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Article

Positive Solutions of a Fractional Thermostat Model with a Parameter

School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China
*
Author to whom correspondence should be addressed.
Symmetry 2019, 11(1), 122; https://doi.org/10.3390/sym11010122
Submission received: 19 December 2018 / Revised: 14 January 2019 / Accepted: 17 January 2019 / Published: 21 January 2019
(This article belongs to the Special Issue Fractional Differential Equations: Theory, Methods and Applications)

Abstract

:
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.

1. Introduction

In this paper, we investigate a fractional nonlocal boundary value problem (BVP)
D 0 + α c x ( t ) + λ g ( t ) f ( x ( t ) ) = 0 , t ( 0 , 1 ) , x ( 0 ) = 0 , β c D 0 + α - 1 x ( 1 ) + x ( η ) = 0 ,
where 1 < α 2 , β > 0 , 0 η 1 , β Γ ( α ) > ( 1 - η ) α - 1 , c D 0 + α is the Gerasimov–Caputo fractional derivative of order α , λ > 0 is a parameter, f C ( [ 0 , + ) , [ 0 , + ) ) , g C ( ( 0 , 1 ) , [ 0 , + ) ) , and 0 < 0 1 g ( t ) d t < + .
One motivation is that the thermostat model
x ( t ) + g ( t ) f ( t , x ( t ) ) = 0 , t ( 0 , 1 ) , x ( 0 ) = 0 , β x ( 1 ) + x ( η ) = 0 ,
which is a special case with α = 2 and λ = 1 , has been discussed by Infante and Webb [1,2]. They established multiplicity results of BVP (2). These types of problems have been investigated by various scholars, see References [3,4,5,6,7,8,9,10,11,12,13,14,15,16,17].
Recently, the thermostat model was extended to the fractional case
D 0 + α c x ( t ) + f ( t , x ( t ) ) = 0 , t ( 0 , 1 ) , α ( 1 , 2 ] , x ( 0 ) = 0 , β c D 0 + α - 1 x ( 1 ) + x ( η ) = 0 ,
where β > 0 , 0 η 1 , f C ( [ 0 , 1 ] × [ 0 , + ) , [ 0 , + ) ) . Nieto and Pimentel [18] proved the existence of positive solutions based on the Krasnosel’skii fixed point theorem. Cabada and Infante [19] discussed the multiplicity results of positive solutions for BVP (3).
In Reference [20], Shen, Zhou, and Yang studied a fractional thermostat model
D 0 + α c x ( t ) + λ f ( t , x ( t ) ) = 0 , t ( 0 , 1 ) , 1 < α 2 , x ( 0 ) = 0 , β c D 0 + α - 1 x ( 1 ) + x ( η ) = 0 ,
where β > 0 , 0 η 1 , β Γ ( α ) > ( 1 - η ) α - 1 , λ > 0 , f : [ 0 , 1 ] × [ 0 , + ) [ 0 , + ) is continuous. 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].

