(See [1].) The Hadamard derivative of fractional order . for a function . : [1, ∞) → R is defined as

where . = [.] + 1, [.] denotes the integer part of the real number ., and log(·) = loge(·).

Let y C[1, e]. Then the Hadamard fractional multi-point boundary value problems

has a solution, which can take the form u(.) = ∫ e . (t, s).(.)(ds/s., where

Proof. Using Lemmas 2–3 of [27], we have

where c. R (. = 1, 2, 3). Note that from .(1) = δu(1) = 0 we have .. = .. = 0. Consequently, we obtain

From .(e) = Σ.−1 a.u.ξ.) we obtain

This, together with (H0), implies that

Therefore, by (2) we have

This completes the proof.

(See [24].) The function G. has the following inequalities:

Proof. From Lemma 2(i) we have

and

This completes the proof.

Let . := .[1,e], ǁ.ǁ = max._∈[1,e] |.(.)|, and . := {. ∈ .: .(.) ≥ 0 ∀. ∈ [1, e]}. Then (E, ǁ·ǁ) becomes a real Banach space, and . is a cone on .. Moreover, . × . is a Banach space with the norm ǁ(u, v)ǁ = ǁ.ǁ + ǁ.ǁ, and . × . is a cone on . × .. Let

Then from Lemma 1 we obtain that (1) is equivalent to the following system of Hammersteintype integral equations:

Therefore, we can define an operator . : . × . → . × . as follows:

Note that G. and f. (. = 1, 2) are nonnegative continuous functions, so the operators A.. P P P (. = 1, 2) and .: P P P P are three completely continuous operators. Moreover, if (u, v) (P P ) . is a fixed point of ., then (u, v) is a positive solution for (1). Therefore, in what follows, we turn to study the existence of

fixed points of the operator ..

Let η = min._{∈[5.4,3e.4]} .(.). Then A..P, P ) ⊂ .. (. = 1, 2), where

Proof. From Lemma 3 we have

This implies that

Consequently, if . ∈ [5.4, 3e.4], we obtain

and then

On the other hand, using the method of Lemma 3, we also have ..(t, s) ≥ .(.)..(τ, s) for t, s, τ ∈ [1, e], and thus ..(P, P ) ⊂ ... This completes the proof.

Let γ, β, θ be nonnegative continuous convex functionals on . , and let α, ψ be nonnegative continuous concave functionals on . ; then for nonnegative numbers .., .., .., .. and .., convex sets are defined:

(See [6].) Let P be a cone in the real Banach space E. Suppose that α and ψ are nonnegative continuous concave functionals on P and γ, β, θ are nonnegative continuous convex functionals on P such that, for some positive numbers c. and e., ...) ≤ .(.) and y ≤ e..(.) for all y P .γ, c′). Suppose further that T : . (γ, c′) . (γ, c′) is completely continuous and there exist constants h., d., a., and b. ≥ 0 with . < d. < a. such that each of the following is satisfied:

Then T has at least three fixed points y., y., y. . (y, c′) such that β(..) < d., a. < α(..) and d. < β(..) with α(..) < a..

(See [12].) Let E be a Banach space, and A: .→ E be a completely continuous operator. Assume that T : . → E is a bounded linear operator such that . is not an eigenvalue of T and lim.._ǁ→∞ ǁAu . Tuǁ.ǁ.ǁ = 0. Then A has a fixed point

(i) If we use t, s to replace log ., log . in .. of Lemma 1, respectively, we can obtain a function

This function happens to be the Green’s function for the Riemann–Liouville fractional boundary value problem

where . ∈ (2, 3], and D. is the Riemann–Liouville fractional derivative, . ∈ .[0, 1].

For details, please refer to Lemma 3.1 in [24].

(ii) Note that for multi-point boundary value problems, the Green’s functions may be complicated. For example, in [7], Bai studied the fractional three-point boundary value problem

where . ∈ (1, 2], βη.⁻¹, . ∈ (0, 1). The Green’s function is

Note that if . = 0, then (3) reduces to the problema.

