Existence of Solutions for Two-Point Boundary Value Problem of Fractional Differential Equations at Resonance

Definition 1 (see [1].)The Riemann-Liouville fractional integral of order α > 0 of a function $$ is given by

provided that the right-hand side is pointwise defined on (0, ∞).

Definition 2 (see [1].)The Riemann-Liouville fractional derivative of order α > 0 of a continuous function $$ is given by

where n − 1 < α ≤ n, provided that the right-hand side is pointwise defined on (0, ∞).

Lemma 3 (see [1].)Let n − 1 < α ≤ n, u ∈ C(0,1) ⋂ L¹(0,1); then

where $$, i = 1,2, …n.

Lemma 4 (see [1].)If α > 0, $$ and D = d/dx. If the fractional derivatives $$ and $$ exist, then

Lemma 5 (see [1].)The relation

is valid in following cases β > 0, α + β > 0, and f(x) ∈ L₁(a, b).

Theorem 6. Let L be a Fredholm operator of index zero and N be L-compact on $$. Suppose that the following conditions are satisfied:

• (1)

  Lx ≠ λNx for each (x, λ) ∈ [(dom⁡L∖ker⁡L)∩∂Ω] × (0,1);

• (2)

  Nx ∉ Im⁡L for each x ∈ ker⁡L∩∂Ω;

• (3)

  deg⁡(JQN|_ker⁡L, Ω∩ker⁡L, 0) ≠ 0, where Q : Z → Z is a continuous projection as above with Im⁡L = ker⁡Q and J : Im⁡Q → ker⁡L is any isomorphism.

Then the equation Lx = Nx has at least one solution in $$.

Lemma 7. The mapping L : dom⁡(L) ⊂ E is a Fredholm operator of index zero.

Proof. It is clear that

Let x ∈ Im⁡L, so there exists a function u ∈ dom⁡L which satisfies Lu = x. By (11) and Lemma 3, we have By $$, we can obtain c₂ = ⋯ = c_N = 0. Hence Then, we have Taking into account $$, we obtain On the other hand, suppose x satisfy $$. Let $$, we can easily prove u(t) ∈ dom⁡(L).

Thus, we conclude that

Consider the linear operators Q : E → E defined by Take x(t) ∈ E; then We can see Q² = Q.

For x(t) ∈ E in the type x(t) = x(t) − Qx(t) + Qx(t), obviously, x(t) − Qx(t) ∈ Ker⁡(Q) = Im⁡(L) and Qx(t) ∈ Im⁡(Q). That is to say, E = Im⁡(L) + Im⁡(Q). If u ∈ Im⁡(L) ⋂ Im⁡(Q), we have u = c₁; then $$. As a result c₁ = 0, and we get E = Im⁡(L) ⊕ Im⁡(Q).

Note that Ind L = dim⁡ ker⁡L − codim Im⁡L = 0. Then L is a Fredholm mapping of index zero.

Lemma 8. Assume Ω ⊂ Y is an open bounded subset such that dom⁡L ⋂ Y ≠ ∅; then map N is L-compact on $$

Proof. By the continuity of f, we can get that $$ and $$ are bounded. So, in view of the Arzela-Ascoli theorem, we need only to prove that $$ is equicontinuous. From the continuity of f, there exists a constant r > 0, such that |(I − Q)N(u(t))| ≤ r, for all $$, t ∈ [0,1].

For 0 ≤ t₁ ≤ t₂ ≤ 1, u ∈ Ω, we have

Furthermore, we have where i = 1,2, …, N − 1. Since t^α and tⁱ are uniformly continuous on [0,1], we can get that $$ is compact. The proof is completed.

Lemma 9. Ω₁ = {u ∈ dom⁡(L)∖Ker⁡(L)∣Lu = λNu, λ ∈ [0,1]} is bounded.

Proof. For u ∈ Ω₁, λ ≠ 0 and Lu = λNu. By (12), Lu = λNu ∈ Im⁡(L) = Ker⁡(Q); that is,

By the integral mean value theorem, there exits a constant t₀ ∈ [0,1] such that Form (H₂), we can get $$.

Again for u ∈ Ω₁, (I − P)u ∈ dom⁡(L)∖Ker⁡(L) and LPu = 0. From (29), we have

Now by Lemma 4 That is, From (25) and (38), we have Furthermore, it follows from (40) and (H₁) that By the definition ∥u∥_X and (H₄), it is easy to see that $$ and ∥u∥_∞ are bounded. So, Ω₁ is bounded.

Lemma 10. Ω₂ = {u ∈ Ker⁡(L) : Nu ∈ Im⁡(L)} is bounded.

Proof. Let u ∈ Ker⁡(L), so we have u = c₁t^α−1, $$. For Nu ∈ Im⁡(L) = Ker⁡(Q),

By the integral mean value theorem, there exits a constant t₁ ∈ [0,1] such that From (H₂), it follows that |c₁| ≤ A/Γ(α). Hence, Ω₂ is bounded.

Lemma 11. Ω₃ = {u ∈ Ker⁡(L) : λu + (1 − λ)QNu = 0,  λ ∈ [0,1]} is bounded.

Proof. Let u ∈ Ker⁡(L), so we have u = c₁t^α−1, $$. If λ = 0, then |c₁| ≤ D. If λ = 1, we have c₁ = 0.

If λ ≠ 0 and λ ≠ 1, then

It follows that Then we get which, together with (H₃), implies |c₁| ≤ D. Here, Ω₃ is bounded.

Remark 12. If the other parts of (H₃) hold, then the set $$ is bounded.

Theorem 13. Suppose (H₁)–(H₄) hold; then the problem (3) has at least one solution in Y.

Proof. Let Ω be a bounded open set of Y, such that $$. It follows from Lemma 8, N is L-compact on Ω. By Lemmas 9, 10, and 11, we get the following:

• (1)

  Lu ≠ λNu, for every u ∈ [(dom⁡L∖Ker⁡L) ⋂ ∂Ω]×(0,1);

• (2)

  Nu ∉ Im⁡L for every u ∈ Ker⁡L ⋂ ∂Ω;

• (3)

  let H(u, λ) = ±λIu + (1 − λ)JQNu, where I is the identical operator. Via the homotopy property of degree, we obtain that

  $$ (48)

Applying Theorem 6, we conclude that Lu = Nu has at least one solution in $$.

Theorem 14. Suppose the conditions (H₁) in the theorem are replaced by the following conditions.

• (H₁)′

  There exist positive constants a_i, i = 0,1, …, N − 1, such that, for all (x₁, x₂, …, x_N), $$, one has

  $$ (49)

• (H₁)′′

  There exist constants l_i, i = 1,2, …, N − 1, such that for all (x₁, x₂, …, x_N), $$, one has

  $$ (50)

Then, the BVP (3) has a unique solution, provided that

Proof. Let y_i = 0, i = 1,2, …, N, and φ₁ = |f(t, 0, …, 0)|; then the condition (H₁) is satisfied. According to Theorem 13, BVP (3) has at least one solution. Suppose u_i ∈ Y, i = 1,2 are two solutions of (3); then

Note that u = u₁ − u₂, so u satisfy the equation According to Im⁡(L) = Ker⁡(Q), we have By the integral mean value theorem, there exists η ∈ [0,1], such that By (H₁)′′, we have We can have Thus, we can obtain According to (25), (38), and (58), we have From the definition of ∥u∥_X and the assumption (51), we have ∥u∥ = 0, so that u₁ = u₂.