Differential-Equations · Unit 1 · Video 5 · Interactive Practice

Why Integral Curves Never Cross or Touch: Existence and Uniqueness

IKey Formulas

Statement	Name	Hypothesis it costs
Integral curves of $y' = f(x,y)$ cannot cross at a positive angle	Integral Curve Theorem (i)	$f(x,y)$ merely defined in the region
Integral curves cannot even be tangent	Integral Curve Theorem (ii)	$f_y(x,y)$ continuous in the region
Through $(x_0, y_0)$ there is one, and only one, solution	Existence and Uniqueness	$f$ and $\dfrac{\partial f}{\partial y}$ continuous near $(x_0, y_0)$
Integral curves cannot intersect at all	Intersection Principle	$f$ and $\dfrac{\partial f}{\partial y}$ continuous in the region

Key Insight: "One, and only one" is two theorems in one phrase — continuity of $f$ buys at least one solution (existence), continuity of $\partial f/\partial y$ buys at most one (uniqueness). Crossing at an angle is vetoed by the direction field alone; mere tangency is vetoed only by uniqueness.

IIVisualization 1 — One Curve Through Each Point

Here $\partial f/\partial y = -1$ is continuous, so exactly one integral curve of $y' = 1 + x - y$ threads through each point.

x₀ 1 y₀ 3

IIIVisualization 2 — Asymptotic, Never Touching

The gap between a trapped curve and the solution $y = x$ shrinks toward zero but is never zero.

x 1.50

💡 If the gap ever closed, two integral curves would share that point — and only one curve is allowed through each point.

IVVisualization 3 — When Uniqueness Fails

Where $\partial f/\partial y$ blows up, infinitely many solutions can run through the same point.

lift-off point c 2.4

💡 A failed hypothesis never proves non-uniqueness — it only withdraws the guarantee; here uniqueness genuinely fails as well.

VQuiz Questions

Problem 1 · The Derivative That Decides Uniqueness

Given: $y' = f(x,y) = x^2 + y^2$ — find $\dfrac{\partial f}{\partial y}$ , the quantity the uniqueness hypothesis asks about.

✅ Correct!

f_y = 2y

is continuous everywhere, so — together with

f

itself being continuous — exactly one integral curve runs through every point of the plane.

❌ That is $\dfrac{\partial f}{\partial x}$ . The theorem's second hypothesis differentiates with respect to

y

, holding

x

fixed.

❌ Not quite. With

x

held fixed,

x^2

is a constant and contributes

0

— not

2x

❌ The $x^2$ was carried along undifferentiated. A partial derivative in

y

sends every

x

-only term to

0

❌ Not quite. Treat

x

as a constant and differentiate

x^2 + y^2

y

Show solution

Here $f(x,y) = x^2 + y^2$ . Holding $x$ fixed:

\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\big(x^2\big) + \frac{\partial}{\partial y}\big(y^2\big) = 0 + 2y = 2y

Both hypotheses now check out at every point: $f$ is a polynomial, hence continuous, which gives existence; and $f_y = 2y$ is continuous, which gives uniqueness. So through each point of the plane runs exactly one integral curve, and no two of them ever meet.

Problem 2 · Existence Without Uniqueness

Given: $y' = 3y^{2/3}$ and the point $(0, 0)$ — which conclusion does the Existence and Uniqueness Theorem support?

✅ Correct!

f = 3y^{2/3}

is continuous everywhere (existence holds), but

f_y = 2y^{-1/3}

is not continuous at

y = 0

(uniqueness is withdrawn) — and here it genuinely fails:

y \equiv 0

and

y = x^3

both pass through the origin.

❌ Uniqueness needs the second hypothesis.

f_y = 2y^{-1/3}

blows up as

y \to 0

, so it is not continuous near

(0,0)

❌ Existence survives. It only asks that

f = 3y^{2/3}

be continuous near the point, and it is;

y \equiv 0

is already one solution through the origin.

❌ Only the second hypothesis fails.

f

itself is continuous everywhere, so the existence half of the theorem still applies.

❌ Not quite. Test the hypotheses one at a time: is

f

continuous at

(0,0)

? Is

\partial f/\partial y

Show solution

Hypothesis 1 — existence. $f(x,y) = 3y^{2/3}$ is continuous at every point, including $(0,0)$ . At least one solution passes through the origin.

Hypothesis 2 — uniqueness.

\frac{\partial f}{\partial y} = 3 \cdot \tfrac{2}{3} y^{-1/3} = \frac{2}{y^{1/3}} \longrightarrow \infty \quad \text{as } y \to 0

Every neighbourhood of $(0,0)$ contains points with $y = 0$ , so $f_y$ is not continuous near the origin and uniqueness is not guaranteed.

