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The existence of symmetric positive solutions for a seconder-order difference equation with sum form boundary conditions

Abstract

In this paper, we consider the existence of positive solutions for a second-order discrete boundary value problem Δ(g(k1)Δu(k1))+w(k)f(k,u(k))=0 subject to the boundary conditions: au(0)bg(0)Δu(0)= i = 1 n 1 h(i)u(i), au(n)+bg(n1)Δu(n1)= i = 1 n 1 h(i)u(i), where a,b>0, Δu(k)=u(k+1)u(k) for k{0,1,,n1}, g(k)>0 is symmetric on {0,1,,n1}, w(k) is symmetric on {0,1,,n}, f:{0,1,,n}×[0,+) is continuous, f(k,u)=f(nk,u) for all (k,u){0,1,,n}×[0,+), and h(i) is nonnegative and symmetric on {0,1,,n}. By the fixed point theorem and the Hölder inequality, we study the existence of symmetric positive solutions for the above difference equation with sum form boundary conditions.

1 Introduction

A class of boundary value problems (BVPs) with integral boundary conditions arise in thermal conduction problems, semiconductor problems, and hydrodynamic problems [13]. Recently, such problems have been investigated by many authors [410]. The equation ( g ( t ) u ( t ) ) +w(t)f(t,u(t))=0, 0<t<1, describes many phenomena in the fields of gas dynamics, nuclear physics, chemically reacting systems and atomic structures [1115]. In [10], Feng considered the following differential equation BVP with integral boundary conditions:

( g ( t ) u ( t ) ) +w(t)f ( t , u ( t ) ) =0,0<t<1,
(1.1)
au(0)b lim t 0 + g(t) u (t)= 0 1 h(s)u(s)ds,
(1.2)
au(1)+b lim t 1 g(t) u (t)= 0 1 h(s)u(s)ds.
(1.3)

Applying the fixed point index theorem and the Hölder inequality, the author studied the existence of symmetric positive solutions for BVP (1.1)-(1.3).

Motivated by the above works, we will study the following BVP with sum form boundary conditions:

Δ ( g ( k 1 ) Δ u ( k 1 ) ) +w(k)f ( k , u ( k ) ) =0,k{1,,n1},
(1.4)
au(0)bg(0)Δu(0)= i = 1 n 1 h(i)u(i),
(1.5)
au(n)+bg(n1)Δu(n1)= i = 1 n 1 h(i)u(i).
(1.6)

Throughout this paper, the following conditions are assumed:

(A1) a,b>0, w(k) is symmetric on {0,1,,n}, and there exists m>0 such that w(k) m n 1 on {0,1,,n}, g(k)>0 for k{0,1,,n}, and g(k) is symmetric on {0,1,,n1}, h is nonnegative, symmetric on {0,1,,n}, and 0s<a, where s= i = 1 n 1 h(i), f:{0,1,,n}×[0,+) is continuous and f(,u) is symmetric on {0,1,,n} for all u0.

Remark 1 The conditions that g and h are symmetric on the different sets, which can guarantee the symmetry of associated kernel function for BVP (1.4)-(1.6). The kernel functions are then used to obtain the existence of symmetric positive solutions for BVP (1.4)-(1.6) by constructing a suitable operator.

In order to study the existence of symmetric positive solutions of problem (1.4)-(1.6), we need the following lemmas.

Lemma 1.1 [16]

Let P be a cone of the real Banach space E and Ω be a bounded open subset of E and θΩ. Assume A:P Ω ¯ P is a completely continuous operator and satisfies Au=μu, uPΩ, μ<1. Then i(A,PΩ,P)=1.

Lemma 1.2 [16]

Suppose A:P Ω ¯ P is a completely continuous operator, and satisfies

  1. (1)

    inf u P Ω Au>0;

  2. (2)

    Au=μu, uPΩ, μ(0,1].

Then i(A,PΩ,P)=0.

Lemma 1.3 (Hölder)

Suppose u={ u 1 , u 2 ,, u n } is a real-valued column, let

u p = { ( k = 1 n | u k | p ) 1 / p , 0 < p < , sup k { 1 , 2 , , n } | u k | , p = ,

where p, q satisfy the condition 1 p + 1 q =1, which are called conjugate exponents, and q= for p=1. If 1p, then

u v 1 u p v q ,

which can be denoted as

k = 1 n | u k v k | { ( k = 1 n | u k | p ) 1 / p ( k = 1 n | v k | q ) 1 / q , 1 < p < , ( k = 1 n | u k | ) ( sup k { 1 , 2 , , n } | v k | ) , p = 1 , ( sup k { 1 , 2 , , n } | u k | ) ( k = 1 n | v k | ) , p = .

