Existence and uniqueness theorem — Picard-Lindelöf (statement): Visual walkthrough

Step 1 — The problem is a starting dot plus a rule for the slope

WHAT. We are given a point $(x_{0}, y_{0})$ in the plane and a rule $f$ . The rule says: at any point $(x, y)$ , the graph passing through it must have slope $f (x, y)$ .

The two facts written as symbols: $slope of the graph at x y^{'} (x) = the slope the rule demands here f (x, y), graph must pass through the start dot y (x_{0}) = y_{0} .$

WHY. This pair is called an initial value problem (IVP). The slope rule alone has infinitely many curves obeying it; the start dot picks out one thread through the tangle. The whole theorem is about whether "one thread" is truly one.

PICTURE. Look at the slope field: tiny line segments, each drawn at the slope the rule $f$ demands. A solution is a curve that is tangent to every segment it touches, and threads through the amber dot.

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

Step 2 — Fence the problem inside a box $R$

WHAT. We refuse to look at the whole plane. We draw a closed rectangle around the start: $R = {(x, y) : width 2 a ∣ x - x_{0} ∣ \leq a, height 2 b ∣ y - y_{0} ∣ \leq b} .$ Here $a > 0$ and $b > 0$ are chosen by us — how far left/right and up/down we are willing to trust the rule.

WHY. Two reasons. First, $f$ might misbehave far away; inside a small box we can control it. Second, a continuous $f$ on a closed bounded box is bounded: there is a number $M$ with $∣ f (x, y) ∣ \leq M$ everywhere in $R$ . That number $M$ is the fastest slope any solution can ever have inside the box — this is the key ceiling we exploit next.

PICTURE. The amber dot at the centre, the cyan box around it, and the steepest allowed lines (slope $+ M$ and $- M$ ) drawn as a wedge — no solution may be steeper than these.

Step 3 — Why the safe interval is $h = min (a, b / M)$

WHAT. Starting at height $y_{0}$ , if the slope never exceeds $M$ , then after moving a horizontal distance $∣ x - x_{0} ∣$ the graph can have moved vertically by at most $M ∣ x - x_{0} ∣$ : $vertical travel ∣ y (x) - y_{0} ∣ \leq max slope M \cdot horizontal travel ∣ x - x_{0} ∣ .$

WHY. To stay inside the box we need the vertical travel to stay $\leq b$ : $M ∣ x - x_{0} ∣ \leq b ⟹ ∣ x - x_{0} ∣ \leq \frac{b}{M} .$ We also can't run off the sides: $∣ x - x_{0} ∣ \leq a$ . Both must hold, so the tighter one wins: $h = min (a, \frac{b}{M}) .$ Why min and not max? Because a constraint is a wall — you stop at the first wall you hit, the smaller number.

PICTURE. The two slope- $\pm M$ lines fan out of the start dot. Where they punch through the top/bottom of the box gives $b / M$ ; the left/right walls give $a$ . The safe interval $h$ is whichever crossing happens first (drawn amber).

Step 4 — Rewrite the ODE as "find a curve you can't move"

WHAT. Integrate both sides of $y^{'} = f (x, y)$ from $x_{0}$ to $x$ . The left side telescopes by the Fundamental Theorem of Calculus, and $y (x_{0}) = y_{0}$ kills the boundary term: $y (x) = start height y_{0} + accumulated slope = total climb \int_{x_{0}}^{x} f (t, y (t)) d t .$

WHY. Differentiation is fragile (it can amplify small wiggles); integration is robust (it smooths). Trading the differential equation for this integral equation lets us estimate with areas instead of slopes. Define the operator $T$ that eats a candidate curve $y$ and spits out a new curve: $T [y] (x) = y_{0} + \int_{x_{0}}^{x} f (t, y (t)) d t .$ A solution of the IVP is now exactly a curve that $T$ leaves unchanged — a fixed point: $T [y] = y$ . (This is the doorway to the Banach fixed-point theorem (contraction mapping).)

PICTURE. A candidate curve goes in the box on the left; $T$ measures the shaded area (the integral) and rebuilds a new curve on the right that starts at $y_{0}$ and rises by that area. A fixed point is a curve where "in" and "out" coincide.

