Solving this differential equation

AI Thread Summary
The differential equation presented, second derivative of phi = (-g/R) sin phi, cannot be solved using elementary functions due to its non-linear nature. Instead, the solution involves Jacobi elliptic functions, which arise from integrating the transformed equation. By multiplying through by the first derivative and recognizing the resulting expression as a first derivative of a squared term, it leads to an elliptic integral. The discussion highlights the limitations of direct integration, particularly noting the dependence on phi rather than time. Numerical methods and approximations, such as Taylor expansion, are suggested as practical alternatives for solving the equation.
Moneer81
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Hello,

while working on a simple mechanics problem using polar coordinates, I got this equation

second derivative of phi = (-g/R) sin phi

now I need to solve this to get an equation for phi (t) but the books says that I cannot solve this using elementary functions and that the solution will be the more complex Jacobi elliptic function. My question is why can't I integrate twice to get the equation for phi (t) ?

thanks a lot
 
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You can't just straight integrate that differential equation. I'm not even sure how one would attempt it. Here's one way to set up the elliptic integral:

Multiply through by \dot{\phi} and get the differential equation
\ddot{\phi} \dot{\phi} = -g/R \dot{\phi} sin \phi
which we recognize as being the first time derivative of
\dot{\phi}^2 = g/R cos \phi + C
when then leads to the integral:
\int d\phi ~ 1/\sqrt{g/R cos \phi + C} = t - t_0
which is then the elliptic integral left behind. This is why we make things like the small angle approximation, where applicable. If you're working with exact numbers, you could pretty simply write a numerical algorithm
 
MalleusScientiarum ,

why can't we straight integrate it with respect to time ?
 
ok never mind...stupid question. i know why :)
 
Yeah - because it's sine of phi, not sine of t.
 
In post #2, you missed a little "2" when going from \dot{\phi}\ddot{\phi} to the square of the first derivative.

Daniel.
 
i guess theylor expansion is the best choice
 
I advise numerical analysis.
 
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