Positive solutions of Neumann problems for a ... - Tubitak Journals

Turk J Math (2018) 42: 2371 – 2379 © TÜBİTAK doi:10.3906/mat-1805-11

Turkish Journal of Mathematics http://journals.tubitak.gov.tr/math/

Research Article

Positive solutions of Neumann problems for a discrete system coming from models of house burglary

1

Tianlan CHEN1,∗,, Ruyun MA1 ,, Yongwen LIANG2 , Department of Mathematics, Northwest Normal University, Lanzhou, P.R. China 2 Lanzhou Petrochemical Polytechnic, Lanzhou, P.R. China

Received: 07.05.2018

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Accepted/Published Online: 29.06.2018

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Final Version: 27.09.2018

Abstract: We show existence results of positive solutions of Neumann problems for a discrete system: η∆2 (Ak−1 − A0k−1 ) − Ak + A0k + Nk Ak = 0, k ∈ [2, n − 1]Z , ( ∆Ak−1 ) ∆ ∆Nk−1 − 2Nk − Nk Ak + A1k − A0k = 0, k ∈ [2, n − 1]Z , Ak ∆A1 = 0 = ∆An−1 , ∆N1 = 0 = ∆Nn−1 , where the assumptions on η, A0k , and A1k are motivated by some mathematics models for house burglary. Our results are based on the topological degree theory. Key words: Neumann boundary value problems, nonconstant positive solutions, topological degree theory

1. Introduction We will denote the integer set by Z. For any a, b ∈ Z with a < b , setting [a, b]Z := {a, a + 1, · · · , b}. In this paper, we study the existence of positive solutions of the difference systems: η∆2 (Ak−1 − A0k−1 ) − Ak + A0k + Nk Ak = 0, k ∈ [2, n − 1]Z , ( ∆Ak−1 ) ∆ ∆Nk−1 − 2Nk − Nk Ak + A1k − A0k = 0, k ∈ [2, n − 1]Z , Ak

(1)

∆A1 = 0 = ∆An−1 , ∆N1 = 0 = ∆Nn−1 , where η > 0 is a constant, A0 = (A01 , · · · , A0n ) ∈ Rn , A1 = (A11 , · · · , A1n ) ∈ Rn , and A1k > A0k > 0 for each k ∈ [1, n]Z . The homogeneous Neumann boundary condition implies that the system is self-contained with zero flux across the boundary. A solution of (1) is a couple of real vector functions (A, N) ∈ Rn × Rn satisfying the system. We are interested in positive solutions, that is, Ak > 0 and Nk ≥ 0 with Nk ̸≡ 0 for all k ∈ [1, n]Z . ∗Correspondence:

[email protected] 2010 AMS Mathematics Subject Classification: 39A12, 34B18 Supported by the NSFC (No. 11671322), Gansu provincial National Science Foundation of China (No. 1606RJYA232, 11801453) and NWNU-LKQN-15-16.

2371 This work is licensed under a Creative Commons Attribution 4.0 International License.

CHEN et al./Turk J Math

This problem is motivated by Neumann boundary value problems for differential systems: η(A − A0 (x))′′ − A + A0 (x) + N A = 0, x ∈ (0, L), (N ′ − 2N

A′ ′ ) − N A + +A1 (x) − +A0 (x) = 0, x ∈ (0, L), A

(2)

A′ (0) = 0 = A′ (L), N ′ (0) = 0 = N ′ (L), which is a one-dimensional problem associated with a very successful model for house burglaries [17], see also the related papers [2, 4, 11, 12, 16]. In most of these models, η > 0 is the diffusion rate of attractiveness, A0 is the intrinsic (static) component of attractiveness, A1 is the average attractiveness, A is the attractiveness of a house for burglary, and N is the density of burglars. Thus, in the discrete case, the restrictions Ak > 0 and Nk ≥ 0 with Nk ̸≡ 0 for all k ∈ [1, n]Z appear as natural. Note that [5] is the consequence of the discretization of differential problems [11]. For some results on nonlinear difference problems, see [1, 3, 5–10, 15] and the references therein. However, the discrete analogues of (2) have received almost no attention. In this article, we will discuss them in detail. We assume that the following conditions are satisfied: (H1) A0 = (A01 , · · · , A0n ) ∈ Rn , ∆A01 = 0 = ∆A0n−1 and A0k > 0, k ∈ [1, n]Z . (H2) A1 = (A11 , · · · , A1n ) ∈ Rn and A1k > 0, k ∈ [1, n]Z . (H3) A1k > A0k , k ∈ [1, n]Z . If A0 , A1 are positive constants, i.e. A0 = (A0 , · · · , A0 ), A1 = (A1 , · · · , A1 ), system (1) admits the unique positive constant solution A = (A, · · · , A), N = (N, · · · , N ), where A = A1 ,

