LECTURE 7: THE PHASE LINE: SINKS, SOURCES, AND NODES
There is a close analogy between this topic the topic from calculus of finding local extrema. Keep
this in mind throughout the lecture. Today, we are only concerned with autonomous differential
equations.
dy
= f (y)
dt
0.1. The Phase Line. For an autonomous equation, the slope field is determined by the slopes of
along any vertical line t = t0 . To get the slope at any other point (t, y), you just determine the slope
at (t0 , y). The phase line of a differential equation characterizes this line of slopes in terms of their
sign.
Here’s how you draw a phase line. Suppose f (y) is a continuous function. First find the equilibrium points of dy/dty = f (y). There may be infinitely many equilibrium points, so be certain
that you have a complete list. We plot these points on a vertical line. Now, since f is continuous f is either strictly positive or strictly negative between two equilibrium points (note that this
follows from the Intermediate Value Theorem!). If the value is positive between two adjacent equilibrium points, we draw an up arrowhead on that segment. If it is negative, we draw a down arrowhead.
Example: Draw a phase line for y ′ = (2 − y)(1 + y). Equilibrium points are at y = −1, 2. Use the
phase line to sketch several solutions.
2
−1
Example: Draw a phase line for y ′ = sin(y). Equilibrium points are y = nπ, n = . . . , −2, −1, 0, 1, 2, . . ..
Use the phase line to sketch several solutions.
Example: Draw the phase line for y ′ = y 2 (1 − y)(1 + y). Use the phase portrait to sketch several
solutions.
0.2. Sinks, Sources, and Nodes. An equilibrium point of an autonomous differential equation
can be classified as a sink, a source or a node. The phase portrait for each type is given below.
A Sink
A Source
Two Nodes
Examples: For each of the examples above, classify the equilibrium points as sinks, sources, or
1
2
LECTURE 7: THE PHASE LINE: SINKS, SOURCES, AND NODES
nodes.
If f (y) is differentiable, the derivative can help to classify each equilibrium point. At a sink, f (y)
changes sign from + to − as y increases. Hence, f (y) is decreasing at a sink. At a source, f (y)
changes from − to + as y increases. Hence, f (y) is decreasing at a source.
Theorem 1 (Linearization Theorem). (as from text pg. 88) Suppose y0 is an equilibrium point
of y ′ = f (y) and f (y) is continuously differentiable( i.e. is differentiable and the derivative is
continuous). Then:
• if f ′ (y0 ) > 0, then y0 is a source.
• if f ′ (y0 ) < 0, then y0 is a sink.
• if f ′ (y0 ) = 0, then we need more information to determine the type of y0 .
Example: Consider y ′ = (y − 1)(y 5 − 7y 4 + 3y 3 + 8y 2 − 11). Classify the equilibrium point y = 1.
(It’s a sink).
Example:(Fox Squirrels) If the population is too small, the fox squirrels cannot find a mate and the
rate of population is decreasing. As fox squirrels are territorial, a large population is determintal to
the population (fox squirrel wars). A model for this is given by:
P
P
dP
= kP 1 −
−1
dt
N
M
Draw a phase protrait and sketch several solutions.