Existence and Uniqueness Theorem for First-Order Equations

Every method in this book assumed a solution was there to be found. This chapter says when that assumption is safe. Continuity of $f$ guarantees that a solution exists; continuity of $\partial f / \partial y$ as well guarantees that it is unique. Both conclusions are local, holding only on some interval around the starting point, and the second worked example shows what goes wrong when the extra hypothesis fails.

Quick Reference

Item	Statement
The problem	$\frac{d y}{d x} = f (x, y)$ , $y (x_{0}) = y_{0}$
The region	An open rectangle $R = (a, b) \times (c, d)$ containing $(x_{0}, y_{0})$
Existence	$f$ continuous on $R$ $⟹$ at least one solution on some open subinterval of $(a, b)$ containing $x_{0}$
Uniqueness	$f$ and $\frac{\partial f}{\partial y}$ both continuous on $R$ $⟹$ exactly one solution there
Both are local	The interval of existence may be much smaller than $(a, b)$
Geometric meaning	Through each point of $R$ passes exactly one integral curve
Standard failure	y' = 3y^{2/3} at $y = 0$ , where $\partial f / \partial y$ blows up
Linear case	For y' + P(x)y = Q(x) with $P$ , $Q$ continuous on an interval, the solution exists and is unique on that entire interval

The Theorem

Consider the initial value problem

\frac{d y}{d x} = f (x, y), y (x_{0}) = y_{0} .

Let $R$ denote an open rectangle

R = (a, b) \times (c, d) = {(x, y) | a < x < b and c < y < d}

containing the point $(x_{0}, y_{0})$ .

If $f$ is continuous at each point of $R$ , then the initial value problem has at least one solution on some open subinterval of $(a, b)$ containing $x_{0}$ .
If both $f$ and $\frac{\partial f}{\partial y}$ are continuous at each point of $R$ , then the initial value problem has a unique solution on some open subinterval of $(a, b)$ containing $x_{0}$ .

This result is often called the Picard-Lindelöf theorem, and part 1 alone is Peano's existence theorem.

Three things to notice.

The hypotheses are on $f$ , not on the solution. They can be checked by inspecting the right-hand side of the equation, before any attempt at solving.

Existence and uniqueness are separate. Continuity of $f$ alone gives a solution but permits several; the extra hypothesis on $\partial f / \partial y$ is what forces there to be only one.

Both conclusions are local. The theorem promises a solution only on some subinterval containing $x_{0}$ , possibly a very short one, even when $f$ is beautifully behaved on a huge rectangle.

Geometric Meaning

Recall from Section 2.2 that y' = f(x,y) prescribes a slope at every point of the plane. The theorem is the statement that the resulting direction field never branches inside $R$ :

Existence says a solution curve can be traced through every point of $R$ , following the field.
Uniqueness says only one such curve passes through each point.

Consequently, in a region where both hypotheses hold, two distinct integral curves can never cross, touch, or merge. That single fact justifies a great deal of informal reasoning about direction fields, and it is what makes equilibrium solutions act as impassable barriers.

Why Only Local?

Show that y' = y^{2} with $y (0) = 1$ has a solution only on a bounded interval, even though $f (x, y) = y^{2}$ is continuous everywhere.

Solution

Both $f = y^{2}$ and $\frac{\partial f}{\partial y} = 2 y$ are continuous on the entire plane, so the theorem guarantees a unique solution near $x = 0$ . Solving by separation,

\frac{d y}{y^{2}} = d x ⟹ - \frac{1}{y} = x + c ⟹ y = \frac{- 1}{x + c} .

The condition $y (0) = 1$ gives $c = - 1$ , so

y = \frac{1}{1 - x} .

This solution blows up as $x \to 1^{-}$ and exists only on $(- \infty, 1)$ , despite $f$ being perfectly well behaved on all of $ℝ^{2}$ .

