Balanced fractional opial inequalities

Balanced fractional opial inequalities

Chaos, Solitons and Fractals 42 (2009) 1523–1528 Contents lists available at ScienceDirect Chaos, Solitons and Fractals journal homepage: www.elsevi...

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Chaos, Solitons and Fractals 42 (2009) 1523–1528

Contents lists available at ScienceDirect

Chaos, Solitons and Fractals journal homepage: www.elsevier.com/locate/chaos

Balanced fractional opial inequalities George A. Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA

a r t i c l e

i n f o

a b s t r a c t

Article history: Accepted 11 March 2009

Here we present Lp , p > 1, fractional Opial type inequalities subject to high order boundary conditions. They involve the right and left Caputo, Riemann–Liouville fractional derivatives. These derivatives are blended together into the balanced Caputo, Riemann–Liouville, respectively, fractional derivative. This balanced fractional derivative is introduced here for the first time. We give applications to a special case. Ó 2009 Elsevier Ltd. All rights reserved.

1. Preliminaries This article is inspired by the famous theorem of Opial [11], 1960, which follows Theorem 1. Let xðtÞ 2 C 1 ð½0; hÞ be such that xð0Þ ¼ xðhÞ ¼ 0, and xðtÞ > 0 in ð0; hÞ. Then

Z

h

jxðtÞx0 ðt Þjdt 6

0

In (1), the constant

( xðtÞ ¼

h 4

h 4

Z

h

2

ðx0 ðtÞÞ dt:

ð1Þ

0

is the best possible. Inequality (1) holds as equality for the optimal function

ct;

0 6 t 6 h=2;

c ðh  t Þ

h 2

6 t 6 h;

where c > 0 is an arbitrary constant. To prove easier Theorem 1, Beesack [5] proved the following well-known Opial type inequality which is used very commonly. This is another inspiration to our work. Theorem 2. Let xðtÞ be absolutely continuous in [0, a], and xð0Þ ¼ 0. Then

Z 0

a

jxðtÞx0 ðt Þjdt 6

a 2

Z

a

2

ðx0 ðtÞÞ dt:

ð2Þ

0

Inequality (2) is sharp, it is attained by xðtÞ ¼ ct, c > 0 is an arbitrary constant. Opial type inequalities are used a lot in proving uniqueness of solutions to differential equations, also to give upper bounds to their solutions. By themselves have made a great subject of intensive research and there exists a great literature about them. Typical and great sources on them are the monographs [1,2]. We need (see also [3,7–9,12]). E-mail address: [email protected] 0960-0779/$ - see front matter Ó 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2009.03.047

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Definition 3. Let f 2 AC m ð½a; bÞ (space of functions from ½a; b into R with m  1 derivative absolutely continuous function on ½a; b), m 2 N, where m ¼ dae, a > 0 (de the ceiling of the number). We define the right Caputo fractional derivative of order a > 0, by

Dab f ðxÞ ¼

Z

ð1Þm Cðm  aÞ

b

ðf  xÞma1 f ðmÞ ðfÞdf:

ð3Þ

x

We set D0b f ðxÞ ¼ f ðxÞ, 8x 2 ½a; b. Note 4. Let f 2 AC m ð½a; bÞ, m ¼ dae, with a > 0, then f ðm1Þ 2 ACð½a; bÞ, which implies that f ðmÞ exists a.e. on ½a; b and that f ðmÞ 2 L1 ð½a; bÞ. Consequently if f 2 AC m ð½a; bÞ, then Dab f ðxÞ exists a.e. on ½a; b and Dab f 2 L1 ð½a; bÞ, see [3]. Observe that when a ¼ m 2 N, then m ðmÞ Dm ðxÞ; 8x 2 ½a; b: b f ðxÞ ¼ ð1Þ f

ð4Þ

We continue with the right Caputo fractional Taylor formula with integral remainder, see [3]. Theorem 5. Let f 2 AC m ð½a; bÞ, x 2 ½a; b, a > 0, m ¼ dae. Then

f ð xÞ ¼

Z b m1 ðkÞ X f ðbÞ 1 k ðx  bÞ þ ðf  xÞa1 Dab f ðfÞdf: k! CðaÞ x k¼0

ð5Þ

We need also (see [6], p. 38). Definition 6. Let f 2 AC m ð½a; bÞ; m 2 N, where m ¼ dae, a > 0. We define the left Caputo fractional derivative of order a > 0, by

