2
d y n dy [ dy ]
---- + --- * ---- * [ e - ---- ] = 0 , y >= yd
2 y dx [ dx ]
dx
The parameter n is a real variable in the range of about 0.1 to 25.
The parameter e can be 0 or 1, or a real variable in the range
[0,1]. The parameter yd is positive real, in the range [0, 100].
The real variable x is in the range [0, delx] and the real variable y is
in
the range [yd, 15000]. I would like to know the analytical form of y(x),
when this equation is expressed between the boundaries (x1,y1) and
(x2,y2). I can solve it numerically, but knowing the analytic form
would answer a lot more questions. I will be glad to make an
acknowlegement in a research paper for the correct answer, maybe
even co-author for signficant work.
Please respond to: mailto://don....@aquarien.com
Thank you,
Don Baker
Aquarius Engineering
2000 West Maine Street
Fayetteville AR 72701
If y(x) = (f(x))^(1/(1-n)), then
2
1/(1-n) d f df
[f(x)] * ----- + n*e*---- = 0
2 dx
dx
I do know that the solution between (x1,f1) and (x2,f2) will
likely involve a very complicated logistic equation of which
(x2-x1) is a term. And the logistic equation will be none of the
common types. Some of the related work for this problem can be
found on my web site:
http://www.aquarien.com
Look under the Darcian means topic.
Can anyone cite a reference to a list of known solutions to
nonlinear ordinary and partial differential equations?
Thanks in advance,
Don Baker
mailto://don....@aquarien.com
A general solution can be given in terms of an integral. Let n and e be
real parameters. n is not necessarily an integer. y is a real into real
mapping.
Let F be a "new" elementary function (of say z) with three parameters,
defined by: F(e,n,K,z) = Int[t=0,z: K*(t^-n)/(1-K*e*(t^-n))] .
e and n are the parameters used within the ODE. K is one of the two
constants of integration arising from integration of the 2nd order ODE.
The implicit general solution of the ODE is: x - C + F(e,n,K,y(x)) = 0 . C
is the second constant of integration.
With e,n,K fixed, one would adopt a particular root-finder strategy, say
regula falsi, and then supply a value for x, solve, obtain y(x), and repeat
for various other x values. The novelty is that F is not a tabulated or
built-in function but this is not a serious problem since the integral can
easily be evaluated as required. Actually, the integral can be expressed
in elementary functions for negative integer n, and this might be useful
for the first few -n.
The other approach would be to let h represent the inverse of y, that is,
h(y(x)) = x for all x. Rewrite the general solution in terms of the
inverse: h(t) - C = -F(e,n,K,t) [t represents a "y-space" value]. The
function t->h(t)-C can be described in terms of a three-parameter set of
graphs which may provide the wanted analytic insight.
Verification of the general solution follows by two differentiations and
the identity y'=1/h'y, where again h is the inverse of y. To obtain the
solution in the first place, one can apply successively the identity
y''=-h''y/((h'y)^3), and then the identity for y'.
David Ziskind
zis...@ntplx.net
In paragraph six, the last sentence is a comment, and is best disregarded.
Technically, the integral can be evaluated in terms of elementary functions
if n is rational (this holds for either sign of n).
Also, in paragraph three, a preferable form for F would be F(e,n,K,z) =
Int[t=0,z: K/(t^n-K*e)] .
David Ziskind
zis...@ntplx.net