Introduction to fractional calculus (Based on lectures by R. Goren‡o, F. Mainardi and I. Podlubny)
R. Vilela Mendes
July 2008
() July2008 1/44 Contents
- Historical origins of fractional calculus - Fractional integral according to Riemann-Liouville - Caputo fractional derivative - Riesz-Feller fractional derivative - Grünwal-Letnikov - Integral equations - Relaxation and oscillation equations - Fractional di¤usion equation - A nonlinear fractional di¤erential equation. Stochastic solution - Geometrical interpretation of fractional integration
() July2008 2/44 Fractional Calculuswas born in1695 n fd dt n What if the order will be n = ½?
It will lead to a paradox, from which one day useful consequences will be Start drawn. JJ II G.F.A. de L’Hôpital G.W. Leibniz J I 10 / 90 (1661–1704) (1646–1716) Back Full screen Close End G. W. Leibniz (1695–1697) In the letters to J. Wallis and J. Bernulli (in 1697) Leibniz mentioned the possible approach to fractional-order differ- entiation in that sense, that for non-integer values of n the definition could be the following: dnemx = mnemx, n dx Start JJ II J I 11 / 90 Back Full screen Close End L. Euler (1730)
n m d x m n = m(m 1) ... (m n + 1)x − dxn − − Γ(m + 1) = m(m 1) ... (m n + 1) Γ(m n + 1) n m − − − d x Γ(m + 1) m n = x − . dxn Γ(m n + 1) − Euler suggested to use this relationship also for negative or 1 non-integer (rational) values of n. Taking m = 1 and n = 2, Euler obtained: Start d1/2x 4x 2 JJ II = r = x1/2 dx1/2 π √π J I 12 / 90 Back Full screen Close End S. F. Lacroix adopted Euler’s derivation for his success- ful textbook ( Trait´edu Calcul Diff´erentiel et du Calcul Int´egral, Courcier, Paris, t. 3, 1819; pp. 409–410).
Start JJ II J I 13 / 90 Back Full screen Close End J. B. J. Fourier (1820–1822) The first step to generalization of the notion of differentia- tion for arbitrary functions was done by J. B. J. Fourier (Th´eorieAnalytique de la Chaleur, Didot, Paris, 1822; pp. 499–508). After introducing his famous formula
1 ∞ ∞ f(x) = Z f(z)dz Z cos (px pz)dp, 2π − −∞ −∞ Fourier made a remark that Start dnf(x) 1 ∞ ∞ π = Z f(z)dzZ cos (px pz + n )dp, JJ II dxn 2π − 2 J I 14 / 90 −∞ −∞ Back and this relationship could serve as a definition of the n-th Full screen order derivative for non-integer n. Close End RiemannಥLiouville definition
t d n df ττ α tfD = 1 § · )( ta )( ¨ ¸ n+−α n α−Γ )( dt ³ t τ− 1 © ¹ a )( <α≤− nn )1(
Start JJ II G.F.B. Riemann J. Liouville J I (1826–1866) (1809–1882) 21 / 90 Back Full screen Close End Extend to any positive real value by using the Gamma function, (n 1)! = Γ(n) Fractional Integral of order α>0 (right-sided) t α 1 α 1 Ja+ f (t) := (t τ) f (τ) dτ , α R (2) Γ (α) a Z 2 0 0 De…ne Ja+ := I , Ja+ f (t) = f (t)
Fractional integral according to Riemann-Liouville
According to Riemann-Liouville the notion of fractional integral of order α (α > 0) for a function f (t), is a natural consequence of the well known formula (Cauchy-Dirichlet ?), that reduces the calculation of the n fold primitive of a function f (t) to a single integral of convolution type t n 1 n 1 Ja+f (t) := (t τ) f (τ) dτ , n N (1) (n 1)! a 2 Z vanishes at t = a with its derivatives of order 1, 2, ... , n 1 . Require n f (t) and Ja+f (t) to be causal functions, that is, vanishing for t < 0 .
() July2008 3/44 Fractional Integral of order α>0 (right-sided) t α 1 α 1 Ja+ f (t) := (t τ) f (τ) dτ , α R (2) Γ (α) a Z 2 0 0 De…ne Ja+ := I , Ja+ f (t) = f (t)
Fractional integral according to Riemann-Liouville
According to Riemann-Liouville the notion of fractional integral of order α (α > 0) for a function f (t), is a natural consequence of the well known formula (Cauchy-Dirichlet ?), that reduces the calculation of the n fold primitive of a function f (t) to a single integral of convolution type t n 1 n 1 Ja+f (t) := (t τ) f (τ) dτ , n N (1) (n 1)! a 2 Z vanishes at t = a with its derivatives of order 1, 2, ... , n 1 . Require n f (t) and Ja+f (t) to be causal functions, that is, vanishing for t < 0 . Extend to any positive real value by using the Gamma function, (n 1)! = Γ(n)
() July2008 3/44 Fractional integral according to Riemann-Liouville
According to Riemann-Liouville the notion of fractional integral of order α (α > 0) for a function f (t), is a natural consequence of the well known formula (Cauchy-Dirichlet ?), that reduces the calculation of the n fold primitive of a function f (t) to a single integral of convolution type t n 1 n 1 Ja+f (t) := (t τ) f (τ) dτ , n N (1) (n 1)! a 2 Z vanishes at t = a with its derivatives of order 1, 2, ... , n 1 . Require n f (t) and Ja+f (t) to be causal functions, that is, vanishing for t < 0 . Extend to any positive real value by using the Gamma function, (n 1)! = Γ(n) Fractional Integral of order α>0 (right-sided) t α 1 α 1 Ja+ f (t) := (t τ) f (τ) dτ , α R (2) Γ (α) a Z 2 0 0 De…ne Ja+ := I , Ja+ f (t) = f (t)
() July2008 3/44 Let α α J := J0+ Semigroup properties JαJ β = Jα+β , α , β 0 Commutative property J βJα = JαJ β E¤ect on power functions Jαtγ = Γ(γ+1) tγ+α , α > 0 , γ > 1 , t > 0 Γ(γ+1+α) (Natural generalization of the positive integer properties). Introduce the following causal function (vanishing for t < 0 ) α 1 t Φ (t) := + , α > 0 α Γ(α)
Fractional integral according to Riemann-Liouville
Alternatively (left-sided integral) b α 1 α 1 Jb f (t) := (τ t) f (τ) dτ , α R Γ (α) t Z 2 (a = 0, b = +∞) Riemann (a = ∞, b = +∞) Liouville
() July2008 4/44 Introduce the following causal function (vanishing for t < 0 ) α 1 t Φ (t) := + , α > 0 α Γ(α)
Fractional integral according to Riemann-Liouville
Alternatively (left-sided integral) b α 1 α 1 Jb f (t) := (τ t) f (τ) dτ , α R Γ (α) t Z 2 (a = 0, b = +∞) Riemann (a = ∞, b = +∞) Liouville Let α α J := J0+ Semigroup properties JαJ β = Jα+β , α , β 0 Commutative property J βJα = JαJ β E¤ect on power functions Jαtγ = Γ(γ+1) tγ+α , α > 0 , γ > 1 , t > 0 Γ(γ+1+α) (Natural generalization of the positive integer properties).
