6t61x
max ju3 j < a:
06t61
3. Proof of Theorem 2.1 In this section, we will prove Theorem 2.1 and its Corollary. The main tools are the Leggett–Williams fixed point theorem and the Schauder fixed point theorem: Theorem A [6]. Let c > 0 and T : P c ! P c be completely continuous and /ðuÞ be a nonnegative continuous concave functional on P, and such that /ðuÞ 6 kuk, for arbitrary u 2 P c . Suppose there exists 0 < a < b < d 6 c, such that (C1) fu 2 P ð/; b; dÞ : /ðuÞ > bg 6¼ ;, and /ðTuÞ > b for arbitrary u 2 P ð/; b; dÞ; (C2) kTuk < a for kuk 6 a; (C3) /ðTuÞ > b for u 2 P ð/; b; cÞ, with kTuk > d. Then, T has at least three fixed points u1 ; u2 and u3 on P c and satisfying ku1 k < a; b < /ðu2 Þ
and
ku3 k > a with /ðu3 Þ < b:
Theorem B [16]. Let D be a bounded, convex closed set, and operator T : D ! D be completely continuous. Then, T has at least a fixed point in D.
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T Lemma 3.1 ([7,11]). Suppose that conditions ðH1 Þ; ðH2 Þ hold, then uðtÞ 2 C½0; 1 C 2 ð0; 1Þ is a solution of boundary value problem (1.1) and (1.2), if and only if uðtÞ is a solution of the following integral equation: 8 R R R > < uq ba nr aðrÞf ðuðrÞÞdr þ 0t uq sr aðrÞf ðuðrÞÞdr ds; 0 6 t 6 r; R R uðtÞ ¼ R > : uq dc rg aðrÞf ðuðrÞÞdr þ t1 uq rs aðrÞf ðuðrÞÞdr ds; r 6 t 6 1; where r 2 ½n; g ð0; 1Þ and u0 ðrÞ ¼ 0. Proof of Theorem 2.1. We define the operator T : P ! E given by 8 R R R > < uq ba nr aðrÞf ðuðrÞÞdr þ 0t uq sr aðrÞf ðuðrÞÞdr ds; R R ðTuÞðtÞ ¼ R > : uq dc rg aðrÞf ðuðrÞÞdr þ t1 uq rs aðrÞf ðuðrÞÞdr ds;
0 6 t 6 r; ð3:1Þ r 6 t 6 1:
First, it follows from Lemma 3.1 that the operator T is well-defined. Next, because of ( R r uq t aðrÞf ðuðrÞÞdr ; 0 6 t 6 r; 0 ðTuÞ ðtÞ ¼ R t uq r aðrÞf ðuðrÞÞdr ; r 6 t 6 1:
ð3:2Þ
0
It is easy to see that the operator ðTuÞ is a continuous monotone decreasing on [0, 1], and ðTuÞðtÞ P 0, for t 2 ½0; 1, these show Tu 2 P , i.e. T ðP Þ P and ðTuÞ0ðrÞ ¼ 0; ðTuÞðrÞ ¼ kTuk, namely Z r Z r Z r b aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds uq a n 0 s Z g Z 1 Z s d ¼ uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds c r r r Furthermore, we can also prove that the operator T : P ! P is a completely continuous (see [10, Lemma 2.4]), and it follows from Lemma 3.1 that each fixed point of T in P is a positive solution of (1.1) and (1.2). We now verify that the conditions of Theorem A are satisfied. Firstly, we choose u 2 P a . Then kuk 6 a, thus 0 6 u 6 a and assumption ðH3 Þ yields f ðuðtÞÞ < up ka1 , for 0 6 t 6 1. Thus, by Lemmas 2.1 and 3.1, we have Z r Z r Z r b aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds kTuk ¼ ðTuÞðrÞ ¼ uq a n 0 s Z g Z g Z g b 6 uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds a n 0 s Z g Z g Z g a b a < uq aðrÞdr þ uq aðrÞdr ds ¼ k1 ¼ a: k1 a n k 1 0 s Hence, condition ðC2 Þ of Theorem A is satisfied. Secondly, we claim that condition ðH4 Þ guarantees the existence of a number c with c > d and T : P c ! P c:
ð3:3Þ
have (3.3) with e ¼ c. Suppose now It is clear that if ðH4 Þ(i) holds, then from the above proof we immediately that ðH4 Þ(ii) is satisfied, then, there exist l > 0 and e < up k11 , such that f ðuÞ 6e up ðuÞ
for u P l:
ð3:4Þ
Let G ¼ maxu2½0;l f ðuÞ. In view of (3.4), it is easy to see that f ðuÞ 6 G þ eup ðuÞ for u P 0:
ð3:5Þ
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Now, let c be such that 1 1 eÞ : up ðcÞ > max up ðdÞ; Gðup k1
ð3:6Þ
Then, for arbitrary u 2 P c and t 2 ½0; 1, from (3.5) and (3.6) and Lemma 3.1, we can obtain Z r Z r Z r b kTuk ¼ ðTuÞðrÞ ¼ uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds a n 0 s Z g Z g Z g b aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds 6 uq a n 0 s Z g Z g Z g b aðrÞðG þ eup ðuðrÞÞÞdr þ uq aðrÞðG þ eup ðuðrÞÞÞdr ds 6 uq a n 0 s Z g Z g Z g b 1 c aðrÞdr þ uq aðrÞdr ds uq up ðcÞ up < uq e þ eup ðcÞ ¼ k1 ¼ c a n k k 1 1 0 s Hence, (3.3) holds. We shall now show that condition ðC1 Þ of Theorem A is satisfied. For this end, we suppose that uðtÞ
bþd : 2
bþd Note that /ðuðtÞÞ ¼ / bþd ¼ 2 > b, thus fu 2 P ð/; b; dÞ : /ðuÞ > bg 6¼ ;. Next, let u 2 P ð/; b; dÞ. Then 2 /ðuÞ P b, and so b 6 kuk 6 d. It follows from Lemma 2.2 that xb 6 xkuk 6 uðtÞ 6 d; t 2 ½x; 1 x, and con 2b for xb 6 uðtÞ 6 d; t 2 ½x; 1 x. In the following, we shall discuss it from dition ðH5 Þ yields f ðuðtÞÞ > up x‘ three respects. (i) If r 2 ð0; x, by ðH5 Þ and (2.1), (3.1), (3.2), and Lemma 2.1, we have 1 /ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ P Tuð1 xÞ 2 Z g Z 1 Z s Z s Z 1 d ¼ uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds P uq aðrÞf ðuðrÞÞdr ds c r 1x r 1x r Z 1x Z 1x Z 1 2b 2b P 2b > b: > uq aðrÞup aðrÞdr dr ds > xuq x‘ x‘ 1x x x (ii) If r 2 ðx; 1 xÞ, then for arbitrary u 2 P ð/; b; dÞ, similarly, we have 2/ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ Z x Z r Z P uq aðrÞf ðuðrÞÞdr ds þ 0
Z
r
uq
P 0
Z > x uq
uq
1x
s
Z
x
Z
1
x
Z aðrÞf ðuðrÞÞdr ds þ
aðrÞf ðuðrÞÞdr ds
r
Z
1
1x
s
uq r
1x
aðrÞf ðuðrÞÞdr ds
Z 1x 2b 2b aðrÞup aðrÞup ð Þdr dr þ uq x‘ x‘ x r Z r Z 1x 2b > x uq P 2b; aðrÞdr þ uq aðrÞdr x‘ x r r
then /ðTuÞ > b holds.
