Equilibria and the Phase Line
Solving a differential equation can be done in three major ways: analytical, qualitative, and numerical. We have
seen some examples of differential equations solved through analytical techniques (for example: linear,
separable, and Bernoulli equations). Euler's Method (though very primitive) illustrates the use of numerical
techniques in solving differential equations. For the qualitative approach, we have defined the slope field of a
differential equation and showed how this can be helpful when other techniques fail.
When the differential equation is autonomous, more can be said about the solutions using qualitative techniques.
Consider the autonomous differential equation
where f(y) is a function. Clearly we recognize a separable equation. So we know how to solve it via the
analytical technique. First we look for the constant solutions as the roots of the equation
Then to find the nonconstant solutions we separate the variable
and integrate
Note that the integration may be hard if not impossible to carry out. So it may not be possible to use the
analytical technique.
In this case, one may try numerical techniques. Unfortunately, numerical techniques fail if f(y) contains a
parameter and we need to make conclusions relative to this parameter. For example, recall the logistic equation
with harvesting (with a constant rate H)
where y(t) is the population of the given species at time t. In order to find the optimal rate H (which makes
hunters happy as well as ecologists worried about preserving the species), numerical techniques may not be the
best tool to use. Here we will see how new ideas based on a graphical approach will help us to say something
about the solutions. Be aware that these ideas are only valid for autonomous equations.
Equilibria of Autonomous Equations
Consider the autonomous equation
The equilibria or constant solutions of this differential equation are the roots of the equation
Using the existence and uniqueness theorem, these constant solutions will cut the entire plane (where the
solutions live) into independent regions. This means, if an initial condition belongs to one of the regions, then
the solution satisfying the initial condition will stay in that region all the time.
Example. Consider the logistic equation
Its equilibria are y=0 and y=1. Consider the solution y(t) to the IVP
Since 0 <y(0) < 1, then we must have
In the next picture, we draw few solutions associated to initial conditions in the different regions (0 < y < 1, y <
0, and 1 < y)
Note that in this example, any solution y(t) is increasing if 0 < y(0) < 1 is satisfied. So one may wonder whether
a similar conclusion is true in general. Consider again the autonomous equation
Assume that
and
(
) are equilibria (that is
f(y) has a constant sign on the interval
where
and
). Assume moreover that
. Assume that f(y) is positive and consider the solution to the IVP
. We know that we will have
for every t. Since
we must have
for all t. This clearly implies that y(t) is always increasing. So the sign of f(y) helps us
determine the behavior of the solutions (whether they are increasing or decreasing).
Example. Consider the autonomous equation
where the graph of f(y) is given below
From the graph, we can determine the equilibria. They are y=0, y=1, and y=2. Using the discussion above, we
obtain the following information:
if y(t) is a solution satisfying y(0) < 0, then y(t) is always increasing;
if y(t) is a solution satisfying 0<y(0) < 1, then y(t) is always decreasing;
if y(t) is a solution satisfying 1 <y(0) < 2, then y(t) is always increasing;
if y(t) is a solution satisfying 2 <y(0), then y(t) is always decreasing.
In the picture below, we draw some solutions
Remark. Recall that if the function h(t) is increasing or decreasing, then the following hold :
the limit of h(t) when
otherwise we have
exists (or is a number) if and only if the function is bounded above;
the limit of h(t) when
otherwise we have
exists (or is a number) if and only if the function is bounded below;
Note that in this case, if h(t) has a limit (as a number) when
must have
The next image illustrates this conclusion better than words can do.
, and h(t) is a "nice function", then we
Example. Consider the solution to the initial value problem
where the graph of f(y) is given below
Discuss the limit of y(t) when
.
Answer. Using the above results, we know that for the given initial condition, y(t) is decreasing and for every t
we have
The above remark implies that
exist and we have
Since y(t) is not a constant solution (because y(0)=0.5), then we have
On the other hand, since
and y is solution to the autonomous equation y'=f(y), then the only possibilities for both limits are the equilibria
0,1, and 2. Clearly if we put everything together, we conclude that
Summary. Let us write down some general steps to follow when dealing with the autonomous equation
Step 1. Find the constant solutions or equilibria through the algebraic equation
Draw the constant solutions. Note that here we are drawing y versus t.
