Positive Solution to a Fractional Boundary Value Problem

Definition 2.1. If g ∈ C([a, b]) and α > 0, then the Riemann-Liouville fractional integral is defined by

Definition 2.2. Let α ≥ 0, n = [α] + 1. If f ∈ ACⁿ[a, b] then the Caputo fractional derivative of order α of f defined by

exists almost everywhere on [a, b] ([α] is the entire part of α).

Lemma 2.3 (see [15].)Let α, β > 0 and n = [α] + 1, then the following relations hold: $$, β > n and $$, k = 0,1, 2, …, n − 1.

Lemma 2.4 (see [15].)For α > 0, g(t) ∈ C(0,1), the homogenous fractional differential equation

has a solution where, c_i ∈ R, i = 0, …, n, and n = [α] + 1.

Lemma 2.5 (see [16].)Let p, q ≥ 0, f ∈ L₁[a, b]. Then $$ and $$, for all t ∈ a, b].

Lemma 2.6 (see [15].)Let β > α > 0. Then the formula $$, holds almost everywhere on t ∈ a, b], for f ∈ L₁[a, b] and it is valid at any point x ∈ [a, b] if f ∈ C[a, b].

Lemma 2.7. Let 2 < q < 3, 1 < σ < 2 and y ∈ C[0,1]. The unique solution of the fractional boundary value problem

is given by where

Proof. Applying Lemmas 2.4 and 2.5 to (2.5) we get

Differentiating both sides of (2.8) and using Lemma 2.6 it yields The first condition in (2.5) implies c₁ = c₃ = 0, the second one gives $$. Substituting c₂ by its value in (2.8), we obtain that can be written as that is equivalent to where G is defined by (2.7). The proof is complete.

Lemma 3.1. The function u ∈ E is solution of the fractional boundary value problem (P1) if and only if Tu(t) = u(t), for all t ∈ [0,1].

Proof. Let u be solution of (P1) and $$. In view of (2.10) we have

With the help of Lemma 2.6 we obtain It is clear that v satisfies conditions (1.2), then it is a solution for the problem (P1). The proof is complete.

Theorem 3.2. Assume that there exist nonnegative functions g, h ∈ L¹([0,1], ℝ₊) such that for all x, y ∈ ℝ and t ∈ [0,1], one has

where Then the fractional boundary value problem (P1) has a unique solution u in E.

Lemma 3.3. Let q > 0, f ∈ L₁([a, b], ℝ₊). Then, for all t ∈ a, b] we have

Proof. Let f ∈ L₁([a, b], ℝ₊), then

Proof. We transform the fractional boundary value problem to a fixed point problem. By Lemma 3.1, the fractional boundary value problem (P1) has a solution if and only if the operator T has a fixed point in E. Now we will prove that T is a contraction. Let u, v ∈ E, in view of (2.10) we get

with the help of (3.4) we obtain Lemma 3.3 implies In view of (3.5) it yields On the other hand we have where

Therefore

Applying hypothesis (3.4) we get Let us estimate the term $$. We have Consequently (3.16) becomes

With the help of hypothesis (3.5) it yields

Taking into account (3.12)–(3.19) we obtain

from here, the contraction principle ensures the uniqueness of solution for the fractional boundary value problem (P1). This finishes the proof.

Theorem 3.4. Assume that f(t, 0,0) ≠ 0 and there exist nonnegative functions k, h, g ∈ L¹([0,1], ℝ₊), $$ nondecreasing on ℝ₊ and r > 0, such that

where C₁ = max {C_k, C_h, C_g}, C₂ = max {A_k, A_h, A_g}, C_h and C_g are defined as in Theorem 3.2 and Then the fractional boundary value problem (P1) has at least one nontrivial solution u^* ∈ E.

Lemma 3.5 (see [17].)Let F be a Banach space and Ω a bounded open subset of F, 0 ∈ Ω. $$ be a completely continuous operator. Then, either there exists x ∈ ∂Ω, λ > 1 such that T(x) = λx, or there exists a fixed point $$.

Proof. First let us prove that T is completely continuous. It is clear that T is continuous since f and G are continuous. Let B_r = {u ∈ E, ∥u∥ ≤ r} be a bounded subset in E. We shall prove that T(B_r) is relatively compact.

(i) For u ∈ B_r and using (3.21) we get

Since ψ and ϕ are nondecreasing then (3.24) implies using similar techniques as to get (3.12) it yields Hence

Moreover, we have

Using (3.17) we obtain From (3.27) and (3.29) we get then T(B_r) is uniformly bounded.

(ii) T(B_r) is equicontinuous. Indeed for all t₁, t₂ ∈ [0,1], t₁ < t₂, u ∈ B_r, let $$, 0 ≤ t ≤ 1, ∥u∥ < r), therefore

that implies

Let us consider the function Φ(x) = x^q−1 − (q − 1)x, we see that Φ is decreasing on [0,1], consequently $$, from which we deduce

Some computations give On the other hand we have Using (3.17) and (3.28) it yields then when t₁ → t₂, in (3.34) and (3.37) then |Tu(t₁) − Tu(t₂)| and $$ tend to 0. Consequently T(B_r) is equicontinuous. From Arzelá-Ascoli Theorem we deduce that T is completely continuous operator.

Now we apply Leray Schauder nonlinear alternative to prove that T has at least one nontrivial solution in E.

