International Journal of Modern Research in Engineering & Management (IJMREM) ||Volume|| 2||Issue|| 1||Pages|| 26-36 || January 2019|| ISSN: 2581-4540
Lyapunov-type inequalities for a fractional q, -difference equation involving p-Laplacian operator 1, 1,2,
Yawen Yan, 2,Chengmin Hou
(Department of Mathematics, Yanbian University, Yanji, 133002, P.R. China)
----------------------------------------------------ABSTRACT------------------------------------------------------In this paper, we present new Lyapunov-type inequalities for a fractional boundary value problem of fractional q, -difference equation with p-Laplacian operator. The obtained inequalities are used to obtain a lower bound for the eigenvalues of corresponding equations.
KEYWORDS: Lyapunov-type inequality; fractional derivative; eigenvalues; boundary value problem --------------------------------------------------------------------------------------------------------------------------------------Date of Submission: Date, 10 January 2018
Date of Accepted: 14. January 2019
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I.
INTRODUCTION
The p-Laplacian operator arises in different mathematical models that describe physical and natural phenomena (see, for example, [1-6]). In this paper, we present some Lyapunov-type inequalities for a fractional q, -difference equation with p-Laplacian operator. More precisely, we are interested with the nonlinear fractional boundary value problem
a Dq, ( p ( a Dq, u (t ))) + (t ) p (u (t )) = 0, a t b, u (a) = Dq , u (a) = Dq , u (b) = 0, a Dq , u (a)= a Dq , u (b) = 0, where 2
(1.1)
3 , 1 2 , a Dq, , a Dq, are the Riemann-Liouville fractional derivatives of orders
, , p ( s) = s
p −1
, p 1 , and
: a, b → R
is a continuous function. Under certain assumptions
imposed on the function g , we obtain necessary conditions for the existence of nontrivial solutions to (1.1). Some applications to eigenvalue problems are also presented. For completeness, let us recall the standard Lyapunov inequality [7], which states that if u is a nontrivial solution of the problem
u (t ) + (t )u (t ) = 0, a t b, u ( a ) = u (b) = 0, where a b are two consecutive zeros of u , and
b
a
(t ) dt
: a, b → R
4 . b−a
is a continuous function, then
(1.2)
Note that in order to obtain this inequality, it is supposed that a and b are two consecutive zeros of u . In our case, as it will be observed in the proof of our main result, we assume just that u is a nontrivial solution to (1.1). Inequality (1.2) is useful in various applications, including oscillation theory, stability criteria for periodic differential equations, and estimates for intervals of disconjugacy.
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Lyapunov-type inequalities for a fractional q, -difference… Several generalizations and extensions of inequality (1.2) to different boundary value problems exist in the literature. As examples, we refer to [8-13] and the references therein. The rest of this paper is organized as follows. In Section 2, we recall some basic concepts on fractional q, -calculus and establish some preliminary result be used in Section 3, where we state and prove our main result. In Section 4, we present some applications of the obtained Lyapunov-type inequalities to eigenvalue problem.
II.
PRELIMINARIES
For the convenience of the reader, we recall some basic concepts on fractional q, -calculus to make easy the analysis of (1.1). For more details, we refer to [19]. Let C[a, b] be the set of real-valued and continuous functions in [ a , b ] . Let f C[a, b] . We define the fractional q, -derivative of Riemann-Liouville type by
(
(
a
Dq,
)
a I q−, f ( x), 0, f ( x) = f ( x), = 0, Dq, a I q,− f ( x), 0,
)
(
Where
denotes the smallest integer greater or equal to
.
q (m) = qm + (1 − q)a . For any positive integer k , we have
Define a q-shifting operator as
a
qk (m)= a qk −1 ( a q (m))
and
a
)
a
q0 (m) = m .
We also define the new power of q-shifting operator as k −1
(n − m) (a0) = 1 , (n − m) (ak ) = (n − a qi (m)) , k N . i =0
More generally, if R , then
( n − m) with
a
( ) a
n− a qi (m)
i =0
n− a qi + (m)
=
,
q (m) = q m + (1 − q )a , R .
For any
, n R , we have s − 0 ) n − 0 = (n − c) , 1+ i + s − 0 i =0 1− q ( ) n − 0 1 − q i +1 (
(n− 0 q ( s))(0)
and
Dq , ( x − a)0 = [ ]q ( x − a) (0−1) .
