 Riley Quinn

2022-03-15

When solving a first order differential equation
$\frac{dy}{dx}=f\left(y\right)$
for the fixed points (or steady states), we set the differential equation to 0 and solve for the values of y that are fixed points. Why does this guarantee that we will obtain fixed points? Why can't we obtain a point at which the slope is 0 but is not a fixed point? ieuemd0l

Step 1
A fixed point ${y}_{0}$ is a constant solution $y\left(t\right)={y}_{0}$ to the differential equation
${y}^{\prime }=f\left(y\right)$ (1)
First, if ${y}_{0}$ is a fixed point, then the constant function $y\left(t\right)={y}_{0}$ has derivative ${y}^{\prime }\left(t\right)=0$. Using (1), we have
$f\left({y}_{0}\right)=f\left(y\left(t\right)\right)={y}^{\prime }\left(t\right)=0,$
thus $f\left({y}_{0}\right)=0$. On the other hand, if ${y}_{0}$ is a zero of f, then the constant functions $y\left(t\right)={y}_{0}$ satisfies (1) and thus is a fixed point.
Step 2
The following remark might be the reason why you are confused: there might be nonzero constant solutions passing through the zero of $f\left({y}_{0}\right)$: the standard example is the ODE
${y}^{\prime }=3{y}^{\frac{2}{3}},$
where ${y}_{0}=0$ is the zero of $f\left(y\right)=3{y}^{\frac{2}{3}}$, however there is a non-constant solution
$y\left(t\right)={t}^{3}$
passing through ${y}_{0}=0$ and is not a fixed point. That is, there is a solution passing through the zero of f with zero slope. Your concern is real.
But this does not invalidate the claim that the any zero of f contribute to a fixed point. The (more worrying) issue is that there are non-uniqueness of solutions to ODE. This could happen when f is not continuously differentiable (note that in our example, ${\left({y}^{\frac{2}{3}}\right)}^{\prime }={y}^{-\frac{1}{3}}$ is not continuous at $y=0$).
When f is continuously differentiable, the Picard-Lindelof theorem says that there is an unique solution to (1) with a given initial condition $y\left({t}_{0}\right)={y}_{0}$. Since we have checked that the constant function $y\left(t\right)={y}_{0}$ is a solution to (1) with that given initial condition, the theorem implies that the constant solutions are the only possible solution.

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