Avances in Dynamical Systems an Applications ISSN 973-532, Volume 4, Number 2, pp. 243 253 (29) http://campus.mst.eu/asa Multiplicity Results of Positive Solutions for Nonlinear Three-Point Bounary Value Problems on Time Scales Bin Zhao Tianshui Normal University Department of Mathematics an Statistics Tianshui, Gansu 74, P. R. China Hong-Rui Sun Lanzhou University School of Mathematics an Statistics Lanzhou, Gansu 73, P. R. China. hrsun@lzu.eu.cn Abstract Let T be a time scale such that, T T. In this paper, we consier the nonlinear secon-orer three-point bounary value problem u (t) + q(t)f(t, u(t), u (t)) =, t (, T ) T βu() γu () =, αu(η) = u(t ), where β, γ, β + γ >, η (, ρ(t )) T, < α < T/η, an = β(t αη) + γ( α) >. By using a fixe point theorem ue to Avery an Peterson, sufficient conitions are obtaine for the existence of three positive solutions of the above problem. The interesting point is the nonlinear term f which is involve with the first orer erivative explicitly. AMS Subject Classifications: 34B5, 39A. Keywors: Positive solution, fixe point theorem, time scales. Introuction Going back to its founer Stefan Hilger (988), the stuy of ynamic equations on time scales is a fairly new area of mathematics. Motivating the subject is the notion that Receive April 28, 28; Accepte April 3, 29 Communicate by Martin Bohner
244 Bin Zhao an Hong-Rui Sun ynamic equations on time scales can buil briges between ifferential equations an ifference equations. Now, the stuy of time scales theory has le to many important applications, for example, in the stuy of insect population moels, neural networks, heat transfer, quantum mechanics, epiemic, crop harvest an stock market [3, 5, 6, 8]. Atici et al present a ynamic optimization problem from economics [3], by constructing a time scale moel, they tol us time scale calculus coul allow exploration of a variety of situations in economics. Very recently, the existence problems of positive solutions for three-point bounary value problems, especially on time scales, have attracte many authors attention an concern [, 2, 5 7, 9 4]. But the nonlinear term f is not involve with the first-orer elta erivative. Many ifficulties occur when the nonlinear term f is involve with the first-orer elta erivative. In [], Sun an Li consiere the existence of positive solution of the problem u (t) + a(t)f(t, u(t)) =, t (, T ) T, βu() γu () =, αu(η) = u(t ) by using fixe point theorem in cones. The authors obtaine the existence of at least one, two an three positive solutions of the above problems. In [7], by a new fixe point theorem, Guo an Ge gave sufficient conitions for the existence of at least one solution to the three-point bounary value problem u (t) + f(t, u, u ) =, < t <, u() =, u() = αu(η). Motivate by those works, in this paper we stuy the existence of multiple positive solutions for the secon-orer nonlinear three-point ynamic equation on time scales u (t) + q(t)f(t, u(t), u (t)) =, t (, T ) T, (.) βu() γu () =, αu(η) = u(t ), (.2) where β, γ, β + γ >, η (, ρ(t )) T, < α < T/η, an = β(t αη) + γ( α) >. We assume a : [, T ] T [, ) is l-continuous, with a (t ) > for at least one t (η, T ) T ; f : [, T ] T [, ) R [, ) is continuous. We establish the existence of at least three positive solutions for bounary value problem (.) an (.2). To our best knowlege, no work has been one for (.) an (.2) on time scales by using the fixe point theorem of Avery an Peterson. 