Show that the orbits are given by ellipses $\omega^{2}x^{2}+v^{2}=C$, where $C$ is any non- negative constant.

Question

I am working through a text book by Strogatz Nonlinear dynamics and chaos . In chapter 5 question 5.1.1 (a). I have answered the question but would like to check if I have performed the integration step properly.

Question

Show that the orbits are given by ellipses $\omega^{2}x^{2}+v^{2}=C$, where $C$ is any non- negative constant.

Answer

Starting with dividing one equation with the other: \begin{equation} \Rightarrow \frac{\dot{x}}{\dot{v}} = \frac{v}{- \omega^{2} x} \\ -\omega^{2} x \dot{x} = v \dot{v} \\ \omega^{2} x \dot{x} + v \dot{v} = 0 \text{.} \end{equation}

Then integrating (The integral of 0 is constant D): \begin{equation} \int \omega^{2} x \dot{x} \text{ }dx + \int v \dot{v} \text{ }dv = \int 0 \text{ } dx \\ \frac{1}{2} \omega^{2} x^{2} + \frac{1}{2} v^{2} = D \\ \omega^{2} x^{2} + v^{2} = 2 D \\ \omega^{2} x^{2} + v^{2} = C \end{equation}

Summary

Could someone please verify this proof, particularly with regards to the $\int \omega^{2} x \dot{x} \text{ }dx + \int v \dot{v} \text{ }dv = \int 0 \text{ } dx $ line. Did I do this properly?

Answer 1

Your derivatives (with a point above) are with respect to the variable time $ t .$

So, when you integrate, you should do it according to $t$.

$$\int \omega^2xx'dt+\int vv'dt=\int 0dt$$