Step 5 — Manufacture a curve by feeding guesses to $T$ (Picard iteration)

WHAT. Start with the laziest guess: the flat line $y_{0} (x) = y_{0}$ . Then apply $T$ again and again: $y_{n + 1} (x) = y_{0} + \int_{x_{0}}^{x} f (t, y_{n} (t)) d t .$ Each pass bends the previous guess a little closer to obeying the slope rule. This is the Picard iteration (method of successive approximations).

WHY. We can't write the answer down, so we approach it. The flat guess is wrong but safe — Step 6 shows every guess stays inside the box, so $f$ is always evaluated somewhere it's actually defined.

PICTURE. Watch $y^{'} = y, y (0) = 1$ : the flat line $1$ , then $1 + x$ , then $1 + x + \frac{x ^{2}}{2}$ , … each new cyan curve hugging $e^{x}$ (amber) more tightly. The iterates are the partial sums of $e^{x}$ — the machine is literally spelling out the answer.

Step 6 — Define Lipschitz, then show the guesses stay in the box and close in

WHAT (the missing definition — state it before using $L$ ).

WHAT (stay inside). On $∣ x - x_{0} ∣ \leq h$ , since $∣ f ∣ \leq M$ : $∣ y_{n + 1} (x) - y_{0} ∣ \leq \int_{x_{0}}^{x} \leq M ∣ f (t, y_{n}) ∣ d t \leq M h \leq M \cdot \frac{b}{M} = b .$ Every iterate stays within height $b$ — so every $y_{n} \in C_{h}$ , and this is exactly why we chose $h \leq b / M$ in Step 3.

WHAT (one step closer). Subtract consecutive iterates; inside the integral apply the Lipschitz bound with $y_{1} = y_{n} (t), y_{2} = y_{n - 1} (t)$ : $∣ y_{n + 1} (x) - y_{n} (x) ∣ \leq \int_{x_{0}}^{x} ∣ f (t, y_{n}) - f (t, y_{n - 1}) ∣ d t \leq L \int_{x_{0}}^{x} ∣ y_{n} (t) - y_{n - 1} (t) ∣ d t .$

WHY (the factorial bound, proved by induction). Write $d_{n} (x) = ∣ y_{n + 1} (x) - y_{n} (x) ∣$ . We claim $d_{n} (x) \leq M \frac{L ^{n} ∣ x - x _{0} ∣ ^{n + 1}}{( n + 1 )!} .$

Base $n = 0$ : $d_{0} (x) = ∣ y_{1} - y_{0} ∣ \leq \int_{x_{0}}^{x} ∣ f (t, y_{0}) ∣ d t \leq M ∣ x - x_{0} ∣ = M \frac{L ^{0} ∣ x - x _{0} ∣ ^{1}}{1 !}$ . ✔
Induction step: assume it for $n - 1$ . Feed that into the one-step inequality: $d_{n} (x) \leq L \int_{x_{0}}^{x} d_{n - 1} (t) d t \leq L \int_{x_{0}}^{x} M \frac{L ^{n - 1} ∣ t - x _{0} ∣ ^{n}}{n !} d t = M \frac{L ^{n} ∣ x - x _{0} ∣ ^{n + 1}}{( n + 1 )!} .$ The integral of $∣ t - x_{0} ∣^{n}$ raised the power by one and turned $n!$ into $(n + 1)!$ . ✔

What each factor in the denominator does: each application of $T$ contributes one more integration, and integrating $∣ x - x_{0} ∣^{k}$ divides by the new exponent $k + 1$ ; doing this $n + 1$ times multiplies all those divisors into $(n + 1)!$ . That is why the factorial — not merely a power — appears, and why it out-grows the $L^{n}$ on top.

CONSEQUENCE. Summing gaps in the supremum norm: $\sum_{n \geq 0} ∥ y_{n + 1} - y_{n} ∥_{\infty} \leq \sum_{n \geq 0} M \frac{L ^{n} h ^{n + 1}}{( n + 1 )!} = \frac{M}{L} (e^{L h} - 1) < \infty.$ A finite total travel means the curves pile up onto one limit curve $y$ (Weierstrass M-test) — they converge uniformly. That limit satisfies $T [y] = y$ : existence is done, and it already used Lipschitz.

PICTURE. A bar chart of the gap sizes $d_{n}$ : the first bars are chunky (the $L^{n}$ tries to grow them) but the factorial slams them toward zero. The stacked heights (total travel) sum to a finite amber line.

