Multiplicity of Positive Solutions for Fractional Differential Equation with p-Laplacian Boundary Value Problems

Lemma 1. Let $$. If y ∈ C[a, b], then the fractional order BVP

has a unique solution, $$, where here

Proof. Assume that u ∈ C^[α]+1[a, b] is a solution of fractional order BVP (4)-(5) and is uniquely expressed as $$, so that

From u^(j)(a) = 0, 0 ≤ j ≤ n − 2, we have c_n = c_n−1 = c_n−2 = ⋯ = c₂ = 0. Then From $$, we have Therefore Thus, the unique solution of (4)-(5) is where G(t, s) is given in (6).

Lemma 2. If h ∈ C[a, b], then the fractional order differential equation

satisfying (5) and has a unique solution, where Here H(t, s) is Green’s function for

Proof. An equivalent integral equation for (13) is given by

By (14), one can determine c₂ = 0 and $$.

Thus, the unique solution of (13), (2) is

Therefore, $$. Consequently, $$.

Hence,

Lemma 3. Green’s function G(t, s) satisfies the following inequalities:

• (i)

  G(t, s) ≤ G(b, s),   for  all  (t, s)∈[a, b]×[a, b],

• (ii)

  G(t, s)≥((η − a)/(b − a)) ^α−1G(b, s),   for  all  (t, s)∈[η, b]×[a, b].

Proof. Consider Green’s function given by (6).

Let a ≤ t ≤ s ≤ b. Then, we have

Let a ≤ s ≤ t ≤ b. Then, we have Therefore G₁(t, s) is increasing in t, which implies G₁(t, s) ≤ G₁(b, s).

Now we prove

In fact, if s ≥ η, obviously, (23) holds. If s ≤ η, one has That implies that (23) is also true. Therefore, by (6), (21), and (23), we find Therefore G(t, s) is increasing with respect to t ∈ [a, b]. Hence the inequality (i) is proved. Now, we establish the inequality (ii).

On the other hand, if a ≤ s ≤ t ≤ b, we have

If a ≤ t ≤ s ≤ b, we have Therefore From (6) and (28) we have Therefore Hence the inequality (ii) is proved.

Lemma 4. Green’s function H(t, s) satisfies the following inequalities:

• (i)

  H(t, s) ≤ H(s, s),   for  all  (t, s)∈[a, b]×[a, b],

• (ii)

  H(t, s) ≥ γH(s, s),   for  all  (t, s) ∈ I × [a, b],

where I = [(3a + b)/4, (a + 3b)/4] and γ = (1/4) ^β−1.

Theorem 5 (Leggett-Williams [3]). Let $$ be completely continuous and let ϕ be a nonnegative continuous concave functional on P such that ϕ(y) ≤ ‖y‖ for all $$. Suppose that there exist a, b, c, and d with 0 < a < b < d ≤ c such that

•  

  A1 {y ∈ P(ϕ, b, d) : ϕ(y) > b} ≠ ∅ and ϕ(Ty) > b for y ∈ P(ϕ, b, d),

•  

  A2 ‖Ty‖ < a for ‖y‖ ≤ a,

•  

  A3 ϕ(Ty) > b for y ∈ P(ϕ, b, c) with ‖Ty‖ > d. Then T has at least three fixed points y₁, y₂, and y₃ in $$ satisfying ‖y₁‖ < a,   b < ϕ(y₂),   ‖y₃‖ > a, and ϕ(y₃) < b.

Theorem 6 (see [3].)Let $$ be a completely continuous operator and let ϕ be a nonnegative continuous concave functional on P such that ϕ(y) ≤ ‖y‖ for all $$. Suppose that there exist a, b, and c with 0 < a < b < c such that

•  

  B1 {y ∈ P(ϕ, b, c) : ϕ(y) > b} ≠ ∅ and ϕ(Ty) > b for y ∈ P(ϕ, b, c),

•  

  B2 ‖Ty‖ < a for ‖y‖ ≤ a,

•  

  B3 ϕ(Ty) > (b/c)‖Ty‖ for $$ with ‖Ty‖ > c. Then T has at least two fixed points y₁ and y₂ in $$ satisfying ‖y₁‖ < a, ‖y₂‖ > a and ϕ(y₂) < b.

