Three positive solutions for a generalized Laplacian boundary value problem with a parameter

Applied Mathematics and Computation 219 (2013) 4782–4788

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Three positive solutions for a generalized Laplacian boundary value problem with a parameter Dingyong Bai a,b, Yuming Chen c,⇑ a

School of Mathematics and Information Science, Guangzhou University, Guangzhou, Guangdong Province 510006, PR China Key Laboratory of Mathematics and Interdisciplinary Sciences of Guangdong Higher Education Institutes, Guangzhou University, Guangzhou, Guangdong Province 510006, PR China c Department of Mathematics, Wilfrid Laurier University, Waterloo, Ontario, Canada N2L 3C5 b

a r t i c l e

i n f o

Keywords: Generalized Laplacian boundary value problem Leggett–Williams fixed point theorem Positive solution Symmetric solution

a b s t r a c t Concerned is a generalized Laplacian boundary value problem with a positive parameter. First we apply the Leggett–Williams fixed point theorem to establish sufficient conditions on the existence of at least three positive solutions for the parameter belonging to an explicit interval. Then, under a little bit stronger assumptions, we show that there are at least three positive symmetric solutions for the parameter in an open interval. The obtained results are new even for Laplacian boundary value problems and they are illustrated with an example. Crown Copyright Ó 2012 Published by Elsevier Inc. All rights reserved.

1. Introduction Consider the following boundary value problem,

(

ðuðu0 ÞÞ0 þ kaðtÞf ðuÞ ¼ 0; uð0Þ ¼ uð1Þ ¼ 0;

0 < t < 1;

ð1:1Þ

where k is a positive parameter. We always assume that (A1) u : R ! R is an odd and increasing homeomorphism, and there exist two increasing homeomorphisms w1 ; w2 : ð0; 1Þ ! ð0; 1Þ such that

w1 ðxÞuðyÞ 6 uðxyÞ 6 w2 ðxÞuðyÞ for all x and y > 0; (A2) f : ½0; 1Þ ! ½0; 1Þ is continuous; R1 (A3) a 2 Cðð0; 1Þ; ½0; 1ÞÞ and 0 < 0 aðtÞdt < 1. Note that (A3) indicates that the function aðtÞ may have L1 -singularity at t ¼ 0 and/or t ¼ 1. Assumption (A1) on u was introduced first by Wang [32], which is satisfied by two important cases uðu0 Þ ¼ u0 and uðu0 Þ ¼ ju0 jp2 u0 (p > 1). When uðu0 Þ ¼ ju0 jp2 u0 , Eq. (1.1) is the one-dimensional p-Laplacian. For u satisfying (A1), we call (1.1) the generalized p-Laplacian. In Ref. [32], Wang studied the n-dimensional system

ðUðu0 ÞÞ0 þ khðtÞfðuÞ ¼ 0;

0
⇑ Corresponding author. E-mail addresses: [email protected] (D. Bai), [email protected] (Y. Chen). 0096-3003/$ - see front matter Crown Copyright Ó 2012 Published by Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.10.100

ð1:2Þ

D. Bai, Y. Chen / Applied Mathematics and Computation 219 (2013) 4782–4788

4783

with the boundary condition

uð0Þ ¼ uð1Þ ¼ 0;

