Differential equation - Reduction of order

[fineTuner]

Hi,

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

[itex]\ddot{r}+a\dot{r}+\frac{b}{r^{2}}=0[/itex]

with [itex]a,b\geq0[/itex]

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

Can someone explain me what i should do?

Thank you,

J.

 

[HallsofIvy]

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 [itex]\begin{bmatrix}r(t) \\ v(t)\end{bmatrix}[/itex] we have the vector equation
[tex]V'= \begin{bmatrix}r(t) \\ v(t) \end{bmatrix}'= \begin{bmatrix}-v- b/r^2 \\ v\end{bmatrix}[/tex]

 

[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.
Solving it leads to Y(r)
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.

 

[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!

 

[vanhees71]

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 [itex]\begin{bmatrix}r(t) \\ v(t)\end{bmatrix}[/itex] we have the vector equation
[tex]V'= \begin{bmatrix}r(t) \\ v(t) \end{bmatrix}'= \begin{bmatrix}-v- b/r^2 \\ v\end{bmatrix}[/tex]

But shouldn't it read
[tex]V'=\begin{bmatrix} r' \\ v' \end{bmatrix}
=\begin{bmatrix} v \\ r'' \end{bmatrix}
=\begin{bmatrix} v \\ -a v-b/r^2 \end{bmatrix}[/tex]