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Lyapunov-type inequalities for a fractional q, -difference… Definition 2.1. Let v 0 and f be a function defined on [ a , b ] . The fractional integration of Riemann-Liouville type is given by
a
I qv, f (t ) =
1 t (t − 0 q ( s))( v0−1) f ( s)d q , s, v 0, t [a, b]. q (v) a
Theorem 2.2. [19] Let
, R + . The Hahn’s fractional integration has the following semi-group property
a
I q a I q f ( x) =
(
a
(
a
)
I q + f ( x),
)
Dq, a Dq , f ( x) =
(
a
(0 a x b).
)
Dq,+1 f ( x),
(0 a x b).
Lemma 2.3. Let f be a function defined on an interval
(
a
)
Dq, a I q, f ( x) = f ( x) ,
Theorem 2.4. Let
( 0 , b )
and
R + . Then the following is valid
(0 a x b).
( N − 1, N ] . Then for some constants ci R, i = 1,2, , N , the following equality
holds:
(
a
)
I q, a Dq, f ( x) = f ( x) + c1 ( x − a) (0 −1) + c2 ( x − a) (0 −2) + + c N ( x − a) (0 − N ).
Now, in
order to obtain an integral formulation of (1.1), we need the following results. Lemma 2.5. Let
2 3 , and y C[a, b] . Then the problem a Dq, u (t ) + y (t ) = 0, a t b, u (a) = Dq , u (a) = Dq , u (b) = 0.
has a unique solution b
u (t ) = G (t , s ) y ( s )d q , s , a
where
(
)
(
)
b − q ( s ) ( − 2 ) 0 (t − a ) (0 −1) , 1 (b − a )( − 2 ) G (t , s ) = ( − 2 ) q ( ) b − 0 q ( s ) (t − a ) (0 −1) − (t − 0 q ( s ))(0 −1) , (b − a )( − 2 )
a t s b, Proof
a s t b.
from Theorem 2.4 we have
u(t ) =− a I q, y(t ) + c1 (t − a) (0 −1) + c2 (t − a) (0 −2) + c3 (t − a) (0 −3) . The condition u ( a ) = 0 implies that c3 = 0 . Therefore,
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Lyapunov-type inequalities for a fractional q, -difference…
(
)
Dq , u (t ) = − a I q,−1 y (t ) + c1 [ − 1]q (t − a) (0 −2) + c2 [ − 2]q (t − a) (0 −3) t 1 =− (t − 0 q ( s))(0 −2) y ( s)d q , 0 q ( s) q ( − 1) a
(
+ c1[ − 1]q (t − a )0
( − 2 )
The condition
+ c2 [ − 2]q (t − a ) (0 −3) .
Dq, u(a) = 0 implies that c2 = 0 . Then
Dq , u (b) = −
1
( − 2 )
b
(b− ( − 1) 0
a
q
q
( s))
( − 2 )
+ c1 [ − 1]q (b − a) 0 Since
)
0
y ( s)d q , s
.
.
Dq, u(b) = 0 , we get
c1 =
1 q ( )(b − a) (0 − 2)
b
a
(b− 0 q ( s ))
( − 2 )
0
y ( s )d q , s .
Thus, ( −1) t 1 ( t − ( s )) y ( s)d q , s 0 q 0 q ( ) a . ( − 2 ) (t − a) (0 −1) b + (b− 0 q ( s)) y ( s)d q , s. 0 q ( )(b − a) (0 − 2) a
u (t ) = −
For the uniqueness, suppose that
u1 and u 2 are two solutions of the considered problem. Define
u = u1 − u 2 . By linearity, u solves the boundary value problem. a Dq, u (t ) = 0, a t b, u (a) = Dq , u (a) = Dq , u (b) = 0. which
has
a
unique
solution
u = 0 . Therefore, u1 = u 2
and
the
uniqueness
follows.
Lemma 2.6. Let y C[ a, b] ,
2 3 , 1 2 , p 1 , and
a Dq, ( p ( a Dq, u (t ))) + y (t ) = 0, u (a) = Dq , u (a) = Dq , u (b) = 0,
1 1 + = 1 . Then the problem p g
a t b, a Dq , u ( a ) = a Dq , u (b) = 0.
