# Stability of fixed points for a differential equation

Consider the differential equation $$x'=x^2-9$$

a. find the stability type of each fixed point

To find the fixed points, I set this equal to $$0$$, right? Would someone mind explaining why I do this? I don't really understand the concept.

So I get the fixed point as being $$x=3$$ and $$x=-3$$.

Checking the stability:

For $$x=3$$:
$$x=2$$ results in $$-5$$ (left)
$$x=4$$ results in $$7$$ (right)

Unstable?

For $$x=-3$$:
$$x=-4$$ is positive (left)
$$x=-2$$ is negative (right)

Stable?

How do I determine their stability exactly?

b. sketch the phase portrait on the line

c. Sketch the graph of $$x(t)=ϕ(t;x_0)$$ in the $$(t,x)$$-plane for several representative initial conditions $$x_0$$.

• You mean $x' = x^2-9$? Commented Nov 7, 2014 at 16:48
• Fixed points are points where the solution to the differential equation is, well, fixed. That is, it doesn't move (i.e. doesn't change with respect to $t$). If it isn't moving, its derivative is zero. If its derivative $x'$ is zero, this means that $x^2 - 9$ is zero as well. Commented Nov 7, 2014 at 16:51
• You might also say a little more about your thoughts on the other parts of this question; what have you tried, or what similar problems have you completed, and why is this one different? (or maybe you have no idea what a phase portrait is, but you should say so) Commented Nov 7, 2014 at 16:54
• @MathMajor yes both are identical. I used the "prime" notation in my class but I know "dot" notation is fairly standard. Commented Nov 7, 2014 at 16:56
• @MathMajor Also, remember $x$ is a function with input $t$, so you aren't checking $x = -2$, you are checking $t = -2 \implies$ checking $x'(-2)$. Do you know how to sketch a phase portrait? Commented Nov 7, 2014 at 17:01

To find the fixed points, we set $x' = 0$ and solve, yielding:

$$x' = x^2 -9 = 0 \implies x_{1,2} = \pm~3$$

To test stability, you can follow Paul's Online Notes, by picking values around the critical points and noting the sign of the derivative $x'$.

If we draw a phase line, we get (note that $+3$ is unstable and $-3$ is stable):

If we draw a direction field plot and then superimpose solution curves on it, we have (compare the two critical points to the phase line and look at each (purple) solution curve):

Note, for the solution curves (the direction field plot shows many examples), you can take several examples for different ICs, for example (just as the plot shows).

$$x' = x^2 - 9, x(0) = 1 \implies x(t) = \dfrac{6-3 e^{6 t}}{e^{6 t}+2}$$

What happens to $x(t)$ as $t$ approaches infinity? It approaches $-3$.

Remember, $x'(t)$ tells us the rate of change of $x(t)$. You want to set it equal to zero because that indicates a point where $x(t)$ is not changing (that is to say, it is fixed). To determine stability, check values of $x'(t)$ on the left and right of your fixed points. For example, for the fixed point $t = -3$, check the value of $x'(-4)$ and $x'(-2)$. If $x'(t)$ is positive to the left of $x'(-3)$ and negative to the right of $x'(-3)$ then $t=-3$ is stable. If the left and right are the same sign, you have semi-stability. If the left side is negative while the right is positive, you have instability.

Hopefully this picture illustrates that solutions move towards $x(-3)$ and away from $x(3)$