NebluxDepth

Mathematics · Ordinary differential equations · 05

Existence and Uniqueness

From one starting point, is there always a way forward, and only one?

Earlier, exactly one exponential solution passed through each chosen starting point; that conclusion had conditions behind it. This page uses a theorem and counterexamples you can check directly to separate three things: existence, uniqueness, and how far a solution reaches.

Series · 05 / 09

Same initial value, but wait a while before leaving?

Choose the non-unique example and change the waiting point a. The initial condition stays y(0)=0. Predict whether the curves will all have the same height at x=2.

Chosen exact solutiona = 0 (dashed)Zero solution / singularity (dashed)
y′=3(∛y)², y(0)=0 allows both the zero solution and cubic solutions that leave after waiting. y′=y², y(0)=1 has the unique solution 1/(1−x), but only for x<1.
One unit may take up a different length on screen along each axis, so read a slope from the change in coordinates. The plot shows only a finite window; leaving it does not mean the solution ends.
Value at x = 2
Conclusion

Three questions, answered separately

“Is there a solution?”, “Is there only one?” and “How far does it reach?” are three different questions. One answer can be yes while another is no.

Write an initial value problem as y′=f(x,y), y(x₀)=y₀. x₀ and y₀ give the starting point; f is the given rule for the slope. Existence asks whether there is at least one such function. Uniqueness asks whether two different functions could start from the same point. Global existence also asks whether the solution extends over the whole stretch you care about.

We are looking for a function that is differentiable on an open interval containing x₀ and satisfies both the equation and the initial value. Drawing one numerical path proves no uniqueness, and it cannot rule out another curve that was never computed. Each of the three questions needs its own answer, from conditions on the equation or from exact calculation.

A local theorem you can check

The theorem works like a checklist: if the rule is smooth enough around the starting point, there is exactly one solution nearby.

Here is the common sufficient-condition version of the existence and uniqueness theorem; the general theorem is not proved here:

Take an open rectangle with (x₀,y₀) inside it. If f(x,y) and its partial derivative fy(x,y) exist and are continuous on that rectangle, then for some h>0 the initial value problem has exactly one solution on x₀−h<x<x₀+h (with h small enough that the solution stays in the rectangle).

fy means differentiate with respect to y while holding x fixed. It is not the total derivative along a solution curve. For example, f=2xy gives fy=2x, and f=y² gives fy=2y. What you check is a whole region around the starting point, not one number at the point.

If you only know that f is continuous near the start, a local solution is already guaranteed to exist (Peano’s existence theorem), but uniqueness needs something more. A common, more general version asks that f be locally Lipschitz in y: on a small rectangle there is one fixed L with |f(x,u)−f(x,v)|≤L|u−v|. It stops the difference in slope between two nearby heights from growing without bound. A continuous fy is a convenient sufficient condition, not the only way in.

  1. Write a familiar example as f

    y′=y; f(x,y)=y

    f is continuous around every starting point.

  2. Differentiate in y with x held fixed

    fᵧ(x,y)=1

    The partial derivative is continuous everywhere too, so every initial value has a unique local solution.

  3. Check the global part separately

    y(x)=y₀ ex−x₀

    The formula is defined for every finite real x and satisfies the equation. That is what establishes global existence for this example; “local” does not quietly become “forever.”

A counterexample: one point, many solutions

One starting point really can grow several solutions. Nothing was miscalculated; the rule itself allows it.

Go back to the example from the second page and check it fully this time:

y′ = 3(∛y)²,   y(0)=0∛y is the real cube root, defined for negative numbers too; (∛y)² = |y|^(2/3).

The zero function clearly makes both sides 0. Now fix any a≥0 and define a waiting solution:

yₐ(x) = 0, x ≤ ayₐ(x) = (x−a)³, x > aa is a chosen constant, not the independent variable. Every solution with a≥0 passes through (0,0).

  1. Check while it waits

    x<a: yₐ′=0, 3(∛yₐ)²=0

    The two sides agree.

  2. Check after it leaves

    x>a: yₐ′=3(x−a)²

    On the right, the cube root is x−a; squaring and multiplying by 3 gives the same result.