ð1:3Þ

where u ¼ ðu1 ; . . . ; un Þ; Uðu0 Þ ¼ ðuðu01 Þ; . . . ; uðu0n ÞÞ; hðtÞ ¼ diag½h1 ðtÞ; . . . ; hn ðtÞ, and fðuÞ ¼ ðf 1 ðu1 ; . . . ; un Þ; . . . ; f n ðu1 ; . . . ; un ÞÞ. With the help of fixed point theorems in a cone, results on the nonexistence and on the existence of one or two positive solutions for k belonging to some open intervals are established. Then Henderson and Wang [19] proved that the boundary value problem (1.2) and (1.3) has at least one positive solution for certain finite intervals of k when UðuÞ ¼ ðqðtÞuðpðtÞu1 Þ; . . . ; qðtÞuðpðtÞun ÞÞ. For more results on positive solutions of generalized p-Laplacian boundary value problems, see, for example, [4,6,20,23,26] and the references therein. In the aforementioned references, the best result is on the existence of at least two positive solutions. The purpose of this paper is to establish results on the existence of at least three positive solutions to (1.1). Under some additional conditions, there are at least three positive symmetric solutions to (1.1). Our tool is the well-known Leggett–Williams fixed point theorem [24]. This is a very powerful tool for dealing with the existence of at least three solutions for nonlinear equations and it has been widely applied to boundary value problems. To name a few, see [1–3,7,13–17,21,25,27–30,33]. A relatively early and representative work about applications of Leggett–Williams fixed point theorem to boundary value problems is [17], which also motivates this work. Here, Henderson and Thompson studied the following boundary value problem



y00 þ f ðyÞ ¼ 0; yð0Þ ¼ 0 ¼ yð1Þ;

ð1:4Þ

where f : ½0; 1Þ ! ½0; 1Þ is continuous and satisfies the following growth conditions, (i) f ðwÞ < 8a for 0 6 w 6 a; (ii) f ðwÞ P 16b for b 6 w 6 2b; (iii) f ðwÞ 6 8c for 0 6 w 6 c; where 0 < a < b < 2c . Under these assumptions, the existence of at least three symmetric positive solutions for (1.4) was established by applying the Leggett–Williams fixed point theorem. We should mention that there is no parameter attached to (1.4) and the function f satisfies piecewise growth conditions. This is also true for the above mentioned references. Moreover, we know that the compress-expansion fixed point theorem and the fixed point index theorem in cones [11,22] can be well applied to study the nonlinear eigenvalue problems and to investigate intervals of parameters such that problems considered have none, or at least one or two positive solutions [10,18]. However, there are no such results obtained by applying the Leggett–Williams fixed point theorem. The main reason is that parameters would bring some difficulties to priori estimations on possible positive solutions when nonlinear terms satisfy piecewise growth conditions. Therefore, our main interest is to find explicit open intervals of k such that (1.1) has at least three positive solutions. Further assuming that aðtÞ is symmetric, we can show that there exists an open interval of k such that (1.1) has at least three positive symmetric solutions. For the study of symmetric solutions to boundary value problems, see [5,8,9,12,16,17,31] as some examples. We should mention that our results are new even for the two special cases that uðu0 Þ ¼ u0 and uðu0 Þ ¼ ju0 jp2 u0 (p > 1). The remaining part of this paper is organized as follows. In Section 2, we provide some preliminary results and cite the Leggett–Williams fixed point theorem. Then, in Section 3, we present sufficient conditions on the growth of f under which (1.1) can have at least three positive solutions for k belonging to some open intervals. Further assuming that aðtÞ is symmetric in Section 4, we show that (1.1) has at least three positive symmetric solutions for k in an open interval. Finally, we give an example to illustrate our main results. 2. Preliminaries   It follows from assumption (A3) that there exists d 2 0; 12 such that

0<

Z

1d

aðtÞdt < 1:

d

For this given d, define a function q : ½d; 1  d ! R by

qðtÞ ¼

1 2

Z d

t

w1 2

Z s

t

 Z s  Z 1 1d 1 aðrÞdr ds þ w2 aðrÞdr ds 2 t t

ð2:1Þ

for t 2 ½d; 1  d. It is easy to see that qðtÞ is continuous and positive on ½d; 1  d. Denote

L ¼ minfqðtÞjt 2 ½d; 1  dg:

ð2:2Þ

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D. Bai, Y. Chen / Applied Mathematics and Computation 219 (2013) 4782–4788

