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?

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

Beginner2022-03-16Added 9 answers

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.

A fixed point

First, if

thus

Step 2

The following remark might be the reason why you are confused: there might be nonzero constant solutions passing through the zero of

where

passing through

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,

When f is continuously differentiable, the Picard-Lindelof theorem says that there is an unique solution to (1) with a given initial condition

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