Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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A first-order initial value problem is equivalent to its Volterra integral equation

Statement

Let n1, let DR×Rn be open, let F:DRn be continuous, let (t0,x0)D, let IR be order-convex with at least two elements, and let x:IRn be continuous with t0I and (t,x(t))D. A curve solves the IVP if and only if it satisfies the associated Volterra integral equation. Explicitly, the equation is

x(t)=x0+t0tF(s,x(s))ds(tI).

The integral is oriented, so the assertion applies on either side of t0.

Facts & Assumptions

[L1]

If a differentiable f:[a,b]Rm has integrable derivative, then abf=f(b)f(a) (If f:[a,b]Rm is differentiable with integrable f then abf=f(b)f(a); and a bounded derivative makes f Lipschitz).

[L2]

If I is order-convex with at least two elements, g:IR is continuous, and t0I, then tt0tg is a primitive of g on I (Every continuous function on an interval has a primitive; two primitives differ by a constant; and abf=G(b)G(a) for any primitive G).

[L3]

Every continuous real function on a compact interval is Riemann integrable (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

technique · direct
1.1

For the forward direction, the identity is immediate at t=t0; for tt0, each component of x=F(,x) is continuous and hence integrable by [L3], so [L1] on the closed interval between t0 and t, followed by orientation when t<t0, gives x(t)x0=t0tF(s,x(s))ds.

givenL1L3
2.1

For the reverse direction, [L2] applied componentwise differentiates the displayed integral equation and gives x(t)=F(t,x(t)); at t=t0 the oriented integral is 0, so x(t0)=x0.

givenL2algebra

Depends on

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