() July2008 4/44 Fractional integral according to Riemann-Liouville
Alternatively (left-sided integral) b α 1 α 1 Jb f (t) := (τ t) f (τ) dτ , α R Γ (α) t Z 2 (a = 0, b = +∞) Riemann (a = ∞, b = +∞) Liouville Let α α J := J0+ Semigroup properties JαJ β = Jα+β , α , β 0 Commutative property J βJα = JαJ β E¤ect on power functions Jαtγ = Γ(γ+1) tγ+α , α > 0 , γ > 1 , t > 0 Γ(γ+1+α) (Natural generalization of the positive integer properties). Introduce the following causal function (vanishing for t < 0 ) α 1 t Φ (t) := + , α > 0 α Γ(α)
() July2008 4/44 Laplace transform
∞ st f (t) := e f (t) dt = f (s) , s C L f g Z0 2 De…ning the Laplace transform pairs by f (et) f (s) f (s) e Jα f (t) , α > 0 sα e
Fractional integral according to Riemann-Liouville
Φ (t) Φ (t) = Φ + (t) , α , β > 0 α β α β Jα f (t) = Φ (t) f (t) , α > 0 α
() July2008 5/44 De…ning the Laplace transform pairs by f (t) f (s) f (s) e Jα f (t) , α > 0 sα e
Fractional integral according to Riemann-Liouville
Φ (t) Φ (t) = Φ + (t) , α , β > 0 α β α β Jα f (t) = Φ (t) f (t) , α > 0 α Laplace transform
∞ st f (t) := e f (t) dt = f (s) , s C L f g Z0 2 e
() July2008 5/44 Fractional integral according to Riemann-Liouville
Φ (t) Φ (t) = Φ + (t) , α , β > 0 α β α β Jα f (t) = Φ (t) f (t) , α > 0 α Laplace transform
∞ st f (t) := e f (t) dt = f (s) , s C L f g Z0 2 De…ning the Laplace transform pairs by f (et) f (s) f (s) e Jα f (t) , α > 0 sα e
() July2008 5/44 Then, de…ne Dα as a left-inverse to Jα. With a positive integer m , m 1 < α m , de…ne: α m m α Fractional Derivative of order α : D f (t) := D J f (t)
d m 1 t f (τ) α dtm α+1 m dτ , m 1 < α < m D f (t) := Γ(m α) 0 (t τ) d m ( h dtm fR(t) , i α = m
Fractional derivative according to Riemann-Liouville
Denote by Dn with n N , the derivative of order n . Note that 2 Dn Jn = I , Jn Dn = I , n N 6 2 Dn is a left-inverse (not a right-inverse) to Jn . In fact
n 1 tk Jn Dn f (t) = f (t) ∑ f (k)(0+) , t > 0 k=0 k!
() July2008 6/44 Fractional derivative according to Riemann-Liouville
Denote by Dn with n N , the derivative of order n . Note that 2 Dn Jn = I , Jn Dn = I , n N 6 2 Dn is a left-inverse (not a right-inverse) to Jn . In fact
n 1 tk Jn Dn f (t) = f (t) ∑ f (k)(0+) , t > 0 k=0 k! Then, de…ne Dα as a left-inverse to Jα. With a positive integer m , m 1 < α m , de…ne: α m m α Fractional Derivative of order α : D f (t) := D J f (t)
d m 1 t f (τ) α dtm α+1 m dτ , m 1 < α < m D f (t) := Γ(m α) 0 (t τ) d m ( h dtm fR(t) , i α = m
() July2008 6/44 The fractional derivative Dα f is not zero for the constant function f (t) 1 if α N 62 t α Dα1 = , α 0 , t > 0 Γ(1 α) Is 0 for α N, due to the poles of the gamma function 2
Fractional derivative according to Riemann-Liouville
De…ne D0 = J0 = I . Then Dα Jα = I , α 0
α γ Γ(γ + 1) γ α D t = t , α > 0 , γ > 1 , t > 0 Γ(γ + 1 α)
() July2008 7/44 Fractional derivative according to Riemann-Liouville
De…ne D0 = J0 = I . Then Dα Jα = I , α 0
α γ Γ(γ + 1) γ α D t = t , α > 0 , γ > 1 , t > 0 Γ(γ + 1 α) The fractional derivative Dα f is not zero for the constant function f (t) 1 if α N 62 t α Dα1 = , α 0 , t > 0 Γ(1 α) Is 0 for α N, due to the poles of the gamma function 2
() July2008 7/44 A de…nition more restrictive than the one before. It requires the absolute integrability of the derivative of order m. In general α m m α m α m α D f (t) := D J f (t) = J D f (t) := D f (t) 6 unless the function f (t) along with its …rst m 1 derivatives vanishes at t = 0+. In fact, for m 1 < α < m and t > 0 , m 1 tk α Dα f (t) = Dα f (t) + f (k)(0+) ∑ Γ(k α + 1) k=0 and therefore, recalling the fractional derivative of the power functions m 1 tk Dα f (t) ∑ f (k)(0+) = Dα f (t) , Dα1 0 , α > 0 k=0 k! !
Caputo fractional derivative
α m α m D f (t) := J D f (t) with m 1 < α m , namely 1 t f (m)(τ) α α+1 m dτ, m 1 < α < m Γ(m α) 0 (t τ) D f (t) := m d ( dtmRf (t) , α = m
() July2008 8/44 Caputo fractional derivative
α m α m D f (t) := J D f (t) with m 1 < α m , namely 1 t f (m)(τ) α α+1 m dτ, m 1 < α < m Γ(m α) 0 (t τ) D f (t) := m d ( dtmRf (t) , α = m A de…nition more restrictive than the one before. It requires the absolute integrability of the derivative of order m. In general α m m α m α m α D f (t) := D J f (t) = J D f (t) := D f (t) 6 unless the function f (t) along with its …rst m 1 derivatives vanishes at t = 0+. In fact, for m 1 < α < m and t > 0 , m 1 tk α Dα f (t) = Dα f (t) + f (k)(0+) ∑ Γ(k α + 1) k=0 and therefore, recalling the fractional derivative of the power functions m 1 tk Dα f (t) ∑ f (k)(0+) = Dα f (t) , Dα1 0 , α > 0 k=0 k! ! () July2008 8/44 Functions which for t > 0 have the same fractional derivative of order α , with m 1 < α m . (the cj ’sare arbitrary constants) m α α α j D f (t) = D g(t) f (t) = g(t) + ∑ cj t () j=1 m α α m j D f (t) = D g(t) f (t) = g(t) + ∑ cj t () j=1 Formal limit as α (m 1)+ ! + α m m 1 α (m 1) = D f (t) D J f (t) = D f (t) ! ) ! + α m m 1 (m 1) + α (m 1) = D f (t) JD f (t) = D f (t) f (0 ) ! ) !
Riemann versus Caputo
α α 1 D t 0 , α > 0 , t > 0 Dα is not a right-inverse to Jα α α α 1 α α α 1 α 1 J D t 0, but D J t = t , α > 0 , t > 0
() July2008 9/44 Formal limit as α (m 1)+ ! + α m m 1 α (m 1) = D f (t) D J f (t) = D f (t) ! ) ! + α m m 1 (m 1) + α (m 1) = D f (t) JD f (t) = D f (t) f (0 ) ! ) !
