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Positive solutions of some singular m-point boundary value problems at non-resonance
Applied Mathematics and Computation 171 (2005) 433–449 www.elsevier.com/locate/amc
Positive solutions of some singular m-point boundary value problems at non-resonance Zhongli Wei *, Changci Pang Department of Mathematics and Physics, Shandong Institute of Architectural and Engineering, Jinan, Shandong 250014, People’s Republic of China
Abstract This paper investigates the existence of positive solutions for second-order singular m-point boundary value problems at non-resonance. A necessary and sufficient condition for the existence of C[0, 1] as well as C1[0, 1] positive solutions is given by constructing lower and upper solutions and with the maximal theorem. Our nonlinearity f(t, x) may be singular at x,t = 0 and/or t = 1. 2005 Elsevier Inc. All rights reserved. Keywords: Singular m-point boundary value problem; Positive solution; Lower and upper solution; Maximum principle; Non-resonance
1. Introduction The singular ordinary differential equations arise in the fields of gas dynamics, Newtonian fluid mechanics, the theory of boundary layer and so on, the *
Corresponding author. E-mail address: [email protected] (Z. Wei).
0096-3003/$ - see front matter 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.01.043
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Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
theory of singular boundary value problems has become an important area of investigation in recent years (see [1–5] and the references therein). Consider the singular m-point boundary value problems of second-order ordinary differential equation at non-resonance 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ f ðt; xðtÞÞ; t 2 ð0; 1Þ; m2 P ð1:1Þ : xð0Þ ¼ ai xðgi Þ; xð1Þ ¼ 0; i¼1
where q > 0 is such that 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ 0; t 2 ð0; 1Þ; m2 P : xð0Þ ¼ ai xðgi Þ; xð1Þ ¼ 0
ð1:2Þ
i¼1
has only the trivial solution, 0 < ai < 1, i = 1,2, . . . , m 2, 0 < g1 < g2 < < Pm2 gm2 < 1, are constants, i¼1 ai < 1, m P 3 and f, p satisfy the following hypothesis. (H1) p(t) 2 C(0, 1), p(t) P 0, t 2 (0, 1), and Z 1 tð1 tÞpðtÞ dt < 1; also
ð1:3Þ
0
limþ t2 pðtÞ ¼ 0 if
t!0
2
lim ð1 tÞ pðtÞ ¼ 0
t!1
Z
1
ð1 tÞpðtÞ dt ¼ 1; and
ð1:4Þ
0
Z if
1
tpðtÞ dt ¼ 1;
ð1:5Þ
0
(H2) p(t) 2 C(0, 1), p(t) P 0, t 2 (0, 1), and Z 1 pðtÞ dt < 1;
ð1:6Þ
0
(H3) f 2 C((0, 1) · (0, 1),[0, 1)), f(t, 1) > 0, t 2 (0, 1), and there exist constants k, l(1 < k < 0 < l < 1), such that, for t 2 (0, 1) and x 2 (0,+1), cl f ðt; xÞ 6 f ðt; cxÞ 6 ck f ðt; xÞ;
if
0 < c 6 1;
ð1:7Þ
if
c P 1:
ð1:8Þ
Remark. (1.7) implies ck f ðt; xÞ 6 f ðt; cxÞ 6 cl f ðt; xÞ;
Conversely, (1.8) implies (1.7). Typical functions thatP satisfy the above sub-linear hypothesis are those taking the form f ðt; xÞ ¼ nk¼1 pk ðtÞxkk ; here pk(t) 2 C(0, 1), pk(t) > 0 on (0, 1), kk < 1, k = 1,2, . . . , n.
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
435
By singularity we mean that the functions f, p in (1.1) are allowed to be unbounded at the points x = 0, t = 0 and t = 1. A function x(t) 2 C[0, 1] \ C2(0, 1) is called a C[0, 1] (positive) solution of (1.1) if it satisfies (1.1) (x(t) > 0, for t 2 (0, 1)). A C[0, 1] (positive) solution of (1.1) is called a C1[0, 1] (positive) solution if x 0 (0+) and x 0 (1-) both exist (x(t) > 0 for t 2 (0, 1)). When the function p = 0, f 2 C([0, 1] · R · R,R) in (1.1), i.e. f is continuous, problem (1.1) is nonsingular, the existence of solutions of (1.1) has been studied by many authors using nonlinear alternative of Leray-Schauder, Coincidence degree theory and fixed point theorem in cones (see [6–15] and references therein). Very recently, in the special case: p = 0, f(t, x) in (1.1) cannot be singular at x = 0, the existence of positive solutions and multiple positive solutions of singular boundary value problem (1.1) have been studied by papers [16–18] using the fixed point index and approximate process; Zhang and Wang in [19] gave some sufficient conditions for the existence results of a class of singular nonlinear second order three-point boundary value problems by the upper and lower solution method and the monotone iterative technique. In this paper, we shall study the existence of positive solutions for secondorder singular m-point boundary value problems at non-resonance (1.1). A necessary and sufficient condition for the existence of C[0, 1] as well as C1[0, 1] positive solutions is given by constructing lower and upper solutions and with the maximal theorem. Our nonlinearity f(t, x) may be singular at x, t = 0 and/or t = 1.
2. Several lemmas Lemma (see [5]). Suppose (H1) holds. (i) Then 00 x þ qpðtÞx ¼ 0; xð0Þ ¼ 0;
t 2 ð0; 1Þ;
x0ð0Þ ¼ 1
ð2:1Þ
has a unique positive increasing solution e1(t) 2 C[0, 1] \ C1[0, 1). And there exists a unique nonnegative function w1 2 C[0, 1] with Z Z q t s spðsÞw1 ðsÞ ds ds; w1 ðtÞ ¼ 1 þ t 0 0 and e1(t) = tw1(t) 2 C[0, 1] \ C1[0, 1) is a solution of (2.1). (ii) Then 00 x þ qpðtÞx ¼ 0; t 2 ð0; 1Þ; xð1Þ ¼ 0; x0ð1Þ ¼ 1
ð2:2Þ
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Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
has a unique positive decreasing solution e2(t) 2 C[0, 1] \ C1(0, 1]. And there exists a nonnegative function w2 2 C[0, 1] with Z 1Z 1 q w2 ðtÞ ¼ 1 þ ð1 sÞpðsÞw2 ðsÞ ds ds; 1t t s and e2(t) = (1 t)w2(t) 2 C[0, 1] \ C1(0, 1] is a positive decreasing solution of (2.2). In addition, if (H2) holds, then e1(t),e2(t) 2 C1[0, 1]. Remark 1. If p(t) = 0, then e1(t) = t, e2(t) = 1 t, w1(t) = w2(t) = 1. To prove the main result, we need the following maximum principle. Suppose that 0 < gm2 < bn, and ( ) m2 X 2 F n ¼ x 2 C½0; bn \ C ð0; bn Þ; xð0Þ ai xðgi Þ P 0; xðbn Þ P 0 : i¼1
Lemma 2.1 (Maximum principle). If x 2 Fn such that x 0 0 (t) + qp(t)x(t) P 0, t 2 (0, bn), then x(t) P 0, t 2 [0, bn]. Proof. Set x00 ðtÞ þ qpðtÞxðtÞ ¼ rðtÞ; xð0Þ
m2 X
ai xðgi Þ ¼ r1 ;
t 2 ð0; bn Þ;
xðbn Þ ¼ r2 ;
ð2:3Þ ð2:4Þ
i¼1
then r1 P 0, r2 P 0, r(t) P 0, t 2 (0, bn). And the solution (2.3) and (2.4) can be stated Z bn x2n ðtÞ x1n ðtÞ xðtÞ ¼ r2 þ Gn ðt; sÞrðsÞ ds r1 þ Pm2 x1n ðbn Þ x2n ð0Þ i¼1 ai x2n ðgi Þ 0 Z bn m2 X x2n ðtÞ a Gn ðgi ; sÞrðsÞ ds; ð2:5Þ þ Pm2 i ai x2n ðgi Þ i¼1 x2n ð0Þ i¼1 0 where 1 Gn ðt; sÞ ¼ xn
x2n ðtÞe1 ðsÞ; s < t; x2n ðsÞe1 ðtÞ; t 6 s;
x2n ðtÞ x2n 0ðtÞ ¼ x2n ð0Þ ¼ e1 ðbn Þ ¼ constant > 0: xn ¼ e1 ðtÞ e1 0ðtÞ
ð2:6Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
and x1n ðtÞ ¼ e1 ðtÞ þ
1
P1m2 i¼1
Pm2 ai
i¼1
437
ai e1 ðgi Þ 2 C 2 ½an ; bn is a unique increasing
positive solution of the problem 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ 0; t 2 ½0; bn ; m2 P : xð0Þ ¼ ai xðgi Þ; x0 ð0Þ ¼ 1:
ð2:7Þ
i¼1
And x2n(t) 2 C2[0, bn] is a unique decreasing positive solution of the problem 00 x ðtÞ þ qpðtÞxðtÞ ¼ 0; t 2 ½0; bn ; ð2:8Þ xðbn Þ ¼ 0; x0ðbn Þ ¼ 1: By means of (2.5)–(2.8), we can obtain x(t) P 0, t 2 [0, bn]. The proof is complete. h Lemma 2.2. Suppose that (H1) and (H3) hold. Let x(t) be a C[0, 1] positive solution of (1.1). Then xðtÞ P a0 kxk
e2 ðtÞ ; e2 ð0Þ
t 2 ½0; 1:
ð2:9Þ
Here, kxk = maxt2[0,1]jx(t)j, and m2 X e1 ðgi Þ e2 ðgi Þ ai a0 ¼ : e1 ð1Þ e2 ð0Þ i¼1
ð2:10Þ
Proof. Assume that x(t) is a C[0, 1] positive solution of (1.1) and (1.2). Then x(t) can be stated Z 1 m2 X e2 ðtÞ Gðt; sÞf ðs; xðsÞÞ ds þ ai xðtÞ ¼ Pm2 e2 ð0Þ i¼1 ai e2 ðgi Þ i¼1 0 Z 1 Gðgi ; sÞf ðs; xðsÞÞ ds; ð2:11Þ 0
where Gðt; sÞ ¼
1 x
e2 ðtÞe1 ðsÞ; s < t;
ð2:12Þ
e2 ðsÞe1 ðtÞ; t 6 s;
e2 ðtÞ e2 0ðtÞ ¼ e2 ð0Þ ¼ e1 ð1Þ ¼ constant > 0: x ¼ e1 ðtÞ e1 0ðtÞ It is easy to see that xð0Þ > 0;
xðtÞ P
e1 ðtÞ e2 ðtÞ kxk; e1 ð1Þ e2 ð0Þ
t 2 ½0; 1:
ð2:13Þ
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Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
From (2.11), we have that e2 ð0Þ ¼
1 Pm2 i¼1
m2 X
ai e2 ðgi Þ
i¼1
Z ai
1
Gðgi ; sÞf ðs; xðsÞÞ ds
0
m2 1 X ai xðgi Þ; e2 ð0Þ i¼1
ð2:14Þ
For 0 6 t 6 1, we have from (2.13) and (2.14) that m2 m2 e2 ðtÞ X e2 ðtÞ X e1 ðgi Þ e2 ðgi Þ ai xðgi Þ P ai xðtÞ P kxk e2 ð0Þ i¼1 e2 ð0Þ i¼1 e1 ð1Þ e2 ð0Þ ¼ a0 kxk
e2 ðtÞ ; e2 ð0Þ
The proof is complete.
