Try
The initial-value problem y′=y², y(0)=2 has the solution y(t)=2/(1−2t) for 0≤t<1/2. Differentiate it and check the initial value. The field is smooth everywhere in the time-state plane, yet this particular solution becomes unbounded as time approaches 1/2. Local existence and uniqueness do not promise a solution for all future time.
A second problem separates unboundedness from leaving the domain. For y′=1/(1−y), y(0)=0, specify the equation only where y<1. Its solution y(t)=1−sqrt(1−2t) stays between zero and one for 0≤t<1/2. Nevertheless, its limiting state 1 lies outside the domain of the equation. A bounded graph can still approach a boundary where continuation fails.
The purpose of this lesson is to attach a precise region and time direction to every uniqueness claim. The fixed-point construction gives more than uniqueness among numerical iterates, but less than unrestricted global existence.
Predict
Suppose the preceding lesson constructed a solution in a closed rectangle using a time radius h and Lipschitz constant L with Lh<1. A different researcher proposes another differentiable solution with the same initial value. Predict which fact is needed before applying the contraction estimate: the researcher used the same iteration, or the competing graph stays in the rectangle where the bound holds?
It is the second. Every classical solution in that rectangle satisfies the same integral equation by the fundamental theorem of calculus. Its method of discovery is irrelevant. On the other hand, a bound for F on a rectangle cannot automatically be used for a curve leaving that rectangle.
Also predict what happens on a much longer comparison interval where L times its length exceeds one. The estimate D≤L·(interval length)·D no longer forces the supremum difference D to vanish. We must compare on short intervals and propagate equality, rather than pretend the long-interval operator remains a contraction.
Experiment
Take L=3 and compare two proposed solutions in a common rectangle over times from 0 to 1. Both have the same value at time zero. Break the comparison into eight intervals of length 1/8. On each interval the contraction factor is 3/8, while the direct full-length factor would be 3.
| Interval | Known equality at | Supremum difference bound | Conclusion |
|---|
| [0,1/8] | 0 | D₁≤(3/8)D₁ | D₁=0 |
| [1/8,1/4] | 1/8 from the preceding step | D₂≤(3/8)D₂ | D₂=0 |
| Each later eighth | Its left endpoint | Same factor 3/8 | Equality continues |
This is a finite proof pattern, not a numerical simulation. The actual solution values need not be computed. Change the comparison direction: if equality is known at the right endpoint, reverse the integral limits and obtain the same absolute-value estimate. This allows comparison backward too when the field satisfies the same Lipschitz hypotheses.
Understand
Let F be continuous on a rectangle centered at (t₀,y₀), and assume |F(t,u)−F(t,v)|≤L|u−v| for states and times in that rectangle. Suppose u and v are classical solutions with the same initial value and remain inside it on [t₀−h,t₀+h]. A classical solution is C¹ and satisfies its equation at every time. The fundamental theorem gives u(t)=y₀+∫ from t₀ to t of F(s,u(s))ds, and similarly for v.
Subtracting gives |u(t)−v(t)|≤L|t−t₀| ||u−v||∞. For backward times reverse the integral and use its positive length. Hence ||u−v||∞≤Lh||u−v||∞. When Lh<1 the finite nonnegative norm must be zero. This proves uniqueness among all such classical solutions, not only among outputs of the construction algorithm.
If both graphs remain in a common Lipschitz rectangle on a longer compact time interval, choose a segment length ℓ=1/[2(L+1)]. Its factor Lℓ is strictly less than one even when L=0. Finitely many segments cover the comparison interval. Equality on one segment gives equality of values at its far endpoint, so the same integral comparison starts there on the next segment. Proceed in either direction from the shared initial time. This establishes comparison on the full common interval without claiming a single long contraction.
A precise continuation theorem. Let Ω be open in the time-state plane, and suppose F is continuous and locally Lipschitz in the state variable, with a finite common constant on a sufficiently small rectangle about each point. Let a solution exist on [t₀,T), where T is finite. Assume its graph is contained in a compact set K lying inside Ω. Continuity bounds |F| by some M on K. Thus |y(t)−y(s)|≤M|t−s| for times before T. The real Cauchy criterion gives a finite limit y_T as t approaches T from the left.
Take any increasing sequence of times tending to T. Its graph points converge to (T,y_T), which belongs to K because K is closed. In particular the limiting point still lies in Ω. Choose a smaller closed rectangle about it inside Ω, with the required boundedness and Lipschitz constants. The previously proved local Picard construction supplies a solution through that point extending to the right of T.
Join the old and new solutions at T. Continuity of F and convergence of the old graph imply F(t,y(t)) tends to F(T,y_T). Passing the old integral equation to the endpoint, or averaging its continuous integrand over shrinking intervals, gives the left derivative F(T,y_T). The new solution gives the same right derivative. The joined function is C¹, satisfies the equation at the join, and extends past T. Local uniqueness makes the two descriptions agree on any overlap. Merely finding a limit was not enough; we used local existence at an interior domain point and checked the equation at the join.
Apply
Failure of a Lipschitz hypothesis does not decide uniqueness by itself. For y′=3|y|^(2/3), start at zero. For each waiting time τ≥0, set y(t)=0 until τ and y(t)=(t−τ)³ afterward. Before the join both sides of the equation are zero; afterward both equal 3(t−τ)². At the join the difference quotients and derivatives tend to zero, so this is a classical C¹ solution. Different waiting times give actual nonuniqueness. The unbounded quotient 3|y|^(−1/3) explains why the Lipschitz theorem cannot apply, but the verified solutions establish the conclusion.
For a contrasting field, let F(y)=−sign(y)sqrt(|y|), with F(0)=0. It too is not locally Lipschitz at zero. Nevertheless every solution through zero satisfies d(y²)/dt=−2|y|^(3/2)≤0. Forward from zero, the nonnegative square cannot decrease below zero, so the only forward solution is zero.
Backward uniqueness is different: y(t)=t²/4 for negative times and zero for nonnegative times is C¹ and solves this same equation. On negative times its derivative is t/2, equal to −sqrt(t²/4); at zero both derivatives vanish. This differs from the identically zero solution before the initial time. State the time direction when drawing a uniqueness conclusion.
Continue
Return to the earlier ODE existence and uniqueness lesson for the equation-level context. The present proof adds the integral comparison, finite interval propagation and compact-interior continuation mechanism.
For a new problem, record the domain of F, the rectangle supporting its bounds, the interval where compared solutions remain there, and any proposed extension point. A failed sufficient condition is a request for another argument, not an automatic counterexample. A local theorem is not a global existence theorem; a finite trace is not a proof that a solution remains inside a safe compact region forever.