2. Preliminaries

Lemma 1 ([20]).
Given u ( t ) C ( 0 , 1 ) L 1 ( 0 , 1 ) , the solution of the problem
D 0 + α c x ( t ) + u ( t ) = 0 , t ( 0 , 1 ) , x ( 0 ) = 0 , β c D 0 + α - 1 x ( 1 ) + x ( η ) = 0
is
x ( t ) = 0 1 G ( t , s ) u ( s ) d s , t [ 0 , 1 ] ,
where
G ( t , s ) = β - ( t - s ) α - 1 Γ ( α ) + ( η - s ) α - 1 Γ ( α ) , 0 s η , s t , β + ( η - s ) α - 1 Γ ( α ) , 0 s η , s t , β - ( t - s ) α - 1 Γ ( α ) , η s 1 , s t , β , η s 1 , s t ,
and G ( t , s ) satisfies:
(i) 
G ( t , s ) : [ 0 , 1 ] × [ 0 , 1 ] ( 0 , + ) is continuous;
(ii) 
t G ( t , s ) 0 , t , s [ 0 , 1 ] ;
(iii) 
γ G ¯ = G ̲ G ( 1 , s ) G ( t , s ) G ( 0 , s ) G ¯ , t , s [ 0 , 1 ] ,
where
γ = β Γ ( α ) - ( 1 - η ) α - 1 β Γ ( α ) + η α - 1 , G ̲ = β Γ ( α ) - ( 1 - η ) α - 1 Γ ( α ) , G ¯ = β Γ ( α ) + η α - 1 Γ ( α ) .
Denote E = C [ 0 , 1 ] and x = sup t [ 0 , 1 ] | x ( t ) | . We define the cone
P = { x E : x ( t ) 0 , inf t [ 0 , 1 ] x ( t ) γ x } .
For any 0 < r < + , let P r = { x P : x < r } . We define T : ( 0 , + ) × E E as
T ( λ , x ) ( t ) = λ 0 1 G ( t , s ) g ( s ) f ( x ( s ) ) d s , t [ 0 , 1 ] .
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
L x ( t ) = 0 1 G ( t , s ) g ( s ) x ( s ) d s , t [ 0 , 1 ] .
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 T u μ u for any u P r and μ 1 , then i ( T , P r , P ) = 1 . (ii) If T u u and T u u for any u P r , then i ( T , P r , P ) = 0 .
Denote
f 0 = lim s 0 f ( s ) s , f = lim s f ( s ) s , A = 0 1 G ( 0 , s ) g ( s ) d s , l = min s ( 0 , ) f ( s ) s .
We assume that:
(H1)
f is nondecreasing on [ 0 , + ) ;
(H2)
there exists a function ϕ : ( 0 , 1 ] [ 0 , 1 ] continuous nondecreasing, such that f ( κ x ) ϕ ( κ ) f ( x ) for 0 < κ < 1 , x > 0 , and F ( κ ) : = κ ϕ ( κ ) is strictly increasing on ( 0 , 1 ] and F ( 1 ) = 1 .
Lemma 3.
Suppose that ( H 1 ) holds, f 0 = and l > 0 . If 0 < λ 1 < λ 2 < 1 l A , then there exist x 1 , x 2 P { θ } , x 1 x 2 , such that T ( λ 1 , x 1 ) ( t ) = x 1 ( t ) and T ( λ 2 , x 2 ) ( t ) = x 2 ( t ) .
Proof. 
Assume s 0 ( 0 , ) such that f ( s 0 ) = l s 0 . Since 0 < λ 1 < λ 2 < 1 l A , we have l < 1 λ 2 A < 1 λ 1 A . We define
x 0 ( t ) = s 0 A 0 1 G ( t , s ) g ( s ) d s , t [ 0 , 1 ] ,
then
x 0 = x 0 ( 0 ) = s 0 , x 0 ( t ) s 0 A 0 1 γ G ( 0 , s ) g ( s ) d s = γ x 0 , t [ 0 , 1 ] .
Therefore, x 0 P and x 0 = s 0 . Direct computations yield
T ( λ 1 , x 0 ) ( t ) = λ 1 0 1 G ( t , s ) g ( s ) f ( x 0 ( s ) ) d s λ 1 0 1 G ( t , s ) g ( s ) f ( x 0 ) d s = λ 1 l s 0 0 1 G ( t , s ) g ( s ) d s < s 0 A 0 1 G ( t , s ) g ( s ) d s = x 0 ( t ) , t [ 0 , 1 ] .
Define
x 1 1 ( t ) = T ( λ 1 , x 0 ) ( t ) , x 1 j ( t ) = T ( λ 1 , x 1 j - 1 ) ( t ) = T j ( λ 1 , x 0 ) ( t ) , j = 2 , 3 , , t [ 0 , 1 ] .
