(ProQuest: ... denotes non-US-ASCII text omitted.)
You-Hui Su 1 and Can-Yun Huang 2
Recommended by Leonid Shaikhet
1, School of Mathematics and Physical Sciences, Xuzhou Institute of Technology, Xuzhou, Jiangsu 221008, China
2, Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou 730050, China
Received 1 July 2009; Accepted 18 November 2009
1. Introduction
Initiated by Hilger in his Ph.D. thesis [1] in 1988, the theory of time scales has been improved ever since, especially in the unification of the theory of differential equations in the continuous case and the theory of difference equations in the discrete case. For the time being, it remains a field of vitality and attracts attention of many distinguished scholars worldwide. In particular, the theory is also widely applied to biology, heat transfer, stock market, wound healing, epidemic models [2-5], and so forth.
Recent research results indicate that considerable achievement has been made in the existence problems of positive solutions to dynamic equations on time scales. For details, please see [6-13] and the references therein. Symmetry and pseudosymmetry have been widely used in science and engineering [14]. The reason is that symmetry and pseudosymmetry are not only of its theoretical value in studying the metric manifolds [15] and symmetric graph [16, 17], and so forth, but also of its practical value, for example, we can apply this characteristic to study graph structure [18, 19] and chemistry structure [20]. Yet, few literature resource [21, 22] is available concerning the characteristics of positive solutions to p -Laplacian dynamic equations on time scales.
Throughout this paper, we denote the p -Laplacian operator by [straight phi]p (u) , that is, [straight phi]p (u)=|u|p-2 u for p>1 with ([straight phi]p)-1 =[straight phi]q and 1/p+1/q=1.
For convenience, we think of the blanket as an assumption that a,b are points in ..., for an interval (a,b)... we always mean (a,b)∩.... Other type of intervals is defined similarly.
We would like to mention the results of Sun and Li [11, 12]. In [12], Sun and Li considered the two-point BVP
[figure omitted; refer to PDF] and established the existence theory for positive solutions of the above problem. They [11] also considered the m -point boundary value problem with p -Laplacian
[figure omitted; refer to PDF] and gave the existence of single or multiple positive solutions to the above problem. The main tools used in these two papers are some fixed-point theorems [23-25].
It is also noted that the researchers mentioned above [11, 12] only considered the existence of positive solutions. As a results, they failed to further provide characteristics of solutions, such as, symmetry. Naturally, it is quite necessary to consider the characteristics of solutions to p -Laplacian dynamic equations on time scales.
Let ... be a symmetric time scale such that 0,T∈... . we consider the following p -Laplacian boundary value problem on time scales ... of the form:
[figure omitted; refer to PDF] By using symmetric technique, the Krasnosel'skii's fixed point theorem [24], the generalized Avery-Henderson fixed point theorem [26], and Avery-Peterson fixed point theorem [27], we obtain the existence of at least single, twin, triple, or arbitrary odd positive symmetric solutions of problem (1.3). As applications, two examples are given to illustrate the main results and their differences. These results are even new for the special cases of continuous and discrete equations as well as in the general time-scale setting.
The rest of the paper is organized as follows. In Section 2, we present several fixed point results. In Section 3, by using Krasnosel'skii's fixed point theorem, we obtain the existence of at least single or twin positive symmetric solutions to problem (1.3). In Section 4, the existence criteria for at least triple positive or arbitrary odd positive symmetric solutions to problem (1.3) are established. In Section 5, we present two simple examples to illustrate our results.
For convenience, we now give some symmetric definitions.
Definition 1.1.
The interval [0,T]... is said to be symmetric if any given t∈[0,T]... , we have T-t∈[0,T]... .
We note that such a symmetric time scale ... exists. For example, let
[figure omitted; refer to PDF] It is obvious that ... is a symmetric time scale.
Definition 1.2.
A function u:[0,T]... [arrow right]... is said to be symmetric if u is symmetric over the interval [0,T]... . That is, u(t)=u(T-t) , for any given t∈[0,T]... .
Definition 1.3.
We say u is a symmetric solution to problem (1.3) on [0,T]... provided that u is a solution to boundary value problem (1.3) and is symmetric over the interval [0,T]... .
Basic definitions on time scale can be found in [6, 7, 28]. Another excellent sources on dynamical systems on measure chains are the book in [29].
Throughout this paper, it is assumed that
(H1) f:[0,∞)[arrow right][0,∞) is continuous, and does not vanish identically;
(H2) h∈Cld ([0,T]... ,[0,∞)) is symmetric over the interval [0,T]... and does not vanish identically on any closed subinterval of [0,T]... , where Cld ([0,T]... ,[0,∞)) denotes the set of all left dense continuous functions from [0,T]... to [0,∞).
2. Preliminaries
Let E=Cld ([0,T]... ,...) and equip norm
[figure omitted; refer to PDF] then E is a Banach space. Define a cone P⊂E by
[figure omitted; refer to PDF]
Assume that r,η∈(0,T/2)... with η<r . By using the symmetric and concave characters of u∈P and u(0)=u(T)=0 , it is easy to obtain the following results.
