POSITIVE SOLUTIONS TO SINGULAR HIGHER ORDER BOUNDARY VALUE PROBLEMS ON PURELY DISCRETE TIME SCALES

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1 Communications in Applied Analysis , POSITIVE SOLUTIONS TO SINGULAR HIGHER ORDER BOUNDARY VALUE PROBLEMS ON PURELY DISCRETE TIME SCALES CURTIS KUNKEL AND ASHLEY MARTIN 1 Department of Mathematics & Statistics, University of Tennessee Martin Martin, TN USA ckunkel@utm.edu ashnpoor@ut.utm.edu ABSTRACT. We study singular discrete higher order boundary value problems with mixed boundary conditions of the form u n t i n 1 + ft i, ut i,..., u n 1 t i n 1 = 0, u n 1 t 0 = u n 2 t N+1 = u n 3 t N+2 = = u t N+n 2 = ut N+n 1 = 0, over a finite discrete interval T = {t 0, t 1,...,t N+n 2, t N+n 1 }. We prove the existence of a positive solution by means of the lower and upper solutions method and the Brouwer fixed point theorem in conjunction with perturbation methods to approximate regular problems. AMS MOS Subject Classification. 39A10, 34B16, 34B PRELIMINARIES This paper is somewhat of an extension of the work done by Rach unková and Rach unek [23] and the works done by Kunkel [17], [18]. Rach unková and Rach unek studied a second order singular boundary value problem for the discrete p-laplacian, φ p x = x p 2 x, p > 1. In particular, Rach unková and Rach unek dealt with the discrete boundary value problem φ p ut 1 + ft, ut, ut 1 = 0, t [1, T + 1], u0 = ut + 2 = 0, in which ft, x 1, x 2 is singular in x 1. Kunkel s results extended theirs to the third order case, but only for p = 2, i.e. φ 2 x = x. That is, Kunkel s extension focused on the boundary value problem 3 ut 2 + ft, ut, ut 1, 2 ut 2 = 0, t [2, T + 1], 2 u0 = ut + 2 = ut + 3 = 0. Received December 15, $15.00 c Dynamic Publishers, Inc.

2 554 C. KUNKEL AND A. MARTIN Kunkel s other work entails an extension to a second order singular discrete boundary value problem with non-uniform step size what we are calling a purely discrete time scale u t i 1 + ft i, ut i, u t i 1 = 0, t i T, u t 0 = ut n+1 = 0. The methods of this paper rely heavily on upper and lower solutions methods in conjunction with an application of the Brouwer fixed point theorem [26]. We consider only the singular third order boundary value problem, while letting our function range over a discrete interval with non-uniform step size. We will provide definitions of appropriate upper and lower solutions. The upper and lower solutions will be applied to nonsingular perturbations of our nonlinear problem, ultimately giving rise to our boundary value problem by passing to the limit. Upper and lower solutions have been used extensively in establishing solutions of boundary value problems for finite difference equations. In addition to [11], [17], [23], we mention especially the paper by Jiang, et al. [13] in which they dealt with singular discrete boundary value problems using upper and lower solutions methods. For other outstanding results in which upper and lower solutions methods were employed to obtain solutions of boundary value problems for finite difference equations, we refer to [1], [2], [4], [5], [6], [10], [12], [16], [20], [21], [22], [27]. Singular discrete boundary value problems also have received a good deal of attention. For a list of a few representative works, we suggest the references [3], [7], [8], [14], [15], [19], [22], [24], [25], [27],[28]. In this section, we will state the definitions that are used in the remainder of the paper. Definition 1.1. For 0 i N + n 1, let t i R, where t 0 < t 1 < < t N+n 2 < t N+n 1. Define the discrete intervals T := [t 0, t N+n 1 ] = {t 0, t 1,...,t N+n 2, t N+n 1 }, and T := [t n 1, t N ] = {t n 1, t n,...,t N 1, t N }. Definition 1.2. For the function u : T R, define the delta derivative [9], u, by u t i := ut i+1 ut i t i+1 t i, t i T\t N+n 1. We make note that u 2 t i = u t i = u t i. Consider the higher order nonlinear discrete dynamic u n t i n 1 + ft i, ut i,...,u n 1 t i n 1 = 0, t i T, 1.1