2 Preliminaries

Let E={u(k):{0,1,,n}R}. It is well known that E is a real Banach space with the norm defined by u= max k { 0 , 1 , , n } |u(k)|. Let K be a cone of E,

K r = { u K : u r } , K r = { u K : u = r } ,

where r>0.

In our main results, we will use the following lemmas.

Lemma 2.1 Assume that (A1) holds. Then for any yE, the BVP

Δ ( g ( k 1 ) Δ u ( k 1 ) ) =y(k),k{1,,n1},
(2.1)
au(0)bg(0)Δu(0)= i = 1 n 1 h(i)u(i),
(2.2)
au(n)+bg(n1)Δu(n1)= i = 1 n 1 h(i)u(i)
(2.3)

has a unique solution u given by

u(k)= i = 1 n 1 H(k,i)y(i),

where

H(k,i)=G(k,i)+ 1 a s τ = 1 n 1 G(τ,i)h(τ),
(2.4)
G(k,i)= 1 Δ { ( b + a j = k n 1 1 g ( j ) ) ( b + a j = 0 i 1 1 g ( j ) ) , 0 i < k , ( b + a j = i n 1 1 g ( j ) ) ( b + a j = 0 k 1 1 g ( j ) ) , k i n ,
(2.5)

and Δ=2ab+ a 2 j = 0 n 1 1 g ( j ) , s= i = 1 n 1 h(i).

Proof From the properties of the difference operator, it is easy to see that

g(k)Δu(k)+g(k1)Δu(k1)=y(k),

then we have

g ( 1 ) Δ u ( 1 ) + g ( 0 ) Δ u ( 0 ) = y ( 1 ) , g ( 2 ) Δ u ( 2 ) + g ( 1 ) Δ u ( 1 ) = y ( 2 ) , g ( k ) Δ u ( k ) + g ( k 1 ) Δ u ( k 1 ) = y ( k ) .

From the above equalities, we can obtain

g(k)Δu(k)+g(0)Δu(0)= i = 1 k y(i).

Let g(0)Δu(0)=A, then

Δu(k)= 1 g ( k ) A 1 g ( k ) i = 1 k y(i),

that is,

u(k+1)u(k)= 1 g ( k ) A 1 g ( k ) i = 1 k y(i).

So,

u ( 1 ) u ( 0 ) = 1 g ( 0 ) A , u ( 2 ) u ( 1 ) = 1 g ( 1 ) A 1 g ( 1 ) i = 1 1 y ( i ) , u ( 3 ) u ( 2 ) = 1 g ( 2 ) A 1 g ( 2 ) i = 1 2 y ( i ) , u ( k ) u ( k 1 ) = 1 g ( k 1 ) A 1 g ( k 1 ) i = 1 k 1 y ( i ) .

It follows that

u(k)=u(0)+A j = 0 k 1 1 g ( j ) j = 1 k 1 1 g ( j ) i = 1 j y(i).

By the boundary conditions, we get

a u ( 0 ) b A = i = 1 n 1 h ( i ) u ( i ) , a u ( 0 ) + ( b + a j = 0 n 1 1 g ( j ) ) A = i = 1 n 1 h ( i ) u ( i ) + a j = 1 n 1 1 g ( j ) i = 1 j y ( i ) + b i = 1 n 1 y ( i ) .

Then

A = 1 2 b + a j = 0 n 1 1 g ( j ) ( a j = 1 n 1 1 g ( j ) i = 1 j y ( i ) + b i = 1 n 1 y ( i ) ) , u ( 0 ) = b 2 a b + a 2 j = 0 n 1 1 g ( j ) ( a j = 1 n 1 1 g ( j ) i = 1 j y ( i ) + b i = 1 n 1 y ( i ) ) + 1 a i = 1 n 1 h ( i ) u ( i ) .