The Green’s function is

Now, if the three-point problem (3) is considered as a perturbation of the two-point problem (5), we can use (6) to obtain (4), i.e.,

This simple idea motivates our study in Lemma 1.

Combining the above, we do not need to construct new Green’s functions to obtain the equivalent Hammerstein-type integral equations for our problem (1).

Let . < a. < b. < b./η < c., (H0).(H1) and the following conditions hold

Where

And

Then (1) has at least triple positive solutions.

Proof. From Lemma 4 we have

Therefore, for our conclusions, we need to define the nonnegative continuous concave functionals α, ψ and the nonnegative continuous convex functionals β, θ, γ on .. by

where . = [5.4, 3e.4], .. = [3.2, 2]. For any (u, v) ∈ .., we have

We show that .: . (γ, c′) → . (γ, c′ . Indeed, if (u, v) ∈ . (γ, c′), then we have 0 ≤ .(.) + .(.) ≤ .. for . ∈ [1, e]. Consequently, (H4) is used to obtain

Now, (B1) and (B2) of Lemma 5 are to be verified. Note that

Therefore, if (u, v) .(γ, θ, α, b., b./η, c.), then .. ≤ .(.) + .(.) ≤ ../η for t I; if (u, v) .(γ, β, ψ, ηa., a., c.), then ηa. ≤ .(.) + .(.) ≤ .. for t I.. As a result, (H3) enables us to find

Moreover, (H2) implies that

Next, (B3) of Lemma 5 is satisfied. Let (u, v) ∈ . (γ, α, b., c.) with .(.(u, v))(.) > b./η. Therefore, for all . ∈ [1, e], we have

Note that the last line of the above is independent of the variable ., and thus we obtain

Finally, we prove that (B4) holds. Let (u, v) ∈ .(γ, β, a., c.) with .(.(u, v))(.) < ηa.. Note that min... G..t, s) ≥ ηG..τ, s) for τ, s ∈ [1, e], and thus min... G..t, s) ≥ . max._∈[1,e]G..τ, s) for . ∈ [1, e], . = 1, 2. Therefore, we have

Up to now, we have proved that all the assumptions of Lemma 5 are satisfied. There- fore, (1) has at least triple positive solutions, (.., x.), (.., y.), and (.., z.) such that .(.., x.) < a., .. < α(y., y.), and .. < β(.., z.) with .(.., z.) < b.. This completes the proof.

Suppose that (H0) and the following conditions hold: (H1.) f. ∈ .([1, e] × R × R, R), i = 1, 2;

Then (1) has at least one nontrivial solution.

Proof. Define operators T. : . × . → . as follows:

Now, we prove that 1 is not an eigenvalue of T. (. = 1, 2), and we only need to consider the case . = 1 (the case . = 2 can be dealt with a similar method). Argument by contrary. If let . + . = ., then we have

and by Lemma 1 we obtain

where q, δ, a., ξ. (. = 1, 2, . . . , m − 1) satisfy (H0). We distinguish two cases.

Case 1. .. = 0. From (8) and Lemma 1 of [23] we have

where c. R (. = 1, 2, 3). By the boundary conditions in (8) we have .. = .. = 0. Therefore, we find

and (H0) indicates that .. = 0. Consequently, we have .(.) 0 for . [1, e]. This contradicts to the definition of eigenvalue and eigenfunction.

Case 2. .. =/ 0. From (7) we have

This has a contradiction.

Above all, 1 is not an eigenvalue of T. (. = 1, 2) as required. Hence, if we let the operator . : . × . → . × . as follows:

we know . := (1, 1) is not an eigenvalue of . .

From (H5), for all ε > 0, there exist M. > 0 (. = 1, 2) such that

Consequently, there are .. > ., ζ. > 0 such that

As a result, we have

For the arbitrariness of ., we have lim_{ǁ(u,v)ǁ→∞} .(u, v) .(u, v) . (u, v) = 0. Note from (H6) that . = (0,0) is not a fixed point of .. Hence, from Lemma 6 we have that . has a fixed point in ., and this fixed point is nontrivial, i.e., (1) has at least one nontrivial solution. This completes the proof.