t 2 ½0; 1:
ð2:15Þ
h
Lemma 2.3. Suppose that (H2) and (H3) hold. Let x(t) be a C1[0, 1] positive solution of (1.1). Then there are constants 0 < I1 < I2, such that I 1 e2 ðtÞ 6 xðtÞ 6 I 2 e2 ðtÞ;
t 2 ½0; 1:
ð2:16Þ
Proof. Assume that x(t) is a C1[0, 1] positive solution of (1.1). Then (2.9) holds, and both x 0 (0) and x 0 (1) exist, x(t) > 0 for t 2 (0, 1). By integration of (1.1), we have Z
1
f ðt; xðtÞÞdt 6 x0ð1Þ þ x0ð0Þ þ q max jxðtÞj t2½0;1
0
Z
1
pðtÞdt < 1:
ð2:17Þ
0
From (2.11) and (1.17), we have " Z 1 1 1 e1 ðsÞf ðs; xðsÞÞ ds þ xðtÞ 6 e2 ðtÞ Pm2 x 0 e ð0Þ i¼1 ai e2 ðgi Þ #2 Z 1 m2 X ai Gðgi ; sÞf ðs; xðsÞÞ ds :
ð2:18Þ
0
i¼1
Setting I1 = a0kxk/e2(0), Z 1 1 1 I2 ¼ e1 ðsÞf ðs; xðsÞÞ ds þ Pm2 x 0 e2 ð0Þ i¼1 ai e2 ðgi Þ Z 1 m2 X ai Gðgi ; sÞf ðs; xðsÞÞ ds; i¼1
0
then from (2.9) and (2.18), we have (2.16). This completes the proof.
h
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
439
Definition 2.1. Suppose a 2 C[0, 1] \ C2(0, 1), if a satisfies a00 ðtÞ þ qpðtÞaðtÞ 6 f ðt; aðtÞÞ; að0Þ
m2 X
ai aðgi Þ 6 0;
t 2 ð0; 1Þ;
að1Þ 6 0:
i¼1
Then a is called a lower solution of the singular boundary value problem (1.1). Definition 2.2. Suppose b 2 C[0, 1] \ C2(0, 1), if b satisfies b00 ðtÞ þ qpðtÞbðtÞ P f ðt; bðtÞÞ; bð0Þ
m2 X
ai bðgi Þ P 0;
t 2 ð0; 1Þ;
bð1Þ P 0:
i¼1
Then b is called a upper solution of the singular boundary value problem (1.1). Lemma 2.4. Assume that there exist lower and upper solutions of (1.1), respectively a(t) and b(t), such that a (t),b(t) 2 C[0, 1] \ C2(0, 1), 0 < a(t) 6 b(t), for t 2 (0, 1). Then problem (1.1) has at least one C[0, 1] positive solution x(t) such that a(t) 6 x(t) 6 b(t), for t 2 [0, 1]. If, in addition, there exists F(t) 2 L1[0, 1] such that jf ðt; xðtÞÞj 6 F ðtÞ;
for
aðtÞ 6 xðtÞ 6 bðtÞ;
t 2 ð0; 1Þ:
ð2:19Þ
then the solution x(t) of (1.1) is a C1[0, 1] positive solution. Proof. First of all, we define a partial ordering in C[0, 1] \ C2(0, 1) by x 6 y if and only if xðtÞ 6 yðtÞ;
8t 2 ½0; 1:
Then, we shall define an 8 > < f ðt; aðtÞÞ; gðt; xÞ ¼ f ðt; xðtÞÞ; > : f ðt; bðtÞÞ;
ð2:20Þ
auxiliary function. "x(t) 2 C [0, 1] \ C2(0, 1), if
aix
if
a6x6b
if
xib:
ð2:21Þ
By the condition (H), we have g:(0, 1) · R ! [0, + 1) is continuous. Let {bn} be a sequence satisfying 0 < gm2 < b1 < < bn < bn+1 < < 1, and bn ! 1 as n ! 1, and let {rn}, be a sequence satisfying aðbn Þ 6 rn 6 bðbn Þ;
n ¼ 1; 2; . . . ;
ð2:22Þ
440
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
For each n, consider the nonsingular problem 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ gðt; xÞ; t 2 ½0; bn ; m2 P : xð0Þ ai xðgi Þ ¼ 0; xðbn Þ ¼ rn :
ð2:23Þ
i¼1
Obviously, by the proof of the Lemma 2.1, the problem (2.23) is equivalent to the integral equation xðtÞ ¼ An xðtÞ x1n ðtÞ rn þ ¼ x1n ðbn Þ þ
x2n ð0Þ
Z
bn
Gn ðt; sÞgðs; xðsÞÞ ds 0
x2n ðtÞ Pm2 i¼1
m2 X
ai x2n ðgi Þ
i¼1
Z ai
bn
Gn ðgi ; sÞgðs; xðsÞÞ ds;
ð2:24Þ
0
where Gn(t, s) is given by (2.6), x1n(t), x2n(t) are given by (2.7) and (2.8). It is easy to verify that An:Xn ! Xn = C[0, bn] is completely continuous and An(Xn) is a bounded set. Moreover, x 2 C[0, bn] is a solution of (2.23) if and only if Anx = x. Using the SchauderÕs fixed point theorem, we assert that An has at least one fixed point xn 2 C2[0, bn]. We claim that a 6 xn 6 b; that is aðtÞ 6 xn ðtÞ 6 bðtÞ;
t 2 ½0; bn :
ð2:25Þ
and hence xn(t) 2 C2[0,bn] and satisfies x00 ðtÞ þ qpðtÞxðtÞ ¼ f ðt; xðtÞÞ;
t 2 ½0; bn :
ð2:26Þ
Indeed, suppose by contradiction that xn ib. By the definition of g, then gðt; xn ðtÞÞ ¼ f ðt; bðtÞÞ;
t 2 ½0; bn ;
and therefore x00n ðtÞ þ qpðtÞxn ðtÞ ¼ f ðt; bðtÞÞ;
t 2 ½0; bn :
ð2:27Þ
On the other hand, since b is an upper solution of (1.1) and (1.2), We also have b00 ðtÞ þ qpðtÞbðtÞ P f ðt; bðtÞÞ; Then setting zðtÞ ¼ bðtÞ xn ðtÞ;
t 2 ½0; bn :
t 2 ð0; 1Þ:
ð2:28Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
441
By (2.22), (2.23), (2.27) and (2.28), we obtain z00 ðtÞ þ qpðtÞzðtÞ P 0;
t 2 ð0; bn Þ; m2 X ai zðgi Þ P 0; zðbn Þ P 0: z 2 C½0; bn \ C 2 ð0; bn Þ; zð0Þ i¼1
By Lemma 2.1, we can conclude that z(t) P 0, t 2 [0, bn], a contradiction with the assumption xn i b. Therefore xn i b is impossible. Similarly, we can show that a 6 xn. So, we have shown that (2.25) holds. Using the method of [3] and the Theorem 3.2 in [20], we can obtain that there is a C[0, 1] positive solution x(t) of (1.1) such that a 6 x 6 b, and a subsequence of {xn(t)} converges to x(t) on any compact subintervals of (0, 1). In addition, if (2.19) holds, then jx 0 0 (t)j 6 F(t) + qp(t)jx(t)j, and hence x 0 0 (t) is absolutely integrable on [0, 1]. This implies x(t) 2 C1[0, 1], so x(t) is a C1[0, 1] positive solution of the problem (1.1). The proof is complete. h
3. The main results Theorem 3.1. Suppose (H1) and (H3) hold, then a necessary and sufficient condition for problem (1.1) and (1.2) to have C[0, 1] positive solutions is that the following integral conditions hold. Z 1 tð1 tÞf ðt; 1Þ dt < 1; ð3:1Þ 0< 0
Z limþ t
t!0
1
ð1 sÞf ðs; 1Þ ds ¼ 0;
ð3:2Þ
t
lim ð1 tÞ
t!1
Z
t
sf ðs; 1Þ ds ¼ 0:
ð3:3Þ
0
Theorem 3.2. Suppose (H2) and (H3) hold, then a necessary and sufficient condition for problem (1.1) and (1.2) to have C1[0, 1] positive solutions is that the following integral conditions hold. Z 1 0< f ðt; e2 ðtÞÞ dt < 1: ð3:4Þ 0
