When solving a first order differential equation \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={f{{\left({y}\right)}}}\) for

Riley Quinn

Riley Quinn

Answered question


When solving a first order differential equation
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?

Answer & Explanation



Beginner2022-03-16Added 9 answers

Step 1
A fixed point y0 is a constant solution y(t)=y0 to the differential equation
y=f(y) (1)
First, if y0 is a fixed point, then the constant function y(t)=y0 has derivative y(t)=0. Using (1), we have
thus f(y0)=0. On the other hand, if y0 is a zero of f, then the constant functions y(t)=y0 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(y0): the standard example is the ODE
where y0=0 is the zero of f(y)=3y23, however there is a non-constant solution
passing through y0=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, (y23)=y13 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(t0)=y0. Since we have checked that the constant function y(t)=y0 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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