# Is it possible to find extrema of solution of first order ODE without solving ODE?

Hello everyone,

Sorry maybe question quite stupid. I have first order, scalar ODE: $$\frac{du}{dt}=u_{\infty}(t)-u$$ where $u_{\infty}(t)$ is really heavy, so I'm pretty sure, that there is not any way to resolve classical integral. I need to find value of solution in exttrema $t_X$, specifically in maximum. Obviously that in extremal points: $u_{\infty}(t_X)=u(t_X)$, so if we can get $t_X$, we would easy get $u(t_X)$. We can define initial condition as $u(0) = u_{\infty}(0) = E$, where E - some constant.

Do anyone have any idea how to get $t_X$ without resolving $u(t)$?

Just note that, if we will looking for standard solution: $$u(t) = e^{-t}\left( \int\limits_{t_0}^{t} u_{\infty}(x)e^x dx + C\right)$$ at $t_X$: $$u_{\infty}(t_X)e^{t_X} = \int\limits_0^{t_X} u_{\infty}(x)e^x dx + C$$ Can we estimate where function $\left(F(t)=u_{\infty}(t)e^{t} \right)$ intersects its integral value and then estimate $C$ from initial condition?

I'll appreciate any help.

Ruben

-
I'm not sure what you mean by "heavy" or "unresolvable". If closed-form solutions are unavailable, have you tried numerical methods? – Robert Israel Dec 5 '12 at 6:35
The solution and also the extrema will depend on the initial condition for $u$. If you're just looking for an extremal value, then for any $t_X$ you can find $u_0$ such that $u(t)$ is extremal at $t_X$. This can easily be seen by writing down the variation of constants integral solution. – Jaap Eldering Dec 5 '12 at 13:55
to Robert: The point is that: I have to solve it analytically, because it is a checker for my numerical results. 'Heavy' or 'unresolvable' means that, standard solution $$u(t) = e^{-t}\, \int_{-\infty}^{t} u_{\infty}(x)e^{x}dx$$ is hardly resolvable. – rth Dec 5 '12 at 18:41
to Jaap: Thank you for comment, you are absolutely right. Solution should obey to some initial condition. $u(-\infty) = E$, where E is a constant, BTW $u_{\infty}(-\infty) = u_{\infty}(\infty) = E$, where E is the same constant. – rth Dec 5 '12 at 18:47
An "initial condition" at $-\infty$ or $+\infty$ generally does not lead to a well-posed problem. Typically either solutions don't have a limit as $t \to -\infty$, or all solutions will have the same limit. In this case the homogeneous equation has a solution $u(t) = e^{-t}$ which goes to $+\infty$ as $t \to -\infty$; there will be at most one solution to your non-homogeneous equation that goes to a finite value as $t \to -\infty$, but there may be none. – Robert Israel Dec 6 '12 at 7:43