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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```

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