Given an ordinary differential equation, I read the definition of a solution (http://tutorial.math.lamar.edu/Classes/DE/Definitions.aspx)
A solution to a differential equation on an interval $\alpha<t<\beta$ is any function which satisfies the differential equation in question on the interval $\alpha<t<\beta$.
My question is why did we specify in the definition that it would be an interval (and not a disjoint union of intervals for example) and also why the interval is open? is there a problem in defining a solution over a closed interval? thanks for your help!
Best Answer
It's mildly annoying to define exactly what the derivative of a function is at $a$ and $b$, if the function is only defined on $[a, b]$, due to the definition of the derivative involving a (two-sided) limit. You can just take a one-sided limit, but here's a specific type of nuisance that might arise. Say you have some ODE $f'(t)=u(f(t), t)$. And suppose some function $f$ is defined on $(a, b+\epsilon)$, and its restriction to $(a, b]$ is a solution to the ODE, meaning the left-hand derivative of $f$ at $b$ is $u(f(b), b)$. What if $f$ is actually non-differentiable at $b$, because the difference quotient at $b$ has a different right-hand limit to its left-hand limit? What if the right-hand derivative of $f$ at $b$ does not satisfy the ODE? Do we really want to say that $f$ solves the ODE on $(a, b]$?
It's just not worth dealing with this kind of irritation in an intro class. It's easier to just say that $f$ solves the ODE on $(a, b)$, and then if for some particular ODE there is some interesting phenomenon happening at the end-points of the interval, investigating them as a special case.