3 SINGULAR HIGHER ORDER BVPS ON PURELY DISCRETE TIME SCALES 555 with mixed boundary conditions u n 1 t 0 = u n 2 t N+1 = u n 3 t N+2 = = u t N+n 2 = ut N+n 1 = Our goal is to prove the existence of a positive solution of problem 1.1, 1.2. Definition 1.3. By a solution of problem 1.1, 1.2, we mean a function u : T R such that u satisfies the discrete dynamic 1.1 on T and the boundary conditions 1.2. If ut > 0 for t T, we say u is a positive solution of the problem 1.1, 1.2. Definition 1.4. Let D R n. We say that f is continuous on T D if ft i, x 1,...,x n is defined on T for each x 1,...,x n D, and if ft i, x 1,..., x n is continuous on D for each t i T. Definition 1.5. Let f : T D R, where D R n. If D = R n, problem 1.1, 1.2 is called regular. If D = R n and f has singularities on the boundary of D, then problem 1.1, 1.2 is called singular. We will assume throughout this paper that the following hold: A: D = 0, R n 1. B: f is continuous on T D. C: ft i, x 1,..., x n has a singularity at x 1 = 0, i.e. lim sup x ft i, x 1,...,x n = for each t i T and for some x 2,..., x n R n LOWER AND UPPER SOLUTIONS METHOD FOR REGULAR PROBLEMS Let us first consider the regular dynamic equation u n t i n 1 + ht i, ut i,...,u n 1 t i n 1 = 0, t i T, 2.1 where h is continuous on T R n satisfying the boundary conditions 1.2. We establish a lower and upper solutions method for this regular problem 2.1, 1.2. Definition 2.1. α : T R is called a lower solution of 2.1, 1.2 if α n t i n 1 + ht i, αt i,...,α n 1 t i n 1 0, t i T, 2.2 satisfying boundary conditions 1 n 1 α n 1 t 0 0, 1 n 2 α n 2 t N+1 0,. α 2 t 0 α t N+n 2 0 αt N+n

4 556 C. KUNKEL AND A. MARTIN Definition 2.2. β : T R is called an upper solution of 2.1, 1.2 if β n t i n 1 + ht i, βt i,..., β n 1 t i n 1 0, t i T, 2.4 satisfying boundary conditions 1 n 1 β n 1 t 0 0, 1 n 2 β n 2 t N+1 0,. β 2 t 0 β t N+n 2 0 βt N+n Theorem 2.3 Lower and Upper Solutions Method. Let α and β be lower and upper solutions of 2.1, 1.2, respectively, and α β on T. Let ht i, x 1,...,x n be continuous on T R n and nonincreasing in its x n variable. Then 2.1, 1.2 has a solution u satisfying αt ut βt, t T. Proof. We proceed through a sequence of steps involving modifications of h. Step 1. For t i T, x 1,...,x n R n, define e x n x n 1 h t i, x 1,..., x n 1, t i n 1 t i n 2 8 where >< >: «= «h t i, βt i,..., β n 1 t i n 2, β n ti n 1 σt i n 1,x n t i n 1 t i n 2 x n > β n t i n 1, β n ti n 1 x n β n t i n 1 x n+1, h t i, x 1..., x n 1, xn σt i n 1,xn, α n t t i n 1 t i n 1 x n β n t i n 1, i n 2 h t i, αt i...,α n 1 t i n 2, α n «ti n 1 σt i n 1,x n t i n 1 t i n 2 x n < α n t i n 1, + xn α n ti n 1, x n α n t i n >< α n t i n 2, x n > 1 n α n t i n 1, σt i n 1, x n = x n, 1 n β n t i n 1 x n 1 n α n t i n 1, >: β n t i n 2, x n < 1 n β n t i n 1. By its construction, h is continuous on T R n, and there exists M > 0 so that, ht i, x 1,...,x n M, t i T, x 1,...,x n R n. We now study the auxiliary equation, u n t i n 1 + ht i, ut i,...,u n 1 t i n 1 = 0, t i T, 2.6 satisfying boundary conditions 1.2. Our immediate goal is to prove the existence of a solution of 2.6, 1.2.

5 SINGULAR HIGHER ORDER BVPS ON PURELY DISCRETE TIME SCALES 557 Step 2. We lay the foundation to use the Brouwer fixed point theorem. To this end, define the Banach space E by E = {u : T R : u n 1 t 0 = u n 2 t N+1 = u n 3 t N+2 = = u t N+n 2 = ut N+n 1 = 0}, and also define u = max{ ut i : t i T}. Further, we define an operator T : E E by N+n 2 T ut z = 1 n t zn 2 +1 t zn 2 z n 2 =z z 0 +n 1 i=n 1 t i n+1 t i n h By its construction, T is a continuous operator. t i, ut i,...,u n 1 t i n 1. Moreover, from the bounds placed on h in Step 1 and from 2.7, if 2.7 r > t N+n 1 t 0 n M, then T Br Br, where Br = {u E : u < r}. Therefore, by the Brouwer fixed point theorem [26], there exists u Br such that u = T u. Step 3. We now show that u is a fixed point of T iff u is a solution of 2.6, 1.2. First, assume u = T u. Then, u E, and thus, satisfies 1.2. Furthermore, u t z = ut z+1 ut z t z+1 t z [ 1 = t z+1 t z = and we see that 1 n N+n 2 z n 2 =z+1 t zn 2 +1 t zn 2 ht i,... ] N+n 2 1 n t zn 2 +1 t zn 2 h t i,... 1 t z+1 t z z n 2 =z [ 1 n t z+1 t z z n 3 =z = 1 n 1 t zn 3 +1 t zn 3 h t i,..., z n 3 =z t zn 3 +1 t zn 3 ht i,... u t z = 1 n 1 t zn 3 +1 t zn 3 h t i,.... z n 3 =z ]

