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Weakly singular fractional Grönwall inequality

Statement

Let $T > 0$ , $0 < α < 1$ , $A, L \geq 0$ , and let $y \in C ([0, T])$ be nonnegative. If $y (t) \leq A + L I^{α} y (t) (0 \leq t \leq T),$ then for every $t \in [0, T]$ , $y (t) \leq A \sum_{k = 0}^{\infty} \frac{L^{k} t^{k α}}{Γ (k α + 1)} .$ The $k = 0$ term is $1$ , including when $L = 0$ , and the series converges uniformly on $[0, T]$ . In particular, if $A = 0$ then $y \equiv 0$ .

Facts & Assumptions

Given: $T > 0$ , $0 < α < 1$ , $A, L \geq 0$ , $y \in C ([0, T])$ , $y \geq 0$ , and the stated pointwise inequality.

[F1]

Fractional integral and Caputo derivative: for continuous $g$ , $I^{β} g (t) = Γ (β)^{- 1} \int_{0}^{t} (t - s)^{β - 1} g (s) d s$ ; its kernel is nonnegative for $β > 0$ .

[F2]

Semigroup law for fractional integrals: for continuous $g$ , $I^{a} (I^{b} g) = I^{a + b} g$ at every $t$ , and $I^{c} 1 (t) = t^{c} / Γ (c + 1)$ for $a, b, c > 0$ .

[F3]

The real Gamma function by Euler's integral and The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ : for $x > 0$ , $Γ (x + 1) = \int_{0}^{\infty} e^{- s} s^{x} d s > 0$ , and $Γ (1) = 1$ .

[F4]

Real powers for positive bases, with the zero-base positive-exponent convention, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , Continuity and derivatives of positive-base real powers, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, and The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : for $α > 0$ , $x \mapsto x^{α}$ is strictly increasing on $(0, \infty)$ and $(x^{r})^{s} = x^{r s}$ for positive $x$ .

[F5]

The Cauchy-sequence reals are Archimedean: for every real $R$ there is a natural number $k \geq 1$ with $R < k$ .

[F6]

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges and If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ : $\sum_{n = 1}^{\infty} 2^{- n}$ converges, and a nonnegative series dominated termwise by a convergent nonnegative series converges.

Proof

1.1givenF3F4F5F6algebra

For $x > 0$ , restricting the Euler integral [F3] to $[x, x + 1]$ gives $Γ (x + 1) \geq e^{- (x + 1)} x^{x}$ , since the interval has length one and $e^{- s} s^{x} \geq e^{- (x + 1)} x^{x}$ there. With $x = k α$ and $k \geq 1$ , this yields $L^{k} T^{k α} / Γ (k α + 1) \leq e (B / k^{α})^{k}$ , where $B : = L T^{α} e^{α} / α^{α}$ . If $L = 0$ , every term with $k \geq 1$ vanishes. Otherwise $B > 0$ ; apply [F5] to $R = (2 B)^{1 / α}$ to choose $k_{0} > R$ . For $k \geq k_{0}$ , [F4] gives $k^{α} \geq k_{0}^{α} > (2 B)^{(1 / α) α} = 2 B$ , so $(B / k^{α})^{k} < 2^{- k}$ . The comparison test [F6] gives convergence of the scalar majorant tail. For any $p > q \geq k_{0}$ and any $t \in [0, T]$ , the absolute value of the series segment from $q$ to $p$ is at most $e \sum_{k = q}^{p} 2^{- k}$ . This bound tends to zero independently of $t$ , so the series is uniformly Cauchy on $[0, T]$ and hence converges uniformly there.

1.2givenF1F2induction

The hypothesis is the iteration formula for $m = 1$ . If it holds for $m \geq 1$ , positivity of the kernel [F1] allows applying $L^{m} I^{m α}$ to $y \leq A + L I^{α} y$ in its remainder, giving $L^{m} I^{m α} y (t) \leq A L^{m} I^{m α} 1 (t) + L^{m + 1} I^{m α} (I^{α} y) (t) = A L^{m} t^{m α} / Γ (m α + 1) + L^{m + 1} I^{(m + 1) α} y (t)$ by [F2]. Substitution proves by induction that $y (t) \leq A \sum_{k = 0}^{m - 1} L^{k} t^{k α} / Γ (k α + 1) + L^{m} I^{m α} y (t)$ for every $m \geq 1$ .

2.1step 1.1step 1.2F1F2algebra∎

By [F1]–[F2], $0 \leq L^{m} I^{m α} y (t) \leq ∥ y ∥_{\infty} L^{m} T^{m α} / Γ (m α + 1)$ . The last factor tends to zero as the $m$ th term of the convergent series in step 1.1. Letting $m \to \infty$ in step 1.2 proves the claimed bound; if $A = 0$ , then $0 \leq y (t) \leq 0$ for every $t$ , so $y \equiv 0$ .

Depends on

Used by

A local Lipschitz existence theorem for the Caputo Volterra equation Theorem

Dependency tree · two levels

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Sources

Jeffrey R. L. Webb, Initial Value Problems for Caputo Fractional Equations with Singular Nonlinearities, Electronic Journal of Differential Equations 2019(117), §1 and Theorem 8.2 (standard reference, not scraped)