Positive Green's Functions for Boundary Value

Positive Green’s Functions for Boundary Value Problems with Conformable Derivatives Douglas R. Anderson

Abstract We use a newly introduced conformable derivative to formulate several boundary value problems with three or four conformable derivatives, including those with conjugate, right-focal, and Lidstone conditions. With the conformable differential equation and boundary conditions established, we find the corresponding Green’s functions and prove their positivity under appropriate assumptions.

1 Introduction The search for the existence of positive solutions and multiple positive solutions to nonlinear fractional boundary value problems has expanded greatly over the past decade; for some recent examples please see [1,4-9,12,15-20]. In all of these works and the references cited therein, however, the definition of the fractional derivative used is either the Caputo or the Riemann-Liouville fractional derivative, involving an integral expression and the gamma function. Recently [10, 11, 14] a new local, limit-based definition of a conformable derivative has been formulated which is not a fractional derivative. In this paper, we use this conformable derivative of order α for 0 < α ≤ 1 and t ∈ [0, ∞) given by Dα f (t) := lim ε→0

f (teεt

−α

) − f (t) , ε

Dα f (0) = lim Dα f (t), t→0+

(1)

provided the limits exist; note that if f is fully differentiable at t, then Dα f (t) = t 1−α f 0 (t),

(2)

Douglas R. Anderson Concordia College, Department of Mathematics, Moorhead, MN 56562 USA, e-mail: [email protected]

1

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Douglas R. Anderson

where f 0 (t) = limε→0 [ f (t +ε)− f (t)]/ε. A function f is α−differentiable at a point t ≥ 0 if the limit in (1) exists and is finite. Using this new definition of a conformable derivative, we reformulate several common boundary value problems, including those with conjugate, right-focal, and Lidstone conditions. With the conformable differential equation and conformable boundary conditions established, we find the corresponding Green’s functions and prove their positivity under appropriate assumptions. This work thus sets the stage for using fixed point theorems to prove the existence of positive and multiple positive solutions to nonlinear conformable problems based on the local conformable derivative (1) and these boundary value problems, as the kernel of the integral operator is often Green’s function. Recently, the conformable derivative (1) has been used successfully to explore two iterated conformable derivatives in comparison with a conformable derivative of order between 1 and 2 in [2]. There is also some potential application in quantum mechanics of this conformable derivative, as shown in [3]. In this work we will focus on boundary value problems with three or four iterated conformable derivatives.

2 Conformable Derivative Properties Throughout this work we are considering the Katugampola derivative formulation of conformable calculus [10], specifically the operator Dα given in (1), where 0 < α ≤ 1 and t ≥ 0. This definition yields the following results [10, Theorem 2.3] for functions f and g that are α−differentiable at a point t > 0: • • • • • •

Dα [a f + bg] = aDα [ f ] + bDα [g] for all a, b ∈ R; Dα t n = nt n−α for all n ∈ R; Dα c = 0 for all constants c ∈ R; Dα [ f g] = f Dα [g] + gDα [ f ]; α f Dα [g] ; Dα [ f /g] = gD [ f ]− g2 Dα [ f ◦ g](t) = f 0 (g(t))Dα [g](t) for f differentiable at g(t).

3 Three Iterated Conformable Derivatives We begin the main discussion by considering the three iterated conformable derivative nonlinear right-focal problem Dγ Dβ Dα x(t) = f (t, x(t)), α

β

0 ≤ t ≤ 1, α

x(0) = D x(τ) = D D x(1) = 0,

(3) (4)

where α, β , γ ∈ (0, 1], with boundary point τ satisfying 0 < τ < 1. One could impose further the conditions α + β ∈ (1, 2] and α + β + γ ∈ (2, 3] if one wishes to explore

Positive Green’s Functions for Boundary Value Problems with Conformable Derivatives