Riemann versus Caputo
α α 1 D t 0 , α > 0 , t > 0 Dα is not a right-inverse to Jα α α α 1 α α α 1 α 1 J D t 0, but D J t = t , α > 0 , t > 0 Functions which for t > 0 have the same fractional derivative of order α , with m 1 < α m . (the cj ’sare arbitrary constants) m α α α j D f (t) = D g(t) f (t) = g(t) + ∑ cj t () j=1 m α α m j D f (t) = D g(t) f (t) = g(t) + ∑ cj t () j=1
() July2008 9/44 Riemann versus Caputo
α α 1 D t 0 , α > 0 , t > 0 Dα is not a right-inverse to Jα α α α 1 α α α 1 α 1 J D t 0, but D J t = t , α > 0 , t > 0 Functions which for t > 0 have the same fractional derivative of order α , with m 1 < α m . (the cj ’sare arbitrary constants) m α α α j D f (t) = D g(t) f (t) = g(t) + ∑ cj t () j=1 m α α m j D f (t) = D g(t) f (t) = g(t) + ∑ cj t () j=1 Formal limit as α (m 1)+ ! + α m m 1 α (m 1) = D f (t) D J f (t) = D f (t) ! ) ! + α m m 1 (m 1) + α (m 1) = D f (t) JD f (t) = D f (t) f (0 ) ! ) ! () July2008 9/44 For the Caputo fractional derivative
m 1 α α (k) + α 1 k D f (t) s f (s) ∑ f (0 ) s , m 1 < α m k=0 Requires the knowledgee of the (bounded) initial values of the function and of its integer derivatives of order k = 1, 2, m 1 in analogy with the case when α = m
Riemann versus Caputo
The Laplace transform
m 1 α α k (m α) + m 1 k D f (t) s f (s) ∑ D J f (0 ) s , m 1 < α m k=0 Requires thee knowledge of the (bounded) initial values of the m α fractional integral J and of its integer derivatives of order k = 1, 2, m 1
() July2008 10/44 Riemann versus Caputo
The Laplace transform
m 1 α α k (m α) + m 1 k D f (t) s f (s) ∑ D J f (0 ) s , m 1 < α m k=0 Requires thee knowledge of the (bounded) initial values of the m α fractional integral J and of its integer derivatives of order k = 1, 2, m 1 For the Caputo fractional derivative
m 1 α α (k) + α 1 k D f (t) s f (s) ∑ f (0 ) s , m 1 < α m k=0 Requires the knowledgee of the (bounded) initial values of the function and of its integer derivatives of order k = 1, 2, m 1 in analogy with the case when α = m
() July2008 10/44 Symbol of an operator ∞ A^ (k) φ^ (k) = eikx Aφ (x) dx ∞ Z For the Liouville integral x α 1 α 1 J∞+ f (x) : = (x ξ) f (ξ) dξ Γ (α) ∞ Z ∞ α 1 α 1 J∞ f (x) : = (ξ x) f (ξ) dξ , α R Γ (α) x Z 2
Riesz - Feller fractional derivative
For functions with Fourier transform ∞ φ (x) = φ^ (k) = eikx φ (x) dx F f g ∞ Z ∞ 1 ^ 1 ikx ^ φ (k) = φ (x) = e φ (k) dx F 2π ∞ Z
() July2008 11/44 For the Liouville integral x α 1 α 1 J∞+ f (x) : = (x ξ) f (ξ) dξ Γ (α) ∞ Z ∞ α 1 α 1 J∞ f (x) : = (ξ x) f (ξ) dξ , α R Γ (α) x Z 2
Riesz - Feller fractional derivative
For functions with Fourier transform ∞ φ (x) = φ^ (k) = eikx φ (x) dx F f g ∞ Z ∞ 1 ^ 1 ikx ^ φ (k) = φ (x) = e φ (k) dx F 2π ∞ Z Symbol of an operator ∞ A^ (k) φ^ (k) = eikx Aφ (x) dx ∞ Z
() July2008 11/44 Riesz - Feller fractional derivative
For functions with Fourier transform ∞ φ (x) = φ^ (k) = eikx φ (x) dx F f g ∞ Z ∞ 1 ^ 1 ikx ^ φ (k) = φ (x) = e φ (k) dx F 2π ∞ Z Symbol of an operator ∞ A^ (k) φ^ (k) = eikx Aφ (x) dx ∞ Z For the Liouville integral x α 1 α 1 J∞+ f (x) : = (x ξ) f (ξ) dξ Γ (α) ∞ Z ∞ α 1 α 1 J∞ f (x) : = (ξ x) f (ξ) dξ , α R Γ (α) x Z 2
() July2008 11/44 Operator symbols
α^ α i(signk)απ/2 α J∞ = k e = ( ik) j j α^ +α i(signk)απ/2 +α D∞ = k e = ( ik) j j α^ α^ 2 cos (απ/2) J∞+ + J∞ = α k j j De…ne a symmetrized version α α ∞ α J∞+ f + J∞ f 1 α 1 I0 f (x) = = x ξ f (ξ) dξ 2 cos (απ/2) 2Γ (α) cos (απ/2) ∞ j j Z (wth exclusion of odd integers). The operator symbol ^α α is I = k 0 j j
Riesz - Feller fractional derivative
Liouville derivatives (m 1 < α < m) m m α α D J∞ f (x) , m odd D∞ = m m α D J∞ f (x) , m even
() July2008 12/44 De…ne a symmetrized version α α ∞ α J∞+ f + J∞ f 1 α 1 I0 f (x) = = x ξ f (ξ) dξ 2 cos (απ/2) 2Γ (α) cos (απ/2) ∞ j j Z (wth exclusion of odd integers). The operator symbol ^α α is I = k 0 j j
Riesz - Feller fractional derivative
Liouville derivatives (m 1 < α < m) m m α α D J∞ f (x) , m odd D∞ = m m α D J∞ f (x) , m even Operator symbols
α^ α i(signk)απ/2 α J∞ = k e = ( ik) j j α^ +α i(signk)απ/2 +α D∞ = k e = ( ik) j j α^ α^ 2 cos (απ/2) J∞+ + J∞ = α k j j
() July2008 12/44 Riesz - Feller fractional derivative
Liouville derivatives (m 1 < α < m) m m α α D J∞ f (x) , m odd D∞ = m m α D J∞ f (x) , m even Operator symbols
α^ α i(signk)απ/2 α J∞ = k e = ( ik) j j α^ +α i(signk)απ/2 +α D∞ = k e = ( ik) j j α^ α^ 2 cos (απ/2) J∞+ + J∞ = α k j j De…ne a symmetrized version α α ∞ α J∞+ f + J∞ f 1 α 1 I0 f (x) = = x ξ f (ξ) dξ 2 cos (απ/2) 2Γ (α) cos (απ/2) ∞ j j Z (wth exclusion of odd integers). The operator symbol ^α α is I0 = k () j j July2008 12/44 De…ne the Riesz fractional derivative by analytical continuation
α α α ^ D f (k) := I f (k) = k f (k) F f 0 g F 0 j j generalized by Feller α Dθ =Riesz-Feller fractional derivative of order α and skewness θ
Dαf (k) := ψθ (k) f^ (k) F f 0 g α with
ψθ (k) = k α ei(signk)θπ/2, 0 < α 2, θ min α, 2 α α j j j j f g The symbol ψθ (k) is the logarithm of the characteristic function of α a Lévy stable probability distribution with index of stability α and asymmetry parameter θ
Riesz-Feller fractional derivative
α I0 f (x) is called the Riesz potential.
