Existence and uniqueness theorem — Picard-Lindelöf (statement): Worked examples

Before anything, a one-line refresher of the tools we lean on, so no symbol appears unearned:

Recall The three numbers we always extract

Question ::: For a rectangle $R = {∣ x - x_{0} ∣ \leq a, ∣ y - y_{0} ∣ \leq b}$ , what do $M$ , $L$ , $h$ mean? $M$ = biggest value of $∣ f ∣$ on $R$ (the steepest slope the solution can have). ::: $L$ = Lipschitz constant, the cap on how fast $f$ changes as $y$ changes. $h = min (a, b / M)$ = the guaranteed half-width of the interval where a unique solution lives.

The scenario matrix

Every ODE this theorem touches falls into one of these cells. The examples below are labelled with the cell(s) they hit — together they cover the whole table.

Cell	Situation	What could go wrong	Example
A	$f$ Lipschitz everywhere, $h$ limited by $a$	nothing — clean	Ex 1
B	$f$ Lipschitz on $R$ , $h$ limited by $b / M$ (box-height wins)	interval shrinks below $a$	Ex 2
C	Boundary case $L h = 1$	naive contraction fails, factorial bound saves it	Ex 3
D	Continuous but NOT Lipschitz at start	uniqueness lost, many solutions	Ex 4
E	Lipschitz locally but solution escapes to $\infty$ in finite time	interval is genuinely finite	Ex 5
F	Degenerate: $f$ independent of $y$ ( $L = 0$ )	is the theorem even needed?	Ex 6
G	Real-world word problem (cooling / mixing)	translate words → IVP, then apply	Ex 7
H	Exam twist: compare Peano vs Picard on the same $f$	which theorem gives what?	Ex 8

Example 1 — Cell A: clean, interval limited by width $a$

Worked example

$y^{'} = sin (x y)$ , $y (0) = 0$ , on $R = {∣ x ∣ \leq 1, ∣ y ∣ \leq 3}$ Forecast (guess first): will the interval $h$ be limited by the box width $a = 1$ or the box height $b = 3$ ? Jot a guess.

Step 1 — Bound $M$ . Since $∣ sin (anything) ∣ \leq 1$ , we get $∣ f ∣ = ∣ sin (x y) ∣ \leq 1 = M$ . Why this step? $M$ is the steepest the solution can climb; a bounded sine hands it to us instantly.

Step 2 — Find the Lipschitz constant $L$ . Differentiate in $y$ : $\partial_{y} f = x cos (x y)$ , so $∣ \partial_{y} f ∣ = ∣ x ∣∣ cos (x y) ∣ \leq 1 \cdot 1 = 1$ on $R$ (because $∣ x ∣ \leq 1$ ). A bounded $y$ -derivative on a convex box gives Lipschitz with $L = sup ∣ \partial_{y} f ∣ = 1$ by the Mean Value Theorem. Why this step? Lipschitz is the hypothesis that makes the solution unique; without a finite $L$ we would only get Peano's existence.

Step 3 — Compute $h$ . $h = min (a, \frac{b}{M}) = min (1, \frac{3}{1}) = min (1, 3) = 1.$ Why this step? The tighter constraint wins. Here the box height ( $b / M = 3$ ) is generous, so the width $a = 1$ is the bottleneck — that answers the forecast.

Verify: $L h = 1 \cdot 1 = 1$ . The naive contraction wants $L h < 1$ , so this sits exactly at the edge — but the factorial/Bielecki bound still gives convergence. A unique solution exists on $∣ x ∣ \leq 1$ . ✔

Example 2 — Cell B: interval limited by box height $b / M$

Worked example

$y^{'} = x^{2} + y^{2}$ , $y (0) = 0$ , on $R = {∣ x ∣ \leq 2, ∣ y ∣ \leq 1}$ Forecast: we made $a = 2$ big and $b = 1$ small on purpose. Which one throttles $h$ ?

Step 1 — Bound $M$ . On $R$ : $∣ x^{2} + y^{2} ∣ \leq 2^{2} + 1^{2} = 4 + 1 = 5 = M$ . Why this step? Each square is largest at the corner of the box; add the two maxima.

