Existence, Uniqueness, and Maximal Intervals, differential equations lesson

A curve sketched through an initial point of a direction field rests on two hopes, that a solution really starts there and that the same data cannot lead to two different futures. Neither hope is automatic, and even when both hold, the solution may exist only on a limited interval of time. Throughout, $t$ is the independent variable, $y' = dy/dt$ , and the problem is $y' = f(t, y)$ with $y(t_0) = y_0$ . This lesson states the hypotheses that guarantee a unique solution near the initial time and shows what they do not promise further away.

Key terms

Local existence

Definition Let $a > 0$ and $b > 0$ , and let $f$ be continuous on the closed rectangle $Q$ of points $(t, y)$ with $\lvert t - t_0 \rvert \le a$ and $\lvert y - y_0 \rvert \le b$ . Let $M$ be the largest value of $\lvert f \rvert$ on $Q$ . Then the problem $y' = f(t, y)$ , $y(t_0) = y_0$ has at least one solution on the open interval $(t_0 - h, t_0 + h)$ , where

h = \min\left(a, \frac{b}{M}\right),

with $h = a$ when $M = 0$ , and the graph of that solution stays inside $Q$ . This is the Peano existence theorem. Its conclusion is local, because $h$ can be much smaller than the range of times on which $f$ is defined.

In plain words If the right side is continuous near the starting point, at least one solution curve passes through that point and lasts at least a short time on each side of $t_0$ . The theorem says nothing about how long the solution lasts beyond that, or whether it is the only one.

Example For $y' = y^2$ with $y(0) = 1$ , take the rectangle $\lvert t \rvert \le a$ , $\lvert y - 1 \rvert \le b$ . The largest value of $y^2$ on it is $M = (1 + b)^2$ , so the theorem gives a solution for $\lvert t \rvert < h$ with $h = \min\left(a, b/(1 + b)^2\right)$ . The fraction $b/(1 + b)^2$ is largest at $b = 1$ , where it equals $\tfrac{1}{4}$ , so the best interval this estimate can guarantee is $(-\tfrac{1}{4}, \tfrac{1}{4})$ , reached with $b = 1$ and any $a \ge \tfrac{1}{4}$ . The actual solution $y = 1/(1 - t)$ exists on the much longer interval $(-\infty, 1)$ . The theorem gives a lower bound for the interval of existence, not its true size.

Local Lipschitz condition

Definition Let $D$ be an open set in the $ty$ -plane and $f$ a function on $D$ . The function $f$ is locally Lipschitz in $y$ on $D$ if every point of $D$ is the center of a closed rectangle $Q \subseteq D$ that has a constant $L \ge 0$ with

\begin{aligned} &\lvert f(t, y_1) - f(t, y_2) \rvert \\[0.4em] &\qquad \le L \lvert y_1 - y_2 \rvert \end{aligned}

for all points $(t, y_1)$ and $(t, y_2)$ of $Q$ . The constant may change from one rectangle to another, but inside one rectangle it works for every $t$ at once. If $\partial f/\partial y$ exists and is continuous on $D$ , then $f$ is locally Lipschitz in $y$ , because by the mean value theorem $L$ can be taken as the largest value of $\lvert \partial f/\partial y \rvert$ on $Q$ . This derivative test is sufficient but not necessary.

In plain words Compare the slopes that the equation assigns to two nearby states at the same time. The condition says that their difference is at most a fixed multiple of the distance between the states. Near each point, the slopes can change with $y$ only at a bounded rate.

Example The function $f(t, y) = \lvert y \rvert$ has a corner at $y = 0$ , so the derivative test fails there. It is still Lipschitz with $L = 1$ on the whole plane, since

\bigl\lvert \lvert y_1 \rvert - \lvert y_2 \rvert \bigr\rvert \le \lvert y_1 - y_2 \rvert.

By contrast, $f(t, y) = 2\sqrt{\lvert y \rvert}$ is not Lipschitz in $y$ on any rectangle around a point of the $t$ -axis. Comparing a state $y \ne 0$ with the state $0$ gives the ratio

\frac{2\sqrt{\lvert y \rvert}}{\lvert y \rvert} = \frac{2}{\sqrt{\lvert y \rvert}},

which grows without bound as $y \to 0$ , so no single constant $L$ works.

Uniqueness of solutions

Definition Let $f$ be continuous on an open set $D$ and locally Lipschitz in $y$ on $D$ , and let $(t_0, y_0)$ be a point of $D$ . Then the problem $y' = f(t, y)$ , $y(t_0) = y_0$ has a solution on some open interval containing $t_0$ . Moreover, any two solutions of the problem, each defined on an interval containing $t_0$ , agree at every $t$ where both are defined. This is often called the Picard-Lindelöf theorem. As a consequence, two solutions of the equation whose graphs share one point of $D$ agree on the whole common part of their intervals.

