Positive solutions of singular boundary. value problems for second order. impulsive differential equations

Transcription

1 Journal of Applied Mathematics & Bioinformatics, vol.4, no.2, 214, ISSN: (print), (online) Scienpress Ltd, 214 Positive solutions of singular boundary value problems for second order impulsive differential equations Ying He 1 Abstract This paper is devoted to study the positive solutions of nonlinear singular two-point boundary value problems for second-order impulsive differential equations.the existence of positive solutions is established by using the fixed point theorem in cones. Mathematics Subject Classification: 34B15 Keywords: Positive solution; Singular two-point boundary value problem; Second-order impulsive differential equations; Fixed point theorem. 1 Introduction This is the text of the introduction. Impulsive and singular differential equations play a very important role in modern applied mathematics due to 1 School of Mathematics and Statistics, Northeast Petroleum University, Daqing163318, P.R.China. Project supported by Heilongjiang Province education department natural science research item, China. heying @163.com Article Info: Received : April 18, 214. Revised : May 29, 214. Published online : June 15, 214.

2 38 Positive solutions of singular boundary value problems... their deep physical background and broad application. In this paper,we consider the existence of positive solutions of u = g(x, u), x I, u x=xk = I k (u(x k )), k = 1, 2,, m, R 1 (u) = α 1 u() + β 1 u () =, R 2 (u) = α 2 u(1) + β 2 u (1) =, (1.1) here α 1, α 2, β 2, β 1, α β 2 1 >, α β 2 2 >, I = [, 1], I = I \ {x 1, x 2,, x m }, < x 1 < x 2 < < x m < 1, R + = [, + ), g C(I R +, R + ), I k C(R +, R + ), u x=xk = u (x + k ) u (x k ), u (x + k ) (respectively u (x k )) denotes the right limit (respectively left limit) of u (x) at x = x k,g(x, u) may be singular at u =. In recent years, boundary problems of second-order differential equations with impulses have been studied extensively in the literature(see for instance [1-9] and their references).in[1],lin and Jiang studied the second-order impulsive differential equation with no singularity and obtained two positive solutions by using the fixed point index theorems in cone. However they did not consider the case when the function is singular.motivated by the work mentioned above,we study the positive solutions of nonlinear singular two-point boundary value problems for second order impulsive differential equations (1.1) in this paper.our argument is based on the fixed point theorem in cones. Moreover,for the simplicity in the following discussion,we introduce the following hypotheses. (H 1 ) : There exists an ε > such that g(x, u) and I k (u) are nonincreasing in u ε,for each fixed x [, 1] (H 2 ) : For each fixed < θ ε < g(y, ( α 1y β 1 α 2 y )θ)dy < (H 3 ) : φ 1 (x) is the eigenfunction related to the smallest eigenvalue λ 1 of the eigenvalue problem φ = λφ, R 1 (φ) = R 2 (φ) =. mp (H 4 ) : g + R 1 ( α 1 x β 1 I (k)φ 1 (x k ) α 2 x )φ 1 (x)dx < λ 1,

3 Ying He 39 where g = lim sup u + max x [,1] g(x,u), I I (k) = lim sup k (u), u u u + Theorem 1.1. Assume that (H 1 ) (H 4 ) are satisfied. Then problem (1.1) has at least one positive solution u.moreover,there exists a θ > such that u(x) θ ( α 1x β 1 ), x [, 1]. (1) 2 Preliminary Notes In order to define the solution of (1.1) we shall consider the following space. P C (I, R) = {u C(I, R); u (xk,x k+1 ) C(x k, x k+1 ), u (x k ) = u (x k ), u (x + k ), k = 1, 2,, m} with the norm u P C = max{ u, u }, here u = sup u(x), x [,1] u = sup u (x). Then P C (I, R) is a Banach space. x [,1] Definition 2.1. A function u P C (I, R) C 2 (I, R) is a solution of (1.1) if it satisfies the differential equation u + g(x, u) =, x I and the function u satisfies conditions u x=xk = I k (u(x k )) and R 1 (u) = R 2 (u) =. Let Q = I I and Q 1 = {(x, y) Q x y 1}, Q 2 = {(x, y) Q y x 1}. Let G(x, y) is the Green s function of the boundary value problem u =, R 1 (u) = R 2 (u) =. Following from [6], G(x, y) can be written by G(x, y) := { (α1 x β 1 )( α 2 y) ω, (x, y) Q 1, (α 1 y β 1 )( α 2 x) ω, (x, y) Q 2. (2.1) where ω = α 1 ( ) β 1 α 2 > It is easy to verify that G(x, y) has the following properties: (i): G(x, y), (x, y) [, 1] [, 1] (ii): G(x, y) G(y, y), (x, y) [, 1] [, 1]

