Existence and uniqueness theorem — Picard-Lindelöf (statement): Question bank

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

Notation reminders used throughout: the IVP is $y^{'} = f (x, y), y (x_{0}) = y_{0}$ ; $M$ bounds $∣ f ∣$ on the rectangle $R$ ; $L$ is the Lipschitz constant in $y$ ; $h = min (a, b / M)$ is the guaranteed half-width of the interval.

The operator $T$ (which most traps circle around) and the Bielecki weighted norm weight both have their own picture too:

True or false — justify

True or false: If

f

is continuous on the rectangle

R

, the IVP has exactly one solution.

False. Continuity alone gives existence (Peano) but can lose uniqueness; you need the Lipschitz condition in

y

to pin down a single solution. Example:

y^{'} = y^{2/3}

is continuous yet has two solutions through

(0, 0)

True or false: The Lipschitz condition must hold in the variable

x

False. The contraction estimate uses

∣ f (t, y_{1}) - f (t, y_{2}) ∣ \leq L ∣ y_{1} - y_{2} ∣

, comparing the same

t

at two different

y

-values. Lipschitz is needed in $y$ ; mere continuity in

x

suffices.

True or false: If

\partial f / \partial y

exists and is bounded on a convex region, then

f

is Lipschitz in

y

there.

True. By the Mean Value Theorem

f (x, y_{1}) - f (x, y_{2}) = \partial_{y} f (x, ξ) (y_{1} - y_{2})

, so

L = sup ∣ \partial_{y} f ∣

works. This is the everyday way we find

L

True or false: Being Lipschitz in

y

is strictly stronger than being continuous in

y

True. Lipschitz implies (uniform) continuity, but

y^{2/3}

is continuous and not Lipschitz near

0

(its

y

-slope blows up). So the class of Lipschitz functions is a proper subset.

True or false: The theorem guarantees the solution on the entire rectangle

∣ x - x_{0} ∣ \leq a

False. The solution is guaranteed only on

∣ x - x_{0} ∣ \leq h = min (a, b / M)

, because the graph may hit the top/bottom of the box (height

b

) before

x

reaches

\pm a

. The result is local.

True or false: Picard–Lindelöf requires

L h < 1

as a hypothesis.

False.

L h < 1

is a convenience of one proof route (direct Banach in the sup-norm). The factorial estimate or the Bielecki weighted norm makes the operator a contraction on the whole interval for any

h

, so no such hypothesis is imposed.

True or false: Uniqueness in the theorem comes from continuity, existence from Lipschitz.

False (both halves). It is roughly the reverse of the naive slogan and even that reversal is imprecise: in the Picard proof the Lipschitz estimate drives both convergence (existence) and uniqueness; continuity supplies the bound

M

that lets iteration start.

True or false: If

f

is continuous in

x

and globally Lipschitz in

y

on all of

R^{2}

, the solution exists for all

x

True. A global Lipschitz bound gives linear growth

∣ f (x, y) ∣ \leq ∣ f (x, 0) ∣ + L ∣ y ∣

, which by Grönwall rules out finite-time blow-up, so the Maximal interval of existence is all of

R

. (The subtlety is only that "Lipschitz on a rectangle" gives just the local

h

; genuinely global Lipschitz gives the whole line.)

True or false: A larger Lipschitz constant

L

can only shrink the guaranteed interval

h

False.

h = min (a, b / M)

does not contain

L

at all —

L

governs convergence speed and uniqueness, not the length

h

. Only

M

(the size bound) shrinks

h

Spot the error

"Since

∣ f (x, y) ∣ \leq M

, the solution's slope is at most

M

, so after distance

a

it rises at most

M a

— that always fits inside height

b

Error:

M a

need not be

\leq b

. If

M a > b

the graph escapes the box, which is exactly why we shrink to

b / M

and take

h = min (a, b / M)

T [y] = y_{0} + \int_{x_{0}}^{x} f (t, y) d t

satisfies

sup ∣ T [u] - T [v] ∣ \leq L h sup ∣ u - v ∣

, so

T

is a contraction — done."

Error: that is a contraction only when

L h < 1

. For the parent's worked example

y^{'} = x^{2} + y^{2}

(where

L = 2, h = \frac{1}{2}

) we get

L h = 1

, so this step fails; you need the factorial bound or Bielecki weighted norm instead.

f = y^{2/3}

is differentiable in

y

, and differentiable functions are Lipschitz, so uniqueness holds."

Error:

\partial_{y} f = \frac{2}{3} y^{- 1/3}

is unbounded as

y \to 0

, so

f

is not Lipschitz near

0

. Differentiability with an unbounded derivative does not give Lipschitz, and uniqueness indeed fails.

"To prove convergence of the Picard iterates we differentiate the ODE for robustness."

Error: we integrate, not differentiate. Converting

y^{'} = f

into

y = y_{0} + \int f

turns a fragile derivative condition into a robust integral operator whose fixed point we chase.

"Grönwall's inequality proves the iterates converge."

