Lecture 3
1.2 Slope Fields and Equilibrium Solutions
This material is covered in Stewart, Section 9.2.
We have seen that in order to solve \[ \dif{y}{t} = f(t) \] we only need to integrate.
1.2 Slope Fields and Equilibrium Solutions
However, for the more general first-order ODE \[ \dif{y}{t}=f(t,y) \] this no longer works.
Nonetheless, the ODE gives a qualitative picture of the solution by noting that at $(t,y)=(a,b)$ the slope of $y(t)$ is $f(a,b)$.
1.2 Slope Fields and Equilibrium Solutions
Nonetheless, the ODE gives a qualitative picture of the solution by noting that at $(t,y)=(a,b)$ the slope of $y(t)$ is $f(a,b)$.
So what one can do is as follows:
Note the slope field can be generated without having to solve the ODE.
1.2.1 Example: the slope field of $\;y'=2y$
From the slope field of $y'=2y$ we can see that $y=0$ is one of the solution curves. It is a constant or equilibrium solution.
1.2.2 Equilibrium solutions
An equilibrium solution is a constant solution $y(t) = c$ to the ODE \[ \dif{y}{t}=f(t,y). \] The graph of an equilibrium solution is a horizontal line. Such a line has a slope of zero, i.e., $y'=0$. This can only happen if $f(t,y)=0$ has a solution $y=c$ for some real constant $c$.
1.2.2 Equilibrium solutions
Example: Find the equilibrium solutions of $\,y'(t)=-3(y-1)$.
Equilibrium solution occurs when $y'=0.$
$\Ra -3(y-1)=0\,$ $\,\Ra y = 1.$
1.2.2 Equilibrium solutions
Example: Find the equilibrium solutions of $\,y'(t)=-3(y-1)$.
1.2.2 Equilibrium solutions
Example: Find the equilibrium solutions of $\; y'(t)=2t+1.$
Set $y'=0\,$ $\,\Ra 2t +1 =0\,$ $\,\Ra t = \ds -\frac{1}{2}$
But this is not an equilibrium solution! Because we need $\ds \ds\frac{dy}{dt}=0$ for all $t.$
We wanted a constant value of $y,$ not $t.$ Then, there are no equilibrium solutions.
1.2.2 Equilibrium solutions
Example: Find the equilibrium solutions of $\; y'(t)=2t+1.$
1.2.2 Equilibrium solutions
Example: Find the equilibrium solutions of $\;y'=y(1-y).$
Set $y'=0\,$ $\,\Ra y(1-y) =0\,$ $\,y = 0, 1.$
In this case we have two equilibrium solutions.
1.2.2 Equilibrium solutions
Example: Find the equilibrium solutions of $\;y'=y(1-y).$
1.2.3 Stability of equilibrium solutions
A pencil sitting balanced vertically is in an equilibrium state. But make one small perturbation and it will topple over. This is an unstable equilibrium. On the other hand, a pendulum hanging vertically is also in an equilibrium state. But if you perturb it slightly, it will eventually (with friction) return to its equilibrium. This is a stable equilibrium.
From the slope field, you can decide if an equilibrium solution is stable or not by looking at whether solution curves will tend toward the equilibrium solution or away from it as time increases.
1.2.3 Stability of equilibrium solutions
Formally, an equilibrium solution $y(t)=y_0$
to the differential
equation $y'=f(t,y)$ is stable if the initial value problem:
\[
\dif{y}{t}=f(t,y), \qquad y(0)=y_0\pm \epsilon
\]
has a solution $y(t)$
which satisfies $\ds\lim_{t\rightarrow \infty} y(t)=y_0.$
In other words, if you start sufficiently close to a stable
equilibrium solution,
then you will approach that equilibrium solution.
1.2.3 Stability of equilibrium solutions
Example: Here is the slope field for $y' = \sin (4y).$
For this ODE, which equilibrium solutions are stable and which are unstable?
1.2.3 Stability of equilibrium solutions
For this ODE, which equilibrium solutions are stable and which are unstable?
If $y'=0\,$ $\Ra \sin(4y)=0\,$ $\Ra 4y=n\pi,\,$ for $\, n\in \N.$
For $n=0\,$ unstable. For $n = \pm 1$ stable.
📝 Try to find under what conditions of $n$ is $y =\ds n\frac{\pi}{4}$ stable or unstable.
1.2.3 Stability of equilibrium solutions
Important remark: It is an important point that equilibrium solutions cannot be crossed by other solution curves. In fact, no solutions curves can cross each other because this would mean that in some point(s) $y'(t)$ has more than one value. Equilibrium solutions therefore partition the solution space.
1.2.4 Long-term behaviour
Exercise: True or False? Assume the ODE \[ \dif{y}{t}=-3(y-1) \] has initial condition $y(t_0)=c$ with $c\lt1$. Then $y(t)\lt 1$ for all $t\gt t_0$.
Equilibrium solutions at $y=1.$ This means that if $y$ starts less than $1,$ it must remain less than $1.$
1.2.4 Long-term behaviour
Example: Consider the ODE $y'=y^2$. It has one equilibrium solution $y=0$.
For $y>0$, $y'>0$ so that $y(t)$ is increasing and hence any solution with $y>0$ tends away from $y=0$.
For $y\lt0$, $y'\gt 0$ so that $y(t)$ is increasing and hence any solution with $y\lt 0$ approaches $y=0$.
$y'=-y\left(y-3/2\right)\left(y+3/2\right)$
Click on graphic view to show/hide solution curves!
$y'=x^2 + y^2 - 1$
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$y'= \cos(2t)\sin(2y) $
Click on graphic view to show/hide solution curves!
1.2.5 Main points