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Caputo and Volterra formulations are equivalent for absolutely continuous solutions

Statement

Assume the Axioms of Countable Choice and Dependent Choice. Let $T > 0$ , $0 < α < 1$ , let $f \in C ([0, T])$ , and let $u \in A C ([0, T])$ . With $u_{0} = u (0)$ and the Caputo derivative of Fractional integral and Caputo derivative,

$D_{C}^{α} u = f almost everywhere on (0, T)$

if and only if

$u (t) = u_{0} + I^{α} f (t) for every t \in [0, T] .$

The absolute-continuity hypothesis is part of the differential-equation solution class. The equivalence does not assert that an arbitrary continuous solution of the Volterra equation is absolutely continuous.

Facts & Assumptions

Given: The Axioms of Countable Choice and Dependent Choice, $T > 0$ , $0 < α < 1$ , $f \in C ([0, T])$ , and $u \in A C ([0, T])$ .

[A1]

The Axiom of Countable Choice: every sequence of nonempty sets admits a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[A2]

The Axiom of Dependent Choice: every total relation on a nonempty set admits a chain from a specified starting point (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[F1]

Fractional integral and Caputo derivative: for $u \in A C$ , $D_{C}^{α} u = I^{1 - α} u^{'}$ almost everywhere.

[F2]

Semigroup law for fractional integrals: for positive $a, b$ and $g \in L^{1}$ , $I^{a} I^{b} g = I^{a + b} g$ almost everywhere; $I^{a} g \in L^{1}$ .

[F3]

Fractional integrals of continuous data are Hölder continuous: if $g$ is continuous and $0 < a < 1$ , then $I^{a} g$ is continuous.

[F4]

Fundamental theorem of calculus for absolutely continuous functions: under [A1] and [A2], an absolutely continuous $u$ has $u^{'} \in L^{1}$ and $u (t) - u (0) = \int_{0}^{t} u^{'} (s) d s$ for every $t$ .

[F5]

The indefinite integral of an $L^{1}$ function is differentiable almost everywhere: under [A1], the indefinite integral of an $L^{1}$ function has derivative equal to its integrand almost everywhere.

Proof

1.1givenA1A2F1F2F3F4algebra

Suppose $D_{C}^{α} u = f$ almost everywhere and put $v = u^{'} \in L^{1}$ by [F4]. Then $u - u_{0} = I^{1} v$ . Applying $I^{α}$ to the Caputo identity and using [F1] and [F2] gives $u - u_{0} = I^{α} f$ almost everywhere. Both sides are continuous: $u$ is absolutely continuous and [F3] applies to $f$ . If their continuous difference were nonzero at some point, continuity would make its absolute value bounded below by a positive constant on a relative interval of positive length (one-sided at an endpoint), contradicting equality almost everywhere. Thus the Volterra identity holds for every $t \in [0, T]$ .

2.1givenA1F1F2F4F5step 1.1algebra∎

Conversely, suppose $u - u_{0} = I^{α} f$ . Applying $I^{1 - α}$ and using [F2] gives $I^{1 - α} (u - u_{0}) = I^{1} f$ . Since $u - u_{0} = I^{1} v$ by [F4], [F2] also gives $I^{1 - α} (u - u_{0}) = I^{1} (I^{1 - α} v) = I^{1} D_{C}^{α} u$ . Both integrands belong to $L^{1}$ by [F1], [F2], and [F4], so their indefinite integrals are continuous. These primitives agree almost everywhere by the preceding identities, hence everywhere by the continuity argument in step 1.1. Applying [F5] gives $f = D_{C}^{α} u$ almost everywhere.

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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 4–5 (standard reference, not scraped)