An investigation of fractional Bagley-Torvik equation

Definition 3.2

Let (X, ⪯) be a partially ordered set and 𝓖 : X × X → X. We say that 𝓖 has the mixed monotone property if 𝓖(x, y) is monotone non-decreasing in x and is monotone non-increasing in y.

Definition 3.3

We call an element (x, y) ∈ X × X a coupled fixed point of the mapping 𝓖 if

Theorem 3.1

Let (X, ⪯) be a partially ordered set and suppose there exists a metric d on X such that (X, d) is a complete metric space. Let 𝓖 : X × X → X be a mapping having the mixed monotone property on X. Assume that there exists a k ∈ [0, 1) with

Suppose either 𝓖 is continuous or X has the following property:

1. if a non-decreasing sequence {x_n} → x, then x_n ⪯ x for all n,

2. if a non-increasing sequence {y_n} → y, then y ⪯ y_n for all n.

   If there exist x₀, y₀ ∈ X such that x₀ ⪯ 𝓖 (x₀, y₀) and y₀ ⪰ 𝓖 (y₀, x₀), then 𝓖 has a coupled fixed point (x^*, y^*) ∈ X × X.

We define the following partial order on the product space X × X:

Theorem 3.2

In addition to the hypothesis of Theorem 3.1, suppose that for every (x, y), (x̃, ỹ) ∈ X × X, there exists an element (u, v) ∈ X × X that is comparable to (x, y) and (x̃, ỹ), then 𝓖 has a unique coupled fixed point (x^*, y^*).

Theorem 3.3

In addition to the hypothesis of Theorem 3.2, suppose that every pair of elements of X has an upper bound or a lower bound in X. Then x^* = y^*. Moreover,

where 𝓖ⁿ (x₀, y₀) = 𝓖(𝓖ⁿ⁻¹ (x₀, y₀), 𝓖ⁿ⁻¹ (y₀, x₀)).

In view of Lemma 2.6, we transform problem (1.3)-(1.4) as the following integral equation

in the set C¹[0, 1].

Definition 3.4

An element (x₀, y₀) ∈ C¹[0, 1] × C¹[0, 1] is called a coupled lower and upper solution of problem (1.3)-(1.4) if

and

for all t ∈ [0, 1].

Theorem 3.4

Assume that (x₀, y₀) ∈ C¹[0, 1] × C¹[0, 1] is a coupled lower and upper solution of problem (1.3)-(1.4) and k=max{2|B|,2AΓ(3−α)}<1.

1. Then the initial value problem (1.3)-(1.4) has a unique solution x^* ∈ C¹[0, 1].

2. Moreover, there exist two monotone iterative sequences {x_n} and {y_n} such that both sequences converge to x^* in C¹[0, 1].

3. In addition, the following error estimates hold,

   ∥xn−x∗∥C1[0,1]≤12kn1−k∥x1−x0∥C1[0,1]+∥y1−y0∥C1[0,1], (3.4)

   ∥yn−x∗∥C1[0,1]≤12kn1−k∥x1−x0∥C1[0,1]+∥y1−y0∥C1[0,1], (3.5)

   ∥xn−yn∥C1[0,1]≤kn1−k∥x1−x0∥C1[0,1]+∥y1−y0∥C1[0,1]. (3.6)

Proof

In view of (3.1), we define the operator 𝓖 : C¹ [0, 1] × C¹ [0, 1] → C¹ [0, 1] by

Obviously, for any x, y ∈ C¹ [0, 1], we have 𝓖(x, y) ∈ C[0, 1]. On the other hand,

and so the operator 𝓖 is well defined.