Step 2 — Lipschitz $L$ . $\partial_{y} f = 2 y$ , and $∣2 y ∣ \leq 2$ on $R$ , so $L = 2$ .

Step 3 — Compute $h$ . $h = min (2, \frac{1}{5}) = \frac{1}{5} = 0.2.$ Why this step? Now $b / M = 1/5$ is tiny, far below $a = 2$ : the box height is the bottleneck. The solution could climb out of the height- $1$ box quickly, so we only trust it for $∣ x ∣ \leq 0.2$ .

Verify: slope-line check — at rate $M = 5$ , over $∣ x ∣ \leq 0.2$ the solution moves at most $5 \times 0.2 = 1 = b$ , exactly touching the box lid. Consistent. ✔ (Contrast Ex 1 in the parent note where $b = 1, M = 2$ gave $h = 1/2$ — same $f$ , different box, different $h$ .)

Figure — Existence and uniqueness theorem — Picard-Lindelöf (statement)

Example 3 — Cell C: the boundary case $L h = 1$

Worked example

$y^{'} = 2 y$ , $y (0) = 1$ , on $R = {∣ x ∣ \leq \frac{1}{2}, ∣ y ∣ \leq 1}$ around $y_{0} = 1$ Here $R$ means $∣ x ∣ \leq \frac{1}{2}$ and $∣ y - 1∣ \leq 1$ (so $0 \leq y \leq 2$ ). Forecast: will the naive " $T$ is a contraction" argument work, or do we need the heavier machinery?

Step 1 — Bound $M$ . $∣ f ∣ = ∣2 y ∣ \leq 2 \cdot 2 = 4 = M$ (largest $∣ y ∣$ on the box is $2$ ).

Step 2 — Lipschitz $L$ . $\partial_{y} f = 2$ constant, so $L = 2$ exactly, everywhere.

Step 3 — Compute $h$ and test $L h$ . $h = min (\frac{1}{2}, \frac{1}{4}) = \frac{1}{4}, L h = 2 \cdot \frac{1}{4} = \frac{1}{2} < 1.$ Wait — with this box, $L h < 1$ , so Banach does apply directly. To hit the true boundary $L h = 1$ , widen: take $a = \frac{1}{2}$ but bound crudely so $h = \frac{1}{2}$ ; then $L h = 2 \cdot \frac{1}{2} = 1$ . Why this step? This shows $L h$ is not part of the theorem — it depends on how you carve the box, not on the ODE.

Verify: the true solution is $y = e^{2 x}$ . Picard iterates from $y_{0} (x) = 1$ : $y_{1} = 1 + \int_{0}^{x} 2 d t = 1 + 2 x$ , $y_{2} = 1 + \int_{0}^{x} 2 (1 + 2 t) d t = 1 + 2 x + 2 x^{2}$ . These are the partial sums of $e^{2 x} = 1 + 2 x + 2 x^{2} + \dots$ . Convergent for all $x$ despite $L h = 1$ — proof by factorial bound, not Banach. ✔

Example 4 — Cell D: continuous but not Lipschitz → uniqueness lost

Worked example

$y^{'} = 3 y^{2/3}$ , $y (0) = 0$ Forecast: $f = 3 y^{2/3}$ is continuous. Does Picard–Lindelöf apply? Guess how many solutions there are.

Step 1 — Check continuity. $y^{2/3} = (3 y)^{2} \geq 0$ is continuous everywhere. Peano therefore guarantees at least one solution. Why this step? Continuity is the price of admission for existence — but not for uniqueness.

Step 2 — Test the Lipschitz condition near $y = 0$ . $\partial_{y} f = 3 \cdot \frac{2}{3} y^{- 1/3} = 2 y^{- 1/3} \to + \infty$ as $y \to 0^{+}$ . No finite $L$ can cap this, so $f$ is not Lipschitz at the initial point. Why this step? The theorem's uniqueness guarantee is void the moment Lipschitz fails at $(x_{0}, y_{0})$ .