6 558 C. KUNKEL AND A. MARTIN Continuing in this manner, u n 1 t u n z+1 u n 1 t z t z = t z+1 t z [ = 1 t z+1 t z z+n 1 i=n 1 z+n i=n 1 t i n+1 t i n ht i,... = t z+1 t z h t z+n,... t z+1 t z = h t z+n,.... Thus, we see that u n t z + h t z+n,... = 0. On the other hand, let ut solve 2.6, 1.2. Then, This implies that Also, t i n+1 t i n h t i,... u n 1 t u n 1 u n 1 t 0 t 0 = t 1 t 0 u n 1 t 1 = t 1 t 0 = h t n 1, ut n 1,...,u n 1 t 0. u n 1 t 1 = t 1 t 0 h u n 1 t u n 2 u n 1 t 1 t 1 = t 2 t 1 u n 1 t 2 t 1 t 0 h = t 2 t 1 = h t n, ut n,...,u t 1. t n 1, ut n 1,...,u n 1 t 0. ] t n 1, ut n 1,..., u n 1 t 0 This implies that u n 1 t 2 = t 2 t 1 h t n, ut n,...,u n 1 t 1 + t 1 t 0 h t n 1, ut n 1,...,u n 1 t 0. Continuing inductively, we see that u n 1 t z = z t i t i 1 h t i+n 2, ut i+n 2,...,u n 1 t i 1, 1 z N. i=1

7 SINGULAR HIGHER ORDER BVPS ON PURELY DISCRETE TIME SCALES 559 Now, we use similar techniques to see that and that u n 2 t u n 1 N+1 u n 2 t N t N = t N+1 t N u n 2 t N = t N+1 t N N = i=1 u n 2 t N = t N+1 t N t i t i 1 h N t i t i 1 h i=1 Proceeding through the interval, we see that u n 2 t z = N z 0 t z0 +1 t z0 t i t i 1 h z 0 =z i=1 In a similar fashion for j = 3, 4,..., n, we see that and that t i+n 2, ut i+n 2,...,u n 1 t i 1. t i+n 2, ut i+n 2,...,u n 1 t i 1. t i+n 2, ut i+n 2,..., u n 1 t i 1. u n j t u n j 1 N+j 1 u n j t N+j 2 t N+j 2 = t N+j 1 t N+j 2 u n j t N+j 2 = t N+j 1 t N+j 2 N+j 1 = 1 j 1 z j 1 =N+j 2 u n j t N+j 2 = 1 j t N+j 1 t N+j 2 h t i,.... N+j 1 z j 1 =N+j 2 ht i,.... Proceeding through the interval, similar to before, we conclude that And specifically, for j = n, N+j 2 u n j t z = 1 j t zj 2 +1 t zj 2 h t i,.... z j 2 =z N+n 2 ut z = 1 n t zn 2 +1 t zn 2 z n 2 =z z 0 +n 1 i=n 1 t i n+1 t i n h t i, ut i,...,u n 1 t i n 1.

8 560 C. KUNKEL AND A. MARTIN We therefore can conclude that u = T u, and this step of the proof is complete. Step 4. We now show that solutions ut of 2.6, 1.2 satisfy αt ut βt, t T. Consider the case of obtaining ut βt. Let vt = ut βt. For the sake of establishing a contradiction, assume that max{vt : t T} := vt l > 0. From the boundary conditions 1.2 and 2.5, we see that t l T. Thus, vt l 1 vt l and vt l+1 vt l. Consequently, v t l 0 and v t l 1 0. This in turn implies that v t l 1 0. Continuing in this manner we see that that v n t l 0. On the other hand, since h is nonincreasing in its x n variable, we have from 2.1 v n t l = u n t l β n t l h t l+n, ut l+n,...,u n 1 t l h t l+n, βt l+n,...,β n 1 t l = ht l+n, βt l+n,...,β n 1 t l + β n t l u n t l β n t l u n t l ht l+n, βt l+n,...,β n 1 t l = β n t l u n t l β n t l u n t l + 1 > 0. But this is a contradiction. Therefore, vl 0. Which means that ut βt for all t T. A similar argument shows that αt ut for all t T. Thus, the conclusion of the theorem holds and our proof is complete. 3. EXISTENCE RESULT In this section, we make use of Theorem 2.3 to obtain positive solutions of the singular problem 1.1, 1.2. In particular, in applying Theorem 2.3, we deal with a sequence of regular perturbations of 1.1, 1.2. Ultimately, we obtain a desired solution of 1.1, 1.2 by passing to the limit on a sequence of solutions for the perturbations. Theorem 3.1. Assume conditions A, B, and C hold, along with the following: D: There exists c 0, so that ft i, c, 0,, 0 0 for all t T. E: ft i, x 1, x 2, x 3,...,x n is nonincreasing in its x n variable for t i T and x 1 0, c].