Error: Grönwall is used for uniqueness (forcing

ϕ = ∣ y - \tilde{y} ∣ \equiv 0

). Convergence of the iterates comes from the Weierstrass M-test on the factorial bound.

"The bound

∣ y_{n + 1} - y_{n} ∣ \leq M \frac{L ^{n} ∣ x - x _{0} ∣ ^{n + 1}}{( n + 1 )!}

only converges if

L h < 1

Error: factorials beat any exponential — the series sums to

\frac{M}{L} (e^{L h} - 1) < \infty

for every finite

h

. That is precisely why the proof survives

L h \geq 1

Why questions

Why does the proof integrate the ODE instead of working with

y^{'}

directly?

Because integration is a smoothing, contractive operation: the operator

T [y] = y_{0} + \int_{x_{0}}^{x} f

maps continuous functions to continuous functions and its fixed points are exactly the solutions, letting us invoke fixed-point machinery.

Why must

h

be a minimum, not just

b / M

Two independent constraints must both hold — stay inside the box vertically (

∣ x - x_{0} ∣ \leq b / M

) and horizontally (

∣ x - x_{0} ∣ \leq a

). The tighter one wins, so

h = min (a, b / M)

Why is a fixed point of

T

the same thing as a solution of the IVP?

T [y] = y

then

y (x) = y_{0} + \int_{x_{0}}^{x} f (t, y) d t

; differentiating gives

y^{'} = f (x, y)

and setting

x = x_{0}

gives

y (x_{0}) = y_{0}

. The equivalence runs both ways.

Why does uniqueness need Lipschitz and not just continuity?

Continuity lets many candidate graphs squeeze through a point where

f

y

-slope is unbounded (like

y^{2/3}

0

). Lipschitz caps that slope, and via Grönwall two solutions cannot separate.

Why can we sometimes recover the full interval

h = min (a, b / M)

even after shrinking to

h < 1/ L

for the Banach step?

By continuation: solve on a short contraction interval, then restart from its endpoint as a new initial condition and patch the pieces together, extending the solution up to

h

(and toward the Maximal interval of existence).

Why does the Bielecki weighted norm

∥ u ∥ = sup e^{- 2 L ∣ x - x_{0} ∣} ∣ u (x) ∣

help?

The exponential weight discounts values far from

x_{0}

, absorbing the factor that would otherwise force

L h < 1

; under this norm

T

becomes a genuine contraction on the whole interval for any

h

Why is

M

(not

L

) the quantity that controls the length

h

M

bounds the slope, so it dictates how fast the graph can climb toward the top of the box;

L

governs how fast nearby solutions pull together (convergence/uniqueness), which is a separate concern from staying inside

R

Edge cases

What happens if

M = 0

on the rectangle?

Then

f \equiv 0

, so

y^{'} \equiv 0

and

y \equiv y_{0}

— a constant. The formula

b / M

+ \infty

, so

h = min (a, \infty) = a

; the trivial solution exists on the whole width, consistent with the theorem.

What if the starting point

(x_{0}, y_{0})

lies exactly on the boundary of

R

where

f

jumps?

The hypotheses require

f

continuous (hence bounded) on the closed

R

containing

(x_{0}, y_{0})

; if

f

is discontinuous there, no

M

exists and the theorem simply does not apply — you can't even guarantee existence.

At the boundary case

L h = 1

exactly (as with

y^{'} = x^{2} + y^{2}

), is uniqueness lost?

No.

L h = 1

only defeats the naive sup-norm contraction; the factorial/Bielecki route still gives a unique solution. Uniqueness never depended on

L h < 1

f

is Lipschitz with

L = 0

, what does that mean?

∣ f (x, y_{1}) - f (x, y_{2}) ∣ \leq 0

forces

f

to be independent of

y

, i.e.

y^{'} = g (x)

. Then

y = y_{0} + \int_{x_{0}}^{x} g

is the one and only solution — a pure integration, unconditionally unique.

Degenerate rectangle

b = 0

: is the theorem meaningful?

No usable interval —

b / M = 0

gives

h = 0

, so the "interval" is the single point

x_{0}

. A rectangle needs

b > 0

(and

a > 0

) to have any interior for the solution to live in.

What if

f

is Lipschitz on

R

but the solution reaches the edge

∣ y - y_{0} ∣ = b

before

x

hits

a

That is exactly the scenario

h = b / M < a

protects against: the theorem stops guaranteeing at

x_{0} + h

. Beyond it you must re-apply the theorem from a new base point (continuation) if

f

is still nice there.

Can a solution exist and be unique even when Picard–Lindelöf's hypotheses fail?

Yes. The theorem gives sufficient, not necessary, conditions — some non-Lipschitz problems still have unique solutions. Failing the hypotheses means "no guarantee," not "guaranteed to break."

Recall Quick self-test

The single fact that resolves most traps ::: $h = min (a, b / M)$ depends on $M$ , not $L$ ; Lipschitz controls uniqueness/convergence, continuity controls existence and the bound $M$ .