Theorem 7 (see [21].)Let P be a closed convex set in a Banach space E and let Ω be a bounded open set such that Ω_p≔Ω∩P ≠ ∅. Let $$ be a compact map. Suppose that x ≠ Tx for all x ∈ ∂_p:

•  

  (C1) Existence: if i(T, Ω_p, P) ≠ ∅, then T has a fixed point in Ω_p.

•  

  (C2) Normalization: if u ∈ Ω_p, then $$, where $$ for $$.

•  

  (C3) Homotopy: let $$ be a compact map such that x ≠ v(t, x) for x ∈ ∂Ω_p and t ∈ [0,1]. Then i(v(0, ·), Ω_p, P) = i(v(1, ·), Ω_p, P).

•  

  (C4) Additivity: if U1, U2 are disjoint relatively open subsets of Ω_p such that x ≠ Tx for $$, then i(T, Ω_p, P) = i(T, U1, P) + i(T, U2, P), where $$.

Theorem 8 (see [22].)Let P be a cone in a Banach space E. For q > 0, define Ω_q = {x ∈ P : ‖x‖ < q}. Assume that $$ is a compact map such that x ≠ Tx for x ∈ ∂Ω_q. Thus, one has the following conclusions:

•  

  (D1) If ‖x‖ < ‖Tx‖ for x ∈ ∂Ω_q, then i(T, Ω_q, P) = 0.

•  

  (D2) If ‖x‖ ≥ ‖Tx‖ for x ∈ ∂Ω_q, then i(T, Ω_q, P) = 1.

Lemma 9. The operator T defined by (32) is a self-map on P.

Proof. Let u ∈ P. Clearly, Tu(t) ≥ 0, for all t ∈ [a, b] and

so that On the other hand, by Lemma 3, we have Hence Tu ∈ P and so T : P → P. Standard argument involving the Arzela-Ascoli theorem shows that T is completely continuous.

Theorem 10. Let f(t, u) be nonnegative continuous on [a, b]×[0, ∞). Assume that there exist constants a^′, b^′ with b^′ > a^′ > 0 such that the following conditions are satisfied:

•  

  H1 f(t, u(t)) ≥ ϕ_p(Ab^′) for all (t, u)∈[η, b]×[b^′, b^′((b − a)/(η − a)) ^2(α−1)].

•  

  H2 f(t, u(t)) < ϕ_p(Ba^′) for all (t, u)∈[a, b]×[0, a^′].

Then fractional order BVP (1)-(2) has at least two positive solutions u₁ and u₂ satisfying ‖u₁‖ < a^′, min_{t∈[η, b]}u₂(t) < b^′, and ‖u₂‖ > a^′.

Proof. Let θ : P → [0, ∞) be the nonnegative continuous concave functional defined by θ(u) = min_t∈[η,b]u(t),   u ∈ P. Evidently, for each u ∈ P, we have θ(u) ≤ ‖u‖.

It is easy to see that $$ is completely continuous and b^′((b − a)/(η − a)) ^{2(α − 1)} > b^′ > a^′ > 0. We choose u(t) = b^′((b − a)/(η − a)) ^2(α−1); then

