When solving a first order differential equation
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 is a constant solution to the differential equation (1) First, if is a fixed point, then the constant function has derivative . Using (1), we have thus . On the other hand, if is a zero of f, then the constant functions 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 : the standard example is the ODE where is the zero of , however there is a non-constant solution passing through 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, is not continuous at ). When f is continuously differentiable, the Picard-Lindelof theorem says that there is an unique solution to (1) with a given initial condition . Since we have checked that the constant function 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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