3

a conformable problem of order (2, 3]. Our approach to the existence of positive solutions would involve Green’s function for this problem. Once the following three theorems are established, an interested reader could then apply a fixed point theorem to get positive solutions to (3), (4), although the details are omitted here. Theorem 1 (Conformable Right-Focal Problem). Let α, β , γ ∈ (0, 1] and 0 < τ < 1. The corresponding Green’s function for the homogeneous problem Dγ Dβ Dα x(t) = 0 satisfying boundary conditions (4) is given by (   u(t, s) : 0 ≤ t ≤ s ≤ 1   s ∈ [0, τ] :   x(0, s) : 0 ≤ s ≤ t ≤ 1   G(t, s) = (    u(t, τ) :0≤t ≤s≤1    s ∈ [τ, 1] : u(t, τ) + x(t, s) : 0 ≤ s ≤ t ≤ 1

(5)

where u(t, s) =

(α + β )t α sβ − αt α+β αβ (α + β )

(6)

and x(·, ·) is the Cauchy function given by
αt α t β − sβ + β sβ (sα − t α ) x(t, s) = . αβ (α + β )

(7)

Proof. Let h be any continuous function. We will show that Z 1

x(t) =

G(t, s)h(s)sγ−1 ds,

0

for G given by (5), is a solution to the linear boundary value problem Dγ Dβ Dα x(t) = h(t) with boundary conditions (4). First let t ∈ [0, τ]. Then Z t

x(t) = 0

x(0, s)h(s)sγ−1 ds +

Z τ

u(t, s)h(s)sγ−1 ds +

t

clearly x(0) = 0 by (6). Differentiating (8), we have

Z 1 τ

u(t, τ)h(s)sγ−1 ds;

(8)

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Douglas R. Anderson α

D x(t) = x(0,t)h(t)t

γ−1 1−α

t

sβ − t β β

Z τ

+ t

−u(t,t)h(t)t

γ−1 1−α

t

Z 1

+ τ

=

1 β

!

τβ − tβ β

h(s)sγ−1 ds ! h(s)sγ−1 ds

Z   1 1 β sβ − t β h(s)sγ−1 ds + τ − t β h(s)sγ−1 ds. β τ

Z τ t

Clearly the second boundary condition Dα x(τ) = 0 is met. Differentiating again, we have Z Z   1 τ 1 1 Dβ Dα x(t) = −βt β −1t 1−β h(s)sγ−1 ds + −βt β −1t 1−β h(s)sγ−1 ds β t β τ Z t

=

h(s)sγ−1 ds.

1

It follows that Dβ Dα x(1) = 0 and Dγ Dβ Dα x(t) = h(t), proving the claim in this case. Next let t ∈ [τ, 1]. Then Z τ

x(t) =

x(0, s)h(s)sγ−1 ds +

Z t

0

x(t, s)h(s)sγ−1 ds + u(t, τ)

Z 1

τ

h(s)sγ−1 ds;

(9)

τ

again x(0) = 0 by (6). Differentiating (9), we have 1 D x(t) = β α

Z t

β

β

t −s



h(s)s

γ−1

ds +

τ

τβ − tβ β

!Z

1

h(s)sγ−1 ds.

τ

The second boundary condition Dα x(τ) = 0 is clearly met. Differentiating again, we have Dβ Dα x(t) =

Z t

h(s)sγ−1 ds;

1

as in the previous case, Dβ Dα x(1) = 0 and Dγ Dβ Dα x(t) = h(t), proving the claim in this case as well. t u Theorem 2. For G(t, s) given in (5), we have that 0 < G(t, s) ≤ G(τ, s)

(10)

for t ∈ (0, 1] and s ∈ (0, 1], provided the condition u(1, τ) > 0 is met for u in (6), that is to say the inequality  1/β α τ> (11) α +β holds.

Positive Green’s Functions for Boundary Value Problems with Conformable Derivatives

5

Proof. Referring to (5), (6), and (7), we see that u(s, s) = x(0, s),

x(0, τ) = u(t, τ) + x(t, τ),

ensuring that G(t, s) is a well-defined function; we also see that G(t, s) = 0 if t = 0 or s = 0. From (6) specifically, u(0, s) = 0 for all s and t α−1 (sβ − t β ) d u(t, s) = ≥ 0, dt β

s ≥ t.

