[SOLVED]2nd order to first order

dwsmith

Well-known member
Can this second order be changed into a system of first order:
$$x''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^3}x$$

Last edited:

Ackbach

Indicium Physicus
Staff member
Let $x_{1}=x$ and $x_{2}=x'$. Then you have the first-order system
\begin{align*}
x_{1}'&=x_{2}\\
x_{2}'&=-\frac{\mu}{ \left(x_{1}^{2}+y^{2}+z^{2} \right)^{3/2}}.
\end{align*}
What are $y$ and $z$ doing? Are they independent functions of time?

Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Physically, this looks like a central force field with a force that varies inversely with the 3rd power of distance.

Can it be that you actually have: $r''(t)=- {\mu \over r^3}$?

If that is the case you can do the following:

$\ddot r=- {\mu \over r^3}$

Multiply left and right with $2 \dot r$.
$2\ddot r \dot r = -2 {\mu \over r^3} \dot r$

$\dot r^2 = {\mu \over r^2} + C$

Jester

Well-known member
MHB Math Helper
My guess - there are two more equations that go with this one!

dwsmith

Well-known member
My guess - there are two more equations that go with this one!
There but if it was straight forward in changing the first equation, I would have done the other two. Will 2 other coupled equations alter how they are re-written as first order equations?

$$y''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^3}y$$
$$z''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^3}z$$

dwsmith

Well-known member
Let $x_{1}=x$ and $x_{2}=x'$. Then you have the first-order system
\begin{align*}
x_{1}'&=x_{2}\\
x_{2}'&=-\frac{\mu}{ \left(x_{1}^{2}+y^{2}+z^{2} \right)^{3/2}}.
\end{align*}
What are $y$ and $z$ doing? Are they independent functions of time?
I had a typo by the way
$$x''(t) = -\frac{\mu}{(\sqrt{x^2+y^2+z^2})^2}x$$

Ackbach

Indicium Physicus
Staff member
Assuming
$$x''(t)=-\frac{\mu}{(x^{2}+y^{2}+z^{2})^{3/2}}\,x,$$
then you can let $x_{1}$ and $x_{2}$ be defined as above, and let $y_{1}=y$ and $y_{2}=y'$, as well as $z_{1}=z$ and $z_{2}=z'$. Then you get the following system:
\begin{align*}
x_{1}'&=x_{2}\\
y_{1}'&=y_{2}\\
z_{1}'&=z_{2}\\
x_{2}'&=-\frac{\mu}{(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})^{3/2}}\,x_{1}\\
y_{2}'&=-\frac{\mu}{(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})^{3/2}}\,y_{1}\\
z_{2}'&=-\frac{\mu}{(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})^{3/2}}\,z_{1}.
\end{align*}