Riemann-Liouville Fractional Calculus of Coalescence Hidden-variable - PowerPoint PPT Presentation

Riemann-Liouville Fractional Calculus of Coalescence Hidden-variable Fractal Interpolation Functions Srijanani Anurag Prasad Department of Applied Sciences, The NorthCap University, Gurgaon 6 th Cornell Conference on Analysis, Probability, and Mathematical Physics on Fractals June 13-17, 2017 S.A.Prasad (NCU) FC June 13-17, 2017 1 / 31

Outline 1 Introduction 2 Riemann-Liouville fractional integral 3 Riemann-Liouville fractional derivative S.A.Prasad (NCU) FC June 13-17, 2017 2 / 31

Outline 1 Introduction 2 Riemann-Liouville fractional integral 3 Riemann-Liouville fractional derivative S.A.Prasad (NCU) FC June 13-17, 2017 3 / 31

Fractal Interpolation Function (FIF) Fractal Interpolation Function (FIF) : [Barnsley M.F., 1986] Similarities of FIF and traditional methods ∗ Geometrical Character - can be plotted on graph ∗ Represented by formulas Difference between FIF and traditional methods ∗ Fractal Character S.A.Prasad (NCU) FC June 13-17, 2017 4 / 31

Coalescence Hidden-variable Interpolation Functions For simulating curves that exhibit self-affine and non-self-affine nature simultaneously, Coalescence Hidden-variable Fractal Interpolation Function (CHFIF) was introduced by [Chand A.K.B. and Kapoor G.P ., 2007]. 120 180 160 100 140 80 120 100 60 80 40 60 40 20 20 0 0 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100 S.A.Prasad (NCU) FC June 13-17, 2017 5 / 31

Construction of a CHFIF Given data { ( x k , y k ) ∈ R 2 : k = 0 , 1 , . . . , N } Generalized data { ( x k , y k , z k ) ∈ R 3 : k = 0 , 1 , . . . , N } [ x 0 , x N ] = I , [ x k − 1 , x k ] = I k , k = 1 , 2 , . . . , N L k : I → I k L k ( x 0 ) = a k x + b k = x k − x k − 1 ( x − x 0 ) + x k − 1 (1) x N − x 0 S.A.Prasad (NCU) FC June 13-17, 2017 6 / 31

Construction of a CHFIF F k : I × R 2 → R 2 F k ( x , y , z ) = α k y + β k z + p k ( x ) , γ k z + q k ( x ) � � (2) | α k | < 1 , | γ k | < 1 , | β k | + | γ k | < 1 F k ( x 0 , y 0 , z 0 ) = ( y k − 1 , z k − 1 ) F k ( x N , y N , Z N ) = ( y k , z k ) ω k : I × R 2 → I × R 2 ω k ( x , y , z ) = ( L k ( x ) , F k ( x , y , z )) , k = 1 , 2 , . . . N S.A.Prasad (NCU) FC June 13-17, 2017 7 / 31

Construction of a CHFIF Theorem ( [Chand A.K.B. and Kapoor G.P ., 2007]) (1) { I × R 2 ; ω k , k = 1 , 2 , . . . , N } is a hyperbolic IFS with respect to a metric equivalent to Euclidean metric on R 3 . (2) The attractor G ⊆ R 3 such that G = � N k = 1 ω k ( G ) of the above IFS is graph of a continuous function f : I → R 2 such that f ( x k ) = ( y k , z k ) for k = 0 , 1 , . . . , N i.e. G = { ( x , f ( x )) : x ∈ I and f ( x ) = ( y ( x ) , z ( x )) } . S.A.Prasad (NCU) FC June 13-17, 2017 8 / 31

Construction of CHFIF Definition The Coalescence Hidden-variable Fractal Interpolation Function (CHFIF) for the given interpolation data { ( x k , y k ) : k = 0 , 1 , . . . , N } is defined as the continuous function f 1 : I → R , where f 1 is the first component of the continuous function f = ( f 1 , f 2 ) , graph of which is attractor of the hyperbolic IFS. f 2 - AFIF (Self-Affine Fractal Interpolation Function) y k = z k and α k + β k = γ k for all k , f 1 = f 2 is FIF S.A.Prasad (NCU) FC June 13-17, 2017 9 / 31