Thus,

u ( k ) = 1 a i = 1 n 1 h ( i ) u ( i ) + b 2 a b + a 2 j = 0 n 1 1 g ( j ) ( a j = 1 n 1 1 g ( j ) i = 1 j y ( i ) + b i = 1 n 1 y ( i ) ) + j = 0 k 1 1 g ( j ) 1 2 b + a j = 0 n 1 1 g ( j ) ( a j = 1 n 1 1 g ( j ) i = 1 j y ( i ) + b i = 1 n 1 y ( i ) ) j = 1 k 1 1 g ( j ) i = 1 j y ( i ) = 1 a i = 1 n 1 h ( i ) u ( i ) + i = 1 n 1 G ( k , i ) y ( i ) ,

where G(k,i) is defined by (2.5). Multiplying the above equation with h(k), and summing from 1 to n1, we can get

i = 1 n 1 h(i)u(i)= a a k = 2 n 1 h ( k ) k = 1 n 1 h(k) i = 1 n 1 G(k,i)y(i).

One deduces that

u ( k ) = i = 1 n 1 G ( k , i ) y ( i ) + 1 a k = 1 n 1 h ( k ) k = 1 n 1 h ( k ) i = 1 n 1 G ( k , i ) y ( i ) = i = 1 n 1 H ( k , i ) y ( i ) ,

where H(k,i) is defined by (2.4). The proof is complete. □

From the above work, we can prove that H(k,i) and G(k,i) have the following properties.

Proposition 2.1 If (A1) holds, then we have

H(k,i)>0,G(k,i)>0,for k,i{0,1,,n};
(2.6)
G(nk,ni)=G(k,i),H(nk,ni)=H(k,i),for k,i{0,1,,n};
(2.7)
1 Δ b 2 G(k,i)G(i,i) 1 Δ D, 1 Δ a b 2 γH(k,i)H(i,i) 1 Δ aγD,
(2.8)

where D= ( b + a j = 0 n 1 g ( j ) ) 2 , γ= 1 a s , k,i{0,1,,n}.

Proof It is clear that (2.6) holds. Now we prove (2.7) holds.

If i{0,1,,k1}, then nink, from (2.5) and (A1) we get

G ( n k , n i ) = 1 Δ ( b + a j = n i n 1 1 g ( j ) ) ( b + a j = 0 n k 1 1 g ( j ) ) = 1 Δ ( b + a j = n i n 1 1 g ( n 1 j ) ) ( b + a j = 0 n k 1 1 g ( n 1 j ) ) = 1 Δ ( b + a j = 0 i 1 1 g ( j ) ) ( b + a j = k n 1 1 g ( j ) ) = G ( k , i ) , i { 0 , 1 , , k 1 } .

Similarly, we can prove that G(nk,ni)=G(k,i), i{k,,n}. So we have G(nk,ni)=G(k,i), for k,i{0,1,,n}. From (2.4) and (A1), we have

H ( n k , n i ) = G ( n k , n i ) + 1 a s τ = 1 n 1 G ( τ , n i ) h ( τ ) = G ( k , i ) + 1 a s τ = 1 n 1 G ( n τ , i ) h ( n τ ) = G ( k , i ) + 1 a s τ = 1 n 1 G ( τ , i ) h ( τ ) = H ( k , i ) .

So, (2.7) is established. Next we prove (2.8) holds. In fact, for k,i{0,1,,n}, if i{0,1,,k1}, then

G ( k , i ) = 1 Δ ( b + a j = k n 1 1 g ( j ) ) ( b + a j = 0 i 1 1 g ( j ) ) 1 Δ ( b + a j = i n 1 1 g ( j ) ) ( b + a j = 0 i 1 1 g ( j ) ) = G ( i , i ) 1 Δ ( b + a j = 0 n 1 g ( j ) ) ( b + a j = 0 n 1 g ( j ) ) 1 Δ ( b + a j = 0 n 1 g ( j ) ) 2 = 1 Δ D .

Similarly, we can prove that G(k,i)G(i,i) 1 Δ D, for i{k,k+1,,n}. Therefore G(k,i)G(i,i) 1 Δ D. For k,i{0,1,,n}, we can get

H ( k , i ) = G ( k , i ) + 1 a s τ = 1 n 1 G ( τ , i ) h ( τ ) G ( i , i ) + 1 a s τ = 1 n 1 G ( τ , i ) h ( τ ) = H ( i , i ) G ( i , i ) + 1 a s τ = 1 n 1 G ( τ , τ ) h ( τ ) 1 Δ D + 1 Δ D 1 a s τ = 1 n 1 h ( τ ) = 1 Δ ( 1 + 1 a s τ = 1 n 1 h ( τ ) ) D = a Δ ( a s ) D = 1 Δ a γ D .