Proof of Theorem 3.1. (1) Necessity. Let x(t) 2 C[0, 1] be a positive solution of (1.1). Then there is a t0 2 (0, 1) such that x 0 (t0) = 0. Let c0 > 0 be a constant such that c0x(t) 6 1 for t 2 [0, 1] and 1/c0 P 1. From (1.7) and (1.8), we have l f ðt; xðtÞÞ P ð1=c0 Þk f ðt; c0 xðtÞÞ P clk 0 x ðtÞf ðt; 1Þ;
for t 2 ð0; 1Þ:
442
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
According to (1.1), we have c0lk f ðt; 1Þ 6 xl ðtÞx00 ðtÞ þ qpðtÞx1l ðtÞ;
t 2 ð0; 1Þ:
ð3:5Þ
For t 2 (0, t0), by integration of (3.5), we obtain Z t0 Z t0 t0 2 lk 0 l c0 f ðs; 1Þ ds 6 x ðsÞx ðsÞjt þ lxl1 ðsÞ ðx0 ðsÞÞ ds t
þq
Z
t t0
pðsÞx1l ðsÞ ds
t
6 xl ðtÞx0 ðtÞ þ q
Z
t0
pðsÞx1l ðsÞ ds;
t 2 ð0; t0 Þ:
ð3:6Þ
t
Integrating (3.6), we have Z t0 Z t0 Z t0 Z t0 x1l ðt0 Þ x1l ð0Þ þ qK c0lk f ðs; 1Þ ds dt 6 pðsÞ ds dt 1l 0 t 0 t Z t0 x1l ðt0 Þ x1l ð0Þ þ qK ¼ spðsÞ ds < 1; 1l 0 where K = maxt2[0,1]x1l(t), so Z t0 sf ðs; 1Þ ds < 1: 0<
ð3:7Þ
0
Similarly, by integration of (3.5), we obtain Z 1 0< ð1 sÞf ðs; 1Þ ds < 1:
ð3:8Þ
t0
(3.7) and (3.8) imply that (3.1) holds. For t 2 (0, t0), by integration of (3.6), we have Z t Z t0 Z t Z t0 x1l ðtÞ x1l ð0Þ þ qK c0lk f ðs; 1Þ ds ds 6 pðsÞ ds ds 1l s s 0 0 ¼
x1l ðtÞ x1l ð0Þ þ qK 1l Z t Z t Z t0 ds þ pðsÞ ds; 0
s
t
therefore, c0lk t
Z
t0
f ðs; 1Þ ds 6 t
x1l ðtÞ x1l ð0Þ 1l Z t Z þ qK spðsÞ ds þ t 0
t
t0
pðsÞ ds :
ð3:9Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
443
Letting R t t ! 0 in (3.9) and noting condition (H1), we have limt!0þ t t 0 f ðs; 1Þds ¼ 0. These imply that (3.2) holds. Similarly, we can prove (3.3). (2) Sufficiency. Suppose that (3.1)–(3.3) hold. Choose a constant m P 2 such that m(l k) > 1, and define the linear operator A qðtÞ ¼ Af ðt; 1Þ Z 1 ¼ Gðt; sÞf ðs; 1Þ ds þ
e2 ð0Þ
0
Z
m2 X
e2 ðtÞ Pm2 i¼1
ai e2 ðgi Þ
1
ai
Gðgi ; sÞf ðs; 1Þds;
ð3:10Þ
0
i¼1
1=ðmðlkÞÞ
QðtÞ ¼ ½qðtÞ
ð3:11Þ
:
where, G(t,s) is given by (2.12). Then q(t),Q(t) 2 C[0, 1] \ C2(0, 1) satisfying q(t) > 0, Q(t) > 0, t 2 (0, 1), and q00 ðtÞ ¼ f ðt; 1Þ;
Q00 ðtÞ P 0;
for
t 2 ð0; 1Þ
Pm2 and from Pm2(3.1)–(3.3), we have q(1) = Q(1) = 0, qð0Þ i¼1 ai qðgi Þ ¼ 0, Qð0Þ i¼1 ai Qðgi Þ P 0, and Z Z 1 kw1 k kw2 k 1 Gðs; sÞf ðs; 1Þ ds 6 sð1 sÞf ðs; 1Þ ds < þ1; x 0 0 0 " #1 1 Z m2 1 X ai e2 ðgi Þ A Gðs; sÞf ðs; 1Þ ds < þ1: qðtÞ 6 @1 þ e2 ð0Þ e2 ð0Þ 0
i¼1
e2 ðtÞ
Z
t
e1 ðsÞQðlkÞ ðsÞf ðs; 1Þ ds 0
1=m Z e2 ðsÞ s e1 ðsÞf ðs; 1Þ ds f ðs; 1Þ ds x 0 0 Z s 1=m Z t 11=m 1=m x e1 ðsÞ e1 ðsÞf ðs; 1Þ ds f ðs; 1Þ ds 6 ðe2 ðtÞÞ 6 e2 ðtÞ
Z
t
e1 ðsÞ
0 1=m
¼x
0
1
11=m
ð1 1=mÞ ðe2 ðtÞÞ
Z
11=m
t
e1 ðsÞf ðs; 1Þ ds 0
6 x1=m ð1 1=mÞ
1
Z 0
11=m
1
e1 ðsÞe2 ðsÞf ðs; 1Þ ds
< 1:
ð3:12Þ
444
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
Similarly, we have Z 1 e1 ðtÞ e2 ðsÞQðlkÞ ðsÞf ðs; 1Þ ds t
6x
1=m
ð1 1=mÞ
1
Z
11=m
1
e1 ðsÞe2 ðsÞf ðs; 1Þ ds
< 1:
ð3:13Þ
0
Let c1 > 0 such that (1/c1)Q(t) 6 1, c1 P 1. From (1.7) and (1.8), we have k
k
Ql f ðt; QÞ 6 Ql ðQ=c1 Þ f ðt; c1 Þ 6 Ql ðQ=c1 Þ cl1 f ðt; 1Þ kl ¼ clk f ðt; 1Þ: 1 Q
Let
h1 ðtÞ ¼ A
e2 ðtÞ e2 ð0Þ
ð3:14Þ
l f ðt; 1Þ;
t 2 ½0; 1;
h2 ðtÞ ¼ AQl ðtÞf ðt; QðtÞÞ þ QðtÞ;
t 2 ½0; 1:
Thus, (3.12)–(3.14) imply that 0 6 h1 ðtÞ < 1; 0 6 h2 ðtÞ < 1;
t 2 ½0; 1: P One can check that hj 2 C[0, 1] \ C (0, 1), h1 ð0Þ m2 i¼1 ai h1 ðgi Þ ¼ 0, h2 ð0Þ Pm2 a h ðg Þ P 0, and i¼1 i 2 i e2 ðtÞ a0 kh1 k 6 h1 ðtÞ 6 kh1 k; QðtÞ 6 h2 ðtÞ 6 kh2 k; t 2 ½0; 1: e2 ð0Þ for
2
h001 ðtÞ þ qpðtÞh1 ðtÞ ¼
e2 ðtÞ e2 ð0Þ
l f ðt; 1Þ;
t 2 ð0; 1Þ;
ð3:15Þ
h002 ðtÞ þ qpðtÞh2 ðtÞ P Ql ðtÞf ðt; QðtÞÞ;
t 2 ð0; 1Þ:
ð3:16Þ
Here, khjk = max06t61jhj(t)j. Let a(t) = k1h1(t), b(t) = k2h2(t), t 2 [0, 1]; here k1, k2 are constants satisfying 0 < k1 6 1 6 k2 and will be determined later. Suppose c2, c3 are positive constants such that c2kh1k 6 1, c2 6 1, c3 P 1. From (1.7),(1.8), we have k
lk
f ðt; aðtÞÞ P ð1=c2 Þ f ðt; c2 aðtÞÞ P ðc2 Þ al ðtÞf ðt; 1Þ l e2 ðtÞ P ðc2 Þlk ðk 1 a0 kh1 kÞl f ðt; 1Þ e2 ð0Þ P a00 ðtÞ þ qpðtÞaðtÞ; t 2 ð0; 1Þ; l bðtÞ f ðt; QðtÞÞ f ðt; bðtÞÞ 6 ðc3 Þ QðtÞ lk l l 6 ðc3 Þ ðk 2 kh2 kÞ Q ðtÞf ðt; QðtÞÞ 6 b00 ðtÞ þ qpðtÞbðtÞ; t 2 ð0; 1Þ: lk
ð3:17Þ
ð3:18Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
445
By virtue of (1.3), (1.4), we can find a k0 such that f(t, Q(t)) P k0Ql(t)f(t,1) and hence, from the definitions of h1(t), h2(t), we have, when k > k 1 0 , h1(t) 6 k 1=ð1lÞ
h2(t) for n t 2 [0, 1]. Now we chooseo k 1 ¼ min f1; ðða0 kh1 kÞ l c2lk Þ g and l lk 1=ð1lÞ 2 1 k 2 ¼ max 1; ; k 0 ; kh2 k c3 . Then a(t),b(t) 2 C[0, 1] \ C (0, 1), 0 < Pm2 Pm2 a(t) 6 b(t), for t 2 (0, 1), að0Þ ¼ i¼1 ai aðgi Þ, bð0Þ i¼1 ai bðgi Þ P 0, a(1) = b(1) = 0. From (3.15)–(3.18), we obtain that for such choice of k1 and k2, a(t) and b(t) are lower and upper solutions of (1.1), respectively. From the first conclusion of Lemma 2.4, we conclude that the problem (1.1) has at least a C[0, 1] positive solution x(t) satisfying a (t) 6 x(t) 6 b(t) for t 2 [0, 1]. This completes the proof of Theorem 3.1. h Proof of Theorem 3.2. (1) Necessity. Suppose that x(t) is a C1[0, 1] positive solution of (1.1). Then both x 0 (0) > 0 and x 0 (1) < 0 exist. By Lemma 2, there are constants I1 and I2, 0 < I1 < I2 such that I 1 e2 ðtÞ 6 xðtÞ 6 I 2 e2 ðtÞ 6 I 2 e2 ð0Þð1 tÞ;
t 2 ½0; 1:
ð3:19Þ
Let c4 be a constant satisfying c4I2 6 1, 1/c4 P 1. Then (1.7), (1.8) and (3.21), lead to k
f ðt; xðtÞÞ P ð1=c4 Þ f ðt; c4 xðtÞe2 ðtÞ=e2 ðtÞÞ P ðc4 Þ P ðc4 Þ