Daa f ðxÞ ¼

1 Cðm  aÞ

Z

x

ðx  tÞma1 f ðmÞ ðt Þdt;

ð6Þ

a

8x 2 ½a; b. We set D0a f ðxÞ ¼ f ðxÞ, 8x 2 ½a; b. Again here Daa f exists a.e. on ½a; b and Daa f 2 L1 ð½a; bÞ, see [6], p. 13, 37 and 38. When a ¼ m 2 N, then ðm Þ Dm ðxÞ; 8x 2 ½a; b: a f ðxÞ ¼ f

ð7Þ

We continue with the left Caputo fractional Taylor formula with integral remainder, see [6], p. 40. Theorem 7. Let f 2 AC m ð½a; bÞ, m 2 N, where m ¼ dae, a > 0, x 2 ½a; b. Then

f ð xÞ ¼

Z x m1 ðkÞ X f ðaÞ 1 ðx  sÞa1 Daa f ðsÞds: ðx  aÞk þ k! C ð a Þ a k¼0

ð8Þ

Above C is the gamma function,

CðaÞ ¼

Z

1

et t a1 dt; a > 0:

0

We introduce the following new balanced Caputo fractional derivative. Definition 8. Let f 2 AC m ð½a; bÞ; m 2 N, m ¼ dae, a > 0, x 2 ½a; b. We define

( a

D f ðxÞ :¼

6 x 6 b; Dab f ðxÞ; for aþb 2 Daa f ðxÞ;

fora 6 x < aþb : 2

ð9Þ

In this article we establish Lp , p > 1, Opial type inequalities involving the balanced Caputo fractional derivative subject to high order boundary conditions, more precisely by assuming that

f ðkÞ ðaÞ ¼ f ðkÞ ðbÞ ¼ 0;

k ¼ 0; 1; . . . ; m  1:

We extend our results to Riemann–Liouville fractional derivatives.

2. Results We present our main result. Theorem 9. Let f 2 AC m ð½a; bÞ; m 2 N, m ¼ dae, a > 0. Assume

ð10Þ

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G.A. Anastassiou / Chaos, Solitons and Fractals 42 (2009) 1523–1528

f ðkÞ ðaÞ ¼ f ðkÞ ðbÞ ¼ 0; 1 1 p; q > 1 : þ ¼ 1; p q

k ¼ 0; 1; . . . ; m  1; 1 and a > : q

(i) Case of 1 < q 6 2. Then

Z

b

  jf ðxÞjDa f ðxÞdx 6

a

pða1Þþ2 p

2ðaþpÞ ðb  aÞð 1

Z

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

b

 a  D f ðxÞq dx

!2=q :

ð11Þ

:

ð12Þ

a

(ii) Case of q > 2. Then

Z

b

  jf ðxÞjDa f ðxÞdx 6

a

pða1Þþ2 p

2ðaþqÞ ðb  aÞð 1

Z

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

(iii) When p ¼ q ¼ 2, a > 12, then

Z

b

  jf ðxÞjDa f ðxÞdx 6

a

2ðaþ2Þ ðb  aÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CðaÞ 2að2a  1Þ 1

a

Z

b



a

2

b

 a  D f ðxÞq dx

!2=q

a

!

D f ðxÞ dx :

ð13Þ

a

Remark 10. Let us say that a ¼ 1, then by (13) we obtain

Z a

b

ðb  aÞ jf ðxÞjjf ðxÞjdx 6 4 0

Z

!

b 0

2

ðf ðxÞÞ dx ;

ð14Þ

a

that is reproving and recovering Opial’s inequality (1), see [11], see also Olech’s result [10]. Proof of Theorem 9. Let x 2 ½a; b. We have by assumption f ðkÞ ðaÞ ¼ 0, k ¼ 0; 1; . . . ; m  1 and Theorem 7 that

f ðxÞ ¼

1 CðaÞ

Z

x

a

ðx  sÞa1 Daa f ðsÞds;

ð15Þ

and by assumption f ðkÞ ðbÞ ¼ 0, k ¼ 0; 1; . . . ; m  1 and Theorem 5 that

f ðxÞ ¼

1 CðaÞ

Z

b

x

ðs  xÞa1 Dab f ðsÞds:

ð16Þ

Using Hölder’s inequality on (15) we get

j f ð xÞ j 6

1 CðaÞ

Z a

x

  ðx  sÞa1 Daa f ðsÞds 6 pða1Þþ1

1 ðx  a Þ p ¼ CðaÞ ðpða  1Þ þ 1Þ1=p

Z a

x

1 CðaÞ

Z

 a  D f ðsÞq ds a

x

 p 1=p Z ðx  sÞa1 ds

a

a

x

 a  D f ðsÞq ds a

1=q

1=q :

ð17Þ

Set

zðxÞ :¼

Z a

x

 a  D f ðsÞq ds; a

ðzðaÞ ¼ 0Þ:

Then

 q z0 ðxÞ ¼ Daa f ðxÞ ; and

 a  D f ðxÞ ¼ ðz0 ðxÞÞ1=q ; a

all a 6 x 6 b:

Therefore by (17) we have

  jf ðxÞjDaa f ðxÞ 6 all a 6 x 6 x.

pða1Þþ1

1 ðx  aÞ p 1=q ðzðxÞz0 ðxÞÞ ; CðaÞ ðpða  1Þ þ 1Þ1=p

ð18Þ

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Next working similarly with (16) we obtain

jf ðxÞj 6

1 CðaÞ

Z x

b

  ðs  xÞa1 Dab f ðsÞds 6 Z

pða1Þþ1 p

¼ Set

kðxÞ :¼

1 ð b  xÞ CðaÞ ðpða  1Þ þ 1Þ1=p Z

b

x

 a  D f ðsÞq ds ¼  b

Z

x

x b

b

Z

1 CðaÞ

b

 p ðs  xÞa1 ds

!1=p Z

x

 a  D f ðsÞq ds b

 a  D f ðsÞq ds; b

b

x

 a  D f ðsÞq ds b

!1=q

!1=q ð19Þ

:

ðkðbÞ ¼ 0Þ:

Then

 q k0 ðxÞ ¼ Dab f ðxÞ and

 a  D f ðxÞ ¼ ðk0 ðxÞÞ1=q ; b

all a 6 x 6 b:

Therefore by (19) we have

  jf ðxÞjDab f ðxÞ 6

pða1Þþ1

1 ðb  xÞ p 1=q ðkðxÞk0 ðxÞÞ ; CðaÞ ðpða  1Þ þ 1Þ1=p

ð20Þ

all x 6 x 6 b. Next we integrate (18) over ½a; x to obtain

Z

x

a

  jf ðxÞjDaa f ðxÞdx 6 6

¼

¼

Z

1

CðaÞðpða  1Þ þ 1Þ

1=p

x

pða1Þþ1 p

ðx  aÞ

Z

1

CðaÞðpða  1Þ þ 1Þ1=p

x

ðx  aÞpða1Þþ1 dx

1=q

1=p Z

a

1

dx

CðaÞðpða  1Þ þ 1Þ

x

zðxÞz0 ðxÞdx

1=q

a

ðx  aÞ 1=p

21=q ðx  aÞ

ðzðxÞz0 ðxÞÞ

a

pða1Þþ2 p

ðpða  1Þ þ 2Þ

2=q

zðxÞ

1=p

21=q

Z

pða1Þþ2 p

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

 a  D f ðxÞq dx a

x

a

2=q :

ð21Þ

2=q  a  D f ðxÞq dx ; a

ð22Þ

So we have proved

Z

x

a

  jf ðxÞjDaa f ðxÞdx 6

21=q ðx  aÞ

Z

pða1Þþ2 p

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

x

a

for all a 6 x 6 b. By (22) we get

Z

aþb 2

a

  jf ðxÞjDaa f ðxÞdx 6

ðb  aÞ

ðpða1Þþ2Þ p

ðpða1Þþ2Þ 1 þq p

2 ½

Z



CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

aþb 2

a

 a  D f ðxÞq dx a

!2=q ð23Þ

:

Similarly we integrate (20) over ½x; b to obtain

Z

b

x

  jf ðxÞjDab f ðxÞdx 6

6

¼

Z

1

CðaÞðpða  1Þ þ 1Þ

1=p

b

ð b  xÞ

pða1Þþ1 p

ðkðxÞk0 ðxÞÞ

1=q

dx

x

Z

1

CðaÞðpða  1Þ þ 1Þ1=p

b

ðb  xÞ

pða1Þþ1

dx

!1=p Z

x

kðxÞk ðxÞdx

ðb  xÞ

ðkðxÞÞ2=q

CðaÞðpða  1Þ þ 1Þ1=p ðpða  1Þ þ 2Þ1=p

21=q

x

b

  jf ðxÞjDab f ðxÞdx 6

for all a 6 x 6 b.