Moreover, from (7) we have x(s, s) = 0 and d t α−1 (t β − sβ ) x(t, s) = ≥ 0, dt β

t ≥ s,

so that G(t, s) is non-decreasing on [0, τ] and non-increasing on [τ, 1]. Thus (10) will hold if G(1, τ) > 0 holds, which occurs if u(1, τ) > 0. This is equivalent to (11), completing the proof. t u Theorem 3. For all t, s ∈ [0, 1], g(t)G(τ, s) ≤ G(t, s) ≤ G(τ, s) where

 g(t) := min

(12)

i 1−t  tα h β β (α + β )τ − αt , . 1−τ β τ α+β

(13)

Proof. From the preceeding theorem, we have G(t, s) ≤ G(τ, s) for all t, s ∈ [0, 1]. For the lower bound, we proceed by cases on the branches of the Green’s function (5), that is we use (6) and (7). (i) 0 ≤ t ≤ s ≤ τ: Here G(t, s) = u(t, s), G(τ, s) = x(0, s) = t, s we have

1 α+β . α(α+β ) s

For these

i u(t, s) u(t, τ) tα h β β (α + β )τ − αt ≥ ≥ x(0, s) x(0, τ) β τ α+β which implies G(t, s) ≥

i tα h (α + β )τ β − αt β G(τ, s). α+β βτ

(ii) 0 ≤ t ≤ τ ≤ s ≤ 1: In this case G(t, s) = u(t, τ) and G(τ, s) = u(τ, τ), so again we have i tα h β β G(t, s) ≥ (α + β )τ − αt G(τ, s). β τ α+β (iii) 0 ≤ s ≤ t ≤ τ or 0 ≤ s ≤ τ ≤ t ≤ 1: Since G(t, s) = G(τ, s) = follows that

1 α+β , α(α+β ) s

it

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Douglas R. Anderson

G(t, s) = G(τ, s). (iv) τ ≤ t ≤ s ≤ 1: As in (ii), G(t, s) = u(t, τ) and G(τ, s) = u(τ, τ). Define   1−t w(t) := u(t, τ) − u(τ, τ) 1−τ   1−t = G(t, s) − G(τ, s). 1−τ

(14)

Now w(τ) = 0, w0 (τ) > 0, and w(1) = G(1, s) > 0 by (11). Since w is concave down, w(t) ≥ 0 on [τ, 1], hence   1−t G(t, s) ≥ G(τ, s). 1−τ (v) τ ≤ s ≤ t ≤ 1: Note that G(τ, s) = u(τ, τ), while G(t, s) = u(t, τ) + x(t, s) ≥ u(t, τ); consequently, the employment of w as in (14) yields   1−t G(t, s) ≥ G(τ, s). 1−τ t u

This completes the proof.

4 Four Iterated Conformable Derivatives In this section we consider four iterated conformable derivatives in the differential operator, with several types of boundary conditions. First, consider the nonlinear two-point cantilever beam eigenvalue problem Dδ Dγ Dβ Dα x(t) = λ a(t) f (x), α

β

α

0 ≤ t ≤ 1, γ

β

α

x(0) = D x(0) = D D x(1) = D D D x(1) = 0,

(15) (16)

where α, β , γ, δ ∈ (0, 1], and with α + β ∈ (1, 2], α + β + γ ∈ (2, 3], and α + β + γ + δ ∈ (3, 4] if one wishes to explore this problem as a conformable order between 3 and 4. The theme throughout this work has been to approach the existence of positive solutions to nonlinear equations such as (15) with boundary conditions (16) by involving Green’s function for this problem. Please note: to obtain symmetry in Green’s function below we must take γ = α in the conformable differential equation, but we will maintain the more general form (γ not necessarily equal to α) in the proofs to follow. Theorem 4 (Conformable Cantilever Beam). Let α, β , γ, δ ∈ (0, 1]. The corresponding Green’s function for the homogeneous problem Dδ Dγ Dβ Dα x(t) = 0

(17)

Positive Green’s Functions for Boundary Value Problems with Conformable Derivatives

satisfying boundary conditions (16) is given by  α+β   sγ tγ t   − : 0 ≤ t ≤ s ≤ 1,  γ  β (α + β ) (β + γ)(α + β + γ) G(t, s) =  tα sα sβ +γ   : 0 ≤ s ≤ t ≤ 1. − α β (β + γ) (α + β )(α + β + γ)

7

(18)

