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Journal of Computational and Applied Mathematics 141 (2002) 65–73 www.elsevier.com/locate/cam

Existence of three positive solutions to a second-order boundary value problem on a measure chain Richard I. Averya , Douglas R. Andersonb;∗

b

a College of Natural Sciences, Dakota State University, Madison, SD 57042, USA Department of Mathematics and Computer Science, Concordia College, Moorhead, MN 56562, USA

Received 24 August 2000; received in revised form 18 December 2000

Abstract Growth conditions are imposed on f such that the boundary value problem −x11 (t) = f(x (t)), t ∈ [t1 ; t2 ], x(t1 ) − x (t1 ) = 0 and x((t2 )) + x1 ((t2 )) = 0, where t1 ¡ t2 from a measure chain T, has at least three positive solutions c 2002 Elsevier Science B.V. All rights reserved. by way of the 6ve functionals 6xed point theorem.  1

MSC: 39A10; 39A99; 34B99 Keywords: Measure chains; Boundary value problem

1. Introduction Since Hilger’s [24] initial paper unifying continuous and discrete calculus, attention is being given to di=erential equations on measure chains (time scales). To facilitate this, the calculus on measure chains has been developed by Agarwal and Bohner [1], Aulbach and Hilger [5], and Erbe and Hilger [15]. Additionally, there is much current attention paid to questions of positive solutions of boundary value problems for ordinary di=erential equations, as well as for 6nite di=erence equations. Many of these works have used Krasnosel’skii [25] 6xed point theorems, the Leggett and Williams [26] multiple 6xed point theorem, or recently the so-called 6ve functionals 6xed point theorem [7], to obtain positive solutions or multiple positive solutions inside a cone. The recent book by Agarwal et al. [2] gives a good overview for much of that work and the methods used. Other papers that have been a driving force behind this interest on positive solutions are [3,4,6,8,9,12,13,16,20,21,27]. ∗

Corresponding author. E-mail addresses: [email protected] (R.I. Avery), [email protected] (D.R. Anderson). c 2002 Elsevier Science B.V. All rights reserved. 0377-0427/02/$ - see front matter  PII: S 0 3 7 7 - 0 4 2 7 ( 0 1 ) 0 0 4 3 6 - 8

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There has been some natural merging of methods in seeking positive solutions of boundary value problems for di=erential equations on measure chains; see Chyan and Henderson [10,11], Henderson [23], and Erbe and Peterson [17–19]. In this paper we are concerned with the existence of at least three positive solutions for a second-order di=erential equation on a measure chain satisfying Sturm-Liouville-like boundary conditions. The results are new for the special cases of di=erence equations and di=erential equations, as well as in the general time scale setting. Before introducing this boundary value problem, we present some background material on the 6ve functionals 6xed point theorem. 2. Five functionals theorem Denition 1. Let E be a real Banach space. A nonempty closed convex set P ⊂ E is called a cone if it satis6es the following two conditions: (i) x ∈ P;  ¿ 0 implies x ∈ P; (ii) x ∈ P; −x ∈ P implies x = 0. Every cone P ⊂ E induces an ordering in E given by x6y

if and only if y − x ∈ P:

Denition 2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets. Denition 3. A map  is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if  : P → [0; ∞) is continuous and (tx + (1 − t)y) ¿ t(x) + (1 − t)(y) for all x; y ∈ P and t ∈ [0; 1]. Similarly we say the map  is a nonnegative continuous convex functional on a cone P of a real Banach space E if  : P → [0; ∞) is continuous and (tx + (1 − t)y) 6 t(x) + (1 − t)(y) for all x; y ∈ P and t ∈ [0; 1]. Let ; ;  be nonnegative continuous convex functionals on P and , be nonnegative continuous concave functionals on P. For nonnegative real numbers h, a, b, k and c we de6ne the following