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421
(iii) If r 2 ½1 x; 1Þ, then for arbitrary u 2 P ð/; b; dÞ, similarly, we have Z x Z r 1 /ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ P TuðxÞ P uq aðrÞf ðuðrÞÞdr ds 2 0 s Z 1x Z x Z 1x 2b 2b P 2b > b > uq aðrÞup aðrÞdr dr ds > xuq x‘ x‘ 0 x x Hence, for arbitrary u 2 P ð/; b; dÞ; /ðTuÞ > b, thus, condition ðC1 Þ of Theorem A is satisfied. Finally, we show that condition ðC3 Þ of Theorem A also holds. For this end, let u 2 P ð/; b; cÞ with kTuk > d. Then, it follows from Lemma 2.2 and d P xb that 1 /ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ P xkTuk > xd P b: 2 Consequently, condition ðC3 Þ of Theorem A holds. Thus, it follows from Theorem A that the boundary value problem (1.1) and (1.2) has at least three positive solutions u1 ; u2 , and u3 , such that 0 < ku1 k < a;
/ðu2 Þ > b;
ku3 k > a;
The Proof of Theorem 2.1 is complete.
and
/ðu3 Þ < b:
h
Proof of Corollary 2.1. Due to the mathematical induction, when n ¼ 1, from the proof process of Theorem 2.1 and condition ðH6 Þ, we can obtain P a1 ! P a1 P a1 , and T is completely continuous. Thus, thanks to Theorem B, we can show that T has at least a fixed point u1 2 P a1 , i.e. the boundary value problem (1.1) and (1.2) has at least a positive solution. When n ¼ 2, let a ¼ a1 ; b ¼ b1 ; d ¼ d 1 ; e ¼ a2 , then, it follows from conditions ðH6 Þ and ðH7 Þ that all conditions of Theorem 2.1 are satisfied. Hence, the boundary value problem (1.1) and (1.2) has at least three positive solutions u1 ; u2 and u3 satisfying 0 < ku1 k < a1 ; /ðu2 Þ > b1 ; ku3 k > a1 and /ðu3 Þ < b1 . In turn, we continuously apply the proof method above, thus Corollary 2.1 holds. h Similarly, we can prove Corollary 2.2, here we omit it. Remark 3.1. For k2 , these proofs are similar to the above Theorem 2.1 and Corollary 2.1, we omit it here. In order to illustrate the above results, we present an example as follows: Example. Consider the singular boundary value problem with p-Laplacian ( 1 0 ðup ðu0 ÞÞ þ t2 f ðuÞ ¼ 0; 0 < t < 1; 0 1 2up ðuð0ÞÞ up u 4 ¼ 0; up ðuð1ÞÞ þ dup u0 12 ¼ 0;
ð3:7Þ
where 3 p¼ ; 2
a ¼ 2;
b ¼ c ¼ 1;
d P 0;
1 n¼ ; 4
and 8 2 0 6 u 6 40; u; > < 4 f ðuÞ ¼ 1592 þ 5 ð50 uÞ; 40 < u 6 50; > : 2uþ15820 pffiffiffiffi ; u > 50: 2u Then, (3.7) has at least three positive solutions. In fact, we have pffiffiffi pffiffiffi 11 6 2 ‘ ¼ 2 3; k1 ¼ : 12
1 g¼ ; 2
1 x¼ ; 4
1
aðtÞ ¼ t2 ;
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X. Li / Applied Mathematics and Computation 202 (2008) 413–426
Obviously, conditions ðH1 Þ and ðH2 Þ are satisfied. Let a ¼ 1; b ¼ 40; d ¼ 200, then 0 < a < xb < b < b=x < d. Thus, we have a 1 f ðuÞ 6 max f ðuÞ ¼ 1 < up ¼ up for 0 6 u 6 1; 06u61 k1 k1 2b f ðuÞ > min f ðuÞ ¼ 100 > up for 10 6 u 6 200; 106u6200 x‘ and lim sup u!1
f ðuÞ 1
u2
pffiffiffi 1 ¼ 2 < up : k1
Consequently, conditions ðH3 Þ; ðH4 ÞðiiÞ; ðH5 Þ of Theorem 2.1 are satisfied. Then by Theorem 2.1, the singular boundary value problem with p-Laplacian (3.7) has at least three positive solutions u1 ; u2 , and u3 , such that 0 < ku1 k < a;
/ðu2 Þ > b;
ku3 k > a;
and
/ðu3 Þ < b:
Remark 3.2. Problem (3.7) is singular at t ¼ 0. 4. Proof of Theorem 2.2 In this section, we will prove Theorem 2.2. The main tools are the Avery–Peterson fixed point theorem below. Theorem C [1]. Let P be a cone in a real Banach space E. Let c and h be nonnegative continuous convex functional on P ; a be a nonnegative continuous concave functional on P, and w be a nonnegative continuous functional on P satisfying wðkuÞ 6 kwðuÞ for 0 6 k 6 1, such that for some positive numbers K and d aðuÞ 6 wðuÞ
and
kuk 6 KcðuÞ;
ð4:1Þ
for all u 2 P ðc; dÞ. Suppose T : P ðc; dÞ ! P ðc; dÞ is a completely continuous and there exist positive numbers a; b and c with a < b, such that (C1) fu 2 P ðc; h; a; b; c; dÞ : aðuÞ > bg 6¼ ;, and aðTuÞ > b for u 2 P ðc; h; a; b; c; dÞ; (C2) aðTuÞ > b for u 2 P ðc; a; b; dÞ with hðTuÞ > c; (C3) 0 62 Rðc; w; a; dÞ and wðTuÞ < a for u 2 Rðc; w; a; dÞ with wðuÞ ¼ a. Then, T has at least three fixed points u1 ; u2 ; u3 2 P ðc; dÞ, such that cðui Þ 6 d a < wðu2 Þ;
for i ¼ 1; 2; 3;
b < aðu1 Þ;
with aðu2 Þ < b;
wðu3 Þ < a:
T Lemma 4.1. Suppose that conditions ðH1 Þ; ðH2 Þ hold, then uðtÞ 2 C½0; 1 C 2 ð0; 1Þ is a solution of boundary value problem (1.1)0 and (1.2), if and only if uðtÞ 2 E is a solution of the following integral equation: 8 R R R r r t > < uq ba n aðrÞgðuðrÞ; u0 ðrÞÞdr þ 0 uq s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; 0 6 t 6 r; uðtÞ ¼ R R R > : uq d g aðrÞgðuðrÞ; u0 ðrÞÞdr þ 1 uq s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; r 6 t 6 1; t r c r where r 2 ½n; g ð0; 1Þ and u0 ðrÞ ¼ 0. Proof. The proof is similar to [10, Lemma 2.3], Here we omit it.
h
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
Proof of Theorem 2.2. We define the operator T : P 1 ! E1 given by 8 R R R r t b r > < uq a n aðrÞgðuðrÞ; u0 ðrÞÞdr þ 0 uq s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; ðTuÞðtÞ ¼ R R R > : u d g aðrÞgðuðrÞ; u0 ðrÞÞdr þ 1 u s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; q c r q r t
423
0 6 t 6 r; ð4:2Þ r 6 t 6 1:
First, it follows from Lemma 4.1 that the operator T is well-defined. Next, because of ( R r uq t aðrÞgðuðrÞ; u0 ðrÞÞdr ðP 0Þ; 0 6 t 6 r ðTuÞ0 ðtÞ ¼ R t uq r aðrÞgðuðrÞ; u0 ðrÞÞdr ð6 0Þ; r 6 t 6 1
ð4:3Þ
then, it is obvious that the operator ðTuÞ0 is a continuous monotone decreasing on [0, 1], and ðTuÞ0 ðrÞ ¼ 0. Meanwhile, it follows from the definition of operator T that for each u 2 P 1 ; Tu 2 E1 is a nonnegative continuous, and by (4.2) and (4.3), we can obtain 0
aup ððTuÞð0ÞÞ bup ððTuÞ ðnÞÞ ¼ 0;
0
cup ððTuÞð1ÞÞ þ dup ððTuÞ ðgÞÞ ¼ 0:
These show Tu 2 P 1 ; i:e: T ðP 1 Þ P 1 , and we can obtain ðTuÞðrÞ ¼ kTuk1 with Z r Z r Z r b 0 0 uq aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds a n 0 s Z g Z 1 Z s d ¼ uq aðrÞgðuðrÞ; u0 ðrÞÞdr þ uq aðrÞgðuðrÞ; u0 ðrÞÞdr ds: c r r r Furthermore, we can also obtain that the operator T : P 1 ! P 1 is a completely continuous (see [10, Lemma 2.4]), and it follows from Lemma 4.1 that each fixed point of T in P 1 is a positive solution of (1.1)0 and (1.2). We now show that the conditions of Theorem C are satisfied. 0 Firstly, we choose u 2 P 1 ðc1 ; dÞ, then c1 ðuÞ ¼ max 06t61 ju ðtÞj 6 d. By Lemma 2.3, there is max06t61 juðtÞj 6 Ld, then assumption ðH10 Þ yields gðuðtÞ; u0 ðtÞÞ 6 up Nd , for arbitrary 0 6 t 6 1. Note that max06t61 jðTuÞ0 ðtÞj ¼ maxfðTuÞ0 ð0Þ; jðTuÞ0 ð1Þjg, and from (4.3), we have Z r Z 1 0 0 0 c1 ðTuÞ ¼ max jðTuÞ ðtÞj ¼ max uq aðrÞgðuðrÞ; u ðrÞÞdr ; uq aðrÞgðuðrÞ; u ðrÞÞdr 06t61
0
Z
6 max uq
g
r
Z aðrÞgðuðrÞ; u0 ðrÞÞdr ; uq
0
6
d max uq N
1
aðrÞgðuðrÞ; u0 ðrÞÞdr
n
Z
Z g aðrÞdr ; uq
0
1
aðrÞdr
¼ d:
n
This proves that T : P 1 ðc1 ; dÞ ! P 1 ðc1 ; dÞ, and by the above proof, we know that T : P 1 ðc1 ; dÞ ! P 1 ðc1 ; dÞ is completely continuous. In addition, due to Lemmas 2.2 and 2.3 and the definition of a1 ; c1 ; h1 ; w1 , we have xh1 ðuÞ 6 a1 ðuÞ 6 h1 ðuÞ ¼ w1 ðuÞ; kuk1 ¼ maxfh1 ðuÞ; c1 ðuÞg 6 Lc1 ðuÞ
ð4:4Þ for all u 2 P 1 :
and w1 ðkuÞ ¼ max jkuðtÞj ¼ k max juðtÞj ¼ kw1 ðuÞ; 06t61
06t61
Therefore, condition (4.1) of Theorem C is satisfied.