pða1Þþ2 p

21=q ðb  xÞ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

Z

b x

ð24Þ

:

We have proved that

Z

0

x pða1Þþ2 p

1

!1=q

b

 a  D f ðxÞq dx b

!2=q ;

ð25Þ

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G.A. Anastassiou / Chaos, Solitons and Fractals 42 (2009) 1523–1528

By (25) we get

Z

b

aþb 2

  jf ðxÞjDab f ðxÞdx 6

ðpða1Þþ2Þ p

ðb  aÞ

2½

ðpða1Þþ2Þ 1 þq p

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

Adding (23) and (26) we get

Z

b

  jf ðxÞjDa f ðxÞdx 6

a

Z



pða1Þþ2 p

2ðaþpÞ ðb  aÞð 1

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

2 4

b

aþb 2

Z

aþb 2

a

 a  D f ðxÞq dx b

 a  D f ðxÞq dx a

!2=q ð26Þ

:

!2=q þ

Z

b aþb 2

 a  D f ðxÞq dx b

!2=q 3 5 ¼: ðÞ: ð27Þ

Assume 1 < q 6 2, then Therefore we get

2 q

P 1.

2ðaþpÞ ðb  aÞð 1

ðÞ 6

pða1Þþ2 p

"Z

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p Z

pða1Þþ2 1 2ðaþpÞ ðb  aÞð p Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

b

aþb 2

a

 a  D f ðxÞq dx þ a

Z

b

aþb 2

!2=q  a  D f ðxÞq dx :

 a  D f ðxÞq dx b

#2=q ¼

ð28Þ

ð29Þ

a

So for 1 < q 6 2 we have proved (11). Assume now q > 2, then 0 < 2q < 1. Therefore we get pða1Þþ2 p

2ðaþpÞ ðb  aÞð 1

ðÞ 6

¼

"Z

Þ 212q

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p 2ð

pða1Þþ2 p

Þ ðb  aÞð

aþ1q

a

Z

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ

aþb 2

1=p

b

 a  D f ðxÞq dx þ a

 a  D f ðxÞq dx

Z

b

aþb 2

!2=q

 a  D f ðxÞq dx b

#2=q

ð30Þ

:

a

So when q > 2 we have established (12). (iii) The case of p ¼ q ¼ 2, see (13), is obvious, it derives from (11) immediately. We need (see [3,6–8], p. 22).



Definition 11. Let a > 0, m ¼ dae, f 2 AC m ð½a; bÞ. We define the right Riemann–Liouville fractional derivative by

Dab f ðxÞ :¼

m Z b ð1Þm d ðt  xÞma1 f ðt Þdt; Cðm  aÞ dx x

ð31Þ

D0b f ðxÞ :¼ IðxÞðthe identity operatorÞ: We also define the left Riemann–Liouville fractional derivative by

Daaþ f ðxÞ :¼

m Z x 1 d ðx  tÞma1 f ðt Þdt; Cðm  aÞ dx a

ð32Þ

D0aþ f ðxÞ :¼ IðxÞ: We further define the new balanced Riemann–Liouville fractional derivative

( a

D f ðxÞ :¼

Dab f ðxÞ; for a

aþb 2

6 x 6 b;

Daþ f ðxÞ; for a 6 x < aþb : 2

ð33Þ

Remark 12. Let now f 2 C m ð½a; bÞ, m ¼ dae, a > 0. In [4] we have proved that Dab f ðxÞ, Daa f ðxÞ are continuous functions in x 2 ½a; b. Of course C m ð½a; bÞ  AC m ð½a; bÞ, so that f 2 AC m ð½a; bÞ. Thus by Theorem 9 of [3], we obtain that also Dab f ðxÞ exists and continuous for every x 2 ½a; b. Furthermore if f ðkÞ ðbÞ ¼ 0, k ¼ 0; 1; . . . ; m  1 we get

Dab f ðxÞ ¼ Dab f ðxÞ;

ð34Þ

8x 2 ½a; b. Similarly, by [6], p. 39, we get that Daaþ f ðxÞ exists and continuous in x 2 ½a; b. Furthermore if f ðkÞ ðaÞ ¼ 0, k ¼ 0; 1; . . . ; m  1 we get

Daaþ f ðxÞ ¼ Daa f ðxÞ;

8x 2 ½a; b.

