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Ordinary Differential Equations 12 | Picard–Lindelöf Theorem thumbnail

Ordinary Differential Equations 12 | Picard–Lindelöf Theorem

5 min read

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TL;DR

Rewrite the initial value problem ẋ = V(t, x), x(0) = x0 as the integral equation x(t) = x0 + ∫[0 to t] V(s, x(s)) ds.

Briefing

Picard–Lindelöf’s existence result for ordinary differential equations hinges on turning an initial value problem into a fixed-point problem on a carefully chosen space of functions. For a system written as ẋ = V(t, x) with x(0) = x0, the key requirement is that V is locally Lipschitz in x. Under that condition, one can guarantee not only uniqueness (handled earlier), but also the existence of a solution—by applying the Banach fixed point theorem.

The construction starts with the integral form of the initial value problem. Define a map F on candidate functions α by F(α)(t) = x0 + ∫[0 to t] V(s, α(s)) ds. A fixed point α = F(α) corresponds exactly to a function satisfying the original initial value problem, because differentiating the integral equation recovers ẋ = V(t, x) and the initial condition is built in through the x0 term.

To use Banach’s theorem, the argument builds a complete metric space X of functions. Since solutions may only exist on a short time interval, the domain is restricted to t ∈ [−ε, ε]. The set X consists of continuous functions α : [−ε, ε] → R^N that satisfy the initial condition α(0) = x0 and take values in the domain where V is defined. The metric D on X is the supremum norm: for α, β ∈ X, D(α, β) = sup_{t ∈ [−ε, ε]} ||α(t) − β(t)||, using the standard norm in R^N. With boundedness ensured by choosing ε small enough (so functions don’t “blow up” on the interval), this function space becomes complete, meaning every Cauchy sequence converges within X.

The next step is proving F is a contraction on X. Take two candidate functions α and β. When comparing F(α)(t) and F(β)(t), the x0 terms cancel, leaving only the integral of V(s, α(s)) − V(s, β(s)). Using the triangle inequality to move the norm inside the integral, the integral is bounded by the length of the time interval times a supremum bound. Since t lies in [−ε, ε], the interval length contributes a factor at most ε. Then the local Lipschitz property of V supplies a constant L such that ||V(s, α(s)) − V(s, β(s))|| ≤ L ||α(s) − β(s)||. Combining these estimates yields D(F(α), F(β)) ≤ ε L · D(α, β). By choosing ε small enough so that εL < 1, F becomes a contraction.

Banach’s fixed point theorem then guarantees a unique fixed point α in X. That fixed point is the local solution to the initial value problem on [−ε, ε]. The final theorem statement: if V is locally Lipschitz and x0 lies in the domain U ⊂ R^N, then there exists ε > 0 and a unique solution α(t) to ẋ = V(t, x), x(0) = x0, valid at least for t in a neighborhood of 0. The method is notable because it produces existence by pure functional-analytic machinery, with the Lipschitz condition supplying the contraction needed for Banach’s theorem.

Cornell Notes

Picard–Lindelöf’s theorem proves local existence (and, with earlier work, uniqueness) for ẋ = V(t, x), x(0) = x0 when V is locally Lipschitz in x. The proof rewrites the ODE as an integral equation and defines a function operator F(α)(t) = x0 + ∫[0 to t] V(s, α(s)) ds. Candidate solutions live in a complete metric space of continuous functions on a short interval [−ε, ε] with the metric given by the supremum norm. Using the local Lipschitz condition, one shows F shrinks distances by a factor ≤ εL, so choosing ε small enough makes F a contraction. Banach’s fixed point theorem then guarantees a fixed point, which corresponds to a solution of the initial value problem.

How does the ODE ẋ = V(t, x), x(0) = x0 turn into a fixed-point problem?

