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Non-Seperable ODE Help

  1. Nov 23, 2006 #1
    Hey,

    I've been teaching myself some DEs that I can use for physics and whatnot. I am comfertable with seperable equations, but I can't figure out how to solve this problem.

    Let's assume we have some rocket with thrust F(t) and drag r(v), plus acceleration due to gravity, g=9.8 m/s/s.

    Overall acceleration: (m is the mass, assume constant)

    [tex]a(t)=\frac{F(t)}{m} - \frac{r(v)}{m} - g[/tex]

    [tex]\frac{dv}{dt}=\frac{F(t)}{m} - \frac{r(v)}{m} - g[/tex]

    Now as you can see, we can't move dt over to the other side, because there are multiple terms there. Can we simply distribute it across them, and get:

    [tex]\int{dv}=\int{\frac{F(t)dt}{m}} - \int{\frac{r(v)dt}{m}} - \int{gdt}[/tex]

    Also, we neet to relate v to t in the r(v) term, but we don't have a v(t)...

    Thanks,

    -Jack Carrozzo
    jack _{at}_ crepinc.com
    http://www.crepinc.com/
     
    Last edited: Nov 23, 2006
  2. jcsd
  3. Nov 23, 2006 #2
    You effectively have a second-order equation, with a generalized function of the first derivative. You will need to be more specific about r(v) if you want to make any progress. If r(v) is linear, you're in luck - otherwise it may be quite tough to solve.
     
  4. Nov 23, 2006 #3
    [tex]r(v)=(\frac{1}{2}rAc)v^2[/tex]

    or simply

    [tex]r(v)=kv^2[/tex]

    which is not linear, per se.

    How would I go about this?

    Thanks,

    -Jack Carrozzo
    jack _{at}_ crepinc.com
    http://www.crepinc.com/
     
    Last edited: Nov 23, 2006
  5. Nov 23, 2006 #4
    I'd wager that you could probably use an integrating factor type equation or something along those lines.
     
  6. Nov 23, 2006 #5
  7. Nov 23, 2006 #6
    Sorry - my bad. What you actually have (if r(v) is proportional to v^2) is a Riccati equation of the form

    [tex]v^{\prime} + v^2 + f(t) = 0 [/tex]

    in v (give or take a constant or two). This is not generally soluble for any f(t). One thing you can do is make the transformation

    [tex]v = \frac{u^{\prime}}{u}[/tex]

    and sub, giving you the Hill equation

    [tex]u^{\prime \prime} + f(t) u = 0 [/tex]

    which only helps if you actually know a solution and can therefore work backwards.

    There are solutions of this Riccati equation for specific forms of f(t) - perhaps if you could give us some clue what the thrust term might be, we could explore it further.

    sorry for the first (slightly misleading) post.

    edit: as a guide to what sort of Riccati equations are soluble, you can check out

    http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc1.htm

    as they have some special cases listed.
     
    Last edited: Nov 23, 2006
  8. Nov 23, 2006 #7
    Thanks, I need to let that sink in for a bit before I can try to apply it.

    The thrust equation is for the most part a type of step function: x newtons for 0.5s, y newtons for 2 sec, and z newtons for 4 secs. That's a bit hard to express in an equation unless we use peice-wise.

    Which come to think of it may work.... if we solve the problem with T constant from t=0 to 0.5, then 0.5 to 2.5 and 6.5, each time carrying v0, perhaps we could solve it.

    What do you think?

    -Jack Carrozzo
    jack _{at}_ crepinc.com
    http://www.crepinc.com/
     
  9. Nov 23, 2006 #8
    By the way - the equation is slightly doubtful - for a rocket you need to include the loss of mass due to fuel expenditure. Therefore m = m(t). You'll have to think about how this fits in - i.e., is the rate of fuel loss proportional to time?

    Basically, you have to resort to a momentum equation, e.g.

    [tex] \frac{d (m(t)v)}{dt} = F(t) - r(v) - m(t)g [/tex]
     
    Last edited: Nov 23, 2006
  10. Nov 23, 2006 #9
    I was using the assumption that the mass lost compared to the total mass of the vehicle was negligable in order to make calculations easier, but if I were to include it, it would be some function of the form

    [tex]m(t)=m_{v} - km_{p0}t[/tex]

    or the mass of the vehicle plus the initial mass of the propellant decreasing linearly with time.

    -Jack Carrozzo
    jack _{at}_ crepinc.com
    http://www.crepinc.com/
     
  11. Nov 23, 2006 #10
    ^actually you'll have

    dp/dt=dm/dt v +dv/dt m

    so you can plug in the terms that give you the force.

    F(t)-r(v)=v dm/dt + m dv/dt

    so

    F(t)- r(v)- v dm/dt = m dv/dt

    so

    F(t)/m(t) -r(v)/m(t) -v/m(t) dm/dt =a

    as I recall rocketry problems are usually solved by a series of applications of newtons laws and conservation of energy, ie no plug and play differential equations.
     
  12. Nov 23, 2006 #11
    out of curiosityin this problem is it allowed for us to assume that the rocket is incredibly massive in comparison to its fuel, thus assuming that mass is a constant?

    also this appears to form an integral equation with


    integral r(v) dt= v + q(t)

    where q(v) is all of the other t terms grouped together.


    I think an equation of this form can usually be solved with a laplace transform correct?
     
  13. Nov 23, 2006 #12
    truthfully I've never heard of laplace transforms, so I'll take your word for it.

    I've only ever seen the problem addressed as a numerical method wherein each part is solved for some small dt. I wanted to do it with diff EQs if possible.

    Thanks guys

    -Jack Carrozzo
    jack _{at}_ crepinc.com
    http://www.crepinc.com/
     
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