From a Direction Field to a Phase Line
A direction field displays slopes at many points in the time-state plane. For some differential equations, however, the slope depends only on the state \(y\), not on the time \(t\). In that case, the same state always has the same slope, whatever the time. This makes it possible to summarize the direction of motion on a single vertical line.
That summary is called a phase line. To construct one, find the values of \(y\) where the slope is zero, then determine the sign of the slope between successive zero values. Positive slopes indicate motion toward larger \(y\)-values as time increases; negative slopes indicate motion toward smaller \(y\)-values.
We will assume that \(g\) is continuous on the intervals being considered. An equilibrium value is a number \(y_*\) for which \(g(y_*)=0\). As in the earlier material on equilibrium solutions in direction fields, substituting this value gives a constant solution \(y(t)=y_*\), since its derivative is zero and the right-hand side is also zero.
How to Read a Phase Line
Draw a vertical line to represent possible values of \(y\), with larger values higher up. Mark each equilibrium value on the line. These marked values divide the line into intervals. Choose one test value in each interval and evaluate \(g\) there. If \(g(y)>0\), draw an upward arrow; if \(g(y)<0\), draw a downward arrow.
Only the sign of \(g\) is needed to determine an arrow. The magnitude of \(g(y)\) gives the slope, and hence information about the rate of change, but it does not change which direction the solution moves. A phase line therefore compresses some of the information in a direction field while making the direction of motion easier to inspect.
Worked Example: A Quadratic Right-Hand Side
Worked Example: A Quadratic Right-Hand Side
Construct a phase line for \(y'=y(2-y)\), and identify its equilibrium solutions.
Here \(g(y)=y(2-y)\). First find its zeros:
These values divide the state line into three intervals: \(y<0\), \(0<y<2\), and \(y>2\). Test one value in each interval:
Thus the arrows point down below \(0\), up between \(0\) and \(2\), and down above \(2\). The equilibria give constant solutions: substituting \(y(t)=0\) yields \(y'=0=g(0)\), and substituting \(y(t)=2\) yields \(y'=0=g(2)=2(2-2)=0\).
The sign checks also verify the arrows: at the chosen point in each interval, the right-hand side has exactly the sign indicated. The complete phase line is described by the ordered marks \(0,2\) and the arrows \(\downarrow,\uparrow,\downarrow\) from bottom to top.
Why the Sign Determines Direction
The connection between the sign of \(g\) and an arrow follows directly from the differential equation. Suppose \(g\) is continuous on an interval \(I\) and has no zero there. Then it cannot take both positive and negative values in \(I\): if it did, the Intermediate Value Theorem would imply that it was zero somewhere between those values. Consequently, its sign is constant throughout \(I\).
Now consider a solution \(y(t)\) during a time interval when its values remain in \(I\). At every such time, the differential equation gives \(y'(t)=g(y(t))\). If \(g\) is positive on \(I\), then \(y'(t)>0\) throughout that time interval, so the solution increases as time increases. If \(g\) is negative on \(I\), then \(y'(t)<0\), so the solution decreases.
This argument justifies each arrow without requiring an explicit formula for a solution. It also explains why the zeros must be marked first: the sign of \(g\) can change only by passing through a zero, provided \(g\) is continuous. The phase line separates the state values into regions where the direction stays fixed.
Worked Example: A Repeated Zero
Worked Example: A Repeated Zero
Construct a phase line for \(y'=(y+1)^2(y-2)\), and find its equilibrium solutions.
Set the right-hand side equal to zero. The product is zero when either factor is zero:
The intervals are \(y<-1\), \(-1<y<2\), and \(y>2\). Evaluate \(g(y)=(y+1)^2(y-2)\) at one point in each interval:
The arrows are therefore down below \(-1\), down between \(-1\) and \(2\), and up above \(2\). The equilibrium values give constant solutions: \(g(-1)=0\) because \((-1+1)^2=0\), and \(g(2)=0\) because \(2-2=0\). In each case a constant function has derivative zero, so it satisfies the equation.
