Existence and Uniqueness (Picard)

The promise and its fine print

For a first-order ODE $y' = f(t, y)$, we usually expect exactly one solution through a given point $(t_0, y_0)$. This is the content of the Picard–Lindelöf theorem (sometimes called the existence and uniqueness theorem, or the Picard theorem):

If $f$ and $\partial f / \partial y$ are continuous in a rectangle around $(t_0, y_0)$, then the IVP $y' = f(t, y),\ y(t_0) = y_0$ has a unique solution on some interval containing $t_0$.

Two hypotheses:

1. $f(t, y)$ is continuous.
2. $\partial f / \partial y$ is continuous (equivalently, $f$ is Lipschitz in $y$).

Today's problem shows what happens when hypothesis 2 fails.

The given IVP

$y' = \sqrt{y}, \qquad y(0) = 0.$

So $f(t, y) = \sqrt{y}$. Continuous on $y \ge 0$, so hypothesis 1 holds. But $\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y}},$ which blows up as $y \to 0^+$. At the initial point $y = 0$, $\partial f / \partial y$ is not even defined, let alone continuous. Hypothesis 2 fails — Picard says nothing, and indeed uniqueness fails.

Step-by-step — find multiple solutions

Solution A: $y \equiv 0$.

Check: $y' = 0$ and $\sqrt{y} = \sqrt{0} = 0$. ✓ The zero function is a perfectly valid solution.

Solution B: $y = t^2 / 4$ for $t \ge 0$.

Check: $y' = t/2$ and $\sqrt{y} = \sqrt{t^2/4} = t/2$ (for $t \ge 0$). ✓

And $y(0) = 0$. ✓

So we've found two solutions to the same IVP.

Solution C (and infinitely many more): the "peel-off" family.

For any $t^\star \ge 0$, define $y_{t^\star}(t) = \begin{cases} 0 & \text{if } 0 \le t \le t^\star \\ (t - t^\star)^2 / 4 & \text{if } t > t^\star \end{cases}$

This stays at zero until time $t^\star$, then "peels off" and grows as a shifted parabola. Check that it is continuously differentiable at the peel-off point:

• At $t = t^\star$: $y(t^\star) = 0$ from both pieces ✓.
• Derivative from the left: $0$. From the right: $y' = (t - t^\star)/2 \to 0$ as $t \to {t^\star}^+$. ✓
• And $y' = \sqrt{y}$ on each piece.

So $y_{t^\star}$ is a valid solution for every $t^\star \ge 0$. Uncountably many solutions share the initial condition.

Geometric interpretation — why the peel-off is possible

The slope field of $y' = \sqrt{y}$ is very flat near the $t$-axis ($y$ small ⇒ slope $\sqrt{y}$ small). A solution crawling along $y = 0$ can sit there indefinitely — the "force pulling it off" is proportional to $\sqrt{y}$, which is zero when $y = 0$. Unlike a Lipschitz-bounded vector field, where a slight kick away from an equilibrium would grow exponentially and force uniqueness, here the slow $\sqrt{\cdot}$ growth is too weak to pin the solution.

By contrast, $y' = y$ also has the equilibrium $y = 0$. But Picard applies ($\partial_y y = 1$ is continuous), and the only solution through $(0, 0)$ is $y \equiv 0$. You can't peel off. Linear pull makes all the difference.

Where uniqueness failure matters in the real world

Physics models usually obey Picard, but when they don't, it's often a red flag:

• Control systems: if the dynamics near a fixed point have $\sqrt{y}$-like slow restoring force, the system's response to a perturbation is non-unique — a tell that your model is missing something (e.g. a higher-order term).
• Hydrodynamics / dam-break problems: $y' = \sqrt{y}$-type equations describe a self-similar flow; the non-uniqueness reflects genuine physical indeterminacy in idealised models (the actual system requires extra information to pick the "correct" solution — often a viscosity term).
• Renormalisation of chaotic systems: some RG flows have non-unique trajectories that must be resolved by physical input.

So uniqueness failure isn't a math bug — it's a warning that the model is incomplete.

The Osgood criterion — a refinement

Picard's hypothesis is sufficient but not necessary. A weaker condition for uniqueness is Osgood's criterion: there exists a function $\omega$ with $\int_{0^+} du / \omega(u) = \infty$ such that $|f(t, y_1) - f(t, y_2)| \le \omega(|y_1 - y_2|).$

For $f = \sqrt{y}$, the modulus is $|\sqrt{y_1} - \sqrt{y_2}| \le \omega(|y_1 - y_2|)$ with $\omega(u) = \sqrt{u}$. But $\int_{0^+} du/\sqrt{u} < \infty$, so Osgood doesn't save us either — and indeed uniqueness fails.

A theorem about the peel-off family

The most "eager" solution (starts growing at $t = 0$) is $y = t^2/4$. The most "lazy" is $y \equiv 0$. Every solution is sandwiched: $0 \le y(t) \le t^2/4$ for all $t \ge 0$. This is the Peano funnel — the maximum and minimum solutions form an envelope; every IVP solution lies inside.

Common mistakes

• Thinking uniqueness always holds. It holds under mild hypotheses, and those hypotheses are satisfied for most textbook ODEs. But the counterexamples exist and they're important.
• Assuming initial conditions pin the solution uniquely. For well-behaved ODEs yes, but not in general. Always check Picard.
• Confusing existence with uniqueness. Peano's theorem guarantees existence from mere continuity of $f$, with no condition on $\partial f / \partial y$. But existence alone doesn't imply uniqueness.
• Dropping the absolute value under $\sqrt{\cdot}$. $\sqrt{y}$ is only real for $y \ge 0$. For $y < 0$ the original ODE isn't defined at all.

Why this matters pedagogically

It's the simplest, cleanest cautionary tale about the limits of ODE theory. Students often take existence and uniqueness for granted after seeing it invoked once; showing that $y' = \sqrt{y}$ has continuum many solutions through the origin makes the lesson stick.

Try it in the visualization

Plot both $y \equiv 0$ (flat along the t-axis) and $y = t^2 / 4$ (the rising parabola) through the origin. Add a slider for the "peel-off time" $t^\star$ from 0 to 3 and watch the hybrid solution $y_{t^\star}$ — flat until $t^\star$, then parabolic. Overlay a visualization of $\partial f/\partial y = 1/(2\sqrt{y})$ blowing up to $\infty$ along the $t$-axis, which is why Picard fails.