Step 7 — Only one thread: uniqueness via Grönwall

WHAT. Suppose two curves $y$ and $\tilde{y}$ both solve the IVP. Let $ϕ (x) = ∣ y (x) - \tilde{y} (x) ∣ \geq 0$ be the gap between them. Both satisfy the integral equation, so subtracting and using Lipschitz: $ϕ (x) \leq L \int_{x_{0}}^{x} ϕ (t) d t .$ Read it aloud: the gap now is bounded by $L$ times the total gap accumulated so far.

WHY. Grönwall's inequality says a non-negative function bounded by a constant times its own running integral — with no head-start term — must be identically zero. Since $ϕ \geq 0$ and $ϕ \leq L \int ϕ$ with nothing added, $ϕ \equiv 0$ . So $y = \tilde{y}$ : the two threads were the same thread all along. Notice this needed only Lipschitz — no shrinking, no $L h < 1$ .

PICTURE. Two candidate solution curves start glued at the amber dot. The Grönwall squeeze (a shrinking cyan envelope) forces them to stay glued — they can never peel apart.

Step 8 — The degenerate & edge cases (never leave the reader stranded)

The figure below is a four-panel breakdown — one panel per corner case, each annotated separately.

Panel A — $M = 0$ (flat rule). If $f \equiv 0$ in the box, no vertical travel ever happens: $b / M = \infty$ , so $h = a$ . The solution is the horizontal line $y = y_{0}$ . No wedge, no shrinking — the widest possible interval.
Panel B — $L h \geq 1$ . The naive contraction fails, but Steps 6–7 (factorial bound / Bielecki norm / Grönwall) never needed $L h < 1$ . Existence and uniqueness still hold — as in the parent's Example 1 where $L h = 1$ exactly.
Panel C — $f$ continuous but NOT Lipschitz. Convergence-and-uniqueness machinery breaks. Existence may survive via Peano existence theorem, but uniqueness can die: $y^{'} = y^{2/3}, y (0) = 0$ has both $y \equiv 0$ and $y = (x /3)^{3}$ . Two threads through one dot.
Panel D — Beyond $h$ . The theorem is local; to grow the interval you re-solve from the endpoint and glue — continuation up to the Maximal interval of existence.

The one-picture summary

Everything on one blueprint: the box $R$ , the $\pm M$ wedge fixing $h = min (a, b / M)$ , the Picard iterates spiralling into the unique fixed-point curve, and the Grönwall clamp keeping any rival curve from splitting off.

Recall Feynman retelling — say it back in plain words

We had a starting dot and a rule for the slope everywhere. First we fenced off a box so the rule couldn't misbehave; inside the box the slope can't exceed some ceiling $M$ . That ceiling means the solution can't climb out of the box faster than the $\pm M$ wedge, which fixes a safe width $h$ — whichever wall (side $a$ or top $b / M$ ) we hit first. Then we turned "match the slope" into "match the area": a solution is a curve that a certain area-machine $T$ (living on the continuous functions that stay in the box, measured by the biggest vertical gap) leaves unchanged. We fed $T$ a lazy flat guess, then its own output, over and over; each pass bent the curve closer, and because each pass adds an integration that divides by a bigger number, a factorial crushes the errors, so the guesses pile onto one curve — that's existence, and it already used the Lipschitz grip $L$ . Finally, if two curves both solved it, their gap obeyed "gap $\leq L \times$ its own running total," and Grönwall says such a gap is zero — so the two curves were one. Edge cases: flat rule gives the widest interval; a merely-continuous-but-not-Lipschitz rule like $y^{2/3}$ can split into two threads; and past $h$ you re-solve and glue.

Recall Quick self-test

Why is the safe interval a min and not a max? ::: Because two constraints (side wall $a$ , top/bottom $b / M$ ) are both walls; you stop at the first one, the smaller number. What exactly is the Lipschitz constant $L$ ? ::: The smallest $L \geq 0$ with $∣ f (x, y_{1}) - f (x, y_{2}) ∣ \leq L ∣ y_{1} - y_{2} ∣$ on $R$ — a ceiling on how fast $f$ changes as $y$ moves. Which hypothesis makes the iterates converge, and which pins down one answer? ::: Lipschitz does both — via the factorial bound (convergence) and via Grönwall (uniqueness); continuity supplies the bound $M$ that starts things. What breaks for $y^{'} = y^{2/3}, y (0) = 0$ ? ::: $f$ is continuous but not Lipschitz at $y = 0$ , so uniqueness fails: $y \equiv 0$ and $y = (x /3)^{3}$ both solve it.