0 6 k 6 1:
ð4:5Þ
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X. Li / Applied Mathematics and Computation 202 (2008) 413–426
b Secondly, we take uðtÞ 2x for t 2 ½0; 1, then, we have
b > b; 2x c1 ðuðtÞÞ ¼ max ju ðtÞj ¼ 0 < d;
a1 ðuðtÞÞ ¼ min juðtÞj ¼ x6t61x 0
06t61
h1 ðuðtÞÞ ¼ max juðtÞj ¼ 06t61
b b < ; 2x x
b thus, uðtÞ 2x 2 P 1 ðc1 ; h1 ; a1 ; b; xb ; dÞ and a1 ðuðtÞÞ > b. Hence, fu 2 P 1 ðc1 ; h1 ; a1 ; b; xb ; dÞ : a1 ðuÞ > bg 6¼ ;, and for any u 2 P 1 ðc1 ; h1 ; a1 ; b; xb ; dÞ, there is b 6 uðtÞ 6 b 0 x ; ju ðtÞj 6 d for t 2 ½x; 1 x. b By condition ðH11 Þ, we have gðuðtÞ; u0 ðtÞÞ > up xm for t 2 ½x; 1 x, and by Lemma 2.2 and the cone P 1 , we have
a1 ðTuÞ ¼ min jðTuÞðtÞj P xh1 ððTuÞðtÞÞ ¼ xðTuÞðrÞ x6t61x Z r Z r Z r b 0 0 ¼ x uq aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds a n 0 s Z g Z 1 Z s d ¼ x uq aðrÞgðuðrÞ; u0 ðrÞÞdr þ uq aðrÞgðuðrÞ; u0 ðrÞÞdr ds c r r r Z n Z n Z 1 Z s 0 0 P x min uq aðrÞgðuðrÞ; u ðrÞÞdr ds; uq aðrÞgðuðrÞ; u ðrÞÞdr ds 0
>x
b min xm
Z
s n
Z
Z
aðrÞdr ds;
uq 0
n
s
Z
1
uq g
P 1 c1 ; h1 ; a1 ; b; xb ; d
g s
g
aðrÞdr ds ¼ b:
g
Consequently, a1 ðTuÞ > b, for all u 2 . Thereupon, condition ðC1 Þ of Theorem C is satisfied. Thirdly, we claim that conditionðC2 Þ of Theorem C is satisfied. For this end, we choose u 2 P 1 ðc1 ; a1 ; b; dÞ with h1 ðTuÞ > xb , and it follows from Lemma 2.2 that b a1 ðTuÞ P xh1 ðTuÞ > x ¼ b: x Thus, condition ðC2 Þ of Theorem C is satisfied. Finally, we show that condition ðC3 Þ of Theorem C is also satisfied. It is easy to see that w1 ð0Þ ¼ 0 < a, hence 0 62 Rðc1 ; w1 ; a; dÞ. Now assume that u 2 Rðc1 ; w1 ; a; dÞ with w1 ðuÞ ¼ a, then, by condition ðH12 Þ, we have w1 ðTuÞ ¼ max jðTuÞðtÞj ¼ ðTuÞðrÞ 06t61 Z r Z r Z r b 0 0 ¼ uq aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds a n 0 s Z g Z 1 Z s d 0 0 aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds ¼ uq c r r r Z g Z g Z g b 0 0 6 max uq aðrÞgðuðrÞ; u ðrÞÞdr ds þ uq aðrÞgðuðrÞ; u ðrÞÞdr ; a n 0 s Z g Z 1 Z s d uq aðrÞgðuðrÞ; u0 ðrÞÞdr ds þ uq aðrÞgðuðrÞ; u0 ðrÞÞdr c n n n Z g Z g Z g Z 1 Z s a b max < uq aðrÞdr ds þ uq aðrÞdr ; uq aðrÞdr ds M a n 0 s n n Z g d þuq aðrÞdr ¼a c n
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
425
Consequently, condition ðC3 Þ of Theorem C is satisfied. Thus, it now follows from Theorem C that the boundary value problem (1.1)0 and (1.2) has at least three positive solutions u1 ; u2 , and u3 satisfying max ju0i ðtÞj 6 d
06t61
for i ¼ 1; 2; 3;
b < min ju1 ðtÞj;
max ju1 ðtÞj 6 Ld;
x6t61x
a < max ju2 ðtÞj 06t61
06t61
with
min ju2 ðtÞj < b;
x6t61x
max ju3 j < a:
06t61
The Proof of Theorem 2.2 is complete.
h
In order to illustrate our result, we present an example as follows: Example. Consider the singular boundary value problem with p-Laplacian ( 1 0 ðup ðu0 ðtÞÞÞ þ 12 t2 gðuðtÞ; u0 ðtÞÞ ¼ 0; 0 < t < 1; aup ðuð0ÞÞ bup u0 ð14Þ ¼ 0; cup ðuð1ÞÞ þ dup u0 ð12Þ ¼ 0:
ð4:6Þ
where 3 p¼ ; 2
a ¼ b;
and
0
gðu; u Þ ¼
c ¼ d;
1 n¼ ; 4
8 5 u0 > u þ sin ; > 100 >2 > > 5 4 sin u < u þ 0; 2 100
1 x¼ ; 4
1 1 aðtÞ ¼ t2 ; 2
ðu; u0 Þ 2 ½0; 1 ð1; 1Þ; ðu; u0 Þ 2 ½1; 2 ð1; 1Þ;
0
u > ; 40 þ sin > 100 > > > 4 0 : 40 þ u10 pffiffi þ sin u ; 100
u
1 g¼ ; 2
ðu; u0 Þ 2 ½2; 104 ð1; 1Þ; ðu; u0 Þ 2 ½104 ; þ1Þ ð1; þ1Þ:
Then, (4.6) has at least three positive solutions. In fact, we have pffiffiffi 1 79 48 2 1 L ¼ 2; N ¼ ; M ¼ ; m¼ : 2 96 96 Clearly, conditions ðH1 Þ and ðH2 Þ are satisfied. Let a ¼ 1; b ¼ 2; d ¼ 5000, thus, we have 1 d < up ð Þ ¼ 100 for ðu; u0 Þ 2 ½0; 10000 ½5000; 5000; 100 N pffiffiffi 1 b 0 0 > up gðu; u Þ > min gðu; u Þ ¼ 40 þ ¼ 16 3 for ðu; u0 Þ 2 ½2; 8 ½5000; 5000; 100 xm a 5 1 < up gðu; u0 Þ < max gðu; u0 Þ ¼ þ for ðu; u0 Þ 2 ½0; 1 ½5000; 5000: 2 100 M gðu; u0 Þ 6 max gðu; u0 Þ ¼ 40 þ
Consequently, conditions ðH10 Þ; ðH11 Þ; and ðH12 Þ of Theorem 2.2 are satisfied. Thus, by Theorem 2.2, the singular boundary value problem with p-Laplacian (4.6) has at least three positive solutions u1 ; u2 , and u3 , such that max ju0i ðtÞj 6 d
06t61
for
i ¼ 1; 2; 3;
b < min ju1 ðtÞj; max ju1 ðtÞj 6 Ld; x6t61x
1 < max ju2 ðtÞj 06t61
max ju3 j < a:
06t61
06t61
with
min ju2 ðtÞj < b;
x6t61x
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Acknowledgements The author is very grateful to the reviewer and the referee for their important comments and suggestions in improving this manuscript. References [1] R.I. Avery, A.C. Peterson, Three positive fixed points of nonlinear operators on ordered Banach spaces, Comput. Math. Appl. 42 (2001) 313–322. [2] C.Z. Bai, J.X. Fang, Existence of multiple positive solutions for nonlinear m-point boundary-value problems, Appl. Math. Comput. 140 (2003) 297–305. [3] X.M. He, Double positive solutions for a three-point boundary value problem for one-dimensional p-Laplacian, Appl. Math. Lett. 17 (2004) 867–873. [4] D.H. Ji, W.G. Ge, Multiple positive solutions for some p-Laplacian boundary value problems, Appl. Math. Comput. 187 (2007) 1315–1325. [5] Dehong Ji, Weigao Ge, Existence of multiple positive solutions for Sturm–Liouville-like four-point boundary value problem with p-Laplacian, Nonlinear Anal., in press, doi:10.1016/j.na.2007.02.010. [6] R.W. Leggett, L.R. Williams, Multiple positive fixed points of nonlinear operators on ordered Banach spaces, Indiana Univ. Math. J. 28 (1979) 673–688. [7] Ruixi Liang, Jun Peng, Jianhua Shen, Double positive solutions for a nonlinear four-point boundary value problem with a p-Laplacian operator, Nonlinear Anal., in press, doi:10.1016/j.na.2007.01.058. [8] B. Liu, Positive solutions three-point boundary value problems for one-dimensional p-Laplacian with infinitely many singularities, Appl. Math. Lett. 17 (2004) 655–661. [9] D.X. Ma, Z.J. Du, W.G. Ge, Existence and iteration of monotone positive solutions for multipoint boundary value problem with p-Laplacian operator, Comput. Math. Appl. 50 (2005) 729–739. [10] H. Su, Z.L. Wei, B.H. Wang, The existence of positive solutions for a nonlinear four-point singular boundary value problem with a p-Laplacian operator, Nonlinear Anal. 66 (2007) 2204–2217. [11] H. Su, Z.L. Wei, F.Y. Xu, The existence of positive solutions for nonlinear singular boundary value problem with a p-Laplacian, Appl. Math. Comput. 181 (2006) 826–836. [12] Y.Y. Wang, W.G. Ge, Multiple positive solutions for multipoint boundary value problems with one-dimensional p-Laplacian, J. Math. Anal. Appl. 327 (2007) 1381–1395. [13] Y.Y. Wang, C.M. Hou, Existence of multiple positive solutions for one-dimensional p-Laplacian, J. Math. Anal. Appl. 315 (2006) 144–153. [14] Z.Y. Wang, J.H. Zhang, Positive solutions for one-dimensional p-Laplacian boundary value problems with dependence on the first order derivative, J. Math. Anal. Appl. 314 (2006) 618–630. [15] D.X. Zhao, H.Z. Wang, W.G. Ge, Existence of triple positive solutions to a class of p-Laplacian boundary value problems, J. Math. Anal. Appl. 328 (2007) 972–983. [16] E. Zeidler, Nonlinear Functional Analysis and its Applications, vol. 2/B, Springer, Berlin, 1990.