(1.1) has
a
unique solution
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Lyapunov-type inequalities for a fractional q, -difference… b b u (t ) = − G(t , s) H ( s, ) y( )d q , d q , s , a g a
(
)
(
)
b − q ( s ) ( −1) 0 0 (t − a ) ( 0 −1) , ( −1) 1 (b − a )0 H (t , s ) = ( −1) q ( ) b − 0 q ( s ) 0 (t − a ) ( 0 −1) − (t − 0 q ( s ))( 0 −1) , ( −1) (b − a )0
a t s b, Proof
a s t b.
from Theorem 2.4 we have
p ( a Dq, u(t )) =− a I q, y(t ) + c1 (t − a) ( −1) + c2 (t − a)( −2) , 0
0
where ci , i = 1,2 , are real constants. The condition conditions
c1 =
a
a
Dq, u (a) = 0 implies that p
Dq, u (b) = 0 implies that p 1 ( -1)
(b − a ) 0
( )
b
a
q
(
a
(
a
)
Dq, u(a) = 0 . which yields c2 = 0 . The
)
Dq, u (b) = 0 . which yields
(b − 0 q ( s ))( 0 -1) y ( s )d q , s .
Therefore,
p ( a Dq, u(t )) = − +
t 1 (t − 0 q ( s))( 0 -1) y ( s)d q , s q ( ) a
(t − a) ( 0 -1) ( -1)
(b − a) 0
b
(b− ( ) q
0
a
q
( s))( 0 -1) y ( s)d q , s ,
that is,
p ( a Dq, u (t ) ) = H (t , s ) y ( s )d q , s . b
a
Then we have a
b Dq, u (t ) − g H (t , s) y ( s)d q , s = 0 . a
Setting b ~ y = − g H (t , s) y ( s)d q , s . a
We obtain
a Dq, u (t ) + ~ y (t ) = 0, a t b, u (a) = Dq , u (a) = Dq , u (b) = 0. Finally, we applying Theorem2.4, we obtain the desired result.
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Lyapunov-type inequalities for a fractional q, -difference… The following estimates will be useful later. Lemma 2.7. We have
0 G(t , s) G(0 q (s), s),
(t , s) [a, b] [a, b].
Proof From the respect of G (t , s ) , set
(b−
g1 (t , s) =
0
q ( s) )( −2)
(b − a )
( − 2 )
(t − a) (0 −1) , a t (s) b , 0 q
and
(b−
g 2 (t , s) =
0
q ( s) )( −2)
(b − a )
( − 2 )
(t − a) (0 −1) − (t − 0 q ( s))(0 −1) , a (s) t b . 0 q
Clearly,
g1 (t , s) 0 ,
a t 0 q (s) b .
On the other hand,
(b−
0
q ( s) )( −2)
(b − a )
( − 2 )
(t − a) (0 −1) − (t − 0 q ( s))(0 −1) 0 , a s t b,
which yields
g 2 (t , s) 0 , a s t b . So G (t , s ) 0 for all (t , s ) [ a, b] [ a, b] , which yields
0 G(t , s) G(0 q (s), s),
(t , s) [a, b] [a, b] .
The proof is complete. Lemma 2.8. We have
0 H (t , s) H (0 q (s), s), Proof
(t , s) [a, b] [a, b] .
(
)
(
)
b − q ( s ) ( − 2 ) 0 (t − a ) (0 − 2 ) , [ − 1] q (b − a )( − 2 ) ( − 2 ) t Dq , G (t , s ) = q ( ) b − 0 q ( s ) (t − a ) (0 − 2 ) − (t − 0 q ( s ))(0 − 2 ) , (b − a )( − 2 ) e that H (t , s ) = G (t , s ) , for
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a t s b, Observ
a s t b.
− 2 = − 1 , Then, from the proof of Lemma 2.8 we have IJMREM
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Lyapunov-type inequalities for a fractional q, -difference…
H (t , s) 0, (t , s) [a, b] [a, b] . On the other hand, for all s [ a, b] , we have
H (t , s) H (0 q (s), s) . Now, we are already to state and prove our main result.
III. MAIN RESULT Our main result is following Lyapunov-type inequality. Theorem 3.1. Suppose that
2 3 , 1 2 , p 1 , and : a, b → R is a continuous function.
If (1.1) has a nontrivial continuous solution, then b
(b−
( s))( 0 -1) ( 0 q ( s ) − a) ( -1) ( s ) d q , s ( -1) 1− p (3.1) p −1 (b − a ) 0 . b ( - 2) ( -1) q ( ) q ( ) ( b − ( s )) ( ( s ) − a ) d s 0 q 0 0 q 0 q , (b − a) (0 -2) a 0
a
q
u
Proof We endow the set C[a, b] with the Chebyshev norm
u
given by
= maxu(t ) : a t b, u C[a, b] .
