COMPLETELY MONOTONE FUNCTIONS IN THE STUDY OF A CLASS OF FRACTIONAL - - PowerPoint PPT Presentation

COMPLETELY MONOTONE FUNCTIONS IN THE STUDY OF A CLASS OF FRACTIONAL EVOLUTION EQUATIONS Emilia Bazhlekova Joint Seminar of Analysis, Geometry and Topology Department Institute of Mathematics and Informatics Bulgarian Academy of Sciences

Completely monotone functions (CMF)

Bulgarian contributions: N. Obreshkov, Y. Tagamlitski, Bl. Sendov, H. Sendov A function f : (0, ∞) → R is called completely monotone if it is of class C∞ and (−1)nf (n)(t) ≥ 0, for all t > 0, n = 0, 1, ... Elementary examples: e−λt; t−1; (λ + µt)−ν; ln

• b + µt−1

; ef(t), f ∈ CMF; where λ, µ, ν > 0, b ≥ 1. Bernstein’s theorem: f(t) ∈ CMF iff f(t) = ∞ e−tx dg(x), where g(x) is nondecreasing and the integral converges for 0 < t < ∞.

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Bernstein functions (BF) and some useful properties

A C∞ function f : (0, ∞) → R is called a Bernstein function if f(t) ≥ 0 and f ′(t) ∈ CMF. Proposition: (a) The class CMF is closed under pointwise addition and multiplication; The class BF is closed under pointwise addition, but, in general not under multiplication; (b) If f ∈ CMF and ϕ ∈ BF, then the composite function f(ϕ) ∈ CMF; (c) If f ∈ BF, then f(t)/t ∈ CMF; (d) Let f ∈ L1

loc(R+) be a nonnegative and nonincreasing function, such that

limt→+∞ f(t) = 0. Then ϕ(s) = s f(s) ∈ BF; (e) If f ∈ L1

loc(R+) and f ∈ CMF, then

f(s) admits analytic extension to the sector | arg s| < π and | arg f(s)| ≤ | arg s|, | arg s| < π.

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The operators of fractional integration and differentiation

Jα

t - the Riemann-Liouville fractional integral of order α > 0:

Jα

t f(t) :=

1 Γ(α) t (t − τ)α−1f(τ) dτ, α > 0, where Γ(·) is the Gamma function. Dα

t - the Riemann-Liouville fractional derivative CDα t - the Caputo fractional derivative

D1

t = CD1 t = d/dt; CDα t = J1−α t

D1

t,

Dα

t = D1 tJ1−α t

, α ∈ (0, 1).

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Mittag-Leffler function

Fractional relaxation equation (λ > 0, 0 < α ≤ 1):

CDα t u(t) + λu(t) = f(t),

t > 0, u(0) = c0. The solution is given by: u(t) = c0Eα(−λtα) + t τ α−1Eα,α(−λτ α)f(t − τ) dτ. Mittag-Leffler function (α, β ∈ R, α > 0): Eα,β(−t) =

∞

• k=0

(−t)k Γ(αk + β), Eα(−t) = Eα,1(−t). E1(−t) = e−t ∈ CMF Eα(−t) ∈ CMF, iff 0 < α < 1 (Pollard, 1948) Eα,β(−t) ∈ CMF, iff 0 ≤ α ≤ 1, α ≤ β (Schneider, 1996; Miller, 1999)

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Plots of Eα(−tα) for different values of α

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Plots of tα−1Eα,α(−tα) for different values of α

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Fractional evolution equation of distributed order

Two alternative forms: 1 µ(β)CDβ

t u(t) dβ = Au(t),

t > 0, (1) and u′(t) = 1 µ(β)Dβ

t Au(t) dβ,

t > 0, (2) A - closed linear unbounded operator densely defined in a Banach space X Initial condition: u(0) = a ∈ X Reference: E. Bazhlekova, Completely monotone functions and some classes of fractional evolution equations, preprint, 2015, arXiv:1502.04647

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Two cases for the weight function µ:

• discrete distribution

µ(β) = δ(β − α) +

m

• j=1

bjδ(β − αj), (3) where 1 > α > α1... > αm > 0, bj > 0, j = 1, ..., m, m ≥ 0, and δ is the Dirac delta function;

• continuous distribution

µ ∈ C[0, 1], µ(β) ≥ 0, β ∈ [0, 1], (4) and µ(β) = 0 on a set of a positive measure.