Applied Mathematics and Computation 202 (2008) 413–426 www.elsevier.com/locate/amc
Multiple positive solutions for some four-point boundary value problems with p-Laplacian q Xiangfeng Li Department of Mathematics, Longdong University, Qingyang, Gansu, 745000, PR China
Abstract This paper is devoted to the existence of multiple positive solutions for some nonlinear singular four-point boundary value problems with p-Laplacian. Under appropriate assumptions on the nonlinear terms, by applying several known fixed point theorems, sufficient conditions for the existence of at least three positive solutions are established. Ó 2008 Elsevier Inc. All rights reserved. Keywords: p-Laplacian operator; Multiple positive solutions; Four-point boundary value problem; Fixed point theorem
1. Introduction In this paper, we study the existence of multiple positive solutions for the following quasilinear singular four-point boundary value problems: 0
ðup ðu0 ÞÞ þ aðtÞf ðuðtÞÞ ¼ 0;
0 < t < 1;
ð1:1Þ
and 0
ðup ðu0 ÞÞ þ aðtÞgðuðtÞ; u0 ðtÞÞ ¼ 0;
0 < t < 1;
ð1:10 Þ
subject to the nonlinear boundary conditions aup ðuð0ÞÞ bup ðu0 ðnÞÞ ¼ 0;
cup ðuð1ÞÞ þ dup ðu0 ðgÞÞ ¼ 0;
ð1:2Þ
where up ðxÞ ¼ jxjp2 x; p > 1; uq ðxÞ ¼ jxjq2 x is the inverse function to up ; 1p þ 1q ¼ 1; a > 0; b P 0; c > 0; d P 0; n; g 2 ð0; 1Þ is prescribed and n < g. aðtÞ may be singular at t ¼ 0 and/or t ¼ 1. In recent years, the study of the existence of positive solutions for boundary value problems with p-Laplacian operator has been an interesting topic; there are much current attentions focused on the study of nonlinear multi-point (at least three point) boundary value problem, see [2,4,9,12,13] and references therein. The q
This work is sponsored by the Tutorial Scientific Research Foundation of Education Department of Gansu Province (0710-04). E-mail address: [email protected]
0096-3003/$ - see front matter Ó 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2008.02.006
414
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
methods and techniques employed in these papers involve the use of the fixed point theorem in cones [2,13], the Avery–Peterson fixed point theorem [4,12], and the monotone iterative technique [9]. Recently, when the nonlinear term f does not depend on the first-order derivative, Eq. (1.1) together with some multi-point boundary conditions has been studied by researchers, for example, (see [3,5,7,8,10,11] and references therein). In [8], Liu studied the existence of positive solution of Eq. (1.1) subject to the three-point boundary value conditions u0 ð0Þ ¼ 0;
uð1Þ ¼ buðgÞ;
by applying the fixed point index theory. In [3], using a three-functional fixed point theorem due to Avery and Henderson, considered the existence of double positive solutions of Eq. (1.1) subject to one of the following two pairs of nonlinear three-point boundary value conditions: uð0Þ B0 ðu0 ðgÞÞ ¼ 0;
uð1Þ B1 ðu0 ð1ÞÞ ¼ 0;
uð0Þ B0 ðu0 ð0ÞÞ ¼ 0;
uð1Þ B1 ðu0 ðgÞÞ ¼ 0:
In [11], Su et al. studied the existence of multiple positive solutions of Eq. (1.1) subject to the nonlinear twopoint boundary value conditions aup ðuð0ÞÞ bup ðu0 ð0ÞÞ ¼ 0;
cup ðuð1ÞÞ þ dup ðu0 ð1ÞÞ ¼ 0:
The main tool is the fixed point theorem of cone expansion and compression of norm type. Very recently in [5,7,10], Ji et al., Liang et al. and Su et al. discussed the existence of a positive solution and double positive solutions of Eqs. (1.1) and (1.2), respectively. The main tools is the fixed point index theory and the Avery and Henderson fixed point theorem. More recently, when the nonlinear term f depends on the first-order derivative, Eq. (1.1)0 subject to the boundary value conditions u0 ð0Þ ¼ uð1Þ ¼ 0
or
uð0Þ ¼ ð1Þ ¼ 0;
has been studied by Wang et al. and Zhao et al., by means of the Leggett–Williams fixed point theorem and the fixed point theorem in cones, respectively, see [14,15]. However, for Eqs. (1.1), (1.2) and (1.1)0 , (1.2), there are currently few papers dealing with the existence of multiple (at least three) positive solutions. Motivated by the papers mentioned above, in this paper we shall consider the existence of multiple positive solutions for quasilinear singular boundary value problem (1.1), (1.2) and (1.1)0 , (1.2) by applying the Leggett–Williams fixed point theorem, the Schauder fixed point theorem and the Avery–Peterson fixed point theorem, respectively. In the rest of the paper, we make the following assumptions: (H1) f : Cð½0; 1Þ ! ½0; 1ÞÞ; g 2 Cð½0; R 1þ1Þ ð1; þ1Þ; ½0; þ1ÞÞ; (H2) aðtÞ 2 Cðð0; 1Þ; ½0; 1ÞÞ and 0 < 0 aðtÞdt < 1. Furthermore, aðtÞ is not an identical zero on any compact subinterval of ð0; 1Þ. 2. Preliminaries and main results In this section, we provide some background materials from the theory of cones in Banach spaces and Lemmas, then give out the main results of this paper. Definition 2.1. Let ðE; k kÞ be a real Banach space. A nonempty, closed, convex set P E is called a cone if it satisfies the following two conditions: (i) u 2 P ; k P 0, implies ku 2 P ; (ii) u 2 P ; u 2 P , implies u ¼ 0.