In plain words With continuity and a local Lipschitz bound, the initial data determine the solution both forward and backward in time, for as long as it exists. Two different solution curves can never cross or touch inside $D$ .

Example For $y' = y^2$ the partial derivative $\partial f/\partial y = 2y$ is continuous on the whole plane, so every initial-value problem for this equation has exactly one solution near its initial time. Since $y = 0$ is a solution, the solution with $y(0) = 1$ can never reach the value $0$ , and it stays positive on its whole interval. The theorem also applies to $y' = \lvert y \rvert$ with $y(0) = 0$ . The derivative test fails at the starting point, but the Lipschitz condition holds, so the zero function is the only solution.

Maximal interval of existence

Definition Let $f$ satisfy the hypotheses of the uniqueness theorem on the open set $D$ , and let $(t_0, y_0)$ be in $D$ . Then the problem has a solution $y$ on an open interval $(\alpha, \beta)$ containing $t_0$ , where $-\infty \le \alpha < t_0 < \beta \le \infty$ , such that every solution of the problem is the restriction of this one to a subinterval containing $t_0$ . This interval is the maximal interval of existence. If $\beta < \infty$ , then for every closed and bounded set $K \subseteq D$ there is a time $t_K < \beta$ such that the point $(t, y(t))$ lies outside $K$ for every $t$ in $(t_K, \beta)$ . The same holds near $\alpha$ when $\alpha > -\infty$ . When $D$ is the whole plane, a finite $\beta$ therefore forces $\lvert y(t) \rvert \to \infty$ as $t \to \beta^-$ .

In plain words The maximal interval belongs to the initial-value problem, not to the equation alone and not to a formula. A solution with a finite endpoint stops because, as the endpoint approaches, it becomes unbounded or comes arbitrarily close to the edge of the region where the hypotheses hold. A solution that stays in a closed and bounded set inside that region can always be continued past a finite time.

Example On the half plane $D$ where $y > 0$ , the right side of $y' = -t/y$ is continuous and has the continuous partial derivative $t/y^2$ . The problem with $y(0) = 1$ has the solution $y = \sqrt{1 - t^2}$ , because

y' = \frac{-t}{\sqrt{1 - t^2}} = -\frac{t}{y}.

Its maximal interval is $(-1, 1)$ . The solution stays bounded, but as $t \to 1^-$ or $t \to -1^+$ it approaches the line $y = 0$ , the edge of $D$ , where the right side is undefined. So a finite endpoint can occur without any blowup.

Finite-time blowup

Definition A solution blows up in finite time if its maximal interval has a finite right end $\beta$ with $\lvert y(t) \rvert \to \infty$ as $t \to \beta^-$ , or a finite left end $\alpha$ with $\lvert y(t) \rvert \to \infty$ as $t \to \alpha^+$ . This can happen even when $f$ is smooth on the whole plane, because a locally Lipschitz right side need not satisfy one Lipschitz bound for all $y$ . For $f = y^2$ and $y_1 \ne y_2$ ,

\frac{\lvert y_1^2 - y_2^2 \rvert}{\lvert y_1 - y_2 \rvert} = \lvert y_1 + y_2 \rvert,

and this quotient is unbounded on the plane. By contrast, for a linear equation $y' = p(t)y + q(t)$ with $p$ and $q$ continuous on an open interval $I$ , every initial-value problem with $t_0$ in $I$ has its solution defined on all of $I$ , so no blowup happens inside $I$ .

In plain words The solution runs off to infinity at a finite time, and that time depends on the initial value, not only on the equation. A smooth right side guarantees exactly one solution near each starting point, but it does not promise that the solution lasts forever.

Example For $y' = y^2$ with $y(0) = 1$ the solution stays positive, as the uniqueness example showed. Dividing by $y^2$ is therefore allowed, and $y'/y^2 = 1$ says that $(-1/y)' = 1$ , so $-1/y = t + c$ for a constant $c$ . The initial value gives $c = -1$ , and

y(t) = \frac{1}{1 - t}.

Its derivative $1/(1 - t)^2$ equals $y^2$ , and $y(0) = 1$ . As $t \to 1^-$ the denominator shrinks to $0$ and $y(t) \to \infty$ , so the maximal interval is $(-\infty, 1)$ and the solution blows up at $t = 1$ . The same formula defines a function for $t > 1$ , but that branch solves a different problem, such as $y(2) = -1$ , and no finite value at $t = 1$ joins it to the first one. More generally, the start $y(0) = y_0 > 0$ leads to $y = y_0/(1 - y_0 t)$ , which blows up at $t = 1/y_0$ , so a larger start blows up sooner.