It really does fail. Both of these solve the equation and pass through $(0,0)$ :

$y \equiv 0$ : then $y' = 0$ and $3y^{2/3} = 0$ ✓
$y = x^3$ : then $y' = 3x^2$ and $3\big(x^3\big)^{2/3} = 3x^2$ ✓

Two integral curves through one point — exactly what the theorem forbids wherever its hypotheses hold.

Problem 3 · The Corridor Curve Through $(1, 3)$

Given: every solution of $y' = 1 + x - y$ has the form $y = x + Ce^{-x}$ — find the constant $C$ for the solution through $(1, 3)$ , then decide whether that curve ever meets the line $y = x$ .

What is $C$ ?

Does that curve ever meet the line?

✅ Correct!

C = 2e

, so

y - x = 2e^{\,1-x}

— it decays by a factor

e

per unit of

x

and never reaches

0

. The curve is asymptotic to

y = x

, never touching it.

❌ Check the exponential factor. At

x = 1

the term is

Ce^{-1}

, not

C

: solving

Ce^{-1} = 2

multiplies by

e

❌ You used $y_0 = 3$ where the equation needs $y_0 - x_0$ . Substituting gives

3 = 1 + Ce^{-1}

, so

Ce^{-1} = 2

❌ Sign slip in the exponent. From

Ce^{-1} = 2

you multiply by

e

, giving

C = 2e

, not

2/e

❌ Check the substitution. Put

x = 1

y = 3

into

y = x + Ce^{-x}

and solve for

C

❌ A limit is not an arrival.

2e^{\,1-x}

approaches

0

but is never equal to

0

— there is no value of

x

at which the curves meet.

❌ The gap is $2e^{\,1-x}$ , which is strictly decreasing — it has no smallest value. At

x = 1

the gap is

2

, hardly a touch. And tangency is forbidden here anyway.

❌ Right answer, wrong reason.

y = x

is a solution:

y' = 1

and

1 + x - x = 1

. That is exactly why no other integral curve may touch it.

❌ Not quite. Compute

y - x

for your curve and ask whether that expression can equal zero.

Show solution

Step 1 — find $C$ . Substitute $(x, y) = (1, 3)$ into $y = x + Ce^{-x}$ :

3 = 1 + Ce^{-1} \quad\Longrightarrow\quad Ce^{-1} = 2 \quad\Longrightarrow\quad C = 2e

Step 2 — measure the gap.

y - x = 2e \cdot e^{-x} = 2e^{\,1-x} > 0 \quad \text{for every } x

An exponential is never zero, so the curve stays strictly above $y = x$ forever, though the gap falls by a factor of $e$ for each unit of $x$ : at $x = 5$ it is $2e^{-4} \approx 0.037$ , at $x = 10$ about $0.00025$ .

Why it could not have been otherwise. $y = x$ is itself a solution ( $y' = 1$ and $1 + x - x = 1$ ). Here $f = 1 + x - y$ and $\partial f/\partial y = -1$ are continuous everywhere, so exactly one integral curve passes through each point. If our curve ever reached $y = x$ , two integral curves would share that point.

Problem 4 · A Constant Solution as a Wall

Given: $y' = f(x,y)$ with $f$ and $\partial f/\partial y$ continuous on the whole plane, and the constant function $y \equiv 2$ is one of its solutions. Another solution satisfies $y(0) = 1$ . What is true of that solution for every $x$ ?

✅ Correct! The horizontal line

y = 2

is an integral curve, and the hypotheses hold everywhere, so no other integral curve may share a point with it. Starting below at

x = 0

, the solution is trapped below

2

forever.

❌ Touching is intersecting. Equality at even one value of

x

would put two integral curves through that point — forbidden exactly as strongly as a crossing.

❌ That claims too much. Staying below

2

says nothing about where the curve goes; it might fall away from the line instead of approaching it.

❌ Matching slopes is precisely tangency. The direction field raises no objection, but uniqueness does: one curve per point.

❌ Not quite. Remember that

y \equiv 2

is itself an integral curve, and that curves may not share a point.

Show solution

Since $f$ and $\partial f/\partial y$ are continuous everywhere, the Intersection Principle applies on the whole plane: no two integral curves may share a point.

The line $y = 2$ is an integral curve. Our solution starts at $(0, 1)$ , strictly below it. If $y(x_1) = 2$ for some $x_1$ , then the point $(x_1, 2)$ would lie on two different integral curves — our solution and the constant solution — which uniqueness forbids.

So $y(x) \ne 2$ for every $x$ , and since the solution is continuous and begins below $2$ , it stays below:

y(x) < 2 \quad \text{for every } x

This is why constant solutions act as walls: they partition the plane into bands that no other solution can leave.

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