@Carl Love I didn't know about InertForm:-Parse.  Thanks for pointing it out.

@vv I like your approach very much.  It's very clever and insightful.

In your procedure I think the definition of c may be shortened to
c := 1/eval(F, [X,Y]);
because the proc is fundamentally based on the identity
F(x,y) = F(a,y)*F(x,b)/F(a,b)
which holds for any function F(x,y)=f(x)*g(y), and any a and b.

It's a pity that sometimes the user needs to supply very special a and b values to produce the simplest results.  I am still impressed by Mathematica's Simplify which does not require such a hint.

@Kitonum I have no idea at all how Mathematica figures that out.  It is very impressive that a machine can be so clever.

It is likely that you already know that there is no analytic solution to your equations, that is, no solution in terms of elementary function.  You may expect a solution expressed in the Fourier series, or a numerical solution.

Currently Maple has no capability for solving Laplace's equation either numerically or in Fourier series.  Maybe in a future version it will.

However you may solve your problem by implementing your own solver.  The Fourier series solution is not difficult; it is done pretty much in every textbook on introductory PDEs.  A numerical solution with finite differences is also quite straightforward.  Doing one or the other, (or both!) can be good learning exercises.

If you attempt to do these, I suggest that you do a simpler problem first.  Solve the PDE with a given right-hand side such as diff(u(x,y),x,x) + diff(u(x,y),y,y) = -1, and zero boundary conditions on all four sides.

You seem to know the difference between a colon ":" and a semicolon ";" for terminating Maple statements, but just in case you don't, let me explain.

Statements terminated by either a colon or semicolon are evaluated fully in Maple.  The only difference is that the result of the evaluation is not shown when the statement is terminated with a colon.

Is there a reason why you have terminated two of your statements with colons?  Don't you want to see the result? Why not? The results can be very informative.

For instance, it seems that you expect your yy to be a pair of partial differential equations.  Is it?

In Maple 2016 I get the expected answer.  Perhaps you have other stuff in your worksheet that interferes with the calculation.  Try it on a fresh worksheet.

The question as stated is incomplete, to say the least.

1. One solves a PDE on a prescribed domain.  What is the domain of your function u(x,t)?  You need to say something like x is in the interval [-1,1] and t > 0.
2. You need to specify boundary conditions.
3. You need to specify an initial condition.
4. I can make no sense of your condition x=0.  What is it saying?
5. I can make no sense of your statement that t is a step function. If t is a function, then it must be a function of something.  What is that something?

@nm OK, your latest version looks good.

As to your comment: "Maybe mapping the interval first from -1..1 to 0..2 and then mapping again solution back to -1..1 at the end is a simpler method", yes, it is.  Not only the calculation is significantly easier (by hand, or by Maple), the solution is more compact, and the distinction between odd and even indices does not arise.  It is simply:

where

@nm As Preben has observed, the problem on an arbitrary interval [a,b] may be translated to a problem on the interval [0,b-a].  Since Maple knows how to solve the latter problem, the solution to the former may be obtained by a simple substituion.  That's what I meant by writing that "it requires no extra effort".  The attached worksheet illustrates this.

heat-equation.mw

@nm Your analysis in report.pdf is correct up to equation (4).  The subsequent argument that leads to "So let us assume B = 0" is incorrect and therefore the rest of the calculations are invalid.

But you don't need to examine the calculations in order to tell that the final result is wrong.  You can tell that just by looking at your infinite series for u(x,t), a sample of which you have included in your post.  According to your calculations, u(x,t) is a sum of cosines regardless of the initial condition.  Since cosine is an even function, your u(x,t) is even in x.  But the correct solution need not be even in x because the initial condition can be anything.

I stand by my assertion that "Solving the PDE over an arbitrary interval [a,b] requires no extra effort".  I will write up the justification of that assertion and post it over the weekend.

@Gillee Thaks for the pointer.  Watching what Maple does with infolevel turned on, we see that is aware of the Fourier series but does not actually apply it.

@nm Maple knows how to solve the heat equation with Fourier series on the interval [0,L].  Solving it over an arbitrary interval [a,b] requires no extra effort.  I am puzzled why it's not programmed to do that.

@maxwell With these modified boundary conditions, Maple produces the exact solution through
pdsolve({pde, ic});
Adding boundary conditions confuses Maple.  I don't know why.

@nm The u(t,x) that you have proposed does not satisfy the boundary conditions.  It would have worked if the x domain were [-Pi/2 .. Pi/2].

@9009134 As I said before, if I were faced with solving such a problem, I will try first to find a simpler version of it which I am able to solve.  By a simpler version I don't mean a different set of parameters.  Rather, I mean a fundamentally reduced model, for example one that involves second order derivatives instead of the current model that involve sixth order derivatives.

I cannot do that for you because I have no idea what is behind these equations. Is it a physics problem?  If so, then it would be worthwhile to examine the assumptions that go into its mathematical model.  Generally, you may expect a coarser model to lead to lower order differential equations.  Do that one first, and only then attempt to solve the more complex version.

Anyway, that's what I would do if I were solving such a problem.