Why Signs Reveal Stability
For an autonomous differential equation \(y'=g(y)\), the rate of change depends on the current value of \(y\), not directly on time. If \(g(r)=0\), then \(y(t)=r\) is a constant solution: the system remains at that value if it starts there. Such a value is an equilibrium.
The key question is what happens to solutions that start near the equilibrium. If \(g(y)>0\), then \(y\) is increasing; if \(g(y)<0\), then \(y\) is decreasing. These directions tell us whether nearby solutions move toward or away from \(r\). A phase line records this information using arrows, and the signs on either side of \(r\) determine the classification.
In this tutorial, we use unstable for an equilibrium that repels nearby solutions on both sides, and semistable for one that attracts solutions on one side and repels them on the other. A semistable equilibrium is not stable in the formal sense above: solutions on its repelling side can move away.
Reading the Phase Line
On an interval where \(g(y)>0\), solutions move upward through \(y\)-values as time increases. On an interval where \(g(y)<0\), they move downward. Thus, an arrow on the phase line points right for a positive sign and left for a negative sign. At an equilibrium \(r\), compare each arrow with the location of \(r\): an arrow pointing toward \(r\) indicates attraction from that side, while one pointing away indicates repulsion.
| Sign to the left of \(r\) | Sign to the right of \(r\) | Behavior at \(r\) |
|---|---|---|
| Positive: solutions move up toward \(r\) | Negative: solutions move down toward \(r\) | Attracts from both sides; stable and asymptotically stable |
| Negative: solutions move down away from \(r\) | Positive: solutions move up away from \(r\) | Repels from both sides; unstable |
| Positive: solutions move toward \(r\) | Positive: solutions move away from \(r\) | Semistable; attracts from the left |
| Negative: solutions move away from \(r\) | Negative: solutions move toward \(r\) | Semistable; attracts from the right |
The sign on the left is interpreted relative to the equilibrium: a positive rate moves an initial value below \(r\) upward toward it, whereas a negative rate moves it farther down. On the right, a negative rate moves a value above \(r\) downward toward it, whereas a positive rate moves it farther up.
Why the Sign Test Works
The sign chart is useful because it translates the differential equation into a local description of motion. Suppose \(g\) is continuous and locally Lipschitz near an isolated equilibrium \(r\). These conditions ensure that solutions exist locally and that a solution starting away from the equilibrium follows the indicated direction as long as it remains in the interval under consideration. The strict sign on each side prevents a solution from reversing direction within that interval.
For example, if \(g(y)>0\) for \(y<r\) near \(r\), a solution starting on the left is increasing. It cannot move farther left while it remains in that neighborhood. If \(g(y)<0\) for \(y>r\), a solution starting on the right is decreasing and cannot move farther right. Both sides therefore move toward \(r\). The reverse pair of signs sends both sides away. If the signs agree, the two sides behave differently: one approaches and the other departs.
Worked Example: A Stable and an Unstable Equilibrium
Worked Example: A Stable and an Unstable Equilibrium
Classify the equilibria of \(y'=(y-1)(y+2)\).
First solve \(g(y)=0\):
The roots divide the number line into three intervals. Test one value in each interval:
Thus \(g\) is positive below \(-2\), negative between \(-2\) and \(1\), and positive above \(1\). At \(r=-2\), the signs are positive on the left and negative on the right. Both sides move toward \(-2\), so this equilibrium is stable and asymptotically stable. At \(r=1\), the signs are negative on the left and positive on the right. Both sides move away, so this equilibrium is unstable.
Check the equilibrium values in the original rate function:
Both proposed equilibria do satisfy \(g(r)=0\), and the tested signs support the classifications.
Worked Example: A Semistable Equilibrium
Worked Example: A Semistable Equilibrium
Classify the equilibria of \(y'=-(y+1)^2(y-2)\).