ð35Þ

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G.A. Anastassiou / Chaos, Solitons and Fractals 42 (2009) 1523–1528

So if f ðkÞ ðaÞ ¼ f ðkÞ ðbÞ ¼ 0, k ¼ 0; 1; . . . ; m  1 we obtain that

Da f ðxÞ ¼ Da f ðxÞ;

ð36Þ

8x 2 ½a; b. So by Theorem 9 we obtain the corresponding results for the balanced Riemann–Liouville fractional derivative Theorem 13. Let f 2 C m ð½a; bÞ, m 2 N, m ¼ dae, a > 0. Assume f ðkÞ ðaÞ ¼ f ðkÞ ðbÞ ¼ 0, k ¼ 0; 1; . . . ; m  1; p, q > 1 : 1p þ 1q ¼ 1, and a > 1q. (i) Case of 1 < q 6 2. Then

Z

pða1Þþ2 p

2ðaþpÞ ðb  aÞð 1

b

a

jf ðxÞjjD f ðxÞjdx 6

a

Z

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

b

!2=q a

q

a

q

jD f ðxÞj dx

:

ð37Þ

:

ð38Þ

a

(ii) Case of q > 2. Then

Z

pða1Þþ2 p

2ðaþqÞ ðb  aÞð 1

b

a

jf ðxÞjjD f ðxÞjdx 6

a

Z

Þ

CðaÞ½ðpða  1Þ þ 1Þðpða  1Þ þ 2Þ1=p

b

!2=q jD f ðxÞj dx

a

(iii) When p ¼ q ¼ 2; a > 12, then

Z

b

2ðaþ2Þ ðb  aÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CðaÞ 2að2a  1Þ 1

jf ðxÞjjDa f ðxÞjdx 6

a

a

Z

b

! 2

ðDa f ðxÞÞ dx :

ð39Þ

a

3. Conclusion According to the monographs [1,2], our presented method of involving balanced fractional derivatives into Opial type inequalities, subject to boundary conditions, could be expanded to all possible directions, by producing interesting results and applications. Especially all these results proved here, and similar to be proved, are expected to have wide applications to fractional differential equations. References [1] [2] [3] [4] [5] [6] [7] [8] [9]

Agarwal RP, Pang PYH. Opial inequalities with applications in differential and difference equations. Dordrecht, London: Kluwer; 1995. Anastassiou GA. Fractional differentiation inequalities, research monograph, accepted. Berlin, NY: Springer; 2009. Anastassiou GA. On right fractional calculus. Chaos Soliton Fract 2009;42:365–76. Anastassiou GA. Fractional Korovkin Theory. Chaos Soliton Fract 2009, accepted for publication. Beesack PR. On an integral inequality of Z. Opial. Trans Am Math Soc 1962;104:470–5. Kai Diethelm. Fractional differential equations. Available from: http://www.tu-bs.de/~diethelm/lehre/f-dgl02/fde-skript.ps.gz. El-Sayed AMA, Gaber M. On the finite Caputo and finite Riesz derivatives. Electron J Theor Phys 2006;3(12):81–95. Frederico GS, Torres DFM. Fractional optimal control in the sense of Caputo and the fractional Noether’s theorem. Int Math Forum 2008;3(10):479–93. Gorenflo R, Mainardi F. Essentials of fractional calculus, Maphysto Center 2000. Available from: http://www.maphysto.dk/oldpages/events/ LevyCAC2000/MainardiNotes/fm2k0a.ps. [10] Olech C. A simple proof of a certain result of Z. Opial. Ann Polon Math 1960;8:61–3. [11] Opial Z. Sur une inegalite. Ann Polon Math 1960;8:29–32. [12] Samko SG, Kilbas AA, Marichev OI. Fractional integrals and derivatives, theory and applications. Amsterdam: Gordon and Breach; 1993 [English translation from the Russian, integrals and derivatives of fractional order and some of their applications. Minsk: Nauka i Tekhnika; 1987].