Step 2. Find the sign of the function f(y) through its graph (versus y).
Step 3. In the region between any two equilibria, the solutions will be all increasing (if f(y) is positive)
or decreasing (if f(y) is negative).
Step 4. If a solution is increasing and bounded above by a constant solution y = L, then the limit of the
solution when
is the number L. Otherwise the limit is
.
If this solution is bounded below by a constant solution y = l, then the limit of the solution when
is the number l. Otherwise the limit is
.
Step 5. If a solution is decreasing and bounded below by a constant solution y = L, then the limit of the
solution when
is the number L. Otherwise the limit is
.
If this solution is bounded above by a constant solution y = l, then the limit of the solution when
is the number l. Otherwise the limit is
Step 6. Draw some solutions.
.
Remark. You must be very careful here, since two graphs will be drawn: one for the function f(y) (versus y)
and another one for the solutions, where this time y is on the vertical axis while t is on the horizontal axis. Keep
in mind that the second graph is the most important one, since it deals with what we are looking for: the
solutions of the differential equation.
Example. Draw some solutions for the equation
Answer. Note that this equation models the logistic growth with threshold. The equilibria or constant solutions
are given by
or y=0, y=2, and y=5. The graph of the function f(y) = .2y(5-y)(y-2) is given in the next picture
So we have:
the solution which satisfies the initial condition
have
, with
, is increasing. Moreover we
the solution which satisfies the initial condition
Moreover we have
, with
, is decreasing.
the solution which satisfies the initial condition
Moreover we have
, with
, is increasing.
the solution which satisfies the initial condition
have
, with
, is decreasing. Moreover we
In the picture below we draw some solutions
Phase Line
Let us reconsider the above example. Let us focus our attention on the solutions which satisfy the initial
condition
and
. We already know that
It is somehow amazing that this conclusion is valid regardless whether is close to 5 or close to 2. In this case
we say that the equilibrium point 5 is attractive from below. In fact, in this example, this equilibrium point is
attractive from below and above. On the other hand, the solutions get away from the equilibrium point 2 from
below and above. We will say that the equilibrium point 2 is repelling. Note that there is no general agreement
on the words used to describe these phenomena.
Definition. Classification of Equilibrium Points.
Consider the autonomous equation
Assume that
is an equilibrium point (that is
).
The equilibrium point is called a sink if and only if any solution y(t) to the autonomous equation such
that y(0) is close to , then we have
The equilibrium point
is called a source if and only if any solution y(t) to the autonomous equation
such that y(0) is (or not) close to
The equilibrium point
, then y(t) will move away from
when
.
is called a node if and only if it is neither a source nor a sink.
Example. Classify the equilibrium points of the equation
as source, sink, or node.
Answer. The equilibrium points are 0, 2, and 5. Using the above results, we see that 0 and 5 are sinks while 2 is
a source.
Example. Find and classify the equilibrium points of the equation
as source, sink, or node.
Answer. It is easy to see that the only equilibrium point is 0. Since
solutions will always be increasing. Using previous results, we conclude:
is always positive, then the
if y(t) is a solution satisfying y(0) < 0, then we have
if y(t) is a solution satisfying y(0) > 0, then we have
Hence, the equilibrium point 0 is a node. In the picture below we draw some solutions.
There is a very nice way to put all this information together. Indeed, draw a vertical line (where the variable
describing it is y) and start by marking the equilibrium points of the equation
on the y-axis.
Then using the sign of f(y), we draw arrows pointing upward in a region where f(y) is positive, and downward in
a region where f(y) is negative. This vertical line is called the phase line of the equation.
An equilibrium point is a sink, if the arrows on both sides point towards the equilibrium point, and it is a source,
if both arrows point away from it.
Example. Draw the phase line of the equations
and
Answer. We will use our previous knowledge to get the two phase lines
and
You can readily see, that an equilibrium point is a sink, if the arrows on both sides point towards the
equilibrium point, and that it is a source, if both arrows point away from it.
Source: sosmath.com