Autonomous Differential Equations

Autonomous: no explicit dependence on time

If the rate of change depends only on how much there is now, not on what time it is, the equation is called autonomous. The same state always moves the same way.

A first-order autonomous differential equation has the form y′=g(y): the slope is set by the current y alone, and x does not appear explicitly on the right. If the independent variable is time t, write dy/dt=g(y). Autonomous does not mean y stays the same; it means the same state meets the same rule of change at any time.

The familiar y′=y is autonomous. y′=2xy is not: even at the same y, the slope can change with x. A first-order autonomous equation is also a kind of separable equation (with f(x)=1), but there is no hurry to integrate yet.

If g(y*)=0, the constant function y(t)=y* satisfies the equation and is called an equilibrium solution. Here y* is a fixed value; the star only marks an equilibrium and has nothing to do with derivatives. At any one height the slope field has the same slope everywhere, so all of its information can be squeezed into a single phase line along the y axis.

The logistic model: units first

A model has to say what each letter stands for and what units it has. Only then do the numbers it produces mean something.

rN is proportional growth at low density. The dimensionless factor (1−N/K) makes each individual’s net growth rate fall as density rises. This page fixes r and K and leaves out seasons, migration, whole-number individuals and time delays; the physical interpretation is limited to N≥0.

Both sides have units of individuals per day. When N≪K the factor is close to 1, so for a short time N′≈rN; near K, net growth approaches 0. This is an idealized model, not a law for every real population, and one smooth curve is not a field forecast.

K is not a hard wall that can’t be passed. The starting value can be greater than K, and the model predicts a decline. A program that simply caps N at K has changed the equation.

Read the direction without solving

You can tell whether N goes up or down without solving for a formula: just mark the sign of the rate of change on a number line.

1. Set the right side to zero

   rN(1−N/K)=0 ⇒ N=0 or N=K

   Since r>0 and K>0, these are the two equilibria.

2. Test between the two equilibria

   0<N<K ⇒ N′>0

   r, N and 1−N/K are all positive, so N increases over time. The arrow on the phase line points toward larger N.

3. Test above K

   N>K ⇒ N′<0

   1−N/K is negative and everything else is positive, so N decreases, and the arrow points toward K.

4. Check the equilibria themselves

   N₀=0 ⇒ N(t)=0; N₀=K ⇒ N(t)=K

   A solution does not leave an equilibrium just because there are arrows next to it. This polynomial right side gives local uniqueness, which stops other solutions from crossing an equilibrium solution.

The phase line shows only the direction of the state. It is not the full curve of N against time, and it does not tell you how long it takes to get near K. The length of an arrow says nothing about speed. The highlighted segment marks only the current starting state; the trajectory in time above is what links t to N.

Check the phase line by separation

The phase line already shows the behavior. Checking with a formula also gives the value at any chosen time.

Keep N=0 and N=K aside first. For the other solutions, let z=N/K be the dimensionless state; since K is constant, z′=N′/K=r z(1−z). That way the input of the logarithm stays dimensionless too.

1. Separate the non-equilibrium solutions

   z′/[z(1−z)]=r

   This step works only on stretches where z≠0, 1.

2. Split the fraction

   1/[z(1−z)] = 1/z + 1/(1−z)

   Over a common denominator, the numerator is (1−z)+z=1.

3. Integrate, minding the sign of the second term

   ln|z/(1−z)|=rt+C

   ∫1/(1−z) dz=−ln|1−z|. Combine into the logarithm of a ratio.

4. Restore the sign and rearrange

   z/(1−z)=B e^rt

   B can be positive or negative; it absorbs the fixed sign of the ratio on one stretch of solution, and B≠0. Solving gives z=B e^(rt)/(1+B e^(rt)).

5. Put in the starting value and rewrite

   N(t)=K/[1+(K/N₀−1)e^−rt]

   Here N₀=N(0)>0. For N₀≠K this follows from the previous line; N₀=K was excluded by the division, but substituting confirms that the same final formula still works. N₀=0 is the zero solution, listed separately.

Take K=100, r=1, N₀=20. Then K/N₀−1=4, so

1. Check the starting value

   N(0)=100/(1+4)=20

   It matches the chosen population.

2. Check the equation

   N′=400e^−t/(1+4e^−t)²

   The right side, N(1−N/100), gives the same expression. Differentiating the negative exponent and the denominator each brings a minus sign, so the result is positive.

3. Where it holds, and its limit

   t≥0: denominator > 0; t→∞: N(t)→100

   For any N₀>0, let A=K/N₀−1>−1; then 1+A e^(−rt) stays positive for t≥0. So every nonnegative starting value on this page extends into the future.

Even when N₀>K, A only lies in (−1,0), the denominator stays positive in the future, and the curve approaches K from above. Extended into the past, some starting values reach a singularity at a finite time. This page’s physical reading and sliders are limited to t≥0, and it does not claim that every starting value works on the whole real line.

Stable and approaching are two claims

Stable means that starting close enough, you stay close. Only if you also approach the equilibrium is it called asymptotically stable.

An equilibrium is stable if solutions that start close enough stay close in the future; if they also tend to it over time, it is asymptotically stable. Both always refer to time moving forward, within a stated range of states.

In the logistic model with N≥0, the arrows on both sides of K point toward it, and the formula confirms that every N₀>0 tends to K. Solutions near K stay between their starting value and K, so K is asymptotically stable. 0 is unstable: however small a positive starting value is, the solution eventually moves away from zero toward K. But the solution that starts exactly at 0 stays at 0 forever.

As a helper, g′(N)=r(1−2N/K), so g′(0)=r>0 and g′(K)=−r<0. This derivative test gives a simple conclusion only for an equilibrium where g is smooth enough and g′≠0; if g′=0, go back to the signs and other information. For example, for y′=−y³ both sides of the zero equilibrium point toward zero, and although g′(0)=0, it can still be shown to be asymptotically stable.

Real populations may oscillate because of delays or outside factors. In this first-order, smooth, autonomous model, positive solutions approach K monotonically and never cross back and forth over K. That conclusion depends on the model’s assumptions.

Phase line and formula, checking each other

The phase line gives the direction; the formula gives the numbers. Two tools that check each other are safer than trusting either one alone.

Take “look at the state” to substitution

Carry the habit of reading the state first and computing second over to the next kind of equation, one that needs a substitution.

Autonomous equations let us read equilibria and long-run behavior from local slopes. On the next page the slope depends not on y alone but on the ratio y/x. Treating that ratio as a new unknown brings a new kind of equation back to separation of variables, which you already know.

Sources and scope

Definitions and the conditions of theorems were checked against the authors and university materials below. The explanations, worked examples and interactive plots here are written independently. Decimals on the plots are rounded displays; the mathematical conclusions rest on derivation and substitution. This page does not prove the general existence and uniqueness theorems it cites.

• Jiří Lebl · Autonomous equations: phase lines, equilibria, stability and the logistic model.
• Jiří Lebl · Separable equations: separating non-equilibrium solutions and checking zero factors.