lk l I 1 f ðt; e2 ðtÞÞ;
Consequently, Z 1 Z kl l f ðt; e2 ðtÞÞ dt 6 ðc4 Þ I 1 0
lk
l
ðxðtÞ=e2 ðtÞÞ f ðt; e2 ðtÞÞ
t 2 ð0; 1Þ:
1
f ðt; xðtÞÞ dt 0
Z 0 0 x ð0Þ x ð1Þ þ I qe ð0Þ ¼ ðc4 Þkl I l 2 2 1
1
pðtÞtð1 tÞ dt
0
< 1: Thus (3.4) holds. (2) Sufficiency. Suppose that (3.4) holds. Let hðtÞ ¼ Af ðt; e2 ðtÞÞ; 1
t 2 ½0; 1: 2
ð3:20Þ
Then h(t) 2 C [0, 1] \ C (0, 1) and (3.19) holds if x(t) is replaced by h(t), and I1 = a0khk/e2(0), " #1 Z m2 X 1 1 I2 ¼ e1 ðsÞf ðs; e2 ðsÞÞ ds þ e2 ð0Þ ai e2 ðgi Þ x 0 i¼1 Z 1 m2 X ai Gðgi ; sÞf ðs; e2 ðsÞÞ ds: i¼1
0
446
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
Suppose that constants c5 and c6 satisfy c5I2 6 1, 1/c5 P 1, c6I1 P 1, 1/ c6 6 1. Let a(t) = k1h(t), nb(t) = k2h(t), to2 [0, 1]; here k 1 ¼ min f1; l lk 1=ð1lÞ 1=ð1lÞ g and k 2 ¼ max 1; I l2 clk . A similar argument to that I 1 c5 6 we have checked in the sufficiency proof of Theorem 3.1 yields that for t 2 (0, 1), k
f ðt; aðtÞÞ P ð1=c5 Þ f ðt; c5 aðtÞÞ P ðc5 Þ P ðc5 Þ
lk
lk
f ðt; bðtÞÞ 6 ðc6 Þ
lk
l
ðaðtÞ=e2 ðtÞÞ f ðt; e2 ðtÞÞ
l
ðk 1 I 1 Þ f ðt; e2 ðtÞÞ P a00 ðtÞ þ qpðtÞaðtÞ;
bðtÞ e2 ðtÞ
l
t 2 ð0; 1Þ; ð3:21Þ
f ðt; e2 ðtÞÞ 6 ðc6 Þlk ðk 2 I 2 Þl f ðt; e2 ðtÞÞ
6 b00 ðtÞ þ qpðtÞbðtÞ;
t 2 ð0; 1Þ:
ð3:22Þ
So, a(t),b(t) 2 C1[0, 1] \ C2(0, 1) are, respectively, lower and upper solutions Pm2 of (1.1)P satisfying 0 < a(t) 6 b(t), for t 2 (0, 1), and að0Þ ¼ i¼1 ai aðgi Þ, m2 bð0Þ ¼ i¼1 ai bðgi Þ, a (1) = b(1) = 0. Additionally, when t 2 (0, 1) and a(t) 6 x 6 b(t), we have k k l k1 c6 x k1 c6 x 0 6 f ðt; xÞ 6 e2 ðtÞ 6 f t; f ðt; e2 ðtÞÞ k 1 e2 ðtÞ k 1 e2 ðtÞ c6 c6 kl k1 l ðk 2 I 2 Þ f ðt; e2 ðtÞÞ ¼ F ðtÞ: 6 c6 R1 From (3.4), we have 0 F ðtÞdt < 1. By Lemma 2.4, we assert that problem (1.1) and (1.2) admits a positive solution x(t) 2 C1[0, 1] \ C2(0, 1) such that a(t) 6 x(t) 6 b(t) for t 2 [0, 1]. The proof of Theorem 3.2 is complete. h Remark 2. Consider the singular m-point boundary value problem 8 00 < x þ qpðtÞx ¼ f ðt; xÞ; t 2 ð0; 1Þ; m2 P ai xðgi Þ: : xð0Þ ¼ 0; xð1Þ ¼
ð3:23Þ
i¼1
By analogous methods, we have the following results. Assume that x(t) is a C[0, 1] positive solution of problem (3.23). Then x(t) can be stated Z 1 e1 ðtÞ xðtÞ ¼ Gðt; sÞf ðs; xðsÞÞ ds þ Pm2 e1 ð1Þ i¼1 ai e1 ðgi Þ 0 Z 1 m2 X ai Gðgi ; sÞf ðs; xðsÞÞ ds: ð3:24Þ i¼1
0
where G(t,s) is given by (2.12).
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
447
Theorem A. Suppose (H1) and (H3) hold, then a necessary and sufficient condition for problem (3.23) to have C[0, 1] positive solutions is that the integral conditions (3.1)–(3.3) hold. Theorem B. Suppose (H2) and (H3) hold, then a necessary and sufficient condition for problem (3.23) to have C1[0, 1] positive solutions is that the following integral conditions hold. Z 1 0< f ðt; e1 ðtÞÞdt < 1: ð3:25Þ 0
4. Some examples Example 1. Consider the following singular three-point boundary value problem 00 b k x ðtÞ ¼ ta ð1 tÞ x ðtÞ; t 2 ð0; 1Þ; ð4:1Þ 1 1 xð0Þ ¼ 2 x 2 ; xð1Þ ¼ 0: where a,b,k 2 R, and k < 1. Evidently, 00 x ðtÞ ¼ 0; t 2 ð0; 1Þ; xð0Þ ¼ 12 x 12 ; xð1Þ ¼ 0:
ð4:2Þ
has only the trivial solution. m = 3, a1 ¼ 12, g1 ¼ 12, p(t) = 0, f(t, x) = ta(1 t)bxk, and from Remark 1, we have e1(t) = t, e2(t) = 1 t, (H1)–(H3) holds. So, from Theorems 3.1 and 3.2, we obtain that Conclusion 4.1. A necessary and sufficient condition for problem (4.1) to have C[0, 1] positive solutions is a > 2, b > 2. Conclusion 4.2. A necessary and sufficient condition for problem (4.1) to have C1[0, 1] positive solutions is a > 1, b + k > 1. Similarly, from Remark 2, for singular three-point boundary value problem ( b x00 ðtÞ ¼ ta ð1 tÞ xk ðtÞ; t 2 ð0; 1Þ; ð4:3Þ xð0Þ ¼ 0; xð1Þ ¼ 12 x 12 : where a,b,k 2 R, and k < 1, we have Conclusion 4.3. A necessary and sufficient condition for problem (4.3) to have C[0, 1] positive solutions is a > 2, b > 2.
448
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
Conclusion 4.4. A necessary and sufficient condition for problem (4.3) to have C1[0, 1] positive solutions is b > 1, a + k > 1. Example 2. Consider singular three-point boundary value problem (
b
x00 ðtÞ þ xðtÞ ¼ ta ð1 tÞ xk ðtÞ; xð0Þ ¼ 12 x 12 ; xð1Þ ¼ 0:
t 2 ð0; 1Þ;
ð4:4Þ
where a,b,k 2 R, and k < 1. Evidently, (
x00 ðtÞ þ xðtÞ ¼ 0; t 2 ð0; 1Þ; xð0Þ ¼ 12 x 12 ; xð1Þ ¼ 0:
ð4:5Þ
has only the trivial solution, m = 3, a1 ¼ 12, g1 ¼ 12, p(t) = 1, f(t,x) = ta(1t)bxk, t t 1t t1 and e1 ðtÞ ¼ e e , e2 ðtÞ ¼ e e , (H1)–(H3) holds. So, from Theorem 3.1 and 2 2 Theorem 3.2, we obtain that Conclusion 4.5. A necessary and sufficient condition for problem (4.4) to have C[0, 1] positive solutions is a > 2, b > 2. Conclusion 4.6. A necessary and sufficient condition for problem (4.4) to have C1[0, 1] positive solutions is a > 1, b + k > 1. Similarly, from Remark 2, for singular three-point boundary value problem ( 00 b x ðtÞ þ xðtÞ ¼ ta ð1 tÞ xk ðtÞ; t 2 ð0; 1Þ; ð4:6Þ xð0Þ ¼ 0; xð1Þ ¼ 12 x 12 : where a, b, k 2 R, and k < 1, we have Conclusion 4.7. A necessary and sufficient condition for problem (4.6) to have C[0, 1] positive solutions is a > 2, b > 2. Conclusion 4.8. A necessary and sufficient condition for problem (4.6) to have C1[0, 1] positive solutions is b > 1, a + k > 1.
Acknowledgment Research supported by NNSF-China (10471077), NSF of Shandong Province (Y2004A01) and XNF of SDAI (040101).
