Direction Fields, Equilibria, and Uniqueness, calculus lesson

Most differential equations cannot be solved by a formula. The equation still tells you the slope of the solution at every point, and that is often enough to see where solutions go, which values they settle on and which ones they run away from. This lesson is about reading a differential equation as a picture.

Key terms

Slope fields

Definition Let $f$ be defined on a region $R$ of the plane. The slope field (or direction field) of $y' = f(x, y)$ assigns to each point $(x, y)$ of $R$ a short line segment through that point with slope $f(x, y)$ . In practice one draws the segments at the points of a grid.

In plain words At each point the equation tells you how steep a solution through that point must be. Draw that steepness as a tiny dash.

Example For $y' = x - y$ , a few slopes are below.

Point $(x, y)$	Slope $x - y$
$(0, 0)$	$0$
$(1, 0)$	$1$
$(0, 1)$	$-1$
$(2, 1)$	$1$
$(1, 2)$	$-1$

Along every line $y = x - c$ the slope is the same number $c$ , so the dashes on such a line are all parallel.

Solution curves

Definition A solution curve (integral curve) of $y' = f(x, y)$ is the graph of a solution $y = \varphi(x)$ on an interval. At each of its points $(x, \varphi(x))$ the curve has slope $\varphi'(x) = f(x, \varphi(x))$ , so it is tangent to the segment of the slope field there.

In plain words A solution curve follows the dashes, like a leaf floating along the current.

Example For $y' = x - y$ , the line $y = x - 1$ has slope $1$ , and on it $x - y = x - (x - 1) = 1$ . So the line is a solution curve. The general solution is $y = x - 1 + Ce^{-x}$ , and every other solution curve approaches this line as $x \to \infty$ .

Autonomous equations

Definition A first order equation is autonomous if the right side does not depend on the independent variable, that is, it has the form

y' = g(y)

for a function $g$ of $y$ alone.

In plain words The rule for how $y$ changes depends only on where $y$ is now, not on the time. So the slope field looks the same along every horizontal line.

Example The equation $y' = y^2 - 4$ is autonomous. The equation $y' = x - y$ is not, because the slope at height $y = 0$ is $0$ at $x = 0$ but $1$ at $x = 1$ .

Equilibrium solutions

Definition A constant function $y(x) = c$ is an equilibrium solution of $y' = f(x, y)$ if $f(x, c) = 0$ for every $x$ in the interval under study. For an autonomous equation $y' = g(y)$ , the equilibrium solutions are exactly $y = c$ with $g(c) = 0$ .

In plain words An equilibrium is a value where the rate of change is zero, so staying at that value forever is a solution.

Example For $y' = y^2 - 4$ the equilibria are the roots of $y^2 - 4 = 0$ , namely $y = -2$ and $y = 2$ .

Phase line

Definition The phase line of an autonomous equation $y' = g(y)$ with $g$ continuous is the $y$ -axis with the zeros of $g$ marked and an arrow on each open interval into which these zeros cut the axis (here $g$ is assumed to have finitely many zeros). The arrow points up where $g(y) > 0$ and down where $g(y) < 0$ , because there every solution increases or decreases.

In plain words It is the slope field squeezed into a single vertical line. It shows which way $y$ moves from each starting value.

Example For $y' = y^2 - 4$ , test one value in each interval.

Interval	Test value	Sign of $y^2 - 4$	Arrow
$y > 2$	$3$	$+$	up
$-2 < y < 2$	$0$	$-$	down
$y < -2$	$-3$	$+$	up

So solutions starting between $-2$ and $2$ fall toward $-2$ , those starting above $2$ rise, and those below $-2$ rise toward $-2$ .

Stability

Definition Let $c$ be an equilibrium of $y' = g(y)$ with $g$ continuous, and write the independent variable as $t$ . It is stable if for every $\varepsilon > 0$ there is a $\delta > 0$ such that every solution with $\lvert y(t_0) - c \rvert < \delta$ exists for all $t \ge t_0$ and satisfies $\lvert y(t) - c \rvert < \varepsilon$ for all $t \ge t_0$ . It is asymptotically stable if moreover $y(t) \to c$ as $t \to \infty$ for every solution starting close enough to $c$ . It is unstable if it is not stable. If $g$ is differentiable at $c$ , then

g'(c) < 0 \implies c \text{ is asymptotically stable}, \qquad g'(c) > 0 \implies c \text{ is unstable},

and $g'(c) = 0$ decides nothing.