Letting Ω = {u ∈ E : ∥u∥ < r}, for any u ∈ ∂Ω, such that u = λTu, 0 < λ < 1, we get, with the help of (3.27),

Taking into account (3.29) we obtain From (3.38), (3.39), and (3.22) we deduce that this contradicts the fact that u ∈ ∂Ω. Lemma 3.5 allows us to conclude that the operator T has a fixed point $$ and then the fractional boundary value problem (P1) has a nontrivial solution u^* ∈ E. The proof is complete.

Lemma 4.1. Let G(t, s) be the function defined by (2.7). If α ≥ 1 then G(t, s) has the following properties:

• (i)

  G(t, s) ∈ C([0,1] × [0,1]), G(t, s) > 0, for all t, s ∈ ]0,1[.

• (ii)

  If t, s ∈ (τ, 1), τ > 0, then

Proof. (i) It is obvious that G(t, s) ∈ C([0,1] × [0,1]), moreover, we have

which is positive if α ≥ 1. Hence G(t, s) is nonnegative for all t, s ∈ ]0,1[.

(ii) Let t, s ∈ (τ, 1), it is easy to see that G(s, s) ≠ 0, then we have

Now we look for lower bounds of G(t, s) Finally, since G(s, s) is nonnegative we obtain 0 < τG(s, s) ≤ G(t, s) ≤ (2/τ)G(s, s).

Definition 4.2. A function u(t) is called positive solution of the fractional boundary value problem (P1) if u(t) ≥ 0, for all t ∈ [0,1].

Lemma 4.3. If u ∈ E⁺ and α ≥ 1, then the solution of the fractional boundary value problem (P1) is positive and satisfies

Proof. First let us remark that under the assumptions on u and f, the function $$ is nonnegative. From Lemma 3.1 we have

Applying the right-hand side of inequality (4.1) we get Moreover, (4.1) gives Combining (4.7) and (4.8) yields hence In view of the-left hand side of (4.1), we obtain for all t ∈ (τ, 1) on the other hand we have

From (4.11) and (4.12) we get

with the help of (4.10) we deduce

The proof is complete.

Theorem 4.4. Under the assumption of Lemma 4.3, the fractional boundary value problem (P1) has at least one positive solution in the both cases superlinear as well as sublinear.

Theorem 4.5 (see [18].)Let E be a Banach space, and let K ⊂ E, be a cone. Assume Ω₁ and Ω₂ are open subsets of E with 0 ∈ Ω₁, $$ and let

be a completely continuous operator such that

• (i)

  ||𝒜u|| ≤ ||u||, u ∈ K∩∂Ω₁, and ||𝒜u|| ≥ ||u||, u ∈ K∩∂Ω₂, or

• (ii)

  ||𝒜u|| ≥ ||u||, u ∈ K∩∂Ω₁, and ||𝒜u|| ≤ ||u||, u ∈ K∩∂Ω₂.

Then 𝒜 has a fixed point in $$.

Proof. To prove Theorem 4.4 we define the cone K by

It is easy to check that K is a nonempty closed and convex subset of E, hence it is a cone. Using Lemma 4.3 we see that TK ⊂ K. From the prove of Theorem 3.4, we know that T is completely continuous in E.

Let us prove the superlinear case. First, since A₀ = 0, for any ε > 0, there exists R₁ > 0, such that

for 0 < |u| + |v| ≤ R₁. Letting Ω₁ = {u ∈ E, ∥u∥ < R₁}, for any u ∈ K∩∂Ω₁, it yields Moreover, we have From (4.19) and (4.20) we conclude

In view of hypothesis (H2), one can choose ε such that

The inequalities (4.21) and (4.22) imply that ||Tu|| ≤ ||u||, for all u ∈ K∩∂Ω₁. Second, in view of A_∞ = ∞, then for any M > 0, there exists R₂ > 0, such that f₁(u, v) ≥ M(|u| + |v|) for |u| + |v| ≥ R₂. Let R = max {2R₁, (2R₂/τ²)} and denote by Ω₂ the open set {u ∈ E/||u|| < R}. If u ∈ K∩∂Ω₂ then Using the left-hand side of (4.1) and Lemma 4.3, we obtain Moreover, we get with the help of (4.12) In view of (4.26) and (4.24) we can write Let us choose M such that then we get $$. Hence, The first part of Theorem 4.5 implies that T has a fixed point in $$ such that R₂ ≤ ||u|| ≤ R. To prove the sublinear case we apply similar techniques. The proof is complete.

Example 4.6. The fractional boundary value problem

has a unique solution in E.

Proof. In this case we have f(t, x, y) = ((t−1)/10)³x + t²y + ln t, 2 < q = 5/2 < 3, σ = (5/4) < 2, α = −1/2 and

then g(t) = ((1 − t)/10) ³ and h(t) = t². Some calculus give Thus Theorem 3.2 implies that fractional boundary value problem (4.29) has a unique in E.

Example 4.7. The fractional boundary value problem

has at least one nontrivial solution in E.

Proof. We apply Theorem 3.4 to prove that the fractional boundary value problem (4.32) has at least one nontrivial solution. We have q = 7/3, σ = 6/5, α = 3/2, and

where k(t) = h(t) = g(t) = (1−t)², ψ(x) = (x/10)², $$, f(t, 0,0) ≠ 0. Let us find r such that (3.22) holds, for this we have We see that (3.22) is equivalent to 1.2664(((r²)/100) + (ln (1 + r²)/9) + 1) − r which is negative for r = 6.