Proof. Let h be any continuous function. We will show that Z 1

x(t) =

G(t, s)h(s)sδ −1 ds,

0

for G given by (18), is a solution to the linear boundary value problem Dδ Dγ Dβ Dα x(t) = h(t) with boundary conditions (16). For any t ∈ [0, 1], using the branches of (18) we have Z t β +γ  s

 tα sα x(t) = − h(s)sδ −1 ds α β (β + γ) (α + β )(α + β + γ) 0  Z  t α+β t tγ sγ − − h(s)sδ −1 ds; γ (β + γ)(α + β + γ) 1 β (α + β ) clearly x(0) = 0. Taking the α-conformable derivative yields 1 D x(t) = β (β + γ) α

+

Z t

β +γ+δ −1

s 0

t β +γ γ(β + γ)

Z t

tβ h(s)ds − βγ

Z t

sγ+δ −1 h(s)ds

1

sδ −1 h(s)ds.

1

It is easy to see that Dα x(0) = 0. Taking the β -conformable derivative of the αconformable derivative yields Dβ Dα x(t) =

1 γ

Z t

(t γ − sγ ) sδ −1 h(s)ds,

1

so that Dβ Dα x(1) = 0. Next, γ

β

α

Z t

D D D x(t) =

sδ −1 h(s)ds,

1

from which we have Dγ Dβ Dα x(1) = 0, and Dδ Dγ Dβ Dα x(t) = h(t). This finishes the proof.

t u

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Douglas R. Anderson

Theorem 5. For all t, s ∈ [0, 1], the cantilever beam Green’s function given by (18) satisfies 0 ≤ G(t, s) ≤ G(1, s). Proof. First observe from (18) that d G(t, s) = dt

(

t α+β −1 γ β γ(β +γ) [γs sβ +γ t α−1 β (β +γ)

+ β (sγ − t γ )] : t ≤ s, : s ≤ t.

(19)

Now fix s ∈ [0, 1]. By the first boundary condition G(0, s) = 0, and dtd G(t, s) as given above implies that G(·, s) is monotone increasing on (0, 1]. In particular, G(1, s) ≥ G(t, s) ≥ 0 for all t ∈ [0, 1]. t u The next Green’s function we consider is for the conformable (2, 2)−conjugate problem Dα Dα Dα Dα x(t) = f (x), α

x(0) = 0 = D x(0),

0 ≤ t ≤ 1,

x(1) = 0 = Dα x(1),

where α ∈ (0, 1]. Theorem 6 (Conformable (2, 2)−Conjugate). Let α ∈ (0, 1]. The symmetric Green’s function for the homogeneous problem Dα Dα Dα Dα x(t) = 0

(20)

satisfying the conformable (2, 2)−conjugate boundary conditions x(0) = 0 = Dα x(0),

x(1) = 0 = Dα x(1)

(21)

is given by ( u(t, s) : 0 ≤ t ≤ s ≤ 1, G(t, s) = u(s,t) : 0 ≤ s ≤ t ≤ 1, where u(t, s) =

t 2α (1 − sα )2 α α [s − t + 2sα (1 − t α )] . 6α 3

(22)

(23)

Proof. In this proof we construct Green’s function from scratch, modifying the classical approach, found for example in [13, Chapter 5]. Here Green’s function takes the form of  u(t, s) : t ≤ s,  α α 3 G(t, s) = u(t, s) + 1 t −s : s ≤ t, 3! α where u(t, s) satisfies (20) and the two boundary conditions in (21) set at t = 0. Thus u(t, s) = a(s)t 2α + b(s)t 3α ;

Positive Green’s Functions for Boundary Value Problems with Conformable Derivatives

9

the two boundary conditions at t = 1 are satisfied when a(s) =

sα (1 − sα )2 2α 3

and b(s) =

(1 − sα )2 (1 + 2sα ) . −6α 3

Combining terms and simplifying leads to (22) and (23).

