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
 

1. What is a non-solvable integral?

A non-solvable integral is an integral that cannot be solved using standard integration techniques. This means that it does not have a closed-form solution and requires more advanced methods to solve.

2. How is a non-solvable integral different from a solvable integral?

A solvable integral has a closed-form solution that can be found using standard integration techniques. On the other hand, a non-solvable integral does not have a closed-form solution and may require more advanced methods, such as numerical integration, to solve.

3. Why are non-solvable integrals important?

Non-solvable integrals are important because they arise in many real-world applications, particularly in physics, engineering, and economics. They also provide interesting mathematical challenges and can lead to new discoveries and techniques in integration.

4. How can non-solvable integrals be approximated?

Non-solvable integrals can be approximated using numerical integration methods, such as the trapezoidal rule or Simpson's rule. These methods break the integral into smaller, solvable pieces and use numerical calculations to approximate the overall result.

5. Is there a general method for solving non-solvable integrals?

There is no single, general method for solving all non-solvable integrals. However, there are various techniques and approaches that can be used depending on the specific form of the integral. These may include substitution, integration by parts, or the use of special functions.

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