@tomleslie Vote up. The square brackets is the issue. If you'd gone further, you'd've seen that the plot works correctly with its five legends, but is, as you say, boring.

@Mac Dude What difference does that make?

@  Dr Venkat Subramanian

The issue is getting a good-looking (and, of course, accurate) plot. It is not necessary to know or to extract the points used internally by dsolve to do this. Yes, plot will by default use 201 points, unless its adaptive scheme decides that more are necessary. Each value of the independent variable chosen by plot is passed to the solution procedure(s) returned by dsolve. Those procedures may (and very likely do) use intermediate values of the independent variable to generate intermediate points. There is no need for those points to appear in the final output. Indeed, there are cases where using fewer points produces a smoother-looking plot. (Using plots:-implicitplot with the resolution option is a good way to generate an example of this.)

@Markiyan Hirnyk The issue is that the OP wants to generate a good-looking plot (no inaccurate sharp turns) without needing to select the points in an ad hoc way. To use the output= Array option, one needs to know a priori which values of the independent variable to use.

@shakuntala This back-and-forth with you not understanding is getting very frustrating, so you need to "come clean" about what exactly you're trying to do. Are you working on a course assignment? Are you working from a book? Please present the assignment or book problem in full.

What is your primary spoken language? If this question is part of a course, what is the language of instruction? If you're working from a book, what language is it written in?

@shakuntala In your originally posted code, you have the line blt:= 10. I assumed that this was still in effect. If it's not, then you must be solving a different ODE problem, in which case X21 and X11 wouldn't make sense either.

@acer If the OP is performing the NLPSolve inside a loop, as claimed, I'd more strongly recommend against using assign. It's still possible to use assign combined with some method of unassigning (either command unassign or assignments such as th:= 'th'), but that quickly turns into a mess. (Of course, I know that you know all the above; I just think that you may have missed the OP's "in a loop" issue.)

@shakuntala Yes, it's easy to combine any number of plots into one. One way is

plot([X11, X21], 0..blt);

The command plots:-display can be used in more-complicated cases:

plots:-display(
     plot(X11, 0..blt),
     plot(X21, 0..blt)
);

The inner plot commands can be any, as long as they all have the same dimension (two or three).

@Abdoulaye No, it doesn't help. There's a discontinuity in the link up of the two pieces at t = 11.72530194.

@jojokaila You can simplify the input of column vectors by using angle brackets:

A:= < 1/sqrt(2), 0, 0, 1/sqrt(2) >;

@jojokaila Oh, okay. I think that I understand your issue. Maple by default doesn't display vectors or matrices with more than 10 rows or columns. It only displays some information about them such as their dimensions. The computation has however still been done. If you want to see the results, then give the command

interface(rtablesize= 25);

before doing the computation. This command only needs to be given once per session, and it can be put in an initialization file. The 25 can be replaced by any positive integer or infinity. Personally, I have mine set at 25. Maple can easily handle computation of 1000 x 1000 matrices, but you wouldn't want one to accidentally display on your screen! That's why the rtablesize option exists.

@Abdoulaye No derivative. I just extracted the bounds of the second period from the original piecewise. The left boundary, op([9, 1, 1], F), is 11.72530194, and the right boundary, op([-2,-1,-1], F), is 17.46028352. I needed to add the 1e-7 because your intervals overlap at the their endpoints because you used <= everywhere instead of alternating <= with <.

@Abdoulaye Except for the horizontal pieces, the individual pieces contain polynomial terms and are certainly not periodic. It is only in the piecewise context that the function can be periodic.

Here's how you can approximate the error:

plot(abs(F - eval(F, t= t-2*Pi)), t= op([9,1,1], F)+1e-7..op([-2,-1,-1], F));

The plot shows that the maximum error is about 1.2e-7.

@Markiyan Hirnyk Here's a plot illustrating the corollary. Adjust n to any positive integer. The convergence of the maximum modulus to 2 is very rapid.

n:= 6:
R:= fsolve(z^n = add(z^k, k= 0..n-1), complex):
plots:-display([
     plot([Re,Im]~([R]), style= point, symbol= diagonalcross, symbolsize= 16),
     plot([1,2], t= 0..2*Pi, coords= polar, thickness= 0, color= blue)
     ], scaling= constrained, axes= boxed, view= [(-2.1..2.1)$2]
      , labels= [Re,Im]
);

I also modified the code in the Answer to show clearly what I meant by union~.

@Abdoulaye This plot command shows that the function is close enough to being 2*Pi periodic that there is no visible gap between the plots:

Fu:= eval(F, piecewise= piecewise[undefined]):
plot([Fu, eval(Fu, t= t-2*Pi)], t= op([1,1,1], F)..op([-2,-1,-1], F)+2*Pi);

On the other hand, clearly it is not exactly periodic because the horizontal piece 0.4977821578 isn't an exact match to the horizontal piece 0.4977820960.

But I wonder what is your true goal here. Are you actually trying to construct a periodic function by duplicating one fundamental period? If so, then that is easy to do, and I can show you how. Or are you truly trying to prove that a given function is periodic?