N=

A1 − A0 A1

under the condition that A1 > A0 . A natural question is whether or not nonconstant positive solutions still exist when A0 , A1 are no longer constants? To this end, under the assumptions (H1)–(H3), we obtain the following result. Theorem 1 Let η > 0 be a constant. Under the assumptions (H1)–(H3), system (1) admits at least one nonconstant positive solution. The purpose of this paper was to show that analogues of existence results of solutions for differential problems proved in [12] hold for the corresponding difference systems. However, some basic ideas from differential calculus are not necessarily available in the field of difference equations such as the intermediate value theorem, the mean value theorem, and Rolle’s theorem. Thus, new challenges are faced and innovation is required. The proof is elementary and relies on Brouwer degree theory [13, 14]. Throughout this paper, we use the following notations and conventions. Given n ∈ N(n ≥ 4) and (x1 , x2 , · · · , xn ) ∈ Rn . Define (∆x1 , · · · , ∆xn−1 ) ∈ Rn−1 by ∆xm = xm+1 − xm , m ∈ [1, n − 1]Z . For every ∑l l, m ∈ N with m > l , we set k=m xk = 0 . Let us introduce the vector space V n−2 = {x ∈ Rn : ∆x1 = 0 = ∆xn−1 }, 2372

(3)


endowed with the orientation of Rn . Its elements can be associated with the coordinates (x2 , · · · , xn−1 ) and correspond to the elements of Rn of the form: (x2 , x2 , x3 , · · · , xn−2 , xn−1 , xn−1 ). We use the norm ∥x∥ :=

max k∈[2,n−1]Z

|xk | in V n−2 , and

max k∈[1,n−2]Z

|xk | in Rn−2 .

2. A priori estimates Let us consider the homotopy corresponding to system (1) for λ ∈ (0, 1], η∆2 (Ak−1 − A0k−1 ) = λ(Ak − A0k − Nk Ak ), k ∈ [2, n − 1]Z , ∆A1 = 0 = ∆An−1 ,

(4)

( ∆Ak−1 ) ∆ ∆Nk−1 − 2λNk = λ(Nk Ak − A1k + A0k ), k ∈ [2, n − 1]Z , ∆N1 = 0 = ∆Nn−1 . Ak

(5)

Obviously, for λ = 1 , (4) and (5) reduce to (1). Setting min B := min Bk , max B := max Bk and B = k∈[1,p]Z

k∈[1,p]Z

n−1 1 ∑ Bk , n−2

B ∈ Rp .

k=2

Lemma 1 Let (A, N) be a solution of (4) and (5) for some λ ∈ (0, 1]. Then, A = A1 .

(6)

Proof Adding the two equations in (4) and (5), and then summing from k = 2 to n − 1, we are led to A = A1 by the Neumann boundary conditions. 2 Lemma 2 Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, NA = A1 − A0 . Proof

(7)

Summing the equation of (4) from k = 2 to n − 1 , we have that n−1 ∑ k=2

(Ak −

A0k )

=

n−1 ∑

Nk Ak ,

k=2

2

which combines with Lemma 1 yields (7).

Lemma 3 Let (H3) hold. Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, for any k ∈ [2, n − 1]Z , |∆Ak | ≤ (n − 2) max |∆2 A0 | + Ak ≤ A1 + (n − 2)2 max |∆2 A0 | +

2(n − 2) 1 A , η

2(n − 2)2 1 A := C1 . η

(8)

(9) 2373


Proof

According to (4), it is apparent that

|∆2 Ak−1 | ≤ |∆2 A0k−1 | +

λ 1 |Ak − A0k − Nk Ak | ≤ |∆2 A0k−1 | + (Ak + A0k + Nk Ak ), η η

k ∈ [2, n − 1]Z .

Combining this with (6) and (7) implies that |∆Ak | =|

k ∑

∆2 Ai−1 | ≤

i=2

n−1 ∑

|∆2 Ai−1 |

i=2

≤(n − 2) max |∆2 A0 | +

n−2 1 (A + A0 + A1 − A0 ) η

=(n − 2) max |∆2 A0 | +

2(n − 2) 1 A . η

(10)

On the other hand, there exists k1 ∈ [2, n − 1]Z such that min A = Ak1 , then Ak1 ≤ A. Thus, by virtue of (6) and (10), it follows that for every k ∈ [2, n − 1]Z , Ak =Ak1 +

k−1 ∑ i=k1

∆Ai ≤ A +

n−1 ∑

∆Ai

i=2

≤A1 + (n − 2)2 max |∆2 A0 | +

2(n − 2)2 1 A . η 2

Lemma 4 Assume (H3). Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, for any k ∈ [2, n − 1]Z , Ak ≥ min A0 := C2 . Proof

(11)

Since there exists k1 ∈ [2, n − 1]Z such that min(A − A0 ) = Ak1 − A0k1 , then 0 ≤ η∆2 (Ak1 − A0k1 ) = λ(Ak1 − A0k1 − Nk1 Ak1 ) ≤ Ak1 − A0k1 .

