1、 No.4 S.G. Kang, etc., TRIPLE POSITIVE SOLUTIONS 401 Now, using the fact that the functions t and t1 are uniformly continuous on the interval 0, 1, we conclude that A(P c is equicontinuous. Therefore, A is completely continuous on P c. We now assert that (H2 of Theorem 2.1 holds. Indeed, 1 1 (Au(t =

2、 0 G(t, sh(sf (u(sds f ( u 0 1 G(t, sh(sds (20 f (d 0 G(t, sh(sds 2 Af (d < d, for all u P d , where the last inequality follows from (15. In addition, we can show that condition (H1 of Theorem 2.1 holds. Obviously (u is a concave continuous function on P with (u u for u P c . We notice that if u

3、(t = (a + b/2 for t 0, 1, then u u P (, a, b : (u > a which implies that u P (, a, b : (u > a is nonempty. For u P (, a, b, we have (u = min u(t a t0,1 and u b. In view of (S1-(S3, we have 1 (Au = min t0,1 G(t, sh(sf (u(sds 0 1 f (a min t0,1 G(t, sh(sds f (aB > 1 f (aB = a, 0 (21 for all u

4、P (, a, b. Finally, we prove that condition (H3 in Theorem 2.1 holds. Let u P (, a, c with Au > b. We notice that (16 implies 1 Au M 0 h(sf (u(sds. (22 Thus 1 1 (Au = min t0,1 G(t, sh(sf (u(sds m 0 0 h(sf (u(sds m m Au > b > a. (23 M M An application of Theorem 2.1 yields our proof. Theorem

5、 3.2 Suppose that (S1-(S3 hold such that 0 < l1 < l2 and there is a number a > 0 such that f (a > 0 and 0 < l1 < l2 < Bf (a/(Aa. For each (a/(Bf (a, 1/(Al2 , choose ve positive numbers , c, b, a and d such that a 1 1 < < , Bf (a A(l2 + A0 l2 aM a > d > 0, cb> m f

6、(c < l2 + c and f (d < l1 + . d (24 (25 (26 402 ANN. OF DIFF. EQS. Vol.28 Then for this , equation (1 has at least three nonnegative solutions u1 , u2 and u3 such that t0,1 max u1 (t < d, a < min u2 (t < max u2 (t < c and d < max u3 (t c, t0,1 t0,1 t0,1 t0,1 min u3 (t < a. Pr

7、oof First of all, A, B > 0. If f (a > 0 then by (S1, f (c > 0 for any c greater than a. Let 1 = a/(Bf (a and 2 = 1/(Al2 . Then 1 , 2 > 0. Furthermore, 0 < 1 < 2 in view of the condition 0 < l1 < l2 < Bf (a/(Aa. For any positive that satises (24, in view of (S3, there is a

8、c with c b > a such that (25 holds, and in view of (S2, there is a d (0, a such that (26 holds. We now dene a continuous mapping A : P P for each (1 , 2 by (16 and a functional : P 0, by (17. As in the proof of Theorem 3.1, it is easy to see that A is completely continuous on P c and maps P c int

9、o P c . For all u P c , we have 1 1 (Au(t = 0 G(t, sh(sf (u(sds f ( u 0 1 G(t, sh(sds (27 f (c 0 G(t, sh(sds f (cA < A(l2 + c c. Furthermore, condition (H2 of Theorem 2.1 holds. Indeed, for u P d , we have 1 1 (Au(t = 0 G(t, sh(sf (u(sds f ( u 0 1 G(t, sh(sds (28 f (d 0 G(t, sh(sds f (dA A(l1 + d

10、 < d. In addition, we can show that condition (H1 of Theorem 2.1 holds. From the proof of Theorem 3.1 we can get that (u is a concave continuous function on P with (u u for u P c . We notice that if u(t = (a + b/2 for t 0, 1, then u u P (, a, b : (u > a which implies that u P (, a, b : (u >

11、 a is nonempty. For u P (, a, b, we have (u = min u(t a and u b. In view of (S1-(S3, we have t0,1 1 1 (Au = min t0,1 G(t, sh(sf (u(sds f (a min 0 t0,1 G(t, sh(sds 0 f (aB > 1 f (aB = a. (29 Finally, we prove condition (H3 of Theorem 2.1. Let u P (, a, c with Au > b. (16 implies 1 Au M 0 h(sf (

12、u(sds. 1 (30 Thus 1 (Au = min t0,1 G(t, sh(sf (u(sds m 0 0 h(sf (u(sds m m Au > b > a. (31 M M An application of Theorem 2.1 yields our proof. Theorem 3.3 Suppose that (S1 holds and f (0 > 0. There are numbers a > 0, c, b and d such that f (a > 0 and (14 and (15 hold. Then for each (a

13、/(Bf (a, c/(Af (a, equation (1 has at least three positive solutions u1 , u2 and u3 such that No.4 S.G. Kang, etc., TRIPLE POSITIVE SOLUTIONS 403 t0,1 max u1 (t < d, a < min u2 (t max u2 (t < c and d < max u3 (t c, t0,1 t0,1 t0,1 t0,1 min u3 (t < a. The proof is similar to that of The

14、orem 3.1 and hence is omitted. 4 An Example Example 4.1 Consider the following fractional boundary value problem 2 u(t + h(tf (u(t = 0, D0+ 7 (32 u(0 = u (0 = u (0 = u (1 = 0, is the standard Riemann-Liouville fractional where D0+ u ln(1 + u, f (u = eu e + ln 2, u 6 + e6 e + ln 2, derivative and 0 u

15、 < 1, 1 u < 6, u 6. Letting h(t = 1, d = 3/2, a = 5, b = 101 and c = 5.0135 × 1041 , we easily check that A = 0.1146, B = 0.0023 and all the conditions of Theorem 3.1 hold. Thus by Theorem 3.1, the boundary value problem (32 has three positive solutions. References 1 Shuqin Zhang, The exi

16、stence of a positive solution for a nonlinear fractional dierential equation, J. Math. Anal. Appl., 252(2000,804-812. 2 Shuqin Zhang, Existence of positive solution for some class of nonlinear fractional dierential equations, J. Math. Anal. Appl., 278:1(2003,136-148. 3 S. Zhang, Positive solutions f

17、or boundary value problems of nonlinear fractional dierential equations, Electronic Journal of Dierential Equations, 2006:36(2006,1-12. 4 C. Bai, Positive solutions for nonlinear fractional dierential equations with coecient that changes sign, Nonlinear Anal., 64(2006,677-685. 5 S. Zhang, Existence

18、of solution for a boundary value problem of fractional order, Acta Math. Sci., 26(2006,220-228. 6 Zhanbing Bai, Haishen Lv, Positive solutions for boundary value problem of nonlinear fractional dierential equation, J. Math. Anal. Appl., 311(2005,495-505. 7 M. El-Shahed, Positive solutions for boundary value problem of nonlinear fracti