Construction of CHFIF CHFIF : if x k − 1 ≤ x ≤ x k then f 1 ( x ) = α k f 1 ( L − 1 k ( x )) + β k f 2 ( L − 1 k ( x )) + p k ( L − 1 k ( x )) FIF : if x k − 1 ≤ x ≤ x k then f 2 ( x ) = γ k f 2 ( L − 1 k ( x )) + q k ( L − 1 k ( x )) S.A.Prasad (NCU) FC June 13-17, 2017 10 / 31

Outline 1 Introduction 2 Riemann-Liouville fractional integral 3 Riemann-Liouville fractional derivative S.A.Prasad (NCU) FC June 13-17, 2017 11 / 31

Riemann-Liouville fractional integral Definition Let −∞ < a < x < b < ∞ . The Riemann-Liouville fractional integral of order ν > 0 with lower limit a is defined for locally integrable functions f : [ a , b ] → R as x 1 � ( x − t ) ν − 1 f ( t ) dt I ν a + f ( x ) = (3) Γ( ν ) a for x > a . S.A.Prasad (NCU) FC June 13-17, 2017 12 / 31

Riemann-Liouville fractional integral Given data { ( x k , y k ) ∈ R 2 : k = 0 , 1 , . . . , N } Generalized data { ( x k , y k , z k ) ∈ R 3 : k = 0 , 1 , . . . , N } x k − 1 1 � ( L k ( x ) − t ) ν − 1 f 1 ( t ) dt p ν k ( x ) = a ν k I ν x 0 + p k ( x ) + (4) Γ( ν ) x 0 and x k − 1 1 � ( L k ( x ) − t ) ν − 1 f 2 ( t ) dt . q ν k ( x ) = a ν k I ν x 0 + q k ( x ) + (5) Γ( ν ) x 0 S.A.Prasad (NCU) FC June 13-17, 2017 13 / 31

Riemann-Liouville fractional integral F ν k ( x , y , z ) = F ν k , 1 ( x , y , z ) , F ν k , 2 ( x , z ) � � a ν k α k y + a ν k β k z + p ν k ( x ) , a ν k γ k z + q ν k ( x ) � � = (6) Define ω ν k ( x , y , z ) = ( L k ( x ) , F ν k ( x , y , z )) ; (7) 0 = 0 = z ν y ν 0 , q ν N ( x N ) z ν N = , 1 − a ν N γ N a ν p ν N ( x N ) N β N y ν z ν N = N + , 1 − a ν 1 − a ν N α N N α N z ν k = a ν k γ k z ν N + q ν k ( x N ) = q ν k + 1 ( x 0 ) k + 1 ( x 0 ) , k = 1 , 2 , . . . , N − 1 . y ν k = a ν k α k y ν N + a ν k β k z ν N + p ν k ( x N ) = p ν and (8) S.A.Prasad (NCU) FC June 13-17, 2017 14 / 31

Riemann-Liouville fractional integral of FIF Proposition Let f 2 be a FIF passing through the interpolation data given by { ( x k , z k ) ∈ R 2 : k = 0 , 1 , . . . , N } . Then, Riemann-Liouville fractional integral of a FIF of order ν is also a FIF passing through the data k ) ∈ R 2 : k = 0 , 1 , . . . , N } , where z ν { ( x k , z ν k are given by (8) . S.A.Prasad (NCU) FC June 13-17, 2017 15 / 31

Riemann-Liouville fractional integral of FIF Theorem ( [S.A.P, 2017]) Let f 1 be the CHFIF passing through the interpolation data given by { ( x k , y k ) ∈ R 2 : k = 0 , 1 , . . . , N } and f 2 be the corresponding FIF passing through the data { ( x k , z k ) ∈ R 2 : k = 0 , 1 , . . . , N } . Then, Riemann-Liouville fractional integral of a CHFIF of order ν given by (3) is also a CHFIF passing through the data k ) ∈ R 2 : k = 0 , 1 , . . . , N } , where y ν { ( x k , y ν k are given by (8) . S.A.Prasad (NCU) FC June 13-17, 2017 16 / 31