Lemma 2.1.
Assume that r,η∈(0,T/2)... with η<r . If u∈P, then
(i) u(η)≥(η/r)u(r);
(ii) (T/2)u(r)≥ru(T/2).
From the previous lemma we know that ||u||=u(T/2) for u∈P.
The operator A:P[arrow right]E is defined by
[figure omitted; refer to PDF] It is obvious that A is completely continuous operator and all the fixed points of A are the solutions to the boundary value problem (1.3).
In addition, it is easy to see that the operator A is symmetric. In fact, for t∈[0,T/2]... , we have T-t∈[T/2,T]... , by using the integral transform, we have
[figure omitted; refer to PDF] Hence, A is symmetric.
Now, we provide some background material from the theory of cones in Banach spaces [24, 26, 27, 30], and then state several fixed point theorems needed later.
Firstly, we list the Krasnosel'skii's fixed point theorem [24].
Lemma 2.2 (see [24]).
Let P be a cone in a Banach space E. Assume that Ω1 , Ω2 are open subsets of E with 0∈Ω1 , Ω¯1 ⊂Ω2 . If A:P∩(Ω¯2 \Ω1 )[arrow right]P is a completely continuous operator such that either
(i) ||Ax||≤||x|| , for all x∈P∩∂Ω1 and ||Ax||≥||x|| , for all x∈P∩∂Ω2 or
(ii) ||Ax||≥||x|| , for all x∈P∩∂Ω1 and ||Ax||≤||x|| , for all x∈P∩∂Ω2 ,
then A has a fixed point in P∩(Ω2 ¯\Ω1 ).
Given a nonnegative continuous functional γ on a cone P of a real Banach space E, we define, for each d>0, the set P(γ,d)={x∈P:γ(x)<d}.
Secondly, we state the generalized Avery-Henderson fixed point theorem [26].
Lemma 2.3 (see [26]).
Let P be a cone in a real Banach space E. Let α , β , and γ be increasing, nonnegative continuous functional on P such that for some c>0 and H>0, γ(x)≤β(x)≤α(x) and ||x||≤Hγ(x) for all x∈P(γ,c)¯. Suppose that there exist positive numbers a and b with a<b<c and A:P(γ,c)¯[arrow right]P is a completely continuous operator such that
(i) γ(Ax)<c for all x∈∂P(γ,c);
(ii) β(Ax)>b for all x∈∂P(β,b);
(iii): P(α,a)≠∅ and α(Ax)<a for x∈∂P(α,a) ,
then A has at least three fixed points x1 , x2 , and x3 belonging to P(γ,c)¯ such that [figure omitted; refer to PDF]
The following lemma can be found in [21].
Lemma 2.4 (see [21]).
Let P be a cone in a real Banach space E. Let α , β , and γ be increasing, nonnegative continuous functional on P such that for some c>0 and H>0, γ(x)≤β(x)≤α(x) and ||x||≤Hγ(x) for all x∈P(γ,c)¯. Suppose that there exist positive numbers a and b with a<b<c and A:P(γ,c)¯[arrow right]P is a completely continuous operator such that:
(i) γ(Ax)>c for all x∈∂P(γ,c);
(ii) β(Ax)<b for all x∈∂P(β,b);
(iii): P(α,a)≠∅ and α(Ax)>a for x∈∂P(α,a) ,
then A has at least three fixed points x1 ,x2 , and x3 belonging to P(γ,c)¯ such that [figure omitted; refer to PDF]
Let β and [varphi] be nonnegative continuous convex functionals on P , λ is a nonnegative continuous concave functional on P , and [straight phi] is a nonnegative continuous functional, respectively on P. We define the following convex sets:
[figure omitted; refer to PDF] and a closed set R([varphi],[straight phi],a,d)={x∈P:a≤[straight phi](x),[varphi](x)≤d}.
Finally, we list the fixed point theorem due to Avery-Peterson [27].
Lemma 2.5 (see [27]).
Let P be a cone in a real Banach space E and β, [varphi], λ, [straight phi] defined as above, moreover, [straight phi] satisfies [straight phi](λ[variant prime] x)≤λ[variant prime] [straight phi](x) for 0≤λ[variant prime] ≤1 such that, for some positive numbers h and d , [figure omitted; refer to PDF] for all x∈P([varphi],d)¯. Suppose that A:P([varphi],d)¯[arrow right]P([varphi],d)¯ is completely continuous and there exist positive real numbers a , b , c , with a<b such that
(i) {x∈P([varphi],β,λ,b,c,d):λ(x)>b}≠∅ and λ(A(x))>b for x∈P([varphi],β,λ,b,c,d);
(ii) λ(A(x))>b for x∈P([varphi],λ,b,d) with β(A(x))>c;
(iii): 0∉R([varphi],[straight phi],a,d) and λ(A(x))<a for all x∈R([varphi],[straight phi],a,d) with [straight phi](x)=a,
then A has at least three fixed points x1 ,x2 ,x3 ∈P([varphi],d)¯ such that [figure omitted; refer to PDF]
3. Single or Twin Solutions
Let
[figure omitted; refer to PDF] We define i0 = number of zeros in the set {f0 ,f∞ } and i∞ = number of infinities in the set {f0 ,f∞ } . Clearly, i0 , i∞ =0,1, or 2 and there exist six possible cases: (i) i0 =1 and i∞ =1 ; (ii) i0 =0 and i∞ =0 ; (iii) i0 =0 and i∞ =1 ; (iv) i0 =0 and i∞ =2 ; (v) i0 =1 and i∞ =0 ; (vi) i0 =2 and i∞ =0 . In the following, by using Krasnosel'skii's fixed point theorem in a cone, we study the existence of positive symmetric solutions to problem (1.3) under the above six possible cases.