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
415
Definition 2.2. A map a is said to be a nonnegative continuous concave functional on cone P of real Banach space E if a : P ! ½0; 1Þ; is continuous and aðtu þ ð1 tÞvÞ P taðuÞ þ ð1 tÞaðvÞ; for all u; v 2 P and t 2 ½0; 1. Similarly, we say that the map b is a nonnegative continuous convex functional on a cone P of real Banach space E if b : P ! ½0; 1Þ; is continuous and bðtu þ ð1 tÞvÞ 6 tbðuÞ þ ð1 tÞbðvÞ; for all u; v 2 P and t 2 ½0; 1. Let 0 < a < b; r > 0 be real numbers, aðuÞ be a nonnegative continuous concave functional on P, then we define the following convex sets: P r ¼ fu 2 P : kuk < rg; P r ¼ fu 2 P : kuk 6 rg; oP r ¼ fu 2 P : kuk ¼ rg; P ða; a; bÞ ¼ fu 2 P : a 6 aðuÞ; kuk 6 bg: Let c and h be nonnegative continuous convex functional on P ; a be a nonnegative continuous concave functional on P, and w be a nonnegative continuous functional on P. Then, for positive real numbers a; b; c; and d, we define the following convex sets: P ðc; dÞ ¼ fu 2 P : cðuÞ < dg; P ðc; dÞ ¼ fu 2 P : cðuÞ 6 dg; P ðc; a; b; dÞ ¼ fu 2 P : b 6 aðuÞ; cðuÞ 6 dg; P ðc; h; a; b; c; dÞ ¼ fu 2 P : b 6 aðuÞ; hðuÞ 6 c; cðuÞ 6 dg; and a convex closed set Rðc; w; a; dÞ ¼ fu 2 P : a 6 wðuÞ; cðuÞ 6 dg: Lemma 2.1. Suppose condition ðH2 Þ holds, then there exists a constant x 2 ð0; 12Þ such that Z 1x 0< aðtÞdt < 1: x
Furthermore, the function Z t Z AðtÞ ¼ uq aðtÞdt þ uq x
1x
aðtÞdt ;
x 6 t 6 1 x;
t
is a positive continuous function on ½x; 1 x, and therefore AðtÞ has a minimum on ½x; 1 x; we suppose that there exists ‘ > 0 such that ‘ ¼ mint2½x;1x AðtÞ. Proof. By condition ðH2 Þ, it is easily seen that Z 1x 0< aðtÞdt < 1; x
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X. Li / Applied Mathematics and Computation 202 (2008) 413–426
is satisfied, and it is easy to prove that AðtÞ is continuous on ½x; 1 x. Next, let Z t Z 1x aðtÞdt ; A2 ðtÞ ¼ uq aðtÞdt : A1 ðtÞ ¼ uq x
t
Then, it follows from condition ðH2 Þ that the function A1 ðtÞ is strictly monotone increasing on ½x; 1 x and A1 ðxÞ ¼ 0; A2 ðtÞ is strictly monotone decreasing on ½x; 1 x and A2 ð1 xÞ ¼ 0; this shows ‘ ¼ mint2½x;1x AðtÞ > 0. The proof is complete. h In the sequel, let E ¼ C½0; 1 be endowed with the maximum norm kuk ¼ max juðtÞj: 06t61
1
E1 ¼ C ½0; 1 be endowed with the maximum norm kuk1 ¼ max max juðtÞj; max ju0 ðtÞj : 06t61
06t61
Clearly E; E1 are two Banach spaces. From the fact 0
ðup ðu0 ðtÞÞÞ ¼ aðtÞf ðuðtÞÞ 6 0;
0
ðup ðu0 ðtÞÞÞ ¼ aðtÞgðuðtÞ; u0 ðtÞÞ 6 0;
we know that u is concave on [0, 1]. So, we define the cone P E by P ¼ fu 2 E : uðtÞ P 0;
uðtÞ is the concave function on ½0; 1g:
we define the cone P 1 E1 by P 1 ¼ fu 2 E : uðtÞ P 0; aup ðuð0ÞÞ bup ðu0 ðnÞÞ ¼ 0; cup ðuð1ÞÞ þ dup ðu0 ðgÞÞ ¼ 0; uðtÞ is the concave function on ½0; 1g; Obviously, E1 E; P 1 P . Let the nonnegative continuous concave functional a1 , the nonnegative continuous convex functional h1 ; c1 , and the nonnegative continuous functional w1 be defined on the cone P 1 by c1 ðuÞ ¼ max ju0 ðtÞj;
w1 ðuÞ ¼ h1 ðuÞ ¼ max juðtÞj; 06t61 1 a1 ðuÞ ¼ min juðtÞj; x 2 0; : x6t6ð1xÞ 2 06t61
Lemma 2.2 [10]. Let u 2 P and x be as in Lemma 2.1, then uðtÞ P xkuk;
t 2 ½x; 1 x:
Apparently, if u 2 P 1 , then Lemma 2.2 also holds. Lemma 2.3. Let u 2 P 1 , then there exists a positive constant L, such that max juðtÞj 6 L max ju0 ðtÞj:
06t61
06t61
Proof. By uðtÞ uð0Þ ¼
Rt 0
u0 ðsÞds, we have
max juðtÞj 6 juð0Þj þ max ju0 ðtÞj:
06t61
06t61
On the other hand, by the first equation of (1.2), we have b b 0 ju ðnÞj 6 uq max ju0 ðtÞj: juð0Þj ¼ uq a a 06t61 Thus, we have
b max juðtÞj 6 1 þ uq max ju0 ðtÞj: 06t61 06t61 a
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417
Similarly, by the second equation of (1.2), we have d max juðtÞj 6 1 þ uq max ju0 ðtÞj: 06t61 06t61 c Therefore, setting b d L ¼ min 1 þ uq ; 1 þ uq ; a c h
the proof of Lemma 2.3 is complete. Next, we define
1 ð2:1Þ /ðuÞ ¼ ðuðxÞ þ uð1 xÞÞ for all u 2 P : 2 Obviously, /ðuÞ is the concave functional, and it is easy to prove that /ðuÞ 6 kuk, for arbitrary u 2 P . For convenience, we introduce the following constants: Z g Z g Z g b k1 ¼ u q aðrÞdr þ uq aðrÞdr ds; a n 0 s Z g Z 1 Z s d aðrÞdr þ uq aðrÞdr ds: k2 ¼ u q c n n n Obviously, k1 ; k2 > 0 Z N ¼ max uq
g
Z aðrÞdr ; uq
0
1
aðrÞdr
;
n
Z g Z g Z g b aðrÞdr þ uq aðrÞdr ds; M ¼ max uq a n 0 s Z g Z 1 Z s d uq aðrÞdr þ uq aðrÞdr ds ; c n n n Z n Z n Z 1 Z s uq aðrÞdr ds; uq aðrÞdr ds : m ¼ min 0
g
s
g
Here are the main results of this paper. Theorem 2.1. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist positive constants a; b; d, such that 0 < a < xb < b < b=x 6 d. Also assume that f satisfies (H3) f ðuÞ < up kai ; i ¼ 1; 2 for 0 6 u 6 a; (H4) one of the following conditions holds: e (i) there exists constant e > d, such that f ðuÞ 6 u p ki ; i ¼ 1; 2 for 0 6 u 6 e or, f ðuÞ (ii) lim supu!1 up ðuÞ < up k1i ; i ¼ 1; 2. (H5) f ðuÞ > up
2b x‘
for xb 6 u 6 d.
Then, the boundary value problem (1.1) and (1.2) has at least three positive solutions u1 ; u2 , and u3 satisfying 0 < ku1 k < a;
/ðu2 Þ > b;
ku3 k > a;
and
/ðu3 Þ < b:
Corollary 2.1. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist some positive constants ai ; bi ; d i ; i 2 N ; N is the positive integer number set, such that 0 < a1 < xb1 < b1 < b1 =x 6 d 1 < a2 < xb2 < b2 < b2 =x 6 d 2 < < an , and assume that f satisfies
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X. Li / Applied Mathematics and Computation 202 (2008) 413–426
(H6) f ðuÞ < up kaji ; j ¼ 1; 2; for 0 6 u 6 ai ; 1 6 i 6 n; i (H7) f ðuÞ > up 2b for xbi 6 u 6 d i ; 1 6 i 6 n 1. x‘ Then, the boundary value problem (1.1) and (1.2) has at least 2n 1 positive solutions. Corollary 2.2. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist some positive constants ai ; bi ; d i :i 2 N ; N is the positive integer number set, such that 0 < a1 < xb1 < b1 < d 1 < a2 < xb2 < b2 < d 2 < < an < , and assume that f satisfies (H8) f ðuÞ < up ðkaji Þ; j ¼ 1; 2 for 0 6 u 6 ai ; i 2 N ; i (H9) f ðuÞ > up ð2b Þ for xbi 6 u 6 d i ; i 2 N . x‘
Then, the boundary value problem (1.1) and (1.2) has infinitely many positive solutions. Theorem 2.2. Assume that ðH1 Þ; ðH2 Þ hold, and suppose that there exist positive constants a; b; d, such that 0 < a < b < xLd. Also assume that g satisfies the following conditions: (H10) ðgðu; u0 Þ 6 up Nd for ðu; u0 Þ 2 ½0; Ld ½d; d;
b for ðu; u0 Þ 2 b; xb ½d; d; (H11) gðu; u0 Þ > up xm (H12) gðu; u0 Þ < up Ma for ðu; u0 Þ 2 ½0; a ½d; d. Then, the boundary value problem (1.1)0 and (1.2) has at least three positive solutions u1 ; u2 , and u3 satisfying max ju0i ðtÞj 6 d
06t61
for i ¼ 1; 2; 3;
b < min ju1 ðtÞj; x6t61x
b 1 < max ju2 ðtÞj < ; 06t61 x
with max ju1 ðtÞj 6 Ld; 06t61
with d min ju2 ðtÞj < b; x6t61x
max ju3 j < a:
06t61
3. Proof of Theorem 2.1 In this section, we will prove Theorem 2.1 and its Corollary. The main tools are the Leggett–Williams fixed point theorem and the Schauder fixed point theorem: Theorem A [6]. Let c > 0 and T : P c ! P c be completely continuous and /ðuÞ be a nonnegative continuous concave functional on P, and such that /ðuÞ 6 kuk, for arbitrary u 2 P c . Suppose there exists 0 < a < b < d 6 c, such that (C1) fu 2 P ð/; b; dÞ : /ðuÞ > bg 6¼ ;, and /ðTuÞ > b for arbitrary u 2 P ð/; b; dÞ; (C2) kTuk < a for kuk 6 a; (C3) /ðTuÞ > b for u 2 P ð/; b; cÞ, with kTuk > d. Then, T has at least three fixed points u1 ; u2 and u3 on P c and satisfying ku1 k < a; b < /ðu2 Þ
and
ku3 k > a with /ðu3 Þ < b:
Theorem B [16]. Let D be a bounded, convex closed set, and operator T : D ! D be completely continuous. Then, T has at least a fixed point in D.