This implies that Ak ≥ A0k for k ∈ [2, n − 1]Z . Now, we may deduce that Ak ≥ min A ≥ min A0 . 2 Corollary 1 Assume (H3). Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1], then, for any k ∈ [2, n − 1]Z ,

∆A (n − 2) max |∆2 A0 | + 2(n − 2)A1 k := C3 . ≤ Ak ηC2

(12)

Lemma 5 Assume that (H3) holds. Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, A1 − A0 A1 − A0 ≤N≤ . C1 C2 2374

(13)


Proof

From (7), (9), (11), and the inequalities (min A)

n−1 ∑

Nk ≤

k=2

n−1 ∑

Nk Ak ≤ (max A)

k=2

n−1 ∑

Nk ,

k=2

2

we may get the desired result. Lemma 6 Assume (H3). Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, ∆Ak ∆Nk − 2λNk+1 ≤ 2(n − 2)(A1 − A0 ), Ak+1 Proof

k ∈ [2, n − 1]Z .

(14)

Owing to (5), it can be easily seen that |∆(∆Nk−1 − 2λNk

∆Ak−1 )| ≤ |Nk Ak + A1k − A0k | = Nk Ak + A1k − A0k , Ak

k ∈ [2, n − 1]Z .

Hence, using the boundary conditions and (7), it is not difficult to prove that for every k ∈ [2, n − 1]Z , k ∆Ak ∑ ∆Ai−1 − 2λN = ∆(∆Ni−1 − 2λNi ) ∆Nk k+1 Ak+1 Ai i=2

≤

n−1 ∑

(Nk Ak + A1k − A0k ) ≤ (n − 2)(A1 − A0 + A1 − A0 )

k=2

= 2(n − 2)(A1 − A0 ). 2 Lemma 7 Assume (H3). Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, [C ] 3 |Nk − N| ≤ 2(n − 2)(A1 − A0 ) + (n − 2) := C4 , C2 Proof

k ∈ [2, n − 1]Z .

(15)

From (14), we have ∆A k |∆Nk | ≤ 2Nk+1 + 2(n − 2)(A1 − A0 ), Ak+1

k ∈ [2, n − 1]Z .

Let now k1 ∈ [2, n − 1]Z be a minimum point of N , then N ≥ Nk1 . Then it follows from (12) and (13) that for any k ∈ [2, n − 1]Z , |Nk − N| ≤|Nk − Nk1 | = |

k−1 ∑

∆Ni |

i=k1

≤

n−1 ∑ k=2

|∆Nk | ≤ 2C3

n−1 ∑

Nk + 2(n − 2)2 (A1 − A0 )

k=2

[C ] 3 ≤2(n − 2)(A1 − A0 ) + (n − 2) . C2 2 2375


Corollary 2 Assume (H3). Let (A, N) be a positive solution of (4) and (5) for some λ ∈ (0, 1]. Then, for any k ∈ [2, n − 1]Z , Nk ≤ C 4 +

A1 − A0 := C5 . C2

(16)

3. Proof of the main result This section is devoted to the existence of nonconstant positive solutions to system (1). For our purpose, we need some preliminary results. Lemma 8 For any λ ∈ (0, 1], (A, N) is a solution of (4) and (5) if and only if (A, N) is a solution of the following system: λ ∑∑ (Ai − A0i − Ni Ai ), k ∈ [1, n]Z , η j=2 i=2

(17)

j k−1 ∑∑ ∆Ai−1 +λ (Ni Ai − A1i + A0i ), k ∈ [1, n]Z . Ai j=2 i=2

(18)

k−1

Ak = A2 + A0k − A02 − (A − A0 − NA) +

Nk = N2 − (NA − A1 + A0 ) + λ

k ∑

2Ni

i=2

Proof that

j

Suppose (A, N) is a solution of the system (17) and (18). By taking k = 2 in both equations, we find (A − A0 − NA) = 0, (NA − A1 + A0 ) = 0.

