Positive Solutions of a Fractional Thermostat Model with a Parameter

We study the existence, multiplicity, and uniqueness results of positive solutions for a fractional thermostat model. Our approach depends on the fixed point index theory, iterative method, and nonsymmetry property of the Green function. The properties of positive solutions depending on a parameter are also discussed.

In Reference [20], Shen, Zhou, and Yang studied a fractional thermostat model The authors obtained intervals of parameter λ that correspond to at least one and no positive solutions. Similar fractional thermostat problems have been studied in References [21][22][23][24].
In this paper, we deal with positive solutions for the fractional thermostat model (1). The existence, multiplicity, and uniqueness results are established by the fixed point index theory and iterative method. The properties of positive solutions depending on a parameter are also discussed. Some of the ideas in this paper are from References [25,26]. Let us remark that the definition of the Gerasimov-Caputo derivative was first introduced and applied by Gerasimov in 1947 and then by Caputo in 1967, see for example, the overview by Novozhenova in Reference [27]. For details on the theory and applications of the fractional derivatives and integrals and fractional differential equations, see References [28][29][30][31].

Preliminaries
Lemma 1 ([20]). Given u(t) ∈ C(0, 1) ∩ L 1 (0, 1), the solution of the problem and G(t, s) satisfies: Denote E = C[0, 1] and x = sup t∈[0,1] |x(t)|. We define the cone For any 0 < r < +∞, let P r = {x ∈ P : x < r}. We define T : (0, +∞) × E → E as It is obvious from Lemma 1 that if x ∈ P is a fixed point of operator T, then x is a positive solution of Problem (1). By regularity arguments, we can show that T is completely continuous and T(P) ⊂ P.
Define the linear operator L : E → E by By the Krein-Rutman theorem, we see that the spectral radius r(L) of the operator L is positive, and L has positive eigenfunction ϕ 1 corresponding to its first eigenvalue µ 1 = (r(L)) −1 .

Lemma 2 ([32]
). Let P be a cone in Banach space E. Suppose that T : P → P is a completely continuous operator. (i) If Tu = µu for any u ∈ ∂P r and µ ≥ 1, then i(T, P r , P) = 1. (ii) If Tu = u and Tu ≥ u for any u ∈ ∂P r , then i(T, P r , P) = 0.
We assume that: Therefore, x 0 ∈ P and x 0 = s 0 . Direct computations yield j=1 is decreasing and bounded from below, lim j→∞ x j 1 (t) exists and convergence is uniform for The contradiction shows that x 1 ∈ P \ {θ} and µγA . Since lim n→∞ x n = ∞, there exists N 0 > 0 such that x n > X γ for n > N 0 , and x n (t) ≥ γ x n > X, t ∈ [0, 1]. Then, for any n > N 0 , we obtain which is absurd, and hence S µ is bounded.
Next, we prove that i(T, P R , P) = 0 for some R > r. In fact, f ∞ = ∞ implies that f (s) > Ms for some large R 1 > 0 and Hence, i(T, P R , P) = 0, and i(T, P R \ P r , P) = −1. Therefore, T admits a fixed point x * ∈ P R \ P r . Theorem 2. Assume that (H 1 ) holds, and that f 0 = f ∞ = ∞. Then, BVP (1) has at least one and two positive solutions for λ = 1 l A and λ ∈ (0, 1 l A ), respectively.
Proof. By Lemma 5, BVP (1) admits a positive solution for λ = 1 l A . For λ ∈ (0, 1 l A ), by Lemmas 3 and 5, there existx, x λ ∈ P \ {θ}, x λ ≤x such that This contradiction shows that x λ <x. Define where r > 0 is the same as in the first part of Theorem 1. For any x ∈ P ∩ ∂Ω 1 , we obtain x = x , and As in the proof in Theorem 1, there is R > 0 large enough such that where Ω 2 = {x ∈ E : x < R}. By compression expansion fixed point theorem, we see that T has a fixed point x λ ∈ P ∩ (Ω 2 \ Ω 1 ). Since x λ ∈ Ω 1 , x λ = x λ , problem (1) has a second positive solution.
Author Contributions: Both authors have contributed equally to this paper. Writing-original draft, X.H. and L.Z.; Writing-review and editing, X.H. and L.Z.

Conflicts of Interest:
The authors declare no conflict of interest.