2010 Right Focal Boundary Value Problems for Difference Equations Johnny Henderson Baylor University Xueyan Liu Baylor University Jeffrey W.. NSUWorks Citation Henderson, Johnny; Liu, Xu
Trang 12010
Right Focal Boundary Value Problems for
Difference Equations
Johnny Henderson
Baylor University
Xueyan Liu
Baylor University
Jeffrey W Lyons
Baylor University,jlyons@nova.edu
Jeffrey T Neugebauer
Baylor University
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NSUWorks Citation
Henderson, Johnny; Liu, Xueyan; Lyons, Jeffrey W.; and Neugebauer, Jeffrey T., "Right Focal Boundary Value Problems for Difference
Equations" (2010) Mathematics Faculty Articles 103.
https://nsuworks.nova.edu/math_facarticles/103
Trang 2Opuscula Mathematica • Vol 30 • No 4 • 2010
RIGHT FOCAL BOUNDARY VALUE PROBLEMS
FOR DIFFERENCE EQUATIONS
Johnny Henderson, Xueyan Liu, Jeffrey W Lyons, Jeffrey T Neugebauer
Abstract An application is made of a new Avery et al fixed point theorem of compression and expansion functional type in the spirit of the original fixed point work of Leggett and Williams, to obtain positive solutions of the second order right focal discrete boundary value problem In the application of the fixed point theorem, neither the entire lower nor entire upper boundary is required to be mapped inward or outward A nontrivial example is also provided
Keywords: difference equation, boundary value problem, right focal, fixed point theorem, positive solution
Mathematics Subject Classification: 39A10
1 INTRODUCTION
For well over a decade, substantial results have been obtained for positive solutions and multiple positive solutions for boundary value problems for finite difference equa-tions; see, for example [2, 5, 10, 11, 13, 15, 16, 20–23, 25–28]
Many of those results have been motivated by the applicability of a number of new fixed point theorems and multiple fixed point theorems as applied to certain discrete boundary value problems; such as the classical fixed point theorems of Guo and Krasnosel’skii [14,17] or Leggett and Williams [19], along with several newer fixed point theorems by Avery et al [1, 3, 6–9], and the fixed point theorem of Ge [12] Recently, Avery, Anderson and Henderson [4] gave a topological proof in obtaining
a Leggett-Williams type of fixed point theorem, which requires only that certain subsets of both boundaries of a subset of a cone for which kxk > b and α(x) = a, where α is a concave positive functional on the cone, be mapped inward and outward, respectively This is an expansion result which is dramatically different from the Leggett-Williams fixed point theorem, which is in itself only a compression result Moreover, this new fixed point theorem [4] is more general than those obtained by
447 http://dx.doi.org/10.7494/OpMath.2010.30.4.447
Trang 3using Guo-Krasnosel’skii compression-expansion results which mapped at least one boundary inward or outward [1,8,14,19,24], or the topological generalizations of fixed point theorems introduced by Kwong [18] which require boundaries to be mapped inward or outward (invariance-like conditions) Moreover, conditions involving the norm in the original Leggett-Williams fixed point theorem were replaced in this recent fixed point theorem [4] by more general conditions on a convex functional
In this paper, we give a first application of the Avery et al fixed point theorem [4] to right focal boundary problems for finite difference equations, by demonstrating
a technique that takes advantage of the flexibility of the new fixed point theorem in obtaining at least one positive solution for
∆2u(k) + f (u(k)) = 0, k ∈ {0, 1, , N }, (1.1)
where f : [0, ∞) → [0, ∞) is continuous In Section 2, we provide some background definitions and we state the new fixed point theorem In Section 3, we apply the fixed point theorem to obtain a positive solution to (1.1), (1.2), and in Section 3, we provide a nontrivial example of the existence result of Section 2
2 BACKGROUND AND A FIXED POINT THEOREM
In this section, we present some definitions used for the remainder of the paper In addition, we include a new fixed point theorem statement whose application, in the next section, will yield a solution of (1.1), (1.2)
