Please post your actual code. The parametric curve expression that you gave is not at all like the plot in your Question. This is the plot of the curve that you gave:

@vidocq wrote:

I have exported second file to mpl and tried read "file.mpl" but it was not work.

No. Export the first file to .mpl and place the read command at the beginning of the second file.

@dena I'm using Maple 18 also.

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@dena Preben's code works for me. You should post your executed code so that we can diagnose what you are doing wrong. Here's my worksheet of it:

Download dsolve.mw

@casperyc asked:

First of all, I have to say that 'myfun' does call a lot of other functions and procedures and returns a single value. Does that mean I have to check every single one of them?

Yes, you have to check them all. But the list is alphabetical, so it's fairly easy.

If just to be safe, say I assume it does not qualify for multithreading and use the Grid option, is there going to be less efficient? Broadly speaking, how much difference?

The difference is huge. The Grid option starts a separate mserver on every processor and copies the entire memory of your base process to each one. Threads, on the other hand, does everything in the same memory space.

@Carl Love Here is a much smoother animation which shows that the analytic and numeric solutions are in perfect agreement and which eliminates the spurious "epi-waves" caused by the default coarse grid spacing.

restart:
PDE:= diff(u(x,t), t$2) = diff(u(x,t), x$2):
IBCs:= u(x,0)=sin(Pi/2*x)*exp(-x), D[2](u)(x,0)=0, u(0,t)=0, u(4,t)=0:
Sol:= pdsolve({PDE, IBCs}):
Sol:= subs(op([2,2,1], Sol)= n, Sol):  #Change _Z1~ to n, just for aesthetics
rhs(Sol);

Sol:= subs(infinity= 20, rhs(Sol)):
A1:= plots:-animate(plot, [Sol, x= 0..4], t= 0..8, frames= 200, thickness= 3):

Numeric solution:
Sol:= pdsolve({PDE}, {IBCs}, numeric, spacestep= .02):
A2:= Sol:-animate(t= 0..8, frames= 200, thickness= 3):
plots:-display(A1,A2);
By displaying the two animations together, we can see that they are in perfect agreement because there is only ever one line.

@Joe Riel Note that the eigenvalues of your Cx are [0, 1, 1] regardless of your choice c12 or c13.

@casperyc Okay, I understand your question now. No, there is no predefined way to do that. It seems very complicated, akin to the identify command.

@casperyc Sorry, I don't understand that any better than the original. Can you give an actual example rather than an explanation?

@Mac Dude Preben's solution does not strictly require eval but it does require evaluation before passing to inttrans[fourier]. This is because assuming does not work with variables in functions. So, using evaluation, this works:

i[beam]:= phi-> exp(-(phi-phi0)^2/2/sigma^2):
ex:= i[beam](phi):
i[beamspec]:= inttrans[fourier](ex, phi, omega) assuming sigma>0;

Please see my Answer below, which you seem to have overlooked.

@Kitonum The animation quality can be improved dramatically by including spacestep= .02 in the call to pdsolve. Note that setting higher values of numpoints does not change the discretization error. Using spacestep=.02 gives 200 points on the interval x= 0..4.

@Markiyan Hirnyk Axel's code works for me. I get the same numeric answer in 10 seconds total (mostly for the integration). I am one of the upvoters.

@Carl Love Using Preben's code, I now understand why f(1) is -x (actually x-> -x) and f(x) is -1. What was confusing me was that I thought that P(x) was always 1+x+x^2.

@Alejandro Jakubi While we cannot construct a patterned matrix of unspecified (abstract) size in Maple, all the OP wants is a procedure that constructs it for specified sizes. This is quite doable in Maple.

@xcyborg You can change the options on this plot, of course. I just made some stuff up.

plot(
     [zip(`[]`, W, M), zip(`[]`, W, B)],
     style= point, symbol= [diamond, box],
     symbolsize= 15, color= [brown, green]
);