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convex sets: P(; c) = {x ∈ P: (x) ¡ c}; P(; ; b; c) = {x ∈ P: b 6 (x); (x) 6 c}; Q(; ; a; c) = {x ∈ P: (x) 6 a; (x) 6 c}; P(; ; ; b; k; c) = {x ∈ P: b 6 (x); (x) 6 k; (x) 6 c}; and Q(; ; ; h; a; c) = {x ∈ P: h 6 (x); (x) 6 a; (x) 6 c}: The following theorem is the Five Functionals Fixed Point Theorem [7], which is a generalization of the Leggett-Williams Fixed Point Theorem [26]. Theorem 4. Let P be a cone in a real Banach space E. Suppose there exist positive numbers c and M; nonnegative continuous concave functionals  and on P; and nonnegative continuous convex functionals ; ; and  on P; with (x) 6 (x)

and

x 6 M(x)

for all x ∈ P(; c). If A : P(; c) → P(; c) is completely continuous and there exist nonnegative numbers h; a; k; b with 0 ¡ a ¡ b such that: (i) (ii) (iii) (iv)

{x ∈ P(; ; ; b; k; c): (x) ¿ b} = ∅ and (Ax) ¿ b for x ∈ P(; ; ; b; k; c); {x ∈ Q(; ; ; h; a; c): (x) ¡ a} = ∅ and (Ax) ¡ a for x ∈ Q(; ; ; h; a; c); (Ax) ¿ b for x ∈ P(; ; b; c) with (Ax) ¿ k; (Ax) ¡ a for x ∈ Q(; ; a; c) with (Ax) ¡ h.

Then A has at least three 7xed points x1 ; x2 ; x3 ∈ P(; c) such that (x1 ) ¡ a; b ¡ (x2 ); and a ¡ (x3 )

with (x3 ) ¡ b:

3. Second-order BVP Throughout the remainder of the paper, let the measure chain T be a nonempty closed subset of R with [a; b]:={t ∈ T: a 6 t 6 b}, and let T have the subspace topology inherited from the Euclidean topology on R. We are interested in the second-order two-point boundary value problem −x11 (t) = f(x (t));

t ∈ [t1 ; t2 ];

(1)

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x(t1 ) − x1 (t1 ) = 0

(2)

x((t2 )) + x1 ((t2 )) = 0

where t1 ¡ t2 ∈ T. The Green’s function [17] for the related homogeneous problem −x11 (t) = 0 with boundary conditions (2) is given by 1 {(t − t1 ) + }{((t2 ) − (s)) + }; t 6 s; d (3) G(t; s) = 1 {((s) − t ) + }{((t ) − t) + }; (s) 6 t; 1 2 d for t ∈ [t1 ; 2 (t2 )] and s ∈ [t1 ; t2 ], where ; ; ; ¿ 0 and d:= +  + ((t2 ) − t1 ) ¿ 0: Note that solutions of (1) and (2) are 6xed points of the operator A de6ned by  (t2 ) Ax(t) = G(t; s)f(x (s))1s: t1

Applying a technique similar to the technique of Avery and Peterson in [9] one can show that if y is a 6xed point of the operator B de6ned by  (t2 )  By(t) = f G((t); s)y(s)1s t1

then

 x(t) =

(t2 )

t1

G(t; s)y(s)1s

is a 6xed point of the operator A, hence is a solution of (1) and (2). Let E = Crd ([t1 ; (t2 )]) denote the set of right-dense continuous functions on [t1 ; (t2 )] endowed with the supremum norm, y =

sup

t1 6t 6(t2 )

|y(t)|;

and de6ne the cone P ⊂ E by P = {y ∈ E: y is nonnegative valued on [t1 ; (t2 )]}: Suppose %1 ; %2 ; %3 ; %4 ∈ (t1 ; t2 ) with %1 ¡ %2 ¡ %3 ¡ %4 : De6ne the nonnegative continuous concave functionals , functionals ; ;  on the cone P by (y):=y; (y):= min y(t); t ∈[%1 ;%4 ]

and the nonnegative continuous convex

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69

(y):= max y(t); t ∈[%1 ;%4 ]