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
449
References [1] D. OÕRegan, Theory of Singular Boundary Value Problems, World Scientific Press, Singapore, 1994. [2] S.D. Taliaferro, A nonlinear singular boundary value problem, Nonlinear Anal. TMA 3 (1979) 897–904. [3] Y. Zhang, Positive solutions of singular sub-linear Emden-Fowler boundary value problems, J. Math. Anal. Appl. 185 (1994) 215–222. [4] D. OÕRegan, Singular Dirichlet boundary value problems-I. Super-linear and non-resonance case, Nonlinear Anal. TMA 29 (2) (1997) 221–245. [5] Zhongli Wei, Changci Pang, Positive solutions of non-resonant singular boundary value problem of second order differential equations, Nagoya Math. J. 162 (2001) 127–148. [6] W. Feng, J.R.L. Webb, Solvability of a m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. [7] W. Feng, On a a m-point boundary value problem, Nonlinear Anal. TMA 30 (6) (1997) 5374– 6369. [8] C.P. Gupta, Solvability of a three-point boundary value problems for a second order ordinary differential equation, J. Math. Anal. Appl. 168 (1992) 540–551. [9] C.P. Gupta, A sharper condition for Solvability of a three-point boundary value problem, J. Math. Anal. Appl. 205 (1997) 586–597. [10] C.P. Gupta, A generalized multi-point boundary value problem for second order ordinary differential equations, Appl. Math. Comput. 89 (1998) 133–146. [11] Ruyun Ma, Positive solutions of a nonlinear three-point boundary value problem, Electron. J. Differential Equations 34 (1999) 1–8. [12] Ruyun Ma, Positive solutions for second order three-point boundary value problem, Appl. Math. Lett. 14 (2001) 1–5. [13] Ruyun Ma, Existence Theorems for a second order m-point boundary value problem, J. Math. Anal. Appl. 211 (1997) 545–555. [14] Ruyun Ma, Existence of solution of nonlinear m-point boundary value problem, J. Math. Anal. Appl. 256 (2001) 556–567. [15] Ruyun Ma, Positive solutions of a nonlinear m-point boundary value problem, Comput. Math. Appl. 42 (2001) 755–765. [16] Xian Xu, Positive solutions for singular m-point boundary value problems with positive parameter, J. Math. Anal. Appl. 291 (2004) 352–367. [17] Xian Xu, Multiplicity results for positive solutions of some semi-positone three-point boundary value problems, J. Math. Anal. Appl. 291 (2004) 673–689. [18] Guowei Zhang, Jingxian Sun, Positive solutions of m-point boundary value problems, J. Math. Anal. Appl. 291 (2004) 406–418. [19] Zhongxin Zhang, Junyu Wang, The upper and lower solution method for a class of singular nonlinear second order three-point boundary value problems, J. Comput. Appl. Math. 147 (2002) 41–52. [20] P. Hartman, Ordinary Differential Equations, second ed., Birkhauser, Boston, 1982.
Positive solutions of some singular m-point boundary value problems at non-resonance Zhongli Wei *, Changci Pang Department of Mathematics and Physics, Shandong Institute of Architectural and Engineering, Jinan, Shandong 250014, People’s Republic of China
Abstract This paper investigates the existence of positive solutions for second-order singular m-point boundary value problems at non-resonance. A necessary and sufficient condition for the existence of C[0, 1] as well as C1[0, 1] positive solutions is given by constructing lower and upper solutions and with the maximal theorem. Our nonlinearity f(t, x) may be singular at x,t = 0 and/or t = 1. 2005 Elsevier Inc. All rights reserved. Keywords: Singular m-point boundary value problem; Positive solution; Lower and upper solution; Maximum principle; Non-resonance
1. Introduction The singular ordinary differential equations arise in the fields of gas dynamics, Newtonian fluid mechanics, the theory of boundary layer and so on, the *
Corresponding author. E-mail address: [email protected] (Z. Wei).
0096-3003/$ - see front matter 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.01.043
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Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
theory of singular boundary value problems has become an important area of investigation in recent years (see [1–5] and the references therein). Consider the singular m-point boundary value problems of second-order ordinary differential equation at non-resonance 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ f ðt; xðtÞÞ; t 2 ð0; 1Þ; m2 P ð1:1Þ : xð0Þ ¼ ai xðgi Þ; xð1Þ ¼ 0; i¼1
where q > 0 is such that 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ 0; t 2 ð0; 1Þ; m2 P : xð0Þ ¼ ai xðgi Þ; xð1Þ ¼ 0
ð1:2Þ
i¼1
has only the trivial solution, 0 < ai < 1, i = 1,2, . . . , m 2, 0 < g1 < g2 < < Pm2 gm2 < 1, are constants, i¼1 ai < 1, m P 3 and f, p satisfy the following hypothesis. (H1) p(t) 2 C(0, 1), p(t) P 0, t 2 (0, 1), and Z 1 tð1 tÞpðtÞ dt < 1; also
ð1:3Þ
0
limþ t2 pðtÞ ¼ 0 if
t!0
2
lim ð1 tÞ pðtÞ ¼ 0
t!1
Z
1
ð1 tÞpðtÞ dt ¼ 1; and
ð1:4Þ
0
Z if
1
tpðtÞ dt ¼ 1;
ð1:5Þ
0
(H2) p(t) 2 C(0, 1), p(t) P 0, t 2 (0, 1), and Z 1 pðtÞ dt < 1;
ð1:6Þ
0
(H3) f 2 C((0, 1) · (0, 1),[0, 1)), f(t, 1) > 0, t 2 (0, 1), and there exist constants k, l(1 < k < 0 < l < 1), such that, for t 2 (0, 1) and x 2 (0,+1), cl f ðt; xÞ 6 f ðt; cxÞ 6 ck f ðt; xÞ;
if
0 < c 6 1;
ð1:7Þ
if
c P 1:
ð1:8Þ
Remark. (1.7) implies ck f ðt; xÞ 6 f ðt; cxÞ 6 cl f ðt; xÞ;
Conversely, (1.8) implies (1.7). Typical functions thatP satisfy the above sub-linear hypothesis are those taking the form f ðt; xÞ ¼ nk¼1 pk ðtÞxkk ; here pk(t) 2 C(0, 1), pk(t) > 0 on (0, 1), kk < 1, k = 1,2, . . . , n.
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
435
By singularity we mean that the functions f, p in (1.1) are allowed to be unbounded at the points x = 0, t = 0 and t = 1. A function x(t) 2 C[0, 1] \ C2(0, 1) is called a C[0, 1] (positive) solution of (1.1) if it satisfies (1.1) (x(t) > 0, for t 2 (0, 1)). A C[0, 1] (positive) solution of (1.1) is called a C1[0, 1] (positive) solution if x 0 (0+) and x 0 (1-) both exist (x(t) > 0 for t 2 (0, 1)). When the function p = 0, f 2 C([0, 1] · R · R,R) in (1.1), i.e. f is continuous, problem (1.1) is nonsingular, the existence of solutions of (1.1) has been studied by many authors using nonlinear alternative of Leray-Schauder, Coincidence degree theory and fixed point theorem in cones (see [6–15] and references therein). Very recently, in the special case: p = 0, f(t, x) in (1.1) cannot be singular at x = 0, the existence of positive solutions and multiple positive solutions of singular boundary value problem (1.1) have been studied by papers [16–18] using the fixed point index and approximate process; Zhang and Wang in [19] gave some sufficient conditions for the existence results of a class of singular nonlinear second order three-point boundary value problems by the upper and lower solution method and the monotone iterative technique. In this paper, we shall study the existence of positive solutions for secondorder singular m-point boundary value problems at non-resonance (1.1). A necessary and sufficient condition for the existence of C[0, 1] as well as C1[0, 1] positive solutions is given by constructing lower and upper solutions and with the maximal theorem. Our nonlinearity f(t, x) may be singular at x, t = 0 and/or t = 1.
2. Several lemmas Lemma (see [5]). Suppose (H1) holds. (i) Then 00 x þ qpðtÞx ¼ 0; xð0Þ ¼ 0;
t 2 ð0; 1Þ;
x0ð0Þ ¼ 1
ð2:1Þ
has a unique positive increasing solution e1(t) 2 C[0, 1] \ C1[0, 1). And there exists a unique nonnegative function w1 2 C[0, 1] with Z Z q t s spðsÞw1 ðsÞ ds ds; w1 ðtÞ ¼ 1 þ t 0 0 and e1(t) = tw1(t) 2 C[0, 1] \ C1[0, 1) is a solution of (2.1). (ii) Then 00 x þ qpðtÞx ¼ 0; t 2 ð0; 1Þ; xð1Þ ¼ 0; x0ð1Þ ¼ 1
ð2:2Þ
436
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
has a unique positive decreasing solution e2(t) 2 C[0, 1] \ C1(0, 1]. And there exists a nonnegative function w2 2 C[0, 1] with Z 1Z 1 q w2 ðtÞ ¼ 1 þ ð1 sÞpðsÞw2 ðsÞ ds ds; 1t t s and e2(t) = (1 t)w2(t) 2 C[0, 1] \ C1(0, 1] is a positive decreasing solution of (2.2). In addition, if (H2) holds, then e1(t),e2(t) 2 C1[0, 1]. Remark 1. If p(t) = 0, then e1(t) = t, e2(t) = 1 t, w1(t) = w2(t) = 1. To prove the main result, we need the following maximum principle. Suppose that 0 < gm2 < bn, and ( ) m2 X 2 F n ¼ x 2 C½0; bn \ C ð0; bn Þ; xð0Þ ai xðgi Þ P 0; xðbn Þ P 0 : i¼1
Lemma 2.1 (Maximum principle). If x 2 Fn such that x 0 0 (t) + qp(t)x(t) P 0, t 2 (0, bn), then x(t) P 0, t 2 [0, bn]. Proof. Set x00 ðtÞ þ qpðtÞxðtÞ ¼ rðtÞ; xð0Þ
m2 X
ai xðgi Þ ¼ r1 ;
t 2 ð0; bn Þ;
xðbn Þ ¼ r2 ;
ð2:3Þ ð2:4Þ
i¼1
then r1 P 0, r2 P 0, r(t) P 0, t 2 (0, bn). And the solution (2.3) and (2.4) can be stated Z bn x2n ðtÞ x1n ðtÞ xðtÞ ¼ r2 þ Gn ðt; sÞrðsÞ ds r1 þ Pm2 x1n ðbn Þ x2n ð0Þ i¼1 ai x2n ðgi Þ 0 Z bn m2 X x2n ðtÞ a Gn ðgi ; sÞrðsÞ ds; ð2:5Þ þ Pm2 i ai x2n ðgi Þ i¼1 x2n ð0Þ i¼1 0 where 1 Gn ðt; sÞ ¼ xn
x2n ðtÞe1 ðsÞ; s < t; x2n ðsÞe1 ðtÞ; t 6 s;
x2n ðtÞ x2n 0ðtÞ ¼ x2n ð0Þ ¼ e1 ðbn Þ ¼ constant > 0: xn ¼ e1 ðtÞ e1 0ðtÞ
ð2:6Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
and x1n ðtÞ ¼ e1 ðtÞ þ
1
P1m2 i¼1
Pm2 ai
i¼1
437
ai e1 ðgi Þ 2 C 2 ½an ; bn is a unique increasing
positive solution of the problem 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ 0; t 2 ½0; bn ; m2 P : xð0Þ ¼ ai xðgi Þ; x0 ð0Þ ¼ 1:
ð2:7Þ
i¼1
And x2n(t) 2 C2[0, bn] is a unique decreasing positive solution of the problem 00 x ðtÞ þ qpðtÞxðtÞ ¼ 0; t 2 ½0; bn ; ð2:8Þ xðbn Þ ¼ 0; x0ðbn Þ ¼ 1: By means of (2.5)–(2.8), we can obtain x(t) P 0, t 2 [0, bn]. The proof is complete. h Lemma 2.2. Suppose that (H1) and (H3) hold. Let x(t) be a C[0, 1] positive solution of (1.1). Then xðtÞ P a0 kxk
e2 ðtÞ ; e2 ð0Þ
t 2 ½0; 1:
ð2:9Þ
Here, kxk = maxt2[0,1]jx(t)j, and m2 X e1 ðgi Þ e2 ðgi Þ ai a0 ¼ : e1 ð1Þ e2 ð0Þ i¼1
ð2:10Þ
Proof. Assume that x(t) is a C[0, 1] positive solution of (1.1) and (1.2). Then x(t) can be stated Z 1 m2 X e2 ðtÞ Gðt; sÞf ðs; xðsÞÞ ds þ ai xðtÞ ¼ Pm2 e2 ð0Þ i¼1 ai e2 ðgi Þ i¼1 0 Z 1 Gðgi ; sÞf ðs; xðsÞÞ ds; ð2:11Þ 0
where Gðt; sÞ ¼
1 x
e2 ðtÞe1 ðsÞ; s < t;
ð2:12Þ
e2 ðsÞe1 ðtÞ; t 6 s;
e2 ðtÞ e2 0ðtÞ ¼ e2 ð0Þ ¼ e1 ð1Þ ¼ constant > 0: x ¼ e1 ðtÞ e1 0ðtÞ It is easy to see that xð0Þ > 0;
xðtÞ P
e1 ðtÞ e2 ðtÞ kxk; e1 ð1Þ e2 ð0Þ
t 2 ½0; 1:
ð2:13Þ
438
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
From (2.11), we have that e2 ð0Þ ¼
1 Pm2 i¼1
m2 X
ai e2 ðgi Þ
i¼1
Z ai
1
Gðgi ; sÞf ðs; xðsÞÞ ds
0
m2 1 X ai xðgi Þ; e2 ð0Þ i¼1
ð2:14Þ
For 0 6 t 6 1, we have from (2.13) and (2.14) that m2 m2 e2 ðtÞ X e2 ðtÞ X e1 ðgi Þ e2 ðgi Þ ai xðgi Þ P ai xðtÞ P kxk e2 ð0Þ i¼1 e2 ð0Þ i¼1 e1 ð1Þ e2 ð0Þ ¼ a0 kxk
e2 ðtÞ ; e2 ð0Þ
The proof is complete.