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419
T Lemma 3.1 ([7,11]). Suppose that conditions ðH1 Þ; ðH2 Þ hold, then uðtÞ 2 C½0; 1 C 2 ð0; 1Þ is a solution of boundary value problem (1.1) and (1.2), if and only if uðtÞ is a solution of the following integral equation: 8 R R R > < uq ba nr aðrÞf ðuðrÞÞdr þ 0t uq sr aðrÞf ðuðrÞÞdr ds; 0 6 t 6 r; R R uðtÞ ¼ R > : uq dc rg aðrÞf ðuðrÞÞdr þ t1 uq rs aðrÞf ðuðrÞÞdr ds; r 6 t 6 1; where r 2 ½n; g ð0; 1Þ and u0 ðrÞ ¼ 0. Proof of Theorem 2.1. We define the operator T : P ! E given by 8 R R R > < uq ba nr aðrÞf ðuðrÞÞdr þ 0t uq sr aðrÞf ðuðrÞÞdr ds; R R ðTuÞðtÞ ¼ R > : uq dc rg aðrÞf ðuðrÞÞdr þ t1 uq rs aðrÞf ðuðrÞÞdr ds;
0 6 t 6 r; ð3:1Þ r 6 t 6 1:
First, it follows from Lemma 3.1 that the operator T is well-defined. Next, because of ( R r uq t aðrÞf ðuðrÞÞdr ; 0 6 t 6 r; 0 ðTuÞ ðtÞ ¼ R t uq r aðrÞf ðuðrÞÞdr ; r 6 t 6 1:
ð3:2Þ
0
It is easy to see that the operator ðTuÞ is a continuous monotone decreasing on [0, 1], and ðTuÞðtÞ P 0, for t 2 ½0; 1, these show Tu 2 P , i.e. T ðP Þ P and ðTuÞ0ðrÞ ¼ 0; ðTuÞðrÞ ¼ kTuk, namely Z r Z r Z r b aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds uq a n 0 s Z g Z 1 Z s d ¼ uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds c r r r Furthermore, we can also prove that the operator T : P ! P is a completely continuous (see [10, Lemma 2.4]), and it follows from Lemma 3.1 that each fixed point of T in P is a positive solution of (1.1) and (1.2). We now verify that the conditions of Theorem A are satisfied. Firstly, we choose u 2 P a . Then kuk 6 a, thus 0 6 u 6 a and assumption ðH3 Þ yields f ðuðtÞÞ < up ka1 , for 0 6 t 6 1. Thus, by Lemmas 2.1 and 3.1, we have Z r Z r Z r b aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds kTuk ¼ ðTuÞðrÞ ¼ uq a n 0 s Z g Z g Z g b 6 uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds a n 0 s Z g Z g Z g a b a < uq aðrÞdr þ uq aðrÞdr ds ¼ k1 ¼ a: k1 a n k 1 0 s Hence, condition ðC2 Þ of Theorem A is satisfied. Secondly, we claim that condition ðH4 Þ guarantees the existence of a number c with c > d and T : P c ! P c:
ð3:3Þ
have (3.3) with e ¼ c. Suppose now It is clear that if ðH4 Þ(i) holds, then from the above proof we immediately that ðH4 Þ(ii) is satisfied, then, there exist l > 0 and e < up k11 , such that f ðuÞ 6e up ðuÞ
for u P l:
ð3:4Þ
Let G ¼ maxu2½0;l f ðuÞ. In view of (3.4), it is easy to see that f ðuÞ 6 G þ eup ðuÞ for u P 0:
ð3:5Þ
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Now, let c be such that 1 1 eÞ : up ðcÞ > max up ðdÞ; Gðup k1
ð3:6Þ
Then, for arbitrary u 2 P c and t 2 ½0; 1, from (3.5) and (3.6) and Lemma 3.1, we can obtain Z r Z r Z r b kTuk ¼ ðTuÞðrÞ ¼ uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds a n 0 s Z g Z g Z g b aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds 6 uq a n 0 s Z g Z g Z g b aðrÞðG þ eup ðuðrÞÞÞdr þ uq aðrÞðG þ eup ðuðrÞÞÞdr ds 6 uq a n 0 s Z g Z g Z g b 1 c aðrÞdr þ uq aðrÞdr ds uq up ðcÞ up < uq e þ eup ðcÞ ¼ k1 ¼ c a n k k 1 1 0 s Hence, (3.3) holds. We shall now show that condition ðC1 Þ of Theorem A is satisfied. For this end, we suppose that uðtÞ
bþd : 2
bþd Note that /ðuðtÞÞ ¼ / bþd ¼ 2 > b, thus fu 2 P ð/; b; dÞ : /ðuÞ > bg 6¼ ;. Next, let u 2 P ð/; b; dÞ. Then 2 /ðuÞ P b, and so b 6 kuk 6 d. It follows from Lemma 2.2 that xb 6 xkuk 6 uðtÞ 6 d; t 2 ½x; 1 x, and con 2b for xb 6 uðtÞ 6 d; t 2 ½x; 1 x. In the following, we shall discuss it from dition ðH5 Þ yields f ðuðtÞÞ > up x‘ three respects. (i) If r 2 ð0; x, by ðH5 Þ and (2.1), (3.1), (3.2), and Lemma 2.1, we have 1 /ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ P Tuð1 xÞ 2 Z g Z 1 Z s Z s Z 1 d ¼ uq aðrÞf ðuðrÞÞdr þ uq aðrÞf ðuðrÞÞdr ds P uq aðrÞf ðuðrÞÞdr ds c r 1x r 1x r Z 1x Z 1x Z 1 2b 2b P 2b > b: > uq aðrÞup aðrÞdr dr ds > xuq x‘ x‘ 1x x x (ii) If r 2 ðx; 1 xÞ, then for arbitrary u 2 P ð/; b; dÞ, similarly, we have 2/ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ Z x Z r Z P uq aðrÞf ðuðrÞÞdr ds þ 0
Z
r
uq
P 0
Z > x uq
uq
1x
s
Z
x
Z
1
x
Z aðrÞf ðuðrÞÞdr ds þ
aðrÞf ðuðrÞÞdr ds
r
Z
1
1x
s
uq r
1x
aðrÞf ðuðrÞÞdr ds
Z 1x 2b 2b aðrÞup aðrÞup ð Þdr dr þ uq x‘ x‘ x r Z r Z 1x 2b > x uq P 2b; aðrÞdr þ uq aðrÞdr x‘ x r r
then /ðTuÞ > b holds.