On the other hand, from (2.5), we have

G(k,i) 1 Δ ( b + a j = n n 1 1 g ( j ) ) ( b + a j = 0 1 1 g ( j ) ) = 1 Δ b 2 .

So, by (2.4), for k,i{0,1,,n}, we can obtain

H ( k , i ) = G ( k , i ) + 1 a s τ = 1 n 1 G ( τ , i ) h ( τ ) 1 Δ b 2 + b 2 ( a s ) Δ τ = 1 n 1 h ( τ ) 1 Δ b 2 ( 1 + 1 a s τ = 1 n 1 h ( τ ) ) = 1 Δ b 2 a a s = 1 Δ b 2 a γ .

Thus,

1 Δ b 2 G ( k , i ) G ( i , i ) 1 Δ D , 1 Δ b 2 a γ H ( k , i ) H ( i , i ) 1 Δ D a γ .

The proof is completed. □

Remark 2 The symmetry of g(k) on {0,1,,n1} can guarantee that G(k,i) is symmetric for k,i{0,1,,n}, and the symmetry of h(k) on {0,1,,n} can guarantee that H(k,i) is symmetric for k,i{0,1,,n}.

Next, we can construct a cone in E by

K = { u E : u 0 , u ( k )  is symmetric on  { 0 , 1 , , n } , Δ ( g ( k ) Δ u ( k ) ) 0 , k { 0 , 1 , , n 2 } ,  and  min k { 0 , 1 , , n } u ( k ) δ u } ,

where δ = 1 D b 2 . Then we define an operator

(Tu)(k)= i = 1 n 1 H(k,i)w(i)f ( i , u ( i ) ) .
(2.9)

It can be observed that u is a solution of problem (1.4)-(1.6) if and only if u is a fixed point of operator T.

We can get the following lemma from Lemma 2.1.

Lemma 2.2 Suppose (A1) holds. If u is a solution of the equation

u(k)=Tu(k)= i = 1 n 1 H(k,i)w(i)f ( i , u ( i ) ) ,

then u is a solution of BVP (1.4)-(1.6).

Lemma 2.3 Assume (A1) holds. Then T(K)K and T:KK is completely continuous.

Proof For uK, from (2.9), we obtain Δ(g(k1)ΔTu(k1))=w(k)f(k,u(k))0. By Proposition 2.1, it is to see that (Tu)(k)0, for k{0,1,,n}. Using the fact that w, u, f(k,u) are symmetric on {0,1,,n}, we have

( T u ) ( n k ) = i = 1 n 1 H ( n k , i ) w ( i ) f ( i , u ( i ) ) = i = 1 n 1 H ( k , n i ) w ( n i ) f ( n i , u ( n i ) ) = i = 1 n 1 H ( k , i ) w ( i ) f ( i , u ( i ) ) = ( T u ) ( k ) ,

then Tu is symmetric on {0,1,,n} for k{0,1,,n}. And from (2.8) we can see

(Tu)(k)= i = 1 n 1 H(k,i)w(i)f ( i , u ( i ) ) 1 Δ aγD i = 1 n 1 w(i)f ( i , u ( i ) ) .

Thus,

Tu 1 Δ aγD i = 1 n 1 w(i)f ( i , u ( i ) ) .

Similarly, by (2.8) we obtain

( T u ) ( k ) = i = 1 n 1 H ( k , i ) w ( i ) f ( i , u ( i ) ) 1 Δ a b 2 γ i = 1 n 1 w ( i ) f ( i , u ( i ) ) = 1 Δ a δ D γ i = 1 n 1 w ( i ) f ( i , u ( i ) ) δ T u .

Thus, TuK and T(K)K. It is clear that T:KK is completely continuous. □

Remark 3 The symmetry of the kernel function H(k,i) for k,i{0,1,,n} can guarantee that Tu is symmetric on {0,1,,n} for uK.

3 Main results

In this section, we will establish that problem (1.4)-(1.6) has at least one positive solution with Lemma 1.1 and Lemma 1.2. We need consider the following situations: p>1, p=1, p=. Next, we will prove a theorem for p>1. At first, we define

H= sup i { 1 , 2 , , n 1 } | H ( i , i ) | , H p = ( i = 1 n 1 | H ( i , i ) | p ) 1 / p .