2 Preliminaries In a cone P, let φ an θ be nonnegative continuous convex functionals, ϕ a nonnegative continuous concave functional an ψ a nonnegative continuous functional. Then for
Positive Solutions for BVP on Time Scales 245 positive real numbers m, m 2, m 3, m 4, we efine the following convex sets: P (φ, m 4 ) = {u P φ(u) < m 4 }, P (φ, ϕ, m 2, m 4 ) = {u P m 2 ϕ(u), φ(u) m 4 }, P (φ, θ, ϕ, m 2, m 3, m 4 ) = {u P m 2 ϕ(u), θ(u) m 3, φ(u) m 4 }, an a close set R(φ, ψ, m, m 4 ) = {u P m ψ(u), φ(u) m 4 }. Lemma 2. (see [4]). Let P be a cone in a real Banach space E. Let θ an φ be nonnegative continuous convex functionals, ϕ a nonnegative continuous concave functional an ψ a nonnegative continuous functional satisfying ψ(λu) λψ(u) for λ, such that for some positive numbers M an m 4, for all u P (φ, m 4 ). Suppose that ψ(u) ψ(u), u Mφ(u) A : P (φ, m 4 ) P (φ, m 4 ) is completely continuous an there exist positive numbers m, m 2 an m 3 with m < m 2 such that (S) {u P (φ, θ, ϕ, m 2, m 3, m 4 ) ϕ(u) > m 2 } =, an ϕ(au) > m 2 for u P (φ, θ, ϕ, m 2, m 3, m 4 ); (S2) ϕ(au) > m 2 for u P (φ, ϕ, m 2, m 4 ) with θ(au) > m 3 ; (S3) / R(φ, ψ, m, m 4 ) an ψ(au) < m for u P (φ, ψ, m, m 4 ) with ψ(u) = m. Then A has at least three fixe points u, u 2, u 3 P (φ, m 4 ), such that φ(u i ) m 4 for i =, 2, 3; m 2 < ϕ(u ); m < ψ(u 2 ) with ϕ(u 2 ) < m 2 ; ψ(u 3 ) < m. 3 Main Results Now we efine the real Banach space E = C ([, T ] T ) as the set of all -ifferentiable functions with continuous -erivative on [, T ] T with the norm u = max{ u, u }, u E,
246 Bin Zhao an Hong-Rui Sun where u := sup{ u(t) : t [, T ] T }, u := sup{ u (t) : t [, T ] T κ}. We efine the cone P E by P = {u E : u(t) is nonnegative an concave on [, T ] T, βu() γu () =, αu(η) = u(t ) }. The following lemma will play an important role in the proof of our main results. Lemma 3.. For u P, there exists a constant M > such that max u(t) M max u (t). t [,T ] T t [,T ] T κ Proof. We first consier the case β >. In this case, by the concavity of u, we have u(t) u() tu () T max u (t). t [,T ] T κ From the bounary conition βu() γu () =, we know u() = γ β u (), thus max t [,T ] T u(t) γ β u () + T Next we consier the case β =. First, by we have ( ) γ max u (t) t [,T ] T κ β + T u(t) = u(t ) t u (s) s, max u (t). t [,T ] T κ max u(t) u(t ) + T max u (t). (3.) t [,T ] T t [,T ] T κ From the bounary conition αu(η) = u(t ) an = γ( α) >, we get ( α)u(t ) = αu(η) αu(t ) α(t η) max u (t). (3.2) t [,T ] T κ In view of (3.) an (3.2), we obtain ( ) α(t η) max u(t) t [,T ] T α + T The proof is complete. max u (t) = t [,T ] T κ ( ) T αη α max u (t). t [,T ] T κ