Let X ¼ Cð½0; 1; RÞ. Equipped with the sup-norm kuk ¼ maxt2½0;1 juðtÞj for u 2 X; X is a Banach space. Define

K ¼ fu 2 XjuðtÞ is nonnegative and concave on ½0; 1g: Then K is a cone in X. In order to apply the Leggett–Williams fixed point theorem, we first introduce a mapping C : K ! X defined by

ðCuÞðtÞ ¼

Z

t

u1

Z

0

t

 Z kaðrÞf ðuðrÞÞdr ds 

s

1

u1

Z

t

s

 kaðrÞf ðuðrÞÞdr ds

t

for t 2 ½0; 1 and u 2 K. With the help of C, for any k > 0, we define an operator T k : K ! X as

8Rt R  < 0 u1 sr kaðrÞf ðuðrÞÞdr ds; ðT k uÞðtÞ ¼ R : 1 u1 R s kaðrÞf ðuðrÞÞdr ds; t r

0 6 t 6 r;

r 6 t 6 1;

ð2:3Þ

for t 2 ½0; 1 and u 2 K, where r 2 ð0; 1Þ is a solution of the equation ðCuÞðtÞ ¼ 0. Arguing as in [32], one can show that T k is well defined. Obviously, ðT k uÞð0Þ ¼ ðT k uÞð1Þ ¼ 0 for u 2 X. Since

8 R < u1 ð tr kaðsÞf ðuðsÞÞdsÞ; 0 6 t 6 r; ðT k uÞ ðtÞ ¼ : u1 ðR t kaðsÞf ðuðsÞÞdsÞ; r 6 t 6 1; r 0

is continuous and non-increasing on ð0; 1Þ and ðT k uÞ0 ðrÞ ¼ 0, we have

 

u ðT k uÞ0 ðtÞ

0

¼ kaðtÞf ðuðtÞÞ

for a.e. t 2 ð0; 1Þ. Therefore, T k ðKÞ  K and each fixed point of T k in K is a positive solution of (1.1). Moreover, it is easy to check that T k : K ! K is compact and continuous. The following result follows easily from the concavity. Lemma 2.1. Assume a < b and u 2 Cð½a; b; RÞ is a nonnegative and concave function with uðaÞ ¼ uðbÞ ¼ 0. Then, for any fixed number c 2 ða; aþb 2 Þ, we have

uðtÞ P

ca jjujj½a;b ba

for t 2 ½c; b þ a  c:

Here jjujj½a;b stands for the sup-norm on Cð½a; b; RÞ. In our discussions, we also need the following result of Wang [32]. Lemma 2.2 [32]. Assume that (A1) holds. Then 1 1 w1 for all x; y 2 ð0; 1Þ: 2 ðxÞy 6 u ðxuðyÞÞ 6 w1 ðxÞy

Finally, we cite the well-known Leggett–Williams fixed point theorem [24]. Let X ¼ ðX; jj  jjÞ be a Banach space and K  X be a cone. Here X and K may be different from those above. A functional / on K is said to be a concave nonnegative continuous functional if / : K ! ½0; 1Þ is continuous and satisfies

/ðku þ ð1  kÞv Þ P k/ðuÞ þ ð1  kÞ/ðv Þ for all k 2 ½0; 1 and u; v 2 K. Given positive constants a; b, and c, for a concave nonnegative continuous functional /, we define

K a ¼ fu 2 K : jjujj < ag and

Kð/; b; cÞ ¼ fu 2 K : /ðuÞ P b and kjujj 6 cg: Now, we are ready to state the Leggett–Williams fixed point theorem. Lemma 2.3 (Leggett–Williams fixed point theorem [24]). Let X ¼ ðX; jj  jjÞ be a Banach space, K  X be a cone, and c4 > 0 be a constant. Suppose there exists a concave nonnegative continuous functional / on K with /ðuÞ 6 jjujj for u 2 K c4 and T : K c4 ! K c4 is a continuous compact mapping. Assume there are numbers c1 ; c2 , and c3 with 0 < c1 < c2 < c3 6 c4 such that (i) fu 2 Kð/; c2 ; c3 Þ : /ðuÞ > c2 g – ; and /ðTuÞ > c2 for all u 2 Kð/; c2 ; c3 Þ; (ii) jjTujj < c1 for all u 2 K c1 ; (iii) /ðTuÞ > c2 for all u 2 Kð/; c2 ; c4 Þ with jjTujj > c3 .