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Discrete distribution:

Multi-term time-fractional equations in the Caputo sense

CDα t u(t) + m

• j=1

bj

CD αj t u(t) = Au(t),

t > 0, (5) and in the Riemann-Liouville sense u′(t) = Dα

t Au(t) + m

• j=1

bj D

αj t Au(t),

t > 0 (6) If m = 0 (single-term equations): problem (5) is equivalent to (6) with α replaced by 1 − α. All problems are generalizations of the classical abstract Cauchy problem u′(t) = Au(t), t > 0; u(0) = a ∈ X. (7)

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Solution u(t) of (5) with A = −1 for: m = 1, α = 0.75, α1 = 0.25, m = 0, α = 0.25 m = 0, α = 0.75.

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Solution u(t) of (5) with A = −1 for: m = 2, α = 0.75, α1 = 0.5, α2 = 0.25 m = 1, α = 0.75, α1 = 0.25, m = 0, α = 0.25 m = 0, α = 0.75.

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Unified approach to the four problems

Rewrite problems (1) and (2) as an abstract Volterra integral equation u(t) = a + t k(t − τ)Au(τ) dτ, t ≥ 0; a ∈ X, where

• k1(s) = (h(s))−1 ,

k2(s) = h(s)/s, In the continuous distribution case: h(s) = 1 µ(β)sβ dβ. In the discrete distribution case: h(s) = sα +

m

• j=1

bjsαj. Define gi(s) = 1/ ki(s), i = 1, 2.

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Particular cases

In the single-term case: k1(t) = tα−1 Γ(α), k2(t) = t−α Γ(1 − α), g1(s) = sα, g2(s) = s1−α, In the double-term case: k1(t) = tα−1Eα−α1,α(−b1tα−α1), k2(t) = t−α Γ(1 − α) + b1 t−α1 Γ(1 − α1), g1(s) = sα + b1sα1, g2(s) = s sα + b1sα1= s k1(s)!!! In the case of continuous distribution in its simplest form: µ(β) ≡ 1. g1(s) = s − 1 log s , g2(s) = s log s s − 1 .

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Properties of the kernels

• Theorem. Let µ(β) be either of the form (3) or of the form (4) with the additional

assumptions µ ∈ C3[0, 1], µ(1) = 0, and µ(0) = 0 or µ(β) = aβν as β → 0, where a, ν > 0. Then for i = 1, 2,: (a) ki ∈ L1

loc(R+) and limt→+∞ ki(t) = 0;

(b) ki(t) ∈ CMF for t > 0; (c) k1 ∗ k2 ≡ 1; (d) gi(s) ∈ BF for s > 0; (e) gi(s)/s ∈ CMF for s > 0; (f) gi(s) admits analytic extension to the sector | arg s| < π and | arg gi(s)| ≤ | arg s|, | arg s| < π. In the discrete distribution case a stronger inequality holds: | arg gi(s)| ≤ α| arg s|, | arg s| < π.

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The classical abstract Cauchy problem: u′(t) = Au(t), t > 0; u(0) = a ∈ X. Main result: Assume that the classical Cauchy problem is well-posed with solution u(t) satisfying u(t) ≤ Ma, t ≥ 0. Then any of the problems 1 µ(β)CDβ

t u(t) dβ = Au(t),

t > 0, u(0) = a ∈ X, u′(t) = 1 µ(β)Dβ

t Au(t) dβ,

t > 0, u(0) = a ∈ X is well-posed with solution satisfying the same estimate.

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The classical abstract Cauchy problem:

u′(t) = Au(t), t > 0; u(0) = a ∈ X. T(t) - solution operator (defined by T(t)a = u(t), t ≥ 0); R(s, A) - resolvent operator of A: R(s, A) = (s − A)−1 = ∞ e−stT(t) dt, s > 0, The Hille-Yosida theorem states that the classical Cauchy problem is well-posed with solution operator T(t) such that T(t) ≤ M, t ≥ 0 iff R(s, A) is well defined for s ∈ (0, ∞) and R(s, A)n ≤ M/sn, s > 0, n ∈ N.