Positive Solutions for BVP on Time Scales 247 For the sake of applying Lemma 2., we let the nonnegative continuous concave functional ϕ, the nonnegative continuous convex functional θ, φ, an the nonnegative continuous functional ψ be efine on the cone P by ϕ(u) = min u(t), φ(u) = max u (t), ψ(u) = θ(u) = max u(t), u P. t [,T ] T κ t [,T ] T For convenience, we introuce the following notations. { β T S = max (T s) q(s) s, q(s) s + αβ η } (η s) q(s) s, L = min{, a}(βη + γ) η (T s) q(s) s, N = βt + γ { α(t η) r = min T αη, αη T, η T }, a = (T s) q(s) s, min{, α}(t η)(βηt + γη), b = at 2 + a [ β(t 2 αη 2 ) + γ( α)σ() ] T +a [ γ(β(t 2 αη 2 ) γ(t αη)σ() ] + r, c = (T + σ(t )) + β(t 2 αη 2 ) + γ( α)σ(). Theorem 3.2. If there exist positive numbers m, m 2 an m 4 with m < m 2 m 4 /c such that (C) f(t, u, v) m 4 /S for (t, u, v) [, T ] T [, Mm 4 ] [ m 4, m 4 ]; (C2) f(t, u, v) > m 2 /L for (t, u, v) [η, T ] T [m 2, bm 2 ] [ m 4, m 4 ]; (C3) f(t, u, v) < m /N for (t, u, v) [, T ] T [, m ] [ m 4, m 4 ], then the problem (.) an (.2) has at least three positive solutions u, u 2, u 3 satisfying max u i (t) m 4 for i =, 2, 3; t [,T ] T κ m 2 < min u (t); m < max t [,T ] T u 2 (t), with min u 2 (t) < m 2 ; max t [,T ] T u 3 (t) < m.
248 Bin Zhao an Hong-Rui Sun Proof. Define an operator A : P E by Au (t) = t (t s) q(s)f(s, u(s), u (s)) s + βt + γ α(βt + γ) η (η s) q(s)f(s, u(s), u (s)) s. From [, Lemma 2.2], we have Au P. By a stanar argument, it is easy to see that A : P P is completely continuous. We now show that all the conitions of Lemma 2. are satisfie. Firstly, we show that if (C) is satisfie, then A : P (φ, m 4 ) P (φ, m 4 ). (3.3) For u P (φ, m 4 ), we have φ(u) = max t [,T ] T κ u (t) m 4. With Lemma 3., we have max u(t) Mm 4. Then conition (C) implies that f(t, u(t), u (t)) m 4 /S for t [,T ] T t [, T ] T. On the other han, for u P, we have Au P. Then T u is concave on [, T ] T, so we have γ(au) = max (Au) (t) t [,T ] T κ = max { (Au) (), (Au) (T ) } { β T = max αβ η (η s) q(s)f(s, u(s), u (s)) s, q(s)f(s, u(s), u (s)) s + β αβ η } (η s) q(s)f(s, u(s), u (s)) s max { β, q(s)f(s, u(s), u (s)) s + αβ m { 4 β T S max (T s) q(s) s, = m 4. Therefore, (3.3) hols. η } (η s) q(s)f(s, u(s), u (s)) s q(s) s + αβ η } (η s) q(s) s
Positive Solutions for BVP on Time Scales 249 We choose u(t) = am 2 t 2 + a [ β(t 2 αη 2 ) + γ( α)σ() ] m 2 t +a [ γ(t 2 αη 2 ) γ(t αη)σ() ] m 2 for t [, T ] T. It is easy to see that βu() γu () =, αu(η) = u(t ), u(t) an is concave on [, T ] T, so u P. Again ϕ(u) = min u(t) = min{, α}(t η)(βηt + γt + γη γσ())am 2 > m 2, θ(u) = max t [,T ] T u(t) am 2 T 2 + a [ β(t 2 αη 2 ) + γ( α)σ() ] m 2 T +a [ γ(t 2 αη 2 ) γ(t αη)σ() ] m 2 = bm 2, φ(u) [ (T + σ(t )) + β(t 2 αη 2 ) + γ( α)σ() ] m 2 = cm 2 m 4. Hence, u P (φ, θ, ϕ, m 2, bm 2, m 4 ) an {u P (φ, θ, ϕ, m 2, bm 2, m 4 ) ϕ(u) > m 2 }. For u P (φ, θ, ϕ, m 2, bm 2, m 4 ), we have m 2 u(t) bm 2 an u (t) m 4 for t [η, T ] T. Hence by conition (C2), one has that f(t, u(t), u (t)) > m 2 /L for t [η, T ] T. So by the efinition of the functional ϕ, we see that ϕ(au) = min Au(t) = min{au(η), Au(T )} ( βη + γ T = min{, α} βt + γ η ) (η s) q(s)f(s, u(s), u (s)) s ( βη + γ T > min{, α} βt + γ η ) (η s) q(s)f(s, u(s), u (s)) s T η η (βs + γ)q(s)f(s, u(s), u (s)) s ( βη + γ T = min{, α} βη + γ η ) = > min{, a}(βη + γ) min{, a}(βη + γ) η η (T s) q(s) m 2 L s = m 2.