3.1. For the Case i0 =1 and i∞ =1
In this subsection, we discuss the existence of single positive symmetric solution of the problem (1.3) under i0 =1 and i∞ =1 .
Theorem 3.1.
Problem (1.3) has at least one positive symmetric solution in the case i0 =1 and i∞ =1.
Proof.
We divide the proof into two cases.
Case 1 (f0 =0 and f∞ =∞ ).
In view of f0 =0 , there exists an H1 >0 such that f(u)≤[straight phi]p ([straight epsilon])[straight phi]p (u)=[straight phi]p ([straight epsilon]u) for u∈(0,H1 ] , where [straight epsilon] arbitrary small and satisfies 0<([straight epsilon]T/2)[straight phi]q (∫0T/2 h(r)∇r)≤1 .
If u∈P with ||u||=H1 , then [figure omitted; refer to PDF] We let ΩH1 ={u∈E:||u||<H1 }, then ||Au||≤||u|| for u∈P∩∂ΩH1 .
From f∞ =∞, there exists an H2[variant prime] >0 such that f(u)≥[straight phi]p (k)[straight phi]p (u)=[straight phi]p (ku) for u∈[H2[variant prime] ,∞) , where k>0, and satisfies the following inequality: [figure omitted; refer to PDF] Set [figure omitted; refer to PDF] If u∈P with ||u||=H2 , then, by Lemma 2.1, one has [figure omitted; refer to PDF] For u∈P∩∂ΩH2 , in terms of (3.3) and (3.5), we get [figure omitted; refer to PDF] Thus, by (i) of Lemma 2.2, problem (1.3) has at least single positive symmetric solution u in P∩(Ω¯H2 \ΩH1 ) with H1 ≤||u||≤H2 .
Case 2 (f0 =∞ and f∞ =0 ).
Since f0 =∞, there exists an H3 >0 such that f(u)≥[straight phi]p (m)[straight phi]p (u)=[straight phi]p (mu) for u∈(0,H3 ], where m is such that [figure omitted; refer to PDF] If u∈P with ||u||=H3 , then, by (3.7), one has [figure omitted; refer to PDF] If we let ΩH3 ={u∈E:||u||<H3 }, then ||Au||≥||u|| for u∈P∩∂ΩH3 .
Now, we consider f∞ =0. By definition, there exists H4[variant prime] >0 such that [figure omitted; refer to PDF] where δ>0 satisfies [figure omitted; refer to PDF]
Suppose that f is bounded, then f(u)≤[straight phi]p (K) for all u∈[0,∞) and some constant K>0. Pick [figure omitted; refer to PDF] If u∈P with ||u||=H4 , then [figure omitted; refer to PDF]
Suppose that f is unbounded. From f∈C([0,+∞),[0,+∞)), we have f(u)≤C3 for arbitrary u∈[0,C4 ], here C3 and C4 are arbitrary positive constants. This implies that f(u)[arrow right]+∞ if u[arrow right]+∞ . Hence, it is easy to know that there exists H4 ≥max {2H3 ,(T/2η)H4[variant prime] } such that f(u)≤f(H4 ) for u∈[0,H4 ]. If u∈P with ||u||=H4 , then by using (3.9) and (3.10), we have [figure omitted; refer to PDF] Consequently, in either case, if we take ΩH4 ={u∈E:||u||<H4 }, then, for u∈P∩∂ΩH4 , we have ||Au||≤||u|| . Thus, condition (ii) of Lemma 2.2 is satisfied. Consequently, problem (1.3) has at least single positive symmetric solution u in P∩(Ω¯H4 \ΩH3 ) with H3 ≤||u||≤H4 . The proof is complete.
3.2. For the Case i0 =0 and i∞ =0
In this subsection, we discuss the existence of positive symmetric solutions to problems (1.3) under i0 =0 and i∞ =0 .
First, we will state and prove the following main result of problem (1.3).
Theorem 3.2.