The equation \(g(y)=0\) gives the equilibria \(y=-1\) and \(y=2\). Test one value in each interval:
The signs are positive for \(y<-1\), positive for \(-1<y<2\), and negative for \(y>2\). At \(-1\), a solution on the left moves up toward the equilibrium, while a solution on the right moves up away from it. Therefore \(-1\) is semistable, attracting from the left and repelling from the right. At \(2\), a solution on the left moves up toward it and a solution on the right moves down toward it. Thus \(2\) is stable and asymptotically stable.
Substitution verifies both roots:
The squared factor makes \(g\) have the same sign on both sides of \(-1\). That same sign does not mean the same behavior relative to the equilibrium: left-side motion is toward \(-1\), but right-side motion is away.
Worked Example: Semistability in Both Directions
Worked Example: Semistability in Both Directions
Classify the equilibria of \(y'=y^2(y-1)\).
Setting \(g(y)=0\) gives \(y=0\) and \(y=1\). Choose test values from the three intervals:
The sign is negative below \(1\), except that the rate is zero at \(0\), and positive above \(1\). At \(0\), the signs immediately to the left and right are both negative. A solution on the left moves down, away from \(0\); a solution on the right moves down, toward \(0\). Hence \(0\) is semistable, attracting from the right. At \(1\), the signs are negative to the left and positive to the right, so solutions move away on both sides. Thus \(1\) is unstable.
Finally, check the equilibrium condition:
The classifications follow from the signs on both sides, not from whether the root is simple or repeated by itself.
A Reliable Sign-Chart Procedure
For an autonomous equation with isolated equilibria and a continuous, locally Lipschitz rate function near them, use the following procedure. The local regularity assumptions support the interpretation of the signs as solution behavior.
Solve \(g(y)=0\). These values divide the \(y\)-axis into intervals.
Choose a test value in each interval and evaluate \(g\), or use the factors of \(g\) to determine its sign.
Use an upward arrow where \(g(y)>0\) and a downward arrow where \(g(y)<0\).
Arrows toward the equilibrium on both sides mean stable; arrows away on both sides mean unstable; one toward and one away mean semistable.
Substitute each proposed equilibrium into \(g\) and check that the result is zero.
Common Pitfall: Treating a Zero as Automatically Stable
A common error is to see \(g(r)=0\) and conclude that \(r\) is stable because the slope there is zero. The zero only establishes that the constant function \(y(t)=r\) is a solution. It says nothing by itself about solutions with initial values slightly above or below \(r\).
For instance, for \(y'=y^2(y-1)\), one might reason incorrectly that the equilibrium \(y=0\) is stable because the rate contains the factor \(y^2\) and vanishes at zero. The sign calculation instead gives
A value just below zero moves downward, away from zero. A value just above zero moves downward, toward zero. So the equilibrium is semistable, not stable. The correction is to check both sides of every equilibrium before classifying it.
Why the Classification Matters
Equilibria describe steady states, while their stability describes how sensitive those states are to small changes in the initial value. If an equilibrium attracts from both sides, small disturbances move the system back toward it. If it repels from both sides, even a small disturbance moves the system farther away. A semistable equilibrium behaves differently depending on which side the disturbance occurs.
This information can be obtained without solving the differential equation explicitly. The rate function's sign determines the direction of motion, and the sign pattern around a zero reveals whether that motion is restorative, departing, or one-sided.
Check Your Understanding
For each question, use the signs of the rate function on the intervals around the equilibrium.
- For \(y'=y(3-y)\), find the equilibria and classify each one.
- If \(g(y)>0\) immediately to both sides of an equilibrium \(r\), is \(r\) stable, unstable, or semistable? Which side attracts?
- What sign pattern gives an equilibrium that attracts solutions from both sides?
- Why does \(g(r)=0\) not determine whether \(r\) is stable?
- For \(y'=-(y-2)^2\), classify the equilibrium at \(y=2\) and state its attracting side.