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D. Bai, Y. Chen / Applied Mathematics and Computation 219 (2013) 4782–4788

Then T has at least three c2 g; u3 2 K c4 n ðKð/; c2 ; c4 Þ [ K c1 Þ.

fixed

points

u1 ; u2

and

u3

in

K c4

with

u1 2 K c1 ; u2 2 fu 2 Kð/; c2 ; c4 Þ : /ðuÞ >

3. Existence of at least three positive solutions For the sake of simplicity, let

f ðuÞ

f 0 ¼ lim sup u!0

uðuÞ

þ

;

f ðuÞ

f 1 ¼ lim sup

uðuÞ

u!1

;

and

2  : A ¼ R1  R1 Rs 1 w1 ð s aðtÞdtÞ þ w1 1 ð 0 aðtÞdtÞ ds 0

ð3:1Þ

Theorem 3.1. Let (A1)–(A3) hold and f 1 < 1. Assume there exist numbers 0 < c1 < c2 such that (A4) f ðuÞ 6 uðuÞ for 0 6 u 6 c1 ; (A5) for all c2 6 u 6 cd2 ; f ðuÞ P uðauÞ holds for some a > dLw1 1ðw 2

1 ðAÞÞ

.

Then, for k 2 ðw2 ða1dLÞ; w1 ðAÞÞ, the boundary value problem (1.1) has at least three positive solutions u1 ; u2 , and u3 such that jju1 jj < c1 ; mint2½d;1d u2 ðtÞ > c2 , and jju3 jj > c1 with mint2½d;1d u3 ðtÞ < c2 . Proof. It is easy to see that the interval above for the parameter k is not empty. Since f 1 < 1, there exist 0 < a0 < 1 and n > 0 such that f ðuÞ 6 a0 uðuÞ for u P n. Let b ¼ max06u6n f ðuÞ. Then 0 6 f ðuÞ 6 a0 uðuÞ þ b for all u P 0. Choose

c4 > max

   c2 1 b : ;u 1  a0 d

The choice of c4 implies that a0 uðc4 Þ þ b < uðc4 Þ and hence u1 ða0 uðc4 Þ þ bÞ < c4 . Clearly, ðT k uÞðrÞ is the maximum value of ðT k uÞðtÞ on ½0; 1. Then for u 2 K c4 , we have

kT k uk ¼

1 2

Z r

u1

0

Z r

 Z kaðtÞf ðuðtÞÞdt ds þ

1

u1

r

s

Z

s

  kaðtÞf ðuðtÞÞdt ds :

r

We estimate the two terms on the right hand side one by one. First,

Z r

u1

Z r

0

 Z kaðtÞf ðuðtÞÞdt ds 6

s

1

u1

0

Z

1 1

u

6 0

Z

1

u1

6 0

Z

1

6 0

w1 1

Z Z Z

Z

1

 Z kaðtÞf ðuðtÞÞdt ds 6

s

1

u1

0

Z

1

 kaðtÞða0 uðuðtÞÞ þ bÞdt ds

s



1

aðtÞdtw1 ðw1 1 ðkÞÞ

1

uðu ða0 uðjjujjÞ þ bÞÞ ds

s 1 s 1

s

   1 aðtÞdt u w1 ðkÞ u ð a u ðjjujjÞ þ bÞ ds 0 1  Z 1 aðtÞdt dsw1 1 ðkÞu ða0 uðc 4 Þ þ bÞ 6