Suppose that the following conditions hold:
(i) there exists constant p[variant prime] >0 such that f(u)≤[straight phi]p (p[variant prime]Λ1 ) for u∈[0,p[variant prime] ], where Λ1 ={(T/2)[straight phi]q (∫0T/2 h(r)∇r)}-1 ;
(ii) there exists constant q[variant prime] >0 such that f(u)≥[straight phi]p (q[variant prime]Λ2 ) for u∈[(2η/T)q[variant prime] ,q[variant prime] ], where Λ2 ={η[straight phi]q (∫ηT/2 h(r)∇r)}-1 , furthermore, p[variant prime] ≠q[variant prime] ,
then problem (1.3) has at least one positive symmetric solution u such that ||u|| lies between p[variant prime] and q[variant prime] .
Proof.
Without loss of generality, we may assume that p[variant prime] <q[variant prime] .
Let Ωp[variant prime] ={u∈E:||u||<p[variant prime] }. For any u∈P∩∂Ωp[variant prime] , in view of condition (i), we have [figure omitted; refer to PDF] which yields [figure omitted; refer to PDF]
Now, set Ωq[variant prime] ={u∈E:||u||<q[variant prime] }. For u∈P∩∂Ωq[variant prime] , Lemma 2.1 implies that [figure omitted; refer to PDF] Hence, by condition (ii) we get [figure omitted; refer to PDF] So, if we take Ωq[variant prime] ={u∈E:||u||<q[variant prime] }, then [figure omitted; refer to PDF] Consequently, in view of p[variant prime] <q[variant prime] , (3.15) and (3.18), it follows from Lemma 2.2 that problem (1.3) has a positive symmetric solution u in P∩(Ω¯q[variant prime] \Ωp[variant prime] ). The proof is complete.
3.3. For the Case i0 =1 and i∞ =0 or i0 =0 and i∞ =1
In this subsection, under the conditions i0 =1 and i∞ =0 or i0 =0 and i∞ =1, we discuss the existence of positive symmetric solutions to problem (1.3).
Theorem 3.3.
Suppose that f0 ∈[0,[straight phi]p (Λ1 )) and f∞ ∈([straight phi]p ((T/2η)Λ2 ),∞) hold. Then problem (1.3) has at least one positive symmetric solution.
Proof.
It is easy to see that under the assumptions, conditions (i) and (ii) in Theorem 3.2 are satisfied. So the proof is easy and we omit it here.
Theorem 3.4.
Suppose that f0 ∈([straight phi]p ((T/2η)Λ2 ),∞) and f∞ ∈[0,[straight phi]p (Λ1 )) hold, then problem (1.3) has at least one positive symmetric solution.
Proof.
Firstly, let [straight epsilon]1 =f0 -[straight phi]p ((T/2η)Λ2 )>0, there exists a sufficiently small q[variant prime] >0 that satisfies [figure omitted; refer to PDF] Thus, u∈[(2η/T)q[variant prime] ,q[variant prime] ], we have [figure omitted; refer to PDF] which implies that condition (ii) in Theorem 3.2 holds.
Nextly, for [straight epsilon]2 =[straight phi]p (Λ1 )-f∞ >0, there exists a sufficiently large p[variant prime][variant prime] (>q[variant prime] ) such that [figure omitted; refer to PDF] We consider two cases.
Case 1.
Assume that f is bounded, that is, [figure omitted; refer to PDF] here K1 >0 some constant. If we take sufficiently large p[variant prime] such that p[variant prime] ≥max {K1 /Λ1 ,p[variant prime][variant prime] }, then [figure omitted; refer to PDF] Consequently, from the above inequality, condition (i) of Theorem 3.2 is true.
Case 2.
Assume that f is unbounded.
From f∈C([0,∞),[0,∞)), there exists p[variant prime] >p[variant prime][variant prime] such that [figure omitted; refer to PDF] Since p[variant prime] >p[variant prime][variant prime] , by (3.21), we get f(p[variant prime] )≤[straight phi]p (Λ1p[variant prime] ), hence [figure omitted; refer to PDF] Thus, condition (i) of Theorem 3.2 is fulfilled.
Consequently, Theorem 3.2 implies that the conclusion of this theorem holds.
From the proof of Theorems 3.1 and 3.2, respectively, we have the following two results.
Corollary 3.5.
Suppose that f0 =0 and condition (ii) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.
Corollary 3.6.
Suppose that f∞ =0 and condition (ii) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.
Theorem 3.7.
Suppose that f0 ∈(0,[straight phi]p (Λ1 )) and f∞ =∞ hold, then problem (1.3) has at least one positive symmetric solution.
Proof.
First, in view of f∞ =∞, then by inequality (3.7), we have ||Au||≥||u|| for u∈P∩∂ΩH2 . Next, by f0 ∈(0,[straight phi]p (Λ1 )), for [straight epsilon]3 =[straight phi]p (Λ1 )-f0 >0, there exists a sufficiently small p[variant prime] ∈(0,H2 ) such that [figure omitted; refer to PDF] which implies that (i) of Theorem 3.2 holds, that is, (3.14) is true. Hence, we obtain ||Au||≤||u|| for u∈P∩∂Ωp[variant prime] . The result is obtained and the proof is complete.
