A Non solvable integral? (dx/dt)^2 dt

Tomder
Messages
4
Reaction score
1
TL;DR Summary
I have a non linear system to which I implement a PD controller, but when applying the kinetic-work theorem I can't solve an integral.
The integral is (dx/dt)^2 dt, where x=x(t) so it can't be just x + C.

The non linear system for whom wants to know how did I get to that point is:

d(dx/dt)/dt = sqrt(a^2+b^2)*sin(x+alfa+phi) - Kd*(dx/dt); where alfa = atan(a/b), phi = constant angle, Kd = constant coefficient.
After applying the kinetic work theorem by multiplying both sides by dx/dt I get:

d(dx/dt)/dt *dx/dt = sqrt(a^2+b^2)*sin(x+alfa+phi)*dx/dt - Kd*(dx/dt)*dx/dt ; So, by integration by dt I get to:

1/2*(dx/dt)^2 = - sqrt(a^2+b^2)*cos(x+alfa`phi) - INTEGRAL[(Kd*(dx/dt)^2]dt + C ;

Rearranging terms:

1/2*(dx/dt)^2 + sqrt(a^2+b^2)*cos(x+alfa`phi) = C - INTEGRAL[(Kd*(dx/dt)^2]dt ;And by this without the Kd term i could get the total energy of the system and the velocity at every point but I don't know how to proceed with the Kd term.
I'm sure maybe some of the theory may be wrong explained so I say sorry in advance.
For further explanation, my system doesn't lose energy when Kd = 0 because the total energy would be constant but with Kd term I assume it is like a frictional component that takes the energy out and the system would slowly stop oscillating in the equilibrium point.
 
Physics news on Phys.org
Simplifying the constants and shifting the origin of x, your system is \ddot x = A \sin x - k\dot x. This is the equation of motion of a pendulum in a resistive medium. The resistive force -k\dot x effectively removes energy from the pendulum as it does work in moving through the resistive medium.

You can write this as the 2D system <br /> \begin{split}<br /> \dot x &amp;= y \\<br /> \dot y &amp;= A\sin x - ky \end{split} and use Dulac's criterion to show that no non-constant periodic solutions exist.
 
Yep, sorry I will reupload this post using the proper notation, thank you for the advice
 
Back
Top