Step 3 — Exhibit two solutions.

The flat one: $y \equiv 0$ satisfies $y^{'} = 0 = 3 \cdot 0^{2/3}$ . ✔
The rising one: try $y = x^{3}$ . Then $y^{'} = 3 x^{2}$ and $3 y^{2/3} = 3 (x^{3})^{2/3} = 3 x^{2}$ . ✔ Both pass through $(0, 0)$ . Why this step? Concretely demonstrates the failure the theorem warns about.

Verify: at $x = 2$ : solution 1 gives $y = 0$ ; solution 2 gives $y = 2^{3} = 8$ . Two different values from one IVP — uniqueness genuinely broken. ✔

Example 5 — Cell E: Lipschitz locally but blow-up in finite time

Worked example

$y^{'} = 1 + y^{2}$ , $y (0) = 0$ Forecast: $f = 1 + y^{2}$ is a smooth polynomial in $y$ . Does the solution live forever, or does the guaranteed interval have a hard finite edge?

Step 1 — Local Lipschitz. $\partial_{y} f = 2 y$ ; on any bounded box $∣ y ∣ \leq b$ we have $∣2 y ∣ \leq 2 b < \infty$ . So $f$ is Lipschitz on every finite box — the theorem applies locally at every point. Why this step? Local Lipschitz gives a unique solution near the start; it says nothing about how far it extends.

Step 2 — Solve exactly to see the edge. This is separable: $\frac{d y}{1 + y ^{2}} = d x \Rightarrow arctan y = x + C$ . With $y (0) = 0$ , $C = 0$ , so $y = tan x$ . Why this step? $arctan$ (the inverse of $tan$ ) is the tool that "undoes" $1 + y^{2}$ under the integral $\int \frac{d y}{1 + y ^{2}} = arctan y$ — the only elementary antiderivative that fits.

Step 3 — Find the maximal interval. $tan x \to + \infty$ as $x \to \frac{π}{2}^{-}$ . The solution exists and is unique on $(- \frac{π}{2}, \frac{π}{2})$ and cannot be continued past $x = \frac{π}{2} \approx 1.5708$ . Why this step? This is the honest meaning of "local": nice $f$ , yet a finite lifespan because $y$ runs off to infinity.

Verify: check $y = tan x$ solves it: $\frac{d}{d x} tan x = sec^{2} x = 1 + tan^{2} x = 1 + y^{2}$ . ✔ And $y (0) = tan 0 = 0$ . ✔

Example 6 — Cell F: degenerate case, $f$ independent of $y$ ( $L = 0$ )

Worked example

$y^{'} = cos x$ , $y (0) = 2$ Forecast: $f = cos x$ has no $y$ in it at all. What is the Lipschitz constant, and does the theorem even do any work here?

Step 1 — Lipschitz constant. $∣ f (x, y_{1}) - f (x, y_{2}) ∣ = ∣ cos x - cos x ∣ = 0 \leq 0 \cdot ∣ y_{1} - y_{2} ∣$ , so $L = 0$ . Why this step? $L = 0$ is the strongest possible Lipschitz condition — $f$ literally doesn't move when $y$ moves. Uniqueness is automatic.

Step 2 — What $h$ becomes. The bound $M = max ∣ cos x ∣ = 1$ still limits $h = min (a, b / M)$ , but the contraction constant $L h = 0 < 1$ trivially — Banach applies with room to spare. Why this step? Shows the machinery collapses to the obvious: integrate once.

Step 3 — Solve directly. $y = y_{0} + \int_{0}^{x} cos t d t = 2 + sin x - sin 0 = 2 + sin x$ . Why this step? When $f$ has no $y$ , the Picard operator gives the answer in one step — no iteration needed. This is just antidifferentiation, the base case of the whole theory.

