Picard iterates and error certificates

Try

A differential equation asks for an entire function, not just its next plotted point. For y′=1+y² with y(0)=0, the equivalent integral equation is y(t)=∫₀ᵗ(1+y(s)²)ds. Start with the continuous guess u₀(t)=0. Replace the unknown function inside the integral by that guess, integrate exactly, and repeat. This operation is called Picard iteration.

The first replacement gives u₁(t)=t. The next gives u₂(t)=t+t³/3. Before calculating the third, predict whether merely appending t⁵/5 could work. It cannot: the whole previous polynomial must be squared. In particular, its cross term matters:

(t+t³/3)²=t²+(2/3)t⁴+t⁶/9.

Integrating 1 plus that square from 0 to t gives u₃(t)=t+t³/3+2t⁵/15+t⁷/63. This is exact polynomial algebra. It does not yet prove that the iterates converge, or that a displayed decimal approximates the solution to a specified tolerance.

Predict

Choose the symmetric interval [−1/4,1/4] and the closed function ball X of continuous functions satisfying ||u||∞≤1. The supremum norm ||u||∞ is the largest absolute value across the whole interval. For u in this ball, |1+u²|≤2. Therefore ||Tu||∞≤(1/4)·2=1/2≤1: the operator returns a function inside the ball.

For u,v in X, |u²−v²|=|u−v||u+v|≤2||u−v||∞. Integration over an interval of length at most 1/4 gives ||Tu−Tv||∞≤(1/2)||u−v||∞. The contraction factor is q=1/2. The preceding construction supplies a unique fixed point y in X and convergence of these exact iterates to y.

Predict which information gives the better guarantee for u₂: the size of the first change, ||u₁−u₀||=1/4, or its actual last change, ||u₂−u₁||=1/192. Both are valid ingredients, but they enter different bounds. An upper bound is not necessarily the actual error; a weaker bound for a later iterate can exceed a sharper bound for an earlier iterate.

Experiment

Use the explorer with b=1 and h=1/4 to match the separate worked example below. Reveal one iteration at a time. Before revealing the next polynomial, square the current expression and integrate its coefficients on paper. Then change b while retaining h. The iteration formula does not depend on the chosen rectangle, but the proof constants do.

Build a function, then check its certificate

For y′=1+y² and y(0)=0, start with u₀(t)=0 and repeatedly integrate 1+uₙ(t)² from 0 to t. Each step constructs a whole polynomial on [−h,h]; it is not an Euler time step.

Rectangle radius b0.512 Time half-width h0.1250.250.50.75 Compute next Picard iterate Restart iterations Reset

Iteration n=0. M=1+b²=2, L=2b=2, q=hL=0.5.

• Self-map: hM=0.5 ≤ b=1: passes.
• Strict contraction: q=0.5 < 1: passes.

Both hypotheses pass. The analytic bound is ‖uₙ−y‖∞ ≤ qⁿh/(1−q) ≈ 0.5. Here ‖u₁−u₀‖∞=h; n=0 is allowed in this a priori estimate.

Dashed outline: proposed ball ‖u‖∞≤b on [−h,h]. Solid curve: u0. Polynomial values are computed with floating-point coefficients, not exact rational arithmetic. A smooth-looking plot does not verify either theorem hypothesis.

For these odd polynomials with nonnegative coefficients, the maximum absolute value is attained at an endpoint: ‖u0‖∞ ≈ 0. This uses the polynomial structure, not a maximum over sampled pixels.

Inspect polynomial coefficients and endpoint values

Coefficient cₖ multiplies tᵏ; omitted coefficients are zero. Values are rounded.

Computed nonzero coefficients

Power k	Coefficient cₖ
0	0 (zero polynomial)

u0(−h)≈0; u0(0)=0; u0(h)≈0.

Predict u₁ and u₂ before computing them. Compare b=1 and b=0.5 at h=0.25: which valid certificate is sharper? Then try b=1, h=0.5, where the self-map condition holds but q=1. Controls are temporary and do not award completion.

