@TomM I tried out your three execution groups (in Maple 2015.1) which altogether produce 4 plots.  The first is a thumbnail.  The last is full-sized.

Curiously, the two middle ones are variable-sized -- they grow and shrink as I resize the Maple window.  When I maximize the window, the two middle ones are actually larger than the fourth plot.  When I reduce the window size, the two middle one can shrink to be as small as a thumbnail.

This does not bother me, and until you brought this up, I was oblivious to the issue.

As to your last paragraph where you write "What I do not need is a lot of space being taken up by the thumbnail just to get this verification", I don't quite see what you are saying.  If you want verification, use semicolon.  If you don't, use colon.  But perhaps you have something else in mind.

As far as I can tell, what you are asking is not possible in Maple, unfortunately.  See this old discussion on the issue.

I think I recall seeing a behavior such as you have described, but I can no longer reproduce it.  Perhaps that is because I have upgraded from Maple 2015 to Maple 2015.1.

With Maple 2015.1's default configuration, p := plot(...); displays a thumbnail.  Subsequently, p; displays a normal-sized plot.

If you don't want to see the thumbnail, replace the semicolon with a colon, as in p := plot(...):

@max125 We know that the (-1)^x is +1 if x is even, and -1 if x is odd.  Now, what do you expect to get if x is neither odd nor even? Think of x = Pi.

@rit I don't understand the question at all.  Perhaps someone else who can understand it may respond.

I have no idea what it is that you are asking.  The expressions that you have shown are correct Maple syntax.  What do you want to know about them?

@Z1493 pdsolve computes its solutions by searching among expressions fitting certain special simple patterns. Those patterns are far from spanning the complete solution space in general. In your case, for instance, the functions

f1(x,t) = A*sin(x+t) + B*cos(x+t),   f2(x,t) = A*cos(x+t) - B*sin(x+t)

are also solutions of the PDEs for any constant A and B, but they are not captured by pdsolve. Note that, in partiular, A=1 and B=-1, produces your "initial conditions", indicating that your ics is insufficient to identify a solution uniquely.

@Z1493 Partial differential equations are much more subtle than ordinary differential equations.  You cannot just throw together a random set of equations and expect a solution.  Seemingly small changes, such as going from

d^2 u/ dx^2 + d^2/dy^2 = 0

to

d^2 u/ dx^2 - d^2/dy^2 = 0

can have drastic effects on what kinds of boundary or initial conditions may be prescribed, and how the solutions bechave.

What you have called ics in your message is not in any way meaningful in the context of the corresponding equations, so you should not expect a solution.  In practice, partial differential equations often arise from physical models which provides a guide to the appropriate initial and/or boundary conditions, and that's where I suggest that you should begin.

Is there a reason to expect anything other than the zero solution?

@Carl Love Yes, the expression you have obtained is good and correct.  However the solution produced by solving the ODE looks better. In particular, it makes it clear that y(x) = x + an oscillation.

By the way, the equation tan(y) = a*tan(x) arises in the modeling of a universal joint, shown as an animation near the top of the web page <https://en.wikipedia.org/wiki/Universal_joint>.

@Raluca84 As to "the whole article is useful for a good understanding", that's an understatement.  To understand that article, one needs to read a bunch of other references first.  For instance, the article does not explain the meaning of the article's very first equation, (2.1).  Rather, it refers to the articles [1-4].  I am not interested enough in the subject to chase those references.

Here is something you can do to help.  Consider the recurrence equation (2.1), which is:

x(t+1) = A(t) x(t),   t in N^m,

where 1 = (1,1,...,1).  I think x is in R^n and A(t) is an nxn matrix, although the article does not say.

Now, suppose m=3 and n=1.  Can you explain what is meant by solving that recurrence?  Can you provide an algorithm?

Carl and Kitonum, your solutions are becoming cleverer and cleverer, as Alice would have said.

Here is a semi-clever solution, inspired by Carl's, but with map instead of map2:

plot(map(((y,x)->[x,y])~, [y1, y2], x));

@Kitonum That's quite clever.  Actually I thought of doing the same thing with zip, which is more verbose than yours:

plot([
  zip((x,y)->[x,y], x, y1),
  zip((x,y)->[x,y], x, y2)]
);

But then I posted the one with pointplot because perhaps that's easier to grasp.

@Carl Love I have attached the outputs of showstat(`convert/list`) from Maple 11.2 and Maple 2015.1.  I haven't analyzed them in any depth, but it's likely that the several calls to eval in the Maple 11 version validate your statement.

convert_list_maple11.txt

convert_list_maple2015.txt

@Raluca84 Your response does not quite answer Carl's question: Could you indicate a specific part of your paper? An equation number will help.

Better yet, don't refer to the paper at all.  Make an attempt to distill your question to a bare minimum.  If you post a well-formed and self-contained question here, there will be many more people who will be willing to help.