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Abstract Volterra integral equation

u(t) = a + t k(t − τ)Au(τ) dτ, t ≥ 0; a ∈ X, The Laplace transform of the solution operator S(t) H(s) = ∞ e−stS(t) dt, s > 0 is given by H(s) = g(s) s R(g(s), A), g(s) = 1/ k(s). The Generation Theorem (Pruss, 1993) states that the integral equation is well- posed with solution operator S(t) satisfying S(t) ≤ M, t ≥ 0, iff H(n)(s) ≤ M n! sn+1, for all s > 0, n ∈ N0.

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Main result

Theorem. Suppose that the classical Cauchy problem is well-posed with solution u(t) satisfying u(t) ≤ Ma, t ≥ 0. Then problems (1) and (2) are well-posed and their solutions satisfy the same estimate. Proof: We know R(s, A)n ≤ M/sn, s > 0, n ∈ N. We have to prove H(n)(s) ≤ M n! sn+1, for all s > 0, n ∈ N0, where H(s) = g(s) s R(g(s), A), and g(s) = 1/ k(s), R(s, A) = (s − A)−1.

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By the Leibniz rule: H(n)(s) =

n

• k=0
• n

k g(s) s (n−k) w(k)(s), w(s) = R(g(s), A). (8) Formula for the k-th derivative of a composite function (P.Todorov, Pacific J. Math., 1981): w(k)(s) =

k

• p=1

ak,p(s)(−1)pp!(R(g(s), A))p+1, (9) where the functions ak,p(s) are defined by ak+1,p(s) = ak,p−1(s)g′(s) + a′

k,p(s),

1 ≤ p ≤ k + 1, k ≥ 1, (10) ak,0 = ak,k+1 ≡ 0, a1,1(s) = g′(s). g(s) ∈ BF ⇒ (−1)k+pak,p(s) ∈ CMF. (11) Proof: by induction.

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So far: (−1)nH(n)(s) =

n

• k=0

k

• p=1

bn,k,p(s)(R(g(s), A))p+1 (12) where bn,k,p(s) = (−1)n+p

• n

k g(s) s (n−k) ak,p(s)p! Positivity? (−1)k+pak,p(s) ≥ 0, g(s) ∈ BF ⇒ g(s)/s ∈ CMF, s > 0. (13) ⇒ bn,k,p(s) = (−1)n+p

• n

k g(s) s (n−k) ak,p(s)p! =

• n

k

• (−1)n−k

g(s) s (n−k) (−1)k+pak,p(s)p! ≥ 0 (−1)nH(n)(s) =

n

• k=0

k

• p=1

bn,k,p(s)(R(g(s), A))p+1

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⇒ H(n)(s) ≤

n

• k=0

k

• p=1

bn,k,p(s)(R(g(s), A))p+1 ≤ M

n

• k=0

k

• p=1

bn,k,p(s)((g(s))−(p+1) = M(−1)n(s−1)(n) = Mn!s−(n+1), s > 0. where we have used that for A ≡ 0: (−1)n(s−1)(n) =

n

• k=0

k

• p=1

bn,k,p(s)(g(s))−(p+1). Therefore, the conditions of the Generation Theorem are satisfied and the problems are well-posed with bounded solution operators S(t), satisfying S(t) ≤ M, t ≥ 0.

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Subordination formula

T(t) - the solution operator of the classical Cauchy problem. Under the assumptions of the previous theorem, the solution operator S(t) of problem (1), resp. (2), satisfies the subordination identity S(t) = ∞ ϕ(t, τ)T(τ) dτ, t > 0, (14) with function ϕ(t, τ) defined by ϕ(t, τ) = 1 2πi γ+i∞

γ−i∞

est−τg(s) g(s) s ds, γ, t, τ > 0, (15) The function ϕ(t, τ) is a probability density function, i.e. it satisfies the properties ϕ(t, τ) ≥ 0, ∞ ϕ(t, τ) dτ = 1. (16) Hint: take function ϕ(t, τ) such that Lt{ϕ}(s, τ) = g(s)

s e−τg(s),

s, τ > 0.

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