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Section Classifying Equilibrium Solutions
Equilibrium solutions mark the places where a system comes to rest. But not all equilibrium solutions are alike, some attract nearby solutions, others repel them, and some do a bit of both. In this section, weβll learn how to classify these points by examining the slope field, the sign of
\(f(y)\text{,}\) and a tool called a
phase line .
Subsection Types of Equilibria
Suppose
\(y = c\) is an equilibrium solution of an autonomous equation
\(y' = f(y)\text{.}\) If you nudge a solution slightly above or below
\(c\text{,}\) it might drift back, move away, or react differently on each side. What it does determines the type of equilibrium it is.
There are three common types:
Stable (sink): Solutions move toward the equilibrium from both sides.
Unstable (source): Solutions move away from the equilibrium on both sides.
Semi-stable (node): Solutions move toward the equilibrium on one side and away from it on the other.
In a slope field, a sink looks like arrows converging toward a horizontal line, a source shows arrows diverging away, and a node is a mix: converging on one side, diverging on the other. Next, weβll look at different ways to determine these behaviors.
Subsection Classification via the Phase Line
Slope fields show a lot of information at once, but for autonomous equations we can simplify. Since the slope depends only on
\(y\text{,}\) we can βcompressβ the slope field into a simple vertical diagram of just
\(y\) -values. This is called a
phase line .
Equilibria are marked with solid dots.
Regions where \(f(y) > 0\) get upward arrows.
Regions where \(f(y) < 0\) get downward arrows.
The arrows summarize how
\(y(t)\) changes: whether solutions are rising or falling. Follow the arrows up or down and youβll see where solutions eventually settleβor whether theyβre pushed away.
Figure 99. Slope Field (left) and Phase Line (right) for \(y'= 1 - y^2\)
π Example 100 . Growth and Decay.
Find and classify the equilibrium solutions of the autonomous equation:
\begin{equation*}
\frac{dy}{dt} = y^2 - 4y
\end{equation*}
Solution .
First, set \(f(y)=0\text{:}\)
\begin{equation*}
y^2 - 4y = y(y - 4) = 0,
\end{equation*}
giving equilibrium solutions at \(y = 0\) and \(y = 4\text{.}\)
These values split the
\(y\) -axis into three regions. We test the sign of
\(f(y)\) in each:
Region
Test Point
\(f(y) = y(y - 4)\) Sign of \(f(y)\)
\((-\infty, 0)\) \(y = -1\) \((-1)(-5) = 5\) Positive
\((0,4)\) \(y = 1\) \((1)(-3) = -3\) Negative
\((4,\infty)\) \(y = 5\) \((5)(1) = 5\) Positive
\begin{equation*}
\rightarrow
\end{equation*}
Now we can classify the equilibria based on the phase line:
Solutions move away from \(y(t) = 0\text{,}\) so it is a source .
Solutions move toward \(y(t) = 4\text{,}\) so it is a sink .
Checkpoint 101 . πβ Match the Equilibria.
Suppose an autonomous equation has three equilibrium solutions at
\begin{equation*}
y=A,\ y=B,\ \text{and}\ \ y=C\text{.}
\end{equation*}
Based on the provided phase line, drag each equilibrium solution to its corresponding classification type.
\begin{equation*}
\rightarrow
\end{equation*}
\begin{equation*}
\begin{array}{c} A\\ \ds\vphantom{\int} B\\ C \end{array}
\end{equation*}
\begin{equation*}
A
\end{equation*}
Sink
\begin{equation*}
B
\end{equation*}
Source
\begin{equation*}
C
\end{equation*}
Node
None of these
Subsection Classification via the Linearization Method
Thereβs another way to classify equilibria: use a quick calculus check on
\(f(y)\text{.}\)
Notice in
FigureΒ 102 that the
sign of
\(f(y)\) flips as you pass through an equilibrium. The pattern of that flip tells you what kind of point it is:
At the source , \(y=-1\text{,}\) \(f(y)\) goes from negative β positive (increases).
At the sink , \(y=1\text{,}\) \(f(y)\) goes from positive β negative (decreases).
The important observation is that
\(f(y)\) increases through sources and decreases through sinks. This is true in general, so we can classify equilibrium solutions by looking at the sign of
\(f'(y)\) at the equilibrium points.
Figure 102. Figure 103. If
\(f(y_0) = 0\text{,}\) then
\(y_0\) is an equilibrium. To classify it:
If \(f'(y_0) < 0\text{,}\) itβs a sink (stable).
If \(f'(y_0) > 0\text{,}\) itβs a source (unstable).
If \(f'(y_0) = 0\text{,}\) the test is inconclusiveβyouβll need more investigation.
This is called the
linearization method , because it looks at the linear (slope) behavior of
\(f(y)\) near the equilibrium.
π Example 104 .
Consider the autonomous equation:
\begin{equation*}
\frac{dy}{dt} = y^2 - 4y + 3.
\end{equation*}
Find and classify the equilibrium solutions using linearization.
Solution .
First, solve \(f(y)=0\text{:}\)
\begin{equation*}
y^2 - 4y + 3 = (y - 1)(y - 3) = 0,
\end{equation*}
so \(y=1\) and \(y=3\) are equilibria.
Then compute the derivative:
\begin{equation*}
f'(y) = 2y - 4.
\end{equation*}
At each equilibrium:
π€ Wrap-Up.
ποΈ Key Takeaways...
Equilibrium solutions occur where \(y' = f(y) = 0\text{.}\)
Sinks pull solutions in, sources push them out, and nodes do one of each.
Phase lines show this behavior simply and clearly.
The derivative \(f'(y)\) can classify stability quickly via the linearization method.
Check Your Understanding.
Checkpoint 105 . πβ Classifying Equilibrium Solutions.
(a) πβ Stability Check.
The autonomous equation
\begin{equation*}
\frac{dy}{dt} = (y - 3)(y + 1).
\end{equation*}
has equilibrium solutions at \(y = -1\) and \(y = 3\text{.}\) Create a simple phase line to classify these equilibria.
\(y = -1\) is a sink, \(y = 2\) source
\(y = -1\) is a source, \(y = 2\) sink
Both are sinks
Both are sources
\(y = -1\) is a node, \(y = 2\) sink
\(y = -1\) is a source, \(y = 2\) node
(b) πβ Comparing Equilibria.
The equation
\begin{equation*}
\frac{dy}{dt} = y^2-3y
\end{equation*}
has equilibrium solutions at \(y=0\) and \(y=3\text{.}\)
Compute
\(f'(y)\) and use the linearization method to determine which of these equilibria
pulls nearby solutions toward it .
\(y=0\) Correct. Since \(f'(y) = 2y-3\) and \(f'(0) = -3 \lt 0\text{,}\) then \(y=0\) is a sink and pulls solutions toward it.
\(y=3\) Incorrect. Since \(f'(y) = 2y-3\) and \(f'(3) = +3 \gt 0\text{,}\) then \(y=3\) is a source and repels solutions.
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