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| LordBeotian... |
 Posted: Fri Oct 30, 2009 6:58 am |
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I would like to solve the following ODE:
f ' (t) = - f(t) ^ 2 + g(t)
where g(t)=(1/2)(b-t/(2sqrt(3)))^-2
I already know a particular solution:
f(t)=(1/sqrt(3))(b-t/(2sqrt(3)))^-1
is there any tecnique which allows to obtain analytic expressions for
the general solution? |
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| LordBeotian... |
| Posted: Fri Oct 30, 2009 8:14 am |
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On 30 Ott, 18:53, "Rod" <rodrodrod... at (no spam) hotmail.com> wrote:
[quote]I would like to solve the following ODE:
f ' (t) = - f(t) ^ 2 + g(t)
where g(t)=(1/2)(b-t/(2sqrt(3)))^-2
I already know a particular solution:
f(t)=(1/sqrt(3))(b-t/(2sqrt(3)))^-1
is there any tecnique which allows to obtain analytic expressions for
the general solution?
Wolfram alpha gives
f(t) = -(3/5 (t-2 sqrt(3) b)^2-(2 c_1)/(t-2 sqrt(3) b)^3)/(1/5 (2 sqrt(3)
b-t)^3-c_1/(t-2 sqrt(3) b)^2)
[/quote]
Surprisingly this "general" solution is not completely general: my
solution given above is obtained only for c->infinity. |
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| Ray Vickson... |
| Posted: Fri Oct 30, 2009 9:02 am |
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On Oct 30, 11:14 am, LordBeotian <pokips... at (no spam) yahoo.it> wrote:
[quote]On 30 Ott, 18:53, "Rod" <rodrodrod... at (no spam) hotmail.com> wrote:
I would like to solve the following ODE:
f ' (t) = - f(t) ^ 2 + g(t)
where g(t)=(1/2)(b-t/(2sqrt(3)))^-2
I already know a particular solution:
f(t)=(1/sqrt(3))(b-t/(2sqrt(3)))^-1
is there any tecnique which allows to obtain analytic expressions for
the general solution?
Wolfram alpha gives
f(t) = -(3/5 (t-2 sqrt(3) b)^2-(2 c_1)/(t-2 sqrt(3) b)^3)/(1/5 (2 sqrt(3)
b-t)^3-c_1/(t-2 sqrt(3) b)^2)
Surprisingly this "general" solution is not completely general: my
solution given above is obtained only for c->infinity.
[/quote]
Maple 9.5 gives: f(t) = fN/fD, where
fN := 2+C*(-30*3^(1/2)*b*t^4+3*t^5-720*3^(1/2)*b^3*t^2-864*3^(1/2)
*b^5+360*b^2*t^3+2160*b^4*t)
and
fD := 2*3^(1/2)*b-t+C*(-12*3^(1/2)*b*t^5+t^6-480*3^(1/2)
*b^3*t^3-1728*3^(1/2)*b^5*t+180*b^2*t^4+2160*b^4*t^2+1728*b^6)
For C = 0 we get f(t) = 2/(2*3^(1/2)*b-t), which is equivalent to your
particular solution.
R.G. Vickson |
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| Rod... |
| Posted: Fri Oct 30, 2009 11:53 am |
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"LordBeotian" <pokipsy76 at (no spam) yahoo.it> wrote in message
news:a100b1e1-d302-46b6-b690-67a76597f2d2 at (no spam) a31g2000yqn.googlegroups.com...
[quote]I would like to solve the following ODE:
f ' (t) = - f(t) ^ 2 + g(t)
where g(t)=(1/2)(b-t/(2sqrt(3)))^-2
I already know a particular solution:
f(t)=(1/sqrt(3))(b-t/(2sqrt(3)))^-1
is there any tecnique which allows to obtain analytic expressions for
the general solution?
[/quote]
Wolfram alpha gives
f(t) = -(3/5 (t-2 sqrt(3) b)^2-(2 c_1)/(t-2 sqrt(3) b)^3)/(1/5 (2 sqrt(3)
b-t)^3-c_1/(t-2 sqrt(3) b)^2) |
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| rancid moth... |
| Posted: Sat Oct 31, 2009 4:00 pm |
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"LordBeotian" <pokipsy76 at (no spam) yahoo.it> wrote in message
news:a100b1e1-d302-46b6-b690-67a76597f2d2 at (no spam) a31g2000yqn.googlegroups.com...
[quote]I would like to solve the following ODE:
f ' (t) = - f(t) ^ 2 + g(t)
where g(t)=(1/2)(b-t/(2sqrt(3)))^-2
I already know a particular solution:
f(t)=(1/sqrt(3))(b-t/(2sqrt(3)))^-1
is there any tecnique which allows to obtain analytic expressions for
the general solution?
[/quote]
Yes. utilise Lie symmetries. Obviously i can't go into huge amount of
detail here and you should read the books by Olver and Bluman and many
others: but first you will need to compute the infintesimal generators of
the symmetries of the ODE either by hand or by a computer package. MuLIE by
Alan Head is free and very good for most things. However given that your
ODE is first order, doing it by hand isnt too much of a chore:
i will take your ODE in the following form to make things easier to read:
f'(t)+f(t)^2 = 1/(b-t)^2
I get one generator being X =n(t,f) * d/df where
n(t,f) = ( (t-b)^2*f^2 - (t-b)*f -1 )/(t-b)
Threfore the cononical coordinates that will hopefully transform this thing
down to something easily solved satisfy the following
dt/0 = df / [( (t-b)^2*f^2 - (t-b)*f -1 )/(t-b)] = ds
to fully understand all this you need to know where the PDE comes from that
generates the equations for the characteristics. however moving on...
the first constant is t=constant=r. then solving s(t) = integrate 1/n(t,f)
df = -2/sqrt(5) arctanh((2*(t-b)^2*f-(t-b))/(sqrt(5)*(t-b)) and
solving for f(t) we have
f(t) = 1/(2*(t-b)) - sqrt(5)*tanh(sqrt(5)/2*s(t))/(2*(t-b))
therefore df/dt = -(1-sqrt(5)*tanh(sqrt(5)*s(t)/2)/(2*(t-b)^2) -
5/(4*(t-b))*(1-tanh(sqrt(5)/2*s(t))^2) ds/dt
recalling that t=r, and that df/dt = -f^2 +1/(b-t)^2, throwing it all in
the mix and doing a lot of simplification one gets your ODE transformed to
ds/dr = 1/(b-r)
this is trivial and has a solution
s(r) = s(0) - ln(1-r/b)
where i have taken r=t=0 as the lower limit. now all we do is invert the
cononical cords...i get...
f(t) = 1/(2*(t-b)) - sqrt(5)/(2*(t-b)) * tanh( sqrt(5)/2 * (s(0) -
ln(1-t/b)) )
where s(0) = is given by that horrible thing above.
cheers
moth |
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| Ken Pledger... |
| Posted: Sun Nov 01, 2009 2:41 pm |
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In article
<a100b1e1-d302-46b6-b690-67a76597f2d2 at (no spam) a31g2000yqn.googlegroups.com>,
LordBeotian <pokipsy76 at (no spam) yahoo.it> wrote:
[quote]I would like to solve the following ODE:
f ' (t) = - f(t) ^ 2 + g(t)
where g(t)=(1/2)(b-t/(2sqrt(3)))^-2
I already know a particular solution:....
[/quote]
In that case there's a standard method to reduce the d.e. to first
order. Look up "Riccati equation".
Ken Pledger. |
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