Need help with an integral -- How to integrate velocity squared?

[Tomder]

TL;DR

Problem with a implicit function integral.

The integral is this one:

$\int (\dot x)^2 \, dt,$

With $x=x(t).$

I don't know how to solve that integral and I haven't find nothing to read about on how to proceed with this kind of (implicit function?) integrals without having the initial function.

 

[fresh_42]

This:
https://www.wolframalpha.com/input?i=integral+(y'(x))^2+dx+=
suggests that it is likely there exists no closed form for the integral.

What's also suspicious: if you take standard units, then you ask for a quantity measured in $\dfrac{m^2}{s}$ and I cannot think of where this might make sense.

 

[pasmith]

TL;DR Summary: Problem with a implicit function integral.

The integral is this one:

$\int (\dot x)^2 \, dt,$

With $x=x(t).$

I don't know how to solve that integral and I haven't find nothing to read about on how to proceed with this kind of (implicit function?) integrals without having the initial function.

By the chain rule: \int \dot x^2\,dt = \int \dot x \frac{dx}{dt}\,dt = \int \dot x \,dx. But this only helps you if you know \dot x in terms of x, ie. your system satisfies a first order ODE. But you originally asked this question in the context of a second-order nonlinear ODE <br /> \ddot x = -f(x) - k\dot x. In that case you do not have \dot x in terms of x; you have \ddot x in terms of x and \dot x. You can't evaluate\int \dot x^2 \,dt unless you already know x(t) and \dot x(t). What you can say is that <br /> \frac{d}{dt} \left( \frac12 \dot x^2 + \int f(x)\,dx \right) = -k\dot x^2 \leq 0, ie. the system dissipates energy.

 

[Tomder]

By the chain rule: \int \dot x^2\,dt = \int \dot x \frac{dx}{dt}\,dt = \int \dot x \,dx. But this only helps you if you know \dot x in terms of x, ie. your system satisfies a first order ODE. But you originally asked this question in the context of a second-order nonlinear ODE <br /> \ddot x = -f(x) - k\dot x. In that case you do not have \dot x in terms of x; you have \ddot x in terms of x and \dot x. You can't evaluate\int \dot x^2 \,dt unless you already know x(t) and \dot x(t). What you can say is that <br /> \frac{d}{dt} \left( \frac12 \dot x^2 + \int f(x)\,dx \right) = -k\dot x^2 \leq 0, ie. the system dissipates energy.

Thanks, I thought about the same process but was unsure of its veracity, guess I‘ll try to work in my system with the idea of energy dissipation term. Thank you very much for your answer.