9 SINGULAR HIGHER ORDER BVPS ON PURELY DISCRETE TIME SCALES 561 F: lim x ft i, x 1,...,x n = for t i T. Then, 1.1, 1.2 has a solution u satisfying 0 < ut c, t T. Proof. Again for the proof, we proceed through a sequence of steps. Step 1. For k N, t T, x 1,...,x n R n, define f t i, x 1, x 2,...,x n, x 1 1 k f k t i, x 1,..., x n = f t i, 1, x k 2,...,x n, x1 < 1 k Then, f k is continuous on T R n and nonincreasing for t i T, x 1 [ c, c]. Assumption F implies that there exists k 0 such that for all k k 0, f k t i, 0,...,0 = f t i, 1k, 0,..., 0 > 0, t i T. Consider, for each k k 0, u n t i n 1 + f k t i, ut i,...,u n 1 t i n 1 = 0, t i T. 3.1k Define αt = 0 and βt = c. Then, α and β are lower and upper solutions for 3.1k, 1.2, and αt βt on T. Thus, by Theorem 2.3, there exists u k a solution of 3.1k, 1.2 satisfying 0 u k t i c, t i T, k k 0. Consequently, u k t i c t i+1 t i, t i T. 3.2 Step 2. Let k N, k k 0. Since u k t solves 3.1k, we get from work similar to that exhibited in Theorem 2.3, u k t z = 1 n 1 t zn 3 +1 t zn 3 f k t i,..., t z T. 3.3 z n 3 =z By assumption F, there exists ε 1 0, 1k0 such that if k 1 ε 1, f k t 1, x 1,...,x n > c t 2 t 1 n, x 1 0, ε 1 ]. 3.4

10 562 C. KUNKEL AND A. MARTIN For the sake of establishing a contradiction, assume that for k 1 ε 1, we have u k t 1 < ε 1. Then, by 3.3 and 3.4, u k t 1 = 1 n 1 t zn 3 +1 t zn 3 f k t i,... z n 3 =1 = 1 n 1 t 2 t 1 n 1 f k t i,... + t 2 t 1 n 1 f k t i,... c > t 2 t 1. z n 3 =2 But this contradicts 3.2. Hence u k t 1 ε 1 for all k 1 ε 1. t zn 3 +1 t zn 3 f k t i,... Proceeding across the interval, we get a sequence of epsilons where 0 < ε N+n 1 < < ε 2 < ε 1 such that u k t i ε N+n 1 for t i T. Hence u k t N+n 1 ε N+n 1 for all k 1 ε N+n 1. Therefore, by letting ε = ε N+n 1 2, we get 0 < ε u k t i c, t i T, k 1 ε. 3.5 Since u k t i satisfies 3.5 and 1.2, we can choose a subsequence {u kl t} {u k t} such that lim l u kl t = ut, t T, ut E, where E is the Banach space as defined in Step 2 of Theorem 2.3. Moreover, 3.3 yields for each sufficiently large l, u k l t z = 1 n 1 t zn 3 +1 t zn 3 f k t i,..., z n 3 =z and so by letting l and from the continuity of f, we get that u t z = 1 n 1 t zn 3 +1 t zn 3 f t i,.... z n 3 =z Consequently, we also get, via similar methods exhibited in Step 3 of Theorem 2.3, u n t z = f t z+n, ut z+n,...,u n 1 t z. Therefore, u solves 1.1, 1.2, and by 3.5, our theorem holds. REFERENCES [1] R. P. Agarwal, A. Cabada and V. Otero-Espinar, Existence and uniqueness results for n-th order nonlinear difference equations in presence of lower and upper solutions, Arch. Inequal. Appl , no. 3-4, [2] R. P. Agarwal, A. Cabada, V. Otero-Espinar and S. Dontha, Existence and uniqueness of solutions for anti-periodic difference equations, Arch. Inequal. Appl , no. 4,

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