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
421
(iii) If r 2 ½1 x; 1Þ, then for arbitrary u 2 P ð/; b; dÞ, similarly, we have Z x Z r 1 /ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ P TuðxÞ P uq aðrÞf ðuðrÞÞdr ds 2 0 s Z 1x Z x Z 1x 2b 2b P 2b > b > uq aðrÞup aðrÞdr dr ds > xuq x‘ x‘ 0 x x Hence, for arbitrary u 2 P ð/; b; dÞ; /ðTuÞ > b, thus, condition ðC1 Þ of Theorem A is satisfied. Finally, we show that condition ðC3 Þ of Theorem A also holds. For this end, let u 2 P ð/; b; cÞ with kTuk > d. Then, it follows from Lemma 2.2 and d P xb that 1 /ðTuÞ ¼ ðTuðxÞ þ Tuð1 xÞÞ P xkTuk > xd P b: 2 Consequently, condition ðC3 Þ of Theorem A holds. Thus, it follows from Theorem A that the boundary value problem (1.1) and (1.2) has at least three positive solutions u1 ; u2 , and u3 , such that 0 < ku1 k < a;
/ðu2 Þ > b;
ku3 k > a;
The Proof of Theorem 2.1 is complete.
and
/ðu3 Þ < b:
h
Proof of Corollary 2.1. Due to the mathematical induction, when n ¼ 1, from the proof process of Theorem 2.1 and condition ðH6 Þ, we can obtain P a1 ! P a1 P a1 , and T is completely continuous. Thus, thanks to Theorem B, we can show that T has at least a fixed point u1 2 P a1 , i.e. the boundary value problem (1.1) and (1.2) has at least a positive solution. When n ¼ 2, let a ¼ a1 ; b ¼ b1 ; d ¼ d 1 ; e ¼ a2 , then, it follows from conditions ðH6 Þ and ðH7 Þ that all conditions of Theorem 2.1 are satisfied. Hence, the boundary value problem (1.1) and (1.2) has at least three positive solutions u1 ; u2 and u3 satisfying 0 < ku1 k < a1 ; /ðu2 Þ > b1 ; ku3 k > a1 and /ðu3 Þ < b1 . In turn, we continuously apply the proof method above, thus Corollary 2.1 holds. h Similarly, we can prove Corollary 2.2, here we omit it. Remark 3.1. For k2 , these proofs are similar to the above Theorem 2.1 and Corollary 2.1, we omit it here. In order to illustrate the above results, we present an example as follows: Example. Consider the singular boundary value problem with p-Laplacian ( 1 0 ðup ðu0 ÞÞ þ t2 f ðuÞ ¼ 0; 0 < t < 1; 0 1 2up ðuð0ÞÞ up u 4 ¼ 0; up ðuð1ÞÞ þ dup u0 12 ¼ 0;
ð3:7Þ
where 3 p¼ ; 2
a ¼ 2;
b ¼ c ¼ 1;
d P 0;
1 n¼ ; 4
and 8 2 0 6 u 6 40; u; > < 4 f ðuÞ ¼ 1592 þ 5 ð50 uÞ; 40 < u 6 50; > : 2uþ15820 pffiffiffiffi ; u > 50: 2u Then, (3.7) has at least three positive solutions. In fact, we have pffiffiffi pffiffiffi 11 6 2 ‘ ¼ 2 3; k1 ¼ : 12
1 g¼ ; 2
1 x¼ ; 4
1
aðtÞ ¼ t2 ;
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X. Li / Applied Mathematics and Computation 202 (2008) 413–426
Obviously, conditions ðH1 Þ and ðH2 Þ are satisfied. Let a ¼ 1; b ¼ 40; d ¼ 200, then 0 < a < xb < b < b=x < d. Thus, we have a 1 f ðuÞ 6 max f ðuÞ ¼ 1 < up ¼ up for 0 6 u 6 1; 06u61 k1 k1 2b f ðuÞ > min f ðuÞ ¼ 100 > up for 10 6 u 6 200; 106u6200 x‘ and lim sup u!1
f ðuÞ 1
u2
pffiffiffi 1 ¼ 2 < up : k1
Consequently, conditions ðH3 Þ; ðH4 ÞðiiÞ; ðH5 Þ of Theorem 2.1 are satisfied. Then by Theorem 2.1, the singular boundary value problem with p-Laplacian (3.7) has at least three positive solutions u1 ; u2 , and u3 , such that 0 < ku1 k < a;
/ðu2 Þ > b;
ku3 k > a;
and
/ðu3 Þ < b:
Remark 3.2. Problem (3.7) is singular at t ¼ 0. 4. Proof of Theorem 2.2 In this section, we will prove Theorem 2.2. The main tools are the Avery–Peterson fixed point theorem below. Theorem C [1]. Let P be a cone in a real Banach space E. Let c and h be nonnegative continuous convex functional on P ; a be a nonnegative continuous concave functional on P, and w be a nonnegative continuous functional on P satisfying wðkuÞ 6 kwðuÞ for 0 6 k 6 1, such that for some positive numbers K and d aðuÞ 6 wðuÞ
and
kuk 6 KcðuÞ;
ð4:1Þ
for all u 2 P ðc; dÞ. Suppose T : P ðc; dÞ ! P ðc; dÞ is a completely continuous and there exist positive numbers a; b and c with a < b, such that (C1) fu 2 P ðc; h; a; b; c; dÞ : aðuÞ > bg 6¼ ;, and aðTuÞ > b for u 2 P ðc; h; a; b; c; dÞ; (C2) aðTuÞ > b for u 2 P ðc; a; b; dÞ with hðTuÞ > c; (C3) 0 62 Rðc; w; a; dÞ and wðTuÞ < a for u 2 Rðc; w; a; dÞ with wðuÞ ¼ a. Then, T has at least three fixed points u1 ; u2 ; u3 2 P ðc; dÞ, such that cðui Þ 6 d a < wðu2 Þ;
for i ¼ 1; 2; 3;
b < aðu1 Þ;
with aðu2 Þ < b;
wðu3 Þ < a:
T Lemma 4.1. Suppose that conditions ðH1 Þ; ðH2 Þ hold, then uðtÞ 2 C½0; 1 C 2 ð0; 1Þ is a solution of boundary value problem (1.1)0 and (1.2), if and only if uðtÞ 2 E is a solution of the following integral equation: 8 R R R r r t > < uq ba n aðrÞgðuðrÞ; u0 ðrÞÞdr þ 0 uq s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; 0 6 t 6 r; uðtÞ ¼ R R R > : uq d g aðrÞgðuðrÞ; u0 ðrÞÞdr þ 1 uq s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; r 6 t 6 1; t r c r where r 2 ½n; g ð0; 1Þ and u0 ðrÞ ¼ 0. Proof. The proof is similar to [10, Lemma 2.3], Here we omit it.
h
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
Proof of Theorem 2.2. We define the operator T : P 1 ! E1 given by 8 R R R r t b r > < uq a n aðrÞgðuðrÞ; u0 ðrÞÞdr þ 0 uq s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; ðTuÞðtÞ ¼ R R R > : u d g aðrÞgðuðrÞ; u0 ðrÞÞdr þ 1 u s aðrÞgðuðrÞ; u0 ðrÞÞdr ds; q c r q r t
423
0 6 t 6 r; ð4:2Þ r 6 t 6 1:
First, it follows from Lemma 4.1 that the operator T is well-defined. Next, because of ( R r uq t aðrÞgðuðrÞ; u0 ðrÞÞdr ðP 0Þ; 0 6 t 6 r ðTuÞ0 ðtÞ ¼ R t uq r aðrÞgðuðrÞ; u0 ðrÞÞdr ð6 0Þ; r 6 t 6 1
ð4:3Þ
then, it is obvious that the operator ðTuÞ0 is a continuous monotone decreasing on [0, 1], and ðTuÞ0 ðrÞ ¼ 0. Meanwhile, it follows from the definition of operator T that for each u 2 P 1 ; Tu 2 E1 is a nonnegative continuous, and by (4.2) and (4.3), we can obtain 0
aup ððTuÞð0ÞÞ bup ððTuÞ ðnÞÞ ¼ 0;
0
cup ððTuÞð1ÞÞ þ dup ððTuÞ ðgÞÞ ¼ 0:
These show Tu 2 P 1 ; i:e: T ðP 1 Þ P 1 , and we can obtain ðTuÞðrÞ ¼ kTuk1 with Z r Z r Z r b 0 0 uq aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds a n 0 s Z g Z 1 Z s d ¼ uq aðrÞgðuðrÞ; u0 ðrÞÞdr þ uq aðrÞgðuðrÞ; u0 ðrÞÞdr ds: c r r r Furthermore, we can also obtain that the operator T : P 1 ! P 1 is a completely continuous (see [10, Lemma 2.4]), and it follows from Lemma 4.1 that each fixed point of T in P 1 is a positive solution of (1.1)0 and (1.2). We now show that the conditions of Theorem C are satisfied. 0 Firstly, we choose u 2 P 1 ðc1 ; dÞ, then c1 ðuÞ ¼ max 06t61 ju ðtÞj 6 d. By Lemma 2.3, there is max06t61 juðtÞj 6 Ld, then assumption ðH10 Þ yields gðuðtÞ; u0 ðtÞÞ 6 up Nd , for arbitrary 0 6 t 6 1. Note that max06t61 jðTuÞ0 ðtÞj ¼ maxfðTuÞ0 ð0Þ; jðTuÞ0 ð1Þjg, and from (4.3), we have Z r Z 1 0 0 0 c1 ðTuÞ ¼ max jðTuÞ ðtÞj ¼ max uq aðrÞgðuðrÞ; u ðrÞÞdr ; uq aðrÞgðuðrÞ; u ðrÞÞdr 06t61
0
Z
6 max uq
g
r
Z aðrÞgðuðrÞ; u0 ðrÞÞdr ; uq
0
6
d max uq N
1
aðrÞgðuðrÞ; u0 ðrÞÞdr
n
Z
Z g aðrÞdr ; uq
0
1
aðrÞdr
¼ d:
n
This proves that T : P 1 ðc1 ; dÞ ! P 1 ðc1 ; dÞ, and by the above proof, we know that T : P 1 ðc1 ; dÞ ! P 1 ðc1 ; dÞ is completely continuous. In addition, due to Lemmas 2.2 and 2.3 and the definition of a1 ; c1 ; h1 ; w1 , we have xh1 ðuÞ 6 a1 ðuÞ 6 h1 ðuÞ ¼ w1 ðuÞ; kuk1 ¼ maxfh1 ðuÞ; c1 ðuÞg 6 Lc1 ðuÞ
ð4:4Þ for all u 2 P 1 :
and w1 ðkuÞ ¼ max jkuðtÞj ¼ k max juðtÞj ¼ kw1 ðuÞ; 06t61
06t61
Therefore, condition (4.1) of Theorem C is satisfied.
