Differential equation - Reduction of order

1. Nov 2, 2013

fineTuner

Hi,

i have to reduce the order of a 2nd order differential equation, to solve it with a numerical method.
The equation is:

$\ddot{r}+a\dot{r}+\frac{b}{r^{2}}=0$

with $a,b\geq0$

I tried to reduce it substituting $\dot{r}=v$, but i don't know what to do with the term $\frac{b}{r^{2}}$ .

Can someone explain me what i should do?

Thank you,

J.

2. Nov 2, 2013

HallsofIvy

Staff Emeritus
In general, you can't just reduce the order of a differential equation. What you can do is convert a second order equation to two first order differential equations or, equivalently to a two dimensional vector equation. Here, if you let v= r', then r''= v' so your first equation becomes, as you saw, v'+ av+ b/r^2= 0 and your second equation is r'= v. Equivalently, if we think of vector, V, as being $\begin{bmatrix}r(t) \\ v(t)\end{bmatrix}$ we have the vector equation
$$V'= \begin{bmatrix}r(t) \\ v(t) \end{bmatrix}'= \begin{bmatrix}-v- b/r^2 \\ v\end{bmatrix}$$

3. Nov 2, 2013

JJacquelin

Hi !
Another manner to present it :
Let dr/dt = Y(r)
d²r/dt² = d(Y(r))/dt = (dY/dr)*(dr/dt) = (dY/dr)*Y
r''+a*r'+b/r² = (dY/dr)*Y+a*Y+b/r²
Y' Y + a Y = -b/r² is an first order ODE, where Y(r) is the unknown function.
Then dt = dr/Y(r)
But the analytic integration of 1/Y(r) will probably be too arduous. That is the raison why numerical solving of the first order ODE is more realisic in practice.

4. Nov 3, 2013

fineTuner

Hi!
Thank you JJacquelin and HallsofIvy for your answers. The most useful form is the first one, in the matrix form. I will put it into a f90 program to simulate a body orbiting around a planet with air drag... now it's time to find the right units to avoid problems with my pc, let's see what will happen!

5. Nov 4, 2013

vanhees71

$$V'=\begin{bmatrix} r' \\ v' \end{bmatrix} =\begin{bmatrix} v \\ r'' \end{bmatrix} =\begin{bmatrix} v \\ -a v-b/r^2 \end{bmatrix}$$