(19)

On the other hand, according to (18) and (19), it becomes apparent that for any k ∈ [1, n − 1]Z , λ∑ (Ai − A0i − Ni Ai ), η i=2

(20)

∆Ak ∑ + (Ni Ai − A1i + A0i )]. Ak+1 i=2

(21)

k

∆Ak = ∆A0k +

k

∆Nk = λ[2Nk+1

It should be noted that we obtain ∆A1 = 0 = ∆N1 if k = 1 and ∆An−1 = 0 = ∆Nn−1 if k = n − 1 by using (19). Finally, it is not difficult to verify that ∆2 (Ak−1 − A0k−1 ) =

λ (Ak − A0k − Nk Ak ), k ∈ [2, n − 1]Z , η

and [ ( ] ∆Ak−1 ) ∆2 Nk−1 = λ ∆ 2Nk + (Nk Ak − A1k + A0k ) , k ∈ [2, n − 1]Z , Ak which is equivalent to systems (4) and (5). Similarly, the proof of the converse is valid.

2

Let us now take 0 < R2 < C2 ≤ C1 < R1 , C3 C1 < R3 and C5 < R5 , where C1 , C2 , C3 and C5 are respectively given in (9),(11),(12), and (16). 2376


Define the vector space E = V n−2 × V n−2 by the usual norm ∥(A, N)∥E = ∥A∥ + ∥∆A∥ + ∥N∥. Set the bounded set Ω ⊂ E , Ω = {(A, N) ∈ E : R2 < Ak < R1 , |∆Ak | < R3 , 0 ≤ Nk < R5 , k ∈ [2, n − 1]Z }.

(22)

Define the operator F : Ω × [0, 1] → E, F(A, N, λ) ) ( ∑k−1 ∑j A2 + A0k − A02 − (A − A0 − NA) + λη j=2 i=2 (Ai − A0i − Ni Ai ), . = ∑k ∑k−1 ∑j N2 − (NA − A1 + A0 ) + λ i=2 2Ni ∆AAi−1 + λ j=2 i=2 (Ni Ai − A1i + A0i ) i

(23)

By Lemma 8, we deduce that (A, N) is a positive solution of systems (4) and (5) if and only if (A, N) is a fixed point of F . ¯ is a fixed point of F(·, 0) if and only if A − A0 and N are constants, set A = A0 + B Lemma 9 (A, N) ∈ Ω and B = (B, · · · , B), N = (N, · · · , N ), where (B, N) satisfies the algebraic system B − N A0 − N B = 0,

N A0 + N B − A1 + A0 = 0,

(24)

whose unique solution is given by: B = A1 − A0 , Proof

N=

A1 − A0 A1

.

(25)

¯ is a fixed point of F(·, 0) if and only if Since (A, N) ∈ Ω Ak = A2 + A0k − A02 − (A − A0 − NA),

Nk = N2 − (NA − A1 + A0 ),

i.e. if and only if (A − A0 − NA) = 0, (NA − A1 + A0 ) = 0,

(26)

and Bk = Ak − A0k ≡ A2 − A02 and Nk ≡ N2 for any k ∈ [2, n − 1]Z , that is, B and N are constants, combining this with (26), we have that (24) and (25) hold.

2

¯. Proof of Theorem 1 It is easy to see that solving (1) is equivalent to finding a fixed point of F(·, 1) in Ω Furthermore, from the definition of Ω and Lemmas 3 and 4, Corollaries 1 and 2, we have that F(·, λ) has no fixed point in ∂Ω for all λ ∈ (0, 1]. By using the homotopy invariance of degree, we can conclude that deg[I − F(·, 1), Ω, 0] = deg[I − F(·, 0), Ω, 0].

(27)

On the other hand, by Lemma 9, any fixed point of F(·, 0) has the form (A0 + B, N) with B and N are constants satisfying (25), so that R2 − A0k < Ak − A0k < R1 − A0k and 0 ≤ Nk < R5 . Set Ak = A0k + B . Now the invariance of the topological degree by translation yields that deg[I − F(·, 0), Ω, 0] = deg[(I − F)(A0 + ·, ·, 0), Ω − (A0 , 0), 0], 2377


where (I − F)(A0 + B, N, 0) = (Bk − B2 + B − NA0 − NB, Nk − N2 + NA0 + NB − A1 + A0 ) ˜ =: (I − F)(B, N). Then it follows from the reduction theorem of the topological degree that ˜ Ω − (A0 , 0), 0] = deg[(I − F)| ˜ R2 , (Ω − (A0 , 0)) ∩ R2 , 0] deg[I − F, = deg[G, (R2 − max A0 , R1 − min A0 ) × [0, R5 ), 0], where G : [R2 − max A0 , R1 − min A0 ] × [0, R5 ] → R is defined by G(B, N ) = (B − N A0 − N B, N A0 + N B − A1 + A0 ). Since the unique zero of G is given by (25). Therefore, deg[G, (R2 − max A0 , R1 − min A0 ) × (0, R5 ), 0] = 1. Consequently, deg[I − F(·, 1), Ω, 0] = 1, and hence, the existence property of degree theory implies that F(·, 1) has at least one fixed point in Ω , i.e. system (1) admits at least one nonconstant positive solution.

2

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