0 6 k 6 1:
ð4:5Þ
424
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
b Secondly, we take uðtÞ 2x for t 2 ½0; 1, then, we have
b > b; 2x c1 ðuðtÞÞ ¼ max ju ðtÞj ¼ 0 < d;
a1 ðuðtÞÞ ¼ min juðtÞj ¼ x6t61x 0
06t61
h1 ðuðtÞÞ ¼ max juðtÞj ¼ 06t61
b b < ; 2x x
b thus, uðtÞ 2x 2 P 1 ðc1 ; h1 ; a1 ; b; xb ; dÞ and a1 ðuðtÞÞ > b. Hence, fu 2 P 1 ðc1 ; h1 ; a1 ; b; xb ; dÞ : a1 ðuÞ > bg 6¼ ;, and for any u 2 P 1 ðc1 ; h1 ; a1 ; b; xb ; dÞ, there is b 6 uðtÞ 6 b 0 x ; ju ðtÞj 6 d for t 2 ½x; 1 x. b By condition ðH11 Þ, we have gðuðtÞ; u0 ðtÞÞ > up xm for t 2 ½x; 1 x, and by Lemma 2.2 and the cone P 1 , we have
a1 ðTuÞ ¼ min jðTuÞðtÞj P xh1 ððTuÞðtÞÞ ¼ xðTuÞðrÞ x6t61x Z r Z r Z r b 0 0 ¼ x uq aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds a n 0 s Z g Z 1 Z s d ¼ x uq aðrÞgðuðrÞ; u0 ðrÞÞdr þ uq aðrÞgðuðrÞ; u0 ðrÞÞdr ds c r r r Z n Z n Z 1 Z s 0 0 P x min uq aðrÞgðuðrÞ; u ðrÞÞdr ds; uq aðrÞgðuðrÞ; u ðrÞÞdr ds 0
>x
b min xm
Z
s n
Z
Z
aðrÞdr ds;
uq 0
n
s
Z
1
uq g
P 1 c1 ; h1 ; a1 ; b; xb ; d
g s
g
aðrÞdr ds ¼ b:
g
Consequently, a1 ðTuÞ > b, for all u 2 . Thereupon, condition ðC1 Þ of Theorem C is satisfied. Thirdly, we claim that conditionðC2 Þ of Theorem C is satisfied. For this end, we choose u 2 P 1 ðc1 ; a1 ; b; dÞ with h1 ðTuÞ > xb , and it follows from Lemma 2.2 that b a1 ðTuÞ P xh1 ðTuÞ > x ¼ b: x Thus, condition ðC2 Þ of Theorem C is satisfied. Finally, we show that condition ðC3 Þ of Theorem C is also satisfied. It is easy to see that w1 ð0Þ ¼ 0 < a, hence 0 62 Rðc1 ; w1 ; a; dÞ. Now assume that u 2 Rðc1 ; w1 ; a; dÞ with w1 ðuÞ ¼ a, then, by condition ðH12 Þ, we have w1 ðTuÞ ¼ max jðTuÞðtÞj ¼ ðTuÞðrÞ 06t61 Z r Z r Z r b 0 0 ¼ uq aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds a n 0 s Z g Z 1 Z s d 0 0 aðrÞgðuðrÞ; u ðrÞÞdr þ uq aðrÞgðuðrÞ; u ðrÞÞdr ds ¼ uq c r r r Z g Z g Z g b 0 0 6 max uq aðrÞgðuðrÞ; u ðrÞÞdr ds þ uq aðrÞgðuðrÞ; u ðrÞÞdr ; a n 0 s Z g Z 1 Z s d uq aðrÞgðuðrÞ; u0 ðrÞÞdr ds þ uq aðrÞgðuðrÞ; u0 ðrÞÞdr c n n n Z g Z g Z g Z 1 Z s a b max < uq aðrÞdr ds þ uq aðrÞdr ; uq aðrÞdr ds M a n 0 s n n Z g d þuq aðrÞdr ¼a c n
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
425
Consequently, condition ðC3 Þ of Theorem C is satisfied. Thus, it now follows from Theorem C that the boundary value problem (1.1)0 and (1.2) has at least three positive solutions u1 ; u2 , and u3 satisfying max ju0i ðtÞj 6 d
06t61
for i ¼ 1; 2; 3;
b < min ju1 ðtÞj;
max ju1 ðtÞj 6 Ld;
x6t61x
a < max ju2 ðtÞj 06t61
06t61
with
min ju2 ðtÞj < b;
x6t61x
max ju3 j < a:
06t61
The Proof of Theorem 2.2 is complete.
h
In order to illustrate our result, we present an example as follows: Example. Consider the singular boundary value problem with p-Laplacian ( 1 0 ðup ðu0 ðtÞÞÞ þ 12 t2 gðuðtÞ; u0 ðtÞÞ ¼ 0; 0 < t < 1; aup ðuð0ÞÞ bup u0 ð14Þ ¼ 0; cup ðuð1ÞÞ þ dup u0 ð12Þ ¼ 0:
ð4:6Þ
where 3 p¼ ; 2
a ¼ b;
and
0
gðu; u Þ ¼
c ¼ d;
1 n¼ ; 4
8 5 u0 > u þ sin ; > 100 >2 > > 5 4 sin u < u þ 0; 2 100
1 x¼ ; 4
1 1 aðtÞ ¼ t2 ; 2
ðu; u0 Þ 2 ½0; 1 ð1; 1Þ; ðu; u0 Þ 2 ½1; 2 ð1; 1Þ;
0
u > ; 40 þ sin > 100 > > > 4 0 : 40 þ u10 pffiffi þ sin u ; 100
u
1 g¼ ; 2
ðu; u0 Þ 2 ½2; 104 ð1; 1Þ; ðu; u0 Þ 2 ½104 ; þ1Þ ð1; þ1Þ:
Then, (4.6) has at least three positive solutions. In fact, we have pffiffiffi 1 79 48 2 1 L ¼ 2; N ¼ ; M ¼ ; m¼ : 2 96 96 Clearly, conditions ðH1 Þ and ðH2 Þ are satisfied. Let a ¼ 1; b ¼ 2; d ¼ 5000, thus, we have 1 d < up ð Þ ¼ 100 for ðu; u0 Þ 2 ½0; 10000 ½5000; 5000; 100 N pffiffiffi 1 b 0 0 > up gðu; u Þ > min gðu; u Þ ¼ 40 þ ¼ 16 3 for ðu; u0 Þ 2 ½2; 8 ½5000; 5000; 100 xm a 5 1 < up gðu; u0 Þ < max gðu; u0 Þ ¼ þ for ðu; u0 Þ 2 ½0; 1 ½5000; 5000: 2 100 M gðu; u0 Þ 6 max gðu; u0 Þ ¼ 40 þ
Consequently, conditions ðH10 Þ; ðH11 Þ; and ðH12 Þ of Theorem 2.2 are satisfied. Thus, by Theorem 2.2, the singular boundary value problem with p-Laplacian (4.6) has at least three positive solutions u1 ; u2 , and u3 , such that max ju0i ðtÞj 6 d
06t61
for
i ¼ 1; 2; 3;
b < min ju1 ðtÞj; max ju1 ðtÞj 6 Ld; x6t61x
1 < max ju2 ðtÞj 06t61
max ju3 j < a:
06t61
06t61
with
min ju2 ðtÞj < b;
x6t61x
426
X. Li / Applied Mathematics and Computation 202 (2008) 413–426
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