This is why the theorem cannot promise more than a local interval. A solution can escape to infinity in finite time, and nothing in the equation's coefficients warns of it. Note also that the interval depends on the initial value: starting from $y (0) = 10$ gives $y = \frac{10}{1 - 10 x}$ , which fails already at $x = \frac{1}{10}$ .

Contrast with linear equations. For a linear equation y' + P(x)y = Q(x) with $P$ and $Q$ continuous on an interval $I$ , the solution exists and is unique on the whole of $I$ . There is no local restriction, and the interval of validity can be read off in advance from the coefficients. This is one of the deep advantages of linearity, and the reason interval-of-validity questions in Chapter 2 arose only for nonlinear equations.

When Uniqueness Fails

Show that the initial value problem

\frac{d y}{d x} = 3 y^{2 / 3}, y (0) = 0

has infinitely many solutions.

Solution

Check the hypotheses. The function $f (x, y) = 3 y^{2 / 3}$ is continuous everywhere, so part 1 applies and a solution exists. But

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

is undefined at $y = 0$ , and no rectangle containing the initial point $(0, 0)$ can avoid the line $y = 0$ . So part 2 does not apply, and uniqueness is not guaranteed.

Find the solutions. First, the constant function $y \equiv 0$ satisfies the equation, since both sides are zero, and it meets the initial condition.

Second, separating variables for $y \neq 0$ ,

\frac{d y}{y^{2 / 3}} = 3 d x ⟹ 3 y^{1 / 3} = 3 x + c ⟹ y = (x + k)^{3} .

Taking $k = 0$ gives $y = x^{3}$ , which also satisfies $y (0) = 0$ .

Infinitely many more. For any $k \geq 0$ , the function

y(x) = \begin{cases} 0 & x \leq k \\ (x-k)^{3} & x > k \end{cases}

is a solution. It is differentiable at $x = k$ , where both pieces have value $0$ and derivative $0$ , and it satisfies the equation on each piece. So the solution may stay at zero for any length of time and then take off.

The picture is that infinitely many integral curves pass through the origin, in direct violation of the "no branching" property. Only the failure of $\partial f / \partial y$ to be continuous makes this possible.

This example is worth remembering as the standard cautionary tale. Physically it corresponds to a system whose future is not determined by its present state, which is why the extra hypothesis matters far beyond its technical role in the proof.

Applying the Theorem

Determine whether the theorem guarantees a unique solution near the given point.

y' = x^{2} + y^{2}, $y (0) = 1$
y' = \dfrac{y}{x}, $y (0) = 2$
y' = \sqrt{y}, $y (0) = 4$
y' = \sqrt{y}, $y (0) = 0$

Solution

Yes. Both $f = x^{2} + y^{2}$ and $\frac{\partial f}{\partial y} = 2 y$ are continuous everywhere, so a unique solution exists near $x = 0$ . Note that this equation has no elementary solution, yet the theorem still applies: existence does not require solvability in closed form.
No. Here $f = \frac{y}{x}$ is not even defined at $x = 0$ , so no rectangle around $(0, 2)$ works and the theorem says nothing. Indeed the general solution is $y = c x$ , and every member passes through the origin but none through $(0, 2)$ : this problem has no solution at all.
Yes. With $f = \sqrt{y}$ and $\frac{\partial f}{\partial y} = \frac{1}{2 \sqrt{y}}$ , both are continuous on any rectangle with $y > 0$ , and $(0, 4)$ has such a neighborhood. A unique solution exists near $x = 0$ .
No. The same $\partial f / \partial y$ blows up at $y = 0$ , and every rectangle containing $(0, 0)$ meets that line. As in the previous example, uniqueness genuinely fails: both $y \equiv 0$ and $y = \frac{x^{2}}{4}$ solve this problem.

Case 3 versus case 4 shows how sharply the conclusion can depend on the initial value alone, with the equation unchanged.

Exercises

Does the theorem guarantee a unique solution of y' = \dfrac{1}{y-2} with $y (0) = 3$ ? With $y (0) = 2$ ?