25 Bin Zhao an Hong-Rui Sun Therefore, we get ϕ(au) > m 2 for u P (φ, θ, ϕ, m 2, bm 2, m 4 ), an conition (S) in Lemma 2. is satisfie. Next, it follows from [, Lemma 2.4] that for any u P (φ, ϕ, m 2, m 4 ) with θ(au) > bm 2, we have ϕ(au) = min A(u) r max A(u) rθ(au) > m 2. t [,T ] T Thus, conition (S2) of Lemma 2. is satisfie. Finally, we assert that (S3) in Lemma 2. also hols. Clearly, as ψ() = < m, so / R(φ, ψ, m, m 4 ). Assume that u P (φ, ψ, m, m 4 ) with ψ(u) = m. Then by the conition (C3), we obtain that ψ(au) = max (Au)(t) t [,T ] T t = max (t s) q(s)f(s, u(s), u (s)) s + βt + γ α(βt + γ) η (η s) q(s)f(s, u(s), u (s)) s max βt + γ T t [,T ] T < βt + γ (T s) q(s) m N s = m. Therefore, an application of Lemma 2. implies that the BVP (.) an (.2) has at least three positive solutions u, u 2 an u 3 such that max u i (t) m 4 for i =, 2, 3; m 2 < min u (t); t [,T ] T κ m < max t [,T ] T u 2 (t) with The proof is complete. min u 2 (t) < m 2 ; max u 3 (t) < m. t [,T ] T 4 Example [ Let T =, ] { 2 2 + }, n =, 2, 3, β =, γ = 2n 2, α = 2, T =, η = 2. Now we consier the BVP u (t) + f(t, u(t), u (t)) =, t (, ) T, (4.)
Positive Solutions for BVP on Time Scales 25 where u() 2 u () =, ( ) 2 u = u(), (4.2) 2 t + 3u3 + sin v, u 2, f(t, u, v) = sin v t + 22464 +, u > 2. It is easy to see by calculating that =, M = 2, r = 4 an S = s + 2 2 L = min{, 2 }( 2 + 2 ) N = 2 2 + 2 2 ( ) 2 s 2 s = 7 6, ( s) s = 2, ( s) s = 2. If we choose m = 4, m 2 =, m 4 = 5, then f(t, u, v) satisfies () f(t, u, v) < 6 7 5 = m 4 S for (t, u, v) [, ] T [, 2 5 ] [ 5, 5 ]; (2) f(t, u, v) 2 + 3 > 2 = m 2 L for ] (t, u, v) [ 2, T [, 49 3 (3) f(t, u, v) + 3 64 + < 3 = m N for (t, u, v) [, ] T [, ] 4 ] [ 5, 5 ]; [ 5, 5 ]. Then all assumptions of Theorem 3.2 hol. Thus by Theorem 3.2, the problem (4.) an (4.2) has at least three positive solutions u, u 2 an u 3 such that max u i (t) 5 for i =, 2, 3, < min u (t); t [,] T κ t [ 2,] T 2 < max t [,] T u 2 (t) with min u 2 (t) < ; t [ 2,] T max t [,] T u 3 (t) < 2.
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