Nonuniqueness

Definition An initial-value problem has nonunique solutions if it has two solutions, both defined on an interval containing $t_0$ , that differ at some time in that interval. Continuity of $f$ alone allows this. By the uniqueness theorem it cannot happen when $f$ is also locally Lipschitz in $y$ , and at the point where two such solutions separate, $f$ fails to be Lipschitz in $y$ on every rectangle around that point.

In plain words The same starting data can lead to several different futures when the slopes change too sharply with $y$ near the starting state. A failed derivative test does not prove that this happens. It only means the theorem is silent, and the equation itself has to be examined.

Example For $y' = 2\sqrt{\lvert y \rvert}$ with $y(0) = 0$ , the right side is continuous and the zero function is a solution. For every waiting time $a \ge 0$ , so is

y_a(t) = \begin{cases} 0, & t \le a, \\ (t - a)^2, & t \ge a. \end{cases}

For $t > a$ the derivative is $2(t - a)$ , and since $t - a > 0$ , also $2\sqrt{\lvert y_a(t) \rvert} = 2(t - a)$ . At $t = a$ both pieces have value $0$ and slope $0$ , so $y_a$ is differentiable there and satisfies the equation. All these curves start at $(0, 0)$ and separate along the $t$ -axis, exactly where the Lipschitz ratio computed above is unbounded.

Check yourself

Solve $y' = y^2$ with $y(0) = -1$ and find the maximal interval. In which direction of time does the solution blow up?

Show the answer

The solution cannot meet the solution $y = 0$ , so it stays negative and dividing by $y^2$ is allowed. As before $-1/y = t + c$ , and $y(0) = -1$ gives $c = 1$ , so

y(t) = -\frac{1}{1 + t}.

Its derivative $1/(1 + t)^2$ equals $y^2$ , and $y(0) = -1$ . The denominator vanishes at $t = -1$ , so the maximal interval is $(-1, \infty)$ . Forward in time the solution rises toward $0$ and exists for every $t > 0$ . Backward in time it blows up, with $y(t) \to -\infty$ as $t \to -1^+$ .

Check that $y = \tan t$ solves $y' = 1 + y^2$ with $y(0) = 0$ , and find the maximal interval. The function $\tan t$ is also defined on $(\pi/2, 3\pi/2)$ . Why is that piece not part of the answer?

Show the answer

Since $(\tan t)' = \sec^2 t = 1 + \tan^2 t$ and $\tan 0 = 0$ , the function solves the problem on $(-\pi/2, \pi/2)$ . The right side $1 + y^2$ has the continuous partial derivative $2y$ , so this is the only solution. As $t$ approaches $\pi/2$ from below, $\tan t \to \infty$ , and as $t$ approaches $-\pi/2$ from above, $\tan t \to -\infty$ . So the maximal interval is $(-\pi/2, \pi/2)$ , and the solution blows up in both directions. The piece on $(\pi/2, 3\pi/2)$ solves a different problem, $y(\pi) = 0$ , and no finite value at $\pi/2$ connects it to the data at $t = 0$ .

Show that $y' = y^{1/3}$ with $y(0) = 0$ , where $y^{1/3}$ is the real cube root, has a solution other than $y = 0$ . Which hypothesis of the uniqueness theorem fails?

Show the answer

Let $y(t) = 0$ for $t \le 0$ and $y(t) = (2t/3)^{3/2}$ for $t \ge 0$ . For $t > 0$ ,

\frac{d}{dt}\left(\frac{2t}{3}\right)^{3/2} = \left(\frac{2t}{3}\right)^{1/2},

which is the cube root of $(2t/3)^{3/2}$ , so the equation holds. At $t = 0$ both pieces have value $0$ and slope $0$ , so $y$ is differentiable there and satisfies the equation too. Its negative $-y$ is a third solution, and delayed starts give infinitely many. The right side is continuous, so existence is guaranteed, but it is not Lipschitz in $y$ near $y = 0$ . For $y \ne 0$ the ratio

\frac{\lvert y^{1/3} \rvert}{\lvert y \rvert} = \lvert y \rvert^{-2/3}

is unbounded as $y \to 0$ . The local Lipschitz hypothesis is the one that fails.

Existence, Uniqueness, and Maximal Intervals

Key terms

Local existence

Local Lipschitz condition

Uniqueness of solutions

Maximal interval of existence

Finite-time blowup

Nonuniqueness

Check yourself

How this lesson connects

Builds on

Used later in