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Re: Methods of solving nonlinear equation

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  • Subject: [mg2489] Re: Methods of solving nonlinear equation
  • From: "Brian J. Albright" <albright at>
  • Date: Wed, 15 Nov 1995 01:57:06 -0500
  • Organization: UCLA Department of Physics and Astronomy

rdm at (Bob Morrison) wrote:
>I'm a non-expert with Mathematica and mathematics, so this is probably
>kind of a stupid question:  when I tried to solve a common nonlinear
>equation D^2 y[t] + k/(y[t]^2) == 0, I seem to be unable to get solutions.
>On paper, there is a degenerate solution at (a +/-b t)^(2/3), and
>also on paper, you can derive a polynomial solution (a0 + a1 t +
> a2 t^2 + ....) which shows two degrees of freedom (a0 and a1 are
>independently unconstrained).  However, the polynomial solution's
>region of convergence is too small to be very useful to me, and the
>boundary conditions of my problem disallow the degenerate solution.
>  I tried using Mathematica to get solutions using DSolve and NDSolve,
>but am too inexperienced to get anything useful.  It turns out if you
>assume that y is of form (f(t))^(2/3) you get another nonlinear equation
>for f(t) which appears to be just as unsolvable.  I tried representing
>y in Exp[f(t)] form, and failed here.  I tried separation
>of variables and didn't get anywhere (to be honest, I can't remember
>why, it's been a little while since I tried it).  Can anybody 
>provide a reference or maybe a little hint how to do this?  Ideally,
>I would like a closed form solution, of course, but a limiting function
>would also be useful.  I created a difference equation which correctly
>shows a coarse version of the solution, but in the region of interest,
>the dt portion is too large to correctly solve the problem.  I think
>the solutions will involve an elliptic integral, but I'm just guessing
>here.  I freely admit this is not my area of expertise and may have 
>overlooked the obvious solution, but I don't see it...
>Thanks, Bob Morrison, rdm at

Hi Bob.

I think the standard trick for doing something like y'' + k / y^2 = 0 
is to multiply both sides by y' and then integrate w.r.t. x.  This gives you
something like y' = [ c + 2 k / y ]^(1/2).  (c is just an integration
Rearranging, you get  x = x0 + Integrate[ 1 / Sqrt[ c + 2 k / y ], y],
which you can either do by hand or have mma do for you.  This will prob. 
give you some kind of trancendental relationship betw. x and y which you 
can look at in whatever limit you desire.  The solution will be parametrized
by the two integration constants, x0 and c.

Hope this helps.


Brian J. Albright                            |
Department of Physics and Astronomy, UCLA    |      To err is human...
albright at                    |         to err really big       |             is government.

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