Definition 2.1 Let E be a real Banach space A nonempty closed convex set P ⊂ E
is called a cone if it satisfies the following two conditions:
(i) x ∈ P, λ ≥ 0 implies λx ∈ P ;
(ii) x ∈ P , −x ∈ P implies x = 0
Definition 2.2 A map α is said to be a nonnegative continuous concave functional
on a cone P of a real Banach space E if
α : P → [0, ∞)
is continuous and
α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y) for all x, y ∈ P and t ∈ [0, 1] Similarly we say the map β is a nonnegative continuous convex functional on a cone P of a real Banach space E if
β : P → [0, ∞)
is continuous and
β(tx + (1 − t)y) ≤ tβ(x) + (1 − t)β(y) for all x, y ∈ P and t ∈ [0, 1]
Trang 4Right focal boundary value problems for difference equations 449
Let ψ and δ be nonnegative continuous functionals on a cone P ; then, for positive real numbers a and b, we define the sets:
and
P (ψ, δ, a, b) := {x ∈ P : a ≤ ψ(x) and δ(x) ≤ b} (2.2) The following theorem [4] is the new fixed point theorem of compression-expansion and functional type
Theorem 2.3 Suppose P is a cone in a real Banach space E, α is a nonnegative continuous concave functional on P , β is a nonnegative continuous convex functional
on P and T : P → P is a completely continuous operator Assume there exist nonnegative numbers a, b, c and d such that:
(A1) {x ∈ P : a < α(x) and β(x) < b} 6= ∅;
(A2) if x ∈ P with β(x) = b and α(x) ≥ a, then β(T x) < b;
(A3) if x ∈ P with β(x) = b and α(T x) < a ,then β(T x) < b;
(A4) {x ∈ P : c < α(x) and β(x) < d} 6= ∅;
(A5) if x ∈ P with α(x) = c and β(x) ≤ d, then α(T x) > c;
(A6) if x ∈ P with α(x) = c and β(T x) > d, then α(T x) > c
If
(H1) a < c, b < d, {x ∈ P : b < β(x) and α(x) < c} 6= ∅, P (β, b) ⊂ P (α, c),
and P (α, c) is bounded,
then T has a fixed point x∗ in P (β, α, b, c)
If
(H2) c < a, d < b, {x ∈ P : a < α(x) and β(x) < d} 6= ∅, P (α, a) ⊂ P (β, d), and P (β, d) is bounded,
then T has a fixed point x∗ in P (α, β, a, d)
3 SOLUTIONS OF (1.1), (1.2)
In this section, we impose growth conditions on f such that the right focal boundary value problem for the finite difference equation, (1.1), (1.2), has a solution as a con-sequence of Theorem 2.3 We note that from the nonnegativity of f , a solution u of (1.1), (1.2) is both nonnegative and concave on {0, 1, , N + 2} In our application
of Theorem 2.3, we will deal with a completely continuous summation operator whose kernel is the Green’s function, H(k, `), for
and satisfying (1.2) In particular, for (k, `) ∈ {0, , N + 2} × {0, , N },
N + 2
k, k ∈ {0, , `},
` + 1, k ∈ {` + 1, , N + 2}
Trang 5We observe that H(k, `) is nonnegative, and for each fixed ` ∈ {0, , N }, H(k, `) is nondecreasing as a function of k In addition, it is straightforward that, for y, w ∈ {0, , N + 2} with y ≤ w,
Next, let E = {v : {0, , N + 2} → R} be endowed with the norm, kvk = maxk∈{0, ,N +2}|v(k)| Choose
τ ∈ {1, , N − 1}, and define the cone P ⊂ E by
P = {v ∈ E : v is nondecreasing and nonnegative-valued on {0, , N + 2},
∆2v(k) ≤ 0, k ∈ {0, , N }, and (N + 2)v(τ ) ≥ τ v(N + 2)
We note that, for any u ∈ P and y, w ∈ {0, , N + 2} with y ≤ w,
For v ∈ P , we define a nonnegative concave functional α on P by
k∈{τ, ,N +2}
v(k) = v(τ ),
and a nonnegative, convex functional β on P by
k∈{0, ,N +2}v(k) = v(N + 2)
We note that for v ∈ P , in terms of the functionals,
(N + 2)α(v) ≥ τ β(v)
Now, we put growth conditions on f such that (1.1), (1.2) has at least one solution
u∗ ∈ P (β, α, b, c), as a consequence of Theorem 2.3 under the expansive condition (H1)
Theorem 3.1 If τ ∈ {1, , N − 1} is fixed, b and c are positive real numbers with 3b ≤ c, and f : [0, ∞) → [0, ∞) is a continuous function such that:
(i) f (w) > τ (N −τ )c(N +2), for w ∈ [c,c(N +2)τ ],
(ii) f (w) is decreasing, for w ∈ [0,N +2bτ ], with f (N +2bτ ) ≥ f (w), for w ∈ [N +2bτ , b], and (iii) Pτ
`=0
(`+1)
N +2f (N +2b` ) < b − f (N +2bτ )[(N +1)(N +2)−(τ +1)(τ +2)2(N +2) ],
then the discrete right-focal problem (1.1), (1.2) has at least one positive solutions
u∗∈ P (β, α, b, c)
Trang 6Right focal boundary value problems for difference equations 451 Proof First, we let
N + 2 and d =
c(N + 2)
Then we have,
3(N + 2) < c and
b ≤ c
dτ 3(N + 2) < d.
