Prove that solutions to autonomous first order DE can't have local maxima

  • #1
zenterix
423
60
Homework Statement
Suppose we have an autonomous first order ordinary differential equation ##\dot{y}=f(y)##.

How do we prove that solutions can't have local maxima or minima?
Relevant Equations
This is the type of question that I can guess the answer to with arguments that seem simple at first but then seem hard to actually construct in full.

I am interested in different proofs that the answer is no.
Here is one argument.

Suppose we have a solution ##y## such that ##y(t_0)=y_0## and ##y'(t_0)=f(y_0)=0##.

##y(t)=y_0## is a solution since ##\dot{y}(t)=0## and so ##\dot{y}=f(y)=f(y_0)=0##.

I am aware of uniqueness theorems for linear differential equations. I don't remember seeing such a theorem for non-linear equations. Is there one?

If there is a unique solution such that ##y(t_0)=y_0## then this solution is the constant function ##y(t)=y_0##.

Thus, if this is the case, we cannot have local maxima in solutions to ##\dot{y}=f(y)## since at the maximum the function would coincide with another solution, namely the constant solution with the value of ##y## at the critical point.

Here is an attempt at another argument.
For autonomous differential equations, all solutions are horizontal shifts of any other solution.

If we have another function ##z(t)=y(t-\Delta t)## then ##\dot{z}=\dot{y}(t-\Delta t)=f(y(t-\Delta t))=f(z(t))##.

Thus, ##z## is also a solution.

If a solution is continuous and a local maximum occurs at ##t_0##, then there are points ##t_1## and ##t_2## such that ##t_1<t_0<t_2## and ##f(t_1)=f(t_2)<f(t_0)##.

If we shift the solution ##y(t)## by ##t_2-t_1## then what we end up with is a new solution that at ##t_1## has the same value as the original solution but a different slope.

Again, this would be precluded by a uniqueness theorem.

Finally, I just want to confirm: under no circumstances can we have solutions that have an inflection point (all for the same ##y_0## but different ##t##) where the slope is zero.

Ie, solutions that look like, say, ##(t-t_0)^3## for different ##t_0##. Now, clearly this specific function doesn't work, since it doesn't satisfy an autonomous DE. But I just cite it here to demonstrate the general shape of the solutions I am referring to.
 
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  • #3
If [itex]y(t)[/itex] has a local extremum at [itex]t = t_0[/itex], you can show that there exists a [itex]t_1 < t_0[/itex] and a [itex]t_2 > t_0[/itex] for which [itex]y(t_1) = y(t_2)[/itex] but [itex]y'(t_1)[/itex] and [itex]y'(t_2)[/itex] have opposite sign. This is impossible if [itex]y'[/itex] is a function of [itex]y[/itex].
 

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