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2d
comment Integro-differential equation
do you have an algorithm in mind? or are you asking about an algorithm?
2d
comment Plotting Contour Family of Curves
@user44840 did you look at all in the docs?
2d
comment eigenvector bug also for matrix with numeric value
Also please provide some way to produce the matrix. Otherwise, the moment you delete the file this question and answer become useless to everybody else. For example, doesn't m = #/Total[#] & /@ (RandomReal[{-1, 1}, {100, 100}]); produce a suitable matrix?
2d
revised eigenvector bug also for matrix with numeric value
added 113 characters in body
2d
answered eigenvector bug also for matrix with numeric value
2d
comment eigenvector bug also for matrix with numeric value
Well Max@Chop@Table[m.evecs[[i]] - evals[[i]]*evecs[[i]],{i, 1, Length@m}] gives 0 so they're correct. Maybe they're degenerate. OK just re-read your post and looked at the evals. The two evecs are degenerate.
2d
comment Plotting Contour Family of Curves
OK, with your edit I see you don't actually want Te. Well don't use D[Te,Plocal] but use eg dtedplocal or something (you'll need to define it)
2d
comment Plotting Contour Family of Curves
After removing some stuff it appears that you expect mathematica to see 1 - 2*Pprobe*Abs[\[CapitalGamma]]*D[Abs[\[CapitalGamma]], Plocal] == Geb*D[Te, Plocal] and, from that, work out what Te is, then insert it into ParametricPlot[{D[Te, Plocal], Plocal}, {Plocal, 0, 10^-13}] (which appears in plt). But you never did tell it to do that. Basically, it appears that you want to solve for Te, maybe like this DSolve[ 1 - 2*Pprobe*Abs[\[CapitalGamma]]*D[Abs[\[CapitalGamma]], Plocal] == Geb*D[Te[Plocal], Plocal], Te, Plocal] (you'll need to take care of the constant).
2d
comment Four dimensional phase space
@belisarius you know, I haven't watched a single episode of that... I don't know how I managed. But, as this is narrated by the Anorak, it seems worth locating and viewing.
2d
comment Numerical solution of the hyperbolic equation
Maybe someone more familiar with the solver for PDEs can. In any case I think the way to go with these things is to start small and build up, not write a big piece of code and then start trying to weed out various syntax errors, only to encounter algorithmic errors.
2d
comment Numerical solution of the hyperbolic equation
Also Derivative[f,{t,2}] isn't correct syntax
2d
comment Numerical solution of the hyperbolic equation
Are you sure you want = and not == in the initial conditions?
2d
comment Implicitly differentiate an equation, then solve the resulting equation
So you'd like mma to differentiate, obtaining $\partial_x y=f(x,z)$, then set $x=3$ and obtain some function of $z$ for the rhs and then what? What's the next step?
2d
comment Implicitly differentiate an equation, then solve the resulting equation
@user44840 it's probably useful to explain what "doesn't work" means.
2d
comment Using NDSolve to solve Equation of Motion in cylindrical coordinates
Maybe this helps with the glitch in the plot.
Jul
7
comment Using NDSolve to solve Equation of Motion in cylindrical coordinates
That's a great answer. Probably worth showing both $r$ and $\phi$, Plot[sol0[[1]][t], {t, 0, 10}, Axes -> True] Plot[sol0[[2]][t], {t, 0, 10}, Axes -> True], to explicitly show $\phi$ rapidly rotating when the radial component hits the centrifugal barrier (so the particle zooms by the origin).
Jul
7
comment Using NDSolve to solve Equation of Motion in cylindrical coordinates
I have no idea, but this may help visualising things: upt = 500; ct = cylindricalTrajectory[1, 0, 2, 1, 0, 1, 1, 1, 1, upt]; traj = Flatten[#, 1] &@ Table[{r[t], phi[t], z[t]} /. ct, {t, 0, upt, .1}]; trajcart = {#[[1]]*Cos[#[[2]]], #[[1]]*Cos[#[[2]]], #[[3]]} & /@ traj; Manipulate[ Graphics3D[ { Arrow[{{0, 0, 0}, #}] &@trajcart[[i]], Point[trajcart[[1 ;; i]]] }, PlotRange -> {{-5, 5}, {-5, 5}, {-5, 5}} ], {i, 1, Length@trajcart, 1} ]
Jul
7
comment Four dimensional phase space
@Dave I... see :)
Jul
7
comment Four dimensional phase space
Well, it's a 4d system with two conserved quantities (energy and momentum), so any trajectory will be restricted onto on 4-2=2 dimensional surfaces. Since it's separable into action-angle variables, there will be two oscillatory motions, at two incommensurate frequencies. So, the trajectory will lie on a torus in 4d space. Please do not take this the wrong way (I mean it as advice, not an insult), but you need to understand how something works before you can get mma to do anything useful with it.
Jul
7
comment Four dimensional phase space
@Dave claimed what? That there is a way to show a shape in 4d space in a 3d space? Of course it's possible but there are an infinite number of ways of plotting it (imagine a 3d shape projected onto a 2d surface, ie, a shadow--obviously there's an infinite number of them depending on the projection you choose). The thing you have is a torus because it is a superposition of two oscillations of different frequencies; this is independent of how you visualise it.