Uniqueness Theorem — Definition, Formula & Examples

A uniqueness theorem states that under certain conditions, a differential equation has at most one solution passing through a given initial point. It tells you that if a solution exists, no other solution can satisfy the same initial condition.

For the initial value problem $y' = f(t, y)$ , $y(t_0) = y_0$ , the Picard-Lindelöf (existence and uniqueness) theorem guarantees a unique solution in a neighborhood of $t_0$ provided that $f$ is continuous in both variables and Lipschitz continuous in $y$ on a rectangle containing $(t_0, y_0)$ . The Lipschitz condition requires a constant $L > 0$ such that $|f(t, y_1) - f(t, y_2)| \leq L|y_1 - y_2|$ for all points in that rectangle.

Key Formula

|f(t, y_1) - f(t, y_2)| \leq L\,|y_1 - y_2|

Where:

$f(t, y)$ = The right-hand side of the ODE y' = f(t, y)
$L$ = Lipschitz constant, a positive real number
$y_1, y_2$ = Two values of the dependent variable in the region of interest

How It Works

To apply the uniqueness theorem, you check two conditions on the function

f(t, y)

y' = f(t,y)

. First, verify that

f

is continuous near the initial point

(t_0, y_0)

. Second, confirm that

\partial f / \partial y

exists and is continuous near that point — this is a convenient sufficient condition for Lipschitz continuity. If both conditions hold, exactly one solution curve passes through

(t_0, y_0)

. When either condition fails, multiple solutions may exist through the same point.

Worked Example

Problem: Determine whether the IVP y' = y^(1/3), y(0) = 0 has a unique solution.

Identify f(t, y): Here f(t, y) = y^(1/3). This function is continuous everywhere, so the existence condition is satisfied.

f(t, y) = y^{1/3}

Check the Lipschitz / partial derivative condition: Compute the partial derivative with respect to y.

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

Evaluate at the initial point: At y = 0, the partial derivative is undefined (it blows up to infinity). The Lipschitz condition fails near (0, 0), so the uniqueness theorem does not guarantee a unique solution.

\left.\frac{\partial f}{\partial y}\right|_{y=0} = \frac{1}{3}(0)^{-2/3} \to \infty

Answer: The uniqueness theorem does not apply. In fact, both y(t) = 0 and y(t) = (2t/3)^(3/2) satisfy this IVP, confirming that the solution is not unique.

Why It Matters

In engineering and physics, knowing a model has a unique solution means the system's future is fully determined by its current state. Without uniqueness, a physical model would predict multiple contradictory outcomes from identical starting conditions, signaling the model needs refinement. You encounter uniqueness arguments throughout courses in ODEs, PDEs, and real analysis.

Common Mistakes

Mistake: Assuming continuity of f alone guarantees uniqueness.

Correction: Continuity of f ensures existence (by Peano's theorem), but uniqueness requires the additional Lipschitz condition on f with respect to y. Always check both conditions.