Problem with a 1st order IV ODE (nonlinear)
- To: mathgroup at smc.vnet.net
- Subject: [mg102752] Problem with a 1st order IV ODE (nonlinear)
- From: garrido at ruth.upc.edu
- Date: Wed, 26 Aug 2009 07:42:50 -0400 (EDT)
Hi Virgil,
h'[t] == -k +
1/((2 R - h[t]) h[t]) /. {h'[t] -> dh/dt, h[t] -> h} // Together
dh/dt == (-1 - h^2 k + 2 h k R)/(h (h - 2 R))
If k = 1/R^2,
dh/dt == (-1 - h^2 k + 2 h k R)/(h (h - 2 R)) /. k -> 1/R^2 // Together
dh/dt == -((-h^2 + 2 h R - R^2)/(h R^2 (-h + 2 R)))
Integrate[-((h R^2 (-h + 2 R))/(-h^2 + 2 h R - R^2)), h] //
FullSimplify // Expand
-h R^2 - R^4/(h - R)
(I) General Solution : -h R^2 - R^4/(h - R) + cte == t
VerifySolution :
1/D[-h R^2 - R^4/(h - R) + cte, h] + (-h^2 + 2 h R - R^2)/(
h R^2 (-h + 2 R)) // FullSimplify
0
Solve[-h R^2 - R^4/(h - R) + cte == t, h] // FullSimplify
{{h -> (cte + R^3 + Sqrt[(cte - 3 R^3 - t) (cte + R^3 - t)] - t)/(
2 R^2)}, {h -> (
cte + R^3 - Sqrt[(cte - 3 R^3 - t) (cte + R^3 - t)] - t)/(2 R^2)}}
Particulars Solutions :
Solve[-h R^2 - R^4/(h - R) + cte == t /. {t -> 0, h -> 0}, cte]
{{cte -> -R^3}}
-h R^2 - R^4/(h - R) + cte == t /. cte -> -R^3 // Together
-((h^2 R^2)/(h - R)) == t
Solve[(h^2 R^2)/(-h + R) == t, h]
{{h -> (-t + Sqrt[t] Sqrt[4 R^3 + t])/(
2 R^2)}, {h -> -((t + Sqrt[t] Sqrt[4 R^3 + t])/(2 R^2))}}
If k != 1/R^2,
Integrate[(-h^2 + 2 h R)/(1 + h^2 k - 2 h k R), h]
-(h/k) - ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2])
Verify Solution :
1/D[-(h/k) - ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) + cte, h] - (-k + 1/(
h (-h + 2 R))) // FullSimplify
0
(II) General solution : -(h/k) -
ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) + cte == t
Particular solution :
Solve[-(h/k) - ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) + cte == t /. {t -> 0, h -> 0}, cte]
{{cte -> ArcTan[(Sqrt[k] R)/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2])}}
-(h/k) - ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) + cte /.
cte -> ArcTan[(Sqrt[k] R)/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) // Together // FullSimplify
-((h Sqrt[k] Sqrt[1 - k R^2] - ArcTan[(Sqrt[k] R)/Sqrt[1 - k R^2]] +
ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]])/(
k^(3/2) Sqrt[1 - k R^2]))
Particular solution curve :
{ -((h Sqrt[k] Sqrt[1 - k R^2] -
ArcTan[(Sqrt[k] R)/Sqrt[1 - k R^2]] +
ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]])/(
k^(3/2) Sqrt[1 - k R^2])), h}
-(h/k) - ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) + cte // Together
(-h Sqrt[k] Sqrt[1 - k R^2] + cte k^(3/2) Sqrt[1 - k R^2] -
ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]])/(k^(3/2) Sqrt[1 - k R^2])
If in (II)
-(h/k) - ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]]/(
k^(3/2) Sqrt[1 - k R^2]) + cte // Together
(-h Sqrt[k] Sqrt[1 - k R^2] + cte k^(3/2) Sqrt[1 - k R^2] -
ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]])/(k^(3/2) Sqrt[1 - k R^2]),
we do h -> 1/R^2 (Hôpital's rule) we obtain (I)
(*III*)Limit[
D[-h Sqrt[k] Sqrt[1 - k R^2] + cte k^(3/2) Sqrt[1 - k R^2] -
ArcTan[(Sqrt[k] (-h + R))/Sqrt[1 - k R^2]], k]/
D[k^(3/2) Sqrt[1 - k R^2], k], k -> 1/R^2,
Direction -> -1] // FullSimplify
cte + h^3 + h^2 R + R^3 + h^4/(-h + R)
(*I*) -h R^2 - R^4/(h - R) + cte ==
cte + h^3 + h^2 R + R^3 + h^4/(-h + R )(*III*)// FullSimplify
True
Regards,
J.L. Garrido
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