4 4 Positive solutions of singular boundary value problems... (iii): G(x, y) min{ α 1x β 1, α 2+β 2 α 2 x ( α 1x β 1 Consider the linear problem } G(y, y) α 2 x ) G(y, y) (x, y) [, 1] [, 1] u (x) = λu(x), R 1 (u) = R 2 (u) =. By the theory of ordinary differential equations, we know that there exists an eigenfunction φ 1 (x) with respect to the first eigenvalue λ 1 > such that φ 1 (x) > for x (, 1). Lemma 2.2. If u is a solution of the equation u(x) = G(x, y)g(y, u(y))dy + then u is a solution of (1.1). G(x, x k )I k (u(x k )), x I. (2.2) In fact by using inequalities (i), (ii) we have that and u u(x) ( α 1x β 1 G(y, y)g(y, u(y))dy + + ( α 1x β 1 ( α 1x β 1 G(x k, x k )I k (u(x k )) ) ) G(y, y)g(y, u(y))dy G(x k, x k )I k (u(x k )) ) u, x [, 1]. 3 Main Results Lemma 3.1. Let E = (E, ) be a Banach space and let K E be a cone in E, and be increasing with respect to K. Also, r, R are constants with < r < R. Suppose that Φ : ( Ω R \Ω r ) K K ( Ω R = {u E, u < R} )

5 Ying He 41 is a continuous, compact map and assume that the conditions are satisfied: (i) Φu > x, for u Ω r K (ii) u µφ(u), for µ [, 1) and u Ω R K Then Φ has a fixed point in K {u E : r u R}. Proof. In applications below, we take E = C(I, R) and define K = {x C(I, R) : x(t) σ x, t [, 1]}. One may readily verify that K is a cone in E. Now,let r > be such that r < min{ε, G( 1 2, y)g(y, ε )dy + G( 1 2, x k)i k (ε )} (3.1) and let R > r be chosen large enough later. Let us define an operator Φ : ( Ω R \Ω r ) K K by (Φu)(x) = G(x, y)g(y, u(y))dy + G(x, x k )I k (u(x k )), x I. First we show that Φ is well defined.to see this, notice that if u ( Ω R \Ω r ) K then r u R and u(x) ( α 1x β 1 x 1. Also notice by (H 1 ) that g(x, u(x)) g(x, ( α 1x β 1 and α 2 x ) u ( α 1x β 1 α 2 x )r, )r), when u(x) r, g(x, u(x)) max max g(x, u) when r u(x) R. r u R x 1 These inequalities with (H 2 ) guarantee that Φ : ( Ω R \Ω r ) K K is well defined. have Next we show that Φ : ( Ω R \Ω r ) K K.If u ( Ω R \Ω r ) K, then we Φu (Φu)(x) ( α 1x β 1 G(y, y)g(y, u(y))dy + + ( α 1x β 1 ( α 1x β 1 ) ) G(x k, x k )I k (u(x k )) G(y, y)g(y, u(y))dy G(x k, x k )I k (u(x k )) ) Φu, x [, 1].

6 42 Positive solutions of singular boundary value problems... i.e.φu K so Φ : ( Ω R \Ω r ) K K. It is clear that Φ is continuous and completely continuous. We now show that Φu > u, for u Ω r K (3.2) To see that,let u Ω r K,then u = r and u(x) ( α 1 x β 1 for x [, 1]. So by (H 1 ) and (3.1) we have (Φu)( 1 2 ) = G( 1 m 2, y)g(y, u(y))dy + G( 1 2, x k)i k (u(x k )) so (3.2) is satisfied. > r = u. G( 1 m 2, y)g(y, r)dy + G( 1 2, x k)i k (r) G( 1 2, y)g(y, ε )dy + G( 1 2, x k)i k (ε ) α 2 x )r On the other hand, from (H 4 ), there exist < ε < λ 1 f and H > p such that (λ 1 ε g ) ( α 1x β 1 )φ 1 (x)dx > (I (k) + ε)φ 1 (x k ); g(x, u) (g + ε)u, I k (u) (I (k) + ε)u x [, 1], u H. (3.3) Let C = max max g(x, u) + m max I k(u), it is clear that r u H x 1 r u H g(x, u) g(x, ( α 1x β 1 )r) + C + (g + ε)u, I k (u) I k (( α 1x β 1 )r) + C + (I (k) + ε)u, x [, 1], u. Next we show that if R is large enough,then µφu u for any u K Ω R and µ < 1. If this is not true,then there exist u K Ω R and µ < 1 such that µ Φu = u. Thus u = R > r and u (x) ( α 1x β 1 α 2 x )R.Note that u (x) satisfies u (x) + µ g(x, u (x)) =, x I, u x=xk = µ I k (u (x k )), k = 1, 2,, m, α 1 u () + β 1 u () = α 2 u (1) + β 2 u (1) =. (3.4)