Theorem 3.8.
Suppose that f0 =∞ and f∞ ∈(0,[straight phi]p (Λ1 )) hold, then problem (1.3) has at least one positive symmetric solution.
Proof.
On one hand, since f0 =∞, by inequality (3.9), one gets ||Au||≥||u|| , u∈P∩∂ΩH3 . On the other hand, since f∞ ∈(0,[straight phi]p (Λ1 )), from the technique similar to the second part proof in Theorem 3.4, one obtains that condition (i) of Theorem 3.2 is satisfied, that is, inequality (3.14) holds, one has ||Au||≤||u|| , u∈P∩∂Ωp[variant prime] , where p[variant prime] >H3 . Hence, problem (1.3) has at least one positive symmetric solution. The proof is complete.
From Theorems 3.7 and 3.8, respectively, it is easy to obtain the following two corollaries.
Corollary 3.9.
Assume that f∞ =∞ and condition (i) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.
Corollary 3.10.
Assume that f0 =∞ and condition (i) in Theorem 3.2 hold, then problem (1.3) has at least one positive symmetric solution.
3.4. For the Case i0 =0 and i∞ =2 or i0 =2 and i∞ =0
In this subsection, under i0 =0 and i∞ =2 or i0 =2 and i∞ =0, we study the existence of multiple positive solutions to problems (1.3).
Combining the proofs of Theorems 3.1 and 3.2, it is easy to prove the following two theorems.
Theorem 3.11.
Suppose that i0 =0 and i∞ =2 and condition (i) of Theorem 3.2 hold, then problem (1.3) has at least two positive solutions u1 ,u2 ∈P such that 0<||u1 ||<p[variant prime] <||u2 ||.
Theorem 3.12.
Suppose that i0 =2 and i∞ =0 and condition (ii) of Theorem 3.2 hold, then problem (1.3) has at least two positive solutions u1 ,u2 ∈P such that 0<||u1 ||<q[variant prime] <||u2 ||.
4. Triple Solutions
In the previous section, we have obtained some results on the existence of at least single or twin positive symmetric solutions to problem (1.3). In this section, we will further discuss the existence of positive symmetric solutions to problem (1.3) by using two different methods. And the conclusions we will arrive at are different with their own distinctive advantages.
Based on the obtained symmetric solution position and local properties, we can only get some local properties of solutions by using method one; however, the position of solutions is not determined. In contrast, by means of method two, we cannot only get some local properties of solutions but also give the position of all solutions, with regard to some subsets of the cone, which has to meet some conditions which are stronger than those of method one. Obviously, the local properties of obtained solutions are different by using the two different methods. Hence, it is convenient for us to comprehensively comprehend the solutions of the models by using the two different techniques.
In Section 5, two examples are given to illustrate the differences of the results obtained by the two different methods.
For the notational convenience, we denote
[figure omitted; refer to PDF]
4.1. Result 1
In this subsection, in view of the generalized Avery-Henderson fixed-point theorem [26], the existence criteria for at least triple and arbitrary odd positive symmetric solutions to problems (1.3) are established.
For u∈P, we define the nonnegative, increasing, continuous functionals γ, β , and α by
[figure omitted; refer to PDF] It is obvious that γ(u)≤β(u)≤α(u) for each u∈P. By Lemma 2.1, one obtains ||u||≤C* γ(u) for all u∈P , here C* =T/2η .
We now present the results in this subsection.
Theorem 4.1.
If there are positive numbers a[variant prime] , b[variant prime] , c[variant prime] such that a[variant prime] <(2r/T)b[variant prime] <(2r/T)(c[variant prime]Nξ /Mξ ). In addition, f(u) satisfies the following conditions:
(i) f(u)<[straight phi]p (c[variant prime] /Mξ ) for u∈[0,(T/2η)c[variant prime] ];
(ii) f(u)>[straight phi]p (b[variant prime] /Nξ ) for u∈[b[variant prime] ,(T/2η)b[variant prime] ];
(iii): f(u)<[straight phi]p (a[variant prime] /Lξ ) for u∈[0,(T/2r)a[variant prime] ].
Then problem (1.3) has at least three positive symmetric solutions u1 , u2 , and u3 such that [figure omitted; refer to PDF]
Proof.
By the definition of completely continuous operator A and its properties, it has to be demonstrated that all the conditions of Lemma 2.3 hold with respect to A. It is easy to obtain that A:P(γ,c)¯[arrow right]P.
Firstly, we verify that if u∈∂P(γ,c[variant prime] ) , then γ(Au)<c[variant prime] .
If u∈∂P(γ,c[variant prime] ), then [figure omitted; refer to PDF] Lemma 2.1 implies that [figure omitted; refer to PDF] we have [figure omitted; refer to PDF] Thus, by condition (i), one has [figure omitted; refer to PDF]
Secondly, we show that β(Au)>b[variant prime] for u∈∂P(β,b[variant prime] ).