() July2008 13/44 α Dθ =Riesz-Feller fractional derivative of order α and skewness θ
Dαf (k) := ψθ (k) f^ (k) F f 0 g α with
ψθ (k) = k α ei(signk)θπ/2, 0 < α 2, θ min α, 2 α α j j j j f g The symbol ψθ (k) is the logarithm of the characteristic function of α a Lévy stable probability distribution with index of stability α and asymmetry parameter θ
Riesz-Feller fractional derivative
α I0 f (x) is called the Riesz potential. De…ne the Riesz fractional derivative by analytical continuation
α α α ^ D f (k) := I f (k) = k f (k) F f 0 g F 0 j j generalized by Feller
() July2008 13/44 The symbol ψθ (k) is the logarithm of the characteristic function of α a Lévy stable probability distribution with index of stability α and asymmetry parameter θ
Riesz-Feller fractional derivative
α I0 f (x) is called the Riesz potential. De…ne the Riesz fractional derivative by analytical continuation
α α α ^ D f (k) := I f (k) = k f (k) F f 0 g F 0 j j generalized by Feller α Dθ =Riesz-Feller fractional derivative of order α and skewness θ
Dαf (k) := ψθ (k) f^ (k) F f 0 g α with
ψθ (k) = k α ei(signk)θπ/2, 0 < α 2, θ min α, 2 α α j j j j f g
() July2008 13/44 Riesz-Feller fractional derivative
α I0 f (x) is called the Riesz potential. De…ne the Riesz fractional derivative by analytical continuation
α α α ^ D f (k) := I f (k) = k f (k) F f 0 g F 0 j j generalized by Feller α Dθ =Riesz-Feller fractional derivative of order α and skewness θ
Dαf (k) := ψθ (k) f^ (k) F f 0 g α with
ψθ (k) = k α ei(signk)θπ/2, 0 < α 2, θ min α, 2 α α j j j j f g The symbol ψθ (k) is the logarithm of the characteristic function of α a Lévy stable probability distribution with index of stability α and asymmetry parameter θ
() July2008 13/44 GrínwaldಥLetnikovdefinition
ª −at º ¬« h ¼» α α α− j § · ta htfD −= ¨ ¸ − jhtf )()1(lim)( →0h ¦ j j=0 © ¹ [x] – integer part of x
Start JJ II J I A.K. Grünwald A.V. Letnikov 22 / 90 Back Full screen Close End the Grünwal-Letnikov fractional derivatives are
[(x a)/h] α 1 k α GLDa+ = lim ( 1) φ (x kh) h 0 hα ∑ k ! k=0 [(b x )/h] 1 α Dα = lim 1 k x + kh GL b α ∑ ( ) φ ( ) h 0 h k ! k=0 [ ] denotes the integer part
Grünwal - Letnikov
From φ (x) φ (x h) Dφ (x) = lim h 0 h ! 1 n n Dn = lim ( 1)k φ (x kh) h 0 hn ∑ k ! k=0
() July2008 14/44 Grünwal - Letnikov
From φ (x) φ (x h) Dφ (x) = lim h 0 h ! 1 n n Dn = lim ( 1)k φ (x kh) h 0 hn ∑ k ! k=0 the Grünwal-Letnikov fractional derivatives are
[(x a)/h] α 1 k α GLDa+ = lim ( 1) φ (x kh) h 0 hα ∑ k ! k=0 [(b x )/h] 1 α Dα = lim 1 k x + kh GL b α ∑ ( ) φ ( ) h 0 h k ! k=0 [ ] denotes the integer part () July2008 14/44 The mechanical problem of the tautochrone, that is, determining a curve in the vertical plane, such that the time required for a particle to slide down the curve to its lowest point is independent of its initial placement on the curve. Found many applications in diverse …elds: - Evaluation of spectroscopic measurements of cylindrical gas discharges - Study of the solar or a planetary atmosphere - Star densities in a globular cluster - Inversion of travel times of seismic waves for determination of terrestrial sub-surface structure - Inverse boundary value problems in partial di¤erential equations
Integral equations
Abel’sequation (1st kind) 1 t u(τ) 1 dτ = f (t) , 0 < α < 1 Γ (α) 0 (t τ) α Z
() July2008 15/44 Found many applications in diverse …elds: - Evaluation of spectroscopic measurements of cylindrical gas discharges - Study of the solar or a planetary atmosphere - Star densities in a globular cluster - Inversion of travel times of seismic waves for determination of terrestrial sub-surface structure - Inverse boundary value problems in partial di¤erential equations
Integral equations
Abel’sequation (1st kind) 1 t u(τ) 1 dτ = f (t) , 0 < α < 1 Γ (α) 0 (t τ) α Z The mechanical problem of the tautochrone, that is, determining a curve in the vertical plane, such that the time required for a particle to slide down the curve to its lowest point is independent of its initial placement on the curve.