Riemann-Liouville fractional integral of CHFIF Sketch of Proof: Let x such that x k − 1 < x < x k for some k ∈ { 1 , 2 , . . . , N } . Then, x 1 � ( x − t ) ν − 1 f 1 ( t ) dt I ν x 0 + f 1 ( x ) = Γ( ν ) x 0  x k − 1  x 1   � ( x − t ) ν − 1 f 1 ( t ) dt + � ( x − t ) ν − 1 f 1 ( t ) dt   = Γ( ν )   x 0 x k − 1   S.A.Prasad (NCU) FC June 13-17, 2017 17 / 31

Riemann-Liouville fractional integral of CHFIF x k − 1  1 � ( x − t ) ν − 1 f 1 ( t ) dt  I ν x 0 + f 1 ( x ) = Γ( ν )  x 0 L − 1 ( x )  k  � k ( x ) − t ) ν − 1 f 1 ( L k ( t )) dt ( L − 1  + a ν k  x 0  x 0 + f 1 ( L − 1 x 0 + f 2 ( L − 1 = a ν k α k I ν k ( x )) + a ν k β k I ν k ( x )) x k − 1   1 � ( x − t ) ν − 1 f 1 ( t ) dt x 0 + p k ( L − 1   + a ν k I ν k ( x )) + Γ( ν )   x 0 S.A.Prasad (NCU) FC June 13-17, 2017 18 / 31

Outline 1 Introduction 2 Riemann-Liouville fractional integral 3 Riemann-Liouville fractional derivative S.A.Prasad (NCU) FC June 13-17, 2017 19 / 31

Riemann-Liouville fractional derivative Definition Let −∞ < a < x < b < ∞ , 0 < ν , f ∈ L 1 ([ a , b ]) and I n − ν f ∈ W 1 , 1 , where n is the smallest integer greater than ν . The Riemann-Liouville fractional derivative of order ν with lower limit a is defined as a + f )( x ) = d n dx n ( I n − ν ( D ν f )( x ) a + a + f )( x ) = f ( x ) when ν = 0 . and ( D ν S.A.Prasad (NCU) FC June 13-17, 2017 20 / 31

Riemann-Liouville fractional derivative of FIF x k − 1   a − n d n � f 2 ( t )( L k ( x ) − t ) n − ν − 1 dt q d ν k k ( x ) = a − ν k D ν q k ( x ) +  (9) dx n Γ( n − ν )  x 0 and x k − 1   a − n d n � k D ν p k ( x ) + f 1 ( t )( L k ( x ) − t ) n − ν − 1 dt p d ν k k ( x ) = a − ν  . dx n Γ( n − ν )  x 0 (10) S.A.Prasad (NCU) FC June 13-17, 2017 21 / 31

Riemann-Liouville fractional derivative of FIF Proposition Let f 2 be a FIF passing through the interpolation data { ( x k , z k ) ∈ R 2 : k = 0 , 1 , . . . , N } and | γ k | < a ν k for some fixed ν > 0 . Then Riemann-Liouville fractional derivative of a FIF of order ν is also a FIF provided (9) is satisfied. S.A.Prasad (NCU) FC June 13-17, 2017 22 / 31

Riemann-Liouville fractional derivative of CHFIF Theorem ( [S.A.P, 2017]) Let f 1 be the CHFIF passing through the interpolation data given by { ( x k , y k ) ∈ R 2 : k = 0 , 1 , . . . , N } and f 2 be the corresponding FIF passing through the data { ( x k , z k ) ∈ R 2 : k = 0 , 1 , . . . , N } . For a fixed ν > 0 , if the free variables and constrained variables are such that | α k | < a ν k , | γ k | < a ν k and | β k | + | γ k | < a ν k then Riemann-Liouville fractional derivative of a CHFIF of order ν is also a CHFIF provided (9) and (10) are satisfied. S.A.Prasad (NCU) FC June 13-17, 2017 23 / 31