Extension of Variation of Parameters to First Order Non-Linear ODE?

[Rruffpaw]

The equation of motion of a rocket with mass depletion during ascent and subject to drag forces can be written as

M(t) dV/dt = A - M(t)g - BV^2 (Eq. 1)

with initial condition V(t=0) = 0 (V is velocity and t is time)

Let us assume a linear mass depletion according to

M(t) = Mo - Kt (Eq. 2)

In Eqs 1 and 2, A, B, g, Mo and K are all constants >/= 0

For the limiting case B = 0, Eq. 1 reduces to a nicely separable form and can easily be solved. For B =/= 0, Eq. 1 can be cast as

dV/dt + p(t)V^2 = q(t) (Eq. 3)

where

p(t) = B/(M(t)

and

q(t) = A/M(t) - g

However, the non-linearity (in V) of Eq. 3 leaves me stuck, since the traditional Variation of Parameters approach does not apply.

Any ideas on how to proceed (in the context of an analytical or at least a semi-analytical solution) would be deeply appreciated.

-Sharat

 

[JJacquelin]

In attachment it is shown how to transform this non-linear EDO to a second order linear EDO.
(which is related to Bessel functions)

 

[Rruffpaw]

In attachment it is shown how to transform this non-linear EDO to a second order linear EDO.
(which is related to Bessel functions)

Thanks! This looks neat. But what about the initial conditions for f and f'?

I could of course solve the 2nd order linear ODE and substitute f and f' back into the expression for V, but I'll have two constants and only one initial condition.

Or do I apply the condition (f'=0 at T=0 which follows from V(0)=0 ) to evaluate one constant, substitute back into V and apply the same condition V=0 to evaluate the other constant?

Also, I notice that the 2nd order linear ODE is not exactly in the form of Bessel's Differential Eqn. There must be some subtlety involved here, that I can look up, but the issue of the initial condition is still a bit confusing.

Many thanks again.

 

[JJacquelin]

Also, I notice that the 2nd order linear ODE is not exactly in the form of Bessel's Differential Eqn. There must be some subtlety involved here, that I can look up, but the issue of the initial condition is still a bit confusing.

Of course, it isn't a standard Bessel equation. There are a number of variants and the solutions generally are combinaisons of sum and products of Bessel functions with some elementary functions.

I could of course solve the 2nd order linear ODE and substitute f and f' back into the expression for V, but I'll have two constants and only one initial condition.
Or do I apply the condition (f'=0 at T=0 which follows from V(0)=0 ) to evaluate one constant, substitute back into V and apply the same condition V=0 to evaluate the other constant?

The substitution V = ((Mo-k*t)/(B*f))(df/dt) and condition V(0)=0 imply not only (df/dt)=0 but also introduce un supplementary condition :
at t=0 Mo²(d²f/dt²)=B*(A-Mo*g)*f
obtained from the second order linear ODE in which df/dt=0
Preferably, solve the second order linear ODE with two arbitrary coefficients. Then subsitute back in order to obtain the solutions V(t) depending on the two coefficients. Apply the condition dV/dt=0 at t=0 and bring back the found functions V(t) in the initial non-linear ODE. Then eliminate the non consistent solutions, which will determine the appropriate values of the coefficients.

 

[Rruffpaw]

JJacquelin,

Many thanks again for your help. Could you kindly review the attached where I've re-derived the last step (I think you had a k in the coefficient of df/dT in your last equation that should not be there) and compared it to a transformed version of Bessel's Differential Equation from which I get a solution in terms of Bessel Functions of the First Kind.

I apologise for any confusion due to my introducing the variable $$\xi$$ = -T

I was thinking of using df/d$$\xi$$ =0 to eliminate one of the constants (C2=Q*C1, where Q is a constant), substitute back into the expression for V whereupon C1 vanishes. Is there something inconsistent about this procedure, that I should be aware of?

Thanks

-Sharat

While it may be obvious, I forgot the = 0 in the RHS of the transformed ODE.

 

[JJacquelin]

I apologise for any confusion due to my introducing the variable = -T

No problem

Could you kindly review the attached where I've re-derived the last step (I think you had a k in the coefficient of df/dT in your last equation that should not be there) and compared it to a transformed version of Bessel's Differential Equation from which I get a solution in terms of Bessel Functions of the First Kind.

You are right. There are mistakes in my last equation. In your equation too (see attachment)
Better, check once again.

 

[JJacquelin]

I was thinking of using df/d =0 to eliminate one of the constants (C2=Q*C1, where Q is a constant), substitute back into the expression for V whereupon C1 vanishes. Is there something inconsistent about this procedure, that I should be aware of?

I am not sure that it's so simple.
V(0)=0 implies f '/f=0 which isn't exactly the same as f '=0 since f might also tends to 0 when t tends to 0.
I think that it should be better to keep the two constants, compute f ' and express f '/f as a serie development of t. Then determine the relationship between tne constants so that the limit be 0 for t=0. May be I am too cautious...

 

[Rruffpaw]

Thanks again! You are right in that I too missed the k^2 in the denominator (amongst other typos)

I will recheck everything and formally derive the solution which I will post.

Cheers,

-Sharat