Existence–uniqueness, Picard, Lipschitz

In one sentence

Continuity of $f$ gives a solution, a Lipschitz condition in $y$ makes it unique, and neither gives a solution that lasts forever.

Why the exam asks it

Because three separate things — existence, uniqueness, and how long the solution lives — get compressed into one expectation, and each fails independently with a small explicit example. The options are built by mixing them up.

The idea

The initial value problem is $y' = f(x,y)$, $y(x_0) = y_0$.

Peano. $f$ continuous near $(x_0,y_0)$ $\Rightarrow$ a solution exists on some interval.

Picard–Lindelöf. $f$ continuous, and Lipschitz in $y$ — meaning $|f(x,y_1) - f(x,y_2)| \le L|y_1 - y_2|$ — $\Rightarrow$ the solution exists and is unique near $x_0$.

Both are local. Neither says anything about how far the solution extends.

Continuity is not enough for uniqueness

$$y' = y^{1/3}, \qquad y(0) = 0.$$

$f(y) = y^{1/3}$ is continuous everywhere, so Peano applies. But $y \equiv 0$ and $y = (2x/3)^{3/2}$ both solve it — check the second: its derivative is $(2x/3)^{1/2}$, which is exactly its own cube root.

The Lipschitz condition fails at $y = 0$ because $\frac{d}{dy}y^{1/3} = \frac13 y^{-2/3}$ is unbounded there. Infinitely many solutions exist in fact: stay at zero for any length of time, then leave.

A practical test: if $\partial f/\partial y$ exists and is continuous, $f$ is locally Lipschitz, which covers almost everything the paper hands you. Look for a fractional power or an absolute value to spot the exception.

Smoothness is not enough for global existence

$$y' = y^2, \qquad y(0) = 1.$$

Here $f(y) = y^2$ is a polynomial — smooth, locally Lipschitz, as well behaved as anything. The solution is $y = \dfrac{1}{1-x}$, which escapes to infinity as $x \to 1^-$ and does not exist beyond.

So the maximal interval of existence is $(-\infty, 1)$, determined by the solution rather than by the equation's domain. The general principle: a solution continues until it leaves every compact subset of the domain — either by blowing up or by reaching the boundary.

Linear equations are the exception. For $y' = a(x)y + b(x)$ with $a, b$ continuous on an interval $I$, the solution exists on all of $I$. Linearity rules out blow-up, which is why the distinction rarely arises in the next chapter.

Picard iteration

The proof is a construction worth knowing, since it is sometimes asked directly:

$$y_{n+1}(x) = y_0 + \int_{x_0}^{x} f(t, y_n(t))\,dt, \qquad y_0(x) = y_0 .$$

The Lipschitz condition makes this a contraction on a suitable complete space, and Banach's fixed point theorem — from the metric spaces chapter — gives a unique fixed point, which is the solution.

Gronwall's inequality is the other standard tool: if $u(x) \le C + \int_{x_0}^x K u(t)\,dt$ then $u(x) \le Ce^{K(x-x_0)}$. It gives uniqueness directly and bounds how fast solutions can separate.

Where intuition breaks

"$f$ continuous, so the solution is unique." $y' = y^{1/3}$. Continuity is Peano, which gives existence only.

"$f$ is smooth, so the solution exists for all $x$." $y' = y^2$. Smoothness gives local uniqueness, never global existence.

"The solution exists as long as $f$ is defined." $f(y) = y^2$ is defined on all of $\mathbb{R}^2$ and the solution dies at $x = 1$. The interval is a property of the solution.

"Lipschitz must hold everywhere." Local suffices, and locally Lipschitz follows from a continuous $\partial f/\partial y$. Options demand a global constant that is not needed.

"Non-uniqueness means two solutions." $y' = y^{1/3}$ from the origin has infinitely many — wait at zero for any duration, then leave.

"Uniqueness fails, so no solution exists." Peano still gives existence. The two conclusions are independent, and options trade one for the other.

"A linear equation can blow up in finite time." It cannot, on an interval where the coefficients are continuous. That is what makes linear systems tractable.

The exam's angle

1. Check the Lipschitz condition by differentiating in $y$. Unbounded there means uniqueness may fail, and the culprit is nearly always a fractional power.
2. For non-uniqueness, look at the initial point. The trouble is at $y_0$, and the standard picture is a solution leaving the equilibrium after an arbitrary wait.
3. Solve explicitly when separable. Most existence questions in this topic are answerable by solving and looking at the interval.
4. If the equation is linear, existence is global on the coefficients' interval. Say so and move on.
5. Distinguish the three claims — exists, unique, global — before reading the options, since each is separately true or false.

The night before

• Peano: $f$ continuous $\Rightarrow$ existence, locally. No uniqueness.
• Picard–Lindelöf: continuous $+$ Lipschitz in $y$ $\Rightarrow$ existence and uniqueness, locally.
• $\partial f/\partial y$ continuous $\Rightarrow$ locally Lipschitz. Fractional powers are where it fails.
• $y' = y^{1/3}$, $y(0)=0$: infinitely many solutions, including $0$ and $(2x/3)^{3/2}$.
• $y' = y^2$, $y(0)=1$: $y = 1/(1-x)$, blows up at $x = 1$ though $f$ is a polynomial.
• Maximal interval is a property of the solution; it ends when the solution leaves every compact set.
• Linear equations have global solutions on the interval where the coefficients are continuous.
• Picard iteration is Banach's fixed point theorem. Gronwall bounds separation.