Suppose that u C[ a, b] is a nontrivial solution of (1.1). From Lemma 2.8 , Lemma 2.9 we have b b u (t ) = − G(t , s) H ( s, ) ( ) p (u ( ))d q , d q , s , t a, b . a g a
Let
t a, b be fixed. We have b b u (t ) G (t , s ) H ( s, ) ( ) p (u ( ))d q , d q , s a g a b
G (t , s ) a b
b
a
H ( s, ) ( ) p (u ( ))d q ,
g −1
d q , s
G (t , s ) ( s )d q , s. a
where
(s)= H (s, ) ( ) u( ) b
a
p −1
d q ,
g −1
,s
a, b.
Using Lemma 2.8 and Lemma 2.9, we obtain b b u (t ) u G( 0 q ( s), s)d q , s H ( 0 q ( s), s) ( s) d q, s a a
g −1
.
Since the last inequality holds for every t [ a, b] , we obtain
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Lyapunov-type inequalities for a fractional q, -difference… b b 1 G( 0 q ( s), s)d q , s H ( 0 q ( s), s) (s) d q , s a a
g −1
,
which yields the desired result.
2 3 , 1 2 , p 1 , and : a, b → R is a continuous function.
Corollary3.2. Suppose that
If (1.1) has a nontrivial continuous solution, then
b
( s) d q , s (b − 0 )
a
1−
( 0 − a)
1−
q ( ) q ( )
p −1
(b − a) ( 0 -1) (b − a) (0 -2)
b (b− 0 q ( s))(0 -2) ( 0 q ( s) − a) (0 -1) d q , s a
Proof
1− p
(3.2)
Let
(s) = (b− q (s))( -1) ( q (s) − a) ( -1) 0
0
0
s − 0 s − 0 ) 1 − q i +1 ( ) b − 0 a − 0 −1 −1 = (b − 0 ) ( 0 − a ) s − 0 1+ i + s − 0 i =0 i =0 1− q ( ) 1 − q 1+i + ( ) b − 0 a − 0 (b − 0 ) −1 ( 0 − a ) −1
Observe that the function
1 − q i +1 (
has a maximum. That is,
= (b − 0 ) −1 (0 − a) −1 , s [a, b] .
The desired result follows immediately from the last equality and inequality (3.1). For p = 2 , problem (1.1) becomes
a Dq, ( a Dq, u (t )) + (t )u (t ) = 0, u (a) = Dq , u (a) = Dq , u (b) = 0,
a Dq , u ( a ) = a Dq , u (b) = 0,
(3.3)
2 3 , 1 2 , and : a, b → R is a continuous function. In this case, taking p = 2
where
in Theorem 3.1, we obtain the following result. Corollary3.3. Suppose that
2 3 , 1 2 , p 1 , and : a, b → R is a continuous function.
If (3.3) has a nontrivial continuous solution, b
(b−
( s))( 0 -1) ( 0 q ( s) − a) ( -1) ( s) d q , s −1 then (b − a) ( 0 -1) b ( - 2) ( -1) q ( )q ( ) ( b − ( s )) ( ( s ) − a ) d s 0 q 0 0 q 0 q , . (b − a) (0 -2) a a
0
q
Taking p = 2 in Corollary3.2, we obtain the following result.
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Lyapunov-type inequalities for a fractional q, -difference…
2 3 , 1 2 , p 1 , and : a, b → R is a continuous function.
Corollary3.4. Suppose that
If (3.3) has a nontrivial continuous solution, then
b
a
( s) d q , s (b − 0 )
−1
( 0 − a)
−1
q ( )q ( )
(b − a) ( 0 -1) (b − a) (0 -2)
−1
b (b− 0 q ( s))(0 -2) ( 0 q ( s) − a) (0 -1) d q , s . a
IV.
APPLICATIONS TO EIGENVALUE PROBLEMS
In this section, we present some applications of the obtained results to eigenvalue problems.
Corollary 4.1. Let
be an eigenvalue of the problem
a Dq, ( p ( a Dq, u (t ))) + p (u (t )) = 0, 0 t 1, u (0) = Dq , u (0) = Dq , u (1) = 0, a Dq , u (0)= a Dq , u (1) = 0,
(4.1)
2 3 , 1 2 , and p 1 . Then
where
q (2 ) q ( )q ( + 1) q ( ) q ( − 1) Proof Let
p −1
.