  3. Don’t skip the joining point

    x=a: both sides have value 0 and derivative 0

    The right-hand difference quotient h³/h=h² tends to 0, and the left-hand one is 0. So the function is differentiable at the join and still satisfies the equation.

  4. Confirm: different, with the same initial value

    a≥0 ⇒ yₐ(0)=0

    For example, at x=2 the solutions with a=0 and a=1 are 8 and 1. That proves directly that they are different.

Here f is continuous, so it is no surprise that a solution exists. But fy has no finite value at y=0. More directly, |f(x,ε)−f(x,0)|/|ε|=3ε−1/3 (ε>0), which is unbounded as ε→0, so no single L works.

Uniqueness does not fail just because the theorem cannot be used. The real evidence of non-uniqueness is that we wrote down different solutions and checked each piece. This family is enough for a counterexample; it does not claim to list every solution.

A failed condition is not the opposite conclusion

Failing the checklist only means this method cannot give a guarantee. It does not mean there must be many solutions.

Compare y′=|y|, y(0)=0. Here too fy does not exist at zero, but the reverse triangle inequality gives

||u|−|v|| ≤ |u−v|

So f satisfies the Lipschitz condition in y with L=1. The more general theorem still guarantees uniqueness, and since y=0 is already a solution, it is the only solution with this initial value. This example separates “one way of checking failed” from “uniqueness really fails.”

There may also be no solution at all. Take x y′=1, required to hold on an interval containing x=0: putting in x=0 immediately gives 0=1, a contradiction. Dividing to get y′=1/x shows a singularity, but that cannot be used to quietly move the starting point.

A smooth rule can still blow up

However smooth the rule, the guarantee is only for “nearby.” A solution can still run off to infinity at a finite position.

Consider y′=y², y(0)=1. f=y² and fy=2y are continuous everywhere, so local uniqueness is not in doubt. The candidate solution, and its check:

  1. Write it down and check the initial value

    y(x)=1/(1−x), y(0)=1

    How to find it comes later, on the pages about separable equations and intervals; here we only check it.

  2. Check the derivative

    y′(x)=1/(1−x)² = y(x)²

    This holds for x≠1. The connected interval containing the starting point 0 is (−∞,1).

  3. Look at the right end

    x→1⁻ ⇒ y(x)→+∞

    No finite y(1) can extend it continuously, let alone differentiably. Nothing is wrong with the equation; this solution has left every finite height.

The same formula draws another curve for x>1, but it does not pass through (0,1), and it cannot be joined on across the blow-up. The local theorem guarantees a start, not an endless journey. “Separable Differential Equations” begins solving, and “Implicit Solutions and Solution Intervals” takes the question of intervals all the way apart.

What is guaranteed, and what is not yet known

When you answer, keep apart what the theorem guarantees and what only looks true on the plot.

1. If f and fᵧ are continuous on the whole plane, does every solution exist for all real x?

See the full reasoning

No. y′=y², y(0)=1 is a counterexample: the solution blows up at x=1. The conditions guarantee local existence and uniqueness at every finite point, but not that a solution’s height stays finite forever.

2. What is the derivative of the waiting solution with a=2 at its joining point? Does it satisfy y(0)=0?

See the full reasoning

The derivative at the join is 0: the constant on the left has derivative 0, and 3(x−2)² on the right also tends to 0 as x→2⁺; the difference quotients prove it is differentiable as well. Since 0≤2, y₂(0)=0, so the initial value holds.

3. If the plot draws only one curve, can you declare the solution unique?

See the full reasoning

No. The program may simply have picked one branch. For example, Euler’s method started at the zero point of the non-unique example sees slope 0 at every step, so it draws only the zero solution. Uniqueness needs the theorem’s conditions or another proof; pixels are no evidence that other solutions are absent.

Sources and scope

Definitions and the conditions of theorems were checked against the authors and university materials below. The explanations, worked examples and interactive plots here are written independently. Decimals on the plots are rounded displays; the mathematical conclusions rest on derivation and substitution. This page does not prove the general existence and uniqueness theorems it cites.