Existence of positive solutions for fourth-order semipositone multi-point boundary value problems with a sign-changing n
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Existence of positive solutions for fourth-order semipositone multi-point boundary value problems with a sign-changing nonlinear term Yan Sun Correspondence: [email protected]. cn Department of Mathematics, Shanghai Normal University, Shanghai 200234, People’s Republic of China
Abstract In this article, some new sufficient conditions are obtained by making use of fixed point index theory in cone and constructing some available integral operators together with approximating technique. They guarantee the existence of at least one positive solution for nonlinear fourth-order semipositone multi-point boundary value problems. The interesting point is that the nonlinear term f not only involve with the first-order and the second-order derivatives explicitly, but also may be allowed to change sign and may be singular at t = 0 and/or t = 1. Moreover, some stronger conditions that common nonlinear term f ≥ 0 will be modified. Finally, two examples are given to demonstrate the validity of our main results. 2000 Mathematics Subject Classification: 34B10; 34B18; 47N20. Keywords: semipositone, positive solutions, multi-point boundary value problems
1 Introduction In this article, we consider the existence of positive solutions to the following nonlinear fourth-order semipositone multi-point boundary value problems with derivatives ⎧ (4) 0 < t < 1, ⎨ y (t) + λf (t, y(t), y (t), y (t)) = 0, m−2 m−2 αi y (ξi ), y (0) = βi y (ξi ), ⎩ y(0) = y (0) = 0, y (1) = i=1
(1:1)
i=1
where f Î C((0, 1) × R×R×R, R) satisfies f(t, y1 y2, y3) ≥ -p(t), p Î L1 ((0,1), (0, +∞)). l > 0, ξi Î (0, 1) with 0 0.
4 Examples Example 4.1. Consider the following singular fourth-order semipositone boundary value problem: ⎧ 8 4λ ⎪ ⎪ y(4) (t) + " sin (|y(t)| + |y (t)|) + e|y(t)|+|y (t)|+|y (t)| ⎪ ⎪ 3 2 ⎪ ⎪ 3 (1 − t) ⎨ 1 2 + (|y(t)| + |y (t)| + |y (t)|) 3 ] − √ = 0, t ∈ (0, 1) ⎪ ⎪ ⎪ t ⎪ ⎪ ⎪ ⎩ y(0) = y (0) = 0, y (1) = 1 y ( 1 ), y (0) = 1 y ( 1 ) 3 2 4 2
(4:1)
Proof. Let # $ 1 4 2 f (t, u1 , u2 , u3 ) = " sin8 (|u1 | + |u2 |) + e|u1 |+|u2 |+|u3 | + (|u1 | + |u2 | + |u3 |) 3 − √ 3 t 2 3 (1 − t)
Then −p(t) ≤ f (t, u1 , u2 , u3 ) ≤ q(t)g(u1 , u2 , u3 ) and
lim
(|u1 |+|u2 |+|u3 |)→+∞
where
f (t, u1 , u2 , u3 ) = +∞ |u1 | + |u2 | + |u3 |
p(t) = 1
1 √ , q(t) t
4 = √ 3 3
(1−t)2
,
g(u1 , u2 , u3 ) = sin8 (|u1 | + |u2 |) + e|u1 |+|u2 |+|u3 | + (|u1 | + |u2 | + |u3 |) 3, which implies that
(H1)-(H3) hold. Since α = 13 , β = 14 , ξ = 12, then we know that
Sun Boundary Value Problems 2012, 2012:12 http://www.boundaryvalueproblems.com/content/2012/1/12
1
1 (p(s) + q(s))ds =
0
0
Page 14 of 16
⎛
⎞
4 ⎜ 2 ⎟ ⎝√ + " ⎠ ds = 8. s 3 3 (1 − s)2
Take r = 2, then M1 = =
max
{g(|u1 |, |u2 |, |u3 |)} + 1 # $% 1 sin8 (|u1 | + |u2 |) + e|u1 |+|u2 |+|u3 | + (|u1 | + |u2 | + |u3 |) 3
(|u1 |,|u2 |,|u3 |)∈[0,2]×[0,2]×[0,2]
max √ 3 +1 = 2 + e6 + 6.
(|u1 |,|u2 |,|u3 |)∈[0,2]×[0,2]×[0,2]
⎛
i=1
m−2 βi 1 − αi ξi
⎞
m−2 1 − α ξ i i ⎟ ⎜ 1 115 ⎟ ⎜ i=1 i=1 i=1 G=⎜ + . = ⎟
m−2 m−2 m−2
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