t 2 ½0; 1:
ð2:15Þ
h
Lemma 2.3. Suppose that (H2) and (H3) hold. Let x(t) be a C1[0, 1] positive solution of (1.1). Then there are constants 0 < I1 < I2, such that I 1 e2 ðtÞ 6 xðtÞ 6 I 2 e2 ðtÞ;
t 2 ½0; 1:
ð2:16Þ
Proof. Assume that x(t) is a C1[0, 1] positive solution of (1.1). Then (2.9) holds, and both x 0 (0) and x 0 (1) exist, x(t) > 0 for t 2 (0, 1). By integration of (1.1), we have Z
1
f ðt; xðtÞÞdt 6 x0ð1Þ þ x0ð0Þ þ q max jxðtÞj t2½0;1
0
Z
1
pðtÞdt < 1:
ð2:17Þ
0
From (2.11) and (1.17), we have " Z 1 1 1 e1 ðsÞf ðs; xðsÞÞ ds þ xðtÞ 6 e2 ðtÞ Pm2 x 0 e ð0Þ i¼1 ai e2 ðgi Þ #2 Z 1 m2 X ai Gðgi ; sÞf ðs; xðsÞÞ ds :
ð2:18Þ
0
i¼1
Setting I1 = a0kxk/e2(0), Z 1 1 1 I2 ¼ e1 ðsÞf ðs; xðsÞÞ ds þ Pm2 x 0 e2 ð0Þ i¼1 ai e2 ðgi Þ Z 1 m2 X ai Gðgi ; sÞf ðs; xðsÞÞ ds; i¼1
0
then from (2.9) and (2.18), we have (2.16). This completes the proof.
h
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
439
Definition 2.1. Suppose a 2 C[0, 1] \ C2(0, 1), if a satisfies a00 ðtÞ þ qpðtÞaðtÞ 6 f ðt; aðtÞÞ; að0Þ
m2 X
ai aðgi Þ 6 0;
t 2 ð0; 1Þ;
að1Þ 6 0:
i¼1
Then a is called a lower solution of the singular boundary value problem (1.1). Definition 2.2. Suppose b 2 C[0, 1] \ C2(0, 1), if b satisfies b00 ðtÞ þ qpðtÞbðtÞ P f ðt; bðtÞÞ; bð0Þ
m2 X
ai bðgi Þ P 0;
t 2 ð0; 1Þ;
bð1Þ P 0:
i¼1
Then b is called a upper solution of the singular boundary value problem (1.1). Lemma 2.4. Assume that there exist lower and upper solutions of (1.1), respectively a(t) and b(t), such that a (t),b(t) 2 C[0, 1] \ C2(0, 1), 0 < a(t) 6 b(t), for t 2 (0, 1). Then problem (1.1) has at least one C[0, 1] positive solution x(t) such that a(t) 6 x(t) 6 b(t), for t 2 [0, 1]. If, in addition, there exists F(t) 2 L1[0, 1] such that jf ðt; xðtÞÞj 6 F ðtÞ;
for
aðtÞ 6 xðtÞ 6 bðtÞ;
t 2 ð0; 1Þ:
ð2:19Þ
then the solution x(t) of (1.1) is a C1[0, 1] positive solution. Proof. First of all, we define a partial ordering in C[0, 1] \ C2(0, 1) by x 6 y if and only if xðtÞ 6 yðtÞ;
8t 2 ½0; 1:
Then, we shall define an 8 > < f ðt; aðtÞÞ; gðt; xÞ ¼ f ðt; xðtÞÞ; > : f ðt; bðtÞÞ;
ð2:20Þ
auxiliary function. "x(t) 2 C [0, 1] \ C2(0, 1), if
aix
if
a6x6b
if
xib:
ð2:21Þ
By the condition (H), we have g:(0, 1) · R ! [0, + 1) is continuous. Let {bn} be a sequence satisfying 0 < gm2 < b1 < < bn < bn+1 < < 1, and bn ! 1 as n ! 1, and let {rn}, be a sequence satisfying aðbn Þ 6 rn 6 bðbn Þ;
n ¼ 1; 2; . . . ;
ð2:22Þ
440
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
For each n, consider the nonsingular problem 8 00 < x ðtÞ þ qpðtÞxðtÞ ¼ gðt; xÞ; t 2 ½0; bn ; m2 P : xð0Þ ai xðgi Þ ¼ 0; xðbn Þ ¼ rn :
ð2:23Þ
i¼1
Obviously, by the proof of the Lemma 2.1, the problem (2.23) is equivalent to the integral equation xðtÞ ¼ An xðtÞ x1n ðtÞ rn þ ¼ x1n ðbn Þ þ
x2n ð0Þ
Z
bn
Gn ðt; sÞgðs; xðsÞÞ ds 0
x2n ðtÞ Pm2 i¼1
m2 X
ai x2n ðgi Þ
i¼1
Z ai
bn
Gn ðgi ; sÞgðs; xðsÞÞ ds;
ð2:24Þ
0
where Gn(t, s) is given by (2.6), x1n(t), x2n(t) are given by (2.7) and (2.8). It is easy to verify that An:Xn ! Xn = C[0, bn] is completely continuous and An(Xn) is a bounded set. Moreover, x 2 C[0, bn] is a solution of (2.23) if and only if Anx = x. Using the SchauderÕs fixed point theorem, we assert that An has at least one fixed point xn 2 C2[0, bn]. We claim that a 6 xn 6 b; that is aðtÞ 6 xn ðtÞ 6 bðtÞ;
t 2 ½0; bn :
ð2:25Þ
and hence xn(t) 2 C2[0,bn] and satisfies x00 ðtÞ þ qpðtÞxðtÞ ¼ f ðt; xðtÞÞ;
t 2 ½0; bn :
ð2:26Þ
Indeed, suppose by contradiction that xn ib. By the definition of g, then gðt; xn ðtÞÞ ¼ f ðt; bðtÞÞ;
t 2 ½0; bn ;
and therefore x00n ðtÞ þ qpðtÞxn ðtÞ ¼ f ðt; bðtÞÞ;
t 2 ½0; bn :
ð2:27Þ
On the other hand, since b is an upper solution of (1.1) and (1.2), We also have b00 ðtÞ þ qpðtÞbðtÞ P f ðt; bðtÞÞ; Then setting zðtÞ ¼ bðtÞ xn ðtÞ;
t 2 ½0; bn :
t 2 ð0; 1Þ:
ð2:28Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
441
By (2.22), (2.23), (2.27) and (2.28), we obtain z00 ðtÞ þ qpðtÞzðtÞ P 0;
t 2 ð0; bn Þ; m2 X ai zðgi Þ P 0; zðbn Þ P 0: z 2 C½0; bn \ C 2 ð0; bn Þ; zð0Þ i¼1
By Lemma 2.1, we can conclude that z(t) P 0, t 2 [0, bn], a contradiction with the assumption xn i b. Therefore xn i b is impossible. Similarly, we can show that a 6 xn. So, we have shown that (2.25) holds. Using the method of [3] and the Theorem 3.2 in [20], we can obtain that there is a C[0, 1] positive solution x(t) of (1.1) such that a 6 x 6 b, and a subsequence of {xn(t)} converges to x(t) on any compact subintervals of (0, 1). In addition, if (2.19) holds, then jx 0 0 (t)j 6 F(t) + qp(t)jx(t)j, and hence x 0 0 (t) is absolutely integrable on [0, 1]. This implies x(t) 2 C1[0, 1], so x(t) is a C1[0, 1] positive solution of the problem (1.1). The proof is complete. h
3. The main results Theorem 3.1. Suppose (H1) and (H3) hold, then a necessary and sufficient condition for problem (1.1) and (1.2) to have C[0, 1] positive solutions is that the following integral conditions hold. Z 1 tð1 tÞf ðt; 1Þ dt < 1; ð3:1Þ 0< 0
Z limþ t
t!0
1
ð1 sÞf ðs; 1Þ ds ¼ 0;
ð3:2Þ
t
lim ð1 tÞ
t!1
Z
t
sf ðs; 1Þ ds ¼ 0:
ð3:3Þ
0
Theorem 3.2. Suppose (H2) and (H3) hold, then a necessary and sufficient condition for problem (1.1) and (1.2) to have C1[0, 1] positive solutions is that the following integral conditions hold. Z 1 0< f ðt; e2 ðtÞÞ dt < 1: ð3:4Þ 0