Direct calculations show that x 0 > x 1 1 > x 1 2 > > x 1 j > x 1 j + 1 > θ . Hence, sequence { x 1 j } j = 1 is decreasing and bounded from below, lim j x 1 j ( t ) exists and convergence is uniform for t [ 0 , 1 ] . Assume that lim j x 1 j = x 1 , we claim that x 1 ( t ) > 0 . Otherwise, since x 1 P , x 1 ( t ) = 0 , i.e., lim j x 1 j ( t ) = 0 , t [ 0 , 1 ] , and hence from x 1 j P , we deduce x 1 j 0 . Since f 0 = , for any H > 1 λ 1 γ A , there is integral Z > 0 such that for j > Z , we have f ( x 1 j ( t ) ) > H x 1 j ( t ) , and hence
x 1 j + 1 ( 0 ) = λ 1 0 1 G ( 0 , s ) g ( s ) f ( x 1 j ( s ) ) d s > λ 1 H γ 0 1 G ( 0 , s ) g ( s ) x 1 j d s x 1 j ( 0 ) λ 1 H γ A > x 1 j ( 0 ) .
The contradiction shows that x 1 P { θ } and x 1 = T ( λ 1 , x 1 ) .
Similarly, from x 2 1 ( t ) = T ( λ 2 , x 0 ) ( t ) and x 2 j ( t ) = T ( λ 2 , x 2 j - 1 ) ( t ) , j = 2 , 3 , , we deduce
x 0 > x 2 1 > x 2 2 > > x 2 j > x 2 j + 1 > θ ,
lim j x 2 j = x 2 P { θ } , and x 2 = T ( λ 2 , x 2 ) . It follows from x 1 1 = T ( λ 1 , x 0 ) < T ( λ 2 , x 0 ) = x 2 1 and the monotonicity of f that x 1 j x 2 j , j = 2 , 3 , . Therefore, x 1 x 2 . ☐
Lemma 4.
If f = , then for any μ > 0 , the set S μ = { x P : T ( λ , x ) = x , λ [ μ , ) } is bounded.
Proof. 
Otherwise, there exists x n S μ corresponding to λ n [ μ , ) such that
T ( λ n , x n ) = x n , lim n x n = .
Because f = , there is X > 0 such that f ( s ) > H s for s > X , where H > 1 μ γ A . 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
x n > λ n 0 1 G ( 0 , s ) g ( s ) H x n ( s ) d s > μ H γ x n A > x n ,
which is absurd, and hence S μ is bounded.  ☐
Lemma 5.
Assume that ( H 1 ) holds, and that f 0 = f = . Then, T admits a fixed point for λ = 1 l A .
Proof. 
Choosing a sequence 0 < λ 1 < λ 2 < < λ n < λ n + 1 < < 1 l A such that lim n λ n = 1 l A . By Lemma 3, there exists a nondecreasing sequence { x n } n = 1 P { θ } such that x n = T ( λ n , x n ) . By Lemma 4, we know that { x n } n = 1 is uniformly bounded and equicontinuous. By using the Arzela–Ascoli theorem, we can prove that there exists { x n k } k = 1 { x n } n = 1 such that x n k x ˜ E uniformly on [ 0 , 1 ] . Therefore, x n k satisfies
x n k ( t ) = T ( λ n k , x n k ) ( t ) = λ n k 0 1 G ( t , s ) g ( s ) f ( x n k ( s ) ) d s , t [ 0 , 1 ] .
Passing to the limit as k , we obtain
x ˜ ( t ) = 1 l A 0 1 G ( t , s ) g ( s ) f ( x ˜ ( s ) ) d s , t [ 0 , 1 ] .
Hence, x ˜ = T 1 l A , x ˜ . ☐
Lemma 6.
Assume that ( H 1 ) holds, and that f ( 0 ) > 0 . Then, for any x P , there exist U x V > 0 such that
V K λ T ( λ , x ) ( t ) U x K λ , t [ 0 , 1 ] ,
where
K λ = λ 0 1 g ( t ) d t .
Proof. 
By ( H 1 ) , for any x P and t [ 0 , 1 ] , we have
T ( λ , x ) ( t ) G ̲ f ( 0 ) λ 0 1 g ( t ) d t : = V K λ ,
and
T ( λ , x ) ( t ) G ¯ f ( x ) λ 0 1 g ( t ) d t : = U x K λ .
 ☐