If we choose u∈∂P(β,b[variant prime] ), then β(u)=min t∈[η,T/2]... u(t)=b[variant prime] . In view of Lemma 2.1, we have [figure omitted; refer to PDF] So [figure omitted; refer to PDF] Using condition (ii), we get [figure omitted; refer to PDF]
Finally, we prove that P(α,a[variant prime] )≠∅ and α(Au)<a[variant prime] for all u∈∂P(α,a[variant prime] ).
In fact, the constant function a[variant prime] /2∈P(α,a[variant prime] ). Moreover, for u∈∂P(α,a[variant prime] ), we have α(u)=max t∈[0,r]... u(t)=a[variant prime] , which implies 0≤u(t)≤a[variant prime] for t∈[0,r]... . Hence, ||u||≤(T/2r)u(r). Therefore [figure omitted; refer to PDF] By using assumption (iii), one has [figure omitted; refer to PDF] Thus, all the conditions in Lemma 2.3 are satisfied. From (H1) and (H2), we have that the solutions to problem (1.3) do not vanish identically on any closed subinterval of [0,T]... . Consequently, problem (1.3) has at least three positive symmetric solutions u1 , u2 , and u3 belonging to P(γ,c[variant prime] )¯, and satisfying (4.3). The proof is complete.
From Theorem 4.1, we see that, when assumptions as (i), (ii), and (iii) are imposed appropriately on f, we can establish the existence of an arbitrary odd number of positive symmetric solutions to problem (1.3).
Theorem 4.2.
Let l=1,2,...,n. Suppose that there exist positive numbers asl [variant prime] , bsl [variant prime] , csl [variant prime] such that [figure omitted; refer to PDF] In addition, f(u) satisfies the following conditions:
(i) f(u)<[straight phi]p (csl [variant prime] /Mξ ) for u∈[0,(T/2η)csl [variant prime] ];
(ii) f(u)>[straight phi]p (bsl [variant prime] /Nξ ) for u∈[bsl [variant prime] ,(T/2η)bsl [variant prime] ];
(iii): f(u)<[straight phi]p (asl [variant prime] /Lξ ) for u∈[0,(T/2r)asl [variant prime] ].
Then problem (1.3) has at least 2l+1 positive symmetric solutions.
Proof.
When l=1, it is clear that Theorem 4.1 holds. Then we can obtain at least three positive symmetric solutions u1 ,u2 , and u3 satisfying [figure omitted; refer to PDF] Following this way, we finish the proof by induction. The proof is complete.
Using Lemma 2.4, it is easy to have the following results.
Theorem 4.3.
Suppose that there are positive numbers a[variant prime] , b[variant prime] , c[variant prime] such that a[variant prime] <(Lθ /Mξ )b[variant prime] <(2η/T)(Lθ /Mξ )c[variant prime] . In addition, f(u) satisfies the following conditions:
(i) f(u)>[straight phi]p (c[variant prime] /Nξ ) for u∈[c[variant prime] ,(T/2η)c[variant prime] ];
(ii) f(u)<[straight phi]p (b[variant prime] /Mξ ) for u∈[0,(T/2η)b[variant prime] ];
(iii): f(u)>[straight phi]p (a[variant prime] /Lθ ) for u∈[a[variant prime] ,(T/2r)a[variant prime] ].
Then problem (1.3) has at least three positive symmetric solutions u1 , u2 , and u3 such that [figure omitted; refer to PDF]
From Theorem 4.3, we can obtain Theorem 4.4 and Corollary 4.5.
Theorem 4.4.
Let l=1,2,...,n. Suppose that there existence positive numbers aλl [variant prime] , bλl [variant prime] , cλl [variant prime] such that [figure omitted; refer to PDF] In addition, f(u) satisfies the following conditions:
(i) f(u)>[straight phi]p (cλl [variant prime] /Nξ ) for u∈[cλl [variant prime] ,(T/2η)cλl [variant prime] ];
(ii) f(u)<[straight phi]p (bλl [variant prime] /Mξ ) for u∈[0,(T/2η)bλl [variant prime] ];
(iii): f(u)>[straight phi]p (aλl [variant prime] /Lθ ) for u∈[aλl [variant prime] ,(T/2r)aλl [variant prime] ].
Then problem (1.3) has at least 2l+1 positive symmetric solutions.
Corollary 4.5.
Assume that f satisfies the following conditions:
(i) f0 =∞,f∞ =∞,
(ii) there exists c0 >0 such that f(u)<[straight phi]p ((2η/T)(c0 /Mξ )) for u∈[0,c0 ],
then problem (1.3) has at least three positive symmetric solutions.
Proof.
First, by condition (ii), let b[variant prime] =(2η/T)c0 , one gets [figure omitted; refer to PDF] which implies that (ii) of Theorem 4.3 holds.