(4.2)
be an eigenvalue of (4.1). Then there exists a nontrivial solution u = u to (4.1). Using
Theorem 3.1 with ( a, b) = (0,1) and
(s) = , we obtain
1
(1− q ( s))( -.1) ( q ( s)) ( -1) d q , s 0
0
0
q ( ) q ( )
0
p −1
1− p (1 − 0) ( 0 -.1) 1 ( - 2) ( -1) . ( 1 − ( s )) ( ( s )) d s 0 q 0 . 0 q 0 . q , (1 − 0) (0 -.2) 0
Observe that 1
(1− 0
0
q
( s ))( 0 -.1) ( 0 q ( s )) ( -1) d q , s = Bq ( , ) ,
q
( s ))(0 -.2) ( 0 q ( s ))(0 -.1) d q , s = Bq ( - 1, ) ,
and 1
(1− 0
Where
0
Bq is the beta function defined by? 1
B q ( x, y ) =
0
(
0
q ( s) )x −1 (1− q ( s) )y −1 d q , s, x, y 0. 0
Using the identity
Bq ( x, y) =
q ( x)q ( y) q ( x + y)
,
We get the desired result.
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Lyapunov-type inequalities for a fractional q, -difference…
Corollary 4.2. Let
be an eigenvalue of the problem
a Dq, ( a Dq, u (t )) + u (t ) = 0, 0 t 1 u (0) = Dq , u (0) = Dq , u (1) = 0, a Dq , u (0)= a Dq , u (1) = 0, where
2 3 , 1 2 , and p = 2 . Then
q (2 ) q ( )q ( + 1) q ( )
q ( − 1)
.
Proof It follows from inequality (4.2) by taking p = 2 .
REFERENCE 1.
Bognàr, G, Rontó, M: Numerical-analytic investigation of the radially symmetric solutions for some nonlinear PDEs.Comput. Math. Appl. 50, 983-991 (2005)
2.
Glowinski, R, Rappaz, J: Approximation of a nonlinear elliptic problem arising in a non-Newtonian fluid flow model in glaciology. Math. Model. Numer. Anal. 37(1), 175-186 (2003)
3.
Kawohl, B: On a family of torsional creep problems. J. Reine Angew. Math. 410, 1-22 (1990)
4.
Liu, C: Weak solutions for a viscous p-Laplacian equation. Electron. J. Differ. Equ. 2003, 63 (2003)
5.
Oruganti, S, Shi, J, Shivaji, R: Diffusive logistic equation with constant yield harvesting, I: steady-states. Trans. Am.Math. Soc. 354(9), 3601-3619 (2002)
6.
Ramaswamy, M, Shivaji, R: Multiple positive solutions for classes of p-Laplacian equations. Differ. Integral Equ.17(11-12), 1255-1261 (2004)
7.
Liapunov, AM: Probleme general de la stabilite du mouvement. Ann. Fac. Sci. Univ. Toulouse 2,203-407 (1907)
8.
Aktas, MF: Lyapunov-type inequalities for a certain class of n-dimensional quasilinear systems.Electron. J. Differ. Equ.2013, 67 (2013)
9.
Cakmak, D: On Lyapunov-type inequality for a class of nonlinear systems. Math. Inequal. Appl.16, 101 一 108 (2013)
10. Liapunov, AM: Problème général de la stabilité du mouvement. Ann. Fac. Sci. Univ. Toulouse 2, 203-407 (1907) 11. Aktas, MF: Lyapunov-type inequalities for a certain class of n-dimensional quasilinear systems. Electron. J. Differ. Equ.2013, 67 (2013) 12. Çakmak, D: On Lyapunov-type inequality for a class of nonlinear systems. Math. Inequal. Appl. 16, 101-108 (2013) 13. Hartman, P, Wintner, A: On an oscillation criterion of Liapounoff. Am. J. Math. 73, 885-890 (1951) 14. Arifi, Nassir Al, et al. "Lyapunov-type inequalities for a fractional p-Laplacian equation."Journal of Inequalities&Applications2016.1(2016):189. 15. Cabrera, I, Sadarangani, K, Samet, B: Hartman-Wintner-type inequalities for a class of nonlocal fractional boundaryvalue problems. Math. Methods Appl. Sci. (2016). doi:10.1002/mma.3972 16. Jleli, M, Samet, B: A Lyapunov-type inequality for a fractional q-difference boundary value problem. J. Nonlinear Sci.Appl. 9, 1965 一 1976 (2016)
www.ijmrem.com
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Lyapunov-type inequalities for a fractional q, -difference… 17. O'Regan, D, Samet, B: Lyapunov-type inequalities for a class of fractional differential equations. J. Inequal. Appl. 2015,247 (2015) 18. Malinowska A B, Torres D F M. The Hahn Quantum Variational Calculus. J. Journal of Optimization Theory & Applications, 2010, 147(3):419-442. 19. Yizhu Wang,Yiding Liu,Chengmin Hou. New concepts of fractional Hahn's q, -derivative of Riemann-Louville type and Caputo type and applications. J. Advances in Difference Equations. December 2018, 2018:292
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