Picard's Existence and Uniqueness Theorem

A first-order differential equation $y^{'} = F (x, y)$ with an initial condition $y (x_{0}) = y_{0}$ asks for a function whose slope at each point is dictated by $F$ . When $F$ depends only on $x$ , the fundamental theorem integrates it. When $F$ depends on $y$ as well, the unknown appears on both sides, and existence of a solution is no longer automatic. Picard's theorem gives a clean sufficient condition, and its proof is a fixed-point argument run in the space of continuous functions.¹

The main tool is the contraction mapping principle, proved first in full generality and then specialized.

The contraction mapping principle

A map that shrinks all distances has a single fixed point, and iterating the map from any start converges to it.

Both hypotheses are needed. Completeness supplies the limit of the iterates, and the strict inequality $k < 1$ is what collapses the distance between two candidate fixed points. A merely $1$ -Lipschitz map, such as $x \mapsto x + 1$ on $R$ , can have no fixed point at all.

Iterating a contraction from any start walks the values in toward the single fixed point where the graph meets the diagonal; slope below one keeps the steps shrinking.

The proof is constructive: it names the fixed point as a limit of iterates and bounds the distance to it after $n$ steps by $d (x, x_{n}) \leq \frac{k ^{n}}{1 - k} d (x_{1}, x_{0})$ . This is the basis of Newton's method and many numerical solvers.

The space of continuous functions

Running the fixed-point theorem on functions needs a complete metric space made of them. The uniform norm supplies exactly one.

Completeness of $C ([a, b])$ is what the Banach theorem requires when the metric space is a space of functions; the uniform norm, the uniform Cauchy criterion, and continuity of the uniform limit are just what its hypotheses demand. A closed subset of $C ([a, b])$ is again complete in the subspace metric, and that is where the solution will be sought.

Picard's theorem

Rewrite the differential equation as an integral equation. If $f$ solves $f^{'} (x) = F (x, f (x))$ with $f (x_{0}) = y_{0}$ , then integrating both sides gives

f (x) = y_{0} + x_{0} \int x F (t, f (t)) d t,

and conversely any continuous $f$ satisfying this equation is differentiable with the right derivative and initial value. The right-hand side defines an operator $T$ on functions,

T (f) (x) = y_{0} + x_{0} \int x F (t, f (t)) d t,

and a solution is precisely a fixed point of $T$ . Solving the differential equation becomes finding $f$ with $T (f) = f$ .

The integral operator

T

sends a candidate

f

to a new function; a solution of the differential equation is exactly a fixed point of this loop.

Proof sketch. Assume $x_{0} = 0$ . Since $I \times J$ is compact and $F$ is continuous, $F$ is bounded, say $∣ F ∣ \leq M$ . Pick $α > 0$ with $[- α, α] \subset I$ and $[y_{0} - α, y_{0} + α] \subset J$ , and set

h = min {α, \frac{α}{M + Lα}} .

Let $Y = {f \in C ([- h, h]) : f ([- h, h]) \subset J}$ , a closed subset of the complete space $C ([- h, h])$ , hence itself complete. The bound $∣ F ∣ \leq M$ makes $T$ map $Y$ into $Y$ : for $∣ x ∣ \leq h$ ,

∣ T (f) (x) - y_{0} ∣ = 0 \int x F (t, f (t)) d t \leq M h \leq M \cdot \frac{α}{M + Lα} \leq α .

The Lipschitz condition makes $T$ a contraction on $Y$ : for $f, g \in Y$ ,

∣ T (f) (x) - T (g) (x) ∣ \leq 0 \int x ∣ F (t, f (t)) - F (t, g (t)) ∣ d t \leq L h d (f, g) \leq \frac{Lα}{M + Lα} d (f, g),

and $C := \frac{Lα}{M + Lα} < 1$ ; taking the supremum over $x$ gives $d (T (f), T (g)) \leq C d (f, g)$ . The Banach theorem supplies a unique $f \in Y$ with $T (f) = f$ , and the fundamental theorem makes $f$ differentiable with $f^{'} = F (x, f (x))$ and $f (0) = y_{0}$ . $□$

The theorem is local. The interval half-width $h$ shrinks as the bound $M$ or the Lipschitz constant $L$ grows, so a fast-changing field yields only a short guarantee. The constant $C < 1$ is engineered by the choice of $h$ : making the interval short enough forces the operator to contract.

Picard iteration

The proof of the Banach theorem is constructive, so the same iteration builds the solution. Starting from the constant $f_{0} (x) = y_{0}$ , the Picard iterates are

f_{k} (x) = y_{0} + x_{0} \int x F (t, f_{k - 1} (t)) d t,

and they converge uniformly to the solution.

The Picard iterates for

y^{'} = y

y (0) = 1

are the exponential's Taylor partial sums; each iterate matches the solution to one more order and hugs it over a wider range.

Necessity of the hypotheses

The Lipschitz condition and continuity of $F$ each rule out a specific failure.

Without Lipschitz, uniqueness fails. For $f^{'} (x) = 2 ∣ f (x) ∣$ , $f (0) = 0$ , the field $F (x, y) = 2 ∣ y ∣$ is continuous but not Lipschitz in $y$ near $0$ . Both $f (x) = 0$ and $f (x) = x^{2}$ (for $x \geq 0$ ) solve it, so the initial value does not determine the solution. The infinite slope of $∣ y ∣$ at $y = 0$ is what prevents a contraction estimate.
Without continuity, existence fails. For $y^{'} = φ (x)$ with $φ = 0$ on the rationals and $φ = 1$ elsewhere, no solution exists: a solution's derivative would be $φ$ , but $φ$ has no intermediate value property, which every derivative must have.
Lipschitz holds but only locally, so $h$ is finite. For $f^{'} (x) = f (x)^{2}$ , $f (0) = 1$ , the solution $f (x) = \frac{1}{1 - x}$ exists only on $(- \infty, 1)$ and blows up at $x = 1$ . The map $y \mapsto y^{2}$ is Lipschitz on any bounded $J$ but not on all of $R$ , and the interval of existence is necessarily short near the blow-up.

Dropping the Lipschitz condition costs uniqueness: from the same start at the origin, both the zero solution and a parabola satisfy the same equation.

Summary

Ingredient	Supplied by	Role in Picard's proof
Uniform norm	pointwise/uniform lesson	the metric on $C ([a, b])$
Uniform Cauchy $\Rightarrow$ converges	same	Cauchy iterates have a limit
Uniform limit is continuous	interchange lesson	the limit is a genuine function in $C ([a, b])$
Fundamental theorem	integration module	integral equation $\Leftrightarrow$ differential equation
Banach fixed point	the contraction principle above	existence and uniqueness of the solution

The differential equation is recast as a fixed-point problem, the space of continuous functions is complete because uniform limits stay continuous, and the Lipschitz condition turns integration into a contraction. The solution is not only shown to exist and to be unique; the iteration that proves it also computes it.

Lebl, §6.3 and §7.6. The two sections prove the same theorem twice — once by hand with the uniform norm and once through the abstract fixed-point theorem — and the treatment here follows the second, cleaner route while keeping the explicit iterates and examples of the first. Rosenlicht, Ch. 8, places the existence-uniqueness result at the head of the theory of differential equations. ↩

The contraction mapping principle

The space of continuous functions

Picard's theorem

Picard iteration

Necessity of the hypotheses

Summary

Footnotes