So {u ∈ P(θ, b^′, b^′((b − a)/(η − a)) ^{2(α − 1)})∣θ(u) > b^′} ≠ ∅. Hence, if u ∈ P(θ, b^′, b^′((b − a)/(η − a)) ^2(α−1)), then b^′ ≤ u(t) ≤ b^′((b − a)/(η − a)) ^2(α−1) for t ∈ [η, b]. Thus for t ∈ [η, b], from assumption (H1), we have Consequently, for η ≤ t ≤ b,   b^′ ≤ u(t) ≤ b^′((b − a)/(η − a)) ^2(α−1). That is, Therefore, condition (B1) of Theorem 6 is satisfied. Now if $$, then ‖u‖ ≤ a^′. By assumption (H2), we have which shows that $$, that is, ‖Tu‖ < a^′ for $$. This shows that condition (B2) of Theorem 6 is satisfied. Finally, we show that (B3) of Theorem 6 also holds. Assume that $$ with ‖Tu‖ > b^′((b − a)/(η − a)) ^2(α−1); then by the definition of cone P, we have So condition (B3) of Theorem 6 is satisfied. Thus using Theorem 6, T has at least two fixed points. Consequently, boundary value problem (1)-(2) has at least two positive solutions u₁ and u₂ in $$ satisfying

Theorem 11. Let f(t, u) be nonnegative continuous on [a, b]×[0, ∞). Assume that there exist constants a^′, b^′, c^′ with ((η − a)/(b − a)) ^{2(α − 1)}  c^′ > b^′ > a^′ > 0 such that

•  

  H3 f(t, u(t)) < ϕ_p(Ba^′) for all (t, u)∈[a, b]×[0, a^′],

•  

  H4 f(t, u(t)) ≥ ϕ_p(Ab^′) for all (t, u)∈[η, b]×[b^′, b^′((b − a)/(η − a)) ^2(α−1)],

•  

  H5 f(t, u(t)) ≤ ϕ_p(Bc^′) for all (t, u)∈[a, b]×[0, c^′].

Then fractional order BVP (1)-(2) has at least three positive solutions u₁, u₂, and u₃ with ‖u₁‖ < a^′, min_{t∈[η, b]}u₂(t) > b^′, ‖u₃‖ > a^′, and min_{t∈[η, b]}u₃(t) < b^′.

Proof. If $$, then ‖u‖ ≤ c^′. By assumption (H5), we have

This shows that $$. Using the same arguments as in the proof of Theorem 10, we can show that $$ is a completely continuous operator. It follows from conditions (H3) and (H4) in Theorem 11 that c^′ > b^′((b − a)/(η − a)) ^{2(α − 1)} > b^′ > a^′. Similarly to the proof of Theorem 10, we have $$ and for all u ∈ P(θ, b^′, b^′((b − a)/(η − a)) ^{2(α − 1)}).

Moreover, for u ∈ P(θ, b^′, c^′) and ‖Tu‖ > b^′((b − a)/(η − a)) ^2(α−1), we have

So all the conditions of Theorem 5 are satisfied. Thus using Theorem 5, T has at least three fixed points. So, th boundary value problem (1)-(2) has at least three positive solutions u₁, u₂, and u₃ with ‖u₁‖ < a^′, min_{t∈[η, b]}u₂(t) > b^′, ‖u₃‖(t) > a^′, and min_{t∈[η, b]}‖u₃‖(t) < b^′.

Theorem 12. Let f(t, u) be nonnegative continuous on [a, b]×[0, ∞). If the following assumptions are satisfied:

•  

  H6 f₀ = f_∞ = ∞;

•  

  H7 there exists a constant μ₁ > 0 such that f(t, u) < ϕ_p(Bμ₁), for (t, u)∈[a, b]×[0, μ₁], then boundary value problem (1)-(2) has at least two positive solutions u₁ and u₂ such that 0 < ‖u₁‖ < μ₁ < ‖u₂‖.

