Non solvable integral? (dx/dt)^2 dt

In summary, the nonlinear 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,
  • #1
Tomder
4
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.
 
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  • #3
Simplifying the constants and shifting the origin of [itex]x[/itex], your system is [tex]\ddot x = A \sin x - k\dot x.[/tex] This is the equation of motion of a pendulum in a resistive medium. The resistive force [itex]-k\dot x[/itex] effectively removes energy from the pendulum as it does work in moving through the resistive medium.

You can write this as the 2D system [tex]
\begin{split}
\dot x &= y \\
\dot y &= A\sin x - ky \end{split}[/tex] and use Dulac's criterion to show that no non-constant periodic solutions exist.
 
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  • #4
Yep, sorry I will reupload this post using the proper notation, thank you for the advice
 

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