# Peeter Joot's (OLD) Blog.

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## Integrating the equation of motion for a one dimensional problem.

Posted by peeterjoot on January 2, 2010

# Motivation.

While linear approximations, such as the small angle approximation for the pendum, are often used to understand the dynamics of non-linear systems, exact solutions may be possible in some cases. Walk through the setup for such an exact solution.

# Guts

The equation to consider solutions of has the form

\begin{aligned}\frac{d}{dt} \left( m \frac{dx}{dt} \right) = -\frac{\partial {U(x)}}{\partial {x}}.\end{aligned} \hspace{\stretch{1}}(2.1)

We have an unpleasant mix of space and time derivatives, preventing any sort of antidifferentiation. Assuming constant mass $m$, and employing the chain rule a refactoring in terms of velocities is possible.

\begin{aligned}\frac{d}{dt} \left( \frac{dx}{dt} \right) &= \frac{dx}{dt} \frac{d}{dx} \left( \frac{dx}{dt} \right) \\ &= \frac{1}{{2}} \frac{d}{dx} \left( \frac{dx}{dt} \right)^2 \\ \end{aligned}

The one dimensional Newton’s law (2.1) now takes the form

\begin{aligned}\frac{d}{dx} \left( \frac{dx}{dt} \right)^2 &= -\frac{2}{m} \frac{\partial {U(x)}}{\partial {x}}.\end{aligned} \hspace{\stretch{1}}(2.2)

This can now be antidifferentiated for

\begin{aligned}\left( \frac{dx}{dt} \right)^2 &= \frac{2}{m} (E - U(x)).\end{aligned} \hspace{\stretch{1}}(2.3)

Taking roots and rearranging produces an implicit differential form $x$ in terms of time

\begin{aligned}dt = \frac{dx}{\sqrt{ \frac{2}{m} (E - U(x)) } }.\end{aligned} \hspace{\stretch{1}}(2.4)

One can concievably integrate this and invert to solve for position as a function of time, but substitution of a more specific potential is required to go further.

\begin{aligned}t(x) = t(x_0) + \int_{y=x_0}^{x} \frac{dy}{\sqrt{ \frac{2}{m} (E - U(y)) } }.\end{aligned} \hspace{\stretch{1}}(2.5)

TODO: doing stuff with this.

EDIT: This was a stupid way to do this. It is nothing more than rearranging the Hamiltonian for $\dot{x}^2$.