Let

F β = lim u β sup max k { 0 , 1 , , n } f ( k , u ) u , F β = lim u β inf min k { 0 , 1 , , n } f ( k , u ) u ,

where β denotes 0 or ∞, and

N 1 = max { H p ( i = 1 n 1 | w ( i ) | q ) 1 / q , ( i = 1 n 1 | H ( i , i ) | ) ( sup i { 1 , 2 , , n 1 } | w ( i ) | ) , H ( i = 1 n 1 | w ( i ) | ) } , L 1 = 1 Δ δ a γ m b 2 .

Theorem 3.1 Assume that conditions (A1) hold. In addition, suppose that

(A2) 0< F 0 <N, and L< F <, or

(A3) 0< F <N, and L< F 0 <

are satisfied. Then problem (1.4)-(1.6) has at least one symmetric positive solution.

Proof We only consider (A2) case, (A3) is similar to (A2). If 0< F 0 <N, then there exist r>0, ε 0 >0 such that N ε 0 >0 and for all 0<ur, we have

f(k,u)(N ε 0 )u(N ε 0 )r,k{0,1,,n}.
(3.1)

For all u K r , from Lemma 1.3 we obtain

( T u ) ( k ) = i = 1 n 1 H ( k , i ) w ( i ) f ( i , u ( i ) ) i = 1 n 1 H ( k , i ) w ( i ) ( N ε 0 ) r i = 1 n 1 H ( i , i ) w ( i ) ( N ε 0 ) r ( i = 1 n 1 | H ( i , i ) | p ) 1 / p ( i = 1 n 1 | w ( i ) | q ) 1 / q ( N ε 0 ) r N 1 ( N ε 0 ) r r .

So Tuλu, for u K r , λ1. From Lemma 1.1, we can get i(T, K r ,K)=1. Next, we prove it satisfies Lemma 1.2. Because L< F <, there exist R> δ r>0, ε 1 >0 such that

f(k,u)(L+ ε 1 )u,uR,k{0,1,,n}.

Let r = δ 1 R, then r >r, and

min k { 0 , 1 , , n } u(k) δ u=R,u K r .

Now we prove that Tuλu, u K r , 0<λ1. If not, then there exist u 0 K r and 0< λ 0 1 such that T u 0 = λ 0 u 0 ; thus we have

r u 0 ( k ) = λ 0 1 ( T u 0 ) ( k ) = λ 0 1 i = 1 n 1 H ( k , i ) w ( i ) f ( i , u ( i ) ) 1 Δ a b 2 γ ( L + ε 1 ) i = 1 n 1 w ( i ) u ( i ) 1 Δ a b 2 γ ( L + ε 1 ) R i = 1 n 1 w ( i ) = 1 Δ a b 2 γ ( L + ε 1 ) δ r i = 1 n 1 w ( i ) 1 Δ a b 2 γ ( L + ε 1 ) δ r m = L 1 ( L + ε 1 ) r = r ( 1 + ε 1 L ) > r ,

i.e., r > r , which is a contradiction. In addition, because (Tu)(k) r (1+ ε 1 L )> r , so inf u K r Tu r >0, from Lemma 1.2 we have i(T, K r ,K)=0. On the other hand, from the above work with the additivity of the fixed point index, we get

i(T, K r K r ¯ ,K)=i(T, K r ,K)i(T, K r ,K)=01=1.

So, T has at least one fixed point u on K r K r ¯ . Then it follows that problem (1.4)-(1.6) has a symmetric positive solution u . The proof is complete. □

Remark 4 From the proof of Theorem 3.1, we can establish that problem (1.4)-(1.6) has another nonnegative solution u , u K r .

The following corollary deals with the case p=1.

Corollary 3.1 Suppose that (A1), (A2) hold. Then problem (1.4)-(1.6) has at least one symmetric positive solution.

Proof It is similar to the proof of Theorem 3.1. Let ( i = 1 n 1 |H(i,i)|)( sup i { 1 , , n 1 } |w(i)|) replace ( i = 1 n 1 | H ( i , i ) | p ) 1 / p ( i = 1 n 1 | w ( i ) | q ) 1 / q and repeat the argument of Theorem 3.1. □

Finally, we consider the case of p=.

Corollary 3.2 Assume that (A1), (A2) hold. Then problem (1.4)-(1.6) has at least one symmetric positive solution.

Proof It is similar to the proof of Theorem 3.1. For all u K r , we have

( T u ) ( k ) = i = 1 n 1 H ( k , i ) w ( i ) f ( i , u ( i ) ) i = 1 n 1 H ( i , i ) w ( i ) ( N ε 0 ) r ( sup i { 1 , 2 , , n 1 } | H ( i , i ) | ) ( i = 1 n 1 | w ( i ) | ) ( N ε 0 ) r N 1 ( N ε 0 ) r < r .