Proof. From Lemma 1, we obtain T : P → P being completely continuous. In view of f₀ = ∞, there exists σ₁ ∈ (0, μ₁) such that f(t, u) ≥ ϕ_p(η₁u), for a ≤ t ≤ b, 0 < u ≤ σ₁, where η₁ ∈ (A/2, ∞). Let $$. Then, for any $$, we have

which implies ‖Tu‖ > ‖u‖ for $$. Hence, Theorem 8 implies On the other hand, since f_∞ = ∞, there exists σ₃ > μ₁ such that f(t, u) ≥ ϕ_p(η₂u), for u ≥ σ₃, where η₂ ∈ (A/2, ∞). Let σ₂ > max⁡{σ₃((b − a)/(η − a)) ^α−1, μ₁} and $$. Then min_t∈[η,b]u(t)≥((b − a)/(η − a)) ^α−1‖u‖ > σ₃, for any $$. By using the method to get (48), we obtain Tu(η)>(2η₁/A)‖u‖ > ‖u‖, which implies ‖Tu‖ > ‖u‖ for u ∈ ∂Ω₂. Thus, from Theorem 8, we have Finally, let $$. Then, for any $$, by (H7), we then get which implies ‖Tu‖ < ‖u‖ for $$. Using Theorem 8 again, we get Note that σ₁ < μ₁ < σ₂, by the additivity of fixed point index and (48)–(51); we obtain Hence, T has a fixed point u₁ in $$, and it has a fixed point u₂ in $$. Clearly, u₁ and u₂ are positive solutions of boundary value problem (1)-(2) and 0 < ‖u₁‖ < μ₁ < ‖u₂‖.

Theorem 13. Let f(t, u) be nonnegative continuous on [a, b]×[0, ∞). If the following assumptions are satisfied:

•  

  H8 f₀ = f_∞ = 0;

•  

  H9 there exists a constant μ₂ > 0 such that f(t, u) > ϕ_p(Aμ₂), for (t, u)∈[η, b]×[((η − a)/(b − a)) ^α−1μ₂, μ₂], then boundary value problem (1)-(2) has at least two positive solutions u₁ and u₂ such that 0 < ‖u₁‖ < μ₂ < ‖u₂‖.

Proof. From Lemma 9, we obtain T : P → P being completely continuous. In view of f₀ = 0, there exists δ₁ ∈ (0, μ₂) such that f(t, u) ≤ ϕ_p(η₂u), for a ≤ t ≤ b, 0 < u ≤ δ₁, where η₂ ∈ (0, B). Let $$. Then, for any $$, we have

which implies ‖Tu‖ < ‖u‖ for $$. Hence, Theorem 8 implies Next, since f_∞ = 0, there exists δ₃ > μ₂ such that f(t, u) ≤ ϕ_p(η₃u), for u ≥ δ₃, where η₃ ∈ (0, B). We consider two cases.

Case (i). Suppose that f is bounded, which implies that there exists N > 0 such that f(t, u) ≤ ϕ_p(N) for all t ∈ [a, b] and u ∈ [0, ∞). Take δ₄ > max⁡{N/B, δ₃}. Then, for u ∈ P with ‖u‖ = δ₄, we get

Case (ii). Suppose that f is unbounded. In view of f : [a, b]×[0, ∞)→[0, ∞) being continuous, there exist t^⋆ ∈ [a, b] and δ₅ > max⁡{((b − a)/(η − a)) ^α−1δ₃, μ₂} such that f(t, u) ≤ f(t^⋆, δ₅), for a ≤ t ≤ b, 0 ≤ u ≤ δ₅. Then, for u ∈ P with ‖u‖ = δ₅, we obtain

So, in either case, if we always choose $$, then we have ‖Tu‖ < ‖u‖, for u ∈ ∂Ω₂. Thus, from Theorem 8, we have Finally, let $$. Then, for any $$, min_t∈[η,b]u(t)≥((η − a)/(b − a)) ^α−1‖u‖ = ((η − a)/(b − a)) ^α−1μ₂, by (H9), and we then obtain which implies ‖Tu‖ > ‖u‖ for $$. An application of Theorem 8 again shows that Note that δ₁ < μ₂ < δ₂ by the additivity of fixed point index and (54)–(59); we obtain Hence, T has a fixed point u₁ in $$, and it has a fixed point u₂ in $$. Consequently, u₁ and u₂ are positive solutions of boundary value problem (1)-(2) and 0 < ‖u₁‖ < μ₂ < ‖u₂‖.