Do you know why maple doesn't want to numerically solve the non-linear equation:

restart;

ode := (diff(x(t), t))^2-(3/4)*(diff(x(t), t))*sin(x(t)) = -(3*9.81)*sin(x(t))+29;

dsolve({ode, x(0) = 0}, x(t), numeric)

Error, (in DEtools/convertsys) unable to convert to an explicit first-order system

@Preben Alsholm It should be 2 traveling solitons:

http://www.vidup.de/v/ih75K/

well, it runs, but how do I get along with the stiffness?

restart;

pde2 := diff(u(x, t), t)+3*(diff(u(x, t)^2, x))+diff(u(x, t), `$`(x, 3)) = 0;

IBC := {u(0, t) = u(2, t), u(x, 0) = sech(50*(x-1/2))^2+2*sech(30*(x-1))^2, (D[1](u))(0, t) = (D[1](u))(2, t), (D[1, 1](u))(0, t) = (D[1, 1](u))(2, t)};

pds := pdsolve(pde2, IBC, numeric, time = t, range = 0 .. 2, timestep = 1/1000, spacestep = 1/1000);

pds:-animate(t = 1, frames = 100, view = -1 .. 2, numpoints = 100)

it seems to do not what it's supposed to do. :-/

@Preben Alsholm Maybe what you actually meant was  D[1,1](u)(0, t) = D[1,1](u)(2, t), which involves the second derivatives w.r.t. the first variable x.

ah that's what I mixed up. Thanks

@Thomas Richard :-/ Thanks anyway..If you or anyone has an idea I'll still be quite interested.

Regarding f1 and f2 I sorta verified till order 300 with series about some point.

@Preben Alsholm Hey, series is not too bad. Do you know how to get the series about s=0 working?

@Carl Love Ah, too late. Yeah thats what I meant. Thanks as well..

@Markiyan Hirnyk yeah, thanks it works. Is this the only possibility, or could one extract via listprocedure the solution and formally integrate it? Like treating it as a list.

@Adri van der Meer 

"Besides: how do I eval r(t) like eval(r(t),t=0) for example."

Simply by R(0);

Yeah thanks^^ this obviously works if I have only 1 r(t)

But I have a function f(r(t)) in which I dont want to set each r(t) seperately.

I could define (haven't tried) a new function f->t->f(r(t)) but I thought eval(...,t=x) should work

@nm Hey I don't get why you can't just download the maple worksheet but here is the plain text:

restart;

Digits := 64;

assume(r > 0);

epsilon := 0.167e-1; AE := 149597870700; mu := 1; v_T := 30290; r_0 := AE; GM := 132712440018*10^9; L_0 := sqrt(GM*mu^2*r_0); d := 86400; t0 := 365.25; M0 := 10^10; M_0 := M0*Heaviside(t0-t); M_1 := M0*Heaviside(t-t0);

ode := diff(r(t), t, t)+(-(M_0*t+M_1*t0+L_0)^2/(mu^2*r(t)^3)+GM/r(t)^2)*d^2 = 0

sol := dsolve({ode, r(0) = r_0, (D(r))(0) = 0}, r(t), numeric);

sol(0);

rt := proc (t) options operator, arrow; rhs(op(2, sol(t))) end proc;

rpt := proc (t) options operator, arrow; rhs(op(3, sol(t))) end proc

plot({rt/AE}, 0 .. 1000)

Now as I said, this works now since I missed one term

But I still would like to know how to differentiate the function r(t) which is numeric

Besides: how do I eval r(t) like eval(r(t),t=0) for example.

For me it says: improper subscript selector

@digerdiga 

oh and why can't I evaluate like eval(rt(t),t=0) ??

Ok sry, first problem was just my fault: I missed one term:Kepler_Gravitatio.mw

Still I'd like to know how to differential r(t)

Download Kepler_Gravitatio.mw

oddly enough I get quite different results with both methods:

X := [51, 61, 71, 81, 101, 121, 141, 161, 201]

Y := [1.041, 1.034, 1.0296, 1.0259, 1.021, 1.017, 1.015, 1.013, 1.010]

f := Fit(1+a*n^b, X, Y, n, output = [leastsquaresfunction])

gives:

a=2.1045

b=1.0016

linear I get:

a=2.225

b=1.014

So which one is more accurate?

PS: ok as I have increased the decimal accuracy (since all values are ~0 in the logarithm rounding errors become large) of my Y values the results tend to that of the nonlinear fit.

oddly enough I get quite different results with both methods:

X := [51, 61, 71, 81, 101, 121, 141, 161, 201]

Y := [1.041, 1.034, 1.0296, 1.0259, 1.021, 1.017, 1.015, 1.013, 1.010]

f := Fit(1+a*n^b, X, Y, n, output = [leastsquaresfunction])

gives:

a=2.1045

b=1.0016

linear I get:

a=2.225

b=1.014

So which one is more accurate?

PS: ok as I have increased the decimal accuracy (since all values are ~0 in the logarithm rounding errors become large) of my Y values the results tend to that of the nonlinear fit.

This works.

Thanks