So Tuλu, u K r , λ1. By Lemma 1.1, we can get i(T, K r ,K)=1. This together with i(T, K r ,K)=0 in the proof of Theorem 3.1 completes the proof. □

References

  1. Cannon JR: The solution of the heat equation subject to the specification of energy. Q. Appl. Math. 1963, 21: 155-160.

    MathSciNet  MATH  Google Scholar 

  2. Ionkin NI: Solution of boundary value problem in heat conduction theory with nonlocal boundary conditions. Differ. Equ. 1977, 13: 294-304.

    MathSciNet  Google Scholar 

  3. Chegis RY: Numerical solution of heat conduction problem with an integral boundary condition. Liet. Mat. Rink. 1984, 24: 209-215.

    MathSciNet  MATH  Google Scholar 

  4. Boucherif A: Second-order boundary value problems with integral boundary conditions. Nonlinear Anal. 2009, 70: 364-371. 10.1016/j.na.2007.12.007

    MathSciNet  Article  MATH  Google Scholar 

  5. Infante, G: Eigenvalues and positive solutions of ODEs involving integral boundary conditions. Discrete Contin. Dyn. Syst. suppl., 436-442 (2005)

  6. Yang Z: Positive solutions of a second order integral boundary value problem. J. Math. Anal. Appl. 2006, 321: 751-765. 10.1016/j.jmaa.2005.09.002

    MathSciNet  Article  MATH  Google Scholar 

  7. Ahmad B, Alsaedi A, Alghamdi BS: Analytic approximation of solutions of the forced Duffing equation with integral boundary conditions. Nonlinear Anal., Real World Appl. 2008, 9: 1727-1740. 10.1016/j.nonrwa.2007.05.005

    MathSciNet  Article  MATH  Google Scholar 

  8. Ahmad B, Alsaedi A: Existence of approximate solutions of the forced Duffing equation with discontinuous type integral boundary conditions. Nonlinear Anal., Real World Appl. 2009, 10: 358-367. 10.1016/j.nonrwa.2007.09.004

    MathSciNet  Article  MATH  Google Scholar 

  9. Feng M, Zhang X, Ge W: New existence results for higher-order nonlinear fractional differential equation with integral boundary conditions. Bound. Value Probl. 2011., 2011: Article ID 720702

    Google Scholar 

  10. Feng M: Existence of symmetric positive solutions for a boundary value problem with integral boundary conditions. Appl. Math. Lett. 2011, 24: 1419-1427. 10.1016/j.aml.2011.03.023

    MathSciNet  Article  MATH  Google Scholar 

  11. Csavinszky P: Universal approximate solution of the Thomas-Fermi equation for ions. Phys. Rev. A 1973, 8: 1688-1701. 10.1103/PhysRevA.8.1688

    Article  Google Scholar 

  12. Granas, A, Guenther, RB, Lee, JW: Nonlinear boundary value problems for ordinary differential equations. Diss. Math. 244, 128 pp. (1985)

  13. Granas A, Guenther RB, Lee JW: A note on the Thomas-Fermi equations. Z. Angew. Math. Mech. 1981, 61: 240-241.

    MathSciNet  MATH  Google Scholar 

  14. Luning CD, Perry WL: Positive solutions of negative exponent generalized Emden-Fowler boundary value problems. SIAM J. Math. Anal. 1981, 12: 874-879. 10.1137/0512073

    MathSciNet  Article  MATH  Google Scholar 

  15. Wong JSW: On the generalized Emden-Fowler equations. SIAM Rev. 1975, 17: 339-360. 10.1137/1017036

    MathSciNet  Article  MATH  Google Scholar 

  16. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, New York; 1988.

    MATH  Google Scholar 

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Acknowledgements

The authors express their sincere thanks to the referees for the careful and details reading of the manuscript and very helpful suggestions. The project was supported by the Natural Science Foundation of China (11371120), the Natural Science Foundation of Hebei Province (A2013208147) and the Education Department of Hebei Province Science and Technology Research Project (Z2014095).

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Guo, Y., Ji, Y. & Lv, X. The existence of symmetric positive solutions for a seconder-order difference equation with sum form boundary conditions. Adv Differ Equ 2014, 237 (2014). https://doi.org/10.1186/1687-1847-2014-237

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Keywords

  • difference equation
  • sum form boundary conditions
  • symmetric positive solutions