@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.

I copied and pasted your code to a Maple worksheet and it worked fine for me.  Add the missing semicolon at the end of the display(P) command and try again.

Aside: Your subject line says "plot using loop" but there is no loop in the code that you have shown.  Perhaps you meant to ask for something else.

@Carl Love Thank your very much for your thorough analysis of the issue.  I have a better understanding of the evaluation rules now after reading the rtable_eval help page.  As you have noted, it does not quite explain the reason for the different outputs of Maple 11 and Maple 2015, especially considering that the rtable_eval help page is virtually identical between the two versions.  I am content with regarding the different behaviors as not a bug but an incompatibility.

@Mac Dude I agree with you -- the 2D math is quite useful for documentation, and I use it that way quite often.  In my remarks on the uselessness of the 2D math I was thinking of 2D math for computation.

@nm You wrote: If Maplesoft is determined to make less and less people use Maple, they are really doing a very good job at it. 

I agree.  I have no idea whom the 2D input is targetted to.  It may appeal to high school students, but I doubt that high school students form a large subset of Maple users.

What is worse, the beginner is more confused than helped by the 2D input, as a plethora of questions in this forum demonstrates.  Hobbling the beginner with the arcane intricacies of the useless 2D input is a disservice.

It seems to me that the vast majority of regular Maple users prefer the 1D input.  I think it will be an improvement if Maplesoft makes the 1D input the default and focuses on 1D in help pages and presentations.  If someone at Maplesoft is overlyl enamored with the 2D input, he may consider keeping it on the side, and out of the way, as a weird curio.

@Kitonum You are right; my z was not as it was supposed to be.  I was careless.

Have a look at this:

http://stackoverflow.com/help/how-to-ask