Second, choose K3 sufficiently large to satisfy [figure omitted; refer to PDF] Since f0 =∞, there exists r1[variant prime] >0 sufficiently small such that [figure omitted; refer to PDF] Without loss of generality, suppose r1[variant prime] ≤(Lθ T/2rMξ )b[variant prime] . Choose a[variant prime] >0 such that a[variant prime] <(2r/T)r1[variant prime] . For a[variant prime] ≤u≤(T/2r)a[variant prime] , we have u≤r1[variant prime] and a[variant prime] <(Lθ /Mξ )b[variant prime] . Thus, by (4.18) and (4.19), we have [figure omitted; refer to PDF] this implies that (iii) of Theorem 4.3 is true.
Third, choose K2 sufficiently large such that [figure omitted; refer to PDF] Since f∞ =∞, there exists r2[variant prime] >0 sufficiently large such that [figure omitted; refer to PDF] Without loss of generality, suppose r2[variant prime] >(T/2η)b[variant prime] . Choose c[variant prime] =r2[variant prime] . Then [figure omitted; refer to PDF] which means that (i) of Theorem 4.3 holds.
From above analysis, we get [figure omitted; refer to PDF] then, all conditions in Theorem 4.3 are satisfied. Hence, problem (1.3) has at least three positive symmetric solutions.
In terms of Theorem 4.1, we also have the following corollary.
Corollary 4.6.
Assume that f satisfies conditions
(i) f0 =0,f∞ =0;
(ii) there exists c0 >0 such that f(u)>[straight phi]p ((2η/T)(c0 /Nξ )) for u∈[(2η/T)c0 ,c0 ],
then problem (1.3) has at least three positive symmetric solutions.
4.2. Result 2
In this subsection, the existence criteria for at least triple positive or arbitrary odd positive symmetric solutions to problems (1.3) are established by using the Avery-Peterson fixed point theorem [27].
Define the nonnegative continuous convex functionals [varphi] and β , nonnegative continuous concave functional λ, and nonnegative continuous functional [straight phi] , respectively, on P by
[figure omitted; refer to PDF]
Now, we list and prove the results in this subsection.
Theorem 4.7.
Suppose that there exist constants a* , b* , d* such that 0<a* <(2η/T)b* <(2η/T)(Nξd* /Wξ ). In addition, suppose that Wξ >[straight phi]q (∫ηT/2 h(s)∇s) holds, f satisfies the following conditions:
(i) f(u)≤[straight phi]p (d* /Wξ ) for u∈[0,d* ];
(ii) f(u)>[straight phi]p (b* /Nξ ) for u∈[b* ,d* ];
(iii): f(u)<[straight phi]p (a* /Mξ ) for u∈[0,(T/2η)a* ],
then problem (1.3) has at least three positive symmetric solutions u1 ,u2 ,u3 such that [figure omitted; refer to PDF]
Proof.
By the definition of completely continuous operator A and its properties, it suffices to show that all the conditions of Lemma 2.5 hold with respect to A.
For all u∈P, λ(u)=[straight phi](u) and ||u||=u(T/2)=[varphi](u) . Hence, condition (2.8) is satisfied.
Firstly, we show that A:P([varphi],d* )¯[arrow right]P([varphi],d* )¯.
For any u∈P([varphi],d* )¯, in view of [varphi](u)=||u||≤d* and assumption (i), one has [figure omitted; refer to PDF] From the above analysis, it remains to show that (i)-(iii) of Lemma 2.5 hold.
Secondly, we verify that condition (i) of Lemma 2.5 holds, let u(t)≡kb* with k=Wξ /Nξ >1. From the definitions of Nξ ,Wξ , and β(u) , respectively, it is easy to see that u(t)=kb* >b* and β(u)=0<kb* . In addition, in view of b* <(Nξ /Wξ )d* , we have [varphi](u)=kb* <d* . Thus [figure omitted; refer to PDF] For any u∈P([varphi],β,λ,b* ,kb* ,d* ), then we get b* ≤u(t)≤d* for all t∈[η,T/2]... . Hence, by assumption (ii), we have [figure omitted; refer to PDF]
Thirdly, we prove that condition (ii) of Lemma 2.5 holds. For any u∈P([varphi],λ,b* ,d* ) with β(Au)>kb* , that is, [figure omitted; refer to PDF] So, in view of k=Wξ /Nξ , Wξ >[straight phi]q (∫ηT/2 h(s)∇s) and (4.30), one has [figure omitted; refer to PDF]
Finally, we check condition (iii) of Lemma 2.5. Clearly, since [straight phi](0)=0<a* , we have 0∉R([varphi],[straight phi],a* ,d* ). If u∈R([varphi],[straight phi],a* ,d* ) with [straight phi](u)=min t∈[η,T/2]... u(t)=a* , then Lemma 2.1 implies that [figure omitted; refer to PDF] This yields 0≤u(t)≤(T/2η)a* for all t∈[0,T/2]... . Hence, by assumption (iii), we have [figure omitted; refer to PDF] Consequently, all conditions of Lemma 2.5 are satisfied. The proof is completed.
We remark that condition (i) in Theorem 4.7 can be replaced by the following condition (i[variant prime] ):
[figure omitted; refer to PDF] which is a special case of (i).
Corollary 4.8.