() July2008 15/44 Integral equations
Abel’sequation (1st kind) 1 t u(τ) 1 dτ = f (t) , 0 < α < 1 Γ (α) 0 (t τ) α Z The mechanical problem of the tautochrone, that is, determining a curve in the vertical plane, such that the time required for a particle to slide down the curve to its lowest point is independent of its initial placement on the curve. Found many applications in diverse …elds: - Evaluation of spectroscopic measurements of cylindrical gas discharges - Study of the solar or a planetary atmosphere - Star densities in a globular cluster - Inversion of travel times of seismic waves for determination of terrestrial sub-surface structure - Inverse boundary value problems in partial di¤erential equations
() July2008 15/44 Then,
x 2 /[4(t τ)] 1 t p(τ) u(x, t) = e dτ , x > 0 , t > 0 pπ 0 pt τ Z
Abel’sequation
- Heating (or cooling) of a semi-in…nite rod by in‡ux (or e• ux) of heat across the boundary into (or from) its interior
ut uxx = 0 , u = u(x, t) in the semi-in…nite intervals 0 < x < ∞ and 0 < t < ∞. Assume initial temperature, u(x, 0) = 0 for 0 < x < ∞ and given in‡ux across the boundary x = 0 from x < 0 to x > 0 ,
ux (0, t) = p(t)
() July2008 16/44 Abel’sequation
- Heating (or cooling) of a semi-in…nite rod by in‡ux (or e• ux) of heat across the boundary into (or from) its interior
ut uxx = 0 , u = u(x, t) in the semi-in…nite intervals 0 < x < ∞ and 0 < t < ∞. Assume initial temperature, u(x, 0) = 0 for 0 < x < ∞ and given in‡ux across the boundary x = 0 from x < 0 to x > 0 ,
ux (0, t) = p(t) Then,
x 2 /[4(t τ)] 1 t p(τ) u(x, t) = e dτ , x > 0 , t > 0 pπ 0 pt τ Z
() July2008 16/44 Is Jα u(t) = f (t) and consequently is solved by
u(t) = Dα f (t)
using Dα Jα = I . Let us now solve using the Laplace transform
u(s) = f (s) = u(s) = sα f (s) sα ) The solution is obtained by the inverse Laplace transform: Two possibilities :
Abel’sequation (1st kind)
1 t u(τ) 1 dτ = f (t) , 0 < α < 1 Γ (α) 0 (t τ) α Z
() July2008 17/44 The solution is obtained by the inverse Laplace transform: Two possibilities :
Abel’sequation (1st kind)
1 t u(τ) 1 dτ = f (t) , 0 < α < 1 Γ (α) 0 (t τ) α Z Is Jα u(t) = f (t) and consequently is solved by
u(t) = Dα f (t)
using Dα Jα = I . Let us now solve using the Laplace transform
u(s) = f (s) = u(s) = sα f (s) sα )
() July2008 17/44 Abel’sequation (1st kind)
1 t u(τ) 1 dτ = f (t) , 0 < α < 1 Γ (α) 0 (t τ) α Z Is Jα u(t) = f (t) and consequently is solved by
u(t) = Dα f (t)
using Dα Jα = I . Let us now solve using the Laplace transform
u(s) = f (s) = u(s) = sα f (s) sα ) The solution is obtained by the inverse Laplace transform: Two possibilities :
() July2008 17/44 2) 1 f (0+) u(s) = [sf (s) f (0+)] + 1 α 1 α s s 1 t f (τ) t α u(t) = 0 dτ + f (0+) Γ (1 α) 0 (t τ)α Γ(1 a) Z Solutions expressed in terms of the fractional derivatives Dα and Dα , respectively
Abel’sequation (1st kind)
1) f (s) u(s) = s 1 α 0 s 1 @ A 1 d t f (τ) u(t) = dτ Γ (1 α) dt 0 (t τ)α Z
() July2008 18/44 Abel’sequation (1st kind)
1) f (s) u(s) = s 1 α 0 s 1 @ A 1 d t f (τ) u(t) = dτ Γ (1 α) dt 0 (t τ)α Z 2) 1 f (0+) u(s) = [sf (s) f (0+)] + 1 α 1 α s s 1 t f (τ) t α u(t) = 0 dτ + f (0+) Γ (1 α) 0 (t τ)α Γ(1 a) Z Solutions expressed in terms of the fractional derivatives Dα and Dα , respectively
() July2008 18/44 In terms of the fractional integral operator (1 + λ Jα) u(t) = f (t) solved as ∞ α 1 n αn u(t) = (1 + λJ ) f (t) = 1 + ∑ ( λ) J f (t) n=1 ! Noting that αn 1 αn t+ J f (t) = Φ n(t) f (t) = f (t) α Γ(αn) ∞ αn 1 t u(t) = f (t) + ∑ ( λ)n + f (t) n=1 Γ(αn) !
Abel’sequation (2nd kind)
λ t u(τ) u(t) + 1 dτ = f (t) , α > 0 , λ C Γ(α) 0 (t τ) α 2 Z
() July2008 19/44 Noting that αn 1 αn t+ J f (t) = Φ n(t) f (t) = f (t) α Γ(αn) ∞ αn 1 t u(t) = f (t) + ∑ ( λ)n + f (t) n=1 Γ(αn) !
Abel’sequation (2nd kind)
λ t u(τ) u(t) + 1 dτ = f (t) , α > 0 , λ C Γ(α) 0 (t τ) α 2 Z In terms of the fractional integral operator (1 + λ Jα) u(t) = f (t) solved as ∞ α 1 n αn u(t) = (1 + λJ ) f (t) = 1 + ∑ ( λ) J f (t) n=1 !
() July2008 19/44 Abel’sequation (2nd kind)
λ t u(τ) u(t) + 1 dτ = f (t) , α > 0 , λ C Γ(α) 0 (t τ) α 2 Z In terms of the fractional integral operator (1 + λ Jα) u(t) = f (t) solved as ∞ α 1 n αn u(t) = (1 + λJ ) f (t) = 1 + ∑ ( λ) J f (t) n=1 ! Noting that αn 1 αn t+ J f (t) = Φ n(t) f (t) = f (t) α Γ(αn) ∞ αn 1 t u(t) = f (t) + ∑ ( λ)n + f (t) n=1 Γ(αn) !
() July2008 19/44 Finally,
u(t) = f (t) + eα0 (t; λ)
Abel’sequation (2nd kind)
Relation to the Mittag-Le• er functions
∞ α n α ( λ t ) eα(t; λ) := Eα( λ t ) = ∑ , t > 0 , α > 0 , λ C n=0 Γ(αn + 1) 2
∞ αn 1 n t+ d α ∑ ( λ) = Eα( λt ) = eα0 (t; λ) , t > 0 n=1 Γ(αn) dt
() July2008 20/44 Abel’sequation (2nd kind)
Relation to the Mittag-Le• er functions
∞ α n α ( λ t ) eα(t; λ) := Eα( λ t ) = ∑ , t > 0 , α > 0 , λ C n=0 Γ(αn + 1) 2
∞ αn 1 n t+ d α ∑ ( λ) = Eα( λt ) = eα0 (t; λ) , t > 0 n=1 Γ(αn) dt Finally,
u(t) = f (t) + eα0 (t; λ)
() July2008 20/44 For the oscillation di¤erential equation
u00(t) = u(t) + q(t) + the solution, under the initial conditions u(0 ) = c0 and + u0(0 ) = c1 , is t u(t) = c0 cos t + c1 sin t + q(t τ) sin τ dτ Z0
Fractional di¤erential equations
Relaxation and oscillation equations. Integer order
u0(t) = u(t) + q(t) + the solution, under the initial condition u(0 ) = c0 , is t t τ u(t) = c0e + q(t τ)e dτ Z0
() July2008 21/44 Fractional di¤erential equations
Relaxation and oscillation equations. Integer order
u0(t) = u(t) + q(t) + the solution, under the initial condition u(0 ) = c0 , is t t τ u(t) = c0e + q(t τ)e dτ Z0 For the oscillation di¤erential equation
u00(t) = u(t) + q(t) + the solution, under the initial conditions u(0 ) = c0 and + u0(0 ) = c1 , is t u(t) = c0 cos t + c1 sin t + q(t τ) sin τ dτ Z0
() July2008 21/44 Relaxation and oscillation equations
Fractional version m 1 tk Dα u(t) = Dα (u(t) ∑ u(k)(0+) = u(t) + q(t) , t > 0 k=0 k! ! (k) + m 1 < α m , initial values u (0 ) = ck , k = 0, ... , m 1 .When α is the integer m m 1 t u(t) = ∑ ck uk (t) + q(t τ) uδ(τ) dτ k=0 Z0 k (h) + uk (t) = J u0(t) , u (0 ) = δk h , h, k = 0, , m 1, u (t) = u0 (t) k δ 0 The uk (t)’sare the fundamental solutions, linearly independent solutions of the homogeneous equation satisfying the initial conditions. The function uδ(t) , which is convoluted with q(t), is the impulse-response solution of the inhomogeneous equation with ck 0 , k = 0, ... , m 1 , t q(t) = δ(t) .For ordinary relaxation and oscillation, u0(t) = e = uδ(t) and u0(t) = cos t , u1(t) = J u0(t) = sin t = cos (t π/2) = u (t) . δ () July2008 22/44 Introducing the Mittag-Le• er type functions sα 1 e (t) e (t; 1) := E ( tα) α α α sα + 1
Relaxation and oscillation equations
Solution of the fractional equation by Laplace transform Applying the operator Jα to the fractional equation m 1 tk u(t) = ∑ Jα u(t) + Jα q(t) k=0 k! Laplace transforming yields m 1 1 1 1 u(s) = u(s) + q(s) ∑ k+1 α α k=0 s s s hence m 1 α k 1 s u(s) = + q(s) ∑ α k=0 s + 1
() July2008 23/44 Relaxation and oscillation equations
Solution of the fractional equation by Laplace transform Applying the operator Jα to the fractional equation m 1 tk u(t) = ∑ Jα u(t) + Jα q(t) k=0 k! Laplace transforming yields m 1 1 1 1 u(s) = u(s) + q(s) ∑ k+1 α α k=0 s s s hence m 1 α k 1 s u(s) = + q(s) ∑ α k=0 s + 1 Introducing the Mittag-Le• er type functions sα 1 e (t) e (t; 1) := E ( tα) α α α sα + 1
() July2008 23/44 When α is not integer, m 1 represents the integer part of α ([α]) and m the number of initial conditions necessary and su¢ cient to ensure the uniqueness of the solution u(t). The m functions k uk (t) = J eα(t) with k = 0, 1, ... , m 1 represent those particular solutions of the homogeneous equation which satisfy the initial conditions (h) + u (0 ) = δk h , h, k = 0, 1, ... , m 1 k and therefore they represent the fundamental solutions of the
fractional equation Furthermore, the function uδ(t) = eα0 (t) represents the impulse-response solution.