Proof of Theorem 3.1. (1) Necessity. Let x(t) 2 C[0, 1] be a positive solution of (1.1). Then there is a t0 2 (0, 1) such that x 0 (t0) = 0. Let c0 > 0 be a constant such that c0x(t) 6 1 for t 2 [0, 1] and 1/c0 P 1. From (1.7) and (1.8), we have l f ðt; xðtÞÞ P ð1=c0 Þk f ðt; c0 xðtÞÞ P clk 0 x ðtÞf ðt; 1Þ;
for t 2 ð0; 1Þ:
442
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
According to (1.1), we have c0lk f ðt; 1Þ 6 xl ðtÞx00 ðtÞ þ qpðtÞx1l ðtÞ;
t 2 ð0; 1Þ:
ð3:5Þ
For t 2 (0, t0), by integration of (3.5), we obtain Z t0 Z t0 t0 2 lk 0 l c0 f ðs; 1Þ ds 6 x ðsÞx ðsÞjt þ lxl1 ðsÞ ðx0 ðsÞÞ ds t
þq
Z
t t0
pðsÞx1l ðsÞ ds
t
6 xl ðtÞx0 ðtÞ þ q
Z
t0
pðsÞx1l ðsÞ ds;
t 2 ð0; t0 Þ:
ð3:6Þ
t
Integrating (3.6), we have Z t0 Z t0 Z t0 Z t0 x1l ðt0 Þ x1l ð0Þ þ qK c0lk f ðs; 1Þ ds dt 6 pðsÞ ds dt 1l 0 t 0 t Z t0 x1l ðt0 Þ x1l ð0Þ þ qK ¼ spðsÞ ds < 1; 1l 0 where K = maxt2[0,1]x1l(t), so Z t0 sf ðs; 1Þ ds < 1: 0<
ð3:7Þ
0
Similarly, by integration of (3.5), we obtain Z 1 0< ð1 sÞf ðs; 1Þ ds < 1:
ð3:8Þ
t0
(3.7) and (3.8) imply that (3.1) holds. For t 2 (0, t0), by integration of (3.6), we have Z t Z t0 Z t Z t0 x1l ðtÞ x1l ð0Þ þ qK c0lk f ðs; 1Þ ds ds 6 pðsÞ ds ds 1l s s 0 0 ¼
x1l ðtÞ x1l ð0Þ þ qK 1l Z t Z t Z t0 ds þ pðsÞ ds; 0
s
t
therefore, c0lk t
Z
t0
f ðs; 1Þ ds 6 t
x1l ðtÞ x1l ð0Þ 1l Z t Z þ qK spðsÞ ds þ t 0
t
t0
pðsÞ ds :
ð3:9Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
443
Letting R t t ! 0 in (3.9) and noting condition (H1), we have limt!0þ t t 0 f ðs; 1Þds ¼ 0. These imply that (3.2) holds. Similarly, we can prove (3.3). (2) Sufficiency. Suppose that (3.1)–(3.3) hold. Choose a constant m P 2 such that m(l k) > 1, and define the linear operator A qðtÞ ¼ Af ðt; 1Þ Z 1 ¼ Gðt; sÞf ðs; 1Þ ds þ
e2 ð0Þ
0
Z
m2 X
e2 ðtÞ Pm2 i¼1
ai e2 ðgi Þ
1
ai
Gðgi ; sÞf ðs; 1Þds;
ð3:10Þ
0
i¼1
1=ðmðlkÞÞ
QðtÞ ¼ ½qðtÞ
ð3:11Þ
:
where, G(t,s) is given by (2.12). Then q(t),Q(t) 2 C[0, 1] \ C2(0, 1) satisfying q(t) > 0, Q(t) > 0, t 2 (0, 1), and q00 ðtÞ ¼ f ðt; 1Þ;
Q00 ðtÞ P 0;
for
t 2 ð0; 1Þ
Pm2 and from Pm2(3.1)–(3.3), we have q(1) = Q(1) = 0, qð0Þ i¼1 ai qðgi Þ ¼ 0, Qð0Þ i¼1 ai Qðgi Þ P 0, and Z Z 1 kw1 k kw2 k 1 Gðs; sÞf ðs; 1Þ ds 6 sð1 sÞf ðs; 1Þ ds < þ1; x 0 0 0 " #1 1 Z m2 1 X ai e2 ðgi Þ A Gðs; sÞf ðs; 1Þ ds < þ1: qðtÞ 6 @1 þ e2 ð0Þ e2 ð0Þ 0
i¼1
e2 ðtÞ
Z
t
e1 ðsÞQðlkÞ ðsÞf ðs; 1Þ ds 0
1=m Z e2 ðsÞ s e1 ðsÞf ðs; 1Þ ds f ðs; 1Þ ds x 0 0 Z s 1=m Z t 11=m 1=m x e1 ðsÞ e1 ðsÞf ðs; 1Þ ds f ðs; 1Þ ds 6 ðe2 ðtÞÞ 6 e2 ðtÞ
Z
t
e1 ðsÞ
0 1=m
¼x
0
1
11=m
ð1 1=mÞ ðe2 ðtÞÞ
Z
11=m
t
e1 ðsÞf ðs; 1Þ ds 0
6 x1=m ð1 1=mÞ
1
Z 0
11=m
1
e1 ðsÞe2 ðsÞf ðs; 1Þ ds
< 1:
ð3:12Þ
444
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
Similarly, we have Z 1 e1 ðtÞ e2 ðsÞQðlkÞ ðsÞf ðs; 1Þ ds t
6x
1=m
ð1 1=mÞ
1
Z
11=m
1
e1 ðsÞe2 ðsÞf ðs; 1Þ ds
< 1:
ð3:13Þ
0
Let c1 > 0 such that (1/c1)Q(t) 6 1, c1 P 1. From (1.7) and (1.8), we have k
k
Ql f ðt; QÞ 6 Ql ðQ=c1 Þ f ðt; c1 Þ 6 Ql ðQ=c1 Þ cl1 f ðt; 1Þ kl ¼ clk f ðt; 1Þ: 1 Q
Let
h1 ðtÞ ¼ A
e2 ðtÞ e2 ð0Þ
ð3:14Þ
l f ðt; 1Þ;
t 2 ½0; 1;
h2 ðtÞ ¼ AQl ðtÞf ðt; QðtÞÞ þ QðtÞ;
t 2 ½0; 1:
Thus, (3.12)–(3.14) imply that 0 6 h1 ðtÞ < 1; 0 6 h2 ðtÞ < 1;
t 2 ½0; 1: P One can check that hj 2 C[0, 1] \ C (0, 1), h1 ð0Þ m2 i¼1 ai h1 ðgi Þ ¼ 0, h2 ð0Þ Pm2 a h ðg Þ P 0, and i¼1 i 2 i e2 ðtÞ a0 kh1 k 6 h1 ðtÞ 6 kh1 k; QðtÞ 6 h2 ðtÞ 6 kh2 k; t 2 ½0; 1: e2 ð0Þ for
2
h001 ðtÞ þ qpðtÞh1 ðtÞ ¼
e2 ðtÞ e2 ð0Þ
l f ðt; 1Þ;
t 2 ð0; 1Þ;
ð3:15Þ
h002 ðtÞ þ qpðtÞh2 ðtÞ P Ql ðtÞf ðt; QðtÞÞ;
t 2 ð0; 1Þ:
ð3:16Þ
Here, khjk = max06t61jhj(t)j. Let a(t) = k1h1(t), b(t) = k2h2(t), t 2 [0, 1]; here k1, k2 are constants satisfying 0 < k1 6 1 6 k2 and will be determined later. Suppose c2, c3 are positive constants such that c2kh1k 6 1, c2 6 1, c3 P 1. From (1.7),(1.8), we have k
lk
f ðt; aðtÞÞ P ð1=c2 Þ f ðt; c2 aðtÞÞ P ðc2 Þ al ðtÞf ðt; 1Þ l e2 ðtÞ P ðc2 Þlk ðk 1 a0 kh1 kÞl f ðt; 1Þ e2 ð0Þ P a00 ðtÞ þ qpðtÞaðtÞ; t 2 ð0; 1Þ; l bðtÞ f ðt; QðtÞÞ f ðt; bðtÞÞ 6 ðc3 Þ QðtÞ lk l l 6 ðc3 Þ ðk 2 kh2 kÞ Q ðtÞf ðt; QðtÞÞ 6 b00 ðtÞ þ qpðtÞbðtÞ; t 2 ð0; 1Þ: lk
ð3:17Þ
ð3:18Þ
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
445
By virtue of (1.3), (1.4), we can find a k0 such that f(t, Q(t)) P k0Ql(t)f(t,1) and hence, from the definitions of h1(t), h2(t), we have, when k > k 1 0 , h1(t) 6 k 1=ð1lÞ
h2(t) for n t 2 [0, 1]. Now we chooseo k 1 ¼ min f1; ðða0 kh1 kÞ l c2lk Þ g and l lk 1=ð1lÞ 2 1 k 2 ¼ max 1; ; k 0 ; kh2 k c3 . Then a(t),b(t) 2 C[0, 1] \ C (0, 1), 0 < Pm2 Pm2 a(t) 6 b(t), for t 2 (0, 1), að0Þ ¼ i¼1 ai aðgi Þ, bð0Þ i¼1 ai bðgi Þ P 0, a(1) = b(1) = 0. From (3.15)–(3.18), we obtain that for such choice of k1 and k2, a(t) and b(t) are lower and upper solutions of (1.1), respectively. From the first conclusion of Lemma 2.4, we conclude that the problem (1.1) has at least a C[0, 1] positive solution x(t) satisfying a (t) 6 x(t) 6 b(t) for t 2 [0, 1]. This completes the proof of Theorem 3.1. h Proof of Theorem 3.2. (1) Necessity. Suppose that x(t) is a C1[0, 1] positive solution of (1.1). Then both x 0 (0) > 0 and x 0 (1) < 0 exist. By Lemma 2, there are constants I1 and I2, 0 < I1 < I2 such that I 1 e2 ðtÞ 6 xðtÞ 6 I 2 e2 ðtÞ 6 I 2 e2 ð0Þð1 tÞ;
t 2 ½0; 1:
ð3:19Þ
Let c4 be a constant satisfying c4I2 6 1, 1/c4 P 1. Then (1.7), (1.8) and (3.21), lead to k
f ðt; xðtÞÞ P ð1=c4 Þ f ðt; c4 xðtÞe2 ðtÞ=e2 ðtÞÞ P ðc4 Þ P ðc4 Þ
lk l I 1 f ðt; e2 ðtÞÞ;