7 Ying He 43 Multiply equation (3.4) by φ 1 (x) and integrate from to 1, using integration by parts in the left side, notice that x1 φ 1 (x)u (x)dx = φ 1 (x)du (x) + = φ 1 (x 1 )u (x 1 ) φ 1 ()u () + x1 m 1 xk+1 u (x)φ 1(x)dx m 1 [φ 1 (x k+1 )u (x k+1 ) φ 1 (x k )u (x k + ) + φ 1 (1)u (1) φ 1 (x m )u (x m + ) = u (x k )φ 1 (x k ) x k φ 1 (x)du (x) + xk+1 x k x m u (x)φ 1(x)dx u (x)φ 1(x)dx] φ 1(x)u (x)dx + φ 1 (1)u (1) φ 1 ()u (). x m φ 1 (x)du (x) Also notice that φ 1(x)u (x)dx = φ 1(x)du (x) = φ 1(1)u (1) φ 1()u () u (x)φ 1(x)dx = φ 1(1)u (1) φ 1()u () + λ 1 u (x)φ 1 (x)dx. thus,by the boundary conditions, we have φ 1 (x)u (x)dx = u (x k )φ 1 (x k ) φ 1(1)u (1) + φ 1()u () λ 1 u (x)φ 1 (x)dx + φ 1 (1)u (1) φ 1 ()u () = = u (x k )φ 1 (x k ) λ 1 u (x)φ 1 (x)dx µ I k (u (x k ))φ 1 (x k ) λ 1 u (x)φ 1 (x)dx.

8 44 Positive solutions of singular boundary value problems... So we obtain + + C λ 1 u (x)φ 1 (x)dx = µ m (I (k) + ε)φ 1 (x k )u (x k ) + C I k (u (x k ))φ 1 (x k ) + µ g(x, u (x))φ 1 (x)dx φ 1 (x k ) I k ( α 1x β 1 )r)φ 1 (x k ) + (g + ε) φ 1 (x)dx + Consequently,we obtain that + (λ 1 g ε) φ 1 (x)g(x, ( α 1x β 1 φ 1 (x)g(x, ( α 1x β 1 u (x)φ 1 (x)dx )r)dx φ 1 (x)u (x)dx (I (k) + ε)φ 1 (x k )u (x k ) )r)dx + C( φ 1 (x k ) + + I k ( α 1x β 1 )r)φ 1 (x k ) u (I (k) + ε)φ 1 (x k ) + φ 1 (x)g(x, ( α 1x β 1 + C( φ 1 (x k ) + φ 1 (x)dx) + I k ( α 1x β 1 We also have Thus u =: R. u (x)φ 1 (x)dx u R 1 φ 1(x)g(x,( α 1 x β 1 ( α 1x β 1 φ 1 (x)dx) )r)dx )r)φ 1 (x k ) )φ 1 (x)dx α 2 x P )r)dx+c( m R 1 P φ α 2 +β 1 (x k )+ 2 φ 1(x)dx)+ m I k ( α 1 x β 1 α 2 x )r)φ α 1 β 1 α 2 +β 1 (x k ) 2 (λ 1 g ε) R 1 ( α 1 x β 1 α 2 x P )φ 1 (x)dx m (I (k)+ε)φ 1 (x k ) Let R > max{r, H}, then for any u K Ω R and µ < 1, we have µφu u. Hence all the assumptions of Lemma 3.1 are satisfied,then Φ has a fixed point u in K {u E : r u R}, u(x) ( α 1x β 1 x [, 1]. Let θ := r, this we complete the proof of Theorem 1. α 2 x )r

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