Relaxation and oscillation equations
α k 1 k s uk (t) := J e (t) , k = 0, 1, ... , m 1 α sα + 1 we …nd m 1 t u(t) = ∑ uk (t) q(t τ) u0 (τ) dτ k=0 Z0
() July2008 24/44 Relaxation and oscillation equations
α k 1 k s uk (t) := J e (t) , k = 0, 1, ... , m 1 α sα + 1 we …nd m 1 t u(t) = ∑ uk (t) q(t τ) u0 (τ) dτ k=0 Z0 When α is not integer, m 1 represents the integer part of α ([α]) and m the number of initial conditions necessary and su¢ cient to ensure the uniqueness of the solution u(t). The m functions k uk (t) = J eα(t) with k = 0, 1, ... , m 1 represent those particular solutions of the homogeneous equation which satisfy the initial conditions (h) + u (0 ) = δk h , h, k = 0, 1, ... , m 1 k and therefore they represent the fundamental solutions of the
fractional equation Furthermore, the function uδ(t) = eα0 (t) represents the impulse-response solution.
() July2008 24/44 Space-fractional di¤usion 0 < α 2 , β = 1 f g Time-fractional di¤usion α = 2 , 0 < β 2 f g Neutral-fractional di¤usion 0 < α = β 2 f g Riesz-Feller space-fractional derivative
α θ x D f (x); κ = ψ (κ) f (κ) F f θ g α ψθ (κ) = κ α ei(signk)θπ/2 , 0 < α 2 , θ min α, 2 α α j j jbj f g
Fractional di¤usion equation
Fractional di¤usion equation, obtained from the standard di¤usion equation by replacing the second-order space derivative with a Riesz-Feller derivative of order α (0, 2] and skewness θ and the 2 …rst-order time derivative with a Caputo derivative of order β (0, 2] 2 α β + x Dθ u(x, t) = t D u(x, t) , x R , t R 2 2 0 < α 2 , θ min α, 2 α , 0 < β 2 j j f g
() July2008 25/44 Riesz-Feller space-fractional derivative
α θ x D f (x); κ = ψ (κ) f (κ) F f θ g α ψθ (κ) = κ α ei(signk)θπ/2 , 0 < α 2 , θ min α, 2 α α j j jbj f g
Fractional di¤usion equation
Fractional di¤usion equation, obtained from the standard di¤usion equation by replacing the second-order space derivative with a Riesz-Feller derivative of order α (0, 2] and skewness θ and the 2 …rst-order time derivative with a Caputo derivative of order β (0, 2] 2 α β + x Dθ u(x, t) = t D u(x, t) , x R , t R 2 2 0 < α 2 , θ min α, 2 α , 0 < β 2 j j f g Space-fractional di¤usion 0 < α 2 , β = 1 f g Time-fractional di¤usion α = 2 , 0 < β 2 f g Neutral-fractional di¤usion 0 < α = β 2 f g
() July2008 25/44 Fractional di¤usion equation
Fractional di¤usion equation, obtained from the standard di¤usion equation by replacing the second-order space derivative with a Riesz-Feller derivative of order α (0, 2] and skewness θ and the 2 …rst-order time derivative with a Caputo derivative of order β (0, 2] 2 α β + x Dθ u(x, t) = t D u(x, t) , x R , t R 2 2 0 < α 2 , θ min α, 2 α , 0 < β 2 j j f g Space-fractional di¤usion 0 < α 2 , β = 1 f g Time-fractional di¤usion α = 2 , 0 < β 2 f g Neutral-fractional di¤usion 0 < α = β 2 f g Riesz-Feller space-fractional derivative
α θ x D f (x); κ = ψ (κ) f (κ) F f θ g α ψθ (κ) = κ α ei(signk)θπ/2 , 0 < α 2 , θ min α, 2 α α j j jbj f g
() July2008 25/44 Cauchy problem
u(x, 0) = ϕ(x) , x R , u( ∞, t) = 0 , t > 0 2 +∞ θ θ uα,β(x, t) = Gα,β(ξ, t) ϕ(x ξ) dξ ∞ Z θ γ θ γ Gα,β(ax , bt) = b Gα,β(ax/b , t) , γ = β/α Similarity variable x/tγ
θ γ θ γ Gα,β(x, t) = t Kα,β(x/t ) , γ = β/α
Fractional di¤usion equation
Caputo time-fractional derivative
1 t f (m)(τ) α α+1 m dτ, m 1 < α < m Γ(m α) 0 (t τ) D f (t) := m d ( dtmRf (t) , α = m
() July2008 26/44 Similarity variable x/tγ
θ γ θ γ Gα,β(x, t) = t Kα,β(x/t ) , γ = β/α
Fractional di¤usion equation
Caputo time-fractional derivative
1 t f (m)(τ) α α+1 m dτ, m 1 < α < m Γ(m α) 0 (t τ) D f (t) := m d ( dtmRf (t) , α = m Cauchy problem
u(x, 0) = ϕ(x) , x R , u( ∞, t) = 0 , t > 0 2 +∞ θ θ uα,β(x, t) = Gα,β(ξ, t) ϕ(x ξ) dξ ∞ Z θ γ θ γ Gα,β(ax , bt) = b Gα,β(ax/b , t) , γ = β/α
() July2008 26/44 Fractional di¤usion equation
Caputo time-fractional derivative
1 t f (m)(τ) α α+1 m dτ, m 1 < α < m Γ(m α) 0 (t τ) D f (t) := m d ( dtmRf (t) , α = m Cauchy problem
u(x, 0) = ϕ(x) , x R , u( ∞, t) = 0 , t > 0 2 +∞ θ θ uα,β(x, t) = Gα,β(ξ, t) ϕ(x ξ) dξ ∞ Z θ γ θ γ Gα,β(ax , bt) = b Gα,β(ax/b , t) , γ = β/α Similarity variable x/tγ
θ γ θ γ Gα,β(x, t) = t Kα,β(x/t ) , γ = β/α
() July2008 26/44 Inverse Laplace transform
∞ n θ θ β z Gα,β (k, t) = Eβ ψα(κ) t , Eβ(z) := ∑ n=0 Γ(β n + 1) h i d +∞ θ 1 ikx θ β Gα,β(x , t) = e Eβ ψα(κ) t dk 2π ∞ Z h i
Fractional di¤usion equation
Solution by Fourier transform for the space variable and the Laplace transform for the time variable