Now we shall show that 𝓖 has the mixed monotone property. Let x, x ∈ C¹ [0, 1] with x ⪯ x. Using the monotonicity of integral operator, we have

and

for every t ∈ [0, 1]. Hence, 𝓖(x, y)(t) ≤ 𝓖(x, y) (t) and 𝓖′(x, y)(t) ≤ 𝓖′(x, y) (t) for every t ∈ [0, 1], that is, 𝓖(x, y) ⪯ 𝓖(x, y). Similarly, let y, y ∈ C¹ [0, 1] with y ⪯ y. From the monotonicity of Riemann-Liouville fractional integral operator, we have

and

for every t ∈ [0, 1]. Hence, 𝓖(x, y)(t) ≥ 𝓖(x, y) (t) and 𝓖′(x, y)(t) ≥ 𝓖′(x, y) (t) for every t ∈ [0, 1], that is, 𝓖(x, y) ⪰ 𝓖(x, y). Thus, 𝓖 (x, y) is monotone non-decreasing in x and monotone non-increasing in y.

Now, for x, y, x, y ∈ C¹[0, 1] with x ⪰ x, y ⪯ y, we have

and

Therefore,

Furthermore, it is easy to see that, if {x_n} is a monotone non-decreasing sequence in C¹ [0, 1] that converges to x ∈ C¹ [0, 1] and {y_n} is a monotone non-increasing sequence in C¹ [0, 1] that converges to y ∈ C¹ [0, 1], then x_n ⪯ x and y ⪯ y_n, for all n.

On the other hand, from (3.2), we have

for every t ∈ [0, 1]. Now by applying the integral operator I¹ on both sides of inequality (3.9) and using x(0) ≤ a, we deduce

for every t ∈ [0, 1]. Furthermore, using (3.3) and applying similar calculation, we get y0′ (t) ≥ 𝓖′(y₀, x₀)(t) and y₀ (t) ≥ 𝓖(y₀, x₀)(t) for every t ∈ [0, 1]. Therefore,

Consequently, Theorem 3.1 yields the existence of coupled fixed point (x^*, y^*) ∈ C¹ [0, 1] × C¹ [0, 1] for the oparator 𝓖.

Also, C¹ [0, 1] × C¹ [0, 1] is a partially ordered set if we define the following order relation in C¹ [0, 1] × C¹ [0, 1] :

Now, if for every (x, y), (x, y) ∈ C¹ [0, 1] × C¹ [0, 1], we define

and

then (x̃, ỹ) ∈ C¹ [0, 1] × C¹ [0, 1] is comparable to (x, y) and (x, y). The uniqueness of the coupled fixed point (x^*, y^*) therefore follows from Theorem 3.2. Finally, an application of Theorem 3.3, together with Lemma 3.1, yields x^* = y^* . This establishes the first assertion.

For the second assertion, we define iterative sequences {x_n} and {y_n} as follows

where x₀ and y₀ are the coupled lower and upper solutions of problem (1.3)-(1.4). We intend to prove by induction on n that {x_n} and {y_n} are non-decreasing and non-increasing sequences, respectively. The case n = 1 being immediate from (3.10). Now we take an arbitrary positive integer n and we assume that x_n−1 ⪯ x_n and y_n ⪯ y_n−1. Then, using the mixed monotone property of 𝓖, we have

and

Moreover, both sequences {x_n} and {y_n} converge to x^* which follow from Theorem 3.3. This proves assertion (ii).

Now we prove the error estimates. From (3.8) it follows that

Again employing (3.8) and using (3.12) and (3.13), we deduce

By a mathematical induction, we obtain

Then for any m ≥ n ≥ 1,

Letting m → ∞ in both sides of (3.16), we can obtain the error estimate (3.4). A similar argument can also be used to prove error estimate (3.5). Finally, (3.6) follows immediately from (3.4) and (3.5). □

Remark 3.1

In view of Theorem 3.4, the sequences defined by (3.11) generate a sequence {(x_n, y_n)} which each of its elements is a coupled lower and upper solution of problem (1.3)-(1.4).