Verify: $y (0) = 2 + sin 0 = 2$ ✔ and $y^{'} = cos x$ ✔. Unique because $L = 0$ . ✔

Example 7 — Cell G: real-world word problem (Newton cooling)

Worked example A cup of coffee at

$9 0^{\circ}$ C sits in a $2 0^{\circ}$ C room; it cools at a rate proportional to the temperature gap with constant $k = 0.1$ per minute. Is the temperature history unique? Forecast: physics feels like it must have one answer. Can Picard–Lindelöf confirm it?

Step 1 — Translate words to an IVP. "Rate of cooling $\propto$ gap" gives $y^{'} = - k (y - 20)$ with $y (0) = 90$ , $k = 0.1$ . Here $y (t)$ is temperature, $t$ in minutes. Why this step? We can only apply the theorem after the sentence becomes $y^{'} = f (t, y)$ .

Step 2 — Extract $f$ , check hypotheses. $f (t, y) = - 0.1 (y - 20)$ is continuous, and $\partial_{y} f = - 0.1$ constant, so $L = 0.1$ everywhere. Linear ODE, globally Lipschitz. Why this step? Constant bounded $y$ -slope ⇒ Picard–Lindelöf gives existence AND uniqueness on the whole time axis.

Step 3 — Solve and read off numbers. Standard linear first-order solution: $y (t) = 20 + (90 - 20) e^{- 0.1 t} = 20 + 70 e^{- 0.1 t}$ . At $t = 10$ : $y = 20 + 70 e^{- 1} = 20 + 70 (0.367879 \dots) = 45.7515 \dots^{\circ}$ C. Why this step? $e^{- 0.1 t}$ is the exponential decay tool — the unique function whose derivative is proportional to itself, exactly what "rate $\propto$ gap" demands.

Verify: units — $[k] = min^{- 1}$ , so $k (y - 20)$ has units $^{\circ}$ C/min = temperature rate. ✔ At $t = 0$ : $20 + 70 = 9 0^{\circ}$ C matches the start. ✔ As $t \to \infty$ : $y \to 2 0^{\circ}$ C (room temperature), physically correct. Uniqueness confirmed since $L = 0.1 < \infty$ . ✔

Example 8 — Cell H: exam twist — Peano vs Picard on the same $f$

Worked example For

$y^{'} = ∣ y ∣$ , $y (0) = 0$ : state precisely what each theorem gives. Forecast: one theorem will say "yes", the other "I can't promise". Which is which?

Step 1 — Continuity. $f = ∣ y ∣$ is continuous everywhere (absolute value inside a root removes the sign problem). So Peano guarantees a solution exists. Why this step? Peano needs only continuity — it is the weaker but broader net.

Step 2 — Lipschitz test at the origin. For $y > 0$ : $\partial_{y} f = \frac{1}{2 y} \to \infty$ as $y \to 0^{+}$ . Not Lipschitz at $(0, 0)$ . So Picard–Lindelöf cannot conclude uniqueness here. Why this step? Picard–Lindelöf is the stronger claim (uniqueness) but demands the stronger hypothesis (Lipschitz), which fails.

Step 3 — Confirm the ambiguity with explicit solutions. $y \equiv 0$ works. Also $y = \frac{x ^{2}}{4}$ for $x \geq 0$ : then $y^{'} = \frac{x}{2}$ and $∣ y ∣ = x^{2} /4 = \frac{x}{2}$ (for $x \geq 0$ ). ✔ Both pass through $(0, 0)$ . Why this step? Cements the exam point: Peano says "≥1 solution", Picard says "exactly 1 only when Lipschitz holds" — and here it doesn't.

Verify: at $x = 2$ : solution 1 gives $0$ ; solution 2 gives $2^{2} /4 = 1$ . Two answers ⇒ genuinely non-unique, matching the Peano-yes / Picard-no verdict. ✔

Recall Self-test: which cell?

Question ::: You have $y^{'} = e^{y}$ , $y (0) = 0$ . Lipschitz locally? Finite lifespan? $\partial_{y} e^{y} = e^{y}$ bounded on any finite box ⇒ locally Lipschitz (unique locally). But solving, $y = - ln (1 - x)$ blows up at $x = 1$ ⇒ Cell E, finite maximal interval $(- \infty, 1)$ . :::