Picard’s operator and invariant function balls

Try

Consider y′=1+y² with y(0)=0. Instead of guessing a solution formula, start with the entire constant curve u₀(t)=0. Use its values to propose a new curve by integration: u₁(t)=∫₀ᵗ(1+u₀(s)²)ds=t. Repeat with that curve: u₂(t)=t+t³/3. These are Picard integral iterates, not Euler steps advancing along a time grid.

Predict whether repeatedly producing curves is already a proof that a solution exists. It is not. We must keep every curve inside the region where our estimates apply, make them converge in a complete space, and prove the limiting curve satisfies the differential equation.

For a concrete safe region choose |t|≤1/4 and |y|≤1. Throughout the larger rectangle |t|≤1, |y|≤1, the rate has magnitude at most 2. Also |(1+u²)−(1+v²)|=|u−v||u+v|≤2|u−v|. These two bounds do different jobs.

Here h is the chosen time half-width, a is the available time half-width, b is the state radius, M bounds the rate, and L bounds its sensitivity to state. The symbols do not denote interchangeable error estimates.

Predict

Increase h to 1/2 while retaining the same state radius and rate bounds. The self-map inequality still holds, but the contraction factor becomes 1. Predict the correct conclusion: this particular strict-contraction certificate fails. It does not prove that the equation has no solution on that larger interval.

Now increase h to 1. The self-map estimate fails too. This is not just an inconclusive bound: the allowed constant input u(t)=1 produces Tu(t)=2t, which leaves the radius-1 ball. A proof must specify the function space on which its map acts, not merely announce that continuous functions form a complete space.

Experiment

Draw the rectangle |t|≤1/4, |y|≤1. Plot u₀=0, u₁=t and u₂=t+t³/3 at the following exact inputs. Join the points according to the supplied formulas, rather than assuming a sampled polyline is the function itself.

The values illustrate the construction but cannot certify every intermediate point or every later iterate. The invariant-ball proof below gives that infinite guarantee. As a manual change, use y′=t+y² and y(1)=1 with |t−1|≤1/2 and |y−1|≤1/2. Compute M=15/4 and L=3. A choice h=1/10 satisfies all conditions. Its first iterate is 1+(t²−1)/2+(t−1). At h=1/4, contraction is strict but the self-map bound fails. Explain why checking only one inequality would miss this problem.

Understand

Let F be continuous on the closed rectangle R=[t₀−a,t₀+a]×[y₀−b,y₀+b], where a,b>0. Assume |F(t,y)|≤M and

|F(t,u)−F(t,v)|≤L|u−v|

for all inputs in the same time slice of R, with finite M,L≥0. Continuity on the compact rectangle supplies some finite M. A bounded continuous partial derivative with respect to y is one sufficient way to supply L, by the mean value theorem; Lipschitz control may also be proved directly, as in the worked difference-of-squares estimate.

Choose h>0 satisfying h≤a, hM≤b and hL<1. Such a choice exists without dividing by zero: for example take a positive h no larger than min(a,b/(M+1),1/(2(L+1))). Let I=[t₀−h,t₀+h] and let X consist of continuous real functions u on I with ||u−y₀||∞≤b. This is a closed ball in C(I), centered at the constant function y₀.

X is nonempty because it contains that constant function. To prove completeness, take a supremum-distance Cauchy sequence in X. The uniform Cauchy criterion supplies a uniform limit u; the uniform-limit theorem makes u continuous. Taking limits in |uₙ(t)−y₀|≤b for each t preserves that bound, so u belongs to X. The closed inequality is essential. The open ball excludes its possible limit: constants y₀+b(1−1/n) converge to y₀+b.

Define the operator

Tu(t)=y₀+∫ from t₀ to t of F(s,u(s)) ds.

For u∈X, every pair (s,u(s)) lies in R, so the integrand is defined and continuous. Its integral is a continuous function of t. For either side of t₀, reversing limits if necessary gives

|Tu(t)−y₀|≤M|t−t₀|≤Mh≤b.

Thus T maps X into itself. For u,v∈X, the same integral estimate and the uniform state-Lipschitz condition give

|Tu(t)−Tv(t)|≤L|t−t₀| ||u−v||∞≤hL ||u−v||∞.

Taking the supremum proves that T is a contraction with q=hL<1 on the complete space X. The already proved contraction theorem now gives a unique fixed point u∈X. If M=0, F vanishes on the rectangle; if L=0, F is independent of the state on each time slice. In either case the estimates remain valid and q=0 is allowed when L=0.

A fixed point is initially a continuous function satisfying an integral equation. To obtain a classical differential-equation solution, observe that F(t,u(t)) is continuous. The fundamental theorem of calculus then gives u′(t)=F(t,u(t)) on the interior of I, with one-sided derivative statements at its endpoints, and u(t₀)=y₀. The derivative is continuous as well. Conversely, integrate the derivative equation of any C¹ solution with that initial value; it satisfies the same fixed-point equation. If its values stay in the chosen state ball throughout I, it lies in X and equals u.

This last condition states the immediate uniqueness domain precisely. We have proved uniqueness among solutions staying in the certified rectangle on I. Extension to other intervals or trajectories requires another argument; local construction does not assert global existence.

Apply

Audit a proposed proof in this order: is the whole graph of every candidate function in the rectangle, is the function space complete, does the operator preserve it, and is its contraction constant strictly below 1? Then verify the fixed-point/solution equivalence. These are distinct steps, even if the same numerical h appears in several inequalities.

For a changed certificate in the original equation y′=1+y², use state radius b=1/2. Then M=5/4 and L=1. Choosing h=1/4 gives hM=5/16≤1/2 and hL=1/4<1. This smaller state ball gives a better contraction bound on the same time interval. It does not change the IVP. Both constructed solutions lie in the larger certified ball, so its uniqueness result proves they agree.

Finally, explain why a numerical sequence appearing stable at a few inputs cannot replace these bounds. The operator acts on entire continuous curves. The estimates must hold for every curve in X and every input in I, not only the first two iterates or a finite list of samples. Later iteration certificates quantify the distance to the actual fixed point using this established operator theorem.

Continue

Reuse the contraction theorem and its error bounds rather than reproving Banach’s theorem here. Return to the uniform Cauchy criterion and uniform-limit continuity for the complete-ball step. The fundamental theorem of calculus turns the integral fixed point into a differentiable solution.

Keep the numerical rectangle and its three checks with your solution record. The next work can study iteration errors and uniqueness beyond one fixed ball without confusing a sufficient local certificate with the maximal lifetime of a solution.