3. Main Results

Theorem 1.
Assume that f = and 0 < f 0 < . Then, for any 0 < λ < μ 1 f 0 , BVP (1) admits a positive solution.
Proof. 
Since 0 < λ < μ 1 f 0 , there exist ε > 0 small enough and r > 0 such that λ ( f 0 + ε ) < μ 1 , and f ( s ) s < f 0 + ε for s ( 0 , r ] . We claim that
T ( λ , x ) μ x , x P r , μ 1 .
Otherwise, there exist x 0 P r and μ 0 1 such that T ( λ , x 0 ) = μ 0 x 0 . Since 0 < γ r x 0 ( t ) x 0 = r , we have
μ 0 x 0 ( t ) λ ( f 0 + ε ) 0 1 G ( t , s ) g ( s ) x 0 ( s ) d s = λ ( f 0 + ε ) L x 0 ( t ) ,
then L x 0 ( t ) μ 0 λ ( f 0 + ε ) x 0 ( t ) . Thus, r ( L ) μ 0 λ ( f 0 + ε ) 1 λ ( f 0 + ε ) . It follows that μ 1 λ ( f 0 + ε ) , which is a contradiction. Then, i ( T , P r , P ) = 1 .
Next, we prove that i ( T , P R , P ) = 0 for some R > r . In fact, f = implies that f ( s ) > M s for some large R 1 > 0 and s R 1 , where M > ( λ γ A ) - 1 . Let R > max { r , R 1 γ } . For x P R , we have x ( t ) γ x = γ R > R 1 , t [ 0 , 1 ] , then
T ( λ , x ) λ M 0 1 G ( 0 , s ) g ( s ) x ( s ) d s λ M γ x A > x .
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 exist x ˜ , x λ P { θ } , x λ x ˜ such that
T 1 l A , x ˜ ( t ) = x ˜ ( t ) , T ( λ , x λ ) ( t ) = x λ ( t ) , t [ 0 , 1 ] .
If x λ = x ˜ , we have
T ( λ , x λ ) = x λ = x ˜ = T 1 l A , x ˜ = T 1 l A , x λ .
This contradiction shows that x λ < x ˜ .
Define Ω 1 = { x E : - r < x ( t ) < x ˜ ( t ) , t [ 0 , 1 ] } , 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
T ( λ , x ) < 1 l A 0 1 G ( 0 , s ) g ( s ) f ( x ˜ ( s ) ) d s = x ˜ ( 0 ) = x ˜ .
Therefore,
T ( λ , x ) < x , x P Ω 1 .
As in the proof in Theorem 1, there is R > 0 large enough such that
T ( λ , x ) > x , x P Ω 2 ,
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. ☐
Theorem 3.
Assume that ( H 1 ) and ( H 2 ) hold, and that f ( 0 ) > 0 . Then, for any λ ( 0 , ) , BVP (1) admits a unique positive solution x ˙ λ ( t ) , and x ˙ λ ( t ) satisfies:
(i) 
x ˙ λ ( t ) is nondecreasing with respect to λ;
(ii) 
lim λ 0 + x ˙ λ = 0 , lim λ x ˙ λ = ;
(iii) 
x ˙ λ - x ˙ λ 0 0 as λ λ 0 .
Proof. 
Since T is nondecreasing, for u P , we have
T ( λ , κ x ) ( t ) ϕ ( κ ) λ 0 1 G ( t , s ) g ( s ) f ( x ( s ) ) d s = ϕ ( κ ) T ( λ , x ) ( t ) , t [ 0 , 1 ] .
Define x ^ ( t ) = K λ , where K λ is given by Lemma 6, then x ^ P and V K λ T ( λ , x ^ ) ( t ) U x K λ . Denote
V ¯ = sup { μ : μ K λ T ( λ , x ^ ) ( t ) } , U ¯ = inf { μ : μ K λ T ( λ , x ^ ) ( t ) } ,
then V ¯ V and U ¯ U x . Select V ˜ and U ˜ so that