Next, we define the summation operator T : E → E by
T u(k) =
N X
`=0 H(k, `)f (u(`)), u ∈ E, k ∈ {0, , N + 2}
It is immediate that T is completely continuous, and it is well known that u ∈ E is a solution of (1.1), (1.2) if, and only if u is a fixed point of T We now show that the conditions of Theorem 2.3 are satisfied with respect to T
So, if we let u ∈ P , then T u(k) = PN
`=0H(k, `)f (u(`)) ≥ 0 on {0, , N + 2} Moreover, ∆2(T u)(k) = −f (u(k)) ≤ 0, and so ∆(T u)(k) is nonincreasing on {0, , N + 1} From properties of H(k, `), ∆(T u)(N + 1) = 0, and so ∆(T u)(k) ≥ 0
on {0, , N + 1} Thus, (T u)(k) is nondecreasing on {0, , N + 2} Moreover,
T u(τ ) =
N
X
`=0
H(k, `)f (u(`)) ≥ τ
N + 2
N X
`=0
H(N + 2, `)f (u(`)) = τ
N + 2T u(N + 2). Therefore, we have T : P → P
We next proceed to verify properties (A1) and (A4) of Theorem 2.3 are satisfied First, for any L ∈ (2N +3−τ2b ,N +12b ), the function uL defined by
uL(k) :=
N
X
`=0
2(N + 2)(2N + 3 − k) ∈ {u ∈ P : a < α(u) and β(u) < b}, since
α(uL) = uL(τ ) = Lτ
2(N + 2)(2N + 3 − τ ) >
bτ
and
β(uL) = uL(N + 2) = L(N + 2)
2(N + 2)(2N + 3 − (N + 2)) < b.
Similarly, for any J ∈ (τ (2N +3−τ )2c(N +2) ,2c(N +2)τ (N +1)), the function uJ defined by
uJ(k) :=
N
X
`=0
J H(k, `) = J k
2(N + 2)(2N + 3 − k) ∈ {u ∈ P : c < α(u) and β(u) < d},
Trang 7α(uJ) = uJ(τ ) = J τ
2(N + 2)(2N + 3 − τ ) > c and
β(uJ) = uJ(N + 2) =J (N + 2)
2(N + 2)(2N + 3 − (N + 2)) =
J (N + 1)
c(N + 2)
Hence we have
{u ∈ P : a < α(u) andβ(u) < b} 6= ∅, and
{u ∈ P : c < α(u) andβ(u) < d} 6= ∅
Therefore conditions (A1) and (A4) of Theorem 2.3 are satisfied
Turning to (A2) of Theorem 2.3, let u ∈ P with β(u) = b and α(u) ≥ a By the concavity of u, for ` ∈ {0, , τ }, we have
u(`) ≥ u(τ )
τ
N + 2 and for all ` ∈ {τ, , N + 2}, we have
bτ
N + 2 ≤ u(`) ≤ b
Hence by (ii) and (iii), it follows that
β(T u) =
N X
`=0 H(N + 2, `)f (u(`)) =
N X
`=0
(` + 1)
N + 2f (u(`)) ≤
≤
τ X
`=0
(` + 1)
N + 2f (
b`
N + 2) +
N X
`=τ +1
(` + 1)
N + 2f (
bτ
N + 2) <
< b −f (
bτ
N +2)
N + 2
(N + 1)(N + 2) − (τ + 1)(τ + 2)
2
+
+f (
bτ
N +2)
N + 2
(N + 1)(N + 2) − (τ + 1)(τ + 2)
2
= b, and so (A2) is satisfied
Next, we establish (A3) of Theorem 2.3, and so we let u ∈ P with β(u) = b and α(T u) < a By the properties of H(k, `),
β(T u) =