t u

Theorem 7. For all t, s ∈ (0, 1), the (2, 2)−conjugate Green’s function given by (22) satisfies   3  α 2  s  2 1−sα : 0 ≤ s ≤ 2−1/α , 3 α 3−2sα  α 3   0 < G(t, s) ≤ 2  1−sα 2 s : 2−1/α ≤ s ≤ 1. 3 α 1+2sα Proof. From (23), it is evident that u(0, s) = u(t, 1) = 0, and u(t, s) ≥ 0 for 0 ≤ t ≤ 2sα s ≤ 1. For fixed s, u(t, s) has a local maximum when t α = 1+2s α , and u(s,t) has 1 α a local maximum when t = 3−2sα . After substitution and comparison, the result holds. t u Finally, we will end the present discussion by considering another common set of boundary conditions, namely the so-called Lidstone conditions. Unlike the argument used below, one could also approach this problem as the conjunction of two conjugate problems, whose Green’s function is given in [2, (2.4)]. Theorem 8 (Conformable Lidstone). Let α, β ∈ (0, 1]. The symmetric Green’s function for the homogeneous problem Dβ Dα Dβ Dα x(t) = 0

(24)

satisfying the Lidstone-type boundary conditions x(0) = 0 = Dβ Dα x(0),

x(1) = 0 = Dβ Dα x(1)

(25)

is given by ( u(t, s) : 0 ≤ t ≤ s ≤ 1, G(t, s) = u(s,t) : 0 ≤ s ≤ t ≤ 1,

(26)

where u(t, s) =

h    i tα 2αsα 1 − sβ − β (1 − sα ) t α+β + sα+β . αβ (α + β )(2α + β ) (27)

Proof. Let x(t, s) be the Cauchy function associated with (17), namely a function satisfying x(s, s) = Dα x(s, s) = Dβ Dα x(s, s) = 0,

Dγ Dβ Dα x(t, s) = 1.

(Note that we will use γ for now, and take γ = α for symmetry purposes at the conclusion.) Using (2) at each step, it is easy to verify that here the Cauchy function

10

Douglas R. Anderson

is given by x(t, s) =

1 γ

Z tZ τ s

(ξ γ − sγ ) ξ β −1 τ α−1 dξ dτ,

(28)

s

and again Green’s function takes the form of ( u(t, s) : t ≤ s, G(t, s) = u(t, s) + x(t, s) : s ≤ t, where u(t, s) satisfies (17) and the two boundary conditions set at t = 0. Thus u(t, s) = a(s) + b(s)t α + c(s)t α+β + d(s)t α+β +γ ; the Lidstone boundary conditions force a(s) = c(s) = 0. The two boundary conditions at t = 1 are satisfied by (u + x) for x given in (28). In particular, we use Dβ Dα [u(t, s) + x(t, s)]|t→1 = 0 to solve for d, leading to d(s) =

sγ − 1 , γ(β + γ)(α + β + γ)

and then u(1, s) + x(1, s) = 0 to solve for b, which yields b(s) =

sα+β +γ sβ +γ − α(α + β )(α + β + γ) αβ (β + γ)   1 1 sγ − . + γ β (α + β ) (β + γ)(α + β + γ)

Altogether we have sα+β α + 2β + γ + α(α + β )(α + β + γ) β (α + β )(β + γ)(α + β + γ) ! sβ (sγ − 1)t α+β +γ − + . αβ (β + γ) γ(β + γ)(α + β + γ)

u(t, s) = sγ t α

To achieve symmetry and to satisfy (24), we take γ = α and arrive at (27).

t u

Theorem 9. For all t, s ∈ (0, 1), the Lidstone Green’s function given by (26) satisfies G(t, s) > 0. Proof. Clearly u(0, s) = u(t, 1) = 0 from (27), and u(t, s) ≥

h   i 2sα t α α 1 − sβ − β (1 − sα ) sβ αβ (α + β )(2α + β )

Positive Green’s Functions for Boundary Value Problems with Conformable Derivatives

11

for 0 ≤ t ≤ s. Define   k(s) := α 1 − sβ − β (1 − sα ) sβ = β sα+β − (α + β )sβ + α. Then k(0) = α > 0, k(1) = 0, and k0 (s) = β (α + β )sβ −1 (sα − 1) < 0. Thus, k is strictly decreasing, so that k(s) > 0 for s ∈ [0, 1), forcing u(t, s) ≥ 0 for t, s ∈ [0, 1]. Therefore the result holds. t u

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