In plain words A stable equilibrium is a valley. Nudge the solution a little and it rolls back. An unstable one is a hilltop, where the smallest nudge sends it away.

Example For $g(y) = y^2 - 4$ we have $g'(y) = 2y$ . Since $g'(-2) = -4 < 0$ , the equilibrium $-2$ is asymptotically stable, and since $g'(2) = 4 > 0$ , the equilibrium $2$ is unstable. This matches the phase line. For $y' = y^2$ the test fails at $0$ since $g'(0) = 0$ . The phase line has arrows up on both sides, so solutions below $0$ approach it and solutions above leave it. This equilibrium is unstable and is often called semistable.

Existence and uniqueness

Definition Let $R$ be an open set in the plane and $f$ continuous on $R$ . Suppose $f$ is locally Lipschitz in $y$ , uniformly in $x$ , meaning every point $(a, b)$ of $R$ has a rectangle $\lvert x - a \rvert \le \alpha$ , $\lvert y - b \rvert \le \beta$ inside $R$ and a constant $K$ with

\lvert f(x, y_1) - f(x, y_2) \rvert \le K \lvert y_1 - y_2 \rvert

for all $(x, y_1)$ and $(x, y_2)$ in that rectangle. Then for every $(x_0, y_0)$ in $R$ there is an $h > 0$ such that the problem $y' = f(x, y)$ , $y(x_0) = y_0$ has exactly one solution on $(x_0 - h, x_0 + h)$ . Moreover two solutions on any common interval containing $x_0$ agree there. The Lipschitz condition holds automatically when $\frac{\partial f}{\partial y}$ exists and is continuous on $R$ .

In plain words If the equation is continuous and does not change too violently in the $y$ direction, then through each starting point there is one solution curve and no other. In particular two solution curves cannot cross or touch.

Example Uniqueness can fail without the Lipschitz condition. For $y' = 3y^{2/3}$ , $y(0) = 0$ , both $y = 0$ and $y = x^3$ are solutions on $\mathbb{R}$ , since

\left(x^3\right)' = 3x^2 = 3\left(x^3\right)^{2/3}.

Here the difference quotient

\frac{\lvert f(x, y) - f(x, 0) \rvert}{\lvert y - 0 \rvert} = 3\lvert y \rvert^{-1/3}, \qquad y \neq 0,

is unbounded near $y = 0$ , so no Lipschitz constant works there. By contrast, for $y' = y^2 - 4$ the partial derivative $2y$ is continuous everywhere, so solutions never cross the equilibria. A solution starting at $y(0) = 0$ therefore stays strictly between $-2$ and $2$ for all time.

Check yourself

Find the equilibria of $y' = y(3 - y)$ and classify each one.

Show the answer

The function $g(y) = 3y - y^2$ has zeros $0$ and $3$ , and $g'(y) = 3 - 2y$ . Since $g'(0) = 3 > 0$ , the equilibrium $0$ is unstable. Since $g'(3) = -3 < 0$ , the equilibrium $3$ is asymptotically stable.

For $y' = y^2 - 4$ with $y(0) = 1$ , what does $y(t)$ approach as $t \to \infty$ ?

Show the answer

It approaches $-2$ . The starting value lies between $-2$ and $2$ , where $y^2 - 4 < 0$ , so the solution decreases, cannot cross $-2$ by uniqueness, and approaches the stable equilibrium $-2$ .

Does the existence and uniqueness theorem apply to $y' = \sqrt{y}$ at the starting point $(0, 0)$ ? At $(0, 4)$ ?

Show the answer

Not at $(0, 0)$ , because $f$ is not even defined for $y < 0$ and $\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y}}$ is unbounded near $y = 0$ . In fact $y = 0$ and $y = \frac{x^2}{4}$ for $x \ge 0$ (with $y = 0$ for $x < 0$ ) both solve it. At $(0, 4)$ it applies, using the open set $y > 0$ where $\frac{\partial f}{\partial y}$ is continuous.

Direction Fields, Equilibria, and Uniqueness

Key terms

Slope fields

Solution curves

Autonomous equations

Equilibrium solutions

Phase line

Stability

Existence and uniqueness

Check yourself

How this lesson connects

Builds on

Used later in