If condition (i) in Theorem 4.7 is replaced by (i[variant prime] ), then the conclusion of Theorem 4.7 also holds.
Proof.
By Theorem 4.7, we only need to prove that (i') implies that (i) holds, that is, if (i') holds, then there is a number d* ≥max {(T/2η)a* ,(Wξ /Nξ )b* } such that f(u)≤[straight phi]p (d* /Wξ ) for u∈[0,d* ].
Suppose on the contrary that for any d* ≥max {(T/2η)a* ,(Wξ /Nξ )b* }, there exists uc ∈[0,d* ] such that f(uc )>[straight phi]p (d* /Wξ ). Hence, if we choose [figure omitted; refer to PDF] with cn[variant prime] [arrow right]∞, then there exist un ∈[0,cn[variant prime] ] such that [figure omitted; refer to PDF] and so [figure omitted; refer to PDF] Since condition (i[variant prime] ) holds, then there exists τ>0 such that [figure omitted; refer to PDF] Hence, we have un ≤τ. Otherwise, if un >τ, then it follows from (4.38) that [figure omitted; refer to PDF] which contradicts (4.36).
Let W=max u∈[0,τ] f(u), then f(un )≤W (n=1,2,...), which also contradicts (4.37). The proof is complete.
Theorem 4.9.
Let l=1,2,...,n. Suppose that there exist constants al* , bl* , dl* such that [figure omitted; refer to PDF] In addition, suppose that Wξ >[straight phi]q (∫ηT/2 h(s)∇s) holds, then f satisfies the following conditions:
(i) f(u)<[straight phi]p (dl* /Wξ ) for u∈[0,dl* ];
(ii) f(u)>[straight phi]p (bl* /Nξ ) for u∈[bl* ,dl* ];
(iii): f(u)<[straight phi]p (al* /Mξ ) for u∈[0,(T/2η)al* ],
then problem (1.3) has at least 2l+1 positive symmetric solutions.
Proof.
Similar to the proof of Theorem 4.2, we omit it here.
5. Examples
In this section, we give two simple examples to illustrate that the conclusions we will arrive at are different with their own distinctive advantages.
Example 5.1.
Let [figure omitted; refer to PDF]
Consider the following boundary value problem with p=3 : [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Note that T=1 . If we choose η=0.1 , r=0.25 , then direct calculation shows that [figure omitted; refer to PDF] and Nξ [approximate]0.0489,Lξ =0.125 ,Wξ =0.2500. If we take a[variant prime] =0.3 , b[variant prime] =1 , c[variant prime] =12, then [figure omitted; refer to PDF] Furthermore, [figure omitted; refer to PDF] By Theorem 4.1 we see that the boundary value problem (5.2) has at least three positive symmetric solutions u1 ,u2 , and u3 such that [figure omitted; refer to PDF] Yet, [straight phi]q (∫0.11/2 h(s)∇s)[approximate]0.489>Wξ [approximate]0.25 , hence, the existence of positive solutions of boundary value problem (5.2) is not obtained by using Theorem 4.7.
Example 5.2.
Let [figure omitted; refer to PDF]
Consider the following boundary value problem: [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Note that T=1. If we choose η=0.45 , r=0.48 , then a direct calculation shows that [figure omitted; refer to PDF] Let [straight epsilon] be an arbitrary small positive number, a* ,b* , and d* are arbitrary positive numbers with [figure omitted; refer to PDF] That is [figure omitted; refer to PDF] where l(u) satisfy l(a* )=[straight phi]p (a* /0.225)-[straight epsilon] , l(b* )=[straight phi]p (b* /0.0981)+[straight epsilon] , (lΔ (u))Δ =0 , u∈[a* ,b* ].
It is obvious that (i), (ii), and (iii) in Theorem 4.7 are satisfied. By Theorem 4.7, we see that the boundary value problem (5.9) has at least three positive solutions u1 ,u2 , and u3 such that [figure omitted; refer to PDF]
However, for arbitrary positive numbers a* , b* , d* , condition (iii) of Theorem 4.1 is not satisfied. Therefore, Theorem 4.1 is not fit to study the boundary value problem (5.9).
Acknowledgments
This work was supported by the Grant of Department of Education Jiangsu Province (09KJD110006) and Science Foundation of Lanzhou University of Technology (BS10200903).
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Abstract
This paper makes a study on the existence of positive solution to p -Laplacian dynamic equations on time scales ... . Some new sufficient conditions are obtained for the existence of at least single or twin positive solutions by using Krasnosel'skii's fixed point theorem and new sufficient conditions are also obtained for the existence of at least triple or arbitrary odd number positive solutions by using generalized Avery-Henderson fixed point theorem and Avery-Peterson fixed point theorem. As applications, two examples are given to illustrate the main results and their differences. These results are even new for the special cases of continuous and discrete equations, as well as in the general time-scale setting.
You have requested "on-the-fly" machine translation of selected content from our databases. This functionality is provided solely for your convenience and is in no way intended to replace human translation. Show full disclaimer
Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer