Non-Seperable ODE Help

crepincdotcom

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/

Matthew Rodman

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.

crepincdotcom

[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/

StatMechGuy

I'd wager that you could probably use an integrating factor type equation or something along those lines.

crepincdotcom

Hm... I looked that up at http://mathworld.wolfram.com/IntegratingFactor.html

Any idea what sort of function I'm looking for to make that integrable?

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

Matthew Rodman

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.

crepincdotcom

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/

Matthew Rodman

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]

crepincdotcom

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/

CPL.Luke

^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.

CPL.Luke

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?

crepincdotcom

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/