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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 $x_0$.

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2 Answers

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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 $-3$.

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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 $x(3)$

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