0 < V ˜ < min { 1 , F - 1 ( V ¯ ) } , 0 < 1 U ˜ < min 1 , F - 1 1 U ¯ .
We define
x 1 ( t ) = V ˜ K λ , x k + 1 ( t ) = T ( λ , x k ) ( t ) , t [ 0 , 1 ] , k = 1 , 2 , ,
y 1 ( t ) = U ˜ K λ , y k + 1 ( t ) = T ( λ , y k ) ( t ) , t [ 0 , 1 ] , k = 1 , 2 , .
Combining the properties of T and (4), we get
V ˜ K λ = x 1 ( t ) x 2 ( t ) x k ( t ) y k ( t ) y 2 ( t ) y 1 ( t ) = U ˜ K λ .
Let d = V ˜ U ˜ , obviously 0 < d < 1 . We claim that
x k ( t ) ϕ k - 1 ( d ) y k ( t ) , t [ 0 , 1 ] , k = 1 , 2 , ,
where ϕ 0 ( d ) = d , ϕ k ( d ) = ϕ ( ϕ k - 1 ( d ) ) , k = 1 , 2 , . In fact, x 1 ( t ) = d y 1 ( t ) = ϕ 0 ( d ) y 1 ( t ) , t [ 0 , 1 ] . Suppose x n ( t ) ϕ n - 1 ( d ) y n ( t ) for t [ 0 , 1 ] , then
x n + 1 ( t ) T ( λ , ϕ n - 1 ( d ) y n ) ( t ) ϕ ( ϕ n - 1 ( d ) ) T ( λ , y n ) ( t ) = ϕ n ( d ) y n + 1 ( t ) .
Hence, it follows by induction that (6) is true. According to (5) and (6), one has
0 x n + m ( t ) - x n ( t ) y n ( t ) - x n ( t ) ( 1 - ϕ n - 1 ( d ) ) y 1 ( t ) = ( 1 - ϕ n - 1 ( d ) ) U ˜ K λ ,
where m 0 is an integer. Thus,
x n + m - x n y n - x n ( 1 - ϕ n - 1 ( d ) ) U ˜ K λ .
We claim that lim n ϕ n ( d ) = 1 . From ( H 2 ) and 0 < d < 1 , we see that ϕ ( d ) ( d , 1 ) and d = ϕ 0 ( d ) < ϕ 1 ( d ) < < ϕ n ( d ) < < 1 . Sequence { ϕ n ( d ) } n = 1 is increasing and bounded, there is p [ d , 1 ] such that lim n ϕ n ( d ) = p . By the continuity of ϕ and ϕ n ( d ) = ϕ ( ϕ n - 1 ( d ) ) , we conclude that p = ϕ ( p ) , i.e., F ( p ) = 1 . It follows that p = 1 . Inequality (7) implies that there exists x ¯ P such that lim n x n ( t ) = lim n y n ( t ) = x ¯ ( t ) for t [ 0 , 1 ] . Clearly, x ¯ ( t ) is a positive solution of problem (1).
Suppose that x ¯ 1 ( t ) and x ¯ 2 ( t ) are positive solutions of problem (1), then T ( λ , x ¯ 1 ) ( t ) = x ¯ 1 ( t ) and T ( λ , x ¯ 2 ) ( t ) = x ¯ 2 ( t ) , t [ 0 , 1 ] . Define δ ˜ = sup { δ : x ¯ 1 ( t ) δ x ¯ 2 ( t ) } , then x ¯ 1 ( t ) δ ˜ x ¯ 2 ( t ) . We claim that δ ˜ 1 . Otherwise, δ ˜ < 1 . Assumption ( H 2 ) implies that f ( δ ˜ x ¯ 2 ( t ) ) > φ ( δ ˜ ) f ( x ¯ 2 ( t ) ) , t [ 0 , 1 ] . Since f is nondecreasing,
x ¯ 1 ( t ) = T ( λ , x ¯ 1 ) ( t ) T ( λ , δ ˜ x ¯ 2 ) ( t ) > ϕ ( δ ˜ ) T ( λ , x ¯ 2 ) ( t ) > δ ˜ x ¯ 2 ( t ) , t [ 0 , 1 ] ,
a contradiction. Then, x ¯ 1 ( t ) x ¯ 2 ( t ) for t [ 0 , 1 ] . Similarly, x ¯ 2 ( t ) x ¯ 1 ( t ) . Therefore, x ¯ 1 ( t ) = x ¯ 2 ( t ) , t [ 0 , 1 ] . This proves the uniqueness result.
Next, we show that ( i ) - ( i i i ) hold. Let
( H x ) ( t ) = 0 1 G ( t , s ) g ( s ) f ( x ( s ) ) d s , t [ 0 , 1 ] ,