Notice that the arrows on the two sides of \(-1\) point in the same direction. The phase line records this fact without requiring us to make any further claim about what happens to a solution over a long time.
Worked Example: Several Equilibria
Worked Example: Several Equilibria
Construct a phase line for \(y'=(y+2)(y-1)(y-3)\), and identify the equilibrium solutions.
The right-hand side is zero when any factor is zero:
These values split the state line into four intervals. Use one test value in each:
Recheck the multiplication: \((-1)(-4)=4\), and \(4(-6)=-24\). Thus \(g(-3)=-24<0\). Continuing with the other intervals:
Therefore the arrows, from bottom to top, are down on \(y<-2\), up on \(-2<y<1\), down on \(1<y<3\), and up on \(y>3\). Each marked value is an equilibrium: \(g(-2)=0\) because the first factor vanishes, \(g(1)=0\) because the second factor vanishes, and \(g(3)=0\) because the third factor vanishes. Hence the constant functions \(y(t)=-2\), \(y(t)=1\), and \(y(t)=3\) satisfy the equation.
The arithmetic at the first test value illustrates why it is useful to show the intermediate products. The three factors there are all negative, so their product must be negative; evaluating them gives \(-24\), confirming the downward arrow.
A Reliable Construction Process
For a polynomial or another easily analyzed function \(g\), the following sequence gives a phase line systematically. The procedure concerns the right-hand side as a function of state; it does not require solving the differential equation.
Write the equation in the form \(y'=g(y)\), making clear which expression determines the slope.
Solve \(g(y)=0\). Mark each resulting value on a vertical state line; each gives a constant equilibrium solution.
Choose a convenient value between each pair of consecutive marks, and evaluate \(g\) there. Include the intervals below the lowest mark and above the highest mark when they are in the domain.
Use an upward arrow where the test value gives \(g(y)>0\), and a downward arrow where it gives \(g(y)<0\).
Confirm that each arrow agrees with its test calculation and that every equilibrium value is a zero of \(g\).
Common Pitfall: Testing the Wrong Expression
A phase line concerns the sign of the derivative \(y'=g(y)\), not the sign of \(y\) alone. The state \(y\) may be negative while its derivative is positive, or \(y\) may be positive while its derivative is negative. Inferring arrows from whether a test value lies above or below zero can therefore give an incorrect diagram.
Worked Example: Correcting a Sign Error
For \(y'=y(2-y)\), suppose we test \(y=3\). A mistaken argument might say, “\(3\) is positive, so the arrow points up.” This uses the sign of \(y\), but the arrow depends on the sign of \(y(2-y)\).
Evaluate the right-hand side:
The correct arrow points down. Substituting \(y=3\) into the differential equation confirms the slope is \(-3\), not a positive number. The correction is to evaluate \(g\) at a test value in every interval, rather than infer a direction from the test value itself.
What a Phase Line Does—and Does Not—Show
A phase line is useful when the question is about the direction of change in an autonomous equation. It reduces a two-dimensional direction field to a one-dimensional picture by using the fact that the slope depends only on \(y\). It also makes the equilibrium values visible and organizes the state values into intervals with consistent signs.
The diagram does not give an explicit solution \(y(t)\), the numerical value of the slope at every state, or the time needed to move between states. For those questions, the phase line alone is not enough. Its purpose here is narrower: identify the zero values and show the direction of motion on the intervals between them.
A practical check is to return to the original differential equation. At every marked equilibrium, verify \(g(y)=0\). On each open interval, verify an arrow with at least one test value. These checks catch missing zeros, incorrect interval signs, and arrows drawn from the sign of \(y\) instead of the sign of \(g(y)\).
Check Your Understanding
For each question, use the sign of the right-hand side to support your answer.
- What makes an equation of the form \(y'=g(y)\) autonomous?
- For \(y'=(y-1)(y+2)\), what values should be marked as equilibria on the phase line?
- For the equation in question 2, choose a test value in each interval and determine the arrow directions.
- Why is the sign of \(g(y)\), rather than the sign of \(y\), used to draw an arrow?
- What information does a phase line summarize, and name one kind of information it does not provide.