Remark 3.2

Let (x₀, y₀) be a coupled lower and upper solution of problem (1.3)-(1.4) such that x₀ ⪯ y₀ and let {x_n} and {y_n} be the sequences defined by (3.11) such that ∥x_n − x^*∥_{C¹[0, 1]} → 0 and ∥y_n − x^*∥_{C¹[0, 1]} → 0. Then

Example 3.1

Let us consider the following problem

where f(t)=14t2−14t−85tπ−2 and the exact solution is x(t) = t(1 − t). Observe that, we have k = max{2|B|,2AΓ(3−α)}=0.90267<1. Now, define 𝓖 : C¹[0, 1] × C¹[0, 1] → C¹[0, 1] by setting

A relatively simple calculation, with the help of Maple, shows that (x₀(t), y₀(t)) = (−3t, 3t) is a coupled lower and upper solution of problem (3.18). Therefore, all the assumptions of Theorem 3.4 hold and consequently, problem (3.18) has a unique solution in C¹[0, 1]. Moreover, the unique solution of (3.18) can be obtained as lim_{n → ∞}𝓖ⁿ (x₀, y₀) where 𝓖ⁿ (x₀, y₀) = 𝓖(𝓖ⁿ⁻¹ (x₀, y₀), 𝓖^{n − 1}(y₀, x₀)). For simplicity, we set x_n = 𝓖 (x_n−1, y_n−1) and and y_n = 𝓖 (y_n−1, x_n−1). Using simple calculation, with the help of Maple, we can now form the first few successive approximations as follows

Therefore, we have

and

Similarly,

and

It is interesting to point out that (x_n, y_n) = (𝓖ⁿ (x₀, y₀), 𝓖ⁿ(y₀, x₀)), n = 1, 2, 3 serve as an approximation to the unique coupled fixed point of 𝓖 of increasing accuracy as n → ∞. On the other hand, from Theorem 3.3, the unique solution of (3.18) can be obtained as

The graphs of x_n and y_n, for n = 0, 1, 6 are shown in Figure 1. Furthermore, the graphs of xn′andyn′, for n = 0, 1, 6 are shown in Figure 2.

Figure 1

Graphs of x_n and y_n

Figure 2

Graphs of xn′andyn′

Remark 3.3

To derive the same results for the case in which A < 0 and B > 0, we can apply the same technique as mentioned above. The main difference is the structure of the function 𝓖 and consequently the definition of coupled lower and upper solution of the problem (1.3)-(1.4). We could carry out a similar argument to prove the existence, uniqueness and approximation results.

Theorem 3.5

Let (X, ⪯) be a partially ordered set such that every pair x, y ∈ X has a lower bound and an upper bound. Furthermore, let d be a metric on X such that (X, d) is a complete metric space and 𝓖 is monotone (i.e., either order-preserving or order-reversing) map from X into X such that

Suppose also that either 𝓖 is continuous or X is such that if x_n → x is a sequence in X whose consecutive terms are comparable, then there exists a subsequence {x_nk} of {x_n} such that every term is comparable to the limit x. Then 𝓖 has a unique fixed point x^*. Moreover, for every x ∈ X, lim_{n → ∞} 𝓖ⁿ (x) = x^*.

In view of Lemma 2.6, we transform problem (1.3)-(1.4) as x = 𝓖(x) where 𝓖 : C¹[0, 1] → C¹[0, 1] is defined by

Definition 3.5

An element x₀ ∈ C¹[0, 1] is called a lower solution of problem (1.3)-(1.4) if

for all t ∈ [0, 1] and it is an upper solution of (1.3)-(1.4) if the above inequalities are reversed.

Theorem 3.6

Assume that x₀ ∈ C¹[0, 1] is a lower solution of problem (1.3)-(1.4) and k=|B|+|A|Γ(3−α)<1.

1. Then the initial value problem (1.3)-(1.4) has a unique solution x^* ∈ C¹[0, 1].

2. Moreover, the iterative sequence {x_n} defined by

   xn(t)=a+bt+AΓ(4−α)t3−α−AΓ(3−α)∫0t(t−τ)2−αxn−1′(τ)dτ+∫0t(t−τ)f(τ)−Bxn−1(τ)dτ,

   converges to x^* in C¹[0, 1].

3. In addition, the following error estimates hold,

   ∥xn−x∗∥C1[0,1]≤kn1−k∥x1−x0∥C1[0,1].∥xn+1−xn∥C1[0,1]≤kn∥x1−x0∥C1[0,1].

Proof

The proof is similar to that for Theorem 3.4. It suffices to define 𝓖 : C¹[0, 1] → C¹[0, 1] by