θ θ β θ β 1 ψ (κ)G = s G s α α,β α,β gd s βgd1 G θ = α,β β θ s + ψα(κ) gd
() July2008 27/44 Fractional di¤usion equation
Solution by Fourier transform for the space variable and the Laplace transform for the time variable
θ θ β θ β 1 ψ (κ)G = s G s α α,β α,β gd s βgd1 G θ = α,β β θ s + ψα(κ) Inverse Laplace transformgd
∞ n θ θ β z Gα,β (k, t) = Eβ ψα(κ) t , Eβ(z) := ∑ n=0 Γ(β n + 1) h i d +∞ θ 1 ikx θ β Gα,β(x , t) = e Eβ ψα(κ) t dk 2π ∞ Z h i
() July2008 27/44 Fractional di¤usion equation
Particular cases α = 2 , β = 1 (Standard di¤usion) f g 0 1/2 1 2 G (x, t) = t exp[ x /(4t)] 2,1 2pπ
0 10 a=2 b=1 q=0
•1 10
•2 10
•3 10 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 28/44 Fractional di¤usion equation
0 < α 2 , β = 1 (Space fractional di¤usion) Thef Mittag-Le• er functiong reduces to the exponential function and we θ obtain a characteristic function of the class Lα(x) of Lévy strictly stable densities f g θ θ θ ψα(κ) θ tψα(κ) Lα(κ) = e , Gα,1(κ, t) = e The Green function of the space-fractional di¤usion equation can be interpreted as ab Lévy strictly stable pdfd, evolving in time θ 1/α θ 1/α G (x, t) = t L (x/t ) , ∞ < x < +∞ , t 0 α,1 α Particular cases: α = 1/2 , θ = 1/2 , Lévy-Smirnov 3/2 s 1/2 1/2 x 1/(4x ) e L L (x) = e , x 0 $ 1/2 2pπ α = 1 , θ = 0 , Cauchy
κ 0 1 1 e j j F L (x) = , ∞ < x < +∞ $ 1 π x2 + 1 () July2008 29/44 Fractional di¤usion equation
0 0 10 10 a=0.50 a=0.50 b=1 b=1 q=0 q=•0.50
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 30/44 Fractional di¤usion equation
0 0 10 10 a=1 a=1 b=1 b=1 q=0 q=•0.99
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 31/44 Fractional di¤usion equation
0 0 10 10 a=1.50 a=1.50 b=1 b=1 q=0 q=•0.50
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 32/44 Fractional di¤usion equation
α = 2 , 0 < β < 2 (Time-fractional di¤usion) f g G 0 (κ, t) = E κ2 t β , κ R , t 0 2,β β 2 or with the equivalentd Laplace transform
0 1 β/2 1 x s β/2 G (x, s) = s e j j , ∞ < x < +∞ , (s) > 0 2,β 2 < with solutiong
0 1 β/2 β/2 G (x, t) = t M x /t , ∞ < x < +∞ , t 0 2,β 2 β/2 j j Mβ/2 is a function of Wright type of order β/2 de…ned for any order ν (0, 1) by 2 ∞ ( z)n 1 ∞ ( z)n 1 M (z) = = Γ(νn) sin(πνn) ν ∑ n! Γ[ νn + (1 ν)] π ∑ (n 1)! n=0 n=1
() July2008 33/44 Fractional di¤usion equation
0 0 10 10 a=2 a=2 b=0.25 b=0.50 q=0 q=0
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 34/44 Fractional di¤usion equation
0 0 10 10 a=2 a=2 b=0.75 b=1.25 q=0 q=0
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 35/44 Fractional di¤usion equation
0 0 10 10 a=2 a=2 b=1.50 b=1.75 q=0 q=0
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 36/44 Fractional di¤usion equation
Space-time fractional di¤usion equation. Some examples
0 0 10 10 a=1.50 a=1.50 b=1.50 b=1.50 q=0 q=•0.49
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 37/44 Fractional di¤usion equation
0 0 10 10 a=1.50 a=1.50 b=1.25 b=1.25 q=0 q=•0.50
•1 •1 10 10
•2 •2 10 10
•3 •3 10 10 •5 •4 •3 •2 •1 0 1 2 3 4 5 •5 •4 •3 •2 •1 0 1 2 3 4 5
() July2008 38/44 Physically it describes a nonlinear di¤usion with growing mass and in our fractional generalization it would represent the same phenomenon taking into account memory e¤ects in time and long range correlations in space.
A fractional nonlinear equation. Stochastic solution
A fractional version of the KPP equation, studied by McKean
α 1 β 2 t D u (t, x) = 2 x Dθ u (t, x) + u (t, x) u (t, x) α t D is a Caputo derivative of order α 1 t f (m)(τ)d τ α α+1 m m 1 < α < m Γ(m β) 0 (t τ) t D f (t) = m d ( dtm f (t)R α = m β x Dθ is a Riesz-Feller derivative de…ned through its Fourier symbol β θ x D f (x) (k) = ψ (k) f (x) (k) F θ β F f g n o with ψθ (k) = k β ei(signk)θπ/2. β j j
() July2008 39/44 A fractional nonlinear equation. Stochastic solution
A fractional version of the KPP equation, studied by McKean
α 1 β 2 t D u (t, x) = 2 x Dθ u (t, x) + u (t, x) u (t, x) α t D is a Caputo derivative of order α 1 t f (m)(τ)d τ α α+1 m m 1 < α < m Γ(m β) 0 (t τ) t D f (t) = m d ( dtm f (t)R α = m β x Dθ is a Riesz-Feller derivative de…ned through its Fourier symbol β θ x D f (x) (k) = ψ (k) f (x) (k) F θ β F f g θ n β i(signko)θπ/2 with ψβ (k) = k e . Physically it describesj j a nonlinear di¤usion with growing mass and in our fractional generalization it would represent the same phenomenon taking into account memory e¤ects in time and long range correlations in space.