1

0

w1 1

Z

1

s

 aðtÞdt dsw1 1 ðkÞc 4 :

Second,

Z

1

u1

Z

r

s

 Z kaðtÞf ðuðtÞÞdt ds 6

r

0

1

w1 1

Z 0

s

 Z 1 aðtÞdt dsw1 1 ðkÞu ða0 uðc 4 Þ þ bÞ 6

Therefore, for u 2 K c4 , we obtain

jjT k ujj 6

1 2

Z

1

0

Z w1 1

1 s

 Z s  aðtÞdt þ w1 aðtÞdt ds w1 1 1 ðkÞc 4 6 c 4 0

which implies that T k : K c4 ! K c4 . For u 2 K c1 , using (A4) and similar arguments as above, one can obtain

jjT k ujj 6

1 2

Z 0

1

Z w1 1

1 s

 Z s  aðtÞdt þ w1 aðtÞdt ds w1 1 1 ðkÞc 1 6 c 1 : 0

0

1

w1 1

Z 0

s

 aðtÞdt dsw1 1 ðkÞc 4 :

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D. Bai, Y. Chen / Applied Mathematics and Computation 219 (2013) 4782–4788

Now we define a concave nonnegative continuous functional / on K by /ðuÞ ¼ mint2½d;1d uðtÞ. Let c3 ¼ cd2 and u0 ðtÞ  c0 , where c0 is any given number satisfying c2 < c0 < c3 . Obviously, u0 2 fu : u 2 Kð/; c2 ; c3 Þ; /ðuÞ > c2 g. For u 2 Kð/; c2 ; c3 Þ, we show that /ðT k uÞ > c2 . Since T k u 2 K, we have by Lemma 2.1 that

/ðT k uÞ ¼ min T k uðtÞ P djjT k ujj: t2½d;1d

In the following, we estimate jjT k ujj for u 2 Kð/; c2 ; c3 Þ. Since mint2½d;1d uðtÞ ¼ /ðuÞ P c2 and jjujj 6 c3 , assumption (A5) gives

f ðuðtÞÞ P uðauðtÞÞ P uðac2 Þ for t 2 ½d; 1  d: We distinguish three cases to finish estimating kT k uk. First, assume

jjT k ujj P

Z r 1 2

1 P 2 1 P 2 1 P 2 Second, assume

jjT k ujj P

u1

s

d

Z r

u1 1

u

Z r s

d

Z r

Z r s

d

Z r

Z r

w1 2

Z r

 Z kaðrÞf ðuðrÞÞdr ds þ

u1

Z

r

 Z aðrÞdrw2 ðw1 2 ðkÞÞuðac2 Þ ds ðw1 2 ðkÞ

aðrÞdr u

 Z aðrÞdr ds þ

w1 2

Z

u1 u

s

r

1

r 2 ½d; 1  d. Then

  kaðrÞf ðuðrÞÞd ds

r

1d

r

s

1d

r

 Z ac2 Þ ds

1d

r

s

d

1d

Z r

Z r

s

s

  aðrÞdrw2 ðw1 2 ðkÞÞuðac2 Þ ds

  aðrÞdr uðw1 ðkÞ a c Þ ds 2 2

  1 aðrÞdr ds w1 2 ðkÞac 2 P Lw2 ðkÞac 2 :

r > 1  d. Then

Z

1d

u1

Z

1d

s

d

 Z 1d  Z 1 1d 1 1 kaðrÞf ðuðrÞÞdr ds P w2 aðrÞdr dsw1 2 ðkÞac 2 ¼ qð1  dÞw2 ðkÞac 2 2 d s

P Lw1 2 ðkÞac 2 : Finally, assume

jjT k ujj P

r < d. Then Z

1d

u1

d

Z d

s

 Z s  Z 1 1d 1 1 1 kaðrÞf ðuðrÞÞdr ds P w2 aðrÞdr dsw1 2 ðkÞac 2 ¼ qðdÞw2 ðkÞac 2 P Lw2 ðkÞac2 : 2 d d