N X
`=0 H(N + 2, `)f (u(`)) ≤
τ
N X
`=0
H(τ, `)f (u(`)) = N + 2
τ α(T u) <
a(N + 2)
and (A3) also holds
Trang 8Right focal boundary value problems for difference equations 453
In dealing with (A5), let u ∈ P with α(u) = c and β(u) ≤ d Then for ` ∈ {τ, , N + 2}, we have
c ≤ u(`) ≤ d = c(N + 2)
By Property (i),
α(T u) =
N X
`=0 H(τ, `)f (u(`)) ≥
N X
`=τ +1 H(τ, `)f (u(`)) =
=
N X
`=τ +1
τ
N + 2f (u(`)) >
N X
`=τ +1
c
N − τ = c, and so (A5) is valid
And now we address (A6) So, let u ∈ P with α(u) = c and β(T u) > d Again by the properties of H,
α(T u) =
N X
`=0 H(τ, `)f (u(`)) ≥
N + 2
N X
`=0 H(N + 2, `)f (u(`)) =
N + 2β(T u) >
τ d
N + 2 = c, and so (A6) of Theorem 2.3 also holds
Finally, we show that the conditions of (H1) are also in effect To that end, if
u ∈ P (α, c), then
τ
N + 2β(u) ≤ α(u) ≤ c, and hence
kxk = β(u) ≤ α(u)(N + 2)
Thus P (α, c) is a bounded subset of P Also, if u ∈ P (β, b), then
α(u) ≤ β(u) ≤ b < c, and hence P (β, b) ⊂ P (α, c)
In addtion, for any M ∈ (N +12b ,N +1c ), the function uM defined by
uM(k) :=
N
X
`=0
M H(k, `) =
k−1 X
`=0
M (` + 1)
N X
`=k
M k
M k 2(N + 2)(2N + 3 − k) belongs to the set P (β, α, b, c), since
α(uM) = uM(τ ) = M τ
2(N + 2)(2N + 3 − τ ) <
cτ 2(N + 1)(N + 2)(2N + 3 − τ ) < c,
Trang 9β(uM) = uM(N + 2) = M (N + 2)
2(N + 2) (2N + 3 − (N + 2)) =
2 (N + 1) >
2b 2(N + 1)(N + 1) = b.
Thus, we also have that {u ∈ P : b < β(u) and α(u) < c} 6= ∅ Hence the conditions
of (H1) are met
It follows from Theorem 2.3 that T has a fixed point u∗ ∈ P (β, α, b, c), and as such u∗ is a desired solution of (1.1), (1.2) The proof is complete
Example Let N = 8, τ = 1, b = 1, and c = 3 Notice that τ (N −τ )c(N +2) = 307, c(N +2)τ = 30, and N +2bτ =101 We define a continuous f : [0, ∞) → [0, ∞) by
f (w) =
−8w + 1, 0 ≤ w ≤ 19, 1
22
9w −73, w ≥ 1
Then:
(i) f (w) > 307, for w ∈ [3, 30],
(ii) f (w) is decreasing on [0,101], and f 101 ≥ f (w), for w ∈ [1
10, 1], and (iii)
1
X
`=0
` + 1
10 f
` 10
100 <
16
100 = 1 − f
1 10
9 · 10 − 2 · 3
2 · 10
Therefore, by Theorem 3.1, the right focal boundary value problem,
∆2u(k) + f (u(k)) = 0, k ∈ {0, , 8},
u(0) = 0 = ∆u(9), has at least one positive solution, u∗, with
1 ≤ u∗(10) and u∗(1) ≤ 3
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