Approximation	First-change bound	Actual last-change bound
u₁=t	1/4	1/4
u₂=t+t³/3	1/8	1/192
u₃=t+t³/3+2t⁵/15+t⁷/63	1/16	1/7680+1/1032192

These entries bound uniform error relative to the fixed point, not merely error at t=1/4. The polynomial differences are odd polynomials with positive coefficients, so their absolute maxima on this symmetric interval occur at its endpoints. This justifies using endpoint arithmetic here; arbitrary functions do not have that property. Decimal displays and finite plot samples remain numerical illustrations of the exact reasoning.

Understand

Here is the general error mechanism. Let T contract distances by a factor q with 0<q<1 on a nonempty complete metric space, let uₙ=Tⁿu₀, and write D=||u₁−u₀||. Repeated contraction gives ||uⱼ₊₁−uⱼ||≤qʲD. For m greater than n, the triangle inequality bounds ||uₘ−uₙ|| by the finite sum of qʲD from j=n to m−1. Let m tend to infinity, using convergence to the fixed point y. The geometric tail yields

||y−uₙ||≤qⁿD/(1−q).

This is an a priori estimate: once D and q are known, it predicts a bound before computing the later iterates. It applies also at n=0 when q is positive. If the latest actual change Δₙ=||uₙ−uₙ₋₁|| is known for n≥1, start the same argument at that change instead. The future changes are bounded by qΔₙ,q²Δₙ,…, giving

||y−uₙ||≤qΔₙ/(1−q).

This is a posteriori: it uses information from the computed iterates. The extra q matters because the error of the current iterate involves future changes, not the already completed change. If q=0, T sends every input to one common point; that point is fixed. Thus u₁=y and all later errors vanish. At n=0 the bound is simply D, avoiding an undefined-looking 0⁰ expression.

For the Riccati equation on [−h,h] with ||u||∞≤b, our sufficient estimates are h(1+b²)≤b for self-mapping and q=2bh<1 for contraction. Both are required before invoking these certificates. On h=1/4, choosing b=1/2 gives self-map estimate 5/16≤1/2 and q=1/4. This improves the a priori error to 4^(−n)/3 even though every exact iterate is unchanged. It reaches 1/1000 at n=5, whereas b=1 first reaches that threshold at n=9.

Choosing b=2 on the same interval makes q=1. That does not disprove existence: the smaller valid rectangles have already proved it. It means this particular strict-contraction estimate is unavailable. Likewise, an iterate that happens to stay in a plotted box is not a proof that T maps every function in that box back into it.

A related certificate uses an arbitrary candidate v in X rather than a known iterate. If ||Tv−v||≤ρ uniformly, then

||v−y||≤||v−Tv||+||Tv−Ty||≤ρ+q||v−y||.

Hence ||v−y||≤ρ/(1−q). Only the next candidate Tv gains the factor q. For v=t in our worked rectangle, ρ=1/192, so the bound for v is 1/96 and that for Tv=u₂ is 1/192. Mixing a next residual with a previous-change formula would attach the wrong bound to the wrong function.

Apply

Change the equation to y′=2+y, y(0)=1 and use [−1/5,1/5] with the ball ||u−1||∞≤1. Find the uniform bound on |2+u|, prove self-mapping, and derive the contraction factor. Then compute three exact iterates starting at the constant 1. Check that the initial value remains 1 at every iteration. The practice supplies a complete review of the resulting 3/500 a priori certificate for u₃.

Audit a numerical claim too. Suppose a curve agrees with u₂ at finitely many nodes, including zero. Adding K times the product of (t−node) over those nodes preserves every displayed value and the initial condition. At any other point the product is nonzero, and choosing K large creates an arbitrarily large discrepancy. Finite agreement cannot establish a supremum bound or even membership in the invariant ball. A rigorous numerical certificate needs an error-controlled representation and uniform residual bound, not just more attractive plotting.

Continue

Keep three objects separate in your record: the exact integral operator, its exact function iterates, and the rounded or sampled representation shown by a computer. State the interval and rectangle, check both operator inequalities, identify which iterate a bound concerns, and include any numerical evaluation error before claiming a certified tolerance. These local certificates establish a solution in the stated function ball; they do not by themselves prove global continuation beyond the interval or uniqueness among functions outside the hypotheses.