To summarize, since k > w2 ða1dLÞ, we have

/ðT k uÞ P djjTujj P dLw1 2 ðkÞac 2 > c 2

for u 2 Kð/; c2 ; c3 Þ:

Last, for u 2 Kð/; c2 ; c4 Þ with jjT k ujj > c3 , we can see that

/ðT k uÞ ¼ min T k uðtÞ P djjT k ujj > dc3 ¼ d  d6t61d

c2 ¼ c2 : d

So far, we have verified all the assumptions of Lemma 2.3 and hence (1.1) has at least three positive solutions. This completes the proof. h From the proof of Theorem 3.1, one can easily see that (A4) can be replaced with f 0 < 1. 1 0 Theorem  3.2. Let (A1)–(A3) hold, f < 1, and f < 1. Suppose that there exists c2 > 0 such that (A5) is satisfied. Then, for k 2 ðw2 a1dL ; w1 ðAÞÞ, the boundary value problem (1.1) has at least three positive solutions.

4. Existence of at least three positive symmetric solutions As mentioned in Introduction, existence of symmetric solutions to boundary value problems has also attracted the attention of many researchers. A function uðtÞ is said to be symmetric on ½0; 1 (respectively, on ð0; 1Þ) if uðtÞ ¼ uð1  tÞ for all t 2 ½0; 1 (respectively, for all t 2 ð0; 1Þ). In this section, we consider the existence of positive symmetric solutions under a little bit stronger assumptions. For this purpose, we make the following assumption. ðA3 Þ a 2 Cðð0; 1Þ; ½0; 1ÞÞ is symmetric on ð0; 1Þ and 0 <

R1 0

aðtÞdt < 1.

A quite simple function satisfying ðA3 Þ is aðtÞ ¼ ðtð1  tÞÞ1 . The following result plays an important role in the coming discussion. Lemma 4.1. Let uðtÞ be continuous and symmetric on ½0; 1. Then C1 uðtÞ ¼ C2 uð1  tÞ for t 2 ½0; 1and C uð12Þ ¼ 0, where C1 ; C2 , and C : Cð½0; 1; RÞ ! Cð½0; 1; RÞ are defined as follows,

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D. Bai, Y. Chen / Applied Mathematics and Computation 219 (2013) 4782–4788

 uðrÞdr ds; 0 s Z s  Z 1  1 ðC2 uÞðtÞ ¼ u uðrÞdr ds;

ðC1 uÞðtÞ ¼

Z

t

u1

Z

t

t

t

ðC uÞðtÞ ¼ C1 uðtÞ  C2 uðtÞ; for t 2 ½0; 1. Proof. Let r ¼ 1  s and s ¼ 1  h. Then we can see that, for t 2 ½0; 1,

C1 uðtÞ ¼

Z

t

u1

Z

0

t

 Z uðrÞdr ds ¼

s

1t

u1

Z

1 

Furthermore, it follows that C uðtÞ ¼ C completes the proof. h

1t

 Z uðsÞds dh ¼

 tÞ  C

 1 uð1

u1

1t

h

 2 uð1

1

Z

h

1t

 uðsÞds dh ¼ C2 uð1  tÞ:



 tÞ ¼ C uð1  tÞ. Taking t ¼ 12 immediately gives C uð12Þ ¼ 0. This

To obtain the main result of this section, we define a cone K  in X as follows,

K  ¼ fu 2 XjuðtÞ is nonnegative; symmetric; and concave on ½0; 1g; where X is the Banach space defined in Section 2. For k > 0, let T k : K  ! X be defined by (2.3) with r ¼ 12, namely, for u 2 K  and t 2 ½0; 1,