By using the integral form: any solution x(t) must satisfy x(t) = x0 + ∫[0 to t] V(s, x(s)) ds. On the space of candidate functions α, define F(α)(t) = x0 + ∫[0 to t] V(s, α(s)) ds. A fixed point α = F(α) means α(t) = x0 + ∫[0 to t] V(s, α(s)) ds, which is exactly the integral equation corresponding to the original initial value problem.

Why restrict attention to a small time interval [−ε, ε]?

Solutions may fail to exist globally in time. The proof constructs a local solution by working on t ∈ [−ε, ε], choosing ε small enough that the function space is well-behaved (e.g., functions stay in the domain where V is defined and remain bounded). This also lets the contraction estimate produce a factor εL < 1.

What metric makes the function space complete, and why does completeness matter?

The metric is the supremum norm: D(α, β) = sup_{t ∈ [−ε, ε]} ||α(t) − β(t)||. With bounded continuous functions on a closed interval, this yields a complete metric space. Completeness is crucial because Banach’s fixed point theorem requires a complete metric space to ensure that iterating the contraction converges to a fixed point.

Where does the contraction inequality D(F(α), F(β)) ≤ εL·D(α, β) come from?

Start by comparing F(α)(t) and F(β)(t); x0 cancels, leaving an integral of V(s, α(s)) − V(s, β(s)). Apply the triangle inequality to bound the norm of the integral by the integral of the norm. Then bound the integral by the interval length (≤ ε) times a supremum over s. Finally, use local Lipschitz continuity: ||V(s, α(s)) − V(s, β(s))|| ≤ L||α(s) − β(s)||. The supremum of ||α(s) − β(s)|| is exactly D(α, β), producing the factor εL.

How does local Lipschitz continuity of V translate into existence of solutions?

Local Lipschitz continuity provides a uniform bound L on how much V can change with respect to x over the relevant region. That bound is what turns the operator F into a contraction when ε is chosen small enough. Once F is a contraction, Banach’s fixed point theorem guarantees a fixed point, which corresponds to an actual solution of the initial value problem on [−ε, ε].

Review Questions

  1. In what way does the operator F(α)(t) encode the initial condition x(0) = x0?
  2. Which estimate introduces the factor ε in the contraction proof, and why can ε be chosen to make εL < 1?
  3. How does the supremum norm metric D(α, β) relate to the contraction property of F?

Key Points

  1. 1

    Rewrite the initial value problem ẋ = V(t, x), x(0) = x0 as the integral equation x(t) = x0 + ∫[0 to t] V(s, x(s)) ds.

  2. 2

    Define an operator F on candidate functions by F(α)(t) = x0 + ∫[0 to t] V(s, α(s)) ds; fixed points of F correspond to solutions.

  3. 3

    Choose a complete metric space X of continuous functions on a short interval [−ε, ε] with α(0) = x0, using the supremum norm metric D(α, β) = sup ||α(t) − β(t)||.

  4. 4

    Show F maps X into itself, so Banach’s theorem applies within the same function space.

  5. 5

    Use the triangle inequality and integral bounds to estimate D(F(α), F(β)) in terms of the interval length and the Lipschitz constant.

  6. 6

    Apply local Lipschitz continuity of V to obtain D(F(α), F(β)) ≤ εL·D(α, β), then pick ε small enough that εL < 1.

  7. 7

    Banach’s fixed point theorem then guarantees a fixed point, yielding a local solution (and, together with prior results, uniqueness).

Highlights

The proof converts an ODE into a fixed-point problem via F(α)(t) = x0 + ∫[0 to t] V(s, α(s)) ds.
A complete metric space of continuous functions on [−ε, ε] with the supremum norm is the setting for Banach’s theorem.
Local Lipschitz continuity supplies a constant L that, combined with a small ε, forces F to be a contraction.
The contraction estimate hinges on bounding the integral by interval length (≤ ε) times a supremum norm bound.
A fixed point of F is exactly a solution of the original initial value problem on a neighborhood of t = 0.