() July2008 39/44 A fractional nonlinear equation
The …rst step towards a probabilistic formulation is the rewriting as an integral equation.Take the Fourier transform ( ) in space and the Laplace transform ( ) in time F L s s s ^ ^ 1 ^ ^ ∞ sαu (s, k) = sα 1u 0+, k ψθ (k) u (s, k) u (s, k) + dte st u2 2 β Z0 F where ^ ∞ u (t, k) = (u (t, x)) = eikx u (t, x) F Z ∞ ∞ s st u (s, x) = (u (t, x)) = e u (t, x) L Z0 ∂ + This equation holds for 0 < α 1 or for 0 < α 2 with dt u (0 , x) = 0. s ^ Solving for u (s, k) one obtains an integral equation
s α 1 ∞ st ^ s ^ + e 2 u (s, k) = 1 u 0 , k + dt 1 u (t, x) sα θ k 0 sα θ k + 2 ψβ ( ) Z + 2 ψβ ( )F () July2008 40/44 A fractional nonlinear equation
Taking the inverse Fourier and Laplace transforms u (t, x) E 1 + 1 θ (k) tα α ∞ α,1 2 ψβ = Eα,1 ( t ) dy 1 x y u 0, y α ( ) ( ) ∞ F 0 Eα,1 ( t ) 1 Z t α 1 @ α A (t τ) E , (t τ) + dτ α α Z0 1 θ α ∞ Eα,α 1 + ψ (k) (t τ) 1 2 β 2 dy α (x y) u (τ, y) ∞ F 0 Eα,α (t τ) 1 Z @ A ∞ z j Eα,ρ is the generalized Mittag-Le• er function Eα,ρ (z) = ∑j=0 Γ(αj+ρ) t α α 1 α Eα,1 ( t ) + dτ (t τ) Eα,α (t τ) = 1 Z0 () July2008 41/44 A fractional nonlinear equation
We de…ne the following propagation kernel
1 θ α Eα,ρ 1 + 2 ψβ (k) t G β t, x = 1 x α,ρ ( ) α ( ) F 0 E , ( t ) 1 α ρ @ A u (t, x) ∞ β E ( tα) G (t, x y) + = α,1 dy α,1 u 0 , y ∞ Z t α 1 α (t τ) E , (t τ) + dτ α α Z0 ∞ β dyGα,α (t τ, x y)u2 (τ, y) ∞ Z α α 1 α E ,1 ( t ) and (t τ) E , (t τ) = survival probability up to α α α time t and the probability density for the branching at time τ (branching process Bα) () July2008 42/44 A fractional nonlinear equation
The propagation kernels satisfy the conditions to be the Green’sfunctions of stochastic processes in R:
+ + + u(t, x) = Ex u(0 , x + ξ )u(0 , x + ξ ) u(0 , x + ξ ) 1 2 n β β Denote the processes associated to Gα,1 (t, x) and Gα,α (t, x), respectively β β by Πα,1 and Πα,α Theorem: The nonlinear fractional partial di¤erential equation, with 0 < α 1, has a stochastic solution, the coordinates x + ξi in the arguments of the initial condition obtained from the exit values of a propagation and branching process, the branching being ruled by the β β process Bα and the propagation by Πα,1 for the …rst particle and by Πα,α for all the remaining ones. A su¢ cient condition for the existence of the solution is
u(0+, x) 1
() July2008 43/44 A fractional nonlinear equation
() July2008 44/44 Geometric interpretation of fractional integration: shadows on the walls
t α 1 α 1 0It f(t) = Z f(τ)(t τ) − dτ, t 0, Γ(α) − ≥ 0
t α 0It f(t) = Z f(τ) dgt(τ), 0 1 Start g (τ) = tα (t τ)α . t Γ(α + 1) JJ II { − − } J I 69 / 90 Back For t1 = kt, τ1 = kτ (k > 0) we have: Full screen g (τ ) = g (kτ) = kαg (τ). Close t1 1 kt t End 10
8
6 f(t) 4
2
0 0
2 0 4 2 Start 6 4 6 JJ II 8 8 J I 10 10 τ 70 / 90 g (τ) t, t Back 1 α “Live fence” and its shadows: 0It f(t) a 0It f(t), Full screen for α = 0.75, f(t) = t + 0.5 sin(t), 0 t 10. Close ≤ ≤ End 0
1
2
3 0 4 1 2 5 3 6 4 5 Start 7 6 JJ II 8 7 8 J I 9 g (τ) 71 / 90 t 9 t, τ 10 10 Back “Live fence”: basis shape is changing Full screen α Close for 0It f(t), α = 0.75, 0 t 10. ≤ ≤ End 10
9
8
7
6
5 F(t)
4
3
2
Start 1 JJ II 0 0 1 2 3 4 5 6 7 8 9 10 J I G (τ) 72 / 90 t Back Snapshots of the changing “shadow” of changing “fence” for Full screen α 0It f(t), α = 0.75, f(t) = t + 0.5 sin(t), with the time Close interval ∆t = 0.5 between the snapshops. End Right-sided R-L integral
b α 1 α 1 tI0 f(t) = Z f(τ)(τ t) − dτ, t b, Γ(α) − ≤ t
0
1
2 3 0 Start 4 1 2 5 3 JJ II 6 4 5 J I 7 6 73 / 90 8 7 8 9 Back g (τ) t 9 t, τ 10 10 Full screen α Close tI10f(t), α = 0.75, 0 t 10 ≤ ≤ End Riesz potential
b α 1 α 1 0Rb f(t) = Z f(τ) τ t − dτ, 0 t b, Γ(α) | − | ≤ ≤ 0
0
1
2 3 0 Start 4 1 2 5 3 JJ II 6 4 5 J I 7 6 74 / 90 8 7 8 9 Back g (τ) t 9 t, τ 10 10 Full screen α Close 0R10f(t), α = 0.75, 0 t 10 ≤ ≤ End References
- F. Gorenflo and F. Mainardi; Fractional Calculus: Integral and Differential Equations of Fractional Order, in Fractals and Fractional Calculus in Continuum Mechanics, pp. 223-276, A. Carpinteri and F. Mainardi (Eds.), Springer, New York 1997 - F. Mainardi, Y. Luchko and G. Pagnini; The fundamental solution of the space-time fractional diffusion equation, Fractional Calculus and Applied Analysis 4 (2001) 153-192. - I. Podlubny; Geometric and physical interpretation of fractional integration and fractional differentiation, Fractional Calculus and Applied Analysis 5 (2002) 367-386. - V. Pipiras and M. S. Taqqu; Fractional Calculus and its connections to fractional Brownian motion, in Theory and Applications of Long-Range Dependence, pp. 165-201, P. Doukhan, G. Oppenheim and M. S. Taqqu (Eds.) Birkhäuser, Boston 2003. - RVM and L. Vazquez; The dynamical nature of a backlash system with and without fluid friction, Nonlinear Dynamics 47 (2007) 363-366. - F. Cipriano, H. Ouerdiane and RVM; Stochastic solution of a KPP-type nonlinear fractional differential equation, arXiv:0803.4457, Fractional Calculus and Applied Analysis, to appear