8R R 1  > < 0t u1 s2 kaðrÞf ðuðrÞÞdr ds;    ðT k uÞðtÞ ¼ R > 1 u1 R s kaðrÞf ðuðrÞÞdr ds; : 1 t 2

0 6 t 6 12 ; 1 2

6 t 6 1:

Then it follows from Lemma 4.1 and (A3⁄) that, for u 2 K  and t 2 ½0; 1; ðT k uÞðtÞ ¼ ðT k uÞð1  tÞ. Therefore, T k ðK  Þ  K  and each fixed point of T k in K  is a symmetric positive solution of (1.1). Now, the following results can be shown with similar arguments in Section 3 and the detail is omitted here. Theorem 4.1. Suppose that the condition (A3) in Theorem 3.1 or Theorem 3.2 is replaced by (A3⁄). Then, for k 2 ðw2 ða1dLÞ; w1 ðAÞÞ, the boundary value problem (1.1) has at least three positive symmetric solutions.

5. An example In this section, we illustrate our main results with an example. Consider

8 0 < ðu0 Þ13 þ kf ðuÞ ¼ 0; :

0 < t < 1;

ð5:1Þ

uð0Þ ¼ uð1Þ ¼ 0

with

f ðuÞ ¼

8 1 > 3 0 6 u 6 18 ; > > c1 u ;   1  > 28 > > < 12 c1 þ 2 3  c1 u3  12 ; 18 6 u 6 1; 25 > > 23 ; >   > > 25 1 > : 2 3 þ c2 u13  43 ;

1 6 u 6 4; u P 4; 1

where c1 and c2 are any given numbers in ð0; 1Þ. Here uðuÞ ¼ u3 . 1 1 It easy to see that (A1) is satisfied with wðuÞ ¼ 12 u3 and w2 ðuÞ ¼ 2u3 . Take d ¼ 14. After some simple calculations, we get 4 L ¼ 2113 and A ¼ 12. Take c1 ¼ 18 ; c2 ¼ 1, and a ¼ 223 . Then we can easily see that a > Lw1 ðw 2

1 3

1 ðAÞÞ

¼ 222 and f ðuÞ P uðauÞ for

1 3

u 2 ½c2 ; 4c2 . Moreover, f ðuÞ ¼ c1 u 6 uðuÞ ¼ u for 0 6 u 6 c1 and

f1

 1  25 1 2 3 þ c2 u3  43 f ðuÞ 1 ¼ lim ¼ < 1: ¼ lim 1 u!1 uðuÞ u!1 2 3 u

   5  4 It follows from Theorem 4.1 that, for w2 a1dL < k < w1 ðAÞ or 12 3 < k < 12 3 , the boundary value problem (5.1) has at least three positive symmetric solutions u1 ; u2 , and u3 such that jju1 jj < 18 ; mint2½1;3 u2 ðtÞ > 1, and jju3 jj > 18 with mint2½1;3 u3 ðtÞ < 1. 44

44

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D. Bai, Y. Chen / Applied Mathematics and Computation 219 (2013) 4782–4788

Acknowledgements This work was done when Bai is visiting the Department of Mathematics, Wilfrid Laurier University. He would like to thank the Department for its hospitality. The research is supported partially by the Research Fund for the Doctoral Program of Higher Education of China (No. 20104410120001), by the Natural Science and Engineering Council of Canada (NSERC), and by the Earlier Researcher Award (ERA) Program of Ontario. References [1] R.P. Agarwal, D. O’Regan, Triple solutions to boundary value problems on time scales, Appl. Math. Lett. 13 (2000) 7–11. [2] R.P. Agarwal, D. O’Regan, A multiplicity result for second order impulsive differential equations via the Leggett Williams fixed point theorem, Appl. Math. Comput. 161 (2005) 433C439. [3] D.R. Anderson, Nonlinear triple-point problems on time scales, Electron. J. Differ. Eqs. 47 (2004) p.12. [4] E. Arrázola, P. 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