Solution

Here $f = \frac{1}{y - 2}$ and $\frac{\partial f}{\partial y} = \frac{- 1}{(y - 2)^{2}}$ , both continuous except on the line $y = 2$ .

For $y (0) = 3$ : a rectangle such as $(- 1, 1) \times (2.5, 3.5)$ contains the point and avoids the bad line, so yes, a unique solution exists near $x = 0$ .

For $y (0) = 2$ : the initial point lies on the excluded line, and $f$ is not even defined there. The theorem gives no information, and in fact y' would have to be infinite at the start.

The problem y' = 2\sqrt{\left|y\right|}, $y (0) = 0$ has more than one solution. Exhibit two of them.

Solution

The constant $y \equiv 0$ is one, since both sides vanish.

For $y > 0$ , separating gives $\frac{d y}{2 \sqrt{y}} = d x$ , so $\sqrt{y} = x + c$ and $y = x^{2}$ with $c = 0$ . Checking: y' = 2x and $2 \sqrt{| x^{2} |} = 2 | x |$ , which agree for $x \geq 0$ . So

y(x) = \begin{cases} 0 & x \leq 0 \\ x^{2} & x > 0 \end{cases}

is a second solution.

Uniqueness fails because $\frac{\partial f}{\partial y} = \frac{1}{\sqrt{y}}$ is unbounded at $y = 0$ , exactly the situation of the worked example.

Find the largest interval on which the theorem guarantees a solution of the linear problem \left(x-3\right)y' + \left(\ln x\right)y = 2x, $y (1) = 5$ .

Solution

In standard form,

y' + \frac{\ln x}{x-3}y = \frac{2x}{x-3} .

The coefficients are continuous where $x > 0$ , required by $\ln x$ , and $x \neq 3$ . The initial point $x = 1$ lies in the interval $(0, 3)$ , so the solution exists and is unique on all of

0 < x < 3 .

Note the contrast with the nonlinear case: because the equation is linear, the interval can be determined in advance from the coefficients alone, with no solving and no local restriction.

Frequently Asked Questions

What does the existence and uniqueness theorem actually say?

That if $f$ is continuous near the initial point, at least one solution passes through it, and if $\partial f / \partial y$ is continuous there as well, exactly one does. Both conclusions hold on some interval around $x_{0}$ , not necessarily a large one.

Why is continuity of

\partial f / \partial y

needed for uniqueness?

It controls how fast $f$ can change with $y$ , which prevents two solutions that start together from separating. Technically it supplies a Lipschitz condition in $y$ , which is what the proof actually uses. In y' = 3y^{2/3} that derivative blows up at $y = 0$ , and uniqueness fails precisely there.

What does "local" mean here?

That the guaranteed interval may be much smaller than the rectangle $R$ . The equation y' = y^{2} with $y (0) = 1$ has $f$ continuous on the whole plane, yet its solution $\frac{1}{1 - x}$ escapes to infinity at $x = 1$ . Nothing in the hypotheses can prevent that, so nothing in the conclusion can promise a longer interval.

Does failing the hypotheses mean the solution is not unique?

No. The conditions are sufficient, not necessary. Failing them means the theorem gives no verdict, and uniqueness may still hold for other reasons. The most you can conclude is that this particular theorem does not apply.

Why is the linear case better?

Because a linear equation's solution is given by an explicit integrating-factor formula, valid wherever the coefficients are continuous. There is no possibility of blow-up in finite time from the equation itself, so existence and uniqueness hold on the entire interval of continuity rather than on some unknown subinterval.

How does this connect to direction fields?

Directly. Uniqueness is the statement that integral curves never cross inside the region. It is what allows equilibrium solutions to act as barriers that trap other solutions in horizontal strips, and it is why the qualitative reasoning of Section 2.2 is trustworthy wherever the hypotheses hold.

Existence and Uniqueness of Solutions of Nonlinear Equations

Quick Reference

The Theorem

Geometric Meaning

Why Only Local?

When Uniqueness Fails

Applying the Theorem

Exercises

Frequently Asked Questions

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