then T ( λ , x ) = λ H x . Since P o = { x P : x ( t ) > 0 , t [ 0 , 1 ] } is nonempty, the operator H : P o P o is increasing, and H ( κ x ) ϕ ( κ ) H x for 0 < κ < 1 . Let ω = 1 λ . We now write H x ω = ω x ω instead of λ H x λ = x λ . Assume 0 < ω 1 < ω 2 , then x ω 1 x ω 2 . Indeed, denote ω ¯ = sup { t > 0 : x ω 1 t x ω 2 } , then ω ¯ 1 . Otherwise 0 < ω ¯ < 1 . Direct computations yield ω 1 x ω 1 = H x ω 1 H ( ω ¯ x ω 2 ) ϕ ( ω ¯ ) H x ω 2 = ϕ ( ω ¯ ) ω 2 x ω 2 , then x ω 1 ω 2 ω 1 ϕ ( ω ¯ ) x ω 2 > ω ¯ x ω 2 . This is a contradiction to the definition of ω ¯ . Thus, ω ¯ 1 , x ω 1 x ω 2 , and further
x ω 1 = 1 ω 1 H x ω 1 1 ω 1 H x ω 2 = ω 2 ω 1 x ω 2 x ω 2 , 0 < ω 1 < ω 2 .
Then, x ω ( t ) is strong decreasing in ω , that is, x λ ( t ) is strong increasing in λ . Let ω 2 = ω and fix ω 1 in (8), for ω > ω 1 , we have x ω 1 ω ω 1 x ω , and
x ω N ω 1 ω x ω 1 ,
where N > 0 is a normal constant of cone P. Because ω = 1 λ , then lim λ 0 + x λ = 0 . Let ω 1 = ω and fix ω 2 in (8), we obtain lim λ + x λ = + .
Finally, for given ω 0 , by (8), we have
x ω x ω 0 , ω > ω 0 .
Let t ω = sup { t > 0 : x ω t x ω 0 , ω > ω 0 } , then 0 < t ω < 1 and x ω t ω x ω 0 . Direct computations yield ω x ω = H x ω H ( t ω x ω 0 ) ϕ ( t ω ) H x ω 0 = ϕ ( t ω ) ω 0 x ω 0 . By the definition of t ω , we have ω 0 ω ϕ ( t ω ) t ω , and
t ω F - 1 ω 0 ω , ω > ω 0 .
Combining (9) with (10), one has that
x ω 0 - x ω N 1 - F - 1 ω 0 ω x ω 0 0 , ω ω 0 + 0 .
Similarly, x ω 0 - x ω 0 , ω ω 0 - 0 . Hence, x ω 0 - x ω 0 as ω ω 0 .  ☐

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.

Funding

Supported financially by the National Natural Science Foundation of China (11501318, 11871302), the China Postdoctoral Science Foundation (2017M612230), the Natural Science Foundation of Shandong Province of China (ZR2017MA036) and the International Cooperation Program of Key Professors by Qufu Normal University.

Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Hao, X.; Zhang, L. Positive Solutions of a Fractional Thermostat Model with a Parameter. Symmetry 2019, 11, 122. https://doi.org/10.3390/sym11010122

AMA Style

Hao X, Zhang L. Positive Solutions of a Fractional Thermostat Model with a Parameter. Symmetry. 2019; 11(1):122. https://doi.org/10.3390/sym11010122

Chicago/Turabian Style

Hao, Xinan, and Luyao Zhang. 2019. "Positive Solutions of a Fractional Thermostat Model with a Parameter" Symmetry 11, no. 1: 122. https://doi.org/10.3390/sym11010122

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