(y):= min y(t); t ∈[%2 ;%3 ]

(y):= max y(t): t ∈[%2 ;%3 ]

We will make use of the following constants  (t2 ) G((t); s)1s; M0 := max t ∈[t1 ;(t2 )]



M1 := min

t ∈[%2 ;%3 ]

M2 := max

t ∈[%2 ;%3 ]

t1

%3

%2



%3

%2

 M3 := max

t ∈[%2 ;%3 ]

M4 := min

t ∈[%1 ;%4 ]

M5 := max

t ∈[%1 ;%4 ]



%4

%4

%1

 M6 := max

t ∈[%1 ;%4 ]

G((t); s)1s; %2

t1

%1



G((t); s)1s;

t1

 G((t); s)1s +

(t2 )

%3

 G((t); s)1s ;

G((t); s)1s; G((t); s)1s; %1

 G((t); s)1s +

(t2 )

%4

 G((t); s)1s :

We are now ready to apply the Five Functionals Fixed Point Theorem to the operator B to give suLcient conditions for the existence of at least three positive solutions to (1) and (2). Theorem 5. Suppose there exist nonnegative numbers h; a; b; k and c such that 0 ¡ h ¡ a ¡ b ¡ k 6 c; and suppose f satis7es the following conditions: (i) (ii) (iii) (iv)

0 6 f(w) 6 c for all w ∈ [0; cM0 ]; f(w) ¡ a for w ∈ [hM4 ; aM5 + cM6 ]; f(w) ¿ b for w ∈ [bM1 ; kM2 + cM3 ]; −f(w0 ) −b | f(ww11)− | 6 cMk0 − for all w0 ; w1 ∈ [bM1 ; cM0 ] with w0 = w1 ; w0 bM1

−f(w2 ) h | 6 aMa5− for all w2 ; w3 ∈ [0; aM5 + cM6 ] with w2 = w3 . (v) | f(ww33)− w2 +cM6

Then; the second order boundary value problem (1) and (2) has three positive solutions x1 ; x2 ; x3 such that  (t2 ) G(t; s)yi (s)1s xi (t) = t1

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where yi ∈ P(; c) for i = 1; 2; 3 with (y1 ) ¡ a (y2 ) ¿ b; and (y3 ) ¿ a

with (y3 ) ¡ b:

Proof. Let y ∈ P(; c) then  (t2 )  06 G((t); s)y(s)1s 6 c t1

(t2 )

t1

G((t); s)1s 6 M0 c

hence by condition (i) for all t ∈ [t1 ; (t2 )] we have 0 6 By(t) 6 c thus B : P(; c) → P(; c) and clearly B is a completely continuous operator. Let b+k a+h and yQ (t) ≡ yP (t) ≡ 2 2 for all t ∈ [t1 ; (t2 )], then yP ∈ {y ∈ P(; ; ; b; k; c): (y) ¿ b}

and

yQ ∈ {y ∈ Q(; ; ; h; a; c): (y) ¡ a}

hence both sets are nonempty. Suppose y ∈ P(; ; ; b; k; c) then  %3 bM1 = b min G((t); s)1s t ∈[%2 ;%3 ]

 6 max

t ∈[%2 ;%3 ]

%2

(t2 )

t1

 6 k max

t ∈[%2 ;%3 ]

%3

%2

G((t); s)y(s)1s 

G((t); s)1s + c max

= kM2 + cM3 hence by condition (iii) By(t) ¿ b: If y ∈ P(; ; b; c) with (By) ¿ k then let  (t2 ) w1 := G((z1 ); s)y(s)1s t1

t ∈[%2 ;%3 ]

t1

%2

 G((t); s)1s +

(t2 )