Consequently, Z 1 Z kl l f ðt; e2 ðtÞÞ dt 6 ðc4 Þ I 1 0
lk
l
ðxðtÞ=e2 ðtÞÞ f ðt; e2 ðtÞÞ
t 2 ð0; 1Þ:
1
f ðt; xðtÞÞ dt 0
Z 0 0 x ð0Þ x ð1Þ þ I qe ð0Þ ¼ ðc4 Þkl I l 2 2 1
1
pðtÞtð1 tÞ dt
0
< 1: Thus (3.4) holds. (2) Sufficiency. Suppose that (3.4) holds. Let hðtÞ ¼ Af ðt; e2 ðtÞÞ; 1
t 2 ½0; 1: 2
ð3:20Þ
Then h(t) 2 C [0, 1] \ C (0, 1) and (3.19) holds if x(t) is replaced by h(t), and I1 = a0khk/e2(0), " #1 Z m2 X 1 1 I2 ¼ e1 ðsÞf ðs; e2 ðsÞÞ ds þ e2 ð0Þ ai e2 ðgi Þ x 0 i¼1 Z 1 m2 X ai Gðgi ; sÞf ðs; e2 ðsÞÞ ds: i¼1
0
446
Z. Wei, C. Pang / Appl. Math. Comput. 171 (2005) 433–449
Suppose that constants c5 and c6 satisfy c5I2 6 1, 1/c5 P 1, c6I1 P 1, 1/ c6 6 1. Let a(t) = k1h(t), nb(t) = k2h(t), to2 [0, 1]; here k 1 ¼ min f1; l lk 1=ð1lÞ 1=ð1lÞ g and k 2 ¼ max 1; I l2 clk . A similar argument to that I 1 c5 6 we have checked in the sufficiency proof of Theorem 3.1 yields that for t 2 (0, 1), k
f ðt; aðtÞÞ P ð1=c5 Þ f ðt; c5 aðtÞÞ P ðc5 Þ P ðc5 Þ
lk
lk
f ðt; bðtÞÞ 6 ðc6 Þ
lk
l
ðaðtÞ=e2 ðtÞÞ f ðt; e2 ðtÞÞ
l
ðk 1 I 1 Þ f ðt; e2 ðtÞÞ P a00 ðtÞ þ qpðtÞaðtÞ;
bðtÞ e2 ðtÞ
l
t 2 ð0; 1Þ; ð3:21Þ
f ðt; e2 ðtÞÞ 6 ðc6 Þlk ðk 2 I 2 Þl f ðt; e2 ðtÞÞ
6 b00 ðtÞ þ qpðtÞbðtÞ;
t 2 ð0; 1Þ:
ð3:22Þ
So, a(t),b(t) 2 C1[0, 1] \ C2(0, 1) are, respectively, lower and upper solutions Pm2 of (1.1)P satisfying 0 < a(t) 6 b(t), for t 2 (0, 1), and að0Þ ¼ i¼1 ai aðgi Þ, m2 bð0Þ ¼ i¼1 ai bðgi Þ, a (1) = b(1) = 0. Additionally, when t 2 (0, 1) and a(t) 6 x 6 b(t), we have k k l k1 c6 x k1 c6 x 0 6 f ðt; xÞ 6 e2 ðtÞ 6 f t; f ðt; e2 ðtÞÞ k 1 e2 ðtÞ k 1 e2 ðtÞ c6 c6 kl k1 l ðk 2 I 2 Þ f ðt; e2 ðtÞÞ ¼ F ðtÞ: 6 c6 R1 From (3.4), we have 0 F ðtÞdt < 1. By Lemma 2.4, we assert that problem (1.1) and (1.2) admits a positive solution x(t) 2 C1[0, 1] \ C2(0, 1) such that a(t) 6 x(t) 6 b(t) for t 2 [0, 1]. The proof of Theorem 3.2 is complete. h Remark 2. Consider the singular m-point boundary value problem 8 00 < x þ qpðtÞx ¼ f ðt; xÞ; t 2 ð0; 1Þ; m2 P ai xðgi Þ: : xð0Þ ¼ 0; xð1Þ ¼
ð3:23Þ
i¼1
By analogous methods, we have the following results. Assume that x(t) is a C[0, 1] positive solution of problem (3.23). Then x(t) can be stated Z 1 e1 ðtÞ xðtÞ ¼ Gðt; sÞf ðs; xðsÞÞ ds þ Pm2 e1 ð1Þ i¼1 ai e1 ðgi Þ 0 Z 1 m2 X ai Gðgi ; sÞf ðs; xðsÞÞ ds: ð3:24Þ i¼1
0
where G(t,s) is given by (2.12).
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Theorem A. Suppose (H1) and (H3) hold, then a necessary and sufficient condition for problem (3.23) to have C[0, 1] positive solutions is that the integral conditions (3.1)–(3.3) hold. Theorem B. Suppose (H2) and (H3) hold, then a necessary and sufficient condition for problem (3.23) to have C1[0, 1] positive solutions is that the following integral conditions hold. Z 1 0< f ðt; e1 ðtÞÞdt < 1: ð3:25Þ 0
4. Some examples Example 1. Consider the following singular three-point boundary value problem 00 b k x ðtÞ ¼ ta ð1 tÞ x ðtÞ; t 2 ð0; 1Þ; ð4:1Þ 1 1 xð0Þ ¼ 2 x 2 ; xð1Þ ¼ 0: where a,b,k 2 R, and k < 1. Evidently, 00 x ðtÞ ¼ 0; t 2 ð0; 1Þ; xð0Þ ¼ 12 x 12 ; xð1Þ ¼ 0:
ð4:2Þ
has only the trivial solution. m = 3, a1 ¼ 12, g1 ¼ 12, p(t) = 0, f(t, x) = ta(1 t)bxk, and from Remark 1, we have e1(t) = t, e2(t) = 1 t, (H1)–(H3) holds. So, from Theorems 3.1 and 3.2, we obtain that Conclusion 4.1. A necessary and sufficient condition for problem (4.1) to have C[0, 1] positive solutions is a > 2, b > 2. Conclusion 4.2. A necessary and sufficient condition for problem (4.1) to have C1[0, 1] positive solutions is a > 1, b + k > 1. Similarly, from Remark 2, for singular three-point boundary value problem ( b x00 ðtÞ ¼ ta ð1 tÞ xk ðtÞ; t 2 ð0; 1Þ; ð4:3Þ xð0Þ ¼ 0; xð1Þ ¼ 12 x 12 : where a,b,k 2 R, and k < 1, we have Conclusion 4.3. A necessary and sufficient condition for problem (4.3) to have C[0, 1] positive solutions is a > 2, b > 2.
448
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Conclusion 4.4. A necessary and sufficient condition for problem (4.3) to have C1[0, 1] positive solutions is b > 1, a + k > 1. Example 2. Consider singular three-point boundary value problem (
b
x00 ðtÞ þ xðtÞ ¼ ta ð1 tÞ xk ðtÞ; xð0Þ ¼ 12 x 12 ; xð1Þ ¼ 0:
t 2 ð0; 1Þ;
ð4:4Þ
where a,b,k 2 R, and k < 1. Evidently, (
x00 ðtÞ þ xðtÞ ¼ 0; t 2 ð0; 1Þ; xð0Þ ¼ 12 x 12 ; xð1Þ ¼ 0:
ð4:5Þ
has only the trivial solution, m = 3, a1 ¼ 12, g1 ¼ 12, p(t) = 1, f(t,x) = ta(1t)bxk, t t 1t t1 and e1 ðtÞ ¼ e e , e2 ðtÞ ¼ e e , (H1)–(H3) holds. So, from Theorem 3.1 and 2 2 Theorem 3.2, we obtain that Conclusion 4.5. A necessary and sufficient condition for problem (4.4) to have C[0, 1] positive solutions is a > 2, b > 2. Conclusion 4.6. A necessary and sufficient condition for problem (4.4) to have C1[0, 1] positive solutions is a > 1, b + k > 1. Similarly, from Remark 2, for singular three-point boundary value problem ( 00 b x ðtÞ þ xðtÞ ¼ ta ð1 tÞ xk ðtÞ; t 2 ð0; 1Þ; ð4:6Þ xð0Þ ¼ 0; xð1Þ ¼ 12 x 12 : where a, b, k 2 R, and k < 1, we have Conclusion 4.7. A necessary and sufficient condition for problem (4.6) to have C[0, 1] positive solutions is a > 2, b > 2. Conclusion 4.8. A necessary and sufficient condition for problem (4.6) to have C1[0, 1] positive solutions is b > 1, a + k > 1.
Acknowledgment Research supported by NNSF-China (10471077), NSF of Shandong Province (Y2004A01) and XNF of SDAI (040101).
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