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A local Lipschitz existence theorem for the Caputo Volterra equation

Statement

Let $T_{0} > 0$ , $u_{0} \in R$ , $r > 0$ , and $0 < α < 1$ . Suppose $f : [0, T_{0}] \times [u_{0} - r, u_{0} + r] \to R$ is continuous and is Lipschitz in its second variable with constant $L \geq 0$ . Let $M \geq 0$ be a finite bound for $∣ f (t, y) ∣$ on this rectangle. Choose $T \in (0, T_{0}]$ such that $\frac{M T^{α}}{Γ (α + 1)} \leq r .$ Then the Volterra equation $u (t) = u_{0} + I^{α} (f (\cdot, u (\cdot))) (t), 0 \leq t \leq T,$ has exactly one continuous solution in the closed ball ${u \in C ([0, T]) : ∥ u - u_{0} ∥_{\infty} \leq r} .$ Such a $T$ exists because $T^{α} \to 0$ as $T ↓ 0$ . This is a local existence theorem for the integral (mild) solution. It does not claim that this continuous solution is absolutely continuous under only the stated hypotheses.

Facts & Assumptions

Given: $T_{0} > 0$ , $u_{0} \in R$ , $r > 0$ , $0 < α < 1$ , continuous $f$ on the stated rectangle, a Lipschitz constant $L \geq 0$ , a finite bound $M$ for $∣ f ∣$ there, and $T \in (0, T_{0}]$ satisfying the displayed bound.

[F1]

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

[F2]

Fractional integrals of continuous data are Hölder continuous: $I^{α} g$ is continuous and $∥ I^{α} g ∥_{\infty} \leq T^{α} ∥ g ∥_{\infty} / Γ (α + 1)$ .

[F3]

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

[F4]

Weakly singular fractional Grönwall inequality: for every $K \geq 0$ the series $\sum_{n = 0}^{\infty} K^{n} t^{n α} / Γ (n α + 1)$ converges uniformly on $[0, T]$ ; if nonnegative continuous $y$ satisfies $y \leq K I^{α} y$ , then $y \equiv 0$ .

[F5]

Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact and Open cover, subcover, compact metric space, and compact subset of a metric space: $[0, T]$ is a compact metric space and is nonempty since it contains $0$ . $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ : $C ([0, T], R)$ is complete for $d_{\infty} (v, w) = \sup_{t \in [0, T]} ∣ v (t) - w (t) ∣$ , which equals $∥ v - w ∥_{\infty}$ . Banach space and Complete metric space: every Cauchy sequence converges in the space identify this as completeness of its norm metric.

[F6]

The real Gamma function by Euler's integral: $Γ (α + 1) > 0$ .

[F7]

Continuity and derivatives of positive-base real powers: since $α > 0$ , $T^{α} \to 0$ as $T ↓ 0$ .

Proof

1.1givenF6F7choose

By [F6]–[F7], $M T^{α} / Γ (α + 1) \to 0$ as $T ↓ 0$ ; since $T_{0} > 0$ and $r > 0$ , choose $T \in (0, T_{0}]$ for which the displayed bound holds.

2.1givenF1F2step 1.1construct

Let $B = {u \in C ([0, T]) : ∥ u - u_{0} ∥_{\infty} \leq r}$ and define $Φ u = u_{0} + I^{α} (f (\cdot, u (\cdot)))$ . For $u \in B$ , $s \mapsto (s, u (s))$ and then $s \mapsto f (s, u (s))$ are continuous, with $∣ f (s, u (s)) ∣ \leq M$ ; [F1]–[F2] therefore make $Φ u$ continuous and give $∥ Φ u - u_{0} ∥_{\infty} \leq M T^{α} / Γ (α + 1) \leq r$ . Thus $Φ$ maps $B$ into itself.

3.1step 2.1givenF3construct

Set $u^{(0)} (t) = u_{0}$ and recursively $u^{(n + 1)} = Φ u^{(n)}$ ; step 2.1 keeps every iterate in $B$ . For $n \geq 1$ put $d_{n} (t) = ∣ u^{(n)} (t) - u^{(n - 1)} (t) ∣$ . The bound on $f$ and [F3] give $d_{1} (t) \leq M I^{α} 1 (t) = M t^{α} / Γ (α + 1)$ .

4.1step 3.1givenF1F3induction

For $n \geq 1$ , Lipschitz continuity and the positive kernel [F1] give $d_{n + 1} \leq L I^{α} d_{n}$ . Starting with step 3.1, induction and the pointwise semigroup identity [F3] yield $d_{n} (t) \leq M L^{n - 1} I^{n α} 1 (t) = M L^{n - 1} t^{n α} / Γ (n α + 1)$ for every $n \geq 1$ , with the coefficient $L^{0}$ at $n = 1$ defined to be $1$ .

5.1step 4.1F4F5algebra

Put $K = 1 + L$ ; then $L^{n - 1} \leq K^{n}$ for $n \geq 1$ , so step 4.1 and the uniform series bound [F4] show $\sum_{n = 1}^{\infty} ∥ d_{n} ∥_{\infty} < \infty$ . The iterates are uniformly Cauchy; completeness [F5] gives a uniform limit $u \in C ([0, T])$ , and since every iterate lies in $B$ , the norm limit also lies in $B$ .

6.1givenF2step 5.1algebra

Lipschitz continuity and [F2] give $∥ Φ u^{(n)} - Φ u ∥_{\infty} \leq (L T^{α} / Γ (α + 1)) ∥ u^{(n)} - u ∥_{\infty} \to 0$ . Since $u^{(n + 1)} = Φ u^{(n)}$ and $u^{(n + 1)} \to u$ , it follows that $u = Φ u$ , proving existence in $B$ .

7.1givenF1F4algebra∎

If $u, v \in B$ are two solutions, $y = ∣ u - v ∣$ is continuous and nonnegative; their equations, Lipschitz continuity, and the positive kernel [F1] give $y (t) \leq L I^{α} y (t)$ for every $t$ . Applying [F4] with $A = 0$ gives $y \equiv 0$ , hence $u = v$ .

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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 8 (standard reference, not scraped)