%3

 G((t); s)1s

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where z1 ∈ [%2 ; %3 ] with (By) = By(z1 ) and similarly let  (t2 ) w0 := G((z0 ); s)y(s)1s t1

where z0 ∈ [%2 ; %3 ] with (By) = By(z0); here the existence of z0 and z1 follows from a continuity (t ) argument for the parameter integral t1 2 G((·); s)y(s)1s. Then we have w0 ; w1 ∈ [bM1 ; cM0 ]. If w0 = w1 then (By) ¿ k ¿ b, and if w0 = w1 then (By) − (By) = |f(w1 ) − f(w0 )|    f(w1 ) − f(w0 )   (cM0 − bM1 ) 6   w1 − w 0 6k −b by condition (iv). Consequently 0 ¡ (By) − k 6 (By) − b; implying (By) ¿ b: Now suppose y ∈ Q(; ; ; h; a; c); then  %4 hM4 = h min G((t); s)1s t ∈[%1 ;%4 ]

 6 max

t ∈[%1 ;%4 ]

%1

(t2 )

t1

 6 a max

t ∈[%1 ;%4 ]

%4

%1

G((t); s)y(s)1s 

G((t); s)1s + c max

t ∈[%1 ;%4 ]

t1

%1

 G((t); s)1s +

(t2 )

%4

 G((t); s)1s

= aM5 + cM6 hence by condition (ii) By(t) ¡ a: If y ∈ Q(; ; a; c) with (By) ¡ h then let  (t2 ) w3 := G((z3 ); s)y(s)1s; t1

where z3 ∈ [%1 ; %4 ] with (By) = By(z3 ) and similarly let  (t2 ) w2 := G((z2 ); s)y(s)1s; t1

By(z2 ); the existence of z2 and z3 follows from a continuity argument where z2 ∈ [%1 ; %4 ] with (By)  (t= 2) for the parameter integral t1 G((·); s)y(s)1s. Then we have w2 ; w3 ∈ [0; aM5 + cM6 ]. If w2 = w3

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then (By) ¡ h ¡ a, and if w2 = w3 then (By) − (By) = |f(w3 ) − f(w2 )|    f(w3 ) − f(w2 )   (aM5 + cM6 ) 6   w3 − w 2 6a−h by condition (v). As a result (By) − a 6 (By) − h ¡ 0 which yields (By) ¡ a: Thus the hypotheses of the Five Functionals Fixed Point Theorem are satis6ed hence there exists 6xed points y1 ; y2 ; y3 ∈ P(; c) for B with (y1 ) ¡ a (y2 ) ¿ b and (y3 ) ¿ a with (y3 ) ¡ b: Therefore the boundary value problem (1), (2) has at least three positive solutions of the form  (t2 ) xi (t) = G(t; s)yi (s)1s t1

for i = 1; 2; 3. Uncited References [14,22] References [1] R.P. Agarwal, M. Bohner, Basic calculus on time scales and some of its applications, Results Math. 35 (1999) 3–22. [2] R.P. Agarwal, D. O’Regan, P.J.Y. Wong, Positive Solutions of Di=erential, Di=erence and Integral Equations, Kluwer Academic, Dordrecht, 1999. [3] D. Anderson, R.I. Avery, A.C. Peterson, Three positive solutions to a discrete focal boundary value problem, J. Comput. Appl. Math. 88 (1998) 103–118. [4] D. Anderson, R.I. Avery, Multiple positive solutions to a third order discrete focal boundary value problem, Comput. Math., Appl. 42 (2001) 333–340. [5] B. Aulbach, S. Hilger, Linear dynamic processes with inhomogeneous time scale, in: Nonlinear Dynamics and Quantum Dynamical Systems, Mathematical Research, vol. 59, Akademie Verlag, Berlin, 1990. [6] R.I. Avery, Multiple Positive Solutions to Boundary Value Problems, Dissertation, University of Nebraska-Lincoln, 1997.

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